SEVENTH EDITION

1496T_fm_i-xxvi 1/6/06 02:56 Page iii Materials Science and Engineering An Introduction 1496T_fm_i-xxvi 1/6/06 22:25 Page v SEVENTH EDITION Materials Science and Engineering An Introduction William D. Callister, Jr. Department of Metallurgical Engineering The University of Utah with special contributions by David G. Rethwisch The University of Iowa John Wiley & Sons, Inc. 1496T_fm_i-xxvi 1/11/06 23:05 Page vi Front Cover: A unit cell for diamond (blue-gray spheres represent carbon atoms), which is positioned above the temperature-versus-logarithm pressure phase diagram for carbon; highlighted in blue is the region for which diamond is the stable phase. Back Cover: Atomic structure for graphite; here the gray spheres depict carbon atoms. The region of graphite stability is highlighted in orange on the pressure-temperature phase diagram for carbon, which is situated behind this graphite structure. ACQUISITIONS EDITOR MARKETING DIRECTOR SENIOR PRODUCTION EDITOR SENIOR DESIGNER COVER ART TEXT DESIGN SENIOR ILLUSTRATION EDITOR COMPOSITOR ILLUSTRATION STUDIO Joseph Hayton Frank Lyman Ken Santor Kevin Murphy Roy Wiemann Michael Jung Anna Melhorn Techbooks/GTS, York, PA Techbooks/GTS, York, PA This book was set in 10/12 Times Ten by Techbooks/GTS, York, PA and printed and bound by Quebecor Versailles. The cover was printed by Quebecor. This book is printed on acid free paper. Copyright © 2007 John Wiley & Sons, Inc. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, scanning or otherwise, except as permitted under Sections 107 or 108 of the 1976 United States Copyright Act, without either the prior written permission of the Publisher, or authorization through payment of the appropriate per-copy fee to the Copyright Clearance Center, 222 Rosewood Drive, Danvers, MA 01923, (508)750-8400, fax (508)750-4470. Requests to the Publisher for permission should be addressed to the Permissions Department, John Wiley & Sons, Inc., 605 Third Avenue, New York, NY 10158-0012, (212) 850-6011, fax (212) 850-6008, E-Mail: [email protected]. To order books or for customer service please call 1(800)225-5945. Library of Congress Cataloging-in-Publication Data Callister, William D., 1940Materials science and engineering : an introduction / William D. Callister, Jr.—7th ed. p. cm. Includes bibliographical references and index. ISBN-13: 978-0-471-73696-7 (cloth) ISBN-10: 0-471-73696-1 (cloth) 1. Materials. I. Title. TA403.C23 2007 620.1’1—dc22 2005054228 Printed in the United States of America 10 9 8 7 6 5 4 3 2 1 1496T_fm_i-xxvi 1/6/06 02:56 Page vii Dedicated to my colleagues and friends in Brazil and Spain 1496T_fm_i-xxvi 1/6/06 02:56 Page viii 1496T_fm_i-xxvi 1/6/06 03:19 Page xv Contents LIST OF SYMBOLS xxiii 1. Introduction 1.1 1.2 1.3 1.4 1.5 1.6 1 Learning Objectives 2 Historical Perspective 2 Materials Science and Engineering 3 Why Study Materials Science and Engineering? 5 Classification of Materials 5 Advanced Materials 11 Modern Materials’ Needs 12 References 13 2. Atomic Structure and Interatomic Bonding 2.1 15 Learning Objectives 16 Introduction 16 ATOMIC STRUCTURE 16 2.2 2.3 2.4 Fundamental Concepts 16 Electrons in Atoms 17 The Periodic Table 23 2.5 2.6 2.7 2.8 Bonding Forces and Energies 24 Primary Interatomic Bonds 26 Secondary Bonding or van der Waals Bonding 30 Molecules 32 ATOMIC BONDING IN SOLIDS 24 Summary 34 Important Terms and Concepts 34 References 35 Questions and Problems 35 3. The Structure of Crystalline Solids 3.1 Learning Objectives 39 Introduction 39 3.2 3.3 3.4 3.5 3.6 Fundamental Concepts 39 Unit Cells 40 Metallic Crystal Structures 41 Density Computations 45 Polymorphism and Allotropy 46 CRYSTAL STRUCTURES 38 39 • xv 1496T_fm_i-xxvi 1/6/06 02:56 Page xvi xvi • Contents 3.7 Crystal Systems 46 CRYSTALLOGRAPHIC POINTS, DIRECTIONS, PLANES 49 3.8 3.9 3.10 3.11 3.12 Point Coordinates 49 Crystallographic Directions 51 Crystallographic Planes 55 Linear and Planar Densities 60 Close-Packed Crystal Structures AND 3.17 61 Single Crystals 63 Polycrystalline Materials 64 Anisotropy 64 X-Ray Diffraction: Determination of Crystal Structures 66 Noncrystalline Solids 71 Summary 72 Important Terms and Concepts 73 References 73 Questions and Problems 74 4. Imperfections in Solids 80 6. Mechanical Properties of Metals 6.1 6.2 6.3 6.4 6.5 Learning Objectives 81 Introduction 81 6.9 POINT DEFECTS 6.10 4.2 4.3 4.4 Vacancies and Self-Interstitials 81 Impurities in Solids 83 Specification of Composition 85 81 88 4.5 4.6 4.7 4.8 Dislocations–Linear Defects 88 Interfacial Defects 92 Bulk or Volume Defects 96 Atomic Vibrations 96 4.9 4.10 4.11 General 97 Microscopic Techniques 98 Grain Size Determination 102 MICROSCOPIC EXAMINATION 5.1 5.2 5.3 AND DESIGN/SAFETY Variability of Material Properties 161 Design/Safety Factors 163 Summary 165 Important Terms and Concepts 166 References 166 Questions and Problems 166 Design Problems 172 97 Summary 104 Important Terms and Concepts 105 References 105 Questions and Problems 106 Design Problems 108 5. Diffusion 143 Tensile Properties 144 True Stress and Strain 151 Elastic Recovery after Plastic Deformation 154 Compressive, Shear, and Torsional Deformation 154 Hardness 155 PROPERTY VARIABILITY FACTORS 161 6.11 6.12 137 Stress-Strain Behavior 137 Anelasticity 140 Elastic Properties of Materials 141 PLASTIC DEFORMATION 6.6 6.7 6.8 131 Learning Objectives 132 Introduction 132 Concepts of Stress and Strain 133 ELASTIC DEFORMATION 4.1 MISCELLANEOUS IMPERFECTIONS Nonsteady-State Diffusion 114 Factors That Influence Diffusion 118 Other Diffusion Paths 125 Summary 125 Important Terms and Concepts 126 References 126 Questions and Problems 126 Design Problems 129 CRYSTALLINE AND NONCRYSTALLINE MATERIALS 63 3.13 3.14 3.15 3.16 5.4 5.5 5.6 109 Learning Objectives 110 Introduction 110 Diffusion Mechanisms 111 Steady-State Diffusion 112 7. Dislocations and Strengthening Mechanisms 174 7.1 Learning Objectives 175 Introduction 175 DISLOCATIONS DEFORMATION 7.2 7.3 7.4 7.5 7.6 7.7 PLASTIC 175 AND Basic Concepts 175 Characteristics of Dislocations 178 Slip Systems 179 Slip in Single Crystals 181 Plastic Deformation of Polycrystalline Materials 185 Deformation by Twinning 185 1496T_fm_i-xxvi 01/10/06 22:13 Page xvii Contents • xvii MECHANISMS OF STRENGTHENING METALS 188 IN 7.8 7.9 7.10 Strengthening by Grain Size Reduction 188 Solid-Solution Strengthening 190 Strain Hardening 191 RECOVERY, RECRYSTALLIZATION, GROWTH 194 7.11 7.12 7.13 AND 9.2 9.3 9.4 9.5 9.6 BINARY PHASE DIAGRAMS GRAIN 9.7 9.8 9.9 Recovery 195 Recrystallization 195 Grain Growth 200 Summary 201 Important Terms and Concepts 202 References 202 Questions and Problems 202 Design Problems 206 9.10 9.11 9.12 9.13 8. Failure 8.1 207 Learning Objectives 208 Introduction 208 FRACTURE 8.2 8.3 8.4 8.5 8.6 9.14 9.15 208 Fundamentals of Fracture 208 Ductile Fracture 209 Brittle Fracture 211 Principles of Fracture Mechanics 215 Impact Fracture Testing 223 FATIGUE 227 8.7 8.8 8.9 8.10 8.11 Cyclic Stresses 228 The S–N Curve 229 Crack Initiation and Propagation 232 Factors That Affect Fatigue Life 234 Environmental Effects 237 8.12 8.13 8.14 8.15 Generalized Creep Behavior 238 Stress and Temperature Effects 239 Data Extrapolation Methods 241 Alloys for High-Temperature Use 242 CREEP 9.16 9.17 9.18 9.19 9.20 Summary 302 Important Terms and Concepts 303 References 303 Questions and Problems 304 10. Phase Transformations in Metals: Development of Microstructure and Alteration of Mechanical Properties 311 10.1 9.1 10.2 10.3 252 Learning Objectives 253 Introduction 253 DEFINITIONS AND BASIC CONCEPTS 10.4 253 290 The Iron–Iron Carbide (Fe–Fe3C) Phase Diagram 290 Development of Microstructure in Iron–Carbon Alloys 293 The Influence of Other Alloying Elements 301 Learning Objectives 312 Introduction 312 PHASE TRANSFORMATIONS 9. Phase Diagrams 258 Binary Isomorphous Systems 258 Interpretation of Phase Diagrams 260 Development of Microstructure in Isomorphous Alloys 264 Mechanical Properties of Isomorphous Alloys 268 Binary Eutectic Systems 269 Development of Microstructure in Eutectic Alloys 276 Equilibrium Diagrams Having Intermediate Phases or Compounds 282 Eutectic and Peritectic Reactions 284 Congruent Phase Transformations 286 Ceramic and Ternary Phase Diagrams 287 The Gibbs Phase Rule 287 THE IRON–CARBON SYSTEM 238 Summary 243 Important Terms and Concepts 245 References 246 Questions and Problems 246 Design Problems 250 Solubility Limit 254 Phases 254 Microstructure 255 Phase Equilibria 255 One-Component (or Unary) Phase Diagrams 256 312 Basic Concepts 312 The Kinetics of Phase Transformations 313 Metastable versus Equilibrium States 324 1496T_fm_i-xxvi 01/10/06 22:13 Page xviii xviii • Contents MICROSTRUCTURAL AND PROPERTY CHANGES IRON–CARBON ALLOYS 324 10.5 10.6 10.7 10.8 10.9 Isothermal Transformation Diagrams 325 Continuous Cooling Transformation Diagrams 335 Mechanical Behavior of Iron–Carbon Alloys 339 Tempered Martensite 343 Review of Phase Transformations and Mechanical Properties for Iron–Carbon Alloys 346 Summary 350 Important Terms and Concepts 351 References 352 Questions and Problems 352 Design Problems 356 11. Applications and Processing of Metal Alloys 358 11.1 OF METAL ALLOYS OF METALS 382 11.4 11.5 11.6 Forming Operations 383 Casting 384 Miscellaneous Techniques 386 11.7 11.8 11.9 Annealing Processes 388 Heat Treatment of Steels 390 Precipitation Hardening 402 THERMAL PROCESSING OF METALS Summary 453 Important Terms and Concepts 454 References 454 Questions and Problems 455 Design Problems 459 13. Applications and Processing of Ceramics 460 13.1 12.2 12.3 12.4 12.5 12.6 Learning Objectives 461 Introduction 461 Glasses 461 Glass–Ceramics 462 Clay Products 463 Refractories 464 Abrasives 466 Cements 467 Advanced Ceramics 468 OF 13.9 Fabrication and Processing of Glasses and Glass–Ceramics 471 13.10 Fabrication and Processing of Clay Products 476 13.11 Powder Pressing 481 13.12 Tape Casting 484 Summary 484 Important Terms and Concepts 486 References 486 Questions and Problems 486 Design Problem 488 Learning Objectives 415 Introduction 415 14. Polymer Structures CERAMIC STRUCTURES 14.1 14.2 14.3 14.4 Crystal Structures 415 Silicate Ceramics 426 Carbon 430 Imperfections in Ceramics 434 Diffusion in Ionic Materials 438 OF FABRICATION AND PROCESSING CERAMICS 471 12. Structures and Properties of Ceramics 414 415 442 12.8 Brittle Fracture of Ceramics 442 12.9 Stress–Strain Behavior 447 12.10 Mechanisms of Plastic Deformation 449 12.11 Miscellaneous Mechanical Considerations 451 387 Summary 407 Important Terms and Concepts 409 References 409 Questions and Problems 410 Design Problems 411 12.1 Ceramic Phase Diagrams 439 MECHANICAL PROPERTIES 13.2 13.3 13.4 13.5 13.6 13.7 13.8 359 Ferrous Alloys 359 Nonferrous Alloys 372 FABRICATION 12.7 TYPES AND APPLICATIONS CERAMICS 461 Learning Objectives 359 Introduction 359 TYPES 11.2 11.3 IN 14.5 489 Learning Objectives 490 Introduction 490 Hydrocarbon Molecules 490 Polymer Molecules 492 The Chemistry of Polymer Molecules 493 Molecular Weight 497 1496T_fm_i-xxvi 01/10/06 22:13 Page xix Contents • xix 14.6 14.7 14.8 14.9 14.10 14.11 14.12 14.13 14.14 Molecular Shape 500 Molecular Structure 501 Molecular Configurations 503 Thermoplastic and Thermosetting Polymers 506 Copolymers 507 Polymer Crystallinity 508 Polymer Crystals 512 Defects in Polymers 514 Diffusion in Polymeric Materials 515 Summary 517 Important Terms and Concepts 519 References 519 Questions and Problems 519 15.2 15.3 15.4 15.5 15.6 OF POLYMERS MECHANISMS OF DEFORMATION AND FOR STRENGTHENING OF POLYMERS 535 15.7 15.8 15.9 Deformation of Semicrystalline Polymers 535 Factors That Influence the Mechanical Properties of Semicrystalline Polymers 538 Deformation of Elastomers 541 CRYSTALLIZATION, MELTING, AND GLASS TRANSITION PHENOMENA IN POLYMERS 544 15.10 Crystallization 544 15.11 Melting 545 15.12 The Glass Transition 545 15.13 Melting and Glass Transition Temperatures 546 15.14 Factors That Influence Melting and Glass Transition Temperatures 547 POLYMER TYPES 15.15 15.16 15.17 15.18 15.19 16.1 560 Polymerization 561 Polymer Additives 563 Forming Techniques for Plastics 565 Fabrication of Elastomers 567 Fabrication of Fibers and Films 568 577 Learning Objectives 578 Introduction 578 16.4 16.5 16.6 16.7 16.8 16.9 16.10 16.11 16.12 16.13 580 Large-Particle Composites 580 Dispersion-Strengthened Composites 584 FIBER-REINFORCED COMPOSITES 585 Influence of Fiber Length 585 Influence of Fiber Orientation and Concentration 586 The Fiber Phase 595 The Matrix Phase 596 Polymer-Matrix Composites 597 Metal-Matrix Composites 603 Ceramic-Matrix Composites 605 Carbon–Carbon Composites 606 Hybrid Composites 607 Processing of Fiber-Reinforced Composites 607 STRUCTURAL COMPOSITES 610 16.14 Laminar Composites 610 16.15 Sandwich Panels 611 Summary 613 Important Terms and Concepts 615 References 616 Questions and Problems 616 Design Problems 619 17. Corrosion and Degradation of Materials 621 549 Plastics 549 Elastomers 552 Fibers 554 Miscellaneous Applications 555 Advanced Polymeric Materials 556 PROCESSING PARTICLE-REINFORCED COMPOSITES 524 Stress–Strain Behavior 524 Macroscopic Deformation 527 Viscoelastic Deformation 527 Fracture of Polymers 532 Miscellaneous Mechanical Characteristics 533 AND Summary 569 Important Terms and Concepts 571 References 571 Questions and Problems 572 Design Questions 576 16.2 16.3 Learning Objectives 524 Introduction 524 MECHANICAL BEHAVIOR 15.20 15.21 15.22 15.23 15.24 16. Composites 15. Characteristics, Applications, and Processing of Polymers 523 15.1 POLYMER SYNTHESIS 17.1 Learning Objectives 622 Introduction 622 17.2 17.3 Electrochemical Considerations 623 Corrosion Rates 630 CORROSION OF METALS 622 1496T_fm_i-xxvi 01/10/06 22:13 Page xx xx • Contents 17.4 Prediction of Corrosion Rates 631 17.5 Passivity 638 17.6 Environmental Effects 640 17.7 Forms of Corrosion 640 17.8 Corrosion Environments 648 17.9 Corrosion Prevention 649 17.10 Oxidation 651 CORROSION OF DEGRADATION CERAMIC MATERIALS OF POLYMERS 654 655 18.20 Types of Polarization 708 18.21 Frequency Dependence of the Dielectric Constant 709 18.22 Dielectric Strength 711 18.23 Dielectric Materials 711 OTHER ELECTRICAL CHARACTERISTICS MATERIALS 711 18.24 Ferroelectricity 711 18.25 Piezoelectricity 712 Summary 713 Important Terms and Concepts 715 References 715 Questions and Problems 716 Design Problems 720 17.11 Swelling and Dissolution 655 17.12 Bond Rupture 657 17.13 Weathering 658 Summary 659 Important Terms and Concepts 660 References 661 Questions and Problems 661 Design Problems 644 18. Electrical Properties 18.1 665 Learning Objectives 666 Introduction 666 ELECTRICAL CONDUCTION 18.2 18.3 18.4 18.5 18.6 18.7 18.8 18.9 666 Ohm’s Law 666 Electrical Conductivity 667 Electronic and Ionic Conduction 668 Energy Band Structures in Solids 668 Conduction in Terms of Band and Atomic Bonding Models 671 Electron Mobility 673 Electrical Resistivity of Metals 674 Electrical Characteristics of Commercial Alloys 677 SEMICONDUCTIVITY 679 18.10 Intrinsic Semiconduction 679 18.11 Extrinsic Semiconduction 682 18.12 The Temperature Dependence of Carrier Concentration 686 18.13 Factors That Affect Carrier Mobility 688 18.14 The Hall Effect 692 18.15 Semiconductor Devices 694 ELECTRICAL CONDUCTION POLYMERS 700 IN IONIC CERAMICS AND IN 18.16 Conduction in Ionic Materials 701 18.17 Electrical Properties of Polymers 701 DIELECTRIC BEHAVIOR 702 18.18 Capacitance 703 18.19 Field Vectors and Polarization 704 OF 19. Thermal Properties 19.1 19.2 19.3 19.4 19.5 W1 Learning Objectives W2 Introduction W2 Heat Capacity W2 Thermal Expansion W4 Thermal Conductivity W7 Thermal Stresses W12 Summary W14 Important Terms and Concepts W15 References W15 Questions and Problems W15 Design Problems W17 20. Magnetic Properties W19 Learning Objectives W20 Introduction W20 Basic Concepts W20 Diamagnetism and Paramagnetism W24 20.4 Ferromagnetism W26 20.5 Antiferromagnetism and Ferrimagnetism W28 20.6 The Influence of Temperature on Magnetic Behavior W32 20.7 Domains and Hysteresis W33 20.8 Magnetic Anisotropy W37 20.9 Soft Magnetic Materials W38 20.10 Hard Magnetic Materials W41 20.11 Magnetic Storage W44 20.12 Superconductivity W47 20.1 20.2 20.3 Summary W50 Important Terms and Concepts W52 References W52 Questions and Problems W53 Design Problems W56 1496T_fm_i-xxvi 01/10/06 22:13 Page xxi Contents • xxi 21. 21.1 Optical Properties Learning Objectives W58 Introduction W58 BASIC CONCEPTS 21.2 21.3 21.4 22.8 22.9 W57 ARTIFICIAL TOTAL HIP REPLACEMENT W58 Electromagnetic Radiation W58 Light Interactions with Solids W60 Atomic and Electronic Interactions W61 OPTICAL PROPERTIES OF METALS OPTICAL PROPERTIES OF NONMETALS W62 W63 21.5 Refraction W63 21.6 Reflection W65 21.7 Absorption W65 21.8 Transmission W68 21.9 Color W69 21.10 Opacity and Translucency in Insulators W71 APPLICATIONS OF OPTICAL PHENOMENA W72 Summary W82 Important Terms and Concepts W83 References W84 Questions and Problems W84 Design Problem W85 22. Materials Selection and Design Considerations W86 Learning Objectives W87 Introduction W87 MATERIALS SELECTION FOR A TORSIONALLY STRESSED CYLINDRICAL SHAFT W87 22.2 22.3 Strength Considerations–Torsionally Stressed Shaft W88 Other Property Considerations and the Final Decision W93 AUTOMOTIVE VALVE SPRING 22.4 22.5 22.6 22.7 W108 22.10 Anatomy of the Hip Joint W108 22.11 Material Requirements W111 22.12 Materials Employed W112 CHEMICAL PROTECTIVE CLOTHING W115 22.13 Introduction W115 22.14 Assessment of CPC Glove Materials to Protect Against Exposure to Methylene Chloride W115 MATERIALS FOR INTEGRATED CIRCUIT PACKAGES W119 21.11 Luminescence W72 21.12 Photoconductivity W72 21.13 Lasers W75 21.14 Optical Fibers in Communications W79 22.1 Testing Procedure and Results W102 Discussion W108 22.15 22.16 22.17 22.18 22.19 22.20 Introduction W119 Leadframe Design and Materials W120 Die Bonding W121 Wire Bonding W124 Package Encapsulation W125 Tape Automated Bonding W127 Summary W129 References W130 Design Questions and Problems W131 23. Economic, Environmental, and Societal Issues in Materials Science and Engineering W135 23.1 Learning Objectives W136 Introduction W136 ECONOMIC CONSIDERATIONS 23.2 23.3 23.4 W136 Component Design W137 Materials W137 Manufacturing Techniques W137 ENVIRONMENTAL AND SOCIETAL CONSIDERATIONS W137 23.5 Recycling Issues in Materials Science and Engineering W140 Summary W143 References W143 Design Question W144 W94 Mechanics of Spring Deformation W94 Valve Spring Design and Material Requirements W95 One Commonly Employed Steel Alloy W98 Appendix A The International System of Units A1 FAILURE OF AN AUTOMOBILE REAR AXLE W101 B.1 B.2 B.3 Introduction W101 Appendix B Properties of Selected Engineering Materials A3 Density A3 Modulus of Elasticity A6 Poisson’s Ratio A10 1496T_fm_i-xxvi 01/10/06 22:13 Page xxii xxii • Contents B.4 B.5 B.6 B.7 B.8 B.9 B.10 Strength and Ductility A11 Plane Strain Fracture Toughness A16 Linear Coefficient of Thermal Expansion A17 Thermal Conductivity A21 Specific Heat A24 Electrical Resistivity A26 Metal Alloy Compositions A29 Appendix E Glass Transition and Melting Temperatures for Common Polymeric Materials A41 Glossary Answers to Selected Problems Index Appendix C Costs and Relative Costs for Selected Engineering Materials A31 Appendix D Repeat Unit Structures for Common Polymers A37 G0 I1 S1 1496T_fm_i-xxvi 1/6/06 02:56 Page ix Preface Imaterials n this Seventh Edition I have retained the objectives and approaches for teaching science and engineering that were presented in previous editions. The first, and primary, objective is to present the basic fundamentals on a level appropriate for university/college students who have completed their freshmen calculus, chemistry, and physics courses. In order to achieve this goal, I have endeavored to use terminology that is familiar to the student who is encountering the discipline of materials science and engineering for the first time, and also to define and explain all unfamiliar terms. The second objective is to present the subject matter in a logical order, from the simple to the more complex. Each chapter builds on the content of previous ones. The third objective, or philosophy, that I strive to maintain throughout the text is that if a topic or concept is worth treating, then it is worth treating in sufficient detail and to the extent that students have the opportunity to fully understand it without having to consult other sources; also, in most cases, some practical relevance is provided. Discussions are intended to be clear and concise and to begin at appropriate levels of understanding. The fourth objective is to include features in the book that will expedite the learning process. These learning aids include: • Numerous illustrations, now presented in full color, and photographs to help visualize what is being presented; • Learning objectives; • “Why Study . . .” and “Materials of Importance” items that provide relevance to topic discussions; • Key terms and descriptions of key equations highlighted in the margins for quick reference; • End-of-chapter questions and problems; • Answers to selected problems; • A glossary, list of symbols, and references to facilitate understanding the subject matter. The fifth objective is to enhance the teaching and learning process by using the newer technologies that are available to most instructors and students of engineering today. FEATURES THAT ARE NEW TO THIS EDITION New/Revised Content Several important changes have been made with this Seventh Edition. One of the most significant is the incorporation of a number of new sections, as well • ix 1496T_fm_i-xxvi 01/10/06 22:13 Page x x • Preface as revisions/amplifications of other sections. New sections/discussions are as follows: • One-component (or unary) phase diagrams (Section 9.6) • Compacted graphite iron (in Section 11.2, “Ferrous Alloys”) • Lost foam casting (in Section 11.5, “Casting”) • Temperature dependence of Frenkel and Schottky defects (in Section 12.5, “Imperfections in Ceramics”) • Fractography of ceramics (in Section 12.8, “Brittle Fracture of Ceramics”) • Crystallization of glass-ceramics, in terms of isothermal transformation and continuous cooling transformation diagrams (in Section 13.3, “Glass-Ceramics”) • Permeability in polymers (in Section 14.14, “Diffusion in Polymeric Materials”) • Magnetic anisotropy (Section 20.8) • A new case study on chemical protective clothing (Sections 22.13 and 22.14). Those sections that have been revised/amplified, include the following: • Treatments in Chapter 1 (“Introduction”) on the several material types have been enlarged to include comparisons of various property values (as bar charts). • Expanded discussions on crystallographic directions and planes in hexagonal crystals (Sections 3.9 and 3.10); also some new related homework problems. • Comparisons of (1) dimensional size ranges for various structural elements, and (2) resolution ranges for the several microscopic examination techniques (in Section 4.10, “Microscopic Techniques”). • Updates on hardness testing techniques (Section 6.10). • Revised discussion on the Burgers vector (Section 7.4). • New discussion on why recrystallization temperature depends on the purity of a metal (Section 7.12). • Eliminated some detailed discussion on fracture mechanics—i.e., used “Concise Version” from sixth edition (Section 8.5). • Expanded discussion on nondestructive testing (Section 8.5). • Used Concise Version (from sixth edition) of discussion on crack initiation and propagation (for fatigue, Section 8.9), and eliminated section on crack propagation rate. • Refined terminology and representations of polymer structures (Sections 14.3 through 14.8). • Eliminated discussion on fringed-micelle model (found in Section 14.12 of the sixth edition). • Enhanced discussion on defects in polymers (Section 14.13). • Revised the following sections in Chapter 15 (“Characteristics, Applications, and Processing of Polymers”): fracture of polymers (Section 15.5), deformation of semicrystalline polymers (Section 15.7), adhesives (in Section 15.18), polymerization (Section 15.20), and fabrication of fibers and films (Section 15.24). • Revised treatment of polymer degradation (Section 17.12). 1496T_fm_i-xxvi 01/10/06 22:13 Page xi Preface • xi Materials of Importance One new feature that has been incorporated in this edition is “Materials of Importance” pieces; in these we discuss familiar and interesting materials/applications of materials. These pieces lend some relevance to topical coverage, are found in most chapters in the book, and include the following: • • • • • • • • • • • • • • • • • Carbonated Beverage Containers Water (Its Volume Expansion Upon Freezing) Tin (Its Allotropic Transformation) Catalysts (and Surface Defects) Aluminum for Integrated Circuit Interconnects Lead-Free Solders Shape-Memory Alloys Metal Alloys Used for Euro Coins Carbon Nanotubes Piezoelectric Ceramics Shrink-Wrap Polymer Films Phenolic Billiard Balls Nanocomposites in Tennis Balls Aluminum Electrical Wires Invar and Other Low-Expansion Alloys An Iron-Silicon Alloy That is Used in Transformer Cores Light-Emitting Diodes Concept Check Another new feature included in this seventh edition is what we call a “Concept Check,” a question that tests whether or not a student understands the subject matter on a conceptual level. Concept check questions are found within most chapters; many of them appeared in the end-of-chapter Questions and Problems sections of the previous edition. Answers to these questions are on the book’s Web site, www.wiley.com/college/callister (Student Companion Site). And, finally, for each chapter, both the Summary and the Questions and Problems are organized by section; section titles precede their summaries and questions/problems. Format Changes There are several other major changes from the format of the sixth edition. First of all, no CD-ROM is packaged with the in-print text; all electronic components are found on the book’s Web site (www.wiley.com/college/callister). This includes the last five chapters in the book—viz. Chapter 19, “Thermal Properties;” Chapter 20, “Magnetic Properties;” Chapter 21, “Optical Properties;” Chapter 22, “Materials Selection and Design Considerations;” and Chapter 23, “Economic, Environmental, and Societal Issues in Materials Science and Engineering.” These chapters are in Adobe Acrobat® pdf format and may be downloaded. Furthermore, only complete chapters appear on the Web site (rather than selected sections for some chapters per the sixth edition). And, in addition, for all sections of the book there is only one version—for the two-version sections of the sixth edition, in most instances, the detailed ones have been retained. 1496T_fm_i-xxvi 01/10/06 22:13 Page xii xii • Preface Six case studies have been relegated to Chapter 22, “Materials Selection and Design Considerations,” which are as follows: • • • • • • Materials Selection for a Torsionally Stressed Cylindrical Shaft Automobile Valve Spring Failure of an Automobile Rear Axle Artificial Total Hip Replacement Chemical Protective Clothing Materials for Integrated Circuit Packages References to these case studies are made in the left-page margins at appropriate locations in the other chapters. All but “Chemical Protective Clothing” appeared in the sixth edition; it replaces the “Thermal Protection System on the Space Shuttle Orbiter” case study. STUDENT LEARNING RESOURCES (WWW.WILEY.COM/COLLEGE/CALLISTER) Also found on the book’s Web site (under “Student Companion Site”) are several important instructional elements for the student that complement the text; these include the following: 1. VMSE: Virtual Materials Science and Engineering. This is essentially the same software program that accompanied the previous edition, but now browserbased for easier use on a wider variety of computer platforms. It consists of interactive simulations and animations that enhance the learning of key concepts in materials science and engineering, and, in addition, a materials properties/cost database. Students can access VMSE via the registration code included with all new copies. Throughout the book, whenever there is some text or a problem that is supplemented by VMSE, a small “icon” that denotes the associated module is included in one of the margins. These modules and their corresponding icons are as follows: Metallic Crystal Structures and Crystallography Phase Diagrams Ceramic Crystal Structures Diffusion Repeat Unit and Polymer Structures Tensile Tests Dislocations Solid-Solution Strengthening 2. Answers to the Concept Check questions. 3. Direct access to online self-assessment exercises. This is a Web-based assessment program that contains questions and problems similar to those found in the text; these problems/questions are organized and labeled according to textbook sections. An answer/solution that is entered by the user in response to a question/problem is graded immediately, and comments are offered for incorrect responses. The student may use this electronic resource to review course material, and to assess his/her mastery and understanding of topics covered in the text. 1496T_fm_i-xxvi 1/6/06 02:56 Page xiii Preface • xiii 4. Additional Web resources, which include the following: • Index of Learning Styles. Upon answering a 44-item questionnaire, a user’s learning style preference (i.e., the manner in which information is assimilated and processed) is assessed. • Extended Learning Objectives. A more extensive list of learning objectives than is provided at the beginning of each chapter. • Links to Other Web Resources. These links are categorized according to general Internet, software, teaching, specific course content/activities, and materials databases. INSTRUCTORS’ RESOURCES The “Instructor Companion Site” (www.wiley.com/college/callister) is available for instructors who have adopted this text. Resources that are available include the following: 1. Detailed solutions of all end-of-chapter questions and problems (in both Microsoft Word® and Adobe Acrobat® PDF formats). 2. Photographs, illustrations, and tables that appear in the book (in PDF and JPEG formats); an instructor can print them for handouts or prepare transparencies in his/her desired format. 3. A set of PowerPoint® lecture slides developed by Peter M. Anderson (The Ohio State University) and David G. Rethwisch (The University of Iowa). These slides follow the flow of topics in the text, and include materials from the text and other sources as well as illustrations and animations. Instructors may use the slides as is or edit them to fit their teaching needs. 4. A list of classroom demonstrations and laboratory experiments that portray phenomena and/or illustrate principles that are discussed in the book; references are also provided that give more detailed accounts of these demonstrations. 5. Suggested course syllabi for the various engineering disciplines. WileyPLUS WileyPLUS gives you, the instructor, the technology to create an environment where students reach their full potential and experience academic success that will last a lifetime! With WileyPLUS, students will come to class better prepared for your lectures, get immediate feedback and context-sensitive help on assignments and quizzes, and have access to a full range of interactive learning resources including a complete online version of their text. WileyPLUS gives you a wealth of presentation and preparation tools, easy-to-navigate assessment tools including an online gradebook, and a complete system to administer and manage your course exactly as you wish. Contact your local Wiley representative for details on how to set up your WileyPLUS course, or visit the website at www.wiley.com/college/wileyplus. FEEDBACK I have a sincere interest in meeting the needs of educators and students in the materials science and engineering community, and therefore would like to solicit feedback on this seventh edition. Comments, suggestions, and criticisms may be submitted to me via e-mail at the following address: [email protected]. ACKNOWLEDGMENTS Appreciation is expressed to those who have made contributions to this edition. I am especially indebted to David G. Rethwisch, who, as a special contributor, 1496T_fm_i-xxvi 01/10/06 22:13 Page xiv xiv • Preface provided invaluable assistance in updating and upgrading important material in a number of chapters. In addition, I sincerely appreciate Grant E. Head’s expert programming skills, which he used in developing the Virtual Materials Science and Engineering software. Important input was also furnished by Carl Wood of Utah State University and W. Roger Cannon of Rutgers University, to whom I also give thanks. In addition, helpful ideas and suggestions have been provided by the following: Tarek Abdelsalam, East Carolina University Keyvan Ahdut, University of the District of Columbia Mark Aindow, University of Connecticut (Storrs) Pranesh Aswath, University of Texas at Arlington Mir Atiqullah, St. Louis University Sayavur Bakhtiyarov, Auburn University Kristen Constant, Iowa State University Raymond Cutler, University of Utah Janet Degrazia, University of Colorado Mark DeGuire, Case Western Reserve University Timothy Dewhurst, Cedarville University Amelito Enriquez, Canada College Jeffrey Fergus, Auburn University Victor Forsnes, Brigham Young University (Idaho) Paul Funkenbusch, University of Rochester Randall German, Pennsylvania State University Scott Giese, University of Northern Iowa Brian P. Grady, University of Oklahoma Theodore Greene, Wentworth Institute of Technology Todd Gross, University of New Hampshire Jamie Grunlan, Texas A & M University Masanori Hara, Rutgers University Russell Herlache, Saginaw Valley State University Susan Holl, California State University (Sacramento) Zhong Hu, South Dakota State University Duane Jardine, University of New Orleans Jun Jin, Texas A & M University at Galveston Paul Johnson, Grand Valley State University Robert Johnson, University of Texas at Arlington Robert Jones, University of Texas (Pan American) Maureen Julian, Virginia Tech James Kawamoto, Mission College Edward Kolesar, Texas Christian University Stephen Krause, Arizona State University (Tempe) Robert McCoy, Youngstown State University Scott Miller, University of Missouri (Rolla) Devesh Misra, University of Louisiana at Lafayette Angela L. Moran, U.S. Naval Academy James Newell, Rowan University Toby Padilla, Colorado School of Mines Timothy Raymond, Bucknell University Alessandro Rengan, Central State University Bengt Selling, Royal Institute of Technology (Stockholm, Sweden) Ismat Shah, University of Delaware Patricia Shamamy, Lawrence Technological University Adel Sharif, California State University at Los Angeles Susan Sinnott, University of Florida Andrey Soukhojak, Lehigh University Erik Spjut, Harvey Mudd College David Stienstra, Rose-Hulman Institute of Technology Alexey Sverdlin, Bradley University Dugan Um, Texas State University Raj Vaidyanatha, University of Central Florida Kant Vajpayee, University of Southern Mississippi Kumar Virwani, University of Arkansas (Fayetteville) Mark Weaver, University of Alabama (Tuscaloosa) Jason Weiss, Purdue University (West Lafayette) I am also indebted to Joseph P. Hayton, Sponsoring Editor, and to Kenneth Santor, Senior Production Editor at Wiley for their assistance and guidance on this revision. Since I undertook the task of writing my first text on this subject in the early80’s, instructors and students too numerous to mention have shared their input and contributions on how to make this work more effective as a teaching and learning tool. To all those who have helped, I express my sincere “Thanks!” Last, but certainly not least, the continual encouragement and support of my family and friends is deeply and sincerely appreciated. WILLIAM D. CALLISTER, JR. Salt Lake City, Utah January 2006 1496T_fm_i-xxvi 01/10/06 22:13 Page xxiii List of Symbols T he number of the section in which a symbol is introduced or explained is given in parentheses. A  area Å  angstrom unit Ai  atomic weight of element i (2.2) APF  atomic packing factor (3.4) a  lattice parameter: unit cell x-axial length (3.4) a  crack length of a surface crack (8.5) at%  atom percent (4.4) B  magnetic flux density (induction) (20.2) Br  magnetic remanence (20.7) BCC  body-centered cubic crystal structure (3.4) b  lattice parameter: unit cell y-axial length (3.7) b  Burgers vector (4.5) C  capacitance (18.18) Ci  concentration (composition) of component i in wt% (4.4) Ci  concentration (composition) of component i in at% (4.4) Cv, Cp  heat capacity at constant volume, pressure (19.2) CPR  corrosion penetration rate (17.3) CVN  Charpy V-notch (8.6) %CW  percent cold work (7.10) c  lattice parameter: unit cell z-axial length (3.7) c  velocity of electromagnetic radiation in a vacuum (21.2) D  diffusion coefficient (5.3) D  dielectric displacement (18.19) DP  degree of polymerization (14.5) d  diameter d  average grain diameter (7.8) dhkl  interplanar spacing for planes of Miller indices h, k, and l (3.16) E  energy (2.5) E  modulus of elasticity or Young’s modulus (6.3) e  electric field intensity (18.3) Ef  Fermi energy (18.5) Eg  band gap energy (18.6) Er(t)  relaxation modulus (15.4) %EL  ductility, in percent elongation (6.6) e  electric charge per electron (18.7) e  electron (17.2) erf  Gaussian error function (5.4) exp  e, the base for natural logarithms F  force, interatomic or mechanical (2.5, 6.3) f  Faraday constant (17.2) FCC  face-centered cubic crystal structure (3.4) G  shear modulus (6.3) H  magnetic field strength (20.2) Hc  magnetic coercivity (20.7) HB  Brinell hardness (6.10) HCP  hexagonal close-packed crystal structure (3.4) HK  Knoop hardness (6.10) HRB, HRF  Rockwell hardness: B and F scales (6.10) • xxiii 1496T_fm_i-xxvi 01/11/06 0:06 Page xxiv xxiv • List of Symbols HR15N, HR45W  superficial Rockwell hardness: 15N and 45W scales (6.10) HV  Vickers hardness (6.10) h  Planck’s constant (21.2) (hkl)  Miller indices for a crystallographic plane (3.10) I  electric current (18.2) I  intensity of electromagnetic radiation (21.3) i  current density (17.3) iC  corrosion current density (17.4) J  diffusion flux (5.3) J  electric current density (18.3) Kc  fracture toughness (8.5) KIc  plane strain fracture toughness for mode I crack surface displacement (8.5) k  Boltzmann’s constant (4.2) k  thermal conductivity (19.4) l  length lc  critical fiber length (16.4) ln  natural logarithm log  logarithm taken to base 10 M  magnetization (20.2) Mn  polymer number-average molecular weight (14.5) Mw  polymer weight-average molecular weight (14.5) mol%  mole percent N  number of fatigue cycles (8.8) NA  Avogadro’s number (3.5) Nf  fatigue life (8.8) n  principal quantum number (2.3) n  number of atoms per unit cell (3.5) n  strain-hardening exponent (6.7) n  number of electrons in an electrochemical reaction (17.2) n  number of conducting electrons per cubic meter (18.7) n  index of refraction (21.5) n  for ceramics, the number of formula units per unit cell (12.2) ni  intrinsic carrier (electron and hole) concentration (18.10) P  dielectric polarization (18.19) P–B ratio  Pilling–Bedworth ratio (17.10) p  number of holes per cubic meter (18.10) Q  activation energy Q  magnitude of charge stored (18.18) R  atomic radius (3.4) R  gas constant %RA  ductility, in percent reduction in area (6.6) r  interatomic distance (2.5) r  reaction rate (17.3) rA, rC  anion and cation ionic radii (12.2) S  fatigue stress amplitude (8.8) SEM  scanning electron microscopy or microscope T  temperature Tc  Curie temperature (20.6) TC  superconducting critical temperature (20.12) Tg  glass transition temperature (13.9, 15.12) Tm  melting temperature TEM  transmission electron microscopy or microscope TS  tensile strength (6.6) t  time tr  rupture lifetime (8.12) Ur  modulus of resilience (6.6) [uvw]  indices for a crystallographic direction (3.9) V  electrical potential difference (voltage) (17.2, 18.2) 1496T_fm_i-xxvi 01/10/06 22:13 Page xxv List of Symbols • xxv VC  unit cell volume (3.4) VC  corrosion potential (17.4) VH  Hall voltage (18.14) Vi  volume fraction of phase i (9.8) v  velocity vol%  volume percent Wi  mass fraction of phase i (9.8) wt%  weight percent (4.4) x  length x  space coordinate Y  dimensionless parameter or function in fracture toughness expression (8.5) y  space coordinate z  space coordinate   lattice parameter: unit cell y–z interaxial angle (3.7) , ,   phase designations l  linear coefficient of thermal expansion (19.3)   lattice parameter: unit cell x–z interaxial angle (3.7)   lattice parameter: unit cell x–y interaxial angle (3.7)   shear strain (6.2)   precedes the symbol of a parameter to denote finite change   engineering strain (6.2)   dielectric permittivity (18.18) r  dielectric constant or relative permittivity (18.18) s  steady-state creep rate (8.12) T  true strain (6.7)   viscosity (12.10)   overvoltage (17.4)   Bragg diffraction angle (3.16) D  Debye temperature (19.2)  wavelength of electromagnetic radiation (3.16)  magnetic permeability (20.2) B  Bohr magneton (20.2) r  relative magnetic permeability (20.2) e  electron mobility (18.7) h  hole mobility (18.10) n  Poisson’s ratio (6.5) n  frequency of electromagnetic radiation (21.2)  density (3.5)  electrical resistivity (18.2) t  radius of curvature at the tip of a crack (8.5)  engineering stress, tensile or compressive (6.2)  electrical conductivity (18.3) *  longitudinal strength (composite) (16.5) c  critical stress for crack propagation (8.5) fs  flexural strength (12.9) m  maximum stress (8.5) m  mean stress (8.7) m  stress in matrix at composite failure (16.5) T  true stress (6.7) w  safe or working stress (6.12) y  yield strength (6.6)  shear stress (6.2) c  fiber–matrix bond strength/matrix shear yield strength (16.4) crss  critical resolved shear stress (7.5) m  magnetic susceptibility (20.2) SUBSCRIPTS c  composite cd  discontinuous fibrous composite cl  longitudinal direction (aligned fibrous composite) ct  transverse direction (aligned fibrous composite) f  final f  at fracture f  fiber i  instantaneous m  matrix m, max  maximum min  minimum 0  original 0  at equilibrium 0  in a vacuum 1496T_fm_i-xxvi 1/6/06 02:56 Page xxvi 1496T_c01_01-14 12/20/05 7:11 Page 1 2nd REVISE PAGES Chapter A 1 Introduction familiar item that is fabricated from three different material types is the beverage container. Beverages are marketed in aluminum (metal) cans (top), glass (ceramic) bottles (center), and plastic (polymer) bottles (bottom). (Permission to use these photographs was granted by the Coca-Cola Company. Coca-Cola, Coca-Cola Classic, the Contour Bottle design and the Dynamic Ribbon are registered trademarks of The Coca-Cola Company and used with its express permission.) • 1 1496T_c01_01-14 12/20/05 7:11 Page 2 2nd REVISE PAGES Learning Objectives After careful study of this chapter you should be able to do the following: 1. List six different property classifications of 4. (a) List the three primary classifications of solid materials that determine their applicability. materials, and then cite the distinctive 2. Cite the four components that are involved in chemical feature of each. the design, production, and utilization of (b) Note the two types of advanced materials materials, and briefly describe the interrelationand, for each, its distinctive feature(s). ships between these components. 5. (a) Briefly define “smart material/system.” 3. Cite three criteria that are important in the ma(b) Briefly explain the concept of “nanotechterials selection process. nology” as it applies to materials. 1.1 HISTORICAL PERSPECTIVE Materials are probably more deep-seated in our culture than most of us realize. Transportation, housing, clothing, communication, recreation, and food production— virtually every segment of our everyday lives is influenced to one degree or another by materials. Historically, the development and advancement of societies have been intimately tied to the members’ ability to produce and manipulate materials to fill their needs. In fact, early civilizations have been designated by the level of their materials development (Stone Age, Bronze Age, Iron Age).1 The earliest humans had access to only a very limited number of materials, those that occur naturally: stone, wood, clay, skins, and so on. With time they discovered techniques for producing materials that had properties superior to those of the natural ones; these new materials included pottery and various metals. Furthermore, it was discovered that the properties of a material could be altered by heat treatments and by the addition of other substances. At this point, materials utilization was totally a selection process that involved deciding from a given, rather limited set of materials the one best suited for an application by virtue of its characteristics. It was not until relatively recent times that scientists came to understand the relationships between the structural elements of materials and their properties. This knowledge, acquired over approximately the past 100 years, has empowered them to fashion, to a large degree, the characteristics of materials. Thus, tens of thousands of different materials have evolved with rather specialized characteristics that meet the needs of our modern and complex society; these include metals, plastics, glasses, and fibers. The development of many technologies that make our existence so comfortable has been intimately associated with the accessibility of suitable materials. An advancement in the understanding of a material type is often the forerunner to the stepwise progression of a technology. For example, automobiles would not have been possible without the availability of inexpensive steel or some other comparable substitute. In our contemporary era, sophisticated electronic devices rely on components that are made from what are called semiconducting materials. 1 The approximate dates for the beginnings of Stone, Bronze, and Iron Ages were 2.5 million 3500 BC and 1000 BC, respectively. BC, 1496T_c01_01-14 11/9/05 17:02 Page 3 REVISED PAGES 1.2 Materials Science and Engineering • 3 1.2 MATERIALS SCIENCE AND ENGINEERING Sometimes it is useful to subdivide the discipline of materials science and engineering into materials science and materials engineering subdisciplines. Strictly speaking, “materials science” involves investigating the relationships that exist between the structures and properties of materials. In contrast, “materials engineering” is, on the basis of these structure–property correlations, designing or engineering the structure of a material to produce a predetermined set of properties.2 From a functional perspective, the role of a materials scientist is to develop or synthesize new materials, whereas a materials engineer is called upon to create new products or systems using existing materials, and/or to develop techniques for processing materials. Most graduates in materials programs are trained to be both materials scientists and materials engineers. “Structure” is at this point a nebulous term that deserves some explanation. In brief, the structure of a material usually relates to the arrangement of its internal components. Subatomic structure involves electrons within the individual atoms and interactions with their nuclei. On an atomic level, structure encompasses the organization of atoms or molecules relative to one another. The next larger structural realm, which contains large groups of atoms that are normally agglomerated together, is termed “microscopic,” meaning that which is subject to direct observation using some type of microscope. Finally, structural elements that may be viewed with the naked eye are termed “macroscopic.” The notion of “property” deserves elaboration. While in service use, all materials are exposed to external stimuli that evoke some type of response. For example, a specimen subjected to forces will experience deformation, or a polished metal surface will reflect light. A property is a material trait in terms of the kind and magnitude of response to a specific imposed stimulus. Generally, definitions of properties are made independent of material shape and size. Virtually all important properties of solid materials may be grouped into six different categories: mechanical, electrical, thermal, magnetic, optical, and deteriorative. For each there is a characteristic type of stimulus capable of provoking different responses. Mechanical properties relate deformation to an applied load or force; examples include elastic modulus and strength. For electrical properties, such as electrical conductivity and dielectric constant, the stimulus is an electric field. The thermal behavior of solids can be represented in terms of heat capacity and thermal conductivity. Magnetic properties demonstrate the response of a material to the application of a magnetic field. For optical properties, the stimulus is electromagnetic or light radiation; index of refraction and reflectivity are representative optical properties. Finally, deteriorative characteristics relate to the chemical reactivity of materials. The chapters that follow discuss properties that fall within each of these six classifications. In addition to structure and properties, two other important components are involved in the science and engineering of materials—namely, “processing” and “performance.” With regard to the relationships of these four components, the structure of a material will depend on how it is processed. Furthermore, a material’s performance will be a function of its properties. Thus, the interrelationship between processing, structure, properties, and performance is as depicted in the schematic illustration shown in Figure 1.1. Throughout this text we draw attention to the 2 Throughout this text we draw attention to the relationships between material properties and structural elements. 1496T_c01_01-14 12/20/05 7:11 Page 4 2nd REVISE PAGES 4 • Chapter 1 / Introduction Processing Structure Properties Performance Figure 1.1 The four components of the discipline of materials science and engineering and their interrelationship. relationships among these four components in terms of the design, production, and utilization of materials. We now present an example of these processing-structure-properties-performance principles with Figure 1.2, a photograph showing three thin disk specimens placed over some printed matter. It is obvious that the optical properties (i.e., the light transmittance) of each of the three materials are different; the one on the left is transparent (i.e., virtually all of the reflected light passes through it), whereas the disks in the center and on the right are, respectively, translucent and opaque. All of these specimens are of the same material, aluminum oxide, but the leftmost one is what we call a single crystal—that is, it is highly perfect—which gives rise to its transparency. The center one is composed of numerous and very small single crystals that are all connected; the boundaries between these small crystals scatter a portion of the light reflected from the printed page, which makes this material optically translucent. Finally, the specimen on the right is composed not only of many small, interconnected crystals, but also of a large number of very small pores or void spaces. These pores also effectively scatter the reflected light and render this material opaque. Thus, the structures of these three specimens are different in terms of crystal boundaries and pores, which affect the optical transmittance properties. Furthermore, each material was produced using a different processing technique. And, of course, if optical transmittance is an important parameter relative to the ultimate in-service application, the performance of each material will be different. Figure 1.2 Photograph of three thin disk specimens of aluminum oxide, which have been placed over a printed page in order to demonstrate their differences in light-transmittance characteristics. The disk on the left is transparent (that is, virtually all light that is reflected from the page passes through it), whereas the one in the center is translucent (meaning that some of this reflected light is transmitted through the disk). And, the disk on the right is opaque—i.e., none of the light passes through it. These differences in optical properties are a consequence of differences in structure of these materials, which have resulted from the way the materials were processed. (Specimen preparation, P. A. Lessing; photography by S. Tanner.) 1496T_c01_01-14 11/9/05 17:02 Page 5 REVISED PAGES 1.4 Classification of Materials • 5 1.3 WHY STUDY MATERIALS SCIENCE AND ENGINEERING? Why do we study materials? Many an applied scientist or engineer, whether mechanical, civil, chemical, or electrical, will at one time or another be exposed to a design problem involving materials. Examples might include a transmission gear, the superstructure for a building, an oil refinery component, or an integrated circuit chip. Of course, materials scientists and engineers are specialists who are totally involved in the investigation and design of materials. Many times, a materials problem is one of selecting the right material from the many thousands that are available. There are several criteria on which the final decision is normally based. First of all, the in-service conditions must be characterized, for these will dictate the properties required of the material. On only rare occasions does a material possess the maximum or ideal combination of properties. Thus, it may be necessary to trade off one characteristic for another. The classic example involves strength and ductility; normally, a material having a high strength will have only a limited ductility. In such cases a reasonable compromise between two or more properties may be necessary. A second selection consideration is any deterioration of material properties that may occur during service operation. For example, significant reductions in mechanical strength may result from exposure to elevated temperatures or corrosive environments. Finally, probably the overriding consideration is that of economics: What will the finished product cost? A material may be found that has the ideal set of properties but is prohibitively expensive. Here again, some compromise is inevitable. The cost of a finished piece also includes any expense incurred during fabrication to produce the desired shape. The more familiar an engineer or scientist is with the various characteristics and structure–property relationships, as well as processing techniques of materials, the more proficient and confident he or she will be to make judicious materials choices based on these criteria. 1.4 CLASSIFICATION OF MATERIALS Solid materials have been conveniently grouped into three basic classifications: metals, ceramics, and polymers. This scheme is based primarily on chemical makeup and atomic structure, and most materials fall into one distinct grouping or another, although there are some intermediates. In addition, there are the composites, combinations of two or more of the above three basic material classes. A brief explanation of these material types and representative characteristics is offered next.Another classification is advanced materials—those used in high-technology applications— viz. semiconductors, biomaterials, smart materials, and nanoengineered materials; these are discussed in Section 1.5. Metals Materials in this group are composed of one or more metallic elements (such as iron, aluminum, copper, titanium, gold, and nickel), and often also nonmetallic elements (for example, carbon, nitrogen, and oxygen) in relatively small amounts.3 Atoms in metals and their alloys are arranged in a very orderly manner (as discussed in Chapter 3), and in comparison to the ceramics and polymers, are relatively dense (Figure 1.3).With 3 The term metal alloy is used in reference to a metallic substance that is composed of two or more elements. 1496T_c01_01-14 11/9/05 17:02 Page 6 REVISED PAGES Figure 1.3 Bar-chart of roomtemperature density values for various metals, ceramics, polymers, and composite materials. Density (g/cm3) (logarithmic scale) 6 • Chapter 1 / Introduction 40 Metals 20 Platinum Silver 10 8 6 Copper Iron/Steel Titanium 4 Aluminum 2 Magnesium Ceramics ZrO2 Al2O3 SiC,Si3N4 Glass Concrete Polymers PTFE Composites GFRC CFRC PVC PS PE Rubber 1.0 0.8 0.6 Woods 0.4 0.2 0.1 regard to mechanical characteristics, these materials are relatively stiff (Figure 1.4) and strong (Figure 1.5), yet are ductile (i.e., capable of large amounts of deformation without fracture), and are resistant to fracture (Figure 1.6), which accounts for their widespread use in structural applications. Metallic materials have large numbers of nonlocalized electrons; that is, these electrons are not bound to particular atoms. Many properties of metals are directly attributable to these electrons. For example, metals are extremely good conductors of electricity (Figure 1.7) and heat, and are not transparent to visible light; a polished metal surface has a lustrous appearance. In addition, some of the metals (viz., Fe, Co, and Ni) have desirable magnetic properties. Figure 1.8 is a photograph that shows several common and familiar objects that are made of metallic materials. Furthermore, the types and applications of metals and their alloys are discussed in Chapter 11. Ceramics Figure 1.4 Bar-chart of roomtemperature stiffness (i.e., elastic modulus) values for various metals, ceramics, polymers, and composite materials. Stiffness [Elastic (or Young’s) Modulus (in units of gigapascals)] (logarithmic scale) Ceramics are compounds between metallic and nonmetallic elements; they are most frequently oxides, nitrides, and carbides. For example, some of the common ceramic 1000 100 10 1.0 Metals Tungsten Iron/Steel Titanium Aluminum Magnesium Ceramics Composites SiC AI2O3 Si3N4 ZrO2 Glass Concrete CFRC GFRC Polymers PVC PS, Nylon PTFE PE 0.1 Rubbers 0.01 0.001 Woods 1496T_c01_01-14 12/20/05 7:11 Page 7 2nd REVISE PAGES 1.4 Classification of Materials • 7 Metals Strength (Tensile Strength, in units of megapascals) (logarithmic scale) Composites Ceramics 1000 Steel alloys Cu,Ti alloys 100 Aluminum alloys Gold CFRC Si3N4 Al2O3 GFRC SiC Polymers Glass Nylon PVC PS PE Woods PTFE 10 materials include aluminum oxide (or alumina,Al2O3), silicon dioxide (or silica, SiO2), silicon carbide (SiC), silicon nitride (Si3N4), and, in addition, what some refer to as the traditional ceramics—those composed of clay minerals (i.e., porcelain), as well as cement, and glass. With regard to mechanical behavior, ceramic materials are relatively stiff and strong—stiffnesses and strengths are comparable to those of the metals (Figures 1.4 and 1.5). In addition, ceramics are typically very hard. On the other hand, they are extremely brittle (lack ductility), and are highly susceptible to fracture (Figure 1.6). These materials are typically insulative to the passage of heat and electricity (i.e., have low electrical conductivities, Figure 1.7), and are more resistant to high temperatures and harsh environments than metals and polymers. With regard to optical characteristics, ceramics may be transparent, translucent, or opaque (Figure 1.2), and some of the oxide ceramics (e.g., Fe3O4) exhibit magnetic behavior. Metals Resistance to Fracture (Fracture Toughness, in units of MPa m) (logarithmic scale) Figure 1.5 Bar-chart of roomtemperature strength (i.e., tensile strength) values for various metals, ceramics, polymers, and composite materials. 100 Steel alloys Composites Titanium alloys Aluminum alloys 10 CFRC GFRC Ceramics Polymers Si3N4 Al2O3 SiC 1.0 Nylon Polystyrene Polyethylene Wood Glass Polyester Concrete 0.1 Figure 1.6 Bar-chart of room-temperature resistance to fracture (i.e., fracture toughness) for various metals, ceramics, polymers, and composite materials. (Reprinted from Engineering Materials 1: An Introduction to Properties, Applications and Design, third edition, M. F. Ashby and D. R. H. Jones, pages 177 and 178, Copyright 2005, with permission from Elsevier.) 1496T_c01_01-14 12/20/05 7:11 Page 8 2nd REVISE PAGES 8 • Chapter 1 / Introduction Metals 108 Semiconductors Electrical Conductivity (in units of reciprocal ohm-meters) (logarithmic scale) Figure 1.7 Bar-chart of roomtemperature electrical conductivity ranges for metals, ceramics, polymers, and semiconducting materials. 104 1 10–4 10–8 Ceramics Polymers 10–12 10–16 10–20 Several common ceramic objects are shown in the photograph of Figure 1.9. The characteristics, types, and applications of this class of materials are discussed in Chapters 12 and 13. Polymers Polymers include the familiar plastic and rubber materials. Many of them are organic compounds that are chemically based on carbon, hydrogen, and other nonmetallic elements (viz. O, N, and Si). Furthermore, they have very large molecular structures, often chain-like in nature that have a backbone of carbon atoms. Some of the common and familiar polymers are polyethylene (PE), nylon, poly(vinyl chloride) (PVC), polycarbonate (PC), polystyrene (PS), and silicone rubber. These materials typically have low densities (Figure 1.3), whereas their mechanical characteristics are generally dissimilar to the metallic and ceramic materials—they are not as stiff nor as strong as these other material types (Figures 1.4 and 1.5). However, on the basis of their low densities, many times their stiffnesses and strengths on a per mass Figure 1.8 Familiar objects that are made of metals and metal alloys: (from left to right) silverware (fork and knife), scissors, coins, a gear, a wedding ring, and a nut and bolt. (Photograpy by S. Tanner.) 1496T_c01_01-14 12/20/05 7:11 Page 9 2nd REVISE PAGES 1.4 Classification of Materials • 9 Figure 1.9 Common objects that are made of ceramic materials: scissors, a china tea cup, a building brick, a floor tile, and a glass vase. (Photography by S. Tanner.) basis are comparable to the metals and ceramics. In addition, many of the polymers are extremely ductile and pliable (i.e., plastic), which means they are easily formed into complex shapes. In general, they are relatively inert chemically and unreactive in a large number of environments. One major drawback to the polymers is their tendency to soften and/or decompose at modest temperatures, which, in some instances, limits their use. Furthermore, they have low electrical conductivities (Figure 1.7) and are nonmagnetic. The photograph in Figure 1.10 shows several articles made of polymers that are familiar to the reader. Chapters 14 and 15 are devoted to discussions of the structures, properties, applications, and processing of polymeric materials. Figure 1.10 Several common objects that are made of polymeric materials: plastic tableware (spoon, fork, and knife), billiard balls, a bicycle helmet, two dice, a lawnmower wheel (plastic hub and rubber tire), and a plastic milk carton. (Photography by S. Tanner.) 1496T_c01_01-14 11/9/05 17:02 Page 10 REVISED PAGES 10 • Chapter 1 / Introduction MATERIALS OF IMPORTANCE Carbonated Beverage Containers O ne common item that presents some interesting material property requirements is the container for carbonated beverages. The material used for this application must satisfy the following constraints: (1) provide a barrier to the passage of carbon dioxide, which is under pressure in the container; (2) be nontoxic, unreactive with the beverage, and, preferably be recyclable; (3) be relatively strong, and capable of surviving a drop from a height of several feet when containing the beverage; (4) be inexpensive and the cost to fabricate the final shape should be relatively low; (5) if optically transparent, retain its optical clarity; and (6) capable of being produced having different colors and/or able to be adorned with decorative labels. All three of the basic material types—metal (aluminum), ceramic (glass), and polymer (polyester plastic)—are used for carbonated beverage containers (per the chapter-opening photographs for this chapter).All of these materials are nontoxic and unreactive with beverages. In addition, each material has its pros and cons. For example, the aluminum alloy is relatively strong (but easily dented), is a very good barrier to the diffusion of carbon dioxide, is easily recycled, beverages are cooled rapidly, and labels may be painted onto its surface. On the other hand, the cans are optically opaque, and relatively expensive to produce. Glass is impervious to the passage of carbon dioxide, is a relatively inexpensive material, may be recycled, but it cracks and fractures easily, and glass bottles are relatively heavy. Whereas the plastic is relatively strong, may be made optically transparent, is inexpensive and lightweight, and is recyclable, it is not as impervious to the passage of carbon dioxide as the aluminum and glass. For example, you may have noticed that beverages in aluminum and glass containers retain their carbonization (i.e., “fizz”) for several years, whereas those in two-liter plastic bottles “go flat” within a few months. Composites A composite is composed of two (or more) individual materials, which come from the categories discussed above—viz., metals, ceramics, and polymers.The design goal of a composite is to achieve a combination of properties that is not displayed by any single material, and also to incorporate the best characteristics of each of the component materials. A large number of composite types exist that are represented by different combinations of metals, ceramics, and polymers. Furthermore, some naturally-occurring materials are also considered to be composites—for example, wood and bone. However, most of those we consider in our discussions are synthetic (or man-made) composites. One of the most common and familiar composites is fiberglass, in which small glass fibers are embedded within a polymeric material (normally an epoxy or polyester).4 The glass fibers are relatively strong and stiff (but also brittle), whereas the polymer is ductile (but also weak and flexible). Thus, the resulting fiberglass is relatively stiff, strong, (Figures 1.4 and 1.5) flexible, and ductile. In addition, it has a low density (Figure 1.3). Another of these technologically important materials is the “carbon fiberreinforced polymer” (or “CFRP”) composite—carbon fibers that are embedded within a polymer. These materials are stiffer and stronger than the glass fiber-reinforced materials (Figures 1.4 and 1.5), yet they are more expensive. The CFRP composites 4 Fiberglass is sometimes also termed a “glass fiber-reinforced polymer” composite, abbreviated “GFRP.” 1496T_c01_01-14 11/9/05 17:02 Page 11 REVISED PAGES 1.5 Advanced Materials • 11 are used in some aircraft and aerospace applications, as well as high-tech sporting equipment (e.g., bicycles, golf clubs, tennis rackets, and skis/snowboards). Chapter 16 is devoted to a discussion of these interesting materials. 1.5 ADVANCED MATERIALS Materials that are utilized in high-technology (or high-tech) applications are sometimes termed advanced materials. By high technology we mean a device or product that operates or functions using relatively intricate and sophisticated principles; examples include electronic equipment (camcorders, CD/DVD players, etc.), computers, fiber-optic systems, spacecraft, aircraft, and military rocketry.These advanced materials are typically traditional materials whose properties have been enhanced, and, also newly developed, high-performance materials. Furthermore, they may be of all material types (e.g., metals, ceramics, polymers), and are normally expensive. Advanced materials include semiconductors, biomaterials, and what we may term “materials of the future” (that is, smart materials and nanoengineered materials), which we discuss below. The properties and applications of a number of these advanced materials—for example, materials that are used for lasers, integrated circuits, magnetic information storage, liquid crystal displays (LCDs), and fiber optics—are also discussed in subsequent chapters. Semiconductors Semiconductors have electrical properties that are intermediate between the electrical conductors (viz. metals and metal alloys) and insulators (viz. ceramics and polymers)—Figure 1.7. Furthermore, the electrical characteristics of these materials are extremely sensitive to the presence of minute concentrations of impurity atoms, for which the concentrations may be controlled over very small spatial regions. Semiconductors have made possible the advent of integrated circuitry that has totally revolutionized the electronics and computer industries (not to mention our lives) over the past three decades. Biomaterials Biomaterials are employed in components implanted into the human body for replacement of diseased or damaged body parts. These materials must not produce toxic substances and must be compatible with body tissues (i.e., must not cause adverse biological reactions). All of the above materials—metals, ceramics, polymers, composites, and semiconductors—may be used as biomaterials. For example, some of the biomaterials that are utilized in artificial hip replacements are discussed in Section 22.12. Materials of the Future Smart Materials Smart (or intelligent) materials are a group of new and state-of-the-art materials now being developed that will have a significant influence on many of our technologies.The adjective “smart” implies that these materials are able to sense changes in their environments and then respond to these changes in predetermined manners— traits that are also found in living organisms. In addition, this “smart” concept is being extended to rather sophisticated systems that consist of both smart and traditional materials. Components of a smart material (or system) include some type of sensor (that detects an input signal), and an actuator (that performs a responsive and adaptive 1496T_c01_01-14 12/20/05 7:11 Page 12 2nd REVISE PAGES 12 • Chapter 1 / Introduction function). Actuators may be called upon to change shape, position, natural frequency, or mechanical characteristics in response to changes in temperature, electric fields, and/or magnetic fields. Four types of materials are commonly used for actuators: shape memory alloys, piezoelectric ceramics, magnetostrictive materials, and electrorheological/magnetorheological fluids. Shape memory alloys are metals that, after having been deformed, revert back to their original shapes when temperature is changed (see the Materials of Importance piece following Section 10.9). Piezoelectric ceramics expand and contract in response to an applied electric field (or voltage); conversely, they also generate an electric field when their dimensions are altered (see Section 18.25). The behavior of magnetostrictive materials is analogous to that of the piezoelectrics, except that they are responsive to magnetic fields. Also, electrorheological and magnetorheological fluids are liquids that experience dramatic changes in viscosity upon the application of electric and magnetic fields, respectively. Materials/devices employed as sensors include optical fibers (Section 21.14), piezoelectric materials (including some polymers), and microelectromechanical devices (MEMS, Section 13.8). For example, one type of smart system is used in helicopters to reduce aerodynamic cockpit noise that is created by the rotating rotor blades. Piezoelectric sensors inserted into the blades monitor blade stresses and deformations; feedback signals from these sensors are fed into a computer-controlled adaptive device, which generates noise-canceling antinoise. Nanoengineered Materials Until very recent times the general procedure utilized by scientists to understand the chemistry and physics of materials has been to begin by studying large and complex structures, and then to investigate the fundamental building blocks of these structures that are smaller and simpler. This approach is sometimes termed “topdown” science. However, with the advent of scanning probe microscopes (Section 4.10), which permit observation of individual atoms and molecules, it has become possible to manipulate and move atoms and molecules to form new structures and, thus, design new materials that are built from simple atomic-level constituents (i.e., “materials by design”). This ability to carefully arrange atoms provides opportunities to develop mechanical, electrical, magnetic, and other properties that are not otherwise possible. We call this the “bottom-up” approach, and the study of the properties of these materials is termed “nanotechnology”; the “nano” prefix denotes that the dimensions of these structural entities are on the order of a nanometer (109 m)—as a rule, less than 100 nanometers (equivalent to approximately 500 atom diameters).5 One example of a material of this type is the carbon nanotube, discussed in Section 12.4. In the future we will undoubtedly find that increasingly more of our technological advances will utilize these nanoengineered materials. 1.6 MODERN MATERIALS’ NEEDS In spite of the tremendous progress that has been made in the discipline of materials science and engineering within the past few years, there still remain technological challenges, including the development of even more sophisticated and specialized 5 One legendary and prophetic suggestion as to the possibility of nanoengineering materials was offered by Richard Feynman in his 1960 American Physical Society lecture that was entitled “There is Plenty of Room at the Bottom.” 1496T_c01_01-14 11/9/05 17:02 Page 13 REVISED PAGES References • 13 materials, as well as consideration of the environmental impact of materials production. Some comment is appropriate relative to these issues so as to round out this perspective. Nuclear energy holds some promise, but the solutions to the many problems that remain will necessarily involve materials, from fuels to containment structures to facilities for the disposal of radioactive waste. Significant quantities of energy are involved in transportation. Reducing the weight of transportation vehicles (automobiles, aircraft, trains, etc.), as well as increasing engine operating temperatures, will enhance fuel efficiency. New highstrength, low-density structural materials remain to be developed, as well as materials that have higher-temperature capabilities, for use in engine components. Furthermore, there is a recognized need to find new, economical sources of energy and to use present resources more efficiently. Materials will undoubtedly play a significant role in these developments. For example, the direct conversion of solar into electrical energy has been demonstrated. Solar cells employ some rather complex and expensive materials. To ensure a viable technology, materials that are highly efficient in this conversion process yet less costly must be developed. The hydrogen fuel cell is another very attractive and feasible energy-conversion technology that has the advantage of being non-polluting. It is just beginning to be implemented in batteries for electronic devices, and holds promise as the power plant for automobiles. New materials still need to be developed for more efficient fuel cells, and also for better catalysts to be used in the production of hydrogen. Furthermore, environmental quality depends on our ability to control air and water pollution. Pollution control techniques employ various materials. In addition, materials processing and refinement methods need to be improved so that they produce less environmental degradation—that is, less pollution and less despoilage of the landscape from the mining of raw materials. Also, in some materials manufacturing processes, toxic substances are produced, and the ecological impact of their disposal must be considered. Many materials that we use are derived from resources that are nonrenewable— that is, not capable of being regenerated. These include polymers, for which the prime raw material is oil, and some metals. These nonrenewable resources are gradually becoming depleted, which necessitates: (1) the discovery of additional reserves, (2) the development of new materials having comparable properties with less adverse environmental impact, and/or (3) increased recycling efforts and the development of new recycling technologies. As a consequence of the economics of not only production but also environmental impact and ecological factors, it is becoming increasingly important to consider the “cradle-to-grave” life cycle of materials relative to the overall manufacturing process. The roles that materials scientists and engineers play relative to these, as well as other environmental and societal issues, are discussed in more detail in Chapter 23. REFERENCES Ashby, M. F. and D. R. H. Jones, Engineering Materials 1, An Introduction to Their Properties and Applications, 3rd edition, ButterworthHeinemann, Woburn, UK, 2005. Ashby, M. F. and D. R. H. Jones, Engineering Materials 2, An Introduction to Microstructures, Pro- cessing and Design, 3rd edition, ButterworthHeinemann, Woburn, UK, 2005. Askeland, D. R. and P. P. Phulé, The Science and Engineering of Materials, 5th edition, Nelson (a division of Thomson Canada), Toronto, 2006. 1496T_c01_01-14 11/9/05 17:02 Page 14 REVISED PAGES 14 • Chapter 1 / Introduction Baillie, C. and L. Vanasupa, Navigating the Materials World, Academic Press, San Diego, CA, 2003. Flinn, R. A. and P. K. Trojan, Engineering Materials and Their Applications, 4th edition, John Wiley & Sons, New York, 1994. Jacobs, J. A. and T. F. Kilduff, Engineering Materials Technology, 5th edition, Prentice Hall PTR, Paramus, NJ, 2005. Mangonon, P. L., The Principles of Materials Selection for Engineering Design, Prentice Hall PTR, Paramus, NJ, 1999. McMahon, C. J., Jr., Structural Materials, Merion Books, Philadelphia, 2004. Murray, G. T., Introduction to Engineering Materials—Behavior, Properties, and Selection, Marcel Dekker, Inc., New York, 1993. Ralls, K. M., T. H. Courtney, and J. Wulff, Introduction to Materials Science and Engineering, John Wiley & Sons, New York, 1976. Schaffer, J. P., A. Saxena, S. D. Antolovich, T. H. Sanders, Jr., and S. B. Warner, The Science and Design of Engineering Materials, 2nd edition, WCB/McGraw-Hill, New York, 1999. Shackelford, J. F., Introduction to Materials Science for Engineers, 6th edition, Prentice Hall PTR, Paramus, NJ, 2005. Smith, W. F. and J. Hashemi, Principles of Materials Science and Engineering, 4th edition, McGraw-Hill Book Company, New York, 2006. Van Vlack, L. H., Elements of Materials Science and Engineering, 6th edition, Addison-Wesley Longman, Boston, MA, 1989. White, M. A., Properties of Materials, Oxford University Press, New York, 1999. 1496T_c02_15-37 12/20/05 7:19 Page 15 Chapter T 2nd REVISE PAGES 2 Atomic Structure and Interatomic Bonding his photograph shows the underside of a gecko. Geckos, harmless tropical lizards, are extremely fascinating and extraordinary animals. They have very sticky feet that cling to virtually any surface. This characteristic makes it possible for them to rapidly run up vertical walls and along the undersides of horizontal surfaces. In fact, a gecko can support its body mass with a single toe! The secret to this remarkable ability is the presence of an extremely large number of microscopically small hairs on each of their toe pads. When these hairs come in contact with a surface, weak forces of attraction (i.e., van der Waals forces) are established between hair molecules and molecules on the surface. The fact that these hairs are so small and so numerous explains why the gecko grips surfaces so tightly. To release its grip, the gecko simply curls up its toes, and peels the hairs away from the surface. Another interesting feature of these toe pads is that they are self-cleaning—that is, dirt particles don’t stick to them. Scientists are just beginning to understand the mechanism of adhesion for these tiny hairs, which may lead to the development of synthetic self-cleaning adhesives. Can you image duct tape that never looses its stickiness, or bandages that never leave a sticky residue? (Photograph courtesy of Professor Kellar Autumn, Lewis & Clark College, Portland, Oregon.) WHY STUDY Atomic Structure and Interatomic Bonding? An important reason to have an understanding of interatomic bonding in solids is that, in some instances, the type of bond allows us to explain a material’s properties. For example, consider carbon, which may exist as both graphite and diamond. Whereas graphite is relatively soft and has a “greasy” feel to it, diamond is the hardest known material. This dramatic disparity in properties is directly attributable to a type of interatomic bonding found in graphite that does not exist in diamond (see Section 12.4). • 15 1496T_c02_15-37 12/20/05 7:19 Page 16 2nd REVISE PAGES Learning Objectives After careful study of this chapter you should be able to do the following: 1. Name the two atomic models cited, and note (b) Note on this plot the equilibrium separation the differences between them. and the bonding energy. 2. Describe the important quantum-mechanical 4. (a) Briefly describe ionic, covalent, metallic, principle that relates to electron energies. hydrogen, and van der Waals bonds. 3. (a) Schematically plot attractive, repulsive, and (b) Note which materials exhibit each of these net energies versus interatomic separation bonding types. for two atoms or ions. 2.1 INTRODUCTION Some of the important properties of solid materials depend on geometrical atomic arrangements, and also the interactions that exist among constituent atoms or molecules. This chapter, by way of preparation for subsequent discussions, considers several fundamental and important concepts—namely, atomic structure, electron configurations in atoms and the periodic table, and the various types of primary and secondary interatomic bonds that hold together the atoms comprising a solid. These topics are reviewed briefly, under the assumption that some of the material is familiar to the reader. A t o m i c St r u c t u r e 2.2 FUNDAMENTAL CONCEPTS atomic number isotope atomic weight atomic mass unit Each atom consists of a very small nucleus composed of protons and neutrons, which is encircled by moving electrons. Both electrons and protons are electrically charged, the charge magnitude being 1.60  1019 C, which is negative in sign for electrons and positive for protons; neutrons are electrically neutral. Masses for these subatomic particles are infinitesimally small; protons and neutrons have approximately the same mass, 1.67  1027 kg, which is significantly larger than that of an electron, 9.11  1031 kg. Each chemical element is characterized by the number of protons in the nucleus, or the atomic number (Z).1 For an electrically neutral or complete atom, the atomic number also equals the number of electrons. This atomic number ranges in integral units from 1 for hydrogen to 92 for uranium, the highest of the naturally occurring elements. The atomic mass (A) of a specific atom may be expressed as the sum of the masses of protons and neutrons within the nucleus. Although the number of protons is the same for all atoms of a given element, the number of neutrons (N) may be variable. Thus atoms of some elements have two or more different atomic masses, which are called isotopes. The atomic weight of an element corresponds to the weighted average of the atomic masses of the atom’s naturally occurring isotopes.2 The atomic mass unit (amu) may be used for computations of atomic weight. A 1 Terms appearing in boldface type are defined in the Glossary, which follows Appendix E. The term “atomic mass” is really more accurate than “atomic weight” inasmuch as, in this context, we are dealing with masses and not weights. However, atomic weight is, by convention, the preferred terminology and will be used throughout this book. The reader should note that it is not necessary to divide molecular weight by the gravitational constant. 2 1496T_c02_15-37 11/10/05 10:42 Page 17 REVISED PAGES 2.3 Electrons in Atoms • 17 scale has been established whereby 1 amu is defined as 121 of the atomic mass of the most common isotope of carbon, carbon 12 1 12C2 1A  12.000002. Within this scheme, the masses of protons and neutrons are slightly greater than unity, and AZN mole (2.1) The atomic weight of an element or the molecular weight of a compound may be specified on the basis of amu per atom (molecule) or mass per mole of material. In one mole of a substance there are 6.023  1023 (Avogadro’s number) atoms or molecules. These two atomic weight schemes are related through the following equation: 1 amu/atom 1or molecule2  1 g/mol For example, the atomic weight of iron is 55.85 amu/atom, or 55.85 g/mol. Sometimes use of amu per atom or molecule is convenient; on other occasions g (or kg)/mol is preferred. The latter is used in this book. Concept Check 2.1 Why are the atomic weights of the elements generally not integers? Cite two reasons. [The answer may be found at www.wiley.com/college/callister (Student Companion Site).] 2.3 ELECTRONS IN ATOMS Atomic Models quantum mechanics Bohr atomic model During the latter part of the nineteenth century it was realized that many phenomena involving electrons in solids could not be explained in terms of classical mechanics. What followed was the establishment of a set of principles and laws that govern systems of atomic and subatomic entities that came to be known as quantum mechanics. An understanding of the behavior of electrons in atoms and crystalline solids necessarily involves the discussion of quantum-mechanical concepts. However, a detailed exploration of these principles is beyond the scope of this book, and only a very superficial and simplified treatment is given. One early outgrowth of quantum mechanics was the simplified Bohr atomic model, in which electrons are assumed to revolve around the atomic nucleus in discrete orbitals, and the position of any particular electron is more or less well defined in terms of its orbital. This model of the atom is represented in Figure 2.1. Another important quantum-mechanical principle stipulates that the energies of electrons are quantized; that is, electrons are permitted to have only specific values of energy. An electron may change energy, but in doing so it must make a quantum jump either to an allowed higher energy (with absorption of energy) or to a lower energy (with emission of energy). Often, it is convenient to think of these allowed electron energies as being associated with energy levels or states. These states do not vary continuously with energy; that is, adjacent states are separated by finite energies. For example, allowed states for the Bohr hydrogen atom are represented in Figure 2.2a. These energies are taken to be negative, whereas the zero reference is the unbound or free electron. Of course, the single electron associated with the hydrogen atom will fill only one of these states. 1496T_c02_15-37 12/20/05 13:51 Page 18 2nd REVISE PAGES 18 • Chapter 2 / Atomic Structure and Interatomic Bonding Figure 2.1 Schematic representation of the Bohr atom. Orbital electron Nucleus 0 0 –1.5 n=3 –3.4 n=2 3d 3p 3s 2p 2s –1 × 10–18 –10 –2 × 10–18 n=1 –13.6 –15 (a) 1s (b) Energy (J) –5 Energy (eV) wave-mechanical model Thus, the Bohr model represents an early attempt to describe electrons in atoms, in terms of both position (electron orbitals) and energy (quantized energy levels). This Bohr model was eventually found to have some significant limitations because of its inability to explain several phenomena involving electrons. A resolution was reached with a wave-mechanical model, in which the electron is considered to exhibit both wave-like and particle-like characteristics. With this model, an electron is no longer treated as a particle moving in a discrete orbital; rather, position is considered to be the probability of an electron’s being at various locations around the nucleus. In other words, position is described by a probability distribution or electron cloud. Figure 2.3 compares Bohr and wavemechanical models for the hydrogen atom. Both these models are used throughout the course of this book; the choice depends on which model allows the more simple explanation. Figure 2.2 (a) The first three electron energy states for the Bohr hydrogen atom. (b) Electron energy states for the first three shells of the wave-mechanical hydrogen atom. (Adapted from W. G. Moffatt, G. W. Pearsall, and J. Wulff, The Structure and Properties of Materials, Vol. I, Structure, p. 10. Copyright © 1964 by John Wiley & Sons, New York. Reprinted by permission of John Wiley & Sons, Inc.) 1496T_c02_15-37 12/20/05 7:19 Page 19 2nd REVISE PAGES 2.3 Electrons in Atoms • 19 Figure 2.3 Comparison of the (a) Bohr and (b) wavemechanical atom models in terms of electron distribution. (Adapted from Z. D. Jastrzebski, The Nature and Properties of Engineering Materials, 3rd edition, p. 4. Copyright © 1987 by John Wiley & Sons, New York. Reprinted by permission of John Wiley & Sons, Inc.) Probability 1.0 0 Distance from nucleus Orbital electron Nucleus (a) (b) Quantum Numbers quantum number Using wave mechanics, every electron in an atom is characterized by four parameters called quantum numbers. The size, shape, and spatial orientation of an electron’s probability density are specified by three of these quantum numbers. Furthermore, Bohr energy levels separate into electron subshells, and quantum numbers dictate the number of states within each subshell. Shells are specified by a principal quantum number n, which may take on integral values beginning with unity; sometimes these shells are designated by the letters K, L, M, N, O, and so on, which correspond, respectively, to n  1, 2, 3, 4, 5, . . . , as indicated in Table 2.1. Note also that this quantum number, and it only, is also associated with the Bohr model. This quantum number is related to the distance of an electron from the nucleus, or its position. The second quantum number, l, signifies the subshell, which is denoted by a lowercase letter—an s, p, d, or f; it is related to the shape of the electron subshell. In addition, the number of these subshells is restricted by the magnitude of n. Allowable subshells for the several n values are also presented in Table 2.1. The number of energy states for each subshell is determined by the third quantum number, ml. For an s subshell, there is a single energy state, whereas for p, d, and f subshells, three, five, and seven states exist, respectively (Table 2.1). In the absence of an external magnetic field, the states within each subshell are identical. However, when a magnetic field is applied these subshell states split, each state assuming a slightly different energy. 1496T_c02_15-37 11/10/05 10:42 Page 20 REVISED PAGES 20 • Chapter 2 / Atomic Structure and Interatomic Bonding Table 2.1 The Number of Available Electron States in Some of the Electron Shells and Subshells Principal Quantum Number n Shell Designation 1 Number of Electrons Subshells Number of States Per Subshell Per Shell K s 1 2 2 2 L s p 1 3 2 6 8 3 M s p d 1 3 5 2 6 10 18 N s p d f 1 3 5 7 2 6 10 14 32 4 Energy Associated with each electron is a spin moment, which must be oriented either up or down. Related to this spin moment is the fourth quantum number, ms, for which two values are possible ( 12 and 12), one for each of the spin orientations. Thus, the Bohr model was further refined by wave mechanics, in which the introduction of three new quantum numbers gives rise to electron subshells within each shell. A comparison of these two models on this basis is illustrated, for the hydrogen atom, in Figures 2.2a and 2.2b. A complete energy level diagram for the various shells and subshells using the wave-mechanical model is shown in Figure 2.4. Several features of the diagram are worth noting. First, the smaller the principal quantum number, the lower the energy level; for example, the energy of a 1s state is less than that of a 2s state, which in turn is lower than the 3s. Second, within each shell, the energy of a subshell level increases with the value of the l quantum number. For example, the energy of a 3d state is greater than a 3p, which is larger than 3s. Finally, there may be overlap in f d f d p s f d p s d p s Figure 2.4 Schematic representation of the relative energies of the electrons for the various shells and subshells. (From K. M. Ralls, T. H. Courtney, and J. Wulff, Introduction to Materials Science and Engineering, p. 22. Copyright © 1976 by John Wiley & Sons, New York. Reprinted by permission of John Wiley & Sons, Inc.) p s d p s p s s 1 2 3 4 5 Principal quantum number, n 6 7 1496T_c02_15-37 11/10/05 13:32 Page 21 REVISED PAGES 2.3 Electrons in Atoms • 21 energy of a state in one shell with states in an adjacent shell, which is especially true of d and f states; for example, the energy of a 3d state is greater than that for a 4s. Electron Configurations Pauli exclusion principle ground state electron configuration valence electron The preceding discussion has dealt primarily with electron states—values of energy that are permitted for electrons.To determine the manner in which these states are filled with electrons, we use the Pauli exclusion principle, another quantum-mechanical concept. This principle stipulates that each electron state can hold no more than two electrons, which must have opposite spins. Thus, s, p, d, and f subshells may each accommodate, respectively, a total of 2, 6, 10, and 14 electrons; Table 2.1 summarizes the maximum number of electrons that may occupy each of the first four shells. Of course, not all possible states in an atom are filled with electrons. For most atoms, the electrons fill up the lowest possible energy states in the electron shells and subshells, two electrons (having opposite spins) per state. The energy structure for a sodium atom is represented schematically in Figure 2.5. When all the electrons occupy the lowest possible energies in accord with the foregoing restrictions, an atom is said to be in its ground state. However, electron transitions to higher energy states are possible, as discussed in Chapters 18 and 21. The electron configuration or structure of an atom represents the manner in which these states are occupied. In the conventional notation the number of electrons in each subshell is indicated by a superscript after the shell–subshell designation. For example, the electron configurations for hydrogen, helium, and sodium are, respectively, 1s1, 1s2, and 1s22s22p63s1. Electron configurations for some of the more common elements are listed in Table 2.2. At this point, comments regarding these electron configurations are necessary. First, the valence electrons are those that occupy the outermost shell. These electrons are extremely important; as will be seen, they participate in the bonding between atoms to form atomic and molecular aggregates. Furthermore, many of the physical and chemical properties of solids are based on these valence electrons. In addition, some atoms have what are termed “stable electron configurations”; that is, the states within the outermost or valence electron shell are completely filled. Normally this corresponds to the occupation of just the s and p states for the outermost shell by a total of eight electrons, as in neon, argon, and krypton; one exception is helium, which contains only two 1s electrons. These elements (Ne, Ar, Kr, and He) are the inert, or noble, gases, which are virtually unreactive chemically. Some atoms of the elements that have unfilled valence shells assume stable electron configurations by gaining or losing electrons to form charged ions, 3p 3s Increasing energy electron state 2p 2s 1s Figure 2.5 Schematic representation of the filled and lowest unfilled energy states for a sodium atom. 1496T_c02_15-37 11/10/05 10:42 Page 22 REVISED PAGES 22 • Chapter 2 / Atomic Structure and Interatomic Bonding Table 2.2 A Listing of the Expected Electron Configurations for Some of the Common Elementsa Element Hydrogen Helium Lithium Beryllium Boron Carbon Nitrogen Oxygen Fluorine Neon Sodium Magnesium Aluminum Silicon Phosphorus Sulfur Chlorine Argon Potassium Calcium Scandium Titanium Vanadium Chromium Manganese Iron Cobalt Nickel Copper Zinc Gallium Germanium Arsenic Selenium Bromine Krypton Symbol Atomic Number H He Li Be B C N O F Ne Na Mg Al Si P S Cl Ar K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 Electron Configuration 1s1 1s2 1s22s1 1s22s2 1s22s22p1 1s22s22p2 1s22s22p3 1s22s22p4 1s22s22p5 1s22s22p6 1s22s22p63s1 1s22s22p63s2 1s22s22p63s23p1 1s22s22p63s23p2 1s22s22p63s23p3 1s22s22p63s23p4 1s22s22p63s23p5 1s22s22p63s23p6 1s22s22p63s23p64s1 1s22s22p63s23p64s2 1s22s22p63s23p63d14s2 1s22s22p63s23p63d24s2 1s22s22p63s23p63d34s2 1s22s22p63s23p63d54s1 1s22s22p63s23p63d54s2 1s22s22p63s23p63d64s2 1s22s22p63s23p63d74s2 1s22s22p63s23p63d84s2 1s22s22p63s23p63d104s1 1s22s22p63s23p63d104s2 1s22s22p63s23p63d104s24p1 1s22s22p63s23p63d104s24p2 1s22s22p63s23p63d104s24p3 1s22s22p63s23p63d104s24p4 1s22s22p63s23p63d104s24p5 1s22s22p63s23p63d104s24p6 a When some elements covalently bond, they form sp hybrid bonds. This is especially true for C, Si, and Ge. or by sharing electrons with other atoms. This is the basis for some chemical reactions, and also for atomic bonding in solids, as explained in Section 2.6. Under special circumstances, the s and p orbitals combine to form hybrid spn orbitals, where n indicates the number of p orbitals involved, which may have a value of 1, 2, or 3. The 3A, 4A, and 5A group elements of the periodic table (Figure 2.6) are those that most often form these hybrids. The driving force for the formation of hybrid orbitals is a lower energy state for the valence electrons. For carbon the sp3 hybrid is of primary importance in organic and polymer chemistries. The shape of the sp3 hybrid is what determines the 109 (or tetrahedral) angle found in polymer chains (Chapter 14). 1496T_c02_15-37 11/10/05 10:42 Page 23 REVISED PAGES 2.4 The Periodic Table • 23 Concept Check 2.2 Give electron configurations for the Fe3 and S2 ions. [The answer may be found at www.wiley.com/college/callister (Student Companion Site).] 2.4 THE PERIODIC TABLE All the elements have been classified according to electron configuration in the periodic table (Figure 2.6). Here, the elements are situated, with increasing atomic number, in seven horizontal rows called periods. The arrangement is such that all elements arrayed in a given column or group have similar valence electron structures, as well as chemical and physical properties. These properties change gradually, moving horizontally across each period and vertically down each column. The elements positioned in Group 0, the rightmost group, are the inert gases, which have filled electron shells and stable electron configurations. Group VIIA and VIA elements are one and two electrons deficient, respectively, from having stable structures. The Group VIIA elements (F, Cl, Br, I, and At) are sometimes termed the halogens. The alkali and the alkaline earth metals (Li, Na, K, Be, Mg, Ca, etc.) are labeled as Groups IA and IIA, having, respectively, one and two electrons in excess of stable structures. The elements in the three long periods, Groups IIIB through IIB, are termed the transition metals, which have partially filled d electron states and in some cases one or two electrons in the next higher energy shell. Groups IIIA, IVA, and VA (B, Si, Ge, As, etc.) display characteristics that are intermediate between the metals and nonmetals by virtue of their valence electron structures. periodic table Metal IA Key 1 29 Atomic number H Cu Symbol 1.0080 3 IIA 63.54 4 Li Be 6.941 11 9.0122 12 0 Nonmetal 2 He Atomic weight Intermediate VIII IIIA IVA VA VIA VIIA 5 6 7 8 9 4.0026 10 B C N O F Ne 10.811 13 12.011 14 14.007 15 15.999 16 18.998 17 20.180 18 Na Mg 22.990 19 24.305 20 IIIB IVB VB VIB 21 22 23 24 25 26 27 28 29 30 K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr 39.098 40.08 44.956 47.87 50.942 51.996 54.938 55.845 58.933 58.69 63.54 65.41 69.72 72.64 74.922 78.96 79.904 83.80 VIIB IB IIB Al Si P S Cl Ar 26.982 31 28.086 32 30.974 33 32.064 34 35.453 35 39.948 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe 85.47 55 87.62 56 88.91 91.22 72 92.91 73 95.94 74 (98) 75 101.07 76 102.91 77 106.4 78 107.87 79 112.41 80 114.82 81 118.71 82 121.76 83 127.60 84 126.90 85 131.30 86 Cs Ba 132.91 87 137.34 88 Fr Ra (223) (226) Rare earth series Actinide series Rare earth series Actinide series Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po At Rn 178.49 104 180.95 105 183.84 106 186.2 107 190.23 108 192.2 109 195.08 110 196.97 200.59 204.38 207.19 208.98 (209) (210) (222) Rf Db Sg Bh Hs Mt Ds (261) (262) (266) (264) (277) (268) (281) 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 La Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu 138.91 140.12 140.91 144.24 (145) 150.35 151.96 157.25 158.92 162.50 164.93 167.26 168.93 173.04 174.97 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 Ac Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No Lr (227) 232.04 231.04 238.03 (237) (244) (243) (247) (247) (251) (252) (257) (258) (259) (262) Figure 2.6 The periodic table of the elements. The numbers in parentheses are the atomic weights of the most stable or common isotopes. 1496T_c02_15-37 11/10/05 10:42 Page 24 REVISED PAGES 24 • Chapter 2 / Atomic Structure and Interatomic Bonding 0 IA 1 2 H He 2.1 3 IIA IIIA IVA VA VIA VIIA 4 5 6 7 8 9 – 10 Li Be B C N O F Ne 1.0 11 1.5 12 2.0 13 2.5 14 3.0 15 3.5 16 4.0 17 – 18 Na Mg 0.9 19 1.2 20 VIII IIIB IVB VB VIB VIIB 21 22 23 24 25 26 27 28 IB IIB 29 30 Al Si P S Cl Ar 1.5 31 1.8 32 2.1 33 2.5 34 3.0 35 – 36 Kr K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br 0.8 1.0 1.3 1.5 1.6 1.6 1.5 1.8 1.8 1.8 1.9 1.6 1.6 1.8 2.0 2.4 2.8 – 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe 0.8 55 1.0 56 1.2 57–71 1.4 72 1.6 73 1.8 74 1.9 75 2.2 76 2.2 77 2.2 78 1.9 79 1.7 80 1.7 81 1.8 82 1.9 83 2.1 84 2.5 85 – 86 Cs Ba La–Lu Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po At Rn 0.7 87 0.9 88 1.1–1.2 89–102 1.3 1.5 1.7 1.9 2.2 2.2 2.2 2.4 1.9 1.8 1.8 1.9 2.0 2.2 – Fr Ra Ac–No 0.7 0.9 1.1–1.7 Figure 2.7 The electronegativity values for the elements. (Adapted from Linus Pauling, The Nature of the Chemical Bond, 3rd edition. Copyright 1939 and 1940, 3rd edition copyright © 1960, by Cornell University. Used by permission of the publisher, Cornell University Press.) electropositive electronegative As may be noted from the periodic table, most of the elements really come under the metal classification. These are sometimes termed electropositive elements, indicating that they are capable of giving up their few valence electrons to become positively charged ions. Furthermore, the elements situated on the right-hand side of the table are electronegative; that is, they readily accept electrons to form negatively charged ions, or sometimes they share electrons with other atoms. Figure 2.7 displays electronegativity values that have been assigned to the various elements arranged in the periodic table. As a general rule, electronegativity increases in moving from left to right and from bottom to top. Atoms are more likely to accept electrons if their outer shells are almost full, and if they are less “shielded” from (i.e., closer to) the nucleus. Atomic Bonding in Solids 2.5 BONDING FORCES AND ENERGIES An understanding of many of the physical properties of materials is predicated on a knowledge of the interatomic forces that bind the atoms together. Perhaps the principles of atomic bonding are best illustrated by considering the interaction between two isolated atoms as they are brought into close proximity from an infinite separation. At large distances, the interactions are negligible, but as the atoms approach, each exerts forces on the other. These forces are of two types, attractive and repulsive, and the magnitude of each is a function of the separation or interatomic distance. The origin of an attractive force FA depends on the particular type of bonding that exists between the two atoms. The magnitude of the attractive force varies with the distance, as represented schematically in Figure 2.8a. Ultimately, the outer electron shells of the two atoms begin to overlap, and a strong repulsive force FR comes into play. The net force FN between the two atoms is just the sum of both attractive and repulsive components; that is, FN  FA  FR (2.2) 1496T_c02_15-37 11/10/05 10:42 Page 25 REVISED PAGES 2.5 Bonding Forces and Energies • 25 + Figure 2.8 (a) The dependence of repulsive, attractive, and net forces on interatomic separation for two isolated atoms. (b) The dependence of repulsive, attractive, and net potential energies on interatomic separation for two isolated atoms. Force F Attraction Attractive force FA 0 Repulsion Interatomic separation r r0 Repulsive force FR Net force FN – (a) Repulsion Repulsive energy ER Interatomic separation r 0 Net energy EN Attraction Potential energy E + E0 Attractive energy EA – (b) which is also a function of the interatomic separation, as also plotted in Figure 2.8a. When FA and FR balance, or become equal, there is no net force; that is, FA  FR  0 (2.3) Then a state of equilibrium exists. The centers of the two atoms will remain separated by the equilibrium spacing r0, as indicated in Figure 2.8a. For many atoms, r0 is approximately 0.3 nm. Once in this position, the two atoms will counteract any attempt to separate them by an attractive force, or to push them together by a repulsive action. Sometimes it is more convenient to work with the potential energies between two atoms instead of forces. Mathematically, energy (E) and force (F) are related as Force-potential energy relationship for two atoms E  F dr (2.4) dr (2.5) or, for atomic systems, r EN  F N q   r FA dr  q  EA  ER r F R dr (2.6) q (2.7) 1496T_c02_15-37 11/10/05 10:42 Page 26 REVISED PAGES 26 • Chapter 2 / Atomic Structure and Interatomic Bonding bonding energy primary bond in which EN, EA, and ER are respectively the net, attractive, and repulsive energies for two isolated and adjacent atoms. Figure 2.8b plots attractive, repulsive, and net potential energies as a function of interatomic separation for two atoms. The net curve, which is again the sum of the other two, has a potential energy trough or well around its minimum. Here, the same equilibrium spacing, r0, corresponds to the separation distance at the minimum of the potential energy curve. The bonding energy for these two atoms, E0, corresponds to the energy at this minimum point (also shown in Figure 2.8b); it represents the energy that would be required to separate these two atoms to an infinite separation. Although the preceding treatment has dealt with an ideal situation involving only two atoms, a similar yet more complex condition exists for solid materials because force and energy interactions among many atoms must be considered. Nevertheless, a bonding energy, analogous to E0 above, may be associated with each atom. The magnitude of this bonding energy and the shape of the energy-versusinteratomic separation curve vary from material to material, and they both depend on the type of atomic bonding. Furthermore, a number of material properties depend on E0, the curve shape, and bonding type. For example, materials having large bonding energies typically also have high melting temperatures; at room temperature, solid substances are formed for large bonding energies, whereas for small energies the gaseous state is favored; liquids prevail when the energies are of intermediate magnitude. In addition, as discussed in Section 6.3, the mechanical stiffness (or modulus of elasticity) of a material is dependent on the shape of its force-versusinteratomic separation curve (Figure 6.7). The slope for a relatively stiff material at the r  r0 position on the curve will be quite steep; slopes are shallower for more flexible materials. Furthermore, how much a material expands upon heating or contracts upon cooling (that is, its linear coefficient of thermal expansion) is related to the shape of its E0-versus-r0 curve (see Section 19.3). A deep and narrow “trough,” which typically occurs for materials having large bonding energies, normally correlates with a low coefficient of thermal expansion and relatively small dimensional alterations for changes in temperature. Three different types of primary or chemical bond are found in solids—ionic, covalent, and metallic. For each type, the bonding necessarily involves the valence electrons; furthermore, the nature of the bond depends on the electron structures of the constituent atoms. In general, each of these three types of bonding arises from the tendency of the atoms to assume stable electron structures, like those of the inert gases, by completely filling the outermost electron shell. Secondary or physical forces and energies are also found in many solid materials; they are weaker than the primary ones, but nonetheless influence the physical properties of some materials. The sections that follow explain the several kinds of primary and secondary interatomic bonds. 2.6 PRIMARY INTERATOMIC BONDS Ionic Bonding ionic bonding Ionic bonding is perhaps the easiest to describe and visualize. It is always found in compounds that are composed of both metallic and nonmetallic elements, elements that are situated at the horizontal extremities of the periodic table. Atoms of a metallic element easily give up their valence electrons to the nonmetallic atoms. In the process all the atoms acquire stable or inert gas configurations and, in addition, an electrical charge; that is, they become ions. Sodium chloride (NaCl) is the classic ionic material. A sodium atom can assume the electron structure of neon (and a net single 1496T_c02_15-37 11/10/05 10:42 Page 27 REVISED PAGES 2.6 Primary Interatomic Bonds • 27 Figure 2.9 Schematic representation of ionic bonding in sodium chloride (NaCl). Coulombic bonding force coulombic force Na+ Cl– Na+ Cl– Na+ Cl– Na+ Cl– Na+ Cl– Na+ Cl– Na+ Cl– Na+ Cl– Na+ Cl– Na+ Cl– positive charge) by a transfer of its one valence 3s electron to a chlorine atom. After such a transfer, the chlorine ion has a net negative charge and an electron configuration identical to that of argon. In sodium chloride, all the sodium and chlorine exist as ions. This type of bonding is illustrated schematically in Figure 2.9. The attractive bonding forces are coulombic; that is, positive and negative ions, by virtue of their net electrical charge, attract one another. For two isolated ions, the attractive energy EA is a function of the interatomic distance according to3 Attractive energyinteratomic separation relationship EA   A r (2.8) An analogous equation for the repulsive energy is Repulsive energyinteratomic separation relationship ER  B rn (2.9) In these expressions, A, B, and n are constants whose values depend on the particular ionic system. The value of n is approximately 8. Ionic bonding is termed nondirectional; that is, the magnitude of the bond is equal in all directions around an ion. It follows that for ionic materials to be stable, all positive ions must have as nearest neighbors negatively charged ions in a threedimensional scheme, and vice versa. The predominant bonding in ceramic materials is ionic. Some of the ion arrangements for these materials are discussed in Chapter 12. Bonding energies, which generally range between 600 and 1500 kJ/mol (3 and 8 eV/atom), are relatively large, as reflected in high melting temperatures.4 Table 2.3 contains bonding energies and melting temperatures for several ionic materials. 3 The constant A in Equation 2.8 is equal to 1 1Z e21Z2e2 4p0 1 where 0 is the permittivity of a vacuum 18.85  1012 F/m2, Z1 and Z2 are the valences of the two ion types, and e is the electronic charge 11.602  1019 C2. 4 Sometimes bonding energies are expressed per atom or per ion. Under these circumstances the electron volt (eV) is a conveniently small unit of energy. It is, by definition, the energy imparted to an electron as it falls through an electric potential of one volt. The joule equivalent of the electron volt is as follows: 1.602  1019 J  1 eV. 1496T_c02_15-37 11/10/05 10:42 Page 28 REVISED PAGES 28 • Chapter 2 / Atomic Structure and Interatomic Bonding Table 2.3 Bonding Energies and Melting Temperatures for Various Substances Bonding Energy Bonding Type Substance Melting Temperature (°C) kJ/mol eV/Atom, Ion, Molecule 640 1000 3.3 5.2 801 2800 Ionic NaCl MgO Covalent Si C (diamond) 450 713 4.7 7.4 1410 3550 Metallic Hg Al Fe W 68 324 406 849 0.7 3.4 4.2 8.8 39 660 1538 3410 van der Waals Ar Cl2 7.7 31 0.08 0.32 189 101 Hydrogen NH3 H2O 35 51 0.36 0.52 78 0 Ionic materials are characteristically hard and brittle and, furthermore, electrically and thermally insulative. As discussed in subsequent chapters, these properties are a direct consequence of electron configurations and/or the nature of the ionic bond. Covalent Bonding covalent bonding In covalent bonding, stable electron configurations are assumed by the sharing of electrons between adjacent atoms. Two atoms that are covalently bonded will each contribute at least one electron to the bond, and the shared electrons may be considered to belong to both atoms. Covalent bonding is schematically illustrated in Figure 2.10 for a molecule of methane 1CH4 2 . The carbon atom has four valence electrons, whereas each of the four hydrogen atoms has a single valence electron. Each hydrogen atom can acquire a helium electron configuration (two 1s valence electrons) when the carbon atom shares with it one electron. The carbon now has four additional shared electrons, one from each hydrogen, for a total of eight valence electrons, and the electron structure of neon. The covalent bond is directional; that is, it is between specific atoms and may exist only in the direction between one atom and another that participates in the electron sharing. H Shared electron from carbon Shared electron from hydrogen H C H H Figure 2.10 Schematic representation of covalent bonding in a molecule of methane 1CH4 2 . 1496T_c02_15-37 11/10/05 10:42 Page 29 REVISED PAGES 2.6 Primary Interatomic Bonds • 29 Many nonmetallic elemental molecules 1H2, Cl2, F2, etc.2 as well as molecules containing dissimilar atoms, such as CH4, H2O, HNO3, and HF, are covalently bonded. Furthermore, this type of bonding is found in elemental solids such as diamond (carbon), silicon, and germanium and other solid compounds composed of elements that are located on the right-hand side of the periodic table, such as gallium arsenide (GaAs), indium antimonide (InSb), and silicon carbide (SiC). The number of covalent bonds that is possible for a particular atom is determined by the number of valence electrons. For N¿ valence electrons, an atom can covalently bond with at most 8  N¿ other atoms. For example, N¿  7 for chlorine, and 8  N¿  1, which means that one Cl atom can bond to only one other atom, as in Cl2. Similarly, for carbon, N¿  4, and each carbon atom has 8  4, or four, electrons to share. Diamond is simply the three-dimensional interconnecting structure wherein each carbon atom covalently bonds with four other carbon atoms. This arrangement is represented in Figure 12.15. Covalent bonds may be very strong, as in diamond, which is very hard and has a very high melting temperature, 7 3550C 16400F2, or they may be very weak, as with bismuth, which melts at about 270C 1518F2 . Bonding energies and melting temperatures for a few covalently bonded materials are presented in Table 2.3. Polymeric materials typify this bond, the basic molecular structure being a long chain of carbon atoms that are covalently bonded together with two of their available four bonds per atom. The remaining two bonds normally are shared with other atoms, which also covalently bond. Polymeric molecular structures are discussed in detail in Chapter 14. It is possible to have interatomic bonds that are partially ionic and partially covalent, and, in fact, very few compounds exhibit pure ionic or covalent bonding. For a compound, the degree of either bond type depends on the relative positions of the constituent atoms in the periodic table (Figure 2.6) or the difference in their electronegativities (Figure 2.7).The wider the separation (both horizontally—relative to Group IVA—and vertically) from the lower left to the upper-right-hand corner (i.e., the greater the difference in electronegativity), the more ionic the bond. Conversely, the closer the atoms are together (i.e., the smaller the difference in electronegativity), the greater the degree of covalency. The percentage ionic character of a bond between elements A and B (A being the most electronegative) may be approximated by the expression % ionic character  51  exp310.2521XA  XB 2 2 4 6  100 (2.10) where XA and XB are the electronegativities for the respective elements. Metallic Bonding metallic bonding Metallic bonding, the final primary bonding type, is found in metals and their alloys. A relatively simple model has been proposed that very nearly approximates the bonding scheme. Metallic materials have one, two, or at most, three valence electrons. With this model, these valence electrons are not bound to any particular atom in the solid and are more or less free to drift throughout the entire metal. They may be thought of as belonging to the metal as a whole, or forming a “sea of electrons” or an “electron cloud.” The remaining nonvalence electrons and atomic nuclei form what are called ion cores, which possess a net positive charge equal in magnitude to the total valence electron charge per atom. Figure 2.11 is a schematic illustration of metallic bonding. The free electrons shield the positively charged ion cores from mutually repulsive electrostatic forces, which they would otherwise exert upon one another; consequently the metallic bond is nondirectional in character. In addition, these free electrons act as a “glue” to hold the ion cores together. Bonding energies 1496T_c02_15-37 12/20/05 7:19 Page 30 2nd REVISE PAGES 30 • Chapter 2 / Atomic Structure and Interatomic Bonding Ion cores + + – + – + – + + + + – + – + + – – – + + Figure 2.11 Schematic illustration of metallic bonding. + – + + Sea of valence electrons and melting temperatures for several metals are listed in Table 2.3. Bonding may be weak or strong; energies range from 68 kJ/mol (0.7 eV/atom) for mercury to 850 kJ/mol (8.8 eV/atom) for tungsten. Their respective melting temperatures are 39 and 3410C 138 and 6170F2 . Metallic bonding is found in the periodic table for Group IA and IIA elements and, in fact, for all elemental metals. Some general behaviors of the various material types (i.e., metals, ceramics, polymers) may be explained by bonding type. For example, metals are good conductors of both electricity and heat, as a consequence of their free electrons (see Sections 18.5, 18.6 and 19.4). By way of contrast, ionically and covalently bonded materials are typically electrical and thermal insulators, due to the absence of large numbers of free electrons. Furthermore, in Section 7.4 we note that at room temperature, most metals and their alloys fail in a ductile manner; that is, fracture occurs after the materials have experienced significant degrees of permanent deformation. This behavior is explained in terms of deformation mechanism (Section 7.2), which is implicitly related to the characteristics of the metallic bond. Conversely, at room temperature ionically bonded materials are intrinsically brittle as a consequence of the electrically charged nature of their component ions (see Section 12.10). Concept Check 2.3 Offer an explanation as to why covalently bonded materials are generally less dense than ionically or metallically bonded ones. [The answer may be found at www.wiley.com/college/callister (Student Companion Site).] 2.7 SECONDARY BONDING OR VAN DER WAALS BONDING secondary bond van der Waals bond Secondary, van der Waals, or physical bonds are weak in comparison to the primary or chemical ones; bonding energies are typically on the order of only 10 kJ/mol (0.1 eV/atom). Secondary bonding exists between virtually all atoms or molecules, 1496T_c02_15-37 12/20/05 7:19 Page 31 2nd REVISE PAGES 2.7 Secondary Bonding or van der Waals Bonding • 31 + – + – Figure 2.12 Schematic illustration of van der Waals bonding between two dipoles. Atomic or molecular dipoles dipole hydrogen bonding but its presence may be obscured if any of the three primary bonding types is present. Secondary bonding is evidenced for the inert gases, which have stable electron structures, and, in addition, between molecules in molecular structures that are covalently bonded. Secondary bonding forces arise from atomic or molecular dipoles. In essence, an electric dipole exists whenever there is some separation of positive and negative portions of an atom or molecule. The bonding results from the coulombic attraction between the positive end of one dipole and the negative region of an adjacent one, as indicated in Figure 2.12. Dipole interactions occur between induced dipoles, between induced dipoles and polar molecules (which have permanent dipoles), and between polar molecules. Hydrogen bonding, a special type of secondary bonding, is found to exist between some molecules that have hydrogen as one of the constituents. These bonding mechanisms are now discussed briefly. Fluctuating Induced Dipole Bonds A dipole may be created or induced in an atom or molecule that is normally electrically symmetric; that is, the overall spatial distribution of the electrons is symmetric with respect to the positively charged nucleus, as shown in Figure 2.13a. All atoms are experiencing constant vibrational motion that can cause instantaneous and short-lived distortions of this electrical symmetry for some of the atoms or molecules, and the creation of small electric dipoles, as represented in Figure 2.13b. One of these dipoles can in turn produce a displacement of the electron distribution of an adjacent molecule or atom, which induces the second one also to become a dipole that is then weakly attracted or bonded to the first; this is one type of van der Waals bonding. These attractive forces may exist between large numbers of atoms or molecules, which forces are temporary and fluctuate with time. The liquefaction and, in some cases, the solidification of the inert gases and other electrically neutral and symmetric molecules such as H2 and Cl2 are realized because of this type of bonding. Melting and boiling temperatures are extremely low in materials for which induced dipole bonding predominates; of all possible intermolecular bonds, these are the weakest. Bonding energies and melting temperatures for argon and chlorine are also tabulated in Table 2.3. Polar Molecule-Induced Dipole Bonds Permanent dipole moments exist in some molecules by virtue of an asymmetrical arrangement of positively and negatively charged regions; such molecules are Atomic nucleus Atomic nucleus Electron cloud Electron cloud + (a) – (b) Figure 2.13 Schematic representations of (a) an electrically symmetric atom and (b) an induced atomic dipole. 1496T_c02_15-37 11/10/05 10:42 Page 32 REVISED PAGES 32 • Chapter 2 / Atomic Structure and Interatomic Bonding H Cl + polar molecule Figure 2.14 Schematic representation of a polar hydrogen chloride (HCl) molecule. – termed polar molecules. Figure 2.14 is a schematic representation of a hydrogen chloride molecule; a permanent dipole moment arises from net positive and negative charges that are respectively associated with the hydrogen and chlorine ends of the HCl molecule. Polar molecules can also induce dipoles in adjacent nonpolar molecules, and a bond will form as a result of attractive forces between the two molecules. Furthermore, the magnitude of this bond will be greater than for fluctuating induced dipoles. Permanent Dipole Bonds Van der Waals forces will also exist between adjacent polar molecules. The associated bonding energies are significantly greater than for bonds involving induced dipoles. The strongest secondary bonding type, the hydrogen bond, is a special case of polar molecule bonding. It occurs between molecules in which hydrogen is covalently bonded to fluorine (as in HF), oxygen (as in H2O), and nitrogen (as in NH3). For each H—F, H—O, or H—N bond, the single hydrogen electron is shared with the other atom. Thus, the hydrogen end of the bond is essentially a positively charged bare proton that is unscreened by any electrons. This highly positively charged end of the molecule is capable of a strong attractive force with the negative end of an adjacent molecule, as demonstrated in Figure 2.15 for HF. In essence, this single proton forms a bridge between two negatively charged atoms. The magnitude of the hydrogen bond is generally greater than that of the other types of secondary bonds and may be as high as 51 kJ/mol (0.52 eV/molecule), as shown in Table 2.3. Melting and boiling temperatures for hydrogen fluoride and water are abnormally high in light of their low molecular weights, as a consequence of hydrogen bonding. 2.8 MOLECULES Many of the common molecules are composed of groups of atoms that are bound together by strong covalent bonds; these include elemental diatomic molecules 1F2, O2, H2, etc.2 as well as a host of compounds 1H2O, CO2, HNO3, C6H6, CH4, etc.2. In the condensed liquid and solid states, bonds between molecules are weak secondary ones. Consequently, molecular materials have relatively low melting and boiling temperatures. Most of those that have small molecules composed of a few atoms are gases at ordinary, or ambient, temperatures and pressures. On the other hand, many of the modern polymers, being molecular materials composed of extremely large molecules, exist as solids; some of their properties are strongly dependent on the presence of van der Waals and hydrogen secondary bonds. H F H Hydrogen bond F Figure 2.15 Schematic representation of hydrogen bonding in hydrogen fluoride (HF). 1496T_c02_15-37 11/10/05 10:43 Page 33 REVISED PAGES 2.8 Molecules • 33 MATERIAL OF IMPORTANCE Water (Its Volume Expansion Upon Freezing) U pon freezing (i.e., transforming from a liquid to a solid upon cooling), most substances experience an increase in density (or, correspondingly, a decrease in volume). One exception is water, which exhibits the anomalous and familiar expansion upon freezing—approximately 9 volume percent expansion. This behavior may be explained on the basis of hydrogen bonding. Each H2O molecule has two hydrogen atoms that can bond to oxygen atoms; in addition, its single O atom can bond to two hydrogen atoms of other H2O molecules. Thus, for solid ice, each water molecule participates in four hydrogen bonds as shown in the three-dimensional schematic of Figure 2.16a; here hydrogen bonds are denoted by dashed lines, and each water molecule has 4 nearest-neighbor molecules. This is a relatively open structure—i.e., the molecules are not closely packed together—and, as a result, the density is comparatively low. Upon melting, this structure is partially destroyed, such that the water molecules become more closely packed together (Figure 2.16b)—at room temperature the average number of nearest-neighbor water molecules has increased to approximately 4.5; this leads to an increase in density. Consequences of this anomalous freezing phenomenon are familiar. This phenomenon explains why icebergs float, why, in cold climates, it is necessary to add antifreeze to an automobile’s cooling system (to keep the engine block from cracking), and why freeze-thaw cycles break up the pavement in streets and cause potholes to form. H H O Hydrogen bond H H O O H H O H O H H (a) H H H O H H H H H H H O H O O H H H O O O H H H H H O O A watering can that ruptured along a side panelbottom panel seam. Water that was left in the can during a cold late-autumn night expanded as it froze and caused the rupture. (Photography by S. Tanner.) O H (b) Figure 2.16 The arrangement of water 1H2O2 molecules in (a) solid ice, and (b) liquid water. 1496T_c02_15-37 11/10/05 10:43 Page 34 REVISED PAGES 34 • Chapter 2 / Atomic Structure and Interatomic Bonding SUMMARY Electrons in Atoms The Periodic Table This chapter began with a survey of the fundamentals of atomic structure, presenting the Bohr and wave-mechanical models of electrons in atoms. Whereas the Bohr model assumes electrons to be particles orbiting the nucleus in discrete paths, in wave mechanics we consider them to be wave-like and treat electron position in terms of a probability distribution. Electron energy states are specified in terms of quantum numbers that give rise to electron shells and subshells. The electron configuration of an atom corresponds to the manner in which these shells and subshells are filled with electrons in compliance with the Pauli exclusion principle.The periodic table of the elements is generated by arrangement of the various elements according to valence electron configuration. Bonding Forces and Energies Primary Interatomic Bonds Atomic bonding in solids may be considered in terms of attractive and repulsive forces and energies. The three types of primary bond in solids are ionic, covalent, and metallic. For ionic bonds, electrically charged ions are formed by the transference of valence electrons from one atom type to another; forces are coulombic.There is a sharing of valence electrons between adjacent atoms when bonding is covalent. With metallic bonding, the valence electrons form a “sea of electrons” that is uniformly dispersed around the metal ion cores and acts as a form of glue for them. Secondary Bonding or van der Waals Bonding Both van der Waals and hydrogen bonds are termed secondary, being weak in comparison to the primary ones. They result from attractive forces between electric dipoles, of which there are two types—induced and permanent. For the hydrogen bond, highly polar molecules form when hydrogen covalently bonds to a nonmetallic element such as fluorine. I M P O R TA N T T E R M S A N D C O N C E P T S Atomic mass unit (amu) Atomic number Atomic weight Bohr atomic model Bonding energy Coulombic force Covalent bond Dipole (electric) Electron configuration Electron state Electronegative Electropositive Ground state Hydrogen bond Ionic bond Isotope Metallic bond Mole Pauli exclusion principle Periodic table Polar molecule Primary bonding Quantum mechanics Quantum number Secondary bonding Valence electron van der Waals bond Wave-mechanical model Note: In each chapter, most of the terms listed in the “Important Terms and Concepts” section are defined in the Glossary, which follows Appendix E. The others are important enough to warrant treatment in a full section of the text and can be referenced from the table of contents or the index. 1496T_c02_15-37 11/10/05 10:43 Page 35 REVISED PAGES Questions and Problems • 35 REFERENCES Most of the material in this chapter is covered in college-level chemistry textbooks.Two are listed here as references. Brady, J. E., and F. Senese, Chemistry: Matter and Its Changes, 4th edition, John Wiley & Sons, Inc., Hoboken, NJ, 2004. Ebbing, D. D., S. D. Gammon, and R. O. Ragsdale, Essentials of General Chemistry, 2nd edition, Houghton Mifflin Company, Boston, 2006. QUESTIONS AND PROBLEMS Fundamental Concepts Electrons in Atoms 2.1 (a) Cite the difference between atomic mass and atomic weight. 2.2 Silicon has three naturally-occurring isotopes: 92.23% of 28Si, with an atomic weight of 27.9769 amu, 4.68% of 29Si, with an atomic weight of 28.9765 amu, and 3.09% of 30Si, with an atomic weight of 29.9738 amu. On the basis of these data, confirm that the average atomic weight of Si is 28.0854 amu. 2.3 (a) How many grams are there in one amu of a material? (b) Mole, in the context of this book, is taken in units of gram-mole. On this basis, how many atoms are there in a pound-mole of a substance? 2.4 (a) Cite two important quantum-mechanical concepts associated with the Bohr model of the atom. (b) Cite two important additional refinements that resulted from the wave-mechanical atomic model. 2.5 Relative to electrons and electron states, what does each of the four quantum numbers specify? 2.6 Allowed values for the quantum numbers of electrons are as follows: n  1, 2, 3, . . . l  0, 1, 2, 3, . . . , n  1 ml  0, 1, 2, 3, . . . , l ms   12 The relationships between n and the shell designations are noted in Table 2.1. Relative to the subshells, l l l l     0 1 2 3 corresponds corresponds corresponds corresponds to to to to an s subshell a p subshell a d subshell an f subshell For the K shell, the four quantum numbers for each of the two electrons in the 1s state, in the order of nlmlms, are 1001 12 2 and 100112 2. Write the four quantum numbers for all of the electrons in the L and M shells, and note which correspond to the s, p, and d subshells. 2.7 Give the electron configurations for the following ions: P5, P3, Sn4, Se2, I, and Ni2. 2.8 Potassium iodide (KI) exhibits predominantly ionic bonding. The K and I ions have electron structures that are identical to which two inert gases? The Periodic Table 2.9 With regard to electron configuration, what do all the elements in Group IIA of the periodic table have in common? 2.10 To what group in the periodic table would an element with atomic number 112 belong? 2.11 Without consulting Figure 2.6 or Table 2.2, determine whether each of the electron configurations given below is an inert gas, a halogen, an alkali metal, an alkaline earth metal, or a transition metal. Justify your choices. 1496T_c02_15-37 11/10/05 10:43 Page 36 REVISED PAGES 36 • Chapter 2 / Atomic Structure and Interatomic Bonding (a) 1s22s22p63s23p5 (b) 1s22s22p63s23p63d74s2 (c) 1s22s22p63s23p63d104s24p6 (d) 1s22s22p63s23p64s1 (e) 1s22s22p63s23p63d104s24p64d55s2 (f) 1s22s22p63s2 2.12 (a) What electron subshell is being filled for the rare earth series of elements on the periodic table? (b) What electron subshell is being filled for the actinide series? Bonding Forces and Energies 2.13 Calculate the force of attraction between a Ca2 and an O2 ion the centers of which are separated by a distance of 1.25 nm. 2.14 The net potential energy between two adjacent ions, EN, may be represented by the sum of Equations 2.8 and 2.9; that is, EN   B A  n r r (2.11) Calculate the bonding energy E0 in terms of the parameters A, B, and n using the following procedure: 1. Differentiate EN with respect to r, and then set the resulting expression equal to zero, since the curve of EN versus r is a minimum at E0. 2. Solve for r in terms of A, B, and n, which yields r0, the equilibrium interionic spacing. 3. Determine the expression for E0 by substitution of r0 into Equation 2.11. 2.15 For an NaCl ion pair, attractive and repulsive energies EA and ER, respectively, depend on the distance between the ions r, according to EA   ER  1.436 r 7.32  106 r8 For these expressions, energies are expressed in electron volts per Na Cl pair, and r is the distance in nanometers. The net energy EN is just the sum of the two expressions above. (a) Superimpose on a single plot EN , ER, and EA versus r up to 1.0 nm. (b) On the basis of this plot, determine (i) the equilibrium spacing r0 between the Na and Cl ions, and (ii) the magnitude of the bonding energy E0 between the two ions. (c) Mathematically determine the r0 and E0 values using the solutions to Problem 2.14 and compare these with the graphical results from part (b). 2.16 Consider a hypothetical X  Y ion pair for which the equilibrium interionic spacing and bonding energy values are 0.38 nm and 5.37 eV, respectively. If it is known that n in Equation 2.11 has a value of 8, using the results of Problem 2.14, determine explicit expressions for attractive and repulsive energies EA and ER of Equations 2.8 and 2.9. 2.17 The net potential energy EN between two adjacent ions is sometimes represented by the expression EN   r C  D exp a b r r (2.12) in which r is the interionic separation and C, D, and r are constants whose values depend on the specific material. (a) Derive an expression for the bonding energy E0 in terms of the equilibrium interionic separation r0 and the constants D and r using the following procedure: 1. Differentiate EN with respect to r and set the resulting expression equal to zero. 2. Solve for C in terms of D, r, and r0. 3. Determine the expression for E0 by substitution for C in Equation 2.12. (b) Derive another expression for E0 in terms of r0, C, and r using a procedure analogous to the one outlined in part (a). Primary Interatomic Bonds 2.18 (a) Briefly cite the main differences between ionic, covalent, and metallic bonding. (b) State the Pauli exclusion principle. 2.19 Compute the percentage ionic character of the interatomic bond for each of the following compounds: MgO, GaP, CsF, CdS, and FeO. 1496T_c02_15-37 11/10/05 10:43 Page 37 REVISED PAGES Questions and Problems • 37 2.20 Make a plot of bonding energy versus melting temperature for the metals listed in Table 2.3. Using this plot, approximate the bonding energy for molybdenum, which has a melting temperature of 2617C. 2.21 Using Table 2.2, determine the number of covalent bonds that are possible for atoms of the following elements: silicon, bromine, nitrogen, sulfur, and neon. 2.22 What type(s) of bonding would be expected for each of the following materials: solid xenon, calcium fluoride (CaF2), bronze, cadmium telluride (CdTe), rubber, and tungsten? Secondary Bonding or van der Waals Bonding 2.23 Explain why hydrogen fluoride (HF) has a higher boiling temperature than hydrogen chloride (HCl) (19.4 vs. 85C), even though HF has a lower molecular weight. 1496T_c03_38-79 12/20/05 7:38 Page 38 Chapter 3 2nd REVISE PAGES The Structure of Crystalline Solids Diffracted beams Incident beam Single crystal X-ray source Lead screen Photographic plate (a) (b) (a) X-ray diffraction photograph [or Laue photograph (Section 3.16)] for a single crystal of magnesium. (b) Schematic diagram illustrating how the spots (i.e., the diffraction pattern) in (a) are produced. The lead screen blocks out all beams generated from the x-ray source, except for a narrow beam traveling in a single direction. This incident beam is diffracted by individual crystallographic planes in the single crystal (having different orientations), which gives rise to the various diffracted beams that impinge on the photographic plate. Intersections of these beams with the plate appear as spots when the film is developed. The large spot in the center of (a) is from the incident beam, which is parallel to a [0001] crystallographic direction. It should be noted that the hexagonal symmetry of magnesium’s hexagonal close-packed crystal structure is indicated by the diffraction spot pattern that was generated. [Figure (a) courtesy of J. G. Byrne, Department of Metallurgical Engineering, University of Utah. Figure (b) from J. E. Brady and F. Senese, Chemistry: Matter and Its Changes, 4th edition. Copyright © 2004 by John Wiley & Sons, Hoboken, NJ. Reprinted by permission of John Wiley & Sons, Inc.] WHY STUDY The Structure of Crystalline Solids? The properties of some materials are directly related to their crystal structures. For example, pure and undeformed magnesium and beryllium, having one crystal structure, are much more brittle (i.e., fracture at lower degrees of deformation) than are pure and undeformed metals such as gold and silver that have yet another crystal structure (see Section 7.4). 38 • Furthermore, significant property differences exist between crystalline and noncrystalline materials having the same composition. For example, noncrystalline ceramics and polymers normally are optically transparent; the same materials in crystalline (or semicrystalline) form tend to be opaque or, at best, translucent. 1496T_c03_38-79 12/20/05 7:38 Page 39 2nd REVISE PAGES Learning Objectives After studying this chapter you should be able to do the following: 1. Describe the difference in atomic/molecular 5. Given three direction index integers, sketch the structure between crystalline and noncrystalline direction corresponding to these indices within a materials. unit cell. 2. Draw unit cells for face-centered cubic, body6. Specify the Miller indices for a plane that has centered cubic, and hexagonal close-packed been drawn within a unit cell. crystal structures. 7. Describe how face-centered cubic and hexagonal 3. Derive the relationships between unit cell edge close-packed crystal structures may be generated length and atomic radius for face-centered cubic by the stacking of close-packed planes of atoms. and body-centered cubic crystal structures. 8. Distinguish between single crystals and polycrys4. Compute the densities for metals having facetalline materials. centered cubic and body-centered cubic crystal 9. Define isotropy and anisotropy with respect to structures given their unit cell dimensions. material properties. 3.1 INTRODUCTION Chapter 2 was concerned primarily with the various types of atomic bonding, which are determined by the electron structures of the individual atoms. The present discussion is devoted to the next level of the structure of materials, specifically, to some of the arrangements that may be assumed by atoms in the solid state. Within this framework, concepts of crystallinity and noncrystallinity are introduced. For crystalline solids the notion of crystal structure is presented, specified in terms of a unit cell. The three common crystal structures found in metals are then detailed, along with the scheme by which crystallographic points, directions, and planes are expressed. Single crystals, polycrystalline, and noncrystalline materials are considered. The final section of this chapter briefly describes how crystal structures are determined experimentally using x-ray diffraction techniques. C r ys t a l St r u c t u r e s 3.2 FUNDAMENTAL CONCEPTS crystalline crystal structure Solid materials may be classified according to the regularity with which atoms or ions are arranged with respect to one another. A crystalline material is one in which the atoms are situated in a repeating or periodic array over large atomic distances; that is, long-range order exists, such that upon solidification, the atoms will position themselves in a repetitive three-dimensional pattern, in which each atom is bonded to its nearest-neighbor atoms. All metals, many ceramic materials, and certain polymers form crystalline structures under normal solidification conditions. For those that do not crystallize, this long-range atomic order is absent; these noncrystalline or amorphous materials are discussed briefly at the end of this chapter. Some of the properties of crystalline solids depend on the crystal structure of the material, the manner in which atoms, ions, or molecules are spatially arranged. There is an extremely large number of different crystal structures all having longrange atomic order; these vary from relatively simple structures for metals to exceedingly complex ones, as displayed by some of the ceramic and polymeric materials. The present discussion deals with several common metallic crystal structures. Chapters 12 and 14 are devoted to crystal structures for ceramics and polymers, respectively. 1496T_c03_38-79 11/11/05 19:05 Page 40 REVISED PAGES 40 • Chapter 3 / The Structure of Crystalline Solids (a) ( b) Figure 3.1 For the facecentered cubic crystal structure, (a) a hard sphere unit cell representation, (b) a reduced-sphere unit cell, and (c) an aggregate of many atoms. [Figure (c) adapted from W. G. Moffatt, G. W. Pearsall, and J. Wulff, The Structure and Properties of Materials, Vol. I, Structure, p. 51. Copyright © 1964 by John Wiley & Sons, New York. Reprinted by permission of John Wiley & Sons, Inc.] (c) lattice When describing crystalline structures, atoms (or ions) are thought of as being solid spheres having well-defined diameters. This is termed the atomic hard sphere model in which spheres representing nearest-neighbor atoms touch one another. An example of the hard sphere model for the atomic arrangement found in some of the common elemental metals is displayed in Figure 3.1c. In this particular case all the atoms are identical. Sometimes the term lattice is used in the context of crystal structures; in this sense “lattice” means a three-dimensional array of points coinciding with atom positions (or sphere centers). 3.3 UNIT CELLS unit cell The atomic order in crystalline solids indicates that small groups of atoms form a repetitive pattern. Thus, in describing crystal structures, it is often convenient to subdivide the structure into small repeat entities called unit cells. Unit cells for most crystal structures are parallelepipeds or prisms having three sets of parallel faces; one is drawn within the aggregate of spheres (Figure 3.1c), which in this case happens to be a cube. A unit cell is chosen to represent the symmetry of the crystal structure, wherein all the atom positions in the crystal may be generated by translations of the unit cell integral distances along each of its edges. Thus, the unit cell is the basic structural unit or building block of the crystal structure and defines the crystal structure by virtue of its geometry and the atom positions within. 1496T_c03_38-79 11/11/05 19:05 Page 41 REVISED PAGES 3.4 Metallic Crystal Structures • 41 Convenience usually dictates that parallelepiped corners coincide with centers of the hard sphere atoms. Furthermore, more than a single unit cell may be chosen for a particular crystal structure; however, we generally use the unit cell having the highest level of geometrical symmetry. 3.4 METALLIC CRYSTAL STRUCTURES The atomic bonding in this group of materials is metallic and thus nondirectional in nature. Consequently, there are minimal restrictions as to the number and position of nearest-neighbor atoms; this leads to relatively large numbers of nearest neighbors and dense atomic packings for most metallic crystal structures. Also, for metals, using the hard sphere model for the crystal structure, each sphere represents an ion core. Table 3.1 presents the atomic radii for a number of metals. Three relatively simple crystal structures are found for most of the common metals: facecentered cubic, body-centered cubic, and hexagonal close-packed. The Face-Centered Cubic Cr ystal Structure face-centered cubic (FCC) The crystal structure found for many metals has a unit cell of cubic geometry, with atoms located at each of the corners and the centers of all the cube faces. It is aptly called the face-centered cubic (FCC) crystal structure. Some of the familiar metals having this crystal structure are copper, aluminum, silver, and gold (see also Table 3.1). Figure 3.1a shows a hard sphere model for the FCC unit cell, whereas in Figure 3.1b the atom centers are represented by small circles to provide a better perspective of atom positions. The aggregate of atoms in Figure 3.1c represents a section of crystal consisting of many FCC unit cells. These spheres or ion cores touch one another across a face diagonal; the cube edge length a and the atomic radius R are related through Unit cell edge length for face-centered cubic Crystal Systems and Unit Cells for Metals a  2R12 (3.1) This result is obtained in Example Problem 3.1. For the FCC crystal structure, each corner atom is shared among eight unit cells, whereas a face-centered atom belongs to only two. Therefore, one-eighth of each of the eight corner atoms and one-half of each of the six face atoms, or a total of four whole atoms, may be assigned to a given unit cell. This is depicted in Figure 3.1a, where only sphere portions are represented within the confines of the cube. The Table 3.1 Atomic Radii and Crystal Structures for 16 Metals Metal Aluminum Cadmium Chromium Cobalt Copper Gold Iron () Lead Crystal Structurea Atomic Radiusb (nm) FCC HCP BCC HCP FCC FCC BCC FCC 0.1431 0.1490 0.1249 0.1253 0.1278 0.1442 0.1241 0.1750 Metal Crystal Structure Atomic Radius (nm) Molybdenum Nickel Platinum Silver Tantalum Titanium () Tungsten Zinc BCC FCC FCC FCC BCC HCP BCC HCP 0.1363 0.1246 0.1387 0.1445 0.1430 0.1445 0.1371 0.1332 FCC  face-centered cubic; HCP  hexagonal close-packed; BCC  body-centered cubic. A nanometer (nm) equals 109 m; to convert from nanometers to angstrom units (Å), multiply the nanometer value by 10. a b 1496T_c03_38-79 11/11/05 19:05 Page 42 REVISED PAGES 42 • Chapter 3 / The Structure of Crystalline Solids coordination number atomic packing factor (APF) Definition of atomic packing factor cell comprises the volume of the cube, which is generated from the centers of the corner atoms as shown in the figure. Corner and face positions are really equivalent; that is, translation of the cube corner from an original corner atom to the center of a face atom will not alter the cell structure. Two other important characteristics of a crystal structure are the coordination number and the atomic packing factor (APF). For metals, each atom has the same number of nearest-neighbor or touching atoms, which is the coordination number. For face-centered cubics, the coordination number is 12. This may be confirmed by examination of Figure 3.1a; the front face atom has four corner nearest-neighbor atoms surrounding it, four face atoms that are in contact from behind, and four other equivalent face atoms residing in the next unit cell to the front, which is not shown. The APF is the sum of the sphere volumes of all atoms within a unit cell (assuming the atomic hard sphere model) divided by the unit cell volume—that is APF  volume of atoms in a unit cell total unit cell volume (3.2) For the FCC structure, the atomic packing factor is 0.74, which is the maximum packing possible for spheres all having the same diameter. Computation of this APF is also included as an example problem. Metals typically have relatively large atomic packing factors to maximize the shielding provided by the free electron cloud. The Body-Centered Cubic Cr ystal Structure body-centered cubic (BCC) Another common metallic crystal structure also has a cubic unit cell with atoms located at all eight corners and a single atom at the cube center. This is called a body-centered cubic (BCC) crystal structure. A collection of spheres depicting this crystal structure is shown in Figure 3.2c, whereas Figures 3.2a and 3.2b are diagrams of BCC unit cells with the atoms represented by hard sphere and reduced-sphere models, respectively. Center and corner atoms touch one another along cube diagonals, and unit cell length a and atomic radius R are related through Unit cell edge length for body-centered cubic (a) a ( b) 4R 13 (3.3) (c) Figure 3.2 For the body-centered cubic crystal structure, (a) a hard sphere unit cell representation, (b) a reduced-sphere unit cell, and (c) an aggregate of many atoms. [Figure (c) from W. G. Moffatt, G. W. Pearsall, and J. Wulff, The Structure and Properties of Materials, Vol. I, Structure, p. 51. Copyright © 1964 by John Wiley & Sons, New York. Reprinted by permission of John Wiley & Sons, Inc.] 1496T_c03_38-79 11/11/05 21:57 Page 43 REVISED PAGES 3.4 Metallic Crystal Structures • 43 H F G E J c D B C A a (a) (b) Figure 3.3 For the hexagonal close-packed crystal structure, (a) a reduced-sphere unit cell (a and c represent the short and long edge lengths, respectively), and (b) an aggregate of many atoms. [Figure (b) from W. G. Moffatt, G. W. Pearsall, and J. Wulff, The Structure and Properties of Materials, Vol. I, Structure, p. 51. Copyright © 1964 by John Wiley & Sons, New York. Reprinted by permission of John Wiley & Sons, Inc.] Crystal Systems and Unit Cells for Metals Chromium, iron, tungsten, as well as several other metals listed in Table 3.1 exhibit a BCC structure. Two atoms are associated with each BCC unit cell: the equivalent of one atom from the eight corners, each of which is shared among eight unit cells, and the single center atom, which is wholly contained within its cell. In addition, corner and center atom positions are equivalent. The coordination number for the BCC crystal structure is 8; each center atom has as nearest neighbors its eight corner atoms. Since the coordination number is less for BCC than FCC, so also is the atomic packing factor for BCC lower—0.68 versus 0.74. The Hexagonal Close-Packed Cr ystal Structure hexagonal closepacked (HCP) Crystal Systems and Unit Cells for Metals Not all metals have unit cells with cubic symmetry; the final common metallic crystal structure to be discussed has a unit cell that is hexagonal. Figure 3.3a shows a reduced-sphere unit cell for this structure, which is termed hexagonal closepacked (HCP); an assemblage of several HCP unit cells is presented in Figure 3.3b.1 The top and bottom faces of the unit cell consist of six atoms that form regular hexagons and surround a single atom in the center. Another plane that provides three additional atoms to the unit cell is situated between the top and bottom planes. The atoms in this midplane have as nearest neighbors atoms in both of the adjacent two planes. The equivalent of six atoms is contained in each unit cell; one-sixth of each of the 12 top and bottom face corner atoms, one-half of each of the 2 center face atoms, and all 3 midplane interior atoms. If a and c represent, respectively, the short and long unit cell dimensions of Figure 3.3a, the ca ratio should be 1.633; however, for some HCP metals this ratio deviates from the ideal value. 1 Alternatively, the unit cell for HCP may be specified in terms of the parallelepiped defined by atoms labeled A through H in Figure 3.3a. As such, the atom denoted J lies within the unit cell interior. 1496T_c03_38-79 11/11/05 19:05 Page 44 REVISED PAGES 44 • Chapter 3 / The Structure of Crystalline Solids The coordination number and the atomic packing factor for the HCP crystal structure are the same as for FCC: 12 and 0.74, respectively. The HCP metals include cadmium, magnesium, titanium, and zinc; some of these are listed in Table 3.1. EXAMPLE PROBLEM 3.1 Determination of FCC Unit Cell Volume Calculate the volume of an FCC unit cell in terms of the atomic radius R. Solution In the FCC unit cell illustrated, R a 4R a the atoms touch one another across a face-diagonal the length of which is 4R. Since the unit cell is a cube, its volume is a3, where a is the cell edge length. From the right triangle on the face, a2  a2  14R2 2 or, solving for a, a  2R 12 (3.1) The FCC unit cell volume VC may be computed from VC  a3  12R122 3  16R3 12 (3.4) EXAMPLE PROBLEM 3.2 Computation of the Atomic Packing Factor for FCC Show that the atomic packing factor for the FCC crystal structure is 0.74. Solution The APF is defined as the fraction of solid sphere volume in a unit cell, or APF  VS volume of atoms in a unit cell  total unit cell volume VC Both the total atom and unit cell volumes may be calculated in terms of the atomic radius R. The volume for a sphere is 43 pR3, and since there are four 1496T_c03_38-79 11/11/05 19:05 Page 45 REVISED PAGES 3.5 Density Computations • 45 atoms per FCC unit cell, the total FCC atom (or sphere) volume is VS  142 43 pR3  163 pR3 From Example Problem 3.1, the total unit cell volume is VC  16R3 12 Therefore, the atomic packing factor is APF  1 163 2 pR3 VS   0.74 VC 16R3 12 3.5 DENSITY COMPUTATIONS A knowledge of the crystal structure of a metallic solid permits computation of its theoretical density r through the relationship Theoretical density for metals r nA VC NA (3.5) where n  number of atoms associated with each unit cell A  atomic weight VC  volume of the unit cell NA  Avogadro’s number 16.023  1023 atoms/mol2 EXAMPLE PROBLEM 3.3 Theoretical Density Computation for Copper Copper has an atomic radius of 0.128 nm, an FCC crystal structure, and an atomic weight of 63.5 g/mol. Compute its theoretical density and compare the answer with its measured density. Solution Equation 3.5 is employed in the solution of this problem. Since the crystal structure is FCC, n, the number of atoms per unit cell, is 4. Furthermore, the atomic weight ACu is given as 63.5 g/mol. The unit cell volume VC for FCC was determined in Example Problem 3.1 as 16R3 12, where R, the atomic radius, is 0.128 nm. Substitution for the various parameters into Equation 3.5 yields r  nACu nACu  VCNA 116R3 122 NA 14 atoms/unit cell2163.5 g/mol2 3161211.28  108 cm2 3/unit cell4 16.023  1023 atoms/mol2  8.89 g/cm3 The literature value for the density of copper is 8.94 g/cm3, which is in very close agreement with the foregoing result. 1496T_c03_38-79 11/11/05 19:05 Page 46 REVISED PAGES 46 • Chapter 3 / The Structure of Crystalline Solids 3.6 POLYMORPHISM AND ALLOTROPY polymorphism allotropy Some metals, as well as nonmetals, may have more than one crystal structure, a phenomenon known as polymorphism. When found in elemental solids, the condition is often termed allotropy. The prevailing crystal structure depends on both the temperature and the external pressure. One familiar example is found in carbon: graphite is the stable polymorph at ambient conditions, whereas diamond is formed at extremely high pressures. Also, pure iron has a BCC crystal structure at room temperature, which changes to FCC iron at 912C (1674F). Most often a modification of the density and other physical properties accompanies a polymorphic transformation. 3.7 CRYSTAL SYSTEMS Crystal Systems and Unit Cells for Metals lattice parameters crystal system Since there are many different possible crystal structures, it is sometimes convenient to divide them into groups according to unit cell configurations and/or atomic arrangements. One such scheme is based on the unit cell geometry, that is, the shape of the appropriate unit cell parallelepiped without regard to the atomic positions in the cell. Within this framework, an x, y, z coordinate system is established with its origin at one of the unit cell corners; each of the x, y, and z axes coincides with one of the three parallelepiped edges that extend from this corner, as illustrated in Figure 3.4. The unit cell geometry is completely defined in terms of six parameters: the three edge lengths a, b, and c, and the three interaxial angles a, b, and g. These are indicated in Figure 3.4, and are sometimes termed the lattice parameters of a crystal structure. On this basis there are seven different possible combinations of a, b, and c, and a, b, and g, each of which represents a distinct crystal system. These seven crystal systems are cubic, tetragonal, hexagonal, orthorhombic, rhombohedral,2 monoclinic, and triclinic. The lattice parameter relationships and unit cell sketches for each are represented in Table 3.2.The cubic system, for which a  b  c and a  b  g  90, has the greatest degree of symmetry. Least symmetry is displayed by the triclinic system, since a  b  c and a  b  g. From the discussion of metallic crystal structures, it should be apparent that both FCC and BCC structures belong to the cubic crystal system, whereas HCP falls within hexagonal. The conventional hexagonal unit cell really consists of three parallelepipeds situated as shown in Table 3.2. z c Figure 3.4 A unit cell with x, y, and z coordinate axes, showing axial lengths (a, b, and c) and interaxial angles 1a, b, and g2 . ␣ ␤ y ␥ a b x 2 Also called trigonal. 1496T_c03_38-79 11/11/05 19:05 Page 47 REVISED PAGES 3.7 Crystal Systems • 47 Table 3.2 Lattice Parameter Relationships and Figures Showing Unit Cell Geometries for the Seven Crystal Systems Crystal System Cubic Axial Relationships abc Interaxial Angles Unit Cell Geometry a  b  g  90 a a Hexagonal abc a  b  90, g  120 a c a Tetragonal abc a  b  g  90 c a Rhombohedral (Trigonal) abc Orthorhombic abc a  b  g  90 a  b  g  90 aa a ␣ a c a Monoclinic abc a a a  g  90  b b c a ␤ b Triclinic abc a  b  g  90 c ␣ ␤ ␥ b a 1496T_c03_38-79 11/11/05 19:05 Page 48 REVISED PAGES 48 • Chapter 3 / The Structure of Crystalline Solids MATERIAL OF IMPORTANCE Tin (Its Allotropic Transformation) A nother common metal that experiences an allotropic change is tin. White (or b) tin, having a body-centered tetragonal crystal structure at room temperature, transforms, at 13.2C (55.8F), to gray (or a) tin, which has a crystal structure similar to diamond (i.e., the diamond cubic crystal structure); this transformation is represented schematically as follows: 13.2°C Cooling White (␤) tin The rate at which this change takes place is extremely slow; however, the lower the temperature (below 13.2C) the faster the rate. Accompanying this white tin-to-gray tin transformation is an increase in volume (27 percent), and, accordingly, a decrease in density (from 7.30 g/cm3 to 5.77 g/cm3). Consequently, this volume expansion results in the disintegration of the white tin metal into a coarse powder of the gray allotrope. For normal subambient temperatures, there is no need to worry about this disintegration process for tin products, due to the very slow rate at which the transformation occurs. This white-to-gray-tin transition produced some rather dramatic results in 1850 in Russia.The winter that year was particularly cold, and record low temperatures persisted for extended periods of time.The uniforms of some Russian soldiers had tin buttons, many of which crumbled due to these extreme cold conditions, as did also many of the tin church organ pipes. This problem came to be known as the “tin disease.” Gray (␣) tin Specimen of white tin (left). Another specimen disintegrated upon transforming to gray tin (right) after it was cooled to and held at a temperature below 13.2C for an extended period of time. (Photograph courtesy of Professor Bill Plumbridge, Department of Materials Engineering, The Open University, Milton Keynes, England.) 1496T_c03_38-79 11/11/05 19:05 Page 49 REVISED PAGES 3.8 Point Coordinates • 49 Concept Check 3.1 What is the difference between crystal structure and crystal system? [The answer may be found at www.wiley.com/college/callister (Student Companion Site).] It is important to note that many of the principles and concepts addressed in previous discussions in this chapter also apply to crystalline ceramic and polymeric systems (Chapters 12 and 14). For example, crystal structures are most often described in terms of unit cells, which are normally more complex than those for FCC, BCC, and HCP. In addition, for these other systems, we are often interested in determining atomic packing factors and densities, using modified forms of Equations 3.2 and 3.5. Furthermore, according to unit cell geometry, crystal structures of these other material types are grouped within the seven crystal systems. C r ys t a l l o g r a p h i c Po i n t s , Directions, and Planes When dealing with crystalline materials, it often becomes necessary to specify a particular point within a unit cell, a crystallographic direction, or some crystallographic plane of atoms. Labeling conventions have been established in which three numbers or indices are used to designate point locations, directions, and planes. The basis for determining index values is the unit cell, with a right-handed coordinate system consisting of three (x, y, and z) axes situated at one of the corners and coinciding with the unit cell edges, as shown in Figure 3.4. For some crystal systems—namely, hexagonal, rhombohedral, monoclinic, and triclinic—the three axes are not mutually perpendicular, as in the familiar Cartesian coordinate scheme. 3.8 POINT COORDINATES The position of any point located within a unit cell may be specified in terms of its coordinates as fractional multiples of the unit cell edge lengths (i.e., in terms of a, b, and c). To illustrate, consider the unit cell and the point P situated therein as shown in Figure 3.5. We specify the position of P in terms of the generalized coordinates z Figure 3.5 The manner in which the q, r, and s coordinates at point P within the unit cell are determined. The q coordinate (which is a fraction) corresponds to the distance qa along the x axis, where a is the unit cell edge length. The respective r and s coordinates for the y and z axes are determined similarly. b a P qrs c sc y qa rb x 1496T_c03_38-79 11/11/05 19:05 Page 50 REVISED PAGES 50 • Chapter 3 / The Structure of Crystalline Solids q, r, and s where q is some fractional length of a along the x axis, r is some fractional length of b along the y axis, and similarly for s. Thus, the position of P is designated using coordinates q r s with values that are less than or equal to unity. Furthermore, we have chosen not to separate these coordinates by commas or any other punctuation marks (which is the normal convention). EXAMPLE PROBLEM 3.4 Location of Point Having Specified Coordinates For the unit cell shown in the accompanying sketch (a), locate the point having coordinates 14 1 12. z z 0.46 nm m 8n 0.4 1 1 1 4 2 P 0.20 nm 0.12 nm M N O 0.46 nm 0.40 nm y x (a) x y (b) Solution From sketch (a), edge lengths for this unit cell are as follows: a  0.48 nm, b  0.46 nm, and c  0.40 nm. Furthermore, in light of the above discussion, fractional lengths are q  14, r  1, and s  12. Therefore, first we move from the origin of the unit cell (point M) qa  14 (0.48 nm)  0.12 nm units along the x axis (to point N), as shown in the (b) sketch. Similarly, we proceed rb  (1)(0.46 nm)  0.46 nm parallel to the y axis, from point N to point O. Finally, we move from this position, sc  12 (0.40 nm)  0.20 nm units parallel to the z axis to point P as noted again in sketch (b). This point P then corresponds to the 14 1 12 point coordinates. EXAMPLE PROBLEM 3.5 Specification of Point Coordinates Specify point coordinates for all atom positions for a BCC unit cell. Solution For the BCC unit cell of Figure 3.2, atom position coordinates correspond to the locations of the centers of all atoms in the unit cell—that is, the eight corner atoms and single center atom. These positions are noted (and also numbered) in the following figure. 1496T_c03_38-79 11/11/05 19:05 Page 51 REVISED PAGES 3.9 Crystallographic Directions • 51 z a a 9 6 8 7 5 a 4 1 2 y 3 x Point coordinates for position number 1 are 0 0 0; this position is located at the origin of the coordinate system, and, as such, the fractional unit cell edge lengths along the x, y, and z axes are, respectively, 0a, 0a, and 0a. Furthermore, for position number 2, since it lies one unit cell edge length along the x axis, its fractional edge lengths are a, 0a, and 0a, respectively, which yield point coordinates of 1 0 0. The following table presents fractional unit cell lengths along the x, y, and z axes, and their corresponding point coordinates for each of the nine points in the above figure. Point Number 1 2 3 4 5 6 7 8 9 Fractional Lengths x axis y axis z axis 0 1 1 0 0 0 1 1 0 0 0 0 1 2 1 2 1 2 0 1 1 0 0 0 1 1 1 1 1 1 Point Coordinates 0 1 1 0 0 0 1 1 0 0 0 0 111 222 0 1 1 0 0 0 1 1 1 1 1 1 3.9 CRYSTALLOGRAPHIC DIRECTIONS A crystallographic direction is defined as a line between two points, or a vector. The following steps are utilized in the determination of the three directional indices: Crystallographic Directions 1. A vector of convenient length is positioned such that it passes through the origin of the coordinate system. Any vector may be translated throughout the crystal lattice without alteration, if parallelism is maintained. 2. The length of the vector projection on each of the three axes is determined; these are measured in terms of the unit cell dimensions a, b, and c. 3. These three numbers are multiplied or divided by a common factor to reduce them to the smallest integer values. 4. The three indices, not separated by commas, are enclosed in square brackets, thus: [uvw]. The u, v, and w integers correspond to the reduced projections along the x, y, and z axes, respectively. 1496T_c03_38-79 11/11/05 19:05 Page 52 REVISED PAGES 52 • Chapter 3 / The Structure of Crystalline Solids z Figure 3.6 The [100], [110], and [111] directions within a unit cell. [111] y [110] [100] x For each of the three axes, there will exist both positive and negative coordinates. Thus negative indices are also possible, which are represented by a bar over the appropriate index. For example, the [111] direction would have a component in the y direction. Also, changing the signs of all indices produces an antiparallel direction; that is, [111] is directly opposite to [111]. If more than one direction (or plane) is to be specified for a particular crystal structure, it is imperative for the maintaining of consistency that a positive–negative convention, once established, not be changed. The [100], [110], and [111] directions are common ones; they are drawn in the unit cell shown in Figure 3.6. EXAMPLE PROBLEM 3.6 Determination of Directional Indices Determine the indices for the direction shown in the accompanying figure. z Projection on x axis (a/2) Projection on y axis (b) y c a b x Solution The vector, as drawn, passes through the origin of the coordinate system, and therefore no translation is necessary. Projections of this vector onto the x, y, and z axes are, respectively, a 2, b, and 0c, which become 12, 1, and 0 in terms of the unit cell parameters (i.e., when the a, b, and c are dropped). Reduction of these numbers to the lowest set of integers is accompanied by multiplication of each by the factor 2. This yields the integers 1, 2, and 0, which are then enclosed in brackets as [120]. 1496T_c03_38-79 11/11/05 19:05 Page 53 REVISED PAGES 3.9 Crystallographic Directions • 53 This procedure may be summarized as follows: Projections Projections (in terms of a, b, and c) Reduction Enclosure x y z a2 b 1 2 [120] 0c 0 0 1 2 1 EXAMPLE PROBLEM 3.7 Construction of Specified Cr ystallographic Direction Draw a [110] direction within a cubic unit cell. Solution First construct an appropriate unit cell and coordinate axes system. In the accompanying figure the unit cell is cubic, and the origin of the coordinate system, point O, is located at one of the cube corners. z a O –y [110] Direction P –a +y a a a x This problem is solved by reversing the procedure of the preceding example. For this [110] direction, the projections along the x, y, and z axes are a, a, and 0a, respectively. This direction is defined by a vector passing from the origin to point P, which is located by first moving along the x axis a units, and from this position, parallel to the y axis a units, as indicated in the figure. There is no z component to the vector, since the z projection is zero. For some crystal structures, several nonparallel directions with different indices are actually equivalent; this means that the spacing of atoms along each direction is the same. For example, in cubic crystals, all the directions represented by the following indices are equivalent: [100], [100], [010], [010], [001], and [001]. As a convenience, equivalent directions are grouped together into a family, which are enclosed in angle brackets, thus: 81009. Furthermore, directions in cubic crystals having the same indices without regard to order or sign, for example, [123] and [213], are equivalent. This is, in general, not true for other crystal systems. For example, for crystals of tetragonal symmetry, [100] and [010] directions are equivalent, whereas [100] and [001] are not. 1496T_c03_38-79 11/11/05 21:57 Page 54 REVISED PAGES 54 • Chapter 3 / The Structure of Crystalline Solids z Figure 3.7 Coordinate axis system for a hexagonal unit cell (Miller–Bravais scheme). a2 a3 120° a1 Hexagonal Cr ystals A problem arises for crystals having hexagonal symmetry in that some crystallographic equivalent directions will not have the same set of indices. This is circumvented by utilizing a four-axis, or Miller–Bravais, coordinate system as shown in Figure 3.7. The three a1, a2, and a3 axes are all contained within a single plane (called the basal plane) and are at 120 angles to one another. The z axis is perpendicular to this basal plane. Directional indices, which are obtained as described above, will be denoted by four indices, as [uvtw]; by convention, the first three indices pertain to projections along the respective a1, a2, and a3 axes in the basal plane. Conversion from the three-index system to the four-index system, 3u¿v¿w¿ 4 ¡ 3uvtw4 is accomplished by the following formulas: u 1 12u¿  v¿2 3 (3.6a) v 1 12v¿  u¿2 3 (3.6b) t  1u  v2 w  w¿ (3.6c) (3.6d) where primed indices are associated with the three-index scheme and unprimed with the new Miller–Bravais four-index system. (Of course, reduction to the lowest set of integers may be necessary, as discussed above.) For example, the [010] direction becomes [1210]. Several different directions are indicated in the hexagonal unit cell (Figure 3.8a). z z [0001] (1010) (1011) a2 [1120] a3 a3 a1 [1100] (a) a1 (0001) (b) Figure 3.8 For the hexagonal crystal system, (a) [0001], [1100], and [1120] directions, and (b) the (0001), (1011), and (1010) planes. 1496T_c03_38-79 11/11/05 19:05 Page 55 REVISED PAGES 3.10 Crystallographic Planes • 55 EXAMPLE PROBLEM 3.8 Determination of Directional Indices for a Hexagonal Unit Cell Determine the indices for the direction shown in the hexagonal unit cell of sketch (a) below. z z H F G E a2 a2 c D B a3 a3 C a a1 (a) a A a1 (b) Solution In sketch (b), one of the three parallelepipeds comprising the hexagonal cell is delineated—its corners are labeled with letters A through H, with the origin of the a1-a2-a3-z axes coordinate system located at the corner labeled C. We use this unit cell as a reference for specifying the directional indices. It now becomes necessary to determine projections of the direction vector on the a1, a2, and z axes. These respective projections are a (a1 axis), a (a2 axis) and c (z axis), which become 1, 1, and 1 in terms of the unit cell parameters. Thus, u¿  1 v¿  1 w¿  1 Also, from Equations 3.6a, 3.6b, 3.6c, and 3.6d u 1 1 1 12u¿  v¿2  3 122112  14  3 3 3 v 1 1 1 12v¿  u¿2  3 122112  14  3 3 3 1 1 2 t  1u  v2  a  b   3 3 3 w  w¿  1 Multiplication of the above indices by 3 reduces them to the lowest set, which yields values for u, v, t, and w of 1, 1, 2 and 3, respectively. Hence, the direction shown in the figure is [1123]. 3.10 CRYSTALLOGRAPHIC PLANES Miller indices The orientations of planes for a crystal structure are represented in a similar manner. Again, the unit cell is the basis, with the three-axis coordinate system as represented in Figure 3.4. In all but the hexagonal crystal system, crystallographic planes are specified by three Miller indices as (hkl). Any two planes parallel to each other 1496T_c03_38-79 11/11/05 19:05 Page 56 REVISED PAGES 56 • Chapter 3 / The Structure of Crystalline Solids are equivalent and have identical indices. The procedure employed in determination of the h, k, and l index numbers is as follows: 1. If the plane passes through the selected origin, either another parallel plane must be constructed within the unit cell by an appropriate translation, or a new origin must be established at the corner of another unit cell. 2. At this point the crystallographic plane either intersects or parallels each of the three axes; the length of the planar intercept for each axis is determined in terms of the lattice parameters a, b, and c. 3. The reciprocals of these numbers are taken. A plane that parallels an axis may be considered to have an infinite intercept, and, therefore, a zero index. 4. If necessary, these three numbers are changed to the set of smallest integers by multiplication or division by a common factor.3 5. Finally, the integer indices, not separated by commas, are enclosed within parentheses, thus: (hkl). Crystallographic Planes An intercept on the negative side of the origin is indicated by a bar or minus sign positioned over the appropriate index. Furthermore, reversing the directions of all indices specifies another plane parallel to, on the opposite side of and equidistant from, the origin. Several low-index planes are represented in Figure 3.9. One interesting and unique characteristic of cubic crystals is that planes and directions having the same indices are perpendicular to one another; however, for other crystal systems there are no simple geometrical relationships between planes and directions having the same indices. EXAMPLE PROBLEM 3.9 Determination of Planar (Miller) Indices Determine the Miller indices for the plane shown in the accompanying sketch (a). z z z⬘ c/2 c y O O O⬘ y a (012) Plane b x x (a) 3 –b x⬘ (b) On occasion, index reduction is not carried out (e.g., for x-ray diffraction studies that are described in Section 3.16); for example, (002) is not reduced to (001). In addition, for ceramic materials, the ionic arrangement for a reduced-index plane may be different from that for a nonreduced one. 1496T_c03_38-79 11/11/05 19:05 Page 57 REVISED PAGES 3.10 Crystallographic Planes • 57 Solution Since the plane passes through the selected origin O, a new origin must be chosen at the corner of an adjacent unit cell, taken as O¿ and shown in sketch (b). This plane is parallel to the x axis, and the intercept may be taken as q a. The y and z axes intersections, referenced to the new origin O¿ , are b and c 2, respectively. Thus, in terms of the lattice parameters a, b, and c, these intersections are q, 1, and 12. The reciprocals of these numbers are 0, 1, and 2; and since all are integers, no further reduction is necessary. Finally, enclosure in parentheses yields 10122. These steps are briefly summarized below: Intercepts Intercepts (in terms of lattice parameters) Reciprocals Reductions (unnecessary) Enclosure z x y z qa q 0 b 1 1 c2 1 2 2 10122 (001) Plane referenced to the origin at point O z (110) Plane referenced to the origin at point O y O y O Other equivalent (001) planes x Other equivalent (110) planes x (a) z (b) (111) Plane referenced to the origin at point O y O Other equivalent (111) planes x (c) Figure 3.9 Representations of a series each of (a) (001), (b) (110), and (c) (111) crystallographic planes. 1496T_c03_38-79 12/20/05 7:38 Page 58 2nd REVISE PAGES 58 • Chapter 3 / The Structure of Crystalline Solids EXAMPLE PROBLEM 3.10 Construction of Specified Cr ystallographic Plane Construct a (011) plane within a cubic unit cell. z z e (011) Plane c f b O –y g y y a Point of intersection along y axis h b x x (a) (b) Solution To solve this problem, carry out the procedure used in the preceding example in reverse order. To begin, the indices are removed from the parentheses, and reciprocals are taken, which yields q , 1, and 1. This means that the particular plane parallels the x axis while intersecting the y and z axes at b and c, respectively, as indicated in the accompanying sketch (a). This plane has been drawn in sketch (b). A plane is indicated by lines representing its intersections with the planes that constitute the faces of the unit cell or their extensions. For example, in this figure, line ef is the intersection between the (011) plane and the top face of the unit cell; also, line gh represents the intersection between this same (011) plane and the plane of the bottom unit cell face extended. Similarly, lines eg and fh are the intersections between (011) and back and front cell faces, respectively. Atomic Arrangements Planar Atomic Arrangements The atomic arrangement for a crystallographic plane, which is often of interest, depends on the crystal structure. The (110) atomic planes for FCC and BCC crystal structures are represented in Figures 3.10 and 3.11; reduced-sphere unit cells are also included. Note that the atomic packing is different for each case. The circles represent atoms lying in the crystallographic planes as would be obtained from a slice taken through the centers of the full-sized hard spheres. A “family” of planes contains all those planes that are crystallographically equivalent—that is, having the same atomic packing; and a family is designated by C A B A B C D E F E F D (a) (b) Figure 3.10 (a) Reducedsphere FCC unit cell with (110) plane. (b) Atomic packing of an FCC (110) plane. Corresponding atom positions from (a) are indicated. 1496T_c03_38-79 12/20/05 7:38 Page 59 2nd REVISE PAGES 3.10 Crystallographic Planes • 59 Figure 3.11 (a) Reduced-sphere BCC unit cell with (110) plane. (b) Atomic packing of a BCC (110) plane. Corresponding atom positions from (a) are indicated. B⬘ A⬘ A⬘ B⬘ C⬘ C⬘ D⬘ E⬘ E⬘ D⬘ (a) (b) indices that are enclosed in braces—such as {100}. For example, in cubic crystals the (111), (111), (111), (111), (111), (111), (111), and (111) planes all belong to the {111} family. On the other hand, for tetragonal crystal structures, the {100} family would contain only the (100), (100), (010), and (010) since the (001) and (001) planes are not crystallographically equivalent. Also, in the cubic system only, planes having the same indices, irrespective of order and sign, are equivalent. For example, both (123) and (312) belong to the {123} family. Hexagonal Cr ystals For crystals having hexagonal symmetry, it is desirable that equivalent planes have the same indices; as with directions, this is accomplished by the Miller–Bravais system shown in Figure 3.7. This convention leads to the four-index (hkil) scheme, which is favored in most instances, since it more clearly identifies the orientation of a plane in a hexagonal crystal. There is some redundancy in that i is determined by the sum of h and k through i  1h  k2 (3.7) Otherwise the three h, k, and l indices are identical for both indexing systems. Figure 3.8b presents several of the common planes that are found for crystals having hexagonal symmetry. EXAMPLE PROBLEM 3.11 Determination of Miller–Bravais Indices for a Plane Within a Hexagonal Unit Cell Determine the Miller–Bravais indices for the plane shown in the hexagonal unit cell. z H F G E a2 c D a3 a B C a A a1 1496T_c03_38-79 11/11/05 19:05 Page 60 REVISED PAGES 60 • Chapter 3 / The Structure of Crystalline Solids Solution To determine these Miller–Bravais indices, consider the plane in the figure referenced to the parallelepiped labeled with the letters A through H at its corners. This plane intersects the a1 axis at a distance a from the origin of the a1-a2-a3-z coordinate axes system (point C). Furthermore, its intersections with the a2 and z axes are a and c, respectively. Therefore, in terms of the lattice parameters, these intersections are 1, 1, and 1. Furthermore, the reciprocals of these numbers are also 1, 1, and 1. Hence h1 k  1 l1 and, from Equation 3.7 i  1h  k2  11  12  0 Therefore the (hkil) indices are (1101). Notice that the third index is zero (i.e., its reciprocal  q ), which means that this plane parallels the a3 axis. Upon inspection of the above figure, it may be noted that this is indeed the case. 3.11 LINEAR AND PLANAR DENSITIES The two previous sections discussed the equivalency of nonparallel crystallographic directions and planes. Directional equivalency is related to linear density in the sense that, for a particular material, equivalent directions have identical linear densities. The corresponding parameter for crystallographic planes is planar density, and planes having the same planar density values are also equivalent. Linear density (LD) is defined as the number of atoms per unit length whose centers lie on the direction vector for a specific crystallographic direction; that is, LD  number of atoms centered on direction vector length of direction vector (3.8) Of course, the units of linear density are reciprocal length (e.g., nm1, m1). For example, let us determine the linear density of the [110] direction for the FCC crystal structure. An FCC unit cell (reduced sphere) and the [110] direction therein are shown in Figure 3.12a. Represented in Figure 3.12b are those five atoms that lie on the bottom face of this unit cell; here the [110] direction vector passes from the center of atom X, through atom Y, and finally to the center of atom Z. With regard to the numbers of atoms, it is necessary to take into account the sharing of atoms with adjacent unit cells (as discussed in Section 3.4 relative to atomic packing factor computations). Each of the X and Z corner atoms are also shared with one other adjacent unit cell along this [110] direction (i.e., one-half of each of these atoms belongs to the unit cell being considered), while atom Y lies entirely within the unit cell. Thus, there is an equivalence of two atoms along the [110] direction vector in the unit cell. Now, the direction vector length is equal to 4R (Figure 3.12b); thus, from Equation 3.8, the [110] linear density for FCC is LD110  2 atoms 1  4R 2R (3.9) 1496T_c03_38-79 12/20/05 7:38 Page 61 2nd REVISE PAGES 3.12 Close-Packed Crystal Structures • 61 R X Y X Z Y [110] Z (a) Figure 3.12 (a) Reducedsphere FCC unit cell with the [110] direction indicated. (b) The bottom face-plane of the FCC unit cell in (a) on which is shown the atomic spacing in the [110] direction, through atoms labeled X, Y, and Z. (b) In an analogous manner, planar density (PD) is taken as the number of atoms per unit area that are centered on a particular crystallographic plane, or PD  number of atoms centered on a plane area of plane (3.10) The units for planar density are reciprocal area (e.g., nm2, m2). For example, consider the section of a (110) plane within an FCC unit cell as represented in Figures 3.10a and 3.10b. Although six atoms have centers that lie on this plane (Figure 3.10b), only one-quarter of each of atoms A, C, D, and F, and onehalf of atoms B and E, for a total equivalence of just 2 atoms are on that plane. Furthermore, the area of this rectangular section is equal to the product of its length and width. From Figure 3.10b, the length (horizontal dimension) is equal to 4R, whereas the width (vertical dimension) is equal to 2R12, since it corresponds to the FCC unit cell edge length (Equation 3.1). Thus, the area of this planar region is 14R212R 122  8R2 12, and the planar density is determined as follows: PD110  2 atoms 1  2 2 8R 12 4R 12 (3.11) Linear and planar densities are important considerations relative to the process of slip—that is, the mechanism by which metals plastically deform (Section 7.4). Slip occurs on the most densely packed crystallographic planes and, in those planes, along directions having the greatest atomic packing. 3.12 CLOSE-PACKED CRYSTAL STRUCTURES Close-Packed Structures (Metals) You may remember from the discussion on metallic crystal structures that both face-centered cubic and hexagonal close-packed crystal structures have atomic packing factors of 0.74, which is the most efficient packing of equal-sized spheres or atoms. In addition to unit cell representations, these two crystal structures may be described in terms of close-packed planes of atoms (i.e., planes having a maximum atom or sphere-packing density); a portion of one such plane is illustrated in Figure 3.13a. Both crystal structures may be generated by the stacking of these close-packed planes on top of one another; the difference between the two structures lies in the stacking sequence. Let the centers of all the atoms in one close-packed plane be labeled A. Associated with this plane are two sets of equivalent triangular depressions formed by three adjacent atoms, into which the next close-packed plane of atoms may rest. Those having the triangle vertex pointing up are arbitrarily designated as B positions, while the remaining depressions are those with the down vertices, which are marked C in Figure 3.13a. 1496T_c03_38-79 12/20/05 7:38 Page 62 2nd REVISE PAGES 62 • Chapter 3 / The Structure of Crystalline Solids B C B C B C B C B C C B B C C C (a) C C C C C C C C Figure 3.13 (a) A portion of a close-packed plane of atoms; A, B, and C positions are indicated. (b) The AB stacking sequence for closepacked atomic planes. (Adapted from W. G. Moffatt, G. W. Pearsall, and J. Wulff, The Structure and Properties of Materials, Vol. I, Structure, p. 50. Copyright © 1964 by John Wiley & Sons, New York. Reprinted by permission of John Wiley & Sons, Inc.) C (b) A second close-packed plane may be positioned with the centers of its atoms over either B or C sites; at this point both are equivalent. Suppose that the B positions are arbitrarily chosen; the stacking sequence is termed AB, which is illustrated in Figure 3.13b. The real distinction between FCC and HCP lies in where the third close-packed layer is positioned. For HCP, the centers of this layer are aligned directly above the original A positions. This stacking sequence, ABABAB . . . , is repeated over and over. Of course, the ACACAC . . . arrangement would be equivalent. These close-packed planes for HCP are (0001)-type planes, and the correspondence between this and the unit cell representation is shown in Figure 3.14. For the face-centered crystal structure, the centers of the third plane are situated over the C sites of the first plane (Figure 3.15a). This yields an ABCABCABC . . . A B A B A Figure 3.14 Close-packed plane stacking sequence for hexagonal close-packed. (Adapted from W. G. Moffatt, G. W. Pearsall, and J. Wulff, The Structure and Properties of Materials, Vol. I, Structure, p. 51. Copyright © 1964 by John Wiley & Sons, New York. Reprinted by permission of John Wiley & Sons, Inc.) 1496T_c03_38-79 12/20/05 7:38 Page 63 2nd REVISE PAGES 3.13 Single Crystals • 63 B A C B A C B A (a) (b) Figure 3.15 (a) Close-packed stacking sequence for face-centered cubic. (b) A corner has been removed to show the relation between the stacking of close-packed planes of atoms and the FCC crystal structure; the heavy triangle outlines a (111) plane. [Figure (b) from W. G. Moffatt, G. W. Pearsall, and J. Wulff, The Structure and Properties of Materials, Vol. I, Structure, p. 51. Copyright © 1964 by John Wiley & Sons, New York. Reprinted by permission of John Wiley & Sons, Inc.] stacking sequence; that is, the atomic alignment repeats every third plane. It is more difficult to correlate the stacking of close-packed planes to the FCC unit cell. However, this relationship is demonstrated in Figure 3.15b. These planes are of the (111) type; an FCC unit cell is outlined on the upper left-hand front face of Figure 3.15b, in order to provide a perspective. The significance of these FCC and HCP close-packed planes will become apparent in Chapter 7. The concepts detailed in the previous four sections also relate to crystalline ceramic and polymeric materials, which are discussed in Chapters 12 and 14. We may specify crystallographic planes and directions in terms of directional and Miller indices; furthermore, on occasion it is important to ascertain the atomic and ionic arrangements of particular crystallographic planes. Also, the crystal structures of a number of ceramic materials may be generated by the stacking of close-packed planes of ions (Section 12.2). C r ys t a l l i n e a n d N o n c r ys t a l l i n e M a t e r i a l s 3.13 SINGLE CRYSTALS single crystal For a crystalline solid, when the periodic and repeated arrangement of atoms is perfect or extends throughout the entirety of the specimen without interruption, the result is a single crystal. All unit cells interlock in the same way and have the same 1496T_c03_38-79 11/11/05 19:05 Page 64 REVISED PAGES 64 • Chapter 3 / The Structure of Crystalline Solids Figure 3.16 Photograph of a garnet single crystal that was found in Tongbei, Fujian Province, China. (Photograph courtesy of Irocks.com, Megan Foreman photo.) orientation. Single crystals exist in nature, but they may also be produced artificially. They are ordinarily difficult to grow, because the environment must be carefully controlled. If the extremities of a single crystal are permitted to grow without any external constraint, the crystal will assume a regular geometric shape having flat faces, as with some of the gem stones; the shape is indicative of the crystal structure. A photograph of a garnet single crystal is shown in Figure 3.16. Within the past few years, single crystals have become extremely important in many of our modern technologies, in particular electronic microcircuits, which employ single crystals of silicon and other semiconductors. 3.14 POLYCRYSTALLINE MATERIALS grain polycrystalline grain boundary Most crystalline solids are composed of a collection of many small crystals or grains; such materials are termed polycrystalline. Various stages in the solidification of a polycrystalline specimen are represented schematically in Figure 3.17. Initially, small crystals or nuclei form at various positions. These have random crystallographic orientations, as indicated by the square grids. The small grains grow by the successive addition from the surrounding liquid of atoms to the structure of each. The extremities of adjacent grains impinge on one another as the solidification process approaches completion. As indicated in Figure 3.17, the crystallographic orientation varies from grain to grain. Also, there exists some atomic mismatch within the region where two grains meet; this area, called a grain boundary, is discussed in more detail in Section 4.6. 3.15 ANISOTROPY anisotropy isotropic The physical properties of single crystals of some substances depend on the crystallographic direction in which measurements are taken. For example, the elastic modulus, the electrical conductivity, and the index of refraction may have different values in the [100] and [111] directions. This directionality of properties is termed anisotropy, and it is associated with the variance of atomic or ionic spacing with crystallographic direction. Substances in which measured properties are independent of the direction of measurement are isotropic. The extent and magnitude of 1496T_c03_38-79 11/11/05 19:05 Page 65 REVISED PAGES 3.15 Anisotropy • 65 (a) (b) (c) (d) Figure 3.17 Schematic diagrams of the various stages in the solidification of a polycrystalline material; the square grids depict unit cells. (a) Small crystallite nuclei. (b) Growth of the crystallites; the obstruction of some grains that are adjacent to one another is also shown. (c) Upon completion of solidification, grains having irregular shapes have formed. (d) The grain structure as it would appear under the microscope; dark lines are the grain boundaries. (Adapted from W. Rosenhain, An Introduction to the Study of Physical Metallurgy, 2nd edition, Constable & Company Ltd., London, 1915.) anisotropic effects in crystalline materials are functions of the symmetry of the crystal structure; the degree of anisotropy increases with decreasing structural symmetry—triclinic structures normally are highly anisotropic. The modulus of elasticity values at [100], [110], and [111] orientations for several materials are presented in Table 3.3. For many polycrystalline materials, the crystallographic orientations of the individual grains are totally random. Under these circumstances, even though each grain may be anisotropic, a specimen composed of the grain aggregate behaves isotropically. Also, the magnitude of a measured property represents some average of the directional values. Sometimes the grains in polycrystalline materials have a preferential crystallographic orientation, in which case the material is said to have a “texture.” The magnetic properties of some iron alloys used in transformer cores are anisotropic—that is, grains (or single crystals) magnetize in a 81009-type direction easier than any other crystallographic direction. Energy losses in transformer cores are minimized by utilizing polycrystalline sheets of these alloys into which have been introduced a “magnetic texture”: most of the grains in each sheet have a 1496T_c03_38-79 11/11/05 19:05 Page 66 REVISED PAGES 66 • Chapter 3 / The Structure of Crystalline Solids Table 3.3 Modulus of Elasticity Values for Several Metals at Various Crystallographic Orientations Modulus of Elasticity (GPa) Metal [100] [110] [111] Aluminum Copper Iron Tungsten 63.7 66.7 125.0 384.6 72.6 130.3 210.5 384.6 76.1 191.1 272.7 384.6 Source: R. W. Hertzberg, Deformation and Fracture Mechanics of Engineering Materials, 3rd edition. Copyright © 1989 by John Wiley & Sons, New York. Reprinted by permission of John Wiley & Sons, Inc. 81009-type crystallographic direction that is aligned (or almost aligned) in the same direction, which direction is oriented parallel to the direction of the applied magnetic field. Magnetic textures for iron alloys are discussed in detail in the Materials of Importance piece in Chapter 20 following Section 20.9. 3.16 X-RAY DIFFRACTION: DETERMINATION OF CRYSTAL STRUCTURES Historically, much of our understanding regarding the atomic and molecular arrangements in solids has resulted from x-ray diffraction investigations; furthermore, x-rays are still very important in developing new materials. We will now give a brief overview of the diffraction phenomenon and how, using x-rays, atomic interplanar distances and crystal structures are deduced. The Diffraction Phenomenon diffraction Diffraction occurs when a wave encounters a series of regularly spaced obstacles that (1) are capable of scattering the wave, and (2) have spacings that are comparable in magnitude to the wavelength. Furthermore, diffraction is a consequence of specific phase relationships established between two or more waves that have been scattered by the obstacles. Consider waves 1 and 2 in Figure 3.18a which have the same wavelength 1l2 and are in phase at point O–O¿ . Now let us suppose that both waves are scattered in such a way that they traverse different paths. The phase relationship between the scattered waves, which will depend upon the difference in path length, is important. One possibility results when this path length difference is an integral number of wavelengths. As noted in Figure 3.18a, these scattered waves (now labeled 1¿ and 2¿ ) are still in phase. They are said to mutually reinforce (or constructively interfere with) one another; and, when amplitudes are added, the wave shown on the right side of the figure results. This is a manifestation of diffraction, and we refer to a diffracted beam as one composed of a large number of scattered waves that mutually reinforce one another. Other phase relationships are possible between scattered waves that will not lead to this mutual reinforcement. The other extreme is that demonstrated in Figure 3.18b, wherein the path length difference after scattering is some integral number of half wavelengths. The scattered waves are out of phase—that is, corresponding amplitudes cancel or annul one another, or destructively interfere (i.e., the 1496T_c03_38-79 11/11/05 19:05 Page 67 REVISED PAGES 3.16 X-Ray Diffraction: Determination of Crystal Structures • 67 O Wave 1 ␭ Scattering event Wave 1' ␭ ␭ Amplitude A A 2A ␭ ␭ A + A Wave 2 O' Wave 2' Position (a) P Wave 3 ␭ Scattering event A Amplitude Figure 3.18 (a) Demonstration of how two waves (labeled 1 and 2) that have the same wavelength l and remain in phase after a scattering event (waves 1¿ and 2¿ ) constructively interfere with one another. The amplitudes of the scattered waves add together in the resultant wave. (b) Demonstration of how two waves (labeled 3 and 4) that have the same wavelength and become out of phase after a scattering event (waves 3¿ and 4¿ ) destructively interfere with one another. The amplitudes of the two scattered waves cancel one another. Wave 3' ␭ A + ␭ A A ␭ Wave 4 Wave 4' P' Position (b) resultant wave has zero amplitude), as indicated on the extreme right side of the figure. Of course, phase relationships intermediate between these two extremes exist, resulting in only partial reinforcement. X-Ray Diffraction and Bragg’s Law X-rays are a form of electromagnetic radiation that have high energies and short wavelengths—wavelengths on the order of the atomic spacings for solids. When a beam of x-rays impinges on a solid material, a portion of this beam will be scattered in all directions by the electrons associated with each atom or ion that lies within the beam’s path. Let us now examine the necessary conditions for diffraction of x-rays by a periodic arrangement of atoms. Consider the two parallel planes of atoms A–A¿ and B–B¿ in Figure 3.19, which have the same h, k, and l Miller indices and are separated by the interplanar spacing dhkl. Now assume that a parallel, monochromatic, and coherent (in-phase) beam of x-rays of wavelength l is incident on these two planes at an angle u. Two rays in this beam, labeled 1 and 2, are scattered by atoms P and Q. Constructive interference of the scattered rays 1¿ and 2¿ occurs also at an angle u to the planes, if the path length difference between 1–P–1¿ and 2–Q–2¿ (i.e., SQ  QT ) is equal to a whole number, n, of wavelengths. That is, the condition for diffraction is nl  SQ  QT (3.12) 1496T_c03_38-79 11/11/05 19:05 Page 68 REVISED PAGES 68 • Chapter 3 / The Structure of Crystalline Solids Figure 3.19 Diffraction of x-rays by planes of atoms (A–A¿ and B–B¿ ). 1 Incident beam 2 A ␭ ␭ P ␪ ␪ B ␪ ␪ S 1' Diffracted beam 2' A' dhkl T B' Q Bragg’s law— relationship among x-ray wavelength, interatomic spacing, and angle of diffraction for constructive interference Bragg’s law Interplanar separation for a plane having indices h, k, and l or nl  dhkl sin u  dhkl sin u  2dhkl sin u (3.13) Equation 3.13 is known as Bragg’s law; also, n is the order of reflection, which may be any integer (1, 2, 3, . . . ) consistent with sin u not exceeding unity. Thus, we have a simple expression relating the x-ray wavelength and interatomic spacing to the angle of the diffracted beam. If Bragg’s law is not satisfied, then the interference will be nonconstructive in nature so as to yield a very low-intensity diffracted beam. The magnitude of the distance between two adjacent and parallel planes of atoms (i.e., the interplanar spacing dhkl) is a function of the Miller indices (h, k, and l) as well as the lattice parameter(s). For example, for crystal structures that have cubic symmetry, a (3.14) dhkl  2h2  k2  l 2 in which a is the lattice parameter (unit cell edge length). Relationships similar to Equation 3.14, but more complex, exist for the other six crystal systems noted in Table 3.2. Bragg’s law, Equation 3.13, is a necessary but not sufficient condition for diffraction by real crystals. It specifies when diffraction will occur for unit cells having atoms positioned only at cell corners. However, atoms situated at other sites (e.g., face and interior unit cell positions as with FCC and BCC) act as extra scattering centers, which can produce out-of-phase scattering at certain Bragg angles. The net result is the absence of some diffracted beams that, according to Equation 3.13, should be present. For example, for the BCC crystal structure, h  k  l must be even if diffraction is to occur, whereas for FCC, h, k, and l must all be either odd or even. 1496T_c03_38-79 11/11/05 19:05 Page 69 REVISED PAGES 3.16 X-Ray Diffraction: Determination of Crystal Structures • 69 Figure 3.20 Schematic diagram of an x-ray diffractometer; T  x-ray source, S  specimen, C  detector, and O  the axis around which the specimen and detector rotate. O S ␪ 0° T 20° 160 ° 2␪ ° 40 14 0° C 0° 60 ° 80° 100° 12 Concept Check 3.2 For cubic crystals, as values of the planar indices h, k, and l increase, does the distance between adjacent and parallel planes (i.e., the interplanar spacing) increase or decrease? Why? [The answer may be found at www.wiley.com/college/callister (Student Companion Site).] Diffraction Techniques One common diffraction technique employs a powdered or polycrystalline specimen consisting of many fine and randomly oriented particles that are exposed to monochromatic x-radiation. Each powder particle (or grain) is a crystal, and having a large number of them with random orientations ensures that some particles are properly oriented such that every possible set of crystallographic planes will be available for diffraction. The diffractometer is an apparatus used to determine the angles at which diffraction occurs for powdered specimens; its features are represented schematically in Figure 3.20. A specimen S in the form of a flat plate is supported so that rotations about the axis labeled O are possible; this axis is perpendicular to the plane of the page. The monochromatic x-ray beam is generated at point T, and the intensities of diffracted beams are detected with a counter labeled C in the figure. The specimen, x-ray source, and counter are all coplanar. The counter is mounted on a movable carriage that may also be rotated about the O axis; its angular position in terms of 2u is marked on a graduated scale.4 Carriage and specimen are mechanically coupled such that a rotation of the specimen through u is accompanied by a 2u rotation of the counter; this assures that the incident and reflection angles are maintained equal to one another (Figure 3.20). 4 Note that the symbol u has been used in two different contexts for this discussion. Here, u represents the angular locations of both x-ray source and counter relative to the specimen surface. Previously (e.g., Equation 3.13), it denoted the angle at which the Bragg criterion for diffraction is satisfied. 1496T_c03_38-79 11/11/05 19:05 Page 70 REVISED PAGES 70 • Chapter 3 / The Structure of Crystalline Solids (111) Intensity Figure 3.21 Diffraction pattern for powdered lead. (Courtesy of Wesley L. Holman.) 0.0 (311) (200) (220) (222) (400) (331) (420) 10.0 20.0 30.0 40.0 50.0 60.0 70.0 80.0 90.0 (422) 100.0 Diffraction angle 2␪ Collimators are incorporated within the beam path to produce a well-defined and focused beam. Utilization of a filter provides a near-monochromatic beam. As the counter moves at constant angular velocity, a recorder automatically plots the diffracted beam intensity (monitored by the counter) as a function of 2u; 2u is termed the diffraction angle, which is measured experimentally. Figure 3.21 shows a diffraction pattern for a powdered specimen of lead. The high-intensity peaks result when the Bragg diffraction condition is satisfied by some set of crystallographic planes. These peaks are plane-indexed in the figure. Other powder techniques have been devised wherein diffracted beam intensity and position are recorded on a photographic film instead of being measured by a counter. One of the primary uses of x-ray diffractometry is for the determination of crystal structure. The unit cell size and geometry may be resolved from the angular positions of the diffraction peaks, whereas arrangement of atoms within the unit cell is associated with the relative intensities of these peaks. X-rays, as well as electron and neutron beams, are also used in other types of material investigations. For example, crystallographic orientations of single crystals are possible using x-ray diffraction (or Laue) photographs. In the (a) chapter-opening photograph for this chapter is shown a photograph that was generated using an incident x-ray beam that was directed on a magnesium crystal; each spot (with the exception of the darkest one near the center) resulted from an x-ray beam that was diffracted by a specific set of crystallographic planes. Other uses of x-rays include qualitative and quantitative chemical identifications and the determination of residual stresses and crystal size. EXAMPLE PROBLEM 3.12 Interplanar Spacing and Diffraction Angle Computations For BCC iron, compute (a) the interplanar spacing, and (b) the diffraction angle for the (220) set of planes. The lattice parameter for Fe is 0.2866 nm. Also, assume that monochromatic radiation having a wavelength of 0.1790 nm is used, and the order of reflection is 1. Solution (a) The value of the interplanar spacing dhkl is determined using Equation 3.14, with a  0.2866 nm, and h  2, k  2, and l  0, since we are considering the (220) planes. Therefore, dhkl   a 2h  k2  l 2 2 0.2866 nm 2122 2  122 2  102 2  0.1013 nm 1496T_c03_38-79 11/11/05 19:05 Page 71 REVISED PAGES 3.17 Noncrystalline Solids • 71 (b) The value of u may now be computed using Equation 3.13, with n  1, since this is a first-order reflection: sin u  11210.1790 nm2 nl   0.884 2dhkl 12210.1013 nm2 u  sin1 10.8842  62.13 The diffraction angle is 2u, or 2u  122162.132  124.26 3.17 NONCRYSTALLINE SOLIDS noncrystalline amorphous It has been mentioned that noncrystalline solids lack a systematic and regular arrangement of atoms over relatively large atomic distances. Sometimes such materials are also called amorphous (meaning literally without form), or supercooled liquids, inasmuch as their atomic structure resembles that of a liquid. An amorphous condition may be illustrated by comparison of the crystalline and noncrystalline structures of the ceramic compound silicon dioxide (SiO2), which may exist in both states. Figures 3.22a and 3.22b present two-dimensional schematic diagrams for both structures of SiO2. Even though each silicon ion bonds to three oxygen ions for both states, beyond this, the structure is much more disordered and irregular for the noncrystalline structure. Whether a crystalline or amorphous solid forms depends on the ease with which a random atomic structure in the liquid can transform to an ordered state during solidification. Amorphous materials, therefore, are characterized by atomic or molecular structures that are relatively complex and become ordered only with some difficulty. Furthermore, rapidly cooling through the freezing temperature favors the formation of a noncrystalline solid, since little time is allowed for the ordering process. Metals normally form crystalline solids, but some ceramic materials are crystalline, whereas others, the inorganic glasses, are amorphous. Polymers may be completely Silicon atom Oxygen atom (a) (b) Figure 3.22 Two-dimensional schemes of the structure of (a) crystalline silicon dioxide and (b) noncrystalline silicon dioxide. 1496T_c03_38-79 11/11/05 19:05 Page 72 REVISED PAGES 72 • Chapter 3 / The Structure of Crystalline Solids noncrystalline and semicrystalline consisting of varying degrees of crystallinity. More about the structure and properties of amorphous ceramics and polymers is contained in Chapters 12 and 14. Concept Check 3.3 Do noncrystalline materials display the phenomenon of allotropy (or polymorphism)? Why or why not? [The answer may be found at www.wiley.com/college/callister (Student Companion Site).] SUMMARY Fundamental Concepts Unit Cells Atoms in crystalline solids are positioned in orderly and repeated patterns that are in contrast to the random and disordered atomic distribution found in noncrystalline or amorphous materials. Atoms may be represented as solid spheres, and, for crystalline solids, crystal structure is just the spatial arrangement of these spheres. The various crystal structures are specified in terms of parallelepiped unit cells, which are characterized by geometry and atom positions within. Metallic Crystal Structures Most common metals exist in at least one of three relatively simple crystal structures: face-centered cubic (FCC), body-centered cubic (BCC), and hexagonal closepacked (HCP). Two features of a crystal structure are coordination number (or number of nearest-neighbor atoms) and atomic packing factor (the fraction of solid sphere volume in the unit cell). Coordination number and atomic packing factor are the same for both FCC and HCP crystal structures, each of which may be generated by the stacking of close-packed planes of atoms. Point Coordinates Crystallographic Directions Crystallographic Planes Crystallographic points, directions, and planes are specified in terms of indexing schemes. The basis for the determination of each index is a coordinate axis system defined by the unit cell for the particular crystal structure. The location of a point within a unit cell is specified using coordinates that are fractional multiples of the cell edge lengths. Directional indices are computed in terms of the vector projection on each of the coordinate axes, whereas planar indices are determined from the reciprocals of axial intercepts. For hexagonal unit cells, a four-index scheme for both directions and planes is found to be more convenient. Linear and Planar Densities Crystallographic directional and planar equivalencies are related to atomic linear and planar densities, respectively.The atomic packing (i.e., planar density) of spheres 1496T_c03_38-79 11/11/05 19:05 Page 73 REVISED PAGES References • 73 in a crystallographic plane depends on the indices of the plane as well as the crystal structure. For a given crystal structure, planes having identical atomic packing yet different Miller indices belong to the same family. Single Crystals Polycrystalline Materials Single crystals are materials in which the atomic order extends uninterrupted over the entirety of the specimen; under some circumstances, they may have flat faces and regular geometric shapes. The vast majority of crystalline solids, however, are polycrystalline, being composed of many small crystals or grains having different crystallographic orientations. Crystal Systems Polymorphism and Allotropy Anisotropy Other concepts introduced in this chapter were: crystal system (a classification scheme for crystal structures on the basis of unit cell geometry), polymorphism (or allotropy) (when a specific material can have more than one crystal structure), and anisotropy (the directionality dependence of properties). X-Ray Diffraction: Determination of Crystal Structures X-ray diffractometry is used for crystal structure and interplanar spacing determinations. A beam of x-rays directed on a crystalline material may experience diffraction (constructive interference) as a result of its interaction with a series of parallel atomic planes according to Bragg’s law. Interplanar spacing is a function of the Miller indices and lattice parameter(s) as well as the crystal structure. I M P O R TA N T T E R M S A N D C O N C E P T S Allotropy Amorphous Anisotropy Atomic packing factor (APF) Body-centered cubic (BCC) Bragg’s law Coordination number Crystal structure Crystal system Crystalline Diffraction Face-centered cubic (FCC) Grain Grain boundary Hexagonal close-packed (HCP) Isotropic Lattice Lattice parameters Miller indices Noncrystalline Polycrystalline Polymorphism Single crystal Unit cell REFERENCES Azaroff, L. F., Elements of X-Ray Crystallography, McGraw-Hill, New York, 1968. Reprinted by TechBooks, Marietta, OH, 1990. Buerger, M. J., Elementary Crystallography, Wiley, New York, 1956. Cullity, B. D., and S. R. Stock, Elements of X-Ray Diffraction, 3rd edition, Prentice Hall, Upper Saddle River, NJ, 2001. 1496T_c03_38-79 11/11/05 19:05 Page 74 REVISED PAGES 74 • Chapter 3 / The Structure of Crystalline Solids QUESTIONS AND PROBLEMS Fundamental Concepts 3.1 What is the difference between atomic structure and crystal structure? Unit Cells Metallic Crystal Structures 3.2 If the atomic radius of lead is 0.175 nm, calculate the volume of its unit cell in cubic meters. 3.3 Show for the body-centered cubic crystal structure that the unit cell edge length a and the atomic radius R are related through a  4R 13. 3.4 For the HCP crystal structure, show that the ideal ca ratio is 1.633. 3.5 Show that the atomic packing factor for BCC is 0.68. 3.6 Show that the atomic packing factor for HCP is 0.74. Density Computations 3.7 Molybdenum has a BCC crystal structure, an atomic radius of 0.1363 nm, and an atomic weight of 95.94 g/mol. Compute and compare its theoretical density with the experimental value found inside the front cover. 3.8 Calculate the radius of a palladium atom, given that Pd has an FCC crystal structure, a density of 12.0 g/cm3, and an atomic weight of 106.4 g/mol. 3.9 Calculate the radius of a tantalum atom, given that Ta has a BCC crystal structure, a density of 16.6 g/cm3, and an atomic weight of 180.9 g/mol. 3.10 Some hypothetical metal has the simple cubic crystal structure shown in Figure 3.23. If its atomic weight is 74.5 g/mol and the atomic radius is 0.145 nm, compute its density. 3.11 Titanium has an HCP crystal structure and a density of 4.51 g/cm3. (a) What is the volume of its unit cell in cubic meters? (b) If the c a ratio is 1.58, compute the values of c and a. 3.12 Using atomic weight, crystal structure, and atomic radius data tabulated inside the front cover, compute the theoretical densities of aluminum, nickel, magnesium, and tungsten, and then compare these values with the measured densities listed in this same table. The ca ratio for magnesium is 1.624. 3.13 Niobium has an atomic radius of 0.1430 nm and a density of 8.57 g/cm3. Determine whether it has an FCC or BCC crystal structure. 3.14 Below are listed the atomic weight, density, and atomic radius for three hypothetical alloys. For each determine whether its crystal structure is FCC, BCC, or simple cubic and then justify your determination. A simple cubic unit cell is shown in Figure 3.23. Alloy A B C Atomic Weight (g/mol) Density (g/cm3) Atomic Radius (nm) 43.1 184.4 91.6 6.40 12.30 9.60 0.122 0.146 0.137 3.15 The unit cell for uranium has orthorhombic symmetry, with a, b, and c lattice parameters of 0.286, 0.587, and 0.495 nm, respectively. If its density, atomic weight, and atomic radius are 19.05 g/cm3, 238.03 g/mol, and 0.1385 nm, respectively, compute the atomic packing factor. Figure 3.23 Hard-sphere unit cell representation of the simple cubic crystal structure. 1496T_c03_38-79 12/20/05 7:38 Page 75 2nd REVISE PAGES Questions and Problems • 75 3.16 Indium has a tetragonal unit cell for which the a and c lattice parameters are 0.459 and 0.495 nm, respectively. (a) If the atomic packing factor and atomic radius are 0.693 and 0.1625 nm, respectively, determine the number of atoms in each unit cell. (b) The atomic weight of indium is 114.82 g/mol; compute its theoretical density. 3.17 Beryllium has an HCP unit cell for which the ratio of the lattice parameters c a is 1.568. If the radius of the Be atom is 0.1143 nm, (a) determine the unit cell volume, and (b) calculate the theoretical density of Be and compare it with the literature value. 3.18 Magnesium has an HCP crystal structure, a ca ratio of 1.624, and a density of 1.74 g/cm3. Compute the atomic radius for Mg. 3.19 Cobalt has an HCP crystal structure, an atomic radius of 0.1253 nm, and a ca ratio of 1.623. Compute the volume of the unit cell for Co. Crystal Systems 3.20 Below is a unit cell for a hypothetical metal. (a) To which crystal system does this unit cell belong? (b) What would this crystal structure be called? (c) Calculate the density of the material, given that its atomic weight is 141 g/mol. 3.23 List the point coordinates of both the sodium and chlorine ions for a unit cell of the sodium chloride crystal structure (Figure 12.2). 3.24 List the point coordinates of both the zinc and sulfur atoms for a unit cell of the zinc blende crystal structure (Figure 12.4). 3.25 Sketch a tetragonal unit cell, and within that cell indicate locations of the 1 1 21 and 21 41 12 point coordinates. 3.26 Using the Molecule Definition Utility found in both “Metallic Crystal Structures and Crystallography” and “Ceramic Crystal Structures” modules of VMSE, located on the book’s web site [www.wiley.com/college/callister (Student Companion Site)], generate (and print out) a three-dimensional unit cell for b tin given the following: (1) the unit cell is tetragonal with a  0.583 nm and c  0.318 nm, and (2) Sn atoms are located at the following point coordinates: 000 011 1 3 100 2 0 4 1 3 110 2 1 4 1 1 124 010 0 21 14 001 1 1 1 2 2 2 101 111 Crystallographic Directions 3.27 Draw an orthorhombic unit cell, and within that cell a [211] direction. 3.28 Sketch a monoclinic unit cell, and within that cell a [101] direction. 3.29 What are the indices for the directions indicated by the two vectors in the sketch below? +z 90° +z 0.45 nm O 90° 90° +y 0.35 nm +x 0.4 nm Direction 2 0.35 nm 3.21 Sketch a unit cell for the face-centered orthorhombic crystal structure. +y 0.3 nm Point Coordinates 3.22 List the point coordinates for all atoms that are associated with the FCC unit cell (Figure 3.1). +x Direction 1 0.5 nm 1496T_c03_38-79 12/20/05 7:38 Page 76 2nd REVISE PAGES 76 • Chapter 3 / The Structure of Crystalline Solids 3.30 Within a cubic unit cell, sketch the following directions: (a) [101], (e) [111], (b) [211], (f) [212], (c) [102], (g) [312], (d) [313], (h) [301]. 3.31 Determine the indices for the directions shown in the following cubic unit cell: 3.35 Determine the indices for the directions shown in the following hexagonal unit cell: z a2 A +z B a3 D C A a1 1 2 C 1, 1 2 2 1 2 B D +y 3.36 Using Equations 3.6a, 3.6b, 3.6c, and 3.6d, derive expressions for each of the three primed indices set (u¿, v¿, and w¿ ) in terms of the four unprimed indices (u, v, t, and w). Crystallographic Planes +x 3.32 Determine the indices for the directions shown in the following cubic unit cell: +z 3.37 (a) Draw an orthorhombic unit cell, and within that cell a (021) plane. (b) Draw a monoclinic unit cell, and within that cell a (200) plane. 3.38 What are the indices for the two planes drawn in the sketch below? 2 3 Plane 2 1 3 C 1, 1 2 2 1 3 D Plane 1 1 2 +y 1, 1 2 2 +y 0.2 nm A 1 2 +z B 2 3 0.4 nm +x 0.4 nm +x 3.33 For tetragonal crystals, cite the indices of directions that are equivalent to each of the following directions: (a) [011] (b) [100] 3.34 Convert the [110] and [001] directions into the four-index Miller–Bravais scheme for hexagonal unit cells. 3.39 Sketch within a cubic unit cell the following planes: (a) (101), (e) (111), (b) (211), (f) (212), (c) (012), (g) (312), (d) (313), (h) (301). 1496T_c03_38-79 11/11/05 19:05 Page 77 REVISED PAGES Questions and Problems • 77 1 3 1 2 B A +y +x 3.41 Determine the Miller indices for the planes shown in the following unit cell: +z 0.4 nm 1 2 1 2 0.4 nm A 0.4 nm (100) (001) B +y 1 2 +x 3.42 Determine the Miller indices for the planes shown in the following unit cell: 0.55 nm +z and (111) planes, (b) (110) and (110) planes, and (c) (111) and (001) planes. 3.44 Sketch the atomic packing of (a) the (100) plane for the FCC crystal structure, and (b) the (111) plane for the BCC crystal structure (similar to Figures 3.10b and 3.11b). 3.45 Consider the reduced-sphere unit cell shown in Problem 3.20, having an origin of the coordinate system positioned at the atom labeled with an O. For the following sets of planes, determine which are equivalent: (a) (100), (010), and (001) (b) (110), (101), (011), and (101) (c) (111), (111), (111), and (111) 3.46 Here are three different crystallographic planes for a unit cell of a hypothetical metal. The circles represent atoms: 0.55 nm 3.40 Determine the Miller indices for the planes shown in the following unit cell: (110) (a) To what crystal system does the unit cell belong? (b) What would this crystal structure be called? 3.47 Below are shown three different crystallographic planes for a unit cell of some hypothetical metal. The circles represent atoms: 0.32 nm 0.39 nm 1 2 1 2 0.36 nm +z A 0.30 nm 0.20 nm (110) B +y +x 3.43 Cite the indices of the direction that results from the intersection of each of the following pair of planes within a cubic crystal: (a) (110) 0.25 nm (101) (011) (a) To what crystal system does the unit cell belong? (b) What would this crystal structure be called? (c) If the density of this metal is 18.91 g/cm3, determine its atomic weight. 3.48 Convert the (111) and (012) planes into the four-index Miller–Bravais scheme for hexagonal unit cells. 1496T_c03_38-79 12/20/05 7:38 Page 78 2nd REVISE PAGES 78 • Chapter 3 / The Structure of Crystalline Solids 3.49 Determine the indices for the planes shown in the hexagonal unit cells below: z 3.50 Sketch the (0111) and (2110) planes in a hexagonal unit cell. Linear and Planar Densities a2 a3 a1 (a) z a2 a3 a1 (b) z 3.51 (a) Derive linear density expressions for FCC [100] and [111] directions in terms of the atomic radius R. (b) Compute and compare linear density values for these same two planes for copper. 3.52 (a) Derive linear density expressions for BCC [110] and [111] directions in terms of the atomic radius R. (b) Compute and compare linear density values for these same two planes for iron. 3.53 (a) Derive planar density expressions for FCC (100) and (111) planes in terms of the atomic radius R. (b) Compute and compare planar density values for these same two planes for aluminum. 3.54 (a) Derive planar density expressions for BCC (100) and (110) planes in terms of the atomic radius R. (b) Compute and compare planar density values for these same two planes for molybdenum. 3.55 (a) Derive the planar density expression for the HCP (0001) plane in terms of the atomic radius R. (b) Compute the planar density value for this same plane for titanium. Polycrystalline Materials a2 3.56 Explain why the properties of polycrystalline materials are most often isotropic. a3 a1 (c) z a2 a3 (d) a1 X-Ray Diffraction: Determination of Crystal Structures 3.57 Using the data for aluminum in Table 3.1, compute the interplanar spacing for the (110) set of planes. 3.58 Determine the expected diffraction angle for the first-order reflection from the (310) set of planes for BCC chromium when monochromatic radiation of wavelength 0.0711 nm is used. 3.59 Using the data for -iron in Table 3.1, compute the interplanar spacings for the (111) and (211) sets of planes. 1496T_c03_38-79 11/11/05 19:05 Page 79 REVISED PAGES Questions and Problems • 79 Intensity Figure 3.24 Diffraction pattern for powdered tungsten. (Courtesy of Wesley L. Holman.) 0.0 20.0 40.0 60.0 80.0 100.0 Diffraction angle 2␪ 3.60 The metal rhodium has an FCC crystal structure. If the angle of diffraction for the (311) set of planes occurs at 36.12 (first-order reflection) when monochromatic x-radiation having a wavelength of 0.0711 nm is used, compute (a) the interplanar spacing for this set of planes, and (b) the atomic radius for a rhodium atom. 3.61 The metal niobium has a BCC crystal structure. If the angle of diffraction for the (211) set of planes occurs at 75.99 (first-order reflection) when monochromatic x-radiation having a wavelength of 0.1659 nm is used, compute (a) the interplanar spacing for this set of planes, and (b) the atomic radius for the niobium atom. 3.62 For which set of crystallographic planes will a first-order diffraction peak occur at a diffraction angle of 44.53 for FCC nickel when monochromatic radiation having a wavelength of 0.1542 nm is used? 3.63 Figure 3.21 shows an x-ray diffraction pattern for lead taken using a diffractometer and monochromatic x-radiation having a wavelength of 0.1542 nm; each diffraction peak on the pattern has been indexed. Compute the interplanar spacing for each set of planes indexed; also determine the lattice parameter of Pb for each of the peaks. 3.64 The diffraction peaks shown in Figure 3.21 are indexed according to the reflection rules for FCC (i.e., h, k, and l must all be either odd or even). Cite the h, k, and l indices of the first four diffraction peaks for BCC crystals consistent with h  k  l being even. 3.65 Figure 3.24 shows the first five peaks of the x-ray diffraction pattern for tungsten, which has a BCC crystal structure; monochromatic x-radiation having a wavelength of 0.1542 nm was used. (a) Index (i.e., give h, k, and l indices) for each of these peaks. (b) Determine the interplanar spacing for each of the peaks. (c) For each peak, determine the atomic radius for W and compare these with the value presented in Table 3.1. Noncrystalline Solids 3.66 Would you expect a material in which the atomic bonding is predominantly ionic in nature to be more or less likely to form a noncrystalline solid upon solidification than a covalent material? Why? (See Section 2.6.) 1496T_c04_80-108 12/20/05 7:46 Page 80 Chapter 4 2nd REVISE PAGES Imperfections in Solids A scanning probe micrograph (generated using a scanning-tunneling microscope) that shows a (111)-type surface plane* for silicon. The arrow points to the location of a silicon atom that was removed using a tungsten nanotip probe. This site from which an atom is missing is the surface analogue of a vacancy defect— that is, a vacant lattice site within the bulk material. Approximately 20,000,000. (Micrograph courtesy of D. Huang, Stanford University.) WHY STUDY Imperfections in Solids? The properties of some materials are profoundly influenced by the presence of imperfections. Consequently, it is important to have a knowledge about the types of imperfections that exist and the roles they play in affecting the behavior of materials. For example, the mechanical properties of pure metals experience significant alterations when alloyed (i.e., when impurity atoms are added)—for example, brass (70% copper–30% zinc) is much harder and stronger than pure copper (Section 7.9). Also, integrated circuit microelectronic devices found in our computers, calculators, and home appliances function because of highly controlled concentrations of specific impurities that are incorporated into small, localized regions of semiconducting materials (Sections 18.11 and 18.15). *The plane shown here is termed a “Si(111)-77 reconstructed surface.” The twodimensional arrangement of atoms in a surface plane is different than the atomic arrangement for the equivalent plane within the interior of the material (i.e., the surface plane has been “reconstructed” by atomic displacements). The “77” notation pertains to the displacement magnitude. Furthermore, the diamond shape that has been drawn indicates a unit cell for this 77 structure. 80 • 1496T_c04_80-108 12/20/05 7:46 Page 81 2nd REVISE PAGES Learning Objectives After studying this chapter you should be able to do the following: 1. Describe both vacancy and self-interstitial crys5. For each of edge, screw, and mixed dislocations: talline defects. (a) describe and make a drawing of the 2. Calculate the equilibrium number of vacancies dislocation. in a material at some specified temperature, (b) note the location of the dislocation line, and given the relevant constants. (c) indicate the direction along which the dislo3. Name the two types of solid solutions, and procation line extends. vide a brief written definition and/or schematic 6. Describe the atomic structure within the vicinity sketch of each. of (a) a grain boundary, and (b) a twin boundary. 4. Given the masses and atomic weights of two or more elements in a metal alloy, calculate the weight percent and atom percent for each element. 4.1 INTRODUCTION imperfection point defect Thus far it has been tacitly assumed that perfect order exists throughout crystalline materials on an atomic scale. However, such an idealized solid does not exist; all contain large numbers of various defects or imperfections. As a matter of fact, many of the properties of materials are profoundly sensitive to deviations from crystalline perfection; the influence is not always adverse, and often specific characteristics are deliberately fashioned by the introduction of controlled amounts or numbers of particular defects, as detailed in succeeding chapters. By “crystalline defect” is meant a lattice irregularity having one or more of its dimensions on the order of an atomic diameter. Classification of crystalline imperfections is frequently made according to geometry or dimensionality of the defect. Several different imperfections are discussed in this chapter, including point defects (those associated with one or two atomic positions), linear (or one-dimensional) defects, as well as interfacial defects, or boundaries, which are two-dimensional. Impurities in solids are also discussed, since impurity atoms may exist as point defects. Finally, techniques for the microscopic examination of defects and the structure of materials are briefly described. Po i n t D e f e c t s 4.2 VACANCIES AND SELF-INTERSTITIALS vacancy Temperaturedependence of the equilibrium number of vacancies The simplest of the point defects is a vacancy, or vacant lattice site, one normally occupied from which an atom is missing (Figure 4.1). All crystalline solids contain vacancies and, in fact, it is not possible to create such a material that is free of these defects. The necessity of the existence of vacancies is explained using principles of thermodynamics; in essence, the presence of vacancies increases the entropy (i.e., the randomness) of the crystal. The equilibrium number of vacancies Nv for a given quantity of material depends on and increases with temperature according to Nv  N exp a Qv b kT (4.1) 1496T_c04_80-108 11/14/05 9:53 Page 82 REVISED PAGES 82 • Chapter 4 / Imperfections in Solids Self-interstitial Boltzmann’s constant self-interstitial Vacancy Figure 4.1 Two-dimensional representations of a vacancy and a self-interstitial. (Adapted from W. G. Moffatt, G. W. Pearsall, and J. Wulff, The Structure and Properties of Materials, Vol. I, Structure, p. 77. Copyright © 1964 by John Wiley & Sons, New York. Reprinted by permission of John Wiley & Sons, Inc.) In this expression, N is the total number of atomic sites, Qv is the energy required for the formation of a vacancy, T is the absolute temperature1 in kelvins, and k is the gas or Boltzmann’s constant. The value of k is 1.38  10 23 J/atom-K, or 8.62  105 eV/atom-K, depending on the units of Qv.2 Thus, the number of vacancies increases exponentially with temperature; that is, as T in Equation 4.1 increases, so also does the expression exp (QvkT ). For most metals, the fraction of vacancies NvN just below the melting temperature is on the order of 104; that is, one lattice site out of 10,000 will be empty. As ensuing discussions indicate, a number of other material parameters have an exponential dependence on temperature similar to that of Equation 4.1. A self-interstitial is an atom from the crystal that is crowded into an interstitial site, a small void space that under ordinary circumstances is not occupied. This kind of defect is also represented in Figure 4.1. In metals, a self-interstitial introduces relatively large distortions in the surrounding lattice because the atom is substantially larger than the interstitial position in which it is situated. Consequently, the formation of this defect is not highly probable, and it exists in very small concentrations, which are significantly lower than for vacancies. EXAMPLE PROBLEM 4.1 Number of Vacancies Computation at a Specified Temperature Calculate the equilibrium number of vacancies per cubic meter for copper at 1000C. The energy for vacancy formation is 0.9 eV/atom; the atomic weight and density (at 1000C) for copper are 63.5 g/mol and 8.4 g/cm3, respectively. Absolute temperature in kelvins (K) is equal to C  273. Boltzmann’s constant per mole of atoms becomes the gas constant R; in such a case R  8.31 J/mol-K. 1 2 1496T_c04_80-108 11/14/05 9:53 Page 83 REVISED PAGES 4.3 Impurities in Solids • 83 Solution This problem may be solved by using Equation 4.1; it is first necessary, however, to determine the value of N, the number of atomic sites per cubic meter for copper, from its atomic weight ACu, its density r, and Avogadro’s number NA, according to Number of atoms per unit volume for a metal N NA r ACu (4.2) 16.023  1023 atoms/mol218.4 g/cm3 21106 cm3/m3 2 63.5 g/mol 28  8.0  10 atoms/m3  Thus, the number of vacancies at 1000C (1273 K) is equal to Nv  N exp a Qv b kT  18.0  1028 atoms/m3 2 exp c   2.2  1025 vacancies/m3 10.9 eV2 18.62  105 eV/K211273 K2 d 4.3 IMPURITIES IN SOLIDS alloy solid solution solute, solvent A pure metal consisting of only one type of atom just isn’t possible; impurity or foreign atoms will always be present, and some will exist as crystalline point defects. In fact, even with relatively sophisticated techniques, it is difficult to refine metals to a purity in excess of 99.9999%. At this level, on the order of 1022 to 1023 impurity atoms will be present in one cubic meter of material. Most familiar metals are not highly pure; rather, they are alloys, in which impurity atoms have been added intentionally to impart specific characteristics to the material. Ordinarily, alloying is used in metals to improve mechanical strength and corrosion resistance. For example, sterling silver is a 92.5% silver–7.5% copper alloy. In normal ambient environments, pure silver is highly corrosion resistant, but also very soft. Alloying with copper significantly enhances the mechanical strength without depreciating the corrosion resistance appreciably. The addition of impurity atoms to a metal will result in the formation of a solid solution and/or a new second phase, depending on the kinds of impurity, their concentrations, and the temperature of the alloy. The present discussion is concerned with the notion of a solid solution; treatment of the formation of a new phase is deferred to Chapter 9. Several terms relating to impurities and solid solutions deserve mention. With regard to alloys, solute and solvent are terms that are commonly employed.“Solvent” represents the element or compound that is present in the greatest amount; on occasion, solvent atoms are also called host atoms. “Solute” is used to denote an element or compound present in a minor concentration. Solid Solutions A solid solution forms when, as the solute atoms are added to the host material, the crystal structure is maintained, and no new structures are formed. Perhaps it is 1496T_c04_80-108 11/14/05 9:53 Page 84 REVISED PAGES 84 • Chapter 4 / Imperfections in Solids Substitutional impurity atom Figure 4.2 Two-dimensional schematic representations of substitutional and interstitial impurity atoms. (Adapted from W. G. Moffatt, G. W. Pearsall, and J. Wulff, The Structure and Properties of Materials, Vol. I, Structure, p. 77. Copyright © 1964 by John Wiley & Sons, New York. Reprinted by permission of John Wiley & Sons, Inc.) Interstitial impurity atom substitutional solid solution interstitial solid solution useful to draw an analogy with a liquid solution. If two liquids, soluble in each other (such as water and alcohol) are combined, a liquid solution is produced as the molecules intermix, and its composition is homogeneous throughout. A solid solution is also compositionally homogeneous; the impurity atoms are randomly and uniformly dispersed within the solid. Impurity point defects are found in solid solutions, of which there are two types: substitutional and interstitial. For the substitutional type, solute or impurity atoms replace or substitute for the host atoms (Figure 4.2). There are several features of the solute and solvent atoms that determine the degree to which the former dissolves in the latter, as follows: 1. Atomic size factor. Appreciable quantities of a solute may be accommodated in this type of solid solution only when the difference in atomic radii between the two atom types is less than about 15% . Otherwise the solute atoms will create substantial lattice distortions and a new phase will form. 2. Crystal structure. For appreciable solid solubility the crystal structures for metals of both atom types must be the same. 3. Electronegativity. The more electropositive one element and the more electronegative the other, the greater is the likelihood that they will form an intermetallic compound instead of a substitutional solid solution. 4. Valences. Other factors being equal, a metal will have more of a tendency to dissolve another metal of higher valency than one of a lower valency. An example of a substitutional solid solution is found for copper and nickel. These two elements are completely soluble in one another at all proportions. With regard to the aforementioned rules that govern degree of solubility, the atomic radii for copper and nickel are 0.128 and 0.125 nm, respectively, both have the FCC crystal structure, and their electronegativities are 1.9 and 1.8 (Figure 2.7); finally, the most common valences are 1 for copper (although it sometimes can be 2) and 2 for nickel. For interstitial solid solutions, impurity atoms fill the voids or interstices among the host atoms (see Figure 4.2). For metallic materials that have relatively high atomic packing factors, these interstitial positions are relatively small. Consequently, 1496T_c04_80-108 11/14/05 9:53 Page 85 REVISED PAGES 4.4 Specification of Composition • 85 the atomic diameter of an interstitial impurity must be substantially smaller than that of the host atoms. Normally, the maximum allowable concentration of interstitial impurity atoms is low (less than 10%). Even very small impurity atoms are ordinarily larger than the interstitial sites, and as a consequence they introduce some lattice strains on the adjacent host atoms. Problem 4.5 calls for determination of the radii of impurity atoms (in terms of R, the host atom radius) that will just fit into interstitial positions without introducing any lattice strains for both FCC and BCC crystal structures. Carbon forms an interstitial solid solution when added to iron; the maximum concentration of carbon is about 2%. The atomic radius of the carbon atom is much less than that for iron: 0.071 nm versus 0.124 nm. Solid solutions are also possible for ceramic materials, as discussed in Section 12.5. 4.4 SPECIFICATION OF COMPOSITION composition weight percent Computation of weight percent (for a two-element alloy) atom percent It is often necessary to express the composition (or concentration)3 of an alloy in terms of its constituent elements. The two most common ways to specify composition are weight (or mass) percent and atom percent. The basis for weight percent (wt%) is the weight of a particular element relative to the total alloy weight. For an alloy that contains two hypothetical atoms denoted by 1 and 2, the concentration of 1 in wt%, C1, is defined as C1  m1  100 m1  m2 (4.3) where m1 and m2 represent the weight (or mass) of elements 1 and 2, respectively. The concentration of 2 would be computed in an analogous manner. The basis for atom percent (at%) calculations is the number of moles of an element in relation to the total moles of the elements in the alloy. The number of moles in some specified mass of a hypothetical element 1, nm1, may be computed as follows: m1¿ nm1  (4.4) A1 Here, m¿1 and A1 denote the mass (in grams) and atomic weight, respectively, for element 1. Concentration in terms of atom percent of element 1 in an alloy containing 1 and 2 atoms, C¿1, is defined by4 Computation of atom percent (for a two-element alloy) C¿1  nm1  100 nm1  nm2 (4.5) In like manner, the atom percent of 2 may be determined. 3 The terms composition and concentration will be assumed to have the same meaning in this book (i.e., the relative content of a specific element or constituent in an alloy) and will be used interchangeably. 4 In order to avoid confusion in notations and symbols that are being used in this section, we should point out that the prime (as in C¿1 and m¿1) is used to designate both composition, in atom percent, as well as mass of material in units of grams. 1496T_c04_80-108 11/14/05 9:53 Page 86 REVISED PAGES 86 • Chapter 4 / Imperfections in Solids Atom percent computations also can be carried out on the basis of the number of atoms instead of moles, since one mole of all substances contains the same number of atoms. Composition Conversions Sometimes it is necessary to convert from one composition scheme to another— for example, from weight percent to atom percent. We will now present equations for making these conversions in terms of the two hypothetical elements 1 and 2. Using the convention of the previous section (i.e., weight percents denoted by C1 and C2, atom percents by C¿1 and C¿2, and atomic weights as A1 and A2), these conversion expressions are as follows: Conversion of weight percent to atom percent (for a two-element alloy) Conversion of atom percent to weight percent (for a twoelement alloy) C¿1  C1A2  100 C1A2  C2A1 (4.6a) C¿2  C2A1  100 C1A2  C2A1 (4.6b) C1  C¿1A1  100 C¿1A1  C¿2A2 (4.7a) C2  C¿2A2  100 C¿1A1  C¿2A2 (4.7b) Since we are considering only two elements, computations involving the preceding equations are simplified when it is realized that C1  C2  100 C¿1  C¿2  100 (4.8a) (4.8b) In addition, it sometimes becomes necessary to convert concentration from weight percent to mass of one component per unit volume of material (i.e., from units of wt% to kg/m3); this latter composition scheme is often used in diffusion computations (Section 5.3). Concentrations in terms of this basis will be denoted using a double prime (i.e., C–1 and C–2), and the relevant equations are as follows: Conversion of weight percent to mass per unit volume (for a twoelement alloy) C–1  C1  103 C2 ¢ ° C1  r1 r2 (4.9a) C–2  C2  103 C2 ¢ ° C1  r1 r2 (4.9b) For density r in units of g/cm3, these expressions yield C–1 and C–2 in kg/m3. Furthermore, on occasion we desire to determine the density and atomic weight of a binary alloy given the composition in terms of either weight percent or atom 1496T_c04_80-108 11/14/05 9:53 Page 87 REVISED PAGES 4.4 Specification of Composition • 87 percent. If we represent alloy density and atomic weight by rave and Aave, respectively, then Computation of density (for a twoelement metal alloy) rave  rave  Computation of atomic weight (for a two-element metal alloy) Aave  Aave  100 (4.10a) C1 C2  r1 r2 C¿1A1  C¿2A2 C¿1A1 C¿2A2  r1 r2 100 C1 C2  A1 A2 (4.10b) (4.11a) C¿1A1  C¿2A2 100 (4.11b) It should be noted that Equations 4.9 and 4.11 are not always exact. In their derivations, it is assumed that total alloy volume is exactly equal to the sum of the volumes of the individual elements. This normally is not the case for most alloys; however, it is a reasonably valid assumption and does not lead to significant errors for dilute solutions and over composition ranges where solid solutions exist. EXAMPLE PROBLEM 4.2 Derivation of Composition-Conversion Equation Derive Equation 4.6a. Solution To simplify this derivation, we will assume that masses are expressed in units of grams, and denoted with a prime (e.g., m¿1). Furthermore, the total alloy mass (in grams) M¿ is M¿  m¿1  m¿2 (4.12) Using the definition of C¿1 (Equation 4.5) and incorporating the expression for nm1, Equation 4.4, and the analogous expression for nm2 yields C¿1   nm1  100 nm1  nm2 m¿1 A1 m¿1 m¿2  A1 A2  100 (4.13) Rearrangement of the mass-in-grams equivalent of Equation 4.3 leads to m¿1  C1M¿ 100 (4.14) 1496T_c04_80-108 11/14/05 9:53 Page 88 REVISED PAGES 88 • Chapter 4 / Imperfections in Solids Substitution of this expression and its m¿2 equivalent into Equation 4.13 gives C1M¿ 100A1 C¿1   100 C1M¿ C2M¿  100A1 100A2 (4.15) Upon simplification we have C¿1  C1A2  100 C1A2  C2A1 which is identical to Equation 4.6a. EXAMPLE PROBLEM 4.3 Composition Conversion—From Weight Percent to Atom Percent Determine the composition, in atom percent, of an alloy that consists of 97 wt% aluminum and 3 wt% copper. Solution If we denote the respective weight percent compositions as CA1  97 and CCu  3, substitution into Equations 4.6a and 4.6b yields C¿A1   CA1ACu  100 CA1ACu  CCu AA1 1972163.55 g/mol2  100 CCu AA1  100 CCu AA1  CA1ACu 132126.98 g/mol2  100 1972163.55 g/mol2  132126.98 g/mol2  98.7 at% and C¿Cu   132126.98 g/mol2  1972163.55 g/mol2  1.30 at% Miscellaneous Imperfections 4.5 DISLOCATIONS—LINEAR DEFECTS edge dislocation A dislocation is a linear or one-dimensional defect around which some of the atoms are misaligned. One type of dislocation is represented in Figure 4.3: an extra portion of a plane of atoms, or half-plane, the edge of which terminates within the crystal. This is termed an edge dislocation; it is a linear defect that centers around the line that is defined along the end of the extra half-plane of atoms. This is 1496T_c04_80-108 12/20/05 7:46 Page 89 2nd REVISE PAGES 4.5 Dislocations—Linear Defects • 89 Burgers vector b Figure 4.3 The atom positions around an edge dislocation; extra half-plane of atoms shown in perspective. (Adapted from A. G. Guy, Essentials of Materials Science, McGraw-Hill Book Company, New York, 1976, p. 153.) Edge dislocation line dislocation line Edge screw dislocation Screw Mixed mixed dislocation Burgers vector sometimes termed the dislocation line, which, for the edge dislocation in Figure 4.3, is perpendicular to the plane of the page. Within the region around the dislocation line there is some localized lattice distortion. The atoms above the dislocation line in Figure 4.3 are squeezed together, and those below are pulled apart; this is reflected in the slight curvature for the vertical planes of atoms as they bend around this extra half-plane. The magnitude of this distortion decreases with distance away from the dislocation line; at positions far removed, the crystal lattice is virtually perfect. Sometimes the edge dislocation in Figure 4.3 is represented by the symbol , which also indicates the position of the dislocation line. An edge dislocation may also be formed by an extra half-plane of atoms that is included in the bottom portion of the crystal; its designation is a . Another type of dislocation, called a screw dislocation, exists, which may be thought of as being formed by a shear stress that is applied to produce the distortion shown in Figure 4.4a: the upper front region of the crystal is shifted one atomic distance to the right relative to the bottom portion. The atomic distortion associated with a screw dislocation is also linear and along a dislocation line, line AB in Figure 4.4b. The screw dislocation derives its name from the spiral or helical path or ramp that is traced around the dislocation line by the atomic planes of atoms. Sometimes the symbol is used to designate a screw dislocation. Most dislocations found in crystalline materials are probably neither pure edge nor pure screw, but exhibit components of both types; these are termed mixed dislocations. All three dislocation types are represented schematically in Figure 4.5; the lattice distortion that is produced away from the two faces is mixed, having varying degrees of screw and edge character. The magnitude and direction of the lattice distortion associated with a dislocation is expressed in terms of a Burgers vector, denoted by a b. Burgers vectors are indicated in Figures 4.3 and 4.4 for edge and screw dislocations, respectively. Furthermore, the nature of a dislocation (i.e., edge, screw, or mixed) is defined by the relative orientations of dislocation line and Burgers vector. For an edge, they are perpendicular (Figure 4.3), whereas for a screw, they are parallel (Figure 4.4); they are neither perpendicular nor parallel for a mixed dislocation. Also, even though a dislocation changes direction and nature within a crystal (e.g., from edge to mixed to screw), the Burgers vector will be the same at all points along its line. For example, all positions of the curved dislocation in Figure 4.5 will have the Burgers vector shown. For metallic materials, the Burgers vector for a dislocation will point in a close-packed crystallographic direction and will be of magnitude equal to the interatomic spacing. 1496T_c04_80-108 12/20/05 7:46 Page 90 2nd REVISE PAGES 90 • Chapter 4 / Imperfections in Solids Figure 4.4 (a) A screw dislocation within a crystal. (b) The screw dislocation in (a) as viewed from above. The dislocation line extends along line AB. Atom positions above the slip plane are designated by open circles, those below by solid circles. [Figure (b) from W. T. Read, Jr., Dislocations in Crystals, McGrawHill Book Company, New York, 1953.] C A D Dislocation line Burgers vector b (a) A B b D C (b) 1496T_c04_80-108 12/20/05 7:46 Page 91 2nd REVISE PAGES 4.5 Dislocations—Linear Defects • 91 Figure 4.5 (a) Schematic representation of a dislocation that has edge, screw, and mixed character. (b) Top view, where open circles denote atom positions above the slip plane. Solid circles, atom positions below. At point A, the dislocation is pure screw, while at point B, it is pure edge. For regions in between where there is curvature in the dislocation line, the character is mixed edge and screw. [Figure (b) from W. T. Read, Jr., Dislocations in Crystals, McGrawHill Book Company, New York, 1953.] b B A b C (a) B b b C A b (b) 1496T_c04_80-108 11/14/05 9:53 Page 92 REVISED PAGES 92 • Chapter 4 / Imperfections in Solids Figure 4.6 A transmission electron micrograph of a titanium alloy in which the dark lines are dislocations. 51,450. (Courtesy of M. R. Plichta, Michigan Technological University.) As we note in Section 7.4, the permanent deformation of most crystalline materials is by the motion of dislocations. In addition, the Burgers vector is an element of the theory that has been developed to explain this type of deformation. Dislocations can be observed in crystalline materials using electron-microscopic techniques. In Figure 4.6, a high-magnification transmission electron micrograph, the dark lines are the dislocations. Virtually all crystalline materials contain some dislocations that were introduced during solidification, during plastic deformation, and as a consequence of thermal stresses that result from rapid cooling. Dislocations are involved in the plastic deformation of crystalline materials, both metals and ceramics, as discussed in Chapters 7 and 12. They have also been observed in polymeric materials, and are discussed in Section 14.13. 4.6 INTERFACIAL DEFECTS Interfacial defects are boundaries that have two dimensions and normally separate regions of the materials that have different crystal structures and/or crystallographic orientations. These imperfections include external surfaces, grain boundaries, twin boundaries, stacking faults, and phase boundaries. External Surfaces One of the most obvious boundaries is the external surface, along which the crystal structure terminates. Surface atoms are not bonded to the maximum number of nearest neighbors, and are therefore in a higher energy state than the atoms at interior positions. The bonds of these surface atoms that are not satisfied give rise to a surface energy, expressed in units of energy per unit area (J/m2 or erg/cm2). To reduce this energy, materials tend to minimize, if at all possible, the total surface area. For example, liquids assume a shape having a minimum area—the droplets become spherical. Of course, this is not possible with solids, which are mechanically rigid. Grain Boundaries Another interfacial defect, the grain boundary, was introduced in Section 3.14 as the boundary separating two small grains or crystals having different crystallographic 1496T_c04_80-108 11/14/05 9:53 Page 93 REVISED PAGES 4.6 Interfacial Defects • 93 Figure 4.7 Schematic diagram showing smalland high-angle grain boundaries and the adjacent atom positions. Angle of misalignment High-angle grain boundary Small-angle grain boundary Angle of misalignment orientations in polycrystalline materials. A grain boundary is represented schematically from an atomic perspective in Figure 4.7. Within the boundary region, which is probably just several atom distances wide, there is some atomic mismatch in a transition from the crystalline orientation of one grain to that of an adjacent one. Various degrees of crystallographic misalignment between adjacent grains are possible (Figure 4.7). When this orientation mismatch is slight, on the order of a few degrees, then the term small- (or low- ) angle grain boundary is used. These boundaries can be described in terms of dislocation arrays. One simple smallangle grain boundary is formed when edge dislocations are aligned in the manner of Figure 4.8. This type is called a tilt boundary; the angle of misorientation, u, is also indicated in the figure. When the angle of misorientation is parallel to the boundary, a twist boundary results, which can be described by an array of screw dislocations. The atoms are bonded less regularly along a grain boundary (e.g., bond angles are longer), and consequently, there is an interfacial or grain boundary energy similar to the surface energy described above. The magnitude of this energy is a function of the degree of misorientation, being larger for high-angle boundaries. Grain boundaries are more chemically reactive than the grains themselves as a consequence of this boundary energy. Furthermore, impurity atoms often preferentially segregate along these boundaries because of their higher energy state. The total interfacial energy is lower in large or coarse-grained materials than in fine-grained ones, since there is less total boundary area in the former. Grains grow at elevated temperatures to reduce the total boundary energy, a phenomenon explained in Section 7.13. In spite of this disordered arrangement of atoms and lack of regular bonding along grain boundaries, a polycrystalline material is still very strong; cohesive forces within and across the boundary are present. Furthermore, the density of a polycrystalline specimen is virtually identical to that of a single crystal of the same material. 1496T_c04_80-108 11/14/05 9:53 Page 94 REVISED PAGES 94 • Chapter 4 / Imperfections in Solids Figure 4.8 Demonstration of how a tilt boundary having an angle of misorientation u results from an alignment of edge dislocations. b ␪ Twin Boundaries A twin boundary is a special type of grain boundary across which there is a specific mirror lattice symmetry; that is, atoms on one side of the boundary are located in mirror-image positions of the atoms on the other side (Figure 4.9). The region of material between these boundaries is appropriately termed a twin. Twins result from atomic displacements that are produced from applied mechanical shear forces (mechanical twins), and also during annealing heat treatments following deformation (annealing twins). Twinning occurs on a definite crystallographic plane and in a specific direction, both of which depend on the crystal structure. Annealing twins are typically found in metals that have the FCC crystal structure, while mechanical twins are observed in BCC and HCP metals. The role of mechanical twins in the deformation process is discussed in Section 7.7. Annealing twins may be observed Twin plane (boundary) Figure 4.9 Schematic diagram showing a twin plane or boundary and the adjacent atom positions (colored circles). 1496T_c04_80-108 12/20/05 7:46 Page 95 2nd REVISE PAGES 4.6 Interfacial Defects • 95 MATERIALS OF IMPORTANCE Catalysts (and Surface Defects) A catalyst is a substance that speeds up the rate of a chemical reaction without participating in the reaction itself (i.e., it is not consumed). One type of catalyst exists as a solid; reactant molecules in a gas or liquid phase are adsorbed5 onto the catalytic surface, at which point some type of interaction occurs that promotes an increase in their chemical reactivity rate. Adsorption sites on a catalyst are normally surface defects associated with planes of atoms; an interatomic/intermolecular bond is formed between a defect site and an adsorbed molecular species. Several types of surface defects, represented schematically in Figure 4.10, include ledges, kinks, terraces, vacancies, and individual adatoms (i.e., atoms adsorbed on the surface). One important use of catalysts is in catalytic converters on automobiles, which reduce the emission of exhaust gas pollutants such as carbon monoxide (CO), nitrogen oxides (NOx, where x is variable), and unburned hydrocarbons. Air is introduced into the exhaust emissions from the automobile engine; this mixture of gases then passes over the catalyst, which adsorbs on its surface molecules of CO, NOx, and O2. The NOx dissociates into N and O atoms, whereas the O2 dissociates into its atomic species. Pairs of nitrogen atoms combine to form N2 molecules, and carbon monoxide is oxidized to form carbon dioxide (CO2). Furthermore, any unburned hydrocarbons are also oxidized to CO2 and H2O. One of the materials used as a catalyst in this application is (Ce0.5Zr0.5)O2. Figure 4.11 is a highresolution transmission electron micrograph, which shows several single crystals of this material. Individual atoms are resolved in this micrograph as well as some of the defects presented in Figure 4.10. These surface defects act as adsorption sites for the atomic and molecular species noted in the previous paragraph. Consequently, dissociation, combination, and oxidation reactions involving these species are facilitated, such that the content of pollutant species (CO, NOx, and unburned hydrocarbons) in the exhaust gas stream is reduced significantly. Terrace Kink Ledge Adatom Step Vacancy Figure 4.10 Schematic representations of surface defects that are potential adsorption sites for catalysis. Individual atom sites are represented as cubes. (From BOUDART, MICHEL, KINETICS OF HETEROGENEOUS CATALYTIC REACTIONS. © 1984 Princeton University Press. Reprinted by permission of Princeton University Press.) 5 Figure 4.11 High-resolution transmission electron micrograph that shows single crystals of (Ce0.5Zr0.5)O2; this material is used in catalytic converters for automobiles. Surface defects represented schematically in Figure 4.10 are noted on the crystals. [From W. J. Stark, L. Mädler, M. Maciejewski, S. E. Pratsinis, A. Baiker, “Flame-Synthesis of Nanocrystalline Ceria/Zirconia: Effect of Carrier Liquid,” Chem. Comm., 588–589 (2003). Reproduced by permission of The Royal Society of Chemistry.] Adsorption is the adhesion of molecules of a gas or liquid to a solid surface. It should not be confused with absorption which is the assimilation of molecules into a solid or liquid. 1496T_c04_80-108 11/14/05 9:53 Page 96 REVISED PAGES 96 • Chapter 4 / Imperfections in Solids in the photomicrograph of the polycrystalline brass specimen shown in Figure 4.13c. The twins correspond to those regions having relatively straight and parallel sides and a different visual contrast than the untwinned regions of the grains within which they reside. An explanation for the variety of textural contrasts in this photomicrograph is provided in Section 4.10. Miscellaneous Interfacial Defects Other possible interfacial defects include stacking faults, phase boundaries, and ferromagnetic domain walls. Stacking faults are found in FCC metals when there is an interruption in the ABCABCABC . . . stacking sequence of close-packed planes (Section 3.12). Phase boundaries exist in multiphase materials (Section 9.3) across which there is a sudden change in physical and/or chemical characteristics. For ferromagnetic and ferrimagnetic materials, the boundary that separates regions having different directions of magnetization is termed a domain wall, which is discussed in Section 20.7. Associated with each of the defects discussed in this section is an interfacial energy, the magnitude of which depends on boundary type, and which will vary from material to material. Normally, the interfacial energy will be greatest for external surfaces and least for domain walls. Concept Check 4.1 The surface energy of a single crystal depends on crystallographic orientation. Does this surface energy increase or decrease with an increase in planar density. Why? [The answer may be found at www.wiley.com/college/callister (Student Companion Site).] 4.7 BULK OR VOLUME DEFECTS Other defects exist in all solid materials that are much larger than those heretofore discussed. These include pores, cracks, foreign inclusions, and other phases. They are normally introduced during processing and fabrication steps. Some of these defects and their effects on the properties of materials are discussed in subsequent chapters. 4.8 ATOMIC VIBRATIONS atomic vibration Every atom in a solid material is vibrating very rapidly about its lattice position within the crystal. In a sense, these atomic vibrations may be thought of as imperfections or defects. At any instant of time not all atoms vibrate at the same frequency and amplitude, nor with the same energy. At a given temperature there will exist a distribution of energies for the constituent atoms about an average energy. Over time the vibrational energy of any specific atom will also vary in a random manner. With rising temperature, this average energy increases, and, in fact, the temperature of a solid is really just a measure of the average vibrational activity of atoms and molecules. At room temperature, a typical vibrational frequency is on the order of 1013 vibrations per second, whereas the amplitude is a few thousandths of a nanometer. Many properties and processes in solids are manifestations of this vibrational atomic motion. For example, melting occurs when the vibrations are vigorous enough to rupture large numbers of atomic bonds. A more detailed discussion of atomic vibrations and their influence on the properties of materials is presented in Chapter 19. 1496T_c04_80-108 11/14/05 9:53 Page 97 REVISED PAGES 4.9 General • 97 M i c ro s c o p i c E x a m i n a t i o n 4.9 GENERAL microstructure microscopy photomicrograph On occasion it is necessary or desirable to examine the structural elements and defects that influence the properties of materials. Some structural elements are of macroscopic dimensions; that is, they are large enough to be observed with the unaided eye. For example, the shape and average size or diameter of the grains for a polycrystalline specimen are important structural characteristics. Macroscopic grains are often evident on aluminum streetlight posts and also on highway guard rails. Relatively large grains having different textures are clearly visible on the surface of the sectioned lead ingot shown in Figure 4.12. However, in most materials the constituent grains are of microscopic dimensions, having diameters that may be on the order of microns,6 and their details must be investigated using some type of microscope. Grain size and shape are only two features of what is termed the microstructure; these and other microstructural characteristics are discussed in subsequent chapters. Optical, electron, and scanning probe microscopes are commonly used in microscopy. These instruments aid in investigations of the microstructural features of all material types. Some of these techniques employ photographic equipment in conjunction with the microscope; the photograph on which the image is recorded is called a photomicrograph. In addition, many microstructural images are computer generated and/or enhanced. Microscopic examination is an extremely useful tool in the study and characterization of materials. Several important applications of microstructural examinations are as follows: to ensure that the associations between the properties and structure (and defects) are properly understood, to predict the properties of materials once these relationships have been established, to design alloys with new property combinations, to determine whether or not a material has been correctly heat treated, and to ascertain the mode of mechanical fracture. Several techniques that are commonly used in such investigations are discussed next. Figure 4.12 High-purity polycrystalline lead ingot in which the individual grains may be discerned. 0.7. (Reproduced with permission from Metals Handbook, Vol. 9, 9th edition, Metallography and Microstructures, American Society for Metals, Metals Park, OH, 1985.) 6 A micron ( m), sometimes called a micrometer, is 106 m. 1496T_c04_80-108 11/14/05 9:53 Page 98 REVISED PAGES 98 • Chapter 4 / Imperfections in Solids 4.10 MICROSCOPIC TECHNIQUES Optical Microscopy With optical microscopy, the light microscope is used to study the microstructure; optical and illumination systems are its basic elements. For materials that are opaque to visible light (all metals and many ceramics and polymers), only the surface is subject to observation, and the light microscope must be used in a reflecting mode. Contrasts in the image produced result from differences in reflectivity of the various regions of the microstructure. Investigations of this type are often termed metallographic, since metals were first examined using this technique. Normally, careful and meticulous surface preparations are necessary to reveal the important details of the microstructure. The specimen surface must first be ground and polished to a smooth and mirrorlike finish. This is accomplished by using successively finer abrasive papers and powders. The microstructure is revealed by a surface treatment using an appropriate chemical reagent in a procedure termed etching. The chemical reactivity of the grains of some single-phase materials depends on crystallographic orientation. Consequently, in a polycrystalline specimen, etching characteristics vary from grain to grain. Figure 4.13b shows how normally incident light is reflected by three etched surface grains, each having a different orientation. Figure 4.13a depicts the surface structure as it might appear when viewed with the microscope; the luster or texture of each grain depends on its reflectance properties. A photomicrograph of a polycrystalline specimen exhibiting these characteristics is shown in Figure 4.13c. Also, small grooves form along grain boundaries as a consequence of etching. Since atoms along grain boundary regions are more chemically active, they dissolve at a greater rate than those within the grains. These grooves become discernible when viewed under a microscope because they reflect light at an angle different from that of the grains themselves; this effect is displayed in Figure 4.14a. Figure 4.14b is a photomicrograph of a polycrystalline specimen in which the grain boundary grooves are clearly visible as dark lines. When the microstructure of a two-phase alloy is to be examined, an etchant is often chosen that produces a different texture for each phase so that the different phases may be distinguished from each other. Electron Microscopy The upper limit to the magnification possible with an optical microscope is approximately 2000 times. Consequently, some structural elements are too fine or small to permit observation using optical microscopy. Under such circumstances the electron microscope, which is capable of much higher magnifications, may be employed. An image of the structure under investigation is formed using beams of electrons instead of light radiation. According to quantum mechanics, a high-velocity electron will become wave-like, having a wavelength that is inversely proportional to its velocity. When accelerated across large voltages, electrons can be made to have wavelengths on the order of 0.003 nm (3 pm). High magnifications and resolving powers of these microscopes are consequences of the short wavelengths of electron beams. The electron beam is focused and the image formed with magnetic lenses; otherwise the geometry of the microscope components is essentially the same as with optical systems. Both transmission and reflection beam modes of operation are possible for electron microscopes. 1496T_c04_80-108 11/14/05 9:53 Page 99 REVISED PAGES 4.10 Microscopic Techniques • 99 (a) Microscope Polished and etched surface Figure 4.13 (a) Polished and etched grains as they might appear when viewed with an optical microscope. (b) Section taken through these grains showing how the etching characteristics and resulting surface texture vary from grain to grain because of differences in crystallographic orientation. (c) Photomicrograph of a polycrystalline brass specimen. 60. (Photomicrograph courtesy of J. E. Burke, General Electric Co.) (b) (c) Transmission Electron Microscopy transmission electron microscope (TEM) The image seen with a transmission electron microscope (TEM) is formed by an electron beam that passes through the specimen. Details of internal microstructural features are accessible to observation; contrasts in the image are produced by differences in beam scattering or diffraction produced between various elements of the microstructure or defect. Since solid materials are highly absorptive to electron beams, a specimen to be examined must be prepared in the form of a very thin foil; 1496T_c04_80-108 11/14/05 9:53 Page 100 REVISED PAGES 100 • Chapter 4 / Imperfections in Solids Microscope Polished and etched surface Surface groove Figure 4.14 (a) Section of a grain boundary and its surface groove produced by etching; the light reflection characteristics in the vicinity of the groove are also shown. (b) Photomicrograph of the surface of a polished and etched polycrystalline specimen of an ironchromium alloy in which the grain boundaries appear dark. 100. [Photomicrograph courtesy of L. C. Smith and C. Brady, the National Bureau of Standards, Washington, DC (now the National Institute of Standards and Technology, Gaithersburg, MD.)] Grain boundary (a) (b) this ensures transmission through the specimen of an appreciable fraction of the incident beam. The transmitted beam is projected onto a fluorescent screen or a photographic film so that the image may be viewed. Magnifications approaching 1,000,000 are possible with transmission electron microscopy, which is frequently utilized in the study of dislocations. Scanning Electron Microscopy scanning electron microscope (SEM) A more recent and extremely useful investigative tool is the scanning electron microscope (SEM). The surface of a specimen to be examined is scanned with an electron beam, and the reflected (or back-scattered) beam of electrons is collected, then displayed at the same scanning rate on a cathode ray tube (similar to a CRT television screen). The image on the screen, which may be photographed, represents the surface features of the specimen. The surface may or may not be polished and etched, but it must be electrically conductive; a very thin metallic surface coating 1496T_c04_80-108 11/14/05 12:38 Page 101 REVISED PAGES 4.10 Microscopic Techniques • 101 must be applied to nonconductive materials. Magnifications ranging from 10 to in excess of 50,000 times are possible, as are also very great depths of field. Accessory equipment permits qualitative and semiquantitative analysis of the elemental composition of very localized surface areas. Scanning Probe Microscopy scanning probe microscope (SPM) In the past decade and a half, the field of microscopy has experienced a revolution with the development of a new family of scanning probe microscopes. This scanning probe microscope (SPM), of which there are several varieties, differs from the optical and electron microscopes in that neither light nor electrons is used to form an image. Rather, the microscope generates a topographical map, on an atomic scale, that is a representation of surface features and characteristics of the specimen being examined. Some of the features that differentiate the SPM from other microscopic techniques are as follows: • Examination on the nanometer scale is possible inasmuch as magnifications as high as 109  are possible; much better resolutions are attainable than with other microscopic techniques. • Three-dimensional magnified images are generated that provide topographical information about features of interest. • Some SPMs may be operated in a variety of environments (e.g., vacuum, air, liquid); thus, a particular specimen may be examined in its most suitable environment. Scanning probe microscopes employ a tiny probe with a very sharp tip that is brought into very close proximity (i.e., to within on the order of a nanometer) of the specimen surface. This probe is then raster-scanned across the plane of the surface. During scanning, the probe experiences deflections perpendicular to this plane, in response to electronic or other interactions between the probe and specimen surface. The in-surface-plane and out-of-plane motions of the probe are controlled by piezoelectric (Section 18.25) ceramic components that have nanometer resolutions. Furthermore, these probe movements are monitored electronically, and transferred to and stored in a computer, which then generates the three-dimensional surface image. Specific scanning probe microscopic techniques differ from one another with regard to the type of interaction that is monitored. A scanning probe micrograph in which may be observed the atomic structure and a missing atom on the surface of silicon is shown in the chapter-opening photograph for this chapter. These new SPMs, which allow examination of the surface of materials at the atomic and molecular level, have provided a wealth of information about a host of materials, from integrated circuit chips to biological molecules. Indeed, the advent of the SPMs has helped to usher in the era of nanomaterials—materials whose properties are designed by engineering atomic and molecular structures. Figure 4.15a is a bar-chart showing dimensional size ranges for several types of structures found in materials (note that the axes are scaled logarithmically). Likewise, the useful dimensional resolution ranges for the several microscopic techniques discussed in this chapter (plus the naked eye) are presented in the bar-chart of Figure 4.15b. For three of these techniques (viz. SPM, TEM, and SEM), an upper resolution value is not imposed by the characteristics of the microscope, and, therefore, is somewhat arbitrary and not well defined. Furthermore, by comparing Figures 4.15a and 4.15b, it is possible to decide which microscopic technique(s) is (are) best suited for examination of each of the structure types. 1496T_c04_80-108 11/14/05 9:53 Page 102 REVISED PAGES 102 • Chapter 4 / Imperfections in Solids Dimensions of structural feature (m) 1014 1012 1010 108 106 104 102 Subatomic particles Atom/ion diameters Unit cell edge lengths Dislocations (width) Second phase particles Grains Macrostructural features (porosity, voids, cracks) 106 104 102 102 1 104 106 108 Dimensions of structural feature (nm) Useful resolution ranges (m) 1012 1010 108 106 104 102 1 Scanning probe microscopes Transmission electron microscopes Scanning electron microscopes Optical microscopes Naked eye 102 1 102 104 106 108 Useful resolution ranges (nm) Figure 4.15 (a) Bar-chart showing size ranges for several structural features found in materials. (b) Bar-chart showing the useful resolution ranges for four microscopic techniques discussed in this chapter, in addition to the naked eye. (Courtesy of Prof. Sidnei Paciornik, DCMM PUC-Rio, Rio de Janeiro, Brazil, and Prof. Carlos Pérez Bergmann, Federal University of Rio Grande do Sul, Porto Alegre, Brazil.) 4.11 GRAIN SIZE DETERMINATION grain size The grain size is often determined when the properties of a polycrystalline material are under consideration. In this regard, there exist a number of techniques by which size is specified in terms of average grain volume, diameter, or area. Grain size may be estimated by using an intercept method, described as follows. Straight lines all the same length are drawn through several photomicrographs that show the grain structure. The grains intersected by each line segment are counted; the line length is then divided by an average of the number of grains intersected, taken over all the line segments. The average grain diameter is found by dividing this result by the linear magnification of the photomicrographs. Probably the most common method utilized, however, is that devised by the American Society for Testing and Materials (ASTM).7 The ASTM has prepared several standard comparison charts, all having different average grain sizes. To each 7 ASTM Standard E 112, “Standard Methods for Estimating the Average Grain Size for Metals.” 1496T_c04_80-108 11/14/05 9:53 Page 103 REVISED PAGES 4.11 Grain Size Determination • 103 Relationship between ASTM grain size number and number of grains per square inch (at 100) is assigned a number ranging from 1 to 10, which is termed the grain size number. A specimen must be properly prepared to reveal the grain structure, which is photographed at a magnification of 100. Grain size is expressed as the grain size number of the chart that most nearly matches the grains in the micrograph. Thus, a relatively simple and convenient visual determination of grain size number is possible. Grain size number is used extensively in the specification of steels. The rationale behind the assignment of the grain size number to these various charts is as follows. Let n represent the grain size number, and N the average number of grains per square inch at a magnification of 100. These two parameters are related to each other through the expression N  2n1 (4.16) Concept Check 4.2 Does the grain size number (n of Equation 4.16) increase or decrease with decreasing grain size? Why? [The answer may be found at www.wiley.com/college/callister (Student Companion Site).] EXAMPLE PROBLEM 4.4 Computations of ASTM Grain Size Number and Number of Grains Per Unit Area (a) Determine the ASTM grain size number of a metal specimen if 45 grains per square inch are measured at a magnification of 100. (b) For this same specimen, how many grains per square inch will there be at a magnification of 85? Solution (a) In order to determine the ASTM grain size number (n) it is necessary to employ Equation 4.16. Taking logarithms of both sides of this expression leads to log N  1n  12 log 2 And, solving for n yields n log N 1 log 2 From the problem statement, N  45, and, therefore n log 45  1  6.5 log 2 (b) At magnifications other than 100, use of the following modified form of Equation 4.16 is necessary: NM a M 2 b  2n1 100 (4.17) 1496T_c04_80-108 11/14/05 9:53 Page 104 REVISED PAGES 104 • Chapter 4 / Imperfections in Solids In this expression NM  the number of grains per square inch at magnification M. In addition, the inclusion of the 1M1002 2 term makes use of the fact that, while magnification is a length parameter, area is expressed in terms of units of length squared. As a consequence, the number of grains per unit area increases with the square of the increase in magnification. Solving Equation 4.17 for NM, realizing that M  85 and n  6.5 leads to 100 2 b M 100 2 b  62.6 grains/in.2  216.512 a 85 NM  2n1 a SUMMARY Vacancies and Self-Interstitials All solid materials contain large numbers of imperfections or deviations from crystalline perfection. The several types of imperfection are categorized on the basis of their geometry and size. Point defects are those associated with one or two atomic positions, including vacancies (or vacant lattice sites), self-interstitials (host atoms that occupy interstitial sites), and impurity atoms. Impurities in Solids A solid solution may form when impurity atoms are added to a solid, in which case the original crystal structure is retained and no new phases are formed. For substitutional solid solutions, impurity atoms substitute for host atoms, and appreciable solubility is possible only when atomic diameters and electronegativities for both atom types are similar, when both elements have the same crystal structure, and when the impurity atoms have a valence that is the same as or less than the host material. Interstitial solid solutions form for relatively small impurity atoms that occupy interstitial sites among the host atoms. Specification of Composition Composition of an alloy may be specified in weight percent or atom percent. The basis for weight percent computations is the weight (or mass) of each alloy constituent relative to the total alloy weight. Atom percents are calculated in terms of the number of moles for each constituent relative to the total moles of all the elements in the alloy. Dislocations—Linear Defects Dislocations are one-dimensional crystalline defects of which there are two pure types: edge and screw. An edge may be thought of in terms of the lattice distortion along the end of an extra half-plane of atoms; a screw, as a helical planar ramp. For mixed dislocations, components of both pure edge and screw are found. The magnitude and direction of lattice distortion associated with a dislocation are specified by its Burgers vector. The relative orientations of Burgers vector and dislocation line are (1) perpendicular for edge, (2) parallel for screw, and (3) neither perpendicular nor parallel for mixed. 1496T_c04_80-108 11/14/05 9:53 Page 105 REVISED PAGES References • 105 Interfacial Defects Bulk or Volume Defects Atomic Vibrations Other imperfections include interfacial defects [external surfaces, grain boundaries (both small- and high-angle), twin boundaries, etc.], volume defects (cracks, pores, etc.), and atomic vibrations. Each type of imperfection has some influence on the properties of a material. Microscopic Techniques Many of the important defects and structural elements of materials are of microscopic dimensions, and observation is possible only with the aid of a microscope. Both optical and electron microscopes are employed, usually in conjunction with photographic equipment. Transmissive and reflective modes are possible for each microscope type; preference is dictated by the nature of the specimen as well as the structural element or defect to be examined. More recent scanning probe microscopic techniques have been developed that generate topographical maps representing the surface features and characteristics of the specimen. Examinations on the atomic and molecular levels are possible using these techniques. Grain Size Determination Grain size of polycrystalline materials is frequently determined using photomicrographic techniques. Two methods are commonly employed: intercept and standard comparison charts. I M P O R TA N T T E R M S A N D C O N C E P T S Alloy Atom percent Atomic vibration Boltzmann’s constant Burgers vector Composition Dislocation line Edge dislocation Grain size Imperfection Interstitial solid solution Microscopy Microstructure Mixed dislocation Photomicrograph Point defect Scanning electron microscope (SEM) Scanning probe microscope (SPM) Screw dislocation Self-interstitial Solid solution Solute Solvent Substitutional solid solution Transmission electron microscope (TEM) Vacancy Weight percent REFERENCES ASM Handbook, Vol. 9, Metallography and Microstructures, ASM International, Materials Park, OH, 2004. Brandon, D., and W. D. Kaplan, Microstructural Characterization of Materials, Wiley, New York, 1999. DeHoff, R. T., and F. N. Rhines, Quantitative Microscopy, TechBooks, Marietta, OH, 1991. Van Bueren, H. G., Imperfections in Crystals, NorthHolland, Amsterdam (Wiley-Interscience, New York), 1960. Vander Voort, G. F., Metallography, Principles and Practice, ASM International, Materials Park, OH, 1984. 1496T_c04_80-108 11/14/05 9:53 Page 106 REVISED PAGES 106 • Chapter 4 / Imperfections in Solids QUESTIONS AND PROBLEMS Vacancies and Self-Interstitials 4.1 Calculate the fraction of atom sites that are vacant for copper at its melting temperature of 1084C (1357 K). Assume an energy for vacancy formation of 0.90 eV/atom. 4.2 Calculate the number of vacancies per cubic meter in gold at 900C. The energy for vacancy formation is 0.98 eV/atom. Furthermore, the density and atomic weight forAu are 18.63 g/cm3 (at 900°C) and 196.9 g/mol, respectively. 4.3 Calculate the energy for vacancy formation in silver, given that the equilibrium number of vacancies at 800C (1073 K) is 3.6  1023 m3. The atomic weight and density (at 800C) for silver are, respectively, 107.9 g/mol and 9.5 g/cm3. Impurities in Solids 4.4 Below, atomic radius, crystal structure, electronegativity, and the most common valence are tabulated, for several elements; for those that are nonmetals, only atomic radii are indicated. Element Atomic Radius (nm) Ni C H O Ag Al Co Cr Fe Pt Zn 0.1246 0.071 0.046 0.060 0.1445 0.1431 0.1253 0.1249 0.1241 0.1387 0.1332 Crystal Structure Electronegativity Valence FCC 1.8 2 FCC FCC HCP BCC BCC FCC HCP 1.9 1.5 1.8 1.6 1.8 2.2 1.6 1 3 2 3 2 2 2 Which of these elements would you expect to form the following with nickel: (a) A substitutional solid solution having complete solubility (b) A substitutional solid solution of incomplete solubility (c) An interstitial solid solution 4.5 For both FCC and BCC crystal structures, there are two different types of interstitial sites. In each case, one site is larger than the other, and is normally occupied by impurity atoms. For FCC, this larger one is located at the center of each edge of the unit cell; it is termed an octahedral interstitial site. On the other hand, with BCC the larger site type is found at 0 12 14 positions—that is, lying on {100} faces, and situated midway between two unit cell edges on this face and one-quarter of the distance between the other two unit cell edges; it is termed a tetrahedral interstitial site. For both FCC and BCC crystal structures, compute the radius r of an impurity atom that will just fit into one of these sites in terms of the atomic radius R of the host atom. Specification of Composition 4.6 Derive the following equations: (a) Equation 4.7a (b) Equation 4.9a (c) Equation 4.10a (d) Equation 4.11b 4.7 What is the composition, in atom percent, of an alloy that consists of 92.5 wt% Ag and 7.5 wt% Cu? 4.8 What is the composition, in weight percent, of an alloy that consists of 5 at% Cu and 95 at% Pt? 4.9 Calculate the composition, in weight percent, of an alloy that contains 105 kg of iron, 0.2 kg of carbon, and 1.0 kg of chromium. 4.10 What is the composition, in atom percent, of an alloy that contains 33 g copper and 47 g zinc? 4.11 What is the composition, in atom percent, of an alloy that contains 44.5 lbm of silver, 83.7 lbm of gold, and 5.3 lbm of Cu? 4.12 What is the composition, in atom percent, of an alloy that consists of 5.5 wt% Pb and 94.5 wt% Sn? 4.13 Convert the atom percent composition in Problem 4.11 to weight percent. 4.14 Calculate the number of atoms per cubic meter in lead. 4.15 The concentration of silicon in an iron-silicon alloy is 0.25 wt%. What is the concentration in kilograms of silicon per cubic meter of alloy? 1496T_c04_80-108 12/20/05 7:46 Page 107 2nd REVISE PAGES Questions and Problems • 107 4.16 Determine the approximate density of a Ti-6Al-4V titanium alloy that has a composition of 90 wt% Ti, 6 wt% Al, and 4 wt% V. 4.17 Calculate the unit cell edge length for an 80 wt% Ag-20 wt% Pd alloy. All of the palladium is in solid solution, the crystal structure for this alloy is FCC, and the roomtemperature density of Pd is 12.02 g/cm3. 4.18 Some hypothetical alloy is composed of 25 wt% of metal A and 75 wt% of metal B. If the densities of metals A and B are 6.17 and 8.00 g/cm3, respectively, whereas their respective atomic weights are 171.3 and 162.0 g/mol, determine whether the crystal structure for this alloy is simple cubic, face-centered cubic, or body-centered cubic. Assume a unit cell edge length of 0.332 nm. 4.19 For a solid solution consisting of two elements (designated as 1 and 2), sometimes it is desirable to determine the number of atoms per cubic centimeter of one element in a solid solution, N1, given the concentration of that element specified in weight percent, C1. This computation is possible using the following expression: N1  NAC1 C1A1 A1  1100  C1 2 r1 r2 (4.18) where NA  Avogadro’s number r1 and r2  densities of the two elements A1  the atomic weight of element 1 Derive Equation 4.18 using Equation 4.2 and expressions contained in Section 4.4. 4.20 Molybdenum forms a substitutional solid solution with tungsten. Compute the number of molybdenum atoms per cubic centimeter for a molybdenum-tungsten alloy that contains 16.4 wt% Mo and 83.6 wt% W. The densities of pure molybdenum and tungsten are 10.22 and 19.30 g/cm3, respectively. 4.21 Niobium forms a substitutional solid solution with vanadium. Compute the number of niobium atoms per cubic centimeter for a niobium-vanadium alloy that contains 24 wt% Nb and 76 wt% V. The densities of pure niobium and vanadium are 8.57 and 6.10 g/cm3, respectively. 4.22 Sometimes it is desirable to be able to determine the weight percent of one element, C1, that will produce a specified concentration in terms of the number of atoms per cubic centimeter, N1, for an alloy composed of two types of atoms. This computation is possible using the following expression: C1  100 NA r2 r2 1  r1 N1A1 (4.19) where NA  Avogadro’s number r1 and r2  densities of the two elements A1 and A2  the atomic weights of the two elements Derive Equation 4.19 using Equation 4.2 and expressions contained in Section 4.4. 4.23 Gold forms a substitutional solid solution with silver. Compute the weight percent of gold that must be added to silver to yield an alloy that contains 5.5  1021 Au atoms per cubic centimeter. The densities of pure Au and Ag are 19.32 and 10.49 g/cm3, respectively. 4.24 Germanium forms a substitutional solid solution with silicon. Compute the weight percent of germanium that must be added to silicon to yield an alloy that contains 2.43  1021 Ge atoms per cubic centimeter. The densities of pure Ge and Si are 5.32 and 2.33 g/cm3, respectively. 4.25 Iron and vanadium both have the BCC crystal structure, and V forms a substitutional solid solution for concentrations up to approximately 20 wt% V at room temperature. Compute the unit cell edge length for a 90 wt% Fe–10 wt% V alloy. Dislocations—Linear Defects 4.26 Cite the relative Burgers vector–dislocation line orientations for edge, screw, and mixed dislocations. 1496T_c04_80-108 11/14/05 9:53 Page 108 REVISED PAGES 108 • Chapter 4 / Imperfections in Solids Interfacial Defects Grain Size Determination 4.27 For an FCC single crystal, would you expect the surface energy for a (100) plane to be greater or less than that for a (111) plane? Why? (Note: You may want to consult the solution to Problem 3.53 at the end of Chapter 3.) 4.28 For a BCC single crystal, would you expect the surface energy for a (100) plane to be greater or less than that for a (110) plane? Why? (Note: You may want to consult the solution to Problem 3.54 at the end of Chapter 3.) 4.29 (a) For a given material, would you expect the surface energy to be greater than, the same as, or less than the grain boundary energy? Why? (b) The grain boundary energy of a smallangle grain boundary is less than for a highangle one. Why is this so? 4.30 (a) Briefly describe a twin and a twin boundary. (b) Cite the difference between mechanical and annealing twins. 4.31 For each of the following stacking sequences found in FCC metals, cite the type of planar defect that exists: (a) . . . A B C A B C B A C B A . . . (b) . . . A B C A B C B C A B C . . . Now, copy the stacking sequences and indicate the position(s) of planar defect(s) with a vertical dashed line. 4.32 (a) Using the intercept method, determine the average grain size, in millimeters, of the specimen whose microstructure is shown in Figure 4.14(b); use at least seven straight-line segments. (b) Estimate the ASTM grain size number for this material. 4.33 (a) Employing the intercept technique, determine the average grain size for the steel specimen whose microstructure is shown in Figure 9.25(a); use at least seven straight-line segments. (b) Estimate the ASTM grain size number for this material. 4.34 For an ASTM grain size of 6, approximately how many grains would there be per square inch at (a) a magnification of 100, and (b) without any magnification? 4.35 Determine the ASTM grain size number if 30 grains per square inch are measured at a magnification of 250. 4.36 Determine the ASTM grain size number if 25 grains per square inch are measured at a magnification of 75. DESIGN PROBLEMS Specification of Composition 4.D1 Aluminum–lithium alloys have been developed by the aircraft industry to reduce the weight and improve the performance of its aircraft. A commercial aircraft skin material having a density of 2.47 g/cm3 is desired. Compute the concentration of Li (in wt%) that is required. 4.D2 Copper and platinum both have the FCC crystal structure, and Cu forms a substitutional solid solution for concentrations up to approximately 6 wt% Cu at room temperature. Determine the concentration in weight percent of Cu that must be added to platinum to yield a unit cell edge length of 0.390 nm. 1496T_c05_109-130 12/20/05 7:52 Page 109 2nd REVISE PAGES Chapter P 5 Diffusion hotograph of a steel gear that has been “case hardened.” The outer surface layer was selectively hardened by a high-temperature heat treatment during which carbon from the surrounding atmosphere diffused into the surface. The “case” appears as the dark outer rim of that segment of the gear that has been sectioned. Actual size. (Photograph courtesy of Surface Division Midland-Ross.) WHY STUDY Diffusion? Materials of all types are often heat treated to improve their properties. The phenomena that occur during a heat treatment almost always involve atomic diffusion. Often an enhancement of diffusion rate is desired; on occasion measures are taken to reduce it. Heat-treating temperatures and times, and/or cooling rates are often predictable using the mathematics of diffusion and appropriate diffusion constants. The steel gear shown on this page has been case hardened (Section 8.10); that is, its hardness and resistance to failure by fatigue have been enhanced by diffusing excess carbon or nitrogen into the outer surface layer. • 109 1496T_c05_109-130 12/20/05 7:52 Page 110 2nd REVISE PAGES Learning Objectives After studying this chapter you should be able to do the following: 1. Name and describe the two atomic mechanisms 4. Write the solution to Fick’s second law for of diffusion. diffusion into a semi-infinite solid when the 2. Distinguish between steady-state and concentration of diffusing species at the surface nonsteady-state diffusion. is held constant. Define all parameters in this 3. (a) Write Fick’s first and second laws in equaequation. tion form, and define all parameters. 5. Calculate the diffusion coefficient for some (b) Note the kind of diffusion for which each material at a specified temperature, given the of these equations is normally applied. appropriate diffusion constants. 5.1 INTRODUCTION Cu Ni (a) (b) Concentration of Ni, Cu diffusion Many reactions and processes that are important in the treatment of materials rely on the transfer of mass either within a specific solid (ordinarily on a microscopic level) or from a liquid, a gas, or another solid phase. This is necessarily accomplished by diffusion, the phenomenon of material transport by atomic motion. This chapter discusses the atomic mechanisms by which diffusion occurs, the mathematics of diffusion, and the influence of temperature and diffusing species on the rate of diffusion. The phenomenon of diffusion may be demonstrated with the use of a diffusion couple, which is formed by joining bars of two different metals together so that there is intimate contact between the two faces; this is illustrated for copper and nickel in Figure 5.1, which includes schematic representations of atom positions and composition across the interface. This couple is heated for an extended period at an elevated temperature (but below the melting temperature of both metals), and 100 Cu Ni 0 Position (c) Figure 5.1 (a) A copper–nickel diffusion couple before a high-temperature heat treatment. (b) Schematic representations of Cu (red circles) and Ni (blue circles) atom locations within the diffusion couple. (c) Concentrations of copper and nickel as a function of position across the couple. 1496T_c05_109-130 11/14/05 13:07 Page 111 REVISED PAGES 5.2 Diffusion Mechanisms • 111 Diffusion of Cu atoms Cu Cu-Ni alloy Diffusion of Ni atoms (a) Ni Figure 5.2 (a) A copper–nickel diffusion couple after a high-temperature heat treatment, showing the alloyed diffusion zone. (b) Schematic representations of Cu (red circles) and Ni (blue circles) atom locations within the couple. (c) Concentrations of copper and nickel as a function of position across the couple. Concentration of Ni, Cu (b) 100 Cu Ni 0 Position (c) interdiffusion impurity diffusion self-diffusion cooled to room temperature. Chemical analysis will reveal a condition similar to that represented in Figure 5.2—namely, pure copper and nickel at the two extremities of the couple, separated by an alloyed region. Concentrations of both metals vary with position as shown in Figure 5.2c. This result indicates that copper atoms have migrated or diffused into the nickel, and that nickel has diffused into copper. This process, whereby atoms of one metal diffuse into another, is termed interdiffusion, or impurity diffusion. Interdiffusion may be discerned from a macroscopic perspective by changes in concentration which occur over time, as in the example for the Cu–Ni diffusion couple. There is a net drift or transport of atoms from high- to low-concentration regions. Diffusion also occurs for pure metals, but all atoms exchanging positions are of the same type; this is termed self-diffusion. Of course, self-diffusion is not normally subject to observation by noting compositional changes. 5.2 DIFFUSION MECHANISMS From an atomic perspective, diffusion is just the stepwise migration of atoms from lattice site to lattice site. In fact, the atoms in solid materials are in constant motion, rapidly changing positions. For an atom to make such a move, two conditions must be met: (1) there must be an empty adjacent site, and (2) the atom must have sufficient energy to break bonds with its neighbor atoms and then cause some lattice distortion during the displacement. This energy is vibrational in nature (Section 4.8). At a specific temperature some small fraction of the total number of atoms is capable of diffusive motion, by virtue of the magnitudes of their vibrational energies. This fraction increases with rising temperature. Several different models for this atomic motion have been proposed; of these possibilities, two dominate for metallic diffusion. 1496T_c05_109-130 11/14/05 10:12 Page 112 REVISED PAGES 112 • Chapter 5 / Diffusion Figure 5.3 Schematic representations of (a) vacancy diffusion and (b) interstitial diffusion. Motion of a host or substitutional atom Vacancy Vacancy (a ) Position of interstitial atom before diffusion Position of interstitial atom after diffusion (b) Vacancy Diffusion vacancy diffusion One mechanism involves the interchange of an atom from a normal lattice position to an adjacent vacant lattice site or vacancy, as represented schematically in Figure 5.3a. This mechanism is aptly termed vacancy diffusion. Of course, this process necessitates the presence of vacancies, and the extent to which vacancy diffusion can occur is a function of the number of these defects that are present; significant concentrations of vacancies may exist in metals at elevated temperatures (Section 4.2). Since diffusing atoms and vacancies exchange positions, the diffusion of atoms in one direction corresponds to the motion of vacancies in the opposite direction. Both self-diffusion and interdiffusion occur by this mechanism; for the latter, the impurity atoms must substitute for host atoms. Interstitial Diffusion interstitial diffusion The second type of diffusion involves atoms that migrate from an interstitial position to a neighboring one that is empty. This mechanism is found for interdiffusion of impurities such as hydrogen, carbon, nitrogen, and oxygen, which have atoms that are small enough to fit into the interstitial positions. Host or substitutional impurity atoms rarely form interstitials and do not normally diffuse via this mechanism. This phenomenon is appropriately termed interstitial diffusion (Figure 5.3b). In most metal alloys, interstitial diffusion occurs much more rapidly than diffusion by the vacancy mode, since the interstitial atoms are smaller and thus more mobile. Furthermore, there are more empty interstitial positions than vacancies; hence, the probability of interstitial atomic movement is greater than for vacancy diffusion. 5.3 STEADY-STATE DIFFUSION Diffusion is a time-dependent process—that is, in a macroscopic sense, the quantity of an element that is transported within another is a function of time. Often it 1496T_c05_109-130 11/14/05 10:12 Page 113 REVISED PAGES 5.3 Steady-State Diffusion • 113 diffusion flux is necessary to know how fast diffusion occurs, or the rate of mass transfer. This rate is frequently expressed as a diffusion flux (J), defined as the mass (or, equivalently, the number of atoms) M diffusing through and perpendicular to a unit cross-sectional area of solid per unit of time. In mathematical form, this may be represented as Definition of diffusion flux J M At (5.1a) where A denotes the area across which diffusion is occurring and t is the elapsed diffusion time. In differential form, this expression becomes J steady-state diffusion concentration profile concentration gradient 1 dM A dt (5.1b) The units for J are kilograms or atoms per meter squared per second (kg/m2-s or atoms/m2-s). If the diffusion flux does not change with time, a steady-state condition exists. One common example of steady-state diffusion is the diffusion of atoms of a gas through a plate of metal for which the concentrations (or pressures) of the diffusing species on both surfaces of the plate are held constant. This is represented schematically in Figure 5.4a. When concentration C is plotted versus position (or distance) within the solid x, the resulting curve is termed the concentration profile; the slope at a particular point on this curve is the concentration gradient: concentration gradient  dC dx (5.2a) In the present treatment, the concentration profile is assumed to be linear, as depicted in Figure 5.4b, and CA  CB ¢C  concentration gradient  (5.2b) xA  xB ¢x PA > PB and constant Thin metal plate Gas at pressure PB Gas at pressure PA Direction of diffusion of gaseous species Concentration of diffusing species, C Figure 5.4 (a) Steady-state diffusion across a thin plate. (b) A linear concentration profile for the diffusion situation in (a). CA CB xA xB Position, x Area, A (a) (b) 1496T_c05_109-130 11/14/05 10:12 Page 114 REVISED PAGES 114 • Chapter 5 / Diffusion For diffusion problems, it is sometimes convenient to express concentration in terms of mass of diffusing species per unit volume of solid (kg/m3 or g/cm3).1 The mathematics of steady-state diffusion in a single (x) direction is relatively simple, in that the flux is proportional to the concentration gradient through the expression Fick’s first law— diffusion flux for steady-state diffusion (in one direction) diffusion coefficient Fick’s first law driving force J  D dC dx (5.3) The constant of proportionality D is called the diffusion coefficient, which is expressed in square meters per second. The negative sign in this expression indicates that the direction of diffusion is down the concentration gradient, from a high to a low concentration. Equation 5.3 is sometimes called Fick’s first law. Sometimes the term driving force is used in the context of what compels a reaction to occur. For diffusion reactions, several such forces are possible; but when diffusion is according to Equation 5.3, the concentration gradient is the driving force. One practical example of steady-state diffusion is found in the purification of hydrogen gas. One side of a thin sheet of palladium metal is exposed to the impure gas composed of hydrogen and other gaseous species such as nitrogen, oxygen, and water vapor. The hydrogen selectively diffuses through the sheet to the opposite side, which is maintained at a constant and lower hydrogen pressure. EXAMPLE PROBLEM 5.1 Diffusion Flux Computation A plate of iron is exposed to a carburizing (carbon-rich) atmosphere on one side and a decarburizing (carbon-deficient) atmosphere on the other side at 700C (1300F). If a condition of steady state is achieved, calculate the diffusion flux of carbon through the plate if the concentrations of carbon at positions of 5 and 10 mm (5  103 and 102 m) beneath the carburizing surface are 1.2 and 0.8 kg/m3, respectively. Assume a diffusion coefficient of 3  1011 m2/s at this temperature. Solution Fick’s first law, Equation 5.3, is utilized to determine the diffusion flux. Substitution of the values above into this expression yields J  D 11.2  0.82 kg/m3 CA  CB  13  1011 m2/s2 xA  xB 15  103  102 2 m  2.4  109 kg/m2-s 5.4 NONSTEADY-STATE DIFFUSION Most practical diffusion situations are nonsteady-state ones. That is, the diffusion flux and the concentration gradient at some particular point in a solid vary with time, with a net accumulation or depletion of the diffusing species resulting. This is illustrated in Figure 5.5, which shows concentration profiles at three different 1 Conversion of concentration from weight percent to mass per unit volume (in kg/m3) is possible using Equation 4.9. 1496T_c05_109-130 11/14/05 10:12 Page 115 REVISED PAGES Concentration of diffusing species 5.4 Nonsteady-State Diffusion • 115 t3 > t2 > t1 Figure 5.5 Concentration profiles for nonsteady-state diffusion taken at three different times, t1, t2, and t3. t3 t2 t1 Distance diffusion times. Under conditions of nonsteady state, use of Equation 5.3 is no longer convenient; instead, the partial differential equation 0C 0 0C  aD b 0t 0x 0x Fick’s second law Fick’s second law— diffusion equation for nonsteady-state diffusion (in one direction) (5.4a) known as Fick’s second law, is used. If the diffusion coefficient is independent of composition (which should be verified for each particular diffusion situation), Equation 5.4a simplifies to 0C 0 2C D 2 0t 0x (5.4b) Solutions to this expression (concentration in terms of both position and time) are possible when physically meaningful boundary conditions are specified. Comprehensive collections of these are given by Crank, and Carslaw and Jaeger (see References). One practically important solution is for a semi-infinite solid2 in which the surface concentration is held constant. Frequently, the source of the diffusing species is a gas phase, the partial pressure of which is maintained at a constant value. Furthermore, the following assumptions are made: 1. Before diffusion, any of the diffusing solute atoms in the solid are uniformly distributed with concentration of C0. 2. The value of x at the surface is zero and increases with distance into the solid. 3. The time is taken to be zero the instant before the diffusion process begins. These boundary conditions are simply stated as For t  0, C  C0 at 0  x  q For t 7 0, C  Cs 1the constant surface concentration2 at x  0 C  C0 at x  q Application of these boundary conditions to Equation 5.4b yields the solution Solution to Fick’s second law for the condition of constant surface concentration (for a semi-infinite solid) Cx  C0 x  1  erf a b Cs  C0 21Dt 2 (5.5) A bar of solid is considered to be semi-infinite if none of the diffusing atoms reaches the bar end during the time over which diffusion takes place. A bar of length l is considered to be semi-infinite when l 7 101Dt. 1496T_c05_109-130 11/14/05 10:12 Page 116 REVISED PAGES 116 • Chapter 5 / Diffusion Table 5.1 Tabulation of Error Function Values z erf(z) z erf(z) z erf(z) 0 0.025 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50 0 0.0282 0.0564 0.1125 0.1680 0.2227 0.2763 0.3286 0.3794 0.4284 0.4755 0.5205 0.55 0.60 0.65 0.70 0.75 0.80 0.85 0.90 0.95 1.0 1.1 1.2 0.5633 0.6039 0.6420 0.6778 0.7112 0.7421 0.7707 0.7970 0.8209 0.8427 0.8802 0.9103 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2.0 2.2 2.4 2.6 2.8 0.9340 0.9523 0.9661 0.9763 0.9838 0.9891 0.9928 0.9953 0.9981 0.9993 0.9998 0.9999 where Cx represents the concentration at depth x after time t. The expression erf(x21Dt) is the Gaussian error function,3 values of which are given in mathematical tables for various x 21Dt values; a partial listing is given in Table 5.1. The concentration parameters that appear in Equation 5.5 are noted in Figure 5.6, a concentration profile taken at a specific time. Equation 5.5 thus demonstrates the relationship between concentration, position, and time—namely, that Cx, being a function of the dimensionless parameter x 1Dt, may be determined at any time and position if the parameters C0, Cs, and D are known. Suppose that it is desired to achieve some specific concentration of solute, C1, in an alloy; the left-hand side of Equation 5.5 now becomes C1  C0  constant Cs  C0 This being the case, the right-hand side of this same expression is also a constant, and subsequently x  constant (5.6a) 21Dt Figure 5.6 Concentration profile for nonsteady-state diffusion; concentration parameters relate to Equation 5.5. Concentration, C Cs Cs – C0 Cx Cx – C0 C0 Distance from interface, x 3 This Gaussian error function is defined by erf1z2  2 z y2  e dy 1p 0 where x2 1Dt has been replaced by the variable z. 1496T_c05_109-130 11/14/05 10:12 Page 117 REVISED PAGES 5.4 Nonsteady-State Diffusion • 117 or x2  constant Dt (5.6b) Some diffusion computations are thus facilitated on the basis of this relationship, as demonstrated in Example Problem 5.3. EXAMPLE PROBLEM 5.2 Nonsteady-State Diffusion Time Computation I carburizing For some applications, it is necessary to harden the surface of a steel (or ironcarbon alloy) above that of its interior. One way this may be accomplished is by increasing the surface concentration of carbon in a process termed carburizing; the steel piece is exposed, at an elevated temperature, to an atmosphere rich in a hydrocarbon gas, such as methane 1CH4 2. Consider one such alloy that initially has a uniform carbon concentration of 0.25 wt% and is to be treated at 950 C (1750 F). If the concentration of carbon at the surface is suddenly brought to and maintained at 1.20 wt%, how long will it take to achieve a carbon content of 0.80 wt% at a position 0.5 mm below the surface? The diffusion coefficient for carbon in iron at this temperature is 1.6  1011 m2/s; assume that the steel piece is semi-infinite. Solution Since this is a nonsteady-state diffusion problem in which the surface composition is held constant, Equation 5.5 is used. Values for all the parameters in this expression except time t are specified in the problem as follows: C0  0.25 wt% C Cs  1.20 wt% C Cx  0.80 wt% C x  0.50 mm  5  104 m D  1.6  1011 m2/s Thus, 15  104 m2 Cx  C0 0.80  0.25   1  erf c d Cs  C0 1.20  0.25 2211.6  1011 m2/s21t2 62.5 s1 2 b 0.4210  erf a 1t We must now determine from Table 5.1 the value of z for which the error function is 0.4210. An interpolation is necessary, as z erf(z) 0.35 z 0.40 0.3794 0.4210 0.4284 z  0.35 0.4210  0.3794  0.40  0.35 0.4284  0.3794 1496T_c05_109-130 11/14/05 10:12 Page 118 REVISED PAGES 118 • Chapter 5 / Diffusion or z  0.392 Therefore, 62.5 s1 2  0.392 1t and solving for t, ta 62.5 s1 2 2 b  25,400 s  7.1 h 0.392 EXAMPLE PROBLEM 5.3 Nonsteady-State Diffusion Time Computation II The diffusion coefficients for copper in aluminum at 500 and 600 C are 4.8  1014 and 5.3  1013 m2/s, respectively. Determine the approximate time at 500 C that will produce the same diffusion result (in terms of concentration of Cu at some specific point in Al) as a 10-h heat treatment at 600°C. Solution This is a diffusion problem in which Equation 5.6b may be employed. The composition in both diffusion situations will be equal at the same position (i.e., x is also a constant), thus Dt  constant (5.7) at both temperatures. That is, D500 t 500  D600 t 600 or t 500  15.3  1013 m2/s2110 h2 D600 t 600   110.4 h D500 4.8  1014 m2/s 5.5 FACTORS THAT INFLUENCE DIFFUSION Diffusing Species The magnitude of the diffusion coefficient D is indicative of the rate at which atoms diffuse. Coefficients, both self- and interdiffusion, for several metallic systems are listed in Table 5.2. The diffusing species as well as the host material influence the diffusion coefficient. For example, there is a significant difference in magnitude between self-diffusion and carbon interdiffusion in a iron at 500 C, the D value being greater for the carbon interdiffusion (3.0  1021 vs. 2.4  1012 m2/s). This comparison also provides a contrast between rates of diffusion via vacancy and interstitial modes as discussed above. Self-diffusion occurs by a vacancy mechanism, whereas carbon diffusion in iron is interstitial. Temperature Temperature has a most profound influence on the coefficients and diffusion rates. For example, for the self-diffusion of Fe in a-Fe, the diffusion coefficient increases approximately six orders of magnitude (from 3.0  1021 to 1.8  1015 m2/s2 in rising temperature from 500 to 900 C (Table 5.2). The temperature dependence of 1496T_c05_109-130 12/20/05 7:52 Page 119 2nd REVISE PAGES 5.5 Factors That Influence Diffusion • 119 Table 5.2 A Tabulation of Diffusion Data Diffusing Species Host Metal Activation Energy Qd 2 Calculated Values kJ/mol eV/atom T(C ) D(m2/s) 4 D0(m /s) Fe -Fe (BCC) 2.8  10 251 2.60 500 900 3.0  1021 1.8  1015 Fe -Fe (FCC) 5.0  105 284 2.94 900 1100 1.1  1017 7.8  1016 C -Fe 6.2  107 80 0.83 500 900 2.4  1012 1.7  1010 C -Fe 2.3  105 148 1.53 900 1100 5.9  1012 5.3  1011 Cu Cu 7.8  105 211 2.19 500 4.2  1019 Zn Cu 2.4  105 189 1.96 500 4.0  1018 Al Al 4 2.3  10 144 1.49 500 4.2  1014 Cu Al 6.5  105 136 1.41 500 4.1  1014 Mg Al 4 1.2  10 131 1.35 500 1.9  1013 Cu Ni 2.7  105 256 2.65 500 1.3  1022 Source: E. A. Brandes and G. B. Brook (Editors), Smithells Metals Reference Book, 7th edition, ButterworthHeinemann, Oxford, 1992. the diffusion coefficients is Dependence of the diffusion coefficient on temperature D  D0 exp a Qd b RT (5.8) where activation energy D0  a temperature-independent preexponential 1m2/s2 Qd  the activation energy for diffusion (J/mol or eV/atom) R  the gas constant, 8.31 J/mol-K or 8.62  105 eV/atom-K T  absolute temperature (K) The activation energy may be thought of as that energy required to produce the diffusive motion of one mole of atoms. A large activation energy results in a relatively small diffusion coefficient. Table 5.2 also contains a listing of D0 and Qd values for several diffusion systems. Taking natural logarithms of Equation 5.8 yields ln D  ln D0  Qd 1 a b R T (5.9a) Qd 1 a b 2.3 R T (5.9b) or in terms of logarithms to the base 10 log D  log D0  Since D0, Qd, and R are all constants, Equation 5.9b takes on the form of an equation of a straight line: y  b  mx where y and x are analogous, respectively, to the variables log D and 1 T. Thus, if log D is plotted versus the reciprocal of the absolute temperature, a straight line 1496T_c05_109-130 11/14/05 10:12 Page 120 REVISED PAGES 120 • Chapter 5 / Diffusion Figure 5.7 Plot of the logarithm of the diffusion coefficient versus the reciprocal of absolute temperature for several metals. [Data taken from E. A. Brandes and G. B. Brook (Editors), Smithells Metals Reference Book, 7th edition, ButterworthHeinemann, Oxford, 1992.] Temperature (°C) 1500 1200 1000 800 600 500 400 300 10–8 10–10 C in ␣ –Fe Diffusion coefficient (m2/s) C in ␥ –Fe 10–12 Zn in Cu 10–14 Fe in ␥ –Fe 10–16 Al in Al Fe in ␣ –Fe Cu in Cu 10–18 10–20 0.5 1.0 1.5 2.0 Reciprocal temperature (1000/K) should result, having slope and intercept of Qd 2.3R and log D0, respectively. This is, in fact, the manner in which the values of Qd and D0 are determined experimentally. From such a plot for several alloy systems (Figure 5.7), it may be noted that linear relationships exist for all cases shown. Concept Check 5.1 Rank the magnitudes of the diffusion coefficients from greatest to least for the following systems: N in Fe at 700°C Cr in Fe at 700°C N in Fe at 900°C Cr in Fe at 900°C Now justify this ranking. (Note: Both Fe and Cr have the BCC crystal structure, and the atomic radii for Fe, Cr, and N are 0.124, 0.125, and 0.065 nm, respectively. You may also want to refer to Section 4.3.) [The answer may be found at www.wiley.com/college/callister (Student Companion Site).] Concept Check 5.2 Consider the self-diffusion of two hypothetical metals A and B. On a schematic graph of ln D versus 1 T , plot (and label) lines for both metals given that D0 1A2 7 D0 1B2 and also that Qd 1A2 7 Qd 1B2. [The answer may be found at www.wiley.com/college/callister (Student Companion Site).] 1496T_c05_109-130 11/14/05 13:07 Page 121 REVISED PAGES 5.5 Factors That Influence Diffusion • 121 EXAMPLE PROBLEM 5.4 Diffusion Coefficient Determination Using the data in Table 5.2, compute the diffusion coefficient for magnesium in aluminum at 550 C. Solution This diffusion coefficient may be determined by applying Equation 5.8; the values of D0 and Qd from Table 5.2 are 1.2  104 m2/s and 131 kJ/mol, respectively. Thus, 1131,000 J/mol2 D  11.2  104 m2/s2 exp c  d 18.31 J/mol-K21550  273 K2  5.8  1013 m2/s EXAMPLE PROBLEM 5.5 Diffusion Coefficient Activation Energy and Preexponential Calculations In Figure 5.8 is shown a plot of the logarithm (to the base 10) of the diffusion coefficient versus reciprocal of absolute temperature, for the diffusion of copper in gold. Determine values for the activation energy and the preexponential. Solution From Equation 5.9b the slope of the line segment in Figure 5.8 is equal to Qd  2.3R, and the intercept at 1T  0 gives the value of log D0. Thus, the activation energy may be determined as D0 and Qd from Experimental Data Qd  2.3R 1slope2  2.3R  2.3R £ ¢ 1log D2 £ § 1 ¢a b T log D1  log D2 § 1 1  T1 T2 where D1 and D2 are the diffusion coefficient values at 1T1 and 1T2, respectively. Let us arbitrarily take 1T1  0.8  103 1K2 1 and 1T2  1.1  103 1K2 1. We may now read the corresponding log D1 and log D2 values from the line segment in Figure 5.8. [Before this is done, however, a parenthetic note of caution is offered. The vertical axis in Figure 5.8 is scaled logarithmically (to the base 10); however, the actual diffusion coefficient values are noted on this axis. For example, for D  1014 m2/s, the logarithm of D is 14.0 not 1014. Furthermore, this logarithmic scaling affects the readings between decade values; for example, at a location midway between 1014 and 1015, the value is not 5  1015 but, rather, 1014.5  3.2  1015]. Thus, from Figure 5.8, at 1T1  0.8  103 1K2 1, log D1  12.40, while for 1T2  1.1  103 1K2 1, log D2  15.45, and the activation energy, as 1496T_c05_109-130 11/14/05 10:12 Page 122 REVISED PAGES 122 • Chapter 5 / Diffusion Figure 5.8 Plot of the logarithm of the diffusion coefficient versus the reciprocal of absolute temperature for the diffusion of copper in gold. Diffusion coefficient (m2/s) 10–12 10–13 10–14 10–15 10–16 10–17 0.7 0.8 0.9 1.0 1.1 1.2 Reciprocal temperature (1000/K) determined from the slope of the line segment in Figure 5.8 is Qd  2.3R £ log D1  log D2 § 1 1  T1 T2  2.3 18.31 J/mol-K2 c 12.40  115.452 0.8  103 1K2 1  1.1  103 1K2 1  194,000 J/mol  194 kJ/mol d Now, rather than trying to make a graphical extrapolation to determine D0, a more accurate value is obtained analytically using Equation 5.9b, and a specific value of D (or log D) and its corresponding T (or 1T ) from Figure 5.8. Since we know that log D  15.45 at 1 T  1.1  103 (K)1, then Qd 1 log D0  log D  a b 2.3R T 1194,000 J/mol211.1  103 3K 4 1 2  15.45  12.3218.31 J/mol-K2  4.28 Thus, D0  104.28 m2/s  5.2  105 m2/s. DESIGN EXAMPLE 5.1 Diffusion Temperature–Time Heat Treatment Specification The wear resistance of a steel gear is to be improved by hardening its surface. This is to be accomplished by increasing the carbon content within an outer surface layer as a result of carbon diffusion into the steel; the carbon is to be supplied from an external carbon-rich gaseous atmosphere at an elevated and constant temperature. The initial carbon content of the steel is 0.20 wt%, whereas the surface concentration is to be maintained at 1.00 wt%. For this treatment to be effective, a carbon content of 0.60 wt% must be established at a position 0.75 mm below the surface. Specify an appropriate heat treatment in terms of temperature and time for temperatures between 900 C and 1050 C. Use data in Table 5.2 for the diffusion of carbon in g-iron. 1496T_c05_109-130 12/20/05 7:52 Page 123 2nd REVISE PAGES 5.5 Factors That Influence Diffusion • 123 Solution Since this is a nonsteady-state diffusion situation, let us first of all employ Equation 5.5, utilizing the following values for the concentration parameters: C0  0.20 wt% C Cs  1.00 wt% C Cx  0.60 wt% C Therefore Cx  C0 0.60  0.20 x   1  erf a b Cs  C0 1.00  0.20 21Dt and thus 0.5  erf a x b 21Dt Using an interpolation technique as demonstrated in Example Problem 5.2 and the data presented in Table 5.1, x  0.4747 21Dt (5.10) The problem stipulates that x  0.75 mm  7.5  104 m. Therefore 7.5  104 m  0.4747 21Dt This leads to Dt  6.24  107 m2 Furthermore, the diffusion coefficient depends on temperature according to Equation 5.8; and, from Table 5.2 for the diffusion of carbon in g-iron, D0  2.3  105 m2/s and Qd  148,000 J/mol. Hence Dt  D0 exp a 12.3  105 m2/s2 exp c  Qd b 1t2  6.24  107 m2 RT 148,000 J/mol d 1t2  6.24  107 m2 18.31 J/mol-K21T 2 and solving for the time t t 1in s2  0.0271 17,810 exp a b T Thus, the required diffusion time may be computed for some specified temperature (in K). Below are tabulated t values for four different temperatures that lie within the range stipulated in the problem. Time Temperature (C ) s h 900 950 1000 1050 106,400 57,200 32,300 19,000 29.6 15.9 9.0 5.3 1496T_c05_109-130 11/14/05 10:13 Page 124 REVISED PAGES 124 • Chapter 5 / Diffusion MATERIAL OF IMPORTANCE Aluminum for Integrated Circuit Interconnects T he heart of all computers and other electronic devices is the integrated circuit (or IC).4 Each integrated circuit chip is a thin square wafer having dimensions on the order of 6 mm by 6 mm by 0.4 mm; furthermore, literally millions of interconnected electronic components and circuits are embedded in one of the chip faces. The base material for ICs is silicon, to which has been added very specific and extremely minute and controlled concentrations of impurities that are confined to very small and localized regions. For some ICs, the impurities are added using high-temperature diffusion heat treatments. One important step in the IC fabrication process is the deposition of very thin and narrow conducting circuit paths to facilitate the passage of current from one device to another; these paths are called “interconnects,” and several are shown in Figure 5.9, a scanning electron micrograph of an IC chip. Of course the material to be used for interconnects must have a high electrical conductivity—a metal, since, of all materials, metals have the highest conductivities. Table 5.3 cites values for silver, copper, gold, and aluminum, the most conductive metals. Table 5.3 Room-Temperature Electrical Conductivity Values for Silver, Copper, Gold, and Aluminum (the Four Most Conductive Metals) Electrical Conductivity [(ohm-meters)1] Metal 6.8  107 6.0  107 4.3  107 3.8  107 Silver Copper Gold Aluminum On the basis of these conductivities, and discounting material cost, Ag is the metal of choice, followed by Cu, Au, and Al. Once these interconnects have been deposited, it is still necessary to subject the IC chip to other heat treatments, which may run as high as 500 C. If, during these treatments, there is significant diffusion of the interconnect metal into the silicon, the electrical functionality of the IC will be destroyed. Thus, since the extent of diffusion is dependent on the magnitude of the diffusion coefficient, it is necessary to select an interconnect metal that has a small value of D in silicon. Figure 5.10 Interconnects Temperature (°C) 1200 1000 900 800 700 600 500 400 10–12 Cu in Si Diffusion coefficient (m2/s) Au in Si 1014 4 × 1013 2.5 × 1015 Ag in Si 4.2 × 1017 1016 Al in Si 1018 2.5 × 1021 1020 1022 0.6 Figure 5.9 Scanning electron micrograph of an integrated circuit chip, on which is noted aluminum interconnect regions. Approximately 2000. (Photograph courtesy of National Semiconductor Corporation.) 4 0.8 1.0 1.2 Reciprocal temperature (1000/K) 1.4 Figure 5.10 Logarithm of D-versus-1T (K) curves (lines) for the diffusion of copper, gold, silver, and aluminum in silicon. Also noted are D values at 500 C. Integrated circuits, their components and materials, are discussed in Section 18.15 and Sections 22.15 through 22.20. 1496T_c05_109-130 11/14/05 10:13 Page 125 REVISED PAGES Summary • 125 plots the logarithm of D versus 1 T for the diffusion, into silicon, of copper, gold, silver, and aluminum. Also, a dashed vertical line has been constructed at 500 C, from which values of D, for the four metals are noted at this temperature. Here it may be seen that the diffusion coefficient for aluminum in silicon (2.5  1021 m2/s) is at least four orders of magnitude (i.e., a factor of 104) lower than the values for the other three metals. Aluminum is indeed used for interconnects in some integrated circuits; even though its electrical conductivity is slightly lower than the values for silver, copper, and gold, its extremely low diffusion coefficient makes it the material of choice for this application. An aluminum-copper-silicon alloy (Al-4 wt% Cu-1.5 wt% Si) is sometimes also used for interconnects; it not only bonds easily to the surface of the chip, but is also more corrosion resistant than pure aluminum. More recently, copper interconnects have also been used. However, it is first necessary to deposit a very thin layer of tantalum or tantalum nitride beneath the copper, which acts as a barrier to deter diffusion of Cu into the silicon. 5.6 OTHER DIFFUSION PATHS Atomic migration may also occur along dislocations, grain boundaries, and external surfaces. These are sometimes called “short-circuit” diffusion paths inasmuch as rates are much faster than for bulk diffusion. However, in most situations shortcircuit contributions to the overall diffusion flux are insignificant because the crosssectional areas of these paths are extremely small. SUMMARY Diffusion Mechanisms Solid-state diffusion is a means of mass transport within solid materials by stepwise atomic motion. The term “self-diffusion” refers to the migration of host atoms; for impurity atoms, the term “interdiffusion” is used. Two mechanisms are possible: vacancy and interstitial. For a given host metal, interstitial atomic species generally diffuse more rapidly. Steady-State Diffusion Nonsteady-State Diffusion For steady-state diffusion, the concentration profile of the diffusing species is time independent, and the flux or rate is proportional to the negative of the concentration gradient according to Fick’s first law. The mathematics for nonsteady state are described by Fick’s second law, a partial differential equation. The solution for a constant surface composition boundary condition involves the Gaussian error function. Factors That Influence Diffusion The magnitude of the diffusion coefficient is indicative of the rate of atomic motion, being strongly dependent on and increasing exponentially with increasing temperature. 1496T_c05_109-130 11/14/05 10:13 Page 126 REVISED PAGES 126 • Chapter 5 / Diffusion I M P O R TA N T T E R M S A N D C O N C E P T S Activation energy Carburizing Concentration gradient Concentration profile Diffusion Diffusion coefficient Diffusion flux Driving force Fick’s first and second laws Interdiffusion (impurity diffusion) Interstitial diffusion Nonsteady-state diffusion Self-diffusion Steady-state diffusion Vacancy diffusion REFERENCES Gale, W. F. and T. C. Totemeier, (Editors), Smithells Metals Reference Book, 8th edition,ButterworthHeinemann Ltd, Woburn, UK, 2004. Carslaw, H. S. and J. C. Jaeger, Conduction of Heat in Solids, 2nd edition, Oxford University Press, Oxford, 1986. Crank, J., The Mathematics of Diffusion, 2nd edition, Oxford University Press, Oxford, 1980. Glicksman, M., Diffusion in Solids, WileyInterscience, New York, 2000. Shewmon, P. G., Diffusion in Solids, 2nd edition, The Minerals, Metals and Materials Society, Warrendale, PA, 1989. QUESTIONS AND PROBLEMS Introduction 5.1 Briefly explain the difference between selfdiffusion and interdiffusion. 5.2 Self-diffusion involves the motion of atoms that are all of the same type; therefore it is not subject to observation by compositional changes, as with interdiffusion. Suggest one way in which self-diffusion may be monitored. Diffusion Mechanisms 5.3 (a) Compare interstitial and vacancy atomic mechanisms for diffusion. (b) Cite two reasons why interstitial diffusion is normally more rapid than vacancy diffusion. Steady-State Diffusion 5.4 Briefly explain the concept of steady state as it applies to diffusion. 5.5 (a) Briefly explain the concept of a driving force. (b) What is the driving force for steady-state diffusion? 5.6 The purification of hydrogen gas by diffusion through a palladium sheet was discussed in Section 5.3. Compute the number of kilograms of hydrogen that pass per hour through a 6-mm-thick sheet of palladium having an area of 0.25 m2 at 600C. Assume a diffusion coefficient of 1.7  108 m2/s, that the concentrations at the high- and low-pressure sides of the plate are 2.0 and 0.4 kg of hydrogen per cubic meter of palladium, and that steady-state conditions have been attained. 5.7 A sheet of steel 2.5 mm thick has nitrogen atmospheres on both sides at 900C and is permitted to achieve a steady-state diffusion condition. The diffusion coefficient for nitrogen in steel at this temperature is 1.2  1010 m2/s, and the diffusion flux is found to be 1.0  107 kg/m2-s. Also, it is known that the concentration of nitrogen in the steel at the high-pressure surface is 2 kg/m3. How far into the sheet from this highpressure side will the concentration be 0.5 kg/m3? Assume a linear concentration profile. 5.8 A sheet of BCC iron 2 mm thick was exposed to a carburizing gas atmosphere on one side and a decarburizing atmosphere on the other side at 675C. After having reached steady state, the iron was quickly cooled to room temperature.The carbon concentrations at the two surfaces of the sheet were determined to be 0.015 and 0.0068 wt%. Compute the diffusion 1496T_c05_109-130 12/20/05 7:52 Page 127 2nd REVISE PAGES Questions and Problems • 127 coefficient if the diffusion flux is 7.36  109 kg/m2-s. Hint: Use Equation 4.9 to convert the concentrations from weight percent to kilograms of carbon per cubic meter of iron. 5.9 When a-iron is subjected to an atmosphere of nitrogen gas, the concentration of nitrogen in the iron, CN (in weight percent), is a function of hydrogen pressure, pN2(in MPa), and absolute temperature (T) according to CN  4.90  103 1pN2 exp a 37.6 kJ/mol b RT (5.11) Furthermore, the values of D0 and Qd for this diffusion system are 3.0  107 m2/s and 76,150 J/mol, respectively. Consider a thin iron membrane 1.5 mm thick that is at 300C. Compute the diffusion flux through this membrane if the nitrogen pressure on one side of the membrane is 0.10 MPa (0.99 atm), and on the other side 5.0 MPa (49.3 atm). Nonsteady-State Diffusion 5.10 Show that Cx  B x2 exp a b 4Dt 1Dt is also a solution to Equation 5.4b. The parameter B is a constant, being independent of both x and t. 5.11 Determine the carburizing time necessary to achieve a carbon concentration of 0.30 wt% at a position 4 mm into an iron–carbon alloy that initially contains 0.10 wt% C. The surface concentration is to be maintained at 0.90 wt% C, and the treatment is to be conducted at 1100C. Use the diffusion data for g-Fe in Table 5.2. 5.12 An FCC iron–carbon alloy initially containing 0.55 wt% C is exposed to an oxygen-rich and virtually carbon-free atmosphere at 1325 K (1052C). Under these circumstances the carbon diffuses from the alloy and reacts at the surface with the oxygen in the atmosphere; that is, the carbon concentration at the surface position is maintained essentially at 0 wt% C. (This process of carbon depletion is termed decarburization.) At what position will the carbon concentration be 0.25 wt% after a 10-h treatment? The value of D at 1325 K is 4.3  1011 m2/s. 5.13 Nitrogen from a gaseous phase is to be diffused into pure iron at 675C. If the surface concentration is maintained at 0.2 wt% N, what will be the concentration 2 mm from the surface after 25 h? The diffusion coefficient for nitrogen in iron at 675C is 1.9  1011 m2/s. 5.14 Consider a diffusion couple composed of two semi-infinite solids of the same metal, and that each side of the diffusion couple has a different concentration of the same elemental impurity; furthermore, assume each impurity level is constant throughout its side of the diffusion couple. For this situation, the solution to Fick’s second law (assuming that the diffusion coefficient for the impurity is independent of concentration), is as follows: Cx  a C1  C2 C1  C2 x ba b erf a b 2 2 21Dt (5.12) In this expression, when the x  0 position is taken as the initial diffusion couple interface, then C1 is the impurity concentration for x 6 0; likewise, C2 is the impurity content for x 7 0. A diffusion couple composed of two platinum-gold alloys is formed; these alloys have compositions of 99.0 wt% Pt-1.0 wt% Au and 96.0 wt% Pt-4.0 wt% Au. Determine the time this diffusion couple must be heated at 1000C (1273 K) in order for the composition to be 2.8 wt% Au at the 10 mm position into the 4.0 wt% Au side of the diffusion couple. Preexponential and activation energy values for Au diffusion in Pt are 1.3  105 m2/s and 252,000 J/mol, respectively. 5.15 For a steel alloy it has been determined that a carburizing heat treatment of 15 h duration will raise the carbon concentration to 0.35 wt% at a point 2.0 mm from the surface. Estimate the time necessary to achieve the same concentration at a 6.0-mm position for an identical steel and at the same carburizing temperature. Factors That Influence Diffusion 5.16 Cite the values of the diffusion coefficients for the interdiffusion of carbon in both a-iron (BCC) and g-iron (FCC) at 900C. Which is larger? Explain why this is the case. 5.17 Using the data in Table 5.2, compute the value of D for the diffusion of magnesium in aluminum at 400C. 1496T_c05_109-130 12/20/05 7:52 Page 128 2nd REVISE PAGES 128 • Chapter 5 / Diffusion 5.18 At what temperature will the diffusion coefficient for the diffusion of zinc in copper have a value of 2.6  1016 m2/s? Use the diffusion data in Table 5.2. 5.19 The preexponential and activation energy for the diffusion of chromium in nickel are 1.1  104 m2/s and 272,000 J/mol, respectively. At what temperature will the diffusion coefficient have a value of 1.2  1014 m2/s? 5.20 The activation energy for the diffusion of copper in silver is 193,000 J/mol. Calculate the diffusion coefficient at 1200 K (927C), given that D at 1000 K (727C) is 1.0  1014 m2/s. 5.21 The diffusion coefficients for nickel in iron are given at two temperatures: T(K) D(m2/s) 1473 1673 2.2  1015 4.8  1014 (a) Determine the values of D0 and the activation energy Qd. (b) What is the magnitude of D at 1300C (1573 K)? 5.22 The diffusion coefficients for carbon in nickel are given at two temperatures: T(C) D(m2/s) 600 700 5.5  1014 3.9  1013 Diffusion coefficient (m2/s) (a) Determine the values of D0 and Qd. (b) What is the magnitude of D at 850C? 5.23 Below is shown a plot of the logarithm (to the base 10) of the diffusion coefficient 10 –13 10 –14 10 –15 0.8 0.9 1.0 Reciprocal temperature (1000/K) versus reciprocal of the absolute temperature, for the diffusion of gold in silver. Determine values for the activation energy and preexponential. 5.24 Carbon is allowed to diffuse through a steel plate 10 mm thick. The concentrations of carbon at the two faces are 0.85 and 0.40 kg C/cm3 Fe, which are maintained constant. If the preexponential and activation energy are 6.2  107 m2/s and 80,000 J/mol, respectively, compute the temperature at which the diffusion flux is 6.3  1010 kg/m2-s. 5.25 The steady-state diffusion flux through a metal plate is 7.8  108 kg/m2-s at a temperature of 1200C (1473 K) and when the concentration gradient is 500 kg/m4. Calculate the diffusion flux at 1000C (1273 K) for the same concentration gradient and assuming an activation energy for diffusion of 145,000 J/mol. 5.26 At approximately what temperature would a specimen of g-iron have to be carburized for 4 h to produce the same diffusion result as at 1000C for 12 h? 5.27 (a) Calculate the diffusion coefficient for magnesium in aluminum at 450C. (b) What time will be required at 550C to produce the same diffusion result (in terms of concentration at a specific point) as for 15 h at 450C? 5.28 A copper–nickel diffusion couple similar to that shown in Figure 5.1a is fashioned.After a 500-h heat treatment at 1000C (1273 K) the concentration of Ni is 3.0 wt% at the 1.0-mm position within the copper. At what temperature should the diffusion couple be heated to produce this same concentration (i.e., 3.0 wt% Ni) at a 2.0-mm position after 500 h? The preexponential and activation energy for the diffusion of Ni in Cu are 2.7  104 m2/s and 236,000 J/mol, respectively. 5.29 A diffusion couple similar to that shown in Figure 5.1a is prepared using two hypothetical metals A and B. After a 20-h heat treatment at 800C (and subsequently cooling to room temperature) the concentration of B in A is 2.5 wt% at the 5.0-mm position within metal A. If another heat treatment is conducted on an identical diffusion couple, only 1496T_c05_109-130 11/14/05 10:13 Page 129 REVISED PAGES Design Problems • 129 at 1000C for 20 h, at what position will the composition be 2.5 wt% B? Assume that the preexponential and activation energy for the diffusion coefficient are 1.5  104 m2/s and 125,000 J/mol, respectively. 5.30 The outer surface of a steel gear is to be hardened by increasing its carbon content; the carbon is to be supplied from an external carbon-rich atmosphere that is maintained at an elevated temperature. A diffusion heat treatment at 600C (873 K) for 100 min increases the carbon concentration to 0.75 wt% at a position 0.5 mm below the surface. Estimate the diffusion time required at 900C (1173 K) to achieve this same concentration also at a 0.5-mm position. Assume that the surface carbon content is the same for both heat treatments, which is maintained constant. Use the diffusion data in Table 5.2 for C diffusion in a-Fe. 5.31 An FCC iron–carbon alloy initially containing 0.10 wt% C is carburized at an elevated temperature and in an atmosphere wherein the surface carbon concentration is maintained at 1.10 wt%. If after 48 h the concentration of carbon is 0.30 wt% at a position 3.5 mm below the surface, determine the temperature at which the treatment was carried out. DESIGN PROBLEMS Steady-State Diffusion (Factors That Influence Diffusion) 5.D1 It is desired to enrich the partial pressure of hydrogen in a hydrogen–nitrogen gas mixture for which the partial pressures of both gases are 0.1013 MPa (1 atm). It has been proposed to accomplish this by passing both gases through a thin sheet of some metal at an elevated temperature; inasmuch as hydrogen diffuses through the plate at a higher rate than does nitrogen, the partial pressure of hydrogen will be higher on the exit side of the sheet. The design calls for partial pressures of 0.051 MPa (0.5 atm) and 0.01013 MPa (0.1 atm), respectively, for hydrogen and nitrogen. The concentrations of hydrogen and nitrogen (CH and CN, in mol/m3) in this metal are functions of gas partial pressures ( pH2 and pN2, in MPa) and absolute temperature and are given by the following expressions: CH  2.5  103 1pH2 exp a 27.8 kJ/mol b RT (5.13a) 37.6 kJ/mol CN  2.75  103 1pN2 exp a b RT (5.13b) Furthermore, the diffusion coefficients for the diffusion of these gases in this metal are functions of the absolute temperature as follows: 13.4 kJ/mol b DH 1m2/s2  1.4  107 exp a RT (5.14a) 76.15 kJ/mol DN 1m2/s2  3.0  107 exp a b RT (5.14b) Is it possible to purify hydrogen gas in this manner? If so, specify a temperature at which the process may be carried out, and also the thickness of metal sheet that would be required. If this procedure is not possible, then state the reason(s) why. 5.D2 A gas mixture is found to contain two diatomic A and B species (A2 and B2) for which the partial pressures of both are 0.1013 MPa (1 atm). This mixture is to be enriched in the partial pressure of the A species by passing both gases through a thin sheet of some metal at an elevated temperature. The resulting enriched mixture is to have a partial pressure of 0.051 MPa (0.5 atm) for gas A and 0.0203 MPa (0.2 atm) for gas B. The concentrations of A and B (CA and CB, in 1496T_c05_109-130 12/20/05 7:52 Page 130 2nd REVISE PAGES 130 • Chapter 5 / Diffusion mol/m3) are functions of gas partial pressures ( pA2 and pB2, in MPa) and absolute temperature according to the following expressions: CA  1.5  103 1pA2 exp a 20.0 kJ/mol b RT (5.15a) 27.0 kJ/mol CB  2.0  103 1pB2 exp a b RT (5.15b) Furthermore, the diffusion coefficients for the diffusion of these gases in the metal are functions of the absolute temperature as follows: DA 1m2/s2  5.0  107 exp a 13.0 kJ/mol b RT (5.16a) 21.0 kJ/mol DB 1m2/s2  3.0  106 exp a b RT (5.16b) Is it possible to purify the A gas in this manner? If so, specify a temperature at which the process may be carried out, and also the thickness of metal sheet that would be required. If this procedure is not possible, then state the reason(s) why. Nonsteady-State Diffusion (Factors That Influence Diffusion) 5.D3 The wear resistance of a steel shaft is to be improved by hardening its surface. This is to be accomplished by increasing the nitrogen content within an outer surface layer as a result of nitrogen diffusion into the steel; the nitrogen is to be supplied from an external nitrogen-rich gas at an elevated and constant temperature. The initial nitrogen content of the steel is 0.0025 wt%, whereas the surface concentration is to be maintained at 0.45 wt%. For this treatment to be effective, a nitrogen content of 0.12 wt% must be established at a position 0.45 mm below the surface. Specify an appropriate heat treatment in terms of temperature and time for a temperature between 475C and 625C. The preexponential and activation energy for the diffusion of nitrogen in iron are 3  107 m2/s and 76,150 J/mol, respectively, over this temperature range. 5.D4 The wear resistance of a steel gear is to be improved by hardening its surface, as described in Design Example 5.1. However, in this case the initial carbon content of the steel is 0.15 wt%, and a carbon content of 0.75 wt% is to be established at a position 0.65 mm below the surface. Furthermore, the surface concentration is to be maintained constant, but may be varied between 1.2 and 1.4 wt% C. Specify an appropriate heat treatment in terms of surface carbon concentration and time, and for a temperature between 1000C and 1200C. 1496T_c06_131-173 12/21/05 8:26 Page 131 Chapter A 2nd REVISE PAGES 6 Mechanical Properties of Metals modern Rockwell hardness tester. (Photograph courtesy of Wilson Instruments Division, Instron Corporation, originator of the Rockwell® Hardness Tester.) WHY STUDY The Mechanical Properties of Metals? It is incumbent on engineers to understand how the various mechanical properties are measured and what these properties represent; they may be called upon to design structures/components using predetermined materials such that unacceptable levels of deformation and/or failure will not occur. We demonstrate this procedure with respect to the design of a tensiletesting apparatus in Design Example 6.1. • 131 1496T_c06_131-173 12/21/05 7:43 Page 132 2nd REVISE PAGES Learning Objectives After studying this chapter you should be able to do the following: 1. Define engineering stress and engineering strain. 7. Give brief definitions of and the units for 2. State Hooke’s law, and note the conditions modulus of resilience and toughness (static). under which it is valid. 8. For a specimen being loaded in tension, given 3. Define Poisson’s ratio. the applied load, the instantaneous cross4. Given an engineering stress–strain diagram, sectional dimensions, as well as original and determine (a) the modulus of elasticity, instantaneous lengths, be able to compute (b) the yield strength (0.002 strain offset), true stress and true strain values. and (c) the tensile strength, and (d) estimate 9. Name the two most common hardness-testing the percent elongation. techniques; note two differences between 5. For the tensile deformation of a ductile cylinthem. drical specimen, describe changes in specimen 10. (a) Name and briefly describe the two differprofile to the point of fracture. ent microindentation hardness testing tech6. Compute ductility in terms of both percent niques, and (b) cite situations for which these elongation and percent reduction of area techniques are generally used. for a material that is loaded in tension to 11. Compute the working stress for a ductile fracture. material. 6.1 INTRODUCTION Many materials, when in service, are subjected to forces or loads; examples include the aluminum alloy from which an airplane wing is constructed and the steel in an automobile axle. In such situations it is necessary to know the characteristics of the material and to design the member from which it is made such that any resulting deformation will not be excessive and fracture will not occur. The mechanical behavior of a material reflects the relationship between its response or deformation to an applied load or force. Important mechanical properties are strength, hardness, ductility, and stiffness. The mechanical properties of materials are ascertained by performing carefully designed laboratory experiments that replicate as nearly as possible the service conditions. Factors to be considered include the nature of the applied load and its duration, as well as the environmental conditions. It is possible for the load to be tensile, compressive, or shear, and its magnitude may be constant with time, or it may fluctuate continuously. Application time may be only a fraction of a second, or it may extend over a period of many years. Service temperature may be an important factor. Mechanical properties are of concern to a variety of parties (e.g., producers and consumers of materials, research organizations, government agencies) that have differing interests. Consequently, it is imperative that there be some consistency in the manner in which tests are conducted, and in the interpretation of their results. This consistency is accomplished by using standardized testing techniques. Establishment and publication of these standards are often coordinated by professional societies. In the United States the most active organization is the American Society for Testing and Materials (ASTM). Its Annual Book of ASTM Standards (http://www.astm.org) comprises numerous volumes, which are issued and updated yearly; a large number of these standards relate to mechanical testing techniques. Several of these are referenced by footnote in this and subsequent chapters. The role of structural engineers is to determine stresses and stress distributions within members that are subjected to well-defined loads. This may be accomplished 1496T_c06_131-173 11/16/05 17:06 Page 133 REVISED PAGES 6.2 Concepts of Stress and Strain • 133 by experimental testing techniques and/or by theoretical and mathematical stress analyses. These topics are treated in traditional stress analysis and strength of materials texts. Materials and metallurgical engineers, on the other hand, are concerned with producing and fabricating materials to meet service requirements as predicted by these stress analyses. This necessarily involves an understanding of the relationships between the microstructure (i.e., internal features) of materials and their mechanical properties. Materials are frequently chosen for structural applications because they have desirable combinations of mechanical characteristics. The present discussion is confined primarily to the mechanical behavior of metals; polymers and ceramics are treated separately because they are, to a large degree, mechanically dissimilar to metals. This chapter discusses the stress–strain behavior of metals and the related mechanical properties, and also examines other important mechanical characteristics. Discussions of the microscopic aspects of deformation mechanisms and methods to strengthen and regulate the mechanical behavior of metals are deferred to later chapters. 6.2 CONCEPTS OF STRESS AND STRAIN If a load is static or changes relatively slowly with time and is applied uniformly over a cross section or surface of a member, the mechanical behavior may be ascertained by a simple stress–strain test; these are most commonly conducted for metals at room temperature. There are three principal ways in which a load may be applied: namely, tension, compression, and shear (Figures 6.1a, b, c). In engineering practice many loads are torsional rather than pure shear; this type of loading is illustrated in Figure 6.1d. Tension Tests 1 One of the most common mechanical stress–strain tests is performed in tension. As will be seen, the tension test can be used to ascertain several mechanical properties of materials that are important in design. A specimen is deformed, usually to fracture, with a gradually increasing tensile load that is applied uniaxially along the long axis of a specimen. A standard tensile specimen is shown in Figure 6.2. Normally, the cross section is circular, but rectangular specimens are also used. This “dogbone” specimen configuration was chosen so that, during testing, deformation is confined to the narrow center region (which has a uniform cross section along its length), and, also, to reduce the likelihood of fracture at the ends of the specimen. The standard diameter is approximately 12.8 mm (0.5 in.), whereas the reduced section length should be at least four times this diameter; 60 mm (214 in.) is common. Gauge length is used in ductility computations, as discussed in Section 6.6; the standard value is 50 mm (2.0 in.). The specimen is mounted by its ends into the holding grips of the testing apparatus (Figure 6.3). The tensile testing machine is designed to elongate the specimen at a constant rate, and to continuously and simultaneously measure the instantaneous applied load (with a load cell) and the resulting elongations (using an extensometer).A stress–strain test typically takes several minutes to perform and is destructive; that is, the test specimen is permanently deformed and usually fractured. 1 ASTM Standards E 8 and E 8M, “Standard Test Methods for Tension Testing of Metallic Materials.” 1496T_c06_131-173 11/16/05 17:21 Page 134 REVISED PAGES 134 • Chapter 6 / Mechanical Properties of Metals Figure 6.1 (a) Schematic illustration of how a tensile load produces an elongation and positive linear strain. Dashed lines represent the shape before deformation; solid lines, after deformation. (b) Schematic illustration of how a compressive load produces contraction and a negative linear strain. (c) Schematic representation of shear strain g, where g  tan u. (d) Schematic representation of torsional deformation (i.e., angle of twist f) produced by an applied torque T. F F A0 l l0 l0 l A0 F F (a) ( b) T ␾ A0 T F F ␪ F (c ) engineering stress engineering strain (d) The output of such a tensile test is recorded (usually on a computer) as load or force versus elongation. These load–deformation characteristics are dependent on the specimen size. For example, it will require twice the load to produce the same elongation if the cross-sectional area of the specimen is doubled. To minimize these geometrical factors, load and elongation are normalized to the respective parameters of engineering stress and engineering strain. Engineering stress s is defined by the relationship Definition of engineering stress (for tension and compression) Figure 6.2 A standard tensile specimen with circular cross section. s F A0 Reduced section 1 2 " 4 3" Diameter 4 0.505" Diameter 2" Gauge length 3" 8 Radius (6.1) 1496T_c06_131-173 11/16/05 17:06 Page 135 REVISED PAGES 6.2 Concepts of Stress and Strain • 135 Load cell Figure 6.3 Schematic representation of the apparatus used to conduct tensile stress–strain tests. The specimen is elongated by the moving crosshead; load cell and extensometer measure, respectively, the magnitude of the applied load and the elongation. (Adapted from H. W. Hayden, W. G. Moffatt, and J. Wulff, The Structure and Properties of Materials, Vol. III, Mechanical Behavior, p. 2. Copyright © 1965 by John Wiley & Sons, New York. Reprinted by permission of John Wiley & Sons, Inc.) Extensometer Specimen Moving crosshead in which F is the instantaneous load applied perpendicular to the specimen cross section, in units of newtons (N) or pounds force 1lbf 2 , and A0 is the original crosssectional area before any load is applied (m2 or in.2). The units of engineering stress (referred to subsequently as just stress) are megapascals, MPa (SI) (where 1 MPa  106 N/m2), and pounds force per square inch, psi (Customary U.S.).2 Engineering strain  is defined according to Definition of engineering strain (for tension and compression)  li  l0 ¢l  l0 l0 (6.2) in which l0 is the original length before any load is applied, and li is the instantaneous length. Sometimes the quantity li  l0 is denoted as ¢l, and is the deformation elongation or change in length at some instant, as referenced to the original length. Engineering strain (subsequently called just strain) is unitless, but meters per meter or inches per inch are often used; the value of strain is obviously independent of the unit system. Sometimes strain is also expressed as a percentage, in which the strain value is multiplied by 100. Compression Tests 3 Compression stress–strain tests may be conducted if in-service forces are of this type. A compression test is conducted in a manner similar to the tensile test, except that the force is compressive and the specimen contracts along the direction of the stress. Equations 6.1 and 6.2 are utilized to compute compressive stress and strain, respectively. By convention, a compressive force is taken to be negative, which yields a negative stress. Furthermore, since l0 is greater than li, compressive strains computed from Equation 6.2 are necessarily also negative. Tensile tests are more common because they are easier to perform; also, for most materials used in structural applications, very little additional information is obtained from compressive tests. 2 Conversion from one system of stress units to the other is accomplished by the relationship 145 psi  1 MPa. 3 ASTM Standard E 9, “Standard Test Methods of Compression Testing of Metallic Materials at Room Temperature.” 1496T_c06_131-173 11/16/05 17:06 Page 136 REVISED PAGES 136 • Chapter 6 / Mechanical Properties of Metals Compressive tests are used when a material’s behavior under large and permanent (i.e., plastic) strains is desired, as in manufacturing applications, or when the material is brittle in tension. Shear and Torsional Tests 4 For tests performed using a pure shear force as shown in Figure 6.1c, the shear stress t is computed according to Definition of shear stress t F A0 (6.3) where F is the load or force imposed parallel to the upper and lower faces, each of which has an area of A0. The shear strain g is defined as the tangent of the strain angle u, as indicated in the figure. The units for shear stress and strain are the same as for their tensile counterparts. Torsion is a variation of pure shear, wherein a structural member is twisted in the manner of Figure 6.1d; torsional forces produce a rotational motion about the longitudinal axis of one end of the member relative to the other end. Examples of torsion are found for machine axles and drive shafts, and also for twist drills. Torsional tests are normally performed on cylindrical solid shafts or tubes. A shear stress t is a function of the applied torque T, whereas shear strain g is related to the angle of twist, f in Figure 6.1d. Geometric Considerations of the Stress State Stresses that are computed from the tensile, compressive, shear, and torsional force states represented in Figure 6.1 act either parallel or perpendicular to planar faces of the bodies represented in these illustrations. Note that the stress state is a function of the orientations of the planes upon which the stresses are taken to act. For example, consider the cylindrical tensile specimen of Figure 6.4 that is subjected to a tensile stress s applied parallel to its axis. Furthermore, consider also the plane p-p¿ that is oriented at some arbitrary angle u relative to the plane of the specimen end-face. Upon this plane p-p¿ , the applied stress is no longer a pure tensile one. Rather, a more complex stress state is present that consists of a tensile (or normal) stress s¿ that acts normal to the p-p¿ plane and, in addition, a shear stress t¿ that acts parallel to this plane; both of these stresses are represented in the figure. Using mechanics of materials principles,5 it is possible to develop equations for s¿ and t¿ in terms of s and u, as follows: 1  cos 2u s¿  s cos2 u  s a b (6.4a) 2 sin 2u b t¿  s sin u cos u  s a (6.4b) 2 These same mechanics principles allow the transformation of stress components from one coordinate system to another coordinate system that has a different orientation. Such treatments are beyond the scope of the present discussion. 4 ASTM Standard E 143, “Standard Test for Shear Modulus.” See, for example, W. F. Riley, L. D. Sturges, and D. H. Morris, Mechanics of Materials, 5th edition, John Wiley & Sons, New York, 1999. 5 1496T_c06_131-173 11/16/05 17:06 Page 137 REVISED PAGES 6.3 Stress–Strain Behavior • 137 ␴ Figure 6.4 Schematic representation showing normal (s¿ ) and shear (t¿ ) stresses that act on a plane oriented at an angle u relative to the plane taken perpendicular to the direction along which a pure tensile stress (s) is applied. p⬘ ␶⬘ ␴⬘ ␪ p ␴ Elastic Deformation 6.3 STRESS–STRAIN BEHAVIOR The degree to which a structure deforms or strains depends on the magnitude of an imposed stress. For most metals that are stressed in tension and at relatively low levels, stress and strain are proportional to each other through the relationship Hooke’s law— relationship between engineering stress and engineering strain for elastic deformation (tension and compression) modulus of elasticity s  E (6.5) This is known as Hooke’s law, and the constant of proportionality E (GPa or psi)6 is the modulus of elasticity, or Young’s modulus. For most typical metals the magnitude of this modulus ranges between 45 GPa (6.5  106 psi), for magnesium, and 407 GPa (59  106 psi), for tungsten. Modulus of elasticity values for several metals at room temperature are presented in Table 6.1. Table 6.1 Room-Temperature Elastic and Shear Moduli, and Poisson’s Ratio for Various Metal Alloys Modulus of Elasticity Shear Modulus Metal Alloy GPa 10 psi GPa 106 psi Poisson’s Ratio Aluminum Brass Copper Magnesium Nickel Steel Titanium Tungsten 69 97 110 45 207 207 107 407 10 14 16 6.5 30 30 15.5 59 25 37 46 17 76 83 45 160 3.6 5.4 6.7 2.5 11.0 12.0 6.5 23.2 0.33 0.34 0.34 0.29 0.31 0.30 0.34 0.28 6 The SI unit for the modulus of elasticity is gigapascal, GPa, where 1 GPa  109 N/m2  103 MPa. 6 1496T_c06_131-173 12/21/05 7:43 Page 138 2nd REVISE PAGES 138 • Chapter 6 / Mechanical Properties of Metals Figure 6.5 Schematic stress–strain diagram showing linear elastic deformation for loading and unloading cycles. Stress Unload Slope = modulus of elasticity Load 0 0 Strain Metal Alloys Deformation in which stress and strain are proportional is called elastic deformation; a plot of stress (ordinate) versus strain (abscissa) results in a linear relationship, as shown in Figure 6.5. The slope of this linear segment corresponds to the modulus of elasticity E. This modulus may be thought of as stiffness, or a material’s resistance to elastic deformation. The greater the modulus, the stiffer the material, or the smaller the elastic strain that results from the application of a given stress. The modulus is an important design parameter used for computing elastic deflections. Elastic deformation is nonpermanent, which means that when the applied load is released, the piece returns to its original shape. As shown in the stress–strain plot (Figure 6.5), application of the load corresponds to moving from the origin up and along the straight line. Upon release of the load, the line is traversed in the opposite direction, back to the origin. There are some materials (e.g., gray cast iron, concrete, and many polymers) for which this elastic portion of the stress–strain curve is not linear (Figure 6.6); hence, it is not possible to determine a modulus of elasticity as described above. For this nonlinear behavior, either tangent or secant modulus is normally used. Tangent modulus is taken as the slope of the stress–strain curve at some specified level of stress, while secant modulus represents the slope of a secant drawn from the origin to some given point of the s– curve. The determination of these moduli is illustrated in Figure 6.6. Figure 6.6 Schematic stress–strain diagram showing non-linear elastic behavior, and how secant and tangent moduli are determined. ␴2 ⌬␴ ⌬⑀ Stress ␴ elastic deformation = Tangent modulus (at ␴2) ␴1 ⌬␴ ⌬⑀ = Secant modulus (between origin and ␴1) Strain ⑀ 1496T_c06_131-173 11/16/05 17:06 Page 139 REVISED PAGES 6.3 Stress–Strain Behavior • 139 Strongly bonded Force F Figure 6.7 Force versus interatomic separation for weakly and strongly bonded atoms. The magnitude of the modulus of elasticity is proportional to the slope of each curve at the equilibrium interatomic separation r0. dF dr r 0 Separation r 0 Weakly bonded On an atomic scale, macroscopic elastic strain is manifested as small changes in the interatomic spacing and the stretching of interatomic bonds. As a consequence, the magnitude of the modulus of elasticity is a measure of the resistance to separation of adjacent atoms, that is, the interatomic bonding forces. Furthermore, this modulus is proportional to the slope of the interatomic force–separation curve (Figure 2.8a) at the equilibrium spacing: E r a dF b dr r0 (6.6) Figure 6.7 shows the force–separation curves for materials having both strong and weak interatomic bonds; the slope at r0 is indicated for each. Values of the modulus of elasticity for ceramic materials are about the same as for metals; for polymers they are lower (Figure 1.4). These differences are a direct consequence of the different types of atomic bonding in the three materials types. Furthermore, with increasing temperature, the modulus of elasticity diminishes, as is shown for several metals in Figure 6.8. Temperature (°F) –400 0 400 800 1200 1600 70 Tungsten 50 300 40 Steel 200 30 20 100 Aluminum 10 0 –200 0 200 400 Temperature (°C) 600 800 0 Modulus of elasticity (106 psi) 60 400 Modulus of elasticity (GPa) Figure 6.8 Plot of modulus of elasticity versus temperature for tungsten, steel, and aluminum. (Adapted from K. M. Ralls, T. H. Courtney, and J. Wulff, Introduction to Materials Science and Engineering. Copyright © 1976 by John Wiley & Sons, New York. Reprinted by permission of John Wiley & Sons, Inc.) 1496T_c06_131-173 11/16/05 17:06 Page 140 REVISED PAGES 140 • Chapter 6 / Mechanical Properties of Metals As would be expected, the imposition of compressive, shear, or torsional stresses also evokes elastic behavior. The stress–strain characteristics at low stress levels are virtually the same for both tensile and compressive situations, to include the magnitude of the modulus of elasticity. Shear stress and strain are proportional to each other through the expression Relationship between shear stress and shear strain for elastic deformation 6.4 t  Gg (6.7) where G is the shear modulus, the slope of the linear elastic region of the shear stress–strain curve. Table 6.1 also gives the shear moduli for a number of the common metals. ANELASTICITY anelasticity Up to this point, it has been assumed that elastic deformation is time independent— that is, that an applied stress produces an instantaneous elastic strain that remains constant over the period of time the stress is maintained. It has also been assumed that upon release of the load the strain is totally recovered—that is, that the strain immediately returns to zero. In most engineering materials, however, there will also exist a time-dependent elastic strain component. That is, elastic deformation will continue after the stress application, and upon load release some finite time is required for complete recovery. This time-dependent elastic behavior is known as anelasticity, and it is due to time-dependent microscopic and atomistic processes that are attendant to the deformation. For metals the anelastic component is normally small and is often neglected. However, for some polymeric materials its magnitude is significant; in this case it is termed viscoelastic behavior, which is the discussion topic of Section 15.4. EXAMPLE PROBLEM 6.1 Elongation (Elastic) Computation A piece of copper originally 305 mm (12 in.) long is pulled in tension with a stress of 276 MPa (40,000 psi). If the deformation is entirely elastic, what will be the resultant elongation? Solution Since the deformation is elastic, strain is dependent on stress according to Equation 6.5. Furthermore, the elongation ¢l is related to the original length l0 through Equation 6.2. Combining these two expressions and solving for ¢l yields s  E  a ¢l  ¢l bE l0 sl0 E The values of s and l0 are given as 276 MPa and 305 mm, respectively, and the magnitude of E for copper from Table 6.1 is 110 GPa (16  106 psi). Elongation is obtained by substitution into the expression above as ¢l  1276 MPa21305 mm2 110  103 MPa  0.77 mm 10.03 in.2 1496T_c06_131-173 11/16/05 17:06 Page 141 REVISED PAGES 6.5 Elastic Properties of Materials • 141 6.5 ELASTIC PROPERTIES OF MATERIALS Poisson’s ratio When a tensile stress is imposed on a metal specimen, an elastic elongation and accompanying strain z result in the direction of the applied stress (arbitrarily taken to be the z direction), as indicated in Figure 6.9. As a result of this elongation, there will be constrictions in the lateral (x and y) directions perpendicular to the applied stress; from these contractions, the compressive strains x and y may be determined. If the applied stress is uniaxial (only in the z direction), and the material is isotropic, then x  y. A parameter termed Poisson’s ratio n is defined as the ratio of the lateral and axial strains, or Definition of Poisson’s ratio in terms of lateral and axial strains n y x  z z (6.8) The negative sign is included in the expression so that n will always be positive, since x and z will always be of opposite sign. Theoretically, Poisson’s ratio for isotropic materials should be 14 ; furthermore, the maximum value for n (or that value for which there is no net volume change) is 0.50. For many metals and other alloys, values of Poisson’s ratio range between 0.25 and 0.35. Table 6.1 shows n values for several common metallic materials. For isotropic materials, shear and elastic moduli are related to each other and to Poisson’s ratio according to Relationship among elastic parameters— modulus of elasticity, shear modulus, and Poisson’s ratio E  2G 11  n2 (6.9) In most metals G is about 0.4E; thus, if the value of one modulus is known, the other may be approximated. Many materials are elastically anisotropic; that is, the elastic behavior (e.g., the magnitude of E) varies with crystallographic direction (see Table 3.3). For these materials the elastic properties are completely characterized only by the specification ⌬lz 2 ␴z l0x ⌬lx 2 l0z z ⑀z ⌬lz/2 = 2 l0z – ⑀x ⌬lx/2 = 2 l0x ␴z y x Figure 6.9 Axial (z) elongation (positive strain) and lateral (x and y) contractions (negative strains) in response to an imposed tensile stress. Solid lines represent dimensions after stress application; dashed lines, before. 1496T_c06_131-173 11/16/05 17:06 Page 142 REVISED PAGES 142 • Chapter 6 / Mechanical Properties of Metals of several elastic constants, their number depending on characteristics of the crystal structure. Even for isotropic materials, for complete characterization of the elastic properties, at least two constants must be given. Since the grain orientation is random in most polycrystalline materials, these may be considered to be isotropic; inorganic ceramic glasses are also isotropic. The remaining discussion of mechanical behavior assumes isotropy and polycrystallinity because such is the character of most engineering materials. EXAMPLE PROBLEM 6.2 Computation of Load to Produce Specified Diameter Change A tensile stress is to be applied along the long axis of a cylindrical brass rod that has a diameter of 10 mm (0.4 in.). Determine the magnitude of the load required to produce a 2.5  103 mm (104 in.) change in diameter if the deformation is entirely elastic. Solution This deformation situation is represented in the accompanying drawing. F d0 di z li x l0 ⑀z = ⌬l l0 = ⑀ x = ⌬d = d0 li – l0 l0 d i – d0 d0 F When the force F is applied, the specimen will elongate in the z direction and at the same time experience a reduction in diameter, ¢d, of 2.5  103 mm in the x direction. For the strain in the x direction, x  ¢d 2.5  103 mm   2.5  104 d0 10 mm which is negative, since the diameter is reduced. 1496T_c06_131-173 11/16/05 17:06 Page 143 REVISED PAGES 6.5 Elastic Properties of Materials • 143 It next becomes necessary to calculate the strain in the z direction using Equation 6.8. The value for Poisson’s ratio for brass is 0.34 (Table 6.1), and thus z   12.5  104 2 x   7.35  104 n 0.34 The applied stress may now be computed using Equation 6.5 and the modulus of elasticity, given in Table 6.1 as 97 GPa (14  106 psi), as s  zE  17.35  104 2197  103 MPa2  71.3 MPa Finally, from Equation 6.1, the applied force may be determined as F  sA0  s a d0 2 bp 2  171.3  106 N/m2 2a 10  103 m 2 b p  5600 N 11293 lbf 2 2 Plastic Deformation Elastic Plastic Upper yield point ␴y Stress Figure 6.10 (a) Typical stress– strain behavior for a metal showing elastic and plastic deformations, the proportional limit P, and the yield strength sy, as determined using the 0.002 strain offset method. (b) Representative stress–strain behavior found for some steels demonstrating the yield point phenomenon. Stress plastic deformation For most metallic materials, elastic deformation persists only to strains of about 0.005. As the material is deformed beyond this point, the stress is no longer proportional to strain (Hooke’s law, Equation 6.5, ceases to be valid), and permanent, nonrecoverable, or plastic deformation occurs. Figure 6.10a plots schematically the tensile stress–strain behavior into the plastic region for a typical metal. The transition from elastic to plastic is a gradual one for most metals; some curvature results at the onset of plastic deformation, which increases more rapidly with rising stress. P ␴y Lower yield point Strain Strain 0.002 (a) (b) 1496T_c06_131-173 12/21/05 7:43 Page 144 2nd REVISE PAGES 144 • Chapter 6 / Mechanical Properties of Metals From an atomic perspective, plastic deformation corresponds to the breaking of bonds with original atom neighbors and then reforming bonds with new neighbors as large numbers of atoms or molecules move relative to one another; upon removal of the stress they do not return to their original positions. The mechanism of this deformation is different for crystalline and amorphous materials. For crystalline solids, deformation is accomplished by means of a process called slip, which involves the motion of dislocations as discussed in Section 7.2. Plastic deformation in noncrystalline solids (as well as liquids) occurs by a viscous flow mechanism, which is outlined in Section 12.10. 6.6 TENSILE PROPERTIES Yielding and Yield Strength yielding proportional limit yield strength Metal Alloys Most structures are designed to ensure that only elastic deformation will result when a stress is applied. A structure or component that has plastically deformed, or experienced a permanent change in shape, may not be capable of functioning as intended. It is therefore desirable to know the stress level at which plastic deformation begins, or where the phenomenon of yielding occurs. For metals that experience this gradual elastic–plastic transition, the point of yielding may be determined as the initial departure from linearity of the stress–strain curve; this is sometimes called the proportional limit, as indicated by point P in Figure 6.10a. In such cases the position of this point may not be determined precisely. As a consequence, a convention has been established wherein a straight line is constructed parallel to the elastic portion of the stress–strain curve at some specified strain offset, usually 0.002. The stress corresponding to the intersection of this line and the stress–strain curve as it bends over in the plastic region is defined as the yield strength y.7 This is demonstrated in Figure 6.10a. Of course, the units of yield strength are MPa or psi.8 For those materials having a nonlinear elastic region (Figure 6.6), use of the strain offset method is not possible, and the usual practice is to define the yield strength as the stress required to produce some amount of strain (e.g.,   0.005). Some steels and other materials exhibit the tensile stress–strain behavior as shown in Figure 6.10b. The elastic–plastic transition is very well defined and occurs abruptly in what is termed a yield point phenomenon.At the upper yield point, plastic deformation is initiated with an actual decrease in stress. Continued deformation fluctuates slightly about some constant stress value, termed the lower yield point; stress subsequently rises with increasing strain. For metals that display this effect, the yield strength is taken as the average stress that is associated with the lower yield point, since it is well defined and relatively insensitive to the testing procedure.9 Thus, it is not necessary to employ the strain offset method for these materials. The magnitude of the yield strength for a metal is a measure of its resistance to plastic deformation. Yield strengths may range from 35 MPa (5000 psi) for a lowstrength aluminum to over 1400 MPa (200,000 psi) for high-strength steels. 7 “Strength” is used in lieu of “stress” because strength is a property of the metal, whereas stress is related to the magnitude of the applied load. 8 For Customary U.S. units, the unit of kilopounds per square inch (ksi) is sometimes used for the sake of convenience, where 1 ksi  1000 psi 9 Note that to observe the yield point phenomenon, a “stiff” tensile-testing apparatus must be used; by stiff is meant that there is very little elastic deformation of the machine during loading. 1496T_c06_131-173 11/16/05 17:06 Page 145 REVISED PAGES 6.6 Tensile Properties • 145 Concept Check 6.1 Cite the primary differences between elastic, anelastic, and plastic deformation behaviors. [The answer may be found at www.wiley.com/college/callister (Student Companion Site).] Tensile Strength Figure 6.11 Typical engineering stress– strain behavior to fracture, point F. The tensile strength TS is indicated at point M. The circular insets represent the geometry of the deformed specimen at various points along the curve. TS M F Stress tensile strength After yielding, the stress necessary to continue plastic deformation in metals increases to a maximum, point M in Figure 6.11, and then decreases to the eventual fracture, point F. The tensile strength TS (MPa or psi) is the stress at the maximum on the engineering stress–strain curve (Figure 6.11). This corresponds to the maximum stress that can be sustained by a structure in tension; if this stress is applied and maintained, fracture will result. All deformation up to this point is uniform throughout the narrow region of the tensile specimen. However, at this maximum stress, a small constriction or neck begins to form at some point, and all subsequent deformation is confined at this neck, as indicated by the schematic specimen insets in Figure 6.11. This phenomenon is termed “necking,” and fracture ultimately occurs at the neck. The fracture strength corresponds to the stress at fracture. Tensile strengths may vary anywhere from 50 MPa (7000 psi) for an aluminum to as high as 3000 MPa (450,000 psi) for the high-strength steels. Ordinarily, when the strength of a metal is cited for design purposes, the yield strength is used. This is because by the time a stress corresponding to the tensile strength has been applied, often a structure has experienced so much plastic deformation that it is useless. Furthermore, fracture strengths are not normally specified for engineering design purposes. Strain 1496T_c06_131-173 11/16/05 17:06 Page 146 REVISED PAGES 146 • Chapter 6 / Mechanical Properties of Metals EXAMPLE PROBLEM 6.3 Mechanical Property Determinations from Stress–Strain Plot From the tensile stress–strain behavior for the brass specimen shown in Figure 6.12, determine the following: (a) The modulus of elasticity (b) The yield strength at a strain offset of 0.002 (c) The maximum load that can be sustained by a cylindrical specimen having an original diameter of 12.8 mm (0.505 in.) (d) The change in length of a specimen originally 250 mm (10 in.) long that is subjected to a tensile stress of 345 MPa (50,000 psi) Solution (a) The modulus of elasticity is the slope of the elastic or initial linear portion of the stress–strain curve. The strain axis has been expanded in the inset, Figure 6.12, to facilitate this computation. The slope of this linear region is the rise over the run, or the change in stress divided by the corresponding change in strain; in mathematical terms, E  slope  s2  s1 ¢s  2  1 ¢ (6.10) 500 70 Tensile strength 450 MPa (65,000 psi) 60 400 300 50 103 psi MPa 40 40 200 30 Yield strength 250 MPa (36,000 psi) 200 20 100 30 20 10 100 10 0 0 0 0.10 0 0 0.005 0.20 0.30 0 0.40 Strain Figure 6.12 The stress–strain behavior for the brass specimen discussed in Example Problem 6.3. Stress (103 psi) Stress (MPa) A 1496T_c06_131-173 11/16/05 17:06 Page 147 REVISED PAGES 6.6 Tensile Properties • 147 Inasmuch as the line segment passes through the origin, it is convenient to take both s1 and 1 as zero. If s2 is arbitrarily taken as 150 MPa, then 2 will have a value of 0.0016. Therefore, E 1150  02 MPa  93.8 GPa 113.6  106 psi2 0.0016  0 which is very close to the value of 97 GPa (14  106 psi) given for brass in Table 6.1. (b) The 0.002 strain offset line is constructed as shown in the inset; its intersection with the stress–strain curve is at approximately 250 MPa (36,000 psi), which is the yield strength of the brass. (c) The maximum load that can be sustained by the specimen is calculated by using Equation 6.1, in which s is taken to be the tensile strength, from Figure 6.12, 450 MPa (65,000 psi). Solving for F, the maximum load, yields F  sA0  s a d0 2 bp 2  1450  106 N/m2 2 a 12.8  103 m 2 b p  57,900 N 113,000 lbf 2 2 (d) To compute the change in length, ¢l, in Equation 6.2, it is first necessary to determine the strain that is produced by a stress of 345 MPa. This is accomplished by locating the stress point on the stress–strain curve, point A, and reading the corresponding strain from the strain axis, which is approximately 0.06. Inasmuch as l0  250 mm, we have ¢l  l0  10.0621250 mm2  15 mm 10.6 in.2 Ductility Ductility is another important mechanical property. It is a measure of the degree of plastic deformation that has been sustained at fracture. A material that experiences very little or no plastic deformation upon fracture is termed brittle. The tensile stress–strain behaviors for both ductile and brittle materials are schematically illustrated in Figure 6.13. Brittle B Figure 6.13 Schematic representations of tensile stress–strain behavior for brittle and ductile materials loaded to fracture. Ductile Stress ductility B⬘ A C C⬘ Strain 1496T_c06_131-173 11/16/05 17:06 Page 148 REVISED PAGES 148 • Chapter 6 / Mechanical Properties of Metals Ductility may be expressed quantitatively as either percent elongation or percent reduction in area. The percent elongation %EL is the percentage of plastic strain at fracture, or Ductility, as percent elongation %EL  a lf  l0 l0 b  100 (6.11) where lf is the fracture length10 and l0 is the original gauge length as above. Inasmuch as a significant proportion of the plastic deformation at fracture is confined to the neck region, the magnitude of %EL will depend on specimen gauge length. The shorter l0, the greater is the fraction of total elongation from the neck and, consequently, the higher the value of %EL. Therefore, l0 should be specified when percent elongation values are cited; it is commonly 50 mm (2 in.). Percent reduction in area %RA is defined as Ductility, as percent reduction in area %RA  a A0  Af A0 b  100 (6.12) where A0 is the original cross-sectional area and Af is the cross-sectional area at the point of fracture.10 Percent reduction in area values are independent of both l0 and A0. Furthermore, for a given material the magnitudes of %EL and %RA will, in general, be different. Most metals possess at least a moderate degree of ductility at room temperature; however, some become brittle as the temperature is lowered (Section 8.6). A knowledge of the ductility of materials is important for at least two reasons. First, it indicates to a designer the degree to which a structure will deform plastically before fracture. Second, it specifies the degree of allowable deformation during fabrication operations. We sometimes refer to relatively ductile materials as being “forgiving,” in the sense that they may experience local deformation without fracture should there be an error in the magnitude of the design stress calculation. Brittle materials are approximately considered to be those having a fracture strain of less than about 5%. Thus, several important mechanical properties of metals may be determined from tensile stress–strain tests. Table 6.2 presents some typical room-temperature values of Table 6.2 Typical Mechanical Properties of Several Metals and Alloys in an Annealed State Metal Alloy Aluminum Copper Brass (70Cu–30Zn) Iron Nickel Steel (1020) Titanium Molybdenum 10 Yield Strength MPa (ksi) Tensile Strength MPa (ksi) Ductility, %EL [in 50 mm (2 in.)] 35 (5) 69 (10) 75 (11) 130 (19) 138 (20) 180 (26) 450 (65) 565 (82) 90 (13) 200 (29) 300 (44) 262 (38) 480 (70) 380 (55) 520 (75) 655 (95) 40 45 68 45 40 25 25 35 Both lf and Af are measured subsequent to fracture and after the two broken ends have been repositioned back together. 1496T_c06_131-173 11/16/05 17:06 Page 149 REVISED PAGES 6.6 Tensile Properties • 149 120 800 –200°C 100 80 60 –100°C 400 40 Stress (103 psi) 600 Stress (MPa) Figure 6.14 Engineering stress– strain behavior for iron at three temperatures. 25°C 200 20 0 0 0.1 0.2 0.3 0.4 0 0.5 Strain yield strength, tensile strength, and ductility for several of the common metals. These properties are sensitive to any prior deformation, the presence of impurities, and/or any heat treatment to which the metal has been subjected. The modulus of elasticity is one mechanical parameter that is insensitive to these treatments. As with modulus of elasticity, the magnitudes of both yield and tensile strengths decline with increasing temperature; just the reverse holds for ductility—it usually increases with temperature. Figure 6.14 shows how the stress–strain behavior of iron varies with temperature. Resilience resilience Resilience is the capacity of a material to absorb energy when it is deformed elastically and then, upon unloading, to have this energy recovered.The associated property is the modulus of resilience, Ur, which is the strain energy per unit volume required to stress a material from an unloaded state up to the point of yielding. Computationally, the modulus of resilience for a specimen subjected to a uniaxial tension test is just the area under the engineering stress–strain curve taken to yielding (Figure 6.15), or Definition of modulus of resilience Ur   y sd (6.13a) 0 Figure 6.15 Schematic representation showing how modulus of resilience (corresponding to the shaded area) is determined from the tensile stress–strain behavior of a material. Stress σy 0.002 y Strain 1496T_c06_131-173 11/16/05 17:06 Page 150 REVISED PAGES 150 • Chapter 6 / Mechanical Properties of Metals Assuming a linear elastic region, Modulus of resilience for linear elastic behavior Ur  12 syy (6.13b) in which y is the strain at yielding. The units of resilience are the product of the units from each of the two axes of the stress–strain plot. For SI units, this is joules per cubic meter (J/m3, equivalent to Pa), whereas with Customary U.S. units it is inch-pounds force per cubic inch (in.-lbf/in.3, equivalent to psi). Both joules and inch-pounds force are units of energy, and thus this area under the stress–strain curve represents energy absorption per unit volume (in cubic meters or cubic inches) of material. Incorporation of Equation 6.5 into Equation 6.13b yields Modulus of resilience for linear elastic behavior, and incorporating Hooke’s law Ur  12 syy  12 sy a sy E b s2y 2E (6.14) Thus, resilient materials are those having high yield strengths and low moduli of elasticity; such alloys would be used in spring applications. Toughness toughness Toughness is a mechanical term that is used in several contexts; loosely speaking, it is a measure of the ability of a material to absorb energy up to fracture. Specimen geometry as well as the manner of load application are important in toughness determinations. For dynamic (high strain rate) loading conditions and when a notch (or point of stress concentration) is present, notch toughness is assessed by using an impact test, as discussed in Section 8.6. Furthermore, fracture toughness is a property indicative of a material’s resistance to fracture when a crack is present (Section 8.5). For the static (low strain rate) situation, toughness may be ascertained from the results of a tensile stress–strain test. It is the area under the s– curve up to the point of fracture. The units for toughness are the same as for resilience (i.e., energy per unit volume of material). For a material to be tough, it must display both strength and ductility; often, ductile materials are tougher than brittle ones. This is demonstrated in Figure 6.13, in which the stress–strain curves are plotted for both material types. Hence, even though the brittle material has higher yield and tensile strengths, it has a lower toughness than the ductile one, by virtue of lack of ductility; this is deduced by comparing the areas ABC and AB¿C¿ in Figure 6.13. Concept Check 6.2 Of those metals listed in Table 6.3, (a) Which will experience the greatest percent reduction in area? Why? (b) Which is the strongest? Why? (c) Which is the stiffest? Why? [The answer may be found at www.wiley.com/college/callister (Student Companion Site).] 1496T_c06_131-173 11/16/05 17:06 Page 151 REVISED PAGES 6.7 True Stress and Strain • 151 Table 6.3 Tensile Stress–Strain Data for Several Hypothetical Metals to be Used with Concept Checks 6.2 and 6.4 Material A B C D E Yield Strength (MPa) 310 100 415 700 Tensile Strength (MPa) Strain at Fracture 340 120 550 850 Fractures before yielding 0.23 0.40 0.15 0.14 Fracture Strength (MPa) Elastic Modulus (GPa) 265 105 500 720 650 210 150 310 210 350 6.7 TRUE STRESS AND STRAIN true stress Definition of true stress true strain Definition of true strain From Figure 6.11, the decline in the stress necessary to continue deformation past the maximum, point M, seems to indicate that the metal is becoming weaker. This is not at all the case; as a matter of fact, it is increasing in strength. However, the cross-sectional area is decreasing rapidly within the neck region, where deformation is occurring. This results in a reduction in the load-bearing capacity of the specimen. The stress, as computed from Equation 6.1, is on the basis of the original crosssectional area before any deformation, and does not take into account this reduction in area at the neck. Sometimes it is more meaningful to use a true stress–true strain scheme. True stress sT is defined as the load F divided by the instantaneous cross-sectional area Ai over which deformation is occurring (i.e., the neck, past the tensile point), or sT  F Ai (6.15) Furthermore, it is occasionally more convenient to represent strain as true strain T, defined by T  ln li l0 (6.16) If no volume change occurs during deformation—that is, if Aili  A0l0 (6.17) true and engineering stress and strain are related according to Conversion of engineering stress to true stress Conversion of engineering strain to true strain sT  s11  2 (6.18a) T  ln11  2 (6.18b) Equations 6.18a and 6.18b are valid only to the onset of necking; beyond this point true stress and strain should be computed from actual load, cross-sectional area, and gauge length measurements. A schematic comparison of engineering and true stress–strain behaviors is made in Figure 6.16. It is worth noting that the true stress necessary to sustain increasing strain continues to rise past the tensile point M¿. 1496T_c06_131-173 11/16/05 17:06 Page 152 REVISED PAGES 152 • Chapter 6 / Mechanical Properties of Metals Figure 6.16 A comparison of typical tensile engineering stress–strain and true stress–strain behaviors. Necking begins at point M on the engineering curve, which corresponds to M¿ on the true curve. The “corrected” true stress– strain curve takes into account the complex stress state within the neck region. True Stress M⬘ Corrected M Engineering Strain Coincident with the formation of a neck is the introduction of a complex stress state within the neck region (i.e., the existence of other stress components in addition to the axial stress). As a consequence, the correct stress (axial) within the neck is slightly lower than the stress computed from the applied load and neck crosssectional area. This leads to the “corrected” curve in Figure 6.16. For some metals and alloys the region of the true stress–strain curve from the onset of plastic deformation to the point at which necking begins may be approximated by True stress-true strain relationship in plastic region of deformation (to point of necking) sT  KTn (6.19) In this expression, K and n are constants; these values will vary from alloy to alloy, and will also depend on the condition of the material (i.e., whether it has been plastically deformed, heat treated, etc.). The parameter n is often termed the strainhardening exponent and has a value less than unity. Values of n and K for several alloys are contained in Table 6.4. Table 6.4 Tabulation of n and K Values (Equation 6.19) for Several Alloys K Material n MPa psi Low-carbon steel (annealed) 0.21 600 87,000 4340 steel alloy (tempered @ 315C) 0.12 2650 385,000 304 stainless steel (annealed) 0.44 1400 205,000 Copper (annealed) 0.44 530 76,500 Naval brass (annealed) 0.21 585 85,000 2024 aluminum alloy (heat treated—T3) 0.17 780 113,000 AZ-31B magnesium alloy (annealed) 0.16 450 66,000 1496T_c06_131-173 11/16/05 17:06 Page 153 REVISED PAGES 6.7 True Stress and Strain • 153 EXAMPLE PROBLEM 6.4 Ductility and True-Stress-At-Fracture Computations A cylindrical specimen of steel having an original diameter of 12.8 mm (0.505 in.) is tensile tested to fracture and found to have an engineering fracture strength sf of 460 MPa (67,000 psi). If its cross-sectional diameter at fracture is 10.7 mm (0.422 in.), determine: (a) The ductility in terms of percent reduction in area (b) The true stress at fracture Solution (a) Ductility is computed using Equation 6.12, as 10.7 mm 2 12.8 mm 2 bpa bp 2 2 %RA   100 12.8 mm 2 a bp 2 2 128.7 mm  89.9 mm2   100  30% 128.7 mm2 a (b) True stress is defined by Equation 6.15, where in this case the area is taken as the fracture area Af. However, the load at fracture must first be computed from the fracture strength as 1 m2 F  sf A0  1460  106 N/m2 21128.7 mm2 2 a 6 b  59,200 N 10 mm2 Thus, the true stress is calculated as sT  F  Af 59,200 N 1 m2 189.9 mm2 2 a 6 b 10 mm2  6.6  108 N/m2  660 MPa 195,700 psi2 EXAMPLE PROBLEM 6.5 Calculation of Strain-Hardening Exponent Compute the strain-hardening exponent n in Equation 6.19 for an alloy in which a true stress of 415 MPa (60,000 psi) produces a true strain of 0.10; assume a value of 1035 MPa (150,000 psi) for K. Solution This requires some algebraic manipulation of Equation 6.19 so that n becomes the dependent parameter. This is accomplished by taking logarithms and rearranging. Solving for n yields log sT  log K log T log1415 MPa2  log11035 MPa2  0.40  log10.12 n 1496T_c06_131-173 11/16/05 17:06 Page 154 REVISED PAGES 154 • Chapter 6 / Mechanical Properties of Metals 6.8 ELASTIC RECOVERY AFTER PLASTIC DEFORMATION Upon release of the load during the course of a stress–strain test, some fraction of the total deformation is recovered as elastic strain. This behavior is demonstrated in Figure 6.17, a schematic engineering stress–strain plot. During the unloading cycle, the curve traces a near straight-line path from the point of unloading (point D), and its slope is virtually identical to the modulus of elasticity, or parallel to the initial elastic portion of the curve. The magnitude of this elastic strain, which is regained during unloading, corresponds to the strain recovery, as shown in Figure 6.17. If the load is reapplied, the curve will traverse essentially the same linear portion in the direction opposite to unloading; yielding will again occur at the unloading stress level where the unloading began. There will also be an elastic strain recovery associated with fracture. 6.9 COMPRESSIVE, SHEAR, AND TORSIONAL DEFORMATION Of course, metals may experience plastic deformation under the influence of applied compressive, shear, and torsional loads. The resulting stress–strain behavior into the plastic region will be similar to the tensile counterpart (Figure 6.10a: yielding and the associated curvature). However, for compression, there will be no maximum, since necking does not occur; furthermore, the mode of fracture will be different from that for tension. Concept Check 6.3 Make a schematic plot showing the tensile engineering stress–strain behavior for a typical metal alloy to the point of fracture. Now superimpose on this plot a schematic compressive engineering stress–strain curve for the same alloy. Explain any differences between the two curves. [The answer may be found at www.wiley.com/college/callister (Student Companion Site).] Figure 6.17 Schematic tensile stress–strain diagram showing the phenomena of elastic strain recovery and strain hardening. The initial yield strength is designated as sy0; syi is the yield strength after releasing the load at point D, and then upon reloading. D ␴yi ␴y0 Stress Unload Reapply load Strain Elastic strain recovery 1496T_c06_131-173 11/16/05 17:06 Page 155 REVISED PAGES 6.10 Hardness • 155 6.10 HARDNESS hardness Another mechanical property that may be important to consider is hardness, which is a measure of a material’s resistance to localized plastic deformation (e.g., a small dent or a scratch). Early hardness tests were based on natural minerals with a scale constructed solely on the ability of one material to scratch another that was softer. A qualitative and somewhat arbitrary hardness indexing scheme was devised, termed the Mohs scale, which ranged from 1 on the soft end for talc to 10 for diamond. Quantitative hardness techniques have been developed over the years in which a small indenter is forced into the surface of a material to be tested, under controlled conditions of load and rate of application. The depth or size of the resulting indentation is measured, which in turn is related to a hardness number; the softer the material, the larger and deeper is the indentation, and the lower the hardness index number. Measured hardnesses are only relative (rather than absolute), and care should be exercised when comparing values determined by different techniques. Hardness tests are performed more frequently than any other mechanical test for several reasons: 1. They are simple and inexpensive—ordinarily no special specimen need be prepared, and the testing apparatus is relatively inexpensive. 2. The test is nondestructive—the specimen is neither fractured nor excessively deformed; a small indentation is the only deformation. 3. Other mechanical properties often may be estimated from hardness data, such as tensile strength (see Figure 6.19). Rockwell Hardness Tests 11 The Rockwell tests constitute the most common method used to measure hardness because they are so simple to perform and require no special skills. Several different scales may be utilized from possible combinations of various indenters and different loads, which permit the testing of virtually all metal alloys (as well as some polymers). Indenters include spherical and hardened steel balls having diameters of 161 , 18, 14, and 12 in. (1.588, 3.175, 6.350, and 12.70 mm), and a conical diamond (Brale) indenter, which is used for the hardest materials. With this system, a hardness number is determined by the difference in depth of penetration resulting from the application of an initial minor load followed by a larger major load; utilization of a minor load enhances test accuracy. On the basis of the magnitude of both major and minor loads, there are two types of tests: Rockwell and superficial Rockwell. For Rockwell, the minor load is 10 kg, whereas major loads are 60, 100, and 150 kg. Each scale is represented by a letter of the alphabet; several are listed with the corresponding indenter and load in Tables 6.5 and 6.6a. For superficial tests, 3 kg is the minor load; 15, 30, and 45 kg are the possible major load values. These scales are identified by a 15, 30, or 45 (according to load), followed by N, T, W, X, or Y, depending on indenter. Superficial tests are frequently performed on thin specimens. Table 6.6b presents several superficial scales. When specifying Rockwell and superficial hardnesses, both hardness number and scale symbol must be indicated. The scale is designated by the symbol HR 11 ASTM Standard E 18, “Standard Test Methods for Rockwell Hardness and Rockwell Superficial Hardness of Metallic Materials.” Diamond pyramid Knoop microhardness 120° l/b = 7.11 b/t = 4.00 136° d D Side View t Shape of Indentation d1 l d b d1 Top View 15 kg 30 kg ¶ Superficial Rockwell 45 kg 60 kg 100 kg ¶ Rockwell 150 kg P P P Load pD 3 D  2D2  d 2 4 2P HK  14.2P l 2 HV  1.854Pd 21 HB  Formula for Hardness Numbera For the hardness formulas given, P (the applied load) is in kg, while D, d, d1, and l are all in mm. Source: Adapted from H. W. Hayden, W. G. Moffatt, and J. Wulff, The Structure and Properties of Materials, Vol. III, Mechanical Behavior. Copyright © 1965 by John Wiley & Sons, New York. Reprinted by permission of John Wiley & Sons, Inc. a Diamond cone; 1 1 1 1 16 , 8 , 4 , 2 in. diameter steel spheres  Diamond pyramid Vickers microhardness Rockwell and Superficial Rockwell 10-mm sphere of steel or tungsten carbide Indenter Brinell Test Table 6.5 Hardness-Testing Techniques 1496T_c06_131-173 11/16/05 17:06 Page 156 REVISED PAGES 156 • Chapter 6 / Mechanical Properties of Metals 1496T_c06_131-173 11/16/05 17:06 Page 157 REVISED PAGES 6.10 Hardness • 157 Table 6.6a Rockwell Hardness Scales Scale Symbol Indenter Major Load (kg) A B C D E F G H K Diamond 1 16 -in. ball Diamond Diamond 1 8 -in. ball 1 16 -in. ball 1 16 -in. ball 1 8 -in. ball 1 8 -in. ball 60 100 150 100 100 60 150 60 150 Table 6.6b Superficial Rockwell Hardness Scales Scale Symbol Indenter Major Load (kg) 15N 30N 45N 15T 30T 45T 15W 30W 45W Diamond Diamond Diamond 1 16 -in. ball 1 16 -in. ball 1 16 -in. ball 1 8 -in. ball 1 8 -in. ball 1 8 -in. ball 15 30 45 15 30 45 15 30 45 followed by the appropriate scale identification.12 For example, 80 HRB represents a Rockwell hardness of 80 on the B scale, and 60 HR30W indicates a superficial hardness of 60 on the 30W scale. For each scale, hardnesses may range up to 130; however, as hardness values rise above 100 or drop below 20 on any scale, they become inaccurate; and because the scales have some overlap, in such a situation it is best to utilize the next harder or softer scale. Inaccuracies also result if the test specimen is too thin, if an indentation is made too near a specimen edge, or if two indentations are made too close to one another. Specimen thickness should be at least ten times the indentation depth, whereas allowance should be made for at least three indentation diameters between the center of one indentation and the specimen edge, or to the center of a second indentation. Furthermore, testing of specimens stacked one on top of another is not recommended. Also, accuracy is dependent on the indentation being made into a smooth flat surface. The modern apparatus for making Rockwell hardness measurements (see the chapter-opening photograph for this chapter) is automated and very simple to use; hardness is read directly, and each measurement requires only a few seconds. The modern testing apparatus also permits a variation in the time of load application. This variable must also be considered in interpreting hardness data. 12 Rockwell scales are also frequently designated by an R with the appropriate scale letter as a subscript, for example, RC denotes the Rockwell C scale. 1496T_c06_131-173 11/16/05 17:06 Page 158 REVISED PAGES 158 • Chapter 6 / Mechanical Properties of Metals Brinell Hardness Tests 13 In Brinell tests, as in Rockwell measurements, a hard, spherical indenter is forced into the surface of the metal to be tested. The diameter of the hardened steel (or tungsten carbide) indenter is 10.00 mm (0.394 in.). Standard loads range between 500 and 3000 kg in 500-kg increments; during a test, the load is maintained constant for a specified time (between 10 and 30 s). Harder materials require greater applied loads. The Brinell hardness number, HB, is a function of both the magnitude of the load and the diameter of the resulting indentation (see Table 6.5).14 This diameter is measured with a special low-power microscope, utilizing a scale that is etched on the eyepiece. The measured diameter is then converted to the appropriate HB number using a chart; only one scale is employed with this technique. Semiautomatic techniques for measuring Brinell hardness are available. These employ optical scanning systems consisting of a digital camera mounted on a flexible probe, which allows positioning of the camera over the indentation. Data from the camera are transferred to a computer that analyzes the indentation, determines its size, and then calculates the Brinell hardness number. For this technique, surface finish requirements are normally more stringent that for manual measurements. Maximum specimen thickness as well as indentation position (relative to specimen edges) and minimum indentation spacing requirements are the same as for Rockwell tests. In addition, a well-defined indentation is required; this necessitates a smooth flat surface in which the indentation is made. Knoop and Vickers Microindentation Hardness Tests 15 Two other hardness-testing techniques are Knoop (pronounced nup) and Vickers (sometimes also called diamond pyramid). For each test a very small diamond indenter having pyramidal geometry is forced into the surface of the specimen. Applied loads are much smaller than for Rockwell and Brinell, ranging between 1 and 1000 g. The resulting impression is observed under a microscope and measured; this measurement is then converted into a hardness number (Table 6.5). Careful specimen surface preparation (grinding and polishing) may be necessary to ensure a well-defined indentation that may be accurately measured. The Knoop and Vickers hardness numbers are designated by HK and HV, respectively,16 and hardness scales for both techniques are approximately equivalent. Knoop and Vickers are referred to as microindentation-testing methods on the basis of indenter size. Both are well suited for measuring the hardness of small, selected specimen regions; furthermore, Knoop is used for testing brittle materials such as ceramics. The modern microindentation hardness-testing equipment has been automated by coupling the indenter apparatus to an image analyzer that incorporates a computer and software package. The software controls important system functions to include indent location, indent spacing, computation of hardness values, and plotting of data. 13 ASTM Standard E 10, “Standard Test Method for Brinell Hardness of Metallic Materials.” The Brinell hardness number is also represented by BHN. 15 ASTM Standard E 92, “Standard Test Method for Vickers Hardness of Metallic Materials,” and ASTM Standard E 384, “Standard Test for Microhardness of Materials.” 16 Sometimes KHN and VHN are used to denote Knoop and Vickers hardness numbers, respectively. 14 1496T_c06_131-173 12/21/05 7:43 Page 159 2nd REVISE PAGES 6.10 Hardness • 159 Other hardness-testing techniques are frequently employed but will not be discussed here; these include ultrasonic microhardness, dynamic (Scleroscope), durometer (for plastic and elastomeric materials), and scratch hardness tests. These are described in references provided at the end of the chapter. Hardness Conversion The facility to convert the hardness measured on one scale to that of another is most desirable. However, since hardness is not a well-defined material property, and because of the experimental dissimilarities among the various techniques, a comprehensive conversion scheme has not been devised. Hardness conversion data have been determined experimentally and found to be dependent on material type and characteristics. The most reliable conversion data exist for steels, some of which are presented in Figure 6.18 for Knoop, Brinell, and two Rockwell scales; the Mohs scale is also included. Detailed conversion tables for various other metals and Figure 6.18 Comparison of several hardness scales. (Adapted from G. F. Kinney, Engineering Properties and Applications of Plastics, p. 202. Copyright © 1957 by John Wiley & Sons, New York. Reprinted by permission of John Wiley & Sons, Inc.) 10,000 10 Diamond 5,000 2,000 Nitrided steels 1,000 80 1000 800 600 Cutting tools 60 110 40 400 200 100 100 20 200 100 80 0 60 Rockwell C 40 20 0 Knoop hardness 50 20 8 Topaz 7 Quartz 6 Orthoclase 5 Apatite 4 3 Fluorite Calcite 2 Gypsum 1 Talc File hard 500 300 9 Corundum or sapphire Rockwell B Easily machined steels Brasses and aluminum alloys Most plastics 10 5 Brinell hardness Mohs hardness 1496T_c06_131-173 11/16/05 17:06 Page 160 REVISED PAGES 160 • Chapter 6 / Mechanical Properties of Metals alloys are contained in ASTM Standard E 140, “Standard Hardness Conversion Tables for Metals.” In light of the preceding discussion, care should be exercised in extrapolation of conversion data from one alloy system to another. Correlation Between Hardness and Tensile Strength Both tensile strength and hardness are indicators of a metal’s resistance to plastic deformation. Consequently, they are roughly proportional, as shown in Figure 6.19, for tensile strength as a function of the HB for cast iron, steel, and brass. The same proportionality relationship does not hold for all metals, as Figure 6.19 indicates. As a rule of thumb for most steels, the HB and the tensile strength are related according to For steel alloys, conversion of Brinell hardness to tensile strength TS1MPa2  3.45  HB (6.20a) TS1psi2  500  HB (6.20b) Concept Check 6.4 Of those metals listed in Table 6.3, which is the hardest? Why? [The answer may be found at www.wiley.com/college/callister (Student Companion Site).] Rockwell hardness 60 70 80 90 100 HRB 20 30 40 50 HRC 250 1500 150 1000 100 500 Brass Cast iron (nodular) 50 0 0 100 200 300 Brinell hardness number 400 0 500 Tensile strength (103 psi) Tensile strength (MPa) 200 Steels Figure 6.19 Relationships between hardness and tensile strength for steel, brass, and cast iron. [Data taken from Metals Handbook: Properties and Selection: Irons and Steels, Vol. 1, 9th edition, B. Bardes (Editor), American Society for Metals, 1978, pp. 36 and 461; and Metals Handbook: Properties and Selection: Nonferrous Alloys and Pure Metals, Vol. 2, 9th edition, H. Baker (Managing Editor), American Society for Metals, 1979, p. 327.] 1496T_c06_131-173 11/16/05 17:06 Page 161 REVISED PAGES 6.11 Variability of Material Properties • 161 P ro p e r t y Va r i a b i l i t y a n d D e s i g n / S a f e t y Fa c t o r s 6.11 VARIABILITY OF MATERIAL PROPERTIES At this point it is worthwhile to discuss an issue that sometimes proves troublesome to many engineering students—namely, that measured material properties are not exact quantities. That is, even if we have a most precise measuring apparatus and a highly controlled test procedure, there will always be some scatter or variability in the data that are collected from specimens of the same material. For example, consider a number of identical tensile samples that are prepared from a single bar of some metal alloy, which samples are subsequently stress–strain tested in the same apparatus. We would most likely observe that each resulting stress–strain plot is slightly different from the others. This would lead to a variety of modulus of elasticity, yield strength, and tensile strength values. A number of factors lead to uncertainties in measured data. These include the test method, variations in specimen fabrication procedures, operator bias, and apparatus calibration. Furthermore, inhomogeneities may exist within the same lot of material, and/or slight compositional and other differences from lot to lot. Of course, appropriate measures should be taken to minimize the possibility of measurement error, and also to mitigate those factors that lead to data variability. It should also be mentioned that scatter exists for other measured material properties such as density, electrical conductivity, and coefficient of thermal expansion. It is important for the design engineer to realize that scatter and variability of materials properties are inevitable and must be dealt with appropriately. On occasion, data must be subjected to statistical treatments and probabilities determined. For example, instead of asking the question, “What is the fracture strength of this alloy?” the engineer should become accustomed to asking the question, “What is the probability of failure of this alloy under these given circumstances?” It is often desirable to specify a typical value and degree of dispersion (or scatter) for some measured property; such is commonly accomplished by taking the average and the standard deviation, respectively. Computation of Average and Standard Deviation Values An average value is obtained by dividing the sum of all measured values by the number of measurements taken. In mathematical terms, the average x of some parameter x is n a xi Computation of average value x i1 (6.21) n where n is the number of observations or measurements and xi is the value of a discrete measurement. Furthermore, the standard deviation s is determined using the following expression: n Computation of standard deviation s £ 2 a 1xi  x2 i1 n1 § 1 2 (6.22) 1496T_c06_131-173 11/16/05 17:06 Page 162 REVISED PAGES 162 • Chapter 6 / Mechanical Properties of Metals where xi, x, and n are defined above. A large value of the standard deviation corresponds to a high degree of scatter. EXAMPLE PROBLEM 6.6 Average and Standard Deviation Computations The following tensile strengths were measured for four specimens of the same steel alloy: Sample Number Tensile Strength (MPa) 1 2 3 4 520 512 515 522 (a) Compute the average tensile strength. (b) Determine the standard deviation. Solution (a) The average tensile strength 1TS2 is computed using Equation 6.21 with n  4: 4 TS  a 1TS2 i i1 4 520  512  515  522  4  517 MPa 525 525 Tensile strength (MPa) Tensile strength (MPa) 僓僒 TS + s 520 515 520 僓僒 TS 515 僓僒 TS – s 510 1 2 3 4 510 Sample number (a) (b) Figure 6.20 (a) Tensile strength data associated with Example Problem 6.6. (b) The manner in which these data could be plotted. The data point corresponds to the average value of the tensile strength 1TS2; error bars that indicate the degree of scatter correspond to the average value plus and minus the standard deviation 1TS  s2. 1496T_c06_131-173 11/16/05 17:06 Page 163 REVISED PAGES 6.12 Design/Safety Factors • 163 (b) For the standard deviation, using Equation 6.22, 4 s £ 2 a 51TS2 i  TS6 i1 41 § 1 2 1520  5172 2  1512  5172 2  1515  5172 2  1522  5172 2 1 2 d 41  4.6 MPa  c Figure 6.20 presents the tensile strength by specimen number for this example problem and also how the data may be represented in graphical form. The tensile strength data point (Figure 6.20b) corresponds to the average value TS, whereas scatter is depicted by error bars (short horizontal lines) situated above and below the data point symbol and connected to this symbol by vertical lines. The upper error bar is positioned at a value of the average value plus the standard deviation 1TS  s2, whereas the lower error bar corresponds to the average minus the standard deviation 1TS  s2. 6.12 DESIGN/SAFETY FACTORS design stress There will always be uncertainties in characterizing the magnitude of applied loads and their associated stress levels for in-service applications; ordinarily load calculations are only approximate. Furthermore, as noted in the previous section, virtually all engineering materials exhibit a variability in their measured mechanical properties. Consequently, design allowances must be made to protect against unanticipated failure. One way this may be accomplished is by establishing, for the particular application, a design stress, denoted as sd. For static situations and when ductile materials are used, sd is taken as the calculated stress level sc (on the basis of the estimated maximum load) multiplied by a design factor, N¿ ; that is, sd  N¿sc safe stress Computation of safe (or working) stress (6.23) where N¿ is greater than unity. Thus, the material to be used for the particular application is chosen so as to have a yield strength at least as high as this value of sd. Alternatively, a safe stress or working stress, sw, is used instead of design stress. This safe stress is based on the yield strength of the material and is defined as the yield strength divided by a factor of safety, N, or sw  sy N (6.24) Utilization of design stress (Equation 6.23) is usually preferred since it is based on the anticipated maximum applied stress instead of the yield strength of the material; normally there is a greater uncertainty in estimating this stress level than in the specification of the yield strength. However, in the discussion of this text, we are concerned with factors that influence the yield strengths of metal alloys and not in the determination of applied stresses; therefore, the succeeding discussion will deal with working stresses and factors of safety. 1496T_c06_131-173 12/21/05 7:43 Page 164 2nd REVISE PAGES 164 • Chapter 6 / Mechanical Properties of Metals Case Study: “Materials Selection for a Torsionally Stressed Cylindrical Shaft,” Chapter 22, which may be found at www.wiley.com/ college/callister (Student Companion Site). The choice of an appropriate value of N is necessary. If N is too large, then component overdesign will result; that is, either too much material or an alloy having a higher-than-necessary strength will be used. Values normally range between 1.2 and 4.0. Selection of N will depend on a number of factors, including economics, previous experience, the accuracy with which mechanical forces and material properties may be determined, and, most important, the consequences of failure in terms of loss of life and/or property damage. DESIGN EXAMPLE 6.1 Specification of Support Post Diameter A tensile-testing apparatus is to be constructed that must withstand a maximum load of 220,000 N (50,000 lbf). The design calls for two cylindrical support posts, each of which is to support half of the maximum load. Furthermore, plain-carbon (1045) steel ground and polished shafting rounds are to be used; the minimum yield and tensile strengths of this alloy are 310 MPa (45,000 psi) and 565 MPa (82,000 psi), respectively. Specify a suitable diameter for these support posts. Solution The first step in this design process is to decide on a factor of safety, N, which then allows determination of a working stress according to Equation 6.24. In addition, to ensure that the apparatus will be safe to operate, we also want to minimize any elastic deflection of the rods during testing; therefore, a relatively conservative factor of safety is to be used, say N  5. Thus, the working stress sw is just sy sw  N 310 MPa   62 MPa 19000 psi2 5 From the definition of stress, Equation 6.1, d 2 F A0  a b p  sw 2 where d is the rod diameter and F is the applied force; furthermore, each of the two rods must support half of the total force or 110,000 N (25,000 psi). Solving for d leads to F d2 B psw 2 110,000 N B p162  106 N/m2 2  4.75  102 m  47.5 mm 11.87 in.2 Therefore, the diameter of each of the two rods should be 47.5 mm or 1.87 in. 1496T_c06_131-173 11/16/05 17:06 Page 165 REVISED PAGES Summary • 165 SUMMARY Concepts of Stress and Strain Stress–Strain Behavior Elastic Properties of Materials True Stress and Strain A number of the important mechanical properties of materials, predominantly metals, have been discussed in this chapter. Concepts of stress and strain were first introduced. Stress is a measure of an applied mechanical load or force, normalized to take into account cross-sectional area. Two different stress parameters were defined—engineering stress and true stress. Strain represents the amount of deformation induced by a stress; both engineering and true strains are used. Some of the mechanical characteristics of metals can be ascertained by simple stress–strain tests. There are four test types: tension, compression, torsion, and shear. Tensile are the most common. A material that is stressed first undergoes elastic, or nonpermanent, deformation, wherein stress and strain are proportional. The constant of proportionality is the modulus of elasticity for tension and compression, and is the shear modulus when the stress is shear. Poisson’s ratio represents the negative ratio of transverse and longitudinal strains. Tensile Properties The phenomenon of yielding occurs at the onset of plastic or permanent deformation; yield strength is determined by a strain offset method from the stress–strain behavior, which is indicative of the stress at which plastic deformation begins. Tensile strength corresponds to the maximum tensile stress that may be sustained by a specimen, whereas percents elongation and reduction in area are measures of ductility—the amount of plastic deformation that has occurred at fracture. Resilience is the capacity of a material to absorb energy during elastic deformation; modulus of resilience is the area beneath the engineering stress–strain curve up to the yield point. Also, static toughness represents the energy absorbed during the fracture of a material, and is taken as the area under the entire engineering stress–strain curve. Ductile materials are normally tougher than brittle ones. Hardness Hardness is a measure of the resistance to localized plastic deformation. In several popular hardness-testing techniques (Rockwell, Brinell, Knoop, and Vickers) a small indenter is forced into the surface of the material, and an index number is determined on the basis of the size or depth of the resulting indentation. For many metals, hardness and tensile strength are approximately proportional to each other. Variability of Material Properties Measured mechanical properties (as well as other material properties) are not exact and precise quantities, in that there will always be some scatter for the measured data. Typical material property values are commonly specified in terms of averages, whereas magnitudes of scatter may be expressed as standard deviations. Design/Safety Factors As a result of uncertainties in both measured mechanical properties and inservice applied stresses, design or safe stresses are normally utilized for design purposes. For ductile materials, safe stress is the ratio of the yield strength and the factor of safety. 1496T_c06_131-173 11/16/05 17:06 Page 166 REVISED PAGES 166 • Chapter 6 / Mechanical Properties of Metals I M P O R TA N T T E R M S A N D C O N C E P T S Anelasticity Design stress Ductility Elastic deformation Elastic recovery Engineering strain Engineering stress Hardness Modulus of elasticity Plastic deformation Poisson’s ratio Proportional limit Resilience Safe stress Shear Tensile strength Toughness True strain True stress Yielding Yield strength REFERENCES ASM Handbook, Vol. 8, Mechanical Testing and Evaluation, ASM International, Materials Park, OH, 2000. Boyer, H. E. (Editor), Atlas of Stress–Strain Curves, 2nd edition,ASM International, Materials Park, OH, 2002. Chandler, H. (Editor), Hardness Testing, 2nd edition, ASM International, Materials Park, OH, 2000. Courtney, T. H., Mechanical Behavior of Materials, 2nd edition, McGraw-Hill Higher Education, Burr Ridge, IL, 2000. Davis, J. R. (Editor), Tensile Testing, 2nd edition, ASM International, Materials Park, OH, 2004. Dieter, G. E., Mechanical Metallurgy, 3rd edition, McGraw-Hill Book Company, New York, 1986. Dowling, N. E., Mechanical Behavior of Materials, 2nd edition, Prentice Hall PTR, Paramus, NJ, 1998. McClintock, F. A. and A. S. Argon, Mechanical Behavior of Materials, Addison-Wesley Publishing Co., Reading, MA, 1966. Reprinted by CBLS Publishers, Marietta, OH, 1993. Meyers, M. A. and K. K. Chawla, Mechanical Behavior of Materials, Prentice Hall PTR, Paramus, NJ, 1999. QUESTIONS AND PROBLEMS Concepts of Stress and Strain 6.1 Using mechanics of materials principles (i.e., equations of mechanical equilibrium applied to a free-body diagram), derive Equations 6.4a and 6.4b. 6.2 (a) Equations 6.4a and 6.4b are expressions for normal 1s¿2 and shear 1t¿2 stresses, respectively, as a function of the applied tensile stress 1s2 and the inclination angle of the plane on which these stresses are taken (u of Figure 6.4). Make a plot on which is presented the orientation parameters of these expressions (i.e., cos2 u and sin u cos u) versus u. (b) From this plot, at what angle of inclination is the normal stress a maximum? (c) Also, at what inclination angle is the shear stress a maximum? Stress–Strain Behavior 6.3 A specimen of copper having a rectangular cross section 15.2 mm  19.1 mm (0.60 in.  0.75 in.) is pulled in tension with 44,500 N (10,000 lbf) force, producing only elastic deformation. Calculate the resulting strain. 6.4 A cylindrical specimen of a nickel alloy having an elastic modulus of 207 GPa (30  106 psi) and an original diameter of 10.2 mm (0.40 in.) will experience only elastic deformation when a tensile load of 8900 N (2000 lbf) is applied. Compute the maximum length of the specimen before deformation if the maximum allowable elongation is 0.25 mm (0.010 in.). 6.5 An aluminum bar 125 mm (5.0 in.) long and having a square cross section 16.5 mm (0.65 in.) on an edge is pulled in tension with a load 1496T_c06_131-173 12/21/05 7:43 Page 167 2nd REVISE PAGES Questions and Problems • 167 6.8 6.9 Figure 6.21 Tensile stress–strain behavior for an alloy steel. tension. Determine its elongation when a load of 65,250 N (14,500 lbf ) is applied. 6.10 Figure 6.22 shows, for a gray cast iron, the tensile engineering stress–strain curve in the elastic region. Determine (a) the tangent modulus at 25 MPa (3625 psi), and (b) the secant modulus taken to 35 MPa (5000 psi). 6.11 As noted in Section 3.15, for single crystals of some substances, the physical properties are anisotropic; that is, they are dependent on crystallographic direction. One such property is the modulus of elasticity. For cubic single crystals, the modulus of elasticity in a general [uvw] direction, Euvw, is described by the relationship 1 1 1 1   3a  b Euvw EH100I EH100I EH111I 1a2b2  b2g2  g2a2 2 where EH100I and EH111I are the moduli of elasticity in [100] and [111] directions, respectively; a, b, and g are the cosines of the angles between [uvw] and the respective [100], [010], and [001] directions. Verify that the EH110I values for aluminum, copper, and iron in Table 3.3 are correct. 6.12 In Section 2.6 it was noted that the net bonding energy EN between two isolated positive and negative ions is a function of interionic distance r as follows: B A EN    n (6.25) r r 300 2000 Stress (MPa) 103 psi 300 MPa 2000 200 200 1000 1000 100 0 0 0.000 0.020 100 0 0.000 0.040 Strain 0.005 0.010 Strain 0.060 0.015 0.080 Stress (103 psi) 6.7 Stress 6.6 of 66,700 N (15,000 lbf), and experiences an elongation of 0.43 mm (1.7  102 in.). Assuming that the deformation is entirely elastic, calculate the modulus of elasticity of the aluminum. Consider a cylindrical nickel wire 2.0 mm (0.08 in.) in diameter and 3  104 mm (1200 in.) long. Calculate its elongation when a load of 300 N (67 lbf ) is applied. Assume that the deformation is totally elastic. For a brass alloy, the stress at which plastic deformation begins is 345 MPa (50,000 psi), and the modulus of elasticity is 103 GPa (15.0  106 psi). (a) What is the maximum load that may be applied to a specimen with a cross-sectional area of 130 mm2 (0.2 in.2) without plastic deformation? (b) If the original specimen length is 76 mm (3.0 in.), what is the maximum length to which it may be stretched without causing plastic deformation? A cylindrical rod of steel (E  207 GPa, 30  106 psi) having a yield strength of 310 MPa (45,000 psi) is to be subjected to a load of 11,100 N (2500 lbf ). If the length of the rod is 500 mm (20.0 in.), what must be the diameter to allow an elongation of 0.38 mm (0.015 in.)? Consider a cylindrical specimen of a steel alloy (Figure 6.21) 8.5 mm (0.33 in.) in diameter and 80 mm (3.15 in.) long that is pulled in 1496T_c06_131-173 11/16/05 17:06 Page 168 REVISED PAGES 168 • Chapter 6 / Mechanical Properties of Metals Figure 6.22 Tensile stress–strain behavior for a gray cast iron. 60 8 6 40 30 4 Stress (103 psi) Stress (MPa) 50 20 2 10 0 0 0.0002 0.0004 0.0006 0 0.0008 Strain where A, B, and n are constants for the particular ion pair. Equation 6.25 is also valid for the bonding energy between adjacent ions in solid materials. The modulus of elasticity E is proportional to the slope of the interionic force–separation curve at the equilibrium interionic separation; that is, E r a dF b dr r0 Derive an expression for the dependence of the modulus of elasticity on these A, B, and n parameters (for the two-ion system) using the following procedure: 1. Establish a relationship for the force F as a function of r, realizing that F dEN dr 2. Now take the derivative dFdr. 3. Develop an expression for r0, the equilibrium separation. Since r0 corresponds to the value of r at the minimum of the EN-versus-r curve (Figure 2.8b), take the derivative dENdr, set it equal to zero, and solve for r, which corresponds to r0. 4. Finally, substitute this expression for r0 into the relationship obtained by taking dF dr. 6.13 Using the solution to Problem 6.12, rank the magnitudes of the moduli of elasticity for the following hypothetical X, Y, and Z materials from the greatest to the least. The appropriate A, B, and n parameters (Equation 6.25) for these three materials are tabulated below; they yield EN in units of electron volts and r in nanometers: Material X Y Z A 1.5 2.0 3.5 B n 6 7.0  10 1.0  105 4.0  106 8 9 7 Elastic Properties of Materials 6.14 A cylindrical specimen of steel having a diameter of 15.2 mm (0.60 in.) and length of 250 mm (10.0 in.) is deformed elastically in tension with a force of 48,900 N (11,000 lbf ). Using the data contained in Table 6.1, determine the following: (a) The amount by which this specimen will elongate in the direction of the applied stress. (b) The change in diameter of the specimen. Will the diameter increase or decrease? 6.15 A cylindrical bar of aluminum 19 mm (0.75 in.) in diameter is to be deformed elastically by application of a force along the bar axis. Using the data in Table 6.1, determine the force that will produce an elastic reduction of 2.5  103 mm (1.0  104 in.) in the diameter. 6.16 A cylindrical specimen of some metal alloy 10 mm (0.4 in.) in diameter is stressed elastically in tension. A force of 15,000 N (3370 lbf ) produces a reduction in specimen diameter of 7  103 mm (2.8  104 in.). Compute Poisson’s ratio for this material if its elastic modulus is 100 GPa (14.5  106 psi). 1496T_c06_131-173 11/16/05 17:06 Page 169 REVISED PAGES Questions and Problems • 169 6.17 A cylindrical specimen of a hypothetical metal alloy is stressed in compression. If its original and final diameters are 30.00 and 30.04 mm, respectively, and its final length is 105.20 mm, compute its original length if the deformation is totally elastic. The elastic and shear moduli for this alloy are 65.5 and 25.4 GPa, respectively. 6.18 Consider a cylindrical specimen of some hypothetical metal alloy that has a diameter of 10.0 mm (0.39 in.). A tensile force of 1500 N (340 lbf ) produces an elastic reduction in diameter of 6.7  104 mm (2.64  105 in.). Compute the elastic modulus of this alloy, given that Poisson’s ratio is 0.35. 6.19 A brass alloy is known to have a yield strength of 240 MPa (35,000 psi), a tensile strength of 310 MPa (45,000 psi), and an elastic modulus of 110 GPa (16.0  106 psi). A cylindrical specimen of this alloy 15.2 mm (0.60 in.) in diameter and 380 mm (15.0 in.) long is stressed in tension and found to elongate 1.9 mm (0.075 in.). On the basis of the information given, is it possible to compute the magnitude of the load that is necessary to produce this change in length? If so, calculate the load. If not, explain why. 6.20 A cylindrical metal specimen 15.0 mm (0.59 in.) in diameter and 150 mm (5.9 in.) long is to be subjected to a tensile stress of 50 MPa (7250 psi); at this stress level the resulting deformation will be totally elastic. (a) If the elongation must be less than 0.072 mm (2.83  103 in.), which of the metals in Table 6.1 are suitable candidates? Why? (b) If, in addition, the maximum permissible diameter decrease is 2.3  103 mm (9.1  105 in.) when the tensile stress of 50 MPa is applied, which of the metals that satisfy the criterion in part (a) are suitable candidates? Why? 6.21 Consider the brass alloy for which the stress–strain behavior is shown in Figure 6.12. A cylindrical specimen of this material 10.0 mm (0.39 in.) in diameter and 101.6 mm (4.0 in.) long is pulled in tension with a force of 10,000 N (2250 lbf ). If it is known that this alloy has a value for Poisson’s ratio of 0.35, compute (a) the specimen elongation, and (b) the reduction in specimen diameter. 6.22 A cylindrical rod 120 mm long and having a diameter of 15.0 mm is to be deformed using a tensile load of 35,000 N. It must not experience either plastic deformation or a diameter reduction of more than 1.2  102 mm. Of the materials listed below, which are possible candidates? Justify your choice(s). Material Aluminum alloy Titanium alloy Steel alloy Magnesium alloy Modulus of Elasticity (GPa) Yield Strength (MPa) Poisson’s Ratio 70 105 205 45 250 850 550 170 0.33 0.36 0.27 0.35 6.23 A cylindrical rod 500 mm (20.0 in.) long, having a diameter of 12.7 mm (0.50 in.), is to be subjected to a tensile load. If the rod is to experience neither plastic deformation nor an elongation of more than 1.3 mm (0.05 in.) when the applied load is 29,000 N (6500 lbf ), which of the four metals or alloys listed below are possible candidates? Justify your choice(s). Material Aluminum alloy Brass alloy Copper Steel alloy Modulus of Elasticity (GPa) Yield Strength (MPa) Tensile Strength (MPa) 70 100 110 207 255 345 210 450 420 420 275 550 Tensile Properties 6.24 Figure 6.21 shows the tensile engineering stress–strain behavior for a steel alloy. (a) What is the modulus of elasticity? (b) What is the proportional limit? (c) What is the yield strength at a strain offset of 0.002? (d) What is the tensile strength? 6.25 A cylindrical specimen of a brass alloy having a length of 100 mm (4 in.) must elongate only 5 mm (0.2 in.) when a tensile load of 100,000 N (22,500 lbf) is applied. Under these circumstances what must be the radius of the specimen? Consider this brass alloy to have the stress–strain behavior shown in Figure 6.12. 1496T_c06_131-173 11/16/05 17:06 Page 170 REVISED PAGES 170 • Chapter 6 / Mechanical Properties of Metals 6.26 A load of 140,000 N (31,500 lbf) is applied to a cylindrical specimen of a steel alloy (displaying the stress–strain behavior shown in Figure 6.21) that has a cross-sectional diameter of 10 mm (0.40 in.). (a) Will the specimen experience elastic and/ or plastic deformation? Why? (b) If the original specimen length is 500 mm (20 in.), how much will it increase in length when this load is applied? 6.27 A bar of a steel alloy that exhibits the stress–strain behavior shown in Figure 6.21 is subjected to a tensile load; the specimen is 375 mm (14.8 in.) long and of square cross section 5.5 mm (0.22 in.) on a side. (a) Compute the magnitude of the load necessary to produce an elongation of 2.25 mm (0.088 in.). (b) What will be the deformation after the load has been released? 6.28 A cylindrical specimen of stainless steel having a diameter of 12.8 mm (0.505 in.) and a gauge length of 50.800 mm (2.000 in.) is pulled in tension. Use the load–elongation characteristics tabulated below to complete parts (a) through (f). Load N 0 12,700 25,400 38,100 50,800 76,200 89,100 92,700 102,500 107,800 119,400 128,300 149,700 159,000 160,400 159,500 151,500 124,700 Length lbf 0 2,850 5,710 8,560 11,400 17,100 20,000 20,800 23,000 24,200 26,800 28,800 33,650 35,750 36,000 35,850 34,050 28,000 Fracture mm in. 50.800 50.825 50.851 50.876 50.902 50.952 51.003 51.054 51.181 51.308 51.562 51.816 52.832 53.848 54.356 54.864 55.880 56.642 2.000 2.001 2.002 2.003 2.004 2.006 2.008 2.010 2.015 2.020 2.030 2.040 2.080 2.120 2.140 2.160 2.200 2.230 (a) Plot the data as engineering stress versus engineering strain. (b) Compute the modulus of elasticity. (c) Determine the yield strength at a strain offset of 0.002. (d) Determine the tensile strength of this alloy. (e) What is the approximate ductility, in percent elongation? (f) Compute the modulus of resilience. 6.29 A specimen of magnesium having a rectangular cross section of dimensions 3.2 mm  19.1 mm (18 in.  34 in.) is deformed in tension. Using the load–elongation data tabulated as follows, complete parts (a) through (f). Load Length lbf N 0 310 625 1265 1670 1830 2220 2890 3170 3225 3110 2810 0 1380 2780 5630 7430 8140 9870 12,850 14,100 14,340 13,830 12,500 in. 2.500 2.501 2.502 2.505 2.508 2.510 2.525 2.575 2.625 2.675 2.725 2.775 Fracture mm 63.50 63.53 63.56 63.62 63.70 63.75 64.14 65.41 66.68 67.95 69.22 70.49 (a) Plot the data as engineering stress versus engineering strain. (b) Compute the modulus of elasticity. (c) Determine the yield strength at a strain offset of 0.002. (d) Determine the tensile strength of this alloy. (e) Compute the modulus of resilience. (f) What is the ductility, in percent elongation? 6.30 A cylindrical metal specimen having an original diameter of 12.8 mm (0.505 in.) and gauge length of 50.80 mm (2.000 in.) is pulled in tension until fracture occurs. The diameter at the point of fracture is 8.13 mm (0.320 in.), and the fractured gauge length is 74.17 mm (2.920 in.). Calculate the ductility in terms of percent reduction in area and percent elongation. 6.31 Calculate the moduli of resilience for the materials having the stress–strain behaviors shown in Figures 6.12 and 6.21. 1496T_c06_131-173 11/16/05 17:06 Page 171 REVISED PAGES Questions and Problems • 171 6.32 Determine the modulus of resilience for each of the following alloys: Yield Strength Material MPa psi 830 380 275 690 120,000 55,000 40,000 100,000 Steel alloy Brass alloy Aluminum alloy Titanium alloy Use modulus of elasticity values in Table 6.1. 6.33 A steel alloy to be used for a spring application must have a modulus of resilience of at least 2.07 MPa (300 psi). What must be its minimum yield strength? True Stress and Strain 6.34 Show that Equations 6.18a and 6.18b are valid when there is no volume change during deformation. 6.35 Demonstrate that Equation 6.16, the expression defining true strain, may also be represented by A0 T  ln a b Ai when specimen volume remains constant during deformation. Which of these two expressions is more valid during necking? Why? 6.36 Using the data in Problem 6.28 and Equations 6.15, 6.16, and 6.18a, generate a true stress–true strain plot for stainless steel. Equation 6.18a becomes invalid past the point at which necking begins; therefore, measured diameters are given below for the last three data points, which should be used in true stress computations. Load Length Diameter N lbf mm in. mm in. 159,500 151,500 124,700 35,850 34,050 28,000 54.864 55.880 56.642 2.160 2.200 2.230 12.22 11.80 10.65 0.481 0.464 0.419 6.37 A tensile test is performed on a metal specimen, and it is found that a true plastic strain of 0.16 is produced when a true stress of 500 MPa (72,500 psi) is applied; for the same metal, the value of K in Equation 6.19 is 825 MPa (120,000 psi). Calculate the true strain that results from the application of a true stress of 600 MPa (87,000 psi). 6.38 For some metal alloy, a true stress of 345 MPa (50,000 psi) produces a plastic true strain of 0.02. How much will a specimen of this material elongate when a true stress of 415 MPa (60,000 psi) is applied if the original length is 500 mm (20 in.)? Assume a value of 0.22 for the strain-hardening exponent, n. 6.39 The following true stresses produce the corresponding true plastic strains for a brass alloy: True Stress ( psi) True Strain 60,000 70,000 0.15 0.25 What true stress is necessary to produce a true plastic strain of 0.21? 6.40 For a brass alloy, the following engineering stresses produce the corresponding plastic engineering strains, prior to necking: Engineering Stress (MPa) Engineering Strain 315 340 0.105 0.220 On the basis of this information, compute the engineering stress necessary to produce an engineering strain of 0.28. 6.41 Find the toughness (or energy to cause fracture) for a metal that experiences both elastic and plastic deformation. Assume Equation 6.5 for elastic deformation, that the modulus of elasticity is 103 GPa (15  106 psi), and that elastic deformation terminates at a strain of 0.007. For plastic deformation, assume that the relationship between stress and strain is described by Equation 6.19, in which the values for K and n are 1520 MPa (221,000 psi) and 0.15, respectively. Furthermore, plastic deformation occurs between strain values of 0.007 and 0.60, at which point fracture occurs. 6.42 For a tensile test, it can be demonstrated that necking begins when dsT  sT dT (6.26) 1496T_c06_131-173 11/16/05 17:06 Page 172 REVISED PAGES 172 • Chapter 6 / Mechanical Properties of Metals Using Equation 6.19, determine the value of the true strain at this onset of necking. 6.43 Taking the logarithm of both sides of Equation 6.19 yields log sT  log K  n log T (6.27) Thus, a plot of log sT versus log T in the plastic region to the point of necking should yield a straight line having a slope of n and an intercept (at log sT  0) of log K. Using the appropriate data tabulated in Problem 6.28, make a plot of log sT versus log T and determine the values of n and K. It will be necessary to convert engineering stresses and strains to true stresses and strains using Equations 6.18a and 6.18b. Elastic Recovery After Plastic Deformation 6.44 A cylindrical specimen of a brass alloy 10.0 mm (0.39 in.) in diameter and 120.0 mm (4.72 in.) long is pulled in tension with a force of 11,750 N (2640 lbf); the force is subsequently released. (a) Compute the final length of the specimen at this time. The tensile stress–strain behavior for this alloy is shown in Figure 6.12. (b) Compute the final specimen length when the load is increased to 23,500 N (5280 lbf) and then released. 6.45 A steel alloy specimen having a rectangular cross section of dimensions 19 mm  3.2 mm (34 in.  18 in.) has the stress–strain behavior shown in Figure 6.21. If this specimen is subjected to a tensile force of 110,000 N (25,000 lbf) then (a) Determine the elastic and plastic strain values. (b) If its original length is 610 mm (24.0 in.), what will be its final length after the load in part (a) is applied and then released? Hardness 6.46 (a) A 10-mm-diameter Brinell hardness indenter produced an indentation 2.50 mm in diameter in a steel alloy when a load of 1000 kg was used. Compute the HB of this material. (b) What will be the diameter of an indentation to yield a hardness of 300 HB when a 500-kg load is used? 6.47 Estimate the Brinell and Rockwell hardnesses for the following: (a) The naval brass for which the stress–strain behavior is shown in Figure 6.12. (b) The steel alloy for which the stress–strain behavior is shown in Figure 6.21. 6.48 Using the data represented in Figure 6.19, specify equations relating tensile strength and Brinell hardness for brass and nodular cast iron, similar to Equations 6.20a and 6.20b for steels. Variability of Material Properties 6.49 Cite five factors that lead to scatter in measured material properties. 6.50 Below are tabulated a number of Rockwell G hardness values that were measured on a single steel specimen. Compute average and standard deviation hardness values. 47.3 52.1 45.6 49.9 47.6 50.4 48.7 50.0 46.2 48.3 51.1 46.7 47.1 50.4 45.9 46.4 48.5 49.7 Design/Safety Factors 6.51 Upon what three criteria are factors of safety based? 6.52 Determine working stresses for the two alloys that have the stress–strain behaviors shown in Figures 6.12 and 6.21. DESIGN PROBLEMS 6.D1 A large tower is to be supported by a series of steel wires; it is estimated that the load on each wire will be 13,300 N (3000 lbf). Determine the minimum required wire diameter, assuming a factor of safety of 2 and a yield strength of 860 MPa (125,000 psi) for the steel. 6.D2 (a) Gaseous hydrogen at a constant pressure of 0.658 MPa (5 atm) is to flow within the inside of a thin-walled cylindrical tube of nickel that has a radius of 0.125 m. The temperature of the tube is to be 350C and the pressure of hydrogen outside of the tube will 1496T_c06_131-173 11/16/05 17:06 Page 173 REVISED PAGES Design Problems • 173 be maintained at 0.0127 MPa (0.125 atm). Calculate the minimum wall thickness if the diffusion flux is to be no greater than 1.25  107 mol/m2-s. The concentration of hydrogen in the nickel, CH (in moles hydrogen per m3 of Ni) is a function of hydrogen pressure, PH2 (in MPa) and absolute temperature (T) according to 12.3 kJ/mol CH  30.81pH2 exp a b (6.28) RT Furthermore, the diffusion coefficient for the diffusion of H in Ni depends on temperature as 39.56 kJ/mol DH 1m2/s2  4.76  107 exp a b RT (6.29) (b) For thin-walled cylindrical tubes that are pressurized, the circumferential stress is a function of the pressure difference across the wall ( ¢p), cylinder radius (r), and tube thickness ( ¢x) as r¢p s (6.30) 4¢x Compute the circumferential stress to which the walls of this pressurized cylinder are exposed. (c) The room-temperature yield strength of Ni is 100 MPa (15,000 psi) and, furthermore, sy diminishes about 5 MPa for every 50C rise in temperature. Would you expect the wall thickness computed in part (b) to be suitable for this Ni cylinder at 350C? Why or why not? (d) If this thickness is found to be suitable, compute the minimum thickness that could be used without any deformation of the tube walls. How much would the diffusion flux increase with this reduction in thickness? On the other hand, if the thickness determined in part (c) is found to be unsuitable, then specify a minimum thickness that you would use. In this case, how much of a diminishment in diffusion flux would result? 6.D3 Consider the steady-state diffusion of hydrogen through the walls of a cylindrical nickel tube as described in Problem 6.D2. One design calls for a diffusion flux of 2.5  108 mol/m2-s, a tube radius of 0.100 m, and inside and outside pressures of 1.015 MPa (10 atm) and 0.01015 MPa (0.1 atm), respectively; the maximum allowable temperature is 300C. Specify a suitable temperature and wall thickness to give this diffusion flux and yet ensure that the tube walls will not experience any permanent deformation. 1496T_c07_174-206 12/21/05 8:27 Page 174 Chapter 7 2nd REVISE PAGES Dislocations and Strengthening Mechanisms I n this photomicrograph of a lithium fluoride (LiF) single crystal, the small pyramidal pits repre- sent those positions at which dislocations intersect the surface. The surface was polished and then chemically treated; these “etch pits” result from localized chemical attack around the dislocations and indicate the distribution of the dislocations. 750. (Photomicrograph courtesy of W. G. Johnston, General Electric Co.) WHY STUDY Dislocations and Strengthening Mechanisms? With a knowledge of the nature of dislocations and the role they play in the plastic deformation process, we are able to understand the underlying mechanisms of the techniques that are used to strengthen and 174 • harden metals and their alloys. Thus, it becomes possible to design and tailor the mechanical properties of materials—for example, the strength or toughness of a metal–matrix composite. 1496T_c07_174-206 12/21/05 7:48 Page 175 2nd REVISE PAGES Learning Objectives After studying this chapter you should be able to do the following: 1. Describe edge and screw dislocation motion 6. Describe and explain solid-solution strengthenfrom an atomic perspective. ing for substitutional impurity atoms in terms of 2. Describe how plastic deformation occurs by the lattice strain interactions with dislocations. motion of edge and screw dislocations in re7. Describe and explain the phenomenon of strain sponse to applied shear stresses. hardening (or cold working) in terms of disloca3. Define slip system and cite one example. tions and strain field interactions. 4. Describe how the grain structure of a polycrys8. Describe recrystallization in terms of both the talline metal is altered when it is plastically alteration of microstructure and mechanical deformed. characteristics of the material. 5. Explain how grain boundaries impede dislocation 9. Describe the phenomenon of grain growth from motion and why a metal having small grains is both macroscopic and atomic perspectives. stronger than one having large grains. 7.1 INTRODUCTION Chapter 6 explained that materials may experience two kinds of deformation: elastic and plastic. Plastic deformation is permanent, and strength and hardness are measures of a material’s resistance to this deformation. On a microscopic scale, plastic deformation corresponds to the net movement of large numbers of atoms in response to an applied stress. During this process, interatomic bonds must be ruptured and then reformed. In crystalline solids, plastic deformation most often involves the motion of dislocations, linear crystalline defects that were introduced in Section 4.5. This chapter discusses the characteristics of dislocations and their involvement in plastic deformation. Twinning, another process by which some metals plastically deform, is also treated. In addition, and probably most importantly, several techniques are presented for strengthening single-phase metals, the mechanisms of which are described in terms of dislocations. Finally, the latter sections of this chapter are concerned with recovery and recrystallization—processes that occur in plastically deformed metals, normally at elevated temperatures—and, in addition, grain growth. Dislocations and Plastic Deformation Early materials studies led to the computation of the theoretical strengths of perfect crystals, which were many times greater than those actually measured. During the 1930s it was theorized that this discrepancy in mechanical strengths could be explained by a type of linear crystalline defect that has since come to be known as a dislocation. It was not until the 1950s, however, that the existence of such dislocation defects was established by direct observation with the electron microscope. Since then, a theory of dislocations has evolved that explains many of the physical and mechanical phenomena in metals [as well as crystalline ceramics (Section 12.10)]. 7.2 BASIC CONCEPTS Edge and screw are the two fundamental dislocation types. In an edge dislocation, localized lattice distortion exists along the end of an extra half-plane of atoms, which also defines the dislocation line (Figure 4.3). A screw dislocation may be thought of 1496T_c07_174-206 12/21/05 7:48 Page 176 2nd REVISE PAGES 176 • Chapter 7 / Dislocations and Strengthening Mechanisms Edge slip as resulting from shear distortion; its dislocation line passes through the center of a spiral, atomic plane ramp (Figure 4.4). Many dislocations in crystalline materials have both edge and screw components; these are mixed dislocations (Figure 4.5). Plastic deformation corresponds to the motion of large numbers of dislocations. An edge dislocation moves in response to a shear stress applied in a direction perpendicular to its line; the mechanics of dislocation motion are represented in Figure 7.1. Let the initial extra half-plane of atoms be plane A. When the shear stress is applied as indicated (Figure 7.1a), plane A is forced to the right; this in turn pushes the top halves of planes B, C, D, and so on, in the same direction. If the applied shear stress is of sufficient magnitude, the interatomic bonds of plane B are severed along the shear plane, and the upper half of plane B becomes the extra halfplane as plane A links up with the bottom half of plane B (Figure 7.1b). This process is subsequently repeated for the other planes, such that the extra half-plane, by discrete steps, moves from left to right by successive and repeated breaking of bonds and shifting by interatomic distances of upper half-planes. Before and after the movement of a dislocation through some particular region of the crystal, the atomic arrangement is ordered and perfect; it is only during the passage of the extra halfplane that the lattice structure is disrupted. Ultimately this extra half-plane may emerge from the right surface of the crystal, forming an edge that is one atomic distance wide; this is shown in Figure 7.1c. The process by which plastic deformation is produced by dislocation motion is termed slip; the crystallographic plane along which the dislocation line traverses is the slip plane, as indicated in Figure 7.1. Macroscopic plastic deformation simply corresponds to permanent deformation that results from the movement of dislocations, or slip, in response to an applied shear stress, as represented in Figure 7.2a. Dislocation motion is analogous to the mode of locomotion employed by a caterpillar (Figure 7.3). The caterpillar forms a hump near its posterior end by pulling in its last pair of legs a unit leg distance. The hump is propelled forward by repeated lifting and shifting of leg pairs. When the hump reaches the anterior end, Shear stress Shear stress A B C D Shear stress A B C D A B Slip plane C D Unit step of slip Edge dislocation line (a) ( b) (c) Figure 7.1 Atomic rearrangements that accompany the motion of an edge dislocation as it moves in response to an applied shear stress. (a) The extra half-plane of atoms is labeled A. (b) The dislocation moves one atomic distance to the right as A links up to the lower portion of plane B; in the process, the upper portion of B becomes the extra half-plane. (c) A step forms on the surface of the crystal as the extra half-plane exits. (Adapted from A. G. Guy, Essentials of Materials Science, McGraw-Hill Book Company, New York, 1976, p. 153.) 1496T_c07_174-206 12/21/05 7:48 Page 177 2nd REVISE PAGES 7.2 Basic Concepts • 177 Figure 7.2 The formation of a step on the surface of a crystal by the motion of (a) an edge dislocation and (b) a screw dislocation. Note that for an edge, the dislocation line moves in the direction of the applied shear stress t; for a screw, the dislocation line motion is perpendicular to the stress direction. (Adapted from H. W. Hayden, W. G. Moffatt, and J. Wulff, The Structure and Properties of Materials,Vol. III, Mechanical Behavior, p. 70. Copyright © 1965 by John Wiley & Sons, New York. Reprinted by permission of John Wiley & Sons, Inc.) ␶ Direction of motion ␶ (a) ␶ Direction of motion ␶ (b) Screw Mixed dislocation density the entire caterpillar has moved forward by the leg separation distance. The caterpillar hump and its motion correspond to the extra half-plane of atoms in the dislocation model of plastic deformation. The motion of a screw dislocation in response to the applied shear stress is shown in Figure 7.2b; the direction of movement is perpendicular to the stress direction. For an edge, motion is parallel to the shear stress. However, the net plastic deformation for the motion of both dislocation types is the same (see Figure 7.2). The direction of motion of the mixed dislocation line is neither perpendicular nor parallel to the applied stress, but lies somewhere in between. All metals and alloys contain some dislocations that were introduced during solidification, during plastic deformation, and as a consequence of thermal stresses that result from rapid cooling. The number of dislocations, or dislocation density in a material, is expressed as the total dislocation length per unit volume or, equivalently, the number of dislocations that intersect a unit area of a random section. The units of dislocation density are millimeters of dislocation per cubic millimeter or just per square millimeter. Dislocation densities as low as 103 mm2 are typically found in carefully solidified metal crystals. For heavily deformed metals, the density may run as high as 109 to 1010 mm2. Heat treating a deformed metal specimen Figure 7.3 Representation of the analogy between caterpillar and dislocation motion. 1496T_c07_174-206 11/17/05 13:32 Page 178 REVISED PAGES 178 • Chapter 7 / Dislocations and Strengthening Mechanisms can diminish the density to on the order of 105 to 106 mm2. By way of contrast, a typical dislocation density for ceramic materials is between 102 and 104 mm2; also, for silicon single crystals used in integrated circuits the value normally lies between 0.1 and 1 mm2. 7.3 CHARACTERISTICS OF DISLOCATIONS lattice strain Several characteristics of dislocations are important with regard to the mechanical properties of metals. These include strain fields that exist around dislocations, which are influential in determining the mobility of the dislocations, as well as their ability to multiply. When metals are plastically deformed, some fraction of the deformation energy (approximately 5%) is retained internally; the remainder is dissipated as heat. The major portion of this stored energy is as strain energy associated with dislocations. Consider the edge dislocation represented in Figure 7.4.As already mentioned, some atomic lattice distortion exists around the dislocation line because of the presence of the extra half-plane of atoms. As a consequence, there are regions in which compressive, tensile, and shear lattice strains are imposed on the neighboring atoms. For example, atoms immediately above and adjacent to the dislocation line are squeezed together. As a result, these atoms may be thought of as experiencing a compressive strain relative to atoms positioned in the perfect crystal and far removed from the dislocation; this is illustrated in Figure 7.4. Directly below the half-plane, the effect is just the opposite; lattice atoms sustain an imposed tensile strain, which is as shown. Shear strains also exist in the vicinity of the edge dislocation. For a screw dislocation, lattice strains are pure shear only. These lattice distortions may be considered to be strain fields that radiate from the dislocation line. The strains extend into the surrounding atoms, and their magnitude decreases with radial distance from the dislocation. The strain fields surrounding dislocations in close proximity to one another may interact such that forces are imposed on each dislocation by the combined interactions of all its neighboring dislocations. For example, consider two edge dislocations that have the same sign and the identical slip plane, as represented in Figure 7.5a. The compressive and tensile strain fields for both lie on the same side of the slip plane; the strain field interaction is such that there exists between these two isolated dislocations a mutual repulsive force that tends to move them apart. On the other hand, two dislocations of opposite sign and having the same slip plane will be attracted to one another, as indicated in Figure 7.5b, and dislocation Compression Tension Figure 7.4 Regions of compression (green) and tension (yellow) located around an edge dislocation. (Adapted from W. G. Moffatt, G. W. Pearsall, and J. Wulff, The Structure and Properties of Materials, Vol. I, Structure, p. 85. Copyright © 1964 by John Wiley & Sons, New York. Reprinted by permission of John Wiley & Sons, Inc.) 1496T_c07_174-206 11/17/05 13:32 Page 179 REVISED PAGES 7.4 Slip Systems • 179 Figure 7.5 (a) Two edge dislocations of the same sign and lying on the same slip plane exert a repulsive force on each other; C and T denote compression and tensile regions, respectively. (b) Edge dislocations of opposite sign and lying on the same slip plane exert an attractive force on each other. Upon meeting, they annihilate each other and leave a region of perfect crystal. (Adapted from H. W. Hayden, W. G. Moffatt, and J. Wulff, The Structure and Properties of Materials, Vol. III, Mechanical Behavior, p. 75. Copyright © 1965 by John Wiley & Sons, New York. Reprinted by permission of John Wiley & Sons.) C C Repulsion T C T (a) T Dislocation annihilation Attraction ; + = (Perfect crystal) T C (b) annihilation will occur when they meet. That is, the two extra half-planes of atoms will align and become a complete plane. Dislocation interactions are possible between edge, screw, and/or mixed dislocations, and for a variety of orientations.These strain fields and associated forces are important in the strengthening mechanisms for metals. During plastic deformation, the number of dislocations increases dramatically. We know that the dislocation density in a metal that has been highly deformed may be as high as 1010 mm2. One important source of these new dislocations is existing dislocations, which multiply; furthermore, grain boundaries, as well as internal defects and surface irregularities such as scratches and nicks, which act as stress concentrations, may serve as dislocation formation sites during deformation. 7.4 SLIP SYSTEMS slip system Dislocations do not move with the same degree of ease on all crystallographic planes of atoms and in all crystallographic directions. Ordinarily there is a preferred plane, and in that plane there are specific directions along which dislocation motion occurs. This plane is called the slip plane; it follows that the direction of movement is called the slip direction. This combination of the slip plane and the slip direction is termed the slip system. The slip system depends on the crystal structure of the metal and is such that the atomic distortion that accompanies the motion of a dislocation is a minimum. For a particular crystal structure, the slip plane is the plane that has the most dense atomic packing—that is, has the greatest planar density. The slip direction corresponds to the direction, in this plane, that is most closely packed with atoms—that is, has the highest linear density. Planar and linear atomic densities were discussed in Section 3.11. Consider, for example, the FCC crystal structure, a unit cell of which is shown in Figure 7.6a.There is a set of planes, the {111} family, all of which are closely packed. A (111)-type plane is indicated in the unit cell; in Figure 7.6b, this plane is positioned within the plane of the page, in which atoms are now represented as touching nearest neighbors. 1496T_c07_174-206 12/21/05 7:48 Page 180 2nd REVISE PAGES 180 • Chapter 7 / Dislocations and Strengthening Mechanisms A Figure 7.6 (a) A {111} H110I slip system shown within an FCC unit cell. (b) The (111) plane from (a) and three H110I slip directions (as indicated by arrows) within that plane comprise possible slip systems. A C B B C F E D E F D (a) (b) Slip occurs along H110I-type directions within the {111} planes, as indicated by arrows in Figure 7.6. Hence, {111}H110I represents the slip plane and direction combination, or the slip system for FCC. Figure 7.6b demonstrates that a given slip plane may contain more than a single slip direction. Thus, several slip systems may exist for a particular crystal structure; the number of independent slip systems represents the different possible combinations of slip planes and directions. For example, for face-centered cubic, there are 12 slip systems: four unique {111} planes and, within each plane, three independent H110I directions. The possible slip systems for BCC and HCP crystal structures are listed in Table 7.1. For each of these structures, slip is possible on more than one family of planes (e.g., {110}, {211}, and {321} for BCC). For metals having these two crystal structures, some slip systems are often operable only at elevated temperatures. Metals with FCC or BCC crystal structures have a relatively large number of slip systems (at least 12). These metals are quite ductile because extensive plastic deformation is normally possible along the various systems. Conversely, HCP metals, having few active slip systems, are normally quite brittle. The Burgers vector concept was introduced in Section 4.5, and denoted by a b for edge, screw, and mixed dislocations in Figures 4.3, 4.4, and 4.5, respectively. With regard to the process of slip, a Burgers vector’s direction corresponds to a dislocation’s slip direction, whereas its magnitude is equal to the unit slip distance (or interatomic separation in this direction). Of course, both the direction and the magnitude of b will depend on crystal structure, and it is convenient to specify a Burgers vector in terms of unit cell edge length (a) and crystallographic direction indices. Table 7.1 Slip Systems for Face-Centered Cubic, Body-Centered Cubic, and Hexagonal Close-Packed Metals Metals Cu, Al, Ni, Ag, Au a-Fe, W, Mo a-Fe, W a-Fe, K Cd, Zn, Mg, Ti, Be Ti, Mg, Zr Ti, Mg Slip Plane Face-Centered Cubic 51116 Body-Centered Cubic 51106 52116 53216 Hexagonal Close-Packed 500016 510106 510116 Slip Direction Number of Slip Systems H110I 12 H111I H111I H111I 12 12 24 H1120I H1120I H1120I 3 3 6 1496T_c07_174-206 11/17/05 13:32 Page 181 REVISED PAGES 7.5 Slip in Single Crystals • 181 Burgers vectors for face-centered cubic, body-centered cubic, and hexagonal closepacked crystal structures are given as follows: a H110I 2 a b1BCC2  H111I 2 a b1HCP2  H1120I 3 b1FCC2  (7.1a) (7.1b) (7.1c) Concept Check 7.1 Which of the following is the slip system for the simple cubic crystal structure? Why? 51006H110I 51106H110I 51006H010I 51106H111I (Note: a unit cell for the simple cubic crystal structure is shown in Figure 3.23.) [The answer may be found at www.wiley.com/college/callister (Student Companion Site).] 7.5 SLIP IN SINGLE CRYSTALS resolved shear stress Resolved shear stress—dependence on applied stress and orientation of stress direction relative to slip plane normal and slip direction A further explanation of slip is simplified by treating the process in single crystals, then making the appropriate extension to polycrystalline materials. As mentioned previously, edge, screw, and mixed dislocations move in response to shear stresses applied along a slip plane and in a slip direction. As noted in Section 6.2, even though an applied stress may be pure tensile (or compressive), shear components exist at all but parallel or perpendicular alignments to the stress direction (Equation 6.4b).These are termed resolved shear stresses, and their magnitudes depend not only on the applied stress, but also on the orientation of both the slip plane and direction within that plane. Let f represent the angle between the normal to the slip plane and the applied stress direction, and l the angle between the slip and stress directions, as indicated in Figure 7.7; it can then be shown that for the resolved shear stress tR tR  s cos f cos l where s is the applied stress. In general, f  l  90, since it need not be the case that the tensile axis, the slip plane normal, and the slip direction all lie in the same plane. A metal single crystal has a number of different slip systems that are capable of operating. The resolved shear stress normally differs for each one because the orientation of each relative to the stress axis (f and l angles) also differs. However, one slip system is generally oriented most favorably—that is, has the largest resolved shear stress, tR(max): tR 1max2  s1cos f cos l2 max critical resolved shear stress (7.2) (7.3) In response to an applied tensile or compressive stress, slip in a single crystal commences on the most favorably oriented slip system when the resolved shear stress reaches some critical value, termed the critical resolved shear stress tcrss; it represents the minimum shear stress required to initiate slip, and is a property of the 1496T_c07_174-206 11/17/05 13:32 Page 182 REVISED PAGES 182 • Chapter 7 / Dislocations and Strengthening Mechanisms Figure 7.7 Geometrical relationships between the tensile axis, slip plane, and slip direction used in calculating the resolved shear stress for a single crystal. F ␾ A ␭ Normal to slip plane Slip direction F material that determines when yielding occurs. The single crystal plastically deforms or yields when tR(max)  tcrss, and the magnitude of the applied stress required to initiate yielding (i.e., the yield strength sy) is Yield strength of a single crystal— dependence on the critical resolved shear stress and the orientation of most favorably oriented slip system sy  tcrss 1cos f cos l2 max (7.4) The minimum stress necessary to introduce yielding occurs when a single crystal is oriented such that f  l  45; under these conditions, sy  2tcrss (7.5) For a single-crystal specimen that is stressed in tension, deformation will be as in Figure 7.8, where slip occurs along a number of equivalent and most favorably oriented planes and directions at various positions along the specimen length. This slip deformation forms as small steps on the surface of the single crystal that are parallel to one another and loop around the circumference of the specimen as indicated in Figure 7.8. Each step results from the movement of a large number of Figure 7.8 Macroscopic slip in a single crystal. Direction of force Slip plane 1496T_c07_174-206 11/17/05 13:32 Page 183 REVISED PAGES 7.5 Slip in Single Crystals • 183 Figure 7.9 Slip in a zinc single crystal. (From C. F. Elam, The Distortion of Metal Crystals, Oxford University Press, London, 1935.) dislocations along the same slip plane. On the surface of a polished single crystal specimen, these steps appear as lines, which are called slip lines. A zinc single crystal that has been plastically deformed to the degree that these slip markings are discernible is shown in Figure 7.9. With continued extension of a single crystal, both the number of slip lines and the slip step width will increase. For FCC and BCC metals, slip may eventually begin along a second slip system, the system that is next most favorably oriented with the tensile axis. Furthermore, for HCP crystals having few slip systems, if the stress axis for the most favorable slip system is either perpendicular to the slip direction (l  90) or parallel to the slip plane (f  90), the critical resolved shear stress will be zero. For these extreme orientations the crystal ordinarily fractures rather than deforming plastically. Concept Check 7.2 Explain the difference between resolved shear stress and critical resolved shear stress. [The answer may be found at www.wiley.com/college/callister (Student Companion Site).] EXAMPLE PROBLEM 7.1 Resolved Shear Stress and Stress-to-Initiate-Yielding Computations Consider a single crystal of BCC iron oriented such that a tensile stress is applied along a [010] direction. (a) Compute the resolved shear stress along a (110) plane and in a [111] direction when a tensile stress of 52 MPa (7500 psi) is applied. 1496T_c07_174-206 11/17/05 13:32 Page 184 REVISED PAGES 184 • Chapter 7 / Dislocations and Strengthening Mechanisms (b) If slip occurs on a (110) plane and in a [111] direction, and the critical resolved shear stress is 30 MPa (4350 psi), calculate the magnitude of the applied tensile stress necessary to initiate yielding. Solution (a) A BCC unit cell along with the slip direction and plane as well as the direction of the applied stress are shown in the accompanying diagram. In order to solve this problem we must use Equation 7.2. However, it is first necessary to determine values for f and l, where, from the diagram below, f is the angle between the normal to the (110) slip plane (i.e., the [110] direction) and the [010] direction, and l represents the angle between [111] and [010] directions. In general, for cubic unit cells, an angle u between directions 1 and 2, represented by [u1v1w1] and [u2v2w2], respectively, is equal to u  cos1 c u1u2  v1v2  w1w2 21u21  v21  w21 21u22  v22  w22 2 d (7.6) For the determination of the value of f, let [u1v1w1]  [110] and [u2v2w2]  [010] such that 112102  112112  102102 f  cos1 e f 23 112 2  112 2  102 2 4 3 102 2  112 2  102 2 4  cos1 a 1 b  45 12 z Slip plane (110) ␾ Normal to slip plane Slip direction [111] ␭ y ␴ Direction of applied stress [010] x However, for l, we take [u1v1w1]  [111] and [u2v2w2]  [010], and l  cos1 c  cos1 a 112102  112112  112102 2 3 112  112 2  112 2 4 3 102 2  112 2  102 2 4 2 1 b  54.7 13 Thus, according to Equation 7.2, tR  s cos f cos l  152 MPa21cos 4521cos 54.72 1 1  152 MPa2a ba b 12 13  21.3 MPa 13060 psi2 d 1496T_c07_174-206 11/17/05 13:32 Page 185 REVISED PAGES 7.7 Deformation by Twinning • 185 (b) The yield strength sy may be computed from Equation 7.4; f and l will be the same as for part (a), and sy  30 MPa  73.4 MPa 110,600 psi2 1cos 4521cos 54.72 7.6 PLASTIC DEFORMATION OF POLYCRYSTALLINE MATERIALS Deformation and slip in polycrystalline materials is somewhat more complex. Because of the random crystallographic orientations of the numerous grains, the direction of slip varies from one grain to another. For each, dislocation motion occurs along the slip system that has the most favorable orientation, as defined above. This is exemplified by a photomicrograph of a polycrystalline copper specimen that has been plastically deformed (Figure 7.10); before deformation the surface was polished. Slip lines1 are visible, and it appears that two slip systems operated for most of the grains, as evidenced by two sets of parallel yet intersecting sets of lines. Furthermore, variation in grain orientation is indicated by the difference in alignment of the slip lines for the several grains. Gross plastic deformation of a polycrystalline specimen corresponds to the comparable distortion of the individual grains by means of slip. During deformation, mechanical integrity and coherency are maintained along the grain boundaries; that is, the grain boundaries usually do not come apart or open up. As a consequence, each individual grain is constrained, to some degree, in the shape it may assume by its neighboring grains. The manner in which grains distort as a result of gross plastic deformation is indicated in Figure 7.11. Before deformation the grains are equiaxed, or have approximately the same dimension in all directions. For this particular deformation, the grains become elongated along the direction in which the specimen was extended. Polycrystalline metals are stronger than their single-crystal equivalents, which means that greater stresses are required to initiate slip and the attendant yielding. This is, to a large degree, also a result of geometrical constraints that are imposed on the grains during deformation. Even though a single grain may be favorably oriented with the applied stress for slip, it cannot deform until the adjacent and less favorably oriented grains are capable of slip also; this requires a higher applied stress level. 7.7 DEFORMATION BY TWINNING In addition to slip, plastic deformation in some metallic materials can occur by the formation of mechanical twins, or twinning. The concept of a twin was introduced in Section 4.6; that is, a shear force can produce atomic displacements such that on one side of a plane (the twin boundary), atoms are located in mirror-image positions of atoms on the other side. The manner in which this is accomplished is demonstrated 1 These slip lines are microscopic ledges produced by dislocations (Figure 7.1c) that have exited from a grain and appear as lines when viewed with a microscope. They are analogous to the macroscopic steps found on the surfaces of deformed single crystals (Figures 7.8 and 7.9). 1496T_c07_174-206 11/17/05 13:32 Page 186 REVISED PAGES 186 • Chapter 7 / Dislocations and Strengthening Mechanisms Figure 7.10 Slip lines on the surface of a polycrystalline specimen of copper that was polished and subsequently deformed. 173. [Photomicrograph courtesy of C. Brady, National Bureau of Standards (now the National Institute of Standards and Technology, Gaithersburg, MD).] Figure 7.11 Alteration of the grain structure of a polycrystalline metal as a result of plastic deformation. (a) Before deformation the grains are equiaxed. (b) The deformation has produced elongated grains. 170. (From W. G. Moffatt, G. W. Pearsall, and J. Wulff, The Structure and Properties of Materials, Vol. I, Structure, p. 140. Copyright © 1964 by John Wiley & Sons, New York. Reprinted by permission of John Wiley & Sons, Inc.) 1496T_c07_174-206 11/17/05 13:32 Page 187 REVISED PAGES 7.7 Deformation by Twinning • 187 Polished surface ␶ ␶ Twin plane Twin plane (a) (b) Figure 7.12 Schematic diagram showing how twinning results from an applied shear stress t. In (b), open circles represent atoms that did not change position; dashed and solid circles represent original and final atom positions, respectively. (From G. E. Dieter, Mechanical Metallurgy, 3rd edition. Copyright © 1986 by McGraw-Hill Book Company, New York. Reproduced with permission of McGraw-Hill Book Company.) in Figure 7.12. Here, open circles represent atoms that did not move, and dashed and solid circles represent original and final positions, respectively, of atoms within the twinned region. As may be noted in this figure, the displacement magnitude within the twin region (indicated by arrows) is proportional to the distance from the twin plane. Furthermore, twinning occurs on a definite crystallographic plane and in a specific direction that depend on crystal structure. For example, for BCC metals, the twin plane and direction are (112) and [111], respectively. Slip and twinning deformations are compared in Figure 7.13 for a single crystal that is subjected to a shear stress t. Slip ledges are shown in Figure 7.13a, the formation of which was described in Section 7.5; for twinning, the shear deformation is homogeneous (Figure 7.13b). These two processes differ from each other in several respects. First, for slip, the crystallographic orientation above and below the slip plane is the same both before and after the deformation; for twinning, there will be a reorientation across the twin plane. In addition, slip occurs in distinct atomic spacing multiples, whereas the atomic displacement for twinning is less than the interatomic separation. Mechanical twinning occurs in metals that have BCC and HCP crystal structures, at low temperatures, and at high rates of loading (shock loading), conditions under which the slip process is restricted; that is, there are few operable slip systems. Twin planes ␶ ␶ Slip planes Twin ␶ ␶ (a) (b) Figure 7.13 For a single crystal subjected to a shear stress t, (a) deformation by slip; (b) deformation by twinning. 1496T_c07_174-206 11/17/05 13:49 Page 188 REVISED PAGES 188 • Chapter 7 / Dislocations and Strengthening Mechanisms The amount of bulk plastic deformation from twinning is normally small relative to that resulting from slip. However, the real importance of twinning lies with the accompanying crystallographic reorientations; twinning may place new slip systems in orientations that are favorable relative to the stress axis such that the slip process can now take place. M e c h a n i s m s o f St r e n g t h e n i n g i n M e t a l s Metallurgical and materials engineers are often called on to design alloys having high strengths yet some ductility and toughness; ordinarily, ductility is sacrificed when an alloy is strengthened. Several hardening techniques are at the disposal of an engineer, and frequently alloy selection depends on the capacity of a material to be tailored with the mechanical characteristics required for a particular application. Important to the understanding of strengthening mechanisms is the relation between dislocation motion and mechanical behavior of metals. Because macroscopic plastic deformation corresponds to the motion of large numbers of dislocations, the ability of a metal to plastically deform depends on the ability of dislocations to move. Since hardness and strength (both yield and tensile) are related to the ease with which plastic deformation can be made to occur, by reducing the mobility of dislocations, the mechanical strength may be enhanced; that is, greater mechanical forces will be required to initiate plastic deformation. In contrast, the more unconstrained the dislocation motion, the greater is the facility with which a metal may deform, and the softer and weaker it becomes. Virtually all strengthening techniques rely on this simple principle: restricting or hindering dislocation motion renders a material harder and stronger. The present discussion is confined to strengthening mechanisms for single-phase metals, by grain size reduction, solid-solution alloying, and strain hardening. Deformation and strengthening of multiphase alloys are more complicated, involving concepts beyond the scope of the present discussion; Chapter 10 and Section 11.9 treat techniques that are used to strengthen multiphase alloys. 7.8 STRENGTHENING BY GRAIN SIZE REDUCTION The size of the grains, or average grain diameter, in a polycrystalline metal influences the mechanical properties. Adjacent grains normally have different crystallographic orientations and, of course, a common grain boundary, as indicated in Figure 7.14. During plastic deformation, slip or dislocation motion must take place Figure 7.14 The motion of a dislocation as it encounters a grain boundary, illustrating how the boundary acts as a barrier to continued slip. Slip planes are discontinuous and change directions across the boundary. (From Van Vlack, A Textbook of Materials Technology, 1st edition, © 1973, p. 53. Adapted by permission of Pearson Education, Inc., Upper Saddle River, NJ.) Grain boundary Slip plane Grain A Grain B 1496T_c07_174-206 11/17/05 13:32 Page 189 REVISED PAGES 7.8 Strengthening by Grain Size Reduction • 189 across this common boundary—say, from grain A to grain B in Figure 7.14. The grain boundary acts as a barrier to dislocation motion for two reasons: 1. Since the two grains are of different orientations, a dislocation passing into grain B will have to change its direction of motion; this becomes more difficult as the crystallographic misorientation increases. 2. The atomic disorder within a grain boundary region will result in a discontinuity of slip planes from one grain into the other. It should be mentioned that, for high-angle grain boundaries, it may not be the case that dislocations traverse grain boundaries during deformation; rather, dislocations tend to “pile up” (or back up) at grain boundaries. These pile-ups introduce stress concentrations ahead of their slip planes, which generate new dislocations in adjacent grains. A fine-grained material (one that has small grains) is harder and stronger than one that is coarse grained, since the former has a greater total grain boundary area to impede dislocation motion. For many materials, the yield strength sy varies with grain size according to sy  s0  kyd1 2 (7.7) In this expression, termed the Hall-Petch equation, d is the average grain diameter, and s0 and ky are constants for a particular material. Note that Equation 7.7 is not valid for both very large (i.e., coarse) grain and extremely fine grain polycrystalline materials. Figure 7.15 demonstrates the yield strength dependence on grain size for a brass alloy. Grain size may be regulated by the rate of solidification from the liquid phase, and also by plastic deformation followed by an appropriate heat treatment, as discussed in Section 7.13. It should also be mentioned that grain size reduction improves not only strength, but also the toughness of many alloys. Small-angle grain boundaries (Section 4.6) are not effective in interfering with the slip process because of the slight crystallographic misalignment across the boundary. On the other hand, twin boundaries (Section 4.6) will effectively block Grain size, d (mm) –1 10 10 –2 5 × 10 –3 30 200 150 20 100 10 50 0 0 4 8 d–1/2 (mm–1/ 2) 12 16 Yield strength (ksi) Yield strength (MPa) Hall-Petch equation— dependence of yield strength on grain size Figure 7.15 The influence of grain size on the yield strength of a 70 Cu–30 Zn brass alloy. Note that the grain diameter increases from right to left and is not linear. (Adapted from H. Suzuki, “The Relation Between the Structure and Mechanical Properties of Metals,” Vol. II, National Physical Laboratory, Symposium No. 15, 1963, p. 524.) 1496T_c07_174-206 1/9/06 12:56 Page 190 2nd REVISE PAGES 190 • Chapter 7 / Dislocations and Strengthening Mechanisms slip and increase the strength of the material. Boundaries between two different phases are also impediments to movements of dislocations; this is important in the strengthening of more complex alloys.The sizes and shapes of the constituent phases significantly affect the mechanical properties of multiphase alloys; these are the topics of discussion in Sections 10.7, 10.8, and 16.1. 7.9 SOLID-SOLUTION STRENGTHENING Another technique to strengthen and harden metals is alloying with impurity atoms that go into either substitutional or interstitial solid solution. Accordingly, this is called solid-solution strengthening. High-purity metals are almost always softer and weaker than alloys composed of the same base metal. Increasing the concentration of the impurity results in an attendant increase in tensile and yield strengths, as indicated in Figures 7.16a and 7.16b for nickel in copper; the dependence of ductility on nickel concentration is presented in Figure 7.16c. Alloys are stronger than pure metals because impurity atoms that go into solid solution ordinarily impose lattice strains on the surrounding host atoms. Lattice strain field interactions between dislocations and these impurity atoms result, and, consequently, dislocation movement is restricted. For example, an impurity atom that is smaller than a host atom for which it substitutes exerts tensile strains on the surrounding crystal lattice, as illustrated in Figure 7.17a. Conversely, a larger substitutional solid-solution strengthening 180 25 60 40 Yield strength (MPa) 300 Tensile strength (ksi) Tensile strength (MPa) 50 140 20 120 15 100 80 200 0 10 20 30 40 30 50 10 60 0 10 20 30 Nickel content (wt%) Nickel content (wt%) (a) (b) 40 Elongation (% in 2 in.) 60 50 40 30 20 0 10 20 30 Nickel content (wt%) (c) 40 50 Figure 7.16 Variation with nickel content of (a) tensile strength, (b) yield strength, and (c) ductility (%EL) for copper–nickel alloys, showing strengthening. 50 Yield strength (ksi) 160 400 1496T_c07_174-206 1/9/06 12:56 Page 191 2nd REVISE PAGES 7.10 Strain Hardening • 191 Figure 7.17 (a) Representation of tensile lattice strains imposed on host atoms by a smaller substitutional impurity atom. (b) Possible locations of smaller impurity atoms relative to an edge dislocation such that there is partial cancellation of impurity–dislocation lattice strains. (a) (b) atom imposes compressive strains in its vicinity (Figure 7.18a).These solute atoms tend to diffuse to and segregate around dislocations in a way so as to reduce the overall strain energy—that is, to cancel some of the strain in the lattice surrounding a dislocation. To accomplish this, a smaller impurity atom is located where its tensile strain will partially nullify some of the dislocation’s compressive strain. For the edge dislocation in Figure 7.17b, this would be adjacent to the dislocation line and above the slip plane. A larger impurity atom would be situated as in Figure 7.18b. The resistance to slip is greater when impurity atoms are present because the overall lattice strain must increase if a dislocation is torn away from them. Furthermore, the same lattice strain interactions (Figures 7.17b and 7.18b) will exist between impurity atoms and dislocations that are in motion during plastic deformation. Thus, a greater applied stress is necessary to first initiate and then continue plastic deformation for solid-solution alloys, as opposed to pure metals; this is evidenced by the enhancement of strength and hardness. 7.10 STRAIN HARDENING strain hardening cold working Percent cold work— dependence on original and deformed crosssectional areas Strain hardening is the phenomenon whereby a ductile metal becomes harder and stronger as it is plastically deformed. Sometimes it is also called work hardening, or, because the temperature at which deformation takes place is “cold” relative to the absolute melting temperature of the metal, cold working. Most metals strain harden at room temperature. It is sometimes convenient to express the degree of plastic deformation as percent cold work rather than as strain. Percent cold work (%CW) is defined as %CW  a A0  Ad b  100 A0 (7.8) where A0 is the original area of the cross section that experiences deformation, and Ad is the area after deformation. Figure 7.18 (a) Representation of compressive strains imposed on host atoms by a larger substitutional impurity atom. (b) Possible locations of larger impurity atoms relative to an edge dislocation such that there is partial cancellation of impurity–dislocation lattice strains. (a) (b) 1496T_c07_174-206 11/17/05 13:32 Page 192 REVISED PAGES 192 • Chapter 7 / Dislocations and Strengthening Mechanisms 140 120 900 800 1040 Steel 120 1040 Steel 800 80 500 Brass 400 60 700 100 600 Brass 80 500 Tensile strength (ksi) 600 Tensile strength (MPa) 100 Yield strength (ksi) Yield strength (MPa) 700 Copper 300 40 60 400 Copper 200 300 40 20 100 0 10 20 30 40 50 60 70 200 0 10 20 30 40 Percent cold work Percent cold work (a) (b) 50 60 70 70 60 Ductility (%EL) 50 40 Brass 30 20 1040 Steel 10 Copper 0 0 10 20 30 40 Percent cold work (c) 50 60 70 Figure 7.19 For 1040 steel, brass, and copper, (a) the increase in yield strength, (b) the increase in tensile strength, and (c) the decrease in ductility (%EL) with percent cold work. [Adapted from Metals Handbook: Properties and Selection: Irons and Steels, Vol. 1, 9th edition, B. Bardes (Editor), American Society for Metals, 1978, p. 226; and Metals Handbook: Properties and Selection: Nonferrous Alloys and Pure Metals, Vol. 2, 9th edition, H. Baker (Managing Editor), American Society for Metals, 1979, pp. 276 and 327.] Figures 7.19a and 7.19b demonstrate how steel, brass, and copper increase in yield and tensile strength with increasing cold work. The price for this enhancement of hardness and strength is in the ductility of the metal. This is shown in Figure 7.19c, in which the ductility, in percent elongation, experiences a reduction with increasing percent cold work for the same three alloys. The influence of cold work on the 1496T_c07_174-206 11/17/05 13:32 Page 193 REVISED PAGES 7.10 Strain Hardening • 193 Figure 7.20 The influence of cold work on the stress–strain behavior of a low-carbon steel; curves are shown for 0%CW, 4%CW, and 24%CW. 24%CW 600 4%CW 500 Stress (MPa) 0%CW 400 300 200 100 0 0 0.05 0.1 0.15 0.2 0.25 Strain stress–strain behavior of a low-carbon steel is shown in Figure 7.20; here stress–strain curves are plotted at 0%CW, 4%CW, and 24%CW. Strain hardening is demonstrated in a stress–strain diagram presented earlier (Figure 6.17). Initially, the metal with yield strength sy0 is plastically deformed to point D. The stress is released, then reapplied with a resultant new yield strength, syi. The metal has thus become stronger during the process because syi is greater than sy0. The strain-hardening phenomenon is explained on the basis of dislocation– dislocation strain field interactions similar to those discussed in Section 7.3. The dislocation density in a metal increases with deformation or cold work, due to dislocation multiplication or the formation of new dislocations, as noted previously. Consequently, the average distance of separation between dislocations decreases—the dislocations are positioned closer together. On the average, dislocation–dislocation strain interactions are repulsive. The net result is that the motion of a dislocation is hindered by the presence of other dislocations. As the dislocation density increases, this resistance to dislocation motion by other dislocations becomes more pronounced. Thus, the imposed stress necessary to deform a metal increases with increasing cold work. Strain hardening is often utilized commercially to enhance the mechanical properties of metals during fabrication procedures. The effects of strain hardening may be removed by an annealing heat treatment, as discussed in Section 11.7. In passing, for the mathematical expression relating true stress and strain, Equation 6.19, the parameter n is called the strain-hardening exponent, which is a measure of the ability of a metal to strain harden; the larger its magnitude, the greater the strain hardening for a given amount of plastic strain. Concept Check 7.3 When making hardness measurements, what will be the effect of making an indentation very close to a preexisting indentation? Why? [The answer may be found at www.wiley.com/college/callister (Student Companion Site).] 1496T_c07_174-206 11/17/05 13:32 Page 194 REVISED PAGES 194 • Chapter 7 / Dislocations and Strengthening Mechanisms Concept Check 7.4 Would you expect a crystalline ceramic material to strain harden at room temperature? Why or why not? [The answer may be found at www.wiley.com/college/callister (Student Companion Site).] EXAMPLE PROBLEM 7.2 Tensile Strength and Ductility Determinations for Cold-Worked Copper Compute the tensile strength and ductility (%EL) of a cylindrical copper rod if it is cold worked such that the diameter is reduced from 15.2 mm to 12.2 mm (0.60 in. to 0.48 in.). Solution It is first necessary to determine the percent cold work resulting from the deformation. This is possible using Equation 7.8: 15.2 mm 2 12.2 mm 2 bpa bp 2 2 %CW   100  35.6% 15.2 mm 2 a bp 2 a The tensile strength is read directly from the curve for copper (Figure 7.19b) as 340 MPa (50,000 psi). From Figure 7.19c, the ductility at 35.6%CW is about 7%EL. In summary, we have just discussed the three mechanisms that may be used to strengthen and harden single-phase metal alloys: strengthening by grain size reduction, solid-solution strengthening, and strain hardening. Of course they may be used in conjunction with one another; for example, a solid-solution strengthened alloy may also be strain hardened. It should also be noted that the strengthening effects due to grain size reduction and strain hardening can be eliminated or at least reduced by an elevatedtemperature heat treatment (Sections 7.12 and 7.13). Conversely, solid-solution strengthening is unaffected by heat treatment. Re c ove r y, Re c r ys t a l l i z a t i o n , a n d G r a i n G row t h As outlined in the preceding paragraphs of this chapter, plastically deforming a polycrystalline metal specimen at temperatures that are low relative to its absolute melting temperature produces microstructural and property changes that include (1) a change in grain shape (Section 7.6), (2) strain hardening (Section 7.10), and (3) an increase in dislocation density (Section 7.3). Some fraction of the energy expended in deformation is stored in the metal as strain energy, which is associated with tensile, compressive, and shear zones around the newly created dislocations (Section 7.3). 1496T_c07_174-206 11/17/05 13:32 Page 195 REVISED PAGES 7.12 Recrystallization • 195 Furthermore, other properties such as electrical conductivity (Section 18.8) and corrosion resistance may be modified as a consequence of plastic deformation. These properties and structures may revert back to the precold-worked states by appropriate heat treatment (sometimes termed an annealing treatment). Such restoration results from two different processes that occur at elevated temperatures: recovery and recrystallization, which may be followed by grain growth. 7.11 RECOVERY recovery During recovery, some of the stored internal strain energy is relieved by virtue of dislocation motion (in the absence of an externally applied stress), as a result of enhanced atomic diffusion at the elevated temperature. There is some reduction in the number of dislocations, and dislocation configurations (similar to that shown in Figure 4.8) are produced having low strain energies. In addition, physical properties such as electrical and thermal conductivities and the like are recovered to their precold-worked states. 7.12 RECRYSTALLIZATION recrystallization Even after recovery is complete, the grains are still in a relatively high strain energy state. Recrystallization is the formation of a new set of strain-free and equiaxed grains (i.e., having approximately equal dimensions in all directions) that have low dislocation densities and are characteristic of the precold-worked condition. The driving force to produce this new grain structure is the difference in internal energy between the strained and unstrained material. The new grains form as very small nuclei and grow until they completely consume the parent material, processes that involve short-range diffusion. Several stages in the recrystallization process are represented in Figures 7.21a to 7.21d; in these photomicrographs, the small (a) ( b) Figure 7.21 Photomicrographs showing several stages of the recrystallization and grain growth of brass. (a) Cold-worked (33%CW) grain structure. (b) Initial stage of recrystallization after heating 3 s at 580C (1075F); the very small grains are those that have recrystallized. (c) Partial replacement of cold-worked grains by recrystallized ones (4 s at 580C). (d) Complete recrystallization (8 s at 580C). (e) Grain growth after 15 min at 580C. ( f ) Grain growth after 10 min at 700C (1290F). All photomicrographs 75. (Photomicrographs courtesy of J. E. Burke, General Electric Company.) 1496T_c07_174-206 11/17/05 13:32 Page 196 REVISED PAGES 196 • Chapter 7 / Dislocations and Strengthening Mechanisms (c) (d) (e) (f ) Figure 7.21 (continued) speckled grains are those that have recrystallized. Thus, recrystallization of coldworked metals may be used to refine the grain structure. Also, during recrystallization, the mechanical properties that were changed as a result of cold working are restored to their precold-worked values; that is, the metal becomes softer, weaker, yet more ductile. Some heat treatments are designed to allow recrystallization to occur with these modifications in the mechanical characteristics (Section 11.7). Recrystallization is a process the extent of which depends on both time and temperature. The degree (or fraction) of recrystallization increases with time, as may be noted in the photomicrographs shown in Figures 7.21a–d. The explicit 1496T_c07_174-206 11/17/05 13:32 Page 197 REVISED PAGES 7.12 Recrystallization • 197 Annealing temperature (°F) 400 600 800 1000 1200 600 60 Tensile strength 500 40 400 Ductility (%EL) Tensile strength (MPa) 50 30 Ductility 20 300 Recovery Recrystallization Grain growth Cold-worked and recovered grains Grain size (mm) Figure 7.22 The influence of annealing temperature (for an annealing time of 1 h) on the tensile strength and ductility of a brass alloy. Grain size as a function of annealing temperature is indicated. Grain structures during recovery, recrystallization, and grain growth stages are shown schematically. (Adapted from G. Sachs and K. R. Van Horn, Practical Metallurgy, Applied Metallurgy and the Industrial Processing of Ferrous and Nonferrous Metals and Alloys, American Society for Metals, 1940, p. 139.) New grains 0.040 0.030 0.020 0.010 100 200 300 400 500 600 700 Annealing temperature (°C) recrystallization temperature time dependence of recrystallization is addressed in more detail near the end of Section 10.3. The influence of temperature is demonstrated in Figure 7.22, which plots tensile strength and ductility (at room temperature) of a brass alloy as a function of the temperature and for a constant heat treatment time of 1 h. The grain structures found at the various stages of the process are also presented schematically. The recrystallization behavior of a particular metal alloy is sometimes specified in terms of a recrystallization temperature, the temperature at which recrystallization just reaches completion in 1 h. Thus, the recrystallization temperature for the brass alloy of Figure 7.22 is about 450C 1850F2. Typically, it is between one-third and one-half of the absolute melting temperature of a metal or alloy and depends on several factors, including the amount of prior cold work and the purity of the alloy. Increasing the percentage of cold work enhances the rate of recrystallization, with the result that the recrystallization temperature is lowered, and approaches a constant or limiting value at high deformations; this effect is shown in Figure 7.23. Furthermore, it is this limiting or minimum recrystallization temperature that is normally specified in the literature. There exists some critical degree of cold work below which recrystallization cannot be made to occur, as shown in the figure; normally, this is between 2% and 20% cold work. Recrystallization proceeds more rapidly in pure metals than in alloys. During recrystallization, grain-boundary motion occurs as the new grain nuclei form and then grow. It is believed that impurity atoms preferentially segregate at and interact 1496T_c07_174-206 11/17/05 13:32 Page 198 REVISED PAGES 198 • Chapter 7 / Dislocations and Strengthening Mechanisms Figure 7.23 The variation of recrystallization temperature with percent cold work for iron. For deformations less than the critical (about 5%CW), recrystallization will not occur. 900 1600 1400 700 1200 600 1000 500 800 400 300 0 10 20 Critical deformation 30 40 Percent cold work 50 60 Recrystallization temperature (F) Recrystallization temperature (C) 800 600 70 with these recrystallized grain boundaries so as to diminish their (i.e., grain boundary) mobilities; this results in a decrease of the recrystallization rate and raises the recrystallization temperature, sometimes quite substantially. For pure metals, the recrystallization temperature is normally 0.3Tm, where Tm is the absolute melting temperature; for some commercial alloys it may run as high as 0.7Tm. Recrystallization and melting temperatures for a number of metals and alloys are listed in Table 7.2. Plastic deformation operations are often carried out at temperatures above the recrystallization temperature in a process termed hot working, described in Section 11.4. The material remains relatively soft and ductile during deformation because it does not strain harden, and thus large deformations are possible. Concept Check 7.5 Briefly explain why some metals (i.e., lead and tin) do not strain harden when deformed at room temperature. [The answer may be found at www.wiley.com/college/callister (Student Companion Site).] Table 7.2 Recrystallization and Melting Temperatures for Various Metals and Alloys Recrystallization Temperature Metal Lead Tin Zinc Aluminum (99.999 wt%) Copper (99.999 wt%) Brass (60 Cu–40 Zn) Nickel (99.99 wt%) Iron Tungsten Melting Temperature C F C F 4 4 10 80 120 475 370 450 1200 25 25 50 176 250 887 700 840 2200 327 232 420 660 1085 900 1455 1538 3410 620 450 788 1220 1985 1652 2651 2800 6170 1496T_c07_174-206 11/17/05 13:32 Page 199 REVISED PAGES 7.12 Recrystallization • 199 Concept Check 7.6 Would you expect it to be possible for ceramic materials to experience recrystallization? Why or why not? [The answer may be found at www.wiley.com/college/callister (Student Companion Site).] DESIGN EXAMPLE 7.1 Description of Diameter Reduction Procedure A cylindrical rod of noncold-worked brass having an initial diameter of 6.4 mm (0.25 in.) is to be cold worked by drawing such that the cross-sectional area is reduced. It is required to have a cold-worked yield strength of at least 345 MPa (50,000 psi) and a ductility in excess of 20%EL; in addition, a final diameter of 5.1 mm (0.20 in.) is necessary. Describe the manner in which this procedure may be carried out. Solution Let us first consider the consequences (in terms of yield strength and ductility) of cold working in which the brass specimen diameter is reduced from 6.4 mm (designated by d0) to 5.1 mm (di ). The %CW may be computed from Equation 7.8 as d0 2 di 2 bpa bp 2 2 %CW   100 2 d0 a bp 2 5.1 mm 2 6.4 mm 2 bpa bp a 2 2  100  36.5%CW  6.4 mm 2 a bp 2 a From Figures 7.19a and 7.19c, a yield strength of 410 MPa (60,000 psi) and a ductility of 8%EL are attained from this deformation. According to the stipulated criteria, the yield strength is satisfactory; however, the ductility is too low. Another processing alternative is a partial diameter reduction, followed by a recrystallization heat treatment in which the effects of the cold work are nullified. The required yield strength, ductility, and diameter are achieved through a second drawing step. Again, reference to Figure 7.19a indicates that 20%CW is required to give a yield strength of 345 MPa. On the other hand, from Figure 7.19c, ductilities greater than 20%EL are possible only for deformations of 23%CW or less. Thus during the final drawing operation, deformation must be between 20%CW and 23%CW. Let’s take the average of these extremes, 21.5%CW, and then calculate the final diameter for the first drawing d¿0, which becomes the original diameter for the second drawing. Again, using Equation 7.8, d¿0 2 5.1 mm 2 b pa b p 2 2 21.5%CW   100 d¿0 2 a b p 2 a 1496T_c07_174-206 12/21/05 7:48 Page 200 2nd REVISE PAGES 200 • Chapter 7 / Dislocations and Strengthening Mechanisms Now, solving for d¿0 from the expression above gives d0¿  5.8 mm 10.226 in.2 7.13 GRAIN GROWTH grain growth For grain growth, dependence of grain size on time After recrystallization is complete, the strain-free grains will continue to grow if the metal specimen is left at the elevated temperature (Figures 7.21d–f); this phenomenon is called grain growth. Grain growth does not need to be preceded by recovery and recrystallization; it may occur in all polycrystalline materials, metals and ceramics alike. An energy is associated with grain boundaries, as explained in Section 4.6. As grains increase in size, the total boundary area decreases, yielding an attendant reduction in the total energy; this is the driving force for grain growth. Grain growth occurs by the migration of grain boundaries. Obviously, not all grains can enlarge, but large ones grow at the expense of small ones that shrink. Thus, the average grain size increases with time, and at any particular instant there will exist a range of grain sizes. Boundary motion is just the short-range diffusion of atoms from one side of the boundary to the other. The directions of boundary movement and atomic motion are opposite to each other, as shown in Figure 7.24. For many polycrystalline materials, the grain diameter d varies with time t according to the relationship d n  d n0  Kt (7.9) where d0 is the initial grain diameter at t  0, and K and n are time-independent constants; the value of n is generally equal to or greater than 2. The dependence of grain size on time and temperature is demonstrated in Figure 7.25, a plot of the logarithm of grain size as a function of the logarithm of time for a brass alloy at several temperatures. At lower temperatures the curves are linear. Furthermore, grain growth proceeds more rapidly as temperature increases; that is, the curves are displaced upward to larger grain sizes. This is explained by the enhancement of diffusion rate with rising temperature. Atomic diffusion across boundary Direction of grain boundary motion Figure 7.24 Schematic representation of grain growth via atomic diffusion. (From Van Vlack, Lawrence H., Elements of Materials Science and Engineering, © 1989, p. 221. Adapted by permission of Pearson Education, Inc., Upper Saddle River, NJ.) 1496T_c07_174-206 11/17/05 13:32 Page 201 REVISED PAGES Summary • 201 850C 1.0 Grain diameter (mm) (Logarithmic scale) Figure 7.25 The logarithm of grain diameter versus the logarithm of time for grain growth in brass at several temperatures. (From J. E. Burke, “Some Factors Affecting the Rate of Grain Growth in Metals.” Reprinted with permission from Metallurgical Transactions, Vol. 180, 1949, a publication of The Metallurgical Society of AIME, Warrendale, Pennsylvania.) 800C 700C 600C 0.1 500C 0.01 1 10 102 Time (min) (Logarithmic scale) 103 104 The mechanical properties at room temperature of a fine-grained metal are usually superior (i.e., higher strength and toughness) to those of coarse-grained ones. If the grain structure of a single-phase alloy is coarser than that desired, refinement may be accomplished by plastically deforming the material, then subjecting it to a recrystallization heat treatment, as described above. SUMMARY Basic Concepts Slip Systems On a microscopic level, plastic deformation corresponds to the motion of dislocations in response to an externally applied shear stress, a process termed “slip.” Slip occurs on specific crystallographic planes and within these planes only in certain directions. A slip system represents a slip plane–slip direction combination, and operable slip systems depend on the crystal structure of the material. Slip in Single Crystals The critical resolved shear stress is the minimum shear stress required to initiate dislocation motion; the yield strength of a single crystal depends on both the magnitude of the critical resolved shear stress and the orientation of slip components relative to the direction of the applied stress. Plastic Deformation of Polycrystalline Materials For polycrystalline materials, slip occurs within each grain along the slip systems that are most favorably oriented with the applied stress; furthermore, during deformation, grains change shape in such a manner that coherency at the grain boundaries is maintained. Deformation by Twinning Under some circumstances limited plastic deformation may occur in BCC and HCP metals by mechanical twinning. Normally, twinning is important to the degree that accompanying crystallographic reorientations make the slip process more favorable. 1496T_c07_174-206 11/17/05 13:32 Page 202 REVISED PAGES 202 • Chapter 7 / Dislocations and Strengthening Mechanisms Characteristics of Dislocations Strengthening by Grain Size Reduction Solid-Solution Strengthening Strain Hardening Since the ease with which a material is capable of plastic deformation is a function of dislocation mobility, restricting dislocation motion increases hardness and strength. On the basis of this principle, three different strengthening mechanisms were discussed. Grain boundaries serve as barriers to dislocation motion; thus refining the grain size of a polycrystalline material renders it harder and stronger. Solid-solution strengthening results from lattice strain interactions between impurity atoms and dislocations. Finally, as a material is plastically deformed, the dislocation density increases, as does also the extent of repulsive dislocation–dislocation strain field interactions; strain hardening is just the enhancement of strength with increased plastic deformation. Recovery Recrystallization Grain Growth The microstructural and mechanical characteristics of a plastically deformed metal specimen may be restored to their predeformed states by an appropriate heat treatment, during which recovery, recrystallization, and grain growth processes are allowed to occur. During recovery there is a reduction in dislocation density and alterations in dislocation configurations. Recrystallization is the formation of a new set of grains that are strain free; in addition, the material becomes softer and more ductile. Grain growth is the increase in average grain size of polycrystalline materials, which proceeds by grain boundary motion. I M P O R TA N T T E R M S A N D C O N C E P T S Cold working Critical resolved shear stress Dislocation density Grain growth Lattice strain Recovery Recrystallization Recrystallization temperature Resolved shear stress Slip Slip system Solid-solution strengthening Strain hardening REFERENCES Hirth, J. P., and J. Lothe, Theory of Dislocations, 2nd edition, Wiley-Interscience, New York, 1982. Reprinted by Krieger, Melbourne, FL, 1992. Hull, D., Introduction to Dislocations, 3rd edition, Butterworth-Heinemann, Woburn, UK, 1984. Read, W. T., Jr., Dislocations in Crystals, McGrawHill, New York, 1953. Weertman, J., and J. R.Weertman, Elementary Dislocation Theory, Macmillan, New York, 1964. Reprinted by Oxford University Press, New York, 1992. QUESTIONS AND PROBLEMS Basic Concepts Characteristics of Dislocations 7.1 To provide some perspective on the dimensions of atomic defects, consider a metal spec- imen that has a dislocation density of 105 mm2. Suppose that all the dislocations in 1000 mm3 11 cm3 2 were somehow removed and linked end to end. How far (in miles) 1496T_c07_174-206 11/17/05 13:32 Page 203 REVISED PAGES Questions and Problems • 203 would this chain extend? Now suppose that the density is increased to 109 mm2 by cold working. What would be the chain length of dislocations in 1000 mm3 of material? 7.2 Consider two edge dislocations of opposite sign and having slip planes that are separated by several atomic distances as indicated in the diagram. Briefly describe the defect that results when these two dislocations become aligned with each other. 7.3 Is it possible for two screw dislocations of opposite sign to annihilate each other? Explain your answer. 7.4 For each of edge, screw, and mixed dislocations, cite the relationship between the direction of the applied shear stress and the direction of dislocation line motion. Slip Systems 7.5 (a) Define a slip system. (b) Do all metals have the same slip system? Why or why not? 7.6 (a) Compare planar densities (Section 3.11 and Problem 3.53) for the (100), (110), and (111) planes for FCC. (b) Compare planar densities (Problem 3.54) for the (100), (110), and (111) planes for BCC. 7.7 One slip system for the BCC crystal structure is 51106H111I. In a manner similar to Figure 7.6b, sketch a {110}-type plane for the BCC structure, representing atom positions with circles. Now, using arrows, indicate two different H111I slip directions within this plane. 7.8 One slip system for the HCP crystal structure is 500016H1120I. In a manner similar to Figure 7.6b, sketch a {0001}-type plane for the HCP structure and, using arrows, indicate three different H1120I slip directions within this plane. You might find Figure 3.8 helpful. 7.9 Equations 7.1a and 7.1b, expressions for Burgers vectors for FCC and BCC crystal structures, are of the form b a HuvwI 2 where a is the unit cell edge length. Also, since the magnitudes of these Burgers vectors may be determined from the following equation: b  a 2 1u  v2  w2 2 1 2 2 (7.10) determine values of  b for copper and iron. You may want to consult Table 3.1. 7.10 (a) In the manner of Equations 7.1a, 7.1b, and 7.1c, specify the Burgers vector for the simple cubic crystal structure. Its unit cell is shown in Figure 3.23. Also, simple cubic is the crystal structure for the edge dislocation of Figure 4.3, and for its motion as presented in Figure 7.1. You may also want to consult the answer to Concept Check 7.1. (b) On the basis of Equation 7.10, formulate an expression for the magnitude of the Burgers vector,  b , for simple cubic. Slip in Single Crystals 7.11 Sometimes cos  cos  in Equation 7.2 is termed the Schmid factor. Determine the magnitude of the Schmid factor for an FCC single crystal oriented with its [120] direction parallel to the loading axis. 7.12 Consider a metal single crystal oriented such that the normal to the slip plane and the slip direction are at angles of 60 and 35, respectively, with the tensile axis. If the critical resolved shear stress is 6.2 MPa (900 psi), will an applied stress of 12 MPa (1750 psi) cause the single crystal to yield? If not, what stress will be necessary? 7.13 A single crystal of zinc is oriented for a tensile test such that its slip plane normal makes an angle of 65 with the tensile axis.Three possible slip directions make angles of 30, 48, and 78 with the same tensile axis. (a) Which of these three slip directions is most favored? (b) If plastic deformation begins at a tensile stress of 2.5 MPa (355 psi), determine the critical resolved shear stress for zinc. 7.14 Consider a single crystal of nickel oriented such that a tensile stress is applied along a [001] direction. If slip occurs on a (111) plane and in a [101] direction, and is initiated at an 1496T_c07_174-206 11/17/05 13:32 Page 204 REVISED PAGES 204 • Chapter 7 / Dislocations and Strengthening Mechanisms 7.15 7.16 7.17 7.18 applied tensile stress of 13.9 MPa (2020 psi), compute the critical resolved shear stress. A single crystal of a metal that has the FCC crystal structure is oriented such that a tensile stress is applied parallel to the [100] direction. If the critical resolved shear stress for this material is 0.5 MPa, calculate the magnitude(s) of applied stress(es) necessary to cause slip to occur on the (111) plane in each of the [110], [101], and [011] directions. (a) A single crystal of a metal that has the BCC crystal structure is oriented such that a tensile stress is applied in the [100] direction. If the magnitude of this stress is 4.0 MPa, compute the resolved shear stress in the [111] direction on each of the (110), (011), and (101) planes. (b) On the basis of these resolved shear stress values, which slip system(s) is (are) most favorably oriented? Consider a single crystal of some hypothetical metal that has the BCC crystal structure and is oriented such that a tensile stress is applied along a [121] direction. If slip occurs on a (101) plane and in a [111] direction, compute the stress at which the crystal yields if its critical resolved shear stress is 2.4 MPa. The critical resolved shear stress for copper is 0.48 MPa (70 psi). Determine the maximum possible yield strength for a single crystal of Cu pulled in tension. Deformation by Twinning 7.19 List four major differences between deformation by twinning and deformation by slip relative to mechanism, conditions of occurrence, and final result. Strengthening by Grain Size Reduction 7.20 Briefly explain why small-angle grain boundaries are not as effective in interfering with the slip process as are high-angle grain boundaries. 7.21 Briefly explain why HCP metals are typically more brittle than FCC and BCC metals. 7.22 Describe in your own words the three strengthening mechanisms discussed in this chapter (i.e., grain size reduction, solid-solution strengthening, and strain hardening). Be sure to explain how dislocations are involved in each of the strengthening techniques. 7.23 (a) From the plot of yield strength versus (grain diameter)1 2 for a 70 Cu–30 Zn cartridge brass, Figure 7.15, determine values for the constants s0 and ky in Equation 7.7. (b) Now predict the yield strength of this alloy when the average grain diameter is 2.0  103 mm. 7.24 The lower yield point for an iron that has an average grain diameter of 1  102 mm is 230 MPa (33,000 psi). At a grain diameter of 6  103 mm, the yield point increases to 275 MPa (40,000 psi). At what grain diameter will the lower yield point be 310 MPa (45,000 psi)? 7.25 If it is assumed that the plot in Figure 7.15 is for noncold-worked brass, determine the grain size of the alloy in Figure 7.19; assume its composition is the same as the alloy in Figure 7.15. Solid-Solution Strengthening 7.26 In the manner of Figures 7.17b and 7.18b, indicate the location in the vicinity of an edge dislocation at which an interstitial impurity atom would be expected to be situated. Now briefly explain in terms of lattice strains why it would be situated at this position. Strain Hardening 7.27 (a) Show, for a tensile test, that %CW  a  b  100 1 if there is no change in specimen volume during the deformation process (i.e., A0l0  Adld). (b) Using the result of part (a), compute the percent cold work experienced by naval brass (for which the stress–strain behavior is shown in Figure 6.12) when a stress of 415 MPa (60,000 psi) is applied. 7.28 Two previously undeformed cylindrical specimens of an alloy are to be strain hardened by reducing their cross-sectional areas (while maintaining their circular cross sections). For one specimen, the initial and deformed radii are 15 mm and 12 mm, respectively. The second specimen, with an initial radius of 11 mm, 1496T_c07_174-206 11/17/05 13:32 Page 205 REVISED PAGES Questions and Problems • 205 must have the same deformed hardness as the first specimen; compute the second specimen’s radius after deformation. 7.29 Two previously undeformed specimens of the same metal are to be plastically deformed by reducing their cross-sectional areas. One has a circular cross section, and the other is rectangular; during deformation the circular cross section is to remain circular, and the rectangular is to remain as such. Their original and deformed dimensions are as follows: Original dimensions Deformed dimensions Circular (diameter, mm) Rectangular (mm) 18.0 15.9 20  50 13.7  55.1 Which of these specimens will be the hardest after plastic deformation, and why? 7.30 A cylindrical specimen of cold-worked copper has a ductility (%EL) of 15%. If its coldworked radius is 6.4 mm (0.25 in.), what was its radius before deformation? 7.31 (a) What is the approximate ductility (%EL) of a brass that has a yield strength of 345 MPa (50,000 psi)? (b) What is the approximate Brinell hardness of a 1040 steel having a yield strength of 620 MPa (90,000 psi)? 7.32 Experimentally, it has been observed for single crystals of a number of metals that the critical resolved shear stress tcrss is a function of the dislocation density rD as tcrss  t0  A2rD where t0 and A are constants. For copper, the critical resolved shear stress is 0.69 MPa (100 psi) at a dislocation density of 104 mm2. If it is known that the value of t0 for copper is 0.069 MPa (10 psi), compute the tcrss at a dislocation density of 106 mm2. Recovery Recrystallization Grain Growth 7.33 Briefly cite the differences between recovery and recrystallization processes. 7.34 Estimate the fraction of recrystallization from the photomicrograph in Figure 7.21c. 7.35 Explain the differences in grain structure for a metal that has been cold worked and one that has been cold worked and then recrystallized. 7.36 (a) What is the driving force for recrystallization? (b) For grain growth? 7.37 (a) From Figure 7.25, compute the length of time required for the average grain diameter to increase from 0.03 to 0.3 mm at 600C for this brass material. (b) Repeat the calculation at 700C. 7.38 The average grain diameter for a brass material was measured as a function of time at 650C, which is tabulated below at two different times: Time (min) 40 100 Grain Diameter (mm) 5.6  102 8.0  102 (a) What was the original grain diameter? (b) What grain diameter would you predict after 200 min at 650C? 7.39 An undeformed specimen of some alloy has an average grain diameter of 0.050 mm. You are asked to reduce its average grain diameter to 0.020 mm. Is this possible? If so, explain the procedures you would use and name the processes involved. If it is not possible, explain why. 7.40 Grain growth is strongly dependent on temperature (i.e., rate of grain growth increases with increasing temperature), yet temperature is not explicitly given as a part of Equation 7.9. (a) Into which of the parameters in this expression would you expect temperature to be included? (b) On the basis of your intuition, cite an explicit expression for this temperature dependence. 7.41 An uncold-worked brass specimen of average grain size 0.01 mm has a yield strength of 150 MPa (21,750 psi). Estimate the yield strength of this alloy after it has been heated to 500C for 1000 s, if it is known that the value of s0 is 25 MPa (3625 psi). 1496T_c07_174-206 11/17/05 13:32 Page 206 REVISED PAGES 206 • Chapter 7 / Dislocations and Strengthening Mechanisms DESIGN PROBLEMS Strain Hardening Recrystallization 7.D1 Determine whether or not it is possible to cold work steel so as to give a minimum Brinell hardness of 240, and at the same time have a ductility of at least 15%EL. Justify your decision. 7.D2 Determine whether or not it is possible to cold work brass so as to give a minimum Brinell hardness of 150, and at the same time have a ductility of at least 20%EL. Justify your decision. 7.D3 A cylindrical specimen of cold-worked steel has a Brinell hardness of 240. (a) Estimate its ductility in percent elongation. (b) If the specimen remained cylindrical during deformation and its original radius was 10 mm (0.40 in.), determine its radius after deformation. 7.D4 It is necessary to select a metal alloy for an application that requires a yield strength of at least 310 MPa (45,000 psi) while maintaining a minimum ductility (%EL) of 27%. If the metal may be cold worked, decide which of the following are candidates: copper, brass, and a 1040 steel. Why? 7.D5 A cylindrical rod of 1040 steel originally 11.4 mm (0.45 in.) in diameter is to be cold worked by drawing; the circular cross section will be maintained during deformation. A cold-worked tensile strength in excess of 825 MPa (120,000 psi) and a ductility of at least 12%EL are desired. Furthermore, the final diameter must be 8.9 mm (0.35 in.). Explain how this may be accomplished. 7.D6 A cylindrical rod of brass originally 10.2 mm (0.40 in.) in diameter is to be cold worked by drawing; the circular cross section will be maintained during deformation. A coldworked yield strength in excess of 380 MPa (55, 000 psi) and a ductility of at least 15%EL are desired. Furthermore, the final diameter must be 7.6 mm (0.30 in.). Explain how this may be accomplished. 7.D7 A cylindrical brass rod having a minimum tensile strength of 450 MPa (65,000 psi), a ductility of at least 13%EL, and a final diameter of 12.7 mm (0.50 in.) is desired. Some 19.0 mm (0.75 in.) diameter brass stock that has been cold worked 35% is available. Describe the procedure you would follow to obtain this material. Assume that brass experiences cracking at 65%CW. 1496T_c08_207-251 12/21/05 8:28 Page 207 2nd REVISE PAGES Chapter A 8 Failure n oil tanker that fractured in a brittle manner by crack propagation around its girth. (Photography by Neal Boenzi. Reprinted with permission from The New York Times.) WHY STUDY Failure? The design of a component or structure often calls upon the engineer to minimize the possibility of failure. Thus, it is important to understand the mechanics of the various failure modes—i.e., fracture, fatigue, and creep—and, in addition, be familiar with appro- priate design principles that may be employed to prevent in-service failures. For example, we discuss in Sections 22.4 through 22.6 material selection and processing issues relating to the fatigue of an automobile valve spring. • 207 1496T_c08_207-251 12/21/05 7:51 Page 208 2nd REVISE PAGES Learning Objectives After studying this chapter you should be able to do the following: 1. Describe the mechanism of crack propagation 6. Define fatigue and specify the conditions under for both ductile and brittle modes of fracture. which it occurs. 2. Explain why the strengths of brittle materials are 7. From a fatigue plot for some material, determuch lower than predicted by theoretical calcumine (a) the fatigue lifetime (at a specified lations. stress level), and (b) the fatigue strength (at a 3. Define fracture toughness in terms of (a) a brief specified number of cycles). statement, and (b) an equation; define all pa8. Define creep and specify the conditions under rameters in this equation. which it occurs. 4. Make a distinction between fracture toughness 9. Given a creep plot for some material, determine and plane strain fracture toughness. (a) the steady-state creep rate, and (b) the 5. Name and describe the two impact fracture rupture lifetime. testing techniques. 8.1 INTRODUCTION The failure of engineering materials is almost always an undesirable event for several reasons; these include human lives that are put in jeopardy, economic losses, and the interference with the availability of products and services. Even though the causes of failure and the behavior of materials may be known, prevention of failures is difficult to guarantee. The usual causes are improper materials selection and processing and inadequate design of the component or its misuse. It is the responsibility of the engineer to anticipate and plan for possible failure and, in the event that failure does occur, to assess its cause and then take appropriate preventive measures against future incidents. The following topics are addressed in this chapter: simple fracture (both ductile and brittle modes), fundamentals of fracture mechanics, impact fracture testing, the ductile-to-brittle transition, fatigue, and creep. These discussions include failure mechanisms, testing techniques, and methods by which failure may be prevented or controlled. Concept Check 8.1 Cite two situations in which the possibility of failure is part of the design of a component or product. [The answer may be found at www.wiley.com/college/callister (Student Companion Site).] Fr a c t u r e 8.2 FUNDAMENTALS OF FRACTURE ductile, brittle fracture Simple fracture is the separation of a body into two or more pieces in response to an imposed stress that is static (i.e., constant or slowly changing with time) and at temperatures that are low relative to the melting temperature of the material. The applied stress may be tensile, compressive, shear, or torsional; the present discussion will be confined to fractures that result from uniaxial tensile loads. For engineering materials, two fracture modes are possible: ductile and brittle. Classification is based on the ability of a material to experience plastic deformation. Ductile materials typically exhibit substantial plastic deformation with high energy absorption before fracture. On the other hand, there is normally little or no plastic deformation with 1496T_c08_207-251 11/18/05 11:50 Page 209 REVISED PAGES 8.3 Ductile Fracture • 209 low energy absorption accompanying a brittle fracture. The tensile stress–strain behaviors of both fracture types may be reviewed in Figure 6.13. “Ductile”and “brittle” are relative terms; whether a particular fracture is one mode or the other depends on the situation. Ductility may be quantified in terms of percent elongation (Equation 6.11) and percent reduction in area (Equation 6.12). Furthermore, ductility is a function of temperature of the material, the strain rate, and the stress state. The disposition of normally ductile materials to fail in a brittle manner is discussed in Section 8.6. Any fracture process involves two steps—crack formation and propagation—in response to an imposed stress. The mode of fracture is highly dependent on the mechanism of crack propagation. Ductile fracture is characterized by extensive plastic deformation in the vicinity of an advancing crack. Furthermore, the process proceeds relatively slowly as the crack length is extended. Such a crack is often said to be stable. That is, it resists any further extension unless there is an increase in the applied stress. In addition, there will ordinarily be evidence of appreciable gross deformation at the fracture surfaces (e.g., twisting and tearing). On the other hand, for brittle fracture, cracks may spread extremely rapidly, with very little accompanying plastic deformation. Such cracks may be said to be unstable, and crack propagation, once started, will continue spontaneously without an increase in magnitude of the applied stress. Ductile fracture is almost always preferred for two reasons. First, brittle fracture occurs suddenly and catastrophically without any warning; this is a consequence of the spontaneous and rapid crack propagation. On the other hand, for ductile fracture, the presence of plastic deformation gives warning that fracture is imminent, allowing preventive measures to be taken. Second, more strain energy is required to induce ductile fracture inasmuch as ductile materials are generally tougher. Under the action of an applied tensile stress, most metal alloys are ductile, whereas ceramics are notably brittle, and polymers may exhibit both types of fracture. 8.3 DUCTILE FRACTURE Ductile fracture surfaces will have their own distinctive features on both macroscopic and microscopic levels. Figure 8.1 shows schematic representations for two characteristic macroscopic fracture profiles. The configuration shown in Figure 8.1a is found for extremely soft metals, such as pure gold and lead at room temperature, and other metals, polymers, and inorganic glasses at elevated temperatures. These highly ductile materials neck down to a point fracture, showing virtually 100% reduction in area. The most common type of tensile fracture profile for ductile metals is that represented in Figure 8.1b, where fracture is preceded by only a moderate amount of Figure 8.1 (a) Highly ductile fracture in which the specimen necks down to a point. (b) Moderately ductile fracture after some necking. (c) Brittle fracture without any plastic deformation. (a) (b) (c) 1496T_c08_207-251 11/18/05 11:50 Page 210 REVISED PAGES 210 • Chapter 8 / Failure (a) (b) (c) Shear Fibrous (d) Figure 8.2 Stages in the cup-and-cone fracture. (a) Initial necking. (b) Small cavity formation. (c) Coalescence of cavities to form a crack. (d) Crack propagation. (e) Final shear fracture at a 45 angle relative to the tensile direction. (From K. M. Ralls, T. H. Courtney, and J. Wulff, Introduction to Materials Science and Engineering, p. 468. Copyright © 1976 by John Wiley & Sons, New York. Reprinted by permission of John Wiley & Sons, Inc.) (e) necking. The fracture process normally occurs in several stages (Figure 8.2). First, after necking begins, small cavities, or microvoids, form in the interior of the cross section, as indicated in Figure 8.2b. Next, as deformation continues, these microvoids enlarge, come together, and coalesce to form an elliptical crack, which has its long axis perpendicular to the stress direction. The crack continues to grow in a direction parallel to its major axis by this microvoid coalescence process (Figure 8.2c). Finally, fracture ensues by the rapid propagation of a crack around the outer perimeter of the neck (Figure 8.2d), by shear deformation at an angle of about 45 with the tensile axis—this is the angle at which the shear stress is a maximum. Sometimes a fracture having this characteristic surface contour is termed a cup-and-cone fracture because one of the mating surfaces is in the form of a cup, the other like a cone. In this type of fractured specimen (Figure 8.3a), the central interior region of the surface has an irregular and fibrous appearance, which is indicative of plastic deformation. Fractographic Studies Much more detailed information regarding the mechanism of fracture is available from microscopic examination, normally using scanning electron microscopy. Studies (a) (b) Figure 8.3 (a) Cup-and-cone fracture in aluminum. (b) Brittle fracture in a mild steel. 1496T_c08_207-251 11/18/05 11:50 Page 211 REVISED PAGES 8.4 Brittle Fracture • 211 Figure 8.4 (a) Scanning electron fractograph showing spherical dimples characteristic of ductile fracture resulting from uniaxial tensile loads. 3300. (b) Scanning electron fractograph showing parabolic-shaped dimples characteristic of ductile fracture resulting from shear loading. 5000. (From R. W. Hertzberg, Deformation and Fracture Mechanics of Engineering Materials, 3rd edition. Copyright © 1989 by John Wiley & Sons, New York. Reprinted by permission of John Wiley & Sons, Inc.) of this type are termed fractographic. The scanning electron microscope is preferred for fractographic examinations since it has a much better resolution and depth of field than does the optical microscope; these characteristics are necessary to reveal the topographical features of fracture surfaces. When the fibrous central region of a cup-and-cone fracture surface is examined with the electron microscope at a high magnification, it will be found to consist of numerous spherical “dimples” (Figure 8.4a); this structure is characteristic of fracture resulting from uniaxial tensile failure. Each dimple is one half of a microvoid that formed and then separated during the fracture process. Dimples also form on the 45 shear lip of the cup-and-cone fracture. However, these will be elongated or C-shaped, as shown in Figure 8.4b. This parabolic shape may be indicative of shear failure. Furthermore, other microscopic fracture surface features are also possible. Fractographs such as those shown in Figures 8.4a and 8.4b provide valuable information in the analyses of fracture, such as the fracture mode, the stress state, and the site of crack initiation. 8.4 BRITTLE FRACTURE Brittle fracture takes place without any appreciable deformation, and by rapid crack propagation. The direction of crack motion is very nearly perpendicular to the direction of the applied tensile stress and yields a relatively flat fracture surface, as indicated in Figure 8.1c. Fracture surfaces of materials that failed in a brittle manner will have their own distinctive patterns; any signs of gross plastic deformation will be absent. For example, in some steel pieces, a series of V-shaped “chevron” markings may form near the center of the fracture cross section that point back toward the crack initiation site (Figure 8.5a). Other brittle fracture surfaces contain lines or ridges that radiate from the origin of the crack in a fanlike pattern (Figure 8.5b). Often, both of these marking patterns will be sufficiently coarse to be discerned with the naked eye. For very hard and fine-grained metals, there will be no discernible fracture pattern. Brittle fracture in amorphous materials, such as ceramic glasses, yields a relatively shiny and smooth surface. 1496T_c08_207-251 12/21/05 7:51 Page 212 2nd REVISE PAGES 212 • Chapter 8 / Failure (a) (b) Figure 8.5 (a) Photograph showing V-shaped “chevron” markings characteristic of brittle fracture. Arrows indicate origin of crack. Approximately actual size. (b) Photograph of a brittle fracture surface showing radial fan-shaped ridges. Arrow indicates origin of crack. Approximately 2. [(a) From R. W. Hertzberg, Deformation and Fracture Mechanics of Engineering Materials, 3rd edition. Copyright © 1989 by John Wiley & Sons, New York. Reprinted by permission of John Wiley & Sons, Inc. Photograph courtesy of Roger Slutter, Lehigh University. (b) Reproduced with permission from D. J. Wulpi, Understanding How Components Fail, American Society for Metals, Materials Park, OH, 1985.] transgranular fracture intergranular fracture For most brittle crystalline materials, crack propagation corresponds to the successive and repeated breaking of atomic bonds along specific crystallographic planes (Figure 8.6a); such a process is termed cleavage. This type of fracture is said to be transgranular (or transcrystalline), because the fracture cracks pass through the grains. Macroscopically, the fracture surface may have a grainy or faceted texture (Figure 8.3b), as a result of changes in orientation of the cleavage planes from grain to grain. This cleavage feature is shown at a higher magnification in the scanning electron micrograph of Figure 8.6b. In some alloys, crack propagation is along grain boundaries (Figure 8.7a); this fracture is termed intergranular. Figure 8.7b is a scanning electron micrograph 1496T_c08_207-251 11/18/05 13:43 Page 213 REVISED PAGES 8.4 Brittle Fracture • 213 SEM Micrograph Path of crack propagation Grains (a) (b) Figure 8.6 (a) Schematic cross-section profile showing crack propagation through the interior of grains for transgranular fracture. (b) Scanning electron fractograph of ductile cast iron showing a transgranular fracture surface. Magnification unknown. [Figure (b) from V. J. Colangelo and F. A. Heiser, Analysis of Metallurgical Failures, 2nd edition. Copyright © 1987 by John Wiley & Sons, New York. Reprinted by permission of John Wiley & Sons, Inc.] 1496T_c08_207-251 11/18/05 13:43 Page 214 REVISED PAGES 214 • Chapter 8 / Failure SEM Micrograph Grain boundaries Path of crack propagation (a) (b) Figure 8.7 (a) Schematic cross-section profile showing crack propagation along grain boundaries for intergranular fracture. (b) Scanning electron fractograph showing an intergranular fracture surface. 50. [Figure (b) reproduced with permission from ASM Handbook, Vol. 12, Fractography, ASM International, Materials Park, OH, 1987.] 1496T_c08_207-251 11/18/05 11:50 Page 215 REVISED PAGES 8.5 Principles of Fracture Mechanics • 215 showing a typical intergranular fracture, in which the three-dimensional nature of the grains may be seen. This type of fracture normally results subsequent to the occurrence of processes that weaken or embrittle grain boundary regions. 8.5 PRINCIPLES OF FRACTURE MECHANICS Brittle fracture of normally ductile materials, such as that shown in the chapteropening photograph for this chapter, has demonstrated the need for a better understanding of the mechanisms of fracture. Extensive research endeavors over the past several decades have led to the evolution of the field of fracture mechanics. This subject allows quantification of the relationships between material properties, stress level, the presence of crack-producing flaws, and crack propagation mechanisms. Design engineers are now better equipped to anticipate, and thus prevent, structural failures. The present discussion centers on some of the fundamental principles of the mechanics of fracture. fracture mechanics Stress Concentration The measured fracture strengths for most brittle materials are significantly lower than those predicted by theoretical calculations based on atomic bonding energies. This discrepancy is explained by the presence of very small, microscopic flaws or cracks that always exist under normal conditions at the surface and within the interior of a body of material. These flaws are a detriment to the fracture strength because an applied stress may be amplified or concentrated at the tip, the magnitude of this amplification depending on crack orientation and geometry. This phenomenon is demonstrated in Figure 8.8, a stress profile across a cross section containing an internal crack. As indicated by this profile, the magnitude of this localized stress diminishes with distance away from the crack tip. At positions far σ0 σm Stress ρt a X X' x 2a x' σ0 x x' Position along X–X' (a) σ0 (b) Figure 8.8 (a) The geometry of surface and internal cracks. (b) Schematic stress profile along the line X–X ¿ in (a), demonstrating stress amplification at crack tip positions. 1496T_c08_207-251 11/18/05 11:50 Page 216 REVISED PAGES 216 • Chapter 8 / Failure stress raiser removed, the stress is just the nominal stress s0, or the applied load divided by the specimen cross-sectional area (perpendicular to this load). Due to their ability to amplify an applied stress in their locale, these flaws are sometimes called stress raisers. If it is assumed that a crack is similar to an elliptical hole through a plate, and is oriented perpendicular to the applied stress, the maximum stress, sm, occurs at the crack tip and may be approximated by For tensile loading, computation of maximum stress at a crack tip a 1 2 sm  2s0 a b rt (8.1) where s0 is the magnitude of the nominal applied tensile stress, rt is the radius of curvature of the crack tip (Figure 8.8a), and a represents the length of a surface crack, or half of the length of an internal crack. For a relatively long microcrack that has a small tip radius of curvature, the factor 1art 2 1 2 may be very large. This will yield a value of sm that is many times the value of s0. Sometimes the ratio sms0 is denoted as the stress concentration factor Kt: Kt  sm a 1 2  2a b s0 rt (8.2) which is simply a measure of the degree to which an external stress is amplified at the tip of a crack. By way of comment, it should be said that stress amplification is not restricted to these microscopic defects; it may occur at macroscopic internal discontinuities (e.g., voids), at sharp corners, and at notches in large structures. Furthermore, the effect of a stress raiser is more significant in brittle than in ductile materials. For a ductile material, plastic deformation ensues when the maximum stress exceeds the yield strength. This leads to a more uniform distribution of stress in the vicinity of the stress raiser and to the development of a maximum stress concentration factor less than the theoretical value. Such yielding and stress redistribution do not occur to any appreciable extent around flaws and discontinuities in brittle materials; therefore, essentially the theoretical stress concentration will result. Using principles of fracture mechanics, it is possible to show that the critical stress sc required for crack propagation in a brittle material is described by the expression Critical stress for crack propagation in a brittle material sc  a 2Egs 1 2 b pa (8.3) where E  modulus of elasticity gs  specific surface energy a  one half the length of an internal crack All brittle materials contain a population of small cracks and flaws that have a variety of sizes, geometries, and orientations. When the magnitude of a tensile stress at the tip of one of these flaws exceeds the value of this critical stress, a crack forms and then propagates, which results in fracture. Very small and virtually defect-free metallic and ceramic whiskers have been grown with fracture strengths that approach their theoretical values. 1496T_c08_207-251 11/18/05 11:50 Page 217 REVISED PAGES 8.5 Principles of Fracture Mechanics • 217 EXAMPLE PROBLEM 8.1 Maximum Flaw Length Computation A relatively large plate of a glass is subjected to a tensile stress of 40 MPa. If the specific surface energy and modulus of elasticity for this glass are 0.3 J/m2 and 69 GPa, respectively, determine the maximum length of a surface flaw that is possible without fracture. Solution To solve this problem it is necessary to employ Equation 8.3. Rearrangement of this expression such that a is the dependent variable, and realizing that s  40 MPa, gs  0.3 J/m2, and E  69 GPa leads to a  2Egs ps2 122169  109 N/m2 210.3 N/m2 p 140  106 N/m2 2 2  8.2  106 m  0.0082 mm  8.2 mm Fracture Toughness Furthermore, using fracture mechanical principles, an expression has been developed that relates this critical stress for crack propagation (sc) and crack length (a) as Fracture toughness— dependence on critical stress for crack propagation and crack length fracture toughness Kc  Ysc 1pa (8.4) In this expression Kc is the fracture toughness, a property that is a measure of a material’s resistance to brittle fracture when a crack is present. Worth noting is that Kc has the unusual units of MPa 1m or psi 1in. (alternatively, ksi 1in.). Furthermore, Y is a dimensionless parameter or function that depends on both crack and specimen sizes and geometries, as well as the manner of load application. Relative to this Y parameter, for planar specimens containing cracks that are much shorter than the specimen width, Y has a value of approximately unity. For example, for a plate of infinite width having a through-thickness crack (Figure 8.9a), Figure 8.9 Schematic representations of (a) an interior crack in a plate of infinite width, and (b) an edge crack in a plate of semi-infinite width. 2a (a) a (b) 1496T_c08_207-251 11/18/05 11:50 Page 218 REVISED PAGES 218 • Chapter 8 / Failure Figure 8.10 The three modes of crack surface displacement. (a) Mode I, opening or tensile mode; (b) mode II, sliding mode; and (c) mode III, tearing mode. (a) plane strain plane strain fracture toughness Plane strain fracture toughness for mode I crack surface displacement (b) (c) Y  1.0; whereas for a plate of semi-infinite width containing an edge crack of length a (Figure 8.9b), Y  1.1. Mathematical expressions for Y have been determined for a variety of crack-specimen geometries; these expressions are often relatively complex. For relatively thin specimens, the value of Kc will depend on specimen thickness. However, when specimen thickness is much greater than the crack dimensions, Kc becomes independent of thickness; under these conditions a condition of plane strain exists. By plane strain we mean that when a load operates on a crack in the manner represented in Figure 8.9a, there is no strain component perpendicular to the front and back faces. The Kc value for this thick-specimen situation is known as the plane strain fracture toughness KIc; furthermore, it is also defined by KIc  Ys2pa (8.5) KIc is the fracture toughness cited for most situations. The I (i.e., Roman numeral “one”) subscript for KIc denotes that the plane strain fracture toughness is for mode I crack displacement, as illustrated in Figure 8.10a.1 Brittle materials, for which appreciable plastic deformation is not possible in front of an advancing crack, have low KIc values and are vulnerable to catastrophic failure. On the other hand, KIc values are relatively large for ductile materials. Fracture mechanics is especially useful in predicting catastrophic failure in materials having intermediate ductilities. Plane strain fracture toughness values for a number of different materials are presented in Table 8.1 (and Figure 1.6); a more extensive list of KIc values is contained in Table B.5, Appendix B. The plane strain fracture toughness KIc is a fundamental material property that depends on many factors, the most influential of which are temperature, strain rate, and microstructure. The magnitude of KIc diminishes with increasing strain rate and decreasing temperature. Furthermore, an enhancement in yield strength wrought by solid solution or dispersion additions or by strain hardening generally produces a corresponding decrease in KIc. Furthermore, KIc normally increases with reduction in grain size as composition and other microstructural variables are maintained constant. Yield strengths are included for some of the materials listed in Table 8.1. Several different testing techniques are used to measure KIc.2 Virtually any specimen size and shape consistent with mode I crack displacement may be utilized, 1 Two other crack displacement modes denoted by II and III and as illustrated in Figures 8.10b and 8.10c are also possible; however, mode I is most commonly encountered. 2 See for example ASTM Standard E 399, “Standard Test Method for Plane Strain Fracture Toughness of Metallic Materials.” 1496T_c08_207-251 11/18/05 11:50 Page 219 REVISED PAGES 8.5 Principles of Fracture Mechanics • 219 Table 8.1 Room-Temperature Yield Strength and Plane Strain Fracture Toughness Data for Selected Engineering Materials Yield Strength Material MPa KIc ksi MPa 1m ksi 1in . Metals Aluminum Alloya (7075-T651) Aluminum Alloya (2024-T3) Titanium Alloya (Ti-6Al-4V) Alloy Steela (4340 tempered @ 260C) Alloy Steela (4340 tempered @ 425C) 495 72 24 22 345 50 44 40 910 132 55 50 1640 238 50.0 45.8 1420 206 87.4 80.0 — — — 0.2–1.4 0.7–0.8 2.7–5.0 0.18–1.27 0.64–0.73 2.5–4.6 Ceramics Concrete Soda-Lime Glass Aluminum Oxide — — — Polystyrene (PS) Poly(methyl methacrylate) (PMMA) Polycarbonate (PC) — — 0.7–1.1 0.64–1.0 53.8–73.1 7.8–10.6 0.7–1.6 0.64–1.5 62.1 9.0 2.2 2.0 Polymers a Source: Reprinted with permission, Advanced Materials and Processes, ASM International, © 1990. and accurate values will be realized provided that the Y scale parameter in Equation 8.5 has been properly determined. Design Using Fracture Mechanics According to Equations 8.4 and 8.5, three variables must be considered relative to the possibility for fracture of some structural component—namely, the fracture toughness (Kc) or plane strain fracture toughness (KIc), the imposed stress (s), and the flaw size (a)—assuming, of course, that Y has been determined. When designing a component, it is first important to decide which of these variables are constrained by the application and which are subject to design control. For example, material selection (and hence Kc or KIc) is often dictated by factors such as density (for lightweight applications) or the corrosion characteristics of the environment. Or, the allowable flaw size is either measured or specified by the limitations of available flaw detection techniques. It is important to realize, however, that once any combination of two of the above parameters is prescribed, the third becomes fixed (Equations 8.4 and 8.5). For example, assume that KIc and the magnitude of a are specified by application constraints; therefore, the design (or critical) stress sc must be Computation of design stress sc  KIc Y1pa (8.6) 1496T_c08_207-251 11/18/05 11:50 Page 220 REVISED PAGES 220 • Chapter 8 / Failure Table 8.2 A List of Several Common Nondestructive Testing (NDT) Techniques Technique Scanning electron microscopy (SEM) Dye penetrant Ultrasonics Optical microscopy Visual inspection Acoustic emission Radiography (X-ray/gamma ray) Defect Location Defect Size Sensitivity (mm) Testing Location Surface 0.001 Laboratory Surface Subsurface Surface Surface Surface/subsurface Subsurface 0.025–0.25 0.050 0.1–0.5 0.1 0.1 2% of specimen thickness Laboratory/in-field Laboratory/in-field Laboratory Laboratory/in-field Laboratory/in-field Laboratory/in-field On the other hand, if stress level and plane strain fracture toughness are fixed by the design situation, then the maximum allowable flaw size ac is Computation of maximum allowable flaw length ac  1 KIc 2 a b p sY (8.7) A number of nondestructive test (NDT) techniques have been developed that permit detection and measurement of both internal and surface flaws.3 Such techniques are used to examine structural components that are in service for defects and flaws that could lead to premature failure; in addition, NDTs are used as a means of quality control for manufacturing processes. As the name implies, these techniques must not destroy the material/structure being examined. Furthermore, some testing methods must be conducted in a laboratory setting; others may be adapted for use in the field. Several commonly employed NDT techniques and their characteristics are listed in Table 8.2. One important example of the use of NDT is for the detection of cracks and leaks in the walls of oil pipelines in remote areas such as Alaska. Ultrasonic analysis is utilized in conjunction with a “robotic analyzer” that can travel relatively long distances within a pipeline. DESIGN EXAMPLE 8.1 Material Specification for a Pressurized Spherical Tank Consider the thin-walled spherical tank of radius r and thickness t (Figure 8.11) that may be used as a pressure vessel. (a) One design of such a tank calls for yielding of the wall material prior to failure as a result of the formation of a crack of critical size and its subsequent rapid propagation. Thus, plastic distortion of the wall may be observed and the pressure within the tank released before the occurrence of catastrophic failure. Consequently, materials having large critical crack lengths are desired. On the basis 3 Sometimes the terms nondestructive evaluation (NDE) and nondestructive inspection (NDI) are also used for these techniques. 1496T_c08_207-251 11/18/05 11:50 Page 221 REVISED PAGES 8.5 Principles of Fracture Mechanics • 221 ␴ 2a p p p t r p p p Figure 8.11 Schematic diagram showing the cross section of a spherical tank that is subjected to an internal pressure p, and that has a radial crack of length 2a in its wall. ␴ p p of this criterion, rank the metal alloys listed in Table B.5,Appendix B, as to critical crack size, from longest to shortest. (b) An alternative design that is also often utilized with pressure vessels is termed leak-before-break. Using principles of fracture mechanics, allowance is made for the growth of a crack through the thickness of the vessel wall prior to the occurrence of rapid crack propagation (Figure 8.11). Thus, the crack will completely penetrate the wall without catastrophic failure, allowing for its detection by the leaking of pressurized fluid. With this criterion the critical crack length ac (i.e., one-half of the total internal crack length) is taken to be equal to the pressure vessel thickness t. Allowance for ac  t instead of ac  t2 assures that fluid leakage will occur prior to the buildup of dangerously high pressures. Using this criterion, rank the metal alloys in Table B.5, Appendix B as to the maximum allowable pressure. For this spherical pressure vessel, the circumferential wall stress s is a function of the pressure p in the vessel and the radius r and wall thickness t according to pr s (8.8) 2t For both parts (a) and (b) assume a condition of plane strain. Solution (a) For the first design criterion, it is desired that the circumferential wall stress be less than the yield strength of the material. Substitution of sy for s in Equation 8.5, and incorporation of a factor of safety N leads to KIc  Y a sy N b 1pac (8.9) where ac is the critical crack length. Solving for ac yields the following expression: ac  N 2 KIc 2 a b Y 2p sy (8.10) Therefore, the critical crack length is proportional to the square of the KIc-sy ratio, which is the basis for the ranking of the metal alloys in Table B.5. The 1496T_c08_207-251 11/18/05 11:50 Page 222 REVISED PAGES 222 • Chapter 8 / Failure Table 8.3 Ranking of Several Metal Alloys Relative to Critical Crack Length (Yielding Criterion) for a ThinWalled Spherical Pressure Vessel Material Medium carbon (1040) steel AZ31B magnesium 2024 aluminum (T3) Ti-5Al-2.5Sn titanium 4140 steel (tempered @ 482C) 4340 steel (tempered @ 425C) Ti-6Al-4V titanium 17-7PH steel 7075 aluminum (T651) 4140 steel (tempered @ 370C) 4340 steel (tempered @ 260C) a KIc Sy 2 b (mm) 43.1 19.6 16.3 6.6 5.3 3.8 3.7 3.4 2.4 1.6 0.93 ranking is provided in Table 8.3, where it may be seen that the medium carbon (1040) steel with the largest ratio has the longest critical crack length, and, therefore, is the most desirable material on the basis of this criterion. (b) As stated previously, the leak-before-break criterion is just met when onehalf of the internal crack length is equal to the thickness of the pressure vessel— that is, when a  t. Substitution of a  t into Equation 8.5 gives KIc  Ys1pt (8.11) and, from Equation 8.8 t pr 2s (8.12) The stress is replaced by the yield strength, inasmuch as the tank should be designed to contain the pressure without yielding; furthermore, substitution of Equation 8.12 into Equation 8.11, after some rearrangement, yields the following expression: K 2Ic 2 (8.13) p 2 a b Y pr sy Hence, for some given spherical vessel of radius r, the maximum allowable pressure consistent with this leak-before-break criterion is proportional to K 2Ic sy. The same several materials are ranked according to this ratio in Table 8.4; as may be noted, the medium carbon steel will contain the greatest pressures. Of the 11 metal alloys that are listed in Table B.5, the medium carbon steel ranks first according to both yielding and leak-before-break criteria. For these reasons, many pressure vessels are constructed of medium carbon steels, when temperature extremes and corrosion need not be considered. 1496T_c08_207-251 11/18/05 11:50 Page 223 REVISED PAGES 8.6 Impact Fracture Testing • 223 Table 8.4 Ranking of Several Metal Alloys Relative to Maximum Allowable Pressure (Leak-Before-Break Criterion) for a Thin-Walled Spherical Pressure Vessel KIc2 Material Medium carbon (1040) steel 4140 steel (tempered @ 482C) Ti-5Al-2.5Sn titanium 2024 aluminum (T3) 4340 steel (tempered @ 425C) 17-7PH steel AZ31B magnesium Ti-6Al-4V titanium 4140 steel (tempered @ 370C) 4340 steel (tempered @ 260C) 7075 aluminum (T651) Sy (MPa-m) 11.2 6.1 5.8 5.6 5.4 4.4 3.9 3.3 2.4 1.5 1.2 8.6 IMPACT FRACTURE TESTING Prior to the advent of fracture mechanics as a scientific discipline, impact testing techniques were established so as to ascertain the fracture characteristics of materials. It was realized that the results of laboratory tensile tests could not be extrapolated to predict fracture behavior; for example, under some circumstances normally ductile metals fracture abruptly and with very little plastic deformation. Impact test conditions were chosen to represent those most severe relative to the potential for fracture—namely, (1) deformation at a relatively low temperature, (2) a high strain rate (i.e., rate of deformation), and (3) a triaxial stress state (which may be introduced by the presence of a notch). Impact Testing Techniques Charpy, Izod tests impact energy Two standardized tests,4 the Charpy and Izod, were designed and are still used to measure the impact energy, sometimes also termed notch toughness. The Charpy V-notch (CVN) technique is most commonly used in the United States. For both Charpy and Izod, the specimen is in the shape of a bar of square cross section, into which a V-notch is machined (Figure 8.12a). The apparatus for making V-notch impact tests is illustrated schematically in Figure 8.12b. The load is applied as an impact blow from a weighted pendulum hammer that is released from a cocked position at a fixed height h. The specimen is positioned at the base as shown. Upon release, a knife edge mounted on the pendulum strikes and fractures the specimen at the notch, which acts as a point of stress concentration for this high-velocity 4 ASTM Standard E 23, “Standard Test Methods for Notched Bar Impact Testing of Metallic Materials.” 1496T_c08_207-251 11/18/05 11:50 Page 224 REVISED PAGES 224 • Chapter 8 / Failure Figure 8.12 (a) Specimen used for Charpy and Izod impact tests. (b) A schematic drawing of an impact testing apparatus. The hammer is released from fixed height h and strikes the specimen; the energy expended in fracture is reflected in the difference between h and the swing height h¿. Specimen placements for both Charpy and Izod tests are also shown. [Figure (b) adapted from H. W. Hayden, W. G. Moffatt, and J. Wulff, The Structure and Properties of Materials, Vol. III, Mechanical Behavior, p. 13. Copyright © 1965 by John Wiley & Sons, New York. Reprinted by permission of John Wiley & Sons, Inc.] 8 mm (0.32 in.) (a) 10 mm (0.39 in.) 10 mm (0.39 in.) Notch Scale Charpy Izod Pointer Starting position Hammer End of swing h Specimen h' Anvil (b) impact blow. The pendulum continues its swing, rising to a maximum height h¿, which is lower than h. The energy absorption, computed from the difference between h and h¿, is a measure of the impact energy. The primary difference between the Charpy and Izod techniques lies in the manner of specimen support, as illustrated in Figure 8.12b. Furthermore, these are termed impact tests in light of the manner of load application. Variables including specimen size and shape as well as notch configuration and depth influence the test results. 1496T_c08_207-251 11/18/05 11:50 Page 225 REVISED PAGES 8.6 Impact Fracture Testing • 225 Both plane strain fracture toughness and these impact tests determine the fracture properties of materials. The former are quantitative in nature, in that a specific property of the material is determined (i.e., KIc). The results of the impact tests, on the other hand, are more qualitative and are of little use for design purposes. Impact energies are of interest mainly in a relative sense and for making comparisons— absolute values are of little significance. Attempts have been made to correlate plane strain fracture toughnesses and CVN energies, with only limited success. Plane strain fracture toughness tests are not as simple to perform as impact tests; furthermore, equipment and specimens are more expensive. Ductile-to-Brittle Transition Temperature (°F) –40 0 40 80 120 160 200 240 280 100 A 80 100 Impact energy 60 80 Shear fracture 60 40 B 40 20 20 0 0 –40 –20 0 20 40 60 Temperature (°C) 80 100 120 140 Shear fracture (%) Figure 8.13 Temperature dependence of the Charpy V-notch impact energy (curve A) and percent shear fracture (curve B) for an A283 steel. (Reprinted from Welding Journal. Used by permission of the American Welding Society.) One of the primary functions of Charpy and Izod tests is to determine whether or not a material experiences a ductile-to-brittle transition with decreasing temperature and, if so, the range of temperatures over which it occurs. The ductile-to-brittle transition is related to the temperature dependence of the measured impact energy absorption. This transition is represented for a steel by curve A in Figure 8.13. At higher temperatures the CVN energy is relatively large, in correlation with a ductile mode of fracture. As the temperature is lowered, the impact energy drops suddenly over a relatively narrow temperature range, below which the energy has a constant but small value; that is, the mode of fracture is brittle. Alternatively, appearance of the failure surface is indicative of the nature of fracture and may be used in transition temperature determinations. For ductile fracture this surface appears fibrous or dull (or of shear character), as in the steel specimen of Figure 8.14 that was tested at 79C. Conversely, totally brittle surfaces have a granular (shiny) texture (or cleavage character) (the 59C specimen, Figure 8.14). Over the ductile-to-brittle transition, features of both types will exist (in Figure 8.14, displayed by specimens tested at 12C, 4C, 16C, and 24C). Frequently, the percent shear fracture is plotted as a function of temperature—curve B in Figure 8.13. Impact energy (J) ductile-to-brittle transition 1496T_c08_207-251 11/18/05 11:50 Page 226 REVISED PAGES 226 • Chapter 8 / Failure 59 12 4 16 24 79 Figure 8.14 Photograph of fracture surfaces of A36 steel Charpy V-notch specimens tested at indicated temperatures (in C). (From R. W. Hertzberg, Deformation and Fracture Mechanics of Engineering Materials, 3rd edition, Fig. 9.6, p. 329. Copyright © 1989 by John Wiley & Sons, Inc., New York. Reprinted by permission of John Wiley & Sons, Inc.) For many alloys there is a range of temperatures over which the ductile-to-brittle transition occurs (Figure 8.13); this presents some difficulty in specifying a single ductile-to-brittle transition temperature. No explicit criterion has been established, and so this temperature is often defined as that temperature at which the CVN energy assumes some value (e.g., 20 J or 15 ft-lbf), or corresponding to some given fracture appearance (e.g., 50% fibrous fracture). Matters are further complicated inasmuch as a different transition temperature may be realized for each of these criteria. Perhaps the most conservative transition temperature is that at which the fracture surface becomes 100% fibrous; on this basis, the transition temperature is approximately 110C (230F) for the steel alloy that is the subject of Figure 8.13. Structures constructed from alloys that exhibit this ductile-to-brittle behavior should be used only at temperatures above the transition temperature, to avoid brittle and catastrophic failure. Classic examples of this type of failure occurred, with disastrous consequences, during World War II when a number of welded transport ships, away from combat, suddenly and precipitously split in half.The vessels were constructed of a steel alloy that possessed adequate ductility according to room-temperature tensile tests.The brittle fractures occurred at relatively low ambient temperatures, at about 4C (40F), in the vicinity of the transition temperature of the alloy. Each fracture crack originated at some point of stress concentration, probably a sharp corner or fabrication defect, and then propagated around the entire girth of the ship. In addition to the ductile-to-brittle transition represented in Figure 8.13, two other general types of impact energy-versus-temperature behavior have been observed; these are represented schematically by the upper and lower curves of Figure 8.15. Here it Impact energy Low-strength (FCC and HCP) metals Low-strength steels (BCC) High-strength materials Temperature Figure 8.15 Schematic curves for the three general types of impact energy-versus-temperature behavior. 1496T_c08_207-251 11/18/05 13:43 Page 227 REVISED PAGES 8.6 Impact Fracture Testing • 227 Temperature (°F) 0 200 –200 400 240 300 0.01 0.11 160 200 0.22 120 0.31 0.43 80 100 Impact energy (ft-lbf) 200 Impact energy (J) Figure 8.16 Influence of carbon content on the Charpy V-notch energy-versustemperature behavior for steel. (Reprinted with permission from ASM International, Metals Park, OH 44073-9989, USA; J. A. Reinbolt and W. J. Harris, Jr., “Effect of Alloying Elements on Notch Toughness of Pearlitic Steels,” Transactions of ASM, Vol. 43, 1951.) 0.53 0.63 40 0.67 0 –200 –100 0 100 Temperature (°C) 200 0 may be noted that low-strength FCC metals (some aluminum and copper alloys) and most HCP metals do not experience a ductile-to-brittle transition (corresponding to the upper curve of Figure 8.15), and retain high impact energies (i.e., remain ductile) with decreasing temperature. For high-strength materials (e.g., high-strength steels and titanium alloys), the impact energy is also relatively insensitive to temperature (the lower curve of Figure 8.15); however, these materials are also very brittle, as reflected by their low impact energy values. And, of course, the characteristic ductile-to-brittle transition is represented by the middle curve of Figure 8.15. As noted, this behavior is typically found in low-strength steels that have the BCC crystal structure. For these low-strength steels, the transition temperature is sensitive to both alloy composition and microstructure. For example, decreasing the average grain size results in a lowering of the transition temperature. Hence, refining the grain size both strengthens (Section 7.8) and toughens steels. In contrast, increasing the carbon content, while increasing the strength of steels, also raises the CVN transition of steels, as indicated in Figure 8.16. Most ceramics and polymers also experience a ductile-to-brittle transition. For ceramic materials, the transition occurs only at elevated temperatures, ordinarily in excess of 1000C (1850F). This behavior as related to polymers is discussed in Section 15.6. Fa t i g u e fatigue Fatigue is a form of failure that occurs in structures subjected to dynamic and fluctuating stresses (e.g., bridges, aircraft, and machine components). Under these circumstances it is possible for failure to occur at a stress level considerably lower than the tensile or yield strength for a static load. The term “fatigue” is used because this type of failure normally occurs after a lengthy period of repeated stress or strain cycling. Fatigue is important inasmuch as it is the single largest cause of failure in metals, estimated to comprise approximately 90% of all metallic failures; polymers and 1496T_c08_207-251 11/18/05 11:50 Page 228 REVISED PAGES 228 • Chapter 8 / Failure ceramics (except for glasses) are also susceptible to this type of failure. Furthermore, fatigue is catastrophic and insidious, occurring very suddenly and without warning. Fatigue failure is brittlelike in nature even in normally ductile metals, in that there is very little, if any, gross plastic deformation associated with failure. The process occurs by the initiation and propagation of cracks, and ordinarily the fracture surface is perpendicular to the direction of an applied tensile stress. 8.7 CYCLIC STRESSES max + 0 – Stress Compression Tension The applied stress may be axial (tension-compression), flexural (bending), or torsional (twisting) in nature. In general, three different fluctuating stress–time modes are possible. One is represented schematically by a regular and sinusoidal time dependence in Figure 8.17a, wherein the amplitude is symmetrical about a mean zero stress level, for example, alternating from a maximum tensile stress (smax) to a minimum compressive stress (smin) of equal magnitude; this is referred to as a reversed stress cycle. min Time (a) a r + m 0 – min Time (b) + – Stress Compression Tension Stress Compression Tension max Time (c) Figure 8.17 Variation of stress with time that accounts for fatigue failures. (a) Reversed stress cycle, in which the stress alternates from a maximum tensile stress (+) to a maximum compressive stress () of equal magnitude. (b) Repeated stress cycle, in which maximum and minimum stresses are asymmetrical relative to the zero-stress level; mean stress sm, range of stress sr, and stress amplitude sa are indicated. (c) Random stress cycle. 1496T_c08_207-251 11/18/05 13:43 Page 229 REVISED PAGES 8.8 The S–N Curve • 229 Another type, termed repeated stress cycle, is illustrated in Figure 8.17b; the maxima and minima are asymmetrical relative to the zero stress level. Finally, the stress level may vary randomly in amplitude and frequency, as exemplified in Figure 8.17c. Also indicated in Figure 8.17b are several parameters used to characterize the fluctuating stress cycle. The stress amplitude alternates about a mean stress sm, defined as the average of the maximum and minimum stresses in the cycle, or Mean stress for cyclic loading—dependence on maximum and minimum stress levels Computation of range of stress for cyclic loading sm  smax  smin 2 (8.14) Furthermore, the range of stress sr is just the difference between smax and smin — namely, sr  smax  smin (8.15) Stress amplitude sa is just one half of this range of stress, or Computation of stress amplitude for cyclic loading sa  smax  smin sr  2 2 (8.16) Finally, the stress ratio R is just the ratio of minimum and maximum stress amplitudes: Computation of stress ratio R smin smax (8.17) By convention, tensile stresses are positive and compressive stresses are negative. For example, for the reversed stress cycle, the value of R is 1. Concept Check 8.2 Make a schematic sketch of a stress-versus-time plot for the situation when the stress ratio R has a value of 1. [The answer may be found at www.wiley.com/college/callister (Student Companion Site).] Concept Check 8.3 Using Equations 8.16 and 8.17, demonstrate that increasing the value of the stress ratio R produces a decrease in stress amplitude sa. [The answer may be found at www.wiley.com/college/callister (Student Companion Site).] 8.8 THE S–N CURVE As with other mechanical characteristics, the fatigue properties of materials can be determined from laboratory simulation tests.5 A test apparatus should be designed to duplicate as nearly as possible the service stress conditions (stress level, time frequency, stress pattern, etc.). A schematic diagram of a rotating-bending test 5 See ASTM Standard E 466, “Standard Practice for Conducting Constant Amplitude Axial Fatigue Tests of Metallic Materials,” and ASTM Standard E 468, “Standard Practice for Presentation of Constant Amplitude Fatigue Test Results for Metallic Materials.” 1496T_c08_207-251 11/18/05 13:43 Page 230 REVISED PAGES 230 • Chapter 8 / Failure Flexible coupling Counter Specimen – Bearing housing ing g hous Bearin High-speed motor + Load Load Figure 8.18 Schematic diagram of fatigue-testing apparatus for making rotatingbending tests. (From Keyser, Materials Science in Engineering, 4th Edition, © 1986, p. 88. Adapted by permission of Pearson Education, Inc., Upper Saddle River, NJ.) fatigue limit fatigue strength fatigue life apparatus, commonly used for fatigue testing, is shown in Figure 8.18; the compression and tensile stresses are imposed on the specimen as it is simultaneously bent and rotated. Tests are also frequently conducted using an alternating uniaxial tension-compression stress cycle. A series of tests are commenced by subjecting a specimen to the stress cycling at a relatively large maximum stress amplitude (smax), usually on the order of twothirds of the static tensile strength; the number of cycles to failure is counted. This procedure is repeated on other specimens at progressively decreasing maximum stress amplitudes. Data are plotted as stress S versus the logarithm of the number N of cycles to failure for each of the specimens. The values of S are normally taken as stress amplitudes (sa, Equation 8.16); on occasion, smax or smin values may be used. Two distinct types of S–N behavior are observed, which are represented schematically in Figure 8.19. As these plots indicate, the higher the magnitude of the stress, the smaller the number of cycles the material is capable of sustaining before failure. For some ferrous (iron base) and titanium alloys, the S–N curve (Figure 8.19a) becomes horizontal at higher N values; or there is a limiting stress level, called the fatigue limit (also sometimes the endurance limit), below which fatigue failure will not occur. This fatigue limit represents the largest value of fluctuating stress that will not cause failure for essentially an infinite number of cycles. For many steels, fatigue limits range between 35% and 60% of the tensile strength. Most nonferrous alloys (e.g., aluminum, copper, magnesium) do not have a fatigue limit, in that the S–N curve continues its downward trend at increasingly greater N values (Figure 8.19b). Thus, fatigue will ultimately occur regardless of the magnitude of the stress. For these materials, the fatigue response is specified as fatigue strength, which is defined as the stress level at which failure will occur for some specified number of cycles (e.g., 107 cycles). The determination of fatigue strength is also demonstrated in Figure 8.19b. Another important parameter that characterizes a material’s fatigue behavior is fatigue life Nf. It is the number of cycles to cause failure at a specified stress level, as taken from the S–N plot (Figure 8.19b). Unfortunately, there always exists considerable scatter in fatigue data—that is, a variation in the measured N value for a number of specimens tested at the same stress level. This variation may lead to significant design uncertainties when fatigue life and/or fatigue limit (or strength) are being considered. The scatter in results is a consequence of the fatigue sensitivity to a number of test and material parameters that are impossible to control precisely. These parameters include specimen fabrication and surface preparation, metallurgical variables, specimen alignment in the apparatus, mean stress, and test frequency. 1496T_c08_207-251 12/21/05 7:51 Page 231 2nd REVISE PAGES Fatigue limit 103 104 105 106 107 108 109 1010 109 1010 Cycles to failure, N (logarithmic scale) (a) Stress amplitude, S Figure 8.19 Stress amplitude (S) versus logarithm of the number of cycles to fatigue failure (N) for (a) a material that displays a fatigue limit, and (b) a material that does not display a fatigue limit. Stress amplitude, S 8.8 The S–N Curve • 231 S1 Fatigue strength at N1 cycles 103 104 107 Fatigue life at stress S1 N1 108 Cycles to failure, N (logarithmic scale) (b) Fatigue S–N curves similar to those shown in Figure 8.19 represent “best fit” curves that have been drawn through average-value data points. It is a little unsettling to realize that approximately one-half of the specimens tested actually failed at stress levels lying nearly 25% below the curve (as determined on the basis of statistical treatments). Several statistical techniques have been developed to specify fatigue life and fatigue limit in terms of probabilities. One convenient way of representing data treated in this manner is with a series of constant probability curves, several of which are plotted in Figure 8.20. The P value associated with each curve represents the probability of failure. For example, at a stress of 200 MPa (30,000 psi), we would expect 1% of the specimens to fail at about 106 cycles and 50% to fail at about 2  107 cycles, and so on. Remember that S–N curves represented in the literature are normally average values, unless noted otherwise. The fatigue behaviors represented in Figures 8.19a and 8.19b may be classified into two domains. One is associated with relatively high loads that produce not only elastic strain but also some plastic strain during each cycle. Consequently, fatigue lives are relatively short; this domain is termed low-cycle fatigue and occurs at less 1496T_c08_207-251 11/18/05 11:50 Page 232 REVISED PAGES 232 • Chapter 8 / Failure 70 60 P = 0.99 50 P = 0.90 300 P = 0.50 40 P = 0.01 200 30 P = 0.10 Stress (103 psi) 400 Stress, S (MPa) Figure 8.20 Fatigue S–N probability of failure curves for a 7075-T6 aluminum alloy; P denotes the probability of failure. (From G. M. Sinclair and T. J. Dolan, Trans. ASME, 75, 1953, p. 867. Reprinted with permission of the American Society of Mechanical Engineers.) 20 100 10 104 105 106 107 108 109 Cycles to failure, N (logarithmic scale) than about 104 to 105 cycles. For lower stress levels wherein deformations are totally elastic, longer lives result. This is called high-cycle fatigue inasmuch as relatively large numbers of cycles are required to produce fatigue failure. High-cycle fatigue is associated with fatigue lives greater than about 104 to 105 cycles. 8.9 CRACK INITIATION AND PROPAGATION The process of fatigue failure is characterized by three distinct steps: (1) crack initiation, wherein a small crack forms at some point of high stress concentration; (2) crack propagation, during which this crack advances incrementally with each stress cycle; and (3) final failure, which occurs very rapidly once the advancing crack has reached a critical size. Cracks associated with fatigue failure almost always initiate (or nucleate) on the surface of a component at some point of stress concentration. Crack nucleation sites include surface scratches, sharp fillets, keyways, threads, dents, and the like. In addition, cyclic loading can produce microscopic surface discontinuities resulting from dislocation slip steps that may also act as stress raisers, and therefore as crack initiation sites. The region of a fracture surface that formed during the crack propagation step may be characterized by two types of markings termed beachmarks and striations. Both of these features indicate the position of the crack tip at some point in time and appear as concentric ridges that expand away from the crack initiation site(s), frequently in a circular or semicircular pattern. Beachmarks (sometimes also called “clamshell marks”) are of macroscopic dimensions (Figure 8.21), and may be observed with the unaided eye. These markings are found for components that experienced interruptions during the crack propagation stage—for example, a machine that operated only during normal work-shift hours. Each beachmark band represents a period of time over which crack growth occurred. On the other hand, fatigue striations are microscopic in size and subject to observation with the electron microscope (either TEM or SEM). Figure 8.22 is an electron fractograph that shows this feature. Each striation is thought to represent the advance distance of a crack front during a single load cycle. Striation width depends on, and increases with, increasing stress range. At this point it should be emphasized that although both beachmarks and striations are fatigue fracture surface features having similar appearances, they are 1496T_c08_207-251 12/21/05 7:51 Page 233 2nd REVISE PAGES 8.9 Crack Initiation and Propagation • 233 Figure 8.21 Fracture surface of a rotating steel shaft that experienced fatigue failure. Beachmark ridges are visible in the photograph. (Reproduced with permission from D. J. Wulpi, Understanding How Components Fail, American Society for Metals, Materials Park, OH, 1985.) nevertheless different, both in origin and size. There may be literally thousands of striations within a single beachmark. Often the cause of failure may be deduced after examination of the failure surfaces. The presence of beachmarks and/or striations on a fracture surface confirms that the cause of failure was fatigue. Nevertheless, the absence of either or both does not exclude fatigue as the cause of failure. One final comment regarding fatigue failure surfaces: Beachmarks and striations will not appear on that region over which the rapid failure occurs. Rather, the Figure 8.22 Transmission electron fractograph showing fatigue striations in aluminum. Magnification unknown. (From V. J. Colangelo and F. A. Heiser, Analysis of Metallurgical Failures, 2nd edition. Copyright © 1987 by John Wiley & Sons, New York. Reprinted by permission of John Wiley & Sons, Inc.) 1496T_c08_207-251 11/18/05 11:50 Page 234 REVISED PAGES 234 • Chapter 8 / Failure Region of slow crack propagation Figure 8.23 Fatigue failure surface. A crack formed at the top edge. The smooth region also near the top corresponds to the area over which the crack propagated slowly. Rapid failure occurred over the area having a dull and fibrous texture (the largest area). Approximately 0.5. [Reproduced by permission from Metals Handbook: Fractography and Atlas of Fractographs, Vol. 9, 8th edition, H. E. Boyer (Editor), American Society for Metals, 1974.] Region of rapid failure rapid failure may be either ductile or brittle; evidence of plastic deformation will be present for ductile, and absent for brittle, failure. This region of failure may be noted in Figure 8.23. Concept Check 8.4 Surfaces for some steel specimens that have failed by fatigue have a bright crystalline or grainy appearance. Laymen may explain the failure by saying that the metal crystallized while in service. Offer a criticism for this explanation. [The answer may be found at www.wiley.com/college/callister (Student Companion Site).] 8.10 FACTORS THAT AFFECT FATIGUE LIFE As mentioned in Section 8.8, the fatigue behavior of engineering materials is highly sensitive to a number of variables. Some of these factors include mean stress level, geometrical design, surface effects, and metallurgical variables, as well as the environment. This section is devoted to a discussion of these factors and, in addition, to measures that may be taken to improve the fatigue resistance of structural components. Mean Stress The dependence of fatigue life on stress amplitude is represented on the S–N plot. Such data are taken for a constant mean stress sm, often for the reversed cycle 1496T_c08_207-251 11/18/05 11:50 Page 235 REVISED PAGES 8.10 Factors That Affect Fatigue Life • 235 Figure 8.24 Demonstration of the influence of mean stress sm on S–N fatigue behavior. Stress amplitude, ␴a ␴ m3 > ␴ m2 > ␴ m1 ␴ m1 ␴ m2 ␴ m3 Cycles to failure, N (logarithmic scale) situation (sm  0). Mean stress, however, will also affect fatigue life; this influence may be represented by a series of S–N curves, each measured at a different sm, as depicted schematically in Figure 8.24. As may be noted, increasing the mean stress level leads to a decrease in fatigue life. Surface Effects For many common loading situations, the maximum stress within a component or structure occurs at its surface. Consequently, most cracks leading to fatigue failure originate at surface positions, specifically at stress amplification sites. Therefore, it has been observed that fatigue life is especially sensitive to the condition and configuration of the component surface. Numerous factors influence fatigue resistance, the proper management of which will lead to an improvement in fatigue life. These include design criteria as well as various surface treatments. Design Factors The design of a component can have a significant influence on its fatigue characteristics. Any notch or geometrical discontinuity can act as a stress raiser and fatigue crack initiation site; these design features include grooves, holes, keyways, threads, and so on. The sharper the discontinuity (i.e., the smaller the radius of curvature), the more severe the stress concentration. The probability of fatigue failure may be reduced by avoiding (when possible) these structural irregularities, or by making design modifications whereby sudden contour changes leading to sharp corners are eliminated—for example, calling for rounded fillets with large radii of curvature at the point where there is a change in diameter for a rotating shaft (Figure 8.25). Fillet (a) ( b) Figure 8.25 Demonstration of how design can reduce stress amplification. (a) Poor design: sharp corner. (b) Good design: fatigue lifetime improved by incorporating rounded fillet into a rotating shaft at the point where there is a change in diameter. 1496T_c08_207-251 11/18/05 13:54 Page 236 REVISED PAGES 236 • Chapter 8 / Failure Figure 8.26 Schematic S–N fatigue curves for normal and shot-peened steel. Stress amplitude Shot peened Normal Cycles to failure (logarithmic scale) Surface Treatments case hardening Two Case Studies: “Automobile Valve Spring,” and “Failure of an Automobile Rear Axle,” Chapter 22, may be found at www.wiley.com/ college/callister (Student Companion Site). During machining operations, small scratches and grooves are invariably introduced into the workpiece surface by cutting tool action. These surface markings can limit the fatigue life. It has been observed that improving the surface finish by polishing will enhance fatigue life significantly. One of the most effective methods of increasing fatigue performance is by imposing residual compressive stresses within a thin outer surface layer. Thus, a surface tensile stress of external origin will be partially nullified and reduced in magnitude by the residual compressive stress. The net effect is that the likelihood of crack formation and therefore of fatigue failure is reduced. Residual compressive stresses are commonly introduced into ductile metals mechanically by localized plastic deformation within the outer surface region. Commercially, this is often accomplished by a process termed shot peening. Small, hard particles (shot) having diameters within the range of 0.1 to 1.0 mm are projected at high velocities onto the surface to be treated. The resulting deformation induces compressive stresses to a depth of between one-quarter and one-half of the shot diameter. The influence of shot peening on the fatigue behavior of steel is demonstrated schematically in Figure 8.26. Case hardening is a technique by which both surface hardness and fatigue life are enhanced for steel alloys. This is accomplished by a carburizing or nitriding process whereby a component is exposed to a carbonaceous or nitrogenous atmosphere at an elevated temperature. A carbon- or nitrogen-rich outer surface layer (or “case”) is introduced by atomic diffusion from the gaseous phase. The case is normally on the order of 1 mm deep and is harder than the inner core of material. (The influence of carbon content on hardness for Fe–C alloys is demonstrated in Figure 10.29a.) The improvement of fatigue properties results from increased hardness within the case, as well as the desired residual compressive stresses the formation of which attends the carburizing or nitriding process. A carbon-rich outer case may be observed for the gear shown in the chapter-opening photograph for Chapter 5; it appears as a dark outer rim within the sectioned segment. The increase in case hardness is demonstrated in the photomicrograph appearing in Figure 8.27. The dark and elongated diamond shapes are Knoop microhardness indentations. The upper indentation, lying within the carburized layer, is smaller than the core indentation. 1496T_c08_207-251 11/18/05 11:50 Page 237 REVISED PAGES 8.11 Environmental Effects • 237 Case Core region Figure 8.27 Photomicrograph showing both core (bottom) and carburized outer case (top) regions of a casehardened steel. The case is harder as attested by the smaller microhardness indentation. 100. (From R. W. Hertzberg, Deformation and Fracture Mechanics of Engineering Materials, 3rd edition. Copyright © 1989 by John Wiley & Sons, New York. Reprinted by permission of John Wiley & Sons, Inc.) 8.11 ENVIRONMENTAL EFFECTS thermal fatigue Thermal stress— dependence on coefficient of thermal expansion, modulus of elasticity, and temperature change corrosion fatigue Environmental factors may also affect the fatigue behavior of materials. A few brief comments will be given relative to two types of environment-assisted fatigue failure: thermal fatigue and corrosion fatigue. Thermal fatigue is normally induced at elevated temperatures by fluctuating thermal stresses; mechanical stresses from an external source need not be present. The origin of these thermal stresses is the restraint to the dimensional expansion and/or contraction that would normally occur in a structural member with variations in temperature. The magnitude of a thermal stress developed by a temperature change ¢T is dependent on the coefficient of thermal expansion al and the modulus of elasticity E according to s  alE ¢T (8.18) (The topics of thermal expansion and thermal stresses are discussed in Sections 19.3 and 19.5.) Of course, thermal stresses will not arise if this mechanical restraint is absent. Therefore, one obvious way to prevent this type of fatigue is to eliminate, or at least reduce, the restraint source, thus allowing unhindered dimensional changes with temperature variations, or to choose materials with appropriate physical properties. Failure that occurs by the simultaneous action of a cyclic stress and chemical attack is termed corrosion fatigue. Corrosive environments have a deleterious influence and produce shorter fatigue lives. Even the normal ambient atmosphere will affect the fatigue behavior of some materials. Small pits may form as a result of chemical reactions between the environment and material, which serve as points of stress concentration and therefore as crack nucleation sites. In addition, crack propagation rate is enhanced as a result of the corrosive environment. The nature of the stress cycles will influence the fatigue behavior; for example, lowering the load application frequency leads to longer periods during which the opened crack is in contact with the environment and to a reduction in the fatigue life. Several approaches to corrosion fatigue prevention exist. On one hand, we can take measures to reduce the rate of corrosion by some of the techniques discussed in Chapter 17—for example, apply protective surface coatings, select a 1496T_c08_207-251 11/18/05 11:50 Page 238 REVISED PAGES 238 • Chapter 8 / Failure more corrosion-resistant material, and reduce the corrosiveness of the environment. And/or it might be advisable to take actions to minimize the probability of normal fatigue failure, as outlined above—for example, reduce the applied tensile stress level and impose residual compressive stresses on the surface of the member. Creep creep Materials are often placed in service at elevated temperatures and exposed to static mechanical stresses (e.g., turbine rotors in jet engines and steam generators that experience centrifugal stresses, and high-pressure steam lines). Deformation under such circumstances is termed creep. Defined as the time-dependent and permanent deformation of materials when subjected to a constant load or stress, creep is normally an undesirable phenomenon and is often the limiting factor in the lifetime of a part. It is observed in all materials types; for metals it becomes important only for temperatures greater than about 0.4Tm (Tm  absolute melting temperature). Amorphous polymers, which include plastics and rubbers, are especially sensitive to creep deformation as discussed in Section 15.4. 8.12 GENERALIZED CREEP BEHAVIOR A typical creep test6 consists of subjecting a specimen to a constant load or stress while maintaining the temperature constant; deformation or strain is measured and plotted as a function of elapsed time. Most tests are the constant load type, which yield information of an engineering nature; constant stress tests are employed to provide a better understanding of the mechanisms of creep. Figure 8.28 is a schematic representation of the typical constant load creep behavior of metals. Upon application of the load there is an instantaneous deformation, as indicated in the figure, which is mostly elastic. The resulting creep curve consists of three regions, each of which has its own distinctive strain–time feature. Primary or transient creep occurs first, typified by a continuously decreasing creep rate; that is, the slope of the curve diminishes with time. This suggests that the material is experiencing an increase in creep resistance or strain hardening (Section 7.10)—deformation becomes more difficult as the material is strained. For secondary creep, sometimes termed steady-state creep, the rate is constant; that is, the plot becomes linear. This is often the stage of creep that is of the longest duration. The constancy of creep rate is explained on the basis of a balance between the competing processes of strain hardening and recovery, recovery (Section 7.11) being the process whereby a material becomes softer and retains its ability to experience deformation. Finally, for tertiary creep, there is an acceleration of the rate and ultimate failure. This failure is frequently termed rupture and results from microstructural and/or metallurgical changes; for example, grain boundary separation, and the formation of internal cracks, cavities, and voids. Also, for tensile loads, a neck may form at some point within the deformation region. These all lead to a decrease in the effective cross-sectional area and an increase in strain rate. 6 ASTM Standard E 139, “Standard Practice for Conducting Creep, Creep-Rupture, and Stress-Rupture Tests of Metallic Materials.” 1496T_c08_207-251 11/18/05 13:43 Page 239 REVISED PAGES 8.13 Stress and Temperature Effects • 239 Creep strain, ⑀ Rupture Primary Δt Δ⑀ Tertiary Figure 8.28 Typical creep curve of strain versus time at constant stress and constant elevated temperature. The minimum creep rate ¢ ¢t is the slope of the linear segment in the secondary region. Rupture lifetime tr is the total time to rupture. Secondary Instantaneous deformation Time, t tr For metallic materials most creep tests are conducted in uniaxial tension using a specimen having the same geometry as for tensile tests (Figure 6.2). On the other hand, uniaxial compression tests are more appropriate for brittle materials; these provide a better measure of the intrinsic creep properties inasmuch as there is no stress amplification and crack propagation, as with tensile loads. Compressive test specimens are usually right cylinders or parallelepipeds having length-to-diameter ratios ranging from about 2 to 4. For most materials creep properties are virtually independent of loading direction. Possibly the most important parameter from a creep test is the slope of the secondary portion of the creep curve ( ¢  ¢t in Figure 8.28); this is often called the # minimum or steady-state creep rate s. It is the engineering design parameter that is considered for long-life applications, such as a nuclear power plant component that is scheduled to operate for several decades, and when failure or too much strain are not options. On the other hand, for many relatively short-life creep situations (e.g., turbine blades in military aircraft and rocket motor nozzles), time to rupture, or the rupture lifetime tr , is the dominant design consideration; it is also indicated in Figure 8.28. Of course, for its determination, creep tests must be conducted to the point of failure; these are termed creep rupture tests. Thus, a knowledge of these creep characteristics of a material allows the design engineer to ascertain its suitability for a specific application. Concept Check 8.5 Superimpose on the same strain-versus-time plot schematic creep curves for both constant tensile stress and constant tensile load, and explain the differences in behavior. [The answer may be found at www.wiley.com/college/callister (Student Companion Site).] 8.13 STRESS AND TEMPERATURE EFFECTS Both temperature and the level of the applied stress influence the creep characteristics (Figure 8.29). At a temperature substantially below 0.4Tm, and after the initial deformation, the strain is virtually independent of time. With either increasing stress or temperature, the following will be noted: (1) the instantaneous strain at the time of stress application increases, (2) the steady-state creep rate is increased, and (3) the rupture lifetime is diminished. 1496T_c08_207-251 11/18/05 11:50 Page 240 REVISED PAGES 240 • Chapter 8 / Failure Figure 8.29 Influence of stress s and temperature T on creep behavior. T3 > T2 > T1 3 > 2 > 1 Creep strain T3 or 3 T2 or 2 T1 or 1 T < 0.4Tm Time Dependence of creep strain rate on stress The results of creep rupture tests are most commonly presented as the logarithm of stress versus the logarithm of rupture lifetime. Figure 8.30 is one such plot for a nickel alloy in which a linear relationship can be seen to exist at each temperature. For some alloys and over relatively large stress ranges, nonlinearity in these curves is observed. Empirical relationships have been developed in which the steady-state creep rate as a function of stress and temperature is expressed. Its dependence on stress can be written # s  K1sn (8.19) # where K1 and n are material constants. A plot of the logarithm of s versus the logarithm of s yields a straight line with slope of n; this is shown in Figure 8.31 for a nickel alloy at three temperatures. Clearly, a straight line segment is drawn at each temperature. Now, when the influence of temperature is included, Dependence of creep strain rate on stress and temperature (in K) Qc # s  K2 sn exp a b RT (8.20) where K2 and Qc are constants; Qc is termed the activation energy for creep. 60 400 300 40 30 200 20 100 80 538°C (1000°F) 10 8 60 649°C (1200°F) 40 6 30 4 20 3 2 102 103 104 Rupture lifetime (h) 105 Stress (103 psi) 427°C (800°F) Stress (MPa) Figure 8.30 Stress (logarithmic scale) versus rupture lifetime (logarithmic scale) for a low carbon–nickel alloy at three temperatures. [From Metals Handbook: Properties and Selection: Stainless Steels, Tool Materials and Special-Purpose Metals, Vol. 3, 9th edition, D. Benjamin (Senior Editor), American Society for Metals, 1980, p. 130.] 1496T_c08_207-251 11/18/05 11:50 Page 241 REVISED PAGES 8.14 Data Extrapolation Methods • 241 30 200 10 8 427°C (800°F) 60 40 6 30 4 538°C (1000°F) 3 20 Stress (103 psi) 20 100 80 Stress (MPa) Figure 8.31 Stress (logarithmic scale) versus steady-state creep rate (logarithmic scale) for a low carbon–nickel alloy at three temperatures. [From Metals Handbook: Properties and Selection: Stainless Steels, Tool Materials and Special-Purpose Metals, Vol. 3, 9th edition, D. Benjamin (Senior Editor), American Society for Metals, 1980, p. 131.] 2 10 8 649°C (1200°F) 10–7 10–6 10–5 1 Steady-state creep rate (h–1) Several theoretical mechanisms have been proposed to explain the creep behavior for various materials; these mechanisms involve stress-induced vacancy diffusion, grain boundary diffusion, dislocation motion, and grain boundary sliding. Each leads to a different value of the stress exponent n in Equation 8.19. It has been possible to elucidate the creep mechanism for a particular material by comparing its experimental n value with values predicted for the various mechanisms. In addition, correlations have been made between the activation energy for creep (Qc) and the activation energy for diffusion (Qd, Equation 5.8). Creep data of this nature are represented pictorially for some well-studied systems in the form of stress–temperature diagrams, which are termed deformation mechanism maps. These maps indicate stress–temperature regimes (or areas) over which various mechanisms operate. Constant strain rate contours are often also included. Thus, for some creep situation, given the appropriate deformation mechanism map and any two of the three parameters—temperature, stress level, and creep strain rate—the third parameter may be determined. 8.14 DATA EXTRAPOLATION METHODS The need often arises for engineering creep data that are impractical to collect from normal laboratory tests. This is especially true for prolonged exposures (on the order of years). One solution to this problem involves performing creep and/or creep rupture tests at temperatures in excess of those required, for shorter time periods, and at a comparable stress level, and then making a suitable extrapolation to the in-service condition. A commonly used extrapolation procedure employs the Larson–Miller parameter, defined as The Larson–Miller parameter—in terms of temperature and rupture lifetime T 1C  log tr 2 (8.21) where C is a constant (usually on the order of 20), for T in Kelvin and the rupture lifetime tr in hours. The rupture lifetime of a given material measured at some specific stress level will vary with temperature such that this parameter remains constant. Or, the data may be plotted as the logarithm of stress versus the Larson–Miller parameter, as shown in Figure 8.32. Utilization of this technique is demonstrated in the following design example. 1496T_c08_207-251 11/18/05 11:50 Page 242 REVISED PAGES 242 • Chapter 8 / Failure 103 T(20 + log tr)(°R–h) 25 30 35 40 45 Figure 8.32 Logarithm stress versus the Larson–Miller parameter for an S-590 iron. (From F. R. Larson and J. Miller, Trans. ASME, 74, 765, 1952. Reprinted by permission of ASME.) 50 1000 100 10 Stress (103 psi) Stress (MPa) 100 10 1 12 16 20 24 28 103 T(20 + log tr)(K–h) DESIGN EXAMPLE 8.2 Rupture Lifetime Prediction Using the Larson–Miller data for S-590 iron shown in Figure 8.32, predict the time to rupture for a component that is subjected to a stress of 140 MPa (20,000 psi) at 800C 11073 K2. Solution From Figure 8.32, at 140 MPa (20,000 psi) the value of the Larson–Miller parameter is 24.0  103, for T in K and tr in h; therefore, 24.0  103  T 120  log tr 2  1073120  log tr 2 and, solving for the time, 22.37  20  log tr tr  233 h 19.7 days2 8.15 ALLOYS FOR HIGH-TEMPERATURE USE There are several factors that affect the creep characteristics of metals. These include melting temperature, elastic modulus, and grain size. In general, the higher the melting temperature, the greater the elastic modulus, and the larger the grain size, the better is a material’s resistance to creep. Relative to grain size, smaller grains permit more grain-boundary sliding, which results in higher creep rates. This 1496T_c08_207-251 11/18/05 11:50 Page 243 REVISED PAGES Summary • 243 Figure 8.33 (a) Polycrystalline turbine blade that was produced by a conventional casting technique. High-temperature creep resistance is improved as a result of an oriented columnar grain structure (b) produced by a sophisticated directional solidification technique. Creep resistance is further enhanced when single-crystal blades (c) are used. (Courtesy of Pratt & Whitney.) (a) Conventional casting (b) (c) Columnar grain Single crystal effect may be contrasted to the influence of grain size on the mechanical behavior at low temperatures [i.e., increase in both strength (Section 7.8) and toughness (Section 8.6)]. Stainless steels (Section 11.2), the refractory metals, and the superalloys (Section 11.3) are especially resilient to creep and are commonly employed in hightemperature service applications. The creep resistance of the cobalt and nickel superalloys is enhanced by solid-solution alloying, and also by the addition of a dispersed phase that is virtually insoluble in the matrix. In addition, advanced processing techniques have been utilized; one such technique is directional solidification, which produces either highly elongated grains or single-crystal components (Figure 8.33). Another is the controlled unidirectional solidification of alloys having specially designed compositions wherein two-phase composites result. SUMMARY Fundamentals of Fracture Ductile Fracture Fracture, in response to tensile loading and at relatively low temperatures, may occur by ductile and brittle modes, both of which involve the formation and propagation of cracks. For ductile fracture, evidence will exist of gross plastic deformation at the fracture surface. In tension, highly ductile metals will neck down to essentially a point fracture; cup-and-cone mating fracture surfaces result for moderate ductility. Cracks in ductile materials are said to be stable (i.e., resist extension without an increase in applied stress); and inasmuch as fracture is noncatastrophic, this fracture mode is almost always preferred. Brittle Fracture For brittle fracture, cracks are unstable, and the fracture surface is relatively flat and perpendicular to the direction of the applied tensile load. Chevron and ridgelike patterns are possible, which indicate the direction of crack propagation. Transgranular (through-grain) and intergranular (between-grain) fractures are found in brittle polycrystalline materials. 1496T_c08_207-251 11/18/05 11:50 Page 244 REVISED PAGES 244 • Chapter 8 / Failure Principles of Fracture Mechanics The significant discrepancy between actual and theoretical fracture strengths of brittle materials is explained by the existence of small flaws that are capable of amplifying an applied tensile stress in their vicinity, leading ultimately to crack formation. Fracture ensues when the theoretical cohesive strength is exceeded at the tip of one of these flaws. The fracture toughness of a material is indicative of its resistance to brittle fracture when a crack is present. It depends on specimen thickness, and, for relatively thick specimens (i.e., conditions of plane strain), is termed the plane strain fracture toughness. This parameter is the one normally cited for design purposes; its value is relatively large for ductile materials (and small for brittle ones), and is a function of microstructure, strain rate, and temperature. With regard to designing against the possibility of fracture, consideration must be given to material (its fracture toughness), the stress level, and the flaw size detection limit. Impact Fracture Testing Qualitatively, the fracture behavior of materials may be determined using Charpy and Izod impact testing techniques. On the basis of the temperature dependence of measured impact energy (or appearance of the fracture surface), it is possible to ascertain whether or not a material experiences a ductile-to-brittle transition and the temperature range over which such a transition occurs. Low-strength steel alloys typify this behavior, and, for structural applications, should be used at temperatures in excess of the transition range. Furthermore, low-strength FCC metals, most HCP metals, and high-strength materials do not experience this ductileto-brittle transition. Cyclic Stresses (Fatigue) The S–N Curve Fatigue is a common type of catastrophic failure wherein the applied stress level fluctuates with time. Test data are plotted as stress versus the logarithm of the number of cycles to failure. For many metals and alloys, stress diminishes continuously with increasing number of cycles at failure; fatigue strength and fatigue life are the parameters used to characterize the fatigue behavior of these materials. On the other hand, for other metals/alloys, at some point, stress ceases to decrease with, and becomes independent of, the number of cycles; the fatigue behavior of these materials is expressed in terms of fatigue limit. Crack Initiation and Propagation The processes of fatigue crack initiation and propagation were discussed. Cracks normally nucleate on the surface of a component at some point of stress concentration. Two characteristic fatigue surface features are beachmarks and striations. Beachmarks form on components that experience applied stress interruptions; they normally may be observed with the naked eye. Fatigue striations are of microscopic dimensions, and each is thought to represent the crack tip advance distance over a single load cycle. Factors That Affect Fatigue Life Measures that may be taken to extend fatigue life include: (1) reducing the mean stress level; (2) eliminating sharp surface discontinuities; (3) improving the surface 1496T_c08_207-251 11/18/05 11:50 Page 245 REVISED PAGES Important Terms and Concepts • 245 finish by polishing; (4) imposing surface residual compressive stresses by shot peening; and (5) case hardening by using a carburizing or nitriding process. Environmental Effects The fatigue behavior of materials may also be affected by the environment. Thermal stresses may be induced in components that are exposed to elevated temperature fluctuations and when thermal expansion and/or contraction is restrained; fatigue for these conditions is termed thermal fatigue. The presence of a chemically active environment may lead to a reduction in fatigue life for corrosion fatigue. Generalized Creep Behavior The time-dependent plastic deformation of materials subjected to a constant load (or stress) and temperatures greater than about 0.4Tm is termed creep. A typical creep curve (strain versus time) will normally exhibit three distinct regions. For transient (or primary) creep, the rate (or slope) diminishes with time. The plot becomes linear (i.e., creep rate is constant) in the steady-state (or secondary) region. And finally, deformation accelerates for tertiary creep, just prior to failure (or rupture). Important design parameters available from such a plot include the steady-state creep rate (slope of the linear region) and rupture lifetime. Stress and Temperature Effects Both temperature and applied stress level influence creep behavior. Increasing either of these parameters produces the following effects: (1) an increase in the instantaneous initial deformation; (2) an increase in the steady-state creep rate; and (3) a diminishment of the rupture lifetime. Analytical expressions were presented # which relate s to both temperature and stress. Data Extrapolation Methods Extrapolation of creep test data to lower temperature—longer time regimes is possible using the Larson–Miller parameter. Alloys for High-Temperature Use Metal alloys that are especially resistant to creep have high elastic moduli and melting temperatures; these include the superalloys, the stainless steels, and the refractory metals. Various processing techniques are employed to improve the creep properties of these materials. I M P O R TA N T T E R M S A N D C O N C E P T S Brittle fracture Case hardening Charpy test Corrosion fatigue Creep Ductile fracture Ductile-to-brittle transition Fatigue Fatigue life Fatigue limit Fatigue strength Fracture mechanics Fracture toughness Impact energy Intergranular fracture Izod test Plane strain Plane strain fracture toughness Stress raiser Thermal fatigue Transgranular fracture 1496T_c08_207-251 11/18/05 11:50 Page 246 REVISED PAGES 246 • Chapter 8 / Failure REFERENCES ASM Handbook, Vol. 11, Failure Analysis and Prevention, ASM International, Materials Park, OH, 1986. ASM Handbook, Vol. 12, Fractography, ASM International, Materials Park, OH, 1987. ASM Handbook, Vol. 19, Fatigue and Fracture, ASM International, Materials Park, OH, 1996. Boyer, H. E. (Editor), Atlas of Creep and Stress–Rupture Curves, ASM International, Materials Park, OH, 1988. Boyer, H. E. (Editor), Atlas of Fatigue Curves, ASM International, Materials Park, OH, 1986. Colangelo, V. J. and F. A. Heiser, Analysis of Metallurgical Failures, 2nd edition, Wiley, New York, 1987. Collins, J.A., Failure of Materials in Mechanical Design, 2nd edition, Wiley, New York, 1993. Courtney, T. H., Mechanical Behavior of Materials, 2nd edition, McGraw-Hill, New York, 2000. Dieter, G. E., Mechanical Metallurgy, 3rd edition, McGraw-Hill, New York, 1986. Esaklul, K. A., Handbook of Case Histories in Failure Analysis, ASM International, Materials Park, OH, 1992 and 1993. In two volumes. Fatigue Data Book: Light Structural Alloys, ASM International, Materials Park, OH, 1995. Hertzberg, R. W., Deformation and Fracture Mechanics of Engineering Materials, 4th edition, Wiley, New York, 1996. Stevens, R. I., A. Fatemi, R. R. Stevens, and H. O. Fuchs, Metal Fatigue in Engineering, 2nd edition, Wiley, New York, 2000. Tetelman, A. S. and A. J. McEvily, Fracture of Structural Materials, Wiley, New York, 1967. Reprinted by Books on Demand, Ann Arbor, MI. Wulpi, D. J., Understanding How Components Fail, 2nd edition, ASM International, Materials Park, OH, 1999. QUESTIONS AND PROBLEMS Principles of Fracture Mechanics 8.1 What is the magnitude of the maximum stress that exists at the tip of an internal crack having a radius of curvature of 1.9  104 mm (7.5  106 in.) and a crack length of 3.8  102 mm (1.5  103 in.) when a tensile stress of 140 MPa (20,000 psi) is applied? 8.2 Estimate the theoretical fracture strength of a brittle material if it is known that fracture occurs by the propagation of an elliptically shaped surface crack of length 0.5 mm (0.02 in.) and having a tip radius of curvature of 5  103 mm (2  104 in.), when a stress of 1035 MPa (150,000 psi) is applied. 8.3 If the specific surface energy for aluminum oxide is 0.90 J/m2, using data contained in Table 12.5, compute the critical stress required for the propagation of an internal crack of length 0.40 mm. 8.4 An MgO component must not fail when a tensile stress of 13.5 MPa (1960 psi) is applied. Determine the maximum allowable surface crack length if the surface energy of MgO is 1.0 J/m2. Data found in Table 12.5 may prove helpful. 8.5 A specimen of a 4340 steel alloy with a plane strain fracture toughness of 54.8 MPa 1m (50 ksi 1in.) is exposed to a stress of 1030 MPa (150,000 psi). Will this specimen experience fracture if it is known that the largest surface crack is 0.5 mm (0.02 in.) long? Why or why not? Assume that the parameter Y has a value of 1.0. 8.6 Some aircraft component is fabricated from an aluminum alloy that has a plane strain fracture toughness of 40 MPa 1m (36.4 ksi1in.). It has been determined that fracture results at a stress of 300 MPa (43,500 psi) when the maximum (or critical) internal crack length is 4.0 mm (0.16 in.). For this same component and alloy, will fracture occur at a stress level of 260 MPa (38,000 psi) when the maximum internal crack length is 6.0 mm (0.24 in.)? Why or why not? 8.7 Suppose that a wing component on an aircraft is fabricated from an aluminum alloy that has a plane strain fracture toughness of 26 MPa 1m (23.7 ksi 1in.). It has been determined that fracture results at a stress of 112 MPa (16,240 psi) when the maximum internal crack length is 8.6 mm (0.34 in.). For this same 1496T_c08_207-251 11/18/05 13:43 Page 247 REVISED PAGES Questions and Problems • 247 8.8 8.9 8.10 8.11 component and alloy, compute the stress level at which fracture will occur for a critical internal crack length of 6.0 mm (0.24 in.). A large plate is fabricated from a steel alloy that has a plane strain fracture toughness of 82.4 MPa1m (75.0 ksi 1in.). If, during service use, the plate is exposed to a tensile stress of 345 MPa (50,000 psi), determine the minimum length of a surface crack that will lead to fracture. Assume a value of 1.0 for Y. Calculate the maximum internal crack length allowable for a Ti-6Al-4V titanium alloy (Table 8.1) component that is loaded to a stress one-half of its yield strength. Assume that the value of Y is 1.50. A structural component in the form of a wide plate is to be fabricated from a steel alloy that has a plane strain fracture toughness of 98.9 MPa1m (90 ksi 1in.) and a yield strength of 860 MPa (125,000 psi). The flaw size resolution limit of the flaw detection apparatus is 3.0 mm (0.12 in.). If the design stress is onehalf of the yield strength and the value of Y is 1.0, determine whether or not a critical flaw for this plate is subject to detection. After consultation of other references, write a brief report on one or two nondestructive test techniques that are used to detect and measure internal and/or surface flaws in metal alloys. Impact Fracture Testing 8.12 Following is tabulated data that were gathered from a series of Charpy impact tests on a tempered 4340 steel alloy. Temperature (C) Impact Energy (J) 0 25 50 75 100 113 125 150 175 200 105 104 103 97 63 40 34 28 25 24 (a) Plot the data as impact energy versus temperature. (b) Determine a ductile-to-brittle transition temperature as that temperature corresponding to the average of the maximum and minimum impact energies. (c) Determine a ductile-to-brittle transition temperature as that temperature at which the impact energy is 50 J. 8.13 Following is tabulated data that were gathered from a series of Charpy impact tests on a commercial low-carbon steel alloy. Temperature (C) 50 40 30 20 10 0 10 20 30 40 Impact Energy (J) 76 76 71 58 38 23 14 9 5 1.5 (a) Plot the data as impact energy versus temperature. (b) Determine a ductile-to-brittle transition temperature as that temperature corresponding to the average of the maximum and minimum impact energies. (c) Determine a ductile-to-brittle transition temperature as that temperature at which the impact energy is 20 J. Cyclic Stresses (Fatigue) The S–N Curve 8.14 A fatigue test was conducted in which the mean stress was 70 MPa (10,000 psi), and the stress amplitude was 210 MPa (30,000 psi). (a) Compute the maximum and minimum stress levels. (b) Compute the stress ratio. (c) Compute the magnitude of the stress range. 8.15 A cylindrical 1045 steel bar (Figure 8.34) is subjected to repeated compression-tension stress cycling along its axis. If the load amplitude is 66,700 N (15,000 lbf), compute the minimum allowable bar diameter to ensure that fatigue failure will not occur. Assume a safety factor of 2.0. 1496T_c08_207-251 11/18/05 11:50 Page 248 REVISED PAGES 248 • Chapter 8 / Failure 80 500 70 60 400 50 1045 steel 300 40 2014–T6 aluminum alloy 30 200 20 Stress amplitude (10 3 psi) Stress amplitude, S (MPa) Figure 8.34 Stress magnitude S versus the logarithm of the number N of cycles to fatigue failure for red brass, an aluminum alloy, and a plain carbon steel. (Adapted from H. W. Hayden, W. G. Moffatt, and J. Wulff, The Structure and Properties of Materials, Vol. III, Mechanical Behavior, p. 15. Copyright © 1965 by John Wiley & Sons, New York. Reprinted by permission of John Wiley & Sons, Inc. Also adapted from ASM Handbook, Vol. 2, Properties and Selection: Nonferrous Alloys and Special-Purpose Materials, 1990. Reprinted by permission of ASM International.) 100 10 Red brass 0 0 103 104 106 105 107 108 109 1010 Cycles to failure, N 8.16 A 6.4 mm (0.25 in.) diameter cylindrical rod fabricated from a 2014-T6 aluminum alloy (Figure 8.34) is subjected to reversed tensioncompression load cycling along its axis. If the maximum tensile and compressive loads are 5340 N (1200 lbf) and 5340 N (1200 lbf), respectively, determine its fatigue life.Assume that the stress plotted in Figure 8.34 is stress amplitude. 8.17 A 15.2 mm (0.60 in.) diameter cylindrical rod fabricated from a 2014-T6 aluminum alloy (Figure 8.34) is subjected to a repeated tensioncompression load cycling along its axis. Compute the maximum and minimum loads that will be applied to yield a fatigue life of 1.0  108 cycles. Assume that the stress plotted on the vertical axis is stress amplitude, and data were taken for a mean stress of 35 MPa (5000 psi). 8.18 The fatigue data for a brass alloy are given as follows: Stress Amplitude (MPa) Cycles to Failure 170 148 130 114 92 80 74 3.7  104 1.0  105 3.0  105 1.0  106 1.0  107 1.0  108 1.0  109 (a) Make an S–N plot (stress amplitude versus logarithm cycles to failure) using these data. (b) Determine the fatigue strength at 4  106 cycles. (c) Determine the fatigue life for 120 MPa. 8.19 Suppose that the fatigue data for the brass alloy in Problem 8.18 were taken from bending-rotating tests, and that a rod of this alloy is to be used for an automobile axle that rotates at an average rotational velocity of 1800 revolutions per minute. Give the maximum torsional stress amplitude possible for each of the following lifetimes of the rod: (a) 1 year, (b) 1 month, (c) 1 day, and (d) 1 hour. 8.20 The fatigue data for a steel alloy are given as follows: Stress Amplitude [MPa (ksi)] Cycles to Failure 470 (68.0) 440 (63.4) 390 (56.2) 350 (51.0) 310 (45.3) 290 (42.2) 290 (42.2) 290 (42.2) 104 3  104 105 3  105 106 3  106 107 108 (a) Make an S–N plot (stress amplitude versus logarithm cycles to failure) using these data. 1496T_c08_207-251 11/18/05 13:43 Page 249 REVISED PAGES Questions and Problems • 249 (b) What is the fatigue limit for this alloy? (c) Determine fatigue lifetimes at stress amplitudes of 415 MPa (60,000 psi) and 275 MPa (40,000 psi). (d) Estimate fatigue strengths at 2  104 and 6  105 cycles. 8.21 Suppose that the fatigue data for the steel alloy in Problem 8.20 were taken for bendingrotating tests, and that a rod of this alloy is to be used for an automobile axle that rotates at an average rotational velocity of 600 revolutions per minute. Give the maximum lifetimes of continuous driving that are allowable for the following stress levels: (a) 450 MPa (65,000 psi), (b) 380 MPa (55,000 psi), (c) 310 MPa (45,000 psi), and (d) 275 MPa (40,000 psi). 8.22 Three identical fatigue specimens (denoted A, B, and C) are fabricated from a nonferrous alloy. Each is subjected to one of the maximumminimum stress cycles listed below; the frequency is the same for all three tests. Specimen max (MPa) min (MPa) A B C 450 300 500 150 300 200 (a) Rank the fatigue lifetimes of these three specimens from the longest to the shortest. (b) Now justify this ranking using a schematic S–N plot. 8.23 Cite five factors that may lead to scatter in fatigue life data. Crack Initiation and Propagation Factors That Affect Fatigue Life 8.24 Briefly explain the difference between fatigue striations and beachmarks both in terms of (a) size and (b) origin. 8.25 List four measures that may be taken to increase the resistance to fatigue of a metal alloy. Generalized Creep Behavior 8.26 Give the approximate temperature at which creep deformation becomes an important consideration for each of the following metals: tin, molybdenum, iron, gold, zinc, and chromium. 8.27 The following creep data were taken on an aluminum alloy at 480C (900F) and a constant stress of 2.75 MPa (400 psi). Plot the data as strain versus time, then determine the steady-state or minimum creep rate. Note: The initial and instantaneous strain is not included. Time (min) Strain Time (min) Strain 0 2 4 6 8 10 12 14 16 0.00 0.22 0.34 0.41 0.48 0.55 0.62 0.68 0.75 18 20 22 24 26 28 30 32 34 0.82 0.88 0.95 1.03 1.12 1.22 1.36 1.53 1.77 Stress and Temperature Effects 8.28 A specimen 1015 mm (40 in.) long of a low carbon–nickel alloy (Figure 8.31) is to be exposed to a tensile stress of 70 MPa (10,000 psi) at 427C (800F). Determine its elongation after 10,000 h. Assume that the total of both instantaneous and primary creep elongations is 1.3 mm (0.05 in.). 8.29 For a cylindrical low carbon–nickel alloy specimen (Figure 8.31) originally 19 mm (0.75 in.) in diameter and 635 mm (25 in.) long, what tensile load is necessary to produce a total elongation of 6.44 mm (0.25 in.) after 5000 h at 538C (1000F)? Assume that the sum of instantaneous and primary creep elongations is 1.8 mm (0.07 in.). 8.30 If a component fabricated from a low carbon–nickel alloy (Figure 8.30) is to be exposed to a tensile stress of 31 MPa (4500 psi) at 649C (1200F), estimate its rupture lifetime. 8.31 A cylindrical component constructed from a low carbon–nickel alloy (Figure 8.30) has a diameter of 19.1 mm (0.75 in.). Determine the maximum load that may be applied for it to survive 10,000 h at 538C (1000F). # 8.32 From Equation 8.19, if the logarithm of s is plotted versus the logarithm of s, then a straight line should result, the slope of which is the stress exponent n. Using Figure 8.31, determine the value of n for the low 1496T_c08_207-251 11/18/05 11:50 Page 250 REVISED PAGES 250 • Chapter 8 / Failure carbon–nickel alloy at each of the three temperatures. 8.33 (a) Estimate the activation energy for creep (i.e., Qc in Equation 8.20) for the low carbon– nickel alloy having the steady-state creep behavior shown in Figure 8.31. Use data taken at a stress level of 55 MPa (8000 psi) and temperatures of 427C and 538C. Assume that the stress exponent n is independent of tem# perature. (b) Estimate s at 649C (922 K). 8.34 Steady-state creep rate data are given here for some alloy taken at 200C (473 K): # s (h1) 3 2.5  10 2.4  102 [MPa (psi)] 55 (8000) 69 (10,000) If it is known that the activation energy for creep is 140,000 J/mol, compute the steady- state creep rate at a temperature of 250C (523 K) and a stress level of 48 MPa (7000 psi). 8.35 Steady-state creep data taken for an iron at a stress level of 140 MPa (20,000 psi) are given here: # s (h1) 4 6.6  10 8.8  102 T (K) 1090 1200 If it is known that the value of the stress exponent n for this alloy is 8.5, compute the steady-state creep rate at 1300 K and a stress level of 83 MPa (12,000 psi). Alloys for High-Temperature Use 8.36 Cite three metallurgical/processing techniques that are employed to enhance the creep resistance of metal alloys. DESIGN PROBLEMS 8.D1 Each student (or group of students) is to obtain an object/structure/component that has failed. It may come from your home, an automobile repair shop, a machine shop, etc. Conduct an investigation to determine the cause and type of failure (i.e., simple fracture, fatigue, creep). In addition, propose measures that can be taken to prevent future incidents of this type of failure. Finally, submit a report that addresses the above issues. Principles of Fracture Mechanics 8.D2 (a) For the thin-walled spherical tank discussed in Design Example 8.1, on the basis of critical crack size criterion [as addressed in part (a)], rank the following polymers from longest to shortest critical crack length: nylon 6,6 (50% relative humidity), polycarbonate, poly(ethylene terephthalate), and poly(methyl methacrylate). Comment on the magnitude range of the computed values used in the ranking relative to those tabulated for metal alloys as provided in Table 8.3. For these computations, use data contained in Tables B.4 and B.5 in Appendix B. (b) Now rank these same four polymers relative to maximum allowable pressure according to the leak-before-break criterion, as described in the (b) portion of Design Example 8.1. As above, comment on these values in relation to those for the metal alloys that are tabulated in Table 8.4. Data Extrapolation Methods 8.D3 An S-590 iron component (Figure 8.32) must have a creep rupture lifetime of at least 20 days at 650C (923 K). Compute the maximum allowable stress level. 8.D4 Consider an S-590 iron component (Figure 8.32) that is subjected to a stress of 55 MPa (8000 psi). At what temperature will the rupture lifetime be 200 h? 8.D5 For an 18-8 Mo stainless steel (Figure 8.35), predict the time to rupture for a component that is subjected to a stress of 100 MPa (14,500 psi) at 600C (873 K). 8.D6 Consider an 18-8 Mo stainless steel component (Figure 8.35) that is exposed to a temperature of 650C (923 K). What is the maximum allowable stress level for a rupture lifetime of 1 year? 15 years? 1496T_c08_207-251 11/18/05 11:50 Page 251 REVISED PAGES Design Problems • 251 Figure 8.35 Logarithm stress versus the Larson–Miller parameter for an 18-8 Mo stainless steel. (From F. R. Larson and J. Miller, Trans. ASME, 74, 765, 1952. Reprinted by permission of ASME.) 103 T(20 + log tr)(°R–h) 25 30 35 40 45 50 100 10 10 12 16 20 24 103 T(20 + log tr)(K–h) 28 1 Stress (103 psi) Stress (MPa) 100 1496T_c09_252-310 12/21/05 8:30 Page 252 Chapter A 9 2nd REVISE PAGES Phase Diagrams scanning electron micrograph showing the microstructure of a plain carbon steel that contains 0.44 wt% C. The large dark areas are proeutectoid ferrite. Regions having the alternating light and dark lamellar structure are pearlite; the dark and light layers in the pearlite correspond, respectively, to ferrite and cementite phases. During etching of the surface prior to examination, the ferrite phase was preferentially dissolved; thus, the pearlite appears in topographical relief with cementite layers being elevated above the ferrite layers. 3000. (Micrograph courtesy of Republic Steel Corporation.) WHY STUDY Phase Diagrams? One reason that a knowledge and understanding of phase diagrams is important to the engineer relates to the design and control of heat-treating procedures; some properties of materials are functions of their microstructures, and, consequently, of their thermal histories. Even though most phase diagrams represent stable (or equilibrium) states and microstructures, they 252 • are nevertheless useful in understanding the development and preservation of nonequilibrium structures and their attendant properties; it is often the case that these properties are more desirable than those associated with the equilibrium state. This is aptly illustrated by the phenomenon of precipitation hardening (Section 11.9). 1496T_c09_252-310 12/21/05 7:56 Page 253 2nd REVISE PAGES Learning Objectives After studying this chapter you should be able to do the following: 1. (a) Schematically sketch simple isomorphous (b) write reactions for all these transformations and eutectic phase diagrams. for either heating or cooling. (b) On these diagrams label the various phase 4. Given the composition of an iron–carbon alloy regions. containing between 0.022 wt% C and 2.14 wt% (c) Label liquidus, solidus, and solvus lines. C, be able to 2. Given a binary phase diagram, the composition (a) specify whether the alloy is hypoeutectoid or of an alloy, its temperature, and assuming that hypereutectoid, the alloy is at equilibrium, determine (b) name the proeutectoid phase, (a) what phase(s) is (are) present, (c) compute the mass fractions of proeutectoid (b) the composition(s) of the phase(s), and phase and pearlite, and (c) the mass fraction(s) of the phase(s). (d) make a schematic diagram of the microstruc3. For some given binary phase diagram, do the ture at a temperature just below the following: eutectoid. (a) locate the temperatures and compositions of all eutectic, eutectoid, peritectic, and congruent phase transformations; and 9.1 INTRODUCTION The understanding of phase diagrams for alloy systems is extremely important because there is a strong correlation between microstructure and mechanical properties, and the development of microstructure of an alloy is related to the characteristics of its phase diagram. In addition, phase diagrams provide valuable information about melting, casting, crystallization, and other phenomena. This chapter presents and discusses the following topics: (1) terminology associated with phase diagrams and phase transformations; (2) pressure–temperature phase diagrams for pure materials; (3) the interpretation of phase diagrams; (4) some of the common and relatively simple binary phase diagrams, including that for the iron–carbon system; and (5) the development of equilibrium microstructures, upon cooling, for several situations. Definitions and Basic Concepts component system It is necessary to establish a foundation of definitions and basic concepts relating to alloys, phases, and equilibrium before delving into the interpretation and utilization of phase diagrams. The term component is frequently used in this discussion; components are pure metals and/or compounds of which an alloy is composed. For example, in a copper–zinc brass, the components are Cu and Zn. Solute and solvent, which are also common terms, were defined in Section 4.3. Another term used in this context is system, which has two meanings. First, “system” may refer to a specific body of material under consideration (e.g., a ladle of molten steel). Or it may relate to the series of possible alloys consisting of the same components, but without regard to alloy composition (e.g., the iron–carbon system). The concept of a solid solution was introduced in Section 4.3. By way of review, a solid solution consists of atoms of at least two different types; the solute atoms occupy either substitutional or interstitial positions in the solvent lattice, and the crystal structure of the solvent is maintained. 1496T_c09_252-310 11/29/05 11:33 Page 254 REVISED PAGES 254 • Chapter 9 / Phase Diagrams 9.2 SOLUBILITY LIMIT solubility limit For many alloy systems and at some specific temperature, there is a maximum concentration of solute atoms that may dissolve in the solvent to form a solid solution; this is called a solubility limit. The addition of solute in excess of this solubility limit results in the formation of another solid solution or compound that has a distinctly different composition. To illustrate this concept, consider the sugar– water (C12H22O11–H2O) system. Initially, as sugar is added to water, a sugar–water solution or syrup forms. As more sugar is introduced, the solution becomes more concentrated, until the solubility limit is reached, or the solution becomes saturated with sugar. At this time the solution is not capable of dissolving any more sugar, and further additions simply settle to the bottom of the container. Thus, the system now consists of two separate substances: a sugar–water syrup liquid solution and solid crystals of undissolved sugar. This solubility limit of sugar in water depends on the temperature of the water and may be represented in graphical form on a plot of temperature along the ordinate and composition (in weight percent sugar) along the abscissa, as shown in Figure 9.1. Along the composition axis, increasing sugar concentration is from left to right, and percentage of water is read from right to left. Since only two components are involved (sugar and water), the sum of the concentrations at any composition will equal 100 wt%. The solubility limit is represented as the nearly vertical line in the figure. For compositions and temperatures to the left of the solubility line, only the syrup liquid solution exists; to the right of the line, syrup and solid sugar coexist. The solubility limit at some temperature is the composition that corresponds to the intersection of the given temperature coordinate and the solubility limit line. For example, at 20C the maximum solubility of sugar in water is 65 wt%. As Figure 9.1 indicates, the solubility limit increases slightly with rising temperature. 9.3 PHASES Also critical to the understanding of phase diagrams is the concept of a phase. A phase may be defined as a homogeneous portion of a system that has uniform physical and chemical characteristics. Every pure material is considered to be a phase; Figure 9.1 The solubility of sugar (C12H22O11) in a sugar–water syrup. 100 200 150 60 Liquid solution + solid sugar Liquid solution (syrup) 40 100 20 50 Sugar Water 0 0 20 40 60 80 100 100 80 60 40 20 0 Composition (wt%) Temperature (°F) Solubility limit 80 Temperature (°C) phase 1496T_c09_252-310 11/29/05 11:33 Page 255 REVISED PAGES 9.5 Phase Equilibria • 255 so also is every solid, liquid, and gaseous solution. For example, the sugar–water syrup solution just discussed is one phase, and solid sugar is another. Each has different physical properties (one is a liquid, the other is a solid); furthermore, each is different chemically (i.e., has a different chemical composition); one is virtually pure sugar, the other is a solution of H2O and C12H22O11. If more than one phase is present in a given system, each will have its own distinct properties, and a boundary separating the phases will exist across which there will be a discontinuous and abrupt change in physical and/or chemical characteristics. When two phases are present in a system, it is not necessary that there be a difference in both physical and chemical properties; a disparity in one or the other set of properties is sufficient. When water and ice are present in a container, two separate phases exist; they are physically dissimilar (one is a solid, the other is a liquid) but identical in chemical makeup. Also, when a substance can exist in two or more polymorphic forms (e.g., having both FCC and BCC structures), each of these structures is a separate phase because their respective physical characteristics differ. Sometimes, a single-phase system is termed “homogeneous.” Systems composed of two or more phases are termed “mixtures” or “heterogeneous systems.” Most metallic alloys and, for that matter, ceramic, polymeric, and composite systems are heterogeneous. Ordinarily, the phases interact in such a way that the property combination of the multiphase system is different from, and more attractive than, either of the individual phases. 9.4 MICROSTRUCTURE Many times, the physical properties and, in particular, the mechanical behavior of a material depend on the microstructure. Microstructure is subject to direct microscopic observation, using optical or electron microscopes; this topic was touched on in Sections 4.9 and 4.10. In metal alloys, microstructure is characterized by the number of phases present, their proportions, and the manner in which they are distributed or arranged. The microstructure of an alloy depends on such variables as the alloying elements present, their concentrations, and the heat treatment of the alloy (i.e., the temperature, the heating time at temperature, and the rate of cooling to room temperature). The procedure of specimen preparation for microscopic examination was briefly outlined in Section 4.10.After appropriate polishing and etching, the different phases may be distinguished by their appearance. For example, for a two-phase alloy, one phase may appear light and the other phase dark, as in the chapter-opening photograph for this chapter. When only a single phase or solid solution is present, the texture will be uniform, except for grain boundaries that may be revealed (Figure 4.12b). 9.5 PHASE EQUILIBRIA equilibrium free energy Equilibrium is another essential concept that is best described in terms of a thermodynamic quantity called the free energy. In brief, free energy is a function of the internal energy of a system, and also the randomness or disorder of the atoms or molecules (or entropy). A system is at equilibrium if its free energy is at a minimum under some specified combination of temperature, pressure, and composition. In a macroscopic sense, this means that the characteristics of the system do not change with time but persist indefinitely; that is, the system is stable. A change in temperature, pressure, and/or composition for a system in equilibrium will result in an increase in the free energy and in a possible spontaneous change to another state whereby the free energy is lowered. 1496T_c09_252-310 11/29/05 12:46 Page 256 REVISED PAGES 256 • Chapter 9 / Phase Diagrams phase equilibrium metastable The term phase equilibrium, often used in the context of this discussion, refers to equilibrium as it applies to systems in which more than one phase may exist. Phase equilibrium is reflected by a constancy with time in the phase characteristics of a system. Perhaps an example best illustrates this concept. Suppose that a sugar–water syrup is contained in a closed vessel and the solution is in contact with solid sugar at 20C. If the system is at equilibrium, the composition of the syrup is 65 wt% C12H22O1135 wt% H2O (Figure 9.1), and the amounts and compositions of the syrup and solid sugar will remain constant with time. If the temperature of the system is suddenly raised—say, to 100C—this equilibrium or balance is temporarily upset in that the solubility limit has been increased to 80 wt% C12H22O11 (Figure 9.1). Thus, some of the solid sugar will go into solution in the syrup. This will continue until the new equilibrium syrup concentration is established at the higher temperature. This sugar–syrup example illustrates the principle of phase equilibrium using a liquid–solid system. In many metallurgical and materials systems of interest, phase equilibrium involves just solid phases. In this regard the state of the system is reflected in the characteristics of the microstructure, which necessarily include not only the phases present and their compositions but, in addition, the relative phase amounts and their spatial arrangement or distribution. Free energy considerations and diagrams similar to Figure 9.1 provide information about the equilibrium characteristics of a particular system, which is important; but they do not indicate the time period necessary for the attainment of a new equilibrium state. It is often the case, especially in solid systems, that a state of equilibrium is never completely achieved because the rate of approach to equilibrium is extremely slow; such a system is said to be in a nonequilibrium or metastable state. A metastable state or microstructure may persist indefinitely, experiencing only extremely slight and almost imperceptible changes as time progresses. Often, metastable structures are of more practical significance than equilibrium ones. For example, some steel and aluminum alloys rely for their strength on the development of metastable microstructures during carefully designed heat treatments (Sections 10.5 and 11.9). Thus not only is an understanding of equilibrium states and structures important, but also the speed or rate at which they are established and the factors that affect the rate must be considered. This chapter is devoted almost exclusively to equilibrium structures; the treatment of reaction rates and nonequilibrium structures is deferred to Chapter 10 and Section 11.9. Concept Check 9.1 What is the difference between the states of phase equilibrium and metastability? [The answer may be found at www.wiley.com/college/callister (Student Companion Site).] 9.6 ONE-COMPONENT (OR UNARY) PHASE DIAGRAMS phase diagram Much of the information about the control of the phase structure of a particular system is conveniently and concisely displayed in what is called a phase diagram, also often termed an equilibrium diagram. Now, there are three externally controllable 1496T_c09_252-310 11/29/05 12:46 Page 257 REVISED PAGES 9.6 One-Component (or Unary) Phase Diagrams • 257 parameters that will affect phase structure—viz. temperature, pressure, and composition—and phase diagrams are constructed when various combinations of these parameters are plotted against one another. Perhaps the simplest and easiest type of phase diagram to understand is that for a one-component system, in which composition is held constant (i.e., the phase diagram is for a pure substance); this means that pressure and temperature are the variables. This one-component phase diagram (or unary phase diagram) [sometimes also called a pressure–temperature (or P–T ) diagram] is represented as a two-dimensional plot of pressure (ordinate, or vertical axis) versus temperature (abscissa, or horizontal axis). Most often, the pressure axis is scaled logarithmically. We illustrate this type of phase diagram and demonstrate its interpretation using as an example the one for H2O, which is shown in Figure 9.2. Here it may be noted that regions for three different phases—solid, liquid, and vapor—are delineated on the plot. Each of the phases will exist under equilibrium conditions over the temperature–pressure ranges of its corresponding area. Furthermore, the three curves shown on the plot (labeled aO, bO, and cO) are phase boundaries; at any point on one of these curves, the two phases on either side of the curve are in equilibrium (or coexist) with one another. That is, equilibrium between solid and vapor phases is along curve aO—likewise for the solid-liquid, curve bO, and the liquidvapor, curve cO. Also, upon crossing a boundary (as temperature and/or pressure is altered), one phase transforms to another. For example, at one atmosphere pressure, during heating the solid phase transforms to the liquid phase (i.e., melting occurs) at the point labeled 2 on Figure 9.2 (i.e., the intersection of the dashed horizontal line with the solid-liquid phase boundary); this point corresponds to a temperature of 0C. Of course, the reverse transformation (liquid-to-solid, or solidification) takes place at the same point upon cooling. Similarly, at the intersection of the dashed line with the liquid-vapor phase boundary [point 3 (Figure 9.2), at 100C] the liquid transforms to the vapor phase (or vaporizes) upon heating; 1,000 b Pressure (atm) 100 Liquid (Water) Solid (Ice) 10 c 3 2 1.0 0.1 Vapor (Steam) O 0.01 a 0.001 20 0 20 40 60 80 100 120 Temperature (°C) Figure 9.2 Pressure–temperature phase diagram for H2O. Intersection of the dashed horizontal line at 1 atm pressure with the solid-liquid phase boundary (point 2) corresponds to the melting point at this pressure (T  0C). Similarly, point 3, the intersection with the liquid-vapor boundary, represents the boiling point (T  100C). 1496T_c09_252-310 11/29/05 12:46 Page 258 REVISED PAGES 258 • Chapter 9 / Phase Diagrams condensation occurs for cooling. And, finally, solid ice sublimes or vaporizes upon crossing the curve labeled aO. As may also be noted from Figure 9.2, all three of the phase boundary curves intersect at a common point, which is labeled O (and for this H2O system, at a temperature of 273.16 K and a pressure of 6.04  103 atm). This means that at this point only, all of the solid, liquid, and vapor phases are simultaneously in equilibrium with one another. Appropriately, this, and any other point on a P–T phase diagram where three phases are in equilibrium, is called a triple point; sometimes it is also termed an invariant point inasmuch as its position is distinct, or fixed by definite values of pressure and temperature. Any deviation from this point by a change of temperature and/or pressure will cause at least one of the phases to disappear. Pressure-temperature phase diagrams for a number of substances have been determined experimentally, which also have solid, liquid, and vapor phase regions. In those instances when multiple solid phases (i.e., allotropes, Section 3.6) exist, there will appear a region on the diagram for each solid phase, and also other triple points. The pressure-temperature phase diagram for carbon (which includes phase regions for diamond and graphite) is shown on the front and back covers of this book. Binary Phase Diagrams Another type of extremely common phase diagram is one in which temperature and composition are variable parameters, and pressure is held constant—normally 1 atm. There are several different varieties; in the present discussion, we will concern ourselves with binary alloys—those that contain two components. If more than two components are present, phase diagrams become extremely complicated and difficult to represent. An explanation of the principles governing and the interpretation of phase diagrams can be demonstrated using binary alloys even though most alloys contain more than two components. Binary phase diagrams are maps that represent the relationships between temperature and the compositions and quantities of phases at equilibrium, which influence the microstructure of an alloy. Many microstructures develop from phase transformations, the changes that occur when the temperature is altered (ordinarily upon cooling). This may involve the transition from one phase to another, or the appearance or disappearance of a phase. Binary phase diagrams are helpful in predicting phase transformations and the resulting microstructures, which may have equilibrium or nonequilibrium character. 9.7 BINARY ISOMORPHOUS SYSTEMS Possibly the easiest type of binary phase diagram to understand and interpret is the type that is characterized by the copper–nickel system (Figure 9.3a). Temperature is plotted along the ordinate, and the abscissa represents the composition of the alloy, in weight percent (bottom) and atom percent (top) of nickel. The composition ranges from 0 wt% Ni (100 wt% Cu) on the left horizontal extremity to 100 wt% Ni (0 wt% Cu) on the right. Three different phase regions, or fields, appear on the diagram, an alpha (a) field, a liquid (L) field, and a two-phase a  L field. Each region is defined by the phase or phases that exist over the range of temperatures and compositions delimited by the phase boundary lines. The liquid L is a homogeneous liquid solution composed of both copper and nickel. The a phase is a substitutional solid solution consisting of both Cu and Ni atoms, and 1496T_c09_252-310 11/29/05 11:33 Page 259 REVISED PAGES 9.7 Binary Isomorphous Systems • 259 Composition (at% Ni) 0 20 40 60 80 100 1600 2800 1500 Liquid 1453°C 2600 Solidus line Liquidus line 1300 2400 ␣ +L B 1200 2200 ␣ A 1100 Temperature (°F) 1400 Temperature (°C) Figure 9.3 (a) The copper–nickel phase diagram. (b) A portion of the copper–nickel phase diagram for which compositions and phase amounts are determined at point B. (Adapted from Phase Diagrams of Binary Nickel Alloys, P. Nash, Editor, 1991. Reprinted by permission of ASM International, Materials Park, OH.) 2000 1085°C 1000 0 40 20 (Cu) 60 80 100 Composition (wt% Ni) (Ni) (a) 1300 Liquid Temperature (°C) Tie line B ␣ + Liquid ␣ + Liquid ␣ 1200 R S ␣ 20 40 30 CL C0 Composition (wt% Ni) 50 C␣ (b) isomorphous having an FCC crystal structure. At temperatures below about 1080C, copper and nickel are mutually soluble in each other in the solid state for all compositions. This complete solubility is explained by the fact that both Cu and Ni have the same crystal structure (FCC), nearly identical atomic radii and electronegativities, and similar valences, as discussed in Section 4.3. The copper–nickel system is termed isomorphous because of this complete liquid and solid solubility of the two components. 1496T_c09_252-310 11/29/05 11:33 Page 260 REVISED PAGES 260 • Chapter 9 / Phase Diagrams A couple of comments are in order regarding nomenclature. First, for metallic alloys, solid solutions are commonly designated by lowercase Greek letters (a, b, g, etc.). Furthermore, with regard to phase boundaries, the line separating the L and a  L phase fields is termed the liquidus line, as indicated in Figure 9.3a; the liquid phase is present at all temperatures and compositions above this line.The solidus line is located between the a and a  L regions, below which only the solid a phase exists. For Figure 9.3a, the solidus and liquidus lines intersect at the two composition extremities; these correspond to the melting temperatures of the pure components. For example, the melting temperatures of pure copper and nickel are 1085C and 1453C, respectively. Heating pure copper corresponds to moving vertically up the left-hand temperature axis. Copper remains solid until its melting temperature is reached. The solid-to-liquid transformation takes place at the melting temperature, and no further heating is possible until this transformation has been completed. For any composition other than pure components, this melting phenomenon will occur over the range of temperatures between the solidus and liquidus lines; both solid a and liquid phases will be in equilibrium within this temperature range. For example, upon heating an alloy of composition 50 wt% Ni–50 wt% Cu (Figure 9.3a), melting begins at approximately 1280C (2340F); the amount of liquid phase continuously increases with temperature until about 1320C (2410F), at which the alloy is completely liquid. 9.8 INTERPRETATION OF PHASE DIAGRAMS For a binary system of known composition and temperature that is at equilibrium, at least three kinds of information are available: (1) the phases that are present, (2) the compositions of these phases, and (3) the percentages or fractions of the phases. The procedures for making these determinations will be demonstrated using the copper–nickel system. Phases Present Bi-Sb Ge-Si The establishment of what phases are present is relatively simple. One just locates the temperature–composition point on the diagram and notes the phase(s) with which the corresponding phase field is labeled. For example, an alloy of composition 60 wt% Ni–40 wt% Cu at 1100C would be located at point A in Figure 9.3a; since this is within the a region, only the single a phase will be present. On the other hand, a 35 wt% Ni–65 wt% Cu alloy at 1250C (point B) will consist of both a and liquid phases at equilibrium. Determination of Phase Compositions Bi-Sb Ge-Si tie line The first step in the determination of phase compositions (in terms of the concentrations of the components) is to locate the temperature–composition point on the phase diagram. Different methods are used for single- and two-phase regions. If only one phase is present, the procedure is trivial: the composition of this phase is simply the same as the overall composition of the alloy. For example, consider the 60 wt% Ni–40 wt% Cu alloy at 1100C (point A, Figure 9.3a). At this composition and temperature, only the a phase is present, having a composition of 60 wt% Ni–40 wt% Cu. For an alloy having composition and temperature located in a two-phase region, the situation is more complicated. In all two-phase regions (and in two-phase regions only), one may imagine a series of horizontal lines, one at every temperature; each of these is known as a tie line, or sometimes as an isotherm. These tie 1496T_c09_252-310 11/29/05 11:33 Page 261 REVISED PAGES 9.8 Interpretation of Phase Diagrams • 261 lines extend across the two-phase region and terminate at the phase boundary lines on either side. To compute the equilibrium concentrations of the two phases, the following procedure is used: 1. A tie line is constructed across the two-phase region at the temperature of the alloy. 2. The intersections of the tie line and the phase boundaries on either side are noted. 3. Perpendiculars are dropped from these intersections to the horizontal composition axis, from which the composition of each of the respective phases is read. For example, consider again the 35 wt% Ni–65 wt% Cu alloy at 1250C, located at point B in Figure 9.3b and lying within the a  L region. Thus, the problem is to determine the composition (in wt% Ni and Cu) for both the a and liquid phases. The tie line has been constructed across the a  L phase region, as shown in Figure 9.3b. The perpendicular from the intersection of the tie line with the liquidus boundary meets the composition axis at 31.5 wt% Ni–68.5 wt% Cu, which is the composition of the liquid phase, CL. Likewise, for the solidus–tie line intersection, we find a composition for the a solid-solution phase, Ca, of 42.5 wt% Ni–57.5 wt% Cu. Determination of Phase Amounts Bi-Sb Ge-Si lever rule The relative amounts (as fraction or as percentage) of the phases present at equilibrium may also be computed with the aid of phase diagrams. Again, the single- and two-phase situations must be treated separately.The solution is obvious in the singlephase region: Since only one phase is present, the alloy is composed entirely of that phase; that is, the phase fraction is 1.0 or, alternatively, the percentage is 100%. From the previous example for the 60 wt% Ni–40 wt% Cu alloy at 1100C (point A in Figure 9.3a), only the a phase is present; hence, the alloy is completely or 100% a. If the composition and temperature position is located within a two-phase region, things are more complex. The tie line must be utilized in conjunction with a procedure that is often called the lever rule (or the inverse lever rule), which is applied as follows: 1. The tie line is constructed across the two-phase region at the temperature of the alloy. 2. The overall alloy composition is located on the tie line. 3. The fraction of one phase is computed by taking the length of tie line from the overall alloy composition to the phase boundary for the other phase, and dividing by the total tie line length. 4. The fraction of the other phase is determined in the same manner. 5. If phase percentages are desired, each phase fraction is multiplied by 100. When the composition axis is scaled in weight percent, the phase fractions computed using the lever rule are mass fractions—the mass (or weight) of a specific phase divided by the total alloy mass (or weight). The mass of each phase is computed from the product of each phase fraction and the total alloy mass. In the employment of the lever rule, tie line segment lengths may be determined either by direct measurement from the phase diagram using a linear scale, preferably graduated in millimeters, or by subtracting compositions as taken from the composition axis. 1496T_c09_252-310 11/29/05 11:33 Page 262 REVISED PAGES 262 • Chapter 9 / Phase Diagrams Consider again the example shown in Figure 9.3b, in which at 1250C both a and liquid phases are present for a 35 wt% Ni–65 wt% Cu alloy. The problem is to compute the fraction of each of the a and liquid phases. The tie line has been constructed that was used for the determination of a and L phase compositions. Let the overall alloy composition be located along the tie line and denoted as C0, and mass fractions be represented by WL and Wa for the respective phases. From the lever rule, WL may be computed according to S RS (9.1a) Ca  C0 Ca  CL (9.1b) WL  or, by subtracting compositions, Lever rule expression for computation of liquid mass fraction (per Figure 9.3b) WL  Composition need be specified in terms of only one of the constituents for a binary alloy; for the computation above, weight percent nickel will be used (i.e., C0  35 wt% Ni, Ca  42.5 wt% Ni, and CL  31.5 wt% Ni), and WL  42.5  35  0.68 42.5  31.5 Wa  R RS (9.2a)  C0  CL Ca  CL (9.2b)  35  31.5  0.32 42.5  31.5 Similarly, for the a phase, Lever rule expression for computation of a-phase mass fraction (per Figure 9.3b) Of course, identical answers are obtained if compositions are expressed in weight percent copper instead of nickel. Thus, the lever rule may be employed to determine the relative amounts or fractions of phases in any two-phase region for a binary alloy if the temperature and composition are known and if equilibrium has been established. Its derivation is presented as an example problem. It is easy to confuse the foregoing procedures for the determination of phase compositions and fractional phase amounts; thus, a brief summary is warranted. Compositions of phases are expressed in terms of weight percents of the components (e.g., wt% Cu, wt% Ni). For any alloy consisting of a single phase, the composition of that phase is the same as the total alloy composition. If two phases are present, the tie line must be employed, the extremities of which determine the compositions of the respective phases. With regard to fractional phase amounts (e.g., mass fraction of the a or liquid phase), when a single phase exists, the alloy is completely that phase. For a two-phase alloy, on the other hand, the lever rule is utilized, in which a ratio of tie line segment lengths is taken. 1496T_c09_252-310 11/29/05 11:33 Page 263 REVISED PAGES 9.8 Interpretation of Phase Diagrams • 263 Concept Check 9.2 A copper–nickel alloy of composition 70 wt% Ni–30 wt% Cu is slowly heated from a temperature of 1300C (2370F). (a) (b) (c) (d) At what temperature does the first liquid phase form? What is the composition of this liquid phase? At what temperature does complete melting of the alloy occur? What is the composition of the last solid remaining prior to complete melting? [The answer may be found at www.wiley.com/college/callister (Student Companion Site).] Concept Check 9.3 Is it possible to have a copper–nickel alloy that, at equilibrium, consists of an a phase of composition 37 wt% Ni–63 wt% Cu, and also a liquid phase of composition 20 wt% Ni–80 wt% Cu? If so, what will be the approximate temperature of the alloy? If this is not possible, explain why. [The answer may be found at www.wiley.com/college/callister (Student Companion Site).] EXAMPLE PROBLEM 9.1 Lever Rule Derivation Derive the lever rule. Solution Consider the phase diagram for copper and nickel (Figure 9.3b) and alloy of composition C0 at 1250C, and let Ca, CL, Wa, and WL represent the same parameters as above. This derivation is accomplished through two conservationof-mass expressions. With the first, since only two phases are present, the sum of their mass fractions must be equal to unity; that is, Wa  WL  1 (9.3) For the second, the mass of one of the components (either Cu or Ni) that is present in both of the phases must be equal to the mass of that component in the total alloy, or WaCa  WLCL  C0 (9.4) Simultaneous solution of these two equations leads to the lever rule expressions for this particular situation, Equations 9.1b and 9.2b: Ca  C0 Ca  CL C0  CL Wa  Ca  CL WL  (9.1b) (9.2b) 1496T_c09_252-310 11/29/05 12:46 Page 264 REVISED PAGES 264 • Chapter 9 / Phase Diagrams For multiphase alloys, it is often more convenient to specify relative phase amount in terms of volume fraction rather than mass fraction. Phase volume fractions are preferred because they (rather than mass fractions) may be determined from examination of the microstructure; furthermore, the properties of a multiphase alloy may be estimated on the basis of volume fractions. For an alloy consisting of a and b phases, the volume fraction of the a phase, Va, is defined as a phase volume fraction— dependence on volumes of a and b phases Va  va va  vb (9.5) where va and vb denote the volumes of the respective phases in the alloy. Of course, an analogous expression exists for Vb, and, for an alloy consisting of just two phases, it is the case that Va  Vb  1. On occasion conversion from mass fraction to volume fraction (or vice versa) is desired. Equations that facilitate these conversions are as follows: Va  Conversion of mass fractions of a and b phases to volume fractions Wa ra Wb Wa  ra rb (9.6a) Wb Vb  rb Wb Wa  ra rb (9.6b) and Wa  Conversion of volume fractions of a and b phases to mass fractions Wb  Vara Vara  Vbrb Vbrb Vara  Vbrb (9.7a) (9.7b) In these expressions, ra and rb are the densities of the respective phases; these may be determined approximately using Equations 4.10a and 4.10b. When the densities of the phases in a two-phase alloy differ significantly, there will be quite a disparity between mass and volume fractions; conversely, if the phase densities are the same, mass and volume fractions are identical. 9.9 DEVELOPMENT OF MICROSTRUCTURE IN ISOMORPHOUS ALLOYS Equilibrium Cooling At this point it is instructive to examine the development of microstructure that occurs for isomorphous alloys during solidification. We first treat the situation in which the cooling occurs very slowly, in that phase equilibrium is continuously maintained. 1496T_c09_252-310 11/29/05 11:33 Page 265 REVISED PAGES 9.9 Development of Microstructure in Isomorphous Alloys • 265 Figure 9.4 Schematic representation of the development of microstructure during the equilibrium solidification of a 35 wt% Ni–65 wt% Cu alloy. L L (35 Ni) L (35 Ni) ␣ (46 Ni) 1300 a ␣ + L L (32 Ni) b ␣ (46 Ni) Temperature (°C) c ␣ (43 Ni) ␣ (43 Ni) L (24 Ni) d ␣ ␣ L (32 Ni) ␣ ␣ 1200 L (24 Ni) e ␣ (35 Ni) ␣ ␣ ␣ ␣ ␣ ␣ ␣ ␣ ␣ ␣ ␣ ␣ ␣ ␣ ␣ (35 Ni) 1100 20 ␣ 30 40 ␣ ␣ 50 Composition (wt% Ni) Bi-Sb Ge-Si Let us consider the copper–nickel system (Figure 9.3a), specifically an alloy of composition 35 wt% Ni–65 wt% Cu as it is cooled from 1300C. The region of the Cu–Ni phase diagram in the vicinity of this composition is shown in Figure 9.4. Cooling of an alloy of the above composition corresponds to moving down the vertical dashed line. At 1300C, point a, the alloy is completely liquid (of composition 35 wt% Ni–65 wt% Cu) and has the microstructure represented by the circle inset in the figure. As cooling begins, no microstructural or compositional changes will be realized until we reach the liquidus line (point b, 1260C). At this point, the first solid a begins to form, which has a composition dictated by the tie line drawn at this temperature [i.e., 46 wt% Ni–54 wt% Cu, noted as a(46 Ni)]; the composition of liquid is still approximately 35 wt% Ni–65 wt% Cu [L(35 Ni)], which is different from that of the solid a. With continued cooling, both compositions and relative amounts of each of the phases will change. The compositions of the liquid and a phases will follow the liquidus and solidus lines, respectively. Furthermore, the fraction of the a phase will increase with continued cooling. Note that the overall alloy composition (35 wt% Ni–65 wt% Cu) remains unchanged during cooling even though there is a redistribution of copper and nickel between the phases. At 1250C, point c in Figure 9.4, the compositions of the liquid and a phases are 32 wt% Ni–68 wt% Cu [L(32 Ni)] and 43 wt% Ni–57 wt% Cu [a(43 Ni)], respectively. 1496T_c09_252-310 11/29/05 11:33 Page 266 REVISED PAGES 266 • Chapter 9 / Phase Diagrams The solidification process is virtually complete at about 1220C, point d; the composition of the solid a is approximately 35 wt% Ni–65 wt% Cu (the overall alloy composition) while that of the last remaining liquid is 24 wt% Ni–76 wt% Cu. Upon crossing the solidus line, this remaining liquid solidifies; the final product then is a polycrystalline a-phase solid solution that has a uniform 35 wt% Ni–65 wt% Cu composition (point e, Figure 9.4). Subsequent cooling will produce no microstructural or compositional alterations. Nonequilibrium Cooling Conditions of equilibrium solidification and the development of microstructures, as described in the previous section, are realized only for extremely slow cooling rates. The reason for this is that with changes in temperature, there must be readjustments in the compositions of the liquid and solid phases in accordance with the phase diagram (i.e., with the liquidus and solidus lines), as discussed. These readjustments are accomplished by diffusional processes—that is, diffusion in both solid and liquid phases and also across the solid–liquid interface. Inasmuch as diffusion is a time-dependent phenomenon (Section 5.3), to maintain equilibrium during cooling, sufficient time must be allowed at each temperature for the appropriate compositional readjustments. Diffusion rates (i.e., the magnitudes of the diffusion coefficients) are especially low for the solid phase and, for both phases, decrease with diminishing temperature. In virtually all practical solidification situations, cooling rates are much too rapid to allow these compositional readjustments and maintenance of equilibrium; consequently, microstructures other than those previously described develop. Some of the consequences of nonequilibrium solidification for isomorphous alloys will now be discussed by considering a 35 wt% Ni–65 wt% Cu alloy, the same composition that was used for equilibrium cooling in the previous section. The portion of the phase diagram near this composition is shown in Figure 9.5; in addition, microstructures and associated phase compositions at various temperatures upon cooling are noted in the circular insets. To simplify this discussion it will be assumed that diffusion rates in the liquid phase are sufficiently rapid such that equilibrium is maintained in the liquid. Let us begin cooling from a temperature of about 1300C; this is indicated by point a¿ in the liquid region. This liquid has a composition of 35 wt% Ni–65 wt% Cu [noted as L(35 Ni) in the figure], and no changes occur while cooling through the liquid phase region (moving down vertically from point a¿ ). At point b¿ (approximately 1260C), a-phase particles begin to form, which, from the tie line constructed, have a composition of 46 wt% Ni–54 wt% Cu [a(46 Ni)]. Upon further cooling to point c¿ (about 1240C), the liquid composition has shifted to 29 wt% Ni–71 wt% Cu; furthermore, at this temperature the composition of the a phase that solidified is 40 wt% Ni–60 wt% Cu [a(40 Ni)]. However, since diffusion in the solid a phase is relatively slow, the a phase that formed at point b has not changed composition appreciably—that is, it is still about 46 wt% Ni—and the composition of the a grains has continuously changed with radial position, from 46 wt% Ni at grain centers to 40 wt% Ni at the outer grain perimeters. Thus, at point c¿, the average composition of the solid a grains that have formed would be some volume weighted average composition, lying between 46 and 40 wt% Ni. For the sake of argument, let us take this average composition to be 42 wt% Ni–58 wt% Cu [a(42 Ni)]. Furthermore, we would also find that, on the basis of lever-rule computations, a greater proportion of liquid is present for these 1496T_c09_252-310 1/10/06 23:53 Page 267 2nd REVISE PAGES 9.9 Development of Microstructure in Isomorphous Alloys • 267 Figure 9.5 Schematic representation of the development of microstructure during the nonequilibrium solidification of a 35 wt% Ni–65 wt% Cu alloy. L L (35 Ni) L (35 Ni) 1300 ␣ (46 Ni) ␣ +L a ␣ b ␣ (40 Ni) L (29 Ni) ␣ (46 Ni) Temperature (°C) c L (24 Ni) ␣ (42 Ni) d L (21 Ni) ␣ (35 Ni) e 1200 L (29 Ni) ␣ (46 Ni) ␣ (40 Ni) ␣ (38 Ni) ␣ (31 Ni) L (24 Ni) ␣ (46 Ni) ␣ (40 Ni) ␣ (35 Ni) f ␣ (46 Ni) ␣ (40 Ni) ␣ (35 Ni) ␣ (31 Ni) L (21 Ni) ␣ (46 Ni) ␣ (40 Ni) ␣ (35 Ni) ␣ (31 Ni) 1100 20 30 40 Composition (wt% Ni) 50 60 nonequilibrium conditions than for equilibrium cooling. The implication of this nonequilibrium solidification phenomenon is that the solidus line on the phase diagram has been shifted to higher Ni contents—to the average compositions of the a phase (e.g., 42 wt% Ni at 1240C)—and is represented by the dashed line in Figure 9.5. There is no comparable alteration of the liquidus line inasmuch as it is assumed that equilibrium is maintained in the liquid phase during cooling because of sufficiently rapid diffusion rates. At point d¿ (1220C) and for equilibrium cooling rates, solidification should be completed. However, for this nonequilibrium situation, there is still an appreciable proportion of liquid remaining, and the a phase that is forming has a composition of 35 wt% Ni [a(35 Ni)]; also the average a-phase composition at this point is 38 wt% Ni [a(38 Ni)]. Nonequilibrium solidification finally reaches completion at point e¿ (1205C). The composition of the last a phase to solidify at this point is about 31 wt% Ni; the average composition of the a phase at complete solidification is 35 wt% Ni. The inset at point f ¿ shows the microstructure of the totally solid material. The degree of displacement of the nonequilibrium solidus curve from the equilibrium one will depend on rate of cooling. The slower the cooling rate, the smaller 1496T_c09_252-310 11/29/05 11:33 Page 268 REVISED PAGES 268 • Chapter 9 / Phase Diagrams this displacement; that is, the difference between the equilibrium solidus and average solid composition is lower. Furthermore, if the diffusion rate in the solid phase is increased, this displacement will be diminished. There are some important consequences for isomorphous alloys that have solidified under nonequilibrium conditions. As discussed above, the distribution of the two elements within the grains is nonuniform, a phenomenon termed segregation; that is, concentration gradients are established across the grains that are represented by the insets of Figure 9.5. The center of each grain, which is the first part to freeze, is rich in the high-melting element (e.g., nickel for this Cu–Ni system), whereas the concentration of the low-melting element increases with position from this region to the grain boundary. This is termed a cored structure, which gives rise to less than the optimal properties. As a casting having a cored structure is reheated, grain boundary regions will melt first inasmuch as they are richer in the low-melting component. This produces a sudden loss in mechanical integrity due to the thin liquid film that separates the grains. Furthermore, this melting may begin at a temperature below the equilibrium solidus temperature of the alloy. Coring may be eliminated by a homogenization heat treatment carried out at a temperature below the solidus point for the particular alloy composition. During this process, atomic diffusion occurs, which produces compositionally homogeneous grains. 9.10 MECHANICAL PROPERTIES OF ISOMORPHOUS ALLOYS We shall now briefly explore how the mechanical properties of solid isomorphous alloys are affected by composition as other structural variables (e.g., grain size) are held constant. For all temperatures and compositions below the melting temperature of the lowest-melting component, only a single solid phase will exist. Therefore, each component will experience solid-solution strengthening (Section 7.9), or an increase in strength and hardness by additions of the other component. This effect is demonstrated in Figure 9.6a as tensile strength versus composition for the 50 300 40 30 200 0 (Cu) 20 40 60 Composition (wt% Ni) (a) 80 100 (Ni) Tensile strength (ksi) Tensile strength (MPa) 60 400 Elongation (% in 50 mm [2 in.]) 60 50 40 30 20 0 (Cu) 20 40 60 Composition (wt% Ni) 80 100 (Ni) (b) Figure 9.6 For the copper–nickel system, (a) tensile strength versus composition, and (b) ductility (%EL) versus composition at room temperature. A solid solution exists over all compositions for this system. 1496T_c09_252-310 11/29/05 11:33 Page 269 REVISED PAGES 9.11 Binary Eutectic Systems • 269 copper–nickel system at room temperature; at some intermediate composition, the curve necessarily passes through a maximum. Plotted in Figure 9.6b is the ductility (%EL)–composition behavior, which is just the opposite of tensile strength; that is, ductility decreases with additions of the second component, and the curve exhibits a minimum. 9.11 BINARY EUTECTIC SYSTEMS Another type of common and relatively simple phase diagram found for binary alloys is shown in Figure 9.7 for the copper–silver system; this is known as a binary eutectic phase diagram. A number of features of this phase diagram are important and worth noting. First, three single-phase regions are found on the diagram: a, b, and liquid. The a phase is a solid solution rich in copper; it has silver as the solute component and an FCC crystal structure. The -phase solid solution also has an FCC structure, but copper is the solute. Pure copper and pure silver are also considered to be a and b phases, respectively. Thus, the solubility in each of these solid phases is limited, in that at any temperature below line BEG only a limited concentration of silver will dissolve in copper (for the a phase), and similarly for copper in silver (for the b phase). The solubility limit for the a phase corresponds to the boundary line, labeled CBA, between the a(a  b) and a(a  L) phase regions; it increases with temperature to Composition (at% Ag) 0 20 40 60 80 100 2200 1200 A 2000 Liquidus 1000 Liquid 1800 F ␣ +L ␣ 800 1600 779°C (TE) B ␤+L E 8.0 (C␣ E) 71.9 (CE) 91.2 (C␤ E) G 1400 ␤ 1200 600 1000 Solvus ␣ +␤ 800 400 C 600 H 200 0 (Cu) 20 40 60 Composition (wt% Ag) 80 400 100 (Ag) Figure 9.7 The copper–silver phase diagram. [Adapted from Binary Alloy Phase Diagrams, 2nd edition, Vol. 1, T. B. Massalski (Editor-in-Chief), 1990. Reprinted by permission of ASM International, Materials Park, OH.] Temperature (°F) Temperature (°C) Solidus 1496T_c09_252-310 11/29/05 11:33 Page 270 REVISED PAGES 270 • Chapter 9 / Phase Diagrams solvus line solidus line liquidus line invariant point The eutectic reaction (per Figure 9.7) a maximum [8.0 wt% Ag at 779C (1434F)] at point B, and decreases back to zero at the melting temperature of pure copper, point A [1085C (1985F)]. At temperatures below 779C (1434F), the solid solubility limit line separating the a and a  b phase regions is termed a solvus line; the boundary AB between the a and a  L fields is the solidus line, as indicated in Figure 9.7. For the b phase, both solvus and solidus lines also exist, HG and GF, respectively, as shown. The maximum solubility of copper in the b phase, point G (8.8 wt% Cu), also occurs at 779C (1434F). This horizontal line BEG, which is parallel to the composition axis and extends between these maximum solubility positions, may also be considered a solidus line; it represents the lowest temperature at which a liquid phase may exist for any copper–silver alloy that is at equilibrium. There are also three two-phase regions found for the copper–silver system (Figure 9.7): a  L, b  L, and a  b. The a- and b-phase solid solutions coexist for all compositions and temperatures within the a  b phase field; the a  liquid and b  liquid phases also coexist in their respective phase regions. Furthermore, compositions and relative amounts for the phases may be determined using tie lines and the lever rule as outlined previously. As silver is added to copper, the temperature at which the alloys become totally liquid decreases along the liquidus line, line AE; thus, the melting temperature of copper is lowered by silver additions. The same may be said for silver: the introduction of copper reduces the temperature of complete melting along the other liquidus line, FE. These liquidus lines meet at the point E on the phase diagram, through which also passes the horizontal isotherm line BEG. Point E is called an invariant point, which is designated by the composition CE and temperature TE; for the copper–silver system, the values of CE and TE are 71.9 wt% Ag and 779C (1434F), respectively. An important reaction occurs for an alloy of composition CE as it changes temperature in passing through TE; this reaction may be written as follows: L1CE 2 Δ a 1CaE 2  b1CbE 2 cooling heating eutectic reaction (9.8) Or, upon cooling, a liquid phase is transformed into the two solid a and b phases at the temperature TE; the opposite reaction occurs upon heating. This is called a eutectic reaction (eutectic means easily melted), and CE and TE represent the eutectic composition and temperature, respectively; CaE and CbE are the respective compositions of the a and b phases at TE. Thus, for the copper–silver system, the eutectic reaction, Equation 9.8, may be written as follows: cooling L171.9 wt% Ag2 Δ a18.0 wt%Ag2  b191.2 wt% Ag2 heating Often, the horizontal solidus line at TE is called the eutectic isotherm. The eutectic reaction, upon cooling, is similar to solidification for pure components in that the reaction proceeds to completion at a constant temperature, or isothermally, at TE. However, the solid product of eutectic solidification is always two solid phases, whereas for a pure component only a single phase forms. Because of this eutectic reaction, phase diagrams similar to that in Figure 9.7 are termed eutectic phase diagrams; components exhibiting this behavior comprise a eutectic system. 1496T_c09_252-310 11/29/05 11:33 Page 271 REVISED PAGES 9.11 Binary Eutectic Systems • 271 In the construction of binary phase diagrams, it is important to understand that one or at most two phases may be in equilibrium within a phase field. This holds true for the phase diagrams in Figures 9.3a and 9.7. For a eutectic system, three phases (a, b, and L) may be in equilibrium, but only at points along the eutectic isotherm. Another general rule is that single-phase regions are always separated from each other by a two-phase region that consists of the two single phases that it separates. For example, the a  b field is situated between the a and b singlephase regions in Figure 9.7. Another common eutectic system is that for lead and tin; the phase diagram (Figure 9.8) has a general shape similar to that for copper–silver. For the lead–tin system the solid solution phases are also designated by a and b; in this case, a represents a solid solution of tin in lead and, for b, tin is the solvent and lead is the solute. The eutectic invariant point is located at 61.9 wt% Sn and 183C (361F). Of course, maximum solid solubility compositions as well as component melting temperatures will be different for the copper–silver and lead–tin systems, as may be observed by comparing their phase diagrams. On occasion, low-melting-temperature alloys are prepared having near-eutectic compositions. A familiar example is the 60–40 solder, containing 60 wt% Sn and 40 wt% Pb. Figure 9.8 indicates that an alloy of this composition is completely molten at about 185C (365F), which makes this material especially attractive as a low-temperature solder, since it is easily melted. Composition (at% Sn) 0 20 40 60 80 100 327°C 600 300 Liquid 500 Temperature (°C) ␣ +L ␣ 200 ␤ +L 183°C 400 ␤ 18.3 61.9 97.8 300 100 ␣ + ␤ 200 100 0 0 (Pb) 20 40 60 Composition (wt% Sn) 80 100 (Sn) Figure 9.8 The lead–tin phase diagram. [Adapted from Binary Alloy Phase Diagrams, 2nd edition, Vol. 3, T. B. Massalski (Editor-in-Chief), 1990. Reprinted by permission of ASM International, Materials Park, OH.] Temperature (°F) 232°C 1496T_c09_252-310 11/29/05 11:33 Page 272 REVISED PAGES 272 • Chapter 9 / Phase Diagrams Concept Check 9.4 At 700C (1290F), what is the maximum solubility (a) of Cu in Ag? (b) Of Ag in Cu? [The answer may be found at www.wiley.com/college/callister (Student Companion Site).] Concept Check 9.5 Below is a portion of the H2O–NaCl phase diagram: 10 50 Liquid (brine) 40 10 30 Salt  Liquid (brine) Ice  Liquid (brine) 20 10 Temperature (°C) Temperature (°C) 0 0 20 10 Ice  Salt NaCl H2O 30 0 100 20 10 90 20 80 30 70 Composition (wt%) (a) Using this diagram, briefly explain how spreading salt on ice that is at a temperature below 0C (32F) can cause the ice to melt. (b) At what temperature is salt no longer useful in causing ice to melt? [The answer may be found at www.wiley.com/college/callister (Student Companion Site).] EXAMPLE PROBLEM 9.2 Determination of Phases Present and Computation of Phase Compositions For a 40 wt% Sn–60 wt% Pb alloy at 150C (300F), (a) What phase(s) is (are) present? (b) What is (are) the composition(s) of the phase(s)? Solution (a) Locate this temperature–composition point on the phase diagram (point B in Figure 9.9). Inasmuch as it is within the a  b region, both a and b phases will coexist. 1496T_c09_252-310 12/21/05 7:56 Page 273 2nd REVISE PAGES 9.11 Binary Eutectic Systems • 273 600 300 Liquid +L 200 +L  B 300 + 100 400  Temperature (°F) Temperature (°C) 500 200 C 100 0 0 (Pb) 20 60 C 80 C1 100 (Sn) Composition (wt% Sn) Figure 9.9 The lead–tin phase diagram. For a 40 wt% Sn–60 wt% Pb alloy at 150C (point B), phase compositions and relative amounts are computed in Example Problems 9.2 and 9.3. (b) Since two phases are present, it becomes necessary to construct a tie line across the a  b phase field at 150C, as indicated in Figure 9.9. The composition of the a phase corresponds to the tie line intersection with the a (a  b) solvus phase boundary—about 10 wt% Sn–90 wt% Pb, denoted as Ca. Similarly for the b phase, which will have a composition of approximately 98 wt% Sn–2 wt% Pb (Cb). EXAMPLE PROBLEM 9.3 Relative Phase Amount Determinations—Mass and Volume Fractions For the lead–tin alloy in Example Problem 9.2, calculate the relative amount of each phase present in terms of (a) mass fraction and (b) volume fraction. At 150C take the densities of Pb and Sn to be 11.23 and 7.24 g/cm3, respectively. Solution (a) Since the alloy consists of two phases, it is necessary to employ the lever rule. If C1 denotes the overall alloy composition, mass fractions may be computed by subtracting compositions, in terms of weight percent tin, as follows: Cb  C1 98  40   0.66 Cb  Ca 98  10 C1  Ca 40  10 Wb    0.34 Cb  Ca 98  10 Wa  1496T_c09_252-310 11/29/05 11:33 Page 274 REVISED PAGES 274 • Chapter 9 / Phase Diagrams (b) To compute volume fractions it is first necessary to determine the density of each phase using Equation 4.10a. Thus ra  CSn1a2 rSn 100 CPb1a2  rPb where CSn(a) and CPb(a) denote the concentrations in weight percent of tin and lead, respectively, in the a phase. From Example Problem 9.2, these values are 10 wt% and 90 wt%. Incorporation of these values along with the densities of the two components lead to ra  100  10.64 g/cm3 10 90  7.24 g/cm3 11.23 g/cm3 Similarly for the b phase: rb  C Sn1b2 rSn  100 CPb1b2  rPb 100 98 2  7.24 g/cm3 11.23 g/cm3  7.29 g/cm3 Now it becomes necessary to employ Equations 9.6a and 9.6b to determine Va and Vb as Va  Wa ra Wb Wa  ra rb 0.66 10.64 g/cm3   0.57 0.66 0.34  10.64 g/cm3 7.29 g/cm3 Wb Vb  rb Wa Wb  ra rb 0.34 7.29 g/cm3   0.43 0.66 0.34  10.64 g/cm3 7.29 g/cm3 1496T_c09_252-310 12/21/05 7:56 Page 275 2nd REVISE PAGES 9.11 Binary Eutectic Systems • 275 MATERIALS OF IMPORTANCE Lead-Free Solders olders are metal alloys that are used to bond or join two or more components (usually other metal alloys). They are used extensively in the electronics industry to physically hold assemblies together; furthermore, they must allow expansion and contraction of the various components, must transmit electrical signals, and also dissipate any heat that is generated. The bonding action is accomplished by melting the solder material, allowing it to flow among and make contact with the components to be joined (which do not melt), and, finally, upon solidification, forming a physical bond with all of these components. In the past, the vast majority of solders have been lead-tin alloys. These materials are reliable, inexpensive, and have relatively low melting temperatures. The most common lead–tin solder has a composition of 63 wt% Sn–37 wt% Pb. According to the lead–tin phase diagram, Figure 9.8, this composition is near the eutectic and has melting temperature of about 183C, the lowest temperature possible with the existence of a liquid phase (at equilibrium) for the lead–tin system. It follows that this alloy is often called a “eutectic lead-tin solder.” Unfortunately, lead is a mildly toxic metal, and there is serious concern about the environmental impact of discarded lead-containing products that can leach into groundwater from landfills or pollute the air if incinerated. Consequently, in some countries legislation has been enacted that bans the use of lead-containing solders. This has forced the development of lead-free solders that, among other things, must have relatively low melting temperatures (or temperature ranges). Some of these are ternary alloys (i.e., composed of three metals), to include tin–silver–copper and tin–silver–bismuth solders. The compositions of several lead-free solders are listed in Table 9.1. Of course, melting temperatures (or temperature ranges) are important in the development and selection of these new solder alloys, information that is available from phase diagrams. For example, the tin-bismuth phase diagram is presented in Figure 9.10. Here it may be noted that a eutectic Table 9.1 Compositions, Solidus Temperatures, and Liquidus Temperatures for Five Lead-Free Solders Composition (wt%) Solidus Temperature (C ) Liquidus Temperature (C ) 118 139 211 118 139 213 217 217 227 227 52 In/48 Sn* 57 Bi/43 Sn* 91.8 Sn/3.4 Ag/ 4.8 Bi 95.5 Sn/3.8 Ag/ 0.7 Cu* 99.3 Sn/0.7 Cu* *The compositions of these alloys are eutectic compositions; therefore, their solidus and liquidus temperatures are identical. Source: Adapted from E. Bastow, “Solder Families and How They Work,”Advanced Materials & Processes, Vol. 161, No. 12, M. W. Hunt (Editor-in-chief), ASM International, 2003, p. 28. Reprinted by permission of ASM International, Materials Park, OH. exists at 57 wt% Bi and 139C, which are indeed the composition and melting temperature of the Bi–Sn soldier in Table 9.1 Composition (at% Bi) 300 0 20 40 60 80 100 271°C L 232°C Temperature (°C) S 200 Bi + L +L 139°C  Sn 21 57 100  Sn + Bi Bi  Sn 13°C 0 0 20 40 60 (Sn) 80 100 (Bi) Composition (wt% Bi) Figure 9.10 The tin–bismuth phase diagram. [Adapted from ASM Handbook, Vol. 3, Alloy Phase Diagrams, H. Baker (Editor), ASM International, 1992, p. 2.106. Reprinted by permission of ASM International, Materials Park, OH.] 1496T_c09_252-310 11/29/05 11:33 Page 276 REVISED PAGES 276 • Chapter 9 / Phase Diagrams 9.12 DEVELOPMENT OF MICROSTRUCTURE IN EUTECTIC ALLOYS Depending on composition, several different types of microstructures are possible for the slow cooling of alloys belonging to binary eutectic systems. These possibilities will be considered in terms of the lead–tin phase diagram, Figure 9.8. The first case is for compositions ranging between a pure component and the maximum solid solubility for that component at room temperature [20C (70F)]. For the lead–tin system, this includes lead-rich alloys containing between 0 and about 2 wt% Sn (for the a phase solid solution), and also between approximately 99 wt% Sn and pure tin (for the b phase). For example, consider an alloy of composition C1 (Figure 9.11) as it is slowly cooled from a temperature within the liquid-phase region, say, 350C; this corresponds to moving down the dashed vertical line ww¿ in the figure. The alloy remains totally liquid and of composition C1 until we cross the liquidus line at approximately 330C, at which time the solid a phase begins to form. While passing through this narrow a  L phase region, solidification proceeds in the same manner as was described for the copper–nickel alloy in the preceding section; that is, with continued cooling more of the solid a forms. Furthermore, liquid- and solid-phase compositions are different, which follow along the liquidus and solidus phase boundaries, respectively. Solidification reaches completion at the point where ww¿ crosses the solidus line. The resulting alloy is polycrystalline with a uniform composition of C1, and no subsequent changes will occur Figure 9.11 Schematic representations of the equilibrium microstructures for a lead–tin alloy of composition C1 as it is cooled from the liquid-phase region. 400 L w (C1 wt% Sn)  L a b L 300 Liquidus c Temperature (°C)     +L  Solidus 200 (C1 wt% Sn)  100  + w 0 10 C1 20 Composition (wt% Sn) 30 1496T_c09_252-310 11/29/05 15:20 Page 277 REVISED PAGES 9.12 Development of Microstructure in Eutectic Alloys • 277 upon cooling to room temperature. This microstructure is represented schematically by the inset at point c in Figure 9.11. The second case considered is for compositions that range between the room temperature solubility limit and the maximum solid solubility at the eutectic temperature. For the lead–tin system (Figure 9.8), these compositions extend from about 2 wt% Sn to 18.3 wt% Sn (for lead-rich alloys) and from 97.8 wt% Sn to approximately 99 wt% Sn (for tin-rich alloys). Let us examine an alloy of composition C2 as it is cooled along the vertical line xx¿ in Figure 9.12. Down to the intersection of xx¿ and the solvus line, changes that occur are similar to the previous case, as we pass through the corresponding phase regions (as demonstrated by the insets at points d, e, and f ). Just above the solvus intersection, point f, the microstructure consists of a grains of composition C2. Upon crossing the solvus line, the a solid solubility is exceeded, which results in the formation of small b-phase particles; these are indicated in the microstructure inset at point g. With continued cooling, these particles will grow in size because the mass fraction of the b phase increases slightly with decreasing temperature. The third case involves solidification of the eutectic composition, 61.9 wt% Sn (C3 in Figure 9.13). Consider an alloy having this composition that is cooled from a temperature within the liquid-phase region (e.g., 250C) down the vertical line yy¿ x d Figure 9.12 Schematic representations of the equilibrium microstructures for a lead–tin alloy of composition C2 as it is cooled from the liquid-phase region. L L (C2 wt% Sn) 300 ␣ L e Temperature (°C) ␣ ␣ ␣ +L ␣ 200 ␣ ␣ C2 wt% Sn f ␤ Solvus line g ␣ 100 ␣ +␤ x 0 10 20 30 C2 Composition (wt% Sn) 40 50 1496T_c09_252-310 1/9/06 13:00 Page 278 2nd REVISE PAGES 278 • Chapter 9 / Phase Diagrams 300 600 y L L (61.9 wt% Sn) 500 h 200 ␣ ␤+L 183°C 97.8 i 18.3 ␤ 400 300 ␣+␤ 100 Temperature (°F) ␣ +L Temperature (°C) Figure 9.13 Schematic representations of the equilibrium microstructures for a lead–tin alloy of eutectic composition C3 above and below the eutectic temperature. 200 ␣ (18.3 wt% Sn) ␤ (97.8 wt% Sn) 100 y 0 0 20 40 60 C3 (61.9) (Pb) 80 100 (Sn) Composition (wt%Sn) Eutectic, Pb-Sn in Figure 9.13. As the temperature is lowered, no changes occur until we reach the eutectic temperature, 183C. Upon crossing the eutectic isotherm, the liquid transforms to the two a and b phases. This transformation may be represented by the reaction cooling L161.9 wt% Sn2 Δ a118.3 wt% Sn2  b197.8 wt% Sn2 heating eutectic structure (9.9) in which the a- and b-phase compositions are dictated by the eutectic isotherm end points. During this transformation, there must necessarily be a redistribution of the lead and tin components, inasmuch as the a and b phases have different compositions neither of which is the same as that of the liquid (as indicated in Equation 9.9). This redistribution is accomplished by atomic diffusion. The microstructure of the solid that results from this transformation consists of alternating layers (sometimes called lamellae) of the a and b phases that form simultaneously during the transformation. This microstructure, represented schematically in Figure 9.13, point i, is called a eutectic structure and is characteristic of this reaction. A photomicrograph of this structure for the lead–tin eutectic is shown in Figure 9.14. Subsequent cooling of the alloy from just below the eutectic to room temperature will result in only minor microstructural alterations. The microstructural change that accompanies this eutectic transformation is represented schematically in Figure 9.15; here is shown the a-b layered eutectic growing into and replacing the liquid phase. The process of the redistribution of lead and tin occurs by diffusion in the liquid just ahead of the eutectic–liquid interface. The arrows indicate the directions of diffusion of lead and tin atoms; lead atoms diffuse toward the a-phase layers since this a phase is lead-rich (18.3 wt% Sn–81.7 wt% Pb); conversely, the direction of diffusion of tin is in the direction of the b, tin-rich (97.8 wt% Sn–2.2 wt% Pb) layers. The eutectic structure forms in 1496T_c09_252-310 1/9/06 13:00 Page 279 2nd REVISE PAGES 9.12 Development of Microstructure in Eutectic Alloys • 279 Figure 9.14 Photomicrograph showing the microstructure of a lead–tin alloy of eutectic composition. This microstructure consists of alternating layers of a leadrich a-phase solid solution (dark layers), and a tin-rich b-phase solid solution (light layers). 375. (Reproduced with permission from Metals Handbook, 9th edition, Vol. 9, Metallography and Microstructures, American Society for Metals, Materials Park, OH, 1985.) eutectic phase primary phase these alternating layers because, for this lamellar configuration, atomic diffusion of lead and tin need only occur over relatively short distances. The fourth and final microstructural case for this system includes all compositions other than the eutectic that, when cooled, cross the eutectic isotherm. Consider, for example, the composition C4, Figure 9.16, which lies to the left of the eutectic; as the temperature is lowered, we move down the line zz¿, beginning at point j. The microstructural development between points j and l is similar to that for the second case, such that just prior to crossing the eutectic isotherm (point l), the a and liquid phases are present having compositions of approximately 18.3 and 61.9 wt% Sn, respectively, as determined from the appropriate tie line. As the temperature is lowered to just below the eutectic, the liquid phase, which is of the eutectic composition, will transform to the eutectic structure (i.e., alternating a and b lamellae); insignificant changes will occur with the a phase that formed during cooling through the a  L region. This microstructure is represented schematically by the inset at point m in Figure 9.16. Thus, the a phase will be present both in the eutectic structure and also as the phase that formed while cooling through the a  L phase field. To distinguish one a from the other, that which resides in the eutectic structure is called eutectic a, while the other that formed prior to crossing the eutectic isotherm is termed primary a; both are labeled in Figure 9.16. The photomicrograph in Figure 9.17 is of a lead–tin alloy in which both primary a and eutectic structures are shown. ␤ ␣ Pb Sn ␤ ␣ ␤ Figure 9.15 Schematic representation of the formation of the eutectic structure for the lead–tin system. Directions of diffusion of tin and lead atoms are indicated by blue and red arrows, respectively. Liquid Pb Sn Pb Eutectic growth direction 1496T_c09_252-310 1/9/06 13:00 Page 280 2nd REVISE PAGES 280 • Chapter 9 / Phase Diagrams L (C4 wt% Sn) z ␣ j 300 600 L ␣ +L L 200 k ␣ ␤ +L l 400 ␤ m L (61.9 wt% Sn) 100 Eutectic structure 300 Temperature (°F) Temperature (°C) ␣ (18.3 wt% Sn) 500 Primary ␣ (18.3 wt% Sn) ␣ + ␤ 200 Eutectic ␣ (18.3 wt% Sn) ␤ (97.8 wt% Sn) 100 z 0 0 (Pb) 20 60 C4 (40) 80 100 (Sn) Composition (wt% Sn) Figure 9.16 Schematic representations of the equilibrium microstructures for a lead–tin alloy of composition C4 as it is cooled from the liquid-phase region. microconstituent In dealing with microstructures, it is sometimes convenient to use the term microconstituent—that is, an element of the microstructure having an identifiable and characteristic structure. For example, in the point m inset, Figure 9.16, there are two microconstituents—namely, primary a and the eutectic structure. Thus, the eutectic structure is a microconstituent even though it is a mixture of two phases, because it has a distinct lamellar structure, with a fixed ratio of the two phases. Figure 9.17 Photomicrograph showing the microstructure of a lead–tin alloy of composition 50 wt% Sn–50 wt% Pb. This microstructure is composed of a primary lead-rich a phase (large dark regions) within a lamellar eutectic structure consisting of a tin-rich b phase (light layers) and a lead-rich a phase (dark layers). 400. (Reproduced with permission from Metals Handbook, 9th edition, Vol. 9, Metallography and Microstructures, American Society for Metals, Materials Park, OH, 1985.) 1496T_c09_252-310 11/29/05 11:33 Page 281 REVISED PAGES 9.12 Development of Microstructure in Eutectic Alloys • 281 300 Temperature (°C) L  +L 200  +L  Q P  Figure 9.18 The lead–tin phase diagram used in computations for relative amounts of primary a and eutectic microconstituents for an alloy of composition C¿4. R 100 0 (Pb) (Sn) 18.3 C4 61.9 97.8 Composition (wt% Sn) It is possible to compute the relative amounts of both eutectic and primary a microconstituents. Since the eutectic microconstituent always forms from the liquid having the eutectic composition, this microconstituent may be assumed to have a composition of 61.9 wt% Sn. Hence, the lever rule is applied using a tie line between the a–(a  b) phase boundary (18.3 wt% Sn) and the eutectic composition. For example, consider the alloy of composition C¿4 in Figure 9.18. The fraction of the eutectic microconstituent We is just the same as the fraction of liquid WL from which it transforms, or Lever rule expression for computation of eutectic microconstituent and liquid phase mass fractions (composition C¿4, Figure 9.18) Lever rule expression for computation of primary a phase mass fraction We  WL   P PQ C¿4  18.3 C¿4  18.3  61.9  18.3 43.6 (9.10) Furthermore, the fraction of primary a, Wa¿, is just the fraction of the a phase that existed prior to the eutectic transformation or, from Figure 9.18, Wa¿   Q PQ 61.9  C¿4 61.9  C¿4  61.9  18.3 43.6 (9.11) The fractions of total a, Wa (both eutectic and primary), and also of total b, Wb, are determined by use of the lever rule and a tie line that extends entirely across the a  b phase field. Again, for an alloy having composition C¿4, Lever rule expression for computation of total a phase mass fraction Wa   QR PQR 97.8  C¿4 97.8  C¿4  97.8  18.3 79.5 (9.12) 1496T_c09_252-310 11/29/05 11:33 Page 282 REVISED PAGES 282 • Chapter 9 / Phase Diagrams and Lever rule expression for computation of total b phase mass fraction Wb   P PQR C¿4  18.3 C¿4  18.3  97.8  18.3 79.5 (9.13) Analogous transformations and microstructures result for alloys having compositions to the right of the eutectic (i.e., between 61.9 and 97.8 wt% Sn). However, below the eutectic temperature, the microstructure will consist of the eutectic and primary b microconstituents because, upon cooling from the liquid, we pass through the b  liquid phase field. When, for case 4 (represented in Figure 9.16), conditions of equilibrium are not maintained while passing through the a (or b)  liquid phase region, the following consequences will be realized for the microstructure upon crossing the eutectic isotherm: (1) grains of the primary microconstituent will be cored, that is, have a nonuniform distribution of solute across the grains; and (2) the fraction of the eutectic microconstituent formed will be greater than for the equilibrium situation. 9.13 EQUILIBRIUM DIAGRAMS HAVING INTERMEDIATE PHASES OR COMPOUNDS terminal solid solution intermediate solid solution intermetallic compound The isomorphous and eutectic phase diagrams discussed thus far are relatively simple, but those for many binary alloy systems are much more complex. The eutectic copper–silver and lead–tin phase diagrams (Figures 9.7 and 9.8) have only two solid phases, a and b; these are sometimes termed terminal solid solutions, because they exist over composition ranges near the concentration extremities of the phase diagram. For other alloy systems, intermediate solid solutions (or intermediate phases) may be found at other than the two composition extremes. Such is the case for the copper–zinc system. Its phase diagram (Figure 9.19) may at first appear formidable because there are some invariant points and reactions similar to the eutectic that have not yet been discussed. In addition, there are six different solid solutions—two terminal (a and h) and four intermediate (b, g, d, and ). (The b¿ phase is termed an ordered solid solution, one in which the copper and zinc atoms are situated in a specific and ordered arrangement within each unit cell.) Some phase boundary lines near the bottom of Figure 9.19 are dashed to indicate that their positions have not been exactly determined. The reason for this is that at low temperatures, diffusion rates are very slow and inordinately long times are required for the attainment of equilibrium. Again, only single- and twophase regions are found on the diagram, and the same rules outlined in Section 9.8 are utilized for computing phase compositions and relative amounts. The commercial brasses are copper-rich copper–zinc alloys; for example, cartridge brass has a composition of 70 wt% Cu–30 wt% Zn and a microstructure consisting of a single a phase. For some systems, discrete intermediate compounds rather than solid solutions may be found on the phase diagram, and these compounds have distinct chemical formulas; for metal–metal systems, they are called intermetallic compounds. For example, consider the magnesium–lead system (Figure 9.20). The compound Mg2Pb 0 (Cu) 200 400 600 800 1000 20  20 +L 40  +  40   +   +  Composition (wt% Zn)  +    +L Composition (at% Zn) 60 60   +  +  +  + L 80 +L  Liquid 80   + + L 600 800 1000 (Zn) 400 100  +L 1200 1400 1600 1800 2000 100 2200 Figure 9.19 The copper–zinc phase diagram. [Adapted from Binary Alloy Phase Diagrams, 2nd edition, Vol. 2, T. B. Massalski (Editor-in-Chief), 1990. Reprinted by permission of ASM International, Materials Park, OH.] Temperature (°C) 1200 0 1496T_c09_252-310 11/29/05 11:33 Page 283 REVISED PAGES 9.13 Equilibrium Diagrams Having Intermediate Phases or Compounds • 283 Temperature (°F) 1496T_c09_252-310 11/29/05 11:33 Page 284 REVISED PAGES 284 • Chapter 9 / Phase Diagrams Composition (at% Pb) 0 5 10 700 20 30 40 70 100 L L + Mg2Pb 600  +L 1200 M 1000  400  + L L + Mg2Pb 600 300 200   + Mg2Pb  + Mg2Pb 100 Mg2Pb 0 0 (Mg) 20 40 60 80 Composition (wt% Pb) 800 Temperature (°F) Temperature (°C) 500 400 200 100 (Pb) Figure 9.20 The magnesium–lead phase diagram. [Adapted from Phase Diagrams of Binary Magnesium Alloys, A. A. Nayeb-Hashemi and J. B. Clark (Editors), 1988. Reprinted by permission of ASM International, Materials Park, OH.] has a composition of 19 wt% Mg–81 wt% Pb (33 at% Pb), and is represented as a vertical line on the diagram, rather than as a phase region of finite width; hence, Mg2Pb can exist by itself only at this precise composition. Several other characteristics are worth noting for this magnesium–lead system. First, the compound Mg2Pb melts at approximately 550C (1020F), as indicated by point M in Figure 9.20. Also, the solubility of lead in magnesium is rather extensive, as indicated by the relatively large composition span for the a-phase field. On the other hand, the solubility of magnesium in lead is extremely limited. This is evident from the very narrow b terminal solid-solution region on the right or lead-rich side of the diagram. Finally, this phase diagram may be thought of as two simple eutectic diagrams joined back to back, one for the Mg–Mg2Pb system and the other for Mg2Pb–Pb; as such, the compound Mg2Pb is really considered to be a component. This separation of complex phase diagrams into smaller-component units may simplify them and, furthermore, expedite their interpretation. 9.14 EUTECTOID AND PERITECTIC REACTIONS In addition to the eutectic, other invariant points involving three different phases are found for some alloy systems. One of these occurs for the copper–zinc system (Figure 9.19) at 560C (1040F) and 74 wt% Zn–26 wt% Cu. A portion of the 1496T_c09_252-310 11/29/05 11:33 Page 285 REVISED PAGES 9.14 Eutectoid and Peritectic Reactions • 285 +L  + +L  600   P L 598°C  + 560°C E 60 1000  + 500 1200 Temperature (°F) 700 Temperature (°C) Figure 9.21 A region of the copper–zinc phase diagram that has been enlarged to show eutectoid and peritectic invariant points, labeled E (560C, 74 wt% Zn) and P (598C, 78.6 wt% Zn), respectively. [Adapted from Binary Alloy Phase Diagrams, 2nd edition, Vol. 2, T. B. Massalski (Editor-in-Chief), 1990. Reprinted by permission of ASM International, Materials Park, OH.] 70  80 +L 90 Composition (wt% Zn) phase diagram in this vicinity appears enlarged in Figure 9.21. Upon cooling, a solid d phase transforms into two other solid phases (g and ) according to the reaction The eutectoid reaction (per point E, Figure 9.21) eutectoid reaction peritectic reaction The peritectic reaction (per point P, Figure 9.21) cooling d Δ g heating (9.14) The reverse reaction occurs upon heating. It is called a eutectoid (or eutectic-like) reaction, and the invariant point (point E, Figure 9.21) and the horizontal tie line at 560C are termed the eutectoid and eutectoid isotherm, respectively. The feature distinguishing “eutectoid” from “eutectic” is that one solid phase instead of a liquid transforms into two other solid phases at a single temperature. A eutectoid reaction is found in the iron–carbon system (Section 9.18) that is very important in the heat treating of steels. The peritectic reaction is yet another invariant reaction involving three phases at equilibrium. With this reaction, upon heating, one solid phase transforms into a liquid phase and another solid phase. A peritectic exists for the copper–zinc system (Figure 9.21, point P) at 598C (1108F) and 78.6 wt% Zn–21.4 wt% Cu; this reaction is as follows: cooling dL Δ  heating (9.15) The low-temperature solid phase may be an intermediate solid solution (e.g.,  in the above reaction), or it may be a terminal solid solution. One of the latter peritectics exists at about 97 wt% Zn and 435C (815F) (see Figure 9.19), wherein the h phase, when heated, transforms to  and liquid phases. Three other peritectics are found for the Cu–Zn system, the reactions of which involve b, d, and g intermediate solid solutions as the low-temperature phases that transform upon heating. 1496T_c09_252-310 11/29/05 11:33 Page 286 REVISED PAGES 286 • Chapter 9 / Phase Diagrams Composition (at% Ti) 30 40 50 60 70 1500 2600 L 1400 1310°C 44.9 wt% Ti Temperature (°C) ␤ +L ␥ +L 1200 2200 ␥ +L 1100 1000 Temperature (°F) 2400 1300 Figure 9.22 A portion of the nickel–titanium phase diagram on which is shown a congruent melting point for the g-phase solid solution at 1310C and 44.9 wt% Ti. [Adapted from Phase Diagrams of Binary Nickel Alloys, P. Nash (Editor), 1991. Reprinted by permission of ASM International, Materials Park, OH.] 2000 ␥ ␤ +␥ 1800 ␥ +␦ ␦ 900 30 40 50 60 70 Composition (wt% Ti) 9.15 CONGRUENT PHASE TRANSFORMATIONS congruent transformation Phase transformations may be classified according to whether or not there is any change in composition for the phases involved. Those for which there are no compositional alterations are said to be congruent transformations. Conversely, for incongruent transformations, at least one of the phases will experience a change in composition. Examples of congruent transformations include allotropic transformations (Section 3.6) and melting of pure materials. Eutectic and eutectoid reactions, as well as the melting of an alloy that belongs to an isomorphous system, all represent incongruent transformations. Intermediate phases are sometimes classified on the basis of whether they melt congruently or incongruently. The intermetallic compound Mg2Pb melts congruently at the point designated M on the magnesium–lead phase diagram, Figure 9.20. Also, for the nickel–titanium system, Figure 9.22, there is a congruent melting point for the g solid solution that corresponds to the point of tangency for the pairs of liquidus and solidus lines, at 1310C and 44.9 wt% Ti. Furthermore, the peritectic reaction is an example of incongruent melting for an intermediate phase. Concept Check 9.6 The figure on the next page is the hafnium–vanadium phase diagram, for which only single-phase regions are labeled. Specify temperature–composition points at which all eutectics, eutectoids, peritectics, and congruent phase transformations occur. Also, for each, write the reaction upon cooling. [Phase diagram from ASM Handbook, Vol. 3, Alloy Phase Diagrams, H. Baker (Editor), 1992, p. 2.244. Reprinted by permission of ASM International, Materials Park, OH.] 1496T_c09_252-310 12/21/05 7:56 Page 287 2nd REVISE PAGES 9.17 The Gibbs Phase Rule • 287 2200 Hf L Temperature (°C) 2000 1800 1600 V 1400 HfV2 1200 Hf 1000 0 20 (Hf) 40 60 Composition (wt% V) 80 100 (V) [The answer may be found at www.wiley.com/college/callister (Student Companion Site).] 9.16 CERAMIC AND TERNARY PHASE DIAGRAMS It need not be assumed that phase diagrams exist only for metal–metal systems; in fact, phase diagrams that are very useful in the design and processing of ceramic systems have been experimentally determined for quite a number of these materials. Ceramic phase diagrams are discussed in Section 12.7. Phase diagrams have also been determined for metallic (as well as ceramic) systems containing more than two components; however, their representation and interpretation may be exceedingly complex. For example, a ternary, or threecomponent, composition–temperature phase diagram in its entirety is depicted by a three-dimensional model. Portrayal of features of the diagram or model in two dimensions is possible but somewhat difficult. 9.17 THE GIBBS PHASE RULE Gibbs phase rule General form of the Gibbs phase rule The construction of phase diagrams as well as some of the principles governing the conditions for phase equilibria are dictated by laws of thermodynamics. One of these is the Gibbs phase rule, proposed by the nineteenth-century physicist J. Willard Gibbs. This rule represents a criterion for the number of phases that will coexist within a system at equilibrium, and is expressed by the simple equation PFCN (9.16) where P is the number of phases present (the phase concept is discussed in Section 9.3). The parameter F is termed the number of degrees of freedom or the number of externally controlled variables (e.g., temperature, pressure, composition) which must be specified to completely define the state of the system. Expressed another 1496T_c09_252-310 11/29/05 11:33 Page 288 REVISED PAGES 288 • Chapter 9 / Phase Diagrams way, F is the number of these variables that can be changed independently without altering the number of phases that coexist at equilibrium. The parameter C in Equation 9.16 represents the number of components in the system. Components are normally elements or stable compounds and, in the case of phase diagrams, are the materials at the two extremities of the horizontal compositional axis (e.g., H2O and C12H22O11, and Cu and Ni for the phase diagrams shown in Figures 9.1 and 9.3a, respectively). Finally, N in Equation 9.16 is the number of noncompositional variables (e.g., temperature and pressure). Let us demonstrate the phase rule by applying it to binary temperature– composition phase diagrams, specifically the copper–silver system, Figure 9.7. Since pressure is constant (1 atm), the parameter N is 1—temperature is the only noncompositional variable. Equation 9.16 now takes the form PFC1 (9.17) Furthermore, the number of components C is 2 (viz. Cu and Ag), and PF213 or F3P Consider the case of single-phase fields on the phase diagram (e.g., a, b, and liquid regions). Since only one phase is present, P  1 and F3P 312 This means that to completely describe the characteristics of any alloy that exists within one of these phase fields, we must specify two parameters; these are composition and temperature, which locate, respectively, the horizontal and vertical positions of the alloy on the phase diagram. For the situation wherein two phases coexist, for example, a  L, b  L, and a  b phase regions, Figure 9.7, the phase rule stipulates that we have but one degree of freedom since F3P 321 Thus, it is necessary to specify either temperature or the composition of one of the phases to completely define the system. For example, suppose that we decide to specify temperature for the a  L phase region, say, T1 in Figure 9.23. The compositions of the a and liquid phases (Ca and CL) are thus dictated by the extremities of the tie line constructed at T1 across the a  L field. Note that only the nature of the phases is important in this treatment and not the relative phase amounts. This is to say that the overall alloy composition could lie anywhere along this tie line constructed at temperature T1 and still give Ca and CL compositions for the respective a and liquid phases. The second alternative is to stipulate the composition of one of the phases for this two-phase situation, which thereby fixes completely the state of the system. For example, if we specified Ca as the composition of the a phase that is in equilibrium with the liquid (Figure 9.23), then both the temperature of the alloy (T1) and the composition of the liquid phase (CL) are established, again by the tie line drawn across the a  L phase field so as to give this Ca composition. 1496T_c09_252-310 11/29/05 11:33 Page 289 REVISED PAGES 9.17 The Gibbs Phase Rule • 289 Figure 9.23 Enlarged copper-rich section of the Cu–Ag phase diagram in which the Gibbs phase rule for the coexistence of two phases (i.e., a and L) is demonstrated. Once the composition of either phase (i.e., Ca or CL) or the temperature (i.e., T1) is specified, values for the two remaining parameters are established by construction of the appropriate tie line. L 1000 T1 Temperature (°C) C 800 ␣+L CL  600 400 0 (Cu) 20 40 Composition (wt% Ag) 60 For binary systems, when three phases are present, there are no degrees of freedom, since F3P 330 This means that the compositions of all three phases as well as the temperature are fixed. This condition is met for a eutectic system by the eutectic isotherm; for the Cu–Ag system (Figure 9.7), it is the horizontal line that extends between points B and G. At this temperature, 779C, the points at which each of the a, L, and b phase fields touch the isotherm line correspond to the respective phase compositions; namely, the composition of the a phase is fixed at 8.0 wt% Ag, that of the liquid at 71.9 wt% Ag, and that of the b phase at 91.2 wt% Ag. Thus, three-phase equilibrium will not be represented by a phase field, but rather by the unique horizontal isotherm line. Furthermore, all three phases will be in equilibrium for any alloy composition that lies along the length of the eutectic isotherm (e.g., for the Cu–Ag system at 779C and compositions between 8.0 and 91.2 wt% Ag). One use of the Gibbs phase rule is in analyzing for nonequilibrium conditions. For example, a microstructure for a binary alloy that developed over a range of temperatures and consisting of three phases is a nonequilibrium one; under these circumstances, three phases will exist only at a single temperature. Concept Check 9.7 For a ternary system, three components are present; temperature is also a variable. What is the maximum number of phases that may be present for a ternary system, assuming that pressure is held constant? [The answer may be found at www.wiley.com/college/callister (Student Companion Site).] 1496T_c09_252-310 11/29/05 11:33 Page 290 REVISED PAGES 290 • Chapter 9 / Phase Diagrams T h e I ro n – C a r b o n S ys t e m Of all binary alloy systems, the one that is possibly the most important is that for iron and carbon. Both steels and cast irons, primary structural materials in every technologically advanced culture, are essentially iron–carbon alloys. This section is devoted to a study of the phase diagram for this system and the development of several of the possible microstructures. The relationships among heat treatment, microstructure, and mechanical properties are explored in Chapters 10 and 11. 9.18 THE IRON–IRON CARBIDE (Fe–Fe3C) PHASE DIAGRAM A portion of the iron–carbon phase diagram is presented in Figure 9.24. Pure iron, upon heating, experiences two changes in crystal structure before it melts. At room temperature the stable form, called ferrite, or a iron, has a BCC crystal structure. Ferrite experiences a polymorphic transformation to FCC austenite, or g iron, at 912C (1674F). This austenite persists to 1394C (2541F), at which temperature the FCC austenite reverts back to a BCC phase known as d ferrite, which finally ferrite austenite Composition (at% C) 1600 0 5 10 15 20 25 1538°C 1493°C L  1400 2500 +L 1394°C 1147°C 2.14 , Austenite 4.30 2000 1000  + Fe3C 912°C 800  +  Temperature (°F) Temperature (°C) 1200 1500 727°C 0.76 0.022 600  + Fe3C , Ferrite Cementite (Fe3C) 400 0 (Fe) 1 2 3 4 Composition (wt% C) 5 6 1000 6.70 Figure 9.24 The iron–iron carbide phase diagram. [Adapted from Binary Alloy Phase Diagrams, 2nd edition, Vol. 1, T. B. Massalski (Editor-in-Chief), 1990. Reprinted by permission of ASM International, Materials Park, OH.] 1496T_c09_252-310 11/29/05 11:33 Page 291 REVISED PAGES 9.18 The Iron–Iron Carbide (Fe–Fe3C) Phase Diagram • 291 Figure 9.25 Photomicrographs of (a) a ferrite (90) and (b) austenite (325). (Copyright 1971 by United States Steel Corporation.) (a) cementite (b) melts at 1538C (2800F). All these changes are apparent along the left vertical axis of the phase diagram.1 The composition axis in Figure 9.24 extends only to 6.70 wt% C; at this concentration the intermediate compound iron carbide, or cementite (Fe3C), is formed, which is represented by a vertical line on the phase diagram. Thus, the iron–carbon system may be divided into two parts: an iron-rich portion, as in Figure 9.24, and the other (not shown) for compositions between 6.70 and 100 wt% C (pure graphite). In practice, all steels and cast irons have carbon contents less than 6.70 wt% C; therefore, we consider only the iron–iron carbide system. Figure 9.24 would be more appropriately labeled the Fe–Fe3C phase diagram, since Fe3C is now considered to be a component. Convention and convenience dictate that composition still be expressed in “wt% C” rather than “wt% Fe3C”; 6.70 wt% C corresponds to 100 wt% Fe3C. Carbon is an interstitial impurity in iron and forms a solid solution with each of a and d ferrites, and also with austenite, as indicated by the a, d, and g singlephase fields in Figure 9.24. In the BCC a ferrite, only small concentrations of carbon are soluble; the maximum solubility is 0.022 wt% at 727C (1341F). The limited solubility is explained by the shape and size of the BCC interstitial positions, which make it difficult to accommodate the carbon atoms. Even though present in relatively low concentrations, carbon significantly influences the mechanical properties of ferrite. This particular iron–carbon phase is relatively soft, may be made magnetic at temperatures below 768C (1414F), and has a density of 7.88 g/cm3. Figure 9.25a is a photomicrograph of a ferrite. 1 The reader may wonder why no b phase is found on the Fe–Fe3C phase diagram, Figure 9.24 (consistent with the a, b, g, etc. labeling scheme described previously). Early investigators observed that the ferromagnetic behavior of iron disappears at 768C and attributed this phenomenon to a phase transformation; the “b” label was assigned to the high-temperature phase. Later it was discovered that this loss of magnetism did not result from a phase transformation (see Section 20.6) and, therefore, the presumed b phase did not exist. 1496T_c09_252-310 11/29/05 11:33 Page 292 REVISED PAGES 292 • Chapter 9 / Phase Diagrams The austenite, or g phase of iron, when alloyed with carbon alone, is not stable below 727C (1341F), as indicated in Figure 9.24. The maximum solubility of carbon in austenite, 2.14 wt%, occurs at 1147C (2097F). This solubility is approximately 100 times greater than the maximum for BCC ferrite, since the FCC interstitial positions are larger (see the results of Problem 4.5), and, therefore, the strains imposed on the surrounding iron atoms are much lower. As the discussions that follow demonstrate, phase transformations involving austenite are very important in the heat treating of steels. In passing, it should be mentioned that austenite is nonmagnetic. Figure 9.25b shows a photomicrograph of this austenite phase. The d ferrite is virtually the same as a ferrite, except for the range of temperatures over which each exists. Since the d ferrite is stable only at relatively high temperatures, it is of no technological importance and is not discussed further. Cementite (Fe3C) forms when the solubility limit of carbon in a ferrite is exceeded below 727C (1341F) (for compositions within the a  Fe3C phase region). As indicated in Figure 9.24, Fe3C will also coexist with the g phase between 727 and 1147C (1341 and 2097F). Mechanically, cementite is very hard and brittle; the strength of some steels is greatly enhanced by its presence. Strictly speaking, cementite is only metastable; that is, it will remain as a compound indefinitely at room temperature. However, if heated to between 650 and 700C (1200 and 1300F) for several years, it will gradually change or transform into a iron and carbon, in the form of graphite, which will remain upon subsequent cooling to room temperature. Thus, the phase diagram in Figure 9.24 is not a true equilibrium one because cementite is not an equilibrium compound. However, inasmuch as the decomposition rate of cementite is extremely sluggish, virtually all the carbon in steel will be as Fe3C instead of graphite, and the iron–iron carbide phase diagram is, for all practical purposes, valid.As will be seen in Section 11.2, addition of silicon to cast irons greatly accelerates this cementite decomposition reaction to form graphite. The two-phase regions are labeled in Figure 9.24. It may be noted that one eutectic exists for the iron–iron carbide system, at 4.30 wt% C and 1147C (2097F); for this eutectic reaction, Eutectic reaction for the iron-iron carbide system cooling L Δ g  Fe3C heating (9.18) the liquid solidifies to form austenite and cementite phases. Of course, subsequent cooling to room temperature will promote additional phase changes. It may be noted that a eutectoid invariant point exists at a composition of 0.76 wt% C and a temperature of 727C (1341F).This eutectoid reaction may be represented by Eutectoid reaction for the iron-iron carbide system g10.76 wt% C2 Δ a10.022 wt% C2  Fe3C 16.7 wt%C2 cooling heating (9.19) or, upon cooling, the solid g phase is transformed into a iron and cementite. (Eutectoid phase transformations were addressed in Section 9.14.) The eutectoid phase changes described by Equation 9.19 are very important, being fundamental to the heat treatment of steels, as explained in subsequent discussions. Ferrous alloys are those in which iron is the prime component, but carbon as well as other alloying elements may be present. In the classification scheme of ferrous alloys based on carbon content, there are three types: iron, steel, and cast iron. Commercially pure iron contains less than 0.008 wt% C and, from the phase diagram, is composed almost exclusively of the ferrite phase at room temperature. The 1496T_c09_252-310 11/29/05 11:33 Page 293 REVISED PAGES 9.19 Development of Microstructure in Iron–Carbon Alloys • 293 iron–carbon alloys that contain between 0.008 and 2.14 wt% C are classified as steels. In most steels the microstructure consists of both a and Fe3C phases. Upon cooling to room temperature, an alloy within this composition range must pass through at least a portion of the g-phase field; distinctive microstructures are subsequently produced, as discussed below. Although a steel alloy may contain as much as 2.14 wt% C, in practice, carbon concentrations rarely exceed 1.0 wt%. The properties and various classifications of steels are treated in Section 11.2. Cast irons are classified as ferrous alloys that contain between 2.14 and 6.70 wt% C. However, commercial cast irons normally contain less than 4.5 wt% C. These alloys are discussed further also in Section 11.2. 9.19 DEVELOPMENT OF MICROSTRUCTURE IN IRON–CARBON ALLOYS Several of the various microstructures that may be produced in steel alloys and their relationships to the iron–iron carbon phase diagram are now discussed, and it is shown that the microstructure that develops depends on both the carbon content and heat treatment. This discussion is confined to very slow cooling of steel alloys, in which equilibrium is continuously maintained. A more detailed exploration of the influence of heat treatment on microstructure, and ultimately on the mechanical properties of steels, is contained in Chapter 10. Phase changes that occur upon passing from the g region into the a  Fe3C phase field (Figure 9.24) are relatively complex and similar to those described for the eutectic systems in Section 9.12. Consider, for example, an alloy of eutectoid composition (0.76 wt% C) as it is cooled from a temperature within the g phase region, say, 800C—that is, beginning at point a in Figure 9.26 and moving down the 1100 ␥ 1000 ␥ + Fe3C 900 ␥ Temperature (°C) x ␥ 800 a ␣ +␥ ␥ 727°C ␣ b 700 ␥ ␣ 600 Fe3C ␣ + Fe3C 500 x 400 0 1.0 Composition (wt% C) 2.0 Figure 9.26 Schematic representations of the microstructures for an iron–carbon alloy of eutectoid composition (0.76 wt% C) above and below the eutectoid temperature. 1496T_c09_252-310 11/29/05 11:33 Page 294 REVISED PAGES 294 • Chapter 9 / Phase Diagrams Figure 9.27 Photomicrograph of a eutectoid steel showing the pearlite microstructure consisting of alternating layers of a ferrite (the light phase) and Fe3C (thin layers most of which appear dark). 500. (Reproduced with permission from Metals Handbook, 9th edition, Vol. 9, Metallography and Microstructures, American Society for Metals, Materials Park, OH, 1985.) pearlite vertical line xx¿. Initially, the alloy is composed entirely of the austenite phase having a composition of 0.76 wt% C and corresponding microstructure, also indicated in Figure 9.26. As the alloy is cooled, there will occur no changes until the eutectoid temperature (727C) is reached. Upon crossing this temperature to point b, the austenite transforms according to Equation 9.19. The microstructure for this eutectoid steel that is slowly cooled through the eutectoid temperature consists of alternating layers or lamellae of the two phases (a and Fe3C) that form simultaneously during the transformation. In this case, the relative layer thickness is approximately 8 to 1. This microstructure, represented schematically in Figure 9.26, point b, is called pearlite because it has the appearance of mother of pearl when viewed under the microscope at low magnifications. Figure 9.27 is a photomicrograph of a eutectoid steel showing the pearlite. The pearlite exists as grains, often termed “colonies”; within each colony the layers are oriented in essentially the same direction, which varies from one colony to another. The thick light layers are the ferrite phase, and the cementite phase appears as thin lamellae most of which appear dark. Many cementite layers are so thin that adjacent phase boundaries are so close together that they are indistinguishable at this magnification, and, therefore, appear dark. Mechanically, pearlite has properties intermediate between the soft, ductile ferrite and the hard, brittle cementite. The alternating a and Fe3C layers in pearlite form as such for the same reason that the eutectic structure (Figures 9.13 and 9.14) forms—because the composition of the parent phase [in this case austenite (0.76 wt% C)] is different from either of the product phases [ferrite (0.022 wt% C) and cementite (6.7 wt% C)], and the phase transformation requires that there be a redistribution of the carbon by diffusion. Figure 9.28 illustrates schematically microstructural changes that accompany this eutectoid reaction; here the directions of carbon diffusion are indicated by arrows. Carbon atoms diffuse away from the 0.022 wt% ferrite regions and to the 6.7 wt% cementite layers, as the pearlite extends from the grain boundary into the unreacted austenite grain. The layered pearlite forms because carbon atoms need diffuse only minimal distances with the formation of this structure. Furthermore, subsequent cooling of the pearlite from point b in Figure 9.26 will produce relatively insignificant microstructural changes. 1496T_c09_252-310 12/21/05 7:56 Page 295 2nd REVISE PAGES 9.19 Development of Microstructure in Iron–Carbon Alloys • 295 Figure 9.28 Schematic representation of the formation of pearlite from austenite; direction of carbon diffusion indicated by arrows. Austenite grain boundary  Ferrite () Austenite ( ) Ferrite () Austenite ( ) Ferrite () Cementite (Fe3C) Growth direction of pearlite Ferrite ()  Carbon diffusion Hypoeutectoid Alloys 1100   1000     + Fe3C  y  M 900  c Temperature (°C) hypoeutectoid alloy Microstructures for iron–iron carbide alloys having other than the eutectoid composition are now explored; these are analogous to the fourth case described in Section 9.12 and illustrated in Figure 9.16 for the eutectic system. Consider a composition C0 to the left of the eutectoid, between 0.022 and 0.76 wt% C; this is termed a hypoeutectoid (less than eutectoid) alloy. Cooling an alloy of this composition is represented by moving down the vertical line yy¿ in Figure 9.29. At about 875C, point c, the microstructure will consist entirely of grains of the g phase, as shown   800 d  e N Te   O f 700   Pearlite 600 Fe3C Proeutectoid  Eutectoid   + Fe3C 500 y 400 0 1.0 C0 Composition (wt% C) 2.0 Figure 9.29 Schematic representations of the microstructures for an iron–carbon alloy of hypoeutectoid composition C0 (containing less than 0.76 wt% C) as it is cooled from within the austenite phase region to below the eutectoid temperature. 1496T_c09_252-310 11/29/05 11:33 Page 296 REVISED PAGES 296 • Chapter 9 / Phase Diagrams proeutectoid ferrite Figure 9.30 Photomicrograph of a 0.38 wt% C steel having a microstructure consisting of pearlite and proeutectoid ferrite. 635. (Photomicrograph courtesy of Republic Steel Corporation.) schematically in the figure. In cooling to point d, about 775C, which is within the a  g phase region, both these phases will coexist as in the schematic microstructure. Most of the small a particles will form along the original g grain boundaries. The compositions of both a and g phases may be determined using the appropriate tie line; these compositions correspond, respectively, to about 0.020 and 0.40 wt% C. While cooling an alloy through the a  g phase region, the composition of the ferrite phase changes with temperature along the a  1a  g2 phase boundary, line MN, becoming slightly richer in carbon. On the other hand, the change in composition of the austenite is more dramatic, proceeding along the 1a  g2  g boundary, line MO, as the temperature is reduced. Cooling from point d to e, just above the eutectoid but still in the a  g region, will produce an increased fraction of the a phase and a microstructure similar to that also shown: the a particles will have grown larger. At this point, the compositions of the a and g phases are determined by constructing a tie line at the temperature Te; the a phase will contain 0.022 wt% C, while the g phase will be of the eutectoid composition, 0.76 wt% C. As the temperature is lowered just below the eutectoid, to point f, all the g phase that was present at temperature Te (and having the eutectoid composition) will transform to pearlite, according to the reaction in Equation 9.19. There will be virtually no change in the a phase that existed at point e in crossing the eutectoid temperature—it will normally be present as a continuous matrix phase surrounding the isolated pearlite colonies. The microstructure at point f will appear as the corresponding schematic inset of Figure 9.29. Thus the ferrite phase will be present both in the pearlite and also as the phase that formed while cooling through the a  g phase region. The ferrite that is present in the pearlite is called eutectoid ferrite, whereas the other, that formed above Te, is termed proeutectoid (meaning pre- or before eutectoid) ferrite, as labeled in Figure 9.29. Figure 9.30 is a photomicrograph of a 0.38 wt% C steel; large, white regions correspond to the proeutectoid ferrite. For pearlite, the spacing between the a and Fe3C layers varies from grain to grain; some of the pearlite appears dark because the many closespaced layers are unresolved at the magnification of the photomicrograph. The Proeutectoid ferrite Pearlite 1496T_c09_252-310 11/29/05 11:33 Page 297 REVISED PAGES 9.19 Development of Microstructure in Iron–Carbon Alloys • 297 chapter-opening photograph for this chapter is a scanning electron micrograph of a hypoeutectoid (0.44 wt% C) steel in which may also be seen both pearlite and proeutectoid ferrite, only at a higher magnification. Note also that two microconstituents are present in these micrographs—proeutectoid ferrite and pearlite— which will appear in all hypoeutectoid iron–carbon alloys that are slowly cooled to a temperature below the eutectoid. The relative amounts of the proeutectoid a and pearlite may be determined in a manner similar to that described in Section 9.12 for primary and eutectic microconstituents. We use the lever rule in conjunction with a tie line that extends from the a  (a  Fe3C) phase boundary (0.022 wt% C) to the eutectoid composition (0.76 wt% C), inasmuch as pearlite is the transformation product of austenite having this composition. For example, let us consider an alloy of composition C0¿ in Figure 9.31. Thus, the fraction of pearlite, Wp, may be determined according to Lever rule expression for computation of pearlite mass fraction (composition C0¿, Figure 9.31) Wp   T TU C0¿  0.022 C0¿  0.022  0.76  0.022 0.74 (9.20) Furthermore, the fraction of proeutectoid a, Wa¿, is computed as follows: Lever rule expression for computation of proeutectoid ferrite mass fraction Wa¿   U TU 0.76  C0¿ 0.76  C0¿  0.76  0.022 0.74 (9.21) Of course, fractions of both total a (eutectoid and proeutectoid) and cementite are determined using the lever rule and a tie line that extends across the entirety of the a  Fe3C phase region, from 0.022 to 6.7 wt% C. ␥ ␥ + Fe3C Temperature Figure 9.31 A portion of the Fe–Fe3C phase diagram used in computations for relative amounts of proeutectoid and pearlite microconstituents for hypoeutectoid (C0¿) and hypereutectoid (C1¿) compositions. ␣ T U V X ␣ + Fe3C 6.70 0.022 C0 0.76 C1 Composition (wt% C) 1496T_c09_252-310 12/21/05 7:56 Page 298 2nd REVISE PAGES 298 • Chapter 9 / Phase Diagrams 1100 Figure 9.32 Schematic representations of the microstructures for an iron–carbon alloy of hypereutectoid composition C1 (containing between 0.76 and 2.14 wt% C), as it is cooled from within the austenite phase region to below the eutectoid temperature. P  + Fe3C  1000 z    g  900 Fe3C Temperature (°C)  800   h   + O 700 i  Pearlite 600  Proeutectoid Eutectoid Fe3C Fe3C 500  + Fe3C z' 400 0 1.0 2.0 C1 Composition (wt% C) Hypereutectoid Alloys hypereutectoid alloy proeutectoid cementite Analogous transformations and microstructures result for hypereutectoid alloys, those containing between 0.76 and 2.14 wt% C, which are cooled from temperatures within the g phase field. Consider an alloy of composition C1 in Figure 9.32 that, upon cooling, moves down the line zz¿. At point g only the g phase will be present with a composition of C1; the microstructure will appear as shown, having only g grains. Upon cooling into the g  Fe3C phase field—say, to point h—the cementite phase will begin to form along the initial g grain boundaries, similar to the a phase in Figure 9.29, point d. This cementite is called proeutectoid cementite—that which forms before the eutectoid reaction. Of course, the cementite composition remains constant (6.70 wt% C) as the temperature changes. However, the composition of the austenite phase will move along line PO toward the eutectoid. As the temperature is lowered through the eutectoid to point i, all remaining austenite of eutectoid composition is converted into pearlite; thus, the resulting microstructure consists of pearlite and proeutectoid cementite as microconstituents (Figure 9.32). In the photomicrograph of a 1.4 wt% C steel (Figure 9.33), note that the proeutectoid cementite appears light. Since it has much the same appearance as proeutectoid ferrite (Figure 9.30), there is some difficulty in distinguishing between hypoeutectoid and hypereutectoid steels on the basis of microstructure. Relative amounts of both pearlite and proeutectoid Fe3C microconstituents may be computed for hypereutectoid steel alloys in a manner analogous to that for hypoeutectoid materials; the appropriate tie line extends between 0.76 and 6.70 1496T_c09_252-310 11/29/05 11:33 Page 299 REVISED PAGES 9.19 Development of Microstructure in Iron–Carbon Alloys • 299 Figure 9.33 Photomicrograph of a 1.4 wt% C steel having a microstructure consisting of a white proeutectoid cementite network surrounding the pearlite colonies. 1000. (Copyright 1971 by United States Steel Corporation.) Proeutectoid cementite Pearlite wt% C. Thus, for an alloy having composition C¿1 in Figure 9.31, fractions of pearlite Wp and proeutectoid cementite WFe3C¿ are determined from the following lever rule expressions: 6.70  C1¿ 6.70  C1¿ X (9.22) Wp    VX 6.70  0.76 5.94 and C1¿  0.76 C1¿  0.76 V (9.23)   WFe3C¿  VX 6.70  0.76 5.94 Concept Check 9.8 Briefly explain why a proeutectoid phase (ferrite or cementite) forms along austenite grain boundaries. Hint: Consult Section 4.6. [The answer may be found at www.wiley.com/college/callister (Student Companion Site).] EXAMPLE PROBLEM 9.4 Determination of Relative Amounts of Ferrite, Cementite, and Pearlite Microconstituents For a 99.65 wt% Fe–0.35 wt% C alloy at a temperature just below the eutectoid, determine the following: (a) The fractions of total ferrite and cementite phases (b) The fractions of the proeutectoid ferrite and pearlite (c) The fraction of eutectoid ferrite 1496T_c09_252-310 11/29/05 11:33 Page 300 REVISED PAGES 300 • Chapter 9 / Phase Diagrams Solution (a) This part of the problem is solved by application of the lever rule expressions employing a tie line that extends all the way across the a  Fe3C phase field. Thus, C¿0 is 0.35 wt% C, and Wa  6.70  0.35  0.95 6.70  0.022 and WFe3C  0.35  0.022  0.05 6.70  0.022 (b) The fractions of proeutectoid ferrite and pearlite are determined by using the lever rule and a tie line that extends only to the eutectoid composition (i.e., Equations 9.20 and 9.21). Or Wp  0.35  0.022  0.44 0.76  0.022 Wa¿  0.76  0.35  0.56 0.76  0.022 and (c) All ferrite is either as proeutectoid or eutectoid (in the pearlite). Therefore, the sum of these two ferrite fractions will equal the fraction of total ferrite; that is, Wa¿  Wae  Wa where Wae denotes the fraction of the total alloy that is eutectoid ferrite.Values for Wa and Wa¿ were determined in parts (a) and (b) as 0.95 and 0.56, respectively. Therefore, Wae  Wa  Wa¿  0.95  0.56  0.39 Nonequilibrium Cooling In this discussion on the microstructural development of iron–carbon alloys it has been assumed that, upon cooling, conditions of metastable equilibrium2 have been continuously maintained; that is, sufficient time has been allowed at each new temperature for any necessary adjustment in phase compositions and relative amounts as predicted from the Fe–Fe3C phase diagram. In most situations these cooling rates are impractically slow and really unnecessary; in fact, on many occasions nonequilibrium conditions are desirable. Two nonequilibrium effects of practical importance are (1) the occurrence of phase changes or transformations at temperatures other than those predicted by phase boundary lines on the phase diagram, and (2) the 2 The term “metastable equilibrium” is used in this discussion inasmuch as Fe3C is only a metastable compound. 1496T_c09_252-310 11/29/05 11:33 Page 301 REVISED PAGES 9.20 The Influence of Other Alloying Elements • 301 Eutectoid temperature (°C) Mo 1200 W 2200 Si 2000 1000 1800 Cr 1600 800 1400 Mn Eutectoid temperature (°F) 2400 Ti Figure 9.34 The dependence of eutectoid temperature on alloy concentration for several alloying elements in steel. (From Edgar C. Bain, Functions of the Alloying Elements in Steel, American Society for Metals, 1939, p. 127.) 1200 600 Ni 0 2 4 6 8 10 12 1000 14 Concentration of alloying elements (wt%) existence at room temperature of nonequilibrium phases that do not appear on the phase diagram. Both are discussed in the next chapter. 9.20 THE INFLUENCE OF OTHER ALLOYING ELEMENTS Additions of other alloying elements (Cr, Ni, Ti, etc.) bring about rather dramatic changes in the binary iron–iron carbide phase diagram, Figure 9.24. The extent of these alterations of the positions of phase boundaries and the shapes of the phase fields depends on the particular alloying element and its concentration. One of the important changes is the shift in position of the eutectoid with respect to temperature and to carbon concentration. These effects are illustrated in Figures 9.34 and 9.35, which plot the eutectoid temperature and eutectoid composition (in wt% C) as a function of concentration for several other alloying elements. Thus, other alloy additions alter not only the temperature of the eutectoid reaction but also the relative fractions of pearlite and the proeutectoid phase that form. Steels are normally alloyed for other reasons, however—usually either to improve their corrosion resistance or to render them amenable to heat treatment (see Section 11.8). Figure 9.35 The dependence of eutectoid composition (wt% C) on alloy concentration for several alloying elements in steel. (From Edgar C. Bain, Functions of the Alloying Elements in Steel, American Society for Metals, 1939, p. 127.) Eutectoid composition (wt% C) 0.8 Ni 0.6 Cr 0.4 Si Mo W 0.2 0 Mn Ti 0 2 4 6 8 10 12 14 Concentration of alloying elements (wt%) 1496T_c09_252-310 11/29/05 11:33 Page 302 REVISED PAGES 302 • Chapter 9 / Phase Diagrams SUMMARY Phase Equilibria One-Component (or Unary) Phase Diagrams Binary Phase Diagrams Interpretation of Phase Diagrams Equilibrium phase diagrams are a convenient and concise way of representing the most stable relationships between phases in alloy systems. This discussion began by considering the unary (or pressure–temperature) phase diagram for a onecomponent system. Solid-, liquid-, and vapor-phase regions are found on this type of phase diagram. For binary systems, temperature and composition are variables, whereas external pressure is held constant. Areas, or phase regions, are defined on these temperature-versus-composition plots within which either one or two phases exist. For an alloy of specified composition and at a known temperature, the phases present, their compositions, and relative amounts under equilibrium conditions may be determined. Within two-phase regions, tie lines and the lever rule must be used for phase composition and mass fraction computations, respectively. Binary Isomorphous Systems Development of Microstructure in Isomorphous Alloys Mechanical Properties of Isomorphous Alloys Several different kinds of phase diagram were discussed for metallic systems. Isomorphous diagrams are those for which there is complete solubility in the solid phase; the copper–nickel system displays this behavior. Also discussed for alloys belonging to isomorphous systems were the development of microstructure for both cases of equilibrium and nonequilibrium cooling, and the dependence of mechanical characteristics on composition. Binary Eutectic Systems Development of Microstructure in Eutectic Alloys In a eutectic reaction, as found in some alloy systems, a liquid phase transforms isothermally to two different solid phases upon cooling. Such a reaction is noted on the copper–silver and lead–tin phase diagrams. Complete solid solubility for all compositions does not exist; instead, solid solutions are terminal—there is only a limited solubility of each component in the other. Four different kinds of microstructures that may develop for the equilibrium cooling of alloys belonging to eutectic systems were discussed. Equilibrium Diagrams Having Intermediate Phases or Compounds Eutectoid and Peritectic Reactions Congruent Phase Transformations Other equilibrium phase diagrams are more complex, having intermediate compounds and/or phases, possibly more than a single eutectic, and other reactions including eutectoid, peritectic, and congruent phase transformations. These are found for copper–zinc and magnesium–lead systems. 1496T_c09_252-310 11/29/05 11:33 Page 303 REVISED PAGES References • 303 The Gibbs Phase Rule The Gibbs phase rule was introduced; it is a simple equation that relates the number of phases present in a system at equilibrium with the number of degrees of freedom, the number of components, and the number of noncompositional variables. The Iron–Iron Carbide (Fe–Fe3C) Phase Diagram Development of Microstructure in Iron–Carbon Alloys Considerable attention was given to the iron–carbon system, and specifically, the iron–iron carbide phase diagram, which technologically is one of the most important. The development of microstructure in many iron–carbon alloys and steels depends on the eutectoid reaction in which the FCC austenite phase of composition 0.76 wt% C transforms isothermally to the BCC a ferrite phase (0.022 wt% C) and the intermetallic compound, cementite (Fe3C). The microstructural product of an iron–carbon alloy of eutectoid composition is pearlite, a microconstituent consisting of alternating layers of ferrite and cementite. The microstructures of alloys having carbon contents less than the eutectoid (hypoeutectoid) are comprised of a proeutectoid ferrite phase in addition to pearlite. On the other hand, pearlite and proeutectoid cementite constitute the microconstituents for hypereutectoid alloys—those with carbon contents in excess of the eutectoid composition. I M P O R TA N T T E R M S A N D C O N C E P T S Austenite Cementite Component Congruent transformation Equilibrium Eutectic phase Eutectic reaction Eutectic structure Eutectoid reaction Ferrite Free energy Gibbs phase rule Hypereutectoid alloy Hypoeutectoid alloy Intermediate solid solution Intermetallic compound Invariant point Isomorphous Lever rule Liquidus line Metastable Microconstituent Pearlite Peritectic reaction Phase Phase diagram Phase equilibrium Primary phase Proeutectoid cementite Proeutectoid ferrite Solidus line Solubility limit Solvus line System Terminal solid solution Tie line REFERENCES ASM Handbook, Vol. 3, Alloy Phase Diagrams, ASM International, Materials Park, OH, 1992. ASM Handbook, Vol. 9, Metallography and Microstructures, ASM International, Materials Park, OH, 2004. Hansen, M. and K. Anderko, Constitution of Binary Alloys, 2nd edition, McGraw-Hill, New York, 1958. First Supplement (R. P. Elliott), 1965. Second Supplement (F. A. Shunk), 1969. Reprinted by Genium Publishing Corp., Schenectady, NY. Massalski,T. B., H. Okamoto, P. R. Subramanian, and L. Kacprzak (Editors), Binary Phase Diagrams, 2nd edition, ASM International, Materials Park, OH, 1990. Three volumes. Also on CD with updates. Okamoto, H., Desk Handbook: Phase Diagrams for Binary Alloys, ASM International, Materials Park, OH, 2000. 1496T_c09_252-310 11/29/05 11:33 Page 304 REVISED PAGES 304 • Chapter 9 / Phase Diagrams Petzow, G., Ternary Alloys, A Comprehensive Compendium of Evaluated Constitutional Data and Phase Diagrams, Wiley, New York, 1988–1995. Fifteen volumes. Villars, P., A. Prince, and H. Okamoto (Editors), Handbook of Ternary Alloy Phase Diagrams, ASM International, Materials Park, OH, 1995. Ten volumes. Also on CD. QUESTIONS AND PROBLEMS Solubility Limit 9.1 Consider the sugar–water phase diagram of Figure 9.1. (a) How much sugar will dissolve in 1000 g of water at 80C (176F)? (b) If the saturated liquid solution in part (a) is cooled to 20C (68F), some of the sugar will precipitate out as a solid. What will be the composition of the saturated liquid solution (in wt% sugar) at 20C? (c) How much of the solid sugar will come out of solution upon cooling to 20C? 9.2 At 100C, what is the maximum solubility (a) of Pb in Sn? (b) of Sn in Pb? Microstructure 9.3 Cite three variables that determine the microstructure of an alloy. Phase Equilibria 9.4 What thermodynamic condition must be met for a state of equilibrium to exist? One-Component (or Unary) Phase Diagrams 9.5 Consider a specimen of ice that is at 25C and 10 atm pressure. Using Figure 9.2, the pressure–temperature phase diagram for H2O, determine the pressure to which the specimen must be raised or lowered to cause it (a) to melt, and (b) to sublime. 9.6 At a pressure of 0.1 atm, determine (a) the melting temperature for ice, and (b) the boiling temperature for water. Binary Isomorphous Systems 9.7 Given here are the solidus and liquidus temperatures for the copper–gold system. Construct the phase diagram for this system and label each region. Composition (wt% Au) Solidus Temperature (C) Liquidus Temperature (C) 0 20 40 60 80 90 95 100 1085 1019 972 934 911 928 974 1064 1085 1042 996 946 911 942 984 1064 Interpretation of Phase Diagrams (Binary Isomorphous Systems) (Binary Eutectic Systems) (Equilibrium Diagrams Having Intermediate Phases or Compounds) 9.8 Cite the phases that are present and the phase compositions for the following alloys: (a) 15 wt% Sn–85 wt% Pb at 100C (212F) (b) 25 wt% Pb–75 wt% Mg at 425C (800F) (c) 85 wt% Ag–15 wt% Cu at 800C (1470F) (d) 55 wt% Zn–45 wt% Cu at 600C (1110F) (e) 1.25 kg Sn and 14 kg Pb at 200C (390F) (f) 7.6 lbm Cu and 144.4 lbm Zn at 600C (1110F) (g) 21.7 mol Mg and 35.4 mol Pb at 350C (660F) (h) 4.2 mol Cu and 1.1 mol Ag at 900C (1650F) 9.9 Is it possible to have a copper–silver alloy that, at equilibrium, consists of a b phase of composition 92 wt% Ag–8 wt% Cu, and also a liquid phase of composition 76 wt% Ag–24 wt% Cu? If so, what will be the approximate temperature of the alloy? If this is not possible, explain why. 9.10 Is it possible to have a copper–silver alloy that, at equilibrium, consists of an a phase of composition 4 wt% Ag–96 wt% Cu, and also a b phase of composition 95 wt% Ag–5 wt% 1496T_c09_252-310 11/29/05 11:33 Page 305 REVISED PAGES Questions and Problems • 305 Cu? If so, what will be the approximate temperature of the alloy? If this is not possible, explain why. 9.11 A lead–tin alloy of composition 30 wt% Sn–70 wt% Pb is slowly heated from a temperature of 150C (300F). (a) At what temperature does the first liquid phase form? (b) What is the composition of this liquid phase? (c) At what temperature does complete melting of the alloy occur? (d) What is the composition of the last solid remaining prior to complete melting? 9.12 A 50 wt% Ni–50 wt% Cu alloy is slowly cooled from 1400C (2550F) to 1200C (2190F). (a) At what temperature does the first solid phase form? (b) What is the composition of this solid phase? (c) At what temperature does the liquid solidify? (d) What is the composition of this last remaining liquid phase? 9.13 For an alloy of composition 52 wt% Zn–48 wt% Cu, cite the phases present and their mass fractions at the following temperatures: 1000C, 800C, 500C, and 300C. 9.14 Determine the relative amounts (in terms of mass fractions) of the phases for the alloys and temperatures given in Problem 9.8. 9.15 A 2.0-kg specimen of an 85 wt% Pb–15 wt% Sn alloy is heated to 200C (390F); at this temperature it is entirely an a-phase solid solution (Figure 9.8). The alloy is to be melted to the extent that 50% of the specimen is liquid, the remainder being the a phase. This may be accomplished by heating the alloy or changing its composition while holding the temperature constant. (a) To what temperature must the specimen be heated? (b) How much tin must be added to the 2.0-kg specimen at 200C to achieve this state? 9.16 A magnesium–lead alloy of mass 7.5 kg consists of a solid a phase that has a composition just slightly below the solubility limit at 300C (570F). (a) What mass of lead is in the alloy? (b) If the alloy is heated to 400C (750F), how much more lead may be dissolved in the a phase without exceeding the solubility limit of this phase? 9.17 A 65 wt% Ni–35 wt% Cu alloy is heated to a temperature within the   liquid-phase region. If the composition of the a phase is 70 wt% Ni, determine: (a) The temperature of the alloy (b) The composition of the liquid phase (c) The mass fractions of both phases 9.18 A 40 wt% Pb–60 wt% Mg alloy is heated to a temperature within the   liquid-phase region. If the mass fraction of each phase is 0.5, then estimate: (a) The temperature of the alloy (b) The compositions of the two phases 9.19 For alloys of two hypothetical metals A and B, there exist an a, A-rich phase and a b, B-rich phase. From the mass fractions of both phases for two different alloys provided in the table below, (which are at the same temperature), determine the composition of the phase boundary (or solubility limit) for both a and b phases at this temperature. Alloy Composition 70 wt% A–30 wt% B 35 wt% A–65 wt% B Fraction  Phase Fraction  Phase 0.78 0.36 0.22 0.64 9.20 A hypothetical A–B alloy of composition 40 wt% B–60 wt% A at some temperature is found to consist of mass fractions of 0.66 and 0.34 for the a and b phases, respectively. If the composition of the a phase is 13 wt% B–87 wt% A, what is the composition of the b phase? 9.21 Is it possible to have a copper–silver alloy of composition 20 wt% Ag–80 wt% Cu that, at equilibrium, consists of a and liquid phases having mass fractions Wa  0.80 and WL  0.20? If so, what will be the approximate temperature of the alloy? If such an alloy is not possible, explain why. 1496T_c09_252-310 11/29/05 11:33 Page 306 REVISED PAGES 306 • Chapter 9 / Phase Diagrams 9.22 For 5.7 kg of a magnesium–lead alloy of composition 50 wt% Pb–50 wt% Mg, is it possible, at equilibrium, to have a and Mg2Pb phases with respective masses of 5.13 and 0.57 kg? If so, what will be the approximate temperature of the alloy? If such an alloy is not possible, then explain why. 9.23 Derive Equations 9.6a and 9.7a, which may be used to convert mass fraction to volume fraction, and vice versa. 9.24 Determine the relative amounts (in terms of volume fractions) of the phases for the alloys and temperatures given in Problems 9.8a, b, and d. Given here are the approximate densities of the various metals at the alloy temperatures: Metal Temperature (C) Density (g/cm3) Cu Mg Pb Pb Sn Zn 600 425 100 425 100 600 8.68 1.68 11.27 10.96 7.29 6.67 Development of Microstructure in Eutectic Alloys 9.28 Briefly explain why, upon solidification, an alloy of eutectic composition forms a microstructure consisting of alternating layers of the two solid phases. 9.29 What is the difference between a phase and a microconstituent? 9.30 Is it possible to have a magnesium–lead alloy in which the mass fractions of primary a and total a are 0.60 and 0.85, respectively, at 460C (860F)? Why or why not? 9.31 For 2.8 kg of a lead–tin alloy, is it possible to have the masses of primary b and total b of 2.21 kg and 2.53 kg, respectively, at 180C (355F)? Why or why not? 9.32 For a lead–tin alloy of composition 80 wt% Sn–20 wt% Pb and at 180C (355F) do the following: (a) Determine the mass fractions of a and b phases. (b) Determine the mass fractions of primary b and eutectic microconstituents. (c) Determine the mass fraction of eutectic b. Development of Microstructure in Isomorphous Alloys 9.25 (a) Briefly describe the phenomenon of coring and why it occurs. (b) Cite one undesirable consequence of coring. Mechanical Properties of Isomorphous Alloys 9.26 It is desirable to produce a copper–nickel alloy that has a minimum noncold-worked tensile strength of 380 MPa (55,000 psi) and a ductility of at least 45%EL. Is such an alloy possible? If so, what must be its composition? If this is not possible, then explain why. Binary Eutectic Systems 9.27 A 60 wt% Pb–40 wt% Mg alloy is rapidly quenched to room temperature from an elevated temperature in such a way that the hightemperature microstructure is preserved. This microstructure is found to consist of the a phase and Mg2Pb, having respective mass fractions of 0.42 and 0.58. Determine the approximate temperature from which the alloy was quenched. 9.33 The microstructure of a copper–silver alloy at 775C (1425F) consists of primary a and eutectic structures. If the mass fractions of these two microconstituents are 0.73 and 0.27, respectively, determine the composition of the alloy. 9.34 Consider the hypothetical eutectic phase diagram for metals A and B, which is similar to that for the lead–tin system, Figure 9.8. Assume that: (l) a and b phases exist at the A and B extremities of the phase diagram, respectively; (2) the eutectic composition is 36 wt% A–64 wt% B; and (3) the composition of the a phase at the eutectic temperature is 88 wt% A–12 wt% B. Determine the composition of an alloy that will yield primary b and total b mass fractions of 0.367 and 0.768, respectively. 9.35 For a 64 wt% Zn–36 wt% Cu alloy, make schematic sketches of the microstructure that would be observed for conditions of very slow cooling at the following temperatures: 900C (1650F), 820C (1510F), 750C (1380F), and 600C (1100F). Label all phases and indicate their approximate compositions. 1496T_c09_252-310 11/29/05 11:33 Page 307 REVISED PAGES Questions and Problems • 307 9.36 For a 76 wt% Pb–24 wt% Mg alloy, make schematic sketches of the microstructure that would be observed for conditions of very slow cooling at the following temperatures: 575C (1070F), 500C (930F), 450C (840F), and 300C (570F). Label all phases and indicate their approximate compositions. 9.37 For a 52 wt% Zn–48 wt% Cu alloy, make schematic sketches of the microstructure that would be observed for conditions of very slow cooling at the following temperatures: 950C (1740F), 860C (1580F), 800C (1470F), and 600C (1100F). Label all phases and indicate their approximate compositions. 9.38 On the basis of the photomicrograph (i.e., the relative amounts of the microconstituents) for the lead–tin alloy shown in Figure 9.17 and the Pb–Sn phase diagram (Figure 9.8), estimate the composition of the alloy, and then compare this estimate with the composition given in the figure legend of Figure 9.17. Make the following assumptions: (1) the area fraction of each phase and microconstituent in the photomicrograph is equal to its volume fraction; (2) the densities of the a and b phases as well as the eutectic structure are 11.2, 7.3, and 8.7 g/cm3, respectively; and (3) this photomicrograph represents the equilibrium microstructure at 180C (355F). 9.39 The room-temperature tensile strengths of pure copper and pure silver are 209 MPa and 125 MPa, respectively. (a) Make a schematic graph of the roomtemperature tensile strength versus composition for all compositions between pure copper and pure silver. (Hint: you may want to consult Sections 9.10 and 9.11, as well as Equation 9.24 in Problem 9.64.) (b) On this same graph schematically plot tensile strength versus composition at 600C. (c) Explain the shapes of these two curves, as well as any differences between them. Equilibrium Diagrams Having Intermediate Phases or Compounds 9.40 Two intermetallic compounds, A3B and AB3, exist for elements A and B. If the compositions for A3B and AB3 are 91.0 wt% A–9.0 wt% B and 53.0 wt% A–47.0 wt% B, respectively, and element A is zirconium, identify element B. Congruent Phase Transformations Binary Eutectic Systems Equilibrium Diagrams Having Intermediate Phases or Compounds Eutectoid and Peritectic Reactions 9.41 What is the principal difference between congruent and incongruent phase transformations? 9.42 Figure 9.36 is the tin–gold phase diagram, for which only single-phase regions are labeled. Specify temperature–composition points at which all eutectics, eutectoids, peritectics, and congruent phase transformations occur. Also, for each, write the reaction upon cooling. 9.43 Figure 9.37 is a portion of the copper– aluminum phase diagram for which only single-phase regions are labeled. Specify temperature–composition points at which all eutectics, eutectoids, peritectics, and congruent phase transformations occur. Also, for each, write the reaction upon cooling. 9.44 Construct the hypothetical phase diagram for metals A and B between room temperature (20C) and 700C given the following information: • The melting temperature of metal A is 480C. • The maximum solubility of B in A is 4 wt% B, which occurs at 420C. • The solubility of B in A at room temperature is 0 wt% B. • One eutectic occurs at 420C and 18 wt% B–82 wt% A. • A second eutectic occurs at 475C and 42 wt% B–58 wt% A. • The intermetallic compound AB exists at a composition of 30 wt% B–70 wt% A, and melts congruently at 525C. • The melting temperature of metal B is 600C. • The maximum solubility of A in B is 13 wt% A, which occurs at 475C. • The solubility of A in B at room temperature is 3 wt% A. 1496T_c09_252-310 11/29/05 11:33 Page 308 REVISED PAGES 308 • Chapter 9 / Phase Diagrams 1000 800 L Temperature (°C) Figure 9.36 The tin–gold phase diagram. (Adapted with permission from Metals Handbook, 8th edition, Vol. 8, Metallography, Structures and Phase Diagrams, American Society for Metals, Metals Park, OH, 1973.) 600  400 200   0 (Sn) 40 60 Composition (wt% Au) 80 100 (Au) 1100  1000 L 1 900  1 800 Temperature (°C) Figure 9.37 The copper–aluminum phase diagram. (Adapted with permission from Metals Handbook, 8th edition, Vol. 8, Metallography Structures and Phase Diagrams, American Society for Metals, Metals Park, OH, 1973.) 20    700 2 2 600  1 1 500 2 2 400 0 (Cu) 4 8 12 16 Composition (wt% Al) 20 24 28 1496T_c09_252-310 11/29/05 11:33 Page 309 REVISED PAGES Questions and Problems • 309 Figure 9.38 Logarithm pressureversus-temperature phase diagram for H2O. 10,000 Ice III 1,000 Pressure (atm) 100 Liquid 10 1.0 A Ice I 0.1 Vapor B 0.01 C 0.001 –20 0 20 40 60 Temperature (°C) The Gibbs Phase Rule 9.45 In Figure 9.38 is shown the pressure– temperature phase diagram for H2O. Apply the Gibbs phase rule at points A, B, and C; that is, specify the number of degrees of freedom at each of the points—that is, the number of externally controllable variables that need be specified to completely define the system. 9.51 The Iron–Iron Carbide (Fe–Fe3C) Phase Diagram Development of Microstructure in Iron–Carbon Alloys 9.46 Compute the mass fractions of a ferrite and cementite in pearlite. 9.47 (a) What is the distinction between hypoeutectoid and hypereutectoid steels? (b) In a hypoeutectoid steel, both eutectoid and proeutectoid ferrite exist. Explain the difference between them. What will be the carbon concentration in each? 9.48 What is the carbon concentration of an iron– carbon alloy for which the fraction of total cementite is 0.10? 9.49 What is the proeutectoid phase for an iron– carbon alloy in which the mass fractions of total ferrite and total cementite are 0.86 and 0.14, respectively? Why? 9.50 Consider 3.5 kg of austenite containing 0.95 wt% C, cooled to below 727C (1341F). (a) What is the proeutectoid phase? 9.52 9.53 9.54 9.55 80 100 120 (b) How many kilograms each of total ferrite and cementite form? (c) How many kilograms each of pearlite and the proeutectoid phase form? (d) Schematically sketch and label the resulting microstructure. Consider 6.0 kg of austenite containing 0.45 wt% C, cooled to below 727C (1341F). (a) What is the proeutectoid phase? (b) How many kilograms each of total ferrite and cementite form? (c) How many kilograms each of pearlite and the proeutectoid phase form? (d) Schematically sketch and label the resulting microstructure. Compute the mass fractions of proeutectoid ferrite and pearlite that form in an iron– carbon alloy containing 0.35 wt% C. The microstructure of an iron–carbon alloy consists of proeutectoid ferrite and pearlite; the mass fractions of these two microconstituents are 0.174 and 0.826, respectively. Determine the concentration of carbon in this alloy. The mass fractions of total ferrite and total cementite in an iron–carbon alloy are 0.91 and 0.09, respectively. Is this a hypoeutectoid or hypereutectoid alloy? Why? The microstructure of an iron–carbon alloy consists of proeutectoid cementite and pearlite; the mass fractions of these microconstituents 1496T_c09_252-310 11/29/05 11:33 Page 310 REVISED PAGES 310 • Chapter 9 / Phase Diagrams 9.56 9.57 9.58 9.59 9.60 9.61 9.62 are 0.11 and 0.89, respectively. Determine the concentration of carbon in this alloy. Consider 1.5 kg of a 99.7 wt% Fe–0.3 wt% C alloy that is cooled to a temperature just below the eutectoid. (a) How many kilograms of proeutectoid ferrite form? (b) How many kilograms of eutectoid ferrite form? (c) How many kilograms of cementite form? Compute the maximum mass fraction of proeutectoid cementite possible for a hypereutectoid iron–carbon alloy. Is it possible to have an iron–carbon alloy for which the mass fractions of total cementite and proeutectoid ferrite are 0.057 and 0.36, respectively? Why or why not? Is it possible to have an iron–carbon alloy for which the mass fractions of total ferrite and pearlite are 0.860 and 0.969, respectively? Why or why not? Compute the mass fraction of eutectoid cementite in an iron–carbon alloy that contains 1.00 wt% C. The mass fraction of eutectoid cementite in an iron–carbon alloy is 0.109. On the basis of this information, is it possible to determine the composition of the alloy? If so, what is its composition? If this is not possible, explain why. The mass fraction of eutectoid ferrite in an iron–carbon alloy is 0.71. On the basis of this information, is it possible to determine the composition of the alloy? If so, what is its composition? If this is not possible, explain why. 9.63 For an iron–carbon alloy of composition 3 wt% C–97 wt% Fe, make schematic sketches of the microstructure that would be observed for conditions of very slow cooling at the following temperatures: 1250C (2280F), 1145C (2095F), and 700C (1290F). Label the phases and indicate their compositions (approximate). 9.64 Often, the properties of multiphase alloys may be approximated by the relationship E 1alloy2  EaVa  EbVb (9.24) where E represents a specific property (modulus of elasticity, hardness, etc.), and V is the volume fraction.The subscripts a and b denote the existing phases or microconstituents. Employ the relationship above to determine the approximate Brinell hardness of a 99.75 wt% Fe–0.25 wt% C alloy. Assume Brinell hardnesses of 80 and 280 for ferrite and pearlite, respectively, and that volume fractions may be approximated by mass fractions. The Influence of Other Alloying Elements 9.65 A steel alloy contains 95.7 wt% Fe, 4.0 wt% W, and 0.3 wt% C. (a) What is the eutectoid temperature of this alloy? (b) What is the eutectoid composition? (c) What is the proeutectoid phase? Assume that there are no changes in the positions of other phase boundaries with the addition of W. 9.66 A steel alloy is known to contain 93.65 wt% Fe, 6.0 wt% Mn, and 0.35 wt% C. (a) What is the approximate eutectoid temperature of this alloy? (b) What is the proeutectoid phase when this alloy is cooled to a temperature just below the eutectoid? (c) Compute the relative amounts of the proeutectoid phase and pearlite. Assume that there are no alterations in the positions of other phase boundaries with the addition of Mn. 1496T_c10_311-357 12/31/05 9:55 Page 311 Chapter 2nd REVISE PAGES 10 Phase Transformations in Metals: Development of Microstructure and Alteration of Mechanical Properties A photomicrograph of a pearlitic steel that has partially transformed to spheroidite. 2000. (Courtesy of United States Steel Corporation.) WHY STUDY Phase Transformations in Metals? The development of a set of desirable mechanical characteristics for a material often results from a phase transformation that is wrought by a heat treatment. The time and temperature dependencies of some phase transformations are conveniently represented on modified phase diagrams. It is important to know how to use these diagrams in order to design a heat treat- ment for some alloy that will yield the desired roomtemperature mechanical properties. For example, the tensile strength of an iron–carbon alloy of eutectoid composition (0.76 wt% C) can be varied between approximately 700 MPa (100,000 psi) and 2000 MPa (300,000 psi) depending on the heat treatment employed. • 311 1496T_c10_311-357 12/31/05 9:55 Page 312 2nd REVISE PAGES Learning Objectives After careful study of this chapter you should be able to do the following: 1. Make a schematic fraction transformationpearlite, coarse pearlite, spheroidite, bainite, versus-logarithm of time plot for a typical martensite, and tempered martensite. Now, in solid–solid transformation; cite the equation terms of microstructure (or crystal structure), that describes this behavior. briefly explain these behaviors. 2. Briefly describe the microstructure for each of 4. Given the isothermal transformation (or continthe following microconstituents that are found uous cooling transformation) diagram for some in steel alloys: fine pearlite, coarse pearlite, iron–carbon alloy, design a heat treatment that spheroidite, bainite, martensite, and tempered will produce a specified microstructure. martensite. 3. Cite the general mechanical characteristics for each of the following microconstituents: fine 10.1 INTRODUCTION transformation rate One reason for the versatility of metallic materials lies in the wide range of mechanical properties they possess, which are accessible to management by various means.Three strengthening mechanisms were discussed in Chapter 7—namely, grain size refinement, solid-solution strengthening, and strain hardening. Additional techniques are available wherein the mechanical properties are reliant on the characteristics of the microstructure. The development of microstructure in both single- and two-phase alloys ordinarily involves some type of phase transformation—an alteration in the number and/or character of the phases. The first portion of this chapter is devoted to a brief discussion of some of the basic principles relating to transformations involving solid phases. Inasmuch as most phase transformations do not occur instantaneously, consideration is given to the dependence of reaction progress on time, or the transformation rate. This is followed by a discussion of the development of two-phase microstructures for iron–carbon alloys. Modified phase diagrams are introduced that permit determination of the microstructure that results from a specific heat treatment. Finally, other microconstituents in addition to pearlite are presented, and, for each, the mechanical properties are discussed. P h a s e Tr a n s f o r m a t i o n s 10.2 BASIC CONCEPTS phase transformation A variety of phase transformations are important in the processing of materials, and usually they involve some alteration of the microstructure. For purposes of this discussion, these transformations are divided into three classifications. In one group are simple diffusion-dependent transformations in which there is no change in either the number or composition of the phases present. These include solidification of a pure metal, allotropic transformations, and, recrystallization and grain growth (see Sections 7.12 and 7.13). In another type of diffusion-dependent transformation, there is some alteration in phase compositions and often in the number of phases present; the final microstructure ordinarily consists of two phases. The eutectoid reaction, described by Equation 9.19, is of this type; it receives further attention in Section 10.5. 1496T_c10_311-357 11/30/05 7:37 Page 313 REVISED PAGES 10.3 The Kinetics of Phase Transformations • 313 The third kind of transformation is diffusionless, wherein a metastable phase is produced. As discussed in Section 10.5, a martensitic transformation, which may be induced in some steel alloys, falls into this category. 10.3 THE KINETICS OF PHASE TRANSFORMATIONS nucleation, growth With phase transformations, normally at least one new phase is formed that has different physical/chemical characteristics and/or a different structure than the parent phase. Furthermore, most phase transformations do not occur instantaneously. Rather, they begin by the formation of numerous small particles of the new phase(s), which increase in size until the transformation has reached completion. The progress of a phase transformation may be broken down into two distinct stages: nucleation and growth. Nucleation involves the appearance of very small particles, or nuclei of the new phase (often consisting of only a few hundred atoms), which are capable of growing. During the growth stage these nuclei increase in size, which results in the disappearance of some (or all) of the parent phase. The transformation reaches completion if the growth of these new phase particles is allowed to proceed until the equilibrium fraction is attained. We now discuss the mechanics of these two processes, and how they relate to solid-state transformations. Nucleation There are two types of nucleation: homogeneous and heterogeneous. The distinction between them is made according to the site at which nucleating events occur. For the homogeneous type, nuclei of the new phase form uniformly throughout the parent phase, whereas for the heterogeneous type, nuclei form preferentially at structural inhomogeneities, such as container surfaces, insoluble impurities, grain boundaries, dislocations, and so on. We begin by discussing homogeneous nucleation because its description and theory are simpler to treat. These principles are then extended to a discussion of the heterogeneous type. Homogeneous Nucleation free energy A discussion of the theory of nucleation involves a thermodynamic parameter called free energy (or Gibbs free energy), G. In brief, free energy is a function of other thermodynamic parameters, of which one is the internal energy of the system (i.e., the enthalpy, H), and another is a measurement of the randomness or disorder of the atoms or molecules (i.e., the entropy, S). It is not our purpose here to provide a detailed discussion of the principles of thermodynamics as they apply to materials systems. However, relative to phase transformations, an important thermodynamic parameter is the change in free energy ¢G; a transformation will occur spontaneously only when ¢G has a negative value. For the sake of simplicity, let us first consider the solidification of a pure material, assuming that nuclei of the solid phase form in the interior of the liquid as atoms cluster together so as to form a packing arrangement similar to that found in the solid phase. Furthermore, it will be assumed that each nucleus is spherical in geometry and has a radius r. This situation is represented schematically in Figure 10.1. There are two contributions to the total free energy change that accompany a solidification transformation. The first is the free energy difference between the solid and liquid phases, or the volume free energy, ¢Gv. Its value will be negative if the temperature is below the equilibrium solidification temperature, and the magnitude of its contribution is the product of ¢Gv and the volume of the spherical nucleus (i.e., 43 pr 3). The second energy contribution results from the formation of the 1496T_c10_311-357 12/31/05 9:55 Page 314 2nd REVISE PAGES 314 • Chapter 10 / Phase Transformations in Metals Volume = 4 3 r 3 Liquid Figure 10.1 Schematic diagram showing the nucleation of a spherical solid particle in a liquid. r Solid Area = 4r2 Solid-liquid interface solid–liquid phase boundary during the solidification transformation. Associated with this boundary is a surface free energy, g, which is positive; furthermore, the magnitude of this contribution is the product of g and the surface area of the nucleus (i.e., 4pr2). Finally, the total free energy change is equal to the sum of these two contributions—that is, ¢G  43 pr3 ¢Gv  4pr2g (10.1) These volume, surface, and total free energy contributions are plotted schematically as a function of nucleus radius in Figures 10.2a and 10.2b. Here (Figure 10.2a) it will be noted that for the curve corresponding to the first term on the right-hand side of Equation 10.1, the free energy (which is negative) decreases with the third power of r. Furthermore, for the curve resulting from the second term in Equation 10.1, energy values are positive and increase with the square of the radius. Consequently, the curve associated with the sum of both terms (Figure 10.2b) first increases, passes through a maximum, and finally decreases. In a physical sense, this means that as a solid particle begins to form as atoms in the liquid cluster together, its free energy first increases. If this cluster reaches a size corresponding to the critical radius r*, then growth will continue with the accompaniment of a 4  r2 0 + radius, r 4 3  r Gv 3 – (a) Free energy change, G + Free energy change, G Total free energy change for a solidification transformation r* G* 0 radius, r – (b) Figure 10.2 (a) Schematic curves for volume free energy and surface free energy contributions to the total free energy change attending the formation of a spherical embryo/nucleus during solidification. (b) Schematic plot of free energy versus embryo/nucleus radius, on which is shown the critical free energy change (¢G*) and the critical nucleus radius (r*). 1496T_c10_311-357 11/30/05 7:37 Page 315 REVISED PAGES 10.3 The Kinetics of Phase Transformations • 315 decrease in free energy. On the other hand, a cluster of radius less than the critical will shrink and redissolve. This subcritical particle is an embryo, whereas the particle of radius greater than r* is termed a nucleus. A critical free energy, ¢G*, occurs at the critical radius and, consequently, at the maximum of the curve in Figure 10.2b. This ¢G* corresponds to an activation free energy, which is the free energy required for the formation of a stable nucleus. Equivalently, it may be considered an energy barrier to the nucleation process. Since r* and ¢G* appear at the maximum on the free energy-versus-radius curve of Figure 10.2b, derivation of expressions for these two parameters is a simple matter. For r*, we differentiate the ¢G equation (Equation 10.1) with respect to r, set the resulting expression equal to zero, and then solve for r ( r*). That is, d1¢G2  43 p¢Gv 13r2 2  4pg12r2  0 dr (10.2) which leads to the result For homogeneous nucleation, critical radius of a stable solid particle nucleus r*   2g ¢Gv (10.3) Now, substitution of this expression for r* into Equation 10.1 yields the following expression for ¢G*: For homogeneous nucleation, activation free energy required for the formation of a stable nucleus ¢G*  16pg3 31¢Gv 2 2 (10.4) This volume free energy change ¢Gv is the driving force for the solidification transformation, and its magnitude is a function of temperature. At the equilibrium solidification temperature Tm, the value of ¢Gv is zero, and with diminishing temperature its value becomes increasingly more negative. It can be shown that ¢Gv is a function of temperature as ¢Gv  ¢Hf 1Tm  T 2 Tm (10.5) where ¢Hf is the latent heat of fusion (i.e., the heat given up during solidification), and Tm and the temperature T are in Kelvin. Substitution of this expression for ¢Gv into Equations 10.3 and 10.4 yields Dependence of critical radius on surface free energy, latent heat of fusion, melting temperature, and transformation temperature Activation free energy expression r*  a 2gTm 1 ba b ¢Hf Tm  T (10.6) 16pg3T m2 1 b 3¢H 2f 1Tm  T 2 2 (10.7) and ¢G*  a Thus, from these two equations, both the critical radius r* and the activation free energy ¢G* decrease as temperature T decreases. (The g and ¢Hf parameters in these expressions are relatively insensitive to temperature changes.) Figure 10.3, a schematic ¢G-versus-r plot that shows curves for two different temperatures, illustrates these relationships. Physically, this means that with a lowering of temperature at temperatures below the equilibrium solidification temperature (Tm), nucleation occurs more 1496T_c10_311-357 11/30/05 7:37 Page 316 REVISED PAGES 316 • Chapter 10 / Phase Transformations in Metals + r*1 T 2 > T1 r*2 G G*1 at T1 G* 2 0 radius, r Figure 10.3 Schematic freeenergy-versus-embryo/nucleusradius curves for two different temperatures. The critical free energy change (¢G*) and critical nucleus radius (r*) are indicated for each temperature. at T2 – readily. Furthermore, the number of stable nuclei n* (having radii greater than r*) is a function of temperature as n*  K1 exp a ¢G* b kT (10.8) where the constant K1 is related to the total number of nuclei of the solid phase. For the exponential term of this expression, changes in temperature have a greater effect on the magnitude of the ¢G* term in the numerator than the T term in the denominator. Consequently, as the temperature is lowered below Tm the exponential term in Equation 10.8 also decreases such that the magnitude of n* increases. This temperature dependence (n* versus T) is represented in the schematic plot of Figure 10.4a. There is another important temperature-dependent step that is involved in and also influences nucleation: the clustering of atoms by short-range diffusion during the formation of nuclei. The influence of temperature on the rate of diffusion (i.e., magnitude of the diffusion coefficient, D) is given in Equation 5.8. Furthermore, this diffusion effect is related to the frequency at which atoms from the liquid attach themselves to the solid nucleus, nd. Or, the dependence of nd on temperature is the same as for the diffusion coefficient—namely, nd  K2 exp a (10.9) Tm Tm Tm G* kT exp – Qd kT d Temperature exp – Temperature T Temperature Figure 10.4 For solidification, schematic plots of (a) number of stable nuclei versus temperature, (b) frequency of atomic attachment versus temperature, and (c) nucleation rate versus temperature (also shown are curves for parts a and b). Qd b kT . N n* . Number of stable nuclei, n* Frequency of attachment, d n*, d, N (a) (b) (c) 1496T_c10_311-357 11/30/05 7:37 Page 317 REVISED PAGES 10.3 The Kinetics of Phase Transformations • 317 where Qd is a temperature-independent parameter—the activation energy for diffusion—and K2 is a temperature-independent constant. Thus, from Equation 10.9 a diminishment of temperature results in a reduction in nd. This effect, represented by the curve shown in Figure 10.4b, is just the reverse of that for n* as discussed above. The principles and concepts just developed are now extended # to a discussion of another important nucleation parameter, the nucleation rate N (which has units of nuclei per unit volume per second). This rate is simply proportional to the product of n* (Equation 10.8) and nd (Equation 10.9); that is, Nucleation rate expression for homogeneous nucleation # Qd ¢G* N  K3n*nd  K1K2K3 c exp a b exp a b d kT kT (10.10) Here K3 is the number of atoms on a nucleus surface. Figure 10.4c schematically plots nucleation rate as a function of temperature and, in addition, the curves of # Figures 10.4a and 10.4b from which the N curve is derived. Note (Figure 10.4c) that, with a lowering of temperature from below Tm, the nucleation rate first increases, achieves a maximum, and # subsequently diminishes. The shape of this N curve is explained as # follows: for the upper region of the curve (a sudden and dramatic increase in N with decreasing T ), ¢G* is greater than Qd, which means that the exp(¢G*kT ) term of Equation 10.10 is much smaller than exp (Qd kT). In other words, the nucleation rate is suppressed at high temperatures due to a small activation driving force. With continued diminishment of temperature, there comes a point at which ¢G* becomes smaller than the temperature-independent Qd with the result that exp(QdkT ) 6 exp(¢G*kT), or that, at lower temperatures, a low atomic mobility suppresses the nucleation rate. This# accounts for the shape of the lower curve segment (a precipitous reduc# tion of N with a continued diminishment of temperature). Furthermore, the N curve of Figure 10.4c necessarily passes through a maximum over the intermediate temperature range where values for ¢G* and Qd are of approximately the same magnitude. Several qualifying comments are in order regarding the above discussion. First, although we assumed a spherical shape for nuclei, this method may be applied to any shape with the same final result. Furthermore, this treatment may be utilized for types of transformations other than solidification (i.e., liquid–solid)—for example, solid–vapor and solid–solid. However, magnitudes of ¢Gv and g, in addition to diffusion rates of the atomic species, will undoubtedly differ among the various transformation types. In addition, for solid–solid transformations, there may be volume changes attendant to the formation of new phases. These changes may lead to the introduction of microscopic strains, which must be taken into account in the ¢G expression of Equation 10.1, and, consequently, will affect the magnitudes of r* and ¢G*. From Figure 10.4c it is apparent that during the cooling of a liquid, an appreciable nucleation rate (i.e., solidification) will begin only after the temperature has been lowered to below the equilibrium solidification (or melting) temperature (Tm). This phenomenon is termed supercooling (or undercooling), and the degree of supercooling for homogeneous nucleation may be significant (on the order of several hundred degrees Kelvin) for some systems. In Table 10.1 is tabulated, for several materials, typical degrees of supercooling for homogeneous nucleation. 1496T_c10_311-357 11/30/05 7:37 Page 318 REVISED PAGES 318 • Chapter 10 / Phase Transformations in Metals Table 10.1 Degree of Supercooling (T ) Values (Homogeneous Nucleation) for Several Metals Metal Antimony Germanium Silver Gold Copper Iron Nickel Cobalt Palladium T (C ) 135 227 227 230 236 295 319 330 332 Source: D. Turnbull and R. E. Cech, “Microscopic Observation of the Solidification of Small Metal Droplets,” J. Appl. Phys., 21, 808 (1950). EXAMPLE PROBLEM 10.1 Computation of Critical Nucleus Radius and Activation Free Energy (a) For the solidification of pure gold, calculate the critical radius r* and the activation free energy ¢G* if nucleation is homogeneous. Values for the latent heat of fusion and surface free energy are 1.16  109 J/m3 and 0.132 J/m2, respectively. Use the supercooling value found in Table 10.1. (b) Now calculate the number of atoms found in a nucleus of critical size. Assume a lattice parameter of 0.413 nm for solid gold at its melting temperature. Solution (a) In order to compute the critical radius, we employ Equation 10.6, using the melting temperature of 1064C for gold, assuming a supercooling value of 230C (Table 10.1), and realizing that ¢Hf is negative. Hence r*  a  c 2gTm 1 ba b ¢Hf Tm  T 12210.132 J/m2 211064  273 K2 1.16  10 J/m 9 3 da 1 b 230 K  1.32  109 m  1.32 nm For computation of the activation free energy, Equation 10.7 is employed. Thus 16pg3Tm2 1 b ¢G*  a 2 3¢Hf 1Tm  T 2 2 11621p210.132 J/m2 2 3 11064  273 K2 2 1  c dc d 13211.16  109 J/m3 2 2 1230 K2 2  9.64  1019 J 1496T_c10_311-357 11/30/05 7:37 Page 319 REVISED PAGES 10.3 The Kinetics of Phase Transformations • 319 (b) In order to compute the number of atoms in a nucleus of critical size (assuming a spherical nucleus of radius r*), it is first necessary to determine the number of unit cells, which we then multiply by the number of atoms per unit cell. The number of unit cells found in this critical nucleus is just the ratio of critical nucleus and unit cell volumes. Inasmuch as gold has the FCC crystal structure (and a cubic unit cell), its unit cell volume is just a3, where a is the lattice parameter (i.e., unit cell edge length); its value is 0.413 nm, as cited in the problem statement. Therefore, the number of unit cells found in a radius of critical size is just 4 pr*3 critical nucleus volume 3 # unit cells/particle   (10.11) unit cell volume a3 4 a b1p211.32 nm2 3 3  137 unit cells  10.413 nm2 3 Inasmuch as there is the equivalence of four atoms per FCC unit cell (Section 3.4), the total number of atoms per critical nucleus is just (137 unit cells/critical nucleus)(4 atoms/unit cell)  548 atoms/critical nucleus Heterogeneous Nucleation For heterogeneous nucleation of a solid particle, relationship among solid-surface, solid-liquid, and liquid-surface interfacial energies and the wetting angle Although levels of supercooling for homogeneous nucleation may be significant (on occasion several hundred degrees Celsius), in practical situations they are often on the order of only several degrees Celsius. The reason for this is that the activation energy (i.e., energy barrier) for nucleation ( ¢G* of Equation 10.4) is lowered when nuclei form on preexisting surfaces or interfaces, since the surface free energy (g of Equation 10.4) is reduced. In other words, it is easier for nucleation to occur at surfaces and interfaces than at other sites. Again, this type of nucleation is termed heterogeneous. In order to understand this phenomenon, let us consider the nucleation, on a flat surface, of a solid particle from a liquid phase. It is assumed that both the liquid and solid phases “wet” this flat surface, that is, both of these phases spread out and cover the surface; this configuration is depicted schematically in Figure 10.5. Also noted in the figure are three interfacial energies (represented as vectors) that exist at two-phase boundaries—gSL, gSI, and gIL—as well as the wetting angle u (the angle between the gSI and gSL vectors). Taking a surface tension force balance in the plane of the flat surface leads to the following expression: gIL  gSI  gSL cos u Liquid Solid  SL   IL  SI Surface or interface (10.12) Figure 10.5 Heterogeneous nucleation of a solid from a liquid. The solid–surface (gSI), solid–liquid (gSL), and liquid–surface (gIL) interfacial energies are represented by vectors. The wetting angle (u) is also shown. 1496T_c10_311-357 11/30/05 7:37 Page 320 REVISED PAGES 320 • Chapter 10 / Phase Transformations in Metals Figure 10.6 Schematic freeenergy-versus-embryo/nucleusradius plot on which are presented curves for both homogeneous and heterogeneous nucleation. Critical free energies and the critical radius are also shown. r* G*hom G G*het 0 r Now, using a somewhat involved procedure similar to the one presented above for homogeneous nucleation (which we have chosen to omit), it is possible to derive equations for r* and ¢G*; these are as follows: For heterogeneous nucleation, critical radius of a stable solid particle nucleus For heterogeneous nucleation, activation free energy required for the formation of a stable nucleus r*   ¢G*  a 2gSL ¢Gv 16pg3SL b S1u2 3¢Gv2 (10.13) (10.14) The S(u) term of this last equation is a function only of u (i.e., the shape of the nucleus), which will have a numerical value between zero and unity.1 From Equation 10.13, it is important to note that the critical radius r* for heterogeneous nucleation is the same as for homogeneous, inasmuch as gSL is the same surface energy as g in Equation 10.3. It is also evident that the activation energy barrier for heterogeneous nucleation (Equation 10.14) is smaller than the homogeneous barrier (Equation 10.4) by an amount corresponding to the value of this S(u) function, or *  ¢Ghom ¢Ghet * S1u2 (10.15) Figure 10.6, a schematic graph of ¢G versus nucleus radius, plots curves for both types of nucleation, and indicates the difference in the magnitudes of ¢G*het and ¢G*hom, in addition to the constancy of r*. This lower ¢G* for heterogeneous means that a smaller energy must be overcome during the nucleation process (than for homogeneous), and, therefore, heterogeneous nucleation occurs more readily (Equa# tion 10.10). In terms of the nucleation rate, the N versus T curve (Figure 10.4c) is shifted to higher temperatures for heterogeneous. This effect is represented in Figure 10.7, which also shows that a much smaller degree of supercooling (¢T) is required for heterogeneous nucleation. Growth The growth step in a phase transformation begins once an embryo has exceeded the critical size, r*, and becomes a stable nucleus. Note that nucleation will continue to occur simultaneously with growth of the new phase particles; of course, nucleation 1 For example, for u angles of 30 and 90, values of S(u) are approximately 0.01 and 0.5, respectively. 1496T_c10_311-357 12/31/05 9:55 Page 321 2nd REVISE PAGES 10.3 The Kinetics of Phase Transformations • 321 Figure 10.7 Nucleation rate versus temperature for both homogeneous and heterogeneous nucleation. Degree of supercooling (¢T ) for each is also shown. Tm T het T hom Temperature . Nhet . Nhom Nucleation rate cannot occur in regions that have already transformed to the new phase. Furthermore, the growth process will cease in any region where particles of the new phase meet, since here the transformation will have reached completion. Particle growth occurs by long-range atomic diffusion, which normally involves several steps—for example, diffusion through the parent phase, across a phase # boundary, and then into the nucleus. Consequently, the growth rate G is determined by the rate of diffusion, and its temperature dependence is the same as for the diffusion coefficient (Equation 5.8)—namely, Dependence of particle growth rate on the activation energy for diffusion and temperature # Q G  C exp a b kT (10.16) where Q (the activation energy) and C (a preexponential) are independent of tem# perature.2 The temperature dependence of G is represented# by one of the curves in Figure 10.8; also shown is a curve for the nucleation rate, N (again, almost always the rate for heterogeneous nucleation). Now, at a specific temperature, the overall # # transformation rate is equal to some product of N and G. The third curve of Tm . Temperature Growth rate, G Figure 10.8 Schematic plot# showing curves for nucleation rate (N ), growth rate # (G ), and overall transformation rate versus temperature. Overall transformation rate . Nucleation rate, N Rate # Processes the rates of which depend on temperature as G in Equation (10.16) are sometimes termed thermally activated. Also, a rate equation of this form (i.e., having the exponential temperature dependence) is termed an Arrhenius rate equation. 2 thermally activated transformation 1496T_c10_311-357 12/31/05 9:55 Page 322 2nd REVISE PAGES 322 • Chapter 10 / Phase Transformations in Metals Figure 10.8, which is for the total rate, represents this combined effect. The general shape of this curve is the same as for the nucleation rate, # in that it has a peak or maximum that has been shifted upward relative to the N curve. Whereas this treatment on transformations has been developed for solidification, the same general principles also apply to solid–solid and solid–gas transformations. As we shall see below, the rate of transformation and the time required for the transformation to proceed to some degree of completion (e.g., time to 50% reaction completion, t0.5) are inversely proportional to one another (Equation 10.18). Thus, if the logarithm of this transformation time (i.e., log t0.5) is plotted versus temperature, a curve having the general shape shown in Figure 10.9b results. This “C-shaped” curve is a virtual mirror image (through a vertical plane) of the transformation rate curve of Figure 10.8, as demonstrated in Figure 10.9. It is often the case that the kinetics of phase transformations are represented using logarithm time- (to some degree of transformation) versus-temperature plots (for example, see Section 10.5). Several physical phenomena may be explained in terms of the transformation rate-versus-temperature curve of Figure 10.8. First, the size of the product phase particles will depend on transformation temperature. For example, for transformations that occur at temperatures near to Tm, corresponding to low nucleation and high growth rates, few nuclei form that grow rapidly. Thus, the resulting microstructure will consist of few and relatively large phase particles (e.g., coarse grains). Conversely, for transformations at lower temperatures, nucleation rates are high and growth rates low, which results in many small particles (e.g., fine grains). Also, from Figure 10.8, when a material is cooled very rapidly through the temperature range encompassed by the transformation rate curve to a relatively low temperature where the rate is extremely low, it is possible to produce nonequilibrium phase structures (for example, see Sections 10.5 and 11.9). Kinetic Considerations of Solid-State Transformations Te Te Temperature Figure 10.9 Schematic plots of (a) transformation rate versus temperature, and (b) logarithm time [to some degree (e.g., 0.5 fraction) of transformation] versus temperature. The curves in both (a) and (b) are generated from the same set of data—i.e., for horizontal axes, the time [scaled logarithmically in the (b) plot] is just the reciprocal of the rate from plot (a). Temperature kinetics The previous discussion of this section has centered on the temperature dependences of nucleation, growth, and transformation rates. The time dependence of rate (which is often termed the kinetics of a transformation) is also an important consideration, often in the heat treatment of materials. Also, since many transformations of interest to materials scientists and engineers involve only solid phases, we have decided to devote the following discussion to the kinetics of solid-state transformations. With many kinetic investigations, the fraction of reaction that has occurred is measured as a function of time while the temperature is maintained constant. Transformation progress is usually ascertained by either microscopic examination or measurement of some physical property (such as electrical conductivity) the magnitude Rate 1 t0.5 (a) Time (t0.5) (logarithmic scale) (b) 1496T_c10_311-357 11/30/05 7:37 Page 323 REVISED PAGES 10.3 The Kinetics of Phase Transformations • 323 Fraction of transformation, y 1.0 Figure 10.10 Plot of fraction reacted versus the logarithm of time typical of many solid-state transformations in which temperature is held constant. 0.5 t0.5 0 Nucleation Growth Logarithm of heating time, t of which is distinctive of the new phase. Data are plotted as the fraction of transformed material versus the logarithm of time; an S-shaped curve similar to that in Figure 10.10 represents the typical kinetic behavior for most solid-state reactions. Nucleation and growth stages are also indicated in the figure. For solid-state transformations displaying the kinetic behavior in Figure 10.10, the fraction of transformation y is a function of time t as follows: Transformation rate—reciprocal of the halfway-tocompletion transformation time y  1  exp1kt n 2 (10.17) where k and n are time-independent constants for the particular reaction. The above expression is often referred to as the Avrami equation. By convention, the rate of a transformation is taken as the reciprocal of time required for the transformation to proceed halfway to completion, t0.5, or rate  1 (10.18) t0.5 Temperature will have a profound influence on the kinetics and thus on the rate of a transformation. This is demonstrated in Figure 10.11, where y-versus-log t 100 Percent recrystallized Avrami equation— dependence of fraction of transformation on time 80 135C 60 119C 113C 102C 88C 43C 40 20 0 1 10 102 Time (min) (Logarithmic scale) 104 Figure 10.11 Percent recrystallization as a function of time and at constant temperature for pure copper. (Reprinted with permission from Metallurgical Transactions, Vol. 188, 1950, a publication of The Metallurgical Society of AIME, Warrendale, PA. Adapted from B. F. Decker and D. Harker, “Recrystallization in Rolled Copper,” Trans. AIME, 188, 1950, p. 888.) 1496T_c10_311-357 11/30/05 7:37 Page 324 REVISED PAGES 324 • Chapter 10 / Phase Transformations in Metals S-shaped curves at several temperatures for the recrystallization of copper are shown. A detailed discussion on the influence of both temperature and time on phase transformations is provided in Section 10.5. 10.4 METASTABLE VERSUS EQUILIBRIUM STATES supercooling superheating Phase transformations may be wrought in metal alloy systems by varying temperature, composition, and the external pressure; however, temperature changes by means of heat treatments are most conveniently utilized to induce phase transformations. This corresponds to crossing a phase boundary on the composition-temperature phase diagram as an alloy of given composition is heated or cooled. During a phase transformation, an alloy proceeds toward an equilibrium state that is characterized by the phase diagram in terms of the product phases, their compositions, and relative amounts.As the previous section noted, most phase transformations require some finite time to go to completion, and the speed or rate is often important in the relationship between the heat treatment and the development of microstructure. One limitation of phase diagrams is their inability to indicate the time period required for the attainment of equilibrium. The rate of approach to equilibrium for solid systems is so slow that true equilibrium structures are rarely achieved. When phase transformations are induced by temperature changes, equilibrium conditions are maintained only if heating or cooling is carried out at extremely slow and unpractical rates. For other than equilibrium cooling, transformations are shifted to lower temperatures than indicated by the phase diagram; for heating, the shift is to higher temperatures. These phenomena are termed supercooling and superheating, respectively. The degree of each depends on the rate of temperature change; the more rapid the cooling or heating, the greater the supercooling or superheating. For example, for normal cooling rates the iron–carbon eutectoid reaction is typically displaced 10 to 20C (18 to 36F) below the equilibrium transformation temperature.3 For many technologically important alloys, the preferred state or microstructure is a metastable one, intermediate between the initial and equilibrium states; on occasion, a structure far removed from the equilibrium one is desired. It thus becomes imperative to investigate the influence of time on phase transformations. This kinetic information is, in many instances, of greater value than a knowledge of the final equilibrium state. M i c ro s t r u c t u r a l a n d P ro p e r t y C h a n ge s i n I ro n – C a r b o n A l l oys Some of the basic kinetic principles of solid-state transformations are now extended and applied specifically to iron–carbon alloys in terms of the relationships among heat treatment, the development of microstructure, and mechanical properties. This 3 It is important to note that the treatments relating to the kinetics of phase transformations in Section 10.3 are constrained to the condition of constant temperature. By way of contrast, the discussion of this section pertains to phase transformations that occur with changing temperature. This same distinction exists between Sections 10.5 (Isothermal Transformation Diagrams) and 10.6 (Continuous Cooling Transformation Diagrams). 1496T_c10_311-357 11/30/05 7:37 Page 325 REVISED PAGES 10.5 Isothermal Transformation Diagrams • 325 system has been chosen because it is familiar and because a wide variety of microstructures and mechanical properties are possible for iron–carbon (or steel) alloys. 10.5 ISOTHERMAL TRANSFORMATION DIAGRAMS Pearlite Consider again the iron–iron carbide eutectoid reaction Eutectoid reaction for the iron-iron carbide system cooling g10.76 wt% C2 Δ a10.022 wt% C2  Fe3C16.70 wt% C2 heating (10.19) which is fundamental to the development of microstructure in steel alloys. Upon cooling, austenite, having an intermediate carbon concentration, transforms to a ferrite phase, having a much lower carbon content, and also cementite, with a much higher carbon concentration. Pearlite is one microstructural product of this transformation (Figure 9.27), and the mechanism of pearlite formation was discussed previously (Section 9.19) and demonstrated in Figure 9.28. Temperature plays an important role in the rate of the austenite-to-pearlite transformation. The temperature dependence for an iron–carbon alloy of eutectoid composition is indicated in Figure 10.12, which plots S-shaped curves of the percentage transformation versus the logarithm of time at three different temperatures. For each curve, data were collected after rapidly cooling a specimen composed of 100% austenite to the temperature indicated; that temperature was maintained constant throughout the course of the reaction. A more convenient way of representing both the time and temperature dependence of this transformation is in the bottom portion of Figure 10.13. Here, the vertical and horizontal axes are, respectively, temperature and the logarithm of time. Two solid curves are plotted; one represents the time required at each temperature for the initiation or start of the transformation; the other is for the transformation conclusion. The dashed curve corresponds to 50% of transformation completion. These curves were generated from a series of plots of the percentage transformation versus the logarithm of time taken over a range of temperatures. The S-shaped curve [for 675C (1247F)], in the upper portion of Figure 10.13, illustrates how the data transfer is made. In interpreting this diagram, note first that the eutectoid temperature [727C (1341F)] is indicated by a horizontal line; at temperatures above the eutectoid and Percent pearlite 600°C 50 650°C 0 1 10 Time (s) 675°C 102 50 100 103 Percent austenite 0 100 Figure 10.12 For an iron–carbon alloy of eutectoid composition (0.76 wt% C), isothermal fraction reacted versus the logarithm of time for the austenite-to-pearlite transformation. 1496T_c10_311-357 11/30/05 7:37 Page 326 REVISED PAGES Percent of austenite transformed to pearlite 326 • Chapter 10 / Phase Transformations in Metals 100 Transformation ends Transformation temperature 675°C 50 Transformation begins 0 1 102 10 103 104 105 Time (s) Eutectoid temperature Austenite (stable) Austenite (unstable) 1200 Pearlite 600 50% Completion curve 1000 Completion curve (~100% pearlite) 500 Begin curve (~ 0% pearlite) 400 1 10 Temperature (°F) Temperature (°C) 700 1400 Figure 10.13 Demonstration of how an isothermal transformation diagram (bottom) is generated from percentage transformation-versuslogarithm of time measurements (top). [Adapted from H. Boyer, (Editor), Atlas of Isothermal Transformation and Cooling Transformation Diagrams, American Society for Metals, 1977, p. 369.] 800 102 103 104 105 Time (s) isothermal transformation diagram for all times, only austenite will exist, as indicated in the figure.The austenite-to-pearlite transformation will occur only if an alloy is supercooled to below the eutectoid; as indicated by the curves, the time necessary for the transformation to begin and then end depends on temperature. The start and finish curves are nearly parallel, and they approach the eutectoid line asymptotically. To the left of the transformation start curve, only austenite (which is unstable) will be present, whereas to the right of the finish curve, only pearlite will exist. In between, the austenite is in the process of transforming to pearlite, and thus both microconstituents will be present. According to Equation 10.18, the transformation rate at some particular temperature is inversely proportional to the time required for the reaction to proceed to 50% completion (to the dashed line in Figure 10.13). That is, the shorter this time, the higher is the rate. Thus, from Figure 10.13, at temperatures just below the eutectoid (corresponding to just a slight degree of undercooling) very long times (on the order of 105 s) are required for the 50% transformation, and therefore the reaction rate is very slow. The transformation rate increases with decreasing temperature such that at 540C (1000F) only about 3 s is required for the reaction to go to 50% completion. Several constraints are imposed on using diagrams like Figure 10.13. First, this particular plot is valid only for an iron–carbon alloy of eutectoid composition; for other compositions, the curves will have different configurations. In addition, these plots are accurate only for transformations in which the temperature of the alloy is held constant throughout the duration of the reaction. Conditions of constant temperature are termed isothermal; thus, plots such as Figure 10.13 are referred to as isothermal transformation diagrams, or sometimes as time–temperature–transformation (or T–T–T ) plots. 1496T_c10_311-357 12/31/05 9:55 Page 327 2nd REVISE PAGES 10.5 Isothermal Transformation Diagrams • 327 1s 700 ␥ Temperature (°C) ␥ ␥ ␥ 1 day Eutectoid temperature Austenite (stable) ␥ 727°C 1h 1400 ␥ ␣ Ferrite ␥ Coarse pearlite 1200 C B D 600 Fe3C Fine pearlite Temperature (°F) A 1 min 1000 500 Austenite → pearlite transformation Denotes that a transformation is occurring 800 1 10 102 103 104 105 Time (s) Figure 10.14 Isothermal transformation diagram for a eutectoid iron–carbon alloy, with superimposed isothermal heat treatment curve (ABCD). Microstructures before, during, and after the austenite-to-pearlite transformation are shown. [Adapted from H. Boyer (Editor), Atlas of Isothermal Transformation and Cooling Transformation Diagrams, American Society for Metals, 1977, p. 28.] coarse pearlite fine pearlite An actual isothermal heat treatment curve (ABCD) is superimposed on the isothermal transformation diagram for a eutectoid iron–carbon alloy in Figure 10.14. Very rapid cooling of austenite to a temperature is indicated by the near-vertical line AB, and the isothermal treatment at this temperature is represented by the horizontal segment BCD. Of course, time increases from left to right along this line. The transformation of austenite to pearlite begins at the intersection, point C (after approximately 3.5 s), and has reached completion by about 15 s, corresponding to point D. Figure 10.14 also shows schematic microstructures at various times during the progression of the reaction. The thickness ratio of the ferrite and cementite layers in pearlite is approximately 8 to 1. However, the absolute layer thickness depends on the temperature at which the isothermal transformation is allowed to occur. At temperatures just below the eutectoid, relatively thick layers of both the a-ferrite and Fe3C phases are produced; this microstructure is called coarse pearlite, and the region at which it forms is indicated to the right of the completion curve on Figure 10.14. At these temperatures, diffusion rates are relatively high, such that during the transformation illustrated in Figure 9.28 carbon atoms can diffuse relatively long distances, which results in the formation of thick lamellae. With decreasing temperature, the carbon diffusion rate decreases, and the layers become progressively thinner. The thin-layered structure produced in the vicinity of 540C is termed fine pearlite; this is also indicated in Figure 10.14. To be discussed in Section 10.7 is the dependence of mechanical properties on lamellar thickness. Photomicrographs of coarse and fine pearlite for a eutectoid composition are shown in Figure 10.15. 1496T_c10_311-357 11/30/05 7:37 Page 328 REVISED PAGES 328 • Chapter 10 / Phase Transformations in Metals Figure 10.15 Photomicrographs of (a) coarse pearlite and (b) fine pearlite. 3000. (From K. M. Ralls et al., An Introduction to Materials Science and Engineering, p. 361. Copyright © 1976 by John Wiley & Sons, New York. Reprinted by permission of John Wiley & Sons, Inc.) For iron–carbon alloys of other compositions, a proeutectoid phase (either ferrite or cementite) will coexist with pearlite, as discussed in Section 9.19. Thus additional curves corresponding to a proeutectoid transformation also must be included on the isothermal transformation diagram. A portion of one such diagram for a 1.13 wt% C alloy is shown in Figure 10.16. Bainite In addition to pearlite, other microconstituents that are products of the austenitic transformation exist; one of these is called bainite. The microstructure of bainite consists of ferrite and cementite phases, and thus diffusional processes are involved 900 1600 A 800 Eutectoid temperature + 700 C A A + 600 1400 1200 P P 1000 500 1 10 102 Time (s) 103 104 Temperature (°F) A Temperature (°C) bainite Figure 10.16 Isothermal transformation diagram for a 1.13 wt% C iron–carbon alloy: A, austenite; C, proeutectoid cementite; P, pearlite. [Adapted from H. Boyer (Editor), Atlas of Isothermal Transformation and Cooling Transformation Diagrams, American Society for Metals, 1977, p. 33.] 1496T_c10_311-357 11/30/05 7:37 Page 329 REVISED PAGES 10.5 Isothermal Transformation Diagrams • 329 Figure 10.17 Transmission electron micrograph showing the structure of bainite. A grain of bainite passes from lower left to upper right-hand corners, which consists of elongated and needle-shaped particles of Fe3C within a ferrite matrix. The phase surrounding the bainite is martensite. (Reproduced with permission from Metals Handbook, 8th edition, Vol. 8, Metallography, Structures and Phase Diagrams, American Society for Metals, Materials Park, OH, 1973.) Martensite Cementite Ferrite in its formation. Bainite forms as needles or plates, depending on the temperature of the transformation; the microstructural details of bainite are so fine that their resolution is possible only using electron microscopy. Figure 10.17 is an electron micrograph that shows a grain of bainite (positioned diagonally from lower left to upper right); it is composed of a ferrite matrix and elongated particles of Fe3C; the various phases in this micrograph have been labeled. In addition, the phase that surrounds the needle is martensite, the topic to which a subsequent section is addressed. Furthermore, no proeutectoid phase forms with bainite. The time–temperature dependence of the bainite transformation may also be represented on the isothermal transformation diagram. It occurs at temperatures below those at which pearlite forms; begin-, end-, and half-reaction curves are just extensions of those for the pearlitic transformation, as shown in Figure 10.18, the isothermal transformation diagram for an iron–carbon alloy of eutectoid composition that has been extended to lower temperatures. All three curves are C-shaped and have a “nose” at point N, where the rate of transformation is a maximum. As may be noted, whereas pearlite forms above the nose [i.e., over the temperature range of about 540 to 727C (1000 to 1341F)], at temperatures between about 215 and 540C (420 and 1000F), bainite is the transformation product. It should also be noted that pearlitic and bainitic transformations are really competitive with each other, and once some portion of an alloy has transformed to either pearlite or bainite, transformation to the other microconstituent is not possible without reheating to form austenite. Spheroidite spheroidite If a steel alloy having either pearlitic or bainitic microstructures is heated to, and left at, a temperature below the eutectoid for a sufficiently long period of time—for example, at about 700C (1300F) for between 18 and 24 h—yet another microstructure will form. It is called spheroidite (Figure 10.19). Instead of the alternating ferrite and cementite lamellae (pearlite), or the microstructure observed for bainite, the Fe3C phase appears as sphere-like particles embedded in a continuous  phase matrix. This transformation has occurred by additional carbon diffusion with no change in the compositions or relative amounts of ferrite and cementite 1496T_c10_311-357 11/30/05 7:37 Page 330 REVISED PAGES 330 • Chapter 10 / Phase Transformations in Metals 800 A Eutectoid temperature 1400 700 A 1200 A + P 1000 N 500 A+B 800 B 400 A 600 300 50% 200 100 10–1 1 10 102 103 400 104 Temperature (°F) Temperature (°C) 600 P Figure 10.18 Isothermal transformation diagram for an iron–carbon alloy of eutectoid composition, including austenite-to-pearlite (A–P) and austeniteto-bainite (A–B) transformations. [Adapted from H. Boyer (Editor), Atlas of Isothermal Transformation and Cooling Transformation Diagrams, American Society for Metals, 1977, p. 28.] 105 Time (s) phases. The chapter-opening photograph for this chapter is a photomicrograph that shows a pearlitic steel that has partially transformed to spheroidite. The driving force for this transformation is the reduction in a–Fe3C phase boundary area. The kinetics of spheroidite formation are not included on isothermal transformation diagrams. Concept Check 10.1 Which is the more stable, the pearlitic or the spheroiditic microstructure? Why? [The answer may be found at www.wiley.com/college/callister (Student Companion Site).] Figure 10.19 Photomicrograph of a steel having a spheroidite microstructure. The small particles are cementite; the continuous phase is a ferrite. 1000. (Copyright 1971 by United States Steel Corporation.) 1496T_c10_311-357 11/30/05 7:37 Page 331 REVISED PAGES 10.5 Isothermal Transformation Diagrams • 331 Martensite martensite Yet another microconstituent or phase called martensite is formed when austenitized iron–carbon alloys are rapidly cooled (or quenched) to a relatively low temperature (in the vicinity of the ambient). Martensite is a nonequilibrium single-phase structure that results from a diffusionless transformation of austenite. It may be thought of as a transformation product that is competitive with pearlite and bainite. The martensitic transformation occurs when the quenching rate is rapid enough to prevent carbon diffusion. Any diffusion whatsoever will result in the formation of ferrite and cementite phases. The martensitic transformation is not well understood. However, large numbers of atoms experience cooperative movements, in that there is only a slight displacement of each atom relative to its neighbors. This occurs in such a way that the FCC austenite experiences a polymorphic transformation to a body-centered tetragonal (BCT) martensite. A unit cell of this crystal structure (Figure 10.20) is simply a body-centered cube that has been elongated along one of its dimensions; this structure is distinctly different from that for BCC ferrite. All the carbon atoms remain as interstitial impurities in martensite; as such, they constitute a supersaturated solid solution that is capable of rapidly transforming to other structures if heated to temperatures at which diffusion rates become appreciable. Many steels, however, retain their martensitic structure almost indefinitely at room temperature. The martensitic transformation is not, however, unique to iron–carbon alloys. It is found in other systems and is characterized, in part, by the diffusionless transformation. Since the martensitic transformation does not involve diffusion, it occurs almost instantaneously; the martensite grains nucleate and grow at a very rapid rate—the velocity of sound within the austenite matrix. Thus the martensitic transformation rate, for all practical purposes, is time independent. Martensite grains take on a plate-like or needle-like appearance, as indicated in Figure 10.21. The white phase in the micrograph is austenite (retained austenite) that did not transform during the rapid quench. As already mentioned, martensite as well as other microconstituents (e.g., pearlite) can coexist. Being a nonequilibrium phase, martensite does not appear on the iron–iron carbide phase diagram (Figure 9.24).The austenite-to-martensite transformation is, however, represented on the isothermal transformation diagram. Since the martensitic transformation is diffusionless and instantaneous, it is not depicted in this diagram as the pearlitic and bainitic reactions are. The beginning of this transformation is represented by a horizontal line designated M(start) (Figure 10.22). Two other horizontal and dashed lines, labeled M(50%) and M(90%), indicate percentages of the austenite-to-martensite transformation. The temperatures at which these lines are Figure 10.20 The body-centered tetragonal unit cell for martensitic steel showing iron atoms (circles) and sites that may be occupied by carbon atoms (crosses). For this tetragonal unit cell, c 7 a. c a a 1496T_c10_311-357 11/30/05 7:37 Page 332 REVISED PAGES 332 • Chapter 10 / Phase Transformations in Metals Figure 10.21 Photomicrograph showing the martensitic microstructure. The needleshaped grains are the martensite phase, and the white regions are austenite that failed to transform during the rapid quench. 1220. (Photomicrograph courtesy of United States Steel Corporation.) athermal transformation plain carbon steel alloy steel located vary with alloy composition but, nevertheless, must be relatively low because carbon diffusion must be virtually nonexistent.4 The horizontal and linear character of these lines indicates that the martensitic transformation is independent of time; it is a function only of the temperature to which the alloy is quenched or rapidly cooled. A transformation of this type is termed an athermal transformation. Consider an alloy of eutectoid composition that is very rapidly cooled from a temperature above 727C (1341F) to, say, 165C (330F). From the isothermal transformation diagram (Figure 10.22) it may be noted that 50% of the austenite will immediately transform to martensite; and as long as this temperature is maintained, there will be no further transformation. The presence of alloying elements other than carbon (e.g., Cr, Ni, Mo, and W) may cause significant changes in the positions and shapes of the curves in the isothermal transformation diagrams. These include (1) shifting to longer times the nose of the austenite-to-pearlite transformation (and also a proeutectoid phase nose, if such exists), and (2) the formation of a separate bainite nose. These alterations may be observed by comparing Figures 10.22 and 10.23, which are isothermal transformation diagrams for carbon and alloy steels, respectively. Steels in which carbon is the prime alloying element are termed plain carbon steels, whereas alloy steels contain appreciable concentrations of other elements, including those cited in the preceding paragraph. Section 11.2 tells more about the classification and properties of ferrous alloys. Concept Check 10.2 Cite two major differences between martensitic and pearlitic transformations. [The answer may be found at www.wiley.com/college/callister (Student Companion Site).] 4 The alloy that is the subject of Figure 10.21 is not an iron-carbon alloy of eutectoid composition; furthermore, its 100% martensite transformation temperature lies below the ambient. Since the photomicrograph was taken at room temperature, some austenite (i.e., the retained austenite) is present, having not transformed to martensite. 1496T_c10_311-357 11/30/05 7:37 Page 333 REVISED PAGES 10.5 Isothermal Transformation Diagrams • 333 Figure 10.22 The complete isothermal transformation diagram for an iron–carbon alloy of eutectoid composition: A, austenite; B, bainite; M, martensite; P, pearlite. 800 A 1400 Eutectoid temperature 700 A 1200 A + 600 P P B 800 A 400 + B A 300 Temperature (°F) Temperature (°C) 1000 500 600 M(start) 200 50% M+A M(50%) 400 M(90%) 100 200 0 10–1 1 102 10 103 104 105 Time (s) 800 A 1400 Eutectoid temperature 700 A+F F+P 1200 A+F +P A 600 1000 800 400 A+B 50% B 600 M(start) 300 M+A M(50%) M(90%) 400 200 M 100 200 0 1 10 102 103 Time (s) 104 105 106 Temperature (°F) 500 Temperature (°C) Figure 10.23 Isothermal transformation diagram for an alloy steel (type 4340): A, austenite; B, bainite; P, pearlite; M, martensite; F, proeutectoid ferrite. [Adapted from H. Boyer (Editor), Atlas of Isothermal Transformation and Cooling Transformation Diagrams, American Society for Metals, 1977, p. 181.] 1496T_c10_311-357 11/30/05 7:37 Page 334 REVISED PAGES 334 • Chapter 10 / Phase Transformations in Metals EXAMPLE PROBLEM 10.2 Microstructural Determinations for Three Isothermal Heat Treatments Using the isothermal transformation diagram for an iron–carbon alloy of eutectoid composition (Figure 10.22), specify the nature of the final microstructure (in terms of microconstituents present and approximate percentages) of a small specimen that has been subjected to the following time–temperature treatments. In each case assume that the specimen begins at 760C (1400F) and that it has been held at this temperature long enough to have achieved a complete and homogeneous austenitic structure. (a) Rapidly cool to 350C (660F), hold for 104 s, and quench to room temperature. (b) Rapidly cool to 250C (480F), hold for 100 s, and quench to room temperature. (c) Rapidly cool to 650C (1200F), hold for 20 s, rapidly cool to 400C (750F), hold for 103 s, and quench to room temperature. Solution The time–temperature paths for all three treatments are shown in Figure 10.24. In each case the initial cooling is rapid enough to prevent any transformation from occurring. (a) At 350C austenite isothermally transforms to bainite; this reaction begins after about 10 s and reaches completion at about 500 s elapsed time. Therefore, by 104 s, as stipulated in this problem, 100% of the specimen is bainite, and no further transformation is possible, even though the final quenching line passes through the martensite region of the diagram. (b) In this case it takes about 150 s at 250C for the bainite transformation to begin, so that at 100 s the specimen is still 100% austenite. As the specimen is cooled through the martensite region, beginning at about 215C, progressively more of the austenite instantaneously transforms to martensite. This transformation is complete by the time room temperature is reached, such that the final microstructure is 100% martensite. (c) For the isothermal line at 650C, pearlite begins to form after about 7 s; by the time 20 s has elapsed, only approximately 50% of the specimen has transformed to pearlite. The rapid cool to 400C is indicated by the vertical line; during this cooling, very little, if any, remaining