Mecanica dos Materiais Hibbeler 8a Edicao - Livro Completo

MECHANICS OF MATERIALS EIGHTH EDITION R C HIBBELER MECHANICS OF MATERIALS This page intentionally left blank MECHANICS OF MATERIALS EIGHTH EDITION R C HIBBELER Prentice Hall Vice President and Editorial Director ECS Marcia Horton Senior Acquisitions Editor Tacy Quinn Editorial Assistant Coleen McDonald Executive Marketing Manager Tim Galligan Senior Managing Editor Scott Disanno Project Manager Rose Kernan Senior Operations Supervisor Alan Fischer Operations Specialist Lisa McDowell Art Director Kenny Beck Text and Cover Designer Kenny Beck Photo Researcher Marta Samsel Cover Images High rise crane Martin MetteShutterstock close up of crane with heavy load Mack7777Shutterstock close up of hoisting rig and telescopic arm of mobile crane 36clicksShutterstock Media Director Daniel Sandin Credits and acknowledgments borrowed from other sources and reproduced with permission in this textbook appear on appropriate page within text or on page xvii Copyright 2011 2008 2005 2003 2001 by R C Hibbeler Published by Pearson Prentice Hall All rights reserved Manufactured in the United States of America This publication is protected by Copyright and permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise To obtain permissions to use material from this work please submit a written request to Pearson Education Inc Permissions Department 1 Lake Street Upper Saddle River NJ 07458 Many of the designations by manufacturers and seller to distinguish their products are claimed as trademarks Where those designations appear in this book and the publisher was aware of a trademark claim the designations have been printed in initial caps or all caps 10 9 8 7 6 5 4 3 2 1 ISBN 10 0136022308 ISBN 13 9780136022305 To the Student With the hope that this work will stimulate an interest in Engineering Mechanics and provide an acceptable guide to its understanding This page intentionally left blank It is intended that this book provide the student with a clear and thorough presentation of the theory and application of the principles of mechanics of materials To achieve this objective over the years this work has been shaped by the comments and suggestions of hundreds of reviewers in the teaching profession as well as many of the authors students The eighth edition has been significantly enhanced from the previous edition and it is hoped that both the instructor and student will benefit greatly from these improvements New to This Edition Updated Content Some portions of the text have been rewritten in order to enhance clarity and be more succinct In this regard some new examples have been added and others have been modified to provide more emphasis on the application of important concepts Also the artwork has been improved throughout the book to support these changes New Photos The relevance of knowing the subject matter is reflected by the realworld applications depicted in over 44 new or updated photos placed throughout the book These photos generally are used to explain how the relevant principles apply to realworld situations and how materials behave under load Fundamental Problems These problem sets are located just after each group of example problems They offer students simple applications of the concepts covered in each section and therefore provide them with the chance to develop their problemsolving skills before attempting to solve any of the standard problems that follow The fundamental problems may be considered as extended examples since the key equations and answers are all listed in the back of the book Additionally when assigned these problems offer students an excellent means of preparing for exams and they can be used at a later time as a review when studying for the Fundamentals of Engineering Exam Conceptual Problems Throughout the text usually at the end of each chapter there is a set of problems that involve conceptual situations related to the application of the principles contained in the chapter These analysis and design problems are intended to engage the students in thinking through a reallife situation as depicted in a photo They can be assigned after the students have developed some expertise in the subject matter and they work well either for individual or team projects New Problems There are approximately 35 or about 550 new problems added to this edition which involve applications to many different fields of engineering Also this new edition now has approximately 134 more problems than in the previous edition PREFACE viii PREFACE Problems with Hints With the additional homework problems in this new edition every problem indicated with a bullet before the problem number includes a suggestion key equation or additional numerical result that is given along with the answer in the back of the book These problems further encourage students to solve problems on their own by providing them with additional checks to the solution Contents The subject matter is organized into 14 chapters Chapter 1 begins with a review of the important concepts of statics followed by a formal definition of both normal and shear stress and a discussion of normal stress in axially loaded members and average shear stress caused by direct shear In Chapter 2 normal and shear strain are defined and in Chapter 3 a discussion of some of the important mechanical properties of materials is given Separate treatments of axial load torsion and bending are presented in Chapters 4 5 and 6 respectively In each of these chapters both linearelastic and plastic behavior of the material are considered Also topics related to stress concentrations and residual stress are included Transverse shear is discussed in Chapter 7 along with a discussion of thinwalled tubes shear flow and the shear center Chapter 8 includes a discussion of thinwalled pressure vessels and provides a partial review of the material covered in the previous chapters such that the state of stress results from combined loadings In Chapter 9 the concepts for transforming multiaxial states of stress are presented In a similar manner Chapter 10 discusses the methods for strain transformation including the application of various theories of failure Chapter 11 provides a means for a further summary and review of previous material by covering design applications of beams and shafts In Chapter 12 various methods for computing deflections of beams and shafts are covered Also included is a discussion for finding the reactions on these members if they are statically indeterminate Chapter 13 provides a discussion of column buckling and lastly in Chapter 14 the problem of impact and the application of various energy methods for computing deflections are considered Sections of the book that contain more advanced material are indicated by a star Time permitting some of these topics may be included in the course Furthermore this material provides a suitable reference for basic principles when it is covered in other courses and it can be used as a basis for assigning special projects Alternative Method of Coverage Some instructors prefer to cover stress and strain transformations first before discussing specific applications of axial load torsion bending and shear One possible method for doing this would be first to cover stress and its transformation Chapter 1 and Chapter 9 followed by strain and its transformation Chapter 2 and the first part of Chapter 10 The discussion and example problems in these later chapters have been PREFACE ix styled so that this is possible Also the problem sets have been subdivided so that this material can be covered without prior knowledge of the intervening chapters Chapters 3 through 8 can then be covered with no loss in continuity Hallmark Elements Organization and Approach The contents of each chapter are organized into welldefined sections that contain an explanation of specific topics illustrative example problems and a set of homework problems The topics within each section are placed into subgroups defined by titles The purpose of this is to present a structured method for introducing each new definition or concept and to make the book convenient for later reference and review Chapter Contents Each chapter begins with a fullpage illustration that indicates a broadrange application of the material within the chapter The Chapter Objectives are then provided to give a general overview of the material that will be covered Procedures for Analysis Found after many of the sections of the book this unique feature provides the student with a logical and orderly method to follow when applying the theory The example problems are solved using this outlined method in order to clarify its numerical application It is to be understood however that once the relevant principles have been mastered and enough confidence and judgment have been obtained the student can then develop his or her own procedures for solving problems Photographs Many photographs are used throughout the book to enhance conceptual understanding and explain how the principles of mechanics of materials apply to realworld situations Important Points This feature provides a review or summary of the most important concepts in a section and highlights the most significant points that should be realized when applying the theory to solve problems Example Problems All the example problems are presented in a concise manner and in a style that is easy to understand Homework Problems Numerous problems in the book depict realistic situations encountered in engineering practice It is hoped that this realism will both stimulate the students interest in the subject and provide a means for developing the skill to reduce any such problem from its physical description to a model or a symbolic representation to which principles may be applied Throughout the book there is an approximate balance of problems using either SI or FPS units Furthermore in any set an attempt has been made to arrange the problems in order of increasing difficulty The answers to all but every fourth problem are listed in the back of the book To alert the user to a x PREFACE problem without a reported answer an asterisk is placed before the problem number Answers are reported to three significant figures even though the data for material properties may be known with less accuracy Although this might appear to be a poor practice it is done simply to be consistent and to allow the student a better chance to validate his or her solution A solid square is used to identify problems that require a numerical analysis or a computer application Appendices The appendices of the book provide a source for review and a listing of tabular data Appendix A provides information on the centroid and the moment of inertia of an area Appendices B and C list tabular data for structural shapes and the deflection and slopes of various types of beams and shafts Accuracy Checking The Eighth Edition has undergone our rigorous Triple Accuracy Checking review In addition to the authors review of all art pieces and pages the text was checked by the following individuals Scott Hendricks Virginia Polytechnic University Karim Nohra University of South Florida Kurt Norlin Laurel Tech Integrated Publishing Services Kai Beng Yap Engineering Consultant Acknowledgments Over the years this text has been shaped by the suggestions and comments of many of my colleagues in the teaching profession Their encouragement and willingness to provide constructive criticism are very much appreciated and it is hoped that they will accept this anonymous recognition A note of thanks is given to the reviewers Akthem AlManaseer San Jose State University Yabin Liao Arizona State University Cliff Lissenden Penn State Gregory M Odegard Michigan Technological University John Oyler University of Pittsburgh Roy Xu Vanderbilt University Paul Ziehl University of South Carolina There are a few people that I feel deserve particular recognition A long time friend and associate Kai Beng Yap was of great help to me in checking the entire manuscript and helping to prepare the problem solutions A special note of thanks also goes to Kurt Norlin of Laurel Tech Integrated Publishing Services in this regard During the production process I am thankful for the assistance of Rose Kernan my production editor for many years and to my wife Conny and daughter PREFACE xi Mary Ann for their help in proofreading and typing that was needed to prepare the manuscript for publication I would also like to thank all my students who have used the previous edition and have made comments to improve its contents I would greatly appreciate hearing from you if at any time you have any comments or suggestions regarding the contents of this edition Russell Charles Hibbeler hibbelerbellsouthnet xii PREFACE Resources for Instructors Instructors Solutions Manual An instructors solutions manual was prepared by the author The manual includes homework assignment lists and was also checked as part of the accuracy checking program Presentation Resources All art from the text is available in PowerPoint slide and JPEG format These files are available for download from the Instructor Resource Center at httpwww pearsonhighered com If you are in need of a login and password for this site please contact your local Pearson Prentice Hall representative Video Solutions Developed by Professor Edward Berger University of Virginia video solutions located on the Companion Website offer stepbystep solution walkthroughs of representative homework problems from each section of the text Make efficient use of class time and office hours by showing students the complete and concise problem solving approaches that they can access anytime and view at their own pace The videos are designed to be a flexible resource to be used however each instructor and student prefers A valuable tutorial resource the videos are also helpful for student selfevaluation as students can pause the videos to check their understanding and work alongside the video Access the videos at httpwww pearsonhigheredcomhibbeler and follow the links for the Mechanics of Materials text Resources for Students Companion WebsiteThe Companion Website located at httpwwwpearsonhigheredcomhibbeler includes opportunities for practice and review including Video SolutionsComplete stepbystep solution walkthroughs of representative homework problems from each section Videos offer Fully Worked SolutionsShowing every step of representative homework problems to help students make vital connections between concepts SelfPaced InstructionStudents can navigate each problem and select play rewind fastforward stop and jumptosections within each problems solution 247 AccessHelp whenever students need it with over 20 hours of helpful review An access code for the Mechanics of Materials Eighth Edition website was included with this text To redeem the code and gain access to the site go to httpwwwpearsonhigheredcomhibbeler and follow the directions on the access code card Access can also be purchased directly from the site CONTENTS 1 Stress 3 Chapter Objectives 3 11 Introduction 3 12 Equilibrium of a Deformable Body 4 13 Stress 22 14 Average Normal Stress in an Axially Loaded Bar 24 15 Average Shear Stress 32 16 Allowable Stress 46 17 Design of Simple Connections 47 2 Strain 65 Chapter Objectives 65 21 Deformation 65 22 Strain 66 3 Mechanical Properties of Materials 81 Chapter Objectives 81 31 The Tension and Compression Test 81 32 The StressStrain Diagram 83 33 StressStrain Behavior of Ductile and Brittle Materials 87 34 Hookes Law 90 35 Strain Energy 92 36 Poissons Ratio 102 37 The Shear StressStrain Diagram 104 38 Failure of Materials Due to Creep and Fatigue 107 4 Axial Load 119 Chapter Objectives 119 41 SaintVenants Principle 119 42 Elastic Deformation of an Axially Loaded Member 122 43 Principle of Superposition 136 44 Statically Indeterminate Axially Loaded Member 137 45 The Force Method of Analysis for Axially Loaded Members 143 46 Thermal Stress 151 47 Stress Concentrations 158 48 Inelastic Axial Deformation 162 49 Residual Stress 164 5 Torsion 179 Chapter Objectives 179 51 Torsional Deformation of a Circular Shaft 179 52 The Torsion Formula 182 53 Power Transmission 190 54 Angle of Twist 200 55 Statically Indeterminate TorqueLoaded Members 214 56 Solid Noncircular Shafts 221 57 ThinWalled Tubes Having Closed Cross Sections 224 58 Stress Concentration 234 59 Inelastic Torsion 237 510 Residual Stress 239 9 Stress Transformation 437 Chapter Objectives 437 91 PlaneStress Transformation 437 92 General Equations of PlaneStress Transformation 442 93 Principal Stresses and Maximum InPlane Shear Stress 445 94 Mohrs CirclePlane Stress 461 95 Absolute Maximum Shear Stress 473 10 Strain Transformation 485 Chapter Objectives 485 101 Plane Strain 485 102 General Equations of PlaneStrain Transformation 486 103 Mohrs CirclePlane Strain 494 104 Absolute Maximum Shear Strain 502 105 Strain Rosettes 504 106 MaterialProperty Relationships 508 107 Theories of Failure 520 11 Design of Beams and Shafts 537 Chapter Objectives 537 111 Basis for Beam Design 537 112 Prismatic Beam Design 540 113 Fully Stressed Beams 554 114 Shaft Design 558 xiv CONTENTS 6 Bending 255 Chapter Objectives 255 61 Shear and Moment Diagrams 255 62 Graphical Method for Constructing Shear and Moment Diagrams 262 63 Bending Deformation of a Straight Member 281 64 The Flexure Formula 285 65 Unsymmetric Bending 302 66 Composite Beams 312 67 Reinforced Concrete Beams 315 68 Curved Beams 319 69 Stress Concentrations 326 610 Inelastic Bending 335 7 Transverse Shear 359 Chapter Objectives 359 71 Shear in Straight Members 359 72 The Shear Formula 361 73 Shear Flow in BuiltUp Members 378 74 Shear Flow in ThinWalled Members 387 75 Shear Center For Open ThinWalled Members 392 8 Combined Loadings 405 Chapter Objectives 405 81 ThinWalled Pressure Vessels 405 82 State of Stress Caused by Combined Loadings 412 CONTENTS xv 14 Energy Methods 715 Chapter Objectives 715 141 External Work and Strain Energy 715 142 Elastic Strain Energy for Various Types of Loading 720 143 Conservation of Energy 733 144 Impact Loading 740 145 Principle of Virtual Work 751 146 Method of Virtual Forces Applied to Trusses 755 147 Method of Virtual Forces Applied to Beams 762 148 Castiglianos Theorem 771 149 Castiglianos Theorem Applied to Trusses 773 1410 Castiglianos Theorem Applied to Beams 776 Appendices A Geometric Properties of an Area 784 A1 Centroid of an Area 784 A2 Moment of Inertia for an Area 787 A3 Product of Inertia for an Area 791 A4 Moments of Inertia for an Area about Inclined Axes 794 A5 Mohrs Circle for Moments of Inertia 797 B Geometric Properties of Structural Shapes 800 C Slopes and Deflections of Beams 808 Fundamental Problems Partial Solutions and Answers 810 Answers to Selected Problems 828 Index 854 12 Deflection of Beams and Shafts 569 Chapter Objectives 569 121 The Elastic Curve 569 122 Slope and Displacement by Integration 573 123 Discontinuity Functions 593 124 Slope and Displacement by the MomentArea Method 604 125 Method of Superposition 619 126 Statically Indeterminate Beams and Shafts 627 127 Statically Indeterminate Beams and ShaftsMethod of Integration 628 128 Statically Indeterminate Beams and ShaftsMomentArea Method 633 129 Statically Indeterminate Beams and ShaftsMethod of Superposition 639 13 Buckling of Columns 657 Chapter Objectives 657 131 Critical Load 657 132 Ideal Column with Pin Supports 660 133 Columns Having Various Types of Supports 666 134 The Secant Formula 678 135 Inelastic Buckling 684 136 Design of Columns for Concentric Loading 692 137 Design of Columns for Eccentric Loading 703 This page intentionally left blank CREDITS Chapter 1 Close up of iron girders Jack SullivanAlamy Images Chapter 2 Photoelastic phenomena tension in a screw mount Alfred PasiekaAlamy Images Chapter 3 A woman stands near a collapsed bridge in one of the worst earthquakehit areas of Yingxiu town in Wenchuan county in Chinas southwestern province of Sichuan on June 2 2008 UN Secretary of State Condoleezza Rice on June 29 met children made homeless by the devastating earthquake that hit southwest China last month and praised the countrys response to the disaster LIU JINStringerGetty Images IncAFP Chapter 3 text Cup and cone steelAlamy Images Chapter 4 Rotary bit on portable oil drilling rig Lowell Georgia CORBIS All Rights Reserved Chapter 5 Steam rising from soils and blurred spinning hollow stem augerAlamy Images Chapter 6 Steel framework at construction site Corbis RF Chapter 7 Train wheels on track Jill StephensonAlamy Images Chapter 7 text Highway flyover Gari Wyn WilliamsAlamy Images Chapter 8 Ski lift with snow covered mountain in background Shutterstock Chapter 9 Turbine blades Chris PearsallAlamy Images Chapter 10 Complex stresses developed within an airplane wing Courtesy of Measurements Group Inc Raleigh North Carolina 27611 USA Chapter 11 Metal frame and yellow crane Stephen FinnAlamy Images Chapter 12 Man pole vaulting in desert Patrick GiardinoCORBIS All Rights Reserved Chapter 13 Water storage tower John DoradoShutterstock Chapter 14 Shot of jackuppiledriver and floating crane John MacCooeyAlamy Images Other images provided by the author This page intentionally left blank MECHANICS OF MATERIALS The bolts used for the connections of this steel framework are subjected to stress In this chapter we will discuss how engineers design these connections and their fasteners 1 3 CHAPTER OBJECTIVES In this chapter we will review some of the important principles of statics and show how they are used to determine the internal resultant loadings in a body Afterwards the concepts of normal and shear stress will be introduced and specific applications of the analysis and design of members subjected to an axial load or direct shear will be discussed 11 Introduction Mechanics of materials is a branch of mechanics that studies the internal effects of stress and strain in a solid body that is subjected to an external loading Stress is associated with the strength of the material from which the body is made while strain is a measure of the deformation of the body In addition to this mechanics of materials includes the study of the bodys stability when a body such as a column is subjected to compressive loading A thorough understanding of the fundamentals of this subject is of vital importance because many of the formulas and rules of design cited in engineering codes are based upon the principles of this subject Stress Historical Development The origin of mechanics of materials dates back to the beginning of the seventeenth century when Galileo performed experiments to study the effects of loads on rods and beams made of various materials However at the beginning of the eighteenth century experimental methods for testing materials were vastly improved and at that time many experimental and theoretical studies in this subject were undertaken primarily in France by such notables as SaintVenant Poisson Lamé and Navier Over the years after many of the fundamental problems of mechanics of materials had been solved it became necessary to use advanced mathematical and computer techniques to solve more complex problems As a resultthis subject expanded into other areas of mechanicssuch as the theory of elasticity and the theory of plasticity Research in these fields is ongoing in order to meet the demands for solving more advanced problems in engineering 12 Equilibrium of a Deformable Body Since statics has an important role in both the development and application of mechanics of materials it is very important to have a good grasp of its fundamentals For this reason we will review some of the main principles of statics that will be used throughout the text External Loads A body is subjected to only two types of external loads namely surface forces or body forces Fig 11 Surface Forces Surface forces are caused by the direct contact of one body with the surface of another In all cases these forces are distributed over the area of contact between the bodies If this area is small in comparison with the total surface area of the body then the surface force can be idealized as a single concentrated force which is applied to a point on the body For example the force of the ground on the wheels of a bicycle can be considered as a concentrated forceIf the surface loading is applied along a narrow strip of area the loading can be idealized as a linear distributed load ws Here the loading is measured as having an intensity of forcelength along the strip and is represented graphically by a series of arrows along the line s The resultant force of ws is equivalent to the area under the distributed loading curve and this resultant acts through the centroid C or geometric center of this area The loading along the length of a beam is a typical example of where this idealization is often applied FR 4 CHAPTER 1 STRESS 1 Fig 11 ws Concentrated force idealization Linear distributed load Surface force Body force s C G FR W Body Forces Abody forceisdevelopedwhenonebodyexertsaforceon another body without direct physical contact between the bodiesExamples include the effects caused by the earths gravitation or its electromagnetic fieldAlthough body forces affect each of the particles composing the body these forces are normally represented by a single concentrated force acting on the body In the case of gravitation this force is called the weight of the body and acts through the bodys center of gravity Support Reactions The surface forces that develop at the supports or points of contact between bodies are called reactions For two dimensional problems ie bodies subjected to coplanar force systems the supports most commonly encountered are shown in Table 11 Note carefully the symbol used to represent each support and the type of reactions it exerts on its contacting member As a general rule if the support prevents translation in a given direction then a force must be developed on the member in that direction Likewise if rotation is prevented a couple moment must be exerted on the member For example the roller support only prevents translation perpendicular or normal to the surface Hence the roller exerts a normal force F on the member at its point of contact Since the member can freely rotate about the roller a couple moment cannot be developed on the member 12 EQUILIBRIUM OF A DEFORMABLE BODY 5 1 F F Type of connection Reaction Cable Roller One unknown F One unknown F F Smooth support One unknown F External pin Internal pin Fx Fy Fx Fy Two unknowns Fx Fy Fx Fy M Fixed support Three unknowns Fx Fy M Two unknowns Fx Fy Type of connection Reaction u u u Many machine elements are pin connected in order to enable free rotation at their connectionsThese supports exert a force on a member but no moment TABLE 11 6 CHAPTER 1 STRESS Equations of Equilibrium Equilibrium of a body requires both a balance of forces to prevent the body from translating or having accelerated motion along a straight or curved path and a balance of moments to prevent the body from rotating These conditions can be expressed mathematically by two vector equations 11 Here represents the sum of all the forces acting on the body and is the sum of the moments of all the forces about any point O either on or off the body If an x y z coordinate system is established with the origin at point O the force and moment vectors can be resolved into components along each coordinate axis and the above two equations can be written in scalar form as six equations namely 12 Often in engineering practice the loading on a body can be represented as a system of coplanar forces If this is the case and the forces lie in the xy plane then the conditions for equilibrium of the body can be specified with only three scalar equilibrium equations that is 13 Here all the moments are summed about point O and so they will be directed along the z axis Successful application of the equations of equilibrium requires complete specification of all the known and unknown forces that act on the body and so the best way to account for all these forces is to draw the bodys freebody diagram Fx 0 Fy 0 MO 0 Fx 0 Fy 0 Fz 0 Mx 0 My 0 Mz 0 MO F F 0 MO 0 1 In order to design the horizontal members of this building frame it is first necessary to find the internal loadings at various points along their length 12 EQUILIBRIUM OF A DEFORMABLE BODY 7 Internal Resultant Loadings In mechanics of materials statics is primarily used to determine the resultant loadings that act within a body For example consider the body shown in Fig 12a which is held in equilibrium by the four external forces In order to obtain the internal loadings acting on a specific region within the body it is necessary to pass an imaginary section or cut through the region where the internal loadings are to be determined The two parts of the body are then separated and a freebody diagram of one of the parts is drawn Fig 12b Notice that there is actually a distribution of internal force acting on the exposed area of the section These forces represent the effects of the material of the top part of the body acting on the adjacent material of the bottom part Although the exact distribution of this internal loading may be unknown we can use the equations of equilibrium to relate the external forces on the bottom part of the body to the distributions resultant force and moment and at any specific point O on the sectioned area Fig 12c It will be shown in later portions of the text that point O is most often chosen at the centroid of the sectioned area and so we will always choose this location for O unless otherwise statedAlso if a member is long and slender as in the case of a rod or beam the section to be considered is generally taken perpendicular to the longitudinal axis of the member This section is referred to as the cross section MRO FR 1 The bodys weight is not shown since it is assumed to be quite small and therefore negligible compared with the other loads Fig 12 section F4 F2 a F1 F3 F1 F2 b FR F1 F2 O MRO c 8 CHAPTER 1 STRESS Three Dimensions Later in this text we will show how to relate the resultant loadings and to the distribution of force on the sectioned area and thereby develop equations that can be used for analysis and design To do this however the components of and acting both normal and perpendicular to the sectioned area must be considered Fig 12d Four different types of resultant loadings can then be defined as follows Normal force N This force acts perpendicular to the area It is developed whenever the external loads tend to push or pull on the two segments of the body Shear force V The shear force lies in the plane of the area and it is developed when the external loads tend to cause the two segments of the body to slide over one another Torsional moment or torque T This effect is developed when the external loads tend to twist one segment of the body with respect to the other about an axis perpendicular to the area Bending moment M The bending moment is caused by the external loads that tend to bend the body about an axis lying within the plane of the area In this text note that graphical representation of a moment or torque is shown in three dimensions as a vector with an associated curlBy the right hand rule the thumb gives the arrowhead sense of this vector and the fingers or curl indicate the tendency for rotation twisting or bending MRO FR MRO FR 1 d O F1 F2 N T M V Torsional Moment Bending Moment Shear Force MRO FR Normal Force O c MRO F1 F2 FR Fig 12 cont 12 EQUILIBRIUM OF A DEFORMABLE BODY 9 Coplanar Loadings If the body is subjected to a coplanar system of forcesFig13athen only normalforceshearforceand bending moment components will exist at the section Fig 13b If we use the x y z coordinate axes as shown on the left segment then N can be obtained by applying and V can be obtained from Finally the bending moment can be determined by summing moments about point O the z axis in order to eliminate the moments caused by the unknowns N and V MO 0 MO Fy 0 Fx 0 1 Fig 13 section F4 F3 F2 F1 a O V MO N x y Bending Moment Shear Force Normal Force b F2 F1 Important Points Mechanics of materials is a study of the relationship between the external loads applied to a body and the stress and strain caused by the internal loads within the body External forces can be applied to a body as distributed or concentrated surface loadings or as body forces that act throughout the volume of the body Linear distributed loadings produce a resultant force having a magnitude equal to the area under the load diagram and having a location that passes through the centroid of this area A support produces a force in a particular direction on its attached member if it prevents translation of the member in that direction and it produces a couple moment on the member if it prevents rotation The equations of equilibrium and must be satisfied in order to prevent a body from translating with accelerated motion and from rotating When applying the equations of equilibrium it is important to first draw the freebody diagram for the body in order to account for all the terms in the equations The method of sections is used to determine the internal resultant loadings acting on the surface of the sectioned body In general these resultants consist of a normal force shear force torsional moment and bending moment M 0 F 0 10 CHAPTER 1 STRESS 1 The following examples illustrate this procedure numerically and also provide a review of some of the important principles of statics Procedure for Analysis The resultant internal loadings at a point located on the section of a body can be obtained using the method of sections This requires the following steps Support Reactions First decide which segment of the body is to be considered If the segment has a support or connection to another body then before the body is sectioned it will be necessary to determine the reactions acting on the chosen segmentTo do this draw the free body diagram of the entire body and then apply the necessary equations of equilibrium to obtain these reactions FreeBody Diagram Keep all external distributed loadings couple moments torques and forces in their exact locations before passing an imaginary section through the body at the point where the resultant internal loadings are to be determined Draw a freebody diagram of one of the cut segments and indicate the unknown resultants N V M and T at the section These resultants are normally placed at the point representing the geometric center or centroid of the sectioned area If the member is subjected to a coplanar system of forces only N V and M act at the centroid Establish the x y z coordinate axes with origin at the centroid and show the resultant internal loadings acting along the axes Equations of Equilibrium Moments should be summed at the section about each of the coordinate axes where the resultants act Doing this eliminates the unknown forces N and V and allows a direct solution for M and T If the solution of the equilibrium equations yields a negative value for a resultant the assumed directional sense of the resultant is opposite to that shown on the freebody diagram EXAMPLE 11 Determine the resultant internal loadings acting on the cross section at C of the cantilevered beam shown in Fig 14a SOLUTION Support Reactions The support reactions at A do not have to be determined if segment CB is considered FreeBody Diagram The freebody diagram of segment CB is shown in Fig 14b It is important to keep the distributed loading on the segment until after the section is made Only then should this loading be replaced by a single resultant force Notice that the intensity of the distributed loading at C is found by proportion ie from Fig 14a w6 m 270 Nm9 m w 180 Nm The magnitude of the resultant of the distributed load is equal to the area under the loading curve triangle and acts through the centroid of this area Thus F 12180 Nm6 m 540 N which acts 136 m 2 m from C as shown in Fig 14b Equations of Equilibrium Applying the equations of equilibrium we have ΣFx 0 NC 0 NC 0 Ans ΣFy 0 VC 540 N 0 VC 540 N Ans ΣMC 0 MC 540 N2 m 0 MC 1080 Nm Ans NOTE The negative sign indicates that MC acts in the opposite direction to that shown on the freebody diagram Try solving this problem using segment AC by first obtaining the support reactions at A which are given in Fig 14c 12 CHAPTER 1 STRESS 1 Determine the resultant internal loadings acting on the cross section at C of the machine shaft shown in Fig 15a The shaft is supported by journal bearings at A and Bwhich only exert vertical forces on the shaft EXAMPLE 12 Fig 15 c 40 N 1875 N 0250 m 0025 m MC VC C A NC 225 N C D 200 mm 100 mm 100 mm 50 mm 50 mm 800 Nm B a A 0275 m 0125 m 800 Nm0150 m 120 N 0100 m 225 N Ay By B b SOLUTION We will solve this problem using segment AC of the shaft Support Reactions The freebody diagram of the entire shaft is shown in Fig 15b Since segment AC is to be considered only the reaction at A has to be determinedWhy The negative sign indicates that acts in the opposite sense to that shown on the freebody diagram FreeBody Diagram The freebody diagram of segment AC is shown in Fig 15c Equations of Equilibrium Ans Ans Ans NOTE The negative signs for and indicate they act in the opposite directions on the freebody diagram As an exercise calculate the reaction at B and try to obtain the same results using segment CBD of the shaft MC VC MC 569 N m MC 40 N10025 m2 1875 N10250 m2 0 d MC 0 VC 588 N 1875 N 40 N VC 0 c Fy 0 NC 0 Fx 0 Ay Ay 1875 N d MB 0 Ay10400 m2 120 N10125 m2 225 N10100 m2 0 12 EQUILIBRIUM OF A DEFORMABLE BODY 13 1 The 500kg engine is suspended from the crane boom in Fig 16a Determine the resultant internal loadings acting on the cross section of the boom at point E SOLUTION Support Reactions We will consider segment AE of the boom so we must first determine the pin reactions at A Notice that member CD is a twoforce member The freebody diagram of the boom is shown in Fig 16bApplying the equations of equilibrium FreeBody Diagram The freebody diagram of segment AE is shown in Fig 16c Equations of Equilibrium Ans Ans Ans ME 24525 N m 245 kN m ME 124525 N211 m2 0 dME 0 VE 24525 N 245 kN VE 24525 N 0 c Fy 0 NE 9810 N 981 kN NE 9810 N 0 Fx 0 Ay 24525 N Ay 112 2625 N2A3 5B 50019812 N 0 cFy 0 Ax 9810 N Ax 112 2625 N2A4 5B 0 Fx 0 FCD 12 2625 N FCDA3 5B12 m2 50019812 N13 m2 0 dMA 0 EXAMPLE 13 A 1 m 2 m 500981 N Ay Ax FCD b 3 4 5 9810 N 24525 N VE ME NE c E A 1 m A 1 m 1 m 1 m 15 m E C B D a Fig 16 EXAMPLE 14 Determine the resultant internal loadings acting on the cross section at G of the beam shown in Fig 17a Each joint is pin connected SOLUTION Support Reactions Here we will consider segment AG The freebody diagram of the entire structure is shown in Fig 17b Verify the calculated reactions at E and C In particular note that BC is a twoforce member since only two forces act on it For this reason the force at C must act along BC which is horizontal as shown Since BA and BD are also twoforce members the freebody diagram of joint B is shown in Fig 17c Again verify the magnitudes of forces FBA and FBD FreeBody Diagram Using the result for FBA the freebody diagram of segment AG is shown in Fig 17d Equations of Equilibrium ΣFx 0 7750 lb 45 NG 0 NG 6200 lb Ans ΣFy 0 1500 lb 7750 lb 35 VG 0 VG 3150 lb Ans ΣMG 0 MG 7750 lb352 ft 1500 lb2 ft 0 MG 6300 lbft Ans EXAMPLE 15 Determine the resultant internal loadings acting on the cross section at B of the pipe shown in Fig 18a The pipe has a mass of 2 kgm and is subjected to both a vertical force of 50 N and a couple moment of 70 Nm at its end A It is fixed to the wall at C SOLUTION The problem can be solved by considering segment AB so we do not need to calculate the support reactions at C FreeBody Diagram The x y z axes are established at B and the freebody diagram of segment AB is shown in Fig 18b The resultant force and moment components at the section are assumed to act in the positive coordinate directions and to pass through the centroid of the crosssectional area at B The weight of each segment of pipe is calculated as follows WBD 2 kgm05 m981 Nkg 981 N WAD 2 kgm125 m981 Nkg 24525 N These forces act through the center of gravity of each segment Equations of Equilibrium Applying the six scalar equations of equilibrium we have ΣFx 0 FBx 0 Ans ΣFy 0 FBy 0 Ans ΣFz 0 FBz 981 N 24525 N 50 N 0 FBz 843 N Ans ΣMBx 0 MBx 70 Nm 50 N 05 m 24525 N 05 m 981 N 025 m 0 MBx 303 Nm Ans ΣMBy 0 MBy 24525 N 0625 m 50 N 125 m 0 MBy 778 Nm Ans ΣMBz 0 MBz 0 Ans NOTE What do the negative signs for MBx and MBy indicate Note that the normal force NB FBy 0 whereas the shear force is VB 02 8432 843 N Also the torsional moment is TB MBy 778 Nm and the bending moment is MB 3032 02 303 Nm The magnitude of each moment about an axis is equal to the magnitude of each force times the perpendicular distance from the axis to the line of action of the force The direction of each moment is determined using the righthand rule with positive moments thumb directed along the positive coordinate axes 1 FUNDAMENTAL PROBLEMS F11 Determine the internal normal force shear force and bending moment at point C in the beam F14 Determine the internal normal force shear force and bending moment at point C in the beam 60 kNm 10 kN 10 kNm 3 m 3 m 2 m 1 m 1 m 2 m F11 F14 F12 Determine the internal normal force shear force and bending moment at point C in the beam F15 Determine the internal normal force shear force and bending moment at point C in the beam 200 Nm 100 Nm 3 ft 3 ft 3 ft 15 m 15 m F12 F15 F13 Determine the internal normal force shear force and bending moment at point C in the beam F16 Determine the internal normal force shear force and bending moment at point C in the beam 20 kNm 5 kNm 3 m 2 m 2 m 2 m 2 m 3 m F13 F16 PROBLEMS 11 Determine the resultant internal normal force acting on the cross section through point A in each column In a segment BC weighs 180 lbft and segment CD weighs 250 lbft In b the column has a mass of 200 kgm 13 Determine the resultant internal torque acting on the cross sections through points B and C 5 kip 8 kN 6 kN 6 kN 45 kN 45 kN 200 mm 200 mm 200 mm 200 mm 10 ft 8 in 8 in 3 kip 3 kip 4 ft 4 ft 3 m 1 m Prob 11 Prob 13 12 Determine the resultant internal torque acting on the cross sections through points C and D The support bearings at A and B allow free turning of the shaft 250 Nm 150 Nm 400 Nm 150 mm 150 mm 150 mm 200 mm 200 mm 300 mm Prob 12 14 A force of 80 N is supported by the bracket as shown Determine the resultant internal loadings acting on the section through point A 03 m 01 m 30 45 80 N Prob 14 15 Determine the resultant internal loadings in the beam at cross sections through points D and E Point E is just to the right of the 3kip load 3 kip 15 kipft 6 ft 6 ft 4 ft 4 ft Prob 15 16 Determine the normal force shear force and moment at a section through point C Take P 8 kN 17 The cable will fail when subjected to a tension of 2 kN Determine the largest vertical load P the frame will support and calculate the internal normal force shear force and moment at the cross section through point C for this loading 18 Determine the resultant internal loadings on the cross section through point C Assume the reactions at the supports A and B are vertical 19 Determine the resultant internal loadings on the cross section through point D Assume the reactions at the supports A and B are vertical 110 The boom DF of the jib crane and the column DE have a uniform weight of 50 lbft If the hoist and load weigh 300 lb determine the resultant internal loadings in the crane on cross sections through points A B and C 111 The force F 80 lb acts on the gear tooth Determine the resultant internal loadings on the root of the tooth ie at the centroid point A of section aa 112 The sky hook is used to support the cable of a scaffold over the side of a building If it consists of a smooth rod that contacts the parapet of a wall at points A B and C determine the normal force shear force and moment on the cross section at points D and E 113 The 800lb load is being hoisted at a constant speed using the motor M which has a weight of 90 lb Determine the resultant internal loadings acting on the cross section through point B in the beam The beam has a weight of 40 lbft and is fixed to the wall at A 114 Determine the resultant internal loadings acting on the cross section through points C and D of the beam in Prob 113 115 Determine the resultant internal loading on the cross section through point C of the pliers There is a pin at A and the jaws at B are smooth 116 Determine the resultant internal loading on the cross section through point D of the pliers 117 Determine resultant internal loadings acting on section aa and section bb Each section passes through the centerline at point C 118 The bolt shank is subjected to a tension of 80 lb Determine the resultant internal loadings acting on the cross section at point C 119 Determine the resultant internal loadings acting on the cross section through point C Assume the reactions at the supports A and B are vertical 120 Determine the resultant internal loadings acting on the cross section through point D Assume the reactions at the supports A and B are vertical 121 The forged steel clamp exerts a force of F 900 N on the wooden block Determine the resultant internal loadings acting on section aa passing through point A 122 The floor crane is used to lift a 600kg concrete pipe Determine the resultant internal loadings acting on the cross section at G 123 The floor crane is used to lift a 600kg concrete pipe Determine the resultant internal loadings acting on the cross section at H 124 The machine is moving with a constant velocity It has a total mass of 20 Mg and its center of mass is located at G excluding the front roller If the front roller has a mass of 5 Mg determine the resultant internal loadings acting on point C of each of the two side members that support the roller Neglect the mass of the side members The front roller is free to roll 125 Determine the resultant internal loadings acting on the cross section through point B of the signpost The post is fixed to the ground and a uniform pressure of 7 lbft² acts perpendicular to the face of the sign 126 The shaft is supported at its ends by two bearings A and B and is subjected to the forces applied to the pulleys fixed to the shaft Determine the resultant internal loadings acting on the cross section located at point C The 300N forces act in the z direction and the 500N forces act in the x direction The journal bearings at A and B exert only x and z components of force on the shaft Prob 126 127 The pipe has a mass of 12 kgm If it is fixed to the wall at A determine the resultant internal loadings acting on the cross section at B Neglect the weight of the wrench CD Prob 127 128 The brace and drill bit is used to drill a hole at O If the drill bit jams when the brace is subjected to the forces shown determine the resultant internal loadings acting on the cross section of the drill bit at A Prob 128 129 The curved rod has a radius r and is fixed to the wall at B Determine the resultant internal loadings acting on the cross section through A which is located at an angle θ from the horizontal Prob 129 130 A differential element taken from a curved bar is shown in the figure Show that dNdθ V dVdθ N dMdθ T and dTdθ M Prob 130 13 Stress It was stated in Section 12 that the force and moment acting at a specified point O on the sectioned area of the body Fig 19 represents the resultant effects of the actual distribution of loading acting over the sectioned area Fig 110a Obtaining this distribution is of primary importance in mechanics of materials To solve this problem it is necessary to establish the concept of stress We begin by considering the sectioned area to be subdivided into small areas such as ΔA shown in Fig 110a As we reduce ΔA to a smaller and smaller size we must make two assumptions regarding the properties of the material We will consider the material to be continuous that is to consist of a continuum or uniform distribution of matter having no voids Also the material must be cohesive meaning that all portions of it are connected together without having breaks cracks or separations A typical finite yet very small force ΔF acting on ΔA is shown in Fig 110a This force like all the others will have a unique direction but for further discussion we will replace it by its three components namely ΔFx ΔFy and ΔFz which are taken tangent tangent and normal to the area respectively As ΔA approaches zero so do ΔF and its components however the quotient of the force and area will in general approach a finite limit This quotient is called stress and as noted it describes the intensity of the internal force acting on a specific plane area passing through a point Fig 19 Fig 110 Normal Stress The intensity of the force acting normal to ΔA is defined as the normal stress σ sigma Since ΔFz is normal to the area then σz lim ΔFzΔA0 ΔFΔA 14 If the normal force or stress pulls on ΔA as shown in Fig 110a it is referred to as tensile stress whereas if it pushes on ΔA it is called compressive stress Shear Stress The intensity of force acting tangent to ΔA is called the shear stress τ tau Here we have shear stress components τzx lim ΔA0 ΔFxΔA τzy lim ΔA0 ΔFyΔA 15 Note that in this subscript notation z specifies the orientation of the area ΔA Fig 111 and x and y indicate the axes along which each shear stress acts General State of Stress If the body is further sectioned by planes parallel to the xz plane Fig 110b and the yz plane Fig 110c we can then cut out a cubic volume element of material that represents the state of stress acting around the chosen point in the body This state of stress is then characterized by three components acting on each face of the element Fig 112 Units Since stress represents a force per unit area in the International Standard or SI system the magnitudes of both normal and shear stress are specified in the basic units of newtons per square meter Nm² This unit called a pascal 1 Pa 1 Nm² is rather small and in engineering work prefixes such as kilo 10³ symbolized by k mega 106 symbolized by M or giga 109 symbolized by G are used to represent larger more realistic values of stress Likewise in the FootPoundSecond system of units engineers usually express stress in pounds per square inch psi or kilopounds per square inch ksi where 1 kilopound kip 1000 lb Sometimes stress is expressed in units of Nmm² where 1 mm 10³ m However in the SI system prefixes are not allowed in the denominator of a fraction and therefore it is better to use the equivalent 1 Nmm² 1 MNm² 1 MPa Fig 111 Fig 112 14 Average Normal Stress in an Axially Loaded Bar In this section we will determine the average stress distribution acting on the crosssectional area of an axially loaded bar such as the one shown in Fig 113a This bar is prismatic since all cross sections are the same throughout its length When the load P is applied to the bar through the centroid of its crosssectional area then the bar will deform uniformly throughout the central region of its length as shown in Fig 113b Homogeneous material has the same physical and mechanical properties throughout its volume and isotropic material has these same properties in all directions Many engineering materials may be approximated as being both homogeneous and isotropic as assumed here Steel for example contains thousands of randomly oriented crystals in each cubic millimeter of its volume and since most problems involving this material have a physical size that is very much larger than a single crystal the above assumption regarding its material composition is quite realistic Note that anisotropic materials such as wood have different properties in different directions and although this is the case like wood if the anisotropy is oriented along the bars axis then the bar will also deform uniformly when subjected to the axial load P Average Normal Stress Distribution If we pass a section through the bar and separate it into two parts then equilibrium requires the resultant normal force at the section to be P Fig 113c Due to the uniform deformation of the material it is necessary that the cross section be subjected to a constant normal stress distribution Fig 113d As a result each small area ΔA on the cross section is subjected to a force ΔF σ ΔA and the sum of these forces acting over the entire crosssectional area must be equivalent to the internal resultant force P at the section If we let ΔA dA and therefore ΔF dF then recognizing σ is constant we have FRz ΣFz dF A σ dA P σ A σ PA 16 Here σ average normal stress at any point on the crosssectional area P internal resultant normal force which acts through the centroid of the crosssectional area P is determined using the method of sections and the equations of equilibrium A crosssectional area of the bar where σ is determined Since the internal load P passes through the centroid of the crosssection the uniform stress distribution will produce zero moments about the x and y axes passing through this point Fig 113d To show this we require the moment of P about each axis to be equal to the moment of the stress distribution about the axes namely MRx Σ Mx 0 A y dF A yσ dA σ A y dA MRy Σ My 0 A x dF A xσ dA σ A x dA These equations are indeed satisfied since by definition of the centroid y dA 0 and x dA 0 See Appendix A Equilibrium It should be apparent that only a normal stress exists on any small volume element of material located at each point on the cross section of an axially loaded bar If we consider vertical equilibrium of the element Fig 114 then apply the equation of force equilibrium ΣFz 0 σΔA σΔA 0 σ σ In other words the two normal stress components on the element must be equal in magnitude but opposite in direction This is referred to as uniaxial stress The previous analysis applies to members subjected to either tension or compression as shown in Fig 115 As a graphical interpretation the magnitude of the internal resultant force P is equivalent to the volume under the stress diagram that is P σ A volume height base Furthermore as a consequence of the balance of moments this resultant passes through the centroid of this volume Although we have developed this analysis for prismatic bars this assumption can be relaxed somewhat to include bars that have a slight taper For example it can be shown using the more exact analysis of the theory of elasticity that for a tapered bar of rectangular cross section for which the angle between two adjacent sides is 15 the average normal stress as calculated by σ PA is only 22 less than its value found from the theory of elasticity Maximum Average Normal Stress In our analysis both the internal force P and the crosssectional area A were constant along the longitudinal axis of the bar and as a result the normal stress σ PA is also constant throughout the bars length Occasionally however the bar may be subjected to several external loads along its axis or a change in its crosssectional area may occur As a result the normal stress within the bar could be different from one section to the next and if the maximum average normal stress is to be determined then it becomes important to find the location where the ratio PA is a maximum To do this it is necessary to determine the internal force P at various sections along the bar Here it may be helpful to show this variation by drawing an axial or normal force diagram Specifically this diagram is a plot of the normal force P versus its position x along the bars length As a sign convention P will be positive if it causes tension in the member and negative if it causes compression Once the internal loading throughout the bar is known the maximum ratio of PA can then be identified This steel tie rod is used as a hanger to suspend a portion of a staircase and as a result it is subjected to tensile stress Fig 113 Fig 114 Fig 115 Tension Compression 14 AVERAGE NORMAL STRESS IN AN AXIALLY LOADED BAR 27 1 Important Points When a body subjected to external loads is sectioned there is a distribution of force acting over the sectioned area which holds each segment of the body in equilibrium The intensity of this internal force at a point in the body is referred to as stress Stress is the limiting value of force per unit area as the area approaches zero For this definition the material is considered to be continuous and cohesive The magnitude of the stress components at a point depends upon the type of loading acting on the body and the orientation of the element at the point When a prismatic bar is made from homogeneous and isotropic material and is subjected to an axial force acting through the centroid of the crosssectional area then the center region of the bar will deform uniformly As a result the material will be subjected only to normal stress This stress is uniform or averaged over the crosssectional area Procedure for Analysis The equation gives the average normal stress on the cross sectional area of a member when the section is subjected to an internal resultant normal force P For axially loaded members application of this equation requires the following steps Internal Loading Section the member perpendicular to its longitudinal axis at the point where the normal stress is to be determined and use the necessary freebody diagram and force equation of equilibrium to obtain the internal axial force P at the section Average Normal Stress Determine the members crosssectional area at the section and calculate the average normal stress It is suggested that be shown acting on a small volume element of the material located at a point on the section where stress is calculated To do this first draw on the face of the element coincident with the sectioned area A Here acts in the same direction as the internal force P since all the normal stresses on the cross section develop this resultant The normal stress on the other face of the element acts in the opposite direction s s s s s PA s PA EXAMPLE 16 The bar in Fig 116a has a constant width of 35 mm and a thickness of 10 mm Determine the maximum average normal stress in the bar when it is subjected to the loading shown SOLUTION Internal Loading By inspection the internal axial forces in regions AB BC and CD are all constant yet have different magnitudes Using the method of sections these loadings are determined in Fig 116b and the normal force diagram which represents these results graphically is shown in Fig 116c The largest loading is in region BC where PBC 30 kN Since the crosssectional area of the bar is constant the largest average normal stress also occurs within this region of the bar Average Normal Stress Applying Eq 16 we have σBC PBC A 30103 N 0035 m0010 m 857 MPa Ans NOTE The stress distribution acting on an arbitrary cross section of the bar within region BC is shown in Fig 116d Graphically the volume or block represented by this distribution of stress is equivalent to the load of 30 kN that is 30 kN 857 MPa35 mm10 mm EXAMPLE 17 The 80kg lamp is supported by two rods AB and BC as shown in Fig 117a If AB has a diameter of 10 mm and BC has a diameter of 8 mm determine the average normal stress in each rod SOLUTION Internal Loading We must first determine the axial force in each rod A freebody diagram of the lamp is shown in Fig 117b Applying the equations of force equilibrium Fx 0 FBC45 FBA cos 60 0 Fy 0 FBC35 FBA sin 60 7848 N 0 FBC 3952 N FBA 6324 N By Newtons third law of action equal but opposite reaction these forces subject the rods to tension throughout their length Average Normal Stress Applying Eq 16 σBC FBC ABC 3952 N π0004 m2 786 MPa Ans σBA FBA ABA 6324 N π0005 m2 805 MPa Ans NOTE The average normal stress distribution acting over a cross section of rod AB is shown in Fig 117c and at a point on this cross section an element of material is stressed as shown in Fig 117d 30 CHAPTER 1 STRESS 1 The casting shown in Fig 118a is made of steel having a specific weight of Determine the average compressive stress acting at points A and B gst 490 lbft3 EXAMPLE 18 075 ft 075 ft 275 ft y z x a A B 075 ft 04 ft Fig 118 275 ft b A P c 936 psi B Wst SOLUTION Internal Loading A freebody diagram of the top segment of the casting where the section passes through points A and B is shown in Fig 118b The weight of this segment is determined from Thus the internal axial force P at the section is Average Compressive Stress The crosssectional area at the sec tion is and so the average compressive stress becomes Ans NOTE The stress shown on the volume element of material in Fig 118c is representative of the conditions at either point A or B Notice that this stress acts upward on the bottom or shaded face of the element since this face forms part of the bottom surface area of the section and on this surface the resultant internal force P is pushing upward 936 psi s 13475 lbft2 11 ft2144 in22 s P A 2381 lb p1075 ft22 13475 lbft2 A p1075 ft22 P 2381 lb P 1490 lbft321275 ft2p1075 ft22 0 P Wst 0 c Fz 0 Wst gst Vst 14 AVERAGE NORMAL STRESS IN AN AXIALLY LOADED BAR 31 1 EXAMPLE 19 b x 3 kN A 200 mm FAB FC x A B C 200 mm a 3 kN Fig 119 SOLUTION Internal Loading The forces at A and C can be related by considering the freebody diagram for member AC Fig 119b There are three unknowns namely and x To solve this problem we will work in units of newtons and millimeters 1 2 Average Normal Stress A necessary third equation can be written that requires the tensile stress in the bar AB and the compressive stress at C to be equivalent ie Substituting this into Eq 1 solving for then solving for we obtain The position of the applied load is determined from Eq 2 Ans NOTE as required 0 6 x 6 200 mm x 124 mm FC 1857 N FAB 1143 N FC FAB FC 1625FAB s FAB 400 mm2 FC 650 mm2 3000 N1x2 FC1200 mm2 0 dMA 0 FAB FC 3000 N 0 c Fy 0 FC FAB Member AC shown in Fig 119a is subjected to a vertical force of 3 kN Determine the position x of this force so that the average compressive stress at the smooth support C is equal to the average tensile stress in the tie rod AB The rod has a crosssectional area of 400 mm2 and the contact area at C is 650 mm2 32 CHAPTER 1 STRESS 15 Average Shear Stress Shear stress has been defined in Section 13 as the stress component that acts in the plane of the sectioned area To show how this stress can develop consider the effect of applying a force F to the bar in Fig 120a If the supports are considered rigid and F is large enough it will cause the material of the bar to deform and fail along the planes identified by AB and CDA freebody diagram of the unsupported center segment of the bar Fig 120b indicates that the shear force must be applied at each section to hold the segment in equilibrium The average shear stress distributed over each sectioned area that develops this shear force is defined by 17 Here average shear stress at the section which is assumed to be the same at each point located on the section internal resultant shear force on the section determined from the equations of equilibrium area at the section The distribution of average shear stress acting over the sections is shown in Fig 120c Notice that is in the same direction as V since the shear stress must create associated forces all of which contribute to the internal resultant force V at the section The loading case discussed here is an example of simple or direct shear since the shear is caused by the direct action of the applied load F This type of shear often occurs in various types of simple connections that use bolts pins welding material etc In all these cases however application of Eq 17 is only approximate A more precise investigation of the shearstress distribution over the section often reveals that much larger shear stresses occur in the material than those predicted by this equation Although this may be the case application of Eq 17 is generally acceptable for many problems in engineering design and analysis For example engineering codes allow its use when considering design sizes for fasteners such as bolts and for obtaining the bonding strength of glued joints subjected to shear loadings tavg A V tavg tavg V A V F2 1 b c F F V V tavg F a B D A C Fig 120 Shear Stress Equilibrium Figure 121a shows a volume element of material taken at a point located on the surface of a sectioned area which is subjected to a shear stress τzy Force and moment equilibrium requires the shear stress acting on this face of the element to be accompanied by shear stress acting on three other faces To show this we will first consider force equilibrium in the y direction Then Fy 0 force stress area τzyΔx Δy τzy Δx Δy 0 τzy τzy In a similar manner force equilibrium in the z direction yields τyz τyz Finally taking moments about the x axis Mx 0 moment force arm stress area τzyΔx Δy Δz τyzΔx Δz Δy 0 τzy τyz so that τzy τzy τyz τyz τ In other words all four shear stresses must have equal magnitude and be directed either toward or away from each other at opposite edges of the element Fig 121b This is referred to as the complementary property of shear and under the conditions shown in Fig 121 the material is subjected to pure shear 34 CHAPTER 1 STRESS 1 Important Points If two parts are thin or small when joined together the applied loads may cause shearing of the material with negligible bending If this is the case it is generally assumed that an average shear stress acts over the crosssectional area When shear stress acts on a plane then equilibrium of a volume element of material at a point on the plane requires associated shear stress of the same magnitude act on three adjacent sides of the element t Procedure for Analysis The equation is used to determine the average shear stress in the materialApplication requires the following steps Internal Shear Section the member at the point where the average shear stress is to be determined Draw the necessary freebody diagram and calculate the internal shear force V acting at the section that is necessary to hold the part in equilibrium Average Shear Stress Determine the sectioned area A and determine the average shear stress It is suggested that be shown on a small volume element of material located at a point on the section where it is determined To do this first draw on the face of the element coincident with the sectioned area A This stress acts in the same direction as V The shear stresses acting on the three adjacent planes can then be drawn in their appropriate directions following the scheme shown in Fig 121 tavg tavg tavg VA tavg VA EXAMPLE 110 Determine the average shear stress in the 20mmdiameter pin at A and the 30mmdiameter pin at B that support the beam in Fig 122a SOLUTION Internal Loadings The forces on the pins can be obtained by considering the equilibrium of the beam Fig 122b ΣMA 0 FB456 m 30 kN2 m 0 FB 125 kN ΣFx 0 125 kN35 Ax 0 Ax 750 kN ΣFy 0 Ay 125 kN45 30 kN 0 Ay 20 kN Thus the resultant force acting on pin A is FA Ax² Ay² 750 kN² 20 kN² 2136 kN The pin at A is supported by two fixed leaves and so the freebody diagram of the center segment of the pin shown in Fig 122c has two shearing surfaces between the beam and each leaf The force of the beam 2136 kN acting on the pin is therefore supported by shear force on each of these surfaces This case is called double shear Thus VA FA2 2136 kN2 1068 kN In Fig 122a note that pin B is subjected to single shear which occurs on the section between the cable and beam Fig 122d For this pin segment VB FB 125 kN Average Shear Stress τAavg VAAA 1068103 N π4002 m² 340 MPa Ans τBavg VBAB 125103 N π4003 m² 177 MPa Ans Fig 122 FUNDAMENTAL PROBLEMS F113 Rods AC and BC are used to suspend the 200kg mass If each rod is made of a material for which the average normal stress can not exceed 150 MPa determine the minimum required diameter of each rod to the nearest mm F114 The frame supports the loading shown The pin at A has a diameter of 025 in If it is subjected to double shear determine the average shear stress in the pin F115 Determine the maximum average shear stress developed in each 34indiameter bolt F116 If each of the three nails has a diameter of 4 mm and can withstand an average shear stress of 60 MPa determine the maximum allowable force P that can be applied to the board F117 The strut is glued to the horizontal member at surface AB If the strut has a thickness of 25 mm and the glue can withstand an average shear stress of 600 kPa determine the maximum force P that can be applied to the strut F118 Determine the maximum average shear stress developed in the 30mmdiameter pin EXAMPLE 111 If the wood joint in Fig 123a has a width of 150 mm determine the average shear stress developed along shear planes aa and bb For each plane represent the state of stress on an element of the material Fig 123 SOLUTION Internal Loadings Referring to the freebody diagram of the member Fig 123b ΣFx 0 6 kN F F 0 F 3 kN Now consider the equilibrium of segments cut across shear planes aa and bb shown in Figs 123c and 123d ΣFx 0 Va 3 kN 0 Va 3 kN ΣFx 0 3 kN Vb 0 Vb 3 kN Average Shear Stress τaavg VaAa 3103 N01 m015 m 200 kPa Ans τbavg VbAb 3103 N0125 m015 m 160 kPa Ans The state of stress on elements located on sections aa and bb is shown in Figs 123c and 123d respectively 15 AVERAGE SHEAR STRESS 37 1 The inclined member in Fig 124a is subjected to a compressive force of 600 lb Determine the average compressive stress along the smooth areas of contact defined by AB and BC and the average shear stress along the horizontal plane defined by DB EXAMPLE 112 1 in 3 4 5 600 lb 15 in 3 in 2 in A C B D a Fig 124 SOLUTION Internal Loadings The freebody diagram of the inclined member is shown in Fig 124b The compressive forces acting on the areas of contact are Also from the freebody diagram of the top segment ABD of the bottom member Fig 124c the shear force acting on the sectioned horizontal plane DB is Average Stress The average compressive stresses along the horizontal and vertical planes of the inclined member are Ans Ans These stress distributions are shown in Fig 124d The average shear stress acting on the horizontal plane defined by DB is Ans This stress is shown uniformly distributed over the sectioned area in Fig 124e tavg 360 lb 13 in2115 in2 80 psi sBC FBC ABC 480 lb 12 in2115 in2 160 psi sAB FAB AAB 360 lb 11 in2115 in2 240 psi V 360 lb Fx 0 FBC 600 lbA4 5B 0 FBC 480 lb c Fy 0 FAB 600 lbA3 5B 0 FAB 360 lb Fx 0 b 3 4 5 600 lb FAB FBC c V 360 lb d 3 4 5 600 lb 160 psi 240 psi e 360 lb 80 psi 38 CHAPTER 1 STRESS 1 FUNDAMENTAL PROBLEMS F17 The uniform beam is supported by two rods AB and CD that have crosssectional areas of and respectively Determine the intensity w of the distributed load so that the average normal stress in each rod does not exceed 300 kPa 15 mm2 10 mm2 F110 If the 600kN force acts through the centroid of the cross section determine the location of the centroid and the average normal stress developed on the cross section Also sketch the normal stress distribution over the cross section y w A C B D 6 m F17 F18 Determine the average normal stress developed on the cross section Sketch the normal stress distribution over the cross section 300 kN 100 mm 80 mm F18 F19 Determine the average normal stress developed on the cross section Sketch the normal stress distribution over the cross section 4 in 1 in 1 in 4 in 1 in 15 kip F19 80 mm 300 mm 60 mm y 80 mm 600 kN x y 60 mm F110 F111 Determine the average normal stress developed at points A B and C The diameter of each segment is indicated in the figure 2 kip 3 kip 8 kip 9 kip 1 in 05 in 05 in A B C F111 F112 Determine the average normal stress developed in rod AB if the load has a mass of 50 kg The diameter of rod AB is 8 mm 8 mm A D B C 5 4 3 F112 15 AVERAGE SHEAR STRESS 39 1 131 The column is subjected to an axial force of 8 kN which is applied through the centroid of the crosssectional area Determine the average normal stress acting at section aa Show this distribution of stress acting over the areas cross section 133 The bar has a crosssectional area A and is subjected to the axial load P Determine the average normal and average shear stresses acting over the shaded section which is oriented at from the horizontal Plot the variation of these stresses as a function of u 10 u 902 u PROBLEMS 20 N 20 N 250 mm 250 mm 12 mm A B Prob 132 8 kN a a 75 mm 10 mm 10 mm 10 mm 75 mm 70 mm 70 mm Prob 131 132 The lever is held to the fixed shaft using a tapered pin AB which has a mean diameter of 6 mm If a couple is applied to the lever determine the average shear stress in the pin between the pin and lever P u P A Prob 133 134 The builtup shaft consists of a pipe AB and solid rod BCThe pipe has an inner diameter of 20 mm and outer diameter of 28 mm The rod has a diameter of 12 mm Determine the average normal stress at points D and E and represent the stress on a volume element located at each of these points C E D A 4 kN 8 kN B 6 kN 6 kN Prob 134 135 The bars of the truss each have a crosssectional area of Determine the average normal stress in each member due to the loading State whether the stress is tensile or compressive 136 The bars of the truss each have a crosssectional area of If the maximum average normal stress in any bar is not to exceed 20 ksi determine the maximum magnitude P of the loads that can be applied to the truss 125 in2 P 8 kip 125 in2 3 ft 4 ft 4 ft P 075 P E D A B C Probs 13536 137 The plate has a width of 05 m If the stress distribution at the support varies as shown determine the force P applied to the plate and the distance d to where it is applied Prob 137 138 The two members used in the construction of an aircraft fuselage are joined together using a 30 fishmouth weld Determine the average normal and average shear stress on the plane of each weld Assume each inclined plane supports a horizontal force of 400 lb Prob 138 139 If the block is subjected to the centrally applied force of 600 kN determine the average normal stress in the material Show the stress acting on a differential volume element of the material Prob 139 140 The pins on the frame at B and C each have a diameter of 025 in If these pins are subjected to double shear determine the average shear stress in each pin 141 Solve Prob 140 assuming that pins B and C are subjected to single shear 142 The pins on the frame at D and E each have a diameter of 025 in If these pins are subjected to double shear determine the average shear stress in each pin 143 Solve Prob 142 assuming that pins D and E are subjected to single shear Probs 140414243 144 A 175lb woman stands on a vinyl floor wearing stiletto highheel shoes If the heel has the dimensions shown determine the average normal stress she exerts on the floor and compare it with the average normal stress developed when a man having the same weight is wearing flatheeled shoes Assume the load is applied slowly so that dynamic effects can be ignored Also assume the entire weight is supported only by the heel of one shoe Prob 144 145 The truss is made from three pinconnected members having the crosssectional areas shown in the figure Determine the average normal stress developed in each member when the truss is subjected to the load shown State whether the stress is tensile or compressive 146 Determine the average normal stress developed in links AB and CD of the smooth twotine grapple that supports the log having a mass of 3 Mg The crosssectional area of each link is 400 mm2 147 Determine the average shear stress developed in pins A and B of the smooth twotine grapple that supports the log having a mass of 3 Mg Each pin has a diameter of 25 mm and is subjected to double shear 148 The beam is supported by a pin at A and a short link BC If P 15 kN determine the average shear stress developed in the pins at A B and C All pins are in double shear as shown and each has a diameter of 18 mm149 The beam is supported by a pin at A and a short link BC Determine the maximum magnitude P of the loads the beam will support if the average shear stress in each pin is not to exceed 80 MPa All pins are in double shear as shown and each has a diameter of 18 mm 150 The block is subjected to a compressive force of 2 kN Determine the average normal and average shear stress developed in the wood fibers that are oriented along section aa at 30 with the axis of the block 42 CHAPTER 1 STRESS 151 During the tension test the wooden specimen is subjected to an average normal stress of 2 ksi Determine the axial force P applied to the specimen Also find the average shear stress developed along section aa of the specimen 1 P P 1 in 2 in 4 in 4 in a a Prob 151 152 If the joint is subjected to an axial force of determine the average shear stress developed in each of the 6mm diameter bolts between the plates and the members and along each of the four shaded shear planes 153 The average shear stress in each of the 6mm diameter bolts and along each of the four shaded shear planes is not allowed to exceed 80 MPa and 500 kPa respectively Determine the maximum axial force P that can be applied to the joint P 9 kN P P 100 mm 100 mm Probs 15253 154 The shaft is subjected to the axial force of 40 kN Determine the average bearing stress acting on the collar C and the normal stress in the shaft 40 kN 30 mm 40 mm C Prob 154 155 Rods AB and BC each have a diameter of 5 mm If the load of is applied to the ring determine the average normal stress in each rod if 156 Rods AB and BC each have a diameter of 5 mm Determine the angle of rod BC so that the average normal stress in rod AB is 15 times that in rod BC What is the load P that will cause this to happen if the average normal stress in each rod is not allowed to exceed 100 MPa u u 60 P 2 kN u C B P A Probs 15556 157 The specimen failed in a tension test at an angle of 52 when the axial load was 1980 kip If the diameter of the specimen is 05 in determine the average normal and average shear stress acting on the area of the inclined failure plane Also what is the average normal stress acting on the cross section when failure occurs 158 The anchor bolt was pulled out of the concrete wall and the failure surface formed part of a frustum and cylinder This indicates a shear failure occurred along the cylinder BC and tension failure along the frustum AB If the shear and normal stresses along these surfaces have the magnitudes shown determine the force P that must have been applied to the bolt 159 The open square butt joint is used to transmit a force of 50 kip from one plate to the other Determine the average normal and average shear stress components that this loading creates on the face of the weld section AB 160 If P 20 kN determine the average shear stress developed in the pins at A and C The pins are subjected to double shear as shown and each has a diameter of 18 mm161 Determine the maximum magnitude P of the load the beam will support if the average shear stress in each pin is not to allowed to exceed 60 MPa All pins are subjected to double shear as shown and each has a diameter of 18 mm 162 The crimping tool is used to crimp the end of the wire E If a force of 20 lb is applied to the handles determine the average shear stress in the pin at A The pin is subjected to double shear and has a diameter of 02 in Only a vertical force is exerted on the wire 163 Solve Prob 162 for pin B The pin is subjected to double shear and has a diameter of 02 in 164 The triangular blocks are glued along each side of the joint A Cclamp placed between two of the blocks is used to draw the joint tight If the glue can withstand a maximum average shear stress of 800 kPa determine the maximum allowable clamping force F165 The triangular blocks are glued along each side of the joint A Cclamp placed between two of the blocks is used to draw the joint tight If the clamping force is F 900 N determine the average shear stress developed in the glued shear plane 166 Determine the largest load P that can be a applied to the frame without causing either the average normal stress or the average shear stress at section aa to exceed σ 150 MPa and τ 60 MPa respectively Member CB has a square cross section of 25 mm on each side 167 The prismatic bar has a crosssectional area A If it is subjected to a distributed axial loading that increases linearly from w 0 at x 0 to w w0 at x a and then decreases linearly to w 0 at x 2a determine the average normal stress in the bar as a function of x for 0 x a 168 The prismatic bar has a crosssectional area A If it is subjected to a distributed axial loading that increases linearly from w 0 at x 0 to w w0 at x a and then decreases linearly to w 0 at x 2a determine the average normal stress in the bar as a function of x for a x 2a 169 The tapered rod has a radius of r 2 x6 in and is subjected to the distributed loading of w 60 40x lbin Determine the average normal stress at the center of the rod B 170 The pedestal supports a load P at its center If the material has a mass density ρ determine the radial dimension r as a function of z so that the average normal stress in the pedestal remains constant The cross section is circular 171 Determine the average normal stress at section aa and the average shear stress at section bb in member AB The cross section is square 05 in on each side 172 Consider the general problem of a bar made from m segments each having a constant crosssectional area Am and length Lm If there are n loads on the bar as shown write a computer program that can be used to determine the average normal stress at any specified location x Show an application of the program using the values L1 4 ft d1 2 ft P1 400 lb A1 3 in2 L2 2 ft d2 6 ft P2 300 lb A2 1 in2 46 CHAPTER 1 STRESS 16 Allowable Stress To properly design a structural member or mechanical element it is necessary to restrict the stress in the material to a level that will be safe To ensure this safety it is therefore necessary to choose an allowable stress that restricts the applied load to one that is less than the load the member can fully support There are many reasons for doing this For example the load for which the member is designed may be different from actual loadings placed on it The intended measurements of a structure or machine may not be exact due to errors in fabrication or in the assembly of its component parts Unknown vibrations impact or accidental loadings can occur that may not be accounted for in the design Atmospheric corrosion decay or weathering tend to cause materials to deteriorate during service And lastly some materials such as wood concrete or fiberreinforced composites can show high variability in mechanical properties One method of specifying the allowable load for a member is to use a number called the factor of safety The factor of safety FS is a ratio of the failure load to the allowable load Here is found from experimental testing of the material and the factor of safety is selected based on experience so that the above mentioned uncertainties are accounted for when the member is used under similar conditions of loading and geometry Stated mathematically 18 If the load applied to the member is linearly related to the stress developed within the member as in the case of using and then we can also express the factor of safety as a ratio of the failure stress or to the allowable stress or that is 19 or 110 FS tfail tallow FS sfail sallow tallow sallow tfail sfail tavg VA s PA FS Ffail Fallow Ffail Fallow Ffail 1 In some cases such as columns the applied load is not linearly related to stress and therefore only Eq 18 can be used to determine the factor of safety See Chapter 13 In any of these equations the factor of safety must be greater than 1 in order to avoid the potential for failure Specific values depend on the types of materials to be used and the intended purpose of the structure or machine For example the FS used in the design of aircraft or spacevehicle components may be close to 1 in order to reduce the weight of the vehicle Or in the case of a nuclear power plant the factor of safety for some of its components may be as high as 3 due to uncertainties in loading or material behavior In many cases the factor of safety for a specific case can be found in design codes and engineering handbooks These values are intended to form a balance of ensuring public and environmental safety and providing a reasonable economic solution to design 17 Design of Simple Connections By making simplifying assumptions regarding the behavior of the material the equations σ PA and τavg VA can often be used to analyze or design a simple connection or mechanical element In particular if a member is subjected to normal force at a section its required area at the section is determined from A P σallow 111 On the other hand if the section is subjected to an average shear force then the required area at the section is A V τallow 112 As discussed in Sec 16 the allowable stress used in each of these equations is determined either by applying a factor of safety to the materials normal or shear failure stress or by finding these stresses directly from an appropriate design code Three examples of where the above equations apply are shown in Fig 125 The area of the column base plate B is determined from the allowable bearing stress for the concrete The embedded length l of this rod in concrete can be determined using the allowable shear stress of the bonding glue The area of the bolt for this lap joint is determined from the shear stress which is largest between the plates 48 CHAPTER 1 STRESS 1 Important Point Design of a member for strength is based on selecting an allowable stress that will enable it to safely support its intended load Since there are many unknown factors that can influence the actual stress in a member then depending upon the intended use of the member a factor of safety is applied to obtain the allowable load the member can support Procedure for Analysis When solving problems using the average normal and shear stress equations a careful consideration should first be made as to choose the section over which the critical stress is acting Once this section is determined the member must then be designed to have a sufficient area at the section to resist the stress that acts on it This area is determined using the following steps Internal Loading Section the member through the area and draw a freebody diagram of a segment of the memberThe internal resultant force at the section is then determined using the equations of equilibrium Required Area Provided the allowable stress is known or can be determined the required area needed to sustain the load at the section is then determined from or A Vtallow A Psallow Appropriate factors of safety must be considered when designing cranes and cables used to transfer heavy loads 17 DESIGN OF SIMPLE CONNECTIONS 49 1 The control arm is subjected to the loading shown in Fig 126a Determine to the nearest the required diameter of the steel pin at C if the allowable shear stress for the steel is tallow 8 ksi 1 4 in EXAMPLE 113 Fig 126 3 5 4 2 in 3 in 8 in Cx 3 kip 5 kip FAB Cy b C a C 3 5 4 2 in 3 in 8 in A C 3 kip 5 kip B SOLUTION Internal Shear Force A freebody diagram of the arm is shown in Fig 126b For equilibrium we have The pin at C resists the resultant force at C which is Since the pin is subjected to double shear a shear force of 3041 kip acts over its crosssectional area between the arm and each supporting leaf for the pin Fig 126c Required Area We have Use a pin having a diameter of Ans d 3 4 in 0750 in d 0696 in pa d 2 b 2 03802 in2 A V tallow 3041 kip 8 kipin2 03802 in2 FC 211 kip22 16 kip22 6082 kip Cy 3 kip 5 kip A3 5B 0 Cy 6 kip c Fy 0 3 kip Cx 5 kip A4 5B 0 Cx 1 kip Fx 0 FAB 3 kip FAB18 in2 3 kip 13 in2 5 kip A3 5B15 in2 0 dMC 0 c 3041 kip 3041 kip 6082 kip Pin at C 50 CHAPTER 1 STRESS 1 The suspender rod is supported at its end by a fixedconnected circular disk as shown in Fig127aIf the rod passes through a 40mmdiameter hole determine the minimum required diameter of the rod and the minimum thickness of the disk needed to support the 20kN load The allowable normal stress for the rod is and the allowable shear stress for the disk is tallow 35 MPa sallow 60 MPa EXAMPLE 114 Fig 127 20 kN A allow b 40 mm t 20 kN t d a 40 mm SOLUTION Diameter of Rod By inspection the axial force in the rod is 20 kN Thus the required crosssectional area of the rod is so that Ans Thickness of Disk As shown on the freebody diagram in Fig 127b the material at the sectioned area of the disk must resist shear stress to prevent movement of the disk through the hole If this shear stress is assumed to be uniformly distributed over the sectioned area then since we have Ans t 45511032 m 455 mm 2p1002 m21t2 2011032 N 3511062 Nm2 A V tallow V 20 kN d 00206 m 206 mm p 4 d2 2011032 N 6011062 Nm2 A P sallow 17 DESIGN OF SIMPLE CONNECTIONS 51 1 The shaft shown in Fig 128a is supported by the collar at C which is attached to the shaft and located on the right side of the bearing at B Determine the largest value of P for the axial forces at E and F so that the bearing stress on the collar does not exceed an allowable stress of and the average normal stress in the shaft does not exceed an allowable stress of 1st2allow 55 MPa 1sb2allow 75 MPa EXAMPLE 115 P 2 P b 3P 80 mm 60 mm P A F 2P C E B a 20 mm c Axial Force Position 3P 2P SOLUTION To solve the problem we will determine P for each possible failure conditionThen we will choose the smallest valueWhy Normal Stress Using the method of sections the axial load within region FE of the shaft is 2P whereas the largest axial force 3P occurs within region EC Fig 128b The variation of the internal loading is clearly shown on the normalforce diagram Fig 128c Since the cross sectional area of the entire shaft is constant region EC is subjected to the maximum average normal stressApplying Eq111we have Ans Bearing Stress As shown on the freebody diagram in Fig 128d the collar at C must resist the load of 3P which acts over a bearing area of Thus By comparison the largest load that can be applied to the shaft is since any load larger than this will cause the allowable normal stress in the shaft to be exceeded NOTE Here we have not considered a possible shear failure of the collar as in Example 114 P 518 kN P 550 kN 219911032 m2 3P 7511062 Nm2 A P sallow Ab p1004 m22 p1003 m22 219911032 m2 P 518 kN p1003 m22 3P 5511062 Nm2 A P sallow 3P d C Fig 128 52 CHAPTER 1 STRESS 1 The rigid bar AB shown in Fig 129a is supported by a steel rod AC having a diameter of 20 mm and an aluminum block having a cross sectional area of The 18mmdiameter pins at A and C are subjected to single shear If the failure stress for the steel and aluminum is and respectively and the failure shear stress for each pin is determine the largest load P that can be applied to the barApply a factor of safety of SOLUTION Using Eqs 19 and 110 the allowable stresses are The freebody diagram of the bar is shown in Fig 129b There are three unknownsHere we will apply the moment equations of equilibrium in order to express and in terms of the applied load PWe have 1 2 We will now determine each value of P that creates the allowable stress in the rod block and pins respectively Rod AC This requires Using Eq 1 P 11068 kN212 m2 125 m 171 kN FAC 1sst2allow1AAC2 34011062 Nm2 p1001 m22 1068 kN FB12 m2 P1075 m2 0 dMA 0 P1125 m2 FAC12 m2 0 dMB 0 FB FAC tallow tfail FS 900 MPa 2 450 MPa 1sal2allow 1sal2fail FS 70 MPa 2 35 MPa 1sst2allow 1sst2fail FS 680 MPa 2 340 MPa FS 2 tfail 900 MPa 1sal2fail 70 MPa 1sst2fail 680 MPa 1800 mm2 EXAMPLE 116 Block B In this case FB 1sal2allow AB 3511062 Nm2 1800 mm2 11062 m2mm2 630 kN Using Eq 2 Pin A or C Due to single shear From Eq 1 By comparisonas P reaches its smallest value 168 kNthe allowable normal stress will first be developed in the aluminum block Hence Ans P 168 kN P 1145 kN 12 m2 125 m 183 kN FAC V tallow A 45011062 Nm2 p10009 m22 1145 kN P 1630 kN212 m2 075 m 168 kN Fig 129 2 m A 075 m a Aluminum Steel P B C b A 075 m P 125 m B FB FAC 54 CHAPTER 1 STRESS 1 F119 If the eyebolt is made of a material having a yield stress of MPa determine the minimum required diameter d of its shank Apply a factor of safety against yielding FS 15 sY 250 30 kN d F119 F120 F120 If the bar assembly is made of a material having a yield stress of ksi determine the minimum required dimensions and to the nearest Apply a factor of safety against yielding Each bar has a thickness of 05 in FS 15 18 in h2 h1 sY 50 A B C 15 kip 15 kip 30 kip h1 h2 F122 80 kN F121 50 mm 60 mm 120 mm a a P 40 mm Section aa F121 Determine the maximum force P that can be applied to the rod if it is made of material having a yield stress of MPa Consider the possibility that failure occurs in the rod and at section aaApply a factor of safety of against yielding FS 2 sY 250 F122 The pin is made of a material having a failure shear stress of Determine the minimum required diameter of the pin to the nearest mm Apply a factor of safety of against shear failure FS 25 tfail 100 MPa F123 40 mm 75 mm 80 mm 30 mm P F123 If the bolt head and the supporting bracket are made of the same material having a failure shear stress of determine the maximum allowable force P that can be applied to the bolt so that it does not pull through the plate Apply a factor of safety of against shear failure FS 25 tfail 120 MPa F124 A B 9 ft 300 lbft F124 Six nails are used to hold the hanger at A against the column Determine the minimum required diameter of each nail to the nearest if it is made of material having Apply a factor of safety of against shear failure FS 2 tfail 16 ksi 116 in 17 DESIGN OF SIMPLE CONNECTIONS 55 1 173 Member B is subjected to a compressive force of 800 lb If A and B are both made of wood and are thick determine to the nearest the smallest dimension h of the horizontal segment so that it does not fail in shear The average shear stress for the segment is tallow 300 psi 1 4 in 3 8 in 176 The lapbelt assembly is to be subjected to a force of 800 N Determine a the required thickness t of the belt if the allowable tensile stress for the material is b the required lap length if the glue can sustain an allowable shear stress of and c the required diameter of the pin if the allowable shear stress for the pin is tallowp 30 MPa dr tallowg 075 MPa dl stallow 10 MPa PROBLEMS 800 lb B h A 12 5 13 Prob 173 174 The lever is attached to the shaft A using a key that has a width d and length of 25 mm If the shaft is fixed and a vertical force of 200 N is applied perpendicular to the handle determine the dimension d if the allowable shear stress for the key is tallow 35 MPa 500 mm 20 mm d a a A 200 N Prob 174 175 The joint is fastened together using two bolts Determine the required diameter of the bolts if the failure shear stress for the bolts is Use a factor of safety for shear of FS 25 tfail 350 MPa 80 kN 40 kN 30 mm 30 mm 40 kN Prob 175 800 N 800 N t dr dl 45 mm Prob 176 177 The wood specimen is subjected to the pull of 10 kN in a tension testing machine If the allowable normal stress for the wood is and the allowable shear stress is determine the required dimensions b and t so that the specimen reaches these stresses simultaneously The specimen has a width of 25 mm tallow 12 MPa stallow 12 MPa 10 kN 10 kN A t b Prob 177 178 Member B is subjected to a compressive force of 600 lb If A and B are both made of wood and are 15 in thick determine to the nearest 18 in the smallest dimension a of the support so that the average shear stress along the blue line does not exceed τallow 50 psi Neglect friction 181 The tension member is fastened together using two bolts one on each side of the member as shown Each bolt has a diameter of 03 in Determine the maximum load P that can be applied to the member if the allowable shear stress for the bolts is τallow 12 ksi and the allowable average normal stress is σallow 20 ksi 179 The joint is used to transmit a torque of T 3 kN m Determine the required minimum diameter of the shear pin A if it is made from a material having a shear failure stress of τfail 150 MPa Apply a factor of safety of 3 against failure 180 Determine the maximum allowable torque T that can be transmitted by the joint The shear pin A has a diameter of 25 mm and it is made from a material having a failure shear stress of τfail 150 MPa Apply a factor of safety of 3 against failure 182 The three steel wires are used to support the load If the wires have an allowable tensile stress of σallow 165 MPa determine the required diameter of each wire if the applied load is P 6 kN 183 The three steel wires are used to support the load If the wires have an allowable tensile stress of σallow 165 MPa and wire AB has a diameter of 6 mm BC has a diameter of 5 mm and BD has a diameter of 7 mm determine the greatest force P that can be applied before one of the wires fails 184 The assembly consists of three disks A B and C that are used to support the load of 140 kN Determine the smallest diameter d1 of the top disk the diameter d2 within the support space and the diameter d3 of the hole in the bottom disk The allowable bearing stress for the material is σallowb 350 MPa and allowable shear stress is τallow 125 MPa 187 The 60 mm 60 mm oak post is supported on the pine block If the allowable bearing stresses for these materials are σoak 43 MPa and σpine 25 MPa determine the greatest load P that can be supported If a rigid bearing plate is used between these materials determine its required area so that the maximum load P can be supported What is this load 185 The boom is supported by the winch cable that has a diameter of 025 in and an allowable normal stress of σallow 24 ksi Determine the greatest load that can be supported without causing the cable to fail when θ 30 and ϕ 45 Neglect the size of the winch 186 The boom is supported by the winch cable that has an allowable normal stress of σallow 24 ksi If it is required that it be able to slowly lift 5000 lb from θ 20 to θ 50 determine the smallest diameter of the cable to the nearest 116 in The boom AB has a length of 20 ft Neglect the size of the winch Set d 12 ft 188 The frame is subjected to the load of 4 kN which acts on member ABD at D Determine the required diameter of the pins at D and C if the allowable shear stress for the material is τallow 40 MPa Pin C is subjected to double shear whereas pin D is subjected to single shear 189 The eye bolt is used to support the load of 5 kip Determine its diameter d to the nearest 18 in and the required thickness h to the nearest 18 in of the support so that the washer will not penetrate or shear through it The allowable normal stress for the bolt is σallow 21 ksi and the allowable shear stress for the supporting material is τallow 5 ksi 190 The softride suspension system of the mountain bike is pinned at C and supported by the shock absorber BD If it is designed to support a load P 1500 N determine the required minimum diameter of pins B and C Use a factor of safety of 2 against failure The pins are made of material having a failure shear stress of τfail 150 MPa and each pin is subjected to double shear 191 The softride suspension system of the mountain bike is pinned at C and supported by the shock absorber BD If it is designed to support a load of P 1500 N determine the factor of safety of pins B and C against failure if they are made of a material having a shear failure stress of τfail 150 MPa Pin B has a diameter of 75 mm and pin C has a diameter of 65 mm Both pins are subjected to double shear 192 The compound wooden beam is connected together by a bolt at B Assuming that the connections at A B C and D exert only vertical forces on the beam determine the required diameter of the bolt at B and the required outer diameter of its washers if the allowable tensile stress for the bolt is σtallow 150 MPa and the allowable bearing stress for the wood is σballow 28 MPa Assume that the hole in the washers has the same diameter as the bolt 193 The assembly is used to support the distributed loading of w 500 lbft Determine the factor of safety with respect to yielding for the steel rod BC and the pins at B and C if the yield stress for the steel in tension is σy 36 ksi and in shear τy 18 ksi The rod has a diameter of 040 in and the pins each have a diameter of 030 in 194 If the allowable shear stress for each of the 030indiameter steel pins at A B and C is τallow 125 ksi and the allowable normal stress for the 040indiameter rod is σallow 22 ksi determine the largest intensity w of the uniform distributed load that can be suspended from the beam 17 DESIGN OF SIMPLE CONNECTIONS 59 1 195 If the allowable bearing stress for the material under the supports at A and B is determine the size of square bearing plates and required to support the load Dimension the plates to the nearest mmThe reactions at the supports are verticalTake P 100 kN B A 1sb2allow 15 MPa 3 m P A B A B 40 kNm 15 m 15 m Probs 19596 196 If the allowable bearing stress for the material under the supports at A and B is determine the maximum load P that can be applied to the beam The bearing plates and have square cross sections of and respectively 250 mm 250 mm 150 mm 150 mm B A 1sb2allow 15 MPa B A D C 4 kN 6 kN 5 kN 3 m 2 m 2 m 3 m Prob 197 197 The rods AB and CD are made of steel having a failure tensile stress of Using a factor of safety of for tension determine their smallest diameter so that they can support the load shown The beam is assumed to be pin connected at A and C FS 175 sfail 510 MPa 198 The aluminum bracket A is used to support the centrally applied load of 8 kip If it has a constant thickness of 05 in determine the smallest height h in order to prevent a shear failure The failure shear stress is Use a factor of safety for shear of FS 25 tfail 23 ksi 8 kip h A Prob 198 20 mm 75 mm 10 mm a a b t P 375 mm 375 mm Probs 199100 199 The hanger is supported using the rectangular pin Determine the magnitude of the allowable suspended load P if the allowable bearing stress is MPa the allowable tensile stress is MPa and the allowable shear stress is Take and 1100 The hanger is supported using the rectangular pin Determine the required thickness t of the hanger and dimensions a and b if the suspended load is The allowable tensile stress is the allowable bearing stress is and the allowable shear stress is tallow 125 MPa sballow 290 MPa stallow 150 MPa P 60 kN b 25 mm a 5 mm t 6 mm tallow 130 MPa stallow 150 sballow 220 CHAPTER REVIEW The internal loadings in a body consist of a normal force shear force bending moment and torsional moment They represent the resultants of both a normal and shear stress distribution that acts over the cross section To obtain these resultants use the method of sections and the equations of equilibrium ΣFₓ 0 ΣFᵧ 0 ΣF𝓏 0 ΣMₓ 0 ΣMᵧ 0 ΣM𝓏 0 If a bar is made from homogeneous isotropic material and it is subjected to a series of external axial loads that pass through the centroid of the cross section then a uniform normal stress distribution will act over the cross section This average normal stress can be determined from σ PA where P is the internal axial load at the section The average shear stress can be determined using τavg VA where V is the shear force acting on the crosssectional area A This formula is often used to find the average shear stress in fasteners or in parts used for connections The design of any simple connection requires that the average stress along any cross section not exceed an allowable stress of σallow or τallow These values are reported in codes and are considered safe on the basis of experiments or through experience Sometimes a factor of safety is reported provided the ultimate stress is known FS σfailσallow τfailτallow CONCEPTUAL PROBLEMS 61 1 P11 Here hurricane winds caused the failure of this highway sign Assuming the wind creates a uniform pressure on the sign of 2 kPa use reasonable dimensions for the sign and determine the resultant shear and moment at the two connections where the failure occurred P13 The hydraulic cylinder H applies a horizontal force F on the pin at A Draw the freebody diagram of the pin and show the forces acting on it Using the method of sections explain why the average shear stress in the pin is largest at sections through the gaps D and E and not at some intermediate section P12 The two structural tubes are connected by the pin which passes through them If the vertical load being supported is 100 kN draw a freebody diagram of the pin and then use the method of sections to find the maximum average shear force in the pin If the pin has a diameter of 50 mm what is the maximum average shear stress in the pin P14 The vertical load on the hook is 1000 lb Draw the appropriate freebody diagrams and determine the maximum average shear force on the pins at A B and C Note that due to symmetry four wheels are used to support the loading on the railing CONCEPTUAL PROBLEMS E H A D A B C P11 P13 P12 P14 REVIEW PROBLEMS 1101 The 200mmdiameter aluminum cylinder supports a compressive load of 300 kN Determine the average normal and shear stress acting on section aa Show the results on a differential element located on the section 1103 Determine the required thickness of the member BC and the diameter of the pins at A and B if the allowable normal stress for member BC is σallow 29 ksi and the allowable shear stress for the pins is τallow 10 ksi 1102 The long bolt passes through the 30mmthick plate If the force in the bolt shank is 8 kN determine the average normal stress in the shank the average shear stress along the cylindrical area of the plate defined by the section lines aa and the average shear stress in the bolt head along the cylindrical area defined by the section lines bb 1104 Determine the resultant internal loadings acting on the cross sections located through points D and E of the frame 1105 The pulley is held fixed to the 20mmdiameter shaft using a key that fits within a groove cut into the pulley and shaft If the suspended load has a mass of 50 kg determine the average shear stress in the key along section aa The key is 5 mm by 5 mm square and 12 mm long 1107 The yokeandrod connection is subjected to a tensile force of 5 kN Determine the average normal stress in each rod and the average shear stress in the pin A between the members 1106 The bearing pad consists of a 150 mm by 150 mm block of aluminum that supports a compressive load of 6 kN Determine the average normal and shear stress acting on the plane through section aa Show the results on a differential volume element located on the plane 1108 The cable has a specific weight γ weightvolume and crosssectional area A If the sag s is small so that its length is approximately L and its weight can be distributed uniformly along the horizontal axis determine the average normal stress in the cable at its lowest point C When the bolt causes compression of these two transparent plates it produces strains in the material that shows up as a spectrum of colors when displayed under polarized light These strains can be related to the stress in the material 2 65 CHAPTER OBJECTIVES In engineering the deformation of a body is specified using the concepts of normal and shear strain In this chapter we will define these quantities and show how they can be determined for various types of problems 21 Deformation Whenever a force is applied to a body it will tend to change the bodys shape and size These changes are referred to as deformation and they may be either highly visible or practically unnoticeable For example a rubber band will undergo a very large deformation when stretched whereas only slight deformations of structural members occur when a building is occupied by people walking about Deformation of a body can also occur when the temperature of the body is changed A typical example is the thermal expansion or contraction of a roof caused by the weather In a general sense the deformation of a body will not be uniform throughout its volume and so the change in geometry of any line segment within the body may vary substantially along its length Hence to study deformational changes in a more uniform manner we will consider line segments that are very short and located in the neighborhood of a point Realize however that these changes will also depend on the orientation of the line segment at the point For example a line segment may elongate if it is oriented in one direction whereas it may contract if it is oriented in another direction Strain Note the before and after positions of three different line segments on this rubber membrane which is subjected to tensionThe vertical line is lengthened the horizontal line is shortened and the inclined line changes its length and rotates 22 Strain In order to describe the deformation of a body by changes in length of line segments and the changes in the angles between them we will develop the concept of strain Strain is actually measured by experiments and once the strain is obtained it will be shown in the next chapter how it can be related to the stress acting within the body Normal Strain If we define the normal strain as the change in length of a line per unit length then we will not have to specify the actual length of any particular line segment Consider for example the line AB which is contained within the undeformed body shown in Fig 21a This line lies along the n axis and has an original length of Δs After deformation points A and B are displaced to A and B and the line becomes a curve having a length of Δs Fig 21b The change in length of the line is therefore Δs Δs If we define the average normal strain using the symbol εavg epsilon then εavg Δs Δs Δs 21 As point B is chosen closer and closer to point A the length of the line will become shorter and shorter such that Δs 0 Also this causes B to approach A such that Δs 0 Consequently in the limit the normal strain at point A and in the direction of n is ε lim BA along n Δs Δs Δs 22 Hence when ε or εavg is positive the initial line will elongate whereas if ε is negative the line contracts Note that normal strain is a dimensionless quantity since it is a ratio of two lengths Although this is the case it is sometimes stated in terms of a ratio of length units If the SI system is used then the basic unit for length is the meter m Ordinarily for most engineering applications ε will be very small so measurements of strain are in micrometers per meter μmm where 1 μm 106 m In the FootPoundSecond system strain is often stated in units of inches per inch inin Sometimes Undeformed body n t A B C a p 2 Deformed body B A C b u 22 STRAIN 67 2 Fig 22 for experimental work strain is expressed as a percent eg As an example a normal strain of can be reported as or 00480Also one can state this answer as simply 480 micros Shear Strain Deformations not only cause line segments to elongate or contract but they also cause them to change direction If we select two line segments that are originally perpendicular to one another then the change in angle that occurs between these two line segments is referred to as shear strain This angle is denoted by gamma and is always measured in radians rad which are dimensionless For example consider the line segments AB and AC originating from the same point A in a body and directed along the perpendicular n and t axes Fig 22a After deformation the ends of both lines are displaced and the lines themselves become curves such that the angle between them at A is Fig 22b Hence the shear strain at point A associated with the n and t axes becomes 23 Notice that if is smaller than the shear strain is positive whereas if is larger than p2 the shear strain is negative u p2 u gnt p 2 lim u B A along n C A along t u g 480 m 48011062 inin 480 mmm 48011062 0001 mm 01 Cartesian Strain Components Using the definitions of normal and shear strain we will now show how they can be used to describe the deformation of the body in Fig 23a To do so imagine the body is subdivided into small elements such as the one shown in Fig 23b This element is rectangular has undeformed dimensions Δx Δy and Δz and is located in the neighborhood of a point in the body Fig 23a If the elements dimensions are very small then its deformed shape will be a parallelepiped Fig 23c since very small line segments will remain approximately straight after the body is deformed In order to achieve this deformed shape we will first consider how the normal strain changes the lengths of the sides of the rectangular element and then how the shear strain changes the angles of each side For example Δx elongates εxΔx so its new length is Δx εxΔx Therefore the approximate lengths of the three sides of the parallelepiped are 1 εx Δx 1 εy Δy 1 εz Δz And the approximate angles between these sides are π2 γxy π2 γyz π2 γxz Notice that the normal strains cause a change in volume of the element whereas the shear strains cause a change in its shape Of course both of these effects occur simultaneously during the deformation In summary then the state of strain at a point in a body requires specifying three normal strains εx εy εz and three shear strains γxy γyz γxz These strains completely describe the deformation of a rectangular volume element of material located at the point and oriented so that its sides are originally parallel to the x y z axes Provided these strains are defined at all points in the body then the deformed shape of the body can be determined 22 STRAIN 69 2 The rubber bearing support under this concrete bridge girder is subjected to both normal and shear strain The normal strain is caused by the weight and bridge loads on the girder and the shear strain is caused by the horizontal movement of the girder due to temperature changes Small Strain Analysis Most engineering design involves applications for which only small deformations are allowed In this text therefore we will assume that the deformations that take place within a body are almost infinitesimal In particular the normal strains occurring within the material are very small compared to 1 so that This assumption has wide practical application in engineering and it is often referred to as a small strain analysis It can be used for example to approximate and provided is very small u tan u u cos u 1 sin u u P V 1 Important Points Loads will cause all material bodies to deform and as a result points in a body will undergo displacements or changes in position Normal strain is a measure per unit length of the elongation or contraction of a small line segment in the bodywhereas shear strain is a measure of the change in angle that occurs between two small line segments that are originally perpendicular to one another The state of strain at a point is characterized by six strain components three normal strains and three shear strains These components depend upon the original orientation of the line segments and their location in the body Strain is the geometrical quantity that is measured using experimental techniques Once obtained the stress in the body can then be determined from material property relations as discussed in the next chapter Most engineering materials undergo very small deformationsand so the normal strain This assumption of small strain analysis allows the calculations for normal strain to be simplified since first order approximations can be made about their size P V 1 gxy gyz gxz Px Py Pz 70 CHAPTER 2 STRAIN 2 The slender rod shown in Fig 24 is subjected to an increase of temperature along its axis which creates a normal strain in the rod of where z is measured in meters Determine a the displacement of the end B of the rod due to the temperature increase and b the average normal strain in the rod Pz 4011032z12 EXAMPLE 21 Fig 24 200 mm A z dz B SOLUTION Part a Since the normal strain is reported at each point along the rod a differential segment dz located at position z Fig 24 has a deformed length that can be determined from Eq 21 that is The sum of these segments along the axis yields the deformed length of the rod ie The displacement of the end of the rod is therefore Ans Part b The average normal strain in the rod is determined from Eq 21 which assumes that the rod or line segment has an original length of 200 mm and a change in length of 239 mm Hence Ans Pavg s s s 239 mm 200 mm 00119 mmmm B 020239 m 02 m 000239 m 239 mm T 020239 m Cz 4011032 2 3 z32D ƒ0 02 m z L 02 m 0 C1 4011032z12D dz dz C1 4011032z12D dz dz dz Pz dz EXAMPLE 22 When force P is applied to the rigid lever arm ABC in Fig 25a the arm rotates counterclockwise about pin A through an angle of 005 Determine the normal strain developed in wire BD SOLUTION I Geometry The orientation of the lever arm after it rotates about point A is shown in Fig 25b From the geometry of this figure α tan1400 mm 300 mm 531301 Then φ 90 α 005 90 531301 005 3692 For triangle ABD the Pythagorean theorem gives LAD 300 mm2 400 mm2 500 mm Using this result and applying the law of cosines to triangle ABD LBD L2AD L2AB 2LADLAB cos φ 500 mm2 400 mm2 2500 mm400 mm cos 3692 3003491 mm Normal Strain εBD LBD LBD LBD 3003491 mm 300 mm 300 mm 000116 mmmm Ans SOLUTION II Since the strain is small this same result can be obtained by approximating the elongation of wire BD as ΔLBD shown in Fig 25b Here ΔLBD θLAB 005 180π rad 400 mm 03491 mm Therefore εBD ΔLBD LBD 03491 mm 300 mm 000116 mmmm Ans Due to a loading the plate is deformed into the dashed shape shown in Fig 26a Determine a the average normal strain along the side AB and b the average shear strain in the plate at A relative to the x and y axes 72 CHAPTER 2 STRAIN 2 EXAMPLE 23 Fig 26 300 mm a C A B y x 250 mm 3 mm 2 mm b A B 250 mm 3 mm 2 mm B c C A u gxy B y x 250 mm 3 mm 2 mm B SOLUTION Part a Line AB coincident with the y axis becomes line after deformation as shown in Fig 26bThe length of is The average normal strain for AB is therefore Ans The negative sign indicates the strain causes a contraction of AB Part b As noted in Fig 26c the once 90 angle BAC between the sides of the plate at A changes to due to the displacement of B to Since then is the angle shown in the figure Thus Ans gxy tan1 a 3 mm 250 mm 2 mm b 00121 rad gxy gxy p2 u B u 79311032 mmmm 1PAB2avg AB AB AB 248018 mm 250 mm 250 mm AB 21250 mm 2 mm22 13 mm22 248018 mm AB AB 22 STRAIN 73 2 The plate shown in Fig 27a is fixed connected along AB and held in the horizontal guides at its top and bottom AD and BC If its right side CD is given a uniform horizontal displacement of 2 mm determine a the average normal strain along the diagonal AC and b the shear strain at E relative to the x y axes SOLUTION Part a When the plate is deformed the diagonal AC becomes Fig 27b The length of diagonals AC and can be found from the Pythagorean theoremWe have Therefore the average normal strain along the diagonal is Ans Part b To find the shear strain at E relative to the x and y axes it is first necessary to find the angle after deformation Fig 27b We have Applying Eq 23 the shear strain at E is therefore Ans The negative sign indicates that the angle is greater than 90 NOTE If the x and y axes were horizontal and vertical at point E then the 90 angle between these axes would not change due to the deformation and so at point E gxy 0 u gxy p 2 158404 rad 00132 rad u 90759 a p 180 b1907592 158404 rad tan a u 2 b 76 mm 75 mm u 000669 mmmm 1PAC2avg AC AC AC 021355 m 021213 m 021213 m AC 210150 m22 10152 m22 021355 m AC 210150 m22 10150 m22 021213 m AC AC EXAMPLE 24 Fig 27 150 mm a 2 mm y 150 mm x D B C A E 76 mm b 75 mm E D B C A 75 mm 76 mm u 74 CHAPTER 2 STRAIN 2 FUNDAMENTAL PROBLEMS F21 When force P is applied to the rigid arm ABC point B displaces vertically downward through a distance of 02 mm Determine the normal strain developed in wire CD F24 The triangular plate is deformed into the shape shown by the dashed line Determine the normal strain developed along edge BC and the average shear strain at corner A with respect to the x and y axes A B C D 400 mm 200 mm 300 mm P F21 F22 If the applied force P causes the rigid arm ABC to rotate clockwise about pin A through an angle of 002 determine the normal strain developed in wires BD and CE A B C D 600 mm 600 mm 600 mm E 400 mm P F22 F23 The rectangular plate is deformed into the shape of a rhombus shown by the dashed line Determine the average shear strain at corner A with respect to the x and y axes y x A B D C 300 mm 2 mm 4 mm 400 mm F23 A C B y x 300 mm 3 mm 5 mm 400 mm F24 F25 The square plate is deformed into the shape shown by the dashed line Determine the average normal strain along diagonal AC and the shear strain of point E with respect to the x and y axes y x E A B D C 300 mm 300 mm 3 mm 3 mm 4 mm F25 PROBLEMS 21 An airfilled rubber ball has a diameter of 6 in If the air pressure within it is increased until the balls diameter becomes 7 in determine the average normal strain in the rubber 22 A thin strip of rubber has an unstretched length of 15 in If it is stretched around a pipe having an outer diameter of 5 in determine the average normal strain in the strip 23 The rigid beam is supported by a pin at A and wires BD and CE If the load P on the beam causes the end C to be displaced 10 mm downward determine the normal strain developed in wires CE and BD 24 The two wires are connected together at A If the force P causes point A to be displaced horizontally 2 mm determine the normal strain developed in each wire 25 The rigid beam is supported by a pin at A and wires BD and CE If the distributed load causes the end C to be displaced 10 mm downward determine the normal strain developed in wires CE and BD 26 Nylon strips are fused to glass plates When moderately heated the nylon will become soft while the glass stays approximately rigid Determine the average shear strain in the nylon due to the load P when the assembly deforms as indicated 27 If the unstretched length of the bowstring is 355 in determine the average normal strain in the string when it is stretched to the position shown 76 CHAPTER 2 STRAIN 2 18 in 6 in 18 in Prob 27 28 Part of a control linkage for an airplane consists of a rigid member CBD and a flexible cable AB If a force is applied to the end D of the member and causes it to rotate by determine the normal strain in the cable Originally the cable is unstretched 29 Part of a control linkage for an airplane consists of a rigid member CBD and a flexible cable AB If a force is applied to the end D of the member and causes a normal strain in the cable of determine the displacement of point D Originally the cable is unstretched 00035 mmmm u 03 400 mm 300 mm A B D P 300 mm C u Probs 289 210 The corners B and D of the square plate are given the displacements indicated Determine the shear strains at A and B 211 The corners B and D of the square plate are given the displacements indicated Determine the average normal strains along side AB and diagonal DB 3 mm 3 mm 16 mm 16 mm 16 mm 16 mm y x A B C D Probs 21011 212 The piece of rubber is originally rectangular Determine the average shear strain at A if the corners B and D are subjected to the displacements that cause the rubber to distort as shown by the dashed lines 213 The piece of rubber is originally rectangular and subjected to the deformation shown by the dashed lines Determine the average normal strain along the diagonal DB and side AD gxy 300 mm 400 mm D A y x 3 mm 2 mm B C Probs 21213 214 Two bars are used to support a load When unloaded AB is 5 in long AC is 8 in long and the ring at A has coordinates 0 0 If a load P acts on the ring at A the normal strain in AB becomes εAB 002 inin and the normal strain in AC becomes εAC 0035 inin Determine the coordinate position of the ring due to the load 215 Two bars are used to support a load P When unloaded AB is 5 in long AC is 8 in long and the ring at A has coordinates 0 0 If a load is applied to the ring at A so that it moves it to the coordinate position 025 in 073 in determine the normal strain in each bar 216 The square deforms into the position shown by the dashed lines Determine the average normal strain along each diagonal AB and CD Side DB remains horizontal 217 The three cords are attached to the ring at B When a force is applied to the ring it moves it to point B such that the normal strain in AB is εAB and the normal strain in CB is εCB Provided these strains are small determine the normal strain in DB Note that AB and CB remain horizontal and vertical respectively due to the roller guides at A and C 218 The piece of plastic is originally rectangular Determine the shear strain γxy at corners A and B if the plastic distorts as shown by the dashed lines 219 The piece of plastic is originally rectangular Determine the shear strain γxy at corners D and C if the plastic distorts as shown by the dashed lines 220 The piece of plastic is originally rectangular Determine the average normal strain that occurs along the diagonals AC and DB 221 The force applied to the handle of the rigid lever arm causes the arm to rotate clockwise through an angle of 3 about pin A Determine the average normal strain developed in the wire Originally the wire is unstretched 222 A square piece of material is deformed into the dashed position Determine the shear strain γxy at A 223 A square piece of material is deformed into the dashed parallelogram Determine the average normal strain that occurs along the diagonals AC and BD 224 A square piece of material is deformed into the dashed position Determine the shear strain γxy at C 225 The guy wire AB of a building frame is originally unstretched Due to an earthquake the two columns of the frame tilt θ 2 Determine the approximate normal strain in the wire when the frame is in this position Assume the columns are rigid and rotate about their lower supports 226 The material distorts into the dashed position shown Determine a the average normal strains along sides AC and CD and the shear strain γxy at F and b the average normal strain along line BE 227 The material distorts into the dashed position shown Determine the average normal strain that occurs along the diagonals AD and CF 228 The wire is subjected to a normal strain that is defined by ε xex2 where x is in millimeters If the wire has an initial length L determine the increase in its length 229 The curved pipe has an original radius of 2 ft If it is heated nonuniformly so that the normal strain along its length is ε 005 cos θ determine the increase in length of the pipe 230 Solve Prob 229 if ε 008 sin θ 231 The rubber band AB has an unstretched length of 1 ft If it is fixed at B and attached to the surface at point A determine the average normal strain in the band The surface is defined by the function y x2 ft where x is in feet 232 The bar is originally 300 mm long when it is flat If it is subjected to a shear strain defined by γxy 002x where x is in meters determine the displacement Δy at the end of its bottom edge It is distorted into the shape shown where no elongation of the bar occurs in the x direction 233 The fiber AB has a length L and orientation θ If its ends A and B undergo very small displacements uA and vB respectively determine the normal strain in the fiber when it is in position AB 234 If the normal strain is defined in reference to the final length that is εn lim pp Δs Δs Δs instead of in reference to the original length Eq 22 show that the difference in these strains is represented as a secondorder term namely εn εn εnεn Horizontal ground displacements caused by an earthquake produced excessive strains in these bridge piers until they fractured The material properties of the concrete and steel reinforcement must be known so that engineers can properly design this structure and thereby avoid such failures 3 81 CHAPTER OBJECTIVES Having discussed the basic concepts of stress and strain we will in this chapter show how stress can be related to strain by using experimental methods to determine the stressstrain diagram for a specific material The behavior described by this diagram will then be discussed for materials that are commonly used in engineering Also mechanical properties and other tests that are related to the development of mechanics of materials will be discussed 31 The Tension and Compression Test The strength of a material depends on its ability to sustain a load without undue deformation or failure This property is inherent in the material itself and must be determined by experiment One of the most important tests to perform in this regard is the tension or compression test Although several important mechanical properties of a material can be determined from this test it is used primarily to determine the relationship between the average normal stress and average normal strain in many engineering materials such as metals ceramics polymers and composites Mechanical Properties of Materials To perform a tension or compression test a specimen of the material is made into a standard shape and size It has a constant circular cross section with enlarged ends so that failure will not occur at the grips Before testing two small punch marks are placed along the specimens uniform length Measurements are taken of both the specimens initial crosssectional area A0 and the gaugelength distance L0 between the punch marks For example when a metal specimen is used in a tension test it generally has an initial diameter of d0 05 in 13 mm and a gauge length of L0 2 in 50 mm Fig 31 In order to apply an axial load with no bending of the specimen the ends are usually seated into ballandsocket joints A testing machine like the one shown in Fig 32 is then used to stretch the specimen at a very slow constant rate until it fails The machine is designed to read the load required to maintain this uniform stretching At frequent intervals during the test data is recorded of the applied load P as read on the dial of the machine or taken from a digital readout Also the elongation δ L L0 between the punch marks on the specimen may be measured using either a caliper or a mechanical or optical device called an extensometer This value of δ delta is then used to calculate the average normal strain in the specimen Sometimes however this measurement is not taken since it is also possible to read the strain directly by using an electricalresistance strain gauge which looks like the one shown in Fig 33 The operation of this gauge is based on the change in electrical resistance of a very thin wire or piece of metal foil under strain Essentially the gauge is cemented to the specimen along its length If the cement is very strong in comparison to the gauge then the gauge is in effect an integral part of the specimen so that when the specimen is strained in the direction of the gauge the wire and specimen will experience the same strain By measuring the electrical resistance of the wire the gauge may be calibrated to read values of normal strain directly 32 THE STRESSSTRAIN DIAGRAM 83 32 The StressStrain Diagram It is not feasible to prepare a test specimen to match the size and of each structural member Rather the test results must be reported so they apply to a member of any size To achieve this the load and corresponding deformation data are used to calculate various values of the stress and corresponding strain in the specimenA plot of the results produces a curve called the stressstrain diagramThere are two ways in which it is normally described Conventional StressStrain Diagram We can determine the nominal or engineering stress by dividing the applied load P by the specimens original crosssectional area This calculation assumes that the stress is constant over the cross section and throughout the gauge lengthWe have 31 Likewise the nominal or engineering strain is found directly from the strain gauge reading or by dividing the change in the specimens gauge length by the specimens original gauge length Here the strain is assumed to be constant throughout the region between the gauge points Thus 32 If the corresponding values of and are plotted so that the vertical axis is the stress and the horizontal axis is the strain the resulting curve is called a conventional stressstrain diagram Realize however that two stressstrain diagrams for a particular material will be quite similar but will never be exactly the sameThis is because the results actually depend on variables such as the materials composition microscopic imperfections the way it is manufactured the rate of loading and the temperature during the time of the test We will now discuss the characteristics of the conventional stressstrain curve as it pertains to steel a commonly used material for fabricating both structural members and mechanical elements Using the method described above the characteristic stressstrain diagram for a steel specimen is shown in Fig 34 From this curve we can identify four different ways in which the material behaves depending on the amount of strain induced in the material P s P d L0 L0 d s P A0 A0 L0 A0 3 84 CHAPTER 3 MECHANICAL PROPERTIES OF MATERIALS Elastic Behavior Elastic behavior of the material occurs when the strains in the specimen are within the light orange region shown in Fig 34 Here the curve is actually a straight line throughout most of this region so that the stress is proportional to the strain The material in this region is said to be linear elastic The upper stress limit to this linear relationship is called the proportional limit If the stress slightly exceeds the proportional limit the curve tends to bend and flatten out as shown This continues until the stress reaches the elastic limit Upon reaching this point if the load is removed the specimen will still return back to its original shape Normally for steel however the elastic limit is seldom determined since it is very close to the proportional limit and therefore rather difficult to detect Yielding A slight increase in stress above the elastic limit will result in a breakdown of the material and cause it to deform permanently This behavior is called yielding and it is indicated by the rectangular dark orange region of the curveThe stress that causes yielding is called the yield stress or yield point and the deformation that occurs is called plastic deformation Although not shown in Fig 34 for low carbon steels or those that are hot rolled the yield point is often distinguished by two values The upper yield point occurs first followed by a sudden decrease in loadcarrying capacity to a lower yield point Notice that once the yield point is reached then as shown in Fig 34 the specimen will continue to elongate strain without any increase in loadWhen the material is in this state it is often referred to as being perfectly plastic sY spl 3 elastic region yielding strain hardening necking elastic behavior plastic behavior elastic limit yield stress ultimate stress true fracture stress fracture stress Conventional and true stressstrain diagrams for ductile material steel not to scale P sf sf sY spl su s proportional limit Fig 34 32 THE STRESSSTRAIN DIAGRAM 85 Strain Hardening When yielding has ended an increase in load can be supported by the specimen resulting in a curve that rises continuously but becomes flatter until it reaches a maximum stress referred to as the ultimate stress The rise in the curve in this manner is called strain hardening and it is identified in Fig 34 as the region in light green Necking Up to the ultimate stress as the specimen elongates its crosssectional area will decreaseThis decrease is fairly uniform over the specimens entire gauge length however just after at the ultimate stress the crosssectional area will begin to decrease in a localized region of the specimen As a result a constriction or neck tends to form in this region as the specimen elongates further Fig 35a This region of the curve due to necking is indicated in dark green in Fig 34 Here the stressstrain diagram tends to curve downward until the specimen breaks at the fracture stress Fig 35b True StressStrain Diagram Instead of always using the original crosssectional area and specimen length to calculate the engineering stress and strain we could have used the actual crosssectional area and specimen length at the instant the load is measured The values of stress and strain found from these measurements are called true stress and true strain and a plot of their values is called the true stressstrain diagram When this diagram is plotted it has a form shown by the lightblue curve in Fig 34 Note that the conventional and true diagrams are practically coincident when the strain is small The differences between the diagrams begin to appear in the strainhardening range where the magnitude of strain becomes more significant In particular there is a large divergence within the necking region Here it can be seen from the conventional diagram that the specimen actually supports a decreasing load since is constant when calculating engineering stress However from the true diagram the actual area A within the necking region is always decreasing until fracture and so the material actually sustains increasing stress since s PA sf sP s PA0 A0 sP sP sf su 3 Typical necking pattern which has occurred on this steel specimen just before fracture Necking Failure of a ductile material Fig 35 a b This steel specimen clearly shows the necking that occurred just before the specimen failed This resulted in the formation of a cupcone shape at the fracture location which is characteristic of ductile materials Although the true and conventional stressstrain diagrams are different most engineering design is done so that the material supports a stress within the elastic range This is because the deformation of the material is generally not severe and the material will restore itself when the load is removed The true strain up to the elastic limit will remain small enough so that the error in using the engineering values of σ and ε is very small about 01 compared with their true values This is one of the primary reasons for using conventional stressstrain diagrams The above concepts can be summarized with reference to Fig 36 which shows an actual conventional stressstrain diagram for a mild steel specimen In order to enhance the details the elastic region of the curve has been shown in light blue color using an exaggerated strain scale also shown in light blue Tracing the behavior the proportional limit is reached at σpl 35 ksi 241 MPa where εpl 00012 inin This is followed by an upper yield point of σYu 38 ksi 262 MPa then suddenly a lower yield point of σYl 36 ksi 248 MPa The end of yielding occurs at a strain of εY 0030 inin which is 25 times greater than the strain at the proportional limit Continuing the specimen undergoes strain hardening until it reaches the ultimate stress of σu 63 ksi 434 MPa then it begins to neck down until a fracture occurs σf 47 ksi 324 MPa By comparison the strain at failure εf 0380 inin is 317 times greater than εpl 33 StressStrain Behavior of Ductile and Brittle Materials Materials can be classified as either being ductile or brittle depending on their stressstrain characteristics Ductile Materials Any material that can be subjected to large strains before it fractures is called a ductile material Mild steel as discussed previously is a typical example Engineers often choose ductile materials for design because these materials are capable of absorbing shock or energy and if they become overloaded they will usually exhibit large deformation before failing One way to specify the ductility of a material is to report its percent elongation or percent reduction in area at the time of fracture The percent elongation is the specimens fracture strain expressed as a percent Thus if the specimens original gauge length is L0 and its length at fracture is Lf then Percent elongation Lf L0L0 100 33 As seen in Fig 36 since εf 0380 this value would be 38 for a mild steel specimen The percent reduction in area is another way to specify ductility It is defined within the region of necking as follows Percent reduction of area A0 AfA0 100 34 Here A0 is the specimens original crosssectional area and Af is the area of the neck at fracture Mild steel has a typical value of 60 Besides steel other metals such as brass molybdenum and zinc may also exhibit ductile stressstrain characteristics similar to steel whereby they undergo elastic stressstrain behavior yielding at constant stress strain hardening and finally necking until fracture In most metals however constant yielding will not occur beyond the elastic range One metal for which this is the case is aluminum Actually this metal often does not have a welldefined yield point and consequently it is standard practice to define a yield strength using a graphical procedure called the offset method Normally a 02 strain 0002 inin is chosen and from this point on the ε axis a line parallel to the initial straightline portion of the stressstrain diagram is drawn The point where this line intersects the curve defines the yield strength An example of the construction for determining the yield strength for an aluminum alloy is shown in Fig 37 From the graph the yield strength is σYS 51 ksi 352 MPa 88 CHAPTER 3 MECHANICAL PROPERTIES OF MATERIALS σ ksi 20 15 10 05 0 2 4 6 8 10 ε inin σε diagram for natural rubber Fig 38 σ ksi σf 22 20 001 006 005 004 003 002 001 A B 20 40 60 80 100 120 C ε inin σε diagram for gray cast iron Fig 39 Concrete used for structural purposes must be routinely tested in compression to be sure it provides the necessary design strength for this bridge deck The concrete cylinders shown are compression tested for ultimate stress after curing for 30 days Realize that the yield strength is not a physical property of the material since it is a stress that causes a specified permanent strain in the material In this text however we will assume that the yield strength yield point elastic limit and proportional limit all coincide unless otherwise stated An exception would be natural rubber which in fact does not even have a proportional limit since stress and strain are not linearly related Instead as shown in Fig 38 this material which is known as a polymer exhibits nonlinear elastic behavior Wood is a material that is often moderately ductile and as a result it is usually designed to respond only to elastic loadings The strength characteristics of wood vary greatly from one species to another and for each species they depend on the moisture content age and the size and arrangement of knots in the wood Since wood is a fibrous material its tensile or compressive characteristics will differ greatly when it is loaded either parallel or perpendicular to its grain Specifically wood splits easily when it is loaded in tension perpendicular to its grain and consequently tensile loads are almost always intended to be applied parallel to the grain of wood members 89 33 STRESSSTRAIN BEHAVIOR OF DUCTILE AND BRITTLE MATERIALS Tension failure of a brittle material a Compression causes material to bulge out b Fig 310 σ ksi σtmax 04 2 0 00030 00025 00020 00015 00010 00005 0 00005 ε inin 2 4 σcmax 5 6 σε diagram for typical concrete mix Fig 311 Brittle Materials Materials that exhibit little or no yielding before failure are referred to as brittle materials Gray cast iron is an example having a stressstrain diagram in tension as shown by portion AB of the curve in Fig 39 Here fracture at σf 22 ksi 152 MPa took place initially at an imperfection or microscopic crack and then spread rapidly across the specimen causing complete fracture Since the appearance of initial cracks in a specimen is quite random brittle materials do not have a welldefined tensile fracture stress Instead the average fracture stress from a set of observed tests is generally reported A typical failed specimen is shown in Fig 310a Compared with their behavior in tension brittle materials such as gray cast iron exhibit a much higher resistance to axial compression as evidenced by portion AC of the curve in Fig 39 For this case any cracks or imperfections in the specimen tend to close up and as the load increases the material will generally bulge or become barrel shaped as the strains become larger Fig 310b Like gray cast iron concrete is classified as a brittle material and it also has a low strength capacity in tension The characteristics of its stressstrain diagram depend primarily on the mix of concrete water sand gravel and cement and the time and temperature of curing A typical example of a complete stressstrain diagram for concrete is given in Fig 311 By inspection its maximum compressive strength is almost 125 times greater than its tensile strength σcmax 5 ksi 345 MPa versus σtmax 040 ksi 276 MPa For this reason concrete is almost always reinforced with steel bars or rods whenever it is designed to support tensile loads It can generally be stated that most materials exhibit both ductile and brittle behavior For example steel has brittle behavior when it contains a high carbon content and it is ductile when the carbon content is reduced Also at low temperatures materials become harder and more brittle whereas when the temperature rises they become softer and more ductile This effect is shown in Fig 312 for a methacrylate plastic Steel rapidly loses its strength when heated For this reason engineers often require main structural members to be insulated in case of fire σ ksi 9 8 7 6 5 4 3 2 1 001 002 003 004 005 006 ε inin 40F 110F 160F σε diagrams for a methacrylate plastic Fig 312 90 CHAPTER 3 MECHANICAL PROPERTIES OF MATERIALS 34 Hookes Law As noted in the previous section the stressstrain diagrams for most engineering materials exhibit a linear relationship between stress and strain within the elastic region Consequently an increase in stress causes a proportionate increase in strain This fact was discovered by Robert Hooke in 1676 using springs and is known as Hookes law It may be expressed mathematically as 35 Here E represents the constant of proportionality which is called the modulus of elasticity or Youngs modulus named after Thomas Young who published an account of it in 1807 Equation 35 actually represents the equation of the initial straight lined portion of the stressstrain diagram up to the proportional limit Furthermore the modulus of elasticity represents the slope of this line Since strain is dimensionless from Eq 35 E will have the same units as stress such as psi ksi or pascals As an example of its calculation consider the stressstrain diagram for steel shown in Fig 36 Here and so that As shown in Fig 313 the proportional limit for a particular type of steel alloy depends on its carbon content however most grades of steel from the softest rolled steel to the hardest tool steel have about the E spl Ppl 35 ksi 00012 inin 2911032 ksi Ppl 00012 inin spl 35 ksi s EP 0002 0004 0006 0008 001 160 140 120 100 80 60 40 20 180 structural steel 02 carbon soft steel 01 carbon machine steel 06 carbon hard steel 06 carbon heat treated spring steel 1 carbon s ksi P inin Fig 313 3 34 HOOKES LAW 91 same modulus of elasticity generally accepted to be or 200 GPa Values of E for other commonly used engineering materials are often tabulated in engineering codes and reference books Representative values are also listed on the inside back cover of this book It should be noted that the modulus of elasticity is a mechanical property that indicates the stiffness of a material Materials that are very stiff such as steel have large values of E or 200 GPa whereas spongy materials such as vulcanized rubber may have low values or 070 MPa The modulus of elasticity is one of the most important mechanical properties used in the development of equations presented in this text It must always be remembered though that E can be used only if a material has linear elastic behavior Also if the stress in the material is greater than the proportional limit the stressstrain diagram ceases to be a straight line and so Eq 35 is no longer valid Strain Hardening If a specimen of ductile material such as steel is loaded into the plastic region and then unloaded elastic strain is recovered as the material returns to its equilibrium state The plastic strain remains however and as a result the material is subjected to a permanent set For example a wire when bent plastically will spring back a little elastically when the load is removed however it will not fully return to its original position This behavior can be illustrated on the stressstrain diagram shown in Fig 314a Here the specimen is first loaded beyond its yield point A to point Since interatomic forces have to be overcome to elongate the specimen elastically then these same forces pull the atoms back together when the load is removed Fig 314a Consequently the modulus of elasticity E is the same and therefore the slope of line is the same as line OA If the load is reappliedthe atoms in the material will again be displaced until yielding occurs at or near the stress and the stressstrain diagram continues along the same path as before Fig 314b It should be noted however that this new stressstrain diagram defined by now has a higher yield point a consequence of strainhardening In other words the material now has a greater elastic region however it has less ductility a smaller plastic region than when it was in its original state 1A2 OAB A OA A Er 010 ksi Est 2911032 ksi Est 2911032 ksi 3 a permanent set elastic recovery elastic region plastic region load unload A A B O O E E s P b O O elastic region plastic region A B P s Fig 314 This pin was made from a hardened steel alloy that is one having a high carbon content It failed due to brittle fracture 35 Strain Energy As a material is deformed by an external loading it tends to store energy internally throughout its volume Since this energy is related to the strains in the material it is referred to as strain energy To obtain this strain energy consider a volume element of material from a tension test specimen It is subjected to uniaxial stress as shown in Fig 315 This stress develops a force F σ A σx y on the top and bottom faces of the element after the element of length z undergoes a vertical displacement ϵ z By definition work is determined by the product of the force and displacement in the direction of the force Since the force is increased uniformly from zero to its final magnitude F when the displacement ϵ z is attained the work done on the element by the force is equal to the average force magnitude F2 times the displacement ϵ z This external work on the element is equivalent to the internal work or strain energy stored in the elementassuming that no energy is lost in the form of heat Consequently the strain energy U is U 12 F ϵ z 12 σ x y ϵ z Since the volume of the element is V x y z then U 12 σ ϵ V For applications it is sometimes convenient to specify the strain energy per unit volume of material This is called the strainenergy density and it can be expressed as u ΔUΔV 12 σε 36 If the material behavior is linear elastic then Hookes law applies σ Eϵ and therefore we can express the elastic strainenergy density in terms of the uniaxial stress as u 12 σ2E 37 Modulus of Resilience In particular when the stress σ reaches the proportional limit the strainenergy density as calculated by Eq 36 or 37 is referred to as the modulus of resilience ie ur 12 σpl ϵpl 12 σpl2E 38 From the elastic region of the stressstrain diagram Fig 316a notice that ur is equivalent to the shaded triangular area under the diagram Physically a materials resilience represents the ability of the material to absorb energy without any permanent damage to the material 35 STRAIN ENERGY 93 Modulus of Toughness Another important property of a material is the modulus of toughness This quantity represents the entire area under the stressstrain diagram Fig 316b and therefore it indicates the strainenergy density of the material just before it fractures This property becomes important when designing members that may be accidentally overloadedAlloying metals can also change their resilience and toughness For example by changing the percentage of carbon in steel the resulting stressstrain diagrams in Fig 317 show how the degrees of resilience and toughness can be changed ut 3 b ut Modulus of toughness ut P s Fig 316 cont soft steel 01 carbon most ductile hard steel 06 carbon highest strength structural steel 02 carbon toughest P s Fig 317 This nylon specimen exhibits a high degree of toughness as noted by the large amount of necking that has occurred just before fracture Important Points A conventional stressstrain diagram is important in engineering since it provides a means for obtaining data about a materials tensile or compressive strength without regard for the materials physical size or shape Engineering stress and strain are calculated using the original crosssectional area and gauge length of the specimen A ductile material such as mild steel has four distinct behaviors as it is loadedThey are elastic behavioryieldingstrain hardening and necking A material is linear elastic if the stress is proportional to the strain within the elastic regionThis behavior is described by Hookes law where the modulus of elasticity E is the slope of the line Important points on the stressstrain diagram are the proportional limit elastic limit yield stress ultimate stress and fracture stress The ductility of a material can be specified by the specimens percent elongation or the percent reduction in area If a material does not have a distinct yield pointa yield strength can be specified using a graphical procedure such as the offset method Brittle materials such as gray cast iron have very little or no yielding and so they can fracture suddenly Strain hardening is used to establish a higher yield point for a materialThis is done by straining the material beyond the elastic limit then releasing the load The modulus of elasticity remains the same however the materials ductility decreases Strain energy is energy stored in a material due to its deformation This energy per unit volume is called strainenergy density If it is measured up to the proportional limit it is referred to as the modulus of resilience and if it is measured up to the point of fracture it is called the modulus of toughness It can be determined from the area under the diagram sP s EP EXAMPLE 31 A tension test for a steel alloy results in the stressstrain diagram shown in Fig 318 Calculate the modulus of elasticity and the yield strength based on a 02 offset Identify on the graph the ultimate stress and the fracture stress SOLUTION Modulus of Elasticity We must calculate the slope of the initial straightline portion of the graph Using the magnified curve and scale shown in blue this line extends from point O to an estimated point A which has coordinates of approximately 00016 inin 50 ksi Therefore E 50 ksi00016 inin 312103 ksi Ans Note that the equation of line OA is thus σ 312103ϵ Yield Strength For a 02 offset we begin at a strain of 02 or 00020 inin and graphically extend a dashed line parallel to OA until it intersects the σϵ curve at A The yield strength is approximately σYS 68 ksi Ans Ultimate Stress This is defined by the peak of the σϵ graph point B in Fig 318 σu 108 ksi Ans Fracture Stress When the specimen is strained to its maximum of ϵf 023 inin it fractures at point C Thus σf 90 ksi Ans EXAMPLE 32 The stressstrain diagram for an aluminum alloy that is used for making aircraft parts is shown in Fig 319 If a specimen of this material is stressed to 600 MPa determine the permanent strain that remains in the specimen when the load is released Also find the modulus of resilience both before and after the load application SOLUTION Permanent Strain When the specimen is subjected to the load it strainhardens until point B is reached on the σϵ diagram The strain at this point is approximately 0023 mmmm When the load is released the material behaves by following the straight line BC which is parallel to line OA Since both lines have the same slope the strain at point C can be determined analytically The slope of line OA is the modulus of elasticity ie E 450 MPa0006 mmmm 750 GPa From triangle CBD we require E BDCD 750109 Pa 600106 PaCD CD 0008 mmmm This strain represents the amount of recovered elastic strain The permanent strain ϵOC is thus ϵOC 0023 mmmm 0008 mmmm 00150 mmmm Ans Note If gauge marks on the specimen were originally 50 mm apart then after the load is released these marks will be 50 mm 0015050 mm 5075 mm apart Modulus of Resilience Applying Eq 38 we have urinitial 12 σpl ϵpl 12450 MPa0006 mmmm 135 MJm3 Ans urfinal 12 σpl ϵpl 12600 MPa0008 mmmm 240 MJm3 Ans NOTE By comparison the effect of strainhardening the material has caused an increase in the modulus of resilience however note that the modulus of toughness for the material has decreased since the area under the original curve OABF is larger than the area under curve CBF Work in the SI system of units is measured in joules where 1 J 1 Nm EXAMPLE 33 An aluminum rod shown in Fig 320a has a circular cross section and is subjected to an axial load of 10 kN If a portion of the stressstrain diagram is shown in Fig 320b determine the approximate elongation of the rod when the load is applied Take Eal 70 GPa SOLUTION For the analysis we will neglect the localized deformations at the point of load application and where the rods crosssectional area suddenly changes These effects will be discussed in Sections 41 and 47 Throughout the midsection of each segment the normal stress and deformation are uniform In order to find the elongation of the rod we must first obtain the strain This is done by calculating the stress then using the stressstrain diagram The normal stress within each segment is σAB PA 10103 N π 001 m2 3183 MPa σBC PA 10103 N π 00075 m2 5659 MPa From the stressstrain diagram the material in segment AB is strained elastically since σAB σY 40 MPa Using Hookes law εAB σAB Eal 3183106 Pa 70109 Pa 00004547 mmmm The material within segment BC is strained plastically since σBC σY 40 MPa From the graph for σBC 5659 MPa εBC 0045 mmmm The approximate elongation of the rod is therefore δ ΣεL 00004547 600 mm 00450 400 mm 183 mm Ans 35 STRAIN ENERGY 97 3 FUNDAMENTAL PROBLEMS F31 Define homogeneous material F32 Indicate the points on the stressstrain diagram which represent the proportional limit and the ultimate stress F310 The material for the 50mmlong specimen has the stressstrain diagram shown If determine the elongation of the specimen F311 The material for the 50mmlong specimen has the stressstrain diagram shown If is applied and then released determine the permanent elongation of the specimen P 150 kN P 100 kN A B C E P D s F32 F33 Define the modulus of elasticity E F34 At room temperature mild steel is a ductile materialTrue or false F35 Engineering stress and strain are calculated using the actual crosssectional area and length of the specimen True or false F36 As the temperature increases the modulus of elasticity will increaseTrue or false F37 A 100mm long rod has a diameter of 15 mm If an axial tensile load of 100 kN is applied determine its change is length F38 A bar has a length of 8 in and crosssectional area of 12 in2 Determine the modulus of elasticity of the material if it is subjected to an axial tensile load of 10 kip and stretches 0003 in The material has linearelastic behavior F39 A 10mmdiameter brass rod has a modulus of elasticity of If it is 4 m long and subjected to an axial tensile load of 6 kN determine its elongation E 100 GPa E 200 GPa P P 450 000225 003 P mmmm 500 20 mm s MPa F31011 F312 If the elongation of wire BC is 02 mm after the force P is applied determine the magnitude of P The wire is A36 steel and has a diameter of 3 mm A B C 400 mm 200 mm 300 mm P F312 98 CHAPTER 3 MECHANICAL PROPERTIES OF MATERIALS 3 31 A concrete cylinder having a diameter of 600 in and gauge length of 12 in is tested in compressionThe results of the test are reported in the table as load versus contraction Draw the stressstrain diagram using scales of and From the diagram determine approximately the modulus of elasticity 1 in 0211032 inin 1 in 05 ksi 34 A tension test was performed on a specimen having an original diameter of 125 mm and a gauge length of 50 mmThe data are listed in the table Plot the stressstrain diagram and determine approximately the modulus of elasticity the ultimate stress and the fracture stress Use a scale of and Redraw the linearelastic region using the same stress scale but a strain scale of 35 A tension test was performed on a steel specimen having an original diameter of 125 mm and gauge length of 50 mm Using the data listed in the table plot the stressstrain diagram and determine approximately the modulus of toughness Use a scale of and 20 mm 005 mmmm 20 mm 50 MPa 20 mm 0001 mmmm 20 mm 005 mmmm 20 mm 50 MPa PROBLEMS 0 50 95 165 205 255 300 345 385 465 500 530 0 00006 00012 00020 00026 00034 00040 00045 00050 00062 00070 00075 Load kip Contraction in Prob 31 32 Data taken from a stressstrain test for a ceramic are given in the tableThe curve is linear between the origin and the first point Plot the diagram and determine the modulus of elasticity and the modulus of resilience 33 Data taken from a stressstrain test for a ceramic are given in the table The curve is linear between the origin and the first point Plot the diagram and determine approximately the modulus of toughnessThe rupture stress is sr 534 ksi 0 332 455 494 515 534 0 00006 00010 00014 00018 00022 S ksi P inin Probs 323 0 111 319 378 409 436 534 623 645 623 588 0 00175 00600 01020 01650 02490 10160 30480 63500 88900 119380 Load kN Elongation mm Probs 345 36 A specimen is originally 1 ft long has a diameter of 05 in and is subjected to a force of 500 lb When the force is increased from 500 lb to 1800 lb the specimen elongates 0009 in Determine the modulus of elasticity for the material if it remains linear elastic 37 A structural member in a nuclear reactor is made of a zirconium alloy If an axial load of 4 kip is to be supported by the member determine its required crosssectional area Use a factor of safety of 3 relative to yielding What is the load on the member if it is 3 ft long and its elongation is 002 in ksi ksi The material has elastic behavior sY 575 Ezr 14103 38 The strut is supported by a pin at C and an A36 steel guy wire AB If the wire has a diameter of 02 in determine how much it stretches when the distributed load acts on the strut 39 The σε diagram for a collagen fiber bundle from which a human tendon is composed is shown If a segment of the Achilles tendon at A has a length of 65 in and an approximate crosssectional area of 0229 in2 determine its elongation if the foot supports a load of 125 lb which causes a tension in the tendon of 34375 lb 310 The stressstrain diagram for a metal alloy having an original diameter of 05 in and a gauge length of 2 in is given in the figure Determine approximately the modulus of elasticity for the material the load on the specimen that causes yielding and the ultimate load the specimen will support 311 The stressstrain diagram for a steel alloy having an original diameter of 05 in and a gauge length of 2 in is given in the figure If the specimen is loaded until it is stressed to 90 ksi determine the approximate amount of elastic recovery and the increase in the gauge length after it is unloaded 312 The stressstrain diagram for a steel alloy having an original diameter of 05 in and a gauge length of 2 in is given in the figure Determine approximately the modulus of resilience and the modulus of toughness for the material 313 A bar having a length of 5 in and crosssectional area of 07 in2 is subjected to an axial force of 8000 lb If the bar stretches 0002 in determine the modulus of elasticity of the material The material has linearelastic behavior 314 The rigid pipe is supported by a pin at A and an A36 steel guy wire BD If the wire has a diameter of 025 in determine how much it stretches when a load of P 600 lb acts on the pipe 315 The rigid pipe is supported by a pin at A and an A36 guy wire BD If the wire has a diameter of 025 in determine the load P if the end C is displaced 0075 in downward 316 Determine the elongation of the square hollow bar when it is subjected to the axial force P 100 kN If this axial force is increased to P 360 kN and released find the permanent elongation of the bar The bar is made of a metal alloy having a stressstrain diagram which can be approximated as shown 317 A tension test was performed on an aluminum 2014T6 alloy specimen The resulting stressstrain diagram is shown in the figure Estimate a the proportional limit b the modulus of elasticity and c the yield strength based on a 02 strain offset method 318 A tension test was performed on an aluminum 2014T6 alloy specimen The resulting stressstrain diagram is shown in the figure Estimate a the modulus of resilience and b modulus of toughness 319 The stressstrain diagram for a bone is shown and can be described by the equation ε 045106σ 0361012σ3 where σ is in kPa Determine the yield strength assuming a 03 offset 320 The stressstrain diagram for a bone is shown and can be described by the equation ε 045106σ 0361012σ3 where σ is in kPa Determine the modulus of toughness and the amount of elongation of a 200mmlong region just before it fractures if failure occurs at ε 012 mmmm 35 STRAIN ENERGY 101 323 By adding plasticizers to polyvinyl chloride it is possible to reduce its stiffness The stressstrain diagrams for three types of this material showing this effect are given below Specify the type that should be used in the manufacture of a rod having a length of 5 in and a diameter of 2 in that is required to support at least an axial load of 20 kip and also be able to stretch at most 1 4 in 321 The stressstrain diagram for a polyester resin is given in the figure If the rigid beam is supported by a strut AB and post CD both made from this material and subjected to a load of determine the angle of tilt of the beam when the load is applied The diameter of the strut is 40 mm and the diameter of the post is 80 mm 322 The stressstrain diagram for a polyester resin is given in the figure If the rigid beam is supported by a strut AB and post CD made from this material determine the largest load P that can be applied to the beam before it ruptures The diameter of the strut is 12 mm and the diameter of the post is 40 mm P 80 kN 3 0 tension compression 001 002 003 004 95 80 100 70 60 50 40 322 20 0 075 m B C D A P 075 m 05 m 2 m P mmmm s MPa Probs 32122 s ksi 0 15 inin 010 020 030 P P flexible plasticized unplasticized copolymer P 10 5 0 Prob 323 s ksi P 106 01 02 03 04 05 80 60 40 20 Prob 324 324 The stressstrain diagram for many metal alloys can be described analytically using the RambergOsgood three parameter equation where E k and n are determined from measurements taken from the diagram Using the stressstrain diagram shown in the figure take and determine the other two parameters k and n and thereby obtain an analytical expression for the curve E 3011032 ksi P sE ksn 102 CHAPTER 3 MECHANICAL PROPERTIES OF MATERIALS 36 Poissons Ratio When a deformable body is subjected to an axial tensile force not only does it elongate but it also contracts laterally For example if a rubber band is stretched it can be noted that both the thickness and width of the band are decreasedLikewisea compressive force acting on a body causes it to contract in the direction of the force and yet its sides expand laterally Consider a bar having an original radius r and length L and subjected to the tensile force P in Fig 321 This force elongates the bar by an amount and its radius contracts by an amount Strains in the longitudinal or axial direction and in the lateral or radial direction are respectively In the early 1800sthe French scientist SDPoisson realized that within the elastic range the ratio of these strains is a constant since the deformations and are proportional This constant is referred to as Poissons ratio nu and it has a numerical value that is unique for a particular material that is both homogeneous and isotropic Stated mathematically it is 39 The negative sign is included here since longitudinal elongation positive strain causes lateral contraction negative strain and vice versa Notice that these strains are caused only by the axial or longitudinal force P ie no force or stress acts in a lateral direction in order to strain the material in this direction Poissons ratio is a dimensionless quantity and for most nonporous solids it has a value that is generally between and Typical values of for common engineering materials are listed on the inside back cover For an ideal material having no lateral deformation when it is stretched or compressed Poissons ratio will be 0 Furthermore it will be shown in Sec 106 that the maximum possible value for Poissons ratio is 05 Therefore 0 n 05 n 1 3 1 4 n Plat Plong n d d Plong d L and Plat d r d d 3 P P r Final Shape L Original Shape Tension d2 d d2 When the rubber block is compressed negative strain its sides will expand positive strain The ratio of these strains remains constant Fig 321 EXAMPLE 34 A bar made of A36 steel has the dimensions shown in Fig 322 If an axial force of P 80 kN is applied to the bar determine the change in its length and the change in the dimensions of its cross section after applying the load The material behaves elastically SOLUTION The normal stress in the bar is σz PA 80103N 01 m005 m 160106 Pa From the table on the inside back cover for A36 steel Est 200 GPa and so the strain in the z direction is εz σz Est 160106 Pa 200109 Pa 80106 mmmm The axial elongation of the bar is therefore δz εzLz 8010615 m 120 μm Ans Using Eq 39 where νst 032 as found from the inside back cover the lateral contraction strains in both the x and y directions are εx εy νstεz 03280106 256 μmm Thus the changes in the dimensions of the cross section are δx εxLx 25610601 m 256 μm Ans δy εyLy 256106005 m 128 μm Ans 37 The Shear StressStrain Diagram In Sec 15 it was shown that when a small element of material is subjected to pure shear equilibrium requires that equal shear stresses must be developed on four faces of the element These stresses τxy must be directed toward or away from diagonally opposite corners of the element as shown in Fig 323a Furthermore if the material is homogeneous and isotropic then this shear stress will distort the element uniformly Fig 323b As mentioned in Sec 22 the shear strain γxy measures the angular distortion of the element relative to the sides originally along the x and y axes The behavior of a material subjected to pure shear can be studied in a laboratory using specimens in the shape of thin tubes and subjecting them to a torsional loading If measurements are made of the applied torque and the resulting angle of twist then by the methods to be explained in Chapter 5 the data can be used to determine the shear stress and shear strain and a shear stressstrain diagram plotted An example of such a diagram for a ductile material is shown in Fig 324 Like the tension test this material when subjected to shear will exhibit linearelastic behavior and it will have a defined proportional limit τpl Also strain hardening will occur until an ultimate shear stress τu is reached And finally the material will begin to lose its shear strength until it reaches a point where it fractures τf For most engineering materials like the one just described the elastic behavior is linear and so Hookes law for shear can be written as τ Gγ 310 Here G is called the shear modulus of elasticity or the modulus of rigidity Its value represents the slope of the line on the τγ diagram that is G τpl γpl Typical values for common engineering materials are listed on the inside back cover Notice that the units of measurement for G will be the same as those for τ Pa or psi since γ is measured in radians a dimensionless quantity It will be shown in Sec 106 that the three material constants E ν and G are actually related by the equation G E 21ν 311 Provided E and G are known the value of ν can then be determined from this equation rather than through experimental measurement For example in the case of A36 steel Est 29103 ksi and Gst 110103 ksi so that from Eq 311 νst 032 EXAMPLE 35 A specimen of titanium alloy is tested in torsion and the shear stressstrain diagram is shown in Fig 325a Determine the shear modulus G the proportional limit and the ultimate shear stress Also determine the maximum distance d that the top of a block of this material shown in Fig 325b could be displaced horizontally if the material behaves elastically when acted upon by a shear force V What is the magnitude of V necessary to cause this displacement SOLUTION Shear Modulus This value represents the slope of the straightline portion OA of the τγ diagram The coordinates of point A are 0008 rad 52 ksi Thus G 52 ksi 0008 rad 6500 ksi Ans The equation of line OA is therefore τ Gγ 6500γ which is Hookes law for shear Proportional Limit By inspection the graph ceases to be linear at point A Thus τpl 52 ksi Ans Ultimate Stress This value represents the maximum shear stress point B From the graph τu 73 ksi Ans Maximum Elastic Displacement and Shear Force Since the maximum elastic shear strain is 0008 rad a very small angle the top of the block in Fig 325b will be displaced horizontally tan0008 rad 0008 rad d 2 in d 0016 in Ans The corresponding average shear stress in the block is τpl 52 ksi Thus the shear force V needed to cause the displacement is τavg V A 52 ksi V 3 in4 in V 624 kip Ans EXAMPLE 36 An aluminum specimen shown in Fig 326 has a diameter of and a gauge length of If a force of 165 kN elongates the gauge length 120 mm determine the modulus of elasticityAlso determine by how much the force causes the diameter of the specimen to contract Take and SOLUTION Modulus of Elasticity The average normal stress in the specimen is and the average normal strain is Since the material behaves elastically The modulus of elasticity is therefore Ans Contraction of Diameter First we will determine Poissons ratio for the material using Eq 311 Since then by Eq 39 The contraction of the diameter is therefore Ans 00416 mm d 10001662125 mm2 Plat 000166 mmmm 0347 Plat 000480 mmmm n Plat Plong Plong 000480 mmmm n 0347 26 GPa 700 GPa 211 n2 G E 211 n2 Eal s P 336111062 Pa 000480 700 GPa s 6 sY 440 MPa P d L 120 mm 250 mm 000480 mmmm s P A 16511032 N 1p4210025 m22 3361 MPa sY 440 MPa Gal 26 GPa L0 250 mm d0 25 mm 106 CHAPTER 3 MECHANICAL PROPERTIES OF MATERIALS 3 d0 L0 165 kN 165 kN Fig 326 108 CHAPTER 3 MECHANICAL PROPERTIES OF MATERIALS In general the creep strength will decrease for higher temperatures or for higher applied stresses For longer periods of time extrapolations from the curves must be madeTo do this usually requires a certain amount of experience with creep behavior and some supplementary knowledge about the creep properties of the material Once the materials creep strength has been determined however a factor of safety is applied to obtain an appropriate allowable stress for design Fatigue When a metal is subjected to repeated cycles of stress or strain it causes its structure to break down ultimately leading to fracture This behavior is called fatigue and it is usually responsible for a large percentage of failures in connecting rods and crankshafts of engines steam or gas turbine blades connections or supports for bridges railroad wheels and axles and other parts subjected to cyclic loading In all these cases fracture will occur at a stress that is less than the materials yield stress The nature of this failure apparently results from the fact that there are microscopic imperfections usually on the surface of the member where the localized stress becomes much greater than the average stress acting over the cross sectionAs this higher stress is cycled it leads to the formation of minute cracks Occurrence of these cracks causes a further increase of stress at their tips or boundaries which in turn causes a further extension of the cracks into the material as the stress continues to be cycled Eventually the crosssectional area of the member is reduced to the point where the load can no longer be sustained and as a result sudden fracture occurs The material even though known to be ductile behaves as if it were brittle In order to specify a safe strength for a metallic material under repeated loading it is necessary to determine a limit below which no evidence of failure can be detected after applying a load for a specified number of cycles This limiting stress is called the endurance or fatigue limit Using a testing machine for this purpose a series of specimens are each subjected to a specified stress and cycled to failure The results are plotted as a graph representing the stress S or on the vertical axis and the number of cyclestofailure N on the horizontal axis This graph is called an SN diagram or stresscycle diagram and most often the values of N are plotted on a logarithmic scale since they are generally quite large Examples of SN diagrams for two common engineering metals are shown in Fig 328 The endurance limit is usually identified as the stress for which the SN graph becomes horizontal or asymptotic As noted it has a welldefined value of 186 MPa for steel For aluminum however the endurance limit is not well defined and so it is normally specified as the stress having a limit of 500 million cycles 131 MPa Once a particular value is obtained it is often assumed that for any stress below this value the fatigue life is infinite and therefore the number of cycles to failure is no longer given consideration 1Sel2al 19 ksi 1Sel2st 27 ksi s 3 Engineers must account for possible fatigue failure of the moving parts of this oil pumping rig The design of members used for amusement park rides requires careful consideration of cyclic loadings that can cause fatigue 112 CHAPTER 3 MECHANICAL PROPERTIES OF MATERIALS 333 The support consists of three rigid plates which are connected together using two symmetrically placed rubber pads If a vertical force of 5 N is applied to plate A determine the approximate vertical displacement of this plate due to shear strains in the rubber Each pad has crosssectional dimensions of 30 mm and 20 mm Gr 020 MPa 3 4 in u y x 5 in Prob 330 331 The shear stressstrain diagram for a steel alloy is shown in the figure If a bolt having a diameter of 075 in is made of this material and used in the double lap joint determine the modulus of elasticity E and the force P required to cause the material to yieldTake n 03 P 000545 60 grad tksi P2 P2 Prob 331 332 A shear spring is made by bonding the rubber annulus to a rigid fixed ring and a plug When an axial load P is placed on the plug show that the slope at point y in the rubber is For small angles we can write Integrate this expression and evaluate the constant of integration using the condition that at From the result compute the deflection y d of the plug r ro y 0 dydr P12phGr2 tan1P12phGr22 dydr tan g P y ro ri y r h d Prob 332 C B 40 mm 40 mm A 5 N Prob 333 P h a a A d Prob 334 334 A shear spring is made from two blocks of rubber each having a height h width b and thickness a The blocks are bonded to three plates as shown If the plates are rigid and the shear modulus of the rubber is G determine the displacement of plate A if a vertical load P is applied to this plate Assume that the displacement is small so that d a tan g L ag 330 The block is made of titanium Ti6A14V and is subjected to a compression of 006 in along the y axis and its shape is given a tilt of Determine Px Py and gxy u 897 CHAPTER REVIEW 113 3 One of the most important tests for material strength is the tension test The results found from stretching a specimen of known size are plotted as normal stress on the vertical axis and normal strain on the horizontal axis Many engineering materials exhibit initial linear elastic behavior whereby stress is proportional to strain defined by Hookes law Here E called the modulus of elasticity is the slope of this straight line on the stressstrain diagram s EP s EP When the material is stressed beyond the yield point permanent deformation will occur In particular steel has a region of yielding whereby the material will exhibit an increase in strain with no increase in stress The region of strain hardening causes further yielding of the material with a corresponding increase in stress Finally at the ultimate stress a localized region on the specimen will begin to constrict forming a neck It is after this that the fracture occurs Ductile materials such as most metals exhibit both elastic and plastic behavior Wood is moderately ductile Ductility is usually specified by the permanent elongation to failure or by the percent reduction in the crosssectional area Percent reduction of area A0 Af A0 11002 Percent elongation Lf L0 L0 11002 CHAPTER REVIEW s P P s E ductile material elastic region yielding strain hardening necking elastic behavior plastic behavior elastic limit yield stress ultimate stress fracture stress P sf sY spl su s proportional limit 114 CHAPTER 3 MECHANICAL PROPERTIES OF MATERIALS 3 Brittle materials exhibit little or no yielding before failure Cast iron concrete and glass are typical examples The yield point of a material at A can be increased by strain hardening This is accomplished by applying a load that causes the stress to be greater than the yield stress then releasing the load The larger stress becomes the new yield point for the material A When a load is applied to a member the deformations cause strain energy to be stored in the material The strain energy per unit volume or strain energy density is equivalent to the area under the stressstrain curve This area up to the yield point is called the modulus of resilienceThe entire area under the stress strain diagram is called the modulus of toughness s P brittle material permanent set elastic recovery elastic region plastic region load unload E E s P A A ur Modulus of resilience Ppl spl s P ut Modulus of toughness s P CHAPTER REVIEW 115 3 Poissons ratio is a dimensionless material property that relates the lateral strain to the longitudinal strain Its range of values is 0 n 05 n n Plat Plong Shear stress versus shear strain diagrams can also be established for a material Within the elastic region where G is the shear modulus found from the slope of the line The value of can be obtained from the relationship that exists between G E and n n t Gg When materials are in service for long periods of time considerations of creep become important Creep is the time rate of deformation which occurs at high stress andor high temperature Design requires that the stress in the material not exceed an allowable stress which is based on the materials creep strength Fatigue can occur when the material undergoes a large number of cycles of loading This effect will cause microscopic cracks to form leading to a brittle failureTo prevent fatigue the stress in the material must not exceed a specified endurance or fatigue limit P P r Final Shape L Original Shape Tension d2 d d2 t g g t G G E 211 n2 116 CHAPTER 3 MECHANICAL PROPERTIES OF MATERIALS 3 335 The elastic portion of the tension stressstrain diagram for an aluminum alloy is shown in the figure The specimen used for the test has a gauge length of 2 in and a diameter of 05 in When the applied load is 9 kip the new diameter of the specimen is 049935 in Compute the shear modulus for the aluminum 336 The elastic portion of the tension stressstrain diagram for an aluminum alloy is shown in the figure The specimen used for the test has a gauge length of 2 in and a diameter of 05 in If the applied load is 10 kip determine the new diameter of the specimen The shear modulus is Gal 3811032 ksi Gal 338 A short cylindrical block of 6061T6 aluminum having an original diameter of 20 mm and a length of 75 mm is placed in a compression machine and squeezed until the axial load applied is 5 kN Determine a the decrease in its length and b its new diameter 339 The rigid beam rests in the horizontal position on two 2014T6 aluminum cylinders having the unloaded lengths shown If each cylinder has a diameter of 30 mm determine the placement x of the applied 80kN load so that the beam remains horizontal What is the new diameter of cylinder A after the load is applied nal 035 REVIEW PROBLEMS 000614 70 sksi P inin Probs 33536 337 The diagram for elastic fibers that make up human skin and muscle is shown Determine the modulus of elasticity of the fibers and estimate their modulus of toughness and modulus of resilience sP 3 m 210 mm 220 mm x A B 80 kN Prob 339 340 The head H is connected to the cylinder of a compressor using six steel bolts If the clamping force in each bolt is 800 lb determine the normal strain in the bolts Each bolt has a diameter of If and what is the strain in each bolt when the nut is unscrewed so that the clamping force is released Est 2911032 ksi sY 40 ksi 3 16 in H L C Prob 340 2 1 225 11 55 Pinin spsi Prob 337 REVIEW PROBLEMS 117 343 The 8mmdiameter bolt is made of an aluminum alloy It fits through a magnesium sleeve that has an inner diameter of 12 mm and an outer diameter of 20 mm If the original lengths of the bolt and sleeve are 80 mm and 50 mm respectively determine the strains in the sleeve and the bolt if the nut on the bolt is tightened so that the tension in the bolt is 8 kN Assume the material at A is rigid Emg 45 GPa Eal 70 GPa 341 The stone has a mass of 800 kg and center of gravity at G It rests on a pad at A and a roller at BThe pad is fixed to the ground and has a compressed height of 30 mm a width of 140 mm and a length of 150 mm If the coefficient of static friction between the pad and the stone is determine the approximate horizontal displacement of the stone caused by the shear strains in the pad before the stone begins to slip Assume the normal force at A acts 15 m from G as shown The pad is made from a material having MPa and n 035 E 4 ms 08 3 04 m 125 m 15 m 03 m P B A G Prob 341 2 ft 3 ft 4 ft 3 ft D A B E C W Prob 342 342 The bar DA is rigid and is originally held in the horizontal position when the weight W is supported from C If the weight causes B to be displaced downward 0025 in determine the strain in wires DE and BC Also if the wires are made of A36 steel and have a crosssectional area of 0002 in2 determine the weight W 50 mm 30 mm A Prob 343 400 mm A B P 400 mm u Prob 344 344 The A36 steel wire AB has a crosssectional area of and is unstretched when Determine the applied load P needed to cause u 449 u 450 10 mm2 The string of drill pipe suspended from this traveling block on an oil rig is subjected to extremely large loadings and axial deformations 4 119 CHAPTER OBJECTIVES In Chapter 1 we developed the method for finding the normal stress in axially loaded members In this chapter we will discuss how to determine the deformation of these members and we will also develop a method for finding the support reactions when these reactions cannot be determined strictly from the equations of equilibrium An analysis of the effects of thermal stress stress concentrations inelastic deformations and residual stress will also be discussed 41 SaintVenants Principle In the previous chapters we have developed the concept of stress as a means of measuring the force distribution within a body and strain as a means of measuring a bodys deformation We have also shown that the mathematical relationship between stress and strain depends on the type of material from which the body is made In particular if the material behaves in a linear elastic manner then Hookes law applies and there is a proportional relationship between stress and strain Axial Load 120 CHAPTER 4 AXIAL LOAD 4 When section cc is so located the theory of elasticity predicts the maximum stress to be smax 102savg Using this idea consider the manner in which a rectangular bar will deform elastically when the bar is subjected to a force P applied along its centroidal axis Fig 41a Here the bar is fixed connected at one end with the force applied through a hole at its other end Due to the loading the bar deforms as indicated by the once horizontal and vertical grid lines drawn on the bar Notice how the localized deformation that occurs at each end tends to even out and become uniform throughout the midsection of the bar If the material remains elastic then the strains caused by this deformation are directly related to the stress in the bar As a result the stress will be distributed more uniformly throughout the crosssectional area when the section is taken farther and farther from the point where any external load is applied For example consider a profile of the variation of the stress distribution acting at sections aa bb and cc each of which is shown in Fig 41b By comparison the stress tends to reach a uniform value at section cc which is sufficiently removed from the end since the localized deformation caused by P vanishes The minimum distance from the bars end where this occurs can be determined using a mathematical analysis based on the theory of elasticity It has been found that this distance should at least be equal to the largest dimension of the loaded cross section Hence section cc should be located at a distance at least equal to the width not the thickness of the bar a P ab c ab c Lines located away from the load and support remain straight Load distorts lines located near load Load distorts lines located near support Fig 41 122 CHAPTER 4 AXIAL LOAD 4 42 Elastic Deformation of an Axially Loaded Member Using Hookes law and the definitions of stress and strain we will now develop an equation that can be used to determine the elastic displacement of a member subjected to axial loads To generalize the development consider the bar shown in Fig 42a which has a cross sectional area that gradually varies along its length LThe bar is subjected to concentrated loads at its ends and a variable external load distributed along its length This distributed load could for example represent the weight of the bar if it does not remain horizontal or friction forces acting on the bars surface Here we wish to find the relative displacement delta of one end of the bar with respect to the other end as caused by this loading We will neglect the localized deformations that occur at points of concentrated loading and where the cross section suddenly changes From SaintVenants principle these effects occur within small regions of the bars length and will therefore have only a slight effect on the final result For the most part the bar will deform uniformly so the normal stress will be uniformly distributed over the cross section Using the method of sectionsa differential element or wafer of length dx and crosssectional area Ax is isolated from the bar at the arbitrary position x The freebody diagram of this element is shown in Fig 42b The resultant internal axial force will be a function of x since the external distributed loading will cause it to vary along the length of the bar This load Px will deform the element into the shape indicated by the dashed outline and therefore the displacement of one end of the element with respect to the other end is The stress and strain in the element are Provided the stress does not exceed the proportional limit we can apply Hookes law ie dd P1x2 dx A1x2E P1x2 A1x2 Ea dd dx b s EP s P1x2 A1x2 and P dd dx dd d dx dd b Px Px P2 P1 x dx L a d Fig 42 42 ELASTIC DEFORMATION OF AN AXIALLY LOADED MEMBER 123 4 For the entire length L of the bar we must integrate this expression to find This yields 41 where displacement of one point on the bar relative to the other point original length of bar internal axial force at the section located a distance x from one end crosssectional area of the bar expressed as a function of x modulus of elasticity for the material Constant Load and CrossSectional Area In many cases the bar will have a constant crosssectional area A and the material will be homogeneous so E is constant Furthermore if a constant external force is applied at each end Fig 43 then the internal force P throughout the length of the bar is also constantAs a result Eq 41 can be integrated to yield 42 If the bar is subjected to several different axial forces along its lengthor the crosssectional area or modulus of elasticity changes abruptly from one region of the bar to the next the above equation can be applied to each segment of the bar where these quantities remain constant The displacement of one end of the bar with respect to the other is then found from the algebraic addition of the relative displacements of the ends of each segmentFor this general case 43 d a PL AE d PL AE E A1x2 P1x2 L d d L L 0 P1x2 dx A1x2E d The vertical displacement at the top of these building columns depends upon the loading applied on the roof and to the floor attached to their midpoint P d P x L Fig 43 42 ELASTIC DEFORMATION OF AN AXIALLY LOADED MEMBER 125 4 Important Points SaintVenants principle states that both the localized deformation and stress which occur within the regions of load application or at the supports tend to even out at a distance sufficiently removed from these regions The displacement of one end of an axially loaded member relative to the other end is determined by relating the applied internal load to the stress using and relating the displacement to the strain using Finally these two equations are combined using Hookes law which yields Eq 41 Since Hookes law has been used in the development of the displacement equation it is important that no internal load causes yielding of the material and that the material is homogeneous and behaves in a linear elastic manner s EP P dddx s PA Procedure for Analysis The relative displacement between any two points A and B on an axially loaded member can be determined by applying Eq 41 or Eq 42Application requires the following steps Internal Force Use the method of sections to determine the internal axial force P within the member If this force varies along the members length due to an external distributed loading a section should be made at the arbitrary location x from one end of the member and the force represented as a function of x ie Px If several constant external forces act on the member the internal force in each segment of the member between any two external forces must be determined For any segment an internal tensile force is positive and an internal compressive force is negative For convenience the results of the internal loading can be shown graphically by constructing the normalforce diagram Displacement When the members crosssectional area varies along its length the area must be expressed as a function of its position x ie Ax If the crosssectional area the modulus of elasticity or the internal loading suddenly changes then Eq 42 should be applied to each segment for which these quantities are constant When substituting the data into Eqs 41 through 43 be sure to account for the proper sign for the internal force PTensile loadings are positive and compressive loadings are negativeAlso use a consistent set of units For any segment if the result is a positive numerical quantity it indicates elongation if it is negative it indicates a contraction 42 ELASTIC DEFORMATION OF AN AXIALLY LOADED MEMBER 129 4 EXAMPLE 44 A member is made from a material that has a specific weight and modulus of elasticity E If it is in the form of a cone having the dimensions shown in Fig 49a determine how far its end is displaced due to gravity when it is suspended in the vertical position SOLUTION Internal Force The internal axial force varies along the member since it is dependent on the weight Wy of a segment of the member below any section Fig 49b Hence to calculate the displacement we must use Eq 41 At the section located a distance y from its free end the radius x of the cone as a function of y is determined by proportion ie The volume of a cone having a base of radius x and height y is Since the internal force at the section becomes Displacement The area of the cross section is also a function of position y Fig 49bWe have Applying Eq 41 between the limits of and yields Ans NOTE As a partial check of this result notice how the units of the terms when canceled give the displacement in units of length as expected gL2 6E g 3E L L 0 y dy d L L 0 P1y2 dy A1y2E L L 0 C1gpr0 23L22 y3D dy C1pr0 2L22 y2D E y L y 0 A1y2 px2 pr0 2 L2 y2 P1y2 gpr0 2 3L2 y3 c Fy 0 W gV V 1 3 pyx2 pr0 2 3L2 y3 x y r0 L x r0 L y g Fig 49 y L x r0 a y y x Wy b Py x 42 ELASTIC DEFORMATION OF AN AXIALLY LOADED MEMBER 133 4 414 The post is made of Douglas fir and has a diameter of 60 mm If it is subjected to the load of 20 kN and the soil provides a frictional resistance that is uniformly distributed along its sides of determine the force F at its bottom needed for equilibriumAlsowhat is the displacement of the top of the post A with respect to its bottom B Neglect the weight of the post 415 The post is made of Douglas fir and has a diameter of 60 mm If it is subjected to the load of 20 kN and the soil provides a frictional resistance that is distributed along its length and varies linearly from at to at determine the force F at its bottom needed for equilibrium Also what is the displacement of the top of the post A with respect to its bottom B Neglect the weight of the post y 2 m 3 kNm w y 0 w 0 w 4 kNm w y A 2 m 20 kN B F Probs 41415 416 The linkage is made of two pinconnected A36 steel members each having a crosssectional area of If a vertical force of is applied to point A determine its vertical displacement at A 417 The linkage is made of two pinconnected A36 steel members each having a crosssectional area of Determine the magnitude of the force P needed to displace point A 0025 in downward 15 in2 P 50 kip 15 in2 15 ft 15 ft C A B 2 ft P Probs 41617 418 The assembly consists of two A36 steel rods and a rigid bar BD Each rod has a diameter of 075 in If a force of 10 kip is applied to the bar as shown determine the vertical displacement of the load 419 The assembly consists of two A36 steel rods and a rigid bar BD Each rod has a diameter of 075 in If a force of 10 kip is applied to the bar determine the angle of tilt of the bar 075 ft 3 ft 125 ft 10 kip A E F C B D 1 ft 2 ft Probs 41819 420 The rigid bar is supported by the pinconnected rod CB that has a crosssectional area of and is made of A36 steel Determine the vertical displacement of the bar at B when the load is applied 500 mm2 Prob 420 4 m 3 m B 45 kNm A C 134 CHAPTER 4 AXIAL LOAD 4 421 A springsupported pipe hanger consists of two springs which are originally unstretched and have a stiffness of three 304 stainless steel rods AB and CD which have a diameter of 5 mm and EF which has a diameter of 12 mm and a rigid beam GH If the pipe and the fluid it carries have a total weight of 4 kN determine the displacement of the pipe when it is attached to the support 422 A springsupported pipe hanger consists of two springs which are originally unstretched and have a stiffness of three 304 stainless steel rods AB and CD which have a diameter of 5 mm and EF which has a diameter of 12 mm and a rigid beam GH If the pipe is displaced 82 mm when it is filled with fluid determine the weight of the fluid k 60 kNm k 60 kNm A C D B F G H E 025 m 025 m 075 m k k 075 m Probs 42122 423 The rod has a slight taper and length L It is suspended from the ceiling and supports a load P at its end Show that the displacement of its end due to this load is Neglect the weight of the material The modulus of elasticity is E d PL1pEr2r12 P L r2 r1 Prob 423 424 Determine the relative displacement of one end of the tapered plate with respect to the other end when it is subjected to an axial load P P t d1 d2 h P Prob 424 425 Determine the elongation of the A36 steel member when it is subjected to an axial force of 30 kN The member is 10 mm thick Use the result of Prob 424 30 kN 30 kN 05 m 20 mm 75 mm Prob 425 426 The casting is made of a material that has a specific weight and modulus of elasticity E If it is formed into a pyramid having the dimensions shown determine how far its end is displaced due to gravity when it is suspended in the vertical position g b0 b0 L Prob 426 44 STATICALLY INDETERMINATE AXIALLY LOADED MEMBER 137 4 44 Statically Indeterminate Axially Loaded Member Consider the bar shown in Fig 411a which is fixed supported at both of its ends From the freebody diagram Fig 411b equilibrium requires This type of problem is called statically indeterminate since the equilibrium equations are not sufficient to determine the two reactions on the bar In order to establish an additional equation needed for solution it is necessary to consider how points on the bar displace Specifically an equation that specifies the conditions for displacement is referred to as a compatibility or kinematic condition In this case a suitable compatibility condition would require the displacement of one end of the bar with respect to the other end to be equal to zero since the end supports are fixed Hence the compatibility condition becomes This equation can be expressed in terms of the applied loads by using a loaddisplacement relationship which depends on the material behavior For example if linearelastic behavior occurs can be used Realizing that the internal force in segment AC is and in segment CB the internal force is Fig 411c the above equation can be written as Assuming that AE is constant then so that using the equilibrium equation the equations for the reactions become Since both of these results are positive the direction of the reactions is shown correctly on the freebody diagram FA PLCB L and FB P LAC L FA FB1LCBLAC2 FALAC AE FBLCB AE 0 FB FA d PLAE dAB 0 FB FA P 0 c F 0 LAC P C LCB L A B a P c FB FA FA FA FB FB b Fig 411 138 CHAPTER 4 AXIAL LOAD 4 Most concrete columns are reinforced with steel rodsand since these two materials work together in supporting the applied load the forces in each material become statically indeterminate Important Points The principle of superposition is sometimes used to simplify stress and displacement problems having complicated loadings This is done by subdividing the loading into components then algebracially adding the results Superposition requires that the loading be linearly related to the stress or displacement and the loading does not significantly change the original geometry of the member A problem is statically indeterminate if the equations of equilibrium are not sufficient to determine all the reactions on a member Compatibility conditions specify the displacement constraints that occur at the supports or other points on a member Procedure for Analysis The support reactions for statically indeterminate problems are determined by satisfying equilibrium compatibility and force displacement requirements for the member Equilibrium Draw a freebody diagram of the member in order to identify all the forces that act on it The problem can be classified as statically indeterminate if the number of unknown reactions on the freebody diagram is greater than the number of available equations of equilibrium Write the equations of equilibrium for the member Compatibility Consider drawing a displacement diagram in order to investigate the way the member will elongate or contract when subjected to the external loads Express the compatibility conditions in terms of the displacements caused by the loading Use a loaddisplacement relation such as to relate the unknown displacements to the reactions Solve the equilibrium and compatibility equations for the reactions If any of the results has a negative numerical value it indicates that this force acts in the opposite sense of direction to that indicated on the freebody diagram d PLAE 142 CHAPTER 4 AXIAL LOAD 4 EXAMPLE 48 The bolt shown in Fig 415a is made of 2014T6 aluminum alloy and is tightened so it compresses a cylindrical tube made of Am 1004T61 magnesium alloyThe tube has an outer radius of and it is assumed that both the inner radius of the tube and the radius of the bolt are The washers at the top and bottom of the tube are considered to be rigid and have a negligible thickness Initially the nut is hand tightened snugly then using a wrench the nut is further tightened onehalf turn If the bolt has 20 threads per inch determine the stress in the bolt SOLUTION Equilibrium The freebody diagram of a section of the bolt and the tube Fig 415b is considered in order to relate the force in the bolt to that in the tube Equilibrium requires 1 Compatibility When the nut is tightened on the bolt the tube will shorten and the bolt will elongate Fig 415c Since the nut undergoes onehalf turn it advances a distance of along the boltThus the compatibility of these displacements requires Taking the moduli of elasticity from the table on the inside back cover and applying Eq 42 yields 2 Solving Eqs 1 and 2 simultaneously we get The stresses in the bolt and tube are therefore Ans These stresses are less than the reported yield stress for each material and see the inside back cover and therefore this elastic analysis is valid 1sY2mg 22 ksi 1sY2al 60 ksi st Ft At 1122 kip p105 in22 1025 in22 191 ksi sb Fb Ab 1122 kip p1025 in22 572 ksi Fb Ft 1122 kip 078595Ft 25 14414Fb 0025 in Fb13 in2 p1025 in2210611032 ksi Ft13 in2 p105 in22 1025 in2264811032 ksi dt 0025 in db 1 c2 0025 in 1 2 1 20 in db dt Fb Ft 0 c Fy 0 Ft Fb 1 4 in 1 2 in Fig 415 c 0025 in Initial position Final position dt db b Ft Fb 3 in in a in 1 2 1 4 45 THE FORCE METHOD OF ANALYSIS FOR AXIALLY LOADED MEMBERS 145 4 431 The column is constructed from highstrength concrete and six A36 steel reinforcing rods If it is subjected to an axial force of 30 kip determine the average normal stress in the concrete and in each rod Each rod has a diameter of 075 in 432 The column is constructed from highstrength concrete and six A36 steel reinforcing rods If it is subjected to an axial force of 30 kip determine the required diameter of each rod so that onefourth of the load is carried by the concrete and threefourths by the steel 434 The 304 stainless steel post A has a diameter of and is surrounded by a red brass C83400 tube B Both rest on the rigid surface If a force of 5 kip is applied to the rigid cap determine the average normal stress developed in the post and the tube 435 The 304 stainless steel post A is surrounded by a red brass C83400 tube B Both rest on the rigid surface If a force of 5 kip is applied to the rigid cap determine the required diameter d of the steel post so that the load is shared equally between the post and tube d 2 in PROBLEMS 3 ft 30 kip 4 in Probs 43132 500 mm 80 kN Prob 433 433 The steel pipe is filled with concrete and subjected to a compressive force of 80 kN Determine the average normal stress in the concrete and the steel due to this loading The pipe has an outer diameter of 80 mm and an inner diameter of 70 mm Ec 24 GPa Est 200 GPa 5 kip d 05 in 8 in A 3 in B A B Probs 43435 436 The composite bar consists of a 20mmdiameter A36 steel segment AB and 50mmdiameter red brass C83400 end segments DA and CB Determine the average normal stress in each segment due to the applied load 437 The composite bar consists of a 20mmdiameter A36 steel segment AB and 50mmdiameter red brass C83400 end segments DA and CB Determine the displacement of A with respect to B due to the applied load 50 mm 20 mm D C 75 kN 75 kN 100 kN 100 kN A B 500 mm 250 mm 250 mm Probs 43637 146 CHAPTER 4 AXIAL LOAD 4 438 The A36 steel column having a crosssectional area of is encased in highstrength concrete as shown If an axial force of 60 kip is applied to the column determine the average compressive stress in the concrete and in the steel How far does the column shorten It has an original length of 8 ft 439 The A36 steel column is encased in highstrength concrete as shown If an axial force of 60 kip is applied to the column determine the required area of the steel so that the force is shared equally between the steel and concrete How far does the column shorten It has an original length of 8 ft 18 in2 60 kip 9 in 8 ft 16 in Probs 43839 440 The rigid member is held in the position shown by three A36 steel tie rods Each rod has an unstretched length of 075 m and a crosssectional area of Determine the forces in the rods if a turnbuckle on rod EF undergoes one full turn The lead of the screw is 15 mm Neglect the size of the turnbuckle and assume that it is rigid Note The lead would cause the rod when unloaded to shorten 15 mm when the turnbuckle is rotated one revolution 125 mm2 05 m B A D C 05 m 075 m 075 m F E Prob 440 441 The concrete post is reinforced using six steel reinforcing rods each having a diameter of 20 mm Determine the stress in the concrete and the steel if the post is subjected to an axial load of 900 kN 442 The post is constructed from concrete and six A36 steel reinforcing rods If it is subjected to an axial force of 900 kN determine the required diameter of each rod so that onefifth of the load is carried by the steel and fourfifths by the concrete Ec 25 GPa Est 200 GPa Ec 25 GPa Est 200 GPa 900 kN 375 mm 250 mm Probs 44142 443 The assembly consists of two red brass C83400 copper alloy rods AB and CD of diameter 30 mm a stainless 304 steel alloy rod EF of diameter 40 mm and a rigid cap G If the supports at A C and F are rigid determine the average normal stress developed in rods AB CD and EF 40 kN 40 kN 300 mm 450 mm 30 mm 30 mm 40 mm A B C D E F G Prob 443 148 CHAPTER 4 AXIAL LOAD 4 449 The tapered member is fixed connected at its ends A and B and is subjected to a load at Determine the reactions at the supports The material is 2 in thick and is made from 2014T6 aluminum 450 The tapered member is fixed connected at its ends A and B and is subjected to a load P Determine the location x of the load and its greatest magnitude so that the average normal stress in the bar does not exceed The member is 2 in thick sallow 4 ksi x 30 in P 7 kip 60 in 3 in x A B 6 in P Probs 44950 451 The rigid bar supports the uniform distributed load of 6 Determine the force in each cable if each cable has a crosssectional area of and 452 The rigid bar is originally horizontal and is supported by two cables each having a crosssectional area of and Determine the slight rotation of the bar when the uniform load is applied E 3111032 ksi 005 in2 E 3111032 ksi 005 in2 kipft 453 The press consists of two rigid heads that are held together by the two A36 steel rods A 6061 T6solidaluminum cylinder is placed in the press and the screw is adjusted so that it just presses up against the cylinder If it is then tightened onehalf turn determine the average normal stress in the rods and in the cylinder The singlethreaded screw on the bolt has a lead of 001 in Note The lead represents the distance the screw advances along its axis for one complete turn of the screw 454 The press consists of two rigid heads that are held together by the two A36 steel rods A 6061 T6solidaluminum cylinder is placed in the press and the screw is adjusted so that it just presses up against the cylinder Determine the angle through which the screw can be turned before the rods or the specimen begin to yield The singlethreaded screw on the bolt has a lead of 001 in Note The lead represents the distance the screw advances along its axis for one complete turn of the screw 1 2indiameter 1 2indiameter 455 The three suspender bars are made of A36 steel and have equal crosssectional areas of Determine the average normal stress in each bar if the rigid beam is subjected to the loading shown 450 mm2 3 ft A D C B 3 ft 6 kipft 3 ft 6 ft Probs 45152 12 in 10 in 2 in Probs 45354 B A D C F E 2 m 50 kN 80 kN 1 m 1 m 1 m 1 m Prob 455 45 THE FORCE METHOD OF ANALYSIS FOR AXIALLY LOADED MEMBERS 149 4 456 The rigid bar supports the 800lb load Determine the normal stress in each A36 steel cable if each cable has a crosssectional area of 457 The rigid bar is originally horizontal and is supported by two A36 steel cables each having a cross sectional area of Determine the rotation of the bar when the 800lb load is applied 004 in2 004 in2 458 The horizontal beam is assumed to be rigid and supports the distributed load shown Determine the vertical reactions at the supports Each support consists of a wooden post having a diameter of 120 mm and an unloaded original length of 140 mTake 459 The horizontal beam is assumed to be rigid and supports the distributed load shown Determine the angle of tilt of the beam after the load is applied Each support consists of a wooden post having a diameter of 120 mm and an unloaded original length of 140 mTake Ew 12 GPa Ew 12 GPa 461 The distributed loading is supported by the three suspender bars AB and EF are made of aluminum and CD is made of steel If each bar has a crosssectional area of determine the maximum intensity of the distributed loading so that an allowable stress of in the steel and in the aluminum is not exceeded Assume ACE is rigid Est 200 GPa Eal 70 GPa 1sallow2al 94 MPa 180 MPa 1sallow2st w 450 mm2 460 The assembly consists of two posts AD and CF made of A36 steel and having a crosssectional area of and a 2014T6 aluminum post BE having a cross sectional area of If a central load of 400 kN is applied to the rigid cap determine the normal stress in each post There is a small gap of 01 mm between the post BE and the rigid member ABC 1500 mm2 1000 mm2 5 ft 5 ft 6 ft A D C B 800 lb 12 ft Probs 45657 140 m A B C 1 m 2 m 18 kNm Probs 45859 C D E F A B 400 kN 05 m 05 m 04 m Prob 460 2 m B D F A C E 15 m 15 m al st al w Prob 461 150 CHAPTER 4 AXIAL LOAD 4 462 The rigid link is supported by a pin at A a steel wire BC having an unstretched length of 200 mm and cross sectional area of and a short aluminum block having an unloaded length of 50 mm and crosssectional area of If the link is subjected to the vertical load shown determine the average normal stress in the wire and the block 463 The rigid link is supported by a pin at A a steel wire BC having an unstretched length of 200 mm and cross sectional area of and a short aluminum block having an unloaded length of 50 mm and crosssectional area of If the link is subjected to the vertical load shown determine the rotation of the link about the pin A Report the answer in radians Eal 70 GPa Est 200 GPa 40 mm2 225 mm2 Eal 70 GPa Est 200 GPa 40 mm2 225 mm2 464 The center post B of the assembly has an original length of 1247 mm whereas posts A and C have a length of 125 mm If the caps on the top and bottom can be considered rigid determine the average normal stress in each post The posts are made of aluminum and have a crosssectional area of Eal 70 GPa 400 mm2 465 The assembly consists of an A36 steel bolt and a C83400 red brass tube If the nut is drawn up snug against the tube so that then turned an additional amount so that it advances 002 mm on the bolt determine the force in the bolt and the tubeThe bolt has a diameter of 7 mm and the tube has a crosssectional area of 466 The assembly consists of an A36 steel bolt and a C83400 red brass tube The nut is drawn up snug against the tube so that Determine the maximum additional amount of advance of the nut on the bolt so that none of the material will yield The bolt has a diameter of 7 mm and the tube has a crosssectional area of 100 mm2 L 75 mm 100 mm2 L 75 mm 467 The three suspender bars are made of the same material and have equal crosssectional areas A Determine the average normal stress in each bar if the rigid beam ACE is subjected to the force P C B A 200 mm 150 mm 100 mm 150 mm D 50 mm 450 N Probs 46263 125 mm 100 mm 100 mm A B C 800 kNm 800 kNm Prob 464 L Probs 46566 L P d B A D C F E d 2 d 2 Prob 467 46 THERMAL STRESS 151 4 46 Thermal Stress A change in temperature can cause a body to change its dimensions Generally if the temperature increases the body will expand whereas if the temperature decreases it will contract Ordinarily this expansion or contraction is linearly related to the temperature increase or decrease that occurs If this is the case and the material is homogeneous and isotropic it has been found from experiment that the displacement of a member having a length L can be calculated using the formula 44 where a property of the material referred to as the linear coefficient of thermal expansionThe units measure strain per degree of temperatureThey are Fahrenheit in the FPS system and Celsius or Kelvin in the SI systemTypical values are given on the inside back cover the algebraic change in temperature of the member the original length of the member the algebraic change in the length of the member The change in length of a statically determinate member can easily be calculated using Eq 44 since the member is free to expand or contract when it undergoes a temperature change However in a statically indeterminate member these thermal displacements will be constrained by the supports thereby producing thermal stresses that must be considered in design Determining these thermal stresses is possible using the methods outlined in the previous sections The following examples illustrate some applications dT L T 1K 1C 1F a dT a TL Most traffic bridges are designed with expansion joints to accommodate the thermal movement of the deck and thus avoid any thermal stress Long extensions of ducts and pipes that carry fluids are subjected to variations in climate that will cause them to expand and contract Expansion joints such as the one shown are used to mitigate thermal stress in the material 152 CHAPTER 4 AXIAL LOAD 4 EXAMPLE 410 The A36 steel bar shown in Fig 418a is constrained to just fit between two fixed supports when If the temperature is raised to determine the average normal thermal stress developed in the bar SOLUTION Equilibrium The freebody diagram of the bar is shown in Fig 418b Since there is no external load the force at A is equal but opposite to the force at B that is The problem is statically indeterminate since this force cannot be determined from equilibrium Compatibility Since the thermal displacement at A that occurs Fig 418c is counteracted by the force F that is required to push the bar back to its original position The compatibility condition at A becomes Applying the thermal and loaddisplacement relationships we have Thus from the data on the inside back cover Since F also represents the internal axial force within the bar the average normal compressive stress is thus Ans NOTE From the magnitude of F it should be apparent that changes in temperature can cause large reaction forces in statically indeterminate members s F A 2871 kip 105 in22 115 ksi 2871 kip 66011062F1120F 60F2105 in222911032 kipin2 F aTAE 0 aTL FL AE dAB 0 dT dF 1 c2 dF dT dAB 0 FA FB F c Fy 0 T2 120F T1 60F Fig 418 b F F c dT dF 2 ft 05 in 05 in A B a 154 CHAPTER 4 AXIAL LOAD 4 EXAMPLE 412 A 2014T6 aluminum tube having a crosssectional area of is used as a sleeve for an A36 steel bolt having a crosssectional area of Fig 420a When the temperature is the nut holds the assembly in a snug position such that the axial force in the bolt is negligible If the temperature increases to determine the force in the bolt and sleeve SOLUTION Equilibrium The freebody diagram of a top segment of the assembly is shown in Fig 420b The forces and are produced since the sleeve has a higher coefficient of thermal expansion than the bolt and therefore the sleeve will expand more when the temperature is increased It is required that 1 Compatibility The temperature increase causes the sleeve and bolt to expand and Fig 420c However the redundant forces and elongate the bolt and shorten the sleeve Consequently the end of the assembly reaches a final position which is not the same as its initial position Hence the compatibility condition becomes Applying Eqs 42 and 44 and using the mechanical properties from the table on the inside back cover we have Using Eq 1 and solving gives Ans NOTE Since linear elastic material behavior was assumed in this analysis the average normal stresses should be checked to make sure that they do not exceed the proportional limits for the material Fs Fb 203 kN Fs10150 m2 1600 mm221106 m2mm2273111092 Nm2 2311062C180C 15C210150 m2 Fb10150 m2 1400 mm221106 m2mm2220011092 Nm2 1211062C180C 15C210150 m2 d 1db2T 1db2F 1ds2T 1ds2F 1 T2 Fs Fb 1db2T 1ds2T Fs Fb c Fy 0 Fs Fb T2 80C T1 15C 400 mm2 600 mm2 150 mm a b Fb Fs dsF dsT dbT dbF d Initial position Final position c Fig 420 156 CHAPTER 4 AXIAL LOAD 4 478 The A36 steel rod has a diameter of 50 mm and is lightly attached to the rigid supports at A and B when If the temperature becomes and an axial force of is applied to its center determine the reactions at A and B 479 The A36 steel rod has a diameter of 50 mm and is lightly attached to the rigid supports at A and B when Determine the force P that must be applied to the collar at its midpoint so that when the reaction at B is zero T2 30C T1 50C P 200 kN T2 20C T1 80C 40 ft d d Prob 475 A C E B D 15 in 025 in 3 in Prob 476 475 The 40ftlongA36 steel rails on a train track are laid with a small gap between them to allow for thermal expansionDetermine the required gap so that the rails just touch one another when the temperature is increased from to Using this gapwhat would be the axial force in the rails if the temperature were to rise to The crosssectional area of each rail is 510 in2 T3 110F T2 90F T1 20F d 476 The device is used to measure a change in temper ature Bars AB and CD are made of A36 steel and 2014T6 aluminum alloy respectively When the temperature is at 75F ACE is in the horizontal position Determine the vertical displacement of the pointer at E when the temperature rises to 150F 477 The bar has a crosssectional area A length L modulus of elasticity E and coefficient of thermal expansion The temperature of the bar changes uniformly along its length from at A to at B so that at any point x along the bar Determine the force the bar exerts on the rigid walls Initially no axial force is in the bar and the bar has a temperature of TA T TA x1TB TA2L TB TA a 480 The rigid block has a weight of 80 kip and is to be supported by posts A and B which are made of A36 steel and the post C which is made of C83400 red brass If all the posts have the same original length before they are loaded determine the average normal stress developed in each post when post C is heated so that its temperature is increased by 20F Each post has a crosssectional area of 8 in2 x TA TB A B Prob 477 05 m P 05 m A B C Probs 47879 A C B 3 ft 3 ft Prob 480 158 CHAPTER 4 AXIAL LOAD 4 47 Stress Concentrations In Sec 41 it was pointed out that when an axial force is applied to a member it creates a complex stress distribution within the localized region of the point of load application Not only do complex stress distributions arise just under a concentrated loading they can also arise at sections where the members crosssectional area changes For example consider the bar in Fig 421a which is subjected to an axial force P Here the once horizontal and vertical grid lines deflect into an irregular pattern around the hole centered in the bar The maximum normal stress in the bar occurs on section aa which is taken through the bars smallest crosssectional area Provided the material behaves in a linearelastic manner the stress distribution acting on this section can be determined either from a mathematical analysis using the theory of elasticity or experimentally by measuring the strain normal to section aa and then calculating the stress using Hookes law Regardless of the method used the general shape of the stress distribution will be like that shown in Fig 421b In a similar manner if the bar has a reduction in its cross section achieved using shoulder fillets as in Fig 422a then again the maximum normal stress in the bar will occur at the smallest crosssectional area section aa and the stress distribution will look like that shown in Fig 422b s EP a a Distorted a P Actual stress distribution b Average stress distribution c smax savg Undistorted P P P P P Fig 421 This saw blade has grooves cut into it in order to relieve both the dynamic stress that develops within it as it rotates and the thermal stress that develops as it heats up Note the small circles at the end of each groove These serve to reduce the stress concentrations that develop at the end of each groove 47 STRESS CONCENTRATIONS 159 4 In both of these cases force equilibrium requires the magnitude of the resultant force developed by the stress distribution to be equal to P In other words 45 This integral graphically represents the total volume under each of the stressdistribution diagrams shown in Fig 421b or Fig 422b The resultant P must act through the centroid of each volume In engineering practice the actual stress distributions in Fig 421b and Fig 422b do not have to be determined Instead only the maximum stress at these sections must be known and the member is then designed to resist this stress when the axial load P is applied Specific values of this maximum normal stress can be determined by experimental methods or by advanced mathematical techniques using the theory of elasticity The results of these investigations are usually reported in graphical form using a stressconcentration factor K We define K as a ratio of the maximum stress to the average normal stress acting at the cross section ie 46 Provided K is known and the average normal stress has been calculated from where A is the smallest crosssectional area Figs 421c and 422c then the maximum normal stress at the cross section is smax K1PA2 savg PA K smax savg P LA s dA P a Distorted a P Actual stress distribution b a smax savg Average stress distribution c Undistorted P P P P Fig 422 Stress concentrations often arise at sharp corners on heavy machinery Engineers can mitigate this effect by using stiffeners welded to the corners 160 CHAPTER 4 AXIAL LOAD 4 Specific values of K are generally reported in handbooks related to stress analysis Examples are given in Figs 424 and 425 Note that K is independent of the bars material properties rather it depends only on the bars geometry and the type of discontinuity As the size r of the discontinuity is decreased the stress concentration is increased For example if a bar requires a change in cross section it has been determined that a sharp corner Fig 423a produces a stress concentration factor greater than 3 In other words the maximum normal stress will be three times greater than the average normal stress on the smallest cross section However this can be reduced to say 15 by introducing a fillet Fig 423b A further reduction can be made by means of small grooves or holes placed at the transition Fig 423c and 423d In all of these cases these designs help to reduce the rigidity of the material surrounding the corners so that both the strain and the stress are more evenly spread throughout the bar The stressconcentration factors given in Figs 424 and 425 were determined on the basis of a static loading with the assumption that the stress in the material does not exceed the proportional limit If the material is very brittle the proportional limit may be at the fracture stress and so for this material failure will begin at the point of stress concentration Essentially a crack begins to form at this point and a higher stress concentration will develop at the tip of this crack This in turn causes the crack to propagate over the cross section resulting in sudden fracture For this reason it is very important to use stress concentration factors in design when using brittle materials On the other hand if the material is ductile and subjected to a static load it is often not necessary to use stressconcentration factors since any stress that exceeds the proportional limit will not result in a crack Instead the material will have reserve strength due to yielding and strain hardening In the next section we will discuss the effects caused by this phenomenon Stress concentrations are also responsible for many failures of structural members or mechanical elements subjected to fatigue loadings For these cases a stress concentration will cause the material to crack if the stress exceeds the materials endurance limit whether or not the material is ductile or brittle Here the material localized at the tip of the crack remains in a brittle state and so the crack continues to grow leading to a progressive fractureAs a result one must seek ways to limit the amount of damage that can be caused by fatigue P P a P P b P P c P P d Fig 423 See Lipson C and R C Juvinall Handbook of Stress and Strength Macmillan 162 CHAPTER 4 AXIAL LOAD 4 48 Inelastic Axial Deformation Up to this point we have only considered loadings that cause the material of a member to behave elastically Sometimes however a member may be designed so that the loading causes the material to yield and thereby permanently deform Such members are often made from a highly ductile metal such as annealed lowcarbon steel having a stressstrain diagram that is similar to that of Fig 36 and for simplicity can be modeled as shown in Fig 426b A material that exhibits this behavior is referred to as being elastic perfectly plastic or elastoplastic To illustrate physically how such a material behaves consider the bar in Fig 426a which is subjected to the axial load P If the load causes an elastic stress to be developed in the bar then applying Eq 45 equilibrium requires A Furthermore the stress causes the bar to strain as indicated on the stressstrain diagram Fig 426b If P is now increased to such that it causes yielding of the material that is then again The load is called the plastic load since it represents the maximum load that can be supported by an elastoplastic material For this case the strains are not uniquely defined Instead at the instant is attained the bar is first subjected to the yield strain Fig 426b after which the bar will continue to yield or elongate such that the strains then etc are generated Since our model of the material exhibits perfectly plastic material behavior this elongation will continue indefinitely with no increase in load In reality however the material will after some yielding actually begin to strainharden so that the extra strength it attains will stop any further straining As a result any design based on this behavior will be safe since strainhardening provides the potential for the material to support an additional load if necessary P3 P2 PY sY Pp Pp 1sY dA sYA s sY Pp P1 s1 P 1s1 dA s1 s s1 b s1 sY s P1 P2 P3 P PY P a s Fig 426 48 INELASTIC AXIAL DEFORMATION 163 4 Consider now the case of a bar having a hole through it as shown in Fig 427a As the magnitude of P is increased a stress concentration occurs in the material at the edge of the hole on section aa The stress here will reach a maximum value of say and have a corresponding elastic strain of Fig 427b The stresses and corresponding strains at other points on the cross section will be smaller as indicated by the stress distribution shown in Fig 427c Equilibrium requires In other words P is geometrically equivalent to the volume contained within the stress distribution If the load is now increased to so that then the material will begin to yield outward from the hole until the equilibrium condition is satisfied Fig 427dAs shown this produces a stress distribution that has a geometrically greater volume than that shown in Fig 427cA further increase in load will cause the material over the entire cross section to yield eventually When this happens no greater load can be sustained by the bar This plastic load is shown in Fig 427e It can be calculated from the equilibrium condition where A is the bars crosssectional area at section aa The following examples illustrate numerically how these concepts apply to other types of problems for which the material has elastoplastic behavior Pp LA sY dA sYA Pp P 1s dA smax sY P P 1s dA P1 smax s1 P a P a a b s sY s1 P P1 PY c P s1 s1 d P sY sY e PP sY sY Fig 427 164 CHAPTER 4 AXIAL LOAD 4 49 Residual Stress If an axially loaded member or group of such members forms a statically indeterminate system that can support both tensile and compressive loads then excessive external loadings which cause yielding of the material will create residual stresses in the members when the loads are removedThe reason for this has to do with the elastic recovery of the material that occurs during unloadingTo show this consider a prismatic member made from an elastoplastic material having the stressstrain diagram shown in Fig 428 If an axial load produces a stress in the material and a corresponding plastic strain then when the load is removed the material will respond elastically and follow the line CD in order to recover some of the plastic strain A recovery to zero stress at point will be possible only if the member is statically determinate since the support reactions for the member must be zero when the load is removed Under these circumstances the member will be permanently deformed so that the permanent set or strain in the member is If the member is statically indeterminate however removal of the external load will cause the support forces to respond to the elastic recovery CD Since these forces will constrain the member from full recovery they will induce residual stresses in the member To solve a problem of this kind the complete cycle of loading and then unloading of the member can be considered as the superposition of a positive load loading on a negative load unloadingThe loading O to C results in a plastic stress distribution whereas the unloading along CD results only in an elastic stress distribution Superposition requires the loads to cancel however the stress distributions will not cancel and so residual stresses will remain PO O PC sY Fig 428 D PO C A B O O PC s sY P 49 RESIDUAL STRESS 165 4 EXAMPLE 413 The bar in Fig 429a is made of steel that is assumed to be elastic perfectly plastic with Determine a the maximum value of the applied load P that can be applied without causing the steel to yield and b the maximum value of P that the bar can support Sketch the stress distribution at the critical section for each case SOLUTION Part a When the material behaves elastically we must use a stressconcentration factor determined from Fig 424 that is unique for the bars geometry Here From the figure The maximum load without causing yielding occurs when The average normal stress is Using Eq46we have Ans This load has been calculated using the smallest cross section The resulting stress distribution is shown in Fig 429b For equilibrium the volume contained within this distribution must equal 914 kN Part b The maximum load sustained by the bar will cause all the material at the smallest cross section to yield Therefore as P is increased to the plastic load it gradually changes the stress distribution from the elastic state shown in Fig 429b to the plastic state shown in Fig 429cWe require Ans Here equals the volume contained within the stress distribution which in this case is Pp sYA Pp Pp 160 kN 25011062 Pa Pp 10002 m210032 m2 sY Pp A Pp PY 914 kN 25011062 Pa 175c PY 10002 m210032 m2 d smax Ksavg sY Ka PY A b savg PA smax sY K L 175 w h 40 mm 140 mm 8 mm2 125 r h 4 mm 140 mm 8 mm2 0125 sY 250 MPa Fig 429 PP c sY PY b sY P 2 mm 40 mm 4 mm P 4 mm a 170 CHAPTER 4 AXIAL LOAD 495 The resulting stress distribution along section AB for the bar is shown From this distribution determine the approximate resultant axial force P applied to the barAlso what is the stressconcentration factor for this geometry 4 498 The bar has a crosssectional area of and is made of a material that has a stressstrain diagram that can be approximated by the two line segments shown Determine the elongation of the bar due to the applied loading 05 in2 08 in 06 in 06 in P 02 in 6 ksi 36 ksi 05 in A B Prob 495 20 mm 80 mm 5 MPa 30 MPa B A 10 mm P Prob 496 496 The resulting stress distribution along section AB for the bar is shown From this distribution determine the approximate resultant axial force P applied to the barAlso what is the stressconcentration factor for this geometry 497 The 300kip weight is slowly set on the top of a post made of 2014T6 aluminum with an A36 steel core If both materials can be considered elastic perfectly plastic determine the stress in each material Aluminum Steel 2 in 1 in Prob 497 5 kip 8 kip A B C 5 ft 2 ft 40 20 0001 0021 P inin sksi Prob 498 499 The rigid bar is supported by a pin at A and two steel wires each having a diameter of 4 mm If the yield stress for the wires is and determine the intensity of the distributed load that can be placed on the beam and will just cause wire EB to yield What is the displacement of point G for this case For the calculation assume that the steel is elastic perfectly plastic 4100 The rigid bar is supported by a pin at A and two steel wires each having a diameter of 4 mm If the yield stress for the wires is and determine a the intensity of the distributed load that can be placed on the beam that will cause only one of the wires to start to yield and b the smallest intensity of the distributed load that will cause both wires to yield For the calculation assume that the steel is elastic perfectly plastic w Est 200 GPa sY 530 MPa w Est 200 GPa sY 530 MPa 400 mm 250 mm 150 mm w A 800 mm E B D C G Probs 499100 172 CHAPTER 4 AXIAL LOAD 4108 The rigid beam is supported by the three posts A B and C of equal length Posts A and C have a diameter of 75 mm and are made of aluminum for which and Post B has a diameter of 20 mm and is made of brass for which and Determine the smallest magnitude of P so that a only rods A and C yield and b all the posts yield 4109 The rigid beam is supported by the three posts A B and C Posts A and C have a diameter of 60 mm and are made of aluminum for which and Post B is made of brass for which and If determine the largest diameter of post B so that all the posts yield at the same time P 130 kN 1sY2br 590 MPa Ebr 100 GPa 1sY2al 20 MPa Eal 70 GPa 1sY2br 590 MPa Ebr 100 GPa 1sY2al 20 MPa Eal 70 GPa 4 4110 The wire BC has a diameter of 0125 in and the material has the stressstrain characteristics shown in the figure Determine the vertical displacement of the handle at D if the pull at the grip is slowly increased and reaches a magnitude of a b P 600 lb P 450 lb al 2 m 2 m 2 m A B P P C br al 2 m Probs 4108109 A B C D P 50 in 30 in 40 in 0007 012 70 80 s ksi P inin Prob 4110 4111 The bar having a diameter of 2 in is fixed connected at its ends and supports the axial load P If the material is elastic perfectly plastic as shown by the stressstrain diagram determine the smallest load P needed to cause segment CB to yield If this load is released determine the permanent displacement of point C 4112 Determine the elongation of the bar in Prob 4111 when both the load P and the supports are removed 2 ft A C 20 0001 P B 3 ft P inin s ksi Probs 4111112 4113 A material has a stressstrain diagram that can be described by the curve Determine the deflection of the end of a rod made from this material if it has a length L crosssectional area A and a specific weight g d s cP12 L A d s P Prob 4113 174 CHAPTER 4 AXIAL LOAD 4 A change in temperature can cause a member made of homogeneous isotropic material to change its length by If the member is confined this change will produce thermal stress in the member d aTL Holes and sharp transitions at a cross section will create stress concentrations For the design of a member made of brittle material one obtains the stress concentration factor K from a graph which has been determined from experimentThis value is then multiplied by the average stress to obtain the maximum stress at the cross section smax Ksavg If the loading in a bar made of ductile material causes the material to yield then the stress distribution that is produced can be determined from the strain distribution and the stressstrain diagramAssuming the material is perfectly plastic yielding will cause the stress distribution at the cross section of a hole or transition to even out and become uniform If a member is constrained and an external loading causes yielding then when the load is released it will cause residual stress in the member P s1 s1 PP sY sY CONCEPTUAL PROBLEMS 175 4 CONCEPTUAL PROBLEMS A P41 P41 The concrete footing A was poured when this column was put in place Later the rest of the foundation slab was poured Can you explain why the 45 cracks occurred at each corner Can you think of a better design that would avoid such cracks P42 P42 The row of bricks along with mortar and an internal steel reinforcing rod was intended to serve as a lintel beam to support the bricks above this ventilation opening on an exterior wall of a building Explain what may have caused the bricks to fail in the manner shown REVIEW PROBLEMS 177 4 4119 The assembly consists of two bars AB and CD of the same material having a modulus of elasticity and coefficient of thermal expansion and a bar EF having a modulus of elasticity and coefficient of thermal expansion All the bars have the same length L and crosssectional area A If the rigid beam is originally horizontal at temperature determine the angle it makes with the horizontal when the temperature is increased to T2 T1 a2 E2 a1 E1 L d B A F E D C d Prob 4119 5 in 4 in 12 in 6 in 350 lb A B C Prob 4120 4120 The rigid link is supported by a pin at A and two A36 steel wires each having an unstretched length of 12 in and crosssectional area of Determine the force developed in the wires when the link supports the vertical load of 350 lb 00125 in2 The torsional stress and angle of twist of this soil auger depend upon the output of the machine turning the bit as well as the resistance of the soil in contact with the shaft 179 CHAPTER OBJECTIVES In this chapter we will discuss the effects of applying a torsional loading to a long straight member such as a shaft or tube Initially we will consider the member to have a circular cross section We will show how to determine both the stress distribution within the member and the angle of twist when the material behaves in a linear elastic manner and also when it is inelastic Statically indeterminate analysis of shafts and tubes will also be discussed along with special topics that include those members having noncircular cross sections Lastly stress concentrations and residual stress caused by torsional loadings will be given special consideration 51 Torsional Deformation of a Circular Shaft Torque is a moment that tends to twist a member about its longitudinal axis Its effect is of primary concern in the design of axles or drive shafts used in vehicles and machinery We can illustrate physically what happens when a torque is applied to a circular shaft by considering the shaft to be made of a highly deformable material such as rubber Fig 51a When the torque is applied the circles and longitudinal grid lines originally marked on the shaft tend to distort into the pattern shown in Fig 51b Note that twisting causes the circles to remain circles and each longitudinal grid line deforms into a helix that intersects the circles at equal angles Also the cross sections from the ends along the shaft will remain flatthat is they do not warp or bulge in or outand radial lines remain straight during the deformation Fig 51b From these observations we can assume that if the angle of twist is small the length of the shaft and its radius will remain unchanged Torsion 5 180 CHAPTER 5 TORSION If the shaft is fixed at one end and a torque is applied to its other end the dark green shaded plane in Fig 52 will distort into a skewed form as shown Here a radial line located on the cross section at a distance x from the fixed end of the shaft will rotate through an angle The angle so defined is called the angle of twist It depends on the position x and will vary along the shaft as shown In order to understand how this distortion strains the material we will now isolate a small element located at a radial distance rho from the axis of the shaft Fig 53 Due to the deformation as noted in Fig 52 the front and rear faces of the element will undergo a rotationthe back face by and the front face by As a result the difference in these rotations causes the element to be subjected to a shear strain To calculate this strain note that before deformation the angle between the edges AB and AC is 90 after deformation however the edges of the element are AD and AC and the angle between them is From the definition of shear strain Eq 24 we have g p 2 u u f f1x2 f f1x2 r f1x2 f1x2 Before deformation a After deformation b Longitudinal lines become twisted Circles remain circular Radial lines remain straight T T Fig 51 T x y x The angle of twist fx increases as x increases Undeformed plane Deformed plane z fx Fig 52 5 Notice the deformation of the rectangular element when this rubber bar is subjected to a torque 182 CHAPTER 5 TORSION 52 The Torsion Formula When an external torque is applied to a shaft it creates a corresponding internal torque within the shaft In this section we will develop an equation that relates this internal torque to the shear stress distribution on the cross section of a circular shaft or tube If the material is linearelastic then Hookes law applies and consequently a linear variation in shear strain as noted in the previous section leads to a corresponding linear variation in shear stress along any radial line on the cross section Hence will vary from zero at the shafts longitudinal axis to a maximum value at its outer surface This variation is shown in Fig 55 on the front faces of a selected number of elements located at an intermediate radial position and at the outer radius c Due to the proportionality of triangles we can write 53 This equation expresses the shearstress distribution over the cross section in terms of the radial position of the elementUsing itwe can now apply the condition that requires the torque produced by the stress distribution over the entire cross section to be equivalent to the resultant internal torque T at the section which holds the shaft in equilibrium Fig 55 r t a r c btmax r tmax t t Gg 5 tmax tmax tmax t t t Shear stress varies linearly along each radial line of the cross section T c dA r Fig 55 52 THE TORSION FORMULA 183 Specifically each element of area dA located at is subjected to a force of The torque produced by this force is We therefore have for the entire cross section 54 Since is constant 55 The integral depends only on the geometry of the shaft It represents the polar moment of inertia of the shafts crosssectional area about the shafts longitudinal axis We will symbolize its value as J and therefore the above equation can be rearranged and written in a more compact form namely 56 Here the maximum shear stress in the shaft which occurs at the outer surface the resultant internal torque acting at the cross section Its value is determined from the method of sections and the equation of moment equilibrium applied about the shafts longitudinal axis the polar moment of inertia of the crosssectional area the outer radius of the shaft Combining Eqs 53 and 56 the shear stress at the intermediate distance can be determined from 57 Either of the above two equations is often referred to as the torsion formula Recall that it is used only if the shaft is circular and the material is homogeneous and behaves in a linear elastic manner since the derivation is based on Hookes law t Tr J r c J T tmax tmax Tc J T tmax c LA r2 dA tmaxc T LA r1t dA2 LA ra r c btmax dA dT r1t dA2 dF t dA r 5 184 CHAPTER 5 TORSION 5 dr c r Fig 56 a T t tmax tmax Shear stress varies linearly along each radial line of the cross section b tmax Fig 57 T T Failure of a wooden shaft due to torsion Fig 58 Solid Shaft If the shaft has a solid circular cross section the polar moment of inertia J can be determined using an area element in the form of a differential ring or annulus having a thickness and circumference Fig 56 For this ring and so 58 Note that J is a geometric property of the circular area and is always positive Common units used for its measurement are or The shear stress has been shown to vary linearly along each radial line of the cross section of the shaft However if an element of material on the cross section is isolated then due to the complementary property of shear equal shear stresses must also act on four of its adjacent faces as shown in Fig 57a Hence not only does the internal torque T develop a linear distribution of shear stress along each radial line in the plane of the crosssectional area but also an associated shearstress distribution is developed along an axial plane Fig 57b It is interesting to note that because of this axial distribution of shear stress shafts made from wood tend to split along the axial plane when subjected to excessive torque Fig 58 This is because wood is an anisotropic material Its shear resistance parallel to its grains or fibers directed along the axis of the shaft is much less than its resistance perpendicular to the fibers directed in the plane of the cross section in4 mm4 J p 2 c4 J LA r2 dA L c 0 r212pr dr2 2p L c 0 r3 dr 2pa 1 4 br4 0 c dA 2pr dr 2pr dr 186 CHAPTER 5 TORSION 5 Important Points When a shaft having a circular cross section is subjected to a torque the cross section remains plane while radial lines rotate This causes a shear strain within the material that varies linearly along any radial line from zero at the axis of the shaft to a maximum at its outer boundary For linear elastic homogeneous material the shear stress along any radial line of the shaft also varies linearly from zero at its axis to a maximum at its outer boundary This maximum shear stress must not exceed the proportional limit Due to the complementary property of shear the linear shear stress distribution within the plane of the cross section is also distributed along an adjacent axial plane of the shaft The torsion formula is based on the requirement that the resultant torque on the cross section is equal to the torque produced by the shear stress distribution about the longitudinal axis of the shaft It is required that the shaft or tube have a circular cross section and that it is made of homogeneous material which has linearelastic behavior Procedure for Analysis The torsion formula can be applied using the following procedure Internal Loading Section the shaft perpendicular to its axis at the point where the shear stress is to be determined and use the necessary freebody diagram and equations of equilibrium to obtain the internal torque at the section Section Property Calculate the polar moment of inertia of the crosssectional area For a solid section of radius c and for a tube of outer radius and inner radius Shear Stress Specify the radial distance measured from the center of the cross section to the point where the shear stress is to be found Then apply the torsion formula or if the maximum shear stress is to be determined use When substituting the data make sure to use a consistent set of units The shear stress acts on the cross section in a direction that is always perpendicular to The force it creates must contribute a torque about the axis of the shaft that is in the same direction as the internal resultant torque T acting on the section Once this direction is established a volume element located at the point where is determined can be isolated and the direction of acting on the remaining three adjacent faces of the element can be shown t t r tmax TcJ t TrJ r J p1co 4 ci 422 ci co J pc42 52 THE TORSION FORMULA 187 5 EXAMPLE 51 The solid shaft of radius c is subjected to a torque T Fig 510a Determine the fraction of T that is resisted by the material contained within the outer region of the shaft which has an inner radius of and outer radius c SOLUTION The stress in the shaft varies linearly such that Eq 53 Therefore the torque on the ring area located within the lightershaded region Fig 510b is For the entire lightershaded area the torque is So that 1 This torque can be expressed in terms of the applied torque T by first using the torsion formula to determine the maximum stress in the shaftWe have or Substituting this into Eq 1 yields Ans NOTE Here approximately 94 of the torque is resisted by the lightershaded region and the remaining 6 or of T is resisted by the inner core of the shaft to As a result the material located at the outer region of the shaft is highly effective in resisting torque which justifies the use of tubular shafts as an efficient means for transmitting torque and thereby saving material r c2 r 0 1 16 T 15 16 T tmax 2T pc3 tmax Tc J Tc 1p22c4 T T 15p 32 tmaxc3 2ptmax c 1 4 r4 c2 c T 2ptmax c L c c2 r3 dr dT r1t dA2 r1rc2tmax12pr dr2 dT t 1rc2tmax c2 a T c c2 c2 b c dr r tmax t Fig 510 190 CHAPTER 5 TORSION 5 53 Power Transmission Shafts and tubes having circular cross sections are often used to transmit power developed by a machine When used for this purpose they are subjected to a torque that depends on the power generated by the machine and the angular speed of the shaft Power is defined as the work performed per unit of timeAlso the work transmitted by a rotating shaft equals the torque applied times the angle of rotationTherefore if during an instant of time dt an applied torque T causes the shaft to rotate then the instantaneous power is Since the shafts angular velocity is we can express the power as 510 In the SI system power is expressed in watts when torque is measured in newtonmeters and is in radians per second In the FPS system the basic units of power are footpounds per second however horsepower hp is often used in engineering practice where For machinery the frequency of a shafts rotation f is often reported This is a measure of the number of revolutions or cycles the shaft makes per second and is expressed in hertz Since then and so the above equation for power becomes 511 Shaft Design When the power transmitted by a shaft and its frequency of rotation are known the torque developed in the shaft can be determined from Eq 511 that is Knowing T and the allowable shear stress for the material we can determine the size of the shafts cross section using the torsion formula provided the material behavior is linear elastic Specifically the design or geometric parameter becomes 512 For a solid shaft and thus upon substitution a unique value for the shafts radius c can be determined If the shaft is tubular so that design permits a wide range of possibilities for the solution This is because an arbitrary choice can be made for either or ci and the other radius can then be determined from Eq 512 co J 1p221co 4 ci 42 J 1p22c4 J c T tallow Jc tallow T P2pf P 2pfT v 2pf 1 cycle 2p rad 11 Hz 1 cycles2 1 hp 550 ft lbs 1ft lbs2 1rads2 11 W 1 N ms2 v 1N m2 P Tv v dudt P T du dt du The chain drive transmits the torque developed by the electric motor to the shaft The stress developed in the shaft depends upon the power transmitted by the motor and the rate of rotation of the connecting shaft P Tv 53 POWER TRANSMISSION 191 5 EXAMPLE 54 A solid steel shaft AB shown in Fig 513 is to be used to transmit 5 hp from the motor M to which it is attached If the shaft rotates at and the steel has an allowable shear stress of determine the required diameter of the shaft to the nearest 1 8 in tallow 145 ksi v 175 rpm M A B v Fig 513 SOLUTION The torque on the shaft is determined from Eq 510 that is Expressing P in footpounds per second and in we have Thus Applying Eq 512 yields Since select a shaft having a diameter of Ans d 7 8 in 0875 in 2c 0858 in c 0429 in c 2T ptallow 13 211501 ft lb2112 inft2 p114 500 lbin22 13 J c p 2 c4 c T tallow T 1501 ft lb 2750 ft lbs T11833 rads2 P Tv v 175 rev min a 2p rad 1 rev b a 1 min 60 s b 1833 rads P 5 hp a 550 ft lbs 1 hp b 2750 ft lbs radianssecond v P Tv 198 CHAPTER 5 TORSION 533 The gear motor can develop 2 hp when it turns at If the shaft has a diameter of 1 in determine the maximum shear stress developed in the shaft 534 The gear motor can develop 3 hp when it turns at If the allowable shear stress for the shaft is determine the smallest diameter of the shaft to the nearest that can be used 1 8 in tallow 12 ksi 150 revmin 450 revmin 537 A ship has a propeller drive shaft that is turning at while developing 1800 hp If it is 8 ft long and has a diameter of 4 in determine the maximum shear stress in the shaft caused by torsion 538 The motor A develops a power of 300 W and turns its connected pulley at Determine the required diameters of the steel shafts on the pulleys at A and B if the allowable shear stress is tallow 85 MPa 90 revmin 1500 revmin 5 Probs 53334 Probs 53536 60 mm 150 mm 90 revmin A BB Prob 538 A F C E D 4 kW 5 kW 12 kW 25 mm 3 kW B Probs 53940 535 The 25mmdiameter shaft on the motor is made of a material having an allowable shear stress of If the motor is operating at its maximum power of 5 kW determine the minimum allowable rotation of the shaft 536 The drive shaft of the motor is made of a material having an allowable shear stress of If the outer diameter of the tubular shaft is 20 mm and the wall thickness is 25 mm determine the maximum allowable power that can be supplied to the motor when the shaft is operating at an angular velocity of 1500 revmin tallow 75 MPa tallow 75 MPa 539 The solid steel shaft DF has a diameter of 25 mm and is supported by smooth bearings at D and E It is coupled to a motor at F which delivers 12 kW of power to the shaft while it is turning at If gears A B and C remove 3 kW 4 kW and 5 kW respectively determine the maximum shear stress developed in the shaft within regions CF and BC The shaft is free to turn in its support bearings D and E 540 Determine the absolute maximum shear stress developed in the shaft in Prob 539 50 revs 53 POWER TRANSMISSION 199 544 The drive shaft AB of an automobile is made of a steel having an allowable shear stress of If the outer diameter of the shaft is 25 in and the engine delivers 200 hp to the shaft when it is turning at determine the minimum required thickness of the shafts wall 545 The drive shaft AB of an automobile is to be designed as a thinwalled tube The engine delivers 150 hp when the shaft is turning at Determine the minimum thickness of the shafts wall if the shafts outer diameter is 25 inThe material has an allowable shear stress of tallow 7 ksi 1500 revmin 1140 revmin tallow 8 ksi 541 The A36 steel tubular shaft is 2 m long and has an outer diameter of 50 mm When it is rotating at 40 rad s it transmits 25 kW of power from the motor M to the pump P Determine the smallest thickness of the tube if the allowable shear stress is 542 The A36 solid tubular steel shaft is 2 m long and has an outer diameter of 60 mm It is required to transmit 60 kW of power from the motor M to the pump P Determine the smallest angular velocity the shaft can have if the allowable shear stress is tallow 80 MPa tallow 80 MPa 5 M P Probs 54142 25 in d Prob 543 A B Probs 54445 543 A steel tube having an outer diameter of 25 in is used to transmit 35 hp when turning at Determine the inner diameter d of the tube to the nearest if the allowable shear stress is tallow 10 ksi 1 8 in 2700 revmin 546 The motor delivers 15 hp to the pulley at A while turning at a constant rate of 1800 rpm Determine to the nearest in the smallest diameter of shaft BC if the allowable shear stress for steel is The belt does not slip on the pulley tallow 12 ksi 1 8 3 in A B C 15 in Prob 546 200 CHAPTER 5 TORSION 54 Angle of Twist Occasionally the design of a shaft depends on restricting the amount of rotation or twist that may occur when the shaft is subjected to a torque Furthermore being able to compute the angle of twist for a shaft is important when analyzing the reactions on statically indeterminate shafts In this section we will develop a formula for determining the angle of twist phi of one end of a shaft with respect to its other endThe shaft is assumed to have a circular cross section that can gradually vary along its length Fig 514a Also the material is assumed to be homogeneous and to behave in a linearelastic manner when the torque is applied Like the case of an axially loaded bar we will neglect the localized deformations that occur at points of application of the torques and where the cross section changes abruptly By SaintVenants principle these effects occur within small regions of the shafts length and generally they will have only a slight effect on the final result Using the method of sections a differential disk of thickness dx located at position x is isolated from the shaft Fig 514b The internal resultant torque is Tx since the external loading may cause it to vary along the axis of the shaft Due to Tx the disk will twist such that the relative rotation of one of its faces with respect to the other face is Fig 514b As a result an element of material located at an arbitrary radius within the disk will undergo a shear strain The values of and are related by Eq 51 namely 513 df g dx r df g g r df f Fig 514 5 Oil wells are commonly drilled to depths exceeding a thousand metersAs a result the total angle of twist of a string of drill pipe can be substantial and must be determined x dx T3 T2 x y z T1 a b c Tx dx df df g g gmax r r 54 ANGLE OF TWIST 201 Since Hookes law applies and the shear stress can be expressed in terms of the applied torque using the torsion formula then Substituting this into Eq 513 the angle of twist for the disk is Integrating over the entire length L of the shaft we obtain the angle of twist for the entire shaft namely 514 Here the angle of twist of one end of the shaft with respect to the other end measured in radians the internal torque at the arbitrary position x found from the method of sections and the equation of moment equilibrium applied about the shafts axis the shafts polar moment of inertia expressed as a function of position x the shear modulus of elasticity for the material Constant Torque and CrossSectional Area Usually in engineering practice the material is homogeneous so that G is constant Also the shafts crosssectional area and the external torque are constant along the length of the shaft Fig 515 If this is the case the internal torque the polar moment of inertia and Eq 514 can be integrated which gives 515 The similarities between the above two equations and those for an axially loaded bar and d PLAE should be noted d 1P1x2 dxA1x2E f TL JG J1x2 J T1x2 T G J1x2 T1x2 f f L L 0 T1x2 dx J1x2G df T1x2 J1x2G dx g T1x2rJ1x2G t T1x2rJ1x2 g tG 5 When computing both the stress and the angle of twist of this soil auger it is necessary to consider the variable torsional loading which acts along its length f L T T Fig 515 202 CHAPTER 5 TORSION Equation 515 is often used to determine the shear modulus of elasticity G of a material To do so a specimen of known length and diameter is placed in a torsion testing machine like the one shown in Fig 516 The applied torque T and angle of twist are then measured along the length L Using Eq 515 Usually to obtain a more reliable value of G several of these tests are performed and the average value is used Multiple Torques If the shaft is subjected to several different torques or the crosssectional area or shear modulus changes abruptly from one region of the shaft to the next Eq 515 can be applied to each segment of the shaft where these quantities are all constantThe angle of twist of one end of the shaft with respect to the other is then found from the vector addition of the angles of twist of each segment For this case 516 Sign Convention In order to apply this equation we must develop a sign convention for both the internal torque and the angle of twist of one end of the shaft with respect to the other endTo do this we will use the righthand rule whereby both the torque and angle will be positive provided the thumb is directed outward from the shaft when the fingers curl to give the tendency for rotation Fig 517 To illustrate the use of this sign convention consider the shaft shown in Fig 518a The angle of twist of end A with respect to end D is to be determined Three segments of the shaft must be considered since the f a TL JG G TLJf f 5 Load dial Motor controls Load range selector Torque strain recorder Fixed head Specimen Motor Turning head Movable unit on rails Fig 516 204 CHAPTER 5 TORSION 5 Important Point When applying Eq 514 to determine the angle of twist it is important that the applied torques do not cause yielding of the material and that the material is homogeneous and behaves in a linear elastic manner Procedure for Analysis The angle of twist of one end of a shaft or tube with respect to the other end can be determined using the following procedure Internal Torque The internal torque is found at a point on the axis of the shaft by using the method of sections and the equation of moment equilibrium applied along the shafts axis If the torque varies along the shafts length a section should be made at the arbitrary position x along the shaft and the internal torque represented as a function of x ie Tx If several constant external torques act on the shaft between its ends the internal torque in each segment of the shaft between any two external torques must be determined The results can be represented graphically as a torque diagram Angle of Twist When the circular crosssectional area of the shaft varies along the shafts axis the polar moment of inertia must be expressed as a function of its position x along the axis Jx Ifthepolarmomentofinertiaortheinternaltorquesuddenlychanges between the ends of the shaft then or must be applied to each segment for which J G and T are continuous or constant When the internal torque in each segment is determined be sure to use a consistent sign convention for the shaft such as the one discussed in Fig 517 Also make sure that a consistent set of units is used when substituting numerical data into the equations f TLJG f 11T1x2J1x2G2 dx 214 CHAPTER 5 TORSION 5 55 Statically Indeterminate TorqueLoaded Members A torsionally loaded shaft may be classified as statically indeterminate if the moment equation of equilibrium applied about the axis of the shaft is not adequate to determine the unknown torques acting on the shaftAn example of this situation is shown in Fig522aAs shown on the freebody diagram Fig 522b the reactive torques at the supports A and B are unknownWe require that In order to obtain a solution we will use the method of analysis discussed in Sec 44 The necessary condition of compatibility or the kinematic condition requires the angle of twist of one end of the shaft with respect to the other end to be equal to zero since the end supports are fixed Therefore Provided the material is linear elastic we can apply the loaddisplacement relation to express the compatibility condition in terms of the unknown torquesRealizing that the internal torque in segment AC is and in segment CB it is Fig 522c we have TALAC JG TBLBC JG 0 TB TA f TLJG fAB 0 T TA TB 0 Mx 0 b c T TA TA TA TB TB TB LAC LBC L C a T A B Fig 522 55 STATICALLY INDETERMINATE TORQUELOADED MEMBERS 215 5 Solving the above two equations for the reactions realizing that we get and TB TLAC L TA TLBC L L LAC LBC Procedure for Analysis The unknown torques in statically indeterminate shafts are determined by satisfying equilibrium compatibility and torquedisplacement requirements for the shaft Equilibrium Draw a freebody diagram of the shaft in order to identify all the external torques that act on it Then write the equation of moment equilibrium about the axis of the shaft Compatibility Write the compatibility equation between two points along the shaft Give consideration as to how the supports constrain the shaft when it is twisted Express the angles of twist in the compatibility condition in terms of the torques using a torquedisplacement relation such as Solve the equilibrium and compatibility equations for the unknown reactive torques If any of the magnitudes have a negative numerical value it indicates that this torque acts in the opposite sense of direction to that shown on the freebody diagram f TLJG The shaft of this cutting machine is fixed at its ends and subjected to a torque at its center allowing it to act as a torsional spring 222 CHAPTER 5 TORSION 5 Notice the deformation of the square element when this rubber bar is subjected to a torque The results of the analysis for square cross sections along with other results from the theory of elasticity for shafts having triangular and elliptical cross sections are reported in Table 51 In all cases the maximum shear stress occurs at a point on the edge of the cross section that is closest to the center axis of the shaft In Table 51 these points are indicated as dots on the cross sections Also given are formulas for the angle of twist of each shaft By extending these results to a shaft having an arbitrary cross section it can also be shown that a shaft having a circular cross section is most efficient since it is subjected to both a smaller maximum shear stress and a smaller angle of twist than a corresponding shaft having a noncircular cross section and subjected to the same torque The drill shaft is connected to the soil auger using a shaft having a square cross section Shape of cross section Tmax Ellipse b b a a 2 T pa3b3G a2 b2TL Square a a T a3 481 T TL a4G 710 TL Equilateral triangle a a a3 20 T a4G 46 TL a F pab2 TABLE 51 224 CHAPTER 5 TORSION 5 The terminology flow is used since q is analogous to water flowing through a tube of rectangular cross section having a constant depth and variable width Although the waters velocity at each point along the tube will be different like the flow will be constant q vw tavg v w 57 ThinWalled Tubes Having Closed Cross Sections Thinwalled tubes of noncircular cross section are often used to construct lightweight frameworks such as those used in aircraft In some applications they may be subjected to a torsional loading In this section we will analyze the effects of applying a torque to a thinwalled tube having a closed cross section that is a tube that does not have any breaks or slits along its length Such a tube having a constant yet arbitrary cross sectional shape and variable thickness t is shown in Fig 528a Since the walls are thin we will obtain the average shear stress by assuming that this stress is uniformly distributed across the thickness of the tube at any given point Before we do this however we will first discuss some preliminary concepts regarding the action of shear stress over the cross section Shear Flow Shown in Figs 528a and 528b is a small element of the tube having a finite length s and differential width dx At one end the element has a thickness and at the other end the thickness is Due to the internal torque T shear stress is developed on the front face of the element Specifically at end A the shear stress is and at end B it is These stresses can be related by noting that equivalent shear stresses and must also act on the longitudinal sides of the element Since these sides have a constant width dx the forces acting on them are and Equilibrium requires these forces to be of equal magnitude but opposite direction so that This important result states that the product of the average shear stress times the thickness of the tube is the same at each point on the tubes crosssectional areaThis product is called shear flow q and in general terms we can express it as 517 Since q is constant over the cross section the largest average shear stress must occur where the tubes thickness is the smallest q tavgt tAtA tBtB dFB tB1tB dx2 dFA tA1tA dx2 tB tA tB tA tB tA T a x dx s t O tA tB dx s b A B tA tB tB tA Fig 528 226 CHAPTER 5 TORSION 5 Solving for we have 518 Here the average shear stress acting over a particular thickness of the tube the resultant internal torque at the cross section the thickness of the tube where is to be determined the mean area enclosed within the boundary of the centerline of the tubes thickness is shown shaded in Fig 528f Since then the shear flow throughout the cross section becomes 519 Angle of Twist The angle of twist of a thinwalled tube of length L can be determined using energy methods and the development of the necessary equation is given as a problem later in the text If the material behaves in a linear elastic manner and G is the shear modulus then this angle given in radians can be expressed as 520 Here again the integration must be performed around the entire boundary of the tubes crosssectional area f TL 4Am 2 G C ds t f q T 2Am q tavg t Am Am tavg t T tavg tavg T 2tAm tavg See Prob 1412 Important Points Shear flow q is the product of the tubes thickness and the average shear stressThis value is the same at all points along the tubes cross section As a result the largest average shear stress on the cross section occurs where the thickness is smallest Both shear flow and the average shear stress act tangent to the wall of the tube at all points and in a direction so as to contribute to the resultant internal torque 57 THINWALLED TUBES HAVING CLOSED CROSS SECTIONS 227 5 EXAMPLE 511 Calculate the average shear stress in a thinwalled tube having a circular cross section of mean radius and thickness t which is subjected to a torque T Fig 529a Also what is the relative angle of twist if the tube has a length L SOLUTION Average Shear Stress The mean area for the tube is Applying Eq 518 gives Ans We can check the validity of this result by applying the torsion formula In this case using Eq 59 we have Since and so that Ans which agrees with the previous result The average shearstress distribution acting throughout the tubes cross section is shown in Fig 529b Also shown is the shearstress distribution acting on a radial line as calculated using the torsion formulaNotice how each acts in a direction such that it contributes to the resultant torque T at the section As the tubes thickness decreasesthe shear stress throughout the tube becomes more uniform Angle of Twist Applying Eq 520 we have The integral represents the length around the centerline boundary which is Substituting the final result is Ans Show that one obtains this same result using Eq 515 f TL 2prm 3 Gt 2prm f TL 4Am 2 G C ds t TL 41prm 2 22Gt C ds tavg tavg Trm J Trm 2prm 3 t T 2ptrm 2 J p 2 12rm 2 212rm2t 2prm 3 t t ro ri rm L ro L ri p 2 1ro 2 ri 221ro ri21ro ri2 p 2 1ro 2 ri 221ro 2 ri 22 J p 2 1ro 4 ri 42 tavg T 2tAm T 2ptrm 2 Am prm 2 rm t T rm L a T T Actual shearstress distribution torsion formula Average shearstress distribution thinwall approximation b rm tavg tavg tmax Fig 529 238 CHAPTER 5 TORSION 5 If the internal torque produces the maximum elastic shear strain at the outer boundary of the shaft then the maximum elastic torque that produces this distribution can be found from the torsion formula so that 524 Furthermore the angle of twist can be determined from Eq 513 namely 525 If the applied torque increases in magnitude above it will begin to cause yielding First at the outer boundary of the shaft and then as the maximum shear strain increases to say in Fig 535a the yielding boundary will progress inward toward the shafts center Fig 535bAs shown this produces an elastic core where by proportion the radius of the core is Also the outer portion of the material forms a plastic annulus or ring since the shear strains within this region are greater than The corresponding shearstress distribution along a radial line of the shaft is shown in Fig 535c It is established by taking successive points on the shearstrain distribution in Fig 535b and finding the corresponding value of shear stress from the diagram Fig 535a For example at gives and at also gives etc Since in Fig 535c can now be expressed as a function of we can apply Eq 523 to determine the torqueWe have 526 ptY 6 14c3 rY 3 2 p 2rY tYrY 4 2p 3 tY1c3 rY 3 2 2p rY tYL rY 0 r3 dr 2ptYL c rY r2 dr 2p L rY 0 tY r rY r2 dr 2p L c rY tYr2 dr T 2p L c 0 tr2 dr r t tY r rY gY tY r c g tg gY g rY 1gYg2c g r c TY df g dx r TY p 2 tYc3 tY TYc3p2c44 TY gY Fig 535 c b Shearstrain distribution Plastic annulus Elastic core rY gY g a gY g g tY t c c Shearstress distribution T rY tY tY 248 CHAPTER 5 TORSION 5 t ksi g rad 10 0004 T T 3 ft 6 in 3 in Probs 5138139 35 mm 30 mm 210 0003 g rad t MPa T Prob 5140 5139 The tube is made of elasticperfectly plastic material which has the diagram shown Determine the torque T that just causes the inner surface of the shaft to yield Also find the residual shearstress distribution in the shaft when the torque is removed tg 5140 The 2mlong tube is made of an elasticperfectly plastic material as shown Determine the applied torque T that subjects the material at the tubes outer edge to a shear strain of What would be the permanent angle of twist of the tube when this torque is removed Sketch the residual stress distribution in the tube gmax 0006 rad 5136 The tubular shaft is made of a strainhardening material having a diagram as shown Determine the torque T that must be applied to the shaft so that the maximum shear strain is 001 rad tg 5137 The shear stressstrain diagram for a solid 50mmdiameter shaft can be approximated as shown in the figure Determine the torque T required to cause a maximum shear stress in the shaft of 125 MPa If the shaft is 15 m long what is the corresponding angle of twist T 075 in 10 0005 g rad t ksi 15 001 05 in Prob 5136 50 00025 g rad t MPa 125 0010 15 m T T Prob 5137 5138 A tube is made of elasticperfectly plastic material which has the diagram shown If the radius of the elastic core is determine the applied torque T Also find the residual shearstress distribution in the shaft and the permanent angle of twist of one end relative to the other when the torque is removed rY 225 in tg CHAPTER REVIEW 251 5 Am T t T tmax c T tY tY rY Solid noncircular shafts tend to warp out of plane when subjected to a torque Formulas are available to determine the maximum elastic shear stress and the twist for these cases The average shear stress in thinwalled tubes is determined by assuming the shear stress across each thickness t of the tube is constant Its value is determined from Stress concentrations occur in shafts when the cross section suddenly changes The maximum shear stress is determined using a stress concentration factor K which is determined from experiment and represented in graphical form Once obtained If the applied torque causes the material to exceed the elastic limit then the stress distribution will not be proportional to the radial distance from the centerline of the shaft Instead the internal torque is related to the stress distribution using the shearstressshearstrain diagram and equilibrium If a shaft is subjected to a plastic torque which is then released it will cause the material to respond elastically thereby causing residual shear stress to be developed in the shaft tmax Ka Tc J b tavg T 2tAm REVIEW PROBLEMS 253 5 2 ft 15 ft B D C A E F 500 lb ft Prob 5148 5148 The motor A develops a torque at gear B of 500 which is applied along the axis of the 2in diameter A36 steel shaft CD This torque is to be transmitted to the pinion gears at E and F If these gears are temporarily fixed determine the maximum shear stress in segments CB and BD of the shaftAlso what is the angle of twist of each of these segments The bearings at C and D only exert force reactions on the shaft lb ft 5149 The coupling consists of two disks fixed to separate shafts each 25 mm in diameterThe shafts are supported on journal bearings that allow free rotation In order to limit the torque T that can be transmitted a shear pin P is used to connect the disks together If this pin can sustain an average shear force of 550 N before it fails determine the maximum constant torque T that can be transmitted from one shaft to the other Also what is the maximum shear stress in each shaft when the shear pin is about to fail 5150 The rotating flywheel and shaft is brought to a sudden stop at D when the bearing freezes This causes the flywheel to oscillate clockwisecounterclockwise so that a point A on the outer edge of the flywheel is displaced through a 10mm arc in either direction Determine the maximum shear stress developed in the tubular 304 stainless steel shaft due to this oscillation The shaft has an inner diameter of 25 mm and an outer diameter of 35 mm The journal bearings at B and C allow the shaft to rotate freely 5151 If the solid shaft AB to which the valve handle is attached is made of C83400 red brass and has a diameter of 10 mm determine the maximum couple forces F that can be applied to the handle just before the material starts to fail Take What is the angle of twist of the handle The shaft is fixed at A tallow 40 MPa 25 mm P 25 mm 130 mm T T Prob 5149 2 m C B D A 80 mm Prob 5150 F 150 mm 150 mm F A 150 mm B Prob 5151 Beams are important structural members used in building construction Their design is often based upon their ability to resist bending stress which forms the subject matter of this chapter Simply supported beam Cantilevered beam Overhanging beam Fig 61 255 CHAPTER OBJECTIVES Beams and shafts are important structural and mechanical elements in engineering In this chapter we will determine the stress in these members caused by bending The chapter begins with a discussion of how to establish the shear and moment diagrams for a beam or shaft Like the normalforce and torque diagrams the shear and moment diagrams provide a useful means for determining the largest shear and moment in a member and they specify where these maximums occur Once the internal moment at a section is determined the bending stress can then be calculated First we will consider members that are straight have a symmetric cross section and are made of homogeneous linear elastic material Afterward we will discuss special cases involving unsymmetric bending and members made of composite materials Consideration will also be given to curved members stress concentrations inelastic bending and residual stresses 61 Shear and Moment Diagrams Members that are slender and support loadings that are applied perpendicular to their longitudinal axis are called beams In general beams are long straight bars having a constant crosssectional area Often they are classified as to how they are supported For example a simply supported beam is pinned at one end and roller supported at the other Fig 61 a cantilevered beam is fixed at one end and free at the other and an overhanging beam has one or both of its ends freely extended over the supports Beams are considered among the most important of all structural elements They are used to support the floor of a building the deck of a bridge or the wing of an aircraft Also the axle of an automobile the boom of a crane even many of the bones of the body act as beams Bending 6 256 CHAPTER 6 BENDING 6 Because of the applied loadings beams develop an internal shear force and bending moment that in general vary from point to point along the axis of the beam In order to properly design a beam it therefore becomes necessary to determine the maximum shear and moment in the beam One way to do this is to express V and M as functions of their arbitrary position x along the beams axis These shear and moment functions can then be plotted and represented by graphs called shear and moment diagrams The maximum values of V and M can then be obtained from these graphs Also since the shear and moment diagrams provide detailed information about the variation of the shear and moment along the beams axis they are often used by engineers to decide where to place reinforcement materials within the beam or how to proportion the size of the beam at various points along its length In order to formulate V and M in terms of x we must choose the origin and the positive direction for x Although the choice is arbitrary most often the origin is located at the left end of the beam and the positive direction is to the right In general the internal shear and moment functions of x will be discontinuous or their slope will be discontinuous at points where a distributed load changes or where concentrated forces or couple moments are applied Because of this the shear and moment functions must be determined for each region of the beam between any two discontinuities of loading For example coordinates and will have to be used to describe the variation of V and M throughout the length of the beam in Fig 62 These coordinates will be valid only within the regions from A to B for from B to C for and from C to D for Beam Sign Convention Before presenting a method for determining the shear and moment as functions of x and later plotting these functions shear and moment diagrams it is first necessary to establish a sign convention so as to define positive and negative values for V and M Although the choice of a sign convention is arbitrary here we will use the one often used in engineering practice and shown in Fig 63 The positive directions are as follows the distributed load acts upward on the beam the internal shear force causes a clockwise rotation of the beam segment on which it acts and the internal moment causes compression in the top fibers of the segment such that it bends the segment so that it holds water Loadings that are opposite to these are considered negative x3 x2 x1 x3 x2 x1 P D B C A w0 x1 x2 x3 Fig 62 V Positive external distributed load Positive internal shear Positive internal moment V M M Beam sign convention wx Fig 63 61 SHEAR AND MOMENT DIAGRAMS 257 6 Procedure for Analysis The shear and moment diagrams for a beam can be constructed using the following procedure Support Reactions Determine all the reactive forces and couple moments acting on the beam and resolve all the forces into components acting perpendicular and parallel to the beams axis Shear and Moment Functions Specify separate coordinates x having an origin at the beams left end and extending to regions of the beam between concentrated forces andor couple moments or where there is no discontinuity of distributed loading Section the beam at each distance x and draw the freebody diagram of one of the segments Be sure V and M are shown acting in their positive sense in accordance with the sign convention given in Fig 63 The shear is obtained by summing forces perpendicular to the beams axis To eliminate V the moment is obtained directly by summing moments about the sectioned end of the segment Shear and Moment Diagrams Plot the shear diagram V versus x and the moment diagram M versus x If numerical values of the functions describing V and M are positive the values are plotted above the x axis whereas negative values are plotted below the axis Generally it is convenient to show the shear and moment diagrams below the freebody diagram of the beam Important Points Beams are long straight members that are subjected to loads perpendicular to their longitudinal axis They are classified according to the way they are supported eg simply supported cantilevered or overhanging In order to properly design a beam it is important to know the variation of the internal shear and moment along its axis in order to find the points where these values are a maximum Using an established sign convention for positive shear and moment the shear and moment in the beam can be determined as a function of its position x on the beam and then these functions can be plotted to form the shear and moment diagrams 62 GRAPHICAL METHOD FOR CONSTRUCTING SHEAR AND MOMENT DIAGRAMS 277 6 621 The beam is subjected to the uniform distributed load shown Draw the shear and moment diagrams for the beam 622 Draw the shear and moment diagrams for the overhang beam 623 Draw the shear and moment diagrams for the beam It is supported by a smooth plate at A which slides within the groove and so it cannot support a vertical force although it can support a moment and axial load B A C 2 m 15 m 1 m 2 kNm 624 Determine the placement distance a of the roller support so that the largest absolute value of the moment is a minimum Draw the shear and moment diagrams for this condition 4 kNm 3 m 3 m A B 625 The beam is subjected to the uniformly distributed moment m Draw the shear and moment diagrams for the beam momentlength L A B w 626 Consider the general problem of a cantilevered beam subjected to n concentrated loads and a constant distributed loading w Write a computer program that can be used to determine the internal shear and moment at any specified location x along the beam and plot the shear and moment diagrams for the beam Show an application of the program using the values w 800 Nm a1 2 m a2 4 m L 4 m P1 4 kN d1 2 m L A m P1 L w a1 d1 d2 dn a2 P2 Pn Prob 621 Prob 622 Prob 623 Prob 626 Prob 625 Prob 624 a w L A B 63 BENDING DEFORMATION OF A STRAIGHT MEMBER 281 6 63 Bending Deformation of a Straight Member In this section we will discuss the deformations that occur when a straight prismatic beam made of homogeneous material is subjected to bendingThe discussion will be limited to beams having a crosssectional area that is symmetrical with respect to an axis and the bending moment is applied about an axis perpendicular to this axis of symmetry as shown in Fig 618 The behavior of members that have unsymmetrical cross sections or are made from several different materials is based on similar observations and will be discussed separately in later sections of this chapter By using a highly deformable material such as rubber we can illustrate what happens when a straight prismatic member is subjected to a bending moment Consider for example the undeformed bar in Fig 619a which has a square cross section and is marked with longitudinal and transverse grid lines When a bending moment is applied it tends to distort these lines into the pattern shown in Fig 619b Notice that the longitudinal lines become curved and the vertical transverse lines remain straight and yet undergo a rotation The bending moment causes the material within the bottom portion of the bar to stretch and the material within the top portion to compress Consequently between these two regions there must be a surface called the neutral surface in which longitudinal fibers of the material will not undergo a change in length Fig 618 Before deformation a M M After deformation b Horizontal lines become curved Vertical lines remain straight yet rotate x y z M Axis of symmetry Longitudinal axis Neutral surface Fig 619 Fig 618 64 THE FLEXURE FORMULA 285 6 Fig 624 64 The Flexure Formula In this section we will develop an equation that relates the stress distribution in a beam to the internal resultant bending moment acting on the beams cross section To do this we will assume that the material behaves in a linearelastic manner and therefore a linear variation of normal strain Fig 624a must then be the result of a linear variation in normal stress Fig 624b Hence like the normal strain variation will vary from zero at the members neutral axis to a maximum value a distance c farthest from the neutral axis Because of the proportionality of triangles Fig 623b or by using Hookes law and Eq 68 we can write 69 This equation describes the stress distribution over the crosssectional area The sign convention established here is significant For positive M which acts in the direction positive values of y give negative values for that is a compressive stress since it acts in the negative x direction Similarly negative y values will give positive or tensile values for If a volume element of material is selected at a specific point on the cross section only these tensile or compressive normal stresses will act on it For example the element located at is shown in Fig 624c We can locate the position of the neutral axis on the cross section by satisfying the condition that the resultant force produced by the stress distribution over the crosssectional area must be equal to zero Noting that the force acts on the arbitrary element dA in Fig 624c we require smax c LA y dA LA a y c bsmax dA 0 LA dF LA s dA FR Fx dF s dA y s s z s a y c bsmax s EP smax s y x c y Normal strain variation profile view a y x y M Bending stress variation profile view b c P Pmax smax s This wood specimen failed in bending due to its fibers being crushed at its top and torn apart at its bottom 286 CHAPTER 6 BENDING 6 Since is not equal to zero then 610 In other words the first moment of the members crosssectional area about the neutral axis must be zeroThis condition can only be satisfied if the neutral axis is also the horizontal centroidal axis for the cross section Consequently once the centroid for the members crosssectional area is determined the location of the neutral axis is known We can determine the stress in the beam from the requirement that the resultant internal moment M must be equal to the moment produced by the stress distribution about the neutral axis The moment of dF in Fig 624c about the neutral axis is Since using Eq 69 we have for the entire cross section or 611 M smax c LA y2 dA M LA y dF LA y1s dA2 LA yy c smax dA 1MR2z Mz dF s dA dM y dF LA y dA 0 smaxc y c Bending stress variation c x z dA M dF y smax s s s Fig 624 cont Recall that the location for the centroid of the crosssectional area is defined from the equation If then and so the centroid lies on the reference neutral axis See Appendix A y 0 1y dA 0 y 1y dA1dA y 64 THE FLEXURE FORMULA 287 6 The integral represents the moment of inertia of the crosssectional area about the neutral axis We will symbolize its value as I Hence Eq 611 can be solved for and written as 612 Here maximum normal stress in the member which occurs at a point on the crosssectional area farthest away from the neutral axis resultant internal moment determined from the method of sections and the equations of equilibrium and calculated about the neutral axis of the cross section perpendicular distance from the neutral axis to a point farthest away from the neutral axisThis is where acts moment of inertia of the crosssectional area about the neutral axis Since Eq 69 the normal stress at the intermediate distance y can be determined from an equation similar to Eq 612 We have 613 Note that the negative sign is necessary since it agrees with the established x y z axes By the righthand rule M is positive along the axis y is positive upward and therefore must be negative compressive since it acts in the negative x direction Fig 624c Either of the above two equations is often referred to as the flexure formula It is used to determine the normal stress in a straight member having a cross section that is symmetrical with respect to an axis and the moment is applied perpendicular to this axisAlthough we have assumed that the member is prismatic we can in most cases of engineering design also use the flexure formula to determine the normal stress in members that have a slight taper For example using a mathematical analysis based on the theory of elasticity a member having a rectangular cross section and a length that is tapered 15 will have an actual maximum normal stress that is about 54 less than that calculated using the flexure formula s z s My I smaxc sy I the smax c the M the smax the smax Mc I smax 288 CHAPTER 6 BENDING 6 Important Points The cross section of a straight beam remains plane when the beam deforms due to bending This causes tensile stress on one portion of the cross section and compressive stress on the other portion In between these portions there exists the neutral axis which is subjected to zero stress Due to the deformation the longitudinal strain varies linearly from zero at the neutral axis to a maximum at the outer fibers of the beam Provided the material is homogeneous and linear elastic then the stress also varies in a linear fashion over the cross section The neutral axis passes through the centroid of the crosssectional areaThis result is based on the fact that the resultant normal force acting on the cross section must be zero The flexure formula is based on the requirement that the resultant internal moment on the cross section is equal to the moment produced by the normal stress distribution about the neutral axis Procedure for Analysis In order to apply the flexure formula the following procedure is suggested Internal Moment Section the member at the point where the bending or normal stress is to be determined and obtain the internal moment M at the sectionThe centroidal or neutral axis for the cross section must be known since M must be calculated about this axis If the absolute maximum bending stress is to be determined then draw the moment diagram in order to determine the maximum moment in the member Section Property Determine the moment of inertia of the crosssectional area about the neutral axis Methods used for its calculation are discussed in Appendix A and a table listing values of I for several common shapes is given on the inside front cover Normal Stress Specify the distance y measured perpendicular to the neutral axis to the point where the normal stress is to be determined Then apply the equation or if the maximum bending stress is to be calculated use When substituting the data make sure the units are consistent The stress acts in a direction such that the force it creates at the point contributes a moment about the neutral axis that is in the same direction as the internal moment M Fig 624c In this manner the stress distribution acting over the entire cross section can be sketched or a volume element of the material can be isolated and used to graphically represent the normal stress acting at the point smax McI s MyI 64 THE FLEXURE FORMULA 293 6 200 mm M 150 mm 150 mm 50 mm 30 mm 30 mm 50 mm 30 mm 30 mm 150 mm 25 mm M A 150 mm 50 mm 50 mm 50 mm 25 mm FUNDAMENTAL PROBLEMS F615 If the beam is subjected to a bending moment of determine the maximum bending stress in the beam M 20 kN m F618 If the beam is subjected to a bending moment of determine the maximum bending stress in the beam M 10 kN m 300 mm 20 mm 20 mm 20 mm M 200 mm 150 mm 150 mm 300 mm M 200 mm 300 mm 20 mm 20 mm 20 mm M F616 If the beam is subjected to a bending moment of sketch the bending stress distribution over the beams cross section M 50 kN m F617 If the beam is subjected to a bending moment of determine the maximum bending stress in the beam M 50 kN m F619 If the beam is subjected to a bending moment of determine the bending stress developed at point A M 5 kN m F615 F616 F617 F618 F619 294 CHAPTER 6 BENDING 6 200 mm 150 mm z y x M 50 mm 30 mm A B C 5 mm 5 mm 5 mm 5 mm 5 mm 7 mm 7 mm 10 mm 647 A member having the dimensions shown is used to resist an internal bending moment of Determine the maximum stress in the member if the moment is applied a about the z axis as shown b about the y axis Sketch the stress distribution for each case M 90 kN m 648 Determine the moment M that will produce a maximum stress of 10 ksi on the cross section 649 Determine the maximum tensile and compressive bending stress in the beam if it is subjected to a moment of M 4 kip ft 652 The beam is subjected to a moment M Determine the percentage of this moment that is resisted by the stresses acting on both the top and bottom boards A and B of the beam 653 Determine the moment M that should be applied to the beam in order to create a compressive stress at point D of Also sketch the stress distribution acting over the cross section and compute the maximum stress developed in the beam sD 30 MPa 650 The channel strut is used as a guide rail for a trolley If the maximum moment in the strut is determine the bending stress at points A B and C 651 The channel strut is used as a guide rail for a trolley If the allowable bending stress for the material is determine the maximum bending moment the strut will resist sallow 175 MPa M 30 N m PROBLEMS 150 mm 25 mm 25 mm 150 mm M 25 mm 25 mm B A D Prob 647 Probs 65051 Probs 65253 3 in D A B 05 in M 05 in 3 in C 10 in 05 in 05 in Probs 64849 64 THE FLEXURE FORMULA 297 6 669 Two designs for a beam are to be considered Determine which one will support a moment of with the least amount of bending stress What is that stress 150 kN m M 670 The simply supported truss is subjected to the central distributed load Neglect the effect of the diagonal lacing and determine the absolute maximum bending stress in the trussThe top member is a pipe having an outer diameter of 1 in and thickness of and the bottom member is a solid rod having a diameter of 1 2 in 3 16 in 671 The axle of the freight car is subjected to wheel loadings of 20 kipIf it is supported by two journal bearings at C and D determine the maximum bending stress developed at the center of the axle where the diameter is 55 in 200 mm 300 mm a b 15 mm 30 mm 15 mm 200 mm 300 mm 30 mm 15 mm 30 mm 6 ft 575 in 6 ft 6 ft 100 lbft C D A B 20 kip 20 kip 10 in 10 in 60 in 672 The steel beam has the crosssectional area shown Determine the largest intensity of distributed load that it can support so that the maximum bending stress in the beam does not exceed 673 The steel beam has the crosssectional area shown If determine the maximum bending stress in the beam w0 05 kipft smax 22 ksi w0 674 The boat has a weight of 2300 lb and a center of gravity at G If it rests on the trailer at the smooth contact A and can be considered pinned at B determine the absolute maximum bending stress developed in the main strut of the trailer Consider the strut to be a boxbeam having the dimensions shown and pinned at C 675 The shaft is supported by a smooth thrust bearing at A and smooth journal bearing at D If the shaft has the cross section shown determine the absolute maximum bending stress in the shaft 10 in 8 in 030 in 12 ft 12 ft 030 in 03 in w0 1 ft 3 ft D A B C 1 ft 5 ft 4 ft G 175 in 3 in 175 in 15 in A C D B 3 kN 3 kN 075 m 075 m 15 m 40 mm 25 mm Prob 669 Probs 67273 Prob 674 Prob 675 Prob 670 Prob 671 64 THE FLEXURE FORMULA 299 6 686 Determine the absolute maximum bending stress in the 2indiameter shaft which is subjected to the concentrated forces The journal bearings at A and B only support vertical forces 687 Determine the smallest allowable diameter of the shaft which is subjected to the concentrated forces The journal bearings at A and B only support vertical forces The allowable bending stress is sallow 22 ksi 688 If the beam has a square cross section of 9 in on each side determine the absolute maximum bending stress in the beam 693 The man has a mass of 78 kg and stands motionless at the end of the diving board If the board has the cross section shown determine the maximum normal strain developed in the board The modulus of elasticity for the material is Assume A is a pin and B is a roller E 125 GPa 694 The two solid steel rods are bolted together along their length and support the loading shown Assume the support at A is a pin and B is a rollerDetermine the required diameter d of each of the rods if the allowable bending stress is 695 Solve Prob 694 if the rods are rotated so that both rods rest on the supports at A pin and B roller 90 sallow 130 MPa 689 If the compound beam in Prob 642 has a square cross section determine its dimension a if the allowable bending stress is 690 If the beam in Prob 628 has a rectangular cross section with a width b and a height h determine the absolute maximum bending stress in the beam sallow 150 MPa 691 Determine the absolute maximum bending stress in the 80mmdiameter shaft which is subjected to the concentrated forces The journal bearings at A and B only support vertical forces 692 Determine the smallest allowable diameter of the shaft which is subjected to the concentrated forces The journal bearings at A and B only support vertical forces The allowable bending stress is sallow 150 MPa 15 in 15 in B A 800 lb 30 in 600 lb A B 8 ft 8 ft 800 lbft 1200 lb 05 m 06 m 04 m 20 kN A B 12 kN B C A 15 m 25 m 350 mm 20 mm 30 mm 10 mm 10 mm 10 mm B A 2 m 80 kN 20 kNm 2 m Probs 68687 Probs 69192 Prob 688 Prob 693 Probs 69495 300 CHAPTER 6 BENDING 6 1 in 3 in a a A 180 lb 25 in 05 in 8 in 696 The chair is supported by an arm that is hinged so it rotates about the vertical axis at A If the load on the chair is 180 lb and the arm is a hollow tube section having the dimensions shown determine the maximum bending stress at section aa 699 If the beam has a square cross section of 6 in on each side determine the absolute maximum bending stress in the beam 6100 The steel beam has the crosssectional area shown Determine the largest intensity of the distributed load that it can support so that the maximum bending stress in the beam does not exceed 6101 The steel beam has the crosssectional area shown If determine the maximum bending stress in the beam w0 2 kipft sallow 22 ksi w0 697 A portion of the femur can be modeled as a tube having an inner diameter of 0375 in and an outer diameter of 125 in Determine the maximum elastic static force P that can be applied to its center Assume the bone to be roller supported at its ends The diagram for the bone mass is shown and is the same in tension as in compression sP 698 If the beam in Prob 618 has a rectangular cross section with a width of 8 in and a height of 16 in determine the absolute maximum bending stress in the beam 4 in 4 in 230 125 002 005 P s ksi P inin 8 in 16 in A B 6 ft 6 ft 400 lbft 12 in 9 in 025 in 9 ft 9 ft 025 in 025 in w0 Prob 696 Prob 699 Probs 6100101 Prob 697 Prob 698 65 UNSYMMETRIC BENDING 303 6 As shown in Sec 64 Eq 614 is satisfied since the z axis passes through the centroid of the area Also since the z axis represents the neutral axis for the cross section the normal stress will vary linearly from zero at the neutral axis to a maximum at Fig 630b Hence the stress distribution is defined by When this equation is substituted into Eq 616 and integrated it leads to the flexure formula When it is substituted into Eq 615 we get which requires This integral is called the product of inertia for the area As indicated in Appendix A it will indeed be zero provided the y and z axes are chosen as principal axes of inertia for the area For an arbitrarily shaped area the orientation of the principal axes can always be determined using either the inertia transformation equations or Mohrs circle of inertia as explained in Appendix A Secs A4 and A5 If the area has an axis of symmetry however the principal axes can easily be established since they will always be oriented along the axis of symmetry and perpendicular to it For example consider the members shown in Fig 631 In each of these cases y and z must define the principal axes of inertia for the cross section in order to satisfy Eqs 614 through 616 In Fig 631a the principal axes are located by symmetry and in Figs 631b and 631c their orientation is determined using the methods of Appendix A Since M is applied about one of the principal axes z axis the stress distribution is determined from the flexure formula and is shown for each case s MyIz LA yz dA 0 0 smax c LA yz dA smax McI s ycs max y c Fig 631 z M y a M y x b z M y x c z 306 CHAPTER 6 BENDING 6 EXAMPLE 615 The rectangular cross section shown in Fig 633a is subjected to a bending moment of Determine the normal stress developed at each corner of the section and specify the orientation of the neutral axis SOLUTION Internal Moment Components By inspection it is seen that the y and z axes represent the principal axes of inertia since they are axes of symmetry for the cross sectionAs required we have established the z axis as the principal axis for maximum moment of inertia The moment is resolved into its y and z components where Section Properties The moments of inertia about the y and z axes are Bending Stress Thus Iz 1 12 102 m2104 m23 106711032 m4 Iy 1 12 104 m2102 m23 0266711032 m4 Mz 3 5 112 kN m2 720 kN m My 4 5 112 kN m2 960 kN m M 12 kN m Ans Ans Ans Ans sE 72011032 N m102 m2 106711032 m4 96011032 N m101 m2 0266711032 m4 495 MPa sD 72011032 N m102 m2 106711032 m4 96011032 N m101 m2 0266711032 m4 225 MPa sC 72011032 N m102 m2 106711032 m4 96011032 N m101 m2 0266711032 m4 495 MPa sB 72011032 N m102 m2 106711032 m4 96011032 N m101 m2 0266711032 m4 225 MPa s Mzy Iz Myz Iy The resultant normalstress distribution has been sketched using these values Fig 633b Since superposition applies the distribution is linear as shown 312 CHAPTER 6 BENDING 6 66 Composite Beams Beams constructed of two or more different materials are referred to as composite beams For example a beam can be made of wood with straps of steel at its top and bottom Fig 635 Engineers purposely design beams in this manner in order to develop a more efficient means for supporting loads Since the flexure formula was developed only for beams having homogeneous material this formula cannot be applied directly to determine the normal stress in a composite beam In this section however we will develop a method for modifying or transforming a composite beams cross section into one made of a single material Once this has been done the flexure formula can then be used for the stress analysis To explain how to do this we will consider a composite beam made of two materials 1 and 2 which have the crosssectional areas shown in Fig 636a If a bending moment is applied to this beam then like one that is homogeneous the total crosssectional area will remain plane after bending and hence the normal strains will vary linearly from zero at the neutral axis to a maximum in the material located farthest from this axis Fig 636b Provided the material is linear elastic then at any point the normal stress in material 1 is determined from and for material 2 the stress distribution is found from Assuming material 1 is stiffer than material 2 then and so the stress distribution will look like that shown in Fig 636c or 636d In particular notice the jump in stress that occurs at the juncture of the two materials Here the strain is the same but since the modulus of elasticity for the materials suddenly changes so does the stress It is possible to determine the location of the neutral axis and the maximum stress based on a trialanderror procedure E1 7 E2 s E2P s E1P Steel plates M Fig 635 66 COMPOSITE BEAMS 313 6 This requires satisfying the conditions that the stress distribution produces a zero resultant force on the cross section and the moment of the stress distribution about the neutral axis must be equal to M A simpler way to satisfy these two conditions is to use the transformed section method which transforms the beam into one made of a single material For example if the beam is thought to consist entirely of the less stiff material 2 then the cross section will look like that shown in Fig 636e Here the height h of the beam remains the same since the strain distribution in Fig 636b must be preserved However the upper portion of the beam must be widened in order to carry a load equivalent to that carried by the stiffer material 1 in Fig 636dThe necessary width can be determined by considering the force dF acting on an area of the beam in Fig 636a It is Assuming the width of a corresponding element of height dy in Fig 636e is n dz then Equating these forces so that they produce the same moment about the z neutral axis we have or 620 n E1 E2 E1P dz dy E2Pn dz dy dF sdA 1E2P2n dz dy dF s dA 1E1P2 dz dy dA dz dy Less stiff material M dy dz y z y h b Stiff material a x 2 1 M x y Normal strain variation profile view b M x y Bending stress variation profile view c y z x M Bending stress variation d Fig 636 67 REINFORCED CONCRETE BEAMS 315 6 67 Reinforced Concrete Beams All beams subjected to pure bending must resist both tensile and compressive stresses Concrete however is very susceptible to cracking when it is in tension and therefore by itself it will not be suitable for resisting a bending moment In order to circumvent this shortcoming engineers place steel reinforcing rods within a concrete beam at a location where the concrete is in tension Fig 637a To be most effective these rods are located farthest from the beams neutral axis so that the moment created by the forces developed in them is greatest about the neutral axis Furthermore the rods are required to have some concrete coverage to protect them from corrosion or loss of strength in the event of a fire Codes used for actual reinforced concrete design assume the ability of concrete will not support any tensile loading since the possible cracking of concrete is unpredictable As a result the normal stress distribution acting on the crosssectional area of a reinforced concrete beam is assumed to look like that shown in Fig 637b The stress analysis requires locating the neutral axis and determining the maximum stress in the steel and concreteTo do this the area of steel is first transformed into an equivalent area of concrete using the transformation factor This ratio which gives requires a greater amount of concrete to replace the steel The transformed area is and the transformed section looks like that shown in Fig 637c Here d represents the distance from the top of the beam to the transformed steel b is the beams width and is the yet unknown distance from the top of the beam to the neutral axisTo obtain we require the centroid C of the crosssectional area of the transformed section to lie on the neutral axis Fig 637c With reference to the neutral axis therefore the moment of the two areas must be zero since Thus Once is obtained from this quadratic equation the solution proceeds in the usual manner for obtaining the stress in the beam Example 618 numerically illustrates application of this method h b 2 h2 nAsth nAstd 0 bh a h 2 b nAst1d h2 0 y yAA 0 yA h h nAst n 7 1 n EstEconc Ast a b d M A b M Concrete assumed cracked within this region N c A N b d h C n Ast Fig 637 Inspection of its particular stressstrain diagram in Fig 311 reveals that concrete can be 125 times stronger in compression than in tension 68 CURVED BEAMS 319 6 68 Curved Beams The flexure formula applies to a straight member since it was shown that the normal strain within it varies linearly from the neutral axis If the member is curved however this assumption becomes inaccurate and so we must develop another method to describe the stress distribution In this section we will consider the analysis of a curved beam that is a member that has a curved axis and is subjected to bending Typical examples include hooks and chain links In all cases the members are not slender but rather have a sharp curve and their crosssectional dimensions are large compared with their radius of curvature The following analysis assumes that the cross section is constant and has an axis of symmetry that is perpendicular to the direction of the applied moment M Fig 640a Also the material is homogeneous and isotropic and it behaves in a linearelastic manner when the load is applied Like the case of a straight beam we will also assume that the cross sections of the member remain plane after the moment is applied Furthermore any distortion of the cross section within its own plane will be neglected To perform the analysis three radii extending from the center of curvature of the member are identified in Fig 640a Here references the known location of the centroid for the crosssectional area R references the yet unspecified location of the neutral axis and r locates the arbitrary point or area element dA on the cross section r O M M R r y R r e N A y C dA a Centroid Neutral axis Area element dA r r O A A This crane hook represents a typical example of a curved beam Fig 640 68 CURVED BEAMS 321 In order to relate the stress distribution to the resultant bending moment we require the resultant internal moment to be equal to the moment of the stress distribution calculated about the neutral axis From Fig 640a the stress acting on the area element dA and located a distance y from the neutral axis creates a moment about the neutral axis of For the entire cross section we require Since and is defined by Eq 622 we have Expanding realizing that Ek and R are constants then The first integral is equivalent to as determined from Eq 623 and the second integral is simply the crosssectional area A Realizing that the location of the centroid of the cross section is determined from the third integral can be replaced by Thus Finally solving for Ek in Eq 622 substituting into the above equation and solving for we have 624 Here the normal stress in the member the internal moment determined from the method of sections and the equations of equilibrium and calculated about the neutral axis for the cross sectionThis moment is positive if it tends to increase the members radius of curvature ie it tends to straighten out the member the crosssectional area of the member the distance measured from the center of curvature to the neutral axis determined from Eq 623 the distance measured from the center of curvature to the centroid of the cross section the distance measured from the center of curvature to the point where the stress s is to be determined r r R A M s s M1R r2 Ar1r R2 s M EkA1r R2 rA r 1r dAA AR M EkR2 L A dA r 2R LA dA LA r dA M LA 1R r2Eka R r r b dA s y R r M 1ys dA dM y1s dA2 s 6 322 CHAPTER 6 BENDING 6 From Fig 640a Also the constant and usually very small distance between the neutral axis and the centroid is When these results are substituted into Eq 624 we can also write 625 These two equations represent two forms of the socalled curved beam formula which like the flexure formula can be used to determine the normalstress distribution in a curved member This distribution is as previously stated hyperbolic an example is shown in Fig 640c and 640d Since the stress acts along the circumference of the beam it is sometimes called circumferential stress Note that due to the curvature of the beam the circumferential stress will create a corresponding component of radial stress so called since this component acts in the radial direction To show how it is developed consider the freebody diagram of the segment shown in Fig 640e Here the radial stress is necessary since it creates the force that is required to balance the two components of circumferential forces dF which act along the line Sometimes the radial stresses within curved members may become significant especially if the member is constructed from thin plates and has for example the shape of an Isection In this case the radial stress can become as large as the circumferential stress and consequently the member must be designed to resist both stresses For most cases however these stresses can be neglected especially if the member has a solid section Here the curvedbeam formula gives results that are in very close agreement with those determined either by experiment or by a mathematical analysis based on the theory of elasticity The curvedbeam formula is normally used when the curvature of the member is very pronounced as in the case of hooks or rings However if the radius of curvature is greater than five times the depth of the member the flexure formula can normally be used to determine the stress For example for rectangular sections for which this ratio equals 5 the maximum normal stress when determined by the flexure formula will be about 7 less than its value when determined by the curved beam formula This error is further reduced when the radius of curvaturetodepth ratio is more than 5 OB dFr sr s My Ae1R y2 e r R r R y Bending stress variation profile view c M smax d M A N smax e B sr s s dF dF dFr O Fig 640 cont See for example Boresi A P et al Advanced Mechanics of Materials 3rd ed p 333 1978 John Wiley Sons New York 68 CURVED BEAMS 323 6 Important Points The curvedbeam formula should be used to determine the circumferential stress in a beam when the radius of curvature is less than five times the depth of the beam Due to the curvature of the beam the normal strain in the beam does not vary linearly with depth as in the case of a straight beam As a result the neutral axis does not pass through the centroid of the cross section The radial stress component caused by bending can generally be neglected especially if the cross section is a solid section and not made from thin plates Procedure for Analysis In order to apply the curvedbeam formula the following procedure is suggested Section Properties Determine the crosssectional area A and the location of the centroid measured from the center of curvature Find the location of the neutral axis R using Eq 623 or Table 61 If the crosssectional area consists of n composite parts determine for each part Then from Eq 623 for the entire section In all cases Normal Stress The normal stress located at a point r away from the center of curvature is determined from Eq 624 If the distance y to the point is measured from the neutral axis then find and use Eq 625 Since generally produces a very small number it is best to calculate and R with sufficient accuracy so that the subtraction leads to a number e having at least four significant figures If the stress is positive it will be tensile whereas if it is negative it will be compressive The stress distribution over the entire cross section can be graphed or a volume element of the material can be isolated and used to represent the stress acting at the point on the cross section where it has been calculated r r R e r R R 6 r R A11 dAr2 1dAr r 324 CHAPTER 6 BENDING 6 EXAMPLE 619 The curved bar has a crosssectional area shown in Fig 641a If it is subjected to bending moments of determine the maximum normal stress developed in the bar 4 kN m 200 mm 250 mm B A 200 mm 50 mm 30 mm 50 mm 280 mm 4 kNm 4 kNm r a O Fig 641 SOLUTION Internal Moment Each section of the bar is subjected to the same resultant internal moment of Since this moment tends to decrease the bars radius of curvature it is negative Thus Section Properties Here we will consider the cross section to be composed of a rectangle and triangle The total crosssectional area is The location of the centroid is determined with reference to the center of curvature point Fig 641a 023308 m 0225 m1005 m21005 m2 0260 m 1 210050 m210030 m2 325011032 m2 r rA A O A 1005 m22 1 2 1005 m21003 m2 325011032 m2 M 4 kN m 4 kN m 69 STRESS CONCENTRATIONS 327 6 geometry Once K is obtained the maximum bending stress shown in Fig 645 is determined using 626 In the same manner Fig 644 can be used if the discontinuity consists of circular grooves or notches Like axial load and torsion stress concentration for bending should always be considered when designing members made of brittle materials or those that are subjected to fatigue or cyclic loadingsAlso realize that stressconcentration factors apply only when the material is subjected to elastic behavior If the applied moment causes yielding of the material as is the case with ductile materials the stress becomes redistributed throughout the member and the maximum stress that results will be lower than that determined using stressconcentration factors This phenomenon is discussed further in the next section smax K Mc I M M smax smax Fig 645 Stress concentrations caused by bending occur at the sharp corners of this window lintel and are responsible for the crack at the corner Important Points Stress concentrations occur at points of sudden crosssectional change caused by notches and holes because here the stress and strain become nonlinear The more severe the change the larger the stress concentration For design or analysis the maximum normal stress occurs on the smallest crosssectional areaThis stress can be obtained by using a stress concentration factor K that has been determined through experiment and is only a function of the geometry of the member Normally the stress concentration in a ductile material subjected to a static moment will not have to be considered in design however if the material is brittle or subjected to fatigue loading then stress concentrations become important 69 STRESS CONCENTRATIONS 329 6 6127 The composite beam is made of 6061T6 aluminum A and C83400 red brass BDetermine the dimension h of the brass strip so that the neutral axis of the beam is located at the seam of the two metals What maximum moment will this beam support if the allowable bending stress for the aluminum is and for the brass 6128 The composite beam is made of 6061T6 aluminum A and C83400 red brass B If the height determine the maximum moment that can be applied to the beam if the allowable bending stress for the aluminum is and for the brass 1sallow2br 35 MPa 1sallow2al 128 MPa h 40 mm 1sallow2br 35 MPa 1sallow2al 128 MPa 6131 The Douglas fir beam is reinforced with A36 straps at its center and sides Determine the maximum stress developed in the wood and steel if the beam is subjected to a bending moment of Sketch the stress distribution acting over the cross section Mz 750 kip ft PROBLEMS B A 50 mm 150 mm h Probs 6127128 y z 6 in 05 in 05 in 2 in 2 in 05 in Prob 6131 w A B 15 ft 3 in 3 in 3 in Probs 6129130 6129 Segment A of the composite beam is made from 2014T6 aluminum alloy and segment B is A36 steel If determine the absolute maximum bending stress developed in the aluminum and steel Sketch the stress distribution on the cross section 6130 Segment A of the composite beam is made from 2014T6 aluminum alloy and segment B is A36 steel If the allowable bending stress for the aluminum and steel are and determine the maximum allowable intensity w of the uniform distributed load sallowst 22 ksi sallowal 15 ksi w 09 kipft 05 in 6 in 05 in 05 in 12 in M 05 in Probs 6132133 6132 The top plate is made of 2014T6 aluminum and is used to reinforce a Kevlar 49 plastic beam Determine the maximum stress in the aluminum and in the Kevlar if the beam is subjected to a moment of 6133 The top plate made of 2014T6 aluminum is used to reinforce a Kevlar 49 plastic beam If the allowable bending stress for the aluminum is and for the Kevlar determine the maximum moment M that can be applied to the beam sallowk 8 ksi sallowal 40 ksi M 900 lb ft 610 INELASTIC BENDING 335 6 610 Inelastic Bending The equations for determining the normal stress due to bending that have previously been developed are valid only if the material behaves in a linearelastic manner If the applied moment causes the material to yield a plastic analysis must then be used to determine the stress distribution For bending of straight members three conditions must be met Linear NormalStrain Distribution Based only on geometric considerations it was shown in Sec 63 that the normal strains always vary linearly from zero at the neutral axis of the cross section to a maximum at the farthest point from the neutral axis Resultant Force Equals Zero Since there is only a resultant internal moment acting on the cross section the resultant force caused by the stress distribution must be equal to zero Since creates a force on the area dA of Fig647then for the entire crosssectional area A we have 627 This equation provides a means for obtaining the location of the neutral axis Resultant Moment The resultant moment at the section must be equivalent to the moment caused by the stress distribution about the neutral axis Since the moment of the force about the neutral axis is Fig 647 then summing the results over the entire cross section we have 628 These conditions of geometry and loading will now be used to show how to determine the stress distribution in a beam when it is subjected to a resultant internal moment that causes yielding of the material Throughout the discussion we will assume that the material has a stressstrain diagram that is the same in tension as it is in compression For simplicity we will begin by considering the beam to have a crosssectional area with two axes of symmetry in this case a rectangle of height h and width b as shown in Fig 648aTwo cases of loading that are of special interest will be considered M LA y1s dA2 1MR2z Mz dM y1s dA2 dF s dA LA s dA 0 FR Fx dF s dA s y dA y x z M s Fig 647 610 INELASTIC BENDING 337 6 Or using Eq 629 630 As noted in Fig 648e M produces two zones of plastic yielding and an elastic core in the member The boundary between them is located a distance from the neutral axis As M increases in magnitude approaches zero This would render the material entirely plastic and the stress distribution will then look like that shown in Fig 648f From Eq 630 with or by finding the moments of the stress blocks around the neutral axis we can write this limiting value as 631 Using Eq 629 or Eq 630 with we have 632 This moment is referred to as the plastic moment Its value applies only for a rectangular section since the analysis depends on the geometry of the cross section Beams used in steel buildings are sometimes designed to resist a plastic moment When this is the case codes usually list a design property for a beam called the shape factorThe shape factor is defined as the ratio 633 This value specifies the additional moment capacity that a beam can support beyond its maximum elastic moment For example from Eq 632 a beam having a rectangular cross section has a shape factor of Therefore this section will support 50 more bending moment than its maximum elastic moment when it becomes fully plastic k 15 k Mp MY Mp 3 2 MY y 0 Mp 1 4 bh2 sY yY 0 yY yY M 3 2 MYa1 4 3 yY 2 h2 b Plastic moment f b C T sY sY Mp h 2 h 2 Fig 648 cont 340 CHAPTER 6 BENDING 6 3 Determine the volumes enclosed by the tensile and compressive stress blocks As an approximation this may require dividing each block into composite regions Equation 627 requires the volumes of these blocks to be equal since they represent the resultant tensile force T and resultant compressive force C on the section Fig 650eIf these forces are unequalan adjustment as to the location of the neutral axis must be made point of zero strain and the process repeated until Eq 627 is satisfied 4 Once the moments produced by T and C can be calculated about the neutral axis Here the moment arms for T and C are measured from the neutral axis to the centroids of the volumes defined by the stress distributions Fig 650e Equation 628 requires If this equation is not satisfied the slope of the strain distribution must be adjusted and the computations for T and C and the moment must be repeated until close agreement is obtained This trialanderror procedure is obviously very tedious and fortunately it does not occur very often in engineering practice Most beams are symmetric about two axes and they are constructed from materials that are assumed to have similar tensionandcompression stressstrain diagrams Whenever this occurs the neutral axis will pass through the centroid of the cross section and the process of relating the stress distribution to the resultant moment is thereby simplified M Ty Cy T C 1T C2 Assumed location of neutral axis Strain distribution profile view c Assumed slope of strain distribution P1 P2 Stress distribution profile view d M s1 s2 y A e N y T C Fig 650 cont Important Points The normal strain distribution over the cross section of a beam is based only on geometric considerations and has been found to always remain linear regardless of the applied load The normal stress distribution however must be determined from the material behavior or stressstrain diagram once the strain distribution is established The location of the neutral axis is determined from the condition that the resultant force on the cross section is zero The resultant internal moment on the cross section must be equal to the moment of the stress distribution about the neutral axis Perfectly plastic behavior assumes the normal stress distribution is constant over the cross section and the beam will continue to bend with no increase in moment This moment is called the plastic moment 610 INELASTIC BENDING 341 6 EXAMPLE 621 The steel wideflange beam has the dimensions shown in Fig 651a If it is made of an elastic perfectly plastic material having a tensile and compressive yield stress of determine the shape factor for the beam SOLUTION In order to determine the shape factor it is first necessary to calculate the maximum elastic moment and the plastic moment Maximum Elastic Moment The normalstress distribution for the maximum elastic moment is shown in Fig 651b The moment of inertia about the neutral axis is Mp MY sY 36 ksi Applying the flexure formula we have MY 15195 kip in 36 kipin2 MY15 in2 2110 in4 smax Mc I I c 1 12 105 in219 in23 d 2c 1 12 18 in2105 in23 8 in 105 in21475 in22d 2110 in4 Plastic Moment The plastic moment causes the steel over the entire cross section of the beam to yield so that the normalstress distribution looks like that shown in Fig 651c Due to symmetry of the crosssectional area and since the tension and compression stressstrain diagrams are the same the neutral axis passes through the centroid of the cross section In order to determine the plastic moment the stress distribution is divided into four composite rectangular blocks and the force produced by each block is equal to the volume of the blockTherefore we have These forces act through the centroid of the volume for each block Calculating the moments of these forces about the neutral axis we obtain the plastic moment Shape Factor Applying Eq 633 gives Ans NOTE This value indicates that a wideflange beam provides a very efficient section for resisting an elastic moment Most of the moment is developed in the flanges ie in the top and bottom segments whereas the web or vertical segment contributes very little In this particular case only 14 additional moment can be supported by the beam beyond that which can be supported elastically k Mp MY 17325 kip in 15195 kip in 114 Mp 21225 in2181 kip2 21475 in21144 kip2 17325 kip in C2 T2 36 kipin2 105 in218 in2 144 kip C1 T1 36 kipin2 105 in2145 in2 81 kip c A 36 ksi 36 ksi N T2 T1 C1 C2 Mp 8 in 9 in 05 in 05 in 05 in a b A 36 ksi MY 36 ksi N Fig 651 346 CHAPTER 6 BENDING 6 a a a a a a 25 mm 150 mm 150 mm 25 mm 25 mm 25 mm 200 mm 15 mm 15 mm 20 mm 200 mm Mp t b h t t 6165 The beam is made of an elastic plastic material for which Determine the residual stress in the beam at its top and bottom after the plastic moment is applied and then released Mp sY 250 MPa 6166 The wideflange member is made from an elastic plastic material Determine the shape factor 6167 Determine the shape factor for the cross section 6168 The beam is made of elastic perfectly plastic material Determine the maximum elastic moment and the plastic moment that can be applied to the cross section Take and sY 36 ksi a 2 in 6169 The box beam is made of an elastic perfectly plastic material for which Determine the residual stress in the top and bottom of the beam after the plastic moment is applied and then released Mp sY 250 MPa PROBLEMS Prob 6165 Prob 6166 Prob 6167168 Prob 6169 610 INELASTIC BENDING 347 6 200 mm 15 mm 15 mm 20 mm 200 mm Mp 3 in 3 in 15 in 15 in 6 in 6170 Determine the shape factor for the wideflange beam 6171 Determine the shape factor of the beams cross section 6172 The beam is made of elasticperfectly plastic material Determine the maximum elastic moment and the plastic moment that can be applied to the cross section Take sY 36 ksi 6173 Determine the shape factor for the cross section of the Hbeam 3 in 3 in 15 in 15 in 6 in 200 mm Mp 20 mm 20 mm 200 mm 20 mm Prob 6170 Prob 6171 Prob 6172 Prob 6173 348 CHAPTER 6 BENDING 6 200 mm Mp 20 mm 20 mm 200 mm 20 mm 3 in 3 in 3 in 3 in 3 in 3 in 6174 The Hbeam is made of an elasticplastic material for which Determine the residual stress in the top and bottom of the beam after the plastic moment is applied and then released Mp sY 250 MPa 6175 Determine the shape factor of the cross section 6176 The beam is made of elasticperfectly plastic material Determine the maximum elastic moment and the plastic moment that can be applied to the cross section Take sY 36 ksi 6177 Determine the shape factor of the cross section for the tube 3 in 3 in 3 in 3 in 3 in 3 in 6 in 5 in Prob 6174 Prob 6175 Prob 6177 Prob 6176 CHAPTER REVIEW 353 6 A bending moment tends to produce a linear variation of normal strain within a straight beam Provided the material is homogeneous and linear elastic then equilibrium can be used to relate the internal moment in the beam to the stress distribution The result is the flexure formula where I and c are determined from the neutral axis that passes through the centroid of the cross section If the crosssectional area of the beam is not symmetric about an axis that is perpendicular to the neutral axis then unsymmetrical bending will occur The maximum stress can be determined from formulas or the problem can be solved by considering the superposition of bending caused by the moment components and about the principal axes of inertia for the area Mz My smax Mc I s Mzy Iz Myz Iy Beams made from composite materials can be transformed so their cross section is considered as if it were made from a single material To do this the transformation factor n which is a ratio of the moduli of elasticity of the materials is used to change the width b of the beam Once the cross section is transformed then the stress in the beam can be determined in the usual manner using the flexure formula c x M y smax x z y M My Mz M h b 2 1 n E1 E2 354 CHAPTER 6 BENDING 6 Curved beams deform such that the normal strain does not vary linearly from the neutral axis Provided the material is homogeneous and linear elastic and the cross section has an axis of symmetry then the curved beam formula can be used to determine the bending stress or s My Ae1R y2 s M1R r2 Ar1r R2 s max KMc I Stress concentrations occur in members having a sudden change in their cross section caused for example by holes and notches The maximum bending stress at these locations is determined using a stress concentration factor K that is found from graphs determined from experiment If the bending moment causes the stress in the material to exceed its elastic limit then the normal strain will remain linear however the stress distribution will vary in accordance with the stress strain diagramThe plastic and ultimate moments supported by the beam can be determined by requiring the resultant force to be zero and the resultant moment to be equivalent to the moment of the stress distribution If an applied plastic or ultimate moment is released it will cause the material to respond elastically thereby inducing residual stresses in the beam M A smax N M M A N Mp h 2 h 2 sY sY CONCEPTUAL PROBLEMS 355 6 CONCEPTUAL PROBLEMS P61 P61 The steel saw blade passes over the drive wheel of the band saw Using appropriate measurements and data explain how to determine the bending stress in the blade P62 P62 This crane boom on a ship has a moment of inertia that varies along its length Draw the moment diagram for the boom to explain why the boom tapers as shown P63 P63 Hurricane winds caused failure of this highway sign by bending the supporting pipes at their connections with the column Assuming the pipes are made of A36 steel use reasonable dimensions for the sign and pipes and try and estimate the smallest uniform wind pressure acting on the face of the sign that caused the pipes to yield P64 P64 These garden shears were manufactured using an inferior material Using a loading of 50 lb applied normal to the blades and appropriate dimensions for the shears determine the absolute maximum bending stress in the material and show why the failure occurred at the critical location on the handle b a Railroad ties act as beams that support very large transverse shear loadings As a result if they are made of wood they will tend to split at their ends where the shear loads are the largest 7 359 CHAPTER OBJECTIVES In this chapter we will develop a method for finding the shear stress in a beam having a prismatic cross section and made from homogeneous material that behaves in a linearelastic manner The method of analysis to be developed will be somewhat limited to special cases of cross sectional geometry Although this is the case it has many widerange applications in engineering design and analysis The concept of shear flow along with shear stress will be discussed for beams and thinwalled members The chapter ends with a discussion of the shear center 71 Shear in Straight Members In general a beam will support both shear and moment The shear V is the result of a transverse shearstress distribution that acts over the beams cross section Due to the complementary property of shear however this stress will create corresponding longitudinal shear stresses which will act along longitudinal planes of the beam as shown in Fig 71 Transverse Shear Fig 71 Longitudinal shear stress Transverse shear stress V t t 360 CHAPTER 7 TRANSVERSE SHEAR 7 a Before deformation V b After deformation V Fig 73 To illustrate this effect consider the beam to be made from three boards Fig 72a If the top and bottom surfaces of each board are smooth and the boards are not bonded together then application of the load P will cause the boards to slide relative to one another when the beam deflects However if the boards are bonded together then the longitudinal shear stresses acting between the boards will prevent their relative sliding and consequently the beam will act as a single unit Fig 72b As a result of the shear stress shear strains will be developed and these will tend to distort the cross section in a rather complex manner For example consider the short bar in Fig 73a made of a highly deformable material and marked with horizontal and vertical grid lines When a shear V is applied it tends to deform these lines into the pattern shown in Fig 73b This nonuniform shearstrain distribution will cause the cross section to warp P Boards not bonded together a Boards bonded together b P Fig 72 Shear connectors are tack welded to this corrugated metal floor liner so that when the concrete floor is poured the connectors will prevent the concrete slab from slipping on the liner surface The two materials will thus act as a composite slab 366 CHAPTER 7 TRANSVERSE SHEAR 7 Important Points Shear forces in beams cause nonlinear shearstrain distributions over the cross section causing it to warp Due to the complementary property of shear stress the shear stress developed in a beam acts over the cross section of the beam and along its longitudinal planes The shear formula was derived by considering horizontal force equilibrium of the longitudinal shearstress and bendingstress distributions acting on a portion of a differential segment of the beam The shear formula is to be used on straight prismatic members made of homogeneous material that has linear elastic behaviorAlso the internal resultant shear force must be directed along an axis of symmetry for the crosssectional area The shear formula should not be used to determine the shear stress on cross sections that are short or flat at points of sudden crosssectional changes or at a point on an inclined boundary Procedure for Analysis In order to apply the shear formula the following procedure is suggested Internal Shear Section the member perpendicular to its axis at the point where the shear stress is to be determined and obtain the internal shear V at the section Section Properties Determine the location of the neutral axisand determine the moment of inertia I of the entire crosssectional area about the neutral axis Pass an imaginary horizontal section through the point where the shear stress is to be determined Measure the width t of the crosssectional area at this section The portion of the area lying either above or below this width is Determine Q by using Here is the distance to the centroid of measured from the neutral axis It may be helpful to realize that is the portion of the members crosssectional area that is being held onto the member by the longitudinal shear stresses See Fig 74c Shear Stress Using a consistent set of units substitute the data into the shear formula and calculate the shear stress It is suggested that the direction of the transverse shear stress be established on a volume element of material located at the point where it is calculated This can be done by realizing that acts on the cross section in the same direction as V From this the corresponding shear stresses acting on the other three planes of the element can then be established t t t A A y Q yA A 72 THE SHEAR FORMULA 367 7 EXAMPLE 71 b 20 mm 50 mm 4 kN 50 mm 4 kN a Fig 79 The solid shaft and tube shown in Fig 79a are subjected to the shear force of 4 kN Determine the shear stress acting over the diameter of each cross section SOLUTION Section Properties Using the table on the inside front cover the moment of inertia of each section calculated about its diameter or neutral axis is The semicircular area shown shaded in Fig 79b above or below each diameterrepresents Qbecause this area is held onto the member by the longitudinal shear stress along the diameter Shear Stress Applying the shear formula where m for the solid section and for the tube we have Ans Ans NOTE As discussed in the limitations for the shear formula the calculations performed here are valid since the shear stress along the diameter is vertical and therefore tangent to the boundary of the cross section An element of material on the diameter is subjected to pure shear as shown in Fig 79b ttube VQ It 4103 N780106 m3 4783106 m4006 m 109 MPa tsolid VQ It 4103 N8333106 m3 4909106 m401 m 679 kPa t 2003 m 006 m t 01 780106 m3 4005 m 3p ap005 m2 2 b 4002 m 3p ap002 m2 2 b Qtube g yA 4co 3p apc2 o 2 b 4ci 3p apc2 i 2 b Qsolid yA 4c 3p a pc2 2 b 4005 m 3p ap005 m2 2 b 8333 106 m3 Itube 1 4 pc4 o c4 i 1 4p3005 m4 002 m44 4783106 m4 Isolid 1 4 pc4 1 4 p005 m4 4909106 m4 368 CHAPTER 7 TRANSVERSE SHEAR 7 EXAMPLE 72 Determine the distribution of the shear stress over the cross section of the beam shown in Fig 710a c b N Tmax A V dy y Shearstress distribution b y N A A b h 2 h 2 y Fig 710 V a h b SOLUTION The distribution can be determined by finding the shear stress at an arbitrary height y from the neutral axis Fig 710b and then plotting this function Here the dark colored area will be used for Q Hence Applying the shear formula we have 1 This result indicates that the shearstress distribution over the cross section is parabolic As shown in Fig 710c the intensity varies from zero at the top and bottom to a maximum value at the neutral axis Specifically since the area of the cross section is then at we have 2 t max 15 V A y 0 A bh y 0 y h2 t VQ It VA1 2B3h24 y24b A 1 12 bh3Bb 6V bh3 a h2 4 y2b Q yA cy 1 2 a h 2 ybd ah 2 ybb 1 2 a h2 4 y2bb A The area below y can also be used but doing so involves a bit more algebraic manipulation 3A bh2 y4 72 THE SHEAR FORMULA 369 7 This same value for can be obtained directly from the shear formula by realizing that occurs where Q is largest since V I and t are constant By inspection Q will be a maximum when the entire area above or below the neutral axis is considered that is and Thus By comparison is 50 greater than the average shear stress determined from Eq 17 that is Itisimportanttorealizethat alsoactsinthelongitudinaldirection of the beam Fig 710d It is this stress that can cause a timber beam to fail as shown Fig 710e Here horizontal splitting of the wood starts to occur through the neutral axis at the beams ends since there the vertical reactions subject the beam to large shear stress and wood has a low resistance to shear along its grains which are oriented in the longitudinal direction It is instructive to show that when the shearstress distribution Eq 1 is integrated over the cross section it yields the resultant shear V To do this a differential strip of area is chosen Fig 710c and since acts uniformly over this strip we have 6V h3 B h2 4 a h 2 h 2 b 1 3 h3 8 h3 8 R V 6V h3 B h2 4 y 1 3 y3R h2 h2 LA t dA L h2 h2 6V bh3 h2 4 y2b dy t dA b dy t max tavg VA t max t max VQ It Vh4bh2 C 1 12bh3Db 15 V A y h4 A bh2 t max t VQIt t max N A d tmax P e Typical shear failure of this wooden beam occurred at the support and through the approximate center of its cross section Fig 710 cont 72 THE SHEAR FORMULA 371 7 N A 002 m 0100 m 0300 m C 0015 m A d For point C and is the dark shaded area shown in Fig 711d Considering this area to be composed of two rectangles we have Thus NOTE From Fig 711b note that most of the shear stress occurs in the web and is almost uniform throughout its depth varying from 226 MPa to 252 MPa It is for this reason that for design some codes permit the use of calculating the average shear stress on the cross section of the web rather than using the shear formula This will be discussed further in Chapter 11 tC tmax VQC ItC 80103 N073511032 m3 155611062 m410015 m2 252 MPa 073511032 m3 005 m10015 m210100 m2 QC yA 0110 m10300 m21002 m2 A tC 0015 m Fig 711 cont 72 THE SHEAR FORMULA 373 7 FUNDAMENTAL PROBLEMS F71 If the beam is subjected to a shear force of kN determine the shear stress developed at point A Represent the state of stress at A on a volume element V 100 F72 Determine the shear stress at points A and B on the beam if it is subjected to a shear force of kN V 600 F73 Determine the absolute maximum shear stress developed in the beam F74 If the beam is subjected to a shear force of kN determine the maximum shear stress developed in the beam V 20 F75 If the beam is made from four plates and subjected to a shear force of kN determine the maximum shear stress developed in the beam V 20 200 mm 90 mm 300 mm 20 mm 20 mm 20 mm V A F71 200 mm V 150 mm 150 mm 50 mm 30 mm 30 mm 30 mm 30 mm 50 mm F74 150 mm 50 mm 25 mm 25 mm A 150 mm 50 mm 50 mm V F75 A 1 ft 3 in 6 in 1 ft 1 ft 6 kip 3 kip B F73 100 mm 100 mm 100 mm 100 mm 100 mm 100 mm B A V F72 72 THE SHEAR FORMULA 375 7 79 Determine the largest shear force V that the member can sustain if the allowable shear stress is 710 If the applied shear force determine the maximum shear stress in the member V 18 kip tallow 8 ksi 713 Determine the maximum shear stress in the strut if it is subjected to a shear force of 714 Determine the maximum shear force V that the strut can support if the allowable shear stress for the material is tallow 40 MPa V 20 kN 711 The wood beam has an allowable shear stress of Determine the maximum shear force V that can be applied to the cross section tallow 7 MPa 715 Plot the shearstress distribution over the cross section of a rod that has a radius c By what factor is the maximum shear stress greater than the average shear stress acting over the cross section 712 The beam has a rectangular cross section and is made of wood having an allowable shear stress of 200 psi Determine the maximum shear force V that can be developed in the cross section of the beam Also plot the shearstress variation over the cross section tallow 716 A member has a cross section in the form of an equilateral triangle If it is subjected to a shear force V determine the maximum average shear stress in the member using the shear formula Should the shear formula actually be used to predict this value Explain V 3 in 1 in 1 in 1 in 3 in Probs 7910 50 mm 50 mm 200 mm 100 mm 50 mm V 50 mm Prob 711 V 60 mm 12 mm 20 mm 20 mm 80 mm 12 mm Probs 71314 c V y V 12 in 8 in Prob 712 V a h Prob 716 Prob 715 376 CHAPTER 7 TRANSVERSE SHEAR 7 717 Determine the maximum shear stress in the strut if it is subjected to a shear force of 718 Determine the maximum shear force V that the strut can support if the allowable shear stress for the material is 719 Plot the intensity of the shear stress distributed over the cross section of the strut if it is subjected to a shear force of V 600 kN tallow 45 MPa V 600 kN 722 Determine the shear stress at point B on the web of the cantilevered strut at section aa 723 Determine the maximum shear stress acting at section aa of the cantilevered strut 720 The steel rod is subjected to a shear of 30 kip Determine the maximum shear stress in the rod 721 The steel rod is subjected to a shear of 30 kip Determine the shear stress at point A Show the result on a volume element at this point 724 Determine the maximum shear stress in the Tbeam at the critical section where the internal shear force is maximum 725 Determine the maximum shear stress in the Tbeam at point C Show the result on a volume element at this point V 150 mm 30 mm 100 mm 100 mm 100 mm 30 mm Probs 7171819 30 kip 2 in 1 in A Probs 72021 a a 2 kN 4 kN 250 mm 250 mm 300 mm 20 mm 50 mm 70 mm 20 mm B Probs 72223 3 m 15 m 15 m 10 kNm A 150 mm 150 mm 30 mm 30 mm B C Probs 72425 72 THE SHEAR FORMULA 377 7 726 Determine the maximum shear stress acting in the fiberglass beam at the section where the internal shear force is maximum 729 Write a computer program that can be used to determine the maximum shear stress in the beam that has the cross section shown and is subjected to a specified constant distributed load and concentrated force P Show an application of the program using the values t1 15 mm t2 20 mm b 50 mm and h 150 mm a 2 m P 15 kN d1 0 d2 2 m w 400 Nm L 4 m w 727 Determine the shear stress at points C and D located on the web of the beam 728 Determine the maximum shear stress acting in the beam at the critical section where the internal shear force is maximum 730 The beam has a rectangular cross section and is subjected to a load P that is just large enough to develop a fully plastic moment at the fixed support If the material is elasticplastic then at a distance the moment creates a region of plastic yielding with an associated elastic core having a height This situation has been described by Eq 630 and the moment M is distributed over the cross section as shown in Fig 648e Prove that the maximum shear stress developed in the beam is given by where the cross sectional area of the elastic core A 2yb tmax 3 21PA2 2y M Px x 6 L Mp PL 731 The beam in Fig 648f is subjected to a fully plastic moment Prove that the longitudinal and transverse shear stresses in the beam are zeroHint Consider an element of the beam as shown in Fig 74c Mp A 150 lbft D 075 in 075 in 4 in 6 in 2 ft 6 ft 6 ft 05 in 200 lbft 4 in Prob 726 A 3 kipft D D C C B 1 in 1 in 6 in 4 in 4 in 6 ft 6 ft 6 ft 075 in 6 in Probs 72728 P d1 a L d2 w A B t1 t1 t2 b h Prob 729 h b L P x Plastic region 2y Elastic region Prob 730 378 CHAPTER 7 TRANSVERSE SHEAR 7 73 Shear Flow in BuiltUp Members Occasionally in engineering practice members are built up from several composite parts in order to achieve a greater resistance to loads Examples are shown in Fig 713 If the loads cause the members to bend fasteners such as nails bolts welding material or glue may be needed to keep the component parts from sliding relative to one another Fig 72 In order to design these fasteners or determine their spacing it is necessary to know the shear force that must be resisted by the fastener This loading when measured as a force per unit length of beam is referred to as shear flow q The magnitude of the shear flow can be obtained using a development similar to that for finding the shear stress in the beamTo show thiswe will consider finding the shear flow along the juncture where the segment in Fig 714a is connected to the flange of the beamAs shown in Fig 714b three horizontal forces must act on this segment Two of these forces F and are developed by normal stresses caused by the moments M and respectively The third force which for equilibrium equals dF acts at the juncture and it is to be supported by the fastener Realizing that dF is the result of dM then like Eq 71 we have The integral represents Q that is the moment of the segments area in Fig 714b about the neutral axis for the entire cross section Since the segment has a length dx the shear flow or force per unit length along the beam is Hence dividing both sides by dx and noting that Eq 62 we can write 74 Here the shear flow measured as a force per unit length along the beam the internal resultant shear force determined from the method of sections and the equations of equilibrium the moment of inertia of the entire crosssectional area computed about the neutral axis where is the crosssectional area of the segment that is connected to the beam at the juncture where the shear flow is to be calculated and is the distance from the neutral axis to the centroid of A y A yA Q I V q q VQ I V dMdx q dFdx A dF dM I LA y dA M dM F dF Fig 713 The use of the word flow in this terminology will become meaningful as it pertains to the discussion in Sec 75 73 SHEAR FLOW IN BUILTUP MEMBERS 381 7 EXAMPLE 76 A box beam is constructed from four boards nailed together as shown in Fig 717a If each nail can support a shear force of 30 lb determine the maximum spacing s of the nails at B and at C so that the beam will support the force of 80 lb SOLUTION Internal Shear If the beam is sectioned at an arbitrary point along its length the internal shear required for equilibrium is always and so the shear diagram is shown in Fig 717b Section Properties The moment of inertia of the crosssectional area about the neutral axis can be determined by considering a square minus a square The shear flow at B is determined using found from the darker shaded area shown in Fig 717c It is this symmetric portion of the beam that is to be held onto the rest of the beam by nails on the left side and by the fibers of the board on the right side Thus Likewise the shear flow at C can be determined using the symmetric shaded area shown in Fig 717dWe have Shear Flow These values represent the shear force per unit length of the beam that must be resisted by the nails at B and the fibers at Fig 717c and the nails at C and the fibers at Fig 717d respectively Since in each case the shear flow is resisted at two surfaces and each nail can resist 30 lb for B the spacing is Ans And for C Ans sC 30 lb 1705922 lbin 850 in Use sC 85 in sB 30 lb 1117622 lbin 510 in Use sB 5 in C B qC VQC I 80 lb12025 in32 2295 in4 7059 lbin qB VQB I 80 lb13375 in32 2295 in4 1176 lbin QC yA 3 in145 in2115 in2 2025 in3 QB yA 3 in175 in2115 in2 3375 in3 QB I 1 12 175 in2175 in23 1 12 145 in2145 in23 2295 in4 45in 45in 75in 75in V 80 lb Fig 717 a 80 lb s 6 in 15 in 6 in 15 in B C 15 in b V lb x ft 80 c 75 in B B A N 3 in 15 in 45 in C A N C d 3 in 15 in 73 SHEAR FLOW IN BUILTUP MEMBERS 383 7 FUNDAMENTAL PROBLEMS F76 The two identical boards are bolted together to form the beam Determine the maximum allowable spacing s of the bolts to the nearest mm if each bolt has a shear strength of 15 kN The beam is subjected to a shear force of F77 The two identical boards are bolted together to form the beam If the spacing of the bolts is mm and each bolt has a shear strength of 15 kN determine the maximum shear force V the beam can resist s 100 V 50 kN F78 Two identical 20mm thick plates are bolted to the top and bottom flange to form the builtup beam If the beam is subjected to a shear force of determine the allowable maximum spacing s of the bolts to the nearest mm Each bolt has a shear strength of 30 kN V 300 kN F710 The boards are bolted together to form the built up beam If the beam is subjected to a shear force of determine the allowable maximum spacing of the bolts to the nearest in Each bolt has a shear strength of 6 kip 1 8 V 15 kip F79 The boards are bolted together to form the built up beam If the beam is subjected to a shear force of determine the allowable maximum spacing of the bolts to the nearest mm Each bolt has a shear strength of 8 kN V 20 kN 100 mm 100 mm 300 mm V s s F767 200 mm 25 mm 50 mm V 25 mm 50 mm 150 mm 150 mm s s F79 4 in 3 in 3 in 1 in 1 in 4 in 1 in 05 in 05 in V s s F710 20 mm 10 mm 10 mm 200 mm 200 mm 10 mm 20 mm 300 mm s s V F78 384 CHAPTER 7 TRANSVERSE SHEAR 7 732 The beam is constructed from two boards fastened together at the top and bottom with two rows of nails spaced every 6 in If each nail can support a 500lb shear force determine the maximum shear force V that can be applied to the beam 733 The beam is constructed from two boards fastened together at the top and bottom with two rows of nails spaced every 6 in If an internal shear force of is applied to the boards determine the shear force resisted by each nail V 600 lb 736 The beam is fabricated from two equivalent structural tees and two plates Each plate has a height of 6 in and a thickness of 05 in If a shear of is applied to the cross section determine the maximum spacing of the bolts Each bolt can resist a shear force of 15 kip 737 The beam is fabricated from two equivalent structural tees and two plates Each plate has a height of 6 in and a thickness of 05 in If the bolts are spaced at determine the maximum shear force V that can be applied to the cross section Each bolt can resist a shear force of 15 kip s 8 in V 50 kip 734 The beam is constructed from two boards fastened together with three rows of nails spaced If each nail can support a 450lb shear force determine the maximum shear force V that can be applied to the beamThe allowable shear stress for the wood is 735 The beam is constructed from two boards fastened together with three rows of nails If the allowable shear stress for the wood is determine the maximum shear force V that can be applied to the beam Also find the maximum spacing s of the nails if each nail can resist 650 lb in shear tallow 150 psi tallow 300 psi s 2 in apart 738 The beam is subjected to a shear of Determine the average shear stress developed in each nail if the nails are spaced 75 mm apart on each side of the beam Each nail has a diameter of 4 mm V 2 kN PROBLEMS V 2 in 6 in 6 in 6 in 2 in Probs 73233 V 15 in s s 6 in 15 in Probs 73435 3 in 3 in A V 05 in 1 in 05 in 6 in s N Probs 73637 75 mm 75 mm 50 mm 25 mm 200 mm 200 mm 25 mm V Prob 738 73 SHEAR FLOW IN BUILTUP MEMBERS 385 7 739 A beam is constructed from three boards bolted together as shown Determine the shear force developed in each bolt if the bolts are spaced apart and the applied shear is V 35 kN s 250 mm 742 The Tbeam is nailed together as shown If the nails can each support a shear force of 950 lb determine the maximum shear force V that the beam can support and the corresponding maximum nail spacing s to the nearest in The allowable shear stress for the wood is tallow 450 psi 1 8 740 The doubleweb girder is constructed from two plywood sheets that are secured to wood members at its top and bottom If each fastener can support 600 lb in single shear determine the required spacing s of the fasteners needed to support the loading Assume A is pinned and B is a roller 741 The doubleweb girder is constructed from two plywood sheets that are secured to wood members at its top and bottom The allowable bending stress for the wood is and the allowable shear stress is If the fasteners are spaced and each fastener can support 600 lb in single shear determine the maximum load P that can be applied to the beam s 6 in tallow 3 ksi sallow 8 ksi P 3000 lb 743 Determine the average shear stress developed in the nails within region AB of the beamThe nails are located on each side of the beam and are spaced 100 mm apart Each nail has a diameter of 4 mmTake 744 The nails are on both sides of the beam and each can resist a shear of 2 kN In addition to the distributed loading determine the maximum load P that can be applied to the end of the beam The nails are spaced 100 mm apart and the allowable shear stress for the wood is tallow 3 MPa P 2 kN s 250 mm 250 mm 100 mm 25 mm 25 mm 25 mm 350 mm V Prob 739 P B s A 2 in 2 in 10 in 6 in 05 in 05 in 2 in 2 in 4 ft 4 ft Probs 74041 12 in 12 in 2 in 2 in V s s Prob 742 P 15 m 15 m B C A 40 mm 20 mm 20 mm 100 mm 200 mm 200 mm 2 kNm Probs 74344 388 CHAPTER 7 TRANSVERSE SHEAR 7 Having determined the direction of the shear flow in each flange we can now find its distribution along the top right flange of the beam in Fig 720a To do this consider the shear flow q acting on the colored element dx located an arbitrary distance x from the centerline of the cross section Fig 720b Here so that 75 By inspection this distribution varies in a linear manner from at to at The limitation of is possible here since the member is assumed to have thin walls and so the thickness of the web is neglected Due to symmetry a similar analysis yields the same distribution of shear flow for the other flange segments so that the results are as shown in Fig 720d The total force developed in each flange segment can be determined by integration Since the force on the element dx in Fig 720b is then We can also determine this result by finding the area under the triangle in Fig 720d Hence All four of these forces are shown in Fig 720e and we can see from their direction that horizontal force equilibrium of the cross section is maintained Ff 1 2 1qmax2fa b 2 b Vt db2 16I Ff L q dx L b2 0 Vt d 2I a b 2 xb dx Vt db2 16I dF q dx x 0 x 0 1qmax2f Vt db4I x b2 q 0 q VQ I Vd21b2 x2t I Vt d 2I a b 2 xb Q yA d21b2 x2t b t t A V N t a d 2 d 2 b N A x dx t q d 2 b 2 c N A b t t t y q dy d 2 Fig 720 390 CHAPTER 7 TRANSVERSE SHEAR 7 From the foregoing analysis three important points should be observed Firstthe value of q changes over the cross sectionsince Q will be different for each area segment for which it is determined In particular q will vary linearly along segments flanges that are perpendicular to the direction of V and parabolically along segments web that are inclined or parallel to V Second q will always act parallel to the walls of the member since the section on which q is calculated is taken perpendicular to the walls And third the directional sense of q is such that the shear appears to flow through the cross section inward at the beams top flange combining and then flowing downward through the web since it must contribute to the shear force V and then separating and flowing outward at the bottom flange If one is able to visualize this flow it will provide an easy means for establishing not only the direction of q but also the corresponding direction of Other examples of how q is directed along the segments of thinwalled members are shown in Fig 721 In all cases symmetry prevails about an axis that is collinear with VAs a result qflowsin a direction such that it will provide the vertical force V and yet also satisfy horizontal force equilibrium for the cross section t A Fig 721 Shear flow q V V V V Important Points The shear flow formula can be used to determine the distribution of the shear flow throughout a thinwalled member provided the shear V acts along an axis of symmetry or principal centroidal axis of inertia for the cross section If a member is made from segments having thin walls only the shear flow parallel to the walls of the member is important The shear flow varies linearly along segments that are perpendicular to the direction of the shear V The shear flow varies parabolically along segments that are inclined or parallel to the direction of the shear V On the cross section the shear flows along the segments so that it results in the vertical shear force V and yet satisfies horizontal force equilibrium q VQI 74 SHEAR FLOW IN THINWALLED MEMBERS 391 7 EXAMPLE 78 The thinwalled box beam in Fig 722a is subjected to a shear of 10 kip Determine the variation of the shear flow throughout the cross section SOLUTION By symmetry the neutral axis passes through the center of the cross section For thinwalled members we use centerline dimensions for calculating the moment of inertia Only the shear flow at points B C and D has to be determined For point B the area Fig 722b since it can be thought of as being located entirely at point B Alternatively can also represent the entire crosssectional area in which case since Because then For point C the area is shown dark shaded in Fig 722c Here we have used the mean dimensions since point C is on the centerline of each segmentWe have Since there are two points of attachment The shear flow at D is determined using the three darkshaded rectangles shown in Fig 722d Again using centerline dimensions qC 1 2 a VQC I b 1 2 a 10 kip1175 in32 1797 in4 b 0487 kipin QC yA 135 in215 in211 in2 175 in3 A qB 0 QB 0 y 0 QB yA 0 A A L 0 I 1 12 12 in217 in23 2 15 in211 in2135 in22 1797 in4 QD yA 2c35 in 2 d11 in2135 in2 35 in15 in211 in2 2975 in3 Because there are two points of attachment Using these results and the symmetry of the cross section the shearflow distribution is plotted in Fig 722e The distribution is linear along the horizontal segments perpendicular to V and parabolic along the vertical segments parallel to V qD 1 2 a VQD I b 1 2 a 10 kip12975 in32 1797 in4 b 0828 kipin Fig 722 a A N C D 3 in 3 in 1 in 1 in 1 in2 in2 in 1 in 10 kip B b A A N 35 in 4 in 5 in N A 1 in 1 in 4 in c 5 in 35 in N A 35 in N A e 0828 kipin 0487 kipin 0487 kipin d 394 CHAPTER 7 TRANSVERSE SHEAR 7 Important Points The shear center is the point through which a force can be applied which will cause a beam to bend and yet not twist The shear center will always lie on an axis of symmetry of the cross section The location of the shear center is only a function of the geometry of the cross section and does not depend upon the applied loading Procedure for Analysis The location of the shear center for an open thinwalled member for which the internal shear is in the same direction as a principal centroidal axis for the cross section may be determined by using the following procedure ShearFlow Resultants By observation determine the direction of the shear flow through the various segments of the cross section and sketch the force resultants on each segment of the cross section For example see Fig 723c Since the shear center is determined by taking the moments of these force resultants about a point A choose this point at a location that eliminates the moments of as many force resultants as possible The magnitudes of the force resultants that create a moment about A must be calculated For any segment this is done by determining the shear flow q at an arbitrary point on the segment and then integrating q along the segments length Realize that V will create a linear variation of shear flow in segments that are perpendicular to V and a parabolic variation of shear flow in segments that are parallel or inclined to V Shear Center Sum the moments of the shearflow resultants about point A and set this moment equal to the moment of V about A Solve this equation to determine the momentarm or eccentric distance e which locates the line of action of V from A If an axis of symmetry for the cross section exists the shear center lies at the point where this axis intersects the line of action of V 398 CHAPTER 7 TRANSVERSE SHEAR 7 100 mm 90 mm 90 mm 200 mm 200 mm 180 mm 190 mm 10 mm 10 mm V A D C B Probs 75051 752 A shear force of is applied to the symmetric box girder Determine the shear flow at A and B 753 A shear force of is applied to the box girder Determine the shear flow at C V 18 kN V 18 kN C A 150 mm 10 mm 10 mm 100 mm 100 mm 10 mm 10 mm 125 mm 150 mm 10 mm 30 mm B V 10 mm Probs 75253 A V 05 in 05 in 5 in 5 in 05 in 2 in 05 in 8 in B A C D Probs 75657 30 mm 40 mm 30 mm V A B 40 mm 10 mm 10 mm 10 mm 10 mm Probs 75455 756 The beam is subjected to a shear force of Determine the shear flow at points A and B 757 The beam is constructed from four plates and is subjected to a shear force of Determine the maximum shear flow in the cross section V 5 kip V 5 kip 750 A shear force of is applied to the box girder Determine the shear flow at points A and B 751 A shear force of is applied to the box girder Determine the shear flow at points C and D V 450 kN V 300 kN 754 The aluminum strut is 10 mm thick and has the cross section shown If it is subjected to a shear of determine the shear flow at points A and B 755 The aluminum strut is 10 mm thick and has the cross section shown If it is subjected to a shear of determine the maximum shear flow in the strut V 150 N V 150 N PROBLEMS 402 CHAPTER 7 TRANSVERSE SHEAR 7 If the beam is made from thinwalled segments then the shearflow distribution along each segment can be determined This distribution will vary linearly along horizontal segments and parabolically along inclined or vertical segments Provided the shearflow distribution in each element of an open thinwalled section is known then using a balance of moments the location O of the shear center for the cross section can be determined When a load is applied to the member through this pointthe member will bend and not twist Shearflow distribution qmaxf qmaxf qmaxw 2qmaxf 2qmaxf P O e The offset hanger supporting this ski gondola is subjected to the combined loadings of axial force and bending moment 8 405 CHAPTER OBJECTIVES This chapter serves as a review of the stress analysis that has been developed in the previous chapters regarding axial load torsion bending and shear We will discuss the solution of problems where several of these internal loads occur simultaneously on a members cross section Before doing this however the chapter begins with an analysis of stress developed in thinwalled pressure vessels 81 ThinWalled Pressure Vessels Cylindrical or spherical vessels are commonly used in industry to serve as boilers or tanks When under pressure the material of which they are made is subjected to a loading from all directions Although this is the case the vessel can be analyzed in a simple manner provided it has a thin wall In generalthin wall refers to a vessel having an innerradius towallthickness ratio of 10 or more Specifically when the results of a thinwall analysis will predict a stress that is approximately 4 less than the actual maximum stress in the vessel For larger ratios this error will be even smaller Provided the vessel wall is thin the stress distribution throughout its thickness will not vary significantly and so we will assume that it is uniform or constant Using this assumption we will now analyze the state of stress in thinwalled cylindrical and spherical pressure vessels In both cases the pressure in the vessel is understood to be the gauge pressure that is it measures the pressure above atmospheric pressure since atmospheric pressure is assumed to exist both inside and outside the vessels wall before the vessel is pressurized rt rt 10 1rt Ú 102 Combined Loadings Cylindrical pressure vessels such as this gas tank have semispherical end caps rather than flat ones in order to reduce the stress in the tank 406 CHAPTER 8 COMBINED LOADINGS 8 Cylindrical Vessels Consider the cylindrical vessel in Fig 81a having a wall thickness t inner radius r and subjected to a gauge pressure p that developed within the vessel by a contained gas Due to this loading a small element of the vessel that is sufficiently removed from the ends and oriented as shown in Fig 81a is subjected to normal stresses in the circumferential or hoop direction and in the longitudinal or axial direction The hoop stress can be determined by considering the vessel to be sectioned by planes a b and c A freebody diagram of the back segment along with the contained gas is shown in Fig 81b Here only the loadings in the x direction are shown These loadings are developed by the uniform hoop stress acting on the vessels wall and the pressure acting on the vertical face of the gas For equilibrium in the x direction we require 81 The longitudinal stress can be determined by considering the left portion of section b of the cylinder Fig 81a As shown in Fig 81c acts uniformly throughout the wall and p acts on the section of the contained gas Since the mean radius is approximately equal to the vessels inner radius equilibrium in the y direction requires 82 In the above equations the normal stress in the hoop and longitudinal directions respectively Each is assumed to be constant throughout the wall of the cylinder and each subjects the material to tension the internal gauge pressure developed by the contained gas the inner radius of the cylinder t the thickness of the wall 1rt Ú 102 r p s1 s2 s2 pr 2t s212prt2 p1pr22 0 Fy 0 s2 s1 pr t 2s11t dy2 p12r dy2 0 Fx 0 s1 s2 s1 a z y b a c x t r s1 s2 t dy 2r t p b s1 s1 t c p r s2 Fig 81 81 THINWALLED PRESSURE VESSELS 407 8 By comparison note that the hoop or circumferential stress is twice as large as the longitudinal or axial stress Consequently when fabricating cylindrical pressure vessels from rolledformed plates the longitudinal joints must be designed to carry twice as much stress as the circumferential joints Spherical Vessels We can analyze a spherical pressure vessel in a similar manner To do this consider the vessel to have a wall thickness t inner radius r and subjected to an internal gauge pressure p Fig 82a If the vessel is sectioned in half the resulting freebody diagram is shown in Fig 82b Like the cylinder equilibrium in the y direction requires 83 This is the same result as that obtained for the longitudinal stress in the cylindrical pressure vessel Furthermore from the analysis this stress will be the same regardless of the orientation of the hemispheric freebody diagram Consequently a small element of the material is subjected to the state of stress shown in Fig 82a The above analysis indicates that an element of material taken from either a cylindrical or a spherical pressure vessel is subjected to biaxial stress ie normal stress existing in only two directions Actually the pressure also subjects the material to a radial stress which acts along a radial line This stress has a maximum value equal to the pressure p at the interior wall and it decreases through the wall to zero at the exterior surface of the vessel since the gauge pressure there is zero For thin walled vessels however we will ignore this radialstress component since our limiting assumption of results in and being respectively 5 and 10 times higher than the maximum radial stress Finally if the vessel is subjected to an external pressure the compressive stress developed within the thin wall may cause the vessel to become unstable and collapse may occur by buckling rather than causing the material to fracture 1s32max p s1 s2 rt 10 s3 s2 pr 2t s22prt p1pr22 0 Fy 0 Shown is the barrel of a shotgun which was clogged with debris just before firing Gas pressure from the charge increased the circumferential stress within the barrel enough to cause the rupture t a r y x z a s2 s2 t p b r s2 Fig 82 408 CHAPTER 8 COMBINED LOADINGS 8 EXAMPLE 81 A cylindrical pressure vessel has an inner diameter of 4 ft and a thickness of Determine the maximum internal pressure it can sustain so that neither its circumferential nor its longitudinal stress component exceeds 20 ksi Under the same conditions what is the maximum internal pressure that a similarsize spherical vessel can sustain SOLUTION Cylindrical Pressure Vessel The maximum stress occurs in the circumferential direction From Eq 81 we have Ans Note that when this pressure is reached from Eq 82 the stress in the longitudinal direction will be Furthermore the maximum stress in the radial direction occurs on the material at the inner wall of the vessel and is This value is 48 times smaller than the circumferential stress 20 ksi and as stated earlier its effects will be neglected Spherical Vessel Here the maximum stress occurs in any two perpendicular directions on an element of the vessel Fig 82a From Eq 83 we have Ans NOTE Although it is more difficult to fabricate the spherical pressure vessel will carry twice as much internal pressure as a cylindrical vessel p 833 psi 20 kipin2 p124 in2 2A1 2 inB s2 pr 2t 1s32max p 417 psi s2 1 2120 ksi2 10 ksi p 417 psi 20 kipin2 p124 in2 1 2 in s1 pr t 1 2 in 81 THINWALLED PRESSURE VESSELS 409 8 81 A spherical gas tank has an inner radius of If it is subjected to an internal pressure of determine its required thickness if the maximum normal stress is not to exceed 12 MPa 82 A pressurized spherical tank is to be made of 05inthick steel If it is subjected to an internal pressure of determine its outer radius if the maximum normal stress is not to exceed 15 ksi 83 The thinwalled cylinder can be supported in one of two ways as shown Determine the state of stress in the wall of the cylinder for both cases if the piston P causes the internal pressure to be 65 psi The wall has a thickness of 025 in and the inner diameter of the cylinder is 8 in p 200 psi p 300 kPa r 15 m 85 Thesphericalgastankisfabricatedbyboltingtogether two hemispherical thin shells of thickness 30 mm If the gas contained in the tank is under a gauge pressure of 2 MPa determine the normal stress developed in the wall of the tank and in each of the boltsThe tank has an inner diameter of 8 m and is sealed with 900 bolts each 25 mm in diameter 86 The spherical gas tank is fabricated by bolting together two hemispherical thin shells If the 8m inner diameter tank is to be designed to withstand a gauge pressure of 2 MPa determine the minimum wall thickness of the tank and the minimum number of 25mm diameter bolts that must be used to seal itThe tank and the bolts are made from material having an allowable normal stress of 150 MPa and 250 MPa respectively PROBLEMS P a b P 8 in 8 in Prob 83 84 The tank of the air compressor is subjected to an internal pressure of 90 psi If the internal diameter of the tank is 22 in and the wall thickness is 025 in determine the stress components acting at point A Draw a volume element of the material at this point and show the results on the element A Prob 84 Probs 856 87 A boiler is constructed of 8mm thick steel plates that are fastened together at their ends using a butt joint consisting of two 8mm cover plates and rivets having a diameter of 10 mm and spaced 50 mm apart as shown If the steam pressure in the boiler is 135 MPa determine a the circumferential stress in the boilers plate apart from the seamb the circumferential stress in the outer cover plate along the rivet line aaand c the shear stress in the rivets a 8 mm 50 mm a 075 m Prob 87 s s s 4 psi 4 psi 410 CHAPTER 8 COMBINED LOADINGS 8 88 The gas storage tank is fabricated by bolting together two half cylindrical thin shells and two hemispherical shells as shown If the tank is designed to withstand a pressure of 3 MPa determine the required minimum thickness of the cylindrical and hemispherical shells and the minimum required number of longitudinal bolts per meter length at each side of the cylindrical shell The tank and the 25 mm diameter bolts are made from material having an allowable normal stress of 150 MPa and 250 MPa respectively The tank has an inner diameter of 4 m 89 The gas storage tank is fabricated by bolting together two half cylindrical thin shells and two hemispherical shells as shown If the tank is designed to withstand a pressure of 3 MPa determine the required minimum thickness of the cylindrical and hemispherical shells and the minimum required number of bolts for each hemispherical cap The tank and the 25 mm diameter bolts are made from material having an allowable normal stress of 150 MPa and 250 MPa respectivelyThe tank has an inner diameter of 4 m 811 The staves or vertical members of the wooden tank are held together using semicircular hoops having a thickness of 05 in and a width of 2 in Determine the normal stress in hoop AB if the tank is subjected to an internal gauge pressure of 2 psi and this loading is transmitted directly to the hoops Also if 025indiameter bolts are used to connect each hoop together determine the tensile stress in each bolt at A and B Assume hoop AB supports the pressure loading within a 12in length of the tank as shown Probs 889 Prob 810 810 A wood pipe having an inner diameter of 3 ft is bound together using steel hoops each having a cross sectional area of If the allowable stress for the hoops is determine their maximum spacing s along the section of pipe so that the pipe can resist an internal gauge pressure of 4 psi Assume each hoop supports the pressure loading acting along the length s of the pipe sallow 12 ksi 02 in2 6 in 12 in 18 in 12 in 6 in A B Prob 811 812 Two hemispheres having an inner radius of 2 ft and wall thickness of 025 in are fitted together and the inside gauge pressure is reduced to psi If the coefficient of static friction is between the hemispheres determine a the torque T needed to initiate the rotation of the top hemisphere relative to the bottom one b the vertical force needed to pull the top hemisphere off the bottom one and c the horizontal force needed to slide the top hemisphere off the bottom one ms 05 10 2 ft 025 in Prob 812 412 CHAPTER 8 COMBINED LOADINGS 8 82 State of Stress Caused by Combined Loadings In previous chapters we developed methods for determining the stress distributions in a member subjected to either an internal axial force a shear force a bending moment or a torsional moment Most often however the cross section of a member is subjected to several of these loadings simultaneously When this occurs the method of superposition can be used to determine the resultant stress distribution Recall from Sec 43 that the principle of superposition can be used for this purpose provided a linear relationship exists between the stress and the loads Also the geometry of the member should not undergo significant change when the loads are applied These conditions are necessary in order to ensure that the stress produced by one load is not related to the stress produced by any other load This chimney is subjected to the combined loading of wind and weight It is important to investigate the tensile stress in the chimney since masonry is weak in tension Procedure for Analysis The following procedure provides a general means for establishing the normal and shear stress components at a point in a member when the member is subjected to several different types of loadings simultaneously It is assumed that the material is homogeneous and behaves in a linear elastic manner Also SaintVenants principle requires that the point where the stress is to be determined is far removed from any discontinuities in the cross section or points of applied load Internal Loading Section the member perpendicular to its axis at the point where the stress is to be determined and obtain the resultant internal normal and shear force components and the bending and torsional moment components The force components should act through the centroid of the cross section and the moment components should be computed about centroidal axes which represent the principal axes of inertia for the cross section Stress Components Determine the stress component associated with each internal loading For each case represent the effect either as a distribution of stress acting over the entire crosssectional area or show the stress on an element of the material located at a specified point on the cross section 82 STATE OF STRESS CAUSED BY COMBINED LOADINGS 413 8 Problems in this section which involve combined loadings serve as a basic review of the application of the stress equations mentioned above A thorough understanding of how these equations are applied as indicated in the previous chapters is necessary if one is to successfully solve the problems at the end of this section The following examples should be carefully studied before proceeding to solve the problems Normal Force The internal normal force is developed by a uniform normalstress distribution determined from Shear Force The internal shear force in a member is developed by a shearstress distribution determined from the shear formula Special care however must be exercised when applying this equation as noted in Sec 72 Bending Moment For straight members the internal bending moment is developed by a normalstress distribution that varies linearly from zero at the neutral axis to a maximum at the outer boundary of the member This stress distribution is determined from the flexure formula If the member is curved the stress distribution is nonlinear and is determined from Torsional Moment For circular shafts and tubes the internal torsional moment is developed by a shearstress distribution that varies linearly from the central axis of the shaft to a maximum at the shafts outer boundary This stress distribution is determined from the torsional formula ThinWalled Pressure Vessels If the vessel is a thinwalled cylinder the internal pressure p will cause a biaxial state of stress in the material such that the hoop or circumferential stress component is and the longitudinal stress component is If the vessel is a thin walled sphere then the biaxial state of stress is represented by two equivalent components each having a magnitude of Superposition Once the normal and shear stress components for each loading have been calculated use the principle of superposition and determine the resultant normal and shear stress components Represent the results on an element of material located at the point or show the results as a distribution of stress acting over the members crosssectional area s2 pr2t s2 pr2t s1 prt t TrJ s MyAe1R y2 s MyI t VQIt s PA 416 CHAPTER 8 COMBINED LOADINGS 8 EXAMPLE 84 The member shown in Fig 85a has a rectangular cross section Determine the state of stress that the loading produces at point C a 15 m 250 mm 50 mm 15 m 2 m 4 m 125 mm 25 m C C A B 50 kNm c 1645 kN 2193 kN 15 m C M N V 125 kN 9759 kN 3 4 5 1645 kN 2193 kN 3 4 5 4 m 125 m 125 m b Fig 85 SOLUTION Internal Loadings The support reactions on the member have been determined and are shown in Fig 85b If the left segment AC of the member is considered Fig 85c the resultant internal loadings at the section consist of a normal force a shear force and a bending moment Solving N 1645 kN V 2193 kN M 3289 kN m 422 CHAPTER 8 COMBINED LOADINGS 8 FUNDAMENTAL PROBLEMS F81 Determine the normal stress developed at corners A and B of the column F83 Determine the state of stress at point A on the cross section of the beam at section aa 300 kN 500 kN 50 mm 100 mm 100 mm 100 mm 150 mm 150 mm 150 mm 150 mm A z y x B F81 A 2 m 05 m 05 m 05 m 30 kN a a 50 mm 10 mm 10 mm 10 mm 180 mm 100 mm Section aa F83 P P 2 in 2 in 05 in a a F84 300 mm 100 mm 100 mm 05 m Section aa 400 kN a a A F82 F82 Determine the state of stress at point A on the cross section at section aa of the cantilever beam F84 Determine the magnitude of the load P that will cause a maximum normal stress of in the link along section aa s max 30 ksi 82 STATE OF STRESS CAUSED BY COMBINED LOADINGS 423 8 F85 The beam has a rectangular cross section and is subjected to the loading shown Determine the components of stress and txy at point B sy sx F86 Determine the state of stress at point A on the cross section of the pipe assembly at section aa F87 Determine the state of stress at point A on the cross section of the pipe at section aa F88 Determine the state of stress at point A on the cross section of the shaft at section aa 15 in 15 in 10 in 400 lb x y z 1 in B 500 lb 2 in 2 in F85 a A A a 300 mm 300 mm Section a a 50 mm 40 mm 6 kN z y x F87 a a 400 mm 200 mm A A Section a a 20 mm 1500 N 1000 N z y x F86 100 mm a 300 N 300 N 900 N 900 N 100 mm 600 mm 400 mm 300 mm 100 mm A Section a a 25 mm 20 mm A z y x a F88 424 CHAPTER 8 COMBINED LOADINGS 8 818 The vertical force P acts on the bottom of the plate having a negligible weight Determine the shortest distance d to the edge of the plate at which it can be applied so that it produces no compressive stresses on the plate at section aa The plate has a thickness of 10 mm and P acts along the center line of this thickness 821 The coping saw has an adjustable blade that is tightened with a tension of 40 N Determine the state of stress in the frame at points A and B PROBLEMS a 500 mm P a 300 mm 200 mm d Prob 818 819 Determine the maximum and minimum normal stress in the bracket at section aa when the load is applied at 820 Determine the maximum and minimum normal stress in the bracket at section aa when the load is applied at x 300 mm x 0 822 The clamp is made from members AB and AC which are pin connected at A If it exerts a compressive force at C and B of 180 N determine the maximum compressive stress in the clamp at section aaThe screw EF is subjected only to a tensile force along its axis 823 The clamp is made from members AB and AC which are pin connected at A If it exerts a compressive force at C and B of 180 N sketch the stress distribution acting over section aa The screw EF is subjected only to a tensile force along its axis 100 kN x 200 mm 150 mm 15 mm 15 mm a a Probs 81920 180 N 180 N B C F E A a a 30 mm 40 mm 15 mm 15 mm Sectiona a Probs 82223 75 mm 50 mm 8 mm 3 mm 3 mm 8 mm A B 100 mm Prob 821 82 STATE OF STRESS CAUSED BY COMBINED LOADINGS 427 8 838 Since concrete can support little or no tension this problem can be avoided by using wires or rods to prestress the concrete once it is formed Consider the simply supported beam shown which has a rectangular cross section of 18 in by 12 in If concrete has a specific weight of determine the required tension in rod AB which runs through the beam so that no tensile stress is developed in the concrete at its center section aa Neglect the size of the rod and any deflection of the beam 839 Solve Prob 838 if the rod has a diameter of 05 in Use the transformed area method discussed in Sec 66 Ec 36011032 ksi Est 2911032 ksi 150 lbft3 840 Determine the state of stress at point A when the beam is subjected to the cable force of 4 kN Indicate the result as a differential volume element 841 Determine the state of stress at point B when the beam is subjected to the cable force of 4 kN Indicate the result as a differential volume element 842 The bar has a diameter of 80 mm Determine the stress components that act at point A and show the results on a volume element located at this point 843 The bar has a diameter of 80 mm Determine the stress components that act at point B and show the results on a volume element located at this point 844 Determine the normal stress developed at points A and B Neglect the weight of the block 845 Sketch the normal stress distribution acting over the cross section at section aa Neglect the weight of the block 16 in 4 ft 4 ft a a A B 18 in 6 in 6 in 2 in Probs 83839 2 m 075 m 1 m 4 kN G 250 mm 375 mm B C D A 200 mm 20 mm 20 mm 150 mm 15 mm A B 100 mm Probs 84041 a a 6 in 6 kip 12 kip 3 in A B Probs 84445 300 mm 200 mm B A 5 kN 4 3 5 Probs 84243 846 The support is subjected to the compressive load P Determine the absolute maximum and minimum normal stress acting in the material P a2 a2 a2 a2 Prob 846 82 STATE OF STRESS CAUSED BY COMBINED LOADINGS 429 8 853 The masonry pier is subjected to the 800kN load Determine the equation of the line along which the load can be placed without causing a tensile stress in the pier Neglect the weight of the pier 854 The masonry pier is subjected to the 800kN load If and determine the normal stress at each corner A B C D not shown and plot the stress distribution over the cross section Neglect the weight of the pier y 05 m x 025 m y f1x2 852 The hook is used to lift the force of 600 lb Determine the maximum tensile and compressive stresses at section aa The cross section is circular and has a diameter of 1 in Use the curvedbeam formula to compute the bending stress 855 The bar has a diameter of 40 mm If it is subjected to the two force components at its end as shown determine the state of stress at point A and show the results on a differential volume element located at this point 856 Solve Prob 855 for point B a a 25 in 15 in 300 lb 300 lb 600 lb Prob 852 100 mm 150 mm y x z B A 300 N 500 N Probs 85556 B A z y x 500 lb 12 in 8 in 800 lb 600 lb Probs 85758 225 m 225 m 15 m 15 m A B C y x x y 800 kN Probs 85354 857 The 2indiameter rod is subjected to the loads shown Determine the state of stress at point A and show the results on a differential element located at this point 858 The 2indiameter rod is subjected to the loads shown Determine the state of stress at point B and show the results on a differential element located at this point 430 CHAPTER 8 COMBINED LOADINGS 8 859 If determine the maximum normal stress developed on the cross section of the column 860 Determine the maximum allowable force P if the column is made from material having an allowable normal stress of sallow 100 MPa P 60 kN 863 The uniform sign has a weight of 1500 lb and is supported by the pipe AB which has an inner radius of 275 in and an outer radius of 300 in If the face of the sign is subjected to a uniform wind pressure of determine the state of stress at points C and D Show the results on a differential volume element located at each of these points Neglect the thickness of the sign and assume that it is supported along the outside edge of the pipe 864 Solve Prob 863 for points E and F p 150 lbft2 100 mm 15 mm 15 mm 15 mm 75 mm 150 mm 150 mm 100 mm 100 mm P 2P Probs 85960 861 The beveled gear is subjected to the loads shown Determine the stress components acting on the shaft at point A and show the results on a volume element located at this pointThe shaft has a diameter of 1 in and is fixed to the wall at C 862 The beveled gear is subjected to the loads shown Determine the stress components acting on the shaft at point B and show the results on a volume element located at this pointThe shaft has a diameter of 1 in and is fixed to the wall at C C B x z y A 200 lb 125 lb 75 lb 8 in 3 in Probs 86162 3 ft 6 ft 12 ft B A y x z 150 lbft2 C D F E Probs 86364 865 Determine the state of stress at point A on the cross section of the pipe at section aa 866 Determine the state of stress at point B on the cross section of the pipe at section aa 50 lb B A 12 in 10 in a a 60 075 in 1 in Section aa z x y Probs 86566 CONCEPTUAL PROBLEMS 433 8 CONCEPTUAL PROBLEMS B A P81 P81 Explain why failure of this garden hose occurred near its end and why the tear occurred along its length Use numerical values to explain your result Assume the water pressure is 30 psi P83 P83 Unlike the turnbuckle at B which is connected along the axis of the rodthe one at A has been welded to the edges of the rod and so it will be subjected to additional stress Use the same numerical values for the tensile load in each rod and the rods diameter and compare the stress in each rod P84 P84 A constant wind blowing against the side of this chimney has caused creeping strains in the mortar jointssuch that the chimney has a noticeable deformation Explain how to obtain the stress distribution over a section at the base of the chimney and sketch this distribution over the section P82 P82 This openended silo contains granular material It is constructed from wood slats and held together with steel bands Explain using numerical values why the bands are not spaced evenly along the height of the cylinder Also how would you find this spacing if each band is to be subjected to the same stress REVIEW PROBLEMS 435 8 880 The hydraulic cylinder is required to support a force of If the cylinder has an inner diameter of 100 mm and is made from a material having an allowable normal stress of determine the required minimum thickness t of the wall of the cylinder 881 The hydraulic cylinder has an inner diameter of 100 mm and wall thickness of If it is made from a material having an allowable normal stress of 150 MPa determine the maximum allowable force P sallow t 4 mm sallow 150 MPa P 100 kN 882 The screw of the clamp exerts a compressive force of 500 lb on the wood blocks Determine the maximum normal stress developed along section The cross section there is rectangular 075 in by 050 in aa 883 Air pressure in the cylinder is increased by exerting forces on the two pistons each having a radius of 45 mm If the cylinder has a wall thickness of 2 mm determine the state of stress in the wall of the cylinder 884 Determine the maximum force P that can be exerted on each of the two pistons so that the circumferential stress component in the cylinder does not exceed 3 MPa Each piston has a radius of 45 mm and the cylinder has a wall thickness of 2 mm P 2 kN 885 The cap on the cylindrical tank is bolted to the tank along the flanges The tank has an inner diameter of 15 m and a wall thickness of 18 mm If the largest normal stress is not to exceed 150 MPa determine the maximum pressure the tank can sustain Also compute the number of bolts required to attach the cap to the tank if each bolt has a diameter of 20 mm The allowable stress for the bolts is 886 The cap on the cylindrical tank is bolted to the tank along the flanges The tank has an inner diameter of 15 m and a wall thickness of 18 mm If the pressure in the tank is determine the force in each of the 16 bolts that are used to attach the cap to the tank Also specify the state of stress in the wall of the tank p 120 MPa 1sallow2b 180 MPa t 100 mm P Probs 88081 4 in 075 in a a Prob 882 47 mm P P Probs 88384 Probs 88586 These turbine blades are subjected to a complex pattern of stress For design it is necessary to determine where and in what direction the maximum stress occurs 437 CHAPTER OBJECTIVES In this chapter we will show how to transform the stress components that are associated with a particular coordinate system into components associated with a coordinate system having a different orientation Once the necessary transformation equations are established we will then be able to obtain the maximum normal and maximum shear stress at a point and find the orientation of elements upon which they act Planestress transformation will be discussed in the first part of the chapter since this condition is most common in engineering practice At the end of the chapter we will discuss a method for finding the absolute maximum shear stress at a point when the material is subjected to both plane and threedimensional states of stress 91 PlaneStress Transformation It was shown in Sec 13 that the general state of stress at a point is characterized by six independent normal and shear stress components which act on the faces of an element of material located at the point Fig 91a This state of stress however is not often encountered in engineering practice Instead engineers frequently make approximations or simplifications of the loadings on a body in order that the stress produced in a structural member or mechanical element can be analyzed in a single planeWhen this is the casethe material is said to be subjected to plane stress Fig 91b For example if there is no load on the surface of a bodythen the normal and shear stress components will be zero on the face of an element that lies on this surface Consequently the corresponding stress components on the opposite face will also be zero and so the material at the point will be subjected to plane stress This case occurred throughout the previous chapter Stress Transformation 9 458 CHAPTER 9 STRESS TRANSFORMATION 9 933 The clamp bears down on the smooth surface at E by tightening the bolt If the tensile force in the bolt is 40 kN determine the principal stress at points A and B and show the results on elements located at each of these points The crosssectional area at A and B is shown in the adjacent figure 935 The square steel plate has a thickness of 10 mm and is subjected to the edge loading shown Determine the maximum inplane shear stress and the average normal stress developed in the steel 934 Determine the principal stress and the maximum inplane shear stress that are developed at point A in the 2indiameter shaft Show the results on an element located at this point The bearings only support vertical reactions 100 mm 50 mm A E B B A 50 mm 30 mm 25 mm 100 mm 300 mm Prob 933 A 24 in 12 in 12 in 300 lb 3000 lb 3000 lb Prob 934 200 mm 200 mm 50 Nm 50 Nm 936 The square steel plate has a thickness of 05 in and is subjected to the edge loading shown Determine the principal stresses developed in the steel 937 The shaft has a diameter d and is subjected to the loadings shown Determine the principal stress and the maximum inplane shear stress that is developed at point A The bearings only support vertical reactions 4 in 4 in 16 lbin 16 lbin Prob 936 A F F P L 2 L 2 Prob 937 Prob 935 94 MOHRS CIRCLEPLANE STRESS 463 Procedure for Analysis The following steps are required to draw and use Mohrs circle Construction of the Circle Establish a coordinate system such that the horizontal axis represents the normal stress with positive to the right and the vertical axis represents the shear stress with positive downwards Fig 917a Using the positive sign convention for as shown in Fig 917b plot the center of the circle C which is located on the axis at a distance from the origin Fig 917a Plot the reference point A having coordinates This point represents the normal and shear stress components on the elements righthand vertical face and since the axis coincides with the x axis this represents Fig 917a Connect point A with the center C of the circle and determine CA by trigonometry This distance represents the radius R of the circle Fig 917a Once R has been determined sketch the circle Principal Stress The principal stresses and are the coordinates of points B and D where the circle intersects the axis ie where Fig 917a These stresses act on planes defined by angles and Fig 917c They are represented on the circle by angles shown and not shown and are measured from the radial reference line CA to lines CB and CD respectively Using trigonometry only one of these angles needs to be calculated from the circle since and are 90 apart Remember that the direction of rotation on the circle here it happens to be counterclockwise represents the same direction of rotation from the reference axis to the principal plane Fig 917c Maximum InPlane Shear Stress The average normal stress and maximum inplane shear stress components are determined from the circle as the coordinates of either point E or F Fig 917a In this case the angles and give the orientation of the planes that contain these components Fig 917dThe angle is shown in Fig 917a and can be determined using trigonometry Here the rotation happens to be clockwise from CA to CE and so must be clockwise on the element Fig 917d us1 2us1 us2 us1 1x2 1x2 up 2up up2 up1 2up2 2up1 up2 up1 t 0 s s2 1s1 Ú s22 s1 u 0 x A1sx txy2 savg 1sx sy22 s sy txy sx t s 94 MOHRS CIRCLEPLANE STRESS 469 9 951 Solve Prob 94 using Mohrs circle 952 Solve Prob 96 using Mohrs circle 953 Solve Prob 914 using Mohrs circle 954 Solve Prob 916 using Mohrs circle 955 Solve Prob 912 using Mohrs circle 956 Solve Prob 911 using Mohrs circle 957 Mohrs circle for the state of stress in Fig 915a is shown in Fig 915b Show that finding the coordinates of point on the circle gives the same value as the stresstransformation Eqs 91 and 92 958 Determine the equivalent state of stress if an element is oriented 25 counterclockwise from the element shown P1sx txy2 960 Determine the equivalent state of stress if an element is oriented 30 clockwise from the element shown Show the result on the element 961 Determine the equivalent state of stress for an element oriented 60 counterclockwise from the element shown Show the result on the element PROBLEMS 550 MPa Prob 958 959 Determine the equivalent state of stress if an element is oriented 20 clockwise from the element shown 2 ksi 3 ksi 4 ksi Prob 959 9 ksi 4 ksi 6 ksi Prob 960 250 MPa 400 MPa 560 MPa Prob 961 9 962 Determine the equivalent state of stress for an element oriented 30 clockwise from the element shown Show the result on the element 963 Determine the principal stress the maximum inplane shear stressand average normal stressSpecify the orientation of the element in each case 964 Determine the principal stress the maximum inplane shear stress and average normal stress Specify the orientation of the element in each case 965 Determine the principal stress the maximum in plane shear stress and average normal stress Specify the orientation of the element in each case 966 Determine the principal stress the maximum inplane shear stressand average normal stressSpecify the orientation of the element in each case 967 Determine the principal stress the maximum inplane shear stressand average normal stressSpecify the orientation of the element in each case 968 Draw Mohrs circle that describes each of the following states of stress 2 ksi 5 ksi Prob 962 5 ksi 15 ksi Prob 963 20 MPa 30 MPa 80 MPa Prob 964 300 psi 120 psi Prob 965 30 MPa 45 MPa 50 MPa Prob 966 200 MPa 500 MPa 350 MPa Prob 967 700 psi 600 psi a b c 4 ksi 40 MPa Prob 968 470 CHAPTER 9 STRESS TRANSFORMATION 95 ABSOLUTE MAXIMUM SHEAR STRESS 473 9 95 Absolute Maximum Shear Stress When a point in a body is subjected to a general threedimensional state of stress an element of material has a normalstress and two shearstress components acting on each of its faces Fig 921a Like the case of plane stress it is possible to develop stresstransformation equations that can be used to determine the normal and shear stress components and acting on any skewed plane of the elementFig921b Furthermore at the point it is also possible to determine the unique orientation of an element having only principal stresses acting on its faces As shown in Fig 921c in general these principal stresses will have magnitudes of maximum intermediate and minimum intensity ie This is a condition known as triaxial stress A discussion of the transformation of stress in three dimensions is beyond the scope of this text however it is discussed in books related to the theory of elasticity For our purposes we will confine our attention only to the case of plane stress For example consider the smax Ú sint Ú smin t s s t b c triaxial stress smin sint smax a Fig 921 474 CHAPTER 9 STRESS TRANSFORMATION 9 material to be subjected to the inplane principal stresses and shown in Fig 922a where both of these stresses are tensile If we view the element in two dimensions that is in the yz xz and xy planes Fig 922b 922c and 922d then we can use Mohrs circle to determine the maximum inplane shear for each case and from this determine the absolute maximum shear stress in the material For example the diameter of Mohrs circle extends between 0 and for the case shown in Fig 922b From this circle Fig 922e the maximum inplane shear stress is For all three circles it is seen that although the maximum in plane shear stress is this value is not the absolute maximum shear stress Instead from Fig 922e 913 and have the same sign s2 s1 t max abs s1 2 txy s1 s22 tyz s22 s2 s2 s1 z y x xy plane stress a s1 s2 Fig 922 y z b s2 x z c s1 x y d s1 s2 s s1 s2 e Absolute maximum shear stress Maximum inplane shear stress txymax tyzmax txzmax t 0 95 ABSOLUTE MAXIMUM SHEAR STRESS 475 9 If one of the inplane principal stresses has the opposite sign of that of the otherFig923athen the three Mohrs circles that describe the state of stress for element orientations about each coordinate axis are shown in Fig 923b Clearly in this case 914 and have opposite signs Calculation of the absolute maximum shear stress as indicated here is important when designing members made of a ductile material since the strength of the material depends on its ability to resist shear stress This situation will be discussed further in Sec 107 s2 s1 t max abs s1 s2 2 s1 s2 y xy plane stress a z x b Maximum inplane and absolute maximum shear stress s1 s2 s txymax t txzmax tyzmax Fig 923 Important Points The general threedimensional state of stress at a point can be represented by an element oriented so that only three principal stresses act on it In the case of plane stress if the inplane principal stresses both have the same sign the absolute maximum shear stress will occur out of the plane and has a value of This value is greater than the inplane shear stress If the inplane principal stresses are of opposite signs then the absolute maximum shear stress will equal the maximum inplane shear stress that is t max abs 1smax smin22 t max abs smax2 smax sint smin 476 CHAPTER 9 STRESS TRANSFORMATION 9 EXAMPLE 910 The point on the surface of the cylindrical pressure vessel in Fig 924a is subjected to the state of plane stress Determine the absolute maximum shear stress at this point b 8 8 16 16 32 s MPa t MPa s1 s2 a 32 MPa 16 MPa Fig 924 SOLUTION The principal stresses are If these stresses are plotted along the axisthe three Mohrs circles can be constructed that describe the stress state viewed in each of the three perpendicular planes Fig 924b The largest circle has a radius of 16 MPa and describes the state of stress in the plane only containing shown shaded in Fig 924a An orientation of an element 45 within this plane yields the state of absolute maximum shear stress and the associated average normal stress namely Ans This same result for can be obtained from direct application of Eq 913 Ans savg 32 0 2 16 MPa t max abs s1 2 32 2 16 MPa t abs max savg 16 MPa t max abs 16 MPa s1 32 MPa s s2 16 MPa s1 32 MPa By comparison the maximum inplane shear stress can be determined from the Mohrs circle drawn between and Fig 924bThis gives a value of savg 32 16 2 24 MPa 32 16 2 8 MPa t max inplane s2 16 MPa s1 32 MPa 478 CHAPTER 9 STRESS TRANSFORMATION 9 5 ksi 3 ksi a 180 MPa b 140 MPa Prob 984 400 psi 300 psi Prob 985 90 MPa z y x 80 MPa Prob 986 30 psi 70 psi z y x 120 psi Prob 987 984 Draw the three Mohrs circles that describe each of the following states of stress 987 The stress at a point is shown on the element Determine the principal stress and the absolute maximum shear stress 985 Draw the three Mohrs circles that describe the following state of stress 988 The stress at a point is shown on the element Determine the principal stress and the absolute maximum shear stress 986 The stress at a point is shown on the element Determine the principal stress and the absolute maximum shear stress PROBLEMS z y x 2 ksi 8 ksi Prob 988 482 CHAPTER 9 STRESS TRANSFORMATION 9 T 075 m A 075 m A F Probs 99697 996 The solid propeller shaft on a ship extends outward from the hull During operation it turns at when the engine develops 900 kW of power This causes a thrust of on the shaft If the shaft has an outer diameter of 250 mm determine the principal stresses at any point located on the surface of the shaft 997 The solid propeller shaft on a ship extends outward from the hull During operation it turns at when the engine develops 900 kW of power This causes a thrust of on the shaft If the shaft has a diameter of 250 mm determine the maximum inplane shear stress at any point located on the surface of the shaft 998 The steel pipe has an inner diameter of 275 in and an outer diameter of 3 in If it is fixed at C and subjected to the horizontal 20lb force acting on the handle of the pipe wrench at its end determine the principal stresses in the pipe at point A which is located on the surface of the pipe F 123 MN v 15 rads F 123 MN v 15 rads 9100 The clamp exerts a force of 150 lb on the boards at G Determine the axial force in each screw AB and CD and then compute the principal stresses at points E and F Show the results on properly oriented elements located at these points The section through EF is rectangular and is 1 in wide 9101 The shaft has a diameter d and is subjected to the loadings shown Determine the principal stress and the maximum inplane shear stress that is developed anywhere on the surface of the shaft 999 Solve Prob 998 for point B which is located on the surface of the pipe REVIEW PROBLEMS 10 in 20 lb 12 in A C y z x B Probs 99899 A C G E B D 05 in 150 lb 150 lb 4 in 15 in 15 in F Prob 9100 F F T0 T0 Prob 9101 Complex stresses developed within this airplane wing are analyzed from strain gauge data Courtesy of Measurements Group Inc Raleigh North Carolina 27611 USA 485 CHAPTER OBJECTIVES The transformation of strain at a point is similar to the transformation of stress and as a result the methods of Chapter 9 will be applied in this chapter Here we will also discuss various ways for measuring strain and develop some important materialproperty relationships including a generalized form of Hookes law At the end of the chapter a few of the theories used to predict the failure of a material will be discussed 101 Plane Strain As outlined in Sec 22 the general state of strain at a point in a body is represented by a combination of three components of normal strain and three components of shear strain These six components tend to deform each face of an element of the material and like stress the normal and shear strain components at the point will vary according to the orientation of the element The strains at a point are often determined by using strain gauges which measure normal strain in specified directions For both analysis and design however engineers must sometimes transform this data in order to obtain the strain in other directions gxy gxz gyz Py Pz Px Strain Transformation 10 102 GENERAL EQUATIONS OF PLANESTRAIN TRANSFORMATION 487 Normal and Shear Strains In order to develop the strain transformation equation for we must determine the elongation of a line segment that lies along the axis and is subjected to strain components As shown in Fig 103a the components of the line along the x and y axes are 101 When the positive normal strain occurs the line dx is elongated Fig 103b which causes line to elongate Likewise when occurs line dy elongates Fig 103c which causes line to elongate Finally assuming that dx remains fixed in position the shear strain which is the change in angle between dx and dy causes the top of line dy to be displaced to the right as shown in Fig 103d This causes to elongate If all three of these elongations are added together the resultant elongation of is then From Eq 22 the normal strain along the line is Using Eq 101 we therefore have 102 Px Px cos2 u Py sin2 u gxy sin u cos u Px dxdx dx dx Px dx cos u Py dy sin u gxy dy cos u dx gxy dy cos u dx gxy dy gxy Py dy sin u dx Py dy Py Px dx cos u dx Px dx Px dy dx sin u dx dx cos u dx Px Py gxy x dx Px 10 b Normal strain Px dx x y y x dx u Pxdx cosu Pxdx Pxdx sinu x y y x dx Before deformation a dy dx u x y x y Normal strain Py dy dx c Pydy sinu Pydy cosu u u Pydy Fig 103 The rubber specimen is constrained between the two fixed supports and so it will undergo plane strain when loads are applied to it in the horizontal plane x y x dx dy d Shear strain gxy dx gxydy cosu gxydy sinu u gxydy dy gxy y 488 CHAPTER 10 STRAIN TRANSFORMATION The straintransformation equation for can be developed by considering the amount of rotation each of the line segments and undergo when subjected to the strain components First we will consider the rotation of which is defined by the counterclockwise angle shown in Fig 103e It can be determined by the displacement caused by using To obtain consider the following three displacement components acting in the direction one from giving Fig 103b another from giving Fig 103c and the last from giving Fig 103dThus as caused by all three strain components is Dividing each term by and using Eq 101 with we have 103 As shown in Fig 103e the line rotates by an amount We can determine this angle by a similar analysis or by simply substituting for into Eq 103 Using the identities we have Since and represent the rotation of the sides and of a differential element whose sides were originally oriented along the and axes Fig 103e the element is then subjected to a shear strain of 104 gxy a b 21Px Py2 sin u cos u gxy1cos2 u sin2 u2 y x dy dx b a 1Px Py2 cos u sin u gxy cos2 u b 1Px Py2 sin1u 902 cos1u 902 gxy sin21u 902 cos1u 902 sin u sin1u 902 cos u u u 90 b dy a 1Px Py2 sin u cos u gxy sin2 u a dydx dx dy Px dx sin u Py dy cos u gxy dy sin u dy gxy dy sin u gxy Py dy cos u Py Px dx sin u Px y dy a dydx dy a dx Px Py gxy dy dx gxy 10 x y x y dy b a u dx dx dy dy e Fig 103 cont x y x dx dy d Shear strain gxy dx gxydy cosu gxydy sinu u gxydy dy gxy y 102 GENERAL EQUATIONS OF PLANESTRAIN TRANSFORMATION 489 10 Using the trigonometric identities and we can rewrite Eqs 102 and 104 in the final form 105 106 These straintransformation equations give the normal strain in the direction and the shear strain of an element oriented at an angle as shown in Fig 104 According to the established sign convention if is positive the element elongates in the positive direction Fig 104a and if is positive the element deforms as shown in Fig 104b If the normal strain in the direction is required it can be obtained from Eq 105 by simply substituting for The result is 107 The similarity between the above three equations and those for plane stress transformation Eqs 91 92 and 93 should be noted By comparison correspond to and correspond to gxy2 gxy2 txy txy Px Py Px Py sy sx sy sx Py Px Py 2 Px Py 2 cos 2u gxy 2 sin 2u u 1u 902 y gxy x Px u gxy x Px gxy 2 Px Py 2 sin 2u gxy 2 cos 2u Px Px Py 2 Px Py 2 cos 2u gxy 2 sin 2u sin2 u cos2 u 1 1 cos 2u2 cos2 u sin 2u 2 sin u cos u x y x y dy dx Positive normal strain Px a u Positive shear strain gxy b x y x y dy dx u Fig 104 490 CHAPTER 10 STRAIN TRANSFORMATION Principal Strains Like stress an element can be oriented at a point so that the elements deformation is caused only by normal strains with no shear strain When this occurs the normal strains are referred to as principal strains and if the material is isotropic the axes along which these strains occur will coincide with the axes that define the planes of principal stress From Eqs 94 and 95 and the correspondence between stress and strain mentioned above the direction of the axis and the two values of the principal strains and are determined from 108 109 Maximum InPlane Shear Strain Using Eqs 96 97 and 98 the direction of the axis and the maximum inplane shear strain and associated average normal strain are determined from the following equations 1010 1011 1012 Pavg Px Py 2 g max inplane 2 Ba Px Py 2 b 2 a gxy 2 b 2 tan 2us Px Py gxy x P12 Px Py 2 C Px Py 2 2 gxy 2 2 tan 2up gxy Px Py P2 P1 x 10 Important Points In the case of plane stress planestrain analysis may be used within the plane of the stresses to analyze the data from strain gauges Remember though there will be a normal strain that is perpendicular to the gauges due to the Poisson effect When the state of strain is represented by the principal strains no shear strain will act on the element The state of strain at a point can also be represented in terms of the maximum inplane shear strain In this case an average normal strain will also act on the element The element representing the maximum inplane shear strain and its associated average normal strains is 45 from the orientation of an element representing the principal strains Complex stresses are often developed at the joints where the cylindrical and hemispherical vessels are joined together The stresses are determined by making measurements of strain 103 MOHRS CIRCLEPLANE STRAIN 501 10 1012 The state of plane strain on an element is given by and Determine the equivalent state of strain on an element at the same point oriented 45 clockwise with respect to the original element gxy 20011062 Py 30011062 Px 50011062 1013 The state of plane strain on an element is and Determine the equivalent state of strain which represents a the principal strains and b the maximum inplane shear strain and the associated average normal strain Specify the orientation of the corresponding elements for these states of strain with respect to the original element gxy 15011062 Py 0 Px 30011062 1014 The state of strain at the point on a boom of an hydraulic engine crane has components of and Use the strain transformation equations to determine a the inplane principal strains and b the maximum inplane shear strain and average normal strain In each case specify the orientation of the element and show how the strains deform the element within the xy plane gxy 18011062 Py 30011062 Px 25011062 1015 Consider the general case of plane strain where and are known Write a computer program that can be used to determine the normal and shear strain and on the plane of an element oriented from the horizontal Also include the principal strains and the elements orientation and the maximum inplane shear strain the average normal strain and the elements orientation 1016 The state of strain at a point on a support has components of Use the straintransformation equations to determine a the inplane principal strains and b the maximum inplane shear strain and average normal strain In each case specify the orientation of the element and show how the strains deform the element within the xy plane 1017 Solve part a of Prob 104 using Mohrs circle 1018 Solve part b of Prob 104 using Mohrs circle 1019 Solve Prob 108 using Mohrs circle 1020 Solve Prob 1010 using Mohrs circle 1021 Solve Prob 1014 using Mohrs circle gxy 67511062 Py 40011062 Px 35011062 u gxy Px gxy Py Px y x dx dy Pydy Pxdx gxy 2 gxy 2 Prob 1012 y x dx dy Pxdx gxy 2 gxy 2 Prob 1013 Prob 1014 y x 106 MATERIALPROPERTY RELATIONSHIPS 509 10 When these three normal strains are superimposed the normal strain is determined for the state of stress in Fig 1018a Similar equations can be developed for the normal strains in the y and z directions The final results can be written as 1018 These three equations express Hookes law in a general form for a triaxial state of stress For application tensile stresses are considered positive quantities and compressive stresses are negative If a resulting normal strain is positive it indicates that the material elongates whereas a negative normal strain indicates the material contracts If we now apply a shear stress to the element Fig 1019a experimental observations indicate that the material will deform only due to a shear strain that is will not cause other strains in the material Likewise and will only cause shear strains and Figs 1019b and 1019c and so Hookes law for shear stress and shear strain can be written as 1019 gxy 1 G txy gyz 1 G tyz gxz 1 G txz gxz gyz txz tyz txy gxy txy Pz 1 E sz n1sx sy2 Py 1 E sy n1sx sz2 Px 1 E sx n1sy sz2 Px a txy b tyz c tzx Fig 1019 512 CHAPTER 10 STRAIN TRANSFORMATION 10 EXAMPLE 109 The bracket in Example 108 Fig 1023a is made of steel for which nst 03 Determine the principal stresses at point A Est 200 GPa a b a c A SOLUTION I From Example 108 the principal strains have been determined as Since point A is on the surface of the bracket for which there is no loading the stress on the surface is zero and so point A is subjected to plane stressApplying Hookes law with we have 1 2 Solving Eqs 1 and 2 simultaneously yields Ans Ans s2 254 MPa s1 620 MPa 67811062 s2 03s1 33911062 s2 20011092 03 20011092 s1 P2 s2 E n E s1 54411062 s1 03s2 27211062 s1 20011092 03 20011092 s2 P1 s1 E n E s2 s3 0 P2 33911062 P1 27211062 Fig 1023 514 CHAPTER 10 STRAIN TRANSFORMATION 10 EXAMPLE 1010 The copper bar in Fig 1024 is subjected to a uniform loading along its edges as shown If it has a length width and thickness before the load is applied determine its new length width and thickness after application of the load Take ncu 034 Ecu 120 GPa t 20 mm b 50 mm a 300 mm Fig 1024 500 MPa 500 MPa 800 MPa 800 MPa a b t SOLUTION By inspection the bar is subjected to a state of plane stress From the loading we have The associated normal strains are determined from the generalized Hookes law Eq 1018 that is The new bar length width and thickness are therefore Ans Ans Ans t 20 mm 100008502120 mm2 1998 mm b 50 mm 10006432150 mm2 4968 mm a 300 mm 0008081300 mm2 3024 mm 0 034 12011032 MPa 1800 MPa 500 MPa2 0000850 Pz sz E n E 1sx sy2 500 MPa 12011032 MPa 034 12011032 MPa 1800 MPa 02 000643 Py sy E n E 1sx sz2 800 MPa 12011032 MPa 034 12011032 MPa 1500 MPa 02 000808 Px sx E n E 1sy sz2 sx 800 MPa sy 500 MPa txy 0 sz 0 107 THEORIES OF FAILURE 525 10 conditions is then plotted as shown in Fig 1035 These three circles are contained in a failure envelope indicated by the extrapolated colored curve that is drawn tangent to all three circles If a planestress condition at a point is represented by a circle that has a point of tangency with the envelope or if it extends beyond the envelopes boundary then failure is said to occur We may also represent this criterion on a graph of principal stresses and This is shown in Fig 1036 Here failure occurs when the absolute value of either one of the principal stresses reaches a value equal to or greater than or or in general if the state of stress at a point defined by the stress coordinates is plotted on the boundary or outside the shaded area Either the maximumnormalstress theory or Mohrs failure criterion can be used in practice to predict the failure of a brittle material However it should be realized that their usefulness is quite limited A tensile fracture occurs very suddenly and its initiation generally depends on stress concentrations developed at microscopic imperfections of the material such as inclusions or voids surface indentations and small cracks Since each of these irregularities varies from specimen to specimen it becomes difficult to specify fracture on the basis of a single test 1s1 s22 1sult2c 1sult2t s2 s1 Important Points If a material is ductile failure is specified by the initiation of yielding whereas if it is brittle it is specified by fracture Ductile failure can be defined when slipping occurs between the crystals that compose the material This slipping is due to shear stress and the maximumshearstress theory is based on this idea Strain energy is stored in a material when it is subjected to normal stressThe maximumdistortionenergy theory depends on the strain energy that distorts the material and not the part that increases its volume The fracture of a brittle material is caused only by the maximum tensile stress in the material and not the compressive stress This is the basis of the maximumnormalstress theory and it is applicable if the stressstrain diagram is similar in tension and compression If a brittle material has a stressstrain diagram that is different in tension and compression then Mohrs failure criterion may be used to predict failure Due to material imperfections tensile fracture of a brittle material is difficult to predict and so theories of failure for brittle materials should be used with caution Mohrs failure criterion sultt sultt sultc sultc s2 s1 Failure envelope s sultt sultc tult t Fig 1035 Fig 1036 107 THEORIES OF FAILURE 529 10 1071 The components of plane stress at a critical point on an A36 steel shell are shown Determine if failure yielding has occurred on the basis of the maximumshear stress theory 1072 The components of plane stress at a critical point on an A36 steel shell are shown Determine if failure yielding has occurred on the basis of the maximum distortionenergy theory 1075 If the A36 steel pipe has outer and inner diameters of 30 mm and 20 mm respectively determine the factor of safety against yielding of the material at point A according to the maximumshearstress theory 1076 If the A36 steel pipe has an outer and inner diameter of 30 mm and 20 mm respectively determine the factor of safety against yielding of the material at point A according to the maximumdistortionenergy theory 1077 The element is subjected to the stresses shown If determine the factor of safety for the loading based on the maximumshearstress theory 1078 Solve Prob 1077 using the maximumdistortion energy theory sY 36 ksi 1079 The yield stress for heattreated beryllium copper is ksi If this material is subjected to plane stress and elastic failure occurs when one principal stress is 145 ksi what is the smallest magnitude of the other principal stress Use the maximumdistortionenergy theory sY 130 1073 If the 2in diameter shaft is made from brittle material having an ultimate strength of for both tension and compression determine if the shaft fails according to the maximumnormalstress theory Use a factor of safety of 15 against rupture 1074 If the 2in diameter shaft is made from cast iron having tensile and compressive ultimate strengths of and respectively determine if the shaft fails in accordance with Mohrs failure criterion 1sult2c 75 ksi 1sult2t 50 ksi sult 50 ksi 60 MPa 70 MPa 40 MPa Probs 107172 30 kip 4 kip ft Probs 107374 150 mm 100 mm 200 mm 200 mm 900 N 900 N A Probs 107576 Probs 107778 12 ksi 4 ksi 8 ksi CHAPTER REVIEW 533 10 If the material is subjected to triaxial stress then the strain in each direction is influenced by the strain produced by all three stresses Hookes law then involves the material properties E and n Pz 1 E sz n1sx sy2 Py 1 E sy n1sx sz2 Px 1 E sx n1sy sz2 If E and are known then G can be determined n The dilatation is a measure of volumetric strain The bulk modulus is used to measure the stiffness of a volume of material If the principal stresses at a critical point in the material are known then a theory of failure can be used as a basis for design Ductile materials fail in shear and here the maximumshearstress theory or the maximumdistortionenergy theory can be used to predict failure Both of these theories make comparison to the yield stress of a specimen subjected to a uniaxial tensile stress Brittle materials fail in tension and so the maximumnormalstress theory or Mohrs failure criterion can be used to predict failure Here comparisons are made with the ultimate tensile stress developed in a specimen G E 211 n2 k E 311 2n2 e 1 2n E 1sx sy sz2 Beams are important structural members that are used to support roof and floor loadings 11 537 CHAPTER OBJECTIVES In this chapter we will discuss how to design a beam so that it is able to resist both bending and shear loads Specifically methods used for designing prismatic beams and determining the shape of fully stressed beams will be developed At the end of the chapter we will consider the design of shafts based on the resistance of both bending and torsional moments 111 Basis for Beam Design Beams are said to be designed on the basis of strength so that they can resist the internal shear and moment developed along their length To design a beam in this way requires application of the shear and flexure formulas provided the material is homogeneous and has linear elastic behavior Although some beams may also be subjected to an axial force the effects of this force are often neglected in design since the axial stress is generally much smaller than the stress developed by shear and bending Design of Beams and Shafts 538 CHAPTER 11 DESIGN OF BEAMS AND SHAFTS As shown in Fig 111 the external loadings on a beam will create additional stresses in the beam directly under the load Notably a compressive stress will be developed in addition to the bending stress and shear stress discussed previously Using advanced methods of analysis as treated in the theory of elasticity it can be shown that diminishes rapidly throughout the beams depth and for most beam spantodepth ratios used in engineering practice the maximum value of generally represents only a small percentage compared to the bending stress that is Furthermore the direct application of concentrated loads is generally avoided in beam design Instead bearing plates are used to spread these loads more evenly onto the surface of the beam Although beams are designed mainly for strength they must also be braced properly along their sides so that they do not buckle or suddenly become unstable Furthermore in some cases beams must be designed to resist a limited amount of deflection as when they support ceilings made of brittle materials such as plaster Methods for finding beam deflections will be discussed in Chapter 12 and limitations placed on beam buckling are often discussed in codes related to structural or mechanical design Since the shear and flexure formulas are used for beam design we will discuss the general results obtained when these equations are applied to various points in a cantilevered beam that has a rectangular cross section and supports a load P at its end Fig 112a In general at an arbitrary section aa along the beams axis Fig 112b the internal shear V and moment M are developed from a parabolic shearstress distribution and a linear normalstress distribution Fig 112c As a result the stresses acting on elements located at points 1 through 5 along the section will be as shown in Fig 112d Note that elements 1 and 5 are subjected only to the maximum normal stress whereas element 3 which is on the neutral axis is subjected only to the maximum shear stress The intermediate elements 2 and 4 resist both normal and shear stress In each case the state of stress can be transformed into principal stresses using either the stresstransformation equations or Mohrs circleThe results are shown in Fig 112e Here each successive element 1 through 5 undergoes a counterclockwise orientation Specifically relative to element 1 considered to be at the 0 position element 3 is oriented at 45 and element 5 is oriented at 90 sx W sy sx sy sy txy sx sy Fig 111 11 w P y x sy sy txy txy sx sx A Whenever large shear loads occur on a beam it is important to use stiffeners such as at A in order to prevent any localized failure such as crimping of the beam flanges 111 BASIS FOR BEAM DESIGN 539 11 a a a P Fig 112 P Stress trajectories for cantilevered beam Fig 113 b P V M 2 3 4 1 5 Bending stress distribution Shear stress distribution c 1 2 3 4 5 d xy stress components 1 2 3 4 5 Principal stresses e If this analysis is extended to many vertical sections along the beam other than aa a profile of the results can be represented by curves called stress trajectories Each of these curves indicated the direction of a principal stress having a constant magnitude Some of these trajectories are shown for the cantilevered beam in Fig 113 Here the solid lines represent the direction of the tensile principal stresses and the dashed lines represent the direction of the compressive principal stresses As expected the lines intersect the neutral axis at 45 angles like element 3 and the solid and dashed lines will intersect at 90 because the principal stresses are always 90 apart Knowing the direction of these lines can help engineers decide where to reinforce a beam if it is made of brittle material so that it does not crack or become unstable 540 CHAPTER 11 DESIGN OF BEAMS AND SHAFTS 112 Prismatic Beam Design Most beams are made of ductile materials and when this is the case it is generally not necessary to plot the stress trajectories for the beam Instead it is simply necessary to be sure the actual bending stress and shear stress in the beam do not exceed allowable bending and shear stress for the material as defined by structural or mechanical codes In the majority of cases the suspended span of the beam will be relatively long so that the internal moments become large When this occurs the engineer will first consider a design based upon bending and then check the shear strength A bending design requires a determination of the beams section modulus a geometric property which is the ratio of I and c that is Using the flexure formula we have 111 Here M is determined from the beams moment diagram and the allowable bending stress is specified in a design code In many cases the beams as yet unknown weight will be small and can be neglected in comparison with the loads the beam must carry However if the additional moment caused by the weight is to be included in the design a selection for S is made so that it slightly exceeds Once is known if the beam has a simple crosssectional shape such as a square a circle or a rectangle of known widthtoheight proportions its dimensions can be determined directly from since However if the cross section is made from several elements such as a wideflange section then an infinite number of web and flange dimensions can be determined that satisfy the value of In practice however engineers choose a particular beam meeting the requirement that from a handbook that lists the standard shapes available from manufacturers Often several beams that have the same section modulus can be selected from these tables If deflections are not restricted usually the beam having the smallest crosssectional area is chosen since it is made of less material and is therefore both lighter and more economical than the others S 7 Sreqd Sreqd Sreqd Ic Sreqd Sreqd Sreqd sallow Sreqd Mmax sallow s McI S Ic 11 A B The two floor beams are connected to the beam AB which transmits the load to the columns of this building frame For design all the connections can be considered to act as pins 542 CHAPTER 11 DESIGN OF BEAMS AND SHAFTS Builtup Sections A builtup section is constructed from two or more parts joined together to form a single unit Since Sreqd the capacity of the beam to resist a moment will vary directly with its section modulus and since Sreqd then Sreqd is increased if I is increased In order to increase I most of the material should be placed as far away from the neutral axis as practicalThis of course is what makes a deep wideflange beam so efficient in resisting a moment For a very large load however an available rolledsteel section may not have a section modulus great enough to support the load Rather than using several available beams instead engineers will usually build up a beam made from plates and anglesA deep Ishaped section having this form is called a plate girder For example the steel plate girder in Fig 115 has two flange plates that are either welded or using angles bolted to the web plate Wood beams are also built up usually in the form of a box beam section Fig 116a They may be made having plywood webs and larger boards for the flanges For very large spans glulam beams are usedThese members are made from several boards gluelaminated together to form a single unit Fig 116b Just as in the case of rolled sections or beams made from a single piece the design of builtup sections requires that the bending and shear stresses be checked In addition the shear stress in the fasteners such as weld glue nails etc must be checked to be certain the beam acts as a single unitThe principles for doing this were outlined in Sec 74 Ic Msallow 11 Fig 115 Welded Bolted Steel plate girders Wooden box beam a Glulam beam b Fig 116 Important Points Beams support loadings that are applied perpendicular to their axes If they are designed on the basis of strength they must resist allowable shear and bending stresses The maximum bending stress in the beam is assumed to be much greater than the localized stresses caused by the application of loadings on the surface of the beam 112 PRISMATIC BEAM DESIGN 543 11 Procedure for Analysis Based on the previous discussion the following procedure provides a rational method for the design of a beam on the basis of strength Shear and Moment Diagrams Determine the maximum shear and moment in the beam Often this is done by constructing the beams shear and moment diagrams For builtup beams shear and moment diagrams are useful for identifying regions where the shear and moment are excessively large and may require additional structural reinforcement or fasteners Bending Stress If the beam is relatively long it is designed by finding its section modulus using the flexure formula Once is determined the crosssectional dimensions for simple shapes can then be computed since If rolledsteel sections are to be used several possible values of S may be selected from the tables in Appendix B Of these choose the one having the smallest crosssectional area since this beam has the least weight and is therefore the most economical Make sure that the selected section modulus S is slightly greater than so that the additional moment created by the beams weight is considered Shear Stress Normally beams that are short and carry large loads especially those made of wood are first designed to resist shear and then later checked against the allowablebendingstress requirements Using the shear formula check to see that the allowable shear stress is not exceeded that is use If the beam has a solid rectangular cross section the shear formula becomes See Eq 2 of Example 72 and if the cross section is a wide flange it is generally appropriate to assume that the shear stress is constant over the crosssectional area of the beams web so that where is determined from the product of the beams depth and the webs thickness See the note at the end of Example 73 Adequacy of Fasteners The adequacy of fasteners used on builtup beams depends upon the shear stress the fasteners can resist Specifically the required spacing of nails or bolts of a particular size is determined from the allowable shear flow calculated at points on the cross section where the fasteners are located See Sec 73 qallow VQI Aweb tallow Ú VmaxAweb tallow Ú 151VmaxA2 tallow Ú Vmax QIt Sreqd Sreqd Ic Sreqd Sreqd Mmaxsallow 112 PRISMATIC BEAM DESIGN 547 11 Shear Stress Maximum shear stress in the beam depends upon the magnitude of Q and t It occurs at the neutral axis since Q is a maximum there and the neutral axis is in the web where the thickness is smallest for the cross section For simplicity we will use the rectangular area below the neutral axis to calculate Q rather than a twopart composite area above this axis Fig 119cWe have so that OK Nail Spacing From the shear diagram it is seen that the shear varies over the entire span Since the nail spacing depends on the magnitude of shear in the beam for simplicity and to be conservative we will design the spacing on the basis of for region BC and for region CD Since the nails join the flange to the web Fig 119d we have V 1 kN V 15 kN 800103 Pa Ú 15103 N037211032 m3 6012511062 m4 1003 m2 309103 Pa tallow Ú VmaxQ It Q yA a 01575 m 2 b101575 m21003 m2 037211032 m3 t 003 m Q yA 100725 m 0015 m2102 m21003 m2 034511032 m3 The shear flow for each region is therefore One nail can resist 150 kN in shear so the maximum spacing becomes For ease of measuring use Ans Ans sCD 250 mm sBC 150 mm sCD 150 kN 574 kNm 0261 m sBC 150 kN 861 kNm 0174 m qCD VCDQ I 1103 N034511032 m3 6012511062 m4 574 kNm qBC VBCQ I 15103 N034511032 m3 6012511062 m4 861 kNm 01575 m 003 m A N 00725 m c 02 m A N 00725 m d 003 m Fig 119 cont 112 PRISMATIC BEAM DESIGN 549 11 113 The brick wall exerts a uniform distributed load of on the beam If the allowable bending stress is ksi determine the required width b of the flange to the nearest in 1 4 sallow 22 120 kipft 112 The brick wall exerts a uniform distributed load of on the beam If the allowable bending stress is and the allowable shear stress is select the lightest wideflange section with the shortest depth from Appendix B that will safely support the load tallow 12 ksi sallow 22 ksi 120 kipft 115 Select the lightestweight steel wideflange beam from Appendix B that will safely support the machine loading shown The allowable bending stress is and the allowable shear stress is tallow 14 ksi sallow 24 ksi 116 The compound beam is made from two sections which are pinned together at B Use Appendix B and select the lightestweight wideflange beam that would be safe for each section if the allowable bending stress is and the allowable shear stress is The beam supports a pipe loading of 1200 lb and 1800 lb as shown tallow 14 ksi sallow 24 ksi 111 The simply supported beam is made of timber that has an allowable bending stress of and an allowable shear stress of Determine its dimensions if it is to be rectangular and have a heightto width ratio of 125 tallow 500 kPa sallow 65 MPa 114 Draw the shear and moment diagrams for the shaft and determine its required diameter to the nearest if and The bearings at A and D exert only vertical reactions on the shaftThe loading is applied to the pulleys at B C and E tallow 3 ksi sallow 7 ksi 1 4 in PROBLEMS Prob 111 Prob 112 2 m 2 m 4 m 8 kNm 4 ft 6 ft 10 ft 120 kipft Prob 113 4 ft 6 ft 9 in 05 in 05 in 05 in 10 ft 120 kipft b A B 14 in 20 in 15 in 12 in 80 lb 110 lb 35 lb C D E Prob 114 Prob 115 Prob 116 5 kip 2 ft 2 ft 2 ft 2 ft 2 ft 5 kip 5 kip 5 kip 6 ft 6 ft 8 ft 10 ft B A C 1200 lb 1800 lb 550 CHAPTER 11 DESIGN OF BEAMS AND SHAFTS 11 117 If the bearing pads at A and B support only vertical forces determine the greatest magnitude of the uniform distributed loading w that can be applied to the beam tallow 15 MPa sallow 15 MPa 1111 The timber beam is to be loaded as shownIf the ends support only vertical forcesdetermine the greatest magnitude of P that can be applied tallow 700 kPa sallow 25 MPa 1112 Determine the minimum width of the beam to the nearest that will safely support the loading of The allowable bending stress is and the allowable shear stress is tallow 15 ksi sallow 24 ksi P 8 kip 1 4 in 1113 Select the shortest and lightestweight steel wide flange beam from Appendix B that will safely support the loading shownThe allowable bending stress is and the allowable shear stress is tallow 12 ksi sallow 22 ksi 119 Select the lightestweight W12 steel wideflange beam from Appendix B that will safely support the loading shown where The allowable bending stress is and the allowable shear stress is 1110 Select the lightestweight W14 steel wideflange beam having the shortest height from Appendix B that will safely support the loading shown where The allowable bending stress is and the allowable shear stress is tallow 12 ksi sallow 22 ksi P 12 kip tallow 12 ksi sallow 22 ksi P 6 kip 118 The simply supported beam is made of timber that has an allowable bending stress of and an allowable shear stress of Determine its smallest dimensions to the nearest in if it is rectangular and has a heighttowidth ratio of 15 1 8 tallow 100 psi sallow 120 ksi 150 mm 25 mm 25 mm 150 mm A w B 1 m 1 m Prob 117 3 ft 3 ft 12 kipft b 15 b A B Prob 118 6 ft 6 ft 9 ft P P Probs 11910 4 m 150 mm 40 mm 30 mm 120 mm A B 4 m P Prob 1111 Prob 1112 Prob 1113 P 6 ft 6 ft A 6 in B 4 ft 4 kip 10 kip 6 kip B A 4 ft 4 ft 4 ft 112 PRISMATIC BEAM DESIGN 551 11 1114 The beam is used in a railroad yard for loading and unloading cars If the maximum anticipated hoist load is 12 kip select the lightestweight steel wideflange section from Appendix B that will safely support the loading The hoist travels along the bottom flange of the beam and has negligible size Assume the beam is pinned to the column at B and roller supported at A The allowable bending stress is and the allowable shear stress is tallow 12 ksi sallow 24 ksi 1 ft x 25 ft 1116 The simply supported beam is composed of two sections built up as shown Determine the maximum uniform loading w the beam will support if the allowable bending stress is and the allowable shear stress is 1117 The simply supported beam is composed of two sections built up as shown Determine if the beam will safely support a loading of w The allowable bending stress is and the allowable shear stress is tallow 14 ksi sallow 22 ksi 2 kipft W12 22 tallow 14 ksi sallow 22 ksi W12 22 1115 The simply supported beam is made of timber that has an allowable bending stress of and an allowable shear stress of Determine its dimensions if it is to be rectangular and have a height towidth ratio of 125 tallow 75 psi sallow 960 psi 1118 Determine the smallest diameter rod that will safely support the loading shown The allowable bending stress is and the allowable shear stress is 1119 The pipe has an outer diameter of 15 mm Determine the smallest inner diameter so that it will safely support the loading shown The allowable bending stress is and the allowable shear stress is tallow 97 MPa sallow 167 MPa tallow 97 MPa sallow 167 MPa A B C 12 kip 27 ft x 15 ft Prob 1114 6 ft 6 ft 5 kipft b 125 b A B Prob 1115 24 ft w Probs 111617 Probs 111819 15 m 25 Nm 15 m 15 Nm 15 Nm 552 CHAPTER 11 DESIGN OF BEAMS AND SHAFTS 11 1120 Determine the maximum uniform loading w the beam will support if the allowable bending stress is and the allowable shear stress is 1121 Determine if the beam will safely support a loading of The allowable bending stress is and the allowable shear stress is tallow 12 ksi sallow 22 ksi w 15 kipft W14 22 tallow 12 ksi sallow 22 ksi W12 14 1123 The box beam has an allowable bending stress of and an allowable shear stress of Determine the maximum intensity w of the distributed loading that it can safely support Also determine the maximum safe nail spacing for each third of the length of the beam Each nail can resist a shear force of 200 N tallow 775 kPa sallow 10 MPa 1124 The simply supported joist is used in the construction of a floor for a building In order to keep the floor low with respect to the sill beams C and D the ends of the joists are notched as shown If the allowable shear stress for the wood is and the allowable bending stress is determine the height h that will cause the beam to reach both allowable stresses at the same time Also what load P causes this to happen Neglect the stress concentration at the notch 1125 The simply supported joist is used in the construction of a floor for a building In order to keep the floor low with respect to the sill beams C and D the ends of the joists are notched as shown If the allowable shear stress for the wood is psi and the allowable bending stress is psi determine the smallest height h so that the beam will support a load of Also will the entire joist safely support the load Neglect the stress concentration at the notch P 600 lb sallow 1700 tallow 350 sallow 1500 psi tallow 350 psi 1122 Determine the minimum depth h of the beam to the nearest that will safely support the loading shown The allowable bending stress is and the allowable shear stress is The beam has a uniform thickness of 3 in tallow 10 ksi sallow 21 ksi 1 8 in 10 ft 10 ft w Probs 112021 A B h 6 ft 12 ft 4 kipft Prob 1122 6 m 150 mm30 mm 250 mm 30 mm 30 mm w Prob 1123 Probs 112425 15 ft 2 in h 10 in A B C D 15 ft P 112 PRISMATIC BEAM DESIGN 553 11 1126 Select the lightestweight steel wideflange beam from Appendix B that will safely support the loading shown The allowable bending stress is and the allowable shear stress is tallow 12 ksi sallow 22 ksi 1129 The wood beam has a rectangular cross section Determine its height h so that it simultaneously reaches its allowable bending stress of and an allowable shear stress of Also what is the maximum load P that the beam can then support tallow 150 psi sallow 150 ksi 1127 The Tbeam is made from two plates welded together as shown Determine the maximum uniform distributed load w that can be safely supported on the beam if the allowable bending stress is and the allowable shear stress is tallow 70 MPa sallow 150 MPa 1128 The beam is made of a ceramic material having an allowable bending stress of psi and an allowable shear stress of psi Determine the width b of the beam if the height h 2b tallow 400 sallow 735 1130 The beam is constructed from three boards as shown If each nail can support a shear force of 300 lb determine the maximum allowable spacing of the nails s for regions AB BC and CD respectively Also if the allowable bending stress is and the allowable shear stress is determine if it can safely support the load tallow 150 psi sallow 15 ksi s s 5 kip 6 ft 12 ft A B 18 kip ft Prob 1126 15 m 200 mm 20 mm 200 mm 20 mm 15 m w A Prob 1127 b h 6 in 2 in 2 in 6 lbin 10 lb 15 lb P 15 ft 15 ft 3 ft 6 in h A B P Prob 1129 A 2 in 10 in 2 in 4 in 10 in 500 lb s s 1500 lb s 6 ft 6 ft 6 ft B C D Prob 1130 Prob 1128 554 CHAPTER 11 DESIGN OF BEAMS AND SHAFTS Fig 1110 a b Haunched concrete beam c Wideflange beam with cover plates 113 Fully Stressed Beams Since the moment in a beam generally varies along its length the choice of a prismatic beam is usually inefficient since it is never fully stressed at points where the internal moment is less than the maximum moment in the beam In order to reduce the weight of the beam engineers sometimes choose a beam having a variable crosssectional area such that at each cross section along the beam the bending stress reaches its maximum allowable value Beams having a variable crosssectional area are called nonprismatic beams They are often used in machines since they can be readily formed by casting Examples are shown in Fig 1110a In structures such beams may be haunched at their ends as shown in Fig 1110b Also beams may be built up or fabricated in a shop using plates An example is a girder made from a rolledshaped wideflange beam and having cover plates welded to it in the region where the moment is a maximum Fig 1110c The stress analysis of a nonprismatic beam is generally very difficult to perform and is beyond the scope of this text Most often these shapes are analyzed by using a computer or the theory of elasticity The results obtained from such an analysis however do indicate that the assumptions used in the derivation of the flexure formula are approximately correct for predicting the bending stresses in nonprismatic sections provided the taper or slope of the upper or lower boundary of the beam is not too severe On the other hand the shear formula cannot be used for nonprismatic beam design since the results obtained from it are very misleading Although caution is advised when applying the flexure formula to nonprismatic beam design we will show here in principle how this formula can be used as an approximate means for obtaining the beams general shape In this regard the size of the cross section of a nonprismatic beam that supports a given loading can be determined using the flexure formula written as If we express the internal moment M in terms of its position x along the beam then since is a known constant the section modulus S or the beams dimensions become a function of x A beam designed in this manner is called a fully stressed beam Although only bending stresses have been considered in approximating its final shape attention must also be given to ensure that the beam will resist shear especially at points where concentrated loads are applied sallow S M sallow 11 The beam for this bridge pier has a variable moment of inertia This design will reduce material weight and save cost 113 FULLY STRESSED BEAMS 555 11 EXAMPLE 114 Determine the shape of a fully stressed simply supported beam that supports a concentrated force at its center Fig 1111a The beam has a rectangular cross section of constant width b and the allowable stress is sallow Fig 1111 P a x h L 2 L 2 h0 x V M b h b P 2 SOLUTION The internal moment in the beam Fig 1111b expressed as a function of position is Hence the required section modulus is Since then for a crosssectional area h by b we have If at then so that Ans By inspection the depth h must therefore vary in a parabolic manner with the distance x NOTE In practice this shape is the basis for the design of leaf springs used to support the rearend axles of most heavy trucks or train cars as shown in the adjacent photo Note that although this result indicates that at it is necessary that the beam resist shear stress at the supports and so practically speaking it must be required that at the supports Fig 1111a h 7 0 x 0 h 0 h2 2h0 2 L x h0 2 3PL 2sallowb x L2 h h0 h2 3P sallowb x I c 1 12 bh3 h2 P 2sallow x S Ic S M sallow P 2sallow x M P 2 x 0 x 6 L2 113 FULLY STRESSED BEAMS 557 11 Applying the flexure formula we have 1 To determine the position x where the absolute maximum normal stress occurs we must take the derivative of with respect to x and set it equal to zeroThis gives Thus Substituting into Eq 1 and simplifying the absolute maximum normal stress is therefore Ans Note that at the wall B the maximum normal stress is which is 111 smaller than NOTE Recall that the flexure formula was derived on the basis of assuming the beam to be prismatic Since this is not the case here some error is to be expected in this analysis and that of Example 114 A more exact mathematical analysis using the theory of elasticity reveals that application of the flexure formula as in the above example gives only small errors in the normal stress if the tapered angle of the beam is small For example if this angle is 15 the stress calculated from the formula will be about 5 greater than that calculated by the more exact analysis It may also be worth noting that the calculation of was done only for illustrative purposes since by SaintVenants principle the actual stress distribution at the support wall is highly irregular 1smax2B sabs max 1smax2B Mc I PL115h02 C 1 12 b13h023D 2 3 PL bh0 2 sabs max 3 4 PL bh0 2 x 1 2 L L2 4x2 0 4x2 4xL L2 8x2 4xL 0 ds dx 6PL2 bh0 2 112x L22 x12212x L2122 12x L24 0 s s Mc I Px1h22 A 1 12 bh3B 6PL2x bh0 2 12x L22 558 CHAPTER 11 DESIGN OF BEAMS AND SHAFTS Fig 1113 114 Shaft Design Shafts that have circular cross sections are often used in the design of mechanical equipment and machineryAs a result they can be subjected to cyclic or fatigue stress which is caused by the combined bending and torsional loads they must transmit or resist In addition to these loadings stress concentrations may exist on a shaft due to keys couplings and sudden transitions in its crosssectional area Sec 58 In order to design a shaft properly it is therefore necessary to take all of these effects into account Here we will discuss some of the important aspects of the design of shafts required to transmit power These shafts are often subjected to loads applied to attached pulleys and gears such as the one shown in Fig 1113a Since the loads can be applied to the shaft at various angles the internal bending and torsional moments at any cross section can be determined by first replacing the loads by their statically equivalent counterparts and then resolving these loads into components in two perpendicular planes Fig 1113bThe bendingmoment diagrams for the loads in each plane can then be drawn and the resultant internal moment at any section along the shaft is then determined by vector addition Fig 1113c In addition to the moment segments of the shaft are also subjected to different internal torques Fig 1113b To account for this general variation of torque along the shaft a torque diagram may also be drawn Fig 1113d M 2M2 x M2 z 11 a A B P1 P2 b x z Az T T y P1z P1x P2 Ax Bz Bx Moment diagram caused by loads in yz plane y Moment diagram caused by loads in xy plane y c Mx Mz Torque diagram caused by torques applied about the shafts axis d y T Ty 562 CHAPTER 11 DESIGN OF BEAMS AND SHAFTS 11 Prob 1133 x L 2 L 2 h0 h w0 B C A Prob 1134 L 2h0 2 2 L h0 w0 h0 Prob 1136 1131 The tapered beam supports a concentrated force P at its center If it is made from a plate that has a constant width b determine the absolute maximum bending stress in the beam 1132 The beam is made from a plate that has a constant thickness b If it is simply supported and carries a uniform load w determine the variation of its depth as a function of x so that it maintains a constant maximum bending stress throughout its length sallow 1135 The beam is made from a plate that has a constant thickness b If it is simply supported and carries the distributed loading shown determine the maximum bending stress in the beam 1136 Determine the variation of the radius r of the cantilevered beam that supports the uniform distributed load so that it has a constant maximum bending stress throughout its length smax 1133 The beam is made from a plate having a constant thickness t and a width that varies as shown If it supports a concentrated force P at its center determine the absolute maximum bending stress in the beam and specify its location x 0 6 x 6 L2 1134 The beam is made from a plate that has a constant thickness b If it is simply supported and carries the distributed loading shown determine the variation of its depth as a function of x so that it maintains a constant maximum bending stress throughout its length sallow PROBLEMS Prob 1131 Prob 1132 P h0 2h0 h0 L 2 L 2 x y w L 2 L 2 h0 b L2 P P 2 L2 P 2 x t b0 L x r0 w r Prob 1135 CHAPTER REVIEW 565 11 Failure of a beam occurs when the internal shear or moment in the beam is a maximum To resist these loadings it is therefore important that the associated maximum shear and bending stress not exceed allowable values as stated in codes Normally the cross section of a beam is first designed to resist the allowable bending stress sallow Mmaxc I CHAPTER REVIEW Then the allowable shear stress is checked For rectangular sections and for wideflange sections it is appropriate to use In general use tallow VQ It tallow Ú VmaxAweb tallow Ú 151VmaxA2 For builtup beams the spacing of fasteners or the strength of glue or weld is determined using an allowable shear flow qallow VQ I Fully stressed beams are nonprismatic and designed such that each cross section along the beam will resist the allowable bending stress This will define the shape of the beam A mechanical shaft generally is designed to resist both torsion and bending stresses Normally the internal bending moment can be resolved into two planes and so it is necessary to draw the moment diagrams for each bendingmoment component and then select the maximum moment based on vector addition Once the maximum bending and shear stresses are determined then depending upon the type of material an appropriate theory of failure is used to compare the allowable stress to what is required A B P1 P2 If the curvature of this pole is measured it is then possible to determine the bending stress developed within it 569 CHAPTER OBJECTIVES Often limits must be placed on the amount of deflection a beam or shaft may undergo when it is subjected to a load and so in this chapter we will discuss various methods for determining the deflection and slope at specific points on beams and shafts The analytical methods include the integration method the use of discontinuity functions and the method of superposition Also a semigraphical technique called the momentarea method will be presented At the end of the chapter we will use these methods to solve for the support reactions on a beam or shaft that is statically indeterminate 121 The Elastic Curve The deflection of a beam or shaft must often be limited in order to provide integrity and stability of a structure or machine and prevent the cracking of any attached brittle materials such as concrete or glass Furthermore code restrictions often require these members not vibrate or deflect severely in order to safely support their intended loading Most important though deflections at specific points on a beam or shaft must be determined if one is to analyze those that are statically indeterminate Before the slope or the displacement at a point on a beam or shaft is determined it is often helpful to sketch the deflected shape of the beam when it is loaded in order to visualize any computed results and thereby partially check these results The deflection curve of the longitudinal axis that passes through the centroid of each crosssectional area of a beam is called the elastic curve For most beams the elastic curve can be sketched without much difficultyWhen doing so however it is necessary to know how the slope or displacement is restricted at various types of supports In general supports that resist a force such as a pin restrict displacement and those that resist a moment such as a fixed wall restrict rotation or slope as well as displacement With this in mind two typical examples of the elastic curves for loaded beams or shafts sketched to an exaggerated scale are shown in Fig 121 Deflection of Beams and Shafts 12 P P Fig 121 121 THE ELASTIC CURVE 571 12 Fig 125 MomentCurvature Relationship We will now develop an important relationship between the internal moment and the radius of curvature rho of the elastic curve at a point The resulting equation will be used for establishing each of the methods presented in the chapter for finding the slope and displacement at points on the elastic curve The following analysis here and in the next section will require the use of three coordinates As shown in Fig 125a the x axis extends positive to the right along the initially straight longitudinal axis of the beam It is used to locate the differential element having an undeformed width dxThe axis extends positive upward from the x axis It measures the displacement of the elastic curve Lastly a localized y coordinate is used to specify the position of a fiber in the beam element It is measured positive upward from the neutral axis or elastic curve as shown in Fig 125b Recall that this same sign convention for x and y was used in the derivation of the flexure formula To derive the relationship between the internal moment and we will limit the analysis to the most common case of an initially straight beam that is elastically deformed by loads applied perpendicular to the beams x axis and lying in the x plane of symmetry for the beams crosssectional area Due to the loading the deformation of the beam is caused by both the internal shear force and bending moment If the beam has a length that is much greater than its depth the greatest deformation will be caused by bending and therefore we will direct our attention to its effects Deflections caused by shear will be discussed in Chapter 14 v r v r P M x x dx v w a u O ds dx Before deformation After deformation b y y dx ds M M du r r 122 SLOPE AND DISPLACEMENT BY INTEGRATION 573 12 122 Slope and Displacement by Integration The equation of the elastic curve for a beam can be expressed mathematically as To obtain this equation we must first represent the curvature in terms of and x In most calculus books it is shown that this relationship is Substituting into Eq 122 we have 124 This equation represents a nonlinear secondorder differential equation Its solution which is called the elastica gives the exact shape of the elastic curveassumingof coursethat beam deflections occur only due to bending Through the use of higher mathematics elastica solutions have been obtained only for simple cases of beam geometry and loading In order to facilitate the solution of a greater number of deflection problems Eq 124 can be modified Most engineering design codes specify limitations on deflections for tolerance or esthetic purposes and as a result the elastic deflections for the majority of beams and shafts form a shallow curve Consequently the slope of the elastic curve which is determined from will be very smalland its square will be negligible compared with unity Therefore the curvature as defined above can be approximated by Using this simplification Eq 124 can now be written as 125 It is also possible to write this equation in two alternative forms If we differentiate each side with respect to x and substitute Eq 62 we get 126 Differentiating again using Eq 61 yields 127 d2 dx2 EI d2v dx2 w1x2 w dVdx d dx EI d2v dx2 V1x2 V dMdx d2v dx2 M EI 1r d2vdx2 dvdx d2vdx2 1 1dvdx2232 M EI 1 r d2vdx2 1 1dvdx2232 v 11r2 v f1x2 See Example 121 574 CHAPTER 12 DEFLECTION OF BEAMS AND SHAFTS 12 For most problems the flexural rigidity EI will be constant along the length of the beam Assuming this to be the case the above results may be reordered into the following set of three equations 128 129 1210 Solution of any of these equations requires successive integrations to obtain the deflection of the elastic curve For each integration it is necessary to introduce a constant of integration and then solve for all the constants to obtain a unique solution for a particular problem For example if the distributed load w is expressed as a function of x and Eq 128 is used then four constants of integration must be evaluated however if the internal moment M is determined and Eq 1210 is used only two constants of integration must be found The choice of which equation to start with depends on the problem Generally however it is easier to determine the internal moment M as a function of x integrate twice and evaluate only two integration constants Recall from Sec 61 that if the loading on a beam is discontinuous that is consists of a series of several distributed and concentrated loads then several functions must be written for the internal moment each valid within the region between the discontinuities Also for convenience in writing each moment expression the origin for each x coordinate can be selected arbitrarily For example consider the beam shown in Fig 127a The internal moment in regions AB BC and CD can be written in terms of the and coordinates selected as shown in either Fig 127b or Fig 127c or in fact in any manner that will yield in as simple a form as possible Once these functions are integrated twice through the use of Eq 1210 and the constants of integration determined the functions will give the slope and deflection elastic curve for each region of the beam for which they are valid M f1x2 x3 x2 x1 v EI d2v dx2 M1x2 EI d3v dx3 V1x2 EI d4v dx4 w1x2 Fig 127 a B C A D P w A D b P w B C x1 x2 x3 A D c P w B C x1 x2 x3 122 SLOPE AND DISPLACEMENT BY INTEGRATION 577 12 Procedure for Analysis The following procedure provides a method for determining the slope and deflection of a beam or shaft using the method of integration Elastic Curve Draw an exaggerated view of the beams elastic curve Recall that zero slope and zero displacement occur at all fixed supports and zero displacement occurs at all pin and roller supports Establish the x and coordinate axes The x axis must be parallel to the undeflected beam and can have an origin at any point along the beam with a positive direction either to the right or to the left If several discontinuous loads are present establish x coordinates that are valid for each region of the beam between the discontinuities Choose these coordinates so that they will simplify subsequent algebraic work In all cases the associated positive axis should be directed upward Load or Moment Function For each region in which there is an x coordinateexpress the loading or the internal moment M as a function of x In particular always assume that M acts in the positive direction when applying the equation of moment equilibrium to determine Slope and Elastic Curve Provided EI is constant apply either the load equation which requires four integrations to get or the moment equation which requires only two integrations For each integration it is important to include a constant of integration The constants are evaluated using the boundary conditions for the supports Table 121 and the continuity conditions that apply to slope and displacement at points where two functions meet Once the constants are evaluated and substituted back into the slope and deflection equations the slope and displacement at specific points on the elastic curve can then be determined The numerical values obtained can be checked graphically by comparing them with the sketch of the elastic curve Realize that positive values for slope are counterclockwise if the x axis extends positive to the right and clockwise if the x axis extends positive to the left In either of these cases positive displacement is upward EI d2vdx2 M1x2 v v1x2 EI d4vdx4 w1x2 M f1x2 w v v 578 CHAPTER 12 DEFLECTION OF BEAMS AND SHAFTS 12 EXAMPLE 121 The cantilevered beam shown in Fig 1210a is subjected to a vertical load P at its end Determine the equation of the elastic curve EI is constant SOLUTION I Elastic Curve The load tends to deflect the beam as shown in Fig 1210a By inspection the internal moment can be represented throughout the beam using a single x coordinate Moment Function From the freebody diagram with M acting in the positive direction Fig 1210b we have Slope and Elastic Curve Applying Eq 1210 and integrating twice yields 1 2 3 Using the boundary conditions at and at Eqs 2 and 3 become Thus and Substituting these results into Eqs 2 and 3 with we get Ans Maximum slope and displacement occur at for which 4 5 vA PL3 3EI uA PL2 2EI A1x 02 v P 6EI 1x3 3L2x 2L32 u P 2EI 1L2 x22 u dvdx C2 PL33 C1 PL22 0 PL3 6 C1L C2 0 PL2 2 C1 x L v 0 x L dvdx 0 EIv Px3 6 C1x C2 EI dv dx Px2 2 C1 EI d2v dx2 Px M Px P x x B A vA Elastic curve L v a uA Fig 1210 M x b P V 122 SLOPE AND DISPLACEMENT BY INTEGRATION 581 12 Slope and Elastic Curve Using Eq 1210 and integrating twice we have 2 The constants of integration are obtained by applying the boundary condition at and the symmetry condition that at This leads to Hence Determining the maximum deflection at we have Ans SOLUTION II Since the distributed loading acts downward it is negative according to our sign convention Using Eq 1 and applying Eq 128 we have Since at then Integrating again yields Here at so This yields Eq 2 The solution now proceeds as before C2 œ 0 x 0 M 0 EI d2v dx2 M w0 3L x3 w0L 4 x C2 œ EI d3v dx3 V w0 L x2 w0L 4 C1 œ w0L4 x 0 V w0L4 EI d3v dx3 V w0 L x2 C1 œ EI d4v dx4 2w0 L x vmax w0L4 120EI x L2 EIv w0 60L x5 w0L 24 x3 5w0L3 192 x EI dv dx w0 12L x4 w0L 8 x2 5w0L3 192 C1 5w0L3 192 C2 0 x L2 dvdx 0 x 0 v 0 EIv w0 60L x5 w0L 24 x3 C1x C2 EI dv dx w0 12L x4 w0L 8 x2 C1 EI d2v dx2 M w0 3L x3 w0L 4 x 122 SLOPE AND DISPLACEMENT BY INTEGRATION 583 12 The four constants are evaluated using two boundary conditions namely and Also two continuity conditions must be applied at B that is at and at Substitution as specified results in the following four equations x1 x2 2a v1 v2 x1 x2 2a dv1dx1 dv2dx2 x2 3a v2 0 v1 0 x1 0 P 18 12a23 C112a2 C2 2P 3 3 2 a12a22 12a23 6 C312a2 C4 v112a2 v212a2 P 6 12a22 C1 2P 3 3a12a2 12a22 2 C3 dv112a2 dx1 dv212a2 dx2 0 2P 3 3 2 a13a22 13a23 6 C313a2 C4 v2 0 at x2 3a 0 0 0 C2 v1 0 at x1 0 Solving we get Thus Eqs 14 become 5 6 7 8 By inspection of the elastic curve Fig 1212b the maximum deflection occurs at D somewhere within region AB Here the slope must be zero From Eq 5 Substituting into Eq 6 Ans The negative sign indicates that the deflection is downward vmax 0484 Pa3 EI x1 1633a 1 6 x1 2 4 9 a2 0 v2 Pa EI x2 2 P 9EI x2 3 22Pa2 9EI x2 4Pa3 3EI dv2 dx2 2Pa EI x2 P 3EI x2 2 22Pa2 9EI v1 P 18EI x1 3 4Pa2 9EI x1 dv1 dx1 P 6EI x1 2 4Pa2 9EI C3 22 9 Pa2 C4 4 3 Pa3 C1 4 9 Pa2 C2 0 584 CHAPTER 12 DEFLECTION OF BEAMS AND SHAFTS 12 EXAMPLE 124 The beam in Fig 1213a is subjected to a load P at its end Determine the displacement at C EI is constant A C B a a x1 vC P x2 2a b P M1 V1 M2 V2 x1 x2 P 2 Fig 1213 SOLUTION Elastic Curve The beam deflects into the shape shown in Fig 1213a Due to the loading two x coordinates will be considered namely and where is directed to the left from C since the internal moment is easy to formulate Moment Functions Using the freebody diagrams shown in Fig 1213b we have Slope and Elastic Curve Applying Eq 1210 For 1 2 EIv1 P 12 x1 3 C1x1 C2 EI dv1 dx1 P 4 x1 2 C1 EI d2v1 dx1 2 P 2 x1 0 x1 2a M1 P 2 x1 M2 Px2 x2 0 x2 6 a 0 x1 6 2a 122 SLOPE AND DISPLACEMENT BY INTEGRATION 585 12 For 3 4 The four constants of integration are determined using three boundary conditions namely at at and at and one continuity equation Here the continuity of slope at the roller requires at and Why is there a negative sign in this equation Note that continuity of displacement at B has been indirectly considered in the boundary conditions since at and Applying these four conditions yields Solving we obtain Substituting and into Eq 4 gives The displacement at C is determined by setting We get Ans vC Pa3 EI x2 0 v2 P 6EI x2 3 7Pa2 6EI x2 Pa3 EI C4 C3 C1 Pa2 3 C2 0 C3 7 6 Pa2 C4 Pa3 P 4 12a22 C1 a P 2 1a22 C3b dv112a2 dx1 dv21a2 dx2 0 P 6 a3 C3a C4 v2 0 at x2 a 0 P 12 12a23 C112a2 C2 v1 0 at x1 2a 0 0 0 C2 v1 0 at x1 0 x2 a x1 2a v1 v2 0 x2 a x1 2a dv1dx1 dv2dx2 x2 a v2 0 x1 2a x1 0 v1 0 v1 0 EIv2 P 6 x2 3 C3x2 C4 EI dv2 dx2 P 2 x2 2 C3 EI d2v2 dx2 2 Px2 0 x2 a 588 CHAPTER 12 DEFLECTION OF BEAMS AND SHAFTS 129 Determine the equations of the elastic curve using the and x2 coordinates EI is constant x1 127 The beam is made of two rods and is subjected to the concentrated load P Determine the maximum deflection of the beam if the moments of inertia of the rods are and IBC and the modulus of elasticity is E IAB 1210 Determine the maximum slope and maximum deflection of the simply supported beam which is subjected to the couple moment EI is constant M0 1211 Determine the equations of the elastic curve for the beam using the and coordinates Specify the beams maximum deflection EI is constant x2 x1 128 Determine the equations of the elastic curve for the beam using the and x2 coordinates EI is constant x1 12 1212 Determine the equations of the elastic curve for the beam using the and coordinates Specify the slope at A and the maximum displacement of the shaft EI is constant x2 x1 A B C L P l Prob 127 P x1 x2 L 2 L 2 Prob 128 P L A B x1 b a x2 Prob 129 A L B M0 2a A B P x1 x2 a Prob 1210 Prob 1211 A B P P L x1 x2 a a Prob 1212 122 SLOPE AND DISPLACEMENT BY INTEGRATION 589 12 1213 The bar is supported by a roller constraint at B which allows vertical displacement but resists axial load and moment If the bar is subjected to the loading shown determine the slope at A and the deflection at C EI is constant 1214 The simply supported shaft has a moment of inertia of 2I for region BC and a moment of inertia I for regions AB and CD Determine the maximum deflection of the beam due to the load P 1216 The fence board weaves between the three smooth fixed postsIf the posts remain along the same linedetermine the maximum bending stress in the board The board has a width of 6 in and a thickness of 05 in Assume the displacement of each end of the board relative to its center is 3 in E 160103 ksi 1217 Determine the equations of the elastic curve for the shaft using the and coordinates Specify the slope at A and the deflection at C EI is constant x2 x1 1218 Determine the equation of the elastic curve for the beam using the x coordinate Specify the slope at A and the maximum deflection EI is constant 1219 Determine the deflection at the center of the beam and the slope at B EI is constant 1215 Determine the equations of the elastic curve for the shaft using the and coordinates Specify the slope at A and the deflection at the center of the shaft EI is constant x3 x1 P A C B L 2 L 2 Prob 1213 C A D P 4 L 4 L 4 L 4 L B Prob 1214 A B a a P b P x1 x3 Prob 1215 4 ft 4 ft A C B 3 in Prob 1216 A B C L L x1 x2 0 M 2 Prob 1217 A L B M0 M0 x Probs 121819 122 SLOPE AND DISPLACEMENT BY INTEGRATION 591 1227 Wooden posts used for a retaining wall have a diameter of 3 in If the soil pressure along a post varies uniformly from zero at the top A to a maximum of at the bottom B determine the slope and displacement at the top of the post Ew 16103 ksi 300 lbft 1228 Determine the slope at end B and the maximum deflection of the cantilevered triangular plate of constant thickness t The plate is made of material having a modulus of elasticity E 1229 The beam is made of a material having a specific weight Determine the displacement and slope at its end A due to its weight The modulus of elasticity for the material is E g 1230 The beam is made of a material having a specific weight of Determine the displacement and slope at its end A due to its weight The modulus of elasticity for the material is E g 12 1226 Determine the equations of the elastic curve using the coordinates and and specify the slope and deflection at B EI is constant x2 x1 L A B a w x1 x2 C Prob 1226 6 ft A 300 lbft B Prob 1227 L t b 2 b 2 w A B x Prob 1228 b L A h Prob 1229 r A L Prob 1230 592 CHAPTER 12 DEFLECTION OF BEAMS AND SHAFTS 1231 The tapered beam has a rectangular cross section Determine the deflection of its free end in terms of the load P length L modulus of elasticity E and the moment of inertia of its fixed end I0 1232 The beam is made from a plate that has a constant thickness t and a width that varies linearly The plate is cut into strips to form a series of leaves that are stacked to make a leaf spring consisting of n leaves Determine the deflection at its end when loaded Neglect friction between the leaves 12 b L A P Prob 1231 b L P Prob 1232 1233 The tapered beam has a rectangular cross section Determine the deflection of its center in terms of the load P length L modulus of elasticity E and the moment of inertia of its center Ic b L 2 L 2 P Prob 1233 P nb b x x L 2 L 2 Prob 1234 1234 The leaf spring assembly is designed so that it is subjected to the same maximum stress throughout its length If the plates of each leaf have a thickness t and can slide freely between each other show that the spring must be in the form of a circular arc in order that the entire spring becomes flat when a large enough load P is applied What is the maximum normal stress in the spring Consider the spring to be made by cutting the n strips from the diamondshaped plate of thickness t and width bThe modulus of elasticity for the material is E Hint Show that the radius of curvature of the spring is constant 123 Discontinuity Functions The method of integration used to find the equation of the elastic curve for a beam or shaft is convenient if the load or internal moment can be expressed as a continuous function throughout the beams entire length If several different loadings act on the beamhoweverthe method becomes more tedious to apply because separate loading or moment functions must be written for each region of the beam Furthermore integration of these functions requires the evaluation of integration constants using both boundary and continuity conditions For example the beam shown in Fig 1214 requires four moment functions to be written They describe the moment in regions AB BC CD and DE When applying the momentcurvature relationship and integrating each moment equation twice we must evaluate eight constants of integration These involve two boundary conditions that require zero displacement at points A and E and six continuity conditions for both slope and displacement at points B C and D In this section we will discuss a method for finding the equation of the elastic curve for a multiply loaded beam using a single expression either formulated from the loading on the beam or from the beams internal moment If the expression for is substituted into and integrated four times or if the expression for M is substituted into and integrated twice the constants of integration will be determined only from the boundary conditionsSince the continuity equations will not be involvedthe analysis will be greatly simplified Discontinuity Functions In order to express the load on the beam or the internal moment within it using a single expression we will use two types of mathematical operators known as discontinuity functions EI d2vdx2 M1x2 EI d4vdx4 w1x2 w M M1x2 w w1x2 EI d2vdx2 M 123 DISCONTINUITY FUNCTIONS 593 12 A E P w C B D M0 Fig 1214 For safety purposes these cantilevered beams that support sheets of plywood must be designed for both strength and a restricted amount of deflection 123 DISCONTINUITY FUNCTIONS 597 12 Procedure for Analysis The following procedure provides a method for using discontinuity functions to determine a beams elastic curve This method is particularly advantageous for solving problems involving beams or shafts subjected to several loadings since the constants of integration can be evaluated by using only the boundary conditions while the compatibility conditions are automatically satisfied Elastic Curve Sketch the beams elastic curve and identify the boundary conditions at the supports Zero displacement occurs at all pin and roller supports and zero slope and zero displacement occur at fixed supports Establish the x axis so that it extends to the right and has its origin at the beams left end Load or Moment Function Calculate the support reactions at and then use the discontinuity functions in Table 122 to express either the loading or the internal moment M as a function of x Make sure to follow the sign convention for each loading as it applies for this equation Note that the distributed loadings must extend all the way to the beams right end to be valid If this does not occur use the method of superposition which is illustrated in Example 126 Slope and Elastic Curve Substitute into or M into the moment curvature relation and integrate to obtain the equations for the beams slope and deflection Evaluate the constants of integration using the boundary conditions and substitute these constants into the slope and deflection equations to obtain the final results When the slope and deflection equations are evaluated at any point on the beam a positive slope is counterclockwise and a positive displacement is upward EI d2vdx2 M EI d4vdx4 w1x2 w w x 0 123 DISCONTINUITY FUNCTIONS 599 12 Notice how this equation can also be established directly using the results of Table 122 for moment Slope and Elastic Curve Integrating twice yields 1 From Eq 1 the boundary condition at and at gives Solving these equations simultaneously for and we get and Thus 2 3 From Fig 1218a maximum displacement may occur either at C or at D where the slope To obtain the displacement of C set in Eq 3We get Ans The negative sign indicates that the displacement is downward as shown in Fig 1218a To locate point D use Eq 2 with and This gives Solving for the positive root Hence from Eq 3 Comparing this value with we see that vmax vC vC vD 5006 kip ft3 EI EIvD 4 3 120323 1203 1023 133312032 12 000 xD 203 ft xD 2 60xD 1633 0 0 4xD 2 31xD 1022 1333 dvdx 0 x 7 10 ft vC 12 000 kip ft3 EI x 0 dvdx 0 EIv 4 3 x3 8x 1093 1333x 12 000 EI dv dx 4x2 38x 1092 1333 C2 12 000 C1 1333 C2 C1 0 36 000 130 1023 C11302 C2 0 1333 110 1023 C11102 C2 x 30 ft v 0 x 10 ft v 0 EIv 4 3 x3 8x 1093 C1x C2 EI dv dx 4x2 38x 1092 C1 EI d2v dx2 8x 68x 1091 602 CHAPTER 12 DEFLECTION OF BEAMS AND SHAFTS 12 1242 Determine the equation of the elastic curve the slope at A and the maximum deflection of the simply supported beam EI is constant 1243 Determine the maximum deflection of the cantilevered beamThe beam is made of material having an and I 650106 mm6 E 200 GPa 1246 Determine the maximum deflection of the simply supported beam and I 650106 mm4 E 200 GPa 1247 The wooden beam is subjected to the load shown Determine the equation of the elastic curve If determine the deflection and the slope at end B Ew 12 GPa 1248 The beam is subjected to the load shownDetermine the slopes at A and B and the displacement at C EI is constant 1244 The beam is subjected to the load shownDetermine the equation of the elastic curve EI is constant 1245 The beam is subjected to the load shown Determine the displacement at and the slope at A EI is constant x 7 m L 3 L 3 L 3 P P A B Prob 1242 A 30 kNm 15 m 15 m 15kN Prob 1243 B A x 4 m 3 m 3 kNm 50 kN 3 m Probs 124445 15 m 15 m 3 m 15 kNm 20 kN A B Prob 1246 B A x 6 kN 4 kN 3 m 15 m 2 kNm 200 mm 400 mm 15 m Prob 1247 x A C B 3 m 5 m 30 kN 12 kNm Prob 1248 124 SLOPE AND DISPLACEMENT BY THE MOMENTAREA METHOD 617 12 1275 The beam is made of a ceramic material In order to obtain its modulus of elasticity it is subjected to the elastic loading shown If the moment of inertia is I and the beam has a measured maximum deflection determine E The supports at A and D exert only vertical reactions on the beam 1278 The rod is constructed from two shafts for which the moment of inertia of AB is I and of BC is 2I Determine the maximum slope and deflection of the rod due to the loadingThe modulus of elasticity is E 1276 The bar is supported by a roller constraint at B which allows vertical displacement but resists axial load and moment If the bar is subjected to the loading shown determine the slope at A and the deflection at C EI is constant 1277 The bar is supported by the roller constraint at C which allows vertical displacement but resists axial load and moment If the bar is subjected to the loading shown determine the slope and displacement at A EI is constant 1279 Determine the slope at point D and the deflection at point C of the simply supported beamThe beam is made of material having a modulus of elasticity EThe moment of inertia of segments AB and CD of the beam is I while the moment of inertia of segment BC of the beam is 2I A D a a L B C P P Prob 1275 P A B C L 2 L 2 Prob 1278 A B C D L 2 L 4 L 4 P P Prob 1279 L 2 L 2 P A C B Prob 1276 A B C P 2a a Prob 1277 1280 Determine the slope at point A and the maximum deflection of the simply supported beamThe beam is made of material having a modulus of elasticity EThe moment of inertia of segments AB and CD of the beam is I while the moment of inertia of segment BC is 2I A B C D L 2 L 4 L 4 P P Prob 1280 125 METHOD OF SUPERPOSITION 619 12 125 Method of Superposition The differential equation satisfies the two necessary requirements for applying the principle of superposition ie the load is linearly related to the deflection and the load is assumed not to change significantly the original geometry of the beam or shaftAs a result the deflections for a series of separate loadings acting on a beam may be superimposed For example if is the deflection for one load and is the deflection for another load the total deflection for both loads acting together is the algebraic sum Using tabulated results for various beam loadings such as the ones listed in Appendix C or those found in various engineering handbooks it is therefore possible to find the slope and displacement at a point on a beam subjected to several different loadings by algebraically adding the effects of its various component parts The following examples illustrate how to use the method of superposition to solve deflection problems where the deflection is caused not only by beam deformations but also by rigidbody displacements such as those that occur when the beam is supported by springs v1 v2 v2 v1 v1x2 w1x2 EI d4vdx4 w1x2 The resultant deflection at any point on this beam can be determined from the superposition of the deflections caused by each of the separate loadings acting on the beam 620 CHAPTER 12 DEFLECTION OF BEAMS AND SHAFTS 12 EXAMPLE 1213 Determine the displacement at point C and the slope at the support A of the beam shown in Fig 1228a EI is constant a 8 kN 4 m A C B vC 2 kNm 4 m 4 m A C B 2 kNm vC1 vC2 b 4 m A C B c 8 kN 4 m 4 m uA uA1 uA2 Fig 1228 SOLUTION The loading can be separated into two component parts as shown in Figs 1228b and 1228c The displacement at C and slope at A are found using the table in Appendix C for each part For the distributed loading For the 8kN concentrated force The displacement at C and the slope at A are the algebraic sums of these components Hence Ans Ans 1 T2 vC 1vC21 1vC22 139 kN m3 EI T 1b2 uA 1uA21 1uA22 56 kN m2 EI b 1vC22 PL3 48EI 8 kN18 m23 48EI 8533 kN m3 EI T 1uA22 PL2 16EI 8 kN18 m22 16EI 32 kN m2 EI b 1vC21 5wL4 768EI 512 kNm218 m24 768EI 5333 kN m3 EI T 1uA21 3wL3 128EI 312 kNm218 m23 128EI 24 kN m2 EI b 622 CHAPTER 12 DEFLECTION OF BEAMS AND SHAFTS 12 EXAMPLE 1215 Determine the displacement at the end C of the cantilever beam shown in Fig 1230 EI is constant 6 m 2 m A B C 4 kNm vB vC uB SOLUTION Using the table in Appendix C for the triangular loading the slope and displacement at point B are The unloaded region BC of the beam remains straight as shown in Fig 1230 Since is small the displacement at C becomes Ans 2448 kN m3 EI T 1728 kN m3 EI 36 kN m2 EI 12 m2 1 T2 vC vB uB1LBC2 uB vB w0L4 30EI 4 kNm16 m24 30EI 1728 kN m3 EI uB w0L3 24EI 4 kNm16 m23 24EI 36 kN m2 EI Fig 1230 125 METHOD OF SUPERPOSITION 625 12 A B 6 in 3 in 8 kip Prob 1294 4 kNm B A 5 m 20 kN 5 m C Prob 1295 1294 Determine the vertical deflection and slope at the end A of the bracket Assume that the bracket is fixed supported at its base and neglect the axial deformation of segment AB EI is constant 1296 Determine the deflection at end E of beam CDE The beams are made of wood having a modulus of elasticity of E 10 GPa 1295 The simply supported beam is made of A36 steel and is subjected to the loading shown Determine the deflection at its center C I 01457103 m4 1297 The pipe assembly consists of three equalsized pipes with flexibility stiffness EI and torsional stiffness GJ Determine the vertical deflection at point A A C D a a a a E B 15 m 15 m 3 kN 2 m 1 m 75 mm 150 mm Section a a Prob 1296 B C P A L2 L2 L2 Prob 1297 626 CHAPTER 12 DEFLECTION OF BEAMS AND SHAFTS 12 1298 Determine the vertical deflection at the end A of the bracket Assume that the bracket is fixed supported at its base B and neglect axial deflection EI is constant 1299 Determine the vertical deflection and slope at the end A of the bracket Assume that the bracket is fixed supported at its base and neglect the axial deformation of segment AB EI is constant 12100 The framework consists of two A36 steel cantilevered beams CD and BA and a simply supported beam CB If each beam is made of steel and has a moment of inertia about its principal axis of determine the deflection at the center G of beam CB Ix 118 in4 12101 The wideflange beam acts as a cantilever Due to an error it is installed at an angle with the vertical Determine the ratio of its deflection in the x direction to its deflection in the y direction at A when a load P is applied at this point The moments of inertia are and For the solution resolve P into components and use the method of superposition Note The result indicates that large lateral deflections x direction can occur in narrow beams when they are improperly installed in this manner To show this numerically compute the deflections in the x and y directions for an A36 steel with P 15 kip u 10 and L 12 ft W10 15 Iy V Ix Iy Ix u 12102 The simply supported beam carries a uniform load of Code restrictions due to a plaster ceiling require the maximum deflection not to exceed of the span length Select the lightestweight A36 steel wide flange beam from Appendix B that will satisfy this requirement and safely support the load The allowable bending stress is and the allowable shear stress is Assume A is a pin and B a roller support tallow 14 ksi sallow 24 ksi 1360 2 kipft a P A b B Prob 1298 A C B 4 in 3 in 80 lb 20 lbin Prob 1299 16 ft A D 8 ft 8 ft C G B 15 kip Prob 12100 x L y P A Vertical u u Prob 12101 4 ft A B 8 ft 8 kip 4 ft 2 kipft 8 kip Prob 12102 126 STATICALLY INDETERMINATE BEAMS AND SHAFTS 627 12 126 Statically Indeterminate Beams and Shafts The analysis of statically indeterminate axially loaded bars and torsionally loaded shafts has been discussed in Secs 44 and 55 respectively In this section we will illustrate a general method for determining the reactions on statically indeterminate beams and shafts Specifically a member of any type is classified as statically indeterminate if the number of unknown reactions exceeds the available number of equilibrium equations The additional support reactions on the beam or shaft that are not needed to keep it in stable equilibrium are called redundants The number of these redundants is referred to as the degree of indeterminacy For example consider the beam shown in Fig 1232a If the freebody diagram is drawn Fig 1232b there will be four unknown support reactions and since three equilibrium equations are available for solution the beam is classified as being indeterminate to the first degree Either or can be classified as the redundant for if any one of these reactions is removed the beam remains stable and in equilibrium cannot be classified as the redundant for if it were removed would not be satisfied In a similar manner the continuous beam in Fig 1233a is indeterminate to the second degree since there are five unknown reactions and only three available equilibrium equations Fig 1233b Here the two redundant support reactions can be chosen among By Cy and Dy Ay Fx 0 Ax MA By Ay Fig 1232 b P MA Ax Ay By a P1 P2 P3 A D C B a P A B Fig 1233 b P1 P2 P3 Ay By Cy Dy Ax 628 CHAPTER 12 DEFLECTION OF BEAMS AND SHAFTS 12 To determine the reactions on a beam or shaft that is statically indeterminate it is first necessary to specify the redundant reactionsWe can determine these redundants from conditions of geometry known as compatibility conditions Once determined the redundants are then applied to the beam and the remaining reactions are determined from the equations of equilibrium In the following sections we will illustrate this procedure for solution using the method of integration Sec 127 the momentarea method Sec 128 and the method of superposition Sec 129 127 Statically Indeterminate Beams and ShaftsMethod of Integration The method of integration discussed in Sec 122 requires two integrations of the differential equation once the internal moment M in the beam is expressed as a function of position x If the beam is statically indeterminate however M can also be expressed in terms of the unknown redundants After integrating this equation twice there will be two constants of integration along with the redundants to be determinedAlthough this is the case these unknowns can always be found from the boundary andor continuity conditions for the problem The following example problems illustrate specific applications of this method using the procedure for analysis outlined in Sec 122 d2vdx2 MEI An example of a statically indeterminate beam used to support a bridge deck 127 STATICALLY INDETERMINATE BEAMS AND SHAFTSMETHOD OF INTEGRATION 629 12 EXAMPLE 1217 The beam is subjected to the distributed loading shown in Fig 1234a Determine the reaction at A EI is constant SOLUTION Elastic Curve The beam deflects as shown in Fig 1234a Only one coordinate x is needed For convenience we will take it directed to the right since the internal moment is easy to formulate Moment Function The beam is indeterminate to the first degree as indicated from the freebody diagram Fig 1234b We can express the internal moment M in terms of the redundant force at A using the segment shown in Fig 1234c Here Slope and Elastic Curve Applying Eq 1210 we have The three unknowns and are determined from the boundary conditions and Applying these conditions yields Solving Ans NOTE Using the result for the reactions at B can be determined from the equations of equilibrium Fig 1234b Show that By 2w0L5 and MB w0L215 Bx 0 Ay C1 1 120 w0L3 C2 0 Ay 1 10 w0L 0 1 6 AyL3 1 120 w0L4 C1L C2 v 0 x L 0 1 2 AyL2 1 24 w0L3 C1 dv dx 0 x L 0 0 0 0 C2 v 0 x 0 v 0 x L dvdx 0 x L v 0 x 0 C2 C1 Ay EIv 1 6 Ayx3 1 120 w0 x5 L C1x C2 EI dv dx 1 2 Ayx2 1 24 w0 x4 L C1 EI d2v dx2 Ayx 1 6 w0 x3 L M Ayx 1 6 w0 x3 L A x L B w0 a By MB Bx A L w0L b Ay L 2 3 1 3 1 2 M V c Ay x x w0 w0 A 2 3 1 3 1 2 x L x2 L Fig 1234 127 STATICALLY INDETERMINATE BEAMS AND SHAFTSMETHOD OF INTEGRATION 631 12 12104 Determine the value of a for which the maximum positive moment has the same magnitude as the maximum negative moment EI is constant 12105 Determine the reactions at the supports A B and C then draw the shear and moment diagrams EI is constant 12107 Determine the moment reactions at the supports A and B EI is constant 12108 Determine the reactions at roller support A and fixed support B L P a Prob 12104 C A B P P L 2 L 2 L 2 L 2 Prob 12105 L B A w 3 2L 3 Prob 12108 L P P A B a a Prob 12107 L A B P L Prob 12106 A B L M0 Prob 12103 12103 Determine the reactions at the supports A and B then draw the moment diagram EI is constant 12106 Determine the reactions at the supports then draw the shear and moment diagram EI is constant PROBLEMS 632 CHAPTER 12 DEFLECTION OF BEAMS AND SHAFTS 12 12109 Use discontinuity functions and determine the reactions at the supports then draw the shear and moment diagrams EI is constant 12110 Determine the reactions at the supports then draw the shear and moment diagrams EI is constant 12112 Determine the moment reactions at fixed supports A and B EI is constant 12111 Determine the reactions at pin support A and roller supports B and C EI is constant 12113 The beam has a constant and is supported by the fixed wall at B and the rod AC If the rod has a crosssectional area and the material has a modulus of elasticity E2 determine the force in the rod A2 E1I1 12114 The beam is supported by a pin at A a roller at B and a post having a diameter of 50 mm at C Determine the support reactions at A B and C The post and the beam are made of the same material having a modulus of elasticity and the beam has a constant moment of inertia I 255106 mm4 E 200 GPa 8 ft 10 ft 3 kipft C A B Prob 12109 B A L 2 L 2 w0 Prob 12112 A B C w L1 L2 Prob 12113 B C A 15 kNm 6 m 1 m 6 m Prob 12114 B C A L L w0 Prob 12110 A C B w L L Prob 12111 638 CHAPTER 12 DEFLECTION OF BEAMS AND SHAFTS 12 12116 The rod is fixed at A and the connection at B consists of a roller constraint which allows vertical displacement but resists axial load and moment Determine the moment reactions at these supports EI is constant 12117 Determine the value of a for which the maximum positive moment has the same magnitude as the maximum negative moment EI is constant 12119 Determine the reactions at the supports then draw the shear and moment diagrams EI is constant Support B is a thrust bearing 12120 Determine the moment reactions at the supports A and B EI is constant L B A M0 A B w L L a P C A B L M0 M0 L C A B L P L 2 L 2 Prob 12115 Prob 12116 Prob 12117 L2 A B w L2 Prob 12120 Prob 12119 Prob 12118 12115 Determine the moment reactions at the supports A and B then draw the shear and moment diagrams EI is constant 12118 Determine the reactions at the supports then draw the shear and moment diagrams EI is constant PROBLEMS 642 CHAPTER 12 DEFLECTION OF BEAMS AND SHAFTS 12 Procedure for Analysis The following procedure provides a means for applying the method of superposition or the force method to determine the reactions on statically indeterminate beams or shafts Elastic Curve Specify the unknown redundant forces or moments that must be removed from the beam in order to make it statically determinate and stable Using the principle of superposition draw the statically indeterminate beam and show it equal to a sequence of corresponding statically determinate beams The first of these beams the primary beam supports the same external loads as the statically indeterminate beam and each of the other beams added to the primary beam shows the beam loaded with a separate redundant force or moment Sketch the deflection curve for each beam and indicate symbolically the displacement slope at the point of each redundant force moment Compatibility Equations Write a compatibility equation for the displacement slope at each point where there is a redundant force moment Determine all the displacements or slopes using an appropriate method as explained in Secs 122 through 125 Substitute the results into the compatibility equations and solve for the unknown redundants If a numerical value for a redundant is positive it has the same sense of direction as originally assumed Similarly a negative numerical value indicates the redundant acts opposite to its assumed sense of direction Equilibrium Equations Once the redundant forces andor moments have been determined the remaining unknown reactions can be found from the equations of equilibrium applied to the loadings shown on the beams freebody diagram The following examples illustrate application of this procedure For brevity all displacements and slopes have been found using the table in Appendix C 644 CHAPTER 12 DEFLECTION OF BEAMS AND SHAFTS 12 EXAMPLE 1222 The beam in Fig 1245a is fixed supported to the wall at A and pin connected to a rod BC If for both members determine the force developed in the rod due to the loading The moment of inertia of the beam about its neutral axis is I 475 in4 E 2911032 ksi 1 2indiameter SOLUTION I Principle of Superposition By inspection this problem is indeterminate to the first degree Here B will undergo an unknown displacement since the rod will stretch The rod will be treated as the redundant and hence the force of the rod is removed from the beam at B Fig 1245b and then reapplied Fig 1245c Compatibility Equation At point B we require 1 The displacements and are determined from the table in Appendix C is calculated from Eq 42Working in kilopounds and inches we have Thus Eq 1 becomes Ans FBC 178 kip 001686FBC 01045 004181FBC 1 T2 vB œ PL3 3EI FBC110 ft23112 inft23 32911023 kipin21475 in42 004181FBCc vB 5PL3 48EI 518 kip2110 ft23112 inft23 482911032 kipin21475 in42 01045 in T vB fl PL AE FBC18 ft2112 inft2 1p42A1 2 inB22911032 kipin2 001686FBC T vB fl vB œ vB vfl B vB vB œ 1 T2 vB fl 5 ft A B 5 ft v Actual beam and rod 8 kip a C 8 ft B A Redundant FBC removed 8 kip b B vB A Only redundant FBC applied c B FBC vB Fig 1245 129 STATICALLY INDETERMINATE BEAMS AND SHAFTSMETHOD OF SUPERPOSITION 645 12 SOLUTION II Principle of Superposition We can also solve this problem by removing the pin support at C and keeping the rod attached to the beam In this case the 8kip load will cause points B and C to be displaced downward the same amount Fig 1245e since no force exists in rod BCWhen the redundant force is applied at point C it causes the end C of the rod to be displaced upward and the end B of the beam to be displaced upward Fig 1245f The difference in these two displacements represents the stretch of the rod due to so that Hence from Figs 1245d 1245e and 1245f the compatibility of displacement at point C is 2 From Solution I we have Therefore Eq 2 becomes Ans FBC 178 kip 0 01045 1001686FBC 004181FBC2 1 T2 vB œ 004181FBCc vBC vB fl 001686FBCc vC vB 01045 in T 0 vC 1vBC vB œ 2 1 T2 vC œ vBC vB œ FBC vBC vB œ vC œ FBC vC A Redundant FBC removed 8 kip e B C vC A Only redundant FBC applied f B C vC vB vBC FBC A 5 ft Actual beam and rod 8 kip d C B 5 ft Fig 1245 cont 129 STATICALLY INDETERMINATE BEAMS AND SHAFTSMETHOD OF SUPERPOSITION 647 12 Compatibility Equations Referring to the displacement and slope at B we require 1 2 Using the table in Appendix C to calculate the slopes and displacements we have Substituting these values into Eqs 1 and 2 and canceling out the common factor EI we get Solving these equations simultaneously gives Ans MB 1125 kip ft By 3375 kip 0 1134 576By 72MB 1 T2 0 108 72By 12MB 1e2 vB fl ML2 2EI MB112 ft22 2EI 72MB EI T uB fl ML EI MB112 ft2 EI 12MB EI b vB œ PL3 3EI By112 ft23 3EI 576By EI T uB œ PL2 2EI By112 ft22 2EI 72By EI b vB 7wL4 384EI 713 kipft2112 ft24 384EI 1134 kip ft3 EI T uB wL3 48EI 3 kipft 112 ft23 48EI 108 kip ft2 EI b 0 vB vB œ vB fl 1 T2 0 uB uB œ uB fl 1e2 648 CHAPTER 12 DEFLECTION OF BEAMS AND SHAFTS 12 FUNDAMENTAL PROBLEMS F1213 Determine the reactions at the fixed support A and the roller B EI is constant F1216 Determine the reaction at the roller B EI is constant F1214 Determine the reactions at the fixed support A and the roller B EI is constant F1215 Determine the reactions at the fixed support A and the roller B Support B settles 2 mm I 650106 m4 E 200 GPa F1217 Determine the reaction at the roller B EI is constant F1218 Determine the reaction at the roller support B if it settles 5 mm and I 650106 m4 E 200 GPa A B 40 kN 4 m 2 m F1213 A B w0 L F1214 A B 6 m 10 kNm F1215 A B C L L M0 F1216 A B C 4 m 6 m 2 m 50 kN F1217 A B C 6 m 6 m 10 kNm F1218 129 STATICALLY INDETERMINATE BEAMS AND SHAFTSMETHOD OF SUPERPOSITION 649 12 12121 Determine the reactions at the bearing supports A B and C of the shaft then draw the shear and moment diagrams EI is constant Each bearing exerts only vertical reactions on the shaft 12124 The assembly consists of a steel and an aluminum bar each of which is 1 in thick fixed at its ends A and B and pin connected to the rigid short link CD If a horizontal force of 80 lb is applied to the link as shown determine the moments created at A and B Eal 1011032 ksi Est 2911032 ksi PROBLEMS 400 N 1 m 1 m C A B 1 m 1 m 400 N Prob 12121 A B L 2 P L Prob 12122 12122 Determine the reactions at the supports A and B EI is constant 12123 Determine the reactions at the supports A B and C then draw the shear and moment diagrams EI is constant 12125 Determine the reactions at the supports A B and C then draw the shear and moment diagrams EI is constant 12126 Determine the reactions at the supports A and B EI is constant 6 ft 12 ft 3 kipft A B C 6 ft 12 kip Prob 12123 80 lb 30 in C D A B 05 in 1 in Aluminum Steel Prob 12124 3 m A B C 3 m 3 m 3 m 10 kN 10 kN Prob 12125 L A M0 B Prob 12126 650 CHAPTER 12 DEFLECTION OF BEAMS AND SHAFTS 12 12127 Determine the reactions at support C EI is constant for both beams 12128 The compound beam segments meet in the center using a smooth contact roller Determine the reactions at the fixed supports A and B when the load P is applied EI is constant 12130 Determine the reactions at A and B Assume the support at A only exerts a moment on the beam EI is constant 12131 The beam is supported by the bolted supports at its ends When loaded these supports do not provide an actual fixed connection but instead allow a slight rotation before becoming fixed Determine the moment at the connections and the maximum deflection of the beam a 12129 The beam has a constant and is supported by the fixed wall at B and the rod AC If the rod has a cross sectional area and the material has a modulus of elasticity E2 determine the force in the rod A2 E1 I1 12132 The beam is supported by a pin at A a spring having a stiffness k at B and a roller at C Determine the force the spring exerts on the beam EI is constant A C D P B L 2 L 2 Prob 12127 P L A C B L Prob 12128 A L2 L1 B C w Prob 12129 L2 L2 A B P Prob 12130 P L 2 L 2 Prob 12131 A B L L k w C Prob 12132 The columns used to support this water tank are braced at their midheight in order to reduce their chance of buckling 657 CHAPTER OBJECTIVES In this chapter we will discuss the behavior of columns and indicate some of the methods used for their design The chapter begins with a general discussion of buckling followed by a determination of the axial load needed to buckle a socalled ideal column Afterwards a more realistic analysis is considered which accounts for any bending of the column Also inelastic buckling of a column is presented as a special topic At the end of the chapter we will discuss some of the methods used to design both concentrically and eccentrically loaded columns made of common engineering materials 131 Critical Load Whenever a member is designed it is necessary that it satisfy specific strength deflection and stability requirements In the preceding chapters we have discussed some of the methods used to determine a members strength and deflection while assuming that the member was always in stable equilibrium Some members however may be subjected to compressive loadings and if these members are long and slender the loading may be large enough to cause the member to deflect laterally or sidesway To be specific long slender members subjected to an axial compressive force are called columns and the lateral deflection that occurs is called buckling Quite often the buckling of a column can lead to a sudden and dramatic failure of a structure or mechanism and as a result special attention must be given to the design of columns so that they can safely support their intended loadings without buckling Buckling of Columns 13 660 CHAPTER 13 BUCKLING OF COLUMNS Fig 134 even greater load than Unfortunately however this loading may require the column to undergo a large deflection which is generally not tolerated in engineering structures or machines For example it may take only a few newtons of force to buckle a meterstick but the additional load it may support can be applied only after the stick undergoes a relatively large lateral deflection 132 Ideal Column with Pin Supports In this section we will determine the critical buckling load for a column that is pin supported as shown in Fig 134aThe column to be considered is an ideal column meaning one that is perfectly straight before loading is made of homogeneous material and upon which the load is applied through the centroid of the cross section It is further assumed that the material behaves in a linearelastic manner and that the column buckles or bends in a single plane In reality the conditions of column straightness and load application are never accomplished however the analysis to be performed on an ideal column is similar to that used to analyze initially crooked columns or those having an eccentric load application These more realistic cases will be discussed later in this chapter Since an ideal column is straight theoretically the axial load P could be increased until failure occurs by either fracture or yielding of the material However when the critical load is reached the column will be on the verge of becoming unstable so that a small lateral force F Fig 134b will cause the column to remain in the deflected position when F is removed Fig 134c Any slight reduction in the axial load P from will allow the column to straighten out and any slight increase in P beyond Pcr will cause further increases in lateral deflection Pcr Pcr Pcr Some slender pinconnected members used in moving machinery such as this short link are subjected to compressive loads and thus act as columns 13 P a L b Pcr F c Pcr 132 IDEAL COLUMN WITH PIN SUPPORTS 661 Fig 135 Whether or not a column will remain stable or become unstable when subjected to an axial load will depend on its ability to restore itself which is based on its resistance to bending Hence in order to determine the critical load and the buckled shape of the column we will apply Eq 1210 which relates the internal moment in the column to its deflected shapeie 131 Recall that this equation assumes that the slope of the elastic curve is small and that deflections occur only by bending When the column is in its deflected position Fig 135a the internal bending moment can be determined by using the method of sections The freebody diagram of a segment in the deflected position is shown in Fig 135b Here both the deflection and the internal moment M are shown in the positive direction according to the sign convention used to establish Eq 131 Moment equilibrium requires Thus Eq 131 becomes 132 This is a homogeneous secondorder linear differential equation with constant coefficients It can be shown by using the methods of differential equations or by direct substitution into Eq 132 that the general solution is 133 The two constants of integration are determined from the boundary conditions at the ends of the column Since at then And since at then This equation is satisfied if however then which is a trivial solution that requires the column to always remain straight even though the load may cause the column to become unstable The other possibility is for which is satisfied if A P EI L np sinaA P EI Lb 0 v 0 C1 0 C1 sinaA P EI Lb 0 x L v 0 C2 0 x 0 v 0 v C1 sinaA P EI xb C2 cosaA P EI xb d2v dx2 a P EIbv 0 EI d2v dx2 Pv M Pv v EI d2v dx2 M 13 L v v x x P P a P M x b P v 132 IDEAL COLUMN WITH PIN SUPPORTS 663 It is also important to realize that a column will buckle about the principal axis of the cross section having the least moment of inertia the weakest axis For example a column having a rectangular cross section like a meter stick as shown in Fig 136 will buckle about the aa axis not the bb axis As a result engineers usually try to achieve a balance keeping the moments of inertia the same in all directions Geometrically then circular tubes would make excellent columnsAlso square tubes or those shapes having are often selected for columns Summarizing the above discussion the buckling equation for a pinsupported long slender column can be rewritten and the terms defined as follows 135 where critical or maximum axial load on the column just before it begins to buckleThis load must not cause the stress in the column to exceed the proportional limit modulus of elasticity for the material least moment of inertia for the columns crosssectional area unsupported length of the column whose ends are pinned For purposes of design the above equation can also be written in a more useful form by expressing where A is the crosssectional area and r is the radius of gyration of the crosssectional areaThus or 136 Here critical stress which is an average normal stress in the column just before the column bucklesThis stress is an elastic stress and therefore modulus of elasticity for the material unsupported length of the column whose ends are pinned smallest radius of gyration of the column determined from where I is the least moment of inertia of the columnscrosssectional area A The geometric ratio in Eq 136 is known as the slenderness ratio It is a measure of the columns flexibility and as will be discussed later it serves to classify columns as long intermediate or short Lr r 2IA r L E scr sY scr scr p2E 1Lr22 a P A b cr p2E 1Lr22 Pcr p2E1Ar22 L2 I Ar2 L I E Pcr Pcr p2EI L2 Ix L Iy 13 Fig 136 P a a b b Typical interior steel pipe columns used to support the roof of a single story building 132 IDEAL COLUMN WITH PIN SUPPORTS 665 13 EXAMPLE 131 The A36 steel member shown in Fig 138 is to be used as a pinconnected column Determine the largest axial load it can support before it either begins to buckle or the steel yields W8 31 Fig 138 12 ft x x y y SOLUTION From the table in Appendix B the columns crosssectional area and moments of inertia are and By inspection buckling will occur about the yy axis Why Applying Eq 135 we have When fully loaded the average compressive stress in the column is Since this stress exceeds the yield stress 36 ksi the load P is determined from simple compression Ans In actual practice a factor of safety would be placed on this loading P 329 kip 36 ksi P 913 in2 scr Pcr A 512 kip 913 in2 561 ksi Pcr p2EI L2 p22911032 kipin21371 in42 12 ft112 inft22 512 kip Iy 371 in4 A 913 in2 Ix 110 in4 666 CHAPTER 13 BUCKLING OF COLUMNS Fig 139 The tubular columns used to support this water tank have been braced at three locations along their length to prevent them from buckling 133 Columns Having Various Types of Supports The Euler load was derived for a column that is pin connected or free to rotate at its endsOftentimeshowevercolumns may be supported in some other way For example consider the case of a column fixed at its base and free at the top Fig 139a As the column buckles the load displaces and at x the displacement is From the freebody diagram in Fig 139b the internal moment at the arbitrary section is Consequently the differential equation for the deflection curve is 137 Unlike Eq 132 this equation is nonhomogeneous because of the nonzero term on the right side The solution consists of both a complementary and a particular solution namely The constants are determined from the boundary conditions At so that Also At so that The deflection curve is therefore 138 Since the deflection at the top of the column is that is at we require The trivial solution indicates that no buckling occurs regardless of the load P Instead The smallest critical load occurs when so that 139 By comparison with Eq 135 it is seen that a column fixedsupported at its base and free at its top will support only onefourth the critical load that can be applied to a column pinsupported at both ends Pcr p2EI 4L2 n 1 cosaA P EI Lb 0 or A P EI L np 2 n 1 3 5 Á d 0 d cosaA P EI Lb 0 v d x L d v dc1 cosaA P EI xb d C1 0 dvdx 0 x 0 dv dx C1A P EI cosaA P EI xb C2A P EI sinaA P EI xb C2 d v 0 x 0 v C1 sinaA P EI xb C2 cosaA P EI xb d d2v dx2 P EI v P EI d EI d2v dx2 P1d v2 M P1d v2 v d 13 L v v x x P a d x P P M b v d 668 CHAPTER 13 BUCKLING OF COLUMNS 13 EXAMPLE 132 A steel column is 24 ft long and is fixed at its ends as shown in Fig 1311a Its loadcarrying capacity is increased by bracing it about the yy weak axis using struts that are assumed to be pin connected to its midheight Determine the load it can support so that the column does not buckle nor the material exceed the yield stress Take and SOLUTION The buckling behavior of the column will be different about the xx and yy axes due to the bracingThe buckled shape for each of these cases is shown in Figs1311b and 1311cFrom Fig1311bthe effective length for buckling about the xx axis is and from Fig 1311c for buckling about the yy axis The moments of inertia for a are found from the table in Appendix B We have Applying Eq 1311 1 2 By comparison buckling will occur about the yy axis The area of the cross section is so the average compressive stress in the column is Since this stress is less than the yield stress buckling will occur before the material yieldsThus Ans NOTE From Eq 1312 it can be seen that buckling will always occur about the column axis having the largest slenderness ratio since a large slenderness ratio will give a small critical stressThus using the data for the radius of gyration from the table in Appendix B we have Hence yy axis buckling will occur which is the same conclusion reached by comparing Eqs 1 and 2 a KL r b y 1008 in 146 in 690 a KL r b x 144 in 256 in 562 Pcr 263 kip scr Pcr A 2625 kip 443 in2 593 ksi 443 in2 1Pcr2y p2EIy 1KL2y 2 p22911032 ksi932 in4 11008 in22 2625 kip 1Pcr2x p2EIx 1KL2x 2 p22911032 ksi291 in4 1144 in22 4017 kip Iy 932 in4 Ix 291 in4 W6 15 07124 ft22 840 ft 1008 in 1KL2y 1KL2x 05124 ft2 12 ft 144 in sY 60 ksi Est 2911032 ksi W6 15 Fig 1311 a P x x y y 12 ft 12 ft 12 ft b xx axis buckling c yy axis buckling 840 ft 672 CHAPTER 13 BUCKLING OF COLUMNS 135 An A36 steel column has a length of 4 m and is pinned at both ends If the cross sectional area has the dimensions shown determine the critical load 136 Solve Prob 135 if the column is fixed at its bottom and pinned at its top 25 mm 10 mm 10 mm 25 mm 25 mm 25 mm Probs 1356 6 in 025 in 025 in 025 in 025 in 55 in Probs 1378 13 137 A column is made of A36 steel has a length of 20 ft and is pinned at both ends If the crosssectional area has the dimensions shown determine the critical load 138 A column is made of 2014T6 aluminum has a length of 30 ft and is fixed at its bottom and pinned at its top If the crosssectional area has the dimensions shown determine the critical load 139 The column is made of A36 steel and is fixed supported at its base If it is subjected to an axial load of determine the factor of safety with respect to buckling 1310 The column is made of A36 steel Determine the critical load if its bottom end is fixed supported and its top is free to move about the strong axis and is pinned about the weak axis W14 38 P 15 kip W14 38 1311 The A36 steel angle has a crosssectional area of and a radius of gyration about the x axis of and about the y axis of The smallest radius of gyration occurs about the z axis and is If the angle is to be used as a pinconnected 10ftlong column determine the largest axial load that can be applied through its centroid C without causing it to buckle rz 0644 in ry 0879 in rx 126 in A 248 in2 1312 An A36 steel column has a length of 15 ft and is pinned at both ends If the crosssectional area has the dimensions shown determine the critical load 20 ft P Probs 13910 x x y y z z C Prob 1311 8 in 05 in 6 in 05 in 05 in Prob 1312 674 CHAPTER 13 BUCKLING OF COLUMNS 1320 The is made of A36 steel and is used as a column that has a length of 15 ft If its ends are assumed pin supported and it is subjected to an axial load of 100 kip determine the factor of safety with respect to buckling 1321 The is made of A36 steel and is used as a column that has a length of 15 ft If the ends of the column are fixed supported can the column support the critical load without yielding W10 45 W10 45 13 15 ft P P Probs 132021 1322 The structural A36 steel column has a length of 12 ft If its bottom end is fixed supported while its top is free and it is subjected to an axial load of determine the factor of safety with respect to buckling 1323 The structural A36 steel column has a length of 12 ft If its bottom end is fixed supported while its top is free determine the largest axial load it can support Use a factor of safety with respect to buckling of 175 W12 87 P 380 kip W12 87 12 ft P Probs 132223 1324 An L2 tool steel link in a forging machine is pin connected to the forks at its ends as shown Determine the maximum load P it can carry without buckling Use a factor of safety with respect to buckling of Note from the figure on the left that the ends are pinned for buckling whereas from the figure on the right the ends are fixed FS 175 P P P P 24 in 15 in 05 in Prob 1324 1325 The is used as a structural A36 steel column that can be assumed pinned at both of its ends Determine the largest axial force P that can be applied without causing it to buckle W14 30 25 ft P Prob 1325 676 CHAPTER 13 BUCKLING OF COLUMNS 1332 The members of the truss are assumed to be pin connected If member AC is an A36 steel rod of 2 in diameter determine the maximum load P that can be supported by the truss without causing the member to buckle D C B P 3 ft A 4 ft Prob 1332 13 6 m 3 m 4 m y x 40 mm 40 mm C B A 40 mm w 1333 The steel bar AB of the frame is assumed to be pin connected at its ends for yy axis buckling If determine the factor of safety with respect to buckling about the yy axis due to the applied loading sY 360 MPa Est 200 GPa w 3 kNm 1336 If load C has a mass of 500 kg determine the required minimum diameter of the solid L2steel rod AB to the nearest mm so that it will not buckle Use against buckling 1337 If the diameter of the solid L2steel rod AB is 50 mm determine the maximum mass C that the rod can support without buckling Use against buckling FS 2 FS 2 1338 The members of the truss are assumed to be pin connected If member GF is an A36 steel rod having a diameter of 2 in determine the greatest magnitude of load P that can be supported by the truss without causing this member to buckle 1339 The members of the truss are assumed to be pin connected If member AG is an A36 steel rod having a diameter of 2 in determine the greatest magnitude of load P that can be supported by the truss without causing this member to buckle A B D C E 2 m 2 m 15 m P Prob 1333 1334 The members of the truss are assumed to be pin connected If member AB is an A36 steel rod of 40 mm diameter determine the maximum force P that can be supported by the truss without causing the member to buckle 1335 The members of the truss are assumed to be pin connected If member CB is an A36 steel rod of 40 mm diameter determine the maximum load P that can be supported by the truss without causing the member to buckle B C D 45 A 60 4 m Probs 133637 G A B D C F P 16 ft 16 ft 12 ft P 16 ft E H Probs 133839 Probs 133435 134 THE SECANT FORMULA 679 or This equation is similar to Eq 137 and has a general solution consisting of the complementary and particular solutions namely 1314 To evaluate the constants we must apply the boundary conditions At so And at which gives Since and we have Hence the deflection curve Eq 1314 can be written as 1315 Maximum Deflection Due to symmetry of loading both the maximum deflection and maximum stress occur at the columns midpointTherefore when so 1316 Notice that if e approaches zero then approaches zero However if the terms in the brackets approach infinity as e approaches zero then will have a nonzero value Mathematically this would represent the behavior of an axially loaded column at failure when subjected to the critical load Therefore to find we require 1317 which is the same result found from the Euler formula Eq 135 If Eq 1316 is plotted as load P versus deflection for various values of eccentricity e the family of colored curves shown in Fig 1314 results Here the critical load becomes an asymptote to the curves and of vmax Pcr p2EI L2 A Pcr EI L 2 p 2 secA Pcr EI L 2 b q Pcr Pcr vmax vmax vmax ecsecaA P EI L 2 b 1 d v vmax x L2 v ectanaA P EI L 2 b sinaA P EI xb cosaA P EI xb 1 d C1 e tanaA P EI L 2 b 2 sin2PEI L2 cos 2PEI L2 sin2PEI L 1 cos2PEI L 2 sin22PEI L2 C1 e1 cos2PEI L sin2PEI L x L v 0 C2 e v 0 x 0 v C1 sin A P EI x C2 cos A P EI x e d2v dx2 P EI v P EI e 13 135 INELASTIC BUCKLING 687 13 1346 Determine the load P required to cause the A36 steel column to fail either by buckling or by yielding The column is fixed at its base and free at its top W8 15 1349 The tube is made of copper and has an outer diameter of 35 mm and a wall thickness of 7 mm Using a factor of safety with respect to buckling and yielding of determine the allowable eccentric load P The tube is pin supported at its ends GPa 750 MPa 1350 The tube is made of copper and has an outer diameter of 35 mm and a wall thickness of 7 mm Using a factor of safety with respect to buckling and yielding of determine the allowable eccentric load P that it can support without failure The tube is fixed supported at its ends GPa sY 750 MPa Ecu 120 FS 25 sY Ecu 120 FS 25 1351 The wood column is fixed at its base and can be assumed pin connected at its top Determine the maximum eccentric load P that can be applied without causing the column to buckle or yield 1352 The wood column is fixed at its base and can be assumed fixed connected at its top Determine the maximum eccentric load P that can be applied without causing the column to buckle or yield sY 8 ksi Ew 1811032 ksi sY 8 ksi Ew 1811032 ksi PROBLEMS 8 ft 1 in P Prob 1346 1347 The hollow red brass C83400 copper alloy shaft is fixed at one end but free at the other end Determine the maximum eccentric force P the shaft can support without causing it to buckle or yield Also find the corresponding maximum deflection of the shaft 1348 The hollow red brass C83400 copper alloy shaft is fixed at one end but free at the other end If the eccentric force is applied to the shaft as shown determine the maximum normal stress and the maximum deflection P 5 kN P a a 150 mm 2 m 30 mm 20 mm Section a a Probs 134748 P 10 ft 10 in 4 in x y P x y Probs 135152 2 m 14 mm P P Probs 134950 688 CHAPTER 13 BUCKLING OF COLUMNS 1353 The A36steel column is fixed at its base Its top is constrained to rotate about the yy axis and free to move along the yy axis Also the column is braced along the xx axis at its midheight Determine the allowable eccentric force P that can be applied without causing the column either to buckle or yield Use against buckling and against yielding 1354 The A36steel column is fixed at its base Its top is constrained to rotate about the yy axis and free to move along the yy axis Also the column is braced along the xx axis at its midheight If determine the maximum normal stress developed in the column P 25 kN W200 22 FS 15 FS 2 W200 22 1357 The A36steel column is fixed at its base Its top is constrained to rotate about the yy axis and free to move along the yy axis If determine the allowable eccentric force P that can be applied without causing the column either to buckle or yield Use against buckling and against yielding 1358 The A36steel column is fixed at its base Its top is constrained to rotate about the yy axis and free to move along the yy axis Determine the force P and its eccentricity e so that the column will yield and buckle simultaneously W250 28 FS 15 FS 2 e 350 mm W250 28 1355 The wood column is fixed at its base and its top can be considered pinned If the eccentric force is applied to the column investigate whether the column is adequate to support this loading without buckling or yieldingTake and sY 15 MPa E 10 GPa P 10 kN 13 x x y y P 5 m 5 m 100 mm Probs 135354 P 5 m 150 mm x 75 mm 75 mm 25 mm 25 mm xy Probs 135556 x x y y P 6 m e Probs 135758 1356 The wood column is fixed at its base and its top can be considered pinned Determine the maximum eccentric force P the column can support without causing it to either buckle or yield Take and sY 15 MPa E 10 GPa 135 INELASTIC BUCKLING 689 1359 The steel column supports the two eccentric loadings If it is assumed to be pinned at its top fixed at the bottom and fully braced against buckling about the yy axis determine the maximum deflection of the column and the maximum stress in the column 1360 The steel column supports the two eccentric loadings If it is assumed to be fixed at its top and bottom and braced against buckling about the yy axis determine the maximum deflection of the column and the maximum stress in the column sY 360 MPa Est 200 GPa sY 360 MPa Est 200 GPa 1361 The A36steel column is pinned at its top and fixed at its base Also the column is braced along its weak axis at midheight If investigate whether the column is adequate to support this loading Use against buckling and against yielding 1362 The A36steel column is pinned at its top and fixed at its base Also the column is braced along its weak axis at midheight Determine the allowable force P that the column can support without causing it either to buckle or yield Use against buckling and FS 15 against yielding FS 2 W250 45 FS 15 FS 2 P 250 kN W250 45 1363 The structural A36 steel member is used as a 20ftlong column that is assumed to be fixed at its top and fixed at its bottom If the 15kip load is applied at an eccentric distance of 10 in determine the maximum stress in the column 1364 The structural A36 steel member is used as a column that is assumed to be fixed at its top and pinned at its bottom If the 15kip load is applied at an eccentric distance of 10 in determine the maximum stress in the column W14 26 W14 26 13 50 kN 80 mm 6 m 120 mm 130 kN 100 mm 10 mm 10 mm 10 mm 100 mm y y x x Probs 135960 4 m 250 mm 250 mm 4 m P P 4 Probs 136162 15 kip 10 in 20 ft Probs 136364 690 CHAPTER 13 BUCKLING OF COLUMNS 1365 Determine the maximum eccentric load P the 2014T6aluminumalloy strut can support without causing it either to buckle or yield The ends of the strut are pinconnected 1366 The structural A36 steel column is fixed at its bottom and free at its top If it is subjected to the eccentric load of 75 kip determine the factor of safety with respect to either the initiation of buckling or yielding 1367 The structural A36 steel column is fixed at its bottom and pinned at its top If it is subjected to the eccentric load of 75 kip determine if the column fails by yielding The column is braced so that it does not buckle about the yy axis W8 48 W8 48 1370 A column of intermediate length buckles when the compressive stress is 40 ksi If the slenderness ratio is 60 determine the tangent modulus 1371 The 6ftlong column has the cross section shown and is made of material which has a stressstrain diagram that can be approximated as shown If the column is pinned at both ends determine the critical load for the column 1372 The 6ftlong column has the cross section shown and is made of material which has a stressstrain diagram that can be approximated as shown If the column is fixed at both ends determine the critical load for the column Pcr Pcr 1368 Determine the load P required to cause the steel structural A36 steel column to fail either by buckling or by yielding The column is fixed at its bottom and the cables at its top act as a pin to hold it 1369 Solve Prob 1368 if the column is an A36 steel section W12 16 W12 50 13 3 m 100 mm 150 mm 100 mm 50 mm 150 mm a a P P Section a a Prob 1365 12 ft 8 in y y x 75 kip Probs 136667 25 ft 2 in P Probs 136869 ksi P inin 55 25 0001 0004 3 in 5 in 05 in 05 in 05 in s Probs 137172 136 DESIGN OF COLUMNS FOR CONCENTRIC LOADING 693 13 Steel Columns Columns made of structural steel can be designed on the basis of formulas proposed by the Structural Stability Research Council SSRCFactors of safety have been applied to these formulas and adopted as specifications for building construction by the American Institute of Steel Construction AISC Basically these specifications provide two formulas for column designeach of which gives the maximum allowable stress in the column for a specific range of slenderness ratios For long columns the Euler formula is proposed ie Application of this formula requires that a factor of safety be appliedThus for design 1321 As stated this equation is applicable for a slenderness ratio bounded by 200 and A specific value of is obtained by requiring the Euler formula to be used only for elastic material behaviorThrough experiments it has been determined that compressive residual stresses can exist in rolledformed steel sections that may be as much as onehalf the yield stress Consequently if the stress in the Euler formula is greater than the equation will not apply Therefore the value of is determined as follows 1322 Columns having slenderness ratios less than are designed on the basis of an empirical formula that is parabolic and has the form Since there is more uncertainty in the use of this formula for longer columns it is divided by a factor of safety defined as follows Here it is seen that at and increases to at 1KLr2c Hence for design purposes FS 23 12 L 192 KLr 0 FS 5 3 L 167 FS 5 3 3 8 1KLr2 1KLr2c 1KLr23 81KLr2c 3 smax B1 1KLr22 21KLr2c 2RsY 1KLr2c 1 2 sY p2E 1KLr2c 2 or a KL r b c B 2p2E sY 1KLr2c 1 2 sY 1KLr2c 1KLr2c sallow 12p2E 231KLr22 a KL r b c KL r 200 FS 23 12 L 192 p2E1KLr22 smax The current AISC code enables engineers to use one of two methods for design namely Load and Resistance Factor DesignandAllowable Stress DesignThe latter is explained here 06 0261 Eq 1323 Eq 1321 KL r c KL r 0 sallow sY 1323 sallow B1 1KLr22 21KLr2c 2RsY 1532 13821KLr21KLr2c C1KLr2381KLr2c 3D Equations 1321 and 1323 are plotted in Fig1321When applying any of these equations either FPS or SI units can be used for the calculations Fig 1321 694 CHAPTER 13 BUCKLING OF COLUMNS 13 Aluminum Columns Column design for structural aluminum is specified by the Aluminum Association using three equations each applicable for a specific range of slenderness ratios Since several types of aluminum alloy exist there is a unique set of formulas for each type For a common alloy 2014T6 used in building construction the formulas are 1324 1325 1326 These equations are plotted in Fig 1322 As shown the first two represent straight lines and are used to model the effects of columns in the short and intermediate rangeThe third formula has the same form as the Euler formula and is used for long columns Timber Columns Columns used in timber construction are designed on the basis of formulas published by the National Forest Products Association NFPA or the American Institute of Timber Construction AITC For example the NFPA formulas for the allowable stress in short intermediate and long columns having a rectangular cross section of dimensions b and d where d is the smallest dimension of the cross section are 1327 1328 1329 Here wood has a modulus of elasticity of and an allowable compressive stress of 12 ksi parallel to the grain In particular Eq 1329 is simply Eulers equation having a factor of safety of 3These three equations are plotted in Fig 1323 Ew 1811032 ksi sallow 540 ksi 1KLd22 26 6 KL d 50 sallow 120c1 1 3 a KLd 260 b 2 d ksi 11 6 KL d 26 sallow 120 ksi 0 KL d 11 sallow 54 000 ksi 1KLr22 55 KL r sallow c307 023a KL r b d ksi 12 6 KL r 6 55 sallow 28 ksi 0 KL r 12 28 18 12 55 KL r Eq 1324 Eq 1325 Eq 1326 0 sallowksi 12 KL d sallowksi Eq 1328 08 0216 11 26 50 Eq 1327 Eq 1329 0 Fig 1322 Fig 1323 136 DESIGN OF COLUMNS FOR CONCENTRIC LOADING 695 13 Procedure for Analysis Column Analysis When using any formula to analyze a column that is to find its allowable load it is first necessary to calculate the slenderness ratio in order to determine which column formula applies Once the average allowable stress has been calculated the allowable load on the column is determined from Column Design If a formula is used to design a column that is to determine the columns crosssectional area for a given loading and effective length then a trialandcheck procedure generally must be followed when the column has a composite shape such as a wideflange section One possible way to apply a trialandcheck procedure would be to assume the columns crosssectional area and calculate the corresponding stress Also use an appropriate design formula to determine the allowable stress From this calculate the required column area If the design is safe When making the comparison it is practical to require to be close to but greater than usually within 23A redesign is necessary if Whenever a trialandcheck procedure is repeated the choice of an area is determined by the previously calculated required area In engineering practice this method for design is usually shortened through the use of computer software or published tables and graphs A 6 Areqd Areqd A A 7 Areqd Areqd Psallow sallow s PA A P sallowA These timber columns can be considered pinned at their bottom and fixed connected to the beams at their tops 696 CHAPTER 13 BUCKLING OF COLUMNS 13 EXAMPLE 136 An A36 steel member is used as a pinsupported column Fig 1324 Using the AISC column design formulas determine the largest load that it can safely support SOLUTION The following data for a is taken from the table in Appendix B Since for both x and y axis buckling the slenderness ratio is largest if is usedThus From Eq 1322 we have Here so Eq 1323 applies The allowable load P on the column is therefore Ans P 476 kip 1617 kipin2 P 294 in2 sallow P A 1617 ksi 1 17245222112612236 ksi 1532 13821724512612 172452381126123 sallow B1 1KLr22 21KLr2c 2RsY 1532 13821KLr21KLr2c C1KLr2381KLrc23D 0 6 KLr 6 1KLr2c 1261 B 2p22911032 ksi 36 ksi a KL r b c B 2p2E sY KL r 1116 ft2112 inft2 265 in 7245 ry K 1 A 294 in2 rx 460 in ry 265 in W10 100 W10 100 P x x y y 16 ft P Fig 1324 698 CHAPTER 13 BUCKLING OF COLUMNS 13 EXAMPLE 138 A bar having a length of 30 in is used to support an axial compressive load of 12 kip Fig 1326 It is pin supported at its ends and made of a 2014T6 aluminum alloy Determine the dimensions of its cross sectional area if its width is to be twice its thickness SOLUTION Since is the same for both x and y axis buckling the larger slenderness ratio is determined using the smaller radius of gyration ie using 1 Here we must apply Eq 1324 1325 or 1326 Since we do not as yet know the slenderness ratio we will begin by using Eq 1324 Checking the slenderness ratio we have Try Eq 1326 which is valid for Ans From Eq 1 OK NOTE It would be satisfactory to choose the cross section with dimensions 1 in by 2 in KL r 1039 105 993 7 55 b 105 in 12 2b1b2 54 000 11039b22 P A 54 000 ksi 1KLr22 KLr Ú 55 KL r 1039 0463 2245 7 12 b 0463 in 12 kip 2b1b2 28 kipin2 P A 28 ksi KL ry KL 2IyA 11302 2111222b1b322b1b2 1039 b Imin Iy KL 30 in Fig 1326 12 kip 30 in 12 kip 2b b y x 136 DESIGN OF COLUMNS FOR CONCENTRIC LOADING 699 13 EXAMPLE 139 A board having crosssectional dimensions of 55 in by 15 in is used to support an axial load of 5 kip Fig 1327 If the board is assumed to be pin supported at its top and bottom determine its greatest allowable length L as specified by the NFPA Fig 1327 SOLUTION By inspection the board will buckle about the y axis In the NFPA equations Assuming that Eq 1329 applies we have Ans Here Since the solution is valid 26 6 KLd 50 KL d 11448 in2 15 in 298 L 448 in 5 kip 155 in2115 in2 540 ksi 11 L15 in22 P A 540 ksi 1KLd22 d 15 in 5 kip L 55 in 15 in y x 5 kip 700 CHAPTER 13 BUCKLING OF COLUMNS 13 1378 Determine the largest length of a structural A36 steel rod if it is fixed supported and subjected to an axial load of 100 kN The rod has a diameter of 50 mm Use the AISC equations 1379 Determine the largest length of a structural steel column if it is pin supported and subjected to an axial load of 290 kip Use the AISC equations 1380 Determine the largest length of a structural A36 steel section if it is pin supported and is subjected to an axial load of 28 kip Use the AISC equations 1381 Using theAISC equationsselect fromAppendix B the lightestweight structural A36 steel column that is 14 ft long and supports an axial load of 40 kipThe ends are pinned Take 1382 Using the AISC equations select from Appendix B the lightestweight structural A36 steel column that is 12 ft long and supports an axial load of 40 kipThe ends are fixed Take 1383 Using the AISC equations select from Appendix B the lightestweight structural A36 steel column that is 24 ft long and supports an axial load of 100 kip The ends are fixed 1384 Using the AISC equations select from Appendix B the lightestweight structural A36 steel column that is 30 ft long and supports an axial load of 200 kip The ends are fixed 1385 A A36steel column of 30ft length is pinned at both ends and braced against its weak axis at mid height Determine the allowable axial force P that can be safely supported by the column Use the AISC column design formulas 1386 Check if a column can safely support an axial force of The column is 20 ft long and is pinned at both ends and braced against its weak axis at midheight It is made of steel having and sY 50 ksi Use the AISC column design formulas E 29103 ksi P 250 kip W10 39 W8 24 sY 50 ksi sY 50 ksi W10 12 sY 50 ksi Est 29103 ksi W10 45 1387 A 5ftlong rod is used in a machine to transmit an axial compressive load of 3 kip Determine its smallest diameter if it is pin connected at its ends and is made of a 2014T6 aluminum alloy 1388 Check if a column can safely support an axial force of The column is 15 ft long and is pinned at both of its ends It is made of steel having and Use the AISC column design formulas 1389 Using the AISC equations check if a column having the cross section shown can support an axial force of 1500 kNThe column has a length of 4 m is made from A36 steel and its ends are pinned sY 50 ksi E 29103 ksi P 200 kip W10 45 PROBLEMS 350 mm 10 mm 300 mm 20 mm 20 mm Prob 1389 1390 The A36steel tube is pinned at both ends If it is subjected to an axial force of 150 kN determine the maximum length that the tube can safely support using the AISC column design formulas 100 mm 80 mm Prob 1390 136 DESIGN OF COLUMNS FOR CONCENTRIC LOADING 701 13 1391 The bar is made of a 2014T6 aluminum alloy Determine its smallest thickness b if its width is 5b Assume that it is pin connected at its ends 1392 The bar is made of a 2014T6 aluminum alloy Determine its smallest thickness b if its width is 5b Assume that it is fixed connected at its ends 5b b 8 ft 600 lb 600 lb Probs 139192 100 mm 15 mm 170 mm 15 mm 15 mm Probs 139394 1393 The 2014T6 aluminum column of 3m length has the cross section shown If the column is pinned at both ends and braced against the weak axis at its midheight determine the allowable axial force P that can be safely supported by the column 1394 The 2014T6 aluminum column has the cross section shown If the column is pinned at both ends and subjected to an axial force determine the maximum length the column can have to safely support the loading P 100 kN 1395 The 2014T6 aluminum hollow section has the cross section shown If the column is 10 ft long and is fixed at both ends determine the allowable axial force P that can be safely supported by the column 1396 The 2014T6 aluminum hollow section has the cross section shown If the column is fixed at its base and pinned at its top and is subjected to the axial force determine the maximum length of the column for it to safely support the load P 100 kip 4 in 3 in Probs 139596 P 6 in y x y x 6 in P 10 ft Probs 13979899 1397 The tube is 025 in thick is made of a 2014T6 aluminum alloy and is fixed at its bottom and pinned at its top Determine the largest axial load that it can support 1398 The tube is 025 in thick is made of a 2014T6 aluminum alloy and is fixed connected at its ends Determine the largest axial load that it can support 1399 The tube is 025 in thick is made of 2014T6 aluminum alloy and is pin connected at its ends Determine the largest axial load it can support 702 CHAPTER 13 BUCKLING OF COLUMNS 13 13100 A rectangular wooden column has the cross section shown If the column is 6 ft long and subjected to an axial force of determine the required minimum dimension a of its crosssectional area to the nearest so that the column can safely support the loading The column is pinned at both ends 13101 A rectangular wooden column has the cross section shown If and the column is 12 ft long determine the allowable axial force P that can be safely supported by the column if it is pinned at its top and fixed at its base 13102 A rectangular wooden column has the cross section shown If and the column is subjected to an axial force of determine the maximum length the column can have to safely support the load The column is pinned at its top and fixed at its base P 15 kip a 3 in a 3 in 1 16 in P 15 kip 2a a Probs 13100101102 14 ft a Prob 13103 13103 The timber column has a square cross section and is assumed to be pin connected at its top and bottom If it supports an axial load of 50 kip determine its smallest side dimension a to the nearest Use the NFPA formulas 1 2 in 13104 The wooden column shown is formed by gluing together the boards If the column is pinned at both ends and is subjected to an axial load determine the required number of boards needed to form the column in order to safely support the loading P 20 kip 6 in 05 in 9 ft 6 in 05 in P P Prob 13104 P L 4 in 2 in x x y y Probs 13105106 13105 The column is made of wood It is fixed at its bottom and free at its top Use the NFPA formulas to determine its greatest allowable length if it supports an axial load of 13106 The column is made of wood It is fixed at its bottom and free at its top Use the NFPA formulas to determine the largest allowable axial load P that it can support if it has a length L 4 ft P 2 kip 704 CHAPTER 13 BUCKLING OF COLUMNS 13 The total area A for the column needed to resist both the axial load and moment requires that or 1331 Here axial stress caused by the force P and determined from where A is the crosssectional area of the column bending stress caused by an eccentric load or applied moment M is found from where I is the moment of inertia of the crosssectional area calculated about the bending or centroidal axis allowable axial stress as defined by formulas given in Sec 136 or by other design code specifications For this purpose always use the largest slenderness ratio for the column regardless of the axis about which the column experiences bending allowable bending stress as defined by code specifications Notice that if the column is subjected only to an axial load then the bendingstress ratio in Eq 1331 would be equal to zero and the design will be based only on the allowable axial stress Likewise when no axial load is present the axialstress ratio is zero and the stress requirement will be based on the allowable bending stress Hence each stress ratio indicates the contribution of axial load or bending moment Since Eq 1331 shows how these loadings interact this equation is sometimes referred to as the interaction formula This design approach requires a trialandcheck procedure where it is required that the designer pick an available column and then check to see if the inequality is satisfied If it is not a larger section is then picked and the process repeated An economical choice is made when the left side is close to but less than 1 The interaction method is often specified in codes for the design of columns made of steel aluminum or timber In particular for allowable stress design the American Institute of Steel Construction specifies the use of this equation only when the axialstress ratio For other values of this ratio a modified form of Eq 1331 is used sa1sa2allow 015 1sb2allow 1sa2allow sb McI sb sb sa PA sa sa 1sa2allow sb 1sb2allow 1 PA 1sa2allow McAr2 1sb2allow 1 Aa Ab P 1sa2allow Mc 1sb2allowr2 A Ab Mc 1sb2allowr2 Typical example of a column used to support an eccentric roof loading 137 DESIGN OF COLUMNS FOR ECCENTRIC LOADING 705 13 EXAMPLE 1310 The column in Fig 1329 is made of aluminum alloy 2014T6 and is used to support an eccentric load P Determine the maximum magnitude of P that can be supported if the column is fixed at its base and free at its top Use Eq 1330 80 in P 2 in 1 in 2 in 2 in Fig 1329 SOLUTION From Fig 1310b The largest slenderness ratio for the column is therefore By inspection Eq 1326 must be used Thus The maximum compressive stress in the column is determined from the combination of axial load and bendingWe have Assuming that this stress is uniform over the cross section we require Ans P 225 kip 07031 03125P sallow smax 03125P P 2 in14 in2 P11 in212 in2 1112212 in214 in23 smax P A 1Pe2c I sallow 54 000 ksi 1KLr22 54 000 ksi 1277122 07031 ksi 12771 7 552 KL r 2180 in2 21112214 in212 in2312 in2 4 in 2771 K 2 137 DESIGN OF COLUMNS FOR ECCENTRIC LOADING 707 13 EXAMPLE 1312 The timber column in Fig 1331 is made from two boards nailed together so that the cross section has the dimensions shown If the column is fixed at its base and free at its top use Eq 1330 to determine the eccentric load P that can be supported 60 in x y 3 in P 3 in 1 in 3 in Fig 1331 SOLUTION From Fig 1310b Here we must calculate to determine which equation from Eqs 1327 through 1329 should be used Since is determined using the largest slenderness ratio we choose This is done to make this ratio as large as possible and thereby yields the lowest possible allowable axial stress We have Since the allowable axial stress is determined using Eq 1329Thus Applying Eq 1330 with we have Ans P 122 kip 03375 ksi P 3 in16 in2 P14 in213 in2 1112213 in216 in23 sallow P A Mc I sallow smax sallow 540 ksi 1KLd22 540 ksi 14022 03375 ksi 26 6 KLd 6 50 KL d 2160 in2 3 in 40 d 3 in sallow KLd K 2 708 CHAPTER 13 BUCKLING OF COLUMNS 13 12 ft 80 kip y y x x 10 in P z Probs 13107108 12 ft y x y x P P M M Probs 13109110 P 10 ft 40 kip 16 in Probs 13111112 13107 The structural A36 steel column supports an axial load of 80 kip in addition to an eccentric load P Determine the maximum allowable value of P based on the AISC equations of Sec 136 and Eq 1330 Assume the column is fixed at its base and at its top it is free to sway in the xz plane while it is pinned in the yz plane 13108 The structural A36 steel column supports an axial load of 80 kip in addition to an eccentric load of Determine if the column fails based on the AISC equations of Sec 136 and Eq 1330Assume that the column is fixed at its base and at its top it is free to sway in the xz plane while it is pinned in the yz plane P 60 kip W12 45 W14 53 13111 The structural A36 steel column is fixed at its bottom and free at its top Determine the greatest eccentric load P that can be applied using Eq 1330 and the AISC equations of Sec 136 13112 The structural A36 steel column is fixed at its bottom and free at its top If it is subjected to a load of determine if it is safe based on the AISC equations of Sec 136 and Eq 1330 P 2 kip W10 45 W14 43 PROBLEMS 13109 The structural A36 steel column is fixed at its top and bottom If a horizontal load not shown causes it to support end moments of determine the maximum allowable axial force P that can be applied Bending is about the xx axis Use the AISC equations of Sec 136 and Eq 1330 13110 The column is fixed at its top and bottom If a horizontal load not shown causes it to support end moments of determine the maximum allowable axial force P that can be applied Bending is about the xx axis Use the interaction formula with 1sb2allow 24 ksi M 15 kip ft W14 22 M 10 kip ft W14 22 137 DESIGN OF COLUMNS FOR ECCENTRIC LOADING 709 13117 A 16ftlong column is made of aluminum alloy 2014T6 If it is fixed at its top and bottom and a compressive load P is applied at point A determine the maximum allowable magnitude of P using the equations of Sec 136 and Eq 1330 13118 A 16ftlong column is made of aluminum alloy 2014T6 If it is fixed at its top and bottom and a compressive load P is applied at point A determine the maximum allowable magnitude of P using the equations of Sec 136 and the interaction formula with 1sb2allow 20 ksi 13 13113 The A36steel column is fixed at its base Its top is constrained to move along the xx axis but free to rotate about and move along the yy axis Determine the maximum eccentric force P that can be safely supported by the column using the allowable stress method 13114 The A36steel column is fixed at its base Its top is constrained to move along the xx axis but free to rotate about and move along the yy axis Determine the maximum eccentric force P that can be safely supported by the column using an interaction formula The allowable bending stress is 13115 The A36steel column is fixed at its base Its top is constrained to move along the xx axis but free to rotate about and move along the yy axis If the eccentric force is applied to the column investigate if the column is adequate to support the loading Use the allowable stress method 13116 The A36steel column is fixed at its base Its top is constrained to move along the xx axis but free to rotate about and move along the yy axis If the eccentric force is applied to the column investigate if the column is adequate to support the loading Use the interaction formulaThe allowable bending stress is sballow 15 ksi P 15 kip W12 50 P 15 kip W12 50 sballow 15 ksi W10 45 W10 45 x x y y P 24 ft 12 in Probs 13113114115116 8 in 05 in 05 in 425 in y x x 8 in y P A 05 in Probs 13117118 13119 The 2014T6 hollow column is fixed at its base and free at its top Determine the maximum eccentric force P that can be safely supported by the column Use the allowable stress method The thickness of the wall for the section is 13120 The 2014T6 hollow column is fixed at its base and free at its top Determine the maximum eccentric force P that can be safely supported by the column Use the interaction formula The allowable bending stress is The thickness of the wall for the section is t 05 in sballow 30 ksi t 05 in P 6 in 6 in 3 in 8 ft Probs 13119120 710 CHAPTER 13 BUCKLING OF COLUMNS 13 13121 The 10ftlong bar is made of aluminum alloy 2014T6 If it is fixed at its bottom and pinned at the top determine the maximum allowable eccentric load P that can be applied using the formulas in Sec 136 and Eq 1330 13122 The 10ftlong bar is made of aluminum alloy 2014T6 If it is fixed at its bottom and pinned at the top determine the maximum allowable eccentric load P that can be applied using the equations of Sec 136 and the interaction formula with 1sb2allow 18 ksi 3 in 15 in 15 in x x y y 2 in 2 in P Probs 13121122 13123 The rectangular wooden column can be considered fixed at its base and pinned at its top Also the column is braced at its midheight against the weak axis Determine the maximum eccentric force P that can be safely supported by the column using the allowable stress method 13124 The rectangular wooden column can be considered fixed at its base and pinned at its top Also the column is braced at its midheight against the weak axis Determine the maximum eccentric force P that can be safely supported by the column using the interaction formula The allowable bending stress is sballow 15 ksi 5 ft 5 ft P 6 in 6 in 6 in 3 in Probs 13123124 13125 The 10indiameter utility pole supports the transformer that has a weight of 600 lb and center of gravity at G If the pole is fixed to the ground and free at its top determine if it is adequate according to the NFPA equations of Sec 136 and Eq 1330 G 18 ft 15 in Prob 13125 13126 Using the NFPA equations of Sec 136 and Eq 1330 determine the maximum allowable eccentric load P that can be applied to the wood columnAssume that the column is pinned at both its top and bottom 13127 Using the NFPA equations of Sec 136 and Eq 1330 determine the maximum allowable eccentric load P that can be applied to the wood columnAssume that the column is pinned at the top and fixed at the bottom 6 in 12 ft P 075 in 3 in Probs 13126127 CHAPTER REVIEW 711 13 CHAPTER REVIEW Pcr Buckling is the sudden instability that occurs in columns or members that support an axial compressive load The maximum axial load that a member can support just before buckling is called the critical load Pcr The critical load for an ideal column is determined from Eulers formula where for pin supports for fixed supports for a pin and a fixed supportand for a fixed support and a free end K 2 K 07 K 05 K 1 If the axial loading is applied eccentrically to the column then the secant formula can be used to determine the maximum stress in the column When the axial load causes yielding of the material then the tangent modulus should be used with Eulers formula to determine the critical load for the column This is referred to as Engessers equation Empirical formulas based on experimental data have been developed for use in the design of steel aluminum and timber columns smax P A c1 ec r2 seca L 2rA P EA b d Pcr p2EI 1KL22 scr p2Et 1KLr22 712 CHAPTER 13 BUCKLING OF COLUMNS 13 13128 The wood column is 4 m long and is required to support the axial load of 25 kN If the cross section is square determine the dimension a of each of its sides using a factor of safety against buckling of The column is assumed to be pinned at its top and bottom Use the Euler equation and sY 10 MPa Ew 11 GPa FS 25 13130 Determine the maximum intensity w of the uniform distributed load that can be applied on the beam without causing the compressive members of the supporting truss to buckle The members of the truss are made from A36steel rods having a 60mm diameter Use FS 2 against buckling REVIEW PROBLEMS 25 kN 4 m a a Prob 13128 P k k B A C L 2 L 2 Prob 13129 13129 If the torsional springs attached to ends A and C of the rigid members AB and BC have a stiffness k determine the critical load Pcr 2 m 36 m 15 m B C A D w Prob 13130 13131 The steel column supports an axial load of 60 kip in addition to an eccentric load P Determine the maximum allowable value of P based on the AISC equations of Sec 136 and Eq 1330 Assume that in the xz plane and in the yz plane sY 50 ksi Ky 20 Est 2911032 ksi Kx 10 W10 45 10 ft 60 kip y x y x 8 in P z Prob 13131 REVIEW PROBLEMS 713 13135 The A36steel column can be considered pinned at its top and fixed at its base Also the column is braced at its midheight against the weak axis Determine the maximum axial load the column can support without causing it to buckle W200 46 13 13132 The A36steel column can be considered pinned at its top and fixed at its base Also it is braced at its midheight along the weak axis Investigate whether a section can safely support the loading shown Use the allowable stress method 13133 The A36steel column can be considered pinned at its top and fixed at its base Also it is braced at its midheight along the weak axis Investigate whether a section can safely support the loading shown Use the interaction formulaThe allowable bending stress is sballow 100 MPa W250 45 W250 45 P P 5 ft 2 in 05 in Prob 13134 6 m 6 m Prob 13135 20 mm 4 m P A 10 mm 100 mm 100 mm 10 mm 150 mm A 100 mm 10 mm Probs 13136137 45 m 45 m 600 mm 40 kN 10 kN Probs 13132133 13134 The member has a symmetric cross section If it is pin connected at its ends determine the largest force it can support It is made of 2014T6 aluminum alloy 13136 The structural A36 steel column has the cross section shown If it is fixed at the bottom and free at the top determine the maximum force P that can be applied at A without causing it to buckle or yield Use a factor of safety of 3 with respect to buckling and yielding 13137 The structural A36 steel column has the cross section shown If it is fixed at the bottom and free at the top determine if the column will buckle or yield when the load Use a factor of safety of 3 with respect to buckling and yielding P 10 kN As piles are driven in place their ends are subjected to impact loading The nature of impact and the energy derived from it must be understood in order to determine the stress developed within the pile 715 CHAPTER OBJECTIVES In this chapter we will show how to apply energy methods to solve problems involving deflection The chapter begins with a discussion of work and strain energy followed by a development of the principle of conservation of energy Using this principle the stress and deflection of a member are determined when the member is subjected to impact The method of virtual work and Castiglianos theorem are then developed and these methods are used to determine the displacement and slope at points on structural members and mechanical elements 141 External Work and Strain Energy The deflection of joints on a truss or points on a beam or shaft can be determined using energy methods Before developing any of these methods however we will first define the work caused by an external force and couple moment and show how to express this work in terms of a bodys strain energyThe formulations to be presented here and in the next section will provide the basis for applying the work and energy methods that follow throughout the chapter Energy Methods 14 141 EXTERNAL WORK AND STRAIN ENERGY 717 M u Fig 142 Work of a Couple Moment A couple moment M does work when it undergoes an angular displacement along its line of action The work is defined as Fig 142 If the total angular displacement is rad the work becomes 144 As in the case of force if the couple moment is applied to a body having linear elastic material behavior such that its magnitude is increased gradually from zero at to M at then the work is 145 However if the couple moment is already applied to the body and other loadings further rotate the body by an amount then the work is Strain Energy When loads are applied to a body they will deform the material Provided no energy is lost in the form of heat the external work done by the loads will be converted into internal work called strain energy This energy which is always positive is stored in the body and is caused by the action of either normal or shear stress Normal Stress If the volume element shown in Fig 143 is subjected to the normal stress then the force created on the elements top and bottom faces is If this force is applied gradually to the element like the force P discussed previously its magnitude is increased from zero to while the element undergoes an elongation The work done by is therefore Since the volume of the element is we have 146 Notice that is always positive even if is compressive since and will always be in the same direction In general then if the body is subjected only to a uniaxial normal stress the strain energy in the body is then 147 Ui LV sP 2 dV s Pz sz sz dUi dUi 1 2 szPz dV dV dx dy dz 1 2sz dx dyPz dz dUi 1 2 dFz dz dFz dz Pz dz dFz dFz sz dA sz dx dy sz Uœ e Mu u Ue 1 2 Mu u u 0 Ue L u 0 M du u dUe M du du 14 dz dx dy sz Fig 143 718 CHAPTER 14 ENERGY METHODS Also if the material behaves in a linearelastic manner then Hookes law applies and we can express the strain energy in terms of the normal stress as 148 Shear Stress A strainenergy expression similar to that for normal stress can also be established for the material when it is subjected to shear stress Consider the volume element shown in Fig 144 Here the shear stress causes the element to deform such that only the shear force acting on the top face of the element is displaced relative to the bottom face The vertical faces only rotate and therefore the shear forces on these faces do no work Hence the strain energy stored in the element is or since 149 The strain energy stored in the body is therefore 1410 Like the case for normal strain energy shear strain energy is always positive since and are always in the same direction If the material is linear elastic then applying Hookes law we can express the strain energy in terms of the shear stress as 1411 Ui LV t2 2G dV g tG g t Ui LV tg 2 dV dUi 1 2 tg dV dV dx dy dz dUi 1 2 t1dx dy2g dz g dz dF t1dx dy2 Ui LV s2 2E dV dx dy dz t gdz g Fig 144 14 141 EXTERNAL WORK AND STRAIN ENERGY 719 In the next section we will use Eqs 148 and 1411 to obtain formal expressions for the strain energy stored in members subjected to several types of loads Once this is done we will then be able to develop the energy methods necessary to determine the displacement and slope at points on a body Multiaxial Stress The previous development may be expanded to determine the strain energy in a body when it is subjected to a general state of stress Fig 145a The strain energies associated with each of the normal and shear stress components can be obtained from Eqs 146 and 149 Since energy is a scalar the total strain energy in the body is therefore 1412 The strains can be eliminated by using the generalized form of Hookes law given by Eqs 1018 and 1019 After substituting and combining terms we have 1413 If only the principal stresses act on the element Fig 145b this equation reduces to a simpler form namely 1414 This equation was used in Sec 107 as a basis for developing the maximumdistortionenergy theory Ui LV c 1 2E As1 2 s2 2 s3 2B n E 1s1s2 s2s3 s1s32d dV s2 s3 s1 1 2G Atxy 2 tyz 2 txz 2B d dV Ui LV c 1 2E Asx 2 sy 2 sz 2B n E 1sxsy sysz sxsz2 1 2 txygxy 1 2 tyzgyz 1 2 txzgxzd dV Ui LV c 1 2 sxPx 1 2 syPy 1 2 szPz 14 a sz sy sx txz tyz txy b s3 s2 s1 Fig 145 720 CHAPTER 14 ENERGY METHODS 142 Elastic Strain Energy for Various Types of Loading Using the equations for elastic strain energy developed in the previous section we will now formulate the strain energy stored in a member when it is subjected to an axial load bending moment transverse shear and torsional moment Examples will be given to show how to calculate the strain energy in members subjected to each of these loadings Axial Load Consider a bar of variable yet slightly tapered cross section Fig 146The internal axial force at a section located a distance x from one end is N If the crosssectional area at this section is A then the normal stress on the section is Applying Eq 148 we have If we choose an element or differential slice having a volume the general formula for the strain energy in the bar is therefore 1415 For the more common case of a prismatic bar of constant cross sectional area A length L and constant axial load N Fig 147 Eq 1415 when integrated gives 1416 Notice that the bars elastic strain energy will increase if the length of the bar is increased or if the modulus of elasticity or crosssectional area is decreased For example an aluminum rod will store approximately three times as much energy as a steel rod having the same size and subjected to the same load However doubling the crosssectional area of a rod will decrease its ability to store energy by onehalf The following example illustrates this point numerically Est 2911032 ksi Eal 1011032 ksi Ui N2L 2AE Ui L L 0 N2 2AE dx dV A dx Ui LV sx 2 2E dV LV N2 2EA2 dV s NA 14 x x A N s Fig 146 L A N N Fig 147 142 ELASTIC STRAIN ENERGY FOR VARIOUS TYPES OF LOADING 721 14 EXAMPLE 141 One of the two highstrength steel bolts A and B shown in Fig 148 is to be chosen to support a sudden tensile loading For the choice it is necessary to determine the greatest amount of elastic strain energy that each bolt can absorb Bolt A has a diameter of 0875 in for 2 in of its length and a root or smallest diameter of 0731 in within the 025in threaded region Bolt B has upset threads such that the diameter throughout its 225in length can be taken as 0731 in In both cases neglect the extra material that makes up the threads Take sY 44 ksi Est 2911032 ksi 2 in 025 in 0875 in 0731 in A 225 in 0731 in B Fig 148 SOLUTION Bolt A If the bolt is subjected to its maximum tension the maximum stress of will occur within the 025in region This tension force is Applying Eq 1416 to each region of the bolt we have Ans Bolt B Here the bolt is assumed to have a uniform diameter of 0731 inthroughout its 225inlengthAlsofrom the calculation above it can support a maximum tension force of Thus Ans NOTE By comparison bolt B can absorb 36 more elastic energy than bolt A because it has a smaller cross section along its shank Ui N2L 2AE 11847 kip221225 in2 2p10731 in2222911032 ksi 00315 in kip Pmax 1847 kip 00231 in kip 11847 kip2212 in2 2p10875 in2222911032 ksi 11847 kip221025 in2 2p10731 in2222911032 ksi Ui a N2L 2AE Pmax sYA 44 ksi Bpa 0731 in 2 b 2 R 1847 kip sY 44 ksi 722 CHAPTER 14 ENERGY METHODS Recall that the flexure formula as used here can also be used with justifiable accuracy to determine the stress in slightly tapered beams See Sec 64 So in the general sense I in Eq 1417 may also have to be expressed as a function of x Bending Moment Since a bending moment applied to a straight prismatic member develops normal stress in the member we can use Eq 148 to determine the strain energy stored in the member due to bending For example consider the axisymmetric beam shown in Fig 149 Here the internal moment is M and the normal stress acting on the arbitrary element a distance y from the neutral axis is If the volume of the element is where dA is the area of its exposed face and dx is its length the elastic strain energy in the beam is or Realizing that the area integral represents the moment of inertia of the area about the neutral axis the final result can be written as 1417 To evaluate the strain energy therefore we must first express the internal moment as a function of its position x along the beam and then perform the integration over the beams entire length The following examples illustrate this procedure Ui L L 0 M2 dx 2EI Ui L L 0 M2 2EI2 LA y2 dA dx Ui LV s2 2E dV LV 1 2E a My I b 2 dA dx dV dA dx s MyI 14 x x z y y dA M s Fig 149 142 ELASTIC STRAIN ENERGY FOR VARIOUS TYPES OF LOADING 723 14 EXAMPLE 142 Determine the elastic strain energy due to bending of the cantilevered beam in Fig 1410a EI is constant w L a Fig 1410 SOLUTION The internal moment in the beam is determined by establishing the x coordinate with origin at the left sideThe left segment of the beam is shown in Fig 1410bWe have Applying Eq 1417 yields or Ans We can also obtain the strain energy using an x coordinate having its origin at the right side of the beam and extending positive to the left Fig 1410c In this case Applying Eq 1417 we obtain the same result as before however more calculations are involved in this case M wL2 2 wLx wx2 2 M wxa x 2 b wL1x2 wL2 2 0 dMNA 0 Ui w2L5 40EI Ui L L 0 M2 dx 2EI L L 0 w1x2222 dx 2EI w2 8EI L L 0 x4 dx M wx2 2 M wxa x 2 b 0 dMNA 0 M V x wx b x 2 M wL x wx c V x 2 wL2 2 726 CHAPTER 14 ENERGY METHODS 14 EXAMPLE 144 Determine the strain energy in the cantilevered beam due to shear if the beam has a square cross section and is subjected to a uniform distributed load Fig 1414a EI and G are constant w Fig 1414 x wx b M V x 2 L w a a a SOLUTION From the freebody diagram of an arbitrary sectionFig1414bwe have Since the cross section is square the form factor Eq 1420 and therefore Eq 1419 becomes or Ans NOTE Using the results of Example 142 with the ratio of shear to bending strain energy is Since and Sec 106 then as an upper bound so that It can be seen that this ratio will increase as L decreases However even for very short beams where say the contribution due to shear strain energy is only 8 of the bending strain energy For this reason the shear strain energy stored in beams is usually neglected in engineering analysis L 5a 1Ui2s 1Ui2b 2a a L b 2 E 3G n 1 2 G E211 n2 1Ui2s 1Ui2b w2L35Ga2 w2L540EA 1 12 a4B 2 3 a a L b 2 E G I 1 12 a4 A a2 1Ui2s w2L3 5GA 1Ui2s L L 0 6 51wx22 dx 2GA 3w2 5GA L L 0 x2 dx fs 6 5 V wx V wx 0 c Fy 0 Torsional Moment To determine the internal strain energy in a circular shaft or tube due to an applied torsional moment we must apply Eq 1411 Consider the slightly tapered shaft in Fig 1415 A section of the shaft taken a distance x from one end is subjected to an internal torque T The shear stress distribution that causes this torque varies linearly from the center of the shaft On the arbitrary element of area dA and length dx the stress is The strain energy stored in the shaft is thus Since the area integral represents the polar moment of inertia J for the shaft at the section the final result can be written as 1421 The most common case occurs when the shaft or tube has a constant crosssectional area and the applied torque is constant Fig 1416 Integration of Eq 1421 then gives 1422 From this equation we may conclude that like an axially loaded member the energyabsorbing capacity of a torsionally loaded shaft is decreased by increasing the diameter of the shaft since this increases J Ui T2L 2GJ Ui L L 0 T2 2GJ dx L L 0 T2 2GJ2 LA r2 dA dx Ui LV t2 2G dV LV 1 2G a Tr J b 2 dA dx t TrJ 142 ELASTIC STRAIN ENERGY FOR VARIOUS TYPES OF LOADING 727 14 Fig 1415 Fig 1416 T x x dA t r T T L Important Points A force does work when it moves through a displacement When a force is applied to a body and its magnitude is increased gradually from zero to F the work is whereas if the force is constant when the displacement occurs then A couple moment does work when it displaces through a rotation Strain energy is caused by the internal work of the normal and shear stresses It is always a positive quantity The strain energy can be related to the resultant internal loadings N V M and T As the beam becomes longer the strain energy due to bending becomes much larger than the strain energy due to shear For this reason the shear strain energy in beams can generally be neglected U F U 1F22 The following example illustrates how to determine the strain energy in a circular shaft due to a torsional loading 732 CHAPTER 14 ENERGY METHODS 2 m 1 m 8 kN 2 m 1 m 8 kN Prob 1420 L C P L 2 y z x B A Prob 1421 P L h b Prob 1422 14 1420 Determine the bending strain energy in the beam and the axial strain energy in each of the two rods The beam is made of 2014T6 aluminum and has a square cross section 50 mm by 50 mm The rods are made of A36 steel and have a circular cross section with a 20mm diameter 1421 The pipe lies in the horizontal plane If it is subjected to a vertical force P at its end determine the strain energy due to bending and torsion Express the results in terms of the crosssectional properties I and J and the material properties E and G 1422 The beam shown is tapered along its width If a force P is applied to its end determine the strain energy in the beam and compare this result with that of a beam that has a constant rectangular cross section of width b and height h 1423 Determine the bending strain energy in the cantilevered beam due to a uniform load w Solve the problem two ways a Apply Eq 1417 b The load w dx acting on a segment dx of the beam is displaced a distance y where the equation of the elastic curve Hence the internal strain energy in the differential segment dx of the beam is equal to the external work ie Integrate this equation to obtain the total strain energy in the beam EI is constant dUi 1 21w dx21y2 y w1x4 4L3x 3L42124EI2 1424 Determine the bending strain energy in the simply supported beam due to a uniform load w Solve the problem two ways a Apply Eq 1417 b The load w dx acting on the segment dx of the beam is displaced a distance y where the equation of the elastic curve Hence the internal strain energy in the differential segment dx of the beam is equal to the external work ie Integrate this equation to obtain the total strain energy in the beam EI is constant dUi 1 21w dx21y2 y w1x4 2Lx3 L3x2124EI2 L dx x w dx w Prob 1423 L dx x w w dx Prob 1424 143 CONSERVATION OF ENERGY 737 14 1425 Determine the horizontal displacement of joint A Each bar is made of A36 steel and has a crosssectional area of 15 in2 1428 Determine the horizontal displacement of joint D AE is constant PROBLEMS 4 ft C B D A 3 ft 3 ft 2 kip Prob 1425 L P C L B A L Prob 1426 1426 Determine the horizontal displacement of joint C AE is constant 1427 Determine the vertical displacement of joint C AE is constant L P C L B A L Prob 1427 L P A B D C L 08 L 06 Prob 1428 1429 The cantilevered beam is subjected to a couple moment applied at its end Determine the slope of the beam at B EI is constant M0 L B A M0 Prob 1429 B A C 4 in 12 in a a Section a a 100 kip 15 ft 15 ft Prob 1430 1430 Determine the vertical displacement of point C of the simply supported 6061T6 aluminum beam Consider both shearing and bending strain energy 143 CONSERVATION OF ENERGY 739 14 1437 The load P causes the open coils of the spring to make an angle with the horizontal when the spring is stretched Show that for this position this causes a torque and a bending moment at the cross section Use these results to determine the maximum normal stress in the material 1438 The coiled spring has n coils and is made from a material having a shear modulus G Determine the stretch of the spring when it is subjected to the load PAssume that the coils are close to each other so that and the deflection is caused entirely by the torsional stress in the coil u L 0 M PR sin u T PR cos u u 1440 The rod has a circular cross section with a polar moment of inertia J and moment of inertia I If a vertical force P is applied at A determine the vertical displacement at this point Consider the strain energy due to bending and torsionThe material constants are E and G 1439 The pipe assembly is fixed at A Determine the vertical displacement of end C of the assembly The pipe has an inner diameter of 40 mm and outer diameter of 60 mm and is made of A36 steel Neglect the shearing strain energy 1441 Determine the vertical displacement of end B of the frame Consider only bending strain energy The frame is made using two A36 steel wideflange sections W460 68 P P R d u Probs 143738 800 mm 400 mm C B 600 N A Prob 1439 r P x z y A Prob 1440 B A 20 kN 4 m 3 m Prob 1441 144 IMPACT LOADING 749 14 L A L 2c B h W Prob 1459 1459 The wideflange beam has a length of 2L a depth 2c and a constant EI Determine the maximum height h at which a weight W can be dropped on its end without exceeding a maximum elastic stress in the beam s max 1463 The diver weighs 150 lb and while holding himself rigid strikes the end of the wooden diving board Determine the maximum height h from which he can jump onto the board so that the maximum bending stress in the wood does not exceed 6 ksi The board has a thickness of 15 in and width of 15 ft Ew 1811032 ksi C h A B 2 m 4 m Probs 146061 4 ft 10 ft h v Probs 146263 8 ft B A 8 ft 4 ft 3 in 4 in k k Probs 146465 1460 The 50kg block C is dropped from onto the simply supported beam If the beam is an A36 steel wideflange section determine the maximum bending stress developed in the beam 1461 Determine the maximum height h from which the 50kg block C can be dropped without causing yielding in the A36 steel wide flange section when the block strikes the beam W310 39 W250 45 h 15 m 1462 The diver weighs 150 lb and while holding himself rigid strikes the end of a wooden diving board with a downward velocity of Determine the maximum bending stress developed in the board The board has a thickness of 15 in and width of 15 ft sY 8 ksi Ew 1811032 ksi 4 fts 1h 02 1464 The weight of 175 lb is dropped from a height of 4 ft from the top of the A36 steel beam Determine the maximum deflection and maximum stress in the beam if the supporting springs at A and B each have a stiffness of The beam is 3 in thick and 4 in wide 1465 The weight of 175 lb is dropped from a height of 4 ft from the top of the A36 steel beam Determine the load factor n if the supporting springs at A and B each have a stiffness of The beam is 3 in thick and 4 in wide k 300 lbin k 500 lbin 145 PRINCIPLE OF VIRTUAL WORK 753 14 See Engineering Mechanics Statics 12th edition RC Hibbeler Prentice Hall Inc 2009 In a similar manner if the angular displacement or slope of the tangent at a point on the body is to be determined at A Fig 1430b then a virtual couple moment having a unit magnitude is applied at the point Fig 1430a As a result this couple moment causes a virtual load in one of the elements of the bodyAssuming the real loads deform the element an amount dL the angular displacement can be found from the virtualwork equation virtual loadings 1435 real displacements Here virtual unit couple moment acting in the direction of virtual load acting on an element angular displacement in radians caused by the real loads displacement of the element in the direction of caused by the real loads This method for applying the principle of virtual work is often referred to as the method of virtual forces since a virtual force is applied resulting in a determination of an external real displacement The equation of virtual work in this case represents a statement of compatibility requirements for the body Although it is not important here realize that we can also apply the principle of virtual work as a method of virtual displacements In this case virtual displacements are imposed on the body when the body is subjected to real loadings This method can be used to determine the external reactive force on the body or an unknown internal loading When it is used in this manner the equation of virtual work is a statement of the equilibrium requirements for the body Internal Virtual Work The terms on the right side of Eqs 1434 and 1435 represent the internal virtual work developed in the bodyThe real internal displacements dL in these terms can be produced in several different ways For example these displacements may result from geometric fabrication errors from a change in temperature or more commonly from stress In particular no restriction has been placed on the magnitude of the external loading so the stress may be large enough to cause yielding or even strain hardening of the material uu dL internal u external uu internal u M 1 external 1 u uu dL u P1 P2 P3 uu M 754 CHAPTER 14 ENERGY METHODS 14 If we assume that the material behavior is linear elastic and the stress does not exceed the proportional limit we can then formulate the expressions for internal virtual work caused by stress using the equations of elastic strain energy developed in Sec 142 They are listed in the center column of Table 141 Recall that each of these expressions assumes that the internal loading N V M or T was applied gradually from zero to its full value As a result the work done by these resultants is shown in these expressions as onehalf the product of the internal loading and its displacement In the case of the virtualforce method however the full virtual internal loading is applied before the real loads cause displacements and therefore the work of the virtual loading is simply the product of the virtual load and its real displacement Referring to these internal virtual loadings u by the corresponding lowercase symbols n m and t the virtual work due to axial load shear bending moment and torsional moment is listed in the righthand column of Table 141 Using these results the virtualwork equation for a body subjected to a general loading can therefore be written as 1436 In the following sections we will apply the above equation to problems involving the displacement of joints on trusses and points on beams and mechanical elements We will also include a discussion of how to handle the effects of fabrication errors and differential temperature For application it is important that a consistent set of units be used for all the terms For example if the real loads are expressed in kilonewtons and the bodys dimensions are in meters a 1kN virtual force or virtual couple should be applied to the body By doing so a calculated displacement will be in meters and a calculated slope will be in radians 1kN m 1 L nN AE dx L mM EI dx L fsvV GA dx L tT GJ dx v Axial load N Shear V Bending moment M Torsional moment T Deformation caused by Strain energy Internal virtual work mM EI dx L 03 fsvV GA dx L 03 nN EA dx L 03 tT GJ dx L 03 M2 2EI dx L 03 fsV2 2GA dx L 03 N2 2EA dx L 03 T2 2GJ dx L 03 TABLE 141 756 CHAPTER 14 ENERGY METHODS 14 Temperature Change Truss members can change their length due to a change in temperature If is the coefficient of thermal expansion for a member and is the change in temperature the change in length of a member is Eq 44 Hence we can determine the displacement of a selected truss joint due to this temperature change from Eq 1434 written as 1438 Here external virtual unit load acting on the truss joint in the direction of joint displacement caused by the temperature change internal virtual force in a truss member caused by the external virtual unit load coefficient of thermal expansion of material change in temperature of member length of member Fabrication Errors Occasionally errors in fabricating the lengths of the members of a truss may occur If this happens the displacement in a particular direction of a truss joint from its expected position can be determined from direct application of Eq 1434 written as 1439 Here external virtual unit load acting on the truss joint in the direction of joint displacement caused by the fabrication errors internal virtual force in a truss member caused by the external virtual unit load difference in length of the member from its intended length caused by a fabrication error A combination of the righthand sides of Eqs 1437 through 1439 will be necessary if external loads act on the truss and some of the members undergo a temperature change or have been fabricated with the wrong dimensions L n 1 1 n L L T a n 1 1 na TL L a TL T a 146 METHOD OF VIRTUAL FORCES APPLIED TO TRUSSES 757 14 Procedure for Analysis The following procedure provides a method that may be used to determine the displacement of any joint on a truss using the method of virtual forces Virtual Forces n Place the virtual unit load on the truss at the joint where the displacement is to be determined The load should be directed along the line of action of the displacement With the unit load so placed and all the real loads removed from the truss calculate the internal n force in each truss member Assume that tensile forces are positive and compressive forces are negative Real Forces N Determine the N forces in each memberThese forces are caused only by the real loads acting on the truss Again assume that tensile forces are positive and compressive forces are negative VirtualWork Equation Apply the equation of virtual work to determine the desired displacement It is important to retain the algebraic sign for each of the corresponding n and N forces when substituting these terms into the equation If the resultant sum is positive the displacement is in the same direction as the virtual unit load If a negative value results is opposite to the virtual unit load When applying an increase in temperature will be positive whereas a decrease in temperature will be negative For when a fabrication error causes an increase in the length of a member is positive whereas a decrease in length is negative When applying this method attention should be paid to the units of each numerical quantity Notice however that the virtual unit load can be assigned any arbitrary unit pounds kips newtons etc since the n forces will have these same units and as a result the units for both the virtual unit load and the n forces will cancel from both sides of the equation L 1 n L T 1 na TL nNLAE 146 METHOD OF VIRTUAL FORCES APPLIED TO TRUSSES 761 1483 Determine the vertical displacement of joint C Each A36 steel member has a crosssectional area of 1484 Determine the vertical displacement of joint H Each A36 steel member has a crosssectional area of 45 in2 45 in2 14 1479 Determine the horizontal displacement of joint B of the truss Each A36 steel member has a crosssectional area of 1480 Determine the vertical displacement of joint C of the truss Each A36 steel member has a crosssectional area of 400 mm2 400 mm2 Probs 147980 Probs 148182 Probs 148384 Probs 148586 15 m C B 2 m 4 kN A D 5 kN 1481 Determine the vertical displacement of point A Each A36 steel member has a crosssectional area of 1482 Determine the vertical displacement of point B Each A36 steel member has a crosssectional area of 400 mm2 400 mm2 30 kN 20 kN 15 m 15 m 2 m A B E D C E 9 ft A I B 12 ft H C G D 6 kip 12 ft 12 ft 12 ft 8 kip 6 kip F J 1485 Determine the vertical displacement of joint C The truss is made from A36 steel bars having a cross sectional area of 1486 Determine the vertical displacement of joint G The truss is made from A36 steel bars having a crosssectional area of 150 mm2 150 mm2 A G C D E F H B 6 kN 6 kN 2 m 2 m 15 m 15 m 15 m 15 m 12 kN 762 CHAPTER 14 ENERGY METHODS 14 147 Method of Virtual Forces Applied to Beams In this section we will apply the method of virtual forces to determine the displacement and slope at a point on a beam To illustrate the principles the vertical displacement of point A on the beam shown in Fig 1434b will be determined To do this we must place a vertical unit load at this point Fig 1434a so that when the real distributed load w is applied to the beam it will cause the internal virtual work Because the load causes both a shear V and moment M within the beam we must actually consider the internal virtual work due to both of these loadings In Example 147 however it was shown that beam deflections due to shear are negligible compared with those caused by bending particularly if the beam is long and slender Since this type of beam is most often used in practice we will only consider the virtual strain energy due to bending Table 141 Hence the real load causes the element dx to deform so its sides rotate by an angle which causes internal virtual work Applying Eq 1434 the virtual work equation for the entire beam we have 1440 Here external virtual unit load acting on the beam in the direction of displacement caused by the real loads acting on the beam internal virtual moment in the beam expressed as a function of x and caused by the external virtual unit load internal moment in the beam expressed as a function of x and caused by the real loads modulus of elasticity of the material moment of inertia of the crosssectional area about the neutral axis In a similar manner if the slope of the tangent at a point on the beams elastic curve is to be determined a virtual unit couple moment must be applied at the point and the corresponding internal virtual moment has to be determined If we apply Eq 1435 for this case and neglect the effect of shear deformations we have 1441 1 u L L 0 muM EI dx mu u I E M m 1 1 L L 0 mM EI dx m du du 1MEI2dx 1 764 CHAPTER 14 ENERGY METHODS 14 Procedure for Analysis The following procedure provides a method that may be used to determine the displacement and slope at a point on the elastic curve of a beam using the method of virtual forces Virtual Moments m or Place a virtual unit load on the beam at the point and directed along the line of action of the desired displacement If the slope is to be determined place a virtual unit couple moment at the point Establish appropriate x coordinates that are valid within regions of the beam where there is no discontinuity of both real and virtual load With the virtual load in place and all the real loads removed from the beam calculate the internal moment m or as a function of each x coordinate Assume that m or acts in the positive direction according to the established beam sign convention for positive moment Fig 63 Real Moments Using the same x coordinates as those established for m or determine the internal moments M caused by the real loads Since positive m or was assumed to act in the conventional positive directionit is important that positive M acts in this same direction This is necessary since positive or negative internal virtual work depends on the directional sense of both the virtual load defined by or and displacement caused by VirtualWork Equation Apply the equation of virtual work to determine the desired displacement or slope It is important to retain the algebraic sign of each integral calculated within its specified region If the algebraic sum of all the integrals for the entire beam is positive or is in the same direction as the virtual unit load or virtual unit couple moment If a negative value results or is opposite to the virtual unit load or couple moment u u u M mu m mu mu mu mu mU 147 METHOD OF VIRTUAL FORCES APPLIED TO BEAMS 767 14 1487 Determine the displacement at point C EI is constant 1492 Determine the displacement at B of the 15in diameter A36 steel shaft 1493 Determine the slope of the 15indiameter A36 steel shaft at the bearing support A PROBLEMS A P B C a a 2 a 2 a P 3 m A 15 m 4 kNm B C 120 mm 180 mm 15 kN 10 m 5 m 2 kNm B A C 320 lb A 320 lb 140 lb 140 lb 15 ft 2 ft B D 2 ft 3 ft C 1488 The beam is made of southern pine for which Determine the displacement at A Ep 13 GPa 1489 Determine the displacement at C of the A36 steel beam 1490 Determine the slope at A of the A36 steel beam 1491 Determine the slope at B of the A36 steel beam I 7011062 mm4 I 7011062 mm4 I 7011062 mm4 1494 The beam is made of Douglas fir Determine the slope at C 8 kN 15 m A 15 m B C 120 mm 180 mm 15 m 3 m A 3 m 200 mm 400 mm 4 kNm B 1495 The beam is made of oak for which Determine the slope and displacement at A Eo 11 GPa Prob 1487 Prob 1488 Probs 14899091 Probs 149293 Prob 1494 Prob 1495 147 METHOD OF VIRTUAL FORCES APPLIED TO BEAMS 769 14112 The frame is made from two segments each of length L and flexural stiffness EI If it is subjected to the uniform distributed load determine the vertical displacement of point C Consider only the effect of bending 14113 The frame is made from two segments each of length L and flexural stiffness EI If it is subjected to the uniform distributed load determine the horizontal displacement of point B Consider only the effect of bending 14 14108 Determine the slope and displacement of end C of the cantilevered beam The beam is made of a material having a modulus of elasticity of E The moments of inertia for segments AB and BC of the beam are 2I and I respectively Prob 14108 A B C P L 2 L 2 A B C 3 m 12 kNm 6 kNm 3 m 14109 Determine the slope at A of the A36 steel simply supported beam 14110 Determine the displacement at point C of the A36 steel simply supported beam W200 46 W200 46 14111 The simply supported beam having a square cross section is subjected to a uniform load w Determine the maximum deflection of the beam caused only by bending and caused by bending and shearTake E 3G L a a w L L A B C w L L P A 14114 Determine the vertical displacement of point A on the angle bracket due to the concentrated force P The bracket is fixed connected to its support EI is constant Consider only the effect of bending Probs 14109110 Prob 14111 Probs 14112113 Prob 14114 770 CHAPTER 14 ENERGY METHODS 14115 Beam AB has a square cross section of 100 mm by 100 mm Bar CD has a diameter of 10 mm If both members are made of A36 steel determine the vertical displacement of point B due to the loading of 10 kN 14116 Beam AB has a square cross section of 100 mm by 100 mm Bar CD has a diameter of 10 mm If both members are made of A36 steel determine the slope at A due to the loading of 10 kN 14 14119 Determine the vertical displacement of point C The frame is made using A36 steel members Consider only the effects of bending 14120 Determine the horizontal displacement of end B The frame is made using A36 steel members Consider only the effects of bending W250 45 W250 45 3 m 2 m 10 kN A D B C 2 m 14117 Bar ABC has a rectangular cross section of 300 mm by 100 mm Attached rod DB has a diameter of 20 mm If both members are made of A36 steel determine the vertical displacement of point C due to the loading Consider only the effect of bending in ABC and axial force in DB 14118 Bar ABC has a rectangular cross section of 300 mm by 100 mm Attached rod DB has a diameter of 20 mm If both members are made of A36 steel determine the slope at A due to the loading Consider only the effect of bending in ABC and axial force in DB 14121 Determine the displacement at point C EI is constant 3 m 20 kN A B C 4 m D 100 mm 300 mm 3 m A C D B 5 m 25 m 15 kNm 15 kN 25 m A B a a M 0 C Probs 14115116 Probs 14117118 Probs 14119120 Prob 14122 A B a a M 0 C Prob 14121 14122 Determine the slope at B EI is constant 148 CASTIGLIANOS THEOREM 771 14 148 Castiglianos Theorem In 1879 Alberto Castigliano an Italian railroad engineer published a book in which he outlined a method for determining the displacement and slope at a point in a body This method which is referred to as Castiglianos second theorem applies only to bodies that have constant temperature and material with linearelastic behavior If the displacement at a point is to be determined the theorem states that the displacement is equal to the first partial derivative of the strain energy in the body with respect to a force acting at the point and in the direction of displacement In a similar manner the slope of the tangent at a point in a body is equal to the first partial derivative of the strain energy in the body with respect to a couple moment acting at the point and in the direction of the slope angle To derive Castiglianos second theorem consider a body of any arbitrary shape which is subjected to a series of n forces Fig 1438 According to the conservation of energy the external work done by these forces is equal to the internal strain energy stored in the body However the external work is a function of the external loads Eq 141 so the internal work is also a function of the external loadsThus 1442 Now if any one of the external forces say is increased by a differential amount the internal work will also be increased such that the strain energy becomes 1443 Ui dUi Ui 0Ui 0Pj dPj dPj Pj Ui Ue f1P1 P2 Pn2 Ue 1P dx P2 Pn P1 P1 P3 Pn P2 Fig 1438 772 CHAPTER 14 ENERGY METHODS 14 This value however will not depend on the sequence in which the n forces are applied to the body For example we could apply to the body first then apply the loads In this case would cause the body to displace a differential amount in the direction of By Eq 142 the increment of strain energy would be This quantity however is a secondorder differential and may be neglected Further application of the loads causes to move through the displacement so that now the strain energy becomes 1444 Here as above is the internal strain energy in the body caused by the loads and is the additional strain energy caused by In summary Eq 1443 represents the strain energy in the body determined by first applying the loads then Eq 1444 represents the strain energy determined by first applying and then the loads Since these two equations must be equal we require 1445 which proves the theorem ie the displacement in the direction of is equal to the first partial derivative of the strain energy with respect to Castiglianos second theorem Eq 1445 is a statement regarding the bodys compatibility requirements since it is a condition related to displacement Also the above derivation requires that only conservative forces be considered for the analysis These forces can be applied in any order and furthermore they do work that is independent of the path and therefore create no energy loss As long as the material has linearelastic behavior the applied forces will be conservative and the theorem is valid Castiglianos first theorem is similar to his second theorem however it relates the load to the partial derivative of the strain energy with respect to the corresponding displacement that is The proof is similar to that given above This theorem is another way of expressing the equilibrium requirements for the body however it has limited application and therefore it will not be discussed here Pj 0Ui0j Pj Pj Pj j j 0Ui 0Pj P1 P2 Pn dPj dPj P2 Pn P1 dPj dPjj P1 P2 Pn Ui Ui dUi Ui dPj j j dPj P2 Pn P1 1 2 dPj dj Ue 1 2 Pj j dPj dj dPj P1 P2 Pn dPj 149 CASTIGLIANOS THEOREM APPLIED TO TRUSSES 773 14 149 Castiglianos Theorem Applied to Trusses Since a truss member is only subjected to an axial load the strain energy for the member is given by Eq 1416 Substituting this equation into Eq 1445 and omitting the subscript i we have It is generally easier to perform the differentiation prior to summation Also L A and E are constant for a given member and therefore we can write 1446 Here displacement of the truss joint force of variable magnitude applied to the truss joint in the direction of axial force in a member caused by both force P and the actual loads on the truss of a member area of a member of elasticity of the material By comparison Eq 1446 is similar to that used for the method of virtual forces Eq 1437 except that n is replaced by These terms n and are the same since they represent the change of the members axial force with respect to the load P or in other words the axial force per unit load 0N0P 0N0P 11 nNLAE2 E modulus A crosssectional L length N internal P an external a Na 0N 0P b L AE 0 0P a N2L 2AE Ui N2L2AE 774 CHAPTER 14 ENERGY METHODS 14 EXAMPLE 1415 Determine the vertical displacement of joint C of the steel truss shown in Fig 1439a The crosssectional area of each member is and SOLUTION External Force P A vertical force P is applied to the truss at joint C since this is where the vertical displacement is to be determined Fig 1439b Est 200 GPa A 400 mm2 D C B A 2 m 2 m 2 m a 100 kN Procedure for Analysis The following procedure provides a method that may be used to determine the displacement of any joint on a truss using Castiglianos second theorem External Force P Place a force P on the truss at the joint where the displacement is to be determinedThis force is assumed to have a variable magnitude and should be directed along the line of action of the displacement Internal Forces N Determine the force N in each member in terms of both the actual numerical loads and the variable force PAssume that tensile forces are positive and compressive forces are negative Find the respective partial derivative for each member After N and have been determined assign P its numerical value if it has actually replaced a real force on the truss Otherwise set P equal to zero Castiglianos Second Theorem Apply Castiglianos second theorem to determine the desired displacement It is important to retain the algebraic signs for corresponding values of N and when substituting these terms into the equation If the resultant sum is positive is in the same direction as P If a negative value results is opposite to P N10N0P2LAE 0N0P 0N0P 0N0P Fig 1439 776 CHAPTER 14 ENERGY METHODS 14 1410 Castiglianos Theorem Applied to Beams The internal strain energy for a beam is caused by both bending and shear However as pointed out in Example 147 if the beam is long and slender the strain energy due to shear can be neglected compared with that of bending Assuming this to be the case the internal strain energy for a beam is given by Eq 1417 Omitting the subscript i Castiglianos second theorem becomes Rather than squaring the expression for internal moment integrating and then taking the partial derivative it is generally easier to differentiate prior to integration Provided E and I are constant we have 1447 L L 0 Ma 0M 0P b dx EI 0 0P L L 0 M2 dx 2EI i 0Ui0Pi Ui 1M2 dx2EI 14123 Solve Prob 1472 using Castiglianos theorem 14124 Solve Prob 1473 using Castiglianos theorem 14125 Solve Prob 1475 using Castiglianos theorem 14126 Solve Prob 1476 using Castiglianos theorem 14127 Solve Prob 1477 using Castiglianos theorem 14128 Solve Prob 1478 using Castiglianos theorem 14129 Solve Prob 1479 using Castiglianos theorem 14130 Solve Prob 1480 using Castiglianos theorem 14131 Solve Prob 1481 using Castiglianos theorem 14132 Solve Prob 1482 using Castiglianos theorem 14133 Solve Prob 1483 using Castiglianos theorem 14134 Solve Prob 1484 using Castiglianos theorem PROBLEMS 1410 CASTIGLIANOS THEOREM APPLIED TO BEAMS 777 Here of the point caused by the real loads acting on the beam force of variable magnitude applied to the beam at the point and in the direction of moment in the beam expressed as a function of x and caused by both the force P and the actual loads on the beam of elasticity of the material of inertia of crosssectional area about the neutral axis If the slope of the tangent at a point on the elastic curve is to be determined the partial derivative of the internal moment M with respect to an external couple moment acting at the point must be found For this case 1448 The above equations are similar to those used for the method of virtual forces Eqs 1440 and 1441 except m and replace and respectively In addition if axial load shear and torsion cause significant strain energy within the member then the effects of all these loadings should be included when applying Castiglianos theorem To do this we must use the strainenergy functions developed in Sec 142 along with their associated partial derivativesThe result is 0M0M 0M0P mu u L L 0 Ma 0M 0M b dx EI M u I moment E modulus M internal P an external displacement 14 1449 L L 0 Ma 0M 0P b dx EI L L 0 Ta 0T 0P b dx GJ Na 0N 0P b L AE L L 0 fsVa 0V 0P b dx GA The method of applying this general formulation is similar to that used to apply Eqs 1447 and 1448 778 CHAPTER 14 ENERGY METHODS 14 Procedure for Analysis The following procedure provides a method that may be used to apply Castiglianos second theorem External Force P or Couple Moment Place a force P on the beam at the point and directed along the line of action of the desired displacement If the slope of the tangent is to be determined at the point place a couple moment at the point Assume that both P and have a variable magnitude Internal Moments M Establish appropriate x coordinates that are valid within regions of the beam where there is no discontinuity of force distributed load or couple moment Determine the internal moments M as a function of x the actual numerical loads and P or and then find the partial derivatives or for each coordinate x After M and or have been determined assign P or its numerical value if it has actually replaced a real force or couple moment Otherwise set P or equal to zero Castiglianos Second Theorem Apply Eq 1447 or 1448 to determine the desired displacement or It is important to retain the algebraic signs for corresponding values of M and or If the resultant sum of all the definite integrals is positive or is in the same direction as P or If a negative value results or is opposite to P or M u M u 0M0M 0M0P u M M 0M0M 0M0P 0M0M 0M0P M M M Mœ 1410 CASTIGLIANOS THEOREM APPLIED TO BEAMS 779 14 EXAMPLE 1416 Determine the displacement of point B on the beam shown in Fig 1440a EI is constant L A a w B x b L w B A P V c wx M x P x 2 Fig 1440 SOLUTION External Force P A vertical force P is placed on the beam at B as shown in Fig 1440b Internal Moments M A single x coordinate is needed for the solution since there are no discontinuities of loading between A and B Using the method of sections Fig 1440c the internal moment and its partial derivative are determined as follows Setting gives Castiglianos Second Theorem Applying Eq 1447 we have Ans The similarity between this solution and that of the virtualwork method Example 1413 should be noted wL4 8EI B L L 0 Ma 0M 0P b dx EI L L 0 1wx2221x2 dx EI M wx2 2 and 0M 0P x P 0 0M 0P x M wx2 2 Px M wxa x 2 b P1x2 0 dMNA 0 780 CHAPTER 14 ENERGY METHODS 14 EXAMPLE 1417 Determine the slope at point B of the beam shown in Fig 1441a EI is constant SOLUTION External Couple Moment M Since the slope at point B is to be determined an external couple moment is placed on the beam at this point Fig 1441b Internal Moments M Two coordinates and must be used to completely describe the internal moments within the beam since there is a discontinuity at B As shown in Fig 1441b ranges from A to B and ranges from B to C Using the method of sections Fig 1441c the internal moments and the partial derivatives for and are determined as follows Castiglianos Second Theorem Setting and applying Eq 1448 we have M 0 0M2 0M 1 M2 M Pa L 2 x2b dMNA 0 0M1 0M 0 M1 Px1 dMNA 0 x2 x1 x2 x1 M x2 x1 M œ Ans L L2 0 1Px12102 dx1 EI L L2 0 P1L22 x2112 dx2 EI 3PL2 8EI uB L L 0 Ma 0M 0M b dx EI Note the similarity between this solution and that of Example 1414 14135 Solve Prob 1487 using Castiglianos theorem 14136 Solve Prob 1488 using Castiglianos theorem 14137 Solve Prob 1490 using Castiglianos theorem 14138 Solve Prob 1492 using Castiglianos theorem 14139 Solve Prob 1493 using Castiglianos theorem 14140 Solve Prob 1496 using Castiglianos theorem 14141 Solve Prob 1497 using Castiglianos theorem 14142 Solve Prob 1498 using Castiglianos theorem 14143 Solve Prob 14112 using Castiglianos theorem 14144 Solve Prob 14114 using Castiglianos theorem 14145 Solve Prob 14121 using Castiglianos theorem PROBLEMS A C P a B L 2 L 2 A C x1 B P b M x2 Fig 1441 x1 x2 B P P c M M1 V1 M2 V2 L 2 CHAPTER REVIEW 781 14 When a force couple moment acts on a deformable body it will do external work when it displaces rotates The internal stresses produced in the body also undergo displacement thereby creating elastic strain energy that is stored in the material The conservation of energy states that the external work done by the loading is equal to the internal elastic strain energy produced by the stresses in the body CHAPTER REVIEW Ue Ui max nst s max nsst n 1 C1 2 h st The conservation of energy can be used to solve problems involving elastic impact which assumes the moving body is rigid and all the strain energy is stored in the stationary body This leads to use of an impact factor n which is a ratio of the dynamic load to the static load It is used to determine the maximum stress and displacement of the body at the point of impact 1 u L L 0 muM EI dx 1 L L 0 mM EI dx 1 a nNL AE The principle of virtual work can be used to determine the displacement of a joint on a truss or the slope and the displacement of points on a beam It requires placing an external virtual unit force virtual unit couple moment at the point where the displacement rotation is to be determined The external virtual work that is produced by the external loading is then equated to the internal virtual strain energy in the structure u L L 0 Ma 0M 0M b dx EI L L 0 Ma 0M 0P b dx EI aNa 0N 0P b L AE Castiglianos second theorem can also be used to determine the displacement of a joint on a truss or the slope and the displacement at a point on a beam Here a variable force P couple moment M is placed at the point where the displacement slope is to be determined The internal loading is then determined as a function of P M and its partial derivative with respect to P M is determined Castiglianos second theorem is then applied to obtain the desired displacement rotation 782 CHAPTER 14 ENERGY METHODS 14 P a a a P C D h k A B 2 m 4 m 8 in 4 in 6 in 02 in A B 2 m C D B A 15 m 15 m 5 kN 14146 Determine the bending strain energy in the beam due to the loading shown EI is constant 14149 The L2 steel bolt has a diameter of 025 in and the link AB has a rectangular cross section that is 05 in wide by 02 in thick Determine the strain energy in the link AB due to bending and in the bolt due to axial force The bolt is tightened so that it has a tension of 350 lb Neglect the hole in the link REVIEW PROBLEMS 14147 The 200kg block D is dropped from a height onto end C of the A36 steel overhang beam If the spring at B has a stiffness determine the maximum bending stress developed in the beam 14148 Determine the maximum height h from which the 200kg block D can be dropped without causing the A36 steel overhang beam to yield The spring at B has a stiffness k 200 kNm W200 36 k 200 kNm W200 36 h 1 m 14150 Determine the vertical displacement of joint A Each bar is made of A36 steel and has a crosssectional area of Use the conservation of energy 600 mm2 Prob 14146 Probs 14147148 Prob 14149 Prob 14150 REVIEW PROBLEMS 783 14156 Determine the displacement of point B on the aluminum beam Use the conservation of energy Eal 10611032 ksi 14 14151 Determine the total strain energy in the A36 steel assembly Consider the axial strain energy in the two 05indiameter rods and the bending strain energy in the beam for which I 434 in4 3 ft 4 ft 4 ft 500 lb 14152 Determine the vertical displacement of joint E For each member Use the method of virtual work 14153 Solve Prob 14152 using Castiglianos theorem E 200 GPa A 400 mm2 C 15 m A D E F 45 kN 2 m B 2 m 14154 The cantilevered beam is subjected to a couple moment applied at its end Determine the slope of the beam at B EI is constant Use the method of virtual work 14155 Solve Prob 14154 using Castiglianos theorem M0 L B A M0 Prob 14151 Probs 14152153 Prob 14156 Probs 14154155 A 3 kip C B 12 ft 12 ft 1 in 6 in 1 in 3 in 3 in Prob 14157 12 ft 4 ft 14157 A 20lb weight is dropped from a height of 4 ft onto the end of a cantilevered A36 steel beam If the beam is a determine the maximum stress developed in the beam W12 50 784 A1 Centroid of an Area The centroid of an area refers to the point that defines the geometric center for the areaIf the area has an arbitrary shapeas shown in FigA1a the x and y coordinates defining the location of the centroid C are determined using the formulas A1 The numerators in these equations are formulations of the moment of the area element dA about the y and the x axis respectively Fig A1b the denominators represent the total area A of the shape x LA x dA LA dA y LA y dA LA dA Geometric Properties of an Area A APPENDIX Fig A1 a y x y x C A b y x x y dA A2 MOMENT OF INERTIA FOR AN AREA 787 A y x x y dA O r Fig A5 A2 Moment of Inertia for an Area The moment of inertia of an area often appears in formulas used in mechanics of materials It is a geometric property that is calculated about an axis and for the x and y axes shown in FigA5 it is defined as A3 These integrals have no physical meaning but they are so named because they are similar to the formulation of the moment of inertia of a mass which is a dynamical property of matter We can also calculate the moment of inertia of an area about the pole O or z axis FigA5This is referred to as the polar moment of inertia A4 Here r is the perpendicular distance from the pole z axis to the element dAThe relationship between and is possible since FigA5 From the above formulations it is seen that and will always be positive since they involve the product of distance squared and area Furthermore the units for moment of inertia involve length raised to the fourth power eg or Using the above equations the moments of inertia for some common area shapes have been calculated about their centroidal axes and are listed on the inside front cover ParallelAxis Theorem for an Area If the moment of inertia for an area is known about a centroidal axis we can determine the moment of inertia of the area about a corresponding parallel axis using the parallelaxis theorem To derive this theorem consider finding the moment of inertia of the shaded area shown in FigA6 about the x axis In this case a differential element dA is located at the arbitrary distance from the centroidal x axis whereas the fixed distance between the y ft4 in4 mm4 m4 JO Iy Ix r2 x2 y2 Ix Iy JO JO LA r2 dA Ix Iy Iy LA x2 dA Ix LA y2 dA y x y dA O d x x dy dx C y Fig A6 788 APPENDIX A GEOMETRIC PROPERTIES OF AN AREA A parallel x and axes is Since the moment of inertia of dA about the x axis is then for the entire area The first term on the right represents the moment of inertia of the area about the axis The second term is zero since the axis passes through the areas centroid C that is since The final result is therefore A5 A similar expression can be written for that is A6 And finally for the polar moment of inertia about an axis perpendicular to the xy plane and passing through the pole O z axis Fig A6 we have A7 The form of each of the above equations states that the moment of inertia of an area about an axis is equal to the areas moment of inertia about a parallel axis passing through the centroid plus the product of the area and the square of the perpendicular distance between the axes Composite Areas Many crosssectional areas consist of a series of connected simpler shapes such as rectangles triangles and semicircles In order to properly determine the moment of inertia of such an area about a specified axis it is first necessary to divide the area into its composite parts and indicate the perpendicular distance from the axis to the parallel centroidal axis for each part Using the table on the inside front cover of the book the moment of inertia of each part is determined about the centroidal axis If this axis does not coincide with the specified axis the parallelaxis theorem should be used to determine the moment of inertia of the part about the specified axisThe moment of inertia of the entire area about this axis is then determined by summing the results of its composite parts In particular if a composite part has a hole the moment of inertia for the composite is found by subtracting the moment of inertia for the hole from the moment of inertia of the entire area including the hole I I Ad2 JO JC Ad2 Iy Iy Adx 2 Iy Ix Ix Ady 2 y 0 1y dA yA 0 x Ix x Ix LA 1y dy22 dA LA y2 dA 2dy LA y dA dy 2 LA dA dIx 1y dy22 dA dy x A2 MOMENT OF INERTIA FOR AN AREA 789 A EXAMPLE A2 Determine the moment of inertia of the crosssectional area of the Tbeam shown in FigA7a about the centroidal axis x x C 5 in 855 in 10 in 8 in a 2 in 445 in 15 in 15 in Fig A7 3 in 13 in 3 in b x 5 in 855 in 10 in 2 in 445 in 65 in C SOLUTION I The area is segmented into two rectangles as shown in Fig A7a and the distance from the axis and each centroidal axis is determined Using the table on the inside front cover the moment of inertia of a rectangle about its centroidal axis is Applying the parallel axis theorem Eq A5 to each rectangle and adding the results we have Ans SOLUTION II The area can be considered as one large rectangle less two small rectangles shown shaded in FigA7bWe have Ans I 646 in4 2c 1 12 13 in2110 in23 13 in2110 in21855 in 5 in22d c 1 12 18 in2113 in23 18 in2113 in21855 in 65 in22 d I Ix Ady 2 I 646 in4 c 1 12 18 in213 in23 18 in213 in21445 in 15 in22 d c 1 12 12 in2110 in23 12 in2110 in21855 in 5 in22 d I Ix Ady 2 I 1 12 bh3 x 790 APPENDIX A GEOMETRIC PROPERTIES OF AN AREA A EXAMPLE A3 Fig A8 400 mm 400 mm x 100 mm y 100 mm 600 mm 100 mm a 300 mm x 100 mm y 100 mm b D 200 mm B A 250 mm 200 mm 250 mm 300 mm Determine the moments of inertia of the beams crosssectional area shown in FigA8a about the x and y centroidal axes SOLUTION The cross section can be considered as three composite rectangular areas A B and D shown in FigA8b For the calculation the centroid of each of these rectangles is located in the figure From the table on the inside front cover the moment of inertia of a rectangle about its centroidal axis is Hence using the parallelaxis theorem for rectangles A and D the calculations are as follows Rectangle A I 1 12 bh3 19011092 mm4 Iy Iy Adx 2 1 12 1300 mm21100 mm23 1100 mm21300 mm21250 mm22 142511092 mm4 Ix Ix Ady 2 1 12 1100 mm21300 mm23 1100 mm21300 mm21200 mm22 Rectangle B Rectangle D Iy 1 12 1100 mm21600 mm23 18011092 mm4 Ix 1 12 1600 mm21100 mm23 00511092 mm4 The moments of inertia for the entire cross section are thus Ans Ans 56011092 mm4 Iy 19011092 18011092 19011092 29011092 mm4 Ix 142511092 00511092 142511092 19011092 mm4 Iy Iy Adx 2 1 12 1300 mm21100 mm23 1100 mm21300 mm21250 mm22 142511092 mm4 Ix Ix Ady 2 1 12 1100 mm21300 mm23 1100 mm21300 mm21200 mm22 A3 PRODUCT OF INERTIA FOR AN AREA 791 A3 Product of Inertia for an Area In general the moment of inertia for an area is different for every axis about which it is computed In some applications of mechanical or structural design it is necessary to know the orientation of those axes that give respectively the maximum and minimum moments of inertia for the areaThe method for determining this is discussed in SecA4To use this method however one must first determine the product of inertia for the area as well as its moments of inertia for given x y axes The product of inertia for the area A shown in Fig A9 is defined as A8 Like the moment of inertia the product of inertia has units of length raised to the fourth power eg or However since x or y may be a negative quantity while dA is always positive the product of inertia may be positive negative or zero depending on the location and orientation of the coordinate axes For example the product of inertia for an area will be zero if either the x or y axis is an axis of symmetry for the area To show this consider the shaded area in Fig A10 where for every element dA located at point x y there is a corresponding element dA located at Since the products of inertia for these elements are respectively xy dA and their algebraic sum or the integration of all the elements of area chosen in this way will cancel each other Consequently the product of inertia for the total area becomes zero xy dA 1x y2 Ixy ft4 in4 mm4 m4 Ixy LA xy dA A y x x y A dA Fig A9 y x x y dA dA y Fig A10 792 APPENDIX A GEOMETRIC PROPERTIES OF AN AREA A ParallelAxis Theorem Consider the shaded area shown in Fig A11 where and represent a set of centroidal axes and x and y represent a corresponding set of parallel axes Since the product of inertia of dA with respect to the x and y axes is then for the entire area The first term on the right represents the product of inertia of the area with respect to the centroidal axis The second and third terms are zero since the moments of the area are taken about the centroidal axis Realizing that the fourth integral represents the total area A we therefore have A9 The similarity between this equation and the parallelaxis theorem for moments of inertia should be noted In particular it is important that here the algebraic signs for and be maintained when applying EqA9 dy dx Ixy Ixy Adxdy Ixy LA xy dA dxLA y dA dyLA x dA dxdyLA dA Ixy LA 1x dx21y dy2 dA dIxy 1x dx21y dy2 dA y x y x C dA y x x y dy dx Fig A11 A3 PRODUCT OF INERTIA FOR AN AREA 793 A EXAMPLE A4 400 mm 400 mm x 100 mm y 100 mm 600 mm 100 mm a Determine the product of inertia of the beams crosssectional area shown in FigA12a about the x and y centroidal axes SOLUTION As in Example A3 the cross section can be considered as three composite rectangular areas A B and D Fig A12b The coordinates for the centroid of each of these rectangles are shown in the figure Due to symmetry the product of inertia of each rectangle is zero about a set of axes that pass through the rectangles centroid Hence application of the parallelaxis theorem to each of the rectangles yields Rectangle A Rectangle B Rectangle D The product of inertia for the entire cross section is thus Ans 30011092 mm4 Ixy 15011092 mm4 0 15011092 mm4 15011092 mm4 0 1300 mm21100 mm21250 mm21200 mm2 Ixy Ixy Adxdy 0 0 0 Ixy Ixy Adxdy 15011092 mm4 0 1300 mm21100 mm21250 mm21200 mm2 Ixy Ixy Adxdy y x 300 mm x 100 mm y 100 mm b D 200 mm B A 250 mm 200 mm 250 mm 300 mm Fig A12 794 APPENDIX A GEOMETRIC PROPERTIES OF AN AREA A4 Moments of Inertia for an Area about Inclined Axes In mechanical or structural design it is sometimes necessary to calculate the moments and product of inertia and for an area with respect to a set of inclined and axes when the values for and are known As shown in Fig A13 the coordinates to the area element dA from each of the two coordinate systems are related by the transformation equations Using these equations the moments and product of inertia of dA about the and axes become Expanding each expression and integratingrealizing that and we obtain These equations may be simplified by using the trigonometric identities and in which case A10 Ixy Ix Iy 2 sin 2u Ixy cos 2u Iy Ix Iy 2 Ix Iy 2 cos 2u Ixy sin 2u Ix Ix Iy 2 Ix Iy 2 cos 2u Ixy sin 2u cos 2u cos2 u sin2 u sin 2u 2 sin u cos u Ixy Ix sin u cos u Iy sin u cos u Ixy1cos2 u sin2 u2 Iy Ix sin2 u Iy cos2 u 2Ixy sin u cos u Ix Ix cos2 u Iy sin2 u 2Ixy sin u cos u Ixy 1xy dA Iy 1x2 dA Ix 1y2 dA dIxy xy dA 1x cos u y sin u21y cos u x sin u2 dA dIy x2 dA 1x cos u y sin u22 dA dIx y2 dA 1y cos u x sin u22 dA y x y y cos u x sin u x x cos u y sin u Ixy u Ix Iy y x Ixy Iy Ix A y x dA O y x x y y x y cos u x sin u y sin u x cos u u u u Fig A13 A5 MOHRS CIRCLE FOR MOMENTS OF INERTIA 797 A A5 Mohrs Circle for Moments of Inertia Equations A10 through A12 have a semigraphical solution that is convenient to use and generally easy to remember Squaring the first and third of EqA10 and adding it is found that A13 In any given problem and are variables and and are known constants Thus the above equation may be written in compact form as When this equation is plotted the resulting graph represents a circle of radius and having its center located at point a 0 where The circle so constructed is called Mohrs circle Its application is similar to that used for stress and strain transformation developed in Chapters 9 and 10 respectively a 1Ix Iy22 R C Ix Iy 2 2 Ixy 2 1Ix a22 Ixy 2 R2 Ixy Ix Iy Ixy Ix Ix Ix Iy 2 2 Ixy 2 Ix Iy 2 2 Ixy 2 Procedure for Analysis The main purpose for using Mohrs circle here is to have a convenient means of transforming and into the principal moments of inertia for the areaThe following procedure provides a method for doing this Calculate Establish the x y axes for the area with the origin located at the point P of interest usually the centroid and determine and Ixy FigA16a Iy Ix Iy Ixy Ix Ixy Iy Ix x y x y Minor axis for principal moment of inertia Imin Major axis for principal moment of inertia Imax P a up1 Fig A16 800 Geometric Properties of Structural Shapes B APPENDIX WideFlange Sections or W Shapes FPS Units Web Flange Area Depth thickness width thickness xx axis yy axis Designation A d I S r I S r in in in in in in 306 2406 0500 12750 0750 3100 258 101 259 407 291 277 2431 0515 9065 0875 2700 222 987 109 240 198 247 2410 0470 9020 0770 2370 196 979 944 209 195 224 2392 0440 8990 0680 2100 176 969 825 184 192 201 2373 0415 8965 0585 1830 154 955 704 157 187 182 2374 0430 7040 0590 1550 131 923 345 980 138 162 2357 0395 7005 0505 1350 114 911 291 830 134 191 1835 0450 7590 0750 1070 117 749 548 144 169 176 1824 0415 7555 0695 984 108 747 501 133 169 162 1811 0390 7530 0630 890 983 741 449 119 167 147 1799 0355 7495 0570 800 889 738 401 107 165 135 1806 0360 6060 0605 712 788 725 225 743 129 118 1790 0315 6015 0525 612 684 721 191 635 127 103 1770 0300 6000 0425 510 576 704 153 512 122 168 1643 0430 7120 0715 758 922 672 431 121 160 147 1626 0380 7070 0630 659 810 668 372 105 159 133 1613 0345 7035 0565 586 727 665 328 934 157 106 1586 0295 6985 0430 448 565 651 245 700 152 912 1588 0275 5525 0440 375 472 641 124 449 117 768 1569 0250 5500 0345 301 384 626 959 349 112 156 1392 0370 8060 0660 541 778 589 577 143 192 126 1366 0305 7995 0530 428 627 582 452 113 189 112 1410 0310 6770 0515 385 546 587 267 788 155 100 1398 0285 6745 0455 340 486 583 233 691 153 885 1384 0270 6730 0385 291 420 573 196 582 149 769 1391 0255 5025 0420 245 353 565 891 354 108 649 1374 0230 5000 0335 199 290 554 700 280 104 W14 22 W14 26 W14 30 W14 34 W14 38 W14 43 W14 53 W16 26 W16 31 W16 36 W16 45 W16 50 W16 57 W18 35 W18 40 W18 46 W18 50 W18 55 W18 60 W18 65 W24 55 W24 62 W24 68 W24 76 W24 84 W24 94 W24 104 in3 in4 in3 in4 in2 in lbft tf bf tw WIDEFLANGE SECTIONS OR W SHAPES FPS UNITS 801 B WideFlange Sections or W Shapes FPS Units Web Flange Area Depth thickness width thickness xx axis yy axis Designation A d I S r I S r in in in in in in 256 1253 0515 12125 0810 740 118 538 241 397 307 147 1219 0370 8080 0640 394 647 518 563 139 196 132 1206 0335 8045 0575 350 581 515 500 124 194 765 1222 0230 6490 0380 204 334 517 173 534 151 648 1231 0260 4030 0425 156 254 491 466 231 0847 471 1199 0220 3990 0265 103 171 467 282 141 0773 416 1191 0200 3970 0225 886 149 462 236 119 0753 294 1110 0680 10340 1120 623 112 460 207 400 265 158 1009 0370 10030 0615 303 600 437 103 206 256 133 1010 0350 8020 0620 248 491 432 534 133 201 115 992 0315 7985 0530 209 421 427 450 113 198 884 1047 0300 5810 0510 170 324 438 167 575 137 562 1024 0250 4020 0395 963 188 414 429 214 0874 441 999 0230 4000 0270 689 138 395 289 145 0810 354 987 0190 3960 0210 538 109 390 218 110 0785 197 900 0570 8280 0935 272 604 372 886 214 212 171 875 0510 8220 0810 228 520 365 751 183 210 141 850 0400 8110 0685 184 433 361 609 150 208 117 825 0360 8070 0560 146 355 353 491 122 204 913 800 0285 7995 0435 110 275 347 371 927 202 708 793 0245 6495 0400 828 209 342 183 563 161 444 811 0245 4015 0315 480 118 329 341 170 0876 734 638 0320 6080 0455 534 167 270 171 561 152 587 620 0260 6020 0365 414 134 266 133 441 150 474 628 0260 4030 0405 321 102 260 443 220 0966 443 599 0230 5990 0260 291 972 256 932 311 146 355 603 0230 4000 0280 221 731 249 299 150 0918 268 590 0170 3940 0215 164 556 247 219 111 0905 W6 9 W6 12 W6 15 W6 16 W6 20 W6 25 W8 15 W8 24 W8 31 W8 40 W8 48 W8 58 W8 67 W10 12 W10 15 W10 19 W10 30 W10 39 W10 45 W10 54 W10 100 W12 14 W12 16 W12 22 W12 26 W12 45 W12 50 W12 87 in3 in4 in3 in4 in2 in lbft tf bf tw y y x x bf tf tw d ANGLES HAVING EQUAL LEGS FPS UNITS 803 B y y y x z z x x Angles Having Equal Legs FPS Units Size and Weight xx axis yy axis zz axis thickness per foot Area A I S r y I S r x r in lb in in in in in 510 150 890 158 244 237 890 158 244 237 156 389 114 697 122 247 228 697 122 247 228 158 264 775 486 836 250 219 486 836 250 219 159 374 110 355 857 180 186 355 857 180 186 117 287 844 282 666 183 178 282 666 183 178 117 196 575 199 461 186 168 199 461 186 168 118 149 436 154 353 188 164 154 353 188 164 119 236 694 157 453 151 152 157 453 151 152 0975 162 475 113 316 154 143 113 316 154 143 0983 123 361 874 242 156 139 874 242 156 139 0990 185 544 767 281 119 127 767 281 119 127 0778 128 375 556 197 122 118 556 197 122 118 0782 98 286 436 152 123 114 436 152 123 114 0788 66 194 304 105 125 109 304 105 125 109 0795 111 325 364 149 106 106 364 149 106 106 0683 85 248 287 115 107 101 287 115 107 101 0687 58 169 201 0794 109 0968 201 0794 109 0968 0694 94 275 222 107 0898 0932 222 107 0898 0932 0584 72 211 176 0833 0913 0888 176 0833 0913 0888 0587 49 144 124 0577 0930 0842 124 0577 0930 0842 0592 77 225 123 0724 0739 0806 123 0724 0739 0806 0487 59 173 0984 0566 0753 0762 0984 0566 0753 0762 0487 41 119 0703 0394 0769 0717 0703 0394 0769 0717 0491 47 136 0479 0351 0594 0636 0479 0351 0594 0636 0389 319 0938 0348 0247 0609 0592 0348 0247 0609 0592 0391 165 0484 0190 0131 0626 0546 0190 0131 0626 0546 0398 L2 2 1 8 L2 2 1 4 L2 2 3 8 L2 1 2 2 1 2 1 4 L2 1 2 2 1 2 3 8 L2 1 2 2 1 2 1 2 L3 3 1 4 L3 3 3 8 L3 3 1 2 L3 1 2 3 1 2 1 4 L3 1 2 3 1 2 1 2 L3 1 2 3 1 2 1 2 L4 4 1 4 L4 4 3 8 L4 4 1 2 L4 4 3 4 L5 5 3 8 L5 5 1 2 L5 5 3 4 L6 6 3 8 L6 6 1 2 L6 6 3 4 L6 6 1 L8 8 1 2 L8 8 3 4 L8 8 1 in3 in4 in3 in4 in2 804 APPENDIX B GEOMETRIC PROPERTIES OF STRUCTURAL SHAPES B y y x x bf tw tf d WideFlange Sections or W Shapes SI Units Web Flange Area Depth thickness width thickness xx axis yy axis Designation A d I S r I S r mm mm mm mm mm mm 19 800 611 1270 3240 190 1 290 4 220 255 108 667 739 17 900 617 1310 2300 222 1 120 3 630 250 451 392 502 15 900 612 1190 2290 196 985 3 220 249 393 343 497 14 400 608 1120 2280 173 875 2 880 247 343 301 488 12 900 603 1050 2280 149 764 2 530 243 295 259 478 11 800 603 1090 1790 150 646 2 140 234 144 161 349 10 500 599 1000 1780 128 560 1 870 231 121 136 339 12 300 466 1140 1930 190 445 1 910 190 228 236 431 11 400 463 1050 1920 177 410 1 770 190 209 218 428 10 400 460 991 1910 160 370 1 610 189 186 195 423 9 460 457 902 1900 145 333 1 460 188 166 175 419 8 730 459 914 1540 154 297 1 290 184 941 122 328 7 590 455 800 1530 133 255 1 120 183 796 104 324 6 640 450 762 1520 108 212 942 179 634 834 309 10 800 417 1090 1810 182 315 1 510 171 180 199 408 9 510 413 965 1800 160 275 1 330 170 156 173 405 8 560 410 876 1790 144 245 1 200 169 138 154 402 6 820 403 749 1770 109 186 923 165 101 114 385 5 890 403 699 1400 112 156 774 163 514 734 295 4 960 399 635 1400 88 126 632 159 402 574 285 10 100 354 940 2050 168 227 1 280 150 242 236 489 8 150 347 775 2030 135 179 1 030 148 188 185 480 7 200 358 787 1720 131 160 894 149 111 129 393 6 450 355 724 1710 116 141 794 148 968 113 387 5 710 352 686 1710 98 121 688 146 816 954 378 4 960 353 648 1280 107 102 578 143 375 586 275 4 190 349 584 1270 85 829 475 141 291 458 264 W360 33 W360 39 W360 45 W360 51 W360 57 W360 64 W360 79 W410 39 W410 46 W410 53 W410 67 W410 74 W410 85 W460 52 W460 60 W460 68 W460 74 W460 82 W460 89 W460 97 W610 82 W610 92 W610 101 W610 113 W610 125 W610 140 W610 155 103 mm3 106 mm4 103 mm3 106 mm4 mm2 mm kgm tf bf tw WIDEFLANGE SECTIONS OR W SHAPES SI UNITS 805 B WideFlange Sections or W Shapes SI Units Web Flange Area Depth thickness width thickness xx axis yy axis Designation A d I S r I S r mm mm mm mm mm mm 16 500 318 1310 3080 206 308 1940 137 100 649 778 9 480 310 940 2050 163 165 1060 132 234 228 497 8 530 306 851 2040 146 145 948 130 207 203 493 4 930 310 584 1650 97 848 547 131 723 876 383 4 180 313 660 1020 108 650 415 125 192 376 214 3 040 305 559 1010 67 428 281 119 116 230 195 2 680 303 508 1010 57 370 244 117 0986 195 192 19 000 282 1730 2630 284 259 1840 117 862 656 674 10 200 256 940 2550 156 126 984 111 431 338 650 8 560 257 889 2040 157 104 809 110 222 218 509 7 400 252 800 2030 135 873 693 109 188 185 504 5 700 266 762 1480 130 711 535 112 703 95 351 3 620 260 635 1020 100 399 307 105 178 349 222 2 850 254 584 1020 69 288 227 101 122 239 207 2 280 251 483 1010 53 225 179 993 0919 182 201 12 700 229 1450 2100 237 113 987 943 366 349 537 11 000 222 1300 2090 206 947 853 928 314 300 534 9 100 216 1020 2060 174 766 709 917 254 247 528 7 580 210 914 2050 142 612 583 899 204 199 519 5 890 203 724 2030 110 455 448 879 153 151 510 4 570 201 622 1650 102 344 342 868 764 926 409 2 860 206 622 1020 80 200 194 836 142 278 223 4 730 162 813 1540 116 222 274 685 707 918 387 3 790 157 660 1530 93 171 218 672 554 724 382 2 860 152 584 1520 66 121 159 650 387 509 368 3 060 160 660 1020 103 134 168 662 183 359 245 2 290 153 584 1020 71 919 120 633 126 247 235 1 730 150 432 1000 55 684 912 629 0912 182 230 W150 14 W150 18 W150 24 W150 22 W150 30 W150 37 W200 22 W200 36 W200 46 W200 59 W200 71 W200 86 W200 100 W250 18 W250 22 W250 28 W250 45 W250 58 W250 67 W250 80 W250 149 W310 21 W310 24 W310 33 W310 39 W310 67 W310 74 W310 129 103 mm3 106 mm4 103 mm3 106 mm4 mm2 mm kgm tf bf tw y y x x bf tw tf d 806 APPENDIX B GEOMETRIC PROPERTIES OF STRUCTURAL SHAPES B American Standard Channels or C Shapes SI Units Web Flange Area Depth thickness width thickness xx axis yy axis Designation A d I S r I S r mm mm mm mm mm mm 9 480 3810 1820 944 1650 168 882 133 458 618 220 7 610 3810 1320 894 1650 145 761 138 384 551 225 6 430 3810 1020 864 1650 131 688 143 338 509 229 5 690 3050 1300 805 1270 674 442 109 214 338 194 4 740 3050 983 774 1270 599 393 112 186 309 198 3 930 3050 716 747 1270 537 352 117 161 283 202 5 690 2540 1710 770 1110 429 338 868 161 271 170 4 740 2540 1340 733 1110 380 299 895 140 243 172 3 790 2540 963 696 1110 328 258 930 117 216 176 2 900 2540 610 660 1110 281 221 984 0949 190 181 3 790 2290 1140 673 1050 253 221 817 101 192 163 2 850 2290 724 631 1050 212 185 862 0803 167 168 2 540 2290 592 618 1050 199 174 885 0733 158 170 3 550 2030 1240 642 990 183 180 718 0824 165 152 2 610 2030 770 595 990 150 148 758 0637 140 156 2 180 2030 559 574 990 136 134 790 0549 128 159 2 790 1780 1060 584 930 113 127 636 0574 128 143 2 320 1780 798 557 930 101 113 660 0487 115 145 1 850 1780 533 531 930 887 997 692 0403 102 148 2 470 1520 1110 548 870 724 953 541 0437 105 133 1 990 1520 798 517 870 633 833 564 0360 922 135 1 550 1520 508 488 870 545 717 593 0288 804 136 1 700 1270 825 479 810 370 583 467 0263 735 124 1 270 1270 483 445 810 312 491 496 0199 618 125 1 370 1020 815 437 750 191 375 373 0180 562 115 1 030 1020 467 402 750 160 314 394 0133 465 114 1 140 762 904 405 690 0862 226 275 0127 439 106 948 762 655 380 690 0770 202 285 0103 383 104 781 762 432 358 690 0691 181 298 0082 332 102 C75 6 C75 7 C75 9 C100 8 C100 11 C130 10 C130 13 C150 12 C150 16 C150 19 C180 15 C180 18 C180 22 C200 17 C200 20 C200 28 C230 20 C230 22 C230 30 C250 23 C250 30 C250 37 C250 45 C310 31 C310 37 C310 45 C380 50 C380 60 C380 74 103 mm3 106 mm4 103 mm3 106 mm4 mm2 mm kgm tf bf tw y y x x d bf tw tf ANGLES HAVING EQUAL LEGS SI UNITS 807 B Angles Having Equal Legs SI Units Mass per xx axis yy axis zz axis Size and thickness Meter Area I S r y I S r x r mm kg mm mm mm mm mm 759 9 680 369 258 617 601 369 258 617 601 396 579 7 380 289 199 626 578 289 199 626 578 401 393 5 000 202 137 636 555 202 137 636 555 404 557 7 100 146 139 453 472 146 139 453 472 297 427 5 440 116 108 462 450 116 108 462 450 297 292 3 710 822 751 471 427 822 751 471 427 300 222 2 810 635 574 475 415 635 574 475 415 302 351 4 480 654 739 382 387 654 739 382 387 248 241 3 060 468 517 391 364 468 517 391 364 250 183 2 330 364 397 395 353 364 397 395 353 251 275 3 510 323 464 303 324 323 464 303 324 198 190 2 420 234 326 311 302 234 326 311 302 199 146 1 840 184 253 316 290 184 253 316 290 200 98 1 250 128 173 320 279 128 173 320 279 202 165 2 100 152 245 269 269 152 245 269 269 173 126 1 600 120 190 274 258 120 190 274 258 174 86 1 090 0840 130 278 246 0840 130 278 246 176 140 1 770 0915 175 227 236 0915 175 227 236 148 107 1 360 0726 136 231 225 0726 136 231 225 149 73 927 0514 939 235 213 0514 939 235 213 150 115 1 450 0524 121 190 206 0524 121 190 206 124 88 1 120 0420 946 194 195 0420 946 194 195 124 61 766 0300 659 198 182 0300 659 198 182 125 70 877 0202 582 152 162 0202 582 152 162 988 47 605 0146 409 156 151 0146 409 156 151 993 25 312 0080 216 160 139 0080 216 160 139 101 L51 51 32 L51 51 64 L51 51 95 L64 64 64 L64 64 95 L64 64 127 L76 76 64 L76 76 95 L76 76 127 L89 89 64 L89 89 95 L89 89 127 L102 102 64 L102 102 95 L102 102 127 L102 102 190 L127 127 95 L127 127 127 L127 127 190 L152 152 95 L152 152 127 L152 152 190 L152 152 254 L203 203 127 L203 203 190 L203 203 254 106 mm3 106 mm4 106 mm3 106 mm4 mm2 y y y x z z x x 808 Slopes and Deflections of Beams C APPENDIX P vmax umax v L 2 L 2 L x P a b v u2 u1 v L u2 u1 x M0 v x L w vmax umax v x w u1 u2 L 2 L 2 v L x w0 u1 u2 u2 7wL3 384EI u1 3wL3 128EI vmax PL3 48EI umax PL2 16EI Simply Supported Beam Slopes and Deflections Beam Slope Deflection Elastic Curve at v w0x 360EIL 13x4 10L2x2 7L42 at x 05193L u2 w0L3 45EI vmax 000652 w0L4 EI u1 7w0L3 360EI L2 x 6 L at x 04598L 17L2x L32 v wL 384EI 18x3 24Lx2 vmax 0006563 wL4 EI 0 x L2 v wx 384EI 116x3 24Lx2 9L32 v xL2 5wL4 768EI v wx 24EI 1x3 2Lx2 L32 vmax 5wL4 384EI umax wL3 24EI x 05774L u2 M0L 3EI v M0x 6EIL 1L2 x22 vmax M0L2 2243EI u1 M0L 6EI 0 x a v xa Pba 6EIL 1L2 b2 a22 u2 Pab1L a2 6EIL v Pbx 6EIL 1L2 b2 x22 u1 Pab1L b2 6EIL 0 x L2 v Px 48EI 13L2 4x22 CANTILEVERED BEAM SLOPES AND DEFLECTIONS 809 C v L x P vmax umax x P v vmax L 2 L 2 umax L x w v vmax umax L x M0 vmax v umax x w v vmax L 2 L 2 umax L x w0 vmax v umax Cantilevered Beam Slopes and Deflections Beam Slope Deflection Elastic Curve v w0x2 120EIL 110L3 10L2x 5Lx2 x32 vmax w0L4 30EI umax w0L3 24EI L2 x L v wL3 192EI 14x L22 0 x L2 v wx2 24EI Ax2 2Lx 3 2 L2B v M0x2 2EI vmax M0L2 2EI umax M0L EI v wx2 24EI 1x2 4Lx 6L22 vmax wL4 8EI umax wL3 6EI v PL2 24EI A3x 1 2 LB L2 x L vmax 5PL3 48EI umax PL2 8EI v Px2 6EI A3 2 L xB 0 x L2 v Px2 6EI 13L x2 vmax PL3 3EI umax PL2 2EI vmax 7wL4 384EI umax wL3 48EI Chapter 1 11 a b 12 13 15 16 17 19 110 111 113 114 115 117 118 119 121 122 123 125 TBz 525 lb ft MBy 788 lb ft TBz 10505 0 MBx 0 MBy 10575 0 VBx 105 lb VBy 0 NBz 0 MH 412 kN m NH 271 kN VH 206 kN NG 981 kN VG 0 MG 0 90002 Maa 0 Maa 180 N m Naa 779 N Vaa 450 N NC 0 VC 450 kip MC 315 kip ft NC 80 lb VC 0 MC 480 lb in Mbb 375 kN m Nbb 177 kN Vbb 354 kN Vaa 125 kN Maa 375 kN m Naa 375 kN NB 5303 kN NC 0 MC 09 N m VC 60 N MD 157 kip ft MC 618 kip ft ND 0 VD 145 kip NC 04 kip VC 108 kip MB 312 kip ft MB 0162 08425 0415 0 VB 0960 kip NB 04 kip NA 207 lb MA 145 lb in VA 773 lb VC 0 NC 120 kip MC 8125 kip ft NB 0 VB 850 lb MB 6325 kip ft VA 450 lb MA 1125 kip ft NA 0 MD 394 kN m ND 0 VD 1875 kN By 300 kN VC 0533 kN MC 0400 kN m NC 200 kN P 0533 kN MC 600 kN m VC 800 kN NC 300 kN ME 240 kip ft MD 135 kip ft NE 0 VE 900 kip By 600 kip ND 0 VD 0750 kip Ay 300 kip 9004 Ay12 0 TC 500 lb ft TB 150 lb ft TD 0 TC 250 N m FA 349 kN FA 138 kip 126 127 129 131 133 134 135 Joint A Joint E Joint B 137 138 139 141 142 143 145 Joint B Joint A 146 147 tA 138 MPa s 339 MPa sAC 833 psi T sBC 469 psi T 417 psi C sAB FAB AAB 625 15 tEavg 124 ksi tDavg 132 ksi tEavg 622 ksi tDavg 662 ksi tBavg 121 ksi FB FC 59424 lb Cy 150 lb By 150 lb Ex 500 lb Ey 350 lb Dy 650 lb savg 5 MPa t 115 psi s 667 psi P 40 MN d 240 m dF 75106 x12 dx sBD 187 ksi C sBC 235 ksi T sEB 480 ksi T sED 853 ksi C sAE 853 ksi C sAB 107 ksi T sD 133 MPa C sE 707 MPa T tavg P 2A sin 2u s P A sin2 u N P sin u V P cos u s 182 MPa MA Pr1 cos u MA Pr1 cos u 0 VA P sin u 0 VA P sin u NA P cos u P cos u NA 0 MBz 0 TBx 942 N m MBy 623 N m VBy 0 VBz 706 N NBx 0 TCy 0 MCz 138 N m MCx 108 N m VCz 240 N NCy 0 VCx 250 N 828 Answers to Selected Problems 149 150 151 153 154 155 157 Inclined plane Cross section 158 159 161 162 163 165 166 167 169 170 171 173 174 175 177 Shear limitation Tension limitation 178 179 181 182 183 185 186 187 Pmax 155 kN A 619103 m2 P 90 kN d 1 1 16 in T 117810 lb FAB 14429 lb W 431 lb P 443 kN dBC 600 mm dBD 700 mm dAB 650 mm P 326 kip P 339 kip N 2827 kip V 1696 kip dA 276 mm a 6 1 2 in b 333 mm t 167 mm d 135 mm d 571 mm h 23 4 in h 274 in tbb 600 psi saa 139 ksi r r1e argpr1 2 2P b z N 720 lb s 102 psi A 7069 in2 s w0 2aA2a2 x2 P 625 kN V 63640 N tavg 509 kPa tBavg 159 ksi tAavg 371 ksi P 153 kN Ax 1732P Ay P FA 2P s 3125 ksi tavg 180 ksi P 683 kN s 101 ksi tavg 0 s 626 ksi tavg 489 ksi V 1219 kip N 15603 kip savgBC 588 MPa savgAB 118 MPa savgb 318 MPa savgs 566 MPa Vp P4 P 905 kN Vb P4 taaavg 250 psi P 4 kip saaavg 667 kPa taaavg 115 kPa P 370 kN Ax 95263P Ay 55P FA 11P 189 190 191 193 194 195 197 198 199 1101 1102 1103 1105 1106 1107 Chapter 2 21 22 23 25 26 27 29 210 211 213 214 y 0218 in x 0192 in PAD 00281103 mmmm PDB 000680 mmmm DB 4966014 mm DB 500 mm AD 40001125 mm AB 30000667 PavgBD 01875 mmmm PavgAB 00889 mmmm gxyA 0206 rad gxyB 0206 rad D 438 mm AB 500 mm AB 50175 mm Pavg 00689 inin g 0197 rad PavgCE 0005 mmmm dB 4 mm PavgBD 000267 mmmm PCE 000250 mmmm PBD 000107 mmmm P 00472 inin P 0167 inin tavg 509 MPa s40 398 MPa s30 707 MPa saa 200 kPa taa 115 kPa tavg 613 MPa F 367875 N dB 13 16 in dA 11 8 in t 1 4 in tavgb 455 MPa tavga 472 MPa ss 208 MPa saaavg 716 MPa taaavg 413 MPa Naa 25981 kN Vaa 150 kN P 550 kN h 174 in dAB 602 mm dCD 541 mm FCD 670 kN FAB 830 kN aA 130 mm aB 300 mm w 0530 kipft tpins 1179 ksi FSpins 153 srod 1326 ksi FSrod 271 FSB 224 FSC 213 dC 629 mm dB 708 mm h 3 8 in 5103 5103 p1h d 5 8 in 210103 5103 p 4d2 ANSWERS TO SELECTED PROBLEMS 829 215 217 218 219 221 222 223 225 226 227 229 230 231 233 Chapter 3 31 32 33 35 36 37 39 310 311 L 0094 in Elastic Recovery 0003 inin PY 118 kip Pult 196 kip E 300103 ksi s 150 ksi P 0035 inin d 0228 in A 0209 in2 P 162 kip E 883103 ksi utapprox 117 MJm3 urapprox 850 in lb in3 ur 996 in lb in3 E 553103 ksi Eapprox 131 0 00004 0 3275103 ksi PAB yB sin u L uA cos u L PAB c1 2vB sin u uA cos u L d 1 2 1 Pavg 0479 ftft L 016 ft L 0100 ft L L 90 0 005 cos u2 du P 005 cos u PavgCF 00687 mmmm PavgAD 0132 mmmm PavgBE 00635 mmmm gxyF 0245 rad PavgCD 0125 mmmm PavgAC 00112 mmmm PAB 168103 mmmm AB 508416 m AB 500 m PBD 113103 mmmm PAC 167103 mmmm gAxy 524103 rad LBD 06155 m Pavg 00258 mmmm gDxy 116103 rad gCxy 116103 rad gAxy 116103 rad gBxy 116103 rad PDB PAB cos2u PCB sin2u LDB L21 2PAB cos2 u 2PCB sin2 u PAC 00274 inin PAB 0152 inin 830 ANSWERS TO SELECTED PROBLEMS 313 314 315 317 318 319 321 322 323 From the stressstrain diagram the copolymer will satisfy both stress and strain requirements 325 326 a b 327 329 330 331 333 334 335 337 338 339 341 342 343 Chapter 4 41 42 43 45 dA 614 mm dD 0850 mm dAD 0766103 in 364103 mm dA 5001038 p 4042 032200109 Pr 0000884 mmmm Pb 000227 mmmm PBC 000193 inin PDE 000116 inin W 112 lb t 14889 kPa G 1481 MPa dh 302 mm x 153 m dA 30008 mm d 00173 mm d 200016 mm E 55 psi ut 1925 psi ur 11 psi Gal 431103 ksi d Pa 2bhG d 0833 mm tavg 416667 Pa g 002083 rad P 530 kip E 286103 ksi gxy 000524 rad Py 00150 inin Px 000540 inin h 2000176 in Plong 00002667 Plat 00000880 n 0330 n 0300 d 05000673 in d 0577103 in d 0126 mm d 000377 mm s 1697 MPa P 113 kN a 0708 sCD 7958 MPa PCD 0002471 mmmm sAB 3183 MPa PAB 0009885 mmmm s 222 MPa Uitapprox 650103 in lb in3 Uir 88 in lb in3 spl 44 ksi sY 60 ksi E 110 103 ksi P 570 lb dBD 00632 in E 286103 ksi s 1143 ksi P 0000400 inin 46 47 49 410 411 413 414 415 417 418 419 421 422 425 426 427 429 430 431 433 434 435 437 438 439 441 442 443 445 446 FD 204 kN FA 180 kN sb 324 MPa st 345 MPa Fb 1017 103 N Ft 2983 103 N sAB 265 MPa sEF 338 MPa d 246 mm scon 842 MPa sst 673 MPa Pcon 36552 Pst Ast 182 in2 d 000545 in d 00055 in sst 166 ksi scon 0240 ksi FD 10789 kN dAB 0335 mm d 239 in sbr 0341 ksi sst 0654 ksi sst 488 MPa scon 585 MPa Pst 5747 kN Pcon 2253 kN sst 314 ksi scon 0455 ksi d 293 mm p0 250 kNm d 0511P pr0E When y r0 4 u 1448 y r0 sin u dy r0 cos u du A pr2 pr0 cos u2 pr2 0 cos2 u d P 2apr0 2E A1 e2aLB d gL2 6E d 0360 mm W 969 kN dC 05332 mm dtot 339 mm dD 01374 mm dAB 03958 mm u 0439103 rad dF 00230 inT d 04310103P P 464 kip dAB 103 mm dAB 0864 mm gL2 2E PL AE d 1 AE L L 0 gAx P dx dt 00260 in u 000878 dF 00113 in dFE 00020690 in dE 00036782 in dC 00055172 in dA 00110344 in dP 00350 inT dA 00128 in ANSWERS TO SELECTED PROBLEMS 831 447 449 450 451 453 454 455 457 458 459 461 Assume failure of AB and EF Assume failure of CD 462 463 465 466 467 469 470 471 473 474 475 477 478 479 481 482 483 485 486 487 489 490 491 493 Maximum normal stress at fillet Maximum normal stress at the hole 494 P 15 kip K 160 smax 883 MPa K 265 K 14 P 121 kip P 771 kN d 0429 mm K 245 w 249 in smax 190 MPa ss 401 MPa sb 295 MPa F 107 44247 N T 172 C FAC 100 lb FAD 136 lb dA 00407 inc FAD 654 kip FAC FAB 409 kip 620136 75FAB 48FAD P 188 kN FB 183 kN FA 383 kN F aAE 2 TB TA 0 T dF d 0348 in F 195 kip F 760 kip 0 dT dF F 1914A s 191 ksi F 116 kip F 0509 kip F 420 kN 0 T d sAB 7P 12A sCD P 3A sEF P 12A a 0120 mm 002 dt db P 116 kN u 000365 sD 134 MPa sBC 955 MPa w 459 kNm FCD 81 000 N F 42 300 N u 114103 FA 579 kN FB 964 kN FC 116 kN FCD 61473 lb FBC 45469 lb u 00633 sCF 113 MPa sBE 963 MPa sAD 796 MPa u 698 srod 928 ksi scyl 116 ksi Fst 1822 kip Fal 3644 kip TCD 272 kip TCD 906 kip x 289 in P 604 kip y 3 0025x FA 409 kip FB 291 kip TAB 361 lb TAB 289 lb 495 497 498 499 4101 4102 a b 4103 4105 4106 4107 4109 4110 a b 4111 4113 4114 4115 4117 4118 4119 Chapter 5 51 a b 52 a b 53 55 56 57 59 tmax 119 MPa J 2545106 m4 tEFmax 0 tCDmax 217 ksi tBCmax 507 ksi tDEmax 362 ksi tmax 267 MPa tB 604 MPa tA 604 MPa r 0841r r 0841r tr05 in 638105 p 20754 054 800 ksi T 638 kip in T 795 kip in u 3E2LT2 T1a2 a1 d5E2 E1 P 565 kN dBA 00918 mm P 464 kip sAB 145 ksi P 485 kip FB 213 kip FA 214 kip d g3L3 3c2 d 1 A2c2 L L 0 gAx2 dx P 126 kip d 000720 in dD 640 in dD 0375 in dB 178 mm FalY 5655 kN Fst 1469 kN w 109 kipft sA 5333 ksi d 869 in sADr 355 MPa C sBEr 532 MPa T sCFr 177 MPa C FBE 91 84461 N FAD 15 43693 N FCF 122 71846 N sCF 250 MPa T P sYA2 cos u 1 dA sYL E cos u P 314 kN P 262 kN dCD 0324 mm dAB 0649 mm FAB 314 kN FCD 272 kN FCD 1800 N FAB 3600 N w 219 kNm dG 424 mm dC 0432 in sst 360 ksi sal 198 ksi Pal 15691 kip Pst 14309 kip P 168 kip K 129 510 511 513 514 515 517 518 519 521 522 525 526 527 529 530 531 533 534 535 537 538 539 541 542 543 545 t 0104 in ri 11460 in T 52521 lb ft d 21 2 in v 177 rads t 25 mm T 625 N m tmaxBC 726 MPa tmaxCF 125 MPa dA 124 mm dB 168 mm tmax 602 ksi T 630254 lb ft P 990 000 ft lbs v 217 rads d 7 8 in tmax 143 ksi T 28011 lb in P 1100 ft lbs tBCmax 311 MPa tABmax 104 MPa c 298 x mm tmax 2TA tALro pr4 o ri 4 TB 2TA tAL 2 TA 1 2tAL TB 0 tBCmax 159 MPa tABmax 239 MPa tmax T 2pri 2h tabs Tmax 26042 lb ft max 359 ksi d 57 mm d 0 tmax 424 MPa d 09 m tmin 0 TAB 2000x 1200 N m tmaxAB 414 MPa tmaxBC 828 MPa tmax 733 ksi tmax 489 ksi TA 960 lb in J 003125p in4 d 33 mm tmaxabs 102 MPa tEAmax 566 MPa tCDmax 891 MPa F 600 N TA 300 N m tBC 236 ksi tAB 782 ksi n 2r3 Rd2 832 ANSWERS TO SELECTED PROBLEMS 546 547 549 550 551 553 554 555 557 558 559 561 562 563 565 566 567 569 570 573 f 7TL 12pr4G Jx pr4 2L4L x4 fD 142 fE 120 fC 0008952 rad fB 001194 rad T2 328 kN m T1 219 kN m fC fB fCB fA fB fAB fC 230 fA 266 fC 0113 fCB 00001119 rad fB 0001852 rad fC 266 tBAmax 186 ksi tBCmax 102 ksi fA 178 fB 153 fF 002667 rad fE 001778 rad fBD 115 fCD 00661 tabs max 317 ksi d 11 4 in TD 6565 lb ft TM 17507 lb ft TC 10942 lb ft fC 0227 fD 101 fB ƒ574ƒ TDA 90 N m TCD 60 N m TBC 80 N m d 275 in f 443 tmax 283 ksi fAD 0879 TBC 85 N m TAB 85 N m tmax 443 MPa f 119 d 7 8 in ANSWERS TO SELECTED PROBLEMS 833 574 575 577 578 579 581 582 583 585 586 587 589 590 591 593 TA 152 189 T TB 37 189 T Jx pc4 2L4 L x4 t abs 550 ksi max tACmax 217 ksi tBDmax 435 ksi fB 0955 TA 1279 kip ft TE 4412 kip ft F 4412 kip fE 166 TA 556 N m TB 222 N m tabs 133 ksi max TR 300x 25x2 lb in fB 175 f 0338 tBDmax 196 ksi tBCmax 147 ksi gbtmax 172106 rad tbrmax 961 psi gstmax 343106 rad tstmax 395 psi fC 0116 tCD 249 MPa TB 0502 kN m TA 1498 kN m tAC 293 ksi tAC 977 MPa tCBmax 407 MPa tACmax 815 MPa TB 100 N m TA 200 N m f 2Lt0L 3TA 3pro 4 ri 4G f 4PLd 3pr4G to 4 pd L 5117 5118 5119 5121 5122 5123 5125 5127 5129 5130 5131 5133 5134 5135 5137 5138 5139 5141 5142 5145 Eq 57 Eq 518 Eq 515 Eq 520 f 4503 f 4495 tavg 8842 MPa tr006 m 8827 MPa ri 00575 m ro 00625 m tmax 19T 12pr3 Tc 761 kN m Tt 7 TYt Plastic Tt 739 kN m Elastic Tt 925695 N m Tc 574305 N m at 15 in t 378 ksi T 412 kip ft at 3 in t 244 ksi T 392 kip ft fP 0413 f 344 T 327 kN m t2 4109r 25106 t1 8109r rg 000625 m TA 570 kN m TC 93 kN m gmax cotY ciG f tYL ciG TP 2 3ptyAco 3 ci 3B T 110 lb ft TP 168 kN m TY 126 kN m T 144 kip ft fr 122 f 03875 rad G 40 GPa f 344 T 208 kN m TP 279 kip ft T 271 kip ft K 140 r 0075 in tmaxf 506 MPa P 101 kW No it is not possible r 798 mm K 128 tavgA tavgB 357 kPa f 0407m tavg 119 MPa f 0428m T 473 MN m Am 18927 m2 834 ANSWERS TO SELECTED PROBLEMS 594 595 597 598 599 5101 5102 5103 5105 5106 5107 5109 5110 5111 5113 5114 5115 tavgB 104 MPa tavgA 156 MPa f 5134 kip ft f 00536 t 5 16 in Am 795950 in2 qst p 4 qct The factor of increase 285 Factor 166 Am 24002 in2 Am 14498 in2 t 0104 in dF 00303 in tmax 231 ksi T 16632 lb in F 104 lb tmaxA 308 MPa fC 00925 TA 48 lb ft TB 32 lb ft T 280 kN m For segment BC T 11 36694 N m For segment AB T 318086 N m fBC ƒ 00643ƒ tACmax 159 MPa tBCmax 0955 MPa fBA 0207 tACmax 159 MPa tBCmax 0955 MPa Factor of increase in shear stress 1 k2 tmaxc 16T pk2d3 tmaxc 16T pd3 fr 00657 fc 00582 trmax 713 psi tcmax 525 psi TA 3t0L 4 TB 7t0L 12 5146 5147 5149 5150 5151 Chapter 6 61 62 63 65 66 67 69 610 611 613 614 617 618 619 621 622 623 625 627 629 630 631 M w0 3L L x3 V w0 L L x2 M w0 2412x2 18Lx 7L2 V w0 4 3L 4x x 254 V 0 M 346 At x 45 m M 2531 kN m At x 4108 m M 2567 kN m x 411 V 0 M 257 M 000617w0L2 x L3 V w0L18 x L V 0 M 0 x L V wL M 0 x 3 V 10 M 18 Ay 15 kN FBC 75 kN x 075 V 0 M 05625 x 5 V 10 M 25 M 800x 120 kip ft V 800 kip M x2 300x 216 kip ft V 300 2x kip M 300x 5556x3 lb ft V 300 1667x2 lb x 6 V 900 M 3000 x 14 V 115 M 3875 x 3a V P M Pa x 6 V 800 M 4800 x 15 V 150 M 225 x 4 V 333 M 467 x 4 V 6 M 24 V 20 M 16 x 15 V 0 M 9 x 4 x 2 V 8 M 39 x 3 V 2000 M 6000 x 2 V 1 M 2 x 4 V 1 M 6 x 025 V 24 M 6 F 262 N f 186 tmax 820 MPa T 715 N m tmax 233 MPa t 8 mm T 331 N m ANSWERS TO SELECTED PROBLEMS 835 633 634 635 637 638 639 641 642 643 645 646 647 649 650 651 653 654 655 657 658 659 661 662 663 a 168r sB 517 MPa T sA 621 MPa C FRC 118 kip sA 20544 ksi sD 02978 ksi y 93043 in I 109307 in4 M 101 kip ft smaxC 200 ksi C smaxT 238 ksi T smax 494 MPa I 178133106 m4 F 456 kN smax 206 MPa M 365 kN m smax 400 MPa I 9114583106 m4 M 771 N m sC 414 MPa sB 101 MPa sA 681 MPa stmax 372 ksi scmax 178 ksi INA 9173 in4 y 340 in smax 90 MPa smax 120 MPa x 0 V 2w0Lp M w0L2p M w0Lx 12 w0x4 12L2 x 0630L V 0 M 00394w0L2 x 14 V 0 M 24 x 1 V 0 M 250 x 4 V 28 M 24 x L V 23 54 wL M 5 54 wL2 Ax 0 Ay 9375 kip x 45 V 0 M 169 M e 100 9 x3 500x 600 f N m V e 100 3 x2 500 f N V 200 N M 200 x N m x 3 V 115 M 21 M 150 lb ft V 300 lb w 400 lbft 6109 6110 6111 6114 6115 6117 6118 6119 6121 6122 6123 6125 6126 6127 6129 6130 6131 6133 6134 6135 6137 6138 6139 6141 sstmax 183 ksi sconmax 195 ksi I 135878 in4 h 5517 in Ast 23562 in2 Mmax 40 kip ft smaxpvc 153 ksi M 986 kN m d 531 mm smaxal 171 MPa smaxst 154 MPa I 1808106 m4 y 01882 m swmax 770 psi sstmax 140 ksi sstmax 201 MPa M 164 kip ft INA 854170 in4 y 25247 in smaxw 0558 ksi smaxst 851 ksi w 0875 kipft smaxal 133 ksi smaxst 226 ksi I 308991 in4 y 23030 in Mmax 253125 kip ft M 660 kN m h 413 mm sA 210 ksi sA 210 ksi C yA 2828 in zA 1155 in sA 293 kPa C sA 293 kPa C smax 161 MPa Mmax 4272 N m M 1186 kN m a 665 sB 131 MPa C sA 760 MPa T Iy 1334583106 m4 Iz 2844583106 m4 My 6000 N m Mz 103923 N m sB 781 ksi sA 895 ksi a 374 sB 0587 MPa T sA 130 MPa C y 574 mm M 119 kip ft a 651 smax 201 ksi C smax 201 ksi T Iz 1584 in4 Iy 736 in4 Mz 1414 kip ft My 1414 kip ft 836 ANSWERS TO SELECTED PROBLEMS 665 666 667 669 670 671 673 674 675 677 678 679 681 682 683 685 686 687 689 690 691 693 694 695 697 698 699 6101 6102 6103 6105 6106 6107 smaxt 3M b h2 a 2Et 2Ec 2Ec b c h2Ec 2Et 2Ec smax 147 psi b 71 2 in I 2 3 b4 v 1125 kNm sB 133 ksi sA 118 ksi smax 198 ksi I 20484375 in4 smax 560 ksi smax 759 ksi Mmax 2P P 119 lb d 199 mm d 116 mm Pmax 0711103 mmmm I 079925106 m4 y 0012848 m smax 119 MPa smax 23w0L2 36bh2 a 669 mm Mmax 750 kN m d 2 in smax 191 ksi b 531 mm smax 129 MPa t 51 2 in smax 125 ksi w 375 kipft smax 156 ksi smax 668 ksi w 165 kipft I 152344 in4 sallow 528 MPa smax 211 ksi smax 100 ksi I 152344 in4 smax 122 ksi smax 221 ksi smax 747 MPa Ib 036135103 m4 Ia 021645103 m4 smax 158 MPa FR 313 kip I 1863 in4 smax 193 psi s 155 psi 6142 6145 6146 6147 6149 6150 6151 6153 6154 6155 6157 6158 6159 6161 6162 6163 6165 6166 6167 6169 6170 6171 6173 6174 6175 6177 6178 6179 6181 6182 a b 6183 a b P 455 kip P 373 kip w0 228 kipft w0 180 kipft M 11ah2 54 sY d 2 3h k 2 k 16ror3 o r3 i 3pr4 o r4 i MY 8783sY Mp 12133sY k 138 k 171 sT sB 142 MPa MY 0000268sY k 157 Ix 268106 m4 Mp 000042sY k 170 k 117 stop sbottom 671 MPa Mp 289 0625 N m I 9114583106 m4 k 171 k 3h 2 c4bth t th 2t2 bh3 b th 2t3 d stop sbottom 435 MPa Iz 8278333106 m4 Mp 21125 kN m L 950 mm smax 295 ksi K 192 P 122 lb r 500 mm smax 120 ksi K 260 M 150 kip ft P 309 N st 201 MPa T sB 262 MPa C A 0008 m2 LA dA r 6479051103 m r 1235 m sB 127 ksi C sA 106 ksi T smaxc 120 psi C smaxt 204 psi T sC 266 MPa T LA dA r 8348614103 m r 05150 m A 000425 m2 smaxc 544 MPa smaxt 451 MPa P 552 kN M 140 kN m LA dA r 0053049301 m A 00028125p m2 M 975 kip ft ANSWERS TO SELECTED PROBLEMS 837 6185 6186 a b 6187 6189 6190 6191 6193 6194 6195 a b 6197 6198 6199 6201 Chapter 7 71 72 73 75 76 77 79 710 711 713 714 715 717 718 719 twmax 374 MPa tfmax 924 MPa V 723 kN tmax 374 MPa Qmax 109125103 m3 I 0175275103 m4 The factor 4 3 V 190 kN tmax 422 MPa y 0080196 m I 48646106 m4 V 100 kN tmax 448 ksi y 11667 in I 675 in4 V 321 kip tmax 462 MPa tA 199 MPa tB 165 MPa Q 6534 6y2 Vf 382 kip y 330 in INA 39060 in4 Vw 190 kN tmax 346 MPa tA 256 MPa I 02501103 m4 QA 064103 m3 a 45 u 45 ds du 0 s 6M a3 cos u sin u x 06 V 233 M 50 M x2 20x 166 V 20 2x sB 265 kPa T sA 225 kPa C A 625103 m2 LA dA r 0012908358 m smax 0410 MPa smax 0410 MPa M 264 kN m M 149 kN m n 18182 I 0130578103 m4 smaxc 162 MPa C smaxt 343 MPa T FR 588 kN M 735 kip ft s 82 ksi M 251 N m M 598 kip ft M 350 kip ft s 50sd 3500106d 0 M 947 N m 766 767 769 770 771 773 774 775 Chapter 8 81 82 83 Case a Case b 85 86 87 a b c 89 810 811 813 814 815 817 818 819 sL 667 MPa C sR 333 MPa T d 667 mm sw pr t t T wt sfil pr t t T wt p Er2 r3 r2 2 r2 r1 r2 3 r4 r3 dri pr2 i Ero ri dF dT 0 sc 269 ksi sh 432 psi sb 880 ksi s 333 in Pballow 12272103 N ns 308 bolts tc 40 mm ts 20 mm tavgr 322 MPa s1 791 MPa s1 127 MPa t 267 mm n 820 bolts sb 228 MPa s 133 MPa Pb 3556103p N s1 104 ksi s2 520 psi s1 104 ksi s2 0 ro 755 in t 188 mm V 749 lb V 410 kip qC 378 kNm qA 0 qB 121 kNm QA 0 QC 016424103 m3 INA 8693913106 m4 y 008798 m VAB 996 kip e 4r sin a a cos a 2a sin 2a e b6h1h2 3h2b 8h3 1 2h3 6bh2 h 2h13 I t 12 2h3 6bh2 h 2h13 Pe Fh 2Vb e 0 e 223 3 a 838 ANSWERS TO SELECTED PROBLEMS 721 722 723 725 726 727 733 734 735 737 738 739 741 742 743 745 747 750 751 753 754 755 757 758 759 761 762 763 765 e 7 10a Q2 at 2 a 2x Q1 t 2 y2 I 10 3 a3t e 3b2 2 b2 1 h 6b1 b2 t V pR2t2R2 y2 qmax 641 lbin qB 452 lbin qA 196 lbin I 92569 in4 y 28362 in qmax 232 kNm qA 215 kNm qmax 414 lbin I 14598 in4 y 370946 in qmax 163 kNm qA 139 kNm qB 125 kNm qC 386 kNm QC 05375103 m3 INA 12517106 m4 qC 0 qD 601 kNm qB 462 kNm qA 228 kNm s 866 in s 121 in P 660 kN Q 0450103 m3 INA 720106 m4 tnailavg 119 MPa s 11 8 in V 882 kip P 691 kip Qmax 2085 in3 INA 2902 in4 Q 168 in2 F 125 kN tn 352 MPa Q 10125 in3 V 345 kip INA 9325 in4 s 21 8 in V 180 kip V 135 kip I 320 in4 Q 120 in4 F 675 lb tC 143 ksi tD 117 ksi tmax 280 psi Qmax 0216103 m3 tmax 367 MPa VC 1375 kN I 270106 m4 tmax 485 MPa tB 441 MPa tA 239 ksi Q 2 3 4 y232 I 4p in4 821 822 823 825 826 827 829 830 831 833 834 835 837 838 839 841 842 843 845 846 847 849 850 851 853 854 sC 128 kPa C sD 691 kPa C sA 988 kPa T sB 494 kPa C Iy 10125 m4 y 075 15x A 135 m2 Ix 2278125 m4 6ey 18ez 6 5a stmax 837 ksi scmax 695 ksi sA 891 MPa C sB 793 kPa T M 142463 N m R 0080889 m N 24525 N smax 0368P r2 C smin 00796P r2 T smax 133P a2 C smin P 3a2 T sA 100 ksi C sB 300 ksi C Iz 540 in4 A 180 in2 Iy 135 in4 s 239 MPa C t 0796 MPa s 179 MPa C t 106 MPa QB 0 sB 0522 MPa C tB 0 A 900 103 m2 I 828 106 m T 216 kip T 216 kip tB 100 MPa sB 153 MPa C J 03125p109 m4 A 25p106 m2 Iz 015625p109 m4 tB 0869 ksi sA 941 ksi tA 0 sB 269 ksi tE 0 sD 0 tD 667 psi sE 233 ksi T tB 0 sC 625 psi C tC 162 psi QC 4103 in3 sB 556 ksi T I 10667103 in4 QB 0 d 667 mm sA 504 kPa C tA 149 kPa tA 600 psi tB 0 sA 533 psi T sB 1067 psi C I 00078125 in4 A 0375 in2 QA 00234375 in3 P 109 kN w 797 mm sB 535 ksi tB 0 M 175 lb in N 606218 lb V 350 lb smax 107 MPa smax 107 MPa 123 MPa sB 625 MPa sA P A Mc I ANSWERS TO SELECTED PROBLEMS 839 855 857 858 859 861 862 863 865 866 867 869 870 871 873 874 875 877 878 879 881 882 883 885 886 s1 500 MPa s2 250 MPa Fb 133 kN n 113 bolts p 360 MPa Fb 63617106 N s1 707 MPa s2 0 smax 440 ksi T P 942 kN p 12106 MPa F 30103p smax 236 psi C stmax 288 ksi T scmax 240 ksi C stmax 490 ksi T scmax 408 ksi C LA dA r 0035774 in A 0049087 in2 sE 802 kPa tE 698 kPa sA 213 psi sB 122 psi s 162 psi T t 384 psi I 09765625103p in4 QB 00104167 in3 R 174103 in e 00089746 in sB 0 tB 0377 ksi tA 0 sA 302 ksi C sB 217 MPa tB 0 A 1256637 103 m2 QB 0 I 01256637 106 m4 h 6 ey h 12 s 2P bh3 h2 18eyy sB 466 psi C tB 422 psi sA 605 psi T tA 327 psi QzA 038542 in3 J 107379 in4 QyA 0 Mz 43301 lb in Iy Iz 053689 in4 T 51962 lb in My 250 lb in tD 624 ksi tC 524 ksi sC 156 ksi T sD 124 ksi T sB 780 ksi T tB 340 ksi tA 284 ksi QAz 008333 in3 sA 162 ksi T I 0049087 in4 QAx 0 A 07854 in2 J 0098175 in4 smax 710 MPa C s 586 ksi C t 480 ksi t 484 ksi Mx 4800 lb in s 176 ksi T Vx 500 lb Ty 7200 lb in Ny 800 lb Vz 600 lb sA 119 MPa T tA 0318 MPa 840 ANSWERS TO SELECTED PROBLEMS Chapter 9 92 93 95 96 97 99 910 911 913 914 a b 915 917 918 919 921 922 923 sx 0507 MPa txy 0958 MPa I 045103 m4 QA 16875103 m3 up1 781 up2 119 s1 493 MPa s2 111 MPa s1 801 ksi s2 199 ksi sx 51962 ksi txy 30 ksi savg 80 MPa us 168 732 tmax inplane 144 MPa up1 618 up2 282 s1 224 MPa s2 642 MPa txy 102 MPa sx 193 MPa sy 357 MPa savg 25 MPa tmax inplane 112 MPa up1 133 up2 767 s1 137 MPa s2 868 MPa txy 50 MPa sx 125 MPa sy 75 MPa us 565 843 tmax inplane 510 MPa up1 393 up2 507 s1 190 MPa s2 121 MPa us 257 643 savg 15 ksi tmax inplane 192 ksi up2 193 up1 707 s2 342 ksi s1 421 ksi sy 127 psi txy 201 psi sx 277 psi sy 350 psi txy 75 psi u 60 sx 200 psi txy 417 ksi sx 271 ksi txy 417 ksi sx 271 ksi txy 40 MPa sx 5 MPa sy 0 txy 45 MPa u 135 sx 80 MPa txy 348 MPa sx 497 MPa txy 348 MPa sx 497 MPa txy 455 psi sx 388 psi txy 0 u 30 sx 650 psi sy 400 psi txy 415 psi sx 678 psi txy 463 ksi sx 348 ksi 925 926 927 929 930 931 933 934 935 937 938 939 941 942 943 up1 819 up2 811 s1 127 MPa s2 624 MPa tmax inplane 355 ksi s1 297 ksi s2 412 ksi s1 212 MPa s2 0380 MPa up 763 A 375103 m2 y 00991 m I 74862106 m4 sx 823 kPa txy 475 kPa tmax inplane 2 pd2 a2PL d Fb s1 4 pd2 a2PL d Fb s2 0 A p 4 d2 I p 64d4 QA 0 savg 0 tmax inplane 5 kPa us 45 45 tmax inplane 668 psi s1 0 s2 134 ksi up1 150 up2 450 s1 240 s2 240 MPa s1 0 s2 192 MPa QB 9375106 m3 I 03125106 m4 QA 0 up1 134 up2 103 s1 638 MPa s2 0360 MPa Point B s1 00723 ksi s2 0683 ksi Point A s1 150 ksi s2 00235 ksi up1 157 up2 743 s1 649 MPa s2 515 MPa I 49175106 m4 QA 0255103 m3 V 705 kN M 3915 kN m savg 115 ksi us 45 135 tmax inplane 115 ksi s1 0 s2 229 ksi savg 149 ksi us 45 45 tmax inplane 149 ksi s1 298 ksi s2 0 savg 630 MPa tmax inplane 630 MPa s1 0 s2 126 MPa A 01103p m2 I 25109p m4 N 400 N M 100 N m 945 946 947 949 950 951 953 954 955 958 959 961 962 963 a b 965 us 257 tmax inplane 192 psi savg 150 psi s2 421 psi up 193 s1 342 psi R 192094 us 282 Counterclockwise tmax inplane 901 ksi up1 168 Clockwise s2 151 ksi s1 165 ksi savg 750 ksi txy 303 ksi sy 325 ksi sx 0250 ksi sy 111 MPa txy 551 MPa sx 299 MPa savg 155 MPa R 56923 MPa sy 399 ksi txy 146 ksi sx 499 ksi sy 421 MPa txy 354 MPa sx 421 MPa sy 989 ksi sx 199 ksi txy 770 ksi us1 301 clockwise savg 750 MPa tmax inplane 605 MPa up1 149 counterclockwise s2 680 MPa s1 530 MPa tmax inplane 192 ksi savg 15 ksi us2 643 s1 421 ksi s2 342 ksi up2 193 R 1921 ksi sx 388 psi txy 455 psi tmax inplane 10 kPa s1 0 s2 20 kPa tmax inplane 387 MPa s1 0 s2 774 MPa QAy 0 Iy 6875106 m4 Iz 0350103 m4 s1 550 MPa s2 0611 MPa up1 45 up2 45 s1 219 psi s2 219 psi savg 270 ksi us 45 45 tmax inplane 270 ksi I 866667 in4 QA 0 V 2 kip M 13 kip ft 966 a b 967 a b 969 970 971 973 974 975 977 978 979 981 982 983 985 986 987 tabs max 832 psi sint 0 psi smin 822 psi smax 158 psi tabs max 918 MPa s3 468 MPa s1 0 s2 137 MPa smin 300 psi sint 0 smax 400 psi us 45 Counterclockwise tmax inplane 376 MPa s1 752 MPa s2 0 up1 344 Counterclockwise s1 325 MPa s2 0118 MPa up1 608 Counterclockwise s1 918 MPa s2 0104 MPa A 14103 m2 I 17367106 m4 N 900 N V 900 N M 675 N m txy 592 kPa sx 470 kPa R 05984 MPa t 02222 MPa savg 05556 MPa sx 500 MPa txy 167 MPa s1 s2 480 ksi tmax inplane 279 ksi s1 438 ksi s2 120 ksi s1 00723 ksi s2 0683 ksi s1 150 ksi s2 00235 ksi A 180 in2 I 540 in4 QA 10125 in3 s2 206 psi s1 686 psi txy 217 kPa sx 125 kPa txy 226 kPa sx 110 kPa us 144 Clockwise tmax inplane 571 MPa up1 306 Counterclockwise s2 496 MPa s1 646 MPa u2 427 savg 375 MPa tmax inplane 506 MPa s2 131 MPa s1 881 MPa ANSWERS TO SELECTED PROBLEMS 841 989 990 993 994 psi 995 997 998 999 9101 9102 9103 9105 9106 Chapter 10 102 103 105 a b Pavg 100106 gmax inplane 108103 up2 652 up1 248 P2 641106 P1 441106 P 111 hp Py 348106 gxy 233106 Px 248106 txy 132 kPa sx 229 kPa up1 45 up2 45 t 264 kPa s1 264 kPa s2 264 kPa I 20833106 m4 QC 3125106 m3 up2 411 Clockwise s2 430 MPa s1 329 MPa txy 357 MPa sx 633 MPa tmax inplane 2 pd2AF2 64T02 d2 s2 2 pd2 aF AF2 64T0 2 d2 b s1 2 pd2 aF AF2 64T02 d2 b s1 329 psi s2 721 psi s1 119 psi s2 119 psi tmax inplane 232 MPa A 0015625p m2 J 03835103 m4 T0 600103 N m P 0900106 N ms tabs max 546 ksi sint smin 0 smax 109 ksi tabs max 755 psi sint 0 smin 926 smax 582 psi tabs max 50 MPa sint 50 MPa smin 0 smax 100 MPa tabs max 548 ksi s2 0 s3 423 ksi s1 673 ksi tabs max 162 MPa s3 102 MPa s1 222 MPa s2 0 MPa 842 ANSWERS TO SELECTED PROBLEMS 106 107 109 1010 1011 1013 1014 a b 1017 1018 1019 1021 a b 1022 a b c gabs max 773106 gmax inplane 696106 P2 768106 P1 773106 Pavg 275106 us 776 822 g max inplane 187106 P2 182106 up1 528 up2 372 P1 368106 R 93408 gxy 423106 Py 541106 Px 309106 us 317 Pavg 30106 gxy max inplane 335106 up 133 P2 198106 P1 138106 R 16771106 Pavg 275106 us 776 822 g max inplane 187106 up1 528 up2 372 P2 182106 P1 368106 Pavg 150106 us 317 122 gmax inplane 335106 up1 767 up2 133 P2 318106 P1 177106 Pavg 150106 us 225 675 gmax inplane 141106 up1 225 up2 675 P2 221106 P1 793106 gxy 718106 Py 467106 Px 103106 542 Pavg 30106 us 358 gmax inplane 316106 up1 922 up2 808 P2 128106 P1 188106 Py 185106 gxy 248106 Px 215106 Py 155106 Px 145106 gxy 583106 1023 a b c 1025 1026 1027 1033 1034 1035 1037 a b 1038 1039 1041 1042 1043 1045 1046 s1 837 ksi s2 626 ksi th 0206 in P 390 lb Pz 244103 Py 0972103 Px 235103 LCD 6 nM Eh2 LAB 3 nM 2Ebh sz 12 My bh3 Py 12 nMy Ebh3 tabs max 857 MPa tmax inplane 0 r 343 MPa Pmin 910106 Pmax 546106 Pint 364106 Kg 513103 ksi Kr 333 ksi Pint Pmin 107106 Pmax 305106 g abs max 411106 s3 0 s1 102 ksi s2 738 ksi us 125 Clockwise Pavg 75106 gmax inplane 828106 up1 325 Counterclockwise P2 489106 P1 339106 us 168 Clockwise Pavg 0 gmax inplane 902106 up1 282Counterclockwise P1 451106 P2 451106 us 369 Counterclockwise Pavg 100106 gmax inplane 416106 R 20817106 up2 805 Clockwise P2 108106 P1 308106 gabs max gmax inplane 344106 P2 152106 P1 192106 ANSWERS TO SELECTED PROBLEMS 843 1047 1049 and 1050 1051 1053 1054 1057 1058 1059 1061 1062 1063 1065 1066 1067 1069 1070 1071 No 1073 Yes 1074 No 1075 1077 1078 1079 1081 sx 121 ksi s2 sx 2 s2 389 ksi FS 180 FS 159 s1 7314 ksi s2 15314 ksi FS 143 t 3056 ksi s 9549 ksi Me 2M2 T2 Yes s2 50 19723 147 MPa s1 50 19723 247 MPa Me AM2 3 4T2 Te A 4 3M2 T2 a 178 in s 450 a3 d 188 in d 0794 in d 0833 in T 3300 x lb in v 80 p rads s2 x s2 y sxsy 3t2 xy s2 Y k 135 Eeff E 1 n2 Pz 544103 Px Py 0 sz 552 ksi sx sy 700 ksi 0 sz 035sx 035sy 620 0 sy 035sz 035sx 262 0 sx 035sy 035sz 262 G 250 GPa n 0335 E 677 GPa wx 723 kN m wy 184 kN m sy 168 ksi C sx 155 ksi C Py 025103 sz 0 Px 03125103 P3 763106 P1 833106 P2 168106 1082 1083 1085 1086 a b 1089 1090 1091 1093 a b 1094 1095 a b c 1097 1098 a b 1099 10101 10102 10103 us 168 Counterclockwise gmax inplane 361106 up1 282 Clockwise P2 120106 P1 480106 Pavg 300106 Py 120106 gxy 232106 Px 480106 us 292 Counterclockwise Pavg 685106 gmax inplane 622106 up1 158 Clockwise P1 996106 P2 374106 T 736 N m us 978 Clockwise gmax inplane 1593106 up 548 Clockwise P2 713106 P1 880106 Pavg 833106 No s1 35042 MPa s2 6542 MPa gabs 48210 max 6 gmax inplane 313106 P2 168106 P1 482106 P pr 2Et1 n t 195 mm t 225 mm sallow 16667106 Pa d 150 in T 967 kN m T 838 kN m s1 994718T s2 994718T tmaxs 994718T tmaxh 862628T p 2t 23r sY p 1 rsY sY 197 ksi s1 88489 ksi s2 108489 ksi sY 910 ksi sY 943 ksi Chapter 11 111 112 113 115 116 117 119 1110 1111 1113 1114 1115 1117 No the beam fails due to bending stress criteria 1118 1119 1121 Yes 1122 1123 1125 1126 1127 1129 1130 Yes it can support the load 1131 1133 The bending stress is constant throughout the span s 3PL 2b0t2 S b0 r2 3L x smax 3PL 8bh2 0 s 111 2 in s 53 4 in Use s 33 4 in P 432 kip h 720 in P 8333h2 w 108 kNm Use W14 22 Yes the joist will safely support the load h 0643 in smid 502 mm sends 167 mm w 302 kNm Use h 91 8 in smax 1746 ksi di 130 mm d 114 mm S 6523 in3 smax 265 ksi b 155 in Use W14 30 Use W12 26 Sreqd 3273 in3 P 249 kN Use W14 43 Use W12 26 tmax 267 ksi Sreqd 2945 in3 w 612 kNm Member BC Use W6 9 Member AB Use W10 12 Use W12 16 Sreqd 150 in3 Use b 425 in Use W12 22 b 211 mm h 264 mm Qmax 01953125b3 Ix 016276b4 844 ANSWERS TO SELECTED PROBLEMS 1134 1135 1137 1138 1139 1141 1142 1143 1145 1146 1147 1149 1150 1151 1153 1154 1155 Chapter 12 121 122 123 125 126 127 vmax P 3EIAB ea1 IAB IAC bl3 L3 f vmax PL3 8EI v3 P 12EI 2x3 3 9Lx3 3 10L2x3 3L3 v1 Px1 12EIx3 1 L2 v2 P 24EI 4x3 2 7L2x2 3L3 v1 P 12EI x3 1 L2x1 Mx1 P 2 x1 Mx2 Px2 W 113 lb s 582 MPa smax c r E 100 MPa Use d 19 mm Use d 21 mm smax wL2 4bh2 0 S bh2 0 6L2 2x L2 Use W10 12 Use d 41 mm M 127475 N m Use d 44 mm Use d 21 mm d 343 mm Use d 36 mm M 4961 N m c 00176 m Use d 11 4 in T 100 N m Use d 33 mm Use d 29 mm T 100 N m c 001421 Use d 20 mm b b0 L2x2 d d0A x L sallow Px b0d26 sabs max 0155w0L2 bh0 2 h h0 L323L2x 4x312 ANSWERS TO SELECTED PROBLEMS 845 129 1210 1211 1213 1214 1215 1217 1218 1219 1221 1222 vmax 0369 inT umax 000466 rad vC 11wL4 384EI T Mx1 wL 8 x1 Mx2 w 2 x2 2 vƒx L 2 0 uB M0L 6EI vmax 00160 M0L2 EI c vmax 00160 M0L2 EI T v M0 6EIL3Lx2 2x3 L2x uA M0L 6EI vC 7M0L2 24EI c v2 M0 24EI12x2 2 20Lx2 7L2 v1 M0 6EILx3 1 L2x1 uA M0L 6EI Mx1 M0 L x1 Mx2 M0 vC Pab2 8 EI v3 Pax3 2EI x3 b a2 2a 3b v1 P 6EI x3 1 3aa bx1 uA Pab 2EI vmax 3PL3 256EIT vC PL3 6EI uA 3 8 PL2 EI vmax 0484 Pa3 EI T v2 P 18EIx3 2 9ax2 2 19a2x2 3a3 v1 P 9EI x3 1 5a2x1 vmax 23M0L2 27EI umax M0L 3EI v2 Pa 6EIL3x3 2 L x3 2 2L2 a2x2 a2L v1 Pb 6EILAx3 1 L2 b2x1B M1 Pb L x1 M2 Pa a1 x2 L b 1239 1241 1242 1243 1245 1246 1247 1249 vmax 176 in T 125 6 x4 125 6 8x 694 23625xd v 1 EIc400x3 1008x 993 250x2 2508x 692 M 2400x 6008x 99 uB 0705 vB 517 mm vmax 110 mm T 06258x 394 1 248x 395 77625xd v 1 EIc375x3 10 3 8x 1593 vƒx7 m 835 kN m3 EI uA 279 kN m2 EI M 246x 15x2 158x 492 508x 79 vmax 129 mm T 1 68x 1595 5 48x 1594 d v 1 EIc625x3 3375x2 1 6x5 vmax 23PL3 648 EI T 3hx 2 3Li 3 2L2xd v P 18EIc3x3 3hx L 3 i 3 uA PL2 9EI vmax 5M0L2 72EI T 3hx 2 3Li 2 Lxd v M0 6EIc3hx L 3 i 2 M M0hx L 3 i 0 M0hx 2 3Li vmax 133 mm T 258x 493 93333x v 1 EI41667x3 58x 293 846 ANSWERS TO SELECTED PROBLEMS 1223 1225 1226 1227 1229 1230 1231 1233 1234 1235 1237 1238 1838x 4093 4000x lb in3 v 1 EI 167x3 667 8x 2093 vE 0501 mm vD 0698 mm vC vE 0501 mm 1508x 0759 M 180x 1508x 0259 608x 059 vmax 364 mm smax 3PL 2nbt2 vC PL3 32EIC I a2Ic L bx vmax PL3 2EI0 vAƒx0 gL4 6r2E uA gL3 3r2E I a2Lc L bx vAƒx0 gL4 h2E uA 2gL3 h2E Ix bh3 12L3 x3 vA 352 in uA 00611 rad vB wa3 24EI 4L a v2 wa3 24EI 4x2 a v1 w 24EI A x4 1 4ax3 1 6a2x2 1B uB wa3 6EI vmax 2074 kN m3 EI T v 1 EI a6x3 1 60x5 540xb kN m3 uA 540 kN m2 EI Mx a36x 1 3x3b kN m umax w0L3 45EI 1250 1251 1253 1254 1255 1257 1258 1259 1261 max ƒtBCƒ M0L2 8EI uCA M0L 2EI umax uA M0L 2EI vC 1080 kip ft3 EI T uC 240 kip ft2 EI vmax vC 1080 kip ft EI T uA 120 kip ft2 EI B ƒtBAƒ 7PL3 16EI T uB ƒuBAƒ 5PL2 8EI C 50 625 EI uC 39375 EI 256x 2637 kip ft3 0005568x 995 v 1 EI 000556x5 1298x 993 vC 3110 kip ft3 EI uA 302 kip ft2 EI 251x 364 kN m3 0258x 1594 46258x 4593 v 1 EI025x4 02088x 1593 958x6 kN m3 08338x 594 2978x 592 v 1 EI500333x3 00833x4 9586 kN m2 03338x 593 890 8x 592 dv dx 1 EI50100x2 0333x3 ANSWERS TO SELECTED PROBLEMS 847 1262 1263 1265 1266 1267 1269 1270 1271 1273 1274 1275 1277 1278 1279 1281 a 0865L D ƒtADƒ 13M0a2 27EI x 13 3 a uA ƒtBAƒ a M0a 6EI C ƒtCAƒ ƒtBAƒ a L uD PL2 16EI C 5PL3 384EI umax 5PL2 16EI max 3PL3 16EI A ƒtACƒ ƒtBCƒ 7Pa3 3EI T uA ƒuACƒ 5Pa2 2EI tAC 13Pa3 3EI tBC 2Pa3 EI uAC 5Pa2 2EI E Pa 24I3L2 4a2 vmax 3048 kip ft3 EI T M0a2 4EI c C 1 2 tBA ƒtCAƒ uA ƒtBAƒ L 5M0a 12EI tBA 5M0a2 6EI tCA M0a2 6EI max 000802PL3 EI uB PL2 12EI uA PL2 24EI C PL3 12EI 19Pa3 6EI T C tAC uA uAC 5Pa2 2EI F P 4 uA 4PL2 81EI a 0858L x 23 3 a uA ƒtBAƒ a C ƒtCAƒ ƒtBAƒ a L uA 000879 rad uC 4M0L 3EI C 5M0L2 6EI 12102 12103 12105 12106 12107 12109 12110 12111 12113 12114 12115 12117 12118 12119 12121 12122 12123 Cy 14625 kip Ay 2625 kip By 3075 kip Cx 0 MA PL 4 Ay 3P 4 By 7P 4 Ay 125 N Cy 125 N By 550 N vB 13333By m3 EI c vB 36667 N m3 EI T Ay 3P 32 Cy 13P 32 By 11P 16 Bx 0 Cx 0 By 3M0 L Cy 3M0 2L Ay 3M0 2L a 0414L Ay PL a22L a 2L3 tAB2 AyL3 3EI tAB1 PL a22L a 6EI MA M0 2 Ay 3M0 2L By 3M0 2L Ax 0 By 340 kN Ay 340 kN FC 112 kN Ax 0 TAC 3A2E2wL 4 1 8A2E2L 3 1 3E1I1L2 Ma wL1 2 2 TACL1 By 5wL 8 Cy wL 16 Ay 7wL 16 Ax 0 Ay w0L 10 By 4w0L 5 Cy w0L 10 Ax 0 Cy 107 kip T Ay 107 kip By 144 kip 3 2 8x 1092 Mx Cyx By 8x 109 MB Pa L L a MA Pa L L a By 5P 2 Ay 3P 2 MA PL 2 Cy 5 16 P By 11 8 P Ay 5 16 P Mx2 Cyx2 Px2 PL 2 Mx1 Cyx1 MB M0 2 By 3M0 2L Ay 3M0 2L Use W14 34 848 ANSWERS TO SELECTED PROBLEMS 1282 1283 1285 1286 1287 1289 1290 1291 1293 1294 1295 1297 1298 1299 12101 xmax 309 in ymax 0736 in xmax ymax Ix Iy tan u ymax P cos u L3 3EIx xmax P sin u L3 3EIy Av 4000 lb in3 EI T uA 1053 lb in2 EI A Pa23b a 3EI A PL3a 1 12EI 1 8JGb A2 PL3 8JG B PL3 24EI A1 PL3 24EI u PL2 4JG C 232 m T uA 36 EI A 72 EI C 190 in C2 80 EI T 2x Mx 6LEI L2 x2 C1 2560 EI T C 133 mm T uB 000722 rad D wa4 12EI T uA wa3 6EI C wa4 8EI T uC wa3 6EI C3 uB3a wa4 3EI T C2 wa4 8EI T uC2 wa3 6EI C1 wa4 3EI c uC1 uB1 wa3 3EI C 0895 in T uA 0175 C ƒtACƒ 00371 in T uB ƒuBCƒ 000160 rad C a3 24EI 64P 7waT uC a2 6EI12P wa vB 0981 in T uB 000778 rad 13109 13110 13111 13113 13114 13115 The column is not adequate 13117 13118 13119 13121 13122 13123 13125 Yes 13126 13127 13129 13130 13131 13133 Yes 13134 13135 13137 No Chapter 14 141 143 Uia 328 J Ui V 1 2E s2 x s2 y 2nsx sy t2 xy 2G Iz 75125104 m4 Iy 20615278106 m4 x 006722 m Pcr 839 kN Pallow 576 kip aKL r b x 5625 a KL r b y 12821 controls P 250 kip w 463 kNm Pcr 2k L M PL 2 u Bx 0 By 2M L sin u M PL 2 sin u P 344 kip P 169 kip KL d 432 in P 248 kip P 132 kip P 980 kip Iy 320 in4 A 240 in2 Ix 720 in4 P 279 kip P 986 kip P 957 kip A 12 in2 Ix 166 in4 Iy 4275 in4 P 146 kip P 860 kip aKL r b y 10030 aKL r b x 13333 controls P 407 kip P 794 kip P 803 kip saallow 16510 ksi KL ry 69231 ANSWERS TO SELECTED PROBLEMS 851 145 146 147 149 1410 1411 1413 1414 1415 1417 1418 1419 1421 1422 15 times as great as for a uniform cross section 1423 a b 1425 1426 1427 1429 1430 1431 1433 1434 1435 B 117 mm B 267 in uE 315 Ue 150 uE Ui 65 625 EI uB 000100 rad Ui 0726 in kip C 00145 in uB M0L EI Ue 1 2 M0 uB C 2PL AE Ch 2PL AE Ah 00142 in FDC 300 kip T FDB 250 kip C FAB 150 kip C FAD 250 kip T Ui w2L5 40EI Ui w2L5 40EI Ui 3P2L3 bh3E Ui P2L3c 3 16EI 1 8JG d Mx Py Ty PL 2 Ui P2r3 JG a3p 8 1b Uib 293 J Uib 336 J Mx a9x 1 4x3b kN m Uib P2L3 48EI Ui M2 0L 24EI Uiy Uib 31 n 5 a h L b 2 M Px V P Uit 00637 J Uit 7T2L 24pr0 4G T 100 kN m Ui 149 J T 20 kN m T 80 kN m P 375 kN Uia 169 kJ Uia 432 J Ui 0372 J NAB 3 kN NBC 7 kN NCD 3 kN 1478 1479 1481 1482 1483 1485 1486 1487 1489 1490 1491 1493 1494 1495 1497 1498 1499 14101 14102 14103 C 0557 in T D wL4 96EI T uC 13wL3 576EI mux1 x1 L mux2 1 Mx2 w 3L x2 3 Mx1 w 24 11Lx1 12x1 2 uA 000700 rad uA Pa2 6EI uC 5Pa2 6EI mu x1 a mu 1 M1 Px M2 Px2 A 274 mm T uA 575103 rad uC 589103 rad uA 273 mu 1 01176x1 mu 01176x3 M 32706x1 M 65412 4706x2 uB 000595 rad uA 000298 rad C 409 mm T mx1 050x1 mx2 x2 Mx1 250x1 Mx2 x2 2 C 57292 kN m3 EI C 23Pa3 24EI Gv 341 mm T Cv 581 mm T 1 N Cv 17428125103 015103200109 Cv 0163 in Bv 379 mm T Av 623 mm T 1 N Av 498125103 N2 m AE Bh 0367 mm Ev 000966 in T 852 ANSWERS TO SELECTED PROBLEMS 1437 1438 1439 1441 1442 a b 1443 1445 1446 1447 1449 1450 1451 1453 1454 1455 1457 1458 1459 1461 1462 1463 1465 1466 1467 1469 1470 1471 1473 1474 1475 1477 Bv 00124 in T 1 kip Bv 1620 kip2 in AE Cv 204 mm Bv 00931103 in T Bh 0699103 in 1 lb Bh 40 53333 lb2 in AE max 233 mm smax 489 MPa beam 0481 in smax 101 ksi v 575 ms kb 493425106 Nm b 0050342 m h 657 m smax 478 MPa kbeam 17700 kipin n 167 h 373 in smax 588 ksi h 223 m n 20913 h smaxL2 3Ec csmaxI WLc 2 d Amax 154 in sst 15378 psi n 19508 h 116 ft smax 436 ksi smax 207 MPa sst 03123 MPa L 325 mm h 0571 m smax 857 MPa smax 237 MPa max 395 mm st 0613125103 m k 160106 Nm smax 216 MPa h 696 mm st 98139106 m smax 359 MPa d 535 in Ui 331 kJ Ui 452 kJ B 152 mm M2 60103 N m M1 20103x1 C 213 mm 64nPR3 d4G smax 16PR pd3 sin u 1 T PR cos u M PR sin u 14105 14106 14107 14109 14110 14111 Bending and shear Bending only 14113 14114 14115 14117 14118 14119 14121 14122 14123 14125 14126 14127 Bv 00124 in T Dv 488 mm 204 mm Cv 122526103 300106200109 Bh 000191 in uB M0a 3EI C 5M0a2 6EI m 1x m 1x M M0 a M M0 Cv 168 mm T uA 0991103 rad C 179 mm T B 435 mm T Av 4PL3 3EI Bh wL4 4EI mx1 0 mx2 10L 10x4 Mx2 wL2 2 Mx1 w 2 x 2 1 5w 96Ga L a b 4 a w Gb aL a b 2 c a 5 96 b aL a b 2 3 20 d C 167 mm T uA 000927 rad mux2 01667x2 kN m mux1 1 01667x1 kN m Mx2 225x2 3x 2 2 kN m Mx1 315x1 6x1 2 kN m uA 5w0L3 192EI C w0L4 120EI B 65wa4 48EI T mx2 1 2 x2 a mx1 x1 2 Mx1 wax1 Mx2 waa x2 w 2 x2 2 ANSWERS TO SELECTED PROBLEMS 853 14129 14130 14131 14133 14134 14135 14137 14138 14139 14141 14142 14143 14145 14146 14147 14149 Bending strain energy Axial force strain energy 14150 14151 14153 14154 14155 14157 smax 105 ksi uB M0L EI uB M0L EI Ev 23625103 400106200109 295 mm Uit 223 in lb A 0536 mm 0344 in lb Uai 35028 229106p 40252 101 ft lb Ubi 1176106 29106 1 1205023 smax 116 MPa Ui 5P2a3 6EI C 5M0a2 6EI M M0 a x M M0 M Px M Px Cv 5wL4 8EI T uA Pa2 6EI uC 5Pa2 6EI uA 273 B 154 in uA 41667103 200109C70106D 000298 rad C 23Pa3 24EI Hv 0156 in Cv 21 232 4529103 0163 in Av 623 mm T Cv 00375 mm 0367 mm Bh 29375103 400106200109 a Specific values may vary for a particular material due to alloy or mineral composition mechanical working of the specimen or heat treatment For a more exact value reference books for the material should be consulted b The yield and ultimate strengths for ductile materials can be assumed equal for both tension and compression c Measured perpendicular to the grain d Measured parallel to the grain e Deformation measured perpendicular to the grain when the load is applied along the grain Specific Modulus of Modulus of Yield Strength ksi Ultimate Strength ksi Coef of Therm Materials Weight Elasticity E Rigidity G Elongation in Poissons Expansion Tens Compb Shear Tens Compb Shear 2 in specimen Ratio Metallic Aluminum 2014T6 0101 106 39 60 60 25 68 68 42 10 035 128 Wrought Alloys 6061T6 0098 100 37 37 37 19 42 42 27 12 035 131 Cast Iron Gray ASTM 20 0260 100 39 26 97 06 028 670 Alloys Malleable ASTM A197 0263 250 98 40 83 5 028 660 Copper Red Brass C83400 0316 146 54 114 114 35 35 35 035 980 Alloys Bronze C86100 0319 150 56 50 50 95 95 20 034 960 Magnesium Am 1004T61 0066 648 25 22 22 40 40 22 1 030 143 Alloy Structural A36 0284 290 110 36 36 58 58 30 032 660 Steel Stainless 304 0284 280 110 30 30 75 75 40 027 960 Alloys Tool L2 0295 290 110 102 102 116 116 22 032 650 Titanium Ti6Al4V 0160 174 64 134 134 145 145 16 036 520 Alloy Nonmetallic Low Strength 0086 320 18 015 60 Concrete High Strength 0086 420 55 015 60 Plastic Kevlar 49 00524 190 104 70 102 28 034 Reinforced 30 Glass 00524 105 13 19 034 Wood Douglas Fir 0017 190 030c 378d 090d 029e Select Structural Grade White Spruce 0130 140 036c 518d 097d 031e 106F n 103 ksi 103 ksi lbin3 a su sY g Average Mechanical Properties of Typical Engineering Materialsa US Customary Units a Specific values may vary for a particular material due to alloy or mineral composition mechanical working of the specimen or heat treatment For a more exact value reference books for the material should be consulted b The yield and ultimate strengths for ductile materials can be assumed equal for both tension and compression c Measured perpendicular to the grain d Measured parallel to the grain e Deformation measured perpendicular to the grain when the load is applied along the grain Modulus of Modulus of Yield Strength MPa Ultimate Strength MPa Coef of Therm Materials Density Elasticity E Rigidity G Elongation in Poissons Expansion Tens Compb Shear Tens Compb Shear 50 mm specimen Ratio Metallic Aluminum 2014T6 279 731 27 414 414 172 469 469 290 10 035 23 Wrought Alloys 6061T6 271 689 26 255 255 131 290 290 186 12 035 24 Cast Iron Gray ASTM 20 719 670 27 179 669 06 028 12 Alloys Malleable ASTM A197 728 172 68 276 572 5 028 12 Copper Red Brass C83400 874 101 37 700 700 241 241 35 035 18 Alloys Bronze C86100 883 103 38 345 345 655 655 20 034 17 Magnesium Am 1004T61 183 447 18 152 152 276 276 152 1 030 26 Alloy Structural A36 785 200 75 250 250 400 400 30 032 12 Steel Stainless 304 786 193 75 207 207 517 517 40 027 17 Alloys Tool L2 816 200 75 703 703 800 800 22 032 12 Titanium Ti6Al4V 443 120 44 924 924 1000 1000 16 036 94 Alloy Nonmetallic Low Strength 238 221 12 015 11 Concrete High Strength 238 290 38 015 11 Plastic Kevlar 49 145 131 717 483 203 28 034 Reinforced 30 Glass 145 724 90 131 034 Wood Douglas Fir 047 131 21c 26d 62d 029e Select Structural Grade White Spruce 360 965 25c 36d 67d 031e 106C n GPa GPa Mgm3 a su sY r Average Mechanical Properties of Typical Engineering Materialsa SI Units Fundamental Equations of Mechanics of Materials Axial Load Normal Stress Displacement Torsion Shear stress in circular shaft where Power Angle of twist Average shear stress in a thinwalled tube Shear Flow Bending Normal stress Unsymmetric bending s Mzy Iz Myz Iy tan a Iz Iy tan u s My I q tavgt T 2Am tavg T 2tAm f TL JG f L L 0 T1x2dx J1x2G P Tv 2pfT J p 2 1co 4 ci 42 tubular cross section J p 2 c4 solid cross section t Tr J dT a TL d PL AE d L L 0 P1x2dx A1x2E s P A Shear Average direct shear stress Transverse shear stress Shear flow Stress in ThinWalled Pressure Vessel Cylinder Sphere Stress Transformation Equations Principal Stress Maximum inplane shear stress Absolute maximum shear stress savg smax smin 2 tabs max smax smin 2 savg sx sy 2 tmax A a sx sy 2 b 2 t2 xy tan 2us 1sx sy22 txy s12 sx sy 2 A a sx sy 2 b 2 t2 xy tan 2up txy 1sx sy22 txy sx sy 2 sin 2u txy cos 2u sx sx sy 2 sx sy 2 cos 2u txy sin 2u s1 s2 pr 2t s1 pr t s2 pr 2t q tt VQ I t VQ It tavg V A Geometric Properties of Area Elements Material Property Relations Poissons ratio Generalized Hookes Law where Relations Between w V M Elastic Curve Buckling Critical axial load Critical stress Secant formula Energy Methods Conservation of energy Strain energy Ui L L 0 T2dx 2GJ torsional moment Ui L L 0 fsV2dx 2GA transverse shear Ui L L 0 M2dx EI bending moment Ui N2L 2AE constant axial load Ue Ui smax P A c1 ec r2 sec a L 2r A P EAb d scr p2E 1KLr22 r 2IA Pcr p2EI 1KL22 EI d2n dx2 M1x2 EI d3n dx3 V1x2 EI d4n dx4 w1x2 1 r M EI dV dx w1x2 dM dx V G E 211 n2 gxy 1 G txy gyz 1 G tyz gzx 1 G tzx Pz 1 E 3sz n1sx sy24 Py 1 E 3sy n1sx sz24 Px 1 E 3sx n1sy sz24 n Plat Plong x h y A bh b C Rectangular area Ix bh3 1 12 Iy hb3 1 12 Ix bh3 x h A bh b C Triangular area 1 36 1 h 3 1 2 x h A ha b b a C Trapezoidal area h 1 3 2a b a b 1 2 Ix πr4 x y C Semicircular area 1 8 A πr2 2 4r 3π Iy πr4 1 8 r Ix πr4 x y C Circular area 1 4 A πr2 Iy πr4 1 4 r A ab C Semiparabolic area 2 3 2 a 5 3 b 8 a zero slope b Exparabolic area 3 a 4 a 3 b 10 zero slope b C A ab 3