Salc(1)

CHAPTER 1 PROBLEM 1.1 Two solid cylindrical rods AB and BC are welded together at B and loaded as shown. Knowing that the average normal stress must not exceed 175 MPa in rod AB and 150 MPa in rod BC, determine the smallest allowable values of d1 and d2. SOLUTION (a) Rod AB P = 40 + 30 = 70 kN = 70 × 103 N (b) σ AB = P = AAB d1 = 4P P 4P = 2 d π d12 4 1 π = πσ AB (4)(70 × 103 ) = 22.6 × 10−3 m π (175 × 106 ) d1 = 22.6 mm  Rod BC P = 30 kN = 30 × 103 N σ BC = d2 = P = ABC 4P πσ BC 4P P = 2 d π d 22 4 2 π = (4)(30 × 103 ) = 15.96 × 10−3 m π (150 × 106 ) d 2 = 15.96 mm  PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 1.2 Two solid cylindrical rods AB and BC are welded together at B and loaded as shown. Knowing that d1 = 50 mm and d 2 = 30 mm, find the average normal stress at the midsection of (a) rod AB, (b) rod BC. SOLUTION (a) Rod AB P = 40 + 30 = 70 kN = 70 × 103 N A= σ AB = (b) π 4 d12 = π 4 (50) 2 = 1.9635 × 103 mm 2 = 1.9635 × 10−3 m 2 P 70 × 103 = = 35.7 × 106 Pa −3 A 1.9635 × 10 σ AB = 35.7 MPa  Rod BC P = 30 kN = 30 × 103 N A= σ BC = π 4 d 22 = π 4 (30)2 = 706.86 mm 2 = 706.86 × 10−6 m 2 P 30 × 103 = = 42.4 × 106 Pa −6 A 706.86 × 10 σ BC = 42.4 MPa  PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 1.3 Two solid cylindrical rods AB and BC are welded together at B and loaded as shown. Determine the magnitude of the force P for which the tensile stress in rod AB is twice the magnitude of the compressive stress in rod BC. SOLUTION σ AB π (2) 2 = 3.1416 in 2 4 P P = = 3.1416 AAB AAB = = 0.31831 P ABC = σ BC π (3)2 = 7.0686 in 2 4 (2)(30) − P = AAB = 60 − P = 8.4883 − 0.14147 P 7.0686 Equating σ AB to 2σ BC 0.31831 P = 2(8.4883 − 0.14147 P) P = 28.2 kips  PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 1.4 In Prob. 1.3, knowing that P = 40 kips, determine the average normal stress at the midsection of (a) rod AB, (b) rod BC. PROBLEM 1.3 Two solid cylindrical rods AB and BC are welded together at B and loaded as shown. Determine the magnitude of the force P for which the tensile stress in rod AB is twice the magnitude of the compressive stress in rod BC. SOLUTION (a) Rod AB P = 40 kips (tension) AAB = σ AB (b) 2 π d AB = π (2) 2 4 4 40 P = = 3.1416 AAB = 3.1416 in 2 σ AB = 12.73 ksi  Rod BC F = 40 − (2)(30) = −20 kips, i.e., 20 kips compression. ABC = σ BC 2 π d BC = π (3) 2 4 4 F −20 = = ABC 7.0686 = 7.0686 in 2 σ BC = −2.83 ksi  PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 1.5 Two steel plates are to be held together by means of 16-mmdiameter high-strength steel bolts fitting snugly inside cylindrical brass spacers. Knowing that the average normal stress must not exceed 200 MPa in the bolts and 130 MPa in the spacers, determine the outer diameter of the spacers that yields the most economical and safe design. SOLUTION At each bolt location the upper plate is pulled down by the tensile force Pb of the bolt. At the same time, the spacer pushes that plate upward with a compressive force Ps in order to maintain equilibrium. Pb = Ps For the bolt, σb = 4Pb Fb = Ab π db2 or Pb = For the spacer, σs = Ps 4Ps = As π (d s2 − db2 ) or Ps = π 4 π 4 σ bdb2 σ s (d s2 − db2 ) Equating Pb and Ps , π 4 σ bdb2 = ds = π 4 σ s (d s2 − db2 )  σb  1 + d = σs  b  200   1 + 130  (16)   d s = 25.2 mm  PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 1.6 Two brass rods AB and BC, each of uniform diameter, will be brazed together at B to form a nonuniform rod of total length 100 m, which will be suspended from a support at A as shown. Knowing that the density of brass is 8470 kg/m3, determine (a) the length of rod AB for which the maximum normal stress in ABC is minimum, (b) the corresponding value of the maximum normal stress. SOLUTION Areas: AAB = ABC = From geometry, Weights: π 4 π 4 (15 mm)2 = 176.71 mm 2 = 176.71 × 10−6 m 2 (10 mm) 2 = 78.54 mm 2 = 78.54 × 10−6 m 2 b = 100 − a WAB = ρ g AAB AB = (8470)(9.81)(176.71 × 10−6 ) a = 14.683 a WBC = ρ g ABC  BC = (8470)(9.81)(78.54 × 10−6 )(100 − a) = 652.59 − 6.526 a Normal stresses: At A, PA = WAB + WBC = 652.59 + 8.157a σA = At B, (a) PA = 3.6930 × 106 + 46.160 × 103 a AAB PB = WBC = 652.59 − 6.526a σB = (2) PB = 8.3090 × 106 − 83.090 × 103 a ABC Length of rod AB. The maximum stress in ABC is minimum when σ A = σ B or 4.6160 × 106 − 129.25 × 103 a = 0 a = 35.71 m (b) (1)  AB = a = 35.7 m  Maximum normal stress. σ A = 3.6930 × 106 + (46.160 × 103 )(35.71) σ B = 8.3090 × 106 − (83.090 × 103 )(35.71) σ A = σ B = 5.34 × 106 Pa σ = 5.34 MPa  PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 1.7 Each of the four vertical links has an 8 × 36-mm uniform rectangular cross section and each of the four pins has a 16-mm diameter. Determine the maximum value of the average normal stress in the links connecting (a) points B and D, (b) points C and E. SOLUTION Use bar ABC as a free body. ΣM C = 0 : (0.040) FBD − (0.025 + 0.040)(20 × 103 ) = 0 FBD = 32.5 × 103 N Link BD is in tension. 3 ΣM B = 0 : − (0.040) FCE − (0.025)(20 × 10 ) = 0 FCE = −12.5 × 103 N Link CE is in compression. Net area of one link for tension = (0.008)(0.036 − 0.016) = 160 × 10−6 m 2. For two parallel links, (a) σ BD = A net = 320 × 10−6 m 2 FBD 32.5 × 103 = = 101.56 × 106 Anet 320 × 10−6 σ BD = 101.6 MPa  Area for one link in compression = (0.008)(0.036) = 288 × 10−6 m 2. For two parallel links, (b) σ CE = A = 576 × 10−6 m 2 FCE −12.5 × 103 = = −21.70 × 10−6 A 576 × 10−6 σ CE = −21.7 MPa  PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 1.8 Knowing that the link DE is 18 in. thick and 1 in. wide, determine the normal stress in the central portion of that link when (a) θ = 0°, (b) θ = 90°. SOLUTION Use member CEF as a free body. Σ M C = 0 : − 12 FDE − (8)(60 sin θ ) − (16)(60 cos θ ) = 0 FDE = −40 sin θ − 80 cos θ lb. 1 ADE = (1)   = 0.125 in.2 8 F σ DE = DE ADE (a) θ = 0: FDE = −80 lb. σ DE = (b) −80 0.125 σ DE = −640 psi  θ = 90°: FDE = −40 lb. σ DE = −40 0.125 σ DE = −320 psi  PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 1.9 Link AC has a uniform rectangular cross section 161 in. thick and 14 in. wide. Determine the normal stress in the central portion of the link. SOLUTION Free Body Diagram of Plate Note that the two 240-lb forces form a couple of moment (240 lb)(6 in.) = 1440 lb ⋅ in. Σ M B = 0 : 1440 lb ⋅ in − ( FAC cos 30°)(10 in.) = 0 FAC = 166.277 lb. Area of link:  1  1  AAC =  in.  in.  = 0.015625 in.2 16 4    Stress: σ AC = 166.277 FAC = = 10640 psi 0.015625 AAC σ AC = 10.64 ksi  PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 1.10 Three forces, each of magnitude P = 4 kN, are applied to the mechanism shown. Determine the cross-sectional area of the uniform portion of rod BE for which the normal stress in that portion is +100 MPa. SOLUTION Draw free body diagrams of AC and CD. Free Body CD: ΣM D = 0: 0.150P − 0.250C = 0 C = 0.6P Free Body AC: Required area of BE: M A = 0: 0.150 FBE − 0.350P − 0.450P − 0.450C = 0 FBE = 1.07 P = 7.1333 P = (7.133)(4 kN) = 28.533 kN 0.150 σ BE = FBE ABE ABE = FBE σ BE = 28.533 × 103 = 285.33 × 10−6 m 2 100 × 106 ABE = 285 mm 2  PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 1.11 The frame shown consists of four wooden members, ABC, DEF, BE, and CF. Knowing that each member has a 2 × 4-in. rectangular cross section and that each pin has a 1/2-in. diameter, determine the maximum value of the average normal stress (a) in member BE, (b) in member CF. SOLUTION  Add support reactions to figure as shown.   Using entire frame as free body, ΣM A = 0: 40 Dx − (45 + 30)(480) = 0 Dx = 900 lb. Use member DEF as free body. Reaction at D must be parallel to FBE and FCF . Dy = 4 Dx = 1200 lb. 3 4  ΣM F = 0: − (30)  FBE  − (30 + 15) DY = 0 5   FBE = −2250 lb. 4  ΣM E = 0: (30)  FCE  − (15) DY = 0 5  FCE = 750 lb. Stress in compression member BE Area: (a) A = 2 in × 4 in = 8 in 2 σ BE = −2250 FBE = 8 A σ BE = −281 psi  Minimum section area occurs at pin. Amin = (2)(4.0 − 0.5) = 7.0 in 2 Stress in tension member CF (b) σ CF = FCF 750 = Amin 7.0 σ CF =107.1 psi  PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 1.12 For the Pratt bridge truss and loading shown, determine the average normal stress in member BE, knowing that the crosssectional area of that member is 5.87 in2. SOLUTION  Use entire truss as free body. ΣM H = 0: (9)(80) + (18)(80) + (27)(80) − 36 Ay = 0 Ay = 120 kips Use portion of truss to the left of a section cutting members BD, BE, and CE.   + ↑ Σ Fy = 0: 120 − 80 − σ BE = 12 FBE = 0 15 50 kips FBE = A 5.87 in 2 ∴ FBE = 50 kips σ BE = 8.52 ksi   PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 1.13 An aircraft tow bar is positioned by means of a single hydraulic cylinder connected by a 25-mm-diameter steel rod to two identical arm-and-wheel units DEF. The mass of the entire tow bar is 200 kg, and its center of gravity is located at G. For the position shown, determine the normal stress in the rod. SOLUTION  FREE BODY – ENTIRE TOW BAR: W = (200 kg)(9.81 m/s 2 ) = 1962.00 N  Σ M A = 0 : 850 R − 1150(1962.00 N) = 0  R = 2654.5 N FREE BODY – BOTH ARM & WHEEL UNITS: tan α = 100 675 α = 8.4270° Σ M E = 0 : ( FCD cos α )(550) − R(500) = 0 FCD = 500 (2654.5 N) 550 cos 8.4270° = 2439.5 N (comp.)  σ CD = − 2439.5 N FCD =− ACD π (0.0125 m) 2 = −4.9697 × 106 Pa σ CD = −4.97 MPa  PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 1.14 A couple M of magnitude 1500 N ⋅ m is applied to the crank of an engine. For the position shown, determine (a) the force P required to hold the engine system in equilibrium, (b) the average normal stress in the connecting rod BC, which has a 450-mm2 uniform cross section. SOLUTION Use piston, rod, and crank together as free body. Add wall reaction H and bearing reactions Ax and Ay. Σ M A = 0 : (0.280 m) H − 1500 N ⋅ m = 0 H = 5.3571 × 103 N Use piston alone as free body. Note that rod is a two-force member; hence the direction of force FBC is known. Draw the force triangle and solve for P and FBE by proportions. l = 2002 + 602 = 208.81 mm P 200 = H 60  ∴ P = 17.86 × 103 N (a) P = 17.86 kN  208.81 FBC = ∴ FBC = 18.643 × 103 N 60 H Rod BC is a compression member. Its area is 450 mm 2 = 450 × 10−6 m 2 Stress,  σ BC = − FBC −18.643 × 103 = = −41.4 × 106 Pa A 450 × 10−6 (b) σ BC = −41.4 MPa  PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 1.15 When the force P reached 8 kN, the wooden specimen shown failed in shear along the surface indicated by the dashed line. Determine the average shearing stress along that surface at the time of failure. SOLUTION Area being sheared: A = 90 mm × 15 mm = 1350 mm 2 = 1350 × 10−6 m 2 Force: P = 8 × 103 N Shearing stress: τ = P 8 × 103 − = 5.93 × 106 Pa A 1350 × 10−6 τ = 5.93 MPa  PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 1.16 The wooden members A and B are to be joined by plywood splice plates, that will be fully glued on the surfaces in contact. As part of the design of the joint, and knowing that the clearance between the ends of the members is to be 14 in., determine the smallest allowable length L if the average shearing stress in the glue is not to exceed 120 psi. SOLUTION There are four separate areas that are glued. Each of these areas transmits one half the 5.8 kip force. Thus F = 1 1 P = (5.8) = 2.9 kips = 2900 lb. 2 2 Let l = length of one glued area and w = 4 in. be its width. For each glued area, A = lw Average shearing stress: τ = F F = A lw The allowable shearing stress is τ = 120 psi F 2900 = = 6.0417 in. τ w (120)(4) Solving for l, l = Total length L: L = l + (gap) + l = 6.0417 + 1 + 6.0417 4 L = 12.33 in.  PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 1.17 A load P is applied to a steel rod supported as shown by an aluminum plate into which a 0.6-in.-diameter hole has been drilled. Knowing that the shearing stress must not exceed 18 ksi in the steel rod and 10 ksi in the aluminum plate, determine the largest load P that can be applied to the rod. SOLUTION A1 = π dt = π (0.6)(0.4) For steel: = 0.7540 in 2 τ1 = P ∴ P = A1τ1 = (0.7540)(18) A = 13.57 kips A2 = π dt = π (1.6)(0.25) = 1.2566 in 2 For aluminum: τ2 = P ∴ P = A2τ 2 = (1.2566)(10) = 12.57 kips A2 Limiting value of P is the smaller value, so P = 12.57 kips  PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 1.18 Two wooden planks, each 22 mm thick and 160 mm wide, are joined by the glued mortise joint shown. Knowing that the joint will fail when the average shearing stress in the glue reaches 820 kPa, determine the smallest allowable length d of the cuts if the joint is to withstand an axial load of magnitude P = 7.6 kN. SOLUTION Seven surfaces carry the total load P = 7.6 kN = 7.6 × 103. Let t = 22 mm. Each glue area is A = dt τ = P 7A A= P 7.6 × 103 = = 1.32404 × 10−3 m 2 7τ (7)(820 × 103 ) = 1.32404 × 103 mm 2 d = A 1.32404 × 103 = = 60.2 t 22 d = 60.2 mm  PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 1.19 The load P applied to a steel rod is distributed to a timber support by an annular washer. The diameter of the rod is 22 mm and the inner diameter of the washer is 25 mm, which is slightly larger than the diameter of the hole. Determine the smallest allowable outer diameter d of the washer, knowing that the axial normal stress in the steel rod is 35 MPa and that the average bearing stress between the washer and the timber must not exceed 5 MPa. SOLUTION Steel rod: A = π 4 (0.022) 2 = 380.13 × 10−6 m 2 σ = 35 × 106 Pa P = σ A = (35 × 106 )(380.13 × 10−6 ) = 13.305 × 103 N Washer: σ b = 5 × 106 Pa Required bearing area: Ab = But, Ab = π 4 P σb = 13.305 × 103 = 2.6609 × 10−3 m 2 5 × 106 (d 2 − di2 ) d 2 = di2 + 4 Ab π = (0.025)2 + (4)(2.6609 × 10−3 ) π −3 = 4.013 × 10 m d = 63.3 × 10−3 m 2 d = 63.3 mm  PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 1.20 The axial force in the column supporting the timber beam shown is P = 20 kips. Determine the smallest allowable length L of the bearing plate if the bearing stress in the timber is not to exceed 400 psi. SOLUTION Bearing area: Ab = Lw σb = L= P P = Ab Lw 20 × 103 P = = 8.33 in. σ b w (400)(6) L = 8.33 in.  PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 1.21 An axial load P is supported by a short W8 × 40 column of crosssectional area A = 11.7 in.2 and is distributed to a concrete foundation by a square plate as shown. Knowing that the average normal stress in the column must not exceed 30 ksi and that the bearing stress on the concrete foundation must not exceed 3.0 ksi, determine the side a of the plate that will provide the most economical and safe design. SOLUTION For the column σ = P or A P = σ A = (30)(11.7) = 351 kips For the a × a plate, σ = 3.0 ksi A= P σ = 351 = 117 in 2 3.0 Since the plate is square, A = a 2 a = A = 117 a = 10.82 in.  PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 1.22 A 40-kN axial load is applied to a short wooden post that is supported by a concrete footing resting on undisturbed soil. Determine (a) the maximum bearing stress on the concrete footing, (b) the size of the footing for which the average bearing stress in the soil is 145 kPa. SOLUTION (a) Bearing stress on concrete footing. P = 40 kN = 40 × 103 N A = (100)(120) = 12 × 103 mm 2 = 12 × 10−3 m 2 σ = (b) 40 × 103 P = = 3.333 × 106 Pa A 12 × 10−3 σ = 145 kPa = 45 × 103 Pa Footing area. P = 40 × 103 N σ = P A A= P σ = 40 × 103 = 0.27586 m 2 145 × 103 Since the area is square, A = b 2 b= 3.33 MPa  A = 0.27586 = 0.525 m b = 525 mm  PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 1.23 A 58 -in.-diameter steel rod AB is fitted to a round hole near end C of the wooden member CD. For the loading shown, determine (a) the maximum average normal stress in the wood, (b) the distance b for which the average shearing stress is 100 psi on the surfaces indicated by the dashed lines, (c) the average bearing stress on the wood. SOLUTION (a) (b) Maximum normal stress in the wood 5  Anet = (1)  4 −  = 3.375 in.2 8  1500 P = = 444 psi σ = 3.375 Anet σ = 444 psi  Distance b for τ = 100 psi For sheared area see dotted lines. P P = A 2bt 1500 P = = 7.50 in. b= 2tτ (2)(1)(100) τ = (c) b = 7.50 in.  Average bearing stress on the wood σb = 1500 P P = = = 2400 psi 5 Ab dt   (1) 8 σ b = 2400 psi  PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 1.24 Knowing that θ = 40° and P = 9 kN, determine (a) the smallest allowable diameter of the pin at B if the average shearing stress in the pin is not to exceed 120 MPa, (b) the corresponding average bearing stress in member AB at B, (c) the corresponding average bearing stress in each of the support brackets at B. SOLUTION Geometry: Triangle ABC is an isoseles triangle with angles shown here. Use joint A as a free body. Law of sines applied to force triangle P FAB FAC = = sin 20° sin110° sin 50° P sin110° FAB = sin 20° (9)sin110° = = 24.73 kN sin 20° PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 1.24 (Continued) (a) Allowable pin diameter. τ = 2 FAB FAB F = πAB 2 = where FAB = 24.73 × 103 N 2 2 AP πd 24d d2 = 2 FAB πτ = (2)(24.73 × 103 ) = 131.18 × 10−6 m 2 6 π (120 × 10 ) d = 11.45 × 10−3 m (b) 11.45 mm  Bearing stress in AB at A. Ab = td = (0.016)(11.45 × 10−3 ) = 183.26 × 10−6 m 2 σb = (c) FAB 24.73 × 103 = = 134.9 × 106 Ab 183.26 × 10−6 134.9 MPa  Bearing stress in support brackets at B. A = td = (0.012)(11.45 × 10−3 ) = 137.4 × 10−6 m 2 σb = 1 2 FAB A = (0.5)(24.73 × 103 ) = 90.0 × 106 137.4 × 10−6 90.0 MPa  PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 1.25 Determine the largest load P which may be applied at A when θ = 60°, knowing that the average shearing stress in the 10-mm-diameter pin at B must not exceed 120 MPa and that the average bearing stress in member AB and in the bracket at B must not exceed 90 MPa. SOLUTION Geometry: Triangle ABC is an isoseles triangle with angles shown here. Use joint A as a free body. Law of sines applied to force triangle P FAB FAC = = sin 30° sin 120° sin 30° P= FAB sin 30° = 0.57735 FAB sin 120° P= FAC sin 30° = FAC sin 30° PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 1.25 (Continued) If shearing stress in pin at B is critical, A= π 4 d2 = π 4 (0.010) 2 = 78.54 × 10−6 m 2 FAB = 2 Aτ = (2)(78.54 × 10−6 )(120 × 106 ) = 18.850 × 103 N If bearing stress in member AB at bracket at A is critical, Ab = td = (0.016)(0.010) = 160 × 10−6 m 2 FAB = Abσ b = (160 × 10−6 )(90 × 106 ) = 14.40 × 103 N If bearing stress in the bracket at B is critical, Ab = 2td = (2)(0.012)(0.010) = 240 × 10−6 m 2 FAB = Abσ b = (240 × 10−6 )(90 × 106 ) = 21.6 × 103 N Allowable FAB is the smallest, i.e., 14.40 × 103N Then from Statics Pallow = (0.57735)(14.40 × 103 ) = 8.31 × 103 N 8.31 kN  PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 1.26 Link AB, of width b = 50 mm and thickness t = 6 mm, is used to support the end of a horizontal beam. Knowing that the average normal stress in the link is − 140 MPa, and that the average shearing stress in each of the two pins is 80 MPa, determine (a) the diameter d of the pins, (b) the average bearing stress in the link. SOLUTION Rod AB is in compression. A = bt where b = 50 mm and t = 6 mm A = (0.050)(0.006) = 300 × 10−6 m 2 P = −σ A = −(−140 × 106 )(300 × 10−6 ) = 42 × 103 N For the pin, Ap = π Ap = (a) P τ = P Ap 42 × 103 = 525 × 10−6 m 2 80 × 106 Diameter d d = (b) 4 and τ = d2 Bearing stress σb = 4 Ap π = (4)(525 × 10 −6 ) π = 2.585 × 10−3 m P 42 × 103 = = 271 × 106 Pa dt (25.85 × 10−3 )(0.006) d = 25.9 mm  σ b = 271 MPa  PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 1.27 For the assembly and loading of Prob. 1.7, determine (a) the average shearing stress in the pin at B, (b) the average bearing stress at B in member BD, (c) the average bearing stress at B in member ABC, knowing that this member has a 10 × 50-mm uniform rectangular cross section. PROBLEM 1.7 Each of the four vertical links has an 8 × 36-mm uniform rectangular cross section and each of the four pins has a 16-mm diameter. Determine the maximum value of the average normal stress in the links connecting (a) points B and D, (b) points C and E. SOLUTION Use bar ABC as a free body. ΣM C = 0 : (0.040) FBD − (0.025 + 0.040)(20 × 103 ) = 0 FBD = 32.5 × 103 N (a) Shear pin at B τ = where A= τ = (b) Bearing: link BD Bearing in ABC at B π 4 d2 = π 4 (0.016) 2 = 201.06 × 10−6 m 2 32.5 × 103 = 80.8 × 106 (2)(201.06 × 10−6 ) τ = 80.8 MPa  A = dt = (0.016)(0.008) = 128 × 10−6 m 2 σb = (c) FBD for double shear, 2A 1 2 FBD A = (0.5)(32.5 × 103 ) = 126.95 × 106 −6 128 × 10 σ b = 127.0 MPa  A = dt = (0.016)(0.010) = 160 × 10−6 m 2 σb = 32.5 × 103 FBD = = 203 × 106 −6 A 160 × 10 σ b = 203 MPa  PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 1.28 The hydraulic cylinder CF, which partially controls the position of rod DE, has been locked in the position shown. Member BD is 58 in. thick and is connected to the vertical rod by a 83 -in.-diameter bolt. Determine (a) the average shearing stress in the bolt, (b) the bearing stress at C in member BD. SOLUTION Use member BCD as a free body, and note that AB is a two force member. l AB = 82 + 1.82 = 8.2 in.  8   1.8  ΣM C = 0: (4 cos 20°)  FAB  − (4sin 20°)  FAB   8.2   8.2  −(7 cos 20°)(400sin 75°) − (7sin 20°)(400 cos 75°) = 0 3.36678FAB − 2789.35 = 0 ∴ FAB = 828.49 lb 1.8 FAB + Cx + 400cos 75° = 0 8.2 (1.8)(828.49) − 400cos 75° = 78.34 lb Cx = 8.2 ΣFx = 0: − 8 FAB + C y − 400sin 75° = 0 8.2 (8)(828.49) + 400sin 75° = 1194.65 lb Cy = 8.2 ΣFy = 0: − C = Cx2 + C y2 = 1197.2 lb PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 1.28 (Continued) A= π 2 π 3 d =   = 0.11045 in 2 4 4 8 τ = 1197.2 P = = 10.84 × 103 psi = A 0.11045 2 (a) (b) Shearing stress in the bolt: P = 1197.2 lb 10.84 ksi   3  5  Bearing stress at C in member BCD: P = 1197.2 lb Ab = dt =    = 0.234375 in 2  8  8  σb = P 1197.2 = = 5.11 × 103 psi = 5.11 ksi  0.234375 Ab PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 1.29 The 1.4-kip load P is supported by two wooden members of uniform cross section that are joined by the simple glued scarf splice shown. Determine the normal and shearing stresses in the glued splice. SOLUTION P = 1400 lb θ = 90° − 60° = 30° A0 = (5.0)(3.0) = 15 in 2 σ = P cos 2 θ (1400)(cos 30°)2 = A0 15 σ = 70.0 psi  τ = P sin 2θ (1400)sin 60° = 2 A0 (2)(15) τ = 40.4 psi  PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 1.30 Two wooden members of uniform cross section are joined by the simple scarf splice shown. Knowing that the maximum allowable tensile stress in the glued splice is 75 psi, determine (a) the largest load P that can be safely supported, (b) the corresponding shearing stress in the splice. SOLUTION A0 = (5.0)(3.0) = 15 in 2 θ = 90° − 60° = 30° σ = (a) (b) P= P cos 2 θ A0 σ A0 (75)(15) = = 1500 lb 2 cos θ cos 2 30° P = 1.500 kips  P sin 2θ (1500)sin 60° = 2 A0 (2)(15) τ = 43.3 psi  τ = PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 1.31 Two wooden members of uniform rectangular cross section are joined by the simple glued scarf splice shown. Knowing that P = 11 kN, determine the normal and shearing stresses in the glued splice. SOLUTION θ = 90° − 45° = 45° P = 11 kN = 11 × 103 N A0 = (150)(75) = 11.25 × 103 mm 2 = 11.25 × 10−3 m 2 σ = P cos 2 θ (11 × 103 ) cos 2 45° = = 489 × 103 Pa −3 A0 11.25 × 10 σ = 489 kPa  τ = (11 × 103 )(sin 90°) P sin 2θ = = 489 × 103 Pa 2 A0 (2)(11.25 × 10−3 ) τ = 489 kPa  PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 1.32 Two wooden members of uniform rectangular cross section are joined by the simple glued scarf splice shown. Knowing that the maximum allowable shearing stress in the glued splice is 620 kPa, determine (a) the largest load P that can be safely applied, (b) the corresponding tensile stress in the splice. SOLUTION θ = 90° − 45° = 45° A0 = (150)(75) = 11.25 × 103 mm 2 = 11.25 × 10−3 m2 τ = 620 kPa = 620 × 103 Pa P sin 2θ τ = 2 A0 (a) P= 2 A0τ (2)(11.25 × 10−3 )(620 × 103 ) = sin2θ sin 90° = 13.95 × 103 N (b) σ = P cos 2 θ (13.95 × 103 )(cos 45°)2 = A0 11.25 × 10−3 = 620 × 103 Pa P = 13.95 kN  σ = 620 kPa  PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 1.33 A steel pipe of 12-in. outer diameter is fabricated from 14 -in.-thick plate by welding along a helix that forms an angle of 25° with a plane perpendicular to the axis of the pipe. Knowing that the maximum allowable normal and shearing stresses in the directions respectively normal and tangential to the weld are σ = 12 ksi and τ = 7.2 ksi, determine the magnitude P of the largest axial force that can be applied to the pipe. SOLUTION 1 d o = 6 in. 2 ri = ro − t = 6 − 0.25 = 5.75 in. do = 12 in. ro = A0 = π (ro2 − ri2 ) = π (62 − 5.752 ) = 9.228 in 2 θ = 25° Based on σ = 12 ksi: σ = P cos 2 θ A0 P= Based on τ = 7.2 ksi: τ = P= (9.228)(12 × 103 ) A0σ = = 134.8 × 103 lb cos 2 θ cos 2 25° P sin 2θ 2 A0 2 A0τ (2)(9.288)(7.2 × 103 ) = = 174.5 × 103 lb sin 2θ sin 50° The smaller calculated value of P is the allowable value. P = 134.8 × 103 lb P = 134.8 kips  PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 1.34 A steel pipe of 12-in. outer diameter is fabricated from 14 -in.-thick plate by welding along a helix that forms an angle of 25° with a plane perpendicular to the axis of the pipe. Knowing that a 66 kip axial force P is applied to the pipe, determine the normal and shearing stresses in directions respectively normal and tangential to the weld. SOLUTION 1 d o = 6 in. 2 ri = ro − t = 6 − 0.25 = 5.75 in. do = 12 in. ro = A0 = π (ro2 − ri2 ) = π (62 − 5.752 ) = 9.228 in 2 θ = 25° Normal stress: Shearing stress: σ = τ = P cos 2 θ (66 × 103 ) cos 2 25° = = 5875 psi A0 9.228 σ = 5.87 ksi  P sin 2θ (66 × 103 )sin 50° = = 2739 psi 2 A0 (2)(9.228) τ = 2.74 ksi  PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 1.35 A 1060-kN load P is applied to the granite block shown. Determine the resulting maximum value of (a) the normal stress, (b) the shearing stress. Specify the orientation of the plane on which each of these maximum values occurs. SOLUTION A0 = (140 mm)(140 mm) = 19.6 × 103 mm 2 = 19.6 × 10−3 m 2 P = 1060 × 103 N σ = (a) P 1060 × 103 cos 2 θ = cos 2 θ = 54.082 × 106 cos 2 θ −3 A0 19.6 × 10 Maximum tensile stress = 0 at θ = 90°. Maximum compressive stress = 54.1 × 106 at θ = 0°. (b) |σ |max = 54.1 MPa  Maximum shearing stress: τ max = P 1060 × 103 = = 27.0 × 106 Pa at θ = 45°. 2 A0 (2)(19.6 × 10−3 ) τ max = 27.0 MPa  PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 1.36 A centric load P is applied to the granite block shown. Knowing that the resulting maximum value of the shearing stress in the block is 18 MPa, determine (a) the magnitude of P, (b) the orientation of the surface on which the maximum shearing stress occurs, (c) the normal stress exerted on that surface, (d) the maximum value of the normal stress in the block. SOLUTION A0 = (140 mm)(140 mm) = 19.6 × 103 mm 2 = 19.6 × 10−3 m 2 τ max = 18 MPa = 18 × 106 Pa θ = 45° for plane of τ max (a) Magnitude of P. τ max = |P| so P = 2 A0 τ max 2 A0 P = 706 kN  P = (2)(19.6 × 10−3 )(18 × 106 ) = 705.6 × 103 N Orientation. (c) Normal stress at θ = 45°. σ = (d) θ = 45°  sin 2θ is maximum when 2θ = 90° (b) P cos 2 θ (705.8 × 103 ) cos 2 45° = = 18.00 × 106 Pa A0 19.6 × 10−3 Maximum normal stress: σ max = σ max = σ = 18.00 MPa  P A0 705.8 × 103 = 36.0 × 106 Pa 19.6 × 10−3 σ max = 36.0 MPa (compression)  PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 1.37 Link BC is 6 mm thick, has a width w = 25 mm, and is made of a steel with a 480-MPa ultimate strength in tension. What was the safety factor used if the structure shown was designed to support a 16-kN load P? SOLUTION Use bar ACD as a free body and note that member BC is a two-force member. ΣM A = 0: (480) FBC − (600) P = 0 FBC = Ultimate load for member BC: 600 (600)(16 × 103 ) = 20 × 103 N P= 480 480 FU = σ U A FU = (480 × 106 )(0.006)(0.025) = 72 × 103 N Factor of safety: F.S. = 72 × 103 FU = FBC 20 × 103 F.S. = 3.60  PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 1.38 Link BC is 6 mm thick and is made of a steel with a 450-MPa ultimate strength in tension. What should be its width w if the structure shown is being designed to support a 20-kN load P with a factor of safety of 3? SOLUTION Use bar ACD as a free body and note that member BC is a two-force member. ΣM A = 0: (480) FBC − 600 P = 0 FBC = 600 P (600)(20 × 103 ) = = 25 × 103 N 480 480 For a factor of safety F.S. = 3, the ultimate load of member BC is FU = (F.S.)( FBC ) = (3)(25 × 103 ) = 75 × 103 N But FU = σ U A ∴A = For a rectangular section FU σU = 75 × 103 = 166.67 × 10−6 m 2 450 × 106 A = wt or w = A 166.67 × 10−6 = t 0.006 w = 27.8 × 10−3 m or 27.8 mm  PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 1.39 A 34 -in.-diameter rod made of the same material as rods AC and AD in the truss shown was tested to failure and an ultimate load of 29 kips was recorded. Using a factor of safety of 3.0, determine the required diameter (a) of rod AC, (b) of rod AD. SOLUTION Forces in AC and AD. Joint C: ΣFy = 0: Joint D: Ultimate stress. From test on ΣFy = 0: 3 4 σU = -in. rod: σ all = Allowable stress: (a) Diameter of rod AC. (b) Diameter of rod AD. σ all = FAC 1 πd2 4 d2 = d2 = 1 FAC − 10 kips = 0 5 FAC = 22.36 kips T 1 FAD − 10 kips = 0 17 FAD = 41.23 kips T PU 29 kips = 1 3 2 = 65.64 ksi A π (4) 4 σU F .S . 4FAC πσ all 4 FAD πσ all 65.64 ksi = 21.88 ksi 3.0 = = 4(22.36) = 1.301 π (21.88) = 4(41.23) = 2.399 π (21.88) d = 1.141 in.  d = 1.549 in.  PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 1.40 In the truss shown, members AC and AD consist of rods made of the same metal alloy. Knowing that AC is of 1-in. diameter and that the ultimate load for that rod is 75 kips, determine (a) the factor of safety for AC, (b) the required diameter of AD if it is desired that both rods have the same factor of safety. SOLUTION Forces in AC and AD. Joint C: ΣFy = 0: Joint D: ΣFy = 0: F.S. = PU FAC F.S. = 1 FAC − 10 kips = 0 5 FAC = 22.36 kips T 1 FAD − 10 kips = 0 17 FAD = 41.23 kips T 75 kips 22.36 kips (a) Factor of safety for AC. (b) For the same factor of safety in AC and AD, σ AD = σ AC . F.S. = 3.35  FAD F = AC AAD AAC AAD = Required diameter: d AD = 41.23 π 2 FAD (1) = 1.4482 in 2 AAC = 22.36 4 FAC 4 AAD π = (4)(1.4482) π d AD = 1.358 in.  PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 1.41 Link AB is to be made of a steel for which the ultimate normal stress is 450 MPa. Determine the cross-sectional area for AB for which the factor of safety will be 3.50. Assume that the link will be adequately reinforced around the pins at A and B. SOLUTION P = (1.2)(8) = 9.6 kN ΣM D = 0 : −(0.8)( FAB sin 35°) + (0.2)(9.6) + (0.4)(20) = 0 FAB = 21.619 kN = 21.619 × 103 N σ AB = AAB = FAB σ = ult AAB F. S . ( F. S.) FAB σ ult = (3.50)(21.619 × 103 ) 450 × 106 = 168.1 × 10−6 m 2 AAB = 168.1 mm 2  PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 1.42 A steel loop ABCD of length 1.2 m and of 10-mm diameter is placed as shown around a 24-mm-diameter aluminum rod AC. Cables BE and DF, each of 12-mm diameter, are used to apply the load Q. Knowing that the ultimate strength of the steel used for the loop and the cables is 480 MPa and that the ultimate strength of the aluminum used for the rod is 260 MPa, determine the largest load Q that can be applied if an overall factor of safety of 3 is desired. SOLUTION Using joint B as a free body and considering symmetry, 2⋅ 3 6 FAB − Q = 0 Q = FAB 5 5 Using joint A as a free body and considering symmetry, 4 FAB − FAC = 0 5 8 5 3 ⋅ Q − FAC = 0 ∴ Q = FAC 5 6 4 2⋅ Based on strength of cable BF: QU = σ U A = σ U π 4 d 2 = (480 × 106 ) π 4 (0.012) 2 = 54.29 × 103 N Based on strength of steel loop: 6 6 6 π FAB, U = σ U A = σ U d 2 5 5 5 4 π 6 = (480 × 106 ) (0.010) 2 = 45.24 × 103 N 5 4 QU = Based on strength of rod AC: QU = 3 3 3 3 π π FAC , U = σ U A = σ U d 2 = (260 × 106 ) (0.024) 2 = 88.22 × 103 N 4 4 4 4 4 4 Actual ultimate load QU is the smallest, ∴ QU = 45.24 × 103 N Allowable load: Q= QU 45.24 × 103 = = 15.08 × 103 N F. S . 3 Q = 15.08 kN  PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 1.43 Two wooden members shown, which support a 3.6 kip load, are joined by plywood splices fully glued on the surfaces in contact. The ultimate shearing stress in the glue is 360 psi and the clearance between the members is 14 in. Determine the required length L of each splice if a factor of safety of 2.75 is to be achieved. SOLUTION There are 4 separate areas of glue. Let l be the length of each area and w = 5 in. its width. Then the area is A = lw. Each glue area transmits one half of the total load. 1 F =   (3.6 kips) = 1.8 kips 2 Required ultimate load for each glue area: FU = ( F . S .) F = (2.75)(1.8) = 4.95 kips Required length of each glue area: FU = τ U A = τ U lw l = Total length of splice: FU 4.95 × 103 = = 2.75 in. τU w (360)(5) L=l+ 1 in. + l 4 L = 2.75 + 0.25 + 2.75 L = 5.75 in.  PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 1.44 Two plates, each 18 in. thick, are used to splice a plastic strip as shown. Knowing that the ultimate shearing stress of the bonding between the surface is 130 psi, determine the factor of safety with respect to shear when P = 325 lb. SOLUTION Bond area: (See figure) 1 (2.25)(0.75) + (2.25)(0.625) = 2.25 in 2 2 PU = 2 AτU = (2)(2.25)(130) = 585 lb. A= F. S . = 585 PU = = 1.800  P 325 PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 1.45 A load P is supported as shown by a steel pin that has been inserted in a short wooden member hanging from the ceiling. The ultimate strength of the wood used is 60 MPa in tension and 7.5 MPa in shear, while the ultimate strength of the steel is 145 MPa in shear. Knowing that b = 40 mm, c = 55 mm, and d = 12 mm, determine the load P if an overall factor of safety of 3.2 is desired. SOLUTION Based on double shear in pin: PU = 2 AτU = 2 = π 4 π 4 d 2τ U (2)(0.012)2 (145 × 106 ) = 32.80 × 103 N Based on tension in wood: PU = Aσ U = w (b − d )σ U = (0.040)(0.040 − 0.012)(60 × 106 ) = 67.2 × 103 N Based on double shear in the wood: PU = 2 AτU = 2wcτU = (2)(0.040)(0.055)(7.5 × 106 ) = 33.0 × 103 N Use smallest Allowable: PU = 32.8 × 103 N P= 32.8 × 103 PU = = 10.25 × 103 N F .S. 3.2 10.25 kN  PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 1.46 For the support of Prob. 1.45, knowing that the diameter of the pin is d = 16 mm and that the magnitude of the load is P = 20 kN, determine (a) the factor of safety for the pin, (b) the required values of b and c if the factor of safety for the wooden members is the same as that found in part a for the pin. PROBLEM 1.45 A load P is supported as shown by a steel pin that has been inserted in a short wooden member hanging from the ceiling. The ultimate strength of the wood used is 60 MPa in tension and 7.5 MPa in shear, while the ultimate strength of the steel is 145 MPa in shear. Knowing that b = 40 mm, c = 55 mm, and d = 12 mm, determine the load P if an overall factor of safety of 3.2 is desired. SOLUTION P = 20 kN = 20 × 103 N (a) Pin: A= Double shear: τ = π 4 d2 = π 4 (0.016) 2 = 2.01.06 × 10−6 m 2 P P τU = U 2A 2A PU = 2 AτU = (2)(201.16 × 10−6 )(145 × 106 ) = 58.336 × 103 N F .S . = (b) Tension in wood: where w = 40 mm = 0.040 m b = 40.3 mm  PU = 58.336 × 103 N for same F.S. Double shear; each area is A = wc c= PU PU = A w(b − d ) 58.336 × 103 PU = 0.016 + = 40.3 × 10−3 m 6 wσ U (0.040)(60 × 10 ) Shear in wood: F .S . = 2.92  PU = 58.336 × 103 N for same F.S. σU = b=d + PU 58.336 × 103 = P 20 × 103 τU = PU P = U 2 A 2wc 58.336 × 103 PU = = 97.2 × 10−3 m 2wτU (2)(0.040)(7.5 × 106 ) c = 97.2 mm  PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 1.47 Three steel bolts are to be used to attach the steel plate shown to a wooden beam. Knowing that the plate will support a 110-kN load, that the ultimate shearing stress for the steel used is 360 MPa, and that a factor of safety of 3.35 is desired, determine the required diameter of the bolts. SOLUTION 110 = 36.667 kN 3 For each bolt, P= Required: PU = ( F. S.) P = (3.35)(36.667) = 122.83 kN τU = d = PU P 4P = π U 2 = U2 A d πd 4 4PU πτU = (4)(122.83 × 103 ) = 20.8 × 10−3 m π (360 × 106 ) d = 20.8 mm  PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 1.48 Three 18-mm-diameter steel bolts are to be used to attach the steel plate shown to a wooden beam. Knowing that the plate will support a 110-kN load and that the ultimate shearing stress for the steel used is 360 MPa, determine the factor of safety for this design. SOLUTION For each bolt, A= π 4 d2 = π 4 (18) 2 = 254.47 mm 2 = 254.47 × 10−6 m 2 PU = Aτ U = (254.47 × 10−6 )(360 × 106 ) = 91.609 × 103 N For the three bolts, PU = (3)(91.609 × 103 ) = 274.83 × 103 N Factor of safety: F. S . = PU 274.83 × 103 = P 110 × 103 F. S . = 2.50  PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 1.49 A steel plate 165 in. thick is embedded in a horizontal concrete slab and is used to anchor a high-strength vertical cable as shown. The diameter of the hole in the plate is 34 in., the ultimate strength of the steel used is 36 ksi, and the ultimate bonding stress between plate and concrete is 300 psi. Knowing that a factor of safety of 3.60 is desired when P = 2.5 kips, determine (a) the required width a of the plate, (b) the minimum depth b to which a plate of that width should be embedded in the concrete slab. (Neglect the normal stresses between the concrete and the lower end of the plate.) SOLUTION Based on tension in plate: A = (a − d )t PU = σ U A F .S . = σ (a − d )t PU = U P P Solving for a, a =d + ( F .S .) P 3 (3.60)(2.5) = + 4 σU t (36) 165 ( ) (a) a = 1.550 in.  Based on shear between plate and concrete slab, τU = 0.300 ksi A = perimeter × depth = 2(a + t )b PU = τ U A = 2τ U (a + t )b Solving for b, b= F .S. = PU P ( F .S.) P (3.6)(2.5) = 2(a + t )τU (2) 1.550 + 165 (0.300) ( ) (b) b = 8.05 in.  PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 1.50 Determine the factor of safety for the cable anchor in Prob. 1.49 when P = 3 kips, knowing that a = 2 in. and b = 7.5 in. PROBLEM 1.49 A steel plate 165 in, thick is embedded in a horizontal concrete slab and is used to anchor a high-strength vertical cable as shown. The diameter of the hole in the plate is 34 in., the ultimate strength of the steel used is 36 ksi, and the ultimate bonding stress between plate and concrete is 300 psi. Knowing that a factor of safety of 3.60 is desired when P = 2.5 kips, determine (a) the required width a of the plate, (b) the minimum depth b to which a plate of that width should be embedded in the concrete slab. (Neglect the normal stresses between the concrete and the lower end of the plate.) SOLUTION Based on tension in plate: A = (a − d )t 3  5   =  2 −   = 0.3906 in 2 4  16   PU = σ U A = (36)(0.3906) = 14.06 kips F .S . = 14.06 PU = = 4.69 3 P Based on shear between plate and concrete slab: 5  A = perimeter × depth = 2(a + t )b = 2  2 +  (7.5) 16   A = 34.69 in 2 τU = 0.300 ksi PU = τ U A = (0.300)(34.69) = 10.41 kips F .S . = 10.41 PU = = 3.47 3 P Actual factor of safety is the smaller value. F .S . = 3.47  PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 1.51 In the steel structure shown, a 6-mm-diameter pin is used at C and 10-mm-diameter pins are used at B and D. The ultimate shearing stress is 150 MPa at all connections, and the ultimate normal stress is 400 MPa in link BD. Knowing that a factor of safety of 3.0 is desired, determine the largest load P that can be applied at A. Note that link BD is not reinforced around the pin holes. SOLUTION Use free body ABC. ΣM C = 0 : 0.280 P − 0.120 FBD = 0 P= 3 FBD 7 (1) ΣM B = 0 : 0.160 P − 0.120 C = 0 P= 3 C 4 (2) Tension on net section of link BD.  400 × 106  −3 −3 3 Anet =   (6 × 10 )(18 − 10)(10 ) = 6.40 × 10 N 3 F. S .   Shear in pins at B and D. FBD = σ Anet = FBD = τ Apin = σU  150 × 106   π  −3 2 3 d 2 =     (10 × 10 ) = 3.9270 × 10 N 3 4 F. S. 4    τU π Smaller value of FBD is 3.9270 × 103 N. From (1) 3 P =   (3.9270 × 103 ) = 1.683 × 103 N 7 Shear in pin at C. C = 2τ Apin = 2 From (2) 3 P =   (2.8274 × 103 ) = 2.12 × 103 N 4 Smaller value of P is allowable value.  150 × 106   π  −3 2 3 d 2 = (2)     (6 × 10 ) = 2.8274 × 10 N 3 4 F. S. 4    τU π P = 1.683 × 103 N P = 1.683 kN  PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 1.52 Solve Prob. 1.51, assuming that the structure has been redesigned to use 12-mm-diameter pins at B and D and no other change has been made. PROBLEM 1.51 In the steel structure shown, a 6-mm-diameter pin is used at C and 10-mmdiameter pins are used at B and D. The ultimate shearing stress is 150 MPa at all connections, and the ultimate normal stress is 400 MPa in link BD. Knowing that a factor of safety of 3.0 is desired, determine the largest load P that can be applied at A. Note that link BD is not reinforced around the pin holes. SOLUTION Use free body ABC. ΣM C = 0 : 0.280 P − 0.120 FBD = 0 P= 3 FBD 7 (1) ΣM B = 0 : 0.160 P − 0.120 C = 0 P= Tension on net section of link BD. FBD = σ Anet = Shear in pins at B and D. FBD = τ Apin = 3 C 4 (2)  400 × 106  −3 −3 3 Anet =   (6 × 10 )(18 − 12)(10 ) = 4.80 × 10 N 3 F. S.   σU  150 × 106   π  −3 2 3 d 2 =     (12 × 10 ) = 5.6549 × 10 N 3 4 F. S. 4     τU π Smaller value of FBD is 4.80 × 103 N. From (1), 3 P =   (4.80 × 103 ) = 2.06 × 103 N 7 Shear in pin at C. C = 2τ Apin = 2 From (2),  150 × 106   π  −3 2 3 d 2 = (2)     (6 × 10 ) = 2.8274 × 10 N 3 4 F. S. 4     τU π 3 P =   (2.8274 × 103 ) = 2.12 × 103 N 4 Smaller value of P is the allowable value. P = 2.06 × 103 N P = 2.06 kN  PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 1.53 Each of the two vertical links CF connecting the two horizontal members AD and EG has a uniform rectangular cross section 14 in. thick and 1 in. wide, and is made of a steel with an ultimate strength in tension of 60 ksi. The pins at C and F each have a 12 -in. diameter and are made of a steel with an ultimate strength in shear of 25 ksi. Determine the overall factor of safety for the links CF and the pins connecting them to the horizontal members. SOLUTION Use member EFG as free body. ΣM E = 0 : 16 FCF − (26)(2) = 0 FCF = 3.25 kips Failure by tension in links CF. (2 parallel links) Net section area for 1 link: A = (b − d ) t = (1 − 12 )( 14 ) = 0.125 in 2 FU = 2 Aσ U = (2)(0.125)(60) = 15 kips Failure by double shear in pins. A= π 4 2   = 0.196350 in 42 = (2)(0.196350)(25) = 9.8175 kips d = FU = 2 AτU π 1 2 2 Actual ultimate load is the smaller value. FU = 9.8175 kips Factor of safety: F. S . = 9.8175 FU = 3.25 FCF F. S . = 3.02  PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 1.54 Solve Prob. 1.53, assuming that the pins at C and F have been replaced by pins with a 34 -in diameter. PROBLEM 1.53 Each of the two vertical links CF connecting the two horizontal members AD and EG has a uniform rectangular cross section 14 in. thick and 1 in. wide, and is made of a steel with an ultimate strength in tension of 60 ksi. The pins at C and F each have a 12 -in. diameter and are made of a steel with an ultimate strength in shear of 25 ksi. Determine the overall factor of safety for the links CF and the pins connecting them to the horizontal members. SOLUTION Use member EFG as free body. ΣM E = 0 : 16 FCF − (26)(2) = 0 FCF = 3.25 kips Failure by tension in links CF. (2 parallel links) A = (b − d ) t = (1 − 34 )( 14 ) = 0.0625 in 2 Net section area for 1 link: FU = 2 Aσ U = (2)(0.0625)(60) = 7.5 kips Failure by double shear in pins. A= π 4 d2 = π 3 2   = 0.44179 in 44 2 FU = 2 AτU = (2)(0.44179)(25) = 22.09 kips Actual ultimate load is the smaller value. FU = 7.5 kips Factor of safety: F. S . = 7.5 FU = 3.25 FCF F. S . = 2.31  PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 1.55 In the structure shown, an 8-mm-diameter pin is used at A, and 12-mm-diameter pins are used at B and D. Knowing that the ultimate shearing stress is 100 MPa at all connections and that the ultimate normal stress is 250 MPa in each of the two links joining B and D, determine the allowable load P if an overall factor of safety of 3.0 is desired. SOLUTION Statics: Use ABC as free body. ΣM B = 0 : 0.20 FA − 0.18 P = 0 ΣM A = 0 : 0.20 FBD − 0.38 P = 0 Based on double shear in pin A: A = FA = 2τ U A = π 4 10 FA 9 P= d2 = P= π 10 FBD 19 (0.008) 2 = 50.266 × 10−6 m 2 4 (2)(100 × 106 )(50.266 × 10−6 ) = 3.351 × 103 N 3.0 F .S. 10 P= FA = 3.72 × 103 N 9 Based on double shear in pins at B and D: A = FBD = 2τ U A = π 4 d2 = π 4 (0.012) 2 = 113.10 × 10−6 m 2 (2)(100 × 106 )(113.10 × 10−6 ) = 7.54 × 103 N 3.0 F .S. 10 P= FBD = 3.97 × 103 N 19 Based on compression in links BD: For one link, A = (0.020)(0.008) = 160 × 10−6 m 2 2σ U A (2)(250 × 106 )(160 × 10−6 ) = = 26.7 × 103 N 3.0 F .S. 10 P= FBD = 14.04 × 103 N 19 FBD = Allowable value of P is smallest, ∴ P = 3.72 × 103 N P = 3.72 kN  PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 1.56 In an alternative design for the structure of Prob. 1.55, a pin of 10-mm-diameter is to be used at A. Assuming that all other specifications remain unchanged, determine the allowable load P if an overall factor of safety of 3.0 is desired. PROBLEM 1.55 In the structure shown, an 8-mmdiameter pin is used at A, and 12-mm-diameter pins are used at B and D. Knowing that the ultimate shearing stress is 100 MPa at all connections and that the ultimate normal stress is 250 MPa in each of the two links joining B and D, determine the allowable load P if an overall factor of safety of 3.0 is desired. SOLUTION Statics: Use ABC as free body. ΣM B = 0: 0.20 FA − 0.18 P = 0 ΣM A = 0: 0.20 FBD − 0.38 P = 0 Based on double shear in pin A: A = 10 FA 9 10 P= FBD 19 P= π 4 d2 = π 4 (0.010) 2 = 78.54 × 10−6 m 2 2τ U A (2)(100 × 106 )(78.54 × 10−6 ) = = 5.236 × 103 N 3.0 F .S. 10 P= FA = 5.82 × 103 N 9 FA = Based on double shear in pins at B and D: A = π 4 d2 = π 4 (0.012) 2 = 113.10 × 10−6 m 2 2τ U A (2)(100 × 106 )(113.10 × 10−6 ) = = 7.54 × 103 N 3.0 F .S. 10 P= FBD = 3.97 × 103 N 19 FBD = Based on compression in links BD: For one link, A = (0.020)(0.008) = 160 × 10−6 m 2 2σ U A (2)(250 × 106 )(160 × 10−6 ) = = 26.7 × 103 N F .S. 3.0 10 P= FBD = 14.04 × 103 N 19 FBD = Allowable value of P is smallest, ∴ P = 3.97 × 103 N P = 3.97 kN  PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 1.57 The Load and Resistance Factor Design method is to be used to select the two cables that will raise and lower a platform supporting two window washers. The platform weighs 160 lb and each of the window washers is assumed to weigh 195 lb with equipment. Since these workers are free to move on the platform, 75% of their total weight and the weight of their equipment will be used as the design live load of each cable. (a) Assuming a resistance factor φ = 0.85 and load factors γ D = 1.2 and γ L = 1.5, determine the required minimum ultimate load of one cable. (b) What is the conventional factor of safety for the selected cables? SOLUTION γ D PD + γ L PL = φ PU (a) PU = γ D PD + γ L PL φ 1  3  (1.2)  × 160  + (1.5)  × 2 × 195  2 4     = 0.85 PU = 629 lb  Conventional factor of safety. P = PD + PL = (b) F. S . = 629 PU = 372.5 P 1 × 160 + 0.75 × 2 × 195 = 372.5 lb 2 F. S . = 1.689  PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 1.58 A 40-kg platform is attached to the end B of a 50-kg wooden beam AB, which is supported as shown by a pin at A and by a slender steel rod BC with a 12-kN ultimate load. (a) Using the Load and Resistance Factor Design method with a resistance factor φ = 0.90 and load factors γ D = 1.25 and γ L = 1.6, determine the largest load that can be safely placed on the platform. (b) What is the corresponding conventional factor of safety for rod BC? SOLUTION 3 ΣM A = 0 : (2.4) P − 2.4W1 − 1.2W2 5 5 5 ∴ P = W1 + W2 3 6 For dead loading, W1 = (40)(9.81) = 392.4 N, W2 = (50)(9.81) = 490.5 N For live loading, W1 = mg W2 = 0 From which m= Design criterion. 5 5 PD =   (392.4) +   (490.5) = 1.0628 × 103 N 3 6 PL = 5 mg 3 3 PL 5 g γ D PD + γ L PL = φ PU PL = φ PU − γ D PD (0.90)(12 × 103 ) − (1.25)(1.0628 × 10−3 ) = γL 1.6 = 5.920 × 103 N (a) m= Allowable load. 3 5.92 × 103 5 9.81 m = 362 kg  Conventional factor of safety. P = PD + PL = 1.0628 × 103 + 5.920 × 103 = 6.983 × 103 N (b) F. S . = PU 12 × 103 = P 6.983 × 103 F. S . = 1.718  PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 1.59 A strain gage located at C on the surface of bone AB indicates that the average normal stress in the bone is 3.80 MPa when the bone is subjected to two 1200-N forces as shown. Assuming the cross section of the bone at C to be annular and knowing that its outer diameter is 25 mm, determine the inner diameter of the bone’s cross section at C. SOLUTION σ = Geometry: A= π 4 P P ∴ A= σ A (d12 − d 22 ) d 22 = d12 − 4A π = d12 − d 22 = (25 × 10−3 ) 2 − (4)(1200) π (3.80 × 106 ) = 222.9 × 10−6 m 2 d 2 = 14.93 × 10−3 m 4P πσ d 2 = 14.93 mm  PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 1.60 Two horizontal 5-kip forces are applied to pin B of the assembly shown. Knowing that a pin of 0.8-in. diameter is used at each connection, determine the maximum value of the average normal stress (a) in link AB, (b) in link BC. SOLUTION Use joint B as free body. Law of Sines 10 FAB FBC = = sin 45° sin 60° sin 95° FAB = 7.3205 kips FBC = 8.9658 kips Link AB is a tension member. Minimum section at pin. Anet = (1.8 − 0.8)(0.5) = 0.5 in 2 (a) Stress in AB : σ AB = 7.3205 FAB = 0.5 Anet σ AB = 14.64 ksi  Link BC is a compression member. Cross sectional area is A = (1.8)(0.5) = 0.9 in 2 (b) Stress in BC: σ BC = − FBC −8.9658 = 0.9 A σ BC = −9.96 ksi  PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 1.61 For the assembly and loading of Prob. 1.60, determine (a) the average shearing stress in the pin at C, (b) the average bearing stress at C in member BC, (c) the average bearing stress at B in member BC. PROBLEM 1.60 Two horizontal 5-kip forces are applied to pin B of the assembly shown. Knowing that a pin of 0.8-in. diameter is used at each connection, determine the maximum value of the average normal stress (a) in link AB, (b) in link BC. SOLUTION Use joint B as free body. Law of Sines 10 FAB FBC = = sin 45° sin 60° sin 95° (a) Shearing stress in pin at C. τ = FBC = 8.9658 kips FBC 2 AP π 4 d2 = π (0.8) 2 = 0.5026 in 2 4 8.9658 τ = = 8.92 (2)(0.5026) AP = τ = 8.92 ksi  PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 1.61 (Continued) (b) Bearing stress at C in member BC. σ b = FBC A A = td = (0.5)(0.8) = 0.4 in 2 σb = (c) Bearing stress at B in member BC. σ b = 8.9658 = 22.4 0.4 σ b = 22.4 ksi  FBC A A = 2td = 2(0.5)(0.8) = 0.8 in 2 σb = 8.9658 = 11.21 0.8 σ b = 11.21 ksi  PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 1.62 In the marine crane shown, link CD is known to have a uniform cross section of 50 × 150 mm. For the loading shown, determine the normal stress in the central portion of that link. SOLUTION W = (80 Mg)(9.81 m/s 2 ) = 784.8 kN Weight of loading: Free Body: Portion ABC  M A = 0: FCD (15 m) − W (28 m) = 0 28 28 (784.8 kN) W = 15 15 = +1465 kN FCD = FCD σ CD = FCD +1465 × 103 N = = +195.3 × 106 Pa (0.050 m)(0.150 m) A σ CD = +195.3 MPa  PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 1.63 Two wooden planks, each 12 in. thick and 9 in. wide, are joined by the dry mortise joint shown. Knowing that the wood used shears off along its grain when the average shearing stress reaches 1.20 ksi, determine the magnitude P of the axial load that will cause the joint to fail. SOLUTION Six areas must be sheared off when the joint fails. Each of these areas has dimensions being A= 5 8 in. × 1 2 in., its area 5 1 5 2 × = in = 0.3125 in 2 8 2 16 At failure, the force carried by each area is F = τ A = (1.20 ksi)(0.3125 in 2 ) = 0.375 kips Since there are six failure areas, P = 6F = (6)(0.375) P = 2.25 kips  PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 1.64 Two wooden members of uniform rectangular cross section of sides a = 100 mm and b = 60 mm are joined by a simple glued joint as shown. Knowing that the ultimate stresses for the joint are σ U = 1.26 MPa in tension and τU = 1.50 MPa in shear, and that P = 6 kN, determine the factor of safety for the joint when (a) α = 20°, (b) α = 35°, (c) α = 45°. For each of these values of α, also determine whether the joint will fail in tension or in shear if P is increased until rupture occurs. SOLUTION Let θ = 90° − α as shown. From the text book: or σ = P cos 2 θ A0 σ = P sin 2 α A0 (1) τ = P sin α cos α A0 (2) τ = P sin θ cos θ A0 A0 = ab = (100 mm) (60 mm) = 6000 mm 2 = 6 × 10−3 m 2 σ U = 1.26 × 106 Pa τU = 1.50 × 106 Pa Ultimate load based on tension across the joint: ( PU )σ = = σ U A0 (1.26 × 106 )(6 × 10−3 ) = sin 2 α sin 2 α 7560 7.56 = kN 2 sin α sin 2 α Ultimate load based on shear across the joint: ( PU )τ = = τU A0 (1.50 × 106 )(6 × 10−3 ) = sin α cos α sin α cos α 9000 9.00 kN = sin α cos α sin α cos α PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 1.64 (Continued) (a) 7.56 = 64.63 kN sin 2 20° 9.00 = ( PU )τ = = 28.00 kN sin 20° cos 20° α = 20° : ( PU )σ = The smaller value governs. The joint will fail in shear and PU = 28.00 kN. F .S . = (b) α = 35° : 28.00 PU = 6 P F .S . = 4.67  7.56 = 22.98 kN sin 2 35° 9.00 ( PU )τ = = 19.155 kN sin 35° cos 35° ( PU )σ = The joint will fail in shear and PU = 19.155 kN. F .S . = (c) 19.155 PU = 6 P F .S . = 3.19  7.56 = 15.12 kN sin 2 45° 9.00 ( PU )τ = = 18.00 kN sin 45° cos 45° α = 45°: ( PU )σ = The joint will fail in tension and PU = 15.12 kN. F .S . = 15.12 PU = 6 P F .S . = 2.52  PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 1.65 Member ABC, which is supported by a pin and bracket at C and a cable BD, was designed to support the 16-kN load P as shown. Knowing that the ultimate load for cable BD is 100 kN, determine the factor of safety with respect to cable failure. SOLUTION Use member ABC as a free body, and note that member BD is a two-force member. ΣM c = 0 : ( P cos 40°)(1.2) + ( P sin 40°)(0.6) − ( FBD cos 30°)(0.6) − ( FBD sin 30°)(0.4) = 0 1.30493P − 0.71962FBD = 0 FBD = 1.81335 P = (1.81335)(16 × 103 ) = 29.014 × 103 N FU = 100 × 103 N F. S . = 100 × 103 FU = FBD 29.014 × 103 F. S . = 3.45  PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 1.66 The 2000-lb load can be moved along the beam BD to any position between stops at E and F. Knowing that σ all = 6 ksi for the steel used in rods AB and CD, determine where the stops should be placed if the permitted motion of the load is to be as large as possible. SOLUTION Permitted member forces:  π  1  AB : ( FAB ) max = σ all AAB = (6)     4  2  = 1.17810 kips 2  π  5  CD : ( FCD ) max = σ all ACD = (6)     4  8  = 1.84078 kips 2 Use member BEFD as a free body. P = 2000 lb = 2.000 kips ΣM D = 0 : − (60) FAB + (60 − x E) P = 0 60 − xE = 60 FAB (60)(1.17810) = P 2.000 = 35.343 ΣM B = 0 : 60 FCD − x F P = 0  xF = 60FCD (60)(1.84078) = 2.000 P x E = 24.7 in.  x F = 55.2 in.  PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 1.67 Knowing that a force P of magnitude 750 N is applied to the pedal shown, determine (a) the diameter of the pin at C for which the average shearing stress in the pin is 40 MPa, (b) the corresponding bearing stress in the pedal at C, (c) the corresponding bearing stress in each support bracket at C. SOLUTION Draw free body diagram of BCD. Since BCD is a 3-force member, the reaction at C is directed toward Point E, the intersection of the lines of action of the other two forces. From geometry, CE = 3002 + 1252 = 325 mm + ↑ ΣFy = 0 : (a) τ pin = (b) σb = (c) σb = 1C 2 Apin = 1C 2 π d2 4 d = 125 C − P = 0 C = 2.6 P = (2.6)(750) = 1950 N 325 2C πτ pin = (2)(1950) = 5.57 × 10−3 m 6 π (40 × 10 ) 1950 C C = = = 38.9 × 106 Pa Ab dt (5.57 × 10−3 )(9 × 10−3 ) 1C 2 Ab = 1950 C = = 35.0 × 106 Pa 2dt (2)(5.57 × 10−3 )(5 × 10−3 ) d = 5.57 mm  σ b = 38.9 MPa  σ b = 35.0 MPa  PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 1.68 A force P is applied as shown to a steel reinforcing bar that has been embedded in a block of concrete. Determine the smallest length L for which the full allowable normal stress in the bar can be developed. Express the result in terms of the diameter d of the bar, the allowable normal stress σ all in the steel, and the average allowable bond stress τ all between the concrete and the cylindrical surface of the bar. (Neglect the normal stresses between the concrete and the end of the bar.) SOLUTION For shear, A = π dL P = τ all A = τ allπ dL For tension, Equating, Solving for L, A= π 4 d2 π  P = σ all A = σ all  d 2  4   τ allπ dL = σ all π 4 d2 Lmin = σ alld/4τ all  PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 1.69 The two portions of member AB are glued together along a plane forming an angle θ with the horizontal. Knowing that the ultimate stress for the glued joint is 2.5 ksi in tension and 1.3 ksi in shear, determine the range of values of θ for which the factor of safety of the members is at least 3.0. SOLUTION A0 = (2.0)(1.25) = 2.50 in.2 P = 2.4 kips PU = ( F. S .) P = 7.2 kips Based on tensile stress: σU = cos 2 θ = PU cos 2 θ A0 σ U A0 PU cos θ = 0.93169 Based on shearing stress: τU = sin 2θ = 2θ = 64.52° Hence, = (2.5)(2.50) = 0.86806 7.2 θ = 21.3° θ > 21.3° PU P sin θ cos θ = U sin 2θ A0 2 A0 2 A0τU PU = (2)(2.50)(1.3) = 0.90278 7.2 θ = 32.3° θ < 32.3° 21.3° < θ < 32.3°  PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 1.70 The two portions of member AB are glued together along a plane forming an angle θ with the horizontal. Knowing that the ultimate stress for the glued joint is 2.5 ksi in tension and 1.3 ksi in shear, determine (a) the value of θ for which the factor of safety of the member is maximum, (b) the corresponding value of the factor of safety. (Hint: Equate the expressions obtained for the factors of safety with respect to normal stress and shear.) SOLUTION A0 = (2.0)(1.25) = 2.50 in 2 At the optimum angle, ( F. S.)σ = ( F. S .)τ σ = σ U A0 P cos 2 θ ∴ PU ,σ = A0 cos 2 θ Normal stress: ( F. S.)σ = Shearing stress: τ = σ U A0 P cos θ Solving: (b) PU = 2 = P = σ U A0 P cos 2 θ τU A0 P sin θ cos θ ∴ PU ,τ = sin θ cos θ A0 ( F. S.)τ = Equating: PU ,σ PU ,τ P = τU A0 P sin θ cos θ τ U A0 P sin θ cos θ sin θ τ 1.3 = tan θ = U = = 0.520 σU cos θ 2.5 (a) θopt = 27.5°  (12.5)(2.50) σ U A0 = = 7.94 kips 2 cos θ cos 2 27.5° F. S . = 7.94 PU = 2.4 P F. S . = 3.31  PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.