PMT 2014 PromotionRelated papers
The Political Economy of Pulse: Techno-somatic rhythm and real-time data
In the context of ubiquitous data capture and the politics of control, there is growing individual and managerial interest in 'pulse', both in the literal sense of arterial pulse (now monitored through wearable technology) and in a metaphorical sense of real-time tracking (for instance taking the 'pulse of an organisation'). This article uses the category of 'pulse' to explore post-Fordism as a set of techniques for governing rhythms, both of the body and of technologies. It draws on Lefebvre's work to introduce notions of eurhythmia, arrhythmia and 'internal measure' as ways of exploring somatic and organisational life. It then introduces two case studies where the idea and physical nature of 'pulse' are at work. These provide an insight into the real-time nature of post-Fordist life, where a chronic sensing of quantities becomes the basis of cooperation , rather than a judgement via measures.
Ensayo Homeopatía
2024
La homeopatía es una práctica médica alternativa creada a finales del siglo XVIII por el médico alemán Samuel Hahnemann. Surgió como una respuesta crítica a los métodos tradicionales de la época, que Hahnemann consideraba excesivamente invasivos y nocivos; recordemos que la medicina actualmente es considerada un arte y una ciencia pero hace 3 siglos era rudimentaria y los tratamientos al no tener ensayos clínicos provocaban efectos secundarios y en ocasiones el empeoramiento de la salud del paciente. El término "homeopatía" proviene del griego "hómoios" (similar) y "páthos" (sufrimiento), reflejando el principio básico de esta disciplina: "similia similibus curentur" o "lo similar cura lo similar". (1)
Manifesting the Inner Aryan with Runes
2023
Paper presented at the 2023 Old Norse Mythology Conference, Aarhus Denmark. In this paper, I'm tracing the ideological history that forms the basis for contemporary rune magic, and asking the question: what has changed since the early 20th century? How far has contemporary rune magic come from its ariosophic origins?
Caratterizzazione delle prestazioni acustiche di infissi antirumore con misure di laboratorio
2006
Il rumore prodotto dalle sorgenti sonore esterne si trasmette all'interno degli edifici attraverso le superfici dell'involucro edilizio meno resistenti al passaggio delle onde sonore; sono questi componenti ad influenzare pertanto la prestazione acustica globale della facciata. Nella maggioranza dei casi i componenti più deboli sono rappresentati dalle finestre, dagli infissi e dai cassonetti per gli avvolgibili; risulta pertanto estremamente importante la caratterizzazione in laboratorio delle proprietà acustiche di questi elementi, al fine di consentire ai progettisti la scelta del componente più idoneo a garantire le prestazioni acustiche di facciata richieste. camere riverberanti accoppiate nelle quali sono stati testati negli ultimi anni numerosi prototipi di infissi antirumore, con particolare riguardo alla tipologia innovativa di infissi ventilati antirumore, nonché diversi cassonetti per avvolgibili dotati di elevate proprietà di isolamento acustico. Il lavoro presenta la metodologia per la misura del potere fonoisolante R secondo la norma UNI EN ISO 140-3 e dell'indice del potere fonoisolante RW secondo la norma UNI EN ISO 717-1, al fine di definire le proprietà di isolamento acustico globali dei campioni sottoposti a prova. Inoltre, sono illustrate le metodiche per la misurazione dei livelli di intensità sonora con sonda intensimetrica, grazie alle quali è possibile costruire le mappe acustiche in prossimità dei campioni testati ed avere una visione più dettagliata del comportamento dei singoli componenti degli infissi testati, localizzando le fughe sonore. Tali tecniche sono fondamentali in vista del perfezionamento e dell'ottimizzazione delle proprietà acustiche di prototipi di componenti finestrati. Il lavoro è completato dai risultati delle campagne di misura più significative realizzate in questi ultimi anni dal Laboratorio di Acustica dell'Università di Perugia.
Archaeology of Household Food Production
Archaeology of Food: An Encyclopedia
Desde las profundidades de la Historia: la teoría de la Historia en Marx (traducción del inglés)
Versión en español de “En las profundidades de la Historia: La investigación sobre teorías de la historia de Marx” El presente trabajo es un nuevo esfuerzo por presentar las ideas del filósofo chino Chen Xianda. Es muy interesante descubrir en español que el hilo conductor de su trabajo confirma la presencia actual de la metodología del materialismo histórico como fuente de conocimiento insoslayable para interpretar los momentos históricos del mundo de alineación que ha vivido ha vivido la especie humana desde su origen hasta nuestros días del siglo XIX. No existe un trascurrir histórico de la humanidad sin que se explique desde la contradicción de clases que explica la necesidad de un instrumental teórico práctico para trasformar el mundo de su imperio capitalista y postcapitalista. El recorrido al que nos invita el profesor Xianda, está plagado de caminos y contracaminos donde han están las categorías de la practica materialista histórica que nos muestra los cursos de acción que reconocen el protagonismo de la inteligencia socialista articulada por una organización política partidaria que existe para adaptarse a las circunstancias del desarrollo y progreso de la vida inteligente del hombre, pero sin saltarse los códigos que explican las relaciones de producción y distribución de la plusvalía en todos los momentos de la humanidad donde el pilar del desarrollo ha sido la teoría del capitalismo y sus estructura de dominación. Los conceptos de, esencia, humanidad, Libertad, desarrollo y socialismo que se han sucedido de forma alterna en la vida de los hombres solo han apuntado a desnudar la gran verdad de la práctica del materialismo histórico: transformarse en la traducción científica del avance de hacia la esencia libertaria y socialista de la Humanidad. La búsqueda del hombre sin muchas veces comprenderlo está en dotar a su inteligencia de la libertad que lo haga menos esclavo de las formas de estar en el mundo y como desean manejar sus acciones y pasiones. Por ello, la tan anhelado búsqueda de una humanidad donde no exista lucha de clases socioeconómicas es una Utopía racional y marxista, y es la llama de fuego intelectual que está presente en cada ser huma no que aprende que no existe un único modo de ser tratado de forma justa en una sociedad donde no es protagonista de su vida y de su muerte. Este libro de Xianda, es por ello, una obligación metodológica de todo militante de la libertad y del socialismo porque no deja ambigüedad al mostrarnos que la historia a pesar de sus profundidades tiene un sendero que cada uno de nosotros puede recorrer sin perderse en los ideales que han definido la lucha por el cambio de las estructuras del capital durante todo la vida en sociedad, y durante toda la vida de observadores que han debido sopesar quienes no son dueños de los medios de producción. China es una forma de comprender como luego de haber transitado por nuevos senderos del materialismo histórico logra sorprendernos con la tesis del socialismo capitalista del siglo XIX sin desconocer los temas de la esencia del hombre contemporáneo como son la libertad e su inteligencia y su acercamiento a distintos modos de percibir los demás modelos de sociedad. Hoy en China podemos reconocer que existieron crisis y contracrisis, la expresión dialéctica de toda la Humanidad, pero fueron capaces de descubrir en la conciencia de sus ciudadanos que la cultura político nacional ha sido el pilar de su crecimiento como potencia mundial económica, sin perder de vista su estructura epistemológica marxista. 4 Estamos en presencia de una forma de hacer filosofía del hombre marxista que debe ser material de trabajo de todo intelectual de izquierda del siglo XIX. Walterio Barra Cabello Filosofo Traductor Viña del Mar, Chile, 13 de septiembre de 2013
Resumo da Lei 8.666 (Lei de Licitações
Licitação é o procedimento administrativo formal em que a Administração Pública convoca, mediante condições estabelecidas em ato próprio (edital ou convite), empresas interessadas na apresentação de propostas para o oferecimento de bens e serviços. A licitação objetiva garantir a observância do princípio constitucional da isonomia e a selecionar a proposta mais vantajosa para a Administração, de maneira a assegurar oportunidade igual a todos os interessados e possibilitar o comparecimento ao certame ao maior número possível de concorrentes. A Lei nº 8.666 de 1993, ao regulamentar o artigo 37, inciso XXI, da Constituição Federal, estabeleceu normas gerais sobre licitações e contratos administrativos pertinentes a obras, serviços, inclusive de publicidade, compras, alienações e locações no âmbito dos Poderes da União, dos Estados, do Distrito Federal e dos Municípios. De acordo com essa Lei, a celebração de contratos com terceiros na Administração Pública deve ser necessariamente precedida na licitação, ressalvadas as hipóteses de dispensa e de inexigibilidade de licitação. O objeto da licitação são compras, os serviços, as obras, alienações e permissões da Administração Pública. O parágrafo único do art. 1.º da Lei 8666/93 estabelece que são obrigados a lançar mão desse instituto os órgãos da administração direta, os fundos especiais, as Autarquias, as fundações públicas, as empresas públicas, as sociedades de economia mista e demais entidades controladas direta ou indiretamente pela União, Estados, Distrito Federal e Municípios. O artigo 1.º da referida lei dispõe que é de competência da União a criação de legislação acerca do tema, Contudo, podem os Estados e Municípios legislar a respeito, desde que as disposições sejam compatíveis com a norma geral. O Artigo 3º traz os princípios da licitação: Nos procedimentos de licitação, esse princípio vincula os licitantes e a Administração Pública às regras estabelecidas, nas normas e princípios em vigor. Significa dar tratamento igual a todos os interessados. É condição essencial para garantir em todas as fases da licitação. Esse princípio obriga a Administração a observar nas suas decisões critérios objetivos previamente estabelecidos, afastando a discricionariedade e o subjetivismo na condução dos procedimentos da licitação. A conduta dos licitantes e dos agentes públicos tem que ser, além de lícita, compatível com a moral, ética, os bons costumes e as regras da boa administração, para processar e julgar as propostas, a administração deve possuir um comportamento não apenas lícito, mas também consoante com a moral, os bons costumes e os princípios de justiça e equidade. Qualquer interessado deve ter acesso às licitações públicas e seu controle, mediante divulgação dos atos praticados pelos administradores em todas as fases da licitação, deve haver divulgação do processo para o conhecimento de todos os interessados. Além disso, os atos da administração praticados nas várias fases do processo devem ser abertas aos interessados, para assegurar a todos a possibilidade de fiscalizar sua legalidade.
Postmodern feminism vs diversity and equality
Do you think postmodern feminists believe in gender diversity? Yes or no. how it is connected/not connected gender equality?
CONTENTS
1. Introduction
...17
1. Definition. 2. Sub-divisions of Theory of Machines.
3. Fundamental Units. 4. Derived Units. 5. Systems
of Units. 6. C.G.S. Units. 7. F.P.S. Units.
8. M.K.S. Units 9. International System of Units (S.I.
Units). 10. Metre. 11. Kilogram. 12. Second.
13. Presentation of Units and their Values.
14. Rules for S.I. Units. 15. Force. 16. Resultant
Force. 17. Scalars and Vectors. 18. Representation
of Vector Quantities. 19. Addition of Vectors.
20. Subtraction of Vectors.
2.
Kinematics of Motion
...823
1. Introduction. 2. Plane Motion. 3. Rectilinear
Motion. 4. Curvilinear Motion. 5. Linear Displacement.
6. Linear Velocity. 7. Linear Acceleration. 8. Equations
of Linear Motion. 9. Graphical Representation of
Displacement with respect to Time. 10. Graphical
Representation of Velocity with respect to Time.
11. Graphical Representation of Acceleration with
respect to Time. 12. Angular Displacement.
13. Representation of Angular Displacement by a
Vector. 14. Angular Velocity. 15. Angular Acceleration
16. Equations of Angular Motion. 17. Relation between
Linear Motion and Angular Motion. 18. Relation
between Linear and Angular Quantities of Motion.
19. Acceleration of a Particle along a Circular Path.
3.
Kinetics of Motion
1. Introduction. 2. Newton's Laws of Motion.
3. Mass and Weight. 4. Momentum. 5. Force.
6. Absolute and Gravitational Units of Force.
7. Moment of a Force. 8. Couple. 9. Centripetal and
Centrifugal Force. 10. Mass Moment of Inertia.
11. Angular Momentum or Moment of Momentum.
12. Torque. 13. Work. 14. Power. 15. Energy.
16. Principle of Conservation of Energy. 17. Impulse
and Impulsive Force. 18. Principle of Conservation
of Momentum. 19. Energy Lost by Friction Clutch
During Engagement. 20. Torque Required to Accelerate
a Geared System. 21. Collision of Two Bodies.
22. Collision of Inelastic Bodies. 23. Collision of
Elastic Bodies. 24. Loss of Kinetic Energy During
Elastic Impact.
(v)
...2471
4.
Simple Harmonic Motion
... 7293
1. Introduction. 2. Velocity and Acceleration of a
Particle Moving with Simple Harmonic Motion.
3. Differential Equation of Simple Harmonic Motion.
4. Terms Used in Simple Harmonic Motion.
5. Simple Pendulum. 6. Laws of Simple Pendulum.
7. Closely-coiled Helical Spring. 8. Compound
Pendulum. 9. Centre of Percussion. 10. Bifilar
Suspension. 11. Trifilar Suspension (Torsional
Pendulum).
5.
Simple Mechanisms
...94118
1. Introduction. 2. Kinematic Link or Element.
3. Types of Links. 4. Structure. 5. Difference Between
a Machine and a Structure. 6. Kinematic Pair.
7. Types of Constrained Motions. 8. Classification
of Kinematic Pairs. 9. Kinematic Chain. 10. Types of
Joints in a Chain. 11. Mechanism. 12. Number of
Degrees of Freedom for Plane Mechanisms.
13. Application of Kutzbach Criterion to Plane
Mechanisms. 14. Grubler's Criterion for Plane
Mechanisms. 15. Inversion of Mechanism. 16. Types
of Kinematic Chains. 17. Four Bar Chain or Quadric
Cycle Chain. 18. Inversions of Four Bar Chain.
19. Single Slider Crank Chain. 20. Inversions of
Single Slider Crank Chain. 21. Double Slider Crank
Chain. 22. Inversions of Double Slider Crank Chain.
6.
Velocity in Mechanisms
(Instantaneous Centre Method)
...119142
Velocity in Mechanisms
(Relative Velocity Method)
...143173
1. Introduction. 2. Space and Body Centrodes.
3. Methods for Determining the Velocity of a Point
on a Link. 4. Velocity of a Point on a Link by
Instantaneous Centre Method. 5. Properties of the
Instantaneous Centre. 6. Number of Instantaneous
Centres in a Mechanism. 7. Types of Instantaneous
Centres. 8. Location of Instantaneous Centres.
9. Aronhold Kennedy (or Three Centres-in-Line)
Theorem. 10. Method of Locating Instantaneous
Centres in a Mechanism.
7.
1. Introduction. 2. Relative Velocity of Two Bodies
Moving in Straight Lines. 3. Motion of a Link.
4. Velocity of a Point on a Link by Relative Velocity
Method. 5. Velocities in a Slider Crank Mechanism.
6. Rubbing Velocity at a Pin Joint. 7. Forces Acting
in a Mechanism. 8. Mechanical Advantage.
(vi)
8.
Acceleration in Mechanisms
...174231
1. Introduction. 2. Acceleration Diagram for a Link.
3. Acceleration of a Point on a Link.
4. Acceleration in the Slider Crank Mechanism.
5. Coriolis Component of Acceleration.
9.
Mechanisms with Lower Pairs
...232257
1. Introduction 2. Pantograph 3. Straight Line
Mechanism. 4. Exact Straight Line Motion Mechanisms
Made up of Turning Pairs. 5. Exact Straight Line
Motion Consisting of One Sliding Pair (Scott Russels
Mechanism). 6. Approximate Straight Line Motion
Mechanisms. 7. Straight Line Motions for Engine
Indicators. 8. Steering Gear Mechanism. 9. Davis
Steering Gear. 10. Ackerman Steering Gear.
11. Universal or Hookes Joint. 12. Ratio of the
Shafts Velocities. 13. Maximum and Minimum Speeds
of the Driven Shaft. 14. Condition for Equal Speeds
of the Driving and Driven Shafts. 15. Angular
Acceleration of the Driven Shaft. 16. Maximum
Fluctuation of Speed. 17. Double Hookes Joint.
10. Friction
...258324
1. Introduction. 2. Types of Friction. 3. Friction
Between Unlubricated Surfaces. 4. Friction Between
Lubricated Surfaces. 5. Limiting Friction. 6. Laws of
Static Friction. 7. Laws of Kinetic or Dynamic Friction.
8. Laws of Solid Friction. 9. Laws of Fluid Friction.
10. Coefficient of Friction. 11. Limiting Angle of
Friction. 12. Angle of Repose. 13. Minimum Force
Required to Slide a Body on a Rough Horizontal
Plane. 14. Friction of a Body Lying on a Rough
Inclined Plane. 15. Efficiency of Inclined Plane.
16. Screw Friction. 17. Screw Jack. 18. Torque
Required to Lift the Load by a Screw Jack.
19. Torque Required to Lower the Load by a Screw
Jack. 20. Efficiency of a Screw Jack. 21. Maximum
Efficiency of a Screw Jack. 22. Over Hauling and
Self Locking Screws. 23. Efficiency of Self Locking
Screws. 24. Friction of a V-thread. 25. Friction in
Journal Bearing-Friction Circle. 26. Friction of Pivot
and Collar Bearing. 27. Flat Pivot Bearing.
28. Conical Pivot Bearing. 29. Trapezoidal or Truncated
Conical Pivot Bearing. 30. Flat Collar Bearing
31. Friction Clutches. 32. Single Disc or Plate Clutch.
33. Multiple Disc Clutch. 34. Cone Clutch.
35. Centrifugal Clutches.
11. Belt, Rope and Chain Drives
...325381
1. Introduction. 2. Selection of a Belt Drive.
3. Types of Belt Drives. 4. Types of Belts.
5. Material used for Belts. 6. Types of Flat Belt
(vii)
Drives. 7. Velocity Ratio of Belt Drive. 8. Velocity
Ratio of a Compound Belt Drive. 9. Slip of Belt.
10. Creep of Belt. 11. Length of an Open Belt Drive.
12. Length of a Cross Belt Drive. 13. Power Transmitted
by a Belt. 14. Ratio of Driving Tensions for Flat Belt
Drive. 15. Determination of Angle of Contact.
16. Centrifugal Tension. 17. Maximum Tension in
the Belt. 18. Condition for the Transmission of
Maximum Power. 19. Initial Tension in the Belt.
20. V-belt Drive. 21. Advantages and Disadvantages
of V-belt Drive Over Flat Belt Drive. 22. Ratio of
Driving Tensions for V-belt. 23. Rope Drive.
24. Fibre Ropes. 25. Advantages of Fibre Rope
Drives. 26. Sheave for Fibre Ropes. 27. Wire Ropes.
28. Ratio of Driving Tensions for Rope Drive. 29.
Chain Drives. 30. Advantages and Disadvantages of
Chain Drive Over Belt or Rope Drive. 31. Terms
Used in Chain Drive. 32. Relation Between Pitch
and Pitch Circle Diameter. 33. Relation Between
Chain Speed and Angular Velocity of Sprocket.
34. Kinematic of Chain Drive. 35. Classification of
Chains. 36. Hoisting and Hauling Chains. 37. Conveyor
Chains. 38. Power Transmitting Chains. 39. Length
of Chains.
12. Toothed Gearing
...382427
1. Introduction. 2. Friction Wheels. 3. Advantages
and Disadvantages of Gear Drive. 4. Classification
of Toothed Wheels. 5. Terms Used in Gears.
6. Gear Materials. 7. Condition for Constant Velocity
Ratio of Toothed Wheels-Law of Gearing. 8. Velocity
of Sliding of Teeth. 9. Forms of Teeth. 10. Cycloidal
Teeth. 11. Involute Teeth. 12. Effect of Altering the
Centre Distance on the Velocity Ratio For Involute
Teeth Gears. 13. Comparison Between Involute and
Cycloidal Gears. 14. Systems of Gear Teeth.
15. Standard Proportions of Gear Systems. 16. Length
of Path of Contact. 17. Length of Arc of Contact.
18. Contact Ratio (or Number of Pairs of Teeth in
Contact). 19. Interference in Involute Gears.
20. Minimum Number of Teeth on the Pinion in
Order to Avoid Interference. 21. Minimum Number
of Teeth on the Wheel in Order to Avoid Interference.
22. Minimum Number of Teeth on a Pinion for
Involute Rack in Order to Avoid Interference.
23. Helical Gears. 24. Spiral Gears. 25. Centre
Distance for a Pair of Spiral Gears. 26. Efficiency of
Spiral Gears.
13. Gear Trains
...428479
1. Introduction. 2. Types of Gear Trains.
3. Simple Gear Train. 4. Compound Gear Train.
(viii)
5. Design of Spur Gears. 6. Reverted Gear Train.
7. Epicyclic Gear Train. 8. Velocity Ratio of Epicyclic
Gear Train. 9. Compound Epicyclic Gear Train (Sun
and Planet Wheel). 10. Epicyclic Gear Train With
Bevel Gears. 11. Torques in Epicyclic Gear Trains.
14. Gyroscopic Couple and Precessional Motion
...480513
1. Introduction. 2. Precessional Angular Motion.
3. Gyroscopic Couple. 4. Effect of Gyroscopic Couple
on an Aeroplane. 5. Terms Used in a Naval Ship.
6. Effect of Gyroscopic Couple on a Naval Ship
during Steering. 7. Effect of Gyroscopic Couple on
a Naval Ship during Pitching. 8. Effect of Gyroscopic
Couple on a Navel during Rolling. 9. Stability of a
Four Wheel drive Moving in a Curved Path.
10. Stability of a Two Wheel Vehicle Taking a Turn.
11. Effect of Gyroscopic Couple on a Disc Fixed
Rigidly at a Certain Angle to a Rotating Shaft.
15. Inertia Forces in Reciprocating Parts
...514564
1. Introduction. 2. Resultant Effect of a System of
Forces Acting on a Rigid Body. 3. D-Alemberts
Principle. 4. Velocity and Acceleration of the
Reciprocating Parts in Engines. 5. Kliens Construction.
6. Ritterhauss Construction. 7. Bennetts Construction.
8. Approximate Analytical Method for Velocity and
Acceleration of the Piston. 9. Angular Velocity and
Acceleration of the Connecting Rod. 10. Forces on
the Reciprocating Parts of an Engine Neglecting
Weight of the Connecting Rod. 11. Equivalent
Dynamical System. 12. Determination of Equivalent
Dynamical System of Two Masses by Graphical
Method. 13. Correction Couple to be Applied to
Make the Two Mass Systems Dynamically Equivalent.
14. Inertia Forces in a Reciprocating Engine Considering
the Weight of Connecting Rod. 15. Analytical Method
for Inertia Torque.
16. Turning Moment Diagrams and Flywheel
1. Introduction. 2. Turning Moment Diagram for a
Single Cylinder Double Acting Steam Engine.
3. Turning Moment Diagram for a Four Stroke Cycle
Internal Combustion Engine. 4. Turning Moment
Diagram for a Multicylinder Engine. 5. Fluctuation
of Energy. 6. Determination of Maximum Fluctuation
of Energy. 7. Coefficient of Fluctuation of Energy.
8. Flywheel. 9. Coefficient of Fluctuation of Speed.
10. Energy Stored in a Flywheel. 11. Dimensions of
the Flywheel Rim. 12. Flywheel in Punching Press.
(ix)
... 565611
17. Steam Engine Valves and Reversing Gears
...612652
1. Introduction. 2. D-slide Valve. 3. Piston Slide
Valve. 4. Relative Positions of Crank and Eccentric
Centre Lines. 5. Crank Positions for Admission, Cut
off, Release and Compression. 6. Approximate
Analytical Method for Crank Positions at Admission,
Cut-off, Release and Compression. 7. Valve Diagram.
8. Zeuner Valve Diagram. 9. Reuleaux Valve Diagram.
10. Bilgram Valve Diagram. 11. Effect of the Early
Point of Cut-off with a Simple Slide Valve.
12. Meyers Expansion Valve. 13. Virtual or Equivalent
Eccentric for the Meyers Expansion Valve.
14. Minimum Width and Best Setting of the Expansion
Plate for Meyers Expansion Valve. 15. Reversing
Gears. 16. Principle of Link Motions-Virtual Eccentric
for a Valve with an Off-set Line of Stroke.
17. Stephenson Link Motion. 18. Virtual or Equivalent
Eccentric for Stephenson Link Motion. 19. Radial
Valve Gears. 20. Hackworth Valve Gear. 21. Walschaert
Valve Gear.
18. Governors
...653731
1. Introduction. 2. Types of Governors. 3. Centrifugal
Governors. 4. Terms Used in Governors. 5. Watt
Governor. 6. Porter Governor. 7. Proell Governor.
8. Hartnell Governor. 9. Hartung Governor.
10. Wilson-Hartnell Governor. 11. Pickering Governor.
12. Sensitiveness of Governors. 13. Stability of
Governors. 14. Isochronous Governor. 15. Hunting.
16. Effort and Power of a Governor. 17. Effort and
Power of a Porter Governor. 18. Controlling Force.
19. Controlling Force Diagram for a Porter Governor.
20. Controlling Force Diagram for a Spring-controlled
Governor. 21. Coefficient of Insensitiveness.
19. Brakes and Dynamometers
...732773
1. Introduction. 2. Materials for Brake Lining.
3. Types of Brakes. 4. Single Block or Shoe Brake.
5. Pivoted Block or Shoe Brake. 6. Double Block or
Shoe Brake. 7. Simple Band Brake. 8. Differential
Band Brake. 9. Band and Block Brake. 10. Internal
Expanding Brake. 11. Braking of a Vehicle.
12. Dynamometer. 13. Types of Dynamometers.
14. Classification of Absorption Dynamometers.
15. Prony Brake Dynamometer. 16. Rope Brake
Dynamometers. 17. Classification of Transmission
Dynamometers. 18. Epicyclic-train Dynamometers.
19. Belt Transmission Dynamometer-Froude or
Throneycraft Transmission Dynamometer. 20. Torsion
Dynamometer. 21. Bevis Gibson Flash Light Torsion
Dynamometer.
(x)
20. Cams
...774832
1. Introduction. 2. Classification of Followers.
3. Classification of Cams. 4. Terms used in Radial
cams. 5. Motion of the Follower. 6. Displacement,
Velocity and Acceleration Diagrams when the Follower
Moves with Uniform Velocity. 7. Displacement,
Velocity and Acceleration Diagrams when the Follower
Moves with Simple Harmonic Motion. 8. Displacement,
Velocity and Acceleration Diagrams when the Follower
Moves with Uniform Acceleration and Retardation.
9. Displacement, Velocity and Acceleration Diagrams
when the Follower Moves with Cycloidal Motion.
10 Construction of Cam Profiles. 11. Cams with
Specified Contours. 12. Tangent Cam with Reciprocating
Roller Follower. 13. Circular Arc Cam with Flatfaced Follower.
21. Balancing of Rotating Masses
...833857
1. Introduction. 2. Balancing of Rotating Masses.
3. Balancing of a Single Rotating Mass By a Single
Mass Rotating in the Same Plane. 4. Balancing of a
Single Rotating Mass By Two Masses Rotating in
Different Planes. 5. Balancing of Several Masses
Rotating in the Same Plane. 6. Balancing of Several
Masses Rotating in Different Planes.
22. Balancing of Reciprocating Masses
...858908
1. Introduction. 2. Primary and Secondary Unbalanced
Forces of Reciprocating Masses. 3. Partial Balancing
of Unbalanced Primary Force in a Reciprocating
Engine. 4. Partial Balancing of Locomotives.
5. Effect of Partial Balancing of Reciprocating Parts
of Two Cylinder Locomotives. 6. Variation of Tractive
Force. 7. Swaying Couple. 8. Hammer Blow.
9. Balancing of Coupled Locomotives. 10. Balancing
of Primary Forces of Multi-cylinder In-line Engines.
11. Balancing of Secondary Forces of Multi-cylinder
In-line Engines. 12. Balancing of Radial Engines
(Direct and Reverse Crank Method). 13. Balancing
of V-engines.
23. Longitudinal and Transverse Vibrations
1. Introduction. 2. Terms Used in Vibratory Motion.
3. Types of Vibratory Motion. 4. Types of Free
Vibrations. 5. Natural Frequency of Free Longitudinal
Vibrations. 6. Natural Frequency of Free Transverse
Vibrations. 7. Effect of Inertia of the Constraint in
Longitudinal and Transverse Vibrations. 8. Natural
Frequency of Free Transverse Vibrations Due to a
Point Load Acting Over a Simply Supported Shaft.
9. Natural Frequency of Free Transverse Vibrations
Due to Uniformly Distributed Load Over a Simply
(xi)
...909971
Supported Shaft. 10. Natural Frequency of Free
Transverse Vibrations of a Shaft Fixed at Both Ends
and Carrying a Uniformly Distributed Load.
11. Natural Frequency of Free Transverse Vibrations
for a Shaft Subjected to a Number of Point Loads.
12. Critical or Whirling Speed of a Shaft. 13. Frequency
of Free Damped Vibrations (Viscous Damping).
14. Damping Factor or Damping Ratio. 15. Logarithmic
Decrement. 16. Frequency of Underdamped Forced
Vibrations. 17. Magnification Factor or Dynamic
Magnifier. 18. Vibration Isolation and Transmissibility.
24. Torsional Vibrations
...9721001
1. Introduction. 2. Natural Frequency of Free Torsional
Vibrations. 3.Effect of Inertia of the Constraint on
Torsional Vibrations. 4. Free Torsional Vibrations
of a Single Rotor System. 5. Free Torsional Vibrations
of a Two Rotor System. 6. Free Torsional Vibrations
of a Three Rotor System. 7. Torsionally Equivalent
Shaft. 8. Free Torsional Vibrations of a Geared
System.
25. Computer Aided Analysis and Synthesis of
Mechanisms
...10021049
1. Introduction. 2. Computer Aided Analysis for
Four Bar Mechanism (Freudensteins Equation).
3. Programme for Four Bar mechanism. 4. Computer
Aided Analysis for Slider Crank Mechanism.
6. Coupler Curves. 7. Synthesis of Mechanisms.
8. Classifications of Synthesis Problem. 9. Precision
Points for Function Generation. 10. Angle Relationship
for function Generation. 11. Graphical Synthesis of
Four Bar Mechanism. 12. Graphical synthesis of
Slider Crank Mechanism. 13. Computer Aided
(Analytical) synthesis of Four Bar Mechanism.
14. Programme to Co-ordinate the Angular
Displacements of the Input and Output Links. 15. Least
square Technique. 16. Programme using Least Square
Technique. 17. Computer Aided Synthesis of Four
Bar Mechanism With Coupler Point. 18. Synthesis
of Four Bar Mechanism for Body Guidance.
19. Analytical Synthesis for slider Crank Mechanism.
26. Automatic Control
...10501062
1. Introduction. 2. Terms Used in Automatic Control
of Systems. 3. Types of Automatic Control System.
4. Block Diagrams. 5. Lag in Response. 6. Transfer
Function. 7. Overall Transfer Function. 8 Transfer
Function for a system with Viscous Damped Output.
9. Transfer Function of a Hartnell Governor.
10. Open-Loop Transfer Function. 11. Closed-Loop
Transfer Function.
Index
...10631071
(xii)
GO To FIRST
CONTENTS
CONTENTS
Chapter 1 : Introduction
l
1
1
Introduction
Features
1. Definition.
2. Sub-divisions of Theory of
Machines.
3. Fundamental Units.
1.1.
Definition
8. M.K.S. Units.
The subject Theory of Machines may be defined as
that branch of Engineering-science, which deals with the study
of relative motion between the various parts of a machine,
and forces which act on them. The knowledge of this subject
is very essential for an engineer in designing the various parts
of a machine.
9. International System of
Units (S.I. Units).
Note:A machine is a device which receives energy in some
available form and utilises it to do some particular type of work.
4. Derived Units.
5. Systems of Units.
6. C.G.S. Units.
7. F.P.S. Units.
10. Metre.
11. Kilogram.
12. Second.
13. Presentation of Units and
their Values.
14. Rules for S.I. Units.
15. Force.
16. Resultant Force.
17. Scalars and Vectors.
18. Representation of Vector
Quantities.
19. Addition of Vectors.
20. Subtraction of Vectors.
1.2.
Sub-divisions of Theory of Machines
The Theory of Machines may be sub-divided into
the following four branches :
1. Kinematics. It is that branch of Theory of
Machines which deals with the relative motion between the
various parts of the machines.
2. Dynamics. It is that branch of Theory of Machines
which deals with the forces and their effects, while acting
upon the machine parts in motion.
3. Kinetics. It is that branch of Theory of Machines
which deals with the inertia forces which arise from the combined effect of the mass and motion of the machine parts.
4. Statics. It is that branch of Theory of Machines
which deals with the forces and their effects while the machine parts are at rest. The mass of the parts is assumed to be
negligible.
1
CONTENTS
CONTENTS
2
l
1.3.
Theory of Machines
Fundamental Units
The measurement of
physical quantities is one of the
most important operations in
engineering. Every quantity is
measured in terms of some
arbitrary, but internationally
accepted units, called
fundamental units. All
physical quantities, met within
this subject, are expressed in
terms of the following three
fundamental quantities :
1. Length (L or l ),
2. Mass (M or m), and
Stopwatch
Simple balance
3. Time (t).
1.4.
Derived Units
Some units are expressed in terms of fundamental units known as derived units, e.g., the units
of area, velocity, acceleration, pressure, etc.
1.5.
Systems of Units
There are only four systems of units, which are commonly used and universally recognised.
These are known as :
1. C.G.S. units,
1.6.
2. F.P.S. units,
3. M.K.S. units, and
4. S.I. units.
C.G.S. Units
In this system, the fundamental units of length, mass and time are centimetre, gram and
second respectively. The C.G.S. units are known as absolute units or physicist's units.
1.7.
F.P
.S. Units
.P.S.
In this system, the fundamental units of length, mass and time are foot, pound and second
respectively.
1.8.
M.K.S. Units
In this system, the fundamental units of length, mass and time are metre, kilogram and second
respectively. The M.K.S. units are known as gravitational units or engineer's units.
1.9.
Inter
na
tional System of Units (S.I. Units)
Interna
national
The 11th general conference* of weights and measures have recommended a unified and
systematically constituted system of fundamental and derived units for international use. This system
is now being used in many countries. In India, the standards of Weights and Measures Act, 1956 (vide
which we switched over to M.K.S. units) has been revised to recognise all the S.I. units in industry
and commerce.
*
It is known as General Conference of Weights and Measures (G.C.W.M.). It is an international organisation,
of which most of the advanced and developing countries (including India) are members. The conference
has been entrusted with the task of prescribing definitions for various units of weights and measures, which
are the very basic of science and technology today.
Chapter 1 : Introduction
l
3
A man whose mass is 60 kg weighs 588.6 N (60 × 9.81 m/s2) on earth, approximately
96 N (60 × 1.6 m/s2) on moon and zero in space. But mass remains the same everywhere.
In this system of units, the fundamental units are metre (m), kilogram (kg) and second (s)
respectively. But there is a slight variation in their derived units. The derived units, which will be
used in this book are given below :
Density (mass density)
kg/m3
Force
N (Newton)
Pressure
Pa (Pascal) or N/m2 ( 1 Pa = 1 N/m2)
Work, energy (in Joules)
1 J = 1 N-m
Power (in watts)
1 W = 1 J/s
Absolute viscosity
kg/m-s
Kinematic viscosity
m2/s
Velocity
m/s
Acceleration
m/s2
Angular acceleration
rad/s2
Frequency (in Hertz)
Hz
The international metre, kilogram and second are discussed below :
1.10. Metre
The international metre may be defined as the shortest distance (at 0°C) between the two
parallel lines, engraved upon the polished surface of a platinum-iridium bar, kept at the International
Bureau of Weights and Measures at Sevres near Paris.
1.11. Kilogram
The international kilogram may be defined as the mass of the platinum-iridium cylinder,
which is also kept at the International Bureau of Weights and Measures at Sevres near Paris.
1.12. Second
The fundamental unit of time for all the three systems, is second, which is 1/24 × 60 × 60
= 1/86 400th of the mean solar day. A solar day may be defined as the interval of time, between the
4
l
Theory of Machines
instants, at which the sun crosses a meridian on two consecutive days. This value varies slightly
throughout the year. The average of all the solar days, during one year, is called the mean solar day.
1.13. Presentation of Units and their Values
The frequent changes in the present day life are facilitated by an international body known as
International Standard Organisation (ISO) which makes recommendations regarding international
standard procedures. The implementation of ISO recommendations, in a country, is assisted by its
organisation appointed for the purpose. In India, Bureau of Indian Standards (BIS) previously known
as Indian Standards Institution (ISI) has been created for this purpose. We have already discussed that
the fundamental units in
M.K.S. and S.I. units for
length, mass and time is metre,
kilogram and second respectively. But in actual practice, it
is not necessary to express all
lengths in metres, all masses in
kilograms and all times in seconds. We shall, sometimes, use
the convenient units, which are
multiples or divisions of our
basic units in tens. As a typical
example, although the metre is
the unit of length, yet a smaller
length of one-thousandth of a
metre proves to be more con- With rapid development of Information Technology, computers are
playing a major role in analysis, synthesis and design of machines.
venient unit, especially in the
dimensioning of drawings. Such convenient units are formed by using a prefix in front of the basic
units to indicate the multiplier. The full list of these prefixes is given in the following table.
Table 1.1. Prefixes used in basic units
Factor by which the unit
Standard form
Prefix
Abbreviation
1 000 000 000 000
1 000 000 000
1 000 000
1 000
100
10
0.1
0.01
0.001
0. 000 001
0. 000 000 001
1012
109
106
103
102
101
10–1
10–2
10–3
10–6
10–9
tera
giga
mega
kilo
hecto*
deca*
deci*
centi*
milli
micro
nano
T
G
M
k
h
da
d
c
m
µ
n
0. 000 000 000 001
10–12
pico
p
is multiplied
*
These prefixes are generally becoming obsolete probably due to possible confusion. Moreover, it is becoming
a conventional practice to use only those powers of ten which conform to 103x , where x is a positive or
negative whole number.
Chapter 1 : Introduction
l
5
1.14. Rules for S.I. Units
The eleventh General Conference of Weights and Measures recommended only the fundamental and derived units of S.I. units. But it did not elaborate the rules for the usage of the units. Later
on many scientists and engineers held a number of meetings for the style and usage of S.I. units. Some
of the decisions of the meetings are as follows :
1. For numbers having five or more digits, the digits should be placed in groups of three separated by spaces* (instead of commas) counting both to the left and right to the decimal point.
2. In a four digit number,** the space is not required unless the four digit number is used in a
column of numbers with five or more digits.
3. A dash is to be used to separate units that are multiplied together. For example, newton
metre is written as N-m. It should not be confused with mN, which stands for millinewton.
4. Plurals are never used with symbols. For example, metre or metres are written as m.
5. All symbols are written in small letters except the symbols derived from the proper names.
For example, N for newton and W for watt.
6. The units with names of scientists should not start with capital letter when written in full. For
example, 90 newton and not 90 Newton.
At the time of writing this book, the authors sought the advice of various international
authorities, regarding the use of units and their values. Keeping in view the international reputation of
the authors, as well as international popularity of their books, it was decided to present units*** and
their values as per recommendations of ISO and BIS. It was decided to use :
4500
not
4 500
or
4,500
75 890 000
not
75890000
or
7,58,90,000
0.012 55
not
0.01255
or
.01255
30 × 106
not
3,00,00,000
or
3 × 107
The above mentioned figures are meant for numerical values only. Now let us discuss about
the units. We know that the fundamental units in S.I. system of units for length, mass and time are
metre, kilogram and second respectively. While expressing these quantities we find it time consuming to write the units such as metres, kilograms and seconds, in full, every time we use them. As a
result of this, we find it quite convenient to use some standard abbreviations.
We shall use :
m
for metre or metres
km
for kilometre or kilometres
kg
for kilogram or kilograms
t
for tonne or tonnes
s
for second or seconds
min
for minute or minutes
N-m
for newton × metres (e.g. work done )
kN-m
for kilonewton × metres
rev
for revolution or revolutions
rad
for radian or radians
*
In certain countries, comma is still used as the decimal mark.
** In certain countries, a space is used even in a four digit number.
*** In some of the question papers of the universities and other examining bodies, standard values are not used.
The authors have tried to avoid such questions in the text of the book. However, at certain places, the
questions with sub-standard values have to be included, keeping in view the merits of the question from the
reader’s angle.
6
l
Theory of Machines
1.15. Force
It is an important factor in the field of Engineering science, which may be defined as an agent,
which produces or tends to produce, destroy or tends to destroy motion.
1.16. Resultant Force
If a number of forces P,Q,R etc. are acting simultaneously on a particle, then a single force,
which will produce the same effect as that of all the given forces, is known as a resultant force. The
forces P,Q,R etc. are called component forces. The process of finding out the resultant force of the
given component forces, is known as composition of forces.
A resultant force may be found out analytically, graphically or by the following three laws:
1. Parallelogram law of forces. It states, “If two forces acting simultaneously on a particle
be represented in magnitude and direction by the two adjacent sides of a parallelogram taken in order,
their resultant may be represented in magnitude and direction by the diagonal of the parallelogram
passing through the point.”
2. Triangle law of forces. It states, “If two forces acting simultaneously on a particle be
represented in magnitude and direction by the two sides of a triangle taken in order, their resultant
may be represented in magnitude and direction by the third side of the triangle taken in opposite
order.”
3. Polygon law of forces. It states, “If a number of forces acting simultaneously on a particle
be represented in magnitude and direction by the sides of a polygon taken in order, their resultant may
be represented in magnitude and direction by the closing side of the polygon taken in opposite order.”
1.17. Scalars and Vectors
1. Scalar quantities are those quantities, which have magnitude only, e.g. mass, time, volume,
density etc.
2. Vector quantities are those quantities which have magnitude as well as direction e.g. velocity,
acceleration, force etc.
3. Since the vector quantities have both magnitude and direction, therefore, while adding or
subtracting vector quantities, their directions are also taken into account.
1.18. Representation of Vector Quantities
The vector quantities are represented by vectors. A vector is a straight line of a certain length
Chapter 1 : Introduction
l
7
possessing a starting point and a terminal point at which it carries an arrow head. This vector is cut off
along the vector quantity or drawn parallel to the line of action of the vector quantity, so that the
length of the vector represents the magnitude to some scale. The arrow head of the vector represents
the direction of the vector quantity.
1.19. Addition of Vectors
(a)
(b)
Fig. 1.1. Addition of vectors.
Consider two vector quantities P and Q, which are required to be added, as shown in Fig.1.1(a).
Take a point A and draw a line AB parallel and equal in magnitude to the vector P. Through B,
draw BC parallel and equal in magnitude to the vector Q. Join A C, which will give the required sum
of the two vectors P and Q, as shown in Fig. 1.1 (b).
1.20. Subtraction of Vector Quantities
Consider two vector quantities P and Q whose difference is required to be found out as
shown in Fig. 1.2 (a).
(a)
(b)
Fig. 1.2. Subtraction of vectors.
Take a point A and draw a line AB parallel and equal in magnitude to the vector P. Through B,
draw BC parallel and equal in magnitude to the vector Q, but in opposite direction. Join A C, which
gives the required difference of the vectors P and Q, as shown in Fig. 1.2 (b).
GO To FIRST
CONTENTS
CONTENTS
8
l
Theory of Machines
2
Features
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
1ntroduction.
Plane Motion.
Rectilinear Motion.
Curvilinear Motion.
Linear Displacement.
Linear Velocity.
Linear Acceleration.
Equations of Linear Motion.
Graphical Representation of
Displacement with respect to
Time.
Graphical Representation of
Velocity with respect to Time.
Graphical Representation of
Acceleration with respect to
Time.
Angular Displacement.
Representation of Angular
Displacement by a Vector.
Angular Velocity.
Angular Acceleration.
Equations of Angular Motion.
Relation Between Linear
Motion and Angular Motion.
Relation Between Linear and
Angular Quantities of
Motion.
Acceleration of a Particle
along a Circular Path.
Kinematics of
Motion
2.1.
Introduction
We have discussed in the previous Chapter, that the
subject of Theory of Machines deals with the motion and
forces acting on the parts (or links) of a machine. In this chapter, we shall first discuss the kinematics of motion i.e. the
relative motion of bodies without consideration of the forces
causing the motion. In other words, kinematics deal with the
geometry of motion and concepts like displacement, velocity
and acceleration considered as functions of time.
2.2.
Plane Motion
When the motion of a body is confined to only one
plane, the motion is said to be plane motion. The plane motion may be either rectilinear or curvilinear.
2.3.
Rectilinear Motion
It is the simplest type of motion and is along a straight
line path. Such a motion is also known as translatory motion.
2.4.
Curvilinear Motion
It is the motion along a curved path. Such a motion,
when confined to one plane, is called plane curvilinear
motion.
When all the particles of a body travel in concentric
circular paths of constant radii (about the axis of rotation
perpendicular to the plane of motion) such as a pulley rotating
8
CONTENTS
CONTENTS
Chapter 2 : Kinematics of Motion
l
9
about a fixed shaft or a shaft rotating about its
own axis, then the motion is said to be a plane
rotational motion.
Note: The motion of a body, confined to one plane,
may not be either completely rectilinear nor completely
rotational. Such a type of motion is called combined
rectilinear and rotational motion. This motion is discussed in Chapter 6, Art. 6.1.
2.5.
Linear Displacement
It may be defined as the distance moved
by a body with respect to a certain fixed point.
The displacement may be along a straight or a
curved path. In a reciprocating steam engine, all
the particles on the piston, piston rod and crosshead trace a straight path, whereas all particles
on the crank and crank pin trace circular paths,
whose centre lies on the axis of the crank shaft. It will be interesting to know, that all the particles on
the connecting rod neither trace a straight path nor a circular one; but trace an oval path, whose radius
of curvature changes from time to time.
The displacement of a body is a vector quantity, as it has both magnitude and direction.
Linear displacement may, therefore, be represented graphically by a straight line.
2.6. Linear Velocity
Spindle
(axis of rotation)
θ
∆θ
r
Reference
θο line
It may be defined as the rate of
change of linear displacement of a body with
respect to the time. Since velocity is always
expressed in a particular direction, therefore
it is a vector quantity. Mathematically, linear velocity,
v = ds/dt
Notes: 1. If the displacement is along a circular
path, then the direction of linear velocity at any
instant is along the tangent at that point.
Axis of rotation
2. The speed is the rate of change of linear displacement of a body with respect to the time. Since the
speed is irrespective of its direction, therefore, it is a scalar quantity.
2.7.
Linear Acceleration
It may be defined as the rate of change of linear velocity of a body with respect to the time. It
is also a vector quantity. Mathematically, linear acceleration,
a=
dv d
=
dt dt
2
ds d s
=
dt dt 2
Notes: 1. The linear acceleration may also be expressed as follows:
a=
dv
dt
=
ds
dt
×
dv
ds
= v×
dv
ds
2. The negative acceleration is also known as deceleration or retardation.
ds
... 3 v =
dt
10
l
2.8.
Theory of Machines
Equations of Linear Motion
The following equations of linear motion are
important from the subject point of view:
1. v = u + a.t
2. s = u.t +
1
2
a.t2
3. v2 = u2 + 2a.s
4. s =
where
(u + v )
2
× t = vav × t
u = Initial velocity of the body,
v = Final velocity of the body,
a = Acceleration of the body,
s = Displacement of the body in time t seconds, and
vav = Average velocity of the body during the motion.
Notes: 1. The above equations apply for uniform
acceleration. If, however, the acceleration is variable,
then it must be expressed as a function of either t, s
or v and then integrated.
2. In case of vertical motion, the body is
subjected to gravity. Thus g (acceleration due to gravity) should be substituted for ‘a’ in the above equations.
3. The value of g is taken as + 9.81 m/s2 for
downward motion, and – 9.81 m/s2 for upward motion of a body.
4. When a body falls freely from a height h,
then its velocity v, with which it will hit the ground is
given by
v = 2 g .h
2.9.
Gra
phical Repr
esenta
tion of
Graphical
Representa
esentation
Displacement with Respect
to Time
t=0s
v = 0 m/s
t = time
v = velocity (downward)
g = 9.81 m/s2 = acceleration
due to gravity
t =1s
v = 9.81 m/s
t=2s
v = 19.62 m/s
The displacement of a moving body in a given time may be found by means of a graph. Such
a graph is drawn by plotting the displacement as ordinate and the corresponding time as abscissa. We
shall discuss the following two cases :
1. When the body moves with uniform velocity. When the body moves with uniform velocity,
equal distances are covered in equal intervals of time. By plotting the distances on Y-axis and time on
X-axis, a displacement-time curve (i.e. s-t curve) is drawn which is a straight line, as shown in Fig. 2.1
(a). The motion of the body is governed by the equation s = u.t, such that
Velocity at instant 1 = s1 / t1
Velocity at instant 2 = s2 / t2
Since the velocity is uniform, therefore
s1 s2 s3
=
= = tan θ
t1 t2 t3
where tan θ is called the slope of s-t curve. In other words, the slope of the s-t curve at any instant
gives the velocity.
Chapter 2 : Kinematics of Motion
l
11
2. When the body moves with variable velocity. When the body moves with variable velocity,
unequal distances are covered in equal intervals of time or equal distances are covered in unequal intervals
of time. Thus the displacement-time graph, for such a case, will be a curve, as shown in Fig. 2.1 (b).
(a) Uniform velocity.
(b) Variable velocity.
Fig. 2.1. Graphical representation of displacement with respect to time.
Consider a point P on the s-t curve and let this point travels to Q by a small distance δs in a
small interval of time δt. Let the chord joining the points P and Q makes an angle θ with the horizontal.
The average velocity of the moving point during the interval PQ is given by
tan θ = δs / δt
. . . (From triangle PQR )
In the limit, when δt approaches to zero, the point Q will tend to approach P and the chord PQ
becomes tangent to the curve at point P. Thus the velocity at P,
v p = tan θ = ds /dt
where tan θ is the slope of the tangent at P. Thus the slope of the tangent at any instant on the s-t curve
gives the velocity at that instant.
2.10. Graphical Representation of Velocity with Respect to Time
We shall consider the following two cases :
1. When the body moves with uniform velocity. When the body moves with zero acceleration,
then the body is said to move with a uniform
velocity and the velocity-time curve (v-t
curve) is represented by a straight line as
shown by A B in Fig. 2.2 (a).
We know that distance covered by a
body in time t second
= Area under the v-t curve A B
= Area of rectangle OABC
Thus, the distance covered by a
body at any interval of time is given by the
area under the v-t curve.
2. When the body moves with
variable velocity. When the body moves with
constant acceleration, the body is said to move with variable velocity. In such a case, there is equal
variation of velocity in equal intervals of time and the velocity-time curve will be a straight
line AB inclined at an angle θ, as shown in Fig. 2.2 (b). The equations of motion i.e. v = u + a.t, and
s = u.t + 12 a.t2 may be verified from this v-t curve.
12
l
Theory of Machines
Let
u = Initial velocity of a moving body, and
v = Final velocity of a moving body after time t.
Then,
tan θ =
BC v − u Change in velocity
=
=
= Acceleration (a)
AC
t
Time
(a) Uniform velocity.
(b) Variable velocity.
Fig. 2.2. Graphical representation of velocity with respect to time.
Thus, the slope of the v-t curve represents the acceleration of a moving body.
BC v − u
a = tan θ =
=
Now
or
v = u + a.t
AC
t
Since the distance moved by a body is given by the area under the v-t curve, therefore
distance moved in time (t),
s = Area OABD = Area OACD + Area ABC
1
1
2
2
= u.t + (v − u ) t = u.t + a.t 2
... (3 v – u = a.t)
2.11. Graphical Representation of Acceleration with Respect to Time
(a) Uniform velocity.
(b) Variable velocity.
Fig. 2.3. Graphical representation of acceleration with respect to time.
We shall consider the following two cases :
1. When the body moves with uniform acceleration. When the body moves with uniform
acceleration, the acceleration-time curve (a-t curve) is a straight line, as shown in Fig. 2.3(a). Since
the change in velocity is the product of the acceleration and the time, therefore the area under the
a-t curve (i.e. OABC) represents the change in velocity.
2. When the body moves with variable acceleration. When the body moves with variable
acceleration, the a-t curve may have any shape depending upon the values of acceleration at various
instances, as shown in Fig. 2.3(b). Let at any instant of time t, the acceleration of moving body is a.
Mathematically,
a = dv / dt
or
dv = a.dt
Chapter 2 : Kinematics of Motion
l
13
Integrating both sides,
v2
∫v
1
t2
dv = ∫ a.dt or
t1
t2
v2 − v1 = ∫ a.dt
t1
where v 1 and v 2 are the velocities of the moving body at time intervals t1 and t2 respectively.
The right hand side of the above expression represents the area (PQQ1P1) under the a-t curve
between the time intervals t1 and t2 . Thus the area under the a-t curve between any two ordinates
represents the change in velocity of the moving body. If the initial and final velocities of the body are
u and v, then the above expression may be written as
v − u = ∫ a.d t = Area under a-t curve A B = Area OABC
t
0
Example 2.1. A car starts from rest and
accelerates uniformly to a speed of 72 km. p.h. over
a distance of 500 m. Calculate the acceleration and
the time taken to attain the speed.
If a further acceleration raises the speed to
90 km. p.h. in 10 seconds, find this acceleration and
the further distance moved. The brakes are now
applied to bring the car to rest under uniform
retardation in 5 seconds. Find the distance travelled
during braking.
Solution. Given : u = 0 ; v = 72 km. p.h. = 20 m/s ; s = 500 m
First of all, let us consider the motion of the car from rest.
Acceleration of the car
Let
a = Acceleration of the car.
v 2 = u2 + 2 a.s
We know that
∴
(20)2 = 0 + 2a × 500 = 1000 a
or
a = (20)2 / 1000 = 0.4 m/s2 Ans.
Time taken by the car to attain the speed
Let
t = Time taken by the car to attain the speed.
We know that
v = u + a.t
∴
20 = 0 + 0.4 × t
or
t = 20/0.4 = 50 s Ans.
Now consider the motion of the car from 72 km.p.h. to 90 km.p.h. in 10 seconds.
Given : * u = 72 km.p.h. = 20 m/s ; v = 96 km.p.h. = 25 m/s ; t = 10 s
Acceleration of the car
Let
a = Acceleration of the car.
We know that
v = u + a.t
25 = 20 + a × 10
a = (25 – 20)/10 = 0.5 m/s2 Ans.
or
Distance moved by the car
We know that distance moved by the car,
1
1
2
2
s = u.t + a.t 2 = 20 × 10 + × 0.5 (10) 2 = 225 m Ans.
*
It is the final velocity in the first case.
14
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Theory of Machines
Now consider the motion of the car during the application of brakes for brining it to rest in
5 seconds.
Given : *u = 25 m/s ; v = 0 ; t = 5 s
We know that the distance travelled by the car during braking,
u+v
25 + 0
×t =
× 5 = 62.5 m Ans.
2
2
Example 2.2. The motion of a particle is given by a = t3 – 3t2 + 5, where a is the acceleration
2
in m/s and t is the time in seconds. The velocity of the particle at t = 1 second is 6.25 m/s, and the
displacement is 8.30 metres. Calculate the displacement and the velocity at t = 2 seconds.
s=
Solution. Given : a = t3 – 3t2 + 5
We know that the acceleration, a = dv/dt. Therefore the above equation may be written as
dv 3
= t − 3t 2 + 5
dt
Integrating both sides
or
dv = (t 3 − 3t 2 + 5)dt
t 4 3t 3
t4
−
+ 5t + C1 = − t 3 + 5 t + C1
...(i)
4
3
4
where C1 is the first constant of integration. We know that when t = 1 s, v = 6.25 m/s. Therefore
substituting these values of t and v in equation (i),
6.25 = 0.25 – 1 + 5 + C1 = 4.25 + C1
or
C1 = 2
Now substituting the value of C1 in equation (i),
v=
v=
t4 3
− t + 5t + 2
4
...(ii)
Velocity at t = 2 seconds
Substituting the value of t = 2 s in the above equation,
v=
24
− 23 + 5 × 2 + 2 = 8 m/s Ans.
4
Displacement at t = 2 seconds
We know that the velocity, v = ds/dt, therefore equation (ii) may be written as
ds t 4 3
= − t + 5t + 2
dt 4
t4
or ds = − t 3 + 5t + 2 dt
4
Integrating both sides,
s=
t 5 t 4 5t 2
− +
+ 2 t + C2
20 4
2
...(iii)
where C2 is the second constant of integration. We know that when t = 1 s, s = 8.30 m. Therefore
substituting these values of t and s in equation (iii),
8.30 =
*
1 1 5
− + + 2 + C2 = 4.3 + C2
20 4 2
It is the final velocity in the second case.
or
C2 = 4
Chapter 2 : Kinematics of Motion
l
15
Substituting the value of C2 in equation (iii),
t 5 t 4 5t 2
− +
+ 2t + 4
20 4
2
Substituting the value of t = 2 s, in this equation,
s=
25 2 4 5 × 2 2
−
+
+ 2 × 2 + 4 = 15.6 m Ans.
20 4
2
Example 2.3. The velocity of a
train travelling at 100 km/h decreases by
10 per cent in the first 40 s after application of the brakes. Calculate the velocity
at the end of a further 80 s assuming that,
during the whole period of 120 s, the retardation is proportional to the velocity.
s=
Solution. Given : Velocity in the
beginning (i.e. when t = 0), v 0 = 100 km/h
Since the velocity decreases by 10
per cent in the first 40 seconds after the
application of brakes, therefore velocity at the end of 40 s,
v 40 = 100 × 0.9 = 90 km/h
Let
v 120 = Velocity at the end of 120 s (or further 80s).
Since the retardation is proportional to the velocity, therefore,
dv
dv
= k .v
= − k .dt
or
dt
v
where k is a constant of proportionality, whose value may be determined from the given conditions.
Integrating the above expression,
a=−
loge v = – k.t + C
... (i)
where C is the constant of integration. We know that when t = 0, v = 100 km/h. Substituting these
values in equation (i),
or
C = 2.3 log 100 = 2.3 × 2 = 4.6
loge100 = C
We also know that when t = 40 s, v = 90 km/h. Substituting these values in equation (i),
loge 90 = – k × 40 + 4.6
...( 3 C = 4.6 )
2.3 log 90 = – 40k + 4.6
or
or
or
k=
4.6 − 2.3log 90 4.6 − 2.3 × 1.9542
=
= 0.0026
40
40
Substituting the values of k and C in equation (i),
loge v = – 0.0026 × t + 4.6
2.3 log v = – 0.0026 × t + 4.6
... (ii)
Now substituting the value of t equal to 120 s, in the above equation,
2.3 log v 120 = – 0.0026 × 120 + 4.6 = 4.288
log v 120 = 4.288 / 2.3 = 1.864
∴
v 120 = 73.1 km/h Ans.
... (Taking antilog of 1.864)
16
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Theory of Machines
Example 2.4. The acceleration (a) of a slider block and its displacement (s) are related by
the expression, a = k s , where k is a constant. The velocity v is in the direction of the displacement
and the velocity and displacement are both zero when time t is zero. Calculate the displacement,
velocity and acceleration as functions of time.
Solution. Given : a = k s
We know that acceleration,
a = v×
∴
dv
ds
k s = v×
or
dv
dv ds dv
= v×
... 3 = ×
ds
dt dt ds
dv
ds
v × dv = k.s1/2 ds
Integrating both sides,
v 2 k .s 3 / 2
=
+ C1
... (i)
∫0
2
3/ 2
where C1 is the first constant of integration whose value is to be determined from the given conditions
of motion. We know that s = 0, when v = 0. Therefore, substituting the values of s and v in equation (i),
we get C1 = 0.
v
∴
v.dv = k ∫ s1/ 2 ds
or
v2 2 3 / 2
= k .s
2 3
or
v=
4k
× s3 / 4
3
... (ii)
Displacement, velocity and acceleration as functions of time
We know that
4k
ds
=v=
× s3 / 4
3
dt
4k
ds
=
dt
3/ 4
3
s
Integrating both sides,
∴
s
∫0 s−3 / 4 ds =
4k
3
or
... [From equation (ii)]
s −3 / 4 ds =
4k
dt
3
t
∫0 dt
4k
s1/ 4
=
× t + C2
1/ 4
3
...(iii)
where C2 is the second constant of integration. We know that displacement, s = 0 when t = 0. Therefore, substituting the values of s and t in equation (iii), we get C2 = 0.
4k
s1/ 4
k 2 .t 4
=
× t or s =
Ans.
1/ 4
3
144
We know that velocity,
∴
and acceleration,
v=
ds k 2
k 2 .t 3
=
× 4t 3 =
Ans.
36
dt 144
k 2 .t 4
... Differentiating
144
a=
dv k 2
k 2 .t 2
=
× 3 t2 =
Ans.
12
dt 36
k 2 .t 3
... Differentiating
36
Chapter 2 : Kinematics of Motion
l
17
Example 2.5. The cutting stroke of a planing
machine is 500 mm and it is completed in 1 second.
The planing table accelerates uniformly during the first
125 mm of the stroke, the speed remains constant during
the next 250 mm of the stroke and retards uniformly during
the last 125 mm of the stroke. Find the maximum cutting
speed.
Solution. Given : s = 500 mm ; t = 1 s ;
s1 = 125 mm ; s2 = 250 mm ; s3 = 125 mm
Fig. 2.4 shows the acceleration-time and velocity-time graph for the planing table of a planing machine.
Let
v = Maximum cutting speed in mm/s.
Average velocity of the table during acceleration
and retardation,
Planing Machine.
vav = (0 + v ) / 2 = v / 2
Time of uniform acceleration t1 = s1 = 125 = 250 s
vav v / 2
v
Time of constant speed,
t2 =
s2 250
s
=
v
v
and time of uniform retardation,
t3 =
s3 125 250
s
=
=
vav v / 2
v
Fig. 2.4
Since the time taken to complete the stroke is 1 s, therefore
t1 + t2 + t3 = t
250 250 250
+
+
= 1 or v = 750 mm/s Ans.
v
v
v
2.12. Angular Displacement
It may be defined as the angle described by a particle from one point to another, with respect
to the time. For example, let a line OB has its inclination θ radians to the fixed line O A, as shown in
18
l
Theory of Machines
Fig. 2.5. If this line moves from OB to OC, through an angle δθ during
a short interval of time δt, then δθ is known as the angular
displacement of the line OB.
Since the angular displacement has both magnitude and
direction, therefore it is also a vector quantity.
2.13. Representation of Angular Displacement by
a Vector
Fig. 2.5. Angular
displacement.
In order to completely represent an angular displacement, by a vector, it must fix the following three conditions :
1. Direction of the axis of rotation. It is fixed by drawing a line perpendicular to the plane
of rotation, in which the angular displacement takes place. In other words, it is fixed along the axis
of rotation.
2. Magnitude of angular displacement. It is fixed by the length of the vector drawn along
the axis of rotation, to some suitable scale.
3. Sense of the angular displacement. It is fixed by a right hand screw rule. This rule
states that if a screw rotates in a fixed nut in a clockwise direction, i.e. if the angular displacement
is clockwise and an observer is looking along the axis of rotation, then the arrow head will point
away from the observer. Similarly, if the angular displacement is anti-clockwise, then the arrow
head will point towards the observer.
2.14. Angular Velocity
It may be defined as the rate of change of angular displacement with respect to time. It is
usually expressed by a Greek letter ω (omega). Mathematically, angular velocity,
ω = d θ / dt
Since it has magnitude and direction, therefore, it is a vector quantity. It may be represented
by a vector following the same rule as described in the previous article.
Note : If the direction of the angular displacement is constant, then the rate of change of magnitude of the
angular displacement with respect to time is termed as angular speed.
2.15. Angular Acceleration
It may be defined as the rate of change of angular velocity with respect to time. It is usually
expressed by a Greek letter α (alpha). Mathematically, angular acceleration,
d ω d d θ d 2θ
dθ
... 3 ω =
= = 2
dt
dt dt dt dt
It is also a vector quantity, but its direction may not be same as that of angular displacement
and angular velocity.
α=
2.16. Equations of Angular Motion
The following equations of angular motion corresponding to linear motion are important
from the subject point of view :
1
1. ω = ω0 + α.t
2. θ = ω 0 .t + α.t 2
2
ω
+
ω
( 0 )t
2
3. ω 2 = (ω0 ) + 2 α.θ
4. θ =
2
where
ω0 = Initial angular velocity in rad/s,
ω = Final angular velocity in rad/s,
l
Chapter 2 : Kinematics of Motion
19
t = Time in seconds,
θ = Angular displacement in time t seconds, and
α = Angular acceleration in rad / s2.
Note : If a body is rotating at the rate of N r.p.m. (revolutions per minute), then its angular velocity,
ω = 2πΝ / 60 rad/s
2.17. Relation between Linear Motion and Angular Motion
Following are the relations between the linear motion and the angular motion :
Particulars
Linear motion
Angular motion
ω0
ω
α
θ
ω = ω0 + α.t
Initial velocity
Final velocity
Constant acceleration
Total distance traversed
Formula for final velocity
u
v
a
s
v = u + a.t
Formula for distance traversed
s = u.t +
Formula for final velocity
v 2 = u2 + 2 a.s
1
2
θ = ω0. t +
a.t2
1
2
α.t2
ω = (ω0) 2 + 2 α.θ
2.18. Relation between Linear and Angular Quantities of Motion
Consider a body moving along a circular path from A to B as shown in Fig. 2.6.
Let
r = Radius of the circular path,
θ = Angular displacement in radians,
s = Linear displacement,
v = Linear velocity,
ω = Angular velocity,
a = Linear acceleration, and
α = Angular acceleration.
From the geometry of the figure, we know that
s=r.θ
We also know that the linear velocity,
Fig. 2.6. Motion of a body
along a circular path.
v=
ds d ( r.θ)
dθ
=
=r×
= r.ω
dt
dt
dt
... (i)
dv d ( r . ω )
dω
=
=r×
= r. α
... (ii)
dt
dt
dt
Example 2.6. A wheel accelerates uniformly from rest to 2000 r.p.m. in 20 seconds. What is its
angular acceleration? How many revolutions does the wheel make in attaining the speed of 2000 r.p.m.?
and linear acceleration,
a=
Solution. Given : N 0 = 0 or ω = 0 ; N = 2000 r.p.m. or ω = 2π × 2000/60 = 209.5 rad/s ; t = 20s
Angular acceleration
Let
We know that
∴
α = Angular acceleration in rad/s2.
ω = ω0 + α.t
or
209.5 = 0 + α × 20
α = 209.5 / 20 = 10.475 rad/s2 Ans.
20
l
Theory of Machines
Number of revolutions made by the wheel
We know that the angular distance moved by the wheel during 2000 r.p.m. (i.e. when
ω = 209.5 rad/s),
θ=
(ω 0 + ω ) t (0 + 209.5) 20
=
= 2095 rad
2
2
Since the angular distance moved by the wheel during one revolution is 2π radians, therefore
number of revolutions made by the wheel,
n = θ /2π = 2095/2π = 333.4 Ans.
2.19. Acceleration of a Particle along a Circular Path
Consider A and B, the two positions of a particle displaced through an angle δθ in time δt as
shown in Fig. 2.7 (a).
Let
r = Radius of curvature of the circular path,
v = Velocity of the particle at A , and
v + dv = Velocity of the particle at B.
The change of velocity, as the particle moves from A to B may be obtained by drawing the
vector triangle oab, as shown in Fig. 2.7 (b). In this triangle, oa represents the velocity v and ob
represents the velocity v + dv. The change of velocity in time δt is represented by ab.
Fig. 2.7. Acceleration of a particle along a circular path.
Now, resolving ab into two components i.e. parallel and perpendicular to oa. Let ac and cb
be the components parallel and perpendicular to oa respectively.
∴
ac = oc – oa = ob cos δθ – oa = (v + δv) cos δθ – v
and
cb = ob sin δθ = (v + δv) sin δθ
Since the change of velocity of a particle (represented by vector ab) has two mutually
perpendicular components, therefore the acceleration of a particle moving along a circular path has
the following two components of the acceleration which are perpendicular to each other.
1. Tangential component of the acceleration. The acceleration of a particle at any instant
moving along a circular path in a direction tangential to that instant, is known as tangential component
of acceleration or tangential acceleration.
∴ Tangential component of the acceleration of particle at A or tangential acceleration at A,
ac ( v + δv )cos δθ − v
=
δt
δt
In the limit, when δt approaches to zero, then
at =
... (i)
at = dv / dt = α.r
2. Normal component of the acceleration. The acceleration of a particle at any instant moving along a circular path in a direction normal to the tangent at that instant and directed towards the
centre of the circular path (i.e. in the direction from A to O) is known as normal component of the
Chapter 2 : Kinematics of Motion
l
21
acceleration or normal acceleration. It is also called radial or centripetal acceleration.
∴ Normal component of the acceleration of the particle at A or normal (or radial or centripetal) acceleration at A ,
cb ( v + δv ) sin θ
=
δt
δt
In the limit, when δt approaches to zero, then
an =
an = v ×
dθ
v v2
= v.ω = v × =
= ω2 .r
dt
r
r
... (ii)
... [3 d θ / dt = ω, and ω = v / r ]
Since the tangential acceleration (at) and the normal acceleration (an) of the particle at any instant A are perpendicular to each other,
as shown in Fig. 2.8, therefore total acceleration of the particle (a) is
equal to the resultant acceleration of at and an.
∴ Total acceleration or resultant acceleration,
a=
(at )2 + (an )2
Fig. 2.8. Total acceleration
of a particle.
and its angle of inclination with the tangential acceleration is given by
tan θ = an/at or θ = tan–1 (an/at)
The total acceleration or resultant acceleration may also be obtained by the vector sum of at
and an.
Notes : 1. From equations (i) and (ii) we see that the tangential acceleration (at ) is equal to the rate of change of
the magnitude of the velocity whereas the normal or radial or centripetal acceleration (an) depends upon its
instantaneous velocity and the radius of curvature of its path.
2. When a particle moves along a straight path, then the radius of curvature is infinitely great. This
means that v 2/r is zero. In other words, there will be no normal or radial or centripetal acceleration. Therefore,
the particle has only tangential acceleration (in the same direction as its velocity and displacement) whose value
is given by
at = dv/dt = α.r
3. When a particle moves with a uniform velocity, then dv/dt will be zero. In other words, there will be
no tangential acceleration; but the particle will have only normal or radial or centripetal acceleration, whose
value is given by
an = v 2/r = v.ω = ω2 r
Example 2.7. A horizontal bar 1.5 metres long and of small cross-section rotates about
vertical axis through one end. It accelerates uniformly from 1200 r.p.m. to 1500 r.p.m. in an interval
of 5 seconds. What is the linear velocity at the beginning and end of the interval ? What are the
normal and tangential components of the acceleration of the mid-point of the bar after 5 seconds
after the acceleration begins ?
Solution. Given : r = 1.5 m ; N 0 = 1200 r.p.m. or ω0 = 2 π × 1200/60 = 125.7 rad/s ;
N = 1500 r.p.m. or ω = 2 π × 1500/60 = 157 rad/s ; t = 5 s
Linear velocity at the beginning
We know that linear velocity at the beginning,
v 0 = r . ω0 = 1.5 × 125.7 = 188.6 m/s Ans.
Linear velocity at the end of 5 seconds
We also know that linear velocity after 5 seconds,
v 5 = r . ω = 1.5 × 157 = 235.5 m/s Ans.
22
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Theory of Machines
Tangential acceleration after 5 seconds
Let
α = Constant angular acceleration.
We know that
ω = ω0+ α.t
157 = 125.7 + α × 5
or
α = (157 – 125.7) /5 = 6.26 rad/s2
Radius corresponding to the middle point,
r = 1.5 / 2 = 0.75 m
∴ Tangential acceleration = α. r = 6.26 × 0.75 = 4.7 m/s2 Ans.
Radial acceleration after 5 seconds
Radial acceleration = ω2 . r = (157)2 0.75 = 18 487 m/s2 Ans.
EXERCISES
1.
A winding drum raises a cage through a height of 120 m. The cage has, at first, an acceleration
of 1.5 m/s2 until the velocity of 9 m/s is reached, after which the velocity is constant until the
cage nears the top, when the final retardation is 6 m/s2. Find the time taken for the cage to reach
the top.
[ Ans. 17.1s ]
2.
The displacement of a point is given by s = 2t3 + t2 + 6, where s is in metres and t in seconds.
Determine the displacement of the point when the velocity changes from 8.4 m/s to 18 m/s. Find also
the acceleration at the instant when the velocity of the particle is 30 m/s. [ Ans. 6.95 m ; 27 m/s2 ]
3.
A rotating cam operates a follower which moves in a straight line. The stroke of the follower is 20
mm and takes place in 0.01 second from rest to rest. The motion is made up of uniform acceleration
for 1/4 of the time, uniform velocity for 1 2 of the time followed by uniform retardation. Find the
maximum velocity reached and the value of acceleration and retardation.
[ Ans. 2.67 m/s ; 1068 m/s2 ; 1068 m/s2 ]
4.
A cage descends a mine shaft with an acceleration of 0.5 m/s2. After the cage has travelled 25 metres,
a stone is dropped from the top of the shaft. Determine : 1. the time taken by the stone to hit the cage,
and 2. distance travelled by the cage before impact.
[ Ans. 2.92 s ; 41.73 m ]
5.
The angular displacement of a body is a function of time and is given by equation :
θ = 10 + 3 t + 6 t2, where t is in seconds.
Determine the angular velocity, displacement and acceleration when t = 5 seconds. State whether or
not it is a case of uniform angular acceleration.
[Ans. 63 rad/s ; 175 rad ; 12 rad/s2]
6.
A flywheel is making 180 r.p.m. and after 20 seconds it is running at 140 r.p.m. How many revolutions will it make, and what time will elapse before it stops, if the retardation is uniform ?
7.
A locomotive is running at a constant speed of 100 km / h. The diameter of driving wheels is 1.8 m. The
stroke of the piston of the steam engine cylinder of the locomotive is 600 mm. Find the centripetal acceleration of the crank pin relative to the engine frame.
[ Ans. 288 m/s2 ]
[ Ans. 135 rev. ; 90 s ]
DO YOU KNOW ?
1.
2.
3.
4.
5.
6.
Distinguish clearly between speed and velocity. Give examples.
What do you understand by the term ‘acceleration’ ? Define positive acceleration and negative acceleration.
Define ‘angular velocity’ and ‘angular acceleration’. Do they have any relation between them ?
How would you find out the linear velocity of a rotating body ?
Why the centripetal acceleration is zero, when a particle moves along a straight path ?
A particle moving with a uniform velocity has no tangential acceleration. Explain clearly.
l
Chapter 2 : Kinematics of Motion
23
OBJECTIVE TYPE QUESTIONS
1. The unit of linear acceleration is
(a) kg-m
(b) m/s
(c) m/s2
(d) rad/s2
2. The angular velocity (in rad/s) of a body rotating at N r.p.m. is
(a) π N/60
(b) 2 π N/60
(c) π N/120
(d) π N/180
3. The linear velocity of a body rotating at ω rad/s along a circular path of radius r is given by
(a) ω.r
(b) ω/r
(c) ω2.r
(d) ω 2/r
4. When a particle moves along a straight path, then the particle has
(a) tangential acceleration only
(b) centripetal acceleration only
(c) both tangential and centripetal acceleration
5. When a particle moves with a uniform velocity along a circular path, then the particle has
(a) tangential acceleration only
(b) centripetal acceleration only
(c)
both tangential and centripetal acceleration
ANSWERS
1. (c)
2. (b)
3. (a)
4. (a)
5. (b)
GO To FIRST
CONTENTS
CONTENTS
24
Theory of Machines
3
Kinetics of
Motion
Features
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
21.
22.
23.
24.
Introduction.
Newton's Laws of Motion.
Mass and Weight.
Momentum.
Force.
Absolute and Gravitational
Units of Force.
Moment of a Force.
Couple.
Centripetal and Centrifugal
Force.
Mass Moment of Inertia.
Angular Momentum or
Moment of Momentum.
Torque.
Work.
Power.
Energy.
Principle of Conservation of
Energy.
Impulse and Impulsive Force.
Principle of Conservation of
Momentum.
Energy Lost by Friction
Clutch During Engagement.
Torque Required to
Accelerate a Geared System.
Collision of Two Bodies.
Collision of Inelastic Bodies.
Collision of Elastic Bodies.
Loss of Kinetic Energy
During Elastic Impact.
3.1.
Introduction
In the previous chapter we have discussed the
kinematics of motion, i.e. the motion without considering
the forces causing the motion. Here we shall discuss the
kinetics of motion, i.e. the motion which takes into
consideration the forces or other factors, e.g. mass or weight
of the bodies. The force and motion is governed by the three
laws of motion.
3.2.
Newton’s Laws of Motion
Newton has formulated three laws of motion, which
are the basic postulates or assumptions on which the whole
system of kinetics is based. Like other scientific laws, these
are also justified as the results, so obtained, agree with the
actual observations. These three laws of motion are as
follows:
1. Newton’s First Law of Motion. It states, “Every
body continues in its state of rest or of uniform motion in
a straight line, unless acted upon by some external force.”
This is also known as Law of Inertia.
The inertia is that property of a matter, by virtue of
which a body cannot move of itself, nor change the motion
imparted to it.
24
CONTENTS
CONTENTS
Chapter 3 : Kinetics of Motion
25
2. Newton’s Second Law of Motion. It states,
“The rate of change of momentum is directly
proportional to the impressed force and takes place in
the same direction in which the force acts.”
3. Newton’s Third Law of Motion. It states, “To
every action, there is always an equal and opposite
reaction.”
3.3.
Mass and Weight
Sometimes much confu-sion and misunderstanding is created, while using the various systems of units
in the measurements of force and mass. This happens
because of the lack of clear understanding of the
difference between the mass and the weight. The
following definitions of mass and weight should be
clearly understood :
The above picture shows space shuttle.
1. Mass. It is the amount of matter contained in a All space vehicles move based on
given body, and does not vary with the change in its Newton’s third laws.
position on the earth's surface. The mass of a body is
measured by direct comparison with a standard mass by using a lever balance.
2. Weight. It is the amount of pull, which the earth exerts upon a given body. Since the pull
varies with distance of the body from the centre of the earth, therefore the weight of the body will
vary with its position on the earth’s surface (say latitude and elevation). It is thus obvious, that the
weight is a force.
The earth’s pull in metric units at sea
level and 45° latitude has been adopted as one
force unit and named as one kilogram of force.
Thus, it is a definite amount of force. But, unfortunately, it has the same name as the unit of mass.
The weight of a body is measured by the use of a
spring balance which indicates the varying tension in the spring as the body is moved from place
to place.
Note: The confusion in the units of mass and weight
is eliminated, to a great extent, in S.I. units. In this system, the mass is taken in kg and force in newtons.
The relation between the mass (m) and the weight (W) of a body is
W = m.g
or
m = W/g
where W is in newtons, m is in kg and g is acceleration due to gravity.
3.4.
Momentum
It is the total motion possessed by a body. Mathematically,
Momentum = Mass × Velocity
Let
m = Mass of the body,
u = Initial velocity of the body,
v = Final velocity of the body,
a = Constant acceleration, and
t = Time required (in seconds) to change the velocity from u to v.
26
Theory of Machines
Now,
and
initial momentum = m.u
final momentum = m.v
∴ Change of momentum = m.v – m.u
and rate of change of momentum =
3.5.
m.v − m.u m (v − u )
=
= m.a
t
t
v −u
... ∵
= a
t
Force
It is an important factor in the field of Engineering-science, which may be defined as an
agent, which produces or tends to produce, destroy or tends to destroy motion.
W, weight (force)
applied force, F
f, friction force
N, normal force
According to Newton’s Second Law of Motion, the applied force or impressed force is
directly proportional to the rate of change of momentum. We have discussed in Art. 3.4, that the rate
of change of momentum
= m.a
where
m = Mass of the body, and
a = Acceleration of the body.
∴
Force , F ∝ m.a or F = k.m.a
where k is a constant of proportionality.
For the sake of convenience, the unit of force adopted is such that it produces a unit
acceleration to a body of unit mass.
∴
F = m.a = Mass × Acceleration
In S.I. system of units, the unit of force is called newton (briefly written as N). A newton
may be defined as the force while acting upon a mass of one kg produces an acceleration of
1 m/s2 in the direction of which it acts. Thus
1 N = 1 kg × 1 m/s2 = 1 kg-m/s2
Note: A force equal in magnitude but opposite in direction and collinear with the impressed force producing
the acceleration, is known as inertia force. Mathematically,
Inertia force = – m.a
3.6.
Absolute and Gravitational Units of Force
We have already discussed, that when a body of mass 1 kg is moving with an acceleration of
1 m/s2, the force acting on the body is one newton (briefly written as N). Therefore, when the same
body is moving with an acceleration of 9.81 m/s2, the force acting on the body is 9.81 newtons. But
we denote 1 kg mass, attracted towards the earth with an acceleration of 9.81 m/s2 as 1 kilogramforce (briefly written as kgf) or 1 kilogram-weight (briefly written as kg-wt). It is thus obvious that
1 kgf = 1 kg × 9.81 m/s2 = 9.81 kg-m/s2 = 9.81 N ... (∵ 1 N = 1 kg-m/s2 )
The above unit of force i.e. kilogram-force (kgf ) is called gravitational or engineer's unit
Chapter 3 : Kinetics of Motion
l
27
of force, whereas newton is the absolute or scientific or S.I. unit of force. It is thus obvious, that the
gravitational units are ‘g’ times the unit of force in the absolute or S.I. units.
It will be interesting to know that the mass of a body in absolute units is numerically equal
to the weight of the same body in gravitational units.
For example, consider a body whose mass, m = 100 kg.
∴ The force, with which it will be attracted towards the centre of the earth,
F = m.a = m.g = 100 × 9.81 = 981 N
Now, as per definition, we know that the weight of a body is the force, by which it is attracted
towards the centre of the earth. Therefore, weight of the body,
W = 981 N = 981 / 9.81 = 100 kgf
... (∵ 1 kgf = 9.81 N)
In brief, the weight of a body of mass m kg at a place where gravitational acceleration is ‘g’
m/s2 is m.g newtons.
3.7.
Moment of a Force
It is the turning effect produced by a force, on the body, on which it acts. The moment of a
force is equal to the product of the force and the perpendicular distance of the point about which the
moment is required, and the line of action of the force. Mathematically,
Moment of a force = F × l
where
F = Force acting on the body, and
l = Perpendicular distance of the
point and the line of action of
the force, as shown in Fig. 3.1.
Fig. 3.1. Moment of a force.
3.8.
Couple
The two equal and opposite parallel forces, whose lines of
action are different, form a couple, as shown in Fig. 3.2.
The perpendicular distance (x) between the lines of action of
two equal and opposite parallel forces (F) is known as arm of the
couple. The magnitude of the couple (i.e. moment of a couple) is
the product of one of the forces and the arm of the couple.
Mathematically,
Fig. 3.2. Couple.
Moment of a couple = F × x
A little consideration will show, that a couple does not produce any translatory motion (i.e.
motion in a straight line). But, a couple produces a motion of rotation of the body, on which it acts.
3.9.
Centripetal and Centrifugal Force
Consider a particle of mass m moving with a linear velocity
v in a circular path of radius r.
We have seen in Art. 2.19 that the centripetal acceleration,
ac = v 2/r = ω2.r
and
Force = Mass × Acceleration
∴ Centripetal force = Mass × Centripetal acceleration
or
Fc = m.v2/r = m.ω2.r
28
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Theory of Machines
This force acts radially inwards and is essential for circular motion.
We have discussed above that the centripetal force acts radially inwards. According to
Newton's Third Law of Motion, action and reaction are equal and opposite. Therefore, the particle
must exert a force radially outwards of equal magnitude. This force is known as centrifugal force
whose magnitude is given by
Fc= m.v 2/r = m.ω2r
3.10. Mass Moment of Inertia
It has been established since long that a rigid body is
composed of small particles. If the mass of every particle of a
body is multiplied by the square of its perpendicular distance
from a fixed line, then the sum of these quantities(for the whole
body) is known as mass moment of inertia of the body. It is
denoted by I.
Consider a body of total mass m. Let it is composed of
small particles of masses m 1, m2, m 3, m 4 etc. If k 1, k 2, k 3, k 4 are
the distances of these masses from a fixed line, as shown in Fig.
3.3, then the mass moment of inertia of the whole body is given
by
Fig. 3.3. Mass moment of inertia.
I = m 1 (k 1)2 + m 2(k 2)2 + m 3 (k 3)2 + m 4 (k 4)2 +....
If the total mass of body may be assumed to concentrate at one point (known as centre of
mass or centre of gravity), at a distance k from the given axis, such that
m.k2 = m 1(k 1)2 + m 2(k 2)2 + m 3(k 3)2 + m 4 (k 4)2 +...
then
I = m.k2
The distance k is called the radius of gyration. It may be defined as the distance, from a
given reference, where the whole mass of body is assumed to be concentrated to give the same
value of I.
The unit of mass moment of inertia in S.I. units is kg-m2.
Notes : 1. If the moment of inertia of a body about an axis through its centre of gravity is known, then the
moment of inertia about any other parallel axis may be obtained by using a parallel axis theorem i.e. moment
of inertia about a parallel axis,
Ip = IG + m.h2
where
IG = Moment of inertia of a body about an axis through its centre of gravity, and
h = Distance between two parallel axes.
2. The following are the values of I for simple cases :
(a) The moment of inertia of a thin disc of radius r, about an axis through its centre of gravity and
perpendicular to the plane of the disc is
I = m.r2 /2
and moment of inertia about a diameter,
I = m.r2/4
(b) The moment of inertia of a thin rod of length l, about an axis through its centre of gravity and
perpendicular to its length,
IG = m.l2/12
and moment of inertia about a parallel axis through one end of a rod,
Ip = m.l2/3
Chapter 3 : Kinetics of Motion
l
29
3. The moment of inertia of a solid cylinder of radius r and length l, about the longitudinal axis or polar
axis
= m.r2/2
and moment of inertia through its centre perpendicular to longitudinal axis
r2 l2
= +
4 12
3.11. Angular Momentum or Moment of Momentum
Consider a body of total mass m rotating with an angular velocity
of ω rad/s, about the fixed axis O as shown in Fig. 3.4. Since the body
is composed of numerous small particles, therefore let us take one of
these small particles having a mass dm and at a distance r from the axis
Fig. 3.4. Angular
of rotation. Let v is its linear velocity acting tangentially at any instant.
momentum.
We know that momentum is the product of mass and velocity, therefore
momentum of mass dm
= dm × v = dm × ω × r
... (3 v = ω.r)
and moment of momentum of mass dm about O
= dm × ω × r × r = dm × r2 × ω = Im × ω
where
Im = Mass moment of inertia of mass dm about O = dm × r2
∴ Moment of momentum or angular momentum of the whole body about O
= ∫ I m .ω= I .ω
∫ Im =
where
Mass moment of inertia of the
whole body about O.
Thus we see that the angular momentum or the moment of momentum is the product of mass
moment of inertia ( I ) and the angular velocity (ω) of the body.
3.12. Torque
It may be defined as the product of
force and the perpendicular distance of its line
of action from the given point or axis. A little
consideration will show that the torque is
equivalent to a couple acting upon a body.
Torque
The Newton’s Second Law of
Motion, when applied to rotating bodies,
states that the torque is directly proportional
to the rate of change of angular momentum.
Mathematically, Torque,
T∝
Same
force
applied
d ( É .ω)
dt
Since I is constant, therefore
T =I×
dω
= I .α
dt
Double
torque
... 3 d ω = α
dt
Double length
spanner
30
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Theory of Machines
The unit of torque (T ) in S.I. units is N-m when I is in kg-m2 and α in rad/s2.
3.13. Work
Whenever a force acts on a body and the body undergoes a displacement in the direction of the
force, then work is said to be done. For example, if a force F acting on a body causes a displacement x
of the body in the direction of the force, then
Work done = Force × Displacement = F × x
If the force varies linearly from zero to a maximum value of F, then
Work done =
0+F
2
×x=
1
2
×F×x
When a couple or torque ( T ) acting on a body causes the angular displacement (θ) about an
axis perpendicular to the plane of the couple, then
Work done = Torque × Angular displacement = T.θ
The unit of work depends upon the unit of force and displacement.
In S.I. system of units, the practical unit of work is N-m. It is the work done by a force of 1
newton, when it displaces a body through 1 metre. The work of 1 N-m is known as joule (briefly
written as J ) such that 1 N-m = 1 J.
Note: While writing the unit of work, it is general practice to put the unit of force first followed by the unit of
displacement (e.g. N-m).
3.14. Power
It may be defined as the rate of doing work or work done per unit time. Mathematically,
Power =
Work done
Time taken
In S.I. system of units, the unit of power is watt (briefly written as W) which is equal to 1 J/s
or 1 N-m/s. Thus, the power developed by a force of F (in newtons) moving with a velocity v m/s is
F.v watt. Generally a bigger unit of power called kilowatt (briefly written as kW) is used which is
equal to 1000 W.
Notes: 1. If T is the torque transmitted in N-m or J and ω is the angular speed in rad/s, then
Power, P = T.ω = T × 2 π N/60 watts
... (∵ ω = 2 π N/60)
where N is the speed in r.p.m.
2. The ratio of power output to power input is known as efficiency of a machine. It is always less than
unity and is represented as percentage. It is denoted by a Greek letter eta (η). Mathematically,
Efficiency, η =
Power output
Power input
3.15. Energy
It may be defined as the capacity to do work. The energy exists in many forms e.g. mechanical,
electrical, chemical, heat, light etc. But we are mainly concerned with mechanical energy.
The mechanical energy is equal to the work done on a body in altering either its position or
its velocity. The following three types of mechanical energies are important from the subject point
of view.
Chapter 3 : Kinetics of Motion
l
31
1. Potential energy. It is the energy possessed by a body for doing work, by virtue of its
position. For example, a body raised to some height above the ground level possesses potential
energy because it can do some work by falling on earth’s surface.
Let
W = Weight of the body,
m = Mass of the body, and
h = Distance through which the body falls.
Then potential energy,
P.E. = W.h = m.g.h
... (∵ W = m.g)
It may be noted that
(a) When W is in newtons and h in metres, then potential energy will be in N-m.
(b) When m is in kg and h in metres, then the potential energy will also be in N-m as
discussed below :
We know that potential energy,
P.E. = m.g .h = kg ×
m
s2
2. Strain energy. It is the potential energy
stored by an elastic body when deformed. A compressed spring possesses this type of energy, because it can do some work in recovering its original
shape. Thus if a compressed spring of stiffness s
newton per unit deformation (i.e. extension or compression) is deformed through a distance x by a load
W, then
× m = N− m
1kg–m
3 1 N =
s2
1
Strain energy = Work done = W .x
2
=
1
2
s. x 2
...(∵ W = s × x)
In case of a torsional spring of stiffness q N-m per unit angular deformation when twisted
through as angle θ radians, then
1 2
q.θ
2
3. Kinetic energy. It is the energy possessed by a body, for doing work, by virtue of its mass
and velocity of motion. If a body of mass m attains a velocity v from rest in time t, under the
influence of a force F and moves a distance s, then
Strain energy = Work done =
Work done = F.s = m.a.s
∴ Kinetic energy of the body or the kinetic energy of translation,
* v2 1
= m.v 2
K.E. = m.a.s = m × a ×
2a 2
*
We know that, v2 – u2 = 2 a.s
Since u = 0 because the body starts from rest, therefore,
v 2 = 2 a.s or s = v 2/2a
... (∵ F = m.a)
32
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Theory of Machines
It may be noted that when m is in kg and v in m/s, then kinetic energy will be in N-m as
discussed below:
We know that kinetic energy,
K.E. =
1
m 2 kg - m
× m = N-m
m.v 2 = kg × 2 =
2
s
s2
1kg-m
... 3 1N =
s2
Notes : 1. When a body of mass moment of inertia I (about a given axis) is rotated about that axis, with an
angular velocity ω, then it possesses some kinetic energy. In this case,
1
2
Kinetic energy of rotation = I .ω
2
2. When a body has both linear and angular motions e.g. in the locomotive driving wheels and
wheels of a moving car, then the total kinetic energy of the body is equal to the sum of kinetic energies of
translation and rotation.
∴
Total kinetic energy =
1
1
m.v 2 + I .ω2
2
2
Example 3.1. The flywheel of a steam engine has a radius
of gyration of 1 m and mass 2500 kg. The starting torque of the
steam engine is 1500 N-m and may be assumed constant.
Determine : 1. Angular acceleration of the flywheel, and 2. Kinetic
energy of the flywheel after 10 seconds from the start.
Solution. Given : k = 1 m ; m = 2500 kg ; T = 1500 N-m
1. Angular acceleration of the flywheel
Let
α = Angular acceleration of the flywheel.
Flywheel
We know that mass moment of inertia of the flywheel,
I = m.k 2 = 2500 × 12 = 2500 kg-m2
We also know that torque ( T ),
1500 = I .α = 2500 × α
or
α = 1500 / 2500 = 0.6 rad/s2 Ans.
2. Kinetic energy of the flywheel after 10 seconds from start
First of all, let us find the angular speed of the flywheel (ω2 ) after t = 10 seconds from the
start (i.e. ω1 = 0 ).
We know that ω2 = ω1 + α.t = 0 + 0.6 × 10 = 6 rad/s
∴ Kinetic energy of the flywheel,
1
1
I (ω2 ) 2 = × 2500 × 62 = 45 000 J = 45 kJ Ans.
2
2
Example 3.2. A winding drum raises a cage of mass 500 kg through a height of 100 metres.
The mass of the winding drum is 250 kg and has an effective radius of 0.5 m and radius of gyration
is 0.35 m. The mass of the rope is 3 kg/m.
E=
The cage has, at first, an acceleration of 1.5 m/s2 until a velocity of 10 m/s is reached, after
which the velocity is constant until the cage nears the top and the final retardation is 6 m/s2. Find
1. The time taken for the cage to reach the top, 2. The torque which must be applied to the drum at
starting; and 3. The power at the end of acceleration period.
Solution. Given : m C = 500 kg ; s = 100 m ; m D = 250 kg ; r = 0.5 m ; k = 0.35 m,
m = 3 kg/m
Chapter 3 : Kinetics of Motion
l
33
Fig. 3.5
Fig. 3.5 shows the acceleration-time and velocity-time graph for the cage.
1. Time taken for the cage to reach the top
Let
t = Time taken for the cage to reach the top = t1 + t2 + t3
where
t1 = Time taken for the cage from initial velocity of u1 = 0 to final
velocity of v1 = 10 m/s with an acceleration of a1 = 1.5 m/s2,
t2 = Time taken for the cage during constant velocity of v 2 = 10 m/s until the
cage nears the top, and
t3 = Time taken for the cage from initial velocity of u3 = 10 m/s to final velocity
of v 3 = 0 with a retardation of a3 = 6 m/s2.
We know that v1 = u1 + a1.t1
10 = 0 + 1.5 t1 or t1 = 10/1.5 = 6.67 s
and distance moved by the cage during time t1,
s1 =
Similarly,
10 + 0
v1 + u1
× t1 =
× 6.67 = 33.35 m
2
2
v 3 = u3 + a3.t3
0 = 10 – 6 × t3 or t3 = 10/6 = 1.67 s
s3 =
and
0 + 10
v3 + u3
× t3 =
× 1.67 = 8.35 m
2
2
Now, distance travelled during constant velocity of v 2 = 10 m/s,
s2 = s − s1 − s3 = 100 − 33.35 − 8.35 = 58.3m
We know that s2 = v 2.t2
or
t2 = s2/v 2 = 58.3/10 = 5.83 s
∴ Time taken for the cage to reach the top,
t = t1 + t2 + t3 = 6.67 + 5.83 + 1.67 = 14.17 s Ans.
2. Torque which must be applied to the drum at starting
Let
where
T = Torque which must be applied to the drum at starting = T 1 + T2 + T 3,
T 1 = Torque to raise the cage and rope at uniform speed,
T 2 = Torque to accelerate the cage and rope, and
T 3 = Torque to accelerate the drum.
34
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Theory of Machines
Since the mass of rope, m = 3 kg/m, therefore total mass of the rope for 100 metres,
m R = m.s = 3 × 100 = 300 kg
We know that the force to raise cage and rope at uniform speed,
F1 = (m C + m R) g = (500 + 300) 9.81 = 7850 N
∴ Torque to raise cage and rope at uniform speed,
T 1 = F1.r = 7850 × 0.5 = 3925 N-m
Force to accelerate cage and rope,
F2 = (m C + m R) a1 = (500 + 300) 1.5 = 1200 N
∴ Torque to accelerate the cage and rope,
T 2 = F2.r = 1200 × 0.5 = 600 N-m
We know that mass moment of inertia of the drum,
I = m D.k 2 = 250 (0.35)2 = 30.6 kg-m2
and angular acceleration of the drum,
a1 1.5
=
= 3 rad/s2
r 0.5
∴ Torque to accelerate the drum,
α=
T 3 = I.α = 30.6 × 3 = 91.8 N-m
and total torque which must be applied to the drum at starting,
T = T 1 + T 2 + T 3 = 3925 + 600 + 91.8 = 4616.8 N-m Ans.
3. Power at the end of acceleration period
When the acceleration period is just finishing, the drum torque will be reduced because
there will be s1 = 33.35 m of rope less for lifting. Since the mass of rope is 3 kg/m, therefore mass of
33.35 m rope,
m 1 = 3 × 33.35 = 100.05 kg
∴ Reduction of torque,
T 4 = (m 1.g + m 1.a1) r = (100.05 × 9.81 + 100.05 × 1.5) 0.5
= 565.8 N-m
and angular velocity of drum,
ω = v / 2πr = 10 / 2π × 0.5 = 3.18 rad/s
We know that power = T 4.ω = 565.8 × 3.18 = 1799 W = 1.799 kW Ans.
Example 3.3. A riveting machine is driven by a 4 kW motor. The moment of inertia of the
rotating parts of the machine is equivalent to 140 kg-m2 at the shaft on which the flywheel is mounted.
At the commencement of operation, the flywheel is making 240 r.p.m. If closing a rivet occupies
1 second and consumes 10 kN-m of energy, find the reduction of speed of the flywheel. What is the
maximum rate at which the rivets can be closed ?
Solution : Given : P = 4 kW = 4000 W ; I = 140 kg-m2 ; N 1 = 240 r.p.m. or ω1 = 2π ×
240/60 = 25.14 rad/s
Reduction of speed of the flywheel
Let
ω2 = Angular speed of the flywheel immediately after closing a rivet.
Chapter 3 : Kinetics of Motion
35
l
Since the power of motor is 4000 W, therefore energy supplied by motor in 1 second,
E1 = 4000 N-m
... (∵ 1 W = 1 N-m/s)
We know that energy consumed in closing a rivet in 1 second,
E2 = 10 kN-m = 10 000 N-m
∴ Loss of kinetic energy of the flywheel during the operation,
E = E2 – E1 = 10 000 – 4000 = 6000 N-m
We know that kinetic energy of the flywheel at the commencement of operation
=
1
1
I (ω1)2 = × 140 (25.14)2 = 44 240 N-m
2
2
∴ Kinetic energy of the flywheel at the end of operation
= 44 240 – 6000 = 38 240 N-m
... (i)
We also know that kinetic energy of the flywheel at the end of operation
=
1
1
I (ω2)2 = × 140 (ω2)2 = 70 (ω2)2
2
2
... (ii)
Equating equations (i) and (ii),
70 (ω2)2 = 38 240
or
(ω2)2 = 38 240/70 = 546.3 and ω = 23.4 rad/s
∴ Reduction of speed
= ω1 – ω2 = 25.14 – 23.4 = 1.74 rad/s
= 1.74 × 60/2π = 16.6 r.p.m. Ans.
... (∵ ω = 2π N/60)
Maximum rate at which the rivets can be closed
Maximum rate at which the rivets can be closed per minute
=
Energy supplied by motor per min 4000 × 60
=
= 24 Ans.
Energy consumed to close a rivet
10 000
Example 3.4. A wagon of mass 14 tonnes is hauled up an incline of 1 in 20 by a rope which
is parallel to the incline and is being wound round a drum of 1 m diameter. The drum, in turn, is
driven through a 40 to 1 reduction gear by an electric motor. The frictional resistance to the movement of the wagon is 1.2 kN, and the efficiency of the gear drive is 85 per cent. The bearing friction
at the drum and motor shafts may be neglected. The rotating parts of the drum have a mass of 1.25
tonnes with a radius of gyration of 450 mm and the rotating parts on the armature shaft have a mass
of 110 kg with a radius of gyration of 125 mm.
At a certain instant the wagon is moving up the slope with a velocity of 1.8 m/s and an
acceleration of 0.1 m/s2. Find the torque on the motor shaft and the power being developed.
Solution. Given : m = 14 t = 14 000 kg ; Slope = 1 in 20 ;
d = 1m or r = 0.5 m ; F = 1.2 kN = 1200 N ; η = 85% = 0.85 ;
m 1 = 1.25 t = 1250 kg ; k 1 = 450 mm = 0.45 m ; m 2 = 110 kg;
k 2 = 125 mm = 0.125 m; v = 1.8 m/s ; a = 0.1 m/s2
Torque on the motor shaft
We know that tension in the rope,
P1 = Forces opposing the motion as shown in
Fig. 3.6.
Fig. 3.6
36
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Theory of Machines
= Component of the weight down the slope
+ *Inertia force + Frictional resistance
= m. g . ×
=
1
+ m.a + F
20
14 000 × 9.81
+ 14 000 × 0.1 + 1200 = 9467 N
20
∴ Torque on the drum shaft to accelerate load,
T 1 = P1.r = 9467 × 0.5 = 4733.5 N-m
We know that mass moment of inertia of the drum,
I1 = m 1 (k 1)2 = 1250 (0.45)2 = 253 kg-m2
and angular acceleration of the drum,
α = a/r = 0.1/0.5 = 0.2 rad/s
∴ Torque on the drum to accelerate drum shaft,
T 2 = I1.α1 = 253 × 0.2 = 50.6 N-m
Since the drum is driven through a 40 to 1 reduction gear and the efficiency of the gear drive
is 85%, therefore
Torque on the armature to accelerate drum and load,
T3 = (T1 + T2 )
1
1
1
1
×
= (4733.5 + 50.6)
×
= 140.7 N -m
40 0.85
40 0.85
We know that mass moment of inertia of the armature,
I2 = m 2 (k 2)2 = 110 (0.125)2 = 1.72 kg-m2
and angular acceleration of the armature,
α2 =
0.1
a
× 40 =
× 40 = 8 rad / s2
0.5
r
... (∵ Armature rotates 40 times that of drum)
∴ Torque on the armature to accelerate armature shaft,
T 4 = I2.α2 = 1.72 × 8 = 13.76 N-m
and torque on the motor shaft
T = T 3 + T 4 = 140.7 + 13.76 = 154.46 N-m Ans.
Power developed by the motor
We know that angular speed of the motor,
ω=
v
1.8
× 40 =
× 40 = 144 rad/s
r
0.5
∴ Power developed by the motor
= T.ω = 154.46 × 144 = 22 240 W = 22.24 kW Ans.
*
Inertia force is equal and opposite to the accelerating force.
Chapter 3 : Kinetics of Motion
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37
Example 3.5. A road roller has a total
mass of 12 tonnes. The front roller has a mass of
2 tonnes, a radius of gyration of 0.4 m and a
diameter of 1.2 m. The rear axle, together with its
wheels, has a mass of 2.5 tonnes, a radius of
gyration of 0.6 m and a diameter of 1.5 m.
Calculate : 1. Kinetic energy of rotation of the
wheels and axles at a speed of 9 km/h, 2. Total
kinetic energy of road roller, and 3. Braking force
required to bring the roller to rest from 9 km/h in 6
m on the level.
Solution. Given : m = 12 t = 12 000 kg ;
m 1 = 2 t = 2000 kg ; k 1 = 0.4 m ; d1 = 1.2 m or r1 = 0.6 m ; m 2 = 2.5 t = 2500 kg ; k 2 = 0.6 m ; d2 = 1.5
m or r2 = 0.75 m ; v = 9 km/h = 2.5 m/s; s = 6 m
1. Kinetic energy of rotation of the wheels and axles
We know that mass moment of inertia of the front roller,
I1 = m1(k 1)2 = 2000 (0.4)2 = 320 kg-m2
and mass moment of inertia of the rear axle together with its wheels,
I2 = m 2 (k 2)2 = 2500 (0.6)2 = 900 kg -m2
Angular speed of the front roller,
ω1 = v/r1 = 2.5/0.6 = 4.16 rad/s
and angular speed of rear wheels,
ω2 = v/r2 = 2.5/0.75 = 3.3 rad/s
We know that kinetic energy of rotation of the front roller,
1
1
I1 ( ω1 )2 = × 320 (4.16)2 = 2770 N-m
2
2
and kinetic energy of rotation of the rear axle together with its wheels,
E1 =
E2 =
1
1
I 2 (ω2 )2 = × 900(3.3)2 = 4900 N-m
2
2
∴ Total kinetic energy of rotation of the wheels,
E = E1 + E2 = 2770 + 4900 = 7670 N-m Ans.
2. Total kinetic energy of road roller
We know that the kinetic energy of motion (i.e. kinetic energy of translation) of the road roller,
1
1
2
2
E3 = m.v 2 =
× 12 000 (2.5) 2 = 37 500 N-m
This energy includes the kinetic energy of translation of the wheels also, because the total
mass (m) has been considered.
∴ Total kinetic energy of road roller,
E4 = Kinetic energy of translation + Kinetic energy of rotation
= E3 + E = 37 500 + 7670 = 45 170 N-m Ans.
38
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Theory of Machines
3. Braking force required to bring the roller to rest
Let
F = Braking force required to bring the roller to rest, in newtons.
We know that the distance travelled by the road roller,
s=6m
... (Given)
∴ Work done by the braking force
= F × s = 6 F N-m
This work done must be equal to the total kinetic energy of road roller to bring the roller to
rest, i.e.
6 F = 45 170
or
F = 45 170/6 = 7528.3 N Ans.
Example 3.6. A steam engine drop-valve is closed by a spring after the operation of a trip
gear. The stiffness of the spring is such that a force of 4 N is required per mm of compression. The
valve is lifted against the spring, and when fully open the compression is 75 mm. When closed the
compression is 30 mm. The mass of the valve is 5 kg and the resistance may be taken as constant
and equal to 70 N. Find the time taken to close the valve after the operation of the trip.
Solution. Given : s = 4 N/mm = 4000 N/m ; x 1 = 75 mm = 0.075 m ; x 2 = 30 mm = 0.03 m;
m = 5 kg ; R = 70 N
Let
x = Displacement of the valve (in metres) from its highest position in
time t seconds.
When the valve is closed, then the value of x
= x 1 – x 2 = 0.075 – 0.03 = 0.045 m
Since the stiffness of the spring is 4000 N/m ; therefore in any position, the push of the spring
Q = 4000 (0.075 – x ) N
If P is the downward force on the valve, then
P = Q + m.g – R = 4000 (0.075 – x) + 5 × 9.81 – 70 = 279 – 4000 x
Also
Force, P = Mass × Acceleration
279 – 4000 x = 5 ×
or
Let
d 2x
dt 2
d 2 x 279 − 4000 x
=
= 56 − 800 x = − 800( x − 0.07)
5
dt 2
y = x – 0.07
d2 y d2x
= 2 = − 800 y
or
dt 2
dt
The solution of this differential equation is
∴
d2 y
+ 800 y = 0
dt 2
y = a cos 800 t + b sin 800 t
x − 0.07 = a cos 800 t + b sin 800 t
... (i)
where a and b are constants to be determined.
Now when t = 0, x = 0, therefore from equation (i), a = – 0.07
Differentiating equation (i),
dx
= − 800 a sin 800 t + 800 b cos 800 t
dt
... (ii)
Chapter 3 : Kinetics of Motion
Now when t = 0,
l
39
dx
= 0, therefore from equation (ii), b = 0
dt
Substituting the values of a and b in equation (i),
x – 0.07 = – 0.07 cos 800 t or x = 0.07 (1 − cos 800 t )
When x = 0.045 m, then
0.045 = 0.07 (1 − cos 800 t )
1 − cos 800 t = 0.045 / 0.07 = 0.642 or cos 800 t = 1 − 0.642 = 0.358
or
800 t = cos − 1 (0.358) = 69 ° = 69 ×
∴
π
= 1.2 rad
180
t = 1.2 / 800 = 1.2 / 28.3 = 0.0424 s Ans.
3.16. Principle of Conservation of Energy
It states “The energy can neither be created nor destroyed, though it can be transformed
from one form into any of the forms, in which the energy can exist.”
Note : The loss of energy in any one form is always accompanied by an equivalent increase in another form.
When work is done on a rigid body, the work is converted into kinetic or potential energy or is used in overcoming friction. If the body is elastic, some of the work will also be stored as strain energy. Thus we say that the total
energy possessed by a system of moving bodies is constant at every instant, provided that no energy is rejected
to or received from an external source to the system.
3.17. Impulse and Impulsive Force
The impulse is the product of force and time. Mathematically,
Impulse = F × t
where
F = Force, and t = Time.
Now consider a body of mass m. Let a force F changes its velocity from an initial velocity v 1
to a final velocity v 2.
We know that the force is equal to the rate of change of linear momentum, therefore
F=
m (v2 − v1 )
or F × t = m( v2 − v1 )
t
i.e.
Impulse = Change of linear momentum
If a force acts for a very short time, it is then known as impulsive force or blow. The impulsive
force occurs in collisions, in explosions, in the striking of a nail or a pile by a hammer.
Note: When the two rotating gears with angular velocities ω1 and ω2 mesh each other, then an impulsive
torque acts on the two gears, until they are both rotating at speeds corresponding to their velocity ratio. The
impulsive torque,
T.t = I (ω2 – ω1)
3.18. Principle of Conservation of Momentum
It states “The total momentum of a system of masses (i.e. moving bodies) in any one direction remains constant, unless acted upon by an external force in that direction.” This principle is
applied to problems on impact, i.e. collision of two bodies. In other words, if two bodies of masses
m 1 and m 2 with linear velocities v 1 and v 2 are moving in the same straight line, and they collide and
begin to move together with a common velocity v, then
40
l
Theory of Machines
Momentum before impact = Momentum after impact
m1v1 ± m2 v2 = (m1 + m2 )v
i.e.
Notes : 1. The positive sign is used when the two bodies move in the
same direction after collision. The negative sign is used when they move
in the opposite direction after collision.
2. Consider two rotating bodies of mass moment of inertia I1
and I2 are initially apart from each other and are made to engage as in
the case of a clutch. If they reach a common angular velocity ω, after
slipping has ceased, then
I1.ω1 ± I2.ω2 = (I1 + I2) ω
The ± sign depends upon the direction of rotation.
3.19. Energy Lost by Friction Clutch During
Engagement
Consider two collinear shafts A and B connected by
a *friction clutch (plate or disc clutch) as shown in Fig. 3.7.
Let I A and IB = Mass moment of inertias of the
rotors attached to shafts A and B
respectively.
ωA and ωB = Angular speeds of shafts A and B
respectively before engagement of
clutch, and
ω = Common angular speed of shafts
A and B after engagement of clutch.
By the principle of conservation of momentum,
IA.ωA + IB.ωB = (IA + IB) ω
∴
ω=
Fig. 3.7. Friction clutch.
I A .ωA + I B .ωB
IA + IB
... (i)
Total kinetic energy of the system before engagement,
1
1
I (ω ) 2 + I B (ωB )2
I A (ωA ) 2 + I B (ωB ) 2 = A A
2
2
2
Kinetic energy of the system after engagement,
E1 =
2
E2 =
=
I .ω + I .ω
1
1
( I A + I B ) ω2 = ( I A + I B ) A A B B
2
2
IA + IB
( I A .ωA + I B .ωB ) 2
2( I A + I B )
∴ Loss of kinetic energy during engagement,
E = E1 − E2 =
*
(I .ω + I .ω )2
I A (ω A )2 + I B (ω B )2
− A A B B
2
2( I A + I B )
Please refer Chapter 10 (Art. 10.32) on Friction.
l
Chapter 3 : Kinetics of Motion
I A . I B (ωA − ωB )2
2( I A + I B )
=
41
... (ii)
Notes: 1. If the rotor attached to shaft B is at rest, then ωB= 0. Therefore, common angular speed after engagement,
I .ω
ω= A A
... [Substituting ωB = 0 in equation (i)] ... (iii)
IA + IB
E=
and loss of kinetic energy,
I A . I B ( ωA ) 2
2 (IA + I B )
... [Substituting ωB = 0 in equation (ii)] ... (iv)
2. If IB is very small as compared to IA and the rotor B is at rest, then
I .ω
ω = A A = ωA
IA + I B
1
1
I B .ω.ωA = I B .ω2
2
2
= Energy given to rotor B
E=
and
... (Neglecting IB)
... [From equations (iii) and (iv)]
Example 3.7. A haulage rope winds on a drum of radius 500 mm, the free end being
attached to a truck. The truck has a mass of 500 kg and is initially at rest. The drum is equivalent to
a mass of 1250 kg with radius of gyration 450 mm. The rim speed of the drum is 0.75 m/s before the
rope tightens. By considering the change in linear momentum of the truck and in the angular momentum of the drum, find the speed of the truck when the motion becomes steady. Find also the
energy lost to the system.
Solution. Given : r = 500 mm = 0.5 m ; m 1 = 500 kg ; m 2 = 1250 kg ; k = 450 mm = 0.45 m ;
u = 0.75 m/s
We know that mass moment of inertia of drum,
I2 = m 2.k 2 = 1250 (0.45)2 = 253 kg-m2
Speed of the truck
Let
v = Speed of the truck in m/s, and
F = Impulse in rope in N-s.
We know that the impulse is equal to the change of linear momentum of the truck. Therefore
F = m 1.v = 500 v N-s
and
moment of impulse = Change in angular momentum of drum
u − v
F × r = I 2 (ω 2 − ω1 ) = I 2
r
i.e.
0.75 − v
500 v × 0.5 = 253
0.5
∴
250 v + 506 v = 380
or
or
u v u−v
... 3 ω2 − ω1 = − =
250 v = 380 − 506 v
v = 380/756 = 0.502 m/s Ans.
Energy lost to the system
We know that energy lost to the system
= Loss in K.E. of drum – Gain in K.E. of truck
=
1
1
× I 2 (ω2 )2 − ( ω1 ) 2 − × m1 .v 2
2
2
r
r
r
42
l
Theory of Machines
=
u 2 − v2 1
1
2
× I2
− × m1.v
2
2
2
r
=
(0.75) 2 − (0.502) 2 1
1
2
× 253
− × 500(0.502) N-m
2
(0.5) 2
2
= 94 N-m
Ans.
Example 3.8. The two buffers at one end of a truck each require a force of 0.7 MN/m of
compression and engage with similar buffers on a truck which it overtakes on a straight horizontal
track. The truck has a mass of 10 tonnes and its initial speed is 1.8 m/s, while the second truck has
mass of 15 tonnes with initial speed 0.6 m/s, in the same direction.
Find : 1. the common velocity when moving together during impact, 2. the kinetic energy
lost to the system, 3. the compression of each buffer to store the kinetic energy lost, and 4. the
velocity of each truck on separation if only half of the energy offered in the springs is returned.
Solution. Given : s = 0.7 MN/m = 0.7 × 106 N/m ; m = 10 t = 10 × 103 kg ; v 1 = 1.8 m/s;
m 2 = 15 t = 15 × 103 kg ; v 2 = 0.6 m/s
1. Common velocity when moving together during impact
Let
v = Common velocity.
We know that momentum before impact = Momentum after impact
i.e.
m1. v 1 + m 2.v 2 = (m 1 + m 2) v
10 × 103 × 1.8 + 15 × 103 × 0.6 = (10 × 103 + 15 + 103) v
or
v = 27 × 103/25 × 103 = 1.08 m/s Ans.
27 × 103 = 25 × 103 v
2. Kinetic energy lost to the system
Since the kinetic energy lost to the system is the kinetic energy before impact minus the
kinetic energy after impact, therefore
Kinetic energy lost to the system
1
1
1
2
2
= m1v1 + m2 v2 − ( m1 + m2 ) v 2
2
2
2
1
1
3
2
3
2
= × 10 × 10 (1.8) + × 15 × 10 (0.6)
2
2
−
(
)
1
10 × 103 + 15 × 103 (1.08) 2
2
= 4.35 × 103 N-m = 4.35 kN-m
Ans.
3. Compression of each buffer spring to store kinetic energy lost
Let
x = Compression of each buffer spring in metre, and
s = Force required by each buffer spring or stiffness of each spring
... (Given)
= 0.7 MN/m = 0.7 × 106 N/m
Since the strain energy stored in the springs (four in number) is equal to kinetic energy lost
in impact, therefore
4×
1
s. x 2 = 4.35 × 10 3
2
Chapter 3 : Kinetics of Motion
4×
l
43
1
× 0.7 × 106 x 2 = 4.35 × 103
2
1.4 × 106 x 2 = 4.35 × 103
or
∴
x2 = 4.35 × 103/1.4 × 106 = 3.11 × 10–3
or
x = 0.056 m = 56 mm
Ans.
4. Velocity of each truck on separation
Let
v3 = Velocity of separation for 10 tonnes truck, and
v4 = Velocity of separation for 15 tonnes truck.
The final kinetic energy after separation is equal to the kinetic energy at the instant of common velocity plus strain energy stored in the springs. Since it is given that only half of the energy
stored in the springs is returned, therefore
Final kinetic energy after separation
= Kinetic energy at common velocity +
or
1
1
1
1
m1 ( v3 ) 2 + m2 ( v4 ) 2 = ( m1 + m2 ) v 2 +
2
2
2
2
1
Energy stored in springs
2
1
2
4 × s. x
2
1
1
1
1
× 10 × 103 (v3 )2 + × 15 × 103 (v4 )2 = (10 × 103 + 15 × 103 ) (1.08)2 + (4.35 × 103 )
2
2
2
2
1
... 3 4 × s. x2 = 4.35 × 103
2
10(v3 ) 2 + 15(v4 ) 2 = 33.51
... (i)
We know that initial momentum and final momentum must be equal, i.e.
m 1.v 3 + m 2.v 4 = (m 1 + m 2) v
10 ×
103
× v 3 + 15 × 103 × v 4 = (10 × 103 + 15 × 103) 1.08
10v 3 + 15 v 4 = 27
... (ii)
From equations (i) and (ii),
v 3 = 0.6 m/s, and v 4 = 1.4 m/s Ans.
Example 3.9. A mass of 300 kg is allowed to fall vertically through 1 metre on to the top of
a pile of mass 500 kg. Assume that the falling mass and pile remain in contact after impact and that
the pile is moved 150 mm at each blow. Find, allowing for the action of gravity after impact 1. The
energy lost in the blow, and 2. The average resistance against the pile.
Solution. Given : m 1 = 300 kg ; s = 1 m ; m 2 = 500 kg ; x = 150 mm = 0.15 m
1. Energy lost in the blow
First of all, let us find the velocity of mass m 1 with which it hits the pile.
Let
v 1 = Velocity with which mass m 1 hits the pile.
We know that v12 − u 2 = 2 g .s
v12 − 0 = 2 × 9.81 × 1 = 19.62
or
v1 = 4.43 m/s
... (∵ u = 0 )
44
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Theory of Machines
Again, let
v 2 = Velocity of the pile before impact, and
v = Common velocity after impact,
We known that momentum before impact
= Momentum after impact
or
m 1.v 1 + m 2.v 2 = (m 1 + m 2) v
300 × 4.43 + 500 × 0 = (300 + 500) v
1329 = 800 v
∴
v = 1329/800 = 1.66 m/s
Now, kinetic energy before impact
= Potential energy = m 1.g.s
Fig. 3.8
= 300 × 9.81 × 1 = 2943 N-m
and kinetic energy after impact
=
1
1
( m1 + m2 ) v 2 = (300 + 500) (1.66) 2 = 1102 N-m
2
2
∴ Energy lost in the blow
= 2943 – 1102 = 1841 N-m
Ans.
2. Average resistance against the pile
Let
R = Average resistance against the pile in N.
Since the net work done by R, m 1 and m 2 is equal to the kinetic energy after impact, therefore
(R – m 1.g – m 2.g) x = Kinetic energy after impact
(R – 300 × 9.81 – 500 × 9.81) 0.15 = 1102
∴
or
R – 7848 = 1102/0.15 = 7347
R = 7347 + 7848 = 15 195 N = 15.195 kN
Ans.
Example 3.10. A hammer B suspended from pin C, and
an anvil A suspended from pin D, are just touching each other
at E, when both hang freely as shown in Fig. 3.9. The mass of B
is 0.7 kg and its centre of gravity is 250 mm below C and its
radius of gyration about C is 270 mm. The mass of A is 2.4 kg
and its centre of gravity is 175 mm below D and its radius of
gyration about D is 185 mm. The hammer B is rotated 20° to the
position shown dotted and released. Assume that the points of
contact move horizontally at the instant of impact and that their
local relative linear velocity of recoil is 0.8 times their relative
linear velocity of impact. Find the angular velocities of hammer
and of the anvil immediately after impact.
Solution. Given : m 1 = 0.7 kg ; k 1 = 270 mm = 0.27 m ;
m 2 = 2.4 kg ; k2 = 185 mm = 0.185 m
Let
Fig. 3.9
ω = Angular velocity of hammer B just before impact, and
h = Distance from release to impact
= Distance of c.g. of mass B below C = 250 mm = 0.25 m ...(Given)
Chapter 3 : Kinetics of Motion
l
45
We know that K.E. of hammer B
= Loss of P.E. from relase to impact
1 I ω2 = m g h
.
1. .
2 1
or
1
m ( k )2 ω 2 = m1 .g .h
2 1 1
1
× 0.7 (0.27) 2 ω 2 = 0.7 × 9.81 × 0.25 (1 − cos 20°)
2
0.0255 ω2 = 0.1032
∴
ω2 = 0.1032 / 0.0255 = 4.05 or ω = 2.01 rad/s
Let ωA and ωB be the angular velocities of the anvil A and hammer B, in the same direction,
immediately after impact.
∴ Relative linear velocity
= ωA × DL – ωA × CM = ωA × 0.2 – ωB × 0.275
... (DL and CM are taken in metres)
= 0.2 ωA – 0.275 ωB
... (i)
But, relative linear velocity
= 0.8 × Relative linear velocity of impact
... (Given)
= 0.8ω × CM = 0.8 × 2.01 × 0.275 = 0.44
... (ii)
Equating (i) and (ii),
0.2 ωA – 0.275 ωB = 0.44 or ωB = 0.727 ωA – 1.6
... (iii)
Since the linear impulse at E is equal and opposite on A and B, then by moments about D for
A and about C for B, it follows that the ratio
Decrease in angular momentum of B CM 0.275
=
=
DL
Increase in angular momentum of A
0.2
I B ( ω − ωB ) 0.275
=
= 1.375
0.2
I A .ωB
i.e.
m1 (k1 )2 (ω − ωB )
m2 (k 2 )2 ωA
∴
= 1.375
2.01– ωB = 2.21 ωA
or
or
0.7 (0.27)2 (2.01 − ωB )
= 1.375
2.4 (0.185) 2 ωA
ωB = 2.01 – 2.21 ωA
From equations (iii) and (iv), we get
0.727 ωA– 1.6 = 2.01 – 2.21 ωA
0.727 ωA + 2.21 ωA = 2.01 + 1.6
Substituting
or
ωA = 1.23 rad/s
ωA = 1.23 rad/s in equation (iv),
ωB = 2.01 – 2.21 × 1.23 = – 0.71 rad/s
= 0.71 rad/s, in reverse direction
Ans.
Ans.
... (iv)
46
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Theory of Machines
Example 3.11. The pendulum of an Izod impact testing machine has a mass of 30 kg. The
centre of gravity of the pendulum is 1 m from the axis of suspension and the striking knife is 150 mm
below the centre of gravity. The radius of gyration about the point of suspension is 1.1 m, and about
the centre of gravity is 350 mm. In making a test, the pendulum is released from an angle of 60° to
the vertical. Determine : 1. striking velocity of the pendulum, 2. impulse on the pendulum and
sudden change of axis reaction when a specimen giving an impact value of 54 N-m is broken,
3. angle of swing of the pendulum after impact, and 4. average force exerted at the pivot and at the
knife edge if the duration of impact is assumed to be 0.005 second.
Solution. Given : m = 30 kg ; A G = a = 1 m ; GB = b = 150 mm = 0.15 m ; k 1 = 1.1 m;
k 2 = 350 mm = 0.35 m ; θ = 60° ; t = 0.005 s
We know that mass moment of inertia of the pendulum about the point of suspension A ,
IA = m (k 1)2 = 30 (1.1)2 = 36.3 kg-m2
and mass moment of inertia of the pendulum about centre of
gravity G,
IG = m (k 2)2 = 30 (0.35)2
= 3.675 kg-m2
1. Striking velocity of the pendulum
Let
v = Striking velocity of the
pendulum, and
ω = Angular velocity of the
pendulum.
Fig. 3.10
Since the potential energy of the pendulum is converted into angular kinetic energy of the
pendulum, therefore,
m.g .h1 =
1
I .ω2
2 A
30 × 9.81 (1 – 1 cos 60°) =
1
× 36.3 ω2
2
... (3 h1 = a – a cos 60°)
147.15 = 18.15 ω2
or
∴
ω2 = 147.15/18.15 = 8.1 or ω = 2.85 rad/s
v = ω × A B = ω (a + b) = 2.85 (1 + 0.15) = 3.28 m/s Ans.
and
2. Impulse on the pendulum
Let
F1 = Impulse at the pivot A ,
F2 = Impulse at the knife edge B,
ω = Angular velocity of the pendulum just before the breakage of the
specimen, and
ω1 = Angular velocity of the pendulum just after the breakage of the specimen.
Since the loss in angular kinetic energy of the pendulum is equal to the energy used for
breaking the specimen (which is 54 N-m ), therefore
1
I (ω 2 − ω12 ) = 54
2 A
or
1
× 36.3 (2.852 − ω12 ) = 54
2
Chapter 3 : Kinetics of Motion
∴
ω12 = (2.85) 2 −
l
47
54 × 2
= 5.125 or ω1 = 2.26 rad/s
36.3
Let v G and v G′ be the linear velocities of G just before and just after the breakage of specimen.
vG = ω × OG = 2.85 × 1 = 2.85m/s
and
vG ′ = ω1 × OG = 2.26 × 1 = 2.26 m/s
We know that Impulse = Change of linear momentum
F1 + F2 = m (v G – v G′) = 30 (2.85 – 2.26) = 17.7 N
... (i)
Taking moments about G, we get
Impulsive torque = Change of angular momentum
F2 × b – F1 × a = IG (ω – ω1)
F2 × 0.15 – F1 × 1 = 3.675 (2.85 – 2.26) = 2.17
... (ii)
From equations (i) and (ii),
F2 = 17.3 N ; and F1 = 0.4 N Ans.
3. Angle of swing of the pendulum after impact
Let
θ = Angle of swing of the pendulum after impact.
Since work done in raising the pendulum is equal to angular kinetic energy of the pendulum,
therefore
m.g.h1 =
30 × 9.81 (1 – 1 cos θ) =
1
2
1
2
I A (ω1 ) 2
× 36.3 (2.26)2 = 92.7
1 – 1 cos θ = 92.7/30 × 9.81 = 0.315
∴
or
cos θ = 1 – 0.315 = 0.685
θ = 46.76° Ans.
4. Average force exerted at the pivot and at the knife edge
We know that average force exerted at the pivot
=
F1
0.4
=
= 80 N Ans.
0.005
t
and average force exerted at the knife edge
=
F2
17.3
=
= 3460 N Ans.
0.005
t
Example 3.12. A motor drives a machine through a friction clutch which transmits a torque
of 150 N-m, while slip occurs during engagement. The rotor, for the motor, has a mass of 60 kg, with
radius of gyration 140 mm and the inertia of the machine is equivalent to a mass of 20 kg at the
driving shaft with radius of gyration 80 mm. If the motor is running at 750 r.p.m. and the machine
is at rest, find the speed after the engagement of the clutch and the time taken. What will be the
kinetic energy lost during the operation ?
48
l
Theory of Machines
Solution. Given : T = 150 N-m ; m 1 = 60 kg ; k 1 = 140 mm = 0.14 m ; m 2 = 20 kg ;
k 2 = 80 mm = 0.08 m ; N 1 = 750 r.p.m. or ω1 = 2 π × 750/60 = 78.55 rad/s ; N 2 = 0 or ω2 = 0
We know that mass moment of inertia of the rotor on motor,
I1 = m1 (k1 )2 = 60 (0.14)2 = 1.176 kg-m 2
and mass moment of inertia of the parts attached to machine,
I2 = m 2 (k 2)2 = 20 (0.08)2 = 0.128 kg-m2
Speed after the engagement of the clutch and the time taken
Let
ω = Speed after the engagement of the clutch in rad/s,
t = Time taken in seconds, and
α = Angular acceleration during the operation in rad/s2.
We know that the impulsive torque = change of angular momentum
∴
T.t = I1 (ω1 – ω)
or
t=
I1 ( ω1 − ω) 1.176 (78.55 − ω)
s
=
150
T
... (i)
Also
T.t = I2 (ω2 – ω)
or
t=
I 2 (ω − ω2 ) 0.128 × ω
s
=
150
T
... (ii)
Equating equations (i) and (ii),
1.176 (78.55 − ω)
0.128 ω
=
150
150
1.304 ω = 92.4
... (3 ω2 = 0)
or
92.4 – 1.176 ω = 0.128 ω
or
ω = 92.4/1.304 = 70.6 rad/s Ans.
Substituting the value of ω in equation (ii),
0.128 × 70.6
= 0.06 s Ans.
150
Kinetic energy lost during the operation
t=
We know that the kinetic energy lost during the operation,
E=
=
I1.I 2 (ω1 − ω2 ) 2
I .I .ω2
= 1 2 1
2 ( I1 + I 2 )
2 ( I1 + I 2 )
... (3 ω2 = 0)
1.176 × 0.128 (78.55)2 928.8
=
= 356 N-m Ans.
2 (1.176 + 0.128)
2.61
3.20. Torque Required to Accelerate a
Geared System
Consider that the two shafts A and B are geared together
as shown in Fig. 3.11. Let the shaft B rotates G times the speed
of shaft A . Therefore, gear ratio,
G=
NB
NA
where N A and N B are speeds of shafts A and B (in r.p.m.)
respectively.
Since the shaft B turns G times the speed of shaft A ,
therefore the rate of change of angular speed of shaft B with
Fig. 3.11. Torque to accelerate a
geared system.
Chapter 3 : Kinetics of Motion
l
49
respect to time (i.e. angular acceleration of shaft B, αB ) must be equal to G times the rate of change
of angular speed of shaft A with respect to time (i.e. angular acceleration of shaft A , αA ).
∴
αB = G.αA
Let
...(i)
IA and IB = Mass moment of inertia of the masses attached to shafts A and B
respectively.
∴ Torque required on shaft A to accelerate itself only,
T A = IA.αA
and torque required on shaft B to accelerate itself only,
T B = IB.αB = G.IB.αA
... [From equation (i)] ... (ii)
In order to provide a torque T B on the shaft B, the torque applied to shaft A must be G × T B.
Therefore, torque applied to shaft A in order to accelerate shaft B,
T AB = G.TB = G2.IB.αA
... [From equation (ii)] ... (iii)
∴ Total torque which must be applied to shaft A in order to accelerate the geared system,
T = T A + T AB = IA.αA + G2.IB.αA
= (IA + G2.IB) αA = I.αA
... (iv)
G2.
IB and may be regarded as equivalent mass moment of inertia of geared system
where I = IA +
referred to shaft A .
Let the torque T required to accelerate the geared system, as shown in Fig. 3.11, is applied
by means of a force F which acts tangentially to a drum or pulley of radius r.
∴
T = F × r = I.αA
... (v)
We know that the tangential acceleration of the drum,
a = αA.r
∴
F×r = I×
F=
or
where me =
1
(
a
r2
or
αA = a/r
(
)
a
a
= I A + G 2 .I B
r
r
(I
A
... (3 I = I A + G 2 .I B )
)
+ G 2 .I B = a.me
... (vi)
)
I A + G 2 .I B and may be regarded as equivalent mass of the system referred to the
r2
line of action of the accelerating force F.
Notes : 1. If η is the efficiency of the gearing between the two shafts A and B, then the torque applied to shaft
A in order to accelerate shaft B,
G 2 .I B .α A
η
and the total torque applied to shaft A in order to accelerate the geared system,
TAB =
T = TA + TAB = I A .α A +
G 2 .I B .α A
G 2 .I B
= I A +
α A = I .α A
η
η
G 2 .I B
, and may be regarded as the equivalent mass moment of inertia of the geared system
η
referred to shaft A .
where I = I A +
2. If the number of shafts (say A to X ) are geared together in series, then the equivalent mass moment
of inertia referred to shaft A is given by,
50
l
Theory of Machines
Gx2 I x
ηx
Gx = Ratio of speed of shaft X to the speed of shaft A ,
Ix = Mass moment of inertia of mass attached to shaft X, and
I = IA + ∑
where
ηx = Overall efficiency of the gearing from shaft A to shaft X .
3. If each pair of gear wheels is assumed to have the same efficiency η and there are m gear pairs
through which the power is transmitted from shaft A to shaft X, then the overall efficiency from shaft A to X is
given by,
ηx = ηm
4. The total kinetic energy of the geared system,
K.E. =
where
1
I (ω A ) 2
2
I = Equivalent mass moment of inertia of the geared system referred to shaft A , and
ωA = Angular speed of shaft A.
Example 3.13. A mass M of 75 kg is hung from a
rope wrapped round a drum of effective radius of 0.3 metre,
which is keyed to shaft A. The shaft A is geared to shaft B
which runs at 6 times the speed of shaft A. The total mass
moment of inertia of the masses attached to shaft A is 100
kg-m2 and that of shaft B is 5 kg-m2.
Find the acceleration of mass M if 1. it is allowed to fall freely, and 2. when the efficiency of the
gearing system is 90%. The configuration of the system
is shown in Fig. 3.12.
Fig. 3.12
Solution. Given : M = 75 kg ; r = 0.3 m ; N B = 6 N A
or G = N B / N A = 6 ; IA = 100 kg-m2 ; IB = 5 kg-m2; η = 90% = 0.9
Let a = Acceleration of the mass M, in m/s2.
1. When it is allowed to fall freely
We know that equivalent mass of the geared system referred to the circumference of the
drum (or the line of action of the accelerating mass M ),
me =
1
r2
(I
A
)
+ G 2 .I B =
1
(0.3) 2
(100 + 6
2
)
× 5 = 3111 kg
and total equivalent mass to be accelerated,
Me = m e + M = 3111 + 75 = 3186 kg
∴ Force required to accelerate this equivalent mass (M e)
= M e.a = 3186 a N
... (i)
and the accelerating force provided by the pull of gravity on the mass M suspended from the rope
= M.g = 75 × 9.81 = 736 N
... (ii)
From equations (i) and (ii),
3186 a = 736
or
a = 736/3186 = 0.231 m/s2 Ans.
2. When the efficiency of the gearing system is 90%
We know that the equivalent mass of the geared system referred to the circumference of the
drum,
Chapter 3 : Kinetics of Motion
me =
1
G 2 . IB
I
+
A
η
r 2
1
=
2
( 0.3)
51
l
62 × 5
100
+
= 3333 kg
0.9
and total equivalent mass to be accelerated,
M e = m e + M = 3333 + 75 = 3408 kg
∴ Force required to accelerate this equivalent mass (M e)
= M e.a = 3408 a N
... (iii)
and accelerating force provided by the pull of gravity on the mass M suspended from the rope
= M.g = 75 × 9.81 = 736 N
... (iv)
Now equating equations (iii) and (iv),
3408 a = 736
or
a = 736/3408 = 0.216 m/s2 Ans.
Example. 3.14. The motor shaft A exerts a constant torque of 100 N-m and is geared to shaft B as
shown in Fig. 3.13. The moments of inertia of the parts
attached to the motor shaft A is 2 kg-m2 and that of the
parts attached to other shaft B is 32 kg-m2.
Find the gear ratio which gives the maximum
angular acceleration of shaft B and the corresponding
angular acceleration of each shaft.
Solution. Given : T = 100 N-m ; IA = 2 kg-m2 ;
IB = 32 kg-m2
Gear ratio which gives the maximum acceleration
Let
Parallel shaft gear motor.
G = Gear ratio which gives the maximum
acceleration.
αA = Angular acceleration of shaft A , and
αB = Angular acceleration of shaft B.
We know that αA = G.αB
... (i)
∴ Torque required on motor shaft A to accelerate rotating
parts on it,
T A = IA. αA= IA.G. αB
Fig. 3.13
... [From equation (i)]
and torque required on motor shaft A to accelerate rotating parts on shaft B,
I B .α B
G
Assuming that there is no resisting torque and the torque exerted on the motor shaft A is
utilised to overcome the inertia of the geared system.
TAB =
∴
or
T = TA + TAB = I A .G.α B +
αB =
G.T
I A .G 2 + I B
I .G 2 + I B
I B .α B
= αB A
G
G
... (ii)
52
l
Theory of Machines
For maximum angular acceleration of B, differentiate with respect to G and equate to zero, i.e.
dαB
=0
dG
G.T
d
2
.
I
G
+ IB
A
dG
or
=0
( I A .G 2 + I B ) T − G.T ( I A × 2G)
( I A .G + I B )
2
∴
I B = G 2 .I A
2
or
or I A .G 2 + I B − 2G 2 .I A = 0
=0
G=
IB
32
=
= 4 Ans.
2
IA
Angular acceleration of each shaft
Substituting the value of G in equation (ii),
αB =
4 × 100
2 × 4 2 + 32
= 6.25 rad/s 2 Ans.
α A = G.α B = 4 × 6.25 = 25 rad/s 2 Ans.
and
Example 3.15. A motor vehicle of total mass 1500 kg has road wheels of 600 mm effective
diameter. The effective moment of inertia of the four road wheels and of the rear axle together is
8 kg-m2 while that of the engine and flywheels is 1 kg-m2. The transmission efficiency is 85%
and a tractive resistance at a speed of 24 km/h is 300 N. The total available engine torque is
200 N-m. Determine :
1. Gear ratio, engine to back axle, to provide maximum acceleration on an upgrade whose
sine is 0.25, when travelling at 24 km/h,
2. The value of this maximum acceleration, and
3. The speed and power of the engine under these conditions.
Solution. Given : m = 1500 kg ; d = 600 mm = 0.6 m or r = 0.3 m ; IA = 8 kg-m2 ;
IB = 1 kg-m2; η = 85% = 0.85 ; v = 24 km/h ; F = 300 N ; T B = 200 N-m ; sin θ = 0.25
1. Gear ratio, engine to back axle, to provide maximum acceleration
Let
G = Gear ratio, engine to back axle, to provide maximum acceleration.
∴ Torque at road wheels,
TW = η × G × TB = 0.85 × G × 200 = 170 G N-m
and available tangential force at road wheels,
TW 170 G
=
= 567 G N
0.3
r
Let the vehicle travels up the gradient a distance of s metre while its speed changes from u
to v m/s.
P=
We know that work done by the tangential force P
= Change of linear K.E. of vehicle + Change of angular K.E. of road
wheels and axle + Change of angular K.E. of engine and flywheel +
Work done in raising vehicle + Work done in overcoming tractive
resistance
Chapter 3 : Kinetics of Motion
1
or
P×s=
or
s ( P − m.g .sin θ − F ) =
2
m (v 2 − u 2 ) +
1
2
I A (ω22 − ω12 ) +
v2 − u2
2
1
2
53
l
I B .G 2 .η (ω22 − ω12 ) + m.g .s. sin θ + F .s
I A I B .G 2 .η
m
+
+
r2
r2
... (Substituting ω1 = u/r, and ω2 = v/r)
s (567 G − 1500 × 9.81 × 0.25 − 300) =
s (567 G − 3980) =
v2 − u2
2
8
1 × G 2 × 0.85
+
1500 +
0.32
0.32
v2 − u2
(1590 + 9.44 G 2 )
2
... (i)
We know that linear acceleration,
567G − 3980
v2 − u2
=
2s
1590 + 9.44 G 2
a=
... [From equation (i)] ... (ii)
For maximum acceleration, differentiate equation (ii) with respect to G and equate to zero,
i.e.
da
=0
dG
(1590 + 9.44 G 2 ) − (567 G − 3980) (9.44 × 2G)
(1590 + 9.44 G 2 )2
=0
or 901 530 + 5352 G2 – 10 705 G2 + 75 142 G = 0
G2 – 14 G – 168.4 = 0
∴
14 ± (14)2 + 4 × 168.4 14 ± 29.5
=
= 21.75 or 22 Ans.
2
2
G=
... (Taking + ve sign)
2. Value of maximum acceleration
Substituting the value of G = 22 in equation (ii), maximum acceleration,
amax =
567 × 22 − 3980
1590 + 9.44 (22) 2
= 1.38 m/s Ans.
3. Speed and power of the engine
Let
ω = Speed of the engine in rad/s.
We know that the speed of the road wheels,
v = 24 km/h = 6.67 m/s
∴ Angular speed of the road wheels
=
v 6.67
=
= 22.23 rad/s
r 0.3
... (Given)
54
l
Theory of Machines
Since the speed of the engine is G times the speed of the road wheels, therefore
ω = G × 22.23 = 22 × 22.23 = 489 rad/s Ans.
We know that power of the engine
= T B.ω = 200 × 489 = 97 800 W = 97.8 kW Ans.
Example 3.16. A super charged road racing automobile has an engine capable of giving
an output torque of 1 kN-m, this torque being reasonably constant over a speed range from
100 km/h to 275 km/h in top gear. The road wheels are of 0.9 m effective diameter, and the back axle
ratio is 3.3 to 1. When travelling at a steady speed of 170 km/h in top gear on a level road, the
power absorbed is 50 kW. The vehicle has a mass of 1000 kg, the four road wheels each has mass of
40 kg and a radius of gyration of 0.25 m. The moment of inertia of the engine and all parts forward
of the differential is 6 kg-m2.
Assuming that the resistance caused by windage and road drag varies as the square of the
speed, determine the time taken for the speed to rise from 100 km/h to 275 km/h in top gear at full
throttle on an upgrade of 1 in 30.
Solution. Given : T B = 1 kN-m = 1000 N-m ; v 1 = 100 km/h = 27.8 m/s ; v 2 = 275 km/h =
76.4m/s ; d = 0.9 m or r = 0.45 m ; G = 3.3 ; v = 170 km/h = 47.2 m/s ; P = 50 kW = 50 × 103 W ;
M = 1000 kg ; m = 40 kg ; k = 0.25 m ; IB = 1 kg-m2
We know that moment of inertia of four road wheels,
IA = 4 × m.k2 = 4 × 40 (0.25)2 = 10 kg -m2
Let
F = Resistance caused by windage and road drag in newtons.
∴ Power absorbed by the automobile at a steady speed (P),
50 × 103 = F.v = F × 47.2 or F = 50 × 103/47.2 = 1060 N
Since the resistance caused by windage and road drag (F) varies as the square of the speed
(v), therefore
F = k.v 2 or k = F/v 2 = 1060/(47.2)2 = 0.476
∴
F = 0.476 v 2 N
We know that the torque at road wheels,
TW = G × TE = 3.3 × 1000 = 3300 N-m
and available tangential force at road wheels,
FT =
TW 3300
=
= 7333 N
0.45
r
Since the gradient is 1 in 30, therefore proceeding in the same way as discussed in the
previous example, we get the linear acceleration,
dv
a=
=
dt
1000 × 9.81
M .g
7333 − 0.476 v 2 −
30 =
30
10
1 × 3.32
I A I B .G 2
1000 +
+
M+ 2 +
r
r2
(0.45)2 (0.45)2
FT − F −
= 6.65 – 0.43 × 10–3 v 2 – 0.3
∴
dt =
dv
6.65 − 0.43 × 10−3 v 2 − 0.3
=
dv
6.35 − 0.43 × 10−3 v 2
Chapter 3 : Kinetics of Motion
l
55
Integrating the above expression,
Let
∫ dt = ∫
=
∴
t=
dv
6.35 − 0.43 × 10 −3 v 2
103
dv
dv
= 2325 ∫
∫
2
0.43 14 768 − v
(121.5)2 − v 2
2325
121.5 + v
log e
+ C1
2 × 121.5
121.5 − v
... (i)
dv
a + v
1
=
... 3 ∫ 2
log e
2
a − v
2a
a −v
where C1 is the constant of integration. We know that when t = 0, v 1 = 27.8 m/s.
∴
0=
2325
121.5 + 27.8
log e
+ C1
2 × 121.5
121.5 − 27.8
= 9.6 log e
∴
... (Substituting v = v 1)
149.3
+ C1 = 9.6 log e 1.6 + C1
93.7
C1 = − 9.6 log e 1.6 = − 9.6 × 0.47 = − 4.5
Now the expression (i) may be written as
t=
2325
121.5 + v
− 4.5
loge
2 × 121.5
121.5 − v
When v 2 = 76.4 m/s, the time taken for the speed to rise
=
2325
121.5 + 76.4
197.9
− 4.5 = 9.6 loge
− 4.5
loge
2 × 121.5
121.5 − 76.4
45.1
= 9.6 log e 4.38 − 4.5 = 9.6 × 1.48 − 4.5 = 9.7 s Ans.
Example 3.17. An electric motor drives a machine through a speed reducing gear of ratio
9:1. The motor armature, with its shaft and gear wheel, has moment of inertia 0.6 kg-m2. The
rotating part of the driven machine has moment of inertia 45 kg-m2. The driven machine has resisting torque of 100 N-m and the efficiency of reduction gear is 95%. Find
1. The power which the motor must develop to drive the machine at a uniform speed of 160
r.p.m.,
2. The time required for the speed of the machine to increase from zero to 60 r.p.m., when the
torque developed on the motor armature in starting from rest is 30 N-m, and
3. If the gear ratio were altered so as to give the machine the greatest possible angular
acceleration in starting from rest, what would then be the gear ratio ? The starting torque of the
motor is 30 N-m as before.
Solution. Given : G = 9; I A = 0.6 kg-m 2 ; I B = 45 kg-m 2 ; T B = 100 N-m;
η = 95% = 0.95; N = 160 r.p.m. ; N 1 = 0 ; N 2 = 60 r.p.m. ; T A = 30 N-m
A motor driving a machine is shown in Fig. 3.14.
56
l
Theory of Machines
1. Power which the motor must develop
We know that the power which the motor must develop,
P=
2π N .TB 2π × 160 × 100
W
=
60 × η
60 × 0.95
= 1764 W = 1.764 kW Ans.
2. Time required for the speed of the machine to increase from
zero to 60 r.p.m.
Let
Fig. 3.14
t = Time required for the speed of the
machine to increase from zero to 60 r.p.m.
αA = Angular acceleration of motor, and
αB = Angular acceleration of machine.
Since the speed of motor A is G times the speed of machine B, therefore
αA = G.αB = 9 αB
We know that torque developed on motor armature,
TA = 30 N-m
... (Given)
Due to the torque (T A) and efficiency of gearing (η), the torque transmitted to machine B,
T B1 = G.TA.η = 9 × 30 × 0.95 = 256.5 N-m
We know that resisting torque on machine B,
T B = 100 N-m
... (Given)
∴ Net torque on machine B
= T B1 – T B = 256.5 – 100 = 156.5 N-m
... (i)
We know that total torque to be applied to machine B in order to accelerate the geared system
= Torque required on B to accelerate B only + Torque required on B
to accelerate A
= IB.αB + G.T A.η = IB.αB + G.IA.αA.η
... (∵ T A = IA.αA)
...(∵ αA= G.αB)
= IB.αB +
G2.
IA .αB.η
= 45 αB +
92
× 0.6 × αB × 0.95 = 45 αB + 46.2 αB
= 91.2 αB
... (ii)
Equating equations (i) and (ii),
αB = 156.5/91.2 = 1.7 rad/s2
We are given that initial angular speed, ω1 = 0, and final angular speed,
ω2 =
We know that
2 π N 2 2 π × 60
=
= 6.28 rad/s
60
60
... (∵ N 2 = 60 r.p.m.)
ω2 = ω1 + αB.t
6.28 = 0 + 1.7 t = 1.7 t
or
t = 6.28/1.7 = 3.7 s Ans.
3. Gear ratio for maximum angular acceleration of the machine
Let
G1 = Gear ratio for maximum angular acceleration of the machine.
Chapter 3 : Kinetics of Motion
l
57
We know that net torque on machine B
= TB1 – T B = G1.T A.η – T B = G1 × 30 × 0.95 – 100
= 27.5 G1 – 100
...(iii)
We also know that total torque required to be applied to machine B in order to accelerate the
geared system
= IB.αB + (G1)2 αB.IA.η
= 45 × αB + (G1)2 αB × 0.6 × 0.95 = αB [45 + 0.57 (G1)2]
...(iv)
From equations (iii) and (iv),
αB =
27.5G1 − 100
45 + 0.57 (G1 ) 2
For maximum angular acceleration, differentiate the above expression and equate to zero, i.e.
d αB
=0
d G1
[45 + 0.57 (G1 )2 ] (27.5) − (27.5 G1 − 100) (2 × 0.57 G1 )
or
[45 + 0.57 (G1 )2 ]
=0
1237.5 + 15.675 (G1)2 – 31.34 (G1)2 + 114 G1 = 0
15.675 (G1)2 – 114 G1 – 1237.5 = 0
(G1)2 – 7.27 G1 – 78.95 = 0
∴
G1 =
7.27 ± (7.27)2 + 4 × 78.95 7.27 ± 19.2
=
= 13.235 Ans.
2
2
... (Taking + ve sign)
Example 3.18. A hoisting gear, with a 1.5 m diameter drum, operates two cages by ropes
passing from the drum over two guide pulleys of 1 m diameter. One cage (loaded) rises while the
other (empty) descends. The drum is driven by a motor through double reduction gearing. The
particulars of the various parts are as follows :
S.No.
1.
2.
3.
4.
5.
6.
Part
Motor
Intermediate gear
Drum and shaft
Guide pulley (each)
Rising rope and cage
Falling rope and cage
Maximum
Speed (r.p.m.)
900
275
50
–
–
–
Mass (kg)
Radius of
Frictional
200
375
2250
200
1150
650
gyration (mm)
90
225
600
450
–
–
resistance
–
150 N-m
1125 N-m
150 N-m
500 N
300 N
Determine the total motor torque necessary to produce a cage an acceleration of 0.9 m/s2.
Solution. Given : d = 1.5 m or r = 0.750 m ; d1 = 1 m ; N M = 900 r.p.m ; N 1 = 275 r.p.m. ;
N D = 50 r.p.m ; m M = 200 kg; k M = 90 mm = 0.09 m ; m I = 375 kg ; k I = 225 mm = 0.225 m ;
M D = 2250 kg ; kD = 600 mm = 0.6 m ; m P = 200 kg ; k P = 450 mm = 0.45 m ; m 1 = 1150 kg ;
m 2 = 650 kg ; FI = 150 N-m ; FD = 1125 N-m ; FP = 150 N-m ; F1 = 500 N ; F2 = 350 N ; a = 0.9 m/s2
58
l
Theory of Machines
Fig. 3.15
We know that speed of guide pulley (P),
NP = ND ×
1.5
d
= 50 ×
= 75 r.p.m.
1
d1
Gear ratio for the intermediate gear and motor,
G1 = N1 / N M = 275 / 900 = 0.306
Gear ratio for the drum and motor,
G2 = N D / N M = 50 / 900 = 0.055
Gear ratio for the guide pulley and motor,
G3 = N P / N M = 75 / 900 = 0.083
Mass moment of inertia of the motor,
I M = mM (k M )2 = 200 (0.09) 2 = 1.62 kg-m2
Mass moment of inertia of the intermediate gear,
I I = mI (kI )2 = 375 (0.225)2 = 18.98 kg-m2
Mass moment of inertia of the drum and shaft,
I D = mD (kD )2 = 2250 (0.6)2 = 810 kg-m2
Mass moment of inertia of the guide pulley,
I P = mP (k P )2 = 200 (0.45)2 = 40.5 kg-m2
and angular acceleration of the drum,
α D = a / r = 0.9 / 0.75 = 1.2 rad/s 2
Chapter 3 : Kinetics of Motion
l
59
Since the speed of the drum (N D) is 0.055 times the speed of motor (N M), therefore angular
acceleration of the drum (αD),
1.2 = 0.055 αM or
αM = 1.2 / 0.055 = 21.8 rad/s2
We know that the equivalent mass moment of inertia of the system (i.e. motor, intermediate
gear shaft and wheel,drum and two guide pulleys) referred to motor M,
I = IM + (G1)2 II + (G2)2 ID + 2 (G3)2 IP
= 1.62 + (0.306)2 18.98 + (0.055)2 810 + 2 (0.083)2 40.5
= 1.62 + 1.78 + 2.45 + 0.56 = 6.41 kg-m2
∴ Torque at motor to accelerate the system,
T 1 = I.αM = 6.41 × 21.8 = 139.7 N-m
and torque at motor to overcome friction at intermediate gear, drum and two guide pulleys,
T 2 = GI.FI + G2.FD + 2 G3.FP
= 0.306 × 150 + 0.055 × 1125 + 2 × 0.83 × 150 N-m
= 45.9 + 61.8 + 25 = 132.7 N-m
Now for the rising rope and cage as shown in Fig. 3.15, tension in the rope between the
pulley and drum,
Q1 = Weight of rising rope and cage + Force to accelerate rising rope
and cage (inertia force) + Frictional resistance
= m 1.g + m 1.a + F1 = 1150 × 9.81 + 1150 × 0.9 + 500
= 12 816 N
Similarly for the falling rope and cage, as shown in Fig. 3.15, tension in the rope between
the pulley and drum,
Q2 = Weight of falling rope and cage – Force to accelerate falling
rope and cage (inertia force) – Frictional resistance
= m 2.g – m2.a – F2 = 650 × 9.81 – 650 × 0.9 – 350 = 5441 N
∴ Torque at drum, T D = (Q1 – Q2) r = (12 816 – 5441) 0.75 = 5531 N-m
and torque at motor to raise and lower cages and ropes and to overcome frictional resistance
T 3 = G2 × T D = 0.055 × 5531 = 304 N-m
∴
Total motor torque required,
T = T1 + T 2 + T 3 = 139.7 + 132.7 + 304 = 576.4 N-m Ans.
3.21. Collision of Two Bodies
Consider the impact between two bodies which move with different velocities along the
same straight line. It is assumed that the point of the impact lies on the line joining the centers of
gravity of the two bodies. The behaviour of these colliding bodies during the complete period of
impact will depend upon the properties of the materials of which they are made. The material of the
two bodies may be *perfectly elastic or perfectly inelastic.
In either case, the first effect of impact is approximately the same. The parts of each body
adjacent to the point of impact is deformed and the deformation will continue until the centre of
gravity of the two bodies are moving with the same velocity. Assuming that there are no external
forces acting on the system, the total momentum must remain constant.
*
The bodies, which rebound after impact are called elastic bodies and the bodies which does not rebound at
all after its impact are called inelastic bodies.
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Theory of Machines
3.22. Collision of Inelastic Bodies
When two *inelastic bodies A and B, as shown in Fig. 3.16 (a), moving with different velocities, collide with each other as shown in Fig. 3.16 (b), the two bodies will remain together after
impact and will move together with a common velocity.
Let
m 1 = Mass of first body A .
m 2 = Mass of second body B.
u1 and u2 = Velocities of bodies A and B respectively before impact, and
v = Common velocity of bodies A and B after impact.
Point of impact
(b) After impact.
(a) Before impact.
Fig. 3.16. Collision of inelastic bodies.
A little consideration will show that the impact will take place only, if u1 is greater than u2.
Now according to principle of conservation of momentum,
Momentum before impact = Momentum after impact
m1.u1 + m 2.u2 = (m 1 + m 2) v
v=
∴
m1.u1 + m2 .u2
m1 + m2
... (i)
The loss of kinetic energy during impact may be obtained by finding out the kinetic energy
of the two bodies before and after impact. The difference between the two kinetic energies of the
system gives the loss of kinetic energy during impact.
We know that the kinetic energy of the first body, before impact
1
m (u ) 2
2 1 1
and kinetic energy of the second body, before impact
=
1
m ( u )2
2 2 2
∴ Total kinetic energy of the system before impact,
=
1
1
m (u )2 + m2 (u2 )2
2 1 1
2
When the two bodies move with the same velocity v after impact, then
Kinetic energy of the system after impact,
E1 =
1
( m + m2 ) v 2
2 1
∴ Loss of kinetic energy during impact,
E2 =
EL = E1 − E2 =
*
1
1
1
m1 .u12 + m2 .u22 − ( m1 + m2 ) v2
2
2
2
The impact between two lead spheres or two clay spheres is approximately an inelastic impact.
Chapter 3 : Kinetics of Motion
=
m . u + m2 . u2
1
1
1
m1 . u12 + m2 . u12 − (m1 + m2 ) 1 1
2
2
2
m1 + m2
l
61
2
. . . [From equation (i)]
=
(m .u + m2 .u2 )2
1
1
m1 .u12 + m2 .u22 − 1 1
2
2
2 (m1 + m2 )
=
1
+ (m1 + m2 ) (m1 ⋅ u12 + m2 ⋅ u22 ) − (m1 ⋅ u1 + m2 ⋅ u 2 ) 2
2 (m1 + m2 )
. . . [Multiplying the numerator and denominator by (m 1 + m 2)]
=
1
m12 . u12 + m1 . m2 . u22 + m1 . m2 . u12 + m22 . u22
2( m1 + m2 )
− m12 . u12 − m22 . u 22 − 2 m1m2 u1 u2
=
1
m1 . m2 . u22 + m1 . m2 . u12 − 2m1 m2 u1 u2
2( m1 + m2 )
=
.
m1 . m2
u12 + u22 − 2u1 . u2 = m1 m2
(u1 − u2 )2
2(m1 + m2 )
2(m1 + m2 )
This *loss of kinetic energy is used for doing the work in deforming the two bodies and is
absorbed in overcoming internal friction of the material. Since there will be no strain energy stored
up in the material due to elastic deformation, therefore the bodies cannot regain its original shape.
Hence the two bodies will adhere together and will move with reduced kinetic energy after impact.
The reduction of kinetic energy appears as heat energy because of the work done in overcoming the
internal friction during deformation.
3.23. Collision of Elastic Bodies
When two elastic bodies, as shown in Fig. 3.17 (a), collide with each other, they suffer a
change of form. When the bodies first touch, the pressure between them is zero. For a short time
thereafter, the bodies continue to approach each other and the pressure exerted by one body over the
other body increases. Thus the two bodies are compressed and deformed at the surface of contact
due to their mutual pressures.
Fig. 3.17. Collision of elastic bodies.
If one of the bodies is fixed then the other will momentarily come to rest and then rebound.
However, if both the bodies are free to move, then each body will momentarily come to rest relative
to the other. At this instant, the pressure between the two bodies becomes maximum and the
deformation is also a maximum. At this stage the two bodies move with a **common velocity, as
shown in Fig. 3.17 (b).
*
**
According to principle of conservation of energy, the energy cannot be lost.
This common velocity (v) may be calculated as discussed in the previous article.
62
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Theory of Machines
The work done in deforming the two bodies is stored up as strain energy. Since no energy is
absorbed in overcoming internal friction, therefore there will be no conversion of kinetic energy into
heat energy. Thus immediately after the instant at which the two bodies move with same velocity, the
bodies begin to regain their original shape.This process of regaining the original shape is called
restitution.
The strain energy thus stored is reconverted into kinetic energy and the two bodies
ultimately separates as shown in Fig. 3.17 (c). In this case, the change of momentum of each body
during the second phase of impact (i.e. when the bodies are separating) is exactly equal to the
change of momentum during the first phase of impact (i.e. when the bodies are approaching or
colliding).
Let
m 1 = Mass of the first body,
u1 = Velocity of the first body before impact,
v1 = Velocity of the first body after impact,
m 2, u2 and v 2 = Corresponding values for the second body, and
v = Common velocity of the two bodies at the instant when
compression has just ended.
∴ Change of momentum of first body during the second phase of impact
= m 1 (v 1 – v)
and change of momentum of the same body during first phase of impact
= m 1 (v – u1)
∴
m 1 (v 1 – v) = m 1 (v – u1)
or
v 1 = 2 v – u1
... (i)
Similarly, for the second body, change of momentum of the second body during second
phase of impact
= m 2 (v 2 – v)
and change of momentum of the second body during first phase of impact
= m 2 (v – u2)
∴
m 2 (v 2 – v) = m 2 (v – u2)
or v 2 = 2v – u2
... (ii)
Subtracting equation (ii) from equation (i), we get
v 1 – v 2 = (u2 – u1) = – (u1 – u2)
... (iii)
Therefore, we see that the relative velocity of the two bodies after impact is equal and
opposite to the relative velocity of the two bodies before impact. Due to the fact that physical bodies
are not perfectly elastic, the relative velocity of two bodies after impact is always less than the
relative velocity before impact. The ratio of the former to the latter is called coefficient of restitution and is represented by e. Mathematically, coefficient of restitution,
e=
=
Relative velocity after impact
v1 − v2
=
Relative velocity before impact
− (u1 − u2 )
v1 − v2
u2 − u1
or
v2 − v1
u1 − u2
The value of e = 0, for the perfectly inelastic bodies and e = 1 for perfectly elastic bodies. In
case the bodies are neither perfectly inelastic nor perfectly elastic, then the value of e lies between 0
and 1.
Chapter 3 : Kinetics of Motion
l
63
The final velocities of the colliding bodies after impact may be calculated as discussed below:
Since the change of velocity of each body during the second phase of impact is e times the
change of velocity during first phase of impact, therefore for the first body,
v 1 – v = e (v – u1) or
v 1 = v (1 + e) – e.u1
...(iv)
Similarly for the second body,
v 2 – v = e (v – u2)
or
v 2 = v (1 + e) – e.u2
...(v)
When e = 1, the above equations (iv) and (v) reduced to equations (i) and (ii).
Notes : 1. The time taken by the bodies in compression, after the instant of collision, is called the time of
compression or compression period.
2. The period of time from the end of the compression stage to the instant when the bodies separate (i.e.
the time for which the restitution takes place) is called time of restitution or restitution period.
3. The sum of compression period and the restitution period is called period of collision or period of
impact.
4. The velocities of the two bodies at the end of restitution period will be different from their common
velocity at the end of the compression period.
3.24. Loss of Kinetic Energy During Elastic Impact
Consider two bodies 1 and 2 having an elastic impact as shown in Fig. 3.17.
Let
m 1 = Mass of the first body,
u1 = Velocity of the first body before impact,
v 1 = Velocity of the first body after impact,
m 2, u2 and v 2 = Corresponding values for the second body,
e = Coefficient of restitution, and
EL= Loss of kinetic energy during impact.
We know that the kinetic energy of the first body, before impact
=
1
m .u2
2 1 1
Similarly, kinetic energy of the second body, before impact
=
1
m .u 2
2 2 2
∴ Total kinetic energy of the two bodies, before impact,
E1 =
1
1
m1 . u12 + m2 ⋅ u22
2
2
. . . (i)
Similarly, total kinetic energy of the two bodies, after impact
E2 =
1
1
m . v2 + m ⋅ v2
2 1 1 2 2 2
∴ Loss of kinetic energy during impact,
1
1
1
1
EL = E1 − E2 = m1 . u12 + m2 . u22 − m1 . v12 + m2 ⋅ v22
2
2
2
2
. . . (ii)
64
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Theory of Machines
=
(
) (
)
1
m1 . u12 + m2 . u22 − m1 . v12 + m2 ⋅ v22
2
Multiplying the numerator and denominator by (m 1 + m 2),
EL =
=
(
)
(
)
1
( m1 + m2 ) m1 ⋅ u12 + m2 ⋅ u22 − ( m1 + m2 ) m1 . v12 + m2 ⋅ v22
2 ( m1 + m2 )
(
1
m12 ⋅ u12 + m1 . m2 ⋅ u22 + m1 . m2 . u 21 + m22 . u22
2 ( m1 + m2 )
)
(
)
− m12 ⋅ v12 + m1 ⋅ m2 ⋅ v22 + m1 ⋅ m2 ⋅ v12 + m22 ⋅ v22
=
{
}
1
m12 ⋅ u12 + m22 ⋅ u22 + m1 ⋅ m2 (u12 + u 22 )
2 ( m1 + m2 )
− {m12 ⋅ v12 + m22 ⋅ v22 + m1 ⋅ m2 (v12 + v22 }
=
{
−
=
}
1
( m .u + m . u )2 − ( 2m m u u ) + m .m (u − u )2 + ( 2 m m u u )
1 1
2 2
1 2 1 2
1 2 1
2
1 2 1 2
2 ( m1 + m2 )
{(m .v + m .v ) − (2 m m
2
1 1
2
2
1
2
}
2
v1 v2 ) + m1 . m2 (v1 − v2 ) + ( 2 m1 m2 v1 v2 )
{
1
( m .u + m . u )2 + m . m (u − u )2
1 1
2 2
1 2 1
2
2 ( m1 + m2 )
}
{
}
2
2
− ( m1 . v1 + m2 .v2 ) + m1 .m2 (v1 − v2 )
We know that in an elastic impact,
Total momentum before impact = Total momentum after impact
i.e.
m 1.u1 + m 2.u2 = m 1.v 1 + m 2.v 2
or
(m 1.u1 + m 2.u2 )2 = (m 1.v 1 + m 2.v 2 )2
... (Squaring both sides)
∴ Loss of kinetic energy due to impact,
EL =
1
m1 .m2 (u1 − u2 ) 2 − m1 .m2 (v1 − v2 ) 2
2 ( m1 + m2 )
Substituting v 1 – v 2 = e (u1 – u2) in the above equation,
EL =
=
1
m1 .m2 (u1 − u2 ) 2 − m1 .m2 .e2 (u1 − u2 ) 2
2 ( m1 + m2 )
m1.m2
(u1 − u2 )2 (1− e2 )
2 ( m1 + m2 )
Notes : 1. The loss of kinetic energy may be found out by calculating the kinetic energy of the system before
impact, and then by subtracting from it the kinetic energy of the system after impact.
Chapter 3 : Kinetics of Motion
l
65
2. For perfectly inelastic bodies, e = 0, therefore
m1.m2
(u1 − u2 )2
2(m1 + m2 )
EL =
. . . (same as before)
3. For perfectly elastic bodies, e = l, therefore EL = 0.
4. If weights (instead of masses) of the two bodies are given, then the same may be used in all the
relations.
Example 3.19. A sphere of mass 50 kg moving at 3 m/s overtakes and collides with another
sphere of mass 25 kg moving at 1.5 m/s in the same direction. Find the velocities of the two masses
after impact and loss of kinetic energy during impact in the following cases :
1. When the impact is inelastic, 2. When the impact
is elastic, and 3. When coefficient of restitution is 0.6.
Solution. Given : m 1 = 50 kg ; u1 = 3 m/s ;
m 2 = 25 kg ; u2 = 1.5 m/s
1. When the impact is inelastic
In case of inelastic impact, the two spheres adhere
after impact and move with a common velocity. We know
that common velocity after impact,
m1.u1 + m2 .u2 50 × 3 + 25 × 1.5
=
= 2.5m s Ans.
50 + 25
m1 + m2
and loss of kinetic energy during impact,
v=
EL =
50 × 25
m1 .m2
(u1 − u2 )2 =
(3 − 1.5)2 N - m
2(m1 + m2 )
2 (50 + 25)
= 18.75 N-m Ans.
2. When the impact is elastic
Let
v 1 = Velocity of the first sphere immediately after impact, and
v 2 = Velocity of the second sphere immediately after impact.
We know that when the impact is elastic, the common velocity of the two spheres is the same
i.e. common velocity, v = 2.5 m/s.
∴
v 1 = 2v – u1 = 2 × 2.5 – 3 = 2 m/s Ans.
and
v 2 = 2v – u2 = 2 × 2.5 – 1.5 = 3.5 m/s Ans.
We know that during elastic impact, there is no loss of kinetic energy, i.e. EL = 0 Ans.
3. When the coefficient of restitution, e = 0.6
We know that v1 = (1 + e) v – e.u1 = (1 + 0.6) 2.5 – 0.6 × 3 = 2.2 m/s Ans.
and
v 2 = (1 + e) v – e.u2 = (1 + 0.6) 2.5 – 0.6 × 1.5 = 3.1 m/s Ans.
Loss of kinetic energy during impact,
EL =
=
m1.m2
(u1 − u2 ) 2 (1 − e 2 )
2 (m1 + m2 )
50 × 25
(3 − 1.5)2 (1 − 0.62 ) = 12 N - m Ans.
2 (50 + 25)
66
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Theory of Machines
Example 3.20. A loaded railway wagon has a mass of 15 tonnes and moves along a level
track at 20 km/h. It over takes and collides with an empty wagon of mass 5 tonnes, which is moving
along the same track at 12 km/h. If the each wagon is fitted with two buffer springs of stiffness 1000
kN/m, find the maximum deflection of each spring during impact and the speeds of the wagons
immediately after impact ends.
If the coefficient of restitution for the buffer springs is 0.5, how would the final speeds be
affected and what amount of energy will be dissipated during impact ?
Solution. Given : m 1 = 15 t = 15 000 kg ; u1 = 20 km/h = 5.55 m/s ; m 2 = 5 t = 5000 kg ;
u2 = 12 km/h = 3.33 m/s ; s = 1000 kN/m = 1 × 106 N/m ; e = 0.5
During impact when both the wagons are moving at the same speed (v) after impact, the
magnitude of the common speed (v) is given by
v=
m1 .u1 + m2 .u2 15 000 × 5.55 + 5000 × 3.33
=
= 5 m / s Ans.
15 000 + 5000
m1 + m2
Maximum deflection of each spring
Let
x = Maximum deflection of each buffer spring during impact, and
s = Stiffness of the spring = 1000 kN/m = 1 × 106 N/m
... (Given)
∴ Strain energy stored in one spring
1
1
s . x 2 = × 1 × 106 × x 2 = 500 × 103 x 2 N - m
2
2
Since the four buffer springs (two in each wagon) are strained, therefore total strain energy
stored in the springs
= 4 × 500 × 103 x 2 = 2 × 106 x 2 N-m
... (i)
Difference in kinetic energies before impact and during impact
=
=
15 000 × 5000
m1 . m2
(u1 − u2 )2 =
(5.55 − 3.33)2 N - m
2(m1 + m2 )
2 (15 000 + 5000)
= 9240 N-m
The difference between the kinetic energy
before impact and kinetic energy during impact is
absorbed by the buffer springs. Thus neglecting all
losses, it must be equal to strain energy stored in the
springs.
Equating equations (i) and (ii),
2 × 106 x 2 = 9240
or
x 2 = 9240 / 2 × 106 = 0.00 462
∴
x = 0.068 m = 68 mm Ans.
...(ii)
Speeds of the wagons immediately after impact ends
Immediately after impact ends, let v 1 and v 2 be the speeds of the loaded wagon and empty
wagon respectively.
We know that
v 1 = 2v – u1 = 2 × 5 – 5.55 = 4.45 m/s Ans.
and
v 2 = 2v – u2 = 2 × 5 – 3.33 = 6.67 m/s Ans.
When the coefficient of restitution, e = 0.5 is taken into account, then
v1 = (1 + e)v – e.u1 = (1+ 0.5) 5 – 0.5 × 5.55 = 4.725 m/s Ans.
l
Chapter 3 : Kinetics of Motion
67
and
v2 = (1 + e)v – e.u2 = (1 + 0.5)5 – 0.5 × 3.33 = 5.635 m/s Ans.
Amount of energy dissipated during impact
We know that amount of energy dissipated during impact,
EL =
m1.m2
(u1 − u2 ) 2 (1 − e 2 ) = 9240 (1 − 0.52 ) N-m
2 (m1 + m2 )
= 9240 × 0.75 = 6930 N-m Ans.
Example 3.21. Fig. 3.18 shows a flywheel A connected through a torsionally flexible spring
to one element C of a dog clutch. The other element D of the clutch is free to slide on the shaft but
it must revolve with the shaft to which the flywheel B is keyed.
The moment of inertia of A and B are 22.5 kg-m2
and 67.5 kg-m2 and the torsional stiffness of the spring is
225 N-m per radian. When the flywheel A is revolving at 150
r.p.m. and the flywheel B is at rest, the dog clutch is suddenly
engaged. Neglecting all losses, find : 1. strain energy stored
in the spring, 2. the maximum twist of the spring, and 3. the
speed of flywheel when the spring regains its initial unstrained
condition.
Fig. 3.18
Solution. Given : IA = 22.5 kg-m2 ; IB = 67.5 kg-m2 ; q = 225 N-m/rad ; N A = 150 r.p.m. or
ωA = 2π × 150/60 = 15.71 rad/s
Immediately after the clutch is engaged, the element C of the clutch comes to rest
momentarily. But the rotating flywheel A starts to wind up the spring, thus causing equal and opposite torques to act on flywheels A and B. The magnitude of the torque increases continuously until
the speeds of flywheels A and B are equal. During this interval, the strain energy is stored in the
spring. Beyond this, the spring starts to unwind and the strain energy stored in the spring is reconverted into kinetic energy of the flywheels.
Since there is no external torque acting on the system, therefore the angular momentum will
remain constant. Let ω be the angular speed of both the flywheels at the instant their speeds are equal.
22.5 × 15.71
I A . ωA
=
= 3.93 rad/s
I A + I B 22.5 + 67.5
Kinetic energy of the system at this instant (i.e. when speeds are equal),
∴
(IA + IB) ω = IA . ωA
or
ω=
1
1
( I A + I B ) ω2 = (22.5 + 67.5) (3.93)2 = 695 N-m
2
2
and the initial kinetic energy of the flywheel A ,
E2 =
1
1
I A (ωA ) 2 = × 22.5 (15.71) 2 = 2776 N-m
2
2
1. Strain energy stored in the spring
E1 =
We know that strain energy stored in the spring
= E1 – E2 = 2776 – 695 = 2081 N-m Ans.
2. Maximum twist of the spring
Let
θ = Maximum twist of the spring in radians, and
q = Torsional stiffness of spring = 225 N-m/rad
...(Given)
68
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Theory of Machines
We know that the strain energy,
1
1
q.θ2 = × 225 θ2 = 112.5 θ2
2
2
2
θ = 2081/112.5 = 18.5
2081 =
∴
θ = 4.3 rad = 4.3 × 180/π = 246.3° Ans.
or
3. Speed of each flywheel when the spring regains its initial unstrained condition
Let N A1 and N B1 be the speeds of the flywheels A and B respectively, when the spring
regains its initial unstrained condition. We know that
60 ω
60 × 3.93
NA1 = 2 N – NA = 2
− NA = 2
− 150
2π
2π
= 75 – 150 = – 75 r.p.m.
Similarly
N B1 = 2 N − N B = 75 − 0 = 75 r.p.m.
...(3NB = 0)
From above we see that when the spring regains its initial unstrained condition, the flywheel
A will revolve at 75 r.p.m. in the opposite direction to its initial motion and the flywheel B will
revolve at 75 r.p.m. in same direction as the initial motion of flywheel A . Ans.
EXERCISES
1.
A flywheel fitted on the crank shaft of a steam engine has a mass of 1 tonne and a radius of gyration
0.4 m. If the starting torque of the engine is 650 J which may be assumed constant, find 1. Angular
acceleration of the flywheel, and 2. Kinetic energy of the flywheel after 10 seconds from the start.
[Ans. 4.06 rad/s2 ; 131.87 kN-m]
2.
A load of mass 230 kg is lifted by means of a rope which is wound several times round a drum and
which then supports a balance mass of 140 kg. As the load rises, the balance mass falls. The drum has
a diameter of 1.2 m and a radius of gyration of 530 mm and its mass is 70 kg. The frictional resistance
to the movement of the load is 110 N, and that to the movement of the balance mass 90 N. The
frictional torque on the drum shaft is 80 N-m.
Find the torque required on the drum, and also the power required, at the instant when the load has an
upward velocity of 2.5 m/s and an upward acceleration of 1.2 m/s2.
[Ans. 916.2 N-m ; 4.32 kW]
3.
4.
5.
A riveting machine is driven by a 3.5 kW motor. The moment of inertia of the rotating parts of the
machine is equivalent to 67.5 kg-m2 at the shaft on which the flywheel is mounted. At the commencement of an operation, the flywheel is making 240 r.p.m. If closing a rivet occupies 1 second and
corresponds to an expenditure of 9 kN-m of energy, find the reduction of speed of the flywheel. What
is the maximum rate at which rivets can be closed ?
[Ans. 33.2 r.p.m. ; 24 per min ]
The drum of a goods hoist has a mass of 900 kg. It has an effective diameter of 1.5 m and a radius of
gyration of 0.6 m. The loaded cage has a mass of 550 kg and its frictional resistance in the vertical
line of travel is 270 N. A maximum acceleration of 0.9 m/s2 is required. Determine : 1. The necessary
driving torque on the drum, 2. The tension in the rope during acceleration, and 3. The power developed at a steady speed of 3.6 m/s.
[Ans. 4.64 kN-m ; 6.16 kN ; 22.3 kW]
A valve operating in a vertical direction is opened by a cam and closed by a spring and when fully
open the valve is in its lowest position. The mass of the valve is 4 kg and its travel is 12.5 mm and the
constant frictional resistance to the motion of the valve is 10 N. The stiffness of the spring is 9.6 N/
mm and the initial compression when the valve is closed is 35 mm. Determine 1. the time taken
to close the valve from its fully open position, and 2. the velocity of the valve at the moment of
impact.
[Ans. 0.0161 s ; 1.4755 m/s]
Chapter 3 : Kinetics of Motion
6.
7.
8.
9.
l
69
A railway truck of mass 20 tonnes, moving at 6.5 km/h is brought to rest by a buffer stop. The
buffer exerts a force of 22.5 kN initially and this force increases uniformly by 60 kN for each 1 m
compression of the buffer. Neglecting any loss of energy at impact, find the maximum compression
of the buffer and the time required for the truck to be brought to rest. [Ans. 0.73 m ; 0.707 s]
A cage of mass 2500 kg is raised and lowered by a winding drum of 1.5 m diameter. A brake drum
is attached to the winding drum and the combined mass of the drums is 1000 kg and their radius of
gyration is 1.2 m. The maximum speed of descent is 6 m/s and when descending at this speed, the
brake must be capable of stopping the load in 6 m. Find 1. the tension of the rope during stopping
at the above rate, 2. the friction torque necessary at the brake, neglecting the inertia of the rope, and
3. In a descent of 30 m, the load starts from rest and falls freely until its speed is 6 m/s. The brake
is then applied and the speed is kept constant at 6 m/s until the load is 10 m from the bottom. The
brake is then tightened so as to give uniform retardation, and the load is brought to rest at the
bottom. Find the total time of descent.
[Ans. 32 kN ; 29.78 kN-m ; 7.27 s]
A mass of 275 kg is allowed to fall vertically through 0.9 m on to the top of a pile of mass 450 kg.
Assuming that the falling mass and the pile remain in contact after impact and that the pile is
moved 150 mm at each blow, find allowing for the action of gravity after impact, 1. The energy lost
in the blow, and 2. The average resistance against the pile.
[Ans. 13.3 kN ; 1.5 kN-m]
Fig. 3.19 shows a hammer of mass 6 kg and pivoted at A . It falls against a wedge of mass 1 kg
which is driven forward 6 mm, by the impact into a heavy rigid block. The resistance to the wedge
varies uniformly with the distance through which it moves, varying zero to R newtons.
Fig. 3.19
Fig. 3.20
Neglecting the small amount by which the hammer rises after passing through the vertical through
A and assuming that the hammer does not rebound, find the value of R.
[Ans. 8.38 kN]
10.
11.
Fig. 3.20 shows a tilt hammer, hinged at O, with its head A resting on top of the pile B. The
hammer, including the arm O A, has a mass of 25 kg. Its centre of gravity G is 400 mm horizontally
from O and its radius of gyration about an axis through G parallel to the axis of the pin O is 75 mm.
The pile has a mass of 135 kg. The hammer is raised through 45° to the position shown in dotted
lines, and released. On striking the pile, there is no rebound. Find the angular velocity of the
hammer immediately before impact and the linear velocity of the pile immediately after impact.
Neglect any impulsive resistance offered by the earth into which the pile is being driven.
[Ans. 5.8 rad/s, 0.343 m/s]
The tail board of a lorry is 1.5 m long and 0.75 m high. It is hinged along the bottom edge to the
floor of the lorry. Chains are attached to the top corners of the board and to the sides of the lorry
so that when the board is in a horizontal position the chains are parallel and inclined at 45° to the
horizontal. A tension spring is inserted in each chain so as to reduce the shock and these are adjusted
to prevent the board from dropping below the horizontal. Each spring exerts a force of 60 N/mm of
extension.
Find the greatest force in each spring and the resultant force at the hinges when the board falls
freely from the vertical position. Assume that the tail board is a uniform body of mass 30 kg.
[Ans. 3636 N ; 9327 N]
70
l
12.
13.
14.
15.
16.
S.No.
Theory of Machines
A motor drives a machine through a friction clutch which transmits 150 N-m while slip
occurs during engagement. For the motor, the rotor has a mass of 60 kg with radius of gyration 140
mm and the inertia of the machine is equivalent to a mass of 20 kg with radius of gyration 80 mm.
If the motor is running at 750 r.p.m. and the machine is at rest, find the speed after engaging the
clutch and the time taken.
[Ans. 70.87 rad/s ; 0.06 s]
A shaft carrying a rotor of moment of inertia 10 kg-m2 revolves at a speed of 600 r.p.m. and is
engaged by means of a friction clutch to another shaft on the same axis having a moment of inertia
of 15 kg-m2. If the second shaft is initially at rest, find 1. the final speed of rotation of the two
shafts together after slipping has ceased, 2. the time of slip if the torque is constant at 250 N-m
during slipping, and 3. the kinetic energy lost during the operation.
[Ans. 25.136 rad/s ; 1.5 s ; 11.85 kN-m]
A self-propelled truck of total mass 25 tonnes and wheel diameter 750 mm runs on a track for
which the resistance is 180 N per tonne. The engine develops 60 kW at its maximum speed of 2400
r.p.m. and drives the axle through a gear box. Determine : 1. the time to reach full speed from rest
on the level if the gear reduction ratio is 10 to 1. Assume the engine torque to be constant and a
gearing efficiency of 94 per cent, and 2. the gear ratio required to give an acceleration of 0.15 m/s2
on an up gradient of 1 in 70 assuming a gearing efficiency of 90 per cent.
[Ans. 157 s ; 20.5]
A motor vehicle of mass 1000 kg has road wheels of 600 mm rolling diameter. The total moment of
inertia of all four road wheels together with the half shafts is 10 kg-m2, while that of the engine and
clutch is 1 kg-m2. The engine torque is 150 N-m, the transmission efficiency is 90 per cent and the
tractive resistance is constant at 500 N. Determine 1. Gear ratio between the engine and the road
wheels to give maximum acceleration on an upgrade of 1 in 20, and 2. The value of this maximum
acceleration.
[Ans. 13 ; 1.74 m/s2]
In a mine hoist a loaded cage is raised and an empty cage is lowered by means of a single rope. This
rope passes from one cage, over a guide pulley of 1.2 m effective diameter, on to the winding drum
of 2.4 m effective diameter, and then over a second guide pulley, also of 1.2 m effective diameter, to
the other cage. The drum is driven by an electric motor through a double reduction gear.
Determine the motor torque required, at an instant when the loaded cage has an upward acceleration of 0.6 m/s2, given the following data :
Part
Maximum speed
Mass
Radius of
Frictional
(r.p.m.)
(kg)
gyration (mm)
resistance
1
Motor and pinion
N
500
150
–
2.
Intermediate gear shaft
N
5
600
225
45 N-m
3000
900
1500 N-m
125
10 000
5000
450
–
–
30 N-m
2500 N
1500 N
and attached wheel
3.
Drum and attached gear
N
20
4.
5.
6.
Guide pulley, each
Rising rope and cage
Falling rope and cage
–
–
–
[Ans. 4003.46 N-m]
DO YOU KNOW ?
1.
2.
3.
4.
5.
State Newton’s three laws of motion.
What do you understand by mass moment of inertia ? Explain clearly.
What is energy ? Explain the various forms of mechanical energies.
State the law of conservation of momentum.
Show that for a relatively small rotor being started from rest with a large rotor, the energy lost in the
clutch is approximately equal to that given to the rotor.
Chapter 3 : Kinetics of Motion
6.
7.
Prove the relation for the torque required in order to accelerate a geared system.
Discuss the phenomenon of collision of elastic bodies.
8.
Define the term ‘coefficient of restitution’.
l
71
OBJECTIVE TYPE QUESTIONS
1.
The force which acts along the radius of a circle and directed ...... the centre of the circle is known
as centripetal force.
(a)
2.
(a)
3.
5.
9.
kgf-m-s2
(c)
kg-m2
(d)
N-m
force
(b)
work
(c)
power
(d)
none of these
(b)
kinetic energy
(c)
electrical energy
(d)
chemical energy
When a body of mass moment of inertia I (about a given axis) is rotated about that axis with an
angular velocity, then the kinetic energy of rotation is
0.5 I.ω
(b)
I.ω
(c)
0.5 I.ω2
(b)
elastic bodies
(d)
I.ω2
The wheels of a moving car possess
potential energy only
(b)
kinetic energy of translation only
(c)
kinetic energy of rotation only
(d)
kinetic energy of translation and rotation both.
The bodies which rebound after impact are called
inelastic bodies
The coefficient of restitution for inelastic bodies is
(a)
zero
(b)
between zero and one
(c)
one
(d)
more than one
Which of the following statement is correct ?
(a)
10.
(b)
potential energy
(a)
8.
m4
(a)
(a)
7.
towards
The energy possessed by a body, for doing work by virtue of its position, is called
(a)
6.
(b)
Joule is a unit of
(a)
4.
away from
The unit of mass moment of inertia in S.I. units is
The kinetic energy of a body during impact remains constant.
(b)
The kinetic energy of a body before impact is equal to the kinetic energy of a body after impact.
(c)
The kinetic energy of a body before impact is less than the kinetic energy of a body after
impact.
(d)
The kinetic energy of a body before impact is more than the kinetic energy of a body after
impact.
A body of mass m moving with a constant velocity v strikes another body of same mass m moving
with same velocity but in opposite direction. The common velocity of both the bodies after collision
is
(a)
v
(b)
2v
(c)
4v
(d)
8v
ANSWERS
1. (b)
2. (c)
3. (b)
4. (a)
5. (c)
6. (d)
7. (b)
8. (a)
9. (d)
10. (b)
GO To FIRST
CONTENTS
CONTENTS
72
l
Theory of Machines
4
Simple
Harmonic
Motion
Features
1. Introduction.
2. Velocity and Acceleration of
a Particle Moving with
Simple Harmonic Motion.
3. Differential Equation of
Simple Harmonic Motion.
4. Terms Used in Simple
Harmonic Motion.
5. Simple Pendulum.
6. Laws of Simple Pendulum.
7. Closely-coiled Helical
Spring.
8. Compound Pendulum.
9. Centre of Percussion.
10. Bifilar Suspension.
11. Trifilar Suspension
(Torsional Pendulum).
4.1.
Introduction
Consider a particle
moving round the circumference of a circle in an
anticlockwise direction, with
a constant angular velocity,
as shown in Fig. 4.1. Let P
be the position of the particle
at any instant and N be the
projection of P on the diameter X X ′ of the circle.
Fig. 4.1. Simple harmonic
motion.
It will be noticed that when
the point P moves round the
circumference of the circle from X to
Y, N moves from X to O, when P
moves from Y to X ′, N moves from O
to X ′. Similarly when P moves from
X ′ to Y ′, N moves from X ′ to O and
finally when P moves from Y ′ to X, N
moves from O to X . Hence, as P
A clock pendulum
completes one revolution, the point N
executes Simple
completes one vibration about the Harmonic Motion.
72
CONTENTS
CONTENTS
Chapter 4 : Simple Harmonic Motion
l
73
point O. This to and fro motion of N is known as simple
harmonic motion (briefly written as S.H.M.).
4.2.
Velocity and Acceleration of a
Particle Moving with Simple
Harmonic Motion
Consider a particle, moving round the circumference of a circle of radius r, with a uniform angular velocity
ω rad/s, as shown in Fig. 4.2. Let P be any position of the
particle after t seconds and θ be the angle turned by the
particle in t seconds. We know that
θ = ω.t
If N is the projection of P on the diameter X X ′,
then displacement of N from its mean position O is
x = r.cos θ = r.cos ω.t
... (i)
The velocity of N is the component of the velocity
of P parallel to X X ′, i.e.
Movements of a ship up and down in
a vertical plane about transverse axis
(called Pitching) and about longitude
(called rolling) are in Simple
Harmonic Motion.
vN = v sin θ = ω.r sin θ = ω r 2 − x 2
... (ii)
... 3 v = ω.r, and r sin θ = NP = r 2 − x 2
Fig. 4.2. Velocity and acceleration of a particle.
A little consideration will show that velocity is maximum, when x = 0, i.e. when N passes
through O i.e., its mean position.
∴
v max = ω.r
We also know that the acceleration of P is the centripetal acceleration whose magnitude is
ω2.r. The acceleration of N is the component of the acceleration of P parallel to X X ′ and is directed
towards the centre O, i.e.,
aN = ω2 . r cos θ = ω2 . x
... (3 x = r cos θ) ...(iii)
The acceleration is maximum when x = r i.e. when P is at X or X′.
∴
amax = ω2.r
It will also be noticed from equation (iii) that when x = 0, the acceleration is zero i.e. N passes
through O. In other words, the acceleration is zero at the mean position. Thus we see from equation
(iii) that the acceleration of N is proportional to its displacement from its mean position O, and it is
74
l
Theory of Machines
always directed towards the centre O; so that the motion of N is simple harmonic.
In general, a body is said to move or vibrate with simple harmonic motion, if it satisfies the
following two conditions :
1. Its acceleration is always directed towards the centre, known as point of reference or
mean position ;
2. Its acceleration is proportional to the distance from that point.
4.3.
Differential Equation of Simple Harmonic Motion
We have discussed in the previous article that the displacement of N from its mean position O is
x = r.cos θ = r.cos ωt
... (i)
Differentiating equation (i), we have velocity of N,
dx = v = r .ω sin ω t
N
dt
Again differentiating equation (ii), we have acceleration of N,
* d2x
dt 2
... (ii)
= aN = − r.ω. ω cos ω t = − ω2 .r cos ωt = − ω2 .x
... (iii)
... (3 r cos ωt = x)
2
d x + ω2 x = 0
dt 2
This is the standard differential equation for simple harmonic motion of a particle. The
solution of this differential equation is
or
x = A cos ω t + B sin ω t
... (iv)
where A and B are constants to be determined by the initial conditions of the motion.
In Fig. 4.2, when t = 0, x = r i.e. when points P and N lie at X, we have from equation (iv), A = r
Differentiating equation (iv),
dx = − A.ω.sin ω t + B.ω cos ω t
dt
When t = 0, dx = 0, therefore, from the above equation, B = 0. Now the equation (iv) becomes
dt
. . . [Same as equation (i)]
x = r cos ω t
The equations (ii) and (iii) may be written as
dx = v = − ω.r sin ωt = ω.r cos ( ωt + π / 2)
N
dt
and
d 2 x = a = − ω2 .r cos ωt = ω2 .r cos (ωt + π )
N
dt 2
These equations show that the velocity leads the displacement by 90° and acceleration leads
the displacement by 180°.
*
The negative sign shows that the direction of acceleration is opposite to the direction in which x increases,
i.e. the acceleration is always directed towards the point O.
Chapter 4 : Simple Harmonic Motion
4.4.
l
75
Terms Used in Simple Harmonic Motion
The following terms, commonly used in simple harmonic motion, are important from the
subject point of view.
1. Amplitude. It is the maximum displacement of a body from its mean position. In Fig. 4.2,
O X or OX ′ is the amplitude of the particle P. The amplitude is always equal to the radius of the
circle.
2. Periodic time. It is the time taken for one complete revolution of the particle.
∴
Periodic time, tp = 2 π/ω seconds
We know that the acceleration,
a = ω2.x
or
2
t p = π = 2π
ω
∴
ω2 = a
x
x = 2π
a
or ω =
a
x
Displacement
seconds
Acceleration
It is thus obvious, that the periodic time is independent of amplitude.
3. Frequency. It is the number of cycles per second and is the reciprocal of time period, tp .
∴
Frequency, n =
ω = 1 = 1
2π t p 2π
a Hz
x
Notes : 1. In S.I. units, the unit of frequency is hertz (briefly written as Hz) which is equal to one cycle per
second.
2. When the particle moves with angular simple harmonic motion, then the periodic time,
t p = 2π
and
frequency, n =
1
2π
Angular displacement
= 2π
Angular acceleration
θ s
α
α
Hz
θ
Example 4.1. The piston of a steam engine moves with simple harmonic motion. The crank
rotates at 120 r.p.m. with a stroke of 2 metres. Find the velocity and acceleration of the piston, when
it is at a distance of 0.75 metre from the centre.
Solution. Given : N = 120 r.p.m. or ω = 2π × 120/60 = 4π rad/s ; 2r = 2 m or r = 1 m;
x = 0.75 m
Velocity of the piston
We know that velocity of the piston,
v = ω r 2 − x 2 = 4 π 1 − (0.75) 2 = 8.31 m/s Ans.
Acceleration of the piston
We also know that acceleration of the piston,
a = ω2.x = (4π)2 0.75 = 118.46 m/s2 Ans.
Example 4.2. A point moves with simple harmonic motion. When this point is 0.75 metre
from the mid path, its velocity is 11 m/s and when 2 metres from the centre of its path its velocity is
3 m/s. Find its angular velocity, periodic time and its maximum acceleration.
76
l
Theory of Machines
Solution. Given : When x = 0.75 m, v = 11 m/s ; when x = 2 m, v = 3 m/s
Angular velocity
Let
ω = Angular velocity of the particle, and
r = Amplitude of the particle.
We know that velocity of the point when it is 0.75 m from the mid path (v),
11 = ω r 2 − x 2 = ω r 2 − (0.75) 2
. . . (i)
Similarly, velocity of the point when it is 2 m from the centre (v),
3 = ω r 2 − 22
. . . (ii)
Dividing equation (i) by equation (ii),
2
2
11 = ω r − (0.75) =
3
ω r 2 − 22
r 2 − (0.75) 2
r 2 − 22
Squaring both sides,
2
121 r − 0.5625
=
9
r2 − 4
121 r2 – 484 = 9r2 – 5.06
∴
or
112 r2 = 478.94
r2 = 478.94 / 112 = 4.276
or
r = 2.07 m
Substituting the value of r in equation (i),
11 = ω (2.07)2 − (0.75)2 = 1.93 ω
∴
ω = 11/1.93 = 5.7 rad/s Ans.
Periodic time
We know that periodic time,
tp = 2π / ω = 2π / 5.7 = 1.1 s Ans.
Maximum acceleration
We know that maximum acceleration,
amax = ω2.r = (5.7)2 2.07 = 67.25 m/s2 Ans.
4.5.
Simple Pendulum
A simple pendulum, in its simplest form, consists
of heavy bob suspended at the end of a light inextensible
and flexible string. The other end of the string is fixed at
O, as shown in Fig. 4.3.
Let
L = Length of the string,
m = Mass of the bob in kg,
W = Weight of the bob in newtons
= m.g, and
θ = Angle through which the string
is displaced.
Fig 4.3. Simple pendulum.
Chapter 4 : Simple Harmonic Motion
l
77
When the bob is at A , the pendulum is in equilibrium position. If the bob is brought to B or C
and released, it will start oscillating between the two positions B and C, with A as the mean position.
It has been observed that if the angle θ is very small (less than 4° ), the bob will have simple harmonic
motion. Now, the couple tending to restore the bob to the equilibrium position or restoring torque,
T = m.g sin θ × L
Since angle θ is very small, therefore sin θ = θ radians.
∴
T = m.g.L.θ
We know that the mass moment of inertia of the bob about an axis through the point of
suspension,
I = mass × (length)2 = m.L 2
∴ Angular acceleration of the string,
m.g .L.θ g .θ
α=T =
=
I
L
m.L2
or
θ = L
α g
Angular displacement L
=
g
Angular acceleration
i.e.
We know that the periodic time,
Displacement
= 2π
Acceleration
t p = 2π
L
g
... (i)
and frequency of oscillation,
n= 1 = 1
t p 2π
g
L
... (ii)
From above we see that the periodic time and the frequency of oscillation of a simple
pendulum depends only upon its length and acceleration due to gravity. The mass of the bob has no
effect on it.
Notes : 1. The motion of the bob from one extremity to the other (i.e. from B to C or C to B) is known as beat
or swing. Thus one beat = 1 oscillation.
2
∴ Periodic time for one beat = π
L /g
2. A pendulum, which executes one beat per second (i.e. one complete oscillation in two seconds) is
known as a second’s pendulum.
4.6.
Laws of Simple Pendulum
The following laws of a simple pendulum are important from the subject point of view :
1. Law of isochronism. It states, “The time period (tp ) of a simple pendulum does not depend
upon its amplitude of vibration and remains the same, provided the angular amplitude (θ) does not
exceed 4°.”
2. Law of mass. It states, “The time period (tp ) of a simple pendulum does not depend upon
the mass of the body suspended at the free end of the string.”
3. Law of length. It states, “The time period (tp ) of a simple pendulum is directly proportional to
L , where L is the length of the string.”
4. Law of gravity. It states, “The time period (tp ) of a simple pendulum is inversely proportional to
g , where g is the acceleration due to gravity.”
78
l
Theory of Machines
Note: The above laws of a simple pendulum are true from the equation of the periodic time i.e.
t p = 2π L / g
4.7.
Closely-coiled Helical Spring
Consider a closely-coiled helical spring, whose upper end is
fixed, as shown in Fig. 4.4. Let a body be attached to the lower end.
Let A A be the equilibrium position of the spring, after the mass is
attached. If the spring is stretched up to BB and then released, the
mass will move up and down with simple harmonic motion.
Let
m = Mass of the body in kg,
W = Weight of the body in newtons = m.g,
x = Displacement of the load below equilibrium position in metres,
s = Stiffnes of the spring in N/m i.e. restoring
force per unit displacement from the equilibrium position,
a = Acceleration of the body in m/s2.
We know that the deflection of the spring,
Fig. 4.4. Closely-coiled
helical spring.
m.g
s
Then disturbing force = m.a
restoring force = s.x
Equating equations (i) and (ii),
δ=
and
m.a = s.x* or
*
... (i)
... (ii)
x= m
a
s
The differential equation for the motion of the spring is
m
d 2x = −
d 2 x = s.x
s.x or
–
m
dt 2
dt 2
(
... Here ω =
2
The – ve sign indicates that the restoring force s.x is opposite to the direction of disturbing force.
s
m
)
Chapter 4 : Simple Harmonic Motion
l
79
We know that periodic time,
Displacement
= 2π
Acceleration
t p = 2π
= 2π
m = π
2
s
frequency, n = 1 = 1
t p 2π
and
x
a
δ
g
s = 1
m 2π
mg
... 3 δ = s
g
δ
Note: If the mass of the spring (m 1) is also taken into consideration, then the periodic time,
t p = 2π
s
Hz
m + m1 / 3
frequency, n = 1
2π
and
m + m1 / 3
seconds,
s
Example 4.3. A helical spring, of negligible mass, and which is found to extend 0.25 mm
under a mass of 1.5 kg, is made to support a mass of 60 kg. The spring and the mass system is
displaced vertically through 12.5 mm and released. Determine the frequency of natural vibration of
the system. Find also the velocity of the mass, when it is 5 mm below its rest position.
Solution. Given : m = 60 kg ; r = 12.5 mm = 0.0125 m ; x = 5 mm = 0.005 m
Since a mass of 1.5 kg extends the spring by 0.25 mm, therefore a mass of 60 kg will extend
the spring by an amount,
δ = 0.25 × 60 = 10 mm = 0.01 m
1.5
Frequency of the system
We know that frequency of the system,
n= 1
2π
g
= 1
δ 2π
9.81 = 4.98
Hz Ans.
0.01
Velocity of the mass
Let
v = Linear velocity of the mass.
We know that angular velocity,
ω** =
g
=
δ
9.81 =
31.32 rad/s
0.01
and
v = ω r 2 − x 2 = 31.32 (0.0125)2 − (0.005)2 = 0.36 m/s Ans.
4.8.
Compound Pendulum
When a rigid body is suspended vertically, and it oscillates
with a small amplitude under the action of the force of gravity, the
body is known as compound pendulum, as shown in Fig. 4.5.
Let
*
m = Mass of the pendulum in kg,
W = Weight of the pendulum in
newtons = m.g,
We know that periodic time,
tp = 2π / ω or
ω = 2π / tp = 2π × n = 2π × 4.98 = 31.3 rad/s
Fig. 4.5. Compound pendulum.
...(∵ n = 1/tp)
80
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Theory of Machines
kG
h
= Radius of gyration
about an axis
through the centre
of gravity G and
perpendicular to
the plane of
motion, and
= Distance of point
of suspension O
from the centre of
gravity G of the
body.
If the pendulum is given a small
angular displacement θ, then the couple
tending to restore the pendulum to the
equilibrium position O A,
T = mg sin θ × h = mgh sin θ
Since θ is very small, therefore substituting sin θ = θ radians, we get
T = mgh θ
Now, the mass moment of inertia about the axis of suspension O,
(
I = I G + m.h 2 = m kG2 + h 2
)
. . . (By parallel axis theorem)
∴ Angular acceleration of the pendulum,
mgh θ
gh θ
α=T =
= 2
= constant × θ
2
2
I
m ( kG + h ) kG + h 2
We see that the angular acceleration is directly proportional to angular displacement,
therefore the pendulum executes simple harmonic motion.
∴
2
2
θ = kG + h
g .h
α
We know that the periodic time,
t p = 2π
= 2π
Displacement
= 2π
Acceleration
kG2 + h2
g.h
1 = 1
and frequency of oscillation, n =
tp
2π
θ
α
... (i)
g .h
kG2
+ h2
... (ii)
Chapter 4 : Simple Harmonic Motion
l
81
Notes : 1. Comparing this equation with equation (ii) of simple pendulum, we see that the equivalent length of
a simple pendulum, which gives the same frequency as compound pendulum, is
L=
kG2 + h 2 kG2
=
+h
h
h
2. Since the equivalent length of simple pendulum (L) depends upon the distance between the point of
suspension and the centre of gravity (G), therefore L can be changed by changing the position of point of suspension. This will, obviously, change the periodic time of a compound pendulum. The periodic time will be minimum
if L is minimum. For L to be minimum, the differentiation of L with respect to h must be equal to zero, i.e.
∴
dL =
0
dh
or
2
d kG + h =
0
dh h
− kG2
+1 = 0
h2
or
kG = h
Thus the periodic time of a compound pendulum is minimum when the distance between the point of
suspension and the centre of gravity is equal to the radius of gyration of the body about its centre of gravity.
∴ Minimum periodic time of a compound pendulum,
t p ( min ) = 2π
4.9.
2kG
g
. . . [Substituting h = k G in equation (i)]
Centre of Percussion
The centre of oscillation is sometimes termed as centre of percussion. It is defined as that point at which a blow
may be struck on a suspended body so that the reaction at the
support is zero.
Consider the case of a compound pendulum suspended
at O as shown in Fig. 4.6. Suppose the pendulum is at rest in
the vertical position, and a blow is struck at a distance L from
the centre of suspension. Let the magnitude of blow is F newtons. A little consideration will show that this blow will have
the following two effects on the body :
Fig. 4.6. Centre of percusssion.
1. A force (F) acting at C will produce a linear motion
with an acceleration a, such that
F = m.a
... (i)
where m is the mass of the body.
2. A couple with moment equal to (F × l ) which will tend to produce a motion of rotation in
the clockwise direction about the centre of gravity G. Let this turning moment (F × l ) produce an
angular acceleration (α), such that
F × l = IG × α
... (ii)
where IG is the moment of inertia of the body about an axis passing through G and parallel to the axis
of rotation.
From equation (i)
a = F/m
... (iii)
and from equation (ii),
α=
F .l
IG
82
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Theory of Machines
Now corresponding linear acceleration of O,
a0 = α.h = F .l.h = F .l .2h
IG
m.kG
... (iv)
(3 I G = m.kG2 )
where k G is the radius of gyration of the body about the centre of
gravity G.
Since there is no reaction at the support when the body is
struck at the centre of percussion, therefore a should be equal to a0.
Equating equations (iii) and (iv),
F
F .l.h
=
m m.kG2
kG2 = l .h, and l =
or
kG2
h
... (v)
We know that the equivalent length of a simple pendulum,
L=
kG2 + h kG2
=
+h=l+h
h
h
... (vi)
From equations (v) and (vi), it follows that
1. The centre of percussion is below the centre of gravity
A pendulum clock designed
by Galileo. Galileo was the
first to deisgn a clock based
on the relationship between
gravitational force (g), length
of the pendulum (l ) and time
of oscillation (t).
and at a distance kG2 / h.
2. The distance between the centre of suspension and the centre of percussion is equal to the
equivalent length of a simple pendulum.
Note: We know that mass moment of inertia of the body about O,
I O = I G + m.h 2 or m.kG2 = m.kO2 + m.h 2
∴
kO2 = kG2 + h 2 = l.h + h 2 = h (l + h) = OG × OC
... (3 kG2 = l.h)
It is thus obvious that the centre of suspension (O) and the centre of percussion (C) are inter-changeable.
In other words, the periodic time and frequency of oscillation will be same, whether the body is suspended at the
point of suspension or at the centre of percussion.
Example 4.4. A uniform thin rod, as shown in Fig. 4.7, has a mass of 1 kg and carries a
concentrated mass of 2.5 kg at B. The rod is hinged at A and is maintained in the horizontal position
by a spring of stiffness 1.8 kN/m at C.
Find the frequency of oscillation, neglecting the effect of the mass of the spring.
Fig. 4.7
Chapter 4 : Simple Harmonic Motion
l
83
Solution. Given : m = 1 kg ; m 1 = 2.5 kg ; s = 1.8 kN/m = 1.8 × 103 N/m
We know that total length of rod,
l = 300 + 300 = 600 mm = 0.6 m
∴ Mass moment of inertia of the system about A ,
IA = Mass moment of inertia of 1 kg about A + Mass moment of interia of
2.5 kg about A
2
2
1(0.6)
= m.l + m1 .l 2 =
+ 2.5 (0.6)2 = 1.02 kg-m2
3
3
If the rod is given a small angular displacement θ and then released, the extension of the spring,
δ = 0.3 sin θ = 0.3θ m
. . . ( ∵ θ is very small, therefore substituting sin θ = θ )
∴ Restoring force = s.δ = 1.8 ×
103
× 0.3 θ = 540 θ N
and restoring torque about A = 540 θ × 0.3 = 162 θ N-m
... (i)
We know that disturbing torque about A
= IA × α = 1.02α N-m
... (ii)
Equating equations (i) and (ii),
1.02 α = 162 θ
or
α / θ = 162 / 1.02 = 159
We know that frequency of oscillation,
n= 1
2π
α = 1 159 = 2.01 Hz Ans.
θ 2π
Example 4.5. A small flywheel of mass 85 kg is suspended in a vertical plane as a compound
pendulum. The distance of centre of gravity from the knife edge support is 100 mm and the flywheel
makes 100 oscillations in 145 seconds. Find the moment of inertia of the flywheel through the centre
of gravity.
Solution. Given : m = 85 kg ; h = 100 mm = 0.1 m
Since the flywheel makes 100 oscillations in 145 seconds, therefore frequency of oscillation,
n = 100/145 = 0.69 Hz
Let
L = Equivalent length of simple pendulum, and
k G = Radius of gyration through C.G.
We know that frequency of oscillation (n),
0.69 = 1
2π
g
= 1
L 2π
9.81 = 0.5
L
L
L = 0.5/0.69 = 0.7246 or L = 0.525 m
∴
We also know that equivalent length of simple pendulum (L),
0.525 =
kG2
k2
k 2 + (0.1)2
+ h = G + 0.1 = G
0.1
0.1
h
kG2 = 0.525 × 0.1 − (0.1)2 = 0.0425 m2
84
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Theory of Machines
and moment of inertia of the flywheel through the centre of gravity,
2
I = m.kG = 85 × 0.0425 = 3.6 kg-m2 Ans.
Example 4.6. The connecting rod of an oil engine has a mass of 60 kg, the distance between
the bearing centres is 1 metre. The diameter of the big end bearing is 120 mm and of the small end
bearing is 75 mm. When suspended vertically with a knife-edge through the small end, it makes 100
oscillations in 190 seconds and with knife-edge through the big end it makes 100 oscillations in 165
seconds. Find the moment of inertia of the rod in kg-m2 and the distance of C.G. from the small end
centre.
Solution. Given : m = 60 kg ; h1 + h2 = 1 m ; d2* = 102 mm; d1* = 75 mm
Moment of inertia of the rod
First of all, let us find the radius of gyration of the connecting
rod about the centre of gravity (i.e. kG).
Let
h1 and h2 = Distance of centre of gravity from
the small and big end centres respectively,
L1 and L2 = Equivalent length of simple
perdulum when the axis of oscillation coincides with the small and big
end centres respectively
Connecting rod
When the axis of oscillation coincides with the small end centre, then frequency of oscillation,
n1 = 100/190 = 0.526 Hz
When the axis of oscillation coincides with the big end centre, the frequency of oscillation,
n2 = 100/165 = 0.606 Hz
We know that for a simple pendulum,
g
Hz
L1
n1 =
1
2π
∴
L1 =
g
9.81
=
= 0.9 m
(2π n1 )2 (2π × 0.526)2
Similarly
L2 =
g
9.81
=
= 0.67 m
2
(2π n2 )
(2π × 0.606)2
We know that
L1 =
kG2 + ( h1 ) 2
or kG2 = L1 .h1 − (h1 )2
h1
Similarly
kG2 = L2 .h2 − (h2 ) 2
From equations (i) and (ii), we have
L1.h1 – (h1)2 = L2.h2 – (h2)2
*
Superfluous data.
... (i)
... (ii)
Chapter 4 : Simple Harmonic Motion
0.9 × h1 – (h1)2 = 0.67 (1 – h1) – (1 – h1)2
l
85
. . . (3 h1 + h2 = 1 m)
= 0.67 – 0.67 h1 – 1 – (h1)2 + 2h1
0.9 h1 + 0.67 h1 – 2 h1 = – 0.33
∴
or
– 0.43 h1 = – 0.33
h1 = 0.33/0.43 = 0.767 m
Substituting the value of h1 in equation (i), we have
kG2 = 0.9 × 0.767 − (0.767)2 = 0.69 − 0.59 = 0.1 m 2
We know that mass moment of inertia of the rod,
I = m.kG2 = 60 × 0.1 = 6 kg-m2 Ans.
Distance of C.G. from the small end centre
We have calculated above that the distance of C.G. from the small end centre,
h1 = 0.767 m Ans.
Example 4.7. A uniform slender rod 1.2 m long is fitted with a transverse pair of knifeedges, so that it can swing in a vertical plane as a compound pendulum. The position of the knife
edges is variable. Find the time of swing of the rod, if 1. the knife edges are 50 mm from one end of
the rod, and 2. the knife edges are so placed that the time of swing is minimum.
In case (1) find also the maximum angular velocity and the maximum angular acceleration
of the rod if it swings through 3° on either side of the vertical.
Solution. Given : l = 1.2 m ; θ = 3° = 3 × π /180 = 0.052 rad
1. Time of swing of the rod when knife edges are 50 mm
Since the distance between knife edges from one end of the rod is 50 mm = 0.05 m, therefore
distance between the knife edge and C.G. of the rod,
h = 1.2 − 0.05 = 0.55m
2
We know that radius of gyration of the rod about C.G.,
1.2
l
k*G =
=
= 0.35 m
12
12
∴ Time of swing of the rod,
kG2 + h2
= 2π
g .h
= 1.76 s Ans.
t p = 2π
(0.35)2 + (0.55)2
9.81 × 0.55
2. Minimum time of swing
We know that minimum time of swing,
t p ( min ) = 2 π
*
2 kG
= 2π
g
2 × 0.35
= 1.68 s Ans.
9.81
We know that mass moment of inertia of the rod about an axis through C.G.
I = m. l 2/12
Also
I = m.k 2 or
k 2 = I/m = m.l 2/12 × m = l 2/12
or
k= l
12
86
l
Theory of Machines
Maximum angular velocity
In case (1), the angular velocity,
ω = 2π / tp = 2π /1.76 = 3.57 rad/s
We know that maximum angular velocity,
ωmax = ω.θ = 3.57 × 0.052 = 0.1856 rad/s Ans.
Maximum angular acceleration
We know that maximum angular acceleration,
αmax = ω2.θ = (3.57)2 × 0.052 = 0.663 rad/s2 Ans.
Example 4.8. The pendulum of an Izod impact testing machine has a mass of 30 kg. Its
centre of gravity is 1.05 m from the axis of suspension and the striking knife is 150 mm below the
centre of gravity. The time for 20 small free oscillations is 43.5 seconds. In making a test the pendulum is released from an angle of 60° to the vertical. Determine :
1. the position of the centre of percussion relative to the striking knife and the striking velocity of the pendulum, and 2. the impulse on the pendulum and the sudden change of axis reaction
when a specimen giving an impact value of 55 N-m is broken.
Solution. Given : m = 30 kg ; OG = h = 1.05 m ; A G = 0.15 m
Since the time for 20 small free oscillations is 43.5 s, therefore frequency of oscillation,
n = 20 = 0.46 Hz
43.5
1. The position of centre of percussion relative to the striking knife and the striking velocity of the
pendulum
Let
L = Equivalent length of simple pendulum,
kG = Radius of gyration of the pendulum about the centre of gravity, and
kO = Radius of gyration of the pendulum about O.
We know that the frequency of oscillation,
n= 1
2π
or
L=
g
L
g
9.81
=
2
(2π n)
(2π × 0.46) 2
= 1.174 m
∴ Distance of centre of percussion (C) from
the centre of gravity (G),
CG = OC – OG = L – OG
= 1.174 – 1.05 = 0.124 m
Fig. 4.8
and distance of centre of percussion (C) from knife edge A ,
AC = AG – CG = 0.15 – 0.124 = 0.026 m Ans.
We know that
kO2 = (h + l ) h = L.h = 1.174 × 1.05 = 1.233 m2
A little consideration will show that the potential energy of the pendulum is converted into
kinetic energy of the pendulum before it strikes the test piece. Let v and ω be the linear and angular
velocity of the pendulum before it strikes the test piece.
Chapter 4 : Simple Harmonic Motion
∴
m.g.h1 =
1 . 2 = 1 . 2 .ω2
mv
mk
2
2 O
l
87
... (∵ v = k O.ω)
1 × 30 × 1.233 ω2
or 154.5 = 18.5 ω2
2
ω2 = 154.5/18.5 = 8.35 or ω = 2.9 rad/s
30 × 9.81 × 1.05 (1 – cos 60°) =
∴
∴ Velocity of striking = ω × O A = 2.9 (1.05 + 0.15) = 3.48 m/s Ans.
2. Impulse on the pendulum and sudden change of axis reaction
It is given that the impact value of the specimen (i.e. the energy used for breaking the specimen)
is 55 N-m. Let ω1 be the angular velocity of the pendulum immediately after impact. We know that
Loss of kinetic energy =
∴
1 (ω2 − ω2 ) = 1 . 2 (ω2 − ω2 ) = 55
I
m kO
N-m
1
1
2
2
1 × 30 × 1.233 (2.92 − ω2 ) = 55
1
2
18.5 (8.41 – ω12 ) = 55
∴
or
ω12 = 8.41 – 55/18.5 = 5.44
ω1 = 2.33 rad/s
Let P and Q be the impulses at the knife edge A and at the pivot O respectively as shown in
Fig. 4.8.
∴
P + Q = Change of linear momentum
= m.h (ω – ω1) = 30 × 1.05 (2.9 – 2.33) = 17.95
... (i)
Taking moments about G,
0.15 P – 1.05 Q = Change of angular momentum
= m.kG2 (ω − ω1 ) = m (kO2 − h 2 ) (ω − ω1 )
= 30 (1.233 – 1.052) (2.9 – 2.33) = 2.27
... (ii)
From equations (i) and (ii),
P = 17.6 N-s; and Q = 0.35 N-s Ans.
∴ Change in axis reaction when pendulum is vertical
= Change in centrifugal force
= m (ω2 − ω12 ) h = 30 (2.92 − 2.332 ) 1.05 = 94 N Ans.
4.10. Bifilar Suspension
The moment of inertia of a body may be determined experimentally by an apparatus called
bifilar suspension. The body whose moment of inertia is to be determined (say A B) is suspended by
two long parallel flexible strings as shown in Fig. 4.9. When the body is twisted through a small angle
θ about a vertical axis through the centre of gravity G, it will vibrate with simple harmonic motion in
a horizontal plane.
Let
m = Mass of the body,
W = Weight of the body in newtons = m.g,
kG = Radius of gyration about an axis through the centre of gravity,
88
l
Theory of Machines
I = Mass moment of inertia of the
body about a vertical axis
through G = m . kG2 ,
l = Length of each string,
x = Distance of A from G (i.e. AG),
y = Distance of B from G (i.e. BG),
θ = Small angular displacement of
the body from the equilibrium
position in the horizontal plane,
φA and φB = Corresponding angular displacements of the strings, and
α = Angular acceleration towards
the equilibrium position.
When the body is stationary, the tension in the strings are
given by
TA =
m. g. y
m. g. x
, and TB =
x+y
x+y
Fig. 4.9. Bifilar suspension.
...(Taking moments about B and A respectively,)
When the body is displaced from its equilibrium position in a horizontal plane through a
small angle θ, then the angular displacements of the strings are given by
A A′ = φA.l = x.θ ; and BB′ = φB.l = y.θ
∴
y.θ
.
φA = x θ ; and φB =
l
l
Component of tension TA in the horizontal plane, acting normal to A′B′ at A ′ as shown in Fig. 4.9
m.g . y x.θ m.g .x. y.θ
= TA .φ A = x + y × l = l ( x + y )
Component of tension T B in the horizontal plane, acting normal to A ′ B′ at B′ as shown in Fig. 4.9
m.g .x
y.θ m.g .x. y.θ
= TB .φ B = x + y × l = l ( x + y )
These components of tensions T A and T B are equal and opposite in direction, which gives rise
to a couple. The couple or torque applied to each string to restore the body to its initial equilibrium
position, i.e. restoring torque
= TA .φA .x + TB .φB . y
=
m.g .x. y.θ
m.g .x. y.θ
( x + y) =
l ( x + y)
l
... (i)
and accelerating (or disturbing) torque
= I .α = m.kG2 .α
Equating equations (i) and (ii),
2
m.g.x. y.θ
θ = kG .l
= m.kG2 .α or
α g .x. y
l
... (ii)
Chapter 4 : Simple Harmonic Motion
i.e.
l
89
k 2 .l
Angular displacement
= G
Angular acceleration
g. x. y
We know that periodic time,
Angular displacement
= 2π
Angular acceleration
t p = 2π
= 2 π kG
and
frequency, n =
1
1
=
t p 2π kG
kG2 .l
g.x. y
l
g.x. y
g.x. y
l
Note : The bifilar suspension is usually used for finding the moment of inertia of a connecting rod of an engine.
In this case, the wires are attached at equal distances from the centre of gravity of the connecting rod (i.e. x = y)
so that the tension in each wire is same.
Example 4.9. A small connecting rod of mass 1.5 kg is suspended in a horizontal plane by
two wires 1.25 m long. The wires are attached to the rod at points 120 mm on either side of the centre
of gravity. If the rod makes 20 oscillations in 40 seconds, find the radius of gyration and the mass
moment of inertia of the rod about a vertical axis through the centre of gravity.
Solution. Given : m = 1.5 kg ; l = 1.25 m ; x = y = 120 mm = 0.12 m
Since the rod makes 20 oscillations in 40 s, therefore frequency of oscillation,
n = 20/40 = 0.5 Hz
Radius of gyration of the connecting rod
Let
k G = Radius of gyration of the connecting rod.
We know that frequency of oscillation (n),
0.5 =
1
2 π kG
g .x. y
= 1
2 π kG
l
9.81 × 0.12 × 0.12 0.0535
=
1.25
k
∴
k G = 0.0535/0.5 = 0.107 m = 107 mm Ans.
Mass moment of inertia of the connecting rod
We know that mass moment of inertia,
I = m (k G)2 = 1.5 (0.107)2 = 0.017 kg-m2 Ans.
4.11. Trifilar Suspension (Torsional Pendulum)
It is also used to find the moment of inertia of a body experimentally. The body (say a disc or flywheel) whose moment of inertia
is to be determined is suspended by three long flexible wires A, B
and C, as shown in Fig. 4.10. When the body is twisted about its axis
through a small angle θ and then released, it will oscillate with simple
harmonic motion.
Let
m = Mass of the body in kg,
W = Weight of the body in newtons = m.g,
kG = Radius of gyration about an axis
Fig. 4.10. Trifilar suspension.
through c.g.,
I = Mass moment of inertia of the disc about an axis through O and perpendicular to it = m.k2,
90
l
Theory of Machines
l = Length of each wire,
r = Distance of each wire from the axis of the disc,
θ = Small angular displacement of the disc,
φ = Corresponding angular displacement of the wires, and
α = Angular acceleration towards the equilibrium position.
Then, for small displacements,
r. θ = l. φ
or φ = r.θ/l
Since the three wires are attached symmetrically with respect to the axis, therefore the tension in each wire will be one-third of the weight of the body.
∴ Tension in each wire = m.g/3
Component of the tension in each wire perpendicular to r
m.g.sin φ m.g .φ m.g .r .θ
=
=
. . . ( ∵ φ is a small angle, and φ = r.θ/l)
3
3
3l
∴ Torque applied to each wire to restore the body to its initial equilibrium position i.e.
restoring torque
=
m.g .r.θ
m.g .r 2 .θ
×r=
3l
3l
Total restoring torque applied to three wires,
=
T = 3×
m. g.r 2 .θ m. g.r 2 .θ
=
3l
l
... (i)
We know that disturbing torque
= I.α = m.kG2 .α
Equating equations (i) and (ii),
m. g.r 2 .θ
= m.kG2 .α
l
or
... (ii)
2
θ = l.kG
α g.r 2
Angular displacement l.kG2
i.e. Angular acceleration =
g.r 2
We know that periodic time,
t p = 2π
and
Angular displacement
= 2π
Angular acceleration
1 = r
frequency, n =
t p 2 π kG
l.kG2
g.r
2
=
2 π kG
r
l
g
g
l
Example 4.10. In order to find the radius of gyration of a car, it is suspended with its axis
vertical from three parallel wires 2.5 metres long. The wires are attached to the rim at points spaced
120° apart and at equal distances 250 mm from the axis.
It is found that the wheel makes 50 torsional oscillations of small amplitude about its axis in
170 seconds. Find the radius of gyration of the wheel.
Solution. Given : l = 2.5 m ; r = 250 mm = 0.25 m ;
Since the wheel makes 50 torsional oscillations in 170 seconds, therefore frequency of
oscillation,
n = 50/170 = 5/17 Hz
Let
k G = Radius of gyration of the wheel
Chapter 4 : Simple Harmonic Motion
l
91
We know that frequency of oscillation (n),
5
r
=
17 2π kG
∴
0.25
g
=
l 2π kG
9.81 0.079
=
2.5
kG
k G = 0.079 × 17/5 = 0.268 m = 268 mm Ans.
Example 4.11. A connecting rod of mass 5.5 kg is placed on a horizontal platform whose
mass is 1.5 kg. It is suspended by three equal wires, each 1.25 m long, from a rigid support. The wires
are equally spaced round the circumference of a circle of 125 mm radius. When the c.g. of the
connecting rod coincides with the axis of the circle, the platform makes 10 angular oscillations in 30
seconds. Determine the mass moment of inertia about an axis through its c.g.
Solution. Given : m 1 = 5.5 kg ; m 2 = 1.5 kg ; l = 1.25 m ; r = 125 mm = 0.125 m
Since the platform makes 10 angular oscillations in 30 s, therefore frequency of oscillation,
n = 10/30 = 1/3 Hz
Let
k G = Radius of gyration about an axis through the c.g.
We know that frequency of oscillation (n)
1= r
3 2πkG
g
= 0.125
2π kG
l
9.81 = 0.056
1.25
kG
k G = 0.056 × 3 = 0.168 m
∴
and mass moment of inertia about an axis through its c.g.,
I = m.kG2 = (m1 + m2 ) kG2 = (5.5 + 1.5) (0.168)2 kg-m2
= 0.198 kg-m2 Ans.
EXERCISES
1.
A particle, moving with simple harmonic motion, performs 10 complete oscillations per minute and
its speed, when at a distance of 80 mm from the centre of oscillation is 3/5 of the maximum speed.
Find the amplitude, the maximum acceleration and the speed of the particle, when it is 60 mm from
the centre of the oscillation.
[Ans. 100 mm ; 109.6 mm/s2 ; 83.76 mm/s]
2.
A piston, moving with a simple harmonic motion, has a velocity of 8 m/s, when it is 1 metre from the
centre position and a velocity of 4 m/s, when it is 2 metres from the centre. Find : 1. Amplitude, 2.
Periodic time, 3. Maximum velocity, and 4. Maximum acceleration.
[Ans. 2.236 m ; 1.571 s ; 8.94 m/s ; 35.77 m/s2]
3.
The plunger of a reciprocating pump is driven by a crank of radius 250 mm rotating at 12.5 rad/s.
Assuming simple harmonic motion, determine the maximum velocity and maximum acceleration of
the plunger.
[Ans. 3.125 m/s ; 39.1 m/s2]
4.
A part of a machine of mass 4.54 kg has a reciprocating motion which is simple harmonic in character.
It makes 200 complete oscillations in 1 minute. Find : 1. the accelerating force upon it and its velocity
when it is 75 mm, from midstroke ; 2. the maximum accelerating force, and 3. the maximum velocity
if its total stroke is 225 mm i.e. if the amplitude of vibration is 112.5 mm.
5.
A helical spring of negligible mass is required to support a mass of 50 kg. The stiffness of the spring
is 60 kN/m. The spring and the mass system is displaced vertically by 20 mm below the equilibrium
position and then released. Find : 1. the frequency of natural vibration of the system ; 2. the velocity
and acceleration of the mass when it is 10 mm below the rest position.
[Ans. 149.5 N ; 1.76 m/s ; 224 N ; 2.36 m/s]
6.
[Ans. 5.5 Hz ; 0.6 m/s ; 11.95 m/s2]
A spring of stiffness 2 kN/m is suspended vertically and two equal masses of 4 kg each are attached to
the lower end. One of these masses is suddenly removed and the system oscillates. Determine : 1. the
amplitude of vibration, 2. the frequency of vibration, 3. the velocity and acceleration of the mass when
92
7.
8.
9.
10.
11.
12.
l
Theory of Machines
passing through half amplitude position, and 4. kinetic energy of the vibration in joules.
[Ans. 0.019 62 m ; 3.56 Hz ; 0.38 m/s , 4.9 m/s2 ; 0.385 J]
A vertical helical spring having a stiffness of 1540 N/m is clamped at its upper end and carries a mass
of 20 kg attached to the lower end. The mass is displaced vertically through a distance of 120 mm and
released. Find : 1. Frequency of oscillation ; 2. Maximum velocity reached ; 3. Maximum acceleration; and 4. Maximum value of the inertia force on the mass.
[Ans. 1.396 Hz ; 1.053 m/s ; 9.24 m/s2 ; 184.8 N]
A small flywheel having mass 90 kg is suspended in a vertical plane as a compound pendulum. The
distance of centre of gravity from the knife edge support is 250 mm and the flywheel makes 50
oscillations in 64 seconds. Find the moment of inertia of the flywheel about an axis through the centre
of gravity.
[Ans. 3.6 kg-m2]
The connecting rod of a petrol engine has a mass 12 kg. In order to find its moment of inertia it is
suspended from a horizontal edge, which passes through small end and coincides with the small end
centre. It is made to swing in a vertical plane, such that it makes 100 oscillations in 96 seconds. If the
point of suspension of the connecting rod is 170 mm from its c.g., find : 1. radius of gyration about an
axis through its c.g., 2. moment of inertia about an axis through its c.g., and 3. length of the equivalent
simple pendulum.
[Ans. 101 mm ; 0.1224 kg-m2 ; 0.23 m]
A connecting rod of mass 40 kg is suspended vertically as a compound pendulum. The distance between
the bearing centres is 800 mm. The time for 60 oscillations is found to be 92.5 seconds when the axis of
oscillation coincides with the small end centre and 88.4 seconds when it coincides with the big end
centre. Find the distance of the centre of gravity from the small end centre, and the moment of inertia of
the rod about an axis through the centre of gravity.
[Ans. 0.442 m ; 2.6 kg-m2]
The following data were obtained from an experiment to find the moment of inertia of a pulley by
bifilar suspension :
Mass of the pulley = 12 kg ; Length of strings = 3 m ; Distance of strings on either side of centre of
gravity = 150 mm ; Time for 20 oscillations about the vertical axis through c.g. = 46.8 seconds
Calculate the moment of inertia of the pulley about the axis of rotation.
[Ans. 0.1226 kg-m2]
In order to find the moment of inertia of a flywheel, it is suspended in the horizontal plane by three
wires of length 1.8 m equally spaced around a circle of 185 mm diameter. The time for 25 oscillations
in a horizontal plane about a vertical axis through the centre of flywheel is 54 s. Find the radius of
gyration and the moment of inertia of the flywheel if it has a mass of 50 kg.
[Ans. 74.2 mm; 0.275 kg-m2]
DO YOU KNOW ?
1.
2.
3.
4.
5.
6.
7.
Explain the meaning of S.H.M. and give an example of S.H.M.
Define the terms amplitude, periodic time, and frequency as applied to S.H.M.
Show that when a particle moves with simple harmonic motion, its time for a complete
oscillation is independent of the amplitude of its motion.
Derive an expression for the period of oscillation of a mass when attached to a helical spring.
What is a simple pendulum ? Under what conditions its motion is regarded as simple harmonic?
Prove the formula for the frequency of oscillation of a compound pendulum. What is the length of a
simple pendulum which gives the same frequency as compound pendulum ?
Show that the minimum periodic time of a compound pendulum is
t p( min) = 2π
8.
2kG
g
where k G is the radius of gyration about the centre of gravity.
What do you understand by centre of percussion ? Prove that it lies below the centre of gravity of the
body and at a distance kG2 / h , where k G is the radius of gyration about c.g. and h is the distance
between the centre of suspension and centre of gravity.
l
Chapter 4 : Simple Harmonic Motion
9.
10.
93
Describe the method of finding the moment of inertia of a connecting rod by means of bifilar suspension. Derive the relations for the periodic time and frequency of oscillation.
What is a torsional pendulum ? Show that periodic time of a torsional pendulum is
tp =
where
2πkG
r
l
g
k G = Radius of gyration,
l = Length of each wire, and
r = Distance of each wire from the axis of the disc.
OBJECTIVE TYPE QUESTIONS
1. The periodic time (tp) is given by
(a) ω / 2 π
(b) 2 π / ω
(c) 2 π × ω
(d) π/ω
2. The velocity of a particle moving with simple harmonic motion is . . . . at the mean position.
(a) zero
(b) minimum
(c) maximum
3. The velocity of a particle (v) moving with simple harmonic motion, at any instant is given by
(a)
ω r 2 − x2
(b)
ω x2 − r 2
(c)
ω2
r 2 − x2
(d)
4. The maximum acceleration of a particle moving with simple harmonic motion is
(a) ω
(b) ω.r
(c) ω 2.r
(d)
5. The frequency of oscillation for the simple pendulum is
(a)
L
g
1
2π
(b)
g
L
1
2π
(c)
L
g
2π
(d)
ω2
x2 − r 2
ω 2/r
g
L
2π
6. When a rigid body is suspended vertically and it oscillates with a small amplitude under the action of the
force of gravity, the body is known as
(a) simple pendulum
(b) torsional pendulum
(c) compound pendulum
(d) second’s pendulum
7. The frequency of oscillation of a compound pendulum is
(a)
1
2π
g .h
kG2 + h 2
(b)
kG2 + h 2
g .h
1
2π
(c) 2π
g .h
kG2 + h 2
(d) 2π
kG2 + h2
g.h
where
kG = Radius of gyration about the centroidal axis, and
h = Distance between the point of suspension and centre of gravity of the body.
8. The equivalent length of a simple pendulum which gives the same frequency as the compound pendulum
is
(a)
h
2
kG + h 2
(b)
kG2 + h2
h
(c)
h2
kG2 + h 2
(d)
kG2 + h2
h2
9. The centre of percussion is below the centre of gravity of the body and is at a distance equal to
(a)
10.
h / kG
(b)
(c) h2/k G
h.k G
(d)
kG2 / h
(d)
r
2πkG
The frequency of oscillation of a torsional pendulum is
(a)
2π kG
r
g
l
(b)
r
2π kG
g
l
(c)
2πkG
r
l
g
l
g
ANSWERS
1. (b)
6. (c)
2. (c)
7. (a)
3. (a)
8. (b)
4. (c)
9. (d)
5. (b)
10. (b)
GO To FIRST
CONTENTS
CONTENTS
94
l
Theory of Machines
5
Features
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
21.
22.
Introduction.
Kinematic Link or Element.
Types of Links.
Structure.
Difference Between a
Machine and a Structure.
Kinematic Pair.
Types of Constrained
Motions.
Classification of Kinematic
Pairs.
Kinematic Chain.
Types of Joints in a Chain.
Mechanism.
Number of Degrees of
Freedom for Plane
Mechanisms.
Application of Kutzbach
Criterion to Plane
Mechanisms.
Grubler's Criterion for
Plane Mechanisms.
Inversion of Mechanism.
Types of Kinematic Chains.
Four Bar Chain or Quadric
Cycle Chain.
Inversions of Four Bar
Chain.
Single Slider Crank Chain.
Inversions of Single Slider
Crank Chain.
Double Slider Crank Chain.
Inversions of Double Slider
Crank Chain.
Simple
Mechanisms
5.1.
Introduction
We have already discussed that a machine is a device which receives energy and transforms it into some useful work. A machine consists of a number of parts or bodies.
In this chapter, we shall study the mechanisms of the various
parts or bodies from which the machine is assembled. This is
done by making one of the parts as fixed, and the relative
motion of other parts is determined with respect to the fixed
part.
5.2.
Kinematic Link or Element
Each part of a machine, which moves relative to some
other part, is known as a kinematic link (or simply link) or
element. A link may consist of several parts, which are rigidly fastened together, so that they do not move relative to
one another. For example, in a reciprocating steam engine,
as shown in Fig. 5.1, piston, piston rod and crosshead constitute one link ; connecting rod with big and small end bearings constitute a second link ; crank, crank shaft and flywheel
a third link and the cylinder, engine frame and main bearings
a fourth link.
94
CONTENTS
CONTENTS
Chapter 5 : Simple Mechanisms
l
95
Fig. 5.1. Reciprocating steam engine.
A link or element need not to be a
rigid body, but it must be a resistant body. A
body is said to be a resistant body if it is
capable of transmitting the required forces
with negligible deformation. Thus a link
should have the following two characteristics:
1. It should have relative motion, and
2. It must be a resistant body.
5.3.
Types of Links
Piston and piston rod of an IC engine.
In order to transmit motion, the driver
and the follower may be connected by the following three types of links :
1. Rigid link. A rigid link is one which does not undergo any deformation while transmitting
motion. Strictly speaking, rigid links do not exist. However, as the deformation of a connecting rod,
crank etc. of a reciprocating steam engine is not appreciable, they can be considered as rigid links.
2. Flexible link. A flexible link is one which is partly deformed in a manner not to affect the
transmission of motion. For example, belts, ropes, chains and wires are flexible links and transmit
tensile forces only.
3. Fluid link. A fluid link is one which is formed by having a fluid in a receptacle and the
motion is transmitted through the fluid by pressure or compression only, as in the case of hydraulic
presses, jacks and brakes.
5.4.
Structure
It is an assemblage of a number of resistant bodies (known as members) having no relative
motion between them and meant for carrying loads having straining action. A railway bridge, a roof
truss, machine frames etc., are the examples of a structure.
5.5.
Difference Between a Machine and a Structure
The following differences between a machine and a structure are important from the subject
point of view :
1. The parts of a machine move relative to one another, whereas the members of a structure
do not move relative to one another.
2. A machine transforms the available energy into some useful work, whereas in a structure
no energy is transformed into useful work.
3. The links of a machine may transmit both power and motion, while the members of a
structure transmit forces only.
96
5.6.
l
Theory of Machines
Kinematic Pair
The two links or elements of a machine, when in contact with each other, are said to form a
pair. If the relative motion between them is completely or successfully constrained (i.e. in a definite
direction), the pair is known as kinematic pair.
First of all, let us discuss the various types of constrained motions.
5.7.
Types of Constrained Motions
Following are the three types of constrained motions :
1. Completely constrained motion. When the motion between a pair is limited to a definite
direction irrespective of the direction of force applied, then the motion is said to be a completely
constrained motion. For example, the piston and cylinder (in a steam engine) form a pair and the
motion of the piston is limited to a definite direction (i.e. it will only reciprocate) relative to the
cylinder irrespective of the direction of motion of the crank, as shown in Fig. 5.1.
Fig. 5.2. Square bar in a square hole.
Fig. 5.3. Shaft with collars in a circular hole.
The motion of a square bar in a square hole, as shown in Fig. 5.2, and the motion of a shaft
with collars at each end in a circular hole, as shown in Fig. 5.3, are also examples of completely
constrained motion.
2. Incompletely constrained motion. When the motion between a pair can take place in more
than one direction, then the motion is called an incompletely constrained motion. The change in the
direction of impressed force may alter the direction of relative motion between the pair. A circular bar
or shaft in a circular hole, as shown in Fig. 5.4, is an example of an incompletely constrained motion
as it may either rotate or slide in a hole. These both motions have no relationship with the other.
Fig. 5.4. Shaft in a circular hole.
Fig. 5.5. Shaft in a foot step bearing.
3. Successfully constrained motion. When the motion between the elements, forming a pair,is
such that the constrained motion is not completed by itself, but by some other means, then the motion
is said to be successfully constrained motion. Consider a shaft in a foot-step bearing as shown in Fig.
5.5. The shaft may rotate in a bearing or it may move upwards. This is a case of incompletely constrained motion. But if the load is placed on the shaft to prevent axial upward movement of the shaft,
then the motion of the pair is said to be successfully constrained motion. The motion of an I.C. engine
Chapter 5 : Simple Mechanisms
l
97
valve (these are kept on their seat by a spring) and the piston reciprocating inside an engine cylinder
are also the examples of successfully constrained motion.
5.8.
Classification of Kinematic Pairs
The kinematic pairs may be classified according to the following considerations :
1. According to the type of relative motion between the elements. The kinematic pairs according to type of relative motion between the elements may be classified as discussed below:
(a) Sliding pair. When the two elements of a pair are connected in such a way that one can
only slide relative to the other, the pair is known as a sliding pair. The piston and cylinder, cross-head
and guides of a reciprocating steam engine, ram and its guides in shaper, tail stock on the lathe bed
etc. are the examples of a sliding pair. A little consideration will show, that a sliding pair has a
completely constrained motion.
(b) Turning pair. When the two elements of a pair are connected in such a way that one can
only turn or revolve about a fixed axis of another link, the pair is known as turning pair. A shaft with
collars at both ends fitted into a circular hole, the crankshaft in a journal bearing in an engine, lathe
spindle supported in head stock, cycle wheels turning over their axles etc. are the examples of a
turning pair. A turning pair also has a completely constrained motion.
(c) Rolling pair. When the two elements of a pair are connected in such a way that one rolls
over another fixed link, the pair is known as rolling pair. Ball and roller bearings are examples of
rolling pair.
(d) Screw pair. When the two elements of a pair are connected in such a way that one element
can turn about the other by screw threads, the pair is known as screw pair. The lead screw of a lathe
with nut, and bolt with a nut are examples of a screw pair.
(e) Spherical pair. When the two elements of a pair are connected in such a way that one
element (with spherical shape) turns or swivels about the other fixed element, the pair formed is
called a spherical pair. The ball and socket joint, attachment of a car mirror, pen stand etc., are the
examples of a spherical pair.
2. According to the type of contact between the elements. The kinematic pairs according to
the type of contact between the elements may be classified as discussed below :
(a) Lower pair. When the two elements of a pair have a surface contact when relative motion
takes place and the surface of one element slides over the surface of the other, the pair formed is
known as lower pair. It will be seen that sliding pairs, turning pairs and screw pairs form lower pairs.
(b) Higher pair. When the two elements of a pair have a line or point contact when relative
motion takes place and the motion between the two elements is partly turning and partly sliding,then
the pair is known as higher pair. A pair of friction discs, toothed gearing, belt and rope drives, ball and
roller bearings and cam and follower are the examples of higher pairs.
3. According to the type of closure. The kinematic pairs according to the type of closure
between the elements may be classified as discussed below :
(a) Self closed pair. When the two elements of a pair are connected together mechanically in
such a way that only required kind of relative motion occurs, it is then known as self closed pair. The
lower pairs are self closed pair.
(b) Force - closed pair. When the two elements of a pair are not connected mechanically but
are kept in contact by the action of external forces, the pair is said to be a force-closed pair. The cam
and follower is an example of force closed pair, as it is kept in contact by the forces exerted by spring
and gravity.
98
5.9.
l
Theory of Machines
Kinematic Chain
When the kinematic pairs are
coupled in such a way that the last link
is joined to the first link to transmit
definite motion (i.e. completely or
successfully constrained motion), it is
called a kinematic chain. In other
words, a kinematic chain may be defined as a combination of kinematic
pairs, joined in such a way that each
link forms a part of two pairs and the
relative motion between the links or
elements is completely or successfully
constrained. For example, the crankshaft of an engine forms a kinematic
pair with the bearings which are fixed
Lawn-mover is a combination of kinematic links.
in a pair, the connecting rod with the
crank forms a second kinematic pair,
the piston with the connecting rod forms a third pair and the piston with the cylinder forms a fourth
pair. The total combination of these links is a kinematic chain.
If each link is assumed to form two pairs with two adjacent links, then the relation between
the number of pairs ( p ) forming a kinematic chain and the number of links ( l ) may be expressed in
the form of an equation :
l=2p–4
. . . (i)
Since in a kinematic chain each link forms a part of two pairs, therefore there will be as many
links as the number of pairs.
Another relation between the number of links (l) and the number of joints ( j ) which
constitute a kinematic chain is given by the expression :
j= 3l−2
...(ii)
2
The equations (i) and (ii) are applicable only to kinematic chains, in which lower pairs are
used. These equations may also be applied to kinematic chains, in which higher pairs are used. In that
case each higher pair may be taken as equivalent to two lower pairs with an additional element or link.
Let us apply the above equations to the following cases to determine whether each of them is
a kinematic chain or not.
1. Consider the arrangement of three links A B, BC and C A with pin joints at A , B and C as
shown in Fig. 5.6. In this case,
and
Number of links,
l=3
Number of pairs,
p=3
number of joints,
j=3
From equation (i),
l = 2p – 4
or
i.e.
3= 2×3–4=2
L.H.S. > R.H.S.
Fig. 5.6. Arrangement of three links.
Now from equation (ii),
j = 3 l−2
2
or
3 = 3 × 3 − 2 = 2.5
2
Chapter 5 : Simple Mechanisms
i.e.
l
99
L.H.S. > R.H.S.
Since the arrangement of three links, as shown in Fig. 5.6, does not satisfy the equations (i)
and (ii) and the left hand side is greater than the right hand side, therefore it is not a kinematic chain
and hence no relative motion is possible. Such type of chain is called locked chain and forms a rigid
frame or structure which is used in bridges and trusses.
2. Consider the arrangement of four links A B, BC, CD and DA as shown in Fig. 5.7. In this case
l = 4, p = 4, and j = 4
From equation (i),
l=2p–4
4=2×4–4=4
i.e.
L.H.S. = R.H.S.
3
l−2
2
4 = 3 × 4−2=4
2
j=
From equation (ii),
i.e.
Fig. 5.7. Arrangement of four links.
L.H.S. = R.H.S.
Since the arrangement of four links, as shown in Fig. 5.7, satisfy the equations (i) and (ii),
therefore it is a kinematic chain of one degree of freedom.
A chain in which a single link such as A D in Fig. 5.7 is sufficient to define the position of all
other links, it is then called a kinematic chain of one degree of freedom.
A little consideration will show that in Fig. 5.7, if a definite displacement (say θ) is given to
the link A D, keeping the link A B fixed, then the resulting displacements of the remaining two links BC
and CD are also perfectly definite. Thus we see that in a four bar chain, the relative motion is completely constrained. Hence it may be called as a constrained kinematic chain, and it is the basis of all
machines.
3. Consider an arrangement of five links, as shown in Fig. 5.8. In this case,
l = 5, p = 5, and j = 5
From equation (i),
l=2p–4
i.e.
or
5=2×5–4=6
or
5 = 3 × 5 − 2 = 5.5
2
L.H.S. < R.H.S.
From equation (ii),
j= 3l−2
2
i.e.
L.H.S. < R.H.S.
Fig. 5.8. Arrangement of five links.
Since the arrangement of five links, as shown in Fig. 5.8 does not satisfy the equations and
left hand side is less than right hand side, therefore it is not a kinematic chain. Such a type of chain is
called unconstrained chain i.e. the relative motion is not completely constrained. This type of chain
is of little practical importance.
4. Consider an arrangement of six links, as shown in Fig. 5.9. This chain is formed by adding
two more links in such a way that these two links form a pair with the existing links as well as form
themselves a pair. In this case
l = 6, p = 5, and j = 7
100
i.e.
i.e.
l
Theory of Machines
From equation (i),
l=2p–4
L.H.S. = R.H.S.
From equation (ii),
j= 3l−2
2
L.H.S. = R.H.S.
or 6 = 2 × 5 – 4 = 6
7 = 3 × 6− 2 = 7
2
or
Since the arrangement of six links, as shown in Fig.
5.9, satisfies the equations (i.e. left hand side is equal to right
hand side), therefore it is a kinematic chain.
Fig. 5.9. Arrangement of six links.
Note : A chain having more than four links is known as compound kinematic chain.
5.10. Types of Joints in a Chain
The following types of joints are usually found in a chain :
1. Binary joint. When two links are joined at the same connection, the joint is known as
binary joint. For example, a chain as shown in Fig. 5.10, has four links and four binary joins at A , B,
C and D.
In order to determine the nature of chain, i.e. whether
the chain is a locked chain (or structure) or kinematic chain
or unconstrained chain, the following relation between the
number of links and the number of binary joints, as given by
A.W. Klein, may be used :
j + h = 3l −2
2 2
where
... (i)
Fig. 5.10.
Kinematic chain with all
binary joints.
j = Number of binary joints,
h = Number of higher pairs, and
l = Number of links.
When h = 0, the equation (i), may be written as
j = 3 l−2
. . . (ii)
2
Applying this equation to a chain, as shown in Fig. 5.10, where l = 4 and j = 4, we have
3
×4−2=4
2
Since the left hand side is equal to the right hand side, therefore the chain is a kinematic chain
or constrained chain.
4=
2. Ternary joint. When three links are joined at the
same connection, the joint is known as ternary joint. It is equivalent to two binary joints as one of the three links joined carry
the pin for the other two links. For example, a chain, as shown
in Fig. 5.11, has six links. It has three binary joints at A , B and
D and two ternary joints at C and E. Since one ternary joint is
equivalent to two binary joints, therefore equivalent binary joints
in a chain, as shown in Fig. 5.11, are 3 + 2 × 2 = 7
Let us now determine whether this chain is a kinematic
chain or not. We know that l = 6 and j = 7, therefore from
Fig. 5.11. Kinematic chain having
binary and ternary joints.
Chapter 5 : Simple Mechanisms
l
101
equation (ii),
j = 3 l −2
2
7= 3×6−2=7
or
2
Since left hand side is equal to right hand side, therefore the chain, as shown in Fig. 5.11, is
a kinematic chain or constrained chain.
3. Quaternary joint. When four links are joined at the same connection, the joint is called a
quaternary joint. It is equivalent to three binary joints. In general, when l number of links are joined
at the same connection, the joint is equivalent to (l – 1) binary joints.
For example consider a chain having eleven links, as shown in Fig. 5.12 (a). It has one binary
joint at D, four ternary joints at A, B, E and F, and two quaternary joints at C and G. Since one
quaternary joint is equivalent to three binary joints and one ternary joint is equal to two binary joints,
therefore total number of binary joints in a chain, as shown in Fig. 5.12 (a), are
(a) Looked chain having binary, ternary
and quaternary joints.
(b) Kinematic chain having binary
and ternary joints.
Fig. 5.12
1 + 4 × 2 + 2 × 3 = 15
Let us now determine whether the chain, as shown in Fig. 5.12 (a), is a kinematic chain or
not. We know that l = 11 and j = 15. We know that,
15 = 3 × 11 − 2 = 14.5, i.e., L.H.S. > R.H.S.
j = 3 l − 2, or
2
2
Since the left hand side is greater than right hand side, therefore the chain, as shown in Fig.
5.12 (a) , is not a kinematic chain. We have discussed in Art 5.9 , that such a type of chain is called
locked chain and forms a rigid frame or structure.
If the link CG is removed, as shown in Fig. 5.12 (b), it has ten links and has one binary joint
at D and six ternary joints at A, B, C, E, F and G.
Therefore total number of binary joints are 1 + 2 × 6 = 13. We know that
13 = 3 × 10 − 2 = 13 , i.e. L.H.S. = R.H.S.
j = 3 l − 2, or
2
2
Since left hand side is equal to right hand side, therefore the chain, as shown in Fig. 5.12 (b),
is a kinematic chain or constrained chain.
5.11. Mechanism
When one of the links of a kinematic chain is fixed, the chain is known as mechanism. It may
be used for transmitting or transforming motion e.g. engine indicators, typewriter etc.
102
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Theory of Machines
A mechanism with four links is known as simple mechanism, and the mechanism with more
than four links is known as compound mechanism. When a mechanism is required to transmit power
or to do some particular type of work, it then becomes a machine. In such cases, the various links or
elements have to be designed to withstand the forces (both static and kinetic) safely.
A little consideration will show that a mechanism may be regarded as a machine in which
each part is reduced to the simplest form to transmit the required motion.
5.12. Number of Degrees of Freedom for Plane Mechanisms
In the design or analysis of a mechanism, one of the most important concern is the number of
degrees of freedom (also called movability) of the mechanism. It is defined as the number of input
parameters (usually pair variables) which must be independently controlled in order to bring the
mechanism into a useful engineering purpose. It is possible to determine the number of degrees of
freedom of a mechanism directly from the number of links and the number and types of joints which
it includes.
(a) Four bar chain.
(b) Five bar chain.
Fig. 5.13
Consider a four bar chain, as shown in Fig. 5.13 (a). A little consideration will show that only
one variable such as θ is needed to define the relative positions of all the links. In other words, we say
that the number of degrees of freedom of a four bar chain is one. Now, let us consider a five bar chain,
as shown in Fig. 5.13 (b). In this case two variables such as θ1 and θ2 are needed to define completely
the relative positions of all the links. Thus, we say that the number of degrees of freedom is * two.
In order to develop the relationship in general, consider two links A B and CD in a plane
motion as shown in Fig. 5.14 (a).
Fig. 5.14. Links in a plane motion.
The link AB with co-ordinate system OXY is taken as the reference link (or fixed link). The
position of point P on the moving link CD can be completely specified by the three variables, i.e. the
*
The differential of an automobile requires that the angular velocity of two elements be fixed in order to
know the velocity of the remaining elements. The differential mechanism is thus said to have two degrees
of freedom. Many computing mechanisms have two or more degrees of freedom.
Chapter 5 : Simple Mechanisms
l
103
co-ordinates of the point P denoted by x and y and the inclination θ of the link CD with X-axis or link
A B. In other words, we can say that each link of a mechanism has three degrees of freedom before it
is connected to any other link. But when the link CD is connected to the link A B by a turning pair at
A , as shown in Fig. 5.14 (b), the position of link CD is now determined by a single variable θ and thus
has one degree of freedom.
From above, we see that when a link is connected to a fixed link by a turning pair (i.e. lower
pair), two degrees of freedom are destroyed. This may be clearly understood from Fig. 5.15, in which
the resulting four bar mechanism has one degree of freedom (i.e. n = 1 ).
Fig. 5.15. Four bar mechanism.
Now let us consider a plane mechanism with l number of links. Since in a mechanism, one of
the links is to be fixed, therefore the number of movable links will be (l – 1) and thus the total number
of degrees of freedom will be 3 (l – 1) before they are connected to any other link. In general, a
mechanism with l number of links connected by j number of binary joints or lower pairs (i.e. single
degree of freedom pairs) and h number of higher pairs (i.e. two degree of freedom pairs), then the
number of degrees of freedom of a mechanism is given by
n = 3 (l – 1) – 2 j – h
... (i)
This equation is called Kutzbach criterion for the movability of a mechanism having plane
motion.
If there are no two degree of freedom pairs (i.e. higher pairs), then h = 0. Substituting
h = 0 in equation (i), we have
n = 3 (l – 1) – 2 j
... (ii)
5.13. Application of Kutzbach Criterion to Plane Mechanisms
We have discussed in the previous article that Kutzbach criterion for determining the number
of degrees of freedom or movability (n) of a plane mechanism is
n = 3 (l – 1) – 2 j – h
Fig. 5.16. Plane mechanisms.
The number of degrees of freedom or movability (n) for some simple mechanisms having no
higher pair (i.e. h = 0), as shown in Fig. 5.16, are determined as follows :
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Theory of Machines
1. The mechanism, as shown in Fig. 5.16 (a), has three links and three binary joints, i.e.
l = 3 and j = 3.
∴
n = 3 (3 – 1) – 2 × 3 = 0
2. The mechanism, as shown in Fig. 5.16 (b), has four links and four binary joints, i.e.
l = 4 and j = 4.
∴
n = 3 (4 – 1) – 2 × 4 = 1
3. The mechanism, as shown in Fig. 5.16 (c), has five links and five binary joints, i.e.
l = 5, and j = 5.
∴
n = 3 (5 – 1) – 2 × 5 = 2
4. The mechanism, as shown in Fig. 5.16 (d), has five links and six equivalent binary joints
(because there are two binary joints at B and D, and two ternary joints at A and C), i.e.
l = 5 and j = 6.
∴
n = 3 (5 – 1) – 2 × 6 = 0
5. The mechanism, as shown in Fig. 5.16 (e), has six links and eight equivalent binary joints
(because there are four ternary joints at A, B, C and D), i.e. l = 6 and j = 8.
∴
n = 3 (6 – 1) – 2 × 8 = – 1
It may be noted that
(a)
When n = 0, then the mechanism forms a structure and no relative motion between the
links is possible, as shown in Fig. 5.16 (a) and (d).
(b) When n = 1, then the mechanism can be driven by a single input motion, as shown in Fig.
5.16 (b).
(c) When n = 2, then two separate input motions are necessary to produce constrained
motion for the mechanism, as shown in Fig. 5.16 (c).
(d) When n = – 1 or less, then there are redundant constraints in the chain and it forms a
statically indeterminate structure, as shown in Fig. 5.16 (e).
The application of Kutzbach’s criterion applied to mechanisms with a higher pair or two
degree of freedom joints is shown in Fig. 5.17.
Fig. 5.17. Mechanism with a higher pair.
In Fig. 5.17 (a), there are three links, two binary joints and one higher pair, i.e. l = 3, j = 2 and h = 1.
∴
n = 3 (3 – 1) – 2 × 2 – 1 = 1
In Fig. 5.17 (b), there are four links, three binary joints and one higher pair, i.e. l = 4,
j = 3 and h = 1
∴
n = 3 (4 – 1) – 2 × 3 – 1 = 2
Here it has been assumed that the slipping is possible between the links (i.e. between the
wheel and the fixed link). However if the friction at the contact is high enough to prevent slipping, the
joint will be counted as one degree of freedom pair, because only one relative motion will be possible
between the links.
Chapter 5 : Simple Mechanisms
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105
5.14. Grubler’s Criterion for Plane Mechanisms
The Grubler’s criterion applies to mechanisms with only single degree of freedom joints
where the overall movability of the mechanism is unity. Substituting n = 1 and h = 0 in Kutzbach
equation, we have
1 = 3 (l – 1) – 2 j
or
3l – 2j – 4 = 0
This equation is known as the Grubler's criterion for plane mechanisms with constrained
motion.
A little consideration will show that a plane mechanism with a movability of 1 and only
single degree of freedom joints can not have odd number of links. The simplest possible machanisms
of this type are a four bar mechanism and a slider-crank mechanism in which l = 4 and j = 4.
5.15. Inversion of Mechanism
We have already discussed that when one of links is fixed in a kinematic chain, it is called a
mechanism. So we can obtain as many mechanisms as the number of links in a kinematic chain by
fixing, in turn, different links in a kinematic chain. This method of obtaining different mechanisms by
fixing different links in a kinematic chain, is known as inversion of the mechanism.
It may be noted that the relative motions between the various links is not changed in any
manner through the process of inversion, but their absolute motions (those measured with respect to
the fixed link) may be changed drastically.
Note: The part of a mechanism which initially moves with respect to the frame or fixed link is called driver and
that part of the mechanism to which motion is transmitted is called follower. Most of the mechanisms are
reversible, so that same link can play the role of a driver and follower at different times. For example, in a
reciprocating steam engine, the piston is the driver and flywheel is a follower while in a reciprocating air
compressor, the flywheel is a driver.
5.16. Types of Kinematic Chains
The most important kinematic chains are those which consist of four lower pairs, each pair
being a sliding pair or a turning pair. The following three types of kinematic chains with four lower
pairs are important from the subject point of view :
1. Four bar chain or quadric cyclic chain,
2. Single slider crank chain, and
3. Double slider crank chain.
These kinematic chains are discussed, in detail, in the following articles.
5.17. Four Bar Chain or Quadric Cycle Chain
We have already discussed that the kinematic chain is a combination of four or more
kinematic pairs, such that the relative motion between the links or elements is completely constrained.
The simplest and the basic kinematic chain is a four bar chain or quadric cycle chain, as shown in Fig. 5.18. It consists of four links, each of
them forms a turning pair at A, B, C and D. The four links may be of
different lengths. According to Grashof ’s law for a four bar mechanism, the sum of the shortest and longest link lengths should not be
greater than the sum of the remaining two link lengths if there is to be
continuous relative motion between the two links.
A very important consideration in designing a mechanism is to
ensure that the input crank makes a complete revolution relative to the
Fig. 5.18. Four bar chain.
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other links. The mechanism in which no link makes a complete revolution will not be useful. In a four
bar chain, one of the links, in particular the shortest link, will make a complete revolution relative to
the other three links, if it satisfies the Grashof ’s law. Such a link is known as crank or driver. In Fig.
5.18, A D (link 4 ) is a crank. The link BC (link 2) which makes a partial rotation or oscillates is known
as lever or rocker or follower and the link CD (link 3) which connects the crank and lever is called
connecting rod or coupler. The fixed link A B (link 1) is known as frame of the mechanism.
When the crank (link 4) is the driver, the mechanism is transforming rotary motion into
oscillating motion.
5.18. Inversions of Four Bar Chain
Though there are many inversions of the four bar
chain, yet the following are important from the subject
point of view :
1. Beam engine (crank and lever mechanism).
A part of the mechanism of a beam engine (also known as
crank and lever mechanism) which consists of four links,
is shown in Fig. 5.19. In this mechanism, when the crank
rotates about the fixed centre A , the lever oscillates about
a fixed centre D. The end E of the lever CDE is
connected to a piston rod which reciprocates due to the
rotation of the crank. In other words, the purpose of this
mechanism is to convert rotary motion into reciprocating
motion.
Fig. 5.19. Beam engine.
Fig. 5.20. Coupling rod of a locomotive.
2. Coupling rod of a locomotive (Double crank mechanism). The mechanism of a coupling
rod of a locomotive (also known as double crank mechanism) which consists of four links, is shown
in Fig. 5.20.
In this mechanism, the links AD and BC (having equal length) act as cranks and are connected to the respective wheels. The link CD acts as a coupling rod and the link A B is fixed in order
to maintain a constant centre to centre distance between them. This mechanism is meant for transmitting rotary motion from one wheel to the other wheel.
3. Watt’s indicator mechanism (Double lever mechanism). A *Watt’s indicator mechanism
(also known as Watt's straight line mechanism or double lever mechanism) which consists of four
*
Refer Chapter 9, Art. 9.6
Chapter 5 : Simple Mechanisms
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107
links, is shown in Fig. 5.21. The four links are : fixed link
at A , link A C, link CE and link BFD. It may be noted that
BF and FD form one link because these two parts have no
relative motion between them. The links CE and BFD act
as levers. The displacement of the link BFD is directly
proportional to the pressure of gas or steam which acts on
the indicator plunger. On any small displacement of the
mechanism, the tracing point E at the end of the link CE
traces out approximately a straight line.
The initial position of the mechanism is shown in
Fig. 5.21 by full lines whereas the dotted lines show the
position of the mechanism when the gas or steam pressure
acts on the indicator plunger.
Fig. 5.21. Watt’s indicator mechanism.
5.19. Single Slider Crank Chain
A single slider crank chain is a modification of the basic four bar chain. It consist of one
sliding pair and three turning pairs. It is,usually, found in reciprocating steam engine mechanism.
This type of mechanism converts rotary motion into reciprocating motion and vice versa.
In a single slider crank chain, as shown in Fig. 5.22, the links 1 and 2, links 2 and 3, and links
3 and 4 form three turning pairs while the links 4 and 1 form a sliding pair.
Fig. 5.22. Single slider crank chain.
The link 1 corresponds to the frame of the engine, which is fixed. The link 2 corresponds to
the crank ; link 3 corresponds to the connecting rod and link 4 corresponds to cross-head. As the
crank rotates, the cross-head reciprocates in the guides and thus the piston reciprocates in the
cylinder.
5.20. Inversions of Single Slider Crank Chain
We have seen in the previous article that a single slider crank chain is a four-link mechanism.
We know that by fixing, in turn, different links in a kinematic chain, an inversion is obtained and
we can obtain as many mechanisms as the links in a kinematic chain. It is thus obvious, that four
inversions of a single slider crank chain are possible. These inversions are found in the following
mechanisms.
1. Pendulum pump or Bull engine. In this mechanism, the inversion is obtained by fixing the
cylinder or link 4 (i.e. sliding pair), as shown in Fig. 5.23. In this case, when the crank (link 2) rotates,
the connecting rod (link 3) oscillates about a pin pivoted to the fixed link 4 at A and the piston
attached to the piston rod (link 1) reciprocates. The duplex pump which is used to supply feed water
to boilers have two pistons attached to link 1, as shown in Fig. 5.23.
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Theory of Machines
Fig. 5.23. Pendulum pump.
Fig. 5.24. Oscillating cylinder engine.
2. Oscillating cylinder engine. The arrangement of oscillating cylinder engine mechanism, as shown in Fig. 5.24, is used to convert
reciprocating motion into rotary motion. In this
mechanism, the link 3 forming the turning pair is
fixed. The link 3 corresponds to the connecting
rod of a reciprocating steam engine mechanism.
When the crank (link 2) rotates, the piston attached to piston rod (link 1) reciprocates and the
cylinder (link 4) oscillates about a pin pivoted to
the fixed link at A.
3. Rotary internal combustion engine
or Gnome engine. Sometimes back, rotary internal combustion engines were used in aviation.
But now-a-days gas turbines are used in its place.
It consists of seven cylinders in one plane and
all revolves about fixed centre D, as shown in
Fig. 5.25, while the crank (link 2) is fixed. In
this mechanism, when the connecting rod (link
4) rotates, the piston (link 3) reciprocates inside
the cylinders forming link 1.
Rotary engine
Fig. 5.25. Rotary internal combustion engine.
4. Crank and slotted lever quick return motion mechanism. This mechanism is mostly used
in shaping machines, slotting machines and in rotary internal combustion engines.
Chapter 5 : Simple Mechanisms
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In this mechanism, the link AC (i.e. link 3) forming the turning pair is fixed, as shown in Fig.
5.26. The link 3 corresponds to the connecting rod of a reciprocating steam engine. The driving crank
CB revolves with uniform angular speed about the fixed centre C. A sliding block attached to the crank
pin at B slides along the slotted bar AP and thus causes AP to oscillate about the pivoted point A . A
short link PR transmits the motion from AP to the ram which carries the tool and reciprocates along the
line of stroke R 1R 2. The line of stroke of the ram (i.e. R 1R 2) is perpendicular to AC produced.
Fig. 5.26. Crank and slotted lever quick return motion mechanism.
In the extreme positions, AP1 and AP2 are tangential to the circle
and the cutting tool is at the end of the stroke. The forward or cutting
stroke occurs when the crank rotates from the position CB1 to CB2 (or
through an angle β) in the clockwise direction. The return stroke occurs
when the crank rotates from the position CB2 to CB1 (or through angle α)
in the clockwise direction. Since the crank has uniform angular speed,
therefore,
Time of cutting stroke β
360° − α
β
= =
or
α
Time of return stroke α 360° − β
Since the tool travels a distance of R 1 R 2 during cutting and return
stroke, therefore travel of the tool or length of stroke
= R 1R 2 = P1P2 = 2P1Q = 2AP1 sin ∠ P1 A Q
(
)
α
α
= 2 AP1 sin 90° − 2 = 2 AP cos 2
= 2 AP ×
CB1
AC
The Shaping Machine
..... (3AP1 = AP)
... 3 cos α = CB1
2
AC
= 2 AP × CB
... (3 CB1 = CB)
AC
Note: From Fig. 5.26, we see that the angle β made by the forward or cutting stroke is greater than the angle α
described by the return stroke. Since the crank rotates with uniform angular speed, therefore the return stroke is
completed within shorter time. Thus it is called quick return motion mechanism.
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Theory of Machines
5. Whitworth quick return motion mechanism. This mechanism is mostly used in shaping
and slotting machines. In this mechanism, the link CD (link 2) forming the turning pair is fixed, as
shown in Fig. 5.27. The link 2 corresponds to a crank in a reciprocating steam engine. The driving
crank C A (link 3) rotates at a uniform angular speed. The slider (link 4) attached to the crank pin at A
slides along the slotted bar PA (link 1) which oscillates at a pivoted point D. The connecting rod PR
carries the ram at R to which a cutting tool is fixed. The motion of the tool is constrained along the
line RD produced, i.e. along a line passing through D and perpendicular to CD.
Fig. 5.27. Whitworth quick return motion mechanism.
When the driving crank C A moves from the position C A1 to C A2 (or the link DP from the
position DP1 to DP2) through an angle α in the clockwise direction, the tool moves from the left hand
end of its stroke to the right hand end through a distance 2 PD.
Now when the driving crank moves from the position CA2 to CA1 (or the link DP from DP2 to
DP1 ) through an angle β in the clockwise direction, the tool moves back from right hand end of its
stroke to the left hand end.
A little consideration will show that the time taken during the left to right movement of the
ram (i.e. during forward or cutting stroke) will be equal to the time taken by the driving crank to move
from CA1 to CA2. Similarly, the time taken during the right to left movement of the ram (or during the
idle or return stroke) will be equal to the time taken by the driving crank to move from C A2 to C A1.
Since the crank link C A rotates at uniform angular velocity therefore time taken during the
cutting stroke (or forward stroke) is more than the time taken during the return stroke. In other words,
the mean speed of the ram during cutting stroke is less than the mean speed during the return stroke.
The ratio between the time taken during the cutting and return strokes is given by
Time of cutting stroke α
α
= =
β 360° − α
Time of return stroke
or
360° − β
β
Note. In order to find the length of effective stroke R 1 R 2, mark P1 R 1 = P2 R 2 = PR. The length of effective
stroke is also equal to 2 PD.
Example 5.1. A crank and slotted lever mechanism used in a shaper has a centre distance of
300 mm between the centre of oscillation of the slotted lever and the centre of rotation of the crank.
The radius of the crank is 120 mm. Find the ratio of the time of cutting to the time of return stroke.
Solution. Given : A C = 300 mm ; CB1 = 120 mm
The extreme positions of the crank are shown in Fig. 5.28. We know that
Chapter 5 : Simple Mechanisms
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111
sin ∠CAB1 = sin (90° − α / 2)
=
CB1 120
=
= 0.4
AC 300
∠ CAB1 = 90° − α / 2
∴
= sin–1 0.4 = 23.6°
α / 2 = 90° – 23.6° = 66.4°
or
α = 2 × 66.4 = 132.8°
and
Fig. 5.28
We know that
Time of cutting stroke 360° − α 360° − 132.8°
=
=
= 1.72 Ans.
Time of return stroke
132.8°
α
Example 5.2. In a crank and slotted lever quick return motion mechanism, the distance
between the fixed centres is 240 mm and the length of the driving crank is 120 mm. Find the inclination of the slotted bar with the vertical in the extreme position and the time ratio of cutting stroke to
the return stroke.
If the length of the slotted bar is 450 mm, find the length of the stroke if the line of stroke
passes through the extreme positions of the free end of the lever.
Solution. Given : AC = 240 mm ; CB1 = 120 mm ; AP1 = 450 mm
Inclination of the slotted bar with the vertical
Let
∠CAB1 = Inclination of the slotted bar with the vertical.
The extreme positions of the crank are
shown in Fig. 5.29. We know that
(
sin ∠CAB1 = sin 90 ° − α
2
=
)
B1C 120
=
= 0.5
240
AC
α
∴ ∠CAB1 = 90° − 2
= sin −1 0.5 = 30° Ans.
Time ratio of cutting stroke to the return stroke
We know that
90° – α / 2 = 30°
α / 2 = 90° – 30° = 60°
∴
or
α = 2 × 60° = 120°
∴
Fig. 5.29
Time of cutting stroke 360 ° − α 360° − 120°
=
=
= 2 Ans.
Time of return stroke
120°
α
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Theory of Machines
Length of the stroke
We know that length of the stroke,
R 1R 2 = P1P2 = 2 P1Q = 2 AP1 sin (90° – α / 2)
= 2 × 450 sin (90°– 60°) = 900 × 0.5 = 450 mm Ans.
Example 5.3. Fig. 5.30 shows the lay out of a quick return mechanism of the oscillating link
type, for a special purpose machine. The driving crank BC is 30 mm long and time ratio of the
working stroke to the return stroke is to be 1.7. If the length of the working stroke of R is 120 mm,
determine the dimensions of AC and AP.
Solution. Given : BC = 30 mm ; R 1R 2 = 120 mm ; Time ratio of working stroke to the return
stroke = 1.7
Fig. 5.30
Fig. 5.31
We know that
Time of working stroke 360 − α
360 − α
=
or 1.7 =
Time of return stroke
α
α
∴
α = 133.3° or α / 2 = 66.65°
The extreme positions of the crank are shown in Fig. 5.31. From right angled triangle A B1C,
we find that
B1C
B1C
BC
sin (90° – α/2) =
or
AC =
=
AC
sin (90° − α / 2) cos α / 2
... (∵ B 1C = BC)
30
= 30 = 75.7 mm Ans.
cos 66.65° 0.3963
We know that length of stroke,
∴
AC =
R 1R 2 = P1P2 = 2P1Q = 2 AP1 sin (90° – α / 2) = 2 AP1 cos α / 2
120 = 2 AP cos 66.65° = 0.7926 AP
... (∵ AP1 = AP)
∴
AP = 120 / 0.7926 = 151.4 mm Ans.
Example 5.4. In a Whitworth quick return motion mechanism, as shown in Fig. 5.32, the
distance between the fixed centers is 50 mm and the length of the driving crank is 75 mm. The length
of the slotted lever is 150 mm and the length of the connecting rod is 135 mm. Find the ratio of the
time of cutting stroke to the time of return stroke and also the effective stroke.
Chapter 5 : Simple Mechanisms
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113
Solution. Given : CD = 50 mm ; C A = 75 mm ; PA = 150 mm ; PR = 135 mm
Fig. 5.32
Fig. 5.33
The extreme positions of the driving crank are shown in Fig. 5.33. From the geometry of the
figure,
∴
cos β / 2 = CD = 50 = 0.667
75
CA2
β / 2 = 48.2° or β = 96.4°
... (3 C A2 = CA)
Ratio of the time of cutting stroke to the time of return stroke
We know that
Time of cutting stroke 360 − β 360 − 96.4
=
=
= 2.735 Ans.
Time of return stroke
96.4
β
Length of effective stroke
In order to find the length of effective stroke (i.e. R 1R 2), draw the space diagram of the
mechanism to some suitable scale, as shown in Fig. 5.33. Mark P1R 2 = P2R 2 = PR. Therefore by
measurement we find that,
Length of effective stroke = R 1R 2 = 87.5 mm Ans.
5.21. Double Slider Crank Chain
A kinematic chain which consists of two turning pairs and two sliding pairs is known as
double slider crank chain, as shown in Fig. 5.34. We see that the link 2 and link 1 form one turning
pair and link 2 and link 3 form the second turning pair. The link 3 and link 4 form one sliding pair and
link 1 and link 4 form the second sliding pair.
5.22. Inversions of Double Slider Crank Chain
The following three inversions of a double slider crank chain are important from the subject
point of view :
1. Elliptical trammels. It is an instrument used for drawing ellipses. This inversion is obtained
by fixing the slotted plate (link 4), as shown in Fig. 5.34. The fixed plate or link 4 has two straight
grooves cut in it, at right angles to each other. The link 1 and link 3, are known as sliders and form sliding
pairs with link 4. The link A B (link 2) is a bar which forms turning pair with links 1 and 3.
When the links 1 and 3 slide along their respective grooves, any point on the link 2 such as
P traces out an ellipse on the surface of link 4, as shown in Fig. 5.34 (a). A little consideration will
show that AP and BP are the semi-major axis and semi-minor axis of the ellipse respectively. This can
be proved as follows :
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Theory of Machines
(a)
(b)
Fig. 5.34. Elliptical trammels.
Let us take O X and O Y as horizontal and vertical axes and let the link B A is inclined at an
angle θ with the horizontal, as shown in Fig. 5.34 (b). Now the co-ordinates of the point P on the link
B A will be
x = PQ = AP cos θ; and y = PR = BP sin θ
or
y
x =
= sin θ
cos θ ; and
AP
BP
Squaring and adding,
2
x2 + y
= cos 2 θ + sin 2 θ = 1
2
2
( AP)
( BP)
This is the equation of an ellipse. Hence the path traced by point P is an ellipse whose semimajor axis is AP and semi-minor axis is BP.
Note : If P is the mid-point of link B A, then AP = BP. The above equation can be written as
x2
y2
+
=1
or
x2 + y 2 = (AP)2
2
( AP)
( AP) 2
This is the equation of a circle whose radius is AP. Hence if P is the mid-point of link B A, it will trace
a circle.
2. Scotch yoke mechanism. This mechanism is used for converting rotary motion into a
reciprocating motion. The inversion is obtained by fixing either the link 1 or link 3. In Fig. 5.35, link
1 is fixed. In this mechanism, when the link 2 (which
corresponds to crank) rotates about B as centre, the link
4 (which corresponds to a frame) reciprocates. The fixed
link 1 guides the frame.
3. Oldham’s coupling. An oldham's coupling is
used for connecting two parallel shafts whose axes are
at a small distance apart. The shafts are coupled in such
a way that if one shaft rotates, the other shaft also rotates
at the same speed. This inversion is obtained by fixing
the link 2, as shown in Fig. 5.36 (a). The shafts to be
connected have two flanges (link 1 and link 3) rigidly
fastened at their ends by forging.
Fig. 5.35. Scotch yoke mechanism.
Chapter 5 : Simple Mechanisms
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115
The link 1 and link 3 form turning pairs with link 2. These flanges have diametrical slots cut
in their inner faces, as shown in Fig. 5.36 (b). The intermediate piece (link 4) which is a circular disc,
have two tongues (i.e. diametrical projections) T 1 and T 2 on each face at right angles to each other, as
shown in Fig. 5.36 (c). The tongues on the link 4 closely fit into the slots in the two flanges (link 1 and
link 3). The link 4 can slide or reciprocate in the slots in the flanges.
(a)
(b)
(c)
Fig. 5.36. Oldham’s coupling.
When the driving shaft A is rotated, the flange C (link 1) causes the intermediate piece (link
4) to rotate at the same angle through which the flange has rotated, and it further rotates the flange D
(link 3) at the same angle and thus the shaft B rotates. Hence links 1, 3 and 4 have the same angular
velocity at every instant. A little consideration will show, that there is a sliding motion between the
link 4 and each of the other links 1 and 3.
If the distance between the axes of the shafts is constant, the centre of intermediate piece will
describe a circle of radius equal to the distance between the axes of the two shafts. Therefore, the
maximum sliding speed of each tongue along its slot is equal to the peripheral velocity of the centre
of the disc along its circular path.
Let
ω = Angular velocity of each shaft in rad/s, and
r = Distance between the axes of the shafts in metres.
∴ Maximum sliding speed of each tongue (in m/s),
v = ω.r
EXERCISES
1.
In a crank and slotted lever quick return mechanism, the distance between the fixed centres is 150
mm and the driving crank is 75 mm long. Determine the ratio of the time taken on the cutting and
return strokes.
[Ans. 2]
2.
In a crank and slotted lever quick return motion mechanism, the distance between the fixed centres O
and C is 200 mm. The driving crank CP is 75 mm long. The pin Q on the slotted lever, 360 mm from
the fulcrum O, is connected by a link QR 100 mm long, to a pin R on the ram. The line of stroke of R
is perpendicular to OC and intersects OC produced at a point 150 mm from C. Determine the ratio of
times taken on the cutting and return strokes.
[Ans. 1.647]
3.
In a crank and slotted lever quick return mechanism, as shown in Fig. 5.37, the driving crank length is
75 mm. The distance between the fixed centres is 200 mm and the length of the slotted lever is 500
mm. Find the ratio of the times taken on the cutting and idle strokes. Determine the effective stroke
also.
[Ans. 1.67 ; 380 mm]
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Theory of Machines
All dimensions in mm
4.
5.
Fig. 5.37
The Whitworth quick return motion mechanism has the driving crank 150 mm long. The distance
between fixed centres is 100 mm. The line of stroke of the ram passes through the centre of rotation of
the slotted lever whose free end is connected to the ram by a connecting link. Find the ratio of time of
cutting to time of return.
[Ans. 2.735]
A Whitworth quick return motion mechanism, as shown in Fig. 5.38, has the following
particulars :
Fig. 5.38
Time of cutting stroke
Length of stroke = 150 mm ; Driving crank length = 40 mm; Time of return stroke = 2
Find the lengths of CD and PD. Also determine the angles α and β.
[Hint : Length of stroke = R 1R 2 = P1P2 = 2PD]
[Ans. 20 mm, 75 mm; 240°, 120°]
DO YOU KNOW ?
1.
2.
3.
4.
5.
6.
7.
Explain the term kinematic link. Give the classification of kinematic link.
What is a machine ? Giving example, differentiate between a machine and a structure.
Write notes on complete and incomplete constraints in lower and higher pairs, illustrating your answer
with neat sketches.
Explain different kinds of kinematic pairs giving example for each one of them.
Explain the terms : 1. Lower pair, 2. Higher pair, 3. Kinematic chain, and 4. Inversion.
In what way a mechanism differ from a machine ?
What is the significance of degrees of freedom of a kinematic chain when it functions as a mechanism? Give examples.
Chapter 5 : Simple Mechanisms
8.
9.
10.
11.
12.
13.
14.
15.
16.
l
117
Determine the mobility (degrees of freedom) of the mechanism shown in Fig. 5.39 (a) and (b) using
Kutzbach mobility criterion and classify them.
Fig. 5.39
Explain Grubler’s criterion for determining degree of freedom for mechanisms. Using Grubler’s criterion for plane mechanism, prove that the minimum number of binary links in a constrained mechanism with simple hinges is four.
Sketch and explain the various inversions of a slider crank chain.
Sketch and describe the four bar chain mechanism. Why it is considered to be the basic chain?
Show that slider crank mechanism is a modification of the basic four bar mechanism.
Sketch slider crank chain and its various inversions, stating actual machines in which these are used in
practice.
Sketch and describe the working of two different types of quick return mechanisms. Give examples of
their applications. Derive an expression for the ratio of times taken in forward and return stroke for
one of these mechanisms.
Sketch and explain any two inversions of a double slider crank chain.
Identify the kinematic chains to which the following mechanisms belong :
1. Steam engine mechanism ; 2. Beam engine ; 3. Whitworth quick return motion mechanism;
4. Elliptical trammels.
OBJECTIVE TYPE QUESTIONS
1. In a reciprocating steam engine, which of the following forms a kinematic link ?
(a) cylinder and piston
(b) piston rod and connecting rod
(c) crank shaft and flywheel
(d) flywheel and engine frame
2. The motion of a piston in the cylinder of a steam engine is an example of
(a) completely constrained motion
(b) incompletely constrained motion
(c) successfully constrained motion
(d) none of these
3. The motion transmitted between the teeth of gears in mesh is
(a) sliding
(b) rolling
(c) may be rolling or sliding depending upon the shape of teeth
(d) partly sliding and partly rolling
4. The cam and follower without a spring forms a
(a) lower pair
(b) higher pair
(c) self closed pair
(d) force closed pair
5. A ball and a socket joint forms a
(a) turning pair
(b) rolling pair
(c) sliding pair
(d) spherical pair
6. The lead screw of a lathe with nut forms a
(a) sliding pair
(b) rolling pair
(c) screw pair
(d) turning pair
7. When the elements of the pair are kept in contact by the action of external forces, the pair is said to be a
(a) lower pair
(b) higher pair
(c) self closed pair
(d) force closed pair
118
l
Theory of Machines
8. Which of the following is a turning pair ?
(a) Piston and cylinder of a reciprocating steam engine
(b) Shaft with collars at both ends fitted in a circular hole
(c) Lead screw of a lathe with nut
(d) Ball and socket joint
9. A combination of kinematic pairs, joined in such a way that the relative motion between the links is
completely constrained, is called a
(a) structure
(b) mechanism
(c) kinematic chain
(d) inversion
10. The relation between the number of pairs ( p ) forming a kinematic chain and the number of links (l) is
(a) l = 2p – 2
(b) l = 2p – 3
(c) l = 2p – 4
(d) l = 2p – 5
11. The relation between the number of links (l) and the number of binary joints ( j) for a kinematic chain
having constrained motion is given by j =
12.
13.
14.
15.
16.
17.
18.
19.
20.
3
l − 2. If the left hand side of this equation is greater than
2
right hand side, then the chain is
(a) locked chain
(b) completely constrained chain
(c) successfully constrained chain
(d) incompletely constrained chain
In a kinematic chain, a quaternary joint is equivalent to
(a) one binary joint
(b) two binary joints
(c) three binary joints
(d) four binary joints
If n links are connected at the same joint, the joint is equivalent to
(a) (n – 1) binary joints (b) (n – 2) binary joints (c) (2n – 1) binary joints (d) none of these
In a 4 – bar linkage, if the lengths of shortest, longest and the other two links are denoted by s, l, p and
q, then it would result in Grashof’s linkage provided that
(a) l + p < s + q
(b) l + s < p + q
(c) l + p = s + q
(d) none of these
A kinematic chain is known as a mechanism when
(a) none of the links is fixed
(b) one of the links is fixed
(c) two of the links are fixed
(d) all of the links are fixed
The Grubler’s criterion for determining the degrees of freedom (n) of a mechanism having plane motion
is
(a) n = (l – 1) – j
(b) n = 2 (l – 1) – 2j
(c) n = 3 (l – 1) – 2j (d) n = 4 (l – 1) – 3j
where l = Number of links, and j = Number of binary joints.
The mechanism forms a structure, when the number of degrees of freedom (n) is equal to
(a) 0
(b) 1
(c) 2
(d) – 1
In a four bar chain or quadric cycle chain
(a) each of the four pairs is a turning pair
(b) one is a turning pair and three are sliding pairs
(c) three are turning pairs and one is sliding pair (d) each of the four pairs is a sliding pair.
Which of the following is an inversion of single slider crank chain ?
(a) Beam engine
(b) Watt’s indicator mechanism
(c) Elliptical trammels
(d) Whitworth quick return motion mechanism
Which of the following is an inversion of double slider crank chain ?
(a) Coupling rod of a locomotive
(b) Pendulum pump
(c) Elliptical trammels
(d) Oscillating cylinder engine
ANSWERS
1.
6.
11.
16.
(c)
(c)
(a)
(c)
2.
7.
12.
17.
(a)
(d)
(c)
(a)
3.
8.
13.
18.
(d)
(b)
(a)
(a)
4.
9.
14.
19.
(c)
(c)
(b)
(d)
5.
10.
15.
20.
(d)
(c)
(b)
(c)
GO To FIRST
CONTENTS
CONTENTS
6
Features
1. 1ntroduction.
2. Space and Body Centrodes.
3. Methods for Determining the
Velocity of a Point on a Link.
4. Velocity of a Point on a Link
by Instantaneous Centre
Method.
5. Properties of the
Instantaneous Centre.
6. Number of Instantaneous
Centres in a Mechanism.
7. Types of Instantaneous
Centres.
8. Location of Instantaneous
Centres.
9. Aronhold Kennedy (or Three
Centres-in-Line) Theorem.
10. Method of Locating
Instantaneous Centres in a
Mechanism.
Velocity in
Mechanisms
(Instantaneous Centre Method)
6.1.
Introduction
Sometimes, a body has simultaneously a motion of
rotation as well
as translation,
such as wheel of
a car, a sphere
rolling (but not
slipping) on the
ground. Such a
motion will have
the combined effect of rotation
Fig. 6.1. Motion of a link.
and translation.
Consider a rigid link AB, which moves from its initial
position AB to A 1 B 1 as shown in Fig. 6.1 (a). A little
consideration will show that the link neither has wholly a
motion of translation nor wholly rotational, but a combination
of the two motions. In Fig. 6.1 (a), the link has first the motion
of translation from AB to A1B′ and then the motion of rotation
about A1, till it occupies the final position A1 B1. In Fig. 6.1
(b), the link AB has first the motion of rotation from AB to
A B′ about A and then the motion of translation from A B′ to
119
CONTENTS
CONTENTS
120
l
Theory of Machines
A 1 B 1. Such a motion of link A B to A 1
B 1 is an example of combined motion
of rotation and translation, it being
immaterial whether the motion of
rotation takes first, or the motion of
translation.
In actual practice, the motion
of link A B is so gradual that it is
difficult to see the two separate
motions. But we see the two separate
motions, though the point B moves
faster than the point A . Thus, this
Mechanisms on a steam automobile engine.
combined motion of rotation and
translation of the link A B may be assumed to be a motion of pure rotation about some centre I, known
as the instantaneous centre of rotation (also called centro or virtual centre). The position of
instantaneous centre may be located as discussed below:
Since the points A and B of the link has moved to A 1 and B 1
respectively under the motion of rotation (as assumed above), therefore the position of the centre of rotation must lie on the intersection of
the right bisectors of chords A A1 and B B1. Let these bisectors intersect
at I as shown in Fig. 6.2, which is the instantaneous centre of rotation or
virtual centre of the link A B.
From above, we see that the position of the link AB goes on
changing, therefore the centre about which the motion is assumed to
take place (i.e. the instantaneous centre of rotation) also goes on changing. Thus the instantaneous centre of a moving body may be defined as
that centre which goes on changing from one instant to another. The
Fig. 6.2. Instantaneous
locus of all such instantaneous centres is known as centrode. A line
centre of rotation.
drawn through an instantaneous centre and perpendicular to the plane
of motion is called instantaneous axis. The locus of this axis is known as axode.
6.2.
Space and Body Centrodes
A rigid body in plane motion relative to a second rigid body, supposed fixed in space, may be
assumed to be rotating about an instantaneous centre at
that particular moment. In other words, the instantaneous
centre is a point in the body which may be considered
fixed at any particular moment. The locus of the
instantaneous centre in space during a definite motion of
the body is called the space centrode and the locus of the
instantaneous centre relative to the body itself is called
the body centrode. These two centrodes have the
instantaneous centre as a common point at any instant and
during the motion of the body, the body centrode rolls
without slipping over the space centrode.
Fig. 6.3. Space and body centrode.
Let I1 and I2 be the instantaneous centres for the
two different positions A 1 B 1 and A 2 B 2 of the link A 1 B 1
after executing a plane motion as shown in Fig. 6.3. Similarly, if the number of positions of the link
A 1 B 1 are considered and a curve is drawn passing through these instantaneous centres (I1, I2....), then
the curve so obtained is called the space centrode.
Chapter 6 : Velocity in Mechanisms
l
121
Now consider a point C1 to be attached to the body or link A 1 B 1 and moves with it in such
a way that C1 coincides with I1 when the body is in position A 1 B 1. Let C2 be the position of the
point C1 when the link A 1 B 1 occupies the position A 2 B 2. A little consideration will show that the
point C2 will coincide with I2 (when the link is in position A 2 B 2) only if triangles A 1 B 1 C1 and
A 2 B 2 C2 are identical.
∴
A1 C 2 = A 2 I 2
and
B 1 C 2 = B 2 I2
In the similar way, the number of positions of the point C1 can be obtained for different
positions of the link A 1B 1. The curve drawn through these points (C1, C2....) is called the body
centrode.
6.3.
Methods for Determining the Velocity of a Point on a Link
Though there are many methods for determining the velocity of any point on a link in a
mechanism whose direction of motion (i.e. path) and velocity of some other point on the same link
is known in magnitude and direction, yet the following two methods are important from the subject
point of view.
1. Instantaneous centre method, and 2. Relative velocity method.
The instantaneous centre method is convenient and easy to apply in simple mechanisms,
whereas the relative velocity method may be used to any configuration diagram. We shall discuss the
relative velocity method in the next chapter.
6.4.
Velocity of a Point on a Link by
Instantaneous Centre Method
The instantaneous centre method of analysing the motion
in a mechanism is based upon the concept (as discussed in Art.
6.1) that any displacement of a body (or a rigid link) having
motion in one plane, can be considered as a pure rotational
motion of a rigid link as a whole about some centre, known as
instantaneous centre or virtual centre of rotation.
Consider two points A and B on a rigid link. Let v A and
Fig. 6.4. Velocity of a point on
v B be the velocities of points A and B, whose directions are given
a link.
by angles α and β as shown in Fig. 6.4. If v A is known in
magnitude and direction and v B in
direction only, then the magnitude of
v B may be determined by the
instantaneous centre method as
discussed below :
Draw A I and BI perpendiculars to the directions v A and v B respectively. Let these lines intersect at I,
which is known as instantaneous centre or virtual centre of the link. The
complete rigid link is to rotate or turn
about the centre I.
Robots use various mechanisms to perform jobs.
Since A and B are the points
on a rigid link, therefore there cannot
be any relative motion between them
along the line A B.
122
l
Theory of Machines
Now resolving the velocities along A B,
vA cos α = v B cos β
vA
cos β sin (90° – β)
=
=
vB
cos α sin (90° – α)
Applying Lami’s theorem to triangle ABI,
AI
BI
=
sin (90° – β) sin (90° – α)
or
...(i)
AI
sin (90° – β)
=
...(ii)
BI sin (90° – α )
From equation (i) and (ii),
vA
AI
vA
v
=
= B =ω
or
...(iii)
vB BI
AI BI
where
ω = Angular velocity of the rigid link.
If C is any other point on the link, then
vA
v
v
= B = C
...(iv)
AI
BI CI
From the above equation, we see that
1. If v A is known in magnitude and direction and v B in direction only, then velocity of point
B or any other point C lying on the same link may be determined in magnitude and direction.
or
2. The magnitude of velocities of the points on a rigid link is inversely proportional to the
distances from the points to the instantaneous centre and is perpendicular to the line joining the point
to the instantaneous centre.
6.5.
Properties of the Instantaneous Centre
The following properties of the instantaneous centre are important from the subject point of
view :
1. A rigid link rotates instantaneously relative to another link at the instantaneous centre for
the configuration of the mechanism considered.
2. The two rigid links have no linear velocity relative to each other at the instantaneous
centre. At this point (i.e. instantaneous centre), the two rigid links have the same linear velocity
relative to the third rigid link. In other words, the velocity of the instantaneous centre relative to any
third rigid link will be same whether the instantaneous centre is regarded as a point on the first rigid
link or on the second rigid link.
6.6.
Number of Instantaneous Centres in a
Mechanism
The number of instantaneous centres in a constrained
kinematic chain is equal to the number of possible combinations of two links. The number of pairs of links or the number Bar 3
of instantaneous centres is the number of combinations of n
links taken two at a time. Mathematically, number of instantaneous centres,
N =
n (n – 1)
, where n = Number of links.
2
Bar 2
3 2 Bar 1
Revolutes
4
Ground 1
Ground 2 Base
Four bar mechanisms.
Chapter 6 : Velocity in Mechanisms
6.7.
l
123
Types of Instantaneous Centres
The instantaneous centres for a mechanism are
of the following three types :
1. Fixed instantaneous centres, 2. Permanent
instantaneous centres, and 3. Neither fixed nor permanent instantaneous centres.
The first two types i.e. fixed and permanent
instantaneous centres are together known as primary
instantaneous centres and the third type is known as
secondary instantaneous centres.
Consider a four bar mechanism ABCD as
shown in Fig. 6.5. The number of instantaneous centres (N) in a four bar mechanism is given by
Fig. 6.5. Types of instantaneous centres.
n (n – 1) 4(4 – 1)
=
=6
... (∵ n = 4)
2
2
The instantaneous centres I12 and I14 are called the fixed instantaneous centres as they remain in the same place for all configurations of the mechanism. The instantaneous centres I23 and I34
are the permanent instantaneous centres as they move when the mechanism moves, but the joints
are of permanent nature. The instantaneous centres I13 and I24 are neither fixed nor permanent
instantaneous centres as they vary with the configuration of the mechanism.
N =
Note: The instantaneous centre of two links such as link 1 and link 2 is usually denoted by I12 and so on. It is
read as I one two and not I twelve.
6.8.
Location of Instantaneous Centres
The following rules may be used in locating the instantaneous centres in a mechanism :
1. When the two links are connected by a pin joint (or pivot joint), the instantaneous centre
Arm moves to a
track to retrive
information stored
there
Track selector
mechanism
The read/write head
is guided by information stored on the disk
itself
The hard disk is
coated with a
magnetic materials
Computer disk drive mechanisms.
Note : This picture is given as additional information and is not a direct example of the current chapter.
124
l
Theory of Machines
lies on the centre of the pin as shown in Fig. 6.6 (a). Such a instantaneous centre is of permanent
nature, but if one of the links is fixed, the instantaneous centre will be of fixed type.
2. When the two links have a pure rolling contact (i.e. link 2 rolls without slipping upon the
fixed link 1 which may be straight or curved), the instantaneous centre lies on their point of contact,
as shown in Fig. 6.6 (b). The velocity of any point A on the link 2 relative to fixed link 1 will be
perpendicular to I12 A and is proportional to I12 A . In other words
vA I12 A
=
vB I12 B
3. When the two links have a sliding contact, the instantaneous centre lies on the common
normal at the point of contact. We shall consider the following three cases :
(a) When the link 2 (slider) moves on fixed link 1 having straight surface as shown in
Fig. 6.6 (c), the instantaneous centre lies at infinity and each point on the slider have
the same velocity.
(b) When the link 2 (slider) moves on fixed link 1 having curved surface as shown in Fig.
6.6 (d),the instantaneous centre lies on the centre of curvature of the curvilinear path
in the configuration at that instant.
(c) When the link 2 (slider) moves on fixed link 1 having constant radius of curvature as
shown in Fig. 6.6 (e), the instantaneous centre lies at the centre of curvature i.e. the
centre of the circle, for all configuration of the links.
Fig. 6.6. Location of instantaneous centres.
6.9. Aronhold Kennedy (or Three Centres in Line) Theorem
The Aronhold Kennedy’s theorem states that if three bodies move relatively to each other,
they have three instantaneous centres and lie on a straight line.
Consider three kinematic links A , B and C having relative
plane motion. The number of instantaneous centres (N) is given by
N =
where
n (n – 1) 3(3 – 1)
=
=3
2
2
n = Number of links = 3
The two instantaneous centres at the pin joints of B with A ,
and C with A (i.e. Iab and Iac) are the permanent instantaneous centres.
According to Aronhold Kennedy’s theorem, the third instantaneous
centre Ibc must lie on the line joining Iab and Iac. In order to prove this,
Fig. 6.7. Aronhold Kennedy’s
theorem.
l
Chapter 6 : Velocity in Mechanisms
125
let us consider that the instantaneous centre Ibc lies outside the line joining Iab and Iac as shown in Fig. 6.7.
The point Ibc belongs to both the links B and C. Let us consider the point Ibc on the link B. Its velocity
v BC must be perpendicular to the line joining Iab and Ibc. Now consider the point Ibc on the link C. Its
velocity v BC must be perpendicular to the line joining Iac and Ibc.
We have already discussed in Art. 6.5, that the velocity of the instantaneous centre is same
whether it is regarded as a point on the first link or as a point on the second link. Therefore, the velocity
of the point Ibc cannot be perpendicular to both lines Iab Ibc and Iac Ibc unless the point Ibc lies on the line
joining the points Iab and Iac. Thus the three instantaneous centres (Iab, Iac and Ibc) must lie on the same
straight line. The exact location of Ibc on line Iab Iac depends upon the directions and magnitudes of the
angular velocities of B and C relative to A .
Drawing
Pencil
Winding handle to
operate the device
Central ring
Ellipses drawn by
the ellipsograph
The above picture shows ellipsograph which is used to draw ellipses.
Note : This picture is given as additional information and is not a direct example of the current chapter.
6.10. Method of Locating Instantaneous Centres in a Mechanism
Consider a pin jointed four bar mechanism as shown in Fig. 6.8 (a). The following procedure
is adopted for locating instantaneous centres.
1. First of all, determine the number of instantaneous centres (N) by using the relation
N =
n (n – 1)
, where n = Number of links.
2
N =
4(4 – 1)
=6
2
In the present case,
...(∵ n = 4)
2. Make a list of all the instantaneous centres in a mechanism. Since for a four bar mechanism, there are six instantaneous centres, therefore these centres are listed as shown in the following
table (known as book-keeping table).
Links
1
2
3
4
Instantaneous
12
23
34
–
centres
(6 in number)
13
14
24
126
l
Theory of Machines
3. Locate the fixed and permanent instantaneous centres by inspection. In Fig. 6.8 (a), I12
and I14 are fixed instantaneous centres and I23 and I34 are permanent instantaneous centres.
Note. The four bar mechanism has four turning pairs, therefore there are four primary (i.e. fixed and permanent)
instantaneous centres and are located at the centres of the pin joints.
Fig. 6.8. Method of locating instantaneous centres.
4. Locate the remaining neither fixed nor permanent instantaneous centres (or secondary
centres) by Kennedy’s theorem. This is done by circle diagram as shown in Fig. 6.8 (b). Mark points
on a circle equal to the number of links in a mechanism. In the present case, mark 1, 2, 3, and 4 on the
circle.
5. Join the points by solid lines to show that these centres are already found. In the circle
diagram [Fig. 6.8 (b)] these lines are 12, 23, 34 and 14 to indicate the centres I12, I23, I34 and I14.
6. In order to find the other two instantaneous centres, join two such points that the line
joining them forms two adjacent triangles in the circle diagram. The line which is responsible for
completing two triangles, should be a common side to the two triangles. In Fig. 6.8 (b), join 1 and 3
to form the triangles 123 and 341 and the instantaneous centre* I13 will lie on the intersection of I12
I23 and I14 I34, produced if necessary, on the mechanism. Thus the instantaneous centre I13 is located.
Join 1 and 3 by a dotted line on the circle diagram and mark number 5 on it. Similarly the instantaneous centre I24 will lie on the intersection of I12 I14 and I23 I34, produced if necessary, on the mechanism. Thus I24 is located. Join 2 and 4 by a dotted line on the circle diagram and mark 6 on it. Hence
all the six instantaneous centres are located.
Note: Since some of the neither fixed nor permanent instantaneous centres are not required in solving problems,
therefore they may be omitted.
Example 6.1. In a pin jointed four bar mechanism, as shown in Fig. 6.9, AB = 300 mm, BC = CD = 360
mm, and AD = 600 mm. The angle BAD = 60°. The crank
AB rotates uniformly at 100 r.p.m. Locate all the instantaneous centres and find the angular velocity of the link BC.
Solution. Given : NAB = 100 r.p.m or
ωAB = 2 π × 100/60 = 10.47 rad/s
Since the length of crank A B = 300 mm = 0.3 m,
therefore velocity of point B on link A B,
*
Fig. 6.9
We may also say as follows: Considering links 1, 2 and 3, the instantaneous centres will be I12, I23 and I13.
The centres I12 and I23 have already been located. Similarly considering links 1, 3 and 4, the instantaneous
centres will be I13, I34 and I14, from which I14 and I34 have already been located. Thus we see that the centre
I13 lies on the intersection of the lines joining the points I12 I23 and I14 I34.
Chapter 6 : Velocity in Mechanisms
l
127
vB = ωAB × A B = 10.47 × 0.3 = 3.141 m/s
Location of instantaneous centres
The instantaneous centres are located as discussed below:
1. Since the mechanism consists of four links (i.e. n = 4 ), therefore number of instantaneous
centres,
n (n – 1) 4(4 – 1)
=
=6
2
2
2. For a four bar mechanism, the book keeping table may be drawn as discussed in Art. 6.10.
N =
3. Locate the fixed and permanent instantaneous centres by inspection. These centres are I12,
I23, I34 and I14, as shown in Fig. 6.10.
4. Locate the remaining neither fixed nor permanent instantaneous centres by Aronhold
Kennedy’s theorem. This is done by circle diagram as shown in Fig. 6.11. Mark four points (equal to
the number of links in a mechanism) 1, 2, 3, and 4 on the circle.
Fig. 6.10
5. Join points 1 to 2, 2 to 3, 3 to 4 and 4 to 1 to indicate the instantaneous centres already
located i.e. I12, I23, I34 and I14.
6. Join 1 to 3 to form two triangles 1 2 3 and 3 4 1. The side 13, common to both triangles,
is responsible for completing the two triangles. Therefore the instantaneous centre I13 lies on the intersection of the lines joining the points I12
I23 and I34 I14 as shown in Fig. 6.10. Thus centre I13 is located. Mark
number 5 (because four instantaneous centres have already been located)
on the dotted line 1 3.
7. Now join 2 to 4 to complete two triangles 2 3 4 and 1 2 4.
The side 2 4, common to both triangles, is responsible for completing
the two triangles. Therefore centre I24 lies on the intersection of the lines
joining the points I23 I34 and I12 I14 as shown in Fig. 6.10. Thus centre I24
is located. Mark number 6 on the dotted line 2 4. Thus all the six instanFig. 6.11
taneous centres are located.
Angular velocity of the link BC
Let
ωBC = Angular velocity of the link BC.
Since B is also a point on link BC, therefore velocity of point B on link BC,
vB = ωBC × I13 B
128
l
Theory of Machines
By measurement, we find that I13 B = 500 mm = 0.5 m
∴
3.141
vB
=
= 6.282 rad/s Ans.
0.5
I13 B
ωBC =
Example 6.2. Locate all the instantaneous centres of the slider crank mechanism as shown
in Fig. 6.12. The lengths of crank OB and connecting rod AB are 100 mm and 400 mm respectively.
If the crank rotates clockwise with an angular velocity of 10 rad/s, find: 1. Velocity of the slider A,
and 2. Angular velocity of the connecting rod AB.
Fig. 6.12
Solution. Given :
ωOB = 10 rad/ s; OB = 100 mm = 0.1 m
We know that linear velocity of the crank OB,
vOB = v B = ωOB × OB = 10 × 0.1 = 1 m/s
Location of instantaneous centres
The instantaneous centres in a slider crank mechanism are located as discussed below:
1. Since there are four links (i.e. n = 4), therefore the number of instantaneous centres,
N =
n ( n – 1) 4 (4 – 1)
=
=6
2
2
Bearing block
Pin
Slider
Crank
Connecting
rod
Slider crank mechanism.
2. For a four link mechanism, the book keeping table may be drawn as discussed in Art. 6.10.
3. Locate the fixed and permanent instantaneous centres by inspection. These centres are I12,
I23 and I34 as shown in Fig. 6.13. Since the slider (link 4) moves on a straight surface (link 1), therefore the instantaneous centre I14 will be at infinity.
Note: Since the slider crank mechanism has three turning pairs and one sliding pair, therefore there will be three
primary (i.e. fixed and permanent) instantaneous centres.
Chapter 6 : Velocity in Mechanisms
l
129
4. Locate the other two remaining neither fixed nor permanent instantaneous centres, by
Aronhold Kennedy’s theorem. This is done by circle diagram as shown in Fig. 6.14. Mark four points
1, 2, 3 and 4 (equal to the number of links in a mechanism) on the circle to indicate I12, I23, I34 and I14.
Fig. 6.13
Fig. 6.14
5. Join 1 to 3 to form two triangles 1 2 3 and 3 4 1 in the circle diagram. The side 1 3,
common to both triangles, is responsible for completing the two triangles. Therefore the centre I13
will lie on the intersection of I12 I23 and I14 I34, produced if necessary. Thus centre I13 is located. Join
1 to 3 by a dotted line and mark number 5 on it.
6. Join 2 to 4 by a dotted line to form two triangles 2 3 4 and 1 2 4. The side 2 4, common
to both triangles, is responsible for completing the two triangles. Therefore the centre I24 lies on the
intersection of I23 I34 and I12 I14. Join 2 to 4 by a dotted line on the circle diagram and mark number 6
on it. Thus all the six instantaneous centres are located.
By measurement, we find that
I13 A = 460 mm = 0.46 m ; and I13 B = 560 mm = 0.56 m
1. Velocity of the slider A
Let
vA = Velocity of the slider A .
We know that
vA
v
= B
I13 A I13 B
vA = vB ×
or
I13 A
0.46
=1×
= 0.82 m/s Ans.
0.56
I13 B
2. Angular velocity of the connecting rod AB
Let
ωAB = Angular velocity of the connecting rod A B.
We know that
vA
v
= B = ωAB
I13 A I13 B
130
l
Theory of Machines
Exhaust
waste heat
Engine
Hydraulic
rams
Load
The above picture shows a digging machine.
Note : This picture is given as additional information and is not a direct example of the current chapter.
∴
ω AB =
vB
1
=
= 1.78 rad/s Ans.
I13 B 0.56
Note: The velocity of the slider A and angular velocity of the connecting rod A B may also be determined as
follows :
From similar triangles I13 I23 I34 and I12 I23 I24,
and
We know that
I12 I 23 I 23 I 24
=
I13 I 23 I 23 I 34
...(i)
I13 I34 I12 I 24
=
I 34 I 23 I 23 I 24
...(ii)
ωAB =
ω × OB
vB
= OB
I13 B
I13 B
= ωOB ×
Also
...(∵ vB = ωOB × OB)
I12 I 23
I I
= ωOB × 23 24
I13 I 23
I 23 I 34
vA = ωAB × I13 A = ωOB ×
I 23 I 24
× I13 I 34 .
I 23 I34
= ωOB × I12 I24 = ωOB × OD
...[From equation (i)] ...(iii)
...[From equation (iii)]
...[From equation (ii)]
Example 6.3. A mechanism, as shown in Fig. 6.15, has the following dimensions:
OA = 200 mm; AB = 1.5 m; BC = 600 mm; CD = 500 mm and BE = 400 mm. Locate all the
instantaneous centres.
If crank OA rotates uniformly at 120 r.p.m. clockwise, find 1. the velocity of B, C and D,
2. the angular velocity of the links AB, BC and CD.
Chapter 6 : Velocity in Mechanisms
l
131
Solution. Given : NOA = 120 r.p.m. or ωOA = 2 π × 120/60 = 12.57 rad/s
Since the length of crank O A = 200 mm = 0.2 m, therefore linear velocity of crank O A,
vOA = v A = ωOA × O A = 12.57 × 0.2 = 2.514 m/s
Fig. 6.15
Location of instantaneous centres
The instantaneous centres are located as discussed below:
1. Since the mechanism consists of six links (i.e. n = 6), therefore the number of instantaneous centres,
n ( n – 1) 6 (6 – 1)
=
= 15
2
2
2. Make a list of all the instantaneous centres in a mechanism. Since the mechanism has 15
instantaneous centres, therefore these centres are listed in the following book keeping table.
N =
Links
1
2
3
4
5
Instantaneous
centres
(15 in number)
12
13
14
15
16
23
24
25
26
34
35
36
45
46
56
Fig. 6.16
6
132
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Theory of Machines
3. Locate the fixed and permanent instantaneous centres by inspection. These centres are I12 I23, I34, I45, I56, I16 and
I14 as shown in Fig. 6.16.
4. Locate the remaining neither fixed nor permanent
instantaneous centres by Aronhold Kennedy’s theorem. Draw a
circle and mark points equal to the number of links such as 1, 2,
3, 4, 5 and 6 as shown in Fig. 6.17. Join the points 12, 23, 34,
45, 56, 61 and 14 to indicate the centres I12, I23, I34, I45, I56, I16
and I14 respectively.
5. Join point 2 to 4 by a dotted line to form the triangles
1 2 4 and 2 3 4. The side 2 4, common to both triangles, is
Fig. 6.17
responsible for completing the two triangles. Therefore the instantaneous centre I24 lies on the intersection of I12 I14 and I23 I34 produced if necessary. Thus centre
I24 is located. Mark number 8 on the dotted line 24 (because seven centres have already been located).
6. Now join point 1 to 5 by a dotted line to form the triangles 1 4 5 and 1 5 6. The side 1 5,
common to both triangles, is responsible for completing the two triangles. Therefore the instantaneous
centre I15 lies on the intersection of I14 I45 and I56 I16 produced if necessary. Thus centre I15 is located.
Mark number 9 on the dotted line 1 5.
7. Join point 1 to 3 by a dotted line to form the triangles 1 2 3 and 1 3 4. The side 1 3,
common to both triangles, is responsible for completing the two triangles. Therefore the instantaneous centre I13 lies on the intersection I12 I23 and I34 I14 produced if necessary. Thus centre I13 is
located. Mark number 10 on the dotted line 1 3.
8. Join point 4 to 6 by a dotted line to form the triangles 4 5 6 and 1 4 6. The side 4 6,
common to both triangles, is responsible for completing the two triangles. Therefore, centre I46 lies
on the intersection of I45 I56 and I14 I16. Thus centre I46 is located. Mark number 11 on the dotted line
4 6.
9. Join point 2 to 6 by a dotted line to form the triangles 1 2 6 and 2 4 6. The side 2 6,
common to both triangles, is responsible for completing the two triangles. Therefore, centre I26 lies
on the intersection of lines joining the points I12 I16 and I24 I46. Thus centre I26 is located. Mark
number 12 on the dotted line 2 6.
10. In the similar way the thirteenth, fourteenth and fifteenth instantaneous centre (i.e. I35, I25
and I36) may be located by joining the point 3 to 5, 2 to 5 and 3 to 6 respectively.
By measurement, we find that
I13 A = 840 mm = 0.84 m ; I13 B = 1070 mm = 1.07 m ; I14 B = 400 mm = 0.4 m ;
I14 C = 200 mm = 0.2 m ; I15 C = 740 mm = 0.74 m ; I15 D = 500 mm = 0.5 m
1. Velocity of points B, C and D
Let vB, v C and v D = Velocity of the points B, C and D respectively.
We know that
∴
Again,
vA
v
= B
I13 A I13 B
...(Considering centre I13)
vA
2.514
× I13 B =
× 1.07 = 3.2 m/s Ans.
0.84
I13 A
v
vB
= C
...(Considering centre I14)
I14 B I14 C
vB =
Chapter 6 : Velocity in Mechanisms
∴
Similarly,
vC =
l
133
vB
3.2
× I14 C =
× 0.2 = 1.6 m/s Ans.
0.4
I14 B
vC
v
= D
I15 C I15 D
...(Considering centre I15)
vC
1.6
× I5 D =
× 0.5 = 1.08 m/s Ans.
0.74
I15 C
2. Angular velocity of the links AB, BC and CD
Let ωAB, ωBC and ωCD = Angular velocity of the links A B, BC and CD respectively.
∴
We know that
vD =
ω AB =
vA
2.514
=
= 2.99 rad/s Ans.
0.84
I13 A
ω BC =
vB
3.2
=
= 8 rad/s Ans.
I14 B 0.4
vC
1.6
=
= 2.16 rad/s Ans.
I15 C 0.74
Example 6.4. The mechanism of a wrapping machine, as shown in Fig. 6.18, has the following dimensions :
O1A = 100 mm; AC = 700 mm; BC = 200 mm; O 3C = 200 mm; O2E = 400 mm;
O2D = 200 mm and BD = 150 mm.
and
ω CD =
The crank O1A rotates at a uniform speed of 100 rad/s. Find the velocity of the point E of the
bell crank lever by instantaneous centre method.
Fig. 6.18
Solution. Given : ωO1A = 100 rad/s ; O1 A = 100 mm = 0.1 m
We know that the linear velocity of crank O1 A ,
v O1A = v A = ωO1A × O1 A = 100 × 0.1 = 10 m/s
Now let us locate the required instantaneous centres as discussed below :
1. Since the mechanism consists of six links (i.e. n = 6), therefore number of instantaneous
centres,
n (n – 1) 6 (6 – 1)
=
= 15
2
2
2. Since the mechanism has 15 instantaneous centres, therefore these centres may be listed in
the book keeping table, as discussed in Example 6.3.
N =
134
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Theory of Machines
Fig. 6.19
Fig. 6.20
3. Locate the fixed and the permanent instantaneous centres by inspection. These centres are
I12, I23, I34, I35, I14, I56 and I16 as shown in Fig. 6.19.
4. Locate the remaining neither fixed nor permanent instantaneous centres by Aronhold
Kennedy’s theorem. This is done by circle diagram as shown in Fig. 6.20. Mark six points on the
circle (i.e. equal to the number of links in a mechanism), and join 1 to 2, 2 to 3, 3 to 4, 3 to 5, 4 to
1, 5 to 6, and 6 to 1, to indicate the fixed and permanent instantaneous centres i.e. I12, I23, I34, I35,
I14, I56, and I16 respectively.
5. Join 1 to 3 by a dotted line to form two triangles 1 2 3 and 1 3 4. The side 1 3, common
to both triangles, is responsible for completing the two triangles. Therefore the instantaneous centre I13 lies on the intersection of the lines joining the points I12 I23 and I14 I34 produced if necessary.
Thus centre I13 is located. Mark number 8 (because seven centres have already been located) on the
dotted line 1 3.
6. Join 1 to 5 by a dotted line to form two triangles 1 5 6 and 1 3 5. The side 1 5, common to
both triangles, is responsible for completing the two triangles. Therefore the instantaneous centre I15
lies on the intersection of the lines joining the points I16 I56 and I13 I35 produced if necessary. Thus
centre I15 is located. Mark number 9 on the dotted line 1 5.
Note: For the given example, we do not require other instantaneous centres.
By measurement, we find that
I13 A = 910 mm = 0.91 m ; I13 B = 820 mm = 0.82 m ; I15 B = 130 mm = 0.13 m ;
I15 D = 50 mm = 0.05 m ; I16 D = 200 mm = 0.2 m ; I16 E = 400 mm = 0.4 m
Velocity of point E on the bell crank lever
Let
vE = Velocity of point E on the bell crank lever,
vB = Velocity of point B, and
vD = Velocity of point D.
We know that
vA
I13 A
=
vB
I13 B
...(Considering centre I13)
Chapter 6 : Velocity in Mechanisms
∴
and
∴
Similarly,
l
135
vA
10
× I13 B =
× 0.82 = 9.01 m/s Ans.
0.91
I13 A
vB
v
= D
...(Considering centre I15)
I15 B I15 D
vB =
vD =
vB
9.01
× I15 D =
× 0.05 = 3.46 m/s Ans.
0.13
I15 B
vD
v
= E
I16 D I16 E
...(Considering centre I16)
vD
3.46
× I16 E =
× 0.4 = 6.92 m/s Ans.
0.2
I16 D
Example 6.5. Fig. 6.21 shows a sewing needle bar mechanism
O1ABO2CD wherein the different dimensions are as follows:
Crank O1A = 16 mm; ∠β = 45°; Vertical distance between O1 and
O2 = 40 mm; Horizontal distance between O1 and O2 = 13 mm; O2 B = 23
mm; AB = 35 mm; ∠ O2 BC = 90°; BC = 16 mm; CD = 40 mm. D lies
vertically below O1.
Find the velocity of needle at D for the given configuration. The crank
O1A rotates at 400 r.p.m.
Solution. Given : N O1A = 400 r.p.m or ωO1A = 2π × 400/60 =
41.9 rad/s ; O1 A = 16 mm = 0.016 m
We know that linear velocity of the crank O1A ,
vO1A = v A = ωO1A × O1A = 41.9 × 0.016 = 0.67 m/s
Now let us locate the required instantaneous centres as discussed
Fig. 6.21
below :
1. Since the mechanism consists of six links (i.e. n = 6), therefore number of instantaneous
centres,
∴
vE =
n (n – 1) 6(6 – 1)
=
= 15
2
2
2. Since the mechanism has 15 instantaneous centres, therefore these centres may be listed in
the book keeping table, as discussed in Example 6.3.
3. Locate the fixed and permanent instantaneous centres by inspections. These centres are
I12, I23, I34, I45, I56, I16 and I14, as shown in Fig. 6.22.
N =
Fig. 6.22
136
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Theory of Machines
4. Locate the remaining neither fixed nor permanent instantaneous centres by Aronhold Kennedy’s theorem. This is done
by circle diagram as shown in Fig. 6.23. Mark six points on the
circle (i.e. equal to the number of links in a mechanism) and join 1
to 2, 2 to 3, 3 to 4, 4 to 5, 5 to 6, 6 to 1 and 1 to 4 to indicate the
fixed and permanent instantaneous centres i.e. I12, I23, I34, I45, I56,
I16 and I14 respectively.
5. Join 1 to 3 by a dotted line to form two triangles 1 2
3 and 1 3 4. The side 1 3, common to both the triangles, is
responsible for completing the two triangles. Therefore the
instantaneous centre I 13 lies on the intersection of I12 I 23 and
I14 I34 produced if necessary. Thus centre I13 is located. Mark
number 8 (because seven centres have already been located)
on the dotted line 1 3.
Fig. 6.23
6. Join 1 to 5 by a dotted line to form two triangles 1 5 6 and 1 4 5. The side 1 5, common
to both the triangles, is responsible for completing the two triangles. Therefore the instantaneous
centre I15 lies on the intersection of I16 I56 and I14 I45 produced if necessary. Thus centre I15 is
located. Mark number 9 on the dotted line 1 5.
Note: For the given example, we do not require other instantaneous centres.
By measurement, we find that
I13 A = 41 mm = 0.041 m ; I13 B = 50 mm = 0.05 m ; I14 B = 23 mm = 0.023 m ;
I14 C= 28 mm = 0.028 m ; I15 C = 65 mm = 0.065 m ; I15 D = 62 mm = 0.062 m
Let
vB = Velocity of point B,
vC = Velocity of point C, and
vD = Velocity of the needle at D.
We know that
∴
vA
v
= B
I13 A I13 B
vB =
0.67
vA
× I13 B =
× 0.05 = 0.817 m/s
0.041
I13 A
v
vB
= C
I14 B I14 C
and
∴
Similarly,
∴
vC =
...(Considering centre I14)
0.817
vB
× I14 C =
× 0.028 = 0.995 m/s
0.023
I14 B
vC
v
= D
I15 C I15 D
vD =
...(Considering centre I13)
...(Considering centre I15)
vC
0.995
× I15 D =
× 0.062 = 0.95 m/s Ans.
0.065
I15 C
Chapter 6 : Velocity in Mechanisms
l
137
Example 6.6. Fig. 6.24 shows a Whitworth quick
return motion mechanism. The various dimensions in
the mechanism are as follows :
OQ = 100 mm ; OA = 200 mm ; QC = 150 mm ;
and CD = 500 mm.
The crank OA makes an angle of 60° with the
vertical and rotates at 120 r.p.m. in the clockwise
direction.
Locate all the instantaneous centres and find the
velocity of ram D.
Solution : Given. N OA = 120 r.p.m. or ωOA =
2 π × 120 / 60 = 12.57 rad/s
Fig. 6.24
Location of instantaneous centres
The instantaneous centres are located as discussed below :
1. Since the mechanism consists of six links (i.e. n = 6), therefore the number of instantaneous centres,
n (n – 1) 6 (6 – 1)
=
= 15
2
2
2. Make a list of all the instantaneous centres in a mechanism as discussed in Example 6.3.
N =
3. Locate the fixed and permanent instantaneous centres by inspection. These centres are I12,
I23, I34, I45, I56, I16 and I14 as shown in Fig. 6.25.
Fig. 6.25
4. Locate the remaining neither fixed nor permanent instantaneous centres by Aronhold
Kennedy’s theorem. Draw a circle and mark points equal to the number of links such as 1, 2, 3, 4, 5,
138
l
Theory of Machines
and 6 as shown in Fig. 6.26. Join the points 1 2, 2 3, 3 4, 4 5, 5 6,
6 1 and 1 4 to indicate the centres I12, I23, I34, I45, I56, I16 and I14
respectively.
5. Join point 1 to 3 by a dotted line to form two triangles
1 2 3 and 1 3 4. The side 1 3, common to both the triangles, is
responsible for completing the two triangles. Therefore the instantaneous centre I13 lies on the intersection of I12 I23, and I14 I34 produced if necessary. Thus centre I13 is located. Mark number 8 on the
dotted line 1 3 (because seven centres have already been located).
6. Join point 1 to 5 by a dotted line to form two triangles
1 4 5 and 1 5 6. The side 1 5, common to both the triangles, is
Fig. 6.26
responsible for completing the two triangles. Therefore the instantaneous centre I15 lies on the intersection of I14 I45 and I56 I16 produced if necessary. Thus centre I15 is
located. Mark number 9 on the dotted line 1 5.
7. Join point 2 to 4 by a dotted line to form two triangles 1 2 4 and 2 3 4. The side 2 4,
common to both the triangles, is responsible for completing the two triangles. Therefore the instantaneous centre I24 lies on the intersection of I12 I14 and I23 I34 produced if necessary. Thus centre I24 is
located. Mark number 10 on the dotted line 2 4.
8. Join point 2 to 5 by a dotted line to form two triangles 1 2 5 and 2 4 5. The side 2 5,
common to both the triangles, is responsible for completing the two triangles. Therefore the instantaneous centre I25 lies on the intersection of I12 I15 and I24 I45 produced if necessary. Thus centre I25 is
located. Mark number 11 on the dotted line 2 5.
9. Join point 2 to 6 by a dotted line to form two triangles 1 2 6 and 2 5 6. The side 2 6
common to both the triangles, is responsible for completing the two triangles. Therefore the instantaneous centre I26 lies on the intersection of I12 I16 and I25 I56 produced if necessary. Thus centre I26 is
located. Mark number 12 on the dotted line 2 6.
10. Join point 3 to 5 by a dotted line to form two triangles 2 3 5 and 3 4 5. The side 3 5,
common to both the triangles, is responsible for completing the two triangles. Therefore the instantaneous centre I35 lies on the intersection of I23 I25 and I34 I45 produced if necessary. Thus centre I35is
located. Mark number 13 on the dotted line 3 5.
11. Join point 3 to 6 by a dotted line to form two triangles 1 3 6 and 3 5 6. The side 3 6,
common to both the triangles, is responsible for completing the two triangles. Therefore the instantaneous centre I36 lies on the intersection of I13 I16 and I35 I56 produced if necessary. Thus centre I36 is
located. Mark number 14 on the dotted line 3 6.
Note. The centre I36 may also be obtained by considering the two triangles 2 3 6 and 3 4 6.
12. Join point 4 to 6 by a dotted line to form two triangles 1 4 6 and 4 5 6. The side 4 6,
common to both the triangles, is responsible for completing the two triangles. Therefore the instantaneous centre I46 lies on the intersection of I14 I16 and I45 I56 produced if necessary. Thus centre I46 is
located. Mark number 15 on the dotted line 4 6.
Velocity of ram D
By measurement, we find that I12 I26 = 65 mm = 0.065 m
∴ Velocity of ram, vD = ωOA × I12 I26 = 12.57 × 0.065 = 0.817 m/s Ans.
EXERCISES
1.
Locate all the instantaneous centres for a four bar mechanism as shown in Fig. 6.27.
The lengths of various links are : A D = 125 mm ; A B = 62.5 mm ; BC = CD = 75 mm.
If the link A B rotates at a uniform speed of 10 r.p.m. in the clockwise direction, find the angular
velocity of the links BC and CD.
[Ans. 0.63 rad/s ; 0.65 rad/s]
Chapter 6 : Velocity in Mechanisms
Fig. 6.27
2.
l
139
Fig. 6.28
Locate all the instantaneous centres for the crossed four bar mechanism as shown in Fig. 6.28. The dimensions of various links are : CD = 65 mm; C A = 60 mm ; D B = 80 mm ; and
A B = 55 mm.
Find the angular velocities of the links A B and DB, if the crank C A rotates at 100 r.p.m. in the
anticlockwise direction.
[Ans. 50 rad/s ; 27 rad/s]
3.
Locate all the instantaneous centres
of the mechanism as shown in Fig.
6.29. The lengths of various links
are : A B = 150 mm ; BC = 300 mm
; CD = 225 mm ; and CE = 500
mm.
When the crank A B rotates in the
anticlockwise direction at a uniform speed of 240 r.p.m. ; find 1.
Velocity of the slider E, and 2.
Angular velocity of the links BC
and CE.
[Ans. 1.6 m/s ; 2.4 rad/s ; 6.6 rad/s]
4.
Fig. 6.29
The crank O A of a mechanism, as shown in Fig. 6.30, rotates clockwise at 120 r.p.m. The lengths of
various links are : O A = 100 mm ; A B = 500 mm ; A C = 100 mm and CD = 750 mm.
Fig. 6.30
140
l
Theory of Machines
Find, by instantaneous centre method : 1. Velocity of point C ; 2. Velocity of slider D ; and 3. Angular
velocities of the links A B and CD.
[Ans. 0.115 m/s; 0.065 m/s; 3 rad/s; 1.3 rad/s]
5.
A mechanism, as shown in Fig. 6.31, has the following dimensions :
O1 A = 60 mm ; A B = 180 mm ; O2 B = 100 mm ; O2 C = 180 mm and CD = 270 mm.
The crank O1 A rotates clockwise at a uniform speed of 120 r.p.m. The block D moves in vertical
guides. Find, by instantaneous centre method, the velocity of D and the angular velocity of CD.
[Ans. 0.08 m/s ; 1.43 rad/s]
6.
The lengths of various links of a mechanism, as shown in Fig. 6.32, are : O A = 0.3 m ;
A B = 1 m ; CD = 0.8 m ; and AC = CB.
Determine, for the given configuration, the velocity of the slider D if the crank O A rotates at 60 r.p.m.
in the clockwise direction. Also find the angular velocity of the link CD. Use instantaneous centre
method.
[Ans. 480 mm/s ; 2.5 rad/s]
Fig. 6.31
7.
Fig. 6.32
In the mechanism shown in Fig. 6.33, find the instantaneous centres of the links B, C and D.
Fig. 6.33
If the link A rotates clockwise at 10 rad/s, find the angular velocity of link E. The lengths of various
links are as follows:
Link A = 25 mm ; Link B = Link C = 100 mm ; Link D = Link E = 50 mm. The link D is hinged to link
B at 25 mm from the left hand end of link B.
[Ans. 1.94 rad/s]
Chapter 6 : Velocity in Mechanisms
8.
l
141
The dimensions of various links in a mechanism, as shown in Fig. 6.34, are as follows :
Fig. 6.34
A B = 25 mm ; BC = 175 mm ; CD = 60 mm ; A D = 150 mm ; BE = EC ; and EF = FG = 100 mm.
The crank A B rotates at 200 r.p.m. When the angle BAD is 135°, determine by instantaneous centre
method : 1. Velocity of G, 2. Angular velocity of EF, and 3. Velocity of sliding of EF in the swivel
block S.
[Ans. 120 mm/s ; 6.5 rad/s ; 400 mm/s]
DO YOU KNOW ?
1.
What do you understand by the instantaneous centre of rotation (centro) in kinematic of
machines? Answer briefly.
2.
Explain, with the help of a neat sketch, the space centrode and body centrode.
3.
Explain with sketch the instantaneous centre method for determination of velocities of links and
mechanisms.
4.
Write the relation between the number of instantaneous centres and the number of links in a mechanism.
5.
Discuss the three types of instantaneous centres for a mechanism.
6.
State and prove the ‘Aronhold Kennedy’s Theorem’ of three instantaneous centres.
OBJECTIVE TYPE QUESTIONS
1.
2.
The total number of instantaneous centres for a mechanism consisting of n links are
(a)
n
2
(b)
n
(c)
n –1
2
(d)
n (n – 1)
2
According to Aronhold Kennedy’s theorem, if three bodies move relatively to each other, their
instantaneous centres will lie on a
(a) straight line
(b)
parabolic curve
(c) ellipse
(d)
none of these
142
3.
l
Theory of Machines
In a mechanism, the fixed instantaneous centres are those centres which
(a) remain in the same place for all configurations of the mechanism
(b)
vary with the configuration of the mechanism
(c) moves as the mechanism moves, but joints are of permanent nature
(d) none of the above
4.
The instantaneous centres which vary with the configuration of the mechanism, are called
(a) permanent instantaneous centres
(b) fixed instantaneous centres
(c) neither fixed nor permanent instantaneous centres
(d) none of these
5.
When a slider moves on a fixed link having curved surface, their instantaneous centre lies
(a) on their point of contact
(b)
at the centre of curvature
(c) at the centre of circle
(d)
at the pin joint
ANSWERS
1. (d)
2.
(a)
3. (a)
4.
(c)
5.
(b)
GO To FIRST
CONTENTS
CONTENTS
Chapter 7 : Velocity in Mechanisms
1. Introduction.
2. Relative Velocity of Two
Bodies Moving in Straight
Lines.
3. Motion of a Link.
4. Velocity of a Point on a Link
by Relative Velocity Method.
5. Velocities in a Slider Crank
Mechanism.
6. Rubbing Velocity at a Pin
Joint.
7. Forces Acting in a
Mechanism.
8. Mechanical Advantage.
143
7
TOGGLE
PRESS
Features
l
Velocity in
Mechanisms
(Relative Velocity Method)
7.1.
Introduction
We have discussed, in the previous chapter, the instantaneous centre method for finding the velocity of various
points in the mechanisms. In this chapter, we shall discuss
the relative velocity method for determining the velocity of
different points in the mechanism. The study of velocity analysis is very important for determining the acceleration of points
in the mechanisms which is discussed in the next chapter.
7.2.
Relative Velocity of Two Bodies
Moving in Straight Lines
Here we shall discuss the application of vectors for
the relative velocity of two bodies moving along parallel lines
and inclined lines, as shown in Fig. 7.1 (a) and 7.2 (a)
respectively.
Consider two bodies A and B moving along parallel
lines in the same direction with absolute velocities v A and
v B such that v A > v B , as shown in Fig. 7.1 (a). The relative
velocity of A with respect to B,
v AB = Vector difference of v A and v B = vA − vB
...(i)
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CONTENTS
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From Fig. 7.1 (b), the relative velocity of A with respect to B (i.e. v AB) may be written in the
vector form as follows :
ba = oa – ob
Fig. 7.1. Relative velocity of two bodies moving along parallel lines.
Similarly, the relative velocity of B with respect to A ,
vBA = Vector difference of v B and vA = vB – vA
or
ab = ob – oa
Now consider the body B moving in an
inclined direction as shown in Fig. 7.2 (a). The
relative velocity of A with respect to B may be
obtained by the law of parallelogram of velocities or triangle law of velocities. Take any fixed
point o and draw vector oa to represent v A in
magnitude and direction to some suitable scale.
Similarly, draw vector ob to represent v B in magnitude and direction to the same scale. Then vector ba represents the relative velocity of A with
respect to B as shown in Fig. 7.2 (b). In the similar way as discussed above, the relative velocity
of A with respect to B,
vAB = Vector difference of v A and vB = vA – vB
or
ba = oa – ob
Fig. 7.2. Relative velocity of two bodies moving along inclined lines.
Similarly, the relative velocity of B with respect to A ,
vBA = Vector difference of v B and vA = vB – vA
or
ab = ob − oa
...(ii)
Chapter 7 : Velocity in Mechanisms
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145
From above, we conclude that the relative velocity of point A with respect to B (v AB) and the
relative velocity of point B with respect A (v BA) are equal in magnitude but opposite in direction, i.e.
vAB = − vBA
or
ba = − ab
Note: It may be noted that to find v AB, start from point b towards a and for v BA, start from point a towards b.
7.3.
Motion of a Link
Consider two points A and B on a rigid link A B, as
shown in Fig. 7.3 (a). Let one of the extremities (B) of the link
move relative to A , in a clockwise direction. Since the distance from A to B remains the same, therefore there can be no
relative motion between A and B, along the line A B. It is thus
obvious, that the relative motion of B with respect to A must
be perpendicular to A B.
Hence velocity of any point on a link with respect to
another point on the same link is always perpendicular to
the line joining these points on the configuration (or space)
diagram.
Fig. 7.3. Motion of a Link.
The relative velocity of B with respect to A (i.e. v BA) is represented by the vector ab and is
perpendicular to the line A B as shown in Fig. 7.3 (b).
Let
ω = Angular velocity of the link A B about A .
We know that the velocity of the point B with respect to A ,
...(i)
vBA = ab = ω. AB
Similarly, the velocity of any point C on A B with respect to A ,
...(ii)
vCA = ac = ω. AC
From equations (i) and (ii),
vCA ac ω. AC AC
=
=
=
...(iii)
vBA ab ω. AB
AB
Thus, we see from equation (iii), that the point c on the vector ab divides it in the same ratio
as C divides the link A B.
Note: The relative velocity of A with respect to B is represented by ba, although A may be a fixed point. The
motion between A and B is only relative. Moreover, it is immaterial whether the link moves about A in a
clockwise direction or about B in a clockwise direction.
7.4.
Velocity of a Point on a Link by Relative Velocity Method
The relative velocity method is based upon the relative velocity of the various points of the
link as discussed in Art. 7.3.
Consider two points A and B on a link as shown in Fig. 7.4 (a). Let the absolute velocity of the
point A i.e. v A is known in magnitude and direction and the absolute velocity of the point B i.e. v B is
known in direction only. Then the velocity of B may be determined by drawing the velocity diagram
as shown in Fig. 7.4 (b). The velocity diagram is drawn as follows :
1. Take some convenient point o, known as the pole.
2. Through o, draw oa parallel and equal to v A, to some suitable scale.
3. Through a, draw a line perpendicular to A B of Fig. 7.4 (a). This line will represent the
velocity of B with respect to A , i.e. v BA.
4. Through o, draw a line parallel to v B intersecting the line of v BA at b.
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5. Measure ob, which gives the required velocity of point B ( v B), to the scale.
(a) Motion of points on a link.
(b) Velocity diagram.
Fig. 7.4
Notes : 1. The vector ab which represents the velocity of B with respect to A (v BA) is known as velocity of
image of the link A B.
2. The absolute velocity of any point C on A B may be determined by dividing vector ab at c in the same
ratio as C divides A B in Fig. 7.4 (a).
In other words
ac AC
=
ab AB
Join oc. The *vector oc represents the absolute velocity
of point C (v C) and the vector ac represents the velocity of C
with respect to A i.e. v CA.
3. The absolute velocity of any other point D outside
A B, as shown in Fig. 7.4 (a), may also be obtained by completing the velocity triangle abd and similar to triangle ABD,
as shown in Fig. 7.4 (b).
4. The angular velocity of the link A B may be found
by dividing the relative velocity of B with respect to A (i.e.
v BA) to the length of the link A B. Mathematically, angular
velocity of the link A B,
v
ab
ωAB = BA =
AB AB
7.5.
Velocities in Slider Crank Mechanism
In the previous article, we have discused the relative velocity method for the velocity of any
point on a link, whose direction of motion and velocity of some other point on the same link is known.
The same method may also be applied for the velocities in a slider crank mechanism.
A slider crank mechanism is shown in Fig. 7.5 (a). The slider A is attached to the connecting
rod A B. Let the radius of crank OB be r and let it rotates in a clockwise direction, about the point O
with uniform angular velocity ω rad/s. Therefore, the velocity of B i.e. v B is known in magnitude and
direction. The slider reciprocates along the line of stroke A O.
The velocity of the slider A (i.e. v A) may be determined by relative velocity method as
discussed below :
1. From any point o, draw vector ob parallel to the direction of v B (or perpendicular to OB)
such that ob = v B = ω.r, to some suitable scale, as shown in Fig. 7.5 (b).
*
The absolute velocities of the points are measured from the pole (i.e. fixed points) of the velocity diagram.
Chapter 7 : Velocity in Mechanisms
(a) Slider crank mechanism.
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147
(b) Velocity diagram.
Fig. 7.5
2. Since A B is a rigid link, therefore the velocity of A relative to B is perpendicular to A B.
Now draw vector ba perpendicular to A B to represent the velocity of A with respect to B i.e. v AB.
3. From point o, draw vector oa parallel to the path of motion of the slider A (which is along
AO only). The vectors ba and oa intersect at a. Now oa represents the velocity of the slider A i.e. v A,
to the scale.
The angular velocity of the connecting rod A B (ωAB) may be determined as follows:
vBA
ab
=
(Anticlockwise about A)
AB AB
The direction of vector ab (or ba) determines the sense of ωAB which shows that it is
anticlockwise.
ωAB =
Note : The absolute velocity of any other point E on the connecting rod AB may also be found out by dividing
vector ba such that be/ba = BE/BA . This is done by drawing any line b A1 equal in length of B A. Mark bE1 = BE.
Join a A1. From E1 draw a line E1e parallel to a A 1. The vector oe now represents the velocity of E and vector ae
represents the velocity of E with respect to A .
7.6.
Rubbing Velocity at a Pin Joint
The links in a mechanism are mostly connected by means of pin joints. The rubbing velocity
is defined as the algebraic sum between the angular velocities of the two links which are connected
by pin joints, multiplied by the radius of the pin.
Consider two links O A and OB connected by a pin joint at O as shown in Fig. 7.6.
Let
ω1 = Angular velocity of the link O A or
the angular velocity of the point A
with respect to O.
ω2 = Angular velocity of the link OB or
the angular velocity of the point B
with respect to O, and
r = Radius of the pin.
Fig. 7.6. Links connected by pin joints.
According to the definition,
Rubbing velocity at the pin joint O
= (ω1 – ω2) r, if the links move in the same direction
= (ω1 + ω2) r, if the links move in the opposite direction
Note : When the pin connects one sliding member and the other turning member, the angular velocity of the
sliding member is zero. In such cases,
Rubbing velocity at the pin joint = ω.r
where
ω = Angular velocity of the turning member, and
r = Radius of the pin.
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Example 7.1. In a four bar chain ABCD, AD is fixed and is 150 mm long. The crank AB is 40
mm long and rotates at 120 r.p.m. clockwise, while the link CD = 80 mm oscillates about D. BC and
AD are of equal length. Find the angular velocity of link CD when angle BAD = 60°.
Solution. Given : N BA = 120 r.p.m. or ωBA = 2 π × 120/60 = 12.568 rad/s
Since the length of crank A B = 40 mm = 0.04 m, therefore velocity of B with respect to A or
velocity of B, (because A is a fixed point),
vBA = v B = ωBA × A B = 12.568 × 0.04 = 0.503 m/s
(a) Space diagram (All dimensions in mm).
(b) Velocity diagram.
Fig. 7.7
First of all, draw the space diagram to some suitable scale, as shown in Fig. 7.7 (a). Now the
velocity diagram, as shown in Fig. 7.7 (b), is drawn as discussed below :
1. Since the link A D is fixed, therefore points a and d are taken as one point in the velocity
diagram. Draw vector ab perpendicular to B A, to some suitable scale, to represent the velocity of B
with respect to A or simply velocity of B (i.e. v BA or v B) such that
vector ab = v BA = v B = 0.503 m/s
2. Now from point b, draw vector bc perpendicular to CB to represent the velocity of C with
respect to B (i.e. v CB) and from point d, draw vector dc perpendicular to CD to represent the velocity
of C with respect to D or simply velocity of C (i.e. v CD or v C). The vectors bc and dc intersect at c.
By measurement, we find that
v CD = v C = vector dc = 0.385 m/s
We know that
CD = 80 mm = 0.08 m
∴ Angular velocity of link CD,
vCD 0.385
=
= 4.8 rad/s (clockwise about D) Ans.
0.08
CD
Example 7.2. The crank and connecting rod
of a theoretical steam engine are 0.5 m and 2 m long
respectively. The crank makes 180 r.p.m. in the
clockwise direction. When it has turned 45° from the
inner dead centre position, determine : 1. velocity of
piston, 2. angular velocity of connecting rod,
3. velocity of point E on the connecting rod 1.5 m
from the gudgeon pin, 4. velocities of rubbing at the
pins of the crank shaft, crank and crosshead when
the diameters of their pins are 50 mm, 60 mm and 30
mm respectively, 5. position and linear velocity of any
point G on the connecting rod which has the least
velocity relative to crank shaft.
ωCD =
Chapter 7 : Velocity in Mechanisms
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Solution. Given : NBO = 180 r.p.m. or ωBO = 2 π × 180/60 = 18.852 rad/s
Since the crank length OB = 0.5 m, therefore linear velocity of B with respect to O or velocity
of B (because O is a fixed point),
vBO = v B = ωBO × OB = 18.852 × 0.5 = 9.426 m/s
. . . (Perpendicular to BO)
1. Velocity of piston
First of all draw the space diagram, to some suitable scale, as shown in Fig. 7.8 (a). Now the
velocity diagram, as shown in Fig. 7.8 (b), is drawn as discussed below :
1. Draw vector ob perpendicular to BO, to some suitable scale, to represent the velocity of B
with respect to O or velocity of B such that
vector ob = v BO = v B = 9.426 m/s
2. From point b, draw vector bp perpendicular to BP to represent velocity of P with respect
to B (i.e. v PB) and from point o, draw vector op parallel to PO to represent velocity of P with respect
to O (i.e. v PO or simply v P). The vectors bp and op intersect at point p.
By measurement, we find that velocity of piston P,
vP = vector op = 8.15 m/s Ans.
(a) Space diagram.
(b) Velocity diagram.
Fig. 7.8
2. Angular velocity of connecting rod
From the velocity diagram, we find that the velocity of P with respect to B,
vPB = vector bp = 6.8 m/s
Since the length of connecting rod PB is 2 m, therefore angular velocity of the connecting rod,
vPB 6.8
=
= 3.4 rad/s (Anticlockwise) Ans.
2
PB
3. Velocity of point E on the connecting rod
The velocity of point E on the connecting rod 1.5 m from the gudgeon pin (i.e. PE = 1.5 m)
is determined by dividing the vector bp at e in the same ratio as E divides PB in Fig. 7.8 (a). This is
done in the similar way as discussed in Art 7.6. Join oe. The vector oe represents the velocity of E. By
measurement, we find that velocity of point E,
vE = vector oe = 8.5 m/s Ans.
ωPB =
Note : The point e on the vector bp may also be obtained as follows :
BE be
BE × bp
=
or be =
BP bp
BP
4. Velocity of rubbing
We know that diameter of crank-shaft pin at O,
dO = 50 mm = 0.05 m
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Theory of Machines
Diameter of crank-pin at B,
dB = 60 mm = 0.06 m
and diameter of cross-head pin,
dC = 30 mm = 0.03 m
We know that velocity of rubbing at the pin of crank-shaft
dO
0.05
× ωBO =
× 18.85 = 0.47 m/s Ans.
2
2
Velocity of rubbing at the pin of crank
=
=
0.06
dB
(ωBO + ωPB ) =
(18.85 + 3.4) = 0.6675 m/s Ans.
2
2
...(3 ωBO is clockwise and ωPB is anticlockwise.)
and velocity of rubbing at the pin of cross-head
0.03
dC
× ωPB =
× 3.4 = 0.051 m/s Ans.
=
2
2
...(3 At the cross-head, the slider does not rotate and only the connecting rod has angular motion.)
5. Position and linear velocity of point G on the connecting rod which has the least velocity
relative to crank-shaft
The position of point G on the connecting rod which has the least velocity relative to crankshaft is determined by drawing perpendicular from o to vector bp. Since the length of og will be the
least, therefore the point g represents the required position of G on the connecting rod.
By measurement, we find that
vector bg = 5 m/s
The position of point G on the connecting rod is obtained as follows:
bg
bg BG
5
× BP =
× 2 = 1.47 m Ans.
=
or BG =
bp
bp BP
6.8
By measurement, we find that the linear velocity of point G,
vG = vector og = 8 m/s Ans.
Example 7.3. In Fig. 7.9, the angular velocity of
the crank OA is 600 r.p.m. Determine the linear velocity of
the slider D and the angular velocity of the link BD, when
the crank is inclined at an angle of 75° to the vertical. The
dimensions of various links are : OA = 28 mm ; AB = 44 mm ;
BC 49 mm ; and BD = 46 mm. The centre distance between
the centres of rotation O and C is 65 mm. The path of travel
of the slider is 11 mm below the fixed point C. The slider
moves along a horizontal path and OC is vertical.
Solution. Given: N AO = 600 r.p.m. or
Fig. 7.9
ωAO = 2 π × 600/60 = 62.84 rad/s
Since O A = 28 mm = 0.028 m, therefore velocity of
A with respect to O or velocity of A (because O is a fixed point),
vAO = v A = ωAO × O A = 62.84 × 0.028 = 1.76 m/s
. . . (Perpendicular to O A)
Linear velocity of the slider D
First of all draw the space diagram, to some suitable scale, as shown in Fig. 7.10 (a). Now the
velocity diagram, as shown in Fig. 7.10 (b), is drawn as discussed below :
Chapter 7 : Velocity in Mechanisms
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1. Since the points O and C are fixed, therefore these points are marked as one point, in the
velocity diagram. Now from point o, draw vector oa perpendicular to O A, to some suitable scale, to
represent the velocity of A with respect to O or simply velocity of A such that
vector oa = v AO = v A = 1.76 m/s
(a) Space diagram.
(b) Velocity diagram.
Fig. 7.10
2. From point a, draw vector ab perpendicular to A B to represent the velocity of B with
respect A (i.e. v BA) and from point c, draw vector cb perpendicular to CB to represent the velocity of
B with respect to C or simply velocity of B (i.e. v BC or v B). The vectors ab and cb intersect at b.
3. From point b, draw vector bd perpendicular to BD to represent the velocity of D with
respect to B (i.e. v DB) and from point o, draw vector od parallel to the path of motion of the slider D
which is horizontal, to represent the velocity of D (i.e. v D). The vectors bd and od intersect at d.
By measurement, we find that velocity of the slider D,
vD = vector od = 1.6 m/s Ans.
Angular velocity of the link BD
By measurement from velocity diagram, we find that velocity of D with respect to B,
vDB = vector bd = 1.7 m/s
Since the length of link BD = 46 mm = 0.046 m, therefore angular velocity of the link BD,
vDB
1.7
=
= 36.96 rad/s (Clockwise about B) Ans.
BD 0.046
Example 7.4. The mechanism, as shown in Fig. 7.11, has the dimensions of various links as
follows :
AB = DE = 150 mm ; BC = CD = 450 mm ; EF = 375 mm.
ωBD =
Fig. 7.11
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The crank AB makes an angle of 45° with the horizontal and rotates about A in the clockwise
direction at a uniform speed of 120 r.p.m. The lever DC oscillates about the fixed point D, which is
connected to AB by the coupler BC.
The block F moves in the horizontal guides, being driven by the link EF. Determine: 1. velocity of
the block F, 2. angular velocity of DC, and 3. rubbing speed at the pin C which is 50 mm in diameter.
Solution. Given : N BA = 120 r.p.m. or ωBA = 2 π × 120/60 = 4 π rad/s
Since the crank length A B = 150 mm = 0.15 m, therefore velocity of B with respect to A or
simply velocity of B (because A is a fixed point),
vBA = v B = ωBA × AB = 4 π × 0.15 = 1.885 m/s
. . . (Perpendicular to A B)
1. Velocity of the block F
First of all draw the space diagram, to some suitable scale, as shown in Fig. 7.12 (a). Now the
velocity diagram, as shown in Fig. 7.12 (b), is drawn as discussed below:
(a) Space diagram.
(b) Velocity diagram.
Fig. 7.12
1. Since the points A and D are fixed, therefore these points are marked as one point* as
shown in Fig. 7.12 (b). Now from point a, draw vector ab perpendicular to A B, to some suitable scale,
to represent the velocity of B with respect to A or simply velocity of B, such that
vector ab = v BA = v B = 1.885 m/s
2. The point C moves relative to B and D, therefore draw vector bc perpendicular to BC to
represent the velocity of C with respect to B (i.e. v CB), and from point d, draw vector dc perpendicular
to DC to represent the velocity of C with respect to D or simply velocity of C (i.e. v CD or v C). The
vectors bc and dc intersect at c.
3. Since the point E lies on DC, therefore divide vector dc in e in the same ratio as E divides
CD in Fig. 7.12 (a). In other words
ce/cd = CE/CD
The point e on dc may be marked in the same manner as discussed in Example 7.2.
4. From point e, draw vector ef perpendicular to EF to represent the velocity of F with respect
to E (i.e. v FE) and from point d draw vector df parallel to the path of motion of F, which is horizontal,
to represent the velocity of F i.e. v F. The vectors ef and df intersect at f.
By measurement, we find that velocity of the block F,
vF = vector df = 0.7 m/s Ans.
2. Angular velocity of DC
By measurement from velocity diagram, we find that velocity of C with respect to D,
v CD = vector dc = 2.25 m/s
*
When the fixed elements of the mechanism appear at more than one place, then all these points lie at one
place in the velocity diagram.
Chapter 7 : Velocity in Mechanisms
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153
Since the length of link DC = 450 mm = 0.45 m, therefore angular velocity of DC,
ωDC =
vCD 2.25
=
= 5 rad/s
DC 0.45
. . . (Anticlockwise about D)
3. Rubbing speed at the pin C
We know that diameter of pin at C,
dC = 50 mm = 0.05 m or Radius , rC = 0.025 m
From velocity diagram, we find that velocity of C with respect to B,
. . . (By measurement)
v CB = vector bc = 2.25 m/s
Length BC = 450 mm = 0.45 m
∴ Angular velocity of BC,
v
2.25
ωCB = CB =
= 5 rad/s
... (Anticlockwise about B)
BC 0.45
We know that rubbing speed at the pin C
= (ωCB – ωCD) rC = (5 – 5) 0.025 = 0 Ans.
Example 7.5. In a mechanism shown in Fig. 7.13, the crank OA is 100 mm long and rotates
clockwise about O at 120 r.p.m. The connceting rod AB is 400 mm long.
Fig. 7.13.
At a point C on AB, 150 mm from A, the rod CE 350 mm long is attached. This rod CE slides
in a slot in a trunnion at D. The end E is connected by a link EF, 300 mm long to the horizontally
moving slider F.
For the mechanism in the position shown, find 1. velocity of F, 2. velocity of sliding of CE in
the trunnion, and 3. angular velocity of CE.
Solution. Given : v AO = 120 r.p.m. or
ωAO = 2 π × 120/60 = 4 π rad/s
Since the length of crank O A = 100 mm
= 0.1 m, therefore velocity of A with respect to O
or velocity of A (because O is a fixed point),
v AO = v A = ωAO × O A = 4 π × 0.1 = 1.26 m/s
. . . (Perpendicular to A O)
1. Velocity of F
First of all draw the space diagram, to
some suitable scale, as shown in Fig. 7.14 (a).
Now the velocity diagram, as shown in Fig. 7.14
(b), is drawn as discussed below :
An aircraft uses many mechanisms in engine,
power transmission and steering.
Note : This picture is given as additional information
and is not a direct example of the current chapter.
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1. Draw vector oa perpendicular to A O, to some suitable scale, to represent the velocity of A
with respect to O or simply velocity of A (i.e. v AO or v A), such that
vector oa = v AO = v A = 1.26 m/s
2. From point a, draw vector ab perpendicular to A B to represent the velocity of B with respect
to A i.e. v BA, and from point o draw vector ob parallel to the motion of B (which moves along BO
only) to represent the velocity of B i.e. v B . The vectors ab and ob intersect at b.
(a) Space diagram.
(b) Velocity diagram.
Fig. 7.14
3. Since the point C lies on A B, therefore divide vector ab at c in the same ratio as C divides A B
in the space diagram. In other words,
ac/ab = AC/AB
4. From point c, draw vector cd perpendicular to CD to represent the velocity of D with respect
to C i.e. v DC, and from point o draw vector od parallel to the motion of CD, which moves along CD
only, to represent the velocity of D, i.e. v D.
5. Since the point E lies on CD produced, therefore divide vector cd at e in the same ratio as E
divides CD in the space diagram. In other words,
cd/ce = CD/CE
6. From point e, draw vector ef perpendicular to EF to represent the velocity of F with respect
to E i.e. v FE, and from point o draw vector of parallel to the motion of F, which is along FD to
represent the velocity of F i.e. v F.
By measurement, we find that velocity of F,
vF = vector of = 0.53 m/s Ans.
2. Velocity of sliding of CE in the trunnion
Since velocity of sliding of CE in the trunnion is the velocity of D, therefore velocity of sliding
of CE in the trunnion
= vector od = 1.08 m/s Ans.
3. Angular velocity of CE
By measurement, we find that linear velocity of C with respect to E,
vCE = vector ec = 0.44 m/s
Since the length CE = 350 mm = 0.35 m, therefore angular velocity of CE,
ωCE =
vCE 0.44
=
= 1.26 rad/s (Clockwise about E) Ans.
0.35
CE
Chapter 7 : Velocity in Mechanisms
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Example 7.6. In a mechanism as shown in Fig. 7.15, the various dimensions are : OC = 125
mm ; CP = 500 mm ; PA = 125 mm ; AQ = 250 mm and QE = 125 mm.
Fig. 7.15. All dimensions in mm.
The slider P translates along an axis which is 25 mm vertically below point O. The crank OC
rotates uniformly at 120 r.p.m. in the anti-clockwise direction. The bell crank lever AQE rocks about
fixed centre Q.
Draw the velocity diagram and calculate the absolute velocity of point E of the lever.
Solution. Given : N CO = 120 r.p.m. or ω CO = 2 π × 120/60 = 12.57 rad/s ;
OC = 125 mm = 0.125 m
We know that linear velocity of C with respect to O or velocity of C, (because O is
as fixed point)
v CO = v C = ωCO × OC = 12.57 × 0.125 = 1.57 m/s
First of all, draw the space diagram, as shown in Fig. 7.16 (a), to some suitable scale. Now the
velocity diagram, as shown in Fig. 7.16 (b) is drawn as discussed below :
1. Since the points O and Q are fixed, therefore these points are taken as one point in the
velocity diagram. From point o, draw vector oc perpendicular to OC, to some suitable scale, to
represent the velocity of C with respect to O or velocity of C, such that
vector oc = v CO = v C = 1.57 m/s
(a) Space diagram.
(b) Velocity diagram.
Fig. 7.16
2. From point c, draw vector cp perpendicular to CP to represent the velocity of P with
respect to C (i.e. v PC) and from point o, draw vector op parallel to the path of motion of slider P
(which is horizontal) to represent the velocity of P (i.e. v P). The vectors cp and op intersect at p.
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Theory of Machines
3. From point p, draw vector pa perpendicular to PA to represent the velocity of A with
respect to P (i.e. v AP) and from point q, draw vector qa perpendicular to Q A to represent the velocity
of A (i.e. v A ). The vectors pa and qa intersect at a.
4. Now draw vector qe perpendicular to vector qa in such a way that
QE/QA = qe/qa
By measurement, we find that the velocity of
point E,
vE = vector oe = 0.7 m/s Ans.
Example 7.7. A quick return mechanism of
the crank and slotted lever type shaping machine is
shown in Fig. 7.17.
The dimensions of the various links are as
follows :
O1O2 = 800 mm ; O1B = 300 mm ;
O2D = 1300 mm ; DR = 400 mm.
The crank O1B makes an angle of 45° with
the vertical and rotates at 40 r.p.m. in the counter
clockwise direction. Find : 1. velocity of the ram R, or
the velocity of the cutting tool, and 2. angular velocity
of link O2D.
Solution. Given: NBO1 = 40 r.p.m. or ωBO1 =
2 π × 40/60 = 4.2 rad/s
Fig. 7.17. All dimensions in mm.
Since the length of crank O1B = 300 mm = 0.3m, therefore velocity of B with respect to O1 or
simply velocity of B (because O1 is a fixed point),
vBO1 = v B = ωBO1 × O1B = 4.2 × 0.3 = 1.26 m/s
. . . (Perpendicular to O1B)
1. Velocity of the ram R
First of all draw the space diagram, to some suitable scale, as shown in Fig. 7.18 (a). Now the
velocity diagram, as shown in Fig. 7.18 (b), is drawn as discussed below :
1. Since O1 and O2 are fixed points, therefore these points are marked as one point in the
velocity diagram. Draw vector o1b perpendicular to O1B, to some suitable scale, to represent the
velocity of B with respect to O1 or simply velocity of B, such that
vector o1b = v BO1 = v B = 1.26 m/s
2. From point o2, draw vector o2c perpendicular to O2C to represent the velocity of the
coincident point C with respect to O2 or simply velocity of C (i.e. v CO2 or v C), and from point b, draw
vector bc parallel to the path of motion of the sliding block (which is along the link O2D) to represent
the velocity of C with respect to B (i.e. v CB). The vectors o2c and bc intersect at c.
3. Since the point D lies on O2C produced, therefore divide the vector o2c at d in the same
ratio as D divides O2C in the space diagram. In other words,
cd / o2d = CD/O2D
4. Now from point d, draw vector dr perpendicular to DR to represent the velocity of R with
respect to D (i.e. v RD), and from point o1 draw vector o1r parallel to the path of motion of R (which is
horizontal) to represent the velocity of R (i.e. v R). The vectors dr and o1r intersect at r.
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By measurement, we find that velocity of the ram R,
vR = vector o1r = 1.44 m/s Ans.
(a) Space diagram
(b) Velocity diagram.
Fig. 7.18
2. Angular velocity of link O2D
By measurement from velocity diagram, we find that velocity of D with respect to O2 or
velocity of D,
v DO2 = v D = vector o2d = 1.32 m/s
We know that length of link O2D = 1300 mm = 1.3 m. Therefore angular velocity of the link O2D,
v
1.32
ωDO 2 = DO2 =
1.3 = 1.015 rad/s (Anticlockwise about O2) Ans.
O2 D
The above picture shows prototype of an industrial steam engine. Before to the invention of
electricity, steam engines used to provide the power needed to turn wheels in the factories.
Note : This picture is given as additional information and is not a direct example of the current chapter.
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Example 7.8. In the mechanism, as shown in Fig. 7.19, the crank O1A rotates at a speed of
60 r.p.m. in a clockwise direction imparting vertical reciprocating motion to the rack R, by means of
toothed quadrant Q. O1 and O2 are fixed centres and the slotted bar BC and quadrant Q are rocking
on O2.
Fig. 7.19. All dimensions are in mm.
Determine : 1. the linear speed of the rack when the crank makes an angle of 30° to the
horizontal, 2. the ratio of the times of lowering and raising the rack, and 3. the length of the stroke of
the rack.
Solution. Given : N AO1 = 60 r.p.m. or ωAO1 = 2 π × 60/60 = 6.28 rad/s
Since crank length O1 A = 85 mm, therefore velocity of A with respect to O1 or velocity of A ,
(because O1 is a fixed point),
v AO1 = v A = ωAO1 × O1A = 6.28 × 85 = 534 mm/s
. . . (Perpendicular to O1A )
(a) Space diagram.
(b) Velocity diagram.
Fig. 7.20
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1. Linear speed of the rack
First of all draw the space diagram, to some suitable scale, as shown in Fig. 7.20 (a). Now the
velocity diagram, as shown in Fig. 7.20 (b), is drawn as discussed below :
1. Since O1 and O2 are fixed points, therefore they are marked as one point in the velocity
diagram. From point o1, draw vector o1a perpendicular to O1A , to some suitable scale, to represent
the velocity of A with respect to O1 or simply velocity of A , such that
vector o1a = v AO1 = v A = 534 mm/s
2. From point a, draw vector ad parallel to the path of motion of D (which is along the slot in
the link BC) to represent the velocity D with respect to A (i.e. v DA), and from point o2 draw vector o2d
perpendicular to the line joining the points O2 and D (because O2 and D lie on the same link) to
represent the velocity of D (i.e. vDO2 or v D). The vectors ad and o2d intersect at d.
Note : The point A represents the point on the crank as well as on the sliding block whereas the point D
represents the coincident point on the lever O2C.
By measurement, we find that
v DO2 = v D = vector o2d = 410 mm/s, and O2D = 264 mm
We know that angular velocity of the quadrant Q,
ωQ =
vDO2 410
=
= 1.55 rad/s (Clockwise about O2)
O2 D 264
Radius of the quadrant Q,
rQ = 50 mm
Since the rack and the quadrant have a rolling contact, therefore the linear velocity at the
points of contact will be same as that of quadrant.
∴ Linear speed of the rack,
vR = w Q.rQ = 1.55 × 50
= 77.5 mm/s
Ans.
2. Ratio of the times of lowering and raising the rack
The two extreme positions of the rack (or A B) are when
the tangent to the circle with centre O1 is also a tangent to the
circle with centre O2, as shown in Fig. 7.21. The rack will be
raising when the crank moves from A 1 to A 2 through an angle α
and it will be lowering when the crank moves from A 2 to A 1
through an angle β. Since the times of lowering and raising the
rack is directly proportional to their respective angles, therefore
Time of lowering β 240°
= =
= 2 Ans.
α 120°
Time of raising
. . . (By measurement)
3. Length of stroke of the rack
By measurement, we find that angle B1O2B 2 = 60° =
60 × π / 180 = 1.047 rad
We know that length of stroke of the rack
= Radius of the quadrant × Angular rotation of the
quadrant in radians
Fig. 7.21. All dimensions in mm.
= rQ × ∠ B 1O2B 2 in radians = 50 × 1.047 = 52.35 mm Ans.
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Example 7.9. Fig. 7.22 shows the
structure of Whitworth quick return mechanism used in reciprocating machine tools. The
various dimensions of the tool are as follows :
OQ = 100 mm ; OP = 200 mm, RQ =
150 mm and RS = 500 mm.
The crank OP makes an angle of 60°
with the vertical. Determine the velocity of the
slider S (cutting tool) when the crank rotates
at 120 r.p.m. clockwise.
Find also the angular velocity of the
link RS and the velocity of the sliding block T
on the slotted lever QT.
Fig. 7.22
Solution. Given : N PO = 120 r.p.m. or ωPO = 2 π × 120/60 = 12.57 rad/s
Since the crank OP = 200 mm = 0.2 m, therefore velocity of P with respect to O or velocity
of P (because O is a fixed point),
vPO = v P = ωPO × OP = 12.57 × 0.2 = 2.514 m/s
. . . (Perpendicular to PO)
Velocity of slider S (cutting tool )
First of all draw the space diagram, to some suitable scale, as shown in Fig. 7.23 (a). Now the
velocity diagram, as shown in Fig. 7.23 (b) is drawn as discussed below :
1. Since O and Q are fixed points, therefore they are taken as one point in the velocity
diagram. From point o, draw vector op perpendicular to OP, to some suitable scale, to represent the
velocity of P with respect to O or simply velocity of P, such that
vector op = v PO = v P = 2.514 m/s
(a) Space diagram.
(b) Velocity diagram.
Fig. 7.23
2. From point q, draw vector qt perpendicular to QT to represent the velocity of T with
respect to Q or simply velocity of T (i.e. v TQ or v T) and from point p draw vector pt parallel to the path
of motion of T (which is parallel to TQ) to represent the velocity of T with respect to P (i.e. v TP). The
vectors qt and pt intersect at t.
Note : The point T is a coincident point with P on the link QT.
Chapter 7 : Velocity in Mechanisms
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3. Since the point R lies on the link TQ produced, therefore divide the vector tq at r in the
same ratio as R divides TQ, in the space diagram. In other words,
qr/qt = QR/QT
The vector qr represents the velocity of R with respect to Q or velocity of R (i.e.v RQ or v R).
4. From point r, draw vector rs perpendicular to R S to represent the velocity of S with respect
to R and from point o draw vector or parallel to the path of motion of S (which is parallel to QS) to
represent the velocity of S (i.e v S). The vectors rs and os intersect at s.
By measurement, we find that velocity of the slider S (cutting tool),
vS = vector os = 0.8 m/s Ans.
Angular velocity of link RS
From the velocity diagram, we find that the linear velocity of the link R S,
v SR = vector rs = 0.96 m/s
Since the length of link R S = 500 mm = 0.5 m, therefore angular velocity of link R S,
v
0.96
ωRS = SR =
= 0.92 rad/s (Clockwise about R) Ans.
0.5
RS
Velocity of the sliding block T on the slotted lever QT
Since the block T moves on the slotted lever with respect to P, therefore velocity of the
sliding block T on the slotted lever QT,
vTP = vector pt = 0.85 m/s
7.7.
Ans.
. . . (By measurement)
Forces Acting in a Mechanism
Consider a mechanism of a four bar chain, as shown in
Fig. 7.24. Let force FA newton is acting at the joint A in the
direction of the velocity of A (v A m/s) which is perpendicular to
the link D A. Suppose a force FB newton is transmitted to the
joint B in the direction of the velocity of B
(i.e. v B m/s) which is perpendicular to the link CB. If we neglect
the effect of friction and the change of kinetic energy of the link
(i.e.), assuming the efficiency of transmission as 100%), then
by the principle of conservation of energy,
Fig. 7.24. Four bar mechanism.
Input work per unit time
= Output work per unit time
∴ Work supplied to the joint A
= Work transmitted by the joint B
or
FA . vA
. . . (i)
vB
If we consider the effect of friction and assuming the efficiency of transmission as η, then
FA.vA = FB.v B or FB =
η=
Output FB . vB
=
Input
FA . vA
or
FB =
η. FA . vA
vB
. . . (ii)
Notes : 1. If the turning couples due to the forces FA and FB about D and C are denoted by T A (known as driving
torque) and T B (known as resisting torque) respectively, then the equations (i) and (ii) may be written as
TA.ωA = T B.ωB, and η =
TB . ωB
TA . ωA
where ωA and ωB are the angular velocities of the links D A and CB respectively.
. . . (iii)
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2. If the forces FA and FB do not act in the direction of the velocities of the points A and B respectively,
then the component of the force in the direction of the velocity should be used in the above equations.
7.8.
Mechanical Advantage
It is defined as the ratio of the load to the effort. In a four bar mechanism, as shown in Fig.
7.24, the link D A is called the driving link and the link CB as the driven link. The force FA acting at A
is the effort and the force FB at B will be the load or the resistance to overcome. We know from the
principle of conservation of energy, neglecting effect of friction,
FA × v A = FB × v B or
∴ Ideal mechanical advantage,
FB vA
=
FA vB
FB vA
=
FA vB
If we consider the effect of friction, less resistance will be overcome with the given effort.
Therefore the actual mechanical advantage will be less.
Let
η = Efficiency of the mechanism.
∴ Actual mechanical advantage,
M.A.( ideal ) =
M.A.( actual ) = η ×
FB
v
= η× A
FA
vB
Note : The mechanical advantage may also be defined as the ratio of output torque to the input torque.
Let
TA = Driving torque,
T B = Resisting torque,
ωA and ωB = Angular velocity of the driving and driven links respectively.
∴ Ideal mechanical advantage,
M.A.( ideal ) =
TB ωA
=
TA ωB
. . . (Neglecting effect of friction)
and actual mechanical advantage,
M.A.( actual ) = η ×
ω
TB
=η× A
ωB
TA
. . . (Considering the effect of friction)
Example 7.10. A four bar mechanism has the following dimensions :
DA = 300 mm ; CB = AB = 360 mm ; DC = 600 mm. The link DC is fixed and the angle ADC
is 60°. The driving link DA rotates uniformly at a speed of 100 r.p.m. clockwise and the constant
driving torque has the magnitude of 50 N-m. Determine the velocity of the point B and angular
velocity of the driven link CB. Also find the actual mechanical advantage and the resisting torque if
the efficiency of the mechanism is 70 per cent.
Solution. Given : N AD = 100 r.p.m. or ωAD = 2 π × 100/60 = 10.47 rad/s ; T A = 50 N-m
Since the length of driving link, D A = 300 mm = 0.3 m, therefore velocity of A with respect
to D or velocity of A (because D is a fixed point),
v AD = v A = ωAD × D A = 10.47 × 0.3 = 3.14 m/s
. . . (Perpendicular to D A)
Velocity of point B
First of all draw the space diagram, to some suitable scale, as shown in Fig. 7.25 (a). Now the
velocity diagram, as shown in Fig. 7.25 (b), is drawn as discussed below :
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1. Since the link DC is fixed, therefore points d and c are taken as one point in the velocity
diagram. Draw vector da perpendicular to D A, to some suitable scale, to represent the velocity of A
with respect to D or simply velocity of A (i.e. v AD or v A) such that
vector da = v AD = v A = 3.14 m/s
2. Now from point a, draw vector ab perpendicular to A B to represent the velocity of B with
respect to A (i.e. v BA), and from point c draw vector cb perpendicular to CB to represent the velocity
of B with respect to C or simply velocity of B (i.e. v BC or v B). The vectors ab and cb intersect at b.
By measurement, we find that velocity of point B,
vB = v BC = vector cb = 2.25 m/s Ans.
(a) Space diagram.
(b) Velocity diagram.
Fig. 7.25
Angular velocity of the driven link CB
Since CB = 360 mm = 0.36 m, therefore angular velocity of the driven link CB,
ωBC =
vBC
2.25
=
= 6.25 rad/s (Clockwise about C) Ans.
0.36
BC
Actual mechanical advantage
We know that the efficiency of the mechanism,
η = 70% = 0.7
∴Actual mechanical advantage,
M.A. ( actual ) = η ×
. . . (Given)
ωA
10.47
= 0.7 ×
= 1.17 Ans.
6.25
ωB
...(3 ωA = ωAD; and ωB = ωBC)
Resisting torque
Let
TB = Resisting torque.
We know that efficiency of the mechanism (η),
0.7 =
∴
TB . ωB
T × 6.25
= B
= 0.012 TB
TA . ωA 50 × 10.47
T B = 58.3 N–m Ans.
Example 7.11. The dimensions of the various links of a
pneumatic riveter, as shown in Fig. 7.26, are as follows :
OA = 175 mm ; AB = 180 mm ; AD = 500 mm ;
and BC = 325 mm.
Find the velocity ratio between C and ram D when OB
is vertical. What will be the efficiency of the machine if a load
of 2.5 kN on the piston C causes a thrust of 4 kN at the ram D ?
Fig. 7.26
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Solution. Given : WC = 2.5 kN = 2500 N ; W D = 4 kN = 4000 N
Let
N = Speed of crank O A.
∴Angular velocity of crank O A,
ωAO = 2 π N/60 rad/s
Since the length of crank OA = 175 mm = 0.175 m, therefore velocity of A with respect to O
(or velocity of A ) (because O is a fixed point),
2πN
vAO = vA =
× 0.175 = 0.0183 N m/s
. . . (Perpendicular to O A)
60
(a) Space diagram.
(b) Velocity diagram.
Fig. 7.27
Velocity ratio between C and the ram D
First of all draw the space diagram, to some
suitable scale, as shown in Fig. 7.27 (a), Now the
velocity diagram, as shown in Fig. 7.27 (b), is drawn
as discussed below :
1. Draw vector oa perpendicular to O A to
represent the velocity of A (i.e. v A) such that
vector oa = vA = 0.0183 N m/s
Since the speed of crank (N) is not given,
therefore let we take vector oa = 20 mm.
2. From point a, draw a vector ab
perpendicular to A B to represent the velocity of B
with respect to A (i.e. v BA), and from point o draw
vector ob perpendicular to OB to represent the
velocity of B with respect to A or simply velocity of
B (i.e. v BO or v B). The vectors ab and ob intersect at
b.
3. Now from point b, draw vector bc
perpendicular to BC to represent the velocity of C
with respect to B (i.e. v CB) and from point o draw
vector oc parallel to the path of motion of C to
represent the velocity of C (i.e. v C). The vectors bc
and oc intersect at c. We see from Fig. 7.27 (b) that
Chapter 7 : Velocity in Mechanisms
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the points b and c coincide. Therefore velocity of B with respect to C is zero and velocity of B is equal
to velocity of C, i.e.
v BC = 0
. . . (3 b and c coincide)
and
vB = vC
. . . (3 vector ob = vector oc)
4. From point a, draw vector ad perpendicular to A D to represent velocity of D with respect
to A i.e. v DA, and from point o draw vector ob parallel to the path of motion of D to represent the
velocity of D i.e. v D. The vectors ad and od intersect at d.
By measurement from velocity diagram, we find that velocity of C,
vC = vector oc = 35 mm
and
velocity of D, v D = vector od = 21 mm
∴Velocity ratio between C and the ram D
= v C /v D = 35/21 = 1.66 Ans.
Efficiency of the machine
Let
η = Efficiency of the machine,
We know that work done on the piston C or input,
= W C × v C = 2500 v C
and work done by the ram D or output,
= W D × v D = 4000 v D
∴
η=
v
.... 3 C = 1.66
v
D
Output 4000 vD 4000
1
=
=
×
Input
2500 vC 2500 1.66
= 0.96 or 96% Ans.
Example 7.12. In the toggle mechanism, as
shown in Fig. 7.28, the slider D is constrained to move
on a horizontal path. The crank OA is rotating in the
counter-clockwise direction at a speed of 180 r.p.m.
The dimensions of various links are as follows :
OA = 180 mm ; CB = 240 mm ; AB = 360 mm ;
and BD = 540 mm.
For the given configuration, find : 1. Velocity
of slider D, 2. Angular velocity of links AB, CB and BD;
3. Velocities of rubbing on the pins of diameter 30 mm
at A and D, and 4. Torque applied to the crank OA, for
a force of 2 kN at D.
Fig. 7.28
Solution. Given : N AO = 180 r.p.m. or ωAO = 2 π × 180/60 = 18.85 rad/s
Since the crank length O A = 180 mm = 0.18 m, therefore velocity of A with respect to O or
velocity of A (because O is a fixed point),
v AO = v A = ωAO × O A = 18.85 × 0.18 = 3.4 m/s
. . . (Perpendicular to O A)
1. Velocity of slider D
First of all draw the space diagram, to some suitable scale, as shown in Fig. 7.29 (a). Now the
velocity diagram, as shown in Fig. 7.29 (b), is drawn as discussed below :
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1. Draw vector oa perpendicular to O A, to some suitable scale, to represent the velocity of A
with respect to O or velocity of A (i.e. v AO or v A, ) such that
vector oa = v AO = vA = 3.4 m/s
(a) Space diagram.
(b) Velocity diagram.
Fig. 7.29
2. Since point B moves with respect to A and also with respect to C, therefore draw vector ab
perpendicular to A B to represent the velocity of B with respect to A i.e. v BA, and draw vector cb
perpendicular to CB to represent the velocity of B with respect to C, i.e. v BC. The vectors ab and cb
intersect at b.
3. From point b, draw vector bd perpendicular to BD to represent the velocity of D with
respect to B i.e. v DB, and from point c draw vector cd parallel to the path of motion of the slider D
(which is along CD) to represent the velocity of D, i.e. v D. The vectors bd and cd intersect at d.
By measurement, we find that velocity of the slider D,
vD = vector cd = 2.05 m/s Ans.
2. Angular velocities of links AB, CB and BD
By measurement from velocity diagram, we find that
Velocity of B with respect to A ,
vBA = vector ab = 0.9 m/s
Velocity of B with respect to C,
v BC = v B = vector cb = 2.8 m/s
and velocity of D with respect to B,
vDB = vector bd = 2.4 m/s
We know that A B = 360 mm = 0.36 m ; CB = 240 mm = 0.24 m and BD = 540 mm = 0.54 m.
∴ Angular velocity of the link A B,
ωAB =
vBA
0.9
=
= 2.5 rad/s (Anticlockwise about A ) Ans.
AB 0.36
Similarly angular velocity of the link CB,
vBC
2.8
=
= 11.67 rad/s (Anticlockwise about C) Ans.
CB 0.24
and angular velocity of the link BD,
ωCB =
ωBD =
vDB
2.4
=
= 4.44 rad/s (Clockwise about B) Ans.
BD 0.54
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3. Velocities of rubbing on the pins A and D
Given : Diameter of pins at A and D,
DA = DD = 30 mm = 0.03 m
∴
Radius, rA = rD = 0.015 m
We know that relative angular velocity at A
= ωBC – ωBA + ωDB = 11.67 – 2.5 + 4.44 = 13.61 rad/s
and relative angular velocity at D
= ωDB = 4.44 rad/s
∴ Velocity of rubbing on the pin A
= 13.61 × 0.015 = 0.204 m/s = 204 mm/s Ans.
and velocity of rubbing on the pin D
= 4.44 × 0.015 = 0.067 m/s = 67 mm/s Ans.
4. Torque applied to the crank OA
Let
TA = Torque applied to the crank O A, in N-m
∴ Power input or work supplied at A
= T A × ωAO = T A × 18.85 = 18.85 T A N-m
We know that force at D,
FD = 2 kN = 2000 N
. . . (Given)
∴ Power output or work done by D,
= FD × v D = 2000 × 2.05 = 4100 N-m
Assuming 100 per cent efficiency, power input is equal to power output.
∴
18.85 T A = 4100 or T A = 217.5 N-m Ans.
Example 7.13. The dimensions of the mechanism, as shown in Fig. 7.30, are as follows :
AB = 0.45 m; BD = 1.5 m : BC = CE = 0.9 m.
Fig. 7.30
The crank A B turns uniformly at 180 r.p.m. in the clockwise direction and the blocks at D
and E are working in frictionless guides.
Draw the velocity diagram for the mechanism and find the velocities of the sliders D and E
in their guides. Also determine the turning moment at A if a force of 500 N acts on D in the direction
of arrow X and a force of 750 N acts on E in the direction of arrow Y.
Solution. Given : N BA = 180 r.p.m. or ωBA = 2 π × 180/60 = 18.85 rad/s
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Theory of Machines
Since A B = 0.45 m, therefore velocity of B with respect to A or velocity of B (because A is a
fixed point),
vBA = v B = ωBA × A B = 18.85 × 0.45 = 8.5 m/s
. . . (Perpendicular to A B)
Velocities of the sliders D and E
First of all draw the space diagram, to some suitable scale, as shown in Fig. 7.31 (a). Now the
velocity diagram, as shown in Fig. 7.31 (b), is drawn as discussed below :
(a) Space diagram.
(b) Velocity diagram.
Fig. 7.31
1. Draw vector ab perpendicular to A B, to some suitable scale, to represent the velocity of B
with respect to A or simply velocity of B (i.e. v BA or v B), such that
vector ab = v BA = v B = 8.5 m/s
2. From point b, draw vector bd perpendicular to BD to represent the velocity of D with
respect to B (i.e. v DB) and from point a draw vector ad parallel to the motion of D to represent the
velocity of D (v D). The vectors bd and ad intersect at d.
3. Since the point C lies on BD, therefore divide vector bd at c in the same ratio as C divides
BD in the space diagram. In other words,
bc/bd = BC/BD
4. Now from point c, draw vector ce perpendicular to CE to represent the velocity of E with
respect to C (i.e. v EC) and from point a draw vector ae parallel to the path of E to represent the
velocity of E (i.e. v E). The vectors ce and ae intersect at e.
By measurement, we find that
Velocity of slider D, vD = vector ad = 9.5 m/s Ans.
Velocity of slider E, vE = vector ae = 1.7 m/s Ans.
Turning moment at A
Let
TA = Turning moment at A (or at the crank-shaft).
We know that force at D, FD = 500 N
. . . (Given)
and
Force at E, FE = 750 N
. . . (Given)
∴ Power input
= FD × v D – FE × v E
. . . (– ve sign indicates that FE opposes the motion)
= 500 × 9.5 – 750 × 1.7 = 3475 N-m/s
Power output = T A.ωBA = T A × 18.85 T A N-m/s
Neglecting losses, power input is equal to power output.
∴
3475 = 18.85 T A or T A = 184.3 N-m Ans.
Chapter 7 : Velocity in Mechanisms
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EXERCISES
1.
2.
In a slider crank mechanism, the length of crank OB and connecting rod A B are 125 mm and 500 mm
respectively. The centre of gravity G of the connecting rod is 275 mm from the slider A . The crank
speed is 600 r.p.m. clockwise. When the crank has turned 45° from the inner dead centre position,
determine: 1. velocity of the slider A , 2. velocity of the point G, and 3. angular velocity of the
connecting rod A B.
[Ans. 6.45 m/s ; 6.75 m/s ; 10.8 rad/s]
In the mechanism, as shown in Fig. 7.32, O A and OB are two equal cranks at right angles rotating
about O at a speed of 40 r.p.m. anticlockwise. The dimensions of the various links are as follows :
Fig. 7.32
O A = OB = 50 mm ; A C = BD = 175 mm ; DE = CE = 75 mm ; FG = 115 mm and EF = FC.
Draw velocity diagram for the given configuration of the mechanism and find velocity of the slider G.
[Ans. 68 mm/s]
3.
The dimensions of various links in a mechanism, as shown in Fig. 7.33, are as follows :
A B = 60 mm ; BC = 400 mm ; CD = 150 mm ; DE = 115 mm ; and EF = 225 mm.
110
mm
225 mm
B
F
C
30°
A
150 mm
E
90°
D
Fig. 7.33
Find the velocity of the slider F when the crank A B rotates uniformly in clockwise direction at a speed
of 60 r.p.m.
[Ans. 250 mm/s]
4.
In a link work, as shown in Fig. 7.34, the crank A B rotates about A at a uniform speed of 150 r.p.m.
The lever DC oscillates about the fixed point D, being connected to A B by the connecting link BC.
The block F moves, in horizontal guides being driven by the link EF, when the crank A B is at 30°. The
dimensions of the various links are :
A B = 150 mm ; BC = 450 mm ; CE = 300 mm ; DE = 150 mm ; and EF = 350 mm.
Find, for the given configuration, 1. velocity of slider F, 2. angular velocity of DC, and 3. rubbing
speed at pin C which is 50 mm in diameter.
[Ans. 500 mm/s ; 3.5 rad/s ; 2.4 m/s]
170
5.
l
Theory of Machines
Fig. 7.34
The oscillating link OAB of a mechanism,
as shown in Fig. 7.35, is pivoted at O and
is moving at 90 r.p.m. anticlockwise. If OA
= 150 mm ; A B = 75 mm, and
A C = 250 mm, calculate
1.
2.
6.
the velocity of the block C;
the angular velocity of the link A C;
and
3. the rubbing velocities of the pins at
O, A and C, assuming that these pins
are of equal diameters of 20 mm.
[Ans. 1.2 m/s; 1.6 rad/s2 clockwise; 21
200 mm/s, 782 mm/s, 160 mm/s]
The dimensions of the various links of a
mechanism, as shown in Fig. 7.36, are as
follows :
Fig. 7.35
A B = 30 mm ; BC = 80 mm ; CD = 45 mm ; and CE = 120 mm.
Fig. 7.36
The crank A B rotates uniformly in the clockwise direction at 120 r.p.m. Draw the velocity diagram for
the given configuration of the mechanism and determine the velocity of the slider E and angular
velocities of the links BC, CD and CE.
Also draw a diagram showing the extreme top and bottom positions of the crank DC and the corresponding configurations of the mechanism.
Find the length of each of the strokes.
[Ans. 120 mm/s ; 2.8 rad/s ; 5.8 rad/s ; 2 rad/s ; 10 mm ; 23 mm]
Chapter 7 : Velocity in Mechanisms
7.
l
171
Fig. 7.37 shows a mechanism in which the crank O A, 100 mm long rotates clockwise about O at 130
r.p.m. The connecting rod A B is 400 mm long. The rod CE, 350 mm long, is attached to A B at C, 150
mm from A . This rod slides in a slot in a trunnion at D. The end E is connected by a link EF, 300 mm
long, to the horizontally moving slider F.
Fig. 7.37
Determine, for the given configuration : 1. velocity of F, 2. velocity of sliding of CE in the trunnion,
and 3. angular velocity of CE.
[Ans. 0.54 m/s ; 1.2 m/s ; 1.4 rad/s]
8.
Fig. 7.38 shows the mechanism of a quick return motion of the crank and slotted lever type shaping
machine. The dimensions of the various links are as follows :
O A = 200 mm ; A B = 100 mm ; OC = 400 mm ; and CR = 150 mm.
The driving crank A B makes 120° with the vertical and rotates at 60 r.p.m. in the clockwise direction.
Find : 1. velocity of ram R, and 2. angular velocity of the slotted link OC.
[Ans. 0.8 m/s ; 1.83 rad/s]
Fig. 7.38
9.
10.
Fig. 7.39
In a Whitworth quick return motion mechanism, as shown in Fig. 7.39, the dimensions of various
links are as follows :
OQ = 100 mm ; O A = 200 mm ; BQ = 150 mm and BP = 500 mm.
If the crank O A turns at 120 r.p.m. in clockwise direction and makes an angle of 120° with OQ,
Find : 1. velocity of the block P, and 2. angular velocity of the slotted link BQ.
[Ans. 0.63 m/s ; 6.3 rad/s]
A toggle press mechanism, as shown in Fig. 7.40, has the dimensions of various links as follows :
OP = 50 mm ; RQ = R S = 200 mm ; PR = 300 mm.
172
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Theory of Machines
Fig. 7.40
Find the velocity of S when the
crank OP rotates at 60 r.p.m. in
the anticlockwise direction. If the
torque on P is 115 N-m, what
pressure will be exerted at S when
the overall efficiency is 60 percent.
[Ans. 400 m/s ; 3.9 kN]
11.
Fig. 7.41 shows a toggle mechanism in which link D is constained
to move in horizontal direction.
For the given configuration, find
out : 1. velocities of points band
D; and 2. angular velocities of
links A B, BC, and BD.
The rank O A rotates at 60 r.p.m.
in anticlockwise direction.
12.
[Ans. 0.9 m/s; 0.5 m/s; 0.0016 rad/s (anticlockwise)
0.0075 rad/s (anti-clockwise),0.0044 rad/s (anticlockwise)]
A riveter, as shown in Fig. 7.42, is operated by a piston
F acting through the links EB, A B and BC. The ram D
carries the tool. The piston moves in a line perpendicular to the line of motion of D. The length of link
BC is twice the length of link A B. In the position
shown, A B makes an angle of 12° with A C and BE is
at right angle to A C. Find the velocity ratio of E to D.
If, in the same position, the total load on the piston is
2.2 kN, find the thrust exerted by D when the efficiency
of the mechanism is 72 per cent,
Ans. [3.2 ; 5 kN]
Fig. 7.41
Fig. 7.42
DO YOU KNOW ?
1. Describe the method to find the velocity of a point on a link whose direction (or path) is known and
the velocity of some other point on the same link in magnitude and direction is given.
2. Explain how the velocities of a slider and the connecting rod are obtained in a slider crank
mechanism.
Chapter 7 : Velocity in Mechanisms
l
173
3. Define rubbing velocity at a pin joint. What will be the rubbing velocity at pin joint when the two links
move in the same and opposite directions ?
4. What is the difference between ideal mechanical advantage and actual mechanical advantage ?
OBJECTIVE TYPE QUESTIONS
1.
2.
3.
4.
The direction of linear velocity of any point on a link with respect to another point on the same link is
(a) parallel to the link joining the points
(b) perpendicular to the link joining the points
(c) at 45° to the link joining the points
(d) none of these
The magnitude of linear velocity of a point B on a link A B relative to point A is
(a) ω.AB
(b) ω (A B)2
2
(c) ω . A B
(d) (ω . A B)2
where ω = Angular velocity of the link A B.
The two links O A and OB are connected by a pin joint at O. If the link O A turns with angular velocity
ω 1 rad/s in the clockwise direction and the link O B turns with angular velocity
ω2 rad/s in the anti-clockwise direction, then the rubbing velocity at the pin joint O is
(a) ω1.ω2.r
(b) (ω1 – ω2) r
(d) (ω1 – ω2) 2 r
(c) (ω1 + ω2) r
where r = Radius of the pin at O.
In the above question, if both the links O A and OB turn in clockwise direction, then the rubbing
velocity at the pin joint O is
(a) ω1.ω2.r
(b) (ω1 – ω2) r
(c)
5.
(ω1 + ω2) r
(d)
(ω1 – ω2) 2 r
In a four bar mechanism, as shown in Fig. 7.43, if a force FA is
acting at point A in the direction of its velocity v A and a force FB is
transmitted to the joint B in the direction of its velocity v B , then the
ideal mechanical advantage is equal to
(a) FB.v A
(b) FA.v B
(c)
FB
vB
(d)
FB
FA
Fig. 7.43
ANSWERS
1. (b)
2. (a)
3. (c)
4. (b)
5. (d)
GO To FIRST
CONTENTS
CONTENTS
174
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Theory of Machines
8
Warping Machine
Fea
tur
es
eatur
tures
1. Introduction.
2. Acceleration Diagram for a
Link.
3. Acceleration of a Point on a
Link.
4. Acceleration in the Slider
Crank Mechanism.
5. Coriolis Component of
Acceleration.
Acceleration
in Mechanisms
8.1.
Intr
oduction
Introduction
We have discussed in the previous chapter the
velocities of various points in the mechanisms. Now we shall
discuss the acceleration of points in the mechanisms. The
acceleration analysis plays a very important role in the
development of machines and mechanisms.
8.2.
Acceleration Diagram for a Link
Consider two points A and B on a rigid link as shown
in Fig. 8.1 (a). Let the point B moves with respect to A, with
an angular velocity of ω rad/s and let α rad/s2 be the angular
acceleration of the link AB.
(a) Link.
(b) Acceleration diagram.
Fig. 8.1. Acceleration for a link.
174
CONTENTS
CONTENTS
Chapter 8 : Acceleration in Mechanisms
l
175
We have already discussed that acceleration of a particle whose velocity changes both in
magnitude and direction at any instant has the following two components :
1. The centripetal or radial component, which is perpendicular to the velocity of the
particle at the given instant.
2. The tangential component, which is parallel to the velocity of the particle at the given
instant.
Thus for a link A B, the velocity of point B with respect to A (i.e. vBA) is perpendicular to the
link A B as shown in Fig. 8.1 (a). Since the point B moves with respect to A with an angular velocity
of ω rad/s, therefore centripetal or radial component of the acceleration of B with respect to A ,
r
2
/ AB
= ω2 × Length of link AB = ω2 × AB = vBA
aBA
v
... 3 ω = BA
AB
This radial component of acceleration acts perpendicular to the velocity v BA, In other words,
it acts parallel to the link AB.
We know that tangential component of the acceleration of B with respect to A ,
t
= α × Length of the link AB = α × AB
aBA
This tangential component of acceleration acts parallel to the velocity v BA. In other words,
it acts perpendicular to the link A B.
In order to draw the acceleration diagram for a link A B, as shown in Fig. 8.1 (b), from any
point b', draw vector b'x parallel to BA to represent the radial component of acceleration of B with
r
and from point x draw vector xa' perpendicular to B A to represent the tangential
respect to A i.e. aBA
t
component of acceleration of B with respect to A i.e. aBA . Join b' a'. The vector b' a' (known as
acceleration image of the link A B) represents the total acceleration of B with respect to A (i.e. aBA)
r
t
and it is the vector sum of radial component (aBA
) and tangential component (aBA
) of acceleration.
8.3.
Acceleration of a Point on a Link
(a) Points on a Link.
(b) Acceleration diagram.
Fig. 8.2. Acceleration of a point on a link.
Consider two points A and B on the rigid link, as shown in Fig. 8.2 (a). Let the acceleration
of the point A i.e. aA is known in magnitude and direction and the direction of path of B is given. The
acceleration of the point B is determined in magnitude and direction by drawing the acceleration
diagram as discussed below.
1. From any point o', draw vector o'a' parallel to the direction of absolute acceleration at
point A i.e. aA , to some suitable scale, as shown in Fig. 8.2 (b).
176
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Theory of Machines
2. We know that the acceleration of B with
respect to A i.e. a BA has the following two
components:
(i) Radial component of the acceleration
r
of B with respect to A i.e. aBA
, and
(ii) Tangential component of the
t
acceleration B with respect to A i.e. aBA
. These two
components are mutually perpendicular.
3. Draw vector a'x parallel to the link A B
(because radial component of the acceleration of B
with respect to A will pass through AB), such that
r
2
vector a′x = aBA
/ AB
= vBA
where
vBA = Velocity of B with respect to A .
Note: The value of v BA may be obtained by drawing the
velocity diagram as discussed in the previous chapter.
4. From point x , draw vector xb'
perpendicular to A B or vector a'x (because tangential
t
component of B with respect to A i.e. aBA
, is
r
perpendicular to radial component aBA
) and
through o' draw a line parallel to the path of B to
represent the absolute acceleration of B i.e. aB. The
vectors xb' and o' b' intersect at b'. Now the values
A refracting telescope uses mechanisms to
change directions.
Note : This picture is given as additional
information and is not a direct example of the
current chapter.
t
of aB and aBA
may be measured, to the scale.
5. By joining the points a' and b' we may determine the total acceleration of B with respect
to A i.e. aBA. The vector a' b' is known as acceleration image of the link A B.
6. For any other point C on the link, draw triangle a' b' c' similar to triangle ABC. Now
vector b' c' represents the acceleration of C with respect to B i.e. aCB, and vector a' c' represents the
acceleration of C with respect to A i.e. aCA. As discussed above, aCB and aCA will each have two
components as follows :
r
t
(i) aCB has two components; aCB
as shown by triangle b' zc' in Fig. 8.2 (b), in
and aCB
which b' z is parallel to BC and zc' is perpendicular to b' z or BC.
r
t
(ii) aCA has two components ; aCA
as shown by triangle a' yc' in Fig. 8.2 (b), in
and aCA
which a' y is parallel to A C and yc' is perpendicular to a' y or A C.
7. The angular acceleration of the link AB is obtained by dividing the tangential components
t
of the acceleration of B with respect to A (aBA
) to the length of the link. Mathematically, angular
acceleration of the link A B,
t
/ AB
α AB = aBA
8.4.
Acceleration in the Slider Crank Mechanism
A slider crank mechanism is shown in Fig. 8.3 (a). Let the crank OB makes an angle θ with
the inner dead centre (I.D.C) and rotates in a clockwise direction about the fixed point O with
uniform angular velocity ωBO rad/s.
∴ Velocity of B with respect to O or velocity of B (because O is a fixed point),
vBO = vB = ωBO × OB , acting tangentially at B .
Chapter 8 : Acceleration in Mechanisms
l
177
We know that centripetal or radial acceleration of B with respect to O or acceleration of B
(because O is a fixed point),
v2
r
2
= aB = ωBO
× OB = BO
aBO
OB
Note : A point at the end of a link which moves with constant angular velocity has no tangential component
of acceleration.
(a) Slider crank mechanism.
(b) Acceleration diagram.
Fig. 8.3. Acceleration in the slider crank mechanism.
The acceleration diagram, as shown in Fig. 8.3 (b), may now be drawn as discussed below:
r
1. Draw vector o' b' parallel to BO and set off equal in magnitude of aBO
= aB , to some
suitable scale.
2. From point b', draw vector b'x parallel to B A. The vector b'x represents the radial component
of the acceleration of A with respect to B whose magnitude is given by :
r
2
/ BA
= vAB
aAB
Since the point B moves with constant angular velocity, therefore there will be no tangential
component of the acceleration.
3. From point x, draw vector xa' perpendicular to b'x (or A B). The vector xa' represents the
t
tangential component of the acceleration of A with respect to B i.e. aAB
.
Note: When a point moves along a straight line, it has no centripetal or radial component of the acceleration.
4. Since the point A reciprocates along A O, therefore the acceleration must be parallel to
velocity. Therefore from o', draw o' a' parallel to A O, intersecting the vector xa' at a'.
t
Now the acceleration of the piston or the slider A (aA) and aAB
may be measured to the scale.
5. The vector b' a', which is the sum of the vectors b' x and x a', represents the total acceleration
of A with respect to B i.e. aAB. The vector b' a' represents the acceleration of the connecting rod A B.
6. The acceleration of any other point on A B such as E may be obtained by dividing the vector
b' a' at e' in the same ratio as E divides A B in Fig. 8.3 (a). In other words
a' e' / a' b' = AE / AB
7. The angular acceleration of the connecting rod A B may be obtained by dividing the
t
tangential component of the acceleration of A with respect to B ( aAB
) to the length of A B. In other
words, angular acceleration of A B,
t
/ AB (Clockwise about B)
α AB = aAB
Example 8.1. The crank of a slider crank mechanism rotates clockwise at a constant speed
of 300 r.p.m. The crank is 150 mm and the connecting rod is 600 mm long. Determine : 1. linear
velocity and acceleration of the midpoint of the connecting rod, and 2. angular velocity and angular
acceleration of the connecting rod, at a crank angle of 45° from inner dead centre position.
178
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Theory of Machines
Solution. Given : N BO = 300 r.p.m. or ωBO = 2 π × 300/60 = 31.42 rad/s; OB = 150 mm =
0.15 m ; B A = 600 mm = 0.6 m
We know that linear velocity of B with respect to O or velocity of B,
vBO = v B = ωBO × OB = 31.42 × 0.15 = 4.713 m/s
...(Perpendicular to BO)
(a) Space diagram.
(b) Velocity diagram.
(c) Acceleration diagram.
Fig. 8.4
Ram moves
outwards
Ram moves
inwards
Oil pressure on
lower side of
piston
Load moves
outwards
Oil pressure on
upper side of
piston
Load
moves
inwards
Pushing with fluids
Note : This picture is given as additional information and is not a direct example of the current chapter.
1. Linear velocity of the midpoint of the connecting rod
First of all draw the space diagram, to some suitable scale; as shown in Fig. 8.4 (a). Now the
velocity diagram, as shown in Fig. 8.4 (b), is drawn as discussed below:
1. Draw vector ob perpendicular to BO, to some suitable scale, to represent the velocity of
B with respect to O or simply velocity of B i.e. vBO or vB, such that
vector ob = v BO = v B = 4.713 m/s
2. From point b, draw vector ba perpendicular to BA to represent the velocity of A with
respect to B i.e. vAB , and from point o draw vector oa parallel to the motion of A (which is along A O)
to represent the velocity of A i.e. vA. The vectors ba and oa intersect at a.
Chapter 8 : Acceleration in Mechanisms
l
179
By measurement, we find that velocity of A with respect to B,
vAB = vector ba = 3.4 m / s
and
Velocity of A, vA = vector oa = 4 m / s
3. In order to find the velocity of the midpoint D of the connecting rod A B, divide the vector
ba at d in the same ratio as D divides A B, in the space diagram. In other words,
bd / ba = BD/BA
Note: Since D is the midpoint of A B, therefore d is also midpoint of vector ba.
4. Join od. Now the vector od represents the velocity of the midpoint D of the connecting
rod i.e. vD.
By measurement, we find that
vD = vector od = 4.1 m/s Ans.
Acceleration of the midpoint of the connecting rod
We know that the radial component of the acceleration of B with respect to O or the
acceleration of B,
2
vBO
(4.713)2
=
= 148.1 m/s 2
0.15
OB
and the radial component of the acceleraiton of A with respect to B,
r
= aB =
aBO
2
vAB
(3.4)2
=
= 19.3 m/s 2
0.6
BA
Now the acceleration diagram, as shown in Fig. 8.4 (c) is drawn as discussed below:
r
=
aAB
1. Draw vector o' b' parallel to BO, to some suitable scale, to represent the radial component
r
of the acceleration of B with respect to O or simply acceleration of B i.e. aBO
or aB , such that
r
vector o′ b′ = aBO
= aB = 148.1 m/s 2
Note: Since the crank OB rotates at a constant speed, therefore there will be no tangential component of the
acceleration of B with respect to O.
2. The acceleration of A with respect to B has the following two components:
r
(a) The radial component of the acceleration of A with respect to B i.e. aAB
, and
t
(b) The tangential component of the acceleration of A with respect to B i.e. aAB
. These two
components are mutually perpendicular.
r
Therefore from point b', draw vector b' x parallel to A B to represent aAB
= 19.3 m/s 2 and
from point x draw vector xa' perpendicular to vector b' x whose magnitude is yet unknown.
3. Now from o', draw vector o' a' parallel to the path of motion of A (which is along A O) to
represent the acceleration of A i.e. aA . The vectors xa' and o' a' intersect at a'. Join a' b'.
4. In order to find the acceleration of the midpoint D of the connecting rod A B, divide the
vector a' b' at d' in the same ratio as D divides A B. In other words
b′d ′ / b′a ′ = BD / BA
Note: Since D is the midpoint of A B, therefore d' is also midpoint of vector b' a'.
5. Join o' d'. The vector o' d' represents the acceleration of midpoint D of the connecting rod
i.e. aD.
By measurement, we find that
aD = vector o' d' = 117 m/s2 Ans.
180
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Theory of Machines
2. Angular velocity of the connecting rod
We know that angular velocity of the connecting rod A B,
vAB 3.4
=
= 5.67 rad/s2 (Anticlockwise about B ) Ans.
BA 0.6
Angular acceleration of the connecting rod
From the acceleration diagram, we find that
ωAB =
t
= 103 m/s 2
aAB
...(By measurement)
We know that angular acceleration of the connecting rod A B,
t
aAB
103
=
= 171.67 rad/s 2 (Clockwise about B ) Ans.
0.6
BA
Example 8.2. An engine mechanism is shown in Fig. 8.5. The crank CB = 100 mm and the
connecting rod BA = 300 mm with centre of gravity G, 100 mm from B. In the position shown, the
crankshaft has a speed of 75 rad/s and an angular acceleration of 1200 rad/s2. Find:1. velocity of
G and angular velocity of AB, and 2. acceleration of G and angular acceleration of AB.
α AB =
Fig. 8.5
Solution. Given : ωBC = 75 rad/s ; αBC = 1200 rad/s2, CB = 100 mm = 0.1 m; B A = 300 mm
= 0.3 m
We know that velocity of B with respect to C or velocity of B,
vBC = vB = ωBC × CB = 75 × 0.1 = 7.5 m/s
...(Perpendicular to BC)
Since the angular acceleration of the crankshaft, αBC = 1200 rad/s2, therefore tangential
component of the acceleration of B with respect to C,
t
= αBC × CB = 1200 × 0.1 = 120 m/s2
aBC
Note: When the angular acceleration is not given, then there will be no tangential component of the acceleration.
1. Velocity of G and angular velocity of AB
First of all, draw the space diagram, to some suitable scale, as shown in Fig. 8.6 (a). Now the
velocity diagram, as shown in Fig. 8.6 (b), is drawn as discussed below:
1. Draw vector cb perpendicular to CB, to some suitable scale, to represent the velocity of
B with respect to C or velocity of B (i.e. vBC or vB), such that
vector cb = vBC = vB = 7.5 m/s
2. From point b, draw vector ba perpendicular to B A to represent the velocity of A with
respect to B i.e. vAB , and from point c, draw vector ca parallel to the path of motion of A (which is
along A C) to represent the velocity of A i.e. vA.The vectors ba and ca intersect at a.
3. Since the point G lies on A B, therefore divide vector ab at g in the same ratio as G divides
A B in the space diagram. In other words,
ag / ab = AG / AB
The vector cg represents the velocity of G.
By measurement, we find that velocity of G,
v G = vector cg = 6.8 m/s Ans.
Chapter 8 : Acceleration in Mechanisms
l
181
From velocity diagram, we find that velocity of A with respect to B,
vAB = vector ba = 4 m/s
We know that angular velocity of A B,
ωAB =
vAB
4
=
= 13.3 rad/s (Clockwise) Ans.
BA 0.3
(a) Space diagram.
(b) Velocity diagram.
Fig. 8.6
2. Acceleration of G and angular acceleration of AB
We know that radial component of the acceleration of B with
respect to C,
2
2
* a r = vBC = (7.5) = 562.5 m/s2
BC
0.1
CB
and radial component of the acceleration of A with respect to B,
2
vAB
42
=
= 53.3 m/s 2
BA 0.3
Now the acceleration diagram, as shown in Fig. 8.6 (c), is drawn
as discussed below:
1. Draw vector c' b'' parallel to CB, to some suitable scale, to (c) Acceleration diagram.
represent the radial component of the acceleration of B with respect to C,
Fig. 8.6
r
i.e. aBC
, such that
r
=
aAB
r
vector c ′ b′′ = aBC
= 562.5 m/s 2
2. From point b'', draw vector b'' b' perpendicular to vector c' b'' or CB to represent the
t
tangential component of the acceleration of B with respect to C i.e. aBC
, such that
t
... (Given)
vector b′′ b′ = aBC
= 120 m/s 2
3. Join c' b'. The vector c' b' represents the total acceleration of B with respect to C i.e. aBC.
4. From point b', draw vector b' x parallel to B A to represent radial component of the
r
acceleration of A with respect to B i.e. aAB
such that
r
vector b′x = aAB
= 53.3 m/s 2
5. From point x, draw vector xa' perpendicular to vector b'x or B A to represent tangential
t
component of the acceleration of A with respect to B i.e. aAB
, whose magnitude is not yet known.
6. Now draw vector c' a' parallel to the path of motion of A (which is along A C) to represent
the acceleration of A i.e. aA.The vectors xa' and c'a' intersect at a'. Join b' a'. The vector b' a'
represents the acceleration of A with respect to B i.e. aAB.
*
t
r =
When angular acceleration of the crank is not given, then there is no aBC
. In that case, aBC
aBC = aB , as
discussed in the previous example.
182
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Theory of Machines
7. In order to find the acceleratio of G, divide vector a' b' in g' in the same ratio as G divides
B A in Fig. 8.6 (a). Join c' g'. The vector c' g' represents the acceleration of G.
By measurement, we find that acceleration of G,
aG = vector c' g' = 414 m/s2 Ans.
From acceleration diagram, we find that tangential component of the acceleration of A with
respect to B,
t
= vector xa′ = 546 m/s 2
aAB
...(By measurement)
∴ Angular acceleration of A B,
α AB =
t
aAB
546
=
= 1820 rad/s 2 (Clockwise) Ans.
0.3
BA
Example 8.3. In the mechanism shown in Fig. 8.7, the slider C is
moving to the right with a velocity of 1 m/s and an acceleration of 2.5 m/s2.
The dimensions of various links are AB = 3 m inclined at 45° with the
vertical and BC = 1.5 m inclined at 45° with the horizontal. Determine: 1. the
magnitude of vertical and horizontal component of the acceleration of the
point B, and 2. the angular acceleration of the links AB and BC.
Solution. Given : v C = 1 m/s ; aC = 2.5 m/s2; A B = 3 m ; BC = 1.5 m
First of all, draw the space diagram, as shown in Fig. 8.8 (a), to some
suitable scale. Now the velocity diagram, as shown in Fig. 8.8 (b), is drawn as
discussed below:
Fig. 8.7
1. Since the points A and D are fixed points, therefore they lie at one place in the velocity
diagram. Draw vector dc parallel to DC, to some suitable scale, which represents the velocity of
slider C with respect to D or simply velocity of C, such that
vector dc = v CD = v C = 1 m/s
2. Since point B has two motions, one with respect to A and the other with respect to C,
therefore from point a, draw vector ab perpendicular to A B to represent the velocity of B with
respect to A , i.e. vBA and from point c draw vector cb perpendicular to CB to represent the velocity
of B with respect to C i.e. vBC .The vectors ab and cb intersect at b.
(a) Space diagram.
(b) Velocity diagram.
(c) Acceleration diagram.
Fig. 8.8
By measurement, we find that velocity of B with respect to A ,
vBA = vector ab = 0.72 m/s
and velocity of B with respect to C,
vBC = vector cb = 0.72 m/s
Chapter 8 : Acceleration in Mechanisms
l
183
We know that radial component of acceleration of B with respect to C,
2
vBC
(0.72) 2
=
= 0.346 m/s2
1.5
CB
and radial component of acceleration of B with respect to A ,
r
=
aBC
2
vBA
(0.72)2
=
= 0.173 m/s 2
3
AB
Now the acceleration diagram, as shown in Fig. 8.8 (c), is drawn as discussed below:
1. *Since the points A and D are fixed points, therefore they lie at one place in the acceleration
diagram. Draw vector d' c' parallel to DC, to some suitable scale, to represent the acceleration of C
with respect to D or simply acceleration of C i.e. aCD or aC such that
r
=
aBA
vector d ′ c′ = aCD = aC = 2.5 m/s 2
2. The acceleration of B with respect to C will have two components, i.e. one radial component
of B with respect to C
r
(aBC
) and
the other tangential component of B with respect to
r
t
C ( aBC
such that
). Therefore from point c', draw vector c' x parallel to CB to represent aBC
r
vector c ′x = aBC
= 0.346 m/s 2
3. Now from point x, draw vector xb' perpendicular to vector c' x or CB to represent atBC
whose magnitude is yet unknown.
4. The acceleration of B with respect to A will also have two components, i.e. one radial
component of B with respect to A (arBA) and other tangential component of B with respect to A (at BA).
Therefore from point a' draw vector a' y parallel to A B to represent arBA, such that
vector a' y = arBA = 0.173 m/s2
t
5. From point y, draw vector yb' perpendicular to vector a'y or AB to represent aBA
. The
vector yb' intersect the vector xb' at b'. Join a' b' and c' b'. The vector a' b' represents the acceleration
of point B (aB) and the vector c' b' represents the acceleration of B with respect to C.
1. Magnitude of vertical and horizontal component of the acceleration of the point B
Draw b' b'' perpendicular to a' c'. The vector b' b'' is the vertical component of the acceleration
of the point B and a' b'' is the horizontal component of the acceleration of the point B. By measurement,
vector b' b'' = 1.13 m/s2 and vector a' b'' = 0.9 m/s2 Ans.
2. Angular acceleration of AB and BC
By measurement from acceleration diagram, we find that tangential component of acceleration
of the point B with respect to A ,
t
= vector yb′ = 1.41 m/s 2
aBA
and tangential component of acceleration of the point B with respect to C,
t
= vector xb′ = 1.94 m/s 2
aBC
*
If the mechanism consists of more than one fixed point, then all these points lie at the same place in the
velocity and acceleration diagrams.
184
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Theory of Machines
We know that angular acceleration of A B,
α AB =
t
aBA
1.41
=
= 0.47 rad/s 2 Ans.
3
AB
and angular acceleration of BC,
t
aBA
1.94
=
= 1.3 rad/s 2 Ans.
1.5
CB
Example 8.4. PQRS is a four bar chain with link PS fixed. The lengths of the links are PQ
= 62.5 mm ; QR = 175 mm ; RS = 112.5 mm ; and PS = 200 mm. The crank PQ rotates at 10 rad/s
clockwise. Draw the velocity and acceleration diagram when angle QPS = 60° and Q and R lie on
the same side of PS. Find the angular velocity and angular acceleration of links QR and RS.
α BC =
Solution. Given : ωQP = 10 rad/s; PQ = 62.5 mm = 0.0625 m ; QR = 175 mm = 0.175 m ;
R S = 112.5 mm = 0.1125 m ; PS = 200 mm = 0.2 m
We know that velocity of Q with respect to P or velocity of Q,
v QP = v Q = ωQP × PQ = 10 × 0.0625 = 0.625 m/s
...(Perpendicular to PQ)
Angular velocity of links QR and RS
First of all, draw the space diagram of a four bar chain, to some suitable scale, as shown in
Fig. 8.9 (a). Now the velocity diagram as shown in Fig. 8.9 (b), is drawn as discussed below:
(a) Space diagram.
(b) Velocity diagram.
(c) Acceleration diagram.
Fig. 8.9
1. Since P and S are fixed points, therefore these points lie at one place in velocity diagram.
Draw vector pq perpendicular to PQ, to some suitable scale, to represent the velocity of Q with
respect to P or velocity of Q i.e. v QP or v Q such that
vector pq = v QP = v Q = 0.625 m/s
2. From point q, draw vector qr perpendicular to QR to represent the velocity of R with
respect to Q (i.e. vRQ) and from point s, draw vector sr perpendicular to S R to represent the velocity
of R with respect to S or velocity of R (i.e. vRS or v R). The vectors qr and sr intersect at r. By
measurement, we find that
vRQ = vector qr = 0.333 m/s, and v RS = v R = vector sr = 0.426 m/s
We know that angular velocity of link QR,
ωQR =
vRQ
RQ
=
0.333
= 1.9 rad/s (Anticlockwise) Ans.
0.175
Chapter 8 : Acceleration in Mechanisms
l
185
and angular velocity of link R S,
vRS
0.426
=
= 3.78 rad/s (Clockwise) AAns.
SR 0.1125
Angular acceleration of links QR and RS
ωRS =
Since the angular acceleration of the crank PQ is not given, therefore there will be no tangential
component of the acceleration of Q with respect to P.
We know that radial component of the acceleration of Q with respect to P (or the acceleration
of Q),
r
= aQP = aQ =
aQP
2
vQP
PQ
=
(0.625) 2
= 6.25 m/s 2
0.0625
Radial component of the acceleration of R with respect to Q,
2
vRQ
(0.333)2
r
=
=
= 0.634 m/s2
aRQ
0.175
QR
and radial component of the acceleration of R with respect to S (or the acceleration of R),
2
vRS
(0.426)2
=
= 1.613 m/s 2
0.1125
SR
The acceleration diagram, as shown in Fig. 8.9 (c) is drawn as follows :
r
= aRS = aR =
aRS
1. Since P and S are fixed points, therefore these points lie at one place in the acceleration
diagram. Draw vector p'q' parallel to PQ, to some suitable scale, to represent the radial component
r
or aQ such that
of acceleration of Q with respect to P or acceleration of Q i.e aQP
r
vector p′q′ = aQP
= aQ = 6.25 m/s 2
2. From point q', draw vector q' x parallel to QR to represent the radial component of
r
acceleration of R with respect to Q i.e. aRQ
such that
r
vector q ′x = aRQ
= 0.634 m/s 2
3. From point x, draw vector xr' perpendicular to QR to represent the tangential component
t
of acceleration of R with respect to Q i.e aRQ
whose magnitude is not yet known.
4. Now from point s', draw vector s'y parallel to S R to represent the radial component of the
r
acceleration of R with respect to S i.e. aRS
such that
r
vector s ′y = aRS = 1.613 m/s 2
5. From point y, draw vector yr' perpendicular to S R to represent the tangential component
t
of acceleration of R with respect to S i.e. aRS
.
6. The vectors xr' and yr' intersect at r'. Join p'r and q' r'. By measurement, we find that
t
t
= vector xr ′ = 4.1 m/s 2 and aRS
= vector yr ′ = 5.3 m/s 2
aRQ
We know that angular acceleration of link QR,
α QR =
t
aRQ
QR
and angular acceleration of link R S,
α RS =
=
4.1
= 23.43 rad/s 2 (Anticlockwise) Ans.
0.175
t
aRS
5.3
=
= 47.1 rad/s 2 (Anticlockwise) Ans.
SR 0.1125
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Theory of Machines
Example 8.5. The dimensions and
configuration of the four bar mechanism, shown in
Fig. 8.10, are as follows :
P1A = 300 mm; P2B = 360 mm; AB = 360
mm, and P1P2 = 600 mm.
The angle AP1P2 = 60°. The crank P1A has
an angular velocity of 10 rad/s and an angular
acceleration of 30 rad/s 2 , both clockwise.
Determine the angular velocities and angular
accelerations of P2B, and AB and the velocity and
acceleration of the joint B.
Fig. 8.10
Solution. Given : ωAP1 = 10 rad/s ; αAP1 = 30 rad/s2; P1A = 300 mm = 0.3 m ; P2B = A B =
360 mm = 0.36 m
We know that the velocity of A with respect to P1 or velocity of A ,
vAP1 = vA = ωAP1 × P1A = 10 × 0.3 = 3 m/s
Velocity of B and angular velocitites of P2B and AB
First of all, draw the space diagram, to some suitable scale, as shown in Fig. 8.11 (a). Now
the velocity diagram, as shown in Fig. 8.11 (b), is drawn as discussed below:
1. Since P1 and P2 are fixed points, therefore these points lie at one place in velocity diagram.
Draw vector p1 a perpendicular to P1A , to some suitable scale, to represent the velocity of A with
respect to P1 or velocity of A i.e. vAP1 or vA, such that
vector p1a = v A P1 = v A = 3 m/s
2. From point a, draw vector ab perpendicular to AB to represent velocity of B with respect
to A (i.e. vBA) and from point p2 draw vector p2b perpendicular to P2B to represent the velocity of B
with respect to P2 or velocity of B i.e. vBP2 or v B. The vectors ab and p2b intersect at b.
By measurement, we find that
vBP2 = v B = vector p2b = 2.2 m/s Ans.
and
vBA = vector ab = 2.05 m/s
We know that angular velocity of P2B,
ωP2B =
vBP 2
P2 B
=
2.2
= 6.1 rad/s (Clockwise) Ans.
0.36
and angular velocity of A B,
ωAB =
vBA
2.05
=
= 5.7 rad/s (Anticlockwise) Ans.
0.36
AB
Acceleration of B and angular acceleration of P2B and AB
We know that tangential component of the acceleration of A with respect to P1,
atA P1 = α A P1 × P1 A = 30 × 0.3 = 9 m/s 2
Radial component of the acceleration of A with respect to P1,
r
aAP
=
1
vA2 P1
P1 A
2
= ωAP
× P1 A = 102 × 0.3 = 30 m/s2
1
Chapter 8 : Acceleration in Mechanisms
l
187
Radial component of the acceleration of B with respect to A .
r
=
aBA
2
vBA
(2.05) 2
=
= 11.67 m/s 2
0.36
AB
and radial component of the acceleration of B with respect to P2,
r
=
aBP
2
2
vBP
2
P2 B
=
(2.2)2
= 13.44 m/s2
0.36
(a) Space diagram.
(b) Velocity diagram.
Fig. 8.11
The acceleration diagram, as shown in Fig. 8.11 (c), is
drawn as follows:
1. Since P1 and P2 are fixed points, therefore these points
will lie at one place, in the acceleration diagram. Draw vector
p1' x parallel to P1A , to some suitable scale, to represent the
radial component of the acceleration of A with respect to P1,
such that
vector p1′ x = aAr P = 30 m/s 2
1
2. From point x, draw vector xa' perpendicular to P1A to
represent the tangential component of the acceleration of A with
respect to P1, such that
vector xa ′ = aAt P1 = 9 m/s 2
(c) Acceleration diagram
3. Join p1' a'. The vector p1' a' represents the acceleration
of A . By measurement, we find that the acceleration of A ,
Fig. 8.11
aA = aAP1 = 31.6 m/s2
4. From point a', draw vector a' y parallel to A B to represent the radial component of the
acceleration of B with respect to A , such that
r
vector a′y = aBA
= 11.67 m/s 2
5. From point y, draw vector yb' perpendicular to A B to represent the tangential component
t
) whose magnitude is yet unknown.
of the acceleration of B with respect to A (i.e. aBA
′
′
6. Now from point p , draw vector p z parallel to P B to represent the radial component
2
2
of the acceleration B with respect to P2, such that
r
vector p2′ z = aBP
= 13.44 m/s 2
2
2
188
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Theory of Machines
7. From point z, draw vector zb' perpendicular to P2B to represent the tangential component
t
.
of the acceleration of B with respect to P2 i.e. aBP
2
8. The vectors yb' and zb' intersect at b'. Now the vector p2' b' represents the acceleration of
B with respect to P2 or the acceleration of B i.e. aBP2 or aB. By measurement, we find that
aBP2 = aB = vector p2' b' = 29.6 m/s2 Ans.
Also
t
t
= 13.6 m/s 2 , and vector zb′ = aBP
= 26.6 m/s 2
vector yb′ = aBA
2
We know that angular acceleration of P2B,
α
P2B
=
and angular acceleration of A B, α AB =
t
aBP
2
P2 B
=
26.6
= 73.8 rad/s 2 (Anticlockwise) Ans.
0.36
t
aBA
13.6
=
= 37.8 rad/s 2 (Anticlockwise) Ans.
AB 0.36
Bicycle is a common example where simple mechanisms are used.
Note : This picture is given as additional information and is not a direct example of the current chapter.
Example 8.6. In the mechanism, as shown in Fig. 8.12, the crank OA rotates at 20 r.p.m.
anticlockwise and gives motion to the sliding blocks B and D. The dimensions of the various links
are OA = 300 mm; AB = 1200 mm; BC = 450 mm and CD = 450 mm.
Fig. 8.12
For the given configuration, determine : 1. velocities of sliding at B and D, 2. angular
velocity of CD, 3. linear acceleration of D, and 4. angular acceleration of CD.
Solution. Given : NAO = 20 r.p.m. or ωAO = 2 π × 20/60 = 2.1 rad/s ; OA = 300 mm = 0.3 m ;
A B = 1200 mm = 1.2 m ; BC = CD = 450 mm = 0.45 m
Chapter 8 : Acceleration in Mechanisms
l
189
We know that linear velocity of A with respect to O or velocity of A ,
vAO = v A = ωAO × O A = 2.1 × 0.3 = 0.63 m/s
...(Perpendicular to O A)
1. Velocities of sliding at B and D
First of all, draw the space diagram, to some suitable scale, as shown in Fig. 8.13 (a). Now
the velocity diagram, as shown in Fig. 8.13 (b), is drawn as discussed below:
(a) Space diagram.
(b) Velocity diagram.
(c) Acceleration diagram.
Fig. 8.13
1. Draw vector oa perpendicular to O A, to some suitable scale, to represent the velocity of
A with respect to O (or simply velocity of A ), such that
vector oa = v AO = v A = 0.63 m/s
2. From point a, draw vector ab perpendicular to A B to represent the velocity of B with
respect to A (i.e. vBA) and from point o draw vector ob parallel to path of motion B (which is along
BO) to represent the velocity of B with respect to O (or simply velocity of B). The vectors ab and ob
intersect at b.
3. Divide vector ab at c in the same ratio as C divides A B in the space diagram. In other
words,
BC/CA = bc/ca
4. Now from point c, draw vector cd perpendicular to CD to represent the velocity of D with
respect to C (i.e. vDC) and from point o draw vector od parallel to the path of motion of D (which
along the vertical direction) to represent the velocity of D.
By measurement, we find that velocity of sliding at B,
vB = vector ob = 0.4 m/s Ans.
and velocity of sliding at D,
v D = vector od = 0.24 m/s Ans.
2. Angular velocity of CD
By measurement from velocity diagram, we find that velocity of D with respect to C,
vDC = vector cd = 0.37 m/s
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Theory of Machines
∴ Angular velocity of CD,
ωCD =
vDC 0.37
=
= 0.82 rad/s (Anticlockwise). Ans.
CD 0.45
3. Linear acceleration of D
We know that the radial component of the acceleration of A with respect to O or acceleration
of A,
2
vAO
2
= ωAO
× OA = (2.1)2 × 0.3 = 1.323 m/s2
OA
Radial component of the acceleration of B with respect to A ,
r
= aA =
aAO
r
=
aBA
2
vBA
(0.54)2
=
= 0.243 m/s 2
1.2
AB
...(By measurement, v BA = 0.54 m/s)
Radial component of the acceleration of D with respect to C,
2
vDC
(0.37)2
=
= 0.304 m/s 2
0.45
CD
Now the acceleration diagram, as shown in Fig. 8.13 (c), is drawn as discussed below:
1. Draw vector o' a' parallel to O A, to some suitable scale, to represent the radial component
of the acceleration of A with respect to O or simply the acceleration of A , such that
r
=
aDC
r
vector o′a′ = aAO
= aA = 1.323 m/s 2
2. From point a', draw vector a' x parallel to A B to represent the radial component of the
acceleration of B with respect to A , such that
r
vector a ′x = aBA
= 0.243 m/s 2
3. From point x, draw vector xb' perpendicular to A B to represent the tangential component
t
of the acceleration of B with respect to A (i.e. aBA
) whose magnitude is not yet known.
4. From point o', draw vector o' b' parallel to the path of motion of B (which is along BO) to
represent the acceleration of B (aB). The vectors xb' and o' b' intersect at b'. Join a' b'. The vector
a' b' represents the acceleration of B with respect to A .
5. Divide vector a' b' at c' in the same ratio as C divides A B in the space diagram. In other
words,
BC / B A = b' c'/b' a'
6. From point c', draw vector c'y parallel to CD to represent the radial component of the
acceleration of D with respect to C, such that
r
vector c ′y = aDC
= 0.304 m/s 2
7. From point y, draw yd' perpendicular to CD to represent the tangential component of
t
acceleration of D with respect to C (i.e. aDC
) whose magnitude is not yet known.
8. From point o', draw vector o' d' parallel to the path of motion of D (which is along the
vertical direction) to represent the acceleration of D (aD). The vectors yd' and o' d' intersect at d'.
By measurement, we find that linear acceleration of D,
aD = vector o' d' = 0.16 m/s2
Ans.
4. Angular acceleration of CD
From the acceleration diagram, we find that the tangential component of the acceleration of
D with respect to C,
t
= vector yd ′ = 1.28 m/s 2
aDC
...(By measurement)
Chapter 8 : Acceleration in Mechanisms
l
191
∴ Angular acceleration of CD,
t
aDC
1.28
=
= 2.84 rad/s2 (Clockwise) Ans.
0.45
CD
Example 8.7. Find out the acceleration of the slider D and
the angular acceleration of link CD for the engine mechanism shown
in Fig. 8.14.
α CD =
The crank OA rotates uniformly at 180 r.p.m. in clockwise
direction. The various lengths are: OA = 150 mm ; AB = 450 mm;
PB = 240 mm ; BC = 210 mm ; CD = 660 mm.
Solution. Given: N AO = 180 r.p.m., or ωAO = 2π × 180/60 =
18.85 rad/s ; O A = 150 mm = 0.15 m ; A B = 450 mm = 0.45 m ;
PB = 240 mm = 0.24 m ; CD = 660 mm = 0.66 m
We know that velocity of A with respect to O or velocity
of A ,
v AO = v A = ωAO × O A
= 18.85 × 0.15 = 2.83 m/s
...(Perpendicular to O A)
All dimensions in mm.
Fig. 8.14
First of all draw the space diagram, to some suitable scale,
as shown in Fig. 8.15 (a). Now the velocity diagram, as shown in Fig. 8.15 (b), is drawn as discussed
below:
(a) Space diagram.
(b) Velocity diagram.
(c) Acceleration diagram.
Fig. 8.15
1. Since O and P are fixed points, therefore these points lie at one place in the velocity
diagram. Draw vector oa perpendicular to O A, to some suitable scale, to represent the velocity of A
with respect to O or velocity of A (i.e. vAO or vA), such that
vector oa = v AO = v A = 2.83 m/s
2. Since the point B moves with respect to A and also with respect to P, therefore draw
vector ab perpendicular to A B to represent the velocity of B with respect to A i.e. vBA ,and from point
p draw vector pb perpendicular to PB to represent the velocity of B with respect to P or velocity of
B (i.e. vBP or vB). The vectors ab and pb intersect at b.
3. Since the point C lies on PB produced, therefore divide vector pb at c in the same ratio as
C divides PB in the space diagram. In other words, pb/pc = PB/PC.
192
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Theory of Machines
4. From point c, draw vector cd perpendicular to CD to represent the velocity of D with
respect to C and from point o draw vector od parallel to the path of motion of the slider D (which is
vertical), to represent the velocity of D, i.e. v D.
By measurement, we find that velocity of the slider D,
v D = vector od = 2.36 m/s
Velocity of D with respect to C,
vDC = vector cd = 1.2 m/s
Velocity of B with respect to A ,
vBA = vector ab = 1.8 m/s
and velocity of B with respect to P, v BP = vector pb = 1.5 m/s
Acceleration of the slider D
We know that radial component of the acceleration of A with respect to O or acceleration
of A ,
r
2
= aA = ωAO
× AO = (18.85)2 × 0.15 = 53.3 m/s2
aAO
Radial component of the acceleration of B with respect to A ,
2
vBA
(1.8)2
=
= 7.2 m/s 2
0.45
AB
Radial component of the acceleration of B with respect to P,
r
=
aBA
2
vBP
(1.5) 2
=
= 9.4 m/s 2
0.24
PB
Radial component of the acceleration of D with respect to C,
r
=
aBP
2
vDC
(1.2)2
=
= 2.2 m/s 2
0.66
CD
Now the acceleration diagram, as shown in Fig. 8.15 (c), is drawn as discussed below:
r
=
aDC
1. Since O and P are fixed points, therefore these points lie at one place in the acceleration
diagram. Draw vector o' a' parallel to O A, to some suitable scale, to represent the radial component
r
of the acceleration of A with respect to O or the acceleration of A (i.e. aAO
or aA), such that
r
vector o′a′ = aAO
= aA = 53.3 m/s 2
2. From point a', draw vector a' x parallel to A B to represent the radial component of the
r
acceleration of B with respect to A (i.e. aBA
), such that
r
vector a′x = aBA
= 7.2 m/s 2
3. From point x, draw vector xb' perpendicular to the vector a'x or AB to represent the
t
tangential component of the acceleration of B with respect to A i.e. aBA
whose magnitude is yet
unknown.
4. Now from point p', draw vector p' y parallel to PB to represent the radial component of
r
the acceleration of B with respect to P (i.e. aBP
), such that
r
vector p′y = aBP
= 9.4 m/s 2
5. From point y, draw vector yb' perpendicular to vector b'y or PB to represent the tangential
t
component of the acceleration of B, i.e. aBP
. The vectors xb' and yb' intersect at b'. Join p' b'. The
vector p' b' represents the acceleration of B, i.e. aB.
Chapter 8 : Acceleration in Mechanisms
l
193
6. Since the point C lies on PB produced, therefore divide vector p'b' at c' in the same ratio
as C divides PB in the space diagram. In other words, p'b'/p'c' = PB/PC
7. From point c', draw vector c'z parallel to CD to represent the radial component of the
r
acceleration of D with respect to C i.e. aDC
, such that
r
vector c′z = aDC
= 2.2 m/s 2
8. From point z, draw vector zd' perpendicular to vector c'z or CD to represent the tangential
t
component of the acceleration of D with respect to C i.e. aDC
, whose magnitude is yet unknown.
9. From point o', draw vector o' d' parallel to the path of motion of D (which is vertical) to
represent the acceleration of D, i.e. aD. The vectors zd' and o' d' intersect at d'. Join c' d'.
By measurement, we find that acceleration of D,
aD = vector o'd' = 69.6 m/s2 Ans.
Angular acceleration of CD
From acceleration diagram, we find that tangential component of the acceleration of D with
respect to C,
t
= vector zd ′ = 17.4 m/s 2
aDC
...(By measurement)
We know that angular acceleration of CD,
t
aDC
17.4
=
= 26.3 rad / s 2 (Anticlockwise) Ans.
CD 0.66
Example 8.8. In the toggle mechanism shown in Fig. 8.16, the slider D is constrained to
move on a horizontal path. The crank OA is rotating in the counter-clockwise direction at a speed
α CD =
Fig. 8.16
of 180 r.p.m. increasing at the rate of 50 rad/s2. The dimensions of the various links are as follows:
OA = 180 mm ; CB = 240 mm ; AB = 360 mm ; and BD = 540 mm.
For the given configuration, find 1. Velocity of slider D and angular velocity of BD, and
2. Acceleration of slider D and angular acceleration of BD.
Solution. Given : NAO = 180 r.p.m. or ωAO = 2 π × 180/60 = 18.85 rad/s ; O A = 180 mm
= 0.18 m ; CB = 240 mm = 0.24 m ; A B = 360 mm = 0.36 m ; BD = 540 mm = 0.54 m
We know that velocity of A with respect to O or velocity of A ,
vAO = v A = ωAO × O A = 18.85 × 0.18 = 3.4 m/s
...(Perpendicular to O A)
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Theory of Machines
1. Velocity of slider D and angular velocity of BD
First of all, draw the space diagram to some suitable scale, as shown in Fig. 8.17 (a). Now
the velocity diagram, as shown in Fig. 8.17 (b), is drawn as discussed below:
1. Since O and C are fixed points, therefore these points lie at one place in the velocity
diagram. Draw vector oa perpendicular to O A, to some suitable scale, to represent the velocity of A
with respect to O or velocity of A i.e. v AO or v A, such that
vector oa = v AO = v A = 3.4 m/s
(a) Space diagram.
(b) Velocity diagram.
(c) Acceleration diagram.
Fig. 8.17
2. Since B moves with respect to A and also with respect to C, therefore draw vector ab
perpendicular to A B to represent the velocity of B with respect to A i.e. vBA, and draw vector cb
perpendicular to CB to represent the velocity of B with respect to C ie. vBC. The vectors ab and cb
intersect at b.
3. From point b, draw vector bd perpendicular to BD to represent the velocity of D with
respect to B i.e. v DB, and from point c draw vector cd parallel to CD (i.e., in the direction of motion
of the slider D) to represent the velocity of D i.e. v D.
By measurement, we find that velocity of B with respect to A ,
v BA = vector ab = 0.9 m/s
Velocity of B with respect to C,
v BC = vector cb = 2.8 m/s
Velocity of D with respect to B,
v DB = vector bd = 2.4 m/s
and velocity of slider D,
vD = vector cd = 2.05 m/s Ans.
Angular velocity of BD
We know that the angular velocity of BD,
vDB
2.4
=
= 4.5 rad/s Ans.
BD 0.54
2. Acceleration of slider D and angular acceleration of BD
Since the angular acceleration of OA increases at the rate of 50 rad/s2, i.e. αAO = 50 rad/s2,
therefore
Tangential component of the acceleration of A with respect to O,
ωBD =
t
= αAO × OA = 50 × 0.18 = 9 m/s 2
aAO
Chapter 8 : Acceleration in Mechanisms
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195
Radial component of the acceleration of A with respect to O,
2
vAO
(3.4)2
=
= 63.9 m/s 2
0.18
OA
Radial component of the acceleration of B with respect to A,
r
=
aAO
2
vBA
(0.9)2
=
= 2.25 m/s 2
0.36
AB
Radial component of the acceleration of B with respect to C,
r
=
aBA
2
vBC
(2.8)2
=
= 32.5 m/s 2
0.24
CB
and radial component of the acceleration of D with respect to B,
r
=
aBC
2
vDB
(2.4)2
=
= 10.8 m/s 2
0.54
BD
Now the acceleration diagram, as shown in Fig. 8.17 (c), is drawn as discussed below:
1. Since O and C are fixed points, therefore
these points lie at one place in the acceleration
diagram. Draw vector o'x parallel to O A, to some
suitable scale, to represent the radial component
of the acceleration of A with respect to O i.e.
r
, such that
aAO
r
=
aDB
r
vector o′x = aAO
= 63.9 m/s 2
2. From point x , draw vector xa'
perpendicular to vector o'x or O A to represent the
tangential component of the acceleration of A with
t
respect to O i.e. aAO
,such that
An experimental IC engine with crank shaft
and cylinders.
vector x a ′ =
= 9 m/s
Note : This picture is given as additional informa3. Join o'a'. The vector o'a' represents the
tion and is not a direct example of the current
chapter.
total acceleration of A with respect to O or
acceleration of A i.e. aAO or aA.
4. Now from point a', draw vector a'y parallel to A B to represent the radial component of the
r
acceleration of B with respect to A i.e. aBA
, such that
t
aAO
2
r
vector a′y = aBA
= 2.25 m/s 2
5. From point y, draw vector yb' perpendicular to vector a'y or A B to represent the tangential
t
component of the acceleration of B with respect to A i.e. aBA
whose magnitude is yet unknown.
6. Now from point c', draw vector c'z parallel to CB to represent the radial component of the
r
acceleration of B with respect to C i.e. aBC
, such that
r
vector c ′z = aBC
= 32.5 m/s 2
7. From point z, draw vector zb' perpendicular to vector c'z or CB to represent the tangential
t
component of the acceleration of B with respect to C i.e. aBC
. The vectors yb' and zb' intersect at b'.
Join c' b'. The vector c' b' represents the acceleration of B with respect to C i.e. aBC.
8. Now from point b', draw vector b's parallel to BD to represent the radial component of the
r
acceleration of D with respect to B i.e. aDB
, such that
r
vector b′s = aDB
= 10.8 m/s 2
196
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Theory of Machines
9. From point s, draw vector sd' perpendicular to vector b's or BD to represent the tangential
t
component of the acceleration of D with respect to B i.e. aDB
whose magnitude is yet unknown.
10. From point c', draw vector c'd' parallel to the path of motion of D (which is along CD)
to represent the acceleration of D i.e. aD. The vectors sd' and c'd' intersect at d'.
By measurement, we find that acceleration of slider D,
aD = vector c'd' = 13.3 m/s2 Ans.
Angular acceleration of BD
By measurement, we find that tangential component of the acceleration of D with respect
to B,
t
= vector sd ′ = 38.5 m/s 2
aDB
We know that angular acceleration of BD,
t
aDB
38.5
=
= 71.3 rad/s2 (Clockwise) Ans.
BD 0.54
Example 8.9. The mechanism of a warping machine, as shown in Fig. 8.18, has the
dimensions as follows:
αBD =
O1A = 100 mm; AC = 700 mm ; BC = 200 mm ; BD = 150 mm ; O2D = 200 mm ; O2E = 400
mm ; O3C = 200 mm.
Fig. 8.18
The crank O1A rotates at a uniform speed of 100 rad/s. For the given configuration,
determine: 1. linear velocity of the point E on the bell crank lever, 2. acceleration of the points E
and B, and 3. angular acceleration of the bell crank lever.
Solution. Given : ωAO1 = 100 rad/s ; O1A = 100 mm = 0.1 m
We know that linear velocity of A with respect to O1, or velocity of A ,
...(Perpendicular to O1A )
vAO1 = v A = ω AO1 × O1A = 100 × 0.1 = 10 m/s
1. Linear velocity of the point E on bell crank lever
First of all draw the space diagram, as shown in Fig. 8.19 (a), to some suitable scale. Now
the velocity diagram, as shown in Fig. 8.19 (b), is drawn as discussed below:
1. Since O1, O2 and O3 are fixed points, therefore these points are marked as one point in the
velocity diagram. From point o1, draw vector o1a perpendicular to O1A to some suitable scale, to
represent the velocity of A with respect to O or velocity of A , such that
vector o1a = v AO1 = v A = 10 m/s
Chapter 8 : Acceleration in Mechanisms
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197
2. From point a, draw vector ac perpendicular to A C to represent the velocity of C with
respect to A (i.e. vCA) and from point o3 draw vector o3c perpendicular to O3C to represent the
velocity of C with respect to O3 or simply velocity of C (i.e. vC). The vectors ac and o3c intersect at
point c.
(a) Space diagram.
(b) Velocity diagram.
(c) Acceleration diagram.
Fig. 8.19
3. Since B lies on A C, therefore divide
vector ac at b in the same ratio as B divides A C in
the space diagram. In other words, ab/ac = AB/AC
4. From point b, draw vector bd
perpendicular to BD to represent the velocity of D
with respect to B (i.e. vDB), and from point o2 draw
vector o2d perpendicular to O2D to represent the
velocity of D with respect to O2 or simply velocity
of D (i.e. v D). The vectors bd and o2d intersect at d.
5. From point o 2 , draw vector o 2 e
perpendicular to vector o2d in such a way that
o2e/o2d = O2E/O2D
By measurement, we find that velocity of
point C with respect to A ,
v CA = vector ac = 7 m/s
Velocity of point C with respect to O3,
vCO3 = v C = vector o3c = 10 m/s
Velocity of point D with respect to B,
vDB = vector bd = 10.2 m/s
Warping machine uses many mechanisms.
198
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Theory of Machines
Velocity of point D with respect to O2,
v DO2 = v D = vector o2d = 2.8 m/s
and velocity of the point E on the bell crank lever,
v E = v EO2 = vector o2e = 5.8 m/s Ans.
2. Acceleration of the points E and B
Radial component of the acceleration of A with respect to O1 (or acceleration of A ),
2
vAO1
102
=
= 1000 m/s 2
0.1
O1 A
Radial component of the acceleration of C with respect to A ,
r
aAO
2 = aAO1 = aA =
2
vCA
72
=
= 70 m/s 2
AC 0.7
Radial component of the acceleration of C with respect to O3,
r
=
aCA
2
vCO
3
10 2
= 500 m/s 2
0.2
O3C
Radial component of the acceleration of D with respect to B,
r
aCO
3 =
=
2
vDB
(10.2)2
=
= 693.6 m/s 2
0.15
BD
Radial component of the acceleration of D with respect to O2,
vD2 O 2 (2.8) 2
r
aDO
=
=
= 39.2 m/s2
2
O2 D
0.2
Radial component of the acceleration of E with respect to O2,
r
=
aDB
r
=
aEO
2
vE2 O 2
O2 E
=
(5.8) 2
= 84.1 m/s 2
0.4
Now the acceleration diagram, as shown in Fig. 8.19 (c), is drawn as discussed below:
1. Since O1, O2 and O3 are fixed points, therefore these points are marked as one point in the
acceleration diagram. Draw vector o1' a' parallel to O1A , to some suitable scale, to represent the
radial component of the acceleration of A with respect to O1 (or simply acceleration of A ), such that
r
vector o1′ a ′ = aAO
= aA = 1000 m/s 2
1
2. From point a', draw a'x parallel to AC to represent the radial component of the acceleration
r
of C with respect to A (i.e. aCA
), such that
r
vector a′x = aCA
= 70 m/s 2
3. From point x, draw vector xc' perpendicular to A C to represent the tangential component
t
of the acceleration of C with respect to A (i.e. aCA
), the magnitude of which is yet unknown.
4. From point o3', draw vector o3' y parallel to O3C to represent the radial component of the
r
acceleration of C with respect to O3 (i.e. aCO
3 ), such that
r
vector o′ y = aCO3
= 500 m/s2
3
5. From point y, draw vector yc' perpendicular to O3C to represent the tangential component
t
of the acceleration of C with respect to O3 (i.e. aCO3
). The vectors xc' and yc' intersect at c'.
Chapter 8 : Acceleration in Mechanisms
l
199
6. Join a' c'. The vector a' c' represents the acceleration of C with respect to A (i.e. aCA).
7. Since B lies on A C, therefore divide vector a'c' at b' in the same ratio as B divides AC in
the space diagram. In other words, a'b'/a'c' = AB/AC. Join b' o2' which represents the acceleration of
point B with respect to O2 or simply acceleration of B. By measurement, we find that
Acceleration of point B = vector o2' b' = 440 m/s2 Ans.
8. Now from point b', draw vector b' z parallel to BD to represent the radial component of
r
the acceleration of D with respect to B (i.e. aDB
), such that
r
vector b′z = aDB
= 693.6 m/s 2
9. From point z, draw vector zd' perpendicular to BD to represent the tangential component
t
of the acceleration of D with respect to B (i.e. aDB
), whose magnitude is yet unknown.
10. From point o2' , draw vector o2' z1 parallel to O2D to represent the radial component of
r
the acceleration of D with respect to O2 (i.e. aDO 2 ), such that
r
2
vector o2′ z1 = aDO
2 = 39.2 m/s
11. From point z 1, draw vector z 1d' perpendicular to O2D to represent the tangential component
t
of the acceleration of D with respect to O2 (i.e. aDO 2 ). The vectors zd' and z 1d' intersect at d'.
12. Join o2' d'. The vector o2'd' represents the acceleration of D with respect to O2 or simply
acceleration of D (i.e. aDO2 or aD).
13. From point o2', draw vector o2' e' perpendicular to o2' d' in such a way that
o2′e′ / o2′ d ′ = O2 E / O2 D
r
t
Note: The point e' may also be obtained drawing aEO 2 and aEO 2 as shown in Fig. 8.19 (c).
By measurement, we find that acceleration of point E,
aE = aEO2 = vector o' 2 e' = 1200 m/s2 Ans.
3. Angular acceleration of the bell crank lever
By measurement, we find that the tangential component of the acceleration of D with respect
to O2,
aDt O 2 = vector z1 d1′ = 610 m/s 2
∴ Angular acceleration of the bell crank lever
=
t
aDO2
610
=
= 3050 rad/s 2 (Anticlockwise) Ans.
O2 D
0.2
Example 8.10. A pump is driven from an engine
crank-shaft by the mechanism as shown in Fig. 8.20. The
pump piston shown at F is 250 mm in diameter and the
crank speed is 100 r.p.m. The dimensions of various links
are as follows:
OA = 150 mm ; AB = 600 mm ; BC = 350 mm ;
CD = 150 mm; and DE = 500 mm.
Determine for the position shown : 1. The velocity of
the cross-head E, 2. The rubbing velocity of the pins A
and B which are 50 mm diameter. 3. The torque required
at the crank shaft to overcome a presure of 0.35 N/mm2,
and 4. The acceleration of the cross-head E.
All dimensions in mm.
Fig. 8.20
200
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Theory of Machines
Solution. Given : NAO = 100 r.p.m. or ωAO = 2 π × 100/60 = 10.47 rad/s; OA = 150 mm = 0.15 m ;
A B = 600 mm = 0.6 m ; BC = 350 mm = 0.35 m ; CD = 150 mm = 0.15 m ; DE = 500 mm = 0.5 m
We know that velocity of A with respect to O or velocity of A ,
v AO = v A = ωAO × O A = 10.47 × 0.15 = 1.57 m/s
...(Perpendicular to O A)
1. Velocity of the cross-head E
First of all, draw the space diagram, to some suitable scale, as shown in Fig. 8.21 (a). Now
the velocity diagram, as shown in Fig. 8.21 (b), is drawn as discussed below:
(a) Space diagram.
(b) Velocity diagram.
(c) Acceleration diagram.
Fig. 8.21
1. Since O and C are fixed points, therefore these points are marked as one point in the
velocity diagram. Now draw vector oa perpendicular to O A, to some suitable scale, to represent the
velocity of A with respect ot O or the velocity of A , such that
vector oa = v AO = v A = 1.57 m/s
2. From point a, draw vector ab perpendicular to A B to represent the velocity of B with
respect to A (i.e. vBA), and from point c draw vector cb perpendicular to CB to represent the velocity
of B with respect to C (i.e. vBC). The vectors ab and cb intersect at b.
By measurement, we find that
vBA = vector ab = 1.65 m/s
and
v BC = v B = vector cb = 0.93 m/s
3. From point c, draw vector cd perpendicular to CD or vector cb to represent the velocity of
D with respect to C or velocity of D, such that
vector cd : vector cb = CD: CB
or
vDC : v BC = CD : CB
∴
vDC CD
0.15
CD
or vDC = vBC ×
=
= 0.93 ×
= 0.4 m/s
0.35
vBC
CB
CB
4. From point d, draw vector de perpendicular to DE to represent the velocity of E with
respect to D (i.e. vED), and from point o draw vector oe parallel to the path of motion of E (which is
vertical) to represent the velocity of E or F. The vectors oe and de intersect at e.
By measurement, we find that velocity of E with respect to D,
vED = vector de = 0.18 m/s
Chapter 8 : Acceleration in Mechanisms
l
201
and velocity of the cross-head E,
vEO = v E = vector oe = 0.36 m/s Ans.
2. Rubbing velocity of the pins at A and B
We know that angular velocity of A with respect to O,
ωAO = 10.47 rad/s
...(Anticlockwise)
Angular velocity of B with respect to A ,
vBA 1.65
=
= 2.75 rad/s
0.6
AB
and angular velocity of B with respect to C,
ωBA =
...(Anticlockwise)
vBC
0.93
=
= 2.66 rad/s
CB 0.35
We know that diameter of pins at A and B,
ωBC =
...(clockwise)
dA = dB = 50 mm = 0.05 m
or
...(Given)
Radius, rA = rB = 0.025 m
∴ Rubbing velocity of pin at A
= (ωAO – ωBA) rA = (10.47 – 2.75) 0.025 = 0.193 m/s Ans.
and rubbing velocity of pin at B
= (ωBA + ωBC) rB = (2.75 + 2.66) 0.025 = 0.135 m/s Ans.
3. Torque required at the crankshaft
Given: Pressure to overcome by the crankshaft,
pF = 0.35 N/mm2
Diameter of the pump piston
DF = 250 mm
∴ Force at the pump piston at F,
π
π
( DF ) 2 = 0.35 × (250)2 = 17 183 N
4
4
FA = Force required at the crankshaft at A .
FF = Pressure × Area = pF ×
Let
Assuming transmission efficiency as 100 per cent,
Work done at A = Work done at F
FA × vA = FF × vF or FA =
FF × vF 17 183 × 0.36
=
= 3940 N
1.57
vA
...(3 vF = vE )
∴ Torque required at the crankshaft,
T A = FA × O A = 3940 × 0.15 = 591 N-m Ans.
Acceleration of the crosshead E
We know that the radial component of the acceleration of A with respect to O or the
acceleration of A ,
v2
(1.57)2
r
= aA = AO =
= 16.43 m/s2
aAO
0.15
OA
202
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Theory of Machines
Radial component of the acceleration of B with respect to A ,
2
vBA
(1.65)2
=
= 4.54 m / s 2
0.6
AB
Radial component of the acceleration of B with respect to C.
r
=
aBA
2
vBC
(0.93)2
=
= 2.47 m/s 2
0.35
CB
and radial component of the acceleration of E with respect to D,
r
=
aBC
2
vED
(0.18)2
=
= 0.065 m/s 2
0.5
DE
Now the acceleration diagram, as shown in Fig. 8.21 (c), is drawn as discussed below:
1. Since O and C are fixed points, therefore these points are marked as one point in the
acceleration diagram. Draw vector o'a' parallel to O A, to some suitable scale, to represent the radial
component of the acceleration of A with respect to O or the acceleration of A , such that
r
=
aED
r
vector o′a′ = aAO
= aA = 16.43 m/s 2
2. From point a', draw vector a'x parallel to A B to represent the radial component of the
r
acceleration of B with respect to A (i.e. aBA
), such that
r
vector a ′x = aBA
= 4.54 m/s 2
3. From point x, draw vector xb' perpendicular to A B to represent the tangential component
t
of the acceleration of B with respect to A (i.e. aBA
) whose magnitude is yet unknown.
4. Now from point c', draw vector c' y parallel to CB to represent the radial component of
r
the acceleration of B with respect to C (i.e. aBC
), such that
r
vector c′y = aBC
= 2.47 m/s 2
5. From point y, draw vector yb' perpendicular to CB to represent the tangential component
t
of the acceleration of B with respect to C (i.e. aBC
). The vectors yb' and xb' intersect at b'. Join c'b'
and a'b'. The vector c'b' represents the acceleration of B with respect to C (i.e. aBC) or the acceleration
of B (i.e. aB) and vector a'b' represents the acceleration of B with respect to A (i.e. aBA).
By measurement, we find that
aBC = aB = vector c'b' = 9.2 m/s2
and
aBA = vector a'b' = 9 m/s2
6. From point c', draw vector c'd' perpendicular to CD or vector c'b' to represent the
acceleration of D with respect to C or the acceleration of D (i.e. aDC or aD), such that
vector c'd' : vector c'b' = CD : CB or aD : aBC = CD : CB
0.15
aD
CD
CD
or aD = aBC ×
=
= 9.2 ×
= 3.94 m/s 2
∴
0.35
aBC CB
CB
7. Now from point d', draw vector d'z parallel to DE to represent the radial component of E
r
with respect to D (i.e. aED
), such that
r
vector d ′z = aED
= 0.065 m/s 2
Note: Since the magnitude of arED is very small, therefore the points d' and z coincide.
8. From point z, draw vector ze' perpendicular to DE to represent the tangential component
t
of the acceleration of E with respect to D (i.e. aED
) whose magnitude is yet unknown.
9. From point o', draw vector o'e' parallel to the path of motion of E (which is vertical) to
represent the acceleration of E. The vectors ze' and o'e' intersect at e'.
Chapter 8 : Acceleration in Mechanisms
l
203
By measurement, we find that acceleration of the crosshead E,
aE = vector o'e' = 3.8 m/s2 Ans.
Example 8.11. Fig. 8.22 shows the mechanism of a radial valve gear. The crank OA turns
uniformly at 150 r.p.m and is pinned at A to rod AB. The point C in the rod is guided in the circular
path with D as centre and DC as radius. The dimensions of various links are:
OA = 150 mm ; AB = 550 mm ; AC = 450 mm ; DC = 500 mm ; BE = 350 mm.
Determine velocity and acceleration of the ram E for the given position of the mechanism.
All dimensions in mm.
Fig. 8.22
Solution. Given : NAO = 150 r.p.m. or ωAO = 2 π × 150/60 = 15.71 rad/s; OA = 150 mm = 0.15 m;
A B = 550 mm = 0.55 m ; AC = 450 mm = 0.45 m ; DC = 500 mm = 0.5 m ; BE = 350 mm = 0.35 m
We know that linear velocity of A with respect to O or velocity of A ,
v AO = v A = ωAO × O A = 15.71 × 0.15 = 2.36 m/s
...(Perpendicular to O A)
Velocity of the ram E
First of all draw the space diagram, as shown in Fig. 8.23 (a), to some suitable scale. Now
the velocity diagram, as shown in Fig. 8.23 (b), is drawn as discussed below:
1. Since O and D are fixed points, therefore these points are marked as one point in the
velocity diagram. Draw vector oa perpendicular to O A, to some suitable scale, to represent the
velocity of A with respect to O or simply velocity of A , such that
vector oa = vAO = vA = 2.36 m/s
2. From point a, draw vector ac perpendicular to A C to represent the velocity of C with
respect to A (i.e. vCA), and from point d draw vector dc perpendicular to DC to represent the velocity
of C with respect to D or simply velocity of C (i.e. vCD or vC). The vectors ac and dc intersect at c.
3. Since the point B lies on A C produced, therefore divide vector ac at b in the same ratio as
B divides A C in the space diagram. In other words ac:cb = AC:CB. Join ob. The vector ob represents
the velocity of B (i.e. vB)
4. From point b, draw vector be perpendicular to be to represent the velocity of E with
respect to B (i.e. vEB), and from point o draw vector oe parallel to the path of motion of the ram E
(which is horizontal) to represent the velocity of the ram E. The vectors be and oe intersect at e.
By measurement, we find that velocity of C with respect to A ,
v CA = vector ac = 0.53 m/s
Velocity of C with respect to D,
v CD = v C = vector dc = 1.7 m/s
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Theory of Machines
Velocity of E with respect to B,
v EB = vector be = 1.93 m/s
and velocity of the ram E, vE = vector oe = 1.05 m/s Ans.
Acceleration of the ram E
We know that the radial component of the acceleration of A with respect to O or the
acceleration of A ,
2
vAO
(2.36)2
=
= 37.13 m/s 2
0.15
OA
Radial component of the acceleration of C with respect to A ,
v2
(0.53) 2
r
= CA =
= 0.624 m/s 2
aCA
0.45
OA
Radial component of the acceleration of C with respect to D,
v2
(1.7)2
r
= CD =
= 5.78 m/s 2
aCD
0.5
DC
Radial component of the acceleration of E with respect to B,
r
= aA =
aAO
2
vEB
(1.93)2
=
= 10.64 m/s 2
0.35
BE
The acceleration diagram, as shown in Fig. 8.23 (c), is drawn as discussed below:
r
=
aEB
(c) Acceleration diagram.
Fig. 8.23
Chapter 8 : Acceleration in Mechanisms
l
205
1. Since O and D are fixed points, therefore these points are marked as one point in the
acceleration diagram. Draw vector o'a' parallel to O A, to some suitable scale, to represent the radial
component of the acceleration of A with respect to O or simply the acceleration of A, such that
r
vector o′a′ = aAO
= aA = 37.13 m/s 2
2. From point d', draw vector d'x parallel to DC to represent the radial component of the
acceleration of C with respect to D, such that
r
vector d ′x = aCD
= 5.78 m/s 2
3. From point x, draw vector xc' perpendicular to DC to represent the tangential component
t
) whose magnitude is yet unknown.
of the acceleration of C with respect to D (i.e. aCD
4. Now from point a', draw vector a'y parallel to A C to represent the radial component of
the acceleration of C with respect to A , such that
r
vector a ′y = aCA
= 0.624 m/s 2
5. From point y, draw vector yc' perpendicular to AC to represent the tangential component
t
). The vectors xc' and yc' intersect at c'.
of acceleration of C with respect to A (i.e. aCA
6. Join a'c'. The
vector a'c' represents the
acceleration of C with
respect to A (i.e. aCA).
7. Since the point
B lies on A C produced,
therefore divide vector a'c'
at b' in the same ratio as B
divides A C in the space
diagram. In other words, a'
c' : c' b' = A C : CB.
8. From point b',
draw vector b' z parallel to
BE to represent the radial A lathe is a machine for shaping a piece of metal, by rotating it rapidly along
component
of
the
its axis while pressing against a fixed cutting or abrading tool.
acceleration of E with Note : This picture is given as additional information and is not a direct
example of the current chapter.
respect to B, such that
r
vector b′z = aEB
= 10.64 m/s 2
9. From point z, draw vector ze' perpendicular to BE to represent the tangential component
t
of the acceleration of E with respect to B (i.e. aEB
) whose magnitude is yet unknown.
10. From point o', draw vector o'e' parallel to the path of motion of E (which is horizontal)
to represent the acceleration of the ram E. The vectors ze' and o'e' intersect at e'.
By measurement, we find that the acceleration of the ram E,
aE = vector o′e′ = 3.1 m/s 2 Ans.
Example 8.12. The dimensions of the Andreau differential stroke engine mechanism, as
shown in Fig. 8.24, are as follows:
AB = 80 mm ; CD = 40 mm ; BE = DE = 150 mm ; and EP = 200 mm.
The links AB and CD are geared together. The speed of the smaller wheel is 1140 r.p.m.
Determine the velocity and acceleration of the piston P for the given configuration.
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Theory of Machines
Solution. Given: N DC = 1140 r.p.m. or ωDC = 2 π × 1140/60 = 119.4 rad/s ; A B = 80 mm
= 0.08 m ; CD = 40 mm = 0.04 m ; BE = DE = 150 mm = 0.15 m ; EP = 200 mm = 0.2 m
Fig. 8.24
We know that velocity of D with respect to C or velocity of D,
vDC = v D = ωDC × CD = 119.4 × 0.04 = 4.77 m/s
...(Perpendicular to CD)
Since the speeds of the gear wheels are inversely proportional to their diameters, therefore
Angular speed of larger wheel
2CD
ω
= BA =
Angular speed of smaller wheel ωDC
2 AB
∴ Angular speed of larger wheel,
0.04
CD
ωBA = ωDC ×
= 119.4 ×
= 59.7 rad/s
0.08
AB
and velocity of B with respect to A or velocity of B,
vBA = vB = ωBA × AB = 59.7 × 0.08 = 4.77 m/s
...(Perpendicular to A B)
Velocity of the piston P
First of all draw the space diagram, to some suitable scale, as shown in Fig. 8.25 (a). Now
the velocity diagram, as shown in Fig. 8.25 (b), is drawn as discussed below:
1. Since A and C are fixed points, therefore these points are marked as one point in the
velocity diagram. Draw vector cd perpendicular to CD, to some suitable scale, to represent the
velocity of D with respect to C or velocity of D (i.e. vDC or v D), such that
vector cd = v DC = v D = 4.77 m/s
2. Draw vector ab perpendicular to A B to represent the velocity of B with respect to A or
velocity of B (i.e. vBA or v B), such that
vector ab = v BA = v B = 4.77 m/s
3. Now from point b, draw vector be perpendicular to BE to represent the velocity of E with
respect to B (i.e. vEB), and from point d draw vector de perpendicular to DE to represent the velocity
of E with respect to D (i.e. vED). The vectors be and de intersect at e.
4. From point e, draw vector ep perpendicular to EP to represent the velocity of P with
respect to E (i.e. vPE), and from point a draw vector ap parallel to the path of motion of P (which is
horizontal) to represent the velocity of P. The vectors ep and ap intersect at p.
Chapter 8 : Acceleration in Mechanisms
l
By measurement, we find that velocity of E with respect to B,
vEB = vector be = 8.1 m/s
Velocity of E with respect to D,
v ED = vector de = 0.15 m/s
Velocity of P with respect to E,
v PE = vector ep = 4.7 m/s
and
velocity of P, v P = vector ap = 0.35 m/s Ans.
(a) Space diagram.
(b) Velocity diagram.
Fig. 8.25
Acceleration of the piston P
We know that the radial component of the
acceleration of B with respect A (or the acceleration
of B),
r
= aB =
aBA
2
vBA
(4.77)2
=
= 284.4 m/s2
0.08
AB
Radial component of the acceleration of D
with respect to C (or the acceleration of D),
2
vDC
(4.77)2
=
= 568.8 m/s 2
0.04
CD
Radial component of the acceleration of E with
respect to B,
r
= aD =
aDC
(c) Acceleration diagram.
v2
(8.1)2
= EB =
= 437.4 m/s 2
0.15
BE
Radial component of the acceleration of E with respect to D,
r
aEB
2
vED
(0.15)2
=
= 0.15 m/s 2
0.15
DE
and radial component of the acceleration of P with respect to E,
r
=
aED
r
=
aPE
2
vPE
(4.7)2
=
= 110.45 m/s 2
0.2
EP
Fig. 8.25
207
208
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Theory of Machines
Now the acceleration diagram, as shown in Fig. 8.25 (c), is drawn as discussed below:
1. Since A and C are fixed points, therefore these points are marked as one point in the
acceleration diagram. Draw vector a'b' parallel to A B, to some suitable scale, to represent the radial
component of the acceleration of B with respect to A or the acceleration of B, such that
r
vector a′b′ = aBA
= aB = 284.4 m/s2
2. Draw vector c'd' parallel to CD to represent the radial component of the acceleration of D
with respect to C or the acceleration of D, such that
r
vector c′d ′ = aDC
= aD = 568.8 m/s 2
3. Now from point b', draw vector b'x parallel to BE to represent the radial component of the
acceleration of E with respect to B, such that
r
vector b′x ′ = aEB
= 437.4 m/s 2
4. From point x, draw vector xe' perpendicular to BE to represent the tangential component
t
) whose magnitude is yet unknown.
of acceleration of E with respect to B (i.e. aEB
5. From point d', draw vector d'y parallel to DE to represent the radial component of the
acceleration of E with respect to D, such that
r
vector d ′y = aED
= 0.15 m/s 2
r
Note: Since the magnitude of aED
is very small (i.e. 0.15 m/s2), therefore the points d' and y coincide.
6. From point y, draw vector ye' perpendicular to DE to represent the tangential component
t
of the acceleration of E with respect to D (i.e. aED
). The vectors xe' and ye' intersect at e'.
7. From point e', draw vector e'z parallel to EP to represent the radial component of the
acceleration of P with respect to E, such that
r
vector e′z = aPE
= 110.45 m/s 2
8. From point z, draw vector zp' perpendicular to EP to represent the tangential component
t
of the acceleration of P with respect to E (i.e. aPE
) whose magnitude is yet unknown.
9. From point a', draw vector a'p' parallel to the path of motion of P (which is horizontal) to
represent the acceleration of P. The vectors zp' and a'p' intersect at p'.
By measurement, we find that acceleration of the piston P,
aP = vector a'p' = 655 m/s2 Ans.
8.5.
Coriolis Component of Acceleration
When a point on one link is sliding along another rotating link, such as in quick return
motion mechanism, then the coriolis component of the acceleration must be calculated.
Consider a link O A and a slider B as shown in Fig. 8.26 (a). The slider B moves along the
link O A. The point C is the coincident point on the link O A.
Let
ω = Angular velocity of the link OA at time t seconds.
v = Velocity of the slider B along the link OA at time t seconds.
ω.r = Velocity of the slider B with respect to O (perpendicular to the link O A)
at time t seconds, and
Chapter 8 : Acceleration in Mechanisms
l
209
(ω + δω), (v + δv) and (ω + δω) (r + δr)
= Corresponding values at time (t + δt) seconds.
Fig. 8.26. Coriolis component of acceleration.
Let us now find out the acceleration of the slider B
with respect to O and with respect to its coincident point C
lying on the link O A.
Fig. 8.26 (b) shows the velocity diagram when their
velocities v and (v + δv) are considered. In this diagram, the
vector bb1 represents the change in velocity in time δt sec ; the
vector bx represents the component of change of velocity bb1
along O A (i.e. along radial direction) and vector xb1 represents
the component of change of velocity bb 1 in a direction
perpendicular to O A (i.e. in tangential direction). Therefore
bx = ox − ob = (v + δv ) cos δθ − v ↑
Since δθ is very small, therefore substituting
cos δθ = 1, we have
bx = (v + δv − v ) ↑ = δv ↑
...(Acting radially outwards)
xb1 = ( v + δv) sin δθ
Since δθ is very small, therefore substituting sin δθ =
δθ, we have
and
xb1 = ( v + δv) δθ = v.δθ + δv.δθ
Neglecting δv.δθ being very small, therefore
←
xb1 = v. δθ
A drill press has a pointed tool
which is used for boring holes in
hard materials usually by rotating
abrasion or repeated bolows.
Note : This picture is given as additional
information and is not a direct example
of the current chapter.
...(Perpendicular to O A and towards left)
Fig. 8.26 (c) shows the velocity diagram when the velocities ω.r and (ω + δω) (r + δr) are
considered. In this diagram, vector bb1 represents the change in velocity ; vector yb1 represents the
component of change of velocity bb1 along O A (i.e. along radial direction) and vector by represents
the component of change of velocity bb1 in a direction perpendicular to O A (i.e. in a tangential
direction). Therefore
yb1 = (ω + δω) (r + δr ) sin δθ ↓
= (ω.r + ω.δr + δω.r + δω.δr ) sin δθ
210
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Theory of Machines
Since δθ is very small, therefore substituting sin δθ = δθ in the above expression, we have
yb1 = ω.r.δθ + ω.δr.δθ + δω.r.δθ + δω.δr.δθ
= ω.r.δθ ↓ , acting radially inwards
and
...(Neglecting all other quantities)
by = oy – ob = (ω + δω) (r + δr) cos δθ – ω.r
= (ω.r + ω.δr + δω.r + δω.δr) cos δθ – ω.r
Since δθ is small, therefore substituting cos δθ = 1, we have
by = ω.r + ω.δr + δω.r + δω.δr – ω.r= ω.δr + r.δω
...(Neglecting δω.δr)
...(Perpendicular to O A and towards left)
Therefore, total component of change of velocity along radial direction
= bx − yb1 = (δv − ω.r.δθ) ↑
...(Acting radially outwards from O to A )
∴ Radial component of the acceleration of the slider B with respect to O on the link O A,
acting radially outwards from O to A ,
r
= Lt
aBO
δv − ω.r.δθ dv
d θ dv
=
− ω.r ×
=
− ω2 .r ↑
δt
dt
dt
dt
...(i)
...(3 d θ / dt = ω)
Also, the total component of change of velocity along tangential direction,
←
←
= xb1 + by = v. δ θ + ( ω. δr + r. δω)
...(Perpendicular to O A and towards left)
∴ Tangential component of acceleration of the slider B with respect to O on the link O A,
acting perpendicular to O A and towards left,
v.δθ + (ω.δr + r.δω)
dθ
dr
dω
t
aBO
= Lt
=v
+ω
+r
dt
dt
dt
δt
←
= v.ω + ω.v + r.α = (2 v.ω + r.α)
...(ii)
...(3 dr / dt = v, and d ω / dt = α)
Now radial component of acceleration of the coincident point C with respect to O, acting in
a direction from C to O,
r
...(iii)
= ω2 .r ↑
aCO
and tangential component of acceleraiton of the coincident point C with respect to O, acting in a
direction perpendicular to CO and towards left,
←
t
...(iv)
aCO
= α.r ↑
Radial component of the slider B with respect to the coincident point C on the link O A,
acting radially outwards,
dv
dv
r
r
r
= aBO
− aCO
=
↑
aBC
− ω2 .r − ( − ω2 .r ) =
dt
dt
and tangential component of the slider B with respect to the coincident point C on the link O A acting
in a direction perpendicular to O A and towards left,
←
t
t
t
= aBO
− aCO
= ( 2 ω.v + α.r ) − α.r = 2 ω.v
aBC
Chapter 8 : Acceleration in Mechanisms
l
211
This tangential component of acceleration of the slider B with respect to the coincident
point C on the link is known as coriolis component of acceleration and is always perpendicualr to
the link.
∴ Coriolis component of the acceleration of B with respect of C,
c
t
= aBC
= 2 ω.v
aBC
where
ω = Angular velocity of the link O A, and
v = Velocity of slider B with respect to coincident point C.
In the above discussion, the anticlockwise direction for ω and the radially outward direction
for v are taken as positive. It may be noted that the direction of coriolis component of acceleration
changes sign, if either ω or v is reversed in direction. But the direction of coriolis component of
acceleration will not be changed in sign if both ω and v are reversed in direction. It is concluded that
the direction of coriolis component of acceleration is obtained by rotating v, at 90°, about its origin
in the same direction as that of ω.
Fig. 8.27. Direction of coriolis component of acceleration.
The direction of coriolis component of acceleration (2 ω.v) for all four possible cases, is
shown in Fig. 8.27. The directions of ω and v are given.
Example 8.13. A mechanism of a crank and slotted lever quick
return motion is shown in Fig. 8.28. If the crank rotates counter clockwise
at 120 r.p.m., determine for the configuration shown, the velocity and
acceleration of the ram D. Also determine the angular acceleration of
the slotted lever.
Crank, AB = 150 mm ; Slotted arm, OC = 700 mm and link
CD = 200 mm.
Solution. Given : N BA = 120 r.p.m or ωBA = 2 π × 120/60
= 12.57 rad/s ; A B = 150 mm = 0.15 m; OC = 700 mm = 0.7 m;
CD = 200 mm = 0.2 m
We know that velocity of B with respect to A ,
vBA = ωBA × AB
= 12.57 × 0.15 = 1.9 m/s
...(Perpendicular to A B)
Velocity of the ram D
First of all draw the space diagram, to some suitable scale, as
shown in Fig. 8.29 (a). Now the velocity diagram, as shown in Fig. 8.29
Fig. 8.28
212
l
Theory of Machines
(b), is drawn as discussed below:
1. Since O and A are fixed points, therefore these points are marked as one point in velocity
diagram. Now draw vector ab in a direction perpendicular to A B , to some suitable scale, to represent
the velocity of slider B with respect to A i.e.vBA, such that
vector ab = v BA = 1.9 m/s
(a) Space diagram.
(b) Velocity diagram.
(c) Direction of coriolis component.
(d) Acceleration diagram.
Fig. 8.29
2. From point o, draw vector ob' perpendicular to OB' to represent the velocity of coincident
point B' (on the link OC) with respect to O i.e. v B′O and from point b draw vector bb' parallel to the
path of motion of B' (which is along the link OC) to represent the velocity of coincident point B' with
respect to the slider B i.e. vB'B. The vectors ob' and bb' intersect at b'.
Note: Since we have to find the coriolis component of acceleration of the slider B with respect to the coincident
point B', therefore we require the velocity of B with respect to B' i.e. vBB'. The vector b'b will represent vBB'
as shown in Fig. 8.29 (b).
3. Since the point C lies on OB' produced, therefore, divide vector ob' at c in the same ratio
as C divides OB' in the space diagram. In other words,
ob′ / oc = OB ′ / OC
The vector oc represents the velocity of C with respect to O i.e. vCO.
Chapter 8 : Acceleration in Mechanisms
l
213
4. Now from point c, draw vector cd perpendicular to CD to represent the velocity of D with
respect to C i.e. vDC ,and from point o draw vector od parallel to the path of motion of D (which is
along the horizontal) to represent the velocity of D i.e. vD.The vectors cd and od intersect at d.
By measurement, we find that velocity of the ram D,
v D = vector od = 2.15 m/s Ans.
From velocity diagram, we also find that
Velocity of B with respect to B',
v BB' = vector b'b = 1.05 m/s
Velocity of D with respect to C,
v DC = vector cd = 0.45 m/s
Velocity of B' with respect to O
v B′O = vector ob' = 1.55 m/s
Velocity of C with respect to O,
v CO = vector oc = 2.15 m/s
∴ Angular velocity of the link OC or OB',
ωCO = ωB′O =
vCO 2.15
=
= 3.07 rad/s (Anticlockwise)
0.7
OC
Acceleration of the ram D
We know that radial component of the acceleration of B with respect to A ,
r
= ω2BA × AB = (12.57)2 × 0.15 = 23.7 m/s2
aBA
Coriolis component of the acceleration of slider B with respect to the coincident point B',
c
′ = 2ω.v = 2ωCO .vBB′ = 2 × 3.07 × 1.05 = 6.45 m/s 2
aBB
...(3 ω = ωCO and v = vBB′)
Radial component of the acceleration of D with respect to C,
2
vDC
(0.45)2
=
= 1.01 m/s 2
0.2
CD
Radial component of the acceleration of the coincident point B' with respect to O,
r
=
aDC
aBr ′O =
vB2 ′O (1.55)2
=
= 4.62 m/s2
0.52
B ′O
...(By measurement B'O = 0.52 m)
Now the acceleration diagram, as shown in Fig. 8.29 (d), is drawn as discussed below:
1. Since O and A are fixed points, therefore these points are marked as one point in the
acceleration diagram. Draw vector a'b' parallel to A B, to some suitable scale, to represent the radial
r
component of the acceleration of B with respect to A i.e. aBA
or aB, such that
r
vector a ′b′ = aBA
= aB = 23.7 m/s 2
2. The acceleration of the slider B with respect to the coincident point B' has the following
two components :
c
(i) Coriolis component of the acceleration of B with respect to B' i.e. aBB
′ , and
r
(ii) Radial component of the acceleration of B with respect to B' i.e. aBB
′.
These two components are mutually perpendicular. Therefore from point b' draw vector b'x
c
2
perpendicular to B'O i.e. in a direction as shown in Fig. 8.29 (c) to represent aBB
′ = 6.45 m/s . The
214
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Theory of Machines
c
direction of aBB
′ is obtained by rotating v BB′ (represented by vector b'b in velocity diagram) through
90° in the same sense as that of link OC which rotates in the counter clockwise direction. Now from
r
point x, draw vector xb'' perpendicular to vector b'x (or parallel to B'O) to represent aBB
′ whose
magnitude is yet unknown.
3. The acceleration of the coincident point B' with respect to O has also the following two
components:
(i) Radial component of the acceleration of coincident point B' with respect to O i.e.
aBr ′O’ and
(ii) Tangential component of the acceleration of coincident point B' with respect to O,
i.e. aBt ′O .
These two components are mutually perpendicular. Therefore from point o', draw vector o'y
parallel to B'O to represent aBr ′O = 4.62 m/s 2 and from point y draw vector yb'' perpendicular to
vector o'y to represent aBt ′O . The vectors xb'' and yb'' intersect at b''. Join o'b''. The vector o'b''
represents the acceleration of B' with respect to O, i.e. aB′O.
4. Since the point C lies on OB' produced, therefore divide vector o'b'' at c' in the same ratio
as C divides OB' in the space diagram. In other words,
o'b''/o'c' = OB'/OC
5. The acceleration of the ram D with respect to C has also the following two components:
r
(i) Radial component of the acceleration of D with respect to C i.e. aDC
, and
t
(ii) Tangential component of the acceleration of D with respect to C, i.e. aDC
.
The two components are mutually perpendicular. Therefore draw vector c'z parallel to CD
r
t
to represent aDC
, whose
= 1.01 m/s 2 and from z draw zd' perpendicular to vector zc' to represent aDC
magnitude is yet unknown.
6. From point o', draw vector o'd' in the direction of motion of the ram D which is along the
horizontal. The vectors zd' and o'd' intersect at d'. The vector o'd' represents the acceleration of ram
D i.e. aD.
By measurement, we find that acceleration of the ram D,
aD = vector o'd' = 8.4 m/s2 Ans.
Angular acceleration of the slotted lever
By measurement from acceleration diagram, we find that tangential component of the
coincident point B' with respect to O,
aBt ′O = vector yb′′ = 6.4 m/s 2
We know that angular acceleration of the slotted lever,
aBt ′O
6.4
=
= 12.3 rad/s 2 (Anticlockwise) Ans.
OB ′ 0.52
Example 8.14. The driving crank AB of the quick-return mechanism, as shown in Fig. 8.30,
revolves at a uniform speed of 200 r.p.m. Find the velocity and acceleration of the tool-box R, in the
position shown, when the crank makes an angle of 60° with the vertical line of centres PA . What is
the acceleration of sliding of the block at B along the slotted lever PQ ?
=
Chapter 8 : Acceleration in Mechanisms
l
215
Solution. Given : N BA = 200 r.p.m. or ωBA = 2 π × 200/60 = 20.95 rad/s ; A B = 75 mm = 0.075 m
We know that velocity of B with respect to A ,
v BA = ωBA × A B = 20.95 × 0.075 = 1.57 m/s
...(Perpendicular to A B)
All dimensions in mm.
Fig. 8.30
Velocity of the tool-box R
First of all draw the space diagram, to some suitable scale, as shown in Fig. 8.31 (a). Now
the velocity diagram, as shown in Fig. 8.31 (b), is drawn as discussed below:
1. Since A and P are fixed points, therefore these points are marked as one point in the
velocity diagram. Now draw vector ab in a direction perpendicular to A B, to some suitable scale, to
represent the velocity of B with respect to A or simply velocity of B (i.e. vBA or v B), such that
vector ab = v BA = v B = 1.57 m/s
2. From point p, draw vector pb' perpendicular to PB' to represent the velocity of coincident
point B' with respect to P (i.e. vB'P or v B') and from point b, draw vector bb' parallel to the path of
motion of B' (which is along PQ) to represent the velocity of coincident point B' with respect to the
slider B i.e. v B'B. The vectors pb' and bb' intersect at b'.
Note. The vector b'b will represent the velocity of the slider B with respect to the coincident point B' i.e.vBB'.
3. Since the point Q lies on PB' produced, therefore divide vector pb' at q in the same ratio
as Q divides PB'. In other words,
pb'/pq = PB'/PQ
The vector pq represents the velocity of Q with respect to P i.e. vQP.
4. Now from point q, draw vector qr perpendicular to QR to represent the velocity of R with
respect to Q i.e. v RQ, and from point a draw vector ar parallel to the path of motion of the tool-box
R (which is along the horizontal), to represent the velocity of R i.e. vR.The vectors qr and ar intersect
at r.
By measurement, we find that velocity of the tool-box R,
v R = vector ar = 1.6 m/s Ans.
We also find that velocity of B' with respect to B,
v B'B = vector bb' = 1.06 m/s
Velocity of B' with respect to P,
vB'P = vector pb' = 1.13 m/s
216
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Theory of Machines
Velocity of R with respect to Q,
v RQ = vector qr = 0.4 m/s
Velocity of Q with respect to P,
v QP = vector pq = 1.7 m/s
∴ Angular velocity of the link PQ,
vQP
1.7
ωPQ =
=
= 4.53 rad/s
QP 0.375
(a) Space diagram.
...(3 PQ = 0.375 m)
(b) Velocity diagram.
(c) Direction of coriolis component.
(d) Acceleration diagram.
Fig. 8.31
Acceleration of the tool box R
We know that the radial component of the acceleration of B with respect to A ,
r
= ω2BA × AB = (20.95)2 × 0.075 = 32.9 m/s2
aBA
Coriolis component of the acceleration of the slider B with respect to coincident point B′.
c
2
aBB
′ = 2ω.v = 2ωQP × vBB′ = 2 × 4.53 ×1.06 = 9.6 m / s
... (3 ω = ωQP , and v = vBB ′)
Radial component of the acceleration of R with respect to Q ,
2
vRQ
(0.4)2
r
=
=
= 0.32 m/s2
aRQ
0.5
QR
Chapter 8 : Acceleration in Mechanisms
l
217
Radial component of the acceleration of B' with respect to P,
aBr ′P =
vB2 ′P (1.13) 2
=
= 5.15 m/s 2
0.248
PB ′
...(By measurement, PB' = 248 mm = 0.248 m)
Now the acceleration diagram, as shown in Fig. 8.31 (d), is drawn as discussed below:
1. Since A and P are fixed points, therefore these points are marked as one point in the
acceleration diagram. Draw vector a'b' parallel to A B, to some suitable scale, to represent the radial
r
component of the acceleration of B with respect to A i.e. aBA
, or aB such that
r
vector a′b′ = aBA
= aB = 32.9 m/s 2
2. The acceleration of the slider B with respect to the coincident point B' has the following
two components:
c
(i) Coriolis component of the acceleration of B with respect to B' i.e. aBB
′ , and
r
(ii) Radial component of the acceleration of B with respect to B' i.e. aBB
′.
These two components are mutually perpendicular. Therefore from point b', draw vector b'x
c
2
perpendicular to BP [i.e. in a direction as shown in Fig. 8.31 (c)] to represent aBB
′ = 9.6 m/s . The
c
direction of aBB′ is obtained by rotating v BB′ (represented by vector b'b in the velocity diagram)
through 90° in the same sense as that of link PQ which rotates in the clockwise direction. Now from
r
point x, draw vector xb'' perpendicular to vector b'x (or parallel to B'P) to represent aBB
′ whose
magnitude is yet unknown.
3. The acceleration of the coincident point B' with respect to P has also the following two
components:
(i) Radial component of the acceleration of B' with respect to P i.e. aBr ′P , and
(ii) Tangential component of the acceleration of B' with respect to P i.e. aBt ′P .
These two components are mutually perpendicular. Therefore from point p' draw vector p'y
parallel to B'P to represent aBr ′P = 5.15 m/s2, and from point y draw vector yb'' perpendicular to
vector p'y to represent aBt ′P . The vectors xb'' and yb'' intersect at b'', join p'b''. The vector p'b''
represents the acceleration of B' with respect to P i.e. aB'P and the vector b''b' represents the acceleration
of B with respect to B' i.e. aBB' .
4. Since the point Q lies on PB' produced, therefore divide vector p'b'' at q' in the same ratio
as Q divides PB in the space diagram. In other words,
p'b''/p'q' = PB'/PQ
5. The acceleration of the tool-box R with respect to Q has the following two components:
r
, and
(i) Radial component of the acceleration of R with respect to Q i.e. aRQ
t
.
(ii) Tangential component of the acceleration of R with respect to Q i.e. aRQ
These two components are mutually perpendicular. Therefore from point q', draw vector a'z
r
parallel to QR to represent aRQ
= 0.32 m/s 2 . Since the magnitude of this component is very small,
therefore the points q' and z coincide as shown in Fig. 8.31 (d). Now from point z (same as q'), draw
t
vector zr' perpendicular to vector q'z (or QR) to represent aRQ
whose magnitude is yet unknown.
6. From point a' draw vector a'r' parallel to the path of motion of the tool-box R (i.e. along
the horizontal) which intersects the vector zr' at r'. The vector a'r' represents the acceleration of the
tool-box R i.e. aR.
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By measurement, we find that
aR = vector a'r' = 22 m/s2 Ans.
Acceleration of sliding of the block B along the slotted lever PQ
By measurement, we find that the acceleration of sliding of the block B along the slotted
lever PQ
= aBB' = vector b''x = 18 m/s2 Ans.
Example 8.15. In a Whitworth quick return motion, as shown in Fig. 8.32. OA is a crank
rotating at 30 r.p.m. in a clockwise direction. The dimensions of various links are : OA = 150 mm;
OC = 100 mm; CD = 125 mm; and DR = 500 mm.
Determine the acceleraion of the sliding block R and the angular acceleration of the slotted
lever CA.
All dimensions in mm.
Fig. 8.32
Solution. Given : N AO = 30 r.p.m. or ωAO = 2π × 30/60 = 3.142 rad/s ; O A = 150 mm
= 0.15 m; OC = 100 mm = 0.1 m ; CD = 125 mm = 0.125 m ; DR = 500 mm = 0.5 m
We know that velocity of A with respect to O or velocity of A ,
v AO = vA = ωAO × O A = 3.142 × 0.15 = 0.47 m/s
...(Perpendicular to O A)
First of all draw the space diagram, to some suitable scale, as shown in Fig. 8.33 (a). Now
the velocity diagram, as shown in Fig. 8.33 (b), is drawn as discussed below:
1. Since O and C are fixed points, therefore these are marked at the same place in velocity
diagram. Now draw vector ca perpendicular to O A, to some suitable scale, to represent the velocity
of A with respect to O or simply velocity of A i.e. vAO or v A, such that
vector oa = v AO = v A = 0.47 m/s
2. From point c, draw vector cb perpendicular to BC to represent the velocity of the coincident
point B with respect to C i.e. vBC or vB and from point a draw vector ab parallel to the path of motion
of B (which is along BC) to represent the velocity of coincident point B with respect to A i.e.vBA. The
vectors cb and ab intersect at b.
Note: Since we have to find the coriolis component of acceleration of slider A with respect to coincident point
B, therefore we require the velocity of A with respect to B i.e. vAB.The vector ba will represent v AB as shown
in Fig. 8.33 (b).
Chapter 8 : Acceleration in Mechanisms
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219
3. Since D lies on BC produced, therefore divide vector bc at d in the same ratio as D divides
BC in the space diagram. In other words,
bd/bc = BD/BC
(a) Space diagram.
(b) Velocity diagram.
(c) Direction of coriolis component.
(d) Acceleration diagram.
Fig. 8.33
4. Now from point d, draw vector dr perpendicular to DR to represent the velocity of R with
respect to D i.e. v RD, and from point c draw vector cr parallel to the path of motion of R (which is
horizontal) to represent the velocity of R i.e.vR.
By measurement, we find that velocity of B with respect to C,
v BC = vector cb = 0.46 m/s
Velocity of A with respect to B,
v AB = vector ba = 0.15 m/s
and velocity of R with respect to D,
v RD = vector dr = 0.12 m/s
We know that angular velocity of the link BC,
ωBC =
vBC 0.46
=
= 1.92 rad/s (Clockwise)
CB 0.24
...(By measurement, CB = 0.24 m)
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Theory of Machines
Acceleration of the sliding block R
We know that the radial component of the acceleration of A with respect to O,
2
vAO
(0.47)2
=
= 1.47 m/s 2
0.15
OA
Coriolis component of the acceleration of slider A with respect to coincident point B,
r
=
aAO
c
= 2 ωBC × vAB = 2 × 1.92 × 0.15 = 0.576 m/s 2
aAB
Radial component of the acceleration of B with respect to C,
2
vBC
(0.46) 2
=
= 0.88 m/s 2
0.24
CB
Radial component of the acceleration of R with respect to D,
r
=
aBC
2
vRD
(0.12)2
=
= 0.029 m/s2
0.5
DR
Now the acceleration diagram, as shown in Fig. 8.33 (d), is drawn as discussed below:
1. Since O and C are fixed points, therefore these are marked at the same place in the
acceleration diagram. Draw vector o'a' parallel to O A, to some suitable scale, to represent the radial
r
=
aRD
r
, or aA such that
component of the acceleration of A with respect to O i.e. aAO
r
vector o′a′ = aAO
= aA = 1.47 m/s 2
2. The acceleration of the slider A with respect to coincident point B has the following two
components:
c
(i) Coriolis component of the acceleration of A with respect to B i.e. aAB
, and
r
(ii) Radial component of the acceleration of A with respect to B i.e. aAB
.
These two components are mutually perpendicular. Therefore from point a' draw vector a′x
c
perpendicular to BC to represent aAB
= 0.576 m/s2 in a direction as shown in Fig. 8.33 (c), and draw
r
whose magnitude is yet
vector xb' perpendicular to vector a'x (or parallel to BC) to represent aAB
unknown.
c
Note: The direction of aAB
is obtained by rotating v AB (represented by vector ba in velocity diagram) through
90° in the same sense as that of ωBC which rotates in clockwise direction.
3. The acceleration of B with respect to C has the following two components:
r
(i) Radial component of B with respect to C i.e. aBC
, and
t
(ii) Tangential component of B with respect to C i.e. aBC
.
These two components are mutually perpendicular. Therefore, draw vector c'y parallel to
r
BC to represent aBC
= 0.88 m/s2 and from point y draw vector yb' perpendicular to c'y to represent
t
aBC . The vectors xb' and yb' intersect at b'. Join b'c'.
4. Since the point D lies on BC produced, therefore divide vector b'c' at d' in the same ratio
as D divides BC in the space diagram. In other words,
b'd'/b'c' = BD/BC.
5. The acceleration of the sliding block R with respect to D has also the following two
components:
r
(i) Radial component of R with respect to D i.e. aRD
, and
t
(ii) Tangential component of R with respect to D i.e. aRD
.
Chapter 8 : Acceleration in Mechanisms
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221
These two components are mutually perpendicular. Therefore from point d', draw vector
r
= 0.029 m/s2 and from z draw zr' perpendicular to d'z to represent
aRD
t
aRD whose magnitude is yet unknown.
d ′z parallel to DR to represent
6. From point c', draw vector c'r' parallel to the path of motion of R (which is horizontal).
The vector c'r' intersects the vector zr' at r'. The vector c'r' represents the acceleration of the sliding
block R.
By measurement, we find that acceleration of the sliding block R,
aR = vector c'r' = 0.18 m/s2 Ans.
Angular acceleration of the slotted lever CA
By measurement from acceleration diagram, we find that tangential component of B with
respect to C,
t
= vector yb′ = 0.14 m/s 2
aBC
We know that angular acceleration of the slotted lever C A,
t
aCB
0.14
=
= 0.583 rad/s 2 (Anticlockwise) Ans.
BC 0.24
Example 8.16. The kinematic diagram of one of the cylinders of a
rotary engine is shown in Fig. 8.34. The crank OA which is vertical and
fixed , is 50 mm long. The length of the connecting rod AB is 125 mm. The
line of the stroke OB is inclined at 50° to the vertical.
α CA = α BC =
The cylinders are rotating at a uniform speed of 300 r.p.m., in a
clockwise direction, about the fixed centre O. Determine: 1. acceleration
of the piston inside the cylinder, and 2. angular acceleration of the
connecting rod.
Solution. Given: A B = 125 mm = 0.125 m ; N CO = 300 r.p.m.
or ωCO = 2π × 300/60 = 31.4 rad/s
Fig. 8.34
First of all draw the space diagram, as shown in Fig. 8.35 (a), to some suitable scale. By
measurement from the space diagram, we find that
OC = 85 mm = 0.085 m
∴ Velocity of C with respect to O,
v CO = ωCO × OC = 31.4 × 0.85 = 2.7 m/s
...(Perpendicular to CO)
Now the velocity diagram, as shown in Fig. 8.35 (b), is drawn as discussed below:
1. Since O and A are fixed points, therefore these are marked at the same place in the
velocity diagram. Draw vector oc perpendicular to OC to represent the velocity of C with respect to
O i.e. v CO, such that
vector oc = v CO = v C = 2.7 m/s.
2. From point c, draw vector cb parallel to the path of motion of the piston B (which is along
CO) to represent the velocity of B with respect to C i.e. vBC , and from point a draw vector ab
perpendicular to A B to represent the velocity of B with respect to A i.e. vBA or vB.
By measurement, we find that velocity of piston B with respect to coincident point C,
vBC = vector cb = 0.85 m/s
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and velocity of piston B with respect to A ,
vBA = vB = vector ab = 2.85 m/s
(a) Space diagram.
(b) Velocity diagram.
(c) Direction of coriolis component.
(d) Acceleration diagram.
Fig. 8.35
1. Acceleration of the piston inside the cylinder
We know that the radial component of the acceleration of the coincident point C with respect
to O,
2
vCO
(2.7)2
=
= 85.76 m/s2
0.085
OC
Coriolis component of acceleration of the piston B with respect to the cylinder or coincident
point C,
r
=
aCO
c
= 2 ωCO × vBC = 2 × 31.4 × 0.85 = 53.4 m/s 2
aBC
Radial component of acceleration of B with respect to A ,
2
vBA
(2.85)2
=
= 65 m/s 2
0.125
AB
The acceleration diagram, as shown in Fig. 8.35 (d), is drawn as discussed below:
r
=
aBA
Chapter 8 : Acceleration in Mechanisms
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223
1. Since O and A are fixed points, therefore these are marked as one point in the acceleration
diagram. Draw vector o'c' parallel to OC, to some suitable scale, to represent the radial component
r
of the acceleration of C with respect to O i.e., aCO
, such that
r
vector o′c ′ = aCO
= 85.76 m/s 2
2. The acceleration of piston B with respect to coincident point C has the following two
components:
c
(i) Coriolis component of the acceleration of B with respect to C i.e. aBC
, and
r
(ii) Radial component of the acceleration of B with respect to C i.e. aBC
.
These two components are mutually perpendicular. Therefore from point c', draw vector c'x
c
perpendicular to CO to represent aBC
= 53.4 m/s 2 in a direction as shown in Fig. 8.35 (c). The
c
direction of aBC
is obtained by rotating v BC (represented by vector cb in velocity diagram) through
90° in the same sense as that of ωCO which rotates in the clockwise direction. Now from point x,
r
draw vector xb' perpendicular to vector c'x (or parallel to OC) to represent aBC
whose magnitude is
yet unknown.
3. The acceleration of B with respect to A has also the following two components:
r
(i) Radial component of the acceleration of B with respect to A i.e. aBA
, and
t
(ii) Tangential component of the acceleration of B with respect to A i.e. aBA
.
These two components are mutually perpendicular. Therefore from point a', draw vector a'y
r
parallel to A B to represent aBA
= 65 m/s 2 , and from point y draw vector yb' perpendicular to vector
t
a'y to represent aBA
. The vectors xb' and yb' intersect at b'.
4. Join c'b' and a'b'. The vector c'b' represents the acceleration of B with respect to C (i.e.
acceleration of the piston inside the cylinder).
By measurement, we find that acceleration of the piston inside the cylinder,
aBC = vector c'b' = 73.2 m/s2 Ans.
2. Angular acceleration of the connecting rod
By measurement from acceleration diagram, we find that the tangential component of the
acceleration of B with respect to A ,
t
= vector yb′ = 37.6 m/s2
aBA
∴ Angular acceleration of the connecting rod A B,
t
aBA
37.6
=
= 301 rad/s 2 (Clockwise) Ans.
AB 0.125
Example 8.17. In a swivelling joint mechanism, as shown in Fig. 8.36, the driving crank
OA is rotating clockwise at 100 r.p.m. The lengths of various links are : OA = 50 mm ; AB = 350
mm; AD = DB ; DE = EF = 250 mm and CB = 125 mm. The horizontal distance between the fixed
points O and C is 300 mm and the vertical distance between F and C is 250 mm.
α AB =
For the given configuration, determine: 1. Velocity of the slider block F, 2. Angular velocity
of the link DE, 3. Velocity of sliding of the link DE in the swivel block,and 4. Acceleration of sliding
of the link DE in the trunnion.
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Theory of Machines
All dimensions in mm.
Fig. 8.36
Solution. Given: N AO = 100 r.p.m. or ωAO = 2π × 100/60 = 10.47 rad/s ; O A = 50 mm =
0.05 m; A B = 350 mm = 0.35 m ; CB = 125 mm = 0.125 m ; DE = EF = 250 mm = 0.25 m
We know that velocity of A with respect to O or velocity of A ,
v AO = v A = ωAO × O A = 10.47 × 0.05 = 0.523 m/s
...(Perpendicular to O A)
This machine uses swivelling joint.
1. Velocity of slider block F
First of all draw the space diagram, to some suitable scale, as shown in Fig. 8.37 (a). Now
the velocity diagram, as shown in Fig. 8.37 (b), is drawn as discussed below:
1. Since O, C and Q are fixed points, therefore these points are marked at one place in the
velocity diagram. Draw vector oa perpendicular to O A, to some suitable scale, to represent the
velocity of A with respect to O or simply velocity of A , i.e. v AO or v A, such that
vector oa = v AO = v A = 0.523 m/s
Chapter 8 : Acceleration in Mechanisms
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225
2. From point a, draw vector ab perpendicular to A B to represent the velocity of B with
respect to A i.e. vBA,and from point c draw vector cb perpendicular to CB to represent the velocity of
B with respect to C or simply velocity of B i.e. vBC or v B. The vectors ab and cb intersect at b.
(a) Space diagram.
(b) Velocity diagram.
(c) Direction of coriolis component.
(d) Acceleration diagram.
Fig. 8.37
3. Since point D lies on A B, therefore divide vector ab at d in the same ratio as D divides A B
in the space diagram. In other words,
ad/ab = AD/AB
Note: Since point D is mid-point of A B, therefore d is also mid-point of ab.
4. Now from point d, draw vector ds perpendicular to DS to represent the velocity of S with
respect to D i.e. vSD, and from point q draw vector qs parallel to the path of motion of swivel block
Q (which is along DE) to represent the velocity of S with respect to Q i.e. vSQ. The vectors ds and qs
intersect at s.
Note: The vector sq will represent the velocity of swivel block Q with respect to S i.e. v QS.
5. Since point E lies on DS produced, therefore divide vector ds at e in the same ratio as E
divides DS in the space diagram. In other words,
de/ds = DE/DS
6. From point e, draw vector ef perpendicular to EF to represent the velocity of F with
respect to E i.e. vFE , and from point o draw vector of parallel to the path of motion of F (which is
along the horizontal direction) to represent the velocity of F i.e. vF..The vectors ef and of intersect
at f.
By measurement, we find that velocity of B with respect to A ,
v BA = vector ab = 0.4 m/s
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Theory of Machines
Velocity of B with respect to C,
v BC = v B = vector cb = 0.485 m/s
Velocity of S with respect to D,
vSD = vector ds = 0.265 m/s
Velocity of Q with respect to S,
vQS = vector sq = 0.4 m/s
Velocity of E with respect to D,
vED = vector de = 0.73 m/s
Velocity of F with respect to E,
v FE = vector ef = 0.6 m/s
and velocity of the slider block F, v F = vector of = 0.27 m/s Ans.
2. Angular velocity of the link DE
We know that angular velocity of the link DE,
vED 0.73
=
= 2.92 rad/s (Anticlockwise) Ans.
DE 0.25
3. Velocity of sliding of the link DE in the swivel block
ωDE =
The velocity of sliding of the link DE in the swivel block Q will be same as that of velocity
of S i.e. vS.
∴ Velocity of sliding of the link DE in the swivel block,
v S = v SQ = 0.4 m/s Ans.
4. Acceleration of sliding of the link DE in the trunnion
We know that the radial component of the acceleration of A with respect to O or the
acceleration of A ,
2
vAO
(0.523)2
=
= 5.47 m/s 2
0.05
OA
Radial component of the acceleration of B with respect to A ,
r
= aA =
aAO
r
=
aBA
2
vBA
(0.4)2
=
= 0.457 m/s2
0.35
AB
Radial component of the acceleration of B with respect to C,
2
vBC
(0.485) 2
=
= 1.88 m/s2
0.125
CB
Radial component of the acceleration of S with respect to D,
r
=
aBC
r
=
aSD
2
vSD
(0.265) 2
=
= 0.826 m/s 2
0.085
DS
...(By measurement DS = 85 mm = 0.085 m)
Coriolis component of the acceleration of Q with respect to S,
c
= 2 ωDE × vQS = 2 × 2.92 × 0.4 = 2.336 m/s 2
aQS
Chapter 8 : Acceleration in Mechanisms
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227
and radial component of the acceleration of F with respect to E,
2
vFE
(0.6)2
=
= 1.44 m/s2
0.25
EF
Now the acceleration diagram, as shown in Fig. 8.37 (d), is drawn as discussed below:
1. Since O, C and Q are fixed points, therefore these points are marked at one place in the
r
acceleration diagram. Now draw vector o'a' parallel to O A, to some suitable scale, to represent aAO
,
or aA such that
r
=
aFE
r
vector o ′ a ′ = aAO
= aA = 5.47 m/s 2
Note : Since O A rotates with uniform speed, therefore there will be no tangential component of the
acceleration.
2. The acceleration of B with respect to A has the following two components:
r
(i) Radial component of the acceleration of B with respect to A i.e. aBA
, and
t
(ii) Tangential component of the acceleration of B with respect to A i.e. aBA
.
These two components are mutually perpendicular. Therefore from point a', draw vector a'x
r
parallel to A B to represent aBA
= 0.457 m/s2 , and from point x draw vector xb' perpendicular to
t
vector a'x to represent aBA
whose magnitude is yet unknown.
3. The acceleration of B with respect to C has the following two components:
r
(i) Radial component of the acceleration of B with respect to C i.e. aBC
, and
t
(ii) Tangential component of the acceleration of B with respect to C i.e. aBC
.
These two components are mutually perpendicular. Therefore from point c', draw vector c'y
r
parallel to CB to represent aBC
= 1.88 m/s2 and from point y draw vector yb' perpendicular to vector
t
c'y to represent aBC
. The vectors xb' and yb' intersect at b'.
4. Join a'b' and c'b'. The vector a'b' represents the acceleration of B with respect to A i.e.
aBA and the vector c'b' represents the acceleration of B with respect to C or simply the acceleration
of B i.e. aBC or aB, because C is a fixed point.
5. Since the point D lies on A B, therefore divide vector a'b' at d' in the same ratio as D
divides A B in the space diagram. In other words,
a'd'/a'b' = AD/AB
Note: Since D is the mid-point of A B, therefore d' is also mid-point of vector a'd'.
6. The acceleration of S with respect to D has the following two components:
r
(i) Radial component of the acceleration of S with respect to D i.e. aSD
, and
t
(ii) Tangential component of the acceleration of S with respect to D i.e. aSD
.
These two components are mutually perpendicular. Therefore from point d′ , draw vector
d′z parallel to DS to represent arSD = 0.826 m/s2, and from point z draw vector zs′ perpendicular to
vector d′z to represent atSD whose magnitude is yet unknown.
7. The acceleration of Q (swivel block) with respect to S (point on link DE i.e. coincident
point) has the following two components:
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Theory of Machines
c
, and
(i) Coriolis component of acceleration of Q with respect to S i.e. aQS
r
.
(ii) Radial component of acceleration of Q with respect to S, i.e. aQS
These two components are mutually perpendicular. Therefore from point q', draw vector
c
= 2.336 m/s 2 in a direction as shown in Fig. 8.37 (c). The
q'z1, perpendicular to DS to represent aQS
c
direction of aQS
is obtained by rotating v QS (represented by vector sq in velocity diagram) through
90° in the same sense as that of ωDE which rotates in the anticlockwise direction. Now from z 1, draw
r
. The vectors zs' and z 1s'
vector z 1s' perpendicular to vector q'z1 (or parallel to DS) to represent aQS
intersect at s'.
8. Join s'q' and d's'. The vector s'q' represents the acceleration of Q with respect to S i.e. aQS
and vector d's' represents the acceleration of S with respect to D i.e. aSD.
By measurement, we find that the acceleration of sliding the link DE in the trunnion,
r
= aQS
= vector z1 s ′ = 1.55 m/s2 Ans.
EXERCISES
1.
The engine mechanism shown in Fig. 8.38 has crank OB = 50 mm and length of connecting rod A B
= 225 mm. The centre of gravity of the rod is at G which is 75 mm from B. The engine speed is 200
r.p.m.
Fig. 8.38
For the position shown, in which OB is turned 45° from O A, Find 1. the velocity of G and the
angular velocity of A B, and 2. the acceleration of G and angular acceleration of A B.
[Ans. 6.3 m/s ; 22.6 rad/s ; 750 m/s2 ; 6.5 rad/s2]
2.
In a pin jointed four bar mechanism ABCD, the lengths of various links are as follows:
A B = 25 mm ; BC = 87.5 mm ; CD = 50 mm and A D = 80 mm.
The link A D is fixed and the angle BAD = 135°. If the velocity of B is 1.8 m/s in the clockwise
direction, find 1. velocity and acceleration of the mid point of BC, and 2. angular velocity and
angular acceleration of link CB and CD.
[Ans. 1.67 m/s, 110 m/s2 ; 8.9 rad/s, 870 rad/s2 ; 32.4 rad/s, 1040 rad/s2]
3.
In a four bar chain ABCD , link A D is fixed and the crank A B rotates at 10 radians per second
clockwise. Lengths of the links are A B = 60 mm ; BC = CD = 70 mm ; D A = 120 mm. When angle
DAB = 60° and both B and C lie on the same side of A D, find 1. angular velocities (magnitude and
direction) of BC and CD ; and 2. angular acceleration of BC and CD.
[Ans. 6.43 rad/s (anticlockwise), 6.43 rad/s (clockwise) ; 10 rad/s2 105 rad/s2]
4.
In a mechanism as shown in Fig. 8.39, the link AB rotates with a uniform angular velocity of
30 rad/s. The lengths of various links are :
A B = 100 mm ; BC = 300 mm ; BD = 150 mm ; DE = 250 mm ; EF = 200 mm ; DG = 165 mm.
Determine the velocity and acceleration of G for the given configuration.
[Ans. 0.6 m/s ; 66 m/s2]
Chapter 8 : Acceleration in Mechanisms
5.
6.
7.
l
229
Fig. 8.39
Fig. 8.40
In a mechanism as shown in Fig. 8.40, the crank
O A is 100 mm long and rotates in a clockwise
direction at a speed of 100 r.p.m. The straight
rod BCD rocks on a fixed point at C. The links
BC and CD are each 200 mm long and the link
A B is 300 mm long. The slider E, which is
driven by the rod DE is 250 mm long. Find the
velocity and acceleration of E.
[ Ans. 1.26 m/s; 10.5 m/s2]
The dimensions of the various links of a
mechanism, as shown in Fig. 8.41, are as
follows:
Fig. 8.41
O A = 80 mm ; A C = CB = CD = 120 mm
If the crank O A rotates at 150 r.p.m. in the anticlockwise direction, find, for the given configuration: 1. velocity and acceleration of B and D ; 2.
rubbing velocity on the pin at C, if its diameter is 20 mm ; and 3. angular acceleration of the links A B
and CD.
[Ans. 1.1 m/s ; 0.37 m/s ; 20.2 m/s2, 16.3 m/s2 ; 0.15 m/s ; 34.6 rad/s2; 172.5 rad/s2]
In the toggle mechanism, as shown in Fig. 8.42, D is constrained to move on a horizontal path. The
dimensions of various links are : A B = 200 mm; BC = 300 mm ; OC = 150 mm; and BD = 450 mm.
Fig. 8.42
Fig. 8.43
The crank OC is rotating in a counter clockwise direction at a speed of 180 r.p.m., increasing at the
rate of 50 rad/s2. Find, for the given configuration 1. velocity and acceleration of D, and 2. angular
velocity and angular acceleration of BD.
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Theory of Machines
8.
In a quick return mechanism, as shown in Fig. 8.43, the driving crank O A is 60 mm long and rotates
at a uniform speed of 200 r.p.m. in a clockwise direction. For the position shown, find 1. velocity of
the ram R ; 2. acceleration of the ram R, and 3. acceleration of the sliding block A along the slotted
bar CD.
[Ans. 1.3 m/s ; 9 m/s2 ; 15 m/s2]
9.
Fig. 8.44 shows a quick return motion mechanism in which the driving crank O A rotates at 120
r.p.m. in a clockwise direction. For the position shown, determine the magnitude and direction of
1, the acceleration of the block D ; and 2. the angular acceleration of the slotted bar QB.
[Ans. 7.7 m/s2 ; 17 rad/s2]
10.
In the oscillating cylinder mechanism as shown in Fig. 8.45, the crank O A is 50 mm long while the
piston rod A B is 150 mm long. The crank O A rotates uniformly about O at 300 r.p.m.
Fig. 8.44
Fig. 8.45
Determine, for the position shown : 1. velocity of the piston B relative to the cylinder walls, 2.
angular velocity of the piston rod A B, 3. sliding acceleration of the piston B relative to the cylinder
walls, and 4. angular acceleration of the piston rod A B.
[Ans. 1.5 m/s ; 2.2 rad/s (anticlockwise) ; 16.75 m/s2 ; 234 rad/s2]
11.
The mechanism as shown in Fig 8.46 is a marine steering gear, called Rapson’s slide. O2B is the
tiller and A C is the actuating rod. If the velocity of AC is 25 mm/min to the left, find the angular
velocity and angular acceleration of the tiller. Either graphical or analytical technique may be used.
[Ans. 0.125 rad/s; 0.018 rad/s2]
Fig. 8.46
Chapter 8 : Acceleration in Mechanisms
l
231
DO YOU KNOW ?
1.
Explain how the acceleration of a point on a link (whose direction is known) is obtained when the
acceleration of some other point on the same link is given in magnitude and direction.
2.
Draw the acceleration diagram of a slider crank mechanism.
3.
Explain how the coriolis component of acceleration arises when a point is rotating about some other
fixed point and at the same time its distance from the fixed point varies.
4.
Derive an expression for the magnitude and direction of coriolis component of acceleration.
5.
Sketch a quick return motion of the crank and slotted lever type and explain the procedure of drawing
the velocity and acceleration diagram, for any given configuration of the mechanism.
OBJECTIVE TYPE QUESTIONS
1.
2.
The component of the acceleration, parallel to the velocity of the particle, at the given instant is
called
(a)
radial component
(b)
tangential component
(c)
coriolis component
(d)
none of these
A point B on a rigid link A B moves with respect to A with angular velocity ω rad/s. The radial
component of the acceleration of B with respect to A ,
(a)
v BA × A B
where
3.
(c)
vBA
AB
(d)
2
vBA
AB
v BA = Linear velocity of B with respect to A = ω × A B
r
aBA
AB
(b)
t
aBA
AB
(c)
v BA × A B
(d)
2
vBA
AB
A point B on a rigid link A B moves with respect to A with angular velocity ω rad/s. The total
acceleration of B with respect to A will be equal to
(a)
5.
v 2BA × A B
A point B on a rigid link A B moves with respect to A with angular velocity ω rad/s. The angular
acceleration of the link A B is
(a)
4.
(b)
vector sum of radial component and coriolis component
(b)
vector sum of tangential component and coriolis component
(c)
vector sum of radial component and tangential component
(d)
vector difference of radial component and tangential component
The coriolis component of acceleration is taken into account for
(a)
slider crank mechanism
(b)
four bar chain mechanism
(c)
quick return motion mechanism
(d)
none of these
ANSWERS
1. (b)
2. (d)
3. (b)
4. (c)
5. (c)
GO To FIRST
CONTENTS
CONTENTS
232
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Theory of Machines
9
Pantograph
Fea
tur
es
eatur
tures
1. Introduction
2. Pantograph
3. Straight Line Mechanism.
4. Exact Straight Line Motion
Mechanisms Made up of
Turning Pairs.
5. Exact Straight Line Motion
Consisting of One Sliding
Pair (Scott Russel’s
Mechanism).
6. Approximate Straight Line
Motion Mechanisms.
7. Straight Line Motions for
Engine Indicators.
8. Steering Gear Mechanism.
9. Davis Steering Gear.
10. Ackerman Steering Gear.
11. Universal or Hooke’s Joint.
12. Ratio of the Shafts Velocities.
13. Maximum and Minimum
Speeds of the Driven Shaft.
14. Condition for Equal Speeds
of the Driving and Driven
Shafts.
15. Angular Acceleration of the
Driven Shaft.
16. Maximum Fluctuation of
Speed.
17.
Mechanisms
with Lower
Pairs
9.1.
Intr
oduction
Introduction
We have already discussed, that when the two elements of a pair have a surface contact and a relative motion
takes place, the surface of one element slides over the surface of the other, the pair formed is known as lower pair. In
this chapter we shall discuss such mechanisms with lower
pairs.
9.2.
Pantograph
A pantograph is an
instrument used to reproduce to an enlarged or a reduced scale and as exactly
as possible the path described by a given point.
It consists of a
Fig. 9.1. Pantograph.
jointed parallelogram
ABCD as shown in Fig. 9.1.
It is made up of bars connected by turning pairs. The bars BA
and BC are extended to O and E respectively, such that
OA/OB = AD/BE
Double Hooke’s Joint.
232
CONTENTS
CONTENTS
Chapter 9 : Mechanisms with Lower Pairs
l
233
Thus, for all relative positions of the bars,
the triangles OAD and OBE are similar and the points
O, D and E are in one straight line. It may be proved
that point E traces out the same path as described by
point D.
From similar triangles OAD and OBE, we
find that
OD/OE = A D/BE
Let point O be fixed and the points D and E
move to some new positions D′ and E′. Then
OD/OE = OD′/OE′
A little consideration will show that the
Pantograph.
straight line DD′ is parallel to the straight line EE′.
Hence, if O is fixed to the frame of a machine by means of a turning pair and D is attached to a point
in the machine which has rectilinear motion relative to the frame, then E will also trace out a straight
line path. Similarly, if E is constrained to move in a straight line, then D will trace out a straight line
parallel to the former.
A pantograph is mostly used for the reproduction of plane areas and figures such as maps,
plans etc., on enlarged or reduced scales. It is, sometimes, used as an indicator rig in order to reproduce to a small scale the displacement of the crosshead and therefore of the piston of a reciprocating
steam engine. It is also used to guide cutting tools. A modified form of pantograph is used to collect
power at the top of an electric locomotive.
9.3.
Straight Line Mechanisms
One of the most common forms of the constraint mechanisms is that it permits only relative
motion of an oscillatory nature along a straight line. The mechanisms used for this purpose are called
straight line mechanisms. These mechanisms are of the following two types:
1. in which only turning pairs are used, and
2. in which one sliding pair is used.
These two types of mechanisms may produce exact straight line motion or approximate straight
line motion, as discussed in the following articles.
9.4.
Exact Straight Line Motion Mechanisms Made up of Turning Pairs
The principle adopted for a mathematically correct
or exact straight line motion is described in Fig.9.2. Let O
be a point on the circumference of a circle of diameter OP.
Let O A be any chord and B is a point on O A produced, such
that
O A × OB = constant
Then the locus of a point B will be a straight line
perpendicular to the diameter OP. This may be proved as
follows:
Draw BQ perpendicular to OP produced. Join AP.
The triangles OAP and OBQ are similar.
Fig. 9.2. Exact straight line
motion mechanism.
234
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Theory of Machines
OA OQ
=
OP OB
∴
or
OP × OQ = O A × OB
OA × OB
OP
But OP is constant as it is the diameter of a circle, therefore, if O A × OB is constant, then OQ will be constant. Hence
the point B moves along the straight path BQ which is perpendicular to OP.
Following are the two well known types of exact straight
line motion mechanisms made up of turning pairs.
OQ =
or
1. Peaucellier mechanism. It consists of a fixed link
OO1 and the other straight links O1A , OC, OD, A D, DB, BC and
CA are connected by turning pairs at their intersections, as shown
in Fig. 9.3. The pin at A is constrained to move along the circumference of a circle with the fixed diameter OP, by means of
the link O1A . In Fig. 9.3,
AC = CB = BD = D A ; OC = OD ; and OO1 = O1A
It may be proved that the product O A × OB remains
constant, when the link O1A rotates. Join CD to bisect A B at R.
Now from right angled triangles ORC and BRC, we have
and
OC2 = OR2 + RC2
...(i)
BC2
...(ii)
=
RB2
+
RC2
A modified form of pantograph is
used to collect electricity at the
top of electric trains and buses.
Subtracting equation (ii) from (i), we have
OC2 – BC2 = OR2 – RB2
= (OR + RB) (OR – RB)
= OB × O A
Since OC and BC are of constant length, therefore
the product OB × O A remains constant. Hence the point B
traces a straight path perpendicular to the diameter OP.
Fig. 9.3. Peaucellier mechanism.
2. Hart’s mechanism. This mechanism requires only
six links as compared with the eight links required by the
Peaucellier mechanism. It consists of a fixed link OO1 and other straight links O1A , FC, CD, DE and
EF are connected by turning pairs at their points of intersection, as shown in Fig. 9.4. The links FC
and DE are equal in length and the lengths of the links CD and EF are also equal. The points O, A and
B divide the links FC, CD and EF in the same ratio. A little consideration will show that BOCE is a
trapezium and O A and OB are respectively parallel to * FD and CE.
Hence OAB is a straight line. It may be proved now that the product O A × OB is constant.
*
In ∆ FCE, O and B divide FC and EF in the same ratio, i.e.
CO/CF = EB/EF
∴ OB is parallel to CE. Similarly, in triangle FCD, O A is parallel to FD.
Chapter 9 : Mechanisms with Lower Pairs
l
235
From similar triangles CFE and OFB,
CE OB
=
FC OF
or
OB =
CE × OF
FC
...(i)
OA =
FD × OC
FC
...(ii)
and from similar triangles FCD and OCA
FD OA
=
FC OC
or
Fig. 9.4. Hart’s mechanism.
Multiplying equations (i) and (ii), we have
FD × OC CE × OF
OC × OF
OA × OB =
×
= FD × CE ×
FC
FC
FC 2
Since the lengths of OC, OF and FC are fixed, therefore
OA × OB = FD × CE × constant
...(iii)
OC × OF
= constant
... substituting
2
FC
Now from point E, draw EM parallel to CF and EN perpendicular to FD. Therefore
...(∵ CE = FM )
FD × CE = FD × FM
= (FN + ND) (FN – M N) = FN2 – ND2
...(∵ MN = ND)
= (FE2 – NE2) – (ED2 – NE2)
...(From right angled triangles FEN and EDN )
= FE2 – ED2 = constant
...(iv)
...(∵ Length FE and ED are fixed)
From equations (iii) and (iv),
OA × OB = constant
It therefore follows that if the mechanism is pivoted about O as a fixed point and the point A
is constrained to move on a circle with centre O1, then the point B will trace a straight line perpendicular to the diameter OP produced.
Note: This mechanism has a great practical disadvantage that even when the path of B is short, a large amount
of space is taken up by the mechanism.
9.5.
Exact Straight Line Motion Consisting of One Sliding Pair-Scott
Russell’s Mechanism
It consists of a fixed member and moving member P of a sliding pair as shown in Fig. 9.5.
236
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Theory of Machines
The straight link PA Q is connected by turning pairs to the link O A and the link P. The link O A rotates
about O. A little consideration will show that the mechanism OAP is same as that of the reciprocating
engine mechanism in which O A is the crank and PA is the
connecting rod. In this mechanism, the straight line motion is not generated but it is merely copied.
In Fig. 9.5, A is the middle point of PQ and
O A = AP = A Q. The instantaneous centre for the link PA Q
lies at I in O A produced and is such that IP is perpendicular to OP. Join IQ. Then Q moves along the perpendicular
to IQ. Since OPIQ is a rectangle and IQ is perpendicular
to OQ, therefore Q moves along the vertical line OQ for
all positions of QP. Hence Q traces the straight line OQ′. If
Fig. 9.5. Scott Russell’s mechanism.
O A makes one complete revolution, then P will oscillate
along the line OP through a distance 2 O A on each side of O and Q will oscillate along OQ′ through
the same distance 2 O A above and below O. Thus, the locus of Q is a copy of the locus of P.
Note: Since the friction and wear of a sliding pair is much more than those of turning pair, therefore this
mechanism is not of much practical value.
9.6.
Approximate Straight Line Motion Mechanisms
The approximate straight line motion mechanisms are the modifications of the four-bar chain
mechanisms. Following mechanisms to give approximate straight line motion, are important from the
subject point of view :
1. Watt’s mechanism. It is a crossed four bar chain mechanism and was used by Watt for his
early steam engines to guide the piston rod in a cylinder to have an approximate straight line motion.
Fig. 9.6. Watt’s mechanism.
In Fig. 9.6, OBAO1 is a crossed four bar chain in which O and O1 are fixed. In the mean
position of the mechanism, links OB and O1A are parallel and the coupling rod A B is perpendicular to
O1A and OB. The tracing point P traces out an approximate straight line over certain positions of its
movement, if PB/PA = O1A/OB. This may be proved as follows :
A little consideration will show that in the initial mean position of the mechanism, the instantaneous centre of the link B A lies at infinity. Therefore the motion of the point P is along the vertical
line B A . Let OB′ A ′O1 be the new position of the mechanism after the links OB and O1A are displaced
through an angle θ and φ respectively. The instantaneous centre now lies at I. Since the angles θ and
φ are very small, therefore
arc B B′ = arc A A′
or
OB × θ = O1 A × φ
...(i)
Chapter 9 : Mechanisms with Lower Pairs
∴
Also
∴
l
237
OB / O1 A = φ / θ
A ′P′ = IP′ × φ, and B′P′ = IP′ × θ
A ′P′ / B′P′ = φ / θ
...(ii)
From equations (i) and (ii),
OB
A′P′ AP
=
=
O1 A B′P′ BP
or
O1 A PB
=
OB
PA
Thus, the point P divides the link A B into two parts whose lengths are inversely proportional
to the lengths of the adjacent links.
2. Modified Scott-Russel mechanism. This mechanism, as shown in Fig. 9.7, is similar to
Scott-Russel mechanism (discussed in Art. 9.5), but in this case AP is not equal to A Q and the points
P and Q are constrained to move in the horizontal and vertical directions. A little consideration will
show that it forms an elliptical trammel, so that any point A
on PQ traces an ellipse with semi-major axis A Q and semiminor axis AP.
If the point A moves in a circle, then for point Q to
move along an approximate straight line, the length OA must
be equal (AP)2 / A Q. This is limited to only small
displacement of P.
3. Grasshopper mechanism. This mechanism is a
modification of modified Scott-Russel’s mechanism with
the difference that the point P does not slide along a straight
line, but moves in a circular arc with centre O.
Fig. 9.7. Modified Scott-Russel
mechanism.
It is a four bar mechanism and all the pairs are turning pairs as shown in Fig. 9.8. In this
mechanism, the centres O and O1 are fixed. The link O A oscillates about O through an angle A O A1
which causes the pin P to move along a circular arc with
O1 as centre and O1P as radius. For small angular displacements of OP on each side of the horizontal, the point
Q on the extension of the link PA traces out an approximately a straight path QQ′, if the lengths are such that O A
= (AP)2 / A Q.
Note: The Grasshopper mechanism was used in early days as
an engine mechanism which gave long stroke with a very short
crank.
4. Tchebicheff’s mechanism. It is a four bar
mechanism in which the crossed links O A and O1B are of
Fig. 9.8. Grasshopper mechanism.
equal length, as shown in Fig. 9.9. The point P, which is
the mid-point of A B traces out an approximately straight
line parallel to OO1. The proportions of the links are, usually, such that point P is exactly above O or
O1 in the extreme positions of the mechanism i.e. when B A lies along O A or when B A lies along BO1.
It may be noted that the point P will lie on a straight line parallel to OO1, in the two extreme positions
and in the mid position, if the lengths of the links are in proportions A B : OO1 : O A = 1 : 2 : 2.5.
5. Roberts mechanism. It is also a four bar chain mechanism, which, in its mean position, has
the form of a trapezium. The links O A and O1 B are of equal length and OO1 is fixed. A bar PQ is
rigidly attached to the link A B at its middle point P.
238
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Theory of Machines
A little consideration will show that if the mechanism is displaced as shown by the dotted
lines in Fig. 9.10, the point Q will trace out an approximately straight line.
Fig. 9.9. Tchebicheff’s mechanism.
9.7.
Fig. 9.10. Roberts mechanism
Straight Line Motions for Engine Indicators
The application of straight
line motions is mostly found in the
engine indicators. In these
instruments, the cylinder of the
indicator
is
in
direct
communication with the steam or
gas inside the cylinder of an
engine. The indicator piston rises
and falls in response to pressure
variation within the engine
cylinder. The piston is resisted by
a spring so that its displacement is
a direct measure of the steam or
gas pressure acting upon it. The
displacement is communicated to
the pencil which traces the
variation of pressure in the
cylinder (also known as indicator
diagram) on a sheet of paper
wrapped on the indicator drum
which oscillates with angular
motion about its axis, according to
the motion of the engine piston.
The variation in pressure is
recorded to an enlarged scale.
Following are the various engine
indicators which work on the
straight line motion mechanism.
Internal damper
absorbs shock
Hydraulic
cylinder folds
wheels for
storage
Liquid
spring
Tyres absorb some
energy
Airplane’s Landing Gear.
Note : This picture is given as additional information and is not a direct
example of the current chapter.
1. Simplex indicator. It closely resembles to the pantograph copying mechanism, as shown
in Fig. 9.11. It consists of a fixed pivot O attached to the body of the indicator. The links A B, BC, CD
Chapter 9 : Mechanisms with Lower Pairs
l
239
and D A form a parallelogram and are pin jointed. The link BC is extended to point P such that O, D
and P lie in one straight line. The point D is attached to the piston rod of the indicator and moves
along the line of stroke of the piston (i.e. in the vertical direction). A little consideration will show
that the displacement of D is reproduced on an enlarged scale, on the paper wrapped on the indicator
drum, by the pencil fixed at point P which describes the path similar to that of D. In other words,
when the piston moves vertically by a distance DD1, the path traced by P is also a vertical straight line
PP1, as shown in Fig. 9.11.
Fig. 9.11. Simplex indicator.
The magnification may be obtained by the following relation :
OP OB BP
PP1
=
=
=
OD OA BC DD1
From the practical point of view, the following are the serious objections to this mechanism:
(a) Since the accuracy of straight line motion of P depends upon the accuracy of motion of
D, therefore any deviation of D from a straight path involves a proportionate deviation of
P from a straight path.
(b) Since the mechanism has five pin joints at O, A , B, C and D, therefore slackness due to
wear in any one of pin joints destroys the accuracy of the motion of P.
2. Cross-by indicator. It is a modified form
of the pantograph copying mechanism, as shown in
Fig. 9.12.
In order to obtain a vertical straight line
for P, it must satisfy the following two conditions:
1. The point P must lie on the line joining
the points O and A , and
2. The velocity ratio between points P and
A must be a constant.
This can be proved by the instantaneous
centre method as discussed below :
The instantaneous centre I1 of the link A C
is obtained by drawing a horizontal line from A
Fig. 9.12. Cross-by indicator.
to meet the line ED produced at I1. Similarly, the
instantaneous centre I2 of the link BP is obtained by drawing a horizontal line from P to meet the
line BO at I2 . We see from Fig. 9.12, that the points I1 and I2 lie on the fixed pivot O. Let v A, v B,
v C and v P be the velocities of the points A , B, C and P respectively.
vC
I C I C
= 1 = 2
...(i)
We know that
vA
I1 A I 2 A
240
and
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Theory of Machines
vP
I P
= 2
vC I 2 C
Multiplying equations (i) and (ii), we get
vC vP
I C I P
×
= 2 × 2
vA vC
I2 A I2 C
...(ii)
or
vP
I P OP
= 2 =
vA
I 2 A OA
...(iii)
...(∵ O and I2 are same points.)
Since A C is parallel to OB, therefore triangles PA C and POB are similar.
OP BP
=
...(iv)
OA BC
From equations (iii) and (iv),
vP OP BP
=
=
= constant
...(∵ Lengths BP and BC are constant.)
vA OA BC
3. Thompson indicator. It consists of the links OB, BD, DE and EO. The tracing point P lies
on the link BD produced. A little consideration will show that it constitutes a straight line motion of
the Grasshopper type as discussed in Art.9.6. The link BD gets the motion from the piston rod of the
indicator at C which is connected by the link A C at A to the end of the indicator piston rod. The
condition of velocity ratio to be constant between P and A may be proved by the instantaneous centre
method, as discussed below :
∴
Fig. 9.13. Thompson indicator.
Draw the instantaneous centres I1 and I2 of the links BD and A C respectively. The line I1P
cuts the links A C at F. Let v A, v C and v P be the velocities of the points A , C and P respectively.
vC I 2 C
=
∴
...(i)
vA I 2 A
From similar triangles I1CF and I2C A
I 2 C I1 C
=
I 2 A I1 F
or
vC I 2 C I1 C
=
=
vA I 2 A I1 F
...(ii)
...[From equation (i)]
vP
I P
= 1
Also
vC I1 C
Multiplying equations (ii) and (iii), we get
vC vP
I C I P
v
I P
×
= 1 × 1
or P = 1
vA vC I1 F I1 C
vA I1 F
...(iii)
...(iv)
Chapter 9 : Mechanisms with Lower Pairs
l
241
Now if the links A C and OB are parallel, the triangles PCF and PBI1 are similar.
I1 P BP
=
...(v)
∴
I1 F BC
From equations (iv) and (v),
vP
I P BP
= 1 =
= constant
...(∵ Lengths BP and BC are constant)
vA I1 F BC
Note: The links A C and OB can not be exactly parallel, nor the line I1P be exactly perpendicular to the line of
stroke of the piston for all positions of the mechanism. Hence the ratio BP/BC cannot be quite constant. Since
the variations are negligible for all practical purposes, therefore the above relation gives fairly good results.
4. Dobbie Mc Innes indicator. It is similar to Thompson indicator with the difference that the
motion is given to the link DE (instead of BD in Thompson indicator) by the link A C connected to the
indicator piston as shown in Fig. 9.14. Let v A, v C, v D and v P be the velocities of the points A , C, D and
P respectively. The condition of velocity ratio (i.e. v P / vA) to be constant between points P and A may
be determined by instantaneous centre method as discussed in Thompson indicator.
Fig. 9.14. Dobbie McInnes indicator.
Draw the instantaneous centres I1 and I2 of the links BD and A C respectively. The line I1P
cuts the link A C at F. Draw DH perpendicular to I1P. We know that
vC I 2 C
=
∴
...(i)
vA I 2 A
From similar triangles I1CF and I2C A,
vC I 2 C I1 C
I 2 C I1 C
=
=
=
or
vA
I 2 A I1 F
I 2 A I1 F
Again from similar triangles I1CF and I1DH,
I1 C
I D
vC
I D
= 1
= 1
or
I1 F I1 H
vA
I1 H
Since the link ED turns about the centre E, therefore
vD
ED
=
vC
EC
...[From equation (i)] ...(ii)
...[From equation (ii)] ...(iii)
...(iv)
242
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Theory of Machines
Also
vP
I P
= 1
vD I1 D
...(v)
Multiplying equations (iii), (iv) and (v), we get
vC vD vP
I D ED I1 P
×
×
= 1 ×
×
vA vC vD
I1 H EC I1 D
or
vP
I P ED
= 1 ×
vA I1 H EC
...(vi)
From similar triangles I1BP and PDH,
I1 P
PB
=
I1 H BD
∴
vP
PB ED
=
×
= constant
vA
BD EC
...[From equation (vi)]
...[∵ Lengths PB, BD, ED and EC are constant.]
9.8.
Steering Gear Mechanism
The steering gear mechanism is used for
changing the direction of two or more of the wheel
axles with reference to the chassis, so as to move the
automobile in any desired path. Usually the two back
wheels have a common axis, which is fixed in direction with reference to the chassis and the steering is
done by means of the front wheels.
In automobiles, the front wheels are placed
over the front axles, which are pivoted at the points
A and B, as shown in Fig. 9.15. These points are fixed to the chassis. The back wheels are placed
over the back axle, at the two ends of the differential tube. When the vehicle takes a turn, the
front wheels along with the respective axles turn about the respective pivoted points. The back
wheels remain straight and do not turn. Therefore, the steering is done by means of front wheels
only.
Fig. 9.15. Steering gear mechanism.
Chapter 9 : Mechanisms with Lower Pairs
l
243
In order to avoid skidding
(i.e. slipping of the wheels sideways), the two front wheels must
turn about the same instantaneous
centre I which lies on the axis of
the back wheels. If the instantaneous centre of the two front
wheels do not coincide with the instantaneous centre of the back
wheels, the skidding on the front
or back wheels will definitely take place, which will cause more wear and tear of the tyres.
Thus, the condition for correct steering is that all the four wheels must turn about the same
instantaneous centre. The axis of the inner wheel makes a larger turning angle θ than the angle φ
subtended by the axis of outer wheel.
Let
a = Wheel track,
b = Wheel base, and
c = Distance between the pivots A and B of the front axle.
Now from triangle IBP,
BP
cot θ =
IP
and from triangle IAP,
AP AB + BP AB BP c
cot φ =
=
=
+
= + cot θ
IP
IP
IP
IP
b
∴ cot φ – cot θ = c / b
...(∵ IP = b)
This is the fundamental equation for correct steering. If this condition is satisfied, there will
be no skidding of the wheels, when the vehicle takes a turn.
9.9.
Davis Steering Gear
The Davis steering gear is shown in Fig. 9.16. It is an exact steering gear mechanism. The
slotted links A M and BH are attached to the front wheel axle, which turn on pivots A and B respectively. The rod CD is constrained to move in the direction of its length, by the sliding members at P
and Q. These constraints are connected to the slotted link A M and BH by a sliding and a turning pair
at each end. The steering is affected by moving CD to the right or left of its normal position. C ′D′
shows the position of CD for turning to the left.
Let
a = Vertical distance between A B and CD,
b = Wheel base,
d = Horizontal distance between A C and BD,
c = Distance between the pivots A and B of the front axle.
x = Distance moved by A C to A C ′ = CC ′ = DD′, and
α = Angle of inclination of the links A C and BD, to the vertical.
From triangle A A′ C′,
tan (α + φ) =
A′ C ′ d + x
=
A A′
a
...(i)
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Theory of Machines
From triangle A A′C,
tan α =
A′C d
=
A A′ a
...(ii)
B ′D ′ d – x
=
BB ′
a
...(iii)
From triangle BB′D′,
tan ( α – θ) =
Fig. 9.16. Davis steering gear.
We know that
tan (α + φ) =
tan α + tan φ
1 – tan α .tan φ
d +x
d / a + tan φ
d + a tan φ
=
=
a
1 – d / a × tan φ a – d tan φ
or
...[From equations (i) and (ii)]
(d + x) (a – d tan φ) = a (d + a tan φ)
a. d – d tan φ + a. x – d.x tan φ = a.d + a2 tan φ
2
tan φ (a2 + d2 + d.x) = ax
or
tan φ =
a.x
a 2 + d 2 + d .x
d – x
, we get
a
ax
tan θ = 2
a + d 2 – d .x
We know that for correct steering,
c
1
1
c
=
–
or
cot φ – cot θ =
tan φ tan θ b
b
...(iv)
Similarly, from tan (α – θ) =
a2 + d 2 + d . x a2 + d 2 – d . x c
–
=
a.x
a. x
b
...(v)
...[From equations (iv) and (v)]
Chapter 9 : Mechanisms with Lower Pairs
or
∴
2d . x c
=
a. x
b
c
2 tan α =
b
or
or
2d c
=
a
b
c
tan α =
2b
l
245
...(∵ d / a = tan α)
Note: Though the gear is theoretically correct, but due to the presence of more sliding members, the wear will
be increased which produces slackness between the sliding surfaces, thus eliminating the original accuracy.
Hence Davis steering gear is not in common use.
Example 9.1. In a Davis steering gear, the distance between the pivots of the front axle is 1.2
metres and the wheel base is 2.7 metres. Find the inclination of the track arm to the longitudinal axis
of the car, when it is moving along a straight path.
Solution. Given :
c = 1.2 m ; b = 2.7 m
Let
α = Inclination of the track arm to the longitudinal axis.
We know that tan α =
c
1.2
=
= 0.222
2b 2 × 2.7
or
α = 12.5° Ans.
9.10. Ackerman Steering Gear
The Ackerman steering gear mechanism is much simpler than Davis gear. The difference
between the Ackerman and Davis steering gears are :
1. The whole mechanism of the Ackerman steering gear is on back of the front wheels;
whereas in Davis steering gear, it is in front of the wheels.
2. The Ackerman steering gear consists of turning pairs, whereas Davis steering gear
consists of sliding members.
Fig. 9.17. Ackerman steering gear.
In Ackerman steering gear, the mechanism ABCD is a four bar crank chain, as shown in Fig.
9.17. The shorter links BC and A D are of equal length and are connected by hinge joints with front
wheel axles. The longer links A B and CD are of unequal length. The following are the only three
positions for correct steering.
1. When the vehicle moves along a straight path, the longer links A B and CD are parallel and
the shorter links BC and A D are equally inclined to the longitudinal axis of the vehicle, as shown by
firm lines in Fig. 9.17.
2. When the vehicle is steering to the left, the position of the gear is shown by dotted lines in
Fig. 9.17. In this position, the lines of the front wheel axle intersect on the back wheel axle at I, for
correct steering.
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Theory of Machines
3. When the vehicle is steering to the right, the similar position may be obtained.
In order to satisfy the fundamental equation for correct steering, as discussed in Art. 9.8, the
links A D and DC are suitably proportioned. The value of θ and φ may be obtained either graphically
or by calculations.
9.11. Universal or Hooke’s Joint
A *Hooke’s joint is used to connect two shafts, which are intersecting at a small angle, as
shown in Fig. 9.18. The end of each shaft is forked to U-type and each fork provides two bearings
Body 1
Axis 2
Body 2
Axis 1
Fig. 9.18. Universal or Hooke’s joint.
for the arms of a cross. The arms of the cross are perpendicular to each other. The motion is transmitted from the driving shaft to driven shaft through a cross. The inclination of the two shafts may be
constant, but in actual practice it varies, when the motion is transmitted. The main application of the
Universal or Hooke’s joint is found in the transmission from the **gear box to the differential or back
axle of the automobiles. It is also used for transmission of power to different spindles of multiple
drilling machine. It is also used as a knee joint in milling machines.
Universal Joint.
*
**
This joint was first suggested by Da Vinci and was named after English physicist and mathematician
Robert Hooke who first applied it to connect two offset misaligned shafts.
In case of automobiles, we use two Hooke’s joints one at each end of the propeller shaft, connecting the
gear box on one end and the differential on the other end.
Chapter 9 : Mechanisms with Lower Pairs
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247
9.12. Ratio of the Shafts Velocities
The top and front views connecting the two shafts by
a universal joint are shown in Fig. 9.19. Let the initial position of the cross be such that both arms lie in the plane of the
paper in front view, while the arm A B attached to the driving
shaft lies in the plane containing the axes of the two shafts.
Let the driving shaft rotates through an angle θ, so that the
arm A B moves in a circle to a new position A 1 B 1 as shown in
front view. A little consideration will show that the arm CD
will also move in a circle of the same size. This circle when
projected in the plane of paper appears to be an ellipse. Therefore the arm CD takes new position C1D1 on the ellipse, at an
angle θ. But the true angle must be on the circular path. To
find the true angle, project the point C1 horizontally to intersect the circle at C2. Therefore the angle COC2 (equal to φ) is
the true angle turned by the driven shaft. Thus when the driving shaft turns through an angle θ, the driven shaft turns
through an angle φ. It may be noted that it is not necessary
that φ may be greater than θ or less than θ. At a particular
point, it may be equal to θ.
In triangle OC1M, ∠ OC1M = θ
∴
tan θ =
OM
MC1
...(i)
and in triangle OC2 N, ∠ OC2 N = φ
∴
tan φ =
Fig. 9.19. Ratio of
velocities.
ON
ON
=
NC2 MC1
...(3 NC2 = MC1)
shafts
...(ii)
Dividing equation (i) by (ii),
tan θ OM MC1 OM
=
×
=
tan φ MC1 ON
ON
But
OM = ON1 cos α = ON cos α
...(where α = Angle of inclination of the driving and driven shafts)
∴
or
Let
tan θ ON cos α
=
= cos α
ON
tan φ
tan θ = tan φ . cos α
...(iii)
ω = Angular velocity of the driving shaft = dθ / dt
ω1 = Angular velocity of the driven shaft = dφ / dt
Differentiating both sides of equation (iii),
sec2 θ × dθ / dt = cos α . sec2 φ × dφ / dt
sec2 θ × ω = cos α . sec2 φ × ω1
∴
ω1
sec 2 θ
1
=
=
2
2
ω cos α .sec φ cos θ .cos α .sec2 φ
...(iv)
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Theory of Machines
We know that
sec 2 φ = 1 + tan 2 φ = 1 +
=1+
=
=
tan 2 θ
cos2 α
...[From equation (iii)]
sin 2 θ
cos 2 θ .cos 2 α + sin 2 θ
=
cos 2 θ .cos 2 α
cos 2 θ .cos 2 α
cos 2 θ (1 – sin 2 α ) + sin 2 θ
cos 2 θ.cos2 α
=
cos 2 θ – cos 2 θ .sin 2 α + sin 2 θ
cos 2 θ.cos 2 α
1 – cos2 θ .sin 2 α
cos 2 θ .cos 2 α
...(∵ cos2 θ + sin2 θ = 1)
Substituting this value of sec2 φ in equation (iv), we have veloity ratio,
1
cos2 θ .cos2 α
cos α
ω1
=
×
=
ω cos 2 θ .cos α 1 – cos 2 θ .sin 2 α 1 – cos 2 θ .sin 2 α
Note: If
...(v)
N = Speed of the driving shaft in r.p.m., and
N1 = Speed of the driven shaft in r.p.m.
Then the equation (v) may also be written as
N1
cos α
=
.
N 1 – cos2 θ .sin 2 α
9.13. Maximum and Minimum Speeds of Driven Shaft
We have discussed in the previous article that velocity ratio,
ω1
cos α
=
ω 1 – cos 2 θ .sin 2 α
or
ω1 =
ω .cos α
1 – cos2 θ .sin 2 α
...(i)
The value of ω1 will be maximum for a given value of α, if the denominator of equation (i) is
minimum. This will happen, when
cos2 θ = 1,
i.e. when θ = 0°, 180°, 360° etc.
∴ Maximum speed of the driven shaft,
ω1( max) =
or
N1( max) =
ω cos α
1 – sin α
2
=
ω cos α
cos α
2
=
ω
cos α
N
cos α
...(ii)
...(where N and N 1 are in r.p.m.)
Similarly, the value of ω1 is minimum, if the denominator of equation (i) is maximum. This
will happen, when (cos2 θ . sin2 α) is maximum, or
cos2 θ = 0, i.e. when θ = 90°, 270° etc.
∴ Minimum speed of the driven shaft,
ω1 (min) = ω cos α
or
N1 (min) = N cos α
...(where N and N 1 are in r.p.m.)
Fig. 9.20, shows the polar diagram depicting the salient features of the driven shaft speed.
Chapter 9 : Mechanisms with Lower Pairs
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249
From above, we see that
1. For one complete revolution of the driven shaft,
there are two points i.e. at 0° and 180° as shown by points
1 and 2 in Fig. 9.20, where the speed of the driven shaft is
maximum and there are two points i.e. at 90° and 270° as
shown by point 3 and 4 where the speed of the driven shaft
is minimum.
2. Since there are two maximum and two minimum speeds of the driven shaft, therefore there are four
points when the speeds of the driven and driver shaft are
Fig. 9.20. Polar diagram-salient
same. This is shown by points, 5,6,7 and 8 in Fig. 9.20 (See
features of driven shaft
Art 9.14).
speed.
3. Since the angular velocity of the driving shaft is
usually constant, therefore it is represented by a circle of radius ω. The driven shaft has a variation in
angular velocity, the maximum value being ω/cos α and minimum value is ω cos α. Thus it is represented by an ellipse of semi-major axis ω/cos α and semi-minor axis ω cos α, as shown in Fig. 9.20.
Note: Due to the variation in speed of the driven shaft, there will be some vibrations in it, the frequency of
which may be decreased by having a heavy mass (a sort of flywheel) on the driven shaft. This heavy mass of
flywheel does not perform the actual function of flywheel.
9.14. Condition for Equal Speeds of the Driving and Driven Shafts
and
We have already discussed that the ratio of the speeds of the driven and driving shafts is
ω1
cos α
ω1 (1 – cos2 θ .sin 2 α)
=
or
ω
=
2
2
ω 1 – cos θ .sin α
cos α
For equal speeds, ω = ω1, therefore
cos α = 1 – cos2 θ . sin2 α
or
cos2 θ . sin2 α = 1 – cos α
1 – cos α
cos 2 θ =
...(i)
sin 2 α
We know that sin 2 θ = 1 – cos2 θ = 1 –
1 − cos α
=1–
1 – cos α
sin α
1 – cos2 α
1 – cos α
1
cos α
=1–
=1–
=
(1 + cos α) (1 – cos α )
1 + cos α 1 + cos α
Dividing equation (ii) by equation (i),
sin 2 θ
cos α
sin 2 α
=
×
cos 2 θ 1 + cos α 1 – cos α
tan 2 θ =
or
∴
2
...(ii)
cos α sin 2 α cos α .sin 2 α
=
= cos α
1 – cos 2 α
sin 2 α
tan θ = ± cos α
There are two values of θ corresponding to positive sign and two values corresponding to
negative sign. Hence, there are four values of θ, at which the speeds of the driving and driven shafts
are same. This is shown by point 5, 6, 7 and 8 in Fig. 9.20.
9.15. Angular Acceleration of the Driven Shaft
We know that
ω1 =
ω cos α
1 – cos 2 θ .sin 2 α
= ω .cos α (1 – cos2 θ .sin 2 α ) –1
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Theory of Machines
Differentiating the above expression, we have the angular acceleration of the driven shaft,
d ω1
dθ
= ω cos α –1(1 – cos 2 θ sin 2 α) – 2 × (2 cos θ sin θ sin 2 α )
dt
dt
– ω2 cos α × sin 2 θ .sin 2 α
=
...(i)
(1 – cos 2 θ sin 2 α )2
...( 2 cos θ sin θ = sin 2 θ, and dθ/dt = ω)
The negative sign does not show that there is always retardation. The angular acceleration
may be positive or negative depending upon the value of sin 2 θ. It means that during one complete
revolution of the driven shaft, there is an angular acceleration corresponding to increase in speed of
ω1 and retardation due to decrease in speed of ω1.
For angular acceleration to be maximum, differentiate dω1 / dt with respect to θ and equate to
zero. The result is * approximated as
cos 2 θ =
sin 2 α (2 – cos2 2 θ)
2 – sin 2 α
Note: If the value of α is less than 30°, then cos 2 θ may approximately be written as
cos 2 θ =
2sin 2 α
2 – sin 2 α
9.16. Maximum Fluctuation of Speed
We know that the maximum speed of the driven shaft,
ω1 (max) = ω/cos α
and minimum speed of the driven shaft,
ω1 (min) = ω cos α
∴ Maximum fluctuation of speed of the driven shaft,
ω
– ω cos α
cos α
1 – cos 2 α ω sin 2 α
1
– cos α = ω
= ω
cos α = cos α
cos α
= ω tan α . sin α
q = ω1( max ) – ω1( min ) =
Since α is a small angle, therefore substituting cos α = 1, and sin α = α radians.
∴ Maximum fluctuation of speed
= ω . α2
Hence, the maximum fluctuation of speed of the driven shaft approximately varies as the
square of the angle between the two shafts.
Note: If the speed of the driving shaft is given in r.p.m. (i.e. N r.p.m.), then in the above relations ω may be
replaced by N.
9.17. Double Hooke’s Joint
We have seen in the previous articles, that the velocity of the driven shaft is not constant, but
varies from maximum to minimum values. In order to have a constant velocity ratio of the driving and
driven shafts, an intermediate shaft with a Hooke’s joint at each end as shown in Fig. 9.21, is used.
This type of joint is known as double Hooke’s joint.
*
Since the differentiation of dω1/dt is very cumbersome, therefore only the result is given.
Chapter 9 : Mechanisms with Lower Pairs
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251
Let the driving, intermediate and driven shafts, in the same time, rotate through angles θ, φ
and γ from the position as discussed previously in Art. 9.12.
Now for shafts A and B,
tan θ = tan φ . cos α
...(i)
and
tan γ = tan φ . cos α
for shafts B and C,
...(ii)
From equations (i) and (ii), we see that θ = γ or ωA = ωC.
Fig. 9.21. Double Hooke’s joint.
This shows that the speed of the driving and driven shaft is constant. In other words, this joint
gives a velocity ratio equal to unity, if
1. The axes of the driving and driven shafts are in the same plane, and
2. The driving and driven shafts make equal angles with the intermediate shaft.
Example. 9.2. Two shafts with an included angle of 160° are connected by a Hooke’s joint.
The driving shaft runs at a uniform speed of 1500 r.p.m. The driven shaft carries a flywheel of mass
12 kg and 100 mm radius of gyration. Find the maximum angular acceleration of the driven shaft
and the maximum torque required.
Solution. Given : α = 180° – 160° = 20°; N = 1500 r.p.m.; m = 12 kg ; k = 100 mm = 0.1 m
We know that angular speed of the driving shaft,
ω = 2 π × 1500 / 60 = 157 rad/s
and mass moment of inertia of the driven shaft,
I = m.k2 = 12 (0.1)2 = 0.12 kg - m2
Maximum angular acceleration of the driven shaft
Let
dω1 / dt = Maximum angular acceleration of the driven shaft, and
θ = Angle through which the driving shaft turns.
We know that, for maximum angular acceleration of the driven shaft,
cos 2 θ =
∴
and
2sin 2 α
2 – sin 2 α
2θ = 82.9°
=
2sin 2 20°
2 – sin 2 20°
= 0.124
θ = 41.45°
or
d ω1 ω2 cos α .sin 2 θ .sin 2 α
=
dt
(1 – cos 2 θ .sin 2 α )2
=
(157)2 cos 20°× sin 82.9° × sin 2 20°
(1 – cos 41.45° × sin 20°)
2
2
2
= 3090 rad/s2 Ans.
Maximum torque required
We know that maximum torque required
= I × d ω1 / dt = 0.12 × 3090 = 371 N-m Ans.
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Theory of Machines
Example. 9.3. The angle between the axes of two shafts connected by Hooke’s joint is 18°.
Determine the angle turned through by the driving shaft when the velocity ratio is maximum and
unity.
Solution. Given : α = 98°
Let
θ = Angle turned through by the driving shaft.
When the velocity ratio is maximum
We know that velocity ratio,
ω1
cos α
=
ω 1 – cos 2 θ .sin 2 α
The velocity ratio will be maximum when cos2 θ is minimum, i.e. when
cos2 θ = 1
or
when θ = 0° or
180° Ans.
When the velocity ratio is unity
The velocity ratio (ω / ω1) will be unity, when
1 – cos2 θ . sin2 α = cos α
∴
cos θ = ±
=±
∴
cos 2 θ =
or
1 – cos α
sin 2 α
1 – cos α
1 – cos α
1
=±
=±
2
2
1 + cos α
sin α
1 – cos α
1
1
=±
= ± 0.7159
1 + cos18°
1 + 0.9510
θ = 44.3°
or
135.7° Ans.
Example. 9.4. Two shafts are connected by a Hooke’s joint. The driving shaft revolves
uniformly at 500 r.p.m. If the total permissible variation in speed of the driven shaft is not to exceed
± 6% of the mean speed, find the greatest permissible angle between the centre lines of the shafts.
Solution. Given : N = 500 r.p.m. or ω = 2 π × 500 / 60 = 52.4 rad/s
Let α = Greatest permissible angle between the centre lines of the shafts.
Since the variation in speed of the driven shaft is ± 6% of the mean speed (i.e. speed of the
driving speed), therefore total fluctuation of speed of the driven shaft,
q = 12 % of mean speed (ω) = 0.12 ω
We know that maximum or total fluctuation of speed of the driven shaft (q),
1 – cos 2 α
0.12 ω = ω
cos α
and
cos α =
or cos2 α + 0.12 cos α – 1 = 0
– 0.12 ± (0.12)2 + 4 – 0.12 ± 2.0036
=
= 0.9418
2
2
α = 19.64° Ans.
...(Taking + sign)
Example. 9.5. Two shafts are connected by a universal joint. The driving shaft rotates at a
uniform speed of 1200 r.p.m. Determine the greatest permissible angle between the shaft axes so that
the total fluctuation of speed does not exceed 100 r.p.m. Also calculate the maximum and minimum
speeds of the driven shaft.
Chapter 9 : Mechanisms with Lower Pairs
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253
Solution. Given : N = 1200 r.p.m.; q = 100 r.p.m.
Greatest permissible angle between the shaft axes
Let
α = Greatest permissible angle between the shaft axes.
We know that total fluctuation of speed (q),
1 – cos 2 α
1 – cos 2 α
100 = N
1200
=
cos α
cos α
1 – cos 2 α 100
=
= 0.083
cos α
1200
cos2 α + 0.083 cos α – 1 = 0
∴
cos α =
and
∴
– 0.083 ±
(0.083)2 + 4
= 0.9593
2
...(Taking + sign)
α = 16.4° Ans.
Maximum and minimum speed of the driven shaft
We know that maximum speed of the driven shaft,
N1 (max) = N / cos α = 1200 / 0.9593 = 1251 r.p.m. Ans.
and minimum speed of the driven shaft,
N1 (min) = N cos α = 1200 × 0.9593 = 1151 r.p.m. Ans.
Example. 9.6. The driving shaft of a Hooke’s joint runs at a uniform speed of 240 r.p.m. and
the angle α between the shafts is 20°. The driven shaft with attached masses has a mass of 55 kg at
a radius of gyration of 150 mm.
1. If a steady torque of 200 N-m resists rotation of the driven shaft, find the torque required
at the driving shaft, when θ = 45°.
2. At what value of ‘α’will the total fluctuation of speed of the driven shaft be limited to 24
r.p.m ?
Solution. Given : N = 240 r.p.m or ω = 2 π × 240/60 = 25.14 rad/s ; α = 20° ; m = 55 kg ;
k = 150 mm = 0.15 m ; T 1 = 200 N-m ; θ = 45° ; q = 24 r.p.m.
1. Torque required at the driving shaft
Let
T′ = Torque required at the driving shaft.
We know that mass moment inertia of the driven shaft,
I = m.k 2 = 55 (0.15)2 = 1.24 kg-m2
and angular acceleration of the driven shaft,
d ω1 – ω2 cos α .sin 2 θ .sin 2 α – (25.14)2 cos 20°× sin 90°× sin 2 20°
=
=
dt
(1 – cos 2 θ sin 2 α) 2
(1 – cos 2 45° sin 2 20°) 2
= – 78.4 rad / s2
∴ Torque required to accelerate the driven shaft,
T2 = I ×
d ω1
= 1.24 × – 78.4 = – 97.2 N − m
dt
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Theory of Machines
and total torque required on the driven shaft,
T = T 1 + T 2 = 200 – 97.2 = 102.8 N– m
Since the torques on the driving and driven shafts are inversely proportional to their angular
speeds, therefore
T ' . ω = T . ω1
T′=
or
=
T . ω1
T cos α
=
ω
1 – cos 2 θ.sin 2 α
102.8 cos 20°
1 – cos 2 45° sin 2 20°
ω
cos α
... 3 1 =
ω 1 – cos 2 θ .sin α
= 102.6 N-m Ans.
2. Value of α for the total fluctuation of speed to be 24 r.p.m.
We know that the total fluctuation of speed of the driven shaft (q),
or
1 – cos 2 α
1 – cos 2 α
24 = N
240
=
cos α
cos α
1 – cos 2 α
24
=
= 0.1
cos α
240
cos2 α + 0.1 cos α – 1 = 0
cos α =
∴
– 0.1 ± (0.1)2 + 4
= 0.95
2
...(Taking + sign)
α = 18.2° Ans.
Example 9.7. A double universal joint is used to connect two shafts in the same plane. The
intermediate shaft is inclined at an angle of 20° to the driving shaft as well as the driven shaft. Find
the maximum and minimum speed of the intermediate shaft and the driven shaft if the driving shaft
has a constant speed of 500 r.p.m.
Solution. Given α = 20° ; N A = 500 r.p.m.
Maximum and minimum speed of the intermediate shaft
Let A , B and C are the driving shaft, intermediate shaft and driven shaft respectively. We
know that for the driving shaft (A ) and intermediate shaft (B),
Maximum speed of the intermediate shaft,
500
NA
=
= 532.1 r.p.m Ans.
cos α cos 20 °
and minimum speed of the intermediate shaft,
N B ( max ) =
N B (min) = N A cos α = 500 × cos 20° = 469.85 r.p.m. Ans.
Maximum and minimum speed of the driven shaft
We know that for the intermediate shaft (B) and driven shaft (C),
Maximum speed of the driven shaft,
N C(max ) =
N B(max )
cos α
=
NA
cos α
2
=
500
cos 2 20°
= 566.25 r.p.m. Ans.
Chapter 9 : Mechanisms with Lower Pairs
l
255
and minimum speed of the driven shaft,
N C (min) = N B (min) × cos α = N A. cos2 α
= 500 × cos2 20° = 441.5 r.p.m. Ans.
EXERCISES
1.
Fig. 9.22 shows the link GAB which oscillates on a fixed centre at A and the link FD on a fixed
centre at F. The link A B is equal to A C and DB, BE, EC and CD are equal in length.
Fig. 9.22
(a) Find the length of AF and the position of centre F so that the point E may move in a straight line.
(b) If the point E is required to move in a circle passing through centre A , what will be the path
of point D ?
[Ans. AF = FD]
(Hint. The mechanism is similar to Peaucellier’s mechanism)
2.
Fig. 9.23 shows a part of the mechanism of a circuit breaker. A and D are fixed centres and the lengths
of the links are : A B = 110 mm, BC = 105 mm, and CD = 150 mm.
All dimensions in mm.
Fig. 9.23
Fig. 9.24
Find the position of a point P on BC produced that will trace out an approximately straight vertical
path 250 mm long.
3.
The mechanism, as shown in Fig. 9.24, is a four bar kinematic chain of which the centres A and B are
fixed. The lengths are : A B = 600 mm, A C = BD = CD = 300 mm. Find the point G on the centre line
of the cross arm of which the locus is an approximately straight line even for considerable displacements from the position shown in the figure.
[Ans. 400 mm.]
(Hint : It is a Robert’s approximate straight line mechanism. Produce A C and BD to intersect at point
E. Draw a vertical line from E to cut the centre line of cross arm at G. The distance of G from CD is
the required distance).
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Theory of Machines
4.
The distance between the fixed centres O and O1 of a Watt’s straight line motion, as shown in Fig. 9.6,
is 250 mm. The lengths of the three moving links OB, B A and A O1 are 150 mm, 75 mm and 100 mm
respectively. Find the position of a point P on B A which gives the best straight line motion.
5.
A Watt’s parallel motion has two bars O A and O′B pivoted at O and O′ respectively and joined by the
link A B in the form of a crossed four bar mechanism. When the mechanism is in its mean position, the
bars O A and O′B are perpendicular to the link A B. If O A = 75 mm, O′B = 25 mm and A B = 100 mm,
find the position of the tracing point P and also find how far P is from the straight line given by the
mean position of A B, when
1. O A and OB are in one straight line, and 2. O′B and A B are in one straight line.
[Ans. 37.5 mm, 6.5 mm,12 mm]
6.
Design a pantograph for an indicator to obtain the indicator diagram of an engine. The distance from
the tracing point of the indicator is 100 mm. The indicator diagram should represent four times the gas
pressure inside the cylinder of an engine.
7.
In a Davis steering gear, the distance between the pivots of the front axle is 1 metre and the wheel base
is 2.5 metres. Find the inclination of the track arm to the longitudinal axis of the car, when it is moving
along a straight path.
[Ans. 11.17°]
8.
A Hooke’s joint connects two shafts whose axes intersect at 150°. The driving shaft rotates uniformly at 120 r.p.m. The driven shaft operates against a steady torque of 150 N-m and carries a
flywheel whose mass is 45 kg and radius of gyration 150 mm. Find the maximum torque which will be
exerted by the driving shaft.
[Ans. 187 N-m]
(Hint : The maximum torque exerted by the driving shaft is the sum of steady torque and the maximum accelerating torque of the driven shaft).
9.
Two shafts are connected by a Hooke’s joint. The driving shaft revolves uniformly at 500 r.p.m. If the
total permissible variation in speed of a driven shaft is not to exceed 6% of the mean speed, find the
greatest permissible angle between the centre lines of the shafts. Also determine the maximum and
minimum speed of the driven shaft.
[Ans. 19.6° ; 530 r.p.m. ; 470 r.p.m.]
10.
Two inclined shafts are connected by means of a universal joint. The speed of the driving shaft is 1000
r.p.m. If the total fluctuation of speed of the driven shaft is not to exceed 12.5% of this, what is the
maximum possible inclination between the two shafts?
With this angle, what will be the maximum acceleration to which the driven shaft is subjected and
when this will occur ?
[Ans. 20.4° ; 1570 rad/s2 ; 41.28°]
DO YOU KNOW ?
1.
Sketch a pantograph, explain its working and show that it can be used to reproduce to an enlarged
scale a given figure.
2.
A circle has OR as its diameter and a point Q lies on its circumference. Another point P lies on the line
OQ produced. If OQ turns about O as centre and the product OQ × OP remains constant, show that
the point P moves along a straight line perpendicular to the diameter OR.
3.
What are straight line mechanisms ? Describe one type of exact straight line motion mechanism
with the help of a sketch.
4.
Describe the Watt’s parallel mechanism for straight line motion and derive the condition under which
the straight line is traced.
5.
Sketch an intermittent motion mechanism and explain its practical applications.
6.
Give a neat sketch of the straight line motion ‘Hart mechanism.’ Prove that it produces an exact
straight line motion.
7.
(a) Sketch and describe the Peaucellier straight line mechanism indicating clearly the conditions
under which the point P on the corners of the rhombus of the mechanism, generates a straight
line.
(b) Prove geometrically that the above mechanism is capable of producing straight line.
Chapter 9 : Mechanisms with Lower Pairs
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257
8.
Draw the sketch of a mechanism in which a point traces an exact straight line. The mechanism must be
made of only revolute pairs. Prove that the point traces an exact straight line motion.
(Hint. Peaucellier straight line mechanism)
9.
Sketch the Dobbie-McInnes indicator mechanism and show that the displacement of the pencil
which traces the indicator diagram is proportional to the displacement of the indicator piston.
10.
What is the condition for correct steering ? Sketch and show the two main types of steering gears
and discuss their relative advantages.
11.
Explain why two Hooke’s joints are used to transmit motion from the engine to the differential of an
automobile.
12.
Derive an expression for the ratio of shafts velocities for Hooke’s joint and draw the polar diagram
depicting the salient features of driven shaft speed.
OBJECTIVE TYPE QUESTIONS
1. In a pantograph, all the pairs are
(a) turning pairs
(b)
sliding pairs
(c) spherical pairs
(d)
self-closed pairs
2. Which of the following mechanism is made up of turning pairs ?
(a) Scott Russel’s mechanism
(b)
Peaucellier’s mechanism
(c) Hart’s mechanism
(d)
none of these
3. Which of the following mechanism is used to enlarge or reduce the size of a drawing ?
(a) Grasshopper mechanism
(b)
Watt mechanism
(c) Pantograph
(d)
none of these
4. The Ackerman steering gear mechanism is preferred to the Davis steering gear mechanism, because
(a) whole of the mechanism in the Ackerman steering gear is on the back of the front wheels.
(b) the Ackerman steering gear consists of turning pairs
(c) the Ackerman steering gear is most economical
(d) both (a) and (b)
5. The driving and driven shafts connected by a Hooke’s joint will have equal speeds, if
(a) cos θ = sin α
(c)
(b)
tan α
(d) cot θ = cos α
tan θ = ± cos α
where
sin θ = ±
θ = Angle through which the driving shaft turns, and
α = Angle of inclination of the driving and driven shafts.
ANSWERS
1. (a)
2. (b), (c)
3. (c)
4. (d)
5. (c)
GO To FIRST
CONTENTS
CONTENTS
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10
Features (Main)
1. Introduction.
2. Types of Friction.
4. Friction Between Lubricated
Surfaces.
5. Limiting Friction.
8. Laws of Solid Friction.
9. Laws of Fluid Friction.
10. Coefficient of Friction.
11. Limiting Angle of Friction.
12. Angle of Repose.
14. Friction of a Body Lying on
a Rough Inclined Plane.
15. Efficiency of Inclined Plane.
16. Screw Friction.
17. Screw Jack.
18. Torque Required to Lift the
Load by a Screw Jack.
20. Efficiency of a Screw Jack.
21. Maximum Efficiency of a
Screw Jack.
22. Over Hauling and Self
Locking Screws.
23. Efficiency of Self Locking
Screws.
24. Friction of a V-thread.
25. Friction in Journal BearingFriction Circle.
26. Friction of Pivot and Collar
Bearing.
27. Flat Pivot Bearing.
28. Conical Pivot Bearing.
29. Trapezoidal or Truncated
Conical Pivot Bearing.
30. Flat Collar Bearing.
31. Friction Clutches.
32. Single Disc or Plate Clutch.
33. Multiple Disc Clutch.
34. Cone Clutch.
35. Centrifugal Clutches.
Friction
10.1. Introduction
It has been established since long, that the surfaces
of the bodies are never perfectly smooth. When, even a very
smooth surface is viewed under a microscope, it is found to
have roughness and irregularities, which may not be detected
by an ordinary touch. If a block of one substance is placed
over the level surface of the same or of different material, a
certain degree of interlocking of the minutely projecting particles takes place. This does not involve any force, so long
as the block does not move or tends to move. But whenever
one block moves or tends to move tangentially with respect
to the surface, on which it rests, the interlocking property of
the projecting particles opposes the motion. This opposing
force, which acts in the opposite direction of the movement
of the upper block, is called the force of friction or simply
friction. It thus follows, that at every joint in a machine, force
of friction arises due to the relative motion between two parts
and hence some energy is wasted in overcoming the friction.
Though the friction is considered undesirable, yet it plays an
important role both in nature and in engineering e.g. walking on a road, motion of locomotive on rails, transmission of
power by belts, gears etc. The friction between the wheels
and the road is essential for the car to move forward.
10.2. Types of Friction
In general, the friction is of the following two types :
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CONTENTS
CONTENTS
Chapter 10 : Friction
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259
1. Static friction. It is the friction, experienced by a body, when at rest.
2. Dynamic friction. It is the friction, experienced by a body, when in motion. The dynamic
friction is also called kinetic friction and is less than the static friction. It is of the following three
types :
(a) Sliding friction. It is the friction, experienced by a body, when it slides over another
body.
(b) Rolling friction. It is the friction, experienced between the surfaces which has balls or
rollers interposed between them.
(c) Pivot friction. It is the friction, experienced by a body, due to the motion of rotation as
in case of foot step bearings.
The friction may further be classified as :
1. Friction between unlubricated surfaces, and
2. Friction between lubricated surfaces.
These are discussed in the following articles.
10.3. Friction Between Unlubricated Surfaces
The friction experienced between two dry and unlubricated surfaces in contact is known as
dry or solid friction. It is due to the surface roughness. The dry or solid friction includes the sliding
friction and rolling friction as discussed above.
10.4. Friction Between Lubricated Surfaces
When lubricant (i.e. oil or grease) is applied between two surfaces in contact, then the friction
may be classified into the following two types depending upon the thickness of layer of a lubricant.
1. Boundary friction (or greasy friction or non-viscous friction). It is the friction,
experienced between the rubbing surfaces, when the surfaces have a very thin layer of lubricant. The thickness of this very thin layer is of the molecular dimension. In this type of friction, a
thin layer of lubricant forms a bond between the two rubbing surfaces. The lubricant is absorbed on
the surfaces and forms a thin film. This thin film of the lubricant results in less friction between
them. The boundary friction follows the laws of solid friction.
2. Fluid friction (or film friction or viscous friction). It is the friction, experienced between
the rubbing surfaces, when the surfaces have a thick layer of the lubrhicant. In this case, the actual
surfaces do not come in contact and thus do not rub against each other. It is thus obvious that fluid
friction is not due to the surfaces in contact but it is due to the viscosity and oiliness of the lubricant.
Note : The viscosity is a measure of the resistance offered to the sliding one layer of the lubricant over an
adjacent layer. The absolute viscosity of a lubricant may be defined as the force required to cause a plate of unit
area to slide with unit velocity relative to a parallel plate, when the two plates are separated by a layer of
lubricant of unit thickness.
The oiliness property of a lubricant may be clearly understood by considering two lubricants of equal
viscosities and at equal temperatures. When these lubricants are smeared on two different surfaces, it is found
that the force of friction with one lubricant is different than that of the other. This difference is due to the
property of the lubricant known as oiliness. The lubricant which gives lower force of friction is said to have
greater oiliness.
10.5. Limiting Friction
Consider that a body A of weight W is lying on a rough horizontal body B as shown in Fig.
10.1 (a). In this position, the body A is in equilibrium under the action of its own weight W , and the
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normal reaction R N (equal to W ) of B on A . Now if a small horizontal force P1 is applied to the body
A acting through its centre of gravity as shown in Fig. 10.1 (b), it does not move because of the
frictional force which prevents the motion. This shows that the applied force P1 is exactly balanced
by the force of friction F1 acting in the opposite direction.
If we now increase the applied force to P2 as shown in Fig. 10.1 (c), it is still found to be in
equilibrium. This means that the force of friction has also increased to a value F2 = P2. Thus every
time the effort is increased the force of friction also increases, so as to become exactly equal to the
applied force. There is, however, a limit beyond which the force of friction cannot increase as shown
in Fig. 10.1 (d). After this, any increase in the applied effort will not lead to any further increase in the
force of friction, as shown in Fig. 10.1 (e), thus the body A begins to move in the direction of the
applied force. This maximum value of frictional force, which comes into play, when a body just
begins to slide over the surface of the other body, is known as limiting force of friction or simply
limiting friction. It may be noted that when the applied force is less than the limiting friction, the body
remains at rest, and the friction into play is called static friction which may have any value between
zero and limiting friction.
Fig. 10.1. Limiting friction.
10.6. Laws of Static Friction
Following are the laws of static friction :
1. The force of friction always acts in a direction, opposite to that in which the body tends to
move.
2. The magnitude of the force of friction is exactly equal to the force, which tends the body
to move.
3. The magnitude of the limiting friction (F ) bears a constant ratio to the normal reaction
(R N) between the two surfaces. Mathematically
F/R N = constant
Chapter 10 : Friction
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261
4. The force of friction is independent of the area of contact, between the two surfaces.
5. The force of friction depends upon the roughness of the surfaces.
10.7. Laws of Kinetic or Dynamic Friction
Following are the laws of kinetic or dynamic friction :
1. The force of friction always acts in a direction, opposite to that in which the body is
moving.
2. The magnitude of the kinetic friction bears a constant ratio to the normal reaction between
the two surfaces. But this ratio is slightly less than that in case of limiting friction.
3. For moderate speeds, the force of friction remains constant. But it decreases slightly with
the increase of speed.
10.8. Laws of Solid Friction
Following are the laws of solid friction :
1. The force of friction is directly proportional to the normal load between the surfaces.
2. The force of friction is independent of the area of the contact surface for a given normal
load.
3. The force of friction depends upon the material of which the contact surfaces are made.
4. The force of friction is independent of the velocity of sliding of one body relative to the
other body.
10.9. Laws of Fluid Friction
Following are the laws of fluid friction :
1. The force of friction is almost independent of the load.
2. The force of friction reduces with the increase of the temperature of the lubricant.
3. The force of friction is independent of the substances of the bearing surfaces.
4. The force of friction is different for different lubricants.
10.10. Coefficient of Friction
It is defined as the ratio of the limiting friction (F) to the normal reaction (R N) between the
two bodies. It is generally denoted by µ. Mathematically, coefficient of friction,
µ = F/R N
10.11. Limiting Angle of Friction
Consider that a body A of weight (W ) is resting on a horizontal plane B, as shown in Fig. 10.2.
If a horizontal force P is applied to the body, no relative motion will
take place until the applied force P is equal to the force of friction
F, acting opposite to the direction of motion. The magnitude of this
force of friction is F = µ.W = µ.R N, where R N is the normal reaction.
In the limiting case, when the motion just begins, the body will be
in equilibrium under the action of the following three forces :
1. Weight of the body (W ),
2. Applied horizontal force (P), and
3. Reaction (R) between the body A and the plane B.
Fig. 10.2. Limiting angle of
friction.
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The reaction R must, therefore, be equal and opposite to the resultant of W and P and will be
inclined at an angle φ to the normal reaction R N. This angle φ is known as the limiting angle of friction.
It may be defined as the angle which the resultant reaction R makes with the normal reaction R N.
From Fig. 10.2, tan φ = F/RN = µ R N / R N = µ
10.12. Angle of Repose
Consider that a body A of weight (W ) is resting on
an inclined plane B, as shown in Fig. 10.3. If the angle of
inclination α of the plane to the horizontal is such that the
body begins to move down the plane, then the angle α is
called the angle of repose.
A little consideration will show that the body will
begin to move down the plane when the angle of inclination
Fig. 10.3. Angle of repose.
of the plane is equal to the angle of friction (i.e. α = φ). This
may be proved as follows :
The weight of the body (W ) can be resolved into the following two components :
1. W sin α, parallel to the plane B.
This component tends to slide the body down
the plane.
2. W cos α, perpendicular to the plane
B. This component is balanced by the normal
reaction (R N) of the body A and the plane B.
Friction is essential to provide grip between tyres
The body will only begin to move
and road. This is a positive aspect of ‘friction’.
down the plane, when
W sin α = F = µ.R N = µ.W cos α
...(∵ R N = W cos α)
∴
tan α = µ = tan φ or α = φ
...(∵ µ = tan φ)
10.13. Minimum Force Required to Slide a Body on a Rough Horizontal
Plane
Consider that a body A of weight (W ) is resting on a
horizontal plane B as shown in Fig. 10.4. Let an effort P is
applied at an angle θ to the horizontal such that the body A
just moves. The various forces acting on the body are shown
in Fig. 10.4. Resolving the force P into two components, i.e.
P sin θ acting upwards and P cos θ acting horizontally. Now
for the equilibrium of the body A ,
R N + P sin θ = W
or
and
RN = W – P sin θ
P cos θ = F = µ.R N
...(i)
Fig. 10.4. Minimum force required
to slide a body.
...(ii)
...(∵ F = µ.R N)
Substituting the value of R N from equation (i), we have
P cos θ = µ (W – P sin θ) = tan φ (W – P sin θ)
=
sin φ
(W − P sin θ )
cos φ
...(∵ µ = tan φ)
Chapter 10 : Friction
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263
P cos θ .cos φ = W sin φ – P sin θ.sin φ
P cos θ.cos φ + P sin θ.sin φ = W sin φ
P cos (θ – φ) = W sin φ
P=
...[3 cos θ. cos φ + sin θ.sin φ = cos (θ – φ)]
W sin φ
cos (θ − φ)
...(iii)
For P to be minimum, cos (θ – φ) should be maximum, i.e.
cos (θ – φ) = 1
θ – φ = 0°
or
or
θ=φ
In other words, the effort P will be minimum, if its inclination with the horizontal is equal to
the angle of friction.
∴
Pmin = W sin θ
...[From equation (iii)]
Example 10.1. A body, resting on a rough horizontal plane required a pull of 180 N inclined
at 30º to the plane just to move it. It was found that a push of 220 N inclined at 30º to the plane just
moved the body. Determine the weight of the body and the coefficient of friction.
Solution. Given : θ = 30º
Let
W = Weight of the body in newtons,
RN = Normal reaction,
µ = Coefficient of friction, and
F = Force of friction.
First of all, let us consider a pull of 180 N. The force of friction (F) acts towards left as shown
in Fig. 10.5 (a).
Resolving the forces horizontally,
F = 180 cos 30º = 180 × 0.866 = 156 N
Fig. 10.5
Now resolving the forces vertically,
RN = W – 180 sin 30º = W – 180 × 0.5 = (W – 90) N
We know that
F = µ.R N
or 156 = µ (W – 90)
...(i)
Now let us consider a push of 220 N. The force of friction (F) acts towards right as shown in Fig.
10.5 (b).
Resolving the forces horizontally,
F = 220 cos 30º = 220 × 0.866 = 190.5 N
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Theory of Machines
Now resolving the forces vertically,
RN = W + 220 sin 30º = W + 220 × 0.5 = (W + 110) N
We know that
F = µ.R N
From equations (i) and (ii),
or 190.5 = µ (W + 110)
...(ii)
W = 1000 N, and µ = 0.1714 Ans.
10.14. Friction of a Body Lying on a Rough Inclined Plane
Consider that a body of weight (W ) is lying on a plane inclined at an angle α with the horizontal, as shown in Fig. 10.6 (a) and (b).
(a) Angle of inclination less than
angle of friction.
(b) Angle of inclination more than
angle of friction.
Fig. 10.6. Body lying on a rough inclined plane.
A little consideration will show that if the inclination of the plane, with the horizontal, is less
than the angle of friction, the body will be in equilibrium as shown in Fig. 10.6 (a). If,in this condition, the body is required to be moved upwards and downwards, a corresponding force is required for
the same. But, if the inclination of the plane is more than the angle of friction, the body will move
down and an upward force (P) will be required to resist the body from moving down the plane as
shown in Fig. 10.6 (b).
Let us now analyse the various forces which act on a body when it slides either up or down an
inclined plane.
1. Considering the motion of the body up the plane
Let
W = Weight of the body,
α = Angle of inclination of the plane to the horizontal,
φ = Limiting angle of friction for the contact surfaces,
P = Effort applied in a given direction in order to cause the body to slide with
uniform velocity parallel to the plane, considering friction,
P0 = Effort required to move the body up the plane neglecting friction,
θ = Angle which the line of action of P makes with the weight of the body W ,
µ = Coefficient of friction between the surfaces of the plane and the body,
RN = Normal reaction, and
R = Resultant reaction.
Chapter 10 : Friction
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265
When the friction is neglected, the body is in equilibrium under the action of the three forces,
i.e. P0, W and R N, as shown in Fig. 10.7 (a). The triangle of forces is shown in Fig. 10.7 (b). Now
applying sine rule for these three concurrent forces,
P0
W sin α
W
or * P0 =
=
sin (θ − α )
sin α sin (θ − α )
...(i)
(a)
(b)
(c)
Fig. 10.7. Motion of the body up the plane, neglecting friction.
When friction is taken into account, a frictional force F = µ.R N acts in the direction opposite
to the motion of the body, as shown in Fig. 10.8 (a). The resultant reaction R between the plane and
the body is inclined at an angle φ with the normal reaction R N. The triangle of forces is shown in Fig.
10.8 (b). Now applying sine rule,
P
W
=
sin (α + φ) sin[θ − (α + φ)]
(a)
(b)
(c)
Fig. 10.8. Motion of the body up the plane, considering friction.
*
1. The effort P0 or (or P) may also be obtained by applying Lami’s theorem to the three forces, as
shown in Fig. 10.7 (c) and 10.8 (c). From Fig. 10.7 (c),
P0
W
=
sin (180º − α) sin [180º − ( θ − α)]
or
P0
W
=
sin α sin ( θ − α)
...[same as before]
2. The effort P0 (or P) may also be obtained by resolving the forces along the plane and perpendicular to
the plane and then applying ΣH = 0 and ΣV = 0.
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Theory of Machines
P=
∴
W sin(α + φ)
sin[θ− (α + φ)]
...(ii)
Notes : 1. When the effort applied is horizontal, then θ = 90º. In that case, the equations (i) and (ii) may be
written as
and
P0 =
W sin α
W sin α
=
= W tan α
sin (90º − α)
cos α
P=
W sin (α + φ)
W sin (α + φ)
=
= W tan (α + φ)
sin[90º − ( α + φ)
cos ( α + φ)
2. When the effort applied is parallel to the plane, then θ = 90º + α. In that case, the equations (i) and
(ii) may be written as
and
P0 =
W sin α
= W sin α
sin (90º + α − α )
P=
W sin (α + φ)
W sin (α + φ)
=
sin[(90º + α) − (α + φ)]
cos φ
=
W (sin α cos φ + cos α sin φ)
= W (sin α + cos α. tan φ)
cos φ
= W (sin α + µ cos α)
...( ∵ µ = tan φ)
2. Considering the motion of the body down the plane
Neglecting friction, the effort required for the motion down the plane will be same as for the
motion up the plane, i.e.
P0 =
(a)
W sin α
sin (θ − α)
(b)
...(iii)
(c)
Fig. 10.9. Motion of the body down the plane, considering friction.
When the friction is taken into account, the force of friction F = µ.R N will act up the plane and
the resultant reaction R will make an angle φ with R N towards its right as shown in Fig. 10.9 (a). The
triangle of forces is shown in Fig. 10.9 (b). Now from sine rule,
P
W
=
sin (α − φ) sin[θ − (α − φ)]
or
P=
W sin (α − φ)
sin[θ − (α − φ)]
...(iv)
Chapter 10 : Friction
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267
Notes : 1. The value of P may also be obtained either by applying Lami’s theorem to Fig. 10.9 (c), or by
resolving the forces along the plane and perpendicular to the plane and then using ΣH = 0 and ΣV = 0 (See Art.
10.18 and 10.19).
2. When P is applied horizontally, then θ = 90º. In that case, equation (iv) may be written as
W sin (α − φ)
W sin (α − φ)
=
= W tan (α – φ))
sin[90º − (α − φ)]
cos (α – φ)
3. When P is applied parallel to the plane, then θ = 90° + α. In that case, equation (iv) may be
written as
W sin (α − φ)
W sin (α − φ)
=
P=
sin[90º + α ) − (α − φ)]
cos φ
P=
=
W (sin α cos φ − cos α sin φ)
= W (sin α − tan φ cos α )
cos φ
= W (sin α – µ cos α)
...(∵ tan φ = µ)
10.15. Efficiency of Inclined Plane
The ratio of the effort required neglecting friction (i.e. P0) to the effort required considering
friction (i.e. P) is known as efficiency of the inclined plane. Mathematically, efficiency of the inclined
plane,
η = P0 / P
Let us consider the following two cases :
1. For the motion of the body up the plane
Efficiency,
η=
=
P0
P
=
sin[θ − (α + φ)]
W sin α
×
sin (θ − α)
W sin (α + φ)
sin α
sin θ cos (α + φ) − cos θ sin (α + φ)
×
sin θ cos α − cos θ sin α
sin (α + φ)
Multiplying the numerator and denominator by sin (α + φ) sin θ, we get
η=
cot (α + φ) − cot θ
cot α − cot θ
Notes : 1. When effort is applied horizontally, then θ = 90°.
∴
η=
tan α
tan (α + φ)
2. When effort is applied parallel to the plane, then θ = 90º + α.
∴
η=
cot (α + φ) − cot (90º + α ) cot (α + φ) + tan α sin α cos φ
=
=
cot α − cot (90º + α )
cot α + tan α
sin (α + φ)
2. For the motion of the body down the plane
Since the value of P will be less than P0, for the motion of the body down the plane, therefore
in this case,
sin ( θ − α)
P
W sin ( α − φ)
η=
=
×
P0 sin [θ − (α − φ)]
W sin α
=
sin (α − φ)
sin θ cos α − cos θ sin α
×
sin θ cos (α − φ) − cos θ sin (α − φ)
sin α
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Theory of Machines
Multiplying the numerator and denominator by sin (α – φ) sin θ, we get
η=
cot α − cot θ
cot (α − φ) − cot θ
Notes : 1. When effort is applied horizontally, then θ = 90º.
∴
η=
cot α
tan (α − φ)
=
cot (α − φ)
tan α
2. When effort is applied parallel to the plane, then θ = 90º + α.
cot α − cot (90º + α )
cot α + tan α
sin (α − φ)
η=
=
=
∴
cot (α − φ) − cot (90º + α ) cot (α − φ) + tan α sin α cos φ
Example 10.2. An effort of 1500 N is required to just move a certain body up an inclined
plane of angle 12º, force acting parallel to the plane. If the angle of inclination is increased to 15º,
then the effort required is 1720 N. Find the weight of the body and the coefficient of friction.
Solution. Given : P1 = 1500 N ; α1 = 12º ; α2 = 15º ; P2 = 1720 N
Let
W = Weight of the body in newtons, and
µ = Coefficient of friction.
(a)
(b)
Fig. 10.10
First of all, let us consider a body lying on a plane inclined at an angle of 12º with the
horizontal and subjected to an effort of 1500 N parallel to the plane as shown in Fig. 10.10 (a).
Let
RN = Normal reaction, and
1
F1 = Force of friction.
We know that for the motion of the body up the inclined plane, the effort applied parallel to
the plane (P1),
1500 = W (sin α1 + µ cos α1) = W (sin 12º + µ cos 12º)
...(i)
Now let us consider the body lying on a plane inclined at an angle of 15º with the horizontal
and subjected to an effort of 1720 N parallel to the plane as shown in Fig. 10.10 (b).
Let
RN = Normal reaction, and
2
F2 = Force of friction.
We know that for the motion of the body up the inclined plane, the effort applied parallel to
the plane (P2),
1720 = W (sin α2 + µ cos α2) = W (sin 15º + µ cos 15º)
...(ii)
Coefficient of friction
Dividing equation (ii) by equation (i),
1720 W (sin 15º + µ cos 15º )
=
1500 W (sin 12º + µ cos 12º )
Chapter 10 : Friction
l
269
1720 sin 12º + 1720 µ cos 12º = 1500 sin 15º + 1500 µ cos 15º
µ (1720 cos 12º – 1500 cos 15º) = 1500 sin 15º – 1720 sin 12º
µ=
∴
=
1500 sin 15º − 1720 sin 12º 1500 × 0.2588 − 1720 × 0.2079
=
1720 cos 12º − 1500 cos 15º 1720 × 0.9781 − 1500 × 0.9659
388.2 − 357.6
30.6
=
= 0.131 Ans.
1682.3 − 1448.5 233.8
Weight of the body
Substituting the value of µ in equation (i),
1500 = W (sin 12º + 0.131 cos 12º)
= W (0.2079 + 0.131 × 0.9781) = 0.336 W
∴
W = 1500/0.336 = 4464 N Ans.
Jet engine used in Jet aircraft.
Note : This picture is given as additional information and is not a direct example of the current chapter.
10.16. Screw Friction
The screws, bolts, studs, nuts etc. are widely used in various machines and structures for
temporary fastenings. These fastenings have screw threads, which are made by cutting a continuous
helical groove on a cylindrical surface. If the threads are cut on the outer surface of a solid rod, these
are known as external threads. But if the threads are cut on the internal surface of a hollow rod, these
are known as internal threads. The screw threads are mainly of two types i.e. V-threads and square
threads. The V-threads are stronger and offer more frictional resistance to motion than square threads.
Moreover, the V-threads have an advantage of preventing the nut from slackening. In general, the Vthreads are used for the purpose of tightening pieces together e.g. bolts and nuts etc. But the square
threads are used in screw jacks, vice screws etc. The following terms are important for the study of
screw :
1. Helix. It is the curve traced by a particle, while describing a circular path at a uniform
speed and advancing in the axial direction at a uniform rate. In other words, it is the curve traced by
a particle while moving along a screw thread.
270
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Theory of Machines
2. Pitch. It is the distance from a point of a screw to a corresponding point on the next thread,
measured parallel to the axis of the screw.
3. Lead. It is the distance, a screw thread advances axially in one turn.
4. Depth of thread. It is the distance between the top and bottom surfaces of a thread (also
known as crest and root of a thread).
5. Single-threaded screw. If the lead of a screw is equal to its pitch, it is known as single
threaded screw.
6. Multi-threaded screw. If more than one thread is cut in one lead distance of a screw, it is
known as multi-threaded screw e.g. in a double threaded screw, two threads are cut in one lead length.
In such cases, all the threads run independently along the length of the rod. Mathematically,
Lead = Pitch × Number of threads
7. Helix angle. It is the slope or inclination of the thread with
the horizontal. Mathematically,
Lead of screw
Circumference of screw
= p/πd
...(In single-threaded screw)
= n.p/πd
...(In multi-threaded screw)
α = Helix angle,
p = Pitch of the screw,
d = Mean diameter of the screw, and
n = Number of threads in one lead.
tan α =
where
10.17. Screw Jack
The screw jack is a device, for lifting heavy loads, by applying a comparatively smaller effort at its handle. The principle, on
which a screw jack works is similar to that of an inclined plane.
(a) Screw jack.
Screw Jack.
(b) Thrust collar.
Fig. 10.11
Chapter 10 : Friction
l
271
Fig. 10.11 (a) shows a common form of a screw jack, which consists of a square threaded rod
(also called screw rod or simply screw) which fits into the inner threads of the nut. The load, to be
raised or lowered, is placed on the head of the square threaded rod which is rotated by the application
of an effort at the end of the lever for lifting or lowering the load.
10.18. Torque Required to Lift the Load by a Screw Jack
If one complete turn of a screw thread by imagined to be unwound, from the body of the
screw and developed, it will form an inclined plane as shown in Fig. 10.12 (a).
(a) Development of a screw.
(b) Forces acting on the screw.
Fig. 10.12
Let
p = Pitch of the screw,
d = Mean diameter of the screw,
α = Helix angle,
P = Effort applied at the circumference of the screw to lift the
load,
W = Load to be lifted, and
µ = Coefficient of friction, between the screw and nut = tan φ,
where φ is the friction angle.
From the geometry of the Fig. 10.12 (a), we find that
tan α = p/π d
Since the principle on which a screw jack works is similar to that of an inclined plane, therefore the force applied on the lever of a screw jack may be considered to be horizontal as shown in Fig.
10.12 (b).
Since the load is being lifted, therefore the force of friction (F = µ.R N) will act downwards.
All the forces acting on the screw are shown in Fig. 10.12 (b).
Resolving the forces along the plane,
P cos α = W sin α + F = W sin α + µ.R N
...(i)
and resolving the forces perpendicular to the plane,
RN = P sin α + W cos α
Substituting this value of R N in equation (i),
P cos α = W sin α + µ (P sin α + W cos α)
= W sin α + µ P sin α + µ W cos α
or
P cos α – µ P sin α = W sin α + µ W cos α
or
P (cos α – µ sin α) = W (sin α + µ cos α)
...(ii)
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Theory of Machines
sin α + µ cos α
cos α − µ sin α
Substituting the value of µ = tan φ in the above equation, we get
∴
P =W ×
sin α + tan φ cos α
cos α − tan φ sin α
Multiplying the numerator and denominator by cos φ,
sin α cos φ + sin φ cos α
sin (α + φ)
P =W ×
=W ×
cos α cos φ − sin α sin φ
cos (α + φ)
P =W ×
= W tan ( α + φ)
∴ Torque required to overcome friction between the screw and nut,
d
d
T1 = P × = W tan ( α + φ)
2
2
When the axial load is taken up by a thrust collar or a flat surface, as shown in Fig. 10.11 (b),
so that the load does not rotate with the screw, then the torque required to overcome friction at the
collar,
where
R + R2
T2 = µ1 .W 1
= µ1 .W .R
2
R 1 and R 2 = Outside and inside radii of the collar,
R = Mean radius of the collar, and
µ1 = Coefficient of friction for the collar.
∴ Total torque required to overcome friction (i.e. to rotate the screw),
d
T = T1 + T2 = P × + µ1.W .R
2
If an effort P1 is applied at the end of a lever of arm length l, then the total torque required to
overcome friction must be equal to the torque applied at the end of the lever, i.e.
d
T = P × = P1 .l
2
Notes : 1. When the *nominal diameter (d0) and the **core diameter (dc) of the screw thread is given, then the
mean diameter of the screw,
d + dc
p
p
d= 0
= d0 − = dc +
2
2
2
2. Since the mechanical advantage is the ratio of load lifted (W ) to the effort applied (P1) at the end of
the lever, therefore mechanical advantage,
M . A. =
=
W W × 2l
=
P1
p.d
P.d
... 3 P1 =
2l
W × 2l
2l
=
W tan (α + φ) d d .tan (α + φ)
Example 10.3. An electric motor driven power screw moves a nut in a horizontal plane
against a force of 75 kN at a speed of 300 mm/min. The screw has a single square thread of 6 mm
pitch on a major diameter of 40 mm. The coefficient of friction at the screw threads is 0.1. Estimate
power of the motor.
*
**
The nominal diameter of a screw thread is also known as outside diameter or major diameter.
The core diameter of a screw thread is also known as inner diameter or root diameter or minor diameter.
Chapter 10 : Friction
l
273
Solution. Given : W = 75 kN = 75 × 103 N ; v = 300 mm/min ; p = 6 mm ; d0 = 40 mm ;
µ = tan φ = 0.1
We know that mean diameter of the screw,
d = d0 – p/2 = 40 – 6/2 = 37 mm = 0.037 m
tan α =
and
p
6
=
= 0.0516
π d π × 37
∴ Force required at the circumference of the screw,
tan α + tan φ
P = W tan ( α + φ) = W
1 − tan α. tan φ
0.0516 + 0.1
3
= 75 × 103
= 11.43 × 10 N
1
0.0516
0.1
−
×
and torque required to overcome friction,
T = P × d/2 = 11.43 × 103 × 0.037/2 = 211.45 N-m
We know that speed of the screw,
Speed of the nut
300
=
= 50 r.p.m.
Pitch of the screw
6
ω = 2 π × 50/60 = 5.24 rad/s
N =
and angular speed,
∴ Power of the motor
= T.ω = 211.45 × 5.24 = 1108 W = 1.108 kW Ans.
Example 10.4. A turnbuckle, with right
and left hand single start threads, is used to couple
two wagons. Its thread pitch is 12 mm and mean
diameter 40 mm. The coefficient of friction between
the nut and screw is 0.16.
1. Determine the work done in drawing the
wagons together a distance of 240 mm, against a
steady load of 2500 N.
2. If the load increases from 2500 N to 6000
N over the distance of 240 m m, what is the work to
be done?
Solution. Given : p = 12 mm ; d = 40 mm ;
µ = tan φ = 0.16 ; W = 2500 N
Turnbuckle.
1. Work done in drawing the wagons together against a steady load of 2500 N
p
12
=
= 0.0955
πd π × 40
∴ Effort required at the circumference of the screw,
We know that
tan α =
tan α + tan φ
P = W tan (α + φ) = W
1 − tan α . tan φ
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Theory of Machines
0.0955 + 0.16
= 2500
= 648.7 N
1 − 0.0955 × 0.16
and torque required to overcome friction between the screw and nut,
T = P × d / 2 = 648.7 × 40 / 2 = 12 947 N-mm = 12.974 N-m
A little consideration will show that for one complete revolution of the screwed rod, the
wagons are drawn together through a distance equal to 2 p, i.e. 2 × 12 = 24 mm. Therefore in order to
draw the wagons together through a distance of 240 mm, the number of turns required are given by
N = 240/24 = 10
∴
Work done = T × 2 π N = 12.974 × 2 π × 10 = 815.36 N-m Ans.
2. Work done in drawing the wagons together when load increases from 2500 N to 6000 N
For an increase in load from 2500 N to 6000 N,
815.3(6000 − 2500)
= 114.4 N-m Ans.
2500
Example 10.5. A 150 mm diameter valve, against which a steam pressure of 2 MN/m2 is
acting, is closed by means of a square threaded screw 50 mm in external diameter with 6 mm pitch.
If the coefficient of friction is 0.12 ; find the torque required to turn the handle.
Work done =
Solution. Given : D = 150 mm = 0.15 mm = 0.15 m ; Ps = 2 MN/m2 = 2 × 106 N/m2 ;
d0 = 50 mm ; p = 6 mm ; µ = tan φ = 0.12
We know that load on the valve,
π 2
π
6
2
W = Pressure × Area = pS × D = 2 × 10 × (0.15) N
4
4
= 35 400 N
Mean diameter of the screw,
d = d0 – p/2 = 50 – 6/2 = 47 mm = 0.047 m
∴
tan α =
p
6
=
= 0.0406
π d π × 47
We know that force required to turn the handle,
tan α + tan φ
P = W tan ( α + φ) = W
1 − tan α.tan φ
0.0406 + 12
= 35 400
= 5713 N
1 − 0.0406 × 0.12
∴ Torque required to turn the handle,
T = P × d/2 = 5713 × 0.047/2 = 134.2 N-m Ans.
Example 10.6. A square threaded bolt of root diameter 22.5 mm and pitch 5 mm is tightened
by screwing a nut whose mean diameter of bearing surface is 50 mm. If coefficient of friction for nut
and bolt is 0.1 and for nut and bearing surface 0.16, find the force required at the end of a spanner
500 mm long when the load on the bolt is 10 kN.
Solution. Given : dc = 22.5 mm ; p = 5 mm ; D = 50 mm or R = 25 mm ; µ = tan φ = 0.1 ;
µ1 = 0.16 ; l = 500 mm ; W = 10 kN = 10 × 103 N
Let
P1 = Force required at the end of a spanner in newtons.
Chapter 10 : Friction
l
275
We know that mean diameter of the screw,
d = d c + p / 2 = 22.5 + 5 / 2 = 25 mm
tan α =
∴
p
5
=
= 0.0636
π d π × 25
Force requred at the circumference of the screw,
tan α + tan φ
P = W tan (α + φ) = W
1 − tan α. tan φ
0.0636 + 0.1
= 10 × 103
= 1646 N
1 − 0.06363 × 0.1
We know that total torque required,
T =P×
25
d
+ µ1.W .R. = 1646 ×
+ 0.16 × 10 × 103 × 25
2
2
= 60575 N - mm
..(i)
We also know that torque required at the end of a spanner,
T = P1 × l = P1 × 500 = 500 P1 N-mm
...(ii)
Equating equations (i) and (ii),
P1 = 60575/500 = 121.15 N Ans.
Example 10.7. A vertical screw with single start square threads 50 mm mean diameter and
12.5 mm pitch is raised against a load of 10 kN by means of a hand wheel, the boss of which is
threaded to act as a nut. The axial load is taken up by a thrust collar which supports the wheel boss
and has a mean diameter of 60 mm. If the coefficient of friction is 0.15 for the screw and 0.18 for the
collar and the tangential force applied by each hand to the wheel is 100 N ; find suitable diameter of
the hand wheel.
Solution. Given : d = 50 mm ; p = 12.5 mm ; W = 10 kN = 10 × 103 N ; D = 60 mm or
R = 30 mm ; µ = tan φ = 0.15 ; µ1 = 0.18 ; P1 = 100 N
p
12.5
=
= 0.08
π d π × 50
and the tangential force required at the circumference of the screw,
We know that tan α =
tan α + tan φ
P = W tan (α + φ) = W
1 − tan α . tan φ
0.08 + 0.15
= 10 × 103
= 2328 N
1 − 0.08 × 0.15
Also we know that the total torque required to turn the hand wheel,
50
d
+ µ1 .W .R = 2328 ×
+ 0.18 × 10 × 103 × 30
2
2
= 112 200 N-mm
Let
D1 = Diameter of the hand wheel in mm.
We know that the torque applied to the hand wheel,
D
D
T = 2 P1 × 1 = 2 × 100 × 1 = 100 D1 N-mm
2
2
T =P×
...(i)
...(ii)
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Theory of Machines
Equating equations (i) and (ii),
D1 = 112 200/100 = 1222 mm = 1.222 m Ans.
Example 10.8. The cutter of a broaching machine is pulled by square threaded screw of 55
mm external diameter and 10 mm pitch. The operating nut takes the axial load of 400 N on a flat
surface of 60 mm internal diameter and 90 mm external diameter. If the coefficient of firction is 0.15
for all contact surfaces on the nut, determine the power required to rotate the operating nut, when
the cutting speed is 6 m/min.
Solution. Given : d0 = 55 mm ; p = 10 mm = 0.01 m ; W = 400 N ; D2 = 60 mm
R 2 = 30 mm ; D1 = 90 mm or R 1 = 45 mm ; µ = tan φ = µ1 = 0.15
or
We know that mean diameter of the screw,
d = d0 – p/2 = 55 – 10/2 = 50 mm
p
10
=
= 0.0637
π d π × 50
and force required at the circumference of the screw,
∴
tan α =
tan α + tan φ
P = W tan ( α + φ) = W
1 − tan α. tan φ
0.0637 + 0.15
= 400
= 86.4 N
1 − 0.0637 × 0.15
We know that mean radius of the flat surface,
R=
∴ Total torque required,
R1 + R2
2
=
45 + 30
= 37.5 mm
2
50
d
+ µ1 .W .R = 86.4 ×
+ 0.15 × 400 × 37.5 N-mm
2
2
= 4410 N-mm = 4.41 N-m
...(∵ µ1 = µ)
T = P×
Since the cutting speed is 6 m/min, therefore speed of the screw,
N =
and
angular speed,
Cutting speed
6
=
= 600 r.p.m.
Pitch
0.01
ω = 2 π × 600/60 = 62.84 rad/s
We know that power required to operate the nut
= T .ω = 4.41 × 62.84 = 277 W = 0.277 kW Ans.
10.19. Torque Required to Lower the Load by a Screw Jack
We have discussed in Art. 10.18, that the principle on which the screw jack works is similar
to that of an inclined plane. If one complete turn of a screw thread be imagined to be unwound from
the body of the screw and developed, it will form an inclined plane as shown in Fig. 10.13 (a).
Let
p = Pitch of the screw,
d = Mean diameter of the screw,
α = Helix angle,
P = Effort applied at the circumference of the screw to lower the
load,
Chapter 10 : Friction
l
277
W = Weight to be lowered, and
µ = Coefficient of friction between the screw and nut = tan φ,
where φ is the friction angle.
(a)
(b)
Fig. 10.13
From the geometry of the figure, we find that
tan α = p/πd
Since the load is being lowered, therefore the force of friction (F = µ.R N) will act upwards.
All the forces acting on the screw are shown in Fig. 10.13 (b).
Resolving the forces along the plane,
P cos α = F – W sin α = µ.R N – W sin α
...(i)
and resolving the forces perpendicular to the plane,
RN = W cos α – P sin α
...(ii)
Substituting this value of R N in equation (i),
P cos α = µ (W cos α – P sin α) – W sin α
= µ.W cos α – µ.P sin α – W sin α
or
P cos α + µ.P sin α = µ.W cos α – W sin α
or
P (cos α + µ sin α) = W (µ cos α – sin α)
(µ cos α − sin α )
(cos α + µ sin α )
Substituting the value of µ = tan φ in the above equation, we get
∴
P =W ×
(tan φ cos α − sin α )
(cos α + tan φ sin α )
Multiplying the numerator and denominator by cos φ,
(sin φ cos α − sin α cos φ)
sin (φ − α)
P =W ×
=W ×
(cos α cos φ + sin φ sin α)
cos (φ − α)
P =W ×
= W tan (φ − α)
∴ Torque required to overcome friction between the screw and nut,
T =P×
d
d
= W tan ( φ − α)
2
2
Note : When α > φ, then P = tan (α – φ).
Example 10.9. The mean diameter of a square threaded screw jack is 50 mm. The pitch of
the thread is 10 mm. The coefficient of friction is 0.15. What force must be applied at the end of a
0.7 m long lever, which is perpendicular to the longitudinal axis of the screw to raise a load of 20 kN
and to lower it?
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Theory of Machines
Solution. Given : d = 50 mm = 0.05 m ; p = 10 mm ; µ = tan φ = 0.15 ; l = 0.7 m ; W = 20 kN
= 20 × 103 N
We know that
p
10
=
= 0.0637
π d π × 50
P1 = Force required at the end of the lever.
tan α =
Let
Force required to raise the load
We know that force required at the circumference of the screw,
tan α + tan φ
P = W tan ( α + φ) = W
1 − tan α.tan φ
0.0637 + 0.15
= 20 × 103
= 4314 N
1 − 0.0637 × 0.15
Now the force required at the end of the lever may be found out by the relation,
P1 × l = P × d/2
P1 =
∴
P × d 4314 × 0.05
=
= 154 N Ans.
2l
2 × 0.7
Force required to lower the load
We know that the force required at the circumference of the screw,
tan φ − tan α
P = W tan ( φ − α) = W
1 + tan φ.tan α
0.15 − 0.0637
= 20 × 103
= 1710 N
1 + 0.15 × 0.0637
Now the force required at the end of the lever may be found out by the relation,
P1 × l = P ×
d
P × d 1710 × 0.05
or P1 =
=
= 61 N Ans.
2
2l
2 × 0.7
10.20. Efficiency of a Screw Jack
The efficiency of a screw jack may be defined as the ratio between the ideal effort (i.e. the
effort required to move the load, neglecting friction) to the actual effort (i.e. the effort required to
move the load taking friction into account).
We know that the effort required to lift the load (W ) when friction is taken into account,
P = W tan (α + φ)
...(i)
where
α = Helix angle,
φ = Angle of friction, and
µ = Coefficient of friction, between the screw and nut = tan φ.
If there would have been no friction between the screw and the nut, then φ will be equal to
zero. The value of effort P0 necessary to raise the load, will then be given by the equation,
P0 = W tan α
(i.e. Putting φ = 0 in equation (i)]
P
Ideal effort
tan α
W tan α
= 0 =
=
Actual effort P W tan (α + φ) tan (α + φ)
which shows that the efficiency of a screw jack, is independent of the load raised.
∴ Efficiency, η =
Chapter 10 : Friction
l
279
In the above expression for efficiency, only the screw friction is considered. However, if the
screw friction and the collar friction is taken into account, then
Torque required to move the load, neglecting friction
η=
∴
Torque required to move the load, including screw and collar friction
=
T0
T
=
P0 × d / 2
P × d / 2 + µ1.W .R
Note: The efficiency of the screw jack may also be defined as the ratio of mechanical advantage to the
velocity ratio.
We know that mechanical advantage,
and velocity ratio,
M . A. =
2l
W W × 2l
W × 2l
=
=
=
P1
P × d W tan (α + φ)d tan (α + φ) d
V .R. =
Distance moved by the effort (P1 ), in one revolution
Distance moved by the load (W ), in one revolution
2πl
2πl
2l
=
=
=
p
tan α × π d tan α × d
...(Refer Art 10.17)
...( tan α = p/πd)
M . A.
2l
tan α × d
tan × α
=
∴ Efficiency, η = V .R. = tan (α + φ) d ×
2l
tan ( α + φ)
10.21. Maximum Efficiency of a Screw Jack
We have seen in Art. 10.20 that the efficiency of a screw jack,
sin α
tan α
cos α
sin α × cos (α + φ)
η=
=
=
tan (α + θ) sin ( α + φ)
cos α × sin (α + φ )
cos (α + φ)
2 sin α × cos (α + φ)
=
2 cos α × sin (α + φ)
...(i)
...(Multiplying the numerator and denominator by 2)
=
sin (2 α + φ) − sin φ
sin (2 α + φ) + sin φ
...(ii)
3 2 sin A cos B = sin ( A + B) + sin ( A − B)
...
2 cos A sin B = sin ( A + B ) − sin ( A − B )
The efficiency given by equation (ii) is maximum when sin (2α + φ) is maximum, i.e. when
sin (2α + φ) = 1 or when 2α + φ = 90°
∴
2α = 90º – φ
or
α = 45º – φ / 2
Substituting the value of 2 α in equation (ii), we have maximum efficiency,
sin (90º − φ + φ) − sin φ sin 90º − sin φ 1 − sin φ
=
=
sin (90º − φ + φ) + sin φ sin 90º + sin φ 1 + sin φ
Example 10.10. The pitch of 50 mm mean diameter threaded screw of a screw jack is 12.5
mm. The coefficient of friction between the screw and the nut is 0.13. Determine the torque required
on the screw to raise a load of 25 kN, assuming the load to rotate with the screw. Determine the ratio
of the torque required to raise the load to the torque required to lower the load and also the efficiency
of the machine.
ηmax =
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Theory of Machines
Solution. Given : d = 50 mm ; p = 12.5 mm ; µ = tan φ = 0.13 ; W = 25 kN = 25 × 103 N
p
12.5
=
= 0.08
π d π × 50
and force required on the screw to raise the load,
We know that,
tan α =
tan φ − tan α
P = W tan ( α + φ) = W
1 + tan φ.tan α
0.08 + 0.13
= 25 × 103
= 5305 N
1 − 0.08 × 0.13
Torque required on the screw
We know that the torque required on the screw to raise the load,
T 1 = P × d/2 = 5305 × 50/2 = 132 625 N-mm Ans.
Ratio of the torques required to raise and lower the load
We know that the force required on the screw to lower the load,
tan φ − tan α
P = W tan ( φ − α) = W
1 + tan φ.tan α
0.13 + 0.08
= 25 × 103
= 1237 N
1 + 0.13 × 0.08
and torque required to lower the load
T 2 = P × d/2 = 1237 × 50/2 = 30 905 N-mm
∴ Ratio of the torques required,
= T1 / T2 = 132 625 / 30 925 = 4.3 Ans.
Efficiency of the machine
We know that the efficiency,
η=
tan α
tan α (1 − tan α.tan φ) 0.08(1 − 0.08 × 0.13)
=
=
tan (α + φ)
tan α + tan φ
0.08 + 0.13
= 0.377 = 37.7% Ans.
Example 10.11. The mean diameter of the screw jack having pitch of 10 mm is 50 mm. A
load of 20 kN is lifted through a distance of 170 mm. Find the work done in lifting the load and
efficiency of the screw jack when
1. the load rotates with the screw, and
2. the load rests on the loose head which does not rotate with the screw.
The external and internal diameter of the bearing surface of the loose head are 60 mm and
10 mm respectively. The coefficient of friction for the screw as well as the bearing surface may be
taken as 0.08.
Solution. Given : p = 10 mm ; d = 50 mm ; W = 20 kN = 20 × 103 N ; D2 = 60 mm or
R 2 = 30 mm ; D1 = 10 mm or R 1 = 5 mm ; µ = tan φ = µ1 = 0.08
We know that
tan α =
p
10
=
= 0.0637
π d π × 50
Chapter 10 : Friction
l
281
∴ Force required at the circumference of the screw to lift the load,
tan α + tan φ
P = W tan ( α + φ) = W
1 − tan α. tan φ
0.0637 + 0.08
= 20 × 103
= 2890 N
1 − 0.0637 × 0.08
and torque required to overcome friction at the screw,
T = P × d / 2 = 2890 × 50 / 2 = 72 250 N-mm = 72.25 N-m
Since the load is lifted through a vertical distance of 170 mm and the distance moved by the
screw in one rotation is 10 mm (equal to pitch), therefore number of rotations made by the screw,
N = 170/10 = 17
1. When the load rotates with the screw
We know that work done in lifting the load
= T × 2πN = 72.25 × 2π × 17 = 7718 N-m Ans.
and efficiency of the screw jack,
η=
tan α
tan α (1 − tan α.tan φ)
=
tan (α + φ)
tan α + tan α
0.0637 (1 − 0.0637 × 0.08)
= 0.441 or 44.1% Ans.
0.0637 + 0.08
2. When the load does not rotate with the screw
=
We know that mean radius of the bearing surface,
R + R2 30 + 5
R= 1
=
= 17.5 mm
2
2
and torque required to overcome friction at the screw and the collar,
T = P × d / 2 + µ1.W .R
= 2890 × 50 / 2 + 0.08 × 20 ×103 × 17.5 = 100 250 N-mm
= 100.25 N-m
∴ Work done by the torque in lifting the load
= T × 2πN = 100.25 × 2π × 17 = 10 710 N-m Ans.
We know that the torque required to lift the load, neglecting friction,
T0 = P0 × d / 2 = W tan α × d / 2
...(3 P0 = W tan α)
= 20 × 103 × 0.0637 × 50/2 = 31 850 N-mm = 31.85 N-m
∴ Efficiency of the screw jack,
η = T0 / T = 31.85 /100.25 = 0.318 or 31.8% Ans.
10.22. Over Hauling and Self Locking Screws
We have seen in Art. 10.20 that the effort required at the circumference of the screw to lower
the load is
P = W tan (φ – α)
282
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Theory of Machines
and the torque required to lower the load
d
d
T = P × = W tan ( φ − α)
2
2
In the above expression, if φ < α, then torque required to lower the load will be negative. In
other words, the load will start moving downward without the application of any torque. Such a
condition is known as over haulding of screws. If however, φ > α, the torque required to lower the
load will positive, indicating that an effort is applied to lower the load. Such a screw is known as self
locking screw. In other words, a screw will be self locking if the friction angle is greater than helix
angle or coefficient of friction is greater than tangent of helix angle i.e. µ or tan φ > tan α.
10.23. Efficiency of Self Locking Screws
We know that efficiency of the screw,
tan α
η=
tan (α + φ)
and for self locking screws, φ ≥ α or α ≤ φ.
∴ Efficiency of self locking screws,
tan φ
tan φ
tan φ (1 − tan 2 φ)
η≤
≤
≤
tan (φ + φ) tan 2φ
2 tan φ
≤
2 tan φ
... 3 tan 2 φ =
1
− tan 2 φ
1 tan 2 φ
−
2
2
From this expression we see that efficiency of self locking screws is less than
1
2
or 50%. If
the efficiency is more than 50%, then the screw is said to be overhauling,
Note : It can also be proved as follows :
Let
W = Load to be lifted, and
h = Distance through which the load is lifted.
∴
Output = W.h
and
Output W .h
=
η
η
∴ Work lost in over coming friction.
Input =
= Input − Output =
W .h
1
− W .h = W .h − 1
η
η
1
− 1 ≤ W .h
η
For self locking,, W .h
∴
1
1
− 1 ≤ 1 or η ≤
or 50%
η
2
Example 10.12. A load of 10 kN is raised by means of a screw jack, having a square threaded
screw of 12 mm pitch and of mean diameter 50 mm. If a force of 100 N is applied at the end of a lever
to raise the load, what should be the length of the lever used? Take coefficient of friction = 0.15.
What is the mechanical advantage obtained? State whether the screw is self locking.
Solution. Given : W = 10 kN = 10 × 103 N ; p = 12 mm ; d = 50 mm ; P1 = 100 N ;
µ = tan φ = 0.15
Length of the lever
Let
l = Length of the lever.
Chapter 10 : Friction
l
283
p
12
=
= 0.0764
πd π × 50
∴ Effort required at the circumference of the screw to raise the load,
tan α + tan φ
P = W tan ( α + φ) = W
1 − tan α. tan φ
We know that
tan α =
0.0764 + 0.15
= 10 × 103
= 2290 N
1 − 0.0764 × 0.15
and torque required to overcome friction,
T = P × d/2 = 2290 × 50/2 = 57 250 N-mm
...(i)
We know that torque applied at the end of the lever,
T = P1 × l = 100 × l N-mm
...(ii)
Equating equations (i) and (ii)
l = 57 250/100 = 572.5 mm Ans.
Mechanical advantage
We know that mechanical advantage,
M . A. =
Self locking of the screw
W 10 × 103
=
= 100 Ans.
100
P1
We know that efficiency of the screw jack,
η=
=
screw.
tan α
tan α (1 − tan α.tan φ)
=
tan (α + φ)
tan α + tan φ
0.0764(1 − 0.0764 × 0.15) 0.0755
=
= 0.3335 or 33.35%
0.0764 + 0.15
0.2264
Since the efficiency of the screw jack is less than 50%, therefore the screw is a self locking
Ans.
10.24. Friction of a V-thread
We have seen Art. 10.18 that the normal reaction in case of a square threaded screw is
R N = W cos α, where α = Helix angle.
But in case of V-thread (or acme or trapezoidal threads), the normal
reaction between the screw and nut is increased because the axial component of
this normal reaction must be equal to the axial load W , as shown in Fig. 10.14.
Let
∴
and
where
2β = Angle of the V-thread, and
β = Semi-angle of the V-thread.
W
RN =
cos β
W
frictional force, F = µ.RN = µ × cos = µ1 .W
Fig. 10.14. V-thread.
β
µ
= µ1 , known as virtual coefficient of friction.
cos β
284
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Theory of Machines
Notes : 1. When coefficient of friction, µ1 =
thread.
µ
is considered, then the V-thread is equivalent to a square
cos β
2. All the equations of square threaded screw also hold good for V-threads. In case of V-threads, µ1
(i.e. tan φ1) may be substituted in place of µ (i.e. tan φ). Thus for V-threads,
P = W tan (α ± φ1 )
φ1 = Virtual friction angle, such that tan φ1 = µ1.
where
Example 10.13. Two co-axial rods are connected by a turn buckle which consists of a box
nut, the one screw being right handed and the other left handed on a pitch diameter of 22 mm, the
pitch of thread being 3 mm. The included angle of the thread is 60º. Assuming that the rods do not
turn, calculate the torque required on the nut to produce a pull of 40 kN, given that the coefficient of
friction is 0.15.
Solution. Given : d = 22 mm ; p = 3 mm ; 2 β = 60º or β = 30º, W = 40 kN = 40 × 103 N ; µ = 0.15
We know that
tan α =
p
3
=
= 0.0434
πd π × 22
and virtual coefficient of friction
0.15
µ
=
= 0.173
cos β cos 30º
We know that the force required at the circumference of the screw,
tan α + tan φ1
P = W tan (α + φ1 ) = W
1 − tan α.tan φ1
0.0434 + 0.173
= 40 × 103
= 8720 N
1 − 0.0434 × 0.173
µ1 = tan φ1 =
and
torque on one rod,
T = P × d/2 = 8720 × 22/2 = 95 920 N-mm = 95.92 N-m
Since the turn buckle has right and left hand threads and the torque on each rod is T = 95.92
N-m, therefore the torque required on the nut,
T 1 = 2T = 2 × 95.92 = 191.84 N-m Ans.
Example 10.14. The mean diameter of a Whitworth bolt having V-threads is 25 mm. The
pitch of the thread is 5 mm and the angle of V is 55º. The bolt is tightened by screwing a nut whose
mean radius of the bearing surface is 25 mm. If the coefficient of friction for nut and bolt is 0.1 and
for nut and bearing surfaces 0.16 ; find the force required at the end of a spanner 0.5 m long when
the load on the bolt is 10 kN.
Solution. Given : d = 25 mm ; p = 5 mm ; 2 β = 55º or β = 27.5º ; R = 25 mm ; µ = tan φ
= 0.1; µ2 = 0.16 ; l = 0.5 m ; W = 10 kN = 10 × 103 N
We know that virtual coefficient of friction,
µ1 = tan φ1 =
tan α =
and
∴ Force on the screw,
0.1
0.1
µ
=
= 0.113
cos β cos 27.5º 0.887
p
5
=
= 0.064
πd π × 25
tan α + tan φ1
P = W tan (α + φ1 ) = W
1− tan α.tan φ1
Chapter 10 : Friction
l
0.064 + 0.113
= 10 × 103
= 1783 N
1 − 0.064 × 0.113
We know that total torque transmitted,
25
d
T = P × + µ2 .W .R = 1783 ×
+ 0.16 × 10 × 103 × 25 N-mm
2
2
= 62 300 N-mm = 62.3 N-m
Let
P1 = Force required at the end of a spanner.
∴ Torque required at the end of a spanner,
T = P1 × l = P1 × 0.5 = 0.5 P1 N-m
Equating equations (i) and (ii),
285
...(i)
...(ii)
P1 = 62.3/0.5 = 124.6 N Ans.
10.25. Friction in Journal Bearing-Friction Circle
A journal bearing forms a turning pair as shown in Fig. 10.15 (a). The fixed outer element of
a turning pair is called a bearing and that portion of the inner element (i.e. shaft) which fits in the
bearing is called a journal. The journal is slightly less in diameter than the bearing, in order to permit
the free movement of the journal in a bearing.
(a)
(b)
Fig. 10.15. Friction in journal bearing.
When the bearing is not lubricated (or the journal is stationary), then there is a line contact
between the two elements as shown in Fig. 10.15 (a). The load W on the journal and normal reaction
R N (equal to W ) of the bearing acts through the centre. The reaction R N acts vertically upwards at
point A . This point A is known as seat or point of pressure.
Now consider a shaft rotating inside a bearing in clockwise direction as shown in Fig. 10.15
(b). The lubricant between the journal and bearing forms a thin layer which gives rise to a greasy
friction.Therefore, the reaction R does not act vertically upward, but acts at another point of pressure
B. This is due to the fact that when shaft rotates, a frictional force F = µ R N acts at the circumference
of the shaft which has a tendency to rotate the shaft in opposite direction of motion and this shifts the
point A to point B.
In order that the rotation may be maintained, there must be a couple rotating the shaft.
Let
φ = Angle between R (resultant of F and R N) and R N,
µ = Coefficient of friction between the journal and bearing,
T = Frictional torque in N-m, and
r = Radius of the shaft in metres.
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Theory of Machines
For uniform motion, the resultant force acting on the shaft must be zero and the resultant
turning moment on the shaft must be zero. In other words,
R = W , and T = W × OC = W × OB sin φ = W.r sin φ
Since φ is very small, therefore substituting sin φ = tan φ
...(∵ µ = tan φ)
∴
T = W.r tan φ = µ.W.r
If the shaft rotates with angular velocity ω rad/s, then power wasted in friction,
P = T.ω = T × 2πN/60 watts
where
N = Speed of the shaft in r.p.m.
Notes : 1. If a circle is drawn with centre O and radius OC = r sin φ, then this circle is called the friction circle
of a bearing.
2. The force R exerted by one element of a turning pair on the other element acts along a tangent to the
friction circle.
Example 10.15. A 60 mm diameter shaft running in a bearing carries a load of 2000 N. If
the coefficient of friction between the shaft and bearing is 0.03, find the power transmitted when it
runs at 1440 r.p.m.
Solution. Given : d = 60 mm or r = 30 mm = 0.03 m ; W = 2000 N ; µ = 0.03 ; N = 1440 r.p.m.
or ω = 2π × 1440/60 = 150.8 rad/s
We know that torque transmitted,
T = µ.W.r = 0.03 × 2000 × 0.03 = 1.8 N-m
∴ Power transmitted,
P = T.ω = 1.8 × 150.8 = 271.4 W Ans.
10.26. Friction of Pivot and Collar Bearing
The rotating shafts are frequently subjected to axial thrust. The bearing surfaces such as pivot
and collar bearings are used to take this axial thrust of the rotating shaft. The propeller shafts of ships, the
shafts of steam turbines, and vertical machine shafts are examples of shafts which carry an axial thrust.
The bearing surfaces placed at the end of a shaft to take the axial thrust are known as
pivots. The pivot may have a flat surface or conical surface as shown in Fig. 10.16 (a) and (b)
respectively. When the cone is truncated, it is then known as truncated or trapezoidal pivot as
shown in Fig. 10.16 (c).
The collar may have flat bearing surface or conical bearing surface, but the flat surface is
most commonly used. There may be a single collar, as shown in Fig. 10.16 (d) or several collars along
the length of a shaft, as shown in Fig. 10.16 (e) in order to reduce the intensity of pressure.
(a) Flat pivot.
(b) Conical pivot. (c) Truncated pivot.
(d) Single flat
collar.
(e) Multiple flat
collar.
Fig. 10.16. Pivot and collar bearings.
In modern practice, ball and roller thrust bearings are used when power is being transmitted
and when thrusts are large as in case of propeller shafts of ships.
Chapter 10 : Friction
A little consideration will show that in a new bearing, the contact between the shaft and bearing may be good
over the whole surface. In other words, we can say that the
pressure over the rubbing surfaces is uniformly distributed.
But when the bearing becomes old, all parts of the rubbing
surface will not move with the same velocity, because the
velocity of rubbing surface increases with the distance from
the axis of the bearing. This means that wear may be different
at different radii and this causes to alter the distribution of
pressure. Hence, in the study of friction of bearings, it is assumed that
l
287
Collar bearing.
1. The pressure is uniformly distributed throughout the bearing surface, and
2. The wear is uniform throughout the bearing surface.
10.27. Flat Pivot Bearing
When a vertical shaft rotates in a flat pivot bearing
(known as foot step bearing), as shown in Fig. 10.17, the
sliding friction will be along the surface of contact between
the shaft and the bearing.
Let
W = Load transmitted over the bearing surface,
R = Radius of bearing surface,
p = Intensity of pressure per unit area of bearing surface between rubbing surfaces, and
µ = Coefficient of friction.
We will consider the following two cases :
1. When there is a uniform pressure ; and
2. When there is a uniform wear.
Fig. 10.17. Flat pivot or footstep
bearing.
1. Considering unifrom pressure
When the pressure is uniformly distributed over the bearing area, then
p=
W
π R2
Consider a ring of radius r and thickness dr of the bearing area.
∴ Area of bearing surface, A = 2πr.dr
Load transmitted to the ring,
δW = p × A = p × 2 π r.dr
...(i)
Frictional resistance to sliding on the ring acting tangentially at radius r,
Fr = µ.δW = µ p × 2π r.dr = 2π µ.p.r.dr
∴ Frictional torque on the ring,
Tr = Fr × r = 2π µ p r.dr × r = 2 π µ p r2 dr
...(ii)
Integrating this equation within the limits from 0 to R for the total frictional torque on the
pivot bearing.
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Theory of Machines
R
∴ Total frictional torque, T =
∫
0
R
2 πµ p r 2 dr = 2 πµ p ∫ r 2 dr
0
3 R
R
2
= 2πµ p r = 2πµ p ×
= × πµ. p.R3
3
3
3 0
2
2
W
= × πµ ×
× R 3 = × µ.W .R
2
3
3
πR
When the shaft rotates at ω rad/s, then power lost in friction,
3
...(3 ω = 2πN / 60)
P = T.ω = T × 2π N/60
where
W
... 3 p =
πR 2
N = Speed of shaft in r.p.m.
2. Considering uniform wear
We have already discussed that the rate of wear depends upon the intensity of pressure (p) and
the velocity of rubbing surfaces (v). It is assumed that the rate of wear is proportional to the product
of intensity of pressure and the velocity of rubbing surfaces (i.e. p.v..). Since the velocity of rubbing
surfaces increases with the distance (i.e. radius r) from the axis of the bearing, therefore for uniform
wear
p.r = C (a constant)
or
p=C/r
and the load transmitted to the ring,
δW = p × 2πr.dr
...[From equation (i)]
C
× 2π r.dr = 2π C .dr
r
∴ Total load transmitted to the bearing
=
R
W =
∫
2π C.dr = 2π C [r ]0R = 2πC.R or C =
0
W
2πR
We know that frictional torque acting on the ring,
Tr = 2 π µ p r 2 dr = 2 π µ ×
= 2π µ.C.r dr
C
× r 2 dr
r
C
...3 p =
r
...(iii)
∴ Total frictional torque on the bearing,
R
T =
∫
0
R
2
2 π µ.C .r.dr = 2 πµ.C r
2 0
= 2πµ.C ×
R2
= πµ.C.R 2
2
= πµ ×
1
W
× R 2 = × µ.W .R
2π R
2
W
... 3 C =
2πR
Example 10.16. A vertical shaft 150 mm in diameter rotating at 100 r.p.m. rests on a flat end
footstep bearing. The shaft carries a vertical load of 20 kN. Assuming uniform pressure distribution
and coefficient of friction equal to 0.05, estimate power lost in friction.
Solution. Given : D = 150 mm or R = 75 mm = 0.075 m ; N = 100 r.p.m or ω = 2 π × 100/60
= 10.47 rad/s ; W = 20 kN = 20 × 103 N ; µ = 0.05
Chapter 10 : Friction
l
289
We know that for uniform pressure distribution, the total frictional torque,
2
2
T = × µ.W .R = × 0.05 × 20 × 103 × 0.075 = 50 N-m
3
3
∴ Power lost in friction,
P = T .ω = 50 × 10.47 = 523.5 W Ans.
10.28. Conical Pivot Bearing
The conical pivot bearing supporting a shaft carrying a load W is shown in Fig. 10.18.
Let
Pn = Intensity of pressure normal to
the cone,
α = Semi angle of the cone,
µ = Coefficient of friction
between the shaft and the
bearing, and
R = Radius of the shaft.
Consider a small ring of radius r and thickness dr. Let dl is
the length of ring along the cone, such that
dl = dr cosec α
∴ Area of the ring,
A = 2πr.dl = 2πr.dr cosec α
...(∵ dl = dr cosec α)
1. Considering uniform pressure
Fig. 10.18.
Conical pivot bearing.
We know that normal load acting on the ring,
δW n = Normal pressure × Area
= pn × 2πr.dr cosec α
and vertical load acting on the ring,
*δW = Vertical component of δW n = δW n.sin α
=pn × 2πr.dr cosec α. sin α = pn × 2π r.dr
∴ Total vertical load transmitted to the bearing,
R
W =
∫
0
or
R
R2
2
pn × 2 πr.dr = 2 π pn r = 2 π pn ×
= π R 2 . pn
2
2 0
pn = W / πR 2
We know that frictional force on the ring acting tangentially at radius r,
Fr = µ.δWn = µ. pn .2πr.dr cosec α = 2π µ. pn .cosec α.r.dr
and frictional torque acting on the ring,
Tr = Fr × r = 2 πµ. pn .cosec α.r.dr × r = 2πµ. pn cosec α.r 2 .dr
*
The vertical load acting on the ring is also given by
δW = Vertical component of pn × Area of the ring
=pn sin α × 2πr.dr.cosec α = pn × 2πr.dr
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Theory of Machines
Integrating the expression within the limits from 0 to R for the total frictional torque on the
conical pivot bearing.
∴ Total frictional torque,
R
T =
∫
0
R
3
2 π µ. pn cosec α.r 2 dr = 2 πµ. pn .cosec α r
3 0
= 2πµ. pn .cosec α ×
R 3 2π R 3
=
× µ. pn .cosec α
3
3
...(i)
Substituting the value of pn in equation (i),
T =
2πR 3
2
W
× π×
× cosec α = × µ.W .R. cosec α
3
3
πR 2
Note : If slant length (l ) of the cone is known, then
T =
2
× µ.W .l
3
...(3 l = R cosec α)
2. Considering uniform wear
In Fig. 10.18, let pr be the normal intensity of pressure at a distance r from the central axis.
We know that, in case of uniform wear, the intensity of pressure varies inversely with the distance.
∴
pr.r = C (a constant) or pr = C/r
and the load transmitted to the ring,
C
× 2πr.dr = 2πC.dr
r
∴ Total load transmitted to the bearing,
δW = pr × 2πr .dr =
R
W =
∫
2πC.dr = 2πC [r ]0R = 2πC.R or C =
0
W
2πR
We know that frictional torque acting on the ring,
Tr = 2π µ. pr .cosec α.r 2 .dr = 2π µ ×
C
× cosec α.r 2 .dr
r
= 2 π µ.C.cosec α.r.dr
∴ Total frictional torque acting on the bearing,
R
T =
∫
0
R
2
2 π µ.C .cosec α.r.dr = 2 π µ.C .cosec α r
2 0
= 2π µ.C.cosec α ×
R2
= πµ.C.cosec α.R 2
2
Substituting the value of C, we have
T = πµ ×
1
1
W
× cosec α.R 2 = ×µ.W .R cosec α = × µ.W .l
2 πR
2
2
10.29. Trapezoidal or Truncated Conical Pivot Bearing
If the pivot bearing is not conical, but a frustrum of a cone with r1 and r2, the external and
internal radius respectively as shown in Fig. 10.19, then
Chapter 10 : Friction
l
291
Area of the bearing surface,
A = π [(r1 )2 − (r2 )2 ]
∴ Intensity of uniform pressure,
pn =
W
W
=
2
A π [(r1 ) − ( r2 )2 ]
...(i)
1. Considering uniform pressure
The total torque acting on the bearing is obtained by integrating the
value of T r (as discussed in Art. 10.27) within the limits r1 and r2.
Fig.10.19. Trapezoidal
pivot bearing.
∴ Total torque acting on the bearing,
T =
r1
∫
r2
3
2πµ. pn cosec α.r .dr = 2πµ. pn .cosec α r
3
r1
2
r2
( r )3 − ( r )3
2
= 2 π µ. pn .cosec α 1
3
Substituting the value of pn from equation (i),
T = 2π.µ ×
=
(r )3 − (r )3
W
2
× cosec α 1
2
π [(r1 ) − ( r2 ) ]
3
2
( r ) 3 − ( r2 ) 3
2
× µ.W .cosec α 1 2
2
3
( r1 ) − ( r2 )
2. Considering uniform wear
We have discussed in Art. 10.26 that the load transmitted to the ring,
δW = 2πC.dr
∴ Total load transmitted to the ring,
W =
r1
∫
r2
C =
or
r
2π C.dr = 2πC [r ]r1 = 2πC (r1 − r2 )
2
W
2π ( r1 − r2 )
...(ii)
We know that the torque acting on the ring, considering uniform wear, is
T r = 2π µ.C cosec α.r.dr
∴ Total torque acting on the bearing,
T =
r1
∫
r2
2
2π µ. C cosec α.r.dr = 2π µ.C.cosec α r
2
= π µ.C .cosec α ( r1 ) 2 − ( r2 ) 2
r1
r2
292
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Theory of Machines
Substituting the value of C from equation (ii), we get
W
T = πµ ×
× cosec α [( r1 ) 2 − ( r2 ) 2 ]
2π ( r1 − r2 )
1
= × µ.W ( r1 + r2 ) cosec α = µ.W .R cosec α
2
r + r2
where
R = Mean radius of the bearing = 1
2
Example 10.17. A conical pivot supports a load of 20 kN, the cone angle is 120º and the
intensity of normal pressure is not to exceed 0.3 N/mm2. The external diameter is twice the internal
diameter. Find the outer and inner radii of the bearing surface. If the shaft rotates at 200 r.p.m. and
the coefficient of friction is 0.1, find the power absorbed in friction. Assume uniform pressure.
Solution. Given : W = 20 kN = 20 × 103 N ; 2 α = 120º or α = 60º ; pn = 0.3 N/mm2 ;
N = 200 r.p.m. or ω = 2 π × 200/60 = 20.95 rad/s ; µ = 0.1
Outer and inner radii of the bearing surface
Let
r1 and r2 = Outer and inner radii of the bearing surface, in mm.
Since the external diameter is twice the internal diameter, therefore
r1 = 2 r2
We know that intensity of normal pressure ( pn),
0.3 =
∴
and
W
π [( r1 ) 2 − ( r2 ) 2 ]
=
20 × 103
π [(2 r2 ) 2 − ( r2 ) 2 ]
=
2.12 × 103
( r2 ) 2
(r2 )2 = 2.12 × 103 / 0.3 = 7.07 × 103 or r2 = 84 mm Ans.
r1 = 2 r2 = 2 × 84 = 168 mm Ans.
Power absorbed in friction
We know that total frictional torque (assuming uniform pressure),
T =
( r )3 − ( r2 )3
2
× µ.W .cosec α 1 2
2
3
( r1 ) − ( r2 )
(168)3 − (84)3
2
N-mm
= × 0.1 × 20 × 103 × cosec 60º =
2
2
3
(168) − (84)
= 301760 N-mm = 301.76 N-m
∴ Power absorbed in friction,
P = T.ω = 301.76 × 20.95 = 6322 W = 6.322 kW Ans.
Example 10.18. A conical pivot bearing supports a vertical shaft of 200 mm diameter. It is
subjected to a load of 30 kN. The angle of the cone is 120º and the coefficient of friction is 0.025.
Find the power lost in friction when the speed is 140 r.p.m., assuming 1. uniform pressure ; and
2. uniform wear.
Solution. Given : D = 200 mm or R = 100 mm = 0.1 m ; W = 30 kN = 30 × 103 N ; 2 α = 120º
or α = 60º ; µ = 0.025 ; N = 140 r.p.m. or ω = 2 π × 140/160 = 14.66 rad/s
1. Power lost in friction assuming uniform pressure
We know that total frictional torque,
2
T = × µ.W .R. cosec α
3
Chapter 10 : Friction
=
∴ Power lost in friction,
l
293
2
× 0.025 × 30 × 103 × 0.1 × cosec 60º = 57.7 N-m
3
P = T.ω = 57.7 × 14.66 = 846 W Ans.
2. Power lost in friction assuming uniform wear
We know that total frictional torque,
1
T = × µ.W .R . cosec α
2
1
= × 0.025 × 30 × 103 × 0.1 × cosec 60º = 43.3 N-m
2
∴ Power lost in friction, P = T.ω = 43.3 × 14.66 = 634.8 W Ans.
10.30. Flat Collar Bearing
We have already discussed that collar bearings are used to take the axial thrust of the rotating
shafts. There may be a single collar or multiple collar bearings as shown in Fig. 10.20 (a) and (b)
respectively. The collar bearings are also known as thrust bearings. The friction in the collar bearings may be found as discussed below :
(a) Single collar bearing
(b) Multiple collar bearing.
Fig. 10.20. Flat collar bearings.
Consider a single flat collar bearing supporting a shaft as shown in Fig. 10.20 (a).
Let
r1 = External radius of the collar, and
r2 = Internal radius of the collar.
∴ Area of the bearing surface,
A = π [(r1)2 – (r2)2]
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Theory of Machines
1. Considering uniform pressure
When the pressure is uniformly distributed over the bearing surface, then the intensity of
pressure,
p=
W
W
=
A π [r1 )2 − (r2 )2 ]
...(i)
We have seen in Art. 10.25, that the frictional torque on the ring of radius r and thickness dr,
Tr = 2 πµ. p.r 2 .dr
Integrating this equation within the limits from r2 to r1 for the total frictional torque on the
collar.
∴ Total frictional torque,
r
r 1
(r )3 − (r2 )3
T = ∫ 2 πµ. p.r .dr = 2πµ. p 3 = 2πµ. p 1
r2
3
3 r2
r1
2
Substituting the value of p from equation (i),
T = 2πµ ×
=
(r1 )3 − (r2 )3
W
π [(r1 )2 − (r2 )2 ]
3
2
× µ.W
3
( r1 )3 − ( r2 )3
2
2
( r1 ) − ( r2 )
Notes: 1. In order to increase the amount of rubbing surfaces so as to reduce the intensity of pressure, it is better
to use two or more collars, as shown in Fig. 10.20 (b), rather than one larger collar.
2. In case of a multi-collared bearings with, say n collars, the intensity of the uniform pressure,
p=
Load
W
=
No. of collars × Bearing area of one collar n π[( r1) 2 − ( r2 ) 2 ]
3. The total torque transmitted in a multi collared shaft remains constant i.e.
T =
( r ) 3 − ( r2 ) 3
2
× µ.W 1 2
2
3
(r1 ) − (r2 )
2. Considering unifrom wear
We have seen in Art. 10.25 that the load transmitted on the ring, considering uniform wear is,
δW = pr .2 πr.dr =
C
× 2πr .dr = 2 πC .dr
r
∴ Total load transmitted to the collar,
W =
r1
∫r
or
C =
r
2π C.dr = 2π C [r ]r1 = 2π C (r1 − r2 )
2
W
2π (r1 − r2 )
2
...(ii)
Chapter 10 : Friction
l
295
We also know that frictional torque on the ring,
Tr = µ. δW .r = µ × 2 πC .dr .r = 2 π µ.C .r .dr
∴ Total frictional torque on the bearing,
T =
r1
∫
r2
2
2πµ C.r.dr = 2πµ.C r
2
r1
r2
( )2 − (r2 ) 2
= 2πµ.C r1
2
= πµ.C [(r1 )2 − (r2 )2 ]
Substituting the value of C from equation (ii),
T = πµ ×
1
W
[(r1 ) 2 − ( r2 ) 2 ] = × µ.W ( r1 + r2 )
2 π ( r1 − r2 )
2
Example 10.19. A thrust shaft of a ship has 6
collars of 600 mm external diameter and 300 mm internal
diameter. The total thrust from the propeller is 100 kN. If
the coefficient of friction is 0.12 and speed of the engine
90 r.p.m., find the power absorbed in friction at the thrust
block, assuming l. uniform pressure ; and 2. uniform
wear.
Solution. Given : n = 6 ; d1 = 600 mm or r1 = 300
mm ; d2 = 300 mm or r2 = 150 mm ; W = 100 kN
= 100 × 10 3 N ; µ = 0.12 ; N = 90 r.p.m. or
ω = 2 π × 90/60 = 9.426 rad/s
1. Power absorbed in friction, assuming uniform
pressure
We know that total frictional torque transmitted,
T =
=
2
× µ.W
3
( r1 )3 − ( r2 )3
2
2
( r1 ) − ( r2 )
Ship propeller.
(300)3 − (150)3
2
× 0.12 × 100 × 103
= 2800 × 103 N-mm
2
2
3
(300) − (150)
= 2800 N-m
∴ Power absorbed in friction,
P = T .ω = 2800 × 9.426 = 26 400 W = 26.4 kW Ans.
2. Power absorbed in friction assuming uniform wear
We know that total frictional torque transmitted,
T =
1
1
× µ.W (r1 + r2 ) = × 0.12 × 100 × 103 (300 + 150) N-mm
2
2
= 2700 × 103 N-mm = 2700 N-m
∴ Power absorbed in friction,
P = T.ω = 2700 × 9.426 = 25 450 W = 25.45 kW Ans.
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Theory of Machines
Example 10.20. A shaft has a number of a collars integral with it. The external diameter of
the collars is 400 mm and the shaft diemater is 250 mm. If the intensity of pressure is 0.35 N/mm2
(uniform) and the coefficient of friction is 0.05, estimate : 1. power absorbed when the shaft runs at
105 r.p.m. carrying a load of 150 kN ; and 2. number of collars required.
Solution. Given : d1 = 400 mm or r1 = 200 mm ; d2 = 250 mm or r2 = 125 mm ; p = 0.35
N/mm2 ; µ = 0.05 ; N = 105 r.p.m or ω = 2 π × 105/60 = 11 rad/s ; W = 150 kN = 150 × 103 N
1. Power absorbed
We know that for uniform pressure, total frictional torque transmitted,
3
3
( r )3 − ( r2 )3 2
2
3 (200) − (125)
0.05
150
10
N-mm
T = × µ.W 1 2
=
×
×
×
2
2
2
3
( r1 ) − ( r2 ) 3
(200) − (125)
= 5000 × 248 = 1240 × 103 N-mm = 1240 N-m
∴ Power absorbed,
P = T .ω = 1240 × 11 = 13640 W = 13.64 kW Ans.
2. Number of collars required
Let
n = Number of collars required.
We know that the intensity of uniform pressure ( p),
0.35 =
W
n.π [(r1 ) 2 − ( r2 ) 2 ]
=
150 × 103
n.π [(200) 2 − (125) 2 ]
=
1.96
n
n = 1.96 / 0.35 = 5.6 say 6 Ans.
∴
Example 10.21. The thrust of a propeller shaft in a marine engine is taken up by a number
of collars integral with the shaft which is 300 mm in diameter. The thrust on the shaft is 200 kN and
the speed is 75 r.p.m. Taking µ constant and equal to 0.05 and assuming intensity of pressure as
uniform and equal to 0.3 N/mm2, find the external diameter of the collars and the number of collars
required, if the power lost in friction is not to exceed 16 kW.
Solution. Given : d2 = 300 mm or r2 = 150 mm = 0.15 m ; W = 200 kN = 200 × 103 N ;
N = 75 r.p.m. or ω = 2 π × 75/60 = 7.86 rad/s ; µ = 0.05 ; p = 0.3 N/mm2 ; P = 16 kW = 16 × 103 W
Let
T = Total frictional torque transmitted in N-m.
We know that power lost in friction (P),
16 × 103 = T.ω = T × 7.86 or T = 16 × 103/7.86 = 2036 N-m
External diameter of the collar
Let
d1 = External diameter of the collar in metres = 2 r1.
We know that for uniform pressure, total frictional torque transmitted (T ),
2036 =
=
(r )3 − (r )3 2
(r )2 + (r ) 2 + r .r *
2
2
1
2
1 2
× µ.W 1 2
=
×
µ
×
W
2
3
3
+
r
r
(
)
(
)
−
r
r
1
2
1
2
(r )2 + (0.15)2 + r1 × 0.15
2
× 0.05 × 200 × 103 1
3
r1 + 0.15
2036 × 3(r1 + 0.15) = 20 × 103 [(r1 )2 + 0.15 r1 + 0.0225]
*
( r1 )3 − ( r2 )3
( r1 ) − ( r2 )
2
3
=
( r1 − r2 ) [( r1 ) 2 + ( r2 ) 2 + r1.r2 ] ( r1 ) 2 + ( r2 ) 2 + r1.r2
=
( r1 + r2 ) ( r1 − r2 )
r1 + r2
Chapter 10 : Friction
l
297
Dividing throughout by 20 × 103,
0.305 (r1 + 0.15) = (r1)2 + 0.15 r1 + 0.0225
(r1)2 – 0.155 r1 – 0.0233 = 0
Solving this as a quadratic equation,
0.155 ± (0.155) 2 + 4 × 0.0233 0.155 ± 0.342
=
2
2
= 0.2485 m = 248.5 mm
...(Taking + ve sign)
r1 =
∴
d1 = 2 r1 = 2 × 248.5 = 497 mm Ans.
Number of collars
Let
n = Number of collars.
We know that intensity of pressure (p),
0.3 =
∴
W
n π [ r1 ) − ( r2 ) ]
2
2
=
200 × 10 3
n π [(248.5) − (150) ]
2
2
=
1.62
n
n = 1.62 / 0.3 = 5.4 or 6 Ans.
10.31. Friction Clutches
A friction clutch has its principal application in the transmission of power of shafts and
machines which must be started and stopped frequently. Its application is also found in cases in which
power is to be delivered to machines partially or fully loaded. The force of friction is used to start the
driven shaft from rest and gradually brings it up to the proper speed without excessive slipping of the
friction surfaces. In automobiles, friction clutch is used to connect the engine to the driven shaft. In
operating such a clutch, care should be taken so that the friction surfaces engage easily and gradually
brings the driven shaft up to proper speed. The proper alignment of the bearing must be maintained
and it should be located as close to the clutch as possible. It may be noted that
1. The contact surfaces should develop a frictional force that may pick up and hold the load
with reasonably low pressure between the contact surfaces.
2. The heat of friction should be rapidly dissipated and tendency to grab should be at a
minimum.
3. The surfaces should be backed by a material stiff enough to ensure a reasonably uniform
distribution of pressure.
The friction clutches of the following types are important from the subject point of view :
1. Disc or plate clutches (single disc or multiple disc clutch),
2. Cone clutches, and
3. Centrifugal clutches.
We shall now discuss, these clutches, in detail, in the following pages. It may be noted that
the disc and cone clutches are based on the same theory as the pivot and collar bearings.
10.32. Single Disc or Plate Clutch
A single disc or plate clutch, as shown in Fig. 10.21, consists of a clutch plate whose both
sides are faced with a friction material (usually of Ferrodo). It is mounted on the hub which is free to
move axially along the splines of the driven shaft. The pressure plate is mounted inside the clutch
body which is bolted to the flywheel. Both the pressure plate and the flywheel rotate with the engine
298
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Theory of Machines
crankshaft or the driving shaft. The pressure plate
pushes the clutch plate towards the flywheel by a set
of strong springs which are arranged radially inside
the body. The three levers (also known as release
levers or fingers) are carried on pivots suspended
from the case of the body. These are arranged in such
a manner so that the pressure plate moves away from
the flywheel by the inward movement of a thrust
bearing. The bearing is mounted upon a forked shaft
and moves forward when the clutch pedal is pressed.
When the clutch pedal is pressed down, its
Single disc clutch
linkage forces the thrust release bearing to move in
towards the flywheel and pressing the longer ends of the levers inward. The levers are forced to turn
on their suspended pivot and the pressure plate moves away from the flywheel by the knife edges,
thereby compressing the clutch springs. This action removes the pressure from the clutch plate and
thus moves back from the flywheel and the driven shaft becomes stationary. On the other hand, when
the foot is taken off from the clutch pedal, the thrust bearing moves back by the levers. This allows the
springs to extend and thus the pressure plate pushes the clutch plate back towards the flywheel.
Fig. 10.21. Single disc or plate clutch.
The axial pressure exerted by the spring provides a frictional force in the circumferential
direction when the relative motion between the driving and driven members tends to take place. If the
torque due to this frictional force exceeds the torque to be transmitted, then no slipping takes place
and the power is transmitted from the driving shaft to the driven shaft.
Now consider two friction surfaces, maintained in contact by an axial thrust W , as shown in
Fig. 10.22 (a).
Chapter 10 : Friction
Let
l
299
T = Torque transmitted by the clutch,
p = Intensity of axial pressure with which the contact surfaces are held
together,
r1 and r2 = External and internal radii of friction faces, and
µ = Coefficient of friction.
Consider an elementary ring of radius r and thickness dr as shown in Fig. 10.22 (b).
We know that area of contact surface or friction surface,
= 2 π r.dr
∴ Normal or axial force on the ring,
δW = Pressure × Area = p × 2 π r.dr
and the frictional force on the ring acting tangentially at radius r,
F r = µ.δW = µ.p × 2 π r.dr
∴ Frictional torque acting on the ring,
Tr = Fr × r = µ.p × 2 π r.dr × r = 2 π × µ .p.r2 dr
(a)
(b)
Fig. 10.22. Forces on a single disc or plate clutch.
We shall now consider the following two cases :
1. When there is a uniform pressure, and
2. When there is a uniform wear.
1. Considering uniform pressure
When the pressure is uniformly distributed over the entire area of the friction face, then the
intensity of pressure,
W
p=
...(i)
2
π [(r ) − (r )2 ]
1
where
2
W = Axial thrust with which the contact or friction surfaces are held together.
We have discussed above that the frictional torque on the elementary ring of radius r and
thickness dr is
Tr = 2 π µ.p.r2 dr
Integrating this equation within the limits from r2 to r1 for the total frictional torque.
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Theory of Machines
∴ Total frictional torque acting on the friction surface or on the clutch,
T =
r2
∫
r1
r
2 πµ. p.r .dr = 2 πµ p 3
3
r1
2
r2
(r )3 − (r2 )3
= 2 πµ p 1
3
Substituting the value of p from equation (i),
T = 2 πµ ×
=
where
W
π[(r1 ) 2 − ( r2 ) 2 ]
×
(r1 )3 − (r2 )3
3
(r )3 − (r )3
2
2
× µ.W 1 2
= µ.W .R
2
3
(r1 ) − (r2 )
R = Mean radius of friction surface
=
2
3
( r1 )3 − ( r2 )3
2
2
( r1 ) − ( r2 )
2. Considering uniform wear
In Fig. 10.22, let p be the normal intensity of pressure at a distance r from the axis of the
clutch. Since the intensity of pressure varies inversely with the distance, therefore
p.r. = C (a constant) or p = C/r
...(i)
and the normal force on the ring,
C
× 2 π C.dr = 2 π C.dr
r
∴ Total force acting on the friction surface,
δW = p.2π r.dr =
W =
r1
∫
2 π C dr = 2 π C [r ]r1 = 2 π C (r1 − r2 )
r
2
r2
or
C =
W
2 π ( r1 − r2 )
We know that the frictional torque acting on the ring,
C
Tr = 2 πµ. p r 2 .dr = 2 πµ × × r 2 .dr = 2πµ.C.r.dr
r
...(∵ p = C/r)
∴ Total frictional torque on the friction surface,
T =
r1
∫
r2
= πµ.C [( r1 ) 2 − ( r2 ) 2 ] = πµ ×
=
where
r
(r )2 − (r )2
2 1
2
2πµ.C.r .dr = 2 πµ.C r = 2 πµ.C 1
2 r2
2
W
( r ) 2 − ( r2 ) 2
2 π ( r1 − r2 ) 1
1
× µ.W ( r1 + r2 ) = µ.W .R
2
R = Mean radius of the friction surface =
r1 + r2
2
Notes : 1. In general, total frictional torque acting on the friction surface (or on the clutch) is given by
T = n.µ.W.R
Chapter 10 : Friction
where
l
301
n = Number of pairs of friction or contact surfaces, and
R = Mean radius of friction surface
=
=
3
3
2 (r1 ) − (r2 )
2
3 ( r1 ) − ( r2 )2
...(For uniform pressure)
r1 + r2
...(For uniform wear)
2
2. For a single disc or plate clutch, normally both sides of the disc are effective. Therefore, a single disc
clutch has two pairs of surfaces in contact, i.e. n = 2.
3. Since the intensity of pressure is maximum at the inner radius (r2) of the friction or contact surface,
therefore equation (i) may be written as
pmax × r2 = C
or
pmax = C/r2
4. Since the intensity of pressure is minimum at the outer radius (r1) of the friction or contact surface,
therefore equation (i) may be written as
pmin × r1 = C
or
pmin = C/r1
5. The average pressure ( pav) on the friction or contact surface is given by
Total force on friction surface
W
=
Cross-sectional area of friction surface π[( r1) 2 − ( r2 ) 2 ]
6. In case of a new clutch, the intensity of pressure is approximately uniform but in an old clutch the
uniform wear theory is more approximate.
pav =
7. The uniform pressure theory gives a higher frictional torque than the uniform wear theory. Therefore
in case of friction clutches, uniform wear should be considered, unless otherwise stated.
10.33. Multiple Disc Clutch
A multiple disc clutch, as shown in Fig. 10.23, may be used when a large torque is to be
transmitted. The inside discs (usually of steel) are fastened to the driven shaft to permit axial motion
Dual Disc Clutches.
(except for the last disc). The outside discs (usually of bronze) are held by bolts and are fastened to
the housing which is keyed to the driving shaft. The multiple disc clutches are extensively used in
motor cars, machine tools etc.
Let
n1 = Number of discs on the driving shaft, and
n2 = Number of discs on the driven shaft.
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Theory of Machines
∴ Number of pairs of contact surfaces,
n = n1 + n2 – 1
and total frictional torque acting on the friction surfaces or on the clutch,
T = n.µ.W.R
where
R = Mean radius of the friction surfaces
=
=
3
3
2 ( r1 ) − ( r2 )
3 ( r1 ) 2 − ( r2 ) 2
r1 + r2
2
...(For uniform pressure)
...(For uniform wear)
Fig. 10.23. Multiple disc clutch.
Example 10.22. Determine the maximum, minimum and average pressure in plate clutch
when the axial force is 4 kN. The inside radius of the contact surface is 50 mm and the outside radius
is 100 mm. Assume uniform wear.
Solution. Given : W = 4 kN = 4 × 103 N ; r2 = 50 mm ; r1 = 100 mm
Maximum pressure
Let
pmax = Maximum pressure.
Since the intensity of pressure is maximum at the inner radius (r2), therefore
pmax × r2 = C or C = 50 pmax
We know that the total force on the contact surface (W ),
4 × 103 = 2 π C (r1 – r2) = 2 π × 50 pmax (100 – 50) = 15 710 pmax
∴
pmax = 4 × 103/15 710 = 0.2546 N/mm2 Ans.
Minimum pressure
Let
pmin = Minimum pressure.
Since the intensity of pressure is minimum at the outer radius (r1), therefore
pmin × r1 = C or C = 100 pmin
Chapter 10 : Friction
l
303
We know that the total force on the contact surface (W ),
4 × 103 = 2 π C (r1 – r2) = 2π × 100 pmin (100 – 50) = 31 420 pmin
∴
pmin = 4 × 103/31 420 = 0.1273 N/mm2 Ans.
Average pressure
We know that average pressure,
Total normal force on contact surface
Cross-sectional area of contact surfaces
4 × 103
W
=
=
= 0.17 N/mm 2 Ans.
π [( r1 ) 2 − ( r2 ) 2 ] π [(100) 2 − (50) 2 ]
Example 10.23. A single plate clutch, with both sides effective, has outer and inner
diameters 300 mm and 200 mm respectively. The maximum intensity of pressure at any point in the
contact surface is not to exceed 0.1 N/mm2. If the coefficient of friction is 0.3, determine the power
transmitted by a clutch at a speed 2500 r.p.m.
pav =
Solution. Given : d1 = 300 mm or r1 = 150 mm ; d2 = 200 mm or r2 = 100 mm ; p = 0.1 N/mm2 ;
µ = 0.3 ; N = 2500 r.p.m. or ω = 2π × 2500/60 = 261.8 rad/s
Since the intensity of pressure ( p) is maximum at the inner radius (r2), therefore for uniform
wear,
p.r2 = C or C = 0.1 × 100 = 10 N/mm
We know that the axial thrust,
W = 2 π C (r1 – r2) = 2 π × 10 (150 – 100) = 3142 N
and mean radius of the friction surfaces for uniform wear,
r + r2 150 + 100
R= 1
=
= 125 mm = 0.125 m
2
2
We know that torque transmitted,
T = n.µ.W.R = 2 × 0.3 × 3142 × 0.125 = 235.65 N-m
...(3 n = 2,for both sides of plate effective)
∴ Power transmitted by a clutch,
P = T.ω = 235.65 × 261.8 = 61 693 W = 61.693 kW Ans.
Example 10.24. A single plate clutch, effective on both sides, is required to transmit 25 kW
at 3000 r.p.m. Determine the outer and inner radii of frictional surface if the coefficient of friction is
0.255, the ratio of radii is 1.25 and the maximum pressure is not to exceed 0.1 N/mm2. Also determine the axial thrust to be provided by springs. Ass ume the theory of uniform wear.
Solution. Given: n = 2 ; P = 25 kW = 25 × 103 W ; N = 3000 r.p.m. or ω = 2π × 3000/60
= 314.2 rad/s ; µ = 0.255 ; r1/r2 = 1.25 ; p = 0.1 N/mm2
Outer and inner radii of frictional surface
Let
r1 and r2 = Outer and inner radii of frictional surfaces, and
T = Torque transmitted.
Since the ratio of radii (r1/r2) is 1.25, therefore
r1 = 1.25 r2
We know that the power transmitted (P),
25 × 103 = T.ω = T × 314.2
∴
T = 25 × 103/314.2 = 79.6 N-m = 79.6 × 103 N-mm
304
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Theory of Machines
Since the intensity of pressure is maximum at the inner radius (r2), therefore
p.r2 = C
or
C = 0.1 r2 N/mm
and the axial thrust transmitted to the frictional surface,
W = 2 π C (r1 – r2) = 2 π × 0.1 r2 (1.25 r2 – r2) = 0.157 (r2)2
...(i)
We know that mean radius of the frictional surface for uniform wear,
r + r2 1.25 r2 + r2
R= 1
=
= 1.125 r2
2
2
We know that torque transmitted (T),
79.6 × 103 = n.µ.W.R = 2 × 0.255 × 0.157 (r2)2 × 1.125 r2 = 0.09 (r2)3
∴
and
(r2)3 = 79.6 × 103/0.09 = 884 × 103 or r2 = 96 mm Ans.
r1 = 1.25 r2 = 1.25 × 96 = 120 mm Ans.
Axial thrust to be provided by springs
We know that axial thrust to be provided by springs,
W = 2 π C (r1 – r2) = 0.157 (r2)2
= 0.157
(96)2
...[From equation (i)]
= 1447 N Ans.
Example 10.25. A single dry plate clutch transmits 7.5 kW at 900 r.p.m. The axial pressure
is limited to 0.07 N/mm2. If the coefficient of friction is 0.25, find 1. Mean radius and face width of
the friction lining assuming the ratio of the mean radius to the face width as 4, and 2. Outer and
inner radii of the clutch plate.
Solution. Given : P = 7.5 kW = 7.5 × 103 W ; N = 900 r.p.m or ω = 2 π × 900/60 = 94.26 rad/s ;
p = 0.07 N/mm2 ; µ = 0.25
1. Mean radius and face width of the friction lining
Let
R = Mean radius of the friction lining in mm, and
w = Face width of the friction lining in mm,
Ratio of mean radius to the face width,
R/w = 4
...(Given)
We know that the area of friction faces,
A = 2 π R.w
∴ Normal or the axial force acting on the friction faces,
W = A × p = 2 π R.w.p
We know that torque transmitted (considering uniform wear),
T = n.µ.W .R = n.µ (2 π R.w. p ) R
π
R
= n.µ 2 π R × × p R = × n.µ. p.R3
2
4
=
π
× 2 × 0.25 × 0.07 R3 = 0.055 R3 N-mm
2
...(∵ w = R/4)
...(i)
...(∵ n = 2, for single plate clutch)
Chapter 10 : Friction
l
305
We also know that power transmitted (P),
7.5 × 103 = T.ω = T × 94.26
∴
T = 7.5 × 103/94.26 = 79.56 N-m = 79.56 × 103 N-mm
...(ii)
From equations (i) and (ii),
R 3 = 79.56 × 103/0.055 = 1446.5 × 103 or R = 113 mm Ans.
and
w = R/4 = 113/4 = 28.25mm Ans.
2. Outer and inner radii of the clutch plate
Let
r1 and r2 = Outer and inner radii of the clutch plate respectively.
Since the width of the clutch plate is equal to the difference of the outer and inner radii,
therefore
w = r1 – r2 = 28.25 mm
...(iii)
Also for uniform wear, the mean radius of the clutch plate,
R=
r1 + r2
or
2
From equations (iii) and (iv),
r1 + r2 = 2 R = 2 × 113 = 226 mm
...(iv)
r1 = 127.125 mm ; and r2 = 98.875 Ans.
Example 10.26. A dry single plate clutch is to be designed for an automotive vehicle whose
engine is rated to give 100 kW at 2400 r.p.m. and maximum torque 500 N-m. The outer radius of
friction plate is 25% more than the inner radius. The intensity of pressure between the plate is not to
exceed 0.07 N/mm2. The coefficient of friction may be assumed equal to 0.3. The helical springs
required by this clutch to provide axial force necessary to engage the clutch are eight. If each spring
has stiffness equal to 40 N /mm, determine the initial compression in the springs and dimensions of
the friction plate.
Solution. Given : P = 100 kW = 100 × 103 W ; T = 500 N-m = 500 × 103 N-mm ;
p = 0.07 N/mm2 ; µ = 0.3 ; Number of springs = 8 ; Stiffness = 40 N/mm
Dimensions of the friction plate
Let
r1 and r2 = Outer and inner radii of the friction plate respectively.
Since the outer radius of the friction plate is 25% more than the inner radius, therefore
r1 = 1.25 r2
We know that, for uniform wear,
p.r2 = C or C = 0.07 r2 N/mm
and load transmitted to the friction plate,
W = 2 π C (r1 – r2) = 2 π × 0.07 r2 (1.125 r2 – r2) = 0.11 (r2)2 N
...(i)
We know that mean radius of the plate for uniform wear,
R=
r1 + r2
2
∴ Torque transmitted (T ),
=
1.25 r2 + r2
2
= 1.125 r2
500 × 103 = n.µ.W .R = 2 × 0.3 × 0.11 (r2)2 × 1.125 r2 = 0.074 (r2)3
...(∵ n = 2)
∴
(r2
)3
= 500 ×
103/0.074
= 6757 ×
103
or r2 = 190 mm Ans.
306
l
Theory of Machines
and
r1 = 1.25 r2 = 1.25 × 190 = 273.5 mm Ans.
Initial compression of the springs
We know that total stiffness of the springs,
s = Stiffness per spring × No. of springs = 40 × 8 = 320 N/mm
Axial force required to engage the clutch,
W = 0.11 (r2)2 = 0.11 (190)2 = 3970 N
...[From equation (i)]
∴ Initial compression in the springs
= W/s = 3970/320 = 12.5 mm Ans.
Example 10.27. A rotor is driven by a co-axial motor through a single plate clutch, both
sides of the plate being effective. The external and internal diameters of the plate are respectively
220 mm and 160 mm and the total spring load pressing the plates together is 570 N. The motor
armature and shaft has a mass of 800 kg with an effective radius of gyration of 200 mm. The rotor
has a mass of 1300 kg with an effective radius of gyration of 180 mm. The coefficient of friction for
the clutch is 0.35.
The driving motor is brought up to a speed of 1250 r.p.m. when the current is switched off
and the clutch suddenly engaged. Determine
1. The final speed of motor and rotor, 2. The time to reach this speed, and 3. The kinetic
energy lost during the period of slipping.
How long would slipping continue if it is assumed that a constant resisting torque of 60 N-m
were present? If instead of a resisting torque, it is assumed that a constant driving torque of 60 N-m
is maintained on the armature shaft, what would then be slipping time?
Solution. Given : d1 = 220 mm or r1 = 110 mm ; d2 = 160 mm or r2 = 80 mm ; W = 570 N ;
m 1 = 800 kg ; k 1 = 200 mm = 0.2 m ; m 2 = 1300 kg ; k 2 = 180 mm = 0.18 m ; µ = 0.35 ; N 1 = 1250 r.p.m.
or ω1 = π × 1250/60 = 131 rad/s
1. Final speed of the motor and rotor
Let
ω3 = Final speed of the motor and rotor in rad/s.
We know that moment of inertia for the motor armature and shaft,
I1 = m 1 (k 1)2 = 800 (0.2)2 = 32 kg-m2
and moment of inertia for the rotor,
I2 = m 2 (k 2)2 = 1300 (0.18)2 = 42.12 kg-m2
Since the angular momentum before slipping is equal to the angular momentum after slipping, therefore
I1.ω1 + I2.ω2 = (I1 + I2) ω3
32 × 131 + I2 × 0 = (32 + 42.12) ω3 = 74.12 ω3
∴
ω3 = 32 × 131 / 74.12 = 56.56 rad/s Ans.
2. Time to reach this speed
Let
t = Time to reach this speed i.e. 56.56 rad/s.
We know that mean radius of the friction plate,
r + r2 110 + 80
R= 1
=
= 95 mm = 0.095 m
2
2
...(∵ ω2 = 0)
l
Chapter 10 : Friction
307
and total frictional torque,
T = n.µ.W.R = 2 × 0.35 × 570 × 0.095 = 37.9 N-m
...(∵ n = 2)
Considering the rotor, let α2, ωI and ωF be the angular acceleration, initial angular speed and
the final angular speed of the rotor respectively.
We know that the torque (T ),
37.9 = I2.α2 = 42.12 α2
α2 = 37.9/42.12 = 0.9 rad/s2
or
Since the angular acceleration is the rate of change of angular speed, therefore
α2 =
ωF − ωI
t
or
t=
ωF − ωI
α2
=
56.56 − 0
= 62.8 s Ans.
0.9
...(∵ ωF = ω3 = 56.56 rad/s, and ω1 = 0)
3. Kinetic energy lost during the period of slipping
We know that angular kinetic energy before impact,
1
1
1
I1 ( ω1 )2 + I 2 ( ω2 ) 2 = I1 ( ω1 ) 2
2
2
2
1
= × 32 (131) 2 = 274 576 N-m
2
and angular kinetic energy after impact,
E1 =
...(∵ ω2 = 0)
1
1
( I1 + I 2 ) (ω3 )2 = (32 + 42.12) (56.56)2 = 118 556 N-m
2
2
∴ Kinetic energy lost during the period of slipping,
= E1 – E2 = 274 576 – 118 556 = 156 020 N-m Ans.
Time of slipping assuming constant resisting torque
Let
t1 = Time of slipping, and
ω2 = Common angular speed of armature and rotor shaft = 56.56 rad/s
When slipping has ceased and there is exerted a constant torque of 60 N-m on the armature
shaft, then
Torque on armature shaft,
T1 = – 60 – 37.9 = – 97.9 N-m
Torque on rotor shaft,
T2 = T = 37.9 N-m
Considering armature shaft,
E2 =
ω3 = ω1 + α1 .t1 = ω1 +
T1
I1
× t1 = 131 −
97.9
× t1 = 131 − 3.06 t1
32
...(i)
Considering rotor shaft,
ω3 = α 2 .t1 =
T2
I2
× t1 =
37.9
× t = 0.9 t1
42.12 1
From equations (i) and (ii),
131 − 3.06 t1 = 0.9 t1
∴
or
3.96 t1 = 131
t1 = 131/ 3.96 = 33.1 s Ans.
...(ii)
308
l
Theory of Machines
Time of slipping assuming constant driving torque of 60 N-m
In this case,
Since
ω1 +
131 +
∴
T 1 = 60 – 37.9 = 22.1 N-m
T1
I1
× t1 =
T2
I2
× t1 , therefore
22.1
37.9
× t1 =
×t
32
42.12 1
0.9 t1 – 0.69 t1 = 131
or
or
131 + 0.69 t1 = 0.9 t1
t1 = 624 s Ans.
Example 10.28. A multiple disc clutch has five plates having four pairs of active friction
surfaces. If the intensity of pressure is not to exceed 0.127 N/mm2, find the power transmitted at 500
r.p.m. The outer and inner radii of friction surfaces are 125 mm and 75 mm respectively. Assume
uniform wear and take coefficient of friction = 0.3.
Solution. Given : n1 + n2 = 5 ; n = 4 ; p = 0.127 N/mm2 ; N = 500 r.p.m. or ω = 2π × 500/60
= 52.4 rad/s ; r1 = 125 mm ; r2 = 75 mm ; µ = 0.3
Since the intensity of pressure is maximum at the inner radius r2, therefore
p.r2 = C
or
C = 0.127 × 75 = 9.525 N/mm
We know that axial force required to engage the clutch,
W = 2 π C (r1 – r2) = 2 π × 9.525 (125 – 75) = 2990 N
and mean radius of the friction surfaces,
r1 + r2
125 + 75
= 100 mm = 0.1 m
2
2
We know that torque transmitted,
R=
=
T = n.µ.W.R = 4 × 0.3 × 2990 × 0.1 = 358.8 N-m
∴ Power transmitted,
P = T.ω = 358.8 × 52.4 = 18 800 W = 18.8 kW Ans.
Example 10.29. A multi-disc clutch has three discs on the driving shaft and two on the
driven shaft. The outside diameter of the contact surfaces is 240 mm and inside diameter 120 mm.
Assuming uniform wear and coefficient of friction as 0.3, find the maximum axial intensity of pressure between the discs for transmitting 25 kW at 1575 r.p.m.
Solution. Given : n1 = 3 ; n2 = 2 ; d1 = 240 mm or r1 = 120 mm ; d2 = 120 mm or r2 = 60 mm ;
µ = 0.3 ; P = 25 kW = 25 × 103 W ; N = 1575 r.p.m. or ω = 2 π × 1575/60 = 165 rad/s
Let
T = Torque transmitted in N-m, and
W = Axial force on each friction surface.
We know that the power transmitted (P),
25 × 103 = T.ω = T × 165 or T = 25 × 103/165 = 151.5 N-m
Number of pairs of friction surfaces,
n = n1 + n2 – 1 = 3 + 2 – 1 = 4
and mean radius of friction surfaces for uniform wear,
R=
r1 + r2
2
=
120 + 60
= 90 mm = 0.09 m
2
Chapter 10 : Friction
l
309
We know that torque transmitted (T ),
151.5 = n.µ.W.R = 4 × 0.3 × W × 0.09 = 0.108 W
∴
Let
W = 151.5/0.108 = 1403 N
p = Maximum axial intensity of pressure.
Since the intensity of pressure ( p) is maximum at the inner radius (r2 ), therefore for uniform
wear
p.r2 = C or
C = p × 60 = 60 p N/mm
We know that the axial force on each friction surface (W ),
1403 = 2 π.C (r1 – r2) = 2 π × 60 p (120 – 60) = 22 622 p
∴
p = 1403/22 622 = 0.062 N/mm2 Ans.
Example 10.30. A plate clutch has three discs on the driving shaft and two discs on the
driven shaft, providing four pairs of contact surfaces. The outside diameter of the contact surfaces is
240 mm and inside diameter 120 mm. Assuming uniform pressure and µ = 0.3; find the total spring
load pressing the plates together to transmit 25 kW at 1575 r.p.m.
If there are 6 springs each of stiffness 13 kN/m and each of the contact surfaces has worn
away by 1.25 mm, find the maximum power that can be transmitted, assuming uniform wear.
Solution. Given : n1 = 3 ; n2 = 2 ; n = 4 ; d1 = 240 mm or r1 = 120 mm ; d2 = 120 mm or
r2 = 60 mm ; µ = 0.3 ; P = 25 kW = 25 × 103 W ; N = 1575 r.p.m. or ω = 2 π × 1575/60 = 165 rad/s
Total spring load
Let
W = Total spring load, and
T = Torque transmitted.
We know that power transmitted (P),
25 × 103 = T.ω = T × 165 or T = 25 × 103/165 = 151.5 N-m
Mean radius of the contact surface, for uniform pressure,
R=
3
3
2 ( r1 ) − ( r2 ) 2 (120)3 − (60)3
=
= 93.3 mm = 0.0933 m
3 ( r1 ) 2 − ( r2 ) 2 3 (120) 2 − (60) 2
and torque transmitted (T ),
151.5 = n.µ.W.R = 4 × 0.3 W × 0.0933 = 0.112 W
∴
W = 151.5/0.112 = 1353 N Ans.
Maximum power transmitted
Given : No of springs = 6
∴ Contact surfaces of the spring
=8
Wear on each contact surface
= 1.25 mm
∴
Total wear = 8 × 1.25 = 10 mm = 0.01 m
Stiffness of each spring = 13 kN/m = 13 × 103 N/m
∴ Reduction in spring force
= Total wear × Stiffness per spring × No. of springs
= 0.01 × 13 × 103 × 6 = 780 N
310
l
Theory of Machines
∴ New axial load, W = 1353 – 780 = 573 N
We know that mean radius of the contact surfaces for uniform wear,
r + r2 120 + 60
R= 1
=
= 90 mm = 0.09 m
2
2
∴ Torque transmitted,
T = n.µ.W.R. = 4 × 0.3 × 573 × 0.09 = 62 N-m
and maximum power transmitted,
P = T. ω = 62 × 155 = 10 230 W = 10.23 kW Ans.
10.34. Cone Clutch
A cone clutch, as shown in Fig. 10.24, was extensively used in automobiles but now-a-days it
has been replaced completely by the disc clutch.
Fig. 10.24. Cone clutch.
It consists of one pair of friction surface only. In a cone clutch, the driver is keyed to the
driving shaft by a sunk key and has an inside conical surface or face which exactly fits into the outside
conical surface of the driven. The driven member resting on the feather key in the driven shaft, may
be shifted along the shaft by a forked lever provided at B, in order to engage the clutch by bringing the
two conical surfaces in contact. Due to the frictional resistance set up at this contact surface, the
torque is transmitted from one shaft to another. In some cases, a spring is placed around the driven
shaft in contact with the hub of the driven. This spring holds the clutch faces in contact and maintains
the pressure between them, and the forked lever is used only for disengagement of the clutch. The
contact surfaces of the clutch may be metal to metal contact, but more often the driven member is
lined with some material like wood, leather, cork or asbestos etc. The material of the clutch faces (i.e.
contact surfaces) depends upon the allowable normal pressure and the coefficient of friction.
Consider a pair of friction surface as shown in Fig. 10.25 (a). Since the area of contact of a
pair of friction surface is a frustrum of a cone, therefore the torque transmitted by the cone clutch may
be determined in the similar manner as discussed for conical pivot bearings in Art. 10.28.
Let
pn = Intensity of pressure with which the conical friction surfaces are held
together (i.e. normal pressure between contact surfaces),
r1 and r2 = Outer and inner radius of friction surfaces respectively.
Chapter 10 : Friction
R = Mean radius of the friction surface =
l
311
r1 + r2
,
2
α = Semi angle of the cone (also called face angle of the cone) or the
angle of the friction surface with the axis of the clutch,
µ = Coefficient of friction between contact surfaces, and
b = Width of the contact surfaces (also known as face width or clutch
face).
Fig. 10.25. Friction surfaces as a frustrum of a cone.
Consider a small ring of radius r and thickness dr, as shown in Fig. 10.25 (b). Let dl is length
of ring of the friction surface, such that
dl = dr.cosec α
∴ Area of the ring,
A = 2π r.dl = 2πr.dr cosec α
We shall consider the following two cases :
1. When there is a uniform pressure, and
2. When there is a uniform wear.
1. Considering uniform pressure
We know that normal load acting on the ring,
δW n = Normal pressure × Area of ring = pn × 2 π r.dr.cosec α
and the axial load acting on the ring,
δW = Horizontal component of δW n (i.e. in the direction of W )
= δW n × sin α = pn × 2π r.dr. cosec α × sin α = 2π × pn.r.dr
∴ Total axial load transmitted to the clutch or the axial spring force required,
W =
r1
∫
r2
2
2 π pn .r.dr = 2 π pn r
2
r1
r2
(r )2 − (r2 )2
= 2 π pn 1
2
= π pn ( r1 ) − ( r2 )
2
∴
pn =
2
W
π[(r1 ) − (r2 ) 2 ]
2
...(i)
312
l
Theory of Machines
We know that frictional force on the ring acting tangentially at radius r,
Fr = µ.δW n = µ.pn × 2 π r.dr.cosec α
∴ Frictional torque acting on the ring,
Tr = Fr × r = µ.pn × 2 π r.dr. cosec α.r = 2 π µ.pn.cosec α.r2 dr
Integrating this expression within the limits from r2 to r1 for the total frictional torque on the
clutch.
∴ Total frictional torque,
T =
r1
∫
r2
3
2 πµ. pn .cosec α.r .dr = 2 πµ pn .cosec α r
3
r1
2
r2
(r )3 − (r2 )3
= 2π µ pn .cosec α 1
3
Substituting the value of pn from equation (i), we get
T = 2π µ ×
=
(r )3 − (r )3
W
2
cosec
×
α
1
3
π [(r1 ) 2 − (r2 )2 ]
( r )3 − ( r2 )3
2
× µ.W .cosec α 1 2
2
3
( r1 ) − ( r2 )
..(ii)
2. Considering uniform wear
In Fig. 10.25, let pr be the normal intensity of pressure at a distance r from the axis of the
clutch. We know that, in case of uniform wear, the intensity of pressure varies inversely with the
distance.
∴
pr .r = C (a constant) or pr = C / r
We know that the normal load acting on the ring,
δW n = Normal pressure × Area of ring = pr × 2πr.dr cosec α
and the axial load acting on the ring ,
δW = δW n × sin α = pr.2 π r.dr.cosec α .sin α = pr × 2 π r.dr
C
× 2 π r.dr = 2 π C.dr
r
∴ Total axial load transmitted to the clutch,
=
W =
r1
∫
r2
r
2 π C.dr = 2πC [r ]r1 = 2 π C (r1 − r2 )
2
W
2 π ( r1 − r2 )
We know that frictional force acting on the ring,
F r = µ.δW n = µ.pr × 2 π r × dr cosec α
and frictional torque acting on the ring,
Tr = Fr × r = µ.pr × 2 π r.dr.cosec α × r
or
C =
=µ×
...(∵ pr = C / r)
C
× 2 π r 2 .dr.cosec α = 2 πµ.C cosec α × r dr
r
...(iii)
Chapter 10 : Friction
l
313
∴ Total frictional torque acting on the clutch,
T =
r1
∫
r2
2
2 πµ.C.cosec α.r dr = 2 πµ.C.cosec α r
2
r1
r2
( r ) 2 − ( r2 ) 2
= 2 π µ.C .cosec α 1
2
Substituting the value of C from equation (i), we have
T = 2 πµ ×
(r )2 − (r )2
W
2
× cosec α 1
2 π (r1 − r2 )
2
r + r2
= µ.W cosec α 1
= µ.W .R cosec α
2
r1 + r2
= Mean radius of friction surface
2
Since the normal force acting on the friction surface, W n = W /sin α, therefore the equation
(iv) may be written as
where
R=
...(iv)
T = µ.W n.R
...(v)
The forces on a friction surface, for steady operation of the clutch and after the clutch is
engaged, is shown in Fig. 10.26.
Fig. 10.26. Forces on a friction surface.
From Fig. 10.26 (a), we find that
r1 − r2 = b sin α; and R =
r1 + r2
2
or r1 + r2 = 2 R
314
l
Theory of Machines
∴ From equation, (i), normal pressure acting on the friction surface,
pn =
or
where
W
π[(r1 ) − (r2 ) ]
2
2
=
W
W
=
(
)
(
)
2
π r1 + r2 r1 − r2
π R.b.sin α
W = pn × 2 π R.b sin α = W n sin α
W n = Normal load acting on the friction surface = pn × 2 π R.b
Now the equation (iv) may be written as,
T = µ ( pn × 2 π R.b sin α) R cosec α = 2 πµ. pn .R 2b
The following points may be noted for a cone clutch :
1. The above equations are valid for steady operation of the clutch and after the clutch is
engaged.
2. If the clutch is engaged when one member is stationary and the other rotating (i.e. during
engagement of the clutch) as shown in Fig. 10.26 (b), then the cone faces will tend to slide on each
other due to the presence of relative motion. Thus an additional force (of magnitude equal to µ.W n.cos α)
acts on the clutch which resists the engagement and the axial force required for engaging the clutch
increases.
∴ Axial force required for engaging the clutch,
We = W + µ.W n cos α = W n sin α + µ.W n cos α
= W n (sin α + µ cos α)
3. Under steady operation of the clutch, a decrease in the semi-cone angle (α) increases the
torque produced by the clutch (T ) and reduces the axial force (W ). During engaging period, the axial
force required for engaging the clutch (W e) increases under the influence of friction as the angle α
decreases. The value of α can not be decreased much because smaller semi-cone angle (α) requires
larger axial force for its disengagement.
For free disengagement of the clutch, the value of tan α must be greater than µ. In case the
value of tan α is less than µ, the clutch will not disengage itself and the axial force required to
disengage the clutch is given by
Wd = W n (µ cos α – sin α)
Example 10.31. A conical friction clutch is used to transmit 90 kW at 1500 r.p.m. The semicone angle is 20º and the coefficient of friction is 0.2. If the mean diameter of the bearing surface is
375 mm and the intensity of normal pressure is not to exceed 0.25 N/mm2, find the dimensions of the
conical bearing surface and the axial load required.
Solution. Given : P = 90 kW = 90 × 103 W ; N = 1500 r.p.m. or ω = 2 π × 1500/60 = 156
rad/s ; α = 20º ; µ = 0.2 ; D = 375 mm or R = 187.5 mm ; pn = 0.25 N/mm2
Dimensions of the conical bearing surface
Let
r1 and r2 = External and internal radii of the bearing surface respectively,
b = Width of the bearing surface in mm, and
T = Torque transmitted.
We know that power transmitted (P),
90 × 103 = T.ω = T × 156
∴
T = 90 × 103/156 = 577 N-m = 577 × 103 N-mm
Chapter 10 : Friction
l
and the torque transmitted (T),
577 × 103 = 2 π µ pn.R 2.b = 2π × 0.2 × 0.25 (187.5)2 b = 11 046 b
∴
b = 577 × 103/11 046 = 52.2 mm Ans.
We know that r1 + r2 = 2R = 2 × 187.5 = 375 mm
and
r1 – r2 = b sin α = 52.2 sin 20º = 18 mm
From equations (i) and (ii),
r1 = 196.5 mm, and r2 = 178.5 mm Ans.
Axial load required
315
...(i)
...(ii)
Since in case of friction clutch, uniform wear is considered and the intensity of pressure is
maximum at the minimum contact surface radius (r2), therefore
pn.r2 = C (a constant) or C = 0.25 × 178.5 = 44.6 N/mm
We know that the axial load required,
W = 2πC (r1 – r2) = 2π × 44.6 (196.5 – 178.5) = 5045 N Ans.
Example 10.32. An engine developing 45 kW at 1000 r.p.m. is fitted with a cone clutch built
inside the flywheel. The cone has a face angle of 12.5º and a maximum mean diameter of 500 mm.
The coefficient of friction is 0.2. The normal pressure on the clutch face is not to exceed 0.1 N/mm2.
Determine : 1. the axial spring force necessary to engage to clutch, and 2. the face width required.
Solution. Given : P = 45 kW = 45 × 103 W ; N = 1000 r.p.m. or ω = 2π × 1000/60 = 104.7
rad/s ; α = 12.5º ; D = 500 mm or R = 250 mm = 0.25 m ; µ = 0.2 ; pn = 0.1 N/mm2
1. Axial spring force necessary to engage the clutch
First of all, let us find the torque (T ) developed by the clutch and the normal load (W n) acting
on the friction surface.
We know that power developed by the clutch (P),
45 × 103 = T.ω = T × 104.7 or T = 45 × 103/104.7 = 430 N-m
We also know that the torque developed by the clutch (T),
430 = µ.W n.R = 0.2 × W n × 0.25 = 0.05 W n
∴
Wn = 430/0.05 = 8600 N
and axial spring force necessary to engage the clutch,
We = W n (sin α + µ cos α)
= 8600 (sin 12.5º + 0.2 cos 12.5º) = 3540 N Ans.
2. Face width required
Let
b = Face width required.
We know that normal load acting on the friction surface (W n),
8600 = pn × 2 π R.b = 0.1 × 2π × 250 × b = 157 b
∴
b = 8600/157 = 54.7 mm Ans.
Example 10.33. A leather faced conical clutch has a cone angle of 30º. If the intensity of
pressure between the contact surfaces is limited to 0.35 N/mm2 and the breadth of the conical surface
is not to exceed one-third of the mean radius, find the dimensions of the contact surfaces to transmit
22.5 kW at 2000 r.p.m. Assume uniform rate of wear and take coefficient of friction as 0.15.
Solution. Given : 2 α = 30º or α = 15º ; pn = 0.35 N/mm2; b = R/3 ; P = 22.5 kW =
22.5 × 103 W ; N = 2000 r.p.m. or ω = 2 π × 2000/60 = 209.5 rad/s ; µ = 0.15
Let
r1 = Outer radius of the contact surface in mm,
316
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Theory of Machines
r2 = Inner radius of the contact surface in mm,
R = Mean radius of the the contact surface in mm,
b = Face width of the contact surface in mm = R/3, and
T = Torque transmitted by the clutch in N-m.
We know that power transmitted (P),
22.5 × 103 = T.ω = T × 209.5
∴
T = 22.5 × 103/209.5 = 107.4 N-m = 107.4 × 103 N-mm
We also know that torque transmitted (T ),
107.4 × 103 = 2π µ pn.R 2. b = 2π × 0.15 × 0.35 × R 2 × R/3 = 0.11 R 3
∴
R 3 = 107.4 × 103/0.11 = 976.4 × 103 or R = 99 mm Ans.
The dimensions of the contact surface are shown in Fig. 10.27.
Fig. 10.27
From Fig. 10.27, we find that
r1 − r2 = b sin α =
99
R
× sin α =
× sin 15º = 8.54 mm
3
3
r1 + r2 = 2 R = 2 × 99 = 198 mm
and
...(i)
...(ii)
From equations (i) and (ii),
r1 = 103.27 mm, and r2 = 94.73 mm Ans.
Example 10.34. The contact surfaces in a cone clutch have an effective diameter of 75 mm.
The semi-angle of the cone is 15º. The coefficient of friction is 0.3. Find the torque required to
produce slipping of the clutch if an axial force applied is 180 N.
This clutch is employed to connect an electric motor running uniformly at 1000 r.p.m. with a
flywheel which is initially stationary. The flywheel has a mass of 13.5 kg and its radius of gyration is
150 mm. Calculate the time required for the flywheel to attain full speed and also the energy lost in
the slipping of the clutch.
Solution. Given : D = 75 mm or R = 37.5 mm = 0.0375 m ; α = 15º ; µ = 0.3 ; W = 180 N ;
N F = 1000 r.p.m. or ωF = 2π × 1000/60 = 104.7 rad/s ; m = 13.5 kg ; k = 150 mm = 0.15 m
Torque required to produce slipping
We know that torque required to produce slipping,
T = µ.W.R.cosec α = 0.3 × 180 × 0.0375 × cosec 15º = 7.8 N-m Ans.
Time required for the flywheel to attain full speed
Let
tF = Time required for the flywheel to attain full speed in seconds, and
αF = Angular acceleration of the flywheel in rad/s2.
We know that the mass moment of inertia of the flywheel,
IF = m.k2 = 13.5 × (0.15)2 = 0.304 kg-m2
Chapter 10 : Friction
l
317
∴ Torque required (T ),
7.8 = IF.αF = 0.304 αF or αF = 7.8/0.304 = 25.6 rad/s2
and angular speed of the flywheel (ωF),
104.7 = αF.tF = 25.6 tF or tF = 104.7/25.6 = 4.1 s Ans.
Energy lost in slipping of the clutch
We know that the angle turned through by the motor and flywheel (i.e. clutch) in time 4.1 s
from rest,
θ = Average angular velocity × time =
1
1
× WF × tF = × 104.7 × 4.1 = 214.6 rad
2
2
∴ Energy lost in slipping of the clutch,
=T.θ = 7.8 × 214.6 = 1674 N-m Ans.
10.35. Centrifugal Clutch
The centrifugal clutches are usually incorporated into the motor pulleys. It consists of a
number of shoes on the inside of a rim of the pulley, as shown in Fig. 10.28. The outer surface of the
shoes are covered with a friction material. These shoes, which can move radially in guides, are held
Fig. 10.28. Centrifugal clutch.
against the boss (or spider) on the driving shaft by means of
springs. The springs exert a radially inward force which is
assumed constant. The mass of the shoe, when revolving, causes
it to exert a radially outward force (i.e. centrifugal force). The
magnitude of this centrifugal force depends upon the speed at
which the shoe is revolving. A little consideration will show
that when the centrifugal force is less than the spring force, the
shoe remains in the same position as when the driving shaft
was stationary, but when the centrifugal force is equal to the
spring force, the shoe is just floating. When the centrifugal
force exceeds the spring force, the shoe moves outward and
comes into contact with the driven member and presses against
it. The force with which the shoe presses against the driven
member is the difference of the centrifugal force and the spring
force. The increase of speed causes the shoe to press harder
Centrifugal clutch.
318
l
Theory of Machines
and enables more torque to be transmitted.
In order to determine the mass and size of the shoes, the following procedure is adopted :
1. Mass of the shoes
Consider one shoe of a centrifugal clutch as shown in Fig. 10.29.
Let
m = Mass of each shoe,
n = Number of shoes,
r = Distance of centre of gravity of
the shoe from the centre of the
spider,
R = Inside radius of the pulley rim,
N = Running speed of the pulley in
r.p.m.,
ω = Angular running speed of the
pulley in rad/s = 2πN/60 rad/s,
ω1 = Angular speed at which the
engagement begins to take place,
and
Fig. 10.29. Forces on a shoe of
centrifugal clutch.
µ = Coefficient of friction between
the shoe and rim.
We know that the centrifugal force acting on each shoe at the running speed,
*Pc = m.ω2.r
and the inward force on each shoe exerted by the spring at the speed at which engagement begins to
take place,
Ps = m (ω1)2 r
∴ The net outward radial force (i.e. centrifugal force) with which the shoe presses against
the rim at the running speed
= Pc – Ps
and the frictional force acting tangentially on each shoe,
F = µ (Pc – Ps)
∴ Frictional torque acting on each shoe,
= F × R = µ (Pc – Ps) R
and total frictional torque transmitted,
T = µ (Pc – Ps) R × n = n.F.R
From this expression, the mass of the shoes (m) may be evaluated.
2. Size of the shoes
Let
l = Contact length of the shoes,
b = Width of the shoes,
*
The radial clearance between the shoe and the rim being very small as compared to r, therefore it is neglected.
If, however, the radial clearance is given, then the operating radius of the mass centre of the shoe from the axis
of the clutch,
r1 = r + c, where c = Radial clearance.
Then
Pc = m.ω2.r1, and Ps = m (ω1)2 r1
Chapter 10 : Friction
l
319
R = Contact radius of the shoes. It is same as the inside radius of the rim
of the pulley.
θ = Angle subtended by the shoes at the centre of the spider in radians.
p = Intensity of pressure exerted on the shoe. In order to ensure reasonable life, the intensity of pressure may be taken as 0.1 N/mm2.
We know that
θ = l/R rad
or
l = θ.R
∴ Area of contact of the shoe,
A = l.b
and the force with which the shoe presses against the rim
= A × p = l.b.p
Since the force with which the shoe presses against the rim at the running speed is (Pc – Ps),
therefore
l.b.p = Pc – Ps
From this expression, the width of shoe (b) may be obtained.
Example 10.35. A centrifugal clutch is to transmit 15 kW at 900 r.p.m. The shoes are four in
number. The speed at which the engagement begins is 3/4th of the running speed. The inside radius
of the pulley rim is 150 mm and the centre of gravity of the shoe lies at 120 mm from the centre of the
spider. The shoes are lined with Ferrodo for which the coefficient of friction may be taken as 0.25.
Determine : 1. Mass of the shoes, and 2. Size of the shoes, if angle subtended by the shoes at the
centre of the spider is 60º and the pressure exerted on the shoes is 0.1 N/mm2.
Solution. Given : P = 15 kW = 15 × 103 W ; N = 900 r.p.m. or ω = 25 × 900/60 = 94.26 rad/s ;
n = 4 ; R = 150 mm = 0.15 m ; r = 120 mm = 0.12 m ; µ = 0.25
Since the speed at which the engagement begins (i.e. ω1) is 3/4th of the running speed (i.e.
ω), therefore
3
3
ω = × 94.26 = 7 0.7 rad/s
4
4
Let
T = Torque transmitted at the running speed.
We know that power transmitted (P),
15 × 103 = T.ω = T × 94.26 or T = 15 × 103/94.26 = 159 N-m
1. Mass of the shoes
Let
m = Mass of the shoes in kg.
We know that the centrifugal force acting on each shoe,
Pc = m.ω2.r = m (94.26)2 × 0.12 = 1066 m N
and the inward force on each shoe exerted by the spring i.e. the centrifugal force at the engagement
speed ω1,
Ps = m (ω1)2 r = m (70.7)2 × 0.12 = 600 m N
∴ Frictional force acting tangentially on each shoe,
F = µ (Pc – Ps) = 0.25 (1066 m – 600 m) = 116.5 m N
We know that the torque transmitted (T ),
159 = n.F.R = 4 × 116.5 m × 0.15 = 70 m or m = 2.27 kg Ans.
2. Size of the shoes
Let
l = Contact length of shoes in mm,
b = Width of the shoes in mm,
ω1 =
320
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Theory of Machines
θ = Angle subtended by the shoes at the centre of the spider in radians
= 60º = π/3 rad, and
...(Given)
p = Pressure exerted on the shoes in N/mm2 = 0.1 N/mm2
We know that
and
∴
...(Given)
π
× 150 = 157.1 mm
3
l.b.p = Pc – Ps = 1066 m – 600 m = 466 m
l = θ. R =
157.1 × b × 0.1 = 466 × 2.27 = 1058
or
b = 1058/157.1 × 0.1 = 67.3 mm Ans.
Example 10.36. A centrifugal clutch has four shoes which slide radially in a spider keyed to
the driving shaft and make contact with the internal cylindrical surface of a rim keyed to the driven
shaft. When the clutch is at rest, each shoe is pulled against a stop by a spring so as to leave a radial
clearance of 5 mm between the shoe and the rim. The pull exerted by the spring is then 500 N. The
mass centre of the shoe is 160 mm from the axis of the clutch.
If the internal diameter of the rim is 400 mm, the mass of each shoe is 8 kg, the stiffness of
each spring is 50 N/mm and the coefficient of friction between the shoe and the rim is 0.3 ; find the
power transmitted by the clutch at 500 r.p.m.
Solution. Given : n = 4 ; c = 5 mm ; S = 500 N ; r = 160 mm ; D = 400 mm or R = 200 mm
= 0.2 m ; m = 8 kg ; s = 50 N/mm ; µ = 0.3 ; N = 500 r.p.m. or ω = 2 π × 500/60 = 52.37 rad/s
We know that the operating radius,
r1 = r + c = 160 + 5 = 165 mm = 0.165 m
Centrifugal force on each shoe,
Pc = m.ω2.r1 = 8 (52.37)2 × 0.165 = 3620 N
and the inward force exerted by the spring,
P4 = S + c.s = 500 + 5 × 50 = 750 N
∴ Frictional force acting tangentially on each shoe,
F = µ (Pc – Ps) = 0.3 (3620 – 750) = 861 N
We know that total frictional torque transmitted by the clutch,
T = n.F.R = 4 × 861 × 0.2 = 688.8 N-m
∴ Power transmitted,
P = T.ω = 688.8 × 52.37 = 36 100 W = 36.1 kW Ans.
EXERCISES
1.
Find the force required to move a load of 300 N up a rough plane, the force being applied parallel to
the plane. The inclination of the plane is such that a force of 60 N inclined at 30º to a similar smooth
plane would keep the same load in equilibrium. The coefficient of friction is 0.3.
[Ans. 146 N]
2.
A square threaded screw of mean diameter 25 mm and pitch of thread 6 mm is utilised to lift a weight
of 10 kN by a horizontal force applied at the circumference of the screw. Find the magnitude of the
force if the coefficient of friction between the nut and screw is 0.02.
[Ans. 966 N]
3.
A bolt with a square threaded screw has mean diameter of 25 mm and a pitch of 3 mm. It carries an
axial thrust of 10 kN on the bolt head of 25 mm mean radius. If µ = 0.12, find the force required at the
end of a spanner 450 mm long, in tightening up the bolt.
[Ans. 110.8 N]
4.
A turn buckle, with right and left hand threads is used to couple two railway coaches. The threads
which are square have a pitch of 10 mm and a mean diameter of 30 mm and are of single start type.
Taking the coefficient of friction as 0.1, find the work to be done in drawing the coaches together a
distance of 200 mm against a steady load of 20 kN.
[Ans. 3927 N-m]
Chapter 10 : Friction
l
321
5.
A vertical two start square threaded screw of a 100 mm mean diameter and 20 mm pitch supports a
vertical load of 18 kN. The axial thrust on the screw
is taken by a collar bearing of 250 mm outside diameter and 100 mm inside diameter. Find the force required at the end of a lever which is 400 mm long in
order to lift and lower the load. The coefficient of
friction for the vertical screw and nut is 0.15 and
that for collar bearing is 0.20.
[Ans. 1423 N ; 838 N]
6.
A sluice gate weighing 18 kN is raised and lowered
by means of square threaded screws, as shown in
Fig.10.30. The frictional resistance induced by water
pressure against the gate when it is in its lowest
position is 4000 N.
The outside diameter of the screw is 60 mm and pitch
is 10 mm. The outside and inside diameter of washer
is 150 mm and 50 mm respectively. The coefficient
of friction between the screw and nut is 0.1 and for
the washer and seat is 0.12. Find :
Fig. 10.30
1. The maximum force to be exerted at the ends of the lever for raising and lowering the gate, and
2. Efficiency of the arrangement.
[Ans. 114 N ; 50 N ; 15.4%]
7.
The spindle of a screw jack has single start square threads with an outside diameter of 45 mm and a
pitch of 10 mm. The spindle moves in a fixed nut. The load is carried on a swivel head but is not free
to rotate. The bearing surface of the swivel head has a mean diameter of 60 mm. The coefficient of
friction between the nut and screw is 0.12 and that between the swivel head and the spindle is 0.10.
Calculate the load which can be raised by efforts of 100 N each applied at the end of two levers each
of effective length of 350 mm. Also determine the velocity ratio and the efficiency of the lifting
arrangement.
[Ans. 9943 N ; 218.7 N ; 39.6%]
8.
The lead screw of a lathe has acme threads of 50 mm outside diameter and 10 mm pitch. The included
angle of the thread is 29°. It drives a tool carriage and exerts an axial pressure of 2500 N. A collar
bearing with outside diameter 100 mm and inside diameter 50 mm is provided to take up the thrust. If
the lead screw rotates at 30 r.p.m., find the efficiency and the power required to drive the screw. The
coefficient of friction for screw threads is 0.15 and for the collar is 0.12. [Ans. 16.3% ; 75.56 W]
9.
A flat foot step bearing 225 mm in diameter supports a load of 7.5 kN. If the coefficient of friction is
0.09 and r.p.m is 60, find the power lost in friction, assuming 1. Uniform pressure, and 2. Uniform
wear.
[Ans. 318 W ; 239 W]
10.
A conical pivot bearing 150 mm in diameter has a cone angle of 120º. If the shaft supports an axial
load of 20 kN and the coefficient of friction is 0.03, find the power lost in friction when the shaft
rotates at 200 r.p.m., assuming 1. Uniform pressure, and 2. uniform wear.
11.
A vertical shaft supports a load of 20 kN in a conical pivot bearing. The external radius of the cone is
3 times the internal radius and the cone angle is 120º. Assuming uniform intensity of pressure as 0.35
MN/m2, determine the dimensions of the bearing.
[Ans. 727.5 W ; 545.6 W]
If the coefficient of friction between the shaft and bearing is 0.05 and the shaft rotates at 120 r.p.m.,
find the power absorbed in friction.
[Ans. 47.7 mm ; 143 mm ; 1.50 kW]
12.
A plain collar type thrust bearing having inner and outer diameters of 200 mm and 450 mm is subjected to an axial thrust of 40 kN. Assuming coefficient of friction between the thrust surfaces as
0.025, find the power absorbed in overcoming friction at a speed of 120 r.p.m. The rate of wear is
considered to be proportional to the pressure and rubbing speed.
[Ans. 4.1 kW]
13.
The thrust on the propeller shaft of a marine engine is taken up by 8 collars whose external and
internal diameters are 660 mm and 420 mm respectively. The thrust pressure is 0.4 MN/m2 and may
322
14.
15.
16.
17.
18.
19.
20.
l
Theory of Machines
be assumed uniform. The coefficient of friction between the shaft and collars is 0.04. If the shaft
rotates at 90 r.p.m. ; find 1. total thrust on the collars ; and 2. power absorbed by friction at the bearing.
[Ans. 651 kN ; 68 kW]
A shaft has a number of collars integral with it. The external diameter of the collars is 400 mm and the
shaft diameter is 250 mm. If the uniform intensity of pressure is 0.35 N/mm2 and its coefficient of
friction is 0.05, estimate : 1. power absorbed in overcoming friction when the shaft runs at 105 r.p.m.
and carries a load of 150 kN, and 2. number of collars required.
[Ans. 13.4 kW ; 6]
A car engine has its rated output of 12 kW. The maximum torque developed is 100 N-m. The clutch
used is of single plate type having two active surfaces. The axial pressure is not to exceed 85 kN/m2.
The external diameter of the friction plate is 1.25 times the internal diameter. Determine the dimensions of the friction plate and the axial force exerted by the springs. Coefficient of friction = 0.3.
[Ans. 129.5 mm ; 103.6 mm ; 1433 N]
A single plate clutch (both sides effective) is required to transmit 26.5 kW at 1600 r.p.m. The outer
diameter of the plate is limited to 300 mm and intensity of pressure between the plates is not to exceed
68.5 kN/m2. Assuming uniform wear and a coefficient of friction 0.3, show that the inner diameter of
the plates is approximately 90 mm.
A multiplate clutch has three pairs of contact surfaces. The outer and inner radii of the contact surfaces are 100 mm and 50 mm respectively. The maximum axial spring force is limited to 1 kN. If the
coefficient of friction is 0.35 and assuming uniform wear, find the power transmitted by the clutch at
1500 r.p.m.
[Ans. 12.37 kW]
A cone clutch is to transmit 7.5 kW at 900 r.p.m. The cone has a face angle of 12º. The width of the
face is half of the mean radius and the normal pressure between the contact faces is not to exceed 0.09
N/mm2. Assuming uniform wear and the coefficient of friction between contact faces as 0.2, find the
main dimensions of the clutch and the axial force required to engage the clutch.
[Ans. R = 112 mm, b = 56 mm, r1 = 117.8 mm, r2 = 106.2 mm ; 1433 N]
A cone clutch with cone angle 20º is to transmit 7.5 kW at 750 r.p.m. The normal intensity of pressure
between the contact faces is not to exceed 0.12 N/mm2. The coefficient of friction is 0.2. If face width
is 1 5 th of mean diameter, find : 1. the main dimensions of the clutch, and 2. axial force required
while running.
[Ans. R = 117 mm ; b = 46.8 mm ; r1 = 125 mm ; r2 = 109 mm ; 1395 N]
A centrifugal friction clutch has a driving member consisting of a spider carrying four shoes which are
kept from contact with the clutch case by means of flat springs until increase of centrifugal force
overcomes the resistance of the springs and the power is transmitted by friction between the shoes and
the case.
Determine the necessary mass of each shoe if 22.5 kW is to be transmitted at 750 r.p.m. with
engagement beginning at 75% of the running speed. The inside diameter of the drum is 300 mm and
the radial distance of the centre of gravity of each shoe from the shaft axis is 125 mm. Assume
µ = 0.25.
[Ans. 5.66 kg]
DO YOU KNOW ?
1.
Discuss briefly the various types of friction experienced by a body.
2.
State the laws of
3.
(i) Static friction ;
(ii) Dynamic friction ;
(iii) Solid friction ; and
(iv) Fluid friction.
Explain the following :
(i) Limiting friction,
(ii) Angle of friction, and
(iii) Coefficient of friction.
4.
Derive from first principles an expression for the effort required to raise a load with a screw jack
taking friction into consideration.
5.
Neglecting collar friction, derive an expression for mechanical advantage of a square threaded screw
moving in a nut, in terms of helix angle of the screw and friction angle.
Chapter 10 : Friction
l
323
6.
In a screw jack, the helix angle of thread is α and the angle of friction is φ. Show that its efficiency is
maximum, when 2α = (90º – φ).
7.
For a screw jack having the nut fixed, derive the equation ( with usual notations),
η=
8.
Neglecting collar friction, from first principles, prove that the maximum efficiency of a square threaded
screw moving in a nut is
9.
tan α
.
tan ( α + φ) + µ.rm .r
1 − sin φ
, where φ is the friction angle.
1 + sin φ
Write a short note on journal bearing.
10.
What is meant by the expression ‘friction circle’? Deduce an expression for the radius of friction
circle in terms of the radius of the journal and the angle of friction.
11.
From first principles, deduce an expression for the friction moment of a collar thrust bearing, stating
clearly the assumptions made.
12.
Derive an expression for the friction moment for a flat collar bearing in terms of the inner radius r1,
outer radius r2, axial thrust W and coefficient of friction µ. Assume uniform intensity of pressure.
13.
Derive from first principles an expression for the friction moment of a conical pivot assuming
(i) Uniform pressure, and (ii) Uniform wear.
14.
A truncated conical pivot of cone angle φ rotating at speed N supports a load W . The smallest and
largest diameter of the pivot over the contact area are ‘d’ and ‘D’ respectively. Assuming uniform
wear, derive the expression for the frictional torque.
15.
Describe with a neat sketch the working of a single plate friction clutch.
16.
Establish a formula for the maximum torque transmitted by a single plate clutch of external and
internal radii r1 and r2, if the limiting coefficient of friction is µ and the axial spring load is W . Assume
that the pressure intensity on the contact faces is uniform.
17.
Which of the two assumptions-uniform intensity of pressure or uniform rate of wear, would you make
use of in designing friction clutch and why ?
18.
Describe with a neat sketch a centrifugal clutch and deduce an equation for the total torque transmitted.
OBJECTIVE TYPE QUESTIONS
1.
The angle of inclination of the plane, at which the body begins to move down the plane, is called
(a)
2.
angle of friction
(b)
angle of repose
(c)
angle of projection
In a screw jack, the effort required to lift the load W is given by
(a)
P = W tan (α – φ)
(b)
P = W tan (α + φ)
(c)
P = W cos (α – φ)
(d)
P = W cos (α + φ)
where
α = Helix angle, and
φ = Angle of friction.
3.
4.
The efficiency of a screw jack is given by
(a)
tan (α + φ)
tan α
(b)
tan α
tan (α + φ)
(c)
tan (α − φ)
tan α
(d)
tan α
tan ( α − φ)
The radius of a friction circle for a shaft of radius r rotating inside a bearing is
(a)
r sin φ
(b)
r cos φ
(c)
r tan φ
(d)
r cot φ
324
5.
l
Theory of Machines
The efficiency of a screw jack is maximum, when
(a)
6.
φ
2
(b)
α = 45º −
φ
2
(c)
α = 90º + φ
(d)
α = 90º − φ
(c)
1 − tan φ
1 + tan φ
(d)
1 + tan φ
1 − tan φ
The maximum efficiency of a screw jack is
(a)
7.
α = 45º +
1 − sin φ
1 + sin φ
(b)
1 + sin φ
1 − sin φ
The frictional torque transmitted in a flat pivot bearing, considering uniform pressure, is
(a)
where
1
× µ.W .R
2
(b)
2
× µ.W .R
3
(c)
3
× µ.W .R
4
(d)
µ.W .R
µ = Coefficient of friction,
W = Load over the bearing, and
R = Radius of the bearing surface.
8.
The frictional torque transmitted in a conical pivot bearing, considering uniform wear, is
(a)
1
× µ .W . R cosec α
2
(b)
2
× µ .W . R cosec α
3
3
× µ .W . R cosec α
(d) µ . W.R cosec α
4
where R = Radius of the shaft, and
α = Semi-angle of the cone.
The frictional torque transmitted by a disc or plate clutch is same as that of
(c)
9.
10.
(a)
flat pivot bearing
(b)
flat collar bearing
(c)
conical pivot bearing
(d)
trapezoidal pivot bearing
The frictional torque transmitted by a cone clutch is same as that of
(a)
flat pivot bearing
(b)
flat collar bearing
(c)
conical pivot bearing
(d)
trapezoidal pivot bearing
ANSWERS
1. (a)
2. (b)
3. (b)
4. (a)
5. (b)
6. (a)
7. (b)
8. (a)
9. (b)
10. (d)
GO To FIRST
CONTENTS
CONTENTS
Chapter 11 : Belt, Rope and Chain Drives
l
325
11
Belt, Rope
and Chain
Drives
Features (Main)
1.
4.
5.
6.
7.
11.
13.
14.
16.
17.
19.
20.
22.
23.
24.
26.
27.
28.
29.
30.
31.
32.
33.
34.
35.
36.
37.
38.
39.
Introduction.
Types of Belts.
Material used for Belts.
Types of Flat Belt Drives.
Velocity Ratio of Belt Drive.
Length of an Open Belt Drive.
Power Transmitted by a Belt.
Ratio of Driving Tensions for
Flat Belt Drive.
Centrifugal Tension.
Maximum Tension in the Belt.
Initial Tension in the Belt.
V-belt Drive.
Ratio of Driving Tensions for
V-belt.
Rope Drive.
Fibre Ropes.
Sheave for Fibre Ropes.
Wire Ropes.
Ratio of Driving Tensions for
Rope Drive.
Chain Drives.
Advantages and
Disadvantages of Chain
Drive Over Belt or Rope
Drive.
Terms Used in Chain Drive.
Relation Between Pitch and
Pitch Circle Diameter.
Relation Between Chain
Speed and Angular Velocity of
Sprocket.
Kinematic of Chain Drive.
Classification of Chains.
Hoisting and Hauling Chains.
Conveyor Chains.
Power Transmitting Chains.
Length of Chains.
11.1. Introduction
The belts or ropes are used to transmit power from
one shaft to another by means of pulleys which rotate at the
same speed or at different speeds. The amount of power transmitted depends upon the following factors :
1. The velocity of the belt.
2. The tension under which the belt is placed on the
pulleys.
3. The arc of contact between the belt and the smaller
pulley.
4. The conditions under which the belt is used.
It may be noted that
(a) The shafts should be properly in line to insure
uniform tension across the belt section.
(b) The pulleys should not be too close together, in
order that the arc of contact on the smaller pulley may be as large as possible.
(c) The pulleys should not be so far apart as to cause
the belt to weigh heavily on the shafts, thus increasing the friction load on the bearings.
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(d) A long belt tends to swing from side to side, causing the belt to run out of the pulleys,
which in turn develops crooked spots in the belt.
(e) The tight side of the belt should be at the bottom, so that whatever sag is present on the
loose side will increase the arc of contact at the pulleys.
( f ) In order to obtain good results with flat belts, the maximum distance between the shafts
should not exceed 10 metres and the minimum should not be less than 3.5 times the
diameter of the larger pulley.
11.2. Selection of a Belt Drive
Following are the various important factors upon which the selection of a belt drive depends:
1. Speed of the driving and driven shafts,
2. Speed reduction ratio,
3. Power to be transmitted,
4. Centre distance between the shafts,
5. Positive drive requirements,
6.
7. Space available, and
8. Service conditions.
Shafts layout,
11.3. Types of Belt Drives
The belt drives are usually classified into the following three groups :
1. Light drives. These are used to transmit small powers at belt speeds upto about 10 m/s, as
in agricultural machines and small machine tools.
2. Medium drives. These are used to transmit medium power at belt speeds over 10 m/s but
up to 22 m/s, as in machine tools.
3. Heavy drives. These are used to transmit large powers at belt speeds above 22 m/s, as in
compressors and generators.
11.4. Types of Belts
(a) Flat belt.
(b) V-belt.
(c) Circular belt.
Fig. 11.1. Types of belts.
Though there are many types of belts used these days, yet the following are important from
the subject point of view :
1. Flat belt. The flat belt, as shown in Fig. 11.1 (a), is mostly used in the factories and
workshops, where a moderate amount of power is to be transmitted, from one pulley to another when
the two pulleys are not more than 8 metres apart.
2. V-belt. The V-belt, as shown in Fig. 11.1 (b), is mostly used in the factories and workshops, where a moderate amount of power is to be transmitted, from one pulley to another, when the
two pulleys are very near to each other.
3. Circular belt or rope. The circular belt or rope, as shown in Fig. 11.1 (c), is mostly used
in the factories and workshops, where a great amount of power is to be transmitted, from one pulley
to another, when the two pulleys are more than 8 meters apart.
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327
If a huge amount of power is to be transmitted, then a single belt may not be sufficient. In
such a case, wide pulleys (for V-belts or circular belts) with a number of grooves are used. Then a belt
in each groove is provided to transmit the required amount of power from one pulley to another.
11.5. Material used for Belts
The material used for belts and ropes must be strong, flexible, and durable. It must have a
high coefficient of friction. The belts, according to the material used, are classified as follows :
1. Leather belts. The most important material for the belt is leather. The best leather belts are
made from 1.2 metres to 1.5 metres long strips cut from either side of the back bone of the top grade
steer hides. The hair side of the leather is smoother and harder than the flesh side, but the flesh side is
stronger. The fibres on the hair side are perpendicular to the surface, while those on the flesh side are
interwoven and parallel to the surface. Therefore for these reasons, the hair side of a belt should be in
contact with the pulley surface, as shown in Fig. 11.2. This gives a more intimate contact between the
belt and the pulley and places the greatest tensile strength of the belt section on the outside, where the
tension is maximum as the belt passes over the pulley.
(a) Single layer belt.
(b) Double layer belt.
Fig. 11.2. Leather belts.
The leather may be either oak-tanned or mineral salt tanned e.g. chrome tanned. In order to
increase the thickness of belt, the strips are cemented together. The belts are specified according to
the number of layers e.g. single, double or triple ply and according to the thickness of hides used e.g.
light, medium or heavy.
The leather belts must be periodically cleaned and dressed or treated with a compound or
dressing containing neats foot or other suitable oils so that the belt will remain soft and flexible.
2. Cotton or fabric belts. Most of the fabric belts are made by folding canvass or cotton duck
to three or more layers (depending upon the thickness desired) and stitching together. These belts are
woven also into a strip of the desired width and thickness. They are impregnated with some filler like
linseed oil in order to make the belts water proof and to prevent injury to the fibres. The cotton belts
are cheaper and suitable in warm climates, in damp atmospheres and in exposed positions. Since the
cotton belts require little attention, therefore these belts are mostly used in farm machinery, belt
conveyor etc.
3. Rubber belt. The rubber belts are made of layers of fabric impregnated with rubber composition and have a thin layer of rubber on the faces. These belts are very flexible but are quickly
destroyed if allowed to come into contact with heat, oil or grease. One of the principal advantage of
these belts is that they may be easily made endless. These belts are found suitable for saw mills, paper
mills where they are exposed to moisture.
4. Balata belts. These belts are similar to rubber belts except that balata gum is used in place
of rubber. These belts are acid proof and water proof and it is not effected by animal oils or alkalies.
The balata belts should not be at temperatures above 40° C because at this temperature the balata
begins to soften and becomes sticky. The strength of balata belts is 25 per cent higher than rubber
belts.
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11.6. Types of Flat Belt Drives
The power from one pulley to another may be transmitted by any of the following types of
belt drives:
1. Open belt drive. The open belt drive, as shown in Fig. 11.3, is used with shafts arranged
parallel and rotating in the same direction. In this case, the driver A pulls the belt from one side (i.e.
lower side RQ) and delivers it to the other side (i.e. upper side LM). Thus the tension in the lower side
belt will be more than that in the upper side belt. The lower side belt (because of more tension) is
known as tight side whereas the upper side belt (because of less tension) is known as slack side, as
shown in Fig. 11.3.
Fig. 11.3. Open belt drive.
2. Crossed or twist belt drive. The crossed or twist belt drive, as shown in Fig. 11.4, is used
with shafts arranged parallel and rotating in the opposite directions.
Fig. 11.4. Crossed or twist belt drive.
In this case, the driver pulls the belt from one side (i.e. RQ) and delivers it to the other side
(i.e. LM). Thus the tension in the belt RQ will be more than that in the belt LM. The belt RQ (because
of more tension) is known as tight side, whereas the belt LM (because of less tension) is known as
slack side, as shown in Fig. 11.4.
Chapter 11 : Belt, Rope and Chain Drives
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329
A little consideration will show that at a point where the belt crosses, it rubs against each
other and there will be excessive wear and tear. In order to avoid this, the shafts should be placed at
a maximum distance of 20 b, where b is the width of belt and the speed of the belt should be less than
15 m/s.
3. Quarter turn belt drive. The quarter turn belt drive also known as right angle belt drive, as
shown in Fig. 11.5 (a), is used with shafts arranged at right angles and rotating in one definite direction. In order to prevent the belt from leaving the pulley, the width of the face of the pulley should be
greater or equal to 1.4 b, where b is the width of belt.
In case the pulleys cannot be arranged, as shown in Fig. 11.5 (a), or when the reversible
motion is desired, then a quarter turn belt drive with guide pulley, as shown in Fig. 11.5 (b), may be
used.
(a) Quarter turn belt drive.
(b) Quarter turn belt drive with guide pulley.
Fig. 11.5
4. Belt drive with idler pulleys. A belt drive with an idler pulley, as shown in Fig. 11.6 (a), is
used with shafts arranged parallel and when an open belt drive cannot be used due to small angle of
contact on the smaller pulley. This type of drive is provided to obtain high velocity ratio and when the
required belt tension cannot be obtained by other means.
(a) Belt drive with single idler pulley.
(b) Belt drive with many idler pulleys.
Fig. 11.6
When it is desired to transmit motion from one shaft to several shafts, all arranged in parallel,
a belt drive with many idler pulleys, as shown in Fig. 11.6 (b), may be employed.
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5. Compound belt drive. A compound belt drive, as shown in Fig. 11.7, is used when power
is transmitted from one shaft to another through a number of pulleys.
Fig. 11.7. Compound belt brive.
6. Stepped or cone pulley drive. A stepped or cone pulley drive, as shown in Fig. 11.8, is
used for changing the speed of the driven shaft while the main or driving shaft runs at constant speed.
This is accomplished by shifting the belt from one part of the steps to the other.
7. Fast and loose pulley drive. A fast and loose pulley drive, as shown in Fig. 11.9, is used
when the driven or machine shaft is to be started or stopped when ever desired without interfering
with the driving shaft. A pulley which is keyed to the machine shaft is called fast pulley and runs at
the same speed as that of machine shaft. A loose pulley runs freely over the machine shaft and is
incapable of transmitting any power. When the driven shaft is required to be stopped, the belt is
pushed on to the loose pulley by means of sliding bar having belt forks.
Fig. 11.8. Stepped or cone pulley drive.
Fig. 11.9. Fast and loose pulley drive.
11.7. Velocity Ratio of Belt Drive
It is the ratio between the velocities of the driver and the follower or driven. It may be
expressed, mathematically, as discussed below :
Let
d1 = Diameter of the driver,
d2 = Diameter of the follower,
Chapter 11 : Belt, Rope and Chain Drives
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331
N1 = Speed of the driver in r.p.m., and
N2 = Speed of the follower in r.p.m.
∴ Length of the belt that passes over the driver, in one minute
= π d1.N1
Similarly, length of the belt that passes over the
follower, in one minute
= π d2 . N2
Since the length of belt that passes over the
driver in one minute is equal to the length of belt that
passes over the follower in one minute, therefore
π d1 . N1 = π d2 . N2
N2
d
= 1
N1 d 2
When the thickness of the belt (t) is considered,
then velocity ratio,
N2
d +t
= 1
N1 d2 + t
∴ Velocity ratio,
Note:
The velocity ratio of a belt drive may also be obtained as discussed below :
We know that peripheral velocity of the belt on the driving pulley,
π d1 . N 1
v1 =
m/s
60
and peripheral velocity of the belt on the driven or follower pulley,
π d2 . N2
v2 =
m/s
60
When there is no slip, then v1 = v2.
N2 d1
π d1 . N1 π d 2 . N 2
=
=
or
∴
N1 d 2
60
60
11.8. Velocity Ra
tio of a Compound Belt Dr
iv
e
Ratio
Driv
ive
Sometimes the power is transmitted from one shaft to another, through a number of pulleys as
shown in Fig. 11.7. Consider a pulley 1 driving the pulley 2. Since the pulleys 2 and 3 are keyed to the
same shaft, therefore the pulley 1 also drives the pulley 3 which, in turn, drives the pulley 4.
Let
d1 = Diameter of the pulley 1,
N1 = Speed of the pulley 1 in r.p.m.,
d2, d3, d4, and N2, N3, N4 = Corresponding values for pulleys 2, 3 and 4.
We know that velocity ratio of pulleys 1 and 2,
N2
d
= 1
...(i)
N1 d 2
Similarly, velocity ratio of pulleys 3 and 4,
N 4 d3
=
...(ii)
N3 d4
Multiplying equations (i) and (ii),
d
N2 N4
d
×
= 1 × 3
N1 N3 d2 d4
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Theory of Machines
d ×d
N4
= 1 3
N1 d2 × d4
or
...(∵ N 2 = N 3, being keyed to the same shaft)
A little consideration will show, that if there are six pulleys, then
N6
d × d × d5
= 1 3
N1 d 2 × d 4 × d 6
Speed of last driven
Product of diameters of drivers
=
Speed of first driver Product of diameters of drivens
or
11.9. Slip of Belt
In the previous articles, we have discussed the motion
of belts and shafts assuming a firm frictional grip between the
belts and the shafts. But sometimes, the frictional grip becomes
insufficient. This may cause some forward motion of the driver
without carrying the belt with it. This may also cause some
forward motion of the belt without carrying the driven pulley
with it. This is called slip of the belt and is generally expressed
as a percentage.
The result of the belt slipping is to reduce the velocity
ratio of the system. As the slipping of the belt is a common
phenomenon, thus the belt should never be used where a
definite velocity ratio is of importance (as in the case of hour,
minute and second arms in a watch).
Let
s1 % = Slip between the
driver and the belt, and
s2 % = Slip between the belt and the follower.
∴ Velocity of the belt passing over the driver per second
π d1 . N1 π d1 . N1
s
π d1 . N1
s
–
1– 1
× 1 =
60
60
100
60
100
and velocity of the belt passing over the follower per second,
v=
...(i)
π d2 . N2
s
s
= v – v × 2 = v 1 – 2
60
100
100
Substituting the value of v from equation (i),
π d 2 N 2 π d1 N1
s
s
1 – 1 1 – 2
=
60
60
100
100
N2
d
s
s
= 1 1 – 1 – 2
100 100
N1 d2
=
s ×s
... Neglecting 1 2
100 × 100
d1
s1 + s2 d1
s
1 –
1 – 100 =
100
d2
d2
... (where s = s1 + s2, i.e. total percentage of slip)
If thickness of the belt (t) is considered, then
N2
d +t
s
= 1
1 –
100
N1 d2 + t
Chapter 11 : Belt, Rope and Chain Drives
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333
Example 11.1. An engine, running at 150 r.p.m., drives a line shaft by means of a belt. The
engine pulley is 750 mm diameter and the pulley on the line shaft being 450 mm. A 900 mm diameter
pulley on the line shaft drives a 150 mm diameter pulley keyed to a dynamo shaft. Find the speed of
the dynamo shaft, when 1. there is no slip, and 2. there is a slip of 2% at each drive.
Solution. Given : N 1 = 150 r.p.m. ; d1 = 750 mm ; d2 = 450 mm ; d3 = 900 mm ; d4 = 150 mm
The arrangement of belt drive is shown in Fig. 11.10.
N4 = Speed of the dynamo shaft .
Let
Fig. 11.10
1. When there is no slip
We know that
d ×d
N4
= 1 3
N1 d2 × d4
∴
N4 = 150 × 10 = 1500 r.p.m. Ans.
or
750 × 900
N4
=
= 10
150 450 × 150
2. When there is a slip of 2% at each drive
We know that
d ×d
N4
= 1 3
N1 d2 × d4
s1
s2
1 – 100 1 – 100
750 × 900
2
2
N4
=
1 –
1 –
= 9.6
150 450 × 150
100
100
∴
N4 = 150 × 9.6 = 1440 r.p.m. Ans.
11.10. Creep of Belt
When the belt passes from the slack side to the tight side, a certain portion of the belt extends
and it contracts again when the belt passes from the tight side to slack side. Due to these changes of
length, there is a relative motion between the belt and the pulley surfaces. This relative motion is
termed as creep. The total effect of creep is to reduce slightly the speed of the driven pulley or
follower. Considering creep, the velocity ratio is given by
E + σ2
N2
d
= 1 ×
N1 d2 E + σ1
where
σ1 and σ2 = Stress in the belt on the tight and slack side respectively, and
E = Young’s modulus for the material of the belt.
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Theory of Machines
Example 11.2. The power is transmitted from a pulley 1 m diameter running at 200 r.p.m. to
a pulley 2.25 m diameter by means of a belt. Find the speed lost by the driven pulley as a result of
creep, if the stress on the tight and slack side of the belt is 1.4 MPa and 0.5 MPa respectively. The
Young’s modulus for the material of the belt is 100 MPa.
Solution. Given : d1 = 1 m ; N 1 = 200 r.p.m. ; d2 = 2.25 m ; σ1 = 1.4 MPa = 1.4 × 106 N/m2;
σ2 = 0.5 MPa = 0.5 × 106 N/m2 ; E = 100 MPa = 100 × 106 N/m2
Let
N2 = Speed of the driven pulley.
Neglecting creep, we know that
1
d
N2
d
= 88.9 r.p.m.
= 1 or N 2 = N1 × 1 = 200 ×
2.25
d2
N1 d2
Considering creep, we know that
E + σ2
N2
d
= 1 ×
N1 d2 E + σ1
or
100 × 106 + 0.5 × 106
1
×
= 88.7 r.p.m.
2.25 100 × 106 + 1.4 × 106
∴ Speed lost by driven pulley due to creep
N 2 = 200 ×
= 88.9 – 88.7 = 0.2 r.p.m. Ans.
11.11. Length of an Open Belt Drive
Fig. 11.11. Length of an open belt drive.
We have already discussed in Art. 11.6 that in an open belt drive, both the pulleys rotate in the
same direction as shown in Fig. 11.11.
Let
r1 and r2 = Radii of the larger and smaller pulleys,
x = Distance between the centres of two pulleys (i.e. O1 O2), and
L = Total length of the belt.
Let the belt leaves the larger pulley at E and G and the smaller pulley at F and H as shown in
Fig. 11.11. Through O2, draw O2 M parallel to FE.
From the geometry of the figure, we find that O2 M will be perpendicular to O1 E.
Let the angle MO2 O1 = α radians.
Chapter 11 : Belt, Rope and Chain Drives
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335
We know that the length of the belt,
L = Arc GJE + EF + Arc FKH + HG
= 2 (Arc JE + EF + Arc FK)
From the geometry of the figure, we find that
...(i)
O1 M
O E – EM
r–r
= 1
= 1 2
O1 O2
O1 O2
x
Since α is very small, therefore putting
r – r
sin α = α (in radians) = 1 2
x
π
∴
Arc JE = r1 + α
2
π
Similarly
Arc FK = r2 – α
2
sin α =
and
EF = MO2 =
(O1 O2 ) 2 – (O1 M ) 2 =
r – r
= x 1– 1 2
x
...(ii)
...(iii)
...(iv)
x 2 – ( r1 – r2 ) 2
2
Expanding this equation by binomial theorem,
2
(r – r ) 2
1r – r
EF = x 1 – 1 2 + .... = x – 1 2
...(v)
2 x
2x
Substituting the values of arc JE from equation (iii), arc FK from equation (iv) and EF from
equation (v) in equation (i), we get
π
(r – r )2
π
L = 2 r1 + α + x – 1 2 + r2 – α
2x
2
2
(r – r )2
π
π
= 2 r1 × + r1 . α + x – 1 2 + r2 × – r2 . α
2
2x
2
π
(r – r )2
= 2 (r1 + r2 ) + α ( r1 – r2 ) + x – 1 2
2 x
2
(r – r )2
= π (r1 + r2 ) + 2 α (r1 – r2 ) + 2 x – 1 2
x
Substituting the value of α =
r1 – r2
from equation (ii),
x
(r1 – r2 )
(r – r ) 2
× (r1 – r2 ) + 2 x – 1 2
x
x
2(r1 – r2 )2
(r1 – r2 )2
= π (r1 + r2 ) +
+ 2x –
x
x
(r1 – r2 )2
= π (r1 + r2 ) + 2 x +
...(In terms of pulley radii)
x
(d – d 2 )2
π
= (d1 + d 2 ) + 2 x + 1
...(In terms of pulley diameters)
2
4x
L = π (r1 + r2 ) + 2 ×
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Theory of Machines
11.12. Length of a Cross Belt Drive
We have already discussed in Art. 11.6 that in a cross belt drive, both the pulleys rotate in
opposite directions as shown in Fig. 11.12.
Fig. 11.12. Length of a cross belt drive.
Let
r1 and r2 = Radii of the larger and smaller pulleys,
x = Distance between the centres of two pulleys (i.e. O1 O2), and
L = Total length of the belt.
Let the belt leaves the larger pulley at E and G and the smaller pulley at F and H, as shown in
Fig. 11.12. Through O2, draw O2M parallel to FE.
From the geometry of the figure, we find that O2M will be perpendicular to O1E.
Let the angle MO2 O1 = α radians.
We know that the length of the belt,
L = Arc GJE + EF + Arc FKH + HG
= 2 (Arc JE + EF + Arc FK)
...(i)
From the geometry of the figure, we find that
sin α =
O1 M
O E + EM
r+r
= 1
= 1 2
O1 O2
O1 O2
x
Since α is very small, therefore putting
sin α = α (in radians) =
and
r1 + r2
x
...(ii)
∴
π
Arc JE = r1 + α
2
...(iii)
Similarly
π
Arc FK = r2 + α
2
...(iv)
EF = MO2 = (O1O2 )2 – (O1 M )2 =
r + r
= x 1– 1 2
x
2
x2 – (r1 + r2 )2
Chapter 11 : Belt, Rope and Chain Drives
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337
Expanding this equation by binomial theorem,
2
(r + r )2
1r + r
EF = x 1 – 1 2 + ... = x – 1 2
2 x
2x
...(v)
Substituting the values of arc JE from equation (iii), arc FK from equation (iv) and EF from
equation (v) in equation (i), we get
π
(r + r )2
π
L = 2 r1 + α + x – 1 2 + r2 + α
2x
2
2
(r + r ) 2
π
π
= 2 r1 × + r1 . α + x – 1 2 + r2 × + r2 . α
2
2x
2
π
(r + r )2
= 2 (r1 + r2 ) + α ( r1 + r2 ) + x – 1 2
2 x
2
= π (r1 + r2 ) + 2α (r1 + r2 ) + 2 x –
Substituting the value of α =
(r1 + r2 )2
x
r1 + r2
from equation (ii),
x
L = π ( r1 + r2 ) +
= π (r1 + r2 ) +
2 ( r1 + r2 )
(r + r )2
× ( r1 + r2 ) + 2 x – 1 2
x
x
2(r1 + r2 )2
(r + r )2
+ 2x – 1 2
x
x
= π (r1 + r2 ) + 2 x +
(r1 + r2 )2
x
...(In terms of pulley radii)
(d + d 2 )2
π
(d1 + d 2 ) + 2 x + 1
...(In terms of pulley diameters)
2
4x
It may be noted that the above expression is a function of
(r1 + r2). It is thus obvious that if sum of the radii of the two pulleys
be constant, then length of the belt required will also remain constant, provided the distance between centres of the pulleys remain
unchanged.
=
Example 11.3. A shaft which rotates at a constant speed of
160 r.p.m. is connected by belting to a parallel shaft 720 mm apart,
which has to run at 60, 80 and 100 r.p.m. The smallest pulley on the
driving shaft is 40 mm in radius. Determine the remaining radii of
the two stepped pulleys for 1. a crossed belt, and 2. an open belt.
Neglect belt thickness and slip.
Solution. Given : N 1 = N 3 = N 5 = 160 r.p.m. ; x = 720 mm ;
N 2 = 60 r.p.m.; N 4 = 80 r.p.m.; N6 = 100 r.p.m. ; r1 = 40 mm
Fig. 11.13.
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Theory of Machines
Let r2, r3, r4, r5 and r6 be the radii of the pulleys 2, 3, 4, 5, and 6 respectively, as shown in Fig.
11.13.
1. For a crossed belt
We know that for pulleys 1 and 2,
N2
r
= 1
N1 r2
r2 = r1 ×
or
N1
160
= 40 ×
= 106.7 mm Ans.
60
N2
and for pulleys 3 and 4,
N 4 r3
N
160
=
or r4 = r3 × 3 = r3 ×
= 2 r3
N3 r4
80
N4
We know that for a crossed belt drive,
r1 + r2 = r3 + r4 = r5 + r6 = 40 + 106.7 = 146.7 mm
∴
r3 + 2 r3 = 146.7 or
and
...(i)
r3 = 146.7/3 = 48.9 mm Ans.
r4 = 2 r3 = 2 × 48.9 = 97.8 mm Ans.
Now for pulleys 5 and 6,
N
160
N6 r5
or r6 = r5 × 5 = r5 ×
= 1.6 r5
=
100
N6
N5 r6
From equation (i),
r5 + 1.6 r5 = 146.7 or r5 = 146.7/2.6 = 56.4 mm Ans.
and
r6 = 1.6 r5 = 1.6 × 56.4 = 90.2 mm Ans.
2. For an open belt
We know that for pulleys 1 and 2,
N
160
N2
r
= 106.7 mm Ans.
= 1 or r2 = r1 × 1 = 40 ×
60
N2
N1 r2
and for pulleys 3 and 4,
N
160
N 4 r3
= 2 r3
or r4 = r3 × 3 = r3 ×
=
80
N4
N3 r4
We know that length of belt for an open belt drive,
L = π (r1 + r2 ) +
(r2 – r1 )2
+ 2x
x
(106.7 – 40)2
+ 2 × 720 = 1907 mm
720
Since the length of the belt in an open belt drive is constant, therefore for pulleys 3 and 4,
length of the belt (L),
= π (40 + 106.7) +
1907 = π (r3 + r4 ) +
(r4 – r3 ) 2
+ 2x
x
Chapter 11 : Belt, Rope and Chain Drives
l
339
(2 r3 – r3 )2
+ 2 × 720
720
= 9.426 r3 + 0.0014 (r3)2 + 1440
0.0014 (r3)2 + 9.426 r3 – 467 = 0
= π (r3 + 2 r3 ) +
or
r3 =
∴
– 9.426 ± (9.426)2 + 4 × 0.0014 × 467
2 × 0.0014
– 9.426 ± 9.564
= 49.3 mm Ans.
0.0028
r4 = 2 r3 = 2 × 49.3 = 98.6 mm Ans.
=
and
...(Taking +ve sign)
Now for pulleys 5 and 6,
N6 r5
=
or
N5 r6
r6 =
N5
160
× r5 =
× r5 = 1.6 r5
100
N6
and length of the belt (L),
1907 = π (r5 + r6 ) +
(r6 – r5 )2
+ 2x
x
(1.6 r5 – r5 )2
+ 2 × 720
720
= 8.17 r5 + 0.0005 (r5)2 + 1440
= π (r5 + 1.6 r5 ) +
0.0005 (r5)2 + 8.17 r5 – 467 = 0
or
r5 =
∴
=
– 8.17 ± (8.17)2 + 4 × 0.0005 × 467
2 × 0.0005
– 8.17 ± 8.23
= 60 mm Ans.
0.001
...(Taking +ve sign)
and
Milling machine is used for dressing
surfaces by rotary cutters.
Note : This picture is given as additional
information and is not a direct example of the
current chapter.
r6 = 1.6 r5 = 1.6 × 60 = 96 mm
Ans.
11.13. Power Transmitted by a Belt
Fig. 11.14 shows the driving pulley (or driver) A and the driven pulley (or follower) B. We
have already discussed that the driving pulley pulls the belt from one side and delivers the same to the
other side. It is thus obvious that the tension on the former side (i.e. tight side) will be greater than the
latter side (i.e. slack side) as shown in Fig. 11.14.
Let
T 1 and T 2 = Tensions in the tight and slack side of the belt respectively in
newtons,
340
l
Theory of Machines
r1 and r2 = Radii of the driver and follower respectively, and
v = Velocity of the belt in m/s.
Fig. 11.14. Power transmitted by a belt.
The effective turning (driving) force at the circumference of the follower is the difference
between the two tensions (i.e. T 1 – T 2).
∴ Work done per second = (T 1 – T 2) v N-m/s
and power transmitted,
P = (T 1 – T 2) v W
...(∵ 1 N-m/s = 1 W)
A little consideration will show that the torque exerted on the driving pulley is (T 1 – T 2) r1.
Similarly, the torque exerted on the driven pulley i.e. follower is (T 1 – T 2) r2.
11.14. Ratio of Driving Tensions For Flat Belt Drive
Consider a driven pulley rotating in the clockwise direction as shown in Fig. 11.15.
Fig. 11.15. Ratio of driving tensions for flat belt.
Let
T1 = Tension in the belt on the tight side,
T2 = Tension in the belt on the slack side, and
θ = Angle of contact in radians (i.e. angle subtended by the arc A B, along
which the belt touches the pulley at the centre).
Now consider a small portion of the belt PQ, subtending an angle δθ at the centre of the
pulley as shown in Fig. 11.15. The belt PQ is in equilibrium under the following forces :
1. Tension T in the belt at P,
2. Tension (T + δ T) in the belt at Q,
3. Normal reaction R N, and
4. Frictional force, F = µ × R N , where µ is the coefficient of friction between the belt and
pulley.
Chapter 11 : Belt, Rope and Chain Drives
l
341
Resolving all the forces horizontally and equating the same,
δθ
δθ
+ T sin
2
2
Since the angle δθ is very small, therefore putting sin δ θ / 2 = δθ / 2 in equation (i),
RN = (T + δ T ) sin
RN = (T + δT )
δθ
δθ T . δθ δT . δθ T . δθ
+T ×
=
+
+
= T . δθ
2
2
2
2
2
...(i)
...(ii)
δT . δ θ
... Neglecting
2
Now resolving the forces vertically, we have
δθ
δθ
– T cos
2
2
Since the angle δ θ is very small, therefore putting cos δ θ / 2 = 1 in equation (iii),
µ × RN = (T + δT ) cos
µ × R N = T + δT – T = δT or RN =
δT
µ
...(iii)
...(iv)
Equating the values of R N from equations (ii) and (iv),
δT
δT
or
= µ .δ θ
µ
T
Integrating both sides between the limits T 2 and T 1 and from 0 to θ respectively,
T .δ θ =
θ
T
δT
or
...(v)
loge 1 = µ . θ or T1 = eµ . θ
=
µ
∫T
∫δθ
T
2
T
0
T2
2
Equation (v) can be expressed in terms of corresponding logarithm to the base 10, i.e.
T1
i.e.
T
2.3log 1 = µ . θ
T2
The above expression gives the relation between the tight side and slack side tensions, in terms
of coefficient of friction and the angle of contact.
11.15. Determination of Angle of Contact
When the two pulleys of different diameters are connected by means of an open belt as
shown in Fig. 11.16 (a), then the angle of contact or lap (θ) at the smaller pulley must be taken into
consideration.
Let
r1 = Radius of larger pulley,
r2 = Radius of smaller pulley, and
x = Distance between centres of two pulleys (i.e. O1 O2).
From Fig. 11.16 (a),
sin α =
∴ Angle of contact or lap,
O1 M
O E – ME r1 – r2
= 1
=
O1 O2
O1 O2
x
θ = (180° – 2 α)
π
rad
180
...(∵ ME = O2 F = r2)
342
l
Theory of Machines
A little consideration will show that when the two pulleys are connected by means of a crossed
belt as shown in Fig. 11.16 (b), then the angle of contact or lap (θ) on both the pulleys is same. From
Fig. 11.16 (b),
O M
O E + ME r1 + r2
sin α = 1
= 1
=
O1 O2
O1 O2
x
∴ Angle of contact or lap,
θ = (180° + 2 α )
π
rad
180
(a) Open belt drive.
(b) Crossed belt drive.
Fig. 11.16
Example 11.4. Find the power transmitted by a belt running over a pulley of 600 mm
diameter at 200 r.p.m. The coefficient of friction between the belt and the pulley is 0.25, angle of lap
160° and maximum tension in the belt is 2500 N.
Solution. Given : d = 600 mm = 0.6 m ; N = 200 r.p.m. ; µ = 0.25 ; θ = 160° = 160 × π / 180
= 2.793 rad ; T 1 = 2500 N
We know that velocity of the belt,
π d . N π × 0.6 × 200
=
= 6.284 m/s
60
60
T2 = Tension in the slack side of the belt.
v=
Let
We know that
T
2.3log 1 = µ . θ = 0.25 × 2.793 = 0.6982
T2
Chapter 11 : Belt, Rope and Chain Drives
l
343
T 0.6982
log 1 =
= 0.3036
2.3
T2
T1
= 2.01
T2
T
2500
T2 = 1 =
= 1244 N
2.01 2.01
We know that power transmitted by the belt,
∴
and
...(Taking antilog of 0.3036)
P = (T 1 – T 2) v = (2500 – 1244) 6.284 = 7890 W
= 7.89 kW Ans.
Another model of milling machine.
Note : This picture is given as additional information and is not a direct example of the current chapter.
Example 11.5. A casting weighing 9 kN hangs freely from a rope which makes 2.5 turns
round a drum of 300 mm diameter revolving at 20 r.p.m. The other end of the rope is pulled by a man.
The coefficient of friction is 0.25. Determine 1. The force required by the man, and 2. The power to
raise the casting.
Solution. Given : W = T 1 = 9 kN = 9000 N ; d = 300 mm = 0.3 m ; N = 20 r.p.m. ; µ = 0.25
1. Force required by the man
Let
T2 = Force required by the man.
Since the rope makes 2.5 turns round the drum, therefore angle of contact,
θ = 2.5 × 2 π = 5 π rad
344
l
Theory of Machines
We know that
T
2.3log 1 = µ . θ = 0.25 × 5 π = 3.9275
T2
T 3.9275
T
log 1 =
= 1.71 or 1 = 51
2.3
T2
T2
...(Taking antilog of 1.71)
∴
T2 =
T1 9000
= 176.47 N Ans.
=
51
51
2. Power to raise the casting
We know that velocity of the rope,
v=
π d . N π × 0.3 × 20
=
= 0.3142 m/s
60
60
∴ Power to raise the casting,
P = (T 1 – T 2) v = (9000 – 176.47) 0.3142 = 2772 W
= 2.772 kW Ans.
Example 11.6. Two pulleys, one 450 mm diameter and the other 200 mm diameter are on
parallel shafts 1.95 m apart and are connected by a crossed belt. Find the length of the belt required
and the angle of contact between the belt and each pulley.
What power can be transmitted by the belt when the larger pulley rotates at 200 rev/min, if
the maximum permissible tension in the belt is 1 kN, and the coefficient of friction between the belt
and pulley is 0.25 ?
Solution. Given : d1 = 450 mm = 0.45 m or r1 = 0.225 m ; d2 = 200 mm = 0.2 m or
r2 = 0.1 m ; x = 1.95 m ; N 1 = 200 r.p.m. ; T 1 = 1 kN = 1000 N ; µ = 0.25
We know that speed of the belt,
π d1. N1 π × 0.45 × 200
v=
=
= 4.714 m/s
60
60
Length of the belt
We know that length of the crossed belt,
L = π (r1 + r2 ) + 2 x +
(r1 + r2 )2
x
= π (0.225 + 0.1) + 2 × 1.95 +
(0.225 + 0.1)2
= 4.975m Ans.
1.95
Angle of contact between the belt and each pulley
Let
θ = Angle of contact between the belt and each pulley.
We know that for a crossed belt drive,
r1 + r2 0.225 + 0.1
=
= 0.1667 or α = 9.6°
1.95
x
θ = 180° + 2 α = 180° + 2 × 9.6° = 199.2°
sin α =
∴
= 199.2 ×
π
= 3.477 rad Ans.
180
Chapter 11 : Belt, Rope and Chain Drives
l
345
Power transmitted
Let
T2 = Tension in the slack side of the belt.
We know that
T
2.3 log 1 = µ . θ = 0.25 × 3.477 = 0.8692
T2
T 0.8692
log 1 =
= 0.378 or T1 = 2.387
2.3
T
T2
2
...(Taking antilog of 0.378)
T1
1000
=
= 419 N
2.387 2.387
We know that power transmitted,
T2 =
∴
P = (T 1 – T 2) v = (1000 – 419) 4.714 = 2740 W = 2.74 kW Ans.
11.16. Centrifugal Tension
Since the belt continuously runs over the pulleys, therefore, some centrifugal force is caused, whose effect is to increase
the tension on both, tight as well as the slack sides. The tension
caused by centrifugal force is called centrifugal tension. At lower
belt speeds (less than 10 m/s), the centrifugal tension is very
small, but at higher belt speeds (more than 10 m/s), its effect is
considerable and thus should be taken into
account.
Consider a small portion PQ of the belt
subtending an angle dθ the centre of the pulley as shown in Fig.
11.17.
Let
m = Mass of the belt per unit length in kg,
Fig. 11.17. Centrifugal tension.
v = Linear velocity of the belt in m/s,
r = Radius of the pulley over which the belt runs in metres, and
TC = Centrifugal tension acting tangentially at P and Q in newtons.
We know that length of the belt PQ
= r. dθ
and mass of the belt PQ
= m. r. dθ
∴ Centrifugal force acting on the belt PQ,
v2
= m. d θ. v 2
r
The centrifugal tension T C acting tangentially at P and Q keeps the belt in equilibrium.
FC = (m . r . d θ)
Now resolving the forces (i.e. centrifugal force and centrifugal tension) horizontally and
equating the same, we have
dθ
dθ
2
TC sin
+ TC sin
= FC = m . d θ . v
2
2
dθ dθ
Since the angle dθ is very small, therefore, putting sin =
, in the above expression,
2
2
346
l
Theory of Machines
dθ
2 TC = m . d θ . v 2 or T = m . v2
C
2
Notes : 1. When the centrifugal tension is taken into account, then total tension in the tight side,
T t1 = T 1 + T C
and total tension in the slack side,
T t2 = T 2 + T C
2. Power transmitted,
P = (T t1 – T t2) v
= [(T 1 + T C) – (T 2 + T C)] v = (T 1 – T 2) v
...(in watts)
...(same as before)
Thus we see that centrifugal tension has no effect on the power transmitted.
3. The ratio of driving tensions may also be written as
T –T
2.3log t1 C = µ . θ
Tt 2 – TC
T t1 = Maximum or total tension in the belt.
where
11.17. Maximum Tension in the Belt
A little consideration will show that the maximum tension in the belt (T) is equal to the total
tension in the tight side of the belt (T t1).
Let
σ = Maximum safe stress in N/mm2,
b = Width of the belt in mm, and
t = Thickness of the belt in mm.
We know that maximum tension in the belt,
T = Maximum stress × cross-sectional area of belt = σ. b. t
When centrifugal tension is neglected, then
T (or T t1) = T 1, i.e. Tension in the tight side of the belt
and when centrifugal tension is considered, then
T (or T t1) = T 1 + T C
11.18. Condition For the Transmission of Maximum Power
We know that power transmitted by a belt,
P = (T 1 – T 2) v
where
...(i)
T1 = Tension in the tight side of the belt in newtons,
T2 = Tension in the slack side of the belt in newtons, and
v = Velocity of the belt in m/s.
From Art. 11.14, we have also seen that the ratio of driving tensions is
T1
T
= eµ . θ or T2 = 1
T2
eµ .θ
Substituting the value of T 2 in equation (i),
T
1
P = T1 – µ1. θ v = T1 1 – µ .θ v = T1 . v . C
e
e
...(ii)
...(iii)
Chapter 11 : Belt, Rope and Chain Drives
C =1–
where
We know that
where
l
347
1
µ. θ
e
T1 = T – T C
T = Maximum tension to which the belt can be subjected in
newtons, and
TC = Centrifugal tension in newtons.
Substituting the value of T 1 in equation (iii),
P = (T – T C) v.C
= (T – m.v 2) v.C = (T.v – m v 3) C
... (Substituting T C = m. v2)
For maximum power, differentiate the above expression with respect to v and equate to zero,
i.e.
∴
or
dP
d
or
(T .v – mv3 ) C = 0
=0
dv
dv
T – 3 m . v2 = 0
T – 3 T C = 0 or T = 3 T C
...(iv)
It shows that when the power transmitted is maximum, 1/3rd of the maximum tension is
absorbed as centrifugal tension.
Notes : 1. We know that T 1 = T– TC and for maximum power, TC =
T
.
3
T 2T
=
3
3
2. From equation (iv), the velocity of the belt for the maximum power,
∴
T1 = T –
v=
T
3m
Example. 11.7. A shaft rotating at 200 r.p.m. drives another shaft at 300 r.p.m. and transmits
6 kW through a belt. The belt is 100 mm wide and 10 mm thick. The distance between the shafts is 4m.
The smaller pulley is 0.5 m in diameter. Calculate the stress in the belt, if it is 1. an open belt drive,
and 2. a cross belt drive. Take µ = 0.3.
Solution. Given : N 1 = 200 r.p.m. ; N 2 = 300 r.p.m. ; P = 6 kW = 6 × 103 W ; b = 100 mm ;
t = 10 mm ; x = 4 m ; d2 = 0.5 m ; µ = 0.3
Let
σ = Stress in the belt.
1. Stress in the belt for an open belt drive
First of all, let us find out the diameter of larger pulley (d1). We know that
N 2 . d 2 300 × 0.5
N2
d
=
= 0.75m
= 1 or d1 =
200
N1
N1 d 2
π d2 . N 2
π × 0.5 × 300
v=
=
= 7.855 m/s
and velocity of the belt,
60
60
Now let us find the angle of contact on the smaller pulley. We know that, for an open belt
drive,
sin α =
r1 – r2 d1 – d2 0.75 – 0.5
=
=
= 0.03125 or α = 1.8°
2x
2×4
x
348
l
Theory of Machines
∴ Angle of contact,
Let
θ =
=
T1 =
T2 =
180° – 2 α = 180 – 2 × 1.8 = 176.4°
176.4 × π / 180 = 3.08 rad
Tension in the tight side of the belt, and
Tension in the slack side of the belt.
We know that
T
2.3log 1 = µ . θ = 0.3 × 3.08 = 0.924
T2
∴
T 0.924
T
log 1 =
= 0.4017 or 1 = 2.52
2.3
T2
T2
...(i)
...(Taking antilog of 0.4017)
We also know that power transmitted (P),
6 × 103 = (T 1 – T 2) v = (T 1 – T 2) 7.855
∴
T 1 – T 2 = 6 × 103 / 7.855 = 764 N
From equations (i) and (ii),
T1 = 1267 N, and T 2 = 503 N
We know that maximum tension in the belt (T 1),
...(ii)
1267 = σ . b. t = σ × 100 × 10 = 1000 σ
∴
σ = 1267 / 1000 = 1.267 N/mm2 = 1.267 MPa Ans.
...[∵ 1 MPa = 1 MN/m2 = 1 N/mm2]
Stress in the belt for a cross belt drive
We know that for a cross belt drive,
r+r
d + d 2 0.75 + 0.5
sin α = 1 2 = 1
=
= 0.1562 or α = 9°
2x
2×4
x
∴ Angle of contact,
θ = 180° + 2α = 180 + 2 × 9 = 198°
= 198 × π / 180 = 3.456 rad
We know that
T
2.3log 1 = µ . θ = 0.3 × 3.456 = 1.0368
T2
T 1.0368
T
log 1 =
= 0.4508 or 1 = 2.82
2.3
T
T2
2
...(iii)
...(Taking antilog of 0.4508)
From equations (ii) and (iii),
T1 = 1184 N and T 2 = 420 N
We know that maximum tension in the belt (T 1),
1184 = σ. b. t = σ × 100 × 10 = 1000 σ
∴
σ = 1184 / 1000 = 1.184 N/mm2 = 1.184 MPa Ans.
Example 11.8. A leather belt is required to transmit 7.5 kW from a pulley 1.2 m in diameter,
running at 250 r.p.m. The angle embraced is 165° and the coefficient of friction between the belt and
the pulley is 0.3. If the safe working stress for the leather belt is 1.5 MPa, density of leather 1 Mg/m3
and thickness of belt 10 mm, determine the width of the belt taking centrifugal tension into account.
Chapter 11 : Belt, Rope and Chain Drives
l
349
Solution. Given : P = 7.5 kW = 7500 W ; d = 1.2 m ; N = 250 r.p.m. ; θ = 165° = 165 × π / 180
= 2.88 rad ; µ = 0.3 ; σ = 1.5 MPa = 1.5 × 106 * N/m2 ; ρ = 1 Mg/m3 = 1 × 106 g/m3 = 1000 kg/m3;
t = 10 mm = 0.01 m
Let
b = Width of belt in metres,
T1 = Tension in the tight side of the belt in N, and
T2 = Tension in the slack side of the belt in N.
We know that velocity of the belt,
v = π d . N / 60 = π × 1.2 × 250/60 = 15.71 m/s
and power transmitted (P),
7500 = (T 1 – T 2) v = (T 1 – T 2) 15.71
∴
T 1 – T 2 = 7500 / 15.71 = 477.4 N
...(i)
We know that
T
2.3log 1 = µ . θ = 0.3 × 2.88 = 0.864
T2
T 0.864
T
log 1 =
= 0.3756 or 1 = 2.375
2.3
T2
T2
...(ii)
...(Taking antilog of 0.3756)
From equations (i) and (ii),
T1 = 824.6 N, and T 2 = 347.2 N
We know that mass of the belt per metre length,
m = Area × length × density = b.t.l.ρ
= b × 0.01 × 1 × 1000 = 10 b kg
∴ Centrifugal tension,
TC = m. v 2 = 10 b (15.71)2 = 2468 b N
and maximum tension in the belt,
T = σ. b. t = 1.5 × 106 × b × 0.01 = 15 000 b N
We know that
T = T 1 + T C or 15000 b = 824.6 + 2468 b
15 000 b – 2468 b = 824.6 or 12 532 b = 824.6
∴
b = 824.6 / 12532 = 0.0658 m = 65.8 mm Ans.
Example. 11.9. Determine the width of a 9.75 mm thick leather belt required to transmit
15 kW from a motor running at 900 r.p.m. The diameter of the driving pulley of the motor is 300 mm.
The driven pulley runs at 300 r.p.m. and the distance between the centre of two pulleys is 3 metres.
The density of the leather is 1000 kg/m3. The maximum allowable stress in the leather is 2.5 MPa.
The coefficient of friction between the leather and pulley is 0.3. Assume open belt drive and neglect
the sag and slip of the belt.
Solution. Given : t = 9.75 mm = 9.75 × 10–3 m ; P = 15 kW = 15 × 103 W ; N1 = 900 r.p.m. ;
d1 = 300 mm = 0.3 m ; N 2 = 300 r.p.m. ; x = 3m ; ρ = 1000 kg/m3 ; σ = 2.5 MPa = 2.5 × 106 N/m2 ;
µ = 0.3
*
1 MPa = 1 × 106 N/m2
350
l
Theory of Machines
First of all, let us find out the diameter of the driven pulley (d2). We know that
900 × 0.3
N ×d
N2
d
= 1 or d2 = 1 1 =
= 0.9 m
300
N2
N1 d 2
v=
and velocity of the belt,
π d1 N1 π × 0.3 × 900
=
= 14.14 m/s
60
60
For an open belt drive,
sin α =
r2 – r1 d2 – d1 0.9 – 0.3
=
=
= 0.1
2x
2×3
x
...(∵ d2 > d1)
α = 5.74°
or
∴
Angle of lap, θ = 180° – 2 α = 180 – 2 × 5.74 = 168.52°
= 168.52 × π / 180 = 2.94 rad
Let
T1 = Tension in the tight side of the belt, and
T2 = Tension in the slack side of the belt.
We know that
T
2.3 log 1 = µ . θ = 0.3 × 2.94 = 0.882
T2
T 0.882
T
log 1 =
= 0.3835 or 1 = 2.42
2.3
T2
T2
...(i)
... (Taking antilog of 0.3835)
We also know that power transmitted (P),
15 × 103 = (T 1 – T 2) v = (T 1 – T 2) 14.14
∴
T 1 – T 2 = 15 × 103 / 14.14 = 1060 N
.... (ii)
From equations (i) and (ii),
T1 = 1806 N
Let
b = Width of the belt in metres.
We know that mass of the belt per metre length,
m = Area × length × density = b.t.l.ρ
= b × 9.75 × 10–3 × 1 × 1000 = 9.75 b kg
∴ Centrifugal tension,
TC = m.v2 = 9.75 b (14.14)2 = 1950 b N
Maximum tension in the belt,
T = σ. b. t = 2.5 × 106 × b × 9.75 × 10–3 = 24 400 b N
We know that
T = T 1 + T C or T – T C = T1
24 400 b – 1950 b = 1806 or 22 450 b = 1806
∴
b = 1806 / 22 450 = 0.080 m = 80 mm Ans.
Example. 11.10. A pulley is driven by a flat belt, the angle of lap being 120°. The belt is 100
mm wide by 6 mm thick and density1000 kg/m3. If the coefficient of friction is 0.3 and the maximum
stress in the belt is not to exceed 2 MPa, find the greatest power which the belt can transmit and the
corresponding speed of the belt.
Chapter 11 : Belt, Rope and Chain Drives
l
351
Solution. Given : θ = 120° = 120 × π / 180 = 2.1 rad ; b = 100 mm = 0.1 m ; t = 6 mm
= 0.006 m ; ρ = 1000 kg / m3 ; µ = 0.3 ; σ = 2 MPa = 2 × 106 N/m2
Speed of the belt for greatest power
We know that maximum tension in the belt,
T = σ. b. t = 2 × 106 × 0.1 × 0.006 = 1200 N
and mass of the belt per metre length,
m = Area × length × density = b. t. l. ρ
= 0.1 × 0.006 × 1 × 1000 = 0.6 kg/m
∴ Speed of the belt for greatest power,
T
1200
=
= 25.82 m/s
Ans.
3m
3 × 0.6
Greatest power which the belt can transmit
We know that for maximum power to be transmitted, centrifugal tension,
TC = T/3 = 1200/3 = 400 N
and tension in the tight side of the belt,
T1 = T – T C = 1200 – 400 = 800 N
Let
T2 =Tension in the slack side of the belt.
We know that
v=
T
2.3log 1 = µ . θ = 0.3 × 2.1 = 0.63
T2
and
T 0.63
T
log 1 =
= 0.2739 or 1 = 1.88
2.3
T
T
2
2
T1
800
T2 =
=
= 425.5 N
1.88 1.88
∴ Greatest power which the belt can transmit,
...(Taking antilog of 0.2739)
P = (T 1 – T 2) v = (800 – 425.5) 25.82 = 9670 W = 9.67 kW Ans.
Example 11.11. An open belt drive connects two pulleys 1.2 m and 0.5 m diameter, on
parallel shafts 4 metres apart. The mass of the belt is 0.9 kg per metre length and the maximum
tension is not to exceed 2000 N.The coefficient of friction is 0.3. The 1.2 m pulley, which is the driver,
runs at 200 r.p.m. Due to belt slip on one of the pulleys, the velocity of the driven shaft is only 450
r.p.m. Calculate the torque on each of the two shafts, the power transmitted, and power lost in
friction. What is the efficiency of the drive ?
Solution. Given : d1 = 1.2 m or r1 = 0.6 m ; d2 = 0.5 m or r2 = 0.25 m ; x = 4 m ; m = 0.9 kg/m;
T = 2000 N ; µ = 0.3 ; N 1 = 200 r.p.m. ; N 2 = 450 r.p.m.
We know that velocity of the belt,
v=
and centrifugal tension,
π d1 . N1 π × 1.2 × 200
=
= 12.57 m/s
60
60
TC = m.v 2 = 0.9 (12.57)2 = 142 N
∴ Tension in the tight side of the belt,
T1 = T – T C = 2000 – 142 = 1858 N
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Theory of Machines
We know that for an open belt drive,
r1 – r2
0.6 – 0.25
=
= 0.0875 or α = 5.02°
4
x
∴ Angle of lap on the smaller pulley,
sin α =
θ = 180° – 2 α = 180° – 2 × 5.02° = 169.96°
= 169.96 × π / 180 = 2.967 rad
Let
T2 = Tension in the slack side of the belt.
We know that
T
2.3log 1 = µ . θ = 0.3 × 2.967 = 0.8901
T2
T 0.8901
T
log 1 =
= 0.387 or 1 = 2.438
2.3
T
T
2
2
...(Taking antilog of 0.387)
∴
T2 =
T1
1858
=
= 762 N
2.438 2.438
Torque on the shaft of larger pulley
We know that torque on the shaft of larger pulley,
TL = (T 1 – T 2) r1 = (1858 – 762) 0.6 = 657.6 N-m
Ans.
Torque on the shaft of smaller pulley
We know that torque on the shaft of smaller pulley,
TS = (T 1 – T 2) r2 = (1858 – 762) 0.25 = 274 N-m
Ans.
Power transmitted
We know that the power transmitted,
P = (T 1 – T 2) v = (1858 – 762) 12.57 = 13780 W
= 13.78 kW Ans.
Power lost in friction
We know that input power,
P1 =
TL × 2 π N1 657.6 × 2 π × 200
=
= 13 780 W = 13.78 kW
60
60
TS × 2 π N 2
274 × 2 π × 450
=
= 12 910 W = 12.91kW
60
60
∴ Power lost in friction = P1 – P2 = 13.78 – 12.91 = 0.87 kW Ans.
and output power,
P2 =
Efficiency of the drive
We know that efficiency of the drive,
η=
Output power 12.91
=
= 0.937 or 93.7 % Ans.
Input power
13.78
Chapter 11 : Belt, Rope and Chain Drives
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353
11.19. Initial Tension in the Belt
When a belt is wound round the two pulleys (i.e. driver and follower), its two ends are joined
together ; so that the belt may continuously move over the pulleys, since the motion of the belt from
the driver and the follower is governed by a firm grip, due to friction between the belt and the pulleys.
In order to increase this grip, the belt is tightened up. At this stage, even when the pulleys are stationary, the belt is subjected to some tension, called initial tension.
When the driver starts rotating, it pulls the belt from one side (increasing tension in the belt
on this side) and delivers it to the other side (decreasing the tension in the belt on that side). The
increased tension in one side of the belt is called tension in tight side and the decreased tension in the
other side of the belt is called tension in the slack side.
Let
T0 = Initial tension in the belt,
T1 = Tension in the tight side of the belt,
T2 = Tension in the slack side of the belt, and
α = Coefficient of increase of the belt length per unit force.
A little consideration will show that the increase of tension in the tight side
= T1 – T0
and increase in the length of the belt on the tight side
= α (T 1 – T 0)
...(i)
Similarly, decrease in tension in the slack side
= T0 – T2
and decrease in the length of the belt on the slack side
= α (T 0 – T 2)
...(ii)
Assuming that the belt material is perfectly elastic such that the length of the belt remains
constant, when it is at rest or in motion, therefore increase in length on the tight side is equal to
decrease in the length on the slack side. Thus, equating equations (i) and (ii),
α (T 1 – T 0) = α (T 0 – T 2) or T 1 – T 0 = T 0 – T 2
∴
T0 =
T1 + T2
2
...(Neglecting centrifugal tension)
T1 + T2 + 2 TC
...(Considering centrifugal tension)
2
Example. 11.12. In a flat belt drive the initial tension is 2000 N. The coefficient of friction
between the belt and the pulley is 0.3 and the angle of lap on the smaller pulley is 150°. The smaller
pulley has a radius of 200 mm and rotates at 500 r.p.m. Find the power in kW transmitted by the belt.
=
Solution. Given : T0 = 2000 N ; µ0 = 0.3 ; θ = 150° = 150° × π / 180 = 2.618 rad ; r2 = 200 mm
or d2 = 400 mm = 0.4 m ; N 2 = 500 r.p.m.
We know that velocity of the belt,
π d2 . N2
π × 0.4 × 500
=
= 10.47 m/s
60
60
T1 = Tension in the tight side of the belt, and
T2 = Tension in the slack side of the belt.
v=
Let
354
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Theory of Machines
We know that initial tension (T 0),
2000 =
We also know that
T1 + T2
2
or
T1 + T 2 = 4000 N
...(i)
T
2.3log 1 = µ . θ = 0.3 × 2.618 = 0.7854
T2
T 0.7854
log 1 =
= 0.3415
2.3
T2
or
T1
= 2.2 ...(ii)
T2
...(Taking antilog of 0.3415)
From equations (i) and (ii),
T1 = 2750 N ;
and
T2 = 1250 N
∴ Power transmitted, P = (T 1 – T 2) v
A military tank uses chain, belt and gear drives
for its movement and operation.
= (2750 – 1250) 10.47
= 15 700 W =15.7 kW Ans.
Example 11.13. Two parallel shafts whose centre lines are 4.8 m apart, are connected by
open belt drive. The diameter of the larger pulley is 1.5 m and that of smaller pulley 1 m. The initial
tension in the belt when stationary is 3 kN. The mass of the belt is 1.5 kg / m length. The coefficient
of friction between the belt and the pulley is 0.3. Taking centrifugal tension into account, calculate
the power transmitted, when the smaller pulley rotates at 400 r.p.m.
Solution. Given : x = 4.8 m ; d1 = 1.5 m ; d2 = 1 m ; T 0 = 3 kN = 3000 N ; m = 1.5 kg / m ;
µ = 0.3 ; N 2 = 400 r.p.m.
We know that velocity of the belt,
π d2 . N2
π × 1 × 400
v=
=
= 21m/s
60
60
and centrifugal tension,
TC = m.v2 = 1.5 (21)2 = 661.5 N
Let
T1 = Tension in the tight side, and
T2 = Tension in the slack side.
We know that initial tension (T 0),
T1 + T2 + 2 TC T1 + T2 + 2 × 661.5
=
2
2
∴
T 1 + T 2 = 3000 × 2 – 2 × 661.5 = 4677 N
For an open belt drive,
3000 =
r1 – r2 d1 – d2 1.5 – 1
=
=
= 0.0521 or α = 3°
2x
2 × 4.8
x
∴ Angle of lap on the smaller pulley,
sin α =
θ = 180° – 2 α = 180° – 2 × 3° = 174°
= 174° × π / 180 = 3.04 rad
...(i)
Chapter 11 : Belt, Rope and Chain Drives
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355
We know that
T
2.3log 1 = µ . θ = 0.3 × 3.04 = 0.912
T2
T 0.912
T
log 1 =
= 0.3965 or 1 = 2.5
2.3
T
T
2
2
...(ii)
...(Taking antilog of 0.3965)
From equations (i) and (ii),
T1 =3341 N ; and T 2 = 1336 N
∴
Power transmitted,
P = (T 1 – T 2) v = (3341 – 1336) 21 = 42 100 W = 42.1 kW Ans.
Example 11.14. An open flat belt drive connects two parallel shafts 1.2 metres apart. The
driving and the driven shafts rotate at 350 r.p.m. and 140 r.p.m. respectively and the driven pulley is
400 mm in diameter. The belt is 5 mm thick and 80 mm wide. The coefficient of friction between the
belt and pulley is 0.3 and the maximum permissible tension in the belting is 1.4 MN/m2. Determine:
1. diameter of the driving pulley, 2. maximum power that may be transmitted by the belting,
and 3. required initial belt tension.
Solution. Given : x = 1.2 m ; N 1 = 350 r.p.m. ; N 2 = 140 r.p.m. ; d2 = 400 mm = 0.4 m ;
t = 5 mm = 0.005 m ; b = 80 mm = 0.08 m ; µ = 0.3 ; σ = 1.4 MN/m2 = 1.4 × 106 N/m2
1. Diameter of the driving pulley
Let
We know that
d1 = Diameter of the driving pulley.
N .d
140 × 0.4
N2
d
= 0.16 m Ans.
= 1 or d1 = 2 2 =
350
N1
N1 d2
2. Maximum power transmitted by the belting
First of all, let us find the angle of contact of the belt on the smaller pulley (or driving
pulley).
Let
θ = Angle of contact of the belt on the driving pulley.
Fig. 11.18
356
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Theory of Machines
From Fig. 11.18, we find that
sin α =
or
α =
θ =
=
T1 =
T2 =
∴
Let
O2 M
r –r
d – d1 0.4 – 0.16
= 2 1 = 2
=
= 0.1
2x
2 × 1.2
O1 O2
x
5.74°
180° – 2 α = 180° – 2 × 5.74° = 168.52°
168.52 × π / 180 = 2.94 rad
Tension in the tight side of the belt, and
Tension in the slack side of the belt.
We know that
T
2.3log 1 = µ . θ = 0.3 × 2.94 = 0.882
T2
T 0.882
T
log 1 =
= 0.3835 or 1 = 2.42
2.3
T
T
2
2
...(i)
...(Taking antilog of 0.3835)
We know that maximum tension to which the belt can be subjected,
T1 = σ × b × t = 1.4 × 106 × 0.08 × 0.005 = 560 N
T2 =
∴
T1
560
=
= 231.4 N
2.42 2.42
...[From equation (i)]
π d1. N1 π × 0.16 × 350
=
= 2.93 m/s
60
60
P = (T 1 – T 2) v = (560 – 231.4) 2.93 = 963 W = 0.963 kW Ans.
v=
Velocity of the belt,
∴ Power transmitted,
3. Required initial belt tension
We know that the initial belt tension,
T1 + T2 560 + 231.4
=
= 395.7 N Ans.
2
2
Example 11.15. An open belt running over two pulleys 240 mm and 600 mm diameter connects
two parallel shafts 3 metres apart and transmits 4 kW from the smaller pulley that rotates at 300
r.p.m. Coefficient of friction between the belt and the pulley is 0.3 and the safe working tension is
10N per mm width. Determine : 1. minimum width of the belt, 2. initial belt tension, and 3. length of
the belt required.
T0 =
Solution. Given : d2 = 240 mm = 0.24 m ; d1 = 600 mm = 0.6 m ; x = 3 m ; P = 4 kW = 4000 W;
N 2 = 300 r.p.m. ; µ = 0.3 ; T 1 = 10 N/mm width
1. Minimum width of belt
We know that velocity of the belt,
π d2 . N2
π × 0.24 × 300
=
= 3.77 m/s
60
60
T1 = Tension in the tight side of the belt, and
v=
Let
T2 = Tension in the slack side of the belt.
∴ Power transmitted (P),
4000 = (T 1 – T 2) v = (T 1 – T 2) 3.77
or
T 1 – T 2 = 4000 / 3.77 = 1061 N
...(i)
Chapter 11 : Belt, Rope and Chain Drives
l
357
We know that for an open belt drive,
sin α =
r1 – r2 d1 – d2 0.6 – 0.24
=
=
= 0.06 or α = 3.44°
2x
2×3
x
and angle of lap on the smaller pulley,
θ = 180° – 2α = 180° – 2 × 3.44° = 173.12°
= 173.12 × π / 180 = 3.022 rad
We know that
T
2.3log 1 = µ . θ = 0.3 × 3.022 = 0.9066
T2
T 0.9066
T
log 1 =
= 0.3942 or 1 = 2.478
2.3
T2
T2
...(ii)
...(Taking antilog of 0.3942)
From equations (i) and (ii),
T1 = 1779 N, and T 2 = 718 N
Since the safe working tension is 10 N per mm width, therefore minimum width of the belt,
b=
T1 1779
=
= 177.9 mm Ans.
10
10
2. Initial belt tension
We know that initial belt tension,
T0 =
T1 + T2 1779 + 718
= 1248.5 N Ans.
=
2
2
3. Length of the belt required
We know that length of the belt required,
L=
=
(d – d 2 )2
π
(d1 – d 2 ) + 2 x + 1
2
4x
(0.6 – 0.24) 2
π
(0.6 + 0.24) + 2 × 3 +
2
4×3
= 1.32 + 6 + 0.01 = 7.33 m Ans.
Example 11.16. The following data refer to an open belt drive :
Diameter of larger pulley = 400 mm ; Diameter of smaller pulley = 250 mm ; Distance
between two pulleys = 2 m ; Coefficient of friction between smaller pulley surface and belt = 0.4 ;
Maximum tension when the belt is on the point of slipping = 1200 N.
Find the power transmitted at speed of 10 m/s. It is desired to increase the power. Which of
the following two methods you will select ?
1. Increasing the initial tension in the belt by 10 per cent.
2. Increasing the coefficient of friction between the smaller pulley surface and belt by 10 per
cent by the application of suitable dressing on the belt.
Find, also, the percentage increase in power possible in each case.
358
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Theory of Machines
Solution. Given : d1 = 400 mm = 0.4 m ; d2 = 250 mm = 0.25 m ; x = 2 m ; µ = 0.4 ;
T = 1200 N ; v = 10 m/s
Power transmitted
We know that for an open belt drive,
sin α =
r1 – r2 d1 – d 2 0.4 – 0.25
=
=
= 0.0375 or α = 2.15°
2x
2×2
x
∴ Angle of contact,
θ = 180° – 2α = 180° – 2 × 2.15° = 175.7°
= 175.7 × π / 180 = 3.067 rad
Let
T1 = Tension in the tight side of the belt, and
T2 = Tension in the slack side of the belt.
Neglecting centrifugal tension,
T1 = T = 1200 N
...(Given)
We know that
T
2.3log 1 = µ . θ = 0.4 × 3.067 = 1.2268
T2
T 1.2268
T
log 1 =
= 0.5334 or 1 = 3.41
2.3
T
T
2
2
and
...(Taking antilog of 0.5334)
T
1200
T2 = 1 =
= 352 N
3.41 3.41
We know that power transmitted,
P = (T 1 – T 2) v = (1200 – 352) 10 = 8480 W = 8.48 kW Ans.
Power transmitted when initial tension is increased by 10%
We know that initial tension,
T0 =
T1 + T2 1200 + 352
=
= 776 N
2
2
∴ Increased initial tension,
776 × 10
= 853.6 N
100
Let T 1 and T 2 be the corresponding tensions in the tight side and slack side of the belt
respectively.
T0′ = 776 +
∴
or
T1 + T2
2
T 1 + T 2 = 2 T'0 = 2 × 853.6 = 1707.2 N
T0′ =
...(i)
Since the ratio of tensions is constant, therefore
T1
= 3.41
T2
...(ii)
Chapter 11 : Belt, Rope and Chain Drives
l
359
From equations (i) and (ii),
T1 = 1320.2 N ; and T 2 = 387 N
∴ Power transmitted,
P = (T 1 – T 2) v = (1320.2 – 387) 10 = 9332 W = 9.332 kW
Power transmitted when coefficient of friction is increased by 10%
We know that coefficient of friction,
µ = 0.4
∴ Increased coefficient of friction,
10
= 0.44
100
Let T 1 and T 2 be the corresponding tensions in the tight side and slack side respectively.
µ ′ = 0.4 + 0.4 ×
We know that
T
2.3log 1 = µ′ . θ = 0.44 × 3.067 = 1.3495
T2
T 1.3495
T
log 1 =
= 0.5867 or 1 = 3.86
2.3
T
T
2
2
...(iii)
... (Taking antilog of 0.5867)
Here the initial tension is constant, i.e.
T0 =
T1 + T2
2
or T 1 + T 2 = 2 T 0 = 2 × 776 = 1552 N
...(iv)
From equations (iii) and (iv),
T = 1232.7 N and T 2 = 319.3 N
∴ Power transmitted,
P = (T 1 – T 2) v = (1232.7 – 319.3) 10 = 9134 W = 9.134 kW
Since the power transmitted by increasing the initial tension is more, therefore in order to
increase the power transmitted we shall adopt the method of increasing the initial tension. Ans.
Percentage increase in power
We know that percentage increase in power when the initial tension is increased
=
9.332 – 8.48
× 100 = 10.05% Ans.
8.48
and percentage increase in power when coefficient of friction is increased
=
9.134 – 8.48
× 100 = 7.7 % Ans.
8.48
11.20. V-belt drive
We have already discussed that a V-belt is mostly used in factories and workshops where a
great amount of power is to be transmitted from one pulley to another when the two pulleys are very
near to each other.
360
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Theory of Machines
The V-belts are made of fabric and cords moulded in rubber and covered with fabric and
rubber, as shown in Fig. 11.19 (a). These belts are moulded to a trapezoidal shape and are made
endless. These are particularly suitable for short drives i.e. when the shafts are at a short distance
apart. The included angle for the V-belt is usually from 30° – 40°. In case of flat belt drive, the belt
runs over the pulleys whereas in case of V-belt drive, the rim of the pulley is grooved in which the
V-belt runs. The effect of the groove is to increase the frictional grip of the V-belt on the pulley and
thus to reduce the tendency of slipping. In order to have a good grip on the pulley, the V-belt is in
contact with the side faces of the groove and not at the bottom. The power is transmitted by the
*wedging action between the belt and the V-groove in the pulley.
(a) Cross-section of a V-belt.
(b) Cross-section of a V-grooved pulley.
Fig. 11.19. V-belt and V-grooved pulley.
A clearance must be provided at the bottom of the groove, as shown in Fig. 11.19 (b), in order
to prevent touching to the bottom as it becomes narrower from wear. The V-belt drive, may be
inclined at any angle with tight side either at top or bottom. In order to increase the power output,
several V- belts may be operated side by side. It may be noted that in multiple V-belt drive, all the
belts should stretch at the same rate so that the load is equally divided between them. When one of the
set of belts break, the entire set should be replaced at the same time. If only one belt is replaced, the
new unworn and unstressed belt will be more tightly stretched and will move with different velocity.
11.21. Advantages and Disadvantages of V-belt Drive Over Flat Belt Drive
Following are the advantages and disadvantages of the V-belt drive over flat belt drive.
Advantages
1. The V-belt drive gives compactness due to the small distance between the centres of pulleys.
2. The drive is positive, because the slip between the belt and the pulley groove is negligible.
3. Since the V-belts are made endless and there is no joint trouble, therefore the drive is
smooth.
4. It provides longer life, 3 to 5 years.
*
The wedging action of the V-belt in the groove of the pulley results in higher forces of friction. A little
consideration will show that the wedging action and the transmitted torque will be more if the groove angle
of the pulley is small. But a smaller groove angle will require more force to pull the belt out of the groove
which will result in loss of power and excessive belt wear due to friction and heat. Hence a selective groove
angle is a compromise between the two. Usually the groove angles of 32° to 38° are used.
Chapter 11 : Belt, Rope and Chain Drives
l
361
5. It can be easily installed and removed.
6. The operation of the belt and pulley is quiet.
7. The belts have the ability to cushion the shock when machines are started.
8. The high velocity ratio (maximum 10) may be obtained.
9. The wedging action of the belt in the groove gives high value of limiting ratio of tensions.
Therefore the power transmitted by V-belts is more than flat belts for the same coefficient of friction,
arc of contact and allowable tension in the belts.
10. The V-belt may be operated in either direction with tight side of the belt at the top or
bottom. The centre line may be horizontal, vertical or inclined.
Disadvantages
1. The V-belt drive cannot be used with large centre distances.
2. The V-belts are not so durable as flat belts.
3. The construction of pulleys for V-belts is more complicated than pulleys for flat belts.
4. Since the V-belts are subjected to certain amount of creep, therefore these are not suitable
for constant speed application such as synchronous machines, and timing devices.
5. The belt life is greatly influenced with temperature changes, improper belt tension and
mismatching of belt lengths.
6. The centrifugal tension prevents the use of V-belts at speeds below 5 m/s and above 50m/s.
11.22. Ratio of Driving Tensions for V-belt
A V-belt with a grooved pulley is shown in Fig. 11.20.
Let
R1 = Normal reaction between the belt and
sides of the groove.
R = Total reaction in the plane of the groove.
2 β = Angle of the groove.
µ = Coefficient of friction between the belt
and sides of the groove.
Resolving the reactions vertically to the groove,
or
Fig. 11.20.
R = R 1 sin β + R 1 sin β = 2 R 1 sin β
R
R1 =
2sin β
We know that the frictional force
R
µ.R
= 2 µ . R1 = 2 µ ×
=
= µ . R cosec β
2sin β sin β
Consider a small portion of the belt, as in Art. 11.14, subtending an angle δθ at the centre.
The tension on one side will be T and on the other side T + δT. Now proceeding as in Art. 11.14, we
get the frictional resistance equal to µ. R cosec β instead of µ . R. Thus the relation between T 1 and T 2
for the V-belt drive will be
T
2.3 log 1 = µ . θ cosec β
T2
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Theory of Machines
Example 11.17. A belt drive consists of two V-belts in parallel, on grooved pulleys of the
same size. The angle of the groove is 30°. The cross-sectional area of each belt is 750 mm2 and
µ. = 0.12. The density of the belt material is 1.2 Mg/m3 and the maximum safe stress in the material
is 7 MPa. Calculate the power that can be transmitted between pulleys 300 mm diameter rotating at
1500 r.p.m. Find also the shaft speed in r.p.m. at which the power transmitted would be maximum.
Solution. Given : 2 β = 30° or β = 15° ; α = 750 mm2 = 750 × 10–6 m2 ; µ = 0.12 ; ρ = 1.2 Mg/m3
= 1200 kg/m3 ; σ = 7 MPa = 7 × 106 N/m2 ; d = 300 mm = 0.3 m ; N = 1500 r.p.m.
Power transmitted
We know that velocity of the belt,
π d . N π × 0.3 × 1500
=
= 23.56 m/s
60
60
and mass of the belt per metre length,
v=
m = Area × length × density = 750 × 10–6 × 1 × 1200 = 0.9 kg/m
∴ Centrifugal tension,
TC = m.v2 = 0.9 (23.56)2 = 500 N
We know that maximum tension in the belt,
T = Maximum stress × cross-sectional area of belt = σ × a
= 7 × 106 × 750 × 10–6 = 5250 N
∴ Tension in the tight side of the belt,
T1 = T – T C = 5250 – 500 = 4750 N
Let
T2 = Tension in the slack side of the belt.
Since the pulleys are of the same size, therefore angle of contact, θ = 180° = π rad.
We know that
T
2.3log 1 = µ . θ cosec β = 0.12 × π × cosec15° = 1.457
T2
T 1.457
T
log 1 =
= 0.6334 or 1 = 4.3
2.3
T2
T2
...(Taking antilog of 0.6334)
and
T1
4750
=
= 1105 N
4.3
4.3
We know that power transmitted,
T2 =
P = (T 1 – T 2) v × 2
...(∵ No. of belts = 2)
= (4750 – 1105) 23.56 × 2 = 171 752 W = 171.752 kW Ans.
Shaft speed
Let
N 1 = Shaft speed in r.p.m., and
v 1= Belt speed in m/s.
We know that for maximum power, centrifugal tension,
T C = T / 3 or m (v 1)2 = T / 3 or 0.9 (v 1)2 = 5250 / 3 = 1750
∴
(v 1)2 = 1750 / 0.9 = 1944.4 or v 1 = 44.1 m/s
Chapter 11 : Belt, Rope and Chain Drives
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363
We know that belt speed (v 1),
π d . N1 π × 0.3 × N1
=
= 0.0157 N1
60
60
N1 = 44.1 / 0.0157 = 2809 r.p.m. Ans.
44.1 =
∴
Example 11.18. Power is transmitted using a V-belt drive. The included angle of V-groove
is 30°. The belt is 20 mm deep and maximum width is 20 mm. If the mass of the belt is 0.35 kg per
metre length and maximum allowable stress is 1.4 MPa, determine the maximum power transmitted
when the angle of lap is 140°. µ = 0.15.
Solution. Given : 2 β = 30° or β = 15° ; t = 20 mm = 0.02 m ; b = 20 mm = 0.02 m ;
m = 0.35 kg/m ; σ = 1.4 MPa = 1.4 × 106 N/m2 ; θ = 140° = 140° × π / 180 = 2.444 rad ; µ = 0.15
We know that maximum tension in the belt,
T = σ. b. t = 1.4 × 106 × 0.02 × 0.02 = 560 N
and for maximum power to be transmitted, velocity of the belt,
T
560
=
= 23.1 m/s
3m
3 × 0.35
T1 = Tension in the tight side of the belt, and
T2 = Tension in the slack side of the belt.
v=
Let
We know that
T
2.3log 1 = µ . θ cosec β = 0.15 × 2.444 × cosec15° = 1.416
T2
T 1.416
T
log 1 =
= 0.616 or 1 = 4.13
2.3
T2
T2
...(i)
...(Taking antilog of 0.616)
T 560
=
= 187 N
3
3
T1 = T – T C = 560 – 187 = 373 N
Centrifugal tension, TC =
and
T1
373
=
= 90.3 N
4.13 4.13
We know that maximum power transmitted,
T2 =
...[From equation (i)]
P = (T 1 – T 2) v = (373 – 90.3) 23.1 = 6530 W = 6.53 kW Ans.
Example 11.19. A compressor, requiring 90 kW is to run at about 250 r.p.m. The drive is by
V-belts from an electric motor running at 750 r.p.m. The diameter of the pulley on the compressor
shaft must not be greater than 1 metre while the centre distance between the pulleys is limited to 1.75
metre. The belt speed should not exceed 1600 m/min.
Determine the number of V-belts required to transmit the power if each belt has a crosssectional area of 375 mm2, density 1000 kg/m3 and an allowable tensile stress of 2.5 MPa. The
groove angle of the pulley is 35°. The coefficient of friction between the belt and the pulley is 0.25.
Calculate also the length required of each belt.
Solution. Given : P = 90 kW ; N 2 = 250 r.p.m. ; N 1 = 750 r.p.m. ; d2 = 1 m ; x = 1.75 m ;
v = 1600 m/min = 26.67 m/s ; a = 375 mm2 = 375 × 10–6 m2 ; ρ = 1000 kg/m3 ; σ = 2.5 MPa
= 2.5 × 106 N/m2 ; 2 β = 35° or β = 17.5° ; µ = 0.25
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First of all, let us find the diameter of pulley on the motor shaft (d1). We know that
N2
d
250 × 1
N .d
= 1 or d1 = 2 2 =
= 0.33m
N1 d2
750
N1
We know that the mass of the belt per metre length,
m = Area × length × density
= 375 × 10–6 × 1 × 1000 = 0.375 kg
∴ Centrifugal tension, TC = m.v2 = 0.375 (26.67)2 = 267 N
and maximum tension in the belt,
T = σ. a = 2.5 × 106 × 375 × 10–6 = 937.5 N
∴ Tension in the tight side of the belt,
T1 = T – T C = 937.5 – 267 = 670.5 N
Let
T2 = Tension in the slack side of the belt.
For an open belt drive, as shown in Fig. 11.21,
O2 M
r –r
d – d1 1 – 0.33
= 2 1 = 2
=
= 0.1914
2x
2 × 1.75
O1 O2
x
∴
α = 11°
and angle of lap on smaller pulley (i.e. pulley on motor shaft),
θ = 180° – 2α = 180° – 2 × 11° = 158°
= 158 × π / 180 = 2.76 rad
sin α =
Fig. 11.21
We know that
T
2.3log 1 = µ . θ cosec β = 0.25 × 2.76 × cosec17.5° = 2.295
T2
and
Number of V-belts
T1
T 2.295
= 9.954
log 1 =
= 0.998 or
T2
2.3
T2
T1
670.5
T2 =
=
= 67.36 N
9.954 9.954
...(Taking antilog of 0.998)
We know that power transmitted per belt
= (T 1 – T 2) v = (670.5 – 67.36) 26.67 = 16 086 W
= 16.086 kW
∴
Number of V-belts =
Total power transmitted
90
=
= 5.6 or 6 Ans.
Power transmitted per belt 16.086
Chapter 11 : Belt, Rope and Chain Drives
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365
Length of each belt
We know that length of belt for an open belt drive,
L=
(d – d1 )2
π
(d 2 + d1 ) + 2 x + 2
2
4x
(1 – 0.33) 2
π
(1 + 0.33) + 2 × 1.75 +
2
4 × 1.75
= 2.1 + 3.5 + 0.064 = 5.664 m Ans.
=
11.23. Rope Drive
The rope drives are widely used where a large amount of power is to be transmitted, from one
pulley to another, over a considerable distance. It may be noted that the use of flat belts is limited for
the transmission of moderate power from one pulley to another when the two pulleys are not more
than 8 metres apart. If large amounts of power are to be transmitted by the flat belt, then it would
result in excessive belt cross-section. It may be noted that frictional grip in case of rope drives is more
than that in V-drive. One of the main advantage of rope drives is that a number of separate drives may
be taken from the one driving pulley. For example, in many spinning mills, the line shaft on each floor
is driven by ropes passing directly from the main engine pulley on the ground floor.
The rope drives use the following two types of ropes :
1. Fibre ropes, and 2. Wire ropes.
The fibre ropes operate successfully when the pulleys are about 60 metres apart, while the
wire ropes are used when the pulleys are upto 150 metres apart.
11.24. Fibre Ropes
The ropes for transmitting power are usually made from fibrous materials such as hemp,
manila and cotton. Since the hemp and manila fibres are rough, therefore the ropes made from these
fibres are not very flexible and possesses poor mechanical properties. The hemp ropes have less
strength as compared to manila ropes. When the hemp and manila ropes are bent over the sheave (or
pulley), there is some sliding of fibres, causing the rope to wear and chafe internally. In order to
minimise this defect, the rope fibres are lubricated with a tar, tallow or graphite. The lubrication also
makes the rope moisture proof. The hemp ropes are suitable only for hand operated hoisting machinery and as tie ropes for lifting tackle, hooks etc.
The cotton ropes are very soft and smooth. The lubrication of cotton ropes is not necessary.
But if it is done, it reduces the external wear between the rope and the grooves of its sheaves. It may
be noted that manila ropes are more durable and stronger than cotton ropes. The cotton ropes are
costlier than manila ropes.
Note : The diameter of manila and cotton ropes usually ranges from 38 mm to 50 mm. The size of the rope is
usually designated by its circumference or ‘girth’.
11.25. Advantages of Fibre Rope Drives
The fibre rope drives have the following advantages :
1. They give smooth, steady and quiet service.
2. They are little affected by out door conditions.
3. The shafts may be out of strict alignment.
4. The power may be taken off in any direction and in fractional parts of the whole amount.
5. They give high mechanical efficiency.
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11.26. Sheave for Fibre Ropes
The fibre ropes are usually circular in cross-section as shown in Fig. 11.22 (a). The sheave
for the fibre ropes is shown in Fig. 11.22 (b). The groove angle of the pulley for rope drives is usually
45°. The grooves in the pulleys are made narrow at the bottom and the rope is pinched between the
edges of the V-groove to increase the holding power of the rope on the pulley.
(a) Cross-section of a rope.
(b) Sheave (Grooved pulley) for ropes.
Fig. 11.22. Rope and sheave.
11.27. Wire Ropes
When a large amount of power is to be transmitted over long distances from one pulley to
another (i.e. when the pulleys are upto 150 metres apart), then wire ropes are used. The wire ropes are
This electric hoist uses wire ropes.
widely used in elevators, mine hoists, cranes, conveyors, hauling devices and suspension bridges.
The wire ropes run on grooved pulleys but they rest on the bottom of the *grooves and are not wedged
between the sides of the grooves. The wire ropes have the following advantage over cotton ropes.
*
The fibre ropes do not rest at the bottom of the groove.
Chapter 11 : Belt, Rope and Chain Drives
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367
1. These are lighter in weight, 2. These offer silent operation, 3. These can withstand shock
loads, 4. These are more reliable, 5. They do not fail suddenly, 6. These are more durable, 7. The
efficiency is high, and 8. The cost is low.
11.28. Ratio of Driving Tensions for Rope Drive
The ratio of driving tensions for the rope drive may be obtained in the similar way as V-belts.
We have discussed in Art. 11.22, that the ratio of driving tensions is
T
2.3log 1 = µ . θ cosec β
T2
where, µ, θ and β have usual meanings.
Example 11.20. A rope drive transmits 600 kW from a pulley of effective diameter 4 m,
which runs at a speed of 90 r.p.m. The angle of lap is 160° ; the angle of groove 45° ; the coefficient
of friction 0.28 ; the mass of rope 1.5 kg / m and the allowable tension in each rope 2400 N. Find the
number of ropes required.
Solution. Given : P = 600 kW ; d = 4 m ; N = 90 r.p.m. ; θ = 160° = 160 × π / 180 = 2.8 rad;
2 β = 45° or β = 22.5° ; µ = 0.28 ; m = 1.5 kg / m ; T = 2400 N
We know that velocity of the rope,
π d . N π× 4 × 90
=
= 18.85 m/s
60
60
∴ Centrifugal tension, TC = m.v2 = 1.5 (18.85)2 = 533 N
v=
and tension in the tight side of the rope,
T1 = T – T C = 2400 – 533 = 1867 N
Let
T2 = Tension in the slack side of the rope.
We know that
T
2.3log 1 = µ . θ cosec β = 0.28 × 2.8 × cosec 22.5° = 2.05
T2
T 2.05
T
log 1 =
= 0.8913 or 1 = 7.786
2.3
T
T
2
2
and
...(Taking antilog of 0.8913)
T1
1867
T2 =
=
= 240 N
7.786 7.786
We know that power transmitted per rope
= (T 1 – T 2) v = (1867 – 240) 18.85 = 30 670 W = 30.67 kW
∴
Number of ropes =
Total power transmitted
600
=
= 19.56 or 20 Ans.
Power transmitted per rope 30.67
Example 11.21. A pulley used to transmit power by means of ropes has a diameter of 3.6
metres and has 15 grooves of 45° angle. The angle of contact is 170° and the coefficient of friction
between the ropes and the groove sides is 0.28. The maximum possible tension in the ropes is 960 N
and the mass of the rope is 1.5 kg per metre length. What is the speed of pulley in r.p.m. and the power
transmitted if the condition of maximum power prevail ?
Solution. Given : d = 3.6 m ; No. of grooves = 15 ; 2 β = 45° or β = 22.5° ; θ = 170°
= 170 π × 180 = 2.967 rad ; µ = 0.28 ; T = 960 N ; m = 1.5 kg/m
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Speed of the pulley
Let
N = Speed of the pulley in r.p.m.
We know that for maximum power, velocity of the rope or pulley,
T
=
3m
v=
∴
N =
960
= 14.6 m/s
3 × 1.5
v × 60 14.6 × 60
=
= 77.5 r.p.m. Ans.
πd
π × 3.6
πd N
... 3 v =
60
Power transmitted
We know that for maximum power, centrifugal tension,
TC = T / 3 = 960 / 3 = 320 N
∴ Tension in the tight side of the rope,
T1 = T – T C = 960 – 320 = 640 N
Let
T2 = Tension in the slack side of the rope.
We know that
T
2.3log 1 = µ . θ cosec β = 0.28 × 2.967 × cosec 22.5° = 2.17
T2
T 2.17
T
log 1 =
= 0.9438 or 1 = 8.78
2.3
T
T
2
2
...(Taking antilog of 0.9438)
T1
640
=
= 73 N
8.78 8.78
∴ Power transmitted per rope = (T 1 – T 2) v = (640 – 73) 14.6 = 8278 W = 8.278 kW
Since the number of grooves are 15, therefore total power transmitted
= 8.278 × 15 = 124.17 kW Ans.
Example 11.22. Following data is given for a rope pulley transmitting 24 kW :
Diameter of pulley = 400 mm ; Speed = 110 r.p.m.; angle of groove = 45° ; Angle of lap on
smaller pulley = 160° ; Coefficient of friction = 0.28 ; Number of ropes = 10 ; Mass in kg/m length
of ropes = 53 C2 ; and working tension is limited to 122 C2 kN, where C is girth of rope in metres.
Find initial tension and diameter of each rope.
Solution. Given : PT = 24 kW ; d = 400 mm = 0.4 m ; N = 110 r.p.m. ; 2 β = 45° or β = 22.5°;
θ = 160° = 160 × π / 180 = 2.8 rad ; n = 0.28 ; n = 10 ; m = 53 C2 kg/m ; T = 122 C2 kN
= 122 × 103 C2 N
Initial tension
We know that power transmitted per rope,
T2 =
and
P=
and velocity of the rope,
Let
Total power transmitted PT 24
=
=
= 2.4 kW = 2400 W
No. of ropes
10
n
π d . N π × 0.4 × 110
=
= 2.3 m/s
60
60
T = Tension in the tight side of the rope, and
T2 = Tension in the slack side of the rope.
v=
Chapter 11 : Belt, Rope and Chain Drives
We know that power transmitted per rope ( P )
2400 = (T 1 – T 2) v = (T 1 – T 2) 2.3
∴
T 1 – T 2 = 2400 / 2.3 = 1043.5 N
We know that
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369
...(i)
T
2.3log 1 = µ . θ cosec β = 0.28 × 2.8 × cosec 22.5° = 2.05
T2
T 2.05
T
log 1 =
= 0.8913 or 1 = 7.786
2.3
T
T
2
2
...(ii)
...(Taking antilog of 0.8913)
From equations (i) and (ii),
T1 = 1197.3 N, and T 2 = 153.8 N
We know that initial tension in each rope,
T0 =
T1 + T2 1197.3 + 153.8
=
= 675.55 N
2
2
Ans.
Diameter of each rope
Let
d1 = Diameter of each rope,
We know that centrifugal tension,
TC = m.v2 = 53 C2 (2.3)2 = 280.4 C2 N
and working tension (T),
122 × 103 C2 = T 1 + T C = 1197.3 + 280.4 C2
122 ×
103
C2 – 280.4 C2 = 1197.3
∴
C2 = 9.836 × 10–3 or C = 0.0992 m = 99.2 mm
We know that girth (i.e. circumference) of rope (C),
99.2 = π d1 or d1 = 99.2 / π = 31.57 mm Ans.
11.29. Chain Drives
We have seen in belt and rope
drives that slipping may occur. In order
to avoid slipping, steel chains are used.
The chains are made up of rigid links
which are hinged together in order to
provide the necessary flexibility for
warping around the driving and driven
wheels. The wheels have projecting teeth
and fit into the corresponding recesses,
in the links of the chain as shown in Fig.
11.23. The wheels and the chain are thus
constrained to move together without
slipping and ensures perfect velocity
ratio. The toothed wheels are known as
sprocket wheels or simply sprockets.
These wheels resemble to spur gears.
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The chains are mostly used to transmit motion and power from one shaft to another, when the
distance between the centres of the shafts is short such
as in bicycles, motor cycles, agricultural machinery,
road rollers, etc.
11.30. Advantages and Disadvantages
of Chain Drive Over Belt or Rope
Drive
Following are the advantages and disadvantages of chain drive over belt or rope drive :
Fig. 11.23. Sprocket and chain.
Advantages
1. As no slip takes place during chain drive, hence perfect velocity ratio is obtained.
2. Since the chains are made of metal, therefore they occupy less space in width than a belt
or rope drive.
3. The chain drives may be used when the distance between the shafts is less.
4. The chain drive gives a high transmission efficiency (upto 98 per cent).
5. The chain drive gives less load on the shafts.
6. The chain drive has the ability of transmitting motion to several shafts by one chain only.
Disadvantages
1. The production cost of chains is relatively high.
2. The chain drive needs accurate mounting and careful maintenance.
3. The chain drive has velocity fluctuations especially when unduly stretched.
11.31. Terms Used in Chain Drive
The following terms are frequently used in chain drive.
1. Pitch of the chain : It is the distance between the hinge centre of a link and the corresponding hinge centre of the adjacent link as shown in Fig. 11.24. It is usually denoted by p.
Fig. 11.24. Pitch of the chain.
Fig. 11.25. Pitch circle diameter of the chain sprocket.
2. Pitch circle diameter of the chain sprocket. It is the diameter of the circle on which the
hinge centres of the chain lie, when the chain is wrapped round a sprocket as shown in Fig. 11.25. The
points A , B, C, and D are the hinge centres of the chain and the circle drawn through these centres is
called pitch circle and its diameter (d) is known as pitch circle diameter.
Chapter 11 : Belt, Rope and Chain Drives
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371
11.32. Relation Between Pitch and Pitch Circle Diameter
A chain wrapped round the sprocket is shown in Fig. 11.25. Since the links of the chain are
rigid, therefore pitch of the chain does not lie on the arc of the pitch circle. The pitch length becomes
a chord. Consider one pitch length A B of the chain subtending an angle θ at the centre of sprocket (or
pitch circle).
Let
d = Diameter of the pitch circle, and
T = Number of teeth on the sprocket.
From Fig. 11.25, we find that pitch of the chain,
d
θ
θ
θ
p = AB = 2 AO sin = 2 × sin = d sin
2
2
2
2
360°
T
We know that
θ=
∴
360°
180°
p = d sin
= d sin
2T
T
180°
d = p cosec
T
or
11.33. Relation Between
Chain Speed and
Angular Velocity
of Sprocket
Since the links of the chain
are rigid, therefore they will have
different positions on the sprocket
at different instants. The relation
between the chain speed (v) and
angular velocity of the sprocket (ω)
also varies with the angular position of the sprocket. The extreme
positions are shown in Fig. 11.26
(a) and (b).
Fig. 11.26. Relation between chain speed and angular velocity of sprocket.
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For the angular position of the sprocket as shown in Fig. 11.26 (a),
v = ω × OA
and for the angular position of the sprocket as shown in Fig. 11.26 (b),
θ
θ
v = ω × OX = ω × OC cos = ω × OA cos
2
2
...(∵ OC = O A)
11.34. Kinematic of Chain Drive
Fig. 11.27 shows an arrangement of a chain drive in which the smaller or driving sprocket has
6 teeth and the larger or driven sprocket has 9 teeth. Though this is an impracticable case, but this is
considered to bring out clearly the kinematic conditions of a chain drive. Let both the sprockets rotate
anticlockwise and the angle subtended by the chain pitch at the centre of the driving and driven
sprockets be α and φ respectively. The lines A B and A 1B 1 show the positions of chain having minimum and maximum inclination respectively with the line of centres O1O2 of the sprockets. The points
A , B 2 and B are in one straight line and the points A 1,C and B 1 are in one straight line. It may be noted
that the straight length of the chain between the two sprockets must be equal to exact number of
pitches.
Fig. 11.27. Kinematic of chain drive.
Let us now consider the pin centre on the driving sprocket in position A . The length of the
chain A B will remain straight as the sprockets rotate, until A reaches A 1 and B reaches B 1. As the
driving sprocket continues to turn, the link A 1C of the chain turns about the pin centre C and the
straight length of the chain between the two sprockets reduces to CB1. When the pin centre C moves
to the position A 1, the pin centre A 1 moves to the position A 2. During this time, each of the sprockets
rotate from its original position by an angle corresponding to one chain pitch. During the first part of
the angular displacement, the radius O1 A moves to O1 A 1 and the radius O2 B moves to O2 B 1. This
arrangement is kinematically equivalent to the four bar chain O1ABO2.
During the second part of the angular displacement, the radius O1 A 1 moves to O1A 2 and the
radius O2 B 1 moves to O2 B 2. This arrangement is kinematically equivalent to the four bar chain
O1CB1O2. The ratio of the angular velocities, under these circumstances, cannot be constant. This
may be easily shown as discussed below :
First of all, let us find the instantaneous centre for the two links O1 A and O2 B. This lies at
point I which is the intersection of B A and O2 O1 produced as shown in Fig. 11.28. If ω1 is the angular
velocity of the driving sprocket and ω2 is the angular velocity of the driven sprocket, then
ω1 × O1 I = ω2 × O2 I
or
ω1 O2 I O2 O1 + O1 I
O O
=
=
=1+ 2 1 .
ω2 O1 I
O1 I
O1 I
Chapter 11 : Belt, Rope and Chain Drives
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373
The distance between the centres of two sprockets O1 O2 is constant for a given chain drive,
but the distance O1 I varies periodically as the two sprockets rotate. This period corresponds to a
rotation of the driving sprocket by an angle α. It is clear from the figure that the line A B has minimum
inclination with line O1 O2. Therefore the distance O1 I is maximum and thus velocity ratio (ω1 / ω2)
is minimum. When the chain occupies the position A 1 B 1, the inclination of line A 1 B 1 is maximum
with the line O1 O2. Therefore the distance O1 I1 is minimum and thus the velocity ratio (ω1 / ω2) is
maximum.
Fig. 11.28. Angular velocities of the two sprockets.
In actual practice, the smaller sprocket have a minimum of 18 teeth and hence the actual
variation of velocity ratio (ω1/ω2) from the mean value is very small.
11.35. Classification of Chains
The chains, on the basis of their use, are classified into the following three groups :
1. Hoisting and hauling (or crane) chains,
2. Conveyor (or tractive) chains, and
3. Power transmitting (or driving) chains.
These chains are discussed, in detail, in the following pages.
11.36. Hoisting and Hauling Chains
These chains are used for hoisting
and hauling purposes. The hoisting and
hauling chains are of the following two
types :
1. Chain with oval links. The links
of this type of chain are of oval shape, as
shown in Fig. 11.29 (a). The joint of each
link is welded. The sprockets which are used
for this type of chain have receptacles to receive the links. Such type of chains are used
only at low speeds such as in chain hoists and in anchors for marine works.
(a) Chain with oval links.
(b) Chain with square links.
Fig. 11.29. Hoisting and hauling chains.
2. Chain with square links. The links of this type of chain are of square shape, as shown in
Fig. 11.29 (b). Such type of chains are used in hoists, cranes, dredges. The manufacturing cost of this
type of chain is less than that of chain with oval links, but in these chains, the kinking occurs easily on
overloading.
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11.37. Conveyor Chains
These chains are used for elevating and conveying the materials continuously. The conveyor
chains are of the following two types :
1. Detachable or hook joint type chain, as shown in Fig. 11.30 (a), and
2. Closed joint type chain, as shown in Fig. 11.30 (b).
(a) Detachable or hook joint type chain.
(b) Closed joint type chain.
Fig. 11.30. Conveyor chains.
The conveyor chains are usually made of malleable cast iron. These chains do not have
smooth running qualities. The conveyor chains run at slow speeds of about 3 to 12 km.p.h.
11.38. Power Transmitting Chains
These chains are used for transmission of power, when the distance between the centres of
shafts is short. These chains have provision for efficient lubrication. The power transmitting chains
are of the following three types.
1. Block chain. A block chain, as shown in Fig. 11.31, is also known as bush chain. This
type of chain was used in the early stages of development in the power transmission.
Fig. 11.31. Block chain.
It produces noise when approaching or leaving the teeth of the sprocket because of rubbing
between the teeth and the links. Such type of chains are used to some extent as conveyor chain at
small speed.
2. Bush roller chain. A bush roller chain, as shown in Fig. 11.32, consists of outer plates or
pin link plates, inner plates or roller link plates, pins, bushes and rollers. A pin passes through the
bush which is secured in the holes of the roller between the two sides of the chain. The rollers are free
to rotate on the bush which protect the sprocket wheel teeth against wear.
A bush roller chain is extremely strong and simple in construction. It gives good service
under severe conditions. There is a little noise with this chain which is due to impact of the rollers on
the sprocket wheel teeth. This chain may be used where there is a little lubrication. When one of these
chains elongates slightly due to wear and stretching of the parts, then the extended chain is of greater
pitch than the pitch of the sprocket wheel teeth. The rollers then fit unequally into the cavities of the
Chapter 11 : Belt, Rope and Chain Drives
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375
wheel. The result is that the total load falls on one teeth or on a few teeth. The stretching of the parts
increase wear of the surfaces of the roller and of the sprocket wheel teeth.
Fig. 11.32. Bush roller chain.
3. Inverted tooth or silent chain. An inverted
tooth or silent chain is shown in Fig. 11.33. It is designed
to eliminate the evil effects caused by stretching and to
produce noiseless running. When the chain stretches and
the pitch of the chain increases, the links ride on the teeth
of the sprocket wheel at a slightly increased radius. This
automatically corrects the small change in the pitch. There
is no relative sliding between the teeth of the inverted tooth
chain and the sprocket wheel teeth. When properly
lubricated, this chain gives durable service and runs very
smoothly and quietly.
Fig. 11.33. Inverted tooth or silent chain.
11.39. Length of Chain
An open chain drive system connecting the two sprockets is shown in Fig. 11.34. We have
already discussed in Art. 11.11 that the length of belt for an open belt drive connecting the two pulleys
of radii r1 and r2 and a centre distance x, is
L = π (r1 + r2 ) + 2 x +
(r1 – r2 )2
x
Fig. 11.34. Length of chain
(i)
376
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Theory of Machines
If this expression is used for determining the length of chain, the result will be slightly greater
than the required length. This is due to the fact that the pitch lines A B C D E F G and P Q R S of the
sprockets are the parts of a polygon and not that of a circle. The exact length of the chain may be
determined as discussed below :
Let
T1 = Number of teeth on the larger sprocket,
T2 = Number of teeth on the smaller sprocket, and
p = Pitch of the chain.
We have discussed in Art. 11.32, that diameter of the pitch circle,
180°
p
180°
d = p cosec
or r = cosec
T
2
T
∴ For larger sprocket,
and for smaller sprocket,
r1 =
180°
p
cosec
2
T1
r2 =
180°
p
cosec
2
T2
Since the term π (r1 + r2) is equal to half the sum of the circumferences of the pitch circles,
therefore the length of chain corresponding to
p
(T1 + T2 )
2
Substituting the values of r1, r2 and π (r1 + r2) in equation (i), the length of chain is given by
π ( r1 + r2 ) =
p
180° p
180°
cosec
– cosec
2
2
T
p
1
T2
L = (T1 + T2 ) + 2 x +
2
x
2
If x = m.p, then
2
180°
180°
cosec
–
cosec
(T1 + T2 )
T1
T2
+ 2m +
L = p
= p.K
2
4m
where
K = Multiplying factor
2
180°
180°
cosec
– cosec
(T1 + T2 )
T1
T2
=
+ 2m +
2
4m
The value of multiplying factor (K) may not be a complete integer. But the length of the chain
must be equal to an integer number of times the pitch of the chain. Thus, the value of K should be
rounded off to the next higher integral number.
Example 11.23. A chain drive is used for reduction of speed from 240 r.p.m. to 120 r.p.m.
The number of teeth on the driving sprocket is 20. Find the number of teeth on the driven sprocket. If
the pitch circle diameter of the driven sprocket is 600 mm and centre to centre distance between the
two sprockets is 800 mm, determine the pitch and length of the chain.
Chapter 11 : Belt, Rope and Chain Drives
377
l
Solution. Given : N 1 = 240 r.p.m ; N 2 = 120 r.p.m ; T 1 = 20 ; d2 = 600 mm or r2 = 300 mm
= 0.3 m ; x = 800 mm = 0.8 m
Number of teeth on the driven sprocket
Let
T2 = Number of teeth on the driven sprocket.
We know that
N 1. T 1 = N 2. T 2
or
T2 =
N1 .T1 240 × 20
=
= 40 Ans.
120
N2
Pitch of the chain
Let
p = Pitch of the chain.
We know that pitch circle radius of the driven sprocket (r2),
180° p
p
180°
cosec
= cosec
= 6.37 p
2
2
T
40
2
p = 0.3 / 6.37 = 0.0471 m = 47.1 mm Ans.
0.3 =
∴
Length of the chain
We know that pitch circle radius of the driving sprocket,
180° 47.1
p
180°
cosec
cosec
=
= 150.5 mm
2
2
T
20
1
x = m.p
or m = x / p = 800 / 47.1 = 16.985
We know that multiplying factor,
r1 =
and
180°
180°
cosec
– cosec
(T1 + T2 )
T1
T2
K =
+ 2m +
2
4m
2
180°
180°
cosec 20 – cosec 40
(20 + 40)
=
+ 2 × 16.985 +
2
4 × 16.985
= 30 + 33.97 +
2
(6.392 – 12.745)2
= 64.56 say 65
67.94
∴ Length of the chain,
L = p.K = 47.1 × 65 = 3061.5 mm = 3.0615 m Ans.
EXERCISES
1.
An engine shaft running at 120 r.p.m. is required to drive a machine shaft by means of a belt. The
pulley on the engine shaft is of 2 m diameter and that of the machine shaft is 1 m diameter. If the belt
thickness is 5 mm ; determine the speed of the machine shaft, when 1. there is no slip ; and 2. there is
a slip of 3%.
[Ans. 239.4 r.p.m. ; 232.3 r.p.m.]
2.
Two parallel shafts 6 metres apart are provided with 300 mm and 400 mm diameter pulleys and are
connected by means of a cross belt. The direction of rotation of the follower pulley is to be reversed by
changing over to an open belt drive. How much length of the belt has to be reduced ?
[Ans. 203.6 mm]
3.
A pulley is driven by a flat belt running at a speed of 600 m/min. The coefficient of friction
between the pulley and the belt is 0.3 and the angle of lap is 160°. If the maximum tension in the belt
is 700 N ; find the power transmitted by a belt.
[Ans. 3.983 kW]
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Theory of Machines
4.
Find the width of the belt, necessary to transmit 7.5 kW to a pulley 300 mm diameter, if the pulley
makes 1600 r.p.m and the coefficient of friction between the belt and the pulley is 0.22. Assume the
angle of contact as 210° and the maximum tension in the belt is not to exceed 8 N/mm width.
[Ans. 67.4 mm]
5.
An open belt 100 mm wide connects two pulleys mounted on parallel shafts with their centres 2.4 m
apart. The diameter of the larger pulley is 450 mm and that of the smaller pulley 300 mm. The coefficient of friction between the belt and the pulley is 0.3 and the maximum stress in the belt is limited to
14 N/mm width. If the larger pulley rotates at 120 r.p.m., find the maximum power that can be transmitted.
[Ans. 2.39 kW]
6.
A leather belt 125 mm wide and 6 mm thick, transmits power from a pulley 750 mm diameter which
runs at 500 r.p.m. The angle of lap is 150° and µ = 0.3. If the mass of 1 m3 of leather is 1 Mg and the
stress in the belt is not to exceed 2.75 MPa, find the maximum power that can be transmitted.
[Ans. 19 kW]
7.
A flat belt is required to transmit 35 kW from a pulley of 1.5 m effective diameter running at 300
r.p.m. The angle of contact is spread over 11/24 of the circumference and the coefficient of friction
between belt and pulley surface is 0.3. Determine, taking centrifugal tension into account, width of
the belt required. It is given that the belt thickness is 9.5 mm, density of its material is 1.1 Mg/m3 and
the related permissible working stress is 2.5 MPa.
[Ans. 143 mm]
8.
A blower is driven by an electric motor though a belt drive. The motor runs at 750 r.p.m. For this
power transmission, a flat belt of 8 mm thickness and 250 mm width is used. The diameter of the
motor pulley is 350 mm and that of the blower pulley 1350 mm. The centre distance between these
pulleys is 1350 mm and an open belt configuration is adopted. The pulleys are made out of cast iron.
The frictional coefficient between the belt and pulley is 0.35 and the permissible stress for the belt
material can be taken as 2.5 N/mm2 with sufficient factor of safety. The mass of a belt is 2 kg per metre
length. Find the maximum power transmitted without belt slipping in any one of the pulleys.
[Ans. 35.9 kW]
9.
An open belt drive connects two pulleys 1.2 m and 0.5 m diameter on parallel shafts 3.6 m apart. The
belt has a mass of 1 kg/m length and the maximum tension in it is not to exceed 2 kN. The 1.2 m
pulley, which is the driver, runs at 200 r.p.m. Due to the belt slip on one of the pulleys, the velocity of
the driven shaft is only 450 r.p.m. If the coefficient of friction between the belt and the pulley is 0.3,
find : 1. Torque on each of the two shafts, 2. Power transmitted, 3. Power lost in friction, and 4.
Efficiency of the drive.
[Ans. 648.6 N-m, 270.25 N-m ; 13.588 kW ; 0.849 kW ; 93.75%]
10.
The power transmitted between two shafts 3.5 metres apart by a cross belt drive round the two pulleys
600 mm and 300 mm in diameters, is 6 kW. The speed of the larger pulley (driver) is 220 r.p.m. The
permissible load on the belt is 25 N/mm width of the belt which is 5 mm thick. The coefficient of
friction between the smaller pulley surface and the belt is 0.35. Determine : 1. necessary length of the
belt ; 2. width of the belt, and 3. necessary initial tension in the belt.
[Ans. 8.472 m ; 53 mm ; 888 N]
11.
A flat belt, 8 mm thick and 100 mm wide transmits power between two pulleys, running at
1600 m/min. The mass of the belt is 0.9 kg/m length. The angle of lap in the smaller pulley is 165° and
the coefficient of friction between the belt and pulley is 0.3. If the maximum permissible stress in the
belt is 2 MN/m2, find : 1. maximum power transmitted ; and 2. initial tension in the belt
[Ans. 14.83 kW ; 1002 N]
12.
An open belt connects two flat pulleys. The smaller pulley is 400 mm diameter and runs at 200 r.p.m.
The angle of lap on this pulley is 160° and the coefficient of friction between the belt and pulley face
is 0.25. The belt is on the point of slipping when 3 kW is being transmitted. Which of the following
two alternatives would be more effective in order to increase the power :
1. Increasing the initial tension in the belt by 10 per cent, and
2. Increasing the coefficient of friction by 10 per cent by the application of a suitable dressing to the
belt?
[Ans. First method is more effective]
Chapter 11 : Belt, Rope and Chain Drives
l
379
13.
A V-belt drive consists of three V-belts in parallel on grooved pulleys of the same size. The angle of
groove is 30° and the coefficient of friction 0.12. The cross-sectional area of each belt is 800 mm2 and
the permissible safe stress in the material is 3 MPa. Calculate the power that can be transmitted
between two pulleys 400 mm in diameter rotating at 960 r.p.m.
[Ans. 111.12 kW]
14.
Power is transmitted between two shafts by a V-belt whose mass is 0.9 kg/m length. The maximum
permissible tension in the belt is limited to 2.2 kN. The angle of lap is 170° and the groove angle 45°.
If the coefficient of friction between the belt and pulleys is 0.17, find : 1. velocity of the belt for
maximum power ; and 2. power transmitted at this velocity.
[Ans. 28.54 m/s ; 30.7 kW]
15.
Two shafts whose centres are 1 m apart are connected by a V-belt drive. The driving pulley is supplied
with 100 kW and has an effective diameter of 300 mm. It runs at 1000 r.p.m. while the driven pulley
runs at 375 r.p.m. The angle of groove on the pulleys is 40°. The permissible tension in 400 mm 2
cross-sectional area belt is 2.1 MPa. The density of the belt is 1100 kg/m3. The coefficient of friction
between the belt and pulley is 0.28. Estimate the number of belts required.
[Ans. 10]
16.
A rope drive is required to transmit 230 kW from a pulley of 1 metre diameter running at 450 r.p.m.
The safe pull in each rope is 800 N and the mass of the rope is 0.46 kg per metre length. The angle of
lap and the groove angle is 160° and 45° respectively. If the coefficient of friction between the rope
and the pulley is 0.3, find the number of ropes required.
[Ans. 21]
17.
Power is transmitted between two shafts, 3 metres apart by an open wire rope passing round two
pulleys of 3 metres and 2 metres diameters respectively, the groove angle being 40°. If the rope has a
mass of 3.7 kg per metre length and the maximum working tension in rope is 20 kN, determine the
maximum power that the rope can transmit and the corresponding speed of the smaller pulley. The
coefficient of friction being 0.15.
[Ans. 400 kW ; 403.5 r.p.m.]
18.
A rope drive transmits 75 kW through a 1.5 m diameter, 45° grooved pulley rotating at 200 r.p.m. The
coefficient of friction between the ropes and the pulley grooves is 0.3 and the angle of lap is 160°.
Each rope has a mass of 0.6 kg/m and can safely take a pull of 800 N. Taking centrifugal tension into
account determine : 1. the number of ropes required for the drive, and 2. initial rope tension.
[Ans. 9 ; 510.2 N]
19.
The reduction of speed from 360 r.p.m. to 120 r.p.m. is desired by the use of chain drive. The driving
sprocket has 10 teeth. Find the number of teeth on the driven sprocket. If the pitch radius of the driven
sprocket is 250 mm and the centre to centre distance between the two sprocket is 400 mm, find the
pitch and length of the chain.
[Ans. 30 ; 52.25 mm ; 1.93 m]
DO YOU KNOW ?
1.
Discuss briefly the various types of belts used for the transmission of power.
2.
How does the velocity ratio of a belt drive effect, when some slip is taking place between the belt and
the two pulleys ?
3.
Obtain an expression for the length of a belt in 1. an open belt drive ; and 2. a cross belt drive.
4.
Explain the phenomena of ‘slip’ and ‘creep’ in a belt drive.
5.
For a flat belt, prove that
T1
= eµθ , where
T2
T1 = Tension in the tight side of the belt,
T2 = Tension in the slack side of the belt,
µ = Coefficient of friction between the belt and the pulley, and
θ = Angle of contact between the belt and the pulley (in radians.)
6.
What is centrifugal tension in a belt ? How does it affect the power transmitted.
7.
Derive the condition for transmitting the maximum power in a flat belt drive.
380
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Theory of Machines
8.
It is stated that the speed at which a belt or rope should be run to transmit maximum power is that at
which the maximum allowable tension is three times the centrifugal tension in the belt or rope at that
speed. Prove the statement.
9.
Explain what do you understand by ‘initial tension in a belt’.
10.
Derive an expression for the ratio of the driving tensions in a rope drive assuming the angle of the
groove of the pulley to be as 2 β.
11.
Discuss relative merits and demerits of belt, rope and chain drive for transmission of power.
12.
What are different types of chains ? Explain, with neat sketches, the power transmission chains.
13.
Obtain an expression for the length of a chain.
OBJECTIVE TYPE QUESTIONS
1.
The velocity ratio of two pulleys connected by an open belt or crossed belt is
(a) directly proportional to their diameters
(b) inversely proportional to their diameters
(c) directly proportional to the square of their diameters
(d) inversely proportional to the square of their diameters
2.
3.
Two pulleys of diameters d1 and d2 and at distance x apart are connected by means of an open belt
drive. The length of the belt is
( a)
π
(d + d 2 )2
( d1 + d 2 ) + 2 x + 1
2
4x
( b)
π
(d – d 2 ) 2
( d1 – d 2 ) + 2 x + 1
2
4x
(c)
π
(d – d 2 )2
( d1 + d 2 ) + 2 x + 1
2
4x
(d )
π
(d + d 2 ) 2
( d1 – d 2 ) + 2 x + 1
2
4x
In a cone pulley, if the sum of radii of the pulleys on the driving and driven shafts is constant, then
(a) open belt drive is recommended
(b) cross belt drive is recommended
(c) both open belt drive and cross belt drive are recommended
(d) the drive is recommended depending upon the torque transmitted
4.
Due to slip of the belt, the velocity ratio of the belt drive
(a) decreases
5.
(b) increases
(a) larger pulley
6.
(b) smaller pulley (c) average of two pulleys
The power transmitted by a belt is maximum when the maximum tension in the belt (T) is equal
to
(a) T C
where
7.
(c) does not change
When two pulleys of different diameters are connected by means of an open belt drive, then the angle
of contact taken into consideration should be of the
(b) 2T C
(c) 3T C
(d) 4T C
T C = Centrifugal tension.
The velocity of the belt for maximum power is
(a)
T
3m
where
(b)
T
4m
(c)
T
5m
m = Mass of the belt in kg per metre length.
(d)
T
6m
Chapter 11 : Belt, Rope and Chain Drives
8.
l
381
The centrifugal tension in belts
(a) increases power transmitted
(b) decreases power transmitted
(c) have no effect on the power transmitted
(d) increases power transmitted upto a certain speed and then decreases
9.
When the belt is stationary, it is subjected to some tension, known as initial tension. The value of this
tension is equal to the
(a) tension in the tight side of the belt
(b) tension in the slack side of the belt
(c) sum of the tensions in the tight side and slack side of the belt
(d) average tension of the tight side and slack side of the belt
10.
The relation between the pitch of the chain ( p) and pitch circle diameter of the sprocket (d) is given by
60°
(a) p = d sin
T
90°
(b) p = d sin
T
120°
(c) p = d sin
T
180°
(d) p = d sin
T
where
T = Number of teeth on the sprocket.
ANSWERS
1. (b)
2. (c)
3. (b)
4. (a)
6. (c)
7. (a)
8. (c)
9. (d)
5. (b)
10.
(d)
GO To FIRST
CONTENTS
CONTENTS
382
l
Theory of Machines
12
Toothed
Gearing
Fea
tur
es
eatur
tures
1. Introduction.
2. Friction Wheels.
3. Advantages and
Disadvantages of Gear Drive.
4. Classification of Toothed
Wheels.
5. Terms Used in Gears.
6. Gear Materials.
7. Law of Gearing.
8. Velocity of Sliding of Teeth.
9. Forms of Teeth.
10. Cycloidal Teeth.
11. Involute Teeth.
12. Effect of Altering the Centre
Distance.
13. Comparison Between Involute
and Cycloidal Gears.
14. Systems of Gear Teeth.
15. Standard Proportions of Gear
Systems.
16. Length of Path of Contact.
17. Length of Arc of Contact.
18. Contact Ratio
19. Interference in Involute
Gears.
20. Minimum Number of Teeth on
the Pinion.
21. Minimum Number of Teeth on
the Wheel.
22. Minimum Number of Teeth on
a Pinion for Involute Rack in
Order to Avoid Interference.
23. Helical Gears.
24. Spiral Gears.
25. Centre Distance For a Pair of
Spiral Gears.
26. Efficiency of Spiral Gears.
12.1. Intr
oduction
Introduction
We have discussed in the previous chapter, that the
slipping of a belt or rope is a common phenomenon, in the
transmission of motion or power between two shafts. The
effect of slipping is to reduce the velocity ratio of the system.
In precision machines, in which a definite velocity ratio is of
importance (as in watch mechanism), the only positive drive
is by means of gears or toothed wheels. A gear drive is also
provided, when the distance between the driver and the follower is very small.
12.2. Friction Wheels
The motion and power transmitted by gears is kinematically equivalent to that transmitted by friction wheels or
discs. In order to
understand how
the motion can be
transmitted by
two
toothed
wheels, consider
two plain circular
wheels A and B
mounted
on
shafts, having sufficient rough surfaces and pressing against
each other as shown in Fig. 12.1 (a).
382
CONTENTS
CONTENTS
Chapter 12 : Toothed Gearing
l
383
Let the wheel A be keyed to the rotating shaft and the wheel B to the shaft, to be rotated. A
little consideration will show, that when the wheel A is rotated by a rotating shaft, it will rotate the
wheel B in the opposite direction as shown in Fig. 12.1 (a).
The wheel B will be rotated (by the wheel A ) so long as the tangential force exerted by the
wheel A does not exceed the maximum frictional resistance between the two wheels. But when the
tangential force (P) exceeds the *frictional resistance (F), slipping will take place between the two
wheels. Thus the friction drive is not a positive drive.
(a) Friction wheels.
(b) Toothed wheels.
Fig. 12.1
In order to avoid the slipping, a number of projections (called teeth) as shown in
Fig. 12.1 (b), are provided on the periphery of the wheel A , which will fit into the corresponding
recesses on the periphery of the wheel B. A friction wheel with the teeth cut on it is known as toothed
wheel or gear. The usual connection to show the toothed wheels is by their **pitch circles.
Note : Kinematically, the friction wheels running without slip and toothed gearing are identical. But due to the
possibility of slipping of wheels, the friction wheels can only be used for transmission of small powers.
12.3. Advantages and Disadvantages of Gear Drive
The following are the advantages and disadvantages of the gear drive as compared to belt,
rope and chain drives :
Advantages
1. It transmits exact velocity ratio.
2. It may be used to transmit large power.
3. It has high efficiency.
4. It has reliable service.
5. It has compact layout.
Disadvantages
1. The manufacture of gears require special tools and equipment.
2. The error in cutting teeth may cause vibrations and noise during operation.
*
**
The frictional force F is equal to µ. R N, where µ = Coefficient of friction between the rubbing surface of
two wheels, and R N = Normal reaction between the two rubbing surfaces.
For details, please refer to Art. 12.4.
384
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Theory of Machines
12.4. Classification of Toothed Wheels
The gears or toothed wheels may be classified as follows :
1. According to the position of axes of the shafts. The axes of the two shafts between which
the motion is to be transmitted, may be
(a) Parallel, (b) Intersecting, and (c) Non-intersecting and non-parallel.
The two parallel and co-planar shafts connected by the gears is shown in Fig. 12.1. These
gears are called spur gears and the arrangement is known as spur gearing. These gears have teeth
parallel to the axis of the wheel as shown in Fig. 12.1. Another name given to the spur gearing is
helical gearing, in which the teeth are inclined to the axis. The single and double helical gears connecting parallel shafts are shown in Fig. 12.2 (a) and (b) respectively. The double helical gears are
known as herringbone gears. A pair of spur gears are kinematically equivalent to a pair of cylindrical
discs, keyed to parallel shafts and having a line contact.
The two non-parallel or intersecting, but coplanar shafts connected by gears is shown in Fig.
12.2 (c). These gears are called bevel gears and the arrangement is known as bevel gearing. The
bevel gears, like spur gears, may also have their teeth inclined to the face of the bevel, in which case
they are known as helical bevel gears.
The two non-intersecting and non-parallel i.e. non-coplanar shaft connected by gears is shown
in Fig. 12.2 (d). These gears are called skew bevel gears or spiral gears and the arrangement is
known as skew bevel gearing or spiral gearing. This type of gearing also have a line contact, the
rotation of which about the axes generates the two pitch surfaces known as hyperboloids.
Notes : (a) When equal bevel gears (having equal teeth) connect two shafts whose axes are mutually perpendicular, then the bevel gears are known as mitres.
(b) A hyperboloid is the solid formed by revolving a straight line about an axis (not in the same
plane), such that every point on the line remains at a constant distance from the axis.
(c) The worm gearing is essentially a form of spiral gearing in which the shafts are usually at right
angles.
(a) Single helical gear. (b) Double helical gear.
(c) Bevel gear.
(d) Spiral gear.
Fig. 12.2
2. According to the peripheral velocity of the gears. The gears, according to the peripheral
velocity of the gears may be classified as :
(a) Low velocity, (b) Medium velocity, and (c) High velocity.
The gears having velocity less than 3 m/s are termed as low velocity gears and gears having
velocity between 3 and 15 m/s are known as medium velocity gears. If the velocity of gears is more
than 15 m/s, then these are called high speed gears.
Chapter 12 : Toothed Gearing
l
385
Helical Gears
Spiral Gears
Double helical gears
3. According to the type of gearing. The gears, according to the type of gearing may be
classified as :
(a) External gearing, (b) Internal gearing, and (c) Rack and pinion.
In external gearing, the gears of the two shafts mesh externally with each other as shown in Fig.
12.3 (a). The larger of these two wheels is called spur wheel and the smaller wheel is called pinion. In
an external gearing, the motion of the two wheels is always unlike, as shown in Fig. 12.3 (a).
(a) External gearing.
(b) Internal gearing.
Fig. 12.3
Fig. 12.4. Rack and pinion.
In internal gearing, the gears of the two shafts mesh internally with each other as shown in
Fig. 12.3 (b). The larger of these two wheels is called annular wheel and the smaller wheel is called
pinion. In an internal gearing, the motion of the two wheels is always like, as shown in Fig. 12.3 (b).
386
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Theory of Machines
Sometimes, the gear of a shaft meshes externally and internally with the gears in a *straight
line, as shown in Fig. 12.4. Such type of gear is called rack and pinion. The straight line gear is called
rack and the circular wheel is called pinion. A little consideration will show that with the help of a
rack and pinion, we can convert linear motion into rotary motion and vice-versa as shown in Fig.
12.4.
4. According to position of teeth on the gear surface. The teeth on the gear surface may be
(a) straight, (b) inclined, and (c) curved.
We have discussed earlier that the spur gears have straight teeth where as helical gears have
their teeth inclined to the wheel rim. In case of spiral gears, the teeth are curved over the rim surface.
Internal gears
Rack and pinion
12.5. Terms Used in Gears
The following terms, which will be mostly used in this chapter, should be clearly understood
at this stage. These terms are illustrated in Fig. 12.5.
Fig. 12.5. Terms used in gears.
1. Pitch circle. It is an imaginary circle which by pure rolling action, would give the same
motion as the actual gear.
*
A straight line may also be defined as a wheel of infinite radius.
Chapter 12 : Toothed Gearing
l
387
2. Pitch circle diameter. It is the diameter of the pitch circle. The size of the gear is usually
specified by the pitch circle diameter. It is also known as pitch diameter.
3. Pitch point. It is a common point of contact between two pitch circles.
4. Pitch surface. It is the surface of the rolling discs which the meshing gears have replaced
at the pitch circle.
5. Pressure angle or angle of obliquity. It is the angle between the common normal to two
gear teeth at the point of contact and the common tangent at the pitch point. It is usually denoted by φ.
The standard pressure angles are 14 12 ° and 20°.
6. Addendum. It is the radial distance of a tooth from the pitch circle to the top of the tooth.
7. Dedendum. It is the radial distance of a tooth from the pitch circle to the bottom of the tooth.
8. Addendum circle. It is the circle drawn through the top of the teeth and is concentric with
the pitch circle.
9. Dedendum circle. It is the circle drawn through the bottom of the teeth. It is also called
root circle.
Note : Root circle diameter = Pitch circle diameter × cos φ, where φ is the pressure angle.
10. Circular pitch. It is the distance measured on the circumference of the pitch circle from
a point of one tooth to the corresponding point on the next tooth. It is usually denoted by pc.
Mathematically,
pc = π D/T
Circular pitch,
where
D = Diameter of the pitch circle, and
T = Number of teeth on the wheel.
A little consideration will show that the two gears will mesh together correctly, if the two
wheels have the same circular pitch.
Note : If D1 and D2 are the diameters of the two meshing gears having the teeth T 1 and T 2 respectively, then for
them to mesh correctly,
pc =
π D1 π D2
=
T1
T2
or
D1
T
= 1
D2 T2
11. Diametral pitch. It is the ratio of number of teeth to the pitch circle diameter in millimetres.
It is denoted by pd . Mathematically,
Diametral pitch,
where
π
T
=
D pc
T = Number of teeth, and
pd =
πD
... 3 pc =
T
D = Pitch circle diameter.
12. Module. It is the ratio of the pitch circle diameter in millimeters to the number of teeth.
It is usually denoted by m. Mathematically,
Module, m = D /T
Note : The recommended series of modules in Indian Standard are 1, 1.25, 1.5, 2, 2.5, 3, 4, 5, 6, 8, 10, 12, 16,
and 20. The modules 1.125, 1.375, 1.75, 2.25, 2.75, 3.5, 4.5, 5.5, 7, 9, 11, 14 and 18 are of second choice.
13. Clearance. It is the radial distance from the top of the tooth to the bottom of the tooth, in
a meshing gear. A circle passing through the top of the meshing gear is known as clearance circle.
14. Total depth. It is the radial distance between the addendum and the dedendum circles of
a gear. It is equal to the sum of the addendum and dedendum.
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15. Working depth. It is the radial distance from the addendum circle to the clearance circle.
It is equal to the sum of the addendum of the two meshing gears.
16. Tooth thickness. It is the width of the tooth measured along the pitch circle.
17. Tooth space . It is the width of space between the two adjacent teeth measured along the pitch
circle.
18. Backlash. It is the difference between the tooth space and the tooth thickness, as measured along the pitch circle. Theoretically, the backlash should be zero, but in actual practice some
backlash must be allowed to prevent jamming of the teeth due to tooth errors and thermal expansion.
19. Face of tooth. It is the surface of the gear tooth above the pitch surface.
20. Flank of tooth. It is the surface of the gear tooth below the pitch surface.
21. Top land. It is the surface of the top of the tooth.
22. Face width. It is the width of the gear tooth measured parallel to its axis.
23. Profile. It is the curve formed by the face and flank of the tooth.
24. Fillet radius. It is the radius that connects the root circle to the profile of the tooth.
25. Path of contact. It is the path traced by the point of contact of two teeth from the
beginning to the end of engagement.
26. *Length of the path of contact. It is the length of the common normal cut-off by the
addendum circles of the wheel and pinion.
27. ** Arc of contact. It is the path traced by a point on the pitch circle from the beginning
to the end of engagement of a given pair of teeth. The arc of contact consists of two parts, i.e.
(a) Arc of approach. It is the portion of the path of contact from the beginning of the
engagement to the pitch point.
(b) Arc of recess. It is the portion of the path of contact from the pitch point to the end of the
engagement of a pair of teeth.
Note : The ratio of the length of arc of contact to the circular pitch is known as contact ratio i.e. number of pairs
of teeth in contact.
12.6. Gear Materials
The material used for the manufacture of gears depends upon the strength and service conditions like wear, noise etc. The gears may be manufactured from metallic or non-metallic materials.
The metallic gears with cut teeth are commercially obtainable in cast iron, steel and bronze. The nonmetallic materials like wood, raw hide, compressed paper and synthetic resins like nylon are used for
gears, especially for reducing noise.
The cast iron is widely used for the manufacture of gears due to its good wearing properties,
excellent machinability and ease of producing complicated shapes by casting method. The cast iron
gears with cut teeth may be employed, where smooth action is not important.
The steel is used for high strength gears and steel may be plain carbon steel or alloy steel. The
steel gears are usually heat treated in order to combine properly the toughness and tooth hardness.
The phosphor bronze is widely used for worm gears in order to reduce wear of the worms
which will be excessive with cast iron or steel.
12.7. Condition for Constant Velocity Ratio of Toothed Wheels–Law of
Gearing
Consider the portions of the two teeth, one on the wheel 1 (or pinion) and the other on the
*
**
For details, see Art. 12.16.
For details, see Art. 12.17.
Chapter 12 : Toothed Gearing
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389
wheel 2, as shown by thick line curves in Fig. 12.6. Let the two teeth
come in contact at point Q, and the wheels rotate in the directions as
shown in the figure.
Let T T be the common tangent and M N be the
common normal to the curves at the point of contact Q. From the
centres O1 and O2 , draw O1M and O2N perpendicular to M N. A
little consideration will show that the point Q moves in the direction
QC, when considered as a point on wheel 1, and in the direction
QD when considered as a point on wheel 2.
Let v 1 and v 2 be the velocities of the point Q on the wheels
1 and 2 respectively. If the teeth are to remain in contact, then the
components of these velocities along the common normal M N must
be equal.
∴
v1 cos α = v2 cos β
Fig. 12.6. Law of gearing.
( ω1 × O1 Q ) cos α = ( ω2 × O2 Q ) cos β
O M
O N
(ω1 × O1 Q ) 1
or ω1 × O1 M = ω2 × O2 N
= (ω2 × O2 Q ) 2
O1 Q
O2 Q
or
∴
ω1 O2 N
=
ω2 O1M
…(i)
Also from similar triangles O1MP and O2NP,
O2 N O2 P
=
O1 M
O1 P
...(ii)
Combining equations (i) and (ii), we have
ω1 O2 N O2 P
=
=
ω2 O1 M
O1 P
...(iii)
From above, we see that the angular velocity ratio is inversely proportional to the ratio of the
distances of the point P from the centres O1 and O2, or the common normal to the two surfaces at the
point of contact Q intersects the line of centres at point P which divides the centre distance inversely
as the ratio of angular velocities.
Therefore in order to have a constant angular velocity ratio for all positions of the wheels, the
point P must be the fixed point (called pitch point) for the two wheels. In other words, the common
normal at the point of contact between a pair of teeth must always pass through the pitch point.
This is the fundamental condition which must be satisfied while designing the profiles for the teeth of
gear wheels. It is also known as law of gearing.
Notes : 1. The above condition is fulfilled by teeth of involute form, provided that the root circles from which
the profiles are generated are tangential to the common normal.
2. If the shape of one tooth profile is arbitrarily chosen and another tooth is designed to satisfy the
above condition, then the second tooth is said to be conjugate to the first. The conjugate teeth are not in common
use because of difficulty in manufacture, and cost of production.
3. If D1 and D2 are pitch circle diameters of wheels 1 and 2 having teeth T 1 and T 2 respectively, then
velocity ratio,
ω1 O2 P D2 T2
=
=
=
ω2
O1 P
D1 T1
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12.8. Velocity of Sliding of Teeth
The sliding between a pair of teeth in contact at Q occurs along the common tangent T T to
the tooth curves as shown in Fig. 12.6. The velocity of sliding is the velocity of one tooth relative to
its mating tooth along the common tangent at the point of contact.
The velocity of point Q, considered as a point on wheel 1, along the common tangent T T is
represented by EC. From similar triangles QEC and O1MQ,
EC
v
=
= ω1
MQ O1Q
or
EC = ω1. MQ
Similarly, the velocity of point Q, considered as a point on wheel 2, along the common tangent T T is represented by ED. From similar triangles QCD and O2 NQ,
ED
v
= 2 = ω2
QN O2 Q
or
ED = ω2 . QN
Let
vS = Velocity of sliding at Q.
∴
vS = ED − EC = ω2 . QN − ω1. MQ
= ω2 (QP + PN ) − ω1 ( MP − QP )
Since
= ( ω1 + ω2 ) QP + ω2 . PN − ω1. MP
...(i)
ω1 O2 P PN
or ω1 . MP = ω2 . PN , therefore equation (i) becomes
=
=
ω2
O1 P MP
vS = (ω1 + ω2 ) QP
...(ii)
Notes : 1. We see from equation (ii), that the velocity of sliding is proportional to the distance of the point
of contact from the pitch point.
2. Since the angular velocity of wheel 2 relative to wheel 1 is (ω1 + ω2 ) and P is the instantaneous
centre for this relative motion, therefore the value of v s may directly be written as v s (ω1 + ω2 ) QP, without the
above analysis.
12.9. Forms of Teeth
We have discussed in Art. 12.7 (Note 2)
that conjugate teeth are not in common use.
Therefore, in actual practice following are the two
types of teeth commonly used :
1. Cycloidal teeth ; and 2. Involute teeth.
We shall discuss both the above mentioned
types of teeth in the following articles. Both these
forms of teeth satisfy the conditions as discussed
in Art. 12.7.
12.10. Cycloidal Teeth
A cycloid is the curve traced by a point on the circumference of a circle which rolls without
slipping on a fixed straight line. When a circle rolls without slipping on the outside of a fixed circle,
the curve traced by a point on the circumference of a circle is known as epi-cycloid. On the other
hand, if a circle rolls without slipping on the inside of a fixed circle, then the curve traced by a point
on the circumference of a circle is called hypo-cycloid.
Chapter 12 : Toothed Gearing
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391
In Fig. 12.7 (a), the fixed line or pitch line of a rack is shown. When the circle C rolls without
slipping above the pitch line in the direction as indicated in Fig. 12.7 (a), then the point P on the circle
traces epi-cycloid PA . This represents the face of the cycloidal tooth profile. When the circle D rolls
without slipping below the pitch line, then the point P on the circle D traces hypo-cycloid PB, which
represents the flank of the cycloidal tooth. The profile BPA is one side of the cycloidal rack tooth.
Similarly, the two curves P' A' and P'B' forming the opposite side of the tooth profile are traced by
the point P' when the circles C and D roll in the opposite directions.
(a)
(b)
Fig. 12.7. Construction of cycloidal teeth of a gear.
In the similar way, the cycloidal teeth of a gear may be constructed as shown in Fig. 12.7 (b).
The circle C is rolled without slipping on the outside of the pitch circle and the point P on the circle
C traces epi-cycloid PA , which represents the face of the cycloidal tooth. The circle D is rolled on the
inside of pitch circle and the point P on the circle D traces hypo-cycloid PB, which represents the
flank of the tooth profile. The profile BPA is one side of the cycloidal tooth. The opposite side of the
tooth is traced as explained above.
The construction of the two mating cycloidal teeth is shown in Fig. 12.8. A point on the circle
D will trace the flank of the tooth T 1 when circle D rolls without slipping on the inside of pitch circle
of wheel 1 and face of tooth T 2 when the circle D rolls without slipping on the outside of pitch circle
of wheel 2. Similarly, a point on the circle C will trace the face of tooth T 1 and flank of tooth T 2. The
rolling circles C and D may have unequal diameters, but if several wheels are to be interchangeable,
they must have rolling circles of equal diameters.
Fig. 12.8. Construction of two mating cycloidal teeth.
A little consideration will show, that the common normal X X at the point of contact between
two cycloidal teeth always passes through the pitch point, which is the fundamental condition for a
constant velocity ratio.
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12.11. Involute Teeth
An involute of a circle is a plane curve generated by a
point on a tangent, which rolls on the circle without slipping or
by a point on a taut string which in unwrapped from a reel as
shown in Fig. 12.9. In connection with toothed wheels, the circle
is known as base circle. The involute is traced as follows :
Let A be the starting point of the involute. The base
circle is divided into equal number of parts e.g. AP1, P1P2,
P2P3 etc. The tangents at P1, P2, P3 etc. are drawn and the
length P1A 1, P2A 2, P3A 3 equal to the arcs AP1, AP2 and AP3 are
set off. Joining the points A, A 1, A 2, A 3 etc. we obtain the involute
Fig. 12.9. Construction of involute.
curve A R. A little consideration will show that at any instant
A 3, the tangent A 3T to the involute is perpendicular to P3A 3 and P3A 3 is the normal to the involute. In
other words, normal at any point of an involute is a tangent to the circle.
Now, let O1 and O2 be the fixed centres of the two base circles as shown in Fig. 12.10 (a). Let
the corresponding involutes A B and A 1B 1 be in contact at point Q. MQ and NQ are normals to the
involutes at Q and are tangents to base circles. Since the normal of an involute at a given point is the
tangent drawn from that point to the base circle, therefore the common normal M N at Q is also the
common tangent to the two base circles. We see that the common normal M N intersects the line of
centres O1O2 at the fixed point P (called pitch point). Therefore the involute teeth satisfy the
fundamental condition of constant velocity ratio.
(a)
(b)
Fig. 12.10. Involute teeth.
From similar triangles O2NP and O1MP,
O1 M
OP ω
= 1 = 2
O2 N O2 P ω1
... (i)
which determines the ratio of the radii of the two base circles. The radii of the base circles is given by
O1M = O1P cos φ, and O2 N = O2 P cos φ
Also the centre distance between the base circles,
OM
O N O M + O2 N
O1O2 = O1 P + O2 P = 1 + 2 = 1
cos φ cos φ
cos φ
Chapter 12 : Toothed Gearing
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where φ is the pressure angle or the angle of obliquity. It is the angle which the common normal to the
base circles (i.e. MN) makes with the common tangent to the pitch circles.
When the power is being transmitted, the maximum tooth pressure (neglecting friction at the
teeth) is exerted along the common normal through the pitch point. This force may be resolved into
tangential and radial or normal components. These components act along and at right angles to the
common tangent to the pitch circles.
If F is the maximum tooth pressure as shown in Fig. 12.10 (b), then
Tangential force,
FT = F cos φ
and radial or normal force, FR = F sin φ.
∴ Torque exerted on the gear shaft
= FT × r, where r is the pitch circle radius of the gear.
Note : The tangential force provides the driving torque and the radial or normal force produces radial deflection
of the rim and bending of the shafts.
12.12. Effect of Altering the Centre Distance on the Velocity Ratio for
Involute Teeth Gears
In the previous article, we have seen that the velocity ratio for the involute teeth gears is given by
O1 M
OP ω
...(i)
= 1 = 2
O2 N O2 P ω1
Let, in Fig. 12.10 (a), the centre of rotation of one of the gears (say wheel 1) is shifted from
O1 to O1' . Consequently the contact point shifts from Q to Q '. The common normal to the teeth at the
point of contact Q ' is the tangent to the base circle, because it has a contact between two involute
curves and they are generated from the base circle. Let the tangent M' N' to the base circles intersects
O1′ O2 at the pitch point P' . As a result of this, the wheel continues to work* correctly.
Now from similar triangles O2NP and O1MP,
O1 M
OP
= 1
O2 N O2 P
...(ii)
and from similar triangles O2N'P' and O1'M'P',
O1′M ′ O1′P′
=
O2 N ′ O2 P′
...(iii)
But O2N = O2N', and O1M = O1' M'. Therefore from equations (ii) and (iii),
O1P O1′ P ′
=
...[Same as equation (i)]
O2 P O2 P ′
Thus we see that if the centre distance is changed within limits, the velocity ratio remains
unchanged. However, the pressure angle increases (from φ to φ′) with the increase in the centre
distance.
Example 12.1. A single reduction gear of 120 kW with a pinion 250 mm pitch circle diameter
and speed 650 r.p.m. is supported in bearings on either side. Calculate the total load due to the
power transmitted, the pressure angle being 20°.
Solution. Given : P = 120 kW = 120 × 103 W ; d = 250 mm or r = 125 mm = 0.125 m ;
N = 650 r.p.m. or ω = 2π × 650/60 = 68 rad/s ; φ = 20°
*
It is not the case with cycloidal teeth.
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Theory of Machines
Let
T = Torque transmitted in N-m.
We know that power transmitted (P),
120 × 103 = T.ω = T × 68
or
T = 120 × 103/68 = 1765 N-m
and tangential load on the pinion,
FT = T /r = 1765 / 0.125 = 14 120 N
∴ Total load due to power transmitted,
F = FT / cos φ = 14 120 / cos 20° = 15 026 N = 15.026 kN Ans.
12.13. Comparison Between Involute and Cycloidal Gears
In actual practice, the involute gears are more commonly used as compared to cycloidal
gears, due to the following advantages :
Advantages of involute gears
Following are the advantages of involute gears :
1. The most important advantage of the involute gears is that the centre distance for a pair of
involute gears can be varied within limits without changing the velocity ratio. This is not true for
cycloidal gears which requires exact centre distance to be maintained.
2. In involute gears, the pressure angle, from the start of the engagement of teeth to the end
of the engagement, remains constant. It is necessary for smooth running and less wear of gears. But in
cycloidal gears, the pressure angle is maximum at the beginning of engagement, reduces to zero at
pitch point, starts decreasing and again becomes maximum at the end of engagement. This results in
less smooth running of gears.
3. The face and flank of involute teeth are generated by a single curve where as in cycloidal
gears, double curves (i.e. epi-cycloid and hypo-cycloid) are required for the face and flank respectively. Thus the involute teeth are easy to manufacture than cycloidal teeth. In involute system, the
basic rack has straight teeth and the same can be cut with simple tools.
Note : The only disadvantage of the involute teeth is that the interference occurs (Refer Art. 12.19) with pinions
having smaller number of teeth. This may be avoided by altering the heights of addendum and dedendum of the
mating teeth or the angle of obliquity of the teeth.
Advantages of cycloidal gears
Following are the advantages of cycloidal gears :
1. Since the cycloidal teeth have wider flanks, therefore the cycloidal gears are stronger than
the involute gears, for the same pitch. Due to this reason, the cycloidal teeth are preferred specially
for cast teeth.
2. In cycloidal gears, the contact takes place between a convex flank and concave surface,
whereas in involute gears, the convex surfaces are in contact. This condition results in less wear in
cycloidal gears as compared to involute gears. However the difference in wear is negligible.
3. In cycloidal gears, the interference does not occur at all. Though there are advantages of
cycloidal gears but they are outweighed by the greater simplicity and flexibility of the involute
gears.
12.14. Systems of Gear Teeth
The following four systems of gear teeth are commonly used in practice :
1. 14 12 ° Composite system, 2. 14 12 ° Full depth involute system, 3. 20° Full depth involute
system, and 4. 20° Stub involute system.
The 14 12 ° composite system is used for general purpose gears. It is stronger but has no inter-
Chapter 12 : Toothed Gearing
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395
changeability. The tooth profile of this system has cycloidal curves at the top and bottom and involute
curve at the middle portion. The teeth are produced by formed milling cutters or hobs. The tooth
profile of the 14 1 ° full depth involute system was developed for use with gear hobs for spur and
2
helical gears.
The tooth profile of the 20° full depth involute system may be cut by hobs. The increase of
the pressure angle from 14 1 ° to 20° results in a stronger tooth, because the tooth acting as a beam is
2
wider at the base. The 20° stub involute system has a strong tooth to take heavy loads.
12.15. Standard Proportions of Gear Systems
The following table shows the standard proportions in module (m) for the four gear systems
as discussed in the previous article.
Table 12.1. Standard proportions of gear systems.
S. No.
Particulars
14 12 ° composite or full
20° full depth
20° stub involute
1.
2.
3.
4.
5.
6.
7.
Addenddm
Dedendum
Working depth
Minimum total depth
Tooth thickness
Minimum clearance
Fillet radius at root
depth involute system
1m
1.25 m
2m
2.25 m
1.5708 m
0.25 m
0.4 m
involute system
1m
1.25 m
2m
2.25 m
1.5708 m
0.25 m
0.4 m
system
0.8 m
1m
1.60 m
1.80 m
1.5708 m
0.2 m
0.4 m
12.16. Length of Path of Contact
Consider a pinion driving the wheel as shown in Fig. 12.11. When the pinion rotates in
clockwise direction, the contact between a pair of involute teeth begins at K (on the flank near the
base circle of pinion or the outer end of the tooth face on the wheel) and* ends at L (outer end of the
tooth face on the pinion or on the flank near the base circle of wheel). M N is the common normal at
the point of contacts and the common tangent to the base circles. The point K is the intersection of the
addendum circle of wheel and the common tangent. The point L is the intersection of the addendum
circle of pinion and common tangent.
Fig. 12.11. Length of path of contact.
*
If the wheel is made to act as a driver and the directions of motion are reversed, then the contact between
a pair of teeth begins at L and ends at K.
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We have discussed in Art. 12.4 that the length
of path of contact is the length of common normal cutoff by the addendum circles of the wheel and the pinion.
Thus the length of path of contact is KL which is the sum
of the parts of the path of contacts KP and PL. The part
of the path of contact KP is known as path of approach
and the part of the path of contact PL is known as path
of recess.
Let
rA = O1L = Radius of addendum
circle of pinion,
RA = O2K = Radius of addendum
circle of wheel,
Bevel gear
r = O1 P = Radius of pitch circle of
pinion, and
R = O2 P = Radius of pitch circle of
wheel.
From Fig. 12.11, we find that radius of the base circle of pinion,
O1M = O1P cos φ = r cos φ
and radius of the base circle of wheel,
O2N = O2P cos φ = R cos φ
Now from right angled triangle O2K N,
KN =
and
(O2 K )2 − (O2 N )2 =
( RA )2
− R2 cos2 φ
PN = O2 P sin φ = R sin φ
∴ Length of the part of the path of contact, or the path of approach,
KP = KN − PN =
( RA )2
− R2 cos 2 φ − R sin φ
Similarly from right angled triangle O1ML,
and
ML = (O1 L)2 − (O1M )2 = (rA )2 − r 2 cos2 φ
MP = O1 P sin φ = r sin φ
∴ Length of the part of the path of contact, or path of recess,
PL = ML − MP =
( rA ) 2 − r 2 cos 2 φ − r sin φ
∴ Length of the path of contact,
KL = KP + PL = ( RA ) 2 − R 2 cos 2 φ +
(rA ) 2 − r 2 cos 2 φ − ( R + r ) sin φ
12.17. Length of Arc of Contact
We have already defined that the arc of contact is the path traced by a point on the pitch circle
from the beginning to the end of engagement of a given pair of teeth. In Fig. 12.11, the arc of contact
is EPF or GPH. Considering the arc of contact GPH, it is divided into two parts i.e. arc GP and arc
PH. The arc GP is known as arc of approach and the arc PH is called arc of recess. The angles
subtended by these arcs at O1 are called angle of approach and angle of recess respectively.
Chapter 12 : Toothed Gearing
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397
We know that the length of the arc of approach (arc GP)
=
KP
Length of path of approach
=
cos φ
cos φ
and the length of the arc of recess (arc PH)
=
PL
Length of path of recess
=
cos φ
cos φ
Since the length of the arc of contact GPH is equal to the sum of the length of arc of approach
and arc of recess, therefore,
Length of the arc of contact
KP
PL
KL
+
=
cos φ cos φ cos φ
Length of path of contact
=
cos φ
= arc GP + arc PH =
12.18. Contact Ratio (or Number of Pairs of Teeth in Contact)
The contact ratio or the number of pairs of teeth in contact is defined as the ratio of the
length of the arc of contact to the circular pitch. Mathematically,
Contact ratio or number of pairs of teeth in contact
Length of the arc of contact
=
pc
where
pc = Circular pitch = π m, and
m = Module.
Notes : 1. The contact ratio, usually, is not a whole number. For example, if the contact ratio is 1.6, it does not
mean that there are 1.6 pairs of teeth in contact. It means that there are alternately one pair and two pairs of teeth
in contact and on a time basis the average is 1.6.
2. The theoretical minimum value for the contact ratio is one, that is there must always be at least one
pair of teeth in contact for continuous action.
3. Larger the contact ratio, more quietly the gears will operate.
Example 12.2. The number of teeth on each of the two equal spur gears in mesh are 40. The
teeth have 20° involute profile and the module is 6 mm. If the arc of contact is 1.75 times the circular
pitch, find the addendum.
Solution. Given : T = t = 40 ; φ = 20° ; m = 6 mm
We know that the circular pitch,
pc = π m = π × 6 = 18.85 mm
∴ Length of arc of contact
= 1.75 pc = 1.75 × 18.85 = 33 mm
and length of path of contact
= Length of arc of contact × cos φ = 33 cos 20° = 31 mm
Let
R A = rA = Radius of the addendum circle of each wheel.
We know that pitch circle radii of each wheel,
R = r = m.T / 2 = 6 × 40/2 = 120 mm
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and length of path of contact
31 = ( RA ) 2 − R 2 cos 2 φ +
( rA ) 2 − r 2 cos 2 φ − ( R + r ) sin φ
= 2 ( RA )2 − R 2 cos 2 φ − R sin φ ...(∵ R = r, and R A = rA)
31
= ( RA ) 2 − (120) 2 cos 2 20° − 120 sin 20°
2
15.5 =
( RA ) 2 − 12 715 − 41
(15.5 + 41)2 = ( RA )2 − 12 715
3192 + 12 715 = ( RA )2
or
RA = 126.12 mm
We know that the addendum of the wheel,
= RA − R = 126.12 − 120 = 6.12 mm Ans.
Example 12.3. A pinion having 30 teeth drives a
gear having 80 teeth. The profile of the gears is involute
with 20° pressure angle, 12 mm module and 10 mm
addendum. Find the length of path of contact, arc of contact
and the contact ratio.
Solution. Given : t = 30 ; T = 80 ; φ = 20° ;
m = 12 mm ; Addendum = 10 mm
Length of path of contact
We know that pitch circle radius of pinion,
r = m.t / 2 = 12 × 30 / 2 = 180 mm
and pitch circle radius of gear,
R = m.T / 2 = 12 × 80 / 2 = 480 mm
∴ Radius of addendum circle of pinion,
rA = r + Addendum = 180 + 10 = 190 mm
and radius of addendum circle of gear,
R A = R + Addendum = 480 + 10 = 490 mm
We know that length of the path of approach,
KP =
=
( RA ) 2 − R 2 cos 2 φ − R sin φ
Worm.
...(Refer Fig.12.11)
(490)2 − (480) 2 cos2 20° − 480 sin 20° = 191.5 − 164.2 = 27.3 mm
and length of the path of recess,
PL =
=
( rA ) 2 − r 2 cos 2 φ − r sin φ
(190) 2 − (180) 2 cos2 20° − 180 sin 20° = 86.6 − 61.6 = 25 mm
We know that length of path of contact,
KL = KP + PL = 27.3 + 25 = 52.3 mm Ans.
Chapter 12 : Toothed Gearing
l
399
Length of arc of contact
We know that length of arc of contact
=
Length of path of contact
52.3
=
= 55.66 mm Ans.
cos φ
cos 20°
Contact ratio
We know that circular pitch,
pc = π.m = π × 12 = 37.7 mm
∴
Contact ratio =
Length of arc of contact 55.66
=
= 1.5 say 2 Ans.
37.7
pc
Example 12.4. Two involute gears of 20° pressure angle are in mesh. The number of teeth
on pinion is 20 and the gear ratio is 2. If the pitch expressed in module is 5 mm and the pitch line
speed is 1.2 m/s, assuming addendum as standard and equal to one module, find :
1. The angle turned through by pinion when one pair of teeth is in mesh ; and
2. The maximum velocity of sliding.
Solution. Given : φ = 20° ; t = 20; G = T/t = 2; m = 5 mm ; v = 1.2 m/s ; addendum = 1 module
= 5 mm
1. Angle turned through by pinion when one pair of teeth is in mesh
We know that pitch circle radius of pinion,
r = m.t / 2 = 5 × 20 / 2 = 50 mm
and pitch circle radius of wheel,
... (3 T = G.t )
R = m.T / 2 = m.G.t / 2 = 2 × 20 × 5 / 2 = 100 mm
∴ Radius of addendum circle of pinion,
rA = r + Addendum = 50 + 5 = 55 mm
and radius of addendum circle of wheel,
R A = R + Addendum = 100 + 5 = 105 mm
We know that length of the path of approach (i.e. the path of contact when engagement
occurs),
KP =
=
( RA ) 2 − R 2 cos 2 φ − R sin φ
...(Refer Fig. 12.11)
(105) 2 − (100) 2 cos2 20° − 100 sin 20°
= 46.85 − 34.2 = 12.65 mm
and the length of path of recess (i.e. the path of contact when disengagement occurs),
PL =
=
( rA ) 2 − r 2 cos 2 φ − r sin φ
(55)2 − (50)2 cos 2 20° − 50 sin 20° = 28.6 − 17.1 = 11.5 mm
∴ Length of the path of contact,
KL = KP + PL = 12.65 + 11.5 = 24.15 mm
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Theory of Machines
and length of the arc of contact
=
Length of path of contact
24.15
=
= 25.7 mm
cos φ
cos 20°
We know that angle turned through by pinion
=
Length of arc of contact × 360° 25.7 × 360°
=
= 29.45° Ans.
Circumference of pinion
2π × 50
2. Maximum velocity of sliding
Let
ω1 = Angular speed of pinion, and
ω2 = Angular speed of wheel.
We know that pitch line speed,
v = ω1.r = ω2.R
∴
and
ω1 = v/r = 120/5 = 24 rad/s
ω2 = v/R = 120/10 = 12 rad/s
∴Maximum velocity of sliding,
v S = (ω1 + ω2) KP
...(3 KP > PL)
= (24 + 12) 12.65 = 455.4 mm/s Ans.
Example 12.5. A pair of gears, having 40 and 20 teeth respectively, are rotating in mesh,
the speed of the smaller being 2000 r.p.m. Determine the velocity of sliding between the gear teeth
faces at the point of engagement, at the pitch point, and at the point of disengagement if the smaller
gear is the driver. Assume that the gear teeth are 20° involute form, addendum length is 5 mm and the
module is 5 mm.
Also find the angle through which the pinion turns while any pairs of teeth are in contact.
Solution. Given : T = 40 ; t = 20 ; N1 = 2000 r.p.m. ; φ = 20° ; addendum = 5 mm ; m = 5 mm
We know that angular velocity of the smaller gear,
2 π N1 2 π × 2000
=
= 209.5 rad/s
60
60
and angular velocity of the larger gear,
20
t
ω2 = ω1 × = 209.5 ×
= 104.75 rad/s
40
T
Pitch circle radius of the smaller gear,
ω1 =
t
ω
... 3 2 =
ω
T
1
r = m.t / 2 = 5 × 20/2 = 50 mm
and pitch circle radius of the larger gear,
R = m.t / 2 = 5 × 40/2 = 100 mm
∴ Radius of addendum circle of smaller gear,
rA = r + Addendum = 50 + 5 = 55 mm
and radius of addendum circle of larger gear,
R A = R + Addendum = 100 + 5 = 105 mm
The engagement and disengagement of the gear teeth is shown in Fig. 12.11. The point K is
the point of engagement, P is the pitch point and L is the point of disengagement. M N is the common
tangent at the points of contact.
Chapter 12 : Toothed Gearing
l
401
We know that the distance of point of engagement K from the pitch point P or the length of
the path of approach,
KP =
( RA ) 2 − R 2 cos2 φ − R sin φ
=
(105) 2 − (100) 2 cos 2 20° − 100 sin 20°
= 46.85 − 34.2 = 12.65 mm
and the distance of the pitch point P from the point of disengagement L or the length of the path of
recess,
PL = (rA )2 − r 2 cos 2 φ − r sin φ
= (55)2 − (50) 2 cos 2 20° − 50 sin 20° = 28.6 − 17.1 = 11.5 mm
Velocity of sliding at the point of engagement
We know that velocity of sliding at the point of engagement K,
vSK = ( ω1 + ω2 ) KP = (209.5 + 104.75) 12.65 = 3975 mm/s Ans.
Velocity of sliding at the pitch point
Since the velocity of sliding is proportional to the distance of the contact point from the pitch
point, therefore the velocity of sliding at the pitch point is zero. Ans.
Velocity of sliding at the point of disengagement
We know that velocity of sliding at the point of disengagement L,
vSL = ( ω1 + ω2 ) PL = (209.5 + 104.75) 11.5 = 3614 mm/s Ans.
Angle through which the pinion turns
We know that length of the path of contact,
KL = KP + PL = 12.65 + 11.5 = 24.15 mm
KL
24.15
=
= 25.7 mm
cos φ cos 20°
Circumference of the smaller gear or pinion
and length of arc of contact =
= 2 π r = 2π × 50 = 314.2 mm
∴ Angle through which the pinion turns
= Length of arc of contact ×
360°
Circumference of pinion
360°
= 29.45 ° Ans.
314.2
Example 12.6. The following data relate to a pair of 20° involute gears in mesh :
= 25.7 ×
Module = 6 mm, Number of teeth on pinion = 17, Number of teeth on gear = 49 ; Addenda
on pinion and gear wheel = 1 module.
Find : 1. The number of pairs of teeth in contact ; 2. The angle turned through by the pinion
and the gear wheel when one pair of teeth is in contact, and 3. The ratio of sliding to rolling motion
when the tip of a tooth on the larger wheel (i) is just making contact, (ii) is just leaving contact with
its mating tooth, and (iii) is at the pitch point.
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Solution. Given : φ = 20° ; m = 6 mm ; t = 17 ; T = 49 ; Addenda on pinion and gear wheel
= 1 module = 6 mm
1. Number of pairs of teeth in contact
We know that pitch circle radius of pinion,
r = m.t / 2 = 6 × 17 / 2 = 51 mm
and pitch circle radius of gear,
r = m.T / 2 = 6 × 49 / 2 = 147 mm
∴ Radius of addendum circle of pinion,
rA = r + Addendum = 51 + 6 = 57 mm
and radius of addendum circle of gear,
R A = R + Addendum = 147 + 6 = 153 mm
Racks
We know that the length of path of approach (i.e. the path of contact when engagement
occurs),
KP = ( RA )2 − R 2 cos2 φ − R sin φ
...(Refer Fig. 12.11)
= (153)2 − (147)2 cos2 20° − 147 sin 20°
= 65.8 − 50.3 = 15.5 mm
and length of path of recess (i.e. the path of contact when disengagement occurs),
PL =
=
( rA ) 2 − r 2 cos 2 φ − r sin φ
(57)2 − (51)2 cos2 20° − 51 sin 20°
= 30.85 − 17.44 = 13.41 mm
∴ Length of path of contact,
KL = KP + PL = 15.5 + 13.41 = 28.91 mm
Chapter 12 : Toothed Gearing
l
403
Length of path of contact
28.91
=
= 30.8 mm
cos φ
cos 20°
We know that circular pitch,
and length of arc of contact =
pc = π.m = π × 6 = 18.852 mm
∴ Number of pairs of teeth in contact (or contact ratio)
=
Length of arc of contact
30.8
=
= 1.6 say 2 Ans.
Circular pitch
18.852
2. Angle turned through by the pinion and gear wheel when one pair of teeth is in contact
We know that angle turned through by the pinion
=
Length of arc of contact × 360° 30.8 × 360
=
= 34.6° Ans.
Circumference of pinion
2π × 51
and angle turned through by the gear wheel
=
Length of arc of contact × 360° 30.8 × 360
=
= 12° Ans.
Circumference of gear
2π × 147
3. Ratio of sliding to rolling motion
Let
ω1 = Angular velocity of pinion, and
ω2 = Angular velocity of gear wheel.
We know that ω1 / ω2 = T / t
or
ω2 = ω1 × t / T = ω1 × 17 / 49 = 0.347 ω1
vR = ω1 .r = ω2 .R = ω1 × 51 = 51 ω1 mm/s
and rolling velocity,
(i) At the instant when the tip of a tooth on the larger wheel is just making contact with its
mating teeth (i.e. when the engagement commences), the sliding velocity
vS = (ω1 + ω2 ) KP = (ω1 + 0.347 ω1) 15.5 = 20.88 ω1 mm/s
∴ Ratio of sliding velocity to rolling velocity,
vS
20.88 ω1
=
= 0.41 Ans.
51ω1
vR
(ii) At the instant when the tip of a tooth on the larger wheel is just leaving contact with its
mating teeth (i.e. when engagement terminates), the sliding velocity,
vS = (ω1 + ω2 ) PL = (ω1 + 0.347 ω1 ) 13.41 = 18.1 ω1 mm/s
∴ Ratio of sliding velocity to rolling velocity
vS 18.1 ω1
=
= 0.355 Ans.
51 ω1
vR
(iii) Since at the pitch point, the sliding velocity is zero, therefore the ratio of sliding velocity
to rolling velocity is zero. Ans.
Example 12.7. A pinion having 18 teeth engages with an internal gear having 72 teeth. If
the gears have involute profiled teeth with 20° pressure angle, module of 4 mm and the addenda on
pinion and gear are 8.5 mm and 3.5 mm respectively, find the length of path of contact.
Solution. Given : t = 18 ; T = 72 ; φ = 20° ; m = 4 mm ; Addendum on pinion = 8.5 mm ;
Addendum on gear = 3.5 mm
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Fig. 12.12 shows a pinion with centre O1, in mesh with internal gear of centre O2. It may be
noted that the internal gears have the addendum circle and the tooth faces inside the pitch circle.
We know that the length of path of contact is the length of the common tangent to the two
base circles cut by the addendum circles. From Fig. 12.12, we see that the addendum circles cut the
common tangents at points K and L. Therefore the length of path of contact is KL which is equal to the
sum of KP (i.e. path of approach) and PL (i.e. path of recess).
Fig. 12.12
We know that pitch circle radius of the pinion,
r = O1P = mt. / 2 = 4 × 18/ 2 = 36 mm
and pitch circle radius of the gear,
R = O2 P = mT
. / 2 = 4 × 72 / 2 = 144 mm
∴ Radius of addendum circle of the pinion,
rA = O1 L = O1 P + Addendum on pinion = 36 + 8.5 = 44.5 mm
and radius of addendum circle of the gear,
RA = O2 K = O2 P − Addendum on wheel = 144 – 3.5 = 140.5 mm
From Fig. 12.12, radius of the base circle of the pinion,
O1M = O1P cos φ = r cos φ = 36 cos 20° = 33.83 mm
and radius of the base circle of the gear,
O2 N = O2 P cos φ = R cos φ = 144 cos 20° = 135.32 mm
We know that length of the path of approach,
KP = PN − KN = O2 P sin 20° −
= 144 × 0.342 –
(O2 K ) 2 − (O2 N ) 2
(140.5) 2 − (135.32) 2 = 49.25 – 37.8 = 11.45 mm
Chapter 12 : Toothed Gearing
l
405
and length of the path of recess,
PL = ML − MP =
=
(O1 L) 2 − (O1 M )2 − O1 P sin 20°
(44.5) 2 − (33.83)2 − 36 × 0.342 = 28.9 − 12.3 = 16.6 mm
∴ Length of the path of contact,
KL = KP + PL = 11.45 + 16.6 = 28.05 mm Ans.
12.19. Interference in Involute Gears
Fig. 12.13 shows a pinion with centre O1, in mesh with wheel or gear with centre O2. M N is
the common tangent to the base circles and KL is the path of contact between the two mating teeth.
Fig. 12.13. Interference in involute gears.
A little consideration will show, that if the radius of the addendum circle of pinion is
increased to O1N, the point of contact L will move from L to N. When this radius is further increased,
the point of contact L will be on the inside of base circle of wheel and not on the involute profile of
tooth on wheel. The tip of tooth on the pinion will then undercut the tooth on the wheel at the root and
remove part of the involute profile of tooth on the wheel. This effect is known as interference, and
occurs when the teeth are being cut. In brief, the phenomenon when the tip of tooth undercuts the
root on its mating gear is known as interference.
Similarly, if the radius of the addendum circle of the wheel increases beyond O2M, then the
tip of tooth on wheel will cause interference with the tooth on pinion. The points M and N are called
interference points. Obviously, interference may be avoided if the path of contact does not extend
beyond interference points. The limiting value of the radius of the addendum circle of the pinion is
*O1N and of the wheel is O2M.
From the above discussion, we conclude that the interference may only be avoided, if the
point of contact between the two teeth is always on the involute profiles of both the teeth. In other
*
From Fig. 12.13, we see that
O1 N = (O1M )2 + ( MN )2 = (rb ) 2 + [r + R) sin φ]2
where
and
where
rb = Radius of base circle of pinion = O1P cos φ = r cos φ
O2 M = (O2 N )2 + ( MN )2 = ( Rb ) 2 + [r + R) sin φ]2
R b = Radius of base circle of wheel = O2P cos φ = R cos φ
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Theory of Machines
words, interference may only be prevented, if the addendum circles of the two mating gears cut the
common tangent to the base circles between the points of tangency.
When interference is just avoided, the maximum length of path of contact is M N when the
maximum addendum circles for pinion and wheel pass through the points of tangency N and M respectively as shown in Fig. 12.13. In such a case,
Maximum length of path of approach,
MP = r sin φ
and maximum length of path of recess,
PN = R sin φ
∴ Maximum length of path of contact,
MN = MP + PN = r sin φ + R sin φ = (r + R) sin φ
and maximum length of arc of contact
=
(r + R ) sin φ
= (r + R ) tan φ
cos φ
Note : In case the addenda on pinion and wheel is such that the path of approach and path of recess are half of
their maximum possible values, then
Path of approach,
or
( RA )2 − R 2 cos 2 φ − R sin φ =
and path of recess,
or
KP =
PL =
(rA )2 − r 2 cos2 φ − r sin φ =
1
2
MP
r sin φ
2
1
2
PN
R sin φ
2
∴ Length of the path of contact
= KP + PL =
1
2
MP +
1
2
PN =
(r + R ) sin φ
2
Example 12.8. Two mating gears have 20 and 40 involute teeth of module 10 mm and 20°
pressure angle. The addendum on each wheel is to be made of such a length that the line of contact
on each side of the pitch point has half the maximum possible length. Determine the addendum
height for each gear wheel, length of the path of contact, arc of contact and contact ratio.
Solution. Given : t = 20 ; T = 40 ; m = 10 mm ; φ = 20°
Addendum height for each gear wheel
We know that the pitch circle radius of the smaller gear wheel,
r = m.t / 2 = 10 × 20 / 2 = 100 mm
and pitch circle radius of the larger gear wheel,
R = m.T / 2 = 10 × 40 / 2 = 200 mm
Let
R A = Radius of addendum circle for the larger gear wheel, and
rA = Radius of addendum circle for the smaller gear wheel.
Since the addendum on each wheel is to be made of such a length that the line of contact on
each side of the pitch point (i.e. the path of approach and the path of recess) has half the maximum
possible length, therefore
Chapter 12 : Toothed Gearing
Path of approach,
or
or
KP =
( RA ) 2 − R 2 cos 2 φ − R sin φ =
1
2
MP
l
407
...(Refer Fig. 12.13)
r.sin φ
2
( RA ) 2 − (200) 2 cos 2 20° − 200 sin 20 ° =
100 × sin 20°
= 50 sin 20°
2
( RA )2 − 35 320 = 50 sin 20° + 200 sin 20° = 250 × 0.342 = 85.5
( RA )2 − 35 320 = (85.5) 2 = 7310
...(Squaring both sides)
(R A)2 = 7310 + 35 320 = 42 630
or
R A = 206.5 mm
∴ Addendum height for larger gear wheel
= RA − R = 206.5 − 200 = 6.5 mm Ans.
Now path of recess,
PL =
1
2
PN
R.sin φ
2
or
( rA ) 2 − r 2 cos 2 φ − r sin φ =
or
( rA ) 2 − (100) 2 cos 2 20° − 100 sin 20° =
200 sin 20°
= 100 sin 20°
2
(rA ) 2 − (100) 2 cos 2 20° = 100 sin 20° + 100 sin 20° = 200 × 0.342 = 68.4
(rA )2 − 8830 = (68.4)2 = 4680
(rA ) = 4680 + 8830 = 13 510
∴ Addendum height for smaller gear wheel
2
...(Squaring both sides)
or
rA = 116.2 mm
= rA − r = 116.2 − 100 = 6.2 mm Ans.
Length of the path of contact
We know that length of the path of contact
= KP + PL =
=
1
1
( r + R ) sin φ
MP + PN =
2
2
2
(100 + 200) sin 20°
= 51.3 mm Ans.
2
Length of the arc of contact
We know that length of the arc of contact
=
Length of the path of contact
51.3
=
= 54.6 mm Ans.
cos φ
cos 20°
Contact ratio
We know that circular pitch,
Pc = π m = π × 10 = 31.42 mm
∴
Contact ratio =
Length of the path of contact
54.6
=
= 1.74 say 2 Ans.
31.42
pc
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Theory of Machines
12.20. Minimum Number of Teeth on the Pinion in Order to Avoid
Interference
We have already discussed in the previous article that in order to avoid interference, the
addendum circles for the two mating gears must cut the common tangent to the base circles between
the points of tangency. The limiting condition reaches, when the addendum circles of pinion and
wheel pass through points N and M (see Fig. 12.13) respectively.
Let
t = Number of teeth on the pinion,,
T = Number of teeth on the wheel,
m = Module of the teeth,
r = Pitch circle radius of pinion = m.t / 2
G = Gear ratio = T / t = R / r
φ = Pressure angle or angle of obliquity.
From triangle O1NP,
(O1 N ) 2 = (O1 P) 2 + ( PN ) 2 − 2 × O1 P × PN cos O1 PN
= r 2 + R 2 sin 2 φ − 2r .R sin φ cos (90° + φ)
...(3 PN = O2 P sin φ = R sin φ)
= r 2 + R 2 sin 2 φ + 2r .R sin 2 φ
R 2 sin 2 φ 2 R sin 2 φ 2
R R
2
= r 2 1 +
+
= r 1 + + 2 sin φ
2
r
r r
r
∴ Limiting radius of the pinion addendum circle,
O1 N = r 1 +
Let
RR
m.t
2
+ 2 sin φ =
2
rr
1+
T
t
T
2
t + 2 sin φ
AP.m = Addendum of the pinion, where A P is a fraction by which the standard
addendum of one module for the pinion should be multiplied in order
to avoid interference.
We know that the addendum of the pinion
= O1N – O1P
∴
AP .m =
=
AP =
or
∴
m.t
2
1+
T T
m.t
2
+ 2 sin φ −
2
t t
...(3O1P = r = m.t / 2)
m.t
T T
2
1 + + 2 sin φ − 1
2
t t
t
T T
2
1 + + 2 sin φ − 1
2
t t
2 AP
t =
1+
T T
2
+ 2 sin φ − 1
t t
=
2 AP
1 + G (G + 2)sin 2 φ − 1
Chapter 12 : Toothed Gearing
l
409
This equation gives the minimum number of teeth required on the pinion in order to avoid
interference.
Notes : 1. If the pinion and wheel have equal teeth, then G = 1. Therefore the above equation reduces to
2 Ap
t=
1 + 3 sin 2 φ − 1
2. The minimum number of teeth on the pinion which will mesh with any gear (also rack) without
interference are given in the following table :
Table 12.2. Minimum number of teeth on the pinion
S. No.
System of gear teeth
Minimum number of teeth on the pinion
1.
14 12 ° Composite
12
2.
14 12 ° Full
32
3.
20° Full depth involute
18
4.
20° Stub involute
14
depth involute
12.21. Minimum Number of Teeth on the Wheel in Order to Avoid
Interference
Let
T = Minimum number of teeth required on the wheel in order to avoid
interference,
AW.m = Addendum of the wheel, where A W is a fraction by which the standard
addendum for the wheel should be multiplied.
and
Using the same notations as in Art. 12.20, we have from triangle O2MP
(O2 M )2 = (O2 P )2 + ( PM )2 − 2 × O2 P × PM cos O2 PM
= R 2 + r 2 sin 2 φ − 2 R.r sin φ cos (90° + φ)
...(3 PM = O1P sin φ = r )
=
R2
+
r2
sin2
φ+
2R.r sin2
φ
r 2 sin 2 φ 2 r sin 2 φ
2
= R 2 1 +
+
=R
2
R
R
∴ Limiting radius of wheel addendum circle,
O2 M = R 1 +
r r
m.T
2
+ 2 sin φ =
2
R R
r r
2
1 + R R + 2 sin φ
1+
tt
2
+ 2 sin φ
T T
We know that the addendum of the wheel
= O2M – O2P
∴
AW m =
=
or
AW =
m.T
t t
m.T
1 + + 2 sin 2 φ −
2
2
T T
m.T
tt
2
1 + + 2 sin φ − 1
2
T T
T
tt
2
1 + + 2 sin φ − 1
2
T T
...(3 O2 P = R = m.T / 2)
410
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∴
Theory of Machines
2 AW
T =
1+
tt
2
+ 2 sin φ − 1
T T
2 AW
=
11
2
+ 2 sin φ − 1
GG
1+
Notes : 1. From the above equation, we may also obtain the minimum number of teeth on pinion.
t
Multiplying both sides by ,
T
t
2 AW ×
t
T
T × =
T
11
1 + + 2 sin 2 φ − 1
GG
t=
2 AW
11
G 1 + + 2 sin 2 φ − 1
GG
2. If wheel and pinion have equal teeth, then G = 1, and
T =
2 AW
1 + 3 sin 2 φ − 1
Example 12.9. Determine the minimum number of teeth required on a pinion, in order to
avoid interference which is to gear with,
1. a wheel to give a gear ratio of 3 to 1 ; and 2. an equal wheel.
The pressure angle is 20° and a standard addendum of 1 module for the wheel may be
assumed.
Solution. Given : G = T / t = 3 ; φ = 20° ; A W = 1 module
1. Minimum number of teeth for a gear ratio of 3 : 1
We know that minimum number of teeth required on a pinion,
t =
=
2 × AW
1 1
G 1 + + 2 sin 2 φ − 1
GG
2×1
2
=
= 15.04 or 16 Ans.
0.133
11
2
3 1 + + 2 sin 20° − 1
33
2. Minimum number of teeth for equal wheel
We know that minimum number of teeth for equal wheel,
t =
2 × AW
1 + 3 sin φ − 1
2
=
2 ×1
1 + 3 sin 20° − 1
2
=
2
0.162
= 12.34 or 13 Ans.
Example 12.10. A pair of spur gears with involute teeth is to give a gear ratio of 4 : 1. The
arc of approach is not to be less than the circular pitch and smaller wheel is the driver. The angle of
pressure is 14.5°. Find : 1. the least number of teeth that can be used on each wheel, and 2. the
addendum of the wheel in terms of the circular pitch ?
Chapter 12 : Toothed Gearing
l
411
Solution. Given : G = T/t = R/r = 4 ; φ = 14.5°
1. Least number of teeth on each wheel
Let
t = Least number of teeth on the smaller wheel i.e. pinion,
T = Least number of teeth on the larger wheel i.e. gear, and
r = Pitch circle radius of the smaller wheel i.e. pinion.
We know that the maximum length of the arc of approach
=
Maximum length of the path of approach r sin φ
=
= r tan φ
cos φ
cos φ
2πr
t
Since the arc of approach is not to be less than the circular pitch, therefore
pc = π m =
and circular pitch,
2r
...3 m =
t
2πr
2π
2π
=
= 24.3 say 25 Ans.
or t =
t
tan φ tan 14.5°
...(3G = T / t)
T = G.t = 4 × 25 = 100 Ans.
r tan φ =
and
2. Addendum of the wheel
We know that addendum of the wheel
=
m.T
tt
2
1 + + 2 sin φ − 1
2
T T
=
m × 100
25 25
+ 2 sin 2 14.5° − 1
1+
2
100 100
= 50 m × 0.017 = 0.85 m = 0.85 × pc / π = 0.27 pc Ans.
...(3 m = pc / π)
Example 12.11. A pair of involute spur gears with 16° pressure angle and pitch of module
6 mm is in mesh. The number of teeth on pinion is 16 and its rotational speed is 240 r.p.m. When the
gear ratio is 1.75, find in order that the interference is just avoided ; 1. the addenda on pinion and
gear wheel ; 2. the length of path of contact ; and 3. the maximum velocity of sliding of teeth on either
side of the pitch point.
Solution. Given : φ = 16° ; m = 6 mm ; t = 16 ; N 1 = 240 r.p.m. or ω1 = 2π × 240/60
= 25.136 rad/s ; G = T / t = 1.75 or T = G.t = 1.75 × 16 = 28
1. Addenda on pinion and gear wheel
We know that addendum on pinion
m.t
T T
1+
2
tt
6 × 16
28
=
1+
2
16
=
+ 2 sin 2 φ − 1
28
+ 2 sin 2 16° − 1
16
= 48 (1.224 − 1) = 10.76 mm Ans.
and addendum on wheel
=
m.T
t t
2
1 + + 2 sin φ − 1
2
T T
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Theory of Machines
=
6 × 28
16 16
+ 2 sin 2 16° − 1
1+
2
28 28
= 84 (1.054 − 1) = 4.56 mm Ans.
2. Length of path of contact
We know that the pitch circle radius of wheel,
R = m.T / 2 = 6 × 28 / 2 = 84 mm
and pitch circle radius of pinion,
r = m.t / 2 = 6 × 16 / 2 = 48 mm
∴ Addendum circle radius of wheel,
RA = R + Addendum of wheel = 84 + 10.76 = 94.76 mm
and addendum circle radius of pinion,
rA = r + Addendum of pinion = 48 + 4.56 = 52.56 mm
We know that the length of path of approach,
KP = ( RA )2 − R 2 cos2 φ − R sin φ
...(Refer Fig. 12.11)
= (94.76)2 − (84) 2 cos 2 16° − 84 sin16°
= 49.6 − 23.15 = 26.45 mm
and the length of the path of recess,
PL =
=
( rA ) 2 − r 2 cos 2 φ − r sin φ
(52.56)2 − (48)2 cos2 16° − 48 sin16°
= 25.17 − 13.23 = 11.94 mm
∴ Length of the path of contact,
KL = KP + PL = 26.45 + 11.94 = 38.39 mm Ans.
3. Maximum velocity of sliding of teeth on either side of pitch point
Let
ω2 = Angular speed of gear wheel.
We know that
ω1 T
= = 1.75
ω2 t
or
ω2 =
ω1
25.136
=
= 14.28 rad/s
1.75
1.75
∴ Maximum velocity of sliding of teeth on the left side of pitch point i.e. at point K
= (ω1 + ω2 ) KP = (25.136 + 14.28) 26.45 = 1043 mm/s Ans.
and maximum velocity of sliding of teeth on the right side of pitch point i.e. at point L
= (ω1 + ω2 ) PL = (25.136 + 14.28) 11.94 = 471 mm/s Ans.
Example 12.12. A pair of 20° full depth involute spur gears having 30 and 50 teeth respectively of module 4 mm are in mesh. The smaller gear rotates at 1000 r.p.m. Determine : 1. sliding
velocities at engagement and at disengagement of pair of a teeth, and 2. contact ratio.
Solution. Given: φ = 20° ; t = 30 ; T = 50 ; m = 4 ; N 1 = 1000 r.p.m. or ω1 = 2π × 1000/60
= 104.7 rad/s
Chapter 12 : Toothed Gearing
l
413
1. Sliding velocities at engagement and at disengagement of pair of a teeth
First of all, let us find the radius of addendum circles of the smaller gear and the larger gear.
We know that
Addendum of the smaller gear,
=
=
m.t
T T
2
1 + + 2 sin φ − 1
2
t t
4 × 30
50 50
+ 2 sin 2 20° − 1
1+
2
30 30
= 60 (1.31 − 1) = 18.6 mm
and addendum of the larger gear,
m.T
t
1+
2
T
4 × 50
=
1+
2
=
t
2
+ 2 sin φ − 1
T
30 30
+ 2 sin 2 20° − 1
50 50
= 100(1.09 − 1) = 9 mm
Pitch circle radius of the smaller gear,
r = m.t / 2 = 4 × 30 / 2 = 60 mm
∴ Radius of addendum circle of the smaller gear,
rA = r + Addendum of the smaller gear = 60 + 18.6 = 78.6 mm
Pitch circle radius of the larger gear,
R = m.T / 2 = 4 × 50 / 2 = 100 mm
∴ Radius of addendum circle of the larger gear,
R A = R + Addendum of the larger gear = 100 + 9 = 109 mm
We know that the path of approach (i.e. path of contact when engagement occurs),
KP = ( RA )2 − R 2 cos2 φ − R sin φ
=
...(Refer Fig. 12.11)
(109) 2 − (100) 2 cos2 20° − 100 sin 20° = 55.2 − 34.2 = 21 mm
and the path of recess (i.e. path of contact when disengagement occurs),
PL =
=
( rA ) 2 − r 2 cos 2 φ − r sin φ
(78.6) 2 − (60) 2 cos2 20° – 60 sin 20° = 54.76 – 20.52 = 34.24 mm
Let
ω2 = Angular speed of the larger gear in rad/s.
We know that
ω1 T
=
ω2
t
or ω2 =
ω1 × t 10.47 × 30
=
= 62.82 rad/s
50
T
∴ Sliding velocity at engagement of a pair of teeth
= (ω1 + ω2 ) KP = (104.7 + 62.82) 21 = 3518 mm/s
= 3.518 m/s Ans.
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Theory of Machines
and sliding velocity at disengagement of a pair of teeth
= (ω1 + ω2 ) PL = (104.7 + 62.82)34.24 = 5736 mm/s
= 5.736 m/s Ans.
2. Contact ratio
We know that the length of the arc of contact
Length of the path of contact KP + PL 21 + 34.24
=
=
cos φ
cos φ
cos 20°
= 58.78 mm
=
and
Circular pitch
∴
= π × m = 3.142 × 4 = 12.568 mm
Contact ratio =
Length of arc of contact
58.78
=
= 4.67 say5 Ans.
Circular pitch
12.568
Example 12.13. Two gear wheels mesh externally and are to give a velocity ratio of 3 to 1.
The teeth are of involute form ; module = 6 mm, addendum = one module, pressure angle = 20°. The
pinion rotates at 90 r.p.m. Determine : 1. The number of teeth on the pinion to avoid interference on
it and the corresponding number of teeth on the wheel, 2. The length of path and arc of contact,
3.The number of pairs of teeth in contact, and 4. The maximum velocity of sliding.
Solution. Given : G = T / t = 3 ; m = 6 mm ; A P = A W = 1 module = 6 mm ; φ = 20° ;
N 1 = 90 r.p.m. or ω1 = 2π × 90 / 60 = 9.43 rad/s
1. Number of teeth on the pinion to avoid interference on it and the corresponding number of teeth
on the wheel
We know that number of teeth on the pinion to avoid interference,
t=
2 AP
1 + G (G + 2) sin φ − 1
2
=
2×6
1 + 3 (3 + 2) sin 2 20° − 1
= 18.2 say 19 Ans.
and corresponding number of teeth on the wheel,
T = G.t = 3 × 19 = 57 Ans.
2. Length of path and arc of contact
We know that pitch circle radius of pinion,
r = m.t / 2 = 6 × 19/2 = 57 mm
∴ Radius of addendum circle of pinion,
rA = r + Addendum on pinion (A P) = 57 + 6 = 63 mm
and pitch circle radius of wheel,
R = m.T / 2 = 6 × 57 / 2 = 171 mm
∴ Radius of addendum circle of wheel,
RA = R + Addendum on wheel ( AW ) = 171 + 6 = 177 mm
We know that the path of approach (i.e. path of contact when engagement occurs),
KP = ( RA )2 − R 2 cos2 φ − R sin φ
...(Refer Fig. 12.11)
= (177)2 − (171)2 cos 2 20° – 171 sin 20° = 74.2 – 58.5 = 15.7 mm
Chapter 12 : Toothed Gearing
l
415
and the path of recess (i.e. path of contact when disengagement occurs),
PL = (rA ) 2 − r 2 cos 2 φ − r sin φ
= (63)2 − (57)2 cos 2 20° − 57 sin 20° = 33.17 − 19.5 = 13.67 mm
∴ Length of path of contact,
KL = KP + PL = 15.7 + 13.67 = 29.37 mm Ans.
We know that length of arc of contact
=
Length of path of contact
29.37
=
= 31.25 mm Ans.
cos φ
cos 20°
3. Number of pairs of teeth in contact
We know that circular pitch,
pc = π × m = π × 6 = 18.852 mm
∴ Number of pairs of teeth in contact
Length of arc of contact
31.25
=
= 1.66 say 2 Ans.
18.852
pc
4. Maximum velocity of sliding
=
Let
We know that
ω2 = Angular speed of wheel in rad/s.
ω1 T
19
t
or ω2 = ω1 × = 9.43 ×
=
= 3.14 rad/s
57
t
T
ω2
∴ Maximum velocity of sliding,
vS = (ω1 + ω2 ) KP
...(3 KP > PL )
= (9.43 + 3.14) 15.7 = 197.35 mm/s Ans.
12.22. Minimum Number of Teeth on a Pinion for Involute Rack in Order to
Avoid Interference
A rack and pinion in mesh is shown in Fig. 12.14.
Fig. 12.14. Rack and pinion in mesh.
Let
t = Minimum number of teeth on the pinion,
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Theory of Machines
r = Pitch circle radius of the pinion = m.t / 2, and
φ = Pressure angle or angle of obliquity, and
A R.m = Addendum for rack, where A R is the fraction by which the standard
addendum of one module for the rack is to be multiplied.
We know that a rack is a part of toothed wheel of infinite diameter. Therefore its base circle
diameter and the profiles of the involute teeth are straight lines. Since these straight profiles are
tangential to the pinion profiles at the point of contact, therefore they are perpendicular to the tangent
PM. The point M is the interference point.
Addendum for rack,
AR .m = LH = PL sin φ
= (OP sin φ) sin φ = OP sin 2 φ
2
= r sin φ =
∴
t =
...(∵ PL = OP sin φ)
m.t
× sin 2 φ
2
2 AR
sin 2 φ
Example 12.14. A pinion of 20 involute teeth and 125 mm pitch circle diameter drives a
rack. The addendum of both pinion and rack is 6.25 mm. What is the least pressure angle which can
be used to avoid interference ? With this pressure angle, find the length of the arc of contact and the
minimum number of teeth in contact at a time.
Solution. Given : T = 20 ; d = 125 mm or r = OP = 62.5 mm ; LH = 6.25 mm
Least pressure angle to avoid interference
Let
φ = Least pressure angle to avoid interference.
We know that for no interference, rack addendum,
LH 6.25
=
= 0.1
6.25
r
LH = r sin 2 φ
or
sin 2 φ =
∴
sin φ = 0.3162
Length of the arc of contact
or
φ = 18.435° Ans.
We know that length of the path of contact,
KL =
=
(OK ) 2 − (OL) 2
...(Refer Fig. 12.14)
(OP + 6.25)2 − (OP cos φ)2
= (62.5 + 6.25)2 − (62.5 cos 18.435°)2
4726.56 − 3515.62 = 34.8 mm
∴ Length of the arc of contact
=
=
Length of the path of contact
34.8
=
= 36.68 mm Ans.
cos φ
cos18.435°
Minimum number of teeth
We know that circular pitch,
pc = πd / T = π × 125 / 20 = 19.64 mm
Chapter 12 : Toothed Gearing
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417
and the number of pairs of teeth in contact
=
Length of the arc of contact 36.68
=
= 1.87
Circular pitch ( pc )
19.64
∴ Minimum number of teeth in contact
= 2 or one pair Ans.
12.23. Helical Gears
A helical gear has teeth in the form
of helix around the gear. Two such gears
may be used to connect two parallel shafts
in place of spur gear. The helixes may be
right handed on one wheel and left handed
on the other. The pitch surfaces are cylindrical as in spur gearing, but the teeth instead of being parallel to the axis, wind
around the cylinders helically like screw
threads. The teeth of helical gears with
parallel axis have line contact, as in spur
Crossed helical gears.
gearing. This provides gradual engagement and continuous contact of the engaging teeth. Hence helical gears give smooth drive with a
high efficiency of transmission.
We have already discussed that the helical gears may be
of single helical type or double helical type. In case of single
helical gears, there is some axial thrust between the teeth, which
is a disadvantage. In order to eliminate this axial thrust, double
helical gears are used. It is equivalent to two single helical gears,
in which equal and opposite thrusts are produced on each gear
and the resulting axial thrust is zero.
The following definitions may be clearly understood in
connection with a helical gear as shown in Fig. 12.15.
Fig. 12.15. Helical gear.
1. Normal pitch. It is the distance between similar faces of adjacent teeth, along a helix on
the pitch cylinder normal to the teeth. It is denoted by pN.
2. Axial pitch. It is the distance measured parallel to the axis, between similar faces of adjacent teeth. It is the same as circular pitch and is therefore denoted by pc. If α is the helix angle, then
circular pitch,
p
pc = N
cos α
Note : The helix angle is also known as spiral angle of the teeth.
12.24. Spiral Gears
We have already discussed that spiral gears (also known as skew gears or screw gears) are
used to connect and transmit motion between two non-parallel and non-intersecting shafts. The pitch
surfaces of the spiral gears are cylindrical and the teeth have point contact. These gears are only
suitable for transmitting small power. We have seen that helical gears, connected on parallel shafts,
are of opposite hand. But spiral gears may be of the same hand or of opposite hand.
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Theory of Machines
12.25. Centre Distance for a Pair of Spiral Gears
The centre distance, for a pair of spiral gears,
is the shortest distance between the two shafts making
any angle between them. A pair of spiral gears 1 and 2,
both having left hand helixes (i.e. the gears are of the
same hand) is shown in Fig. 12.16. The shaft angle θ is
the angle through which one of the shafts must be rotated
so that it is parallel to the other shaft, also the two shafts
be rotating in opposite directions.
Let
α1 and α2 = Spiral angles of gear
teeth for gears 1 and 2
respectively,
pc1 and pc2 = Circular pitches of gears 1
and 2,
T1 and T 2 = Number of teeth on gears
1 and 2,
d1 and d2= Pitch circle diameters of
gears 1 and 2,
N1 and N 2 = Speed of gears 1 and 2,
T
N
G = Gear ratio = 2 = 1 ,
T1
N2
pN = Normal pitch, and
L = Least centre distance
between the axes of shafts.
Fig. 12.16. Centre distance for
a pair of spiral gears.
Since the normal pitch is same for both the spiral gears, therefore
pN
pN
, and pc 2 =
pc1 =
cos α1
cos α2
Helical gears
Chapter 12 : Toothed Gearing
We know that
and
∴
π d1
,
T1
π d2
,
=
T2
pc1 =
or
pc 2
or
L=
l
419
pc1 × T1
π
p × T2
d2 = c 2
π
d1 =
d1 + d 2 1 pc1 × T1
p × T2
=
+ c2
2
2
π
π
T2 T1 p N
pN
T1
pc1 + pc 2 × T =
cos α + cos α × G
π
2π
2
1
1
2
G
PN × T1 1
+
=
2 π cos α1 cos α 2
=
Notes : 1. If the pair of spiral gears have teeth of the same hand, then
θ = α1 + α 2
and for a pair of spiral gears of opposite hand,
θ = α1 – α2
2. When θ = 90°, then both the spiral gears must have teeth of the same hand.
12.26. Efficiency of Spiral Gears
A pair of spiral gears 1 and 2 in mesh is shown in Fig. 12.17. Let the gear 1 be the driver and
the gear 2 the driven. The forces acting on each of a pair of teeth in contact are shown in Fig. 12.17.
The forces are assumed to act at the centre of the width of each teeth and in the plane tangential to the
pitch cylinders.
Fig. 12.17. Efficiency of spiral gears.
Let
F1 = Force applied tangentially on the driver,
F2 = Resisting force acting tangentially on the driven,
Fa1 = Axial or end thrust on the driver,
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Fa2 = Axial or end thrust on the driven,
R N = Normal reaction at the point of contact,
φ = Angle of friction,
R = Resultant reaction at the point of contact, and
θ = Shaft angle = α1+ α2
...(3 Both gears are of the same hand)
From triangle OPQ, F1 = R cos (α1 – φ)
∴ Work input to the driver
= F1× π d1.N1 = R cos (α1 – φ) π d1.N1
From triangle OST, F2 = R cos (α2 + φ)
∴ Work output of the driven
= F2 × π d2.N 2 = R cos (α2 + φ) π d2.N 2
∴ Efficiency of spiral gears,
η=
=
Work output R cos (α 2 + φ) π d 2 . N 2
=
Work input
R cos (α1 − φ) π d1 . N1
cos (α 2 + φ) d2 . N 2
cos (α1 − φ) d1. N1
...(i)
We have discussed in Art. 12.25, that pitch circle diameter of gear 1,
p × T1
PN
T
=
× 1
d1 = c1
cos α1 π
π
and pitch circle diameter of gear 2,
d2 =
pc 2 × T2
PN
T
=
× 2
cos α 2
π
π
∴
d 2 T2 cos α1
=
d1 T1 cos α 2
...(ii)
We know that
N2 T1
=
N1 T2
...(iii)
Multiplying equations (ii) and (iii), we get,
d 2 . N 2 cos α1
=
cos α 2
d1. N1
Substituting this value in equation (i), we have
η=
=
cos (α 2 + φ) cos α1
cos (α1 − φ) cos α 2
...(iv)
cos (α1 + α 2 + φ) + cos (α1 − α 2 − φ)
cos (α 2 + α1 − φ) + cos (α 2 − α1 + φ)
1
... 3 cos A cos B = [cos ( A + B) + cos ( A − B)]
2
Chapter 12 : Toothed Gearing
=
cos (θ + φ) + cos (α1 − α 2 − φ)
cos (θ − φ) + cos (α 2 − α1 + φ)
l
421
...(v)
...(3θ = α1 + α 2 )
Since the angles θ and φ are constants, therefore the efficiency will be maximum, when
cos (α1 – α2 – φ) is maximum, i.e.
cos (α1 – α2 – φ) = 1
∴
α1 = α2 + φ
or
α1 – α2 – φ = 0
and
α2 = α1 – φ
Since α1 + α2 = θ, therefore
α1 = θ − α 2 = θ − α1 + φ or α1 =
θ+φ
2
θ−φ
2
Substituting α1 = α2 + φ and α2 = α1 – φ, in equation (v), we get
α2 =
Similarly,
ηmax =
Note: From Fig. 12.17, we find that
RN =
cos (θ + φ) + 1
cos (θ − φ) + 1
F1
cos α1
=
...(vi)
F2
cos α 2
∴ Axial thrust on the driver, Fa1 = R N.sin α1 = F1.tan α1
and axial thrust on the driven,
Fa2 = R N.sin α2 = F2.tan α2
Example 12.15. A pair of spiral gears is required to connect two shafts 175 mm apart, the
shaft angle being 70°. The velocity ratio is to be 1.5 to 1, the faster wheel having 80 teeth and a pitch
circle diameter of 100 mm. Find the spiral angles for each wheel. If the torque on the faster wheel is
75 N-m ; find the axial thrust on each shaft, neglecting friction.
Solution. Given : L = 175 mm = 0.175 m ; θ = 70° ; G = 1.5 ; T 2 = 80 ; d2 = 100 mm = 0.1 m
or r2 = 0.05 m ; Torque on faster wheel = 75 N-m
Spiral angles for each wheel
Let
α1 = Spiral angle for slower wheel, and
α2 = Spiral angle for faster wheel.
We know that velocity ratio, G =
N 2 T1
=
= 1.5
N1 T2
∴ No. of teeth on slower wheel,
T 1 = T 2 × 1.5 = 80 × 1.5 = 120
We also know that the centre distance between shafts (L),
d1 + d2 d1 + 0.1
=
2
2
d1 = 2 × 0.175 – 0.1 = 0.25 m
0.175 =
∴
and
0.1
80 cos α1
2 cos α1
d 2 T2 cos α1
or
=
=
=
0.25 120 cos α 2 3 cos α 2
d1 T1 cos α 2
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Theory of Machines
∴
cos α1
0.1 × 3
=
= 0.6
cos α 2 0.25 × 2
We know that,
α1 + α2 = θ = 70° or α2 = 70° – α1
or
cos α1 = 0.6 cos α 2
...(i)
Substituting the value of α2 in equation (i),
cos α1 = 0.6 cos (70° – α1) = 0.6 (cos 70° cos α1 + sin 70° sin α1)
...[3 cos ( A − B ) = cos A cos B + sin A sin B ]
= 0.2052 cos α1 + 0.5638 sin α1
cos α1 – 0.2052 cos α1 = 0.5638 sin α1
0.7948 cos α1 = 0.5638 sin α1
∴
tan α1 =
sin α1 0.7948
=
= 1.4097
cos α1 0.5638
or α1 = 54.65°
α 2 = 70° − 54.65° = 15.35° Ans.
and
Axial thrust on each shaft
We know that
Torque = Tangential force × Pitch circle radius
∴ Tangential force at faster wheel,
Torque on the faster wheel
75
=
= 1500 N
Pitch circle radius (r2 )
0.05
and normal reaction at the point of contact,
F2 =
R N = F2 / cos α2 = 1500/cos 15.35° = 1556 N
We know that axial thrust on the shaft of slower wheel,
Fa1 = R N. sin α1 = 1556 × sin 54.65° = 1269 N Ans.
and axial thrust on the shaft of faster wheel,
Fa2 = R N. sin α2 = 1556 × sin 15.35° = 412 N Ans.
Example 12.16. In a spiral gear drive connecting two shafts, the approximate centre
distance is 400 mm and the speed ratio = 3. The angle between the two shafts is 50° and the normal
pitch is 18 mm. The spiral angle for the driving and driven wheels are equal. Find : 1. Number of
teeth on each wheel, 2. Exact centre distance, and 3. Efficiency of the drive, if friction angle = 6°.
Solution. Given : L = 400 mm = 0.4 m ; G = T 2 / T 1 = 3 ; θ = 50° ; pN = 18 mm ; φ = 6°
1. Number of teeth on each wheel
Let
T 1 = Number of teeth on wheel 1 (i.e. driver), and
T 2 = Number of teeth on wheel 2 (i.e. driven).
Since the spiral angle α1 for the driving wheel is equal to the spiral angle α2 for the driven
wheel, therefore
α1 = α2 = θ/2 = 25°
...(3 α1 + α2 = θ = 50°)
We know that centre distance between two shafts (L),
400 =
G pN .T1 1 + G
pN. T1 1
+
=
...(3 α1 = α 2 )
2 π cos α1 cos α 2
2 π cos α1
Chapter 12 : Toothed Gearing
=
l
423
18 × T1 1 + 3
= 12.64 T1
2π cos 25°
∴
T 1 = 400/12.64 = 31.64 or 32 Ans.
and
T 2 = G.T1 = 3 × 32 = 96 Ans.
2. Exact centre distance
We know that exact centre distance,
L1 =
=
G pN .T1 1 + G
pN .T1 1
cos α + cos α =
2π
2π cos α1
1
2
...(3 α1 = α 2 )
18 × 32 1 + 3
= 404.5 mm Ans.
2π cos 25 °
3. Efficiency of the drive
We know that efficiency of the drive,
η=
cos (α 2 + φ) cos α1 cos (α1 + φ)
=
cos (α1 − φ) cos α 2
cos (α1 − φ)
...(3 α1 = α 2 )
cos (25° + 6°) cos 31° 0.8572
=
=
= 0.907 = 90.7% Ans.
cos (25° − 6°) cos 19° 0.9455
Example 12.17. A drive on a machine tool is to be made by two spiral gear wheels, the
spirals of which are of the same hand and has normal pitch of 12.5 mm. The wheels are of equal
diameter and the centre distance between the axes of the shafts is approximately 134 mm. The angle
between the shafts is 80° and the speed ratio 1.25. Determine : 1. the spiral angle of each wheel,
2. the number of teeth on each wheel, 3. the efficiency of the drive, if the friction angle is 6°, and
4. the maximum efficiency.
=
Solution. Given : pN = 12.5 mm ; L = 134 mm ; θ = 80° ; G = N 2 / N 1 = T 1 / T 2 = 1.25
1. Spiral angle of each wheel
Let
α1 and α2 = Spiral angles of wheels 1 and 2 respectively, and
d1 and d2 = Pitch circle diameter of wheels 1 and 2 respectively.
We know that
d 2 T2 cos α1
=
d1 T1 cos α 2
cos α1 T1
=
= 1.25
cos α 2 T2
We also know that
∴
T1 cos α 2 = T2 cos α1
or
...(3 d1 = d 2 )
cos α1 = 1.25 cos α 2
or
...(i)
α1 + α 2 = θ = 80° or α 2 = 80° − α1
Substituting the value of α2 in equation (i),
cos α1 = 1.25 cos (80° – α1) = 1.25 (cos 80° cos α1 + sin 80° sin α1)
= 1.25 (0.1736 cos α1 + 0.9848 sin α1)
= 0.217 cos α1 + 1.231 sin α1
cos α1 – 0.217 cos α1 = 1.231 sin α1
∴
and
or
0.783 cos α1 = 1.231 sin α1
tan α1 = sin α1 / cos α1 = 0.783 / 1.231 = 0.636
α2 = 80° – 32.46° = 47.54° Ans.
or α1 = 32.46° Ans.
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Theory of Machines
2. Number of teeth on each wheel
Let
T 1 = Number of teeth on wheel 1, and
T 2 = Number of teeth on wheel 2.
We know that centre distance between the two shafts (L),
d1 + d 2
or
2
p .T
p .T
d1 = c1 1 = N 1
π
π cos α1
134 =
We know that
∴
and
d1 = d 2 = 134 mm
...(3 d1 = d2 )
T1 =
πd1.cos α1 π × 134 × cos 32.46°
=
= 28.4 or 30 Ans.
12.5
pN
T2 =
T1
30
=
= 24 Ans.
1.25 1.25
3. Efficiency of the drive
We know that efficiency of the drive,
η=
=
cos (α 2 + φ) cos α1 cos (47.54° + 6°) cos 32.46°
=
cos (α1 − φ) cos α 2 cos (32.46° − 6°) cos 47.54°
0.5943 × 0.8437
= 0.83 or 83% Ans.
0.8952 × 0.6751
4. Maximum efficiency
We know that maximum efficiency,
ηmax =
cos (θ + φ) + 1 cos (80° + 6°) + 1 1.0698
=
=
cos (θ − φ) + 1 cos (80° − 6°) + 1 1.2756
= 0.838 or 83.8% Ans.
EXERCISES
1.
2.
3.
4.
5.
The pitch circle diameter of the smaller of the two spur wheels which mesh externally and have
involute teeth is 100 mm. The number of teeth are 16 and 32. The pressure angle is 20° and the
addendum is 0.32 of the circular pitch. Find the length of the path of contact of the pair of teeth.
[Ans. 29.36 mm]
A pair of gears, having 40 and 30 teeth respectively are of 25° involute form. The addendum length is
5 mm and the module pitch is 2.5 mm. If the smaller wheel is the driver and rotates at 1500 r.p.m., find
the velocity of sliding at the point of engagement and at the point of disengagement.
[Ans. 2.8 m/s ; 2.66 m/s]
Two gears of module 4mm have 24 and 33 teeth. The pressure angle is 20° and each gear has a
standard addendum of one module. Find the length of arc of contact and the maximum velocity of
sliding if the pinion rotates at 120 r.p.m.
[Ans. 20.58 mm ; 0.2147 m/s]
The number of teeth in gears 1 and 2 are 60 and 40 ; module = 3 mm ; pressure angle = 20° and
addendum = 0.318 of the circular pitch. Determine the velocity of sliding when the contact is at the tip
of the teeth of gear 2 and the gear 2 rotates at 800 r.p.m.
[Ans. 1.06 m/s]
Two spur gears of 24 teeth and 36 teeth of 8 mm module and 20° pressure angle are in mesh. Addendum of each gear is 7.5 mm. The teeth are of involute form. Determine : 1. the angle through which the
pinion turns while any pair of teeth are in contact, and 2. the velocity of sliding between the teeth
when the contact on the pinion is at a radius of 102 mm. The speed of the pinion is 450 r.p.m.
[Ans. 20.36°, 1.16 m/s]
Chapter 12 : Toothed Gearing
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
l
425
A pinion having 20 involute teeth of module pitch 6 mm rotates at 200 r.p.m. and transmits 1.5 kW to
a gear wheel having 50 teeth. The addendum on both the wheels is 1/4 of the circular pitch. The angle
of obliquity is 20°. Find (a) the length of the path of approach ; (b) the length of the arc of approach;
(c) the normal force between the teeth at an instant where there is only pair of teeth in contact.
[Ans. 13.27 mm ; 14.12 mm ; 1193 N]
Two mating involute spur gear of 20° pressure angle have a gear ratio of 2. The number of teeth on the
pinion is 20 and its speed is 250 r.p.m. The module pitch of the teeth is 12 mm.
If the addendum on each wheel is such that the path of approach and the path of recess on each side are
half the maximum possible length, find : 1. the addendum for pinion and gear wheel ; 2. the length of
the arc of contact ; and 3. the maximum velocity of sliding during approach and recess.
Assume pinion to be the driver.
[Ans. 19.5 mm, 7.8 mm ; 65.5 mm ; 807.5 mm/s, 1615 mm/s]
Two mating gears have 20 and 40 involute teeth of module 10 mm and 20° pressure angle. If the
addendum on each wheel is such that the path of contact is maximum and interference is just avoided,
find the addendum for each gear wheel, path of contact, arc of contact and contact ratio.
[Ans. 14 mm ; 39 mm ; 102.6 mm ; 109.3 mm ; 4]
A 20° involute pinion with 20 teeth drives a gear having 60 teeth. Module is 8 mm and addendum of
each gear is 10 mm.
1. State whether interference occurs or not. Give reasons.
2. Find the length of path of approach and arc of approach if pinion is the driver.
[Ans. Interference does not occur ; 25.8 mm, 27.45 mm]
A pair of spur wheels with involute teeth is to give a gear ratio of 3 to 1. The arc of approach is not to
be less than the circular pitch and the smaller wheel is the driver. The pressure angle is 20°. What is
the least number of teeth that can be used on each wheel ? What is the addendum of the wheel in terms
of the circular pitch ?
[Ans. 18, 54 ; 0.382 Pc]
Two gear wheels mesh externally and are to give a velocity ratio of 3. The teeth are of involute form
of module 6. The standard addendum is 1 module. If the pressure angle is 18° and pinion rotates at 90
r.p.m., find : 1. the number of teeth on each wheel, so that the interference is just avoided, 2. the length
of the path of contact, and 3. the maximum velocity of sliding between the teeth.
[Ans. 19, 57 ; 31.5 mm ; 213.7 mm/s]
A pinion with 24 involute teeth of 150 mm of pitch circle diameter drives a rack. The addendum of the
pinion and rack is 6 mm. Find the least pressure angle which can be used if under cutting of the teeth
is to be avoided. Using this pressure angle, find the length of the arc of contact and the minimum
number of teeth in contact at one time.
[Ans. 16.8° ; 40 mm ; 2 pairs of teeth]
Two shafts, inclined at an angle of 65° and with a least distance between them of 175 mm are to be
connected by spiral gears of normal pitch 15 mm to give a reduction ratio 3 : 1. Find suitable diameters and numbers of teeth. Determine, also, the efficiency if the spiral angles are determined by the
condition of maximum efficiency. The friction angle is 7°.
[Ans. 88.5 mm ; 245.7 mm ; 15, 45 ; 85.5 %]
A spiral wheel reduction gear, of ratio 3 to 2, is to be used on a machine, with the angle between the
shafts 80°. The approximate centre distance between the shafts is 125 mm. The normal pitch of the
teeth is 10 mm and the wheel diameters are equal. Find the number of teeth on each wheel, pitch circle
diameters and spiral angles. Find the efficiency of the drive if the friction angle is 5°.
[Ans. 24, 36 ; 128 mm ; 53.4°, 26.6° ; 85.5 %]
A right angled drive on a machine is to be made by two spiral wheels. The wheels are of equal
diameter with a normal pitch of 10 mm and the centre distance is approximately 150 mm. If the speed
ratio is 2.5 to 1, find : 1. the spiral angles of the teeth, 2. the number of teeth on each wheel, 3.the exact
centre distance, and 4. transmission efficiency, if the friction angle is 6°.
[Ans. 21.8°, 68.2° ; 18 , 45 ; 154 mm ; 75.8 %]
DO YOU KNOW ?
1.
2.
3.
Explain the terms : (i) Module, (ii) Pressure angle, and (iii) Addendum.
State and prove the law of gearing. Show that involute profile satisfies the conditions for correct
gearing.
Derive an expression for the velocity of sliding between a pair of involute teeth. State the advantages
of involute profile as a gear tooth profile.
426
4.
5.
6.
7.
8.
9.
10.
11.
12.
l
Theory of Machines
Prove that the velocity of sliding is proportional to the distance of the point of contact from the pitch
point.
Prove that for two involute gear wheels in mesh, the angular velocity ratio does not change if the
centre distance is increased within limits, but the pressure angle increases.
Derive an expression for the length of the arc of contact in a pair of meshed spur gears.
What do you understand by the term ‘interference’ as applied to gears?
Derive an expression for the minimum number of teeth required on the pinion in order to avoid
interference in involute gear teeth when it meshes with wheel.
Derive an expression for minimum number of teeth required on a pinion to avoid interference when it
gears with a rack.
Define (i) normal pitch, and (ii) axial pitch relating to helical gears.
Derive an expression for the centre distance of a pair of spiral gears.
Show that, in a pair of spiral gears connecting inclined shafts, the efficiency is maximum when the
spiral angle of the driving wheel is half the sum of the shaft and friction angles.
OBJECTIVE TYPE QUESTIONS
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
The two parallel and coplanar shafts are connected by gears having teeth parallel to the axis of the
shaft. This arrangement is called
(a) spur gearing
(b) helical gearing
(c) bevel gearing (d) spiral gearing
The type of gears used to connect two non-parallel non-intersecting shafts are
(a) spur gears
(b) helical gears
(c) spiral gears
(d) none of these
An imaginary circle which by pure rolling action, gives the same motion as the actual gear, is called
(a) addendum circle (b) dedendum circle (c) pitch circle
(d) clearance circle
The size of a gear is usually specified by
(a) pressure angle
(b) circular pitch
(c) diametral pitch (d) pitch circle diameter
The radial distance of a tooth from the pitch circle to the bottom of the tooth, is called
(a) dedendum
(b) addendum
(c) clearance
(d) working depth
The product of the diametral pitch and circular pitch is equal to
(a) 1
(b) 1/π
(c) π
(d) 2π
The module is the reciprocal of
(a) diametral pitch
(b) circular pitch
(c) pitch diameter (d) none of these
Which is the incorrect relationship of gears?
(a) Circular pitch × Diametral pitch = π
(b) Module = P.C.D/No.of teeth
(c) Dedendum = 1.157 module
(d) Addendum = 2.157 module
If the module of a gear be m, the number of teeth T and pitch circle diameter D, then
(a) m = D/T
(b) D = T/m
(c) m = D/2T
(d) none of these
Mitre gears are used for
(a) great speed reduction
(b) equal speed
(c) minimum axial thrust
(d) minimum backlash
The condition of correct gearing is
(a) pitch line velocities of teeth be same
(b) radius of curvature of two profiles be same
(c) common normal to the pitch surface cuts the line of centres at a fixed point
(d) none of the above
Law of gearing is satisfied if
(a) two surfaces slide smoothly
(b) common normal at the point of contact passes through the pitch point on the line joining the
centres of rotation
(c) number of teeth = P.C.D. / module
(d) addendum is greater than dedendum
Chapter 12 : Toothed Gearing
13.
l
427
Involute profile is preferred to cyloidal because
(a) the profile is easy to cut
(b) only one curve is required to cut
(c) the rack has straight line profile and hence can be cut accurately
(d) none of the above
The contact ratio for gears is
(a) zero
(b) less than one
(c) greater than one
The maximum length of arc of contact for two mating gears, in order to avoid interference, is
(a) (r + R) sin φ
(b) (r + R) cos φ
(c) (r + R) tan φ
(d) none of these
where r = Pitch circle radius of pinion,
R = Pitch circle radius of driver, and
φ = Pressure angle.
When the addenda on pinion and wheel is such that the path of approach and path of recess are half of
their maximum possible values, then the length of the path of contact is given by
14.
15.
16.
( r + R) sin φ
( r + R) cos φ
( r + R) tan φ
(b)
(c)
(d) none of these
2
2
2
Interference can be avoided in involute gears with 20° pressure angle by
(a) cutting involute correctly
(b) using as small number of teeth as possible
(c) using more than 20 teeth
(d) using more than 8 teeth
The ratio of face width to transverse pitch of a helical gear with α as the helix angle is normally
(a) more than 1.15/tan α
(b) more than 1.05/tan α
(c) more than 1/tan α
(d) none of these
The maximum efficiency for spiral gears is
(a)
17.
18.
19.
(a)
sin (θ + φ) + 1
cos (θ − φ) + 1
cos (θ + φ) + 1
(c) cos (θ − φ) + 1
(b)
cos (θ − φ) + 1
sin (θ + φ) + 1
(d)
cos (θ − φ) + 1
cos (θ + φ) + 1
where
θ = Shaft angle, and φ = Friction angle.
For a speed ratio of 100, smallest gear box is obtained by using
(a) a pair of spur gears
(b) a pair of helical and a pair of spur gear compounded
(c) a pair of bevel and a pair of spur gear compounded
(d) a pair of helical and a pair of worm gear compounded
20.
ANSWERS
1. (a)
2. (c)
3. (c)
4. (d)
6. (c)
11. (c)
7. (a)
12. (b)
8. (d)
13. (b)
9. (a)
14. (c)
10.
15.
(b)
(c)
16.
17.
18.
19.
20.
(d)
(a)
(c)
(a)
(c)
5. (a)
GO To FIRST
CONTENTS
CONTENTS
428
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Theory of Machines
13
Fea
tur
es
eatur
tures
1. Introduction.
Gear Trains
2. Types of Gear Trains.
3. Simple Gear Train.
4. Compound Gear Train.
5. Design of Spur Gears.
6. Reverted Gear Train.
7. Epicyclic Gear Train.
8. Velocity Ratio of Epicyclic
Gear Train.
9. Compound Epicyclic Gear
Train (Sun and Planet
Wheel).
13.1. Intr
oduction
Introduction
Sometimes, two or more gears are made to mesh with
each other to transmit power from one shaft to another. Such
a combination is called gear train or train of toothed wheels.
The nature of the train used depends upon the velocity ratio
required and the relative position of the axes of shafts. A
gear train may consist of spur, bevel or spiral gears.
13.2. Types of Gear Trains
10. Epicyclic Gear Train With
Bevel Gears.
Following are the different types of gear trains, depending upon the arrangement of wheels :
11. Torques in Epicyclic Gear
Trains.
1. Simple gear train, 2. Compound gear train, 3. Reverted gear train, and 4. Epicyclic gear train.
In the first three types of gear trains, the axes of the
shafts over which the gears are mounted are fixed relative to
each other. But in case of epicyclic gear trains, the axes of
the shafts on which the gears are mounted may move relative
to a fixed axis.
13.3. Simple Gear Train
When there is only one gear on each shaft, as shown
in Fig. 13.1, it is known as simple gear train. The gears are
represented by their pitch circles.
When the distance between the two shafts is small,
the two gears 1 and 2 are made to mesh with each other to
428
CONTENTS
CONTENTS
Chapter 13 : Gear Trains
l
429
transmit motion from one shaft to the other, as shown in Fig. 13.1 (a). Since the gear 1 drives the gear
2, therefore gear 1 is called the driver and the gear 2 is called the driven or follower. It may be noted
that the motion of the driven gear is opposite to the motion of driving gear.
(a)
(b)
(c)
Fig. 13.1. Simple gear train.
Let
N 1 = Speed of gear 1(or driver) in r.p.m.,
N 2 = Speed of gear 2 (or driven or follower) in r.p.m.,
T 1 = Number of teeth on gear 1, and
T 2 = Number of teeth on gear 2.
Since the speed ratio (or velocity ratio) of gear train is the ratio of the speed of the driver to
the speed of the driven or follower and ratio of speeds of any pair of gears in mesh is the inverse of
their number of teeth, therefore
Speed ratio =
N1 T2
=
N 2 T1
It may be noted that ratio of the speed of the driven or follower to the speed of the driver is
known as train value of the gear train. Mathematically,
Train value =
N 2 T1
=
N1 T2
From above, we see that the train value is the reciprocal of speed ratio.
Sometimes, the distance between the two gears is large. The motion from one gear to another,
in such a case, may be transmitted by either of the following two methods :
1. By providing the large sized gear, or 2. By providing one or more intermediate gears.
A little consideration will show that the former method (i.e. providing large sized gears) is
very inconvenient and uneconomical method ; whereas the latter method (i.e. providing one or more
intermediate gear) is very convenient and economical.
It may be noted that when the number of intermediate gears are odd, the motion of both the
gears (i.e. driver and driven or follower) is like as shown in Fig. 13.1 (b).
But if the number of intermediate gears are even, the motion of the driven or follower will be
in the opposite direction of the driver as shown in Fig. 13.1 (c).
Now consider a simple train of gears with one intermediate gear as shown in Fig. 13.1 (b).
Let
N 1 = Speed of driver in r.p.m.,
N 2 = Speed of intermediate gear in r.p.m.,
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Theory of Machines
N 3 = Speed of driven or follower in r.p.m.,
T 1 = Number of teeth on driver,
T 2 = Number of teeth on intermediate gear, and
T 3 = Number of teeth on driven or follower.
Since the driving gear 1 is in mesh with the intermediate gear 2, therefore speed ratio for
these two gears is
N1
T
= 2
N2
T1
...(i)
Similarly, as the intermediate gear 2 is in mesh with the driven gear 3, therefore speed ratio
for these two gears is
T
N2
...(ii)
= 3
N3
T2
The speed ratio of the gear train as shown in Fig. 13.1 (b) is obtained by multiplying the
equations (i) and (ii).
∴
T
N1
N
T
× 2 = 2 × 3
N2
N3
T1 T2
or
T
N1
= 3
N3
T1
i.e.
Speed ratio =
Speed of driver
No. of teeth on driven
=
Speed of driven
No. of teeth on driver
and
Train value =
Speed of driven
No. of teeth on driver
=
Speed of driver
No. of teeth on driven
Similarly, it can be proved that the
above equation holds good even if there are
any number of intermediate gears. From
above, we see that the speed ratio and the
train value, in a simple train of gears, is independent of the size and number of intermediate gears. These intermediate gears are
called idle gears, as they do not effect the
speed ratio or train value of the system. The
idle gears are used for the following two purposes :
1. To connect gears where a large
centre distance is required, and
2. To obtain the desired direction of
motion of the driven gear (i.e. clockwise or
anticlockwise).
13.4. Compound Gear Train
Gear trains inside a mechanical watch
When there are more than one gear on a shaft, as shown in Fig. 13.2, it is called a compound
train of gear.
We have seen in Art. 13.3 that the idle gears, in a simple train of gears do not effect the speed
ratio of the system. But these gears are useful in bridging over the space between the driver and the
driven.
Chapter 13 : Gear Trains
l
431
But whenever the distance between the driver and the driven or follower has to be bridged
over by intermediate gears and at the same time a great ( or much less ) speed ratio is required, then
the advantage of intermediate gears is intensified by providing compound gears on intermediate shafts.
In this case, each intermediate shaft has two gears rigidly fixed to it so that they may have the same
speed. One of these two gears meshes with the driver and the other with the driven or follower
attached to the next shaft as shown in Fig.13.2.
Fig. 13.2. Compound gear train.
In a compound train of gears, as shown in Fig. 13.2, the gear 1 is the driving gear mounted on
shaft A , gears 2 and 3 are compound gears which are mounted on shaft B. The gears 4 and 5 are also
compound gears which are mounted on shaft C and the gear 6 is the driven gear mounted on shaft D.
Let
N 1 = Speed of driving gear 1,
T 1 = Number of teeth on driving gear 1,
N 2 ,N 3 ..., N 6 = Speed of respective gears in r.p.m., and
T 2 ,T 3..., T 6 = Number of teeth on respective gears.
Since gear 1 is in mesh with gear 2, therefore its speed ratio is
N1 T2
=
N2 T1
Similarly, for gears 3 and 4, speed ratio is
...(i)
N3 T4
=
N4 T3
and for gears 5 and 6, speed ratio is
...(ii)
N5 T6
=
N6 T5
...(iii)
The speed ratio of compound gear train is obtained by multiplying the equations (i), (ii) and (iii),
∴
*
N1 N3 N5 T2 T4 T6
×
×
=
×
×
N 2 N 4 N 6 T1 T3 T5
or
*N
1
N6
=
T2 × T4 × T6
T1 × T3 × T5
Since gears 2 and 3 are mounted on one shaft B, therefore N 2 = N 3. Similarly gears 4 and 5 are mounted on
shaft C, therefore N 4 = N 5.
432
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Theory of Machines
Speed of the first driver
Speed of the last driven or follower
Product of the number of teeth on the drivens
=
Product of the number of teeth on the drivers
i.e.
Speed ratio =
Speed of the last driven or follower
Speed of the first driver
Product of the number of teeth on the drivers
=
Product of the number of teeth on the drivens
The advantage of a compound train over a simple gear train is that a much larger speed
reduction from the first shaft to the last shaft can be obtained with small gears. If a simple gear train
is used to give a large speed reduction, the last gear has to be very large. Usually for a speed reduction
in excess of 7 to 1, a simple train is not used and a compound train or worm gearing is employed.
Train value =
and
Note: The gears which mesh must have the same circular pitch or module. Thus gears 1 and 2 must have the
same module as they mesh together. Similarly gears 3 and 4, and gears
5 and 6 must have the same module.
Example 13.1. The gearing of a machine tool is shown
in Fig. 13.3. The motor shaft is connected to gear A and rotates
at 975 r.p.m. The gear wheels B, C, D and E are fixed to parallel
shafts rotating together. The final gear F is fixed on the output
shaft. What is the speed of gear F ? The number of teeth on
each gear are as given below :
Fig. 13.3
Gear
A
B
C
D
E
F
No. of teeth
20
50
25
75
26
65
Solution. Given : N A = 975 r.p.m. ;
T A = 20 ; T B = 50 ; T C = 25 ; T D = 75 ; T E = 26 ;
T F = 65
From Fig. 13.3, we see that gears A , C
and E are drivers while the gears B, D and F are
driven or followers. Let the gear A rotates in
clockwise direction. Since the gears B and C are
mounted on the same shaft, therefore it is a
compound gear and the direction or rotation of
both these gears is same (i.e. anticlockwise).
Similarly, the gears D and E are mounted on the
same shaft, therefore it is also a compound gear
and the direction of rotation of both these gears
is same (i.e. clockwise). The gear F will rotate in
anticlockwise direction.
Let
Battery Car: Even though it is run by batteries,
the power transmission, gears, clutches,
brakes, etc. remain mechanical in nature.
Note : This picture is given as additional information
and is not a direct example of the current chapter.
N F = Speed of gear F, i.e. last driven or follower.
We know that
Speed of the first driver Product of no. of teeth on drivens
=
Speed of the last driven
Product of no. of teeth on drivers
Chapter 13 : Gear Trains
l
433
50 × 75 × 65
N A TB × TD × TF
=
=
= 18.75
20 × 25 × 26
N F TA × TC × TE
or
∴
NF =
NA
975
=
= 52 r. p. m. Ans.
18.75 18.75
13.5. Design of Spur Gears
Sometimes, the spur gears (i.e. driver and driven) are to be designed for the given velocity
ratio and distance between the centres of their shafts.
Let
x = Distance between the centres of two shafts,
N 1 = Speed of the driver,
T 1 = Number of teeth on the driver,
d1 = Pitch circle diameter of the driver,
N 2 , T 2 and d2 = Corresponding values for the driven or follower, and
pc = Circular pitch.
We know that the distance between the centres of two shafts,
x=
and speed ratio or velocity ratio,
d1 + d 2
2
...(i)
N1 d2 T2
...(ii)
=
=
N2
d1 T1
From the above equations, we can conveniently find out the values of d1 and d2 (or T 1 and T 2)
and the circular pitch ( pc ). The values of T 1 and T 2, as obtained above, may or may not be whole
numbers. But in a gear since the number of its teeth is always a whole number, therefore a slight
alterations must be made in the values of x, d1 and d2, so that the number of teeth in the two gears may
be a complete number.
Example 13.2. Two parallel shafts, about 600 mm apart are to be connected by spur gears.
One shaft is to run at 360 r.p.m. and the other at 120 r.p.m. Design the gears, if the circular pitch is
to be 25 mm.
Solution. Given : x = 600 mm ; N 1 = 360 r.p.m. ; N 2 = 120 r.p.m. ; pc = 25 mm
Let
d1 = Pitch circle diameter of the first gear, and
d2 = Pitch circle diameter of the second gear.
We know that speed ratio,
N1 d 2 360
=
=
=3
N2
d1 120
or
d2 = 3d1
...(i)
and centre distance between the shafts (x),
1
(d1 + d2 )
2
From equations (i) and (ii), we find that
600 =
or
d1 + d2 = 1200
d1 = 300 mm, and d2 = 900 mm
∴ Number of teeth on the first gear,
T1 =
π d 2 π × 300
=
= 37.7
25
pc
...(ii)
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Theory of Machines
and number of teeth on the second gear,
T2 =
π d 2 π × 900
=
= 113.1
25
pc
Since the number of teeth on both the gears are to be in complete numbers, therefore let us
make the number of teeth on the first gear as 38. Therefore for a speed ratio of 3, the number of teeth
on the second gear should be 38 × 3 = 114.
Now the exact pitch circle diameter of the first gear,
T1 × pc 38 × 25
=
= 302.36 mm
π
π
and the exact pitch circle diameter of the second gear,
d1′ =
T × pc 114 × 25
d 2′ = 2
=
= 907.1 mm
π
π
∴ Exact distance between the two shafts,
d1′ + d 2′ 302.36 + 907.1
=
= 604.73 mm
2
2
Hence the number of teeth on the first and second gear must be 38 and 114 and their pitch
circle diameters must be 302.36 mm and 907.1 mm
respectively. The exact distance between the two shafts
must be 604.73 mm. Ans.
x′ =
13.6. Reverted Gear Train
When the axes of the first gear (i.e. first driver)
and the last gear (i.e. last driven or follower) are co-axial,
then the gear train is known as reverted gear train as
shown in Fig. 13.4.
We see that gear 1 (i.e. first driver) drives the
gear 2 (i.e. first driven or follower) in the opposite direction. Since the gears 2 and 3 are mounted on the same
shaft, therefore they form a compound gear and the gear
3 will rotate in the same direction as that of gear 2. The
gear 3 (which is now the second driver) drives the gear 4
(i.e. the last driven or follower) in the same direction as
that of gear 1. Thus we see that in a reverted gear train,
the motion of the first gear and the last gear is like.
Let
T1 = Number of teeth on gear 1,
Fig. 13.4. Reverted gear train.
r1 = Pitch circle radius of gear 1, and
N 1 = Speed of gear 1 in r.p.m.
Similarly,
T2, T 3, T 4 = Number of teeth on respective gears,
r2, r3, r4 = Pitch circle radii of respective gears, and
N2, N 3, N 4 = Speed of respective gears in r.p.m.
Chapter 13 : Gear Trains
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435
Since the distance between the centres of the shafts of gears 1 and 2 as well as gears 3 and 4
is same, therefore
r1 + r2 = r3 + r4
...(i)
Also, the circular pitch or module of all the gears is assumed to be same, therefore number of
teeth on each gear is directly proportional to its circumference or radius.
∴
*T 1 + T 2 = T 3 + T 4
and
Speed ratio =
...(ii)
Product of number of teeth on drivens
Product of number of teeth on drivers
N1 T2 × T4
=
N 4 T1 × T3
or
... (iii)
From equations (i), (ii) and (iii), we can determine the number of teeth on each gear for the
given centre distance, speed ratio and module only when
the number of teeth on one gear is chosen arbitrarily.
The reverted gear trains are used in automotive transmissions, lathe back gears, industrial speed reducers, and in
clocks (where the minute and hour hand shafts are co-axial).
Example 13.3. The speed ratio of the reverted gear
train, as shown in Fig. 13.5, is to be 12. The module pitch of
gears A and B is 3.125 mm and of gears C and D is 2.5 mm.
Calculate the suitable numbers of teeth for the gears. No
gear is to have less than 24 teeth.
Solution. Given : Speed ratio, N A/N D = 12 ;
m A = m B = 3.125 mm ; m C = m D = 2.5 mm
Let
Fig. 13.5
N A = Speed of gear A ,
T A = Number of teeth on gear A ,
rA = Pitch circle radius of gear A ,
N B, N C , N D = Speed of respective gears,
T B, T C , T D = Number of teeth on respective gears, and
rB, rC , rD = Pitch circle radii of respective gears.
*
We know that circular pitch,
∴
pc =
2πr
= πm
T
r1 =
m.T1
m.T2
m.T3
m.T4
; r2 =
; r3 =
; r4 =
2
2
2
2
Now from equation (i),
m.T1 m.T2 m.T3 m.T4
+
=
+
2
2
2
2
T 1 + T 2 = T 3 + T4
or
r=
m.T
, where m is the module.
2
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Theory of Machines
Since the speed ratio between the gears A and B and between the gears C and D are to be
same, therefore
N
* NA
= C = 12 = 3.464
NB
ND
Also the speed ratio of any pair of gears in mesh is the inverse of their number of teeth,
therefore
TB TD
=
= 3.464
TA TC
...(i)
We know that the distance between the shafts
x = rA + rB = rC + rD = 200 mm
or
∴
and
and
and
m .T
... 3 r =
2
m .T
mA .TA
m .T
m .T
+ B B = C C + D D = 200
2
2
2
2
3.125 (T A + T B) = 2.5 (T C + T D) = 400
...(∵ mA = m B, and m C = m D)
T A + T B = 400 / 3.125 = 128
...(ii)
T C + T D = 400 / 2.5 = 160
...(iii)
From equation (i), T B = 3.464 T A. Substituting this value of T B in equation (ii),
T A + 3.464 T A = 128 or
T A = 128 / 4.464 = 28.67 say 28 Ans.
T B = 128 – 28 = 100 Ans.
Again from equation (i), T D = 3.464 T C. Substituting this value of T D in equation (iii),
T C + 3.464 T C = 160
or T C = 160 / 4.464 = 35.84 say 36 Ans.
T D = 160 – 36 = 124 Ans.
Note : The speed ratio of the reverted gear train with the calculated values of number of teeth on each gear is
NA TB × TD 100 × 124
=
=
= 12.3
N D TA × TC
28 × 36
13.7. Epicyclic Gear Train
We have already discussed that in an epicyclic gear train, the axes of the shafts, over which
the gears are mounted, may move relative to a fixed axis. A simple epicyclic gear train is shown in
Fig. 13.6, where a gear A and the arm C have a common axis at O1 about which they can rotate. The
gear B meshes with gear A and has its axis on the arm at O2, about which the gear B can rotate. If the
*
We know that speed ratio
=
Speed of first driver N A
=
= 12
Speed of last driven N D
NA
N
N
= A × C
ND
NB
ND
Also
NA
NC
...(N B = NC, being on the same shaft)
For N and N to be same, each speed ratio should be 12 so that
B
D
N
NA
N
= A × C = 12 × 12 = 12
ND
NB
ND
Chapter 13 : Gear Trains
l
437
arm is fixed, the gear train is simple and gear A can drive gear B
or vice- versa, but if gear A is fixed and the arm is rotated about
the axis of gear A (i.e. O1), then the gear B is forced to rotate
upon and around gear A . Such a motion is called epicyclic and
the gear trains arranged in such a manner that one or more of
their members move upon and around another member are
known as epicyclic gear trains (epi. means upon and cyclic
means around). The epicyclic gear trains may be simple or compound.
The epicyclic gear trains are useful for transmitting
high velocity ratios with gears of moderate size in a comparatively lesser space. The epicyclic gear trains are used in the
back gear of lathe, differential gears of the automobiles, hoists,
pulley blocks, wrist watches etc.
Fig. 13.6. Epicyclic gear train.
13.8. Velocity Ratioz of Epicyclic Gear Train
The following two methods may be used for finding out the velocity ratio of an epicyclic
gear train.
1. Tabular method, and 2. Algebraic method.
These methods are discussed, in detail, as follows :
1. Tabular method. Consider an epicyclic gear train as shown in Fig. 13.6.
Let
T A = Number of teeth on gear A , and
T B = Number of teeth on gear B.
First of all, let us suppose that
the arm is fixed. Therefore the axes of
both the gears are also fixed relative to
each other. When the gear A makes one
revolution anticlockwise, the gear B will
make *T A / T B revolutions, clockwise.
Assuming the anticlockwise rotation as
positive and clockwise as negative, we
may say that when gear A makes + 1
revolution, then the gear B will make
(– T A / T B) revolutions. This statement
of relative motion is entered in the first
row of the table (see Table 13.1).
Secondly, if the gear A makes
+ x revolutions, then the gear B will
Inside view of a car engine.
make – x × T A / T B revolutions. This
statement is entered in the second row
Note : This picture is given as additional information and is not
a direct example of the current chapter.
of the table. In other words, multiply
the each motion (entered in the first row) by x.
Thirdly, each element of an epicyclic train is given + y revolutions and entered in the third
row. Finally, the motion of each element of the gear train is added up and entered in the fourth row.
*
We know that N B / N A = TA / TB. Since N A = 1 revolution, therefore N B = T A / T B.
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Theory of Machines
Table 13.1. Table of motions
Revolutions of elements
Step No.
Conditions of motion
Arm C
Gear A
Gear B
1.
Arm fixed-gear A rotates through + 1
revolution i.e. 1 rev. anticlockwise
0
+1
–
TA
TB
2.
Arm fixed-gear A rotates through + x
revolutions
0
+x
– x×
TA
TB
3.
Add + y revolutions to all elements
+y
+y
+y
4.
Total motion
+y
x +y
y – x×
TA
TB
A little consideration will show that when two conditions about the motion of rotation of any
two elements are known, then the unknown speed of the third element may be obtained by substituting the given data in the third column of the fourth row.
2. Algebraic method. In this method, the motion of each element of the epicyclic train relative
to the arm is set down in the form of equations. The number of equations depends upon the number of
elements in the gear train. But the two conditions are, usually, supplied in any epicyclic train viz. some
element is fixed and the other has specified motion. These two conditions are sufficient to solve all the
equations ; and hence to determine the motion of any element in the epicyclic gear train.
Let the arm C be fixed in an epicyclic gear train as shown in Fig. 13.6. Therefore speed of the
gear A relative to the arm C
= NA – N C
and speed of the gear B relative to the arm C,
= NB – N C
Since the gears A and B are meshing directly, therefore they will revolve in opposite directions.
∴
NB – NC
T
=– A
NA – NC
TB
Since the arm C is fixed, therefore its speed, N C = 0.
∴
NB
T
=– A
NA
TB
If the gear A is fixed, then N A = 0.
N B – NC
T
=– A
0 – NC
TB
or
NB
T
=1+ A
NC
TB
Note : The tabular method is easier and hence mostly used in solving problems on epicyclic gear train.
Example 13.4. In an epicyclic gear train, an arm carries
two gears A and B having 36 and 45 teeth respectively. If the arm
rotates at 150 r.p.m. in the anticlockwise direction about the centre
of the gear A which is fixed, determine the speed of gear B. If the
gear A instead of being fixed, makes 300 r.p.m. in the clockwise
direction, what will be the speed of gear B ?
Solution. Given : T A = 36 ; T B = 45 ; N C = 150 r.p.m.
(anticlockwise)
The gear train is shown in Fig. 13.7.
Fig. 13.7
Chapter 13 : Gear Trains
l
439
We shall solve this example, first by tabular method and then by algebraic method.
1. Tabular method
First of all prepare the table of motions as given below :
Table 13.2. Table of motions.
Revolutions of elements
Step No.
Conditions of motion
Arm C
Gear A
Gear B
1.
Arm fixed-gear A rotates through + 1
revolution (i.e. 1 rev. anticlockwise)
0
+1
–
TA
TB
2.
Arm fixed-gear A rotates through + x
revolutions
0
+x
– x×
TA
TB
3.
Add + y revolutions to all elements
+y
+y
+y
4.
Total motion
+y
x +y
y –x×
TA
TB
Speed of gear B when gear A is fixed
Since the speed of arm is 150 r.p.m. anticlockwise, therefore from the fourth row of the table,
y = + 150 r.p.m.
Also the gear A is fixed, therefore
x+y=0
or
∴ Speed of gear B, NB = y – x ×
x = – y = – 150 r.p.m.
36
TA
= 150 + 150 ×
= + 270 r.p.m.
45
TB
= 270 r.p.m. (anticlockwise) Ans.
Speed of gear B when gear A makes 300 r.p.m. clockwise
Since the gear A makes 300 r.p.m.clockwise, therefore from the fourth row of the table,
x + y = – 300
or
x = – 300 – y = – 300 – 150 = – 450 r.p.m.
∴ Speed of gear B,
NB = y – x ×
36
TA
= 150 + 450 ×
= + 510 r.p.m.
45
TB
= 510 r.p.m. (anticlockwise)
Ans.
2. Algebraic method
Let
N A = Speed of gear A .
N B = Speed of gear B, and
N C = Speed of arm C.
Assuming the arm C to be fixed, speed of gear A relative to arm C
= NA – NC
and speed of gear B relative to arm C = N B – N C
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Theory of Machines
Since the gears A and B revolve in opposite directions, therefore
NB – NC
T
=– A
NA – NC
TB
...(i)
Speed of gear B when gear A is fixed
When gear A is fixed, the arm rotates at 150 r.p.m. in the anticlockwise direction, i.e.
N A = 0,
∴
or
and
N C = + 150 r.p.m.
36
NB – 150
=–
= – 0.8
0 – 150
45
...[From equation (i)]
N B = – 150 × – 0.8 + 150 = 120 + 150 = 270 r.p.m. Ans.
Speed of gear B when gear A makes 300 r.p.m. clockwise
Since the gear A makes 300 r.p.m. clockwise, therefore
N A = – 300 r.p.m.
36
N B – 150
=–
= – 0.8
–300 – 150
45
∴
or
N B = – 450 × – 0.8 + 150 = 360 + 150 = 510 r.p.m. Ans.
Example 13.5. In a reverted epicyclic gear
train, the arm A carries two gears B and C and a
compound gear D - E. The gear B meshes with gear E
and the gear C meshes with gear D. The number of teeth
on gears B, C and D are 75, 30 and 90 respectively.
Find the speed and direction of gear C when gear B is
fixed and the arm A makes 100 r.p.m. clockwise.
Solution. Given : T B = 75 ; T C = 30 ; T D = 90 ;
N A = 100 r.p.m. (clockwise)
Fig. 13.8
The reverted epicyclic gear train is
shown in Fig. 13.8. First of all, let us find the
number of teeth on gear E (T E). Let dB , dC , dD
and dE be the pitch circle diameters of gears B,
C, D and E respectively. From the geometry of
the figure,
dB + dE = dC + dD
Since the number of teeth on each gear,
for the same module, are proportional to their
pitch circle diameters, therefore
TB + T E = T C + T D
∴
TE = T C + T D – T B = 30 + 90 – 75 = 45
The table of motions is drawn as
follows :
A gear-cutting machine is used to cut gears.
Note : This picture is given as additional information
and is not a direct example of the current chapter.
Chapter 13 : Gear Trains
l
441
Table 13.3. Table of motions.
Revolutions of elements
Step
No.
Conditions of motion
Arm A
Compound
gear D-E
Gear B
1.
Arm fixed-compound gear D-E
rotated through + 1 revolution ( i.e.
1 rev. anticlockwise)
0
+1
–
TE
TB
–
TD
TC
2.
Arm fixed-compound gear D-E
rotated through + x revolutions
0
+x
– x×
TE
TB
–x ×
TD
TC
3.
Add + y revolutions to all elements
+y
+y
4.
Total motion
+y
x +y
+y
y –x×
TE
TB
Gear C
+y
y –x ×
TD
TC
Since the gear B is fixed, therefore from the fourth row of the table,
y –x×
∴
TE
=0
TB
or
y –x×
45
=0
75
y – 0.6 = 0
...(i)
Also the arm A makes 100 r.p.m. clockwise, therefore
y = – 100
...(ii)
Substituting y = – 100 in equation (i), we get
– 100 – 0.6 x = 0
or
x = – 100 / 0.6 = – 166.67
Hydraulic or Pneumatic Speed
Change Actuator
Ratio Detection Switches
Round Housing With O-ring
Seated Cooling Jacket
Motor Flange
Hollow Through Bore for
Drawbar Integration
OUTPUT- External Spline to
Spindle
INPUT
Spline to Accept
Motor Shaft
Housing OD Designed to meet
RAM Bore Dia, and Share Motor
Coolant Supply
Model of sun and planet gears.
442
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Theory of Machines
From the fourth row of the table, speed of gear C,
NC = y – x ×
90
TD
= – 100 + 166.67 ×
= + 400 r.p.m.
30
TC
= 400 r.p.m. (anticlockwise) Ans.
13.9. Compound Epicyclic Gear Train—Sun and Planet Gear
A compound epicyclic
gear train is shown in Fig. 13.9.
It consists of two co-axial shafts
S1 and S 2, an annulus gear A which
is fixed, the compound gear (or
planet gear) B-C, the sun gear D
and the arm H. The annulus gear
has internal teeth and the compound gear is carried by the arm
and revolves freely on a pin of the
arm H. The sun gear is co-axial
with the annulus gear and the arm
but independent of them.
Speed Change
Shift Axis
Bearing Housing
Output Belt Pulley
The annulus gear A
meshes with the gear B and the Slide Dog
sun gear D meshes with the gear Clutch
C. It may be noted that when the
Output Sun
Gear
annulus gear is fixed, the sun gear
Sun and Planet gears.
provides the drive and when the
sun gear is fixed, the annulus gear
provides the drive. In both cases, the arm acts as a follower.
Oil
Collector
Planet
Gears
Input Sun
Gear
Motor
Flange
Note : The gear at the centre is called the sun gear and the gears whose axes move are called planet gears.
Fig. 13.9. Compound epicyclic gear train.
Chapter 13 : Gear Trains
l
443
Let T A , T B , T C , and T D be the teeth and N A, N B, N C and N D be the speeds for the gears A , B,
C and D respectively. A little consideration will show that when the arm is fixed and the sun gear D is
turned anticlockwise, then the compound gear B-C and the annulus gear A will rotate in the clockwise
direction.
The motion of rotations of the various elements are shown in the table below.
Table 13.4. Table of motions.
Revolutions of elements
Step
No.
Conditions of motion
Arm
Gear D
Compound gear
B-C
1.
Arm fixed-gear D rotates
through + 1 revolution
0
+1
–
TD
TC
–
2.
Arm fixed-gear D rotates
through + x revolutions
0
+x
– x×
TD
TC
–x ×
3.
Add + y revolutions to all
elements
+y
+y
+y
4.
Total motion
+y
x +y
y–x×
Gear A
TD TB
×
TC TA
TD TB
×
TC TA
+y
TD
TC
y –x×
TD TB
×
TC TA
Note : If the annulus gear A is rotated through one revolution anticlockwise with the arm fixed, then the
compound gear rotates through T A / T B revolutions in the same sense and the sun gear D rotates through
T A / T B × T C / T D revolutions in clockwise direction.
Example 13.6. An epicyclic gear consists of three gears A, B and C as shown in Fig. 13.10.
The gear A has 72 internal teeth and gear C has 32 external teeth. The gear B meshes with both A
and C and is carried on an arm EF which rotates about the centre of A at 18 r.p.m.. If the gear A is
fixed, determine the speed of gears B and C.
Solution. Given : T A = 72 ; T C = 32 ; Speed of arm EF = 18 r.p.m.
Considering the relative motion of rotation as shown in Table 13.5.
Table 13.5. Table of motions.
Revolutions of elements
Step No.
Conditions of motion
Arm EF
Gear C
Gear B
1.
Arm fixed-gear C rotates through
+ 1 revolution (i.e. 1 rev.
anticlockwise)
0
+1
–
TC
TB
2.
Arm fixed-gear C rotates through
+ x revolutions
0
+x
–x×
TC
TB
3.
Add + y revolutions to all
elements
+y
+y
+y
4.
Total motion
+y
x +y
y –x×
Gear A
–
TC TB
T
×
=– C
TB TA
TA
– x×
TC
TA
+y
TC
TB
y –x×
TC
TA
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Theory of Machines
Speed of gear C
We know that the speed of the arm is 18 r.p.m. therefore,
y = 18 r.p.m.
and the gear A is fixed, therefore
y –x×
∴
TC
=0
TA
18 – x ×
or
32
=0
72
x = 18 × 72 / 32 = 40.5
∴ Speed of gear C
= x + y = 40.5 + 18
= + 58.5 r.p.m.
= 58.5 r.p.m. in the direction
of arm. Ans.
Fig. 13.10
Speed of gear B
Let dA, dB and dC be the pitch circle diameters of gears
A , B and C respectively. Therefore, from the geometry of Fig. 13.10,
dB +
dC
d
= A
2
2
or
2 dB + dC = dA
Since the number of teeth are proportional to their pitch circle diameters, therefore
2 TB + T C = TA
∴ Speed of gear B
or
2 T B + 32 = 72
or
T B = 20
TC
32
= 18 – 40.5 ×
= – 46.8 r.p.m.
20
TB
= 46.8 r.p.m. in the opposite direction of arm. Ans.
= y –x×
Example 13.7. An epicyclic train of gears is arranged as shown in
Fig.13.11. How many revolutions does the arm, to which the pinions B and
C are attached, make :
1. when A makes one revolution clockwise and D makes half a
revolution anticlockwise, and
2. when A makes one revolution clockwise and D is stationary ?
The number of teeth on the gears A and D are 40 and 90
respectively.
Fig. 13.11
Solution. Given : T A = 40 ; T D = 90
First of all, let us find the number of teeth on gears B and C (i.e. TB and T C). Let dA, dB, dC
and dD be the pitch circle diameters of gears A , B, C and D respectively. Therefore from the geometry
of the figure,
dA + dB + dC = dD
or
dA + 2 dB = dD
...(3 dB = dC)
Since the number of teeth are proportional to their pitch circle diameters, therefore,
TA + 2 T B = T D
∴
T B = 25,
or
and
40 + 2 T B = 90
T C = 25
...(3 T B = T C)
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Chapter 13 : Gear Trains
445
The table of motions is given below :
Table 13.6. Table of motions.
Revolutions of elements
Step No.
Conditions of motion
Arm
Arm fixed, gear A rotates
through – 1 revolution (i.e. 1
rev. clockwise)
1.
Gear A
Compound
gear B-C
0
–1
+
TA
TB
TA
TB
2.
Arm fixed, gear A rotates
through – x revolutions
0
–x
+ x×
3.
Add – y revolutions to all
elements
–y
–y
–y
4.
Total motion
–y
–x–y
x×
TA
– y
TB
Gear D
+
TA TB
T
×
=+ A
TB TD
TD
+ x×
TA
TD
–y
x×
TA
– y
TD
1. Speed of arm when A makes 1 revolution clockwise and D makes half revolution anticlockwise
Since the gear A makes 1 revolution clockwise, therefore from the fourth row of the table,
– x – y = –1
or
x+y=1
...(i)
Also, the gear D makes half revolution anticlockwise, therefore
x×
∴
1
TA
– y=
2
TD
40 x – 90 y = 45
x×
or
x – 2.25 y = 1.125
From equations (i) and (ii), x = 1.04
∴
40
1
– y=
90
2
or
and
...(ii)
y = – 0.04
Speed of arm = – y = – (– 0.04) = + 0.04
= 0.04 revolution anticlockwise Ans.
2. Speed of arm when A makes 1 revolution clockwise and D is stationary
Since the gear A makes 1 revolution clockwise, therefore from the fourth row of the
table,
–x –y =–1
or
x+y=1
...(iii)
Also the gear D is stationary, therefore
TA
– y=0
TD
or
40 x – 90 y = 0
or
x×
∴
x×
40
– y=0
90
x – 2.25 y = 0
From equations (iii) and (iv),
x = 0.692
∴
and
y = 0.308
Speed of arm = – y = – 0.308 = 0.308 revolution clockwise Ans.
...(iv)
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Theory of Machines
Example 13.8. In an epicyclic gear train, the internal wheels A and B and compound wheels
C and D rotate independently about axis O. The wheels E and F rotate on pins fixed to the arm G. E
gears with A and C and F gears with B and D. All the wheels have
the same module and the number of teeth are : TC = 28; TD = 26;
T E = TF = 18.
1. Sketch the arrangement ; 2. Find the number of teeth on
A and B ; 3. If the arm G makes 100 r.p.m. clockwise and A is fixed,
find the speed of B ; and 4. If the arm G makes 100 r.p.m. clockwise
and wheel A makes 10 r.p.m. counter clockwise ; find the speed of
wheel B.
Solution. Given : T C = 28 ; T D = 26 ; T E = T F = 18
1. Sketch the arrangement
The arrangement is shown in Fig. 13.12.
Fig. 13.12
2. Number of teeth on wheels A and B
Let
TA = Number of teeth on wheel A , and
TB = Number of teeth on wheel B.
If dA , dB , dC , dD , dE and dF are the pitch circle diameters of wheels A , B, C, D, E and F
respectively, then from the geometry of Fig. 13.12,
dA = dC + 2 dE
and
dB = dD + 2 dF
Since the number of teeth are proportional to their pitch circle diameters, for the same
module, therefore
Ans.
TA = T C + 2 T E = 28 + 2 × 18 = 64
and
TB = T D + 2 T F = 26 + 2 × 18 = 62
Ans.
3. Speed of wheel B when arm G makes 100 r.p.m. clockwise and wheel A is fixed
First of all, the table of motions is drawn as given below :
Table 13.7. Table of motions.
Revolutions of elements
Step
No.
Conditions of
motion
Arm Wheel
G
A
1.
Arm fixed- wheel A
rotates through + 1
revolution (i.e. 1 rev.
anticlockwise)
0
2.
Arm fixed-wheel A
rotates through + x
revolutions
0
+x
+ x×
3.
Add + y revolutions
to all elements
+y
+y
+y
4.
Total motion
+y
x+y
+1
Wheel
E
+
TA
TE
Compound
wheel C-D
–
TA TE
×
TE TC
=–
TA
TE
y + x×
– x×
TA
TC
y–x×
+
TA TD
×
TC TF
TA
TC
+y
TA
TE
Wheel F
+
TA TD TF
× ×
TC TF TB
=+
+ x×
TA TD
×
TC TF
+y
TA
TC
Wheel B
y + x×
TA TD
×
TC TF
TA TD
×
TC TB
+ x×
TA TD
×
TC TB
+y
y + x×
TA TD
×
TC TB
Chapter 13 : Gear Trains
447
l
Since the arm G makes 100 r.p.m. clockwise, therefore from the fourth row of the table,
y = – 100
...(i)
Also, the wheel A is fixed, therefore from the fourth row of the table,
x+y=0
or
x = – y = 100
∴ Speed of wheel B = y + x ×
...(ii)
64 26
TA TD
×
= – 100 + 100 ×
×
= – 100 + 95.8 r.p.m.
28 62
TC TB
= – 4.2 r.p.m. = 4.2 r.p.m. clockwise Ans.
4. Speed of wheel B when arm G makes 100 r.p.m. clockwise and wheel A makes 10 r.p.m. counter
clockwise
Since the arm G makes 100 r.p.m. clockwise, therefore from the fourth row of the
table
y = – 100
...(iii)
Also the wheel A makes 10 r.p.m. counter clockwise, therefore from the fourth row of the
table,
x + y = 10 or x = 10 – y = 10 + 100 = 110
...(iv)
∴
Speed of wheel B = y + x ×
64 26
TA TD
×
= – 100 + 110 ×
×
= – 100 +105.4 r.p.m.
28 62
TC TB
= + 5.4 r.p.m. = 5.4 r.p.m. counter clockwise Ans.
Example 13.9. In an epicyclic gear of the ‘sun and planet’ type shown
in Fig. 13.13, the pitch circle diameter of the internally toothed ring is to be
224 mm and the module 4 mm. When the ring D is stationary, the spider A,
which carries three planet wheels C of equal size, is to make one revolution in
the same sense as the sunwheel B for every five revolutions of the driving
spindle carrying the sunwheel B. Determine suitable numbers of teeth for all
the wheels.
Fig. 13.13
Solution. Given : dD = 224 mm ; m = 4 mm ;
NA = NB / 5
Let T B , T C and T D be the number of teeth on the sun wheel B,
planet wheels C and the internally toothed ring D. The table of motions is given below :
Table 13.8. Table of motions.
Revolutions of elements
Step No.
1.
2.
3.
4.
Conditions of motion
Spider A fixed, sun wheel
B rotates through + 1
revolution (i.e. 1 rev.
anticlockwise)
Spider A fixed, sun wheel
B rotates through + x
revolutions
Add + y revolutions to all
elements
Total motion
Spider A
Sun wheel B
Planet wheel C
0
+1
–
0
+x
– x×
+y
+y
+y
+y
x +y
y – x×
TB
TC
Internal gear D
–
TB
TC
TB TC
T
× =– B
TC TD
TD
– x×
TB
TD
+y
TB
TC
y – x×
TB
TD
448
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Theory of Machines
We know that when the sun
wheel B makes + 5 revolutions, the spider A makes + 1 revolution. Therefore
from the fourth row of the table,
y = + 1 ; and x + y = + 5
∴
x=5–y=5–1=4
Main rotor
Drive shaft
Cockpit
Tail boom
Since the internally toothed ring
D is stationary, therefore from the fourth
row of the table,
y – x×
TB
=0
TD
T
1– 4× B =0
TD
or
∴
TB 1
=
TD 4
Tail rotor
Landing skids
Engine, transmission fuel, etc.
Power transmission in a helicopter is essentially through
gear trains.
Note : This picture is given as additional information and is not a
direct example of the current chapter.
or
TD = 4 TB
We know that
T D = dD / m = 224 / 4 = 56 Ans.
∴
T B = T D / 4 = 56 / 4 = 14 Ans.
...(i)
...[From equation (i)]
Let dB, dC and dD be the pitch circle diameters of sun wheel B, planet wheels C and internally
toothed ring D respectively. Assuming the pitch of all the gears to be same, therefore from the geometry of Fig. 13.13,
dB + 2 dC = dD
Since the number of teeth are proportional to their pitch circle diameters, therefore
TB + 2 TC = TD
∴
or
14 + 2 T C = 56
T C = 21 Ans.
Example 13.10. Two shafts A and B are co-axial. A gear C (50 teeth) is rigidly mounted
on shaft A. A compound gear D-E gears with C and an internal gear G. D has 20 teeth and gears
with C and E has 35 teeth and gears with an internal gear G. The gear G is fixed and is concentric with the shaft axis. The compound gear D-E is mounted on a pin which projects from an arm
keyed to the shaft B. Sketch the arrangement and find the number of teeth on internal gear G
assuming that all gears have the same module. If the shaft A rotates at 110 r.p.m., find the speed
of shaft B.
Solution. Given : T C = 50 ; T D = 20 ; T E = 35 ; N A = 110 r.p.m.
The arrangement is shown in Fig. 13.14.
Number of teeth on internal gear G
Let dC , dD , dE and dG be the pitch circle diameters of gears C, D, E and G respectively. From
the geometry of the figure,
dG
d
d
d
= C + D + E
2
2
2
2
or
dG = dC + dD + dE
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Chapter 13 : Gear Trains
449
Let T C , T D , T E and T G be the number of teeth on gears C, D, E and G respectively. Since all
the gears have the same module, therefore number of teeth are proportional to their pitch circle
diameters.
∴
T G = T C + T D + T E = 50 + 20 + 35 = 105 Ans.
Fig. 13.14
Speed of shaft B
The table of motions is given below :
Table 13.9. Table of motions.
Revolutions of elements
Step
No.
Conditions of motion
Arm
Gear C (or
shaft A)
Compound
gear D-E
TC
TD
1.
Arm fixed - gear C rotates through + 1
revolution
0
+1
–
2.
Arm fixed - gear C rotates through + x
revolutions
0
+x
– x×
3.
Add + y revolutions to all elements
+y
+y
+y
4.
Total motion
+y
x+y
y – x×
Gear G
–
TC
TD
TC TE
×
TD TG
– x×
TC TE
×
TD TG
+y
TC
TD
y – x×
TC TE
×
TD TG
Since the gear G is fixed, therefore from the fourth row of the table,
y – x×
∴
TC TE
×
=0
TD TG
y–
5
x=0
6
or
y – x×
50 35
×
=0
20 105
...(i)
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Theory of Machines
Since the gear C is rigidly mounted on shaft A , therefore speed of gear C and shaft A is same.
We know that speed of shaft A is 110 r.p.m., therefore from the fourth row of the table,
x + y = 100
...(ii)
From equations (i) and (ii), x = 60, and y = 50
∴
Speed of shaft B = Speed of arm = + y = 50 r.p.m. anticlockwise Ans.
Example 13.11. Fig. 13.15 shows diagrammatically a compound
epicyclic gear train. Wheels A , D and E are free to rotate independently
on spindle O, while B and C are compound and rotate together on spindle
P, on the end of arm OP. All the teeth on different wheels have the same
module. A has 12 teeth, B has 30 teeth and C has 14 teeth cut externally.
Find the number of teeth on wheels D and E which are cut internally.
If the wheel A is driven clockwise at 1 r.p.s. while D is driven
counter clockwise at 5 r.p.s., determine the magnitude and direction of
the angular velocities of arm OP and wheel E.
Fig. 13.15
Solution. Given : T A = 12 ; T B = 30 ;T C = 14 ; NA = 1 r.p.s. ; N D = 5 r.p.s.
Number of teeth on wheels D and E
Let T D and T E be the number of teeth on wheels D and E respectively. Let dA , dB , dC , dD and dE
be the pitch circle diameters of wheels A , B, C, D and E respectively. From the geometry of the figure,
dE = dA + 2dB
and
dD = dE – (dB – dC)
Since the number of teeth are proportional to their pitch circle diameters for the same module,
therefore
T E = T A + 2T B = 12 + 2 × 30 = 72 Ans.
and
T D = T E – (T B – T C) = 72 – (30 – 14) = 56 Ans.
Magnitude and direction of angular velocities of arm OP and wheel E
The table of motions is drawn as follows :
Table 13.10. Table of motions.
Revolutions of elements
Step
No.
Conditions of motion
Arm Wheel A
1.
Arm fixed A rotated through
– 1 revolution (i.e. 1 revolution clockwise)
0
2.
Arm fixed-wheel A rotated
through – x revolutions
0
–x
+ x×
3.
Add – y revolutions to all elements
–y
–y
–y
4.
Total motion
–y
–x–y
–1
Compound
wheel B-C
+
TA
TB
Wheel D
+
TA TC
×
TB TD
Wheel E
+
TA TB
×
TB TE
=+
x×
TA
TB
TA
– y
TB
+ x×
TA TC
×
TB TD
+ x×
–y
x×
TA TC
×
– y
TB TD
TA
TE
TA
TE
–y
x×
TA
– y
TE
Chapter 13 : Gear Trains
l
451
Since the wheel A makes 1 r.p.s. clockwise, therefore from the fourth row of the table,
– x – y = – 1 or
x+y=1
...(i)
Also, the wheel D makes 5 r.p.s. counter clockwise, therefore
TA TC
12 14
– y=5
×
or
x×
– y=5
×
TB TD
30 56
∴
0.1 x – y = 5
From equations (i) and (ii),
x = 5.45
and
y = – 4.45
∴ Angular velocity of arm OP
= – y = –(– 4.45) = 4.45 r.p.s
= 4.45 × 2 π = 27.964 rad/s (counter clockwise) Ans.
x×
...(ii)
12
TA
– y = 5.45 ×
– (– 4.45) = 5.36 r.p.s.
72
TE
= 5.36 × 2 π = 33.68 rad/s (counter clockwise) Ans.
Example 13.12. An internal wheel B with 80 teeth is keyed to a shaft F. A fixed internal
wheel C with 82 teeth is concentric
with B. A compound wheel D-E
gears with the two internal wheels;
D has 28 teeth and gears with C
while E gears with B. The compound
wheels revolve freely on a pin which
projects from a disc keyed to a shaft
A co-axial with F. If the wheels have
the same pitch and the shaft A makes
800 r.p.m., what is the speed of the
Helicopter
shaft F ? Sketch the arrangement.
and angular velocity of wheel E = x ×
Solution. Given : T B = 80 ; T C
= 82 ; T D = 28 ; N A = 500 r.p.m.
Note : This picture is given as additional information and is not a
direct example of the current chapter.
The arrangement is shown in Fig. 13.16.
Fig. 13.16
First of all, let us find out the number of teeth on wheel E (T E). Let dB , dC , dD and dE be the
pitch circle diameter of wheels B, C, D and E respectively. From the geometry of the figure,
dB = dC – (dD – dE )
452
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Theory of Machines
or
dE = dB + dD – dC
Since the number of teeth are proportional to their pitch circle diameters for the same pitch,
therefore
T E = T B + T D – T C = 80 + 28 – 82 = 26
The table of motions is given below :
Table 13.11. Table of motions.
Revolutions of elements
Step
No.
Conditions of motion
Arm (or
shaft A)
Wheel B (or
shaft F)
Compound
gear D-E
1.
Arm fixed - wheel B rotated
through + 1 revolution (i.e. 1
revolution anticlockwise)
0
+1
+
2.
Arm fixed - wheel B rotated
through + x revolutions
0
+x
+ x×
3.
Add + y revolutions to all
elements
+y
+y
+y
4.
Total motion
+y
x +y
y + x×
TB
TE
Wheel C
+
TB
TE
TB TD
×
TE TC
+ x×
TB TD
×
TE TC
+y
TB
TE
y + x×
TB TD
×
TE TC
Since the wheel C is fixed, therefore from the fourth row of the table,
y+ x×
∴
TB TD
×
=0
TE TC
or
y+ x×
80 28
×
=0
26 82
y + 1.05 x = 0
...(i)
Also, the shaft A (or the arm) makes 800 r.p.m., therefore from the fourth row of the table,
y = 800
...(ii)
From equations (i) and (ii),
x = – 762
∴ Speed of shaft F = Speed of wheel B = x + y = – 762 + 800 = + 38 r.p.m.
= 38 r.p.m. (anticlockwise) Ans.
Example 13.13. Fig. 13.17 shows an epicyclic gear
train known as Ferguson’s paradox. Gear A is fixed to the
frame and is, therefore, stationary. The arm B and gears C
and D are free to rotate on the shaft S. Gears A, C and D have
100, 101 and 99 teeth respectively. The planet gear has 20
teeth. The pitch circle diameters of all are the same so that the
planet gear P meshes with all of them. Determine the
revolutions of gears C and D for one revolution of the arm B.
Solution. Given : T A = 100 ; T C = 101 ; T D = 99 ;
T P = 20
Fig. 13.17
l
Chapter 13 : Gear Trains
453
The table of motions is given below :
Table 13.12. Table of motions.
Revolutions of elements
Step No.
Conditions of motion
Arm B
Gear A
Gear C
TA
TC
1.
Arm B fixed, gear A rotated
through + 1 revolution (i.e. 1
revolution anticlockwise)
0
+1
+
2.
Arm B fixed, gear A rotated
through + x revolutions
0
+x
+ x×
3.
Add + y revolutions to all
elements
+y
+y
+y
4.
Total motion
+y
x +y
y + x×
Gear D
+
TA TC
T
×
=+ A
TC TD
TD
TA
TC
+ x×
TA
TD
+y
TA
TC
y + x×
TA
TD
The arm B makes one revolution, therefore
y=1
Since the gear A is fixed, therefore from the fourth row of the table,
x+y=0
or
x =–y =–1
Let
N C and N D = Revolutions of gears C and D respectively.
From the fourth row of the table, the revolutions of gear C,
T
100
1
=+
NC = y + x × A = 1 – 1 ×
Ans.
101
101
TC
and the revolutions of gear D,
T
100
1
=–
ND = y + x × A = 1 –
Ans.
99
99
TD
From above we see that for one revolution of the arm B, the gear C rotates through 1/101
revolutions in the same direction and the gear D rotates through 1/99 revolutions in the opposite
direction.
Example 13.14. In the gear drive as shown in Fig.
13.18, the driving shaft A rotates at 300 r.p.m. in the clockwise direction, when seen from left hand. The shaft B is the
driven shaft. The casing C is held stationary. The wheels E
and H are keyed to the central vertical spindle and wheel F
can rotate freely on this spindle. The wheels K and L are
rigidly fixed to each other and rotate together freely on a
pin fitted on the underside of F. The wheel L meshes with
internal teeth on the casing C. The numbers of teeth on the
different wheels are indicated within brackets in Fig. 13.18.
Find the number of teeth on wheel C and the speed
and direction of rotation of shaft B.
Solution. Given : N A = 300 r.p.m. (clockwise) ;
T D = 40 ; T B = 30 ; T F = 50 ; T G = 80 ; T H = 40 ; T K = 20 ; T L = 30
Fig. 13.18
In the arrangement shown in Fig. 13.18, the wheels D and G are auxillary gears and do not
form a part of the epicyclic gear train.
454
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Theory of Machines
Speed of wheel E, N E = N A ×
40
TD
= 300 ×
= 400 r.p.m. (clockwise)
30
TE
Number of teeth on wheel C
Let
T C = Number of teeth on wheel C.
Assuming the same module for all teeth and since the pitch circle diameter is proportional to
the number of teeth ; therefore from the geometry of Fig.13.18,
T C = T H + T K + T L = 40 + 20 + 30 = 90 Ans.
Speed and direction of rotation of shaft B
The table of motions is given below. The wheel F acts as an arm.
Table 13.13. Table of motions.
Revolutions of elements
Step
No.
Conditions of motion
Arm or
wheel F
Wheel E
Wheel H
1.
Arm fixed-wheel E
rotated through – 1
revolution (i.e. 1
revolution clockwise)
0
–1
2.
Arm fixed-wheel E
rotated through – x
revolutions
0
–x
–x
+ x×
3.
Add – y revolutions to
all elements
–y
–y
–y
–y
4.
Total motion
–y
–x–y
–x–y
– 1( 3 E and
H are on the
same shaft)
Compound
wheel K-L
+
x×
Wheel C
TH
TK
TH
TK
TH
– y
TK
+
TH TL
×
TK TC
+ x×
TH TL
×
TK TC
–y
x×
TH TL
×
–y
TK TC
Since the speed of wheel E is 400 r.p.m. (clockwise), therefore from the fourth row of the table,
or
– x – y = – 400 or
Also the wheel C is fixed, therefore
T
T
x× H × L – y=0
TK TC
40
30
x×
– y=0
×
20
90
2x
∴
– y=0
3
From equations (i) and (ii),
x = 240
and
x + y = 400
...(i)
...(ii)
y = 160
∴ Speed of wheel F, N F = – y = – 160 r.p.m.
Since the wheel F is in mesh with wheel G, therefore speed of wheel G or speed of shaft B
= – NF ×
50
TF
= – – 160 ×
= 100 r.p.m.
80
TG
...(3 Wheel G will rotate in opposite direction to that of wheel F.)
= 100 r.p.m. anticlockwise i.e. in opposite direction of
shaft A . Ans.
Chapter 13 : Gear Trains
455
l
Example 13.15. Fig. 13.19 shows a compound epicyclic gear in which the casing C contains
an epicyclic train and this casing is inside the larger casing D.
Determine the velocity ratio of the output shaft B to the input shaft A when the casing D is
held stationary. The number of teeth on various wheels are as follows :
Wheel on A = 80 ; Annular wheel on B = 160 ; Annular wheel on C = 100 ; Annular wheel
on D = 120 ; Small pinion on F = 20 ; Large pinion on F = 66.
Fig. 13.19
Solution. Given : T 1 = 80 ; T 8 = 160 ; T 4 = 100; T 3 = 120 ; T 6 = 20 ; T 7 = 66
First of all, let us consider the train of wheel 1 (on A ), wheel 2 (on E), annular wheel 3 (on D)
and the arm i.e. casing C. Since the pitch circle diameters of wheels are proportional to the number of
teeth, therefore from the geometry of Fig. 13.19,
T 1 + 2 T2 = T 3
or 80 + 2 T 2 = 120
∴
T 2 = 20
The table of motions for the train considered is given below :
Table 13.14. Table of motions.
Revolutions of elements
Step No.
Conditons of motion
Arm
Wheel 1
1.
Arm fixed - wheel 1 rotated
through + 1 revolution
(anticlockwise)
0
+1
–
2.
Arm fixed - wheel 1 rotated
through + x revolutions
0
+x
– x×
+y
+y
+y
y
x +y
y – x×
3.
4.
Add + y revolutions to all
elements
Total motion
Wheel 2
T1
T2
Wheel 3
–
T1
T2
T1 T2
T
×
=– 1
T2 T3
T3
– x×
T1
T3
+y
T1
T2
y – x×
T1
T3
456
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Theory of Machines
Let us assume that wheel 1 makes 1 r.p.s. anticlockwise.
∴
x+y=1
...(i)
Also the wheel 3 is stationary, therefore from the fourth row of the table,
y – x×
∴
y –
T1
=0
T3
y – x×
or
80
=0
120
2
x=0
3
...(ii)
From equations (i) and (ii), x = 0.6,
and
y = 0.4
∴ Speed of arm or casing C = y = 0.4 r.p.s.
= y – x×
and speed of wheel 2 or arm E
80
T1
= 0.4 – 0.6 ×
= – 2 r.p.s.
20
T2
= 2 r.p.s. (clockwise)
Let us now consider the train of annular wheel 4 (on C), wheel 5 (on E), wheel 6 (on F) and
arm E. We know that
T6 + 2 T5 = T 4
∴
or
20 + 2 T 5 = 100
T 5 = 40
The table of motions is given below :
Table 13.15. Table of motions.
Revolutions of elements
Step
No.
Conditions of motion
Arm E or
wheel 2
Wheel 6
Wheel 5
1.
Arm fixed, wheel 6 rotated
through + 1 revolution
0
+1
–
2.
Arm fixed, wheel 6 rotated
through + x 1 revolutions
0
x1
– x1 ×
3.
Add + y 1 revolutions to all
elements
+ y1
+ y1
+ y1
4.
Total motion
+ y1
x1 + y1
y1 – x1 ×
T6
T5
Wheel 4
–
T6
T5
T6 T5
T
×
=– 6
T5 T4
T4
– x1 ×
T6
T4
+ y1
T6
T5
y1 – x1 ×
T6
T4
We know that speed of arm E = Speed of wheel 2 in the first train
∴
y1 = – 2
...(iii)
Also speed of wheel 4 = Speed of arm or casing C in the first train
∴
or
y1 – x1 ×
T6
= 0.4
T4
or
x1 = (–2 – 0.4)
–2 – x1 ×
100
= – 12
20
20
= 0.4
100
...(iv)
l
Chapter 13 : Gear Trains
457
∴ Speed of wheel 6 (or F)
= x 1 + y 1 = – 12 – 2 = – 14 r.p.s. = 14 r.p.s. (clockwise)
Now consider the train of wheels 6 and 7 (both on F), annular wheel 8 (on B) and the arm i.e.
casing C. The table of motions is given below :
Table 13.16. Table of motions.
Revolutions of elements
Step No.
Conditions of motion
Arm
Wheel 8
Wheel 7
1.
Arm fixed, wheel 8 rotated through
+ 1 revolution
0
+1
+
2.
Arm fixed, wheel 8 rotated through
+ x 2 revolutions
0
+ x2
+ x2 ×
3.
Add + y 2 revolutions to all
elements
+ y2
+ y2
+ y2
4.
Total motion
y2
x2 + y2
T8
T7
y2 + x2 ×
T8
T7
T8
T7
We know that the speed of C in the first train is 0.4 r.p.s., therefore
y 2 = 0.4
...(v)
Also the speed of wheel 7 is equal to the speed of F or wheel 6 in the second train, therefore
y2 + x2 ×
∴
T8
= –14
T7
or
x2 = ( − 14 − 0.4)
0.4 + x2 ×
160
= – 14
66
...(vi)
66
= − 5.94
160
∴ Speed of wheel 8 or of the shaft B
x 2 + y 2 = – 5.94 + 0.4 = – 5.54 r.p.s. = 5.54 r.p.s. (clockwise)
We have already assumed that the speed of wheel 1 or the shaft A is 1 r.p.s. anticlockwise
∴ Velocity ratio of the output shaft B to the input shaft A
= – 5.54 Ans.
Note : The – ve sign shows that the two shafts A and B rotate in opposite directions.
13.10. Epicyclic Gear Train with Bevel Gears
The bevel gears are used to make a more compact epicyclic system and they permit a very
high speed reduction with few gears. The useful application of the epicyclic gear train with bevel
gears is found in Humpage’s speed reduction gear and differential gear of an automobile as discussed
below :
1. Humpage’s speed reduction gear. The Humpage’s speed reduction gear was originally
designed as a substitute for back gearing of a lathe, but its use is now considerably extended to all
kinds of workshop machines and also in electrical machinery. In Humpage’s speed reduction gear, as
shown in Fig. 13.20, the driving shaft X and the driven shaft Y are co-axial. The driving shaft carries
a bevel gear A and driven shaft carries a bevel gear E. The bevel gear B meshes with gear A (also
known as pinion) and a fixed gear C. The gear E meshes with gear D which is compound with gear B.
458
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Theory of Machines
This compound gear B-D is mounted on the arm or spindle F which is rigidly connected with a hollow
sleeve G. The sleeve revolves freely loose on the axes of the driving and driven shafts.
Fig. 13.20. Humpage’s speed reduction gear.
2. Differential gear of an automobile. The differential gear used in the rear drive of an
automobile is shown in Fig. 13.21. Its function is
(a) to transmit motion from the engine shaft to the rear driving wheels, and
(b) to rotate the rear wheels at different speeds while the automobile is taking a turn.
As long as the automobile is running on a straight path, the rear wheels are driven directly by
the engine and speed of both the wheels is same. But when the automobile is taking a turn, the outer
wheel will run faster than the * inner wheel because at that time the outer rear wheel has to cover more
distance than the inner rear wheel. This is achieved by epicyclic gear train with bevel gears as shown
in Fig. 13.21.
The bevel gear A (known as pinion) is keyed to
the propeller shaft driven from the engine shaft through
universal coupling. This gear A drives the gear B (known
as crown gear) which rotates freely on the axle P. Two
equal gears C and D are mounted on two separate parts P
and Q of the rear axles respectively. These gears, in turn,
mesh with equal pinions E and F which can rotate freely
on the spindle provided on the arm attached to gear B.
When the automobile runs on a straight path, the
gears C and D must rotate together. These gears are rotated
through the spindle on the gear B. The gears E and F do
not rotate on the spindle. But when the automobile is taking
Fig. 13.21. Differential gear of an automobile.
a turn, the inner rear wheel should have lesser speed than
the outer rear wheel and due to relative speed of the inner and outer gears D and C, the gears E and F
start rotating about the spindle axis and at the same time revolve about the axle axis.
Due to this epicyclic effect, the speed of the inner rear wheel decreases by a certain amount
and the speed of the outer rear wheel increases, by the same amount. This may be well understood by
drawing the table of motions as follows :
*
This difficulty does not arise with the front wheels as they are greatly used for steering purposes and are
mounted on separate axles and can run freely at different speeds.
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Chapter 13 : Gear Trains
Table 13.17. Table of motions.
Revolutions of elements
Step No.
1.
Conditions of motion
Gear B
Gear C
Gear B fixed-Gear C rotated
through + 1 revolution (i.e.
1 revolution anticlockwise )
0
+1
Gear E
+
Gear D
TC
TE
–
TC TE
×
= –1
TE TD
(3 TC = TD )
TC
TE
2.
Gear B fixed-Gear C rotated
through + x revolutions
0
+x
3.
Add + y revolutions to all
elements
+y
+y
+y
4.
Total motion
+y
x +y
y + x×
+x ×
–x
+y
TC
TE
y– x
From the table, we see that when the gear B, which derives motion from the engine shaft,
rotates at y revolutions, then the speed of inner gear D (or the rear axle Q) is less than y by x revolutions and the speed of the outer gear C (or the rear axle P) is greater than y by x revolutions. In other
words, the two parts of the rear axle and thus the two wheels rotate at two different speeds. We also
see from the table that the speed of gear B is the mean of speeds of the gears C and D.
Example 13.16. Two bevel gears A and B (having 40 teeth and 30 teeth) are rigidly mounted
on two co-axial shafts X and Y. A bevel gear C (having
50 teeth) meshes with A and B and rotates freely on one
end of an arm. At the other end of the arm is welded a
sleeve and the sleeve is riding freely loose on the axes of
the shafts X and Y. Sketch the arrangement.
If the shaft X rotates at 100 r.p.m. clockwise and
arm rotates at 100 r.p.m.anitclockwise, find the
speed of shaft Y.
Solution. Given : T A = 40 ; T B = 30 ; T C = 50 ; N X
= N A = 100 r.p.m. (clockwise) ; Speed of arm = 100 r.p.m.
(anticlockwise)
The arangement is shown in Fig. 13.22.
The table of motions is drawn as below :
Fig. 13.22
Table 13.18. Table of motions.
Revolutions of elements
Step No.
Arm
Gear A
Gear C
* TA
1.
Arm B fixed, gear A rotated
through + 1 revolution (i.e. 1
revolution anticlockwise)
0
+1
±
2.
Arm B fixed, gear A rotated
through + x revolutions
Add + y revolutions to all
elements
0
+x
± x×
+y
+y
+y
+y
x +y
y ± x×
3.
4.
*
Conditions of motion
Total motion
Gear B
–
TC
TA
TC
TA TC
T
×
=– A
TC TB
TB
– x×
TA
TB
+y
TA
TC
y – x×
TA
TB
The ± sign is given to the motion of the wheel C because it is in a different plane. So we cannot indicate the
direction of its motion specifically, i.e. either clockwise or anticlockwise.
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Theory of Machines
Since the speed of the arm is 100 r.p.m. anticlockwise, therefore from the fourth row of the
table,
y = + 100
Also, the speed of the driving shaft X or gear A is 100 r.p.m. clockwise.
∴
x + y = – 100
or
x = – y – 100 = – 100 – 100 = – 200
∴ Speed of the driven shaft i.e. shaft Y ,
N Y = Speed of gear B = y – x ×
40
TA
= 100 – – 200 ×
30
TB
= + 366.7 r.p.m. = 366.7 r.p.m. (anticlockwise) Ans.
Example 13.17. In a gear train, as
shown in Fig. 13.23, gear B is connected to the
input shaft and gear F is connected to the output
shaft. The arm A carrying the compound wheels
D and E, turns freely on the output shaft. If the
input speed is 1000 r.p.m. counter- clockwise
when seen from the right, determine the speed of
the output shaft under the following conditions :
1. When gear C is fixed, and 2. when
gear C is rotated at 10 r.p.m. counter clockwise.
Solution. Given : T B = 20 ; T C = 80 ;
T D = 60 ; T E = 30 ; T F = 32 ; N B = 1000 r.p.m.
(counter-clockwise)
Fig. 13.23
The table of motions is given below :
Table 13.19. Table of motions.
Revolutions of elements
Step
No.
Conditions of motion
Arm A
Gear B
(or input
shaft)
1.
Arm fixed, gear B rotated
through + 1 revolution (i.e.
1 revolution anticlockwise)
0
+1
2.
Arm fixed, gear B rotated
through + x revolutions
0
+x
3.
Add + y revolutions to all
elements
+y
+y
4.
Total motion
+y
x +y
Compound
wheel D-E
+
TB
TD
Gear C
–
TB TD
×
TD TC
=–
+ x×
TB
TD
+y
y + x×
Gear F (or
output shaft)
–
TB TE
×
TD TF
TB
TC
– x×
TB
TC
– x×
+y
TB
TD
y – x×
TB TE
×
TD TF
+y
TB
TC
y – x×
TB TE
×
TD TF
Chapter 13 : Gear Trains
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461
1. Speed of the output shaft when gear C is fixed
Since the gear C is fixed, therefore from the fourth row of the table,
y –x×
TB
=0
TC
or
y – x×
20
=0
80
∴
y – 0.25 x = 0
...(i)
We know that the input speed (or the speed of gear B) is 1000 r.p.m. counter clockwise,
therefore from the fourth row of the table,
x + y = + 1000
...(ii)
From equations (i) and (ii), x = + 800, and y = + 200
∴ Speed of output shaft = Speed of gear F = y – x ×
TB TE
×
TD TF
20 30
×
= 200 – 187.5 = 12.5 r.p.m.
80 32
= 12.5 r.p.m. (counter clockwise) Ans.
= 200 – 800 ×
2. Speed of the output shaft when gear C is rotated at 10 r.p.m. counter clockwise
Since the gear C is rotated at 10 r.p.m. counter clockwise, therefore from the fourth row of the table,
y – x×
TB
= + 10
TC
∴
y – 0.25 x = 10
From equations (ii) and (iii),
x = 792,
∴ Speed of output shaft
or
y – x×
20
= 10
80
...(iii)
and
y = 208
= Speed of gear F = y – x ×
20 30
TB TE
×
= 208 – 792 ×
×
80 32
TD TF
= 208 – 185.6 = 22.4 r.p.m. = 22.4 r.p.m. (counter clockwise) Ans.
Example 13.18. Fig. 13.24 shows a differential
gear used in a motor car. The pinion A on the propeller
shaft has 12 teeth and gears with the crown gear B which
has 60 teeth. The shafts P and Q form the rear axles to
which the road wheels are attached. If the propeller
shaft rotates at 1000 r.p.m. and the road wheel attached
to axle Q has a speed of 210 r.p.m. while taking a turn,
find the speed of road wheel attached to axle P.
Solution. Given : T A = 12 ; T B = 60 ; N A = 1000
r.p.m. ; N Q = N D = 210 r.p.m.
Since the propeller shaft or the pinion A rotates at
1000 r.p.m., therefore speed of crown gear B,
12
TA
= 1000 ×
60
TB
= 200 r.p.m.
The table of motions is given below :
NB = NA ×
Fig. 13.24
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Table 13.20. Table of motions.
Revolutions of elements
Step No.
Conditions of motion
Gear B
Gear C
Gear E
+
TC
TE
Gear B fixed-Gear C rotated
through + 1 revolution (i.e. 1
revolution anticlockwise)
0
2.
Gear B fixed-Gear C rotated
through + x revolutions
0
+x
+ x×
3.
Add + y revolutions to all
elements
+y
+y
+y
4.
Total motion
+y
x+y
y +x×
1.
+1
Gear D
–
TC TE
×
= –1
TE TD
(3TC = TD )
TC
TE
–x
+y
TC
TE
y –x
Since the speed of gear B is 200 r.p.m., therefore from the fourth row of the table,
y = 200
...(i)
Also, the speed of road wheel attached to axle Q or the speed of gear D is 210 r.p.m., therefore from the fourth row of the table,
y – x = 210
or
x = y – 210 = 200 – 210 = – 10
∴ Speed of road wheel attached to axle P
= Speed of gear C = x + y
= – 10 + 200 = 190 r.p.m. Ans.
13.11. Torques in Epicyclic Gear Trains
Fig. 13.25. Torques in epicyclic gear trains.
When the rotating parts of an epicyclic gear train, as shown in Fig. 13.25, have no angular
acceleration, the gear train is kept in equilibrium by the three externally applied torques, viz.
1. Input torque on the driving member (T 1),
2. Output torque or resisting or load torque on the driven member (T 2),
3. Holding or braking or fixing torque on the fixed member (T 3).
Chapter 13 : Gear Trains
l
463
The net torque applied to the gear train must be zero. In other words,
∴
T 1 + T2 + T 3 = 0
...(i)
F1.r1 + F2.r2 + F3.r3 = 0
...(ii)
where F1, F2 and F3 are the corresponding externally applied forces at radii r1, r2 and r3.
Further, if ω1, ω2 and ω3 are the angular speeds of the driving, driven and fixed members
respectively, and the friction be neglected, then the net kinetic energy dissipated by the gear train
must be zero, i.e.
T 1.ω1 + T 2.ω2 + T 3.ω3 = 0
...(iii)
But, for a fixed member, ω3 = 0
∴
T 1.ω1 + T 2.ω2 = 0
...(iv)
Notes : 1. From equations (i) and (iv), the holding or braking torque T 3 may be obtained as follows :
ω1
ω2
...[From equation (iv)]
T 3 = – (T 1+ T 2 )
...[From equation (i)]
T2 = – T1 ×
and
ω
N
= T1 1 – 1 = T1 1 – 1
ω
N
2
2
2. When input shaft (or driving shaft) and output shaft (or driven shaft) rotate in the same direction,
then the input and output torques will be in opposite directions. Similarly, when the input and output shafts
rotate in opposite directions, then the input and output torques will be in the same direction.
Example 13.19. Fig. 13.26 shows an epicyclic gear train. Pinion
A has 15 teeth and is rigidly fixed to the motor shaft. The wheel B has 20
teeth and gears with A and also with the annular fixed wheel E. Pinion
C has 15 teeth and is integral with B (B, C being a compound gear
wheel). Gear C meshes with annular wheel D, which is keyed to the
machine shaft. The arm rotates about the same shaft on which A is fixed
and carries the compound wheel B, C. If the motor runs at 1000 r.p.m.,
find the speed of the machine shaft. Find the torque exerted on the
machine shaft, if the motor develops a torque of 100 N-m.
Fig. 13.26
Solution. Given : TA = 15 ; TB = 20 ; TC = 15 ; NA = 1000 r.p.m.; Torque developed by motor (or
pinion A) = 100 N-m
First of all, let us find the number of teeth on wheels D and E. Let T D and T E be the number of
teeth on wheels D and E respectively. Let dA, dB, dC, dD and dE be the pitch circle diameters of wheels
A , B, C, D and E respectively. From the geometry of the figure,
dE = dA + 2 dB
and
dD = dE – (dB – dC)
Since the number of teeth are proportional to their pitch circle diameters, therefore,
T E = T A + 2 T B = 15 + 2 × 20 = 55
and
T D = T E – (T B – T C) = 55 – (20 – 15) = 50
Speed of the machine shaft
The table of motions is given below :
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Theory of Machines
Table 13.21. Table of motions.
Revolutions of elements
Step
No.
Conditions of motion
Arm
Pinion
A
Compound
wheel B-C
Wheel D
1.
Arm fixed-pinion A
rotated through + 1
revolution
(anticlockwise)
0
+1
–
TA
TB
–
2.
Arm fixed-pinion A
rotated through + x
revolutions
0
+x
– x×
TA
TB
–x ×
3.
Add + y revolutions to
all elements
+y
+y
4.
Total motion
+y
x +y
+y
y – x×
TA TC
×
TB TD
TA TC
×
TB TD
+y
TA
TB
y –x×
TA TC
×
TB TD
Wheel E
−
TA TB
T
×
=− A
TB TE
TE
− x×
TA
TE
+y
y − x×
TA
TE
We know that the speed of the motor or the speed of the pinion A is 1000 r.p.m.
Therefore
x + y = 1000
...(i)
Also, the annular wheel E is fixed, therefore
y –x×
TA
=0
TE
y = x×
or
15
TA
=x×
= 0.273 x
55
TE
...(ii)
From equations (i) and (ii),
x = 786
and
y = 214
∴ Speed of machine shaft = Speed of wheel D,
ND = y – x ×
TA TC
15 15
×
= 214 – 786 ×
×
= + 37.15 r.p.m.
20 50
TB TD
= 37.15 r.p.m. (anticlockwise) Ans.
Torque exerted on the machine shaft
We know that
Torque developed by motor × Angular speed of motor
= Torque exerted on machine shaft
× Angular speed of machine shaft
or
100 × ωA = Torque exerted on machine shaft × ωD
∴ Torque exerted on machine shaft
= 100 ×
1000
ωA
N
= 100 × A = 100 ×
= 2692 N-m Ans.
37.15
ωD
ND
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Chapter 13 : Gear Trains
465
Example 13.20. An epicyclic gear train consists of a sun wheel
S, a stationary internal gear E and three identical planet wheels P
carried on a star- shaped planet carrier C. The size of different toothed
wheels are such that the planet carrier C rotates at 1/5th of the speed
of the sunwheel S. The minimum number of teeth on any wheel is 16.
The driving torque on the sun wheel is 100 N-m. Determine : 1. number of teeth on different wheels of the train, and 2. torque necessary to
keep the internal gear stationary.
Solution. Given :
NC =
NS
5
Fig. 13.27
1. Number of teeth on different wheels
The arrangement of the epicyclic gear train is shown in Fig. 13.27. Let T S and T E be the
number of teeth on the sun wheel S and the internal gear E respectively. The table of motions is
given below :
Table 13.22. Table of motions.
Revolutions of elements
Step
No.
Conditions of motion
Planet
carrier C
Sun
wheel S
Planet
wheel P
TS
TP
1.
Planet carrier C fixed, sunwheel S
rotates through + 1 revolution (i.e.
1 rev. anticlockwise)
0
+1
–
2.
Planet carrier C fixed, sunwheel S
rotates through + x revolutions
0
+x
– x×
3.
Add + y revolutions to all elements
+y
+y
+y
4.
Total motion
+y
x +y
y –x×
Internal gear E
–
TS TP
T
×
=– S
TP TE
TE
TS
TP
– x×
TS
TE
+y
TS
TP
y –x×
TS
TE
We know that when the sunwheel S makes 5 revolutions, the planet carrier C makes 1
revolution. Therefore from the fourth row of the table,
y = 1,
and
x+y=5
or
x=5–y=5–1=4
Since the gear E is stationary, therefore from the fourth row of the table,
y –x×
∴
TS
=0
TE
or
1–4×
TS
=0
TE
or
TS 1
=
TE 4
T E = 4T S
Since the minimum number of teeth on any wheel is 16, therefore let us take the number of
teeth on sunwheel,
T S = 16
∴
T E = 4 T S = 64 Ans.
Let dS, dP and dE be the pitch circle diameters of wheels S, P and E respectively. Now from the
geometry of Fig. 13.27,
dS + 2 dP = dE
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Theory of Machines
Assuming the module of all the gears to be same, the number of teeth are proportional to their
pitch circle diameters.
TS + 2 T P = TE
or
16 + 2 T P = 64
or
T P = 24 Ans.
2. Torque necessary to keep the internal gear stationary
We know that
Torque on S × Angular speed of S
= Torque on C × Angular speed of C
100 × ωS = Torque on C × ωC
∴
Torque on C = 100 ×
ωS
N
= 100 × S = 100 × 5 = 500 N-m
ωC
NC
∴ Torque necessary to keep the internal gear stationary
= 500 – 100 = 400 N-m Ans.
Example 13.21. In the epicyclic gear train, as
shown in Fig. 13.28, the driving gear A rotating in clockwise direction has 14 teeth and the fixed annular gear C
has 100 teeth. The ratio of teeth in gears E and D is 98 :
41. If 1.85 kW is supplied to the gear A rotating at 1200
r.p.m., find : 1. the speed and direction of rotation of gear
E, and 2. the fixing torque required at C, assuming 100
per cent efficiency throughout and that all teeth have the
same pitch.
Solution. Given : T A = 14 ; T C = 100 ; T E / T D
= 98 / 41 ; PA = 1.85 kW = 1850 W ; N A = 1200 r.p.m.
Fig. 13.28
Let dA, dB and dC be the pitch circle diameters of gears A , B and C respectively. From Fig.
13.28,
dA + 2 dB = dC
Gears are extensively used in trains for power transmission.
467
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Chapter 13 : Gear Trains
Since teeth of all gears have the same pitch and the number of teeth are proportional to their
pitch circle diameters, therefore
T – TA 100 – 14
TB = C
=
= 43
T A + 2T B = T C
or
2
2
The table of motions is now drawn as below :
Table 13.23. Table of motions.
Revolutions of elements
Step
No.
Conditions of motion
Arm
Gear
A
1.
Arm fixed-Gear A rotated
through – 1 revolution (i.e.
1 revolution clockwise)
0
–1
2.
3.
4.
Arm fixed-Gear A rotated
through – x revolutions
Add – y revolutions to all
elements
Total motion
Compound
gear B-D
+
Gear C
TA
TB
+
TA TB
×
TB TC
=+
0
–x
–y
–y
–y
– y–x
+ x×
TA
TB
–y + x ×
+
TA
TC
+x ×
–y
TA
TB
TA TD
×
TB TE
TA
TC
+ x×
–y
Gear E
–y + x ×
TA TD
×
TB TE
–y
TA
TC
–y + x ×
TA TD
×
TB TE
Since the annular gear C is fixed, therefore from the fourth row of the table,
– y +x×
∴
TA
=0
TC
–y + x ×
or
14
=0
100
– y + 0.14 x = 0
...(i)
Also, the gear A is rotating at 1200 r.p.m., therefore
– x – y = 1200
...(ii)
From equations (i) and (ii), x = – 1052.6,
and
y = – 147.4
1. Speed and direction of rotation of gear E
From the fourth row of the table, speed of gear E,
NE = – y + x ×
14 41
TA TD
×
= 147.4 – 1052.6 ×
×
43 98
TB TE
= 147.4 – 143.4 = 4 r.p.m.
= 4 r.p.m. (anticlockwise) Ans.
2. Fixing torque required at C
We know that torque on A =
PA × 60 1850 × 60
=
= 14.7 N-m
2π N A
2π × 1200
Since the efficiency is 100 per cent throughout, therefore the power available at E (PE) will
be equal to power supplied at A (PA).
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Theory of Machines
=
∴ Torque on E
PA × 60 1850 × 60
=
= 4416 N-m
2π × N E
2π × 4
∴ Fixing torque required at C
= 4416 – 14.7 = 4401.3 N-m Ans.
Example 13.22. An over drive for a vehicle consists of an
epicyclic gear train, as shown in Fig. 13.29, with compound planets
B-C. B has 15 teeth and meshes with an annulus A which has 60
teeth. C has 20 teeth and meshes with the sunwheel D which is fixed.
The annulus is keyed to the propeller shaft Y which rotates at 740
rad /s. The spider which carries the pins upon which the planets
revolve, is driven directly from main gear box by shaft X, this shaft
being relatively free to rotate with respect to wheel D. Find the
speed of shaft X, when all the teeth have the same module.
When the engine develops 130 kW, what is the holding
torque on the wheel D ? Assume 100 per cent efficiency
throughout.
Fig. 13.29
Solution. Given : TB = 15 ; TA = 60 ; TC = 20 ; ωY = ωA = 740 rad /s ; P = 130 kW = 130 × 103 W
First of all, let us find the number of teeth on the sunwheel D (T D). Let dA , dB , dC and dD be
the pitch circle diameters of wheels A , B, C and D respectively. From Fig. 13.29,
dD
d
d
d
or
dD + dC + dB = dA
+ C + B = A
2
2
2
2
Since the module is same for all teeth and the number of teeth are proportional to their pitch
circle diameters, therefore
TD + T C + T B = T A
or
T D = T A – (T C + T B) = 60 – (20 + 15) = 25
The table of motions is given below :
Table 13.24. Table of motions.
Revolutions of elements
Step
No.
Conditions of motion
Arm (or
shaft X)
Wheel D
Compound
wheel C-B
TD
TC
1.
Arm fixed-wheel D rotated
through + 1 revolution
(anticlockwise)
0
+1
–
2.
Arm fixed-wheel D rotated
through + x revolutions
0
+x
– x×
3.
Add + y revolutions to all elements
+y
+y
+y
4.
Total motion
+y
x+y
y –x×
Wheel A
(or shaft Y)
–
TD
TC
TD TB
×
TC TA
–x ×
TD TB
×
TC TA
+y
TD
TC
y –x×
TD TB
×
TC TA
Since the shaft Y or wheel A rotates at 740 rad/s, therefore
y –x ×
TD TB
×
= 740
TC TA
y – 0.3125 x = 740
or
y –x×
25 15
×
= 740
20 60
...(i)
Chapter 13 : Gear Trains
l
469
Also the wheel D is fixed, therefore
x +y=0
or
y=–x
...(ii)
From equations (i) and (ii),
x = – 563.8
and
y = 563.8
Speed of shaft X
Since the shaft X will make the same number of revolutions as the arm, therefore
Speed of shaft X , ωX = Speed of arm = y = 563.8 rad/s Ans.
Holding torque on wheel D
We know that torque on A = P/ωA = 130 × 103 / 740 = 175.7 N-m
and
Torque on X = P/ωX = 130 × 103/563.8 = 230.6 N-m
∴ Holding torque on wheel D
= 230.6 – 175.7 = 54.9 N-m Ans.
Example 13.23. Fig. 13.30 shows some details of a compound epicyclic gear drive where I
is the driving or input shaft and O is the driven or output shaft which carries two arms A and B
rigidly fixed to it. The arms carry planet wheels which mesh with annular wheels P and Q and the
sunwheels X and Y. The sun wheel X is a part of Q. Wheels Y and Z are fixed to the shaft I. Z engages
with a planet wheel carried on Q and this planet wheel engages the fixed annular wheel R. The
numbers of teeth on the wheels are :
P = 114, Q = 120, R = 120, X = 36, Y = 24 and Z = 30.
Fig. 13.30.
The driving shaft I makes 1500 r.p.m.clockwise looking from our right and the input at I is
7.5 kW.
1. Find the speed and direction of rotation of the driven shaft O and the wheel P.
2. If the mechanical efficiency of the drive is 80%, find the torque tending to rotate the fixed
wheel R.
Solution. Given : T P =144 ; T Q = 120 ; T R = 120 ; T X = 36 ; T Y = 24 ; T Z = 30 ; N I = 1500
r.p.m. (clockwise) ; P = 7.5 kW = 7500 W ; η = 80% = 0.8
First of all, consider the train of wheels Z,R and Q (arm). The revolutions of various wheels
are shown in the following table.
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Theory of Machines
Table 13.25. Table of motions.
Revolutions of elements
Step No.
Conditions of motion
Q (Arm)
Z (also I)
R (Fixed)
1.
Arm fixed-wheel Z rotates through + 1
revolution (anticlockwise)
0
+1
–
TZ
TR
2.
Arm fixed-wheel Z rotates through + x revolutions
0
+x
– x×
TZ
TR
3.
Add + y revolutions to all elements
+y
+y
4.
Total motion
+y
x +y
+y
y –x×
TZ
TR
Since the driving shaft I as well as wheel Z rotates at 1500 r.p.m. clockwise, therefore
x + y = – 1500
...(i)
Also, the wheel R is fixed. Therefore
y –x×
TZ
=0
TR
y=x×
or
30
TZ
= x×
= 0.25 x
120
TR
...(ii)
From equations (i) and (ii),
x = – 1200,
and
y = – 300
Now consider the train of wheels Y , Q, arm A , wheels P and X . The revolutions of various
elements are shown in the following table.
Table 13.26. Table of motions.
Revolutions of elements
Step
No.
Conditions of motion
Arm A, B
and Shaft O
Wheel Y
Compound
wheel Q-X
TY
TQ
1.
Arm A fixed-wheel Y
rotates through + 1
revolution (anticlockwise)
0
+1
–
2.
Arm A fixed-wheel Y rotates
through + x 1 revolutions
0
+ x1
– x1 ×
3.
Add + y 1 revolutions to all
elements
+ y1
+ y1
4.
Total motion
+ y1
x1 + y 1
+
TY
TQ
+ y1
y1 – x1 ×
or
TY
= y = – 300
TQ
y1 – x1 ×
24
= – 300
120
TY TX
×
TQ TP
+ x1 ×
TY TX
×
TQ TP
+ y1
TY
TQ
Since the speed of compound wheel Q-X is same as that of Q, therefore
y1 – x1 ×
Wheel P
y1 + x1 ×
TY TX
×
TQ TP
Chapter 13 : Gear Trains
∴
y 1 = 0.2 x 1 – 300
Also
∴
l
471
...(iii)
Speed of wheel Y = Speed of wheel Z or shaft I
x 1 + y 1 = x + y = – 1500
x 1 + 0.2 x 1 – 300 = – 1500
...(iv)
...[From equation (iii)]
1.2 x 1= – 1500 + 300 = – 1200
or
x 1 = – 1200/1.2 = – 1000
and
y 1 = – 1500 – x 1 = – 1500 + 1000 = – 500
1. Speed and direction of the driven shaft O and the wheel P
Speed of the driven shaft O,
N O = y 1 = – 500 = 500 r.p.m. clockwise Ans.
and
Speed of the wheel P, N P = y1 + x1 ×
TY TX
24
36
×
= – 500 – 1000 ×
×
120 144
TQ TP
= – 550 = 550 r.p.m. clockwise Ans.
2. Torque tending to rotate the fixed wheel R
We know that the torque on shaft I or input torque
T1 =
7500 × 60
P × 60
=
= 47.74 N-m
2π × N1 2π × 1500
and torque on shaft O or output torque,
T2 =
η × P × 60 0.8 × 7500 × 60
=
= 114.58 N-m
2π × NO
2π × 500
Since the input and output shafts rotate in the same direction (i.e. clockwise), therefore input
and output torques will be in opposite direction.
∴ Torque tending to rotate the fixed wheel R
= T2 – T 1 = 114.58 – 47.74 = 66.84 N-m Ans.
Example 13.24. An epicyclic bevel gear train (known as Humpage’s reduction gear) is shown
in Fig. 13.31. It consists of a fixed wheel C, the
driving shaft X and the driven shaft Y. The compound
wheel B-D can revolve on a spindle F which can
turn freely about the axis X and Y.
Show that (i) if the ratio of tooth numbers
T B / T D is greater than TC / TE , the wheel E will rotate in the same direction as wheel A, and (ii) if the
ratio TB / T D is less than TC / T E, the direction of E is
reversed.
If the numbers of teeth on wheels A, B, C, D
and E are 34, 120, 150, 38 and 50 respectively and
7.5 kW is put into the shaft X at 500 r.p.m., what is
Fig. 13.31
the output torque of the shaft Y, and what are the
forces (tangential to the pitch cones) at the contact
points between wheels D and E and between wheels B and C, if the module of all wheels is 3.5 mm ?
Solution. Given : T A = 34 ; T B = 120 ; T C = 150 ; T D = 38 ; T E = 50 ; PX = 7.5 kW = 7500 W ;
N X = 500 r.p.m. ; m = 3.5 mm
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Theory of Machines
The table of motions is given below :
Table 13.27. Table of motions.
Revolutions of elements
Step
No.
1.
Conditions of motion
Spindle fixed, wheel A
is rotated through + 1
revolution
Spindle
F
Wheel A
(or shaft X)
0
+1
Compound
wheel B-D
+
TA
TB
Wheel C
−
TA TB
×
TB TC
=–
TA
TB
2.
Spindle fixed, wheel A
is rotated through + x
revolutions
0
+x
+ x×
3.
Add + y revolutions to
all elements
+y
+y
+y
4.
Total motion
+y
x +y
y + x×
Wheel E (or
shaft Y)
−
TA TD
×
TB TE
TA
TC
– x×
TA
TC
–x ×
+y
TA
TB
y – x×
TA TD
×
TB TE
+y
TA
TC
y – x×
TA TD
×
TB TE
Let us assume that the driving shaft X rotates through 1 revolution anticlockwise, therefore
the wheel A will also rotate through 1 revolution anticlockwise.
∴
x + y = + 1 or
y=1–x
We also know that the wheel C is fixed, therefore
...(i)
TA
=0
TC
or
(1 – x) – x ×
T
1 – x 1 + A = 0
TC
or
T + TA
x C
=1
TC
y – x×
x=
and
TA
=0
TC
...[From equation (i)]
TC
TC + TA
...(ii)
From equation (i),
y =1– x =1–
We know that speed of wheel E,
NE = y – x ×
=
and the speed of wheel A ,
TA
TC + TA
TC
TA
=
TC + TA TC + TA
...(iii)
TC
TA TD
TA
T
T
–
×
=
× A × D
TB TE TC + TA TC + TA TB TE
TC TD
×
1 –
TB TE
...(iv)
N A = x + y = + 1 revolution
(i) If
TB TC
>
or T B × T E > T C × T D , then the equation (iv) will be positive. Therefore the
TD TE
wheel E will rotate in the same direction as wheel A . Ans.
Chapter 13 : Gear Trains
(ii) If
l
473
TB TC
or T B × T E < T C × T D , then the equation (iv) will be negative. Therefore the
<
TD TE
wheel E will rotate in the opposite direction as wheel A. Ans.
Output torque of shaft Y
We know that the speed of the driving shaft X (or wheel A ) or input speed is 500 r.p.m.,
therefore from the fourth row of the table,
x + y = 500
or
y = 500 – x
...(v)
Since the wheel C is fixed, therefore
TA
=0
TC
or
(500 – x) – x ×
500 – x – 0.227 x = 0
or
x = 500/1.227 = 407.5 r.p.m.
y – x×
∴
and
34
=0
150
...[From equation (v)]
y = 500 – x = 500 – 407.5 = 92.5 r.p.m.
Since the speed of the driven or output shaft Y (i.e. N Y) is equal to the speed of wheel E
(i.e. N E), therefore
34 38
TA TD
×
= 92.5 – 407.5 ×
×
120 50
TB TE
= 92.5 – 87.75 = 4.75 r.p.m.
NY = NE = y – x ×
Assuming 100 per cent efficiency of the gear train, input power PX is equal to output power
(PY), i.e.
PY = PX = 7.5 kW = 7500 W
∴ Output torque of shaft Y ,
=
PY × 60 7500 × 60
=
= 15 076 N-m = 15.076 kN-m Ans.
2π N Y
2π × 4.75
Tangential force between wheels D and E
We know that the pitch circle radius of wheel E,
m × TE 3.5 × 50
=
= 87.5 mm = 0.0875 m
2
2
∴ Tangential force between wheels D and E,
rE =
=
Torque on wheel E
15.076
=
= 172.3 kN Ans.
Pitch circle radius of wheel E 0.0875
...(∴ Torque on wheel E = Torque on shaft Y )
Tangential force between wheels B and C
We know that the input torque on shaft X or on wheel A
=
PX × 60 7500 × 60
=
= 143 N-m
2π N X
2π × 500
∴ Fixing torque on the fixed wheel C
= Torque on wheel E – Torque on wheel A
= 15 076 – 143 = 14 933 N-m = 14.933 kN-m
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Theory of Machines
Pitch circle radius of wheel C,
m × TC 3.5 × 150
=
= 262.5 mm = 0.2625 m
2
2
Tangential force between wheels B and C
rC =
=
Fixing torque on wheel C 14.933
=
= 57 kN Ans.
0.2625
rC
EXERCISES
1.
A compound train consists of six gears. The number of teeth on the gears are as follows :
Gear
:
A
B
C
D
E
F
No. of teeth :
60
40
50
25
30
24
The gears B and C are on one shaft while the gears D and E are on another shaft. The gear A drives gear
B, gear C drives gear D and gear E drives gear F. If the gear A transmits 1.5 kW at 100 r.p.m. and the gear
train has an efficiency of 80 per cent, find the torque on gear F.
[Ans. 30.55 N-m]
2.
3.
4.
Two parallel shafts are to be connected by spur gearing. The approximate distance between the shafts
is 600 mm. If one shaft runs at 120 r.p.m. and the other at 360 r.p.m., find the number of teeth on each
wheel, if the module is 8 mm. Also determine the exact distance apart of the shafts.
[Ans. 114, 38 ; 608 mm]
In a reverted gear train, as shown in Fig. 13.32, two shafts A and B are
in the same straight line and are geared together through an intermediate parallel shaft C. The gears connecting the shafts A and C have a
module of 2 mm and those connecting the shafts C and B have a
module of 4.5 mm. The speed of shaft A is to be about but greater than
12 times the speed of shaft B, and the ratio at each reduction is same.
Find suitable number of teeth for gears. The number of teeth of each
gear is to be a minimum but not less than 16. Also find the exact
velocity ratio and the distance of shaft C from A and B.
[Ans. 36, 126, 16, 56 ; 12.25 ; 162 mm]
Fig. 13.32
In an epicyclic gear train, as shown in Fig.13.33, the number of teeth
on wheels A , B and C are 48, 24 and 50 respectively. If the arm rotates at 400 r.p.m., clockwise,
find : 1. Speed of wheel C when A is fixed, and 2. Speed of wheel A when C is fixed.
[Ans. 16 r.p.m. (clockwise) ; 16.67 (anticlockwise)]
Fig. 13.33
Fig. 13.34
Chapter 13 : Gear Trains
5.
6.
475
In an epicyclic gear train, as shown in Fig. 13.34, the wheel C is keyed to the shaft B and wheel F is
keyed to shaft A . The wheels D and E rotate together on a pin fixed to the arm G. The number of teeth
on wheels C, D, E and F are 35, 65, 32 and 68 respectively.
If the shaft A rotates at 60 r.p.m. and the shaft B rotates at 28 r.p.m. in the opposite direction, find
the speed and direction of rotation of arm G. [Ans. 90 r.p.m., in the same direction as shaft A]
An epicyclic gear train, as shown in Fig. 13.35, is composed of a fixed annular wheel A having 150
teeth. The wheel A is meshing with wheel B which drives wheel D through an idle wheel C, D being
concentric with A . The wheels B and C are carried on an arm which revolves clockwise at 100 r.p.m.
about the axis of A and D. If the wheels B and D have 25 teeth and 40 teeth respectively, find the
number of teeth on C and the speed and sense of rotation of C.
[Ans. 30 ; 600 r.p.m. clockwise]
Fig. 13.35
7.
l
Fig. 13.36
Fig. 13.36, shows an epicyclic gear train with the following details :
A has 40 teeth external (fixed gear) ; B has 80 teeth internal ; C - D is a compound wheel having 20 and
50 teeth (external) respectively, E-F is a compound wheel having 20 and 40 teeth (external) respectively, and G has 90 teeth (external).
The arm runs at 100 r.p.m. in clockwise direction. Determine the speeds for gears C, E, and B.
[Ans. 300 r.p.m. clockwise ; 400 r.p.m. anticlockwise ; 150 r.p.m. clockwise]
8.
An epicyclic gear train, as shown in Fig. 13.37, has a sun wheel S of 30 teeth and two planet wheels
P-P of 50 teeth. The planet wheels mesh with the internal teeth of a fixed annulus A . The driving shaft
carrying the sunwheel, transmits 4 kW at 300 r.p.m. The driven shaft is connected to an arm which
carries the planet wheels. Determine the speed of the driven shaft and the torque transmitted, if the
overall efficiency is 95%.
[Ans. 56.3 r.p.m. ; 644.5 N-m]
Fig. 13.37
9.
Fig. 13.38
An epicyclic reduction gear, as shown in Fig. 13.38, has a shaft A fixed to arm B. The arm B has a pin
fixed to its outer end and two gears C and E which are rigidly fixed, revolve on this pin. Gear C
meshes with annular wheel D and gear E with pinion F. G is the driver pulley and D is kept stationary.
The number of teeth are : D = 80 ; C = 10 ; E = 24 and F = 18.
If the pulley G runs at 200 r.p.m. ; find the speed of shaft A.
[Ans. 17.14 r.p.m. in the same direction as that of G]
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10.
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Theory of Machines
A reverted epicyclic gear train for a hoist block is shown in
Fig. 13.39. The arm E is keyed to the same shaft as the load
drum and the wheel A is keyed to a second shaft which carries a chain wheel, the chain being operated by hand. The
two shafts have common axis but can rotate independently.
The wheels B and C are compound and rotate together on a
pin carried at the end of arm E. The wheel D has internal
teeth and is fixed to the outer casing of the block so that it
does not rotate.
The wheels A and B have 16 and 36 teeth respectively with a
module of 3 mm. The wheels C and D have a module of 4
mm. Find : 1. the number of teeth on wheels C and D when
the speed of A is ten times the speed of arm E, both rotating
in the same sense, and 2. the speed of wheel D when the
wheel A is fixed and the arm E rotates at 450 r.p.m.
anticlockwise.
[Ans. TC = 13 ; TD = 52 ; 500 r.p.m. anticlockwise]
Fig. 13.39
11.
A compound epicyclic gear is shown diagrammatically in Fig. 13.40. The gears A , D and E are free to
rotate on the axis P. The compound gear B and C rotate together on the axis Q at the end of arm F. All
the gears have equal pitch. The number of external teeth on the gears A , B and C are 18, 45 and 21
respectively. The gears D and E are annular gears. The gear A rotates at 100 r.p.m. in the anticlockwise
direction and the gear D rotates at 450 r.p.m. clockwise. Find the speed and direction of the arm and
the gear E.
[Ans. 400 r.p.m. clockwise ; 483.3 r.p.m. clockwise]
12.
In an epicyclic gear train of the ‘sun and planet type’ as shown in Fig. 13.41, the pitch circle diameter
of the internally toothed ring D is to be 216 mm and the module 4 mm. When the ring D is stationary,
the spider A , which carries three planet wheels C of equal size, is to make one revolution in the same
sense as the sun wheel B for every five revolutions of the driving spindle carrying the sunwheel B.
Determine suitable number of teeth for all the wheels and the exact diameter of pitch circle of the ring.
[Ans. TB = 14 , TC = 21 , TD = 56 ; 224 mm]
Fig. 13.40
13.
Fig. 13.41
An epicyclic train is shown in Fig. 13.42. Internal gear A is keyed to the driving shaft and has 30 teeth.
Compound wheel C and D of 20 and 22 teeth respectively are free to rotate on the pin fixed to the arm
P which is rigidly connected to the driven shaft. Internal gear B which has 32 teeth is fixed. If the
driving shaft runs at 60 r.p.m. clockwise, determine the speed of the driven shaft. What is the direction
of rotation of driven shaft with reference to driving shaft?
[Ans. 1980 r.p.m. clockwise]
Chapter 13 : Gear Trains
14.
15.
l
477
Fig. 13.42
Fig. 13.43
A shaft Y is driven by a co-axial shaft X by means of an epicyclic gear train, as shown in Fig. 13.43.
The wheel A is keyed to X and E to Y . The wheels B and D are compound and carried on an arm F
which can turn freely on the common axes of X and Y . The wheel C is fixed. If the numbers of teeth
on A , B, C, D and E are respectively 20, 64, 80, 30 and 50 and the shaft X makes 600 r.p.m.,
determine the speed in r.p.m. and sense of rotation of the shaft Y .
[Ans. 30 r.p.m. in the same sense as shaft X]
An epicyclic bevel gear train, as shown in Fig. 13.44, has fixed gear B meshing with pinion C. The
gear E on the driven shaft meshes with the pinion D. The pinions C and D are keyed to a shaft,
which revolves in bearings on the arm A . The arm A is keyed to the driving shaft. The number of
teeth are : T B = 75, T C = 20, T D = 18, and T E = 70. Find the speed of the driven shaft, if 1. the driving
shaft makes 1000 r.p.m., and 2. the gear B turns in the same sense as the driving shaft at 400
r.p.m., the driving shaft still making 1000 r.p.m.
[Ans. 421.4 r.p.m. in the same direction as driving shaft]
16.
The epicyclic gear train is shown in Fig. 13.45. The wheel D is held stationary by the shaft A and the
arm B is rotated at 200 r.p.m. The wheels E (20 teeth) and F (40 teeth) are fixed together and rotate
freely on the pin carried by the arm. The wheel G (30 teeth) is rigidly attached to the shaft C. Find the
speed of shaft C stating the direction of rotation to that of B.
If the gearing transmits 7.5 kW, what will be the torque required to hold the shaft A stationary, neglecting all friction losses?
[Ans. 466.7 r.p.m. in opposite direction of B; 511.5 N-m in opposite direction of B]
Fig. 13.44
17.
Fig. 13.45
An epicyclic gear train, as shown in Fig. 13.46, consists of two sunwheels A and D with 28 and 24
teeth respectively, engaged with a compound planet wheels B and C with 22 and 26 teeth. The sunwheel
478
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Theory of Machines
D is keyed to the driven shaft and the sunwheel A is a fixed wheel co-axial with the driven shaft. The
planet wheels are carried on an arm E from the driving shaft which is co-axial with the driven shaft.
Find the velocity ratio of gear train. If 0.75 kW is transmitted and input speed being 100 r.p.m.,
determine the torque required to hold the sunwheel A .
[Ans. 2.64 ; 260.6 N-m]
Fig. 13.46
18.
Fig. 13.47
In the epicyclic reduction gear, as shown in Fig. 13.47, the sunwheel D has 20 teeth and is keyed
to the input shaft. Two planet wheels B, each having 50 teeth, gear with wheel D and are carried
by an arm A fixed to the output shaft. The wheels B also mesh with an internal gear C which is
fixed. The input shaft rotates at 2100 r.p.m. Determine the speed of the output shaft and the torque
required to fix C when the gears are transmitting 30 kW.
[Ans. 300 r.p.m. in the same sense as the input shaft ; 818.8 N-m]
19.
An epicyclic gear train for an electric motor is shown in Fig. 13.48. The wheel S has 15 teeth and is
fixed to the motor shaft rotating at 1450 r.p.m. The planet P has 45 teeth, gears with fixed annulus A
and rotates on a spindle carried by an arm which is fixed to the output shaft. The planet P also gears
with the sun wheel S. Find the speed of the output shaft. If the motor is transmitting 1.5 kW, find the
torque required to fix the annulus A .
[Ans. 181.3 r.p.m. ; 69.14 N-m]
Fig. 13.48
20.
Fig. 13.49
An epicyclic gear consists of bevel wheels as shown in Fig. 13.49. The driving pinion A has 20 teeth
and meshes with the wheel B which has 25 teeth. The wheels B and C are fixed together and turn freely
on the shaft F. The shaft F can rotate freely about the main axis X X. The wheel C has 50 teeth and
meshes with wheels D and E, each of which has 60 teeth. Find the speed and direction of E when A
rotates at 200 r.p.m., if
1. D is fixed, and 2. D rotates at 100 r.p.m., in the same direction as A .
In both the cases, find the ratio of the torques transmitted by the shafts of the wheels A and E, the
friction being neglected.
[Ans. 800 r.p.m. in the opposite direction of A ; 300 r.p.m. in the opposite
direction of A ; 4 ; 1.5]
Chapter 13 : Gear Trains
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479
DO YOU KNOW ?
1.
2.
3.
4.
5.
6.
What do you understand by ‘gear train’? Discuss the various types of gear trains.
Explain briefly the differences between simple, compound, and epicyclic gear trains. What are the
special advantages of epicyclic gear trains ?
Explain the procedure adopted for designing the spur wheels.
How the velocity ratio of epicyclic gear train is obtained by tabular method?
Explain with a neat sketch the ‘sun and planet wheel’.
What are the various types of the torques in an epicyclic gear train ?
OBJECTIVE TYPE QUESTIONS
1. In a simple gear train, if the number of idle gears is odd, then the motion of driven gear will
(a) be same as that of driving gear
(b) be opposite as that of driving gear
(c) depend upon the number of teeth on the driving gear
(d) none of the above
2. The train value of a gear train is
(a) equal to velocity ratio of a gear train
(b) reciprocal of velocity ratio of a gear train
(c) always greater than unity
(d) always less than unity
3. When the axes of first and last gear are co-axial, then gear train is known as
(a) simple gear train
(b) compound gear train
(c) reverted gear train
(d) epicyclic gear train
4. In a clock mechanism, the gear train used to connect minute hand to hour hand, is
(a) epicyclic gear train
(b) reverted gear train
(c) compound gear train
(d) simple gear train
5. In a gear train, when the axes of the shafts, over which the gears are mounted, move relative to a fixed
axis, is called
(a) simple gear train
(b) compound gear train
(c) reverted gear train
(d) epicyclic gear train
6. A differential gear in an automobile is a
(a) simple gear train
(b) epicyclic gear train
(c) compound gear train
(d) none of these
7. A differential gear in automobilies is used to
(a) reduce speed
(b) assist in changing speed
(c)
provide jerk-free movement of vehicle
(d)
help in turning
ANSWERS
1. (a)
2. (b)
6. (b)
7. (d)
3. (c)
4. (b)
5. (d)
GO To FIRST
CONTENTS
CONTENTS
480
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Theory of Machines
14
Gyroscopic
Couple and
Precessional
Motion
Fea
tur
es
eatur
tures
1. Introduction.
2. Precessional Angular
Motion.
3. Gyroscopic Couple.
4. Effect of Gyroscopic Couple
on an Aeroplane.
5. Terms Used in a Naval Ship.
6. Effect of Gyroscopic Couple
on a Naval Ship during
Steering.
7. Effect of Gyroscopic Couple
on a Naval Ship during
Pitching.
8. Effect of Gyroscopic Couple
on a Navalship during
Rolling.
9. Stability of a Four Wheel
drive Moving in a Curved
Path.
10. Stability of a Two Wheel
Vehicle Taking a Turn.
11. Effect of Gyroscopic Couple
on a Disc Fixed Rigidly at a
Certain Angle to a Rotating
Shaft.
14.1. Intr
oduction
Introduction
We have already discussed that,
1. When a body moves along a curved path with a
uniform linear velocity, a force in the direction of centripetal
acceleration (known as centripetal force) has to be applied
externally over the body, so that it moves along the required
curved path. This external force applied is known as active
force.
2. When a body, itself, is moving with uniform linear velocity along a circular path, it is subjected to the centrifugal force* radially outwards. This centrifugal force is
called reactive force. The action of the reactive or centrifugal force is to tilt or move the body along radially outward
direction.
Note: Whenever the effect of any force or couple over a moving or
rotating body is to be considered, it should be with respect to the
reactive force or couple and not with respect to active force or
couple.
*
CONTENTS
CONTENTS
Centrifugal force is equal in magnitude to centripetal force but
opposite in direction.
480
Chapter 14 : Gyroscopic Couple and Precessional Motion
Spin axis
l
481
Input axis
Gyorscope
will resist
movement
in these
directions
Wheel
Axle
Output axis
Gimbals
Gyroscopic inertia prevents a spinning top from falling sideways.
14.2. Precessional Angular Motion
We have already discussed that the angular acceleration is the rate of change of angular
velocity with respect to time. It is a vector quantity and may be represented by drawing a vector
diagram with the help of right hand screw rule (see chapter 2, Art. 2.13).
Fig. 14.1. Precessional angular motion.
Consider a disc, as shown in Fig. 14.1 (a), revolving or spinning about the axis O X (known as
axis of spin) in anticlockwise when seen from the front, with an angular velocity ω in a plane at right
angles to the paper.
After a short interval of time δt, let the disc be spinning about the new axis of spin OX ′ (at an
angle δθ) with an angular velocity (ω + δω). Using the right hand screw rule, initial angular velocity
of the disc (ω) is represented by vector ox; and the final angular velocity of the disc (ω + δω) is
represented by vector ox′ as shown in Fig. 14.1 (b). The vector x x′ represents the change of angular
velocity in time δt i.e. the angular acceleration of the disc. This may be resolved into two components,
one parallel to ox and the other perpendicular to ox.
Component of angular acceleration in the direction of ox,
xr or – ox ox ′ cos δθ – ox
αt =
=
=
δt
δt
δt
(ω + δω ) cos δθ – ω ω cos δθ + δω cos δθ – ω
=
δt
δt
Since δθ is very small, therefore substituting cos δθ = 1, we have
ω + δω – ω δω
αt =
=
δt
δt
=
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Theory of Machines
In the limit, when δt → 0 ,
δω d ω
αt = Lt
=
δt → 0 δt
dt
Component of angular acceleration in the direction perpendicular to ox,
rx ′ ox′ sin δθ (ω + δω) sin δθ ω sin δθ + δω .sin δθ
αc =
=
=
=
δt
δt
δt
δt
Since δθ in very small, therefore substituting sin δθ = δθ, we have
ω . δθ + δω . δθ ω . δθ
αc =
=
δt
δt
...(Neglecting δω.δθ, being very small)
In the limit when δt → 0,
dθ
ω . δθ
= ω×
= ω. ωP
dt
δt
∴ Total angular acceleration of the disc
= vector x x′ = vector sum of αt and αc
α c = Lt
δt → 0
... Substituting
dθ
= ωP
dt
dω
dθ dω
+ ω×
=
+ ω . ωP
dt
dt
dt
where dθ/dt is the angular velocity of the axis of spin about a certain axis, which is perpendicular to
the plane in which the axis of spin is going to rotate. This angular velocity of the axis of spin (i.e.
dθ/dt) is known as angular velocity of precession and is denoted by ωP. The axis, about which the
axis of spin is to turn, is known as axis of precession. The angular motion of the axis of spin about the
axis of precession is known as precessional angular motion.
=
Notes:1. The axis of precession is perpendicular to the plane in which the axis of spin is going to rotate.
2. If the angular velocity of the disc remains constant at all positions of the axis of spin, then dθ/dt is
zero; and thus αc is zero.
3. If the angular velocity of the disc changes the direction, but remains constant in magnitude, then
angular acceleration of the disc is given by
αc = ω.dθ/dt = ω.ωP
The angular acceleration αc is known as gyroscopic acceleration.
Evaporators change liquid
hydrogen to gas
Fuel tank
Engine
This experimental car burns hydrogen fuel in an ordinary piston engine. Its exhaust gases cause no pollution,
because they contain only water vapour.
Note : This picture is given as additional information and is not a direct example of the current chapter.
Chapter 14 : Gyroscopic Couple and Precessional Motion
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483
14.3. Gyroscopic Couple
Consider a disc spinning with an angular velocity ω rad/s about the axis of spin OX, in
anticlockwise direction when seen from the front, as shown in Fig. 14.2 (a). Since the plane in which
the disc is rotating is parallel to the plane YOZ, therefore it is called plane of spinning. The plane
XOZ is a horizontal plane and the axis of spin rotates in a plane parallel to the horizontal plane about
an axis O Y. In other words, the axis of spin is said to be rotating or processing about an axis O Y. In
other words, the axis of spin is said to be rotating or processing about an axis OY (which is perpendicular
to both the axes OX and OZ) at an angular velocity ωP rap/s. This horizontal plane XOZ is called
plane of precession and O Y is the axis of precession.
Let
I = Mass moment of inertia of the disc about OX, and
ω = Angular velocity of the disc.
∴ Angular momentum of the disc
= I.ω
Since the angular momentum is a vector quantity, therefore it may be represented by the
→
vector ox , as shown in Fig. 14.2 (b). The axis of spin OX is also rotating anticlockwise when seen
from the top about the axis O Y. Let the axis O X is turned in the plane XOZ through a small angle δθ
radians to the position OX ′ , in time δt seconds. Assuming the angular velocity ω to be constant, the
angular momentum will now be represented by vector ox′.
Fig. 14.2. Gyroscopic couple.
∴ Change in angular momentum
→
→
→
→
→
...(in the direction of xx′ )
= ox ′ – ox = xx ′ = ox . δθ
= I. ω.δθ
and rate of change of angular momentum
δθ
= I .ω ×
dt
Since the rate of change of angular momentum will result by the application of a couple to the
disc, therefore the couple applied to the disc causing precession,
C = Lt I . ω ×
δt →0
dθ
δθ
= I .ω ×
= I . ω . ωP
dt
δt
... Q
dθ
= ωP
dt
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Theory of Machines
where ωP = Angular velocity of precession of the
axis of spin or the speed of rotation of the axis of
spin about the axis of precession O Y.
In S.I. units, the units of C is N-m when I is
in kg-m2.
It may be noted that
1. The couple I.ω.ωp, in the direction of
the vector x x′ (representing the change in angular
momentum) is the active gyroscopic couple, which
has to be applied over the disc when the axis of
spin is made to rotate with angular velocity ωP about
the axis of precession. The vector x x′ lies in the
plane XOZ or the horizontal plane. In case of a very
small displacement δθ, the vector x x′ will be
perpendicular to the vertical plane X O Y. Therefore
the couple causing this change in the angular
Above picture shows an aircraft propeller.
momentum will lie in the plane X O Y. The vector
These rotors play role in gyroscopic couple.
x x′ , as shown in Fig. 14.2 (b), represents an
anticlockwise couple in the plane X O Y. Therefore, the plane XOY is called the plane of active
gyroscopic couple and the axis OZ perpendicular to the plane X O Y, about which the couple acts, is
called the axis of active gyroscopic couple.
2. When the axis of spin itself moves with angular velocity ωP, the disc is subjected to
reactive couple whose magnitude is same (i.e. I. ω.ωP) but opposite in direction to that of active
couple. This reactive couple to which the disc is subjected when the axis of spin rotates about the axis
of precession is known as reactive gyroscopic couple. The axis of the reactive gyroscopic couple is
represented by OZ′ in Fig. 14.2 (a).
3. The gyroscopic couple is usually applied through the bearings which support the shaft.
The bearings will resist equal and opposite couple.
4. The gyroscopic principle is used in an instrument or toy known as gyroscope. The
gyroscopes are installed in ships in order to minimize the rolling and pitching effects of waves. They
are also used in aeroplanes, monorail cars, gyrocompasses etc.
Example 14.1. A uniform disc of diameter 300 mm and of mass 5 kg is mounted on one end
of an arm of length 600 mm. The other end of the arm is free to rotate in a universal bearing. If the
disc rotates about the arm with a speed of 300 r.p.m. clockwise, looking from the front, with what
speed will it precess about the vertical axis?
Solution. Given: d = 300 mm or r = 150 mm = 0.15 m ; m = 5 kg ; l = 600 mm = 0.6 m ;
N = 300 r.p.m. or ω = 2π × 300/60 = 31.42 rad/s
We know that the mass moment of inertia of the disc, about an axis through its centre of
gravity and perpendicular to the plane of disc,
I = m.r2/2 = 5(0.15)2/2 = 0.056 kg-m2
and couple due to mass of disc,
C = m.g.l = 5 × 9.81 × 0.6 = 29.43 N-m
Let
ωP = Speed of precession.
We know that couple (C),
29.43 = I.ω.ωP = 0.056 × 31.42 × ωP = 1.76 ωP
∴
ωP = 29.43/1.76 = 16.7 rad/s Ans.
Chapter 14 : Gyroscopic Couple and Precessional Motion
Example 14.2. A uniform disc of 150 mm diameter has a
mass of 5 kg. It is mounted centrally in bearings which maintain
its axle in a horizontal plane. The disc spins about it axle with a
constant speed of 1000 r.p.m. while the axle precesses uniformly
about the vertical at 60 r.p.m. The directions of rotation are as
shown in Fig. 14.3. If the distance between the bearings is 100
mm, find the resultant reaction at each bearing due to the mass
and gyroscopic effects.
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485
Fig. 14.3
Solution. Given: d = 150 mm or r = 75 mm = 0.075 m ; m = 5 kg ; N = 1000 r.p.m. or
ω = 2π × 1000/60 = 104.7 rad/s (anticlockwise); N P = 60 r.p.m. or ωP = 2π × 60/60 = 6.284 rad/s
(anticlockwise); x = 100 mm = 0.1 m
We know that mass moment of inertia of the disc, about an axis through its centre of gravity
and perpendicular to the plane of disc,
I = m.r2/2 = 5 (0.075)2/2 = 0.014 kg m2
∴ Gyroscopic couple acting on the disc,
C = I. ω. ωP = 0.014 × 104.7 × 6.284 = 9.2 N-m
The direction of the reactive gyroscopic couple is shown in Fig.14.4 (b). Let F be the force at
each bearing due to the gyroscopic couple.
∴
F = C/x = 9.2/0.1 = 92 N
The force F will act in opposite directions at the bearings as shown in Fig. 14.4 (a). Now let
R A and R B be the reaction at the bearing A and B respectively due to the weight of the disc. Since the
disc is mounted centrally in bearings, therefore,
RA = R B = 5/2 = 2.5 kg = 2.5 × 9.81 = 24.5 N
(a)
(b)
Fig. 14.4
Resultant reaction at each bearing
Let
R A1 and R B1 = Resultant reaction at the bearings A and B respectively.
Since the reactive gyroscopic couple acts in clockwise direction when seen from the front,
therefore its effect is to increase the reaction on the left hand side bearing (i.e. A) and to decrease the
reaction on the right hand side bearing (i.e. B).
∴
and
RA1 = F + RA = 92 + 24.5 = 116.5 N (upwards) Ans.
R B1 = F – RB = 92 – 24.5 = 67.5 N (downwards) Ans.
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14.4. Effect of the Gyroscopic Couple on an Aeroplane
The top and front view of an aeroplane are shown in Fig 14.5 (a). Let engine or propeller
rotates in the clockwise direction when seen from the rear or tail end and the aeroplane takes a turn to
the left.
Let
ω = Angular velocity of the engine in rad/s,
m = Mass of the engine and the propeller in kg,
k = Its radius of gyration in metres,
I = Mass moment of inertia of the engine and the propeller in kg-m2
= m.k2,
v = Linear velocity of the aeroplane in m/s,
R = Radius of curvature in metres, and
ωP = Angular velocity of precession =
∴ Gyroscopic couple acting on the aeroplane,
C = I.ω.ωP
Fig. 14.5. Aeroplane taking a left turn.
v
rad/s
R
Chapter 14 : Gyroscopic Couple and Precessional Motion
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487
Before taking the left turn, the angular momentum vector is represented by ox. When it takes
left turn, the active gyroscopic couple will change the direction of the angular momentum vector from
ox to ox′ as shown in Fig. 14.6 (a). The vector x x′, in the limit, represents the change of angular
momentum or the active gyroscopic couple and is perpendicular to ox. Thus the plane of active
gyroscopic couple XOY will be perpendicular to x x′ , i.e. vertical in this case, as shown in Fig 14.5
(b). By applying right hand screw rule to vector x x′, we find that the direction of active gyroscopic
couple is clockwise as shown in the front view of Fig. 14.5 (a). In other words, for left hand turning,
the active gyroscopic couple on the aeroplane in the axis OZ will be clockwise as shown in Fig. 14.5
(b).The reactive gyroscopic couple (equal in magnitude of active gyroscopic couple) will act in the
opposite direction (i.e. in the anticlockwise direction) and the effect of this couple is, therefore, to
raise the nose and dip the tail of the aeroplane.
(a) Aeroplane taking left turn.
(b) Aeroplane taking right turn.
Fig. 14.6. Effect of gyroscopic couple on an aeroplane.
Notes : 1. When the aeroplane takes a right turn under similar conditions as discussed above, the effect of the
reactive gyroscopic couple will be to dip the nose and raise the tail of the aeroplane.
2. When the engine or propeller rotates in anticlockwise direction when viewed from the rear or tail
end and the aeroplane takes a left turn, then the effect of reactive gyroscopic couple will be to dip the nose and
raise the tail of the aeroplane.
3. When the aeroplane takes a right turn under similar conditions as mentioned in note 2 above, the
effect of reactive gyroscopic couple will be to raise the nose and dip the tail of the aeroplane.
4. When the engine or propeller rotates in clockwise direction when viewed from the front and the
aeroplane takes a left turn, then the effect of reactive gyroscopic couple will be to raise the tail and dip the nose
of the aeroplane.
5. When the aeroplane takes a right turn under similar conditions as mentioned in note 4-above, the
effect of reactive gyroscopic couple will be to raise the nose and dip the tail of the aeroplane.
Example 14.3. An aeroplane makes a complete half circle of 50 metres radius, towards left,
when flying at 200 km per hr. The rotary engine and the propeller of the plane has a mass of 400 kg
and a radius of gyration of 0.3 m. The engine rotates at 2400 r.p.m. clockwise when viewed from the
rear. Find the gyroscopic couple on the aircraft and state its effect on it.
Solution. Given : R = 50 m ; v = 200 km/hr = 55.6 m/s ; m = 400 kg ; k = 0.3 m ;
N = 2400 r.p.m. or ω = 2π × 2400/60 = 251 rad/s
We know that mass moment of inertia of the engine and the propeller,
I = m.k2 = 400(0.3)2 = 36 kg-m2
and angular velocity of precession,
ωP = v/R = 55.6/50 = 1.11 rad/s
We know that gyroscopic couple acting on the aircraft,
C = I. ω. ωP = 36 × 251.4 × 1.11 = 100 46 N-m
= 10.046 kN-m Ans.
We have discussed in Art. 14.4 that when the aeroplane turns towards left, the effect of the
gyroscopic couple is to lift the nose upwards and tail downwards. Ans.
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Theory of Machines
14.5. Terms Used in a Naval Ship
The top and front views of a naval ship are shown in Fig 14.7. The fore end of the ship is
called bow and the rear end is known as stern or aft. The left hand and right hand sides of the ship,
when viewed from the stern are called port and star-board respectively. We shall now discuss the
effect of gyroscopic couple on the naval ship in the following three cases:
1. Steering, 2. Pitching, and 3. Rolling.
Fig. 14.7. Terms used in a naval ship.
14.6. Effect of Gyroscopic Couple on a Naval Ship during Steering
Steering is the turning of a complete ship in a curve towards left or right, while it moves
forward. Consider the ship taking a left turn, and rotor rotates in the clockwise direction when viewed
from the stern, as shown in Fig. 14.8. The effect of gyroscopic couple on a naval ship during steering
taking left or right turn may be obtained in the similar way as for an aeroplane as discussed in Art.14.4.
Fig. 14.8. Naval ship taking a left turn.
When the rotor of the ship rotates in the clockwise direction when viewed from the stern, it will have
its angular momentum vector in the direction ox as shown in Fig. 14.9 (a). As the ship steers to the
left, the active gyroscopic couple will change the angular momentum vector from ox to ox′. The
vector x x′ now represents the active gyroscopic couple and is perpendicular to ox. Thus the plane of
active gyroscopic couple is perpendicular to x x′ and its direction in the axis OZ for left hand turn is
clockwise as shown in Fig. 14.8. The reactive gyroscopic couple of the same magnitude will act in the
Chapter 14 : Gyroscopic Couple and Precessional Motion
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489
opposite direction (i.e. in anticlockwise direction). The effect of this reactive gyroscopic couple is to
raise the bow and lower the stern.
Notes: 1. When the ship
steers to the right under similar conditions as discussed
above, the effect of the reactive gyroscopic couple, as
shown in Fig. 14.9 (b), will
be to raise the stern and
lower the bow.
2. When the rotor rates in
Fig. 14.9. Effect of gyroscopic couple on a naval ship during steering.
the anticlockwise direction,
when viewed from the stern and the ship is steering to the
left, then the effect of reactive gyroscopic couple will be
to lower the bow and raise the stern.
3. When the ship is steering to the right under similar
conditions as discussed in note 2 above, then the effect of
reactive gyroscopic couple will be to raise the bow and
lower the stern.
4. When the rotor rotates in the clockwise direction when
viewed from the bow or fore end and the ship is steering
to the left, then the effect of reactive gyroscopic couple will be to raise the stern and lower the bow.
5. When the ship is steering to the right under similar conditions as discussed in note 4 above, then the effect of
reactive gyroscopic couple will be to raise the bow and lower the stern.
6. The effect of the reactive gyroscopic couple on a boat propelled by a turbine taking left or right turn is similar
as discussed above.
14.7. Effect of Gyroscopic Couple on a Naval Ship during Pitching
Pitching is the movement of a complete ship up and down in a vertical plane about transverse
axis, as shown in Fig. 14.10 (a). In this case, the transverse axis is the axis of precession. The pitching
of the ship is assumed to take place with simple harmonic motion i.e. the motion of the axis of spin
about transverse axis is simple harmonic.
Fig. 14.10. Effect of gyroscopic couple on a naval ship during pitching.
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Theory of Machines
Gryroscopic couple plays its role during ship’s turning and pitching.
∴ Angular displacement of the axis of spin from mean position after time t seconds,
θ = φ sin ω1. t
where
φ = Amplitude of swing i.e. maximum angle turned from the mean
position in radians, and
ω1 = Angular velocity of S.H.M.
=
2π
2π
rad/s
=
Time period of S.H.M. in seconds t p
Angular velocity of precession,
ωP =
dθ d
= (φ sin ω1 . t ) = φ ω1 cos ω1 t
dt
dt
The angular velocity of precession will be maximum, if cos ω1.t = 1.
∴ Maximum angular velocity of precession,
ωPmax = φ.ω1 = φ × 2π / tp
...(Substituting cos ω1.t = 1)
Let
I = Moment of inertia of the rotor in kg-m2, and
ω = Angular velocity of the rotor in rad/s.
∴ Mamimum gyroscopic couple,
Cmax = I. ω. ωPmax
When the pitching is upward, the effect of the reactive gyroscopic couple, as shown in Fig. 14.10
(b), will try to move the ship toward star-board. On the other hand, if the pitching is downward, the effect
of the reactive gyroscopic couple, as shown in Fig. 14.10 (c), is to turn the ship towards port side.
Notes : 1. The effect of the gyroscopic couple is always given on specific position of the axis of spin i.e.
whether it is pitching downwards or upwards.
2. The pitching of a ship produces forces on the bearings which act horizontally and perpendicular to
the motion of the ship.
3. The maximum gyroscopic couple tends to shear the holding-down bolts.
4. The angular acceleration during pitching,
d 2θ
= – φ (ω1 ) 2 sin ω1t
dt 2
The angular acceleration is maximum, if sin ω1t = 1.
∴ Maximum angular acceleration during pitching,
α=
αmax = (ω1)2
dθ
with respect to t
... Differentiating
dt
Chapter 14 : Gyroscopic Couple and Precessional Motion
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491
14.8. Effect of Gyroscopic Couple on a Naval Ship during Rolling
We know that, for the effect of gyroscopic couple to occur, the axis of precession should
always be perpendicular to the axis of spin. If, however, the axis of precession becomes parallel to the
axis of spin, there will be no effect of the gyroscopic couple acting on the body of the ship.
In case of rolling of a ship, the axis of precession (i.e. longitudinal axis) is always parallel to
the axis of spin for all positions. Hence, there is no effect of the gyroscopic couple acting on the body
of a ship.
Example 14.4. The turbine rotor of a ship has a mass of 8 tonnes and a radius of gyration
0.6 m. It rotates at 1800 r.p.m. clockwise, when looking from the stern. Determine the gyroscopic
couple, if the ship travels at 100 km/hr and steer to the left in a curve of 75 m radius.
Solution. Given: m = 8 t = 8000 kg ; k = 0.6 m ; N = 1800 r.p.m. or ω = 2π × 1800/60
= 188.5 rad/s ; v = 100 km/h = 27.8 m/s ; R = 75 m
We know that mass moment of inertia of the rotor,
I = m.k2 = 8000 (0.6)2 = 2880 kg-m2
and angular velocity of precession,
ωP = v / R = 27.8 / 75 = 0.37 rad/s
We know that gyroscopic couple,
C = I.ω.ωP= 2880 × 188.5 × 0.37 = 200 866 N-m
= 200.866 kN-m Ans.
We have discussed in Art. 14.6, that when the rotor rotates in clockwise direction when
looking from the stern and the ship steers to the left, the effect of the reactive gyroscopic couple is to
raise the bow and lower the stern.
Example 14.5. The heavy turbine
rotor of a sea vessel rotates at 1500 r.p.m.
clockwise looking from the stern, its mass
being 750 kg. The vessel pitches with an
angular velocity of 1 rad/s. Determine the
gyroscopic couple transmitted to the hull when
bow is rising, if the radius of gyration for the
rotor is 250 mm. Also show in what direction
the couple acts on the hull?
Solution. Given: N = 1500 r.p.m. or
ω = 2π × 1500/60 = 157.1 rad/s; m = 750 kg;
ωP = 1 rad/s; k = 250 mm = 0.25 m
We know that mass moment of inertia
of the rotor,
I = m.k2 = 750 (0.25)2 = 46.875 kg-m2
∴ Gyroscopic couple transmitted to
the hull (i.e. body of the sea vessel),
Ship’s propeller shown as a separate part. A ship’s
propeller is located at backside (stern) of the ship
below the water surface.
C = I.ω.ωP = 46.875 × 157.1 × 1= 7364 N-m = 7.364 kN-m
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Theory of Machines
We have discussed in Art. 14.7, that when the bow is rising i.e. when the pitching is upward,
the reactive gyroscopic couple acts in the clockwise direction which moves the sea vessel towards
star-board.
Example 14.6. The turbine rotor of a ship has a mass of 3500 kg. It has a radius of gyration
of 0.45 m and a speed of 3000 r.p.m. clockwise when looking from stern. Determine the gyroscopic
couple and its effect upon the ship:
1. when the ship is steering to the left on a curve of 100 m radius at a speed of 36 km/h.
2. when the ship is pitching in a simple harmonic motion, the bow falling with its maximum
velocity. The period of pitching is 40 seconds and the total angular displacement between the two
extreme positions of pitching is 12 degrees.
Solution. Given : m = 3500 kg ; k = 0.45 m; N = 3000 r.p.m. or ω = 2π × 3000/60 = 314.2 rad/s
1. When the ship is steering to the left
Given:
R =100 m ; v = km/h = 10 m/s
We know that mass moment of inertia of the rotor,
I = m.k2 = 3500 (0.45)2 = 708.75 kg-m2
and angular velocity of precession,
ωP = v/R = 10/100 = 0.1 rad/s
∴ Gyroscopic couple,
C = I.ω.ωP = 708.75 × 314.2 × 0.1 = 22 270 N-m
= 22.27 kN-m Ans.
We have discussed in Art. 14.6, that when the rotor rotates clockwise when looking from the
stern and the ship takes a left turn, the effect of the reactive gyroscopic couple is to raise the bow and
lower the stern. Ans.
2. When the ship is pitching with the bow falling
Given: tp = 40 s
Since the total angular displacement between the two extreme positions of pitching is 12°
(i.e. 2φ = 12°), therefore amplitude of swing,
φ = 12 / 2 = 6° = 6 × π/180 = 0.105 rad
and angular velocity of the simple harmonic motion,
ω1 = 2π / tp = 2π / 40 = 0.157 rad/s
We know that maximum angular velocity of precession,
ωP = φ.ω1 = 0.105 × 0.157 = 0.0165 rad/s
∴ Gyroscopic couple,
C = I.ω.ωP = 708.75 × 314.2 × 0.0165 = 3675 N-m
= 3.675 kN-m Ans.
We have discussed in Art. 14.7, that when the bow is falling (i.e. when the pitching is downward), the effect of the reactive gyroscopic couple is to move the ship towards port side. Ans.
Example 14.7. The mass of the turbine rotor of a ship is 20 tonnes and has a radius of
gyration of 0.60 m. Its speed is 2000 r.p.m. The ship pitches 6° above and 6° below the horizontal
position. A complete oscillation takes 30 seconds and the motion is simple harmonic. Determine the
following:
Chapter 14 : Gyroscopic Couple and Precessional Motion
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493
1. Maximum gyroscopic couple, 2. Maximum angular acceleration of the ship during pitching, and 3. The direction in which the bow will tend to turn when rising, if the rotation of the rotor is
clockwise when looking from the left.
Solution. Given : m = 20 t = 20 000 kg ; k = 0.6 m ; N = 2000 r.p.m. or ω = 2π × 2000/60 =
209.5 rad/s; φ = 6° = 6 × π/180 = 0.105 rad ; tp = 30 s
1. Maximum gyroscopic couple
We know that mass moment of inertia of the rotor,
I = m.k2 = 20 000 (0.6)2 = 7200 kg-m2
and angular velocity of the simple harmonic motion,
ω1 = 2π / tp = 2π/30 = 0.21 rad/s
∴ Maximum angular velocity of precession,
ωPmax = φ.ω1 = 0.105 × 0.21 = 0.022 rad/s
We know that maximum gyroscopic couple,
Cmax = I.ω.ωPmax = 7200 × 209.5 × 0.022 = 33 185 N-m
= 33.185 kN-m Ans.
2. Maximum angular acceleration during pitching
We know that maximum angular acceleration during pitching
= φ(ω1)2 = 0.105 (0.21)2 = 0.0046 rad/s2
3. Direction in which the bow will tend to turn when rising
We have discussed in Art. 14.7, that when the rotation of the rotor is clockwise when looking
from the left (i.e. rear end or stern) and when the bow is rising (i.e. pitching is upward), then the
reactive gyroscopic couple acts in the clockwise direction which tends to turn the bow towards right
(i.e. towards star-board). Ans.
Example 14.8. A ship propelled by a turbine rotor which has a mass of 5 tonnes and a speed
of 2100 r.p.m. The rotor has a radius of gyration of 0.5 m and rotates in a clockwise direction when
viewed from the stern. Find the gyroscopic effects in the following conditions:
1. The ship sails at a speed of 30 km/h and steers to the left in a curve having 60 m radius.
2. The ship pitches 6 degree above and 6 degree below the horizontal position. The bow is
descending with its maximum velocity. The motion due to pitching is simple harmonic and the periodic
time is 20 seconds.
3. The ship rolls and at a certain instant it has an angular velocity of 0.03 rad/s clockwise
when viewed from stern.
Determine also the maximum angular acceleration during pitching. Explain how the direction
of motion due to gyroscopic effect is determined in each case.
Solution. Given : m = 5 t = 5000 kg ; N = 2100 r.p.m. or ω = 2π × 2100/60 = 220 rad/s ;
k = 0.5 m
1. When the ship steers to the left
Given:
v = 30 km / h = 8.33 m / s ; R = 60 m
We know that angular velocity of precession,
ωP = v/R = 8.33/60 = 0.14 rad/s
and mass moment of inertia of the rotor,
I = m.k2 = 5000(0.5)2 = 1250 kg-m2
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∴ Gyroscopic couple,
C = I.ω.ωP = 1250 × 220 × 0.14 = 38 500 N-m = 38.5 kN-m
We have discussed in Art. 14.6, that when the rotor in a clockwise direction when viewed
from the stern and the ship steers to the left, the effect of reactive gyroscopic couple is to raise the
bow and lower the stern. Ans.
2. When the ship pitches with the bow descending
Given: φ = 6° = 6 × π/180 = 0.105 rad/s ; tp = 20 s
We know that angular velocity of simple harmonic motion,
ω1 = 2π / tp = 2π / 20 = 0.3142 rad/s
and maximum angular velocity of precession,
ωPmax = φ.ω1 = 0.105 × 0.3142 = 0.033 rad/s
∴ Maximum gyroscopic couple,
Cmax = I.ω.ωPmax = 1250 × 220 × 0.033 = 9075 N-m
Since the ship is pitching with the bow descending, therefore the effect of this maximum
gyroscopic couple is to turn the ship towards port side. Ans.
3. When the ship rolls
Since the ship rolls at an angular velocity of 0.03 rad / s, therefore angular velocity of precession
when the ship rolls,
ωP = 0.03 rad /s
∴ Gyroscopic couple,
C = I.ω.ωP = 1250 × 220 × 0.03 = 8250 N-m
In case of rolling of a ship, the axis of precession is always parallel to the axis of spin for all
positions, therefore there is no effect of gyroscopic couple. Ans.
Maximum angular acceleration during pitching
We know that maximum angular acceleration during pitching.
αmax = φ (ω1)2 = 0.105 (0.3142)2 = 0.01 rad/s2 Ans.
Example 14.9. The turbine rotor of a ship has a mass of 2000 kg and rotates at a speed of
3000 r.p.m. clockwise when looking from a stern. The radius of gyration of the rotor is 0.5 m.
Determine the gyroscopic couple and its effects upon the ship when the ship is steering to the
right in a curve of 100 m radius at a speed of 16.1 knots (1 knot = 1855 m/hr).
Calculate also the torque and its effects when the ship is pitching in simple harmonic motion,
the bow falling with its maximum velocity. The period of pitching is 50 seconds and the total angular
displacement between the two extreme positions of pitching is 12°. Find the maximum acceleration
during pitching motion.
Solution. Given : m = 2000 kg ; N = 3000 r.p.m. or ω = 2π × 3000/60 = 314.2 rad/s ;
k = 0.5 m ; R = 100 m ; v = 16.1 knots = 16.1 × 1855 / 3600 = 8.3 m/s
Gyroscopic couple
We know that mass moment of inertia of the rotor,
I = m.k 2 = 2000 (0.5)2 = 500 kg-m2
and angular velocity of precession,
ωP = v/R = 8.3/100 = 0.083 rad /s
Chapter 14 : Gyroscopic Couple and Precessional Motion
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495
∴ Gyroscopic couple,
C = I.ω.ωP = 500 × 314.2 × 0.083 = 13 040 N-m = 13.04 kN-m
We have discussed in Art. 14.6, that when the rotor rotates clockwise when looking from a
stern and the ship steers to the right, the effect of the reactive gyroscopic couple is to raise the stern
and lower the bow. Ans.
Torque during pitching
Given : tp =50 s ; 2 φ = 12° or φ = 6° × π/180 = 0.105 rad
We know that angular velocity of simple harmonic motion,
ω1 = 2π /tp = 2π /50 = 0.1257 rad/s
and maximum angular velocity of precession,
ωPmax = φ.ω1 = 0.105 × 0.1257 = 0.0132 rad/s
∴ Torque or maximum gyroscopic couple during pitching,
Cmax = I.ω.ωP max = 500 × 314.2 × 0.0132 = 2074 N-m Ans.
We have discussed in Art. 14.7, that when the pitching is downwards, the effect of the reactive gyroscopic couple is to turn the ship towards port side.
Maximum acceleration during pitching
We know that maximum acceleration during pitching
αmax = φ (ω1)2 = 0.105 (0.1257)2 = 0.00166 rad/s2 Ans.
14.9. Stability of a Four Wheel Drive Moving in a Curved Path
Consider the four wheels A , B, C and D of an
automobile locomotive taking a turn towards left as shown
in Fig. 14.11. The wheels A and C are inner wheels, whereas
B and D are outer wheels. The centre of gravity (C.G.) of
the vehicle lies vertically above the road surface.
Let m = Mass of the vehicle in kg,
W = Weight of the vehicle in newtons = m.g,
rW = Radius of the wheels in metres,
R = Radius of curvature in metres
(R > rW),
h = Distance of centre of gravity, vertically
above the road surface in metres,
x = Width of track in metres,
IW = Mass moment of inertia of one of the
wheels in kg-m2,
ωW = Angular velocity of the wheels or velocity of spin in rad/s,
IE = Mass moment of inertia of the rotating
parts of the engine in kg-m2,
ωE = Angular velocity of the rotating parts of
the engine in rad/s,
G = Gear ratio = ωE /ωW,
v = Linear velocity of the vehicle in m/s = ωW.rW
Fig. 14.11. Four wheel drive
moving in a curved path.
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Theory of Machines
A little considereation will show,
that the weight of the vehicle (W ) will be
equally distributed over the four wheels
which will act downwards. The reaction
between each wheel and the road surface
of the same magnitude will act upwards.
Therefore
Road reaction over each wheel
= W /4 = m.g /4 newtons
Let us now consider the effect of
the gyroscopic couple and centrifugal couple on the vehicle.
1. Effect of the gyroscopic couple
Since the vehicle takes a turn towards left due to the precession and other rotating parts,
therefore a gyroscopic couple will act.
We know that velocity of precession,
ωP = v/R
∴ Gyroscopic couple due to 4 wheels,
CW = 4 IW.ωW.ωP
and gyroscopic couple due to the rotating parts of the engine,
CE = IE.ωE.ωP = IE.G.ωW.ωP
... ( G = ωE/ωW)
∴ Net gyroscopic couple,
C = CW ± CE = 4 IW.ωW.ωP ± IE.G.ωW.ωP
= ωW.ωP (4 IW ± G.IE)
The positive sign is used when the wheels and rotating parts of the engine rotate in the same
direction. If the rotating parts of the engine revolves in opposite direction, then negative sign is used.
Due to the gyroscopic couple, vertical reaction on the road surface will be produced. The
reaction will be vertically upwards on the outer wheels and vertically downwards on the inner wheels.
Let the magnitude of this reaction at the two outer or inner wheels be P newtons. Then
P × x = C or P = C/x
∴ Vertical reaction at each of the outer or inner wheels,
P /2 = C/ 2x
Note: We have discussed above that when rotating parts of the engine rotate in opposite directions, then –ve
sign is used, i.e. net gyroscopic couple,
C = CW – CE
When CE > CW, then C will be –ve. Thus the reaction will be vertically downwards on the outer wheels
and vertically upwards on the inner wheels.
2. Effect of the centrifugal couple
Since the vehicle moves along a curved path, therefore centrifugal force will act outwardly at
the centre of gravity of the vehicle. The effect of this centrifugal force is also to overturn the vehicle.
We know that centrifugal force,
FC =
m × v2
R
Chapter 14 : Gyroscopic Couple and Precessional Motion
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497
∴ The couple tending to overturn the vehicle or overturning couple,
m.v 2
×h
R
This overturning couple is balanced by vertical reactions, which are vertically upwards on
the outer wheels and vertically downwards on the inner wheels. Let the magnitude of this reaction at
the two outer or inner wheels be Q. Then
CO = FC × h =
CO m.v 2 .h
=
x
R. x
∴ Vertical reaction at each of the outer or inner wheels,
Q × x = CO or Q =
Q m.v 2 .h
=
2
2R.x
∴ Total vertical reaction at each of the outer wheel,
W
P Q
PO =
+ +
4
2
2
and total vertical reaction at each of the inner wheel,
W
P Q
PI =
− −
4
2
2
A little consideration will show that when the vehicle is running at high speeds, PI may be
zero or even negative. This will cause the inner wheels to leave the ground thus tending to overturn
the automobile. In order to have the contact between the inner wheels and the ground, the sum of
P/2 and Q/2 must be less than W /4.
Example 14.10. A four-wheeled trolley car of mass
2500 kg runs on rails, which are 1.5 m apart and travels
around a curve of 30 m radius at 24 km / hr. The rails are at
the same level. Each wheel of the trolley is 0.75 m in diameter
and each of the two axles is driven by a motor running in a
direction opposite to that of the wheels at a speed of five
times the speed of rotation of the wheels. The moment of
inertia of each axle with gear and wheels is 18 kg-m2. Each
motor with shaft and gear pinion has a moment of inertia of
12 kg-m2. The centre of gravity of the car is 0.9 m above the
rail level. Determine the vertical force exerted by each wheel
on the rails taking into consideration the centrifugal and
gyroscopic effects. State the centrifugal and gyroscopic effects
on the trolley.
Solution. Given : m = 2500 kg ; x = 1.5 m ; R = 30 m ;
v = 24 km/h = 6.67 m/s ; dW = 0.75 m or rW = 0.375 m ; G = ωE/ωW = 5 ; IW = 18 kg-m2 ;
IE = 12 kg-m2 ; h = 0.9 m
The weight of the trolley (W = m.g) will be equally distributed over the four wheels, which
will act downwards. The reaction between the wheels and the road surface of the same magnitude will
act upwards.
∴ Road reaction over each wheel
= W /4 = m.g/4 = 2500 × 9.81/4 = 6131.25 N
We know that angular velocity of the wheels,
ωW = v/rW = 6.67/0.375 = 17.8 rad/s
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Theory of Machines
ωP = v/R = 6.67/30 = 0.22 rad/s
and angular velocity of precession,
∴Gyroscopic couple due to one pair of wheels and axle,
CW = 2 IW.ωW.ωP = 2 × 18 × 17.8 × 0.22 = 141 N-m
and gyroscopic couple due to the rotating parts of the motor and gears,
CE = 2 IE.ωE.ωP = 2 IE.G.ωW.ωP
... (Q ωE = G. ωW)
= 2 × 12 × 5 × 17.8 × 0.22 = 470 N-m
∴ Net gyroscopic couple,
C = CW – CE = 141 – 470 = – 329 N-m
... (–ve sign is used due to opposite direction of motor)
Due to this net gyroscopic couple, the vertical reaction on the rails will be produced. Since
CE is greater than CW, therefore the reaction will be vertically downwards on the outer wheels and
vertically upwards on the inner wheels. Let the magnitude of this reaction at each of the outer or inner
wheel be P/2 newton.
∴
P/ 2 = C/2x = 329 / 2 × 1.5 = 109.7 N
We know that centrifugal force,
FC = m.v 2/R = 2500 (6.67)2/30 = 3707 N
∴ Overturning couple,
CO = FC × h = 3707 × 0.9 = 3336.3 N-m
This overturning couple is balanced by the vertical reactions which are vertically upwards on
the outer wheels and vertically downwards on the inner wheels. Let the magnitude of this reaction at
each of the outer or inner wheels be Q/2 newton.
∴
Q/ 2 = CO / 2x = 3336.3 / 2 × 1.5 = 1112.1 N
We know that vertical force exerted on each outer wheel,
W
P Q
− + = 6131.25 – 109.7 + 1112.1 = 7142.65 N Ans.
4
2
2
and vertical force exerted on each inner wheel,
PO =
PI =
W
P Q
+ − = 6131.25 + 109.7 – 1112.1 = 5128.85 N Ans.
4
2
2
Example 14.11. A rear engine automobile is travelling along a track of 100 metres mean
radius. Each of the four road wheels has a moment of inertia of 2.5 kg-m2 and an effective diameter
of 0.6 m. The rotating parts of the engine have a moment of inertia of 1.2 kg-m2. The engine axis is
parallel to the rear axle and the crankshaft rotates in the same sense as the road wheels. The ratio of
engine speed to back axle speed is 3 : 1. The automobile has a mass of 1600 kg and has its centre of
gravity 0.5 m above road level. The width of the track of the vehicle is 1.5 m.
Determine the limiting speed of the vehicle around the curve for all four wheels to maintain
contact with the road surface. Assume that the road surface is not cambered and centre of gravity of
the automobile lies centrally with respect to the four wheels.
Solution. Given : R = 100 m ; IW = 2.5 kg-m2 ; dW = 0.6 m or rW = 0.3 m ; IE = 1.2 kg-m2;
G = ωE/ωW = 3 ; m = 1600 kg ; h = 0.5 m ; x = 1.5 m
The weight of the vehicle (m.g) will be equally distributed over the four wheels which will act
downwards. The reaction between the wheel and the road surface of the same magnitude will act
upwards.
∴ Road reaction over each wheel
= W /4 = m.g / 4 = 1600 × 9.81/4 = 3924 N
Chapter 14 : Gyroscopic Couple and Precessional Motion
Let
l
499
v = Limiting speed of the vehicle in m/s.
We know that angular velocity of the wheels,
ωW =
v
v
=
= 3.33 v rad /s
rW 0.3
and angular velocity of precession,
ωP =
v
v
=
= 0.01 v rad /s
R 100
∴ Gyroscopic couple due to 4 wheels,
CW = 4 IW.ωW.ωP = 4 × 2.5 ×
v
v
×
= 0.33 v 2 N-m
0.3 100
and gyroscopic couple due to rotating parts of the engine,
CE = IE.ωE.ωP = IE.G.ωW.ωP
= 1.2 × 3 × 3.33v × 0.01v = 0.12 v 2 N-m
∴ Total gyroscopic couple,
C = CW + CE = 0.33 v 2 + 0.12 v 2 = 0.45 v 2 N-m
Due to this gyroscopic couple, the vertical reaction on the rails will be produced. The reaction will be vertically upwards on the outer wheels and vertically downwards on the inner wheels. Let
the magnitude of this reaction at each of the outer or inner wheel be P/2 newtons.
∴
P/2 = C/2x = 0.45v 2/2 × 1.5 = 0.15 v 2 N
We know that centrifugal force,
FC = m.v 2/R = 1600 × v 2/100 = 16 v 2 N
∴Overturning couple acting in the outward direction,
CO = FC × h = 16 v 2 × 0.5 = 8 v 2 N-m
This overturning couple is balanced by vertical reactions which are vertically upwards on the
outer wheels and vertically downwards on the inner wheels. Let the magnitude of this reaction at each
of the outer or inner wheels be Q/2 newtons.
∴
Q / 2 = CO / 2x = 8 v 2/ 2 × 1.5 = 2.67 v 2 N
We know that total vertical reaction at each of the outer wheels,
PO =
W
P Q
+ +
4
2
2
...(i)
and total vertical reaction at each of the inner wheels,
W
P Q W P Q
–
–
– +
=
...(ii)
4
2
2
4 2
2
From equation (i), we see that there will always be contact between the outer wheels and
the road surface because W /4, P/2 and Q/2 are vertically upwards. In order to have contact between
the inner wheels and road surface, the reactions should also be vertically upwards, which is only
possible if
PI =
P Q W
+
≤
2
2
4
500
l
i.e.
Theory of Machines
0.15 v 2 + 2.67 v 2 ≤ 3924
or
2.82 v 2 ≤ 3924
∴
v 2 ≤ 3924/2.82 = 1391.5
or
v ≤ 37.3 m/s = 37.3 × 3600 / 1000 = 134.28 km/h Ans.
Example 14.12. A four wheeled motor car of mass 2000 kg has a wheel base 2.5 m, track
width 1.5 m and height of centre of gravity 500 mm above the ground level and lies at 1 metre from
the front axle. Each wheel has an effective diameter of 0.8 m and a moment of inertia of 0.8 kg-m 2.
The drive shaft, engine flywheel and transmission are rotating at 4 times the speed of road wheel, in
a clockwise direction when viewed from the front, and is equivalent to a mass of 75 kg having a
radius of gyration of 100 mm. If the car is taking a right turn of 60 m radius at 60 km/h, find the load
on each wheel.
Solution. Given : m = 2000 kg : b = 2.5 m ; x = 1.5 m ; h = 500 mm = 0.5 m ; L = 1 m ; dW =
0.8 m or rW = 0.4 m ; IW = 0.8 kg-m2 ; G = ωE / ωW = 4 ; m E = 75 kg ; k E = 100 mm = 0.1 m ;
R = 60 m ; v = 60 km/h = 16.67 m/s
Since the centre of gravity of the car lies at 1 m from the front axle and the weight of the car
(W = m.g) lies at the centre of gravity, therefore weight on the front wheels and rear wheels will be
different.
Let
W1 = Weight on the front wheels, and
W2 = Weight on the rear wheels.
Taking moment about the front wheels,
W 2 × 2.5 = W × 1 = m.g × 1 = 2000 × 9.81 × 1 = 19 620
∴
W2 = 19 620 / 2.5 = 7848 N
We know that weight of the car or on the four wheels,
W = W 1 + W 2 = m.g = 2000 × 9.81 = 19 620 N
or
W1 = W – W 2 = 19 620 – 7848 = 11 772 N
∴ Weight on each of the front wheels
= W 1 / 2 = 11 772 / 2 = 5886 N
and weight on each of the rear wheels
= W 2 /2 = 7874 / 2 = 3924 N
Since the weight of the car over the four wheels will act
downwards, therefore the reaction between each wheel and the
road surface of the same magnitude will act upwards as shown
in Fig. 14.12.
Let us now consider the effect of gyroscopic couple
due to four wheels and rotating parts of the engine.
We know angular velocity of wheels,
ωW = v/rW = 16.67 / 0.4 = 41.675 rad /s
and angular velocity of precession,
ωP = v/R = 16.67 / 60 = 0.278 rad /s
Fig. 14.12
Chapter 14 : Gyroscopic Couple and Precessional Motion
l
501
∴ Gyroscopic couple due to four wheels,
CW = 4 IW.ωW.ωP
= 4 × 0.8 × 41.675 × 0.278 = 37.1 N-m
This gyroscopic couple tends to lift the inner wheels and to press the outer wheels. In other
words, the reaction will be vertically downward on the inner wheels (i.e. wheels 1 and 3) and vertically upward on the outer wheels (i.e. wheels 2 and 4) as shown in Fig. 14.12. Let P/2 newtons be the
magnitude of this reaction at each of the inner or outer wheel.
∴
P / 2 = CW / 2x = 37.1 / 2 × 1.5 = 12.37 N
We know that mass moment of inertia of rotating parts of the engine,
IE = m E (k E)2 = 75 (0.1)2 = 0.75 kg-m2
...(∵ I = m.k 2)
∴ Gyroscopic couple due to rotating parts of the engine,
CE = IE.ωE.ωP = m E (k E)2 G. ωW.ωP
= 75 (0.1)2 4 × 41.675 × 0.278 = 34.7 N-m
This gyroscopic couple tends to lift the front wheels and to press the outer wheels. In other
words, the reaction will be vertically downwards on the front wheels and vertically upwards on the
rear wheels as shown in Fig. 14.12. Let F/2 newtons be the magnitude of this reaction on each of the
front and rear wheels.
∴
F / 2 = CE / 2b = 34.7/2 × 2.5 = 6.94 N
Now let us consider the effect of centrifugal couple acting on the car. We know that centrifugal
force,
FC = m.v 2 / R = 2000 (16.67)2/ 60 = 9263 N
∴ Centrifugal couple tending to overturn the car or over turning couple,
CO = FC × h = 9263 × 0.5 = 4631.5 N-m
This overturning couple tends to reduce the pressure on the inner wheels and to increase on
the outer wheels. In other words, the reactions are vertically downward on the inner wheels and
vertically upwards on the outer wheels. Let Q/2 be the magnitude of this reaction on each of the inner
and outer wheels.
∴
Q / 2 = CO / 2x = 4631.5 / 2 × 1.5 = 1543.83 N
From Fig. 14.12, we see that
Load on the front wheel 1
W1
P F Q
–
–
–
= 5886 – 12.37 – 6.94 – 1543.83 = 4322.86 N Ans.
2
2
2
2
Load on the front wheel 2
=
W1
P
F
Q
–
+
+
= 5886 + 12.37 – 6.94 + 1543.83 = 7435.26 N Ans.
2
2
2
2
Load on the rear wheel 3
=
W2
P
F Q
−
+
−
= 3924 – 12.37 + 6.94 – 1543.83 = 2374.74 N Ans.
2
2
2
2
Load on the rear wheel 4
=
=
W2
P
F
Q
+
+
+
= 3924 + 12.37 + 6.94 + 1543.83 = 5487.14 N Ans.
2
2
2
2
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Theory of Machines
Example 14.13. A four-wheeled trolley car of total mass 2000 kg running on rails of 1.6 m
gauge, rounds a curve of 30 m radius at 54 km/h. The track is banked at 8°. The wheels have an
external diameter of 0.7 m and each pair with axle has a mass of 200 kg. The radius of gyration for
each pair is 0.3 m. The height of centre of gravity of the car above the wheel base is 1 m. Determine,
allowing for centrifugal force and gyroscopic couple actions, the pressure on each rail.
Solution. Given : m = 2000 kg ; x = 1.6 m ; R = 30 m ; v = 54 km / h = 15 m / s ; θ = 8° ;
dW = 0.7 m or rW = 0.35 m ; m 1 = 200 kg ; k = 0.3 m ; h = 1 m
First of all, let us find the reactions R A and R B at the wheels A and B respectively. The various
forces acting on the trolley car are shown in Fig. 14.13.
Resolving the forces perpendicular to the track,
R A + R B = W cos θ + FC sin θ = m.g cos θ +
m.v 2
sin θ
R
2000 (15) 2
× sin 8°
30
= 19 620 × 0.9903 + 15 000 × 0.1392 = 21 518 N
= 2000 × 9.81 cos 8° +
Fig. 14.13
Now taking moments about B,
R A × x = (W cos θ + FC sin θ)
∴
x
+ W sin θ × h – FC cos θ × h
2
1
h
m.v 2
m.v 2
sin θ + m.g sin θ –
cos θ
RA = m.g cos θ +
x
R
R
2
1
2000 (15) 2
sin 8°
= 2000 × 9.81 cos 8° +
30
2
1
2000 (15) 2
+ 2000 × 9.81 sin 8° −
cos 8°
30
1.6
1
= (19 620 × 0.9903 + 15 000 × 0.1392)
2
+ (19 620 × 0.1392 – 15 000 × 0.9903)
1
1.6
Chapter 14 : Gyroscopic Couple and Precessional Motion
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503
1
1
+ (2731 – 14 855)
2
1.6
= 10 759 – 7577 = 3182 N
∴
RB = (R A + R B) – R A = 21 518 – 3182 = 18 336 N
We know that angular velocity of wheels,
= (19 430 + 2088)
15
v
=
= 42.86 rad /s
rW 0.35
and angular velocity of precession,
v 15
ωP = =
= 0.5 rad /s
R 30
∴ Gyroscopic couple,
C = * I ωW cos θ × ωP = m I.k 2.ωW cos θ.ωP
...(∵ I = m I.k 2)
2
= 200 (0.3) 42.86 cos 8° × 0.5 = 382 N-m
Due to this gyroscopic couple, the car will tend to overturn about the outer wheels. Let P be
the force at each pair of wheels or each rail due to the gyroscopic couple,
∴
P = C / x = 382 / 1.6 = 238.75 N
We know that pressure (or total reaction) on the inner rail,
PI = R A – P = 3182 – 238.75 = 2943.25 N Ans.
and pressure on the outer rail,
ωW =
PO = R B + P = 18 336 + 238.75 = 18 574.75 N Ans.
Example 14.14. A pair of locomotive driving wheels with the axle, have a moment of inertia
of 180 kg-m2. The diameter of the wheel treads is 1.8 m and the distance between wheel centres is
1.5 m. When the locomotive is travelling on a level track at 95 km/h, defective ballasting causes one
wheel to fall 6 mm and to rise again in a total time of 0.1 s. If the displacement of the wheel takes
place with simple harmonic motion, find : 1. The gyroscopic couple set up, and 2. The reaction
between the wheel and rail due to this couple.
Solution. Given : I = 180 kg-m2 ; D = 1.8 m or R = 0.9 m ; x = 1.5 m ; v = 95 km / h = 26.4 m /s
1. Gyroscopic couple set up
We know that angular velocity of the locomotive,
ω = v/R = 26.4/ 0.9 = 29.3 rad /s
Since the defective ballasting causes one wheel to fall 6 mm and to rise again in a total time
(t) of 0.1 s, therefore
1
1
1
Fall = Rise = × 6 = 3 mm
2
2
2
and maximum velocity while falling,
2π
2π
vmax =
×A=
× 3 = 118.5 mm / s = 0.1885 m / s
t
0.1
∴ Maximum angular velocity of tilt of the axle or angular velocity of precession,
Amplitude, A =
ωP max =
*
vmax
0.1885
=
= 0.126 rad /s
1.5
x
Angular momentum about axle = I.ωW
∴ Angular momentum about horizontal = I.ωW cos θ
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Theory of Machines
We know that gyroscopic couple set up,
C = I.ω.ωP max = 180 × 29.3 × 0.126 = 664.5 N-m Ans.
The gyroscopic couple will act in a horizontal plane and this couple will tend to produce
swerve i.e. it tends to turn the locomotive aside.
2. Reaction between the wheel and rail due to the gyroscopic couple
We know that the reaction between the wheel and rail due to the gyroscopic couple is
P = C / x = 664.5 / 1.5 = 443 N Ans.
14.10. Stability of a Two Wheel Vehicle Taking a Turn
Consider a two wheel vehicle (say a scooter or motor cycle) taking a right turn as shown in
Fig. 14.14 (a).
(a)
(b)
(c)
Fig. 14.14. Stability of a two wheel vehicle taking a turn.
Let
m = Mass of the vehicle and its
rider in kg,
W = Weight of the vehicle and
its rider in newtons = m.g,
h = Height of the centre of
gravity of the vehicle and
rider,
rW = Radius of the wheels,
R = Radius of
curvature,
track
or
Motorcycle taking a turn.
IW = Mass moment of inertia of each wheel,
IE = Mass moment of inertia of the rotating parts of the engine,
ωW = Angular velocity of the wheels,
ωE = Angular velocity of the engine,
G = Gear ratio = ωE / ωW,
Chapter 14 : Gyroscopic Couple and Precessional Motion
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505
v = Linear velocity of the vehicle = ωW × rW,
θ = Angle of heel. It is inclination of the vehicle to the vertical for equilibrium.
Let us now consider the effect of the gyroscopic couple and centrifugal couple on the vehicle,
as discussed below.
1. Effect of gyroscopic couple
We know that
v = ωW × rW
ωE = G.ωW = G ×
and
∴ Total
ωW = v / rW
or
v
rW
(I × ω) = 2 IW × ωW ± IE × ωE
= 2 IW ×
v
v
v
(2 I W ± G.I E )
± IE × G ×
=
rW
rW rW
and velocity of precession, ωP = v /R
A little consideration will show that when the wheels move over the curved path, the vehicle
is always inclined at an angle θ with the vertical plane as shown in Fig. 14.14 (b). This angle is known
as angle of heel. In other words, the axis of spin is inclined to the horizontal at an angle θ, as shown
in Fig. 14.14 (c). Thus the angular momentum vector Iω due to spin is represented by O A inclined to
O X at an angle θ. But the precession axis is vertical. Therefore the spin vector is resolved along O X.
∴ Gyroscopic couple,
C1 = I .ω cos θ × ωP =
=
v
v
(2 I W ± G.I E ) cos θ ×
rW
R
v2
(2 I W ± G.I E ) cos θ
R.rW
Notes : (a) When the engine is rotating in the same direction as that of wheels, then the positive sign is used in
the above expression and if the engine rotates in opposite direction, then negative sign is used.
(b) The gyroscopic couple will act over the vehicle outwards i.e. in the anticlockwise direction
when seen from the front of the vehicle. The tendency of this couple is to overturn the vehicle in outward
direction.
An aircraft of 1920’s model.
506
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Theory of Machines
2. Effect of centrifugal couple
We know that centrifugal force,
m.v 2
R
This force acts horizontally through the centre of gravity (C.G.) along the outward direction.
∴ Centrifugal couple,
FC =
m.v 2
C2 = FC × h cos θ =
h cos θ
R
Since the centrifugal couple has a tendency to overturn the vehicle, therefore
Total overturning couple,
CO = Gyroscopic couple + Centrifugal couple
=
v2
m.v 2
× h cos θ
( 2 I W + G.I E ) cos θ +
R.rW
R
v 2 2 I W + G.I E
+ m.h cos θ
R
rW
We know that balancing couple = m.g.h sin θ
The balancing couple acts in clockwise direction when seen from the front of the vehicle.
Therefore for stability, the overturning couple must be equal to the balancing couple, i.e.
=
v 2 2 I W + G.I E
+ m.h cos θ = m.g .h sin θ
R
rW
From this expression, the value of the angle of heel (θ) may be determined, so that the vehicle
does not skid.
Example 14.15. Find the angle of inclination with respect to the vertical of a two wheeler
negotiating a turn. Given : combined mass of the vehicle with its rider 250 kg ; moment of inertia of
the engine flywheel 0.3 kg-m2 ; moment of inertia of each road wheel 1 kg-m2 ; speed of engine
flywheel 5 times that of road wheels and in the same direction ; height of centre of gravity of rider
with vehicle 0.6 m ; two wheeler speed 90 km/h ; wheel radius 300 mm ; radius of turn 50 m.
ω
Solution. Given : m = 250 kg ; IE = 0.3 kg-m2 ; IW = 1 kg-m2 ; ωE = 5 ωW or G = E = 5 ;
ωW
h = 0.6 m ; v = 90 km/h = 25 m/s ; rW = 300 mm = 0.3 m ; R = 50 m
Let
θ = Angle of inclination with respect to the vertical of a two wheeler.
We know that gyroscopic couple,
(25)2
v2
(2 I W + G.I E ) cos θ =
(2 × 1 + 5 × 0.3) cos θ
50 × 0.3
R × rW
= 146 cos θ N-m
C1 =
250 (25)2
m.v 2
× h cos θ =
× 0.6 cos θ = 1875 cos θ N-m
50
R
∴ Total overturning couple,
= C1 + C2 = 146 cos θ + 1875 cos θ = 2021 cos θ N-m
We know that balancing couple
= m.g.h sin θ = 250 × 9.81 × 0.6 sin θ = 1471.5 sin θ N-m
and centrifugal couple,
C2 =
Chapter 14 : Gyroscopic Couple and Precessional Motion
l
507
Since the overturning couple must be equal to the balancing couple for equilibrium condition, therefore
2021 cos θ = 1471.5 sin θ
∴
tan θ = sin θ / cos θ = 2021 / 1471.5 = 1.3734 or θ = 53.94° Ans.
Example 14.16. A gyrowheel D of mass 0.5 kg, with a radius
of gyration of 20 mm, is mounted in a pivoted frame C as shown in
Fig. 14.15. The axis AB of the pivots passes through the centre of
rotation O of the wheel, but the centre of gravity G of the frame C is
10 mm below O. The frame has a mass of 0.30 kg and the speed of
rotation of the wheel is 3000 r.p.m. in the anticlockwise direction as
shown.
The entire unit is mounted on a vehicle so that the axis AB is
parallel to the direction of motion of the vehicle. If the vehicle travels
at 15 m/s in a curve of 50 metres radius, find the inclination of the
gyrowheel from the vertical, when
Fig. 14.15
1. The vehicle moves in the direction of the arrow ‘X’ taking
a left hand turn along the curve, and
2. The vehicle reverse at the same speed in the direction of arrow ‘Y’ along the same path.
Solution. Given : m 1 = 0.5 kg ; k = 20 mm = 0.02 m ; OG = h = 10 mm = 0.01 m ;
m 2 = 0.3 kg ; N = 3000 r.p.m. or ω = 2 π × 3000 / 60 = 314.2 rad/s ; v = 15 m/s ; R = 50 m
We know that mass moment of inertia of the gyrowheel,
I = m 1.k 2 = 0.5 (0.02)2 = 0.0002 kg-m2
and angular velocity of precession,
ωP = v/R = 15 / 50 = 0.3 rad /s
Let
θ = Angle of inclination of gyrowheel from the vertical.
1. When the vehicle moves in the direction of arrow X taking a left turn along the curve
We know that gyroscopic couple about O,
C1 = I ω.ωP cos θ = 0.0002 × 314.2 × 0.3 cos θ N-m
= 0.019 cos θ N-m (anticlockwise)
and centrifugal couple about O,
C2 =
m2 .v 2
0.3 (15)2
× h cos θ =
× 0.01 cos θ N-m
50
R
= 0.0135 cos θ N-m (anticlockwise)
∴ Total overturning couple
= C1 – C2 = 0.019 cos θ – 0.0135 cos θ
... (– ve sign due to opposite direction)
= 0.0055 cos θ N-m (anticlockwise)
We know that balancing couple due to weight (W 2 = m 2.g) of the frame about O,
= m 2.g.h sin θ = 0.3 × 9.81 × 0.01 sin θ N-m
= 0.029 sin θ N-m (clockwise)
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Theory of Machines
Since the overturning couple must be equal to the balancing couple for equilibrium condition, therefore
0.0055 cos θ = 0.029 sin θ
tan θ = sin θ / cos θ = 0.0055 / 0.029 = 0.1896
or
∴
θ = 10.74° Ans.
Fig. 14.16
2. When the vehicle reverses at the same speed in the direction of arrow Y along the same path
When the vehicle reverses at the same speed in the direction of arrow Y , then the gyroscopic
and centrifugal couples (C1 and C2) will be in clockwise direction about O and the balancing couple
due to weight (W 2 = m 2.g) of the frame about O will be in anticlockwise direction.
∴ Total overturning couple
= C1 + C2 = 0.019 cos θ + 0.0135 cos θ = 0.0325 cos θ N-m
Equating the total overturning couple to the balancing couple, we have
0.0325 cos θ = 0.029 sin θ
tan θ = sin θ / cos θ = 0.0325 / 0.029 = 1.1207
or
∴
θ = 48.26° Ans.
14.11. Effect of Gyroscopic Couple on a Disc Fixed Rigidly at a Certain
Angle to a Rotating Shaft
Consider a disc fixed rigidly to a rotating shaft such that the polar axis of the disc makes an
angle θ with the shaft axis, as shown in Fig. 14.17. Let the shaft rotates with an angular velocity ω
rad/s in the clockwise direction when viewed from the front. A little consideration will show that the
disc will also rotate about O X with the same angular velocity ω rad/s. Let OP be the polar axis and
OD the diametral axis of the disc.
Fig. 14.17. Effect of gyroscopic couple on a disc fixed rigidly at a certain angle to a rotating shaft.
Chapter 14 : Gyroscopic Couple and Precessional Motion
l
509
∴ Angular velocity of the disc about the polar axis OP or the angular velocity of spin
= ω cos θ
... (Component of ω in the direction of OP)
Since the shaft rotates, therefore the point P will move in a plane perpendicular to the plane
of paper. In other words, precession is produced about OD.
∴ Angular velocity of the disc about the diametral axis OD or the angular velocity of precession
= ω sin θ
If IP is the mass moment of inertia of the disc about the polar axis OP, then gyroscopic couple
acting on the disc,
1
CP = IP.ω cos θ.ω sin θ = × IP.ω2 sin 2θ
2
... (∵ 2 sin θ cos θ = sin 2θ)
The effect of this gyroscopic couple is to turn the disc in the anticlockwise when viewed from
the top, about an axis through O in the plane of paper.
Now consider the movement of point D about the polar axis OP. In this case, OD is axis of
spin and OP is the axis of precession.
∴ Angular velocity of disc about OD or angular velocity of spin
= ω sin θ
and angular velocity of D about OP or angular velocity of precession
= ω cos θ
If ID is the mass moment of inertia of the disc about the diametral axis OD, then gyroscopic
couple acting on the disc,
1
CD = ID.ω sin θ.ω cos θ = × ID.ω2 sin 2θ
2
The effect of this couple will be opposite to that of CP.
∴ Resultant gyroscopic couple acting on the disc,
1
× ω2 sin 2θ (IP – ID)
2
This resultant gyroscopic couple will act in the anticlockwise direction as seen from the top.
In other words, the shaft tends to turn in the plane of paper in anticlockwise direction as seen from the
top, as a result the horizontal force is exerted on the shaft bearings.
C = CP – CD =
Notes: 1. The mass moment of inertia of the disc about polar axis OP,
IP = m.r2/2
and mass moment of inertia of the disc about diametral axis OD,
l2 r2
+
I D = m
12 4
where
m = Mass of disc,
r = Radius of disc, and
l = Width of disc.
2. If the disc is thin, l may be neglected. In such a case
ID = m.r2/4
∴
C=
m.r 2 m.r 2 m
1
2 2
× ω2 sin 2θ
–
= × ω .r sin 2θ
2
2
4
8
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Theory of Machines
Example 14.17. A shaft carries a uniform thin disc of 0.6 m diameter and mass 30 kg. The
disc is out of truth and makes an angle of 1° with a plane at right angles to the axis of the shaft. Find
the gyroscopic couple acting on the bearing when the shaft rotates at 1200 r.p.m.
Solution. Given : d = 0.6 m or r = 0.3 m , m = 30 kg ; θ = 1° ; N = 1200 r.p.m. or
ω = 2 π × 1200/60 = 125.7 rad /s
We know that gyroscopic couple acting on the bearings,
C =
30
m
(125.7) 2 (0.3)2 sin 2 ° = 186 N-m Ans.
× ω2 .r 2 sin 2θ =
8
8
EXERCISES
1.
2.
3.
4.
5.
6.
7.
8.
A flywheel of mass 10 kg and radius of gyration 200 mm is spinning about its axis, which is horizontal
and is suspended at a point distant 150 mm from the plane of rotation of the flywheel. Determine the
angular velocity of precession of the flywheel. The spin speed of flywheel is 900 r.p.m.
[Ans. 0.39 rad/s]
A horizontal axle A B, 1 m long, is pivoted at the mid point C. It carries a weight of 20 N at A and a
wheel weighing 50 N at B. The wheel is made to spin at a speed of 600 r.p.m in a clockwise direction
looking from its front. Assuming that the weight of the flywheel is uniformly distributed around the
rim whose mean diameter is 0.6 m, calculate the angular velocity of precession of the system around
the vertical axis through C.
[Ans. 0.52 rad/s]
An aeroplane runs at 600 km / h. The rotor of the engine weighs 4000 N with radius of gyration of
1 metre. The speed of rotor is 3000 r.p.m. in anticlockwise direction when seen from rear side of the
aeroplane.
If the plane takes a loop upwards in a curve of 100 metres radius, find : 1. gyroscopic couple developed; and 2. effect of reaction gyroscopic couple developed on the body of aeroplane.
[Ans. 213.5 kN-m]
An aeroplane makes a complete half circle of 50 metres radius, towards left, when flying at 200 km
per hour. The rotary engine and the propeller of the plane has a mass of 400 kg with a radius of
gyration of 300 mm. The engine runs at 2400 r.p.m. clockwise, when viewed from the rear. Find the
gyroscopic couple on the aircraft and state its effect on it. What will be the effect, if the aeroplane
turns to its right instead of to the left ?
[Ans. 10 kN-m]
Each paddle wheel of a steamer have a mass of 1600 kg and a radius of gyration of 1.2 m. The steamer
turns to port in a circle of 160 m radius at 24 km / h, the speed of the paddles being 90 r.p.m. Find the
magnitude and effect of the gyroscopic couple acting on the steamer.
[Ans. 905.6 N-m]
The rotor of the turbine of a yacht makes 1200 r.p.m. clockwise when viewed from stern. The rotor has
a mass of 750 kg and its radius of gyration is 250 mm. Find the maximum gyroscopic couple transmitted to the hull (body of the yacht) when yacht pitches with maximum angular velocity of 1 rad /s. What is
the effect of this couple ?
[Ans. 5892 N-m]
The rotor of a turbine installed in a boat with its axis along the longitudinal axis of the boat makes
1500 r.p.m. clockwise when viewed from the stern. The rotor has a mass of 750 kg and a radius of
gyration of 300 mm. If at an instant, the boat pitches in the longitudinal vertical plane so that the bow
rises from the horizontal plane with an angular velocity of 1 rad /s, determine the torque acting on the
boat and the direction in which it tends to turn the boat at the instant.
[Ans. 10.6 kN-m]
The mass of a turbine rotor of a ship is 8 tonnes and has a radius of gyration 0.6 m. It rotates at 1800
r.p.m. clockwise when looking from the stern. Determine the gyroscopic effects in the following
cases:
1. If the ship travelling at 100 km / h strees to the left in a curve of 75 m radius, 2. If the ship is pitching
and the bow is descending with maximum velocity. The pitching is simple harmonic, the periodic time
being 20 seconds and the total angular movement between the extreme positions is 10°, and 3. If the
ship is rolling and at a certain instant has an angular velocity of 0.03 rad/s clockwise when looking
from stern.
Chapter 14 : Gyroscopic Couple and Precessional Motion
9.
10.
11.
12.
13.
14.
15.
l
511
In each case, explain clearly how you determine the direction in which the ship tends to move as a
result of the gyroscopic action.
[Ans. 201 kN-m ; 14.87 kN-m ; 16.3 kN-m]
The turbine rotor of a ship has a mass of 20 tonnes and a radius of gyration of 0.75 m. Its speed is 2000
r.p.m. The ship pitches 6° above and below the horizontal position. One complete oscillation takes 18
seconds and the motion is simple harmonic. Calculate :
1. the maximum couple tending to shear the holding down bolts of the turbine, 2. the maximum
angular acceleration of the ship during pitching, and 3. the direction in which the bow will tend to turn
while rising, if the rotation of the rotor is clockwise when looking from rear.
[Ans. 86.26 kN-m ; 0.0128 rad /s2, towards star-board]
A motor car takes a bend of 30 m radius at a speed of 60 km / hr. Determine the magnitudes of
gyroscopic and centrifugal couples acting on the vehicle and state the effect that each of these has on
the road reactions to the road wheels. Assume that :
Each road wheel has a moment of inertia of 3 kg-m2 and an effective road radius of 0.4 m.
The rotating parts of the engine and transmission are equivalent to a flywheel of mass 75 kg with a radius
of gyration of 100 mm. The engine turns in a clockwise direction when viewed from the front.
The back-axle ratio is 4 : 1, the drive through the gear box being direct. The gyroscopic effects of the
half shafts at the back axle are to be ignored.
The car has a mass of 1200 kg and its centre of gravity is 0.6 m above the road wheel.
The turn is in a right hand direction.
If the turn has been in a left hand direction, all other details being unaltered, which answers, if any,
need modification.
[Ans. 347.5 N-m : 6670 N-m]
A rail car has a total mass of 4 tonnes. There are two axles, each of which together with its wheels and
gearing has a total moment of inertia of 30 kg-m2. The centre distance between the two wheels on an
axle is 1.5 metres and each wheel is of 375 mm radius. Each axle is driven by a motor, the speed ratio
between the two being 1 : 3. Each motor with its gear has a moment of inertia of 15 kg-m2 and runs
in a direction opposite to that of its axle. The centre of gravity of the car is 1.05 m above the rails.
Determine the limiting speed for this car, when it rounding a curve of 240 metres radius such that no
wheel leaves the rail. Consider the centrifugal and gyroscopic effects completely. Assume that no cant
is provided for outer rail.
[Ans. 144 km / h]
A racing car weighs 20 kN. It has a wheel base of 2 m, track width 1 m and height of C.G. 300 mm
above the ground level and lies midway between the front and rear axle. The engine flywheel rotates
at 3000 r.p.m. clockwise when viewed from the front. The moment of inertia of the flywheel is
4 kg-m2 and moment of inertia of each wheel is 3 kg-m2. Find the reactions between the wheels and
the ground when the car takes a curve of 15 m radius towards right at 30 km / h, taking into consideration the gyroscopic and the centrifugal effects. Each wheel radius is 400 mm.
[Ans. Front inner wheel = 3341.7 N ; Front outer wheel = 6309.5 N ;
Rear inner wheel = 3690.5 N ; Rear outer wheel = 6658.3 N]
A four wheel trolley car of total mass 2000 kg running on rails of 1 m gauge, rounds a curve of 25 m
radius at 40 km / h. The track is banked at 10°. The wheels have an external diameter of 0.6 m and
each pair of an axle has a mass of 200 kg. The radius of gyration for each pair is 250 mm. The height
of C.G. of the car above the wheel base is 0.95 m. Allowing for centrifugal force and gyroscopic
couple action, determine the pressure on each rail.
[Ans. 4328 N ; 16 704 N]
A 2.2 tonne racing car has a wheel base of 2.4 m and a track of 1.4 m from the rear axle. The equivalent mass of engine parts is 140 kg with radius of gyration of 150 mm. The back axle ratio is 5. The
engine shaft and flywheel rotate clockwise when viewed from the front. Each wheel has a diameter of
0.8 m and a moment of inertia of 0.7 kg-m2. Determine the load distribution on the wheels when the
car is rounding a curve of 100 m radius at a speed of 72 km / h to the left.
A disc has a mass of 30 kg and a radius of gyration about its axis of symmetry 125 mm while its radius
of gyration about a diameter of the disc at right angles to the axis of symmetry is 75 mm. The disc is
pressed on to the shaft but due to incorrect boring, the angle between the axis of symmetry and the
actual axis of rotation is 0.25°, though both these axes pass through the centre of gravity of the disc.
Assuming that the shaft is rigid and is carried between bearings 200 mm apart, determine the bearing
forces due to the misalignment at a speed of 5000 r.p.m.
[Ans. 1810 N]
512
16.
17.
18.
l
Theory of Machines
A wheel of a locomotive, travelling on a level track at 90 km / h, falls in a spot hole 10 mm deep and
rises again in a total time of 0.8 seconds. The displacement of the wheel takes place with simple
harmonic motion. The wheel has a diameter of 3 m and the distance between the wheel centres is 1.75
m. The wheel pair with axle has a moment of inertia of 500 kg-m2. Determine the magnitude and the
effect of gyrocouple produced in this case.
[Ans. 186.6 N-m]
Each road wheel of a motor cycle has a mass moment of inertia of 1.5 kg-m2. The rotating parts of the
engine of the motor cycle have a mass moment of inertia of 0.25 kg-m2. The speed of the engine is 5
times the speed of the wheels and is in the same sense. The mass of the motor cycle with its rider is 250
kg and its centre of gravity is 0.6 m above the ground level.
Find the angle of heel if the cycle is travelling at 50 km / h and is taking a turn of 30 m radius. The
wheel diameter is 0.6 m.
[Ans. 35.7°]
A racing motor cyclist travels at 140 km/h round a curve of 120 m radius measured horizontally. The
cycle and rider have mass of 150 kg and their centre of gravity lies at 0.7 m above the ground level
when the motor cycle is vertical. Each wheel is 0.6 m in diameter and has moment of inertia about its
axis of rotation 1.5 kg-m2. The engine has rotating parts whose moment of inertia about their axis of
rotation is 0.25 kg-m2 and it rotates at five times the wheel speed in the same direction. Find : 1. the
correct angle of banking of the track so that there is no tendency to side slip, and 2. the correct angle
of inclination of the cycle and rider to the vertical.
[Ans. 52.12°; 55.57°]
[Hint. In calculating the angle of banking of the track, neglect the effect of gyroscopic couple]
DO YOU KNOW ?
1.
Write a short note on gyroscope.
2.
What do you understand by gyroscopic couple ? Derive a formula for its magnitude.
3.
Explain the application of gyroscopic principles to aircrafts.
4.
Describe the gyroscopic effect on sea going vessels.
5.
Explain the effect of the gyroscopic couple on the reaction of the four wheels of a vehicle negotiating
a curve.
6.
Discuss the effect of the gyroscopic couple on a two wheeled vehicle when taking a turn.
7.
What will be the effect of the gyroscopic couple on a disc fixed at a certain angle to a rotating shaft ?
OBJECTIVE TYPE QUESTIONS
1.
A disc is spinning with an angular velocity ω rad/s about the axis of spin. The couple applied to the
disc causing precession will be
(a)
1
I .ω2
2
where
(b)
I.ω2
(c)
1
I .ω.ωP
2
(d)
I.ω.ωP
I = Mass moment of inertia of the disc, and
ωP = Angular velocity of precession of the axis of spin.
2.
A disc spinning on its axis at 20 rad/s will undergo precession when a torque 100 N-m is applied about
an axis normal to it at an angular speed, if mass moment of inertia of the disc is the 1 kg-m2
(a) 2 rad/s
3.
(b)
5 rad/s
(c)
10 rad/s
(d)
20 rad/s
The engine of an aeroplane rotates in clockwise direction when seen from the tail end and the aeroplane
takes a turn to the left. The effect of the gyroscopic couple on the aeroplane will be
(a) to raise the nose and dip the tail
(b)
to dip the nose and raise the tail
(c) to raise the nose and tail
(d)
to dip the nose and tail
Chapter 14 : Gyroscopic Couple and Precessional Motion
4.
l
513
The air screw of an aeroplane is rotating clockwise when looking from the front. If it makes a left turn,
the gyroscopic effect will
(a) tend to depress the nose and raise the tail
(b) tend to raise the nose and depress the tail
(c) tilt the aeroplane
(d) none of the above
5.
6.
7.
The rotor of a ship rotates in clockwise direction when viewed from the stern and the ship takes a left
turn. The effect of the gyroscopic couple acting on it will be
(a) to raise the bow and stern
(b)
to lower the bow and stern
(c) to raise the bow and lower the stern
(d)
to lower the bow and raise the stern
When the pitching of a ship is upward, the effect of gyroscopic couple acting on it will be
(a) to move the ship towards port side
(b)
to move the ship towards star-board
(c) to raise the bow and lower the stern
(d)
to raise the stern and lower the bow
In an automobile, if the vehicle makes a left turn, the gyroscopic torque
(a) increases the forces on the outer wheels
(b)
decreases the forces on the outer wheels
(c) does not affect the forces on the outer wheels
(d) none of the above
8.
A motor car moving at a certain speed takes a left turn in a curved path. If the engine rotates in the
same direction as that of wheels, then due to the centrifugal forces
(a) the reaction on the inner wheels increases and on the outer wheels decreases
(b) the reaction on the outer wheels increases and on the inner wheels decreases
(c) the reaction on the front wheels increases and on the rear wheels decreases
(d) the reaction on the rear wheels increases and on the front wheels decreases
ANSWERS
1. (d)
2. (b)
3. (a)
4. (b)
5. (c)
6. (b)
7. (a)
8. (b)
GO To FIRST
CONTENTS
CONTENTS
514
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Theory of Machines
15
Features
1. Introduction.
2. Resultant Effect of a System of
Forces Acting on a Rigid Body.
3. D-Alembert’s Principle.
4. Velocity and Acceleration of
the Reciprocating Parts in
Engines.
5. Klien’s Construction.
6. Ritterhaus’s Construction.
7. Bennett’s Construction.
8. Approximate Analytical
Method for Velocity and
Acceleration of the Piston.
9. Angular
Velocity and
Acceleration of the Connecting
Rod.
10. Forces on the Reciprocating
Parts of an Engine Neglecting
Weight of the Connecting Rod.
11. Equivalent Dynamical System.
12. Determination of Equivalent
Dynamical System of Two
Masses by Graphical Method.
13. Correction Couple to be
Applied to Make the Two Mass
Systems
Dynamically
Equivalent.
14. Inertia
Forces
in
a
Reciprocating
Engine
Considering the Weight of
Connecting Rod.
15. Analytical Method for Inertia
Torque.
Inertia
Forces in
Reciprocating
Parts
15.1. Introduction
The inertia force is an imaginary force, which when
acts upon a rigid body, brings it in an equilibrium position. It
is numerically equal to the accelerating force in magnitude,
but opposite in direction. Mathematically,
Inertia force = – Accelerating force = – m.a
where
m = Mass of the body, and
a = Linear acceleration of the centre
of gravity of the body.
Similarly, the inertia torque is an imaginary torque,
which when applied upon the rigid body, brings it in equilibrium position. It is equal to the accelerating couple in magnitude but opposite in direction.
15.2. Resultant Effect of a System of Forces
Acting on a Rigid Body
Consider a rigid body acted upon by a system of
forces. These forces may be reduced to a single resultant force
514
CONTENTS
CONTENTS
Chapter 15 : Inertia Forces in Reciprocating Parts
l
515
F whose line of action is at a distance h from the centre of
gravity G. Now let us assume two equal and opposite forces
(of magnitude F ) acting through G, and parallel to the
resultant force, without influencing the effect of the
resultant force F, as shown in Fig. 15.1.
A little consideration will show that the body is
now subjected to a couple (equal to F × h) and a force,
equal and parallel to the resultant force F passing through
G. The force F through G causes linear acceleration of the
c.g. and the moment of the couple (F × h) causes angular
acceleration of the body about an axis passing through G
and perpendicular to the point in which the couple acts.
Fig. 15.1. Resultant effect of a system of
forces acting on a rigid body.
α = Angular acceleration of the rigid body due to couple,
Let
h = Perpendicular distance between the force and centre of gravity of the
body,
m = Mass of the body,
k = Least radius of gyration about an axis through G, and
I = Moment of inertia of the body about an axis passing through its centre
of gravity and perpendicular to the point in which the couple acts
= m.k2
We know that
Force,
and
F = Mass × Acceleration = m.a
...(i)
...(Q I = m.k 2 ) ...(ii)
F.h = m.k2.α = I.α
From equations (i) and (ii), we can
find the values of a and α, if the values of F,
m, k, and h are known.
15.3. D-Alembert’s Principle
Consider a rigid body acted upon by
a system of forces. The system may be
reduced to a single resultant force acting on
the body whose magnitude is given by the
product of the mass of the body and the linear
acceleration of the centre of mass of the body.
According to Newton’s second law of
motion,
The above picture shows the reciprocating parts
of a 19th century oil engine.
F = m.a
where
...(i)
F = Resultant force acting on the body,
m = Mass of the body, and
a = Linear acceleration of the centre of mass of the body.
The equation (i) may also be written as:
F – m.a = 0
...(ii)
A little consideration will show, that if the quantity – m.a be treated as a force, equal, opposite
516
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Theory of Machines
and with the same line of action as the resultant force F, and include this force with the system of
forces of which F is the resultant, then the complete system of forces will be in equilibrium. This
principle is known as D-Alembert’s principle. The equal and opposite force – m.a is known as reversed
effective force or the inertia force (briefly written as FI). The equation (ii) may be written as
F + FI = 0
...(iii)
Thus, D-Alembert’s principle states that the resultant force acting on a body together with
the reversed effective force (or inertia force), are in equilibrium.
This principle is used to reduce a dynamic problem into an equivalent static problem.
15.4. Velocity and Acceleration of the Reciprocating Parts in Engines
The velocity and acceleration of the reciprocating parts of the steam engine or internal
combustion engine (briefly called as I.C. engine) may be determined by graphical method or analytical
method. The velocity and acceleration, by graphical method, may be determined by one of the following
constructions:
1. Klien’s construction, 2. Ritterhaus’s construction, and 3. Bennett’s construction.
We shall now discuss these constructions, in detail, in the following pages.
15.5. Klien’s Construction
Let OC be the crank and PC the connecting rod of a reciprocating steam engine, as shown in
Fig. 15.2 (a). Let the crank makes an angle θ with the line of stroke PO and rotates with uniform
angular velocity ω rad/s in a clockwise direction. The Klien’s velocity and acceleration diagrams are
drawn as discussed below:
(a) Klien’s acceleration diagram.
(b) Velocity diagram.
(c) Acceleration diagram.
Fig. 15.2. Klien’s construction.
Klien’s velocity diagram
First of all, draw OM perpendicular to OP; such that it intersects the line PC produced at M.
The triangle OCM is known as Klien’s velocity diagram. In this triangle OCM,
OM may be regarded as a line perpendicular to PO,
CM may be regarded as a line parallel to PC, and
...(Q It is the same line.)
CO may be regarded as a line parallel to CO.
We have already discussed that the velocity diagram for given configuration is a triangle ocp
Chapter 15 : Inertia Forces in Reciprocating Parts
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517
as shown in Fig. 15.2 (b). If this triangle is revolved through 90°, it will be a triangle oc1 p1, in which
oc1 represents v CO (i.e. velocity of C with respect to O or velocity of crank pin C) and is paralel to OC,
op1 represents v PO (i.e. velocity of P with respect to O or velocity of cross-head or piston P)
and is perpendicular to OP, and
c1p1 represents v PC (i.e. velocity of P with respect to C) and is parallel to CP.
A little consideration will show, that the triangles oc1p1 and OCM are similar. Therefore,
oc1
op1
c p
=
= 1 1 = ω (a constant)
OC OM
CM
vCO
v
v
= PO = PC = ω
OC OM
CM
or
∴
vCO = ω × OC ; v PO = ω × OM, and v PC = ω × CM
Thus, we see that by drawing the Klien’s velocity diagram, the velocities of various points
may be obtained without drawing a separate velocity diagram.
Klien’s acceleration diagram
The Klien’s acceleration diagram is drawn as discussed below:
1. First of all, draw a circle
with C as centre and CM as radius.
2. Draw another circle with
PC as diameter. Let this circle intersect the previous circle at K and L.
3. Join KL and produce it to
intersect PO at N. Let KL intersect
PC at Q. This forms the quadrilateral
CQNO, which is known as Klien’s
acceleration diagram.
We have already discussed
that the acceleration diagram for the given configuration is as shown in Fig. 15. 2 (c). We know that
r
(i) o'c' represents aCO
(i.e. radial component of the acceleration of crank pin C with respect
to O ) and is parallel to CO;
r
(ii) c'x represents aPC
(i.e. radial component of the acceleration of crosshead or piston P
with respect to crank pin C) and is parallel to CP or CQ;
t
(i.e. tangential component of the acceleration of P with respect to C )
(iii) xp' represents aPC
and is parallel to QN (because QN is perpendicular to CQ); and
(iv) o'p' represents aPO (i.e. acceleration of P with respect to O or the acceleration of piston
P) and is parallel to PO or NO.
A little consideration will show that the quadrilateral o'c'x p' [Fig. 15.2 (c)] is similar to
quadrilateral CQNO [Fig. 15.2 (a)]. Therefore,
o′c ′ c ′x
xp ′ o′p′
=
=
=
= ω2 (a constant)
OC CQ QN
NO
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Theory of Machines
r
aCO
ar
at
a
= PC = PC = PO = ω2
OC
CQ QN
NO
or
∴
r
r
= ω2 × OC ; aPC
= ω2 × CQ
aCO
t
= ω2 × QN ; and aPO = ω2 × NO
aPC
Thus we see that by drawing the Klien’s acceleration diagram, the acceleration of various
points may be obtained without drawing the separate acceleration diagram.
Notes: 1. The acceleration of piston P with respect to crank pin C (i.e. aPC) may be obtained from:
∴
c′p ′
= ω2
or
CN
aPC = ω2 × CN
aPC
= ω2
CN
2. To find the velocity of any point D on the connecting rod PC, divide CM at D1 in the same ratio as D
divides CP. In other words,
CD1 CD
=
CM
CP
∴
Velocity of D, v D = ω × OD1
3. To find the acceleration of any point D on the connecting rod PC, draw a line from a point D parallel
to PO which intersects CN at D2 .
∴ Acceleration of D, aD = ω2 × OD2
4. If the crank position is such that the point N lies on the right of O instead of to the left as shown in
Fig. 15.2 (a), then the acceleration of the piston is negative. In other words, the piston is under going retardation.
5. The acceleration of the piston P is zero and its velocity is maximum, when N coincides with O.
There is no simple graphical method of finding the corresponding crank position, but it can be shown that for N
and O to coincide, the angle between the crank and the connecting rod must be slightly less than 90°. For most
practical purposes, it is assumed that the acceleration of piston P is zero, when the crank OC and connecting rod
PC are at right angles to each other.
15.6. Ritterhaus’s Construction
Let OC be the crank and PC the connecting rod of a rciprocating steam engine, as shown in
Fig. 15.3. Let the crank makes an angle θ with the line of stroke PO and rotates with uniform angular
velocity ω rad/s in a clockwise direction. The Ritterhaus’s velocity and acceleration diagrams are
drawn as discussed below:
Fig. 15.3. Ritterhaus’s construction.
Ritterhaus’s velocity diagram
Draw OM perpendicular to the line of stroke PO, such that it intersects the line PC produced
at M. The triangle OCM is known as Ritterhaus’s velocity diagram. It is similar to Klien’s velocity
diagram.
Chapter 15 : Inertia Forces in Reciprocating Parts
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519
∴ Velocity of C with respect to O or the velocity of crank pin C,
v CO = v C = ω × OC
Velocity of P with respect to O or the velocity of crosshead or piston P,
vPO = v P = ω × OM
and velocity of P with respect to C,
vPC = ω × CM
Ritterhaus’s acceleration diagram
The Ritterhaus’s acceleration diagram is drawn as discussed below:
1. From point M, draw M K parallel to the line of stroke PO, to interect OC produced at K.
2. Draw KQ parallel to MO. From Q draw QN perpendicular to PC.
3. The quadrilateral CQNO is known as Ritterhaus’s acceleration diagram. This is similar
to Klien’s acceleration diagram.
∴ Radial component of the acceleration of C with respect to O or the acceleration of crank
pin C,
r
= aC = ω2 × OC
aCO
Radial component of the acceleration of the crosshead or piston P with respect to crank
pin C,
r
= ω2 × CQ
aPC
Tangential component of the acceleration of P with respect to C,
t
= ω2 × QN
aPC
and acceleration of P with respect to O or the acceleration of piston P,
aPO = aP = ω2 × NO
Notes : 1. The acceleration of piston P with respect to crank pin C is given by
aPC = ω2 × CN
2. To find the velocity of any point D on the connecting rod PC, divide CM at D1 in the same ratio as D
divides CP. In other words,
CD1 CD
=
CM
CP
∴ Velocity of D
vD = ω × OD1
3. To find the acceleration of any point D on the connecting rod PC, draw DD2 parallel to the line of
stroke PO, which intersects CN at D2. The acceleration of D is given by
aD = ω2 × OD2
15.7. Bennett’s Construction
Let OC be the crank and PC the connecting rod of reciprocating steam engine, as shown in
Fig. 15.4. Let the crank makes an angle θ with the line of stroke PO and rotates with uniform angular
velocity ω rad/s in the clockwise direction. The Bennett’s velocity and acceleration diagrams are
drawn as discussed below:
Bennett’s velocity diagram
When the crank OC is at right angle to the line of stroke, it occupies the postition OC1 and the
crosshead P moves to the position P1, as shown in Fig. 15.4. Now, produce PC to intersect OC1 at M.
The triangle OCM is known as Bennett’s velocity diagram. It is similar to Klien’s velocity diagram.
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Theory of Machines
Fig. 15.4. Bennett’s construction.
∴ Velocity of C with respect to O or the velocity of crank pin C,
vCO = vC = ω × OC
Velocity of P with respect to O or the velocity of crosshead or piston P,
vPO = v P = ω × OM
and velocity of P with respect to C,
vPC = ω × CM
Bennett’s acceleration diagram
The Bennett’s acceleration diagram is drawn as discussed below:
1. From O, draw OL1 perpendicular to P1C1 (i.e. position of connecting rod PC when crank is
at right angle). Mark the position of point L on the connecting rod PC such that CL = C1L 1.
2. From L, draw LK perpendicular to PC and from point K draw KQ perpendicular to the line
of stroke PO. From point C, draw CN perpendicular to the line of stroke PO. Join NQ. A little
consideration will show that NQ is perpendicular to PC.
3. The quadrilateral CQNO is known as Bennett’s acceleration diagram. It is similar to
Klien’s acceleration diagram.
∴ Radial component of the acceleration of C with respect to O or the acceleration of the
crank pin C,
r
= aC = ω2 × OC
aCO
Radial component of the acceleration of the crosshead or piston P with respect to crank
pin C,
r
= ω2 × CQ
aPC
Tangential component of the acceleration of P with respect to C,
t
= ω2 × QN
aPC
and acceleration of P with respect to O or the acceleration of piston P,
aPO = aP = ω2 × NO
Notes : 1. The acceleration of piston P with respect to crank pin C is given by
aPC = ω2 × CN
2. The velocity and acceleration of any point D on the connecting rod PC may be obtained in the similar
way, as discussed in the previous articles, i.e.
Velocity of D, vD = ω × OD1
and
Acceleration of D, aD = ω2 × OD2
Chapter 15 : Inertia Forces in Reciprocating Parts
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521
Example 15.1. The crank and connecting rod of a reciprocating engine are 200 mm and 700
mm respectively. The crank is rotating in clockwise direction at 120 rad/s. Find with the help of
Klein’s construction: 1. Velocity and acceleration of the piston, 2. Velocity and acceleration of the
mid point of the connecting rod, and 3. Angular velocity and angular acceleration of the connecting
rod, at the instant when the crank is at 30° to I.D.C. (inner dead centre).
Solution. Given: OC = 200 mm = 0.2 m ; PC = 700 mm = 0.7 m ; ω = 120 rad/s
Fig. 15.5
The Klein’s velocity diagram OCM and Klein’s acceleration diagram CQNO as shown in Fig.
15.5 is drawn to some suitable scale, in the similar way as discussed in Art. 15.5. By measurement, we
find that
OM = 127 mm = 0.127 m ; CM = 173 mm = 0.173 m ; QN = 93 mm = 0.093 m ; NO = 200 mm
= 0.2 m
1. Velocity and acceleration of the piston
We know that the velocity of the piston P,
v P = ω × OM = 120 × 0.127 = 15.24 m/s Ans.
and acceleration of the piston P,
aP = ω2 × NO = (120)2 × 0.2 = 2880 m/s2 Ans.
2. Velocity and acceleration of the mid-point of the connecting rod
In order to find the velocity of the mid-point D of the connecting rod, divide CM at D1 in the
same ratio as D divides CP. Since D is the mid-point of CP, therefore D1 is the mid-point of CM, i.e.
CD1 = D1M. Join OD1. By measurement,
OD1 = 140 mm = 0.14 m
∴
Velocity of D, v D = ω × OD1 = 120 × 0.14 = 16.8 m/s Ans.
In order to find the acceleration of the mid-point of the connecting rod, draw a line DD2
parallel to the line of stroke PO which intersects CN at D2. By measurement,
OD2 = 193 mm = 0.193 m
∴ Acceleration of D,
aD = ω2 × OD2 = (120)2 × 0.193 = 2779.2 m/s2 Ans.
3. Angular velocity and angular acceleration of the connecting rod
We know that the velocity of the connecting rod PC (i.e. velocity of P with respect to C),
vPC = ω × CM = 120 × 0.173 = 20.76 m/s
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Theory of Machines
∴ Angular acceleration of the connecting rod PC,
ωPC =
vPC
20.76
=
= 29.66 rad/s Ans.
0.7
PC
We know that the tangential component of the acceleration of P with respect to C,
t
= ω2 × QN = (120)2 × 0.093 = 1339.2 m/s2
aPC
∴ Angular acceleration of the connecting rod PC,
t
aPC
1339.2
=
= 1913.14 rad/s2 Ans.
0.7
PC
Example 15.2. In a slider crank mechanism, the length of the crank and connecting rod are
150 mm and 600 mm respectively. The crank position is 60° from inner dead centre. The crank shaft
speed is 450 r.p.m. clockwise. Using Ritterhaus’s construction, determine 1. Velocity and acceleration of the slider, 2. Velocity and acceleration of point D on the connecting rod which is 150 mm from
crank pin C, and 3. angular velocity and angular acceleration of the connecting rod.
α PC =
Solution. Given : OC = 150 mm = 0.15m ; PC = 600 mm = 0.6 m ; CD = 150 mm = 0.15 m ;
N = 450 r.p.m. or ω = 2π × 450/60 = 47.13 rad/s
The Ritterhaus’s velocity diagram OCM and acceleration diagram CQNO, as shown in
Fig. 15.6, is drawn to some suitable scale in the similar way as discussed in Art. 15.6. By measurement, we find that
OM = 145 mm = 0.145 m ; CM = 78 mm = 0.078 m ; QN = 130 mm = 0.13 m ; and
NO = 56 mm = 0.056 m
Fig. 15.6
1. Velocity and acceleration of the slider
We know that the velocity of the slider P,
vP = ω × OM = 47.13 × 0.145 = 6.834 m/s Ans.
and acceleration of the slider P,
aP = ω2 × NO = (47.13)2 × 0.056 = 124.4 m/s2 Ans.
2. Velocity and acceleration of point D on the connecting rod
In order to find the velocity of point D on the connecting rod, divide CM at D1 in the same
ratio as D divides CP. In other words,
CD1 CD
150
CD
or CD1 =
=
× CM =
× 78 = 19.5 mm
600
CM
CP
CP
Join OD1. By measurement, OD1 = 145 mm = 0.145 m
∴ Velocity of point D,
vD = ω × OD1 = 47.13 × 0.145 = 6.834 m/s Ans.
Chapter 15 : Inertia Forces in Reciprocating Parts
l
523
In order to find the acceleration of point D on the connecting rod, draw DD2 parallel to the
line of stroke PO. Join OD2. By measurement, we find that OD2 = 120 mm = 0.12 m.
∴ Acceleration of point D,
aD = ω2 × OD2 = (47.13)2 × 0.12 = 266.55 m/s2 Ans.
3. Angular velocity and angular acceleration of the connecting rod
We know that the velocity of the connecting rod PC (or the velocity of point P with respect
to C ),
vPC = ω × CM = 47.13 × 0.078 = 3.676 m/s
∴ Angular velocity of the connecting rod,
vPC 3.676
=
= 6.127 rad/s Ans.
0.6
PC
We know that the tangential component of the acceleration of P with respect to C,
ωPC =
t
= ω2 × QN = (47.13)2 × 0.13 = 288.76 m/s2
aPC
∴ Angular acceleration of the connecting rod PC,
α PC =
t
aPC
288.76
=
= 481.27 rad/s 2 Ans.
0.6
PC
15.8. Approximate Analytical Method for Velocity and Acceleration of the
Piston
Consider the motion of a crank and connecting rod of a reciprocating steam engine as shown
in Fig. 15.7. Let OC be the crank and PC the connecting rod. Let the crank rotates with angular
velocity of ω rad/s and the crank turns through an angle θ from the inner dead centre (briefly written
as I.D.C). Let x be the displacement of a reciprocating body P from I.D.C. after time t seconds, during
which the crank has turned through an angle θ.
Fig. 15.7. Motion of a crank and connecting rod of a reciprocating steam engine.
Let
l = Length of connecting rod between the centres,
r = Radius of crank or crank pin circle,
φ = Inclination of connecting rod to the line of stroke PO, and
n = Ratio of length of connecting rod to the radius of crank = l/r.
Velocity of the piston
From the geometry of Fig. 15.7,
x = P ′P = OP′ − OP = ( P′C ′ + C ′O) − ( PQ + QO)
= (l + r ) − (l cos φ + r cos θ)
Q PQ = l cos φ ,
...
and QO = r cos θ
524
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Theory of Machines
l
= r (1 − cos θ) + l (1 − cos φ) = r (1 − cos θ) + (1 − cos φ)
r
= r [(1 − cos θ) + n (1 − cos φ)]
...(i)
From triangles CPQ and CQO,
CQ = l sin φ = r sin θ or l/r = sin θ/sin φ
∴
n = sin θ/sin φ or sin φ = sin θ/n
...(ii)
1
sin 2 θ 2
cos φ = (1 − sin 2 φ ) = 1 −
We know that,
n2
Expanding the above expression by binomial theorem, we get
1
2
cos φ = 1 −
or
1 − cos φ =
1 sin 2 θ
×
+ .....
2
n2
...(Neglecting higher terms)
sin 2 θ
...(iii)
2 n2
Substituting the value of (1 – cos φ) in equation (i), we have
sin 2 θ
sin 2 θ
= r (1 − cos θ) +
x = r (1 − cos θ) + n ×
2
2n
2n
Differentiating equation (iv) with respect to θ,
1
sin 2θ
dx
× 2 sin θ. cos θ = r sin θ +
= r sin θ +
2
n
dθ
2 n
..(iv)
... (v)
(Q 2 sin θ. cos θ = sin 2θ)
∴ Velocity of P with respect to O or velocity of the piston P,
dx dx d θ dx
vPO = vP =
=
×
=
×ω
dt
d θ dt
dθ
...(Q Ratio of change of angular velocity = d θ / dt = ω)
Substituting the value of dx/dθ from equation (v), we have
sin 2θ
vPO = vP = ω.r sin θ +
2 n
Note:
...(vi)
We know that by Klien’s construction,
vP = ω × OM
Comparing this equation with equation (vi), we find that
sin 2θ
OM = r sin θ +
2 n
Acceleration of the piston
Since the acceleration is the rate of change of velocity, therefore acceleration of the piston P,
aP =
dvP
dv
d θ dvP
= P ×
=
×ω
dt
dθ
dt
dθ
Chapter 15 : Inertia Forces in Reciprocating Parts
l
525
Differentiating equation (vi) with respect to θ,
cos 2 θ × 2
dvP
cos 2θ
= ω.r cos θ +
= ω.r cos θ +
2
n
dθ
n
Substituting the value of
dvP
in the above equation, we have
dθ
cos 2θ
× ω = ω2 .r
aP = ω.r cos θ +
n
cos 2θ
cos θ + n
...(vii)
Notes: 1. When crank is at the inner dead centre (I.D.C.), then θ = 0°.
cos 0°
1
= ω2 .r 1 +
aP = ω2 .r cos 0° +
n
n
2. When the crank is at the outer dead centre (O.D.C.), then θ = 180°.
∴
∴
cos 2 × 180°
1
aP = ω2.r cos 180° +
= ω2.r −1 +
n
n
As the direction of motion is reversed at the outer dead centre therefore changing the sign of the above
expression,
1
aP = ω2.r 1 −
n
Above picture shows a diesel engine. Steam engine, petrol engine and diesel engine, all
have reciprocating parts such as piston, piston rod, etc.
15.9. Angular Velocity and Acceleration of the Connecting Rod
Consider the motion of a connecting rod and a crank as shown in Fig. 15.7.From the geometry
of the figure, we find that
CQ = l sin φ = r sin θ
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Theory of Machines
r
sin θ
× sin θ =
l
n
Differentiating both sides with respect to time t,
∴
...Q n =
sin φ =
l
r
d φ cos θ d θ cos θ
dθ
... Q
=
×
=
×ω
= ω
dt
dt
n
dt
n
Since the angular velocity of the connecting rod PC is same as the angular velocity of point P
with respect to C and is equal to dφ/dt, therefore angular velocity of the connecting rod
cos φ ×
ωPC =
d φ cos θ
ω
ω cos θ
=
×
= ×
cos φ n cos φ
dt
n
1
We know that, cos φ = (1 − sin 2 φ )
ωPC =
∴
=
ω
×
n
1
2
sin 2 θ 2
= 1 −
n2
cos θ
1 −
1
sin 2 θ 2
n2
=
sin θ
... Q sin φ =
n
cos θ
ω
×
n 1 ( n 2 − sin 2 θ)1/2
n
ω cos θ
...(i)
(n2 − sin2 θ)1/2
Angular acceleration of the connecting rod PC,
αPC = Angular acceleration of P with respect to C =
d (ωPC )
dt
We know that
d (ωPC ) d (ωPC ) d θ d (ωPC )
=
×
=
×ω
dt
dθ
dt
dθ
...(ii)
...(Q d θ / dt = ω)
Now differentiating equation (i), we get
ω cos θ
d (ωPC )
d
=
(n 2 − sin 2 θ)1/2
dθ
dθ
(n2 − sin 2 θ)1/2 (− sin θ)] − [(cos θ) × 1 (n2 − sin 2 θ) –1/ 2 × − 2 sin θ cos θ
2
= ω
n2 − sin 2 θ
(n 2 − sin 2 θ)1/2 (− sin θ) + (n 2 − sin 2 θ) –1/ 2 sin θ cos 2 θ
=ω
n2 − sin 2 θ
2
2
1/2
2
2
–1/
2
2
(n − sin θ) − (n − sin θ)
cos θ
= − ω sin θ
2
2
n − sin θ
(n2 − sin 2 θ) − cos2
= − ω sin θ
2
2 3/ 2
(n − sin θ)
θ
...[Dividing and multiplying by (n2 − sin 2 θ)1/2 ]
Chapter 15 : Inertia Forces in Reciprocating Parts
=
− ω sin θ
(n2 − sin 2 θ)3/ 2
n2 − (sin 2 θ + cos2 θ) =
l
527
−ω sin θ (n2 − 1)
(n2 − sin 2 θ)3/ 2
...(Q sin 2 θ + cos 2 θ = 1)
∴ α PC =
d (ωPC )
− ω2 sin θ (n2 − 1)
×ω=
dθ
(n2 − sin 2 θ)3/ 2
...[From equation (ii)]
...(iii)
The negative sign shows that the sense of the acceleration of the connecting rod is such that it tends to
reduce the angle φ.
Notes: 1. Since sin2 θ is small as compared to n2, therefore it may be neglected. Thus, equations (i) and (iii) are
reduced to
ω cos θ
− ω2 sin θ ( n2 − 1)
, and α PC =
n
n3
2. Also in equation (iii), unity is small as compared to n2, hence the term unity may be neglected.
ωPC =
α PC =
∴
−ω2 sin θ
n
Example 15.3. If the crank and the connecting rod are 300 mm and 1 m long respectively
and the crank rotates at a constant speed of 200 r.p.m., determine:1. The crank angle at which the
maximum velocity occurs, and 2. Maximum velocity of the piston.
Solution. Given : r = 300 mm = 0.3 m ; l = 1 m ; N = 200 r.p.m. or ω = 2 π × 200/60 = 20.95 rad/s
1. Crank angle at which the maximum velocity occurs
θ = Crank angle from the inner dead centre at which the maximum
velocity occurs.
Let
We know that ratio of length of connecting rod to crank radius,
n = l/r = 1/0.3 = 3.33
and velocity of the piston,
sin 2θ
vP = ω.r sin θ +
2 n
For maximum velocity of the piston,
dvP
=0
dθ
or
...(i)
2 cos 2θ
i.e. ω.r cos θ +
=0
2n
n cos θ + 2 cos2 θ – 1 = 0
...(Q cos 2 θ = 2 cos 2 θ − 1)
2 cos2 θ + 3.33 cos θ – 1 = 0
∴
or
cos θ =
−3.33 ± (3.33)2 + 4 × 2 × 1
= 0.26
2×2
...(Taking + ve sign)
θ = 75º Ans.
2. Maximum velocity of the piston
Substituting the value of θ = 75° in equation (i), maximum velocity of the piston,
sin150°
= 20.95 × 0.3
vP( max ) = ω.r sin 75° +
2n
= 6.54 m/s Ans.
0.5
0.966 + 3.33 m/s
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Theory of Machines
Example 15.4. The crank and connecting rod of a steam engine are 0.3 m and 1.5 m in
length. The crank rotates at 180 r.p.m. clockwise. Determine the velocity and acceleration of the
piston when the crank is at 40 degrees from the inner dead centre position. Also determine the
position of the crank for zero acceleration of the piston.
Solution. Given : r = 0.3; l = 1.5 m ; N = 180 r.p.m. or ω = π × 180/60 = 18.85 rad/s; θ = 40°
Velocity of the piston
We know that ratio of lengths of the connecting rod and crank,
n = l/r = 1.5/0.3 = 5
∴ Velocity of the piston,
sin 80°
sin 2θ
vP = ω.r sin θ +
= 18.85 × 0.3 sin 40° + 2 × 5 m/s
2n
= 4.19 m/s Ans.
Acceleration of the piston
We know that acceleration of piston,
cos 2θ
cos80°
2
2
aP = ω2 .r cos θ +
= (18.85) × 0.3 cos 40° +
m/s
5
n
= 85.35 m/s 2 Ans.
Position of the crank for zero acceleration of the piston
Let
θ1 = Position of the crank from the inner dead centre for zero acceleration
of the piston.
We know that acceleration of piston,
cos 2θ1
aP = ω2 .r cos θ1 +
n
or
ω2 .r
(n cos θ1 + cos 2θ1 )
n
∴ n cos θ1 + cos 2θ1 = 0
5 cos θ1 + 2 cos2 θ1 – 1 = 0
∴
or
...(Q aP = 0)
0=
cos θ1 =
or
2 cos2 θ1 + 5 cos θ1 – 1 = 0
− 5 ± 52 + 4 × 1 × 2
= 0.1862
2×2
...(Taking + ve sign)
θ1 = 79.27° or 280.73° Ans.
Example 15.5. In a slider crank mechanism, the length of the crank and connecting rod are
150 mm and 600 mm respectively. The crank position is 60° from inner dead centre. The crank shaft
speed is 450 r.p.m. (clockwise). Using analytical method, determine: 1. Velocity and acceleration of
the slider, and 2. Angular velocity and angular acceleration of the connecting rod.
Solution. Given : r = 150 mm = 0.15 m ; l = 600 mm = 0.6 m ; θ = 60°; N = 400 r.p.m or
ω = π × 450/60 = 47.13 rad/s
1. Velocity and acceleration of the slider
We know that ratio of the length of connecting rod and crank,
n = l / r = 0.6 / 0.15 = 4
Chapter 15 : Inertia Forces in Reciprocating Parts
l
529
∴ Velocity of the slider,
sin 120°
sin 2 θ
vP = ω.r sin θ +
= 47.13 × 0.15 sin 60° + 2 × 4 m/s
2n
= 6.9 m/s Ans.
and acceleration of the slider,
cos 2θ
cos 120°
2
2
aP = ω2 .r cos θ +
= (47.13) × 0.15 cos 60° +
m/s
4
n
= 124.94 m/s 2 Ans.
2. Angular velocity and angular acceleration of the connecting rod
We know that angular velocity of the connecting rod,
ω cos θ 47.13 × cos 60°
=
= 5.9 rad/s Ans.
4
n
and angular acceleration of the connecting rod,
ωPC =
α PC =
ω2 sin θ (47.13)2 × sin 60°
=
= 481 rad/s 2 Ans.
4
n
15.10. Forces on the Reciprocating Parts of an Engine, Neglecting the
Weight of the Connecting Rod
The various forces acting on the reciprocating parts of a horizontal engine are shown in Fig.
15.8. The expressions for these forces, neglecting the weight of the connecting rod, may be derived as
discussed below :
1. Piston effort. It is the net force acting on the piston or crosshead pin, along the line of
stroke. It is denoted by FP in Fig. 15.8.
Fig. 15.8. Forces on the reciprocating parts of an engine.
Let
mR = Mass of the reciprocating parts, e.g. piston, crosshead pin or
gudgeon pin etc., in kg, and
WR = Weight of the reciprocating parts in newtons = m R.g
We know that acceleration of the reciprocating parts,
cos 2θ
aR = aP = ω2 .r cos θ +
n
530
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Theory of Machines
∴ *Accelerating force or inertia force of the reciprocating parts,
cos 2θ
FI = mR . aR = mR .ω 2 .r cos θ +
n
It may be noted that in a horizontal
engine, the reciprocating parts are
accelerated from rest, during the latter half
of the stroke (i.e. when the piston moves
from inner dead centre to outer dead
centre). It is, then, retarded during the latter
half of the stroke (i.e. when the piston
moves from outer dead centre to inner dead
centre). The inertia force due to the
acceleration of the reciprocating parts,
opposes the force on the piston due to the
difference of pressures in the cylinder on
Connecting rod of a petrol engine.
the two sides of the piston. On the other
hand, the inertia force due to retardation of the reciprocating parts, helps the force on the piston.
Therefore,
Piston effort,
FP = Net load on the piston m Inertia force
= FL m FI
...(Neglecting frictional resistance)
= FL m FI − RF
...(Considering frictional resistance)
where
RF = Frictional resistance.
The –ve sign is used when the piston is accelerated, and +ve sign is used when the piston is
retarded.
In a double acting reciprocating steam engine, net load on the piston,
FL = p1A 1 – p2 A 2 = p1 A 1 – p2 (A 1 – a)
where
p1, A 1 = Pressure and cross-sectional area on the back end side of the
piston,
p2, A2 = Pressure and cross-sectional area on the crank end side of the
piston,
a = Cross-sectional area of the piston rod.
Notes : 1. If ‘p’ is the net pressure of steam or gas on the piston and D is diameter of the piston, then
Net load on the piston, FL = Pressure × Area = p ×
π
× D2
4
2. In case of a vertical engine, the weight of the reciprocating parts assists the piston effort during the
downward stroke (i.e. when the piston moves from top dead centre to bottom dead centre) and opposes during
the upward stroke of the piston (i.e. when the piston moves from bottom dead centre to top dead centre).
∴ Piston effort,
FP = FL m FI ± WR − RF
2. Force acting along the connecting rod. It is denoted by FQ in Fig. 15.8. From the geometry of the figure, we find that
F
FQ = P
cos φ
*
The acceleration of the reciprocating parts by Klien’s construction is,
aP = ω2 × NO
∴
FI = m R.. ω2 × NO
Chapter 15 : Inertia Forces in Reciprocating Parts
We know that
∴
cos φ =
l
531
sin 2 θ
1−
n2
FP
FQ =
1−
sin 2 θ
n2
3. Thrust on the sides of the cylinder walls or normal reaction on the guide bars. It is
denoted by FN in Fig. 15.8. From the figure, we find that
FN = FQ sin φ =
F
... Q FQ = P
cos φ
FP
× sin φ = FP tan φ
cos φ
4. Crank-pin effort and thrust on crank shaft bearings. The force acting on the connecting
rod FQ may be resolved into two components, one perpendicular to the crank and the other along the
crank. The component of FQ perpendicular to the crank is known as crank-pin effort and it is denoted
by FT in Fig. 15.8. The component of FQ along the crank produces a thrust on the crank shaft bearings
and it is denoted by FB in Fig. 15.8.
Resolving FQ perpendicular to the crank,
FT = FQ sin (θ + φ) =
FP
× sin (θ + φ)
cos φ
and resolving FQ along the crank,
FB = FQ cos (θ + φ) =
FP
× cos (θ + φ)
cos φ
5. Crank effort or turning moment or torque on the crank shaft. The product of the crankpin effort (FT) and the crank pin radius (r) is known as crank effort or turning moment or torque on
the crank shaft. Mathematically,
Crank effort,
FP sin (θ + φ)
×r
cos φ
F (sin θ cos φ + cos θ sin φ)
= P
×r
cos φ
sin φ
= FP sin θ + cos θ ×
×r
cos φ
T = FT × r =
= FP (sin θ + cos θ tan φ) × r
...(i)
We know that l sin φ = r sin θ
and
∴
sin φ =
r
sin θ
sin θ =
l
n
cos φ =
1 − sin 2 φ =
tan φ =
sin φ sin θ
=
×
cos φ
n
l
...Q n =
r
1−
sin 2 θ
n
2
=
n
n − sin θ
2
2
1
n
=
n2 − sin 2 θ
sin θ
n − sin 2 θ
2
532
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Theory of Machines
Substituting the value of tan φ in equation (i), we have crank effort,
cos θ sin θ
T = FP sin θ +
×r
n2 − sin 2 θ
sin 2θ
= FP × r sin θ +
2
2
2 n − sin θ
...(ii)
...(Q 2 cos θ sin θ = sin 2θ)
Note:
Since
sin2
θ is very small as compared to
n2
therefore neglecting
sin2
θ, we have,
sin 2θ
T = FP × r sin θ +
= FP × OM
2n
We have seen in Art. 15.8, that
Crank effort,
sin 2θ
OM = r sin θ +
2n
Therefore, it is convenient to find OM instead of solving the large expression.
Example 15.6. Find the inertia force for the following data of an I.C. engine.
Bore = 175 mm, stroke = 200 mm, engine speed = 500 r.p.m., length of connecting rod =
400 mm, crank angle = 60° from T.D.C and mass of reciprocating parts = 180 kg.
Solution. Given : *D =175 mm ; L = 200 mm = 0.2 m or r = L / 2 = 0.1 m ; N = 500 r.p.m. or
ω = 2π × 500/60 =52.4 rad/s ; l = 400 mm = 0.4 m ; m R = 180 kg
The inertia force may be calculated by graphical method or analytical method as discussed
below:
1. Graphical method
First of all, draw the Klien’s acceleration diagram OCQN to some suitable scale as shown in
Fig. 15.9. By measurement,
ON = 38 mm = 0.038 m
∴ Acceleration of the reciprocating parts,
aR = ω2 × ON
= (52.4)2 × 0.038 = 104.34 m/s
We know that inertia force,
FI = m R × aR = 180 × 104.34 N
= 18 780 N = 18.78 kN Ans.
2. Analytical method
We know that ratio of lengths of connecting rod and crank,
n = l / r = 0.4 / 0.1 = 4
∴ Inertia force,
cos 2θ
FI = mR .ω2 .r cos θ +
n
Fig. 15.9
cos 120°
= 180 × (52.4)2 × 0.1 cos 60° +
= 18 530 N
4
= 18.53 kN Ans.
*
Superfluous data.
Chapter 15 : Inertia Forces in Reciprocating Parts
l
533
Example 15.7. The crank-pin circle radius of a horizontal engine is 300 mm. The mass of the
reciprocating parts is 250 kg. When the crank has travelled 60° from I.D.C., the difference between
the driving and the back pressures is 0.35 N/mm2. The connecting rod length between centres is 1.2 m
and the cylinder bore is 0.5 m. If the engine runs at 250 r.p.m. and if the effect of piston rod diameter
is neglected, calculate : 1. pressure on slide bars, 2. thrust in the connecting rod, 3. tangential force
on the crank-pin, and 4. turning moment on the crank shaft.
Solution. Given: r = 300 mm = 0.3 m ; m R = 250 kg; θ = 60°; p1 – p2 = 0.35 N/mm2;
l = 1.2 m ; D = 0.5 m = 500 mm ; N = 250 r.p.m. or ω = 2 π × 250/60 = 26.2 rad/s
First of all, let us find out the piston effort (FP).
We know that net load on the piston,
FL = ( p1 − p2 )
π
π
× D 2 = 0.35 × (500) 2 = 68730 N
4
4
...(Q Force = Pressure × Area)
Ratio of length of connecting rod and crank,
n = l / r = 1.2 / 0.3 = 4
and accelerating or inertia force on reciprocating parts,
cos 2θ
FI = mR . ω2 r cos θ +
n
= 250 (26.2) 2 0.3 cos 60° +
∴ Piston effort,
1. Pressure on slide bars
Let
We know that,
cos 120°
= 19306 N
4
FP = FL – FI = 68 730 – 19 306 = 49 424 N = 49.424 kN
φ = Angle of inclination of the connecting rod to the line of stroke.
sin φ =
sin θ sin 60 ° 0.866
=
=
= 0.2165
n
4
4
∴
φ = 12.5°
We know that pressure on the slide bars,
FN = FP tan φ = 49.424 × tan 12.5° = 10.96 kN Ans.
2. Thrust in the connecting rod
We know that thrust in the connecting rod,
FQ =
49.424
FP
=
= 50.62 kN Ans.
cos φ cos 12.5°
3. Tangential force on the crank-pin
We know that tangential force on the crank pin,
FT = FQ sin ( θ + φ) = 50.62 sin (60 ° + 12.5°) = 48.28 kN Ans.
4. Turning moment on the crank shaft
We know that turning moment on the crank shaft,
T = FT × r = 48.28 × 0.3 = 14.484 kN-m Ans.
534
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Theory of Machines
Example 15.8. A vertical double acting steam engine has a cylinder 300 mm diameter and
450 mm stroke and runs at 200 r.p.m. The reciprocating parts has a mass of 225 kg and the piston rod
is 50 mm diameter. The connecting rod is 1.2 m long. When the crank has turned through 125°
from the top dead centre, the steam pressure above the piston is 30 kN/m2 and below the piston is 1.5
kN/m2. Calculate the effective turning moment on the crank shaft.
Solution. Given : D = 300 mm = 0.3 m ; L = 450 mm or r = L/2 = 225 mm = 0.225 m ;
N = 200 r.p.m. or ω = 2 π × 200/60 = 20.95 rad/s ; m R= 225 kg ; d = 50 mm = 0.05 m ; l = 1.2 m ;
θ = 125° ; p1 = 30 kN/m2 = 30 × 103 N/m2 ; p2 = 1.5 kN/m2 = 1.5 × 103 N/m2
We know that area of the piston,
π
π
× D2 = × (0.3)2 = 0.0707 m2
4
4
π
π
2
a = × d = × (0.05)2 = 0.001 96 m2
and area of the piston rod,
4
4
∴ Force on the piston due to steam pressure,
A1 =
FL = p1. A1 − p2 ( A1 − a)
= 30 × 103 × 0.0707 − 1.5 × 103 (0.0707 − 0.001 96) N
= 2121 – 103 = 2018 N
Ratio of lengths of connecting rod and crank,
n = l / r = 1.2 / 0.225 = 5.33
and inertia force on the reciprocating parts,
cos 2θ
F1 = mR .ω 2 .r cos θ +
n
cos 250°
= 225 (20.95) 2 × 0.225 cos 125° +
= − 14 172 N
5.33
We know that for a vertical engine, net force on the piston or piston effort,
FP = FL – FI + m R.g
= 2018 – (– 14 172) + 225 × 9.81 = 18 397 N
Let
φ = Angle of inclination of the connecting rod to the line of stroke.
We know that,
∴
sin θ sin 125° 0.8191
=
=
= 0.1537
n
5.33
5.33
φ = 8.84°
sin φ =
We know that effective turning moment on the crank shaft,
18397 sin (125° + 8.84°)
F × sin (θ + φ)
T = P
×r =
× 0.225 N-m
cos φ
cos 8.84°
= 3021.6 N-m Ans.
Example 15.9. The crank and connecting rod of a petrol engine, running at 1800 r.p.m.are
50 mm and 200 mm respectively. The diameter of the piston is 80 mm and the mass of the reciprocating parts is 1 kg. At a point during the power stroke, the pressure on the piston is 0.7 N/mm2, when it
has moved 10 mm from the inner dead centre. Determine : 1. Net load on the gudgeon pin, 2. Thrust
in the connecting rod, 3. Reaction between the piston and cylinder, and 4. The engine speed at which
the above values become zero.
Solution. Given : N = 1800 r.p.m. or ω = 2π × 1800/60 = 188.52 rad/s ; r = 50 mm = 0.05 m;
l = 200 mm ; D = 80 mm ; m R = 1 kg ; p = 0.7 N/mm2 ; x = 10 mm
Chapter 15 : Inertia Forces in Reciprocating Parts
l
535
1. Net load on the gudgeon pin
We know that load on the piston,
FL =
π 2
π
D × p = × (80)2 × 0.7 = 3520 N
4
4
Fig. 15.10
When the piston has moved 10 mm from the inner dead centre, i.e. when P1P = 10 mm, the
crank rotates from OC1 to OC through an angle θ as shown in Fig. 15.10.
By measurement, we find that *θ = 33°.
We know that ratio of lengths of connecting rod and crank,
n = l/r = 200 /50 = 4
and inertia force on the reciprocating parts,
cos 2θ
FI = mR .aR = mR .ω2 .r cos θ +
n
cos 66°
= 1 × (188.52) 2 × 0.05 cos 33° +
= 1671 N
4
We know that net load on the gudgeon pin,
FP = FL − FI = 3520 − 1671 = 1849 N Ans.
2. Thrust in the connecting rod
φ = Angle of inclination of the connecting rod to the line of
stroke.
Let
We know that,
∴
*
sin φ =
sin θ sin 33° 0.5446
=
=
= 0.1361
n
4
4
φ = 7.82°
The angle θ may also be obtained as follows:
We know that
sin 2 θ
1 − cos2 θ
x = r (1 − cos θ) +
= r (1 − cos θ) +
2n
2n
1 − cos 2 θ 50
10 = 50 (1 − cos θ) +
(8 − 8 cos θ + 1 − cos 2 θ
=
8
2×4
= 50 – 50 cos θ + 6.25 – 6.25 cos2 θ
or
6.25 cos2 θ + 50 cos θ – 56.25 = 0
Solving this quadratic equation, we get θ = 33.14°
536
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Theory of Machines
Twin-cylinder aeroplane engine.
We know that thrust in the connecting rod,
FQ =
1849
FP
=
= 1866.3N Ans.
cos φ cos 7.82°
3. Reaction between the piston and cylinder
We know that reaction between the piston and cylinder,
FN = FP tan φ = 1849 tan 7.82° = 254 N A ns.
4. Engine speed at which the above values will become zero
A little consideration will show that the above values will become zero, if the inertia force on
the reciprocating parts (FI) is equal to the load on the piston (FL). Let ω1 be the speed in rad/s, at
which FI = FL .
∴
cos 2θ π 2
mR (ω1 )2 r cos θ +
= D × p
n 4
cos 66° π
2
2
1 (ω1 )2 × 0.05 cos 33° +
= × (80) × 0.7 or 0.0 47 (ω1 ) = 3520
4 4
∴
(ω1)2 = 3520 / 0.047 = 74 894 or ω1 = 273.6 rad/s
∴ Corresponding speed in r.p.m.,
N 1 = 273.6 × 60 / 2π = 2612 r.p.m. Ans.
Example 15.10. During a trial on steam engine, it is found that the acceleration of the piston
is 36 m/s2 when the crank has moved 30° from the inner dead centre position. The net effective steam
pressure on the piston is 0.5 N/mm2 and the frictional resistance is equivalent to a force of 600 N. The
diameter of the piston is 300 mm and the mass of the reciprocating parts is 180 kg. If the length of
the crank is 300 mm and the ratio of the connecting rod length to the crank length is 4.5, find:
1. Reaction on the guide bars, 2. Thrust on the crank shaft bearings, and 3. Turning moment on the
crank shaft.
Solution. Given : aP = 36 m/s2 ; θ = 30°; p = 0.5 N/mm2 ; R F = 600 N; D = 300 mm ;
m R = 180 kg ; r = 300 mm = 0.3 m ; n = l / r = 4.5
1. Reaction on the guide bars
First of all, let us find the piston effort (FP). We know that load on the piston,
FL = p ×
π
π
× D 2 = 0.5 × × (300) 2 = 35350 N
4
4
Chapter 15 : Inertia Forces in Reciprocating Parts
l
537
and inertia force due to reciprocating parts,
FI = m R × aP = 180 × 36 = 6480 N
∴ Piston effort, FP = FL – FI – R F = 35 350 – 6480 – 600 = 28 270 N = 28.27 kN
Let
φ = Angle of inclination of the connecting rod to the line of stroke.
We know that sin φ = sin θ/n = sin 30°/4.5 = 0.1111
∴
φ = 6.38°
We know that reaction on the guide bars,
FN = FP tan φ = 28.27 tan 6.38° = 3.16 kN Ans.
2. Thrust on the crank shaft bearing
We know that thrust on the crank shaft bearings,
FP cos (θ + φ) 28.27 cos (30° + 6.38°)
=
= 22.9 kN Ans.
cos φ
cos 6.38°
3. Turning moment on the crank shaft
FB =
We know that turning moment on the crank shaft,
T =
28.27 sin (30° + 6.38°)
FP sin (θ + φ)
×r =
× 0.3 kN-m
cos φ
cos 6.38°
= 5.06 kN-m
Example 15.11. A vertical petrol engine 100 mm diameter and 120 mm stroke has a
connecting rod 250 mm long. The mass of the piston is 1.1 kg. The speed is 2000 r.p.m. On the
expansion stroke with a crank 20° from top dead centre, the gas pressure is 700 kN/m2. Determine:
1. Net force on the piston, 2. Resultant load on the gudgeon pin,
3. Thrust on the cylinder walls, and 4. Speed above which, other things remaining same, the gudgeon pin load would be reversed in direction.
Solution. Given: D = 100 mm = 0.1 m ; L = 120 mm = 0.12 m or
r = L/2 = 0.06 m ; l = 250 mm = 0.25 m ; m R = 1.1 kg ; N = 2000 r.p.m. or
ω = 2 π × 2000/60 = 209.5 rad/s ; θ = 20°; p = 700 kN/m2
1. Net force on the piston
The configuration diagram of a vertical engine is shown in Fig. 15.11.
We know that force due to gas pressure,
π
π
× D2 = 700 × × (0.1)2 = 5.5 kN
4
4
= 5500 N
FL = p ×
and ratio of lengths of the connecting rod and crank,
n = l/r = 0.25 /0.06 = 4.17
∴ Inertia force on the piston,
cos 2θ
FI = mR .ω2 .r cos θ +
n
cos 40°
= 1.1 × (209.5)2 × 0.06 × cos 20° +
4.17
= 3254 N
Fig. 15.11
538
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Theory of Machines
We know that for a vertical engine, net force on the piston,
FP = FL − FI + WR = FL − FI + mR .g
= 5500 − 3254 + 1.1 × 9.81 = 2256.8 N Ans.
2. Resultant load on the gudgeon pin
φ = Angle of inclination of the connecting rod to the line of stroke.
Let
We know that,
sin φ = sin θ / n = sin 20°/4.17 = 0.082
∴
φ = 4.7°
We know that resultant load on the gudgeon pin,
2256.8
FP
=
= 2265 N Ans.
cos φ cos 4.7°
3. Thrust on the cylinder walls
FQ =
We know that thrust on the cylinder walls,
FN = FP tan φ = 2256.8 × tan 4.7° = 185.5 N Ans.
4. Speed, above which, the gudgeon pin load would be reversed in direction
Let
N 1 = Required speed, in r.p.m.
The gudgeon pin load i.e. FQ will be reversed in direction, if FQ becomes negative. This is only
possible when FP is negative. Therefore, for FP to be negative, FI must be greater than (FL + W R),
i.e.
cos 2θ
mR (ω1 )2 r cos θ +
> 5500 + 1.1 × 9.81
n
cos 40°
1.1 × (ω1 ) 2 × 0.06 cos 20° +
> 5510.8
4.17
0.074 (ω1 )2 > 5510.8
or
or
(ω1 )2 > 5510.8 / 0.074 or 74 470
ω1 > 273 rad/s
∴ Corresponding speed in r.p.m.,
N1 > 273 × 60 / 2π
or
2606 r.p.m. Ans.
Example 15. 12. A horizontal steam engine running at 120 r.p.m. has a bore of 250 mm and
a stroke of 400 mm. The connecting rod is 0.6 m and mass of the reciprocating parts is 60 kg. When
the crank has turned through an angle of 45° from the inner dead centre, the steam pressure on the
cover end side is 550 kN/m2 and that on the crank end side is 70 kN/m2. Considering the diameter of
the piston rod equal to 50 mm, determine:
1. turning moment on the crank shaft, 2. thrust on the bearings, and 3. acceleration of the
flywheel, if the power of the engine is 20 kW, mass of the flywheel 60 kg and radius of gyration 0.6 m.
Solution. Given : N = 120 r.p.m. or ω = 2π × 120/60 = 12.57 rad/s ; D = 250 mm = 0.25 m ;
L = 400 mm = 0.4 m or r = L/2 = 0.2 m ; l = 0.6 m ; m R = 60 kg ; θ = 45° ; d = 50 mm = 0.05 m ;
p1 = 550 kN/m2 = 550 × 103 N/m2 ; p2 = 70 kN/m2 = 70 × 103 N/m2
Chapter 15 : Inertia Forces in Reciprocating Parts
High-pressure steam in
Valve rod
Cross head
l
539
Exhaust steam
Slide valve
Cross-head guide
Piston Rod
Piston
Cylinder
1. Turning moment on the crankshaft
First of all, let us find the net load on the piston (FP).
We know that area of the piston on the cover end side,
A1 =
π
π
× D 2 = × (0.25) 2 = 0.049 m2
4
4
π
π
× d 2 = × (0.05) 2 = 0.00196 m2
4
4
∴ Net load on the piston,
and area of piston rod,
a=
FL = p1 . A1 − p2 . A2 = p1 . A1 − p2 ( A1 − a )
= 550 × 103 × 0.049 − 70 × 103 (0.049 − 0.001 96) = 23657 N
We know that ratio of lengths of the connecting rod and crank,
n = l/r = 0.6/0.2 = 3
and inertia force on the reciprocating parts,
cos 2θ
FI = mR .ω2.r cos θ +
n
cos 90°
= 60 × (12.57)2 × 0.2 cos 45° +
= 1340 N
3
∴ Net force on the piston or piston effort,
FP = FL − FI = 23657 − 1340 = 22 317 N = 22.317 kN
Let
We know that,
∴
φ = Angle of inclination of the connecting rod to the line of stroke.
sin φ = sin θ/n = sin 45°/3 = 0.2357
φ = 13.6°
We know that turning moment on the crankshaft,
FP sin (θ + φ)
22.317 × sin (45° + 13.6 °)
×r=
× 0.2 kN-m
cos φ
cos 13.6°
= 3.92 kN-m = 3920 N-m Ans.
T =
540
l
Theory of Machines
2. Thrust on the bearings
We know that thrust on the bearings,
FP cos (θ + φ) 22.317 × cos (45° + 13.6°)
=
= 11.96 kN Ans.
cos φ
cos 13.6°
3. Acceleration of the flywheel
Given: P = 20 kW = 20 × 103 W; m = 60 kg ; k = 0.6 m
Let
α = Acceleration of the flywheel in rad/s2.
We know that mass moment of inertia of the flywheel,
I = m.k2 = 60 × (0.6) 2 = 21.6 kg-m2
∴ Accelerating torque, T A = I.α = 21.6 α N-m
FB =
...(i)
2πNT
P × 60 20 × 103 × 60
∴ P =
=
= 1591 N-m
60
2π N
2 π × 120
Since the accelerating torque is equal to the difference of torques on the crankshaft or turning
moment (T) and the resisting torque (T R), therefore, accelerating torque,
T A = T – T R = 3920 – 1591 = 2329 N-m
...(ii)
From equation (i) and (ii),
and resisting torque,
TR =
α = 2329/21.6 = 107.8 rad/s2 Ans.
Example 15.13. A vertical, single cylinder,
single acting diesel engine has a cylinder diameter
300 mm, stroke length 500 mm, and connecting rod
length 4.5 times the crank length. The engine runs
at 180 r.p.m. The mass of the reciprocating parts is
280 kg. The compression ratio is 14 and the pressure
remains constant during the injection of the oil for
1/10th of the stroke. If the compression and
expansion follows the law p.V1.35 = constant, find:
1. Crank-pin effort, 2. Thrust on the bearings, and
3. Turning moment on the crank shaft, when the
crank displacement is 45° from the inner dead centre
position during expansion stroke.
The suction pressure may be taken as
0.1 N/mm2.
Solution. Given : D = 300 mm = 0.3 m ;
Fig. 15.12
L = 500 mm = 0.5 m or r = 0.25 m ; l = 4.5 r or n = l / r = 4.5; N = 180 r.p.m.
or
V1
= 14; θ = 45°; p1 = 0.1 N/mm2
V2
The pressure-volume (i.e. p-V) diagram for a *diesel engine is shown in Fig 15.12, in which
ω = 2π × 180/60 = 18.85 rad/s ; mR = 280 kg ;
1-2 represents the compression, 2-3 represents the injection of fuel, 3-4 represents the expansion, and 4-1 represents the exhaust.
Let p1, p2, p3, and p4 = Pressures corresponding to points 1, 2, 3 and 4 respectively, and
V 1, V 2, V 3, and
*
V4 = Volumes corresponding to points 1, 2, 3 and 4 respectively.
In a diesel engine, the compression and expanssion are isentropic i.e. according to the law p.Vγ = constant.
The injection of fuel takes place at constant pressure and the exhaust is at constant volume.
Chapter 15 : Inertia Forces in Reciprocating Parts
l
541
Since the compression follows the law p.V1.35 = constant, therefore
p1 (V1 )1.35 = p2 (V2 )1.35
1.35
or
V
p2 = p1 1
V2
We know that swept volume,
VS =
= 0.1 × (14)1.35 = 3.526 N/mm 2
π
π
× D 2 × L = × (0.3) 2 × 0.5 = 0.035 m 3
4
4
V + VS
V
compression ratio, = V1 = C
=1+ S
V2
VC
VC
and
∴
14 = 1 +
0.035
VC
or
VC =
...(Q V2 = VC )
0.035
= 0.0027 m3
14 − 1
Since the injection of fuel takes place at constant pressure (i.e. p2 = p3) and continues up to
1/10th of the stroke, therefore volume at the end of the injection of fuel,
1
0.035
× VS = 0.0027 +
= 0.0062 m 3
10
10
When the crank displacement is 45° (i.e. when θ = 45°) from the inner dead centre during
expansion stroke, the corresponding displacement of the piston (marked by point 4' on the p-V diagram)
is given by
V3 = VC +
sin 2 θ
sin 2 45°
x = r (1 − cos θ) +
= r (1 − cos 45°) +
2n
2 × 4.5
0.5
= 0.25 (1 − 0.707) +
= 0.087 m
9
π
π
× D 2 × x = 0.0027 + × (0.3)2 × 0.087 = 0.0088 m 2
4
4
1.35
Since the expansion follows the law p.V = constant, therefore,
∴
V4′ = VC +
p3 (V3 )1.35 = p4′ (V4′ )1.35
1.35
1.35
V3
0.0062
3.526
p
p
=
=
= 2.2 N/mm 2
∴
3
4′
V
0.0088
4′
Difference of pressures on two sides of the piston,
p = p4′ − p1 = 2.2 − 0.1 = 2.1 N/mm 2 = 2.1 × 106 N/m 2
∴ Net load on the piston,
π
π
× D2 = 2.1 × 106 × × (0.3)2 = 148 460 N
4
4
Inertia force on the reciprocating parts,
FL = p ×
cos 2θ
FI = mR .ω2 .r cos θ +
n
cos 90°
= 280 × (18.85)2 × 0.25 cos 45° +
= 17 585 N
4.5
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Theory of Machines
We know that net force on the piston or piston effort,
FP = FL − FI + WR = FL − FI + mR .g
= 148 460 − 17 858 + 280 × 9.81 = 133 622 N
1. Crank-pin effort
Let
φ = Angle of inclination of the connecting rod to the line of stroke.
We know that, sin φ = sin θ/n = sin 45°/4.5 = 0.1571
∴
φ = 9.04°
We know that crank-pin effort,
F sin (θ + φ) 133622 × sin (45° + 9.04°)
FT = P
=
= 109522 N
cos φ
cos 9.04°
= 109.522 kN Ans.
2. Thrust on the bearings
We know that thrust on the bearings,
FP .cos ( θ + φ) 133622 × sin (45° + 9.04°)
=
= 79 456 N
cos φ
cos 9.04°
= 79.956 kN Ans.
FB =
3. Turning moment on the crankshaft
We know that the turning moment on the crankshaft,
T = FT × r = 109.522 × 0.25 = 27.38 kN-m Ans.
Example 15.14. A vertical double acting steam engine
has cylinder diameter 240 mm, length of stroke 360 mm and
length of connecting rod 0.6 m. The crank rotates at 300 r.p.m.
and the mass of the reciprocating parts is 160 kg. The steam is
admitted at a pressure of 8 bar gauge and cut-off takes place at
1/3rd of the stroke. The expansion of steam is hyperbolic. The
exhaust of steam takes place at a pressure of –0.75 bar gauge.
The frictional resistance is equivalent to a force of 500 N. Determine the turning moment on the crankshaft, when the piston is
75° from the top dead centre. Neglect the effect of clearance and
assume the atmospheric presssure as 1.03 bar.
Solution. Given D = 240 mm = 0.24 m ; L = 360 mm
= 0.36 m or r = L /2 = 0.18 m ; l = 0.6 m ; l= 0.6 m ;
N = 300 r.p.m. or ω = 2π × 300/60= 31.42 rad/s; m R = 160 kg ;
pA = 8 + 1.03 = 9.03 bar = 903 × 103 N/m2 ; pE = – 0.75 + 1.03
= 0.28 bar = 28 × 103 N/m2 ; FR = 500 N ; θ = 75°
First of all, let us find the piston effort (FP).
Fig. 15.13
The pressure-volume ( p-V ) diagram for a steam engine,
neglecting clearance, is shown in FIg. 15.13, in which A B
represents the admission of steam, BC the expansion and DE the exhaust of steam. The steam is cutoff at point B.
We know that the stroke volume,
VS =
π
π
× D 2 × L = × (0.24)2 × 0.36 = 0.0163 m3
4
4
Chapter 15 : Inertia Forces in Reciprocating Parts
l
543
Since the admission of steam is cut-off at 1/3rd of the stroke, therefore volume of steam at
cut-off,
VB = V S / 3 = 0.0163/3 = 0.005 43 m3
We know that ratio of the lengths of the connecting rod and crank,
n = l / r = 0.6 / 0.18 = 3.33
When the crank position is 75° from the top dead centre (i.e. when θ = 75°), the displacement
of the piston (marked by point C' on the expansion curve BC) is given by
sin 2 θ
sin 2 75°
0.18
=
x = r (1 − cos θ) +
1
cos
75
−
°
+
2n
2 × 3.33
= 0.1586 m
0.1586
x
VC′ = VS × = 0.0163 ×
= 0.0072 m3
0.36
L
Since the expansion is hyperbolic (i.e. according to the law p V = constant), therefore
∴
or
pB .VB = pC′ .VC′
p × VB 903 × 103 × 0.005 43
pC′ = B
=
= 681 × 103 N/m2
0.0072
′
VC
∴ Difference of pressures on the two sides of the piston,
p = pC′ − pE = 681 × 103 − 28 × 103 = 653 × 103 N/m 2
We know that net load on the piston,
π
π
× D 2 × p = × (0.24) 2 × 653 × 103 = 29545 N
4
4
and inertia force on the reciprocating parts,
FL =
∴ Piston effort,
cos 2θ
FI = mR .ω2 .r cos θ +
n
cos 150°
= 160 × (31.42)2 × 0.18 cos75° +
= − 36 N
3.33
FP = FL − FI + WR − FR
= 29 545 − ( − 36) + 160 × 9.81 − 500 = 30 651 N
Turning moment on the crankshaft
Let
We know that
∴
φ = Angle of inclination of the connecting rod to the line of stroke.
sin φ = sin θ/n = sin 75°/3.33 = 0.29
φ = 16.86°
We know that turning moment on the crankshaft
FP sin ( θ + φ)
30 651 sin (75° + 16.86°)
×r=
× 0.18 N-m
cos φ
cos 16.86 °
= 5762 N-m Ans.
T =
15.11. Equivalent Dynamical System
In order to determine the motion of a rigid body, under the action of external forces, it is
usually convenient to replace the rigid body by two masses placed at a fixed distance apart, in such
a way that,
544
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Theory of Machines
1. the sum of their masses is equal to the total mass of the body ;
2. the centre of gravity of the two masses coincides with that of the body ; and
3. the sum of mass moment of inertia of the masses about their centre of gravity is equal to
the mass moment of inertia of the body.
When these three conditions are satisfied, then it is said to be an equivalent dynamical system.
Consider a rigid body, having its centre of gravity at G, as shown in Fig. 15.14.
Let
m = Mass of the body,
k G = Radius of gyration about
its centre of gravity G,
m 1 and m 2 = Two masses which form a
dynamical equivalent system,
l1 = Distance of mass m 1 from G,
l2 = Distance of mass m 2 from G,
and
L = Total distance between the
masses m 1 and m 2.
Fig. 15.14. Equivalent
dynamical system.
Thus, for the two masses to be dynamically equivalent,
m1 + m2 = m
...(i)
m1.l1 = m2 .l2
...(ii)
m1 (l1 )2 + m2 (l2 )2 = m (kG )2
and
...(iii)
From equations (i) and (ii),
l2 .m
l1 + l2
l .m
m2 = 1
l1 + l2
Substituting the value of m 1 and m 2 in equation (iii), we have
m1 =
and
l2 .m
l .m
(l1 )2 + 1
(l2 )2 = m (kG )2
l1 + l2
l1 + l2
∴
or
...(iv)
...(v)
l1 .l2 (l1 + l2 )
= (k G )2
l1 + l2
l1 .l 2 = (kG ) 2
...(vi)
This equation gives the essential condition of placing the two masses, so that the system
becomes dynamical equivalent. The distance of one of the masses (i.e. either l1 or l2) is arbitrary
chosen and the other distance is obtained from equation (vi).
Note : When the radius of gyration k G is not known, then the position of the second mass may be obtained by
considering the body as a compound pendulum. We have already discussed, that the length of the simple pendulum which gives the same frequency as the rigid body (i.e. compound pendulum) is
L=
We also know that
( kG ) 2 + h 2 ( kG ) 2 + ( l1 ) 2
=
h
l1
..(Replacing h by l1)
l1.l2 = ( k G ) 2
l1.l2 + ( l1 ) 2
= l2 + l1
l1
This means that the second mass is situated at the centre of oscillation or percussion of the
body, which is at a distance of l2 = (k G)2/l1.
∴
L=
Chapter 15 : Inertia Forces in Reciprocating Parts
l
545
15.12. Determination of Equivalent Dynamical System of Two Masses by
Graphical Method
Consider a body of mass m, acting at G as
shown in Fig. 15.15. This mass m, may be replaced
by two masses m 1 and m2 so that the system becomes
dynamical equivalent. The position of mass m 1 may
be fixed arbitrarily at A . Now draw perpendicular
CG at G, equal in length of the radius of gyration of
the body, k G . Then join A C and draw C B
perpendicular to A C intersecting A G produced in
B. The point B now fixes the position of the second
mass m 2.
A little consideration will show that the
triangles ACG and BCG are similar. Therefore,
kG
l
= 2
l1
kG
or
Fig. 15.15. Determination of equivalent
dynamical system by graphical method.
( kG )2 = l1 .l2
...(Same as before)
Example 15.15. The connecting rod of a gasoline engine
is 300 mm long between its centres. It has a mass of 15 kg and
mass moment of inertia of 7000 kg-mm2. Its centre of gravity is at
200 mm from its small end centre. Determine the dynamical
equivalent two-mass system of the connecting rod if one of the
masses is located at the small end centre.
Solution. Given : l = 300 mm ; m = 15 kg; I = 7000 kg-mm2 ;
l1 = 200 mm
The connecting rod is shown in Fig. 15.16.
Let
kG = Radius of gyration of the connecting rod
about an axis passing through its centre of
gravity G.
We know that mass moment of inertia (I),
7000 = m (k G)2 = 15 (k G)2
∴
(k G)2 = 7000/15 = 466.7 mm2 or k G = 21.6 mm
It is given that one of the masses is located at the small end
centre. Let the other mass is placed at a distance l2 from the centre
of gravity G, as shown in Fig. 15.17.
We know that for a dynamical equivalent system,
Fig. 15.16
l1.l2 = (kG ) 2
∴
Let
(kG )2 466.7
=
= 2.33 mm
200
l1
m1 = Mass placed at the small end
centre, and
l2 =
m2 = Mass placed at a distance l2 from
the centre of gravity G.
Fig. 15.17
546
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Theory of Machines
We know that
m1 =
2.33 × 15
l2 .m
=
= 0.17 kg Ans.
l1 + l2 200 + 2.33
200 × 15
l1 .m
=
= 14.83 kg Ans.
l1 + l2 200 + 2.33
Example 15.16. A connecting rod is suspended from a point 25 mm above the centre of
small end, and 650 mm above its centre of gravity, its mass being 37.5 kg. When permitted to oscillate, the time period is found to be 1.87 seconds. Find the dynamical equivalent system constituted of
two masses, one of which is located at the small end centre.
m2 =
and
Solution. Given : h = 650 mm = 0.65 m ; l1 = 650 – 25 = 625 mm
= 0.625 m ; m = 37.5 kg ; tp = 1.87 s
First of all, let us find the radius of gyration (k G) of the connecting rod (considering it is a compound pendulum), about an axis passing
through its centre of gravity, G.
We know that for a compound pendulum, time period of
oscillation (tp),
1.87 = 2 π
( kG ) 2 + h 2
g .h
or
1.87
=
2π
( kG ) 2 + (0.65) 2
9.81 × 0.65
Squaring both sides, we have
0.0885 =
( kG )2 + 0.4225
6.38
(kG ) 2 = 0.0885 × 6.38 − 0.4225 = 0.1425 m2
∴
k G = 0.377 m
It is given that one of the masses is located at the small end centre.
Let the other mass is located at a distance l2 from the centre of gravity G,
as shown in Fig. 15.19. We know that, for a dynamically equivalent system,
Fig. 15.18
l1.l2 = (k G)2
(kG )2 0.1425
=
= 0.228 m
0.625
l1
∴
l2 =
Let
m1 = Mass placed at the small end
centre A , and
m2 = Mass placed at a distance l2 from
G, i.e. at B.
We know that, for a dynamically equivalent system,
and
m1 =
0.228 × 37.5
l2 .m
=
= 10 kg Ans.
l1 + l2 0.625 + 0.228
m2 =
0.625 × 37.5
l1 .m
=
= 27.5 kg Ans.
l1 + l2 0.625 + 0.228
Fig. 15.19
Chapter 15 : Inertia Forces in Reciprocating Parts
l
547
Example 15.17. The following data relate to a connecting rod of a reciprocating engine:
Mass = 55 kg; Distance between bearing centres = 850 mm; Diameter of small end bearing
= 75 mm; Diameter of big end bearing = 100 mm; Time of oscillation when the connecting rod is
suspended from small end = 1.83 s; Time of oscillation when the connecting rod is suspended from
big end = 1.68 s.
Determine: 1. the radius of gyration of the rod about an axis passing through the centre of
gravity and perpendicular to the plane of oscillation; 2. the moment of inertia of the rod about the
same axis; and 3. the dynamically equivalent system for the connecting rod, constituted of two masses,
one of which is situated at the small end centre.
Solution. Given : m = 55 kg ; l = 850 mm = 0.85 m ; d1 = 75 mm = 0.075 m ;
d2 = 100 mm = 0.1 m ; tp1 = 1.83 s ; tp2 = 1.68 s
First of all, let us find the lengths of the equivalent simple
pendulum when suspended
(a) from the top of small end bearing; and
(b) from the top of big end bearing.
Let
L1 = Length of equivalent simple pendulum
when suspended from the top of small
end bearing,
L2 = Length of equivalent simple pendulum
when suspended from the top of big end
bearing,
h1 = Distance of centre of gravity, G, from the
top of small end bearing, and
h2 = Distance of centre of gravity, G, from the
top of big end bearing.
We know that for a simple pendulum
t p1 = 2 π
2
t p1
L1
or
=
2
g
π
L1
g
2
∴
...(Squaring both sides)
2
t
1.83
L1 = g p1 = 9.81
= 0.832 m
2π
2π
2
Similarly,
Fig. 15.20
2
t
1.68 = 0.7 m
L2 = g p 2 = 9.81
2π
2π
1. Radius of gyration of the rod about an axis passing through the centre of gravity and perpendicular to the plane of oscillation
Let
k G = Required radius of gyration of the rod.
We know that the length of equivalent simple pendulum,
( kG ) 2 + h 2
or (kG )2 = L.h − h 2 = h ( L − h)
h
∴ When the rod is suspended from the top of small end bearing,
L=
( kG )2
= h1 ( L1 − h1 )
...(i)
548
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Theory of Machines
and when the rod is suspended from the top of big end bearing,
( kG )2
= h2 ( L2 − h2 )
...(ii)
Also, from the geometry of the Fig. 15.20,
h1 + h2 =
0.075
0.1
d1
d
+l+ 2 =
+ 0.85 +
= 0.9375 m
2
2
2
2
h2 = 0.9375 − h1
∴
...(iii)
From equations (i) and (ii),
h1 ( L1 − h1 ) = h2 ( L2 − h2 )
Substituting the value of h2 from equation (iii),
h1 (0.832 − h1 ) = (0.9375 − h1 ) [0.7 − (0.9375 − h1 )]
0.832 h1 − (h1 )2 = − 0.223 + 1.175 h1 − (h1 )2
0.343 h1 = 0.233 or h1 = 0.223 / 0.343 = 0.65 m
Now from equation (i),
(kG )2 = 0.65 (0.832 − 0.65) = 0.1183 or kG = 0.343 m Ans.
2. Moment of inertia of the rod
We know that moment of inertia of the rod,
I = m (kG ) 2 = 55 × 0.1183 = 6.51 kg-m 2Ans.
3. Dynamically equivalent system for the rod
Since one of the masses (m 1) is situated at the centre of small end bearing, therefore its
distance from the centre of gravity, G, is
l1 = h1 – 0.075 / 2 = 0.65 – 0.0375 = 0.6125 m
Let
m2 = Magnitude of the second mass, and
l2 = Distance of the second mass from the centre of gravity, G,
towards big end bearing.
For a dynamically equivalent system,
l1.l2 = (kG ) 2 or l2 =
(kG )2 0.1183
=
= 0.193 m
0.6125
l1
0.193 × 55
l2 .m
=
= 13.18 kg Ans.
l1 + l2 0.6125 + 0.193
0.6125 × 55
l .m
m2 = 1
=
= 41.82 kg Ans.
l1 + l2 0.6125 + 0.193
We know that m1 =
and
15.13. Correction Couple to be Applied to Make Two Mass System
Dynamically Equivalent
In Art. 15.11, we have discussed the conditions for equivalent dynamical system of two
bodies. A little consideration will show that when two masses are placed arbitrarily*, then the condi*
When considering the inertia forces on the connecting rod in a mechanism, we replace the rod by two
masses arbitrarily. This is discussed in Art. 15.14.
Chapter 15 : Inertia Forces in Reciprocating Parts
l
549
tions (i) and (ii) as given in Art. 15.11 will only be satisfied. But the condition (iii) is not possible to
satisfy. This means that the mass moment of inertia of these two masses placed arbitrarily, will differ
than that of mass moment of inertia of the rigid body.
Fig. 15.21. Correction couple to be applied to make the two-mass system dynamically equivalent.
Consider two masses, one at A and the other at D be placed arbitrarily, as shown in Fig. 15.21.
Let
l3 = Distance of mass placed at D from G,
I1 = New mass moment of inertia of the two masses;
k1 = New radius of gyration;
α = Angular acceleration of the body;
I = Mass moment of inertia of a dynamically equivalent system;
kG = Radius of gyration of a dynamically equivalent system.
We know that the torque required to accelerate the body,
T = I.α = m (k G)2 α
...(i)
Similarly, the torque required to accelerate the two-mass system placed arbitrarily,
T1 = I1.α = m (k 1)2 α
...(ii)
∴ Difference between the torques required to accelerate the two-mass system and the torque
required to accelerate the rigid body,
T' = T 1–T = m (k 1)2 α – m (k G)2 α = m [(k 1)2 – (k G)2] α
...(iv)
The difference of the torques T' is known as correction couple. This couple must be applied,
when the masses are placed arbitrarily to make the system dynamical equivalent. This, of course, will
satisfy the condition (iii) of Art. 15.11.
Note:
We know that
∴ Correction couple,
But
∴
where
(k G)2 = l1.l2 ,
and (k 1)2 = l1.l3
T' = m (l1.l3 – l1.l2) α = m.ll (l3 – l2) α
l3 – l2 = l – L
T' = m.l1 (l – L) α
l = Distance between the two arbitrarily masses, and
L = Distance between the two masses for a true dynamically equivalent
system. It is the equivalent length of a simple pendulum when a body
is suspended from an axis which passes through the position of mass
m, and perpendicular to the plane of rotation of the two mass system.
=
( kG ) 2 + (l1) 2
l1
550
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Theory of Machines
Example 15.18. A connecting rod of an I.C. engine has a mass of 2 kg and the distance
between the centre of gudgeon pin and centre of crank pin is 250 mm. The C.G. falls at a point 100 m m
from the gudgeon pin along the line of centres. The radius of gyration about an axis through the C.G.
perpendicular to the plane of rotation is 110 mm. Find the equivalent dynamical system if only one of
the masses is located at gudgeon pin.
If the connecting rod is replaced by two masses, one at the gudgeon pin and the other at the
crank pin and the angular acceleration of the rod is 23 000 rad/s2 clockwise, determine the correction couple applied to the system to reduce it to a dynamically equivalent system.
Solution. Given : m = 2 kg ; l = 250 mm = 0.25 m ; l1 = 100 mm = 0.1m ; k G = 110 mm = 0.11 m ;
α = 23 000 rad/s2
Equivalent dynamical system
It is given that one of the masses is located at the gudgeon pin. Let the other mass be located
at a distance l2 from the centre of gravity. We know that for an equivalent dynamical system.
l1.l2 = (kG ) 2 or l2 =
Let
(kG ) 2 (0.11)2
=
= 0.121 m
0.1
l1
m1 = Mass placed at the gudgeon pin, and
m2 = Mass placed at a distance l2 from C.G.
We know that
and
0.121 × 2
l2 .m
=
= 1.1 kg Ans.
l1 + l 2 0.1 + 0.121
0.1 × 2
l .m
m2 = 1
=
= 0.9 kg Ans.
l1 + l2 0.1 + 0.121
m1 =
Correction couple
Since the connecting rod is replaced by two masses located at the two centres (i.e. one at the
gudgeon pin and the other at the crank pin), therefore,
l = 0.1 m, and l3 = l – l1 = 0.25 – 0.1 = 0.15 m
Let
We know that
k1 = New radius of gyration.
(k 1)2= l1.l3 = 0.1 × 0.15 = 0.015 m2
∴ Correction couple,
T ′ = m ( k12 − kG2 ) α = 2 0.015 − (0.11) 2 23 000 = 133.4 N-m Ans.
Note : Since T' is positive, therefore, the direction of correction couple is same as that of angular acceleration
i.e. clockwise.
15.14. Inertia Forces in a Reciprocating Engine, Considering the Weight of
Connecting Rod
In a reciprocating engine, let OC be the crank and PC, the connecting rod whose centre of
gravity lies at G. The inertia forces in a reciprocating engine may be obtained graphically as discussed
below:
1. First of all, draw the acceleration diagram OCQN by Klien’s construction. We know that
the acceleration of the piston P with respect to O,
aPO = aP = ω2 × NO,
Chapter 15 : Inertia Forces in Reciprocating Parts
l
551
acting in the direction from N to O. Therefore, the inertia force FI of the reciprocating parts will act in
the opposite direction as shown in Fig. 15.22.
Fig. 15.22. Inertia forces is reciprocating engine, considering the weight of connecting rod.
2. Replace the connecting rod by dynamically equivalent system of two masses as discussed
in Art. 15.12. Let one of the masses be arbitrarily placed at P. To obtain the position of the other mass,
draw GZ perpendicular to CP such that GZ = k, the radius of gyration of the connecting rod. Join PZ
and from Z draw perpendicular to DZ which intersects CP at D. Now, D is the position of the second
mass.
Note: The position of the second mass may also be obtained from the equation,
GP × GD = k2
3. Locate the points G and D on NC which is the acceleration image of the connecting rod.
This is done by drawing parallel lines from G and D to the line of stroke PO. Let these parallel lines
intersect NC at g and d respectively. Join gO and dO. Therefore, acceleration of G with respect to O,
in the direction from g to O,
aGO = aG = ω2 × gO
and acceleration of D with respect to O, in the direction from d to O,
aDO = aD = ω2 × dO
4. From D, draw DE parallel to dO which intersects the line of stroke PO at E. Since the
accelerating forces on the masses at P and D intersect at E, therefore their resultant must also pass
through E. But their resultant is equal to the accelerang force on the rod, so that the line of action of
the accelerating force on the rod, is given by a line drawn through E and parallel to gO, in the direction from g to O. The inertia force of the connecting rod FC therefore acts through E and in the
opposite direction as shown in Fig. 15.22. The inertia force of the connecting rod is given by
FC = m C × ω2 × gO
...(i)
where
mC = Mass of the connecting rod.
A little consideration will show that the forces acting on the connecting rod are :
(a) Inertia force of the reciprocating parts (FI ) acting along the line of stroke PO,
552
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Theory of Machines
(b) The side thrust between the
crosshead and the guide bars
(FN) acting at P and right angles
to line of stroke PO,
(c) The weight of the connecting rod
(W C = m C.g),
(d) Inertia force of the connecting
rod (FC),
(e) The radial force (FR) acting
through O and parallel to the
crank OC,
(f) The force (FT) acting perpendicular to the crank OC.
Now, produce the lines of action of
FR and FN to intersect at a point I, known as
instantaneous centre. From I draw I X and I
Y , perpendicular to the lines of action of FC
and W C. Taking moments about I, we have
Radial engines of a motor cycle.
FT × IC = FI × IP + FC × I X + W C × I Y
...(ii)
The value of FT may be obtained from this equation and from the force polygon as shown in
Fig. 15.22, the forces FN and FR may be calculated. We know that, torque exerted on the crankshaft to
overcome the inertia of the moving parts = FT × OC
Note : When the mass of the reciprocating parts is neglected, then FI is zero.
15.15. Analytical Method for Inertia Torque
The effect of the inertia of the connecting rod on the crankshaft torque may be obtained as
discussed in the following steps:
Fig. 15.23. Analytical method for inertia torque.
1. The mass of the connecting rod (m C) is divided into two masses. One of the mass is placed
at the crosshead pin P and the other at the crankpin C as shown in Fig. 15.23, so that the centre of
gravity of these two masses coincides with the centre of gravity of the rod G.
2. Since the inertia force due to the mass at C acts radially outwards along the crank OC,
therefore the mass at C has no effect on the crankshaft torque.
3. The inertia force of the mass at P may be obtained as follows:
Let
mC = Mass of the connecting rod,
l = Length of the connecting rod,
l1 = Length of the centre of gravity of the connecting rod from P.
Chapter 15 : Inertia Forces in Reciprocating Parts
l
553
∴ Mass of the connecting rod at P,
l − l1
× mC
l
The mass of the reciprocating parts (m R) is also acting at P. Therefore,
=
Total equivalent mass of the reciprocating parts acting at P
l − l1
× mC
l
∴ Total inertia force of the equivalent mass acting at P,
= mR +
l − l1
FI = mR +
× mC aR
l
aR = Acceleration of the reciprocating parts
where
...(i)
cos 2θ
= ω2 .r cos θ +
n
cos 2θ
l − l1
FI = mR +
× mC ω2 .r cos θ +
l
n
and corresponding torque exerted on the crank shaft,
∴
sin 2θ
TI = FI × OM = FI .r sin θ +
2
2
2 n − sin θ
...(ii)
Note : Usually the value of OM is measured by drawing the perpendicular from O on PO which intersects PC
produced at M.
4. In deriving the equation (ii) of the torque exerted on the crankshaft, it is assumed that one
of the two masses is placed at C and the other at P. This assumption does not satisfy the condition for
kinetically equivalent system of a rigid bar. Hence to compensate for it, a correcting torque is necessary whose value is given by
T ′ = mC ( k1 ) 2 − ( kG ) 2 α PC = mC .l1 (l − L) α PC
where
L = Equivalent length of a simple pendulum when swung about an
axis through P
(kG )2 + (l1 )2
l1
= Angular acceleration of the connecting rod PC.
=
αPC
− ω2 sin θ
...(From Art. 15.9)
n
The correcting torque T' may be applied to the system by two equal and opposite forces FY
acting through P and C. Therefore,
F × PN = T' or F = T '/PN
=
Y
Y
and corresponding torque on the crankshaft,
T′
× NO
PN
NO = OC cos θ = r cos θ
TC = FY × NO =
We know that,
and
PN = PC cos φ = l cos φ
...(iii)
554
l
∴
Theory of Machines
NO r cos θ
cos θ
=
=
PN
l cos φ n cos φ
=
cos θ
=
l
...Q n =
r
cos θ
2
... Q cos φ = 1 − sin θ
sin θ
n − sin 2 θ
n 1−
n2
n2
Since sin2θ is very small as compared to n2, therefore neglecting sin2θ, we have
2
2
NO cos θ
=
PN
n
Substituting this value in equation (iii), we have
TC = T ′ ×
cos θ
cos θ
= mC × l1 (l − L) α PC ×
n
n
= − mC × l1 (l − L)
= − mC × l1 (l − L)
ω2 sin θ cos θ
×
n
n
ω2 sin 2θ
2 n2
2
... Q αPC = −ω sin θ
n
... (Q 2 sin θ cos θ = sin 2θ )
5. The equivalent mass of the rod acting at C,
l1
l
∴ Torque exerted on the crank shaft due to mass m 2 ,
m2 = mC ×
TW = − m2 × g × NO = − mC × g ×
l1
l
× NO = − mC × g × 1 × r cos θ
l
l
...(Q NO = r cos θ)
l1
× cos θ
...(Q l / r = n)
n
6. The total torque exerted on the crankshaft due to the inertia of the moving parts is the
algebraic sum of T I , T C and T W.
= − mC × g ×
Example 15.19. The crank and connecting rod lengths of an engine are 125 mm and 500
mm respectively. The mass of the connecting rod is 60 kg and its centre of gravity is 275 mm from the
crosshead pin centre, the radius of gyration about centre of gravity being 150 mm.
If the engine speed is 600 r.p.m. for a crank position of 45° from the inner dead centre,
determine, using Klien’s or any other construction 1. the acceleration of the piston; 2. the magnitude, position and direction of inertia force due to the mass of the connecting rod.
Solution. Given : r = OC = 125 mm ; l = PC = 500 mm; m C = 60 kg ; PG = 275 mm ;
m C = 60 kg ; PG = 275 mm ; k G = 150 mm ; N = 600 r.p.m. or ω = 2π × 600/60 = 62.84 rad/s ; θ = 45°
1. Acceleration of the piston
Let
aP = Acceleration of the piston.
First of all, draw the configuration diagram OCP, as shown in Fig. 15.24, to some suitable
scale, such that
OC = r = 125 mm ; PC = l = 500 mm ; and θ = 45°.
Chapter 15 : Inertia Forces in Reciprocating Parts
l
555
Now, draw the Klien’s acceleration diagram OCQN, as shown in Fig. 15.24, in the same
manner as already discussed. By measurement,
NO = 90 mm = 0.09 m
∴ Acceleration of the piston,
aP = ω2 × NO = (62.84)2 × 0.09 = 355.4 m/s Ans.
Fig. 15.24
2. The magnitude, position and direction of inertia force due to the mass of the connecting rod
The magnitude, postition and direction of the inertia force may be obtained as follows:
(i) Replace the connecting rod by dynamical equivalent system of two masses, assuming that
one of the masses is placed at P and the other mass at D. The position of the point D is obtained as
discussed in Art. 15.12.
(ii) Locate the points G and D on NC which is the acceleration image of the connecting rod.
Let these points are g and d on NC. Join gO and dO. By measurement,
gO = 103 mm = 0.103 m
∴ Acceleration of G, aG = ω2 × gO, acting in the direction from g to O.
(iii) From point D, draw DE parallel to dO. Now E is the point through which the inertia force
of the connecting rod passes. The magnitude of the inertia force of the connecting rod is given by
FC = m C × ω2 × gO = 60 × (62.84)2 × 0.103 = 24 400 N = 24.4 kN Ans.
(iv) From point E, draw a line parallel to gO, which shows the position of the inertia force of
the connecting rod and acts in the opposite direction of gO.
Example 15.20. The following data refer to a steam engine:
Diameter of piston = 240 mm; stroke = 600 mm ; length of connecting rod = 1.5 m ; mass of
reciprocating parts = 300 kg; mass of connecting rod = 250 kg; speed = 125 r.p.m ; centre of gravity
of connecting rod from crank pin = 500 mm ; radius of gyration of the connecting rod about an axis
through the centre of gravity = 650 mm.
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Theory of Machines
Determine the magnitude and direction of the torque exerted on the crankshaft when the
crank has turned through 30° from inner dead centre.
Solution. Given : D = 240 mm = 0.24 m ; L = 600 mm or r = L/2 = 300 mm = 0.3 m ;
l = 1.5 m ; m R = 300 kg ; mC = 250 kg ; N = 125 r.p.m. or ω = 2π × 125/60 = 13.1 rad/s ;
GC = 500 mm = 0.5 m ; kG = 650 mm = 0.65 m ; θ = 30°
The inertia torque on the crankshaft may be determined by graphical method or analytical
method as discussed below:
1. Graphical method
First of all, draw the configuration diagram OCP, as shown in Fig. 15.25, to some suitable
scale, such that
OC = r = 300 mm ; PC = l = 1.5 m ; and angle POC = θ = 30°.
Fig. 15.25
Now draw the Klien’s acceleration diagram OCQN, as shown in Fig. 15.25, and complete the
figure in the similar manner as discussed in Art. 15.14.
By measurement; NO = 0.28 m ; gO = 0.28 m ; IP = 1.03 m ; I X = 0.38 m ; I Y = 0.98 m,
and IC = 1.7 m.
We know that inertia force of reciprocating parts,
FI = mR × ω2 × NO = 300 × (13.1) 2 × 0.28 = 14 415 N
and inertia force of connecting rod,
FC = mC × ω2 × gO = 250 × (13.1) 2 × 0.28 = 12 013 N
Let
FT = Force acting perpendicular to the crank OC.
Taking moments about point I,
FT × IC = FI × IP + WC × IY + FC × IX
Chapter 15 : Inertia Forces in Reciprocating Parts
l
557
FT × 1.7 = 14 415 × 1.03 + 250 × 9.81 × 0.98 + 12013 × 0.38 = 21816
∴
FT = 2.816 /1.7 = 12 833 N
...(Q WC = mC. g )
We know that torque exerted on the crankshaft
= FT × r = 12 833 × 0.3 = 3850 N-m Ans.
2. Analytical method
We know that the distance of centre of gravity (G) of the connecting rod from P, i.e.,
l1 = l – GC = 1.5 – 0.5 = 1 m
∴ Inertia force due to total mass of the reciprocating parts at P,
cos 2θ
l − l1
FI = mR +
× mC ω2 .r cos θ +
5
l
1.5 − 1
cos 60°
= 300 +
× 250 × (13.1)2 × .0.3 cos 30° +
= 19 064 N
1.5
5
∴ Corresponding torque due to FI ,
l 1.5
... Q n = =
= 5
r 0.3
sin 2θ
TI = FI × OM = FI .r sin θ +
2
2
2 n − sin θ
sin 60°
= 19 064 × 0.3 sin 30° +
2
2
2 5 − sin 30°
= 5719.2 × 0.587 = 3357 N-m (anticlockwise)
Equivalent length of a simple pendulum when swung about an axis through P,
L=
∴ Correcting torque,
(kG )2 + (l1 )2 (0.65) 2 + 12
=
= 1.42 m
1
l1
ω2 sin 2θ
TC = mC .l1 (l − L)
2
2n
(13.1) 2 sin 60°
= 250 × 1 (1.5 − 1.42)
= 59.5 N-m (anticlockwise)
2 × 52
Torque due to the weight of the connecting rod at C,
l1
l
× cos θ = mC × g × 1 × cos θ
n
n
1
= 250 × 9.81 × × cos 30° = 424.8 N-m (anticlockwise)
5
TW = WC ×
∴ Total torque exerted on the crankshaft,
= TI + TC + TW
= 3357 + 59.5 + 424.8 = 3841.3 N-m (anticlockwise) Ans.
Note: The slight difference in results arrived at by the above two methods is mainly due to error in measurement
in graphical method.
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Theory of Machines
Example 15.21. A vertical engine running at 1200 r.p.m. with a stroke of 110 mm, has a
connecting rod 250 mm between centres and mass 1.25 kg. The mass centre of the connecting rod is
75 mm from the big end centre and when suspended as a pendulum from the gudgeon pin axis makes
21 complete oscillations in 20 seconds.
1. Calculate the radius of gyration of the connecting rod about an axis through its mass
centre.
2. When the crank is at 40° from the top dead centre and the piston is moving downwards,
find analytically, the acceleration of the piston and the angular acceleration of the connecting rod.
Hence find the inertia torque exerted on the crankshaft. To make the two-mass system to be dynamically equivalent to the connecting rod, necessary correction torque has to be applied and since the
engine is vertical, gravity effects are to be considered.
Solution. Given : N = 1200 r.p.m. or ω = 2π × 1200/60 = 125.7 rad/s ;
L = 110 mm or r = L/2 = 55 mm = 0.055 m ; l = PC = 250 mm = 0.25 m ;
m C = 1.25 kg ; CG = 75 mm = 0.075 m ; θ = 40°
The configuration diagram of the engine is shown in Fig. 15.26.
1. Radius of gyration of the connecting rod about an axis through its mass
centre
Let
kG = Radius of gyration of the connecting rod about
an axis through its mass centre,
l1 = Distance of the centre of gravity from the point
of suspension = PG
= 250 – 75 = 175 mm = 0.175 m
Since the connecting rod makes 21 complete oscillations in 20 seconds,
therefore frequency of oscillation,
21
n=
= 1.05 H Z
20
We know that for a compound pendulum, frequency of oscillation,
Fig. 15.26
g.l1
g.l1
1
1
2
or
n=
n
=
×
2π ( kG ) 2 + (l1 )2
4π2 ( kG ) 2 + (l1 ) 2
...(Squaring both sides)
g .l1
( kG ) 2 =
and
∴
− (l1 ) 2 =
9.81 × 0.175
4π n
4π 2 × (1.05) 2
kG = 0.094 m = 94 mm Ans.
2 2
− (0.175)2 = 0.0088 m2
2. Acceleration of the piston
We know that acceleration of the piston,
cos 80°
cos 2θ
2
aP = ω2 .r cos θ +
= (125.7) 0.055 cos 40° + 0.25 / 0.055
n
= 698.7 m/s2 Ans.
Angular acceleration of the connecting rod
We know that mass of the connecting rod at P,
α PC =
− ω2 sin θ − (125.7)2 sin 40°
=
= − 2234.4 rad/s 2 Ans.
0.25 / 0.055
n
...(Q n = l / r )
Chapter 15 : Inertia Forces in Reciprocating Parts
l
559
Inertia torque exerted on the crankshaft
We know that mass of the connecting rod at P,
l − l1
0.25 − 0.175
× mC =
× 1.25 = 0.375 kg
0.25
l
∴ Vertical inertia force,
m1 =
FI = m1.aP = 0.375 × 698.7 = 262 N
and corresponding torque due to FI,
TI = − FI × OM = − 262 × 0.0425 = − 11.135 N-m
= 11.135 N-m (anticlockwise)
...(By measurement, OM = 0.0425 m)
We know that the equivalent length of a simple pendulum when swung about an axis passing
through P,
L=
(kG ) 2 + (l1 ) 2 (0.094)2 + (0.175) 2
=
= 0.225 m
0.175
l1
∴ Correction couple,
T ′ = − mC .l1 (l − L) α PC = − 1.25 × 0.175 (0.25 − 0.225) 2234.4 = − 12.22 N-m
Corresponding torque on the crankshaft,
T ′ cos θ −12.22 × cos 40°
=
= − 2.06 N-m = 2.06 N-m (anticlockwise)
0.25 / 0.055
n
Torque due to the mass at P,
TC =
TP = m1 × g × OM = 0.375 × 9.81 × 0.0425 = 0.156 N-m (clockwise)
Equivalent mass of the connecting rod at C,
l1
0.175
= 1.25 ×
= 0.875 kg
0.25
l
Torque due to mass at C,
m2 = mC ×
TW = m2 × g × NC = 0.875 × 9.81 × 0.035 = 0.3 N-m (clockwise)
...(By measurement, NC = 0.035 m)
∴ Inertia torque exerted on the crankshaft
= TI + TC − TP − TW
= 11.135 + 2.06 − 0.156 − 0.3 = 12.739 N-m (anticlockwise) Ans.
Example 15.22. The connecting rod of an internal combustion engine is 225 mm long and
has a mass 1.6 kg. The mass of the piston and gudgeon pin is 2.4 kg and the stroke is 150 mm. The
cylinder bore is 112.5 mm. The centre of gravity of the connection rod is 150 mm from the small end.
Its radius of gyration about the centre of gravity for oscillations in the plane of swing of the connecting rod is 87.5 mm. Determine the magnitude and direction of the resultant force on the crank pin
when the crank is at 40° and the piston is moving away from inner dead centre under an effective gas
presure of 1.8 MN/m2. The engine speed is 1200 r.p.m.
Solution. Given : l = PC = 225 mm = 0.225 m; m C = 1.6 kg; m R = 2.4 kg; L = 150 mm or
r = L/2 = 75 mm = 0.075 m ; D = 112.5 mm = 0.1125 m ; PG = 150 mm ; k G = 87.5 mm = 0.0875 m ;
θ = 40° ; p = 1.8 MN/m2 = 1.8 × 106 N/m2 ; N = 1200 r.p.m. or ω = 2π × 1200/60 = 125.7 rad/s
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Theory of Machines
First of all, draw the configuration diagram OCP, as shown in Fig. 15.27 to some suitable
scale, such that OC = r = 75 mm ; PC = l = 225 mm ; and θ = 40°.
Fig. 15.27
Now, draw the Klien’s acceleration diagram OCQN. Complete the diagram in the same manner
as discussed earlier. By measurement,
NO = 0.0625 m ; gO = 0.0685 m ; IC = 0.29 m ; IP = 0.24 m ; I Y = 0.148 m ; and I X = 0.08 m
We know that force due to gas pressure,
π
π
× D 2 × p = × (0.1125) 2 × 1.8 × 10 6 = 17 895 N
4
4
Inertia force due to mass of the reciprocating parts,
FL =
FI = mR × ω2 × NO = 2.4 (125.7)2 × 0.0625 = 2370 N
∴ Net force on the piston,
FP = FL − FI = 17 895 − 2370 = 15 525 N
Inertia force due to mass of the connecting rod,
FC = mC × ω2 × gO = 1.6 × (125.7)2 × 0.0685 = 1732 N
Let
FT = Force acting perpendicular to the crank OC.
Now, taking moments about point I,
FP × IP = WC × IY + FC × IX + FT × IC
15 525 × 0.24 = 1.6 × 9.81 × 0.148 + 1732 × 0.08 + FT × 0.29
∴
FT = 12 362 N
...(Q WC = mC. g )
Let us now find the values of FN and FR in magnitude and direction. Draw the force polygon
as shown in Fig. 15.25.
Chapter 15 : Inertia Forces in Reciprocating Parts
l
561
By measurement, FN = 3550 N; and FR = 7550 N
The magnitude and direction of the resultant force on the crank pin is given by FQ , which is
the resultant of FR and FT.
By measurement, FQ = 13 750 N Ans.
EXERCISES
1.
The crank and connecting rod of a reciprocating engine are 150 mm and 600 mm respectively. The
crank makes an angle of 60° with the inner dead centre and revolves at a uniform speed of 300 r.p.m.
Find, by Klein’s or Ritterhaus’s construction, 1. Velocity and acceleration of the piston, 2. Velocity
and acceleration of the mid-point D of the connecting rod, and 3. Angular velocity and angular acceleration of the connecting rod.[Ans. 4.6 m/s, 61.7 m/s2 ; 4.6 m/s, 93.8 m/s2 ; 4.17 rad/s, 214 rad/s2]
2.
In a slider crank mechanism, the length of the crank and connecting rod are 100 mm and 400 mm
respectively. The crank rotates uniformly at 600 r.p.m. clockwise. When the crank has turned through
45° from the inner dead centre, find, by analytical method : 1. Velocity and acceleration of the slider,
2. Angular velocity and angular acceleration of the connecting rod. Check your result by Klein’s or
Bennett’s construction.
[Ans. 5.2 m/s; 279 m/s2; 11 rad/s; 698 rad/s2]
3.
A petrol engine has a stroke of 120 mm and connecting rod is 3 times the crank length. The crank
rotates at 1500 r.p.m. in clockwise direction. Determine: 1. Velocity and acceleration of the piston,
and 2. Angular velocity and angular acceleration of the connecting rod, when the piston had travelled
one-fourth of its stroke from I.D.C.
[Ans. 8.24 m/s, 1047 m/s2; 37 rad/s, 5816 rad/s2]
4.
The stroke of a steam engine is 600 mm and the length of connecting rod is 1.5 m. The crank rotates
at 180 r.p.m. Determine: 1. velocity and acceleration of the piston when crank has travelled through
an angle of 40° from inner dead centre, and 2. the position of the crank for zero acceleration of the
piston.
[Ans. 4.2 m/s, 85.4 m/s2; 79.3° from I.D.C]
5.
The following data refer to a steam engine :
Diameter of piston = 240 mm; stroke = 600 mm; length of connecting rod = 1.5 m; mass of reciprocating parts = 300 kg; speed = 125 r.p.m.
Determine the magnitude and direction of the inertia force on the crankshaft when the crank has
turned through 30° from inner dead centre.
[Ans. 14.92 kN]
6.
A vertical petrol engine 150 mm diameter and 200 mm stroke has a connecting rod 350 mm long. The
mass of the piston is 1.6 kg and the engine speed is 1800 r.p.m. On the expansion stroke with crank
angle 30° from top dead centre, the gas pressure is 750 kN/m2. Determine the net thrust on the piston.
[Ans. 7535 N]
7.
A horizontal steam engine running at 240 r.p.m. has a bore of 300 mm and stroke 600 mm. The
connecting rod is 1.05 m long and the mass of reciprocating parts is 60 kg. When the crank is 60° past
its inner dead centre, the steam pressure on the cover side of the piston is 1.125 N/mm2 while that on
the crank side is 0.125 N/mm2. Neglecting the area of the piston rod, determine : 1. the force in the
piston rod ; and 2. the turning moment on the crankshaft.
[Ans. 66.6 kN ; 19.86 kN-m]
8.
A steam engine 200 mm bore and 300 mm stroke has a connecting rod 625 mm long. The mass of the
reciprocating parts is 15 kg and the speed is 250 r.p.m.When the crank is at 30° to the inner dead centre
and moving outwards, the difference in steam pressures is 840 kN/m2. If the crank pin radius is 30 mm,
determine: 1. the force on the crankshaft bearing; and 2. the torque acting on the frame.
[Ans. 20.04 kN ; 2253 N-m]
9.
A vertical single cylinder engine has a cylinder diameter of 250 mm and a stroke of 450 mm. The
reciprocating parts have a mass of 180 kg. The connecting rod is 4 times the crank radius and the
speed is 360 r.p.m. When the crank has turned through an angle of 45° from top dead centre, the net
pressure on the piston is 1.05 MN/m2. Calculate the effective turning moment on the crankshaft for
this position.
[Ans. 2368 N-m]
562
10.
11.
12.
13.
14.
15.
16.
l
Theory of Machines
A horizontal, double acting steam engine has a stroke of 300 mm and runs at 240 r.p.m. The cylinder
diameter is 200 mm, connecting rod is 750 mm long and the mass of the reciprocating parts is 70 kg.
The steam is admitted at 600 kN/m2 for one-third of the stroke, after which expansion takes place
according to the hyperbolic law p.V = constant. The exhaust pressure is 20 kN/m2. Neglecting the
effect of clearance and the diameter of the piston rod, find : 1. Thrust in the connecting rod, and 2.
Effective turning moment on the crankshaft when the crank has turned through 120° from inner dead
centre.
[Ans. 11.506 kN; 1322 N-m]
A horizontal steam engine running at 150 r.p.m. has a bore of 200 mm and a stroke of 400 mm. The
connecting rod is 1 m long and the reciprocating parts has a mass of 60 kg. When the
crank has turned through an angle of 30° from inner dead centre, steam pressure on the cover side is
0.6 N/mm2 while on the crankside is 0.1 N/mm2. Neglecting the area of the piston rod, determine:
1. turning moment on the crankshaft, 2. acceleration of the flywheel, if the mean resistance torque is
600 N-m and the moment of inertia is 2.8 kg-m2.
[Ans. 1508 N-m; 324.3 rad/s2]
The ratio of the connecting rod length to crank length for a vertical petrol engine is 4:1. The bore /
stroke is 80/100 mm and mass of the reciprocating parts is 1 kg. The gas pressure on the piston is
0.7N/mm2 when it has moved 10 mm from T.D.C. on its power stroke. Determine the net load on the
gudgeon pin. The engine runs at 1800 r.p.m. At what engine speed will this load be zero?
[Ans. 1862.8 N; 2616 r.p.m.]
A petrol engine 90 mm in diameter and 120 mm stroke has a connecting rod of 240 mm length. The
piston has a mass of 1 kg and the speed is 1800 r.p.m. On the explosion stroke with the crank at 30°
from top dead centre, the gas pressure is 0.5 N/mm2. Find :
1. the resultant load on the gudgeon pin, 2. the thrust on the cylinder walls, and 3. the speed, above
which other things remaining same, the gudgeon pin load would be reserved in direction.
Also calculate the crank effort at the given position of the crank.
[Ans. 1078 N; 136 N ; 2212 r.p.m.; 39.4 N-m]
A single cylinder vertical engine has a bore of 300 mm, storke 360 mm and a connecting rod of length
720 mm. The mass of the reciprocating parts is 130 kg. When the piston is at quarter stroke from top
dead centre and is moving downwards, the net pressure on it is 0.6 MPa. If the speed of the engine is
250 r.p.m., calculate the turning moment on the crankshaft at the instant corresponding to the position
stated above.
[Ans. 6295 N-m]
A horizontal, single cylinder, single acting, otto cycle gas engine has a bore of 300 mm and a stroke of
500 mm. The engine runs at 180 r.p.m. The ratio of compression is 5.5. The maximum explosion
pressure is 3.2 N/mm2 gauge and expansion follows the law p.V1.3 = constant. If the mass of the piston
is 150 kg and the connecting rod is 1.25 m long. Calculate the turning moment on the crankshaft when
the crank has turned through 60° from the inner dead centre. The atmospheric pressure is 0.1 N/mm2.
[Ans. 15.6 kN-m]
A vertical single cylinder, diesel engine running at 300 r.p.m. has a cylinder diameter 250 mm
and stroke 400 mm. The mass of the reciprocating parts is 200 kg. The length of the connecting
rod is 0.8 m. The ratio of compression is 14 and the pressure remains constant during injection of
oil for 1/10th of stroke. If the index of the law of expansion and compression is 1.35, find the torque
on the crankshaft when it makes an angle of 60° with the top dead centre during the expansion stroke.
The suction pressure may be taken as 0.1 N/mm2.
[Ans. 7034 N-m]
17.
A gas engine is coupled to a compressor, the two cylinders being horizontally opposed with the
pistons connected to a common crank pin. The stroke of each piston is 500 mm and the ratio of the
length of the connecting rod to the length of crank is 5. The cylinder diameters are 200 mm and 250 mm
and the masses of reciprocating parts are 130 kg and 150 kg respectively. When the crank has moved
through 60° from inner dead centre on the firing stroke, the pressure of gas on the engine cylinder is
1 N/mm2 gauge and the pressure in the compressor cylinder is 0.1 N/mm2 gauge. If the crank moves
with 200 r.p.m. and the flywheel of radius of gyration 1 m has a mass of 1350 kg, determine the
angular acceleration of the flywheel.
[Ans. 2.4 rad/s2]
18.
The length of a connecting rod of an engine is 500 mm measured between the centres and its mass is 18
kg. The centre of gravity is 125 mm from the crank pin centre and the crank radius is 100 mm.
Chapter 15 : Inertia Forces in Reciprocating Parts
l
563
Determine the dynamically equivalent system keeping one mass at the small end. The frequency of
oscillation of the rod, when suspended from the centre of the small end is 43 vibrations per minute.
[Ans. 4.14 kg; 13.86 kg]
19.
A small connecting rod 220 mm long between centres has a mass of 2 kg and a moment of inertia of
0.02 kg-m2 about its centre of gravity. The centre of gravity is located at a distance of 150 mm from
the small end centre. Determine the dynamically equivalent two mass system when one mass is located at the small end centre.
If the connecting rod is replaced by two masses located at the two centres, find the correction couple
that must be applied for complete dynamical equivalence of the system when the angular acceleration
of the connecting rod is 20 000 rad/s2 anticlockwise.
[Ans. 0.617 kg; 1.383 kg; 20 N-m (anticlockwise)]
20.
The connecting rod of a horizontal reciprocating engine is 400 mm and length of the stroke is 200
mm. The mass of the reciprocating parts is 125 kg and that the connecting rod is 100 kg. The radius
of gyration of the connecting rod about an axis through the centre of gravity is 120 mm and the
distance of centre of gravity of the connecting rod from big end centre is 160 mm. The engine runs at
750 r.p.m. Determine the torque exerted on the crankshaft when the crank has turned 30° from the
inner dead centre.
[Ans. 7078 N-m]
21.
If the crank has turned through 135° from the inner dead centre in the above question, find the torque
on the crankshaft.
[Ans. 5235 N-m]
DO YOU KNOW ?
1.
2.
3.
4.
5.
6.
7.
8.
Define ‘inertia force’ and ‘inertia torque’.
Draw and explain Klien’s construction for determining the velocity and acceleration of the piston in a
slider crank mechanism.
Explain Ritterhaus’s and Bennett’s constructions for determining the acceleration of the piston of a
reciprocating engine.
How are velocity and acceleration of the slider of a single slider crank chain determined analytically?
Derive an expression for the inertia force due to reciprocating mass in reciprocating engine, neglecting the mass of the connecting rod.
What is the difference between piston effort, crank effort and crank-pin effort?
Discuss the method of finding the crank effort in a reciprocating single acting, single cylinder petrol
engine.
The inertia of the connecting rod can be replaced by two masses concentrated at two points and
connected rigidly together. How to determine the two masses so that it is dynamically equivalent to
the connecting rod ? Show this.
9.
Given acceleration image of a link. Explain how dynamical equivalent system can be used to
determine the direction of inertia force on it.
10.
Describe the graphical and analytical method of finding the inertia torque on the crankshaft of a
horizontal reciprocating engine.
11.
Derive an expression for the correction torque to be applied to a crankshaft if the connecting rod of
a reciprocating engine is replaced by two lumped masses at the piston pin and the crank pin
respectively.
OBJECTIVE TYPE QUESTIONS
1.
When the crank is at the inner dead centre, in a horizontal reciprocating steam engine, then the velocity
of the piston will be
(a) zero
(b) minimum
(c) maximum
564
2.
l
Theory of Machines
The acceleration of the piston in a reciprocating steam engine is given by
(a)
sin 2θ
ω.r sin θ +
n
(b)
cos 2θ
ω.r cos θ +
n
(c)
sin 2θ
ω2 .r sin θ +
n
(d)
cos 2θ
ω2 .r cos θ +
n
where
ω = Angular velocity of the crank,
r
= Radius of the crank,
θ = Angle turned by the crank from inner dead centre, and
n = Ratio of length of connecting rod to crank radius.
3.
A rigid body, under the action of external forces, can be replaced by two masses placed at a fixed
distance apart. The two masses form an equivalent dynamical system, if
(a) the sum of two masses is equal to the total mass of the body
(b) the centre of gravity of the two masses coincides with that of the body
(c) the sum of mass moment of inertia of the masses about their centre of gravity is equal to the mass
moment of inertia of the body
(d) all of the above
4.
The essential condition of placing the two masses, so that the system becomes dynamically equivalent
is
(a) l1 .l2 = k G2
where
(b) l1 .l2 = k G
(c)
l1 = k G
(d)
l2 = k G
l1 and l2 = Distance of two masses from the centre of gravity of the body, and
k G = Radius of gyration of the body.
5.
In an engine, the work done by inertia forces in a cycle is
(a) positive
(b) zero
(c)
negative
(d) none of these
4. (a)
5. (a)
ANSWERS
1. (a)
2. (d)
3. (d)
GO To FIRST
CONTENTS
CONTENTS
Chapter 16 : Turning Moment Diagrams and Flywheel
l
565
16
Fea
tur
es
eatur
tures
1. Introduction.
2. Turning Moment Diagram
for a Single Cylinder
Double Acting Steam
Engine.
3. Turning Moment Diagram
for a Four Stroke Cycle
Internal Combustion
Engine.
4. Turning Moment Diagram
for a Multicylinder Engine.
5. Fluctuation of Energy.
6. Determination of Maximum
Fluctuation of Energy.
7. Coefficient of Fluctuation
of Energy.
8. Flywheel.
9. Coefficient of Fluctuation of
Speed.
10. Energy Stored in a
Flywheel.
11. Dimensions of the Flywheel
Rim.
12. Flywheel in Punching Press.
Turning
Moment
Diagrams
and Flywheel
16.1. Intr
oduction
Introduction
The turning moment diagram (also known as crankeffort diagram) is the graphical representation of the turning
moment or crank-effort for various positions of the crank. It is
plotted on cartesian co-ordinates, in which the turning moment
is taken as the ordinate and crank angle as abscissa.
16.2. Tur
ning Moment Diagram ffor
or a Single
urning
Cylinder Double Acting Steam Engine
A turning moment diagram for a single cylinder
double acting steam engine is shown in Fig. 16.1. The vertical
ordinate represents the turning moment and the horizontal
ordinate represents the crank angle.
We have discussed in Chapter 15 (Art. 15.10.) that
the turning moment on the crankshaft,
sin 2 θ
T = FP × r sin θ +
2
2
2 n − sin θ
565
CONTENTS
CONTENTS
566
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Theory of Machines
Fig. 16.1. Turning moment diagram for a single cylinder, double acting steam engine.
where
FP = Piston effort,
r = Radius of crank,
n = Ratio of the connecting rod length and radius of crank, and
θ = Angle turned by the crank from inner dead centre.
From the above expression, we see
that the turning moment (T ) is zero, when the
crank angle (θ) is zero. It is maximum when
the crank angle is 90° and it is again zero when
crank angle is 180°.
This is shown by the curve abc in
Fig. 16.1 and it represents the turning
moment diagram for outstroke. The curve
cde is the turning moment diagram for
instroke and is somewhat similar to the
curve abc.
Since the work done is the product
of the turning moment and the angle turned,
therefore the area of the turning moment
diagram represents the work done per
revolution. In actual practice, the engine is
assumed to work against the mean resisting
torque, as shown by a horizontal line AF.
The height of the ordinate a A represents the
mean height of the turning moment diagram.
Since it is assumed that the work done by
the turning moment per revolution is equal
to the work done against the mean resisting
torque, therefore the area of the rectangle
aAFe is proportional to the work done against
the mean resisting torque.
For flywheel, have a look at your tailor’s manual
sewing machine.
Notes: 1. When the turning moment is positive (i.e. when the engine torque is more than the mean resisting
torque) as shown between points B and C (or D and E) in Fig. 16.1, the crankshaft accelerates and the work
is done by the steam.
Chapter 16 : Turning Moment Diagrams and Flywheel
l
567
2. When the turning moment is negative (i.e. when the engine torque is less than the mean resisting
torque) as shown between points C and D in Fig. 16.1, the crankshaft retards and the work is done on the
steam.
3. If
T
= Torque on the crankshaft at any instant, and
T mean = Mean resisting torque.
Then accelerating torque on the rotating parts of the engine
= T – T mean
4. If (T –T mean) is positive, the flywheel accelerates and if (T – T mean) is negative, then the flywheel retards.
16.3. Turning Moment Diagram for a Four Stroke Cycle Internal
Combustion Engine
A turning moment diagram for a four stroke cycle internal combustion engine is shown in
Fig. 16.2. We know that in a four stroke cycle internal combustion engine, there is one working
stroke after the crank has turned through two revolutions, i.e. 720° (or 4 π radians).
Fig. 16.2. Turning moment diagram for a four stroke cycle internal combustion engine.
Since the pressure inside the engine cylinder is less than the atmospheric pressure during
the suction stroke, therefore a negative loop is formed as shown in Fig. 16.2. During the compression
stroke, the work is done on the gases, therefore a higher negative loop is obtained. During the
expansion or working stroke, the fuel burns and the gases expand, therefore a large positive loop is
obtained. In this stroke, the work is done by the gases. During exhaust stroke, the work is done on
the gases, therefore a negative loop is formed. It may be noted that the effect of the inertia forces on
the piston is taken into account in Fig. 16.2.
16.4. Turning Moment Diagram for a Multi-cylinder Engine
A separate turning moment diagram for a compound steam engine having three cylinders
and the resultant turning moment diagram is shown in Fig. 16.3. The resultant turning moment
diagram is the sum of the turning moment diagrams for the three cylinders. It may be noted that the
first cylinder is the high pressure cylinder, second cylinder is the intermediate cylinder and the third
cylinder is the low pressure cylinder. The cranks, in case of three cylinders, are usually placed at
120° to each other.
568
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Theory of Machines
Fig. 16.3. Turning moment diagram for a multi-cylinder engine.
16.5. Fluctuation of Energy
The fluctuation of energy may be determined by the turning moment diagram for one complete
cycle of operation. Consider the turning moment diagram for a single cylinder double acting steam
engine as shown in Fig. 16.1. We see that the mean resisting torque line AF cuts the turning moment
diagram at points B, C, D and E. When the crank moves from a to p, the work done by the engine is
equal to the area aBp, whereas the energy required is represented by the area aABp. In other words,
the engine has done less work (equal to the area a AB) than the requirement. This amount of energy
is taken from the flywheel and hence the speed of the flywheel decreases. Now the crank moves
from p to q, the work done by the engine is equal to the area pBbCq, whereas the requirement of
energy is represented by the area pBCq. Therefore, the engine has done more work than the
requirement. This excess work (equal to the area BbC) is stored in the flywheel and hence the speed
of the flywheel increases while the crank moves from p to q.
Similarly, when the crank moves from q to r, more work is taken from the engine than is
developed. This loss of work is represented by the area C c D. To supply this loss, the flywheel gives
up some of its energy and thus the speed decreases while the crank moves from q to r. As the crank
moves from r to s, excess energy is again developed given by the area D d E and the speed again
increases. As the piston moves from s to e, again there is a loss of work and the speed decreases. The
variations of energy above and below the mean resisting torque line are called fluctuations of
energy. The areas BbC, CcD, DdE, etc. represent fluctuations of energy.
A little consideration will show that the engine has a maximum speed either at q or at s. This
is due to the fact that the flywheel absorbs energy while the crank moves from p to q and from r to s.
On the other hand, the engine has a minimum speed either at p or at r. The reason is that the flywheel
gives out some of its energy when the crank moves from a to p and q to r. The difference between the
maximum and the minimum energies is known as maximum fluctuation of energy.
16.6. Determination of Maximum Fluctuation of Energy
A turning moment diagram for a multi-cylinder engine is shown by a wavy curve in Fig.
16.4. The horizontal line A G represents the mean torque line. Let a1, a3, a5 be the areas above the
mean torque line and a2, a4 and a6 be the areas below the mean torque line. These areas represent
some quantity of energy which is either added or subtracted from the energy of the moving parts of
the engine.
Chapter 16 : Turning Moment Diagrams and Flywheel
l
569
Let the energy in the flywheel at A = E,
then from Fig. 16.4, we have
Energy at B = E + a1
Energy at C = E + a1– a2
Energy at D = E + a1 – a2 + a3
Energy at E = E + a1 – a2 + a3 – a4
Energy at F = E + a1 – a2 + a3 – a4 + a5
Energy at G = E + a1 – a2 + a3 – a4 + a5 – a6
= Energy at A (i.e. cycle
repeats after G)
Let us now suppose that the greatest of
these energies is at B and least at E. Therefore,
Maximum energy in flywheel
= E + a1
A flywheel stores energy when the supply
is in excess and releases energy when
energy is in deficit.
Minimum energy in the flywheel
= E + a1 – a2 + a3 – a4
∴ Maximum fluctuation of energy,
∆ E = Maximum energy – Minimum energy
= (E + a1) – (E + a1 – a2 + a3 – a4) = a2 – a3 + a4
Fig. 16.4. Determination of maximum fluctuation of energy.
16.7. Coefficient of Fluctuation of Energy
It may be defined as the ratio of the maximum fluctuation of energy to the work done
per cycle. Mathematically, coefficient of fluctuation of energy,
Maximum fluctuation of energy
Work done per cycle
The work done per cycle (in N-m or joules) may be obtained by using the following two
relations :
1. Work done per cycle = T mean × θ
where
T mean = Mean torque, and
θ = Angle turned (in radians), in one revolution.
= 2π, in case of steam engine and two stroke internal combustion
engines
= 4π, in case of four stroke internal combustion engines.
CE =
570
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Theory of Machines
The mean torque (T mean) in N-m may be obtained by using the following relation :
Tmean =
where
P × 60 P
=
ω
2π N
P = Power transmitted in watts,
N = Speed in r.p.m., and
ω = Angular speed in rad/s = 2 πN/60
2. The work done per cycle may also be obtained by using the following relation :
P × 60
n
n = Number of working strokes per minute,
Work done per cycle =
where
= N, in case of steam engines and two stroke internal combustion
engines,
= N /2, in case of four stroke internal combustion engines.
The following table shows the values of coefficient of fluctuation of energy for steam engines
and internal combustion engines.
Table 16.1. Coefficient of fluctuation of energy (CE) for steam and internal
combustion engines.
S.No.
Type of engine
Coefficient of fluctuation
of energy (CE)
1.
Single cylinder, double acting steam engine
0.21
2.
Cross-compound steam engine
0.096
3.
Single cylinder, single acting, four stroke gas engine
1.93
4.
Four cylinders, single acting, four stroke gas engine
0.066
5.
Six cylinders, single acting, four stroke gas engine
0.031
16.8. Flywheel
A flywheel used in machines serves as a reservoir, which stores energy during the period
when the supply of energy is more than the requirement, and releases it during the period when the
requirement of energy is more than the supply.
In case of steam engines, internal combustion engines, reciprocating compressors and pumps,
the energy is developed during one stroke and the engine is to run for the whole cycle on the energy
produced during this one stroke. For example, in internal combustion engines, the energy is developed
only during expansion or power stroke which is much more than the engine load and no energy is
being developed during suction, compression and exhaust strokes in case of four stroke engines and
during compression in case of two stroke engines. The excess energy developed during power stroke
is absorbed by the flywheel and releases it to the crankshaft during other strokes in which no energy
is developed, thus rotating the crankshaft at a uniform speed. A little consideration will show that
when the flywheel absorbs energy, its speed increases and when it releases energy, the speed decreases.
Hence a flywheel does not maintain a constant speed, it simply reduces the fluctuation of speed. In
other words, a flywheel controls the speed variations caused by the fluctuation of the engine
turning moment during each cycle of operation.
Chapter 16 : Turning Moment Diagrams and Flywheel
l
571
In machines where the operation is intermittent like *punching machines, shearing machines,
rivetting machines, crushers, etc., the flywheel stores energy from the power source during the greater
portion of the operating cycle and gives it up during a small period of the cycle. Thus, the energy
from the power source to the machines is supplied practically at a constant rate throughout the
operation.
Note: The function of a **governor in an engine is entirely different from that of a flywheel. It
regulates the mean speed of an engine when there are variations in the load, e.g., when the load on the engine
increases, it becomes necessary to increase the supply of working fluid. On the other hand, when the load
decreases, less working fluid is required. The governor automatically controls the supply of working fluid to
the engine with the varying load condition and keeps the mean speed of the engine within certain limits.
As discussed above, the flywheel does not maintain a constant speed, it simply reduces the fluctuation
of speed. It does not control the speed variations caused by the varying load.
16.9. Coefficient of Fluctuation of Speed
The difference between the maximum and minimum speeds during a cycle is called the
maximum fluctuation of speed. The ratio of the maximum fluctuation of speed to the mean speed is
called the coefficient of fluctuation of speed.
Let
N 1 and N 2 = Maximum and minimum speeds in r.p.m. during the cycle, and
N1 + N 2
N = Mean speed in r.p.m. =
∴ Coefficient of fluctuation of speed,
Cs =
=
=
N1 − N 2
N
ω1 − ω2
=
ω
v1 − v2
v
=
=
(
2 N1 − N 2
2
)
N1 + N 2
(
2 ω1 − ω2
)
ω1 + ω2
(
2 v1 − v2
)
v1 + v2
...(In terms of angular speeds)
...(In terms of linear speeds)
The coefficient of fluctuation of speed is a limiting factor in the design of flywheel. It varies
depending upon the nature of service to which the flywheel is employed.
Note. The reciprocal of the coefficient of fluctuation of speed is known
as coefficient of steadiness and is denoted by m.
∴
m=
N
1
=
Cs N1 − N 2
16.10. Energy Stored in a Flywheel
A flywheel is shown in Fig. 16.5. We have discussed in
Art. 16.5 that when a flywheel absorbs energy, its speed increases
and when it gives up energy, its speed decreases.
Let
m = Mass of the flywheel in kg,
k = Radius of gyration of the
flywheel in metres,
*
**
See Art. 16.12.
See Chapter 18 on Governors.
Fig. 16.5. Flywheel.
572
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Theory of Machines
I = Mass moment of inertia of the flywheel about its axis of rotation
in kg-m2 = m.k 2,
N 1 and N 2 = Maximum and minimum speeds during the cycle in r.p.m.,
ω1 and ω2 = Maximum and minimum angular speeds during the cycle in rad/s,
N + N2
,
N = Mean speed during the cycle in r.p.m. = 1
2
ω + ω2
,
ω = Mean angular speed during the cycle in rad/s = 1
2
N1 − N 2
ω − ω2
or 1
CS = Coefficient of fluctuation of speed, =
N
ω
We know that the mean kinetic energy of the flywheel,
1
1
× I . ω 2 = × m . k 2 . ω2
(in N-m or joules)
2
2
As the speed of the flywheel changes from ω1 to ω2, the maximum fluctuation of energy,
∆E = Maximum K.E. – Minimum K.E.
2
2
2
2
1
1
1
= × I ω1 − × I ω2 = × I ω1 − ω2
2
2
2
1
= × I ω1 + ω2 ω1 − ω2 = I . ω ω1 − ω2
...(i)
2
E=
( )
( )
(
)(
)
( ) ( )
(
)
ω +ω
2
... Q ω = 1
2
ω − ω2
= I . ω2 1
ω
= I.ω2.CS = m.k 2.ω2.CS
... (Multiplying and dividing by ω)
... (∵ I = m.k 2)
= 2.E.CS (in N–m or joules)
...(ii)
1
... Q E = × I . ω2 ... (iii)
2
The radius of gyration (k) may be taken equal to the mean radius of the rim (R), because the
thickness of rim is very small as compared to the diameter of rim. Therefore, substituting k = R, in
equation (ii), we have
∆E = m.R 2.ω2.CS = m.v 2.CS
where
v = Mean linear velocity (i.e. at the mean radius) in m/s = ω.R
Notes. 1. Since ω = 2 π N/60, therefore equation (i) may be written as
∆E = I ×
2 π N 2 π N1 2 π N 2 4 π 2
−
× I × N N1 − N 2
=
60 60
60 3600
(
(
=
π2
× m . k 2 . N N1 − N 2
900
=
π2
× m . k 2 . N 2 . CS
900
)
)
N −N
2
... Q Cs = 1
N
-
Chapter 16 : Turning Moment Diagrams and Flywheel
l
573
2. In the above expressions, only the mass moment of inertia of the flywheel rim (I) is considered
and the mass moment of inertia of the hub and arms is neglected. This is due to the fact that the major portion
of the mass of the flywheel is in the rim and a small portion is in the hub and arms. Also the hub and arms are
nearer to the axis of rotation, therefore the mass moment of inertia of the hub and arms is small.
Example 16.1. The mass of flywheel of an engine is 6.5 tonnes and the radius of gyration
is 1.8 metres. It is found from the turning moment diagram that the fluctuation of energy is
56 kN-m. If the mean speed of the engine is 120 r.p.m., find the maximum and minimum speeds.
Solution. Given : m = 6.5 t = 6500 kg ; k = 1.8 m ; ∆ E = 56 kN-m = 56 × 103 N-m ;
N = 120 r.p.m.
Let
N 1 and N 2 = Maximum and minimum speeds respectively.
We know that fluctuation of energy (∆ E),
56 × 103 =
π2
π2
× m.k 2 . N (N 1 – N 2) =
× 6500 (1.8)2 120 (N 1 – N 2)
900
900
= 27 715 (N 1 – N 2)
∴
N 1 – N 2 = 56 × 103 /27 715 = 2 r.p.m.
...(i)
We also know that mean speed (N),
120 =
N +N
1
2
2
or N1 + N 2 = 120 × 2 = 240 r.p.m.
From equations (i) and (ii),
N1 = 121 r.p.m., and N 2 = 119 r.p.m.
...(ii)
Ans.
Example 16.2. The flywheel of a steam engine has a radius of gyration of 1 m and mass
2500 kg. The starting torque of the steam engine is 1500 N-m and may be assumed constant.
Determine: 1. the angular acceleration of the flywheel, and 2. the kinetic energy of the flywheel
after 10 seconds from the start.
Solution. Given : k = 1 m ; m = 2500 kg ; T = 1500 N-m
1. Angular acceleration of the flywheel
Let
α = Angular acceleration of the flywheel.
We know that mass moment of inertia of the flywheel,
I = m.k 2 = 2500 × 12 = 2500 kg-m2
∴ Starting torque of the engine (T),
1500 = I.α = 2500 × α
or
α = 1500 / 2500 = 0.6 rad /s2 Ans.
2. Kinetic energy of the flywheel
First of all, let us find out the angular speed of the flywheel after 10 seconds from the start
(i.e. from rest), assuming uniform acceleration.
Let
ω1 = Angular speed at rest = 0
ω2 = Angular speed after 10 seconds, and
t = Time in seconds.
We know that
ω2 = ω1 + α t = 0 + 0.6 × 10 = 6 rad /s
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Theory of Machines
∴ Kinetic energy of the flywheel
( )
2
1
1
× I ω2 = × 2500 × 62 = 45 000 N-m = 45 kN-m Ans.
2
2
Example 16.3. A horizontal cross compound steam engine develops 300 k W at 90 r.p.m.
The coefficient of fluctuation of energy as found from the turning moment diagram is to be 0.1 and
the fluctuation of speed is to be kept within ± 0.5% of the mean speed. Find the weight of the
flywheel required, if the radius of gyration is 2 metres.
Solution. Given : P = 300 kW = 300 × 103 W; N = 90 r.p.m.; CE = 0.1; k = 2 m
We know that the mean angular speed,
ω = 2 π N/60 = 2 π × 90/60 = 9.426 rad/s
Let
ω1 and ω2 = Maximum and minimum speeds respectively.
Since the fluctuation of speed is ± 0.5% of mean speed, therefore total fluctuation of speed,
ω1 – ω2 = 1% ω = 0.01 ω
and coefficient of fluctuation of speed,
ω − ω2
Cs = 1
= 0.01
ω
We know that work done per cycle
= P × 60 / N = 300 × 103 × 60 / 90 = 200 × 103 N-m
∴ Maximum fluctuation of energy,
∆E = Work done per cycle × CE = 200 × 103 × 0.1 = 20 × 103 N-m
Let
m = Mass of the flywheel.
We know that maximum fluctuation of energy ( ∆E ),
=
20 × 103 = m.k 2.ω2.CS = m × 22 × (9.426)2 × 0.01 = 3.554 m
∴
m = 20 × 103/3.554 = 5630 kg
Ans.
Example 16.4. The turning moment diagram for a petrol engine is drawn to the following
scales : Turning moment, 1 m m = 5 N-m ; crank angle, 1 m m = 1°. The turning moment diagram
repeats itself at every half revolution of the engine and the areas above and below the mean turning
moment line taken in order are 295, 685, 40, 340, 960, 270 m m2. The rotating parts are equivalent
to a mass of 36 kg at a radius of gyration of 150 m m. Determine the coefficient of fluctuation of
speed when the engine runs at 1800 r.p.m.
Solution. Given : m = 36 kg ; k = 150 mm = 0.15 m ; N = 1800 r.p.m. or ω = 2 π × 1800/60
= 188.52 rad /s
Fig. 16.6
The turning moment diagram is shown in Fig. 16.6.
Chapter 16 : Turning Moment Diagrams and Flywheel
l
575
Since the turning moment scale is 1 mm = 5 N-m and
crank angle scale is 1 mm = 1° = π /180 rad, therefore,
1 mm2 on turning moment diagram
π
π
N-m
=
180 36
Let the total energy at A = E, then referring to
Fig. 16.6,
=5×
Energy at B = E + 295
... (Maximum energy)
Energy at C = E + 295 – 685 = E – 390
Energy at D = E – 390 + 40 = E – 350
Flywheel of an electric motor.
Energy at E = E – 350 – 340 = E – 690 ...(Minimum energy)
Energy at F = E – 690 + 960 = E + 270
Energy at G = E + 270 – 270 = E = Energy at A
We know that maximum fluctuation of energy,
∆ E = Maximum energy – Minimum energy
= (E + 295) – (E – 690) = 985 mm2
π
= 86 N - m = 86 J
36
CS = Coefficient of fluctuation of speed.
= 985 ×
Let
We know that maximum fluctuation of energy (∆ E),
86 = m.k 2 ω2.CS = 36 × (0.15)2 × (188.52)2 CS = 28 787 CS
∴
CS = 86 / 28 787 = 0.003 or 0.3%
Ans.
Example 16.5. The turning moment diagram for a multicylinder engine has been drawn to
a scale 1 m m = 600 N-m vertically and 1 m m = 3° horizontally. The intercepted areas between the
output torque curve and the mean resistance line, taken in order from one end, are as follows :
+ 52, – 124, + 92, – 140, + 85, – 72 and + 107 m m2, when the engine is running at a speed
of 600 r.p.m. If the total fluctuation of speed is not to exceed ± 1.5% of the mean, find the necessary
mass of the flywheel of radius 0.5 m.
Solution. Given : N = 600 r.p.m. or ω = 2 π × 600 / 60 = 62.84 rad / s ; R = 0.5 m
Fig. 16.7
Since the total fluctuation of speed is not to exceed ± 1.5% of the mean speed, therefore
ω1 – ω2 = 3% ω = 0.03 ω
576
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Theory of Machines
and coefficient of fluctuation of speed,
ω − ω2
Cs = 1
= 0.03
ω
The turning moment diagram is shown in Fig. 16.7.
Since the turning moment scale is 1 mm = 600 N-m and crank angle scale is 1 mm = 3°
= 3° × π/180 = π / 60 rad, therefore
1 mm2 on turning moment diagram
= 600 × π/60 = 31.42 N-m
Let the total energy at A = E, then referring to Fig. 16.7,
Energy at B = E + 52
...(Maximum energy)
Energy at C = E + 52 – 124 = E – 72
Energy at D = E – 72 + 92 = E + 20
Energy at E = E + 20 – 140 = E – 120
...(Minimum energy)
Energy at F = E – 120 + 85 = E – 35
Energy at G = E – 35 – 72 = E – 107
Energy at H = E – 107 + 107 = E = Energy at A
We know that maximum fluctuation of energy,
∆ E = Maximum energy – Minimum energy
= (E + 52) – (E – 120) = 172 = 172 × 31.42 = 5404 N-m
Let
m = Mass of the flywheel in kg.
We know that maximum fluctuation of energy (∆ E ),
5404 = m.R 2.ω2.CS = m × (0.5)2 × (62.84)2 × 0.03 = 29.6 m
∴
m = 5404 / 29.6 = 183 kg
Ans.
Example 16.6. A shaft fitted with a flywheel rotates at 250 r.p.m. and drives a machine.
The torque of machine varies in a cyclic manner over a period of 3 revolutions. The torque rises
from 750 N-m to 3000 N-m uniformly during 1/2 revolution and remains constant for the following
revolution. It then falls uniformly to 750 N-m during the next 1/2 revolution and remains constant
for one revolution, the cycle being repeated thereafter.
Determine the power required to drive the machine and percentage fluctuation in speed, if
the driving torque applied to the shaft is constant and the mass of the flywheel is 500 kg with radius
of gyration of 600 m m.
Solution. Given : N = 250 r.p.m. or ω = 2π × 250/60 = 26.2 rad/s ; m = 500 kg ;
k = 600 mm = 0.6 m
The turning moment diagram for the complete cycle is shown in Fig. 16.8.
We know that the torque required for one complete cycle
= Area of figure OABCDEF
= Area OAEF + Area ABG + Area BCHG + Area CDH
1
1
= OF × OA + × AG × BG + GH × CH + × HD × CH
2
2
Chapter 16 : Turning Moment Diagrams and Flywheel
l
577
1
= 6 π × 750 + × π (3000 − 750 ) + 2 π (3000 − 750 )
2
1
+ × π (3000 − 750 )
2
= 11 250 π N-m
...(i)
If T mean is the mean torque in N-m, then torque required for one complete cycle
= T mean × 6 π Ν-m
...(ii)
From equations (i) and (ii),
T mean = 11 250 π / 6 π = 1875 N-m
Fig. 16.8
Power required to drive the machine
We know that power required to drive the machine,
P = T mean × ω = 1875 × 26.2 = 49 125 W = 49.125 kW Ans.
Coefficient of fluctuation of speed
Let
CS = Coefficient of fluctuation of speed.
First of all, let us find the values of L M and NP. From similar triangles ABG and BLM,
LM 3000 − 1875
LM
BM
=
= 0.5
=
or
π
3000 − 750
AG
BG
Now, from similar triangles CHD and CNP,
or
LM = 0.5 π
NP 3000 − 1875
=
= 0.5
π
3000 − 750
or
NP = 0.5 π
NP CN
=
HD CH
From Fig. 16.8, we find that
or
BM = CN = 3000 – 1875 = 1125 N-m
Since the area above the mean torque line represents the maximum fluctuation of energy,
therefore, maximum fluctuation of energy,
∆E = Area LBCP = Area LBM + Area MBCN + Area PNC
=
1
1
× LM × BM + MN × BM + × NP × CN
2
2
578
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Theory of Machines
1
1
× 0.5 π × 1125 + 2 π × 1125 + × 0.5 π × 1125
2
2
= 8837 N - m
We know that maximum fluctuation of energy (∆ E),
8837 = m.k 2.ω2.CS = 500 × (0.6)2 × (26.2)2 × CS = 123 559 CS
=
CS =
8837
= 0.071 Ans.
123559
Flywheel of a pump run by a diesel engine.
Example 16.7. During forward stroke of the piston of the double acting steam engine, the
turning moment has the maximum value of 2000 N-m when the crank makes an angle of 80° with
the inner dead centre. During the backward stroke, the maximum turning moment is 1500 N-m
when the crank makes an angle of 80° with the outer dead centre. The turning moment diagram for
the engine may be assumed for simplicity to be represented by two triangles.
If the crank makes 100 r.p.m. and the radius of gyration of the flywheel is 1.75 m, find the
coefficient of fluctuation of energy and the mass of the flywheel to keep the speed within ± 0.75% of
the mean speed. Also determine the crank angle at which the speed has its minimum and maximum
values.
Solution. Given : N = 100 r.p.m. or ω = 2π × 100/60 = 10.47 rad /s; k = 1.75 m
Since the fluctuation of speed is ± 0.75% of mean speed, therefore total fluctuation of speed,
ω1 – ω2 = 1.5% ω
and coefficient of fluctuation of speed,
ω – ω2
CS = 1
= 1.5% = 0.015
ω
Coefficient of fluctuation of energy
The turning moment diagram for the engine during forward and backward strokes is shown
in Fig. 16.9. The point O represents the inner dead centre (I.D.C.) and point G represents the
outer dead centre (O.D.C). We know that maximum turning moment when crank makes an
angle of 80° (or 80 × π / 180 = 4π/9 rad) with I.D.C.,
∴
AB = 2000 N-m
Chapter 16 : Turning Moment Diagrams and Flywheel
l
579
and maximum turning moment when crank makes an angle of 80° with outer dead centre (O.D.C.) or
180° + 80° = 260° = 260 × π /180 = 13 π / 9 rad with I.D.C.,
LM = 1500 N-m
Let
T mean = EB = QM = Mean resisting torque.
Fig. 16.9
We know that work done per cycle
= Area of triangle OAG + Area of triangle GLS
=
1
1
× OG × AB + × GS × LM
2
2
1
1
× π × 2000 + × π × 1500 = 1750 π N-m
2
2
We also know that work done per cycle
=
...(i)
= T mean × 2 π N-m
...(ii)
From equations (i) and (ii),
T mean = 1750 π / 2 π = 875 N-m
From similar triangles ACD and AOG,
CD OG
=
AE
AB
or
OG
OG
π
× AE =
( AB – EB ) =
(2000 − 875) = 1.764 rad
AB
AB
2000
∴ Maximum fluctuation of energy,
CD =
∆E = Area of triangle ACD =
1
× CD × AE
2
1
1
× CD ( AB – EB ) = × 1.764 ( 2000 − 875 ) = 992 N-m
2
2
We know that coefficient of fluctuation of energy,
=
CE =
Max.fluctuation of energy
992
=
= 0.18 or 18% Ans.
Work done per cycle
1750 π
580
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Theory of Machines
Mass of the flywheel
Let
m = Mass of the flywheel.
We know that maximum fluctuation of energy (∆E),
992 = m.k 2.ω2.CS = m × (1.75)2 × (10.47)2 × 0.015 = 5.03 m
∴
m = 992 / 5.03 = 197.2 kg Ans.
Crank angles for the minimum and
maximum speeds
We know that the speed of
the flywheel is minimum at point C
and maximum at point D (See
Art. 16.5).
Let θC and θD = Crank angles from
I.D.C., for the minimum and maximum
speeds.
From similar triangles ACE and
AOB,
CE
AE
=
OB AB
CE =
or
∴
Flywheel of small steam engine.
AE
AB – EB
2000 − 875 4 π π
× OB =
× OB =
×
= rad
AB
AB
2000
9
4
θC =
4π π 7π
7 π 180
rad =
− =
×
= 35°
9
4 36
36
π
Ans.
Again from similar triangles AED and ABG,
ED AE
=
BG
AB
or
∴
ED =
AE
AB – EB
× BG =
(OG − OB )
AB
AB
=
2000 − 875
4 π 2.8 π
rad
π −
=
2000
9
9
θD =
4 π 2.8 π 6.8 π
6.8 π 180
rad =
+
=
×
= 136° Ans.
9
9
9
9
π
Example 16.8. A three cylinder single acting engine has its cranks set equally at 120° and
it runs at 600 r.p.m. The torque-crank angle diagram for each cycle is a triangle for the power stroke
with a maximum torque of 90 N-m at 60° from dead centre of corresponding crank. The torque on the
return stroke is sensibly zero. Determine : 1. power developed. 2. coefficient of fluctuation of speed,
if the mass of the flywheel is 12 kg and has a radius of gyration of 80 mm, 3. coefficient of fluctuation
of energy, and 4. maximum angular acceleration of the flywheel.
Solution. Given : N = 600 r.p.m. or ω = 2 π × 600/60 = 62.84 rad /s; T max = 90 N-m;
m = 12 kg; k = 80 mm = 0.08 m
Chapter 16 : Turning Moment Diagrams and Flywheel
l
581
The torque-crank angle diagram for the individual cylinders is shown in Fig. 16.10 (a), and
the resultant torque-crank angle diagram for the three cylinders is shown in Fig. 16.10 (b).
Fig. 16.10
1. Power developed
We know that work done/cycle
= Area of three triangles = 3 ×
1
× π × 90 = 424 N-m
2
Work done / cycle
424
=
= 67.5 N - m
Crank angle / cycle
2π
∴ Power developed = T mean × ω = 67.5 × 62.84 = 4240 W = 4.24 kW Ans.
and mean torque,
Tmean =
2. Coefficient of fluctuation of speed
Let
CS = Coefficient of fluctuation of speed.
First of all, let us find the maximum fluctuation of energy (∆E).
From Fig. 16.10 (b), we find that
a1 = Area of triangle AaB =
1
× AB × Aa
2
1 π
× × ( 67.5 − 45) = 5.89 N-m = a7 ...(∵ A B = 30° = π / 6 rad)
2 6
1
a2 = Area of triangle BbC = × BC × bb '
2
1 π
= × (90 − 67.5 ) = 11.78 N - m
...(∵BC = 60° = π/3 rad)
2 3
= a3 = a4 = a5 = a6
=
Now, let the total energy at A = E, then referring to Fig. 16.10 (b),
Energy at B =
Energy at C =
Energy at D =
Energy at E =
Energy at G =
Energy at H =
Energy at J =
E – 5.89
E – 5.89 + 11.78 = E + 5.89
E + 5.89 – 11.78 = E – 5.89
E – 5.89 + 11.78 = E + 5.89
E + 5.89 – 11.78 = E – 5.89
E – 5.89 + 11.78 = E + 5.89
E + 5.89 – 5.89 = E = Energy at A
582
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Theory of Machines
From above we see that maximum energy
= E + 5.89
and minimum energy
= E – 5.89
∴ * Maximum fluctuation of energy,
∆E = (E + 5.89) – (E – 5.89) = 11.78 N-m
We know that maximum fluctuation of energy (∆E),
11.78 = m.k 2.ω2.CS = 12 × (0.08)2 × (62.84)2 × CS = 303.3 CS
∴
CS = 11.78 / 303.3 = 0.04 or 4% Ans.
3. Coefficient of fluctuation of energy
We know that coefficient of fluctuation of energy,
Max. fluctuation of energy 11.78
=
= 0.0278 = 2.78% Ans.
Work done/cycle
424
4. Maximum angular acceleration of the flywheel
CE =
α = Maximum angular acceleration of the flywheel.
Let
We know that,
T max – T mean = I.α = m.k 2.α
90 – 67.5 = 12 × (0.08)2 × α = 0.077 α
90 − 67.5
= 292 rad / s2 Ans.
0.077
Example 16.9. A single cylinder, single acting, four stroke gas engine develops 20 kW at
300 r.p.m. The work done by the gases during the expansion stroke is three times the work done on
the gases during the compression stroke, the work done during the suction and exhaust strokes
being negligible. If the total fluctuation of speed is not to exceed ± 2 per cent of the mean speed and
the turning moment diagram during compression and expansion is assumed to be triangular in
shape, find the moment of inertia of the flywheel.
∴
α=
Solution. Given : P = 20 kW = 20 × 103 W; N = 300 r.p.m. or ω = 2π × 300/60 = 31.42 rad/s
Since the total fluctuation of speed (ω1 – ω2) is not to exceed ± 2 per cent of the mean speed
(ω), therefore
ω1 – ω2 = 4% ω
and coefficient of fluctuation of speed,
ω1 − ω2
= 4% = 0.04
ω
The turning moment-crank angle diagram for a four stroke engine is shown in Fig. 16.11. It
is assumed to be triangular during compression and expansion strokes, neglecting the suction and
exhaust strokes.
CS =
*
Since the area above the mean torque line represents the maximum fluctuation of energy, therefore maximum fluctuation of energy,
∆E = Area Bbc = Area DdE = Area Ggh
=
1 π
× (90 – 67.5) = 11.78 N-m
2 3
Chapter 16 : Turning Moment Diagrams and Flywheel
l
583
We know that for a four stroke engine, number of working strokes per cycle,
n = N/2 = 300 / 2 = 150
∴ Work done/cycle =P × 60/n = 20 × 103 × 60/150 = 8000 N-m
...(i)
Fig. 16.11
Since the work done during suction and exhaust strokes is negligible, therefore net work
done per cycle (during compression and expansion strokes)
WE
2
= WE
... ( ∵ WE = 3WC) ...(ii)
3
3
Equating equations (i) and (ii), work done during expansion stroke,
= WE – WC = WE –
W E = 8000 × 3/2 = 12 000 N-m
We know that work done during expansion stroke (W E),
1
1
× BC × AG = × π × AG
2
2
AG = T max = 12 000 × 2/π = 7638 N-m
12 000 = Area of triangle ABC =
∴
and mean turning moment,
* Tmean = FG =
Work done/cycle
8000
=
= 637 N-m
Crank angle/cycle
4π
∴ Excess turning moment,
Texcess = AF = A G – FG = 7638 – 637 = 7001 N-m
Now, from similar triangles ADE and ABC,
DE
AF
AF
7001
=
or DE =
× BC =
× π = 2.88 rad
BC
AG
AG
7638
Since the area above the mean turning moment line represents the maximum fluctuation of
energy, therefore maximum fluctuation of energy,
∆E = Area of ∆ ADE =
*
1
1
× DE × AF = × 2.88 × 7001 = 10 081 N-m
2
2
The mean turning moment (T mean) may also be obtained by using the following relation :
P = T mean × ω or T mean = P/ω = 20 × 103/31.42 = 637 N-m
584
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Theory of Machines
Let
I = Moment of inertia of the flywheel in kg-m2 .
We know that maximum fluctuation of energy (∆ E),
10 081 = I.ω2.CS = I × (31.42)2 × 0.04 = 39.5 I
∴
I = 10081/ 39.5 = 255.2 kg-m2 Ans.
Example 16.10. The turning moment diagram for a four stroke gas engine may be assumed
for simplicity to be represented by four triangles, the areas of which from the line of zero pressure
are as follows :
Suction stroke = 0.45 × 10–3 m2; Compression stroke = 1.7 × 10–3 m2; Expansion stroke
= 6.8 × 10–3 m2; Exhaust stroke = 0.65 × 10–3 m2. Each m2 of area represents 3 MN-m of energy.
Assuming the resisting torque to be uniform, find the mass of the rim of a flywheel required
to keep the speed between 202 and 198 r.p.m. The mean radius of the rim is 1.2 m.
Solution. Given : a1 = 0.45 × 10 –3 m 2 ; a2 = 1.7 × 10 –3 m 2 ; a3 = 6.8 × 10 –3 m 2 ;
a4 = 0.65 × 10 –3 m 2 ; N 1 = 202 r.p.m; N 2 = 198 r.p.m.; R = 1.2 m
The turning moment crank angle diagram for a four stroke engine is shown in Fig. 16.12.
The areas below the zero line of pressure are taken as negative while the areas above the zero line of
pressure are taken as positive.
∴
Net area = a3 – (a1 + a2 + a4)
= 6.8 × 10–3 – (0.45 × 10–3 + 1.7 × 10–3 + 0.65 × 10–3) = 4 × 10–3 m2
Since the energy scale is 1 m2 = 3 MN-m = 3 × 106 N-m, therefore,
Net work done per cycle = 4 × 10–3 × 3 ×106 = 12 × 103 N-m
. . . (i)
We also know that work done per cycle,
= T mean × 4π N-m
. . . (ii)
From equations (i) and (ii),
T mean = FG = 12 × 103/4π = 955 N-m
Fig. 16.12
Work done during expansion stroke
= a3 × Energy scale = 6.8 × 10–3 × 3 × 106 = 20.4 × 103 N-m ...(iii)
Chapter 16 : Turning Moment Diagrams and Flywheel
l
585
Also, work done during expansion stroke
= Area of triangle ABC
1
1
× BC × AG = × π × AG = 1.571 × AG
. . . (iv)
2
2
From equations (iii) and (iv),
AG = 20.4 × 103/1.571 = 12 985 N-m
∴ Excess torque,
Texcess = AF = A G – FG = 12 985 – 955 = 12 030 N-m
Now from similar triangles ADE and ABC,
AF
12 030
DE
AF
× BC =
× π = 2.9 rad
=
or DE =
AG
12 985
BC
AG
We know that the maximum fluctuation of energy,
=
∆ E = Area of ∆ ADE =
1
1
× DE × AF = × 2.9 × 12 030 N-m
2
2
= 17 444 N-m
Mass of the rim of a flywheel
Let
m = Mass of the rim of a flywheel in kg, and
N = Mean speed of the flywheel
202 + 198
N1 + N 2
=
= 200 r .p.m.
2
2
We know that the maximum fluctuation of energy (∆E ),
=
π2
π2
2
× m.R 2 . N ( N1 – N 2 ) =
× (1.2 ) 200 × (202 – 198 )
900
900
= 12.63 m
17 444 =
∴
m = 17 444 /12.36 = 1381 kg Ans.
Example 16.11. The turning moment curve for an engine is represented by the equation,
T = (20 000 + 9500 sin 2θ – 5700 cos 2θ) N-m, where θ is the angle moved by the crank from
inner dead centre. If the resisting torque is constant, find:
1. Power developed by the engine ; 2. Moment of inertia of flywheel in kg-m2, if the total
fluctuation of speed is not exceed 1% of mean speed which is 180 r.p.m; and 3. Angular acceleration
of the flywheel when the crank has turned through 45° from inner dead centre.
Solution. Given : T = (20 000 + 9500 sin 2θ – 5700 cos 2θ) N-m ; N = 180 r.p.m. or
ω = 2π × 180/60 = 18.85 rad/s
Since the total fluctuation of speed (ω1 – ω2) is 1% of mean speed (ω), therefore coefficient
of fluctuation of speed,
ω – ω2
CS = 1
= 1% = 0.01
ω
1. Power developed by the engine
We know that work done per revolution
2π
2π
= ∫ T d θ = ∫ ( 20 000 + 9500sin 2 θ – 5700 cos 2 θ) d θ
0
0
586
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Theory of Machines
9500 cos 2 θ 5700sin 2 θ 2 π
= 20 000 θ –
–
2
2
0
= 20 000 × 2π = 40 000 π N-m
and mean resisting torque of the engine,
Work done per revolution 40 000
=
= 20000 N-m
2π
2π
We know that power developed by the engine
Tmean =
= T mean . ω = 20 000 × 18.85 = 377 000 W = 377 kW Ans.
2. Moment of inertia of the flywheel
Let
I = Moment of inertia of the flywheel in kg-m2.
The turning moment diagram for one stroke (i.e. half revolution of the crankshaft) is shown
in Fig. 16.13. Since at points B and D, the torque exerted on the crankshaft is equal to the mean
resisting torque on the flywheel, therefore,
T = T mean
20 000 + 9500 sin 2θ – 5700 cos 2θ = 20 000
or
9500 sin 2θ = 5700 cos 2θ
tan 2θ = sin 2θ/cos 2θ = 5700/9500 = 0.6
∴
2θ = 31° or θ = 15.5°
∴
θB = 15.5° and θD = 90° + 15.5° = 105.5°
Fig. 16.13
Maximum fluctuation of energy,
θ
D
(
)
∆ E = ∫ T – Tmean d θ
θ
=
B
105.5 °
∫
(20 000 + 9500 sin 2 θ – 5700 cos 2 θ – 20 000 ) d θ
15.5°
105.5°
9500 cos 2 θ 5700sin 2 θ
= –
= 11 078 N-m
–
2
2
15.5°
Chapter 16 : Turning Moment Diagrams and Flywheel
l
587
We know that maximum fluctuation of energy
(∆ E),
11 078 = I.ω2.CS = I × (18.85)2 × 0.01 = 3.55 I
∴
I =11078/3.55 = 3121 kg-m2 Ans.
3. Angular acceleration of the flywheel
Let α = Angular acceleration of the flywheel,
and
θ = Angle turned by the crank from inner
dead centre = 45°
. . . (Given)
The angular acceleration in the flywheel is
produced by the excess torque over the mean torque.
We know that excess torque at any instant,
T excess = T – T mean
= 20000 + 9500 sin 2θ – 5700 cos 2θ
– 20000
= 9500 sin 2θ – 5700 cos 2θ
∴ Excess torque at 45°
Nowadays steam turbines like this can
be produced entirely by computercontrolled machine tools, directly from the
engineer’s computer.
Note : This picture is given as additional information and is not a direct example of the current
chapter.
= 9500 sin 90° – 5700 cos 90° = 9500 N-m
. . . (i)
We also know that excess torque
= I.α = 3121 × α
. . . (ii)
From equations (i) and (ii),
α = 9500/3121 = 3.044 rad /s2 Ans.
Example 16.12. A certain machine requires a torque of (5000 + 500 sin θ ) N-m to drive it,
where θ is the angle of rotation of shaft measured from certain datum. The machine is directly
coupled to an engine which produces a torque of (5000 + 600 sin 2θ) N-m. The flywheel and the
other rotating parts attached to the engine has a mass of 500 kg at a radius of gyration of 0.4 m. If
the mean speed is 150 r.p.m., find : 1. the fluctuation of energy, 2. the total percentage fluctuation of
speed, and 3. the maximum and minimum angular acceleration of the flywheel and the corresponding
shaft position.
Solution. Given : T 1 = ( 5000 + 500 sin θ) N-m ; T 2 = (5000 + 600 sin 2θ) N-m ;
m = 500 kg; k = 0.4 m ; N = 150 r.p.m. or ω = 2 π × 150/60 = 15.71 rad/s
Fig. 16.14
588
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Theory of Machines
1. Fluctuation of energy
We know that change in torque
= T 2 – T 1 = (5000 + 600 sin 2θ) – (5000 + 500 sin θ)
= 600 sin 2θ – 500 sin θ
This change is zero when
600 sin 2θ = 500 sin θ or
1.2 × 2 sin θ cos θ = sin θ
or
1.2 sin 2θ = sin θ
2.4 sin θ cos θ = sin θ . . . (∵sin 2θ = 2 sin θ cos θ)
∴ Either
sin θ = 0 or cos θ = 1/2.4 = 0.4167
when
sin θ = 0, θ = 0°, 180° and 360°
i.e.
when
i.e.
θA = 0°, θC = 180° and θE = 360°
cos θ = 0.4167, θ = 65.4° and 294.6°
θB = 65.4° and θD = 294.6°
The turning moment diagram is shown in Fig. 16.14. The maximum fluctuation of energy
lies between C and D (i.e. between 180° and 294.6°), as shown shaded in Fig. 16.14.
∴ Maximum fluctuation of energy,
∆E =
294.6°
∫
180°
=
(T
2
)
– T dθ
1
294.6°
∫ (5000 + 600sin 2 θ) – (5000 + 500sin θ) d θ
180°
294.6 °
600 cos 2 θ
= –
+ 500 cos θ
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