Ebook- shigley's mechanical engineering design 8th edition

EBOOK- Shigley's Mechanical Engineering Design 8th Edition
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Unformatted text preview: Mechanical Engineering Shigleys Mechanical Engineering Design, Eighth Edition BudynasNisbett =>? McGraw-Hill McGrawHill Primis ISBN: 0390764876 Text: Shigleys Mechanical Engineering Design, Eighth Edition BudynasNisbett This book was printed on recycled paper. Mechanical Engineering http://www.primisonline.com Copyright ©2006 by The McGrawHill Companies, Inc. All rights reserved. Printed in the United States of America. Except as permitted under the United States Copyright Act of 1976, no part of this publication may be reproduced or distributed in any form or by any means, or stored in a database or retrieval system, without prior written permission of the publisher. This McGrawHill Primis text may include materials submitted to McGrawHill for publication by the instructor of this course. The instructor is solely responsible for the editorial content of such materials. 111 0192GEN ISBN: 0390764876 Mechanical Engineering Contents BudynasNisbett Shigleys Mechanical Engineering Design, Eighth Edition Front Matter 1 Preface List of Symbols 1 5 I. Basics 8 Introduction 1. Introduction to Mechanical Engineering Design 2. Materials 3. Load and Stress Analysis 4. Deflection and Stiffness 8 9 33 72 1 45 II. Failure Prevention 208 Introduction 5. Failures Resulting from Static Loading 6. Fatigue Failure Resulting from Variable Loading 208 2 09 2 60 III. Design of Mechanical Elements 349 Introduction 7. Shafts and Shaft Components 8. Screws, Fasteners, and the Design of Nonpermanent Joints 9. Welding, Bonding, and the Design of Permanent Joints 10. Mechanical Springs 11. RollingContact Bearings 12. Lubrication and Journal Bearings 13. Gears General 14. Spur and Helical Gears 15. Bevel and Worm Gears 16. Clutches, Brakes, Couplings, and Flywheels 17. Flexible Mechanical Elements 18. Power Transmission Case Study 349 3 50 3 98 4 60 IV. Analysis Tools 928 Introduction 19. FiniteElement Analysis 20. Statistical Considerations 928 9 29 9 52 iii 5 01 5 50 5 97 6 52 7 11 7 62 8 02 8 56 9 09 Back Matter 978 Appendix A: Useful Tables Appendix B: Answers to Selected Problems Index iv 9 78 1 034 1039 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition Front Matter Preface © The McGrawHill Companies, 2008 1 Preface Objectives This text is intended for students beginning the study of mechanical engineering design. The focus is on blending fundamental development of concepts with practical specification of components. Students of this text should find that it inherently directs them into familiarity with both the basis for decisions and the standards of industrial components. For this reason, as students transition to practicing engineers, they will find that this text is indispensable as a reference text. The objectives of the text are to: Cover the basics of machine design, including the design process, engineering mechanics and materials, failure prevention under static and variable loading, and characteristics of the principal types of mechanical elements. Offer a practical approach to the subject through a wide range of real-world applications and examples. Encourage readers to link design and analysis. Encourage readers to link fundamental concepts with practical component specication. New to This Edition This eighth edition contains the following signicant enhancements: New chapter on the Finite Element Method. In response to many requests from reviewers, this edition presents an introductory chapter on the nite element method. The goal of this chapter is to provide an overview of the terminology, method, capabilities, and applications of this tool in the design environment. New transmission case study. The traditional separation of topics into chapters sometimes leaves students at a loss when it comes time to integrate dependent topics in a larger design process. A comprehensive case study is incorporated through standalone example problems in multiple chapters, then culminated with a new chapter that discusses and demonstrates the integration of the parts into a complete design process. Example problems relevant to the case study are presented on engineering paper background to quickly identify them as part of the case study. Revised and expanded coverage of shaft design. Complementing the new transmission case study is a signicantly revised and expanded chapter focusing on issues relevant to shaft design. The motivating goal is to provide a meaningful presentation that allows a new designer to progress through the entire shaft design process from general shaft layout to specifying dimensions. The chapter has been moved to immediately follow the fatigue chapter, providing an opportunity to seamlessly transition from the fatigue coverage to its application in the design of shafts. Availability of information to complete the details of a design. Additional focus is placed on ensuring the designer can carry the process through to completion. xv 2 xvi BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition Front Matter Preface © The McGrawHill Companies, 2008 Mechanical Engineering Design By assigning larger design problems in class, the authors have identied where the students lack details. For example, information is now provided for such details as specifying keys to transmit torque, stress concentration factors for keyways and retaining ring grooves, and allowable deections for gears and bearings. The use of internet catalogs and engineering component search engines is emphasized to obtain current component specications. Streamlining of presentation. Coverage of material continues to be streamlined to focus on presenting straightforward concept development and a clear design procedure for student designers. Content Changes and Reorganization A new Part 4: Analysis Tools has been added at the end of the book to include the new chapter on nite elements and the chapter on statistical considerations. Based on a survey of instructors, the consensus was to move these chapters to the end of the book where they are available to those instructors wishing to use them. Moving the statistical chapter from its former location causes the renumbering of the former chapters 2 through 7. Since the shaft chapter has been moved to immediately follow the fatigue chapter, the component chapters (Chapters 8 through 17) maintain their same numbering. The new organization, along with brief comments on content changes, is given below: Part 1: Basics Part 1 provides a logical and unied introduction to the background material needed for machine design. The chapters in Part 1 have received a thorough cleanup to streamline and sharpen the focus, and eliminate clutter. Chapter 1, Introduction. Some outdated and unnecessary material has been removed. A new section on problem specication introduces the transmission case study. Chapter 2, Materials. New material is included on selecting materials in a design process. The Ashby charts are included and referenced as a design tool. Chapter 3, Load and Stress Analysis. Several sections have been rewritten to improve clarity. Bending in two planes is specically addressed, along with an example problem. Chapter 4, Deection and Stiffness. Several sections have been rewritten to improve clarity. A new example problem for deection of a stepped shaft is included. A new section is included on elastic stability of structural members in compression. Part 2: Failure Prevention This section covers failure by static and dynamic loading. These chapters have received extensive cleanup and clarication, targeting student designers. Chapter 5, Failures Resulting from Static Loading. In addition to extensive cleanup for improved clarity, a summary of important design equations is provided at the end of the chapter. Chapter 6, Fatigue Failure Resulting from Variable Loading. Confusing material on obtaining and using the S-N diagram is claried. The multiple methods for obtaining notch sensitivity are condensed. The section on combination loading is rewritten for greater clarity. A chapter summary is provided to overview the analysis roadmap and important design equations used in the process of fatigue analysis. BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition Front Matter Preface 3 © The McGrawHill Companies, 2008 Preface xvii Part 3: Design of Mechanical Elements Part 3 covers the design of specic machine components. All chapters have received general cleanup. The shaft chapter has been moved to the beginning of the section. The arrangement of chapters, along with any signicant changes, is described below: Chapter 7, Shafts and Shaft Components. This chapter is signicantly expanded and rewritten to be comprehensive in designing shafts. Instructors that previously did not specically cover the shaft chapter are encouraged to use this chapter immediately following the coverage of fatigue failure. The design of a shaft provides a natural progression from the failure prevention section into application toward components. This chapter is an essential part of the new transmission case study. The coverage of setscrews, keys, pins, and retaining rings, previously placed in the chapter on bolted joints, has been moved into this chapter. The coverage of limits and ts, previously placed in the chapter on statistics, has been moved into this chapter. Chapter 8, Screws, Fasteners, and the Design of Nonpermanent Joints. The section on setscrews, keys, and pins, has been moved from this chapter to Chapter 7. The coverage of bolted and riveted joints loaded in shear has been returned to this chapter. Chapter 9, Welding, Bonding, and the Design of Permanent Joints. The section on bolted and riveted joints loaded in shear has been moved to Chapter 8. Chapter 10, Mechanical Springs. Chapter 11, Rolling-Contact Bearings. Chapter 12, Lubrication and Journal Bearings. Chapter 13, Gears General. New example problems are included to address design of compound gear trains to achieve specied gear ratios. The discussion of the relationship between torque, speed, and power is claried. Chapter 14, Spur and Helical Gears. The current AGMA standard (ANSI/AGMA 2001-D04) has been reviewed to ensure up-to-date information in the gear chapters. All references in this chapter are updated to reect the current standard. Chapter 15, Bevel and Worm Gears. Chapter 16, Clutches, Brakes, Couplings, and Flywheels. Chapter 17, Flexible Mechanical Elements. Chapter 18, Power Transmission Case Study. This new chapter provides a complete case study of a double reduction power transmission. The focus is on providing an example for student designers of the process of integrating topics from multiple chapters. Instructors are encouraged to include one of the variations of this case study as a design project in the course. Student feedback consistently shows that this type of project is one of the most valuable aspects of a rst course in machine design. This chapter can be utilized in a tutorial fashion for students working through a similar design. Part 4: Analysis Tools Part 4 includes a new chapter on nite element methods, and a new location for the chapter on statistical considerations. Instructors can reference these chapters as needed. Chapter 19, Finite Element Analysis. This chapter is intended to provide an introduction to the nite element method, and particularly its application to the machine design process. 4 xviii BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition Front Matter Preface © The McGrawHill Companies, 2008 Mechanical Engineering Design Chapter 20, Statistical Considerations. This chapter is relocated and organized as a tool for users that wish to incorporate statistical concepts into the machine design process. This chapter should be reviewed if Secs. 513, 617, or Chap. 11 are to be covered. Supplements The 8th edition of Shigleys Mechanical Engineering Design features McGraw-Hills ARIS (Assessment Review and Instruction System). ARIS makes homework meaningfuland manageablefor instructors and students. Instructors can assign and grade text-specic homework within the industrys most robust and versatile homework management system. Students can access multimedia learning tools and benet from unlimited practice via algorithmic problems. Go to aris.mhhe.com to learn more and register! The array of tools available to users of Shigleys Mechanical Engineering Design includes: Student Supplements TutorialsPresentation of major concepts, with visuals. Among the topics covered are pressure vessel design, press and shrink ts, contact stresses, and design for static failure. MATLAB® for machine design. Includes visual simulations and accompanying source code. The simulations are linked to examples and problems in the text and demonstrate the ways computational software can be used in mechanical design and analysis. Fundamentals of engineering (FE) exam questions for machine design. Interactive problems and solutions serve as effective, self-testing problems as well as excellent preparation for the FE exam. Algorithmic Problems. Allow step-by-step problem-solving using a recursive computational procedure (algorithm) to create an innite number of problems. Instructor Supplements (under password protection) Solutions manual. The instructors manual contains solutions to most end-of-chapter nondesign problems. PowerPoint® slides. Slides of important gures and tables from the text are provided in PowerPoint format for use in lectures. BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition Front Matter List of Symbols © The McGrawHill Companies, 2008 5 List of Symbols This is a list of common symbols used in machine design and in this book. Specialized use in a subject-matter area often attracts fore and post subscripts and superscripts. To make the table brief enough to be useful the symbol kernels are listed. See Table 141, pp. 715716 for spur and helical gearing symbols, and Table 151, pp. 769770 for bevel-gear symbols. A A a a ˆ a B Bhn B b ˆ b b C c CDF COV c D d E e F f fom G g H HB HRC h hC R ¯ I i i Area, coefcient Area variate Distance, regression constant Regression constant estimate Distance variate Coefcient Brinell hardness Variate Distance, Weibull shape parameter, range number, regression constant, width Regression constant estimate Distance variate Basic load rating, bolted-joint constant, center distance, coefcient of variation, column end condition, correction factor, specic heat capacity, spring index Distance, viscous damping, velocity coefcient Cumulative distribution function Coefcient of variation Distance variate Helix diameter Diameter, distance Modulus of elasticity, energy, error Distance, eccentricity, efciency, Naperian logarithmic base Force, fundamental dimension force Coefcient of friction, frequency, function Figure of merit Torsional modulus of elasticity Acceleration due to gravity, function Heat, power Brinell hardness Rockwell C-scale hardness Distance, lm thickness Combined overall coefcient of convection and radiation heat transfer Integral, linear impulse, mass moment of inertia, second moment of area Index Unit vector in x-direction xxiii 6 xxiv BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition Front Matter List of Symbols © The McGrawHill Companies, 2008 Mechanical Engineering Design J j K k k L LN l M M m N N n nd P PDF p Q q R R r r S S s T T t U U u V v W W w w X x x Y y y Z z z Mechanical equivalent of heat, polar second moment of area, geometry factor Unit vector in the y-direction Service factor, stress-concentration factor, stress-augmentation factor, torque coefcient Marin endurance limit modifying factor, spring rate k variate, unit vector in the z-direction Length, life, fundamental dimension length Lognormal distribution Length Fundamental dimension mass, moment Moment vector, moment variate Mass, slope, strain-strengthening exponent Normal force, number, rotational speed Normal distribution Load factor, rotational speed, safety factor Design factor Force, pressure, diametral pitch Probability density function Pitch, pressure, probability First moment of area, imaginary force, volume Distributed load, notch sensitivity Radius, reaction force, reliability, Rockwell hardness, stress ratio Vector reaction force Correlation coefcient, radius Distance vector Sommerfeld number, strength S variate Distance, sample standard deviation, stress Temperature, tolerance, torque, fundamental dimension time Torque vector, torque variate Distance, Students t-statistic, time, tolerance Strain energy Uniform distribution Strain energy per unit volume Linear velocity, shear force Linear velocity Cold-work factor, load, weight Weibull distribution Distance, gap, load intensity Vector distance Coordinate, truncated number Coordinate, true value of a number, Weibull parameter x variate Coordinate Coordinate, deection y variate Coordinate, section modulus, viscosity Standard deviation of the unit normal distribution Variate of z BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition Front Matter List of Symbols © The McGrawHill Companies, 2008 List of Symbols α β δ ǫ ε Ŵ γ λ L µ ν ω φ ψ ρ σ σ S σ ˆ τ θ ¢ $ 7 xxv Coefcient, coefcient of linear thermal expansion, end-condition for springs, thread angle Bearing angle, coefcient Change, deection Deviation, elongation Eccentricity ratio, engineering (normal) strain Normal distribution with a mean of 0 and a standard deviation of s True or logarithmic normal strain Gamma function Pitch angle, shear strain, specic weight Slenderness ratio for springs Unit lognormal with a mean of l and a standard deviation equal to COV Absolute viscosity, population mean Poisson ratio Angular velocity, circular frequency Angle, wave length Slope integral Radius of curvature Normal stress Von Mises stress Normal stress variate Standard deviation Shear stress Shear stress variate Angle, Weibull characteristic parameter Cost per unit weight Cost 8 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition PART I. Basics Introduction 1 Basics © The McGrawHill Companies, 2008 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition I. Basics © The McGrawHill Companies, 2008 1. Introduction to Mechanical Engineering Design 9 1 Introduction to Mechanical Engineering Design Chapter Outline 11 Design 12 Mechanical Engineering Design 13 Phases and Interactions of the Design Process 14 Design Tools and Resources 15 The Design Engineers Professional Responsibilities 16 Standards and Codes 17 Economics 18 Safety and Product Liability 19 Stress and Strength 4 5 5 8 10 12 12 15 15 110 Uncertainty 111 Design Factor and Factor of Safety 112 Reliability 113 Dimensions and Tolerances 114 Units 115 Calculations and Signicant Figures 116 Power Transmission Case Study Specications 16 17 18 19 21 22 23 3 10 4 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition I. Basics 1. Introduction to Mechanical Engineering Design © The McGrawHill Companies, 2008 Mechanical Engineering Design Mechanical design is a complex undertaking, requiring many skills. Extensive relationships need to be subdivided into a series of simple tasks. The complexity of the subject requires a sequence in which ideas are introduced and iterated. We rst address the nature of design in general, and then mechanical engineering design in particular. Design is an iterative process with many interactive phases. Many resources exist to support the designer, including many sources of information and an abundance of computational design tools. The design engineer needs not only to develop competence in their eld but must also cultivate a strong sense of responsibility and professional work ethic. There are roles to be played by codes and standards, ever-present economics, safety, and considerations of product liability. The survival of a mechanical component is often related through stress and strength. Matters of uncertainty are ever-present in engineering design and are typically addressed by the design factor and factor of safety, either in the form of a deterministic (absolute) or statistical sense. The latter, statistical approach, deals with a designs reliability and requires good statistical data. In mechanical design, other considerations include dimensions and tolerances, units, and calculations. The book consists of four parts. Part 1, Basics, begins by explaining some differences between design and analysis and introducing some fundamental notions and approaches to design. It continues with three chapters reviewing material properties, stress analysis, and stiffness and deection analysis, which are the key principles necessary for the remainder of the book. Part 2, Failure Prevention, consists of two chapters on the prevention of failure of mechanical parts. Why machine parts fail and how they can be designed to prevent failure are difcult questions, and so we take two chapters to answer them, one on preventing failure due to static loads, and the other on preventing fatigue failure due to time-varying, cyclic loads. In Part 3, Design of Mechanical Elements, the material of Parts 1 and 2 is applied to the analysis, selection, and design of specic mechanical elements such as shafts, fasteners, weldments, springs, rolling contact bearings, lm bearings, gears, belts, chains, and wire ropes. Part 4, Analysis Tools, provides introductions to two important methods used in mechanical design, nite element analysis and statistical analysis. This is optional study material, but some sections and examples in Parts 1 to 3 demonstrate the use of these tools. There are two appendixes at the end of the book. Appendix A contains many useful tables referenced throughout the book. Appendix B contains answers to selected end-of-chapter problems. 11 Design To design is either to formulate a plan for the satisfaction of a specied need or to solve a problem. If the plan results in the creation of something having a physical reality, then the product must be functional, safe, reliable, competitive, usable, manufacturable, and marketable. Design is an innovative and highly iterative process. It is also a decision-making process. Decisions sometimes have to be made with too little information, occasionally with just the right amount of information, or with an excess of partially contradictory information. Decisions are sometimes made tentatively, with the right reserved to adjust as more becomes known. The point is that the engineering designer has to be personally comfortable with a decision-making, problem-solving role. BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition I. Basics © The McGrawHill Companies, 2008 1. Introduction to Mechanical Engineering Design Introduction to Mechanical Engineering Design 11 5 Design is a communication-intensive activity in which both words and pictures are used, and written and oral forms are employed. Engineers have to communicate effectively and work with people of many disciplines. These are important skills, and an engineers success depends on them. A designers personal resources of creativeness, communicative ability, and problemsolving skill are intertwined with knowledge of technology and rst principles. Engineering tools (such as mathematics, statistics, computers, graphics, and languages) are combined to produce a plan that, when carried out, produces a product that is functional, safe, reliable, competitive, usable, manufacturable, and marketable, regardless of who builds it or who uses it. 12 Mechanical Engineering Design Mechanical engineers are associated with the production and processing of energy and with providing the means of production, the tools of transportation, and the techniques of automation. The skill and knowledge base are extensive. Among the disciplinary bases are mechanics of solids and uids, mass and momentum transport, manufacturing processes, and electrical and information theory. Mechanical engineering design involves all the disciplines of mechanical engineering. Real problems resist compartmentalization. A simple journal bearing involves uid ow, heat transfer, friction, energy transport, material selection, thermomechanical treatments, statistical descriptions, and so on. A building is environmentally controlled. The heating, ventilation, and air-conditioning considerations are sufciently specialized that some speak of heating, ventilating, and air-conditioning design as if it is separate and distinct from mechanical engineering design. Similarly, internal-combustion engine design, turbomachinery design, and jet-engine design are sometimes considered discrete entities. Here, the leading string of words preceding the word design is merely a product descriptor. Similarly, there are phrases such as machine design, machine-element design, machine-component design, systems design, and uid-power design. All of these phrases are somewhat more focused examples of mechanical engineering design. They all draw on the same bodies of knowledge, are similarly organized, and require similar skills. 13 Phases and Interactions of the Design Process What is the design process? How does it begin? Does the engineer simply sit down at a desk with a blank sheet of paper and jot down some ideas? What happens next? What factors inuence or control the decisions that have to be made? Finally, how does the design process end? The complete design process, from start to finish, is often outlined as in Fig. 11. The process begins with an identification of a need and a decision to do something about it. After many iterations, the process ends with the presentation of the plans for satisfying the need. Depending on the nature of the design task, several design phases may be repeated throughout the life of the product, from inception to termination. In the next several subsections, we shall examine these steps in the design process in detail. Identication of need generally starts the design process. Recognition of the need and phrasing the need often constitute a highly creative act, because the need may be only a vague discontent, a feeling of uneasiness, or a sensing that something is not right. The need is often not evident at all; recognition is usually triggered by a particular 12 6 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition I. Basics © The McGrawHill Companies, 2008 1. Introduction to Mechanical Engineering Design Mechanical Engineering Design Figure 11 Identification of need The phases in design, acknowledging the many feedbacks and iterations. Definition of problem Synthesis Analysis and optimization Evaluation Iteration Presentation adverse circumstance or a set of random circumstances that arises almost simultaneously. For example, the need to do something about a food-packaging machine may be indicated by the noise level, by a variation in package weight, and by slight but perceptible variations in the quality of the packaging or wrap. There is a distinct difference between the statement of the need and the denition of the problem. The denition of problem is more specic and must include all the specications for the object that is to be designed. The specications are the input and output quantities, the characteristics and dimensions of the space the object must occupy, and all the limitations on these quantities. We can regard the object to be designed as something in a black box. In this case we must specify the inputs and outputs of the box, together with their characteristics and limitations. The specications dene the cost, the number to be manufactured, the expected life, the range, the operating temperature, and the reliability. Specied characteristics can include the speeds, feeds, temperature limitations, maximum range, expected variations in the variables, dimensional and weight limitations, etc. There are many implied specifications that result either from the designers particular environment or from the nature of the problem itself. The manufacturing processes that are available, together with the facilities of a certain plant, constitute restrictions on a designers freedom, and hence are a part of the implied specifications. It may be that a small plant, for instance, does not own cold-working machinery. Knowing this, the designer might select other metal-processing methods that can be performed in the plant. The labor skills available and the competitive situation also constitute implied constraints. Anything that limits the designers freedom of choice is a constraint. Many materials and sizes are listed in suppliers catalogs, for instance, but these are not all easily available and shortages frequently occur. Furthermore, inventory economics requires that a manufacturer stock a minimum number of materials and sizes. An example of a specification is given in Sec. 116. This example is for a case study of a power transmission that is presented throughout this text. The synthesis of a scheme connecting possible system elements is sometimes called the invention of the concept or concept design. This is the rst and most important step in the synthesis task. Various schemes must be proposed, investigated, and BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition I. Basics 1. Introduction to Mechanical Engineering Design © The McGrawHill Companies, 2008 Introduction to Mechanical Engineering Design 13 7 quantied in terms of established metrics.1 As the eshing out of the scheme progresses, analyses must be performed to assess whether the system performance is satisfactory or better, and, if satisfactory, just how well it will perform. System schemes that do not survive analysis are revised, improved, or discarded. Those with potential are optimized to determine the best performance of which the scheme is capable. Competing schemes are compared so that the path leading to the most competitive product can be chosen. Figure 11 shows that synthesis and analysis and optimization are intimately and iteratively related. We have noted, and we emphasize, that design is an iterative process in which we proceed through several steps, evaluate the results, and then return to an earlier phase of the procedure. Thus, we may synthesize several components of a system, analyze and optimize them, and return to synthesis to see what effect this has on the remaining parts of the system. For example, the design of a system to transmit power requires attention to the design and selection of individual components (e.g., gears, bearings, shaft). However, as is often the case in design, these components are not independent. In order to design the shaft for stress and deection, it is necessary to know the applied forces. If the forces are transmitted through gears, it is necessary to know the gear specications in order to determine the forces that will be transmitted to the shaft. But stock gears come with certain bore sizes, requiring knowledge of the necessary shaft diameter. Clearly, rough estimates will need to be made in order to proceed through the process, rening and iterating until a nal design is obtained that is satisfactory for each individual component as well as for the overall design specications. Throughout the text we will elaborate on this process for the case study of a power transmission design. Both analysis and optimization require that we construct or devise abstract models of the system that will admit some form of mathematical analysis. We call these models mathematical models. In creating them it is our hope that we can nd one that will simulate the real physical system very well. As indicated in Fig. 11, evaluation is a signicant phase of the total design process. Evaluation is the nal proof of a successful design and usually involves the testing of a prototype in the laboratory. Here we wish to discover if the design really satises the needs. Is it reliable? Will it compete successfully with similar products? Is it economical to manufacture and to use? Is it easily maintained and adjusted? Can a prot be made from its sale or use? How likely is it to result in product-liability lawsuits? And is insurance easily and cheaply obtained? Is it likely that recalls will be needed to replace defective parts or systems? Communicating the design to others is the nal, vital presentation step in the design process. Undoubtedly, many great designs, inventions, and creative works have been lost to posterity simply because the originators were unable or unwilling to explain their accomplishments to others. Presentation is a selling job. The engineer, when presenting a new solution to administrative, management, or supervisory persons, is attempting to sell or to prove to them that this solution is a better one. Unless this can be done successfully, the time and effort spent on obtaining the solution have been largely wasted. When designers sell a new idea, they also sell themselves. If they are repeatedly successful in selling ideas, designs, and new solutions to management, they begin to receive salary increases and promotions; in fact, this is how anyone succeeds in his or her profession. 1 An excellent reference for this topic is presented by Stuart Pugh, Total DesignIntegrated Methods for Successful Product Engineering, Addison-Wesley, 1991. A description of the Pugh method is also provided in Chap. 8, David G. Ullman, The Mechanical Design Process, 3rd ed., McGraw-Hill, 2003. 14 8 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition I. Basics © The McGrawHill Companies, 2008 1. Introduction to Mechanical Engineering Design Mechanical Engineering Design Design Considerations Sometimes the strength required of an element in a system is an important factor in the determination of the geometry and the dimensions of the element. In such a situation we say that strength is an important design consideration. When we use the expression design consideration, we are referring to some characteristic that inuences the design of the element or, perhaps, the entire system. Usually quite a number of such characteristics must be considered and prioritized in a given design situation. Many of the important ones are as follows (not necessarily in order of importance): 1 2 3 4 5 6 7 8 9 10 11 12 13 Functionality Strength/stress Distortion/deection/stiffness Wear Corrosion Safety Reliability Manufacturability Utility Cost Friction Weight Life 14 15 16 17 18 19 20 21 22 23 24 25 26 Noise Styling Shape Size Control Thermal properties Surface Lubrication Marketability Maintenance Volume Liability Remanufacturing/resource recovery Some of these characteristics have to do directly with the dimensions, the material, the processing, and the joining of the elements of the system. Several characteristics may be interrelated, which affects the conguration of the total system. 14 Design Tools and Resources Today, the engineer has a great variety of tools and resources available to assist in the solution of design problems. Inexpensive microcomputers and robust computer software packages provide tools of immense capability for the design, analysis, and simulation of mechanical components. In addition to these tools, the engineer always needs technical information, either in the form of basic science/engineering behavior or the characteristics of specic off-the-shelf components. Here, the resources can range from science/engineering textbooks to manufacturers brochures or catalogs. Here too, the computer can play a major role in gathering information.2 Computational Tools Computer-aided design (CAD) software allows the development of three-dimensional (3-D) designs from which conventional two-dimensional orthographic views with automatic dimensioning can be produced. Manufacturing tool paths can be generated from the 3-D models, and in some cases, parts can be created directly from a 3-D database by using a rapid prototyping and manufacturing method (stereolithography)paperless manufacturing! Another advantage of a 3-D database is that it allows rapid and accurate calculations of mass properties such as mass, location of the center of gravity, and mass moments of inertia. Other geometric properties such as areas and distances between points are likewise easily obtained. There are a great many CAD software packages available such 2 An excellent and comprehensive discussion of the process of gathering information can be found in Chap. 4, George E. Dieter, Engineering Design, A Materials and Processing Approach, 3rd ed., McGraw-Hill, New York, 2000. BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition I. Basics 1. Introduction to Mechanical Engineering Design © The McGrawHill Companies, 2008 Introduction to Mechanical Engineering Design 15 9 as Aries, AutoCAD, CadKey, I-Deas, Unigraphics, Solid Works, and ProEngineer, to name a few. The term computer-aided engineering (CAE) generally applies to all computerrelated engineering applications. With this denition, CAD can be considered as a subset of CAE. Some computer software packages perform specic engineering analysis and/or simulation tasks that assist the designer, but they are not considered a tool for the creation of the design that CAD is. Such software ts into two categories: engineeringbased and non-engineering-specic. Some examples of engineering-based software for mechanical engineering applicationssoftware that might also be integrated within a CAD systeminclude nite-element analysis (FEA) programs for analysis of stress and deection (see Chap. 19), vibration, and heat transfer (e.g., Algor, ANSYS, and MSC/NASTRAN); computational uid dynamics (CFD) programs for uid-ow analysis and simulation (e.g., CFD++, FIDAP, and Fluent); and programs for simulation of dynamic force and motion in mechanisms (e.g., ADAMS, DADS, and Working Model). Examples of non-engineering-specic computer-aided applications include software for word processing, spreadsheet software (e.g., Excel, Lotus, and Quattro-Pro), and mathematical solvers (e.g., Maple, MathCad, Matlab, Mathematica, and TKsolver). Your instructor is the best source of information about programs that may be available to you and can recommend those that are useful for specic tasks. One caution, however: Computer software is no substitute for the human thought process. You are the driver here; the computer is the vehicle to assist you on your journey to a solution. Numbers generated by a computer can be far from the truth if you entered incorrect input, if you misinterpreted the application or the output of the program, if the program contained bugs, etc. It is your responsibility to assure the validity of the results, so be careful to check the application and results carefully, perform benchmark testing by submitting problems with known solutions, and monitor the software company and user-group newsletters. Acquiring Technical Information We currently live in what is referred to as the information age, where information is generated at an astounding pace. It is difcult, but extremely important, to keep abreast of past and current developments in ones eld of study and occupation. The reference in Footnote 2 provides an excellent description of the informational resources available and is highly recommended reading for the serious design engineer. Some sources of information are: Libraries (community, university, and private). Engineering dictionaries and encyclopedias, textbooks, monographs, handbooks, indexing and abstract services, journals, translations, technical reports, patents, and business sources/brochures/catalogs. Government sources. Departments of Defense, Commerce, Energy, and Transportation; NASA; Government Printing Ofce; U.S. Patent and Trademark Ofce; National Technical Information Service; and National Institute for Standards and Technology. Professional societies. American Society of Mechanical Engineers, Society of Manufacturing Engineers, Society of Automotive Engineers, American Society for Testing and Materials, and American Welding Society. Commercial vendors. Catalogs, technical literature, test data, samples, and cost information. Internet. The computer network gateway to websites associated with most of the categories listed above.3 3 Some helpful Web resources, to name a few, include www.globalspec.com, www.engnetglobal.com, www.efunda.com, www.thomasnet.com, and www.uspto.gov. 16 10 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition I. Basics 1. Introduction to Mechanical Engineering Design © The McGrawHill Companies, 2008 Mechanical Engineering Design This list is not complete. The reader is urged to explore the various sources of information on a regular basis and keep records of the knowledge gained. 15 The Design Engineers Professional Responsibilities In general, the design engineer is required to satisfy the needs of customers (management, clients, consumers, etc.) and is expected to do so in a competent, responsible, ethical, and professional manner. Much of engineering course work and practical experience focuses on competence, but when does one begin to develop engineering responsibility and professionalism? To start on the road to success, you should start to develop these characteristics early in your educational program. You need to cultivate your professional work ethic and process skills before graduation, so that when you begin your formal engineering career, you will be prepared to meet the challenges. It is not obvious to some students, but communication skills play a large role here, and it is the wise student who continuously works to improve these skillseven if it is not a direct requirement of a course assignment! Success in engineering (achievements, promotions, raises, etc.) may in large part be due to competence but if you cannot communicate your ideas clearly and concisely, your technical prociency may be compromised. You can start to develop your communication skills by keeping a neat and clear journal/logbook of your activities, entering dated entries frequently. (Many companies require their engineers to keep a journal for patent and liability concerns.) Separate journals should be used for each design project (or course subject). When starting a project or problem, in the denition stage, make journal entries quite frequently. Others, as well as yourself, may later question why you made certain decisions. Good chronological records will make it easier to explain your decisions at a later date. Many engineering students see themselves after graduation as practicing engineers designing, developing, and analyzing products and processes and consider the need of good communication skills, either oral or writing, as secondary. This is far from the truth. Most practicing engineers spend a good deal of time communicating with others, writing proposals and technical reports, and giving presentations and interacting with engineering and nonengineering support personnel. You have the time now to sharpen your communication skills. When given an assignment to write or make any presentation, technical or nontechnical, accept it enthusiastically, and work on improving your communication skills. It will be time well spent to learn the skills now rather than on the job. When you are working on a design problem, it is important that you develop a systematic approach. Careful attention to the following action steps will help you to organize your solution processing technique. Understand the problem. Problem denition is probably the most signicant step in the engineering design process. Carefully read, understand, and rene the problem statement. Identify the known. From the rened problem statement, describe concisely what information is known and relevant. Identify the unknown and formulate the solution strategy. State what must be determined, in what order, so as to arrive at a solution to the problem. Sketch the component or system under investigation, identifying known and unknown parameters. Create a owchart of the steps necessary to reach the nal solution. The steps may require the use of free-body diagrams; material properties from tables; equations BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition I. Basics 1. Introduction to Mechanical Engineering Design © The McGrawHill Companies, 2008 Introduction to Mechanical Engineering Design 17 11 from rst principles, textbooks, or handbooks relating the known and unknown parameters; experimentally or numerically based charts; specic computational tools as discussed in Sec. 14; etc. State all assumptions and decisions. Real design problems generally do not have unique, ideal, closed-form solutions. Selections, such as choice of materials, and heat treatments, require decisions. Analyses require assumptions related to the modeling of the real components or system. All assumptions and decisions should be identied and recorded. Analyze the problem. Using your solution strategy in conjunction with your decisions and assumptions, execute the analysis of the problem. Reference the sources of all equations, tables, charts, software results, etc. Check the credibility of your results. Check the order of magnitude, dimensionality, trends, signs, etc. Evaluate your solution. Evaluate each step in the solution, noting how changes in strategy, decisions, assumptions, and execution might change the results, in positive or negative ways. If possible, incorporate the positive changes in your nal solution. Present your solution. Here is where your communication skills are important. At this point, you are selling yourself and your technical abilities. If you cannot skillfully explain what you have done, some or all of your work may be misunderstood and unaccepted. Know your audience. As stated earlier, all design processes are interactive and iterative. Thus, it may be necessary to repeat some or all of the above steps more than once if less than satisfactory results are obtained. In order to be effective, all professionals must keep current in their elds of endeavor. The design engineer can satisfy this in a number of ways by: being an active member of a professional society such as the American Society of Mechanical Engineers (ASME), the Society of Automotive Engineers (SAE), and the Society of Manufacturing Engineers (SME); attending meetings, conferences, and seminars of societies, manufacturers, universities, etc.; taking specic graduate courses or programs at universities; regularly reading technical and professional journals; etc. An engineers education does not end at graduation. The design engineers professional obligations include conducting activities in an ethical manner. Reproduced here is the Engineers Creed from the National Society of Professional Engineers (NSPE)4: As a Professional Engineer I dedicate my professional knowledge and skill to the advancement and betterment of human welfare. I pledge: To give the utmost of performance; To participate in none but honest enterprise; To live and work according to the laws of man and the highest standards of professional conduct; To place service before prot, the honor and standing of the profession before personal advantage, and the public welfare above all other considerations. In humility and with need for Divine Guidance, I make this pledge. 4 Adopted by the National Society of Professional Engineers, June 1954. The Engineers Creed. Reprinted by permission of the National Society of Professional Engineers. This has been expanded and revised by NSPE. For the current revision, January 2006, see the website www.nspe.org/ethics/ehl-code.asp, or the pdf le, www.nspe.org/ethics/code-2006-Jan.pdf. 18 12 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition I. Basics 1. Introduction to Mechanical Engineering Design © The McGrawHill Companies, 2008 Mechanical Engineering Design 16 Standards and Codes A standard is a set of specications for parts, materials, or processes intended to achieve uniformity, efciency, and a specied quality. One of the important purposes of a standard is to place a limit on the number of items in the specications so as to provide a reasonable inventory of tooling, sizes, shapes, and varieties. A code is a set of specications for the analysis, design, manufacture, and construction of something. The purpose of a code is to achieve a specied degree of safety, efciency, and performance or quality. It is important to observe that safety codes do not imply absolute safety. In fact, absolute safety is impossible to obtain. Sometimes the unexpected event really does happen. Designing a building to withstand a 120 mi/h wind does not mean that the designers think a 140 mi/h wind is impossible; it simply means that they think it is highly improbable. All of the organizations and societies listed below have established specications for standards and safety or design codes. The name of the organization provides a clue to the nature of the standard or code. Some of the standards and codes, as well as addresses, can be obtained in most technical libraries. The organizations of interest to mechanical engineers are: Aluminum Association (AA) American Gear Manufacturers Association (AGMA) American Institute of Steel Construction (AISC) American Iron and Steel Institute (AISI) American National Standards Institute (ANSI)5 ASM International6 American Society of Mechanical Engineers (ASME) American Society of Testing and Materials (ASTM) American Welding Society (AWS) American Bearing Manufacturers Association (ABMA)7 British Standards Institution (BSI) Industrial Fasteners Institute (IFI) Institution of Mechanical Engineers (I. Mech. E.) International Bureau of Weights and Measures (BIPM) International Standards Organization (ISO) National Institute for Standards and Technology (NIST)8 Society of Automotive Engineers (SAE) 17 Economics The consideration of cost plays such an important role in the design decision process that we could easily spend as much time in studying the cost factor as in the study of the entire subject of design. Here we introduce only a few general concepts and simple rules. 5 In 1966 the American Standards Association (ASA) changed its name to the United States of America Standards Institute (USAS). Then, in 1969, the name was again changed, to American National Standards Institute, as shown above and as it is today. This means that you may occasionally nd ANSI standards designated as ASA or USAS. 6 Formally American Society for Metals (ASM). Currently the acronym ASM is undened. 7 In 1993 the Anti-Friction Bearing Manufacturers Association (AFBMA) changed its name to the American Bearing Manufacturers Association (ABMA). 8 Former National Bureau of Standards (NBS). BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition I. Basics 1. Introduction to Mechanical Engineering Design © The McGrawHill Companies, 2008 Introduction to Mechanical Engineering Design 19 13 First, observe that nothing can be said in an absolute sense concerning costs. Materials and labor usually show an increasing cost from year to year. But the costs of processing the materials can be expected to exhibit a decreasing trend because of the use of automated machine tools and robots. The cost of manufacturing a single product will vary from city to city and from one plant to another because of overhead, labor, taxes, and freight differentials and the inevitable slight manufacturing variations. Standard Sizes The use of standard or stock sizes is a rst principle of cost reduction. An engineer who species an AISI 1020 bar of hot-rolled steel 53 mm square has added cost to the product, provided that a bar 50 or 60 mm square, both of which are preferred sizes, would do equally well. The 53-mm size can be obtained by special order or by rolling or machining a 60-mm square, but these approaches add cost to the product. To ensure that standard or preferred sizes are specied, designers must have access to stock lists of the materials they employ. A further word of caution regarding the selection of preferred sizes is necessary. Although a great many sizes are usually listed in catalogs, they are not all readily available. Some sizes are used so infrequently that they are not stocked. A rush order for such sizes may mean more on expense and delay. Thus you should also have access to a list such as those in Table A17 for preferred inch and millimeter sizes. There are many purchased parts, such as motors, pumps, bearings, and fasteners, that are specied by designers. In the case of these, too, you should make a special effort to specify parts that are readily available. Parts that are made and sold in large quantities usually cost somewhat less than the odd sizes. The cost of rolling bearings, for example, depends more on the quantity of production by the bearing manufacturer than on the size of the bearing. Large Tolerances Among the effects of design specications on costs, tolerances are perhaps most signicant. Tolerances, manufacturing processes, and surface nish are interrelated and inuence the producibility of the end product in many ways. Close tolerances may necessitate additional steps in processing and inspection or even render a part completely impractical to produce economically. Tolerances cover dimensional variation and surface-roughness range and also the variation in mechanical properties resulting from heat treatment and other processing operations. Since parts having large tolerances can often be produced by machines with higher production rates, costs will be significantly smaller. Also, fewer such parts will be rejected in the inspection process, and they are usually easier to assemble. A plot of cost versus tolerance/machining process is shown in Fig. 12, and illustrates the drastic increase in manufacturing cost as tolerance diminishes with finer machining processing. Breakeven Points Sometimes it happens that, when two or more design approaches are compared for cost, the choice between the two depends on a set of conditions such as the quantity of production, the speed of the assembly lines, or some other condition. There then occurs a point corresponding to equal cost, which is called the breakeven point. 20 14 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition I. Basics © The McGrawHill Companies, 2008 1. Introduction to Mechanical Engineering Design Mechanical Engineering Design Figure 12 Costs, % Cost versus tolerance/ machining process. (From David G. Ullman, The Mechanical Design Process, 3rd ed., McGraw-Hill, New York, 2003.) 400 380 360 340 320 300 280 260 240 220 200 180 160 140 120 100 80 60 40 20 Material: steel 0.030 0.015 0.010 0.005 0.003 0.001 0.0005 0.00025 0.063 0.025 0.012 0.006 Semifinish turn Finish turn Grind Nominal tolerances (inches) 0.75 0.50 0.50 0.125 Nominal tolerance (mm) Rough turn Hone Machining operations Figure 13 140 A breakeven point. Breakeven point 120 Cost, $ 100 Automatic screw machine 80 60 Hand screw machine 40 20 0 0 20 40 60 Production 80 100 As an example, consider a situation in which a certain part can be manufactured at the rate of 25 parts per hour on an automatic screw machine or 10 parts per hour on a hand screw machine. Let us suppose, too, that the setup time for the automatic is 3 h and that the labor cost for either machine is $20 per hour, including overhead. Figure 13 is a graph of cost versus production by the two methods. The breakeven point for this example corresponds to 50 parts. If the desired production is greater than 50 parts, the automatic machine should be used. BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition I. Basics © The McGrawHill Companies, 2008 1. Introduction to Mechanical Engineering Design Introduction to Mechanical Engineering Design 21 15 Cost Estimates There are many ways of obtaining relative cost gures so that two or more designs can be roughly compared. A certain amount of judgment may be required in some instances. For example, we can compare the relative value of two automobiles by comparing the dollar cost per pound of weight. Another way to compare the cost of one design with another is simply to count the number of parts. The design having the smaller number of parts is likely to cost less. Many other cost estimators can be used, depending upon the application, such as area, volume, horsepower, torque, capacity, speed, and various performance ratios.9 18 Safety and Product Liability The strict liability concept of product liability generally prevails in the United States. This concept states that the manufacturer of an article is liable for any damage or harm that results because of a defect. And it doesnt matter whether the manufacturer knew about the defect, or even could have known about it. For example, suppose an article was manufactured, say, 10 years ago. And suppose at that time the article could not have been considered defective on the basis of all technological knowledge then available. Ten years later, according to the concept of strict liability, the manufacturer is still liable. Thus, under this concept, the plaintiff needs only to prove that the article was defective and that the defect caused some damage or harm. Negligence of the manufacturer need not be proved. The best approaches to the prevention of product liability are good engineering in analysis and design, quality control, and comprehensive testing procedures. Advertising managers often make glowing promises in the warranties and sales literature for a product. These statements should be reviewed carefully by the engineering staff to eliminate excessive promises and to insert adequate warnings and instructions for use. 19 Stress and Strength The survival of many products depends on how the designer adjusts the maximum stresses in a component to be less than the components strength at specic locations of interest. The designer must allow the maximum stress to be less than the strength by a sufcient margin so that despite the uncertainties, failure is rare. In focusing on the stress-strength comparison at a critical (controlling) location, we often look for strength in the geometry and condition of use. Strengths are the magnitudes of stresses at which something of interest occurs, such as the proportional limit, 0.2 percent-offset yielding, or fracture. In many cases, such events represent the stress level at which loss of function occurs. Strength is a property of a material or of a mechanical element. The strength of an element depends on the choice, the treatment, and the processing of the material. Consider, for example, a shipment of springs. We can associate a strength with a specic spring. When this spring is incorporated into a machine, external forces are applied that result in load-induced stresses in the spring, the magnitudes of which depend on its geometry and are independent of the material and its processing. If the spring is removed from the machine unharmed, the stress due to the external forces will return 9 For an overview of estimating manufacturing costs, see Chap. 11, Karl T. Ulrich and Steven D. Eppinger, Product Design and Development, 3rd ed., McGraw-Hill, New York, 2004. 22 16 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition I. Basics 1. Introduction to Mechanical Engineering Design © The McGrawHill Companies, 2008 Mechanical Engineering Design to zero. But the strength remains as one of the properties of the spring. Remember, then, that strength is an inherent property of a part, a property built into the part because of the use of a particular material and process. Various metalworking and heat-treating processes, such as forging, rolling, and cold forming, cause variations in the strength from point to point throughout a part. The spring cited above is quite likely to have a strength on the outside of the coils different from its strength on the inside because the spring has been formed by a cold winding process, and the two sides may not have been deformed by the same amount. Remember, too, therefore, that a strength value given for a part may apply to only a particular point or set of points on the part. In this book we shall use the capital letter S to denote strength, with appropriate subscripts to denote the type of strength. Thus, Ss is a shear strength, Sy a yield strength, and Su an ultimate strength. In accordance with accepted engineering practice, we shall employ the Greek letters σ (sigma) and τ (tau) to designate normal and shear stresses, respectively. Again, various subscripts will indicate some special characteristic. For example, σ1 is a principal stress, σ y a stress component in the y direction, and σr a stress component in the radial direction. Stress is a state property at a specic point within a body, which is a function of load, geometry, temperature, and manufacturing processing. In an elementary course in mechanics of materials, stress related to load and geometry is emphasized with some discussion of thermal stresses. However, stresses due to heat treatments, molding, assembly, etc. are also important and are sometimes neglected. A review of stress analysis for basic load states and geometry is given in Chap. 3. 110 Uncertainty Uncertainties in machinery design abound. Examples of uncertainties concerning stress and strength include Composition of material and the effect of variation on properties. Variations in properties from place to place within a bar of stock. Effect of processing locally, or nearby, on properties. Effect of nearby assemblies such as weldments and shrink ts on stress conditions. Effect of thermomechanical treatment on properties. Intensity and distribution of loading. Validity of mathematical models used to represent reality. Intensity of stress concentrations. Inuence of time on strength and geometry. Effect of corrosion. Effect of wear. Uncertainty as to the length of any list of uncertainties. Engineers must accommodate uncertainty. Uncertainty always accompanies change. Material properties, load variability, fabrication delity, and validity of mathematical models are among concerns to designers. There are mathematical methods to address uncertainties. The primary techniques are the deterministic and stochastic methods. The deterministic method establishes a BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition I. Basics © The McGrawHill Companies, 2008 1. Introduction to Mechanical Engineering Design Introduction to Mechanical Engineering Design 23 17 design factor based on the absolute uncertainties of a loss-of-function parameter and a maximum allowable parameter. Here the parameter can be load, stress, deection, etc. Thus, the design factor n d is dened as nd = loss-of-function parameter maximum allowable parameter (11) If the parameter is load, then the maximum allowable load can be found from Maximum allowable load = loss-of-function load nd (12) EXAMPLE 11 Consider that the maximum load on a structure is known with an uncertainty of ±20 percent, and the load causing failure is known within ±15 percent. If the load causing failure is nominally 2000 lbf, determine the design factor and the maximum allowable load that will offset the absolute uncertainties. Solution To account for its uncertainty, the loss-of-function load must increase to 1/0.85, whereas the maximum allowable load must decrease to 1/1.2. Thus to offset the absolute uncertainties the design factor should be Answer nd = 1/0.85 = 1.4 1/1.2 From Eq. (12), the maximum allowable load is found to be Answer Maximum allowable load = 2000 = 1400 lbf 1.4 Stochastic methods (see Chap. 20) are based on the statistical nature of the design parameters and focus on the probability of survival of the designs function (that is, on reliability). Sections 513 and 617 demonstrate how this is accomplished. 111 Design Factor and Factor of Safety A general approach to the allowable load versus loss-of-function load problem is the deterministic design factor method, and sometimes called the classical method of design. The fundamental equation is Eq. (11) where nd is called the design factor. All loss-of-function modes must be analyzed, and the mode leading to the smallest design factor governs. After the design is completed, the actual design factor may change as a result of changes such as rounding up to a standard size for a cross section or using off-the-shelf components with higher ratings instead of employing what is calculated by using the design factor. The factor is then referred to as the factor of safety, n. The factor of safety has the same denition as the design factor, but it generally differs numerically. Since stress may not vary linearly with load (see Sec. 319), using load as the loss-of-function parameter may not be acceptable. It is more common then to express 24 18 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition I. Basics © The McGrawHill Companies, 2008 1. Introduction to Mechanical Engineering Design Mechanical Engineering Design the design factor in terms of a stress and a relevant strength. Thus Eq. (11) can be rewritten as nd = S loss-of-function strength = allowable stress σ (or τ ) (13) The stress and strength terms in Eq. (13) must be of the same type and units. Also, the stress and strength must apply to the same critical location in the part. EXAMPLE 12 Solution A rod with a cross-sectional area of A and loaded in tension with an axial force of P 2000 lbf undergoes a stress of σ = P/A. Using a material strength of 24 kpsi and a design factor of 3.0, determine the minimum diameter of a solid circular rod. Using Table A17, select a preferred fractional diameter and determine the rods factor of safety. Since A = πd 2/4, and σ = S / n d , then σ= S P 2 000 24 000 = = = nd 3 A π d 2/4 or, Answer d= 4 Pn d πS 1/2 = 4(2000)3 π (24 000) 1/2 = 0.564 in From Table A17, the next higher preferred size is 5 in 0.625 in. Thus, according to 8 the same equation developed earlier, the factor of safety n is Answer n= π (24 000)0.6252 π Sd 2 = = 3.68 4P 4(2000) Thus rounding the diameter has increased the actual design factor. 112 Reliability In these days of greatly increasing numbers of liability lawsuits and the need to conform to regulations issued by governmental agencies such as EPA and OSHA, it is very important for the designer and the manufacturer to know the reliability of their product. The reliability method of design is one in which we obtain the distribution of stresses and the distribution of strengths and then relate these two in order to achieve an acceptable success rate. The statistical measure of the probability that a mechanical element will not fail in use is called the reliability of that element. The reliability R can be expressed by a number having the range 0 R 1. A reliability of R = 0.90 means that there is a 90 percent chance that the part will perform its proper function without failure. The failure of 6 parts out of every 1000 manufactured might be considered an acceptable failure rate for a certain class of products. This represents a reliability of R =1 or 99.4 percent. 6 = 0.994 1000 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition I. Basics © The McGrawHill Companies, 2008 1. Introduction to Mechanical Engineering Design Introduction to Mechanical Engineering Design 25 19 In the reliability method of design, the designers task is to make a judicious selection of materials, processes, and geometry (size) so as to achieve a specic reliability goal. Thus, if the objective reliability is to be 99.4 percent, as above, what combination of materials, processing, and dimensions is needed to meet this goal? Analyses that lead to an assessment of reliability address uncertainties, or their estimates, in parameters that describe the situation. Stochastic variables such as stress, strength, load, or size are described in terms of their means, standard deviations, and distributions. If bearing balls are produced by a manufacturing process in which a diameter distribution is created, we can say upon choosing a ball that there is uncertainty as to size. If we wish to consider weight or moment of inertia in rolling, this size uncertainty can be considered to be propagated to our knowledge of weight or inertia. There are ways of estimating the statistical parameters describing weight and inertia from those describing size and density. These methods are variously called propagation of error, propagation of uncertainty, or propagation of dispersion. These methods are integral parts of analysis or synthesis tasks when probability of failure is involved. It is important to note that good statistical data and estimates are essential to perform an acceptable reliability analysis. This requires a good deal of testing and validation of the data. In many cases, this is not practical and a deterministic approach to the design must be undertaken. 113 Dimensions and Tolerances The following terms are used generally in dimensioning: Nominal size. The size we use in speaking of an element. For example, we may specify a 1 1 -in pipe or a 1 -in bolt. Either the theoretical size or the actual measured size 2 2 may be quite different. The theoretical size of a 1 1 -in pipe is 1.900 in for the outside 2 diameter. And the diameter of the 1 -in bolt, say, may actually measure 0.492 in. 2 Limits. The stated maximum and minimum dimensions. Tolerance. The difference between the two limits. Bilateral tolerance. The variation in both directions from the basic dimension. That is, the basic size is between the two limits, for example, 1.005 ± 0.002 in. The two parts of the tolerance need not be equal. Unilateral tolerance. The basic dimension is taken as one of the limits, and variation is permitted in only one direction, for example, 1.005 +0.004 0.000 in Clearance. A general term that refers to the mating of cylindrical parts such as a bolt and a hole. The word clearance is used only when the internal member is smaller than the external member. The diametral clearance is the measured difference in the two diameters. The radial clearance is the difference in the two radii. Interference. The opposite of clearance, for mating cylindrical parts in which the internal member is larger than the external member. Allowance. The minimum stated clearance or the maximum stated interference for mating parts. When several parts are assembled, the gap (or interference) depends on the dimensions and tolerances of the individual parts. 26 20 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition I. Basics © The McGrawHill Companies, 2008 1. Introduction to Mechanical Engineering Design Mechanical Engineering Design EXAMPLE 13 A shouldered screw contains three hollow right circular cylindrical parts on the screw before a nut is tightened against the shoulder. To sustain the function, the gap w must equal or exceed 0.003 in. The parts in the assembly depicted in Fig. 14 have dimensions and tolerances as follows: a = 1.750 ± 0.003 in b = 0.750 ± 0.001 in c = 0.120 ± 0.005 in d = 0.875 ± 0.001 in Figure 14 a An assembly of three cylindrical sleeves of lengths a, b, and c on a shoulder bolt shank of length a. The gap w is of interest. b c d w All parts except the part with the dimension d are supplied by vendors. The part containing the dimension d is made in-house. (a) Estimate the mean and tolerance on the gap w . (b) What basic value of d will assure that w 0.003 in? Solution (a) The mean value of w is given by w = a b c d = 1.750 0.750 0.120 0.875 = 0.005 in ¯ ¯¯¯¯ Answer For equal bilateral tolerances, the tolerance of the gap is Answer tw = all t = 0.003 + 0.001 + 0.005 + 0.001 = 0.010 in Then, w = 0.005 ± 0.010, and ¯ wmax = w + tw = 0.005 + 0.010 = 0.015 in wmin = w tw = 0.005 0.010 = 0.005 in ¯ Thus, both clearance and interference are possible. ¯ (b) If wmin is to be 0.003 in, then, w = wmin + tw = 0.003 + 0.010 = 0.013 in. Thus, ¯¯¯¯¯ d = a b c w = 1.750 0.750 0.120 0.013 = 0.867 in Answer The previous example represented an absolute tolerance system. Statistically, gap dimensions near the gap limits are rare events. Using a statistical tolerance system, the probability that the gap falls within a given limit is determined.10 This probability deals with the statistical distributions of the individual dimensions. For example, if the distributions of the dimensions in the previous example were normal and the tolerances, t, were 10 See Chapter 20 for a description of the statistical terminology. BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition I. Basics © The McGrawHill Companies, 2008 1. Introduction to Mechanical Engineering Design Introduction to Mechanical Engineering Design 27 21 given in terms of standard deviations of the dimension distribution, the standard devia¯ tion of the gap w would be tw = t 2 . However, this assumes a normal distribution all for the individual dimensions, a rare occurrence. To nd the distribution of w and/or the probability of observing values of w within certain limits requires a computer simulation in most cases. Monte Carlo computer simulations are used to determine the distribution of w by the following approach: Generate an instance for each dimension in the problem by selecting the value of each dimension based on its probability distribution. 2 Calculate w using the values of the dimensions obtained in step 1. 3 Repeat steps 1 and 2 N times to generate the distribution of w . As the number of trials increases, the reliability of the distribution increases. 1 114 Units In the symbolic units equation for Newtons second law, F ma, 2 F = M LT (14) F stands for force, M for mass, L for length, and T for time. Units chosen for any three of these quantities are called base units. The rst three having been chosen, the fourth unit is called a derived unit. When force, length, and time are chosen as base units, the mass is the derived unit and the system that results is called a gravitational system of units. When mass, length, and time are chosen as base units, force is the derived unit and the system that results is called an absolute system of units. In some English-speaking countries, the U.S. customary foot-pound-second system (fps) and the inch-pound-second system (ips) are the two standard gravitational systems most used by engineers. In the fps system the unit of mass is FT 2 (pound-force)(second)2 M= = = lbf · s2 /ft = slug (15) L foot Thus, length, time, and force are the three base units in the fps gravitational system. The unit of force in the fps system is the pound, more properly the pound-force. We shall often abbreviate this unit as lbf; the abbreviation lb is permissible however, since we shall be dealing only with the U.S. customary gravitational system. In some branches of engineering it is useful to represent 1000 lbf as a kilopound and to abbreviate it as kip. Note: In Eq. (15) the derived unit of mass in the fps gravitational system is the lbf · s2 /ft and is called a slug; there is no abbreviation for slug. The unit of mass in the ips gravitational system is (pound-force)(second)2 FT 2 = = lbf · s2/in M= (16) L inch The mass unit lbf · s2 /in has no ofcial name. The International System of Units (SI) is an absolute system. The base units are the meter, the kilogram (for mass), and the second. The unit of force is derived by using Newtons second law and is called the newton. The units constituting the newton (N) are F= ML (kilogram)(meter) = = kg · m /s2 = N T2 (second)2 (17) The weight of an object is the force exerted upon it by gravity. Designating the weight as W and the acceleration due to gravity as g, we have W = mg (18) 28 22 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition I. Basics 1. Introduction to Mechanical Engineering Design © The McGrawHill Companies, 2008 Mechanical Engineering Design In the fps system, standard gravity is g 32.1740 ft/s2. For most cases this is rounded off to 32.2. Thus the weight of a mass of 1 slug in the fps system is W = mg = (1 slug)(32.2 ft /s2 ) = 32.2 lbf In the ips system, standard gravity is 386.088 or about 386 in/s2. Thus, in this system, a unit mass weighs W = (1 lbf · s2 /in)(386 in/s2 ) = 386 lbf With SI units, standard gravity is 9.806 or about 9.81 m/s. Thus, the weight of a 1-kg mass is W = (1 kg)(9.81 m/s2 ) = 9.81 N A series of names and symbols to form multiples and submultiples of SI units has been established to provide an alternative to the writing of powers of 10. Table A1 includes these prexes and symbols. Numbers having four or more digits are placed in groups of three and separated by a space instead of a comma. However, the space may be omitted for the special case of numbers having four digits. A period is used as a decimal point. These recommendations avoid the confusion caused by certain European countries in which a comma is used as a decimal point, and by the English use of a centered period. Examples of correct and incorrect usage are as follows: 1924 or 1 924 but not 1,924 0.1924 or 0.192 4 but not 0.192,4 192 423.618 50 but not 192,423.61850 The decimal point should always be preceded by a zero for numbers less than unity. 115 Calculations and Signicant Figures The discussion in this section applies to real numbers, not integers. The accuracy of a real number depends on the number of signicant gures describing the number. Usually, but not always, three or four signicant gures are necessary for engineering accuracy. Unless otherwise stated, no less than three signicant gures should be used in your calculations. The number of signicant gures is usually inferred by the number of gures given (except for leading zeros). For example, 706, 3.14, and 0.002 19 are assumed to be numbers with three signicant gures. For trailing zeros, a little more clarication is necessary. To display 706 to four signicant gures insert a trailing zero and display either 706.0, 7.060 × 102 , or 0.7060 × 103. Also, consider a number such as 91 600. Scientic notation should be used to clarify the accuracy. For three signicant gures express the number as 91.6 × 103. For four signicant gures express it as 91.60 × 103. Computers and calculators display calculations to many signicant gures. However, you should never report a number of signicant gures of a calculation any greater than the smallest number of signicant gures of the numbers used for the calculation. Of course, you should use the greatest accuracy possible when performing a calculation. For example, determine the circumference of a solid shaft with a diameter of d = 0.40 in. The circumference is given by C = πd . Since d is given with two signicant gures, C should be reported with only two signicant gures. Now if we used only two signicant gures for π our calculator would give C = 3.1 (0.40) = 1.24 in. This rounds off to two significant gures as C = 1.2 in. However, using π = 3.141 592 654 as programmed in the calculator, C = 3.141 592 654 (0.40) = 1.256 637 061 in. This rounds off to C = 1.3 in, which is 8.3 percent higher than the rst calculation. Note, however, since d is given BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition I. Basics 1. Introduction to Mechanical Engineering Design © The McGrawHill Companies, 2008 Introduction to Mechanical Engineering Design 29 23 with two signicant gures, it is implied that the range of d is 0.40 ± 0.005. This means that the calculation of C is only accurate to within ±0.005/0.40 = ±0.0125 = ±1.25%. The calculation could also be one in a series of calculations, and rounding each calculation separately may lead to an accumulation of greater inaccuracy. Thus, it is considered good engineering practice to make all calculations to the greatest accuracy possible and report the results within the accuracy of the given input. 116 Power Transmission Case Study Specications A case study incorporating the many facets of the design process for a power transmission speed reducer will be considered throughout this textbook. The problem will be introduced here with the denition and specication for the product to be designed. Further details and component analysis will be presented in subsequent chapters. Chapter 18 provides an overview of the entire process, focusing on the design sequence, the interaction between the component designs, and other details pertinent to transmission of power. It also contains a complete case study of the power transmission speed reducer introduced here. Many industrial applications require machinery to be powered by engines or electric motors. The power source usually runs most efciently at a narrow range of rotational speed. When the application requires power to be delivered at a slower speed than supplied by the motor, a speed reducer is introduced. The speed reducer should transmit the power from the motor to the application with as little energy loss as practical, while reducing the speed and consequently increasing the torque. For example, assume that a company wishes to provide off-the-shelf speed reducers in various capacities and speed ratios to sell to a wide variety of target applications. The marketing team has determined a need for one of these speed reducers to satisfy the following customer requirements. Design Requirements Power to be delivered: 20 hp Input speed: 1750 rev/min Output speed: 85 rev/min Targeted for uniformly loaded applications, such as conveyor belts, blowers, and generators Output shaft and input shaft in-line Base mounted with 4 bolts Continuous operation 6-year life, with 8 hours/day, 5 days/wk Low maintenance Competitive cost Nominal operating conditions of industrialized locations Input and output shafts standard size for typical couplings In reality, the company would likely design for a whole range of speed ratios for each power capacity, obtainable by interchanging gear sizes within the same overall design. For simplicity, in this case study only one speed ratio will be considered. Notice that the list of customer requirements includes some numerical specics, but also includes some generalized requirements, e.g., low maintenance and competitive cost. These general requirements give some guidance on what needs to be considered in the design process, but are difcult to achieve with any certainty. In order to pin down these nebulous requirements, it is best to further develop the customer requirements into a set of product specications that are measurable. This task is usually achieved through the work of a team including engineering, marketing, management, and customers. Various tools 30 24 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition I. Basics 1. Introduction to Mechanical Engineering Design © The McGrawHill Companies, 2008 Mechanical Engineering Design may be used (see Footnote 1) to prioritize the requirements, determine suitable metrics to be achieved, and to establish target values for each metric. The goal of this process is to obtain a product specication that identies precisely what the product must satisfy. The following product specications provide an appropriate framework for this design task. Design Specications Power to be delivered: 20 hp Power efciency: >95% Steady state input speed: 1750 rev/min Maximum input speed: 2400 rev/min Steady-state output speed: 8288 rev/min Usually low shock levels, occasional moderate shock Input and output shaft diameter tolerance: ±0.001 in Output shaft and input shaft in-line: concentricity ±0.005 in, alignment ±0.001 rad Maximum allowable loads on input shaft: axial, 50 lbf; transverse, 100 lbf Maximum allowable loads on output shaft: axial, 50 lbf; transverse, 500 lbf Base mounted with 4 bolts Mounting orientation only with base on bottom 100% duty cycle Maintenance schedule: lubrication check every 2000 hours; change of lubrication every 8000 hours of operation; gears and bearing life >12,000 hours; innite shaft life; gears, bearings, and shafts replaceable Access to check, drain, and rell lubrication without disassembly or opening of gasketed joints. Manufacturing cost per unit: < $300 Production: 10,000 units per year Operating temperature range: 10 to 120 F Sealed against water and dust from typical weather Noise: < 85 dB from 1 meter PROBLEMS 11 Select a mechanical component from Part 3 of this book (roller bearings, springs, etc.), go to your universitys library or the appropriate internet website, and, using the Thomas Register of American Manufacturers, report on the information obtained on ve manufacturers or suppliers. 12 Select a mechanical component from Part 3 of this book (roller bearings, springs, etc.), go to the Internet, and, using a search engine, report on the information obtained on ve manufacturers or suppliers. 13 Select an organization listed in Sec. 16, go to the Internet, and list what information is available on the organization. 14 Go to the Internet and connect to the NSPE website (www.nspe.org). Read the full version of the NSPE Code of Ethics for Engineers and briey discuss your reading. 15 Highway tunnel trafc (two parallel lanes in the same direction) experience indicates the average spacing between vehicles increases with speed. Data from a New York tunnel show that between 15 and 35 mi/h, the space x between vehicles (in miles) is x = 0.324/(42.1 v) where v is the vehicles speed in miles per hour. (a) Ignoring the length of individual vehicles, what speed will give the tunnel the largest volume in vehicles per hour? BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition I. Basics © The McGrawHill Companies, 2008 1. Introduction to Mechanical Engineering Design Introduction to Mechanical Engineering Design 31 25 (b) Does including the length of the vehicles cut the tunnel capacity prediction signicantly? Assume the average vehicle length is 10 ft. (c) For part (b), does the optimal speed change much? 16 The engineering designer must create (invent) the concept and connectivity of the elements that constitute a design, and not lose sight of the need to develop ideas with optimality in mind. A useful design attribute can be cost, which can be related to the amount of material used (volume or weight). When you think about it, the weight is a function of the geometry and density. When the design is solidied, nding the weight is a straightforward, sometimes tedious task. The gure depicts a simple bracket frame that has supports that project from a wall column. The bracket supports a chain-fall hoist. Pinned joints are used to avoid bending. The cost of a link can be approximated by $ = ¢ Al γ , where ¢ is the cost of the link per unit weight, A is the cross-sectional area of the prismatic link, l is the pin-to-pin link length, and γ is the specic weight of the material used. To be sure, this is approximate because no decisions have been made concerning the geometric form of the links or their ttings. By investigating cost now in this approximate way, one can detect whether a particular set of proportions of the bracket (indexed by angle θ ) is advantageous. Is there a preferable angle θ ? Show that the cost can be expressed as $= 1 + cos2 θ sin θ cos θ γ ¢W l2 S where W is the weight of the hoist and load, and S is the allowable tensile or compressive stress in the link material (assume S = | Fi / A| and no column buckling action). What is the desirable angle θ corresponding to the minimal cost? l1 Problem 16 F1 (a) A chain-hoist bracket frame. (b) Free body of pin. F2 l2 W (b) (a ) 17 When one knows the true values x1 and x2 and has approximations X 1 and X 2 at hand, one can see where errors may arise. By viewing error as something to be added to an approximation to attain a true value, it follows that the error ei , is related to X i , and xi as xi = X i + ei (a) Show that the error in a sum X 1 + X 2 is (x1 + x2 ) ( X 1 + X 2 ) = e1 + e2 (b) Show that the error in a difference X 1 X 2 is (x1 x2 ) ( X 1 X 2 ) = e1 e2 (c) Show that the error in a product X 1 X 2 is x1 x2 X 1 X 2 = X 1 X 2 e1 e2 + X1 X2 (d ) Show that in a quotient X 1 / X 2 the error is x1 X1 X1 = x2 X2 X2 e2 e1 X1 X2 32 26 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition I. Basics 1. Introduction to Mechanical Engineering Design © The McGrawHill Companies, 2008 Mechanical Engineering Design 18 Use the true values x1 = 5 and x2 = 6 (a) Demonstrate the correctness of the error equation from Prob. 17 for addition if three correct digits are used for X 1 and X 2 . (b) Demonstrate the correctness of the error equation for addition using three-digit signicant numbers for X 1 and X 2 . 19 Convert the following to appropriate SI units: (a) A stress of 20 000 psi. (b) A force of 350 lbf. (c) A moment of 1200 lbf in. (d) An area of 2.4 in2 . (e) A second moment of area of 17.4 in4 . ( f ) An area of 3.6 mi2 . (g ) A modulus of elasticity of 21 Mpsi. (h) A speed of 45 mi/h. (i) A volume of 60 in3 . 110 Convert the following to appropriate ips units: (a) A length of 1.5 m. (b) A stress of 600 MPa. (c) A pressure of 160 kPa. (d) A section modulus of 1.84 (105 ) mm3 . (e) A unit weight of 38.1 N/m. ( f ) A deection of 0.05 mm. (g) A velocity of 6.12 m/s. (h) A unit strain of 0.0021 m/m. (i) A volume of 30 L. 111 Generally, nal design results are rounded to or xed to three digits because the given data cannot justify a greater display. In addition, prexes should be selected so as to limit number strings to no more than four digits to the left of the decimal point. Using these rules, as well as those for the choice of prexes, solve the following relations: (a) σ = M / Z , where M = 200 N · m and Z = 15.3 × 103 mm3. (b) σ = F / A , where F = 42 kN and A = 600 mm2 . (c) y = Fl 3 /3 E I , where F = 1200 N, l = 800 mm, E = 207 GPa, and I = 64 × 103 mm4 . (d ) θ = T l / G J , where J = π d 4 /32, T = 1100 N · m, l = 250 mm, G = 79.3 GPa, and d = 25 mm. Convert results to degrees of angle. 112 Repeat Prob. 111 for the following: (a) σ = F /wt , where F = 600 N, w = 20 mm, and t = 6 mm. (b) I = bh 3 /12, where b = 8 mm and h = 24 mm. (c) I = π d 4 /64, where d = 32 mm. (d ) τ = 16T /π d 3 , where T = 16 N m and d = 25 mm. 113 Repeat Prob. 111 for: (a) τ = F / A , where A = π d 2 /4, F = 120 kN, and d = 20 mm. (b) σ = 32 Fa /π d 3 , where F = 800 N, a = 800 mm, and d = 32 mm. (c) Z = (π/32d )(d 4 di4 ) for d = 36 mm and di = 26 mm. (d ) k = (d 4 G )/(8 D 3 N ) , where d = 1.6 mm, G = 79.3 GPa, D = 19.2 mm, and N = 32 (a dimensionless number). BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition I. Basics © The McGrawHill Companies, 2008 2. Materials 33 2 Materials Chapter Outline 21 Material Strength and Stiffness 22 The Statistical Signicance of Material Properties 23 Strength and Cold Work 24 Hardness 25 Impact Properties 26 Temperature Effects 27 Numbering Systems 28 Sand Casting 41 29 Shell Molding 42 28 32 33 36 37 39 40 210 Investment Casting 211 Powder-Metallurgy Process 212 Hot-Working Processes 213 Cold-Working Processes 214 The Heat Treatment of Steel 215 Alloy Steels 216 Corrosion-Resistant Steels 217 Casting Materials 218 Nonferrous Metals 219 Plastics 220 Composite Materials 221 Materials Selection 42 42 43 44 44 47 48 49 51 54 55 56 27 34 28 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition I. Basics © The McGrawHill Companies, 2008 2. Materials Mechanical Engineering Design The selection of a material for a machine part or a structural member is one of the most important decisions the designer is called on to make. The decision is usually made before the dimensions of the part are established. After choosing the process of creating the desired geometry and the material (the two cannot be divorced), the designer can proportion the member so that loss of function can be avoided or the chance of loss of function can be held to an acceptable risk. In Chaps. 3 and 4, methods for estimating stresses and deections of machine members are presented. These estimates are based on the properties of the material from which the member will be made. For deections and stability evaluations, for example, the elastic (stiffness) properties of the material are required, and evaluations of stress at a critical location in a machine member require a comparison with the strength of the material at that location in the geometry and condition of use. This strength is a material property found by testing and is adjusted to the geometry and condition of use as necessary. As important as stress and deection are in the design of mechanical parts, the selection of a material is not always based on these factors. Many parts carry no loads on them whatever. Parts may be designed merely to ll up space or for aesthetic qualities. Members must frequently be designed to also resist corrosion. Sometimes temperature effects are more important in design than stress and strain. So many other factors besides stress and strain may govern the design of parts that the designer must have the versatility that comes only with a broad background in materials and processes. 21 Material Strength and Stiffness The standard tensile test is used to obtain a variety of material characteristics and strengths that are used in design. Figure 2l illustrates a typical tension-test specimen and its characteristic dimensions.1 The original diameter d0 and the gauge length l0 , used to measure the deections, are recorded before the test is begun. The specimen is then mounted in the test machine and slowly loaded in tension while the load P and deection are observed. The load is converted to stress by the calculation σ= P A0 (21) 2 where A0 = 1 π d0 is the original area of the specimen. 4 d0 P P l0 Figure 21 A typical tension-test specimen. Some of the standard dimensions used for d0 are 2.5, 6.25, and 12.5 mm and 0.505 in, but other sections and sizes are in use. Common gauge lengths l0 used are 10, 25, and 50 mm and 1 and 2 in. 1 See ASTM standards E8 and E-8 m for standard dimensions. BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition I. Basics 35 © The McGrawHill Companies, 2008 2. Materials Materials 29 The deection, or extension of the gage length, is given by l l0 where l is the gauge length corresponding to the load P. The normal strain is calculated from l l0 l0 ǫ= (22) At the conclusion of, or during, the test, the results are plotted as a stress-strain diagram. Figure 22 depicts typical stress-strain diagrams for ductile and brittle materials. Ductile materials deform much more than brittle materials. Point pl in Fig. 22a is called the proportional limit. This is the point at which the curve rst begins to deviate from a straight line. No permanent set will be observable in the specimen if the load is removed at this point. In the linear range, the uniaxial stress-strain relation is given by Hookes law as (23) σ = Eǫ where the constant of proportionality E, the slope of the linear part of the stress-strain curve, is called Youngs modulus or the modulus of elasticity. E is a measure of the stiffness of a material, and since strain is dimensionless, the units of E are the same as stress. Steel, for example, has a modulus of elasticity of about 30 Mpsi (207 GPa) regardless of heat treatment, carbon content, or alloying. Stainless steel is about 27.5 Mpsi (190 GPa). Point el in Fig. 22 is called the elastic limit. If the specimen is loaded beyond this point, the deformation is said to be plastic and the material will take on a permanent set when the load is removed. Between pl and el the diagram is not a perfectly straight line, even though the specimen is elastic. During the tension test, many materials reach a point at which the strain begins to increase very rapidly without a corresponding increase in stress. This point is called the yield point. Not all materials have an obvious yield point, especially for brittle materials. For this reason, yield strength Sy is often dened by an offset method as shown in Fig. 22, where line a y is drawn at slope E. Point a corresponds to a denite or stated amount of permanent set, usually 0.2 percent of the original gauge length (ǫ = 0.002), although 0.01, 0.1, and 0.5 percent are sometimes used. The ultimate, or tensile, strength Su or Sut corresponds to point u in Fig. 22 and is the maximum stress reached on the stress-strain diagram.2 As shown in Fig. 22a, Su Sf Sy u f y el pl Sut Sy u, f y Stress Stress-strain diagram obtained from the standard tensile test (a) Ductile material; (b) brittle material. pl marks the proportional limit; el, the elastic limit; y, the offset-yield strength as dened by offset strain a; u, the maximum or ultimate strength; and f, the fracture strength. = P/A0 Figure 22 Oa y u f a Strain 2 Strain (a ) (b) Usage varies. For a long time engineers used the term ultimate strength, hence the subscript u in Su or Sut . However, in material science and metallurgy the term tensile strength is used. 36 30 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition I. Basics © The McGrawHill Companies, 2008 2. Materials Mechanical Engineering Design some materials exhibit a downward trend after the maximum stress is reached and fracture at point f on the diagram. Others, such as some of the cast irons and high-strength steels, fracture while the stress-strain trace is still rising, as shown in Fig. 22b, where points u and f are identical. As noted in Sec. 19, strength, as used in this book, is a built-in property of a material, or of a mechanical element, because of the selection of a particular material or process or both. The strength of a connecting rod at the critical location in the geometry and condition of use, for example, is the same no matter whether it is already an element in an operating machine or whether it is lying on a workbench awaiting assembly with other parts. On the other hand, stress is something that occurs in a part, usually as a result of its being assembled into a machine and loaded. However, stresses may be built into a part by processing or handling. For example, shot peening produces a compressive stress in the outer surface of a part, and also improves the fatigue strength of the part. Thus, in this book we will be very careful in distinguishing between strength, designated by S, and stress, designated by σ or τ . The diagrams in Fig. 22 are called engineering stress-strain diagrams because the stresses and strains calculated in Eqs. (21) and (22) are not true values. The stress calculated in Eq. (21) is based on the original area before the load is applied. In reality, as the load is applied the area reduces so that the actual or true stress is larger than the engineering stress. To obtain the true stress for the diagram the load and the crosssectional area must be measured simultaneously during the test. Figure 22a represents a ductile material where the stress appears to decrease from points u to f. Typically, beyond point u the specimen begins to neck at a location of weakness where the area reduces dramatically, as shown in Fig. 23. For this reason, the true stress is much higher than the engineering stress at the necked section. The engineering strain given by Eq. (22) is based on net change in length from the original length. In plotting the true stress-strain diagram, it is customary to use a term called true strain or, sometimes, logarithmic strain. True strain is the sum of the incremental elongations divided by the current gauge length at load P, or l ε= l0 dl l = ln l l0 (24) where the symbol ε is used to represent true strain. The most important characteristic of a true stress-strain diagram (Fig. 24) is that the true stress continually increases all the way to fracture. Thus, as shown in Fig. 24, the true fracture stress σ f is greater than the true ultimate stress σu . Contrast this with Fig. 22a, where the engineering fracture strength S f is less than the engineering ultimate strength Su . Compression tests are more difcult to conduct, and the geometry of the test specimens differs from the geometry of those used in tension tests. The reason for this is that the specimen may buckle during testing or it may be difcult to distribute the stresses evenly. Other difculties occur because ductile materials will bulge after yielding. However, the results can be plotted on a stress-strain diagram also, and the same strength denitions can be applied as used in tensile testing. For most ductile materials the compressive strengths are about the same as the tensile strengths. When substantial differences occur between tensile and compressive strengths, however, as is the case with Figure 23 Tension specimen after necking. BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition I. Basics 37 © The McGrawHill Companies, 2008 2. Materials Materials 31 Figure 24 f f True stress-strain diagram plotted in Cartesian coordinates. u True stress u u f True strain the cast irons, the tensile and compressive strengths should be stated separately, Sut , Suc , where Suc is reported as a positive quantity. Torsional strengths are found by twisting solid circular bars and recording the torque and the twist angle. The results are then plotted as a torque-twist diagram. The shear stresses in the specimen are linear with respect to radial location, being zero at the center of the specimen and maximum at the outer radius r (see Chap. 3). The maximum shear stress τmax is related to the angle of twist θ by τmax = Gr θ l0 (25) where θ is in radians, r is the radius of the specimen, l0 is the gauge length, and G is the material stiffness property called the shear modulus or the modulus of rigidity. The maximum shear stress is also related to the applied torque T as τmax = Tr J (26) where J = 1 π r 4 is the polar second moment of area of the cross section. 2 The torque-twist diagram will be similar to Fig. 22, and, using Eqs. (25) and (26), the modulus of rigidity can be found as well as the elastic limit and the torsional yield strength Ssy . The maximum point on a torque-twist diagram, corresponding to point u on Fig. 22, is Tu . The equation Ssu = Tu r J (27) denes the modulus of rupture for the torsion test. Note that it is incorrect to call Ssu the ultimate torsional strength, as the outermost region of the bar is in a plastic state at the torque Tu and the stress distribution is no longer linear. All of the stresses and strengths dened by the stress-strain diagram of Fig. 22 and similar diagrams are specically known as engineering stresses and strengths or nominal stresses and strengths. These are the values normally used in all engineering design calculations. The adjectives engineering and nominal are used here to emphasize that the stresses are computed by using the original or unstressed cross-sectional area of the specimen. In this book we shall use these modiers only when we specically wish to call attention to this distinction. 38 32 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition I. Basics © The McGrawHill Companies, 2008 2. Materials Mechanical Engineering Design The Statistical Signicance of Material Properties 22 There is some subtlety in the ideas presented in the previous section that should be pondered carefully before continuing. Figure 22 depicts the result of a single tension test (one specimen, now fractured). It is common for engineers to consider these important stress values (at points pl, el, y , u , and f) as properties and to denote them as strengths with a special notation, uppercase S, in lieu of lowercase sigma σ , with subscripts added: Spl for proportional limit, Sy for yield strength, Su for ultimate tensile strength ( Sut or Suc , if tensile or compressive sense is important). If there were 1000 nominally identical specimens, the values of strength obtained would be distributed between some minimum and maximum values. It follows that the description of strength, a material property, is distributional and thus is statistical in nature. Chapter 20 provides more detail on statistical considerations in design. Here we will simply describe the results of one example, Ex. 20-4. Consider the following table, which is a histographic report containing the maximum stresses of 1000 tensile tests on a 1020 steel from a single heat. Here we are seeking the ultimate tensile strength Sut . The class frequency is the number of occurrences within a 1 kpsi range given by the class midpoint. Thus, 18 maximum stress values occurred in the range of 57 to 58 kpsi. Class Frequency f i Class Midpoint xi , kpsi 2 18 23 31 83 109 138 151 139 130 82 49 28 11 4 2 56.5 57.5 58.5 59.5 60.5 61.5 62.5 63.5 64.5 65.5 66.5 67.5 68.5 69.5 70.5 71.5 The probability density is dened as the number of occurrences divided by the total sample number. The bar chart in Fig. 25 depicts the histogram of the probability density. If the data is in the form of a Gaussian or normal distribution, the probability density function determined in Ex. 20-4 is f (x ) = 1 1 exp 2 2.594 2π x 63.62 2.594 2 where the mean stress is 63.62 kpsi and the standard deviation is 2.594 kpsi. A plot of f (x ) is included in Fig. 25. The description of the strength Sut is then expressed in terms of its statistical parameters and its distribution type. In this case Sut = N(63.62, 2.594) kpsi. Note that the test program has described 1020 property Sut, for only one heat of one supplier. Testing is an involved and expensive process. Tables of properties are often prepared to be helpful to other persons. A statistical quantity is described by its mean, standard deviation, and distribution type. Many tables display a single number, which is often the mean, minimum, or some percentile, such as the 99th percentile. Always read the foonotes to the table. If no qualication is made in a single-entry table, the table is subject to serious doubt. Since it is no surprise that useful descriptions of a property are statistical in nature, engineers, when ordering property tests, should couch the instructions so the data generated are enough for them to observe the statistical parameters and to identify the distributional characteristic. The tensile test program on 1000 specimens of 1020 steel is a large one. If you were faced with putting something in a table of ultimate tensile strengths and constrained to a single number, what would it be and just how would your footnote read? BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition I. Basics 39 © The McGrawHill Companies, 2008 2. Materials 33 Materials 0.2 Figure 25 Histogram for 1000 tensile tests on a 1020 steel from a single heat. Probability density f(x ) 0.1 0 50 60 70 Ultimate tensile strength, kpsi 23 Strength and Cold Work Cold working is the process of plastic straining below the recrystallization temperature in the plastic region of the stress-strain diagram. Materials can be deformed plastically by the application of heat, as in blacksmithing or hot rolling, but the resulting mechanical properties are quite different from those obtained by cold working. The purpose of this section is to explain what happens to the signicant mechanical properties of a material when that material is cold-worked. Consider the stress-strain diagram of Fig. 26a. Here a material has been stressed beyond the yield strength at y to some point i, in the plastic region, and then the load removed. At this point the material has a permanent plastic deformation ǫ p . If the load corresponding to point i is now reapplied, the material will be elastically deformed by Figure 26 u Pu u i Pi i i f Sy f Py y y Load, P Nominal stress, (a) Stress-strain diagram showing unloading and reloading at point i in the plastic region; (b) analogous load-deformation diagram. Su O p Unit strain, e A0 Ai Af Ai Area deformation (reduction) (a ) (b) 40 34 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition I. Basics © The McGrawHill Companies, 2008 2. Materials Mechanical Engineering Design the amount ǫe . Thus at point i the total unit strain consists of the two components ǫ p and ǫe and is given by the equation ǫ = ǫ p + ǫe (a) This material can be unloaded and reloaded any number of times from and to point i, and it is found that the action always occurs along the straight line that is approximately parallel to the initial elastic line Oy. Thus ǫe = σi E (b) The material now has a higher yield point, is less ductile as a result of a reduction in strain capacity, and is said to be strain-hardened. If the process is continued, increasing ǫ p , the material can become brittle and exhibit sudden fracture. It is possible to construct a similar diagram, as in Fig. 26b, where the abscissa is the area deformation and the ordinate is the applied load. The reduction in area corresponding to the load Pf , at fracture, is dened as R= A0 A f Af =1 A0 A0 (28) where A0 is the original area. The quantity R in Eq. (28) is usually expressed in percent and tabulated in lists of mechanical properties as a measure of ductility. See Appendix Table A20, for example. Ductility is an important property because it measures the ability of a material to absorb overloads and to be cold-worked. Thus such operations as bending, drawing, heading, and stretch forming are metal-processing operations that require ductile materials. Figure 26b can also be used to dene the quantity of cold work. The cold-work factor W is dened as W= A0 Ai A 0 Ai A0 A0 (29) where Ai corresponds to the area after the load Pi has been released. The approximation in Eq. (29) results because of the difculty of measuring the small diametral changes in the elastic region. If the amount of cold work is known, then Eq. (29) can be solved for the area Ai . The result is Ai = A0 (1 W ) (210) Cold working a material produces a new set of values for the strengths, as can be seen from stress-strain diagrams. Datsko3 describes the plastic region of the true stresstrue strain diagram by the equation σ = σ0 εm 3 (211) Joseph Datsko, Solid Materials, Chap. 32 in Joseph E. Shigley, Charles R. Mischke, and Thomas H. Brown, Jr. (eds.), Standard Handbook of Machine Design, 3rd ed., McGraw-Hill, New York, 2004. See also Joseph Datsko, New Look at Material Strength, Machine Design, vol. 58, no. 3, Feb. 6, 1986, pp. 8185. BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition I. Basics 41 © The McGrawHill Companies, 2008 2. Materials 35 Materials where σ = true stress σ0 = a strength coefcient, or strain-strengthening coefcient ε = true plastic strain m = strain-strengthening exponent It can be shown4 that (212) m = εu provided that the load-deformation curve exhibits a stationary point (a place of zero slope). Difculties arise when using the gauge length to evaluate the true strain in the plastic range, since necking causes the strain to be nonuniform. A more satisfactory relation can be obtained by using the area at the neck. Assuming that the change in volume of the material is small, Al = A0 l0 . Thus, l / l0 = A0 / A , and the true strain is given by ε = ln l A0 = ln l0 A (213) Returning to Fig. 26b, if point i is to the left of point u, that is, Pi < Pu , then the new yield strength is Sy = Pi = σ0 εim Ai Pi Pu (214) Because of the reduced area, that is, because Ai < A0 , the ultimate strength also changes, and is Su = Pu Ai (c) Since Pu = Su A0 , we nd, with Eq. (210), that Su = Su Su A0 = A0 (1 W ) 1W εi εu (215) which is valid only when point i is to the left of point u. For points to the right of u, the yield strength is approaching the ultimate strength, and, with small loss in accuracy, . . Su = Sy = σ0 εim εi εu (216) A little thought will reveal that a bar will have the same ultimate load in tension after being strain-strengthened in tension as it had before. The new strength is of interest to us not because the static ultimate load increases, butsince fatigue strengths are correlated with the local ultimate strengthsbecause the fatigue strength improves. Also the yield strength increases, giving a larger range of sustainable elastic loading. 4 See Sec. 52, J. E. Shigley and C. R. Mischke, Mechanical Engineering Design, 6th ed., McGraw-Hill, New York, 2001. 42 36 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition I. Basics © The McGrawHill Companies, 2008 2. Materials Mechanical Engineering Design EXAMPLE 21 Solution An annealed AISI 1018 steel (see Table A22) has Sy = 32.0 kpsi, Su = 49.5 kpsi, σ f = 91.1 kpsi, σ0 = 90 kpsi, m = 0.25, and ε f = 1.05 in/in. Find the new values of the strengths if the material is given 15 percent cold work. From Eq. (212), we nd the true strain corresponding to the ultimate strength to be εu = m = 0.25 The ratio A0 / Ai is, from Eq. (29), A0 1 1 = = 1.176 = Ai 1W 1 0.15 The true strain corresponding to 15 percent cold work is obtained from Eq. (213). Thus εi = ln A0 = ln 1.176 = 0.1625 Ai Since εi < εu , Eqs. (214) and (215) apply. Therefore, Answer Sy = σ0 εim = 90(0.1625)0.25 = 57.1 kpsi Answer Su = 24 49.5 Su = = 58.2 kpsi 1W 1 0.15 Hardness The resistance of a material to penetration by a pointed tool is called hardness. Though there are many hardness-measuring systems, we shall consider here only the two in greatest use. Rockwell hardness tests are described by ASTM standard hardness method E18 and measurements are quickly and easily made, they have good reproducibility, and the test machine for them is easy to use. In fact, the hardness number is read directly from a dial. Rockwell hardness scales are designated as A , B , C , . . . , etc. The indenters are 1 described as a diamond, a 16 -in-diameter ball, and a diamond for scales A, B, and C, respectively, where the load applied is either 60, 100, or 150 kg. Thus the Rockwell B 1 scale, designated R B , uses a 100-kg load and a No. 2 indenter, which is a 16 -in-diameter ball. The Rockwell C scale RC uses a diamond cone, which is the No. 1 indenter, and a load of 150 kg. Hardness numbers so obtained are relative. Therefore a hardness RC = 50 has meaning only in relation to another hardness number using the same scale. The Brinell hardness is another test in very general use. In testing, the indenting tool through which force is applied is a ball and the hardness number HB is found as a number equal to the applied load divided by the spherical surface area of the indentation. Thus the units of HB are the same as those of stress, though they are seldom used. Brinell hardness testing takes more time, since HB must be computed from the test data. The primary advantage of both methods is that they are nondestructive in most cases. Both are empirically and directly related to the ultimate strength of the BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition I. Basics 43 © The McGrawHill Companies, 2008 2. Materials 37 Materials material tested. This means that the strength of parts could, if desired, be tested part by part during manufacture. For steels, the relationship between the minimum ultimate strength and the Brinell hardness number for 200 HB 450 is found to be Su = 0.495 HB 3.41 HB kpsi MPa (217) Similar relationships for cast iron can be derived from data supplied by Krause.5 Data from 72 tests of gray iron produced by one foundry and poured in two sizes of test bars are reported in graph form. The minimum strength, as dened by the ASTM, is found from these data to be Su = 0.23 HB 12.5 kpsi 1.58 HB 86 MPa (218) Walton6 shows a chart from which the SAE minimum strength can be obtained. The result is Su = 0.2375 HB 16 kpsi (219) which is even more conservative than the values obtained from Eq. (218). EXAMPLE 22 Solution It is necessary to ensure that a certain part supplied by a foundry always meets or exceeds ASTM No. 20 specications for cast iron (see Table A24). What hardness should be specied? From Eq. (218), with (Su)min = 20 kpsi, we have Answer HB = 20 + 12.5 Su + 12.5 = = 141 0.23 0.23 If the foundry can control the hardness within 20 points, routinely, then specify 145 < HB < 165. This imposes no hardship on the foundry and assures the designer that ASTM grade 20 will always be supplied at a predictable cost. 25 Impact Properties An external force applied to a structure or part is called an impact load if the time of application is less than one-third the lowest natural period of vibration of the part or structure. Otherwise it is called simply a static load. 5 D. E. Krause, Gray IronA Unique Engineering Material, ASTM Special Publication 455, 1969, pp. 329, as reported in Charles F. Walton (ed.), Iron Castings Handbook, Iron Founders Society, Inc., Cleveland, 1971, pp. 204, 205. 6 Ibid. I. Basics © The McGrawHill Companies, 2008 2. Materials Mechanical Engineering Design A mean trace shows the effect of temperature on impact values. The result of interest is the brittle-ductile transition temperature, often dened as the temperature at which the mean trace passes through the 15 ft · lbf level. The critical temperature is dependent on the geometry of the notch, which is why the Charpy V notch is closely dened. 60 40 20 0 400 200 0 200 400 Temperature, °F Figure 28 100 Inuence of strain rate on tensile properties. 100 80 80 Ratio, Sy /Su Strength, kpsi Ultimate strength, Su 60 Ratio, Sy /Su , % Figure 27 The Charpy (commonly used) and Izod (rarely used) notched-bar tests utilize bars of specied geometries to determine brittleness and impact strength. These tests are helpful in comparing several materials and in the determination of low-temperature brittleness. In both tests the specimen is struck by a pendulum released from a xed height, and the energy absorbed by the specimen, called the impact value, can be computed from the height of swing after fracture, but is read from a dial that essentially computes the result. The effect of temperature on impact values is shown in Fig. 27 for a material showing a ductile-brittle transition. Not all materials show this transition. Notice the narrow region of critical temperatures where the impact value increases very rapidly. In the low-temperature region the fracture appears as a brittle, shattering type, whereas the appearance is a tough, tearing type above the critical-temperature region. The critical temperature seems to be dependent on both the material and the geometry of the notch. For this reason designers should not rely too heavily on the results of notched-bar tests. The average strain rate used in obtaining the stress-strain diagram is about 0.001 in/(in · s) or less. When the strain rate is increased, as it is under impact conditions, the strengths increase, as shown in Fig. 28. In fact, at very high strain rates the yield strength seems to approach the ultimate strength as a limit. But note that the curves show little change in the elongation. This means that the ductility remains about the same. Also, in view of the sharp increase in yield strength, a mild steel could be expected to behave elastically throughout practically its entire strength range under impact conditions. 60 Total elongation 40 40 Yield strength, Sy 20 0 106 20 104 102 1 1 Strain rate, s 10 2 0 10 4 Elongation, % 38 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition Charpy, ft lbf 44 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition I. Basics 45 © The McGrawHill Companies, 2008 2. Materials Materials 39 The Charpy and Izod tests really provide toughness data under dynamic, rather than static, conditions. It may well be that impact data obtained from these tests are as dependent on the notch geometry as they are on the strain rate. For these reasons it may be better to use the concepts of notch sensitivity, fracture toughness, and fracture mechanics, discussed in Chaps. 5 and 6, to assess the possibility of cracking or fracture. 26 Temperature Effects Strength and ductility, or brittleness, are properties affected by the temperature of the operating environment. The effect of temperature on the static properties of steels is typied by the strength versus temperature chart of Fig. 29. Note that the tensile strength changes only a small amount until a certain temperature is reached. At that point it falls off rapidly. The yield strength, however, decreases continuously as the environmental temperature is increased. There is a substantial increase in ductility, as might be expected, at the higher temperatures. Many tests have been made of ferrous metals subjected to constant loads for long periods of time at elevated temperatures. The specimens were found to be permanently deformed during the tests, even though at times the actual stresses were less than the yield strength of the material obtained from short-time tests made at the same temperature. This continuous deformation under load is called creep. One of the most useful tests to have been devised is the long-time creep test under constant load. Figure 210 illustrates a curve that is typical of this kind of test. The curve is obtained at a constant stated temperature. A number of tests are usually run simultaneously at different stress intensities. The curve exhibits three distinct regions. In the rst stage are included both the elastic and the plastic deformation. This stage shows a decreasing creep rate, which is due to the strain hardening. The second stage shows a constant minimum creep rate caused by the annealing effect. In the third stage the specimen shows a considerable reduction in area, the true stress is increased, and a higher creep eventually leads to fracture. When the operating temperatures are lower than the transition temperature (Fig. 27), the possibility arises that a part could fail by a brittle fracture. This subject will be discussed in Chap. 5. Figure 29 1.0 Sut 0.9 ST /SRT A plot of the results of 145 tests of 21 carbon and alloy steels showing the effect of operating temperature on the yield strength Sy and the ultimate strength Sut . The ordinate is the ratio of the strength at the operating temperature to the strength at room temperature. The standard deviations were 0.0442 σ Sy 0.152 for ˆ ˆ Sy and 0.099 σ Sut 0.11 for Sut . (Data source: E. A. Brandes (ed.), Smithells Metal Reference Book, 6th ed., Butterworth, London, 1983 pp. 22128 to 22131.) Sy 0.8 0.7 0.6 0.5 0 RT 200 400 Temperature, °C 600 46 40 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition I. Basics © The McGrawHill Companies, 2008 2. Materials Mechanical Engineering Design Figure 210 Creep deformation Creep-time curve. 1st stage 2nd stage 3rd stage Time Of course, heat treatment, as will be shown, is used to make substantial changes in the mechanical properties of a material. Heating due to electric and gas welding also changes the mechanical properties. Such changes may be due to clamping during the welding process, as well as heating; the resulting stresses then remain when the parts have cooled and the clamps have been removed. Hardness tests can be used to learn whether the strength has been changed by welding, but such tests will not reveal the presence of residual stresses. 27 Numbering Systems The Society of Automotive Engineers (SAE) was the rst to recognize the need, and to adopt a system, for the numbering of steels. Later the American Iron and Steel Institute (AISI) adopted a similar system. In 1975 the SAE published the Unied Numbering System for Metals and Alloys (UNS); this system also contains cross-reference numbers for other material specications.7 The UNS uses a letter prex to designate the material, as, for example, G for the carbon and alloy steels, A for the aluminum alloys, C for the copper-base alloys, and S for the stainless or corrosion-resistant steels. For some materials, not enough agreement has as yet developed in the industry to warrant the establishment of a designation. For the steels, the rst two numbers following the letter prex indicate the composition, excluding the carbon content. The various compositions used are as follows: G10 Plain carbon G46 Nickel-molybdenum G11 Free-cutting carbon steel with more sulfur or phosphorus G48 Nickel-molybdenum G50 Chromium G13 Manganese G51 Chromium G23 Nickel G52 Chromium G25 Nickel G61 Chromium-vanadium G31 Nickel-chromium G86 Chromium-nickel-molybdenum G33 Nickel-chromium G87 Chromium-nickel-molybdenum G40 Molybdenum G92 Manganese-silicon G41 Chromium-molybdenum G94 Nickel-chromium-molybdenum G43 Nickel-chromium-molybdenum 7 Many of the materials discussed in the balance of this chapter are listed in the Appendix tables. Be sure to review these. BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition I. Basics 47 © The McGrawHill Companies, 2008 2. Materials Materials Table 21 Aluminum 99.00% pure and greater Aluminum Alloy Designations 41 Ax1xxx Copper alloys Ax2xxx Manganese alloys Ax3xxx Silicon alloys Ax4xxx Magnesium alloys Ax5xxx Magnesium-silicon alloys Ax6xxx Zinc alloys Ax7xxx The second number pair refers to the approximate carbon content. Thus, G10400 is a plain carbon steel with a nominal carbon content of 0.40 percent (0.37 to 0.44 percent). The fth number following the prex is used for special situations. For example, the old designation AISI 52100 represents a chromium alloy with about 100 points of carbon. The UNS designation is G52986. The UNS designations for the stainless steels, prex S, utilize the older AISI designations for the rst three numbers following the prex. The next two numbers are reserved for special purposes. The rst number of the group indicates the approximate composition. Thus 2 is a chromium-nickel-manganese steel, 3 is a chromium-nickel steel, and 4 is a chromium alloy steel. Sometimes stainless steels are referred to by their alloy content. Thus S30200 is often called an 18-8 stainless steel, meaning 18 percent chromium and 8 percent nickel. The prefix for the aluminum group is the letter A. The first number following the prefix indicates the processing. For example, A9 is a wrought aluminum, while A0 is a casting alloy. The second number designates the main alloy group as shown in Table 21. The third number in the group is used to modify the original alloy or to designate the impurity limits. The last two numbers refer to other alloys used with the basic group. The American Society for Testing and Materials (ASTM) numbering system for cast iron is in widespread use. This system is based on the tensile strength. Thus ASTM A18 speaks of classes; e.g., 30 cast iron has a minimum tensile strength of 30 kpsi. Note from Appendix A-24, however, that the typical tensile strength is 31 kpsi. You should be careful to designate which of the two values is used in design and problem work because of the signicance of factor of safety. 28 Sand Casting Sand casting is a basic low-cost process, and it lends itself to economical production in large quantities with practically no limit to the size, shape, or complexity of the part produced. In sand casting, the casting is made by pouring molten metal into sand molds. A pattern, constructed of metal or wood, is used to form the cavity into which the molten metal is poured. Recesses or holes in the casting are produced by sand cores introduced into the mold. The designer should make an effort to visualize the pattern and casting in the mold. In this way the problems of core setting, pattern removal, draft, and solidication can be studied. Castings to be used as test bars of cast iron are cast separately and properties may vary. 48 42 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition I. Basics 2. Materials © The McGrawHill Companies, 2008 Mechanical Engineering Design Steel castings are the most difcult of all to produce, because steel has the highest melting temperature of all materials normally used for casting. This high temperature aggravates all casting problems. The following rules will be found quite useful in the design of any sand casting: 1 2 3 4 All sections should be designed with a uniform thickness. The casting should be designed so as to produce a gradual change from section to section where this is necessary. Adjoining sections should be designed with generous llets or radii. A complicated part should be designed as two or more simple castings to be assembled by fasteners or by welding. Steel, gray iron, brass, bronze, and aluminum are most often used in castings. The minimum wall thickness for any of these materials is about 5 mm, though with particular care, thinner sections can be obtained with some materials. 29 Shell Molding The shell-molding process employs a heated metal pattern, usually made of cast iron, aluminum, or brass, which is placed in a shell-molding machine containing a mixture of dry sand and thermosetting resin. The hot pattern melts the plastic, which, together with the sand, forms a shell about 5 to 10 mm thick around the pattern. The shell is then baked at from 400 to 700°F for a short time while still on the pattern. It is then stripped from the pattern and placed in storage for use in casting. In the next step the shells are assembled by clamping, bolting, or pasting; they are placed in a backup material, such as steel shot; and the molten metal is poured into the cavity. The thin shell permits the heat to be conducted away so that solidication takes place rapidly. As solidication takes place, the plastic bond is burned and the mold collapses. The permeability of the backup material allows the gases to escape and the casting to air-cool. All this aids in obtaining a ne-grain, stress-free casting. Shell-mold castings feature a smooth surface, a draft that is quite small, and close tolerances. In general, the rules governing sand casting also apply to shell-mold casting. 210 Investment Casting Investment casting uses a pattern that may be made from wax, plastic, or other material. After the mold is made, the pattern is melted out. Thus a mechanized method of casting a great many patterns is necessary. The mold material is dependent upon the melting point of the cast metal. Thus a plaster mold can be used for some materials while others would require a ceramic mold. After the pattern is melted out, the mold is baked or red; when ring is completed, the molten metal may be poured into the hot mold and allowed to cool. If a number of castings are to be made, then metal or permanent molds may be suitable. Such molds have the advantage that the surfaces are smooth, bright, and accurate, so that little, if any, machining is required. Metal-mold castings are also known as die castings and centrifugal castings. 211 Powder-Metallurgy Process The powder-metallurgy process is a quantity-production process that uses powders from a single metal, several metals, or a mixture of metals and nonmetals. It consists essentially of mechanically mixing the powders, compacting them in dies at high pressures, BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition I. Basics 49 © The McGrawHill Companies, 2008 2. Materials Materials 43 and heating the compacted part at a temperature less than the melting point of the major ingredient. The particles are united into a single strong part similar to what would be obtained by melting the same ingredients together. The advantages are (1) the elimination of scrap or waste material, (2) the elimination of machining operations, (3) the low unit cost when mass-produced, and (4) the exact control of composition. Some of the disadvantages are (1) the high cost of dies, (2) the lower physical properties, (3) the higher cost of materials, (4) the limitations on the design, and (5) the limited range of materials that can be used. Parts commonly made by this process are oil-impregnated bearings, incandescent lamp laments, cemented-carbide tips for tools, and permanent magnets. Some products can be made only by powder metallurgy: surgical implants, for example. The structure is different from what can be obtained by melting the same ingredients. 212 Hot-Working Processes By hot working are meant such processes as rolling, forging, hot extrusion, and hot pressing, in which the metal is heated above its recrystallation temperature. Hot rolling is usually used to create a bar of material of a particular shape and dimension. Figure 211 shows some of the various shapes that are commonly produced by the hot-rolling process. All of them are available in many different sizes as well as in different materials. The materials most available in the hot-rolled bar sizes are steel, aluminum, magnesium, and copper alloys. Tubing can be manufactured by hot-rolling strip or plate. The edges of the strip are rolled together, creating seams that are either butt-welded or lap-welded. Seamless tubing is manufactured by roll-piercing a solid heated rod with a piercing mandrel. Extrusion is the process by which great pressure is applied to a heated metal billet or blank, causing it to ow through a restricted orice. This process is more common with materials of low melting point, such as aluminum, copper, magnesium, lead, tin, and zinc. Stainless steel extrusions are available on a more limited basis. Forging is the hot working of metal by hammers, presses, or forging machines. In common with other hot-working processes, forging produces a rened grain structure that results in increased strength and ductility. Compared with castings, forgings have greater strength for the same weight. In addition, drop forgings can be made smoother and more accurate than sand castings, so that less machining is necessary. However, the initial cost of the forging dies is usually greater than the cost of patterns for castings, although the greater unit strength rather than the cost is usually the deciding factor between these two processes. Figure 211 Common shapes available through hot rolling. Round Square Half oval Flat Hexagon (a) Bar shapes Wide flange Channel Angle (b) Structural shapes Tee Zee 50 44 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition I. Basics 2. Materials © The McGrawHill Companies, 2008 Mechanical Engineering Design Figure 212 100 Cold-drawn 80 Strength, kpsi Stress-strain diagram for hot-rolled and cold-drawn UNS G10350 steel. Hot-rolled Yield point 60 Yield point 40 20 0 0 0.2 0.4 0.6 Elongation, in 213 Cold-Working Processes By cold working is meant the forming of the metal while at a low temperature (usually room temperature). In contrast to parts produced by hot working, cold-worked parts have a bright new nish, are more accurate, and require less machining. Cold-nished bars and shafts are produced by rolling, drawing, turning, grinding, and polishing. Of these methods, by far the largest percentage of products are made by the cold-rolling and cold-drawing processes. Cold rolling is now used mostly for the production of wide ats and sheets. Practically all cold-nished bars are made by cold drawing but even so are sometimes mistakenly called cold-rolled bars. In the drawing process, the hot-rolled bars are rst cleaned of scale and then drawn by pulling them 1 1 through a die that reduces the size about 32 to 16 in. This process does not remove material from the bar but reduces, or draws down, the size. Many different shapes of hot-rolled bars may be used for cold drawing. Cold rolling and cold drawing have the same effect upon the mechanical properties. The cold-working process does not change the grain size but merely distorts it. Cold working results in a large increase in yield strength, an increase in ultimate strength and hardness, and a decrease in ductility. In Fig. 212 the properties of a colddrawn bar are compared with those of a hot-rolled bar of the same material. Heading is a cold-working process in which the metal is gathered, or upset. This operation is commonly used to make screw and rivet heads and is capable of producing a wide variety of shapes. Roll threading is the process of rolling threads by squeezing and rolling a blank between two serrated dies. Spinning is the operation of working sheet material around a rotating form into a circular shape. Stamping is the term used to describe punch-press operations such as blanking, coining, forming, and shallow drawing. 214 The Heat Treatment of Steel Heat treatment of steel refers to time- and temperature-controlled processes that relieve residual stresses and/or modies material properties such as hardness (strength), ductility, and toughness. Other mechanical or chemical operations are sometimes grouped under the heading of heat treatment. The common heat-treating operations are annealing, quenching, tempering, and case hardening. BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition I. Basics 2. Materials 51 © The McGrawHill Companies, 2008 Materials 45 Annealing When a material is cold- or hot-worked, residual stresses are built in, and, in addition, the material usually has a higher hardness as a result of these working operations. These operations change the structure of the material so that it is no longer represented by the equilibrium diagram. Full annealing and normalizing is a heating operation that permits the material to transform according to the equilibrium diagram. The material to be annealed is heated to a temperature that is approximately 100°F above the critical temperature. It is held at this temperature for a time that is sufcient for the carbon to become dissolved and diffused through the material. The object being treated is then allowed to cool slowly, usually in the furnace in which it was treated. If the transformation is complete, then it is said to have a full anneal. Annealing is used to soften a material and make it more ductile, to relieve residual stresses, and to rene the grain structure. The term annealing includes the process called normalizing. Parts to be normalized may be heated to a slightly higher temperature than in full annealing. This produces a coarser grain structure, which is more easily machined if the material is a low-carbon steel. In the normalizing process the part is cooled in still air at room temperature. Since this cooling is more rapid than the slow cooling used in full annealing, less time is available for equilibrium, and the material is harder than fully annealed steel. Normalizing is often used as the nal treating operation for steel. The cooling in still air amounts to a slow quench. Quenching Eutectoid steel that is fully annealed consists entirely of pearlite, which is obtained from austenite under conditions of equilibrium. A fully annealed hypoeutectoid steel would consist of pearlite plus ferrite, while hypereutectoid steel in the fully annealed condition would consist of pearlite plus cementite. The hardness of steel of a given carbon content depends upon the structure that replaces the pearlite when full annealing is not carried out. The absence of full annealing indicates a more rapid rate of cooling. The rate of cooling is the factor that determines the hardness. A controlled cooling rate is called quenching. A mild quench is obtained by cooling in still air, which, as we have seen, is obtained by the normalizing process. The two most widely used media for quenching are water and oil. The oil quench is quite slow but prevents quenching cracks caused by rapid expansion of the object being treated. Quenching in water is used for carbon steels and for medium-carbon, low-alloy steels. The effectiveness of quenching depends upon the fact that when austenite is cooled it does not transform into pearlite instantaneously but requires time to initiate and complete the process. Since the transformation ceases at about 800°F, it can be prevented by rapidly cooling the material to a lower temperature. When the material is cooled rapidly to 400°F or less, the austenite is transformed into a structure called martensite. Martensite is a supersaturated solid solution of carbon in ferrite and is the hardest and strongest form of steel. If steel is rapidly cooled to a temperature between 400 and 800°F and held there for a sufcient length of time, the austenite is transformed into a material that is generally called bainite. Bainite is a structure intermediate between pearlite and martensite. Although there are several structures that can be identied between the temperatures given, depending upon the temperature used, they are collectively known as bainite. By the choice of this transformation temperature, almost any variation of structure may be obtained. These range all the way from coarse pearlite to ne martensite. 52 46 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition I. Basics © The McGrawHill Companies, 2008 2. Materials Mechanical Engineering Design Tempering When a steel specimen has been fully hardened, it is very hard and brittle and has high residual stresses. The steel is unstable and tends to contract on aging. This tendency is increased when the specimen is subjected to externally applied loads, because the resultant stresses contribute still more to the instability. These internal stresses can be relieved by a modest heating process called stress relieving, or a combination of stress relieving and softening called tempering or drawing. After the specimen has been fully hardened by being quenched from above the critical temperature, it is reheated to some temperature below the critical temperature for a certain period of time and then allowed to cool in still air. The temperature to which it is reheated depends upon the composition and the degree of hardness or toughness desired.8 This reheating operation releases the carbon held in the martensite, forming carbide crystals. The structure obtained is called tempered martensite. It is now essentially a superne dispersion of iron carbide(s) in ne-grained ferrite. The effect of heat-treating operations upon the various mechanical properties of a low alloy steel is shown graphically in Fig. 213. Figure 213 300 Tensile strength 600 200 150 Brinell hardness Tensile and yield strength, kpsi 250 Yield strength 80 500 Brinell 60 400 40 Reduction area 100 50 300 20 Elongation 200 400 600 800 1000 1200 Percent elongation and reduction in area The effect of thermalmechanical history on the mechanical properties of AISI 4340 steel. (Prepared by the International Nickel Company.) 0 1400 Tempering temperature, °F Condition Yield strength, kpsi Reduction in area, % Elongation in 2 in, % Brinell hardness, Bhn Normalized 200 147 20 10 410 As rolled 190 144 18 9 380 Annealed 8 Tensile strength, kpsi 120 99 43 18 228 For the quantitative aspects of tempering in plain carbon and low-alloy steels, see Charles R. Mischke, The Strength of Cold-Worked and Heat-Treated Steels, Chap. 33 in Joseph E. Shigley, Charles R. Mischke, and Thomas H. Brown, Jr. (eds.), Standard Handbook of Machine Design, 3rd ed., McGraw-Hill, New York, 2004. BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition I. Basics 2. Materials 53 © The McGrawHill Companies, 2008 Materials 47 Case Hardening The purpose of case hardening is to produce a hard outer surface on a specimen of lowcarbon steel while at the same time retaining the ductility and toughness in the core. This is done by increasing the carbon content at the surface. Either solid, liquid, or gaseous carburizing materials may be used. The process consists of introducing the part to be carburized into the carburizing material for a stated time and at a stated temperature, depending upon the depth of case desired and the composition of the part. The part may then be quenched directly from the carburization temperature and tempered, or in some cases it must undergo a double heat treatment in order to ensure that both the core and the case are in proper condition. Some of the more useful case-hardening processes are pack carburizing, gas carburizing, nitriding, cyaniding, induction hardening, and ame hardening. In the last two cases carbon is not added to the steel in question, generally a medium carbon steel, for example SAE/AISI 1144. Quantitative Estimation of Properties of Heat-Treated Steels Courses in metallurgy (or material science) for mechanical engineers usually present the addition method of Crafts and Lamont for the prediction of heat-treated properties from the Jominy test for plain carbon steels.9 If this has not been in your prerequisite experience, then refer to the Standard Handbook of Machine Design, where the addition method is covered with examples.10 If this book is a textbook for a machine elements course, it is a good class project (many hands make light work) to study the method and report to the class. For low-alloy steels, the multiplication method of Grossman11 and Field12 is explained in the Standard Handbook of Machine Design (Secs. 29.6 and 33.6). Modern Steels and Their Properties Handbook explains how to predict the Jominy curve by the method of Grossman and Field from a ladle analysis and grain size.13 Bethlehem Steel has developed a circular plastic slide rule that is convenient to the purpose. 215 Alloy Steels Although a plain carbon steel is an alloy of iron and carbon with small amounts of manganese, silicon, sulfur, and phosphorus, the term alloy steel is applied when one or more elements other than carbon are introduced in sufcient quantities to modify its properties substantially. The alloy steels not only possess more desirable physical properties but also permit a greater latitude in the heat-treating process. Chromium The addition of chromium results in the formation of various carbides of chromium that are very hard, yet the resulting steel is more ductile than a steel of the same hardness produced by a simple increase in carbon content. Chromium also renes the grain structure so that these two combined effects result in both increased toughness and increased hardness. The addition of chromium increases the critical range of temperatures and moves the eutectoid point to the left. Chromium is thus a very useful alloying element. 9 W. Crafts and J. L. Lamont, Hardenability and Steel Selection, Pitman and Sons, London, 1949. 10 Charles R. Mischke, Chap. 33 in Joseph E. Shigley, Charles R. Mischke, and Thomas H. Brown, Jr. (eds.), Standard Handbook of Machine Design, 3rd ed., McGraw-Hill, New York, 2004, p. 33.9. 11 M. A. Grossman, AIME, February 1942. 12 J. Field, Metals Progress, March 1943. 13 Modern Steels and Their Properties, 7th ed., Handbook 2757, Bethlehem Steel, 1972, pp. 4650. 54 48 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition I. Basics 2. Materials © The McGrawHill Companies, 2008 Mechanical Engineering Design Nickel The addition of nickel to steel also causes the eutectoid point to move to the left and increases the critical range of temperatures. Nickel is soluble in ferrite and does not form carbides or oxides. This increases the strength without decreasing the ductility. Case hardening of nickel steels results in a better core than can be obtained with plain carbon steels. Chromium is frequently used in combination with nickel to obtain the toughness and ductility provided by the nickel and the wear resistance and hardness contributed by the chromium. Manganese Manganese is added to all steels as a deoxidizing and desulfurizing agent, but if the sulfur content is low and the manganese content is over 1 percent, the steel is classied as a manganese alloy. Manganese dissolves in the ferrite and also forms carbides. It causes the eutectoid point to move to the left and lowers the critical range of temperatures. It increases the time required for transformation so that oil quenching becomes practicable. Silicon Silicon is added to all steels as a deoxidizing agent. When added to very-low-carbon steels, it produces a brittle material with a low hysteresis loss and a high magnetic permeability. The principal use of silicon is with other alloying elements, such as manganese, chromium, and vanadium, to stabilize the carbides. Molybdenum While molybdenum is used alone in a few steels, it nds its greatest use when combined with other alloying elements, such as nickel, chromium, or both. Molybdenum forms carbides and also dissolves in ferrite to some extent, so that it adds both hardness and toughness. Molybdenum increases the critical range of temperatures and substantially lowers the transformation point. Because of this lowering of the transformation point, molybdenum is most effective in producing desirable oil-hardening and air-hardening properties. Except for carbon, it has the greatest hardening effect, and because it also contributes to a ne grain size, this results in the retention of a great deal of toughness. Vanadium Vanadium has a very strong tendency to form carbides; hence it is used only in small amounts. It is a strong deoxidizing agent and promotes a ne grain size. Since some vanadium is dissolved in the ferrite, it also toughens the steel. Vanadium gives a wide hardening range to steel, and the alloy can be hardened from a higher temperature. It is very difcult to soften vanadium steel by tempering; hence, it is widely used in tool steels. Tungsten Tungsten is widely used in tool steels because the tool will maintain its hardness even at red heat. Tungsten produces a ne, dense structure and adds both toughness and hardness. Its effect is similar to that of molybdenum, except that it must be added in greater quantities. 216 Corrosion-Resistant Steels Iron-base alloys containing at least 12 percent chromium are called stainless steels. The most important characteristic of these steels is their resistance to many, but not all, corrosive conditions. The four types available are the ferritic chromium steels, the BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition I. Basics 2. Materials 55 © The McGrawHill Companies, 2008 Materials 49 austenitic chromium-nickel steels, and the martensitic and precipitation-hardenable stainless steels. The ferritic chromium steels have a chromium content ranging from 12 to 27 percent. Their corrosion resistance is a function of the chromium content, so that alloys containing less than 12 percent still exhibit some corrosion resistance, although they may rust. The quench-hardenability of these steels is a function of both the chromium and the carbon content. The very high carbon steels have good quench hardenability up to about 18 percent chromium, while in the lower carbon ranges it ceases at about 13 percent. If a little nickel is added, these steels retain some degree of hardenability up to 20 percent chromium. If the chromium content exceeds 18 percent, they become difcult to weld, and at the very high chromium levels the hardness becomes so great that very careful attention must be paid to the service conditions. Since chromium is expensive, the designer will choose the lowest chromium content consistent with the corrosive conditions. The chromium-nickel stainless steels retain the austenitic structure at room temperature; hence, they are not amenable to heat treatment. The strength of these steels can be greatly improved by cold working. They are not magnetic unless cold-worked. Their work hardenability properties also cause them to be difcult to machine. All the chromium-nickel steels may be welded. They have greater corrosion-resistant properties than the plain chromium steels. When more chromium is added for greater corrosion resistance, more nickel must also be added if the austenitic properties are to be retained. 217 Casting Materials Gray Cast Iron Of all the cast materials, gray cast iron is the most widely used. This is because it has a very low cost, is easily cast in large quantities, and is easy to machine. The principal objections to the use of gray cast iron are that it is brittle and that it is weak in tension. In addition to a high carbon content (over 1.7 percent and usually greater than 2 percent), cast iron also has a high silicon content, with low percentages of sulfur, manganese, and phosphorus. The resultant alloy is composed of pearlite, ferrite, and graphite, and under certain conditions the pearlite may decompose into graphite and ferrite. The resulting product then contains all ferrite and graphite. The graphite, in the form of thin akes distributed evenly throughout the structure, darkens it; hence, the name gray cast iron. Gray cast iron is not readily welded, because it may crack, but this tendency may be reduced if the part is carefully preheated. Although the castings are generally used in the as-cast condition, a mild anneal reduces cooling stresses and improves the machinability. The tensile strength of gray cast iron varies from 100 to 400 MPa (15 to 60 kpsi), and the compressive strengths are 3 to 4 times the tensile strengths. The modulus of elasticity varies widely, with values extending all the way from 75 to 150 GPa (11 to 22 Mpsi). Ductile and Nodular Cast Iron Because of the lengthy heat treatment required to produce malleable cast iron, engineers have long desired a cast iron that would combine the ductile properties of malleable iron with the ease of casting and machining of gray iron and at the same time would possess these properties in the as-cast conditions. A process for producing such a material using magnesium-containing material seems to fulll these requirements. 56 50 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition I. Basics 2. Materials © The McGrawHill Companies, 2008 Mechanical Engineering Design Ductile cast iron, or nodular cast iron, as it is sometimes called, is essentially the same as malleable cast iron, because both contain graphite in the form of spheroids. However, ductile cast iron in the as-cast condition exhibits properties very close to those of malleable iron, and if a simple 1-h anneal is given and is followed by a slow cool, it exhibits even more ductility than the malleable product. Ductile iron is made by adding MgFeSi to the melt; since magnesium boils at this temperature, it is necessary to alloy it with other elements before it is introduced. Ductile iron has a high modulus of elasticity (172 GPa or 25 Mpsi) as compared with gray cast iron, and it is elastic in the sense that a portion of the stress-strain curve is a straight line. Gray cast iron, on the other hand, does not obey Hookes law, because the modulus of elasticity steadily decreases with increase in stress. Like gray cast iron, however, nodular iron has a compressive strength that is higher than the tensile strength, although the difference is not as great. In 40 years it has become extensively used. White Cast Iron If all the carbon in cast iron is in the form of cementite and pearlite, with no graphite present, the resulting structure is white and is known as white cast iron. This may be produced in two ways. The composition may be adjusted by keeping the carbon and silicon content low, or the gray-cast-iron composition may be cast against chills in order to promote rapid cooling. By either method, a casting with large amounts of cementite is produced, and as a result the product is very brittle and hard to machine but also very resistant to wear. A chill is usually used in the production of gray-iron castings in order to provide a very hard surface within a particular area of the casting, while at the same time retaining the more desirable gray structure within the remaining portion. This produces a relatively tough casting with a wear-resistant area. Malleable Cast Iron If white cast iron within a certain composition range is annealed, a product called malleable cast iron is formed. The annealing process frees the carbon so that it is present as graphite, just as in gray cast iron but in a different form. In gray cast iron the graphite is present in a thin ake form, while in malleable cast iron it has a nodular form and is known as temper carbon. A good grade of malleable cast iron may have a tensile strength of over 350 MPa (50 kpsi), with an elongation of as much as 18 percent. The percentage elongation of a gray cast iron, on the other hand, is seldom over 1 percent. Because of the time required for annealing (up to 6 days for large and heavy castings), malleable iron is necessarily somewhat more expensive than gray cast iron. Alloy Cast Irons Nickel, chromium, and molybdenum are the most common alloying elements used in cast iron. Nickel is a general-purpose alloying element, usually added in amounts up to 5 percent. Nickel increases the strength and density, improves the wearing qualities, and raises the machinability. If the nickel content is raised to 10 to 18 percent, an austenitic structure with valuable heat- and corrosion-resistant properties results. Chromium increases the hardness and wear resistance and, when used with a chill, increases the tendency to form white iron. When chromium and nickel are both added, the hardness and strength are improved without a reduction in the machinability rating. Molybdenum added in quantities up to 1.25 percent increases the stiffness, hardness, tensile strength, and impact resistance. It is a widely used alloying element. BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition I. Basics 2. Materials 57 © The McGrawHill Companies, 2008 Materials 51 Cast Steels The advantage of the casting process is that parts having complex shapes can be manufactured at costs less than fabrication by other means, such as welding. Thus the choice of steel castings is logical when the part is complex and when it must also have a high strength. The higher melting temperatures for steels do aggravate the casting problems and require closer attention to such details as core design, section thicknesses, fillets, and the progress of cooling. The same alloying elements used for the wrought steels can be used for cast steels to improve the strength and other mechanical properties. Cast-steel parts can also be heat-treated to alter the mechanical properties, and, unlike the cast irons, they can be welded. 218 Nonferrous Metals Aluminum The outstanding characteristics of aluminum and its alloys are their strength-weight ratio, their resistance to corrosion, and their high thermal and electrical conductivity. The density of aluminum is about 2770 kg/m3 (0.10 lbf/in3), compared with 7750 kg/m3 (0.28 lbf/in3) for steel. Pure aluminum has a tensile strength of about 90 MPa (13 kpsi), but this can be improved considerably by cold working and also by alloying with other materials. The modulus of elasticity of aluminum, as well as of its alloys, is 71.7 GPa (10.4 Mpsi), which means that it has about one-third the stiffness of steel. Considering the cost and strength of aluminum and its alloys, they are among the most versatile materials from the standpoint of fabrication. Aluminum can be processed by sand casting, die casting, hot or cold working, or extruding. Its alloys can be machined, press-worked, soldered, brazed, or welded. Pure aluminum melts at 660°C (1215°F), which makes it very desirable for the production of either permanent or sand-mold castings. It is commercially available in the form of plate, bar, sheet, foil, rod, and tube and in structural and extruded shapes. Certain precautions must be taken in joining aluminum by soldering, brazing, or welding; these joining methods are not recommended for all alloys. The corrosion resistance of the aluminum alloys depends upon the formation of a thin oxide coating. This lm forms spontaneously because aluminum is inherently very reactive. Constant erosion or abrasion removes this lm and allows corrosion to take place. An extra-heavy oxide lm may be produced by the process called anodizing. In this process the specimen is made to become the anode in an electrolyte, which may be chromic acid, oxalic acid, or sulfuric acid. It is possible in this process to control the color of the resulting lm very accurately. The most useful alloying elements for aluminum are copper, silicon, manganese, magnesium, and zinc. Aluminum alloys are classied as casting alloys or wrought alloys. The casting alloys have greater percentages of alloying elements to facilitate casting, but this makes cold working difcult. Many of the casting alloys, and some of the wrought alloys, cannot be hardened by heat treatment. The alloys that are heattreatable use an alloying element that dissolves in the aluminum. The heat treatment consists of heating the specimen to a temperature that permits the alloying element to pass into solution, then quenching so rapidly that the alloying element is not precipitated. The aging process may be accelerated by heating slightly, which results in even greater hardness and strength. One of the better-known heat-treatable alloys is duraluminum, or 2017 (4 percent Cu, 0.5 percent Mg, 0.5 percent Mn). This alloy hardens in 4 days at room temperature. Because of this rapid aging, the alloy must be stored under 58 52 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition I. Basics 2. Materials © The McGrawHill Companies, 2008 Mechanical Engineering Design refrigeration after quenching and before forming, or it must be formed immediately after quenching. Other alloys (such as 5053) have been developed that age-harden much more slowly, so that only mild refrigeration is required before forming. After forming, they are articially aged in a furnace and possess approximately the same strength and hardness as the 2024 alloys. Those alloys of aluminum that cannot be heat-treated can be hardened only by cold working. Both work hardening and the hardening produced by heat treatment may be removed by an annealing process. Magnesium The density of magnesium is about 1800 kg/m3 (0.065 lb/in3), which is two-thirds that of aluminum and one-fourth that of steel. Since it is the lightest of all commercial metals, its greatest use is in the aircraft and automotive industries, but other uses are now being found for it. Although the magnesium alloys do not have great strength, because of their light weight the strength-weight ratio compares favorably with the stronger aluminum and steel alloys. Even so, magnesium alloys nd their greatest use in applications where strength is not an important consideration. Magnesium will not withstand elevated temperatures; the yield point is denitely reduced when the temperature is raised to that of boiling water. Magnesium and its alloys have a modulus of elasticity of 45 GPa (6.5 Mpsi) in tension and in compression, although some alloys are not as strong in compression as in tension. Curiously enough, cold working reduces the modulus of elasticity. A range of cast magnesium alloys are also available. Titanium Titanium and its alloys are similar in strength to moderate-strength steel but weigh half as much as steel. The material exhibits very good resistence to corrosion, has low thermal conductivity, is nonmagnetic, and has high-temperature strength. Its modulus of elasticity is between those of steel and aluminum at 16.5 Mpsi (114 GPa). Because of its many advantages over steel and aluminum, applications include: aerospace and military aircraft structures and components, marine hardware, chemical tanks and processing equipment, uid handling systems, and human internal replacement devices. The disadvantages of titanium are its high cost compared to steel and aluminum and the difculty of machining it. Copper-Base Alloys When copper is alloyed with zinc, it is usually called brass. If it is alloyed with another element, it is often called bronze. Sometimes the other element is specied too, as, for example, tin bronze or phosphor bronze. There are hundreds of variations in each category. Brass with 5 to 15 Percent Zinc The low-zinc brasses are easy to cold work, especially those with the higher zinc content. They are ductile but often hard to machine. The corrosion resistance is good. Alloys included in this group are gilding brass (5 percent Zn), commercial bronze (10 percent Zn), and red brass (15 percent Zn). Gilding brass is used mostly for jewelry and articles to be gold-plated; it has the same ductility as copper but greater strength, accompanied by poor machining characteristics. Commercial bronze is used for jewelry and for forgings and stampings, because of its ductility. Its machining properties are poor, but it has excellent cold-working properties. Red brass has good corrosion resistance as well as high-temperature strength. Because of this it is used a great deal in the form of tubing or piping to carry hot water in such applications as radiators or condensers. BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition I. Basics 2. Materials 59 © The McGrawHill Companies, 2008 Materials 53 Brass with 20 to 36 Percent Zinc Included in the intermediate-zinc group are low brass (20 percent Zn), cartridge brass (30 percent Zn), and yellow brass (35 percent Zn). Since zinc is cheaper than copper, these alloys cost less than those with more copper and less zinc. They also have better machinability and slightly greater strength; this is offset, however, by poor corrosion resistance and the possibility of cracking at points of residual stresses. Low brass is very similar to red brass and is used for articles requiring deep-drawing operations. Of the copper-zinc alloys, cartridge brass has the best combination of ductility and strength. Cartridge cases were originally manufactured entirely by cold working; the process consisted of a series of deep draws, each draw being followed by an anneal to place the material in condition for the next draw, hence the name cartridge brass. Although the hot-working ability of yellow brass is poor, it can be used in practically any other fabricating process and is therefore employed in a large variety of products. When small amounts of lead are added to the brasses, their machinability is greatly improved and there is some improvement in their abilities to be hot-worked. The addition of lead impairs both the cold-working and welding properties. In this group are low-leaded brass (32 1 percent Zn, 1 percent Pb), high-leaded brass (34 percent Zn, 2 2 2 percent Pb), and free-cutting brass (35 1 percent Zn, 3 percent Pb). The low-leaded 2 brass is not only easy to machine but has good cold-working properties. It is used for various screw-machine parts. High-leaded brass, sometimes called engravers brass, is used for instrument, lock, and watch parts. Free-cutting brass is also used for screwmachine parts and has good corrosion resistance with excellent mechanical properties. Admiralty metal (28 percent Zn) contains 1 percent tin, which imparts excellent corrosion resistance, especially to saltwater. It has good strength and ductility but only fair machining and working characteristics. Because of its corrosion resistance it is used in power-plant and chemical equipment. Aluminum brass (22 percent Zn) contains 2 percent aluminum and is used for the same purposes as admiralty metal, because it has nearly the same properties and characteristics. In the form of tubing or piping, it is favored over admiralty metal, because it has better resistance to erosion caused by highvelocity water. Brass with 36 to 40 Percent Zinc Brasses with more than 38 percent zinc are less ductile than cartridge brass and cannot be cold-worked as severely. They are frequently hot-worked and extruded. Muntz metal (40 percent Zn) is low in cost and mildly corrosion-resistant. Naval brass has the same composition as Muntz metal except for the addition of 0.75 percent tin, which contributes to the corrosion resistance. Bronze Silicon bronze, containing 3 percent silicon and 1 percent manganese in addition to the copper, has mechanical properties equal to those of mild steel, as well as good corrosion resistance. It can be hot- or cold-worked, machined, or welded. It is useful wherever corrosion resistance combined with strength is required. Phosphor bronze, made with up to 11 percent tin and containing small amounts of phosphorus, is especially resistant to fatigue and corrosion. It has a high tensile strength and a high capacity to absorb energy, and it is also resistant to wear. These properties make it very useful as a spring material. Aluminum bronze is a heat-treatable alloy containing up to 12 percent aluminum. This alloy has strength and corrosion-resistance properties that are better than those of brass, and in addition, its properties may be varied over a wide range by cold working, heat treating, 60 54 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition I. Basics © The McGrawHill Companies, 2008 2. Materials Mechanical Engineering Design or changing the composition. When iron is added in amounts up to 4 percent, the alloy has a high endurance limit, a high shock resistance, and excellent wear resistance. Beryllium bronze is another heat-treatable alloy, containing about 2 percent beryllium. This alloy is very corrosion resistant and has high strength, hardness, and resistance to wear. Although it is expensive, it is used for springs and other parts subjected to fatigue loading where corrosion resistance is required. With slight modication most copper-based alloys are available in cast form. 219 Plastics The term thermoplastics is used to mean any plastic that ows or is moldable when heat is applied to it; the term is sometimes applied to plastics moldable under pressure. Such plastics can be remolded when heated. A thermoset is a plastic for which the polymerization process is nished in a hot molding press where the plastic is liqueed under pressure. Thermoset plastics cannot be remolded. Table 22 lists some of the most widely used thermoplastics, together with some of their characteristics and the range of their properties. Table 23, listing some of the Table 22 The Thermoplastics Source: These data have been obtained from the Machine Design Materials Reference Issue, published by Penton/IPC, Cleveland. These reference issues are published about every 2 years and constitute an excellent source of data on a great variety of materials. Name ABS group Su, kpsi 28 E, Mpsi Hardness Rockwell 0.100.37 60110R 350 4060 Acetal group 810 0.410.52 8094M Acrylic 510 0.200.47 92110M Fluoroplastic group 0.507 ··· 5080D Elongation Dimensional Heat Chemical % Stability Resistance Resistance Processing 375 100300 Nylon 814 0.180.45 112120R 10200 Phenylene oxide 718 0.350.92 115R, 106L 560 Good * Fair EMST Excellent Good High M High * Fair EMS High Excellent Excellent MPR Poor Poor Good CEM Excellent Good Fair EFM Polycarbonate 816 0.340.86 6291M 10125 Excellent Excellent Fair EMS Polyester 818 0.281.6 6590M 1300 Excellent Poor Excellent CLMR Polyimide 650 Excellent Excellent Excellent CLMP Good Excellent Excellent M Poor Poor EM Excellent Excellent EFM Poor Poor EFM Polyphenylene sulde ··· 88120M Very low 1419 0.11 122R 1.0 Polystyrene group 1.512 0.140.60 1090M 0.560 Polysulfone 10 0.36 120R 50100 Excellent 6585D 40450 ·· Polyvinyl chloride 1.57.5 0.350.60 ··· *Heat-resistant grades available. With exceptions. C Coatings L Laminates R Resins E Extrusions M Moldings S Sheet F Foams P Press and sinter methods T Tubing BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition I. Basics 61 © The McGrawHill Companies, 2008 2. Materials Materials 55 Table 23 The Thermosets Source: These data have been obtained from the Machine Design Materials Reference Issue, published by Penton/IPC, Cleveland. These reference issues are published about every 2 years and constitute an excellent source of data on a great variety of materials. Name Su, kpsi E, Mpsi Hardness Rockwell 0.050.30 99M* Elongation % Dimensional Stability Heat Resistance Chemical Resistance Excellent Good Fair M Excellent Excellent Excellent CM Good Excellent* Excellent* LR Excellent Excellent CMR Processing Alkyd 39 Allylic 410 Amino group 58 0.130.24 Epoxy 520 0.030.30* 80120M 110 Excellent Phenolics 59 0.100.25 7095E Excellent Good EMR 56 ··· 8090M ··· Excellent Silicones ··· Excellent Excellent CL MR 105120M ··· 110120M ··· ··· 0.300.90 ··· *With exceptions. C Coatings L Laminates R Resins E Extrusions M Moldings S Sheet F Foams P Press and sinter methods T Tubing thermosets, is similar. These tables are presented for information only and should not be used to make a nal design decision. The range of properties and characteristics that can be obtained with plastics is very great. The inuence of many factors, such as cost, moldability, coefcient of friction, weathering, impact strength, and the effect of llers and reinforcements, must be considered. Manufacturers catalogs will be found quite helpful in making possible selections. 220 Composite Materials14 Composite materials are formed from two or more dissimilar materials, each of which contributes to the nal properties. Unlike metallic alloys, the materials in a composite remain distinct from each other at the macroscopic level. Most engineering composites consist of two materials: a reinforcement called a ller and a matrix. The ller provides stiffness and strength; the matrix holds the material together and serves to transfer load among the discontinuous reinforcements. The most common reinforcements, illustrated in Fig. 214, are continuous bers, either straight or woven, short chopped bers, and particulates. The most common matrices are various plastic resins although other materials including metals are used. Metals and other traditional engineering materials are uniform, or isotropic, in nature. This means that material properties, such as strength, stiffness, and thermal conductivity, are independent of both position within the material and the choice of coordinate system. The discontinuous nature of composite reinforcements, though, means that material properties can vary with both position and direction. For example, an 14 For references see I. M. Daniel and O. Ishai, Engineering Mechanics of Composite Materials, Oxford University Press, 1994, and ASM Engineered Materials Handbook: Composites, ASM International, Materials Park, OH, 1988. 62 56 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition I. Basics © The McGrawHill Companies, 2008 2. Materials Mechanical Engineering Design Figure 214 Composites categorized by type of reinforcement. Particulate composite Randomly oriented short fiber composite Unidirectional continuous fiber composite Woven fabric composite epoxy resin reinforced with continuous graphite bers will have very high strength and stiffness in the direction of the bers, but very low properties normal or transverse to the bers. For this reason, structures of composite materials are normally constructed of multiple plies (laminates) where each ply is oriented to achieve optimal structural stiffness and strength performance. High strength-to-weight ratios, up to 5 times greater than those of high-strength steels, can be achieved. High stiffness-to-weight ratios can also be obtained, as much as 8 times greater than those of structural metals. For this reason, composite materials are becoming very popular in automotive, aircraft, and spacecraft applications where weight is a premium. The directionality of properties of composite materials increases the complexity of structural analyses. Isotropic materials are fully dened by two engineering constants: Youngs modulus E and Poissons ratio ν . A single ply of a composite material, however, requires four constants, dened with respect to the ply coordinate system. The constants are two Youngs moduli (the longitudinal modulus in the direction of the bers, E 1 , and the transverse modulus normal to the bers, E 2 ), one Poissons ratio (ν12 , called the major Poissons ratio), and one shear modulus (G 12 ). A fth constant, the minor Poissons ratio, ν21 , is determined through the reciprocity relation, ν21 / E 2 = ν12 / E 1 . Combining this with multiple plies oriented at different angles makes structural analysis of complex structures unapproachable by manual techniques. For this reason, computer software is available to calculate the properties of a laminated composite construction.15 221 Materials Selection As stated earlier, the selection of a material for a machine part or structural member is one of the most important decisions the designer is called on to make. Up to this point in this chapter we have discussed many important material physical properties, various characteristics of typical engineering materials, and various material production processes. The actual selection of a material for a particular design application can be an easy one, say, based on previous applications (1020 steel is always a good candidate because of its many positive attributes), or the selection process can be as involved and daunting as any design problem with the evaluation of the many material physical, economical, and processing parameters. There are systematic and optimizing approaches to material selection. Here, for illustration, we will only look at how to approach some material properties. One basic technique is to list all the important material properties associated with the design, e.g., strength, stiffness, and cost. This can be prioritized by using a weighting measure depending on what properties are more 15 About Composite Materials Software listing, http://composite.about.com/cs/software/index.htm. BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition I. Basics 63 © The McGrawHill Companies, 2008 2. Materials Materials 57 important than others. Next, for each property, list all available materials and rank them in order beginning with the best material; e.g., for strength, high-strength steel such as 4340 steel should be near the top of the list. For completeness of available materials, this might require a large source of material data. Once the lists are formed, select a manageable amount of materials from the top of each list. From each reduced list select the materials that are contained within every list for further review. The materials in the reduced lists can be graded within the list and then weighted according to the importance of each property. M. F. Ashby has developed a powerful systematic method using materials selection charts.16 This method has also been implemented in a software package called CES Edupack.17 The charts display data of various properties for the families and classes of materials listed in Table 24. For example, considering material stiffness properties, a simple bar chart plotting Youngs modulus E on the y axis is shown in Fig. 215. Each vertical line represents the range of values of E for a particular material. Only some of the materials are labeled. Now, more material information can be displayed if the x axis represents another material property, say density. Table 24 Family Short Name Metals (the metals and alloys of engineering) Material Families and Classes Classes Aluminum alloys Al alloys Copper alloys Cu alloys Lead alloys Lead alloys Magnesium alloys Mg alloys Nickel alloys Ni alloys Carbon steels Steels Stainless steels Stainless steels Tin alloys Tin alloys Titanium alloys W alloys Lead alloys Pb alloys Zinc alloys Ceramics Technical ceramics (ne ceramics capable of load-bearing application) Ti alloys Tungsten alloys Zn alloys Alumina AI2 O3 Aluminum nitride AIN Boron carbide B4 C Silicon carbide SiC Silicon nitride Si3 N4 Tungsten carbide Nontechnical ceramics (porous ceramics of construction) WC Brick Brick Concrete Concrete Stone Stone (continued) 16 M. F. Ashby, Materials Selection in Mechanical Design, 3rd ed., Elsevier Butterworth-Heinemann, Oxford, 2005. 17 Produced by Granta Design Limited. See www.grantadesign.com. 64 58 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition I. Basics © The McGrawHill Companies, 2008 2. Materials Mechanical Engineering Design Table 24 (continued) Family Classes Short Name Glasses Soda-lime glass Soda-lime glass Borosilicate glass Borosilicate glass Silica glass Silica glass Glass ceramic Glass ceramic Polymers (the thermoplastics and thermosets of engineering) Acrylonitrile butadiene styrene ABS Cellulose polymers CA lonomers lonomers Epoxies Epoxy Phenolics Phenolics Polyamides (nylons) PA Polycarbonate PC Polyesters Polyester Polyetheretherkeytone PEEK Polyethylene PE Polyethylene terephalate PET or PETE Polymethylmethacrylate PMMA Polyoxymethylene(Acetal) POM Polypropylene PP Polystyrene PTFE Polyvinylchloride Elastomers (engineering rubbers, natural and synthetic) PS Polytetrauorethylene PVC Butyl rubber Butyl rubber EVA EVA lsoprene lsoprene Natural rubber Natural rubber Polychloroprene (Neoprene) Neoprene Polyurethane PU Silicon elastomers Silicones Al-SiC Flexible polymer foams Flexible foams Rigid foams Cork Cork Bamboo Wood Natural materials GFRP Bamboo Foams CFRP Glass-ber reinforced polymers Rigid polymer foams Composites Carbon-ber reinforced polymers SiC reinforced aluminum Hybrids Wood From M. F. Ashby, Materials Selection in Mechanical Design, 3rd ed., Elsevier Butterworth-Heinemann, Oxford, 2005. Table 41, pp. 4950. BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition I. Basics 65 © The McGrawHill Companies, 2008 2. Materials Materials 59 Figure 215 Youngs modulus E for various materials. (Figure courtesy of Prof. Mike Ashby, Granta Design, Cambridge, U.K.) 1000 Tungsten carbides Nickel alloys 100 Cast iron, gray Titanium alloys Low-alloy steel GFRP, epoxy matrix (isotropic) Copper alloys Soda-lime glass 10 Polyester Young's modulus, GPa Wood, typical along grain Wood, typical across grain 1 Acrylonitrile butadiene styrene (ABS) Rigid polymer foam (MD) 0.1 Cork 0.01 Polyurethane 1e-3 Butyl rubber Flexible polymer foam (VLD) 1e-4 Figure 216, called a bubble chart, represents Youngs modulus E plotted against density ρ . The line ranges for each material property plotted two-dimensionally now form ellipses, or bubbles. This plot is more useful than the two separate bar charts of each property. Now, we also see how stiffness/weight for various materials relate. Figure 216 also shows groups of bubbles outlined according to the material families of Table 24. In addition, dotted lines in the lower right corner of the chart indicate ratios of E β /ρ , which assist in material selection for minimum mass design. Lines drawn parallel to these lines represent different values for E β /ρ . For example, several parallel dotted lines are shown in Fig. 216 that represent different values of E /ρ ( β = 1). Since ( E /ρ ) 1/2 represents the speed of sound in a material, each dotted line, E /ρ , represents a different speed as indicated. To see how β ts into the mix, consider the following. The performance metric P of a structural element depends on (1) the functional requirements, (2) the geometry, and (3) the material properties of the structure. That is, P functional geometric material [(requirements F), (parameters G), (properties M)] or, symbolically, P = f ( F, G, M ) (220) 66 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 60 I. Basics © The McGrawHill Companies, 2008 2. Materials Mechanical Engineering Design Figure 216 Youngs modulus E versus density ρ for various materials. (Figure courtesy of Prof. Mike Ashby, Granta Design, Cambridge, U.K.) 1000 Composites 100 Wood n grain Young's modulus E, GPa Natural materials Longitudinal wave speed Metals Polyester o Concrete Zinc alloys Lead alloys y PEEK PET T Epoxies PC E1/3 PTFE Polymers Leather 1 WC W alloys Cu alloys y Wood PS n grain PP PE 104 m/s Rigid polymer foams 10 B4C Al A alloys C CFRP Glass s Mg alloys l GFRP PMMA PA 10 1 Al2O3 Steels Ni alloys Si3N4 SiC Ti alloys Technical ceramics E1/2 Foams 10 2 E EVA 103 m/s Silicone elastomers Polyurethane Cork Guidelines for minimum mass design Isoprene 10 Neoprene 3 Flexible polymer foams 10 4 102 m/s 0.01 Elastomers Butyl rubber MFA C4 0.1 1 10 Density , Mg/m3 If the function is separable, which it often is, we can write Eq. (220) as P = f 1 ( F ) · f 2 (G ) · f 3 ( M ) (221) For optimum design, we desire to maximize or minimize P. With regards to material properties alone, this is done by maximizing or minimizing f 3 ( M ), called the material efciency coefcient. For illustration, say we want to design a light, stiff, end-loaded cantilever beam with a circular cross section. For this we will use the mass m of the beam for the performance metric to minimize. The stiffness of the beam is related to its material and geometry. The stiffness of a beam is given by k = F /δ , where F and δ are the end load and deection, respectively (see Chap. 4). The end deection of an end-loaded cantilever beam is given in Table A9, beam 1, as δ = ymax = ( Fl 3 )/(3 E I ) , where E is Youngs modulus, I the second moment of the area, and l the length of the beam. Thus, the stiffness is given by k= F 3E I =3 δ l (222) From Table A-18, the second moment of the area of a circular cross section is I= π D4 A2 = 64 4π (223) BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition I. Basics 67 © The McGrawHill Companies, 2008 2. Materials Materials 61 where D and A are the diameter and area of the cross section, respectively. Substituting Eq. (223) in (222) and solving for A, we obtain A= 4π kl 3 3E 1/2 (224) The mass of the beam is given by (225) m = Alρ Substituting Eq. (224) into (225) and rearranging yields π 1/2 5/2 ρ (k )(l ) (226) 3 E 1/2 Equation (226) is of the form of Eq. (221). The term 2 π/3 is simply a constant and can be associated with any function, say f 1 ( F ). Thus, f 1 ( F ) = 2 π/3(k 1/2 ) is the functional requirement, stiffness; f 2 (G ) = (l 5/2 ) , the geometric parameter, length; and the material efciency coefcient m=2 f3( M ) = ρ E 1/2 (227) is the material property in terms of density and Youngs modulus. To minimize m we want to minimize f 3 ( M ), or maximize M= E 1/2 ρ (228) where M is called the material index, and β = 1 . Returning to Fig. 216, draw lines of 2 various values of E 1/2 /ρ as shown in Fig. 217. Lines of increasing M move up and to the left as shown. Thus, we see that good candidates for a light, stiff, end-loaded cantilever beam with a circular cross section are certain woods, composites, and ceramics. Other limits/constraints may warrant further investigation. Say, for further illustration, the design requirements indicate that we need a Youngs modulus greater than 50 GPa. Figure 218 shows how this further restricts the search region. This eliminates woods as a possible material. Figure 217 Modulusdensity Search region 100 Young's modulus E, GPa A schematic E versus ρ chart showing a grid of lines for various values the material index M = E1/2 /ρ . (From M. F. Ashby, Materials Selection in Mechanical Design, 3rd ed., Elsevier ButterworthHeinemann, Oxford, 2005.) 1000 Ceramics 3 1 0.3 Metals Composites 0.1 Increasing values of index E1/2/ 10 E1/2/ (GPa)1/2/(Mg/m)3 Woods 1 Polymers Foams 0.1 Elastomers 0.01 0.1 MFA 04 1 10 Density, Mg/m3 100 68 62 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition I. Basics © The McGrawHill Companies, 2008 2. Materials Mechanical Engineering Design Figure 218 1000 Modulusdensity Ceramics Metals Composites Search region 100 Young's modulus E, GPa The search region of Fig. 216 further reduced by restricting E 50 GPa. (From M. F. Ashby, Materials Selection in Mechanical Design, 3rd ed., Elsevier Butterworth-Heinemann, Oxford, 2005.) Index E1/2/r 3 10 Woods E Modulus 50 GPa 1 Polymers 0.1 Foams Elastomers MFA 04 0.01 0.1 1 10 100 Density, Mg/m3 Figure 219 Strength S versus density ρ for various materials. For metals, S is the 0.2 percent offset yield strength. For polymers, S is the 1 percent yield strength. For ceramics and glasses, S is the compressive crushing strength. For composites, S is the tensile strength. For elastomers, S is the tear strength. (Figure courtesy of Prof. Mike Ashby, Granta Design, Cambridge, U.K.) 10000 Strengthdensity 1000 Metals and polymers yield strength Ceramics and glasses MGR Elastomers tensile tear strength Composites tensile failure Strength S, MPa 100 Natural materials Ceramics Si3N4 Ti alloys y Metals Steels s SiC Al2O3 N Ni alloys A Al alloys Tungsten CFRP P alloys Mg alloys Polymers and Tungsten GFRP elastomers carbide PEEK Copper PET T PA alloys PC Wood PMMA to grain Composites Rigid polymer foams 10 Zinc alloys Lead alloys Foams 1 Concrete Butyl Wood rubber Silicone elastomers to grain Guide lines for minimum mass design Cork 0.1 S 3 S2/3 Flexible polymer foams 0.01 0.01 0.1 S1/2 MFA D4 1 Density , Mg/m3 10 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition I. Basics 69 © The McGrawHill Companies, 2008 2. Materials Materials 63 Certainly, in a given design exercise, there will be other considerations such as strength, environment, and cost, and other charts may be necessary to investigate. For example, Fig. 219 represents strength versus density for the material families. Also, we have not brought in the material process selection part of the picture. If done properly, material selection can result in a good deal of bookkeeping. This is where software packages such as CES Edupack become very effective. PROBLEMS 21 Determine the minimum tensile and yield strengths for SAE 1020 cold-drawn steel. 22 Determine the minimum tensile and yield strengths for UNS G10500 hot-rolled steel. 23 For the materials in Probs. 21 and 22, compare the following properties: minimum tensile and yield strengths, ductility, and stiffness. 24 Assuming you were specifying an AISI 1040 steel for an application where you desired to maximize the yield strength, how would you specify it? 25 Assuming you were specifying an AISI 1040 steel for an application where you desired to maximize the ductility, how would you specify it? 26 Determine the yield strength-to-weight density ratios (called specic strength) in units of inches for UNS G10350 hot-rolled steel, 2024-T4 aluminum, Ti-6A1-4V titanium alloy, and ASTM No. 30 gray cast iron. 27 Determine the stiffness-to-weight density ratios (called specic modulus) in units of inches for UNS G10350 hot-rolled steel, 2024-T4 aluminum, Ti-6A1-4V titanium alloy, and ASTM No. 30 gray cast iron. 28 Poissons ratio ν is a material property and is the ratio of the lateral strain and the longitudinal strain for a member in tension. For a homogeneous, isotropic material, the modulus of rigidity G is related to Youngs modulus as G= E 2(1 + ν) Using the tabulated values of G and E, determine Poissons ratio for steel, aluminum, beryllium copper, and gray cast iron. 29 A specimen of medium-carbon steel having an initial diameter of 0.503 in was tested in tension using a gauge length of 2 in. The following data were obtained for the elastic and plastic states: Elastic State Plastic State Load P, lbf Elongation, in Load P, lbf Area Ai, in2 1 000 0.0004 8 800 0.1984 2 000 0.0006 9 200 0.1978 3 000 0.0010 9 100 0.1963 4 000 0.0013 13 200 0.1924 7 000 0.0023 15 200 0.1875 8 400 0.0028 17 000 0.1563 8 800 0.0036 16 400 0.1307 9 200 0.0089 14 800 0.1077 70 64 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition I. Basics © The McGrawHill Companies, 2008 2. Materials Mechanical Engineering Design Note that there is some overlap in the data. Plot the engineering or nominal stress-strain diagram using two scales for the unit strain ǫ, one from zero to about 0.02 in/in and the other from zero to maximum strain. From this diagram nd the modulus of elasticity, the 0.2 percent offset yield strength, the ultimate strength, and the percent reduction in area. 210 Compute the true stress and the logarithmic strain using the data of Prob. 29 and plot the results on log-log paper. Then nd the plastic strength coefcient σ0 and the strain-strengthening exponent m. Find also the yield strength and the ultimate strength after the specimen has had 20 percent cold work. 211 The stress-strain data from a tensile test on a cast-iron specimen are Engineering stress, kpsi 5 16 19 26 32 40 46 49 54 0.20 Engineering strain, ǫ · 103 in/in 10 0.44 0.80 1.0 1.5 2.0 2.8 3.4 4.0 5.0 Plot the stress-strain locus and nd the 0.1 percent offset yield strength, and the tangent modulus of elasticity at zero stress and at 20 kpsi. 212 A straight bar of arbitrary cross section and thickness h is cold-formed to an inner radius R about an anvil as shown in the gure. Some surface at distance N having an original length L A B will remain unchanged in length after bending. This length is π( R + N ) 2 L AB = L AB = The lengths of the outer and inner surfaces, after bending, are Lo = π ( R + h) 2 Li = π R 2 Using Eq. (24), we then nd the true strains to be εo = ln R+h R+N εi = ln R R+N Tests show that |εo | = |εi |. Show that N=R B B h LAB Problem 212 N A R 1+ h R 1/2 1 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition I. Basics 71 © The McGrawHill Companies, 2008 2. Materials Materials 65 and εo = ln 1 + h R 1/2 213 A hot-rolled AISI 1212 steel is given 20 percent cold work. Determine the new values of the yield and ultimate strengths. 214 A steel member has a Brinell of HB = 250. Estimate the ultimate strength of the steel in MPa. 215 Brinell hardness tests were made on a random sample of 10 steel parts during processing. The results were HB values of 252 (2), 260, 254, 257 (2), 249 (3), and 251. Estimate the mean and standard deviation of the ultimate strength in kpsi. 216 Repeat Prob. 215 assuming the material to be cast iron. 217 Toughness is a term that relates to both strength and ductility. The fracture toughness, for examǫ ple, is dened as the total area under the stress-strain curve to fracture, u T = 0 f σ d ǫ . This area, called the modulus of toughness, is the strain energy per unit volume required to cause the material to fracture. A similar term, but dened within the elastic limit of the material, is called ǫ the modulus of resilience, u R = 0 y σ d ǫ, where ǫ y is the strain at yield. If the stress-strain is 2 linear to σ = Sy , then it can be shown that u R = Sy /2 E . For the material in Prob. 29: (a) Determine the modulus of resilience, and (b) Estimate the modulus of toughness, assuming that the last data point corresponds to fracture. 218 What is the material composition of AISI 4340 steel? 219 Search the website noted in Sec. 220 and report your ndings. 220 Research the material Inconel, briey described in Table A5. Compare it to various carbon and alloy steels in stiffness, strength, ductility, and toughness. What makes this material so special? 221 Pick a specic material given in the tables (e.g., 2024-T4 aluminum, SAE 1040 steel), and consult a local or regional distributor (consulting either the Yellow Pages or the Thomas Register) to obtain as much information as you can about cost and availability of the material and in what form (bar, plate, etc.). 222 Consider a tie rod transmitting a tensile force F. The corresponding tensile stress is given by σ = F / A , where A is the area of the cross section. The deection of the rod is given by Eq. (43), which is δ = ( Fl )/( AE ) , where l is the length of the rod. Using the Ashby charts of Figs. 216 and 219, explore what ductile materials are best suited for a light, stiff, and strong tie rod. Hints: Consider stiffness and strength separately. For use of Fig. 216, prove that β = 1 . For use of Fig. 219, relate the applied tensile stress to the material strength. 72 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition I. Basics © The McGrawHill Companies, 2008 3. Load and Stress Analysis 3 Load and Stress Analysis Chapter Outline 31 Equilibrium and Free-Body Diagrams 32 Shear Force and Bending Moments in Beams 33 Singularity Functions 34 Stress 35 Cartesian Stress Components 75 36 Mohrs Circle for Plane Stress 76 37 General Three-Dimensional Stress 38 Elastic Strain 39 Uniformly Distributed Stresses 68 71 73 75 82 83 84 310 Normal Stresses for Beams in Bending 311 Shear Stresses for Beams in Bending 312 Torsion 313 Stress Concentration 314 Stresses in Pressurized Cylinders 315 Stresses in Rotating Rings 316 Press and Shrink Fits 317 Temperature Effects 318 Curved Beams in Bending 319 Contact Stresses 320 Summary 85 90 95 105 107 110 110 111 112 117 121 67 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 68 I. Basics 73 © The McGrawHill Companies, 2008 3. Load and Stress Analysis Mechanical Engineering Design One of the main objectives of this book is to describe how specic machine components function and how to design or specify them so that they function safely without failing structurally. Although earlier discussion has described structural strength in terms of load or stress versus strength, failure of function for structural reasons may arise from other factors such as excessive deformations or deections. Here it is assumed that the reader has completed basic courses in statics of rigid bodies and mechanics of materials and is quite familiar with the analysis of loads, and the stresses and deformations associated with the basic load states of simple prismatic elements. In this chapter and Chap. 4 we will review and extend these topics briey. Complete derivations will not be presented here, and the reader is urged to return to basic textbooks and notes on these subjects. This chapter begins with a review of equilibrium and free-body diagrams associated with load-carrying components. One must understand the nature of forces before attempting to perform an extensive stress or deection analysis of a mechanical component. An extremely useful tool in handling discontinuous loading of structures employs Macaulay or singularity functions. Singularity functions are described in Sec. 33 as applied to the shear forces and bending moments in beams. In Chap. 4, the use of singularity functions will be expanded to show their real power in handling deections of complex geometry and statically indeterminate problems. Machine components transmit forces and motion from one point to another. The transmission of force can be envisioned as a ow or force distribution that can be further visualized by isolating internal surfaces within the component. Force distributed over a surface leads to the concept of stress, stress components, and stress transformations (Mohrs circle) for all possible surfaces at a point. The remainder of the chapter is devoted to the stresses associated with the basic loading of prismatic elements, such as uniform loading, bending, and torsion, and topics with major design ramications such as stress concentrations, thin- and thick-walled pressurized cylinders, rotating rings, press and shrink ts, thermal stresses, curved beams, and contact stresses. 31 Equilibrium and Free-Body Diagrams Equilibrium The word system will be used to denote any isolated part or portion of a machine or structureincluding all of it if desiredthat we wish to study. A system, under this denition, may consist of a particle, several particles, a part of a rigid body, an entire rigid body, or even several rigid bodies. If we assume that the system to be studied is motionless or, at most, has constant velocity, then the system has zero acceleration. Under this condition the system is said to be in equilibrium. The phrase static equilibrium is also used to imply that the system is at rest. For equilibrium, the forces and moments acting on the system balance such that F=0 (31) M=0 (32) which states that the sum of all force and the sum of all moment vectors acting upon a system in equilibrium is zero. 74 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition I. Basics 3. Load and Stress Analysis © The McGrawHill Companies, 2008 Load and Stress Analysis 69 Free-Body Diagrams We can greatly simplify the analysis of a very complex structure or machine by successively isolating each element and studying and analyzing it by the use of free-body diagrams. When all the members have been treated in this manner, the knowledge can be assembled to yield information concerning the behavior of the total system. Thus, free-body diagramming is essentially a means of breaking a complicated problem into manageable segments, analyzing these simple problems, and then, usually, putting the information together again. Using free-body diagrams for force analysis serves the following important purposes: The diagram establishes the directions of reference axes, provides a place to record the dimensions of the subsystem and the magnitudes and directions of the known forces, and helps in assuming the directions of unknown forces. The diagram simplies your thinking because it provides a place to store one thought while proceeding to the next. The diagram provides a means of communicating your thoughts clearly and unambiguously to other people. Careful and complete construction of the diagram claries fuzzy thinking by bringing out various points that are not always apparent in the statement or in the geometry of the total problem. Thus, the diagram aids in understanding all facets of the problem. The diagram helps in the planning of a logical attack on the problem and in setting up the mathematical relations. The diagram helps in recording progress in the solution and in illustrating the methods used. The diagram allows others to follow your reasoning, showing all forces. EXAMPLE 31 Solution Figure 31a shows a simplied rendition of a gear reducer where the input and output shafts AB and C D are rotating at constant speeds ωi and ωo, respectively. The input and output torques (torsional moments) are Ti = 240 lbf · in and To, respectively. The shafts are supported in the housing by bearings at A, B, C, and D. The pitch radii of gears G1 and G2 are r1 = 0.75 in and r2 = 1.5 in, respectively. Draw the free-body diagrams of each member and determine the net reaction forces and moments at all points. First, we will list all simplifying assumptions. 1 2 3 4 5 Gears G1 and G2 are simple spur gears with a standard pressure angle φ = 20° (see Sec. 135). The bearings are self-aligning and the shafts can be considered to be simply supported. The weight of each member is negligible. Friction is negligible. The mounting bolts at E, F, H, and I are the same size. The separate free-body diagrams of the members are shown in Figs. 31bd. Note that Newtons third law, called the law of action and reaction, is used extensively where each member mates. The force transmitted between the spur gears is not tangential but at the pressure angle φ. Thus, N = F tan φ. BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 70 I. Basics Mechanical Engineering Design F i, G1 240 lbf in Ti 5 in D 5 in C RCz x H C RI 4 in RH I 4 in (a) Gear reducer (b) Gear box 1 in RAz r1 G1 F RAz RCy I B y RAy H 1.5 in RBy RF z A RDy D G2 E RBz RDz A 0 B RBy B F RE E T0 RBz 75 © The McGrawHill Companies, 2008 3. Load and Stress Analysis RAy T0 Ti 240 lbf in N F D r2 C A N RDy RDz G2 RCy (c) Input shaft RCz (d ) Output shaft Figure 31 (a) Gear reducer; (b d ) free-body diagrams. Diagrams are not drawn to scale. Summing moments about the x axis of shaft A B in Fig. 31d gives Mx = F (0.75) 240 = 0 F = 320 lbf The normal force is N = 320 tan 20° = 116.5 lbf. Using the equilibrium equations for Figs. 31c and d, the reader should verify that: R Ay = 192 lbf, R Az = 69.9 lbf, R By = 128 lbf, R Bz = 46.6 lbf, RC y = 192 lbf, RC z = 69.9 lbf, R Dy = 128 lbf, R Dz = 46.6 lbf, and To = 480 lbf · in. The direction of the output torque To is opposite ωo because it is the resistive load on the system opposing the motion ωo. Note in Fig. 31b the net force from the bearing reactions is zero whereas the net moment about the x axis is 2.25 (192) + 2.25 (128) = 720 lbf · in. This value is the same as Ti + To = 240 + 480 = 720 lbf · in, as shown in Fig. 31a. The reaction forces R E , R F , R H , and R I , from the mounting bolts cannot be determined from the equilibrium equations as there are too many unknowns. Only three equations are Fy = Fz = Mx = 0. In case you were wondering about assumption available, 5, here is where we will use it (see Sec. 812). The gear box tends to rotate about the x axis because of a pure torsional moment of 720 lbf · in. The bolt forces must provide 76 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition I. Basics © The McGrawHill Companies, 2008 3. Load and Stress Analysis Load and Stress Analysis 71 an equal but opposite torsional moment. The center of rotation relative to the bolts lies at the centroid of the bolt cross-sectional areas. Thus if the bolt areas are equal: the center of rotation is at the center of the four bolts, a distance of (4/2)2 + (5/2)2 = 3.202 in from each bolt; the bolt forces are equal ( R E = R F = R H = R I = R ), and each bolt force is perpendicular to the line from the bolt to the center of rotation. This gives a net torque from the four bolts of 4 R (3.202) = 720. Thus, R E = R F = R H = R I = 56.22 lbf. 32 Shear Force and Bending Moments in Beams Figure 32a shows a beam supported by reactions R1 and R2 and loaded by the concentrated forces F1 , F2 , and F3 . If the beam is cut at some section located at x = x1 and the left-hand portion is removed as a free body, an internal shear force V and bending moment M must act on the cut surface to ensure equilibrium (see Fig. 32b). The shear force is obtained by summing the forces on the isolated section. The bending moment is the sum of the moments of the forces to the left of the section taken about an axis through the isolated section. The sign conventions used for bending moment and shear force in this book are shown in Fig. 33. Shear force and bending moment are related by the equation V= dM dx (33) Sometimes the bending is caused by a distributed load q (x ), as shown in Fig. 34; q (x ) is called the load intensity with units of force per unit length and is positive in the Figure 32 y y Free-body diagram of simplysupported beam with V and M shown in positive directions. F2 F1 F3 F1 V x x1 x1 R2 R1 R1 (a) (b) Figure 33 Sign conventions for bending and shear. Positive bending y Negative bending Positive shear Figure 34 x M Negative shear q (x ) Distributed load on beam. x BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 72 I. Basics 77 © The McGrawHill Companies, 2008 3. Load and Stress Analysis Mechanical Engineering Design positive y direction. It can be shown that differentiating Eq. (33) results in dV d2 M = =q dx dx 2 (34) Normally the applied distributed load is directed downward and labeled w (e.g., see Fig. 36). In this case, w = q . Equations (33) and (34) reveal additional relations if they are integrated. Thus, if we integrate between, say, x A and x B , we obtain VB xB dV = VA (35) q dx = VB V A xA which states that the change in shear force from A to B is equal to the area of the loading diagram between x A and x B . In a similar manner, MB xB dM = MA (36) V dx = M B M A xA which states that the change in moment from A to B is equal to the area of the shearforce diagram between x A and x B . Table 31 Function Singularity (Macaulay ) Functions Concentrated moment (unit doublet) Graph of fn (x) xa Meaning xa 2 xa x 2 2 xa =0 x=a = ± x = a 2 dx = x a 1 a Concentrated force (unit impulse) xa xa 1 xa x 1 1 xa =0 x=a = + x = a 1 dx = x a 0 a Unit step xa 0 xa 1 x 0 = xa 0 1 = 0 x<a 1 xa dx = x a 1 a Ramp xa 1 xa 1 1 x xa 1 0 x<a xa xa dx = xa 2 a W. H. Macaulay, Note on the deection of beams, Messenger of Mathematics, vol. 48, pp. 129130, 1919. 2 78 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition I. Basics © The McGrawHill Companies, 2008 3. Load and Stress Analysis Load and Stress Analysis 33 73 Singularity Functions The four singularity functions dened in Table 31 constitute a useful and easy means of integrating across discontinuities. By their use, general expressions for shear force and bending moment in beams can be written when the beam is loaded by concentrated moments or forces. As shown in the table, the concentrated moment and force functions are zero for all values of x not equal to a. The functions are undened for values of x = a . Note that the unit step and ramp functions are zero only for values of x that are less than a. The integration properties shown in the table constitute a part of the mathematical denition too. The rst two integrations of q (x ) for V (x ) and M (x ) do not require constants of integration provided all loads on the beam are accounted for in q (x ). The examples that follow show how these functions are used. EXAMPLE 32 Derive expressions for the loading, shear-force, and bending-moment diagrams for the beam of Fig. 35. Figure 35 y l q F1 F2 x O a1 R1 Solution Answer a2 R2 Using Table 31 and q (x ) for the loading function, we nd q = R1 x 1 F1 x a1 1 F2 x a2 1 + R2 x l 1 (1) Next, we use Eq. (35) to get the shear force. Answer V= 0 q dx = R1 x F1 x a1 0 F2 x a2 0 + R2 x l 0 (2) + R2 x l 1 (3) Note that V = 0 at x = 0 . A second integration, in accordance with Eq. (36), yields Answer M= V dx = R1 x 1 F1 x a1 1 F2 x a2 1 The reactions R1 and R2 can be found by taking a summation of moments and forces as usual, or they can be found by noting that the shear force and bending moment must be zero everywhere except in the region 0 x l . This means that Eq. (2) should give V = 0 at x slightly larger than l. Thus R1 F1 F2 + R2 = 0 (4) Since the bending moment should also be zero in the same region, we have, from Eq. (3), R1l F1 (l a1 ) F2 (l a2 ) = 0 Equations (4) and (5) can now be solved for the reactions R1 and R2 . (5) BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 74 I. Basics 79 © The McGrawHill Companies, 2008 3. Load and Stress Analysis Mechanical Engineering Design EXAMPLE 33 Solution Figure 36a shows the loading diagram for a beam cantilevered at A with a uniform load of 20 lbf/in acting on the portion 3 in x 7 in, and a concentrated counterclockwise moment of 240 lbf · in at x = 10 in. Derive the shear-force and bendingmoment relations, and the support reactions M1 and R1 . Following the procedure of Example 32, we nd the load intensity function to be q = M1 x 2 + R1 x 1 20 x 3 0 + 20 x 7 0 240 x 10 2 (1) Note that the 20 x 7 0 term was necessary to turn off the uniform load at C. Integrating successively gives Answers V = M1 x M = M1 x 1 0 + R1 x + R1 x 1 0 20 x 3 1 + 20 x 7 1 240 x 10 10 x 3 2 + 10 x 7 2 240 x 10 0 1 (2) (3) The reactions are found by making x slightly larger than 10 in, where both V and M are zero in this region. Equation (2) will then give M1 (0) + R1 (1) 20(10 3) + 20(10 7) 240(0) = 0 Answer which yields R1 = 80 lbf. From Eq. (3) we get M1 (1) + 80(10) 10(10 3)2 + 10(10 7)2 240(1) = 0 Answer which yields M1 = 160 lbf · in. Figures 36b and c show the shear-force and bending-moment diagrams. Note that the impulse terms in Eq. (2), M1 x 1 and 240 x 10 1 , are physically not forces Figure 36 y (a) Loading diagram for a beam cantilevered at A. (b) Shear-force diagram. (c) Bending-moment diagram. q 10 in 7 in 3 in 20 lbf/in 240 lbf in D B A (a) M1 C x R1 V (lbf) 80 (b) Step Ramp x O M (lbf in) 240 Parabolic Step 80 x O (c) 160 Ramp Slope = 80 lbf in/in 80 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition I. Basics 3. Load and Stress Analysis © The McGrawHill Companies, 2008 Load and Stress Analysis 75 and are not shown in the V diagram. Also note that both the M1 and 240 lbf · in moments are counterclockwise and negative singularity functions; however, by the convention shown in Fig. 32 the M1 and 240 lbf · in are negative and positive bending moments, respectively, which is reected in Fig. 36c. 34 Stress When an internal surface is isolated as in Fig. 32b, the net force and moment acting on the surface manifest themselves as force distributions across the entire area. The force distribution acting at a point on the surface is unique and will have components in the normal and tangential directions called normal stress and tangential shear stress, respectively. Normal and shear stresses are labeled by the Greek symbols σ and τ , respectively. If the direction of σ is outward from the surface it is considered to be a tensile stress and is a positive normal stress. If σ is into the surface it is a compressive stress and commonly considered to be a negative quantity. The units of stress in U.S. Customary units are pounds per square inch (psi). For SI units, stress is in newtons per square meter (N/m2 ); 1 N/m2 = 1 pascal (Pa). 35 Cartesian Stress Components The Cartesian stress components are established by defining three mutually orthogonal surfaces at a point within the body. The normals to each surface will establish the x, y, z Cartesian axes. In general, each surface will have a normal and shear stress. The shear stress may have components along two Cartesian axes. For example, Fig. 37 shows an infinitesimal surface area isolation at a point Q within a body where the surface normal is the x direction. The normal stress is labeled σx . The symbol σ indicates a normal stress and the subscript x indicates the direction of the surface normal. The net shear stress acting on the surface is (τx )net which can be resolved into components in the y and z directions, labeled as τx y and τx z , respectively (see Fig. 37). Note that double subscripts are necessary for the shear. The first subscript indicates the direction of the surface normal whereas the second subscript is the direction of the shear stress. The state of stress at a point described by three mutually perpendicular surfaces is shown in Fig. 38a. It can be shown through coordinate transformation that this is sufcient to determine the state of stress on any surface intersecting the point. As the y Figure 37 Stress components on surface normal to x direction. xy ( x)net Q xz z x x BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 76 I. Basics 81 © The McGrawHill Companies, 2008 3. Load and Stress Analysis Mechanical Engineering Design y Figure 38 y y (a) General three-dimensional stress. (b) Plane stress with cross-shears equal. y xy yx yz xy xy x x zy x x x xy zx xz xy y z z (a ) (b) dimensions of the cube in Fig. 38a approach zero, the stresses on the hidden faces become equal and opposite to those on the opposing visible faces. Thus, in general, a complete state of stress is dened by nine stress components, σx , σ y , σz , τx y , τx z , τ yx , τ yz , τzx , and τzy . For equilibrium, in most cases, cross-shears are equal, hence τ yx = τx y τzy = τ yz τx z = τzx (37) This reduces the number of stress components for most three-dimensional states of stress from nine to six quantities, σx , σ y , σz , τx y , τ yz , and τzx . A very common state of stress occurs when the stresses on one surface are zero. When this occurs the state of stress is called plane stress. Figure 38b shows a state of plane stress, arbitrarily assuming that the normal for the stress-free surface is the z direction such that σz = τzx = τzy = 0. It is important to note that the element in Fig. 38b is still a three-dimensional cube. Also, here it is assumed that the cross-shears are equal such that τ yx = τx y , and τ yz = τzy = τx z = τzx = 0. 36 Mohrs Circle for Plane Stress Suppose the d x dy dz element of Fig. 38b is cut by an oblique plane with a normal n at an arbitrary angle φ counterclockwise from the x axis as shown in Fig. 39. This section is concerned with the stresses σ and τ that act upon this oblique plane. By summing the forces caused by all the stress components to zero, the stresses σ and τ are found to be σ= σx + σ y σx σ y + cos 2φ + τx y sin 2φ 2 2 τ = σx σ y sin 2φ + τx y cos 2φ 2 (38) (39) Equations (38) and (39) are called the plane-stress transformation equations. Differentiating Eq. (38) with respect to φ and setting the result equal to zero gives tan 2φ p = 2τx y σx σ y (310) 82 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition I. Basics © The McGrawHill Companies, 2008 3. Load and Stress Analysis Load and Stress Analysis 77 y Figure 39 n x dy dy ds ds dx dx xy xy x y Equation (310) denes two particular values for the angle 2φ p , one of which denes the maximum normal stress σ1 and the other, the minimum normal stress σ2 . These two stresses are called the principal stresses, and their corresponding directions, the principal directions. The angle between the principal directions is 90°. It is important to note that Eq. (310) can be written in the form σx σ y sin 2φ p τx y cos 2φ p = 0 2 (a) Comparing this with Eq. (39), we see that τ = 0, meaning that the surfaces containing principal stresses have zero shear stresses. In a similar manner, we differentiate Eq. (39), set the result equal to zero, and obtain tan 2φs = σx σ y 2τx y (311) Equation (311) denes the two values of 2φs at which the shear stress τ reaches an extreme value. The angle between the surfaces containing the maximum shear stresses is 90°. Equation (311) can also be written as σx σ y cos 2φ p + τx y sin 2φ p = 0 2 (b) Substituting this into Eq. (38) yields σ= σx + σ y 2 (312) Equation (312) tells us that the two surfaces containing the maximum shear stresses also contain equal normal stresses of (σx + σ y )/2. Comparing Eqs. (310) and (311), we see that tan 2φs is the negative reciprocal of tan 2φ p . This means that 2φs and 2φ p are angles 90° apart, and thus the angles between the surfaces containing the maximum shear stresses and the surfaces containing the principal stresses are ±45 . Formulas for the two principal stresses can be obtained by substituting the angle 2φ p from Eq. (310) in Eq. (38). The result is σ1 , σ2 = σx + σ y ± 2 σx σ y 2 2 2 + τx y (313) BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 78 I. Basics 83 © The McGrawHill Companies, 2008 3. Load and Stress Analysis Mechanical Engineering Design In a similar manner the two extreme-value shear stresses are found to be τ1 , τ2 = ± σx σ y 2 2 2 + τx y (314) Your particular attention is called to the fact that an extreme value of the shear stress may not be the same as the actual maximum value. See Sec. 37. It is important to note that the equations given to this point are quite sufcient for performing any plane stress transformation. However, extreme care must be exercised when applying them. For example, say you are attempting to determine the principal state of stress for a problem where σx = 14 MPa, σ y = 10 MPa, and τx y = 16 MPa. Equation (310) yields φ p = 26.57 and 63.43° to locate the principal stress surfaces, whereas, Eq. (313) gives σ1 = 22 MPa and σ2 = 18 MPa for the principal stresses. If all we wanted was the principal stresses, we would be nished. However, what if we wanted to draw the element containing the principal stresses properly oriented relative to the x, y axes? Well, we have two values of φ p and two values for the principal stresses. How do we know which value of φ p corresponds to which value of the principal stress? To clear this up we would need to substitute one of the values of φ p into Eq. (38) to determine the normal stress corresponding to that angle. A graphical method for expressing the relations developed in this section, called Mohrs circle diagram, is a very effective means of visualizing the stress state at a point and keeping track of the directions of the various components associated with plane stress. Equations (38) and (39) can be shown to be a set of parametric equations for σ and τ , where the parameter is 2φ. The relationship between σ and τ is that of a circle plotted in the σ, τ plane, where the center of the circle is located at C = (σ, τ ) = 2 [(σx + σ y )/2, 0] and has a radius of R = [(σx σ y )/2]2 + τx y . A problem arises in the sign of the shear stress. The transformation equations are based on a positive φ being counterclockwise, as shown in Fig. 39. If a positive τ were plotted above the σ axis, points would rotate clockwise on the circle 2φ in the opposite direction of rotation on the element. It would be convenient if the rotations were in the same direction. One could solve the problem easily by plotting positive τ below the axis. However, the classical approach to Mohrs circle uses a different convention for the shear stress. Mohrs Circle Shear Convention This convention is followed in drawing Mohrs circle: Shear stresses tending to rotate the element clockwise (cw) are plotted above the σ axis. Shear stresses tending to rotate the element counterclockwise (ccw) are plotted below the σ axis. For example, consider the right face of the element in Fig. 38b. By Mohrs circle convention the shear stress shown is plotted below the σ axis because it tends to rotate the element counterclockwise. The shear stress on the top face of the element is plotted above the σ axis because it tends to rotate the element clockwise. In Fig. 310 we create a coordinate system with normal stresses plotted along the abscissa and shear stresses plotted as the ordinates. On the abscissa, tensile (positive) normal stresses are plotted to the right of the origin O and compressive (negative) normal stresses to the left. On the ordinate, clockwise (cw) shear stresses are plotted up; counterclockwise (ccw) shear stresses are plotted down. 84 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition I. Basics © The McGrawHill Companies, 2008 3. Load and Stress Analysis Load and Stress Analysis 79 cw Figure 310 x ( y Mohrs circle diagram. x y) x F y B ( 2 y H cw y , xy ) xy 2 E O 2 D C y x 1 xy x 2 y2 + 2 xy ( A x, 2 p ccw xy ) x ccw x + 2 y G Using the stress state of Fig. 38b, we plot Mohrs circle, Fig. 310, by rst looking at the right surface of the element containing σx to establish the sign of σx and the cw or ccw direction of the shear stress. The right face is called the x face where φ = 0 . If σx is positive and the shear stress τx y is ccw as shown in Fig. 38b, we can ccw establish point A with coordinates (σx , τx y ) in Fig. 310. Next, we look at the top y face, where φ = 90 , which contains σ y , and repeat the process to obtain point B with cw coordinates (σ y , τx y ) as shown in Fig. 310. The two states of stress for the element are φ = 90 from each other on the element so they will be 2 φ = 180 from each other on Mohrs circle. Points A and B are the same vertical distance from the σ axis. Thus, AB must be on the diameter of the circle, and the center of the circle C is where AB intersects the σ axis. With points A and B on the circle, and center C, the complete circle can then be drawn. Note that the extended ends of line A B are labeled x and y as references to the normals to the surfaces for which points A and B represent the stresses. The entire Mohrs circle represents the state of stress at a single point in a structure. Each point on the circle represents the stress state for a specic surface intersecting the point in the structure. Each pair of points on the circle 180° apart represent the state of stress on an element whose surfaces are 90° apart. Once the circle is drawn, the states of stress can be visualized for various surfaces intersecting the point being analyzed. For example, the principal stresses σ1 and σ2 are points D and E, respectively, and their values obviously agree with Eq. (313). We also see that the shear stresses are zero on the surfaces containing σ1 and σ2 . The two extreme-value shear stresses, one clockwise and one counterclockwise, occur at F and G with magnitudes equal to the radius of the circle. The surfaces at F and G each also contain normal stresses of (σx + σ y )/2 as noted earlier in Eq. (312). Finally, the state of stress on an arbitrary surface located at an angle φ counterclockwise from the x face is point H. BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 80 I. Basics © The McGrawHill Companies, 2008 3. Load and Stress Analysis Mechanical Engineering Design At one time, Mohrs circle was used graphically where it was drawn to scale very accurately and values were measured by using a scale and protractor. Here, we are strictly using Mohrs circle as a visualization aid and will use a semigraphical approach, calculating values from the properties of the circle. This is illustrated by the following example. EXAMPLE 34 A stress element has σx = 80 MPa and τx y = 50 MPa cw, as shown in Fig. 311a. (a) Using Mohrs circle, nd the principal stresses and directions, and show these on a stress element correctly aligned with respect to the x y coordinates. Draw another stress element to show τ1 and τ2 , nd the corresponding normal stresses, and label the drawing completely. (b) Repeat part a using the transformation equations only. Solution (a) In the semigraphical approach used here, we rst make an approximate freehand sketch of Mohrs circle and then use the geometry of the gure to obtain the desired information. Draw the σ and τ axes rst (Fig. 311b) and from the x face locate σx = 80 MPa along the σ axis. On the x face of the element, we see that the shear stress is 50 MPa in the cw direction. Thus, for the x face, this establishes point A (80, 50cw) MPa. Corresponding to the y face, the stress is σ = 0 and τ = 50 MPa in the ccw direction. This locates point B (0, 50ccw) MPa. The line A B forms the diameter of the required circle, which can now be drawn. The intersection of the circle with the σ axis denes σ1 and σ2 as shown. Now, noting the triangle AC D , indicate on the sketch the length of the legs A D and C D as 50 and 40 MPa, respectively. The length of the hypotenuse AC is Answer τ1 = (50)2 + (40)2 = 64.0 MPa and this should be labeled on the sketch too. Since intersection C is 40 MPa from the origin, the principal stresses are now found to be Answer σ1 = 40 + 64 = 104 MPa and σ2 = 40 64 = 24 MPa The angle 2φ from the x axis cw to σ1 is Answer 2φ p = tan1 50 40 = 51.3 To draw the principal stress element (Fig. 311c), sketch the x and y axes parallel to the original axes. The angle φ p on the stress element must be measured in the same direction as is the angle 2φ p on the Mohr circle. Thus, from x measure 25.7° (half of 51.3°) clockwise to locate the σ1 axis. The σ2 axis is 90° from the σ1 axis and the stress element can now be completed and labeled as shown. Note that there are no shear stresses on this element. The two maximum shear stresses occur at points E and F in Fig. 311b. The two normal stresses corresponding to these shear stresses are each 40 MPa, as indicated. Point E is 38.7° ccw from point A on Mohrs circle. Therefore, in Fig. 311d, draw a stress element oriented 19.3° (half of 38.7°) ccw from x. The element should then be labeled with magnitudes and directions as shown. In constructing these stress elements it is important to indicate the x and y directions of the original reference system. This completes the link between the original machine element and the orientation of its principal stresses. 85 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition I. Basics © The McGrawHill Companies, 2008 3. Load and Stress Analysis 81 Load and Stress Analysis y Figure 311 x cw 1 All stresses in MPa. (80, 50cw) E 50 A 80 64 86 38.7° x 2 p 50 50 y= 2 (a ) 0 51.3° C 40 D 40 1 x= (0, 50ccw) 80 B F 2 ccw y (b) y 2 y = 40 2= 24 2= 64 = 40 F Answer 104 x 1= 25.7° 1= 19.3° E x 64 1 (c) (d ) (b) The transformation equations are programmable. From Eq. (310), φp = 1 tan1 2 2τx y σx σ y = 1 tan1 2 2(50) 80 = 25.7 , 64.3 From Eq. (38), for the rst angle φ p = 25.7 , σ= 80 + 0 80 0 + cos[2(25.7)] + (50) sin[2(25.7)] = 104.03 MPa 2 2 The shear on this surface is obtained from Eq. (39) as τ = 80 0 sin[2(25.7)] + (50) cos[2(25.7)] = 0 MPa 2 which conrms that 104.03 MPa is a principal stress. From Eq. (38), for φ p = 64.3 , σ= 80 + 0 80 0 + cos[2(64.3)] + (50) sin[2(64.3)] = 24.03 MPa 2 2 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 82 I. Basics 3. Load and Stress Analysis © The McGrawHill Companies, 2008 Mechanical Engineering Design Answer Substituting φ p = 64.3 into Eq. (39) again yields τ = 0, indicating that 24.03 MPa is also a principal stress. Once the principal stresses are calculated they can be ordered such that σ1 σ2 . Thus, σ1 = 104.03 MPa and σ2 = 24.03 MPa. Since for σ1 = 104.03 MPa, φ p = 25.7 , and since φ is dened positive ccw in the transformation equations, we rotate clockwise 25.7° for the surface containing σ1 . We see in Fig. 311c that this totally agrees with the semigraphical method. To determine τ1 and τ2 , we rst use Eq. (311) to calculate φs : φs = σx σ y 1 tan1 2 2τx y = 1 80 tan1 2 2(50) = 19.3 , 109.3 For φs = 19.3 , Eqs. (38) and (39) yield Answer σ= 80 + 0 80 0 + cos[2(19.3)] + (50) sin[2(19.3)] = 40.0 MPa 2 2 τ = 80 0 sin[2(19.3)] + (50) cos[2(19.3)] = 64.0 MPa 2 Remember that Eqs. (38) and (39) are coordinate transformation equations. Imagine that we are rotating the x, y axes 19.3° counterclockwise and y will now point up and to the left. So a negative shear stress on the rotated x face will point down and to the right as shown in Fig. 311d. Thus again, results agree with the semigraphical method. For φs = 109.3 , Eqs. (38) and (39) give σ = 40.0 MPa and τ = +64.0 MPa. Using the same logic for the coordinate transformation we nd that results again agree with Fig. 311d. 37 General Three-Dimensional Stress As in the case of plane stress, a particular orientation of a stress element occurs in space for which all shear-stress components are zero. When an element has this particular orientation, the normals to the faces are mutually orthogonal and correspond to the principal directions, and the normal stresses associated with these faces are the principal stresses. Since there are three faces, there are three principal directions and three principal stresses σ1 , σ2 , and σ3 . For plane stress, the stress-free surface contains the third principal stress which is zero. In our studies of plane stress we were able to specify any stress state σx , σ y , and τx y and find the principal stresses and principal directions. But six components of stress are required to specify a general state of stress in three dimensions, and the problem of determining the principal stresses and directions is more difficult. In design, three-dimensional transformations are rarely performed since most maximum stress states occur under plane stress conditions. One notable exception is contact stress, which is not a case of plane stress, where the three principal stresses are given in Sec. 319. In fact, all states of stress are truly three-dimensional, where they might be described one- or two-dimensionally with respect to specific coordinate axes. Here it is most important to understand the relationship amongst the three principal stresses. The process in finding the three principal stresses from the six 87 88 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition I. Basics © The McGrawHill Companies, 2008 3. Load and Stress Analysis Load and Stress Analysis 83 Figure 312 Mohrs circles for threedimensional stress. 1/3 1/2 2/3 1/2 3 2 1 1 2 (a) (b ) stress components σx , σ y , σz , τx y , τ yz , and τzx , involves finding the roots of the cubic equation1 2 2 2 σ 3 (σx + σ y + σz )σ 2 + σx σ y + σx σz + σ y σz τx y τ yz τzx σ 2 2 2 σx σ y σz + 2τx y τ yz τzx σx τ yz σ y τzx σz τx y = 0 (315) In plotting Mohrs circles for three-dimensional stress, the principal normal stresses are ordered so that σ1 σ2 σ3 . Then the result appears as in Fig. 312a. The stress coordinates σ , τ for any arbitrarily located plane will always lie on the boundaries or within the shaded area. Figure 312a also shows the three principal shear stresses τ1/2 , τ2/3 , and τ1/3 .2 Each of these occurs on the two planes, one of which is shown in Fig. 312b. The gure shows that the principal shear stresses are given by the equations τ1/2 = σ1 σ2 2 τ2/3 = σ2 σ3 2 τ1/3 = σ1 σ3 2 (316) Of course, τmax = τ1/3 when the normal principal stresses are ordered (σ1 > σ2 > σ3 ) , so always order your principal stresses. Do this in any computer code you generate and youll always generate τmax . 38 Elastic Strain Normal strain ǫ is dened and discussed in Sec. 2-1 for the tensile specimen and is given by Eq. (22) as ǫ = δ/ l , where δ is the total elongation of the bar within the length l. Hookes law for the tensile specimen is given by Eq. (23) as σ = Eǫ (317) where the constant E is called Youngs modulus or the modulus of elasticity. 1 For development of this equation and further elaboration of three-dimensional stress transformations see: Richard G. Budynas, Advanced Strength and Applied Stress Analysis, 2nd ed., McGraw-Hill, New York, 1999, pp. 4678. 2 Note the difference between this notation and that for a shear stress, say, τx y . The use of the shilling mark is not accepted practice, but it is used here to emphasize the distinction. BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 84 I. Basics 89 © The McGrawHill Companies, 2008 3. Load and Stress Analysis Mechanical Engineering Design When a material is placed in tension, there exists not only an axial strain, but also negative strain (contraction) perpendicular to the axial strain. Assuming a linear, homogeneous, isotropic material, this lateral strain is proportional to the axial strain. If the axial direction is x, then the lateral strains are ǫ y = ǫz = νǫx . The constant of proportionality v is called Poissons ratio, which is about 0.3 for most structural metals. See Table A5 for values of v for common materials. If the axial stress is in the x direction, then from Eq. (317) ǫx = σx E ǫ y = ǫz = ν σx E (318) For a stress element undergoing σx , σ y , and σz simultaneously, the normal strains are given by 1 σ x ν ( σ y + σz ) E 1 σ y ν ( σ x + σz ) ǫy = E 1 σz ν ( σ x + σ y ) ǫz = E ǫx = (319) Shear strain γ is the change in a right angle of a stress element when subjected to pure shear stress, and Hookes law for shear is given by τ = Gγ (320) where the constant G is the shear modulus of elasticity or modulus of rigidity. It can be shown for a linear, isotropic, homogeneous material, the three elastic constants are related to each other by E = 2G (1 + ν) 39 (321) Uniformly Distributed Stresses The assumption of a uniform distribution of stress is frequently made in design. The result is then often called pure tension, pure compression, or pure shear, depending upon how the external load is applied to the body under study. The word simple is sometimes used instead of pure to indicate that there are no other complicating effects. The tension rod is typical. Here a tension load F is applied through pins at the ends of the bar. The assumption of uniform stress means that if we cut the bar at a section remote from the ends and remove one piece, we can replace its effect by applying a uniformly distributed force of magnitude σ A to the cut end. So the stress σ is said to be uniformly distributed. It is calculated from the equation σ= F A (322) This assumption of uniform stress distribution requires that: The bar be straight and of a homogeneous material The line of action of the force contains the centroid of the section The section be taken remote from the ends and from any discontinuity or abrupt change in cross section 90 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition I. Basics © The McGrawHill Companies, 2008 3. Load and Stress Analysis Load and Stress Analysis 85 For simple compression, Eq. (322) is applicable with F normally being considered a negative quantity. Also, a slender bar in compression may fail by buckling, and this possibility must be eliminated from consideration before Eq. (322) is used.3 Use of the equation F τ= (323) A for a body, say, a bolt, in shear assumes a uniform stress distribution too. It is very difcult in practice to obtain a uniform distribution of shear stress. The equation is included because occasions do arise in which this assumption is utilized. 310 Normal Stresses for Beams in Bending The equations for the normal bending stresses in straight beams are based on the following assumptions: 1 2 3 4 5 6 7 The beam is subjected to pure bending. This means that the shear force is zero, and that no torsion or axial loads are present. The material is isotropic and homogeneous. The material obeys Hookes law. The beam is initially straight with a cross section that is constant throughout the beam length. The beam has an axis of symmetry in the plane of bending. The proportions of the beam are such that it would fail by bending rather than by crushing, wrinkling, or sidewise buckling. Plane cross sections of the beam remain plane during bending. In Fig. 313 we visualize a portion of a straight beam acted upon by a positive bending moment M shown by the curved arrow showing the physical action of the moment together with a straight arrow indicating the moment vector. The x axis is coincident with the neutral axis of the section, and the xz plane, which contains the neutral axes of all cross sections, is called the neutral plane. Elements of the beam coincident with this plane have zero stress. The location of the neutral axis with respect to the cross section is coincident with the centroidal axis of the cross section. Figure 313 y Straight beam in positive bending. M z M 3 See Sec. 411. x BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 86 I. Basics 91 © The McGrawHill Companies, 2008 3. Load and Stress Analysis Mechanical Engineering Design y Figure 314 Bending stresses according to Eq. (324). Compression c y Neutral axis, Centroidal axis x Tension The bending stress varies linearly with the distance from the neutral axis, y, and is given by σx = My I (324) where I is the second moment of area about the z axis. That is I= y2d A (325) The stress distribution given by Eq. (324) is shown in Fig. 314. The maximum magnitude of the bending stress will occur where y has the greatest magnitude. Designating σmax as the maximum magnitude of the bending stress, and c as the maximum magnitude of y σmax = Mc I (326a) Equation (324) can still be used to ascertain as to whether σmax is tensile or compressive. Equation (326a) is often written as σmax = M Z (326b) where Z = I/c is called the section modulus. EXAMPLE 35 A beam having a T section with the dimensions shown in Fig. 315 is subjected to a bending moment of 1600 N · m that causes tension at the top surface. Locate the neutral axis and nd the maximum tensile and compressive bending stresses. Solution The area of the composite section is A = 1956 mm2 . Now divide the T section into two rectangles, numbered 1 and 2, and sum the moments of these areas about the top edge. We then have 1956c1 = 12(75)(6) + 12(88)(56) and hence c1 = 32.99 mm. Therefore c2 = 100 32.99 = 67.01 mm. Next we calculate the second moment of area of each rectangle about its own centroidal axis. Using Table A-18, we nd for the top rectangle I1 = 13 1 bh = (75)123 = 1.080 × 104 mm4 12 12 92 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition I. Basics © The McGrawHill Companies, 2008 3. Load and Stress Analysis Load and Stress Analysis Figure 315 87 y Dimensions in millimeters. 75 12 1 c1 z 100 2 c2 12 For the bottom rectangle, we have I2 = 1 (12)883 = 6.815 × 105 mm4 12 We now employ the parallel-axis theorem to obtain the second moment of area of the composite gure about its own centroidal axis. This theorem states Iz = Icg + Ad 2 where Icg is the second moment of area about its own centroidal axis and Iz is the second moment of area about any parallel axis a distance d removed. For the top rectangle, the distance is d1 = 32.99 6 = 26.99 mm and for the bottom rectangle, d2 = 67.01 44 = 23.01 mm Using the parallel-axis theorem for both rectangles, we now nd that I = [1.080 × 104 + 12(75)26.992 ] + [6.815 × 105 + 12(88)23.012 ] = 1.907 × 106 mm4 Finally, the maximum tensile stress, which occurs at the top surface, is found to be Answer σ= 1600(32.99)103 Mc1 = = 27.68(106 ) Pa = 27.68 MPa I 1.907(106 ) Similarly, the maximum compressive stress at the lower surface is found to be Answer σ = 1600(67.01)103 Mc2 = = 56.22(106 ) Pa = 56.22 MPa I 1.907(106 ) BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 88 I. Basics 93 © The McGrawHill Companies, 2008 3. Load and Stress Analysis Mechanical Engineering Design Two-Plane Bending Quite often, in mechanical design, bending occurs in both xy and xz planes. Considering cross sections with one or two planes of symmetry only, the bending stresses are given by σx = My z Mz y + Iz Iy (327) where the rst term on the right side of the equation is identical to Eq. (324), M y is the bending moment in the xz plane (moment vector in y direction), z is the distance from the neutral y axis, and I y is the second area moment about the y axis. For noncircular cross sections, Eq. (327) is the superposition of stresses caused by the two bending moment components. The maximum tensile and compressive bending stresses occur where the summation gives the greatest positive and negative stresses, respectively. For solid circular cross sections, all lateral axes are the same and the plane containing the moment corresponding to the vector sum of Mz and M y contains the maximum bending stresses. For a beam of diameter d the maximum distance from the neutral axis is d/2, and from Table A18, I = π d 4/64. The maximum bending stress for a solid circular cross section is then 2 2 ( M y + Mz ) 1/2 ( d /2) 32 Mc 2 = = ( M 2 + Mz ) 1/2 4 /64 I πd π d3 y σm = EXAMPLE 36 As shown in Fig. 316a, beam OC is loaded in the xy plane by a uniform load of 50 lbf/in, and in the xz plane by a concentrated force of 100 lbf at end C. The beam is 8 in long. Figure 316 (a) Beam loaded in two planes; (b) loading and bending-moment diagrams in xy plane; (c) loading and bending-moment diagrams in xz plane. (328) y y 50 lbf/in x A O 1600 lbf-in 400 lbf 50 lbf/in OB z C Mz (lbf-in) C 1.5 in 0 x x 100 lbf 1600 0.75 in (b ) (a ) 100 lbf 800 lbf-in x O C z 100 lbf My (lbf-in) 800 x 0 (c) 94 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition I. Basics © The McGrawHill Companies, 2008 3. Load and Stress Analysis Load and Stress Analysis 89 (a) For the cross section shown determine the maximum tensile and compressive bending stresses and where they act. (b) If the cross section was a solid circular rod of diameter, d = 1.25 in, determine the magnitude of the maximum bending stress. Solution (a) The reactions at O and the bending-moment diagrams in the xy and xz planes are shown in Figs. 316b and c, respectively. The maximum moments in both planes occur at O where 1 ( M y ) O = 100(8) = 800 lbf-in ( Mz ) O = (50)82 = 1600 lbf-in 2 The second moments of area in both planes are Iz = 1 (0.75)1.53 = 0.2109 in4 12 Iy = 1 (1.5)0.753 = 0.05273 in4 12 The maximum tensile stress occurs at point A, shown in Fig. 316a, where the maximum tensile stress is due to both moments. At A, y A = 0.75 in and z A = 0.375 in. Thus, from Eq. (327) Answer (σx ) A = 1600(0.75) 800(0.375) + = 11 380 psi = 11.38 kpsi 0.2109 0.05273 The maximum compressive bending stress occurs at point B where, y B = 0.75 in and z B = 0.375 in. Thus Answer (σx ) B = 1600(0.75) 800(0.375) + = 11 380 psi = 11.38 kpsi 0.2109 0.05273 (b) For a solid circular cross section of diameter, d = 1.25 in, the maximum bending stress at end O is given by Eq. (328) as Answer σm = 32 8002 + (1600)2 π(1.25)3 1/2 = 9326 psi = 9.329 kpsi Beams with Asymmetrical Sections The relations developed earlier in this section can also be applied to beams having asymmetrical sections, provided that the plane of bending coincides with one of the two principal axes of the section. We have found that the stress at a distance y from the neutral axis is σ = My I (a) Therefore, the force on the element of area d A in Fig. 317 is dF = σ dA = My dA I BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 90 I. Basics 95 © The McGrawHill Companies, 2008 3. Load and Stress Analysis Mechanical Engineering Design Figure 317 y y z M y x dA z Taking moments of this force about the y axis and integrating across the section gives My = z dF = σz dA = M I yz d A (b) We recognize that the last integral in Eq. (b) is the product of inertia I yz . If the bending moment on the beam is in the plane of one of the principal axes, say the x y plane, then I yz = yz d A = 0 (c) With this restriction, the relations developed in Sec. 310 hold for any cross-sectional shape. Of course, this means that the designer has a special responsibility to ensure that the bending loads do, in fact, come onto the beam in a principal plane! 311 Shear Stresses for Beams in Bending Most beams have both shear forces and bending moments present. It is only occasionally that we encounter beams subjected to pure bending, that is to say, beams having zero shear force. The exure formula is developed on the assumption of pure bending. This is done, however, to eliminate the complicating effects of shear force in the development. For engineering purposes, the exure formula is valid no matter whether a shear force is present or not. For this reason, we shall utilize the same normal bendingstress distribution [Eqs. (324) and (326)] when shear forces are also present. In Fig. 318a we show a beam segment of constant cross section subjected to a shear force V and a bending moment M at x. Because of external loading and V, the shear force and bending moment change with respect to x. At x + dx the shear force and bending moment are V + d V and M + d M , respectively. Considering forces in the x direction only, Fig. 318b shows the stress distribution σx due to the bending moments. If dM is positive, with the bending moment increasing, the stresses on the right face, for a given value of y, are larger in magnitude than the stresses on the left face. If we further isolate the element by making a slice at y = y1 (see Fig. 318b), the net force in the x direction will be directed to the left with a value of c y1 (d M ) y dA I as shown in the rotated view of Fig. 318c. For equilibrium, a shear force on the bottom face, directed to the right, is required. This shear force gives rise to a shear stress τ , where, if assumed uniform, the force is τ b d x . Thus c τ b dx = y1 (d M ) y dA I (a) 96 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition I. Basics © The McGrawHill Companies, 2008 3. Load and Stress Analysis Load and Stress Analysis w (x) y x My I x M x dM M x V x dMy I c y1 V My I 91 dV dx dx (b) (a ) A Figure 318 Beam section isolation. Note: Only forces shown in x direction on dx element in (b). y F dx c dM y I y1 b y1 x (c) The term dM/I can be removed from within the integral and b dx placed on the right side of the equation; then, from Eq. (33) with V = d M /dx , Eq. (a) becomes τ= V Ib c yd A (329) y1 In this equation, the integral is the rst moment of the area A with respect to the neutral axis (see Fig. 318c). This integral is usually designated as Q. Thus c Q= y1 yd A = y A ¯ (330) ¯ where, for the isolated area y1 to c, y is the distance in the y direction from the neutral plane to the centroid of the area A . With this, Eq. (329) can be written as τ= VQ Ib (331) In using this equation, note that b is the width of the section at y = y1 . Also, I is the second moment of area of the entire section about the neutral axis. Because cross shears are equal, and area A is nite, the shear stress τ given by Eq. (331) and shown on area A in Fig. 318c occurs only at y = y1 . The shear stress on the lateral area varies with y (normally maximum at the neutral axis where y = 0, and zero at the outer bers of the beam where Q A 0). EXAMPLE 37 A beam 12 in long is to support a load of 488 lbf acting 3 in from the left support, as shown in Fig. 319a. Basing the design only on bending stress, a designer has selected a 3-in aluminum channel with the cross-sectional dimensions shown. If the direct shear is neglected, the stress in the beam may be actually higher than the designer thinks. Determine the principal stresses considering bending and direct shear and compare them with that considering bending only. BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 92 I. Basics © The McGrawHill Companies, 2008 3. Load and Stress Analysis Mechanical Engineering Design Figure 319 y 488 lbf 3 in 9 in 0.273 in x O 0.170 in 3 in 1.410 in R1 = 366 lbf I I = 1.66 in , c = 1.10 in3 4 R2 = 122 lbf (a) y dy dA 488 lbf x O a b y 1.227 in 366 lbf 122 lbf 366 lbf O 122 lbf 1098 lbf in (c ) O (b ) Solution The loading, shear-force, and bending-moment diagrams are shown in Fig. 319b. If the direct shear force is included in the analysis, the maximum stresses at the top and bottom of the beam will be the same as if only bending were considered. The maximum bending stresses are σ =± Mc 1098(1.5) =± = ± 992 psi I 1.66 However, the maximum stress due to the combined bending and direct shear stresses may be maximum at the point (3, 1.227) that is just to the left of the applied load, where the web joins the ange. To simplify the calculations we assume a cross section with square corners (Fig. 319c). The normal stress at section ab, with x = 3 in, is σ = My 1098(1.227) = = 812 psi I 1.66 For the shear stress at section ab, considering the area above ab and using Eq. (330) gives Q = y A = 1.227 + ¯ 0.273 ( 1.410)(0.273) = 0.525 in3 2 97 98 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition I. Basics © The McGrawHill Companies, 2008 3. Load and Stress Analysis Load and Stress Analysis 93 Using Eq. (331) with V = 366 lbf, I = 1.66 in4 , Q = 0.525 in3 , and b = 0.170 in yields τx y = 366(0.525) VQ = = 681 psi Ib 1.66(0.170) The negative sign comes from recognizing that the shear stress is down on an x face of a dx dy element at the location being considered. The principal stresses at the point can now be determined. Using Eq. (313), we nd that at x = 3 in, y = 1.227 in, σ1 , σ2 = = σx + σ y ± 2 812 + 0 ± 2 σx σ y 2 2 2 + τx y 812 0 2 2 + ( 681) 2 = 387, 1200 psi For a point at x = 3 in, y = 1.227 in, the principal stresses are σ1 , σ2 = 1200, 387 psi. Thus we see that the maximum principal stresses are ±1200 psi, 21 percent higher than thought by the designer. Shear Stresses in Standard-Section Beams The shear stress distribution in a beam depends on how Q /b varies as a function of y1. Here we will show how to determine the shear stress distribution for a beam with a rectangular cross section and provide results of maximum values of shear stress for other standard cross sections. Figure 320 shows a portion of a beam with a rectangular cross section, subjected to a shear force V and a bending moment M. As a result of the bending moment, a normal stress σ is developed on a cross section such as A-A, which is in compression above the neutral axis and in tension below. To investigate the shear stress at a distance y1 above the neutral axis, we select an element of area d A at a distance y above the neutral axis. Then, d A = b dy , and so Eq. (330) becomes c Q= y1 c ydA = b y1 y dy = by 2 2 c y1 = b2 2 c y1 2 (a) Substituting this value for Q into Eq. (331) gives V2 2 τ= c y1 (332) 2I This is the general equation for shear stress in a rectangular beam. To learn something about it, let us make some substitutions. From Table A18, the second moment of area for a rectangular section is I = bh 3 /12; substituting h = 2c and A = bh = 2bc gives I= Ac2 3 (b) BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 94 I. Basics 99 © The McGrawHill Companies, 2008 3. Load and Stress Analysis Mechanical Engineering Design Figure 320 y y y A Shear stresses in a rectangular beam. dy b dA max = 3V 2A V M c yy 1 x O z x h O A (a) (b ) (c) y c y1 x (d ) If we now use this value of I for Eq. (332) and rearrange, we get τ= 3V 2A 1 2 y1 c2 (333) We note that the maximum shear stress exists when y1 = 0, which is at the bending neutral axis. Thus τmax = 3V 2A (334) for a rectangular section. As we move away from the neutral axis, the shear stress decreases parabolically until it is zero at the outer surfaces where y1 = ±c , as shown in Fig. 320c. It is particularly interesting and significant here to observe that the shear stress is maximum at the bending neutral axis, where the normal stress due to bending is zero, and that the shear stress is zero at the outer surfaces, where the bending stress is a maximum. Horizontal shear stress is always accompanied by vertical shear stress of the same magnitude, and so the distribution can be diagrammed as shown in Fig. 320d. Figure 320c shows that the shear τ on the vertical surfaces varies with y. We are almost always interested in the horizontal shear, τ in Fig. 320d, which is nearly uniform with constant y. The maximum horizontal shear occurs where the vertical shear is largest. This is usually at the neutral axis but may not be if the width b is smaller somewhere else. Furthermore, if the section is such that b can be minimized on a plane not horizontal, then the horizontal shear stress occurs on an inclined plane. For example, with tubing, the horizontal shear stress occurs on a radial plane and the corresponding vertical shear is not vertical, but tangential. Formulas for the maximum exural shear stress for the most commonly used shapes are listed in Table 32. 100 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition I. Basics © The McGrawHill Companies, 2008 3. Load and Stress Analysis 95 Load and Stress Analysis Table 32 Beam Shape Formula Formulas for Maximum Shear Stress Due to Bending τmax = Beam Shape 3V 2A Formula τmax = 2V A τmax = V A web Rectangular Hollow, thin-walled round τmax = 4V 3A Web Structural I beam (thin-walled) Circular 312 Torsion Any moment vector that is collinear with an axis of a mechanical element is called a torque vector, because the moment causes the element to be twisted about that axis. A bar subjected to such a moment is also said to be in torsion. As shown in Fig. 3 21, the torque T applied to a bar can be designated by drawing arrows on the surface of the bar to indicate direction or by drawing torque-vector arrows along the axes of twist of the bar. Torque vectors are the hollow arrows shown on the x axis in Fig. 321. Note that they conform to the right-hand rule for vectors. The angle of twist, in radians, for a solid round bar is Tl θ= (335) GJ where T = torque l = length G = modulus of rigidity J = polar second moment of area Figure 321 T l A y dx B T C r B' O C' z x BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 96 I. Basics 101 © The McGrawHill Companies, 2008 3. Load and Stress Analysis Mechanical Engineering Design Shear stresses develop throughout the cross section. For a round bar in torsion, these stresses are proportional to the radius ρ and are given by τ= Tρ J (336) Designating r as the radius to the outer surface, we have τmax = Tr J (337) The assumptions used in the analysis are: The bar is acted upon by a pure torque, and the sections under consideration are remote from the point of application of the load and from a change in diameter. Adjacent cross sections originally plane and parallel remain plane and parallel after twisting, and any radial line remains straight. The material obeys Hookes law. Equation (337) applies only to circular sections. For a solid round section, J= π d4 32 (338) where d is the diameter of the bar. For a hollow round section, J= π4 d di4 32 o (339) where the subscripts o and i refer to the outside and inside diameters, respectively. In using Eq. (337) it is often necessary to obtain the torque T from a consideration of the power and speed of a rotating shaft. For convenience when U. S. Customary units are used, three forms of this relation are H= where 2π T n Tn FV = = 33 000 33 000(12) 63 025 (340) H = power, hp T = torque, lbf · in n = shaft speed, rev/min F = force, lbf V = velocity, ft/min When SI units are used, the equation is where H = power, W T = torque, N · m ω = angular velocity, rad/s H = Tω (341) 102 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition I. Basics © The McGrawHill Companies, 2008 3. Load and Stress Analysis 97 Load and Stress Analysis The torque T corresponding to the power in watts is given approximately by T = 9.55 H n (342) where n is in revolutions per minute. There are some applications in machinery for noncircular-cross-section members and shafts where a regular polygonal cross section is useful in transmitting torque to a gear or pulley that can have an axial change in position. Because no key or keyway is needed, the possibility of a lost key is avoided. Saint Venant (1855) showed that the maximum shearing stress in a rectangular b × c section bar occurs in the middle of the longest side b and is of the magnitude τmax = T .T =2 α bc2 bc 3+ 1.8 b/c (343) where b is the longer side, c the shorter side, and α a factor that is a function of the ratio b/c as shown in the following table.4 The angle of twist is given by θ= Tl β bc3 G (344) where β is a function of b/c , as shown in the table. b/c 1.00 1.50 1.75 2.00 2.50 3.00 4.00 6.00 8.00 10 α 0.208 0.231 0.239 0.246 0.258 0.267 0.282 0.299 0.307 0.313 0.333 β 0.141 0.196 0.214 0.228 0.249 0.263 0.281 0.299 0.307 0.313 0.333 In Eqs. (343) and (344) b and c are the width (long side) and thickness (short side) of the bar, respectively. They cannot be interchanged. Equation (343) is also approximately valid for equal-sided angles; these can be considered as two rectangles, each of which is capable of carrying half the torque.5 4 S. Timoshenko, Strength of Materials, Part I, 3rd ed., D. Van Nostrand Company, New York, 1955, p. 290. 5 For other sections see W. C. Young and R. G. Budynas, Roarks Formulas for Stress and Strain, 7th ed., McGraw-Hill, New York, 2002. EXAMPLE 38 Figure 322 shows a crank loaded by a force F = 300 lbf that causes twisting and bending of a 3 -in-diameter shaft xed to a support at the origin of the reference system. 4 In actuality, the support may be an inertia that we wish to rotate, but for the purposes of a stress analysis we can consider this a statics problem. (a) Draw separate free-body diagrams of the shaft A B and the arm B C , and compute the values of all forces, moments, and torques that act. Label the directions of the coordinate axes on these diagrams. (b) Compute the maxima of the torsional stress and the bending stress in the arm BC and indicate where these act. BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 98 I. Basics © The McGrawHill Companies, 2008 3. Load and Stress Analysis Mechanical Engineering Design y Figure 322 1.5 in F C A 3 4 in dia. 1 4 1 2 in 1 14 B in dia. in z 4 in 5 in x (c) Locate a stress element on the top surface of the shaft at A, and calculate all the stress components that act upon this element. (d ) Determine the maximum normal and shear stresses at A. Solution (a) The two free-body diagrams are shown in Fig. 323. The results are F = 300j lbf, TC = 450k lbf · in F = 300j lbf, M1 = 1200i lbf · in, T1 = 450k lbf · in F = 300j lbf, T2 = 1200i lbf · in, M2 = 450k lbf · in F = 300j lbf, MA = 1950k lbf · in, TA = 1200i lbf · in At end C of arm BC: At end B of arm BC: At end B of shaft AB: At end A of shaft AB: Figure 323 F y TC C 4 in B M1 F T1 x z y MA TA A z 5 in F F M2 B T2 x 103 104 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition I. Basics © The McGrawHill Companies, 2008 3. Load and Stress Analysis Load and Stress Analysis 99 (b) For arm BC , the bending moment will reach a maximum near the shaft at B. If we assume this is 1200 lbf · in, then the bending stress for a rectangular section will be Answer σ= M 6(1200) 6M = 18 400 psi = 2= I /c bh 0.25(1.25)2 Of course, this is not exactly correct, because at B the moment is actually being transferred into the shaft, probably through a weldment. For the torsional stress, use Eq. (343). Thus Answer τmax = T bc2 3+ 1.8 b/c = 450 1.8 3+ 1.25(0.252 ) 1.25/0.25 = 19 400 psi This stress occurs at the middle of the 1 1 -in side. 4 (c) For a stress element at A, the bending stress is tensile and is Answer σx = M 32(1950) 32 M = = 47 100 psi = 3 I /c πd π(0.75)3 The torsional stress is Answer τx z = T 16(1200) 16T = = 14 500 psi = J /c π d3 π(0.75)3 where the reader should verify that the negative sign accounts for the direction of τx z . (d ) Point A is in a state of plane stress where the stresses are in the x z plane. Thus the principal stresses are given by Eq. (313) with subscripts corresponding to the x, z axes. Answer The maximum normal stress is then given by σ1 = = Answer σ x + σz + 2 47.1 + 0 + 2 σ x σz 2 2 47.1 0 2 2 + τx z 2 + (14.5)2 = 51.2 kpsi The maximum shear stress at A occurs on surfaces different than the surfaces containing the principal stresses or the surfaces containing the bending and torsional shear stresses. The maximum shear stress is given by Eq. (314), again with modied subscripts, and is given by τ1 = σ x σz 2 2 2 + τx z = 47.1 0 2 2 + (14.5)2 = 27.7 kpsi BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 100 I. Basics 105 © The McGrawHill Companies, 2008 3. Load and Stress Analysis Mechanical Engineering Design EXAMPLE 39 The 1.5-in-diameter solid steel shaft shown in Fig. 324a is simply supported at the ends. Two pulleys are keyed to the shaft where pulley B is of diameter 4.0 in and pulley C is of diameter 8.0 in. Considering bending and torsional stresses only, determine the locations and magnitudes of the greatest tensile, compressive, and shear stresses in the shaft. Solution Figure 324b shows the net forces, reactions, and torsional moments on the shaft. Although this is a three-dimensional problem and vectors might seem appropriate, we will look at the components of the moment vector by performing a two-plane analysis. Figure 324c shows the loading in the x y plane, as viewed down the z axis, where bending moments are actually vectors in the z direction. Thus we label the moment diagram as Mz versus x. For the x z plane, we look down the y axis, and the moment diagram is M y versus x as shown in Fig. 324d. The net moment on a section is the vector sum of the components. That is, M= 2 2 M y + Mz (1) At point B, MB = At point C, 20002 + 80002 = 8246 lbf · in MC = 40002 + 40002 = 5657 lbf · in Thus the maximum bending moment is 8246 lbf · in and the maximum bending stress at pulley B is σ= 32(8246) M d /2 32 M = = = 24 890 psi π d 4 /64 π d3 π(1.53 ) The maximum torsional shear stress occurs between B and C and is 16T 16(1600) T d /2 = = 2414 psi = 4 /32 3 πd πd π(1.53 ) τ= The maximum bending and torsional shear stresses occur just to the right of pulley B at points E and F as shown in Fig. 324e. At point E, the maximum tensile stress will be σ1 given by Answer σ1 = σ + 2 2 σ 2 + τ2 = 24 890 + 2 24 890 2 2 + 24142 = 25 120 psi At point F, the maximum compressive stress will be σ2 given by Answer σ2 = σ 2 σ 2 2 + τ2 = 24 890 2 24 890 2 2 + 24142 = 25 120 psi The extreme shear stress also occurs at E and F and is Answer τ1 = ±σ 2 2 + τ2 = ±24 890 2 2 + 24142 = 12 680 psi 106 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition I. Basics © The McGrawHill Companies, 2008 3. Load and Stress Analysis 101 Load and Stress Analysis y A 10 in z B 200 lbf 10 in C 10 in 1000 lbf D x 100 lbf 500 lbf (a) y 800 lbf A 10 in z 1600 lbf in B 200 lbf 10 in 1600 lbf in 1200 lbf 10 in C D 600 lbf x 400 lbf 400 lbf (b) 1200 lbf y 600 lbf A A C B D C B D 400 lbf 800 lbf 200 lbf Mz (lbf in) z 400 lbf 4000 8000 My (lbf in) 2000 4000 x O x O (d ) (c) Location: at B (x = 10+ ) 8000 lbf in 8246 lbf in F Max. compression and shear 2000 lbf in = tan1 8000 = 76 2000 E Max. tension and shear Figure 324 x x (e) BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 102 I. Basics 107 © The McGrawHill Companies, 2008 3. Load and Stress Analysis Mechanical Engineering Design Figure 325 ds The depicted cross section is elliptical, but the section need not be symmetrical nor of constant thickness. r t dAm = 1 rds 2 Closed Thin-Walled Tubes ( t « r) 6 In closed thin-walled tubes, it can be shown that the product of shear stress times thickness of the wall τ t is constant, meaning that the shear stress τ is inversely proportional to the wall thickness t. The total torque T on a tube such as depicted in Fig. 325 is given by T= τ tr ds = (τ t ) r ds = τ t (2 Am ) = 2 Am t τ where Am is the area enclosed by the section median line. Solving for τ gives τ= T 2 Am t (345) For constant wall thickness t, the angular twist (radians) per unit of length of the tube θ1 is given by θ1 = T Lm 4G A2 t m (346) where L m is the perimeter of the section median line. These equations presume the buckling of the tube is prevented by ribs, stiffeners, bulkheads, and so on, and that the stresses are below the proportional limit. 6 See Sec. 313, F. P. Beer, E. R. Johnston, and J. T. De Wolf, Mechanics of Materials, 4th ed., McGraw-Hill, New York, 2006. EXAMPLE 310 Solution A welded steel tube is 40 in long, has a 1 -in wall thickness, and a 2.5-in by 3.6-in 8 rectangular cross section as shown in Fig. 326. Assume an allowable shear stress of 11 500 psi and a shear modulus of 11.5(106) psi. (a) Estimate the allowable torque T. (b) Estimate the angle of twist due to the torque. (a) Within the section median line, the area enclosed is Am = (2.5 0.125)(3.6 0.125) = 8.253 in2 and the length of the median perimeter is L m = 2[(2.5 0.125) + (3.6 0.125)] = 11.70 in 108 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition I. Basics © The McGrawHill Companies, 2008 3. Load and Stress Analysis Load and Stress Analysis 103 Figure 326 A rectangular steel tube produced by welding. 1 8 in 40 in 2.5 in 3.6 in Answer From Eq. (345) the torque T is T = 2 Am t τ = 2(8.253)0.125(11 500) = 23 730 lbf · in Answer (b) The angle of twist θ from Eq. (346) is θ = θ1l = EXAMPLE 311 Solution 23 730(11.70) T Lm l= (40) = 0.0284 rad = 1.62 4G A2 t 4(11.5 × 106 )(8.2532 )(0.125) m Compare the shear stress on a circular cylindrical tube with an outside diameter of 1 in and an inside diameter of 0.9 in, predicted by Eq. (337), to that estimated by Eq. (345). From Eq. (337), τmax = T (0.5) Tr Tr = = = 14.809T 4 J (π/32)(14 0.94 ) (π/32) do di4 From Eq. (345), τ= T T = = 14.108T 2 Am t 2(π 0.952 /4)0.05 Taking Eq. (337) as correct, the error in the thin-wall estimate is 4.7 percent. Open Thin-Walled Sections When the median wall line is not closed, it is said to be open. Figure 327 presents some examples. Open sections in torsion, where the wall is thin, have relations derived from the membrane analogy theory7 resulting in: τ = G θ1 c = 7 3T Lc2 (347) See S. P. Timoshenko and J. N. Goodier, Theory of Elasticity, 3rd ed., McGraw-Hill, New York, 1970, Sec.109. BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 104 I. Basics 3. Load and Stress Analysis © The McGrawHill Companies, 2008 Mechanical Engineering Design Figure 327 c Some open thin-wall sections. L where τ is the shear stress, G is the shear modulus, θ1 is the angle of twist per unit length, T is torque, and L is the length of the median line. The wall thickness is designated c (rather than t) to remind you that you are in open sections. By studying the table that follows Eq. (3 44) you will discover that membrane theory presumes b/c . Note that open thin-walled sections in torsion should be avoided in design. As indicated in Eq. (3 47), the shear stress and the angle of twist are inversely proportional to c2 and c3 , respectively. Thus, for small wall thickness, stress and twist can become quite large. For example, consider the thin round tube with a slit in Fig. 327. For a ratio of wall thickness of outside diameter of c/do = 0.1, the open section has greater magnitudes of stress and angle of twist by factors of 12.3 and 61.5, respectively, compared to a closed section of the same dimensions. EXAMPLE 312 Solution A 12-in-long strip of steel is 1 in thick and 1 in wide, as shown in Fig. 328. If the 8 allowable shear stress is 11 500 psi and the shear modulus is 11.5(106) psi, nd the torque corresponding to the allowable shear stress and the angle of twist, in degrees, (a) using Eq. (347) and (b) using Eqs. (343) and (344). (a) The length of the median line is 1 in. From Eq. (347), (1)(1/8)2 11 500 Lc2 τ = = 59.90 lbf · in 3 3 11500(12) τl = = 0.0960 rad = 5.5° θ = θ1l = Gc 11.5(106 )(1/8) T= A torsional spring rate kt can be expressed as T /θ : T kt = 59.90/0.0960 = 624 lbf · in/rad 1 in (b) From Eq. (343), T= 1 8 in Figure 328 The cross-section of a thin strip of steel subjected to a torsional moment T. 11 500(1)(0.125)2 τmax bc2 = = 55.72 lbf · in 3 + 1.8/(b/c) 3 + 1.8/(1/0.125) From Eq. (344), with b/c = 1/0.125 = 8, θ= 55.72(12) Tl = = 0.0970 rad = 5.6° β bc3 G 0.307(1)0.1253 (11.5)106 kt = 55.72/0.0970 = 574 lbf · in/rad 109 110 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition I. Basics © The McGrawHill Companies, 2008 3. Load and Stress Analysis 105 Load and Stress Analysis 313 Stress Concentration In the development of the basic stress equations for tension, compression, bending, and torsion, it was assumed that no geometric irregularities occurred in the member under consideration. But it is quite difcult to design a machine without permitting some changes in the cross sections of the members. Rotating shafts must have shoulders designed on them so that the bearings can be properly seated and so that they will take thrust loads; and the shafts must have key slots machined into them for securing pulleys and gears. A bolt has a head on one end and screw threads on the other end, both of which account for abrupt changes in the cross section. Other parts require holes, oil grooves, and notches of various kinds. Any discontinuity in a machine part alters the stress distribution in the neighborhood of the discontinuity so that the elementary stress equations no longer describe the state of stress in the part at these locations. Such discontinuities are called stress raisers, and the regions in which they occur are called areas of stress concentration. The distribution of elastic stress across a section of a member may be uniform as in a bar in tension, linear as a beam in bending, or even rapid and curvaceous as in a sharply curved beam. Stress concentrations can arise from some irregularity not inherent in the member, such as tool marks, holes, notches, grooves, or threads. The nominal stress is said to exist if the member is free of the stress raiser. This denition is not always honored, so check the denition on the stress-concentration chart or table you are using. A theoretical, or geometric, stress-concentration factor Kt or Kts is used to relate the actual maximum stress at the discontinuity to the nominal stress. The factors are dened by the equations Kt = σmax σ0 K ts = τmax τ0 (348) where Kt is used for normal stresses and Kts for shear stresses. The nominal stress σ0 or τ0 is more difcult to dene. Generally, it is the stress calculated by using the elementary stress equations and the net area, or net cross section. But sometimes the gross cross section is used instead, and so it is always wise to double check your source of Kt or Kts before calculating the maximum stress. The subscript t in Kt means that this stress-concentration factor depends for its value only on the geometry of the part. That is, the particular material used has no effect on the value of Kt. This is why it is called a theoretical stress-concentration factor. The analysis of geometric shapes to determine stress-concentration factors is a difcult problem, and not many solutions can be found. Most stress-concentration factors are found by using experimental techniques.8 Though the nite-element method has been used, the fact that the elements are indeed nite prevents nding the true maximum stress. Experimental approaches generally used include photoelasticity, grid methods, brittle-coating methods, and electrical strain-gauge methods. Of course, the grid and strain-gauge methods both suffer from the same drawback as the nite-element method. Stress-concentration factors for a variety of geometries may be found in Tables A15 and A16. 8 The best source book is W. D. Pilkey, Petersons Stress Concentration Factors, 2nd ed., John Wiley & Sons, New York, 1997. BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 106 I. Basics 111 © The McGrawHill Companies, 2008 3. Load and Stress Analysis Mechanical Engineering Design Figure 329 Thin plate in tension or simple compression with a transverse central hole. The net tensile force is F = σ wt, where t is the thickness of the plate. The nominal stress is given by F w σ σ0 = = (w d )t (w d ) 3.0 d 2.8 w 2.6 Kt 2.4 2.2 2.0 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 d /w An example is shown in Fig. 329, that of a thin plate loaded in tension where the plate contains a centrally located hole. In static loading, stress-concentration factors are applied as follows. In ductile (ǫ f 0.05) materials, the stress-concentration factor is not usually applied to predict the critical stress, because plastic strain in the region of the stress is localized and has a strengthening effect. In brittle materials (ǫ f < 0.05), the geometric stressconcentration factor Kt is applied to the nominal stress before comparing it with strength. Gray cast iron has so many inherent stress raisers that the stress raisers introduced by the designer have only a modest (but additive) effect. EXAMPLE 313 Be Alert to Viewpoint On a spade rod end (or lug) a load is transferred through a pin to a rectangular-crosssection rod or strap. The theoretical or geometric stress-concentration factor for this geometry is known as follows, on the basis of the net area A = (w d )t as shown in Fig. 330. d/w Kt 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50 7.4 5.4 4.6 3.7 3.2 2.8 2.6 2.45 As presented in the table, Kt is a decreasing monotone. This rod end is similar to the square-ended lug depicted in Fig. A15-12 of appendix A. σmax = K t σ0 Kt F F = Kt σmax = A (w d )t (a) (b) It is insightful to base the stress concentration factor on the unnotched area, wt . Let F σmax = K t (c) wt By equating Eqs. (b) and (c) and solving for K t we obtain K t = Kt wt F Kt = F (w d )t 1 d /w (d ) 112 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition I. Basics © The McGrawHill Companies, 2008 3. Load and Stress Analysis 107 Load and Stress Analysis A power regression curve-t for the data in the above table in the form K t = a (d /w)b gives the result a = exp(0.204 521 2) = 1.227, b = 0.935, and r 2 = 0.9947. Thus F d A B K t = 1.227 0.935 d w (e) which is a decreasing monotone (and unexciting). However, from Eq. (d ), w t K t = Form another table from Eq. ( f ): F d/w Figure 330 K t A round-ended lug end to a rectangular cross-section rod. The maximum tensile stress in the lug occurs at locations A and B. The net area A = ( w d) t is used in the denition of K t , but there is an advantage to using the total area wt. 0.15 0.20 0.25 1.227 1 d /w 0.30 0.35 d w 0.935 0.40 0.45 (f ) 0.50 0.55 0.60 8.507 6.907 5.980 5.403 5.038 4.817 4.707 4.692 4.769 4.946 which shows a stationary-point minimum for K t . This can be found by differentiating Eq. ( f ) with respect to d /w and setting it equal to zero: (1 d /w)ab(d /w)b1 + a (d /w)b d K t = =0 d (d /w) [1 (d /w)]2 where b = 0.935, from which d w = b 0.935 = = 0.483 b1 0.935 1 with a corresponding K t of 4.687. Knowing the section w × t lets the designer specify the strongest lug immediately by specifying a pin diameter of 0.483w (or, as a rule of thumb, of half the width). The theoretical K t data in the original form, or a plot based on the data using net area, would not suggest this. The right viewpoint can suggest valuable insights. 314 po dr r pi ri ro Stresses in Pressurized Cylinders Cylindrical pressure vessels, hydraulic cylinders, gun barrels, and pipes carrying uids at high pressures develop both radial and tangential stresses with values that depend upon the radius of the element under consideration. In determining the radial stress σr and the tangential stress σt , we make use of the assumption that the longitudinal elongation is constant around the circumference of the cylinder. In other words, a right section of the cylinder remains plane after stressing. Referring to Fig. 331, we designate the inside radius of the cylinder by ri, the outside radius by ro, the internal pressure by pi, and the external pressure by po. Then it can be shown that tangential and radial stresses exist whose magnitudes are9 σt = Figure 331 2 2 pi ri2 po ro ri2ro ( po pi )/ r 2 2 r2 ro i 2 2 pi ri2 po ro + ri2ro ( po pi )/ r 2 σr = 2 ro ri2 A cylinder subjected to both internal and external pressure. 9 (349) See Richard G. Budynas, Advanced Strength and Applied Stress Analysis, 2nd ed., McGraw-Hill, New York, 1999, pp. 348352. BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 108 I. Basics 113 © The McGrawHill Companies, 2008 3. Load and Stress Analysis Mechanical Engineering Design Figure 332 po = 0 po = 0 t Distribution of stresses in a thick-walled cylinder subjected to internal pressure. ro ri pi ri pi r ro (a) Tangential stress distribution (b) Radial stress distribution As usual, positive values indicate tension and negative values, compression. The special case of po = 0 gives σt = ri2 pi 2 ro ri2 r 2 pi σr = 2 i 2 r o ri 1+ 2 ro r2 r2 1 o r2 (350) The equations of set (350) are plotted in Fig. 332 to show the distribution of stresses over the wall thickness. It should be realized that longitudinal stresses exist when the end reactions to the internal pressure are taken by the pressure vessel itself. This stress is found to be σl = pi ri2 2 ro ri2 (351) We further note that Eqs. (349), (350), and (351) apply only to sections taken a signicant distance from the ends and away from any areas of stress concentration. Thin-Walled Vessels When the wall thickness of a cylindrical pressure vessel is about one-twentieth, or less, of its radius, the radial stress that results from pressurizing the vessel is quite small compared with the tangential stress. Under these conditions the tangential stress can be obtained as follows: Let an internal pressure p be exerted on the wall of a cylinder of thickness t and inside diameter di. The force tending to separate two halves of a unit length of the cylinder is pdi . This force is resisted by the tangential stress, also called the hoop stress, acting uniformly over the stressed area. We then have pdi = 2t σt , or pdi (σt )av = (352) 2t This equation gives the average tangential stress and is valid regardless of the wall thickness. For a thin-walled vessel an approximation to the maximum tangential stress is p(di + t ) (σt )max = (353) 2t where di + t is the average diameter. 114 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition I. Basics © The McGrawHill Companies, 2008 3. Load and Stress Analysis Load and Stress Analysis 109 In a closed cylinder, the longitudinal stress σl exists because of the pressure upon the ends of the vessel. If we assume this stress is also distributed uniformly over the wall thickness, we can easily nd it to be pdi σl = (354) 4t EXAMPLE 314 An aluminum-alloy pressure vessel is made of tubing having an outside diameter of 8 in and a wall thickness of 1 in. 4 (a) What pressure can the cylinder carry if the permissible tangential stress is 12 kpsi and the theory for thin-walled vessels is assumed to apply? (b) On the basis of the pressure found in part (a), compute all of the stress components using the theory for thick-walled cylinders. Solution (a) Here di = 8 2(0.25) = 7.5 in, ri = 7.5/2 = 3.75 in, and ro = 8/2 = 4 in. Then 1 t / ri = 0.25/3.75 = 0.067. Since this ratio is greater than 20 , the theory for thin-walled vessels may not yield safe results. We rst solve Eq. (353) to obtain the allowable pressure. This gives Answer p= 2(0.25)(12)(10)3 2t (σt )max = = 774 psi di + t 7.5 + 0.25 Then, from Eq. (354), we nd the average longitudinal stress to be σl = 774(7.5) pdi = = 5810 psi 4t 4(0.25) (b) The maximum tangential stress will occur at the inside radius, and so we use r = ri in the rst equation of Eq. (350). This gives Answer (σt )max = ri2 pi 2 ro ri2 2 ro ri2 1+ = pi 2 ro + ri2 42 + 3.752 = 774 2 = 12 000 psi 2 4 3.752 ro ri2 Similarly, the maximum radial stress is found, from the second equation of Eq. (350) to be Answer σr = pi = 774 psi Equation (351) gives the longitudinal stress as Answer σl = pi ri2 774(3.75)2 =2 = 5620 psi 2 4 3.752 ri 2 ro These three stresses, σt , σr , and σl , are principal stresses, since there is no shear on these surfaces. Note that there is no signicant difference in the tangential stresses in parts (a) and (b), and so the thin-wall theory can be considered satisfactory. BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 110 I. Basics 115 © The McGrawHill Companies, 2008 3. Load and Stress Analysis Mechanical Engineering Design 315 Stresses in Rotating Rings Many rotating elements, such as ywheels and blowers, can be simplied to a rotating ring to determine the stresses. When this is done it is found that the same tangential and radial stresses exist as in the theory for thick-walled cylinders except that they are caused by inertial forces acting on all the particles of the ring. The tangential and radial stresses so found are subject to the following restrictions: The outside radius of the ring, or disk, is large compared with the thickness ro 10t . The thickness of the ring or disk is constant. The stresses are constant over the thickness. The stresses are10 2 ri2ro 1 + 3ν 2 r 2 r 3+ν σt = ρω2 3+ν 8 2 ri2 + ro + σr = ρω2 3+ν 8 r 2r 2 2 ri2 + ro i 2o r 2 r (355) where r is the radius to the stress element under consideration, ρ is the mass density, and ω is the angular velocity of the ring in radians per second. For a rotating disk, use ri = 0 in these equations. 316 Press and Shrink Fits When two cylindrical parts are assembled by shrinking or press tting one part upon another, a contact pressure is created between the two parts. The stresses resulting from this pressure may easily be determined with the equations of the preceding sections. Figure 333 shows two cylindrical members that have been assembled with a shrink t. Prior to assembly, the outer radius of the inner member was larger than the inner radius of the outer member by the radial interference δ . After assembly, an interference contact pressure p develops between the members at the nominal radius R, causing radial stresses σr = p in each member at the contacting surfaces. This pressure is given by11 p= δ 1 R Eo 2 ro 2 ro 2 +R 1 + νo + 2 R Ei R 2 + ri2 νi R 2 ri2 (356) where the subscripts o and i on the material properties correspond to the outer and inner members, respectively. If the two members are of the same material with E o = E i = E , νo = vi , the relation simplies to p= 2 E δ (ro R 2 )( R 2 ri2 ) 2 2 R3 ro ri2 (357) For Eqs. (356) or (357), diameters can be used in place of R, ri , and ro , provided δ is the diametral interference (twice the radial interference). 10 Ibid, pp. 348357. 11 Ibid, pp. 348354. 116 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition I. Basics © The McGrawHill Companies, 2008 3. Load and Stress Analysis Load and Stress Analysis 111 Figure 333 Notation for press and shrink ts. (a) Unassembled parts; (b) after assembly. ro R ri (a ) (b ) With p, Eq. (349) can be used to determine the radial and tangential stresses in each member. For the inner member, po = p and pi = 0, For the outer member, po = 0 and pi = p . For example, the magnitudes of the tangential stresses at the transition radius R are maximum for both members. For the inner member (σt )i r=R = p R 2 + ri2 R 2 ri2 (358) and, for the outer member (σt )o r=R =p 2 ro + R 2 2 ro R 2 (359) Assumptions It is assumed that both members have the same length. In the case of a hub that has been press-tted onto a shaft, this assumption would not be true, and there would be an increased pressure at each end of the hub. It is customary to allow for this condition by employing a stress-concentration factor. The value of this factor depends upon the contact pressure and the design of the female member, but its theoretical value is seldom greater than 2. 317 Temperature Effects When the temperature of an unrestrained body is uniformly increased, the body expands, and the normal strain is ǫx = ǫ y = ǫz = α( T ) (360) where α is the coefcient of thermal expansion and T is the temperature change, in degrees. In this action the body experiences a simple volume increase with the components of shear strain all zero. If a straight bar is restrained at the ends so as to prevent lengthwise expansion and then is subjected to a uniform increase in temperature, a compressive stress will develop because of the axial constraint. The stress is σ = ǫ E = α( T ) E (361) In a similar manner, if a uniform at plate is restrained at the edges and also subjected to a uniform temperature rise, the compressive stress developed is given by the equation σ = α( T ) E 1ν (362) BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 112 I. Basics 117 © The McGrawHill Companies, 2008 3. Load and Stress Analysis Mechanical Engineering Design Table 33 Material Coefcients of Thermal Expansion (Linear Mean Coefcients for the Temperature Range 0 100°C) Celsius Scale (°C 1 Fahrenheit Scale (°F1) ) Aluminum 23.9(10)6 13.3(10)6 Brass, cast 18.7(10)6 10.4(10)6 Carbon steel 6 10.8(10) 6.0(10)6 Cast iron 10.6(10)6 5.9(10)6 Magnesium 25.2(10) 6 14.0(10)6 Nickel steel 13.1(10)6 7.3(10)6 6 9.6(10)6 4.3(10)6 2.4(10)6 Stainless steel 17.3(10) Tungsten The stresses expressed by Eqs. (361) and (362) are called thermal stresses. They arise because of a temperature change in a clamped or restrained member. Such stresses, for example, occur during welding, since parts to be welded must be clamped before welding. Table 33 lists approximate values of the coefcients of thermal expansion. 318 Curved Beams in Bending The distribution of stress in a curved exural member is determined by using the following assumptions: The cross section has an axis of symmetry in a plane along the length of the beam. Plane cross sections remain plane after bending. The modulus of elasticity is the same in tension as in compression. We shall nd that the neutral axis and the centroidal axis of a curved beam, unlike the axes of a straight beam, are not coincident and also that the stress does not vary linearly from the neutral axis. The notation shown in Fig. 334 is dened as follows: ro = radius of outer ber ri = radius of inner ber Figure 334 a Note that y is positive in the direction toward the center of curvature, point O. Centroidal axis b' b co h e y y ci M d M c c' Neutral axis ro rn r d ri rn O O rc 118 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition I. Basics © The McGrawHill Companies, 2008 3. Load and Stress Analysis Load and Stress Analysis 113 h = depth of section co = distance from neutral axis to outer ber ci = distance from neutral axis to inner ber rn = radius of neutral axis rc = radius of centroidal axis e = distance from centroidal axis to neutral axis M = bending moment; positive M decreases curvature Figure 334 shows that the neutral and centroidal axes are not coincident.12 It turns out that the location of the neutral axis with respect to the center of curvature O is given by the equation A dA r rn = (363) The stress distribution can be found by balancing the external applied moment against the internal resisting moment. The result is found to be σ= My Ae(rn y ) (364) where M is positive in the direction shown in Fig. 334. Equation (363) shows that the stress distribution is hyperbolic. The critical stresses occur at the inner and outer surfaces where y = ci and y = co , respectively, and are σi = Mci Aeri σo = Mco Aero (365) These equations are valid for pure bending. In the usual and more general case, such as a crane hook, the U frame of a press, or the frame of a clamp, the bending moment is due to forces acting to one side of the cross section under consideration. In this case the bending moment is computed about the centroidal axis, not the neutral axis. Also, an additional axial tensile or compressive stress must be added to the bending stresses given by Eqs. (364) and (365) to obtain the resultant stresses acting on the section. 12 For a complete development of the relations in this section, see Richard G. Budynas, Advanced Strength and Applied Stress Analysis, 2nd ed., Mcgraw-Hill, New York, 1999, pp. 309317. EXAMPLE 315 Solution Plot the distribution of stresses across section A-A of the crane hook shown in Fig.335a. The cross section is rectangular, with b = 0.75 in and h = 4 in, and the load is F = 5000 lbf. Since A = bh , we have d A = b dr and, from Eq. (363), bh A h = ro rn = =r o dA b ln dr ri r r r i (1) BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 114 I. Basics © The McGrawHill Companies, 2008 3. Load and Stress Analysis Mechanical Engineering Design From Fig. 335b, we see that ri = 2 in, ro = 6 in, rc = 4 in, and A = 3 in2. Thus, from Eq. (1), 4 h rn = = 6 = 3.641 in ln(ro / ri ) ln 2 and so the eccentricity is e = rc rn = 4 3.641 = 0.359 in. The moment M is positive and is M = Frc = 5000(4) = 20 000 lbf · in. Adding the axial component of stress to Eq. (364) gives F My 5000 (20 000)(3.641 r ) σ= + = + (2) A Ae(rn y ) 3 3(0.359)r Substituting values of r from 2 to 6 in results in the stress distribution shown in Fig. 335c. The stresses at the inner and outer radii are found to be 16.9 and 5.63 kpsi, respectively, as shown. Figure 335 3/4 in (a) Plan view of crane hook; (b) cross section and notation; (c) resulting stress distribution. There is no stress concentration. Section A-A rc e rn y r 6-in R. F A 2 in A 0.75 in 2-in R. 4 in 6 in (a) (b) 16.9 kpsi + 4 2 5 3 6 r 5.63 kpsi (c) Note in the hook example, the symmetrical rectangular cross section causes the maximum tensile stress to be 3 times greater than the maximum compressive stress. If we wanted to design the hook to use material more effectively we would use more material at the inner radius and less material at the outer radius. For this reason, trapezoidal, T, or unsymmetric I, cross sections are commonly used. Sections most frequently encountered in the stress analysis of curved beams are shown in Table 34. 119 120 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition I. Basics © The McGrawHill Companies, 2008 3. Load and Stress Analysis Table 34 Formulas for Sections of Curved Beams rc = r i + h 2 h ro rn = rc rn h ln (ro /r i ) ri bo rc = ri + e h ro bi ri rn rc bo c2 ro rn bi A bo b i + [(b i ro bo r i )/h] ln (ro /r i ) rc = r i + e c1 rn = rc rn = b i c 2 + 2bo c1 c2 + bo c 2 1 2 2(bo c2 + b i c1 ) b i c1 + bo c2 b i ln [(r i + c1 )/r i )] + bo ln [ro /(r i + c1 )] ri R rc = r i + R e rn = ri rn rc R2 2 rc bo rc = r i + to e h b i + 2bo 3 b i + bo r2 R2 c 12 ht 2 t i (b i t ) + to (bo t ) + ht t h rn = ro ti rc r n 1 + 2 t i2 (b i t ) + to (bo t )(h to /2) t i (b i t ) + to (bo t ) + hto ri + t ro to ro b i ln + t ln + bo ln ri ri + ti ro to bi ri b rc = r i + to e t 2 t 2 h ro rc r n ti rn = 12 ht 2 1 + 2 t 2 (b t ) + to (b t )(h to /2) i ht + (b t )(t i + to ) (b t )(t i + t o ) + ht ri + ti ro ro to b ln + ln + t ln ri ro to ri + ti ri 115 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 116 I. Basics 121 © The McGrawHill Companies, 2008 3. Load and Stress Analysis Mechanical Engineering Design Alternative Calculations for e Calculating rn and rc mathematically and subtracting the difference can lead to large errors if not done carefully, since rn and rc are typically large values compared to e. Since e is in the denominator of Eqs. (364) and (365), a large error in e can lead to an inaccurate stress calculation. Furthermore, if you have a complex cross section that the tables do not handle, alternative methods for determining e are needed. For a quick and simple approximation of e, it can be shown that13 .I e= rc A (366) . This approximation is good for a large curvature where e is small with rn = rc . Substituting Eq. (366) into Eq. (364), with rn y = r , gives . M y rc σ= Ir (367) . If rn = rc , which it should be to use Eq. (367), then it is only necessary to calculate rc , and to measure y from this axis. Determining rc for a complex cross section can be done easily by most CAD programs or numerically as shown in the before mentioned reference. Observe that as the curvature increases, r rc , and Eq. (367) becomes the straight-beam formulation, Eq. (324). Note that the negative sign is missing because y in Fig. 334 is vertically downward, opposite that for the straight-beam equation. 13 Ibid., pp 317321. Also presents a numerical method. EXAMPLE 316 Solution Consider the circular section in Table 34 with rc = 3 in and R = 1 in. Determine e by using the formula from the table and approximately by using Eq. (366). Compare the results of the two solutions. Using the formula from Table 34 gives rn = R2 2 rc 2 rc R 2 = 12 = 2.91421 in 2 3 32 1 This gives an eccentricity of Answer e = rc rn = 3 2.91421 = 0.08579 in The approximate method, using Eq. (366), yields Answer π R 4 /4 R2 12 .I = = = 0.08333 in e= = rc A rc (π R 2 ) 4rc 4(3) This differs from the exact solution by 2.9 percent. 122 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition I. Basics © The McGrawHill Companies, 2008 3. Load and Stress Analysis 117 Load and Stress Analysis 319 Contact Stresses When two bodies having curved surfaces are pressed together, point or line contact changes to area contact, and the stresses developed in the two bodies are threedimensional. Contact-stress problems arise in the contact of a wheel and a rail, in automotive valve cams and tappets, in mating gear teeth, and in the action of rolling bearings. Typical failures are seen as cracks, pits, or flaking in the surface material. The most general case of contact stress occurs when each contacting body has a double radius of curvature; that is, when the radius in the plane of rolling is different from the radius in a perpendicular plane, both planes taken through the axis of the contacting force. Here we shall consider only the two special cases of contacting spheres and contacting cylinders.14 The results presented here are due to Hertz and so are frequently known as Hertzian stresses. Spherical Contact When two solid spheres of diameters d1 and d2 are pressed together with a force F, a circular area of contact of radius a is obtained. Specifying E 1 , ν1 and E 2 , ν2 as the respective elastic constants of the two spheres, the radius a is given by the equation a= 3 2 2 3 F 1 ν1 E 1 + 1 ν2 8 1/d1 + 1/d2 (368) E2 The pressure distribution within the contact area of each sphere is hemispherical, as shown in Fig. 336b. The maximum pressure occurs at the center of the contact area and is pmax = 3F 2π a 2 (369) Equations (368) and (369) are perfectly general and also apply to the contact of a sphere and a plane surface or of a sphere and an internal spherical surface. For a plane surface, use d = . For an internal surface, the diameter is expressed as a negative quantity. The maximum stresses occur on the z axis, and these are principal stresses. Their values are 1 z (1 + ν) σ1 = σ2 = σx = σ y = pmax 1 tan1 a |z /a | σ3 = σz = 14 pmax z2 1+ 2 a 1 2 2 1+ z a2 (370) (371) A more comprehensive presentation of contact stresses may be found in Arthur P. Boresi and Richard J. Schmidt, Advanced Mechanics of Materials, 6th ed., Wiley, New York, 2003 pp. 589623. BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 118 I. Basics 123 © The McGrawHill Companies, 2008 3. Load and Stress Analysis Mechanical Engineering Design Figure 336 F (a) Two spheres held in contact by force F; (b) contact stress has a hemispherical distribution across contact zone diameter 2a. F x d1 y y 2a d2 F F z z (a ) (b) These equations are valid for either sphere, but the value used for Poissons ratio must correspond with the sphere under consideration. The equations are even more complicated when stress states off the z axis are to be determined, because here the x and y coordinates must also be included. But these are not required for design purposes, because the maxima occur on the z axis. Mohrs circles for the stress state described by Eqs. (370) and (371) are a point and two coincident circles. Since σ1 = σ2 , we have τ1/2 = 0 and τmax = τ1/3 = τ2/3 = σ2 σ3 σ1 σ3 = 2 2 (372) Figure 337 is a plot of Eqs. (370), (371), and (372) for a distance to 3a below the surface. Note that the shear stress reaches a maximum value slightly below the surface. It is the opinion of many authorities that this maximum shear stress is responsible for the surface fatigue failure of contacting elements. The explanation is that a crack originates at the point of maximum shear stress below the surface and progresses to the surface and that the pressure of the lubricant wedges the chip loose. Cylindrical Contact Figure 338 illustrates a similar situation in which the contacting elements are two cylinders of length l and diameters d1 and d2. As shown in Fig. 338b, the area of contact is a narrow rectangle of width 2b and length l , and the pressure distribution is elliptical. The half-width b is given by the equation b= 2 2 2 F 1 ν1 E 1 + 1 ν2 πl 1/d1 + 1/d2 E2 (373) BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition I. Basics © The McGrawHill Companies, 2008 3. Load and Stress Analysis Load and Stress Analysis Magnitude of the stress components below the surface as a function of the maximum pressure of contacting spheres. Note that the maximum shear stress is slightly below the surface at z = 0.48a and is approximately 0.3pmax. The chart is based on a Poisson ratio of 0.30. Note that the normal stresses are all compressive stresses. 119 , Figure 337 1.0 0.8 Ratio of stress to pmax 124 z 0.6 x , y 0.4 max 0.2 0 z 0 0.5a 1.5a a 2a 2.5a 3a Distance from contact surface Figure 338 F (a) Two right circular cylinders held in contact by forces F uniformly distributed along cylinder length l. (b) Contact stress has an elliptical distribution across the contact zone width 2b. F x x d1 l y y 2b d2 F F z z (a ) (b ) The maximum pressure is pmax = 2F π bl (374) Equations (373) and (374) apply to a cylinder and a plane surface, such as a rail, by making d = for the plane surface. The equations also apply to the contact of a cylinder and an internal cylindrical surface; in this case d is made negative for the internal surface. BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 120 I. Basics 125 © The McGrawHill Companies, 2008 3. Load and Stress Analysis Mechanical Engineering Design The stress state along the z axis is given by the equations σx = 2ν pmax 1+ z2 z 2 b b (375) z2 1 + 2 b2 z σ y = pmax 2 2 b z 1+ 2 b pmax σ3 = σz = (376) (377) 1 + z 2 /b2 These three equations are plotted in Fig. 339 up to a distance of 3b below the surface. For 0 z 0.436b, σ1 = σx , and τmax = (σ1 σ3 )/2 = (σx σz )/2. For z 0.436b, σ1 = σ y , and τmax = (σ y σz )/2. A plot of τmax is also included in Fig. 339, where the greatest value occurs at z /b = 0.786 with a value of 0.300 pmax . Hertz (1881) provided the preceding mathematical models of the stress eld when the contact zone is free of shear stress. Another important contact stress case is line of contact with friction providing the shearing stress on the contact zone. Such shearing stresses are small with cams and rollers, but in cams with atfaced followers, wheel-rail contact, and gear teeth, the stresses are elevated above the Hertzian eld. Investigations of the effect on the stress eld due to normal and shear stresses in the contact zone were begun theoretically by Lundberg (1939), and continued by Mindlin (1949), Smith-Liu (1949), and Poritsky (1949) independently. For further detail, see the reference cited in Footnote 14. , Figure 339 1.0 0.8 y Ratio of stress to pmax Magnitude of the stress components below the surface as a function of the maximum pressure for contacting cylinders. The largest value of τmax occurs at z / b = 0.786. Its maximum value is 0.30pmax. The chart is based on a Poisson ratio of 0.30. Note that all normal stresses are compressive stresses. z 0.6 0.4 x max 0.2 0 z 0 0.5b b 1.5b 2b Distance from contact surface 2.5b 3b 126 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition I. Basics 3. Load and Stress Analysis © The McGrawHill Companies, 2008 Load and Stress Analysis 320 121 Summary The ability to quantify the stress condition at a critical location in a machine element is an important skill of the engineer. Why? Whether the member fails or not is assessed by comparing the (damaging) stress at a critical location with the corresponding material strength at this location. This chapter has addressed the description of stress. Stresses can be estimated with great precision where the geometry is sufciently simple that theory easily provides the necessary quantitative relationships. In other cases, approximations are used. There are numerical approximations such as nite element analysis (FEA, see Chap. 19), whose results tend to converge on the true values. There are experimental measurements, strain gauging, for example, allowing inference of stresses from the measured strain conditions. Whatever the method(s), the goal is a robust description of the stress condition at a critical location. The nature of research results and understanding in any field is that the longer we work on it, the more involved things seem to be, and new approaches are sought to help with the complications. As newer schemes are introduced, engineers, hungry for the improvement the new approach promises, begin to use the approach. Optimism usually recedes, as further experience adds concerns. Tasks that promised to extend the capabilities of the nonexpert eventually show that expertise is not optional. In stress analysis, the computer can be helpful if the necessary equations are available. Spreadsheet analysis can quickly reduce complicated calculations for parametric studies, easily handling what if questions relating trade-offs (e.g., less of a costly material or more of a cheaper material). It can even give insight into optimization opportunities. When the necessary equations are not available, then methods such as FEA are attractive, but cautions are in order. Even when you have access to a powerful FEA code, you should be near an expert while you are learning. There are nagging questions of convergence at discontinuities. Elastic analysis is much easier than elastic-plastic analysis. The results are no better than the modeling of reality that was used to formulate the problem. Chapter 19 provides an idea of what nite-element analysis is and how it can be used in design. The chapter is by no means comprehensive in nite-element theory and the application of nite elements in practice. Both skill sets require much exposure and experience to be adept. PROBLEMS 31 The symbol W is used in the various gure parts to specify the weight of an element. If not given, assume the parts are weightless. For each gure part, sketch a free-body diagram of each element, including the frame. Try to get the forces in the proper directions, but do not compute magnitudes. 32 Using the gure part selected by your instructor, sketch a free-body diagram of each element in the gure. Compute the magnitude and direction of each force using an algebraic or vector method, as specied. 33 Find the reactions at the supports and plot the shear-force and bending-moment diagrams for each of the beams shown in the gure on page 123. Label the diagrams properly. BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 127 © The McGrawHill Companies, 2008 3. Load and Stress Analysis Mechanical Engineering Design 1 1 1 1 2 3 2 W 2 1 1 W W (b) (a ) (c ) Problem 31 1 1 3 2 2 1 2 1 W W W (d ) (e) y (f) y 2 0.4 m 0.15m radius B A 1 45° W = 2 kN F = 800 N 0.6 m 60° 30° 2 O (a) x 1 (b ) y Problem 32 y F = 400 N B 30° F = 1.2 kN 0.9 m 122 I. Basics C 3 4 2 2 3 B 60° A O 1 x 60° A 1.9 m D 60° O 1 5 E x 9m (c) (d ) 34 Repeat Prob. 33 using singularity functions exclusively (for reactions as well). 35 Select a beam from Table A9 and nd general expressions for the loading, shear-force, bendingmoment, and support reactions. Use the method specied by your instructor. 128 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition I. Basics © The McGrawHill Companies, 2008 3. Load and Stress Analysis Load and Stress Analysis y y 4 in O 4 in 6 in A 4 in D O x C B R1 2 kN 4 kN/m 60 lbf 40 lbf A C B x 200 mm 150 mm 150 mm R2 30 lbf 123 (b) (a ) y y 1000 lbf Problem 33 6 ft O 1000 lbf 4 ft x A O B R1 R2 2 ft 2000 lbf 6 ft 2 ft A R2 (d ) Hinge y 40 lbf/in y 400 lbf 4 ft 3 ft B R1 B O x R2 C D x A R1 C R3 R2 8 in (e) 36 320 lbf 800 lbf 3 ft A C R1 (c) O x B 5 in 2 in 5 in (f) A beam carrying a uniform load is simply supported with the supports set back a distance a from the ends as shown in the gure. The bending moment at x can be found from summing moments to zero at section x : 1 1 M = M + w(a + x )2 wlx = 0 2 2 or M= w [lx (a + x )2 ] 2 where w is the loading intensity in lbf/in. The designer wishes to minimize the necessary weight of the supporting beam by choosing a setback resulting in the smallest possible maximum bending stress. (a) If the beam is congured with a = 2.25 in, l = 10 in, and w = 100 lbf/in, nd the magnitude of the severest bending moment in the beam. (b) Since the conguration in part (a) is not optimal, nd the optimal setback a that will result in the lightest-weight beam. x w(a + x) w, lbf/in M Problem 36 V a x a l wl 2 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 124 I. Basics 3. Load and Stress Analysis © The McGrawHill Companies, 2008 Mechanical Engineering Design 37 An artist wishes to construct a mobile using pendants, string, and span wire with eyelets as shown in the gure. (a) At what positions w, x , y , and z should the suspension strings be attached to the span wires? (b) Is the mobile stable? If so, justify; if not, suggest a remedy. z y x Problem 37 w l 38 For each of the plane stress states listed below, draw a Mohrs circle diagram properly labeled, nd the principal normal and shear stresses, and determine the angle from the x axis to σ1 . Draw stress elements as in Fig. 311c and d and label all details. (a) σx = 12, σ y = 6, τx y = 4 cw (b) σx = 16, σ y = 9, τx y = 5 ccw (c) σx = 10, σ y = 24, τx y = 6 ccw (d ) σx = 9, σ y = 19, τx y = 8 cw 39 Repeat Prob. 38 for: (a) σx = 4, σ y = 12, τx y = 7 ccw (b) σx = 6, σ y = 5, τx y = 8 ccw (c) σx = 8, σ y = 7, τx y = 6 cw (d ) σx = 9, σ y = 6, τx y = 3 cw 310 Repeat Prob. 38 for: (a) σx = 20, σ y = 10, τx y (b) σx = 30, σ y = 10, τx y (c) σx = 10, σ y = 18, τx y (d ) σx = 12, σ y = 22, τx y = 8 cw = 10 ccw = 9 cw = 12 cw 311 For each of the stress states listed below, nd all three principal normal and shear stresses. Draw a complete Mohrs three-circle diagram and label all points of interest. (a) σx = 10, σ y = 4 (b) σx = 10, τx y = 4 ccw (c) σx = 2, σ y = 8, τx y = 4 cw (d ) σx = 10, σ y = 30, τx y = 10 ccw 312 Repeat Prob. 311 for: (a) σx = 80, σ y = 30, τx y = 20 cw (b) σx = 30, σ y = 60, τx y = 30 cw (c) σx = 40, σz = 30, τx y = 20 ccw (d ) σx = 50, σz = 20, τx y = 30 cw 129 130 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition I. Basics © The McGrawHill Companies, 2008 3. Load and Stress Analysis Load and Stress Analysis 125 313 1 A 2 -in-diameter steel tension rod is 72 in long and carries a load of 2000 lbf. Find the tensile stress, the total deformation, the unit strains, and the change in the rod diameter. 314 Twin diagonal aluminum alloy tension rods 15 mm in diameter are used in a rectangular frame to prevent collapse. The rods can safely support a tensile stress of 135 MPa. If the rods are initially 3 m in length, how much must they be stretched to develop this stress? 315 Electrical strain gauges were applied to a notched specimen to determine the stresses in the notch. The results were ǫx = 0.0021 and ǫ y = 0.00067. Find σx and σ y if the material is carbon steel. 316 An engineer wishes to determine the shearing strength of a certain epoxy cement. The problem is to devise a test specimen such that the joint is subject to pure shear. The joint shown in the gure, in which two bars are offset at an angle θ so as to keep the loading force F centroidal with the straight shanks, seems to accomplish this purpose. Using the contact area A and designating Ssu as the ultimate shearing strength, the engineer obtains F cos θ A Ssu = The engineers supervisor, in reviewing the test results, says the expression should be Ssu = F A 1+ 1 tan2 θ 4 1/2 cos θ Resolve the discrepancy. What is your position? Problem 316 317 318 F F The state of stress at a point is σx = 2, σ y = 6, σz = 4, τx y = 3, τ y z = 2, and τz x = 5 kpsi. Determine the principal stresses, draw a complete Mohrs three-circle diagram, labeling all points of interest, and report the maximum shear stress for this case. Repeat Prob. 317 with σx = 10, σ y = 0, σz = 10, τx y = 20, τ y z = 10 2, and τz x = 0 MPa. 319 Repeat Prob. 317 with σx = 1, σ y = 4, σz = 4, τx y = 2, τ y z = 4, and τz x = 2 kpsi. 320 The Roman method for addressing uncertainty in design was to build a copy of a design that was satisfactory and had proven durable. Although the early Romans did not have the intellectual tools to deal with scaling size up or down, you do. Consider a simply supported, rectangular-crosssection beam with a concentrated load F, as depicted in the gure. F c a Problem 320 R2 h b R1 l BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 126 I. Basics © The McGrawHill Companies, 2008 3. Load and Stress Analysis Mechanical Engineering Design (a) Show that the stress-to-load equation is F= σ bh 2 l 6ac (b) Subscript every parameter with m (for model) and divide into the above equation. Introduce a scale factor, s = am /a = bm /b = cm /c etc. Since the Roman method was to not lean on the material any more than the proven design, set σm /σ = 1. Express Fm in terms of the scale factors and F, and comment on what you have learned. 321 Using our experience with concentrated loading on a simple beam, Prob. 320, consider a uniformly loaded simple beam (Table A97). (a) Show that the stress-to-load equation for a rectangular-cross-section beam is given by W= 4 σ bh 2 3l where W = wl . (b) Subscript every parameter with m (for model) and divide the model equation into the prototype equation. Introduce the scale factor s as in Prob. 320, setting σm /σ = 1. Express Wm and wm in terms of the scale factor, and comment on what you have learned. 322 The Chicago North Shore & Milwaukee Railroad was an electric railway running between the cities in its corporate title. It had passenger cars as shown in the gure, which weighed 104.4 kip, had 32-ft, 8-in truck centers, 7-ft-wheelbase trucks, and a coupled length of 55 ft, 3 1 in. Consider 4 the case of a single car on a 100-ft-long, simply supported deck plate girder bridge. (a) What was the largest bending moment in the bridge? (b) Where on the bridge was the moment located? (c) What was the position of the car on the bridge? (d ) Under which axle is the bending moment? 7 ft 32 ft, 8 in Problem 322 Copyright 1963 by Central Electric Railfans Association, Bull. 107, p. 145, reproduced by permission. 131 132 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition I. Basics © The McGrawHill Companies, 2008 3. Load and Stress Analysis 127 Load and Stress Analysis 323 For each section illustrated, nd the second moment of area, the location of the neutral axis, and the distances from the neutral axis to the top and bottom surfaces. Suppose a positive bending moment of 10 kip · in is applied; nd the resulting stresses at the top and bottom surfaces and at every abrupt change in cross section. D y D 7 8 in 1 4 in 1 4 C in C 3 8 1 4 B B in 60° A 1 4 1 2 in 1 in 60° A in 2 in (a) (b) y y C D C B 3 in Problem 323 1 2 1 2 in B A 1 2 4 in in 30° 30° in A 1 in 2 in 4 in 4 in 4 in (c) (d ) y 1 1 in 4 6 in C y 1 1 in 2 D 1 in B 3 in 1 4 C in B A 1 1 in 2 1 in 1 in A (e) (f) 324 From basic mechanics of materials, in the derivation of the bending stresses, it is found that the radius of curvature of the neutral axis, ρ , is given by ρ = E I / M . Find the x and y coordinates of the center of curvature corresponding to the place where the beam is bent the most, for each beam shown in the gure. The beams are both made of Douglas r (see Table A5) and have rectangular sections. 325 For each beam illustrated in the gure, nd the locations and magnitudes of the maximum tensile bending stress and the maximum shear stress due to V. BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 128 I. Basics 133 © The McGrawHill Companies, 2008 3. Load and Stress Analysis Mechanical Engineering Design y y 50 lbf 50 lbf 50 lbf 1 2 Problem 324 C O 20 in in 1 2 x B 30 in A in 5 in O B 20 in A 50 lbf 5 in C x 2 in 2 in 20 in (a) (b) y y 1000 lbf 1000 lbf 3 4 12 in A 6 in in 1 in B x x O O 8 in A B 8 in 1 1 2 in (a ) 2 in (b) Problem 325 y y 3 4 w = 120 lbf/in in 1 in w = 100 lbf/in x O 5 in A 15 in B x 5 in C O 6 in A 12 in 2 in 2 in (c) 326 B (d ) The gure illustrates a number of beam sections. Use an allowable bending stress of 1.2 kpsi for wood and 12 kpsi for steel and nd the maximum safe uniformly distributed load that each beam can carry if the given lengths are between simple supports. 1 (a) Wood joist 1 2 by 9 1 in and 12 ft long 2 3 (b) Steel tube, 2 in OD by 8 -in wall thickness, 48 in long 3 (c) Hollow steel tube 3 by 2 in, outside dimensions, formed from 16 -in material and welded, 48 in long (d ) Steel angles 3 × 3 × 1 in and 72 in long 4 (e) A 5.4-lb, 4-in steel channel, 72 in long ( f ) A 4-in × 1-in steel bar, 72 in long y z Problem 326 y z y z (a) (b) (c) y y y z z z (d ) (e) (f) 134 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition I. Basics © The McGrawHill Companies, 2008 3. Load and Stress Analysis Load and Stress Analysis 327 129 A pin in a knuckle joint carrying a tensile load F deects somewhat on account of this loading, making the distribution of reaction and load as shown in part b of the gure. The usual designers assumption of loading is shown in part c; others sometimes choose the loading shown in part d. If a = 0.5 in, b = 0.75 in, d = 0.5 in, and F = 1000 lbf, estimate the maximum bending stress and the maximum shear stress due to V for each approximation. F (b) b 2 Problem 327 d a+b (c) a a b F a+b b (d ) (a) 328 The gure illustrates a pin tightly tted into a hole of a substantial member. A usual analysis is one that assumes concentrated reactions R and M at distance l from F. Suppose the reaction is distributed linearly along distance a. Is the resulting moment reaction larger or smaller than the concentrated reaction? What is the loading intensity q? What do you think of using the usual assumption? F l a Problem 328 329 For the beam shown, determine (a) the maximum tensile and compressive bending stresses, (b) the maximum shear stress due to V, and (c) the maximum shear stress in the beam. 3000 lbf 2 in 600 lbf/ft A B C 6 in Problem 329 2 in 5 ft 15 ft 6 in Cross section (enlarged) BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 130 I. Basics © The McGrawHill Companies, 2008 3. Load and Stress Analysis Mechanical Engineering Design 330 Consider a simply supported beam of rectangular cross section of constant width b and variable depth h, so proportioned that the maximum stress σx at the outer surface due to bending is constant, when subjected to a load F at a distance a from the left support and a distance c from the right support. Show that the depth h at location x is given by h= 331 0x a In Prob. 330, h 0 as x 0, which cannot occur. If the maximum shear stress τmax due to direct shear is to be constant in this region, show that the depth h at location x is given by h= 332 6 Fcx lbσmax 3 Fc 2 lbτmax 0x 3 Fcσmax 2 8 lbτmax Consider a simply supported static beam of circular cross section of diameter d, so proportioned by varying the diameter such that the maximum stress σx at the surface due to bending is constant, when subjected to a steady load F located at a distance a from the left support and a distance b from the right support. Show that the diameter d at a location x is given by d= 32 Fbx π l σmax 1/3 0x a 333 Two steel thin-wall tubes in torsion of equal length are to be compared. The rst is of square cross section, side length b, and wall thickness t. The second is a round of diameter b and wall thickness t. The largest allowable shear stress is τall and is to be the same in both cases. How does the angle of twist per unit length compare in each case? 334 Begin with a 1-in-square thin-wall steel tube, wall thickness t = 0.05 in, length 40 in, then introduce corner radii of inside radii ri , with allowable shear stress τall of 11 500 psi, shear modulus of 11.5(106) psi; now form a table. Use a column of inside corner radii in the range 0 ri 0.45 in. Useful columns include median line radius rm , periphery of the median line L m , area enclosed by median curve, torque T, and the angular twist θ . The cross section will vary from square to circular round. A computer program will reduce the calculation effort. Study the table. What have you learned? ri Problem 334 rm t 1 in 1 in 335 An unequal leg angle shown in the gure carries a torque T. Show that T= G θ1 3 τmax = G θ1 cmax L i ci3 135 136 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition I. Basics 3. Load and Stress Analysis © The McGrawHill Companies, 2008 Load and Stress Analysis 131 c1 Problem 335 L1 c2 L2 336 1 In Prob. 335 the angle has one leg thickness 16 in and the other 1 in, with both leg lengths 8 The allowable shear stress is τall = 12 000 psi for this steel angle. (a) Find the torque carried by each leg, and the largest shear stress therein. (b) Find the angle of twist per unit length of the section. 337 Two 12 in long thin rectangular steel strips are placed together as shown. Using a maximum allowable shear stress of 12 000 psi, determine the maximum torque and angular twist, and the 1 torsional spring rate. Compare these with a single strip of cross section 1 in by 8 in. 1 8 1 16 5 8 in. in in Problem 337 1 in T 338 Using a maximum allowable shear stress of 60 MPa, nd the shaft diameter needed to transmit 35 kw when (a) The shaft speed is 2000 rev/min. (b) The shaft speed is 200 rev/min. 339 A 15-mm-diameter steel bar is to be used as a torsion spring. If the torsional stress in the bar is not to exceed 110 MPa when one end is twisted through an angle of 30°, what must be the length of the bar? 340 A 3-in-diameter solid steel shaft, used as a torque transmitter, is replaced with a 3-in hollow shaft 1 having a 4 -in wall thickness. If both materials have the same strength, what is the percentage reduction in torque transmission? What is the percentage reduction in shaft weight? 341 A hollow steel shaft is to transmit 5400 N · m of torque and is to be sized so that the torsional stress does not exceed 150 MPa. (a) If the inside diameter is three-fourths of the outside diameter, what size shaft should be used? Use preferred sizes. (b) What is the stress on the inside of the shaft when full torque is applied? BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 132 I. Basics © The McGrawHill Companies, 2008 3. Load and Stress Analysis Mechanical Engineering Design 342 The gure shows an endless-belt conveyor drive roll. The roll has a diameter of 6 in and is driven at 5 rev/min by a geared-motor source rated at 1 hp. Determine a suitable shaft diameter dC for an allowable torsional stress of 14 kpsi. (a) What would be the stress in the shaft you have sized if the motor starting torque is twice the running torque? (b) Is bending stress likely to be a problem? What is the effect of different roll lengths B on bending? (a ) Problem 342 y dA dB dA dC x A B A C (b ) 343 The conveyer drive roll in the gure for Prob. 342 is 150 mm in diameter and is driven at 8 rev/min by a geared-motor source rated at 1 kW. Find a suitable shaft diameter dC based on an allowable torsional stress of 75 MPa. 344 For the same cross-sectional area A = s 2 = π d 2 /4, for a square cross-sectional area shaft and a circular cross-sectional area shaft, in torsion which has the higher maximum shear stress, and by what multiple is it higher? 345 For the same cross-sectional area A = s 2 = π d 2 /4, for a square cross-sectional area shaft and a circular cross-sectional area shaft, both of length l , in torsion which has the greater angular twist θ , and by what multiple is it greater? 346 In the figure, shaft AB is rotating at 1000 rev/min and transmits 10 hp to shaft CD through a set of bevel gears contacting at point E. The contact force at E on the gear of shaft CD is determined to be (FE)CD 92.8i 362.8j 808.0k lbf. For shaft CD: (a) draw a free-body diagram and determine the reactions at C and D assuming simple supports (assume also that bearing C is a thrust bearing), (b) draw the shear-force and bending-moment diagrams, and (c) assuming that the shaft diameter is 1.25 in, determine the maximum tensile and shear stresses in the beam. 137 138 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition I. Basics © The McGrawHill Companies, 2008 3. Load and Stress Analysis Load and Stress Analysis 133 y 6.50 in 3.90 in 3 in D 3 in Problem 346 x 1.30 in A B E 4 in C 347 Repeat the analysis of Prob. 346 for shaft AB. Let the diameter of the shaft be 1.0 in, and assume that bearing A is a thrust bearing. 348 A torque of T = 1000 lbf · in is applied to the shaft EFG, which is running at constant speed and contains gear F. Gear F transmits torque to shaft ABCD through gear C, which drives the chain sprocket at B, transmitting a force P as shown. Sprocket B, gear C, and gear F have pitch diameters of 6, 10, and 5 in, respectively. The contact force between the gears is transmitted through the pressure angle φ = 20°. Assuming no frictional losses and considering the bearings at A, D, E, and G to be simple supports, locate the point on shaft ABCD that contains the maximum tensile bending and maximum torsional shear stresses. From this, determine the maximum tensile and shear stresses in the shaft. y a E G F 5 in y T T = 1000 lbf in B A D C x Problem 348 z 10 in 1.25-in dia. a P 10 in 3 in P 5 in 6 in View aa 349 If the tension-loaded plate of Fig. 329 is innitely wide, then the stress state anywhere in the plate can be described in polar coordinates as15 σr = d2 1 d2 σ 1 2 + 1 2 2 4r 4r 1 3d 2 4r 2 cos 2θ 15 See R. G. Budynas, Advanced Strength and Applied Stress Analysis, 2nd ed. McGraw-Hill, New York, 1999, pp. 235238. BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 134 I. Basics 139 © The McGrawHill Companies, 2008 3. Load and Stress Analysis Mechanical Engineering Design σθ = d2 1 3 d4 σ 1+ 2 1+ 2 4r 16 r 4 1 d2 1 τr θ = σ 1 2 4 r2 1+ 3 d2 4 r2 cos 2θ sin 2θ for the radial, tangential, and shear components, respectively. Here r is the distance from the center to the point of interest and θ is measured positive counterclockwise from the horizontal axis. (a) Find the stress components at the top and side of the hole for r = d /2. (b) If d = 10 mm, plot a graph of the tangential stress distribution σθ /σ for θ = 90º from r = 5 mm to 20 mm. (c) Repeat part (b) for θ = 0º 350 Considering the stress concentration at point A in the gure, determine the maximum normal and shear stresses at A if F = 200 lbf. y 2 in O A 12 in z Problem 350 1 12 -in dia. 1 8 -in B 2 in C R. 1-in dia. 15 in F 1 x 1 2 -in dia. D 351 Develop the formulas for the maximum radial and tangential stresses in a thick-walled cylinder due to internal pressure only. 352 Repeat Prob. 351 where the cylinder is subject to external pressure only. At what radii do the maximum stresses occur? 353 Develop the stress relations for a thin-walled spherical pressure vessel. 354 A pressure cylinder has a diameter of 150 mm and has a 6-mm wall thickness. What pressure can this vessel carry if the maximum shear stress is not to exceed 25 Mpa? 355 A cylindrical pressure vessel has an outside diameter of 10 in and a wall thickness of internal pressure is 350 psi, what is the maximum shear stress in the vessel walls? 3 8 in. If the 140 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition I. Basics © The McGrawHill Companies, 2008 3. Load and Stress Analysis Load and Stress Analysis 135 356 1 An AISI 1020 cold-drawn steel tube has an ID of 1 4 in and an OD of 1 3 in. What maximum 4 external pressure can this tube take if the largest principal normal stress is not to exceed 80 percent of the minimum yield strength of the material? 357 An AISI 1020 cold-drawn steel tube has an ID of 40 mm and an OD of 50 mm. What maximum internal pressure can this tube take if the largest principal normal stress is not to exceed 80 percent of the minimum yield strength of the material? 358 Find the maximum shear stress in a 10-in circular saw if it runs idle at 7200 rev/min. The saw is 14 gauge (0.0747 in) and is used on a 3 -in arbor. The thickness is uniform. What is the maximum 4 radial component of stress? 359 The maximum recommended speed for a 300-mm-diameter abrasive grinding wheel is 2069 rev/min. Assume that the material is isotropic; use a bore of 25 mm, ν = 0.24, and a mass density of 3320 kg/m3; and nd the maximum tensile stress at this speed. 360 1 An abrasive cutoff wheel has a diameter of 6 in, is 16 in thick, and has a 1-in bore. It weighs 6 oz and is designed to run at 10 000 rev/min. If the material is isotropic and ν = 0.20, nd the maximum shear stress at the design speed. 361 A rotary lawn-mower blade rotates at 3000 rev/min. The steel blade has a uniform cross section 1 1 in thick by 1 4 in wide, and has a 1 -in-diameter hole in the center as shown in the gure. 8 2 Estimate the nominal tensile stress at the central section due to rotation. 12 in 1 8 in 1 1 in 4 Problem 361 24 in 362 to 367 The table lists the maximum and minimum hole and shaft dimensions for a variety of standard press and shrink ts. The materials are both hot-rolled steel. Find the maximum and minimum values of the radial interference and the corresponding interface pressure. Use a collar diameter of 80 mm for the metric sizes and 3 in for those in inch units. Problem Number 362 363 364 365 366 367 Fit Designation* Basic Size 40H7/p6 (1.5 in)H7/p6 40H7/s6 (1.5 in)H7/s6 40H7/u6 (1.5 in)H7/u6 40 mm 1.5 in 40 mm 1.5 in 40 mm 1.5 in Hole Dmax Dmin 40.025 1.5010 40.025 1.5010 40.025 1.5010 40.000 1.5000 40.000 1.5000 40.000 1.5000 Shaft dmax dmin 40.042 1.5016 40.059 1.5023 40.076 1.5030 40.026 1.5010 40.043 1.5017 40.060 1.5024 *Note: See Table 79 for description of ts. 368 to 371 The table gives data concerning the shrink fit of two cylinders of differing materials and dimensional specification in inches. Elastic constants for different materials may be found in Table A5. Identify the radial interference δ , then find the interference pressure p , and the tangential normal stress on both sides of the fit surface. If dimensional tolerances are given at fit surfaces, repeat the problem for the highest and lowest stress levels. BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 136 I. Basics 141 © The McGrawHill Companies, 2008 3. Load and Stress Analysis Mechanical Engineering Design Problem Number Inner Cylinder Material di d0 Outer Cylinder Material Di D0 368 369 Steel Steel 0 0 1.002 1.002 Steel Cast iron 1.000 1.000 2.00 2.00 370 371 Steel Steel 0 0 1.002/1.003 2.005/2.003 Steel Aluminum 1.000/1.001 2.000/2.002 2.00 4.00 372 Force ts of a shaft and gear are assembled in an air-operated arbor press. An estimate of assembly force and torque capacity of the t is needed. Assume the coefcient of friction is f , the t interface pressure is p, the nominal shaft or hole radius is R , and the axial length of the gear bore is l . (a) Show that the estimate of the axial force is Fax = 2π f Rlp. (b) Show the estimate of the torque capacity of the t is T = 2π f R 2 lp. 373 A utility hook was formed from a 1-in-diameter round rod into the geometry shown in the gure. What are the stresses at the inner and outer surfaces at section A- A if the load F is 1000 lbf? F 3 in Problem 373 10 in 1 in 3 in A 374 A F The steel eyebolt shown in the gure is loaded with a force F of 100 lbf. The bolt is formed of 3 1 -in-diameter wire to a 8 -in radius in the eye and at the shank. Estimate the stresses at the inner 4 and outer surfaces at sections A- A and B - B . F 3 8 Problem 374 3 8 1 4 375 -in R. B in B A F A in Shown in the gure is a 12-gauge (0.1094-in) by 3 -in latching spring that supports a load of 4 1 F = 3 lbf. The inside radius of the bend is 8 in. Estimate the stresses at the inner and outer surfaces at the critical section. 142 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition I. Basics © The McGrawHill Companies, 2008 3. Load and Stress Analysis Load and Stress Analysis 137 376 The cast-iron bell-crank lever depicted in the gure is acted upon by forces F1 of 250 lbf and F2 of 333 lbf. The section A- A at the central pivot has a curved inner surface with a radius of ri = 1 in. Estimate the stresses at the inner and outer surfaces of the curved portion of the lever. 377 The crane hook depicted in Fig. 335 has a 1-in-diameter hole in the center of the critical section. For a load of 5 kip, estimate the bending stresses at the inner and outer surfaces at the critical section. 378 A 20-kip load is carried by the crane hook shown in the gure. The cross section of the hook uses two concave anks. The width of the cross section is given by b = 2/ r ,where r is the radius from the center. The inside radius ri is 2 in, and the outside radius ro = 6 in. Find the stresses at the inner and outer surfaces at the critical section. F 4 in A 1 -in 8 A 3 4 R. in Problem 375 Section A-A No. 12 gauge (0.1094 in) F1 8 in Nylon bushing 1 A 1 in 3 2 in 3 8 Problem 376 1-in R. A 6 in 1 1 8 in 7 F2 1 8 in Section A-A Problem 378 4 in 2 in in 1 1 4 in BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 138 I. Basics 3. Load and Stress Analysis © The McGrawHill Companies, 2008 Mechanical Engineering Design 379 An offset tensile link is shaped to clear an obstruction with a geometry as shown in the gure. The cross section at the critical location is elliptical, with a major axis of 4 in and a minor axis of 2 in. For a load of 20 kip, estimate the stresses at the inner and outer surfaces of the critical section. 10-in R. Problem 379 380 8 in A cast-steel C frame as shown in the gure has a rectangular cross section of 1 in by 1.6 in, with a 0.4-in-radius semicircular notch on both sides that forms midank uting as shown. Estimate A, rc , rn , and e, and for a load of 3000 lbf, estimate the inner and outer surface stresses at the throat C . Note: Table 34 can be used to determine rn for this section. From the table, the integral d A / r can be evaluated for a rectangle and a circle by evaluating A / rn for each shape [see Eq. (364)]. Subtracting A / rn of the circle from that of the rectangle yields d A / r for the C frame, and rn can then be evaluated. 0.4-in R. Problem 380 3000 lbf 4 in 1 in 1-in R. 0.4 in 0.4 in 381 Two carbon steel balls, each 25 mm in diameter, are pressed together by a force F . In terms of the force F , nd the maximum values of the principal stress, and the maximum shear stress, in MPa. 382 One of the balls in Prob. 381 is replaced by a at carbon steel plate. If F = 18 N, at what depth does the maximum shear stress occur? 383 An aluminum alloy roller with diameter 1 in and length 2 in rolls on the inside of a cast-iron ring having an inside radius of 4 in, which is 2 in thick. Find the maximum contact force F that can be used if the shear stress is not to exceed 4000 psi. 384 The gure shows a hip prosthesis containing a stem that is cemented into a reamed cavity in the femur. The cup is cemented and fastened to the hip with bone screws. Shown are porous layers of titanium into which bone tissue will grow to form a longer-lasting bond than that afforded by cement alone. The bearing surfaces are a plastic cup and a titanium femoral head. The lip shown in the gures bears against the cutoff end of the femur to transfer the load to the leg from the hip. Walking will induce several million stress uctuations per year for an average person, so there is danger that the prosthesis will loosen the cement bonds or that metal cracks may occur because of the many repetitions of stress. Prostheses like this are made in many different sizes. Typical 143 144 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition I. Basics © The McGrawHill Companies, 2008 3. Load and Stress Analysis Load and Stress Analysis 139 C Offset D Neck length Problem 384 Porous hip prosthesis. (Photograph and drawing courtesy of Zimmer, Inc., Warsaw, Indiana.) B Stem length A distal stem diameter (a ) (b) dimensions are ball diameter 50 mm, stem diameter 15 mm, stem length 155 mm, offset 38 mm, and neck length 39 mm. Develop an outline to follow in making a complete stress analysis of this prosthesis. Describe the material properties needed, the equations required, and how the loading is to be dened. 385 Simplify Eqs. (370), (371), and (372) by setting z = 0 and nding σx / pmax , σ y / pmax , σz / pmax , and τ2/3 / pmax and, for cast iron, check the ordinate intercepts of the four loci in Fig. 337. 386 A 6-in-diameter cast-iron wheel, 2 in wide, rolls on a at steel surface carrying an 800-lbf load. (a) Find the Hertzian stresses σx , σ y , σz , and τ2/3 . (b) What happens to the stresses at a point A that is 0.010 in below the wheel rim surface during a revolution? BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition I. Basics © The McGrawHill Companies, 2008 4. Deflection and Stiffness 4 145 Deection and Stiffness Chapter Outline 41 Spring Rates 42 Tension, Compression, and Torsion 43 Deection Due to Bending 144 44 Beam Deection Methods 146 45 Beam Deections by Superposition 46 Beam Deections by Singularity Functions 47 Strain Energy 48 Castiglianos Theorem 49 Deection of Curved Members 142 143 147 150 156 158 163 410 Statically Indeterminate Problems 411 Compression MembersGeneral 412 Long Columns with Central Loading 413 Intermediate-Length Columns with Central Loading 414 Columns with Eccentric Loading 415 Struts or Short Compression Members 416 Elastic Stability 417 Shock and Impact 418 Suddenly Applied Loading 168 173 173 176 176 180 182 183 184 141 146 142 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition I. Basics © The McGrawHill Companies, 2008 4. Deflection and Stiffness Mechanical Engineering Design All real bodies deform under load, either elastically or plastically. A body can be sufciently insensitive to deformation that a presumption of rigidity does not affect an analysis enough to warrant a nonrigid treatment. If the body deformation later proves to be not negligible, then declaring rigidity was a poor decision, not a poor assumption. A wire rope is exible, but in tension it can be robustly rigid and it distorts enormously under attempts at compressive loading. The same body can be both rigid and nonrigid. Deection analysis enters into design situations in many ways. A snap ring, or retaining ring, must be exible enough to be bent without permanent deformation and assembled with other parts, and then it must be rigid enough to hold the assembled parts together. In a transmission, the gears must be supported by a rigid shaft. If the shaft bends too much, that is, if it is too exible, the teeth will not mesh properly, and the result will be excessive impact, noise, wear, and early failure. In rolling sheet or strip steel to prescribed thicknesses, the rolls must be crowned, that is, curved, so that the nished product will be of uniform thickness. Thus, to design the rolls it is necessary to know exactly how much they will bend when a sheet of steel is rolled between them. Sometimes mechanical elements must be designed to have a particular force-deection characteristic. The suspension system of an automobile, for example, must be designed within a very narrow range to achieve an optimum vibration frequency for all conditions of vehicle loading, because the human body is comfortable only within a limited range of frequencies. The size of a load-bearing component is often determined on deections, rather than limits on stress. This chapter considers distortion of single bodies due to geometry (shape) and loading, then, briey, the behavior of groups of bodies. 41 Spring Rates Elasticity is that property of a material that enables it to regain its original conguration after having been deformed. A spring is a mechanical element that exerts a force when deformed. Figure 41a shows a straight beam of length l simply supported at the ends and loaded by the transverse force F. The deection y is linearly related to the force, as long as the elastic limit of the material is not exceeded, as indicated by the graph. This beam can be described as a linear spring. In Fig. 41b a straight beam is supported on two cylinders such that the length between supports decreases as the beam is deected by the force F. A larger force is required to deect a short beam than a long one, and hence the more this beam is deected, the stiffer it becomes. Also, the force is not linearly related to the deection, and hence this beam can be described as a nonlinear stiffening spring. Figure 41 l l d F (a) A linear spring; (b) a stiffening spring; (c) a softening spring. F F y y y F F F (a) y y y (b) (c) I. Basics © The McGrawHill Companies, 2008 4. Deflection and Stiffness 147 Deection and Stiffness BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 143 Figure 41c is an edge-view of a dish-shaped round disk. The force necessary to atten the disk increases at rst and then decreases as the disk approaches a at conguration, as shown by the graph. Any mechanical element having such a characteristic is called a nonlinear softening spring. If we designate the general relationship between force and deection by the equation F = F (y) (a) then spring rate is dened as F dF = y dy k ( y ) = lim y 0 (41) where y must be measured in the direction of F and at the point of application of F. Most of the force-deection problems encountered in this book are linear, as in Fig. 41a. For these, k is a constant, also called the spring constant; consequently Eq. (41) is written k= F y (42) We might note that Eqs. (41) and (42) are quite general and apply equally well for torques and moments, provided angular measurements are used for y. For linear displacements, the units of k are often pounds per inch or newtons per meter, and for angular displacements, pound-inches per radian or newton-meters per radian. 42 Tension, Compression, and Torsion The total extension or contraction of a uniform bar in pure tension or compression, respectively, is given by δ= Fl AE (43) This equation does not apply to a long bar loaded in compression if there is a possibility of buckling (see Secs. 411 to 415). Using Eqs. (42) and (43), we see that the spring constant of an axially loaded bar is k= AE l (44) The angular deection of a uniform round bar subjected to a twisting moment T was given in Eq. (335), and is θ= Tl GJ (45) where θ is in radians. If we multiply Eq. (45) by 180/π and substitute J = π d 4 /32 for a solid round bar, we obtain θ= 583.6T l Gd 4 (46) where θ is in degrees. Equation (45) can be rearranged to give the torsional spring rate as k= GJ T = θ l (47) 148 144 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition I. Basics © The McGrawHill Companies, 2008 4. Deflection and Stiffness Mechanical Engineering Design 43 Deection Due to Bending The problem of bending of beams probably occurs more often than any other loading problem in mechanical design. Shafts, axles, cranks, levers, springs, brackets, and wheels, as well as many other elements, must often be treated as beams in the design and analysis of mechanical structures and systems. The subject of bending, however, is one that you should have studied as preparation for reading this book. It is for this reason that we include here only a brief review to establish the nomenclature and conventions to be used throughout this book. The curvature of a beam subjected to a bending moment M is given by 1 M = ρ EI (48) where ρ is the radius of curvature. From studies in mathematics we also learn that the curvature of a plane curve is given by the equation d 2 y /dx 2 1 = ρ [1 + (dy /dx )2 ]3/2 (49) where the interpretation here is that y is the lateral deection of the beam at any point x along its length. The slope of the beam at any point x is θ= dy dx (a) For many problems in bending, the slope is very small, and for these the denominator of Eq. (49) can be taken as unity. Equation (48) can then be written d2 y M = EI dx 2 (b) Noting Eqs. (33) and (34) and successively differentiating Eq. (b) yields d3 y V = EI dx 3 (c) d4 y q = EI dx 4 (d ) It is convenient to display these relations in a group as follows: d4 y q = EI dx 4 (410) d3 y V = EI dx 3 (411) d2 y M = EI dx 2 (412) θ= dy dx y = f (x ) (413) (414) I. Basics © The McGrawHill Companies, 2008 4. Deflection and Stiffness Figure 42 149 Deection and Stiffness BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 145 y l = 20 in w x x x Moment, M M0 = Ml = 0 Slope, EI l/2 = 0 x R1 = wl 2 Shear, V V0 = + 800 lbf Vl = 800 lbf x (a) Loading, w w = 80 lbf/in Deflection, EIy y0 = yl = 0 R2 = wl 2 V V0 + Vl (b) M + (c) M0 Ml EI + (d) EI 0 EI l EIy (e) The nomenclature and conventions are illustrated by the beam of Fig. 42. Here, a beam of length l = 20 in is loaded by the uniform load w = 80 lbf per inch of beam length. The x axis is positive to the right, and the y axis positive upward. All quantities loading, shear, moment, slope, and deectionhave the same sense as y; they are positive if upward, negative if downward. The reactions R1 = R2 = +800 lbf and the shear forces V0 = +800 lbf and Vl = 800 lbf are easily computed by using the methods of Chap. 3. The bending moment is zero at each end because the beam is simply supported. For a simplysupported beam, the deections are also zero at each end. EXAMPLE 41 For the beam in Fig. 42, the bending moment equation, for 0 x l, is M= w wl x x2 2 2 Using Eq. (412), determine the equations for the slope and deection of the beam, the slopes at the ends, and the maximum deection. Solution Integrating Eq. (412) as an indenite integral we have EI dy = dx M dx = wl 2 w 3 x x + C1 4 6 (1) where C1 is a constant of integration that is evaluated from geometric boundary conditions. We could impose that the slope is zero at the midspan of the beam, since the beam and 150 146 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition I. Basics © The McGrawHill Companies, 2008 4. Deflection and Stiffness Mechanical Engineering Design loading are symmetric relative to the midspan. However, we will use the given boundary conditions of the problem and verify that the slope is zero at the midspan. Integrating Eq. (1) gives EIy = M dx = w wl 3 x x 4 + C1 x + C2 12 24 (2) The boundary conditions for the simply supported beam are y = 0 at x = 0 and l. Applying the rst condition, y = 0 at x = 0, to Eq. (2) results in C2 = 0. Applying the second condition to Eq. (2) with C2 = 0, E I y (l ) = wl 3 w l l 4 + C1l = 0 12 24 Solving for C1 yields C1 = wl 3 /24. Substituting the constants back into Eqs. (1) and (2) and solving for the deection and slope results in y= wx (2lx 2 x 3 l 3 ) 24 E I (3) θ= w dy = (6lx 2 4x 3 l 3 ) dx 24 E I (4) Comparing Eq. (3) with that given in Table A9, beam 7, we see complete agreement. For the slope at the left end, substituting x = 0 into Eq. (4) yields θ |x =0 = and at x = l, θ |x = l = wl 3 24 E I wl 3 24 E I At the midspan, substituting x = l/2 gives d y /dx = 0, as earlier suspected. The maximum deection occurs where d y /dx = 0. Substituting x = l/2 into Eq. (3) yields ymax = 5wl 4 384 E I which again agrees with Table A97. The approach used in the example is ne for simple beams with continuous loading. However, for beams with discontinuous loading and/or geometry such as a step shaft with multiple gears, ywheels, pulleys, etc., the approach becomes unwieldy. The following section discusses bending deections in general and the techniques that are provided in this chapter. 44 Beam Deection Methods Equations (410) through (414) are the basis for relating the intensity of loading q, vertical shear V, bending moment M, slope of the neutral surface θ, and the transverse deflection y. Beams have intensities of loading that range from q = constant I. Basics © The McGrawHill Companies, 2008 151 Deection and Stiffness BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 147 4. Deflection and Stiffness (uniform loading), variable intensity q (x ), to Dirac delta functions (concentrated loads). The intensity of loading usually consists of piecewise contiguous zones, the expressions for which are integrated through Eqs. (410) to (414) with varying degrees of difculty. Another approach is to represent the deection y (x ) as a Fourier series, which is capable of representing single-valued functions with a nite number of nite discontinuities, then differentiating through Eqs. (414) to (410), and stopping at some level where the Fourier coefcients can be evaluated. A complication is the piecewise continuous nature of some beams (shafts) that are stepped-diameter bodies. All of the above constitute, in one form or another, formal integration methods, which, with properly selected problems, result in solutions for q, V, M, θ, and y. These solutions may be 1 2 3 Closed-form, or Represented by innite series, which amount to closed form if the series are rapidly convergent, or Approximations obtained by evaluating the rst or the rst and second terms. The series solutions can be made equivalent to the closed-form solution by the use of a computer. Roarks1 formulas are committed to commercial software and can be used on a personal computer. There are many techniques employed to solve the integration problem for beam deection. Some of the popular methods include: Superposition (see Sec. 45) The moment-area method2 Singularity functions (see Sec. 46) Numerical integration3 The two methods described in this chapter are easy to implement and can handle a large array of problems. There are methods that do not deal with Eqs. (410) to (414) directly. An energy method, based on Castiglianos theorem, is quite powerful for problems not suitable for the methods mentioned earlier and is discussed in Secs. 47 to 410. Finite element programs are also quite useful for determining beam deections. 45 Beam Deections by Superposition The results of many simple load cases and boundary conditions have been solved and are available. Table A9 provides a limited number of cases. Roarks4 provides a much more comprehensive listing. Superposition resolves the effect of combined loading on a structure by determining the effects of each load separately and adding 1 Warren C. Young and Richard G. Budynas, Roarks Formulas for Stress and Strain, 7th ed., McGraw-Hill, New York, 2002. 2 See Chap. 9, F. P. Beer, E. R. Johnston Jr., and J. T. DeWolf, Mechanics of Materials, 4th ed., McGraw-Hill, New York, 2006. 3 See Sec. 44, J. E. Shigley and C. R. Mischke, Mechanical Engineering Design, 6th ed., McGraw-Hill, New York, 2001. 4 Warren C. Young and Richard G. Budynas, Roarks Formulas for Stress and Strain, 7th ed., McGraw-Hill, New York, 2002. 152 148 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition I. Basics © The McGrawHill Companies, 2008 4. Deflection and Stiffness Mechanical Engineering Design the results algebraically. Superposition may be applied provided: (1) each effect is linearly related to the load that produces it, (2) a load does not create a condition that affects the result of another load, and (3) the deformations resulting from any specific load are not large enough to appreciably alter the geometric relations of the parts of the structural system. The following examples are illustrations of the use of superposition. EXAMPLE 42 Consider the uniformly loaded beam with a concentrated force as shown in Fig. 43. Using superposition, determine the reactions and the deection as a function of x. Solution Considering each load state separately, we can superpose beams 6 and 7 of Table A9. For the reactions we nd Answer R1 = Fb wl + l 2 Answer R2 = wl Fa + l 2 The loading of beam 6 is discontinuous and separate deection equations are given for regions A B and BC. Beam 7 loading is not discontinuous so there is only one equation. Superposition yields Answer y AB = Fbx 2 wx ( x + b2 l 2 ) + (2lx 2 x 3 l 3 ) 6E I l 24 E I Answer y BC = Fa (l x ) 2 wx (x + a 2 2lx ) + (2lx 2 x 3 l 3 ) 6E I l 24 E I y Figure 43 l F b a w C A B R1 x R2 If we wanted to determine the maximum deection in the previous example, we would set dy /dx = 0 and solve for the value of x where the deection is a maximum. If a = l /2, the maximum deection would obviously occur at x = l /2 because of symmetry. However, if a < l /2, where would the maximum deection be? It can be shown that as the force F moves toward the left support, the maximum deection moves toward the left support also, but not as much as F (see Prob. 434). Thus, we would set dy BC /dx = 0 and solve for x. Sometimes it may not be obvious that we can use superposition with the tables at hand, as demonstrated in the next example. I. Basics EXAMPLE 43 Solution © The McGrawHill Companies, 2008 4. Deflection and Stiffness 153 Deection and Stiffness BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 149 Consider the beam in Fig. 44a and determine the deection equations using superposition. For region A B we can superpose beams 7 and 10 of Table A9 to obtain Answer y AB = wx Fax 2 (2lx 2 x 3 l 3 ) + (l x 2 ) 24 E I 6E I l For region B C , how do we represent the uniform load? Considering the uniform load only, the beam deflects as shown in Fig. 44b. Region B C is straight since there is no bending moment due to w . The slope of the beam at B is θB and is obtained by taking the derivative of y given in the table with respect to x and setting x = l . Thus, d w wx dy = (2lx 2 x 3 l 3 ) = (6lx 2 4x 3 l 3 ) dx dx 24 E I 24 E I Substituting x = l gives θB = w wl 3 (6ll 2 4l 3 l 3 ) = 24 E I 24 E I The deection in region B C due to w is θ B (x l ), and adding this to the deection due to F, in BC, yields Answer y BC = y Figure 44 (a) Beam with uniformly distributed load and overhang force; (b) deections due to uniform load only. y a l w F w B A R2 R1 (a) EXAMPLE 44 F (x l ) wl 3 (x l ) + [(x l )2 a (3x l )] 24 E I 6E I C B A x B yBC = B( x l) x C l x (b) Figure 45a shows a cantilever beam with an end load. Normally we model this problem by considering the left support as rigid. After testing the rigidity of the wall it was found that the translational stiffness of the wall was kt force per unit vertical deection, and the rotational stiffness was kr moment per unit angular (radian) deection (see Fig. 45b). Determine the deection equation for the beam under the load F. 154 150 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition I. Basics © The McGrawHill Companies, 2008 4. Deflection and Stiffness Mechanical Engineering Design Solution Here we will superpose the modes of deection. They are: (1) translation due to the compression of spring kt , (2) rotation of the spring kr , and (3) the elastic deformation of the beam given by Table A91. The force in spring kt is R1 = F , giving a deection from Eq. (42) of y1 = F kt (1) The moment in spring kr is M1 = Fl . This gives a clockwise rotation of θ = Fl / kr . Considering this mode of deection only, the beam rotates rigidly clockwise, leading to a deection equation of y2 = Fl x kr (2) Finally, the elastic deformation of the beam from Table A91 is y3 = F x2 (x 3l ) 6E I (3) Adding the deections from each mode yields Answer y= F F x2 Fl (x 3l ) x 6E I kt kr y Figure 45 l F x M1 R1 (a) kr F x kt R1 (b ) 46 Beam Deections by Singularity Functions Introduced in Sec. 33, singularity functions are excellent for managing discontinuities, and their application to beam deection is a simple extension of what was presented in the earlier section. They are easy to program, and as will be seen later, they can greatly simplify the solution of statically indeterminate problems. The following examples illustrate the use of singularity functions to evaluate deections of statically determinate beam problems. I. Basics EXAMPLE 45 Solution © The McGrawHill Companies, 2008 155 Deection and Stiffness BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 151 4. Deflection and Stiffness Consider the beam of Table A96, which is a simply supported beam having a concentrated load F not in the center. Develop the deection equations using singularity functions. First, write the load intensity equation from the free-body diagram, q = R1 x 1 F x a 1 + R2 x l 1 (1) Integrating Eq. (1) twice results in V = R1 x 0 F x a 0 + R2 x l 0 (2) M = R1 x 1 F x a 1 + R2 x l 1 (3) Recall that as long as the q equation is complete, integration constants are unnecessary for V and M; therefore, they are not included up to this point. From statics, setting V = M = 0 for x slightly greater than l yields R1 = Fb/ l and R2 = Fa / l . Thus Eq. (3) becomes M= Fb x l 1 F x a 1 + Fa x l l 1 Integrating Eqs. (412) and (413) as indenite integrals gives EI dy Fb x = dx 2l EIy = Fb x 6l 2 F x a 2 3 F x a 6 2 + Fa x l 2l 3 + Fa x l 6l 2 + C1 3 + C1 x + C2 Note that the rst singularity term in both equations always exists, so x 2 = x 2 and x 3 = x 3 . Also, the last singularity term in both equations does not exist until x = l, where it is zero, and since there is no beam for x > l we can drop the last term. Thus EI Fb 2 F dy x a = x dx 2l 2 EIy = Fb 3 F x a x 6l 6 2 + C1 (4) 3 + C1 x + C2 (5) The constants of integration C1 and C2 are evaluated by using the two boundary conditions y = 0 at x = 0 and y = 0 at x = l. The rst condition, substituted into Eq. (5), gives C2 = 0 (recall that 0 a 3 = 0). The second condition, substituted into Eq. (5), yields 0= Fb3 Fbl 2 Fb 3 F l (l a )3 + C1l = + C1l 6l 6 6 6 Solving for C1 , C1 = Fb 2 (l b2 ) 6l 156 152 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition I. Basics © The McGrawHill Companies, 2008 4. Deflection and Stiffness Mechanical Engineering Design Finally, substituting C1 and C2 in Eq. (5) and simplifying produces F [bx (x 2 + b2 l 2 ) l x a 3 ] 6E I l y= (6) Comparing Eq. (6) with the two deection equations in Table A96, we note that the use of singularity functions enables us to express the deection equation with a single equation. EXAMPLE 46 Determine the deection equation for the simply supported beam with the load distribution shown in Fig. 46. Solution This is a good beam to add to our table for later use with superposition. The load intensity equation for the beam is q = R1 x 1 w x 0 +w x a 0 + R2 x l 1 (1) where the w x a 0 is necessary to turn off the uniform load at x = a. From statics, the reactions are R1 = wa (2l a ) 2l R2 = wa 2 2l (2) For simplicity, we will retain the form of Eq. (1) for integration and substitute the values of the reactions in later. Two integrations of Eq. (1) reveal V = R1 x 0 w x M = R1 x 1 w x 2 1 +w x a 2 + w x a 2 1 + R2 x l 2 + R2 x l 0 (3) 1 (4) As in the previous example, singularity functions of order zero or greater starting at x = 0 can be replaced by normal polynomial functions. Also, once the reactions are determined, singularity functions starting at the extreme right end of the beam can be omitted. Thus, Eq. (4) can be rewritten as M = R1 x y Figure 46 l a w B A R1 C R2 x w2 w x a x+ 2 2 2 (5) I. Basics © The McGrawHill Companies, 2008 4. Deflection and Stiffness 157 Deection and Stiffness BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 153 Integrating two more times for slope and deection gives EI dy R1 2 w 3 w x a = x x+ dx 2 6 6 EIy = 3 R1 3 w w x a x x4 + 6 24 24 + C1 (6) 4 (7) + C1 x + C2 The boundary conditions are y = 0 at x = 0 and y = 0 at x = l. Substituting the rst condition in Eq. (7) shows C2 = 0. For the second condition 0= R1 3 w w l l 4 + (l a )4 + C1l 6 24 24 Solving for C1 and substituting into Eq. (7) yields EIy = R1 w w w x a x (x 2 l 2 ) x (x 3 l 3 ) x (l a )4 + 6 24 24l 24 4 Finally, substitution of R1 from Eq. (2) and simplifying results gives Answer y= w [2ax (2l a )(x 2 l 2 ) xl (x 3 l 3 ) x (l a )4 + l x a 4 ] 24 E I l As stated earlier, singularity functions are relatively simple to program, as they are omitted when their arguments are negative, and the brackets are replaced with ( ) parentheses when the arguments are positive. EXAMPLE 47 The steel step shaft shown in Fig. 47a is mounted in bearings at A and F. A pulley is centered at C where a total radial force of 600 lbf is applied. Using singularity functions evaluate the shaft displacements at 1 - in increments. Assume the shaft is 2 simply supported. Solution The reactions are found to be R1 = 360 lbf and R2 = 240 lbf. Ignoring R2 , using singularity functions, the moment equation is M = 360x 600 x 8 1 (1) This is plotted in Fig. 47b. For simplication, we will consider only the step at D. That is, we will assume section AB has the same diameter as BC and section EF has the same diameter as DE. Since these sections are short and at the supports, the size reduction will not add much to the deformation. We will examine this simplication later. The second area moments for BC and DE are I BC = π 1.54 = 0.2485 in4 64 IDE = π 1.754 = 0.4604 in4 64 158 154 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition I. Basics © The McGrawHill Companies, 2008 4. Deflection and Stiffness Mechanical Engineering Design Figure 47 y 600 lbf 1.000 AB Dimensions in inches. 1.750 1.500 C D 1.000 E F x 0.5 8 R1 8.5 19.5 (a) 2880 lbf-in M R2 20 2760 lbf-in (b ) M/I a b c d (c ) A plot of M / I is shown in Fig. 47c. The values at points b and c, and the step change are M I b M I = 2760 = 11 106.6 lbf/in3 0.2485 M I c = 2760 = 5 994.8 lbf/in3 0.4604 = 5 994.8 11 106.6 = 5 111.8 lbf/in3 The slopes for ab and cd, and the change are m ab = 360 600 = 965.8 lbf/in4 0.2485 5 994.8 = 521.3 lbf/in4 11.5 m cd = m = 521.3 ( 965.8) = 444.5 lbf/in4 Dividing Eq. (1) by I BC and, at x of slope 444.5 lbf/in4 , gives 8.5 in, adding a step of 5 111.8 lbf/in3 and a ramp M = 1 448.7x 2 414.5 x 8 1 5 111.8 x 8.5 0 + 444.5 x 8.5 I 1 (2) Integrating twice gives E dy = 724.35x 2 1207.3 x 8 2 5 111.8 x 8.5 dx +222.3 x 8.5 2 + C1 1 (3) and E y = 241.5x 3 402.4 x 8 3 2 555.9 x 8.5 2 + 74.08 x 8.5 3 + C1 x + C2 (4) At x = 0, y = 0. This gives C2 = 0 (remember, singularity functions do not exist until the argument is positive). At x = 20 in, y = 0, and 0 = 241.5(20) 3 402.4(20 8) 3 2 555.9(20 8.5) 2 + 74.08(20 8.5) 3 + C1 (20) I. Basics © The McGrawHill Companies, 2008 159 Deection and Stiffness BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 155 4. Deflection and Stiffness Solving, gives C1 = 50 565 lbf/in2 . Thus, Eq. (4) becomes, with E = 30(10) 6 psi, y= 1 (241.5x 3 402.4 x 8 3 2 555.9 x 8.5 30(106 ) 2 +74.08 x 8.5 3 50 565x ) (5) When using a spreadsheet, program the following equations: y= 1 (241.5x 3 50 565x ) 30(106 ) 0 x 8 in y= 1 [241.5x 3 402.4( x 8) 3 50 565x ] 30(106 ) 8 x 8.5 in y= 1 [241.5x 3 402.4 ( x 8) 3 2 555.9 ( x 8.5) 2 30(106 ) +74.08 ( x 8.5) 3 50 565x ] 8.5 x 20 in The following table results. x y x y x y x y x y 0 0.000000 4.5 0.006851 9 0.009335 13.5 0.007001 18 0.002377 0.5 0.000842 5 0.007421 9.5 0.009238 14 0.006571 18.5 0.001790 1 0.001677 5.5 0.007931 10 0.009096 14.5 0.006116 19 0.001197 1.5 0.002501 6 0.008374 10.5 0.008909 15 0.005636 19.5 0.000600 20 0.000000 2 0.003307 6.5 0.008745 11 0.008682 15.5 0.005134 2.5 0.004088 7 0.009037 11.5 0.008415 16 0.004613 3 0.004839 7.5 0.009245 12 0.008112 16.5 0.004075 3.5 0.005554 8 0.009362 12.5 0.007773 17 0.003521 4 0.006227 8.5 0.009385 13 0.007403 17.5 0.002954 where x and y are in inches. We see that the greatest deection is at x = 8.5 in, where y = 0.009385 in. Substituting C1 into Eq. (3) the slopes at the supports are found to be θ A = 1.686(103 ) rad = 0.09657 deg, and θ F = 1.198(103 ) rad = 0.06864 deg. You might think these to be insignicant deections, but as you will see in Chap. 7, on shafts, they are not. A nite-element analysis was performed for the same model and resulted in y |x = 8.5 in = 0.009380 in θ A = 0.09653 θ F = 0.06868 Virtually the same answer save some round-off error in the equations. If the steps of the bearings were incorporated into the model, more equations result, but the process is the same. The solution to this model is y |x = 8.5 in = 0.009387 in θ A = 0.09763 θ F = 0.06973 The largest difference between the models is of the order of 1.5 percent. Thus the simplication was justied. 160 156 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition I. Basics © The McGrawHill Companies, 2008 4. Deflection and Stiffness Mechanical Engineering Design In Sec. 49, we will demonstrate the usefulness of singularity functions in solving statically indeterminate problems. 47 Strain Energy The external work done on an elastic member in deforming it is transformed into strain, or potential, energy. If the member is deformed a distance y, and if the force-deection relationship is linear, this energy is equal to the product of the average force and the deection, or U= F2 F y= 2 2k (a) This equation is general in the sense that the force F can also mean torque, or moment, provided, of course, that consistent units are used for k. By substituting appropriate expressions for k, strain-energy formulas for various simple loadings may be obtained. For tension and compression and for torsion, for example, we employ Eqs. (44) and (47) and obtain U= F 2l 2 AE tension and compression (415) U= T 2l 2G J torsion (416) To obtain an expression for the strain energy due to direct shear, consider the element with one side xed in Fig. 48a. The force F places the element in pure shear, and the work done is U = F δ/2. Since the shear strain is γ = δ/ l = τ/ G = F / AG , we have U= F 2l 2 AG direct shear (417) The strain energy stored in a beam or lever by bending may be obtained by referring to Fig. 48b. Here AB is a section of the elastic curve of length d s having a radius of curvature ρ. The strain energy stored in this element of the beam is d U = ( M /2)d θ. Since ρ d θ = ds , we have dU = O Figure 48 F d A F l ds F B dx (a) Pure shear element (b) Beam bending element M ds 2ρ (b) I. Basics © The McGrawHill Companies, 2008 4. Deflection and Stiffness 161 Deection and Stiffness BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 157 We can eliminate ρ by using Eq. (48). Thus dU = M 2 ds 2E I (c) . For small deections, d s = dx . Then, for the entire beam M 2 dx 2E I U= (418) bending Equation (418) is exact only when a beam is subject to pure bending. Even when shear is present, Eq. (418) continues to give quite good results, except for very short beams. The strain energy due to shear loading of a beam is a complicated problem. An approximate solution can be obtained by using Eq. (417) with a correction factor whose value depends upon the shape of the cross section. If we use C for the correction factor and V for the shear force, then the strain energy due to shear in bending is the integral of Eq. (417), or U= C V 2 dx 2 AG bending shear (419) Values of the factor C are listed in Table 41. Table 41 Beam Cross-Sectional Shape Strain-Energy Correction Factors for Shear Source: Richard G. Budynas, Advanced Strength and Applied Stress Analysis, 2nd ed., McGraw-Hill, New York, 1999. Copyright © 1999 The McGraw-Hill Companies. EXAMPLE 48 Solution Factor C Rectangular 1.2 Circular 1.11 Thin-walled tubular, round 2.00 Box sections 1.00 Structural sections 1.00 Use area of web only. Find the strain energy due to shear in a rectangular cross-section beam, simply supported, and having a uniformly distributed load. Using Appendix Table A97, we nd the shear force to be V= wl wx 2 Substituting into Eq. (419), with C = 1.2, gives Answer U= 1.2 2 AG l 0 wl wx 2 2 dx = w2l 3 20 AG 162 158 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition I. Basics © The McGrawHill Companies, 2008 4. Deflection and Stiffness Mechanical Engineering Design EXAMPLE 49 A cantilever has a concentrated load F at the end, as shown in Fig. 49. Find the strain energy in the beam by neglecting shear. Figure 49 l ymax F x Solution At any point x along the beam, the moment is M = F x . Substituting this value of M into Eq. (418), we nd l Answer 48 U= 0 F 2l 3 F 2 x 2 dx = 2E I 6E I Castiglianos Theorem A most unusual, powerful, and often surprisingly simple approach to deection analysis is afforded by an energy method called Castiglianos theorem. It is a unique way of analyzing deections and is even useful for nding the reactions of indeterminate structures. Castiglianos theorem states that when forces act on elastic systems subject to small displacements, the displacement corresponding to any force, in the direction of the force, is equal to the partial derivative of the total strain energy with respect to that force. The terms force and displacement in this statement are broadly interpreted to apply equally to moments and angular displacements. Mathematically, the theorem of Castigliano is δi = U Fi (420) where δi is the displacement of the point of application of the force Fi in the direction of Fi . For rotational displacement Eq. (420) can be written as θi = U Mi (421) where θi is the rotational displacement, in radians, of the beam where the moment Mi exists and in the direction of Mi . As an example, apply Castiglianos theorem using Eqs. (415) and (416) to get the axial and torsional deections. The results are δ= F F 2l 2 AE = Fl AE (a) θ= T T 2l 2G J = Tl GJ (b) I. Basics © The McGrawHill Companies, 2008 163 Deection and Stiffness BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 159 4. Deflection and Stiffness Compare Eqs. (a) and (b) with Eqs. (43) and (45). In Example 48, the bending strain energy for a cantilever having a concentrated end load was found. According to Castiglianos theorem, the deection at the end of the beam due to bending is y= U = F F F 2l 3 6E I = Fl 3 3E I (c) which checks with Table A91. Castiglianos theorem can be used to nd the deection at a point even though no force or moment acts there. The procedure is: 1 2 3 Set up the equation for the total strain energy U by including the energy due to a fictitious force or moment Q i acting at the point whose deflection is to be found. Find an expression for the desired deection δi , in the direction of Q i , by taking the derivative of the total strain energy with respect to Q i . Since Q i is a ctitious force, solve the expression obtained in step 2 by setting Q i equal to zero. Thus, δi = EXAMPLE 410 Solution U Qi (422) Q i =0 The cantilever of Ex. 49 is a carbon steel bar 10 in long with a 1-in diameter and is loaded by a force F = 100 lbf. (a) Find the maximum deection using Castiglianos theorem, including that due to shear. (b) What error is introduced if shear is neglected? (a) From Eq. (419) and Example 49 data, the total strain energy is U= F 2l 3 + 6E I l 0 C V 2 dx 2 AG (1) For the cantilever, the shear force is constant with repect to x, V = F . Also, C = 1.11, from Table 41. Performing the integration and substituting these values in Eq. (1) gives, for the total strain energy, U= 1.11 F 2l F 2l 3 + 6E I 2 AG (2) Then, according to Castiglianos theorem, the deection of the end is y= U Fl 3 1.11 Fl = + F 3E I AG We also nd that I= π d4 π(1)4 = = 0.0491 in4 64 64 A= π d2 π(1)2 = = 0.7854 in2 4 4 (3) 164 160 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition I. Basics © The McGrawHill Companies, 2008 4. Deflection and Stiffness Mechanical Engineering Design Substituting these values, together with F = 100 lbf, l = 10 in, E = 30 Mpsi, and G = 11.5 Mpsi, in Eq. (3) gives Answer Answer y = 0.022 63 + 0.000 12 = 0.022 75 in Note that the result is positive because it is in the same direction as the force F. (b) The error in neglecting shear for this problem is found to be about 0.53 percent. In performing any integrations, it is generally better to take the partial derivative with respect to the load Fi rst. This is true especially if the force is a ctitious force Q i , since it can be set to zero as soon as the derivative is taken. This is demonstrated in the next example. The forms for deection can then be rewritten. Here we will assume, for axial and torsional loading, that material and cross section properties and loading can vary along the length of the members. From Eqs. (415), (416), and (418), δi = U = Fi 1 AE F F Fi θi = U = Mi 1 GJ T T Mi δi = U = Fi 1 EI M M Fi dx dx dx tension and compression (423) torsion (424) bending (425) EXAMPLE 411 Using Castiglianos method, determine the deections of points A and B due to the force F applied at the end of the step shaft shown in Fig. 410. The second area moments for sections AB and BC are I1 and 2 I1 , respectively. Solution With cantilever beams we normally set up the coordinate system such that x starts at the wall and is directed towards the free end. Here, for simplicity, we have reversed that. With the coordinate system of Fig. 410 the bending moment expression is simpler than with the usual coordinate system, and does not require the support reactions. For 0 x l , the bending moment is M = F x (1) Since F is at A and in the direction of the desired deection, the deection at A from Eq. (425) is Figure 410 y l /2 x A F l /2 I1 B Qi 2 I1 C I. Basics © The McGrawHill Companies, 2008 4. Deflection and Stiffness U = F δA = l 0 1 EI M F M 165 Deection and Stiffness BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 161 (2) dx Substituting Eq. (1) into Eq. (2), noting that I = I1 for 0 x l /2, and I = 2 I1 for l /2 x l , we get δA = = Answer 1 E l /2 1 E 0 l 1 ( F x ) (x ) dx + I1 l /2 1 ( F x ) (x ) dx 2 I1 3 Fl 3 Fl3 7 Fl 3 = + 24 I1 48 I1 16 E I1 which is positive, as it is in the direction of F. For B, a ctitious force Q i is necessary at the point. Assuming Q i acts down at B, and x is as before, the moment equation is M = F x M = F x Qi 0 x l /2 l x 2 (3) l /2 x l For Eq. (425), we need M / Q i . From Eq. (3), M =0 Qi 0 x l /2 l M = x Qi 2 l /2 x l (4) Once the derivative is taken, Q i can be set to zero, so from Eq. (3), M = F x for 0 x l , and Eq. (425) becomes l δB = 0 1 EI M M Qi = dx Q i =0 1 E I1 l /2 0 ( F x )(0) dx + 1 E (2 I1 ) l l /2 ( F x ) x l 2 dx Evaluating the last integral gives Answer F δB = 2 E I1 x 3 lx 2 3 4 l l /2 = 5 Fl 3 96 E I1 which again is positive, in the direction of Q i . EXAMPLE 412 For the wire form of diameter d shown in Fig. 411a, determine the deection of point B in the direction of the applied force F (neglect the effect of bending shear). Solution It is very important to include the loading effects on all parts of the structure. Coordinate systems are not important, but loads must be consistent with the problem. Thus 166 162 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition I. Basics © The McGrawHill Companies, 2008 4. Deflection and Stiffness Mechanical Engineering Design Figure 411 G c z F x B D a b C y (a ) MG 2 = MD 2 = Fb G F MD 2 = MD = Fb D MG1 = MD1 = Fa F F B MD1 = TD = Fa MD = Fb C MC = Fa TD = TC = Fa F D F F C TC = MC = Fa (b ) appropriate use of free-body diagrams is essential here. The reader should verify that the reactions as functions of F in elements B C , C D , and G D are as shown in Fig. 411b. The deection of B in the direction of F is given by δB = U F so the partial derivatives in Eqs. (423) to (425) will all be taken with respect to F. Element BC is in bending only so from Eq. (425),5 1 U BC = F EI a 0 ( F y )( y ) dy = Fa 3 3E I (1) Element C D is in bending and in torsion. The torsion is constant so Eq. (424) can be written as U T =T Fi Fi 5 l GJ It is very tempting to mix techniques and try to use superposition also, for example. However, some subtle things can occur that you may visually miss. It is highly recommended that if you are using Castiglianos theorem on a problem, you use it for all parts of the problem. I. Basics © The McGrawHill Companies, 2008 4. Deflection and Stiffness 167 Deection and Stiffness BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 163 where l is the length of the member. So for the torsion in member CD, Fi = F , T = Fa , and l = b. Thus, UCD F torsion = ( Fa )(a ) b Fa 2 b = GJ GJ (2) For the bending in CD, UCD F bending = b 1 EI ( F x )(x ) dx = 0 Fb3 3E I (3) Member DG is axially loaded and is bending in two planes. The axial loading is constant, so Eq. (423) can be written as U = Fi F F Fi l AE where l is the length of the member. Thus, for the axial loading of D G , F = Fi , l = c, and U DG F axial = Fc AE (4) The bending moments in each plane of D G are constant along the length of M y = Fb and Mx = Fa . Considering each one separately in the form of Eq. (425) gives U DG F bending = 1 EI c 0 ( Fb)(b) dz + 1 EI c ( Fa )(a ) dz 0 Fc(a 2 + b2 ) = EI (5) Adding Eqs. (1) to (5), noting that I = π d 4 /64, J = 2 I , A = π d 2 /4, and G = E /[2(1 + ν)], we nd that the deection of B in the direction of F is Answer (δ B ) F = 4F [16(a 3 + b3 ) + 48c(a 2 + b2 ) + 48(1 + ν)a 2 b + 3cd 2 ] 3π Ed 4 Now that we have completed the solution, see if you can physically account for each term in the result. 49 Deection of Curved Members Machine frames, springs, clips, fasteners, and the like frequently occur as curved shapes. The determination of stresses in curved members has already been described in Sec. 318. Castiglianos theorem is particularly useful for the analysis of deections in curved parts too. Consider, for example, the curved frame of Fig. 412a. We are interested in nding the deection of the frame due to F and in the direction of F . The total strain energy consists of four terms, and we shall consider each separately. The rst is due to the bending moment and is6 6 See Richard G. Budynas, Advanced Strength and Applied Stress Analysis, 2nd ed., Sec. 6.7, McGraw-Hill, New York, 1999. 168 164 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition I. Basics © The McGrawHill Companies, 2008 4. Deflection and Stiffness Mechanical Engineering Design Fr d M F h R F F (b) (a ) Figure 412 (a) Curved bar loaded by force F. R = radius to centroidal axis of section; h = section thickness. (b) Diagram showing forces acting on section taken at angle θ. F r = V = shear component of F; F θ is component of F normal to section; M is moment caused by force F. U1 = M2 dθ 2 AeE (426) In this equation, the eccentricity e is (427) e = R rn where rn is the radius of the neutral axis as dened in Sec. 318 and shown in Fig. 334. An approximate result can be obtained by using the equation . U1 = M2 R dθ 2E I R > 10 h (428) which is obtained directly from Eq. (418). Note the limitation on the use of Eq. (428). The strain energy component due to the normal force Fθ consists of two parts, one of which is axial and analogous to Eq. (415). This part is U2 = Fθ2 R d θ 2 AE (429) The force Fθ also produces a moment, which opposes the moment M in Fig. 412b. The resulting strain energy will be subtractive and is U3 = M Fθ d θ AE (430) The negative sign of Eq. (430) can be appreciated by referring to both parts of Fig. 412. Note that the moment M tends to decrease the angle d θ . On the other hand, the moment due to Fθ tends to increase d θ . Thus U3 is negative. If Fθ had been acting in the opposite direction, then both M and Fθ would tend to decrease the angle d θ . The fourth and last term is the shear energy due to Fr . Adapting Eq. (419) gives U4 = C Fr2 R d θ 2 AG (431) where C is the correction factor of Table 41. Combining the four terms gives the total strain energy U= M2 dθ + 2 AeE Fθ2 R d θ 2 AE M Fθ d θ + AE C Fr2 R d θ 2 AG (432) I. Basics © The McGrawHill Companies, 2008 4. Deflection and Stiffness 169 Deection and Stiffness BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 165 The deection produced by the force F can now be found. It is δ= U = F π 0 π M F M AeE 0 π dθ + 0 1 ( M Fθ ) dθ + AE F Fθ F Fθ R AE π 0 dθ Fr F C Fr R AG dθ (433) Using Fig. 412b, we nd M = F R sin θ M = R sin θ F Fθ = F sin θ Fθ = sin θ F MFθ = F 2 R sin2 θ M Fθ = 2 F R sin2 θ F Fr = cos θ F Fr = F cos θ Substituting these into Eq. (433) and factoring yields F R2 AeE π π 0 sin2 θ d θ + FR AE 0 sin2 θ d θ 2F R AE + δ= CFR AG y A R x C + F O z B T axis = Figure 413 Ring ABC in the xy plane subject to force F parallel to the z axis. Corresponding to a ring segment CB at angle θ from the point of application of F, the moment axis is a line BO and the torque axis is a line in the xy plane tangent to the ring at B. Note the positive directions of the T and M axes. sin2 θ d θ 0 π cos2 θ d θ 0 π F R2 πFR πFR πC F R π F R2 πFR πC F R + + = + 2 AeE 2 AE AE 2 AG 2 AeE 2 AE 2 AG (434) Because the rst term contains the square of the radius, the second two terms will be small if the frame has a large radius. Also, if R / h > 10, Eq. (428) can be used. An approximate result then turns out to be . π F R3 δ= 2E I M axis + π (435) The determination of the deection of a curved member loaded by forces at right angles to the plane of the member is more difcult, but the method is the same.7 We shall include here only one of the more useful solutions to such a problem, though the methods for all are similar. Figure 413 shows a cantilevered ring segment having a span angle φ . Assuming R / h > 10, the strain energy neglecting direct shear, is obtained from the equation φ U= 7 0 M2 R dθ + 2E I φ 0 T 2 R dθ 2G J (436) For more solutions than are included here, see Joseph E. Shigley, Curved Beams and Rings, Chap. 38 in Joseph E. Shigley, Charles R. Mischke, and Thomas H. Brown, Jr. (eds.), Standard Handbook of Machine Design, 3rd ed., McGraw-Hill, New York, 2004. 170 166 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition I. Basics © The McGrawHill Companies, 2008 4. Deflection and Stiffness Mechanical Engineering Design The moments and torques acting on a section at B, due to the force F, are M = F R sin θ T = F R (1 cos θ) The deection δ of the ring segment at C and in the direction of F is then found to be δ= F R3 U = F 2 β α + EI GJ (437) where the coefcients α and β are dependent on the span angle φ and are dened as follows: α = φ sin φ cos φ (438) β = 3φ 4 sin φ + sin φ cos φ (438) where φ is in radians. EXAMPLE 413 Deection in a Variable-Cross-Section Punch-Press Frame The general result expressed in Eq. (434), δ= πFR πC F R π F R2 + 2 AeE 2 AE 2 AG is useful in sections that are uniform and in which the centroidal locus is circular. The bending moment is largest where the material is farthest from the load axis. Strengthening requires a larger second area moment I. A variable-depth cross section is attractive, but it makes the integration to a closed form very difcult. However, if you are seeking results, numerical integration with computer assistance is helpful. Consider the steel C frame depicted in Fig. 414a in which the centroidal radius is 32 in, the cross section at the ends is 2 in × 2 in, and the depth varies sinusoidally with an amplitude of 2 in. The load is 1000 lbf. It follows that C = 1.2, G = 11.5(106 ) psi, E = 30(106 ) psi. The outer and inner radii are Rout = 33 + 2sin θ Rin = 31 2sin θ The remaining geometrical terms are h = Rout Rin = 2(1 + 2 sin θ) A = bh = 4(1 + 2 sin θ rn = 2(1 + 2 sin θ) h = ln[( R + h /2)/( R h /2)] ln[(33 + 2 sin θ)/(31 2 sin θ)] e = R rn = 32 rn Note that M = F R sin θ M / F = R sin θ Fθ = F sin θ Fθ / F = sin θ M Fθ = F 2 R sin2 θ Fr = F cos θ M Fθ / F = 2 F R sin2 θ Fr / F = cos θ 1 I. Basics © The McGrawHill Companies, 2008 171 Deection and Stiffness BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 167 4. Deflection and Stiffness Figure 414 (a) A steel punch press has a C frame with a varying-depth rectangular cross section depicted. The cross section varies sinusoidally from 2 in × 2 in at θ = 0 to 2 in × 6 in at θ = 90 , and back to 2 in × 2 in at θ = 180 . Of immediate interest to the designer is the deection in the load axis direction under the load. (b) Finite-element model. 31- in R 1000 lbf 1000 lbf 1000 lbf (a ) (b) Substitution of the terms into Eq. (433) yields three inteqrals (1) δ = I1 + I2 + I3 where the integrals are sin2 θ d θ π I1 = 8.5333(103 ) 0 (1 + 2 sin θ) 32 π I2 = 2.6667(104 ) 0 π I3 = 8.3478(104 ) 0 sin2 θ d θ 1 + 2 sin θ 2(1 + 2 sin θ) 33 + 2 sin θ ln 31 2 sin θ (2) cos2 θ d θ 1 + 2 sin θ (3) (4) The integrals may be evaluated in a number of ways: by a program using Simpsons rule integration,8 by a program using a spreadsheet, or by mathematics software. Using MathCad and checking the results with Excel gives the integrals as I1 = 0.076 615, I2 = 0.000 159, and I3 = 0.000 773. Substituting these into Eq. (1) gives Answer δ = 0.077 23 in Finite-element (FE) programs are also very accessible. Figure 414b shows a simple half-model, using symmetry, of the press consisting of 216 plane-stress (2-D) elements. Creating the model and analyzing it to obtain a solution took minutes. Doubling the results from the FE analysis yielded δ = 0.07790 in, a less than 1 percent variation from the results of the numerical integration. 8 See Case Study 4, p. 203, J. E. Shigley and C. R. Mischke, Mechanical Engineering Design, 6th ed., McGraw-Hill, New York, 2001. 172 168 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition I. Basics © The McGrawHill Companies, 2008 4. Deflection and Stiffness Mechanical Engineering Design 410 Statically Indeterminate Problems A system in which the laws of statics are not sufcient to determine all the unknown forces or moments is said to be statically indeterminate. Problems of which this is true are solved by writing the appropriate equations of static equilibrium and additional equations pertaining to the deformation of the part. In all, the number of equations must equal the number of unknowns. A simple example of a statically indeterminate problem is furnished by the nested helical springs in Fig. 415a. When this assembly is loaded by the compressive force F, it deforms through the distance δ . What is the compressive force in each spring? Only one equation of static equilibrium can be written. It is F = F F1 F2 = 0 (a) which simply says that the total force F is resisted by a force F1 in spring 1 plus the force F2 in spring 2. Since there are two unknowns and only one equation, the system is statically indeterminate. To write another equation, note the deformation relation in Fig. 415b. The two springs have the same deformation. Thus, we obtain the second equation as δ1 = δ2 = δ (b) If we now substitute Eq. (42) in Eq. (b), we have F2 F1 = (c) k1 k2 Now we solve Eq. (c) for F1 and substitute the result in Eq. (a). This gives k2 F k1 F F2 F2 = 0 or F2 = (d ) k2 k1 + k2 This completes the solution, because with F2 known, F1 can be found from Eq. (c). F Figure 415 k1 k2 (a) F1 F2 k1 k2 (b) I. Basics © The McGrawHill Companies, 2008 4. Deflection and Stiffness 173 Deection and Stiffness BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 169 In the spring example, obtaining the necessary deformation equation was very straightforward. However, for other situations, the deformation relations may not be as easy. A more structured approach may be necessary. Here we will show two basic procedures for general statically indeterminate problems. Procedure 1 1 Choose the redundant reaction(s). There may be alternative choices (See Example 414). 2 Write the equations of static equilibrium for the remaining reactions in terms of the applied loads and the redundant reaction(s) of step 1. 3 Write the deection equation(s) for the point(s) at the locations of the redundant reaction(s) of step 1 in terms of the applied loads and the redundant reaction(s) of step 1. Normally the deection(s) is (are) zero. If a redundant reaction is a moment, the corresponding deection equation is a rotational deection equation. 4 The equations from steps 2 and 3 can now be solved to determine the reactions. In step 3 the deection equations can be solved in any of the standard ways. Here we will demonstrate the use of superposition and Castiglianos theorem on a beam problem. EXAMPLE 414 The indeterminate beam of Appendix Table A911 is reproduced in Fig. 416. Determine the reactions using procedure 1. Solution The reactions are shown in Fig. 416b. Without R2 the beam is a statically determinate cantilever beam. Without M1 the beam is a statically determinate simply supported beam. In either case, the beam has only one redundant support. We will rst solve this problem using superposition, choosing R2 as the redundant reaction. For the second solution, we will use Castiglianos theorem with M1 as the redundant reaction. Solution 1 1 2 Choose R2 at B to be the redundant reaction. Using static equilibrium equations solve for R1 and M1 in terms of F and R2 . This results in R1 = F R2 3 Fl R2 l 2 (1) Write the deection equation for point B in terms of F and R2 . Using superposition of Table A91 with F = R2 , and Table A92 with a = l /2, the deection of B, at x = l , is δB = Figure 416 M1 = F (l /2)2 R2 l 2 (l 3l ) + 6E I 6E I y l 3l 2 = R2 l 3 5 Fl 3 =0 3E I 48 E I (2) y F l l 2 F A O (a) B A B x x O M1 R1 x ˆ (b) R2 174 170 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition I. Basics © The McGrawHill Companies, 2008 4. Deflection and Stiffness Mechanical Engineering Design 4 Equation (2) can be solved for R2 directly. This yields Answer 5F 16 R2 = (3) Next, substituting R2 into Eqs. (1) completes the solution, giving Answer R1 = 11 F 16 M1 = 3 Fl 16 (4) Note that the solution agrees with what is given in Table A911. Solution 2 1 2 Choose M1 at O to be the redundant reaction. Using static equilibrium equations solve for R1 and R2 in terms of F and M1 . This results in R1 = 3 M1 F + 2 l R2 = M1 F 2 l (5) Since M1 is the redundant reaction at O, write the equation for the angular deection at point O. From Castiglianos theorem this is θO = U M1 (6) We can apply Eq. (425), using the variable x as shown in Fig. 416b. However, simˆ pler terms can be found by using a variable x that starts at B and is positive to the left. With this and the expression for R2 from Eq. (5) the moment equations are M= F M1 2 l x ˆ M= M1 F 2 l x F x ˆ ˆ l 2 (7) l x l ˆ 2 (8) 0x ˆ l 2 For both equations M x ˆ = M1 l (9) Substituting Eqs. (7) to (9) in Eq. (6), using the form of Eq. (425) where Fi = M1 , gives θO = U 1 = M1 EI l /2 0 F M1 2 l F x ˆ l 2 x ˆ x ˆ l x ˆ l l dx + ˆ F M1 2 l l /2 x ˆ dx = 0 ˆ Canceling 1/ E I l , and combining the rst two integrals, simplies this quite readily to M1 F 2 l l 0 l x2 dx F ˆˆ l /2 x ˆ l x dx = 0 ˆˆ 2 Integrating gives F M1 2 l l3 F3 l 3 3 l 2 3 + Fl 2 l 4 l 2 2 =0 I. Basics © The McGrawHill Companies, 2008 4. Deflection and Stiffness 175 Deection and Stiffness BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 171 which reduces to M1 = 4 3 Fl 16 (10) Substituting Eq. (10) into (5) results in R1 = 11 F 16 R2 = 5F 16 (11) which again agrees with Table A911. For some problems even procedure 1 can be a task. Procedure 2 eliminates some tricky geometric problems that would complicate procedure 1. We will describe the procedure for a beam problem. Procedure 2 1 Write the equations of static equilibrium for the beam in terms of the applied loads and unknown restraint reactions. 2 Write the deection equation for the beam in terms of the applied loads and unknown restraint reactions. 3 Apply boundary conditions consistent with the restraints. 4 Solve the equations from steps 1 and 3. EXAMPLE 415 The rods A D and C E shown in Fig. 417a each have a diameter of 10 mm. The secondarea moment of beam A BC is I = 62.5(103 ) mm4 . The modulus of elasticity of the material used for the rods and beam is E = 200 GPa. The threads at the ends of the rods are single-threaded with a pitch of 1.5 mm. The nuts are rst snugly t with bar A BC horizontal. Next the nut at A is tightened one full turn. Determine the resulting tension in each rod and the deections of points A and C. Solution There is a lot going on in this problem; a rod shortens, the rods stretch in tension, and the beam bends. Lets try the procedure! 1 The free-body diagram of the beam is shown in Fig. 417b. Summing forces, and moments about B, gives FB FA FC = 0 4 FA 3 FC = 0 Figure 417 200 A Dimensions in mm. (1) (2) 150 FA B C 200 A 150 B C x FB 600 800 D E (a) FC (b) Free-body diagram of beam ABC 176 172 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition I. Basics © The McGrawHill Companies, 2008 4. Deflection and Stiffness Mechanical Engineering Design 2 Using singularity functions, we nd the moment equation for the beam is M = FA x + FB x 0.2 1 where x is in meters. Integration yields dy FA = x2 + dx 2 FA E I y = x3 + 6 EI 3 FB x 0.2 2 + C1 2 FB x 0.2 3 + C1 x + C2 6 (3) The term E I = 200(109 ) 62.5(109 ) = 1.25(104 ) N · m2 . The upward deection of point A is ( Fl / AE ) AD N p, where the rst term is the elastic stretch of A D , N is the number of turns of the nut, and p is the pitch of the thread. Thus, the deection of A is FA (0.6) (1)(0.0015) yA = π (0.010)2 (200)(109 ) 4 (4) = 3.8197(108 ) FA 1.5(103 ) The upward deection of point C is ( Fl / AE )C E , or FC (0.8) yC = π = 5.093(108 ) FC (0.010)2 (200)(109 ) 4 (5) Equations (4) and (5) will now serve as the boundary conditions for Eq. (3). At x = 0, y = y A . Substituting Eq. (4) into (3) with x = 0 and E I = 1.25(104 ), noting that the singularity function is zero for x = 0, gives 4.7746(104 ) FA + C2 = 18.75 (6) At x = 0.2 m, y = 0, and Eq. (3) yields 1.3333(103 ) FA + 0.2C1 + C2 = 0 (7) At x = 0.35 m, y = yC . Substituting Eq. (5) into (3) with x = 0.35 m and E I = 1.25(104 ) gives 7.1458(103 ) FA + 5.625(104 ) FB 6.3662(104 ) FC + 0.35C1 + C2 = 0 Equations (1), (2), (6), (7), and (8) are ve equations in Written in matrix form, they are 1 1 1 0 4 0 3 0 0 0 0 4.7746(104 ) 1.3333(103 ) 0 0 0.2 7.1458(103 ) 5.625(104 ) 6.3662(104 ) 0.35 Solving these equations yields Answer FA = 2988 N 2 C1 = 106.54 N · m FB = 6971 N 3 C2 = 17.324 N · m (8) FA , FB , FC , C1 , and C2 . 0 FA 0 0 FB 0 1 FC = 18.75 1 C1 0 1 C2 0 FC = 3983 N I. Basics © The McGrawHill Companies, 2008 4. Deflection and Stiffness 177 Deection and Stiffness BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 173 Equation (3) can be reduced to y = (39.84x 3 92.95 x 0.2 3 8.523x + 1.386)(103 ) At x = 0, y = y A = 1.386(103 ) m = 1.386 mm. Answer Answer At x = 0.35 m, y = yC = [39.84(0.35)3 92.95(0.35 0.2)3 8.523(0.35) + 1.386](103 ) = 0.203(103 ) m = 0.203 mm Note that we could have easily incorporated the stiffness of the support at B if we were given a spring constant. 411 Compression MembersGeneral The analysis and design of compression members can differ signicantly from that of members loaded in tension or in torsion. If you were to take a long rod or pole, such as a meterstick, and apply gradually increasing compressive forces at each end, nothing would happen at rst, but then the stick would bend (buckle), and nally bend so much as to fracture. Try it. The other extreme would occur if you were to saw off, say, a 5-mm length of the meterstick and perform the same experiment on the short piece. You would then observe that the failure exhibits itself as a mashing of the specimen, that is, a simple compressive failure. For these reasons it is convenient to classify compression members according to their length and according to whether the loading is central or eccentric. The term column is applied to all such members except those in which failure would be by simple or pure compression. Columns can be categorized then as: 1 2 3 4 Long columns with central loading Intermediate-length columns with central loading Columns with eccentric loading Struts or short columns with eccentric loading Classifying columns as above makes it possible to develop methods of analysis and design specic to each category. Furthermore, these methods will also reveal whether or not you have selected the category appropriate to your particular problem. The four sections that follow correspond, respectively, to the four categories of columns listed above. 412 Long Columns with Central Loading Figure 418 shows long columns with differing end (boundary) conditions. If the axial force P shown acts along the centroidal axis of the column, simple compression of the member occurs for low values of the force. However, under certain conditions, when P reaches a specic value, the column becomes unstable and bending as shown in Fig. 418 develops rapidly. This force is determined by writing the bending deection equation for the column, resulting in a differential equation where when the boundary conditions are applied, results in the critical load for unstable bending.9 The critical force for the pin-ended column of Fig. 418a is given by Pcr = π2E I l2 9 See F. P. Beer, E. R. Johnston, Jr., and J. T. DeWolf, Mechanics of Materials, 4th ed., McGraw-Hill, New York, 2006, pp. 610613. (439) 174 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition I. Basics © The McGrawHill Companies, 2008 4. Deflection and Stiffness Mechanical Engineering Design (a) Both ends rounded or pivoted; (b) both ends xed; (c) one end free and one end xed; (d ) one end rounded and pivoted, and one end xed. P P Figure 418 P P y l 4 A l 2 l 0.707l 178 l l 4 l A B x (a) C 1 (b) C 4 (c) C 1 4 (d ) C 2 which is called the Euler column formula. Equation (439) can be extended to apply to other end-conditions by writing Pcr = Cπ2 E I l2 (440) where the constant C depends on the end conditions as shown in Fig. 418. Using the relation I = Ak 2 , where A is the area and k the radius of gyration, enables us to rearrange Eq. (440) into the more convenient form Cπ2 E Pcr = A (l / k )2 (441) where l / k is called the slenderness ratio. This ratio, rather than the actual column length, will be used in classifying columns according to length categories. The quantity Pcr / A in Eq. (441) is the critical unit load. It is the load per unit area necessary to place the column in a condition of unstable equilibrium. In this state any small crookedness of the member, or slight movement of the support or load, will cause the column to begin to collapse. The unit load has the same units as strength, but this is the strength of a specic column, not of the column material. Doubling the length of a member, for example, will have a drastic effect on the value of Pcr / A but no effect at all on, say, the yield strength Sy of the column material itself. Equation (441) shows that the critical unit load depends only upon the modulus of elasticity and the slenderness ratio. Thus a column obeying the Euler formula made of high-strength alloy steel is no stronger than one made of low-carbon steel, since E is the same for both. The factor C is called the end-condition constant, and it may have any one of the theoretical values 1 , 1, 2, and 4, depending upon the manner in which the load is 4 applied. In practice it is difcult, if not impossible, to x the column ends so that the factor C = 2 or C = 4 would apply. Even if the ends are welded, some deection will occur. Because of this, some designers never use a value of C greater than unity. However, if liberal factors of safety are employed, and if the column load is accurately known, then a value of C not exceeding 1.2 for both ends xed, or for one end rounded and one end xed, is not unreasonable, since it supposes only partial xation. Of course, the value C = 1 must always be used for a column having one end xed and one end 4 free. These recommendations are summarized in Table 42. I. Basics © The McGrawHill Companies, 2008 4. Deflection and Stiffness Table 42 179 Deection and Stiffness BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 175 End-Condition Constant C Theoretical Value Conservative Value Recommended Value* Fixed-free 1 4 1 4 1 4 Rounded-rounded 1 1 1 Fixed-rounded 2 1 1.2 Fixed-xed End-Condition Constants for Euler Columns [to Be Used with Eq. (440)] Column End Conditions 4 1 1.2 *To be used only with liberal factors of safety when the column load is accurately known. Figure 419 P Euler curve plotted using Eq. (440) with C = 1. Q Unit load Pcr A Sy Parabolic curve T Euler curve R l kQ l k 1 l Slenderness ratio k When Eq. (441) is solved for various values of the unit load Pcr / A in terms of the slenderness ratio l / k , we obtain the curve PQR shown in Fig. 419. Since the yield strength of the material has the same units as the unit load, the horizontal line through Sy and Q has been added to the gure. This would appear to make the gure cover the entire range of compression problems from the shortest to the longest compression member. Thus it would appear that any compression member having an l / k value less than (l / k ) Q should be treated as a pure compression member while all others are to be treated as Euler columns. Unfortunately, this is not true. In the actual design of a member that functions as a column, the designer will be aware of the end conditions shown in Fig. 418, and will endeavor to congure the ends, using bolts, welds, or pins, for example, so as to achieve the required ideal end conditions. In spite of these precautions, the result, following manufacture, is likely to contain defects such as initial crookedness or load eccentricities. The existence of such defects and the methods of accounting for them will usually involve a factor-of-safety approach or a stochastic analysis. These methods work well for long columns and for simple compression members. However, tests show numerous failures for columns with slenderness ratios below and in the vicinity of point Q, as shown in the shaded area in Fig. 419. These have been reported as occurring even when near-perfect geometric specimens were used in the testing procedure. A column failure is always sudden, total, unexpected, and hence dangerous. There is no advance warning. A beam will bend and give visual warning that it is overloaded, but not so for a column. For this reason neither simple compression methods nor the 180 176 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition I. Basics © The McGrawHill Companies, 2008 4. Deflection and Stiffness Mechanical Engineering Design Euler column equation should be used when the slenderness ratio is near (l / k ) Q . Then what should we do? The usual approach is to choose some point T on the Euler curve of Fig. 419. If the slenderness ratio is specied as (l / k )1 corresponding to point T, then use the Euler equation only when the actual slenderness ratio is greater than (l / k )1 . Otherwise, use one of the methods in the sections that follow. See Examples 417 and 418. Most designers select point T such that Pcr / A = Sy /2. Using Eq. (440), we nd the corresponding value of (l / k )1 to be l k 413 1 2π 2 C E Sy = 1/2 (442) Intermediate-Length Columns with Central Loading Over the years there have been a number of column formulas proposed and used for the range of l / k values for which the Euler formula is not suitable. Many of these are based on the use of a single material; others, on a so-called safe unit load rather than the critical value. Most of these formulas are based on the use of a linear relationship between the slenderness ratio and the unit load. The parabolic or J. B. Johnson formula now seems to be the preferred one among designers in the machine, automotive, aircraft, and structural-steel construction elds. The general form of the parabolic formula is l Pcr =ab A k 2 (a) where a and b are constants that are evaluated by tting a parabola to the Euler curve of Fig. 419 as shown by the dashed line ending at T . If the parabola is begun at Sy , then a = Sy . If point T is selected as previously noted, then Eq. (a) gives the value of (l / k )1 and the constant b is found to be b= Sy 2π 2 1 CE (b) Upon substituting the known values of a and b into Eq. (a), we obtain, for the parabolic equation, Pcr = Sy A 414 Sy l 2π k 2 1 CE l k l k (443) 1 Columns with Eccentric Loading We have noted before that deviations from an ideal column, such as load eccentricities or crookedness, are likely to occur during manufacture and assembly. Though these deviations are often quite small, it is still convenient to have a method of dealing with them. Frequently, too, problems occur in which load eccentricities are unavoidable. Figure 420a shows a column in which the line of action of the column forces is separated from the centroidal axis of the column by the eccentricity e. This problem is developed by using Eq. (412) and the free-body diagram of Fig. 420b. I. Basics © The McGrawHill Companies, 2008 4. Deflection and Stiffness 181 Deection and Stiffness BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 177 x Figure 420 P Notation for an eccentrically loaded column. A x P l M y x y y O P Pe e P (a ) (b) This results in the differential equation d2 y Pe P y= + 2 dx EI EI (a) The solution of Eq. (a), for the boundary conditions that y y [( e tan lP 2 EI )( sin P x EI ) ( cos 0 at x P x EI 0, l is )] 1 (b) By substituting x = l /2 in Eq. (b) and using a trigonometric identity, we obtain [( e sec Pl EI 2 )] (444) 1 The maximum bending moment also occurs at midspan and is Mmax = P (e + δ) = Pe sec l 2 P EI (445) The magnitude of the maximum compressive stress at midspan is found by superposing the axial component and the bending component. This gives σc = Mc P Mc P = A I A Ak 2 (c) Substituting Mmax from Eq. (445) yields σc = ec P l 1 + 2 sec A k 2k P EA (446) By imposing the compressive yield strength Syc as the maximum value of σc , we can write Eq. (446) in the form Syc P = 2 ) sec[(l /2k ) P / AE ] A 1 + (ec/ k (447) This is called the secant column formula. The term ec/ k 2 is called the eccentricity ratio. Figure 421 is a plot of Eq. (447) for a steel having a compressive (and tensile) 182 178 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition I. Basics © The McGrawHill Companies, 2008 4. Deflection and Stiffness Mechanical Engineering Design Figure 421 ec/k 2 = 0.1 Unit load P/A Comparison of secant and Euler equations for steel with Sy = 40 kpsi. Sy 0.3 0.6 Euler's curve 1.0 0 50 100 150 200 250 Slenderness ratio l / k yield strength of 40 kpsi. Note how the P / A contours asymptotically approach the Euler curve as l / k increases. Equation (447) cannot be solved explicitly for the load P . Design charts, in the fashion of Fig. 421, can be prepared for a single material if much column design is to be done. Otherwise, a root-nding technique using numerical methods must be used. EXAMPLE 416 Solution Develop specic Euler equations for the sizes of columns having (a) Round cross sections (b) Rectangular cross sections (a) Using A = π d 2 /4 and k = gives Answer I / A = [(π d 4 /64)/(π d 2 /4)]1/2 = d /4 with Eq. (441) d= 1/4 64 Pcrl 2 π 3C E (448) (b) For the rectangular column, we specify a cross section h × b with the restriction that h b . If the end conditions are the same for buckling in both directions, then buckling will occur in the direction of the least thickness. Therefore I= bh 3 12 A = bh k2 = I / A = h2 12 Substituting these in Eq. (441) gives Answer b= 12 Pcrl 2 π 2 C Eh 3 (449) Note, however, that rectangular columns do not generally have the same end conditions in both directions. I. Basics EXAMPLE 417 Solution © The McGrawHill Companies, 2008 4. Deflection and Stiffness 183 Deection and Stiffness BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 179 Specify the diameter of a round column 1.5 m long that is to carry a maximum load estimated to be 22 kN. Use a design factor n d = 4 and consider the ends as pinned (rounded). The column material selected has a minimum yield strength of 500 MPa and a modulus of elasticity of 207 GPa. We shall design the column for a critical load of Pcr = n d P = 4(22) = 88 kN Then, using Eq. (448) with C = 1 (see Table 42) gives 64 Pcrl 2 π 3C E d= 1/4 = 64(88)(1.5)2 π 3 (1)(207) 1/4 1/4 103 109 (103 ) = 37.48 mm Table A17 shows that the preferred size is 40 mm. The slenderness ratio for this size is l 1.5(103 ) l = = = 150 k d /4 40/4 To be sure that this is an Euler column, we use Eq. (548) and obtain l k 1 = 2π 2 C E Sy 1/2 = 2π 2 (1)(207) 500 1/2 109 106 1/2 = 90.4 which indicates that it is indeed an Euler column. So select Answer EXAMPLE 418 Solution d = 40 mm Repeat Ex. 416 for J. B. Johnson columns. (a) For round columns, Eq. (443) yields Sy l 2 Pcr d=2 +2 π Sy π CE Answer 1/2 (450) (b) For a rectangular section with dimensions h b , we nd Answer EXAMPLE 419 b= Pcr h Sy 3l 2 Sy 1 2 π C Eh 2 hb (451) Choose a set of dimensions for a rectangular link that is to carry a maximum compressive load of 5000 lbf. The material selected has a minimum yield strength of 75 kpsi and a modulus of elasticity E = 30 Mpsi. Use a design factor of 4 and an end condition constant C = 1 for buckling in the weakest direction, and design for (a) a length of 15 in, and (b) a length of 8 in with a minimum thickness of 1 in. 2 184 180 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition I. Basics © The McGrawHill Companies, 2008 4. Deflection and Stiffness Mechanical Engineering Design Solution (a) Using Eq. (441), we nd the limiting slenderness ratio to be l k 1 = 2π 2 C E Sy 1/2 = 2π 2 (1)(30)(106 ) 75(10)3 1/2 = 88.9 By using Pcr = n d P = 4(5000) = 20 000 lbf, Eqs. (449) and (451) are solved, using various values of h, to form Table 43. The table shows that a cross section of 5 by 3 8 4 in, which is marginally suitable, gives the least area. (b) An approach similar to that in part (a) is used with l = 8 in. All trial computations are found to be in the J. B. Johnson region of l / k values. A minimum area occurs when the section is a near square. Thus a cross section of 1 by 3 in is found to be suit2 4 able and safe. Table 43 h Table Generated to Solve Ex. 419, part (a) b A l/k Type Eq. No. 0.375 3.46 1.298 139 Euler (449) 0.500 1.46 0.730 104 Euler (449) 0.625 415 0.475 83 Johnson (451) 1.03 0.579 92 Euler (449) Struts or Short Compression Members A short bar loaded in pure compression by a force P acting along the centroidal axis will shorten in accordance with Hookes law, until the stress reaches the elastic limit of the material. At this point, permanent set is introduced and usefulness as a machine member may be at an end. If the force P is increased still more, the material either becomes barrel-like or fractures. When there is eccentricity in the loading, the elastic limit is encountered at smaller loads. A strut is a short compression member such as the one shown in Fig. 422. The magnitude of the maximum compressive stress in the x direction at point B in an intermediate section is the sum of a simple component P / A and a exural component Mc/ I ; that is, P x e B 0.76 0.5625 l c y P Figure 422 Eccentrically loaded strut. σc = ec Mc P Pec A P P 1+ 2 + = + = A I A IA A k (452) where k = ( I / A)1/2 and is the radius of gyration, c is the coordinate of point B, and e is the eccentricity of loading. Note that the length of the strut does not appear in Eq. (452). In order to use the equation for design or analysis, we ought, therefore, to know the range of lengths for which the equation is valid. In other words, how long is a short member? The difference between the secant formula Eq. (447) and Eq. (452) is that the secant equation, unlike Eq. (452), accounts for an increased bending moment due to bending deection. Thus the secant equation shows the eccentricity to be magnied by the bending deection. This difference between the two formulas suggests that one way I. Basics © The McGrawHill Companies, 2008 4. Deflection and Stiffness 185 Deection and Stiffness BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 181 of differentiating between a secant column and a strut, or short compression member, is to say that in a strut, the effect of bending deection must be limited to a certain small percentage of the eccentricity. If we decide that the limiting percentage is to be 1 percent of e, then, from Eq. (444), the limiting slenderness ratio turns out to be l k = 0.282 2 AE P 1/2 (453) This equation then gives the limiting slenderness ratio for using Eq. (452). If the actual slenderness ratio is greater than (l / k )2 , then use the secant formula; otherwise, use Eq. (452). EXAMPLE 420 Figure 423a shows a workpiece clamped to a milling machine table by a bolt tightened to a tension of 2000 lbf. The clamp contact is offset from the centroidal axis of the strut by a distance e = 0.10 in, as shown in part b of the gure. The strut, or block, is steel, 1 in square and 4 in long, as shown. Determine the maximum compressive stress in the block. Solution First we nd A = bh = 1(1) = 1 in2 , I = bh 3 /12 = 1(1)3 /12 = 0.0833 in4 , k 2 = I / A = 0.0833/1 = 0.0833 in2, and l / k = 4/(0.0833)1/2 = 13.9. Equation (453) gives the limiting slenderness ratio as l k 2 = 0.282 1/2 AE P = 0.282 1(30)(106 ) 1000 1/2 = 48.8 Thus the block could be as long as l = 48.8k = 48.8(0.0833)1/2 = 14.1 in before it need be treated by using the secant formula. So Eq. (452) applies and the maximum compressive stress is Answer σc = ec P 1+ 2 A k = 0.1(0.5) 1000 1+ = 1600 psi 1 0.0833 Figure 423 P = 1000 lbf A strut that is part of a workpiece clamping assembly. 1-in square 4 in 0.10 in P (a ) (b) 186 182 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition I. Basics © The McGrawHill Companies, 2008 4. Deflection and Stiffness Mechanical Engineering Design 416 Elastic Stability Section 412 presented the conditions for the unstable behavior of long, slender columns. Elastic instability can also occur in structural members other than columns. Compressive loads/stresses within any long, thin structure can cause structural instabilities (buckling). The compressive stress may be elastic or inelastic and the instability may be global or local. Global instabilities can cause catastrophic failure, whereas local instabilities may cause permanent deformation and function failure but not a catastrophic failure. The buckling discussed in Sec. 412 was global instability. However, consider a wide ange beam in bending. One ange will be in compression, and if thin enough, can develop localized buckling in a region where the bending moment is a maximum. Localized buckling can also occur in the web of the beam, where transverse shear stresses are present at the beam centroid. Recall, for the case of pure shear stress τ , a stress transformation will show that at 45 , a compressive stress of σ = τ exists. If the web is sufciently thin where the shear force V is a maximum, localized buckling of the web can occur. For this reason, additional support in the form of bracing is typically applied at locations of high shear forces.10 Thin-walled beams in bending can buckle in a torsional mode as illustrated in Fig. 424. Here a cantilever beam is loaded with a lateral force, F. As F is increases from zero, the end of the beam will deect in the negative y direction normally according to the bending equation, y = F L 3 /(3 E I ) . However, if the beam is long enough and the ratio of b/h is sufciently small, there is a critical value of F for which the beam will collapse in a twisting mode as shown. This is due to the compression in the bottom bers of the beam which cause the bers to buckle sideways (z direction). There are a great many other examples of unstable structural behavior, such as thinwalled pressure vessels in compression or with outer pressure or inner vacuum, thin-walled open or closed members in torsion, thin arches in compression, frames in compression, and shear panels. Because of the vast array of applications and the complexity of their analyses, further elaboration is beyond the scope of this book. The intent of this section is to make the reader aware of the possibilities and potential safety issues. The key issue is that the designer should be aware that if any unbraced part of a structural member is thin, and/or long, and in compression (directly or indirectly), the possibility of buckling should be investigated.11 Figure 424 y Torsional buckling of a thin-walled beam in bending. z h z y x b F Figure 425 10 Finite-element representation of ange buckling of a channel in compression. 11 See S. P. Timoshenko and J. M. Gere, Theory of Elastic Stability, 2nd ed., McGraw-Hill, New York, 1961. See also, Z. P. Bazant and L. Cedolin, Stability of Structures, Oxford University Press, New York, 1991. See C. G. Salmon and J. E. Johnson, Steel Structures: Design and Behavior, 4th ed., Harper, Collins, New York, 1996. I. Basics © The McGrawHill Companies, 2008 4. Deflection and Stiffness 187 Deection and Stiffness BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 183 For unique applications, the designer may need to revert to a numerical solution such as using nite elements. Depending on the application and the nite-element code available, an analysis can be performed to determine the critical loading (see Fig. 425). 417 Shock and Impact Impact refers to the collision of two masses with initial relative velocity. In some cases it is desirable to achieve a known impact in design; for example, this is the case in the design of coining, stamping, and forming presses. In other cases, impact occurs because of excessive deections, or because of clearances between parts, and in these cases it is desirable to minimize the effects. The rattling of mating gear teeth in their tooth spaces is an impact problem caused by shaft deection and the clearance between the teeth. This impact causes gear noise and fatigue failure of the tooth surfaces. The clearance space between a cam and follower or between a journal and its bearing may result in crossover impact and also cause excessive noise and rapid fatigue failure. Shock is a more general term that is used to describe any suddenly applied force or disturbance. Thus the study of shock includes impact as a special case. Figure 426 represents a highly simplied mathematical model of an automobile in collision with a rigid obstruction. Here m 1 is the lumped mass of the engine. The displacement, velocity, and acceleration are described by the coordinate x1 and its time derivatives. The lumped mass of the vehicle less the engine is denoted by m 2 , and its motion by the coordinate x2 and its derivatives. Springs k1 , k2 , and k3 represent the linear and nonlinear stiffnesses of the various structural elements that compose the vehicle. Friction and damping can and should be included, but is not shown in this model. The determination of the spring rates for such a complex structure will almost certainly have to be performed experimentally. Once these valuesthe ks, ms, damping and frictional coefcientsare obtained, a set of nonlinear differential equations can be written and a computer solution obtained for any impact velocity. Figure 427 is another impact model. Here mass m 1 has an initial velocity v and is just coming into contact with spring k1 . The part or structure to be analyzed is represented by mass m 2 and spring k2 . The problem facing the designer is to nd the maximum deection of m 2 and the maximum force exerted by k2 against m 2 . In the analysis it doesnt matter whether k1 is fastened to m 1 or to m 2 , since we are interested Figure 426 x2 x1 Two-degree-of-freedom mathematical model of an automobile in collision with a rigid obstruction. k1 k2 m1 m2 k3 Figure 427 x1 x2 k1 m1 k2 m2 188 184 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition I. Basics © The McGrawHill Companies, 2008 4. Deflection and Stiffness Mechanical Engineering Design only in a solution up to the point in time for which x2 reaches a maximum. That is, the solution for the rebound isnt needed. The differential equations are not difcult to derive. They are m 1 x 1 + k1 ( x 1 x 2 ) = 0 ¨ m 2 x 2 + k2 x 2 k1 ( x 1 x 2 ) = 0 ¨ (454) The analytical solution of Eq. pair (454) is harmonic and is studied in a course on mechanical vibrations.12 If the values of the ms and ks are known, the solution can be obtained easily using a program such as MATLAB. 418 Suddenly Applied Loading A simple case of impact is illustrated in Fig. 428a. Here a weight W falls a distance h and impacts a cantilever of stiffness EI and length l. We want to nd the maximum deection and the maximum force exerted on the beam due to the impact. Figure 428b shows an abstract model of the system. Using Table A91, we nd the spring rate to be k = F / y = 3 E I / l 3 . The beam mass and damping can be accounted for, but for this example will be considered negligible. The origin of the coordinate y corresponds to the point where the weight is released. Two free-body diagrams, shown in Fig. 428c and d are necessary. The rst corresponds to y h , and the second when y > h to account for the spring force. For each of these free-body diagrams we can write Newtons law by stating that the ¨ inertia force (W /g ) y is equal to the sum of the external forces acting on the weight. We then have W y=W ¨ g yh W y = k ( y h ) + W ¨ g y>h (a) We must also include in the mathematical statement of the problem the knowledge that the weight is released with zero initial velocity. Equation pair (a) constitutes a set of piecewise differential equations. Each equation is linear, but each applies only for a certain range of y . Figure 428 (a) A weight free to fall a distance h to free end of a beam. (b) Equivalent spring model. (c) Free body of weight during fall. (d ) Free body of weight during arrest. W W y h y EI, l y W h y W W k (a) 12 (b ) W k ( y h) (c) y h (d ) y h See William T. Thomson and Marie Dillon Dahleh, Theory of Vibrations with Applications, Prentice Hall, 5th ed., 1998. BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition © The McGrawHill Companies, 2008 189 Deection and Stiffness I. Basics 185 4. Deflection and Stiffness The solution to the set is valid for all values of t, but we are interested in values of y only up until the time that the spring or structure reaches its maximum deection. The solution to the rst equation in the set is y= gt 2 2 (455) yh and you can verify this by direct substitution. Equation (455) is no longer valid after y = h ; call this time t1 . Then t1 = (b) 2h / g Differentiating Eq. (455) to get the velocity gives y = gt ˙ (c) yh and so the velocity of the weight at t = t1 is y1 = gt1 = g 2h /g = ˙ (d) 2gh Having moved from y = 0 to y = h , we then need to solve the second equation of the set (a). It is convenient to dene a new time t = t t1 . Thus t = 0 at the instant the weight strikes the spring. Applying your knowledge of differential equations, you should nd the solution to be y = A cos ωt + B sin ωt + h + W k (e) y>h where ω= kg W (456) is the circular frequency of vibration. The initial conditions for the beam motion at t = 0, are y = h and y = y1 = 2gh (neglecting the mass of the beam, the velocity is ˙ ˙ the same as the weight at t = 0). Substituting the initial conditions into Eq. (e) yields A and B, and Eq. (e) becomes W W 2W h cos ωt + sin ωt + h + y>h (f) k k k 2W h / k = C sin φ , where it can be shown that Let W / k = C cos φ and C = [(W / k )2 + 2W h / k ]1/2 . Substituting this into Eq. ( f ) and using a trigonometric identity gives y= y= W k 2 + 2W h k 1/2 cos[ωt φ ] + h + W k y>h (457) The maximum deection of the spring (beam) occurs when the cosine term in Eq. (457) is unity. We designate this as δ and, after rearranging, nd it to be δ = ymax h = W W 1+ + k k 1/2 2hk W (458) The maximum force acting on the beam is now found to be F = kδ = W + W 1 + 2hk W 1/2 (459) 190 186 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition I. Basics 4. Deflection and Stiffness © The McGrawHill Companies, 2008 Mechanical Engineering Design Note, in this equation, that if h = 0, then F = 2W . This says that when the weight is released while in contact with the spring but is not exerting any force on the spring, the largest force is double the weight. Most systems are not as ideal as those explored here, so be wary about using these relations for nonideal systems. PROBLEMS 41 Structures can often be considered to be composed of a combination of tension and torsion members and beams. Each of these members can be analyzed separately to determine its force-deflection relationship and its spring rate. It is possible, then, to obtain the deflection of a structure by considering it as an assembly of springs having various series and parallel relationships. (a) What is the overall spring rate of three springs in series? (b) What is the overall spring rate of three springs in parallel? (c) What is the overall spring rate of a single spring in series with a pair of parallel springs? 42 The gure shows a torsion bar O A xed at O , simply supported at A, and connected to a cantilever A B . The spring rate of the torsion bar is k T , in newton-meters per radian, and that of the cantilever is kC , in newtons per meter. What is the overall spring rate based on the deection y at point B ? F O B L Problem 42 l A y R 43 A torsion-bar spring consists of a prismatic bar, usually of round cross section, that is twisted at one end and held fast at the other to form a stiff spring. An engineer needs a stiffer one than usual and so considers building in both ends and applying the torque somewhere in the central portion of the span, as shown in the figure. If the bar is uniform in diameter, that is, if d = d1 = d2 , investigate how the allowable angle of twist, the largest torque, and the spring rate depend on the location x at which the torque is applied. Hint: Consider two springs in parallel. d2 T Problem 43 d1 l x I. Basics © The McGrawHill Companies, 2008 4. Deflection and Stiffness 191 Deection and Stiffness BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 187 44 An engineer is forced by geometric considerations to apply the torque on the spring of Prob. 43 at the location x = 0.2l . For a uniform-diameter spring, this would cause the long leg of the span to be underutilized when both legs have the same diameter. If the diameter of the long leg is reduced sufciently, the shear stress in the two legs can be made equal. How would this change affect the allowable angle of twist, the largest torque, and the spring rate? 45 A bar in tension has a circular cross section and includes a conical portion of length l , as shown. The task is to find the spring rate of the entire bar. Equation (44) is useful for the outer portions of diameters d1 and d2 , but a new relation must be derived for the tapered section. If α is the apex half-angle, as shown, show that the spring rate of the tapered portion of the shaft is k= Problem 45 d2 E A1 l 1+ 2l tan α d1 dl l 46 When a hoisting cable is long, the weight of the cable itself contributes to the elongation. If a cable has a weight per unit length of w, a length of l , and a load P attached to the free end, show that the cable elongation is δ= Pl wl 2 + AE 2 AE 47 Use integration to verify the deection equation given for the uniformly loaded cantilever beam of appendix Table A93. 48 Use integration to verify the deection equation given for the end moment loaded cantilever beam of appendix Table A94. 49 When an initially straight beam sags under transverse loading, the ends contract because the neutral surface of zero strain neither extends nor contracts. The length of the deflected neutral surface is the same as the original beam length l . Consider a segment of the initially straight beam s . After bending, the x-direction component is shorter than s , namely, x . The contraction is s x , and these summed for the entire beam gives the end contraction λ. Show that .1 λ= 2 l 0 dy dx 2 dx 410 Using the results of Prob. 49, determine the end contraction of the uniformly loaded cantilever beam of appendix Table A93. 411 Using the results of Prob. 49, determine the end contraction of the uniformly loaded simplysupported beam of appendix Table A97. Assume the left support cannot deect in the x direction, whereas the right support can. 412 1 The gure shows a cantilever consisting of steel angles size 4 × 4 × 2 in mounted back to back. Using superposition, nd the deection at B and the maximum stress in the beam. 192 188 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition I. Basics 4. Deflection and Stiffness © The McGrawHill Companies, 2008 Mechanical Engineering Design y 10 ft 600 lbf Problem 412 7 ft 50 lbf/ft x O B A 413 A simply supported beam loaded by two forces is shown in the gure. Select a pair of structural steel channels mounted back to back to support the loads in such a way that the deec1 tion at midspan will not exceed 16 in and the maximum stress will not exceed 6 kpsi. Use superposition. y 800 lbf 600 lbf Problem 413 3 ft 5 ft 2 ft C O A 414 x B Using superposition, nd the deection of the steel shaft at A in the gure. Find the deection at midspan. By what percentage do these two values differ? y 400 mm 600 mm 1500 N Problem 414 2 kN/m B O x A 40 mm-dia. shaft 415 A rectangular steel bar supports the two overhanging loads shown in the gure. Using superposition, nd the deection at the ends and at the center. y 250 250 500 500 N 500 N Problem 415 Dimensions in millimeters. A B x C O Bar, b = 9, h = 35 416 Using the formulas in Appendix Table A9 and superposition, nd the deection of the cantilever at B if I = 13 in4 and E = 30 Mpsi. I. Basics © The McGrawHill Companies, 2008 4. Deflection and Stiffness 193 Deection and Stiffness BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 189 y 400 lbf 400 lbf Problem 416 3 ft 3 ft O x A B 417 The cantilever shown in the gure consists of two structural-steel channels size 3 in, 5.0 lbf/ft. Using superposition, nd the deection at A. y 48 in 220 lbf Problem 417 10 lbf/in x A O 418 Using superposition, determine the maximum deection of the beam shown in the gure. The material is carbon steel. y 10 in 10 in 10 in 10 in 120 lbf 85 lbf 85 lbf Problem 418 D O A B x C 2-in-dia. shaft 419 Illustrated is a rectangular steel bar with simple supports at the ends and loaded by a force F at the middle; the bar is to act as a spring. The ratio of the width to the thickness is to be about b = 16h , and the desired spring scale is 2400 lbf/in. (a) Find a set of cross-section dimensions, using preferred sizes. (b) What deection would cause a permanent set in the spring if this is estimated to occur at a normal stress of 90 kpsi? F A b Problem 419 A 4 ft 420 h Section AA 1 Illustrated in the gure is a 1 2 -in-diameter steel countershaft that supports two pulleys. Pulley A delivers power to a machine causing a tension of 600 lbf in the tight side of the belt and 80 lbf in 194 190 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition I. Basics © The McGrawHill Companies, 2008 4. Deflection and Stiffness Mechanical Engineering Design the loose side, as indicated. Pulley B receives power from a motor. The belt tensions on pulley B have the relation T1 = 0.125T2 . Find the deection of the shaft in the z direction at pulleys A and B . Assume that the bearings constitute simple supports. y 12 in 21 in O A 15 in Problem 420 T2 z 600 lbf T1 80 lbf C 9-in dia. B 1 1 -in dia. 2 x 12-in dia. 421 The gure shows a steel countershaft that supports two pulleys. Pulley C receives power from a motor producing the belt tensions shown. Pulley A transmits this power to another machine through the belt tensions T1 and T2 such that T1 = 8T2 . y 9 in O T2 z T1 Problem 421 A 11 in 1 1 -in dia. 4 12 in B 10-in dia. C 16-in dia. 50 lbf x 400 lbf (a) Find the deection of the overhanging end of the shaft, assuming simple supports at the bearings. (b) If roller bearings are used, the slope of the shaft at the bearings should not exceed 0.06 for 1 good bearing life. What shaft diameter is needed to conform to this requirement? Use 8 -in increments in any iteration you may make. What is the deection at pulley C now? 422 The structure of a diesel-electric locomotive is essentially a composite beam supporting a deck. Above the deck are mounted the diesel prime mover, generator or alternator, radiators, switch gear, and auxiliaries. Beneath the deck are found fuel and lubricant tanks, air reservoirs, and small auxiliaries. This assembly is supported at bolsters by the trucks that house the I. Basics © The McGrawHill Companies, 2008 4. Deflection and Stiffness 195 Deection and Stiffness BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 191 traction motors and brakes. This equipment is distributed as uniformly as possible in the span between the bolsters. In an approximate way, the loading can be viewed as uniform between the bolsters and simply supported. Because the hoods that shield the equipment from the weather have many rectangular access doors, which are mass-produced, it is important that the hood structure be level and plumb and sit on a flat deck. Aesthetics plays a role too. The center sill beam has a second moment of area of I = 5450 in4 , the bolsters are 36 ft apart, and the deck loading is 5000 lbf/ft. (a) What is the camber of the curve to which the deck will be built in order that the service-ready locomotive will have a at deck? (b) What equation would you give to locate points on the curve of part (a)? 423 The designer of a shaft usually has a slope constraint imposed by the bearings used. This limit will be denoted as ξ . If the shaft shown in the gure is to have a uniform diameter d except in the locality of the bearing mounting, it can be approximated as a uniform beam with simple supports. Show that the minimum diameters to meet the slope constraints at the left and right bearings are, respectively, dL = 32 Fb(l 2 b2 ) 3π El ξ 1/4 dR = 32 Fa (l 2 a 2 ) 3π El ξ 1/4 F a b l Problem 423 y F x 424 A shaft is to be designed so that it is supported by roller bearings. The basic geometry is shown in the gure. The allowable slope at the bearings is 0.001 mm/mm without bearing life penalty. For a design factor of 1.28, what uniform-diameter shaft will support the 3.5-kN load 100 mm from the left bearing without penalty? Use E = 207 GPa. F = 3.5 kN 100 150 Problem 424 Dimensions in millimeters. d 250 425 Determine the maximum deection of the shaft of Prob. 424. 426 For the shaft shown in the gure, let a1 = 4 in, b1 = 12 in, a2 = 10 in, F1 = 100 lbf, F2 = 300 lbf, and E = 30 Mpsi. The shaft is to be sized so that the maximum slope at either bearing A or bearing B does not exceed 0.001 rad. Determine a suitable diameter d. 196 192 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition I. Basics © The McGrawHill Companies, 2008 4. Deflection and Stiffness Mechanical Engineering Design y F1 a1 A Problem 426 b1 z a2 B b2 F2 x 427 If the diameter of the beam for Prob. 426 is 1.375 in, determine the deection of the beam at x = 8 in. 428 See Prob. 426 and the accompanying gure. The loads and dimensions are F1 = 3.5 kN, F2 = 2.7 kN, a1 = 100 mm, b1 = 150 mm, and a2 = 175 mm. Find the uniform shaft diameter necessary to limit the slope at the bearings to 0.001 rad. Use a design factor of n d = 1.5 and E = 207 Gpa. 429 Shown in the gure is a uniform-diameter shaft with bearing shoulders at the ends; the shaft is subjected to a concentrated moment M = 1200 lbf · in. The shaft is of carbon steel and has a = 5 in and l = 9 in. The slope at the ends must be limited to 0.002 rad. Find a suitable diameter d . a b MB Problem 429 B l 430 The rectangular member O AB , shown in the gure, is held horizontal by the round hooked bar AC. The modulus of elasticity of both parts is 10 Mpsi. Use superposition to nd the deection at B due to a force F = 80 lbf. 1 -in 2 dia. C y Problem 430 12 in 2 in 1 -in 4 thick F x A B O 6 in 431 12 in The gure illustrates a torsion-bar spring O A having a diameter d = 12 mm. The actuating cantilever A B also has d = 12 mm. Both parts are of carbon steel. Use superposition and nd the spring rate k corresponding to a force F acting at B. I. Basics © The McGrawHill Companies, 2008 4. Deflection and Stiffness 197 Deection and Stiffness BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 193 y O d x Problem 431 1.5 m A F d z 0.1 m B 432 Consider the simply supported beam with an intermediate load in Appendix A96. Determine the deection equation if the stiffness of the left and right supports are k1 and k2 , respectively. 433 Consider the simply supported beam with a uniform load in Appendix A97. Determine the deection equation if the stiffness of the left and right supports are k1 and k2 , respectively. 434 Prove that for a uniform-cross-section beam with simple supports at the ends loaded by a single concentrated load, the location of the maximum deection will never be outside the range of 0.423l x 0.577l regardless of the location of the load along the beam. The importance of this is that you can always get a quick estimate of ymax by using x = l /2. 435 Solve Prob. 412 using singularity functions. Use statics to determine the reactions. 436 Solve Prob. 413 using singularity functions. Use statics to determine the reactions. 437 Solve Prob. 414 using singularity functions. Use statics to determine the reactions. 438 Consider the uniformly loaded simply supported beam with an overhang as shown. Use singularity functions to determine the deection equation of the beam. Use statics to determine the reactions. w Problem 438 l a 439 Solve Prob. 415 using singularity functions. Since the beam is symmetric, only write the equation for half the beam and use the slope at the beam center as a boundary condition. Use statics to determine the reactions. 440 Solve Prob. 430 using singularity functions. Use statics to determine the reactions. 441 Determine the deection equation for the steel beam shown using singularity functions. Since the beam is symmetric, write the equation for only half the beam and use the slope at the beam center as a boundary condition. Use statics to determine the reactions. w = 200 lbf/in 1.5-in diameter 1.5-in diameter Problem 441 2-in diameter 4 in 12 in 4 in 198 194 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition I. Basics © The McGrawHill Companies, 2008 4. Deflection and Stiffness Mechanical Engineering Design 442 Determine the deection equation for the cantilever beam shown using singularity functions. Evaluate the deections at B and C and compare your results with Example 411. y l /2 Problem 442 A 2 I1 l /2 B I1 x C F 443 Examine the expression for the deection of the cantilever beam, end-loaded, shown in Appendix Table A91 for some intermediate point, x = a , as F1 a 2 (a 3l ) 6E I y |x = a = In Table A92, for a cantilever with intermediate load, the deection at the end is y |x = l = F2 a 2 (a 3l ) 6E I These expressions are remarkably similar and become identical when F1 = F2 = 1. In other words, the deection at x = a (station 1) due to a unit load at x = l (station 2) is the same as the deection at station 2 due to a unit load at station 1. Prove that this is true generally for an elastic body even when the lines of action of the loads are not parallel. This is known as a special case of Maxwells reciprocal theorem. (Hint: Consider the potential energy of strain when the body is loaded by two forces in either order of application.) 444 A steel shaft of uniform 2-in diameter has a bearing span l of 23 in and an overhang of 7 in on which a coupling is to be mounted. A gear is to be attached 9 in to the right of the left bearing and will carry a radial load of 400 lbf. We require an estimate of the bending deection at the coupling. Appendix Table A96 is available, but we cant be sure of how to expand the equation to predict the deection at the coupling. (a) Show how Appendix Table A910 and Maxwells theorem (see Prob. 443) can be used to obtain the needed estimate. (b) Check your work by nding the slope at the right bearing and extending it to the coupling location. 445 Use Castiglianos theorem to verify the maximum deection for the uniformly loaded beam of Appendix Table A97. Neglect shear. 446 Solve Prob. 417 using Castiglianos theorem. Hint: Write the moment equation using a position variable positive to the left starting at the right end of the beam. 447 Solve Prob. 430 using Castiglianos theorem. 448 Solve Prob. 431 using Castiglianos theorem. 449 Determine the deection at midspan for the beam of Prob. 441 using Castiglianos theorem. I. Basics 450 4. Deflection and Stiffness © The McGrawHill Companies, 2008 199 Deection and Stiffness BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 195 Using Castiglianos theorem, determine the deection of point B in the direction of the force F for the bar shown. The solid bar has a uniform diameter, d. Neglect bending shear. l O A Problem 450 a B 4 3 F 451 A cable is made using a 16-gauge (0.0625-in) steel wire and three strands of 12-gauge (0.0801-in) copper wire. Find the stress in each wire if the cable is subjected to a tension of 250 lbf. 452 The gure shows a steel pressure cylinder of diameter 4 in which uses six SAE grade 5 steel bolts having a grip of 12 in. These bolts have a proof strength (see Chap. 8) of 85 kpsi for this size of bolt. Suppose the bolts are tightened to 90 percent of this strength in accordance with some recommendations. (a) Find the tensile stress in the bolts and the compressive stress in the cylinder walls. (b) Repeat part (a), but assume now that a uid under a pressure of 600 psi is introduced into the cylinder. Six 3 8 -in grade 5 bolts t= Problem 452 lc = 11 in 1 4 in D = 4 in lb = 12 in 453 A torsion bar of length L consists of a round core of stiffness (G J )c and a shell of stiffness (G J )s . If a torque T is applied to this composite bar, what percentage of the total torque is carried by the shell? 454 A rectangular aluminum bar 12 mm thick and 50 mm wide is welded to xed supports at the ends, and the bar supports a load W = 3.5 kN, acting through a pin as shown. Find the reactions at the supports. 200 196 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition I. Basics 4. Deflection and Stiffness © The McGrawHill Companies, 2008 Mechanical Engineering Design y B 750 mm Problem 454 50 mm W 12 mm thick A 500 mm x O 455 The steel shaft shown in the gure is subjected to a torque of 50 lbf-in applied at point A. Find the torque reactions at O and B . y 1 50 lbf-in 1 2 -in dia. Problem 455 x O A B 4 in 6 in 456 Repeat Prob. 455 with the diameters of section OA being 1.5 in and section AB being 1.75 in. 457 In testing the wear life of gear teeth, the gears are assembled by using a pretorsion. In this way, a large torque can exist even though the power input to the tester is small. The arrangement shown in the gure uses this principle. Note the symbol used to indicate the location of the shaft bearings used in the gure. Gears A, B , and C are assembled rst, and then gear C is held xed. Gear D is assembled and meshed with gear C by twisting it through an angle of 4 to provide the pretorsion. Find the maximum shear stress in each shaft resulting from this preload. 4 ft C, 6-in dia. Problem 457 B, 6-in dia. 1 1 -in dia. 4 7 8 2 -in dia. 1 D, 2 1 -in dia. 2 A, 2 1 -in dia. 2 458 3 The gure shows a 8 - by 1 1 -in rectangular steel bar welded to xed supports at each end. The 2 bar is axially loaded by the forces FA = 10 kip and FB = 5 kip acting on pins at A and B . Assuming that the bar will not buckle laterally, nd the reactions at the xed supports. Use procedure 1 from Sec. 410. 459 For the beam shown, determine the support reactions using superposition and procedure 1 from Sec. 410. I. Basics © The McGrawHill Companies, 2008 4. Deflection and Stiffness 201 Deection and Stiffness BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 197 y 10 in 20 in Problem 458 A 1 15 in B FA 1 2 in C FB x O 3 8 in thick w Problem 459 B A C a l 460 Solve Prob. 459 using Castiglianos theorem and procedure 1 from Sec. 410. 461 The steel beam A BC D shown is simply supported at A and supported at B and D by steel cables, each having an effective diameter of 12 mm. The second area moment of the beam is I = 8(105 ) mm4 . A force of 20 kN is applied at point C. Using procedure 2 of Sec. 410 determine the stresses in the cables and the deections of B, C, and D. For steel, let E = 209 GPa. E F 1m A Problem 461 B C D 20 kN 500 mm 462 500 mm 500 mm The steel beam A BC D shown is supported at C as shown and supported at B and D by steel bolts 5 each having a diameter of 16 in. The lengths of B E and D F are 2 and 2.5 in, respectively. The beam has a second area moment of 0.050 in4 . Prior to loading, the nuts are just in contact with the horizontal beam. A force of 500 lbf is then applied at point A. Using procedure 2 of Sec. 410, determine the stresses in the bolts and the deections of points A, B, and D. For steel, let E = 30 Mpsi. E 500 lbf A B D C Problem 462 F 3 in 463 3 in 3 in The horizontal deection of the right end of the curved bar of Fig. 412 is given by Eq. (435) for R / h > 10. For the same conditions, determine the vertical deection. 202 198 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition I. Basics 4. Deflection and Stiffness © The McGrawHill Companies, 2008 Mechanical Engineering Design 464 A cast-iron piston ring has a mean diameter of 81 mm, a radial height h = 6 mm, and a thickness b = 4 mm. The ring is assembled using an expansion tool that separates the split ends a distance δ by applying a force F as shown. Use Castiglianos theorem and determine the deection δ as a function of F . Use E = 131 GPa and assume Eq. (428) applies. h = 6 mm F Problem 464 + F 465 For the wire form shown use Castiglianos method to determine the vertical deection of point A. Consider bending only and assume Eq. (428) applies for the curved part. C Problem 465 P R A B l 466 For the wire form shown determine the vertical deections of points A and B. Consider bending only and assume Eq. (428) applies. A C R P Problem 466 B 467 For the wire form shown, determine the deection of point A in the y direction. Assume R / h > 10 and consider the effects of bending and torsion only. The wire is steel with E = 200 GPa, ν = 0.29, and has a diameter of 5 mm. Before application of the 200-N force the wire form is in the x z plane where the radius R is 100 mm. I. Basics © The McGrawHill Companies, 2008 4. Deflection and Stiffness 203 Deection and Stiffness BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 199 y x Problem 467 R z 90° A 200 N 468 For the wire form shown, determine (a) the reactions at points A and B, (b) how the bending moment varies along the wire, and (c) the deection of the load F. Assume that the entire energy is described by Eq. (428). F Problem 468 R A 469 B For the curved beam shown, F = 30 kN. The material is steel with E = 207 GPa and G = 79 GPa. Determine the relative deection of the applied forces. 80 10 50 F F A A 20 40 Problem 469 10 Section AA 100 (All dimensions in millimeters.) 470 Solve Prob. 463 using Eq. (432). 471 A thin ring is loaded by two equal and opposite forces F in part a of the gure. A free-body diagram of one quadrant is shown in part b. This is a statically indeterminate problem, because the moment M A cannot be found by statics. We wish to nd the maximum bending moment in the ring due to the forces F. Assume that the radius of the ring is large so that Eq. (428) can be used. 204 200 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition I. Basics © The McGrawHill Companies, 2008 4. Deflection and Stiffness Mechanical Engineering Design y y F B B ds d R Problem 471 A C A x O O x MA F 2 D F (b ) (a) 472 Find the increase in the diameter of the ring of Prob. 471 due to the forces F and along the y axis. 473 A round tubular column has outside and inside diameters of D and d, respectively, and a diametral ratio of K = d / D . Show that buckling will occur when the outside diameter is 64 Pcr l 2 3 C E (1 K 4 ) π D= 474 For the conditions of Prob. 473, show that buckling according to the parabolic formula will occur when the outside diameter is D=2 475 1/4 Sy l 2 Pcr +2 2) π Sy (1 K π C E (1 + K 2 ) 1/2 Link 2, shown in the gure, is 1 in wide, has 1 -in-diameter bearings at the ends, and is cut from 2 low-carbon steel bar stock having a minimum yield strength of 24 kpsi. The end-condition constants are C = 1 and C = 1.2 for buckling in and out of the plane of the drawing, respectively. (a) Using a design factor n d = 5, nd a suitable thickness for the link. (b) Are the bearing stresses at O and B of any signicance? y 1 Problem 475 x 2 O A 3 3 180 lbf 1 4 ft B 3 ft 476 C 1 2 2 ft Link 3, shown schematically in the gure, acts as a brace to support the 1.2-kN load. For buckling in the plane of the gure, the link may be regarded as pinned at both ends. For out-of-plane buckling, the ends are xed. Select a suitable material and a method of manufacture, such as forging, casting, stamping, or machining, for casual applications of the brace in oil-eld machinery. Specify the dimensions of the cross section as well as the ends so as to obtain a strong, safe, wellmade, and economical brace. I. Basics © The McGrawHill Companies, 2008 4. Deflection and Stiffness 205 Deection and Stiffness BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 201 y B F = 1.2 kN 3 Problem 476 0.9 m 2 O 60° 1 477 A x The hydraulic cylinder shown in the gure has a 3-in bore and is to operate at a pressure of 800 psi. With the clevis mount shown, the piston rod should be sized as a column with both ends rounded for any plane of buckling. The rod is to be made of forged AISI 1030 steel without further heat treatment. d Problem 477 3 in (a) Use a design factor n d = 3 and select a preferred size for the rod diameter if the column length is 60 in. (b) Repeat part (a) but for a column length of 18 in. (c) What factor of safety actually results for each of the cases above? 478 The gure shows a schematic drawing of a vehicular jack that is to be designed to support a maximum mass of 400 kg based on the use of a design factor n d = 2.50. The opposite-handed threads on the two ends of the screw are cut to allow the link angle θ to vary from 15 to 70 . The links are to be machined from AISI 1020 hot-rolled steel bars with a minimum yield strength of 380 MPa. Each of the four links is to consist of two bars, one on each side of the central bearings. The bars are to be 300 mm long and have a bar width of 25 mm. The pinned ends are to be designed to secure an end-condition constant of at least C = 1.4 for out-of-plane buckling. Find a suitable preferred thickness and the resulting factor of safety for this thickness. W l Problem 478 w 206 202 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition I. Basics 4. Deflection and Stiffness © The McGrawHill Companies, 2008 Mechanical Engineering Design 479 If drawn, a gure for this problem would resemble that for Prob. 452. A strut that is a standard 3 hollow right circular cylinder has an outside diameter of 4 in and a wall thickness of 8 in and is compressed between two circular end plates held by four bolts equally spaced on a bolt circle of 5.68-in diameter. All four bolts are hand-tightened, and then bolt A is tightened to a tension of 2000 lbf and bolt C, diagonally opposite, is tightened to a tension of 10 000 lbf. The strut axis of symmetry is coincident with the center of the bolt circles. Find the maximum compressive load, the eccentricity of loading, and the largest compressive stress in the strut. 480 Design link C D of the hand-operated toggle press shown in the gure. Specify the cross-section dimensions, the bearing size and rod-end dimensions, the material, and the method of processing. F A B L l Problem 480 C L = 12 in, l = 4 in, θmin = 0°. l D 481 Find expressions for the maximum values of the spring force and deection y of the impact system shown in the gure. Can you think of a realistic application for this model? W y k Problem 481 h 482 As shown in the gure, the weight W1 strikes W2 from a height h. Find the maximum values of the spring force and the deection of W2 . Name an actual system for which this model might be used. h W1 W2 Problem 482 y k I. Basics 483 © The McGrawHill Companies, 2008 4. Deflection and Stiffness 207 Deection and Stiffness BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 203 Part a of the gure shows a weight W mounted between two springs. If the free end of spring k1 is suddenly displaced through the distance x = a , as shown in part b, what would be the maximum displacement y of the weight? x y k1 k2 W Problem 483 a t x (a ) ( b) 208 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition PART II. Failure Prevention Introduction 2 Failure Prevention © The McGrawHill Companies, 2008 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition II. Failure Prevention © The McGrawHill Companies, 2008 5. Failures Resulting from Static Loading 5 209 Failures Resulting from Static Loading Chapter Outline 51 Static Strength 52 Stress Concentration 53 Failure Theories 54 Maximum-Shear-Stress Theory for Ductile Materials 55 Distortion-Energy Theory for Ductile Materials 56 Coulomb-Mohr Theory for Ductile Materials 57 Failure of Ductile Materials Summary 58 Maximum-Normal-Stress Theory for Brittle Materials 59 Modications of the Mohr Theory for Brittle Materials 208 209 211 510 Failure of Brittle Materials Summary 511 Selection of Failure Criteria 512 Introduction to Fracture Mechanics 513 Stochastic Analysis 514 Important Design Equations 211 213 219 222 226 227 229 230 231 240 246 205 210 206 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition II. Failure Prevention 5. Failures Resulting from Static Loading © The McGrawHill Companies, 2008 Mechanical Engineering Design In Chap. 1 we learned that strength is a property or characteristic of a mechanical element. This property results from the material identity, the treatment and processing incidental to creating its geometry, and the loading, and it is at the controlling or critical location. In addition to considering the strength of a single part, we must be cognizant that the strengths of the mass-produced parts will all be somewhat different from the others in the collection or ensemble because of variations in dimensions, machining, forming, and composition. Descriptors of strength are necessarily statistical in nature, involving parameters such as mean, standard deviations, and distributional identification. A static load is a stationary force or couple applied to a member. To be stationary, the force or couple must be unchanging in magnitude, point or points of application, and direction. A static load can produce axial tension or compression, a shear load, a bending load, a torsional load, or any combination of these. To be considered static, the load cannot change in any manner. In this chapter we consider the relations between strength and static loading in order to make the decisions concerning material and its treatment, fabrication, and geometry for satisfying the requirements of functionality, safety, reliability, competitiveness, usability, manufacturability, and marketability. How far we go down this list is related to the scope of the examples. Failure is the rst word in the chapter title. Failure can mean a part has separated into two or more pieces; has become permanently distorted, thus ruining its geometry; has had its reliability downgraded; or has had its function compromised, whatever the reason. A designer speaking of failure can mean any or all of these possibilities. In this chapter our attention is focused on the predictability of permanent distortion or separation. In strength-sensitive situations the designer must separate mean stress and mean strength at the critical location sufciently to accomplish his or her purposes. Figures 51 to 55 are photographs of several failed parts. The photographs exemplify the need of the designer to be well-versed in failure prevention. Toward this end we shall consider one-, two-, and three-dimensional stress states, with and without stress concentrations, for both ductile and brittle materials. Figure 51 (a) Failure of a truck drive-shaft spline due to corrosion fatigue. Note that it was necessary to use clear tape to hold the pieces in place. (b) Direct end view of failure. BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition II. Failure Prevention 5. Failures Resulting from Static Loading © The McGrawHill Companies, 2008 Failures Resulting from Static Loading Figure 52 Impact failure of a lawnmower blade driver hub. The blade impacted a surveying pipe marker. Figure 53 Failure of an overhead-pulley retaining bolt on a weightlifting machine. A manufacturing error caused a gap that forced the bolt to take the entire moment load. Figure 54 Chain test xture that failed in one cycle. To alleviate complaints of excessive wear, the manufacturer decided to case-harden the material. (a) Two halves showing fracture; this is an excellent example of brittle fracture initiated by stress concentration. (b) Enlarged view of one portion to show cracks induced by stress concentration at the support-pin holes. 211 207 212 208 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition II. Failure Prevention 5. Failures Resulting from Static Loading © The McGrawHill Companies, 2008 Mechanical Engineering Design Figure 55 Valve-spring failure caused by spring surge in an oversped engine. The fractures exhibit the classic 45 shear failure. 51 Static Strength Ideally, in designing any machine element, the engineer should have available the results of a great many strength tests of the particular material chosen. These tests should be made on specimens having the same heat treatment, surface nish, and size as the element the engineer proposes to design; and the tests should be made under exactly the same loading conditions as the part will experience in service. This means that if the part is to experience a bending load, it should be tested with a bending load. If it is to be subjected to combined bending and torsion, it should be tested under combined bending and torsion. If it is made of heat-treated AISI 1040 steel drawn at 500 C with a ground nish, the specimens tested should be of the same material prepared in the same manner. Such tests will provide very useful and precise information. Whenever such data are available for design purposes, the engineer can be assured of doing the best possible job of engineering. The cost of gathering such extensive data prior to design is justied if failure of the part may endanger human life or if the part is manufactured in sufciently large quantities. Refrigerators and other appliances, for example, have very good reliabilities because the parts are made in such large quantities that they can be thoroughly tested in advance of manufacture. The cost of making these tests is very low when it is divided by the total number of parts manufactured. You can now appreciate the following four design categories: 1 2 3 Failure of the part would endanger human life, or the part is made in extremely large quantities; consequently, an elaborate testing program is justied during design. The part is made in large enough quantities that a moderate series of tests is feasible. The part is made in such small quantities that testing is not justied at all; or the design must be completed so rapidly that there is not enough time for testing. BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition II. Failure Prevention 5. Failures Resulting from Static Loading © The McGrawHill Companies, 2008 Failures Resulting from Static Loading 213 209 The part has already been designed, manufactured, and tested and found to be unsatisfactory. Analysis is required to understand why the part is unsatisfactory and what to do to improve it. 4 More often than not it is necessary to design using only published values of yield strength, ultimate strength, percentage reduction in area, and percentage elongation, such as those listed in Appendix A. How can one use such meager data to design against both static and dynamic loads, two- and three-dimensional stress states, high and low temperatures, and very large and very small parts? These and similar questions will be addressed in this chapter and those to follow, but think how much better it would be to have data available that duplicate the actual design situation. 52 Stress Concentration Stress concentration (see Sec. 313) is a highly localized effect. In some instances it may be due to a surface scratch. If the material is ductile and the load static, the design load may cause yielding in the critical location in the notch. This yielding can involve strain strengthening of the material and an increase in yield strength at the small critical notch location. Since the loads are static and the material is ductile, that part can carry the loads satisfactorily with no general yielding. In these cases the designer sets the geometric (theoretical) stress concentration factor K t to unity. The rationale can be expressed as follows. The worst-case scenario is that of an idealized nonstrain-strengthening material shown in Fig. 56. The stress-strain curve rises linearly to the yield strength Sy , then proceeds at constant stress, which is equal to Sy . Consider a lleted rectangular bar as depicted in Fig. A155, where the crosssection area of the small shank is 1 in2. If the material is ductile, with a yield point of 40 kpsi, and the theoretical stress-concentration factor (SCF) K t is 2, A load of 20 kip induces a tensile stress of 20 kpsi in the shank as depicted at point A in Fig. 56. At the critical location in the llet the stress is 40 kpsi, and the SCF is K = σmax /σnom = 40/20 = 2. Figure 56 An idealized stress-strain curve. The dashed line depicts a strain-strengthening material. 50 Tensile stress , kpsi C Sy D E B A 0 Tensile strain, 214 210 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition II. Failure Prevention 5. Failures Resulting from Static Loading © The McGrawHill Companies, 2008 Mechanical Engineering Design A load of 30 kip induces a tensile stress of 30 kpsi in the shank at point B. The llet stress is still 40 kpsi (point D), and the SCF K = σmax /σnom = Sy /σ = 40/30 = 1.33. At a load of 40 kip the induced tensile stress (point C) is 40 kpsi in the shank. At the critical location in the llet, the stress (at point E) is 40 kpsi. The SCF K = σmax /σnom = Sy /σ = 40/40 = 1. For materials that strain-strengthen, the critical location in the notch has a higher Sy . The shank area is at a stress level a little below 40 kpsi, is carrying load, and is very near its failure-by-general-yielding condition. This is the reason designers do not apply K t in static loading of a ductile material loaded elastically, instead setting K t = 1. When using this rule for ductile materials with static loads, be careful to assure yourself that the material is not susceptible to brittle fracture (see Sec. 512) in the environment of use. The usual denition of geometric (theoretical) stress-concentration factor for normal stress K t and shear stress K ts is σmax = K t σnom (a) τmax = K ts τnom (b) Since your attention is on the stress-concentration factor, and the denition of σnom or τnom is given in the graph caption or from a computer program, be sure the value of nominal stress is appropriate for the section carrying the load. Brittle materials do not exhibit a plastic range. A brittle material feels the stress concentration factor K t or K ts , which is applied by using Eq. (a) or (b). An exception to this rule is a brittle material that inherently contains microdiscontinuity stress concentration, worse than the macrodiscontinuity that the designer has in mind. Sand molding introduces sand particles, air, and water vapor bubbles. The grain structure of cast iron contains graphite akes (with little strength), which are literally cracks introduced during the solidication process. When a tensile test on a cast iron is performed, the strength reported in the literature includes this stress concentration. In such cases K t or K ts need not be applied. An important source of stress-concentration factors is R. E. Peterson, who compiled them from his own work and that of others.1 Peterson developed the style of presentation in which the stress-concentration factor K t is multiplied by the nominal stress σnom to estimate the magnitude of the largest stress in the locality. His approximations were based on photoelastic studies of two-dimensional strips (Hartman and Levan, 1951; Wilson and White, 1973), with some limited data from three-dimensional photoelastic tests of Hartman and Levan. A contoured graph was included in the presentation of each case. Filleted shafts in tension were based on two-dimensional strips. Table A15 provides many charts for the theoretical stress-concentration factors for several fundamental load conditions and geometry. Additional charts are also available from Peterson.2 Finite element analysis (FEA) can also be applied to obtain stress-concentration factors. Improvements on K t and K ts for lleted shafts were reported by Tipton, Sorem, and Rolovic.3 1 R. E. Peterson, Design Factors for Stress Concentration, Machine Design, vol. 23, no. 2, February 1951; no. 3, March 1951; no. 5, May 1951; no. 6, June 1951; no. 7, July 1951. 2 Walter D. Pilkey, Petersons Stress Concentration Factors, 2nd ed, John Wiley & Sons, New York, 1997. 3 S. M. Tipton, J. R. Sorem Jr., and R. D. Rolovic, Updated Stress-Concentration Factors for Filleted Shafts in Bending and Tension, Trans. ASME, Journal of Mechanical Design, vol. 118, September 1996, pp. 321327. BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition II. Failure Prevention © The McGrawHill Companies, 2008 5. Failures Resulting from Static Loading Failures Resulting from Static Loading 53 215 211 Failure Theories Section 51 illustrated some ways that loss of function is manifested. Events such as distortion, permanent set, cracking, and rupturing are among the ways that a machine element fails. Testing machines appeared in the 1700s, and specimens were pulled, bent, and twisted in simple loading processes. If the failure mechanism is simple, then simple tests can give clues. Just what is simple? The tension test is uniaxial (thats simple) and elongations are largest in the axial direction, so strains can be measured and stresses inferred up to failure. Just what is important: a critical stress, a critical strain, a critical energy? In the next several sections, we shall show failure theories that have helped answer some of these questions. Unfortunately, there is no universal theory of failure for the general case of material properties and stress state. Instead, over the years several hypotheses have been formulated and tested, leading to todays accepted practices. Being accepted, we will characterize these practices as theories as most designers do. Structural metal behavior is typically classied as being ductile or brittle, although under special situations, a material normally considered ductile can fail in a brittle manner (see Sec. 512). Ductile materials are normally classied such that ε f 0.05 and have an identiable yield strength that is often the same in compression as in tension ( Syt = Syc = Sy ). Brittle materials, ε f < 0.05, do not exhibit an identiable yield strength, and are typically classied by ultimate tensile and compressive strengths, Sut and Suc , respectively (where Suc is given as a positive quantity). The generally accepted theories are: Ductile materials (yield criteria) Maximum shear stress (MSS), Sec. 54 Distortion energy (DE), Sec. 55 Ductile Coulomb-Mohr (DCM), Sec. 56 Brittle materials (fracture criteria) Maximum normal stress (MNS), Sec. 58 Brittle Coulomb-Mohr (BCM), Sec. 59 Modied Mohr (MM), Sec. 59 It would be inviting if we had one universally accepted theory for each material type, but for one reason or another, they are all used. Later, we will provide rationales for selecting a particular theory. First, we will describe the bases of these theories and apply them to some examples. 54 Maximum-Shear-Stress Theory for Ductile Materials The maximum-shear-stress theory predicts that yielding begins whenever the maximum shear stress in any element equals or exceeds the maximum shear stress in a tensiontest specimen of the same material when that specimen begins to yield. The MSS theory is also referred to as the Tresca or Guest theory. Many theories are postulated on the basis of the consequences seen from tensile tests. As a strip of a ductile material is subjected to tension, slip lines (called Lüder lines) form at approximately 45° with the axis of the strip. These slip lines are the 216 212 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition II. Failure Prevention © The McGrawHill Companies, 2008 5. Failures Resulting from Static Loading Mechanical Engineering Design beginning of yield, and when loaded to fracture, fracture lines are also seen at angles approximately 45° with the axis of tension. Since the shear stress is maximum at 45° from the axis of tension, it makes sense to think that this is the mechanism of failure. It will be shown in the next section, that there is a little more going on than this. However, it turns out the MSS theory is an acceptable but conservative predictor of failure; and since engineers are conservative by nature, it is quite often used. Recall that for simple tensile stress, σ = P / A , and the maximum shear stress occurs on a surface 45° from the tensile surface with a magnitude of τmax = σ/2. So the maximum shear stress at yield is τmax = Sy /2. For a general state of stress, three principal stresses can be determined and ordered such that σ1 σ2 σ3 . The maximum shear stress is then τmax = (σ1 σ3 )/2 (see Fig. 312). Thus, for a general state of stress, the maximum-shear-stress theory predicts yielding when τmax = Sy σ1 σ3 2 2 or σ1 σ3 Sy (51) Note that this implies that the yield strength in shear is given by (52) Ssy = 0.5 Sy which, as we will see later is about 15 percent low (conservative). For design purposes, Eq. (51) can be modied to incorporate a factor of safety, n. Thus, τmax = Sy 2n or σ1 σ3 = Sy n (53) Plane stress problems are very common where one of the principal stresses is zero, and the other two, σ A and σ B , are determined from Eq. (313). Assuming that σ A σ B , there are three cases to consider in using Eq. (51) for plane stress: Case 1: σ A σ B 0. For this case, σ1 = σ A and σ3 = 0. Equation (51) reduces to a yield condition of σ A Sy (54) Case 2: σ A 0 σ B . Here, σ1 = σ A and σ3 = σ B , and Eq. (51) becomes σ A σ B Sy (55) Case 3: 0 σ A σ B . For this case, σ1 = 0 and σ3 = σ B , and Eq. (51) gives σ B Sy (56) Equations (54) to (56) are represented in Fig. 57 by the three lines indicated in the σ A , σ B plane. The remaining unmarked lines are cases for σ B σ A , which are not normally used. Equations (54) to (56) can also be converted to design equations by substituting equality for the equal to or greater sign and dividing Sy by n. Note that the rst part of Eq. (5-3), τmax = Sy /2n , is sufcient for design purposes provided the designer is careful in determining τmax . For plane stress, Eq. (314) does not always predict τmax . However, consider the special case when one normal stress is zero in the plane, say σx and τx y have values and σ y = 0. It can be easily shown that this is a Case 2 problem, and the shear stress determined by Eq. (314) is τmax . Shaft design problems typically fall into this category where a normal stress exists from bending and/or axial loading, and a shear stress arises from torsion. BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition II. Failure Prevention © The McGrawHill Companies, 2008 5. Failures Resulting from Static Loading Failures Resulting from Static Loading Figure 57 217 213 B Sy The maximum-shear-stress (MSS) theory for plane stress, where σ A and σ B are the two nonzero principal stresses. Case 1 Sy A Sy Case 2 Sy Case 3 55 Distortion-Energy Theory for Ductile Materials The distortion-energy theory predicts that yielding occurs when the distortion strain energy per unit volume reaches or exceeds the distortion strain energy per unit volume for yield in simple tension or compression of the same material. The distortion-energy (DE) theory originated from the observation that ductile materials stressed hydrostatically exhibited yield strengths greatly in excess of the values given by the simple tension test. Therefore it was postulated that yielding was not a simple tensile or compressive phenomenon at all, but, rather, that it was related somehow to the angular distortion of the stressed element. To develop the theory, note, in Fig. 58a, the unit volume subjected to any three-dimensional stress state designated by the stresses σ1 , σ2 , and σ3 . The stress state shown in Fig. 58b is one of hydrostatic tension due to the stresses σav acting in each of the same principal directions as in Fig. 58a. The formula for σav is simply σ1 + σ2 + σ3 3 σav = (a) Thus the element in Fig. 58b undergoes pure volume change, that is, no angular distortion. If we regard σav as a component of σ1 , σ2 , and σ3 , then this component can be 2 2 av = + 1 3 1 > 2 (a) Triaxial stresses > av 1 av av 3 av 3 (b) Hydrostatic component (c) Distortional component Figure 58 (a) Element with triaxial stresses; this element undergoes both volume change and angular distortion. (b) Element under hydrostatic tension undergoes only volume change. (c) Element has angular distortion without volume change. av 218 214 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition II. Failure Prevention © The McGrawHill Companies, 2008 5. Failures Resulting from Static Loading Mechanical Engineering Design subtracted from them, resulting in the stress state shown in Fig. 58c. This element is subjected to pure angular distortion, that is, no volume change. The strain energy per unit volume for simple tension is u = 1 ǫσ . For the element 2 of Fig. 58a the strain energy per unit volume is u = 1 [ǫ1 σ1 + ǫ2 σ2 + ǫ3 σ3 ]. 2 Substituting Eq. (319) for the principal strains gives u= 1 2 2 σ 2 + σ2 + σ3 2ν(σ1 σ2 + σ2 σ3 + σ3 σ1 ) 2E 1 (b) The strain energy for producing only volume change u v can be obtained by substituting σav for σ1 , σ2 , and σ3 in Eq. (b). The result is uv = 2 3σav (1 2ν) 2E (c) If we now substitute the square of Eq. (a) in Eq. (c) and simplify the expression, we get uv = 1 2ν 2 2 2 σ1 + σ2 + σ3 + 2σ1 σ2 + 2σ2 σ3 + 2σ3 σ1 6E (57) Then the distortion energy is obtained by subtracting Eq. (57) from Eq. (b). This gives ud = u uv = 1+ν 3E (σ1 σ2 )2 + (σ2 σ3 )2 + (σ3 σ1 )2 2 (58) Note that the distortion energy is zero if σ1 = σ2 = σ3 . For the simple tensile test, at yield, σ1 = Sy and σ2 = σ3 = 0, and from Eq. (58) the distortion energy is ud = 1+ν 2 S 3E y (59) So for the general state of stress given by Eq. (58), yield is predicted if Eq. (58) equals or exceeds Eq. (59). This gives (σ1 σ2 )2 + (σ2 σ3 )2 + (σ3 σ1 )2 2 1/2 Sy (510) If we had a simple case of tension σ , then yield would occur when σ Sy . Thus, the left of Eq. (510) can be thought of as a single, equivalent, or effective stress for the entire general state of stress given by σ1 , σ2 , and σ3 . This effective stress is usually called the von Mises stress, σ , named after Dr. R. von Mises, who contributed to the theory. Thus Eq. (510), for yield, can be written as σ Sy (511) where the von Mises stress is σ = (σ1 σ2 )2 + (σ2 σ3 )2 + (σ3 σ1 )2 2 1/2 (512) For plane stress, let σ A and σ B be the two nonzero principal stresses. Then from Eq. (512), we get 2 2 σ = σ A σ AσB + σB 1/2 (513) BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition II. Failure Prevention 219 © The McGrawHill Companies, 2008 5. Failures Resulting from Static Loading Failures Resulting from Static Loading Figure 59 215 B The distortion-energy (DE) theory for plane stress states. This is a plot of points obtained from Eq. (513) with σ = Sy . Sy Sy Sy A Pure shear load line ( Sy A B ) DE MSS Equation (513) is a rotated ellipse in the σ A , σ B plane, as shown in Fig. 59 with σ = Sy . The dotted lines in the gure represent the MSS theory, which can be seen to be more restrictive, hence, more conservative.4 Using xyz components of three-dimensional stress, the von Mises stress can be written as 1 2 2 2 σ = (σx σ y )2 + (σ y σz )2 + (σz σx )2 + 6 τx y + τ yz + τzx 2 1/2 (514) and for plane stress, 2 2 2 σ = σx σx σ y + σ y + 3τx y 1/2 (515) The distortion-energy theory is also called: The von Mises or von MisesHencky theory The shear-energy theory The octahedral-shear-stress theory Understanding octahedral shear stress will shed some light on why the MSS is conservative. Consider an isolated element in which the normal stresses on each surface are equal to the hydrostatic stress σav . There are eight surfaces symmetric to the principal directions that contain this stress. This forms an octahedron as shown in Fig. 510. The shear stresses on these surfaces are equal and are called the octahedral shear stresses (Fig. 510 has only one of the octahedral surfaces labeled). Through coordinate transformations the octahedral shear stress is given by5 τoct = 1 (σ1 σ2 )2 + (σ2 σ3 )2 + (σ3 σ1 )2 3 1/2 (516) 4 The three-dimensional equations for DE and MSS can be plotted relative to three-dimensional σ1 , σ2 , σ3 , coordinate axes. The failure surface for DE is a circular cylinder with an axis inclined at 45° from each principal stress axis, whereas the surface for MSS is a hexagon inscribed within the cylinder. See Arthur P. Boresi and Richard J. Schmidt, Advanced Mechanics of Materials, 6th ed., John Wiley & Sons, New York, 2003, Sec. 4.4. 5 For a derivation, see Arthur P. Boresi, op. cit., pp. 3637. 220 216 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition II. Failure Prevention © The McGrawHill Companies, 2008 5. Failures Resulting from Static Loading Mechanical Engineering Design Figure 510 2 Octahedral surfaces. av oct 1 3 Under the name of the octahedral-shear-stress theory, failure is assumed to occur whenever the octahedral shear stress for any stress state equals or exceeds the octahedral shear stress for the simple tension-test specimen at failure. As before, on the basis of the tensile test results, yield occurs when σ1 = Sy and σ2 = σ3 = 0. From Eq. (516) the octahedral shear stress under this condition is 2 Sy τoct = (517) 3 When, for the general stress case, Eq. (516) is equal or greater than Eq. (517), yield is predicted. This reduces to (σ1 σ2 )2 + (σ2 σ3 )2 + (σ3 σ1 )2 2 1/2 Sy (518) which is identical to Eq. (510), verifying that the maximum-octahedral-shear-stress theory is equivalent to the distortion-energy theory. The model for the MSS theory ignores the contribution of the normal stresses on the 45° surfaces of the tensile specimen. However, these stresses are P /2 A , and not the hydrostatic stresses which are P /3 A . Herein lies the difference between the MSS and DE theories. The mathematical manipulation involved in describing the DE theory might tend to obscure the real value and usefulness of the result. The equations given allow the most complicated stress situation to be represented by a single quantity, the von Mises stress, which then can be compared against the yield strength of the material through Eq. (511). This equation can be expressed as a design equation by σ = Sy n (519) The distortion-energy theory predicts no failure under hydrostatic stress and agrees well with all data for ductile behavior. Hence, it is the most widely used theory for ductile materials and is recommended for design problems unless otherwise specified. One nal note concerns the shear yield strength. Consider a case of pure shear τx y , where for plane stress σx = σ y = 0. For yield, Eq. (511) with Eq. (515) gives 2 3τx y 1/2 = Sy or Sy τx y = = 0.577 Sy 3 (520) BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition II. Failure Prevention 221 © The McGrawHill Companies, 2008 5. Failures Resulting from Static Loading 217 Failures Resulting from Static Loading Thus, the shear yield strength predicted by the distortion-energy theory is Ssy = 0.577 Sy (521) which as stated earlier, is about 15 percent greater than the 0.5 Sy predicted by the MSS theory. For pure shear, τx y the principal stresses from Eq. (313) are σ A = σ B = τx y . The load line for this case is in the third quadrant at an angle of 45o from the σ A , σ B axes shown in Fig. 59. EXAMPLE 51 A hot-rolled steel has a yield strength of Syt = Syc = 100 kpsi and a true strain at fracture of ε f = 0.55. Estimate the factor of safety for the following principal stress states: (a) 70, 70, 0 kpsi. (b) 30, 70, 0 kpsi. (c) 0, 70, 30 kpsi. (d ) 0, 30, 70 kpsi. (e) 30, 30, 30 kpsi. Solution Since ε f > 0.05 and Syc and Syt are equal, the material is ductile and the distortionenergy (DE) theory applies. The maximum-shear-stress (MSS) theory will also be applied and compared to the DE results. Note that cases a to d are plane stress states. (a) The ordered principal stresses are σ A = σ1 = 70, σ B = σ2 = 70, σ3 = 0 kpsi. DE From Eq. (513), σ = [702 70(70) + 702 ]1/2 = 70 kpsi Answer n= Sy 100 = = 1.43 σ 70 MSS Case 1, using Eq. (54) with a factor of safety, Answer n= Sy 100 = 1.43 = σA 70 (b) The ordered principal stresses are σ A = σ1 = 70, σ B = σ2 = 30, σ3 = 0 kpsi. DE Answer σ = [702 70(30) + 302 ]1/2 = 60.8 kpsi n= Sy 100 = = 1.64 σ 60.8 MSS Case 1, using Eq. (54), Answer n= Sy 100 = = 1.43 σA 70 222 218 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition II. Failure Prevention © The McGrawHill Companies, 2008 5. Failures Resulting from Static Loading Mechanical Engineering Design (c) The ordered principal stresses are σ A = σ1 = 70, σ2 = 0, σ B = σ3 = 30 kpsi. σ = [702 70(30) + (30)2 ]1/2 = 88.9 kpsi DE Answer Sy 100 = 1.13 = σ 88.9 n= MSS Case 2, using Eq. (55), Answer n= Sy 100 = 1.00 = σ A σB 70 (30) (d ) The ordered principal stresses are σ1 = 0, σ A = σ2 = 30, σ B = σ3 = 70 kpsi. DE σ = [(70)2 (70)(30) + (30)2 ]1/2 = 60.8 kpsi Answer Sy 100 = = 1.64 σ 60.8 n= MSS Case 3, using Eq. (56), Answer n= Sy 100 = = 1.43 σB 70 (e) The ordered principal stresses are σ1 = 30, σ2 = 30, σ3 = 30 kpsi DE From Eq. (512), σ = (30 30)2 + (30 30)2 + (30 30)2 2 Answer n= 1/2 = 0 kpsi Sy 100 = σ 0 MSS From Eq. (53), Answer n= Sy 100 = σ1 σ3 30 30 A tabular summary of the factors of safety is included for comparisons. (a) (b) (c) (d ) (e) DE 1.43 1.64 1.13 1.64 MSS 1.43 1.43 1.00 1.43 Since the MSS theory is on or within the boundary of the DE theory, it will always predict a factor of safety equal to or less than the DE theory, as can be seen in the table. For each case, except case (e), the coordinates and load lines in the σ A , σ B plane are shown in Fig. 511. Case (e) is not plane stress. Note that the load line for case (a) is BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition II. Failure Prevention © The McGrawHill Companies, 2008 5. Failures Resulting from Static Loading Failures Resulting from Static Loading Figure 511 223 219 B (a) Load lines for Example 51. Sy (b) B A Sy Sy A (c) Sy DE MSS Load lines (d ) the only plane stress case given in which the two theories agree, thus giving the same factor of safety. 56 Coulomb-Mohr Theory for Ductile Materials Not all materials have compressive strengths equal to their corresponding tensile values. For example, the yield strength of magnesium alloys in compression may be as little as 50 percent of their yield strength in tension. The ultimate strength of gray cast irons in compression varies from 3 to 4 times greater than the ultimate tensile strength. So, in this section, we are primarily interested in those theories that can be used to predict failure for materials whose strengths in tension and compression are not equal. Historically, the Mohr theory of failure dates to 1900, a date that is relevant to its presentation. There were no computers, just slide rules, compasses, and French curves. Graphical procedures, common then, are still useful today for visualization. The idea of Mohr is based on three simple tests: tension, compression, and shear, to yielding if the material can yield, or to rupture. It is easier to dene shear yield strength as Ssy than it is to test for it. The practical difculties aside, Mohrs hypothesis was to use the results of tensile, compressive, and torsional shear tests to construct the three circles of Fig. 512 dening a failure envelope, depicted as line ABCDE in the gure, above the σ axis. The failure envelope need not be straight. The argument amounted to the three Mohr circles describing the stress state in a body (see Fig. 312) growing during loading until one of them became tangent to the failure envelope, thereby dening failure. Was the form of the failure envelope straight, circular, or quadratic? A compass or a French curve dened the failure envelope. A variation of Mohrs theory, called the Coulomb-Mohr theory or the internal-friction theory, assumes that the boundary BCD in Fig. 512 is straight. With this assumption only the tensile and compressive strengths are necessary. Consider the conventional ordering of the principal stresses such that σ1 σ2 σ3 . The largest circle connects σ1 and σ3 , as shown in Fig. 513. The centers of the circles in Fig. 513 are C1, C2, and C3. Triangles OBiCi are similar, therefore B3 C3 B1 C1 B2 C2 B1 C1 = OC2 OC1 OC3 OC1 224 220 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition II. Failure Prevention 5. Failures Resulting from Static Loading © The McGrawHill Companies, 2008 Mechanical Engineering Design Figure 512 Three Mohr circles, one for the uniaxial compression test, one for the test in pure shear, and one for the uniaxial tension test, are used to dene failure by the Mohr hypothesis. The strengths Sc and S t are the compressive and tensile strengths, respectively; they can be used for yield or ultimate strength. A B C D E Sc Figure 513 St Coulomb-Mohr failure line Mohrs largest circle for a general state of stress. B3 B2 B1 O Sc 3 C3 C2 1 C1 St or Sc σ1 σ3 St St 2 2=2 2 σ1 + σ3 St St Sc + 2 2 2 2 Cross-multiplying and simplifying reduces this equation to σ3 σ1 =1 St Sc (522) where either yield strength or ultimate strength can be used. For plane stress, when the two nonzero principal stresses are σ A σ B , we have a situation similar to the three cases given for the MSS theory, Eqs. (54) to (56). That is, Case 1: σ A σ B 0. For this case, σ1 = σ A and σ3 = 0. Equation (522) reduces to a failure condition of σ A St (523) Case 2: σ A 0 σ B . Here, σ1 = σ A and σ3 = σ B , and Eq. (522) becomes σB σA 1 St Sc (524) Case 3: 0 σ A σ B . For this case, σ1 = 0 and σ3 = σ B , and Eq. (522) gives σ B Sc (525) BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition II. Failure Prevention © The McGrawHill Companies, 2008 5. Failures Resulting from Static Loading Failures Resulting from Static Loading Figure 514 225 221 B Plot of the Coulomb-Mohr theory of failure for plane stress states. St Sc St A Sc A plot of these cases, together with the normally unused cases corresponding to σ B σ A , is shown in Fig. 514. For design equations, incorporating the factor of safety n, divide all strengths by n. For example, Eq. (522) as a design equation can be written as σ3 1 σ1 = St Sc n (526) Since for the Coulomb-Mohr theory we do not need the torsional shear strength circle we can deduce it from Eq. (522). For pure shear τ, σ1 = σ3 = τ . The torsional yield strength occurs when τmax = Ssy . Substituting σ1 = σ3 = Ssy into Eq. (522) and simplifying gives Ssy = EXAMPLE 52 Solution Syt Syc Syt + Syc (527) A 25-mm-diameter shaft is statically torqued to 230 N · m. It is made of cast 195-T6 aluminum, with a yield strength in tension of 160 MPa and a yield strength in compression of 170 MPa. It is machined to nal diameter. Estimate the factor of safety of the shaft. The maximum shear stress is given by τ= 16T 16(230) = 3 πd π 25 103 3 = 75 106 N/m2 = 75 MPa The two nonzero principal stresses are 75 and 75 MPa, making the ordered principal stresses σ1 = 75, σ2 = 0, and σ3 = 75 MPa. From Eq. (526), for yield, Answer n= 1 1 = 1.10 = σ1 / Syt σ3 / Syc 75/160 (75)/170 Alternatively, from Eq. (527), Ssy = Syt Syc 160(170) = = 82.4 MPa Syt + Syc 160 + 170 226 222 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition II. Failure Prevention © The McGrawHill Companies, 2008 5. Failures Resulting from Static Loading Mechanical Engineering Design and τmax = 75 MPa. Thus, Answer 57 n= Ssy 82.4 = 1.10 = τmax 75 Failure of Ductile Materials Summary Having studied some of the various theories of failure, we shall now evaluate them and show how they are applied in design and analysis. In this section we limit our studies to materials and parts that are known to fail in a ductile manner. Materials that fail in a brittle manner will be considered separately because these require different failure theories. To help decide on appropriate and workable theories of failure, Marin6 collected data from many sources. Some of the data points used to select failure theories for ductile materials are shown in Fig. 515.7 Mann also collected many data for copper and nickel alloys; if shown, the data points for these would be mingled with those already diagrammed. Figure 515 shows that either the maximum-shear-stress theory or the distortion-energy theory is acceptable for design and analysis of materials that would Figure 515 2 Experimental data superposed on failure theories. (From Fig. 7.11, p. 257, Mechanical Behavior of Materials, 2nd ed., N. E. Dowling, Prentice Hall, Englewood Cliffs, N.J., 1999. Modied to show only ductile failures.) /Sc Oct. shear Yielding ( Sc = Sy ) 1.0 Ni-Cr-Mo steel AISI 1023 steel 2024-T4 Al 3S-H Al Max. shear 1.0 0 1.0 1 /Sc 1.0 6 Joseph Marin was one of the pioneers in the collection, development, and dissemination of material on the failure of engineering elements. He has published many books and papers on the subject. Here the reference used is Joseph Marin, Engineering Materials, Prentice-Hall, Englewood Cliffs, N.J., 1952. (See pp. 156 and 157 for some data points used here.) 7 Note that some data in Fig. 515 are displayed along the top horizontal boundary where σ B σ A . This is often done with failure data to thin out congested data points by plotting on the mirror image of the line σB = σ A . BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition II. Failure Prevention © The McGrawHill Companies, 2008 5. Failures Resulting from Static Loading Failures Resulting from Static Loading 227 223 fail in a ductile manner. You may wish to plot other theories using a red or blue pencil on Fig. 515 to show why they are not acceptable or are not used. The selection of one or the other of these two theories is something that you, the engineer, must decide. For design purposes the maximum-shear-stress theory is easy, quick to use, and conservative. If the problem is to learn why a part failed, then the distortion-energy theory may be the best to use; Fig. 515 shows that the plot of the distortion-energy theory passes closer to the central area of the data points, and thus is generally a better predictor of failure. For ductile materials with unequal yield strengths, Syt in tension and Syc in compression, the Mohr theory is the best available. However, the theory requires the results from three separate modes of tests, graphical construction of the failure locus, and tting the largest Mohrs circle to the failure locus. The alternative to this is to use the Coulomb-Mohr theory, which requires only the tensile and compressive yield strengths and is easily dealt with in equation form. EXAMPLE 53 This example illustrates the use of a failure theory to determine the strength of a mechanical element or component. The example may also clear up any confusion existing between the phrases strength of a machine part, strength of a material, and strength of a part at a point. A certain force F applied at D near the end of the 15-in lever shown in Fig. 516, which is quite similar to a socket wrench, results in certain stresses in the cantilevered bar OABC. This bar (OABC) is of AISI 1035 steel, forged and heat-treated so that it has a minimum (ASTM) yield strength of 81 kpsi. We presume that this component would be of no value after yielding. Thus the force F required to initiate yielding can be regarded as the strength of the component part. Find this force. y Figure 516 2 in O A 12 in z 1 1 -in D. 2 B 1 8 -in R. 2 in C 1-in D. 15 in F D x 1 1 -in D. 2 228 224 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition II. Failure Prevention © The McGrawHill Companies, 2008 5. Failures Resulting from Static Loading Mechanical Engineering Design Solution We will assume that lever DC is strong enough and hence not a part of the problem. A 1035 steel, heat-treated, will have a reduction in area of 50 percent or more and hence is a ductile material at normal temperatures. This also means that stress concentration at shoulder A need not be considered. A stress element at A on the top surface will be subjected to a tensile bending stress and a torsional stress. This point, on the 1-in-diameter section, is the weakest section, and governs the strength of the assembly. The two stresses are σx = 32 M M 32(14 F ) = = 142.6 F = 3 I /c πd π(13 ) τzx = 16T Tr 16(15 F ) = = 76.4 F = 3 J πd π(13 ) Employing the distortion-energy theory, we nd, from Eq. (515), that 2 2 σ = σx + 3τzx 1/2 1/2 = [(142.6 F )2 + 3(76.4 F )2 ] = 194.5 F Equating the von Mises stress to Sy , we solve for F and get Answer F= Sy 81 000 = = 416 lbf 194.5 194.5 In this example the strength of the material at point A is Sy = 81 kpsi. The strength of the assembly or component is F = 416 lbf. Let us see how to apply the MSS theory. For a point undergoing plane stress with only one non-zero normal stress and one shear stress, the two nonzero principal stresses σ A and σ B will have opposite signs and hence t case 2 for the MSS theory. From Eq. (313), σ A σB = 2 σx 2 2 1/2 2 + τzx 2 2 = σx + 4τzx 1/2 For case 2 of the MSS theory, Eq. (55) applies and hence 2 2 σx + 4τzx 1/2 2 = Sy [(142.6 F ) + 4(76.4 F )2 ]1/2 = 209.0 F = 81 000 F = 388 lbf which is about 7 percent less than found for the DE theory. As stated earlier, the MSS theory is more conservative than the DE theory. EXAMPLE 54 The cantilevered tube shown in Fig. 517 is to be made of 2014 aluminum alloy treated to obtain a specied minimum yield strength of 276 MPa. We wish to select a stock-size tube from Table A8 using a design factor n d = 4. The bending load is F = 1.75 kN, the axial tension is P = 9.0 kN, and the torsion is T = 72 N · m. What is the realized factor of safety? BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition II. Failure Prevention © The McGrawHill Companies, 2008 5. Failures Resulting from Static Loading Failures Resulting from Static Loading Figure 517 229 225 y 12 0m m F z P T x Solution Since the maximum bending moment is M = 120 F , the normal stress, for an element on the top surface of the tube at the origin, is σx = Mc 9 120(1.75)(do /2) 9 105do P + =+ =+ A I A I A I (1) where, if millimeters are used for the area properties, the stress is in gigapascals. The torsional stress at the same point is 72(do /2) 36do Tr = = J J J τzx = (2) For accuracy, we choose the distortion-energy theory as the design basis. The von Mises stress, as in the previous example, is 2 2 σ = σx + 3τzx 1/2 (3) On the basis of the given design factor, the goal for σ is σ Sy 0.276 = 0.0690 GPa = nd 4 (4) where we have used gigapascals in this relation to agree with Eqs. (1) and (2). Programming Eqs. (1) to (3) on a spreadsheet and entering metric sizes from Table A8 reveals that a 42- × 5-mm tube is satisfactory. The von Mises stress is found to be σ = 0.06043 GPa for this size. Thus the realized factor of safety is Answer n= Sy 0.276 = 4.57 = σ 0.06043 For the next size smaller, a 42- × 4-mm tube, σ = 0.07105 GPa giving a factor of safety of n= Sy 0.276 = 3.88 = σ 0.07105 230 226 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition II. Failure Prevention © The McGrawHill Companies, 2008 5. Failures Resulting from Static Loading Mechanical Engineering Design 58 Maximum-Normal-Stress Theory for Brittle Materials The maximum-normal-stress (MNS) theory states that failure occurs whenever one of the three principal stresses equals or exceeds the strength. Again we arrange the principal stresses for a general stress state in the ordered form σ1 σ2 σ3 . This theory then predicts that failure occurs whenever σ1 Sut or σ3 Suc (528) where Sut and Suc are the ultimate tensile and compressive strengths, respectively, given as positive quantities. For plane stress, with the principal stresses given by Eq. (313), with σ A σ B , Eq. (528) can be written as σ A Sut or σ B Suc (529) which is plotted in Fig. 518a. As before, the failure criteria equations can be converted to design equations. We can consider two sets of equations for load lines where σ A σ B as Figure 518 (a) Graph of maximum-normalstress (MNS) theory of failure for plane stress states. Stress states that plot inside the failure locus are safe. (b) Load line plot. B Sut Suc A Sut Suc (a) B Load line 1 O A Sut Load line 2 Suc Load line 4 (b) Load line 3 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition II. Failure Prevention © The McGrawHill Companies, 2008 5. Failures Resulting from Static Loading Failures Resulting from Static Loading σA = Sut n σ A σB 0 Suc n 227 Load line 1 σ A 0 σB σB = 231 and σ A 0 σB and Suc σB σA Sut Suc σB > σA Sut 0 σ A σB Load line 2 (530a) Load line 3 Load line 4 (530b) where the load lines are shown in Fig. 518b. Before we comment any further on the MNS theory we will explore some modications to the Mohr theory for brittle materials. 59 Modications of the Mohr Theory for Brittle Materials We will discuss two modications of the Mohr theory for brittle materials: the BrittleCoulomb-Mohr (BCM) theory and the modied Mohr (MM) theory. The equations provided for the theories will be restricted to plane stress and be of the design type incorporating the factor of safety. The Coulomb-Mohr theory was discussed earlier in Sec. 56 with Eqs. (523) to (525). Written as design equations for a brittle material, they are: Brittle-Coulomb-Mohr σA = Sut n σA σB 1 = Sut Suc n σB = Suc n (531a) σ A σB 0 (531b) σ A 0 σB (531c) 0 σ A σB On the basis of observed data for the fourth quadrant, the modied Mohr theory expands the fourth quadrant as shown in Fig. 519. Modied Mohr σA = Sut n σ A σB 0 σ A 0 σB σB 1 ( Suc Sut ) σ A = Suc Sut Suc n σB = Suc n and σ A 0 σB 0 σ A σB (532a) σB 1 σA and σB >1 σA (532b) (532c) Data are still outside this extended region. The straight line introduced by the modied Mohr theory, for σ A 0 σ B and |σ B /σ A | > 1, can be replaced by a parabolic relation 232 228 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition II. Failure Prevention © The McGrawHill Companies, 2008 5. Failures Resulting from Static Loading Mechanical Engineering Design Figure 519 B, Biaxial fracture data of gray cast iron compared with various failure criteria. (Dowling, N. E., Mechanical Behavior of Materials, 2/e, 1999, p. 261. Reprinted by permission of Pearson Education, Inc., Upper Saddle River, New Jersey.) MPa 300 Sut max. normal ohr d. M mo Suc 700 Cou lomb - Mo hr Sut 300 300 0 A, MPa Sut r To sio n 300 Gray cast-iron data Suc 700 which can more closely represent some of the data.8 However, this introduces a nonlinear equation for the sake of a minor correction, and will not be presented here. 8 See J. E. Shigley, C. R. Mischke, R. G. Budynas, Mechanical Engineering Design, 7th ed., McGraw-Hill, New York, 2004, p. 275. EXAMPLE 55 Consider the wrench in Ex. 53, Fig. 516, as made of cast iron, machined to dimension. The force F required to fracture this part can be regarded as the strength of the component part. If the material is ASTM grade 30 cast iron, nd the force F with (a) Coulomb-Mohr failure model. (b) Modied Mohr failure model. Solution We assume that the lever DC is strong enough, and not part of the problem. Since grade 30 cast iron is a brittle material and cast iron, the stress-concentration factors K t and K ts are set to unity. From Table A24, the tensile ultimate strength is 31 kpsi and the compressive ultimate strength is 109 kpsi. The stress element at A on the top surface will be subjected to a tensile bending stress and a torsional stress. This location, on the 1-indiameter section llet, is the weakest location, and it governs the strength of the assembly. The normal stress σx and the shear stress at A are given by σx = K t 32(14 F ) M 32 M = Kt = (1) = 142.6 F I /c π d3 π(1)3 τx y = K ts 16(15 F ) Tr 16T = K ts = (1) = 76.4 F 3 J πd π(1)3 From Eq. (313) the nonzero principal stresses σ A and σ B are σ A, σB = 142.6 F + 0 ± 2 142.6 F 0 2 2 + (76.4 F )2 = 175.8 F , 33.2 F BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition II. Failure Prevention © The McGrawHill Companies, 2008 5. Failures Resulting from Static Loading Failures Resulting from Static Loading 233 229 This puts us in the fourth-quadrant of the σ A , σ B plane. (a) For BCM, Eq. (531b) applies with n = 1 for failure. (33.2 F ) σB 175.8 F σA =1 = Sut Suc 31(103 ) 109(103 ) Solving for F yields Answer F = 167 lbf (b) For MM, the slope of the load line is |σ B /σ A | = 33.2/175.8 = 0.189 < 1. Obviously, Eq. (532a) applies. 175.8 F σA =1 = Sut 31(103 ) Answer F = 176 lbf As one would expect from inspection of Fig. 519, Coulomb-Mohr is more conservative. 510 Failure of Brittle Materials Summary We have identied failure or strength of brittle materials that conform to the usual meaning of the word brittle, relating to those materials whose true strain at fracture is 0.05 or less. We also have to be aware of normally ductile materials that for some reason may develop a brittle fracture or crack if used below the transition temperature. Figure 520 shows data for a nominal grade 30 cast iron taken under biaxial Figure 520 A plot of experimental data points obtained from tests on cast iron. Shown also are the graphs of three failure theories of possible usefulness for brittle materials. Note points A, B, C, and D. To avoid congestion in the rst quadrant, points have been plotted for σ A > σ B as well as for the opposite sense. (Source of data: Charles F. Walton (ed.), Iron Castings Handbook, Iron Founders Society, 1971, pp. 215, 216, Cleveland, Ohio.) B Modified Mohr Sut 30 Sut 120 Suc 90 60 30 30 ASTM No. 30 C.I. Sut = 31 kpsi, Suc = 109 kpsi 30 Sut A Sut B Coulomb-Mohr 60 A Maximum-normal-stress 90 B 120 A C D A B 150 Suc = 1 234 230 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition II. Failure Prevention © The McGrawHill Companies, 2008 5. Failures Resulting from Static Loading Mechanical Engineering Design stress conditions, with several brittle failure hypotheses shown, superposed. We note the following: In the rst quadrant the data appear on both sides and along the failure curves of maximum-normal-stress, Coulomb-Mohr, and modied Mohr. All failure curves are the same, and data t well. In the fourth quadrant the modied Mohr theory represents the data best. In the third quadrant the points A, B, C, and D are too few to make any suggestion concerning a fracture locus. 511 Selection of Failure Criteria For ductile behavior the preferred criterion is the distortion-energy theory, although some designers also apply the maximum-shear-stress theory because of its simplicity and conservative nature. In the rare case when Syt = Syc , the ductile Coulomb-Mohr method is employed. For brittle behavior, the original Mohr hypothesis, constructed with tensile, compression, and torsion tests, with a curved failure locus is the best hypothesis we have. However, the difculty of applying it without a computer leads engineers to choose modications, namely, Coulomb Mohr, or modied Mohr. Figure 521 provides a summary owchart for the selection of an effective procedure for analyzing or predicting failures from static loading for brittle or ductile behavior. Figure 521 Brittle behavior Failure theory selection owchart. Ductile behavior < 0.05 0.05 f No Mod. Mohr (MM) Eq. (5-32) Conservative? Yes No Yes · Syt = Syc? Brittle Coulomb-Mohr Ductile Coulomb-Mohr (BCM) (DCM) Eq. (5-31) Eq. (5-26) No Conservative? Distortion-energy (DE) Eqs. (5-15) and (5-19) Yes Maximum shear stress (MSS) Eq. (5-3) BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition II. Failure Prevention © The McGrawHill Companies, 2008 5. Failures Resulting from Static Loading Failures Resulting from Static Loading 512 235 231 Introduction to Fracture Mechanics The idea that cracks exist in parts even before service begins, and that cracks can grow during service, has led to the descriptive phrase damage-tolerant design. The focus of this philosophy is on crack growth until it becomes critical, and the part is removed from service. The analysis tool is linear elastic fracture mechanics (LEFM). Inspection and maintenance are essential in the decision to retire parts before cracks reach catastrophic size. Where human safety is concerned, periodic inspections for cracks are mandated by codes and government ordinance. We shall now briey examine some of the basic ideas and vocabulary needed for the potential of the approach to be appreciated. The intent here is to make the reader aware of the dangers associated with the sudden brittle fracture of so-called ductile materials. The topic is much too extensive to include in detail here and the reader is urged to read further on this complex subject.9 The use of elastic stress-concentration factors provides an indication of the average load required on a part for the onset of plastic deformation, or yielding; these factors are also useful for analysis of the loads on a part that will cause fatigue fracture. However, stress-concentration factors are limited to structures for which all dimensions are precisely known, particularly the radius of curvature in regions of high stress concentration. When there exists a crack, aw, inclusion, or defect of unknown small radius in a part, the elastic stress-concentration factor approaches innity as the root radius approaches zero, thus rendering the stress-concentration factor approach useless. Furthermore, even if the radius of curvature of the aw tip is known, the high local stresses there will lead to local plastic deformation surrounded by a region of elastic deformation. Elastic stress-concentration factors are no longer valid for this situation, so analysis from the point of view of stress-concentration factors does not lead to criteria useful for design when very sharp cracks are present. By combining analysis of the gross elastic changes in a structure or part that occur as a sharp brittle crack grows with measurements of the energy required to produce new fracture surfaces, it is possible to calculate the average stress (if no crack were present) that will cause crack growth in a part. Such calculation is possible only for parts with cracks for which the elastic analysis has been completed, and for materials that crack in a relatively brittle manner and for which the fracture energy has been carefully measured. The term relatively brittle is rigorously dened in the test procedures,10 but it means, roughly, fracture without yielding occurring throughout the fractured cross section. Thus glass, hard steels, strong aluminum alloys, and even low-carbon steel below the ductile-to-brittle transition temperature can be analyzed in this way. Fortunately, ductile materials blunt sharp cracks, as we have previously discovered, so that fracture occurs at average stresses of the order of the yield strength, and the designer is prepared 9 References on brittle fracture include: H. Tada and P. C. Paris, The Stress Analysis of Cracks Handbook, 2nd ed., Paris Productions, St. Louis, 1985. D. Broek, Elementary Engineering Fracture Mechanics, 4th ed., Martinus Nijhoff, London, 1985. D. Broek, The Practical Use of Fracture Mechanics, Kluwar Academic Pub., London, 1988. David K. Felbeck and Anthony G. Atkins, Strength and Fracture of Engineering Solids, Prentice-Hall, Englewood Cliffs, N.J., 1984. Kåre Hellan, Introduction to Fracture Mechanics, McGraw-Hill, New York, 1984. 10 BS 5447:1977 and ASTM E399-78. 236 232 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition II. Failure Prevention © The McGrawHill Companies, 2008 5. Failures Resulting from Static Loading Mechanical Engineering Design for this condition. The middle ground of materials that lie between relatively brittle and ductile is now being actively analyzed, but exact design criteria for these materials are not yet available. Quasi-Static Fracture Many of us have had the experience of observing brittle fracture, whether it is the breaking of a cast-iron specimen in a tensile test or the twist fracture of a piece of blackboard chalk. It happens so rapidly that we think of it as instantaneous, that is, the cross section simply parting. Fewer of us have skated on a frozen pond in the spring, with no one near us, heard a cracking noise, and stopped to observe. The noise is due to cracking. The cracks move slowly enough for us to see them run. The phenomenon is not instantaneous, since some time is necessary to feed the crack energy from the stress eld to the crack for propagation. Quantifying these things is important to understanding the phenomenon in the small. In the large, a static crack may be stable and will not propagate. Some level of loading can render the crack unstable, and the crack propagates to fracture. The foundation of fracture mechanics was rst established by Grifth in 1921 using the stress eld calculations for an elliptical aw in a plate developed by Inglis in 1913. For the innite plate loaded by an applied uniaxial stress σ in Fig. 522, the maximum stress occurs at (±a, 0) and is given by (σ y )max = 1 + 2 a σ b (533) Note that when a = b , the ellipse becomes a circle and Eq. (533) gives a stress concentration factor of 3. This agrees with the well-known result for an innite plate with a circular hole (see Table A151). For a ne crack, b/a 0, and Eq. (534) predicts that (σ y )max . However, on a microscopic level, an innitely sharp crack is a hypothetical abstraction that is physically impossible, and when plastic deformation occurs, the stress will be nite at the crack tip. Grifth showed that the crack growth occurs when the energy release rate from applied loading is greater than the rate of energy for crack growth. Crack growth can be stable or unstable. Unstable crack growth occurs when the rate of change of the energy release rate relative to the crack length is equal to or greater than the rate of change of the crack growth rate of energy. Grifths experimental work was restricted to brittle materials, namely glass, which pretty much conrmed his surface energy hypothesis. However, for ductile materials, the energy needed to perform plastic work at the crack tip is found to be much more crucial than surface energy. Figure 522 y b x a BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition II. Failure Prevention © The McGrawHill Companies, 2008 5. Failures Resulting from Static Loading Failures Resulting from Static Loading 237 233 Figure 523 Crack propagation modes. Mode I Mode II Mode III Crack Modes and the Stress Intensity Factor Three distinct modes of crack propagation exist, as shown in Fig. 523. A tensile stress eld gives rise to mode I, the opening crack propagation mode, as shown in Fig. 523a. This mode is the most common in practice. Mode II is the sliding mode, is due to in-plane shear, and can be seen in Fig. 523b. Mode III is the tearing mode, which arises from out-of-plane shear, as shown in Fig. 523c. Combinations of these modes can also occur. Since mode I is the most common and important mode, the remainder of this section will consider only this mode. Consider a mode I crack of length 2a in the innite plate of Fig. 524. By using complex stress functions, it has been shown that the stress eld on a d x dy element in the vicinity of the crack tip is given by σx = σ θ θ 3θ a 1 sin sin cos 2r 2 2 2 (534a) σy = σ θ 3θ θ a cos 1 + sin sin 2r 2 2 2 (534b) τx y = σ θ θ 3θ a sin cos cos 2r 2 2 2 (534c) 0 ν(σx + σ y ) σz = Figure 524 y Mode I crack model. dx dy r a x (for plane stress) (for plane strain) (534d) 238 234 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition II. Failure Prevention © The McGrawHill Companies, 2008 5. Failures Resulting from Static Loading Mechanical Engineering Design The stress σ y near the tip, with θ = 0, is σ y |θ =0 = σ a 2r (a) As with the elliptical crack, we see that σ y |θ =0 as r 0, and again the concept of an innite stress concentration at the crack tip is inappropriate. The quantity σ y |θ =0 2r = σ a , however, does remain constant as r 0. It is common practice to dene a factor K called the stress intensity factor given by K = σ πa (b) where the units are MPa m or kpsi in. Since we are dealing with a mode I crack, Eq. (b) is written as KI = σ πa (535) The stress intensity factor is not to be confused with the static stress concentration factors K t and K ts dened in Secs. 313 and 52. Thus Eqs. (534) can be rewritten as 3θ KI θ θ 1 sin sin cos σx = 2 2 2 2π r (536a) θ θ 3θ KI cos 1 + sin sin σy = 2 2 2 2π r (536b) θ θ 3θ KI sin cos cos τx y = 2 2 2 2π r (536c) σz = 0 ν(σx + σ y ) (for plane stress) (for plane strain) (536d) The stress intensity factor is a function of geometry, size and shape of the crack, and the type of loading. For various load and geometric congurations, Eq. (535) can be written as K I = βσ π a (537) where β is the stress intensity modication factor. Tables for β are available in the literature for basic congurations.11 Figures 525 to 530 present a few examples of β for mode I crack propagation. Fracture Toughness When the magnitude of the mode I stress intensity factor reaches a critical value, K I c crack propagation initiates. The critical stress intensity factor K I c is a material property that depends on the material, crack mode, processing of the material, temperature, 11 See, for example: H. Tada and P. C. Paris, The Stress Analysis of Cracks Handbook, 2nd ed., Paris Productions, St. Louis, 1985. G. C. Sib, Handbook of Stress Intensity Factors for Researchers and Engineers, Institute of Fracture and Solid Mechanics, Lehigh University, Bethlehem, Pa., 1973. Y. Murakami, ed., Stress Intensity Factors Handbook, Pergamon Press, Oxford, U.K., 1987. W. D. Pilkey, Formulas for Stress, Strain, and Structural Matrices, 2nd ed. John Wiley& Sons, New York, 2005. BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition II. Failure Prevention Figure 525 Off-center crack in a plate in longitudinal tension; solid curves are for the crack tip at A; dashed curves are for the tip at B. © The McGrawHill Companies, 2008 5. Failures Resulting from Static Loading 2.2 239 AA 2.0 2a A A B d 1.8 2b 1.6 0.4 1.4 d b = 1.0 B 0.2 B 0.4 1.2 0.2 1.0 Figure 526 Plate loaded in longitudinal tension with a crack at the edge; for the solid curve there are no constraints to bending; the dashed curve was obtained with bending constraints added. 0 0.2 0.4 a d ratio 0.6 0.8 0.6 0.8 7.0 6.0 h a b h 5.0 4.0 3.0 h b = 0.5 1.0 2.0 1.0 0 0.2 0.4 a b ratio 235 240 236 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition II. Failure Prevention © The McGrawHill Companies, 2008 5. Failures Resulting from Static Loading Mechanical Engineering Design Figure 527 Beams of rectangular cross section having an edge crack. 2.0 h a M M F 1.8 h a F 2 F 2 l l 1.6 Pure bending 1.4 l =4 h 1.2 l =2 h 1.0 0 Figure 528 0.2 0.4 a h ratio 0.6 0.8 3 Plate in tension containing a circular hole with two cracks. 2a r = 0.5 b 2 r 2b r = 0.25 b 1r =0 b 0 0 0.2 0.4 a b ratio 0.6 0.8 loading rate, and the state of stress at the crack site (such as plane stress versus plane strain). The critical stress intensity factor K I c is also called the fracture toughness of the material. The fracture toughness for plane strain is normally lower than that for plane stress. For this reason, the term K I c is typically dened as the mode I, plane strain fracture toughness. Fracture toughness K I c for engineering metals lies in the range 20 K I 200 MPa · m; for engineering polymers and ceramics, 1 K I c c 5 MPa · m. For a 4340 steel, where the yield strength due heat treatment ranges to from 800 to 1600 MPa, K I c decreases from 190 to 40 MPa · m. BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition II. Failure Prevention © The McGrawHill Companies, 2008 5. Failures Resulting from Static Loading Failures Resulting from Static Loading Figure 529 A cylinder loading in axial tension having a radial crack of depth a extending completely around the circumference of the cylinder. 241 237 4.0 ri ro = 0 a a 3.0 0.1 0.4 2.0 1.0 Figure 530 Cylinder subjected to internal pressure p, having a radial crack in the longitudinal direction of depth a. Use Eq. (451) for the tangential stress at r = r 0 . 0 0.8 ro ri 0.2 0.4 a (ro ri ) ratio 0.6 0.8 0.6 0.8 3.4 a 3.0 pi ri ro 2.6 2.2 1.8 ri ro = 0.9 0.75 0.35 1.4 1.0 0 0.2 0.4 a (ro ri ) ratio Table 51 gives some approximate typical room-temperature values of K I c for several materials. As previously noted, the fracture toughness depends on many factors and the table is meant only to convey some typical magnitudes of K I c . For an actual application, it is recommended that the material specied for the application be certied using standard test procedures [see the American Society for Testing and Materials (ASTM) standard E399]. 242 238 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition II. Failure Prevention © The McGrawHill Companies, 2008 5. Failures Resulting from Static Loading Mechanical Engineering Design Table 51 K Ic, MPa m Material Values of KIc for Some Engineering Materials at Room Temperature Aluminum 2024 7075 7178 Sy, MPa 26 24 33 115 55 52100 860 1515 14 Steel 4340 4340 910 1035 99 60 Titanium Ti-6AL-4V Ti-6AL-4V 455 495 490 2070 One of the rst problems facing the designer is that of deciding whether the conditions exist, or not, for a brittle fracture. Low-temperature operation, that is, operation below room temperature, is a key indicator that brittle fracture is a possible failure mode. Tables of transition temperatures for various materials have not been published, possibly because of the wide variation in values, even for a single material. Thus, in many situations, laboratory testing may give the only clue to the possibility of a brittle fracture. Another key indicator of the possibility of fracture is the ratio of the yield strength to the ultimate strength. A high ratio of Sy / Su indicates there is only a small ability to absorb energy in the plastic region and hence there is a likelihood of brittle fracture. The strength-to-stress ratio K I c / K I can be used as a factor of safety as n= KIc KI (538) EXAMPLE 56 A steel ship deck plate is 30 mm thick and 12 m wide. It is loaded with a nominal uniaxial tensile stress of 50 MPa. It is operated below its ductile-to-brittle transition temperature with K I c equal to 28.3 MPa. If a 65-mm-long central transverse crack is present, estimate the tensile stress at which catastrophic failure will occur. Compare this stress with the yield strength of 240 MPa for this steel. Solution For Fig. 525, with d = b, 2a = 65 mm and 2b = 12 m, so that d /b = 1 and a /d = 65/12(103 ) = 0.00542. Since a/d is so small, β = 1, so that K I = σ π a = 50 π (32.5 × 103 ) = 16.0 MPa m From Eq. (538), n= KIc 28.3 = 1.77 = KI 16.0 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition II. Failure Prevention © The McGrawHill Companies, 2008 5. Failures Resulting from Static Loading Failures Resulting from Static Loading 243 239 The stress at which catastrophic failure occurs is Answer σc = KIc 28.3 (50) = 88.4 MPa σ= KI 16.0 The yield strength is 240 MPa, and catastrophic failure occurs at 88.4/240 = 0.37, or at 37 percent of yield. The factor of safety in this circumstance is K I c / K I = 28.3/16 = 1.77 and not 240/50 = 4.8. EXAMPLE 57 A plate of width 1.4 m and length 2.8 m is required to support a tensile force in the 2.8-m direction of 4.0 MN. Inspection procedures will detect only through-thickness edge cracks larger than 2.7 mm. The two Ti-6AL-4V alloys in Table 51 are being considered for this application, for which the safety factor must be 1.3 and minimum weight is important. Which alloy should be used? Solution (a) We elect rst to estimate the thickness required to resist yielding. Since σ = P /wt , we have t = P /wσ. For the weaker alloy, we have, from Table 51, Sy = 910 MPa. Thus, σall = Sy 910 = = 700 MPa n 1.3 Thus t= 4.0(10)3 P = = 4.08 mm or greater wσall 1.4(700) For the stronger alloy, we have, from Table 51, σall = 1035 = 796 MPa 1.3 and so the thickness is Answer t= P 4.0(10)3 = 3.59 mm or greater = wσall 1.4(796) (b) Now let us nd the thickness required to prevent crack growth. Using Fig. 526, we have h 2.8/2 = =1 b 1.4 a 2.7 = = 0.001 93 b 1.4(103 ) . Corresponding to these ratios we nd from Fig. 526 that β = 1.1, and K I = 1.1σ π a . KIc 115 103 KIc n= = σ= , KI 1.1σ π a 1.1n π a 244 240 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition II. Failure Prevention © The McGrawHill Companies, 2008 5. Failures Resulting from Static Loading Mechanical Engineering Design From Table 51, K I c = 115 MPa m for the weaker of the two alloys. Solving for σ with n = 1 gives the fracture stress σ= 115 1.1 π (2.7 × 103 ) = 1135 MPa which is greater than the yield strength of 910 MPa, and so yield strength is the basis for the geometry decision. For the stronger alloy Sy = 1035 MPa, with n = 1 the fracture stress is σ= 55 KIc = 542.9 MPa = nKI 1(1.1) π (2.7 × 103 ) which is less than the yield strength of 1035 MPa. The thickness t is t= P 4.0(103 ) = 6.84 mm or greater = wσall 1.4(542.9/1.3) This example shows that the fracture toughness K I c limits the geometry when the stronger alloy is used, and so a thickness of 6.84 mm or larger is required. When the weaker alloy is used the geometry is limited by the yield strength, giving a thickness of only 4.08 mm or greater. Thus the weaker alloy leads to a thinner and lighter weight choice since the failure modes differ. 513 Stochastic Analysis12 Reliability is the probability that machine systems and components will perform their intended function satisfactorily without failure. Up to this point, discussion in this chapter has been restricted to deterministic relations between static stress, strength, and the design factor. Stress and strength, however, are statistical in nature and very much tied to the reliability of the stressed component. Consider the probability density functions for stress and strength, and S, shown in Fig. 531a. The mean values of stress and strength are µσ and µ S , respectively. Here, the average factor of safety is n= ¯ µS µσ (a) The margin of safety for any value of stress σ and strength S is dened as m = Sσ (b) ¯ The average part will have a margin of safety of m = µ S µσ . However, for the overlap of the distributions shown by the shaded area in Fig. 531a, the stress exceeds the strength, the margin of safety is negative, and these parts are expected to fail. This shaded area is called the interference of and S. Figure 531b shows the distribution of m, which obviously depends on the distributions of stress and strength. The reliability that a part will perform without failure, R, is the area of the margin of safety distribution for m > 0. The interference is the area 12 Review Chap. 20 before reading this section. BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition II. Failure Prevention © The McGrawHill Companies, 2008 5. Failures Resulting from Static Loading Failures Resulting from Static Loading 245 241 S s Stress (a) m f (m) Plot of density functions showing how the interference of S and is used to obtain the stress margin m. (a) Stress and strength distributions. (b) Distribution of interference; the reliability R is the area of the density function for m greater than zero; the interference is the area (1 R). f (s), f ( ) Figure 531 (1 R) R + 0 m Stress margin (b) 1 R where parts are expected to fail. We next consider some typical cases involving stress-strength interference. Normal-Normal Case ˆ ˆ Consider the normal distributions, S = N(µ S , σ S ) and = N(µσ , σσ ) . The stress margin is m = S , and will be normally distributed because the addition or subˆ traction of normals is normal. Thus m = N(µm , σm ) . Reliability is the probability p that m > 0. That is, (539) R = p( S > σ ) = p( S σ > 0) = p(m > 0) To nd the chance that m > 0 we form the z variable of m and substitute m = 0 [See ˆ2 ˆ2 ˆ Eq. (2016)]. Noting that µm = µ S µσ and σm = (σ S + σσ )1/2 , we write z= 0 µm µm µ S µσ m µm = = = 1/2 σm ˆ σm ˆ σm ˆ σ S + σσ ˆ2 ˆ2 (540) Equation (540) is called the normal coupling equation. The reliability associated with z is given by R= x u2 1 du = 1 F = 1 exp 2 2π (z ) (541) The body of Table A10 gives R when z > 0 and (1 R = F ) when z 0. Noting that n = µ S /µσ , square both sides of Eq. (540), and introduce C S and Cσ where ¯ ¯ Cs = σs /µs and Cσ = σσ /µσ . Solve the resulting quadratic for n to obtain ˆ ˆ n= ¯ 1± 2 1 1 z2C S 2 1 z2C S 2 1 z 2 Cσ The plus sign is associated with R > 0.5, and the minus sign with R < 0.5. (542) 246 242 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition II. Failure Prevention © The McGrawHill Companies, 2008 5. Failures Resulting from Static Loading Mechanical Engineering Design LognormalLognormal Case ˆ ˆ Consider the lognormal distributions S = LN(µ S , σ S ) and = LN(µσ , σσ ). If we interfere their companion normals using Eqs. (2018) and (2019), we obtain 2 µln S = ln µ S ln 1 + C S σln S = ˆ (strength) 2 ln 1 + C S and 2 µln σ = ln µσ ln 1 + Cσ σln σ = ˆ (stress) 2 ln 1 + Cσ Using Eq. (540) for interfering normal distributions gives z= µln S µln σ ˆ2 σln S + σln σ ˆ2 1/2 = 2 1 + Cσ 2 1 + CS µS µσ ln 2 ln 1 + C S (543) 2 1 + Cσ The reliability R is expressed by Eq. (541). The design factor n is the random variable that is the quotient of S/ . The quotient of lognormals is lognormal, so pursuing the z variable of the lognormal n, we note µn = µS µσ Cn = 2 2 C S + Cσ 2 1 + Cσ σn = Cn µn ˆ ˆ The companion normal to n = LN(µn , σn ) , from Eqs. (2018) and (2019), has a mean and standard deviation of 2 µ y = ln µn ln 1 + Cn σy = ˆ 2 ln 1 + Cn The z variable for the companion normal y distribution is z= y µy σy ˆ ¯ Failure will occur when the stress is greater than the strength, when n < 1, or when y < 0. z= 2 2 ln µn / 1 + Cn ln µn ln 1 + Cn µy 0 µy = = = ˙ σy ˆ σy 2 2 ln 1 + Cn ln 1 + Cn (544) Solving for µn gives Cn . 2 2 µn = n = exp z ln 1 + Cn + ln 1 + Cn = exp Cn z + ¯ 2 (545) BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition II. Failure Prevention 247 © The McGrawHill Companies, 2008 5. Failures Resulting from Static Loading Failures Resulting from Static Loading 243 Equations (542) and (545) are remarkable for several reasons: ¯ They relate design factor n to the reliability goal (through z) and the coefcients of variation of strength and stress. They are not functions of the means of stress and strength. They estimate the design factor necessary to achieve the reliability goal before decisions involving means are made. The C S depends slightly on the particular material. The Cσ has the coefcient of variation (COV) of the load, and that is generally given. EXAMPLE 58 Solution A round cold-drawn 1018 steel rod has an 0.2 percent yield strength S y = N(78.4, 5.90) kpsi and is to be subjected to a static axial load of P = N(50, 4.1) kip. What value of ¯ the design factor n corresponds to a reliability of 0.999 against yielding (z = 3.09)? Determine the corresponding diameter of the rod. C S = 5.90/78.4 = 0.0753 , and P 4P = A π d2 Since the COV of the diameter is an order of magnitude less than the COV of the load or strength, the diameter is treated deterministically: 4.1 Cσ = C P = = 0.082 50 From Eq. (542), = 1 1 [1 ( 3.09) (0.0753 )][1 2 ( 3.09) (0.082 )] 2 2 2 n 1 2 2 ( 3.09) (0.0753 ) The diameter is found deterministically: Answer d= Check ¯ 4P = ¯¯ π Sy /n 4(50 000) = 1.072 in π(78 400)/1.416 S y = N(78.4, 5.90) kpsi, P = N(50, 4.1) kip, and d = 1.072 in. Then π d2 π(1.0722 ) = = 0.9026 in2 4 4 ¯ (50 000) P = = 55 400 psi σ= ¯ A 0.9026 4.1 = 0.082 C P = Cσ = 50 A= σσ = Cσ σ = 0.082(55 400) = 4540 psi ˆ ¯ From Eq. (540) σ S = 5.90 kpsi ˆ 78.4 55.4 = 3.09 (5.902 + 4.542 )1/2 From Appendix Table A10, R = (3.09) = 0.999. z= 1.416 248 244 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition II. Failure Prevention © The McGrawHill Companies, 2008 5. Failures Resulting from Static Loading Mechanical Engineering Design EXAMPLE 59 Solution Rework Ex. 58 with lognormally distributed stress and strength. C S = 5.90/78.4 = 0.0753, and Cσ = C P = 4.1/50 = 0.082. Then P 4P = A π d2 = 2 2 C S + Cσ = 2 1 + Cσ Cn = 0.07532 + 0.0822 = 0.1110 1 + 0.0822 From Table A10, z = 3.09. From Eq. (545), n = exp (3.09) ln(1 + 0.1112 ) + ln 1 + 0.1112 = 1.416 ¯ d= Check ¯ 4P = ¯¯ π Sy /n 4(50 000) = 1.0723 in π(78 400)/1.416 S y = LN(78.4, 5.90), P = LN (50, 4.1) kip. Then π d2 π(1.07232 ) = = 0.9031 4 4 ¯ 50 000 P = = 55 365 psi σ= ¯ A 0.9031 4.1 = 0.082 Cσ = C P = 50 A= σσ = Cσ µσ = 0.082(55 367) = 4540 psi ˆ From Eq. (543), z= ln 78.4 55.365 2 1 + 0.082 1 + 0.07532 ln[(1 + 0.07532 )(1 + 0.0822 )] = 3.1343 Appendix Table A10 gives R = 0.99950. InterferenceGeneral In the previous segments, we employed interference theory to estimate reliability when the distributions are both normal and when they are both lognormal. Sometimes, however, it turns out that the strength has, say, a Weibull distribution while the stress is distributed lognormally. In fact, stresses are quite likely to have a lognormal distribution, because the multiplication of variates that are normally distributed produces a result that approaches lognormal. What all this means is that we must expect to encounter interference problems involving mixed distributions and we need a general method to handle the problem. It is quite likely that we will use interference theory for problems involving distributions other than strength and stress. For this reason we employ the subscript 1 to BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition II. Failure Prevention © The McGrawHill Companies, 2008 5. Failures Resulting from Static Loading Failures Resulting from Static Loading Figure 532 249 245 f1(S ) (a) PDF of the strength distribution; (b) PDF of the load-induced stress distribution. dF1(x) = f1(x) dx S dx (a) x Cursor f2( ) F2(x) R2(x) (b) designate the strength distribution and the subscript 2 to designate the stress distribution. Figure 532 shows these two distributions aligned so that a single cursor x can be used to identify points on both distributions. We can now write Probability that stress is less than strength = dp(σ < x ) = d R = F2 (x ) d F1 (x ) By substituting 1 R2 for F2 and d R1 for d F1 , we have d R = [1 R2 (x )] d R1 (x ) The reliability for all possible locations of the cursor is obtained by integrating x from to ; but this corresponds to an integration from 1 to 0 on the reliability R1 . Therefore 0 R= 1 [1 R2 (x )] d R1 (x ) which can be written 1 R2 d R1 R =1 (546) f1( S) d S (547) f 2 (σ ) d σ (548) 0 where R1 ( x ) = R2 ( x ) = x x 250 246 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition II. Failure Prevention © The McGrawHill Companies, 2008 5. Failures Resulting from Static Loading Mechanical Engineering Design 1 1 R2 R2 R1 R1 1 (a) 1 (b) Figure 533 Curve shapes of the R 1 R 2 plot. In each case the shaded area is equal to 1 R and is obtained by numerical integration. (a) Typical curve for asymptotic distributions; (b) curve shape obtained from lower truncated distributions such as the Weibull. For the usual distributions encountered, plots of R1 versus R2 appear as shown in Fig. 533. Both of the cases shown are amenable to numerical integration and computer solution. When the reliability is high, the bulk of the integration area is under the right-hand spike of Fig. 533a. 514 Important Design Equations The following equations and their locations are provided as a summary. Maximum Shear Theory τmax = p. 212 Sy σ1 σ3 = 2 2n (53) Distortion-Energy Theory Von Mises stress, p. 214 σ = (σ1 σ2 )2 + (σ2 σ3 )2 + (σ3 σ1 )2 2 1/2 1 2 2 2 p. 215 σ = (σx σ y )2 + (σ y σz )2 + (σz σx )2 + 6(τx y + τ yz + τzx ) 2 (512) 1/2 (514) Plane stress, p. 214 2 2 σ = (σ A σ A σ B + σ B )1/2 p. 215 2 2 2 σ = (σx σx σ y + σ y + 3τx y )1/2 (513) (515) Yield design equation, p. 216 σ = Sy n (519) Shear yield strength, p. 217 Ssy = 0.577 Sy (521) BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition II. Failure Prevention 251 © The McGrawHill Companies, 2008 5. Failures Resulting from Static Loading Failures Resulting from Static Loading 247 Coulomb-Mohr Theory σ1 σ3 1 = St Sc n p. 221 (526) where St is tensile yield (ductile) or ultimate tensile (brittle), and St is compressive yield (ductile) or ultimate compressive (brittle) strengths. Maximum-Normal-Stress Theory σ1 = p. 226 Sut n σ3 = or Suc n (530) Modied Mohr (Plane Stress) Use maximum-normal-stress equations, or p. 227 σB 1 ( Suc Sut )σ A = Suc Sut Suc n σ A 0 σB and σB >1 σA (532b) Failure Theory Flowchart Fig. 521, p. 230 Brittle behavior Ductile behavior < 0.05 0.05 f No Mod. Mohr (MM) Eq. (5-32) Conservative? Yes No Yes · Syt = Syc? Brittle Coulomb-Mohr Ductile Coulomb-Mohr (BCM) (DCM) Eq. (5-31) Eq. (5-26) No Conservative? Distortion-energy (DE) Eqs. (5-15) and (5-19) Yes Maximum shear stress (MSS) Eq. (5-3) Fracture Mechanics p. 234 K I = βσ π a where β is found in Figs. 525 to 530 (pp. 235 to 237) (537) 252 248 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition II. Failure Prevention © The McGrawHill Companies, 2008 5. Failures Resulting from Static Loading Mechanical Engineering Design n= p. 238 KIc KI (538) where K I c is found in Table 51 (p. 238) Stochastic Analysis ¯ Mean factor of safety dened as n = µ S /µσ (µ S and µσ are mean strength and stress, respectively) Normal-Normal Case n= p. 241 1± 2 2 1 (1 z 2 Cs )(1 z 2 Cσ ) 2C 2 1z s (542) ˆ ˆ where z can be found in Table A10, C S = σ S /µ S , and Cσ = σσ /µσ . Lognormal-Lognormal Case p. 242 Cn . 2 2 n = exp z ln(1 + Cn ) + ln 1 + Cn = exp Cn z + 2 (545) where Cn = 2 2 C S + Cσ 2 1 + Cσ (See other denitions in normal-normal case.) PROBLEMS 51 A ductile hot-rolled steel bar has a minimum yield strength in tension and compression of 50 kpsi. Using the distortion-energy and maximum-shear-stress theories determine the factors of safety for the following plane stress states: (a) σx = 12 kpsi, σ y = 6 kpsi (b) σx = 12 kpsi, τx y = 8 kpsi (c) σx = 6 kpsi, σ y = 10 kpsi, τx y = 5 kpsi (d ) σx = 12 kpsi, σ y = 4 kpsi, τx y = 1 kpsi 52 Repeat Prob. 51 for: (a) σ A = 12 kpsi, σ B = 12 kpsi (b) σ A = 12 kpsi, σ B = 6 kpsi (c) σ A = 12 kpsi, σ B = 12 kpsi (d ) σ A = 6 kpsi, σ B = 12 kpsi 53 Repeat Prob. 51 for a bar of AISI 1020 cold-drawn steel and: (a) σx = 180 MPa, σ y = 100 MPa (b) σx = 180 MPa, τx y = 100 MPa (c) σx = 160 MPa, τx y = 100 MPa (d ) τx y = 150 MPa 54 Repeat Prob. 51 for a bar of AISI 1018 hot-rolled steel and: (a) σ A = 100 MPa, σ B = 80 MPa (b) σ A = 100 MPa, σ B = 10 MPa (c) σ A = 100 MPa, σ B = 80 MPa (d ) σ A = 80 MPa, σ B = 100 MPa BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition II. Failure Prevention 5. Failures Resulting from Static Loading © The McGrawHill Companies, 2008 Failures Resulting from Static Loading 253 249 55 Repeat Prob. 53 by rst plotting the failure loci in the σ A , σ B plane to scale; then, for each stress state, plot the load line and by graphical measurement estimate the factors of safety. 56 Repeat Prob. 54 by rst plotting the failure loci in the σ A , σ B plane to scale; then, for each stress state, plot the load line and by graphical measurement estimate the factors of safety. 57 An ASTM cast iron has minimum ultimate strengths of 30 kpsi in tension and 100 kpsi in compression. Find the factors of safety using the MNS, BCM, and MM theories for each of the following stress states. Plot the failure diagrams in the σ A , σ B plane to scale and locate the coordinates of each stress state. (a) σx = 20 kpsi, σ y = 6 kpsi (b) σx = 12 kpsi, τx y = 8 kpsi (c) σx = 6 kpsi, σ y = 10 kpsi, τx y = 5 kpsi (d ) σx = 12 kpsi, τx y = 8 kpsi 58 For Prob. 57, case (d ), estimate the factors of safety from the three theories by graphical measurements of the load line. 59 Among the decisions a designer must make is selection of the failure criteria that is applicable to the material and its static loading. A 1020 hot-rolled steel has the following properties: Sy = 42 kpsi, Sut = 66.2 kpsi, and true strain at fracture ε f = 0.90. Plot the failure locus and, for the static stress states at the critical locations listed below, plot the load line and estimate the factor of safety analytically and graphically. (a) σx = 9 kpsi, σ y = 5 kpsi. (b) σx = 12 kpsi, τx y = 3 kpsi ccw. (c) σx = 4 kpsi, σ y = 9 kpsi, τx y = 5 kpsi cw. (d) σx = 11 kpsi, σ y = 4 kpsi, τx y = 1 kpsi cw. 510 A 4142 steel Q&T at 80 F exhibits Syt = 235 kpsi, Syc = 275 kpsi, and ε f = 0.06. Choose and plot the failure locus and, for the static stresses at the critical locations, which are 10 times those in Prob. 59, plot the load lines and estimate the factors of safety analytically and graphically. 511 For grade 20 cast iron, Table A24 gives Sut = 22 kpsi, Suc = 83 kpsi. Choose and plot the failure locus and, for the static loadings inducing the stresses at the critical locations of Prob. 59, plot the load lines and estimate the factors of safety analytically and graphically. 512 A cast aluminum 195-T6 has an ultimate strength in tension of Sut = 36 kpsi and ultimate strength in compression of Suc = 35 kpsi, and it exhibits a true strain at fracture ε f = 0.045. Choose and plot the failure locus and, for the static loading inducing the stresses at the critical locations of Prob. 59, plot the load lines and estimate the factors of safety analytically and graphically. 513 An ASTM cast iron, grade 30 (see Table A24), carries static loading resulting in the stress state listed below at the critical locations. Choose the appropriate failure locus, plot it and the load lines, and estimate the factors of safety analytically and graphically. (a) σ A = 20 kpsi, σ B = 20 kpsi. (b) τx y = 15 kpsi. (c) σ A = σ B = 80 kpsi. (d ) σ A = 15 kpsi, σ B = 25 kpsi. 514 This problem illustrates that the factor of safety for a machine element depends on the particular point selected for analysis. Here you are to compute factors of safety, based upon the distortion-energy theory, for stress elements at A and B of the member shown in the gure. This bar is made of AISI 1006 cold-drawn steel and is loaded by the forces F = 0.55 kN, P = 8.0 kN, and T = 30 N · m. 254 250 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition II. Failure Prevention © The McGrawHill Companies, 2008 5. Failures Resulting from Static Loading Mechanical Engineering Design y 10 0m m A B Problem 514 F z 20-mm D. P T x 515 The gure shows a crank loaded by a force F = 190 lbf which causes twisting and bending of the 3 -in-diameter shaft xed to a support at the origin of the reference system. In actuality, the 4 support may be an inertia which we wish to rotate, but for the purposes of a strength analysis we can consider this to be a statics problem. The material of the shaft AB is hot-rolled AISI 1018 steel (Table A20). Using the maximum-shear-stress theory, nd the factor of safety based on the stress at point A. y 1 in F C A 3 -in 4 1 -in 2 dia. 1 4 Problem 515 B in 1 14 dia. in z 4 in 5 in x 516 517* 518 Solve Prob. 515 using the distortion energy theory. If you have solved Prob. 515, compare the results and discuss the difference. Design the lever arm CD of Fig. 516 by specifying a suitable size and material. A spherical pressure vessel is formed of 18-gauge (0.05-in) cold-drawn AISI 1018 sheet steel. If the vessel has a diameter of 8 in, estimate the pressure necessary to initiate yielding. What is the estimated bursting pressure? *The asterisk indicates a problem that may not have a unique result or may be a particularly challenging problem. BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition II. Failure Prevention © The McGrawHill Companies, 2008 5. Failures Resulting from Static Loading Failures Resulting from Static Loading 255 251 519 This problem illustrates that the strength of a machine part can sometimes be measured in units other than those of force or moment. For example, the maximum speed that a ywheel can reach without yielding or fracturing is a measure of its strength. In this problem you have a rotating ring made of hot-forged AISI 1020 steel; the ring has a 6-in inside diameter and a 10-in outside diameter and is 1.5 in thick. What speed in revolutions per minute would cause the ring to yield? At what radius would yielding begin? [Note: The maximum radial stress occurs at r = (ro ri )1/2 ; see Eq. (355).] 520 A light pressure vessel is made of 2024-T3 aluminum alloy tubing with suitable end closures. 1 This cylinder has a 3 2 -in OD, a 0.065-in wall thickness, and ν = 0.334. The purchase order species a minimum yield strength of 46 kpsi. What is the factor of safety if the pressure-release valve is set at 500 psi? 521 A cold-drawn AISI 1015 steel tube is 300 mm OD by 200 mm ID and is to be subjected to an external pressure caused by a shrink t. What maximum pressure would cause the material of the tube to yield? 522 What speed would cause fracture of the ring of Prob. 519 if it were made of grade 30 cast iron? 523 The gure shows a shaft mounted in bearings at A and D and having pulleys at B and C. The forces shown acting on the pulley surfaces represent the belt tensions. The shaft is to be made of ASTM grade 25 cast iron using a design factor n d = 2.8. What diameter should be used for the shaft? x 6-in D. 300 lbf 50 lbf y 27 lbf Problem 523 8-in D. z A B 360 lbf D C 6 in 8 in 8 in 524 By modern standards, the shaft design of Prob. 523 is poor because it is so long. Suppose it is redesigned by halving the length dimensions. Using the same material and design factor as in Prob. 523, nd the new shaft diameter. 525 The gear forces shown act in planes parallel to the yz plane. The force on gear A is 300 lbf. Consider the bearings at O and B to be simple supports. For a static analysis and a factor of safety of 3.5, use distortion energy to determine the minimum safe diameter of the shaft. Consider the material to have a yield strength of 60 kpsi. 526 Repeat Prob. 525 using maximum-shear-stress. 527 The gure is a schematic drawing of a countershaft that supports two V-belt pulleys. For each pulley, the belt tensions are parallel. For pulley A consider the loose belt tension is 15 percent of the tension on the tight side. A cold-drawn UNS G10180 steel shaft of uniform diameter is to be selected for this application. For a static analysis with a factor of safety of 3.0, determine the minimum preferred size diameter. Use the distortion-energy theory. 256 252 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition II. Failure Prevention © The McGrawHill Companies, 2008 5. Failures Resulting from Static Loading Mechanical Engineering Design y 20 in O 16 in FC 10 in Problem 525 20° z Gear A 24-in D. B A C FA Gear C 10-in D. x 20° y 300 45° O 400 T2 Problem 527 T1 z 150 Dimensions in millimeters 250 Dia. 300 Dia. A 50 N B C x 270 N 528 Repeat Prob. 527 using maximum shear stress. 529 The clevis pin shown in the gure is 12 mm in diameter and has the dimensions a = 12 mm and b = 18 mm. The pin is machined from AISI 1018 hot-rolled steel (Table A20) and is to be loaded to no more than 4.4 kN. Determine whether or not the assumed loading of gure c yields a factor of safety any different from that of gure d. Use the maximum-shear-stress theory. 530 Repeat Prob. 529, but this time use the distortion-energy theory. 531 1 A split-ring clamp-type shaft collar is shown in the gure. The collar is 2 in OD by 1 in ID by 2 1 in wide. The screw is designated as 4 -28 UNF. The relation between the screw tightening torque T, the nominal screw diameter d, and the tension in the screw Fi is approximately T = 0.2 Fi d . The shaft is sized to obtain a close running t. Find the axial holding force Fx of the collar as a function of the coefcient of friction and the screw torque. BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition II. Failure Prevention © The McGrawHill Companies, 2008 5. Failures Resulting from Static Loading Failures Resulting from Static Loading 257 253 F (b) Problem 529 b 2 d a+b a a (c) F b b (a) a+b (d ) A Problem 531 532 Suppose the collar of Prob. 531 is tightened by using a screw torque of 190 lbf · in. The collar material is AISI 1040 steel heat-treated to a minimum tensile yield strength of 63 kpsi. (a) Estimate the tension in the screw. (b) By relating the tangential stress to the hoop tension, nd the internal pressure of the shaft on the ring. (c) Find the tangential and radial stresses in the ring at the inner surface. (d ) Determine the maximum shear stress and the von Mises stress. (e) What are the factors of safety based on the maximum-shear-stress hypothesis and the distortionenergy theory? 533 In Prob. 531, the role of the screw was to induce the hoop tension that produces the clamping. The screw should be placed so that no moment is induced in the ring. Just where should the screw be located? 534 A tube has another tube shrunk over it. The specications are: Inner Member ID OD Outer Member 1.000 ± 0.002 in 2.000 ± 0.0004 in 1.999 ± 0.0004 in 3.000 ± 0.004 in Both tubes are made of a plain carbon steel. (a) Find the nominal shrink-t pressure and the von Mises stresses at the t surface. (b) If the inner tube is changed to solid shafting with the same outside dimensions, nd the nominal shrink-t pressure and the von Mises stresses at the t surface. 258 254 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition II. Failure Prevention © The McGrawHill Companies, 2008 5. Failures Resulting from Static Loading Mechanical Engineering Design 535 Steel tubes with a Youngs modulus of 207 GPa have the specications: Inner Tube Outer Tube ID 25 ± 0.050 mm 49.98 ± 0.010 mm OD 50 ± 0.010 mm 75 ± 0.10 mm These are shrink-tted together. Find the nominal shrink-t pressure and the von Mises stress in each body at the t surface. 536 Repeat Prob. 535 for maximum shrink-t conditions. 537 A 2-in-diameter solid steel shaft has a gear with ASTM grade 20 cast-iron hub ( E = 14.5 Mpsi) shrink-tted to it. The specications for the shaft are 2.000 + 0.0000 0.0004 in 1 The hole in the hub is sized at 1.999 ± 0.0004 in with an OD of 4.00 ± 32 in. Using the midrange values and the modied Mohr theory, estimate the factor of safety guarding against fracture in the gear hub due to the shrink t. 538 Two steel tubes are shrink-tted together where the nominal diameters are 1.50, 1.75, and 2.00 in. Careful measurement before tting revealed that the diametral interference between the tubes to be 0.00246 in. After the t, the assembly is subjected to a torque of 8000 lbf · in and a bending-moment of 6000 lbf · in. Assuming no slipping between the cylinders, analyze the outer cylinder at the inner and outer radius. Determine the factor of safety using distortion energy with Sy = 60 kpsi. 539 Repeat Prob. 538 for the inner tube. 540 For Eqs. (536) show that the principal stresses are given by θ KI θ 1 + sin σ1 = cos 2 2 2π r σ3 = θ θ KI 1 sin cos σ2 = 2 2 2π r (plane stress) 0 θ 2 ν K I cos πr 2 (plane strain) 541 Use the results of Prob. 540 for plane strain near the tip with θ = 0 and ν = 1 . If the yield 3 strength of the plate is Sy , what is σ1 when yield occurs? (a) Use the distortion-energy theory. (b) Use the maximum-shear-stress theory. Using Mohrs circles, explain your answer. 542 A plate 4 in wide, 8 in long, and 0.5 in thick is loaded in tension in the direction of the length. The plate contains a crack as shown in Fig. 526 with the crack length of 0.625 in. The material is steel with K I c = 70 kpsi · in, and Sy = 160 kpsi. Determine the maximum possible load that can be applied before the plate (a) yields, and (b) has uncontrollable crack growth. 543 A cylinder subjected to internal pressure pi has an outer diameter of 350 mm and a 25-mm wall thickness. For the cylinder material, K I c = 80 MPa · m, Sy = 1200 MPa, and Sut = 1350 MPa. BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition II. Failure Prevention 5. Failures Resulting from Static Loading © The McGrawHill Companies, 2008 Failures Resulting from Static Loading 259 255 If the cylinder contains a radial crack in the longitudinal direction of depth 12.5 mm determine the pressure that will cause uncontrollable crack growth. 544 A carbon steel collar of length 1 in is to be machined to inside and outside diameters, respectively, of Di = 0.750 ± 0.0004 in Do = 1.125 ± 0.002 in This collar is to be shrink-tted to a hollow steel shaft having inside and outside diameters, respectively, of di = 0.375 ± 0.002 in do = 0.752 ± 0.0004 in These tolerances are assumed to have a normal distribution, to be centered in the spread interval, and to have a total spread of ±4 standard deviations. Determine the means and the standard deviations of the tangential stress components for both cylinders at the interface. 545 Suppose the collar of Prob. 544 has a yield strength of S y = N(95.5, 6.59) kpsi. What is the probability that the material will not yield? 546 A carbon steel tube has an outside diameter of 1 in and a wall thickness of 1 in. The tube is to 8 carry an internal hydraulic pressure given as p = N(6000, 500) psi. The material of the tube has a yield strength of S y = N(50, 4.1) kpsi. Find the reliability using thin-wall theory. 260 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition II. Failure Prevention © The McGrawHill Companies, 2008 6. Fatigue Failure Resulting from Variable Loading 6 Fatigue Failure Resulting from Variable Loading Chapter Outline 61 Introduction to Fatigue in Metals 62 Approach to Fatigue Failure in Analysis and Design 63 Fatigue-Life Methods 64 The Stress-Life Method 265 65 The Strain-Life Method 268 66 The Linear-Elastic Fracture Mechanics Method 67 The Endurance Limit 68 Fatigue Strength 69 Endurance Limit Modifying Factors 258 264 265 270 274 275 278 610 Stress Concentration and Notch Sensitivity 611 Characterizing Fluctuating Stresses 612 Fatigue Failure Criteria for Fluctuating Stress 613 Torsional Fatigue Strength under Fluctuating Stresses 614 Combinations of Loading Modes 615 Varying, Fluctuating Stresses; Cumulative Fatigue Damage 616 Surface Fatigue Strength 617 Stochastic Analysis 618 Road Maps and Important Design Equations for the Stress-Life Method 287 292 295 309 309 313 319 322 336 257 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 258 II. Failure Prevention 6. Fatigue Failure Resulting from Variable Loading © The McGrawHill Companies, 2008 Mechanical Engineering Design In Chap. 5 we considered the analysis and design of parts subjected to static loading. The behavior of machine parts is entirely different when they are subjected to timevarying loading. In this chapter we shall examine how parts fail under variable loading and how to proportion them to successfully resist such conditions. 61 Introduction to Fatigue in Metals In most testing of those properties of materials that relate to the stress-strain diagram, the load is applied gradually, to give sufcient time for the strain to fully develop. Furthermore, the specimen is tested to destruction, and so the stresses are applied only once. Testing of this kind is applicable, to what are known as static conditions; such conditions closely approximate the actual conditions to which many structural and machine members are subjected. The condition frequently arises, however, in which the stresses vary with time or they uctuate between different levels. For example, a particular ber on the surface of a rotating shaft subjected to the action of bending loads undergoes both tension and compression for each revolution of the shaft. If the shaft is part of an electric motor rotating at 1725 rev/min, the ber is stressed in tension and compression 1725 times each minute. If, in addition, the shaft is also axially loaded (as it would be, for example, by a helical or worm gear), an axial component of stress is superposed upon the bending component. In this case, some stress is always present in any one ber, but now the level of stress is uctuating. These and other kinds of loading occurring in machine members produce stresses that are called variable, repeated, alternating, or uctuating stresses. Often, machine members are found to have failed under the action of repeated or uctuating stresses; yet the most careful analysis reveals that the actual maximum stresses were well below the ultimate strength of the material, and quite frequently even below the yield strength. The most distinguishing characteristic of these failures is that the stresses have been repeated a very large number of times. Hence the failure is called a fatigue failure. When machine parts fail statically, they usually develop a very large deection, because the stress has exceeded the yield strength, and the part is replaced before fracture actually occurs. Thus many static failures give visible warning in advance. But a fatigue failure gives no warning! It is sudden and total, and hence dangerous. It is relatively simple to design against a static failure, because our knowledge is comprehensive. Fatigue is a much more complicated phenomenon, only partially understood, and the engineer seeking competence must acquire as much knowledge of the subject as possible. A fatigue failure has an appearance similar to a brittle fracture, as the fracture surfaces are at and perpendicular to the stress axis with the absence of necking. The fracture features of a fatigue failure, however, are quite different from a static brittle fracture arising from three stages of development. Stage I is the initiation of one or more microcracks due to cyclic plastic deformation followed by crystallographic propagation extending from two to ve grains about the origin. Stage I cracks are not normally discernible to the naked eye. Stage II progresses from microcracks to macrocracks forming parallel plateau-like fracture surfaces separated by longitudinal ridges. The plateaus are generally smooth and normal to the direction of maximum tensile stress. These surfaces can be wavy dark and light bands referred to as beach marks or clamshell marks, as seen in Fig. 61. During cyclic loading, these cracked surfaces open and close, rubbing together, and the beach mark appearance depends on the changes in the level or frequency of loading and the corrosive nature of the environment. Stage III occurs during the nal stress cycle when the remaining material cannot support the loads, resulting in 261 262 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition II. Failure Prevention 6. Fatigue Failure Resulting from Variable Loading © The McGrawHill Companies, 2008 Fatigue Failure Resulting from Variable Loading 259 Figure 61 Fatigue failure of a bolt due to repeated unidirectional bending. The failure started at the thread root at A, propagated across most of the cross section shown by the beach marks at B, before final fast fracture at C. (From ASM Handbook, Vol. 12: Fractography, ASM International, Materials Park, OH 44073-0002, g 50, p. 120. Reprinted by permission of ASM International ®, www.asminternational.org.) a sudden, fast fracture. A stage III fracture can be brittle, ductile, or a combination of both. Quite often the beach marks, if they exist, and possible patterns in the stage III fracture called chevron lines, point toward the origins of the initial cracks. There is a good deal to be learned from the fracture patterns of a fatigue failure.1 Figure 62 shows representations of failure surfaces of various part geometries under differing load conditions and levels of stress concentration. Note that, in the case of rotational bending, even the direction of rotation inuences the failure pattern. Fatigue failure is due to crack formation and propagation. A fatigue crack will typically initiate at a discontinuity in the material where the cyclic stress is a maximum. Discontinuities can arise because of: Design of rapid changes in cross section, keyways, holes, etc. where stress concentrations occur as discussed in Secs. 313 and 52. Elements that roll and/or slide against each other (bearings, gears, cams, etc.) under high contact pressure, developing concentrated subsurface contact stresses (Sec. 319) that can cause surface pitting or spalling after many cycles of the load. Carelessness in locations of stamp marks, tool marks, scratches, and burrs; poor joint design; improper assembly; and other fabrication faults. Composition of the material itself as processed by rolling, forging, casting, extrusion, drawing, heat treatment, etc. Microscopic and submicroscopic surface and subsurface discontinuities arise, such as inclusions of foreign material, alloy segregation, voids, hard precipitated particles, and crystal discontinuities. Various conditions that can accelerate crack initiation include residual tensile stresses, elevated temperatures, temperature cycling, a corrosive environment, and high-frequency cycling. The rate and direction of fatigue crack propagation is primarily controlled by localized stresses and by the structure of the material at the crack. However, as with crack formation, other factors may exert a signicant inuence, such as environment, temperature, and frequency. As stated earlier, cracks will grow along planes normal to the 1 See the ASM Handbook, Fractography, ASM International, Metals Park, Ohio, vol. 12, 9th ed., 1987. BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 260 II. Failure Prevention Mechanical Engineering Design Figure 62 Schematics of fatigue fracture surfaces produced in smooth and notched components with round and rectangular cross sections under various loading conditions and nominal stress levels. (From ASM Handbook, Vol. 11: Failure Analysis and Prevention, ASM International, Materials Park, OH 44073-0002, g 18, p. 111. Reprinted by permission of ASM International ®, www.asminternational.org.) 6. Fatigue Failure Resulting from Variable Loading © The McGrawHill Companies, 2008 263 264 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition II. Failure Prevention 6. Fatigue Failure Resulting from Variable Loading © The McGrawHill Companies, 2008 Fatigue Failure Resulting from Variable Loading 261 maximum tensile stresses. The crack growth process can be explained by fracture mechanics (see Sec. 66). A major reference source in the study of fatigue failure is the 21-volume ASM Metals Handbook. Figures 61 to 68, reproduced with permission from ASM International, are but a minuscule sample of examples of fatigue failures for a great variety of conditions included in the handbook. Comparing Fig. 63 with Fig. 62, we see that failure occurred by rotating bending stresses, with the direction of rotation being clockwise with respect to the view and with a mild stress concentration and low nominal stress. Figure 63 Fatigue fracture of an AISI 4320 drive shaft. The fatigue failure initiated at the end of the keyway at points B and progressed to nal rupture at C. The nal rupture zone is small, indicating that loads were low. (From ASM Handbook, Vol. 11: Failure Analysis and Prevention, ASM International, Materials Park, OH 44073-0002, g 18, p. 111. Reprinted by permission of ASM International ®, www.asminternational.org.) Figure 64 Fatigue fracture surface of an AISI 8640 pin. Sharp corners of the mismatched grease holes provided stress concentrations that initiated two fatigue cracks indicated by the arrows. (From ASM Handbook, Vol. 12: Fractography, ASM International, Materials Park, OH 44073-0002, g 520, p. 331. Reprinted by permission of ASM International ®, www.asminternational.org.) BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 262 II. Failure Prevention 6. Fatigue Failure Resulting from Variable Loading Mechanical Engineering Design Figure 65 Fatigue fracture surface of a forged connecting rod of AISI 8640 steel. The fatigue crack origin is at the left edge, at the ash line of the forging, but no unusual roughness of the ash trim was indicated. The fatigue crack progressed halfway around the oil hole at the left, indicated by the beach marks, before nal fast fracture occurred. Note the pronounced shear lip in the nal fracture at the right edge. (From ASM Handbook, Vol. 12: Fractography, ASM International, Materials Park, OH 44073-0002, g 523, p. 332. Reprinted by permission of ASM International ®, www.asminternational.org.) Figure 66 Fatigue fracture surface of a 200-mm (8-in) diameter piston rod of an alloy steel steam hammer used for forging. This is an example of a fatigue fracture caused by pure tension where surface stress concentrations are absent and a crack may initiate anywhere in the cross section. In this instance, the initial crack formed at a forging ake slightly below center, grew outward symmetrically, and ultimately produced a brittle fracture without warning. (From ASM Handbook, Vol. 12: Fractography, ASM International, Materials Park, OH 44073-0002, g 570, p. 342. Reprinted by permission of ASM International ®, www.asminternational.org.) © The McGrawHill Companies, 2008 265 266 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition II. Failure Prevention © The McGrawHill Companies, 2008 6. Fatigue Failure Resulting from Variable Loading Fatigue Failure Resulting from Variable Loading 263 Medium-carbon steel (ASTM A186) 30 dia Web Fracture Fracture Tread Flange (1 of 2) (a) Coke-oven-car wheel Figure 67 Fatigue failure of an ASTM A186 steel double-ange trailer wheel caused by stamp marks. (a) Coke-oven car wheel showing position of stamp marks and fractures in the rib and web. (b) Stamp mark showing heavy impression and fracture extending along the base of the lower row of numbers. (c) Notches, indicated by arrows, created from the heavily indented stamp marks from which cracks initiated along the top at the fracture surface. (From ASM Handbook, Vol. 11: Failure Analysis and Prevention, ASM International, Materials Park, OH 440730002, g 51, p. 130. Reprinted by permission of ASM International ®, www.asminternational.org.) Figure 68 Aluminum alloy 7075-T73 landing-gear torque-arm assembly redesign to eliminate fatigue fracture at a lubrication hole. (a) Arm conguration, original and improved design (dimensions given in inches). (b) Fracture surface where arrows indicate multiple crack origins. (From ASM Handbook, Vol. 11: Failure Analysis and Prevention, ASM International, Materials Park, OH 44073-0002, g 23, p. 114. Reprinted by permission of ASM International ®, www.asminternational.org.) 4.94 Aluminum alloy 7075-T73 Rockwell B 85.5 25.5 10.200 Lug (1 of 2) Fracture A Primary-fracture surface Lubrication hole 1.750-in.-dia bushing, 0.090-in. wall Lubrication hole 1 in 3.62 dia Secondary fracture Improved design Original design Detail A (a) BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 264 II. Failure Prevention 6. Fatigue Failure Resulting from Variable Loading © The McGrawHill Companies, 2008 Mechanical Engineering Design 62 Approach to Fatigue Failure in Analysis and Design As noted in the previous section, there are a great many factors to be considered, even for very simple load cases. The methods of fatigue failure analysis represent a combination of engineering and science. Often science fails to provide the complete answers that are needed. But the airplane must still be made to ysafely. And the automobile must be manufactured with a reliability that will ensure a long and troublefree life and at the same time produce prots for the stockholders of the industry. Thus, while science has not yet completely explained the complete mechanism of fatigue, the engineer must still design things that will not fail. In a sense this is a classic example of the true meaning of engineering as contrasted with science. Engineers use science to solve their problems if the science is available. But available or not, the problem must be solved, and whatever form the solution takes under these conditions is called engineering. In this chapter, we will take a structured approach in the design against fatigue failure. As with static failure, we will attempt to relate to test results performed on simply loaded specimens. However, because of the complex nature of fatigue, there is much more to account for. From this point, we will proceed methodically, and in stages. In an attempt to provide some insight as to what follows in this chapter, a brief description of the remaining sections will be given here. Fatigue-Life Methods (Secs. 63 to 66) Three major approaches used in design and analysis to predict when, if ever, a cyclically loaded machine component will fail in fatigue over a period of time are presented. The premises of each approach are quite different but each adds to our understanding of the mechanisms associated with fatigue. The application, advantages, and disadvantages of each method are indicated. Beyond Sec. 66, only one of the methods, the stress-life method, will be pursued for further design applications. Fatigue Strength and the Endurance Limit (Secs. 67 and 68) The strength-life (S-N ) diagram provides the fatigue strength S f versus cycle life N of a material. The results are generated from tests using a simple loading of standard laboratorycontrolled specimens. The loading often is that of sinusoidally reversing pure bending. The laboratory-controlled specimens are polished without geometric stress concentration at the region of minimum area. For steel and iron, the S-N diagram becomes horizontal at some point. The strength at this point is called the endurance limit Se and occurs somewhere between 106 and 107 cycles. The prime mark on Se refers to the endurance limit of the controlled laboratory specimen. For nonferrous materials that do not exhibit an endurance limit, a fatigue strength at a specic number of cycles, S f , may be given, where again, the prime denotes the fatigue strength of the laboratory-controlled specimen. The strength data are based on many controlled conditions that will not be the same as that for an actual machine part. What follows are practices used to account for the differences between the loading and physical conditions of the specimen and the actual machine part. Endurance Limit Modifying Factors (Sec. 69) Modifying factors are dened and used to account for differences between the specimen and the actual machine part with regard to surface conditions, size, loading, temperature, reliability, and miscellaneous factors. Loading is still considered to be simple and reversing. 267 268 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition II. Failure Prevention 6. Fatigue Failure Resulting from Variable Loading © The McGrawHill Companies, 2008 Fatigue Failure Resulting from Variable Loading 265 Stress Concentration and Notch Sensitivity (Sec. 610) The actual part may have a geometric stress concentration by which the fatigue behavior depends on the static stress concentration factor and the component materials sensitivity to fatigue damage. Fluctuating Stresses (Secs. 611 to 613) These sections account for simple stress states from uctuating load conditions that are not purely sinusoidally reversing axial, bending, or torsional stresses. Combinations of Loading Modes (Sec. 614) Here a procedure based on the distortion-energy theory is presented for analyzing combined uctuating stress states, such as combined bending and torsion. Here it is assumed that the levels of the uctuating stresses are in phase and not time varying. Varying, Fluctuating Stresses; Cumulative Fatigue Damage (Sec. 615) The uctuating stress levels on a machine part may be time varying. Methods are provided to assess the fatigue damage on a cumulative basis. Remaining Sections The remaining three sections of the chapter pertain to the special topics of surface fatigue strength, stochastic analysis, and roadmaps with important equations. 63 Fatigue-Life Methods The three major fatigue life methods used in design and analysis are the stress-life method, the strain-life method, and the linear-elastic fracture mechanics method. These methods attempt to predict the life in number of cycles to failure, N, for a specic level of loading. Life of 1 N 103 cycles is generally classied as low-cycle fatigue, whereas high-cycle fatigue is considered to be N > 103 cycles. The stress-life method, based on stress levels only, is the least accurate approach, especially for low-cycle applications. However, it is the most traditional method, since it is the easiest to implement for a wide range of design applications, has ample supporting data, and represents high-cycle applications adequately. The strain-life method involves more detailed analysis of the plastic deformation at localized regions where the stresses and strains are considered for life estimates. This method is especially good for low-cycle fatigue applications. In applying this method, several idealizations must be compounded, and so some uncertainties will exist in the results. For this reason, it will be discussed only because of its value in adding to the understanding of the nature of fatigue. The fracture mechanics method assumes a crack is already present and detected. It is then employed to predict crack growth with respect to stress intensity. It is most practical when applied to large structures in conjunction with computer codes and a periodic inspection program. 64 The Stress-Life Method To determine the strength of materials under the action of fatigue loads, specimens are subjected to repeated or varying forces of specied magnitudes while the cycles or stress reversals are counted to destruction. The most widely used fatigue-testing device BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 266 II. Failure Prevention © The McGrawHill Companies, 2008 6. Fatigue Failure Resulting from Variable Loading Mechanical Engineering Design is the R. R. Moore high-speed rotating-beam machine. This machine subjects the specimen to pure bending (no transverse shear) by means of weights. The specimen, shown in Fig. 69, is very carefully machined and polished, with a nal polishing in an axial direction to avoid circumferential scratches. Other fatigue-testing machines are available for applying uctuating or reversed axial stresses, torsional stresses, or combined stresses to the test specimens. To establish the fatigue strength of a material, quite a number of tests are necessary because of the statistical nature of fatigue. For the rotating-beam test, a constant bending load is applied, and the number of revolutions (stress reversals) of the beam required for failure is recorded. The rst test is made at a stress that is somewhat under the ultimate strength of the material. The second test is made at a stress that is less than that used in the rst. This process is continued, and the results are plotted as an S-N diagram (Fig. 610). This chart may be plotted on semilog paper or on log-log paper. In the case of ferrous metals and alloys, the graph becomes horizontal after the material has been stressed for a certain number of cycles. Plotting on log paper emphasizes the bend in the curve, which might not be apparent if the results were plotted by using Cartesian coordinates. 7 3 16 in 0.30 in 9 7 in R. 8 Figure 69 Test-specimen geometry for the R. R. Moore rotatingbeam machine. The bending moment is uniform over the curved at the highest-stressed portion, a valid test of material, whereas a fracture elsewhere (not at the higheststress level) is grounds for suspicion of material aw. Figure 610 High cycle Finite life Infinite life Sut 100 Fatigue strength Sf , kpsi An S-N diagram plotted from the results of completely reversed axial fatigue tests. Material: UNS G41300 steel, normalized; Sut = 116 kpsi; maximum Sut = 125 kpsi. (Data from NACA Tech. Note 3866, December 1966.) Low cycle 50 100 Se 101 102 103 10 4 10 5 Number of stress cycles, N 106 107 108 269 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition II. Failure Prevention © The McGrawHill Companies, 2008 6. Fatigue Failure Resulting from Variable Loading Fatigue Failure Resulting from Variable Loading Figure 611 S-N bands for representative aluminum alloys, excluding wrought alloys with Sut < 38 kpsi. (From R. C. Juvinall, Engineering Considerations of Stress, Strain and Strength. Copyright © 1967 by The McGraw-Hill Companies, Inc. Reprinted by permission.) 267 80 70 60 Peak alternating bending stress S, kpsi (log) 270 50 40 35 30 25 Wrought 20 18 16 14 12 Permanent mold cast 10 Sand cast 8 7 6 5 103 104 105 106 Life N, cycles (log) 107 108 109 The ordinate of the S-N diagram is called the fatigue strength S f ; a statement of this strength value must always be accompanied by a statement of the number of cycles N to which it corresponds. Soon we shall learn that S-N diagrams can be determined either for a test specimen or for an actual mechanical element. Even when the material of the test specimen and that of the mechanical element are identical, there will be signicant differences between the diagrams for the two. In the case of the steels, a knee occurs in the graph, and beyond this knee failure will not occur, no matter how great the number of cycles. The strength corresponding to the knee is called the endurance limit Se , or the fatigue limit. The graph of Fig. 610 never does become horizontal for nonferrous metals and alloys, and hence these materials do not have an endurance limit. Figure 611 shows scatter bands indicating the S-N curves for most common aluminum alloys excluding wrought alloys having a tensile strength below 38 kpsi. Since aluminum does not have an endurance limit, normally the fatigue strength S f is reported at a specic number of cycles, normally N = 5(108 ) cycles of reversed stress (see Table A24). We note that a stress cycle ( N = 1) constitutes a single application and removal of a load and then another application and removal of the load in the opposite direction. Thus N = 1 means the load is applied once and then removed, which is the case with 2 the simple tension test. The body of knowledge available on fatigue failure from N = 1 to N = 1000 cycles is generally classied as low-cycle fatigue, as indicated in Fig. 610. High-cycle fatigue, then, is concerned with failure corresponding to stress cycles greater than 103 cycles. We also distinguish a nite-life region and an innite-life region in Fig. 610. The boundary between these regions cannot be clearly dened except for a specic material; but it lies somewhere between 106 and 107 cycles for steels, as shown in Fig. 610. As noted previously, it is always good engineering practice to conduct a testing program on the materials to be employed in design and manufacture. This, in fact, is a requirement, not an option, in guarding against the possibility of a fatigue failure. BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 268 II. Failure Prevention 6. Fatigue Failure Resulting from Variable Loading © The McGrawHill Companies, 2008 Mechanical Engineering Design Because of this necessity for testing, it would really be unnecessary for us to proceed any further in the study of fatigue failure except for one important reason: the desire to know why fatigue failures occur so that the most effective method or methods can be used to improve fatigue strength. Thus our primary purpose in studying fatigue is to understand why failures occur so that we can guard against them in an optimum manner. For this reason, the analytical design approaches presented in this book, or in any other book, for that matter, do not yield absolutely precise results. The results should be taken as a guide, as something that indicates what is important and what is not important in designing against fatigue failure. As stated earlier, the stress-life method is the least accurate approach especially for low-cycle applications. However, it is the most traditional method, with much published data available. It is the easiest to implement for a wide range of design applications and represents high-cycle applications adequately. For these reasons the stress-life method will be emphasized in subsequent sections of this chapter. However, care should be exercised when applying the method for low-cycle applications, as the method does not account for the true stress-strain behavior when localized yielding occurs. 65 The Strain-Life Method The best approach yet advanced to explain the nature of fatigue failure is called by some the strain-life method. The approach can be used to estimate fatigue strengths, but when it is so used it is necessary to compound several idealizations, and so some uncertainties will exist in the results. For this reason, the method is presented here only because of its value in explaining the nature of fatigue. A fatigue failure almost always begins at a local discontinuity such as a notch, crack, or other area of stress concentration. When the stress at the discontinuity exceeds the elastic limit, plastic strain occurs. If a fatigue fracture is to occur, there must exist cyclic plastic strains. Thus we shall need to investigate the behavior of materials subject to cyclic deformation. In 1910, Bairstow veried by experiment Bauschingers theory that the elastic limits of iron and steel can be changed, either up or down, by the cyclic variations of stress.2 In general, the elastic limits of annealed steels are likely to increase when subjected to cycles of stress reversals, while cold-drawn steels exhibit a decreasing elastic limit. R. W. Landgraf has investigated the low-cycle fatigue behavior of a large number of very high-strength steels, and during his research he made many cyclic stress-strain plots.3 Figure 612 has been constructed to show the general appearance of these plots for the rst few cycles of controlled cyclic strain. In this case the strength decreases with stress repetitions, as evidenced by the fact that the reversals occur at ever-smaller stress levels. As previously noted, other materials may be strengthened, instead, by cyclic stress reversals. The SAE Fatigue Design and Evaluation Steering Committee released a report in 1975 in which the life in reversals to failure is related to the strain amplitude ε/2.4 2 L. Bairstow, The Elastic Limits of Iron and Steel under Cyclic Variations of Stress, Philosophical Transactions, Series A, vol. 210, Royal Society of London, 1910, pp. 3555. 3 R. W. Landgraf, Cyclic Deformation and Fatigue Behavior of Hardened Steels, Report no. 320, Department of Theoretical and Applied Mechanics, University of Illinois, Urbana, 1968, pp. 8490. 4 Technical Report on Fatigue Properties, SAE J1099, 1975. 271 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition II. Failure Prevention © The McGrawHill Companies, 2008 6. Fatigue Failure Resulting from Variable Loading Fatigue Failure Resulting from Variable Loading Figure 612 True stresstrue strain hysteresis loops showing the rst ve stress reversals of a cyclicsoftening material. The graph is slightly exaggerated for clarity. Note that the slope of the line AB is the modulus of elasticity E. The stress range is σ , ε p is the plastic-strain range, and εe is the elastic strain range. The total-strain range is ε = ε p + εe . 269 1st reversal A 3d 5th 4th 2d B p e Figure 613 A log-log plot showing how the fatigue life is related to the true-strain amplitude for hot-rolled SAE 1020 steel. (Reprinted with permission from SAE J1099_200208 © 2002 SAE International.) 10 0 ' F 101 Strain amplitude, /2 272 c 1.0 102 ' F E Total strain Plastic strain b 1.0 103 Elastic strain 10 4 100 101 10 2 10 3 10 4 10 5 106 Reversals to failure, 2 N The report contains a plot of this relationship for SAE 1020 hot-rolled steel; the graph has been reproduced as Fig. 613. To explain the graph, we rst dene the following terms: Fatigue ductility coefcient εF is the true strain corresponding to fracture in one reversal (point A in Fig. 612). The plastic-strain line begins at this point in Fig. 613. Fatigue strength coefcient σ F is the true stress corresponding to fracture in one reversal (point A in Fig. 612). Note in Fig. 613 that the elastic-strain line begins at σF / E . Fatigue ductility exponent c is the slope of the plastic-strain line in Fig. 613 and is the power to which the life 2N must be raised to be proportional to the true plasticstrain amplitude. If the number of stress reversals is 2N, then N is the number of cycles. BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 270 II. Failure Prevention 6. Fatigue Failure Resulting from Variable Loading 273 © The McGrawHill Companies, 2008 Mechanical Engineering Design Fatigue strength exponent b is the slope of the elastic-strain line, and is the power to which the life 2N must be raised to be proportional to the true-stress amplitude. Now, from Fig. 612, we see that the total strain is the sum of the elastic and plastic components. Therefore the total strain amplitude is half the total strain range εp ε εe = + 2 2 2 (a) The equation of the plastic-strain line in Fig. 613 is εp = εF (2 N )c 2 (61) The equation of the elastic strain line is σ εe = F (2 N )b 2 E (62) Therefore, from Eq. (a), we have for the total-strain amplitude ε σ = F (2 N )b + εF (2 N )c 2 E (63) which is the Manson-Cofn relationship between fatigue life and total strain.5 Some values of the coefcients and exponents are listed in Table A23. Many more are included in the SAE J1099 report.6 Though Eq. (63) is a perfectly legitimate equation for obtaining the fatigue life of a part when the strain and other cyclic characteristics are given, it appears to be of little use to the designer. The question of how to determine the total strain at the bottom of a notch or discontinuity has not been answered. There are no tables or charts of strain concentration factors in the literature. It is possible that strain concentration factors will become available in research literature very soon because of the increase in the use of nite-element analysis. Moreover, nite element analysis can of itself approximate the strains that will occur at all points in the subject structure.7 66 The Linear-Elastic Fracture Mechanics Method The rst phase of fatigue cracking is designated as stage I fatigue. Crystal slip that extends through several contiguous grains, inclusions, and surface imperfections is presumed to play a role. Since most of this is invisible to the observer, we just say that stage I involves several grains. The second phase, that of crack extension, is called stage II fatigue. The advance of the crack (that is, new crack area is created) does produce evidence that can be observed on micrographs from an electron microscope. The growth of 5 J. F. Tavernelli and L. F. Cofn, Jr., Experimental Support for Generalized Equation Predicting Low Cycle Fatigue, and S. S. Manson, discussion, Trans. ASME, J. Basic Eng., vol. 84, no. 4, pp. 533537. 6 See also, Landgraf, Ibid. 7 For further discussion of the strain-life method see N. E. Dowling, Mechanical Behavior of Materials, 2nd ed., Prentice-Hall, Englewood Cliffs, N.J., 1999, Chap. 14. BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition II. Failure Prevention © The McGrawHill Companies, 2008 6. Fatigue Failure Resulting from Variable Loading Fatigue Failure Resulting from Variable Loading 271 the crack is orderly. Final fracture occurs during stage III fatigue, although fatigue is not involved. When the crack is sufciently long that K I = K Ic for the stress amplitude involved, then K Ic is the critical stress intensity for the undamaged metal, and there is sudden, catastrophic failure of the remaining cross section in tensile overload (see Sec. 512). Stage III fatigue is associated with rapid acceleration of crack growth then fracture. Crack Growth Fatigue cracks nucleate and grow when stresses vary and there is some tension in each stress cycle. Consider the stress to be uctuating between the limits of σmin and σmax , where the stress range is dened as σ = σmax σmin . From Eq. (537) the stress intensity is given by K I = βσ π a . Thus, for σ, the stress intensity range per cycle is K I = β(σmax σmin ) π a = β σ π a (64) To develop fatigue strength data, a number of specimens of the same material are tested at various levels of σ. Cracks nucleate at or very near a free surface or large discontinuity. Assuming an initial crack length of ai , crack growth as a function of the number of stress cycles N will depend on σ, that is, K I . For K I below some threshold value ( K I )th a crack will not grow. Figure 614 represents the crack length a as a function of N for three stress levels ( σ )3 > ( σ )2 > ( σ )1 , where ( K I )3 > ( K I )2 > ( K I )1 . Notice the effect of the higher stress range in Fig. 614 in the production of longer cracks at a particular cycle count. When the rate of crack growth per cycle, d a /d N in Fig. 614, is plotted as shown in Fig. 615, the data from all three stress range levels superpose to give a sigmoidal curve. The three stages of crack development are observable, and the stage II data are linear on log-log coordinates, within the domain of linear elastic fracture mechanics (LEFM) validity. A group of similar curves can be generated by changing the stress ratio R = σmin /σmax of the experiment. Here we present a simplied procedure for estimating the remaining life of a cyclically stressed part after discovery of a crack. This requires the assumption that plane strain Figure 614 The increase in crack length a from an initial length of ai as a function of cycle count for three stress ranges, ( σ ) 3 > ( σ) 2 > ( σ) 1. (KI )3 Crack length a 274 (KI )2 (KI )1 da a dN ai Log N Stress cycles N BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 272 II. Failure Prevention 275 © The McGrawHill Companies, 2008 6. Fatigue Failure Resulting from Variable Loading Mechanical Engineering Design Figure 615 Log da dN When da/dN is measured in Fig. 614 and plotted on loglog coordinates, the data for different stress ranges superpose, giving rise to a sigmoid curve as shown. ( K I ) th is the threshold value of K I , below which a crack does not grow. From threshold to rupture an aluminum alloy will spend 85--90 percent of life in region I, 5--8 percent in region II, and 1--2 percent in region III. Region I Region II Crack initiation Crack propagation Region III Crack unstable Increasing stress ratio R Kc (K )th Log K Table 61 Conservative Values of Factor C and Exponent m in Eq. (65) for Various Forms of Steel . (R = 0) C, Material m/cycle m MPa m C, 6.89(1012 ) Ferritic-pearlitic steels 10 Martensitic steels 1.36(10 Austenitic stainless steels in/cycle m kpsi in 3.60(1010 ) 9 5.61(1012 ) ) 6.60(10 ) 3.00(1010 ) m 3.00 2.25 3.25 From J.M. Barsom and S.T. Rolfe, Fatigue and Fracture Control in Structures, 2nd ed., Prentice Hall, Upper Saddle River, NJ, 1987, pp. 288291, Copyright ASTM International. Reprinted with permission. conditions prevail.8 Assuming a crack is discovered early in stage II, the crack growth in region II of Fig. 615 can be approximated by the Paris equation, which is of the form da = C ( K I )m dN (65) where C and m are empirical material constants and K I is given by Eq. (64). Representative, but conservative, values of C and m for various classes of steels are listed in Table 61. Substituting Eq. (64) and integrating gives Nf 0 d N = Nf = 1 C af ai da (β σ π a )m (66) Here ai is the initial crack length, a f is the nal crack length corresponding to failure, and N f is the estimated number of cycles to produce a failure after the initial crack is formed. Note that β may vary in the integration variable (e.g., see Figs. 525 to 530). 8 Recommended references are: Dowling, op. cit.; J. A. Collins, Failure of Materials in Mechanical Design, John Wiley & Sons, New York, 1981; H. O. Fuchs and R. I. Stephens, Metal Fatigue in Engineering, John Wiley & Sons, New York, 1980; and Harold S. Reemsnyder, Constant Amplitude Fatigue Life Assessment Models, SAE Trans. 820688, vol. 91, Nov. 1983. 276 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition II. Failure Prevention © The McGrawHill Companies, 2008 6. Fatigue Failure Resulting from Variable Loading Fatigue Failure Resulting from Variable Loading 273 If this should happen, then Reemsnyder9 suggests the use of numerical integration employing the algorithm δ a j = C ( K I )m (δ N ) j j a j +1 = a j + δ a j (67) N j +1 = N j + δ N j Nf = δ Nj Here δ a j and δ N j are increments of the crack length and the number of cycles. The procedure is to select a value of δ N j , using ai determine β and compute K I , determine δ a j , and then nd the next value of a. Repeat the procedure until a = a f . The following example is highly simplied with β constant in order to give some understanding of the procedure. Normally, one uses fatigue crack growth computer programs such as NASA/FLAGRO 2.0 with more comprehensive theoretical models to solve these problems. 9 Op. cit. EXAMPLE 61 Solution The bar shown in Fig. 616 is subjected to a repeated moment 0 M 1200 lbf in. · The bar is AISI 4430 steel with Sut = 185 kpsi, Sy = 170 kpsi, and K Ic = 73 kpsi in. Material tests on various specimens of this material with identical heat treatment indicate worst-case constants of C = 3.8(1011 ) (in/cycle) (kpsi in)m and m = 3.0. As shown, a nick of size 0.004 in has been discovered on the bottom of the bar. Estimate the number of cycles of life remaining. The stress range σ is always computed by using the nominal (uncracked) area. Thus I bh 2 0.25(0.5)2 = = = 0.010 42 in3 c 6 6 Therefore, before the crack initiates, the stress range is σ= M 1200 = = 115.2(103 ) psi = 115.2 kpsi I /c 0.010 42 which is below the yield strength. As the crack grows, it will eventually become long enough such that the bar will completely yield or undergo a brittle fracture. For the ratio of Sy / Sut it is highly unlikely that the bar will reach complete yield. For brittle fracture, designate the crack length as a f . If β = 1, then from Eq. (537) with K I = K Ic , we approximate a f as af = 1 π K Ic βσmax Figure 616 2 1 4 M M Nick .1 = π 73 115.2 in 1 2 in 2 = 0.1278 in BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 274 II. Failure Prevention © The McGrawHill Companies, 2008 6. Fatigue Failure Resulting from Variable Loading Mechanical Engineering Design From Fig. 527, we compute the ratio a f / h as af 0.1278 = = 0.256 h 0.5 Thus a f / h varies from near zero to approximately 0.256. From Fig. 527, for this range β is nearly constant at approximately 1.07. We will assume it to be so, and re-evaluate a f as af = 1 π 73 1.07(115.2) 2 = 0.112 in Thus, from Eq. (66), the estimated remaining life is Nf = 1 C = 67 af ai 1 da = m 3.8(1011 ) (β σ π a ) 5.047(103 ) a 0.112 0.004 da [1.07(115.2) π a ]3 0.112 0.004 = 64.7 (103 ) cycles The Endurance Limit The determination of endurance limits by fatigue testing is now routine, though a lengthy procedure. Generally, stress testing is preferred to strain testing for endurance limits. For preliminary and prototype design and for some failure analysis as well, a quick method of estimating endurance limits is needed. There are great quantities of data in the literature on the results of rotating-beam tests and simple tension tests of specimens taken from the same bar or ingot. By plotting these as in Fig. 617, it is possible to see whether there is any correlation between the two sets of results. The graph appears to suggest that the endurance limit ranges from about 40 to 60 percent of the tensile strength for steels up to about 210 kpsi (1450 MPa). Beginning at about Sut = 210 kpsi (1450 MPa), the scatter appears to increase, but the trend seems to level off, as sug gested by the dashed horizontal line at Se = 105 kpsi. We wish now to present a method for estimating endurance limits. Note that estimates obtained from quantities of data obtained from many sources probably have a large spread and might deviate signicantly from the results of actual laboratory tests of the mechanical properties of specimens obtained through strict purchase-order specications. Since the area of uncertainty is greater, compensation must be made by employing larger design factors than would be used for static design. For steels, simplifying our observation of Fig. 617, we will estimate the endurance limit as Sut 200 kpsi (1400 MPa) 0.5 Sut Se = 100 kpsi Sut > 200 kpsi (68) 700 MPa Sut > 1400 MPa where Sut is the minimum tensile strength. The prime mark on Se in this equation refers to the rotating-beam specimen itself. We wish to reserve the unprimed symbol Se for the endurance limit of any particular machine element subjected to any kind of loading. Soon we shall learn that the two strengths may be quite different. 277 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition II. Failure Prevention © The McGrawHill Companies, 2008 6. Fatigue Failure Resulting from Variable Loading 275 Fatigue Failure Resulting from Variable Loading 140 0 S 'e = u S Carbon steels Alloy steels Wrought irons 120 Endurance limit S 'e , kpsi 278 0.5 .6 0.4 105 kpsi 100 80 60 40 20 0 0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 Tensile strength Su t , kpsi Figure 617 Graph of endurance limits versus tensile strengths from actual test results for a large number of wrought irons and steels. Ratios of Se/ Sut of 0.60, 0.50, and 0.40 are shown by the solid and dashed lines. Note also the horizontal dashed line for Se = 105 kpsi. Points shown having a tensile strength greater than 210 kpsi have a mean endurance limit of Se = 105 kpsi and a standard deviation of 13.5 kpsi. (Collated from data compiled by H. J. Grover, S. A. Gordon, and L. R. Jackson in Fatigue of Metals and Structures, Bureau of Naval Weapons Document NAVWEPS 00-25-534, 1960; and from Fatigue Design Handbook, SAE, 1968, p. 42.) Steels treated to give different microstructures have different Se / Sut ratios. It appears that the more ductile microstructures have a higher ratio. Martensite has a very brittle nature and is highly susceptible to fatigue-induced cracking; thus the ratio is low. When designs include detailed heat-treating specifications to obtain specific microstructures, it is possible to use an estimate of the endurance limit based on test data for the particular microstructure; such estimates are much more reliable and indeed should be used. The endurance limits for various classes of cast irons, polished or machined, are given in Table A24. Aluminum alloys do not have an endurance limit. The fatigue strengths of some aluminum alloys at 5(108) cycles of reversed stress are given in Table A24. 68 Fatigue Strength As shown in Fig. 610, a region of low-cycle fatigue extends from N = 1 to about 103 cycles. In this region the fatigue strength S f is only slightly smaller than the tensile strength Sut . An analytical approach has been given by Mischke10 for both 10 J. E. Shigley, C. R. Mischke, and T. H. Brown, Jr., Standard Handbook of Machine Design, 3rd ed., McGraw-Hill, New York, 2004, pp. 29.2529.27. BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 276 II. Failure Prevention 279 © The McGrawHill Companies, 2008 6. Fatigue Failure Resulting from Variable Loading Mechanical Engineering Design high-cycle and low-cycle regions, requiring the parameters of the Manson-Coffin equation plus the strain-strengthening exponent m. Engineers often have to work with less information. Figure 610 indicates that the high-cycle fatigue domain extends from 103 cycles for steels to the endurance limit life Ne , which is about 106 to 107 cycles. The purpose of this section is to develop methods of approximation of the S-N diagram in the highcycle region, when information may be as sparse as the results of a simple tension test. Experience has shown high-cycle fatigue data are rectied by a logarithmic transform to both stress and cycles-to-failure. Equation (62) can be used to determine the fatigue strength at 103 cycles. Dening the specimen fatigue strength at a specic number of cycles as ( S f ) N = E εe /2, write Eq. (62) as ( S f ) N = σ F (2 N )b (69) At 103 cycles, ( S f )103 = σ F (2.103 )b = f Sut where f is the fraction of Sut represented by ( S f )103 cycles . Solving for f gives f= σF (2 · 103 )b Sut (610) Now, from Eq. (211), σ F = σ0 εm , with ε = εF . If this true-stresstrue-strain equation is not known, the SAE approximation11 for steels with HB 500 may be used: σ F = Sut + 50 kpsi or σ F = Sut + 345 MPa (611) To nd b, substitute the endurance strength and corresponding cycles, Se and Ne , respectively into Eq. (69) and solving for b b= log σ F / Se log (2 N e ) (612) Thus, the equation S f = σ F (2 N )b is known. For example, if Sut = 105 kpsi and Se = 52.5 kpsi at failure, Eq. (611) σ F = 105 + 50 = 155 kpsi Eq. (612) b= Eq. (610) f= log(155/52.5) = 0.0746 log 2 · 106 155 2 · 103 105 0.0746 = 0.837 and for Eq. (69), with S f = ( S f ) N , S f = 155(2 N )0.0746 = 147 N 0.0746 11 Fatigue Design Handbook, vol. 4, Society of Automotive Engineers, New York, 1958, p. 27. (a) 280 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition II. Failure Prevention © The McGrawHill Companies, 2008 6. Fatigue Failure Resulting from Variable Loading Fatigue Failure Resulting from Variable Loading Figure 618 Fatigue strength fraction, f, of Sut at 103 cycles for Se = Se = 0.5 Sut . f 277 0.9 0.88 0.86 0.84 0.82 0.8 0.78 0.76 70 80 90 100 110 120 130 140 150 160 170 180 190 200 Su t , kpsi The process given for nding f can be repeated for various ultimate strengths. Figure 618 is a plot of f for 70 Sut 200 kpsi. To be conservative, for Sut < 70 kpsi, let f 0.9. For an actual mechanical component, Se is reduced to S e (see Sec. 69) which is less than 0.5 Sut . However, unless actual data is available, we recommend using the value of f found from Fig. 618. Equation (a), for the actual mechanical component, can be written in the form Sf = a N b (613) where N is cycles to failure and the constants a and b are dened by the points 103 , S f 103 and 106 , Se with S f 103 = f Sut . Substituting these two points in Eq. (613) gives a= ( f Sut )2 Se 1 b = log 3 (614) f Sut Se (615) If a completely reversed stress σa is given, setting S f = σa in Eq. (613), the number of cycles-to-failure can be expressed as N= σa a 1/b (616) Low-cycle fatigue is often dened (see Fig. 610) as failure that occurs in a range of 1 N 103 cycles. On a loglog plot such as Fig. 610 the failure locus in this range is nearly linear below 103 cycles. A straight line between 103 , f Sut and 1, Sut (transformed) is conservative, and it is given by S f Sut N (log f )/3 1 N 103 (617) BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 278 II. Failure Prevention © The McGrawHill Companies, 2008 6. Fatigue Failure Resulting from Variable Loading Mechanical Engineering Design EXAMPLE 62 Solution Given a 1050 HR steel, estimate (a) the rotating-beam endurance limit at 106 cycles. (b) the endurance strength of a polished rotating-beam specimen corresponding to 104 cycles to failure (c) the expected life of a polished rotating-beam specimen under a completely reversed stress of 55 kpsi. (a) From Table A20, Sut = 90 kpsi. From Eq. (68), Se = 0.5(90) = 45 kpsi Answer . (b) From Fig. 618, for Sut = 90 kpsi, f = 0.86. From Eq. (614), a= [0.86(90)2 ] = 133.1 kpsi 45 From Eq. (615), 1 0.86(90) b = log = 0.0785 3 45 Thus, Eq. (613) is S f = 133.1 N 0.0785 Answer For 104 cycles to failure, S f = 133.1(104 ) 0.0785 = 64.6 kpsi (c) From Eq. (616), with σa = 55 kpsi, Answer N= 55 133.1 1/0.0785 = 77 500 = 7.75(104 )cycles Keep in mind that these are only estimates. So expressing the answers using three-place accuracy is a little misleading. 69 Endurance Limit Modifying Factors We have seen that the rotating-beam specimen used in the laboratory to determine endurance limits is prepared very carefully and tested under closely controlled conditions. It is unrealistic to expect the endurance limit of a mechanical or structural member to match the values obtained in the laboratory. Some differences include Material: composition, basis of failure, variability Manufacturing: method, heat treatment, fretting corrosion, surface condition, stress concentration Environment: corrosion, temperature, stress state, relaxation times Design: size, shape, life, stress state, stress concentration, speed, fretting, galling 281 282 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition II. Failure Prevention © The McGrawHill Companies, 2008 6. Fatigue Failure Resulting from Variable Loading 279 Fatigue Failure Resulting from Variable Loading Marin12 identied factors that quantied the effects of surface condition, size, loading, temperature, and miscellaneous items. The question of whether to adjust the endurance limit by subtractive corrections or multiplicative corrections was resolved by an extensive statistical analysis of a 4340 (electric furnace, aircraft quality) steel, in which a correlation coefcient of 0.85 was found for the multiplicative form and 0.40 for the additive form. A Marin equation is therefore written as Se = ka kb kc kd ke k f Se where (618) ka = surface condition modication factor kb = size modication factor kc = load modication factor kd = temperature modication factor ke = reliability factor13 kf = miscellaneous-effects modication factor Se = rotary-beam test specimen endurance limit Se = endurance limit at the critical location of a machine part in the geometry and condition of use When endurance tests of parts are not available, estimations are made by applying Marin factors to the endurance limit. Surface Factor ka The surface of a rotating-beam specimen is highly polished, with a nal polishing in the axial direction to smooth out any circumferential scratches. The surface modication factor depends on the quality of the nish of the actual part surface and on the tensile strength of the part material. To nd quantitative expressions for common nishes of machine parts (ground, machined, or cold-drawn, hot-rolled, and as-forged), the coordinates of data points were recaptured from a plot of endurance limit versus ultimate tensile strength of data gathered by Lipson and Noll and reproduced by Horger.14 The data can be represented by b ka = aSut (619) where Sut is the minimum tensile strength and a and b are to be found in Table 62. 12 Joseph Marin, Mechanical Behavior of Engineering Materials, Prentice-Hall, Englewood Cliffs, N.J., 1962, p. 224. 13 Complete stochastic analysis is presented in Sec. 617. Until that point the presentation here is one of a deterministic nature. However, we must take care of the known scatter in the fatigue data. This means that we will not carry out a true reliability analysis at this time but will attempt to answer the question: What is the probability that a known (assumed) stress will exceed the strength of a randomly selected component made from this material population? 14 C. J. Noll and C. Lipson, Allowable Working Stresses, Society for Experimental Stress Analysis, vol. 3, no. 2, 1946, p. 29. Reproduced by O. J. Horger (ed.), Metals Engineering Design ASME Handbook, McGraw-Hill, New York, 1953, p. 102. BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 280 II. Failure Prevention 283 © The McGrawHill Companies, 2008 6. Fatigue Failure Resulting from Variable Loading Mechanical Engineering Design Table 62 Parameters for Marin Surface Modication Factor, Eq. (619) Factor a Surface Finish Sut, kpsi Sut, MPa Ground 1.34 1.58 Machined or cold-drawn 2.70 4.51 Hot-rolled 14.4 As-forged 39.9 57.7 Exponent b 0.085 0.265 0.718 272. 0.995 From C.J. Noll and C. Lipson, Allowable Working Stresses, Society for Experimental Stress Analysis, vol. 3, no. 2, 1946 p. 29. Reproduced by O.J. Horger (ed.) Metals Engineering Design ASME Handbook, McGraw-Hill, New York. Copyright © 1953 by The McGraw-Hill Companies, Inc. Reprinted by permission. EXAMPLE 63 Solution Answer A steel has a minimum ultimate strength of 520 MPa and a machined surface. Estimate ka. From Table 62, a = 4.51 and b = 0.265. Then, from Eq. (619) ka = 4.51(520)0.265 = 0.860 Again, it is important to note that this is an approximation as the data is typically quite scattered. Furthermore, this is not a correction to take lightly. For example, if in the previous example the steel was forged, the correction factor would be 0.540, a signicant reduction of strength. Size Factor kb The size factor has been evaluated using 133 sets of data points.15 The results for bending and torsion may be expressed as (d /0.3)0.107 = 0.879d 0.107 0.91d 0.157 kb = (d /7.62)0.107 = 1.24d 0.107 1.51d 0.157 0.11 d 2 in 2 < d 10 in 2.79 d 51 mm 51 < d 254 mm ( 620) For axial loading there is no size effect, so kb = 1 (621) but see kc . One of the problems that arises in using Eq. (620) is what to do when a round bar in bending is not rotating, or when a noncircular cross section is used. For example, what is the size factor for a bar 6 mm thick and 40 mm wide? The approach to be used 15 Charles R. Mischke, Prediction of Stochastic Endurance Strength, Trans. of ASME, Journal of Vibration, Acoustics, Stress, and Reliability in Design, vol. 109, no. 1, January 1987, Table 3. 284 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition II. Failure Prevention © The McGrawHill Companies, 2008 6. Fatigue Failure Resulting from Variable Loading Fatigue Failure Resulting from Variable Loading 281 here employs an effective dimension de obtained by equating the volume of material stressed at and above 95 percent of the maximum stress to the same volume in the rotating-beam specimen.16 It turns out that when these two volumes are equated, the lengths cancel, and so we need only consider the areas. For a rotating round section, the 95 percent stress area is the area in a ring having an outside diameter d and an inside diameter of 0.95d. So, designating the 95 percent stress area A0.95σ , we have π A0.95σ = [d 2 (0.95d )2 ] = 0.0766d 2 (622) 4 This equation is also valid for a rotating hollow round. For nonrotating solid or hollow rounds, the 95 percent stress area is twice the area outside of two parallel chords having a spacing of 0.95d, where d is the diameter. Using an exact computation, this is A0.95σ = 0.01046d 2 (623) with de in Eq. (622), setting Eqs. (622) and (623) equal to each other enables us to solve for the effective diameter. This gives (624) de = 0.370d as the effective size of a round corresponding to a nonrotating solid or hollow round. A rectangular section of dimensions h × b has A0.95σ = 0.05hb . Using the same approach as before, de = 0.808(hb)1/2 (625) Table 63 provides A0.95σ areas of common structural shapes undergoing nonrotating bending. 16 See R. Kuguel, A Relation between Theoretical Stress Concentration Factor and Fatigue Notch Factor Deduced from the Concept of Highly Stressed Volume, Proc. ASTM, vol. 61, 1961, pp. 732748. EXAMPLE 64 Solution A steel shaft loaded in bending is 32 mm in diameter, abutting a lleted shoulder 38 mm in diameter. The shaft material has a mean ultimate tensile strength of 690 MPa. Estimate the Marin size factor kb if the shaft is used in (a) A rotating mode. (b) A nonrotating mode. (a) From Eq. (620) Answer kb = d 7.62 0.107 = 32 7.62 0.107 = 0.858 (b) From Table 63, de = 0.37d = 0.37(32) = 11.84 mm From Eq. (620), Answer kb = 11.84 7.62 0.107 = 0.954 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 282 II. Failure Prevention 285 © The McGrawHill Companies, 2008 6. Fatigue Failure Resulting from Variable Loading Mechanical Engineering Design Table 63 A0.95σ Areas of Common Nonrotating Structural Shapes A 0.95σ = 0.01046d 2 d de = 0.370d b 2 h 1 A 0 .95 σ = 0.05hb de = 0.808 hb 1 2 a 1 b 2 A 0 .95 σ = 2 0.10at f 0.05ba axis 1-1 t f > 0.025a axis 2-2 tf 1 a 1 x 2 b A 0 .95 σ = 2 tf 0.05ab axis 1-1 0.052xa + 0.1t f ( b x ) axis 2-2 1 Loading Factor kc When fatigue tests are carried out with rotating bending, axial (push-pull), and torsional loading, the endurance limits differ with Sut. This is discussed further in Sec. 617. Here, we will specify average values of the load factor as kc = 1 0.85 0.59 bending axial torsion17 (626) Temperature Factor kd When operating temperatures are below room temperature, brittle fracture is a strong possibility and should be investigated rst. When the operating temperatures are higher than room temperature, yielding should be investigated rst because the yield strength drops off so rapidly with temperature; see Fig. 29. Any stress will induce creep in a material operating at high temperatures; so this factor must be considered too. 17 Use this only for pure torsional fatigue loading. When torsion is combined with other stresses, such as bending, kc = 1 and the combined loading is managed by using the effective von Mises stress as in Sec. 55. Note: For pure torsion, the distortion energy predicts that (kc)torsion = 0.577. 286 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition II. Failure Prevention © The McGrawHill Companies, 2008 6. Fatigue Failure Resulting from Variable Loading Fatigue Failure Resulting from Variable Loading Table 64 Effect of Operating Temperature on the Tensile Strength of Steel.* (ST = tensile strength at operating temperature; SRT = tensile strength at room temperature; ˆ 0.099 σ 0.110) Temperature, °C 20 ST/SRT 1.000 Temperature, °F 70 283 ST/SRT 1.000 50 1.010 100 1.008 100 1.020 200 1.020 150 1.025 300 1.024 200 1.020 400 1.018 250 1.000 500 0.995 300 0.975 600 0.963 350 0.943 700 0.927 400 0.900 800 0.872 450 0.843 900 0.797 500 0.768 1000 0.698 550 0.672 1100 0.567 600 0.549 *Data source: Fig. 29. Finally, it may be true that there is no fatigue limit for materials operating at high temperatures. Because of the reduced fatigue resistance, the failure process is, to some extent, dependent on time. The limited amount of data available show that the endurance limit for steels increases slightly as the temperature rises and then begins to fall off in the 400 to 700°F range, not unlike the behavior of the tensile strength shown in Fig. 29. For this reason it is probably true that the endurance limit is related to tensile strength at elevated temperatures in the same manner as at room temperature.18 It seems quite logical, therefore, to employ the same relations to predict endurance limit at elevated temperatures as are used at room temperature, at least until more comprehensive data become available. At the very least, this practice will provide a useful standard against which the performance of various materials can be compared. Table 64 has been obtained from Fig. 29 by using only the tensile-strength data. Note that the table represents 145 tests of 21 different carbon and alloy steels. A fourthorder polynomial curve t to the data underlying Fig. 29 gives 2 kd = 0.975 + 0.432(103 )TF 0.115(105 )TF 3 4 + 0.104(108 )TF 0.595(1012 )TF ( 627) where 70 TF 1000 F. Two types of problems arise when temperature is a consideration. If the rotatingbeam endurance limit is known at room temperature, then use ST kd = (628) S RT 18 For more, see Table 2 of ANSI/ASME B106. 1M-1985 shaft standard, and E. A. Brandes (ed.), Smithells Metals Reference Book, 6th ed., Butterworth, London, 1983, pp. 22134 to 22136, where endurance limits from 100 to 650°C are tabulated. BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 284 II. Failure Prevention 6. Fatigue Failure Resulting from Variable Loading 287 © The McGrawHill Companies, 2008 Mechanical Engineering Design from Table 64 or Eq. (627) and proceed as usual. If the rotating-beam endurance limit is not given, then compute it using Eq. (68) and the temperature-corrected tensile strength obtained by using the factor from Table 64. Then use kd = 1. EXAMPLE 65 Solution A 1035 steel has a tensile strength of 70 kpsi and is to be used for a part that sees 450°F in service. Estimate the Marin temperature modication factor and ( Se )450 if (a) The room-temperature endurance limit by test is ( Se )70 = 39.0 kpsi. (b) Only the tensile strength at room temperature is known. (a) First, from Eq. (627), kd = 0.975 + 0.432(103 )(450) 0.115(105 )(4502 ) + 0.104(108 )(4503 ) 0.595(1012 )(4504 ) = 1.007 Thus, ( Se )450 = kd ( Se )70 = 1.007(39.0) = 39.3 kpsi Answer (b) Interpolating from Table 64 gives ( ST / S RT )450 = 1.018 + (0.995 1.018) 450 400 = 1.007 500 400 Thus, the tensile strength at 450°F is estimated as ( Sut )450 = ( ST / S RT )450 ( Sut )70 = 1.007(70) = 70.5 kpsi From Eq. (68) then, Answer ( S e )450 = 0.5 ( Sut )450 = 0.5(70.5) = 35.2 kpsi Part a gives the better estimate due to actual testing of the particular material. Reliability Factor ke The discussion presented here accounts for the scatter of data such as shown in . Fig. 617 where the mean endurance limit is shown to be Se / Sut = 0.5, or as given by Eq. (68). Most endurance strength data are reported as mean values. Data presented by Haugen and Wirching19 show standard deviations of endurance strengths of less than 8 percent. Thus the reliability modication factor to account for this can be written as ke = 1 0.08 z a (629) where za is dened by Eq. (2016) and values for any desired reliability can be determined from Table A10. Table 65 gives reliability factors for some standard specied reliabilities. For a more comprehensive approach to reliability, see Sec. 617. 19 E. B. Haugen and P. H. Wirsching, Probabilistic Design, Machine Design, vol. 47, no. 12, 1975, pp. 1014. 288 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition II. Failure Prevention © The McGrawHill Companies, 2008 6. Fatigue Failure Resulting from Variable Loading Fatigue Failure Resulting from Variable Loading Table 65 Reliability Factors ke Corresponding to 8 Percent Standard Deviation of the Endurance Limit Reliability, % Transformation Variate za Reliability Factor ke 50 0 1.000 90 1.288 0.897 95 1.645 0.868 99 2.326 0.814 99.9 3.091 0.753 99.99 3.719 0.702 99.999 4.265 0.659 99.9999 4.753 0.620 Figure 619 The failure of a case-hardened part in bending or torsion. In this example, failure occurs in the core. 285 Se (case) or Case Core Se (core) Miscellaneous-Effects Factor kf Though the factor k f is intended to account for the reduction in endurance limit due to all other effects, it is really intended as a reminder that these must be accounted for, because actual values of k f are not always available. Residual stresses may either improve the endurance limit or affect it adversely. Generally, if the residual stress in the surface of the part is compression, the endurance limit is improved. Fatigue failures appear to be tensile failures, or at least to be caused by tensile stress, and so anything that reduces tensile stress will also reduce the possibility of a fatigue failure. Operations such as shot peening, hammering, and cold rolling build compressive stresses into the surface of the part and improve the endurance limit signicantly. Of course, the material must not be worked to exhaustion. The endurance limits of parts that are made from rolled or drawn sheets or bars, as well as parts that are forged, may be affected by the so-called directional characteristics of the operation. Rolled or drawn parts, for example, have an endurance limit in the transverse direction that may be 10 to 20 percent less than the endurance limit in the longitudinal direction. Parts that are case-hardened may fail at the surface or at the maximum core radius, depending upon the stress gradient. Figure 619 shows the typical triangular stress distribution of a bar under bending or torsion. Also plotted as a heavy line in this gure are the endurance limits Se for the case and core. For this example the endurance limit of the core rules the design because the gure shows that the stress σ or τ, whichever applies, at the outer core radius, is appreciably larger than the core endurance limit. BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 286 II. Failure Prevention 6. Fatigue Failure Resulting from Variable Loading © The McGrawHill Companies, 2008 Mechanical Engineering Design Of course, if stress concentration is also present, the stress gradient is much steeper, and hence failure in the core is unlikely. Corrosion It is to be expected that parts that operate in a corrosive atmosphere will have a lowered fatigue resistance. This is, of course, true, and it is due to the roughening or pitting of the surface by the corrosive material. But the problem is not so simple as the one of nding the endurance limit of a specimen that has been corroded. The reason for this is that the corrosion and the stressing occur at the same time. Basically, this means that in time any part will fail when subjected to repeated stressing in a corrosive atmosphere. There is no fatigue limit. Thus the designers problem is to attempt to minimize the factors that affect the fatigue life; these are: Mean or static stress Alternating stress Electrolyte concentration Dissolved oxygen in electrolyte Material properties and composition Temperature Cyclic frequency Fluid ow rate around specimen Local crevices Electrolytic Plating Metallic coatings, such as chromium plating, nickel plating, or cadmium plating, reduce the endurance limit by as much as 50 percent. In some cases the reduction by coatings has been so severe that it has been necessary to eliminate the plating process. Zinc plating does not affect the fatigue strength. Anodic oxidation of light alloys reduces bending endurance limits by as much as 39 percent but has no effect on the torsional endurance limit. Metal Spraying Metal spraying results in surface imperfections that can initiate cracks. Limited tests show reductions of 14 percent in the fatigue strength. Cyclic Frequency If, for any reason, the fatigue process becomes time-dependent, then it also becomes frequency-dependent. Under normal conditions, fatigue failure is independent of frequency. But when corrosion or high temperatures, or both, are encountered, the cyclic rate becomes important. The slower the frequency and the higher the temperature, the higher the crack propagation rate and the shorter the life at a given stress level. Frettage Corrosion The phenomenon of frettage corrosion is the result of microscopic motions of tightly tting parts or structures. Bolted joints, bearing-race ts, wheel hubs, and any set of tightly tted parts are examples. The process involves surface discoloration, pitting, and eventual fatigue. The frettage factor k f depends upon the material of the mating pairs and ranges from 0.24 to 0.90. 289 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition II. Failure Prevention © The McGrawHill Companies, 2008 6. Fatigue Failure Resulting from Variable Loading Fatigue Failure Resulting from Variable Loading 610 287 Stress Concentration and Notch Sensitivity In Sec. 313 it was pointed out that the existence of irregularities or discontinuities, such as holes, grooves, or notches, in a part increases the theoretical stresses signicantly in the immediate vicinity of the discontinuity. Equation (348) dened a stress concentration factor K t (or K ts ), which is used with the nominal stress to obtain the maximum resulting stress due to the irregularity or defect. It turns out that some materials are not fully sensitive to the presence of notches and hence, for these, a reduced value of Kt can be used. For these materials, the maximum stress is, in fact, σmax = K f σ0 or (630) τmax = K f s τ0 where K f is a reduced value of K t and σ0 is the nominal stress. The factor K f is commonly called a fatigue stress-concentration factor, and hence the subscript f. So it is convenient to think of Kf as a stress-concentration factor reduced from Kt because of lessened sensitivity to notches. The resulting factor is dened by the equation Kf = maximum stress in notched specimen stress in notch-free specimen (a) Notch sensitivity q is dened by the equation q= Kf 1 Kt 1 qshear = or Kfs 1 K ts 1 (631) where q is usually between zero and unity. Equation (631) shows that if q = 0, then K f = 1, and the material has no sensitivity to notches at all. On the other hand, if q = 1, then K f = K t , and the material has full notch sensitivity. In analysis or design work, nd Kt rst, from the geometry of the part. Then specify the material, nd q, and solve for Kf from the equation K f = 1 + q ( K t 1) K f s = 1 + qshear ( K ts 1) or (632) For steels and 2024 aluminum alloys, use Fig. 620 to nd q for bending and axial loading. For shear loading, use Fig. 621. In using these charts it is well to know that the actual test results from which the curves were derived exhibit a large amount of Figure 620 Notch-sensitivity charts for steels and UNS A92024-T wrought aluminum alloys subjected to reversed bending or reversed axial loads. For larger notch radii, use the values of q corresponding to the r = 0.16-in (4-mm) ordinate. (From George Sines and J. L. Waisman (eds.), Metal Fatigue, McGraw-Hill, New York. Copyright © 1969 by The McGraw-Hill Companies, Inc. Reprinted by permission.) Notch radius r, mm 1.0 0 0.5 S ut = 1.0 kpsi 200 15 0.8 2.0 2.5 3.0 3.5 4.0 (0.7) 0 (0.4) 10 0.6 1.5 (1.4 GPa) (1.0) 0 Notch sensitivity q 290 60 0.4 Steels Alum. alloy 0.2 0 0 0.02 0.04 0.06 0.08 0.10 Notch radius r, in 0.12 0.14 0.16 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 288 II. Failure Prevention 291 © The McGrawHill Companies, 2008 6. Fatigue Failure Resulting from Variable Loading Mechanical Engineering Design Figure 621 1.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 0.14 0.16 0.8 Notch sensitivity qshear Notch-sensitivity curves for materials in reversed torsion. For larger notch radii, use the values of qshear corresponding to r = 0.16 in (4 mm). Notch radius r, mm 0 Quenched and drawn steels (Bhn > 200) Annealed steels (Bhn < 200) 0.6 0.4 Aluminum alloys 0.2 0 0 0.02 0.04 0.06 0.08 0.10 0.12 Notch radius r, in scatter. Because of this scatter it is always safe to use K f = K t if there is any doubt about the true value of q. Also, note that q is not far from unity for large notch radii. The notch sensitivity of the cast irons is very low, varying from 0 to about 0.20, depending upon the tensile strength. To be on the conservative side, it is recommended that the value q = 0.20 be used for all grades of cast iron. Figure 620 has as its basis the Neuber equation, which is given by Kf = 1 + Kt 1 1 + a /r (633) where a is dened as the Neuber constant and is a material constant. Equating Eqs. (631) and (633) yields the notch sensitivity equation q= 1 a 1+ r (634) For steel, with Sut in kpsi, the Neuber constant can be approximated by a third-order polynomial t of data as a = 0.245 799 0.307 794(102 ) Sut 2 3 + 0.150 874(104 ) Sut 0.266 978(107 ) Sut (635) To use Eq. (633) or (634) for torsion for low-alloy steels, increase the ultimate strength by 20 kpsi in Eq. (635) and apply this value of a . EXAMPLE 66 A steel shaft in bending has an ultimate strength of 690 MPa and a shoulder with a llet radius of 3 mm connecting a 32-mm diameter with a 38-mm diameter. Estimate Kf using: (a) Figure 620. (b) Equations (633) and (635). 292 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition II. Failure Prevention © The McGrawHill Companies, 2008 6. Fatigue Failure Resulting from Variable Loading Fatigue Failure Resulting from Variable Loading Solution Answer Answer 289 From Fig. A159, using D /d = 38/32 = 1.1875, r /d = 3/32 = 0.093 75, we read . the graph to nd K t = 1.65. . (a) From Fig. 620, for Sut = 690 MPa and r = 3 mm, q = 0.84. Thus, from Eq. (632) . K f = 1 + q ( K t 1) = 1 + 0.84(1.65 1) = 1.55 (b) From Eq. (635) with Sut = 690 MPa = 100 kpsi, a = 0.0622 in = 0.313 mm. Substituting this into Eq. (633) with r = 3 mm gives Kf = 1 + Kt 1 . 1.65 1 =1+ = 1.55 0.313 1 + a /r 1+ 3 For simple loading, it is acceptable to reduce the endurance limit by either dividing the unnotched specimen endurance limit by K f or multiplying the reversing stress by K f . However, in dealing with combined stress problems that may involve more than one value of fatigue-concentration factor, the stresses are multiplied by K f . EXAMPLE 67 Solution Consider an unnotched specimen with an endurance limit of 55 kpsi. If the specimen was notched such that K f = 1.6, what would be the factor of safety against failure for N > 106 cycles at a reversing stress of 30 kpsi? (a) Solve by reducing Se . (b) Solve by increasing the applied stress. (a) The endurance limit of the notched specimen is given by Se = Se 55 = 34.4 kpsi = Kf 1.6 and the factor of safety is Answer n= Se 34.4 = 1.15 = σa 30 (b) The maximum stress can be written as (σa )max = K f σa = 1.6(30) = 48.0 kpsi and the factor of safety is Answer n= Se 55 = 1.15 = K f σa 48 Up to this point, examples illustrated each factor in Marins equation and stress concentrations alone. Let us consider a number of factors occurring simultaneously. BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 290 II. Failure Prevention 6. Fatigue Failure Resulting from Variable Loading © The McGrawHill Companies, 2008 Mechanical Engineering Design EXAMPLE 68 A 1015 hot-rolled steel bar has been machined to a diameter of 1 in. It is to be placed in reversed axial loading for 70 000 cycles to failure in an operating environment of 550°F. Using ASTM minimum properties, and a reliability of 99 percent, estimate the endurance limit and fatigue strength at 70 000 cycles. Solution From Table A20, Sut = 50 kpsi at 70°F. Since the rotating-beam specimen endurance limit is not known at room temperature, we determine the ultimate strength at the elevated temperature rst, using Table 64. From Table 64, ST S RT 550 = 0.995 + 0.963 = 0.979 2 The ultimate strength at 550°F is then ( Sut )550 = ( ST / S RT )550 ( Sut )70 = 0.979(50) = 49.0 kpsi The rotating-beam specimen endurance limit at 550°F is then estimated from Eq. (68) as Se = 0.5(49) = 24.5 kpsi Next, we determine the Marin factors. For the machined surface, Eq. (619) with Table 62 gives b ka = aSut = 2.70(490.265 ) = 0.963 For axial loading, from Eq. (621), the size factor kb = 1, and from Eq. (626) the loading factor is kc = 0.85. The temperature factor kd = 1, since we accounted for the temperature in modifying the ultimate strength and consequently the endurance limit. For 99 percent reliability, from Table 65, ke = 0.814. Finally, since no other conditions were given, the miscellaneous factor is kf = 1. The endurance limit for the part is estimated by Eq. (618) as Se = ka kb kc kd ke k f Se Answer = 0.963(1)(0.85)(1)(0.814)(1)24.5 = 16.3 kpsi For the fatigue strength at 70 000 cycles we need to construct the S-N equation. From p. 277, since Sut = 49 < 70 kpsi, then f 0.9. From Eq. (614) a= ( f Sut )2 [0.9(49)]2 = 119.3 kpsi = Se 16.3 and Eq. (615) 1 b = log 3 f Sut Se 1 0.9(49) = log = 0.1441 3 16.3 Finally, for the fatigue strength at 70 000 cycles, Eq. (613) gives Answer S f = a N b = 119.3(70 000)0.1441 = 23.9 kpsi 293 294 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition II. Failure Prevention © The McGrawHill Companies, 2008 6. Fatigue Failure Resulting from Variable Loading Fatigue Failure Resulting from Variable Loading 291 EXAMPLE 69 Figure 622a shows a rotating shaft simply supported in ball bearings at A and D and loaded by a nonrotating force F of 6.8 kN. Using ASTM minimum strengths, estimate the life of the part. Solution From Fig. 622b we learn that failure will probably occur at B rather than at C or at the point of maximum moment. Point B has a smaller cross section, a higher bending moment, and a higher stress-concentration factor than C, and the location of maximum moment has a larger size and no stress-concentration factor. We shall solve the problem by rst estimating the strength at point B, since the strength will be different elsewhere, and comparing this strength with the stress at the same point. From Table A20 we nd Sut = 690 MPa and Sy = 580 MPa. The endurance limit Se is estimated as Se = 0.5(690) = 345 MPa From Eq. (619) and Table 62, ka = 4.51(690)0.265 = 0.798 From Eq. (620), kb = (32/7.62)0.107 = 0.858 Since kc = kd = ke = k f = 1, Se = 0.798(0.858)345 = 236 MPa To nd the geometric stress-concentration factor K t we enter Fig. A159 with D /d = . 1. 38/32 = 1.1875 and r /d = 3/32 = 0.093 75 and read K t = 65. Substituting Sut = 690/6.89 = 100 kpsi into Eq. (635) yields a = 0.0622 in = 0.313 mm. Substituting this into Eq. (633) gives Kf = 1 + Figure 622 (a) Shaft drawing showing all dimensions in millimeters; all llets 3-mm radius. The shaft rotates and the load is stationary; material is machined from AISI 1050 cold-drawn steel. (b) Bendingmoment diagram. A 1.65 1 Kt 1 =1+ = 1.55 1 + a /r 1 + 0.313/ 3 6.8 kN B 250 75 C 100 125 10 10 32 30 D 35 38 30 R2 R1 (a ) Mmax MB MC A B C (b) D BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 292 II. Failure Prevention © The McGrawHill Companies, 2008 6. Fatigue Failure Resulting from Variable Loading Mechanical Engineering Design The next step is to estimate the bending stress at point B. The bending moment is M B = R1 x = 225(6.8) 225 F 250 = 250 = 695.5 N · m 550 550 Just to the left of B the section modulus is I /c = π d 3 /32 = π 323 /32 = 3.217 (103 )mm3 . The reversing bending stress is, assuming innite life, σ = Kf 695.5 MB = 1.55 (10)6 = 335.1(106 ) Pa = 335.1 MPa I /c 3.217 This stress is greater than Se and less than Sy. This means we have both nite life and no yielding on the rst cycle. For nite life, we will need to use Eq. (616). The ultimate strength, Sut = 690 MPa = 100 kpsi. From Fig. 618, f = 0.844. From Eq. (614) a= [0.844(690)]2 ( f Sut )2 = 1437 MPa = Se 236 and from Eq. (615) 1 b = log 3 f Sut Se 0.844(690) 1 = 0.1308 = log 3 236 From Eq. (616), Answer 611 N= σa a 1/b = 335.1 1437 1/0.1308 = 68(103 ) cycles Characterizing Fluctuating Stresses Fluctuating stresses in machinery often take the form of a sinusoidal pattern because of the nature of some rotating machinery. However, other patterns, some quite irregular, do occur. It has been found that in periodic patterns exhibiting a single maximum and a single minimum of force, the shape of the wave is not important, but the peaks on both the high side (maximum) and the low side (minimum) are important. Thus Fmax and Fmin in a cycle of force can be used to characterize the force pattern. It is also true that ranging above and below some baseline can be equally effective in characterizing the force pattern. If the largest force is Fmax and the smallest force is Fmin , then a steady component and an alternating component can be constructed as follows: Fm = Fmax + Fmin 2 Fa = Fmax Fmin 2 where Fm is the midrange steady component of force, and Fa is the amplitude of the alternating component of force. 295 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition II. Failure Prevention © The McGrawHill Companies, 2008 6. Fatigue Failure Resulting from Variable Loading 293 Fatigue Failure Resulting from Variable Loading Figure 623 Stress a Stress r Time a max m min (a ) O Time Stress (d ) Stress Time a r max a O (b ) min =0 m Time (e) + a Time Stress Some stress-time relations: (a) uctuating stress with highfrequency ripple; (b and c) nonsinusoidal uctuating stress; (d) sinusoidal uctuating stress; (e) repeated stress; (f ) completely reversed sinusoidal stress. Stress 296 Time O r a m (c) =0 (f) Figure 623 illustrates some of the various stress-time traces that occur. The components of stress, some of which are shown in Fig. 623d, are σmin = minimum stress σmax = maximum stress σa = amplitude component σm = midrange component σr = range of stress σs = static or steady stress The steady, or static, stress is not the same as the midrange stress; in fact, it may have any value between σmin and σmax . The steady stress exists because of a xed load or preload applied to the part, and it is usually independent of the varying portion of the load. A helical compression spring, for example, is always loaded into a space shorter than the free length of the spring. The stress created by this initial compression is called the steady, or static, component of the stress. It is not the same as the midrange stress. We shall have occasion to apply the subscripts of these components to shear stresses as well as normal stresses. The following relations are evident from Fig. 623: σmax + σmin 2 σmax σmin σa = 2 σm = (636) BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 294 II. Failure Prevention 297 © The McGrawHill Companies, 2008 6. Fatigue Failure Resulting from Variable Loading Mechanical Engineering Design In addition to Eq. (636), the stress ratio σmin σmax R= (637) σa σm (638) and the amplitude ratio A= are also dened and used in connection with uctuating stresses. Equations (636) utilize symbols σa and σm as the stress components at the location under scrutiny. This means, in the absence of a notch, σa and σm are equal to the nominal stresses σao and σmo induced by loads Fa and Fm , respectively; in the presence of a notch they are K f σao and K f σmo , respectively, as long as the material remains without plastic strain. In other words, the fatigue stress concentration factor K f is applied to both components. When the steady stress component is high enough to induce localized notch yielding, the designer has a problem. The rst-cycle local yielding produces plastic strain and strain-strengthening. This is occurring at the location where fatigue crack nucleation and growth are most likely. The material properties ( Sy and Sut ) are new and difcult to quantify. The prudent engineer controls the concept, material and condition of use, and geometry so that no plastic strain occurs. There are discussions concerning possible ways of quantifying what is occurring under localized and general yielding in the presence of a notch, referred to as the nominal mean stress method, residual stress method, and the like.20 The nominal mean stress method (set σa = K f σao and σm = σmo ) gives roughly comparable results to the residual stress method, but both are approximations. There is the method of Dowling21 for ductile materials, which, for materials with a pronounced yield point and approximated by an elasticperfectly plastic behavior model, quantitatively expresses the steady stress component stress-concentration factor K f m as Kfm = Kf Kfm = Sy K f σao |σmo | Kfm = 0 K f |σmax,o | < Sy K f |σmax,o | > Sy (639) K f |σmax,o σmin,o | > 2 Sy For the purposes of this book, for ductile materials in fatigue, Avoid localized plastic strain at a notch. Set σa = K f σa ,o and σm = K f σmo . When plastic strain at a notch cannot be avoided, use Eqs. (639); or conservatively, set σa = K f σao and use K f m = 1, that is, σm = σmo . 20 R. C. Juvinall, Stress, Strain, and Strength, McGraw-Hill, New York, 1967, articles 14.914.12; R. C. Juvinall and K. M. Marshek, Fundamentals of Machine Component Design, 4th ed., Wiley, New York, 2006, Sec. 8.11; M. E. Dowling, Mechanical Behavior of Materials, 2nd ed., Prentice Hall, Englewood Cliffs, N.J., 1999, Secs. 10.310.5. 21 Dowling, op. cit., p. 437438. BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition II. Failure Prevention © The McGrawHill Companies, 2008 6. Fatigue Failure Resulting from Variable Loading Fatigue Failure Resulting from Variable Loading 612 295 Fatigue Failure Criteria for Fluctuating Stress Now that we have dened the various components of stress associated with a part subjected to uctuating stress, we want to vary both the midrange stress and the stress amplitude, or alternating component, to learn something about the fatigue resistance of parts when subjected to such situations. Three methods of plotting the results of such tests are in general use and are shown in Figs. 624, 625, and 626. The modied Goodman diagram of Fig. 624 has the midrange stress plotted along the abscissa and all other components of stress plotted on the ordinate, with tension in the positive direction. The endurance limit, fatigue strength, or nite-life strength, whichever applies, is plotted on the ordinate above and below the origin. The midrangestress line is a 45 line from the origin to the tensile strength of the part. The modied Goodman diagram consists of the lines constructed to Se (or S f ) above and below the origin. Note that the yield strength is also plotted on both axes, because yielding would be the criterion of failure if σmax exceeded Sy . Another way to display test results is shown in Fig. 625. Here the abscissa represents the ratio of the midrange strength Sm to the ultimate strength, with tension plotted to the right and compression to the left. The ordinate is the ratio of the alternating strength to the endurance limit. The line B C then represents the modied Goodman criterion of failure. Note that the existence of midrange stress in the compressive region has little effect on the endurance limit. The very clever diagram of Fig. 626 is unique in that it displays four of the stress components as well as the two stress ratios. A curve representing the endurance limit for values of R beginning at R = 1 and ending with R = 1 begins at Se on the σa axis and ends at Sut on the σm axis. Constant-life curves for N = 105 and N = 104 cycles Figure 624 + Modied Goodman diagram showing all the strengths and the limiting values of all the stress components for a particular midrange stress. Su Stress Sy max ss tre x. s Ma Se a M id str ran es ge s r a min 45° 0 Sy m M in. str ess Parallel 298 Se Midrange stress Su BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 296 II. Failure Prevention 299 © The McGrawHill Companies, 2008 6. Fatigue Failure Resulting from Variable Loading Mechanical Engineering Design 1.2 Amplitude ratio Sa / Se 1.0 A B 0.8 0.6 0.4 0.2 C 1.2 1.0 0.8 0.6 0.4 0.2 0 0.2 Compression Sm / Suc 0.4 0.6 0.8 1.0 Tension Sm / Sut Midrange ratio Figure 625 Plot of fatigue failures for midrange stresses in both tensile and compressive regions. Normalizing the data by using the ratio of steady strength component to tensile strength Sm / Sut , steady strength component to compressive strength Sm / Suc and strength amplitude component to endurance limit Sa / Se enables a plot of experimental results for a variety of steels. [Data source: Thomas J. Dolan, Stress Range, Sec. 6.2 in O. J. Horger (ed.), ASME HandbookMetals Engineering Design, McGraw-Hill, New York, 1953.] Figure 626 1.5 0.2 A=1 R=0 0.67 0.2 RA 0.43 0.4 0.25 0.6 0.11 0.8 0 1.0 16 0 18 0 2.33 0.4 0 180 ,k m 0 10 tre es ng 80 ra id 60 40 40 i ps ,k 60 a s es str g tin Se M a rn 40 80 lte A 60 20 20 20 kpsi max , 10 0 Maximum stress A 6 10 80 160 14 5 10 0 100 Sut ss 120 4c 10 s y cle 12 0 ps i A= R = 1.0 12 Master fatigue diagram created for AISI 4340 steel having Sut = 158 and Sy = 147 kpsi. The stress components at A are σmin = 20, σmax = 120, σm = 70, and σa = 50, all in kpsi. (Source: H. J. Grover, Fatigue of Aircraft Structures, U.S. Government Printing Ofce, Washington, D.C., 1966, pp. 317, 322. See also J. A. Collins, Failure of Materials in Mechanical Design, Wiley, New York, 1981, p. 216.) 4.0 0.6 120 100 80 60 40 20 0 20 40 Minimum stress 60 min , 80 100 120 140 kpsi have been drawn too. Any stress state, such as the one at A, can be described by the minimum and maximum components, or by the midrange and alternating components. And safety is indicated whenever the point described by the stress components lies below the constant-life line. BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition II. Failure Prevention © The McGrawHill Companies, 2008 6. Fatigue Failure Resulting from Variable Loading Fatigue Failure Resulting from Variable Loading Figure 627 Yield (Langer) line a Fatigue diagram showing various criteria of failure. For each criterion, points on or above the respective line indicate failure. Some point A on the Goodman line, for example, gives the strength Sm as the limiting value of σm corresponding to the strength Sa , which, paired with σm , is the limiting value of σa . 297 Sy Alternating stress 300 Se Gerber line Load line, slope r = Sa /Sm Modified Goodman line Sa A ASME-elliptic line Soderberg line 0 0 Sm Midrange stress Sy Sut m When the midrange stress is compression, failure occurs whenever σa = Se or whenever σmax = Syc , as indicated by the left-hand side of Fig. 625. Neither a fatigue diagram nor any other failure criteria need be developed. In Fig. 627, the tensile side of Fig. 625 has been redrawn in terms of strengths, instead of strength ratios, with the same modied Goodman criterion together with four additional criteria of failure. Such diagrams are often constructed for analysis and design purposes; they are easy to use and the results can be scaled off directly. The early viewpoint expressed on a σa σm diagram was that there existed a locus which divided safe from unsafe combinations of σa and σm . Ensuing proposals included the parabola of Gerber (1874), the Goodman (1890)22 (straight) line, and the Soderberg (1930) (straight) line. As more data were generated it became clear that a fatigue criterion, rather than being a fence, was more like a zone or band wherein the probability of failure could be estimated. We include the failure criterion of Goodman because It is a straight line and the algebra is linear and easy. It is easily graphed, every time for every problem. It reveals subtleties of insight into fatigue problems. Answers can be scaled from the diagrams as a check on the algebra. We also caution that it is deterministic and the phenomenon is not. It is biased and we cannot quantify the bias. It is not conservative. It is a stepping-stone to understanding; it is history; and to read the work of other engineers and to have meaningful oral exchanges with them, it is necessary that you understand the Goodman approach should it arise. Either the fatigue limit Se or the nite-life strength S f is plotted on the ordinate of Fig. 627. These values will have already been corrected using the Marin factors of Eq. (618). Note that the yield strength Sy is plotted on the ordinate too. This serves as a reminder that rst-cycle yielding rather than fatigue might be the criterion of failure. The midrange-stress axis of Fig. 627 has the yield strength Sy and the tensile strength Sut plotted along it. 22 It is difcult to date Goodmans work because it went through several modications and was never published. BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 298 II. Failure Prevention 301 © The McGrawHill Companies, 2008 6. Fatigue Failure Resulting from Variable Loading Mechanical Engineering Design Five criteria of failure are diagrammed in Fig. 627: the Soderberg, the modied Goodman, the Gerber, the ASME-elliptic, and yielding. The diagram shows that only the Soderberg criterion guards against any yielding, but is biased low. Considering the modied Goodman line as a criterion, point A represents a limiting point with an alternating strength Sa and midrange strength Sm. The slope of the load line shown is dened as r = Sa / Sm . The criterion equation for the Soderberg line is Sm Sa + =1 Se Sy (640) Similarly, we nd the modied Goodman relation to be Sa Sm + =1 Se Sut (641) Examination of Fig. 625 shows that both a parabola and an ellipse have a better opportunity to pass among the midrange tension data and to permit quantication of the probability of failure. The Gerber failure criterion is written as Sa + Se 2 Sm Sut (642) =1 and the ASME-elliptic is written as 2 Sa Se 2 Sm Sy + (643) =1 The Langer first-cycle-yielding criterion is used in connection with the fatigue curve: (644) Sa + Sm = Sy The stresses n σa and n σm can replace Sa and Sm , where n is the design factor or factor of safety. Then, Eq. (640), the Soderberg line, becomes σa σm 1 + = Se Sy n Soderberg (645) Equation (641), the modied Goodman line, becomes mod-Goodman σa σm 1 + = Se Sut n (646) Equation (642), the Gerber line, becomes Gerber n σa + Se n σm Sut 2 (647) =1 Equation (643), the ASME-elliptic line, becomes ASME-elliptic n σa Se 2 + n σm Sy 2 =1 (648) We will emphasize the Gerber and ASME-elliptic for fatigue failure criterion and the Langer for rst-cycle yielding. However, conservative designers often use the modied Goodman criterion, so we will continue to include it in our discussions. The design equation for the Langer rst-cycle-yielding is Langer static yield σa + σm = Sy n (649) 302 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition II. Failure Prevention © The McGrawHill Companies, 2008 6. Fatigue Failure Resulting from Variable Loading Fatigue Failure Resulting from Variable Loading 299 The failure criteria are used in conjunction with a load line, r = Sa / Sm = σa /σm . Principal intersections are tabulated in Tables 66 to 68. Formal expressions for fatigue factor of safety are given in the lower panel of Tables 66 to 68. The rst row of each table corresponds to the fatigue criterion, the second row is the static Langer criterion, and the third row corresponds to the intersection of the static and fatigue Table 66 Intersecting Equations Amplitude and Steady Coordinates of Strength and Important Intersections in First Quadrant for Modied Goodman and Langer Failure Criteria Sa Sm + =1 Se Sut Intersection Coordinates Sa = Sa Sm r Se Sut r Sut + Se Sm = Sa Sm + =1 Sy Sy Sa Sm Load line r = Sa r Sa = Load line r = r Sy 1+r Sm = Sy 1+r Sy Se Sut Sut Se Sa Sm + =1 Se Sut Sm = Sa Sm + =1 Sy Sy Sa = Sy Sm , r crit = Sa / Sm Fatigue factor of safety 1 n f = σa σm + Se Sut Table 67 Amplitude and Steady Coordinates of Strength and Important Intersections in First Quadrant for Gerber and Langer Failure Criteria Intersecting Equations Sa + Se Sm Sut Intersection Coordinates 2 2 Se 2 r 2 Sut 1 + 1 + Sa = 2 Se r Sut 2 =1 Load line r = Sm = Sa Sm + =1 Sy Sy Load line r = Sa + Se Sm Sut Sa r Sa = Sa Sm r Sy 1+r Sy 1+r 2 Sut Sm = 1 2 Se Sa Sm Sm = 2 =1 Sa Sm + =1 Sy Sy 1+ 2 Se Sut 2 Sy 1 Se Sa = Sy Sm , r crit = Sa / Sm Fatigue factor of safety 1 nf = 2 Sut σm 2 σa 1 + Se 1+ 2σm Se Sut σa 2 σm > 0 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 300 II. Failure Prevention © The McGrawHill Companies, 2008 6. Fatigue Failure Resulting from Variable Loading Mechanical Engineering Design Table 68 Intersecting Equations Amplitude and Steady Coordinates of Strength and Important Intersections in First Quadrant for ASMEElliptic and Langer Failure Criteria Sa Se 2 + Sm Sy Intersection Coordinates 2 22 r 2 Se Sy Sa = =1 2 2 Se + r 2 Sy Load line r = Sa / Sm Sm = Sa r Sa Sm + =1 Sy Sy Sa = r Sy 1+r Load line r = Sa / Sm Sm = Sy 1+r Sa Se 2 + Sm Sy 2 Sa = 0, =1 Sa Sm + =1 Sy Sy 2 2 Sy Se 2 + S2 Se y Sm = Sy Sa , r crit = Sa / Sm Fatigue factor of safety nf = 1 2 ( σa / Se ) + σm / Sy 2 criteria. The rst column gives the intersecting equations and the second column the intersection coordinates. There are two ways to proceed with a typical analysis. One method is to assume that fatigue occurs first and use one of Eqs. (645) to (648) to determine n or size, depending on the task. Most often fatigue is the governing failure mode. Then follow with a static check. If static failure governs then the analysis is repeated using Eq. (649). Alternatively, one could use the tables. Determine the load line and establish which criterion the load line intersects rst and use the corresponding equations in the tables. Some examples will help solidify the ideas just discussed. EXAMPLE 610 A 1.5-in-diameter bar has been machined from an AISI 1050 cold-drawn bar. This part is to withstand a uctuating tensile load varying from 0 to 16 kip. Because of the ends, and the llet radius, a fatigue stress-concentration factor K f is 1.85 for 106 or larger life. Find Sa and Sm and the factor of safety guarding against fatigue and rst-cycle yielding, using (a) the Gerber fatigue line and (b) the ASME-elliptic fatigue line. Solution We begin with some preliminaries. From Table A20, Sut = 100 kpsi and Sy = 84 kpsi. Note that Fa = Fm = 8 kip. The Marin factors are, deterministically, ka = 2.70(100)0.265 = 0.797: Eq. (619), Table 62, p. 279 kb = 1 (axial loading, see kc ) 303 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition II. Failure Prevention © The McGrawHill Companies, 2008 6. Fatigue Failure Resulting from Variable Loading Fatigue Failure Resulting from Variable Loading 301 kc = 0.85: Eq. (626), p. 282 kd = ke = k f = 1 Se = 0.797(1)0.850(1)(1)(1)0.5(100) = 33.9 kpsi: Eqs. (68), (618), p. 274, p. 279 The nominal axial stress components σao and σmo are σao = 4 Fa 4(8) = = 4.53 kpsi 2 πd π 1.52 σmo = 4 Fm 4(8) = = 4.53 kpsi 2 πd π 1.52 Applying K f to both components σao and σmo constitutes a prescription of no notch yielding: σa = K f σao = 1.85(4.53) = 8.38 kpsi = σm Answer (a) Let us calculate the factors of safety rst. From the bottom panel from Table 67 the factor of safety for fatigue is 1 100 2 8.38 2(8.38)33.9 2 nf = = 3.66 1 + 1 + 2 8.38 33.9 100(8.38) From Eq. (649) the factor of safety guarding against rst-cycle yield is Answer Answer Figure 628 ny = Sy 84 = 5.01 = σa + σm 8.38 + 8.38 Thus, we see that fatigue will occur rst and the factor of safety is 3.68. This can be seen in Fig. 628 where the load line intersects the Gerber fatigue curve rst at point B. If the plots are created to true scale it would be seen that n f = O B / O A. From the rst panel of Table 67, r = σa /σm = 1, 2 2 2 (1) 100 2(33.9) = 30.7 kpsi Sa = 1 + 1 + 2(33.9) (1)100 100 Principal points A, B, C, and D on the designers diagram drawn for Gerber, Langer, and load line. a, kpsi 84 Stress amplitude 304 50 Load line C 42 Langer line 33.9 30.7 B D 20 rcrit Gerber fatigue curve A 8.38 0 0 8.38 30.7 42 50 Midrange stress m, 64 kpsi 84 100 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 302 II. Failure Prevention © The McGrawHill Companies, 2008 6. Fatigue Failure Resulting from Variable Loading Mechanical Engineering Design Answer Sm = 30.7 Sa = = 30.7 kpsi r 1 As a check on the previous result, n f = O B / O A = Sa /σa = Sm /σm = 30.7/8.38 = 3.66 and we see total agreement. We could have detected that fatigue failure would occur rst without drawing Fig. 628 by calculating rcrit . From the third row third column panel of Table 67, the intersection point between fatigue and rst-cycle yield is 1002 2(33.9) 2 84 1 1+ = 64.0 kpsi Sm = 1 2(33.9) 100 33.9 Sa = Sy Sm = 84 64 = 20 kpsi The critical slope is thus Sa 20 = 0.312 = Sm 64 rcrit = which is less than the actual load line of r = 1. This indicates that fatigue occurs before rst-cycle-yield. (b) Repeating the same procedure for the ASME-elliptic line, for fatigue Answer nf = (8.38/33.9) 2 1 = 3.75 + (8.38/84) 2 Again, this is less than n y = 5.01 and fatigue is predicted to occur rst. From the rst row second column panel of Table 68, with r = 1, we obtain the coordinates Sa and Sm of point B in Fig. 629 as Figure 629 100 Principal points A, B, C, and D on the designers diagram drawn for ASME-elliptic, Langer, and load lines. Stress amplitude a, kpsi 84 50 Load line C 42 Langer line B 31.4 D 23.5 ASME-elliptic line A 8.38 0 0 8.38 31.4 42 50 Midrange stress 60.5 kpsi m, 84 100 305 306 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition II. Failure Prevention © The McGrawHill Companies, 2008 6. Fatigue Failure Resulting from Variable Loading Fatigue Failure Resulting from Variable Loading Answer Sa = ( 1) 2 33.92 (84) 2 = 31.4 kpsi, 33.92 + (1) 2 842 Sm = 303 Sa 31.4 = = 31.4 kpsi r 1 To verify the fatigue factor of safety, n f = Sa /σa = 31.4/8.38 = 3.75. As before, let us calculate rcrit . From the third row second column panel of Table 68, 2(84)33.92 = 23.5 kpsi, 33.92 + 842 Sa = rcrit = Sm = Sy Sa = 84 23.5 = 60.5 kpsi Sa 23.5 = = 0.388 Sm 60.5 which again is less than r = 1, verifying that fatigue occurs rst with n f = 3.75. The Gerber and the ASME-elliptic fatigue failure criteria are very close to each other and are used interchangeably. The ANSI/ASME Standard B106.1M1985 uses ASME-elliptic for shafting. EXAMPLE 611 A at-leaf spring is used to retain an oscillating at-faced follower in contact with a plate cam. The follower range of motion is 2 in and xed, so the alternating component of force, bending moment, and stress is xed, too. The spring is preloaded to adjust to various cam speeds. The preload must be increased to prevent follower oat or jump. For lower speeds the preload should be decreased to obtain longer life of cam and follower surfaces. The spring is a steel cantilever 32 in long, 2 in wide, and 1 in thick, 4 as seen in Fig. 630a. The spring strengths are Sut = 150 kpsi, Sy = 127 kpsi, and Se = 28 kpsi fully corrected. The total cam motion is 2 in. The designer wishes to preload the spring by deecting it 2 in for low speed and 5 in for high speed. (a) Plot the Gerber-Langer failure lines with the load line. (b) What are the strength factors of safety corresponding to 2 in and 5 in preload? Solution We begin with preliminaries. The second area moment of the cantilever cross section is I= bh 3 2(0.25)3 = = 0.00260 in4 12 12 Since, from Table A9, beam 1, force F and deection y in a cantilever are related by F = 3 E I y / l 3, then stress σ and deection y are related by σ= where K = Mc 32 Fc 32(3 E I y ) c 96 Ecy = = = = Ky I I l3 I l3 96(30 · 106 )0.125 96 Ec = = 10.99(103 ) psi/in = 10.99 kpsi/in 3 l 323 Now the minimums and maximums of y and σ can be dened by ymin = δ ymax = 2 + δ σmin = K δ σmax = K (2 + δ) BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 304 II. Failure Prevention © The McGrawHill Companies, 2008 6. Fatigue Failure Resulting from Variable Loading Mechanical Engineering Design Figure 630 Cam follower retaining spring. (a) Geometry; (b) designers fatigue diagram for Ex. 611. 1 4 2 in + in 32 in = 2 in + = 2 in preload = 5 in = 5 in preload + (a ) Amplitude stress component a, kpsi 150 100 Langer line 50 Gerber line A 0 A' 11 33 A" 50 65.9 Steady stress component m, 100 kpsi 115.6 127 150 (b) The stress components are thus σa = K (2 + δ) K δ = K = 10.99 kpsi 2 σm = K (2 + δ) + K δ = K (1 + δ) = 10.99(1 + δ) 2 For δ = 0, σa = σm = 10.99 = 11 kpsi 307 308 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition II. Failure Prevention © The McGrawHill Companies, 2008 6. Fatigue Failure Resulting from Variable Loading Fatigue Failure Resulting from Variable Loading For δ = 2 in, σa = 11 kpsi, σm = 10.99(1 + 2) = 33 kpsi For δ = 5 in, 305 σa = 11 kpsi, σm = 10.99(1 + 5) = 65.9 kpsi (a) A plot of the Gerber and Langer criteria is shown in Fig. 630b. The three preload deections of 0, 2, and 5 in are shown as points A, A , and A . Note that since σa is constant at 11 kpsi, the load line is horizontal and does not contain the origin. The intersection between the Gerber line and the load line is found from solving Eq. (642) for Sm and substituting 11 kpsi for Sa : Sm = Sut 1 Sa 11 = 116.9 kpsi = 150 1 Se 28 The intersection of the Langer line and the load line is found from solving Eq. (644) for Sm and substituting 11 kpsi for Sa : Sm = Sy Sa = 127 11 = 116 kpsi The threats from fatigue and rst-cycle yielding are approximately equal. (b) For δ = 2 in, Answer nf = Sm 116.9 = 3.54 = σm 33 ny = 116 = 3.52 33 and for δ = 5 in, Answer EXAMPLE 612 Solution nf = 116.9 = 1.77 65.9 ny = 116 = 1.76 65.9 A steel bar undergoes cyclic loading such that σmax = 60 kpsi and σmin = 20 kpsi. For the material, Sut = 80 kpsi, Sy = 65 kpsi, a fully corrected endurance limit of Se = 40 kpsi, and f = 0.9. Estimate the number of cycles to a fatigue failure using: (a) Modied Goodman criterion. (b) Gerber criterion. From the given stresses, σa = 60 (20) = 40 kpsi 2 σm = 60 + (20) = 20 kpsi 2 From the material properties, Eqs. (614) to (616), p. 277, give a= ( f Sut )2 [0.9(80)]2 = 129.6 kpsi = Se 40 1 b = log 3 N= Sf a 1 0.9(80) = log = 0.0851 3 40 f Sut Se 1/b = Sf 129.6 where S f replaced σa in Eq. (616). 1/0.0851 (1) BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 306 II. Failure Prevention 309 © The McGrawHill Companies, 2008 6. Fatigue Failure Resulting from Variable Loading Mechanical Engineering Design (a) The modied Goodman line is given by Eq. (646), p. 298, where the endurance limit Se is used for innite life. For nite life at S f > Se , replace Se with S f in Eq. (646) and rearrange giving Sf = σa 40 = 53.3 kpsi σm = 20 1 1 Sut 80 Substituting this into Eq. (1) yields Answer N= 53.3 129.6 1/0.0851 . = 3.4(104 ) cycles (b) For Gerber, similar to part (a), from Eq. (647), Sf = σa 1 σm Sut 2 = 40 1 20 80 2 = 42.7 kpsi Again, from Eq. (1), Answer N= 42.7 129.6 1/0.0851 . = 4.6(105 ) cycles Comparing the answers, we see a large difference in the results. Again, the modied Goodman criterion is conservative as compared to Gerber for which the moderate difference in S f is then magnied by a logarithmic S, N relationship. For many brittle materials, the rst quadrant fatigue failure criteria follows a concave upward Smith-Dolan locus represented by Sa 1 Sm / Sut = Se 1 + Sm / Sut (650) n σa 1 n σm / Sut = Se 1 + n σm / Sut (651) or as a design equation, For a radial load line of slope r, we substitute Sa / r for Sm in Eq. (650) and solve for Sa , obtaining Sa = r Sut + Se 1 + 2 1+ 4r Sut Se (r Sut + Se )2 (652) The fatigue diagram for a brittle material differs markedly from that of a ductile material because: Yielding is not involved since the material may not have a yield strength. Characteristically, the compressive ultimate strength exceeds the ultimate tensile strength severalfold. 310 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition II. Failure Prevention © The McGrawHill Companies, 2008 6. Fatigue Failure Resulting from Variable Loading Fatigue Failure Resulting from Variable Loading 307 First-quadrant fatigue failure locus is concave-upward (Smith-Dolan), for example, and as at as Goodman. Brittle materials are more sensitive to midrange stress, being lowered, but compressive midrange stresses are benecial. Not enough work has been done on brittle fatigue to discover insightful generalities, so we stay in the rst and a bit of the second quadrant. The most likely domain of designer use is in the range from Sut σm Sut . The locus in the rst quadrant is Goodman, Smith-Dolan, or something in between. The portion of the second quadrant that is used is represented by a straight line between the points Sut , Sut and 0, Se , which has the equation Sa = Se + Se 1 Sm Sut Sut Sm 0 (for cast iron) (653) Table A24 gives properties of gray cast iron. The endurance limit stated is really ka kb Se and only corrections kc , kd , ke , and k f need be made. The average kc for axial and torsional loading is 0.9. EXAMPLE 613 A grade 30 gray cast iron is subjected to a load F applied to a 1 by 3 -in cross-section 8 link with a 1 -in-diameter hole drilled in the center as depicted in Fig. 631a. The sur4 faces are machined. In the neighborhood of the hole, what is the factor of safety guarding against failure under the following conditions: (a) The load F = 1000 lbf tensile, steady. (b) The load is 1000 lbf repeatedly applied. (c) The load uctuates between 1000 lbf and 300 lbf without column action. Use the Smith-Dolan fatigue locus. Alternating stress, a F Sut 1 in 1 4 r = 1.86 in D. drill Sa = 18.5 kpsi Se 3 8 r=1 in Sa = 7.63 Sm F Sut 9.95 0 Midrange stress (a) 7.63 10 m, 20 30 Sut kpsi (b) Figure 631 The grade 30 cast-iron part in axial fatigue with (a) its geometry displayed and (b) its designers fatigue diagram for the circumstances of Ex. 613. BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 308 II. Failure Prevention © The McGrawHill Companies, 2008 6. Fatigue Failure Resulting from Variable Loading Mechanical Engineering Design Solution Some preparatory work is needed. From Table A24, Sut = 31 kpsi, Suc = 109 kpsi, ka kb Se = 14 kpsi. Since kc for axial loading is 0.9, then Se = (ka kb Se )kc = 14(0.9) = 12.6 kpsi. From Table A151, A = t (w d ) = 0.375(1 0.25) = 0.281 in2 , d /w = 0.25/1 = 0.25, and K t = 2.45. The notch sensitivity for cast iron is 0.20 (see p. 288), so K f = 1 + q ( K t 1) = 1 + 0.20(2.45 1) = 1.29 (a) σa = K f Fa 1.29(0) = =0 A 0.281 σm = K f Fm 1.29(1000) 3 = (10 ) = 4.59 kpsi A 0.281 and Answer n= Sut 31.0 = 6.75 = σm 4.59 Fa = Fm = 1000 F = = 500 lbf 2 2 σa = σm = (b) K f Fa 1.29(500) 3 = (10 ) = 2.30 kpsi A 0.281 r= σa =1 σm From Eq. (652), Sa = (1)31 + 12.6 1 + 2 Answer n= (c) Fa = Fm = 1+ = 7.63 kpsi Sa 7.63 = 3.32 = σa 2.30 1 |300 (1000)| = 650 lbf 2 1 [300 + (1000)] = 350 lbf 2 r= 4(1)31(12.6) [(1)31 + 12.6]2 σa = σm = 1.29(650) 3 (10 ) = 2.98 kpsi 0.281 1.29(350) 3 (10 ) = 1.61 kpsi 0.281 σa 3.0 = 1.86 = σm 1.61 From Eq. (653), Sa = Se + ( Se / Sut 1) Sm and Sm = Sa / r . It follows that Sa = Answer 1 1 r Se Se 1 Sut n= = 12.6 12.6 1 1 1 1.86 31 = 18.5 kpsi Sa 18.5 = 6.20 = σa 2.98 Figure 631b shows the portion of the designers fatigue diagram that was constructed. 311 312 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition II. Failure Prevention © The McGrawHill Companies, 2008 6. Fatigue Failure Resulting from Variable Loading Fatigue Failure Resulting from Variable Loading 613 309 Torsional Fatigue Strength under Fluctuating Stresses Extensive tests by Smith23 provide some very interesting results on pulsating torsional fatigue. Smiths rst result, based on 72 tests, shows that the existence of a torsional steady-stress component not more than the torsional yield strength has no effect on the torsional endurance limit, provided the material is ductile, polished, notch-free, and cylindrical. Smiths second result applies to materials with stress concentration, notches, or surface imperfections. In this case, he nds that the torsional fatigue limit decreases monotonically with torsional steady stress. Since the great majority of parts will have surfaces that are less than perfect, this result indicates Gerber, ASME-elliptic, and other approximations are useful. Joerres of Associated Spring-Barnes Group, conrms Smiths results and recommends the use of the modied Goodman relation for pulsating torsion. In constructing the Goodman diagram, Joerres uses Ssu = 0.67 Sut (654) Also, from Chap. 5, Ssy = 0.577 Syt from distortion-energy theory, and the mean load factor kc is given by Eq. (626), or 0.577. This is discussed further in Chap. 10. 614 Combinations of Loading Modes It may be helpful to think of fatigue problems as being in three categories: Completely reversing simple loads Fluctuating simple loads Combinations of loading modes The simplest category is that of a completely reversed single stress which is handled with the S-N diagram, relating the alternating stress to a life. Only one type of loading is allowed here, and the midrange stress must be zero. The next category incorporates general uctuating loads, using a criterion to relate midrange and alternating stresses (modied Goodman, Gerber, ASME-elliptic, or Soderberg). Again, only one type of loading is allowed at a time. The third category, which we will develop in this section, involves cases where there are combinations of different types of loading, such as combined bending, torsion, and axial. In Sec. 69 we learned that a load factor kc is used to obtain the endurance limit, and hence the result is dependent on whether the loading is axial, bending, or torsion. In this section we want to answer the question, How do we proceed when the loading is a mixture of, say, axial, bending, and torsional loads? This type of loading introduces a few complications in that there may now exist combined normal and shear stresses, each with alternating and midrange values, and several of the factors used in determining the endurance limit depend on the type of loading. There may also be multiple stress-concentration factors, one for each mode of loading. The problem of how to deal with combined stresses was encountered when developing static failure theories. The distortion energy failure theory proved to be a satisfactory method of combining the 23 James O. Smith, The Effect of Range of Stress on the Fatigue Strength of Metals, Univ. of Ill. Eng. Exp. Sta. Bull. 334, 1942. BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 310 II. Failure Prevention 313 © The McGrawHill Companies, 2008 6. Fatigue Failure Resulting from Variable Loading Mechanical Engineering Design multiple stresses on a stress element into a single equivalent von Mises stress. The same approach will be used here. The rst step is to generate two stress elementsone for the alternating stresses and one for the midrange stresses. Apply the appropriate fatigue stress concentration factors to each of the stresses; i.e., apply ( K f ) bending for the bending stresses, ( K f s ) torsion for the torsional stresses, and ( K f ) axial for the axial stresses. Next, calculate an equivalent von Mises stress for each of these two stress elements, σa and σm . Finally, select a fatigue failure criterion (modied Goodman, Gerber, ASME-elliptic, or Soderberg) to complete the fatigue analysis. For the endurance limit, Se , use the endurance limit modiers, ka , kb , and kc , for bending. The torsional load factor, kc = 0.59 should not be applied as it is already accounted for in the von Mises stress calculation (see footnote 17 on page 282). The load factor for the axial load can be accounted for by dividing the alternating axial stress by the axial load factor of 0.85. For example, consider the common case of a shaft with bending stresses, torsional shear stresses, and axial stresses. For this case, 1/2 the von Mises stress is of the form σ = σx 2 + 3τx y 2 . Considering that the bending, torsional, and axial stresses have alternating and midrange components, the von Mises stresses for the two stress elements can be written as σa = ( K f ) bending ( σa ) bending + ( K f ) axial ( σa ) axial 0.85 1/2 2 + 3 ( K f s ) torsion ( τa ) torsion 2 (655) σm = ( K f ) bending ( σm ) bending + ( K f ) axial ( σm ) axial 2 + 3 ( K f s ) torsion ( τm ) torsion 2 1/2 (656) For rst-cycle localized yielding, the maximum von Mises stress is calculated. This would be done by rst adding the axial and bending alternating and midrange stresses to obtain σmax and adding the alternating and midrange shear stresses to obtain τmax . Then substitute σmax and τmax into the equation for the von Mises stress. A simpler and more con. servative method is to add Eq. (655) and Eq. (656). That is, let σmax = σa + σm If the stress components are not in phase but have the same frequency, the maxima can be found by expressing each component in trigonometric terms, using phase angles, and then nding the sum. If two or more stress components have differing frequencies, the problem is difcult; one solution is to assume that the two (or more) components often reach an in-phase condition, so that their magnitudes are additive. EXAMPLE 614 A rotating shaft is made of 42- × 4-mm AISI 1018 cold-drawn steel tubing and has a 6-mm-diameter hole drilled transversely through it. Estimate the factor of safety guarding against fatigue and static failures using the Gerber and Langer failure criteria for the following loading conditions: (a) The shaft is subjected to a completely reversed torque of 120 N · m in phase with a completely reversed bending moment of 150 N · m. (b) The shaft is subjected to a pulsating torque uctuating from 20 to 160 N · m and a steady bending moment of 150 N · m. Solution Here we follow the procedure of estimating the strengths and then the stresses, followed by relating the two. 314 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition II. Failure Prevention © The McGrawHill Companies, 2008 6. Fatigue Failure Resulting from Variable Loading Fatigue Failure Resulting from Variable Loading 311 From Table A20 we nd the minimum strengths to be Sut = 440 MPa and Sy = 370 MPa. The endurance limit of the rotating-beam specimen is 0.5(440) = 220 MPa. The surface factor, obtained from Eq. (619) and Table 62, p. 279 is ka = 4.51 Sut0.265 = 4.51(440)0.265 = 0.899 From Eq. (620) the size factor is kb = d 7.62 0.107 = 42 7.62 0.107 = 0.833 The remaining Marin factors are all unity, so the modied endurance strength Se is Se = 0.899(0.833)220 = 165 MPa (a) Theoretical stress-concentration factors are found from Table A16. Using a / D = 6/42 = 0.143 and d / D = 34/42 = 0.810, and using linear interpolation, we obtain A = 0.798 and K t = 2.366 for bending; and A = 0.89 and K ts = 1.75 for torsion. Thus, for bending, Z net = πA π (0.798) ( D4 d 4) = [(42) 4 (34) 4 ] = 3.31 (103 )mm3 32 D 32(42) and for torsion Jnet = πA 4 π (0.89) ( D d 4) = [(42) 4 (34) 4 ] = 155 (103 )mm4 32 32 Next, using Figs. 620 and 621, pp. 287288, with a notch radius of 3 mm we nd the notch sensitivities to be 0.78 for bending and 0.96 for torsion. The two corresponding fatigue stress-concentration factors are obtained from Eq. (632) as K f = 1 + q ( K t 1) = 1 + 0.78(2.366 1) = 2.07 K f s = 1 + 0.96(1.75 1) = 1.72 The alternating bending stress is now found to be σxa = K f M 150 = 93.8(106 )Pa = 93.8 MPa = 2.07 Z net 3.31(106 ) and the alternating torsional stress is τx ya = K f s 120(42)(103 ) TD = 28.0(106 )Pa = 28.0 MPa = 1.72 2 Jnet 2(155)(109 ) The midrange von Mises component σm is zero. The alternating component σa is given by 2 2 σa = σxa + 3τx ya 1/2 = [93.82 + 3(282 )]1/2 = 105.6 MPa Since Se = Sa , the fatigue factor of safety n f is Answer nf = Sa 165 = 1.56 = σa 105.6 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 312 II. Failure Prevention © The McGrawHill Companies, 2008 6. Fatigue Failure Resulting from Variable Loading Mechanical Engineering Design Figure 632 Designers fatigue diagram for Ex. 614. Von Mises amplitude stress component a , MPa ' 400 300 200 Gerber 165 r = 0.28 100 105.6 85.5 0 305 Von Mises steady stress component m, MPa ' 440 500 The rst-cycle yield factor of safety is Answer ny = Sy 370 = 3.50 = σa 105.6 There is no localized yielding; the threat is from fatigue. See Fig. 632. (b) This part asks us to nd the factors of safety when the alternating component is due to pulsating torsion, and a steady component is due to both torsion and bending. We have Ta = (160 20)/2 = 70 N · m and Tm = (160 + 20) /2 = 90 N · m. The corresponding amplitude and steady-stress components are τx ya = K f s 70(42)(103 ) Ta D = 16.3(106 )Pa = 16.3 MPa = 1.72 2 Jnet 2(155)(109 ) τx ym = K f s Tm D 90(42)(103 ) = 21.0(106 )Pa = 21.0 MPa = 1.72 2 Jnet 2(155)(109 ) The steady bending stress component σxm is σxm = K f 150 Mm = 93.8(106 )Pa = 93.8 MPa = 2.07 Z net 3.31(106 ) The von Mises components σa and σm are σa = [3(16.3)2 ]1/2 = 28.2 MPa σm = [93.82 + 3(21)2 ]1/2 = 100.6 MPa From Table 67, p. 299, the fatigue factor of safety is Answer nf = 1 2 440 100.6 2 28.2 1 + 165 1+ 2(100.6)165 440(28.2) 2 = 3.03 315 316 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition II. Failure Prevention © The McGrawHill Companies, 2008 6. Fatigue Failure Resulting from Variable Loading Fatigue Failure Resulting from Variable Loading 313 From the same table, with r = σa /σm = 28.2/100.6 = 0.280, the strengths can be shown to be Sa = 85.5 MPa and Sm = 305 MPa. See the plot in Fig. 632. The rst-cycle yield factor of safety n y is Answer ny = σa Sy 370 = 2.87 = + σm 28.2 + 100.6 There is no notch yielding. The likelihood of failure may rst come from rst-cycle yielding at the notch. See the plot in Fig. 632. 615 Varying, Fluctuating Stresses; Cumulative Fatigue Damage Instead of a single fully reversed stress history block composed of n cycles, suppose a machine part, at a critical location, is subjected to A fully reversed stress σ1 for n 1 cycles, σ2 for n 2 cycles, . . . , or A wiggly time line of stress exhibiting many and different peaks and valleys. What stresses are signicant, what counts as a cycle, and what is the measure of damage incurred? Consider a fully reversed cycle with stresses varying 60, 80, 40, and 60 kpsi and a second fully reversed cycle 40, 60, 20, and 40 kpsi as depicted in Fig. 633a. First, it is clear that to impose the pattern of stress in Fig. 633a on a part it is necessary that the time trace look like the solid line plus the dashed line in Fig. 633a. Figure 633b moves the snapshot to exist beginning with 80 kpsi and ending with 80 kpsi. Acknowledging the existence of a single stress-time trace is to discover a hidden cycle shown as the dashed line in Fig. 633b. If there are 100 applications of the all-positive stress cycle, then 100 applications of the all-negative stress cycle, the Figure 633 100 100 50 50 0 0 50 50 Variable stress diagram prepared for assessing cumulative damage. (a) (b) BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 314 II. Failure Prevention 317 © The McGrawHill Companies, 2008 6. Fatigue Failure Resulting from Variable Loading Mechanical Engineering Design hidden cycle is applied but once. If the all-positive stress cycle is applied alternately with the all-negative stress cycle, the hidden cycle is applied 100 times. To ensure that the hidden cycle is not lost, begin on the snapshot with the largest (or smallest) stress and add previous history to the right side, as was done in Fig. 633b. Characterization of a cycle takes on a maxminsame max (or minmaxsame min) form. We identify the hidden cycle rst by moving along the dashed-line trace in Fig. 633b identifying a cycle with an 80-kpsi max, a 60-kpsi min, and returning to 80 kpsi. Mentally deleting the used part of the trace (the dashed line) leaves a 40, 60, 40 cycle and a 40, 20, 40 cycle. Since failure loci are expressed in terms of stress amplitude component σa and steady component σm , we use Eq. (636) to construct the table below: Cycle Number max 1 80 60 70 10 2 60 40 10 50 3 20 40 10 30 min a m The most damaging cycle is number 1. It could have been lost. Methods for counting cycles include: Number of tensile peaks to failure. All maxima above the waveform mean, all minima below. The global maxima between crossings above the mean and the global minima between crossings below the mean. All positive slope crossings of levels above the mean, and all negative slope crossings of levels below the mean. A modication of the preceding method with only one count made between successive crossings of a level associated with each counting level. Each local maxi-min excursion is counted as a half-cycle, and the associated amplitude is half-range. The preceding method plus consideration of the local mean. Rain-ow counting technique. The method used here amounts to a variation of the rain-ow counting technique. The Palmgren-Miner24 cycle-ratio summation rule, also called Miners rule, is written ni =c Ni (657) where n i is the number of cycles at stress level σi and Ni is the number of cycles to failure at stress level σi . The parameter c has been determined by experiment; it is usually found in the range 0.7 < c < 2.2 with an average value near unity. 24 A. Palmgren, Die Lebensdauer von Kugellagern, ZVDI, vol. 68, pp. 339341, 1924; M. A. Miner, Cumulative Damage in Fatigue, J. Appl. Mech., vol. 12, Trans. ASME, vol. 67, pp. A159A164, 1945. 318 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition II. Failure Prevention © The McGrawHill Companies, 2008 6. Fatigue Failure Resulting from Variable Loading Fatigue Failure Resulting from Variable Loading 315 Using the deterministic formulation as a linear damage rule we write ni Ni D= (658) where D is the accumulated damage. When D = c = 1, failure ensues. EXAMPLE 615 Solution Given a part with Sut = 151 kpsi and at the critical location of the part, Se = 67.5 kpsi. For the loading of Fig. 633, estimate the number of repetitions of the stress-time block in Fig. 633 that can be made before failure. From Fig. 618, p. 277, for Sut = 151 kpsi, f = 0.795. From Eq. (614), a= ( f Sut )2 [0.795(151)]2 = 213.5 kpsi = Se 67.5 From Eq. (615), 1 b = log 3 f Sut Se 1 0.795(151) = log = 0.0833 3 67.5 So, S f = 213.5 N 0.0833 N= Sf 213.5 1/0.0833 (1), (2) We prepare to add two columns to the previous table. Using the Gerber fatigue criterion, Eq. (647), p. 298, with Se = S f , and n = 1, we can write σa 1 (σm / Sut )2 Se Sf = σm > 0 σm 0 (3) Cycle 1: r = σa /σm = 70/10 = 7, and the strength amplitude from Table 67, p. 299, is 72 1512 2(67.5) 2 = 67.2 kpsi 1 + 1 + Sa = 2(67.5) 7(151) Since σa > Sa , that is, 70 > 67.2, life is reduced. From Eq. (3), Sf = 70 = 70.3 kpsi 1 (10/151)2 and from Eq. (2) N= 70.3 213.5 1/0.0833 = 619(103 ) cycles Cycle 2: r = 10/50 = 0.2, and the strength amplitude is 2 2 2 0.2 151 2(67.5) Sa = 1 + 1 + = 24.2 kpsi 2(67.5) 0.2(151) BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 316 II. Failure Prevention © The McGrawHill Companies, 2008 6. Fatigue Failure Resulting from Variable Loading Mechanical Engineering Design Since σa < Sa , that is 10 < 24.2, then S f = Se and indenite life follows. Thus, . N Cycle 3: r = 10/30 = 0.333, and since σm < 0, S f = Se , indenite life follows and N Cycle Number Sf , kpsi N , cycles 1 70.3 619(103) 2 67.5 3 67.5 From Eq. (658) the damage per block is N 1 1 1 ni = + + =N 3) Ni 619(10 619(103 ) D= Answer Setting D = 1 yields N = 619(103 ) cycles. To further illustrate the use of the Miner rule, let us choose a steel having the prop erties Sut = 80 kpsi, Se,0 = 40 kpsi, and f = 0.9, where we have used the designation Se,0 instead of the more usual Se to indicate the endurance limit of the virgin, or undamaged, material. The log Slog N diagram for this material is shown in Fig. 634 by the heavy solid line. Now apply, say, a reversed stress σ1 = 60 kpsi for n 1 = 3000 cycles. Since σ1 > Se,0 , the endurance limit will be damaged, and we wish to nd the new endurance limit Se,1 of the damaged material using the Miner rule. The equation of the virgin material failure line in Fig. 634 in the 103 to 106 cycle range is S f = a N b = 129.6 N 0.085 091 The cycles to failure at stress level σ1 = 60 kpsi are N1 = Figure 634 σ1 129.6 1/0.085 091 60 129.6 = 1/0.085 091 = 8520 cycles 4.9 Use of the Miner rule to predict the endurance limit of a material that has been overstressed for a nite number of cycles. 0.9Sut 72 4.8 Sf, 0 1 60 Sf, 1 4.7 So kpsi Log S n1 = 3 (10 3) N1 = 8.52(10 3) N1 n1 = 5.52(10 3) 4.6 Se ,0 40 38.6 Sf,2 Se ,1 n 2 = 0.648(106) 4.5 10 3 10 4 10 5 10 6 5 6 N 3 4 Log N 319 320 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition II. Failure Prevention © The McGrawHill Companies, 2008 6. Fatigue Failure Resulting from Variable Loading Fatigue Failure Resulting from Variable Loading 317 Figure 634 shows that the material has a life N1 = 8520 cycles at 60 kpsi, and consequently, after the application of σ1 for 3000 cycles, there are N1 n 1 = 5520 cycles of life remaining at σ1 . This locates the nite-life strength S f ,1 of the damaged material, as shown in Fig. 634. To get a second point, we ask the question: With n 1 and N1 given, how many cycles of stress σ2 = Se,0 can be applied before the damaged material fails? This corresponds to n 2 cycles of stress reversal, and hence, from Eq. (658), we have n2 n1 + =1 N1 N2 (a) or n2 = 1 n1 N1 (b) N2 Then n2 = 1 3(10)3 (106 ) = 0.648(106 ) cycles 8.52(10)3 This corresponds to the nite-life strength S f ,2 in Fig. 634. A line through S f ,1 and S f ,2 is the log Slog N diagram of the damaged material according to the Miner rule. The new endurance limit is Se,1 = 38.6 kpsi. We could leave it at this, but a little more investigation can be helpful. We have two points on the new fatigue locus, N1 n 1 , σ1 and n 2 , σ2 . It is useful to prove that the slope of the new line is still b. For the equation S f = a N b , where the values of a and b are established by two points α and β . The equation for b is b = log σα /σβ log Nα / Nβ (c) Examine the denominator of Eq. (c): log N1 n 1 N1 n 1 N1 Nα = log = log = log Nβ n2 (1 n 1 / N1 ) N2 N2 = log (σ1 /a )1/b σ1 = log 1/b (σ2 /a ) σ2 1/b = σ1 1 log b σ2 Substituting this into Eq. (c) with σα /σβ = σ1 /σ2 gives b = log(σ1 /σ2 ) =b (1/b) log(σ1 /σ2 ) which means the damaged material line has the same slope as the virgin material line; therefore, the lines are parallel. This information can be helpful in writing a computer program for the Palmgren-Miner hypothesis. Though the Miner rule is quite generally used, it fails in two ways to agree with experiment. First, note that this theory states that the static strength Sut is damaged, that is, decreased, because of the application of σ1 ; see Fig. 634 at N = 103 cycles. Experiments fail to verify this prediction. The Miner rule, as given by Eq. (658), does not account for the order in which the stresses are applied, and hence ignores any stresses less than Se,0 . But it can be seen in Fig. 634 that a stress σ3 in the range Se,1 < σ3 < Se,0 would cause damage if applied after the endurance limit had been damaged by the application of σ1 . BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition © The McGrawHill Companies, 2008 6. Fatigue Failure Resulting from Variable Loading Mechanical Engineering Design Figure 635 4.9 Use of the Manson method to predict the endurance limit of a material that has been overstressed for a nite number of cycles. 0.9Sut 72 4.8 Sf, 0 1 60 Sf, 1 Log S n1 = 3 (10 3) 4.7 So kpsi 318 II. Failure Prevention N1 = 8.52(10 3) N1 n1 = 5.52(10 3) 4.6 S'e ,0 40 S'e ,1 34.4 4.5 10 3 10 4 10 5 10 6 5 6 N 3 4 Log N Mansons25 approach overcomes both of the deciencies noted for the PalmgrenMiner method; historically it is a much more recent approach, and it is just as easy to use. Except for a slight change, we shall use and recommend the Manson method in this book. Manson plotted the Slog N diagram instead of a log Slog N plot as is recommended here. Manson also resorted to experiment to nd the point of convergence of the Slog N lines corresponding to the static strength, instead of arbitrarily selecting the intersection of N = 103 cycles with S = 0.9 Sut as is done here. Of course, it is always better to use experiment, but our purpose in this book has been to use the simple test data to learn as much as possible about fatigue failure. The method of Manson, as presented here, consists in having all log Slog N lines, that is, lines for both the damaged and the virgin material, converge to the same point, 0.9 Sut at 103 cycles. In addition, the log Slog N lines must be constructed in the same historical order in which the stresses occur. The data from the preceding example are used for illustrative purposes. The results are shown in Fig. 635. Note that the strength S f ,1 corresponding to N1 n 1 = 5.52(103 ) cycles is found in the same manner as before. Through this point and through 0.9 Sut at 103 cycles, draw the heavy dashed line to meet N = 106 cycles and dene the endurance limit Se,1 of the damaged material. In this case the new endurance limit is 34.4 kpsi, somewhat less than that found by the Miner method. It is now easy to see from Fig. 635 that a reversed stress σ = 36 kpsi, say, would not harm the endurance limit of the virgin material, no matter how many cycles it might be applied. However, if σ = 36 kpsi should be applied after the material was damaged by σ1 = 60 kpsi, then additional damage would be done. Both these rules involve a number of computations, which are repeated every time damage is estimated. For complicated stress-time traces, this might be every cycle. Clearly a computer program is useful to perform the tasks, including scanning the trace and identifying the cycles. 25 S. S. Manson, A. J. Nachtigall, C. R. Ensign, and J. C. Fresche, Further Investigation of a Relation for Cumulative Fatigue Damage in Bending, Trans. ASME, J. Eng. Ind., ser. B, vol. 87, No. 1, pp. 2535, February 1965. 321 322 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition II. Failure Prevention © The McGrawHill Companies, 2008 6. Fatigue Failure Resulting from Variable Loading Fatigue Failure Resulting from Variable Loading 319 Collins said it well: In spite of all the problems cited, the Palmgren linear damage rule is frequently used because of its simplicity and the experimental fact that other more complex damage theories do not always yield a signicant improvement in failure prediction reliability.26 616 Surface Fatigue Strength The surface fatigue mechanism is not denitively understood. The contact-affected zone, in the absence of surface shearing tractions, entertains compressive principal stresses. Rotary fatigue has its cracks grown at or near the surface in the presence of tensile stresses that are associated with crack propagation, to catastrophic failure. There are shear stresses in the zone, which are largest just below the surface. Cracks seem to grow from this stratum until small pieces of material are expelled, leaving pits on the surface. Because engineers had to design durable machinery before the surface fatigue phenomenon was understood in detail, they had taken the posture of conducting tests, observing pits on the surface, and declaring failure at an arbitrary projected area of hole, and they related this to the Hertzian contact pressure. This compressive stress did not produce the failure directly, but whatever the failure mechanism, whatever the stress type that was instrumental in the failure, the contact stress was an index to its magnitude. Buckingham27 conducted a number of tests relating the fatigue at 108 cycles to endurance strength (Hertzian contact pressure). While there is evidence of an endurance limit at about 3(107 ) cycles for cast materials, hardened steel rollers showed no endurance limit up to 4(108 ) cycles. Subsequent testing on hard steel shows no endurance limit. Hardened steel exhibits such high fatigue strengths that its use in resisting surface fatigue is widespread. Our studies thus far have dealt with the failure of a machine element by yielding, by fracture, and by fatigue. The endurance limit obtained by the rotating-beam test is frequently called the exural endurance limit, because it is a test of a rotating beam. In this section we shall study a property of mating materials called the surface endurance shear. The design engineer must frequently solve problems in which two machine elements mate with one another by rolling, sliding, or a combination of rolling and sliding contact. Obvious examples of such combinations are the mating teeth of a pair of gears, a cam and follower, a wheel and rail, and a chain and sprocket. A knowledge of the surface strength of materials is necessary if the designer is to create machines having a long and satisfactory life. When two surfaces roll or roll and slide against one another with sufcient force, a pitting failure will occur after a certain number of cycles of operation. Authorities are not in complete agreement on the exact mechanism of the pitting; although the subject is quite complicated, they do agree that the Hertz stresses, the number of cycles, the surface nish, the hardness, the degree of lubrication, and the temperature all inuence the strength. In Sec. 319 it was learned that, when two surfaces are pressed together, a maximum shear stress is developed slightly below the contacting surface. It is postulated by some authorities that a surface fatigue failure is initiated by this maximum shear stress and then is propagated rapidly to the surface. The lubricant then enters the crack that is formed and, under pressure, eventually wedges the chip loose. 26 J. A. Collins, Failure of Materials in Mechanical Design, John Wiley & Sons, New York, 1981, p. 243. 27 Earle Buckingham, Analytical Mechanics of Gears, McGraw-Hill, New York, 1949. BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 320 II. Failure Prevention 323 © The McGrawHill Companies, 2008 6. Fatigue Failure Resulting from Variable Loading Mechanical Engineering Design To determine the surface fatigue strength of mating materials, Buckingham designed a simple machine for testing a pair of contacting rolling surfaces in connection with his investigation of the wear of gear teeth. Buckingham and, later, Talbourdet gathered large numbers of data from many tests so that considerable design information is now available. To make the results useful for designers, Buckingham dened a load-stress factor, also called a wear factor, which is derived from the Hertz equations. Equations (373) and (374), pp. 118119, for contacting cylinders are found to be 2 2 2 F 1 ν1 / E 1 + 1 ν2 / E 2 πl (1/d1 ) + (1/d2 ) b= pmax = 2F π bl (659) (660) where b = half width of rectangular contact area F = contact force l = length of cylinders ν = Poissons ratio E = modulus of elasticity d = cylinder diameter It is more convenient to use the cylinder radius, so let 2r = d . If we then designate the length of the cylinders as w (for width of gear, bearing, cam, etc.) instead of l and remove the square root sign, Eq. (659) becomes b2 = 2 2 4 F 1 ν1 / E 1 + 1 ν2 / E 2 πw 1/ r1 + 1/ r2 (661) We can dene a surface endurance strength SC using pmax = 2F π bw (662) as SC = 2F π bw (663) which may also be called contact strength, the contact fatigue strength, or the Hertzian endurance strength. The strength is the contacting pressure which, after a specied number of cycles, will cause failure of the surface. Such failures are often called wear because they occur over a very long time. They should not be confused with abrasive wear, however. By squaring Eq. (663), substituting b2 from Eq. (661), and rearranging, we obtain F w 1 1 + r1 r2 2 = π SC 2 2 1 ν2 1 ν1 + = K1 E1 E2 (664) The left expression consists of parameters a designer may seek to control independently. The central expression consists of material properties that come with the material and condition specication. The third expression is the parameter K 1 , Buckinghams loadstress factor, determined by a test xture with values F, w , r1 , r2 and the number of 324 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition II. Failure Prevention © The McGrawHill Companies, 2008 6. Fatigue Failure Resulting from Variable Loading Fatigue Failure Resulting from Variable Loading 321 cycles associated with the rst tangible evidence of fatigue. In gear studies a similar K factor is used: K1 sin φ Kg = (665) 4 2 2 where φ is the tooth pressure angle, and the term [(1 ν1 )/ E 1 + (1 ν2 )/ E 2 ] is dened as 1/(π C 2 ), so that P SC = C P F w 1 1 + r1 r2 (666) Buckingham and others reported K 1 for 108 cycles and nothing else. This gives only one point on the SC N curve. For cast metals this may be sufcient, but for wrought steels, heattreated, some idea of the slope is useful in meeting design goals of other than 108 cycles. Experiments show that K 1 versus N, K g versus N, and SC versus N data are rectied by loglog transformation. This suggests that Kg = a N b K 1 = α1 N β1 SC = α N β The three exponents are given by β1 = log( K 1 / K 2 ) log( N1 / N2 ) b= log( K g1 / K g2 ) log( N1 / N2 ) β= log( SC 1 / SC 2 ) log( N1 / N2 ) (667) Data on induction-hardened steel on steel give ( SC )107 = 271 kpsi and ( SC )108 = 239 kpsi, so β , from Eq. (667), is β= log(271/239) = 0.055 log(107 /108 ) It may be of interest that the American Gear Manufacturers Association (AGMA) uses β 0.056 between 104 < N < 1010 if the designer has no data to the contrary beyond 107 cycles. A longstanding correlation in steels between SC and HB at 108 cycles is ( SC )108 = 0.4 HB 10 kpsi 2.76 HB 70 MPa (668) 0.99 ( SC )107 = 0.327 HB + 26 kpsi (669) AGMA uses Equation (666) can be used in design to nd an allowable surface stress by using a design factor. Since this equation is nonlinear in its stress-load transformation, the designer must decide if loss of function denotes inability to carry the load. If so, then to nd the allowable stress, one divides the load F by the design factor n d : σC = C P F wn d 1 1 + r1 r2 CP = nd F w 1 1 + r1 r2 SC = nd and n d = ( SC /σC )2 . If the loss of function is focused on stress, then n d = SC /σC . It is recommended that an engineer Decide whether loss of function is failure to carry load or stress. Dene the design factor and factor of safety accordingly. Announce what he or she is using and why. Be prepared to defend his or her position. BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 322 II. Failure Prevention 6. Fatigue Failure Resulting from Variable Loading © The McGrawHill Companies, 2008 Mechanical Engineering Design In this way everyone who is party to the communication knows what a design factor (or factor of safety) of 2 means and adjusts, if necessary, the judgmental perspective. 617 Stochastic Analysis28 As already demonstrated in this chapter, there are a great many factors to consider in a fatigue analysis, much more so than in a static analysis. So far, each factor has been treated in a deterministic manner, and if not obvious, these factors are subject to variability and control the overall reliability of the results. When reliability is important, then fatigue testing must certainly be undertaken. There is no other way. Consequently, the methods of stochastic analysis presented here and in other sections of this book constitute guidelines that enable the designer to obtain a good understanding of the various issues involved and help in the development of a safe and reliable design. In this section, key stochastic modications to the deterministic features and equations described in earlier sections are provided in the same order of presentation. Endurance Limit To begin, a method for estimating endurance limits, the tensile strength correlation ¯ method, is presented. The ratio = Se / Sut is called the fatigue ratio.29 For ferrous metals, most of which exhibit an endurance limit, the endurance limit is used as a numerator. For materials that do not show an endurance limit, an endurance strength at a specied number of cycles to failure is used and noted. Gough30 reported the stochastic nature of the fatigue ratio for several classes of metals, and this is shown in Fig. 636. The rst item to note is that the coefcient of variation is of the order 0.10 to 0.15, and the distribution varies for classes of metals. The second item to note is that Goughs data include materials of no interest to engineers. In the absence of testing, engineers use the correlation that represents to estimate the endurance limit Se from ¯ the mean ultimate strength Sut . Goughs data are for ensembles of metals, some chosen for metallurgical interest, and include materials that are not commonly selected for machine parts. Mischke31 analyzed data for 133 common steels and treatments in varying diameters in rotating bending,32 and the result was = 0.445d 0.107 LN(1, 0.138) where d is the specimen diameter in inches and LN(1, 0.138) is a unit lognormal variate with a mean of 1 and a standard deviation (and coefcient of variation) of 0.138. For the standard R. R. Moore specimen, 0.30 = 0.445(0.30)0.107 LN(1, 0.138) = 0.506LN(1, 0.138) 28 Review Chap. 20 before reading this section. 29 From this point, since we will be dealing with statistical distributions in terms of means, standard ¯ deviations, etc. A key quantity, the ultimate strength, will here be presented by its mean value, Sut . This means that certain terms that were dened earlier in terms of the minimum value of Sut will change slightly. 30 In J. A. Pope, Metal Fatigue, Chapman and Hall, London, 1959. 31 Charles R. Mischke, Prediction of Stochastic Endurance Strength, Trans. ASME, Journal of Vibration, Acoustics, Stress, and Reliability in Design, vol. 109, no. 1, January 1987, pp. 113122. 32 Data from H. J. Grover, S. A. Gordon, and L. R. Jackson, Fatigue of Metals and Structures, Bureau of Naval Weapons, Document NAVWEPS 00-2500435, 1960. 325 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition II. Failure Prevention © The McGrawHill Companies, 2008 6. Fatigue Failure Resulting from Variable Loading Fatigue Failure Resulting from Variable Loading Figure 636 The lognormal probability density PDF of the fatigue ratio φb of Gough. 4 5 Class All metals Nonferrous Iron and carbon steels Low alloy steels Special alloy steels 1 2 3 4 5 3 Probability density 326 No. 380 152 111 78 39 2 1 0 323 0.3 5 0.4 0.5 0.6 Rotary bending fatigue ratio 0.7 b Also, 25 plain carbon and low-alloy steels with Sut > 212 kpsi are described by Se = 107LN(1, 0.139) kpsi In summary, for the rotating-beam specimen, ¯ 0.506 Sut LN(1, 0.138) kpsi or MPa Se = 107LN(1, 0.139) kpsi 740LN(1, 0.139) MPa ¯ Sut 212 kpsi (1460 MPa) ¯ Sut > 212 kpsi (670) ¯ut > 1460 MPa S ¯ where Sut is the mean ultimate tensile strength. Equations (670) represent the state of information before an engineer has chosen a material. In choosing, the designer has made a random choice from the ensemble of possibilities, and the statistics can give the odds of disappointment. If the testing is lim¯ ited to nding an estimate of the ultimate tensile strength mean Sut with the chosen material, Eqs. (670) are directly helpful. If there is to be rotary-beam fatigue testing, then statistical information on the endurance limit is gathered and there is no need for the correlation above. ¯ Table 69 compares approximate mean values of the fatigue ratio φ0.30 for several classes of ferrous materials. Endurance Limit Modifying Factors A Marin equation can be written as Se = ka kb kc kd kf Se (671) where the size factor kb is deterministic and remains unchanged from that given in Sec. 69. Also, since we are performing a stochastic analysis, the reliability factor ke is unnecessary here. The surface factor ka cited earlier in deterministic form as Eq. (620), p. 280, is now given in stochastic form by ¯b ka = a Sut LN(1, C ) ¯ ( Sut in kpsi or MPa) where Table 610 gives values of a, b, and C for various surface conditions. (672) BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 324 II. Failure Prevention 327 © The McGrawHill Companies, 2008 6. Fatigue Failure Resulting from Variable Loading Mechanical Engineering Design Table 69 Material Class Wrought steels 0.50 Cast steels 0.40 Powdered steels 0.38 Gray cast iron 0.35 Malleable cast iron 0.40 Normalized nodular cast iron Comparison of Approximate Values of Mean Fatigue Ratio for Some Classes of Metals φ 0.30 0.33 Table 610 ka Parameters in Marin Surface Condition Factor Surface Finish b aSut LN(1, C ) a kpsi Ground 1.34 Machined or Cold-rolled 2.67 Hot-rolled 14.5 As-forged 39.8 MPa Solution 0.086 1.58 0.265 4.45 0.719 58.1 0.995 271 *Due to the wide scatter in ground surface data, an alternate function is ka EXAMPLE 616 b Coefcient of Variation, C 0.120 0.058 0.110 0.145 0.878LN(1, 0.120). Note: Sut in kpsi or MPa. A steel has a mean ultimate strength of 520 MPa and a machined surface. Estimate ka . From Table 610, ka = 4.45(520)0.265 LN(1, 0.058) ¯ ka = 4.45(520)0.265 (1) = 0.848 Answer ¯ σka = C ka = (0.058)4.45(520)0.265 = 0.049 ˆ so ka = LN(0.848, 0.049) . The load factor kc for axial and torsional loading is given by ¯ (kc )axial = 1.23 Sut0.0778 LN(1, 0.125) ¯0 (kc )torsion = 0.328 Sut.125 LN(1, 0.125) (673) (674) ¯ where Sut is in kpsi. There are fewer data to study for axial fatigue. Equation (673) was deduced from the data of Landgraf and of Grover, Gordon, and Jackson (as cited earlier). Torsional data are sparser, and Eq. (674) is deduced from data in Grover et al. Notice the mild sensitivity to strength in the axial and torsional load factor, so kc in these cases is not constant. Average values are shown in the last column of Table 611, and as footnotes to Tables 612 and 613. Table 614 shows the inuence of material classes on the load factor kc . Distortion energy theory predicts (kc )torsion = 0.577 for materials to which the distortion-energy theory applies. For bending, kc = LN(1, 0) . 328 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition II. Failure Prevention © The McGrawHill Companies, 2008 6. Fatigue Failure Resulting from Variable Loading Fatigue Failure Resulting from Variable Loading Table 611 Parameters in Marin Loading Factor kc Mode of Loading MPa Bending 1 1 Axial 1.23 1.43 Torsion Table 612 0.328 0.258 ¯ Sut , kpsi C Average kc 0 1 0.125 0.85 0.125 0.59 50 0.125 0.860 0.832 200 0 0.0778 0.907 150 β k* c 100 Average Marin Loading Factor for Axial Load β αSut LN(1, C) α kpsi 0.814 *Average entry 0.85. Table 613 Average Marin Loading Factor for Torsional Load ¯ Sut , kpsi k* c 50 0.535 100 0.583 150 0.614 200 0.636 *Average entry 0.59. Table 614 Average Marin Torsional Loading Factor kc for Several Materials Range n ¯c k ˆkc σ Wrought steels 0.520.69 31 0.60 0.03 Wrought Al 0.430.74 13 0.55 0.09 Wrought Cu and alloy 0.410.67 7 0.56 0.10 Wrought Mg and alloy 0.490.60 2 0.54 0.08 Material 325 Titanium 0.370.57 3 0.48 0.12 Cast iron 0.791.01 9 0.90 0.07 Cast Al, Mg, and alloy 0.710.91 5 0.85 0.09 Source: The table is an extension of P. G. Forrest, Fatigue of Metals, Pergamon Press, London, 1962, Table 17, p. 110, with standard deviations estimated from range and sample size using Table A1 in J. B. Kennedy and A. M. Neville, Basic Statistical Methods for Engineers and Scientists, 3rd ed., Harper & Row, New York, 1986, pp. 5455. BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 326 II. Failure Prevention 329 © The McGrawHill Companies, 2008 6. Fatigue Failure Resulting from Variable Loading Mechanical Engineering Design EXAMPLE 617 Solution Estimate the Marin loading factor kc for a 1in-diameter bar that is used as follows. (a) In bending. It is made of steel with Sut = 100LN(1, 0.035) kpsi, and the designer ¯ intends to use the correlation Se = 0.30 Sut to predict Se . (b) In bending, but endurance testing gave Se = 55LN(1, 0.081) kpsi. (c) In push-pull (axial) fatigue, Sut = LN(86.2, 3.92) kpsi, and the designer intended to ¯ use the correlation Se = 0.30 Sut . (d) In torsional fatigue. The material is cast iron, and Se is known by test. (a) Since the bar is in bending, Answer kc = (1, 0) (b) Since the test is in bending and use is in bending, Answer kc = (1, 0) (c) From Eq. (673), Answer (kc )ax = 1.23(86.2)0.0778 LN(1, 0.125) ¯ kc = 1.23(86.2)0.0778 (1) = 0.870 ¯ σkc = C kc = 0.125(0.870) = 0.109 ˆ ¯ ˆ (d) From Table 615, kc = 0.90, σkc = 0.07, and Answer Ckc = 0.07 = 0.08 0.90 The temperature factor kd is ¯ kd = kd LN(1, 0.11) (675) ¯ where kd = kd , given by Eq. (627), p. 283. Finally, kf is, as before, the miscellaneous factor that can come about from a great many considerations, as discussed in Sec. 69, where now statistical distributions, possibly from testing, are considered. Stress Concentration and Notch Sensitivity Notch sensitivity q was dened by Eq. (631), p. 287. The stochastic equivalent is q= Kf 1 Kt 1 (676) where K t is the theoretical (or geometric) stress-concentration factor, a deterministic quantity. A study of lines 3 and 4 of Table 206, will reveal that adding a scalar to (or subtracting one from) a variate x will affect only the mean. Also, multiplying (or dividing) by a scalar affects both the mean and standard deviation. With this in mind, we can 330 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition II. Failure Prevention © The McGrawHill Companies, 2008 6. Fatigue Failure Resulting from Variable Loading Fatigue Failure Resulting from Variable Loading a( in) , a( mm) , Notch Type Sut in kpsi Sut in MPa Coefcient of Variation CKf Transverse hole 5/Sut 174/Sut 0.10 Shoulder 4/Sut 139/Sut 0.11 Groove 3/Sut 104/Sut 0.15 Table 615 Heywoods Parameter a and coefcients of variation CKf for steels 327 relate the statistical parameters of the fatigue stress-concentration factor K f to those of notch sensitivity q. It follows that q = LN ¯ ¯ Kf 1 CKf , Kt 1 Kt 1 where C = C K f and q= ¯ σq = ˆ Cq = ¯ Kf 1 Kt 1 ¯ CKf Kt 1 (677) ¯ CKf ¯ Kf 1 The fatigue stress-concentration factor K f has been investigated more in England than in ¯ the United States. For K f , consider a modied Neuber equation (after Heywood33 ), where the fatigue stress-concentration factor is given by Kt (678) 2( K t 1) a 1+ Kt r where Table 615 gives values of a and C K f for steels with transverse holes, shoulders, or grooves. Once K f is described, q can also be quantified using the set Eqs. (677). The modied Neuber equation gives the fatigue stress concentration factor as ¯ Kf = ¯ K f = K f LN 1, C K f (679) 33 R. B. Heywood, Designing Against Fatigue, Chapman & Hall, London, 1962. EXAMPLE 618 Solution Estimate K f and q for the steel shaft given in Ex. 66, p. 288. From Ex. 66, a steel shaft with Sut = 690 Mpa and a shoulder with a llet of 3 mm . was found to have a theoretical stress-concentration-factor of K t = 1.65. From Table 615, 139 139 = 0.2014 mm a= = Sut 690 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 328 II. Failure Prevention 6. Fatigue Failure Resulting from Variable Loading © The McGrawHill Companies, 2008 Mechanical Engineering Design From Eq. (678), Kf = Answer 1.65 Kt = 1.51 = 2(1.65 1) 0.2014 2( K t 1) a 1+ 1+ 1.65 Kt 3 r which is 2.5 percent lower than what was found in Ex. 66. From Table 615, C K f = 0.11. Thus from Eq. (679), K f = 1.51 LN(1, 0.11) From Eq. (677), with K t = 1.65 q= ¯ Cq = 1.51 1 = 0.785 1.65 1 ¯ CK f K f 0.11(1.51) = = 0.326 ¯ 1.51 1 Kf 1 σq = Cq q = 0.326(0.785) = 0.256 ˆ ¯ So, Answer EXAMPLE 619 Solution q = LN(0.785, 0.256) The bar shown in Fig. 637 is machined from a cold-rolled at having an ultimate strength of Sut = LN(87.6, 5.74) kpsi. The axial load shown is completely reversed. The load amplitude is Fa = LN(1000, 120) lbf. (a) Estimate the reliability. (b) Reestimate the reliability when a rotating bending endurance test shows that Se = LN(40, 2) kpsi. ¯ (a) From Eq. (670), Se = 0.506 Sut LN(1, 0.138) = 0.506(87.6)LN(1, 0.138) = 44.3LN(1, 0.138) kpsi From Eq. (672) and Table 610, ¯ ka = 2.67 Sut0.265 LN(1, 0.058) = 2.67(87.6)0.265 LN(1, 0.058) = 0.816LN(1, 0.058) kb = 1 (axial loading) 3 16 Figure 637 1000 lbf 2 1 in 4 in R. 1000 lbf 1 1 in 2 1 4 in 3 4 in D. 331 332 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition II. Failure Prevention © The McGrawHill Companies, 2008 6. Fatigue Failure Resulting from Variable Loading Fatigue Failure Resulting from Variable Loading 329 From Eq. (673), ¯ kc = 1.23 Sut0.0778 LN(1, 0.125) = 1.23(87.6)0.0778 LN(1, 0.125) = 0.869LN(1, 0.125) kd = k f = (1, 0) The endurance strength, from Eq. (671), is Se = ka kb kc kd k f Se Se = 0.816LN(1, 0.058)(1)0.869LN(1, 0.125)(1)(1)44.3LN(1, 0.138) The parameters of Se are ¯ Se = 0.816(0.869)44.3 = 31.4 kpsi C Se = (0.0582 + 0.1252 + 0.1382 )1/2 = 0.195 so Se = 31.4LN(1, 0.195) kpsi. In computing the stress, the section at the hole governs. Using the terminology . of Table A151 we nd d /w = 0.50, therefore K t = 2.18. From Table 615, a = 5/ Sut = 5/87.6 = 0.0571 and Ck f = 0.10. From Eqs. (678) and (679) with r = 0.375 in, 2.18 Kt LN(1, 0.10) LN 1, C K f = 2(2.18 1) 0.0571 2( K t 1) a 1+ 1+ 2.18 0.375 Kt r = 1.98LN(1, 0.10) Kf = The stress at the hole is = Kf 1000LN(1, 0.12) F = 1.98LN(1, 0.10) A 0.25(0.75) σ = 1.98 ¯ 1000 103 = 10.56 kpsi 0.25(0.75) Cσ = (0.102 + 0.122 )1/2 = 0.156 so stress can be expressed as = 10.56LN(1, 0.156) kpsi.34 The endurance limit is considerably greater than the load-induced stress, indicating that nite life is not a problem. For interfering lognormal-lognormal distributions, Eq. (543), p. 242, gives ln z= ¯ Se σ ¯ 2 1 + Cσ 2 1 + C Se 2 ln 1 + C Se 2 1 + Cσ = ln 31.4 10.56 ln[(1 + 0.1952 )(1 + 0.1562 )] From Table A10 the probability of failure p f = reliability is Answer 1 + 0.1562 1 + 0.1952 = 4.37 (4.37) = .000 006 35, and the R = 1 0.000 006 35 = 0.999 993 65 34 Note that there is a simplication here. The area is not a deterministic quantity. It will have a statistical distribution also. However no information was given here, and so it was treated as being deterministic. BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 330 II. Failure Prevention © The McGrawHill Companies, 2008 6. Fatigue Failure Resulting from Variable Loading Mechanical Engineering Design (b) The rotary endurance tests are described by Se = 40LN(1, 0.05) kpsi whose mean ¯ is less than the predicted mean in part a. The mean endurance strength Se is ¯ Se = 0.816(0.869)40 = 28.4 kpsi C Se = (0.0582 + 0.1252 + 0.052 )1/2 = 0.147 so the endurance strength can be expressed as Se = 28.3LN(1, 0.147) kpsi. From Eq. (543), 28.4 1 + 0.1562 ln 10.56 1 + 0.1472 z= = 4.65 ln[(1 + 0.1472 )(1 + 0.1562 )] Using Table A10, we see the probability of failure p f = and (4.65) = 0.000 001 71, R = 1 0.000 001 71 = 0.999 998 29 an increase! The reduction in the probability of failure is (0.000 001 71 0.000 006 35)/0.000 006 35 = 0.73, a reduction of 73 percent. We are analyzing an existing ¯¯ ¯ design, so in part (a) the factor of safety was n = S /σ = 31.4/10.56 = 2.97. In part (b) n = 28.4/ 10.56 = 2.69, a decrease. This example gives you the opportunity to see the role ¯ ¯ ¯ of the design factor. Given knowledge of S, C S, σ , Cσ , and reliability (through z), the mean ¯ ¯ factor of safety (as a design factor) separates S and σ so that the reliability goal is achieved. ¯ ¯ Knowing n alone says nothing about the probability of failure. Looking at n = 2.97 and n = 2.69 says nothing about the respective probabilities of failure. The tests did not reduce ¯ ¯ Se signicantly, but reduced the variation C S such that the reliability was increased. ¯¯ When a mean design factor (or mean factor of safety) defined as Se /σ is said to be silent on matters of frequency of failures, it means that a scalar factor of safety by itself does not offer any information about probability of failure. Nevertheless, some engineers let the factor of safety speak up, and they can be wrong in their conclusions. As revealing as Ex. 619 is concerning the meaning (and lack of meaning) of a design factor or factor of safety, let us remember that the rotary testing associated with part (b) changed nothing about the part, but only our knowledge about the part. The mean endurance limit was 40 kpsi all the time, and our adequacy assessment had to move with what was known. Fluctuating Stresses Deterministic failure curves that lie among the data are candidates for regression models. Included among these are the Gerber and ASME-elliptic for ductile materials, and, for brittle materials, Smith-Dolan models, which use mean values in their presentation. Just as the deterministic failure curves are located by endurance strength and ultimate tensile (or yield) strength, so too are stochastic failure curves located by Se and by Sut or S y . Figure 632, p. 312, shows a parabolic Gerber mean curve. We also need to establish a contour located one standard deviation from the mean. Since stochastic 333 334 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition II. Failure Prevention © The McGrawHill Companies, 2008 6. Fatigue Failure Resulting from Variable Loading Fatigue Failure Resulting from Variable Loading 331 curves are most likely to be used with a radial load line we will use the equation given in Table 67, p. 299, expressed in terms of the strength means as ¯2 ¯2 r 2 Sut 2 Se ¯ 1 + 1 + Sa = (680) ¯ ¯ 2 Se r Sut ¯¯ ¯ Because of the positive correlation between Se and Sut , we increment Se by C Se Se , Sut ¯ ¯ ¯ by C Sut Sut , and Sa by C Sa Sa , substitute into Eq. (680), and solve for C Sa to obtain 2 ¯ 2 Se (1 + C Se ) 1 + 1 + ¯ r Sut (1 + C Sut ) (1 + C Sut )2 1 C Sa = (681) 1 + C Se ¯2 2 Se 1 + 1 + ¯ r Sut Equation (681) can be viewed as an interpolation formula for C Sa , which falls between ¯ C Se and C Sut depending on load line slope r. Note that Sa = Sa LN(1, C Sa ) . Similarly, the ASME-elliptic criterion of Table 68, p. 300, expressed in terms of its means is ¯ Sa = ¯¯ r Sy Se (682) ¯2 ¯2 r 2 Sy + Se ¯¯ ¯ ¯ ¯ ¯ Similarly, we increment Se by C Se Se , Sy by C Sy Sy , and Sa by C Sa Sa , substitute into Eq. (682), and solve for C Sa : C Sa = (1 + C Sy )(1 + C Se ) ¯2 ¯2 r 2 Sy + Se 1 ¯2 ¯2 r 2 Sy (1 + C Sy )2 + Se (1 + C Se )2 (683) Many brittle materials follow a Smith-Dolan failure criterion, written deterministically as n σa 1 n σm / Sut = Se 1 + n σm / Sut (684) ¯¯ ¯ 1 Sm / Sut Sa = ¯ ¯¯ Se 1 + Sm / Sut (685) Expressed in terms of its means, ¯ ¯ ¯ For a radial load line slope of r, we substitute Sa / r for Sm and solve for Sa , obtaining ¯ ¯¯ ¯ r Sut + Se 4r Sut Se ¯ 1 + 1 + Sa = (686) ¯ ¯ 2 (r Sut + Se )2 and the expression for C Sa is ¯ ¯ r Sut (1 + C Sut ) + Se (1 + C Se ) C Sa = ¯ 2 Sa · 1 + 1+ ¯¯ 4r Sut Se (1 + C Se )(1 + C Sut ) ¯ ¯ [r Sut (1 + C Sut ) + Se (1 + C Se )]2 (687) 1 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 332 II. Failure Prevention © The McGrawHill Companies, 2008 6. Fatigue Failure Resulting from Variable Loading Mechanical Engineering Design EXAMPLE 620 Solution A rotating shaft experiences a steady torque T = 1360LN(1, 0.05) lbf · in, and at a shoulder with a 1.1-in small diameter, a fatigue stress-concentration factor K f = 1.50LN(1, 0.11) , K f s = 1.28LN(1, 0.11) , and at that location a bending moment of M = 1260LN(1, 0.05) lbf · in. The material of which the shaft is machined is hot-rolled 1035 with Sut = 86.2LN(1, 0.045) kpsi and S y = 56.0LN(1, 0.077) kpsi. Estimate the reliability using a stochastic Gerber failure zone. Establish the endurance strength. From Eqs. (670) to (672) and Eq. (620), p. 280, Se = 0.506(86.2)LN(1, 0.138) = 43.6LN(1, 0.138) kpsi ka = 2.67(86.2)0.265 LN(1, 0.058) = 0.820LN(1, 0.058) kb = (1.1/0.30)0.107 = 0.870 kc = kd = k f = LN(1, 0) Se = 0.820LN(1, 0.058)0.870(43.6)LN(1, 0.138) ¯ Se = 0.820(0.870)43.6 = 31.1 kpsi C Se = (0.0582 + 0.1382 )1/2 = 0.150 and so Se = 31.1LN(1, 0.150) kpsi. Stress (in kpsi): σa = 32K f Ma 32(1.50)LN(1, 0.11)1.26LN(1, 0.05) = π d3 π(1.1)3 σa = ¯ 32(1.50)1.26 = 14.5 kpsi π(1.1)3 Cσ a = (0.112 + 0.052 )1/2 = 0.121 m = τm = ¯ 16K f s Tm 16(1.28)LN(1, 0.11)1.36LN(1, 0.05) = 3 πd π(1.1)3 16(1.28)1.36 = 6.66 kpsi π(1.1)3 Cτ m = (0.112 + 0.052 )1/2 = 0.121 σa = σa + 3τa ¯ ¯2 ¯2 1/2 = [14.52 + 3(0)2 ]1/2 = 14.5 kpsi σm = σm + 3τm ¯ ¯2 ¯2 1/2 = [0 + 3(6.66)2 ]1/2 = 11.54 kpsi r= σa ¯ 14.5 = = 1.26 σm ¯ 11.54 Strength: From Eqs. (680) and (681), 1.26 86.2 ¯ 1 + Sa = 2(31.1) 2 2 1+ 2(31.1) 1.26(86.2) 2 = 28.9 kpsi 335 336 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition II. Failure Prevention © The McGrawHill Companies, 2008 6. Fatigue Failure Resulting from Variable Loading Fatigue Failure Resulting from Variable Loading C Sa 1 + (1 + 0.045)2 = 1 + 0.150 1+ 1 + 2(31.1)(1 + 0.15) 1.26(86.2)(1 + 0.045) 2(31.1) 1+ 1.26(86.2) 2 333 2 1 = 0.134 Reliability: Since Sa = 28.9LN(1, 0.134) kpsi and a = 14.5LN(1, 0.121) kpsi, Eq. (544), p. 242, gives 2 28.9 1 + 0.1212 ¯a 1 + Cσa S ln ln 2 14.5 1 + 0.1342 σa 1 + C Sa ¯ z= = = 3.83 ln[(1 + 0.1342 )(1 + 0.1212 )] ln 1 + C 2 1 + C 2 Sa σa From Table A10 the probability of failure is p f = 0.000 065, and the reliability is, against fatigue, Answer R = 1 p f = 1 0.000 065 = 0.999 935 The chance of rst-cycle yielding is estimated by interfering S y with max . The ¯ ¯ quantity max is formed from a + m . The mean of max is σa + σm = 14.5 + 11.54 = 26.04 kpsi. The coefcient of variation of the sum is 0.121, since both COVs are 0.121, thus Cσ max = 0.121. We interfere S y = 56LN(1, 0.077) kpsi with max = 26.04LN (1, 0.121) kpsi. The corresponding z variable is 1 + 0.1212 56 ln 26.04 1 + 0.0772 = 5.39 z= ln[(1 + 0.0772 )(1 + 0.1212 )] which represents, from Table A10, a probability of failure of approximately 0.07 358 [which represents 3.58(108 )] of rst-cycle yield in the llet. The probability of observing a fatigue failure exceeds the probability of a yield failure, something a deterministic analysis does not foresee and in fact could lead one to expect a yield failure should a failure occur. Look at the a Sa interference and the max S y interference and examine the z expressions. These control the relative probabilities. A deterministic analysis is oblivious to this and can mislead. Check your statistics text for events that are not mutually exclusive, but are independent, to quantify the probability of failure: p f = p(yield) + p(fatigue) p(yield and fatigue) = p(yield) + p(fatigue) p(yield) p(fatigue) = 0.358(107 ) + 0.65(104 ) 0.358(107 )0.65(104 ) = 0.650(104 ) R = 1 0.650(104 ) = 0.999 935 against either or both modes of failure. BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 334 II. Failure Prevention 337 © The McGrawHill Companies, 2008 6. Fatigue Failure Resulting from Variable Loading Mechanical Engineering Design Figure 638 ea M Designers fatigue diagram for Ex. 620. 50 n La ng cu rv e Amplitude stress component a, kpsi er 40 Load line 30 1 S ig ma c urve Sa Mea nG +1 erbe r cu _ Sa Sig ma rve cur ve 20 a _ a 10 0 0 10 20 30 40 50 Steady stress component 60 m, 70 80 90 kpsi Examine Fig. 638, which depicts the results of Ex. 620. The problem distribution of Se was compounded of historical experience with Se and the uncertainty manifestations due to features requiring Marin considerations. The Gerber failure zone displays this. The interference with load-induced stress predicts the risk of failure. If additional information is known (R. R. Moore testing, with or without Marin features), the stochastic Gerber can accommodate to the information. Usually, the accommodation to additional test information is movement and contraction of the failure zone. In its own way the stochastic failure model accomplishes more precisely what the deterministic models and conservative postures intend. Additionally, stochastic models can estimate the probability of failure, something a deterministic approach cannot address. The Design Factor in Fatigue The designer, in envisioning how to execute the geometry of a part subject to the imposed constraints, can begin making a priori decisions without realizing the impact on the design task. Now is the time to note how these things are related to the reliability goal. The mean value of the design factor is given by Eq. (545), repeated here as 2 2. n = exp z ln 1 + Cn + ln 1 + Cn = exp[Cn (z + Cn /2)] ¯ (688) in which, from Table 206 for the quotient n = S/ , Cn = 2 2 C S + Cσ 2 1 + Cσ where C S is the COV of the signicant strength and Cσ is the COV of the signicant ¯ stress at the critical location. Note that n is a function of the reliability goal (through z) and the COVs of the strength and stress. There are no means present, just measures of variability. The nature of C S in a fatigue situation may be C Se for fully reversed loading, or C Sa otherwise. Also, experience shows C Se > C Sa > C Sut , so C Se can be used as a conservative estimate of C Sa . If the loading is bending or axial, the form of 338 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition II. Failure Prevention © The McGrawHill Companies, 2008 6. Fatigue Failure Resulting from Variable Loading Fatigue Failure Resulting from Variable Loading a 335 might be a = Kf Ma c I respectively. This makes the COV of a or a, = Kf F A namely Cσa , expressible as 2 2 Cσa = C K f + C F 1/2 again a function of variabilities. The COV of Se , namely C Se , is 2 2 2 2 2 C Se = Cka + Ckc + Ckd + Ck f + C Se 1/2 again, a function of variabilities. An example will be useful. EXAMPLE 621 Solution A strap to be made from a cold-drawn steel strip workpiece is to carry a fully reversed axial load F = LN(1000, 120) lbf as shown in Fig. 639. Consideration of adjacent parts established the geometry as shown in the gure, except for the thickness t. Make a decision as to the magnitude of the design factor if the reliability goal is to be 0.999 95, then make a decision as to the workpiece thickness t. Let us take each a priori decision and note the consequence: A Priori Decision Use 1018 CD steel Consequence ¯ Sut 87.6kpsi, CSut 0.0655 Function: Carry axial load Fa = 1000 lbf R 0.999 95 CF z in D. drill Cka Hole critical 3 8 Machined surfaces CKf Ambient temperature Ckd CS e Hole drilled 3 4 Correlation method CSe Cn in n ¯ Fa = 1000 lbf 0.12, Ckc 3.891 0.058 0.10, C a 0 0.138 0.125 (0.102 0.122)1/2 = 0.156 (0.0582 + 0.1252 + 0.1382 ) 1/2 = 0.195 2 2 CSe + Cσ a 1+ 2 Cσ a = 0.1952 + 0.1562 = 0.2467 1 + 0.1562 exp ( 3.891) ln(1 + 0.24672 ) + ln 1 + 0.24672 = 2.65 Figure 639 A strap with a thickness t is subjected to a fully reversed axial load of 1000 lbf. Example 621 considers the thickness necessary to attain a reliability of 0.999 95 against a fatigue failure. ¯ These eight a priori decisions have quantied the mean design factor as n = 2.65. Proceeding deterministically hereafter we write ¯ ¯ Se F ¯ = Kf σa = n ¯ (w d )t from which ¯ ¯¯ Kf nF t= (1) ¯ (w d ) Se BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 336 II. Failure Prevention 6. Fatigue Failure Resulting from Variable Loading 339 © The McGrawHill Companies, 2008 Mechanical Engineering Design ¯ ¯ To evaluate the preceding equation we need Se and K f . The Marin factors are ¯ ka = 2.67 Sut0.265 LN(1, 0.058) = 2.67(87.6)0.265 LN(1, 0.058) ¯ ka = 0.816 kb = 1 ¯ kc = 1.23 Sut0.078 LN(1, 0.125) = 0.868LN(1, 0.125) ¯ kc = 0.868 ¯ ¯ kd = k f = 1 and the endurance strength is ¯ Se = 0.816(1)(0.868)(1)(1)0.506(87.6) = 31.4 kpsi The hole governs. From Table A151 we nd d /w = 0.50, therefore K t = 2.18. From ¯ Table 615 a = 5/ Sut = 5/87.6 = 0.0571, r = 0.1875 in. From Eq. (678) the fatigue stress concentration factor is 2.18 ¯ = 1.91 Kf = 2(2.18 1) 0.0571 1+ 2.18 0.1875 The thickness t can now be determined from Eq. (1) ¯ ¯¯ Kf nF 1.91(2.65)1000 t = 0.430 in = (w d ) Se (0.75 0.375)31 400 Use 1 -in-thick strap for the workpiece. The 1 -in thickness attains and, in the rounding 2 2 to available nominal size, exceeds the reliability goal. The example demonstrates that, for a given reliability goal, the fatigue design factor that facilitates its attainment is decided by the variabilities of the situation. Furthermore, the necessary design factor is not a constant independent of the way the concept unfolds. Rather, it is a function of a number of seemingly unrelated a priori decisions that are made in giving denition to the concept. The involvement of stochastic methodology can be limited to dening the necessary design factor. In particular, in the example, the design factor is not a function of the design variable t; rather, t follows from the design factor. 618 Road Maps and Important Design Equations for the Stress-Life Method As stated in Sec. 615, there are three categories of fatigue problems. The important procedures and equations for deterministic stress-life problems are presented here. Completely Reversing Simple Loading 1 Determine Se either from test data or p. 274 0.5 Sut Se = 100 kpsi 700 MPa Sut 200 kpsi (1400 MPa) Sut > 200 kpsi Sut > 1400 MPa (68) 340 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition II. Failure Prevention © The McGrawHill Companies, 2008 6. Fatigue Failure Resulting from Variable Loading Fatigue Failure Resulting from Variable Loading 337 2 Modify Se to determine Se . Se = ka kb kc kd ke k f Se p. 279 ka = Table 62 Parameters for Marin Surface Modication Factor, Eq. (619) (619) Factor a Surface Finish Sut, kpsi Sut, MPa Ground 1.34 1.58 Machined or cold-drawn 2.70 4.51 Hot-rolled 14.4 As-forged (618) b aSut 39.9 57.7 0.085 0.265 0.718 272. Rotating shaft. For bending or torsion, ( d /0.3) 0.107 = 0.879d 0.107 0.91d 0.157 kb = p. 280 ( d /7.62) 0.107 = 1.24d 0.107 1.51d 0.157 Exponent b 0.995 0.11 d 2 in 2 < d 10 in 2.79 d 51 mm 51 < 254 mm (620) For axial, (621) kb = 1 Nonrotating member. Use Table 63, p. 282, for de and substitute into Eq. (620) for d. bending 1 kc = 0.85 axial p. 282 (626) 0.59 torsion p. 283 Use Table 64 for kd, or 2 kd = 0.975 + 0.432(103 ) TF 0.115(105 ) TF 3 4 + 0.104(108 ) TF 0.595(1012 ) TF (627) pp. 284285, ke Table 65 Reliability Factors ke Corresponding to 8 Percent Standard Deviation of the Endurance Limit Reliability, % Transformation Variate za Reliability Factor ke 50 0 1.000 90 1.288 0.897 95 1.645 0.868 99 2.326 0.814 99.9 3.091 0.753 99.99 3.719 0.702 99.999 4.265 0.659 99.9999 4.753 0.620 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 338 II. Failure Prevention 341 © The McGrawHill Companies, 2008 6. Fatigue Failure Resulting from Variable Loading Mechanical Engineering Design pp. 285286, k f 3 Determine fatigue stress-concentration factor, K f or K f s . First, nd K t or K ts from Table A15. p. 287 K f = 1 + q ( K t 1) K f s = 1 + q ( K ts 1) or (632) Obtain q from either Fig. 620 or 621, pp. 287288. Alternatively, for reversed bending or axial loads, Kf = 1 + p. 288 Kt 1 1 + a /r (633) For Sut in kpsi, a = 0.245 799 0.307 794(102 ) Sut 2 3 +0.150 874(104 ) Sut 0.266 978(107 ) Sut (635) For torsion for low-alloy steels, increase Sut by 20 kpsi and apply to Eq. (635). 4 Apply K f or K f s by either dividing Se by it or multiplying it with the purely reversing stress not both. 5 Determine fatigue life constants a and b. If Sut 70 kpsi, determine f from Fig. 618, p. 277. If Sut < 70 kpsi, let f = 0.9. a = ( f Sut ) 2 / Se (614) b = [log( f Sut / Se )]/3 p. 277 (615) 6 Determine fatigue strength S f at N cycles, or, N cycles to failure at a reversing stress σa (Note: this only applies to purely reversing stresses where σm = 0). Sf = a N b p. 277 (613) N = ( σa /a ) 1/b (616) Fluctuating Simple Loading For Se , K f or K f s , see previous subsection. 1 Calculate σm and σa . Apply K f to both stresses. p. 293 σm = ( σmax + σmin ) /2 σa = |σmax σmin |/2 (636) 2 Apply to a fatigue failure criterion, p. 298 σm 0 Soderburg σa / Se + σm / Sy = 1/ n (645) mod-Goodman σa / Se + σm / Sut = 1/ n (646) n σa / Se + ( n σm / Sut ) = 1 (647) Gerber ASME-elliptic 2 ( σa / Se ) 2 + ( σm / Sut ) 2 = 1/ n 2 σm < 0 p. 297 σa = Se / n (648) 342 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition II. Failure Prevention © The McGrawHill Companies, 2008 6. Fatigue Failure Resulting from Variable Loading 339 Fatigue Failure Resulting from Variable Loading Torsion. Use the same equations as apply for σm 0, except replace σm and σa with τm and τa , use kc = 0.59 for Se , replace Sut with Ssu = 0.67 Sut [Eq. (654), p. 309], and replace Sy with Ssy = 0.577 Sy [Eq. (521), p. 217] 3 Check for localized yielding. (649) p. 298 σa + σm = Sy / n or, for torsion, τa + τm = 0.577 Sy / n 4 For nite-life fatigue strength (see Ex. 612, pp. 305306), mod-Goodman Sf = σa 1 ( σm / Sut ) Gerber Sf = σa 1 ( σm / Sut ) 2 If determining the nite life N with a factor of safety n, substitute S f / n for σa in Eq. (616). That is, N= Sf /n a 1/b Combination of Loading Modes See previous subsections for earlier denitions. 1 Calculate von Mises stresses for alternating and midrange stress states, σa and σm . When determining Se , do not use kc nor divide by K f or K f s . Apply K f and/or K f s directly to each specic alternating and midrange stress. If axial stress is present divide the alternating axial stress by kc = 0.85. For the special case of combined bending, torsional shear, and axial stresses p. 310 σa = ( σa ) axial ( K f ) bending ( σa ) bending + ( K f ) axial 0.85 1/2 2 + 3 ( K f s ) torsion ( τa ) torsion 2 (655) σm = ( K f ) bending ( σm ) bending + ( K f ) axial ( σm ) axial 2 + 3 ( K f s ) torsion ( τm ) torsion 2 1/2 (656) 2 Apply stresses to fatigue criterion [see Eq. (645) to (648), p. 338 in previous subsection]. 3 Conservative check for localized yielding using von Mises stresses. p. 298 σa + σm = Sy / n (649) BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 340 II. Failure Prevention 6. Fatigue Failure Resulting from Variable Loading © The McGrawHill Companies, 2008 Mechanical Engineering Design PROBLEMS Problems 61 to 631 are to be solved by deterministic methods. Problems 632 to 638 are to be solved by stochastic methods. Problems 639 to 646 are computer problems. Deterministic Problems 61 A 1 -in drill rod was heat-treated and ground. The measured hardness was found to be 490 Brinell. 4 Estimate the endurance strength if the rod is used in rotating bending. 62 Estimate Se for the following materials: (a) AISI 1020 CD steel. (b) AISI 1080 HR steel. (c) 2024 T3 aluminum. (d) AISI 4340 steel heat-treated to a tensile strength of 250 kpsi. 63 Estimate the fatigue strength of a rotating-beam specimen made of AISI 1020 hot-rolled steel corresponding to a life of 12.5 kilocycles of stress reversal. Also, estimate the life of the specimen corresponding to a stress amplitude of 36 kpsi. The known properties are Sut = 66.2 kpsi, σ0 = 115 kpsi, m = 0.22, and ε f = 0.90. 64 65 66 Derive Eq. (617). For the specimen of Prob. 63, estimate the strength corresponding to 500 cycles. For the interval 103 N 106 cycles, develop an expression for the axial fatigue strength ( S f )ax for the polished specimens of 4130 used to obtain Fig. 610. The ultimate strength is Sut = 125 kpsi and the endurance limit is ( Se )ax = 50 kpsi. Estimate the endurance strength of a 32-mm-diameter rod of AISI 1035 steel having a machined nish and heat-treated to a tensile strength of 710 MPa. 67 Two steels are being considered for manufacture of as-forged connecting rods. One is AISI 4340 Cr-Mo-Ni steel capable of being heat-treated to a tensile strength of 260 kpsi. The other is a plain carbon steel AISI 1040 with an attainable Sut of 113 kpsi. If each rod is to have a size giving an equivalent diameter de of 0.75 in, is there any advantage to using the alloy steel for this fatigue application? 68 A solid round bar, 25 mm in diameter, has a groove 2.5-mm deep with a 2.5-mm radius machined into it. The bar is made of AISI 1018 CD steel and is subjected to a purely reversing torque of 200 N · m. For the S-N curve of this material, let f = 0.9. (a) Estimate the number of cycles to failure. (b) If the bar is also placed in an environment with a temperature of 450 C, estimate the number of cycles to failure. 69 A solid square rod is cantilevered at one end. The rod is 0.8 m long and supports a completely reversing transverse load at the other end of ±1 kN. The material is AISI 1045 hot-rolled steel. If the rod must support this load for 104 cycles with a factor of safety of 1.5, what dimension should the square cross section have? Neglect any stress concentrations at the support end and assume that f = 0.9. 610 A rectangular bar is cut from an AISI 1018 cold-drawn steel at. The bar is 60 mm wide by 10 mm thick and has a 12-mm hole drilled through the center as depicted in Table A151. The bar is concentrically loaded in push-pull fatigue by axial forces Fa , uniformly distributed across the width. Using a design factor of n d = 1.8, estimate the largest force Fa that can be applied ignoring column action. 611 Bearing reactions R1 and R2 are exerted on the shaft shown in the gure, which rotates at 1150 rev/min and supports a 10-kip bending force. Use a 1095 HR steel. Specify a diameter d using a design factor of n d = 1.6 for a life of 3 min. The surfaces are machined. 343 344 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition II. Failure Prevention © The McGrawHill Companies, 2008 6. Fatigue Failure Resulting from Variable Loading Fatigue Failure Resulting from Variable Loading 341 F = 10 kip 12 in 6 in 6 in d /5 R. Problem 611 d R1 1.5 d d /10 R. R2 d 1 in 612 A bar of steel has the minimum properties Se = 276 MPa, Sy = 413 MPa, and Sut = 551 MPa. The bar is subjected to a steady torsional stress of 103 MPa and an alternating bending stress of 172 MPa. Find the factor of safety guarding against a static failure, and either the factor of safety guarding against a fatigue failure or the expected life of the part. For the fatigue analysis use: (a) Modied Goodman criterion. (b) Gerber criterion. (c) ASME-elliptic criterion. 613 Repeat Prob. 612 but with a steady torsional stress of 138 MPa and an alternating bending stress of 69 MPa. 614 Repeat Prob. 612 but with a steady torsional stress of 103 MPa, an alternating torsional stress of 69 MPa, and an alternating bending stress of 83 MPa. 615 Repeat Prob. 612 but with an alternating torsional stress of 207 MPa. 616 Repeat Prob. 612 but with an alternating torsional stress of 103 MPa and a steady bending stress of 103 MPa. 617 The cold-drawn AISI 1018 steel bar shown in the gure is subjected to an axial load uctuating between 800 and 3000 lbf. Estimate the factors of safety n y and n f using (a) a Gerber fatigue failure criterion as part of the designers fatigue diagram, and (b) an ASME-elliptic fatigue failure criterion as part of the designers fatigue diagram. 1 4 in D. 1 in Problem 617 3 8 in 618 Repeat Prob. 617, with the load uctuating between 800 and 3000 lbf. Assume no buckling. 619 Repeat Prob. 617, with the load uctuating between 800 and 3000 lbf. Assume no buckling. 620 The gure shows a formed round-wire cantilever spring subjected to a varying force. The hardness tests made on 25 springs gave a minimum hardness of 380 Brinell. It is apparent from the mounting details that there is no stress concentration. A visual inspection of the springs indicates 16 in Problem 620 3 8 in D. Fmax = 30 lbf Fmin = 15 lbf BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 342 II. Failure Prevention © The McGrawHill Companies, 2008 6. Fatigue Failure Resulting from Variable Loading Mechanical Engineering Design that the surface nish corresponds closely to a hot-rolled nish. What number of applications is likely to cause failure? Solve using: (a) Modied Goodman criterion. (b) Gerber criterion. 621 The figure is a drawing of a 3- by 18-mm latching spring. A preload is obtained during assembly by shimming under the bolts to obtain an estimated initial deflection of 2 mm. The latching operation itself requires an additional deflection of exactly 4 mm. The material is ground high-carbon steel, bent then hardened and tempered to a minimum hardness of 490 Bhn. The radius of the bend is 3 mm. Estimate the yield strength to be 90 percent of the ultimate strength. (a) Find the maximum and minimum latching forces. (b) Is it likely the spring will fail in fatigue? Use the Gerber criterion. F 100 A A Problem 621 Dimensions in millimeters 18 3 Section AA 622 Repeat Prob. 621, part b, using the modied Goodman criterion. 623 The gure shows the free-body diagram of a connecting-link portion having stress concentration at three sections. The dimensions are r = 0.25 in, d = 0.75 in, h = 0.50 in, w1 = 3.75 in, and w2 = 2.5 in. The forces F uctuate between a tension of 4 kip and a compression of 16 kip. Neglect column action and nd the least factor of safety if the material is cold-drawn AISI 1018 steel. A Problem 623 F F w1 w2 A 624 h r d Section AA The torsional coupling in the gure is composed of a curved beam of square cross section that is welded to an input shaft and output plate. A torque is applied to the shaft and cycles from zero to T. The cross section of the beam has dimensions of 5 by 5 mm, and the centroidal axis of the beam describes a curve of the form r = 20 + 10 θ/π , where r and θ are in mm and radians, respectively (0 θ 4π ). The curved beam has a machined surface with yield and ultimate strength values of 420 and 770 MPa, respectively. (a) Determine the maximum allowable value of T such that the coupling will have an innite life with a factor of safety, n = 3, using the modied Goodman criterion. (b) Repeat part (a) using the Gerber criterion. (c) Using T found in part (b), determine the factor of safety guarding against yield. 345 346 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition II. Failure Prevention © The McGrawHill Companies, 2008 6. Fatigue Failure Resulting from Variable Loading Fatigue Failure Resulting from Variable Loading 343 T 5 T 20 Problem 624 60 (Dimensions in mm) 625 Repeat Prob. 624 ignoring curvature effects on the bending stress. 626 In the gure shown, shaft A, made of AISI 1010 hot-rolled steel, is welded to a xed support and is subjected to loading by equal and opposite forces F via shaft B. A theoretical stress concentration K t s of 1.6 is induced by the 3-mm llet. The length of shaft A from the xed support to the connection at shaft B is 1 m. The load F cycles from 0.5 to 2 kN. (a) For shaft A, nd the factor of safety for innite life using the modied Goodman fatigue failure criterion. (b) Repeat part (a) using the Gerber fatigue failure criterion. F 20 mm 25 Problem 626 mm m mm m 2510 3 mm fillet Shaft B Shaft A F 627 A schematic of a clutch-testing machine is shown. The steel shaft rotates at a constant speed ω. An axial load is applied to the shaft and is cycled from zero to P. The torque T induced by the clutch face onto the shaft is given by f P(D + d) T= 4 where D and d are dened in the gure and f is the coefcient of friction of the clutch face. The shaft is machined with Sy = 800 MPa and Sut = 1000 MPa. The theoretical stress concentration factors for the llet are 3.0 and 1.8 for the axial and torsional loading, respectively. (a) Assume the load variation P is synchronous with shaft rotation. With f = 0.3, nd the maximum allowable load P such that the shaft will survive a minimum of 106 cycles with a factor of safety of 3. Use the modied Goodman criterion. Determine the corresponding factor of safety guarding against yielding. BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 344 II. Failure Prevention © The McGrawHill Companies, 2008 6. Fatigue Failure Resulting from Variable Loading Mechanical Engineering Design (b) Suppose the shaft is not rotating, but the load P is cycled as shown. With f = 0.3, nd the maximum allowable load P so that the shaft will survive a minimum of 106 cycles with a factor of safety of 3. Use the modied Goodman criterion. Determine the corresponding factor of safety guarding against yielding. R=3 d = 30 mm Problem 627 P Friction pad D = 150 mm 628 For the clutch of Prob. 627, the external load P is cycled between 20 kN and 80 kN. Assuming that the shaft is rotating synchronous with the external load cycle, estimate the number of cycles to failure. Use the modied Goodman fatigue failure criteria. 629 A at leaf spring has uctuating stress of σmax = 420 MPa and σmin = 140 MPa applied for 5 (104) cycles. If the load changes to σmax = 350 MPa and σmin = 200 MPa, how many cycles should the spring survive? The material is AISI 1040 CD and has a fully corrected endurance strength of Se = 200 MPa. Assume that f = 0.9. (a) Use Miners method. (b) Use Mansons method. 630 A machine part will be cycled at ±48 kpsi for 4 (103) cycles. Then the loading will be changed to ±38 kpsi for 6 (104) cycles. Finally, the load will be changed to ±32 kpsi. How many cycles of operation can be expected at this stress level? For the part, Sut = 76 kpsi, f = 0.9, and has a fully corrected endurance strength of Se = 30 kpsi. (a) Use Miners method. (b) Use Mansons method. 631 A rotating-beam specimen with an endurance limit of 50 kpsi and an ultimate strength of 100 kpsi is cycled 20 percent of the time at 70 kpsi, 50 percent at 55 kpsi, and 30 percent at 40 kpsi. Let f = 0.9 and estimate the number of cycles to failure. Stochastic Problems 632 Solve Prob. 61 if the ultimate strength of production pieces is found to be Sut = 245LN (1, 0.0508)kpsi. 633 The situation is similar to that of Prob. 610 wherein the imposed completely reversed axial load Fa = 15LN(1, 0.20) kN is to be carried by the link with a thickness to be specied by you, the designer. Use the 1018 cold-drawn steel of Prob. 610 with Sut = 440LN(1, 0.30) MPa and S yt = 370LN(1, 0.061) . The reliability goal must exceed 0.999. Using the correlation method, specify the thickness t. 634 A solid round steel bar is machined to a diameter of 1.25 in. A groove 1 in deep with a radius of 8 1 in is cut into the bar. The material has a mean tensile strength of 110 kpsi. A completely 8 reversed bending moment M = 1400 lbf · in is applied. Estimate the reliability. The size factor should be based on the gross diameter. The bar rotates. 347 348 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition II. Failure Prevention 6. Fatigue Failure Resulting from Variable Loading © The McGrawHill Companies, 2008 Fatigue Failure Resulting from Variable Loading 345 635 Repeat Prob. 634, with a completely reversed torsional moment of T = 1400 lbf · in applied. 636 1 A 1 1 -in-diameter hot-rolled steel bar has a 8 -in diameter hole drilled transversely through it. The 4 bar is nonrotating and is subject to a completely reversed bending moment of M = 1600 lbf · in in the same plane as the axis of the transverse hole. The material has a mean tensile strength of 58 kpsi. Estimate the reliability. The size factor should be based on the gross size. Use Table A16 for K t . 637 Repeat Prob. 636, with the bar subject to a completely reversed torsional moment of 2400 lbf · in. 638 The plan view of a link is the same as in Prob. 623; however, the forces F are completely reversed, the reliability goal is 0.998, and the material properties are Sut = 64LN(1, 0.045) kpsi and S y = 54LN(1, 0.077) kpsi. Treat Fa as deterministic, and specify the thickness h. Computer Problems 639 1 1 A 4 by 1 2 -in steel bar has a 3 -in drilled hole located in the center, much as is shown in 4 Table A151. The bar is subjected to a completely reversed axial load with a deterministic load ¯ of 1200 lbf. The material has a mean ultimate tensile strength of Sut = 80 kpsi. (a) Estimate the reliability. (b) Conduct a computer simulation to conrm your answer to part a. 640 From your experience with Prob. 639 and Ex. 619, you observed that for completely reversed axial and bending fatigue, it is possible to Observe the COVs associated with a priori design considerations. Note the reliability goal. ¯ Find the mean design factor n d which will permit making a geometric design decision that ¯ will attain the goal using deterministic methods in conjunction with n d . ¯ Formulate an interactive computer program that will enable the user to nd n d . While the material properties Sut , S y , and the load COV must be input by the user, all of the COVs associated with 0 .30 , ka , kc , kd , and K f can be internal, and answers to questions will allow C σ and C S , as well ¯ as Cn and n d , to be calculated. Later you can add improvements. Test your program with problems you have already solved. 641 When using the Gerber fatigue failure criterion in a stochastic problem, Eqs. (680) and (681) are useful. They are also computationally complicated. It is helpful to have a computer subroutine or procedure that performs these calculations. When writing an executive program, and it is appropriate to nd Sa and C Sa , a simple call to the subroutine does this with a minimum of effort. Also, once the subroutine is tested, it is always ready to perform. Write and test such a program. 642 Repeat Problem. 641 for the ASME-elliptic fatigue failure locus, implementing Eqs. (682) and (683). 643 Repeat Prob. 641 for the Smith-Dolan fatigue failure locus, implementing Eqs. (686) and (687). 644 Write and test computer subroutines or procedures that will implement ¯ (a) Table 62, returning a, b, C, and ka . (b) Equation (620) using Table 64, returning kb . ¯ (c) Table 611, returning α , β , C, and kc . ¯ (d) Equations (627) and (675), returning kd and Ckd . 645 Write and test a computer subroutine or procedure that implements Eqs. (676) and (677), ¯ˆ returning q , σq , and Cq . 646 Write and test a computer subroutine or procedure that implements Eq. (678) and Table 615, ¯ returning a , C K f , and K f . BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition PART III. Design of Mechanical Elements Introduction 3 © The McGrawHill Companies, 2008 Design of Mechanical Elements 349 350 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition III. Design of Mechanical Elements © The McGrawHill Companies, 2008 7. Shafts and Shaft Components 7 Shafts and Shaft Components Chapter Outline 71 Introduction 72 Shaft Materials 73 Shaft Layout 74 Shaft Design for Stress 75 Deection Considerations 367 76 Critical Speeds for Shafts 371 77 Miscellaneous Shaft Components 78 Limits and Fits 348 348 349 354 376 383 347 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 348 III. Design of Mechanical Elements 7. Shafts and Shaft Components © The McGrawHill Companies, 2008 Mechanical Engineering Design 71 Introduction A shaft is a rotating member, usually of circular cross section, used to transmit power or motion. It provides the axis of rotation, or oscillation, of elements such as gears, pulleys, ywheels, cranks, sprockets, and the like and controls the geometry of their motion. An axle is a nonrotating member that carries no torque and is used to support rotating wheels, pulleys, and the like. The automotive axle is not a true axle; the term is a carry-over from the horse-and-buggy era, when the wheels rotated on nonrotating members. A non-rotating axle can readily be designed and analyzed as a static beam, and will not warrant the special attention given in this chapter to the rotating shafts which are subject to fatigue loading. There is really nothing unique about a shaft that requires any special treatment beyond the basic methods already developed in previous chapters. However, because of the ubiquity of the shaft in so many machine design applications, there is some advantage in giving the shaft and its design a closer inspection. A complete shaft design has much interdependence on the design of the components. The design of the machine itself will dictate that certain gears, pulleys, bearings, and other elements will have at least been partially analyzed and their size and spacing tentatively determined. Chapter 18 provides a complete case study of a power transmission, focusing on the overall design process. In this chapter, details of the shaft itself will be examined, including the following: Material selection Geometric layout Stress and strength Static strength Fatigue strength Deection and rigidity Bending deection Torsional deection Slope at bearings and shaft-supported elements Shear deection due to transverse loading of short shafts Vibration due to natural frequency In deciding on an approach to shaft sizing, it is necessary to realize that a stress analysis at a specic point on a shaft can be made using only the shaft geometry in the vicinity of that point. Thus the geometry of the entire shaft is not needed. In design it is usually possible to locate the critical areas, size these to meet the strength requirements, and then size the rest of the shaft to meet the requirements of the shaft-supported elements. The deection and slope analyses cannot be made until the geometry of the entire shaft has been dened. Thus deection is a function of the geometry everywhere, whereas the stress at a section of interest is a function of local geometry. For this reason, shaft design allows a consideration of stress rst. Then, after tentative values for the shaft dimensions have been established, the determination of the deections and slopes can be made. 72 Shaft Materials Deection is not affected by strength, but rather by stiffness as represented by the modulus of elasticity, which is essentially constant for all steels. For that reason, rigidity cannot be controlled by material decisions, but only by geometric decisions. 351 352 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition III. Design of Mechanical Elements 7. Shafts and Shaft Components © The McGrawHill Companies, 2008 Shafts and Shaft Components 349 Necessary strength to resist loading stresses affects the choice of materials and their treatments. Many shafts are made from low carbon, cold-drawn or hot-rolled steel, such as ANSI 1020-1050 steels. Signicant strengthening from heat treatment and high alloy content are often not warranted. Fatigue failure is reduced moderately by increase in strength, and then only to a certain level before adverse effects in endurance limit and notch sensitivity begin to counteract the benets of higher strength. A good practice is to start with an inexpensive, low or medium carbon steel for the rst time through the design calculations. If strength considerations turn out to dominate over deection, then a higher strength material should be tried, allowing the shaft sizes to be reduced until excess deection becomes an issue. The cost of the material and its processing must be weighed against the need for smaller shaft diameters. When warranted, typical alloy steels for heat treatment include ANSI 1340-50, 3140-50, 4140, 4340, 5140, and 8650. Shafts usually dont need to be surface hardened unless they serve as the actual journal of a bearing surface. Typical material choices for surface hardening include carburizing grades of ANSI 1020, 4320, 4820, and 8620. Cold drawn steel is usually used for diameters under about 3 inches. The nominal diameter of the bar can be left unmachined in areas that do not require fitting of components. Hot rolled steel should be machined all over. For large shafts requiring much material removal, the residual stresses may tend to cause warping. If concentricity is important, it may be necessary to rough machine, then heat treat to remove residual stresses and increase the strength, then finish machine to the final dimensions. In approaching material selection, the amount to be produced is a salient factor. For low production, turning is the usual primary shaping process. An economic viewpoint may require removing the least material. High production may permit a volumeconservative shaping method (hot or cold forming, casting), and minimum material in the shaft can become a design goal. Cast iron may be specied if the production quantity is high, and the gears are to be integrally cast with the shaft. Properties of the shaft locally depend on its historycold work, cold forming, rolling of llet features, heat treatment, including quenching medium, agitation, and tempering regimen.1 Stainless steel may be appropriate for some environments. 73 Shaft Layout The general layout of a shaft to accommodate shaft elements, e.g. gears, bearings, and pulleys, must be specied early in the design process in order to perform a free body force analysis and to obtain shear-moment diagrams. The geometry of a shaft is generally that of a stepped cylinder. The use of shaft shoulders is an excellent means of axially locating the shaft elements and to carry any thrust loads. Figure 71 shows an example of a stepped shaft supporting the gear of a worm-gear speed reducer. Each shoulder in the shaft serves a specic purpose, which you should attempt to determine by observation. 1 See Joseph E. Shigley, Charles R. Mischke, and Thomas H. Brown, Jr. (eds-in-chief), Standard Handbook of Machine Design, 3rd ed., McGraw-Hill, New York, 2004. For cold-worked property prediction see Chap. 29, and for heat-treated property prediction see Chaps. 29 and 33. BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 350 III. Design of Mechanical Elements © The McGrawHill Companies, 2008 7. Shafts and Shaft Components Mechanical Engineering Design Figure 71 A vertical worm-gear speed reducer. (Courtesy of the Cleveland Gear Company.) Figure 72 (a) Choose a shaft conguration to support and locate the two gears and two bearings. (b) Solution uses an integral pinion, three shaft shoulders, key and keyway, and sleeve. The housing locates the bearings on their outer rings and receives the thrust loads. (c) Choose fanshaft conguration. (d) Solution uses sleeve bearings, a straight-through shaft, locating collars, and setscrews for collars, fan pulley, and fan itself. The fan housing supports the sleeve bearings. (a) (b) Fan (c) (d) The geometric conguration of a shaft to be designed is often simply a revision of existing models in which a limited number of changes must be made. If there is no existing design to use as a starter, then the determination of the shaft layout may have many solutions. This problem is illustrated by the two examples of Fig. 72. In Fig. 72a a geared countershaft is to be supported by two bearings. In Fig. 72c a fanshaft is to be congured. The solutions shown in Fig. 72b and 72d are not necessarily the best ones, but they do illustrate how the shaft-mounted devices are xed and located in the axial direction, and how provision is made for torque transfer from one element to another. There are no absolute rules for specifying the general layout, but the following guidelines may be helpful. 353 354 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition III. Design of Mechanical Elements 7. Shafts and Shaft Components © The McGrawHill Companies, 2008 Shafts and Shaft Components 351 Axial Layout of Components The axial positioning of components is often dictated by the layout of the housing and other meshing components. In general, it is best to support load-carrying components between bearings, such as in Fig. 72a, rather than cantilevered outboard of the bearings, such as in Fig. 72c. Pulleys and sprockets often need to be mounted outboard for ease of installation of the belt or chain. The length of the cantilever should be kept short to minimize the deection. Only two bearings should be used in most cases. For extremely long shafts carrying several load-bearing components, it may be necessary to provide more than two bearing supports. In this case, particular care must be given to the alignment of the bearings. Shafts should be kept short to minimize bending moments and deections. Some axial space between components is desirable to allow for lubricant ow and to provide access space for disassembly of components with a puller. Load bearing components should be placed near the bearings, again to minimize the bending moment at the locations that will likely have stress concentrations, and to minimize the deection at the load-carrying components. The components must be accurately located on the shaft to line up with other mating components, and provision must be made to securely hold the components in position. The primary means of locating the components is to position them against a shoulder of the shaft. A shoulder also provides a solid support to minimize deection and vibration of the component. Sometimes when the magnitudes of the forces are reasonably low, shoulders can be constructed with retaining rings in grooves, sleeves between components, or clamp-on collars. In cases where axial loads are very small, it may be feasible to do without the shoulders entirely, and rely on press ts, pins, or collars with setscrews to maintain an axial location. See Fig. 72b and 72d for examples of some of these means of axial location. Supporting Axial Loads In cases where axial loads are not trivial, it is necessary to provide a means to transfer the axial loads into the shaft, then through a bearing to the ground. This will be particularly necessary with helical or bevel gears, or tapered roller bearings, as each of these produces axial force components. Often, the same means of providing axial location, e.g., shoulders, retaining rings, and pins, will be used to also transmit the axial load into the shaft. It is generally best to have only one bearing carry the axial load, to allow greater tolerances on shaft length dimensions, and to prevent binding if the shaft expands due to temperature changes. This is particularly important for long shafts. Figures 73 and 74 show examples of shafts with only one bearing carrying the axial load against a shoulder, while the other bearing is simply press-fit onto the shaft with no shoulder. Providing for Torque Transmission Most shafts serve to transmit torque from an input gear or pulley, through the shaft, to an output gear or pulley. Of course, the shaft itself must be sized to support the torsional stress and torsional deection. It is also necessary to provide a means of transmitting the torque between the shaft and the gears. Common torque-transfer elements are: Keys Splines Setscrews BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 352 III. Design of Mechanical Elements 7. Shafts and Shaft Components © The McGrawHill Companies, 2008 Mechanical Engineering Design Figure 73 Tapered roller bearings used in a mowing machine spindle. This design represents good practice for the situation in which one or more torquetransfer elements must be mounted outboard. (Source: Redrawn from material furnished by The Timken Company.) Figure 74 A bevel-gear drive in which both pinion and gear are straddle-mounted. (Source: Redrawn from material furnished by Gleason Machine Division.) Pins Press or shrink ts Tapered ts In addition to transmitting the torque, many of these devices are designed to fail if the torque exceeds acceptable operating limits, protecting more expensive components. Details regarding hardware components such as keys, pins, and setscrews are addressed in detail in Sec. 77. One of the most effective and economical means of transmitting moderate to high levels of torque is through a key that fits in a groove in the shaft and gear. Keyed components generally have a slip fit onto the shaft, so assembly and disassembly is easy. The key provides for positive angular orientation of the component, which is useful in cases where phase angle timing is important. 355 356 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition III. Design of Mechanical Elements 7. Shafts and Shaft Components © The McGrawHill Companies, 2008 Shafts and Shaft Components 353 Splines are essentially stubby gear teeth formed on the outside of the shaft and on the inside of the hub of the load-transmitting component. Splines are generally much more expensive to manufacture than keys, and are usually not necessary for simple torque transmission. They are typically used to transfer high torques. One feature of a spline is that it can be made with a reasonably loose slip t to allow for large axial motion between the shaft and component while still transmitting torque. This is useful for connecting two shafts where relative motion between them is common, such as in connecting a power takeoff (PTO) shaft of a tractor to an implement. SAE and ANSI publish standards for splines. Stress concentration factors are greatest where the spline ends and blends into the shaft, but are generally quite moderate. For cases of low torque transmission, various means of transmitting torque are available. These include pins, setscrews in hubs, tapered ts, and press ts. Press and shrink ts for securing hubs to shafts are used both for torque transfer and for preserving axial location. The resulting stress-concentration factor is usually quite small. See Sec. 78 for guidelines regarding appropriate sizing and tolerancing to transmit torque with press and shrink ts. A similar method is to use a split hub with screws to clamp the hub to the shaft. This method allows for disassembly and lateral adjustments. Another similar method uses a two-part hub consisting of a split inner member that ts into a tapered hole. The assembly is then tightened to the shaft with screws, which forces the inner part into the wheel and clamps the whole assembly against the shaft. Tapered ts between the shaft and the shaft-mounted device, such as a wheel, are often used on the overhanging end of a shaft. Screw threads at the shaft end then permit the use of a nut to lock the wheel tightly to the shaft. This approach is useful because it can be disassembled, but it does not provide good axial location of the wheel on the shaft. At the early stages of the shaft layout, the important thing is to select an appropriate means of transmitting torque, and to determine how it affects the overall shaft layout. It is necessary to know where the shaft discontinuities, such as keyways, holes, and splines, will be in order to determine critical locations for analysis. Assembly and Disassembly Consideration should be given to the method of assembling the components onto the shaft, and the shaft assembly into the frame. This generally requires the largest diameter in the center of the shaft, with progressively smaller diameters towards the ends to allow components to be slid on from the ends. If a shoulder is needed on both sides of a component, one of them must be created by such means as a retaining ring or by a sleeve between two components. The gearbox itself will need means to physically position the shaft into its bearings, and the bearings into the frame. This is typically accomplished by providing access through the housing to the bearing at one end of the shaft. See Figs. 75 through 78 for examples. Figure 75 Arrangement showing bearing inner rings press-tted to shaft while outer rings oat in the housing. The axial clearance should be sufcient only to allow for machinery vibrations. Note the labyrinth seal on the right. BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 354 III. Design of Mechanical Elements 7. Shafts and Shaft Components © The McGrawHill Companies, 2008 Mechanical Engineering Design Figure 76 Similar to the arrangement of Fig. 7--5 except that the outer bearing rings are preloaded. Figure 77 In this arrangement the inner ring of the left-hand bearing is locked to the shaft between a nut and a shaft shoulder. The locknut and washer are AFBMA standard. The snap ring in the outer race is used to positively locate the shaft assembly in the axial direction. Note the oating right-hand bearing and the grinding runout grooves in the shaft. Figure 78 This arrangement is similar to Fig. 7--7 in that the left-hand bearing positions the entire shaft assembly. In this case the inner ring is secured to the shaft using a snap ring. Note the use of a shield to prevent dirt generated from within the machine from entering the bearing. 74 When components are to be press-t to the shaft, the shaft should be designed so that it is not necessary to press the component down a long length of shaft. This may require an extra change in diameter, but it will reduce manufacturing and assembly cost by only requiring the close tolerance for a short length. Consideration should also be given to the necessity of disassembling the components from the shaft. This requires consideration of issues such as accessibility of retaining rings, space for pullers to access bearings, openings in the housing to allow pressing the shaft or bearings out, etc. Shaft Design for Stress Critical Locations It is not necessary to evaluate the stresses in a shaft at every point; a few potentially critical locations will sufce. Critical locations will usually be on the outer surface, at axial locations where the bending moment is large, where the torque is present, and where stress concentrations exist. By direct comparison of various points along the shaft, a few critical locations can be identied upon which to base the design. An assessment of typical stress situations will help. 357 358 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition III. Design of Mechanical Elements © The McGrawHill Companies, 2008 7. Shafts and Shaft Components Shafts and Shaft Components 355 Most shafts will transmit torque through a portion of the shaft. Typically the torque comes into the shaft at one gear and leaves the shaft at another gear. A free body diagram of the shaft will allow the torque at any section to be determined. The torque is often relatively constant at steady state operation. The shear stress due to the torsion will be greatest on outer surfaces. The bending moments on a shaft can be determined by shear and bending moment diagrams. Since most shaft problems incorporate gears or pulleys that introduce forces in two planes, the shear and bending moment diagrams will generally be needed in two planes. Resultant moments are obtained by summing moments as vectors at points of interest along the shaft. The phase angle of the moments is not important since the shaft rotates. A steady bending moment will produce a completely reversed moment on a rotating shaft, as a specic stress element will alternate from compression to tension in every revolution of the shaft. The normal stress due to bending moments will be greatest on the outer surfaces. In situations where a bearing is located at the end of the shaft, stresses near the bearing are often not critical since the bending moment is small. Axial stresses on shafts due to the axial components transmitted through helical gears or tapered roller bearings will almost always be negligibly small compared to the bending moment stress. They are often also constant, so they contribute little to fatigue. Consequently, it is usually acceptable to neglect the axial stresses induced by the gears and bearings when bending is present in a shaft. If an axial load is applied to the shaft in some other way, it is not safe to assume it is negligible without checking magnitudes. Shaft Stresses Bending, torsion, and axial stresses may be present in both midrange and alternating components. For analysis, it is simple enough to combine the different types of stresses into alternating and midrange von Mises stresses, as shown in Sec. 614, p. 309. It is sometimes convenient to customize the equations specically for shaft applications. Axial loads are usually comparatively very small at critical locations where bending and torsion dominate, so they will be left out of the following equations. The uctuating stresses due to bending and torsion are given by σa = K f Ma c I σm = K f Mm c I (71) Ta c J τm = K f s Tm c J (72) τa = K f s where Mm and Ma are the midrange and alternating bending moments, Tm and Ta are the midrange and alternating torques, and K f and K f s are the fatigue stress concentration factors for bending and torsion, respectively. Assuming a solid shaft with round cross section, appropriate geometry terms can be introduced for c, I, and J resulting in σa = K f 32 Ma π d3 σm = K f 16Ta π d3 τm = K f s τa = K f s 32 Mm π d3 (73) 16Tm π d3 (74) BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 356 III. Design of Mechanical Elements 359 © The McGrawHill Companies, 2008 7. Shafts and Shaft Components Mechanical Engineering Design Combining these stresses in accordance with the distortion energy failure theory, the von Mises stresses for rotating round, solid shafts, neglecting axial loads, are given by σa σm = 2 (σa = 2 (σm + 2 3τa )1/2 + 2 3τm )1/2 2 = 32 K f Ma π d3 2 = 32 K f Mm π d3 16 K f s Ta +3 π d3 16 K f s Tm +3 π d3 2 1/2 (75) 2 1/2 (76) Note that the stress concentration factors are sometimes considered optional for the midrange components with ductile materials, because of the capacity of the ductile material to yield locally at the discontinuity. These equivalent alternating and midrange stresses can be evaluated using an appropriate failure curve on the modied Goodman diagram (See Sec. 612, p. 295, and Fig. 627). For example, the fatigue failure criteria for the modied Goodman line as expressed previously in Eq. (646) is σ 1 σ = a+ m n Se Sut Substitution of σa and σm from Eqs. (75) and (76) results in 1 16 = n π d3 1 4( K f Ma )2 + 3( K f s Ta )2 Se 1/2 + 1 4( K f Mm )2 + 3( K f s Tm )2 Sut 1/2 For design purposes, it is also desirable to solve the equation for the diameter. This results in d= 16n π 1 1/2 4( K f Ma )2 + 3( K f s Ta )2 Se 1 1/2 4( K f Mm )2 + 3( K f s Tm )2 + Sut 1/3 Similar expressions can be obtained for any of the common failure criteria by substituting the von Mises stresses from Eqs. (75) and (76) into any of the failure criteria expressed by Eqs. (645) through (648), p. 298. The resulting equations for several of the commonly used failure curves are summarized below. The names given to each set of equations identifies the significant failure theory, followed by a fatigue failure locus name. For example, DE-Gerber indicates the stresses are combined using the distortion energy (DE) theory, and the Gerber criteria is used for the fatigue failure. DE-Goodman 16 1 = n π d3 1 4( K f Ma )2 + 3( K f s Ta )2 Se 1/2 + 1 4( K f Mm )2 + 3( K f s Tm )2 Sut 1/2 (77) d= 16n π 1 1/2 4( K f Ma )2 + 3( K f s Ta )2 Se 1 1/2 4( K f Mm )2 + 3( K f s Tm )2 + Sut 1/3 (78) 360 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition III. Design of Mechanical Elements © The McGrawHill Companies, 2008 7. Shafts and Shaft Components Shafts and Shaft Components 357 DE-Gerber 1 8A = n π d 3 Se 8n A d= π Se where 1+ 1+ 2 B Se ASut 2 B Se ASut 1+ 1+ 2 1/2 2 1/2 (79) 1/3 (710) A= 4( K f Ma ) 2 + 3( K f s Ta ) 2 B= 4( K f Mm ) 2 + 3( K f s Tm ) 2 DE-ASME Elliptic 1 16 = 4 n π d3 2 K f Ma Se +3 2 K f s Ta Se +4 2 K f Mm Sy +3 K f s Tm Sy 2 1/2 (711) d= 16n π 4 K f Ma Se 2 +3 K f s Ta Se 2 +4 2 K f Mm Sy +3 K f s Tm Sy 1/3 2 1/2 (712) DE-Soderberg 1 16 = n π d3 1 4( K f Ma )2 + 3( K f s Ta )2 Se 1/2 + 1 4( K f Mm )2 + 3( K f s Tm )2 Syt 1/2 (713) d= 16n π + 1 4( K f Ma )2 + 3( K f s Ta )2 Se 1 4( K f Mm )2 + 3( K f s Tm )2 Syt 1/2 1/2 1/3 (714) For a rotating shaft with constant bending and torsion, the bending stress is completely reversed and the torsion is steady. Equations (77) through (714) can be simplied by setting Mm and Ta equal to 0, which simply drops out some of the terms. Note that in an analysis situation in which the diameter is known and the factor of safety is desired, as an alternative to using the specialized equations above, it is always still valid to calculate the alternating and mid-range stresses using Eqs. (75) and (76), and substitute them into one of the equations for the failure criteria, Eqs. (645) through (648), and solve directly for n. In a design situation, however, having the equations pre-solved for diameter is quite helpful. It is always necessary to consider the possibility of static failure in the rst load cycle. The Soderberg criteria inherently guards against yielding, as can be seen by noting that its failure curve is conservatively within the yield (Langer) line on Fig. 627, p. 297. The ASME Elliptic also takes yielding into account, but is not entirely conservative BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 358 III. Design of Mechanical Elements 361 © The McGrawHill Companies, 2008 7. Shafts and Shaft Components Mechanical Engineering Design throughout its entire range. This is evident by noting that it crosses the yield line in Fig. 627. The Gerber and modied Goodman criteria do not guard against yielding, requiring a separate check for yielding. A von Mises maximum stress is calculated for this purpose. σmax = ( σm + σa ) 2 + 3 ( τm + τa ) 2 = 32 K f ( Mm + Ma ) π d3 2 1/2 16 K f s ( Tm + Ta ) +3 π d3 2 1/2 (715) To check for yielding, this von Mises maximum stress is compared to the yield strength, as usual. ny = Sy σmax (716) For a quick, conservative check, an estimate for σmax can be obtained by simply adding σa and σm . (σa + σm ) will always be greater than or equal to σmax , and will therefore be conservative. EXAMPLE 71 At a machined shaft shoulder the small diameter d is 1.100 in, the large diameter D is 1.65 in, and the llet radius is 0.11 in. The bending moment is 1260 lbf · in and the steady torsion moment is 1100 lbf · in. The heat-treated steel shaft has an ultimate strength of Sut = 105 kpsi and a yield strength of Sy = 82 kpsi. The reliability goal is 0.99. (a) Determine the fatigue factor of safety of the design using each of the fatigue failure criteria described in this section. (b) Determine the yielding factor of safety. Solution (a) D /d = 1.65/1.100 = 1.50, r /d = 0.11/1.100 = 0.10, K t = 1.68 (Fig. A159), K ts = 1.42 (Fig. A158), q = 0.85 (Fig. 620), qshear = 0.92 (Fig. 621). From Eq. (632), K f = 1 + 0.85(1.68 1) = 1.58 K f s = 1 + 0.92(1.42 1) = 1.39 Eq. (68): Se = 0.5(105) = 52.5 kpsi Eq. (619): ka = 2.70(105) 0.265 = 0.787 Eq. (620): kb = 1.100 0.30 0.107 kc = kd = k f = 1 = 0.870 362 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition III. Design of Mechanical Elements © The McGrawHill Companies, 2008 7. Shafts and Shaft Components Shafts and Shaft Components 359 ke = 0.814 Table 66: Se = 0.787(0.870)0.814(52.5) = 29.3 kpsi For a rotating shaft, the constant bending moment will create a completely reversed bending stress. Ma = 1260 lbf · in Tm = 1100 lbf · in Mm = Ta = 0 Applying Eq. (77) for the DE-Goodman criteria gives 16 1 = n π (1.1) 3 4 (1.58 · 1260) 2 29 300 Answer n = 1.62 1/2 + 3 (1.39 · 1100) 2 105 000 1/2 = 0.615 DE-Goodman Similarly, applying Eqs. (79), (711), and (713) for the other failure criteria, Answer n = 1.87 DE-Gerber Answer n = 1.88 DE-ASME Elliptic Answer n = 1.56 DE-Soderberg For comparison, consider an equivalent approach of calculating the stresses and applying the fatigue failure criteria directly. From Eqs. (75) and (76), σa 32 · 1.58 · 1260 π (1.1) 3 = σm = 3 16 · 1.39 · 1100 π (1.1) 3 2 1/2 = 15 235 psi 2 1/2 = 10 134 psi Taking, for example, the Goodman failure critera, application of Eq. (646) gives 1 σ 10 134 σ 15 235 = a+ m= + = 0.616 n Se Sut 29 300 105 000 n = 1.62 which is identical with the previous result. The same process could be used for the other failure criteria. (b) For the yielding factor of safety, determine an equivalent von Mises maximum stress using Eq. (715). σmax Answer = 32(1.58) (1260) π (1.1) 3 ny = 2 16(1.39) (1100) +3 π (1.1) 3 2 1/2 = 18 300 psi Sy 82 000 = 4.48 = σmax 18 300 For comparison, a quick and very conservative check on yielding can be obtained by replacing σmax with σa + σm . This just saves the extra time of calculating σmax if σa and σm have already been determined. For this example, BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 360 III. Design of Mechanical Elements © The McGrawHill Companies, 2008 7. Shafts and Shaft Components Mechanical Engineering Design ny = σa Sy 82 000 = 3.23 = + σm 15 235 + 10 134 which is quite conservative compared with ny 4.48. Estimating Stress Concentrations The stress analysis process for fatigue is highly dependent on stress concentrations. Stress concentrations for shoulders and keyways are dependent on size specications that are not known the rst time through the process. Fortunately, since these elements are usually of standard proportions, it is possible to estimate the stress concentration factors for initial design of the shaft. These stress concentrations will be ne-tuned in successive iterations, once the details are known. Shoulders for bearing and gear support should match the catalog recommendation for the specic bearing or gear. A look through bearing catalogs shows that a typical bearing calls for the ratio of D /d to be between 1.2 and 1.5. For a rst approximation, the worst case of 1.5 can be assumed. Similarly, the llet radius at the shoulder needs to be sized to avoid interference with the llet radius of the mating component. There is a signicant variation in typical bearings in the ratio of llet radius versus bore diameter, with r /d typically ranging from around 0.02 to 0.06. A quick look at the stress concentration charts (Figures A158 and A159) shows that the stress concentrations for bending and torsion increase signicantly in this range. For example, with D /d = 1.5 for bending, K t = 2.7 at r /d = 0.02, and reduces to K t = 2.1 at r /d = 0.05, and further down to K t = 1.7 at r /d = 0.1. This indicates that this is an area where some attention to detail could make a signicant difference. Fortunately, in most cases the shear and bending moment diagrams show that bending moments are quite low near the bearings, since the bending moments from the ground reaction forces are small. In cases where the shoulder at the bearing is found to be critical, the designer should plan to select a bearing with generous llet radius, or consider providing for a larger llet radius on the shaft by relieving it into the base of the shoulder as shown in Fig. 79a. This effectively creates a dead zone in the shoulder area that does not Sharp radius Large radius undercut Stress flow Large-radius relief groove Shoulder relief groove Bearing Shaft (a) (b ) (c) Figure 79 Techniques for reducing stress concentration at a shoulder supporting a bearing with a sharp radius. (a) Large radius undercut into the shoulder. (b) Large radius relief groove into the back of the shoulder. (c) Large radius relief groove into the small diameter 363 364 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition III. Design of Mechanical Elements © The McGrawHill Companies, 2008 7. Shafts and Shaft Components Shafts and Shaft Components 361 carry the bending stresses, as shown by the stress ow lines. A shoulder relief groove as shown in Fig. 79b can accomplish a similar purpose. Another option is to cut a large-radius relief groove into the small diameter of the shaft, as shown in Fig. 79c. This has the disadvantage of reducing the cross-sectional area, but is often used in cases where it is useful to provide a relief groove before the shoulder to prevent the grinding or turning operation from having to go all the way to the shoulder. For the standard shoulder fillet, for estimating K t values for the first iteration, an r /d ratio should be selected so K t values can be obtained. For the worst end of the spectrum, with r /d = 0.02 and D /d = 1.5, K t values from the stress concentration charts for shoulders indicate 2.7 for bending, 2.2 for torsion, and 3.0 for axial. A keyway will produce a stress concentration near a critical point where the loadtransmitting component is located. The stress concentration in an end-milled keyseat is a function of the ratio of the radius r at the bottom of the groove and the shaft diameter d. For early stages of the design process, it is possible to estimate the stress concentration for keyways regardless of the actual shaft dimensions by assuming a typical ratio of r /d = 0.02. This gives K t = 2.2 for bending and K ts = 3.0 for torsion, assuming the key is in place. Figures A1516 and A1517 give values for stress concentrations for atbottomed grooves such as used for retaining rings. By examining typical retaining ring specications in vendor catalogs, it can be seen that the groove width is typically slightly greater than the groove depth, and the radius at the bottom of the groove is around 1/10 of the groove width. From Figs. A1516 and A1517, stress concentration factors for typical retaining ring dimensions are around 5 for bending and axial, and 3 for torsion. Fortunately, the small radius will often lead to a smaller notch sensitivity, reducing K f . Table 71 summarizes some typical stress concentration factors for the rst iteration in the design of a shaft. Similar estimates can be made for other features. The point is to notice that stress concentrations are essentially normalized so that they are dependent on ratios of geometry features, not on the specic dimensions. Consequently, by estimating the appropriate ratios, the rst iteration values for stress concentrations can be obtained. These values can be used for initial design, then actual values inserted once diameters have been determined. Table 71 First Iteration Estimates for Stress Concentration Factors Kt. Warning: These factors are only estimates for use when actual dimensions are not yet determined. Do not use these once actual dimensions are available. Bending Shoulder lletsharp (r/d Shoulder lletwell rounded (r/d End-mill keyseat (r/d 0.02) 0.1) Axial 2.7 0.02) Torsional 2.2 3.0 1.7 1.5 1.9 2.2 3.0 Sled runner keyseat 1.7 Retaining ring groove 5.0 3.0 5.0 Missing values in the table are not readily available. BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition © The McGrawHill Companies, 2008 7. Shafts and Shaft Components Mechanical Engineering Design EXAMPLE 72 This example problem is part of a larger case study. See Chap. 18 for the full context. A double reduction gearbox design has developed to the point that the general layout and axial dimensions of the countershaft carrying two spur gears has been proposed, as shown in Fig. 710. The gears and bearings are located and supported by shoulders, and held in place by retaining rings. The gears transmit torque through keys. Gears have been specified as shown, allowing the tangential and radial forces transmitted through the gears to the shaft to be determined as follows. t W23 = 540 lbf t W54 = 2431 lbf r W23 = 197 lbf r W54 = 885 lbf where the superscripts t and r represent tangential and radial directions, respectively; and, the subscripts 23 and 54 represent the forces exerted by gears 2 and 5 (not shown) on gears 3 and 4, respectively. Proceed with the next phase of the design, in which a suitable material is selected, and appropriate diameters for each section of the shaft are estimated, based on providing sufficient fatigue and static stress capacity for infinite life of the shaft, with minimum safety factors of 1.5. Bearing A Bearing B Gear 3 d3 12 Gear 4 d4 2.67 D5 Figure 710 Shaft layout for Example 72. Dimensions in inches. KL MBN 11.50 D7 11.25 J 10.25 I 9.50 9.75 3.50 H 8.50 G D6 7.50 C A D EF 2.75 1.75 2.0 1.25 0.75 D4 D2 10.75 D3 D1 Datum 0.25 362 III. Design of Mechanical Elements 365 366 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition III. Design of Mechanical Elements © The McGrawHill Companies, 2008 7. Shafts and Shaft Components Shafts and Shaft Components Solution r W23 Perform free body diagram analysis to get reaction forces at the bearings. t W23 y r W54 RBy RAy R Az = 115.0 lbf R Ay = 356.7 lbf t W54 A x R Bz = 1776.0 lbf R By = 725.3 lbf G I RAz J B K RBz z From Mx , find the torque in the shaft between the gears, T 3240 t T = W23 ( d3 /2) = 540 (12/2) = 3240 lbf · in Generate shear-moment diagrams for two planes. V 655 115 1776 x-z Plane 3341 M 3996 2220 230 V 357 160 725 1472 x-y Plane 1632 M 713 907 3651 Combine orthogonal planes as vectors to get total moments, e.g. at J, 39962 + 16322 = 4316 lbf · in. 4316 MTOT 2398 749 Start with Point I, where the bending moment is high, there is a stress concentration at the shoulder, and the torque is present. At I , Ma = 3651 lbf-in, Tm = 3240 lbf-in, Mm = Ta = 0 363 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 364 III. Design of Mechanical Elements © The McGrawHill Companies, 2008 7. Shafts and Shaft Components Mechanical Engineering Design Assume generous fillet radius for gear at I. From Table 71, estimate K t = 1.7, K ts = 1.5 . For quick, conservative first pass, assume K f = K t , K f s = K ts . Choose inexpensive steel, 1020 CD, with Sut = 68 kpsi. For Se , b ka = aSut = 2.7(68) 0.265 = 0.883 Eq. (619) Guess kb = 0.9. Check later when d is known. kc = kd = ke = 1 Se = (0.883)(0.9)(0.5)(68) = 27.0 kpsi. Eq. (618) For first estimate of the small diameter at the shoulder at point I, use the DE-Goodman criterion of Eq. (78). This criterion is good for the initial design, since it is simple and conservative. With Mm = Ta = 0, Eq. (78) reduces to 16n 2 K M 3 K f s Tm f a d= + π Se Sut d= 16(1.5) π 2 1/2 1/3 3 [(1.5) (3240)]2 2 (1.7) (3651) + 27 000 68 000 1/2 1/3 d = 1.65 in. All estimates have probably been conservative, so select the next standard size below 1.65 in. and check, d 1.625 in. A typical D/d ratio for support at a shoulder is D/d 1.2, thus, D 1.2(1.625) 1.95 in. Increase to D 2.0 in. A nominal 2 in. cold-drawn shaft diameter can be used. Check if estimates were acceptable. D /d = 2/1.625 = 1.23 = Assume fillet radius r = d /10 0.16 in. r /d = 0.1 K t = 1.6 (Fig. A159), q = 0.82 (Fig. 620) Eq. (632) K f = 1 + 0.82(1.6 1) = 1.49 K ts = 1.35 (Fig. A158), qs = 0.95 (Fig. 621) K f s = 1 + 0.95(1.35 1) = 1.33 ka = 0.883 (no change) Eq. (620) kb = 1.625 0.3 0.107 = 0.835 Se = (0.883)(0.835)(0.5)(68) = 25.1 kpsi Eq. (75) Eq. (76) 32 K f Ma 32(1.49)(3651) = = 12 910 psi 1 3 πd π (1.625) 3 1/2 16 K f s Tm 2 3(16)(1.33)(3240) σm = 3 = = 8859 psi π d3 π (1.625) 3 σa = Using Goodman criterion σ σ 129 10 8859 1 = a+ m= + = 0.645 nf Se Sut 25 100 68 000 n f = 1.55 367 368 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition III. Design of Mechanical Elements © The McGrawHill Companies, 2008 7. Shafts and Shaft Components Shafts and Shaft Components Note that we could have used Eq. (77) directly. Check yielding. ny = Sy Sy 57 000 = 2.62 > = σmax σa + σm 12 910 + 8859 Also check this diameter at the end of the keyway, just to the right of point I, and at the groove at point K. From moment diagram, estimate M at end of keyway to be M 3750 lbf-in. Assume the radius at the bottom of the keyway will be the standard rd 0.02, r 0.02 d 0.02 (1.625) 0.0325 in. K t = 2.14 (Fig. A1518), q 0.65 (Fig. 620) K f = 1 + 0.65(2.14 1) = 1.74 K ts = 3.0 (Fig. A1519), qs = 0.9 (Fig. 621) K f s = 1 + 0.9(3 1) = 2.8 32 K f Ma 32(1.74)(3750) σa = = = 15 490 psi 3 πd π (1.625) 3 K f s Tm 3(16)(2.8)(3240) σm = 3(16) = = 18 650 psi π d3 π (1.625) 3 σ σ 15 490 18 650 1 + = 0.891 = a+ m= nf Se Sut 25 100 68 000 n f = 1.12 The keyway turns out to be more critical than the shoulder. We can either increase the diameter, or use a higher strength material. Unless the deflection analysis shows a need for larger diameters, let us choose to increase the strength. We started with a very low strength, and can afford to increase it some to avoid larger sizes. Try 1050 CD, with Sut = 100 kpsi. Recalculate factors affected by Sut , i.e. ka Se ; q K f σa ka = 2.7(100) 0.265 = 0.797, Se = 0.797(0.835)(0.5)(100) = 33.3 kpsi q = 0.72, K f = 1 + 0.72(2.14 1) = 1.82 32(1.82)(3750) = 16 200 psi π (1.625) 3 1 18 650 16 200 + = 0.673 = nf 33 300 100 000 σa = n f = 1.49 Since the Goodman criterion is conservative, we will accept this as close enough to the requested 1.5. Check at the groove at K, since K t for flat-bottomed grooves are often very high. From the torque diagram, note that no torque is present at the groove. From the moment diagram, Ma = 2398 lbf in, Mm = Ta = Tm = 0 . To quickly check if this location is potentially critical just use K f = K t = 5.0 as an estimate, from Table 71. σa = 32 K f Ma 32(5)(2398) = = 28 460 psi 3 πd π (1.625) 3 nf = Se 33 300 = = 1.17 σa 28 460 365 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 366 III. Design of Mechanical Elements 7. Shafts and Shaft Components © The McGrawHill Companies, 2008 Mechanical Engineering Design This is low. We will look up data for a specific retaining ring to obtain K f more accurately. With a quick on-line search of a retaining ring specification using the website www.globalspec.com, appropriate groove specifications for a retaining ring for a shaft diameter of 1.625 in are obtained as follows: width, a = 0.068 in; depth, t = 0.048 in; and corner radius at bottom of groove, r = 0.01in. From Fig. A1516, with r / t = 0.01/0.048 = 0.208 , and a / t = 0.068/0.048 = 1.42 K t = 4.3 , q = 0.65 (Fig. 620) K f = 1 + 0.65(4.3 1) = 3.15 32 K f Ma 32(3.15)(2398) = = 17 930 psi π d3 π (1.625) 3 Se 33 300 = 1.86 = nf = σa 17 930 σa = Quickly check if point M might be critical. Only bending is present, and the moment is small, but the diameter is small and the stress concentration is high for a sharp fillet required for a bearing. From the moment diagram, Ma = 959 lbf · in, and Mm = Tm = Ta = 0 . Estimate K t = 2.7 from Table 71, d = 1.0 in, and fillet radius r to fit a typical bearing. r /d = 0.02, r = 0.02(1) = 0.02 q = 0.7 (Fig. 620) K f = 1 + (0.7)(2.7 1) = 2.19 32 K f Ma 32(2.19)(959) = = 21 390 psi σa = π d3 π (1) 3 nf = Se 33 300 = 1.56 = σa 21 390 Should be OK. Close enough to recheck after bearing is selected. With the diameters specified for the critical locations, fill in trial values for the rest of the diameters, taking into account typical shoulder heights for bearing and gear support. D1 = D7 = 1.0 in D2 = D6 = 1.4 in D3 = D5 = 1.625 in D4 = 2.0 in The bending moments are much less on the left end of shaft, so D1 , D2 , and D3 could be smaller. However, unless weight is an issue, there is little advantage to requiring more material removal. Also, the extra rigidity may be needed to keep deflections small. 369 370 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition III. Design of Mechanical Elements © The McGrawHill Companies, 2008 7. Shafts and Shaft Components Shafts and Shaft Components Table 72 367 Slopes Typical Maximum Ranges for Slopes and Transverse Deections 0.00050.0012 rad Tapered roller Cylindrical roller 0.00080.0012 rad Deep-groove ball 0.0010.003 rad Spherical ball 0.0260.052 rad Self-align ball 0.0260.052 rad Uncrowned spur gear 0.0005 rad Transverse deections Spur gears with P < 10 teeth/in 0.010 in Spur gears with 11 < P < 19 75 0.005 in Spur gears with 20 < P < 50 0.003 in Deection Considerations Deection analysis at even a single point of interest requires complete geometry information for the entire shaft. For this reason, it is desirable to design the dimensions at critical locations to handle the stresses, and ll in reasonable estimates for all other dimensions, before performing a deection analysis. Deection of the shaft, both linear and angular, should be checked at gears and bearings. Allowable deections will depend on many factors, and bearing and gear catalogs should be used for guidance on allowable misalignment for specic bearings and gears. As a rough guideline, typical ranges for maximum slopes and transverse deections of the shaft centerline are given in Table 72. The allowable transverse deections for spur gears are dependent on the size of the teeth, as represented by the diametral pitch P number of teeth/pitch diameter. In Sec. 44 several beam deection methods are described. For shafts, where the deections may be sought at a number of different points, integration using either singularity functions or numerical integration is practical. In a stepped shaft, the crosssectional properties change along the shaft at each step, increasing the complexity of integration, since both M and I vary. Fortunately, only the gross geometric dimensions need to be included, as the local factors such as llets, grooves, and keyways do not have much impact on deection. Example 47 demonstrates the use of singularity functions for a stepped shaft. Many shafts will include forces in multiple planes, requiring either a three dimensional analysis, or the use of superposition to obtain deections in two planes which can then be summed as vectors. A deection analysis is straightforward, but it is lengthy and tedious to carry out manually, particularly for multiple points of interest. Consequently, practically all shaft deflection analysis will be evaluated with the assistance of software. Any general-purpose finite-element software can readily handle a shaft problem (see Chap. 19). This is practical if the designer is already familiar with using the software and with how to properly model the shaft. Special-purpose software solutions for 3-D shaft analysis are available, but somewhat expensive if only used occasionally. Software requiring very little training is readily available for planar beam analysis, and can be downloaded from the internet. Example 73 demonstrates how to incorporate such a program for a shaft with forces in multiple planes. BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 368 III. Design of Mechanical Elements © The McGrawHill Companies, 2008 7. Shafts and Shaft Components Mechanical Engineering Design EXAMPLE 73 This example problem is part of a larger case study. See Chap. 18 for the full context. In Example 72 a preliminary shaft geometry was obtained on the basis of design for stress. The resulting shaft is shown in Fig. 710, with proposed diameters of D1 = D7 = 1 in D2 = D6 = 1.4 in D3 = D5 = 1.625 in D4 = 2.0 in Check that the deflections and slopes at the gears and bearings are acceptable. If necessary, propose changes in the geometry to resolve any problems. Solution A simple planar beam analysis program will be used. By modeling the shaft twice, with loads in two orthogonal planes, and combining the results, the shaft deflections can readily be obtained. For both planes, the material is selected (steel with E = 30 Mpsi), the shaft lengths and diameters are entered, and the bearing locations are specified. Local details like grooves and keyways are ignored, as they will have insignificant effect on the deflections. Then the tangential gear forces are entered in the horizontal xz plane model, and the radial gear forces are entered in the vertical xy plane model. The software can calculate the bearing reaction forces, and numerically integrate to generate plots for shear, moment, slope, and deflection, as shown in Fig. 711. xy plane xz plane Beam length: 11.5 in Beam length: 11.5 in in Deflection in Deflection deg Slope deg Slope lbf-in Moment lbf-in Moment lbf Shear lbf Shear Figure 711 Shear, moment, slope, and deection plots from two planes. (Source: Beam 2D Stress Analysis, Orand Systems, Inc.) 371 372 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition III. Design of Mechanical Elements © The McGrawHill Companies, 2008 7. Shafts and Shaft Components Shafts and Shaft Components Point of interest xz plane xy plane Total Left bearing slope 0.02263 deg 0.01770 deg 0.02872 deg Right bearing slope 0.05711 deg 0.02599 deg 0.06274 deg Left gear slope 0.02067 deg 0.01162 deg 0.02371 deg Right gear slope 0.02155 deg 0.01149 deg 0.02442 deg Left gear deection 0.0007568 in 0.0005153 in 0.0009155 in Right gear deection 0.0015870 in 0.0007535 in 369 0.0017567 in 0.000501 rad 0.001095 rad 0.000414 rad 0.000426 rad Table 73 Slope and Deection Values at Key Locations The deflections and slopes at points of interest are obtained from the plots, 2 2 and combined with orthogonal vector addition, that is, δ = δx z + δx y . Results are shown in Table 73. Whether these values are acceptable will depend on the specific bearings and gears selected, as well as the level of performance expected. According to the guidelines in Table 72, all of the bearing slopes are well below typical limits for ball bearings. The right bearing slope is within the typical range for cylindrical bearings. Since the load on the right bearing is relatively high, a cylindrical bearing might be used. This constraint should be checked against the specific bearing specifications once the bearing is selected. The gear slopes and deflections more than satisfy the limits recommended in Table 72. It is recommended to proceed with the design, with an awareness that changes that reduce rigidity should warrant another deflection check. Once deections at various points have been determined, if any value is larger than the allowable deection at that point, a new diameter can be found from n d yold 1/4 dnew = dold (717) yall where yall is the allowable deection at that station and n d is the design factor. Similarly, if any slope is larger than the allowable slope θall , a new diameter can be found from n d ( dy /dx ) old 1/4 dnew = dold (718) (slope) all where (slope)all is the allowable slope. As a result of these calculations, determine the largest dnew /dold ratio, then multiply all diameters by this ratio. The tight constraint will be just tight, and all others will be loose. Dont be too concerned about end journal sizes, as their inuence is usually negligible. The beauty of the method is that the deections need to be completed just once and constraints can be rendered loose but for one, with diameters all identied without reworking every deection. BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 370 III. Design of Mechanical Elements 373 © The McGrawHill Companies, 2008 7. Shafts and Shaft Components Mechanical Engineering Design EXAMPLE 74 Solution For the shaft in Example 73, it was noted that the slope at the right bearing is near the limit for a cylindrical roller bearing. Determine an appropriate increase in diameters to bring this slope down to 0.0005 rad. Applying Eq. (717) to the deection at the right bearing gives dnew = dold n d slopeold slopeall 1/4 = 1.0 ( 1)(0.001095) (0.0005) 1/4 = 1.216 in Multiplying all diameters by the ratio dnew 1.216 = 1.216 = dold 1.0 gives a new set of diameters, D1 = D7 = 1.216 in D2 = D6 = 1.702 in D3 = D5 = 1.976 in D4 = 2.432 in Repeating the beam deection analysis of Example 73 with these new diameters produces a slope at the right bearing of 0.0005 in, with all other deections less than their previous values. The transverse shear V at a section of a beam in exure imposes a shearing deection, which is superposed on the bending deection. Usually such shearing deection is less than 1 percent of the transverse bending deection, and it is seldom evaluated. However, when the shaft length-to-diameter ratio is less than 10, the shear component of transverse deection merits attention. There are many short shafts. A tabular method is explained in detail elsewhere2, including examples. For right-circular cylindrical shafts in torsion the angular deection θ is given in Eq. (45). For a stepped shaft with individual cylinder length li and torque Ti , the angular deection can be estimated from θ= θi = Ti li G i Ji (719) or, for a constant torque throughout homogeneous material, from θ= T G li Ji (720) This should be treated only as an estimate, since experimental evidence shows that the actual θ is larger than given by Eqs. (719) and (720).3 2 C.R. Mischke, Tabular Method for Transverse Shear Deection, Sec. 17.3 in Joseph E. Shigley, Charles R. Mischke, and Thomas H. Brown, Jr. (eds.), Standard Handbook of Machine Design, 3rd ed., McGrawHill, New York, 2004. 3 R. Bruce Hopkins, Design Analysis of Shafts and Beams, McGraw-Hill, New York, 1970, pp. 9399. 374 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition III. Design of Mechanical Elements © The McGrawHill Companies, 2008 7. Shafts and Shaft Components 371 Shafts and Shaft Components If torsional stiffness is defined as ki = Ti /θi and, since θi = Ti / ki and θ = θi = ( Ti / ki ), for constant torque θ = T ( 1/ ki ), it follows that the torsional stiffness of the shaft k in terms of segment stiffnesses is 1 1 = (721) k ki 76 Critical Speeds for Shafts When a shaft is turning, eccentricity causes a centrifugal force deection, which is resisted by the shafts exural rigidity E I . As long as deections are small, no harm is done. Another potential problem, however, is called critical speeds: at certain speeds the shaft is unstable, with deections increasing without upper bound. It is fortunate that although the dynamic deection shape is unknown, using a static deection curve gives an excellent estimate of the lowest critical speed. Such a curve meets the boundary condition of the differential equation (zero moment and deection at both bearings) and the shaft energy is not particularly sensitive to the exact shape of the deection curve. Designers seek rst critical speeds at least twice the operating speed. The shaft, because of its own mass, has a critical speed. The ensemble of attachments to a shaft likewise has a critical speed that is much lower than the shafts intrinsic critical speed. Estimating these critical speeds (and harmonics) is a task of the designer. When geometry is simple, as in a shaft of uniform diameter, simply supported, the task is easy. It can be expressed4 as ω1 = π l 2 EI = m π l 2 gE I Aγ (722) where m is the mass per unit length, A the cross-sectional area, and γ the specic weight. For an ensemble of attachments, Rayleighs method for lumped masses gives5 ω1 = g wi yi wi yi2 (723) where wi is the weight of the ith location and yi is the deection at the ith body location. It is possible to use Eq. (723) for the case of Eq. (722) by partitioning the shaft into segments and placing its weight force at the segment centroid as seen in Fig. 712. Figure 712 y (a) A uniform-diameter shaft for Eq. (722). (b) A segmented uniform-diameter shaft for Eq. (723). x (a) y x (b) 4 William T. Thomson and Marie Dillon Dahleh, Theory of Vibration with Applications, Prentice Hall, 5th ed., 1998, p. 273. 5 Thomson, op. cit., p. 357. BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 372 III. Design of Mechanical Elements 375 © The McGrawHill Companies, 2008 7. Shafts and Shaft Components Mechanical Engineering Design y Figure 713 Unit load aj The inuence coefcient δi j is the deection at i due to a unit load at j. bj xi x l Computer assistance is often used to lessen the difculty in nding transverse deections of a stepped shaft. Rayleighs equation overestimates the critical speed. To counter the increasing complexity of detail, we adopt a useful viewpoint. Inasmuch as the shaft is an elastic body, we can use inuence coefcients. An inuence coefcient is the transverse deection at location i on a shaft due to a unit load at location j on the shaft. From Table A96 we obtain, for a simply supported beam with a single unit load as shown in Fig. 713, b j xi l 2 b2 x 2 j i 6E I l δi j = a (l x ) i j 2 2 6 E I l 2lxi a j xi xi ai (724) xi > ai For three loads the inuence coefcients may be displayed as j i 1 2 3 1 δ11 δ12 δ13 2 δ21 δ22 δ23 3 δ31 δ32 δ33 Maxwells reciprocity theorem6 states that there is a symmetry about the main diagonal, composed of δ11 , δ22 , and δ33 , of the form δi j = δ ji . This relation reduces the work of nding the inuence coefcients. From the inuence coefcients above, one can nd the deections y1 , y2 , and y3 of Eq. (723) as follows: y1 = F1 δ11 + F2 δ12 + F3 δ13 y2 = F1 δ21 + F2 δ22 + F3 δ23 y3 = F1 δ31 + F2 δ32 + F3 δ33 (725) The forces Fi can arise from weight attached wi or centrifugal forces m i ω2 yi . The equation set (725) written with inertial forces can be displayed as y1 = m 1 ω2 y1 δ11 + m 2 ω2 y2 δ12 + m 3 ω2 y3 δ13 y2 = m 1 ω2 y1 δ21 + m 2 ω2 y2 δ22 + m 3 ω2 y3 δ23 y3 = m 1 ω2 y1 δ31 + m 2 ω2 y2 δ32 + m 3 ω2 y3 δ33 6 Thomson, op. cit., p. 167. 376 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition III. Design of Mechanical Elements © The McGrawHill Companies, 2008 7. Shafts and Shaft Components Shafts and Shaft Components 373 which can be rewritten as (m 1 δ11 1/ω2 ) y1 + (m 2 δ12 ) y2 + (m 3 δ13 ) y3 = 0 (m 1 δ21 ) y1 + (m 2 δ22 1/ω2 ) y2 + (m 3 δ23 ) y3 = 0 (a ) (m 1 δ31 ) y1 + (m 2 δ32 ) y2 + (m 3 δ33 1/ω2 ) y3 = 0 Equation set (a) is three simultaneous equations in terms of y1 , y2 , and y3 . To avoid the trivial solution y1 = y2 = y3 = 0, the determinant of the coefcients of y1 , y2 , and y3 must be zero (eigenvalue problem). Thus, m 2 δ12 m 3 δ13 (m 1 δ11 1/2 ) 2 =0 m 1 δ21 (m 2 δ22 1/ω ) m 3 δ23 2 m 1 δ31 m 2 δ32 (m 3 δ33 1/ω ) (726) which says that a deection other than zero exists only at three distinct values of ω, the critical speeds. Expanding the determinant, we obtain 1 ω2 3 (m 1 δ11 + m 2 δ22 + m 3 δ33 ) 1 ω2 2 + ··· = 0 (727) 2 2 2 The three roots of Eq. (727) can be expressed as 1/ω1 , 1/ω2 , and 1/ω3 . Thus Eq. (727) can be written in the form 1 1 2 ω2 ω1 1 1 2 ω2 ω2 1 1 2 ω2 ω3 =0 or 1 ω2 3 1 ω2 1 1 1 + 2+ 2 2 ω1 ω2 ω3 2 + ··· = 0 (728) Comparing Eqs. (727) and (728) we see that 1 1 1 + 2 + 2 = m 1 δ11 + m 2 δ22 + m 3 δ33 2 ω1 ω2 ω3 (729) If we had only a single mass m 1 alone, the critical speed would be given by 1/ω2 = m 1 δ11 . Denote this critical speed as ω11 (which considers only m 1 acting alone). Like2 wise for m 2 or m 3 acting alone, we similarly dene the terms 1/ω22 = m 2 δ22 or 2 1/ω33 = m 3 δ33 , respectively. Thus, Eq. (729) can be rewritten as 1 1 1 1 1 1 + 2+ 2= 2 + 2 + 2 2 ω1 ω2 ω3 ω11 ω22 ω33 (730) 2 2 2 If we order the critical speeds such that ω1 < ω2 < ω3 , then 1/ω1 1/ω2 , and 1/ω3 . So the rst, or fundamental, critical speed ω1 can be approximated by 1 1 1.1 = 2+ 2+ 2 2 ω1 ω11 ω22 ω33 (731) This idea can be extended to an n-body shaft: 1. = 2 ω1 n 1=1 1 2 ωii (732) BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 374 III. Design of Mechanical Elements 377 © The McGrawHill Companies, 2008 7. Shafts and Shaft Components Mechanical Engineering Design This is called Dunkerleys equation. By ignoring the higher mode term(s), the rst critical speed estimate is lower than actually is the case. Since Eq. (732) has no loads appearing in the equation, it follows that if each load could be placed at some convenient location transformed into an equivalent load, then the critical speed of an array of loads could be found by summing the equivalent loads, all placed at a single convenient location. For the load at station 1, placed at the center of span, denoted with the subscript c, the equivalent load is found from 2 ω11 = g g = w1 δ11 w1c δcc or w1c = w1 EXAMPLE 75 Solution δ11 δcc Consider a simply supported steel shaft as depicted in Fig. 714, with 1 in diameter and a 31-in span between bearings, carrying two gears weighing 35 and 55 lbf. (a) Find the inuence coefcients. w y and w y 2 and the rst critical speed using Rayleighs equation, (b) Find Eq. (723). (c) From the inuence coefcients, nd ω11 and ω22 . (d) Using Dunkerleys equation, Eq. (732), estimate the rst critical speed. (e) Use superposition to estimate the rst critical speed. ( f ) Estimate the shafts intrinsic critical speed. Suggest a modication to Dunkerleys equation to include the effect of the shafts mass on the rst critical speed of the attachments. I= (a) π(1)4 π d4 = = 0.049 09 in4 64 64 6 E I l = 6(30)106 (0.049 09)31 = 0.2739(109 ) lbf · in3 Figure 714 (733) y w1 = 35 lbf (a) A 1-in uniform-diameter shaft for Ex. 75. (b) Superposing of equivalent loads at the center of the shaft for the purpose of nding the rst critical speed. 7 in w2 = 55 lbf 13 in 11 in x 31 in (a ) y w1c 17.1 lbf w2c 46.1 lbf 15.5 in 15.5 in x (b ) 378 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition III. Design of Mechanical Elements © The McGrawHill Companies, 2008 7. Shafts and Shaft Components Shafts and Shaft Components 375 From Eq. set (724), δ11 = 24(7)(312 242 72 ) = 2.061(104 ) in/lbf 0.2739(109 ) δ22 = 11(20)(312 112 202 ) = 3.534(104 ) in/lbf 0.2739(109 ) δ12 = δ21 = Answer 11(7)(312 112 72 ) = 2.224(104 ) in/lbf 0.2739(109 ) j i 1 1 2.061(10 4) 2 2 2.224(10 4) 4 3.534(10 4) 2.224(10 ) y1 = w1 δ11 + w2 δ12 = 35(2.061)104 + 55(2.224)104 = 0.019 45 in y2 = w1 δ21 + w2 δ22 = 35(2.224)104 + 55(3.534)104 = 0.027 22 in wi yi = 35(0.019 45) + 55(0.027 22) = 2.178 lbf · in (b) wi yi2 = 35(0.019 45)2 + 55(0.027 22)2 = 0.053 99 lbf · in2 Answer Answer ω= 386.1(2.178) = 124.8 rad/s , or 1192 rev/min 0.053 99 (c) w1 1 δ11 = 2 g ω11 Answer ω11 = 386.1 = 231.4 rad/s, or 2210 rev/min 35(2.061)104 ω22 = Answer g = w1 δ11 g = w2 δ22 386.1 = 140.9 rad/s, or 1346 rev/min 55(3.534)104 1. = 2 ω1 (d ) . ω1 = Answer 1 1 1 = + = 6.905(105 ) 2 2 231.4 140.92 ωii 1 = 120.3 rad/s, or 1149 rev/min 6.905(105 ) which is less than part b, as expected. (e) From Eq. (724), 2 2 bcc xcc l 2 bcc xcc 15.5(15.5)(312 15.52 15.52 ) = 6E I l 0.2739(109 ) 4 = 4.215(10 ) in/lbf δcc = (1) BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 376 III. Design of Mechanical Elements 379 © The McGrawHill Companies, 2008 7. Shafts and Shaft Components Mechanical Engineering Design From Eq. (733), w1c = w1 w2c = w2 Answer ω= g δcc wic 2.061(104 ) δ11 = 17.11 lbf = 35 δcc 4.215(104 ) 3.534(104 ) δ22 = 46.11 lbf = 55 δcc 4.215(104 ) 386.1 = 120.4 rad/s, or 1150 rev/min 4.215(104 )(17.11 + 46.11) = which, except for rounding, agrees with part d, as expected. ( f ) For the shaft, E = 30(106 ) psi, γ = 0.282 lbf/in3, and A = π(12 )/4 = 0.7854 in2. Considering the shaft alone, the critical speed, from Eq. (722), is Answer ωs = π l 2 gE I = Aγ π 31 2 386.1(30)106 (0.049 09) 0.7854(0.282) = 520.4 rad/s, or 4970 rev/min 2 We can simply add 1/ωs to the right side of Dunkerleys equation, Eq. (1), to include the shafts contribution, Answer 1 1. = + 6.905(105 ) = 7.274(105 ) 2 520.42 ω1 . ω1 = 117.3 rad/s, or 1120 rev/min which is slightly less than part d, as expected. The shafts rst critical speed ωs is just one more single effect to add to Dunkerleys equation. Since it does not t into the summation, it is usually written up front. Answer 1.1 = 2+ 2 ωs ω1 n i =1 1 2 ωii (734) Common shafts are complicated by the stepped-cylinder geometry, which makes the inuence-coefcient determination part of a numerical solution. 77 Miscellaneous Shaft Components Setscrews Unlike bolts and cap screws, which depend on tension to develop a clamping force, the setscrew depends on compression to develop the clamping force. The resistance to axial motion of the collar or hub relative to the shaft is called holding power. This holding power, which is really a force resistance, is due to frictional resistance of the contacting portions of the collar and shaft as well as any slight penetration of the setscrew into the shaft. 380 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition III. Design of Mechanical Elements © The McGrawHill Companies, 2008 7. Shafts and Shaft Components Shafts and Shaft Components 377 Figure 715 shows the point types available with socket setscrews. These are also manufactured with screwdriver slots and with square heads. Table 74 lists values of the seating torque and the corresponding holding power for inch-series setscrews. The values listed apply to both axial holding power, for Figure 715 Socket setscrews: (a) at point; (b) cup point; (c) oval point; (d ) cone point; (e) half-dog point. L L L T T D T D (a) D (b) (c) L L T T D (d) Table 74 Typical Holding Power (Force) for Socket Setscrews* Source: Unbrako Division, SPS Technologies, Jenkintown, Pa. Size, in P D Seating Torque, lbf . in (e) Holding Power, lbf #0 1.0 50 #1 1.8 65 #2 1.8 85 #3 5 120 #4 5 160 #5 10 200 #6 10 250 #8 20 385 #10 36 540 1 4 5 16 87 1000 165 1500 3 8 290 2000 7 16 430 2500 1 2 620 3000 9 16 620 3500 5 8 1325 4000 3 4 7 8 2400 5000 5200 6000 1 7200 7000 *Based on alloy-steel screw against steel shaft, class 3A coarse or ne threads in class 2B holes, and cup-point socket setscrews. BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 378 III. Design of Mechanical Elements 381 © The McGrawHill Companies, 2008 7. Shafts and Shaft Components Mechanical Engineering Design resisting thrust, and the tangential holding power, for resisting torsion. Typical factors of safety are 1.5 to 2.0 for static loads and 4 to 8 for various dynamic loads. Setscrews should have a length of about half of the shaft diameter. Note that this practice also provides a rough rule for the radial thickness of a hub or collar. Keys and Pins Keys and pins are used on shafts to secure rotating elements, such as gears, pulleys, or other wheels. Keys are used to enable the transmission of torque from the shaft to the shaft-supported element. Pins are used for axial positioning and for the transfer of torque or thrust or both. Figure 716 shows a variety of keys and pins. Pins are useful when the principal loading is shear and when both torsion and thrust are present. Taper pins are sized according to the diameter at the large end. Some of the most useful sizes of these are listed in Table 75. The diameter at the small end is (735) d = D 0.0208 L where d D L diameter at small end, in diameter at large end, in length, in Figure 716 (a) Square key; (b) round key; (c and d ) round pins; (e) taper pin; (f ) split tubular spring pin. The pins in parts (e) and (f ) are shown longer than necessary, to illustrate the chamfer on the ends, but their lengths should be kept smaller than the hub diameters to prevent injuries due to projections on rotating parts. (a) Dimensions at Large End of Some Standard Taper PinsInch Series (c) (d ) Table 75 (b ) (e) ( f) Commercial Precision Size Maximum Minimum Maximum Minimum 4/0 0.1103 0.1083 0.1100 0.1090 2/0 0.1423 0.1403 0.1420 0.1410 0 0.1573 0.1553 0.1570 0.1560 2 0.1943 0.1923 0.1940 0.1930 4 0.2513 0.2493 0.2510 0.2500 6 0.3423 0.3403 0.3420 0.3410 8 0.4933 0.4913 0.4930 0.4920 382 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition III. Design of Mechanical Elements © The McGrawHill Companies, 2008 7. Shafts and Shaft Components Shafts and Shaft Components Table 76 Shaft Diameter Key Size Over w h Keyway Depth 7 16 3 32 3 32 7 16 9 16 1 8 3 32 1 8 1 8 3 64 3 64 1 16 3 16 1 8 1 16 3 16 Source: Joseph E. Shigley, Unthreaded Fasteners, Chap. 24 in Joseph E. Shigley, Charles R. Mischke, and Thomas H. Brown, Jr. (eds.), Standard Handbook of Machine Design, 3rd ed., McGraw-Hill, New York, 2004. To (Incl.) 5 16 Inch Dimensions for Some Standard Squareand Rectangular-Key Applications 379 3 16 3 32 1 4 1 4 5 16 3 16 3 32 1 4 1 4 5 16 1 8 9 16 7 8 11 4 7 8 1 14 3 18 5 16 5 32 3 8 21 4 23 4 1 24 3 24 1 34 3 16 1 2 3 8 3 16 1 2 5 8 7 16 1 4 7 32 5 8 13 4 1 4 3 8 1 2 13 8 3 14 1 8 5 8 5 16 3 4 3 4 1 2 1 4 3 8 3 8 3 4 1 8 For less important applications, a dowel pin or a drive pin can be used. A large variety of these are listed in manufacturers catalogs.7 The square key, shown in Fig. 716a, is also available in rectangular sizes. Standard sizes of these, together with the range of applicable shaft diameters, are listed in Table 76. The shaft diameter determines standard sizes for width, height, and key depth. The designer chooses an appropriate key length to carry the torsional load. Failure of the key can be by direct shear, or by bearing stress. Example 76 demonstrates the process to size the length of a key. The maximum length of a key is limited by the hub length of the attached element, and should generally not exceed about 1.5 times the shaft diameter to avoid excessive twisting with the angular deection of the shaft. Mulo tiple keys may be used as necessary to carry greater loads, typically oriented at 90 from one another. Excessive safety factors should be avoided in key design, since it is desirable in an overload situation for the key to fail, rather than more costly components. Stock key material is typically made from low carbon cold-rolled steel, and is manufactured such that its dimensions never exceed the nominal dimension. This allows standard cutter sizes to be used for the keyseats. A setscrew is sometimes used along with a key to hold the hub axially, and to minimize rotational backlash when the shaft rotates in both directions. 7 See also Joseph E. Shigley, Unthreaded Fasteners, Chap. 24. In Joseph E. Shigley, Charles R. Mischke, and Thomas H. Brown, Jr. (eds.), Standard Handbook of Machine Design, 3rd ed., McGraw-Hill, New York, 2004. BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 380 III. Design of Mechanical Elements © The McGrawHill Companies, 2008 7. Shafts and Shaft Components Mechanical Engineering Design Figure 717 (a) Gib-head key; (b) Woodruff key. 1 Taper 8 " in 12" w w h (a ) D w (b ) The gib-head key, in Fig. 717a, is tapered so that, when rmly driven, it acts to prevent relative axial motion. This also gives the advantage that the hub position can be adjusted for the best axial location. The head makes removal possible without access to the other end, but the projection may be hazardous. The Woodruff key, shown in Fig. 717b, is of general usefulness, especially when a wheel is to be positioned against a shaft shoulder, since the keyslot need not be machined into the shoulder stress-concentration region. The use of the Woodruff key also yields better concentricity after assembly of the wheel and shaft. This is especially important at high speeds, as, for example, with a turbine wheel and shaft. Woodruff keys are particularly useful in smaller shafts where their deeper penetration helps prevent key rolling. Dimensions for some standard Woodruff key sizes can be found in Table 77, and Table 78 gives the shaft diameters for which the different keyseat widths are suitable. Pilkey8 gives values for stress concentrations in an end-milled keyseat, as a function of the ratio of the radius r at the bottom of the groove and the shaft diameter d. For fillets cut by standard milling-machine cutters, with a ratio of r /d = 0.02, Petersons charts give K t = 2.14 for bending and K ts = 2.62 for torsion without the key in place, or K ts = 3.0 for torsion with the key in place. The stress concentration at the end of the keyseat can be reduced somewhat by using a sled-runner keyseat, eliminating the abrupt end to the keyseat, as shown in Fig. 717. It does, however, still have the sharp radius in the bottom of the groove on the sides. The sled-runner keyseat can only be used when denite longitudinal key positioning is not necessary. It is also not as suitable near a shoulder. Keeping the end of a keyseat at least a distance 8 W. D. Pilkey, Petersons Stress Concentration Factors, 2nd ed., John Wiley & Sons, New York, 1997, pp. 408409. 383 384 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition III. Design of Mechanical Elements © The McGrawHill Companies, 2008 7. Shafts and Shaft Components Shafts and Shaft Components Table 77 Dimensions of Woodruff KeysInch Series Table 78 Sizes of Woodruff Keys Suitable for Various Shaft Diameters Key Size Height b Offset e 1 4 3 8 3 8 1 2 5 8 1 2 5 8 3 4 5 8 3 4 7 8 3 4 7 8 0.109 1 0.438 7 8 0.375 1 64 1 64 1 64 3 64 1 16 3 64 1 16 1 16 1 16 1 16 1 16 1 16 1 16 1 16 1 16 1 16 5 64 1 16 5 64 7 64 5 64 7 64 w D 1 16 1 16 3 32 3 32 3 32 1 8 1 8 1 8 5 32 5 32 5 32 3 16 3 16 3 16 1 4 1 4 1 4 5 16 5 16 5 16 3 8 3 8 0.172 0.172 0.203 0.250 0.203 0.250 0.313 0.250 0.313 0.375 0.313 0.375 1 0.438 11 4 0.547 1 0.438 11 4 0.547 11 2 0.641 11 4 0.547 11 2 0.641 Keyseat Width, in 1 16 3 32 1 8 5 32 3 16 1 4 5 16 3 8 Shaft Diameter, in From To (inclusive) 5 16 3 8 3 8 1 2 9 16 11 16 3 4 1 2 7 8 11 2 15 8 1 25 8 2 21 4 23 8 Keyseat Depth Shaft Hub 0.0728 0.0372 0.1358 0.0372 0.1202 0.0529 0.1511 0.0529 0.1981 0.0529 0.1355 0.0685 0.1825 0.0685 0.2455 0.0685 0.1669 0.0841 0.2299 0.0841 0.2919 0.0841 0.2143 0.0997 0.2763 0.0997 0.3393 0.0997 0.2450 0.1310 0.3080 0.1310 0.4170 0.1310 0.2768 0.1622 0.3858 0.1622 0.4798 0.1622 0.3545 0.1935 0.4485 0.1935 381 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 382 III. Design of Mechanical Elements © The McGrawHill Companies, 2008 7. Shafts and Shaft Components Mechanical Engineering Design Figure 718 Typical uses for retaining rings. (a) External ring and (b) its application; (c) internal ring and (d ) its application. Retaining ring Retaining ring (a ) (b ) (c) (d) of d/10 from the start of the shoulder llet will prevent the two stress concentrations from combining with each other.9 Retaining Rings A retaining ring is frequently used instead of a shaft shoulder or a sleeve to axially position a component on a shaft or in a housing bore. As shown in Fig. 718, a groove is cut in the shaft or bore to receive the spring retainer. For sizes, dimensions, and axial load ratings, the manufacturers catalogs should be consulted. Appendix Tables A1516 and A1517 give values for stress concentration factors for at-bottomed grooves in shafts, suitable for retaining rings. For the rings to seat nicely in the bottom of the groove, and support axial loads against the sides of the groove, the radius in the bottom of the groove must be reasonably sharp, typically about one-tenth of the groove width. This causes comparatively high values for stress concentration factors, around 5 for bending and axial, and 3 for torsion. Care should be taken in using retaining rings, particularly in locations with high bending stresses. 9 Ibid, p. 381. EXAMPLE 76 A UNS G10350 steel shaft, heat-treated to a minimum yield strength of 75 kpsi, has 7 a diameter of 1 16 in. The shaft rotates at 600 rev/min and transmits 40 hp through a gear. Select an appropriate key for the gear. Solution A 3 -in square key is selected, UNS G10200 cold-drawn steel being used. The design 8 will be based on a yield strength of 65 kpsi. A factor of safety of 2.80 will be employed in the absence of exact information about the nature of the load. The torque is obtained from the horsepower equation t a F F b r T= 63 025 H (63 025)(40) = = 4200 lbf · in n 600 From Fig. 719, the force F at the surface of the shaft is F= 4200 T = = 5850 lbf r 1.4375/2 By the distortion-energy theory, the shear strength is Figure 719 Ssy = 0.577 Sy = (0.577)(65) = 37.5 kpsi 385 386 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition III. Design of Mechanical Elements © The McGrawHill Companies, 2008 7. Shafts and Shaft Components Shafts and Shaft Components 383 Failure by shear across the area ab will create a stress of τ = F / tl . Substituting the strength divided by the factor of safety for τ gives Ssy F = n tl or 37.5(10) 3 5850 = 2.80 0.375l or l = 1.16 in. To resist crushing, the area of one-half the face of the key is used: Sy F = n tl /2 or 65(10) 3 5850 = 2.80 0.375l /2 and l = 1.34 in. The hub length of a gear is usually greater than the shaft diameter, for stability. If the key, in this example, is made equal in length to the hub, it would 7 therefore have ample strength, since it would probably be 1 16 in or longer. 78 Limits and Fits The designer is free to adopt any geometry of t for shafts and holes that will ensure the intended function. There is sufcient accumulated experience with commonly recurring situations to make standards useful. There are two standards for limits and ts in the United States, one based on inch units and the other based on metric units.10 These differ in nomenclature, denitions, and organization. No point would be served by separately studying each of the two systems. The metric version is the newer of the two and is well organized, and so here we present only the metric version but include a set of inch conversions to enable the same system to be used with either system of units. In using the standard, capital letters always refer to the hole; lowercase letters are used for the shaft. The denitions illustrated in Fig. 720 are explained as follows: Basic size is the size to which limits or deviations are assigned and is the same for both members of the t. Deviation is the algebraic difference between a size and the corresponding basic size. Upper deviation is the algebraic difference between the maximum limit and the corresponding basic size. Lower deviation is the algebraic difference between the minimum limit and the corresponding basic size. Fundamental deviation is either the upper or the lower deviation, depending on which is closer to the basic size. Tolerance is the difference between the maximum and minimum size limits of a part. International tolerance grade numbers (IT) designate groups of tolerances such that the tolerances for a particular IT number have the same relative level of accuracy but vary depending on the basic size. Hole basis represents a system of ts corresponding to a basic hole size. The fundamental deviation is H. 10 Preferred Limits and Fits for Cylindrical Parts, ANSI B4.1-1967. Preferred Metric Limits and Fits, ANSI B4.2-1978. BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 384 III. Design of Mechanical Elements 7. Shafts and Shaft Components © The McGrawHill Companies, 2008 Mechanical Engineering Design Figure 720 Denitions applied to a cylindrical t. Upper deviation, Max. size, dmax u Lower deviation, l Min. size, dmin International tolerance grade, d (IT number) Fundamental deviation, F (letter) Basic size, D(d) Lower deviation, Upper deviation, u International tolerance grade, D (IT number) l Fundamental deviation, F (letter) Min. size, Dmin Max. size, Dmax Shaft basis represents a system of ts corresponding to a basic shaft size. The fundamental deviation is h. The shaft-basis system is not included here. The magnitude of the tolerance zone is the variation in part size and is the same for both the internal and the external dimensions. The tolerance zones are specied in international tolerance grade numbers, called IT numbers. The smaller grade numbers specify a smaller tolerance zone. These range from IT0 to IT16, but only grades IT6 to IT11 are needed for the preferred ts. These are listed in Tables A11 to A13 for basic sizes up to 16 in or 400 mm. The standard uses tolerance position letters, with capital letters for internal dimensions (holes) and lowercase letters for external dimensions (shafts). As shown in Fig. 720, the fundamental deviation locates the tolerance zone relative to the basic size. Table 79 shows how the letters are combined with the tolerance grades to establish a preferred t. The ISO symbol for the hole for a sliding t with a basic size of 32 mm is 32H7. Inch units are not a part of the standard. However, the designation (1 3 in) H7 includes the same information and is recommended for use here. In both 8 cases, the capital letter H establishes the fundamental deviation and the number 7 denes a tolerance grade of IT7. For the sliding t, the corresponding shaft dimensions are dened by the symbol 32g6 [(1 3 in)g6]. 8 The fundamental deviations for shafts are given in Tables A11 and A13. For letter codes c, d, f, g, and h, Upper deviation = fundamental deviation Lower deviation = upper deviation tolerance grade For letter codes k, n, p, s, and u, the deviations for shafts are Lower deviation = fundamental deviation Upper deviation = lower deviation + tolerance grade 387 388 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition III. Design of Mechanical Elements © The McGrawHill Companies, 2008 7. Shafts and Shaft Components 385 Shafts and Shaft Components Table 79 Type of Fit Description Symbol Descriptions of Preferred Fits Using the Basic Hole System Clearance Loose running t: for wide commercial tolerances or allowances on external members H11/c11 Free running t: not for use where accuracy is essential, but good for large temperature variations, high running speeds, or heavy journal pressures H9/d9 Close running t: for running on accurate machines and for accurate location at moderate speeds and journal pressures H8/f7 Sliding t: where parts are not intended to run freely, but must move and turn freely and locate accurately H7/g6 Locational clearance t: provides snug t for location of stationary parts, but can be freely assembled and disassembled H7/h6 Locational transition t for accurate location, a compromise between clearance and interference H7/k6 Locational transition t for more accurate location where greater interference is permissible H7/n6 Locational interference t: for parts requiring rigidity and alignment with prime accuracy of location but without special bore pressure requirements H7/p6 Medium drive t: for ordinary steel parts or shrink ts on light sections, the tightest t usable with cast iron H7/s6 Source: Preferred Metric Limits and Fits, ANSI B4.2-1978. See also BS 4500. Transition Interference Force t: suitable for parts that can be highly stressed H7/u6 or for shrink ts where the heavy pressing forces required are impractical The lower deviation H (for holes) is zero. For these, the upper deviation equals the tolerance grade. As shown in Fig. 720, we use the following notation: D = basic size of hole d = basic size of shaft δu = upper deviation δl = lower deviation δ F = fundamental deviation D = tolerance grade for hole d = tolerance grade for shaft Note that these quantities are all deterministic. Thus, for the hole, Dmax = D + D (736) Dmin = D For shafts with clearance ts c, d, f, g, and h, dmax = d + δ F dmin = d + δ F d (737) d (738) For shafts with interference ts k, n, p, s, and u, dmin = d + δ F dmax = d + δ F + BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 386 III. Design of Mechanical Elements © The McGrawHill Companies, 2008 7. Shafts and Shaft Components Mechanical Engineering Design EXAMPLE 77 Find the shaft and hole dimensions for a loose running t with a 34-mm basic size. Solution From Table 79, the ISO symbol is 34H11/c11. From Table A11, we nd that tolerance grade IT11 is 0.160 mm. The symbol 34H11/c11 therefore says that D = d = 0.160 mm. Using Eq. (736) for the hole, we get Answer Dmax = D + Answer D = 34 + 0.160 = 34.160 mm Dmin = D = 34.000 mm The shaft is designated as a 34c11 shaft. From Table A12, the fundamental deviation is δ F = 0.120 mm. Using Eq. (737), we get for the shaft dimensions Answer Answer dmax = d + δ F = 34 + (0.120) = 33.880 mm dmin = d + δ F d = 34 + (0.120) 0.160 = 33.720 mm EXAMPLE 78 Find the hole and shaft limits for a medium drive t using a basic hole size of 2 in. Solution The symbol for the t, from Table 78, in inch units is (2 in)H7/s6. For the hole, we use Table A13 and nd the IT7 grade to be D = 0.0010 in. Thus, from Eq. (736), Answer Dmax = D + Answer D = 2 + 0.0010 = 2.0010 in Dmin = D = 2.0000 in The IT6 tolerance for the shaft is d = 0.0006 in. Also, from Table A14, the fundamental deviation is δ F = 0.0017 in. Using Eq. (738), we get for the shaft that Answer Answer dmin = d + δ F = 2 + 0.0017 = 2.0017 in dmax = d + δ F + d = 2 + 0.0017 + 0.0006 = 2.0023 in Stress and Torque Capacity in Interference Fits Interference ts between a shaft and its components can sometimes be used effectively to minimize the need for shoulders and keyways. The stresses due to an interference t can be obtained by treating the shaft as a cylinder with a uniform external pressure, and the hub as a hollow cylinder with a uniform internal pressure. Stress equations for these situations were developed in Sec. 316, and will be converted here from radius terms into diameter terms to match the terminology of this section. 389 390 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition III. Design of Mechanical Elements © The McGrawHill Companies, 2008 7. Shafts and Shaft Components Shafts and Shaft Components 387 The pressure p generated at the interface of the interference t, from Eq. (356) converted into terms of diameters, is given by δ p= (739) 2 d d 2 + di2 d do + d 2 + νo + νi 2 E o do d 2 E i d 2 di2 or, in the case where both members are of the same material, p= 2 E δ (do d 2 )(d 2 di2 ) 2 2d 3 do di2 (740) where d is the nominal shaft diameter, di is the inside diameter (if any) of the shaft, do is the outside diameter of the hub, E is Youngs modulus, and v is Poissons ratio, with subscripts o and i for the outer member (hub) and inner member (shaft), respectively. δ is the diametral interference between the shaft and hub, that is, the difference between the shaft outside diameter and the hub inside diameter. δ = dshaft dhub (741) Since there will be tolerances on both diameters, the maximum and minimum pressures can be found by applying the maximum and minimum interferences. Adopting the notation from Fig. 720, we write δmin = dmin Dmax (742) δmax = dmax Dmin (743) where the diameter terms are dened in Eqs. (736) and (738). The maximum interference should be used in Eq. (739) or (740) to determine the maximum pressure to check for excessive stress. From Eqs. (358) and (359), with radii converted to diameters, the tangential stresses at the interface of the shaft and hub are σt , shaft = p σt , hub = p d 2 + di 2 d 2 di 2 do 2 + d 2 do 2 d 2 (744) (745) The radial stresses at the interface are simply σr, shaft = p (746) σr, hub = p (747) The tangential and radial stresses are orthogonal, and should be combined using a failure theory to compare with the yield strength. If either the shaft or hub yields during assembly, the full pressure will not be achieved, diminishing the torque that can be transmitted. The interaction of the stresses due to the interference t with the other stresses in the shaft due to shaft loading is not trivial. Finite-element analysis of the interface would be appropriate when warranted. A stress element on the surface of a rotating shaft will experience a completely reversed bending stress in the longitudinal direction, as well as the steady compressive stresses in the tangential and radial directions. This is a three-dimensional stress element. Shear stress due to torsion in shaft may also be present. Since the stresses due to the press t are compressive, the fatigue situation is usually actually improved. For this reason, it may be acceptable to simplify BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 388 III. Design of Mechanical Elements 7. Shafts and Shaft Components 391 © The McGrawHill Companies, 2008 Mechanical Engineering Design the shaft analysis by ignoring the steady compressive stresses due to the press t. There is, however, a stress concentration effect in the shaft bending stress near the ends of the hub, due to the sudden change from compressed to uncompressed material. The design of the hub geometry, and therefore its uniformity and rigidity, can have a signicant effect on the specic value of the stress concentration factor, making it difcult to report generalized values. For rst estimates, values are typically not greater than 2. The amount of torque that can be transmitted through an interference t can be estimated with a simple friction analysis at the interface. The friction force is the product of the coefcient of friction f and the normal force acting at the interface. The normal force can be represented by the product of the pressure p and the surface area A of interface. Therefore, the friction force Ff is Ff = f N = f ( p A) = f [ p2π(d /2)l ] = f pπ dl (748) where l is the length of the hub. This friction force is acting with a moment arm of d/2 to provide the torque capacity of the joint, so T = Ff d /2 = f pπ dl (d /2) T = (π/2) f pld 2 (749) The minimum interference, from Eq. (742), should be used to determine the minimum pressure to check for the maximum amount of torque that the joint should be designed to transmit without slipping. PROBLEMS 71 A shaft is loaded in bending and torsion such that Ma = 600 lbf · in, Ta = 400 lbf · in, Mm = 500 lbf · in, and Tm = 300 lbf · in. For the shaft, Su = 100 kpsi and Sy = 80 kpsi, and a fully corrected endurance limit of Se = 30 kpsi is assumed. Let K f = 2.2 and K f s = 1.8. With a design factor of 2.0 determine the minimum acceptable diameter of the shaft using the (a) DE-Gerber criterion. (b) DE-elliptic criterion. (c) DE-Soderberg criterion. (d ) DE-Goodman criterion. Discuss and compare the results. 72 The section of shaft shown in the gure is to be designed to approximate relative sizes of d = 0.75 D and r = D /20 with diameter d conforming to that of standard metric rolling-bearing bore sizes. The shaft is to be made of SAE 2340 steel, heat-treated to obtain minimum strengths in the shoulder area of 1226-MPa ultimate tensile strength and 1130-MPa yield strength with a Brinell hardness not less than 368. At the shoulder the shaft is subjected to a completely reversed bending moment of 70 N · m, accompanied by a steady torsion of 45 N · m. Use a design factor of 2.5 and size the shaft for an innite life. Problem 72 Section of a shaft containing a grinding-relief groove. Unless otherwise specied, the diameter at the root of the groove dr = d 2r, and though the section of diameter d is ground, the root of the groove is still a machined surface. r D d 392 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition III. Design of Mechanical Elements © The McGrawHill Companies, 2008 7. Shafts and Shaft Components Shafts and Shaft Components 73 389 The rotating solid steel shaft is simply supported by bearings at points B and C and is driven by a gear (not shown) which meshes with the spur gear at D, which has a 6-in pitch diameter. o The force F from the drive gear acts at a pressure angle of 20 . The shaft transmits a torque to point A of T A = 3000 lbf · in. The shaft is machined from steel with Sy = 60 kpsi and Sut = 80 kpsi. Using a factor of safety of 2.5, determine the minimum allowable diameter of the 10 in section of the shaft based on (a) a static yield analysis using the distortion energy theory and (b) a fatigue-failure analysis. Assume sharp llet radii at the bearing shoulders for estimating stress concentration factors. TA 10 in A F Problem 73 4 in B 20 C D 74 A geared industrial roll shown in the gure is driven at 300 rev/min by a force F acting on a 3-in-diameter pitch circle as shown. The roll exerts a normal force of 30 lbf/in of roll length on the material being pulled through. The material passes under the roll. The coefcient of friction is 0.40. Develop the moment and shear diagrams for the shaft modeling the roll force as (a) a concentrated force at the center of the roll, and (b) a uniformly distributed force along the roll. These diagrams will appear on two orthogonal planes. y O 4 dia. F Problem 74 Material moves under the roll. Dimensions in inches. A z 20° 3 14 3 B 8 3 14 3 24 2 x Gear 4 3 dia. 75 Design a shaft for the situation of the industrial roll of Prob. 74 with a design factor of 2 and a reliability goal of 0.999 against fatigue failure. Plan for a ball bearing on the left and a cylindrical roller on the right. For deformation use a factor of safety of 2. BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 390 III. Design of Mechanical Elements © The McGrawHill Companies, 2008 7. Shafts and Shaft Components Mechanical Engineering Design 76 The gure shows a proposed design for the industrial roll shaft of Prob. 74. Hydrodynamic lm bearings are to be used. All surfaces are machined except the journals, which are ground and polished. The material is 1035 HR steel. Perform a design assessment. Is the design satisfactory? 11 4 Problem 76 Bearing shoulder llets 0.030 in, 1 others 16 in. Sled-runner keyway is 31 in long. Dimensions in inches. 2 keyway A 1 1 1 10 12 77 1 4 1 O 7 8 4 12 In the double-reduction gear train shown, shaft a is driven by a motor attached by a exible coupling attached to the overhang. The motor provides a torque of 2500 lbf · in at a speed of 1200 rpm. The gears have 20o pressure angles, with diameters shown on the gure. Use an AISI 1020 cold-drawn steel. Design one of the shafts (as specied by the instructor) with a design factor of 1.5 by performing the following tasks. (a) Sketch a general shaft layout, including means to locate the gears and bearings, and to transmit the torque. (b) Perform a force analysis to nd the bearing reaction forces, and generate shear and bending moment diagrams. (c) Determine potential critical locations for stress design. (d) Determine critical diameters of the shaft based on fatigue and static stresses at the critical locations. (e) Make any other dimensional decisions necessary to specify all diameters and axial dimensions. Sketch the shaft to scale, showing all proposed dimensions. (f) Check the deection at the gear, and the slopes at the gear and the bearings for satisfaction of the recommended limits in Table 72. (g) If any of the deections exceed the recommended limits, make appropriate changes to bring them all within the limits. 3 8 24 F E c Problem 77 16 Dimensions in inches. 20 4 D C b 8 A B a 12 78 9 2 6 In the gure is a proposed shaft design to be used for the input shaft a in Prob. 77. A ball bearing is planned for the left bearing, and a cylindrical roller bearing for the right. (a) Determine the minimum fatigue factor of safety by evaluating at any critical locations. Use a fatigue failure criteria that is considered to be typical of the failure data, rather than one that is considered conservative. Also ensure that the shaft does not yield in the rst load cycle. 393 394 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition III. Design of Mechanical Elements © The McGrawHill Companies, 2008 7. Shafts and Shaft Components 391 Shafts and Shaft Components (b) Check the design for adequacy with respect to deformation, according to the recommendations in Table 72. 8 3 74 Problem 78 Shoulder llets at bearing seat 0.030-in radius, others 1 -in radius, 8 except right-hand bearing seat 1 transition, 4 in. The material is 1030 HR. Keyways 3 in wide by 8 3 in deep. Dimensions in inches. 16 0.354 0.453 1.875 1.875 1.500 1.574 1.574 9 6 11 79 The shaft shown in the gure is driven by a gear at the right keyway, drives a fan at the left keyway, and is supported by two deep-groove ball bearings. The shaft is made from AISI 1020 cold-drawn steel. At steady-state speed, the gear transmits a radial load of 230 lbf and a tangential load of 633 lbf at a pitch diameter of 8 in. (a) Determine fatigue factors of safety at any potentially critical locations. (b) Check that deections satisfy the suggested minimums for bearings and gears. 12.87 8.50 1.181 2.0 2.20 0.20 0.75 0.485 1.750 2.75 1.70 1.40 1.181 1.000 Problem 79 Dimensions in inches. 2.0 1 16 1 4 710 × 1 8 R. keyway 1 32 1 8 0.15 R. 3 8 R. × 3 16 0.1 R. 1 8 keyway R. 1 32 R. An AISI 1020 cold-drawn steel shaft with the geometry shown in the gure carries a transverse load of 7 kN and a torque of 107 N · m. Examine the shaft for strength and deection. If the largest allowable slope at the bearings is 0.001 rad and at the gear mesh is 0.0005 rad, what 7 kN 155 40 35 30 55 45 40 35 30 20 Problem 710 Dimensions in millimeters. 30 30 60 55 115 85 10 150 375 All fillets 2 mm BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 392 III. Design of Mechanical Elements © The McGrawHill Companies, 2008 7. Shafts and Shaft Components Mechanical Engineering Design is the factor of safety guarding against damaging distortion? What is the factor of safety guarding against a fatigue failure? If the shaft turns out to be unsatisfactory, what would you recommend to correct the problem? 711 A shaft is to be designed to support the spur pinion and helical gear shown in the gure on two bearings spaced 28 in center-to-center. Bearing A is a cylindrical roller and is to take only radial load; bearing B is to take the thrust load of 220 lbf produced by the helical gear and its share of the radial load. The bearing at B can be a ball bearing. The radial loads of both gears are in the same plane, and are 660 lbf for the pinion and 220 lbf for the gear. The shaft speed is 1150 rev/min. Design the shaft. Make a sketch to scale of the shaft showing all llet sizes, keyways, shoulders, and diameters. Specify the material and its heat treatment. C brg L C brg L 2 4 Problem 711 Dimensions in inches. A B 7 712 16 5 A heat-treated steel shaft is to be designed to support the spur gear and the overhanging worm shown in the gure. A bearing at A takes pure radial load. The bearing at B takes the wormthrust load for either direction of rotation. The dimensions and the loading are shown in the gure; note that the radial loads are in the same plane. Make a complete design of the shaft, including a sketch of the shaft showing all dimensions. Identify the material and its heat treatment (if necessary). Provide an assessment of your nal design. The shaft speed is 310 rev/min. 4 4 A B Problem 712 Dimensions in inches. 4 3 14 950 lbf 600 lbf RB 5600 lbf T = 4800 lbf-in T RA 713 RB A bevel-gear shaft mounted on two 40-mm 02-series ball bearings is driven at 1720 rev/min by a motor connected through a exible coupling. The gure shows the shaft, the gear, and the bearings. The shaft has been giving troublein fact, two of them have already failedand the down time on the machine is so expensive that you have decided to redesign the shaft yourself rather than order replacements. A hardness check of the two shafts in the vicinity of the fracture of the two shafts showed an average of 198 Bhn for one and 204 Bhn of the other. As closely as you can estimate the two shafts failed at a life measure between 600 000 and 1 200 000 cycles 395 396 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition III. Design of Mechanical Elements © The McGrawHill Companies, 2008 7. Shafts and Shaft Components Shafts and Shaft Components 393 of operation. The surfaces of the shaft were machined, but not ground. The llet sizes were not measured, but they correspond with the recommendations for the ball bearings used. You know that the load is a pulsating or shock-type load, but you have no idea of the magnitude, because the shaft drives an indexing mechanism, and the forces are inertial. The keyways are 3 in wide 8 3 by 16 in deep. The straight-toothed bevel pinion drives a 48-tooth bevel gear. Specify a new shaft in sufcient detail to ensure a long and trouble-free life. 2 Shaft failed here 1 3 Problem 713 1 2 dia. 1 8 dia. Dimensions in inches. 4 6 1 2 2 4P, 16T 714 A 1-in-diameter uniform steel shaft is 24 in long between bearings. (a) Find the lowest critical speed of the shaft. (b) If the goal is to double the critical speed, nd the new diameter. (c) A half-size model of the original shaft has what critical speed? 715 Demonstrate how rapidly Rayleighs method converges for the uniform-diameter solid shaft of Prob. 714, by partitioning the shaft into rst one, then two, and nally three elements. 716 Compare Eq. (727) for the angular frequency of a two-disk shaft with Eq. (728), and note that the constants in the two equations are equal. (a) Develop an expression for the second critical speed. (b) Estimate the second critical speed of the shaft addressed in Ex. 75, parts a and b. 717 For a uniform-diameter shaft, does hollowing the shaft increase or decrease the critical speed? 718 The shaft shown in the gure carries a 20-lbf gear on the left and a 35-lbf gear on the right. Estimate the rst critical speed due to the loads, the shafts critical speed without the loads, and the critical speed of the combination. 35 lbf 20 lbf 2.000 2.763 2.472 2.000 Problem 718 Dimensions in inches. 1 2 9 14 15 16 719 A transverse drilled and reamed hole can be used in a solid shaft to hold a pin that locates and holds a mechanical element, such as the hub of a gear, in axial position, and allows for the transmission of torque. Since a small-diameter hole introduces high stress concentration, and a larger diameter hole erodes the area resisting bending and torsion, investigate the existence of a pin diameter with minimum adverse affect on the shaft. Then formulate a design rule. (Hint: Use Table A16.) BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 394 III. Design of Mechanical Elements 7. Shafts and Shaft Components © The McGrawHill Companies, 2008 Mechanical Engineering Design 720 A guide pin is required to align the assembly of a two-part xture. The nominal size of the pin is 15 mm. Make the dimensional decisions for a 15-mm basic size locational clearance t. 721 An interference t of a cast-iron hub of a gear on a steel shaft is required. Make the dimensional decisions for a 45-mm basic size medium drive t. 722 A pin is required for forming a linkage pivot. Find the dimensions required for a 50-mm basic size pin and clevis with a sliding t. 723 A journal bearing and bushing need to be described. The nominal size is 1 in. What dimensions are needed for a 1-in basic size with a close running t if this is a lightly loaded journal and bushing assembly? 724 A gear and shaft with nominal diameter of 1.5 in are to be assembled with a medium drive t, as specied in Table 79. The gear has a hub, with an outside diameter of 2.5 in, and an overall length of 2 in. The shaft is made from AISI 1020 CD steel, and the gear is made from steel that has been through hardened to provide Su 100 kpsi and Sy 85 kpsi. (a) Specify dimensions with tolerances for the shaft and gear bore to achieve the desired t. (b) Determine the minimum and maximum pressures that could be experienced at the interface with the specied tolerances. (c) Determine the worst-case static factors of safety guarding against yielding at assembly for the shaft and the gear based on the distortion energy failure theory. (d ) Determine the maximum torque that the joint should be expected to transmit without slipping, i.e., when the interference pressure is at a minimum for the specied tolerances. 397 398 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition III. Design of Mechanical Elements © The McGrawHill Companies, 2008 8. Screws, Fasteners, and the Design of Nonpermanent Joints 8 Screws, Fasteners, and the Design of Nonpermanent Joints Chapter Outline 81 Thread Standards and Denitions 396 82 The Mechanics of Power Screws 400 83 Threaded Fasteners 84 JointsFastener Stiffness 410 85 JointsMember Stiffness 413 86 Bolt Strength 87 Tension JointsThe External Load 88 Relating Bolt Torque to Bolt Tension 89 Statically Loaded Tension Joint with Preload 408 417 421 422 810 Gasketed Joints 811 Fatigue Loading of Tension Joints 812 Bolted and Riveted Joints Loaded in Shear 425 429 429 435 395 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 396 III. Design of Mechanical Elements 8. Screws, Fasteners, and the Design of Nonpermanent Joints © The McGrawHill Companies, 2008 Mechanical Engineering Design The helical-thread screw was undoubtably an extremely important mechanical invention. It is the basis of power screws, which change angular motion to linear motion to transmit power or to develop large forces (presses, jacks, etc.), and threaded fasteners, an important element in nonpermanent joints. This book presupposes a knowledge of the elementary methods of fastening. Typical methods of fastening or joining parts use such devices as bolts, nuts, cap screws, setscrews, rivets, spring retainers, locking devices, pins, keys, welds, and adhesives. Studies in engineering graphics and in metal processes often include instruction on various joining methods, and the curiosity of any person interested in mechanical engineering naturally results in the acquisition of a good background knowledge of fastening methods. Contrary to rst impressions, the subject is one of the most interesting in the entire eld of mechanical design. One of the key targets of current design for manufacture is to reduce the number of fasteners. However, there will always be a need for fasteners to facilitate disassembly for whatever purposes. For example, jumbo jets such as Boeings 747 require as many as 2.5 million fasteners, some of which cost several dollars apiece. To keep costs down, aircraft manufacturers, and their subcontractors, constantly review new fastener designs, installation techniques, and tooling. The number of innovations in the fastener eld over any period you might care to mention has been tremendous. An overwhelming variety of fasteners are available for the designers selection. Serious designers generally keep specic notebooks on fasteners alone. Methods of joining parts are extremely important in the engineering of a quality design, and it is necessary to have a thorough understanding of the performance of fasteners and joints under all conditions of use and design. 81 Thread Standards and Denitions The terminology of screw threads, illustrated in Fig. 81, is explained as follows: The pitch is the distance between adjacent thread forms measured parallel to the thread axis. The pitch in U.S. units is the reciprocal of the number of thread forms per inch N. The major diameter d is the largest diameter of a screw thread. The minor (or root) diameter dr is the smallest diameter of a screw thread. The pitch diameter d p is a theoretical diameter between the major and minor diameters. The lead l, not shown, is the distance the nut moves parallel to the screw axis when the nut is given one turn. For a single thread, as in Fig. 81, the lead is the same as the pitch. A multiple-threaded product is one having two or more threads cut beside each other (imagine two or more strings wound side by side around a pencil). Standardized products such as screws, bolts, and nuts all have single threads; a double-threaded screw has a lead equal to twice the pitch, a triple-threaded screw has a lead equal to 3 times the pitch, and so on. All threads are made according to the right-hand rule unless otherwise noted. The American National (Unied) thread standard has been approved in this country and in Great Britain for use on all standard threaded products. The thread angle is 60 and the crests of the thread may be either at or rounded. Figure 82 shows the thread geometry of the metric M and MJ proles. The M prole replaces the inch class and is the basic ISO 68 prole with 60 symmetric threads. The MJ prole has a rounded llet at the root of the external thread and a 399 400 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition III. Design of Mechanical Elements © The McGrawHill Companies, 2008 8. Screws, Fasteners, and the Design of Nonpermanent Joints Screws, Fasteners, and the Design of Nonpermanent Joints Figure 81 397 Major diameter Pitch diameter Terminology of screw threads. Sharp vee threads shown for clarity; the crests and roots are actually attened or rounded during the forming operation. Minor diameter Pitch p 45° chamfer Root Thread angle 2α Crest Figure 82 Basic prole for metric M and M J threads. d major diameter dr minor diameter dp pitch diameter p itch p H 3 2 H 8 H p 8 5H 8 p 2 p 4 p Internal threads p 2 3H 8 60° H 4 60° H 4 d 30° dp p External threads dr larger minor diameter of both the internal and external threads. This prole is especially useful where high fatigue strength is required. Tables 81 and 82 will be useful in specifying and designing threaded parts. Note that the thread size is specied by giving the pitch p for metric sizes and by giving the number of threads per inch N for the Unied sizes. The screw sizes in Table 82 with diameter under 1 in are numbered or gauge sizes. The second column 4 in Table 82 shows that a No. 8 screw has a nominal major diameter of 0.1640 in. A great many tensile tests of threaded rods have shown that an unthreaded rod having a diameter equal to the mean of the pitch diameter and minor diameter will have the same tensile strength as the threaded rod. The area of this unthreaded rod is called the tensile-stress area At of the threaded rod; values of At are listed in both tables. Two major Unied thread series are in common use: UN and UNR. The difference between these is simply that a root radius must be used in the UNR series. Because of reduced thread stress-concentration factors, UNR series threads have improved fatigue strengths. Unied threads are specied by stating the nominal major diameter, the number of threads per inch, and the thread series, for example, 5 in-18 8 UNRF or 0.625 in-18 UNRF. Metric threads are specied by writing the diameter and pitch in millimeters, in that order. Thus, M12 × 1.75 is a thread having a nominal major diameter of 12 mm and a pitch of 1.75 mm. Note that the letter M, which precedes the diameter, is the clue to the metric designation. BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 398 III. Design of Mechanical Elements 401 © The McGrawHill Companies, 2008 8. Screws, Fasteners, and the Design of Nonpermanent Joints Mechanical Engineering Design Table 81 Diameters and Areas of Coarse-Pitch and FinePitch Metric Threads.* Nominal Major Diameter d mm Coarse-Pitch Series Pitch p mm TensileStress Area At mm2 MinorDiameter Area Ar mm2 1.6 0.35 1.27 0.40 2.07 0.45 3.39 0.5 5.03 4.47 3.5 0.6 6.78 6.00 4 0.7 8.78 7.75 5 0.8 14.2 12.7 6 1 20.1 MinorDiameter Area Ar mm2 2.98 3 TensileStress Area At mm2 1.79 2.5 Pitch p mm 1.07 2 Fine-Pitch Series 17.9 1.25 36.6 32.8 1 39.2 36.0 10 8 1.5 58.0 52.3 1.25 61.2 56.3 12 1.75 76.3 1.25 14 2 16 2 157 144 1.5 167 157 20 2.5 245 225 1.5 272 259 24 3 353 324 2 384 365 30 3.5 561 519 2 621 596 36 4 42 4.5 48 5 1470 1380 2 1670 1630 56 5.5 2030 1910 2 2300 2250 64 6 2680 2520 2 3030 2980 72 6 3460 3280 2 3860 3800 80 6 4340 4140 1.5 4850 4800 90 6 5590 5360 2 6100 6020 100 6 6990 6740 110 84.3 115 104 1.5 92.1 125 86.0 116 817 759 2 915 884 1120 1050 2 1260 1230 2 7560 7470 2 9180 9080 *The equations and data used to develop this table have been obtained from ANSI B1.1-1974 and B18.3.1-1978. The minor diameter was found from the equation dr d 1.226 869p, and the pitch diameter from dp d 0.649 519p. The mean of the pitch diameter and the minor diameter was used to compute the tensile-stress area. Square and Acme threads, shown in Fig. 83a and b, respectively, are used on screws when power is to be transmitted. Table 83 lists the preferred pitches for inchseries Acme threads. However, other pitches can be and often are used, since the need for a standard for such threads is not great. Modications are frequently made to both Acme and square threads. For instance, the square thread is sometimes modied by cutting the space between the teeth so as to have an included thread angle of 10 to 15 . This is not difcult, since these threads are usually cut with a single-point tool anyhow; the modication retains most of the high efciency inherent in square threads and makes the cutting simpler. Acme threads 402 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition III. Design of Mechanical Elements © The McGrawHill Companies, 2008 8. Screws, Fasteners, and the Design of Nonpermanent Joints Screws, Fasteners, and the Design of Nonpermanent Joints 399 Table 82 Diameters and Area of Unied Screw Threads UNC and UNF* Coarse SeriesUNC Size Designation Nominal Major Diameter in Fine SeriesUNF Threads per Inch N MinorDiameter Area Ar in2 TensileStress Area At in2 MinorDiameter Area Ar in2 80 Threads per Inch N TensileStress Area At in2 0.001 80 0.001 51 0 0.0600 1 0.0730 64 0.002 63 0.002 18 72 0.002 78 0.002 37 2 0.0860 56 0.003 70 0.003 10 64 0.003 94 0.003 39 3 0.0990 48 0.004 87 0.004 06 56 0.005 23 0.004 51 4 0.1120 40 0.006 04 0.004 96 48 0.006 61 0.005 66 5 0.1250 40 0.007 96 0.006 72 44 0.008 80 0.007 16 6 0.1380 32 0.009 09 0.007 45 40 0.010 15 0.008 74 8 0.1640 32 0.014 0 0.011 96 36 0.014 74 0.012 85 10 0.1900 24 0.017 5 0.014 50 32 0.020 0 0.017 5 12 0.2160 24 0.024 2 0.020 6 28 0.025 8 0.022 6 1 4 5 16 0.2500 20 0.031 8 0.026 9 28 0.036 4 0.032 6 0.3125 18 0.052 4 0.045 4 24 0.058 0 0.052 4 0.3750 16 0.077 5 0.067 8 24 0.087 8 0.080 9 0.4375 14 0.106 3 0.093 3 20 0.118 7 0.109 0 0.5000 13 0.141 9 0.125 7 20 0.159 9 0.148 6 3 8 7 16 1 2 9 16 0.5625 12 0.182 0.162 18 0.203 0.189 5 8 3 4 7 8 0.6250 11 0.226 0.202 18 0.256 0.240 0.7500 10 0.334 0.302 16 0.373 0.351 0.8750 9 0.462 0.419 14 0.509 0.480 1 1.0000 8 0.606 0.551 12 0.663 0.625 1 14 1 12 1.2500 7 0.969 0.890 12 1.073 1.024 1.5000 6 1.405 1.294 12 1.581 1.521 *This table was compiled from ANSI B1.1-1974. The minor diameter was found from the equation dr mean of the pitch diameter and the minor diameter was used to compute the tensile-stress area. Figure 83 p 1.299 038p, and the pitch diameter from dp p p 2 (a) Square thread; (b) Acme thread. d 29° p 2 d p 2 d dr dr (a ) (b) p 2 d 0.649 519p. The BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 400 III. Design of Mechanical Elements 403 © The McGrawHill Companies, 2008 8. Screws, Fasteners, and the Design of Nonpermanent Joints Mechanical Engineering Design Table 83 d, in Preferred Pitches for Acme Threads 1 4 5 16 3 8 1 2 5 8 3 4 7 8 1 11 4 11 2 13 4 2 21 2 3 p, in 1 16 1 14 1 12 1 10 1 8 1 6 1 6 1 5 1 5 1 4 1 4 1 4 1 3 1 2 are sometimes modied to a stub form by making the teeth shorter. This results in a larger minor diameter and a somewhat stronger screw. 82 The Mechanics of Power Screws A power screw is a device used in machinery to change angular motion into linear motion, and, usually, to transmit power. Familiar applications include the lead screws of lathes, and the screws for vises, presses, and jacks. An application of power screws to a power-driven jack is shown in Fig. 84. You should be able to identify the worm, the worm gear, the screw, and the nut. Is the worm gear supported by one bearing or two? Figure 84 The Joyce worm-gear screw jack. (Courtesy Joyce-Dayton Corp., Dayton, Ohio.) 404 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition III. Design of Mechanical Elements © The McGrawHill Companies, 2008 8. Screws, Fasteners, and the Design of Nonpermanent Joints Screws, Fasteners, and the Design of Nonpermanent Joints 401 Figure 85 Portion of a power screw. dm F p Nut F 2 F 2 Figure 86 Force diagrams: (a) lifting the load; (b) lowering the load. F F fN PR l fN PL N N dm (a) l dm (b) In Fig. 85 a square-threaded power screw with single thread having a mean diameter dm , a pitch p, a lead angle λ, and a helix angle ψ is loaded by the axial compressive force F. We wish to nd an expression for the torque required to raise this load, and another expression for the torque required to lower the load. First, imagine that a single thread of the screw is unrolled or developed (Fig. 86) for exactly a single turn. Then one edge of the thread will form the hypotenuse of a right triangle whose base is the circumference of the mean-thread-diameter circle and whose height is the lead. The angle λ, in Figs. 85 and 86, is the lead angle of the thread. We represent the summation of all the unit axial forces acting upon the normal thread area by F. To raise the load, a force PR acts to the right (Fig. 86a), and to lower the load, PL acts to the left (Fig. 86b). The friction force is the product of the coefcient of friction f with the normal force N, and acts to oppose the motion. The system is in equilibrium under the action of these forces, and hence, for raising the load, we have FH = PR N sin λ f N cos λ = 0 (a) FV = F + f N sin λ N cos λ = 0 In a similar manner, for lowering the load, we have FH = PL N sin λ + f N cos λ = 0 FV = F f N sin λ N cos λ = 0 (b) BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 402 III. Design of Mechanical Elements 405 © The McGrawHill Companies, 2008 8. Screws, Fasteners, and the Design of Nonpermanent Joints Mechanical Engineering Design Since we are not interested in the normal force N, we eliminate it from each of these sets of equations and solve the result for P. For raising the load, this gives PR = F (sin λ + f cos λ) cos λ f sin λ (c) PL = F ( f cos λ sin λ) cos λ + f sin λ (d ) and for lowering the load, Next, divide the numerator and the denominator of these equations by cos λ and use the relation tan λ = l /π dm (Fig. 86). We then have, respectively, PR = F [(l /π dm ) + f ] 1 ( f l /π dm ) (e) PL = F [ f (l /π dm )] 1 + ( f l /π dm ) (f ) Finally, noting that the torque is the product of the force P and the mean radius dm /2, for raising the load we can write TR = Fdm 2 l + π f dm π dm f l (81) where TR is the torque required for two purposes: to overcome thread friction and to raise the load. The torque required to lower the load, from Eq. ( f ), is found to be TL = Fdm 2 π f dm l π dm + f l (82) This is the torque required to overcome a part of the friction in lowering the load. It may turn out, in specic instances where the lead is large or the friction is low, that the load will lower itself by causing the screw to spin without any external effort. In such cases, the torque TL from Eq. (82) will be negative or zero. When a positive torque is obtained from this equation, the screw is said to be self-locking. Thus the condition for self-locking is π f dm > l Now divide both sides of this inequality by π dm . Recognizing that l /π dm = tan λ, we get f > tan λ (83) This relation states that self-locking is obtained whenever the coefcient of thread friction is equal to or greater than the tangent of the thread lead angle. An expression for efciency is also useful in the evaluation of power screws. If we let f = 0 in Eq. (81), we obtain T0 = Fl 2π (g) 406 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition III. Design of Mechanical Elements © The McGrawHill Companies, 2008 8. Screws, Fasteners, and the Design of Nonpermanent Joints Screws, Fasteners, and the Design of Nonpermanent Joints 403 which, since thread friction has been eliminated, is the torque required only to raise the load. The efciency is therefore e= T0 Fl = TR 2π TR (84) The preceding equations have been developed for square threads where the normal thread loads are parallel to the axis of the screw. In the case of Acme or other threads, the normal thread load is inclined to the axis because of the thread angle 2α and the lead angle λ. Since lead angles are small, this inclination can be neglected and only the effect of the thread angle (Fig. 87a) considered. The effect of the angle α is to increase the frictional force by the wedging action of the threads. Therefore the frictional terms in Eq. (81) must be divided by cos α . For raising the load, or for tightening a screw or bolt, this yields TR = Fdm 2 l + π f dm sec α π dm f l sec α (85) In using Eq. (85), remember that it is an approximation because the effect of the lead angle has been neglected. For power screws, the Acme thread is not as efcient as the square thread, because of the additional friction due to the wedging action, but it is often preferred because it is easier to machine and permits the use of a split nut, which can be adjusted to take up for wear. Usually a third component of torque must be applied in power-screw applications. When the screw is loaded axially, a thrust or collar bearing must be employed between the rotating and stationary members in order to carry the axial component. Figure 87b shows a typical thrust collar in which the load is assumed to be concentrated at the mean collar diameter dc . If f c is the coefcient of collar friction, the torque required is Tc = F f c dc 2 (86) For large collars, the torque should probably be computed in a manner similar to that employed for disk clutches. Figure 87 (a) Normal thread force is increased because of angle α ; (b) thrust collar has frictional diameter dc. dc F 2 F cos F F 2 Collar Nut 2= Thread angle F 2 (a) F 2 (b) BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 404 III. Design of Mechanical Elements 407 © The McGrawHill Companies, 2008 8. Screws, Fasteners, and the Design of Nonpermanent Joints Mechanical Engineering Design Nominal body stresses in power screws can be related to thread parameters as follows. The maximum nominal shear stress τ in torsion of the screw body can be expressed as τ= 16T π dr3 (87) The axial stress σ in the body of the screw due to load F is σ= F 4F = A π dr2 (88) in the absence of column action. For a short column the J. B. Johnson buckling formula is given by Eq. (443), which is F A crit = Sy Sy l 2π k 2 1 CE (89) Nominal thread stresses in power screws can be related to thread parameters as follows. The bearing stress in Fig. 88, σ B , is σB = F 2F = π dm n t p/2 π dm n t p (810) where n t is the number of engaged threads. The bending stress at the root of the thread σb is found from I π (π dr n t ) ( p/2)2 = = dr n t p2 c 6 24 M= Fp 4 so M 6F Fp 24 = = 2 I /c 4 π dr n t p π dr n t p σb = (811) The transverse shear stress τ at the center of the root of the thread due to load F is 3V 3 3F F = = 2A 2 π dr n t p/2 π dr n t p τ= (812) and at the top of the root it is zero. The von Mises stress σ at the top of the root plane is found by rst identifying the orthogonal normal stresses and the shear stresses. From dm Figure 88 Geometry of square thread useful in nding bending and transverse shear stresses at the thread root. F z p2 x p 2 dr 408 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition III. Design of Mechanical Elements © The McGrawHill Companies, 2008 8. Screws, Fasteners, and the Design of Nonpermanent Joints Screws, Fasteners, and the Design of Nonpermanent Joints 405 the coordinate system of Fig. 88, we note σx = 6F π dr n t p τ yz = σy = 0 σz = τx y = 0 4F π dr2 16T π dr3 τzx = 0 then use Eq. (514) of Sec. 55. The screw-thread form is complicated from an analysis viewpoint. Remember the origin of the tensile-stress area At , which comes from experiment. A power screw lifting a load is in compression and its thread pitch is shortened by elastic deformation. Its engaging nut is in tension and its thread pitch is lengthened. The engaged threads cannot share the load equally. Some experiments show that the rst engaged thread carries 0.38 of the load, the second 0.25, the third 0.18, and the seventh is free of load. In estimating thread stresses by the equations above, substituting 0.38 F for F and setting n t to 1 will give the largest level of stresses in the thread-nut combination. EXAMPLE 81 A square-thread power screw has a major diameter of 32 mm and a pitch of 4 mm with double threads, and it is to be used in an application similar to that in Fig. 84. The given data include f = f c = 0.08, dc = 40 mm, and F = 6.4 kN per screw. (a) Find the thread depth, thread width, pitch diameter, minor diameter, and lead. (b) Find the torque required to raise and lower the load. (c) Find the efciency during lifting the load. (d ) Find the body stresses, torsional and compressive. (e) Find the bearing stress. ( f ) Find the thread stresses bending at the root, shear at the root, and von Mises stress and maximum shear stress at the same location. Solution (a) From Fig. 83a the thread depth and width are the same and equal to half the pitch, or 2 mm. Also dm = d p/2 = 32 4/2 = 30 mm Answer dr = d p = 32 4 = 28 mm l = np = 2(4) = 8 mm (b) Using Eqs. (81) and (86), the torque required to turn the screw against the load is TR = = Answer Fdm 2 l + π f dm π dm f l + F f c dc 2 6.4(30) 8 + π(0.08)(30) 6.4(0.08)40 + 2 π(30) 0.08(8) 2 = 15.94 + 10.24 = 26.18 N · m BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 406 III. Design of Mechanical Elements © The McGrawHill Companies, 2008 8. Screws, Fasteners, and the Design of Nonpermanent Joints Mechanical Engineering Design Using Eqs. (82) and (86), we nd the load-lowering torque is TL = = Answer π f dm l π dm + f l Fdm 2 + F f c dc 2 6.4(0.08)(40) 6.4(30) π (0.08)30 8 + 2 π(30) + 0.08(8) 2 = 0.466 + 10.24 = 9.77 N · m The minus sign in the rst term indicates that the screw alone is not self-locking and would rotate under the action of the load except for the fact that the collar friction is present and must be overcome, too. Thus the torque required to rotate the screw with the load is less than is necessary to overcome collar friction alone. (c) The overall efciency in raising the load is Answer e= Fl 6.4(8) = = 0.311 2π TR 2π(26.18) (d ) The body shear stress τ due to torsional moment TR at the outside of the screw body is Answer τ= 16TR 16(26.18)(103 ) = = 6.07 MPa π dr3 π(283 ) The axial nominal normal stress σ is Answer σ = 4(6.4)103 4F = = 10.39 MPa π dr2 π(282 ) (e) The bearing stress σ B is, with one thread carrying 0.38 F , Answer σB = 2(0.38 F ) 2(0.38)(6.4)103 = = 12.9 MPa π dm (1) p π(30)(1)(4) ( f ) The thread-root bending stress σb with one thread carrying 0.38 F is σb = 6(0.38 F ) 6(0.38)(6.4)103 = = 41.5 MPa π dr (1) p π(28)(1)4 The transverse shear at the extreme of the root cross section due to bending is zero. However, there is a circumferential shear stress at the extreme of the root cross section of the thread as shown in part (d ) of 6.07 MPa. The three-dimensional stresses, after Fig. 88, noting the y coordinate is into the page, are σx = 41.5 MPa τx y = 0 σy = 0 τ yz = 6.07 MPa σz = 10.39 MPa τzx = 0 Equation (514) of Sec. 55 can be written as Answer 1 σ = {(41.5 0) 2 + [0 ( 10.39)]2 + ( 10.39 41.5) 2 + 6(6.07) 2 }1/2 2 = 48.7 MPa 409 410 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition III. Design of Mechanical Elements © The McGrawHill Companies, 2008 8. Screws, Fasteners, and the Design of Nonpermanent Joints Screws, Fasteners, and the Design of Nonpermanent Joints 407 Alternatively, you can determine the principal stresses and then use Eq. (512) to nd the von Mises stress. This would prove helpful in evaluating τmax as well. The principal stresses can be found from Eq. (315); however, sketch the stress element and note that there are no shear stresses on the x face. This means that σx is a principal stress. The remaining stresses can be transformed by using the plane stress equation, Eq. (313). Thus, the remaining principal stresses are 10.39 ± 2 10.39 2 2 + 6.072 = 2.79, 13.18 MPa Ordering the principal stresses gives σ1 , σ2 , σ3 = 41.5, 2.79, 13.18 MPa. Substituting these into Eq. (512) yields σ = Answer [41.5 2.79]2 + [2.79 (13.18)]2 + [13.18 41.5]2 2 1/2 = 48.7 MPa The maximum shear stress is given by Eq. (316), where τmax = τ1/3 , giving Answer Table 84 Screw Bearing Pressure pb Source: H. A. Rothbart, Mechanical Design and Systems Handbook, 2nd ed., McGraw-Hill, New York, 1985. τmax = σ1 σ3 41.5 (13.18) = = 27.3 MPa 2 2 Screw Material Nut Material Safe pb, psi Notes Steel Bronze 25003500 Low speed Steel Bronze 16002500 10 fpm Cast iron 18002500 Steel 8 fpm 8001400 2040 fpm Cast iron Steel Bronze 6001000 2040 fpm Bronze 150240 50 fpm Ham and Ryan1 showed that the coefcient of friction in screw threads is independent of axial load, practically independent of speed, decreases with heavier lubricants, shows little variation with combinations of materials, and is best for steel on bronze. Sliding coefcients of friction in power screws are about 0.100.15. Table 84 shows safe bearing pressures on threads, to protect the moving surfaces from abnormal wear. Table 85 shows the coefcients of sliding friction for common material pairs. Table 86 shows coefcients of starting and running friction for common material pairs. 1 Ham and Ryan, An Experimental Investigation of the Friction of Screw-threads, Bulletin 247, University of Illinois Experiment Station, Champaign-Urbana, Ill., June 7, 1932. BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 408 III. Design of Mechanical Elements 411 © The McGrawHill Companies, 2008 8. Screws, Fasteners, and the Design of Nonpermanent Joints Mechanical Engineering Design Table 85 Coefcients of Friction f for Threaded Pairs Source: H. A. Rothbart, Mechanical Design and Systems Handbook, 2nd ed., McGraw-Hill, New York, 1985. Table 86 Nut Material Screw Material Steel Bronze Brass Cast Iron Steel, dry 0.150.25 0.150.23 0.150.19 0.150.25 Steel, machine oil 0.110.17 0.100.16 0.100.15 0.110.17 Bronze 0.080.12 0.040.06 0.060.09 Combination Running Starting Thrust-Collar Friction Coefcients Soft steel on cast iron 0.12 0.17 Hard steel on cast iron 0.09 0.15 Source: H. A. Rothbart, Mechanical Design and Systems Handbook, 2nd ed., McGraw-Hill, New York, 1985. Soft steel on bronze 0.08 0.10 Hard steel on bronze 0.06 0.08 83 Threaded Fasteners As you study the sections on threaded fasteners and their use, be alert to the stochastic and deterministic viewpoints. In most cases the threat is from overproof loading of fasteners, and this is best addressed by statistical methods. The threat from fatigue is lower, and deterministic methods can be adequate. Figure 89 is a drawing of a standard hexagon-head bolt. Points of stress concentration are at the llet, at the start of the threads (runout), and at the thread-root llet in the plane of the nut when it is present. See Table A29 for dimensions. The diameter of the washer face is the same as the width across the ats of the hexagon. The thread length of inch-series bolts, where d is the nominal diameter, is LT = 2d + 2d + 1 4 1 2 in in L 6 in (813) L > 6 in and for metric bolts is LT = 2d + 6 2d + 12 2d + 25 L 125 125 < L 200 d 48 (814) L > 200 where the dimensions are in millimeters. The ideal bolt length is one in which only one or two threads project from the nut after it is tightened. Bolt holes may have burrs or sharp edges after drilling. These could bite into the llet and increase stress concentration. Therefore, washers must always be used under the bolt head to prevent this. They should be of hardened steel and loaded onto the bolt so that the rounded edge of the stamped hole faces the washer face of the bolt. Sometimes it is necessary to use washers under the nut too. The purpose of a bolt is to clamp two or more parts together. The clamping load stretches or elongates the bolt; the load is obtained by twisting the nut until the bolt has elongated almost to the elastic limit. If the nut does not loosen, this bolt tension 412 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition III. Design of Mechanical Elements © The McGrawHill Companies, 2008 8. Screws, Fasteners, and the Design of Nonpermanent Joints Screws, Fasteners, and the Design of Nonpermanent Joints Figure 89 H Approx. Hexagon-head bolt; note the washer face, the llet under the head, the start of threads, and the chamfer on both ends. Bolt lengths are always measured from below the head. 1 64 409 W in R 30° Figure 810 Typical cap-screw heads: (a) llister head; (b) at head; (c) hexagonal socket head. Cap screws are also manufactured with hexagonal heads similar to the one shown in Fig. 89, as well as a variety of other head styles. This illustration uses one of the conventional methods of representing threads. A A A 80 to 82° H H H D D D L L l L l l (a) (b ) (c) remains as the preload or clamping force. When tightening, the mechanic should, if possible, hold the bolt head stationary and twist the nut; in this way the bolt shank will not feel the thread-friction torque. The head of a hexagon-head cap screw is slightly thinner than that of a hexagon-head bolt. Dimensions of hexagon-head cap screws are listed in Table A30. Hexagon-head cap screws are used in the same applications as bolts and also in applications in which one of the clamped members is threaded. Three other common capscrew head styles are shown in Fig. 810. A variety of machine-screw head styles are shown in Fig. 811. Inch-series machine screws are generally available in sizes from No. 0 to about 3 in. 8 Several styles of hexagonal nuts are illustrated in Fig. 812; their dimensions are given in Table A31. The material of the nut must be selected carefully to match that of the bolt. During tightening, the rst thread of the nut tends to take the entire load; but yielding occurs, with some strengthening due to the cold work that takes place, and the load is eventually divided over about three nut threads. For this reason you should never reuse nuts; in fact, it can be dangerous to do so. BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 410 III. Design of Mechanical Elements 413 © The McGrawHill Companies, 2008 8. Screws, Fasteners, and the Design of Nonpermanent Joints Mechanical Engineering Design Types of heads used on machine screws. A A D H 80 to 82° Figure 811 D H L (a) Round head 80 to 82° (b) Flat head A A D H L D H L (c) Fillister head L (d) Oval head 5° ±3° A A D D R H L L (e) Truss head ( f ) Binding head D D W W H L H (g) Hex head (trimmed) Figure 812 W Hexagonal nuts: (a) end view, general; (b) washer-faced regular nut; (c) regular nut chamfered on both sides; (d) jam nut with washer face; (e) jam nut chamfered on both sides. 84 (h) Hex head (upset) H 1 Approx. 64 in 30 (a) L H H 30 ( b) ( c) Approx. 1 64 in H 30 30 ( d) (e) JointsFastener Stiffness When a connection is desired that can be disassembled without destructive methods and that is strong enough to resist external tensile loads, moment loads, and shear loads, or a combination of these, then the simple bolted joint using hardened-steel washers is a good solution. Such a joint can also be dangerous unless it is properly designed and assembled by a trained mechanic. 414 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition III. Design of Mechanical Elements © The McGrawHill Companies, 2008 8. Screws, Fasteners, and the Design of Nonpermanent Joints Screws, Fasteners, and the Design of Nonpermanent Joints Figure 813 P 411 P A bolted connection loaded in tension by the forces P. Note the use of two washers. Note how the threads extend into the body of the connection. This is usual and is desired. l is the grip of the connection. l P P Figure 814 Section of cylindrical pressure vessel. Hexagon-head cap screws are used to fasten the cylinder head to the body. Note the use of an O-ring seal. l is the effective grip of the connection (see Table 87). l' A section through a tension-loaded bolted joint is illustrated in Fig. 813. Notice the clearance space provided by the bolt holes. Notice, too, how the bolt threads extend into the body of the connection. As noted previously, the purpose of the bolt is to clamp the two, or more, parts together. Twisting the nut stretches the bolt to produce the clamping force. This clamping force is called the pretension or bolt preload. It exists in the connection after the nut has been properly tightened no matter whether the external tensile load P is exerted or not. Of course, since the members are being clamped together, the clamping force that produces tension in the bolt induces compression in the members. Figure 814 shows another tension-loaded connection. This joint uses cap screws threaded into one of the members. An alternative approach to this problem (of not using a nut) would be to use studs. A stud is a rod threaded on both ends. The stud is screwed into the lower member rst; then the top member is positioned and fastened down with hardened washers and nuts. The studs are regarded as permanent, and so the joint can be disassembled merely by removing the nut and washer. Thus the threaded part of the lower member is not damaged by reusing the threads. The spring rate is a limit as expressed in Eq. (41). For an elastic member such as a bolt, as we learned in Eq. (42), it is the ratio between the force applied to the member and the deection produced by that force. We can use Eq. (44) and the results of Prob. 41 to nd the stiffness constant of a fastener in any bolted connection. The grip l of a connection is the total thickness of the clamped material. In Fig. 813 the grip is the sum of the thicknesses of both members and both washers. In Fig. 814 the effective grip is given in Table 87. The stiffness of the portion of a bolt or screw within the clamped zone will generally consist of two parts, that of the unthreaded shank portion and that of the BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 412 III. Design of Mechanical Elements 415 © The McGrawHill Companies, 2008 8. Screws, Fasteners, and the Design of Nonpermanent Joints Mechanical Engineering Design Table 87 Suggested Procedure for Finding Fastener Stiffness ld h t1 t lt H t t2 d d lt LT l' l LT ld L L (a) (b) Given fastener diameter d and pitch p or number of threads l = Grip is thickness l Washer thickness from Table A32 or A33 Threaded length LT Inch series: LT = Fastener length: L > l H 2d + 2d + 1 4 1 2 in, in, Effective grip h + t2 /2, h + d /2, t2 < d t2 d L 6 in L > 6 in Metric series: 2d + 6 mm, L 125, d 48 mm LT = 2d + 12 mm, 125 < L 200 mm 2d + 25 mm, L > 200 mm Fastener length: L > h 1.5d Round up using Table A17 Length of useful unthreaded portion: ld L LT Length of threaded portion: lt l ld Length of useful portion: ld L Length of useful portion: lt l unthreaded LT threaded ld Area of unthreaded portion: Ad π d 2 4 Area of threaded portion: At, Table 81 or 82 Fastener stiffness: AdAtE kb = A d l t + A t ld *Bolts and cap screws may not be available in all the preferred lengths listed in Table A17. Large fasteners may not be available in fractional inches or in millimeter lengths ending in a nonzero digit. Check with your bolt supplier for availability. 416 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition III. Design of Mechanical Elements © The McGrawHill Companies, 2008 8. Screws, Fasteners, and the Design of Nonpermanent Joints Screws, Fasteners, and the Design of Nonpermanent Joints 413 threaded portion. Thus the stiffness constant of the bolt is equivalent to the stiffnesses of two springs in series. Using the results of Prob. 41, we nd 1 1 1 + = k k1 k2 k= or k1 k2 k1 + k2 (815) for two springs in series. From Eq. (44), the spring rates of the threaded and unthreaded portions of the bolt in the clamped zone are, respectively, kt = where At E lt kd = Ad E ld (816) At = tensile-stress area (Tables 81, 82) lt = length of threaded portion of grip Ad = major-diameter area of fastener ld = length of unthreaded portion in grip Substituting these stiffnesses in Eq. (815) gives kb = Ad At E Ad lt + At ld (817) where kb is the estimated effective stiffness of the bolt or cap screw in the clamped zone. For short fasteners, the one in Fig. 814, for example, the unthreaded area is small and so the rst of the expressions in Eq. (816) can be used to nd kb . For long fasteners, the threaded area is relatively small, and so the second expression in Eq. (816) can be used. Table 87 is useful. 85 JointsMember Stiffness In the previous section, we determined the stiffness of the fastener in the clamped zone. In this section, we wish to study the stiffnesses of the members in the clamped zone. Both of these stiffnesses must be known in order to learn what happens when the assembled connection is subjected to an external tensile loading. There may be more than two members included in the grip of the fastener. All together these act like compressive springs in series, and hence the total spring rate of the members is 1 1 1 1 1 = + + + ··· + km k1 k2 k3 ki (818) If one of the members is a soft gasket, its stiffness relative to the other members is usually so small that for all practical purposes the others can be neglected and only the gasket stiffness used. If there is no gasket, the stiffness of the members is rather difcult to obtain, except by experimentation, because the compression spreads out between the bolt head and the nut and hence the area is not uniform. There are, however, some cases in which this area can be determined. Ito2 has used ultrasonic techniques to determine the pressure distribution at the member interface. The results show that the pressure stays high out to about 1.5 bolt radii. 2 Y. Ito, J. Toyoda, and S. Nagata, Interface Pressure Distribution in a Bolt-Flange Assembly, ASME paper no. 77-WA/DE-11, 1977. BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 414 III. Design of Mechanical Elements 417 © The McGrawHill Companies, 2008 8. Screws, Fasteners, and the Design of Nonpermanent Joints Mechanical Engineering Design Figure 815 D x Compression of a member with the equivalent elastic properties represented by a frustum of a hollow cone. Here, l represents the grip length. y dw t y x l 2 d t d dx x (a) (b ) The pressure, however, falls off farther away from the bolt. Thus Ito suggests the use of Rotschers pressure-cone method for stiffness calculations with a variable cone angle. This method is quite complicated, and so here we choose to use a simpler approach using a xed cone angle. Figure 815 illustrates the general cone geometry using a half-apex angle α . An angle α = 45 has been used, but Little3 reports that this overestimates the clamping stiffness. When loading is restricted to a washer-face annulus (hardened steel, cast iron, or aluminum), the proper apex angle is smaller. Osgood4 reports a range of 25 α 33 for most combinations. In this book we shall use α = 30 except in cases in which the material is insufcient to allow the frusta to exist. Referring now to Fig. 815b, the contraction of an element of the cone of thickness dx subjected to a compressive force P is, from Eq. (43), dδ = P dx EA (a) The area of the element is 2 A = π ro ri2 = π = π x tan α + x tan α + D+d 2 D 2 2 x tan α + d 2 2 Dd 2 (b) Substituting this in Eq. (a) and integrating gives a total contraction of δ= t P πE 0 dx [x tan α + ( D + d )/2][x tan α + ( D d )/2] (c) Using a table of integrals, we nd the result to be δ= P (2t tan α + D d )( D + d ) ln π Ed tan α (2t tan α + D + d )( D d ) (d ) Thus the spring rate or stiffness of this frustum is k= P = δ π Ed tan α (2t tan α + D d )( D + d ) ln (2t tan α + D + d )( D d ) 3 R. E. Little, Bolted Joints: How Much Give? Machine Design, Nov. 9, 1967. 4 C. C. Osgood, Saving Weight on Bolted Joints, Machine Design, Oct. 25, 1979. (819) 418 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition III. Design of Mechanical Elements © The McGrawHill Companies, 2008 8. Screws, Fasteners, and the Design of Nonpermanent Joints Screws, Fasteners, and the Design of Nonpermanent Joints 415 With α = 30 , this becomes k= 0.5774π Ed (1.155t + D d )( D + d ) ln (1.155t + D + d )( D d ) (820) Equation (820), or (819), must be solved separately for each frustum in the joint. Then individual stiffnesses are assembled to obtain km using Eq. (818). If the members of the joint have the same Youngs modulus E with symmetrical frusta back to back, then they act as two identical springs in series. From Eq. (818) we learn that km = k /2. Using the grip as l = 2t and dw as the diameter of the washer face, we nd the spring rate of the members to be km = π Ed tan α (l tan α + dw d ) (dw + d ) 2 ln (l tan α + dw + d ) (dw d ) (821) The diameter of the washer face is about 50 percent greater than the fastener diameter for standard hexagon-head bolts and cap screws. Thus we can simplify Eq. (821) by letting dw = 1.5d . If we also use α = 30 , then Eq. (821) can be written as km = 0.5774π Ed 0.5774l + 0.5d 2 ln 5 0.5774l + 2.5d (822) It is easy to program the numbered equations in this section, and you should do so. The time spent in programming will save many hours of formula plugging. To see how good Eq. (821) is, solve it for km / Ed : km = Ed π tan α (l tan α + dw d ) (dw + d ) 2 ln (l tan α + dw + d ) (dw d ) Earlier in the section use of α = 30 was recommended for hardened steel, cast iron, or aluminum members. Wileman, Choudury, and Green5 conducted a nite element study of this problem. The results, which are depicted in Fig. 816, agree with the α = 30 recommendation, coinciding exactly at the aspect ratio d / l = 0.4. Additionally, they offered an exponential curve-t of the form km = A exp( Bd / l ) Ed (823) with constants A and B dened in Table 88. For standard washer faces and members of the same material, Eq. (823) offers a simple calculation for member stiffness km . For departure from these conditions, Eq. (820) remains the basis for approaching the problem. 5 J.Wileman, M. Choudury, and I. Green, Computation of Member Stiffness in Bolted Connections, Trans. ASME, J. Mech. Design, vol. 113, December 1991, pp. 432437. BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 416 III. Design of Mechanical Elements 419 © The McGrawHill Companies, 2008 8. Screws, Fasteners, and the Design of Nonpermanent Joints Mechanical Engineering Design Figure 816 3.4 3.2 3.0 2.8 2.6 Dimensionless stiffness, k m Ed The dimensionless plot of stiffness versus aspect ratio of the members of a bolted joint, showing the relative accuracy of methods of Rotscher, Mischke, and Motosh, compared to a nite-element analysis (FEA) conducted by Wileman, Choudury, and Green. 2.4 2.2 2.0 1.8 1.6 1.4 1.2 1.0 0.8 0.6 0.4 0.1 0.3 0.5 0.7 0.9 1.1 1.3 1.5 1.7 1.9 Aspect ratio, d l FEA Table 88 Stiffness Parameters of Various Member Materials Source: J. Wileman, M. Choudury, and I. Green, Computation of Member Stiffness in Bolted Connections, Trans. ASME, J. Mech. Design, vol. 113, December 1991, pp. 432437. Rotscher Material Used Mischke 45° Mischke 30° Modulus Mpsi Motosh Poisson Ratio Elastic GPa Steel 0.291 207 30.0 0.787 15 0.628 73 Aluminum 0.334 71 10.3 0.796 70 0.638 16 A B Copper 0.326 119 17.3 0.795 68 0.635 53 Gray cast iron 0.211 100 14.5 0.778 71 0.616 16 0.789 52 0.629 14 General expression EXAMPLE 82 Two 1 -in-thick steel plates with a modulus of elasticity of 30(106 ) psi are clamped 2 by washer-faced 1 -in-diameter UNC SAE grade 5 bolts with a 0.095-in-thick washer 2 under the nut. Find the member spring rate km using the method of conical frusta, and compare the result with the nite element analysis (FEA) curve-t method of Wileman et al. Solution The grip is 0.5 + 0.5 + 0.095 = 1.095 in. Using Eq. (822) with l = 1.095 and d = 0.5 in, we write km = 0.5774π 30(106 )0.5 0.5774(1.095) + 0.5(0.5) 2 ln 5 0.5774(1.095) + 2.5(0.5) = 15.97(106 ) lbf/in BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition III. Design of Mechanical Elements © The McGrawHill Companies, 2008 8. Screws, Fasteners, and the Design of Nonpermanent Joints Screws, Fasteners, and the Design of Nonpermanent Joints 417 From Table 88, A = 0.787 15, B = 0.628 73. Equation (823) gives km = 30(106 )(0.5)(0.787 15) exp[0.628 73(0.5)/1.095] = 15.73(106 ) lbf/in For this case, the difference between the results for Eqs. (822) and (823) is less than 2 percent. 86 Bolt Strength In the specication standards for bolts, the strength is specied by stating ASTM minimum quantities, the minimum proof strength, or minimum proof load, and the minimum tensile strength. The proof load is the maximum load (force) that a bolt can withstand without acquiring a permanent set. The proof strength is the quotient of the proof load and the tensile-stress area. The proof strength thus corresponds roughly to the proportional limit and corresponds to 0.0001 in permanent set in the fastener (rst measurable deviation from elastic behavior). The value of the mean proof strength, the mean tensile strength, and the corresponding standard deviations are not part of the specication codes, so it is the designers responsibility to obtain these values, perhaps by laboratory testing, before designing to a reliability specication. Figure 817 shows the distribution of ultimate tensile strength from a bolt production run. If the ASTM minimum strength equals or exceeds 120 kpsi, the bolts can be offered as SAE grade 5. The designer does not see this histogram. Instead, in Table 89, the designer sees the entry Sut = 120 kpsi under the 1 1-in size in grade 5 bolts. Similarly, minimum 4 strengths are shown in Tables 810 and 811. The SAE specications are found in Table 89. The bolt grades are numbered according to the tensile strengths, with decimals used for variations at the same strength level. Bolts and screws are available in all grades listed. Studs are available in grades 1, 2, 4, 5, 8, and 8.1. Grade 8.1 is not listed. 120 Figure 817 Histogram of bolt ultimate tensile strength based on 539 tests displaying a mean ultimate tensile strength ¯ Sut = 145.1 kpsi and a standard deviation of σ Sut = 10.3 kpsi. ˆ 100 Number of specimens 420 80 60 40 20 0 0 120 130 140 150 160 Tensile strength, Sut , kpsi 170 180 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 418 III. Design of Mechanical Elements © The McGrawHill Companies, 2008 8. Screws, Fasteners, and the Design of Nonpermanent Joints Mechanical Engineering Design Table 89 SAE Specications for Steel Bolts Minimum Proof Strength,* kpsi Minimum Tensile Strength,* kpsi Minimum Yield Strength,* kpsi 1 1 1 1 4 2 33 60 36 Low or medium carbon 2 13 44 55 74 57 Low or medium carbon 7 1 1 8 2 33 60 36 4 1 1 1 4 2 65 115 100 5 1 1 4 85 120 92 1 1 1 1 8 2 74 105 81 5.2 1 1 4 85 120 92 7 1 1 1 4 2 105 133 115 Medium-carbon alloy, Q&T 8 1 1 1 4 2 120 150 130 Medium-carbon alloy, Q&T 8.2 1 1 4 120 150 130 Low-carbon martensite, Q&T SAE Grade No. Size Range Inclusive, in Material Head Marking Medium carbon, cold-drawn Medium carbon, Q&T Low-carbon martensite, Q&T *Minimum strengths are strengths exceeded by 99 percent of fasteners. ASTM specications are listed in Table 810. ASTM threads are shorter because ASTM deals mostly with structures; structural connections are generally loaded in shear, and the decreased thread length provides more shank area. Specications for metric fasteners are given in Table 811. It is worth noting that all specication-grade bolts made in this country bear a manufacturers mark or logo, in addition to the grade marking, on the bolt head. Such marks conrm that the bolt meets or exceeds specications. If such marks are missing, the bolt may be imported; for imported bolts there is no obligation to meet specications. Bolts in fatigue axial loading fail at the llet under the head, at the thread runout, and at the rst thread engaged in the nut. If the bolt has a standard shoulder under 421 422 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition III. Design of Mechanical Elements © The McGrawHill Companies, 2008 8. Screws, Fasteners, and the Design of Nonpermanent Joints Screws, Fasteners, and the Design of Nonpermanent Joints 419 Table 810 ASTM Specications for Steel Bolts ASTM Size DesigRange, nation Inclusive, No. in Minimum Proof Strength,* kpsi Minimum Tensile Strength,* kpsi A307 1 1 1 4 2 33 60 36 Low carbon A325, 1 1 2 85 120 92 Medium carbon, Q&T type 1 1 1 1 1 8 2 74 105 81 A325, 1 1 2 85 120 92 Low-carbon, martensite, type 2 1 1 1 1 8 2 74 105 81 Q&T A325, 1 1 2 85 120 92 Weathering steel, type 3 1 1 1 1 8 2 74 105 81 Q&T A354, 1 2 1 4 2 105 125 109 2 3 4 4 95 115 99 1 4 4 120 150 130 1 1 4 85 120 92 1 1 1 1 8 2 74 105 81 1 3 3 4 55 90 58 1 1 1 2 2 120 150 130 1 1 1 2 2 120 150 130 grade BC A354, Minimum Yield Strength,* kpsi Material Head Marking A325 A325 A325 Alloy steel, Q&T BC Alloy steel, Q&T grade BD A449 A490, Medium-carbon, Q&T Alloy steel, Q&T type 1 A490, type 3 *Minimum strengths are strengths exceeded by 99 percent of fasteners. A490 Weathering steel, Q&T A490 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 420 III. Design of Mechanical Elements © The McGrawHill Companies, 2008 8. Screws, Fasteners, and the Design of Nonpermanent Joints Mechanical Engineering Design Table 811 Metric Mechanical-Property Classes for Steel Bolts, Screws, and Studs* Property Class 4.6 Size Range, Inclusive Minimum Proof Strength, MPa Minimum Tensile Strength, MPa Minimum Yield Strength, MPa Material M5M36 225 400 240 Low or medium carbon Head Marking 4.6 4.8 M1.6M16 310 420 340 Low or medium carbon 4.8 5.8 M5M24 380 520 420 Low or medium carbon 5.8 8.8 M16M36 600 830 660 Medium carbon, Q&T 8.8 9.8 M1.6M16 650 900 720 Medium carbon, Q&T 9.8 10.9 M5M36 830 1040 940 Low-carbon martensite, Q&T 12.9 M1.6M36 970 1220 1100 10.9 Alloy, Q&T 12.9 *The thread length for bolts and cap screws is 2d + 6 L T = 2d + 12 2d + 25 L 125 125 < L 200 L > 200 where L is the bolt length. The thread length for structural bolts is slightly shorter than given above. strengths are strength exceeded by 99 percent of fasteners. Minimum the head, it has a value of K f from 2.1 to 2.3, and this shoulder llet is protected from scratching or scoring by a washer. If the thread runout has a 15 or less halfcone angle, the stress is higher at the rst engaged thread in the nut. Bolts are sized by examining the loading at the plane of the washer face of the nut. This is the weakest part of the bolt if and only if the conditions above are satised (washer protection of the shoulder llet and thread runout 15 ). Inattention to this requirement has led to a record of 15 percent fastener fatigue failure under the head, 20 percent at thread runout, and 65 percent where the designer is focusing attention. It does little good to concentrate on the plane of the nut washer face if it is not the weakest location. 423 424 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition III. Design of Mechanical Elements © The McGrawHill Companies, 2008 8. Screws, Fasteners, and the Design of Nonpermanent Joints Screws, Fasteners, and the Design of Nonpermanent Joints 421 Nuts are graded so that they can be mated with their corresponding grade of bolt. The purpose of the nut is to have its threads deect to distribute the load of the bolt more evenly to the nut. The nuts properties are controlled in order to accomplish this. The grade of the nut should be the grade of the bolt. 87 Tension JointsThe External Load Let us now consider what happens when an external tensile load P, as in Fig. 813, is applied to a bolted connection. It is to be assumed, of course, that the clamping force, which we will call the preload Fi , has been correctly applied by tightening the nut before P is applied. The nomenclature used is: Fi = preload P = external tensile load Pb = portion of P taken by bolt Pm = portion of P taken by members Fb = Pb + Fi = resultant bolt load Fm = Pm Fi = resultant load on members C = fraction of external load P carried by bolt 1 C = fraction of external load P carried by members The load P is tension, and it causes the connection to stretch, or elongate, through some distance δ . We can relate this elongation to the stiffnesses by recalling that k is the force divided by the deection. Thus δ= Pb kb δ= and Pm km (a) or Pm = Pb km kb (b) Since P = Pb + Pm , we have Pb = kb P = CP kb + km (c) and Pm = P Pb = (1 C ) P (d ) where C= kb kb + km (e) is called the stiffness constant of the joint. The resultant bolt load is Fb = Pb + Fi = C P + Fi Fm < 0 (824) BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 422 III. Design of Mechanical Elements 425 © The McGrawHill Companies, 2008 8. Screws, Fasteners, and the Design of Nonpermanent Joints Mechanical Engineering Design Table 812 Stiffnesses, M lbf/in Computation of Bolt and Member Stiffnesses. Steel members clamped using a 1 in-13 NC 2 kb steel bolt. C = Bolt Grip, in kb km C 1 C 2 2.57 12.69 0.168 0.832 3 1.79 11.33 0.136 0.864 4 1.37 10.63 0.114 0.886 kb + km and the resultant load on the connected members is Fm = Pm Fi = (1 C ) P Fi Fm < 0 (825) Of course, these results are valid only as long as some clamping load remains in the members; this is indicated by the qualier in the equations. Table 812 is included to provide some information on the relative values of the stiffnesses encountered. The grip contains only two members, both of steel, and no washers. The ratios C and 1 C are the coefcients of P in Eqs. (824) and (825), respectively. They describe the proportion of the external load taken by the bolt and by the members, respectively. In all cases, the members take over 80 percent of the external load. Think how important this is when fatigue loading is present. Note also that making the grip longer causes the members to take an even greater percentage of the external load. 88 Relating Bolt Torque to Bolt Tension Having learned that a high preload is very desirable in important bolted connections, we must next consider means of ensuring that the preload is actually developed when the parts are assembled. If the overall length of the bolt can actually be measured with a micrometer when it is assembled, the bolt elongation due to the preload Fi can be computed using the formula δ = Fi l /( AE ). Then the nut is simply tightened until the bolt elongates through the distance δ. This ensures that the desired preload has been attained. The elongation of a screw cannot usually be measured, because the threaded end is often in a blind hole. It is also impractical in many cases to measure bolt elongation. In such cases the wrench torque required to develop the specied preload must be estimated. Then torque wrenching, pneumatic-impact wrenching, or the turn-of-the-nut method may be used. The torque wrench has a built-in dial that indicates the proper torque. With impact wrenching, the air pressure is adjusted so that the wrench stalls when the proper torque is obtained, or in some wrenches, the air automatically shuts off at the desired torque. The turn-of-the-nut method requires that we rst dene the meaning of snug-tight. The snug-tight condition is the tightness attained by a few impacts of an impact wrench, or the full effort of a person using an ordinary wrench. When the snug-tight condition is attained, all additional turning develops useful tension in the bolt. The turn-of-the-nut method requires that you compute the fractional number of turns necessary to develop the required preload from the snug-tight condition. For example, for heavy hexagonal structural bolts, the turn-of-the-nut specication states that the nut should be turned a minimum of 180 from the snug-tight condition under optimum 426 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition III. Design of Mechanical Elements © The McGrawHill Companies, 2008 8. Screws, Fasteners, and the Design of Nonpermanent Joints Screws, Fasteners, and the Design of Nonpermanent Joints Table 813 Distribution of Preload Fi for 20 Tests of Unlubricated Bolts Torqued to 90 N · m 423 23.6, 27.6, 28.0, 29.4, 30.3, 30.7, 32.9, 33.8, 33.8, 33.8, 34.7, 35.6, 35.6, 37.4, 37.8, 37.8, 39.2, 40.0, 40.5, 42.7 ˆ *Mean value Fi = 34.3 kN. Standard deviation, σ = 4.91 kN. conditions. Note that this is also about the correct rotation for the wheel nuts of a passenger car. Problems 815 to 817 illustrate the method further. Although the coefcients of friction may vary widely, we can obtain a good estimate of the torque required to produce a given preload by combining Eqs. (85) and (86): T= l + π f dm sec α π dm f l sec α Fi dm 2 + Fi f c dc 2 (a) where dm is the average of the major and minor diameters. Since tan λ = l /π dm , we divide the numerator and denominator of the rst term by π dm and get T= tan λ + f sec α l f tan λ sec α Fi dm 2 + Fi f c dc 2 (b) The diameter of the washer face of a hexagonal nut is the same as the width across ats and equal to 1 1 times the nominal size. Therefore the mean collar diameter is 2 dc = (d + 1.5d )/2 = 1.25d . Equation (b) can now be arranged to give T= dm 2d tan λ + f sec α 1 f tan λ sec α + 0.625 f c Fi d (c) We now dene a torque coefcient K as the term in brackets, and so K= dm 2d tan λ + f sec α 1 f tan λ sec α + 0.625 f c (826) Equation (c) can now be written T = K Fi d (827) The coefcient of friction depends upon the surface smoothness, accuracy, and degree of lubrication. On the average, both f and f c are about 0.15. The interesting . fact about Eq. (826) is that K = 0.20 for f = f c = 0.15 no matter what size bolts are employed and no matter whether the threads are coarse or ne. Blake and Kurtz have published results of numerous tests of the torquing of bolts.6 By subjecting their data to a statistical analysis, we can learn something about the distribution of the torque coefcients and the resulting preload. Blake and Kurtz determined the preload in quantities of unlubricated and lubricated bolts of size 1 in-20 2 UNF when torqued to 800 lbf · in. This corresponds roughly to an M12 × 1.25 bolt torqued to 90 N · m. The statistical analyses of these two groups of bolts, converted to SI units, are displayed in Tables 813 and 814. 6 J. C. Blake and H. J. Kurtz, The Uncertainties of Measuring Fastener Preload, Machine Design, vol. 37, Sept. 30, 1965, pp. 128131. BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 424 III. Design of Mechanical Elements 427 © The McGrawHill Companies, 2008 8. Screws, Fasteners, and the Design of Nonpermanent Joints Mechanical Engineering Design Table 814 30.3, Distribution of Preload Fi for 10 Tests of Lubricated Bolts Torqued to 90 N · m Table 815 32.5, 32.5, 32.9, 33.8, 34.3, 34.7, 37.4, 40.5 ˆ *Mean value, Fi = 34.18 kN. Standard deviation, σ = 2.88 kN. Bolt Condition Torque Factors K for Use with Eq. (827) 32.9, K Nonplated, black nish 0.30 Zinc-plated 0.20 Lubricated 0.18 Cadmium-plated 0.16 With Bowman Anti-Seize 0.12 With Bowman-Grip nuts 0.09 We rst note that both groups have about the same mean preload, 34 kN. The unlubricated bolts have a standard deviation of 4.9 kN and a COV of about 0.15. The lubricated bolts have a standard deviation of 3 kN and a COV of about 0.9. The means obtained from the two samples are nearly identical, approximately 34 kN; using Eq. (827), we nd, for both samples, K = 0.208. Bowman Distribution, a large manufacturer of fasteners, recommends the values shown in Table 815. In this book we shall use these values and use K = 0.2 when the bolt condition is not stated. EXAMPLE 83 Solution A 3 in-16 UNF × 2 1 in SAE grade 5 bolt is subjected to a load P of 6 kip in a ten4 2 sion joint. The initial bolt tension is Fi = 25 kip. The bolt and joint stiffnesses are kb = 6.50 and km = 13.8 Mlbf/in, respectively. (a) Determine the preload and service load stresses in the bolt. Compare these to the SAE minimum proof strength of the bolt. (b) Specify the torque necessary to develop the preload, using Eq. (827). (c) Specify the torque necessary to develop the preload, using Eq. (826) with f = f c = 0.15. From Table 82, At = 0.373 in2. (a) The preload stress is Answer σi = Fi 25 = = 67.02 kpsi At 0.373 The stiffness constant is C= 6.5 kb = = 0.320 kb + km 6.5 + 13.8 428 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition III. Design of Mechanical Elements © The McGrawHill Companies, 2008 8. Screws, Fasteners, and the Design of Nonpermanent Joints Screws, Fasteners, and the Design of Nonpermanent Joints 425 From Eq. (824), the stress under the service load is σb = Answer Fb C P + Fi P = =C + σi At At At = 0.320 6 + 67.02 = 72.17 kpsi 0.373 From Table 89, the SAE minimum proof strength of the bolt is Sp = 85 kpsi. The preload and service load stresses are respectively 21 and 15 percent less than the proof strength. (b) From Eq. (827), the torque necessary to achieve the preload is T = K Fi d = 0.2(25)(103 )(0.75) = 3750 lbf · in Answer (c ) The minor diameter can be determined from the minor area in Table 82. Thus dr = 4 Ar /π = 4(0.351)/π = 0.6685 in. Thus, the mean diameter is dm = (0.75 + 0.6685)/2 = 0.7093 in. The lead angle is λ = tan1 l 1 1 = tan1 = tan1 = 1.6066 π dm π dm N π(0.7093)(16) For α = 30 , Eq. (826) gives T= 0.7093 2(0.75) tan 1.6066 + 0.15(sec 30 ) + 0.625(0.15) 25(103 )(0.75) 1 0.15(tan 1.6066 )(sec 30 ) = 3551 lbf · in which is 5.3 percent less than the value found in part (b). 89 Statically Loaded Tension Joint with Preload Equations (824) and (825) represent the forces in a bolted joint with preload. The tensile stress in the bolt can be found as in Ex. 83 as σb = CP Fi + At At (a) The limiting value of σb is the proof strength Sp . Thus, with the introduction of a load factor n, Eq. (a) becomes Cn P Fi + = Sp At At (b) or n= Sp At Fi CP (828) Here we have called n a load factor rather than a factor of safety, though the two ideas are somewhat related. Any value of n > 1 in Eq. (828) ensures that the bolt stress is less than the proof strength. Another means of ensuring a safe joint is to require that the external load be smaller than that needed to cause the joint to separate. If separation does occur, then BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 426 III. Design of Mechanical Elements 429 © The McGrawHill Companies, 2008 8. Screws, Fasteners, and the Design of Nonpermanent Joints Mechanical Engineering Design the entire external load will be imposed on the bolt. Let P0 be the value of the external load that would cause joint separation. At separation, Fm = 0 in Eq. (825), and so (1 C ) P0 Fi = 0 (c) Let the factor of safety against joint separation be n0 = P0 P (d ) Substituting P0 = n 0 P in Eq. (c), we nd n0 = Fi P (1 C ) (829) as a load factor guarding against joint separation. Figure 818 is the stress-strain diagram of a good-quality bolt material. Notice that there is no clearly dened yield point and that the diagram progresses smoothly up to fracture, which corresponds to the tensile strength. This means that no matter how much preload is given the bolt, it will retain its load-carrying capacity. This is what keeps the bolt tight and determines the joint strength. The pre-tension is the muscle of the joint, and its magnitude is determined by the bolt strength. If the full bolt strength is not used in developing the pre-tension, then money is wasted and the joint is weaker. Good-quality bolts can be preloaded into the plastic range to develop more strength. Some of the bolt torque used in tightening produces torsion, which increases the principal tensile stress. However, this torsion is held only by the friction of the bolt head and nut; in time it relaxes and lowers the bolt tension slightly. Thus, as a rule, a bolt will either fracture during tightening, or not at all. Above all, do not rely too much on wrench torque; it is not a good indicator of preload. Actual bolt elongation should be used whenever possibleespecially with fatigue loading. In fact, if high reliability is a requirement of the design, then preload should always be determined by bolt elongation. Russell, Burdsall & Ward Inc. (RB&W) recommendations for preload are 60 kpsi for SAE grade 5 bolts for nonpermanent connections, and that A325 bolts (equivalent to SAE grade 5) used in structural applications be tightened to proof load or beyond Sut Figure 818 Typical stress-strain diagram for bolt materials showing proof strength Sp, yield strength Sy, and ultimate tensile strength Sut. Sy Stress Sp Strain 430 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition III. Design of Mechanical Elements © The McGrawHill Companies, 2008 8. Screws, Fasteners, and the Design of Nonpermanent Joints Screws, Fasteners, and the Design of Nonpermanent Joints 427 (85 kpsi up to a diameter of 1 in).7 Bowman8 recommends a preload of 75 percent of proof load, which is about the same as the RB&W recommendations for reused bolts. In view of these guidelines, it is recommended for both static and fatigue loading that the following be used for preload: Fi = 0.75 Fp 0.90 Fp for nonpermanent connections, reused fasteners for permanent connections (830) where Fp is the proof load, obtained from the equation (831) Fp = At Sp Here Sp is the proof strength obtained from Tables 89 to 811. For other materials, an approximate value is Sp = 0.85 Sy . Be very careful not to use a soft material in a threaded fastener. For high-strength steel bolts used as structural steel connectors, if advanced tightening methods are used, tighten to yield. You can see that the RB&W recommendations on preload are in line with what we have encountered in this chapter. The purposes of development were to give the reader the perspective to appreciate Eqs. (830) and a methodology with which to handle cases more specically than the recommendations. 7 Russell, Burdsall & Ward Inc., Helpful Hints for Fastener Design and Application, Mentor, Ohio, 1965, p. 42. 8 Bowman DistributionBarnes Group, Fastener Facts, Cleveland, 1985, p. 90. EXAMPLE 84 Solution Figure 819 is a cross section of a grade 25 cast-iron pressure vessel. A total of N bolts are to be used to resist a separating force of 36 kip. (a) Determine kb , km , and C. (b) Find the number of bolts required for a load factor of 2 where the bolts may be reused when the joint is taken apart. (a) The grip is l = 1.50 in. From Table A31, the nut thickness is 2 threads beyond the nut of 11 in gives a bolt length of L= Figure 819 5 8 35 2 + 1.50 + = 2.229 in 64 11 in-11 UNC × 2 1 in grade 5 4 finished hex head bolt No. 25 CI 3 4 in 3 4 in 35 64 in. Adding two BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 428 III. Design of Mechanical Elements 431 © The McGrawHill Companies, 2008 8. Screws, Fasteners, and the Design of Nonpermanent Joints Mechanical Engineering Design From Table A17 the next fraction size bolt is L = 2 1 in. From Eq. (813), the thread 4 length is L T = 2(0.625) + 0.25 = 1.50 in. Thus the length of the unthreaded portion in the grip is ld = 2.25 1.50 = 0.75 in. The threaded length in the grip is lt = l ld = 0.75 in. From Table 82, At = 0.226 in2. The major-diameter area is Ad = π(0.625)2 /4 = 0.3068 in2. The bolt stiffness is then kb = Answer Ad At E 0.3068(0.226)(30) = Ad lt + At ld 0.3068(0.75) + 0.226(0.75) = 5.21 Mlbf/in From Table A24, for no. 25 cast iron we will use E = 14 Mpsi. The stiffness of the members, from Eq. (822), is km = Answer 0.5774π Ed 0.5774l + 0.5d 2 ln 5 0.5774l + 2.5d = 0.5774π(14)(0.625) 0.5774 (1.5) + 0.5 (0.625) 2 ln 5 0.5774 (1.5) + 2.5 (0.625) = 8.95 Mlbf/in If you are using Eq. (823), from Table 88, A = 0.778 71 and B = 0.616 16, and km = Ed A exp( Bd / l ) = 14(0.625)(0.778 71) exp[0.616 16(0.625)/1.5] = 8.81 Mlbf/in which is only 1.6 percent lower than the previous result. From the rst calculation for km , the stiffness constant C is Answer C= kb 5.21 = = 0.368 kb + km 5.21 + 8.95 (b) From Table 89, Sp = 85 kpsi. Then, using Eqs. (830) and (831), we nd the recommended preload to be Fi = 0.75 At Sp = 0.75(0.226)(85) = 14.4 kip For N bolts, Eq. (828) can be written n= Sp At Fi C( P/N ) (1) or N= 0.368(2)(36) Cn P = = 5.52 Sp At Fi 85(0.226) 14.4 With six bolts, Eq. (1) gives n= 85(0.226) 14.4 = 2.18 0.368(36/6) which is greater than the required value. Therefore we choose six bolts and use the recommended tightening preload. 432 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition III. Design of Mechanical Elements © The McGrawHill Companies, 2008 8. Screws, Fasteners, and the Design of Nonpermanent Joints Screws, Fasteners, and the Design of Nonpermanent Joints 810 429 Gasketed Joints If a full gasket is present in the joint, the gasket pressure p is found by dividing the force in the member by the gasket area per bolt. Thus, for N bolts, p= Fm Ag / N (a) With a load factor n, Eq. (825) can be written as Fm = (1 C )n P Fi (b) Substituting this into Eq. (a) gives the gasket pressure as p = [ Fi n P (1 C )] N Ag (832) In full-gasketed joints uniformity of pressure on the gasket is important. To maintain adequate uniformity of pressure adjacent bolts should not be placed more than six nominal diameters apart on the bolt circle. To maintain wrench clearance, bolts should be placed at least three diameters apart. A rough rule for bolt spacing around a bolt circle is 3 π Db 6 Nd (833) where Db is the diameter of the bolt circle and N is the number of bolts. 811 Fatigue Loading of Tension Joints Tension-loaded bolted joints subjected to fatigue action can be analyzed directly by the methods of Chap. 6. Table 816 lists average fatigue stress-concentration factors for the llet under the bolt head and also at the beginning of the threads on the bolt shank. These are already corrected for notch sensitivity and for surface nish. Designers should be aware that situations may arise in which it would be advisable to investigate these factors more closely, since they are only average values. In fact, Peterson9 observes that the distribution of typical bolt failures is about 15 percent under the head, 20 percent at the end of the thread, and 65 percent in the thread at the nut face. Use of rolled threads is the predominant method of thread-forming in screw fasteners, where Table 816 applies. In thread-rolling, the amount of cold work and strainstrengthening is unknown to the designer; therefore, fully corrected (including K f ) axial endurance strength is reported in Table 817. For cut threads, the methods of Chap. 6 are useful. Anticipate that the endurance strengths will be considerably lower. Most of the time, the type of fatigue loading encountered in the analysis of bolted joints is one in which the externally applied load uctuates between zero and some Table 816 SAE Grade Cut Threads 3.6 to 5.8 2.2 2.8 2.1 4 to 8 9 Rolled Threads 0 to 2 Fatigue StressConcentration Factors Kf for Threaded Elements Metric Grade 6.6 to 10.9 3.0 3.8 2.3 Fillet W. D. Pilkey, Petersons Stress Concentration Factors, 2nd ed., John Wiley & Sons, New York, 1997, p. 387. BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 430 III. Design of Mechanical Elements 433 © The McGrawHill Companies, 2008 8. Screws, Fasteners, and the Design of Nonpermanent Joints Mechanical Engineering Design Table 817 Fully Corrected Endurance Strengths for Bolts and Screws with Rolled Threads* Grade or Class Size Range Endurance Strength 1 1 in 4 1 1 1 8 1 2 in 1 1 1 2 in 4 1 1 1 2 in 4 SAE 5 SAE 7 SAE 8 18.6 kpsi 16.3 kpsi 20.6 kpsi 23.2 kpsi ISO 8.8 M16M36 129 MPa ISO 9.8 M1.6M16 140 MPa ISO 10.9 M5M36 162 MPa ISO 12.9 M1.6M36 190 MPa *Repeatedly-applied, axial loading, fully corrected. Figure 820 Se a Load line Alternating stress Designers fatigue diagram showing a Goodman failure line and how a load line is used to dene failure and safety in preloaded bolted joints in fatigue. Point B represents nonfailure; point C, failure. 1 1 C Sa B a A Fi i=A t m D Sm Sut Sa Steady stress m maximum force P. This would be the situation in a pressure cylinder, for example, where a pressure either exists or does not exist. For such cases, Fmax = Fb and Fmin = Fi and the alternating component of the force is Fa = ( Fmax Fmin )/2 = ( Fb Fi )/2. Dividing this by At yields the alternating component of the bolt stress. Employing the notation from Sec. 87 with Eq. (824), we obtain σa = Fb Fi (C P + Fi ) Fi CP = = 2 At 2 At 2 At (834) The mean stress is equal to the alternating component plus the minimum stress, σi = Fi / At , which results in σm = CP Fi + 2 At At (835) On the designers fatigue diagram, shown in Fig. 820, the load line is σm = σa + σi (836) 434 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition III. Design of Mechanical Elements © The McGrawHill Companies, 2008 8. Screws, Fasteners, and the Design of Nonpermanent Joints Screws, Fasteners, and the Design of Nonpermanent Joints 431 The next problem is to nd the strength components Sa and Sm of the fatigue failure line. These depend on the failure criteria: Goodman: Sa Sm + =1 Se Sut (837) Gerber: 2 Sm Sut Sa + Se =1 (838) ASME-elliptic: Sa Se 2 + Sm Sp 2 =1 (839) For simultaneous solution between Eq. (836), as Sm = Sa + σi , and each of Eqs. (837) to (839) gives Goodman: Sa = Se ( Sut σi ) Sut + Se Sm = Sa + σi (840) (841) Gerber: Sa = 1 2 2 Sut Sut + 4 Se ( Se + σi ) Sut 2σi Se 2 Se (842) Sm = Sa + σi ASME-elliptic: Sa = 2 Sp Se 2 2 Sp Sp + Se σi2 σi Se 2 + Se (843) Sm = Sa + σi When using relations of this section, be sure to use Kf for both σa and σm . Otherwise, the slope of the load line will not remain 1 to 1. Examination of Eqs. (837) to (843) shows parametric equations that relate the coordinates of interest to the form of the criteria. The factor of safety guarding against fatigue is given by nf = Sa σa (844) Applying this to the Goodman criterion, for example, with Eqs. (834) and (840) and σi = Fi / At gives nf = 2 Se ( Sut At Fi ) C P ( Sut + Se ) (845) BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 432 III. Design of Mechanical Elements 435 © The McGrawHill Companies, 2008 8. Screws, Fasteners, and the Design of Nonpermanent Joints Mechanical Engineering Design when preload Fi is present. With no preload, C = 1, Fi = 0, and Eq. (845) becomes nf0 = 2 Se Sut At P ( Sut + Se ) (846) Preload is benecial for resisting fatigue when n f / n f 0 is greater than unity. For Goodman, Eqs. (845) and (846) with n f / n f 0 1 puts an upper bound on the preload Fi of Fi (1 C ) Sut At (847) If this cannot be achieved, and nf is unsatisfactory, use the Gerber or ASME-elliptic criterion to obtain a less conservative assessment. If the design is still not satisfactory, additional bolts and/or a different size bolt may be called for. Bolts loosen, as they are friction devices, and cyclic loading and vibration as well as other effects allow the fasteners to lose tension with time. How does one ght loosening? Within strength limitations, the higher the preload the better. A rule of thumb is that preloads of 60 percent of proof load rarely loosen. If more is better, how much more? Well, not enough to create reused fasteners as a future threat. Alternatively, fastener-locking schemes can be employed. After solving Eq. (844), you should also check the possibility of yielding, using the proof strength np = Sp σm + σa (848) EXAMPLE 85 Figure 821 shows a connection using cap screws. The joint is subjected to a uctuating force whose maximum value is 5 kip per screw. The required data are: cap screw, 1 5/8 in-11 NC, SAE 5; hardened-steel washer, tw = 16 in thick; steel cover plate, t1 = 5 5 E = 30 Mpsi; and cast-iron base, t2 = 8 in, E ci = 16 Mpsi. 8 in, s (a) Find kb , km , and C using the assumptions given in the caption of Fig. 821. (b) Find all factors of safety and explain what they mean. Solution (a) For the symbols of Figs. 815 and 821, h = t1 + tw = 0.6875 in, l = h + d /2 = 1 in, and D2 = 1.5d = 0.9375 in. The joint is composed of three frusta; the upper two frusta are steel and the lower one is cast iron. For the upper frustum: t = l /2 = 0.5 in, D = 0.9375 in, and E = 30 Mpsi. Using these values in Eq. (820) gives k1 = 46.46 Mlbf/in. Figure 821 Pressure-cone frustum member model for a cap screw. For this model the signicant sizes are t2 < d h + t 2 /2 l= h + d /2 t2 d D1 = dw + l tan α = 1.5d + 0.577l D2 = dw = 1.5d where l = effective grip. The solutions are for α = 30 and dw = 1.5d . D1 l l 2 t1 t2 d D2 h BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition III. Design of Mechanical Elements © The McGrawHill Companies, 2008 8. Screws, Fasteners, and the Design of Nonpermanent Joints Screws, Fasteners, and the Design of Nonpermanent Joints 433 For the middle frustum: t = h l /2 = 0.1875 in and D = 0.9375 + 2(l h ) tan 30 = 1.298 in. With these and E s = 30 Mpsi, Eq. (820) gives k2 = 197.43 Mlbf/in. The lower frustum has D = 0.9375 in, t = l h = 0.3125 in, and E ci = 16 Mpsi. The same equation yields k3 = 32.39 Mlbf/in. Substituting these three stiffnesses into Eq. (818) gives km = 17.40 Mlbf/in. The cap screw is short and threaded all the way. Using l = 1 in for the grip and At = 0.226 in2 from Table 82, we find the stiffness to be kb = At E / l = 6.78 Mlbf/in. Thus the joint constant is Answer kb 6.78 = = 0.280 kb + km 6.78 + 17.40 C= (b) Equation (830) gives the preload as Fi = 0.75 Fp = 0.75 At Sp = 0.75(0.226)(85) = 14.4 kip where from Table 89, Sp = 85 kpsi for an SAE grade 5 cap screw. Using Eq. (828), we obtain the load factor as Answer n= Sp At Fi 85(0.226) 14.4 = = 3.44 CP 0.280(5) This factor prevents the bolt stress from becoming equal to the proof strength. Next, using Eq. (829), we have Answer Fi 14.4 = = 4.00 P (1 C ) 5(1 0.280) n0 = If the force P gets too large, the joint will separate and the bolt will take the entire load. This factor guards against that event. For the remaining factors, refer to Fig. 822. This diagram contains the modied Goodman line, the Gerber line, the proof-strength line, and the load line. The intersection Figure 822 E Sa D Sa Sa C Sp a B A a Designers fatigue diagram for preloaded bolts, drawn to scale, showing the modied Goodman line, the Gerber line, and the Langer proofstrength line, with an exploded view of the area of interest. The strengths used are Sp = 85 kpsi, Se = 18.6 kpsi, and Sut = 120 kpsi. The coordinates are A, σi = 63.72 kpsi; B, σa = 3.10 kpsi, σm = 66.82 kpsi; C, Sa = 7.55 kpsi, Sm = 71.29 kpsi; D, Sa = 10.64 kpsi, Sm = 74.36 kpsi; E, Sa = 11.32 kpsi, Sm = 75.04 kpsi. L 60 Stress amplitude 436 i m Sm Sm 70 Sm 80 Sp Proof strength line Gerber line L Se Modified Goodman line i Sp Steady stress component Sut m 90 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 434 III. Design of Mechanical Elements 8. Screws, Fasteners, and the Design of Nonpermanent Joints 437 © The McGrawHill Companies, 2008 Mechanical Engineering Design of the load line L with the respective failure lines at points C, D, and E denes a set of strengths Sa and Sm at each intersection. Point B represents the stress state σa , σm . Point A is the preload stress σi . Therefore the load line begins at A and makes an angle having a unit slope. This angle is 45° only when both stress axes have the same scale. The factors of safety are found by dividing the distances AC , AD , and A E by the distance AB. Note that this is the same as dividing Sa for each theory by σa . The quantities shown in the caption of Fig. 822 are obtained as follows: Point A σi = Fi 14.4 = = 63.72 kpsi At 0.226 Point B σa = CP 0.280(5) = = 3.10 kpsi 2 At 2(0.226) σm = σa + σi = 3.10 + 63.72 = 66.82 kpsi Point C This is the modied Goodman criteria. From Table 817, we nd Se = 18.6 kpsi. Then, using Eq. (840), we get Sa = Se ( Sut σi ) 18.6(120 63.72) = = 7.55 kpsi Sut + Se 120 + 18.6 The factor of safety is found to be Answer nf = Sa 7.55 = = 2.44 σa 3.10 Point D This is on the proof-strength line where Sm + Sa = Sp (1) In addition, the horizontal projection of the load line AD is Sm = σi + Sa (2) Solving Eqs. (1) and (2) simultaneously results in Sa = Sp σi 85 63.72 = = 10.64 kpsi 2 2 The factor of safety resulting from this is Answer np = Sa 10.64 = = 3.43 σa 3.10 which, of course, is identical to the result previously obtained by using Eq. (828). A similar analysis of a fatigue diagram could have been done using yield strength instead of proof strength. Though the two strengths are somewhat related, proof strength is a much better and more positive indicator of a fully loaded bolt than is the yield strength. It is also worth remembering that proof-strength values are specied in design codes; yield strengths are not. 438 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition III. Design of Mechanical Elements © The McGrawHill Companies, 2008 8. Screws, Fasteners, and the Design of Nonpermanent Joints 435 Screws, Fasteners, and the Design of Nonpermanent Joints We found n f = 2.44 on the basis of fatigue and the modied Goodman line, and n p = 3.43 on the basis of proof strength. Thus the danger of failure is by fatigue, not by overproof loading. These two factors should always be compared to determine where the greatest danger lies. Point E For the Gerber criterion, from Eq. (842), Sa = = 1 2 2 Sut Sut + 4 Se ( Se + σi ) Sut 2σi Se 2 Se 1 120 1202 + 4(18.6)(18.6 + 63.72) 1202 2(63.72)(18.6) 2(18.6) = 11.33 kpsi Thus for the Gerber criterion the safety factor is Answer nf = Sa 11.33 = = 3.65 σa 3.10 which is greater than n p = 3.43 and contradicts the conclusion earlier that the danger of failure is fatigue. Figure 822 clearly shows the conict where point D lies between points C and E. Again, the conservative nature of the Goodman criterion explains the discrepancy and the designer must form his or her own conclusion. 812 Bolted and Riveted Joints Loaded in Shear10 Riveted and bolted joints loaded in shear are treated exactly alike in design and analysis. Figure 823a shows a riveted connection loaded in shear. Let us now study the various means by which this connection might fail. Figure 823b shows a failure by bending of the rivet or of the riveted members. The bending moment is approximately M = Ft /2, where F is the shearing force and t is the grip of the rivet, that is, the total thickness of the connected parts. The bending stress in the members or in the rivet is, neglecting stress concentration, σ= M I /c (849) where I /c is the section modulus for the weakest member or for the rivet or rivets, depending upon which stress is to be found. The calculation of the bending stress in 10 The design of bolted and riveted connections for boilers, bridges, buildings, and other structures in which danger to human life is involved is strictly governed by various construction codes. When designing these structures, the engineer should refer to the American Institute of Steel Construction Handbook, the American Railway Engineering Association specications, or the Boiler Construction Code of the American Society of Mechanical Engineers. BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 436 III. Design of Mechanical Elements 439 © The McGrawHill Companies, 2008 8. Screws, Fasteners, and the Design of Nonpermanent Joints Mechanical Engineering Design Figure 823 Modes of failure in shear loading of a bolted or riveted connection: (a) shear loading; (b) bending of rivet; (c) shear of rivet; (d) tensile failure of members; (e) bearing of rivet on members or bearing of members on rivet; (f ) shear tear-out; (g) tensile tear-out. (a ) (b ) (e ) ( c) (d ) (f) ( g) this manner is an assumption, because we do not know exactly how the load is distributed to the rivet or the relative deformations of the rivet and the members. Although this equation can be used to determine the bending stress, it is seldom used in design; instead its effect is compensated for by an increase in the factor of safety. In Fig. 823c failure of the rivet by pure shear is shown; the stress in the rivet is τ= F A (850) where A is the cross-sectional area of all the rivets in the group. It may be noted that it is standard practice in structural design to use the nominal diameter of the rivet rather than the diameter of the hole, even though a hot-driven rivet expands and nearly lls up the hole. Rupture of one of the connected membes or plates by pure tension is illustrated in Fig. 823d. The tensile stress is σ= F A (851) where A is the net area of the plate, that is, the area reduced by an amount equal to the area of all the rivet holes. For brittle materials and static loads and for either ductile or brittle materials loaded in fatigue, the stress-concentration effects must be included. It is true that the use of a bolt with an initial preload and, sometimes, a rivet will place the area around the hole in compression and thus tend to nullify the effects of stress concentration, but unless denite steps are taken to ensure that the preload does not relax, it is on the conservative side to design as if the full stress-concentration effect were present. The stress-concentration effects are not considered in structural design, because the loads are static and the materials ductile. 440 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition III. Design of Mechanical Elements © The McGrawHill Companies, 2008 8. Screws, Fasteners, and the Design of Nonpermanent Joints Screws, Fasteners, and the Design of Nonpermanent Joints 437 In calculating the area for Eq. (851), the designer should, of course, use the combination of rivet or bolt holes that gives the smallest area. Figure 823e illustrates a failure by crushing of the rivet or plate. Calculation of this stress, which is usually called a bearing stress, is complicated by the distribution of the load on the cylindrical surface of the rivet. The exact values of the forces acting upon the rivet are unknown, and so it is customary to assume that the components of these forces are uniformly distributed over the projected contact area of the rivet. This gives for the stress σ = F A (852) where the projected area for a single rivet is A = td . Here, t is the thickness of the thinnest plate and d is the rivet or bolt diameter. Edge shearing, or tearing, of the margin is shown in Fig. 823f and g, respectively. In structural practice this failure is avoided by spacing the rivets at least 1 1 2 diameters away from the edge. Bolted connections usually are spaced an even greater distance than this for satisfactory appearance, and hence this type of failure may usually be neglected. In a rivet joint, the rivets all share the load in shear, bearing in the rivet, bearing in the member, and shear in the rivet. Other failures are participated in by only some of the joint. In a bolted joint, shear is taken by clamping friction, and bearing does not exist. When bolt preload is lost, one bolt begins to carry the shear and bearing until yielding slowly brings other fasteners in to share the shear and bearing. Finally, all participate, and this is the basis of most bolted-joint analysis if loss of bolt preload is complete. The usual analysis involves Bearing in the bolt (all bolts participate) Bearing in members (all holes participate) Shear of bolt (all bolts participate eventually) Distinguishing between thread and shank shear Edge shearing and tearing of member (edge bolts participate) Tensile yielding of member across bolt holes Checking member capacity EXAMPLE 86 Two 1- by 4-in 1018 cold-rolled steel bars are butt-spliced with two 1 - by 4-in 1018 2 cold-rolled splice plates using four 3 in-16 UNF grade 5 bolts as depicted in Fig. 4 824. For a design factor of n d = 1.5 estimate the static load F that can be carried if the bolts lose preload. Solution From Table A20, minimum strengths of Sy = 54 kpsi and Sut = 64 kpsi are found for the members, and from Table 89 minimum strengths of Sp = 85 kpsi and Sut = 120 kpsi for the bolts are found. F /2 is transmitted by each of the splice plates, but since the areas of the splice plates are half those of the center bars, the stresses associated with the plates are the same. So for stresses associated with the plates, the force and areas used will be those of the center plates. BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 438 III. Design of Mechanical Elements © The McGrawHill Companies, 2008 8. Screws, Fasteners, and the Design of Nonpermanent Joints Mechanical Engineering Design Figure 824 1 1 2 in 1 1 1 2 in 1 2 in 1 1 2 in 1 1 4 in F 1 1 2 in w F 1 1 4 in (a ) 1 2 3 4 in in - 16 UNF SAE grade 5 1in F 1 2 F in (b ) Bearing in bolts, all bolts loaded: Sp F = σ= 2td nd F= 2(1) 3 85 2td Sp 4 = = 85 kip nd 1.5 Bearing in members, all bolts active: σ= ( Sy )mem F = 2td nd F= 2(1) 3 54 2td ( Sy )mem 4 = = 54 kip nd 1.5 Shear of bolt, all bolts active: If the bolt threads do not extend into the shear planes for four shanks: τ= Sp F = 0.577 4π d 2 /4 nd F = 0.577π d 2 Sp 85 = 0.577π(0.75)2 = 57.8 kip nd 1.5 If the bolt threads extend into a shear plane: τ= Sp F = 0.577 4 Ar nd F= 0.577(4) Ar Sp 0.577(4)0.351(85) = = 45.9 kip nd 1.5 441 442 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition III. Design of Mechanical Elements © The McGrawHill Companies, 2008 8. Screws, Fasteners, and the Design of Nonpermanent Joints Screws, Fasteners, and the Design of Nonpermanent Joints 439 Edge shearing of member at two margin bolts: From Fig. 825, τ= 0.577( Sy )mem F = 4at nd F= 4at 0.577( Sy )mem 4(1.125)(1)0.577(54) = 93.5 kip = nd 1.5 Tensile yielding of members across bolt holes: σ= F= F 42 42 = ( Sy )mem nd 3 4 t 3 4 t ( Sy )mem nd = 42 3 4 (1)54 1.5 = 90 kip Member yield: F= wt ( Sy )mem 4(1)54 = 144 kip = nd 1.5 On the basis of bolt shear, the limiting value of the force is 45.9 kip, assuming the threads extend into a shear plane. However, it would be poor design to allow the threads to extend into a shear plane. So, assuming a good design based on bolt shear, the limiting value of the force is 57.8 kip. For the members, the bearing stress limits the load to 54 kip. Figure 825 Edge shearing of member. Bolt d a Shear Joints with Eccentric Loading Integral to the analysis of a shear joint is locating the center of relative motion between the two members. In Fig. 826 let A1 to A5 be the respective cross-sectional areas of a group of ve pins, or hot-driven rivets, or tight-tting shoulder bolts. Under this assumption the rotational pivot point lies at the centroid of the cross-sectional area pattern of the pins, rivets, or bolts. Using statics, we learn that the centroid G is located ¯ ¯ by the coordinates x and y , where x1 and yi are the distances to the ith area center: x= ¯ A1 x 1 + A2 x 2 + A3 x 3 + A4 x 4 + A5 x 5 = A1 + A2 + A3 + A4 + A5 A1 y1 + A2 y2 + A3 y3 + A4 y4 + A5 y5 y= ¯ = A1 + A2 + A3 + A4 + A5 n 1 Ai x i Ai n 1 n 1 Ai yi n 1 Ai (853) BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 440 III. Design of Mechanical Elements 443 © The McGrawHill Companies, 2008 8. Screws, Fasteners, and the Design of Nonpermanent Joints Mechanical Engineering Design Figure 826 y Centroid of pins, rivets, or bolts. A3 A2 A4 G A1 _ y A5 O x _ x Figure 827 w lbf in M1 (a) Beam bolted at both ends with distributed load; (b) freebody diagram of beam; (c) enlarged view of bolt group centered at O showing primary and secondary resultant shear forces. O M2 V2 V1 (b ) FA ' FB ' FB " w lbf in A O FA " + B rB rA O Beam FC ' rC rD FD ' FD " (a ) C D FC " (c) In many instances the centroid can be located by symmetry. An example of eccentric loading of fasteners is shown in Fig. 827. This is a portion of a machine frame containing a beam subjected to the action of a bending load. In this case, the beam is fastened to vertical members at the ends with specially prepared load-sharing bolts. You will recognize the schematic representation in Fig. 827b as a statically indeterminate beam with both ends xed and with moment and shear reactions at each end. For convenience, the centers of the bolts at the left end of the beam are drawn to a larger scale in Fig. 827c. Point O represents the centroid of the group, and it is assumed in this example that all the bolts are of the same diameter. Note that the forces shown in Fig. 827c are the resultant forces acting on the pins with a net force and moment equal and opposite to the reaction loads V1 and M1 acting at O. The total load taken by each bolt will be calculated in three steps. In the rst step the shear V1 is divided equally among the bolts so that each bolt takes F = V1 / n , where n refers to the number of bolts in the group and the force F is called the direct load, or primary shear. It is noted that an equal distribution of the direct load to the bolts assumes an absolutely rigid member. The arrangement of the bolts or the shape and size of the 444 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition III. Design of Mechanical Elements © The McGrawHill Companies, 2008 8. Screws, Fasteners, and the Design of Nonpermanent Joints Screws, Fasteners, and the Design of Nonpermanent Joints 441 members sometimes justies the use of another assumption as to the division of the load. The direct loads F are shown as vectors on the loading diagram (Fig. 827c). The moment load, or secondary shear, is the additional load on each bolt due to the moment M1 . If r A , r B , rC , etc., are the radial distances from the centroid to the center of each bolt, the moment and moment loads are related as follows: M1 = FA r A + FB r B + FC rC + · · · (a) where the F are the moment loads. The force taken by each bolt depends upon its radial distance from the centroid; that is, the bolt farthest from the centroid takes the greatest load, while the nearest bolt takes the smallest. We can therefore write F F FA = B= C rA rB rC (b) where again, the diameters of the bolts are assumed equal. If not, then one replaces F in Eq. (b) with the shear stresses τ = 4 F /π d 2 for each bolt. Solving Eqs. (a) and (b) simultaneously, we obtain M1 r n 2 2 2 r A + r B + rC + · · · Fn = (854) where the subscript n refers to the particular bolt whose load is to be found. These moment loads are also shown as vectors on the loading diagram. In the third step the direct and moment loads are added vectorially to obtain the resultant load on each bolt. Since all the bolts or rivets are usually the same size, only that bolt having the maximum load need be considered. When the maximum load is found, the strength may be determined by using the various methods already described. EXAMPLE 87 Shown in Fig. 828 is a 15- by 200-mm rectangular steel bar cantilevered to a 250-mm steel channel using four tightly tted bolts located at A, B, C, and D. Figure 828 250 Dimensions in millimeters. 10 15 M16 C 2 bolts F = 16 kN B 60 200 O D 60 A 75 75 50 300 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 442 III. Design of Mechanical Elements © The McGrawHill Companies, 2008 8. Screws, Fasteners, and the Design of Nonpermanent Joints Mechanical Engineering Design For a F = 16 kN load nd (a) The resultant load on each bolt (b) The maximum shear stress in each bolt (c) The maximum bearing stress (d) The critical bending stress in the bar Solution (a) Point O, the centroid of the bolt group in Fig. 828, is found by symmetry. If a free-body diagram of the beam were constructed, the shear reaction V would pass through O and the moment reactions M would be about O. These reactions are V = 16 kN M = 16(425) = 6800 N · m In Fig. 829, the bolt group has been drawn to a larger scale and the reactions are shown. The distance from the centroid to the center of each bolt is (60)2 + (75)2 = 96.0 mm r= The primary shear load per bolt is F = V 16 = = 4 kN n 4 Since the secondary shear forces are equal, Eq. (854) becomes F = Mr M 6800 = = = 17.7 kN 2 4r 4r 4(96.0) The primary and secondary shear forces are plotted to scale in Fig. 829 and the resultants obtained by using the parallelogram rule. The magnitudes are found by measurement Figure 829 y FC " FC B C ' FC FB ' rB rC FB " FB O " FD M V rA rD FD D A ' FA ' FD " FA FA x 445 446 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition III. Design of Mechanical Elements 8. Screws, Fasteners, and the Design of Nonpermanent Joints © The McGrawHill Companies, 2008 Screws, Fasteners, and the Design of Nonpermanent Joints 443 (or analysis) to be Answer FA = FB = 21.0 kN Answer FC = FD = 14.8 kN (b) Bolts A and B are critical because they carry the largest shear load. Does this shear act on the threaded portion of the bolt, or on the unthreaded portion? The bolt length will be 25 mm plus the height of the nut plus about 2 mm for a washer. Table A31 gives the nut height as 14.8 mm. Including two threads beyond the nut, this adds up to a length of 43.8 mm, and so a bolt 46 mm long will be needed. From Eq. (814) we compute the thread length as L T = 38 mm. Thus the unthreaded portion of the bolt is 46 38 = 8 mm long. This is less than the 15 mm for the plate in Fig. 828, and so the bolt will tend to shear across its minor diameter. Therefore the shear-stress area is As = 144 mm2, and so the shear stress is Answer τ= F 21.0(10)3 = = 146 MPa As 144 (c) The channel is thinner than the bar, and so the largest bearing stress is due to the pressing of the bolt against the channel web. The bearing area is Ab = td = 10(16) = 160 mm2. Thus the bearing stress is Answer σ = F 21.0(10)3 = = 131 MPa Ab 160 (d ) The critical bending stress in the bar is assumed to occur in a section parallel to the y axis and through bolts A and B. At this section the bending moment is M = 16(300 + 50) = 5600 N · m The second moment of area through this section is obtained by the use of the transfer formula, as follows: ¯ I = Ibar 2( Iholes + d 2 A) = 15(16)3 15(200)3 2 + (60)2 (15)(16) = 8.26(10)6 mm4 12 12 Then Answer σ= Mc 5600(100) (10)3 = 67.8 MPa = I 8.26(10)6 PROBLEMS 81 A power screw is 25 mm in diameter and has a thread pitch of 5 mm. (a) Find the thread depth, the thread width, the mean and root diameters, and the lead, provided square threads are used. (b) Repeat part (a) for Acme threads. BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 444 III. Design of Mechanical Elements © The McGrawHill Companies, 2008 8. Screws, Fasteners, and the Design of Nonpermanent Joints Mechanical Engineering Design 82 Using the information in the footnote of Table 81, show that the tensile-stress area is At = 83 π (d 0.938 194 p)2 4 Show that for zero collar friction the efciency of a square-thread screw is given by the equation e = tan λ 1 f tan λ tan λ + f Plot a curve of the efciency for lead angles up to 45 . Use f = 0.08. 84 A single-threaded 25-mm power screw is 25 mm in diameter with a pitch of 5 mm. A vertical load on the screw reaches a maximum of 6 kN. The coefcients of friction are 0.05 for the collar and 0.08 for the threads. The frictional diameter of the collar is 40 mm. Find the overall efciency and the torque to raise and lower the load. 85 The machine shown in the gure can be used for a tension test but not for a compression test. Why? Can both screws have the same hand? Motor Bearings Worm Spur gears [ Problem 85 Bronze bushings 2 's C.I. Collar bearing B C 2 [ 's Foot A 86 The press shown for Prob. 85 has a rated load of 5000 lbf. The twin screws have Acme threads, 1 a diameter of 3 in, and a pitch of 2 in. Coefcients of friction are 0.05 for the threads and 0.06 for the collar bearings. Collar diameters are 5 in. The gears have an efciency of 95 percent and a speed ratio of 75:1. A slip clutch, on the motor shaft, prevents overloading. The full-load motor speed is 1720 rev/min. (a) When the motor is turned on, how fast will the press head move? (b) What should be the horsepower rating of the motor? 87 3 A screw clamp similar to the one shown in the gure has a handle with diameter 16 in made 7 of cold-drawn AISI 1006 steel. The overall length is 3 in. The screw is 16 in-14 UNC and is 3 3 5 4 in long, overall. Distance A is 2 in. The clamp will accommodate parts up to 4 16 in high. (a) What screw torque will cause the handle to bend permanently? (b) What clamping force will the answer to part (a) cause if the collar friction is neglected and if the thread friction is 0.075? 447 448 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition III. Design of Mechanical Elements 8. Screws, Fasteners, and the Design of Nonpermanent Joints © The McGrawHill Companies, 2008 Screws, Fasteners, and the Design of Nonpermanent Joints 445 (c) What clamping force will cause the screw to buckle? (d ) Are there any other stresses or possible failures to be checked? Problem 87 A B 88 The C clamp shown in the gure for Prob. 87 uses a 5 in-6 Acme thread. The frictional coef8 cients are 0.15 for the threads and for the collar. The collar, which in this case is the anvil 7 strikers swivel joint, has a friction diameter of 16 in. Calculations are to be based on a max3 imum force of 6 lbf applied to the handle at a radius of 2 4 in from the screw centerline. Find the clamping force. 89 Find the power required to drive a 40-mm power screw having double square threads with a pitch of 6 mm. The nut is to move at a velocity of 48 mm/s and move a load of F = 10 kN. The frictional coefcients are 0.10 for the threads and 0.15 for the collar. The frictional diameter of the collar is 60 mm. 810 A single square-thread power screw has an input power of 3 kW at a speed of 1 rev/s. The screw has a diameter of 36 mm and a pitch of 6 mm. The frictional coefcients are 0.14 for the threads and 0.09 for the collar, with a collar friction radius of 45 mm. Find the axial resisting load F and the combined efciency of the screw and collar. 811 A bolted joint is to have a grip consisting of two 1 -in steel plates and one wide 1 -in American 2 2 Standard plain washer to t under the head of the 1 in-13 × 1.75 in UNC hex-head bolt. 2 (a) What is the length of the thread L T for this diameter inch-series bolt? (b) What is the length of the grip l? (c) What is the height H of the nut? (d ) Is the bolt long enough? If not, round to the next larger preferred length (Table A17). (e) What is the length of the shank and threaded portions of the bolt within the grip? These lengths are needed in order to estimate the bolt spring rate kb . 812 A bolted joint is to have a grip consisting of two 14-mm steel plates and one 14R metric plain washer to t under the head of the M14 × 2 hex-head bolt, 50 mm long. (a) What is the length of the thread L T for this diameter metric coarse-pitch series bolt? (b) What is the length of the grip l? (c) What is the height H of the nut? (d ) Is the bolt long enough? If not, round to the next larger preferred length (Table A17). (e) What is the length of the shank and the threaded portions of the bolt within the grip? These lengths are needed in order to estimate bolt spring rate kb . BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 446 III. Design of Mechanical Elements © The McGrawHill Companies, 2008 8. Screws, Fasteners, and the Design of Nonpermanent Joints Mechanical Engineering Design 813 A blanking disk 0.875 in thick is to be fastened to a spool whose ange is 1 in thick, using eight 1 in-13 × 1.75 in hex-head cap screws. 2 (a) What is the length of threads L T for this cap screw? (b) What is the effective length of the grip l ? (c) Is the length of this cap screw sufcient? If not, round up. (d ) Find the shank length ld and the useful thread length lt within the grip. These lengths are needed for the estimate of the fastener spring rate kb . 814 A blanking disk is 20 mm thick and is to be fastened to a spool whose ange is 25 mm thick, using eight M12 × 40 hex-head metric cap screws. (a) What is the length of the threads L T for this fastener? (b) What is the effective grip length l ? (c) Is the length of this fastener sufcient? If not, round to the next preferred length. (d ) Find the shank length ld and the useful threaded length in the grip lt . These lengths are needed in order to estimate the fastener spring rate kb . 815 A 3 in-16 UNF series SAE grade 5 bolt has a 3 -in ID tube 13 in long, clamped between washer 4 4 faces of bolt and nut by turning the nut snug and adding one-third of a turn. The tube OD is the washer-face diameter dw = 1.5d = 1.5(0.75) = 1.125 in = OD. 3 4 in-16 UNF grade 1.125 in Problem 815 13 in (a) What is the spring rate of the bolt and the tube, if the tube is made of steel? What is the joint constant C? (b) When the one-third turn-of-nut is applied, what is the initial tension Fi in the bolt? (c) What is the bolt tension at opening if additional tension is applied to the bolt external to the joint? 816 From your experience with Prob. 815, generalize your solution to develop a turn-of-nut equation Nt = θ = 360 kb + km kb km Fi N where Nt = turn of the nut from snug tight θ = turn of the nut in degrees N = number of thread/in (1/ p where p is pitch) Fi = initial preload kb , km = spring rates of the bolt and members, respectively Use this equation to nd the relation between torque-wrench setting T and turn-of-nut Nt . (Snug tight means the joint has been tightened to perhaps half the intended preload to atten asperities on the washer faces and the members. Then the nut is loosened and retightened 449 450 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition III. Design of Mechanical Elements © The McGrawHill Companies, 2008 8. Screws, Fasteners, and the Design of Nonpermanent Joints Screws, Fasteners, and the Design of Nonpermanent Joints 447 nger tight, and the nut is rotated the number of degrees indicated by the equation. Properly done, the result is competitive with torque wrenching.) 817 RB&W11 recommends turn-of-nut from snug t to preload as follows: 1/3 turn for bolt grips of 14 diameters, 1/2 turn for bolt grips 48 diameters, and 2/3 turn for grips of 812 diameters. These recommendations are for structural steel fabrication (permanent joints), producing preloads of 100 percent of proof strength and beyond. Machinery fabricators with fatigue loadings and possible joint disassembly have much smaller turns-of-nut. The RB&W recommendation enters the nonlinear plastic deformation zone. Position mark on work surface Position mark on nut Problem 817 Turn-of-nut method Position mark on nut Tighten nut to snug fit Addition turn (a) For Ex. 84, use Eq. (827) with K = 0.2 to estimate the torque necessary to establish the desired preload. Then, using the results from Prob. 816, determine the turn of the nut in degrees. How does this compare with the RB&W recommendations? (b) Repeat part (a) for Ex. 85. 818 Take Eq. (822) and express km /( Ed ) as a function of l /d , then compare with Eq. (823) for d / l = 0.5. 819 A joint has the same geometry as Ex. 84, but the lower member is steel. Use Eq. (823) to nd the spring rate of the members in the grip. Hint: Equation (823) applies to the stiffness of two sections of a joint of one material. If each section has the same thickness, then what is the stiffness of one of the sections? 820 The gure illustrates the connection of a cylinder head to a pressure vessel using 10 bolts and a confined-gasket seal. The effective sealing diameter is 150 mm. Other dimensions are: A = 100, B = 200, C = 300, D = 20, and E = 20, all in millimeters. The cylinder is used to store gas at a static pressure of 6 MPa. ISO class 8.8 bolts with a diameter of 12 mm have been selected. This provides an acceptable bolt spacing. What load factor n results from this selection? C B D Problem 820 E Cylinder head is steel; cylinder is grade 30 cast iron. A 11 Russell, Burdsall & Ward, Inc., Metal Forming Spets, Mentor, Ohio. BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 448 III. Design of Mechanical Elements 8. Screws, Fasteners, and the Design of Nonpermanent Joints © The McGrawHill Companies, 2008 Mechanical Engineering Design 821 The computer can be very helpful to the engineer. In matters of analysis it can take the drudgery out of calculations and improve accuracy. In synthesis, good programming is a matter of organizing decisions that must be made, soliciting them while displaying enough information, accepting them, and doing the number crunching. In either case, one cannot program what one does not understand. Understanding comes from experience with problems executed manually. It is useful to program the protocol of Table 87 because it is so easy to make a mistake in longhand. Focusing on the fastener, recognize two situations: (1) the fastener has been chosen, its diameter and length are known, and the designer needs to know all the pertinent dimensions, including the effective grip of a cap-screw joint and whether the length is adequate; and (2) the fastener diameter, nut, and washers are chosen, and the designer has to make the length decision, after which documentation of pertinent dimensions is in order. Code the protocol of Table 87, bearing in mind that you may wish to embed some of it in a larger program. 822 Figure P820 illustrates the connection of a cylinder head to a pressure vessel using 10 bolts and a conned-gasket seal. The effective sealing diameter is 150 mm. Other dimensions are: A = 100, B = 200, C = 300, D = 20, and E = 25, all in millimeters. The cylinder is used to store gas at a static pressure of 6 MPa. ISO class 8.8 bolts with a diameter of 12 mm have been selected. This provides an acceptable bolt spacing. What load factor n results from this selection? 823 We wish to alter the gure for Prob. 822 by decreasing the inside diameter of the seal to the diameter A = 100 mm. This makes an effective sealing diameter of 120 mm. Then, by using cap screws instead of bolts, the bolt circle diameter B can be reduced as well as the outside diameter C. If the same bolt spacing and the same edge distance are used, then eight 12-mm cap screws can be used on a bolt circle with B = 160 mm and an outside diameter of 260 mm, a substantial savings. With these dimensions and all other data the same as in Prob. 822, nd the load factor. 824 In the gure for Prob. 820, the bolts have a diameter of 1 in and the cover plate is steel, with 2 1 D = 2 in. The cylinder is cast iron, with E = 5 in and a modulus of elasticity of 18 Mpsi. 8 The 1 -in SAE washer to be used under the nut has OD = 1.062 in and is 0.095 in thick. Find 2 the stiffnesses of the bolt and the members and the joint constant C. 825 1 The same as Prob. 824, except that 2 -in cap screws are used with washers (see Fig. 821). 826 In addition to the data of Prob. 824, the dimensions of the cylinder are A = 3.5 in and an effective seal diameter of 4.25 in. The internal static pressure is 1500 psi. The outside diameter of the head is C = 8 in. The diameter of the bolt circle is 6 in, and so a bolt spacing in the range of 3 to 5 bolt diameters would require from 8 to 13 bolts. Select 10 SAE grade 5 bolts and nd the resulting load factor n. 827 A 3 -in class 5 cap screw and steel washer are used to secure a cap to a cast-iron frame of a 8 machine having a blind threaded hole. The washer is 0.065 in thick. The frame has a modulus 1 of elasticity of 14 Mpsi and is 4 in thick. The screw is 1 in long. The material in the frame also has a modulus of elasticity of 14 Mpsi. Find the stiffnesses kb and km of the bolt and members. 828 Bolts distributed about a bolt circle are often called upon to resist an external bending moment as shown in the gure. The external moment is 12 kip · in and the bolt circle has a diameter of 8 in. The neutral axis for bending is a diameter of the bolt circle. What needs to be determined is the most severe external load seen by a bolt in the assembly. (a) View the effect of the bolts as placing a line load around the bolt circle whose intensity Fb , in pounds per inch, varies linearly with the distance from the neutral axis according to the relation Fb = Fb ,max R sin θ . The load on any particular bolt can be viewed as the effect of the line load over the arc associated with the bolt. For example, there are 12 bolts shown 451 452 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition III. Design of Mechanical Elements © The McGrawHill Companies, 2008 8. Screws, Fasteners, and the Design of Nonpermanent Joints Screws, Fasteners, and the Design of Nonpermanent Joints 449 in the gure. Thus each bolt load is assumed to be distributed on a 30° arc of the bolt circle. Under these conditions, what is the largest bolt load? (b) View the largest load as the intensity Fb ,max multiplied by the arc length associated with each bolt and nd the largest bolt load. (c) Express the load on any bolt as F = Fmax sin θ , sum the moments due to all the bolts, and estimate the largest bolt load. Compare the results of these three approaches to decide how to attack such problems in the future. R Problem 828 Bolted connection subjected to bending. M M Neutral axis 829 The gure shows a cast-iron bearing block that is to be bolted to a steel ceiling joist and is to support a gravity load. Bolts used are M20 ISO 8.8 with coarse threads and with 3.4-mmthick steel washers under the bolt head and nut. The joist anges are 20 mm in thickness, and the dimension A, shown in the gure, is 20 mm. The modulus of elasticity of the bearing block is 135 GPa. A Problem 829 B d C (a) Find the wrench torque required if the fasteners are lubricated during assembly and the joint is to be permanent. (b) Determine the load factor for the design if the gravity load is 15 kN. 830 The upside-down steel A frame shown in the gure is to be bolted to steel beams on the ceiling of a machine room using ISO grade 8.8 bolts. This frame is to support the 40-kN radial load as illustrated. The total bolt grip is 48 mm, which includes the thickness of the steel beam, the A-frame feet, and the steel washers used. The bolts are size M20 × 2.5. (a) What tightening torque should be used if the connection is permanent and the fasteners are lubricated? (b) What portion of the external load is taken by the bolts? By the members? 831 If the pressure in Prob. 820 is cycling between 0 and 6 MPa, determine the fatigue factor of safety using the: (a) Goodman criterion. (b) Gerber criterion. (c) ASME-elliptic criterion. BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 450 III. Design of Mechanical Elements 8. Screws, Fasteners, and the Design of Nonpermanent Joints © The McGrawHill Companies, 2008 Mechanical Engineering Design Drill 2 holes for M20 × 2.5 bolts Problem 830 W = 40 kN 832 In the gure for Prob. 820, let A = 0.9 m, B = 1 m, C = 1.10 m, D = 20 mm, and E = 25 mm. The cylinder is made of ASTM No. 35 cast iron ( E = 96 GPa), and the head, of low-carbon steel. There are thirty-six M10 × 1.5 ISO 10.9 bolts tightened to 75 percent of proof load. During use, the cylinder pressure uctuates between 0 and 550 kPa. Find the factor of safety guarding against a fatigue failure of a bolt using the: (a) Goodman criterion. (b) Gerber criterion. (c) ASME-elliptic criterion. 833 A 1-in-diameter hot-rolled AISI 1144 steel rod is hot-formed into an eyebolt similar to that shown in the gure for Prob. 374, with an inner 2-in-diameter eye. The threads are 1 in-12 UNF and are die-cut. (a) For a repeatedly applied load collinear with the thread axis, using the Gerber criterion is fatigue failure more likely in the thread or in the eye? (b) What can be done to strengthen the bolt at the weaker location? (c) If the factor of safety guarding against a fatigue failure is n f = 2, what repeatedly applied load can be applied to the eye? 834 The section of the sealed joint shown in the gure is loaded by a repeated force P = 6 kip. The members have E = 16 Mpsi. All bolts have been carefully preloaded to Fi = 25 kip each. 3 4 in-16 UNF SAE grade 5 Problem 834 1 1 2 in No. 40 CI (a) If hardened-steel washers 0.134 in thick are to be used under the head and nut, what length of bolts should be used? (b) Find kb , km , and C. (c) Using the Goodman criterion, nd the factor of safety guarding against a fatigue failure. (d ) Using the Gerber criterion, nd the factor of safety guarding against a fatigue failure. (e) Find the load factor guarding against overproof loading. 453 454 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition III. Design of Mechanical Elements © The McGrawHill Companies, 2008 8. Screws, Fasteners, and the Design of Nonpermanent Joints Screws, Fasteners, and the Design of Nonpermanent Joints 835 451 Suppose the welded steel bracket shown in the gure is bolted underneath a structural-steel ceiling beam to support a uctuating vertical load imposed on it by a pin and yoke. The bolts 1 are 2 in coarse-thread SAE grade 5, tightened to recommended preload. The stiffnesses have already been computed and are kb = 4.94 Mlb/in and km = 15.97 Mlb/in. A C Problem 835 d B (a) Assuming that the bolts, rather than the welds, govern the strength of this design, determine the safe repeated load P that can be imposed on this assembly using the Goodman criterion and a fatigue design factor of 2. (b) Repeat part (a) using the Gerber criterion. (c) Compute the load factors based on the load found in part (b). 836 Using the Gerber fatigue criterion and a fatigue-design factor of 2, determine the external repeated load P that a 1 1 -in SAE grade 5 coarse-thread bolt can take compared with that for 4 a ne-thread bolt. The joint constants are C = 0.30 for coarse- and 0.32 for ne-thread bolts. 837 An M30 × 3.5 ISO 8.8 bolt is used in a joint at recommended preload, and the joint is subject to a repeated tensile fatigue load of P = 80 kN per bolt. The joint constant is C = 0.33. Find the load factors and the factor of safety guarding against a fatigue failure based on the Gerber fatigue criterion. 838 The gure shows a uid-pressure linear actuator (hydraulic cylinder) in which D = 4 in, t = 3 8 in, L = 12 in, and w = 3 in. Both brackets as well as the cylinder are of steel. The actuator 4 3 has been designed for a working pressure of 2000 psi. Six 8 -in SAE grade 5 coarse-thread bolts are used, tightened to 75 percent of proof load. w Problem 838 t L w D (a) Find the stiffnesses of the bolts and members, assuming that the entire cylinder is compressed uniformly and that the end brackets are perfectly rigid. (b) Using the Goodman fatigue criterion, nd the factor of safety guarding against a fatigue failure. (c) Repeat part (b) using the Gerber fatigue criterion. (d ) What pressure would be required to cause total joint separation? BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 452 III. Design of Mechanical Elements © The McGrawHill Companies, 2008 8. Screws, Fasteners, and the Design of Nonpermanent Joints Mechanical Engineering Design 839 The gure shows a bolted lap joint that uses SAE grade 8 bolts. The members are made of cold-drawn AISI 1040 steel. Find the safe tensile shear load F that can be applied to this connection if the following factors of safety are specied: shear of bolts 3, bearing on bolts 2, bearing on members 2.5, and tension of members 3. 5 8 Problem 839 3 8 in 5 in 16 in-16 UNC 1 1 in 8 5 8 in 1 4 1 1 4 in 840 in The bolted connection shown in the gure uses SAE grade 5 bolts. The members are hot-rolled AISI 1018 steel. A tensile shear load F = 4000 lbf is applied to the connection. Find the factor of safety for all possible modes of failure. 5 8 1 5 8 in 3 8 in 5 8 5 8 1 8 in in 1 4 in in-16 UNC in Problem 840 1 4 841 in A bolted lap joint using SAE grade 5 bolts and members made of cold-drawn SAE 1040 steel is shown in the gure. Find the tensile shear load F that can be applied to this connection if the following factors of safety are specied: shear of bolts 1.8, bearing on bolts 2.2, bearing on members 2.4, and tension of members 2.6. 7 8 1 3 4 in in-9 UNC 1 2 in 3 Problem 841 2 4 in 1 1 2 in 3 in 842 3 4 in The bolted connection shown in the gure is subjected to a tensile shear load of 20 kip. The bolts are SAE grade 5 and the material is cold-drawn AISI 1015 steel. Find the factor of safety of the connection for all possible modes of failure. 455 456 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition III. Design of Mechanical Elements © The McGrawHill Companies, 2008 8. Screws, Fasteners, and the Design of Nonpermanent Joints Screws, Fasteners, and the Design of Nonpermanent Joints 3 3 1 8 in Problem 842 3 3 1 8 in 2 8 in 2 8 in 453 3 4 1 3 in 8 5 8 in in-10 UNC 1 3 in 8 3 4 843 The gure shows a connection that employs three SAE grade 5 bolts. The tensile shear load on the joint is 5400 lbf. The members are cold-drawn bars of AISI 1020 steel. Find the factor of safety for each possible mode of failure. 5 8 5 8 Problem 843 1 1 8 in in 3 8 in 5 in 16 in-16 UNC 1 in 5 8 in 3 116 in 5 in 16 2 3 in 8 844 in A beam is made up by bolting together two cold-drawn bars of AISI 1018 steel as a lap joint, as shown in the gure. The bolts used are ISO 5.8. Ignoring any twisting, determine the factor of safety of the connection. y A 2.8 kN Problem 844 200 50 100 350 10 Dimensions in millimeters. x 50 10 A 845 M10 1.5 Section AA Standard design practice, as exhibited by the solutions to Probs. 839 to 843, is to assume that the bolts, or rivets, share the shear equally. For many situations, such an assumption may lead to an unsafe design. Consider the yoke bracket of Prob. 835, for example. Suppose this bracket is bolted to a wide-ange column with the centerline through the two bolts in the vertical direction. A vertical load through the yoke-pin hole at distance B from the column ange would place a shear load on the bolts as well as a tensile load. The tensile load comes about because the bracket tends to pry itself about the bottom corner, much like a claw hammer, exerting a large tensile load on the upper bolt. In addition, it is almost certain that both the spacing of the bolt holes and their diameters will be slightly different on the column ange from what they are on the yoke bracket. Thus, unless yielding occurs, only one of the bolts will take the shear load. The designer has no way of knowing which bolt this will be. BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 454 III. Design of Mechanical Elements © The McGrawHill Companies, 2008 8. Screws, Fasteners, and the Design of Nonpermanent Joints Mechanical Engineering Design In this problem the bracket is 8 in long, A = 1 in, B = 3 in, C = 6 in, and the column 2 1 ange is 2 in thick. The bolts are 1 in UNC SAE 5. Steel washers 0.095 in thick are used 2 under the nuts. The nuts are tightened to 75 percent of proof load. The vertical yoke-pin load is 3000 lbf. If the upper bolt takes all the shear load as well as the tensile load, how closely does the bolt stress approach the proof strength? 846 The bearing of Prob. 829 is bolted to a vertical surface and supports a horizontal shaft. The bolts used have coarse threads and are M20 ISO 5.8. The joint constant is C = 0.30, and the dimensions are A = 20 mm, B = 50 mm, and C = 160 mm. The bearing base is 240 mm long. The bearing load is 12 kN. If the bolts are tightened to 75 percent of proof load, will the bolt stress exceed the proof strength? Use worst-case loading, as discussed in Prob. 845. 847 A split-ring clamp-type shaft collar such as is described in Prob. 531 must resist an axial load of 1000 lbf. Using a design factor of n = 3 and a coefcient of friction of 0.12, specify an SAE Grade 5 cap screw using ne threads. What wrench torque should be used if a lubricated screw is used? 848 A vertical channel 152 × 76 (see Table A7) has a cantilever beam bolted to it as shown. The channel is hot-rolled AISI 1006 steel. The bar is of hot-rolled AISI 1015 steel. The shoulder bolts are M12 × 1.75 ISO 5.8. For a design factor of 2.8, nd the safe force F that can be applied to the cantilever. 12 F Problem 848 Dimensions in millimeters. A 50 849 O 50 B 50 125 Find the total shear load on each of the three bolts for the connection shown in the gure and compute the signicant bolt shear stress and bearing stress. Find the second moment of area of the 8-mm plate on a section through the three bolt holes, and nd the maximum bending stress in the plate. Holes for M12 1.75 bolts 8 mm thick 36 Problem 849 Dimensions in millimeters. 12 kN 32 64 36 200 Column 457 458 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition III. Design of Mechanical Elements © The McGrawHill Companies, 2008 8. Screws, Fasteners, and the Design of Nonpermanent Joints Screws, Fasteners, and the Design of Nonpermanent Joints 850 A 3 - × 2-in AISI 1018 cold-drawn steel bar is cantilevered to support a static load of 300 lbf 8 as illustrated. The bar is secured to the support using two 1 in-13 UNC SAE 5 bolts. Find the 2 factor of safety for the following modes of failure: shear of bolt, bearing on bolt, bearing on member, and strength of member. 3 8 Problem 850 455 1 in in 14 in 3 in 1 in 300 lbf 851 The gure shows a welded tting which has been tentatively designed to be bolted to a channel so as to transfer the 2500-lbf load into the channel. The channel is made of hot-rolled lowcarbon steel having a minimum yield strength of 46 kpsi; the two tting plates are of hot-rolled stock having a minimum Sy of 45.5 kpsi. The tting is to be bolted using six SAE grade 2 shoulder bolts. Check the strength of the design by computing the factor of safety for all possible modes of failure. 6 holes for 5 8 in-11 NC bolts F = 2500 lbf 1 4 in 4 in 1 in Problem 851 2 5 in 1 4 in 8 in [ 11.5 8 in 3 16 in 7 1 in 2 852 A cantilever is to be attached to the at side of a 6-in, 13.0-lbf/in channel used as a column. The cantilever is to carry a load as shown in the gure. To a designer the choice of a bolt array is usually an a priori decision. Such decisions are made from a background of knowledge of the effectiveness of various patterns. 1 2 Problem 852 in steel plate 6 in 6 in 6 in 2000 lbf BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 456 III. Design of Mechanical Elements 8. Screws, Fasteners, and the Design of Nonpermanent Joints © The McGrawHill Companies, 2008 Mechanical Engineering Design (a) If two fasteners are used, should the array be arranged vertically, horizontally, or diagonally? How would you decide? (b) If three fasteners are used, should a linear or triangular array be used? For a triangular array, what should be the orientation of the triangle? How would you decide? 853 Using your experience with Prob. 852, specify a bolt pattern for Prob. 852, and size the bolts. 854 Determining the joint stiffness of nonsymmetric joints of two or more different materials using a frustum of a hollow cone can be time-consuming and prone to error. Develop a computer program to determine km for a joint composed of two different materials of differing thickness. Test the program to determine km for problems such as Ex. 85 and Probs. 819, 820, 822, 824, and 827. 459 460 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition III. Design of Mechanical Elements © The McGrawHill Companies, 2008 9. Welding, Bonding, and the Design of Permanent Joints 9 Welding, Bonding, and the Design of Permanent Joints Chapter Outline 91 Welding Symbols 92 Butt and Fillet Welds 93 Stresses in Welded Joints in Torsion 94 Stresses in Welded Joints in Bending 95 The Strength of Welded Joints 96 Static Loading 97 Fatigue Loading 98 Resistance Welding 99 Adhesive Bonding 458 460 464 469 471 474 478 480 480 457 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 458 III. Design of Mechanical Elements © The McGrawHill Companies, 2008 9. Welding, Bonding, and the Design of Permanent Joints Mechanical Engineering Design Form can more readily pursue function with the help of joining processes such as welding, brazing, soldering, cementing, and gluingprocesses that are used extensively in manufacturing today. Whenever parts have to be assembled or fabricated, there is usually good cause for considering one of these processes in preliminary design work. Particularly when sections to be joined are thin, one of these methods may lead to signicant savings. The elimination of individual fasteners, with their holes and assembly costs, is an important factor. Also, some of the methods allow rapid machine assembly, furthering their attractiveness. Riveted permanent joints were common as the means of fastening rolled steel shapes to one another to form a permanent joint. The childhood fascination of seeing a cherry-red hot rivet thrown with tongs across a building skeleton to be unerringly caught by a person with a conical bucket, to be hammered pneumatically into its nal shape, is all but gone. Two developments relegated riveting to lesser prominence. The rst was the development of high-strength steel bolts whose preload could be controlled. The second was the improvement of welding, competing both in cost and in latitude of possible form. 91 Welding Symbols A weldment is fabricated by welding together a collection of metal shapes, cut to particular congurations. During welding, the several parts are held securely together, often by clamping or jigging. The welds must be precisely specied on working drawings, and this is done by using the welding symbol, shown in Fig. 91, as standardized by the American Welding Society (AWS). The arrow of this symbol points to the joint to be welded. The body of the symbol contains as many of the following elements as are deemed necessary: Reference line Arrow Groove angle; included angle of countersink for plug welds Length of weld Size; size or strength for resistance welds Pitch (center-to-center spacing) of welds F A R Arrow connecting reference line to arrow side of joint, to grooved member, or both Other side Reference line sides) S LP T Specification; process; or other reference Tail (may be omitted when reference is not used) Basic weld symbol or detail reference Arrow side The AWS standard welding symbol showing the location of the symbol elements. Finish symbol Contour symbol Root opening; depth of filling for plug and slot welds (Both Figure 91 (N) Field weld symbol Weld all around symbol Number of spot or projection welds 461 462 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition III. Design of Mechanical Elements © The McGrawHill Companies, 2008 9. Welding, Bonding, and the Design of Permanent Joints Welding, Bonding, and the Design of Permanent Joints 459 Basic weld symbols as in Fig. 92 Dimensions and other data Supplementary symbols Finish symbols Tail Specication or process The arrow side of a joint is the line, side, area, or near member to which the arrow points. The side opposite the arrow side is the other side. Figures 93 to 96 illustrate the types of welds used most frequently by designers. For general machine elements most welds are llet welds, though butt welds are used a great deal in designing pressure vessels. Of course, the parts to be joined must be arranged so that there is sufcient clearance for the welding operation. If unusual joints are required because of insufcient clearance or because of the section shape, the design may be a poor one and the designer should begin again and endeavor to synthesize another solution. Since heat is used in the welding operation, there are metallurgical changes in the parent metal in the vicinity of the weld. Also, residual stresses may be introduced because of clamping or holding or, sometimes, because of the order of welding. Usually these Figure 92 Arc- and gas-weld symbols. Type of weld Bead Fillet Figure 93 Fillet welds. (a) The number indicates the leg size; the arrow should point only to one weld when both sides are the same. (b) The symbol indicates that the welds are intermittent and staggered 60 mm along on 200-mm centers. Plug or slot Groove Square V Bevel 60 5 (b) Figure 94 The circle on the weld symbol indicates that the welding is to go all around. 5 J 200 60200 (a) U BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 460 III. Design of Mechanical Elements © The McGrawHill Companies, 2008 9. Welding, Bonding, and the Design of Permanent Joints Mechanical Engineering Design Figure 95 60° Butt or groove welds: (a) square butt-welded on both sides; (b) single V with 60° bevel and root opening of 2 mm; (c) double V; (d ) single bevel. 2 2 60° (a) (b) 60° 45° (d ) (c) Figure 96 Special groove welds: (a) T joint for thick plates; (b) U and J welds for thick plates; (c) corner weld (may also have a bead weld on inside for greater strength but should not be used for heavy loads); (d) edge weld for sheet metal and light loads. (a) (b) (c) (d ) residual stresses are not severe enough to cause concern; in some cases a light heat treatment after welding has been found helpful in relieving them. When the parts to be welded are thick, a preheating will also be of benet. If the reliability of the component is to be quite high, a testing program should be established to learn what changes or additions to the operations are necessary to ensure the best quality. 92 Butt and Fillet Welds Figure 97a shows a single V-groove weld loaded by the tensile force F. For either tension or compression loading, the average normal stress is F σ= (91) hl where h is the weld throat and l is the length of the weld, as shown in the gure. Note that the value of h does not include the reinforcement. The reinforcement can be desirable, but it varies somewhat and does produce stress concentration at point A in the gure. If fatigue loads exist, it is good practice to grind or machine off the reinforcement. 463 464 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition III. Design of Mechanical Elements © The McGrawHill Companies, 2008 9. Welding, Bonding, and the Design of Permanent Joints Welding, Bonding, and the Design of Permanent Joints Figure 97 Reinforcement 461 Reinforcement A A typical butt joint. l l F F F F Throat h Throat h (a) Tensile loading Figure 98 (b) Shear loading Throat A transverse llet weld. D A h C h F B 2F h F Figure 99 x Free body from Fig. 98. t h Fs Fn F 90 y The average stress in a butt weld due to shear loading (Fig. 97b) is τ= F hl (92) Figure 98 illustrates a typical transverse llet weld. In Fig. 99 a portion of the welded joint has been isolated from Fig. 98 as a free body. At angle θ the forces on each weldment consist of a normal force Fn and a shear force Fs . Summing forces in the x and y directions gives Fs = F sin θ (a) Fn = F cos θ (b) Using the law of sines for the triangle in Fig. 99 yields 2h h t h = = = θ + 45 ) θ) sin 45 sin(90 sin(135 cos θ + sin θ Solving for the throat length t gives t= h cos θ + sin θ (c) BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 462 III. Design of Mechanical Elements 9. Welding, Bonding, and the Design of Permanent Joints 465 © The McGrawHill Companies, 2008 Mechanical Engineering Design The nominal stresses at the angle θ in the weldment, τ and σ , are τ= Fs F sin θ(cos θ + sin θ) F = = (sin θ cos θ + sin2 θ) A hl hl (d ) σ= Fn F cos θ(cos θ + sin θ) F = = (cos2 θ + sin θ cos θ) A hl hl (e) The von Mises stress σ at angle θ is σ = (σ 2 + 3τ 2 )1/2 = F [(cos2 θ + sin θ cos θ)2 + 3(sin2 θ + sin θ cos θ)2 ]1/2 hl (f ) The largest von Mises stress occurs at θ = 62.5 with a value of σ = 2.16 F /(hl ) . The corresponding values of τ and σ are τ = 1.196 F /(hl ) and σ = 0.623 F /(hl ) . The maximum shear stress can be found by differentiating Eq. ( d ) with respect to θ and equating to zero. The stationary point occurs at θ = 67.5 with a corresponding τmax = 1.207 F /(hl ) and σ = 0.5 F /(hl ) . There are some experimental and analytical results that are helpful in evaluating Eqs. ( d ) through ( f ) and consequences. A model of the transverse llet weld of Fig. 98 is easily constructed for photoelastic purposes and has the advantage of a balanced loading condition. Norris constructed such a model and reported the stress distribution along the sides AB and BC of the weld.1 An approximate graph of the results he obtained is shown as Fig. 910a. Note that stress concentration exists at A and B on the horizontal leg and at B on the vertical leg. Norris states that he could not determine the stresses at A and B with any certainty. Salakian2 presents data for the stress distribution across the throat of a llet weld (Fig. 910b). This graph is of particular interest because we have just learned that it is the throat stresses that are used in design. Again, the gure shows stress concentration at point B. Note that Fig. 910a applies either to the weld metal or to the parent metal, and that Fig. 910b applies only to the weld metal. Equations (a) through ( f ) and their consequences seem familiar, and we can become comfortable with them. The net result of photoelastic and nite element analysis of transverse llet weld geometry is more like that shown in Fig. 910 than those given by mechanics of materials or elasticity methods. The most important concept here is that we have no analytical approach that predicts the existing stresses. The geometry of the llet is crude by machinery standards, and even if it were ideal, the macrogeometry is too abrupt and complex for our methods. There are also subtle bending stresses due to eccentricities. Still, in the absence of robust analysis, weldments must be specied and the resulting joints must be safe. The approach has been to use a simple and conservative model, veried by testing as conservative. The approach has been to Consider the external loading to be carried by shear forces on the throat area of the weld. By ignoring the normal stress on the throat, the shearing stresses are inated sufciently to render the model conservative. 1 C. H. Norris, Photoelastic Investigation of Stress Distribution in Transverse Fillet Welds, Welding J., vol. 24, 1945, p. 557s. 2 A. G. Salakian and G. E. Claussen, Stress Distribution in Fillet Welds: A Review of the Literature, Welding J., vol. 16, May 1937, pp. 124. 466 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition III. Design of Mechanical Elements © The McGrawHill Companies, 2008 9. Welding, Bonding, and the Design of Permanent Joints Welding, Bonding, and the Design of Permanent Joints Figure 910 463 C Stress distribution in llet welds: (a) stress distribution on the legs as reported by Norris; (b) distribution of principal stresses and maximum shear stress as reported by Salakian. + 1 D x ma + + A 0 D B B 2 (a) (b) Figure 911 Parallel llet welds. l F h 2F F Use distortion energy for signicant stresses. Circumscribe typical cases by code. For this model, the basis for weld analysis or design employs τ= F 1.414 F = 0.707hl hl (93) which assumes the entire force F is accounted for by a shear stress in the minimum throat area. Note that this inates the maximum estimated shear stress by a factor of 1.414/1.207 = 1.17. Further, consider the parallel llet welds shown in Fig. 911 where, as in Fig. 98, each weld transmits a force F. However, in the case of Fig. 911, the maximum shear stress is at the minimum throat area and corresponds to Eq. (93). Under circumstances of combined loading we Examine primary shear stresses due to external forces. Examine secondary shear stresses due to torsional and bending moments. Estimate the strength(s) of the parent metal(s). Estimate the strength of deposited weld metal. Estimate permissible load(s) for parent metal(s). Estimate permissible load for deposited weld metal. BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 464 III. Design of Mechanical Elements 467 © The McGrawHill Companies, 2008 9. Welding, Bonding, and the Design of Permanent Joints Mechanical Engineering Design 93 Stresses in Welded Joints in Torsion Figure 912 illustrates a cantilever of length l welded to a column by two llet welds. The reaction at the support of a cantilever always consists of a shear force V and a moment M. The shear force produces a primary shear in the welds of magnitude τ = V A (94) where A is the throat area of all the welds. The moment at the support produces secondary shear or torsion of the welds, and this stress is given by the equation τ = Mr J (95) where r is the distance from the centroid of the weld group to the point in the weld of interest and J is the second polar moment of area of the weld group about the centroid of the group. When the sizes of the welds are known, these equations can be solved and the results combined to obtain the maximum shear stress. Note that r is usually the farthest distance from the centroid of the weld group. Figure 913 shows two welds in a group. The rectangles represent the throat areas of the welds. Weld 1 has a throat width b1 = 0.707h 1 , and weld 2 has a throat width d2 = 0.707h 2 . Note that h 1 and h 2 are the respective weld sizes. The throat area of both welds together is A = A1 + A2 = b1 d1 + b2 d2 (a) This is the area that is to be used in Eq. (94). The x axis in Fig. 913 passes through the centroid G 1 of weld 1. The second moment of area about this axis is Ix = 3 b1 d1 12 Similarly, the second moment of area about an axis through G 1 parallel to the y axis is Iy = 3 d1 b1 12 Figure 912 This is a moment connection; such a connection produces torsion in the welds. F O r ro O l 468 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition III. Design of Mechanical Elements © The McGrawHill Companies, 2008 9. Welding, Bonding, and the Design of Permanent Joints Welding, Bonding, and the Design of Permanent Joints Figure 913 465 y x2 b2 2 G2 1 G r2 x G1 O x1 y2 y r1 d1 d2 M b1 x Thus the second polar moment of area of weld 1 about its own centroid is JG 1 = Ix + I y = 3 3 b1 d1 d1 b1 + 12 12 (b) In a similar manner, the second polar moment of area of weld 2 about its centroid is JG 2 = 3 3 b2 d2 d2 b2 + 12 12 (c) The centroid G of the weld group is located at x= ¯ A1 x 1 + A2 x 2 A y= ¯ A1 y1 + A2 y2 A Using Fig. 913 again, we see that the distances r1 and r2 from G 1 and G 2 to G, respectively, are r1 = [(x x1 )2 + y 2 ]1/2 ¯ ¯ r2 = [( y2 y )2 + (x2 x )2 ]1/2 ¯ ¯ Now, using the parallel-axis theorem, we nd the second polar moment of area of the weld group to be 2 2 J = JG 1 + A1r1 + JG 2 + A2r2 (d ) This is the quantity to be used in Eq. (95). The distance r must be measured from G and the moment M computed about G. The reverse procedure is that in which the allowable shear stress is given and we wish to nd the weld size. The usual procedure is to estimate a probable weld size and then to use iteration. 3 3 Observe in Eqs. (b) and (c) the quantities b1 and d2 , respectively, which are the cubes of the weld widths. These quantities are small and can be neglected. This leaves 3 3 the terms b1 d1 /12 and d2 b2 /12, which make JG 1 and JG 2 linear in the weld width. Setting the weld widths b1 and d2 to unity leads to the idea of treating each llet weld as a line. The resulting second moment of area is then a unit second polar moment of area. The advantage of treating the weld size as a line is that the value of Ju is the same regardless of the weld size. Since the throat width of a llet weld is 0.707h , the relationship between J and the unit value is J = 0.707h Ju (96) BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 466 III. Design of Mechanical Elements © The McGrawHill Companies, 2008 9. Welding, Bonding, and the Design of Permanent Joints Mechanical Engineering Design in which Ju is found by conventional methods for an area having unit width. The transfer formula for Ju must be employed when the welds occur in groups, as in Fig. 912. Table 91 lists the throat areas and the unit second polar moments of area for the most common llet welds encountered. The example that follows is typical of the calculations normally made. Table 91 Torsional Properties of Fillet Welds* Weld Throat Area Ju d 3 /12 Ju = d (3b2 + d 2 ) 6 b2 2(b + d ) Ju = d2 2(b + d ) (b + d )4 6b 2 d 2 12(b + d ) ¯ y= A G Unit Second Polar Moment of Area Location of G ¯ x= b2 2b + d Ju = 8b3 + 6bd 2 + d 3 b4 12 2b + d Ju = (b + d )3 6 Ju 2π r3 ¯ x 0.70 hd 0 ¯ y = d /2 d y b A ¯ x = b/2 1.41 hd ¯ y = d /2 d G y x b A 0.707h(2b d) d G y ¯ x= x b A 0.707h(2b d) ¯ y = d /2 d G y x b A G 1.414h(b d) ¯ x = b/2 ¯ y = d /2 d y x A r 1.414 π hr G *G is centroid of weld group; h is weld size; plane of torque couple is in the plane of the paper; all welds are of unit width. 469 470 BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition III. Design of Mechanical Elements © The McGrawHill Companies, 2008 9. Welding, Bonding, and the Design of Permanent Joints Welding, Bonding, and the Design of Permanent Joints 467 EXAMPLE 91 A 50-kN load is transferred from a welded tting into a 200-mm steel channel as illustrated in Fig. 914. Estimate the maximum stress in the weld. Solution3 (a) Label the ends and corners of each weld by letter. Sometimes it is desirable to label each weld of a set by number. See Fig. 915. (b) Estimate the primary shear stress τ . As shown in Fig. 914, each plate is welded to the channel by means of three 6-mm llet welds. Figure 915 shows that we have divided the load in half and are considering only a single plate. From case 4 of Table 91 we nd the throat area as A = 0.707(6)[2(56) + 190] = 1280 mm2 Then the primary shear stress is V 25(10)3 = = 19.5 MPa A 1280 τ = (c) Draw the τ stress, to scale, at each lettered corner or end. See Fig. 916. (d ) Locate the centroid of the weld pattern. Using case 4 of Table 91, we nd x= ¯ (56)2 = 10.4 mm 2(56) + 190 This is shown as point O on Figs. 915 and 916. Figure 914 6 200 6 Dimensions in millimeters. 50 kN 6 100 6 56 200-mm 190 6 Figure 915 Diagram showing the weld geometry; all dimensions in millimeters. Note that V and M represent loads applied by the welds to the plate. 25 kN 100 110.4 C D V 56 y O 45.6 M B A 95 x 3 We are indebted to Professor George Piotrowski of the University of Florida for the detailed steps, presented here, of his method of weld analysis R.G.B, J.K.N. BudynasNisbett: Shigleys Mechanical Engineering Design, Eighth Edition 468 III. Design of Mechanical Elements © The McGrawHill Companies, 2008 9. Welding, Bonding, and the Design of Permanent Joints Mechanical Engineering Design Figure 916 F D D Free-body diagram of one of the side plates. C A D C rC D A rD O rA rB B A B A C C B B (e) Find the distances ri (see Fig. 916): r A = r B = [(190/2)2 + (56 10.4)2 ]1/2 = 105 mm rC = r D = [(190/2)2 + (10.4)2 ]1/2 = 95.6 mm These distances can also be scaled from the drawing. ( f ) Find J. Using case 4 of Table 91 again, we get J = 0.707(6) (g) Find M: 8(56)3 + 6(56)(190)2 + (190)3 (56)4 12 2(56) + 190 = 7.07(10)6 mm4 M = Fl = 25(100 + 10.4) = 2760 N · m (h) Estimate the secondary shear stresses τ at each lettered end or corner: τ A = τB = Mr 2760(10)3 (105) = 41.0 MPa = J 7.07(10)6 τC = τ D = 2760(10)3 (95.6) = 37.3 MPa 7.07(10)6 (i) Draw the τ stress, to scale, at each corner and end. See Fig. 916. Note that this is a freebody diagram of one of the side plates, and therefore the τ and τ stresses represent what the channel is doing to the plate (through the welds) to hold the plate in equilibrium. ( j) At each letter, combine the two stress components as vectors. This gives τ A = τ B = 37 MPa τC = τ D = 44 MPa (k) Identify the most highly stressed point: Answer τmax = τC = τ D = 44 MPa 471 472 BudynasNisbett: Shigleys Mechanical Engineering Design, E