The am1004 t61 magnesium tube

Fundamental Equations of Mechanics of Materials Axial Load

Normal Stress

s = P A

Displacement

d = L L

0 P(x)dx A (x)E

d = ! PL AE

dT = a "TL

Torsion

Shear stress in circular shaft

t = Tr

J

where

J =

p

2 c4 solid cross section

J =

p

2 (co

4 - ci 4) tubular cross section

Power

P = Tv = 2pf T

Angle of twist

f = L L

0 T(x)dx J(x)G

f = !

TL JG

Average shear stress in a thin-walled tube

tavg =

T 2tA m

Shear Flow

q = tavg t =

T 2Am

Bending

Normal stress

s =

My

I Unsymmetric bending

s = -

Mz y

Iz +

Myz

Iy , tan a =

Iz Iy

tan u

Shear

Average direct shear stress

tavg = V A

Transverse shear stress

t =

VQ It

Shear flow

q = tt =

VQ I

Stress in Thin-Walled Pressure Vessel

Cylinder

s1 =

pr

t s2 =

pr

2t

Sphere

s1 = s2 =

pr

2t

Stress Transformation Equations

sx# = sx + sy

2 +

sx - sy 2

cos 2u + txy sin 2u

tx#y# = - sx - sy

2 sin 2u + txy cos 2u

Principal Stress

tan 2up = txy

(sx - sy)>2 s1,2 =

sx + sy 2

{ A asx - sy2 b2 + txy2 Maximum in-plane shear stress

tan 2us = - (sx - sy)>2

txy

tmax = A asx - sy2 b2 + t2xy savg =

sx + sy 2

Absolute maximum shear stress

tabs max

= smax

2 for smax, smin same sign

tabs max

= smax - smin

2 for smax, smin opposite signs

Geometric Properties of Area Elements Material Property Relations

Poisson’s ratio

n = - Plat

Plong

Generalized Hooke’s Law

ex =

1 E

3sx - n(sy + sz)4

ey = 1 E

3sy - n(sx + sz)4

ez = 1 E

3sz - n(sx + sy)4

gxy = 1 G

txy, gyz = 1 G

tyz, gzx = 1 G

tzx

where

G =

E 2(1 + n)

Relations Between w, V, M

dV dx

= w(x), dM dx

= V

Elastic Curve

1 r

= M EI

EI

d4v

dx4 = w(x)

EI

d3v

dx3 = V (x)

EI

d2v

dx2 = M(x)

Buckling Critical axial load

Pcr =

p2EI

(KL)2

Critical stress

scr =

p2E

(KL >r)2 , r = 2I>A Secant formula

smax =

P A

c 1 + ec r2

sec a L 2r

A PEA b d Energy Methods Conservation of energy

Ue = Ui

Strain energy Ui =

N2L 2AE

constant axial load

Ui = L

L

0

M2dx 2EI

bending moment

Ui = L

L

0

fsV 2dx

2GA transverse shear

Ui = L

L

0

T2dx 2GJ

torsional moment

xh

y A = bh

b

C

Rectangular area

Ix = bh 31

12

Iy = hb 31

12

Ix = bh 3x

h

A = bh

b

C

Triangular area

1 36h13

1 2

xh

A = h(a + b)

b

a

C

Trapezoidal area

h13 2a + b a + b

1 2

Ix = pr 4

x

y

C

Semicircular area

1 8

A = pr 2

2

4r 3p

Iy = pr 41

8

r

Ix = pr 4

x

y

C

Circular area

1 4

A = pr2

Iy = pr 41

4

r

A = ab

C

Semiparabolic area

2 3

a25

b38 a zero slope

b

Exparabolic area

a34 a

b310 zero slope

b C

A = ab3

Average Mechanical Properties of Typical Engineering Materialsa

(SI Units)

Materials Density R (Mg/m 3 )

Moduls of Elasticity E

(GPa)

Modulus of Rigidity G

(GPa)

Yield Strength (MPa) SY

Tens. Comp.b Shear

Ultimate Strength (MPa) Su

Tens. Comp. b Shear

%Elongation in 50 mm specimen

Poisson’s Ratio N

Coef. of Therm. Expansion A

(10–6)/°C

Metallic

Aluminum 2014-T6 Wrought Alloys 6061-T6

2.79 73.1 27 414 414 172 469 469 290 10 0.35 23

2.71 68.9 26 255 255 131 290 290 186 12 0.35 24

Cast Iron Gray ASTM 20 Alloys Malleable ASTM A-197

7.19 67.0 27 – – – 179 669 – 0.6 0.28 12

7.28 172 68 – – – 276 572 – 5 0.28 12

Copper Red Brass C83400 Alloys Bronze C86100

8.74 101 37 70.0 70.0 – 241 241 – 35 0.35 18

8.83 103 38 345 345 – 655 655 – 20 0.34 17

Magnesium [Am 1004-T61]

Alloy 1.83 44.7 18 152 152 – 276 276 152 1 0.30 26

Structural A-36

Steel Structural A992

Alloys Stainless 304

Tool L2

7.85 200 75 250 250 – 400 400 – 30 0.32 12

7.85 200 75 345 345 – 450 450 – 30 0.32 12

7.86 193 75 207 207 – 517 517 – 40 0.27 17

8.16 200 75 703 703 – 800 800 – 22 0.32 12

Titanium [Ti-6Al-4V]

Alloy 4.43 120 44 924 924 – 1,000 1,000 – 16 0.36 9.4

Nonmetallic

Concrete Low Strength

High Strength

2.38 22.1 – – – 12 – – – – 0.15 11

2.37 29.0 – – – 38 – – – – 0.15 11

Plastic Kevlar 49

Reinforced 30% Glass

1.45 131 – – – – 717 483 20.3 2.8 0.34 –

1.45 72.4 – – – – 90 131 – – 0.34 –

Wood Douglas Fir Select Structural White Spruce Grade

0.47 13.1 – – – – 2.1c 26d 6.2 d – 0.29e –

3.60 9.65 – – – – 2.5 c 36 d 6.7 d – 0.31 e –

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.

Average Mechanical Properties of Typical Engineering Materialsa

(U.S. Customary Units)

Materials Specific Weight (lb/in 3 )

Moduls of Elasticity E

(103) ksi

Modulus of Rigidity G (103) ksi

Yield Strength (ksi) SY

Tens. Comp.b Shear

Ultimate Strength (ksi) Su

Tens. Comp. b Shear

%Elongation in 2 in. specimen

Poisson’s Ratio N

Coef. of Therm. Expansion A

(10–6)/°F

Metallic

Aluminum 2014-T6 Wrought Alloys 6061-T6

0.101 10.6 3.9 60 60 25 68 68 42 10 0.35 12.8

0.098 10.0 3.7 37 37 19 42 42 27 12 0.35 13.1

Cast Iron Gray ASTM 20 Alloys Malleable ASTM A-197

0.260 10.0 3.9 – – – 26 96 – 0.6 0.28 6.70

0.263 25.0 9.8 – – – 40 83 – 5 0.28 6.60

Copper Red Brass C83400

Alloys Bronze C86100

0.316 14.6 5.4 11.4 11.4 – 35 35 – 35 0.35 9.80

0.319 15.0 5.6 50 50 – 35 35 – 20 0.34 9.60

Magnesium [Am 1004-T61]

Alloy 0.066 6.48 2.5 22 22 – 40 40 22 1 0.30 14.3

Structural A-36

Steel Structural A992

Alloys Stainless 304

Tool L2

0.284 29.0 11.0 36 36 – 58 58 – 30 0.32 6.60

0.284 29.0 11.0 50 50 – 65 65 – 30 0.32 6.60

0.284 28.0 11.0 30 30 – 75 75 – 40 0.27 9.60

0.295 29.0 11.0 102 102 – 116 116 – 22 0.32 6.50

Titanium [Ti-6Al-4V]

Alloy 0.160 17.4 6.4 134 134 – 145 145 – 16 0.36 5.20

Nonmetallic

Concrete Low Strength

High Strength

0.086 3.20 – – – 1.8 – – – – 0.15 6.0

0.086 4.20 – – – 5.5 – – – – 0.15 6.0

Plastic Kevlar 49

Reinforced 30% Glass

0.0524 19.0 – – – – 104 70 10.2 2.8 0.34 –

0.0524 10.5 – – – – 13 19 – – 0.34 –

Wood Douglas Fir Select Structural White Spruce Grade

0.017 1.90 – – – – 0.30c 3.78d 0.90 d – 0.29e –

0.130 1.40 – – – – 0.36 c 5.18 d 0.97 d – 0.31 e –

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.

MECHANICS OF MATERIALS

MECHANICS OF MATERIALS NINTH EDITION

R. C. HIBBELER

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Credits and acknowledgments borrowed from other sources and reproduced, with permission, in this textbook appear on appropriate page within text (or on page xix).

Copyright © 2014, 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 permission(s) 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

Library of Congress Cataloging-in-Publication Data on File.

ISBN 10: 0-13-325442-9 ISBN 13: 978-0-13-325442-6

Prentice Hall is an imprint of

www.pearsonhighered.com

To the Student With the hope that this work will stimulate

an interest in Engineering Mechanics and provide an acceptable guide to its understanding.

PREFACE

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 author’s 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

• Preliminary Problems. This feature can be found throughout the text, and is given just before the Fundamental Problems. The intent here is to test the student’s conceptual understanding of the theory. Normally the solutions require little or no calculation, and as such, these problems provide a basic understanding of the concepts before they are applied numerically. All the solutions are given in the back of the text.

• Updated Examples. 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. Included is application of the LRFD method of design, and use of A992 steel for structural applications. 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 real-world applications depicted in over 30 new or updated photos placed throughout the book. These photos generally are used to explain how the relevant principles apply to real-world situations and how materials behave under load.

• Additional Fundamental Problems. These problem sets are located just after each group of example problems. In this edition they have been expanded. They offer students simple applications of the concepts covered in each section and, therefore, provide them with the chance to develop their problem-solving 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.

VI I I PREFACE

• Additional 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 real-life 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 31%, or about 460, new problems added to this edition, which involve applications to many different fields of engineering.

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 linear- elastic 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 thin-walled tubes, shear flow, and the shear center. Chapter 8 includes a discussion of thin-walled pressure vessels and provides a partial review of the material covered in the previous chapters, where 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.

PREFACE IX

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 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 well-defined 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 full-page illustration that indicates a broad-range 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 real-world 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. Apart from the preliminary, fundamental, and conceptual problems, there are numerous standard problems in the

X PREFACE

book that depict realistic situations encountered in engineering practice. It is hoped that this realism will both stimulate the student’s 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 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 Ninth Edition has undergone a rigorous Triple Accuracy Checking review. In addition to the author’s 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, LaurelTech 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.

S. Suliman, Penn State C. Valle, Georgia Institute of Tech C. Sulzbach, Colorado School of Mines K. Cook-Chennault, Rutgers University J. Ramirez, Purdue University J. Oyler, University of Pittsburg

PREFACE XI

P. Mokashi, Ohio State Y. Liao, Arizona State University P. 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 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, 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; including those in the teaching profession who have taken the time to e-mail me their comments, notably S. Alghamdi, A. Atai, S. Larwood, D. Kuemmerle, and J. Love.

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 hibbeler@bellsouth.net Mastering Ad to come

your work...

your answer specific feedback

®

XIV PREFACE

Resources for Instructors • MasteringEngineering. This online Tutorial Homework program allows you to integrate dynamic homework with automatic grading and adaptive tutoring. MasteringEngineering allows you to easily track the performance of your entire class on an assignment-by- assignment basis, or the detailed work of an individual student. • Instructor’s Solutions Manual. An instructor’s solutions manual was prepared by the author. The manual includes homework assignment lists and was also checked as part of the accuracy checking program. The Instructor Solutions Manual is available at www.pearsonhighered.com . • 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 www.pearsonhighered.com . If you are in need of a login and password for this site, please contact your local Pearson representative. • Video Solutions. Developed by Professor Edward Berger, University of Virginia, video solutions located on the Companion Website offer step-by-step 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 self-evaluation as students can pause the videos to check their understanding and work alongside the video. Access the videos at www.pearsonhighered.com/ hibbeler and follow the links for the Mechanics of Materials text.

Resources for Students • Mastering Engineering. Tutorial homework problems emulate the instrutor’s office-hour environment. • Companion Website —The Companion Website, located at www.pearsonhighered.com/hibbeler includes opportunities for practice and review including: • Video Solutions —Complete, step-by-step solution walkthroughs of representative homework problems from each section. Videos offer: students need it with over 20 hours helpful review.

An access code for the Mechanics of Materials, Ninth Edition companion website was included with this text. To redeem the code and gain access to the site, go to www.pearsonhighered.com/hibbeler 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 1.1 Introduction 3 1.2 Equilibrium of a Deformable Body 4 1.3 Stress 22 1.4 Average Normal Stress in an Axially

Loaded Bar 24 1.5 Average Shear Stress 32 1.6 Allowable Stress Design 46 1.7 Limit State Design 48

2 Strain 67

Chapter Objectives 67 2.1 Deformation 67 2.2 Strain 68

3 Mechanical Properties of Materials 83

Chapter Objectives 83 3.1 The Tension and Compression Test 83 3.2 The Stress–Strain Diagram 85 3.3 Stress–Strain Behavior of Ductile and

Brittle Materials 89 3.4 Hooke’s Law 92 3.5 Strain Energy 94 3.6 Poisson’s Ratio 104 3.7 The Shear Stress–Strain Diagram 106 *3.8 Failure of Materials Due to Creep

and Fatigue 109

4 Axial Load 121

Chapter Objectives 121 4.1 Saint-Venant’s Principle 121 4.2 Elastic Deformation of an Axially Loaded

Member 124 4.3 Principle of Superposition 138 4.4 Statically Indeterminate Axially Loaded

Member 139 4.5 The Force Method of Analysis for Axially

Loaded Members 145 4.6 Thermal Stress 153 4.7 Stress Concentrations 160 *4.8 Inelastic Axial Deformation 164 *4.9 Residual Stress 166

5 Torsion 181

Chapter Objectives 181 5.1 Torsional Deformation of a Circular

Shaft 181 5.2 The Torsion Formula 184 5.3 Power Transmission 192 5.4 Angle of Twist 204 5.5 Statically Indeterminate Torque-Loaded

Members 218 *5.6 Solid Noncircular Shafts 225 *5.7 Thin-Walled Tubes Having Closed Cross

Sections 228 5.8 Stress Concentration 238 *5.9 Inelastic Torsion 241 *5.10 Residual Stress 243

XVI CONTENTS

6 Bending 259

Chapter Objectives 259 6.1 Shear and Moment Diagrams 259 6.2 Graphical Method for Constructing Shear

and Moment Diagrams 266 6.3 Bending Deformation of a Straight

Member 285 6.4 The Flexure Formula 289 6.5 Unsymmetric Bending 306 *6.6 Composite Beams 316 *6.7 Reinforced Concrete Beams 319 *6.8 Curved Beams 323 6.9 Stress Concentrations 330 *6.10 Inelastic Bending 339

7 Transverse Shear 363

Chapter Objectives 363 7.1 Shear in Straight Members 363 7.2 The Shear Formula 365 7.3 Shear Flow in Built-Up Members 382 7.4 Shear Flow in Thin-Walled Members 391 *7.5 Shear Center for Open Thin-Walled

Members 396

8 Combined Loadings 409

Chapter Objectives 409 8.1 Thin-Walled Pressure Vessels 409 8.2 State of Stress Caused by Combined

Loadings 416

9 Stress Transformation 441

Chapter Objectives 441 9.1 Plane-Stress Transformation 441 9.2 General Equations of Plane-Stress

Transformation 446 9.3 Principal Stresses and Maximum In-Plane

Shear Stress 449 9.4 Mohr’s Circle—Plane Stress 465 9.5 Absolute Maximum Shear Stress 477

10 Strain Transformation 489

Chapter Objectives 489 10.1 Plane Strain 489 10.2 General Equations of Plane-Strain

Transformation 490 *10.3 Mohr’s Circle—Plane Strain 498 *10.4 Absolute Maximum Shear Strain 506 10.5 Strain Rosettes 508 10.6 Material-Property Relationships 512 *10.7 Theories of Failure 524

11 Design of Beams and Shafts 541

Chapter Objectives 541 11.1 Basis for Beam Design 541 11.2 Prismatic Beam Design 544 *11.3 Fully Stressed Beams 558 *11.4 Shaft Design 562

CONTENTS XVI I

12 Deflection of Beams and Shafts 573

Chapter Objectives 573 12.1 The Elastic Curve 573 12.2 Slope and Displacement by

Integration 577 *12.3 Discontinuity Functions 597 *12.4 Slope and Displacement by the Moment-

Area Method 608 12.5 Method of Superposition 623 12.6 Statically Indeterminate Beams

and Shafts 631 12.7 Statically Indeterminate Beams and

Shafts—Method of Integration 632 *12.8 Statically Indeterminate Beams and

Shafts—Moment-Area Method 637 12.9 Statically Indeterminate Beams and

Shafts—Method of Superposition 643

13 Buckling of Columns 661

Chapter Objectives 661 13.1 Critical Load 661 13.2 Ideal Column with Pin Supports 664 13.3 Columns Having Various Types

of Supports 670 *13.4 The Secant Formula 682 *13.5 Inelastic Buckling 688 *13.6 Design of Columns for Concentric

Loading 696 *13.7 Design of Columns for Eccentric

Loading 707

14 Energy Methods 719

Chapter Objectives 719 14.1 External Work and Strain Energy 719 14.2 Elastic Strain Energy for Various Types

of Loading 724 14.3 Conservation of Energy 737 14.4 Impact Loading 744 *14.5 Principle of Virtual Work 755 *14.6 Method of Virtual Forces Applied

to Trusses 759 *14.7 Method of Virtual Forces Applied

to Beams 766 *14.8 Castigliano’s Theorem 775 *14.9 Castigliano’s Theorem Applied

to Trusses 777 *14.10 Castigliano’s Theorem Applied

to Beams 780

Appendix A. Geometric Properties of an Area 788 B. Geometric Properties of Structural

Shapes 804 C. Slopes and Deflections of Beams 812

Solutions and Answers for Preliminary Problems 814

Fundamental Problems Partial Solutions and Answers 825

Selected Answers 844

Index 866

Cover. Construction cranes. Mushakesa / Shutterstock. Chapter 1 Opener. Structural steel framework on new industrial unit. Copyright by Joe Gough/Shutterstock. Chapter 2 Opener. A copper rod after necking and fracture. sciencephotos / Alamy. Chapter 3 Opener. Earthquake damage on Olfusarbru bridge, South Coast, Iceland. ARCTIC IMAGES / Alamy. Page 63. Photo for Prob. C1-3 is from the Deutsches Museum. Used with permission. Chapter 4 Opener. Oil Rig. George Hammerstein /Corbis. Chapter 5 Opener. Piling machine drilling foundation on building site. HAWKEYE / Alamy. Chapter 6 Opener. Fragment view of the road under reconstruction. Lev Kropotov/Shutterstock. Chapter 7 Opener. Edmonton, Alberta, Canada. Design Pics / LJM Photo/LJM Photo/Newscom. Chapter 8 Opener. Tram on mountain, Snowbird Ski Resort, Wasatch Mountains, Utah, United States. Tetra Images / Alamy. Chapter 9 Opener. Industrial texture of power station generator turbine. Yurchyks/Shutterstock. Chapter 10 Opener. Support for Frederick Douglass-Susan B. Anthony Memorial Bridge, Rochester, NY USA. Russell C. Hibbeler. Chapter 11 Opener. Zinc steel structure of light fl oor. Jarous/Shutterstock. Chapter 12 Opener. Pole vaulter in midair. Corbis Flirt / Alamy. Chapter 13 Opener. Water tower against blue sky. VanHart/Shutterstock Chapter 14 Opener. An Olympic worker cleans a diving board as an Olympic athlete dives in practice at the Aquatics Center ahead of competition in the London Olympics in London, Britain, 24 July 2012. BARBARA WALTON/EPA/Newscom.

All other photos provided by the author.

CREDITS

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.

Chapter 1

3

1.1 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 body’s 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.

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.

Stress

4 CHAPTER 1 STRESS

1 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 Saint-Venant, 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 result, this subject expanded into other areas of mechanics, such 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.

1.2 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 and body forces, Fig. 1–1 .

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 force. If the surface loading is applied along a narrow strip of area, the loading can be idealized as a linear distributed load , w ( s ). Here the loading is measured as having an intensity of force/length along the strip and is represented graphically by a series of arrows along the line s . The resultant force FR of w(s) 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.

w(s)

Concentrated force idealization

Linear distributed load

Surface force

Body force

s

C

G

FR W

Fig. 1–1

1.2 EQUILIBRIUM OF A DEFORMABLE BODY 5

1

Many machine elements are pin connected in order to enable free rotation at their connections. These supports exert a force on a member, but no moment.

Body Forces. A body force is developed when one body exerts a force on another body without direct physical contact between the bodies. Examples include the effects caused by the earth’s gravitation or its electromagnetic field. Although 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 body’s 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, i.e., bodies subjected to coplanar force systems, the supports most commonly encountered are shown in Table 1–1 . 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.

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

FyM

Fixed support Three unknowns: Fx, Fy, M

Two unknowns: Fx, Fy

Type of connection Reaction

u u

u

TABLE 1–1

6 CHAPTER 1 STRESS

1 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

!F = 0 !MO = 0

(1–1)

Here, ! F represents the sum of all the forces acting on the body, and ! MO 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,

!Fx = 0 !Fy = 0 !Fz = 0

!Mx = 0 !My = 0 !Mz = 0 (1–2)

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 x – y plane, then the conditions for equilibrium of the body can be specified with only three scalar equilibrium equations; that is,

!Fx = 0 !Fy = 0

!MO = 0 (1–3)

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 body’s free-body diagram .

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.

1.2 EQUILIBRIUM OF A DEFORMABLE BODY 7

1

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. 1–2 a , 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 free-body diagram of one of the parts is drawn, Fig. 1–2 b . 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 distribution’s resultant force and moment , FR and MRO, at any specific point O on the sectioned area, Fig. 1–2 c . 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 stated. Also, 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 .

* The body’s weight is not shown, since it is assumed to be quite small, and therefore negligible compared with the other loads.

section

F4

F2 (a)

F1

F3

F1 F2

(b)

FR

F1 F2

O

MRO

(c)

Fig. 1–2

8 CHAPTER 1 STRESS

1

Three Dimensions. Later in this text we will show how to relate the resultant loadings, FR and MRO, 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 FR and MRO acting both normal and perpendicular to the sectioned area must be considered, Fig. 1–2 d . 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 curl. By 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).

O

(c)

MRO

F1 F2

FR

(d)

O

F1 F2

N

T

M V

Torsional Moment

Bending Moment

Shear Force

MRO

FR

Normal Force

Fig. 1–2 (cont.)

The weight of this sign and the wind loadings acting on it will cause normal and shear forces and bending and torsional moments in the supporting column.

1.2 EQUILIBRIUM OF A DEFORMABLE BODY 9

1

Coplanar Loadings. If the body is subjected to a coplanar system of forces , Fig. 1–3 a , then only normal-force, shear-force, and bending-moment components will exist at the section, Fig. 1–3 b . If we use the x, y, z coordinate axes, as shown on the left segment, then N can be obtained by applying !Fx = 0, and V can be obtained from !Fy = 0. Finally, the bending moment MO can be determined by summing moments about point O (the z axis), !MO = 0, in order to eliminate the moments caused by the unknowns N and V .

section

F4

F3F2

F1

(a)

O

V MO

N x

y

Bending Moment

Shear Force

Normal Force

(b)

F2

F1

Fig. 1–3

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 !F = 0 and !M = 0 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 free-body 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.

10 CHAPTER 1 STRESS

1 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 segment. To do this draw the free-body diagram of the entire body and then apply the necessary equations of equilibrium to obtain these reactions.

Free-Body 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 free-body 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 directional sense of the resultant is opposite to that shown on the free-body diagram.

The following examples illustrate this procedure numerically and also provide a review of some of the important principles of statics.

1.2 EQUILIBRIUM OF A DEFORMABLE BODY 11

1 EXAMPLE 1.1

Determine the resultant internal loadings acting on the cross section at C of the cantilevered beam shown in Fig. 1–4 a .

(a)

A B

C 3 m 6 m

270 N/m

Fig. 1–4

SOLUTION Support Reactions. The support reactions at A do not have to be determined if segment CB is considered.

Free-Body Diagram. The free-body diagram of segment CB is shown in Fig. 1–4 b . 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, i.e., from Fig. 1–4 a , w >6 m = (270 N>m)>9 m, w = 180 N>m. 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 = 12(180 N>m)(6 m) = 540 N, which acts 13(6 m) = 2 m from C as shown in Fig. 1–4 b .

Equations of Equilibrium. Applying the equations of equilibrium we have

S + !Fx = 0; -NC = 0 NC = 0 Ans. + c!Fy = 0; VC - 540 N = 0 VC = 540 N Ans. a+ !MC = 0; -MC - 540 N(2 m) = 0

MC = -1080 N # m Ans. NOTE: The negative sign indicates that MC acts in the opposite direction to that shown on the free-body diagram. Try solving this problem using segment AC , by first obtaining the support reactions at A , which are given in Fig. 1–4 c .

180 N/m

540 N

2 m 4 mVC

MC

NC

(b)

BC

1.5 m 0.5 m

1 m

180 N/m90 N/m

540 N 135 N

VC

MC

NC

(c)

1215 N

3645 N"m CA

12 CHAPTER 1 STRESS

1 EXAMPLE 1.2

The 500-kg engine is suspended from the crane boom in Fig. 1–5 a . 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 two-force member. The free-body diagram of the boom is shown in Fig. 1–5 b . Applying the equations of equilibrium,

a+ !MA = 0; FCD1352(2 m) - [500(9.81) N](3 m) = 0 FCD = 12 262.5 N

S + !Fx = 0; A x - (12 262.5 N)1452 = 0 A x = 9810 N

+ c!Fy = 0; -A y + (12 262.5 N)1352 - 500(9.81) N = 0 A y = 2452.5 N

Free-Body Diagram. The free-body diagram of segment AE is shown in Fig. 1–5 c .

Equations of Equilibrium.

S + !Fx = 0; NE + 9810 N = 0

NE = -9810 N = -9.81 kN Ans.

+ c !Fy = 0; -VE - 2452.5 N = 0

VE = -2452.5 N = -2.45 kN Ans.

a+ !ME = 0; ME + (2452.5 N)( 1 m) = 0

ME = -2452.5 N # m = -2.45 kN # m Ans.

A 1 m1 m1 m

1.5 m

E

C

B

D

(a)

A

1 m2 m

500(9.81) N

Ay

Ax

FCD

(b)

3 4

5

9810 N

2452.5 N

VE

ME

NE

(c)

EA

1 m

Fig. 1–5

1.2 EQUILIBRIUM OF A DEFORMABLE BODY 13

1 EXAMPLE 1.3

Determine the resultant internal loadings acting on the cross section at G of the beam shown in Fig. 1–6 a . Each joint is pin connected.

(a)

300 lb/ft

2 ft 2 ft 6 ft

1500 lb

A

B

G D

C

3 ft E

3 ft

6 ft (6 ft) ! 4 ft

(6 ft)(300 lb/ft) ! 900 lb

1500 lb

Ey ! 2400 lb

Ex ! 6200 lb

FBC ! 6200 lb

(b)

2 3

1 2

6200 lb

3 4

5

(c)

B

FBA ! 7750 lb FBD ! 4650 lb

(d)

NG

MGVG2 ft

3 4

5

7750 lb1500 lb

A G

SOLUTION Support Reactions. Here we will consider segment AG . The free-body diagram of the entire structure is shown in Fig. 1–6 b . Verify the calculated reactions at E and C . In particular, note that BC is a two-force 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 two-force members, the free-body diagram of joint B is shown in Fig. 1–6 c . Again, verify the magnitudes of forces FBA and FBD .

Free-Body Diagram. Using the result for FBA , the free-body diagram of segment AG is shown in Fig. 1–6 d .

Equations of Equilibrium.

S + !Fx = 0; 7750 lb1452 + NG = 0 NG = -6200 lb Ans. + c !Fy = 0; -1500 lb + 7750 lb1352 - VG = 0 VG = 3150 lb Ans.

a+ !MG = 0; MG - (7750 lb)1352(2 ft) + 1500 lb(2 ft) = 0 MG = 6300 lb # ft Ans.

Fig. 1–6

14 CHAPTER 1 STRESS

1 EXAMPLE 1.4

Determine the resultant internal loadings acting on the cross section at B of the pipe shown in Fig. 1–7 a . End A is subjected to a vertical force of 50 N, a horizontal force of 30 N, and a couple moment of 70 N # m . Neglect the pipe’s mass.

0.75 m

50 N

1.25 m

B

A

0.5 m

C

D

70 N!m

(a)

30 N

1.25 m

70 N·m 30 N

(b)

y

A

50 N

0.5 m

x

z

B

(FB)z (MB)z

(MB)x (FB)x

(MB)y

(FB)y

Fig. 1–7

SOLUTION The problem can be solved by considering segment AB , so we do not need to calculate the support reactions at C .

Free-Body Diagram. The x, y, z axes are established at B and the free-body diagram of segment AB is shown in Fig. 1–7 b . 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 cross-sectional area at B .

Equations of Equilibrium. Applying the six scalar equations of equilibrium, we have *

"Fx = 0; (FB)x = 0 Ans.

"Fy = 0; (FB)y + 30 N = 0 (FB)y = -30 N Ans.

"Fz = 0; (FB)z - 50 N = 0 (FB)z = 50 N Ans.

"(MB)x = 0; (MB)x + 70 N # m - 50 N (0.5 m) = 0 (MB)x = -45 N # m Ans. "(MB)y = 0; (MB)y + 50 N (1.25 m) = 0

(MB)y = -62.5 N # m Ans. "(MB)z = 0; (MB)z + (30 N)(1.25) = 0 Ans.

(MB)z = -37.5 N # m NOTE: What do the negative signs for (FB)y, (MB)x , (MB)y and (MB)z indicate? The normal force NB = ! (FB)y ! = 30 N, whereas the shear force is VB = 2(0)2 + (50)2 = 50 N. Also, the torsional moment is TB = ! (MB)y ! = 62.5 N # m and the bending moment is MB = 2(45)2 + (37.5)2 = 58.6 N # m.

* 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 right-hand rule, with positive moments (thumb) directed along the positive coordinate axes.

1.2 EQUILIBRIUM OF A DEFORMABLE BODY 15

1

P1–1. In each case, explain how to find the resultant internal loading acting on the cross section at point A . Draw all necessary free-body diagrams, and indicate the relevant equations of equilibrium. Do not calculate values. The lettered dimensions, angles, and loads are assumed to be known.

P

B Au

2a aa

(a)

C

D

B

a a

w

A C

a

P

(b)

B CA

P

u

a

(c)

a/2 a/2

M

O P

B

r

A

(d)

f

u

(e)

a

a

2a

A

C

B

P

u

a

a a

a 3a

a

P (f)

B

A

D

C

PRELIMINARY PROBLEMS

It is suggested that you test yourself on the solutions to these examples, by covering them over and then trying to think about which equilibrium equations must be used and how they are applied in order to determine the unknowns. Then before solving any of the Problems, build your skills by first trying to solve the Preliminary Problems, which actually require little or no calculations, and then do some of the Fundamental Problems given on the following pages. The solutions and answers to all these problems are given in the back of the book . Doing this throughout the book will help immensely in understanding how to apply the theory, and thereby develop your problem-solving skills.

16 CHAPTER 1 STRESS

1

F1–1. Determine the internal normal force, shear force, and bending moment at point C in the beam.

A B

C

2 m 2 m1 m

60 kN!m

1 m

10 kN

F1–1

F1–2. Determine the internal normal force, shear force, and bending moment at point C in the beam.

A B C

1.5 m 1.5 m

100 N/m 200 N/m

F1–2

F1–3. Determine the internal normal force, shear force, and bending moment at point C in the beam.

A

B

2 m 2 m 2 m

C

20 kN/m

F1–3

F1–4. Determine the internal normal force, shear force, and bending moment at point C in the beam.

A C

B

3 m3 m

10 kN/m

F1–4

F1–5. Determine the internal normal force, shear force, and bending moment at point C in the beam.