Mechanics of Materials - Riley 6e

Mechanics ofMaterials Sixth Edition

WILLIAM F. RILEY LEROY D. STURGES Associate Professor Aerospace Engineering and Engineering Mechanics Iowa State University

DON H. MORRIS Professor Emeritus

Engineering Science and Mechanics Virginia Polytechnic Institute and State University

JOHN WILEY & SONS, INC.

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Preface INTRODUCTION The primary objectives of a course in mechanics of materials are: (1) to develop a working knowledge of the relations between the loads applied to a nonrigid body made of a given material and the resulting deformations of the body; (2) to develop a thorough understanding ofthe relations between the loads applied to a nonrigid body and the stresses produced in the body: (3) to develop a clear insight into the relations between stress and strain for a wide variety of conditions and materials; and (4) to develop adequate procedures for finding the required dimensions ofa member of a specified material to carry a given load subject to stated specifications of stress and deflection. These objectives involve the concepts and skills that form the foundation of all structural and machine design. The principles and methods used to meet the general objectives are drawn largely from prerequisite courses in mechanics, physics, and mathematics together with the basic concepts of the theory of elasticity and the properties of engineering materials. This book is designed to emphasize the required fundamental principles, with numerous applications to demonstrate and develop logical, orderly methods ofprocedure. Instead of deriving numerous formulas for all types of problems, we have stressed the use of fiee-body diagrams and the equations of equilibrium, together with the geometry of the deformed body, and the observed relations between stress and strain, for the analysis of the force system acting on a body. This book is designed for a first course in mechanics of deformable bodies. Because of the extensive subdivision into different topics, the book will provide flexibility in the choice of assignments to cover courses of different length and content. The developments of structural applications include the inelastic as well as the elastic range of stress; however, the material is organized so that the book will be found satisfactory for elastic coverage only.

NEW T0 THIS EDITION Content changes 0 A sign convention for internal forces is established in Chapter I and followed consistently throughout the text.

I Discussion ofthe stress element ir1 Chapter 2 is expanded. I Section 4-6 of the Fifth Edition has been moved and combined with section 5-1 l on Design I Updated coverage of combined loadings provided in multiple chaptcrs—Chapters 5, 6, and 7—to offer continuous reinforcement of this difficult topic ' New and revised example problems and homework problems throughout.

New visualization tool MecMovies, by Tint Philpot of University of MissouriRolla is integrated at appropriate places in the text. Winner of the Premier Award for Excellence in Engineering Education Software, Mecllilovies offers a series of interactive tutorials, quizzes, problems, and games to support lectures and aid student self-study. Icons i.n the margin refer to appropriate sections of MecMovies, and MecMovie Problems and Activities are provided with most end of section problem sets. Available by accessing the companion site www.wiley.comfcollege/riley (student companion site) and using the registration code that accompanies new copies of the text.

ORGANIZATION OF TI-IE TEXT Since most mechanics of materials problems begin with a statics problem (finding the forces in structural members or the forces in pins cormecting structural members), we have included a review of statics in Chapter 1 of this book. The coverage is perhaps more complete and comprehensive than would be necessary for review so that the book could be used for both statics and mechanics of materials if desired. After the review of statics in Chapter 1, Chapters 2 and 3 consist of a thorough discussion ofmaterial stress and strain including principal stresses and principal strains. We choose to present principal stresses and principal strains at this early position to make it easier to talk about maximum stresses in the axial, torsional, and flexural applications that follow. It also allows us to talk about the maximum stresses in combined loading situations in Chapters 5, 6, and 7, rather than waiting until the end ofthe book. The ideas ofprincipal values also allow for continuous reinforcement throughout the book of the state of stress and strain at a point. in

=1:

iv

PREFACE

Material properties and the relationship between stress and strain are presented in Chapter 4. In Chapters 5, 6, and 7, we consider the stresses and strains in axial, torsional, and flexural loading applications. In addition to calculating the stresses in members subjected to axial loading and the stresses in pressure vessels subjected to internal pressure, in Chapter 5 we also calculate the stresses in pressure vessels subjected to axial and pressure loading. In addition to calculating the stresses in circular shafts subjected to torsional loading. In Chapter 6, we also calculate the stresses in circular shafts subjected to axial and torsional loads and in presstne vessels subjected to torsional loads. In Chapter 7, we first calculate the normal and shear stresses in beams subjected to flexural loading. We conclude Chapter 7 with the calculation of stresses in beams and circular shafis subjected to a combination of axial, torsional, and flexural loads. In Chapter 8, we calculate the deflection ofbeams due to various loading situations and also cover the calculation of support reactions for and stresses in statically indeterminate beams. In Chapter 9, we consider the tendency of columns to buckle. Finally, in Chapter 10 we discuss theories of failure and the use of energy methods. Every chapter opens with a brief Introduction and ends with a Summary of important concepts covered in the chapter followed by a set of Review Problems. All principles are illustrated by one or more Example Problems and several Homework Problems. The Homework Problems are graded in difliculty and are separated into groups of Intt'oductory, Intermediate, and Challenging problems. Several sections of Homework Problems also have a set of Computer Problems. While the computation could be accomplished by the student writing a FORTRAN program, the computation could just as easily be carried out using MathCAD, Mathematica, or a spreadsheet program. The important concept of the Computer problems is that they require students to analyze how the solution depends on some parameter of the problem. Most chapters conclude with a section on Design, which includes Example Problems and a set of Homework Problems. The emphasis in these problems is that there are ofien more than just one criteria to be satisfied in a design specification. An acceptable design must satisfy all specified criteria. In addition, standard lumber, pipes, beams, etc. come in specific sizes. The student must choose an appropriate structural member from these standard materials. Since each different choice of a beam or a piece of lumber has a different specific weight and affects the overall problem differently, students are also introduced to the idea that design is an iterative process.

FREE-BODY DIAGRAMS We strongly feel that a proper free-body diagram is just as important in mechanics of materials as it is in statics. It is our approach that, whenever an equation of equilibrium is written, it must be accompanied by a complete, proper fleebody diagram. Furthermore, since the primary purpose of a free-body diagram is to show the forces acting on a body, the fiee-body diagram should not be used for any other purpose. We encourage students to draw separate diagrams to show deformation and compatibility relationships.

PROBLEMS-SOLVING PROCEDURES Students are tnged to develop the ability to reduce problems to a series of simpler component problems that can be easily analyzed and combined to give the solution of the initial problem. Along with an effective methodology for problem decomposition and solution, the ability to present results in a clear, logical, and neat manner is emphasized throughout the text.

HOMEWORK PROBLEMS The illustrative examples and problems have been selected with special attention devoted to problems that require an understanding of the principles of mechanics of materials without demanding excessive time for computational work. A large number of homework problems are included so that problem assignments may be varied from term to term. The problems in each set represent a considerable range of difficulty and are grouped according to this range of difficulty. Mastery, ir1 general, is not achieved by solving a large number of simple but similar problems. ‘While the solution of simple problems is necessary to build a student’s problem-solving skills and confidence, we believe that a student gains mastery of a subject through application of basic theory to the solution of problems that appear somewhat difficult.

SIGNIFICANT FIGURES Results should always be reported as accurately as possible. However, results should not be reported to I0 significant figures merely because the calculator displays that many digits. One of the tasks in all engineering work is to determine the

PREFACE

accuracy of the given data and the expected accuiacy of the final answer. Results should always reflect the accuracy of the given data. In a textbook, however, it is not possible for students to examine or question the accuracy of the given data. It is also impractical, in an introductory course, to give error bounds on every number. Therefore, since an accuracy greater than about 0.2% is seldom possible for practical engineering problems, all given data in Example Problems and Homework Problems, regardless of the number of figures shown, will be assumed suificiently accurate to justify rounding off the final answer to approximately this degree of accuracy (three to four significant figures).

SI VERSUS USCS UNITS U.S. customary units and SI units are used in approximately equal proportions in the text for both Example Problems and Homework Problems. To help the instructor who wants to assign problems of one type or the other, odd-numbered Homework Problems are in U.S. customary units and evennumbered Homework Problems are in SI rmits.

ANSWERS PROVIDED Answers to about half of the Homework Problems are provided on the student companion site: www.wiley.comf college/Riley. Since the convenient designation ofproblems for which answers are provided is ofgreat value to those who make up assignment sheets, the problems for which answers are provided are indicated by means of an asterisk [*) after the problem number.

V

INSTRUCTOR RESOURCES In addition to a firlly worked solutions manual, all figures from the text are available in electronic format for instructors who adopt this book for use in their classes. All resources will be available for download fiom the book’s website: www.wiley.comfcollegelriley.

ACKNOWLEDGMENTS We are grateful for comments and suggestions received from colleagues and from users of the earlier editions of this book. Special thanks go to the following people who provided input and comments: Candace M. Anunerman, Colorado School of Mines, James N. Craddock, Southem Illinois University, Leonard De Rooy, Calvin College, Xin-Lin Gao, Michigan Technological University, John B. Ligon, Michigan Technological University, Charles E. Bakis, Pennsylvania State University, Shashi S. Marikunte, Southern Illinois University, Timothy A. Philpot, University of Missouri-Rolla, Ray Ruichong Zhang, Colorado School of Mines, Jiang Zhe, University of Akron. Final judgments concerning organization of material and emphasis of topics, however, were made by the authors. We will be pleased to receive comments from readers and will attempt to acknowledge all such communications. Comments can be sent by email to sturges@ iastate.edu or to dhmorris @ vt.edu.

William F. Riley Leroy D. Sturges Don H. Morris

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I Contents Chapter I

Chapter 4

Introduction and Review ofStatiw I

Material Properties and Stress-Strain

1-1 1-2 1-3 1-4 1-5

INTRODUCTION l CLASSIFICATION OF FORCES 2 EQUILIBRIUM OF A RIGID BODY 4 EQUILIBRIUM OF A DEFORMABLE BODY 30 INTERNAL FORCES 34 SUMMARY 44

Relationships 153 4-1 4-2 4-3 4-4 4-5

Chapter 2

INTRODUCTION 153 STRESS-STRAIN DIAGRAMS I53 GENERALIZED HOOKE‘S LAW I64 THERLIAL STRAIN 176 STRESS-STRAIN EQUATIONS FOR ORTHOTROPIC MATERIALS 180 SUMMARY I84

Analysis ofStress: Concepts and

Definitions 48 2-1 2-2 2-3 2-4 2-5 2-6 2-7 2-8 2-9 2-10 2-11 2-12

INTRODUCTION 48 NORMAL STRESS UNDER AXIAL LOADING 48 SHEARING STRESS IN CONNECTIONS 49 BEARING STRESS 51 UNITS OF STRESS 51 STRESSES ON AN INCLINED PLANE IN AN AXIALLY LOADED MEMBER 65 STRESS AT A GENERAL POINT IN AN ARBITRARILY LOADED MEMBER 72 TWO-DIMENSIONAL OR PLANE STRESS 74 TI-IE STRESS TRANSFORMATION EQUATIONS FOR PLANE STRESS 75 PRINCIPAL STRESSES AND MAXIMUM SI-IEARING STRESS—PLANE STRESS 85 MOI-IR’S CIRCLE FOR PLANE STRESS 98 GENERAL STATE OF STRESS AT A POINT 108 SUMMARY 1 17

Chapter 3

Analysis ofStrain: Concepts ami Definitions 121 3-1 3-2 3-3 3-4 3-5 3-6 3-7

INTRODUCTION 121 DISPLACEMENT, DEFORMATION, AND STRAIN 121 TIIE STATE OF STRAIN AT A POINT 129 THE STRAIN TRANSFORMATION EQUATIONS FOR PLANE STRAIN 130 PRINCIPAL STRAINS AND MAXIMUM SHEAR STRAIN 135 MOI-IR’S CIRCLE FOR PLANE STRAIN 140 STRAIN MEASUREMENT AND ROSETTE ANALYSIS 142 SUMMARY 148

Chapter 5 Axial Loading Applications and Prmsure Vessels 189 5-1 5-2 5-3 5-4 5-5 5-6 5-7 5-8 5-9 5-10 5-ll

INTRODUCTION 189 DEFORMATION OF AXIALLY LOADED MEMBERS 189 DEFORMATIONS IN A SYSTEM OF AXLALLY LOADED BARS 201 STATICALLY INDETERMINATE AXIALLY LOADED MEMBERS 209 THERMAL EFFECTS 225 STRESS CONCENTRATIONS 234 INELASTIC BEHAVIOR OF AXIALLY LOADED MEMBERS 239 THIN-WALLED PRESSURE VESSELS 246 COMBINED EFFECTS—AXIAL AND PRESSURE LOADS 254 THICK-‘WALLED CYLINDRICAL PRESSURE VESSELS 257 DESIGN 264 SUMMARY 270

Chapter 6

Torsional Loading ofShafls 2 76 6-1 6-2 6-3 6-4 6-5 6-6 6-7 6-8

INTRODUCTION 276 TORSIONAL SI-IEARING STRAIN 277 TORSIONAL SHEARING STRESS—THE ELASTIC TORSION FORMULA 279 TORSIONAL DISPLACEMENTS 281 STRESSES ON OBLIQUE PLANES 295 POWER TRANSMISSION 300 STATICALLY INDETERIVIINATE MEMBERS 303 COMBINED LOADING—AXI.AL, TORSIONAL, AND PRESSURE VESSEL 315

vii

viii 6-9 6-10 6-11 6-12 6-13

CONTENTS STRESS CONCENTRATIONS IN CIRCULAR SHAFT S UNDER TORSIONAL LOADINGS 322 INELASTIC BEHAVIOR OF TORSIONAL MEMBERS 325 TORSION OF NONCIRCULAR SECTIONS 331 TORSION OF THIN-WALLED TUBES—SHEAR FLOW 333 DESIGN PROBLEMS 339 SUMMARY 344

8-6 8-7 8-8 8-9 8-1 0

DEFLECTIONS BY SUPERPOSITION 520 DEFLECTIONS DUE TO SI-IEARING STRESS 530 DEFLECTIONS BY ENERGY METHODSCASTIGLIANOS THEOREM 532 STATICALLY INDETERMINATE BEAMS 542 DESIGN PROBLEMS 567 SUMMARY 574

Chapter 9 Colmnns 5 78 Chapter 7 Flexaral Loading: Stresses in Beams 349 7-1 7-2 7-3 7-4 7-5 7-6 7-7 7-8 7-9 7-10 7-11 7-12 7-13 7-14 7-15 7-16 7-17

INTRODUCTION 349 FLEXURAL STRAINS 352 FLEXURAL STRESSES 354 THE ELASTIC FLEXURE FORMULA 356 SHEAR FORCES AND BENDING MOMENTS IN BEAMS 366 LOAD, SHEAR FORCE, AND BENDING MOMENT RELATIONSI-HPS 376 SHEARING STRESSES IN BEAMS 391 PRINCIPAL STRESSES IN FLEXURAL MEMBERS 405 FLEXURAL STRESSES—UNSYMMETRICAL BENDING 410 STRESS CONCENTRATIONS UNDER FLEXURAL LOADINGS 418 INELASTIC BEHAVIOR OF FLEXURAL MEMBERS 422 SHEARING STRESSES IN TI-]1N-WALLED OPEN SECTIONS—SHEAR CENTER 431 FLEXURAL STRESSES IN BEAMS OF TWO MATERIALS 441 FLEXURAL STRESSES IN REINFORCED CONCRETE BEAMS 445 FLEXURAL STRESSES IN CURVED BEAMS 450 COMBINED LOADING: AXIAL, PRESSURE, FLEXURAL, AND TORSIONAL 457 DESIGN PROBLEMS 475 SUMMARY 480

9-1 9-2 9-3 9-4 9-5 9-6

INTRODUCTION 578 BUCKLING OF LONG, STRAIGHT COLUIVINS 579 EFFECTS OF DIFFERENT IDEALIZED END CONDITIONS 587 EMPIRICAL COLUMN FORMULAS—CENTRIC LOADING 592 ECCENTRICALLY LOADED COLUMNS 600 DESIGN PROBLEMS 606 SUMMARY 610

Chapter 10 10-1 10-2 10-3 10-4

10-5 10-6 10-7

A B

SECOND MOMENTS OF AREA 659 TABLES OF PROPERTIES 683

Flexaral Loading: Beam Deflections 487 8-1 8-2 8-3 8-4 8-5

INTRODUCTION 487 THE DIFFERENTIAL EQUATION OF THE ELASTIC CURVE 487 DEFLECTION BY INTEGRATION 489 DEFLECTIONS BY INTEGRATION OF SHEAR FORCE OR LOAD EQUATIONS 502 SINGULARITY FUNCTIONS 507

,.

INTRODUCTION 614 PART A: ENERGY IVIETHODS 615 STRAIN ENERGY 615 ELASTIC STRAIN ENERGY FOR VARIOUS LOADS 61 7 IMPACT LOADING 624 PART B: TI-IEORIES OF FAILURE FOR STATIC LOADING 637 INTRODUCTION 637 FAILURE THEORIES FOR DUCTILE MATERIALS 637 FAILURE THEORIES FOR BRITILE MATERIALS 650 SUMI\4ARY 654

Appendices Chapter 8

.

Energy Methods and Theories ofFailure 614

Answers‘

Index 705 “Available online at the Wiley website www.wiley.com

Chapter 1 Introduction and Review of Statics

1-1 INTRODUCTION The primary objective of a course in mechanics of materials is the development of relationships between the loads applied to a nonrigid body and the intemal forces and deformations induced in the body. Ever since the time ofGalileo Galilei (1564l642), scientists and engineers have studied the problem of the load-carrying capacity of structural members and machine components, and have developed mathematical and experimental methods of analysis for determining the internal forces and the deformations induced by the applied loads. The experiences and observations of these scientists and engineers of the last three centuries are the heritage of the engineer of today. The fundamental knowledge gained over the last three centuries, together with the theories and analysis techniques developed, permit the modern engineer to design, with complete competence and assurance, structures and machines of unprecedented size and complexity. The subject matter of this book forms the basis for the solution of three general types of problems: 1. Given a certain function to perform (transporting traffic over a river by means of a bridge, conveying scientific instruments to Mars in a space vehicle, conve1'ing water power into electric power), of what materials should the machine or structure be constructed, and what should be the sizes and proportions of the various elements? This is the designers task, and obviously there is no single solution to any given problem. 2. Given the completed design, is it adequate? That is, does it perform the function economically and without excessive deformation? This is the checker’s problem. 3. Given a completed structure or machine, what is its actual load-carrying capacity? The structure may have been designed for some purpose other than the one for which it is now to be used. Is it adequate for the proposed use? For example, a building may have been designed as an office building out is later found to be desirable for use as a warehouse. In such a case, what maximum loading may the floor safely support? This is the rating problem. Because the complete scope of these problems is obviously too comprehensive for mastery in a single course, this book is restricted to a study of individual members and very simple structures or machines. The design courses that follow will consider the entire structure or machine, and will provide essential background for the complete analysis of the three problems.

2

CH.AP'I'ER 1

INTRODUCTION AND REVIEW OF STATICS

The principles and methods used to meet the objective stated at the beginning of this chapter depend to a great extent on prerequisite courses in mathematics and mechanics, supplemented by additional concepts fi'om the theory of elasticity and the properties of engineering materials. The equations of equilibrium from statics are used extensively, with one major change in the free-body diagrams; namely, most free bodies are isolated by cutting through a member instead of simply removing a pin or some other connection. The internal force on the cut section is related to the stresses (force per unit area) generated by the cohesive forces holding the member together. The size and shape of the member must be adjusted to keep the stress below the limiting value for the type of material from which the member is constructed. In some instances, the specified maximum deformation, not the specified maximum stress, will govem the maximum load that a member may carry. In other instances, it may be found that the equations of equilibrium (or motion) are not sufiicient to determine all of the unknown loads or reactions acting on a body. In such cases it is necessary to consider the geometry (the change in size or shape) of the body after the loads are applied. The deformation per unit length in any direction or dimension is called strain. Some knowledge of the physical and mechanical properties of materials is required in order to create a design, to properly evaluate a given design, or even to write the correct relation between an applied load and the resulting deformation of a loaded member. Essential information will be introduced as required, and more complete information can be obtained fi"om textbooks and handbooks on properties of materials.

1-2 CLASSIFICATION OF FORCES

rt, F u

~

1,-‘

Figure 1-1

I

F fir 4

4

1‘

Concentrated loads :M,.=0

(=1)

where EM, = 0 has been replaced with EMA = 0 (pointA is the intersection of the z-axis and the plane ofthe two-dimensional structure). Poir1tA was selected for the moment equation; any other point on or offbody AB could have been selected. There is no particular order ir1 which we write the equations of equilibrium; mathematical convenience usually dictates the order. In this example, we use the order given in Eq. (a).

+->21-",=0 +¢ )3F,, =0 +l EM, = 0;

A,-r cos45°=0 Ay+Tsin 45°-5000=0 (T sin 45=>)(4) — 5000(4) = 0

(b) (C) (d)

Equation (d) is solved for T, which is then substituted into Eqs. (b) and (c) to find the components of the pin forces at A. The results are A, = 5000 lb

Ay = 0.04795 lb

T = 7071 lb

(6)

1-5

EQUILIBRIUM OF A RIGID BODY

Before proceeding further, we examine the results. Why were the forces written with the number of significant figures shown in Eq. (e)? For example, consider the solution of Eq. (d): T=

S111 45

= 7071.067812lb

(f)

Although results should always be reported as accurately as possible, the numbers to the right of the decimal point in Eq. (f ) have meaning only if the original data (dimensions and applied load) are lcnown to the same relative accuracy as the solution for the force T. One of the tasks in all engineering work is to determine the accuracy of the given data and the expected accuracy of the final answer. Results should always reflect the accuracy of the given data. It is not possible, however, for students to examine or question the accuracy of the given data in a textbook. It is also impractical in an introductory course to give error bounds on every nmnber. Therefore, since an accuracy greater than about 0.2 percent is seldom possible for practical engineering problems, all given data in Example Problems and Homework Problems, regardless of the number of figures shown, will be assumed sufliciently accurate to justify rounding off the final answer to approximately this degree of accuracy (three to four significant figures). One commonly used rounding scheme uses the leading digit to determine how many significant figures to keep in the final answer. If the first nonzero digit of the result is a 1, then the answer is reported with four significant figures; otherwise the answer is reported with three significant figures. Of course, all intermediate steps in the solution must maintain more significant figures than are used to represent the final results so as to reduce the effect of roundoff errors on the final results. Using the value of T from Eq. (f ) in Eq. (c) would yield A, =0 instead of the value shown in Eq. (e). The point of this discussion is: Don’t report final results with more accuracy than is justified by the data and don‘t rormd off numbers too much too soon. For this example problem, then, the answer is A, = 5000 lb

A,. = 0 lb

T = 7070 lb

A55-

1 Example Problem 1-2 A 900-kg mass is supported by a roller that

3 m _,

Bm

can move along a beam, as shown in Fig. 1-6a. The beam is supported by a pin at A and a roller at B. (a) Neglect the mass of the beam and determine the reactions at A and B. (b) If the mass of the beam is 8.5 kg/m, determine the reactions at A and B. SOLUTION The beam can be modeled as a rigid member with frictionless pin and roller supports atA and B. A fiee-body diagram for the beam is constructed by “freeing” the beam from its supports atA and B and fiom the roller that supports the 900-kg mass, as shown ir1 Fig. 1-6b. (a) As in Example Problem 1- 1, the components A, and A,. of the pin reaction at A are shown. Rollers exert forces on the beam that are perpendicular to the

J’

I ""

X Figure 1-6(a)

01)

7

8

CHAPIER 1

INTRODUCTION AND REVIEW OF 5'l'ATICS

I;-"= mg = (9o0)(9.s1) N A,

i fizmfiksmi A}, 3

beam. The masses of the bar and roller connecting the 900-kg mass to the beam a.re neglected. The force F in Fig. l-6b is the weight of the 900-kg mass, F = mg = 8829 N. There are three unknown forces (Ax, Ay, and B) shown on the free-body diagram (Fig. l-6b) for the beam. The three equations of equilibrium available to solve for the unknowns are

(5) F=3329N

w=83.39 Nfm

AR

A_,.

3 m#Sm

B

(C) sszs N

EF, =0

EIFy=0

+—>EF,=0:

A,=0

EMA =0

+1‘EF_,=0:

Ay+B—8829=0

+LZMA =0:

B(8)—8829(3)=0

Solving for AJ, and B gives

667.1 N Ay=55l8N

B=33llN

Thus, the reactions at supports A and B are

AX

I

A’.

3m4\i-—%4n1 lm

B

A, =0N

Ay=5520N

B=331UN

Alls-

(4) Figum l-6(b-cl)

(bJ For a beam mass of 8.5 kg/m, the uniformly distributed force on the beam resulting from its weight is w = mg = (8.5)(9.Sl) = 83.39 N/m. A free-body diagram for this beam is shown in Fig. 1-6c. In the equilibrium equations, the distributed force is statically equivalent to a single force whose magnitude is equal to the area under the load diagram (the area of a rectangle,

8 m x 83.39 N/m = 667.1 N) and which acts through the centroid of the load diagram (which is 4 m to the right ofA). The free-body diagram of Fig. 1-6d and the equations of equilibrium give P The maximum difference in the results from parts (a) and (b) is less than 10 percent. In many problems in engineering, we neglect the weight of members as being small when compared to the applied loads. As you gain experience in solving problems, you will be able to judge when you can safely neglect and when youmust include the weights of members of a structure or machine. Of course, if you have any doubts, the safest approach is to include these weights.

+ —> EF, = Q

A, = 0

+1‘ EF,=0:

Ay+B—8829—8(B3.39)=0

+ 5 EMA = 0:

B(8) — 8829(3) — 8(83.39)(4) = 0

Solving gives Ay=5852N

B=3644N

Thus, the reactions atA and B are A, = 0 N

Ay = 5850 N

B = 3640 N

Alls-

EXHIIIPIB Pfflblfilll 1 3 The truss shown m Fig l-"la supports one side of a bridge; an identical truss supports the other side. ‘Floor beams carry

vehicle loads to the truss joints. A 3400-lb car is stopped on the bridge. Assume that the weight ofthe car is evenly distributed among the wheels and that the center

1-5 EQUILIBRIUM or A mom sour 9 of gravity of the car is 16 Pt from support A. Calculate the support reactions and the forces in members BD, DE, and CE of the truss. B§'13fifl3fi1'D

::» _,,p,,,; i: lfifiwi 32fi

32f’(?>

( E15; = (k

28 N

91.0 N

57.0 N 24.6 N I

-28 cos 30.96‘ +l15.1— 91.0 = 0.08925 E 0

+ 1‘ BF), = O:

-28 sin3U.96° — 67.2 + 57.0 + 2.4.6 = —4.307 X 10-3 '5 O

+ L EMF = ll

28(0.3) + 57.0 (0.1667) — 91.0 (0.2777) + 24.6(0.3)

Figure 1-10

75113

= 0.0112 E 0

' '-;sln. - A

The force and moment equations of equilibrium do not exactly equate to zero due to roundoff error.

'F'~

4in.—# --1'

ii C..

1 Example Problem 1-5 The weight OfbO01(S OI1 a shelf bracket is equivalent to a vertical force of 75 lb as shown on Fig. 1-1 la. All members are made of 195-T6 cast aluminum and all pins have 114-in. diameters. Determine all

(I1)

forces acting on all three members of this frame. SOLUTION First draw the free-body diagram of the entire shelfbracket as in Fig. 1-1 lb. Here the “body of interest” is the frame ABC. The pins atA, B, and C remain attached to the frame, and thus the forces that would result from removal of the pins are not shown on the free-body diagram. The bracket at A has been removed from the frame, and the forces that the bracket exerts at A are shown as A, and A, (directions assumed). Similarly, the rocker at C has been removed, and the force of the rocker on the flame is shown as C. The equations of equilibrium are

+LzM,l=0= +->ZF,=0: +1‘EF,.— 0-

sc-s(75)=0 A,,+C=0

75 lb A?

A,

i

5 in.

(b) Figure 1-11

A, -75 =0

Ay = 75.0 lb

TM] A

C = 120.0 lb

A115C

Next, dismember the bracket and draw separate fiee-body diagrams of each

Tac 1

member (Fig. 1-12). Members AC and BC are straight two-force members, and

thus the forces in these members must act along the members. PinA connects a

(0) Figure 1-12(a)

1

B y

2L.

which are solved to get the support reactions A, = —120.0 lb

8 in.

I

14

CHAPTER 1

INTRODUCTION AND REVIEW OF STATICS

1"’ E

B

7511!

8111.

1 4111 -|

TAG

(b)

see

TAC

B

C

i

C

(C) Figure l-l2(b, c)

support and two members. Since memberAC is a two-force member, pinA will be left attached to member AB. The forces that act on pin A are the support reactions

A, and A, and a vertical force due to the two-force member AC. Similarly, pin B connects two members, one of which is a two-force member. Therefore, pin B is left attached to member AB, and the only force on pin B is along the twoforce member BC. Pin C connects a support and two members. Since both of the members are two-force members, pin C is arbitrarily left attached to memberBC, and thus the force on pin C due to memberAC is vertical. Then the equations of equilibrium can be written for member AB (Fig. 1-12b)

+1/EM); = 0; + L EMA = 0:

4(75)+12r,,¢ —12(75.0)= 0 l2[(5/13)B] - 8(75) = 0

fi'om which Tm; = 50.0 lb

B = 130.0 lb

A115-

50. D

E'*~

ip

” lb

130.011:

75.0 lb

4%

130.0 lb AI

120.0 lb A fi

C

50.0 lb

12

in

B

E3 P 0 l—| 1:-

jg. -mvi

(H1

.0Q .- 1:"

l50.0 lb

C

lb)

(Cl

Figure 1-13 It is easily verified that these values also satisfy the equations of equilibrium for the other free-body diagrams. These forces are all shown on the “report diagrams“ of Fig. 1-13.

1-5 EQUILIBRIUM or A RIGID sour

15

1 Example Problem 1-6 Determine the reaction at support A of the pipe system shown in Fig. 1- 1 4a. The 200-N force is parallel to the z-axis. Neglect the weights of the pipe and the wrench. Z

200 N

l80mii1 1*’!

_3,_.

250

15

'~__

“-~—- -J *7 230mm _

@l

x

_""'350mm ___*

‘ \y

(H) Figure l-l4(a)

SOLUTION A free-body diagram of the pipe system is shown in Fig. l-14b. The support at A is modeled as a rigid support that does not translate or rotate. There are three forces A,,A,,, and A, to prevent translation and three couples Mx, M,., and M, to prevent rotation. Couple M, lies in the yz-plane, couple My lies in the xz-plane, and couple M, lies in thexy-plane. Since there are six equations ofequilibrium for a three-dimensional force system, all six unlcnowns can be found. Using Eqs. 1-3 and 1-4 yields

21-", = ax, = BF, = . slug. = EM, = EM, = 9:~>s =>:~>s

=O

=0 - 200 = 0 - 200(0.ss0 + 0.230) = 0 - 200(0.1s0) = 0 = § §3‘*~E“.?* 0 Z

ZUUN

130 *——_ _

A2

.1, My ,

I/J41

|

LT-‘TM,

‘x~—-.

1

250mm

M +_T"'- 350 mm __ (5)

Figure l-14(1))

st-es.

-_

_..;\

230 mm M‘

P The summation EM, is the net tendency of all forces and moments to rotate the pipe about the x-axis through point A. For the 200-N force, this is just the magnitude of the force times the perpendicular distance between the line of action of the force and the x-axis through point A. For a more complicated force system, the tendency to rotate the body about the x-axis would be computed using the x-component of the vector product r >< F.

16

CHAPTER 1

IN'TRODUC'T'ION AND REVIEW OF STATICS

Thus, the reaction at A is

A,,=0N

A,.=0N

A,=200N

Ans-

M,=116.0N-m

M,=36.0N-m

M,=0N-m

ABS-

Since all the reactions are positive, they act in the directions shown on the fi'eebody diagram of Fig. 1-14b.

i Example Problem 1-7 A 1000-lb load is securely fastened to a hoisting cable as shown in Fig. 1-15a. The tension in the flexible cable does not change as it passes arolmd the small frictionless pulley at the right support. The weight of the cable may be neglected. Plot the tensions in the two cables (L13 and P) as a function of the sag distance d (0 5 d 5 10 ft). Determine the minimum sag d for which P is less than (a) Twice the weight of the load.

(b) Four tinles the weight of the load. (c) Eight times the weight of the load.

j

3011

1‘.

4

v

,

1‘

l_

'

'

b

'.E:

J

10ft

.

L

l (Q) Figure 1-15(2)

SOLUTION The ring B holds the wires together, and it will be isolated to generate the fi'ee-body diagram shown in Fig. 1-15b. The tension forces in the cables and the weight of the load are concurrent at the ring B. Writing the x- and y-components of the equilibrium equation for the fi'ee-body diagram of Fig. 1-15b results in +—>-EF,,=0:

T3ccos6;;—T,u,-cos6,;=0

(11)

+1‘ ZF,,=0:

T43 sir19,,+T,,¢;-sin6¢—l000=0

(lb)

Solving Eq. (a) for TA3 gives

T-B =

r -

9 A



" ‘YA \

.

6 I

1-24 A drum ofoil with a mass of200 kg is supported by a pair of frames (the second frame is behind the one shown] as shown in Fig. P1-24. Determine all forces acting on member ACE.

.

lite

O

I. D

Figure P1-21

45°, 1-22 The mass of block A in Fig. Pl-22 is 250 kg. Block A is supported by a small wheel that is free to roll on the continuous cable between supports B and C. The length of the cable is 42 n1. Determine the distance x and the tension T in the cable when the system is in equilibrium.

y

m

E.

‘

yr /H . -gas‘?

Figure P1-24

'J'(: . i';',=;ii~'lr—2.rs ru-I 0.5 tn

Figure P1-50

1-s2 A group of workers propose to raise a uniform 250-kg post AB to a vertical position using the rope and brace arrangement shown in Fig. Pl-52a. Assmne that the weight of the brace can be neglected and that end A acts as a frictionless pin for both the 6-m-long post AB and the 6-m-long brace AC. a. Plot the rope force P and the force E41" in the brace AC as functions ofthe angle 6 (0° 5 6 5 90° ). b. Repeat the problem if two braces are used as shown in Fig. P1-52b. Plot the rope force P and the brace forces E4; and FAD as functions of the angle 6 (0° 5 B 5 90° ). Assume that when the force in brace AD becomes zero, the brace falls out of the way and from that point on the rope is attached to C instead ofD. , c

B

1 51 The crane and boom shown in Fig. P1-51 weigh 12,000 lb and 600 lb, respectively. The pulleys at D and E are small, and the cables attached to them are essentially parallel. a. Plot d, the location of the resultant force of the groimd on the crane relative to point C, as a fimction of the boom angle 6 (0° 5 6 5 80°) when the crane is lifting a 3600-lb load. b. Plot A/3600 and T5,;/3600 as functions of the boom angle 6 (0° 5 6 5 80° ) when the crane is lifting a 3600-lb load (in which A is the magnitude of the reaction force on the pin at A, TED is the tension in the cable raising the boom, and 3600 lb is the weight of the load being lifted.) c. It is desired that the resultant force on the tread always be at least 1 ft behind C to ensure that the crane is never in danger of tipping over. Plot Wm, the maximum load that may be lifted, as a fimction of the boom angle 6 (0° 5 6 5 80° ). (Don't forget to check the tension in cable BD.)

.§0°t/r I

2" I25 mm

220 mm

:'

Figure P1-60

_ AISOM

250 mm

1-61 The reaction between a crutch and the ground is 35 lb, as shown in Fig. Pl-61. Determine the internal forces acting on section a-a.

250 mm

2700 N Figure P1-62 1-63 Apin-connected systemoflevers and bars is used as a toggle for a press as shown in Fig. P1-63. Three members are joined by pin D, as shown in the insert. Determine the internal forces on a section perpendicular to and midway between D and E when P = 1000 lb.

.\ ' \

C I-'5

\\

co

or

./

Pin D BD

‘\

33°

\

P

DE

His

fl/ F=35lb 25° 2“

ta: H U

45?

30 in.

B

Figure P1-61

Figure P1-63 Challenging Problems

1-62 The finnt-wheel suspension of an automobile is shown in Fig. Pl -62. The pavement exerts a vertical force of 2700 N on the tire. Determine the internal forces on a section perpendicular to and midway between C and D.

1-64* A steel shafi 120 mm in diameter is supported in flexible bearings at its ends. Two pulleys, each S00 mm in diameter, are keyed to the shaft. The pulleys carry belts that produce the forces shown in Fig. Pl~64. Determine the internal forces on a vertical section through point A.

1-S

800mm _

Q .

- 400mm -

-

30.101

"

]N'l'li]lNAL FORCES

43

/\ 1270mm /K ‘ C 900mm O .

800mm _

530 mun/' ‘_ F .1 gs

‘ mm

ml; g

5 in MN

/

-

380 mm

16 I

2;;

'

1930 mm

510 mm

Figure Pl-64 Figure Pl-66 l-65* A device for lifiing rectangular objects such as bricks and concrete blocks is shown in Fig. Pl-65. The coeflicients of friction at all vertical contact surfaces are it, = 0.4 and it, = 0.3. The device is to liit two blocks, each weighing 15 lb. Determine the internal forces on a vertical section 4 in. to the left of pin B.

1-67 Determine the internal forces on section a—a of the pipe system shown in Fig. Pl-67.

50 lb

Z

7in.

"U!

;

3

111

10 in.

P .

G

in-i 4in.

4in.

Sin. X

I B

14 in.

B

Figure Pl-67

legs O

1'

Lit. Ty.

J’

1-68" Determine the internal forces on section a—a in bar ABC of the three-bar lrarne shown in Fig. Pl-68.

A

C W

3 kN

Figure Pl-65

/' _'|‘_

-i-ii.-j

l-66 An automobile engine with amass of360 kg is supported by an engine hoist, as shown in Fig. Pl -66. Determine the internal forces on section a-a.

_-l00mm 100mm 100mm 100mm

Figure Pl-68

44 crnmzn 1 INTRODUCTION rum nsvrsw or srmcs Computer Problems

1-71 A 4000-lb cart rolls along a beam as shown in Fig. Pl-71.

1-69 The hook shown in Fig. Pl-69 supports a 10-kip load. Plot P, V, and M, the intemal fgfggg and moment mmgmimgd by 3 section ofthe hook, as a function ofthe angle 6 (0° 5 6 5 150°).

6- Sbflw 11131 the 11111K1lT1l1l‘I1 bending 11101116111 111 the beam 06' curs at the wheel that is closer to the middle of the beam ll P101 lM|mul the l'1111Xi1'f1111T1 bending 11101116111 111 I116 1363111, 35 a function of the cart’s position x (0 ft 5 x 5 15 it).

x—*I

1 4000 lb total

I i}

I‘

V_ A I s

3

/\\

l-

s= 10in.

l

2011

Figure P1-71

10 kip (

Figure P1-74 1-75 The electric motor shown in Fig. P1-75 weighs 25 lb. Due to friction between the belt and pulley, the belt forces have magnitudes of T1 = 21 lb and T; = llb. Determine the support reactions at A and B.

M

Z (e

T

150

E §

E22=‘52E5TE£E5;E£EEE!E::I==¢II=1=

¥ ~ "e.

it

‘T4

;

B V-4100 W Ttt'uu'\ Lv1 mml>75 rn1nl~l00 mm

Figure P1-78

Figure P1-75 1-76" Determine the force P required to push the 135-kg cylinder over the small block shown in Fig. Pl-76.

1-79* A three-bar fiame is loaded and supported as shown in Fig. Pl-79. Determine the internal forces transmitted by a. Section a-a in bar BEF. b. Section b—b in bar ABCD. 400 lb

P iv; 220mm

,-¢75n1.rn

F *

[

1000 lb

4fl

/*\-‘Y 2 s rt -'=_ - - '20"

Figure P1-76 1-77 For the beam shown in Fig. P1-77, determine a. The reactions at supports A and B. b. The internal forces on a transverse cross section 10 it to the right of support A.

1:-£111}

P

\

I \I / _‘\

1

\

-"-1'-1

"=71

Ag!

__E__

Figure 2-4

Fig. 2-5. In this diagram, a transverse cut has been made through the bolt, and the Figure 2-5

2.1

Illllillllllllllllllil P V

Figure 2-6

lower portion ofthe bolt remains in contact with the lefi member. The distribution of shearing force on the transverse cross section of the bolt has been replaced by a resultant shear force V. Since only one cross section of the bolt is used to effect load transfer between the members, the bolt is said to be in single shear; therefore, equilibrium requires that the resultant shear force V equal the applied load P. A free-body diagram for the threaded eyebar at the right end of the connection of Fig. 2-4 is shown in Fig. 2-6. In this diagram, two transverse cuts have been made through the bolt, and the middle portion of the bolt remains in contact with the eyebar. In this case, two transverse cross sections of the pin are used to effect load transfer between members of the connection and the pin is said to be in double shear. As a result, equilibrium requires that the resultant shear force V on each cross section of the pin equals one-half of the applied load P.

From the definition of stress given by Eq. 2-1, an average shearing stress on the transverse cross section of the bolt or pin can be computed as 2.2

rt. =§

EF=F}+270=0 +—t>EF=Fg,+270—245=O

F}=—270kN=270kN(C) F;,=—25kN=25kN(C)

+—>EF=Fn+270—245+200=0

Fn=—225kN=225kN(C)

P An allowable stress is the maximum permissible stress allowed in the design of a member. This will be discussed in more detail in Chapter 4.

54

CHAPTER 2

ANALYSIS OF STRESS: CONCEPTS AND DEFINITIONS

270 kN

-C

270 kN 270 kN 15

F,

mnirlfim-' J“kea)

:1

0- |

\___)

Figure 2-l0(h)

P The analysis of this example ignores the problem of how the three different materials are to be fastened together. The actual stress in the vicinity of the concentrated loads may be considerably different than the average stress computed here. In addition, it can be shown that near abrupt changes in diameter or near holes through the section, the actual stress can be two to three times greater than the average stress at that section. This ef-

An axial-force diagram for bar ABCD is shown in Fig. 2-10c. The crosssectional areas of the bar required to limit the stresses to the specified values are obtained fi'om Eq. 2-2. Thus,

Tension 400 200 0

Position as

B |

“ Axiaforce,-200 kN

fect is called stress concentration and will

be covered in later sections of this book and in structural design courses.

C |

D |

25 kN 270 kN

225 kN

-400 Compression

(C) Figure 2-l|](c)

F = -270 103 (a) A, = Edf = -*1

-12s(10) d, = s2.44(10-3)m 2 52.4mm

Ans.

n F1, -2s(10’) 4=21.3210‘ _ A =-n'1=-=i 221.3 ” 4 b 0;, -10(10°) ” l )m mm n F, -22s(103) (c) A. = Haj = Z = Y6) d,, = 5B.05(l0'3)m 2 ss.1 mm

An. S

4

0,

lb)

Ans.

-I EXHIIIPIB P1'0lIIl8lIl 2-3 A brass tube with an outside diameter of 2.00 in. and a wall thickness of 0.375 in. is connected to a steel tube with an inside diameter of 2.00 in. and a wall thickness of 0.250 in. by using a 0.750-indiameter pin as shown in Fig. 2-1 la. Determine

(a) The shearing stress in the pin when thejoint is carrying an axial load P of 10 kip.

(b) The length of joint required if the pin is replaced by a glued joint and the shearing stress in the glue must be limited to 250 psi.

2-5 uxrrs or srrrsss

P

SS

V-ii

V ‘Ii

(4)

(ll)

Figure 2-l1(a-b) SOLUTION (a) A free-body diagram of the brass tube and pin is shown in Fig. 2-1 lb. Since the pin is in double shear, A = 2(’{)(0.750)2 = 0.8836 in.2 Thus, from Eq. 2-4,

V 10 W , -r=Z:0_8836=ll.3l7ks1_ll.32ksr

Ans.

L -1--1-4-

2.00111.

Tl Tl it 4-,— \

——v

xi

Figure P2-62 (Q)

(3!)

Figure P2-59 2-60* At a point in a stressed body, the stresses on two perpendicular planes are as shown in Fig. P2-60. Determine a. The stresses on plane a—a. b. The stresses on horizontal and vertical planes at the point.

.

‘\/’ /

A //

200 MPa

,,. so MPa '1

2-63* The thin-walled cylindrical pressure vessel shown in Fig. P2-63 was constructed by wrapping a thin steel plate into a helix that forms an angle of 6 = 35° with respect to a transverse plane through the cylinder and butt-welding the resulting seam. In a thin-walled cylindrical pressure vessel, the normal stress 0,. on a horizontal plane through a point on the surface of the vessel is twice as large as the normal stress 0, on a vertical plane through the point, and the shear stresses on both the horizontal and vertical planes are zero. If the stresses in the weld material on the plane of the weld must be limited to 10 ksi in tension and 7 ksi in shear, determine the maximrnn normal stress 0,, permitted in the vessel.

H G)" “X

Figure P2-60 2-61 The stresses on horizontal and vertical planes at a point on the outside surface of a solid circular shaft subjected to an axial load P and a torque T are shown in Fig. P2-61. The normal stress on plane a—a at this point is 8000 psi (T). Determine a. The magnitude of the shearing stresses r,, and r,,. b. The magnitude and direction of the shearing stress on the inclined plane a—a.

't _.;9

Figure P2-63 2-64 Known stresses at point A in a structural member (see Fig. P2-64) are 125-MPa tension and zero shear on plane b-b and 225-MPa compression on plane c—c. Determine a. The stresses on a vertical plane through the point. h. The suesses on a horizontal plane through the point.

c \ \

R U

P

_'\-F

T3,

J R

,,

r

\L>I

,I I rz /

I

I

n

\\

b /43» \

\\

\

"—‘

\

fl

’ 4

\ A ,_< ,1 \\

8000 psi

/' /

Figure P2-61

\

/’

1:‘.

iti‘

U“

Figure P2-64

\

\

\

\ -B-—v

1 MIL!

G

2-10 PRINCIPAL sntsssss AND MAXIMUM SHEARING srttas-rmsr smsss 85 2-65 A steel bar with a 4 x l-in. rectangular cross section is being used to transmit an axial tensile load, as shown in Fig. P2-65. Normal and shear stresses on plane a—b of the bar are 12-ksi tension and 9-ksi shear. Determine the angle 6 and the applied load P.

13 ksi

ll I

rfiksi

Q‘

£1

Figure P2-67 b

Figure P2-65

Computer Problems 2-66 The stresses on horizontal and vertical planes at a point on the outside surface of a solid circular shaft subjected to an axial load P and a torque T are shown in Fig. P2-66.

a. Calculate and graph the normal stress 0,, and the shearing stress rm on the inclined plane a-b as a function ofthe angle 6 (0" 5 6 5180"). b. For what angle 6 is the normal stress a maximum? A minimum? What is the value of the shear stress on these planes? c. For what angle 6 is the shear stress a maximum? A minimum? What is the value ofthe normal stress on these planes?

a. Calculate and graph the normal stress 0,, and the shearing stress r,,, on the inclined plane a—b as a fimction of the angle 6 (0" 5 6 5 180°). b. For what angle 6 is the normal stress a rnaximmn? A miniminn? What is the value of the shear stress on these planes? c. For what angle 6 is the shear stress a maximum? A miniminn? What is the value ofthe normal stress on these planes?

I-1

P

T

I

j

I

—l 40 MPa

T

Figure P2-68

41

-->

10 MPa

so MPa - 40 MPa 1* 40 MPa

Figure P2-66 2-67 The stresses shown in Fig. P2-67 act at a point on the free s|.u~face of a stressed body.

2-68 The stresses shown in Fig. P2-68 act at a point on the free stu-face of a stressed body. Calculate the normal stress 0,, and the shearing stress r,,, on the inclined plane a-b as a function of the angle 6 (0° 5 6 5 180"). For each angle 6, graph the negative of the shearing stress (—r,,,, vertical axis) as a fimction of the normal stress (Um horizontal axis). On your graph, clearly identify the points associated with the angles 6 = 0", 30°, 45°, 60°‘, 90°, 120°, 135", I50“, and 180".

2-10 PRINCIPAL STRESSES AND MAXIMUM SHEARING STRESS—PLANE STRESS The transformation equations for plane stress (0, = r,,, = r,,, = 1:2,, = try, = 0), Eqs. 2.12 and 2-13, provide a means for determining the normal stress 0,, and the shearing stress rm on different planes through a point in a stressed body. As an example, consider the state of stress shown on the stress element in Fig.

2-3 la, which acts at a point on the free surface of a machine component or structural member. As the angle 6 varies (0° 5 6 5 360°), the normal stress 0,, and the

86

CHAPTER 2

ANALYSIS OF STRESS: CONCEPTS AND DEFINITIONS 7 ksi

12 ksi I —>2s ksi

(0) 40 .._

30

/-' on

-:20 a.’

I0

Sl1:, ks 0,, orress

~'sQ

0

_[()

/Int \

\./'

3U°‘\

l

I

60°

90°

\ \ ‘I . _ _ ‘ _ ’ av /'

1

I

-o_

“Q

rt’

,1 120°

.0.‘ l

|

150°

180°

Angle B

-20 (B) Figure 2-31 shearing stress rm on the different planes vary as shown in Fig. 2-31b. Note that, for9 = 0°, an = 0,, and tn, = 1:,_,.; and for6 = 90°, 0,, = 0,, and rm = —'r},, = —r,_v. For design purposes, critical stresses at the point are usually the maximum tensile stress and the maximum shea.ring stress. For a bar under axial load, the planes on which maximum normal stresses and maximum shearing stresses act are lcnown from the results of Section 2-6. For more complicated forms of loading, these stresses can be determined by plotting curves similar to those shown in Fig. 2-31b for each different state of stress encountered, but this process is time—consuming and inefiicient. Therefore, more general methods for finding the critical stresses have been developed. The transformation equations for plane stress developed previously are as follows: For normal stress an,

6,, = w + 2 cos 29 + 1,, sin 29

(2-121»)

For shear stress 1r,,,, 0, — 0

r,,, = - T’ sin 20 + In cos 29

(2-131»)

Maximum and minimum values of 0,, occur at values of 9 for which da,,1'd6 is equal to zero. Differentiation of 0,, with respect to 6 yields do _ T9” = —(cr,, — 0,.) S111 29 + 21,), cos 29

(a)

Z-10

PRINCIPAL STIIESSBS AND MAXIMUM SHEARING STRESS—PLANE STRESS

Setting Eq. (a) equal to zero and solving gives

an 29,, =

21: .

”

(2-14)

ax — 0,,

Note that the expression for dc"/d9 fi'om Eq. (a) is numerically twice the value of the expression for tn, from Eq. 2- 1 3b. Consequently, the shearing stress is zero on planes experiencing maximum and minimum values of normal stress. Planes fi'ee of shear stress are known as principal planes. Normal stresses occurring on principal planes are known as principal stresses. The values of 6;, from Eq. 2-14 give the orientations of two principal planes. A third principal plane for the plane stress state has an outward normal in the z-direction. For a given set of values of 0,, 0),, and r,,_,,, there are two values of 26,, differing by 180” and, consequently, two values of9,, that are 90‘ apart. This proves that the principal planes are normal to each other. When 1:,_,, and (0, — 0,.) have the same sign, tan 26,, is positive and one value of 9,, is between 0" and 90°, with the other value 180“ greater, as shown in Fig. 2-32. Consequently, one value of 6,, is between 0° and 45“, and the other one is 90° greater. In the first case, both sin 26,, and cos 26;, are positive, and in the second case both are negative. When these fimctions of 29,, are substituted into Eq. 2-12b, two in-plane principal stresses cpl and op; are foimd to be 0 +0

0 —cr. 2

¢,,_,,,= *2 yzb‘/(%) +1},

(2-15)

Equation 2-15 gives the two principal stresses in the xy-plane, and the third one is op; = U2 = 0. Equation 2-14 gives the angles 49;, and 6P + 90° between the x- (ory-) plane and the mutually perpendicular planes on which the principal stresses act. In order to determine which of the principal stresses (found using Eq. 2-15) acts on which of the principal planes (found using Eq. 2-14), substitute one of the values of Op into the stress transformation equation Eq. 2- 12a (or Eq. 2-12b). Since Hp is used, the calculated value of 0 ,, must be one of the principal stresses given by Eq. 2-15, and it acts on the surface whose normal points in the direction Bp. Because principal planes are perpendicular, the other principal stress given by Eq. 2-15 acts on the surface whose normal points in the direction Sp zt 90“. This procedure is illustrated in Example Problem 2-11.

~l(‘*e.-P*'i1’+ r2 ,

__.' \‘

1

\

.--

my

'~

29;, + lBO°,."

‘\'~"\.

-iQ

., ,+_

1,29,, »

__,9.i . __o Figure 2-32

88

crrsrrrn 2 ANALYSIS or STRESS: concsrrs AND rmrrsrrroms Note that ifone or both ofthe principal stresses from Eq. 2- 15 is negative, the algebraic maximum stress can have a smaller absolute value than the “minimum” stress. The maximum in-plane shearing stress rp occurs on planes located by values of 6 where a'r,,,/d6 is equal to zero. Differentiation of Eq. 2-13b yields

dr“ de

_ —(a, — a_,.)c0s

29 — 2 tn. sin ' 29

When dr,,,1’d6 is equated to zero, the value of 6, is given by the expression

131129, = _@*;-V) 2r,,y

3» 2.10

(2-16)

where 61- locates the planes of maximum in-plane shearing stress. Comparison of Eqs. 2- 1 6 and 2- 1 4 reveals that the two tangents are negative reciprocals. Therefore, the two angles 26p and 26, differ by 90°, and 6,, and 6, are 45° apart. This means that the planes on which the maximum in-plane shearing stresses occur are 45° from the principal planes. The maximum in-plane shearing stresses are found by substituting values of angle fiinctions obtained from Eq. 2-16 in Eq. 2-13b. The results are 0, — cry 2 Tp = i

T

2 + Txy

Equation 2-17 has the same magnitude as the second term of Eq. 2-15. Equation 2-16 gives tvm perpendicular planes of maximum ir1-plane shearing stress. The shearing stresses on these two planes have the same magnitude but opposite signs (Eq. 2-17). To determine which sign in Eq. 2-17 corresponds to each ofthe surfaces found using Eq. 2-16, substitute one value of 6, into the stress transformation equation for shearing stress (Eq. 2-13a or 2-13b). Since 6, is used, the calculated value of 1:," must be one of the shear stresses given by Eq. 2-17, and it acts on the surface whose normal points in the direction 6,. This procedure is illustrated in Example Problem 2-1 1. A useful relation between the principal stresses and the maximum ir1-plane shearing stress is obtained from Eqs. 2-15 and 2-17 by subtracting the values for the two in-plane principal stresses and substituting the value of the radical from Eq. 2-17. The result is -L-P =

7-'3)

(248)

or, in words, the maximum value of rm (rp) is equal in magnitude to one-half the difference between the two in-plane principal stresses. In general, when stresses act ir1 three directions it can be shown (see Section 2-12 that there are three orthogonal planes on which the shearing stress is zero. These planes are known as the principal planes, and the stresses acting on them (the principal stresses) will have three values: one maximum, one minimum, and a third stress between the other two. The maximum shearing stress, rm, on any plane that could be passed through the point, is one-half the difference between

Z-10

PRINCIPAL STIIESSBS AND MAXIMUM SHEARING STRESS—PLANE STRESS Upl fllld 63,2

same signs

same signs

opposlm slgns

ispli > iOp2i

Io-pli < l0p2i

P2

P2 Gpl

A

P2 U312

B‘ .

C

A

B iC

as

D

cpl

B‘

__‘ p3

Up] fllld GP;

op, and op;

E

P

7C

as

pl

F

A

1

}_____.

P3

(Q)

I

D

\_\ E

pl

F

P

1

}___-_.-

P3

(I5)

Figure 2-33

the maximum and minimum principal stresses and acts on planes that bisect the angles between the planes of the maximum and minimum normal stresses. _ Umin rm, _ Umax g

(2-19)

When a state ofplane stress exists, one of the principal stresses is zero. If the values of 0,1 and op; from Eq. 2-15 have the same sign, then the third principal stress, op; equals zero, will be either the maximum or the minimum normal stress. Thus, the maximum shearing stress may be

(Up — -

0'1,

.

__ _. _- 1

2

re, I

X

Figure 2-34

Application of the formulas and procedures developed in this section are illustrated by the following examples.

J’

8000 psi

-I EXEIIIIPIE PTODICIII 2 - I I

At a point in a structural member subjected

to plane stress there are normal and shearing stresses on horizontal and vertical planes through the point, as shown on the element in Fig. 2-35. 4000 psi iii 10,000 psi

_I,_. Figure 2-as

(a) Determine the principal stresses and the maximum shearing stress at the point. (b) Locate the planes on which these stresses act and show the stresses on the planes on which they act. (c) Sketch a triangular stress element showing the principal stresses and the maximum shearing stress (and the associated normal stress).

SOLUTION (a) On the basis of the axes shown in Fig. 2-35 and the established sign conventions, 03, is positive, whereas 0,, and r,,, are negative. For use in Eqs. 2-14 and

2- 15, the given values are 0,, = +10,000 psi

0,. = -8000 psi

1:,y = -4000 psi

When these values are substituted in Eq. 2- l 5, the principal stresses are found to be

%*i(%)

0 +0

0 -0

2

=10,0001:-8000)i\/(10,000;(-s000))’+F4000), = 1000 :1: 9849

0,1 = 1000 + 9849 = 10,249 psi 2 10,250 psi (r) s-,2 = 1000 - 9349 = -8349 psi 2 ssso psi (c)

Ans. Ans-

UP3 = 0, = 0

A115.

2-10

PRINCIPAL STRESSES AND MAXIMUM SHEARING STRESS—PLANE STRESS

The maximum in-plane shearing stress is given by Eq. 2-17, which is simply the last term in Eq. 2-15, and is rp = 9850 psi. The maximum shearing stress is given by 2_ 1 9 as aw _ cmin 1,-mm, = m 10.349 _ (_3g49)

v N

_

= g = 9849 psi = 9850 psi

Ans.

P The sum ofthe normal stresses on any two orthogonal planes is a constant for planes stress; oj, + :5. = 10,000 + (—8000) =q,1 + :52 = 10,850 + (—8850)=2000 psi. Also, the normal stress on the planes of maxi» mum shear is equal to the average ofthe nor-

ma) stresses 0,1: 1()()0=(¢x + gy),=2=(gp,

+ q,,)r2. Since 0;,| and 0),; have opposite signs, the values of the maximum inplane shearing stress (rp) and the maximum shearing stress (rm) are

equal. (b) Vlfhen the given data are substituted in Eq. 2-14, the results are

tan 29 - 2”-" — 0 4444 "_ s,-9, _10,000-(-3000)‘ ' from which 26,, = -23.96“, 180” + (—23.96°) = -1-156.04", - - and 6;, = —l1.98°, +7802”, -~ ~ E 1l.98° J and 78.0” '1

Ans.

Vlfhich principal stress is associated with which angle (6,, = -11.98“ or =

+ 78.02“) is determined by substituting the angles into the stress transformation equations. When 6 = 6,, = —11.98°, Eq. 2-12a gives

2

_ 2

_

___ __

X ~ - _.__ .11.98°

‘*' 7 - ~. .0

0,,=0, cos 6+0, S111 6+2r,y s1.n6cos6

10,350 psi _____n

= 10,000 cos2(—l1.98°) + (-8000) sin2(—1l.98°) + 2(—4000) sin (—l1.98°) cos (—11.98°) = 0P1 = 10,849 psi '5 10,850 psi (T)

(H) Figum 246(8)

The result is shown in Fig. 2-36a.

When 9 = 9,, = +7s.02 , Eq. 2-129 gives

/n

fipsi

0,, = 0, cos 2 6 + 0,, sin 2 6 +2-tn. sin 6 cos 6

_.

= 10,000 cos2(+78.02°) + (-8000) sin2[+78.02”) +2(-4000) sin(+78.02“) cos(+78.02°)

.

= 0,, = -3849 psi 2 ss50 psi (c)

.

“‘~-78-°° \.|

I

|‘

‘

The result is shown in Fig. 2-36b. The principal stresses shown on the principal planes ir1 Figs. 2-36a and b are shown on a stress element in Fig. 2-36c. The maximum in-plane shear

I (5) Figure 2-36(h)

92

crrrrrrrr 2 xnnrsrs or s'mnss= concnrvrs AND nrrrrsrrrons 8850 psi

10,850 psi

mm, 111.93“

(9) Figure 2-36(c)

stress occ1n's on a surface oriented at 45° to the principal directions. When 6 = -11.98” + 45° = +33.02°, Eqs. 2-12a and 2-13a give 0,, = 0,, cosz 6 + 0,, sinz 6 +213, sin6 cos 6

= 10.000 cos2(-1-33.02”) + (-3000) sin2(+33.02°) + 2(—4000) S111 [+33.02°) COS (+33.02°)

= +999.5 psi 2 1000 psi (T) and /_ n

_ .- ’

rm = —(0, — 0,) sir16 cos 6 + r,,,,[cos2 6 — sing 6]

‘°°° PS‘

= -[10,000 - (-9000)] sin (+33.02=) cos (+33.029)

9850 psi

+ (-4-000)[cos2(+33.02‘) - sin2(+33.O2°)] = -9349 psi 2 -9350 psi

3- C "ll 33.0“

Figure 2-37

These results are shown in Fig. 2-37. Substituting 6 = -+-33.02" + 90° =123.02° into Eqs. 2-12a and 2-13a

gives 0,, = +1000 psi

rm = +9850 psi

That is, the nonnal stresses on the perpendicularplanes ofmaximum shearing stresses are equal both in magnitude and sign, whereas the shearing stresses have equal magnitude but opposite sign. These results are shown on the stress element in Fig. 2-38. fl ,.

1000 psi

/ I000 psi

\‘-._‘ 31020 I

985° P51

Figure 2-as

9250 psi tit

2-10

PRINCIPAL STRESSES AND MAXIMUM SHEARING STRESS—PLANE STRESS

--._,,._11.9s" A I‘" 45° 10,350 psi

1000 psi

"

9850 psi

45°

8850 psi

Figure 2-39 (c) All of these results can be combined into the triangular stress element shown in Fig. 2-39. Note that all planes represented in this problem pass through a single point, and all of the stresses calculated act at that point. Sometimes the planes and stresses are represented individually (as in Figs. 2-36a and b or Fig. 2-37) for clarity. Other times the planes and stresses are combined on a single sketch (as in Figs. 2-36c, 2-38, and 2-39) for brevity.

1 EXHIIIPIB PTODIBIII 2-I2 At a point on the outside surface of a thin-walled pressure vessel there are normal and shearing stresses on horizontal and vertical planes through the point, as shown in Fig. 2-40a and b. (a) Determine the principal stresses and the maximum shearing stress at the point. (b) Locate the planes on which these stresses act and show the stresses on a stress element. SOLUTION (a) On the basis of the axes shown in Fig. 2-4-0b and the established sign conventions, 0X is positive, 0,, is positive, and 11,, is negative. For use in Eqs. 2-14 and 2-15, the given values are 0, = +100 MPa

0,, = +80 MPa

r,,- = -40 MIPa

When these values are substituted in Eq. 2-15, the principal stresses are found tobe 2

100+30

=5

100-30 2

7

+(-40)

2

= 90:l:4-1.23 0,,1= 90+4l.23 = +13l.23 MPa E 131.2M1’a (T) Ans. 0,2 = 90 — 41.23 = +48.77 MPa E 48.8MPa (T)

Ans.

0,03 = 0, = 0

Ans.

94

CH.AP'l'ER 2

ANALYSIS OF STRESS: CONCEPTS AND DEFINITIONS

y

80MIPa _L0 MPa

65.6 MPa 4-1‘ 65.6 MIPa _

100 MPa i,

1

mix

\ *1!\33° 1

48.3 MP3

\\

1 1.-/,

131.2 MPa

'1

11>)

. _ ’’

I

'.\

(J)

\\\

Face A -X

131.2 MPa

,1

T P

Q

i P

-,\

( )

C

T

4s.s MPa 65.6 MPa 5

(Q)

65.6 M1‘a W 131.2 MPa (e) Figure 2-40

Since cpl and op; have the same sign, the maximum shearing stress is given b)/Eq. 2-l9as

- -

151.23 -0

Tmax = NW = e = 65.61 MPa 2 65.6 MPa (b) When the given data are substituted in Eq. 2-14, the results are

211,. 2(-40) 26 =i=-i=-4.000 ta“ P 6,-6, 100-so fi'0m which 26,, = —75.96°,+104.04°, ...

and

9,, = -37.93", + 52.02", . .. E 38.05 J and 52.001

Ans.

2-10 PRINCIPAL snuzssss AND MAXIMUM sunuunc STRESS—PLANE smsss

95

“Then 8 = -37.98“, Eq. 2-12a gives 0,, = ax cosz 0 + 0}. sinz 8 + 21:” sin 9 cos 6

= 100 666%-37.98") + so 616%-37.92“) + 2(-40) 6&6 (-31.93“) 666 (-31.93“) = 6,,1= +131.23 MP6 2151.2 MPa(T) P The maximum shear stress in the xy-plane (the 12-plane) is r,,=(a,,1 — c;,;)f2=41.2 MPa and occurs on the plane midway between the principal planes l and 2. The normal stress on this surface is the average of the normal stresses 0,, = (061 + cr,,2)1'2 = 90 MPa. However, the difierence between the principal stresses cpl and op; is bigger than the difference between the principal stresses Upl and o,,;. Therefore, the maximum shear stress occurs on the surface midway between the principal planes l and 3 and has

Vifhen 8 = +52.02°, Eq. 2-12a gives an = 0,, cos: 6 + 0). sinz 9 + Zr”, sin9 cos 9

= 100 6661(+52.02°) + so sin2(+52.02“) +2(-40) 616 (+52.02

_

C

\ Q29‘,

i

(H) F mm?‘ B

Gr

Ix

ox +0, 2 ox on

op, =1

Figure 2-41

i 1

_..

100 CHAPTER 2 tuwxsrs or snusss coscstrrs AND nrrmnoss Line CV on Mohr’s circle represents the plane (the vertical plane of Fig. 2-41a) through the stressed point from which the angle 6 is measured. The coordinates of each point on the circle represent on and r,,, for one particular plane through the stressed point, the abscissa representing 0,, and the ordinate representing 1:,,,. To demonstrate this statement, draw any radius CF in Fig. 2-41b at an angle 26 counterclockwise from radius CV. From the figure, it is apparent that OF’ = OC + CF cos (26,, — 29) and since CF equals CV, the above equation reduces to OF’ = OC+CVcos28,, cos2t9 +CVsin26_,, sin2t9 Referring to Fig. 2-41b, note that CV cos 26,, = CV’ = (0, —oy)/2 CV sin26,, = VV' = 17,), OC = (ax + ay)/2 = oavg Therefore, OF’: OC+CV’cos 29+ VV’sin26 = ax-gay +(%cos2t9—l—r,_,-sin29 This expression is identical to Eq. 2-12b. Therefore, OF’ is equal to on. in a similar manner, F’F = CF sin (29,, — 29) = CVsin29,, cos29 — CVcos29,, sin26 = V'V cos 26 — CV’ sin 29

= 6,, 665 20 - 6% 3°" 616 29 This expression is identical to Eq. 2-13b. Therefore, F‘F is equal to r,,,. Since the horizontal coordinate of each point on the circle represents the normal stress 0,, on some plane through the point, the maximum normal stress at the point is represented by OD, and its value is 0',,|= OD=OC+CD=OC+CV

- , ./(T _a,+o'y

o,,—o'y2

which agrees with Eq. 2-15. Likewise, the vertical coordinate of each point on the circle represents the shearing stress 1:,,, (called the in-plane shearing stress) on some plane through the point, which means that the maximum in-plane shearing stresses at the point are

2-11 nouns CIRCLE FOR PLANE srnsss represented by CA and CB, and their value is 2

"r_,,=CA=CB=‘i(%) +175}, which agrees with Eq. 2-17. If the two nonzero principal stresses have the same sign, the maximum shearing stress at the point will not be in the plane of the applied stresses. The angle 29,, from CV to CD is counterclockwise orpositive, and its tangent is 1'1}.

29 = -ii

ta“ '° (GI —/2

which is Eq. 2-14. From the derivation of Eq. 2- 15, the angle between the vertical plane and one of the principal planes was 9p. In obtaining the same equation fi'om Mohr’s circle, the angle between the radii representing these same two planes is 29),. In other words, all angles on Mohr’s circle are twice the corresponding angles for the actual stressed body. The angle from the vertical plane to the horizontal plane in Fig. 2-41a is 90°, but in Fig. 2-41b, the angle between line CV (which represents the vertical plane) and line CH (which represents the horizontal plane) on Mohrs circle is 180”. The results obtained from Mohr’s circle have been shown to be identical with the equations derived fi"om the free-body diagram of Fig. 2-27. Thus, Mohr’s circle provides an extremely useful aid for both the visualization ofand the solution of stresses on various planes through a point in a stressed body in terms of the stresses on two mutually perpendicular planes through the point. Although Mohr’s circle can be drawn to scale and used to obtain values of stresses and angles by direct measurements on the figure, it is probably more useful as a pictorial aid to the analyst who is performing analytical determinations of stresses and their directions at the point. When the state of stress at a point is specified by means ofa sketch ofa small element, the procedure for drawing and using Mohr’s circle to obtain specific stress information can be briefly summarized as follows: 1. Choose a set ofx—y reference axes. 2. Identify the stresses ax, oy and 1:,_,l= -5,, and list them with the proper sign. 3. Draw a set of o,,—1:,,, coordinate axes with an and 1:", positive to the right and upward, respectively. 4. Plot the point (0,, —r,,.) and label it point V (vertical plane). 5. Plot the point ((5., 1:,.,) and label it point H (horizontal plane). 6. Draw a line between V and H. This establishes the center C and the radius R of Mohr‘s circle. 7. Draw the circle. 8. An extension of the radius between C and V can be identified as the x-axis or the reference line for angle measurements (i.e., 9 = 0°). By plotting points V and H as (0,, —1r,y) and (ay, ryx), respectively, shear stresses that tend to rotate the stress element clockwise will plot above the 0,,-axis, while those tending to rotate the element counterclockwise will plot below the 0,,-axis. The use of a negative sign with one of the shearing stresses (Ix); or ryx)

101

102

crnrrsn 2 ANALYSIS or snrsss= coucsrrs rum mzrrrrrrross is required for plotting purposes, since for a given state of stress, the shearing stresses (try = r_,,,,) have only one sign (both are positive or both are negative). The use of the negative sign at point V on Mohr’s circle brings the direction of angular measurements 28 on Mohr’s circle into agreement with the direction of angular measurements 6 on the stress element. Once the circle has been drawn, the normal and shearing stresses on an arbitrary inclined plane a—a having an outward normal n that is oriented at an angle 9 with respect to the reference x-axis (see Fig. 2-4-la) can be obtained from the coordinates of point F (see Fig. 2-41b) on the circle that is located at angular position 26 fi'om the reference axis through point V. The coordinates of point F must be interpreted as stresses 0,, and —r,,,. Other points on Mohr’s circle that provide stresses of interest are l. Point D, which provides the principal stress cpl. 2. Point E, which provides the principal stress 05,2. 3. Point A, which provides the maximum in-plane shearing stress —r,, and the accompanying normal stress aavg that acts on the plane.

.V 6 ksi iii ksi \\\

\ eP

\

\

..\ t\

|

\

(H) Figure 2-42(2)

P The point V on Mohr‘s circle represents the state of stress on the vertical surface of the stress element and can appear in any quadrant of the circle. If the shear stress on the vertical surface tends to rotate the stress element counterclockwise (as in this example), then the point V is plotted below the 0,,-axis. If the shear stress on the vertical surface tends to rotate the stress element clockwise, then the point V would be plotted above the 0,,-axis. Ifthe normal stress on the vertical surface is positive, then the point V is plotted to the right ofthe rm-axis (as in this example); if the normal stress is negative, it would be plotted to the left of the rm-axis. If the normal stress on the vertical surface is more positive than the normal stress on the horizontal surface, then the point V is plotted to the right of the point H. If the normal stress on the vertical surface is less positive than the nonnal stress on the horizontal surface, then the point V would be plotted to the left of the point H.

A negative sign must be used when interpreting shearing stresses r,,, and 1:], obtained fi'om the circle since a shearing stress tending to produce a clockwise rotation of the stress element is a negative shearing stress when a right-hand n—t coordinate system is used. Problems of the type presented in Section 2-10 can readily be solved by this semigraphic method, as illustrated in the following examples.

-I Example Pfflblfifll 2 - 13 Atapoint in a structural member subjected to plane stress there are normal and shearing stresses on horizontal and vertical planes through the point, as shown on the stress element in Fig. 2-42a. Determine and show on a sketch: (a) The principal and maximum shearing stresses at the point. (b) The normal and shearing stresses on plane a—a through the point. SOLUTION Mohr‘s circle (see Fig. 2-42b) is constructed fi'om the given data by plotting point V (representing the stresses on the vertical plane) at (8, -4) because the stresses on the vertical plane are 8 ksi (T) and 4 ksi (counterclockwise) shear. Likewise, point H (representing the stresses on the horizontal plane) has the coordinates (-6, 4). Draw line HV, which is a diameter of Mohr’s circle, and note that the center of the circle is at (1, 0). The radius of the circle is

2

_ I ( 51-01‘ 2 CV_ 2 )+1:”, = '\/72 +42 = 8.062 ksi

(a) The principal stresses and the maximum shearing stress at the point are a,,1 = OD = 1+ 8.062 = +9.062 ksi E 9.06 ksi (T)

Ans.

op; = OE = 1 — 8.062 = -7.062 ksi E 7.06 ksi (T)

Ans.

0'_,,3 = 0', = 0

Ans.

2-11 rrornrs cntcu: rorr PLANE snrsss

A

.

Ir ei'

H(-6, 4) -- . r

I1":

103

I

I

Q

0

QVI, P-3 "+.-.

Q»'

\“‘-2‘~O ...‘-FD ,:__ __/'1

D

",;3~ -

v65

‘,3

on ksi

_£‘__

,,(8’_4)

9-t.W

C/' (5)

Figure 2-42(0)

Since op; and op; have opposite signs, the maximum shearing stress is 1:,, = rm, = CA = CB — 8.062 ksi 2 8.06 ksi

Ans.

The principal planes are represented by lines CD and CE, where tan 26,, = 4/7 — 0.5714 which gives

7.06 ksi

20, = +2914“

or

0, = 14.s7= "l

Since the angle 26,, is counterclockwise, the principal planes are counterclockwise from the vertical and horizontal planes of the stress block, as shown in Fig. 2-42c. Actually, the angle 9,, is measured from the x-axis to the outward normal to one of the principal planes. Since the outward normal and the plane are perpendicular, the angle between the x-plane and one of the principal planes is also 9p. To determine which principal stress acts on which plane, note that as the radius ofthe circle rotates counterclockwise, the end of the radius CV moves from V to D, indicating that as the initially vertical plane rotates through 14.87”, the stresses change to 9.06 ksi (T) (normal) and 0 ksi (shear). Note also that the end H or radius CH moves to E, indicating that as the initially horizontal plane rotates through 14.87‘, the stresses change to 7.06 ksi (C) (normal) and 0 ksi (shear). These stresses are shown on the sketch of Fig. 2-42c, in which the surfaces D and E are the principal planes represented by points D and E on Mohr’s circle. Point A on Mohr’s circle represents the state of stress on a surface rotated 45“ counterclockwise fi'om the surface represented by point D on the circle. On this surface, the shear stress is the maximum it can be (in the xy-plane), r,,, = rp = rm, =8.06 ksi, and the normal stress is the average value 0,, = (0, + o_,.)r'2 = ac = 1.000 ksi (the normal stress at the center of the circle). Likewise, point B represents the state of stress on a surface rotated

9.06 ksi 9.06 in __ __ ,. .-

_ , _- -- -_._e,,= r4.sr

1.06 kn

I

(6) Figure 2-42(2)

P Positive 6 is measured counterclockwise from the positive x-axis, whether on the stress element or on Mohr's circle. On a plane ofthe stress element a shear stress that rotates the element counterclockwise is plotted below the 0,,-axis on Mohr’s circle, and a shear stress that rotates the element clockwise is plotted above the 0,,-axis of Mohr’s circle.

104 crnrrrzrr 2 ANALYSIS or s'nrsss= coucsrrrs rum nsrrrvrrross 1.000 ksi ' s.o6 ksi

1.000 ksi s.06 ksi 1 1.000 ksi

_

1.000 ksi

f.

45° counterclockwise from the surface represented by point E. The stresses on this surface are also 0,, = (Tc — 1.000 ksi and rm = rp — 8.06 ksi. These stresses are shown on the sketch of Fig. 2-42d. Planes A and B in Fig. 2-42d are the surfaces represented by points A and B on Mohr’s circle, Fig. 2-4-2b. Note that any two orthogonal surfaces of Fig. 2-42c are sufficient to completely specify the principal stresses. Also, only one of the surfaces of Fig. 2-42d is required to completely specify the maximum shear stress. Therefore, these two separate sketches can be combined as shown on the triangular stress element in Fig. 2-422.

4‘~\_v4s 0 f.

x

__-3, _. __ - * ' ‘ B,,— 14-87°

1.000 ksi

I

9.06 ksi

y

an

8.06 ksi

Figum 2_42(d)

D

B

THTH

‘\\

O

x

\ ‘Jr’

I

“ i‘ -\

\/

45° E _ ‘_ _- ~'

Figure 2-42(e-1')

n

5.00 ksi

=

F

(B)

.,_

\/"

30.0 M1=_X/ 3

~ S‘ -.

on MPa

Q.\\

54.0 MPa

~_r26.6 " ..

__,_

42.0 MPa

A’ 42.0 MPa _.

J

.1

(b) 84.0 MPa

6: = 0 Section A-A

Figure 2-43¢»-tr) The principal planes are represented by lines CD and CE, where tan26l,, = -24/18 = —1.3333

which gives

2 9,, = -s3.13= = 53.13% 0,, = -26.5"/= = 26.5"/"J Since 0,,| and 0],; have the sanre sign, the maximum in-plane shearing stress 17,, is

not the maximum shearing stress r,,,,,,, at the point. The maximum shearing stress at the point is represented on Mohr’s circle by drawing an additional circle (shown dashed in Fig. 2-43b) which has line OD as a diameter. This circle is centered at point F (42, 0) and has a radius FG equal to 42 MPa. This circle represents the combinations of normal and shearing stresses existing on planes obtained by rotating the element about the principal axis associated with the principal stress 0_,,g. A third Mohr’s circle (shown dotted in Fig. 2-43b) has line OE as a diameter. This circle is centered at point J (12, 0) and has a radius JK equal to 12 MPa. This circle represents the combinations of normal and shearing stresses existing on planes obtained by rotating the element about the principal axis associated with the principal stress 0,,|. Thus,

1,, = ca = 30.0 MPa 1...... = FG = 42.0 MPa

Ans.

(4)

(C)

105

106 CHAPI'Elt 2 ANALYSIS or s'rrrsss= coucsrrs AND nsrrnrrross The principal stresses 0,1, 0,2, and 0, = 0,3 = 0, the maximum in-plane shearing stress rp, and the maximum shearing stress rm, at the point are all shown on Figs. 2-43c and 2-43d.

1 PROBLEMS MecMovie Activities and Problems MM2.12 Sketching stress transformation results. Learning tool. Constructing appropriate sketches showing orientation ofprincipal and maximum shear stresses.

the nonnal and shear stresses at this point on plane a—a, and show these stresses on a sketch.

MM2.13 Coach Mohr’s Circle of Stress. Theory; Interactive example; Game. Learn to construct and use Mohr‘s circle to deterrnine principal stresses including the proper orientation of the principal stress planes.

T

a

/ /

.\‘[M2.14 Absolute maximum shear stress. Example; Try one. Investigate a three-dimensional stress state at a point.

I

{I

I

iX

I

/45"

MIM2.15 Mohr‘s circle game—plane stress. Game. Recognize correctly constructed Mohr's circles. .\‘[M2.16 Mohr’s circle garne—principalfmax shear. Game. Given a Mohr’s circle, recognize the corresponding stress element for the principal stress state and the rnaximurn in-plane shear stress

a

L

Figure P2-86

state.

Introductory Problems 2-85* The stresses shown in Fig. P2-85 act at a point on the free surface ofa stressed body. Using Mohr’s circle, determine, and show on a sketch, the normal and shear stresses at this point on plane a—a.

2-87 The stresses shown in Fig. P2-87 act at a point on the free surface of a stressed body. Use Mohr‘s circle to determine the normal and shear stresses at this point on the inclined plane a-b shown in the figure. Show these stresses on a sketch.

i

16 ksi

Y a

\\

1

1

_.' b U

45°\\\

x \\

{IE4

l0 ksi

\

l2ksi—x

4

a

Figure P2-85

Figure P2-87

2-86* At a point in a structural member subjected to plane stress there are stresses on horizontal and vertical planes through the point, as shown in Fig. P2-86. Use Mohr's circle to determine

2-88 The stresses shown in Fig. P2-88 act at a point on the free surface of a stressed body. Use Mohr’s circle to determine the norrnal and shear stresses at this point on the

2-11 MOHR’S cntcu: ron PLANE snusss inclined plane a—b shownin the figure. Show these stresses on a sketch.

Ii’

107

2-91 At a point in a structural member subjected to plane stress there are stresses on horizontal and vertical planes through the point, as shown in Fig. P2-91. Use Mohr’s circle to determine the principal stresses and the maximum shear stress at the point. Show these stresses on the planes on which they act.

45 MPa l0k.si b

If

1, jay

27 MPa —x

T

;

3 ksi

l

15 ksi Ii

Figure P2-88 Figure P2-91

Intermediate Problems 2-89* At a point in a structural member subjected to plane stress there are stresses on horizontal and vertical planes through the

2-92* The stresses shown in Fig. P2-92 act at a point on the free

point, as shown in Fig. P2-S9. Use Mohr's circle to determine

smface of a machine component. Use M0hr’s circle to deter-

the principal stresses andthemaximum shear stress at thepoint. Show these stresses on the planes on which they aet_

mine the principal stresses and the maximum shear stress at the point. Show these stresses on the planes on which they act.

7 ksi 25 MPa

_

12 ksi

I

T

25 ksi

I

-vi-Z

I

50 MP5

‘

Figure P2-92 Figure P2-89 2-90* At a point in a structural member subjected to plane stress there are stresses on horizontal and vertical planes through the point’ as shown in Fig‘ P2‘90' Use Mamas circle to detmnme the principal stressa and the maximum shear stress at the point. Show these stresses on the planes on which they act.

2-93 The stresses shown in Fig. P2-93 act at a point on the free surface of a thin-walled pressure vessel. Use Mohr's circle to determine the principal stresses and the maximum shear stress at the point. Show these stresses on the planes on which they act

20 MPa L

|

12 ksi

ih.

T

I40 MPa

L

P

P

1

I 25 k l

1;

|

SO MP3

U

s1

| T

L 10 ksi i

108 CHAPTER 2 ANALYSIS or s'nu:ss= coucsrrs AND nsrnwrross Challenging Problems

2-96 At a point in a structural member subjected to plane stress

2-94* At a point in a structural member subjected to plane stress there are stresses on horizontal and vertical planes through the point, as shown in Fig. P2-94. Using Mohr’s circle, a. Determine the principal stresses and the maximum shear stress at the point. Show these stresses on a triangular stress element. b. Determine the normal and shear stresses on the inclined plane a-b shown in the figure. Show the results on a sketch of the plane on which these stresses act.

there are normal and shear stresses on horizontal and vertical planes through the point, as shown in Fig. P2-96. Using Mohr's circle a. Determine the principal stresses and the maximum shear stress at the point. Show these stresses on a triangular stress element. b. Determine the normal and shear stresses on the inclined plane a—b shown in the figure. Show the results on a sketch of the plane on which these stresses act. 120 MPa

100 MPa

E \e|-

_u \

b\ 60 MPa

‘t7

a

20 MPa

80 MPa

iv

Figure P2-94

Figure P2-96

2-95* The stresses on horizontal and vertical planes at a point on the outside surface ofa solid circular shaft subjected to an axial load P and a torque T are shown in Fig. P2-95. Using Mohr's circle, a. Determine the principal stresses and the maximum shear stress at the point. Show these stresses on a triangular stress element. b. Determine the normal and shear stresses on the inclined plane a—b shown in the figure. Show the results on a sketch of the plane on which these stresses act.

2-97 At a point in a machine component subjected to plane stress there are normal and shear stresses on horizontal and vertical planes through the point, as shown in Fig. P2-97. Using Mohr's circle. a. Determine the principal stresses and the maximum shear stress at the point. Show these stresses on a triangular stress element. b. Determine the normal and shear stresses on the inclined plane a—b shown in the figure. Show the results on a sketch of the plane on which these stresses act.

6000 psi ‘MR

E

\ D-

“tr

I

\\;,__

I, b r10,o00 psi

\\w,

_-.

“-3

\\

ll

Figure P2-95

/

, /I

\

\

\

\

\

\

8000 psi

/iL60° |

12,000 psi

a I

iv

5000 psi Figure P2-97

2-12 GENERAL STATE OF STRESS AT A POINT The general state of stress at a point was previously illustrated in Fig. 2-24. Expressions for the stresses on any oblique plane through the point in terms of stresses on the reference planes can be developed with the aid of the free-body diagram of Fig. 2-4-4, where the rt-axis is normal to the oblique (shaded) face. The areas of

2-12 csmnttl. srars or sntnss AT A ronrr

5:411:

F = SJA

cxdA.r ‘__§“wA5

3

‘"“*

/ Z

dz ,

- 1qdAz

%\"P

-H ---__.a“-~__ ," . _

//' Fadfly % ‘Y=‘“y

1

_

‘""r~,,~

o}dAy

Figure 2-44 the faces of the element are dA for the oblique face and dA cos 6,, dA cos By, and dA cos 9; for the x-, y-, and z-faces, respectively.3 The resultant force F on the oblique face is S dA, whcre S is thc resultant stress (stress vcctor—see Section 2-7) on the area and is equal to ‘/03 + 1:3,. The forccs on the x-, y-, and z-faces are shown as three components, the magnitude of each being the product of the arca and the appropriate stress. If we use I, m, and n for cos 6,, cos 6)., and cos 9,, respectively; the three force equations of equilibrium in the x-, y-, and z-directions are F, = S,dA = axd/11+ tyxdAm + r,xdAn F} = SydA = aydAm +r,ydAn + r,,ydAl F, = S,dA = o,dAn + TXgdA.l + Tygd/1H1 where I, m, and rt arc direction cosincs. Thus, the thrcc orthogonal components of the resultant stress arc S, = 0,1 + ryxm + run

S). = 1:,,1 + crym + Tzyn

('1)

S, = r,,l + tyzm + czn The normal component 0,, ofthe resultant stress S equals Sxl + Sym + S,n; therefore, fi"om Eq. (a), the following equation for the normal stress on any oblique plane through thc point is obtained: 0,, = 0,12 + aymz + 02:12 + 21.',_,,lm + Zryzmn + 21;, nl

(2-21)

3 These relationships can be established by considering the volume of the tetrahedron shown

in Fig. 2-44. Thus, V = %dn c|'A = %dxd.»-1, = §r1ya.4_..=§¢1zr1,4,.sur.r» = at cos6‘, = dy cos 6,. = dz cos6,; therefore, a’/ix = r1A cos 6;, d/1). = dA cos9_,., dA, = u'A cos 9,.

109

11O

CHAPTER 2 ANALYSIS or smass concarrs AND rmrmrrross 'The shearing stress tn, on the oblique plane can be obtained fi'om the relation SQ = 0,? + rip For a given problem the values of S and 0,, will be obtained from Eqs. (a) and 2-21. A principal plane was previously defined as a plane on which the shear stress rm is zero. The normal stress 0,, on such a plane was defined as a principal stress 0p. If the oblique plane of Fig. 2-44 is a principal plane, then S = 0,, and S, = 0,,l, Sy = 0pm, S, = apn. VVhen these components are substituted into Eq. (0), the equations can be rewritten to produce the following homogeneous linear equations in l, m, and n: (0,, — 0,)! — rpm — run = 0 (0,, — 0,.)m — ray" — try! = O

(b)

(0,, — 0,)n — tn! — tyzm = 0 This set of equations has a nontrivial solution only if the determinant of the coefficients of 1, m, and n is equal to zero. Thus, (Up _ ax)

-710‘ -1,;

—Tyx

—Tzx

(UP _ Gr) —T=.v —1:,., (0,, — 0,)

=0

Expansion of the determinant yields the following cubic equation for determining the principal stresses: 03 — (0, + 0,. + 0,)0,§ + (010,, + 0y0, + 0,0,, — rfy — 1'3, — rf,)0;, 2 — 0,-ta 2 — 0311,, 2 + 21:,,-r,-,1“) = 0 —(0,,0y0, — 0,17,,

(L22)

For given values of 0,, 0,-,. . ., rm Eq. 2-22 gives three values for the principal stresses 0P| , 0P2, 0P3. By substituting these values for 0p, in turn, into Eq. (b) and using the relation

I2+m2+n2=l

(C)

three sets of direction cosines may be determined for the normals to the three principal planes. The foregoing discussion verifies the existence of three mutually perpendicular principal planes for the most general state of stress. In developing equations for maximum and minimum normal stresses, the special case will be considered in which 1:,_,. = ry, = -cu = 0. No loss in generality is introduced by considering this special case since it involves only a reorientation of the reference x-, y-, z-axes to coincide with the principal directions. Since the 1-, y-, z-planes are now principal planes, the stresses 0,, 0),, 0, becomes 0],] , 0P2, and 0P3. Solving Eq. (a) for the direction cosines yields I: Sx/apl

m = Sy/Gp2

n = S2/ap3

By substituting these values into Eq. (c), the following equation is obtained:

52

s?

s?

—3‘ + —§ 5 =1 0102 + 0p3

01,1

Id)

2-12 GENERAL srsrs or smass AT A Ponrr Sy op;

tr’ “'1 '0 (-5'» 5;» 51) §‘~

¢-T \ \\

\ '1

\: _-‘T

1

Tm '-—_ I /$_"""-~_ r_J____.\

-____________

0P1

Sr

ftp; S

Z

cpl > op; > 0P3

Figure 2-45

The plot of Eq. (d) is the ellipsoid shown in Fig. 2-45. It can be observed that the magnitude of 0,, is everywhere less than that of S (since S2 = 0,? + 1:3,) except at the intercepts, where S is 0P1, 0'02, or 0P3. Therefore, it can be concluded that two of the principal stresses (0p1 and 0p; of Fig. 2-45) are the maximum and minimum normal stresses at the point. The third principal stress is intermediate in value and has no particular significance. The discussion above demonstrates that the set of principal stresses includes the maximum and minimum normal stresses at the point. Continuing with the special case where the given stresses 0,, 0y, and 0, are principal stresses, we can develop equations for the maximum shearing stress at the point. The resultant stress on the oblique plane is given by the expression

§=$+$+$ Substitution of values for S,,, Sy and S, from Eq. (a), with zero shearing stresses, yields the expression S2 = 0312 + 031112 + 0,3512

(e)

03 = (axlz + aymz + 0,n2)2

(f)

Also, from Eq. 2-21,

Since S2 = 0,? + 1'3” an expression for the shearing stress rm on the oblique plane is obtained from Eqs. (e) and (f) as rm = ‘/0312 + afmz + 0,2112 — (0112 + 0_,,m2 + 0,,n2)2

(2-23)

The planes on which maximum and minimum shearing stresses occur can be obtained from Eq. 2-23 by differentiating with respect to the direction cosines I, m, and n. One ofthe direction cosines in Eq. 2-23 (rz, for example) canbe eliminated

111

1 12

crnrrsn 2 ANALYSIS or STRESS CONCEPTS AND nnrnvrrross by solving Eq. (0) for n2 and substituting into Eq. 2-23. Thus,

13,, = {(03 — 0,2)I2 + (0,? — 0,2]m2

(g)

+03 — [(0, — 0,)l2 + (0,, — 0,]m2 + 0,]2}1/2

By taking the partial derivatives of Eq. (g), first with respect to I and then with respect to m and equating to zero, the following equations are obtained for determining the direction cosines associated with planes having maximum and minimum shearing stresses:

Z[%(ax — 52) _ (51: _ U2)l2 _ (Uy _ o'z)m2:| : 0

(I1)

m [%(a_v _ U2) _ (ax _ Ur)-Z2 _ (Uy _ cz)-'52:] = 0

One solution of these equations is obviously l= m = 0. Then from Eq. (c), n = :i: 1. But this is just the principal surface whose normal is in the z-direction, and on which the shear stresses are their minimum values, 17,, = tzy — 0. Solutions different from zero are also possible for this set ofequations. Consider first that m = 0; then from Eq. (11),! = :t,,/W and from Eq. (c), n = zt,/l/i2. This is the surface whose normal is 45° relative to both the x- and z-axes and perpendicular to the y-axis. This surface has the largest shear stress of all surfaces having a normal perpendicular to the y-axis. Also if I = O, then from Eq. (i), m = :l:,/U2 and fi'om Eq. (c), n = :l:.,/lfi. This is the smface whose normal is 45° relative to both the y- and z-axes and perpendicular to the x-axis. This surface has the largest shear stress of all surfaces having a normal perpendicular to the x-axis. Repeating the above procedure by eliminating I and m in tlnn from Eq. (g) yields other values for the direction cosines that make the shearing stresses maximum or minimum. All the possible solutions are listed in Table 2-1. In the last line of the table the planes corresponding to the direction cosines in the column above are shown shaded. Note that in each case only one of the two possible planes is shown.

Table 2 1

1 m

Direction Conslnes for Shearing Stresses l

Mmimum 2

3

4

Maximum 5

6

;l:l 0

0 11

0 0

1./—1/2 1./—1/2

i./—1/2 0

0 1,/—1/2

i~/T73

i~/T71

Lenin

2-12 csmzrtu. srxrs or STRESS AT A Ponrr Substituting values for the direction cosines from column 4 of Table 2-1 into Eq. 2-23, with 0,, 0y, 0, replaced with 0P1, Upz, 0P3, yields

Tmax =

1 2

1 2

Z0,“ + ZGP; —

1

1

£0,111 + 551:2

2

_ Up1—Up;_

_

2

Similarly, using the values of the cosines from columns 5 and 6 gives 01—03 and rmax=% 0;—03 rw=LEi

Of these three possible results, the largest magnitude will be the maximum stress; hence, the expression for the maximum shearing stress is

1-max = (

(244)

which verifies the statement in Section 2-10 regarding maximum shearing stress.

1 EXHIIIIPIB Pfflblfilll 2- 15 stresses are

At a point in a stressed body, the known

0, = 14 ksi (T)

0,. = 12 ksi (T)

0, = 10 ksi (T)

Ix, = 8 ksi

1:_,.z = —10 ksi

ru = 6 ksi

Determine (a) The normal and shearing stresses on a plane whose outward nonnal is oriented at angles of 61.3“, 53.1“, and 50.2“ with the x-, y-, and z-axes, respectively. (b) The principal stresses and the maximum shearing stress at the point.

SOLUTION (a) The direction cosines for a plane whose outward normal is oriented at angles of 61 .3”, 53.1”, and 50.2” with the x-, y-, and z-axes, respectively, are

I = cos 61.3“ = 0.4802 m = cos 53.1“ = 0.6004 n = cos 50.2“ = 0.6401

1 13

1 14 crrtrrtn 2 ANALYSIS or s'nu1ss= coscnrrrs AND nnrnvrrross The three orthogonal components of the resultant stress S on the plane are S, = 0,1 + r,.,,m -1- run = 14(0.4-802) + 8(0.6004) + 6(0.64-01) = 15.367 ksi S, = rxyl +0_,,m + Tzyn = 8(0.-4802) + 12(0.6004) — l0(0.64-01) = 4.645 ksi S, = 1:”! + rum + 0,21 = 6(0.4802) — l0(0.6004) + 10(0.64-01) = 3.278 ksi P Stresses are not vectors and generally cannot be manipulated as vectors. However, the total force on the surface having area dA and normal in the n-direction has a n1agnitude of S dA. The magnitudes of the normal and tangential components of this force are 0,, a'A and r,,, dA, respectively. Therefore, dividing the relationship between the force vector components, (S dA)2 = (0,, dA)2 + (r,,, dA),Z by dA2 and rearranging gives the relationship between the stress components If, = S2 — 0: or rm = ,/S2 —’0f-

Once the three orthogonal components of the resultant stress S on the plane are known, the resultant stress S, the normal stress 0,, and the shear stress 1,, on the plane can be determined by using the equations

>56 -i" ~33+ =53

= \/(15.367)2 + 14.645)’ +(3.27s)1 = 16.335 ksi 0,, = Sxl + S,.m + Szn = l5.367(0.4802) + 4-.645(0.6004) + 3.278(0.6401)

A115-

= 12.266 ksi E 12.27 ksi (T)

6,, = ,/6'2 - 6,; = ,/(16.3s5)2 - (12.266? = 10.863 ksi E l0.86ksi (T)

Ans.

(b) The three principal stresses and the maximum shearing stress at the point are determined by using Eqs. 2-22 and 2-24. Substituting the given values of stress at the point into Eq. 2-22 yields

6; - 366; +2286, +1"/52 = 0 which has the solution

6,1 = 6,,,,, = 22.03 ksi 2 22.1 ksi (T)

Ans.

6,, = 6,, = 13.263 ksi = 18.26 ksi (T) 6,,3 = am, = -4.344 1641 2 4.34 ksi (c)

AnsAns.

Finally, Eq. 2-24 yields

am, - 6,1,, rm =

2

22.08 -1- 4.344) =

= 13.212101; 13.211151

2

Ans.

2-12 csmntu. srxrs or STRESS AT A ronrr

I Example Problem 2-16

1 15

At @6161 in a stressed body, the known

stresses are

0,, = 60 MPa (T)

0,. = 40 MPa (T)

0, = 20 MPa (T)

txy = +40 MPa

ry, = +20 l\/IPa

I“ = +30 MPa

Determine (a) The principal stresses and the maximum shearing stress at the point. (b) The orientation of the plane on which the maximum tensile stress acts.

SOLUTION (a) The three principal stresses and the maximum shearing stress at the point are determined by using Eqs. 2-22 and 2-24. Substituting the given values of stress at the point into Eq. 2-22 yeilds

aj - 1200; + 15006, - 4000 = 0 which has the solution

6,1 = 6,,,,, = 106.23 MPa 2 106.2 MPa (T)

Ans.

0p; = 0m = 10.000 MPa E 10.00 MPa (T)

An

6,3 = am, = 3.365 l‘vIPa 2 3."/7 MPa (T)

Ans.

Equation 2-24 yields

_ am, - 6,,-m _ 106.23 - 3.165 Tmax

_

2

—

2

=51.23 MIPaE5l.2MPa

Ans

(b) The orientation of the plane on which the maximum tensile stress acts obtained by substituting the stress value 0'01 = 0 max = 106.23 MPa into Eqs. (b) and solving for I, m, and n after noting that F + m2 + n2 = 1. Thus,

(46.23)1- 40m - 30» = 0 (66.23)m - 20» - 401 = 0 (86.23)n - 301- 20m = 0 From Eqs. 1 and 2 m = 0.76231’ From Eqs. 2 and 3 n = 0.52471

(1) (2) (3)

P These three homogeneous equations are not independent because the determinate of their coeflicients is zero (by construction). Therefore, these three equations cannot give unique values for the three direction consines, 1, m, and n; they can only give relationships between the direction cosines. The fourth equation, 12 + mi + :11 = 1, must be used along with these three equations in order to get values for the three direction cosines.

1 1 6 CHAPTER 2 ANALYSIS or STRESS: coscarrrs AND narnvrrross Therefore,

1* + (0.76231)” +(0.s24v1)1 = 1 which yields

1 = 0.7339

0,, = 42.793 2 42.88

Ans.

m = 0.5595 ” = 0.3850

0,, = 55.988 2 56.08 0,, = 67.368 2 67.48

AnsAns.

1 PROBIEMS Introductory Problems 2-98* At a point in a stressed body, the known stresses are 0, = 40 It/[Pa (T), 0,. =20 MPa (C), 0, = 20 It/[Pa (T), tn. = +40 MPa, r,., =0 MPa, and r,, = +30 MPa. Determine the nonnal and shear stresses on a plane whose outward normal is oriented at angles of 40°, 75°, and 54* with the x-, y-, and z-axes, respectively. 2-99* At a point in a stressed body, the known stresses are q,= 14 ksi (T), 0,=12 ksi (T), Q: 10 ksi (T), r,, = +4 ksi, r_,., = -4 ksi, and r,, =0 ksi. Determine the normal and shear stresses on a plane whose outward normal is oriented at angles of 403, 60", and 66.2"’ with the x-, y-, and z-axes, respectively. 2-100 At a point in a stressed body, the known stresses are 0, = 60 It/[Pa (T),0_,=90 l\/[Pa (T),0,= 60 MPa (T), r,, =+120 MPa, 1:,-Z = +75 MPa, and ti, = +90 MPa. Determine the normal and shear stresses on a plane whose outward normal is oriented at angles of 60°, 70°, and 37.3“ with the x-, y-, and z-axes, respectively.

2-101* At a point in a stressed body, the known stresses are o5,=0 ksi, 0,.=0 ksi, 0,=0 ksi, r_,,,= +6 ksi, r,,,= +10 ksi, and ta = +8 ksi. Determine the normal and shear stresses on a plane whose outward normal makes equal angles with the x-, y-, and z-axes. 2-102 At a point in a stressed body, the known stresses are 0, = 72 It/[Pa('I‘),0,.= 32 MPa(C),0,= 0MPa,r,,.= +21 MPa,r,,,=0 MPa, and ta = +21 MPa. Determine the normal and shear stresses on a plane whose outward normal makes equal angles with the x-, y-, and z-axes.

Intermediate Problems 2-103* At a point in a stressed body, the known stresses are q,= 12 ksi (T), o_r,= 10 ksi (C), 0,=8 ksi (T), r,_,=+8 ksi, i:_,,=-10 ksi, and r,,=+12 ksi. Determine the principal stresses and the maximum shearing stress at the point.

2-104* At a point in a stressed body, the known stresses are 0, = 40 MPa (T), c5, = 20 MPa (C), ca = 20 MPa (T), Txy = +40 MPa, 1,, = 0 M1-"a, and ta = +30 MPa. Determine the principal stresses and the maximum shearing stress at the point. 2-105 At a point in a stressed body the known stresses are 05, = 14 ksi (T), 0,. = 12 ksi (T), 0, =10 ksi (T), r_,, = +4 ksi, r_,, = -4 ksi, and rz, = 0 ksi. Determine the principal stresses and the maximum shearing stress at the point. 2-106* At a point in a stressed body, the known stresses are 0, = 60 MPa (T),o_1, = 90 MPa (T), 0;. = 60 MPa(T), rm. = +120 MPa, ry, = +75 MPa, and ru = +90 MPa. Determine the principal stresses and the maximum shearing stress at the point. 2-107 At a point in a stressed body, the known stresses are 0, = 0 ksi, 0,=0 ksi, 0,=0 ksi, r,,,=+6 ksi, r,._,=+10 ksi, and tn = +8 ksi. Determine the principal stresses and the maximum shearing stress at the point. 2-108 At a point in a stressed body, the known stresses are 05, = 72 MPa(T), 0, = 32 MPa (C),0, = 0 MPa,1:,,,.= +21 MPa, r,._, =0 MPa, and T11» = +21 MPa. Determine the principal stresses and the maximum shearing stress at the point.

Challenging Problems 2-109* At a point in a stressed body, the known stresses are 0,, =18 ksi (C), 05,- =15 ksi (C), 0, =12 ksi (C), Ix), = -15 ksi, ryz = +12 ksi, and tn = -9 ksi. Determine a. The principal stresses and the maximum shearing stress at the point. b. The orientation of the plane on which the maximum compressive stress acts. 2-110* At a point in a stressed body, the known stresses are 0, = 75 MPa (T), :5. = 35 MPa (T), 0, = 55 MPa (T), r,_,. = +45 MPa, r_,, = +28 MPa, and rm, = +36 MPa. Determine a. The principal stresses and the maximum shearing stress at the point.

SUMMARY b. The orientation of the plane on which the maximum tensile stress acts. 2-111 At a point in a stressed body, the known stresses are cg, = 18 ksi (T), 05. = 12 ksi (T), 0, = 6 ksi (T), 1:,,. = +12 ksi, r,., = -6 ksi, and 1:1, = +9 ksi. Determine a. The principal stresses and the maximum shearing stress at the point. b. The orientation of the plane on which the maximum tensile

2-112 At a point in a stressed body, the known stresses are 0,-=100 MPa (T), 05. = 100 MPa (C), 0,=80 MPa (T), r,,. = +50 MPa, r_,., = -70 MPa, and ta = -64 MPa. Determine a. The principal stresses and the maximum shaming stress at the point. b. The orientation of the plane on which the maximum compressive stress acts.

SITBSS 3615.

SUMMARY Application of the equations of equilibrium is usually just the first step in the solution of engineering problems. Using these equations, an engineer can determine the forces exerted on a structure by its supports, the forces on bolts and rivets that connect parts of a machine, or the intemal forces in cables or rods that either support the structure or are a part of the structure. A second and equally important step is detennining the internal effect of the forces on the structure or machine. In the simplest qualitative terms, stress is the intensity of internal force. A body must be able to withstand the intensity of intemal force; if not, the body may rupture or deform excessively. Force intensity (stress) is force divided by the area over which the force is distributed

F Stress = E

117

(2-1)

Experimental studies indicate that materials respond differently to forces that tend to pull surfaces apart than to forces that tend to slide surfaces relative to each other. Therefore, the resultant intemal force is usually resolved into normal and tangential (shear) components. These are the components that are used to compute the stresses on an intemal surface. The ratio of the normal force N and the area on which it acts is the normal stress 0. Positive normal stresses are tensile normal stresses—they tend to stretch the material. The ratio of the tangential (shear) component of force V and the area on which it acts is the shear stress 1:. The positive direction for shear stress is in a direction 90’ counterclockwise fiom the outward normal direction to the surface. In general, the stress at a given point on a transverse cross section of an axially loaded bar will not be the same as the average stress computed by dividing the total force F by the total cross-sectional area A. For long, slender, axially loaded members such as those found in trusses and similar structures, however, it is generally assumed that the normal stresses are uniformly distributed except in the vicinity of the points of application of the loads. It can be shown that the shear stress camrot be uniformly distributed over the area. Therefore, the actual shear stress at any particular point and the maximum shear stress on a cross section will generally be different than the average shear stress. However, the design of simple comrections is usually based on average stress considerations. The stress at a point depends on the orientation of the surface (the area A) used to compute it. For example, in an axially loaded member the normal stress and the shear stress on a surface oriented at an angle 9 to the axis are given by

1 18

crntrrsa 2 ANALYSIS or STRESS coscarrs AND narmrrross 0,, = §(1 + cos 29) and r,, = ;—: sin 26, respectively. That is, the normal stress is a maximum (0 mu = P/A) and the shear stress is zero on a transverse surface (6 = 0"), and the shear stress is a maximum (rm = Pf2A = 0m,,J2) on a surface oriented at an angle of 6 = 45" to the axial direction. Although the normal stress and the shear stress at a point both depend on the orientation of the surface on which they act, the state of stress at a point is completely determined by the normal stress and shear stress on three mutually orthogonal smfaces through the point. For two-dimensional or plane stress (for which 0, = 1'2, = 13,, = rm = r_,, = 0), the normal and shear stresses on a plane oriented at an angle 9 relative to the x-axis are given by

,, + . + %os — . 6,, = ? 29 + 17,, 511129

(2-126)

1:,,, = - %i sin 20 + 1,, 665 26

(2-136)

respectively. For design purposes, the critical stresses at a point are usually the maximum normal stress and the maximum shearing stress. The maximum (and minimum) values of normal stress are called the principal stresses and always occur on planes free of shear stress, which are called principal planes. At every point in a stressed body, there exist three principal stresses acting on mutually orthogonal planes. In plane stress situations, two principal stresses are in the xy-plane 2

6,,1.p1 = U‘ :0’ 1 ,/

+ 1},

(2-15)

and the third principal stress is 0,3 = 0, = 0. The principal directions are oriented at an angle 6P given by

tan26,, =

2 T"

0, — 0,,

(2-14)

relative to the x-axis. The maximum in-plane shearing stress can be found from the principal stresses tp =

(248)

The maximum shear stress at the point may be (0),) — 0,,2)f2, (0,,1 — 0)f2, or (0 — 0,,2)f2, depending on the relative magnitudes and signs ofthe principal stresses. The planes associated with the maximum shear stress bisect the angles between the planes experiencing maximum and minimum normal stresses. The transformation equations for plane stress have a simple graphical representation called Mohr’s circle. Normal stresses are plotted on the horizontal axis, with tensile stresses (positive) plotted to the right of the origin and compressive stresses (negative) plotted to the left of the origin. Shearing stresses are plotted on the vertical axis with those tending to produce a clockwise rotation of the stress element plotted above the 0-axis and those tending to produce a counterclockwise rotation of the stress element plotted below the 0-axis. Each point on Mohr’s circle

REVIEWPROBLEMS

119

represents the state of stress on some surface through the given point. The angular separation of two points on Mohr‘s circle is ir1 the same direction but twice as large as the physical angular separation of the two surfaces represented by the points. Since Mohr‘s circle is simply a graphical representation ofthe transformation equations, every problem that can be solved using Mohr‘s circle can also be solved using the stress transformation equations. Mohr's circle is simply an alternative

(graphical) method of representing and working with the stress transformation equations. Although Mohr's circle can be drawn to scale and used to obtain values of stresses and angles by direct measurements on the figure, it is more useful as a pictorial aid to the analyst who is performing analytical determinations of stresses and their directions at the point.

1 REVIEW PROBLEMS 2-113* A farmer is extracting a post fi'om the ground using the structure shown in Fig. P2-1 13. Ifthe force required to remove the post is 2000 lb, determine the normal stresses in the 0.25in.-diameter cables CE and AB.

A

,@

E .

\|60°

4*,

. k ~ .. .\

|.'.|

_. '

"

it E ._ i

Q V _

L

15.

'75 kN FlgureP2-114

_. ..

=2

_,

15' I

B '

I I/if’

A. C D

L

2-115 A pin-connected truss is loaded and supported as shown in Fig. P2-115. Determine a. The normal stress in member AC if it has a cross-sectional area of 1.477 in? b. The minimum cross-sectional area for member CD if the axial stress must he limited to 3500 psi.

Figure P2-113

A

2-114* Two tie rods are used to support a 75-kN load as shown in Fig. P2-l l4. Determine a. The minimum cross-sectional area required for each of the rods if the normal stress in each rod must be limited to 75 MPa. b. The minimum diameters required for the pins at A and C if the shear stress in each pin must be limited to 100 MPa. Both pins are in single shear.

5' '

Ci

7500 lb

__ as D '

9000 lb Figure P2415

5 Pl

120 cmwrnx 2 xnursrs or s'nu:ss= coscnvrs xrrn DEFINITIONS 2-116* Determine the maximum axial load P that can be applied to the 150 x 180-mm wood compression block shown in Fig. P2-116 if the shear stress parallel to the grain must not exceed 1.40 MPa and the normal stress perpendicular to the grain must not exceed l2 MPa.

45 MPa

1

35 MPa

i

L18 MPa ii,

P

/= Rigid Ir

25° '

.\

plate

7

..-

-

_

2-119* At a point in a structural member subjected to plane stress there are normal and shear stresses on horizontal and vertical planes through the point, as shown in Fig. P2-1 19. Using Mohr's circle

_ _- .' . 180 nun

Grain

— e

_ - .

Figure P2-1 18

‘

‘ Figure P2416

a. Determine the principal stresses and the maximum shear stress at the point. Show these stresses on a triangular stress element. b. Determine the normal and shear stresses on the inclined plane a-b shown in the figure. Show these stresses on a sketch of the plane on which they act.

2-111 The stresses shown in Fig. P2-117 act at a point on the free

7 ks‘

surface of a stressed body. Determine the normal and shear stresses at this point on the inclined plane a-b shown in the figure. Show these stresses on a sketch of the plane on which they act.

“ \\ | mg‘

l \b

13 ksi 4 ksi

28 ksi

Figure P2-1 19 a\

I 6:2 I \ \b

7 ksi

i

12 ksi

'_ F1211" P2417

2-118 The stresses shown in Fig. P2-118 act at a point on the free surface of a stressed body. Determine the principal stresses and the maximum shear stress at the point. Show these stresses on a triangular stress element.

2-120 At a point in a stressed body, the known stresses are 05, = 53 MPa (T), qr, = 28 MPa (C), at = 36 MPa (T), r,,, = +24 MPa, r_,,,=-18 MPa, and rg=+46MPaDetermine a. The normal and shear stresses on a plane whose outward normal is oriented at angles of40°, 75°, and 54° with the x-, y-, and z-axes, respectively. b. The principal stresses and the maximum shear stress at the point. c. The orientation of the plane on which the maximinn tensile stress acts. 2-121 Demonstrate that Eq. 2-22 reduces to Eq. 2-15 for the state of plane stress.

Chapter 3 Analysis of Strain: Concepts and Definitions

3-1 INTRODUCTION Relationships were developed in Chapter 2 between forces and stresses and between stresses on planes having different orientations at a point using equilib1'ium considerations. No assumptions involving deformations or materials used in fabricating the body were made; therefore, the results are valid for an idealized rigid body or for a real deformable body. In the design of structural elements or machine components, the deformations experienced by the body, as a result of the applied loads, ofien represent as important a design consideration as the stresses. For this reason, the nature of the deformations experienced by a real deformable body as a result of internal force or stress distributions will be studied, and methods to measure or compute deformations will be established.

3-2 DISPLACEMENT, DEFORMATION, AND STRAIN 3-2-1 DiSpl2lC€I1l€1lt When a system of loads is applied to a machine component or structural element, individual points of the body generally move. This movement of a point with respect to some convenient reference system of axes is a vector quantity known as a displacement. For example, consider a body where displacements are restricted to the x—y plane. In Fig. 3-1 the solid lines represent a body before loads are applied, and the dashed lines represent the body after loads are applied and the body has deformed. Line AB of initial length L; has deformed into a line of final length Ly. Point A is displaced to point A’. The vector from A to A’ is called the displacement of point A. For this twodimensional example, the scalar components of the displacement vector are 14,; in the x-direction and v,, in the y-direction. If L,» = Ly, the body is rigid; and if L; qé Lf, the body has deformed. In some instances, displacements are associated with a translation andfor a rotation of the body as a whole and neither the size nor the shape of the body is changed. A study of displacements in which neither the size nor the shape of the body is changed is the concern of courses in rigidbody mechanics. When displacements induced by applied loads cause the size and/or shape of a body to be altered, individual points of the body move relative to one another. The change in any dimension associated with these relative displacements is known as a deformation and will be designated by the Greek letter delta (5).

122

crrsrrra 5 ANALYSIS or S'l’llAlN CONCEPTS AND DEFINITIONS ___ /’

,1

/

__\\

~. \

1’

\

\

' _£f__ __ _. OB.

A

5?"

=-I-it

E

23 L 2L

l

L

*8

fig»I-

2L

I

L_ ,J '

-

I 2P

_¢,-

1;‘ 7 ".1 - -~ 5

i

J—>"=§

1

P

25

P

(_._->-;'___..-»"'

_,,-.-;;;..::11»4""'

r_____~—

B

__ _ __

I

‘(lg

_ -In C"

k'

\|J

c

X

(c)

151

1 52

cmrr'r"s1r 5 ANALYSIS or srrrsnr coscarvrs AND rmrrrvmoss Since the strains (6,, 6,, yxy, and 6,,) and the angles ¢,, and 1,6 are all small [but the angle 6 is not necessarily small), .

71'

,._,

sm(5 + 3/U) = cos yxy = l sin[6 + (Q5, — 111)] = sin 6 cos(¢,, — 1,0‘) + cos 6 sin(¢,, — 1,0) E sin6+(¢., -1,//) cos6 and Eq. (c) can be written (1 +6y)dy E (1 + 6,,)dn [sin 6 + (¢,, — tit) cos 6]

(d)

where dy = dn sin 6 [see Fig. 3-7a). Therefore, Eq. (d) can be reduced to

(éy — 6,.) @1119 E (¢,. — tit) cos 9 +€n(¢n — 11¢) W5 9 E’ (¢,. — W) COS 9

(Q)

since 6,,, Q5“, and all are all small. Substituting Eq. (3-7a) into Eq. (e) and solving for 45,, yields ¢,, = —(6,, — 6,.) sin 6 cos 6 — y,,. sin: 6 + 1,0

(f)

Equation gives the counterclockwise rotation of a line which makes an angle of 6 with the x-axis initially. Thinking of ¢,, as 45(6), the cormterclockwise rotation of the t-axis can be written

a -w + a

= —(6, -6,) sin(6 +

cos(6 +

— yxy sing (6 +

=(6,,—6,.)cos6 sin6—y,,).cos26+1,b

+16 (g)

Finally, the shearing strain y,,, is the decrease in the right angle between the orthogonal n- and t-directions or the difference between the rotations ¢,, and 4),. Therefore, ]/m = 65:1 — ¢r

= —2(6, — 6y) sin 6 cos 6 + J/,_,,(cos2 6 — sin: 6)

(3-Ba)

or in terms of the double angle y,,, = —(6,, — 6,.) sin 26 + 31,}. cos 26

(3-Sb)

Equations 3-8a and 3-Sb are the strain transformation equations for shearing strain when the state of strain is two-dimensional or plane strain. Note the similarity between the shear strain transformation equations (Eqs. 3-Ba and 3-Sb) and the shear stress transformation equations (Eqs. 2-13a and 2-13b). Equations 3-7 and 3-8 provide a means for determining the normal strain 6,, associated with a line oriented in an arbitrary rt-direction in the x—y plane and

5-4 ms STRAIN TRANSFORMATION squsnoss sort PLANE snnrs

153

the shearing strain y,,, associated with any two orthogonal lines oriented in the

11- and t-directions in the x—y plane when the strains 6,, 6,, and y,,. associated with the coordinate directions are known. When these equations a.re used, the sign conventions used in their development must be rigorously followed. The sign conventions used are as follows:

1. Tensile strains are positive; compressive strains are negative. 2. Shearing strains that decrease the angle between the two lines at the origin of coordinates are positive. 3. Angles measured counterclockwise fi'om the reference x-axis are positive. 4. The (n, I, 2) axes have the same order as the (x, y, 2) axes. Both sets of axes form a right-hand coordinate system.

J1’ I

1 EXHIIIPIB PTOIJIBIII 5-4 The strain components at a point are 6,, = +300;r, 6,. = —l000;.t, and y,,_,, = —600,rt. Determine the strain components 6,,, 6,, and y,,, if the xy- and nt-axes are oriented as shown in Fig. 3-9.

If-.\

1

I

1

\~\

-\ _.1_z

\~ \»'e=30°1 /' \--i

SOLUTION The rt-axis is located at an angle 6,, = —30° with respect to the x-axis; therefore, the strain 6,, is given by Eq. 3-Ta as

\

.. \ tvr

\n

Figure 3-9

6,, = 6,, cosz 6,, + 6,, sinz 6,, + 11,), sin 6,, cos 6,, = 800 cos2(—30"') + (-1000) sin2(—30°) + (-600) sin(—30°) cos(—30“) = 609.8_|.t E 6l0_r.t

Ans. 3.1

The t-axis is located at an angle 6, = +60“ with respect to the x-axis; therefore, the strain 6, is given by Eq. 3-7a as

6, = 6,, cos2 6, +6, sing 6, + y,,,, sin 6, cos 6,

= 800 cos2(+60°) + (-1000) sin2(+60") + (—e00) sin(+60°) cos(+60”) = -809.8,, E -srou Ans.

3.2

P As a check of the small angle assumptions, note that when y,,, = 0.0012588 rad, sin y,,, = 0.00l25900 2 y,,, and cos y,, = 0.9999992 Z 1.

In a similar manner, the shearing strain Ym is given by Eq. 3-Ba as

y,,, = -2(€, - 6,) sin 6,, cos 0.. + y,_.. (cosz 0,, - sin: 0,)

= -2[s00 - (-1000)] sin(—30") cos(—30") +(—600)[cos2(—30‘) - sin2(—30°)] = 1250.8 ,u.rad 2 1259 _u.rad

'3»

Ans.

Therefore, a line element in the n-direction has increased in length, a line element in the t-direction has decreased in length, and the angle at the origin of the nt-axes is now less than 90°.

3.3

134 CHAPTER 5 anuvsrs or s'r'ruu1r= coscsrrrs urn nsnrvrrross 1 PROBLEMS Introductory Problems

Intermediate Problems

3-16* The thin rectangular plate shown in Fig. P3-16 is uniformly deformed such that 6, = -2000 um/rn, 6, = -1500 um/m, and y,,- = + 1250 rtrad. Determine the normal strain 6,, in the plate.

3-19* A thin square plate 30 in. on a side is unifonnly deformed into the rectangle indicated by the dashed lines shown in Fig. P3-19. Determine

I

a. The normal strains 6,, and 6, and the shearing strain yg. b. The normal strain 6,,.

i

J’

i as" 150mm

n\

0.03 in.

\\

I

1t_

Figure P3-16 3-17* The thin rectangular plate shown in Fig. P3-17 is uniformly deformed such that 6,, = +880 /.1.in./in., 6, = +960 ;.r.infin., and yg. = -750 nrad. Determine a. The normal strain 6,,¢ along diagonal AC of the plate. b. The normal strain 63,; along diagonal BD of the plate.

hi 30

Figure P3-19

J"

Di

C \\‘

3-20* A plate undergoing plane strain is deformed into the dashed shape shown in Fig. P3-20. Determine the strains 6,, 6,., and

III

-.__

,/ I

Ir \ I

\I

I

A /,

Yo‘ -

2in.

I\ '~.

0.06 in.

\\\

I

ii 4 mi» B

J-’

Figure P3-17

6

3-18 The thin square plate shown in Fig. P3-18 is uniformly deformed such that 6,, = +1750 p.n1/m, 6,. = -2200 um/m, and yg = -800 rrrad. Determine a. The normal strains 6,, and 6, in the plate. b. The shearing strain y,,, in the plate.

L

100mm '

T r----

7 0.0lmml 50mm

-_.___

J" I

I‘

l-0.02mm ““--

_

— x __________ _l___Q--__,0.lrmn 0.05 mm

Figure P3-20 I

-— 300 mm —i

Figure P3-18

,,

3-21 The strain components at a point in a body rmdergoing plane strain are 6, = -800 rein/‘in., 6, = +640 ,uin./in., and y,,, = -960 ,urad. Determine the strain components 6,, 6,, and y,,,

5-5 PRINCIPAL srrmns AND MAXIMUM srraur STRAIN when the nt-axes are oriented at 6 = 42"‘ counterclockwise from the any-axes as shown in Fig. P3-21.

155

y

5|

Y

/

A

\

/1

I

//

\\

1/

240 mm

n

\\

\ /

\/ \\

/1 \

\

,

/

A

8

\\ /'

ix

Figure P3-21

3-22 The strain components at a point in a body undergoing plane strain are 6, = + 720 rrmfm, 6,. = -480 um/rn, and y,,. = +360 rtrad. Determine the strain components 6,, 6,, and 1.-,, when the nt-axes are oriented at 6 = 30" clockwise from the xy-axes as shown in Fig. P3-22.

C -200mm--200rrr1n

D

x

'

Figure P3-24 3-25* The thin rectangular plate shown in Fig. P3-25 is unifonnly deformed such that6, = + 1575 ,uin./in., 6, = + 1350 ,uin./in., and 6, = + 1250 ,u.in.!in.. Determine a. The shearing strain y,,, in the plate. b. The normal strain 6J, in the plate. t\

/n Q

R

45° 45° 30 in. x

P r~ I

xx‘

I

Figure P3-25

\\

7 I

'~ Pk

I I -\

\‘) '~.

-.,

|' , ,-B ‘n

\\\

\\

1

,

_r‘x 1' I

3-26 The thin rectangular plate shown in Fig. P3-26 is unifonnly deformed such that 6,, = +1950 rtmfm, 6,. = -1625 umlni, and 6,, = - 1275 urn/m. Determine a. The shearing strain ya in the plate. b. The normal strain 69,, along the diagonal QR of the plate.

\\‘,,

ll

J‘

Figure P3 -22

, l ms /" .

Challenging Problems

//

I

If /

3-23 Using the small strain approximation, show that Eq. (b) of Section 3-4 reduces to Eq. 3-7a of Section 3-4.

/

300 mm

/I

‘

I’,

3-24* The thinrectangularplate shown in Fig. P3-24 is uniformly deformed such that 6A, = -1200 ,um/rn, 6 3,, = +750 ,u.m./rn, and 6,,,, = -600 nmfm. Determine the nonnal strain 6,. and the shearing strain yg. in the plate.

P ' i

Figure P3-26

3-5 PRINCIPAL STRAINS AND MAXIMUM SHEAR STRAIN The similarity between Eqs. 3-7 and 3-8 for plane strain and Eqs. 2-12 and 2-13

for plane stress indicates that all the equations developed for plane stress can be applied to plane strain by substituting 6, for 0,, 6y for cry, and y,,,/2 for 1:xy. Thus,

—-— x 400 rnrn

iQ

136 CHAPTER 5 ANALYSIS or STIIAIN: c ONCEPTS AND DEFINITIONS fi'om Eqs. 2-14, 2-15, and 2-17, expressions are obtained for determining in-plane principal directions, in-plane principal strains, and the maximum in-plane shear

strain. Thus, tt1l'126p =

+

yxy

(3_9)

6, — 6,

—

2

1

6,1-as = ex 2 Q iv (ex 2 Q) + (1/ii)

(3-101

V” 2 ”»~--2 "‘_" T 2 1%?)

6,.) is shown in Fig. 3-13. It is apparent that the sign convention for shear strain needs to be extended to cover the construction of Mohr’s circle. Observe for a positive shear strain (indicated in Fig. 3-7) that the edge of the element parallel to the x-axis tends to rotate counterclockwise while the edge parallel to the y-axis tends to rotate clockwise. For Mohrs circle construction, the clockwise rotation will be designated positive and the counterclockwise rotation will be designated negative. This is consistent with the sign convention for shear stresses given in Section 2-7. A Mohr's circle solution for Example Problem 3-5 is shown in Fig. 3-14.

1 PROBLEMS In Problems 3-47 through 3-58 certain strains and angles are given for a point in a body subjected to plane strain. Use Mohr‘s circle to determine the unknown quantities for each problem and prepare a sketch showing the angle 8,, the principal strain deformations, and the maximum shearing strain distortions. In some problems there may be more than one possible value of HP, depending on the sign of y,,..

MecMovie Activities and Problems MM3.4 Coach Mohr‘s Circle of Strain. Theory; Interactive Example; Game. Learn to construct and use Mohr’s circle to determine principal strains including the proper orientation of the principal strain planes.

142

cnsrrss 5 ANALYSIS or sraxne coscsrrs AND DEFINITIONS

Introductory Problems Problem

6,

PS'1

-e _fl'\

l'l\

5,02

"U

i

Vp

Vrnax

use + 1s.43'=

+ 703,1

-600... l -7ssn| -104n|

-114;.

-903.“

+19.2s@

3-47*

+40011

3-48*

+ 945pt

3-49 3-50

-1-16.85"

-34.10“

Intermediate Problems Problem

l"f\ in

m ‘<

*2‘Q

- M -.1Ul ‘R

3-51*

'1"\O Ur @ ‘F

3-52‘

+ too0 7;

-225 1; -333 ‘F

3-53

'i'“-Jut @ ‘P

+Lu 8 ‘P

3-54

+600,u.

+480,u,

m E.

m -1"u:onM Ti

m E F-J

R3

ymax

GP?

VP

yrnax

cs ‘m

ii.

is 38 1:‘:

Challenging Problems Problem

l'f\ is

3-55"

-680 '2

-i-320 R

+ ii -rt‘R

3-56"

'1"J5- Ur @ 1;

-1"Go ‘F

+ -I@ o ‘R

I'h

m

'<

1’-\'.v

i<

mE

*-

3-57

*1"Lu @ @ 1;

-1"-r{II o F

+ 120;.

3-ss

-300..

+ox S 1‘

—-450p.

m

hm

3-7 STRAIN MEASUREMENT AND ROSETTE ANALYSIS In most experimental work involving strain measurement, the strains are measured on a free surface of a member where a state of plane stress exists. If the outward normal to the surface is taken as the z-axis, then 0, = 1'2, = 1:2,, = 0. Since this state of stress offers no restraint to out-of-plane deformation, a normal strain 6, J’

develops in addition to the in-plane strains 6,, 6}|, and yxy. The shear strains 1/,1 and 11,}. remain zero; therefore, the normal strain 6, is aprincipal strain. In Section 3-5, expressions were developed for the plane strain case relating the in-plane principal strains (6_,,| and 6_,,;) and their orientations to the in-plane strains 6,, 6_,,, and y,,_,,.

,.+\ J 3'

1'" A

iv

+

r—|—Li|

“V

For the plane stress case, which involves the normal strain 6, in addition to the in-plane strains, similar expressions are needed. As an illustration of the effects of an out-of-plane displacement on the deformation (change in length) ofa line segment originally located in the x—y plane,

3-

be

I wflfie’ H

consider line AB of Fig. 3-15. As a result of the loads imposed on the member, the

/_____ _ _ _ ;_ _

u+du

X

line AB is displaced and extended into the line A’B' . The displacements associated with point A‘ are u in the x-direction, v ir1 the y-direction, and win the z-direction.

Point B’ displaces u + du in the x-direction, v + dv in the y-direction, and w + dw z

Figure 3-15

in the z-direction. The deformation 8,“; is obtained from the original length of the line, and the displacements du, dv, and dw of point B’ with respect to A’. Thus,

3-1 srruus MEASUREMEl\T no aossrrs ANALYSIS

(A’B')2 = (L + 5,18? = (L + du)2 + (mi + (dw)2 and afier squaring the terms on both sides, the result is

L2 + 2La,,,. + sf”, = L2 + 2L(du) + (du)2 + (dv)2 + (dw)2 If the deformations are small, the second-degree terms can be neglected; hence, 5,13 = du This indicates that the normal strain along AB (3.45 divided by L) is not affected by the presence ofthe out-of-plane displacements. In fact, none ofthe in-plane strains is affected; therefore, Eqs. 3-7 and 3-8 are valid not only for the plane strain case but also for the plane stress case present when strain measurements are made on a free surface. Electrical resistance strain gages have been developed to provide accurate measurements ofnormal strain. The gage may be an etched foil conductor mounted on epoxy or polyimide backing (see Fig. 3-16). The foil gage is cemented to the material for which the strain is to be determined. As the material is strained, the wires are lengthened or shortened; this changes the electrical resistance of the gage. The change in resistance is measured by means of a Wheatstone bridge, which may be calibrated to directly read strain. Shear strains are more difficult to measure directly than normal strains. The electrical resistance strain gages are sensitive only to normal strains and cannot respond to shear strains. Instead, shear strains are often obtained by measuring nonnal strains in two or three different directions. The shearing strain yxy can be computed from the normal strain data by using Eq. 3-7a. For example, consider the most general case of three arbitrary normal strain measurements, as shown in

5". to l'-'

I.

E.

K. I

;,_: - .1‘.

*"h\.-1",

/ Figure 3-16

E

I

143

144

C

CHAPTER 3

ANALYSIS OF STRAIN: CONCEPTS AND DEFINITIONS

Y

Fig. 3-17. From Eq. 3-7a, b

6,, = 6, cos: 9,, + 6,, sing 9,, + )/1,. sin 9, cos 9,, .

6,, = 6, cosz 9;, + 6,, sinz 9;, + 7/xy sin 9;, cos 6;,

Q

(3-12)

\

ix \\

6, = 6, cosz 9,, + 6,, sinz 19¢ + J/,,. sin 9,, cos 9,,

'\

,

\K 9 \|v|ec

l||6b

ll

II

I

Figure 3-17

\ ~

‘tr v

1 '

F

‘

/‘1’\

-1

-

L‘

i

in

Figure 3-18

From the measured values of60, 6,5, and 6,, and a knowledge ofthe gage orientations 6,, 9,, and 6, with respect to the reference x-axis, the values of 6,, 6,, and y,_,, can be determined by simultaneous solution of the three equations. In practice the angles 6,, 95, and 9,, are selected to simplify the calculations. Multiple-element strain gages used for this type of measurement are known as strain rosettes. Two rosette configurations that are marketed conmrercially are shown in Fig. 3-18. It should be noted that the strain rosettes shown in Fig. 3- 1 9 are all equivalent. The choice of which to use is often determined by the geometry of the machine part and the point at which the strains are to be determined. The rosettes shown in Figs. 3-19a and 3-19b could be used to determine strains at a point on the fi'ee surface of a shaft, pressure vessel, or other type of machine component. The point of interest would be located at the center of the triangular arrangement ofgages for the rosette shown in Fig. 3-19a or at the intersection of the three gage lines for the rosette shown ir1 Fig. 3-19b. Neither of these rosettes could be used to determine the strains at a point near the edge ofa hole or other type of boundary in a machine component. Since the intersections of the gage lines for the rosettes shown in Figs. 3-19c and 3-19d are outside the regions occupied by the gages, they can be used to determine the strains at a point near the edge of a hole or other type of boundary ir1 a machine component. In this book the angles used to identify the normal strain directions of the various elements of a rosette will always be measured cotmterclockwise from the reference x-axis. Once 6,, 6_,,, and y,,,. have been determined, Eqs. 3-9, 3-10, and 3-1 1 can be used to determine the in-plane principal strains, their orientations, and the maximum ir1-plane shear strain at the point. In Section 4-3 , it will be shown that for plane stress v 6, = 6p; = —:(61 + 6,.)

(3-13)

where v is Poisson‘s ratio (a property of the material used in fabricating the member), which is defined in Section 4-2. For the case of plane stress, this out-of-plane

_

0&3

W g

0%

@ (a )

Figure 3-19

(b)

(v)

(4)

3-1 STRAIN MEASUREMENI‘ AND noserrs mursrs

145

principal strain is important because the maximum shear strain at the point may be (epl — Gpg), (e_,,1 — Epg), or (€p3 — 6P3), depending on the relative magnitudes and signs of the principal strains at the point. Strain-measuring transducers such as electrical resistance strain gages measure the average normal strain under the sensing foil element of the gage (along some gage length) and not the strain at a point. So long as the gage length is kept small, errors associated with such measurements can be kept within acceptable liniits. The following example illustrates the application of Eqs. 3-12 to principal strain and maximum shear strain determinations under conditions of plane stress. 666

1 Example Problem 3-7 A mmrosette, composed ofthree electrical resistance strain gages making angles of 0°, 60°, and 120° with the x-axis (see Fig. 3-20) was mounted on the free surface of a material for which Poisson’s ratio is 1/3. Under load, the following strains were measured: 6,, = +l000,u

Eb = +750,tz

Gage 1.

../

"0-t

Gage b

60f/

o0

LI

C‘)Hm-HCD a

sf = —650,u Figure 3-20

Determine the principal strains and the maximum shear strain. Show the directions of the in-plane principal strains on a sketch. SOLUTION The given data are substituted in Eqs. 3-12 to yield ea = so = ex = +l000p. (measured) 6,, = 660 = —650;1. = 1000,11. cosl 60° + ey sin: 60° + yxy sin 60° cos 60° eb = em] = +750p. = l000,u. cos: 120° + 6}. sing 120° + yn, sin 120° cosl20° from which 6), = —266.7,u.

and

y,,_,, = —l6l6.61.t

The in-plane principal strains em and ep; are given by Eq. 3-10 as 6,, +6

6, — e

2

= 2 ’ it/(Ty) * 1000 — 266.7

(e,.1.e,.2>(10°) = — 1 2

11,. 2

1000 + 266.7 2

—l616.6 2

— + — 2 2

= 366.7 5: 1026.9 6p| = (366.7 + l026.9)[10_6) = l393.6(l0"’) E 13941.1.

Ans.

s_,,;; = (366.7 — l026.9)(l0‘°) = —660.2(l0“°) E —660,u.

Ans.

The third principal strain 6P3 = ez is given by Eq. 3-13 as

t. = =

H,»

= - i(l000,u. — 266.7,u)

1 - (1 /3) = -366.1“ 2 -36'/it

Ans.

P Note that gage b could be considered oriented 120"‘ above the x-axis or 60“ below the x-axis because cos: 120“ = cos2(—60°), sinz 120" = sin2(—60°), and cos 120" sin 120° = cos(—60°‘) sin(—60"’). Similarly, gage c could be considered oriented 60" above the x-axis or 120"" below the x-axis.

CHAPTER 3

ANALYSIS OF S'I'IUiIN: CONCEPTS AND DEFINITIONS

Since em is less compressive than the in-plane principal strain GP], the maximum shear strain at the point is the maximum in-plane shear strain yp. Thus,

Ymax = VP =€.v1 —€P2 = l393.6,u. + 660.2u = 2053.8,u. "5 2050_u.

.1’

550“

Ans.

The in-plane principal strain directions are given by Eq. 3-9 as

.- 26.0” ,

”

y,,. = -1616.6 =_ 1.2763 e,-6, 10o0+26e."/ 29,, = -51.92“ 9,, = -25.960 E -2e.0=

1I8Il29p=

Imp Figure 3-21

Ans.

The required sketch is given in Fig. 3-21.

1 PROBLEMS MecMovie Activities and Problems MM3.5 Strain measurement with rosettes. Example; Try One. Calculating strain states with rosette strain data.

a. The strain components ex, ey, and yr, at the point. b. The principal strains and the maximum shearing strain at the point. A sketch is not required.

Introductory Problems 3-59* At a point on the free surface of a steel (v = 0.30) machine part, the strain rosette shown in Fig. P3-59 was used to obtain the following normal strain data: ea = +750 pin./in., ab = —125 uin.fin., and st = —250 _u.in.;'in. Determine

Gageb

it ___lY__120'~=

a. The strain components ex, e_.., and ya. at the point. b. The principal strains and the maximum shearing strain at the point. A sketch is not required.

izosii

.-.__\_- L

J’

I

'./‘.'

" “ 120°

X

G

age G

Gage c iGagec

@ 45°

Gage b

Figure P3-60 Gage a

Figure P3-59 3-60* At a point on the free surface of a steel (v — 0.30) machine part, the strain rosette shown in Fig. P3-60 was used to obtain the following normal strain data: ea = —555 um/m, 6;, = +925 urn/m, and ec = +740 ,um/m. Determine

3-61 The strain rosette shown in Fig. P3-61 is attached to a point on the free siuface ofan aluminum (v = 0.33) machine part. The following normal strain data were taken: ea = +800 p.in..-‘in., ei, = +950 uin./in., and ec = +600 ,u.in.,/in. Determine

3-7

STIIAIN MEASUREMENT‘ AND ROSEITE ANALYSIS

a. The strain components ex, e,., and y,,. at the point. h. The principal strains and the maximum shearing strain at the point. A sketch is not required.

J’

Gages

\ y

W

/

Gage c

é

4

3

'

Gags @-

_4§@ G flaw

_ (11)

I

(bu

Figure P3 63

Gage b

»

Gage a

as Rx

r

Figure P3-61

3-64 The strain rosette shown in Fig. P3-64 was used to obtain normal strain data at a point on the free surface of a 2024T4 aluminum alloy (v = 0.30) machine part. The gage readings were ea = +525 um/rn, e.-, = +450 uni/rn, and efi = +1425 ,um/m. Determine the principal strains at this point, and show the principal strain deformations on a sketch.

Intermediate Problems 3 62* At a point on the free surface of an aluminum alloy (v = 0.33) machine part, the strain rosette shown in Fig. P3-62 was used to obtain the following normal strain data: e.-i = +780 um/m, ab = +345 urnlm, and er = -332 um/in. a. Determine the strain components e,,, ey, and yg at the point. b. Determine the principal strains and the maximtun shearing strain at the point. A sketch is not required.

Gagec .-"

T

Gageb

2°? I

"~t_ ‘E45 O

l

x

Gage a

Figure P3-64

Challenging Problems

G55 Gage

Gage b J’

it

.60“

I

Gagea

is ~x

60°./’

3-65* At a point on the free surface ofa steel (v = 0.30) machine part, the strain rosette shown in Fig. P3-65 was used to obtain the following normal strain data: ea = +875 ,uin./in., 6,, = +700 ,uinJin., and e, = +350 ;.tin.fin. a. Determine the principal strains and the maximum shearing strain at the point. Prepare a sketch showing all of these strains. b. Determine the normal strain in the n-direction at the point.

Figure P3-62

y 4 3-63* At a point on the outside surface of a steel (v = 0.30) thin-walled pressure vessel the strain rosette shown in Fig. P3-63 indicates the normal strains ea = +36 prin./in., eh = +310 uin.!in., and e, = +150 ,uin./"m. Gages a and c are oriented in the axial and hoop directions of the vessel, respectively. Determine the principal strains and the maximum shearing strain at this point. Show the principal strain deformations and maximum shear strain distortion on a sketch.

H

Gage c

3 Gage b

501,...-

+

Figure P3-65

Gagea

I

148 CHAPTERS iuwursrs or STIIAIN: CONCEPTSANDDEPINITIONS 3-66 At a point on the free surface of an aluminum alloy (v = 0.33) machine part, the strain rosette shown in Fig. P3-66 was used to obtain the following normal strain data: ea = +875 urn/m, ei, = +700 um/m, and ac = -650 um/m.

3-67* At a point on the free surface of an aluminum alloy (v = 0.33) machine part, the strain rosette shown in Fig. P3-67 was used to obtain the following normal strain data: en = +800 ,uin./in., er, = +950 ,u.in.1'in., and e, = +600 uinjin.

a. Determine the principal strains and the maximum shearing strain at the point. Prepare a sketch showing all of these strains. b. Determine the shearing strain y,., at the point.

a. Determine the principal strains and the maximum shearing strain at the point. Prepare a sketch showing all of these strains. b. Determine the normal strain in the n-direction at the point.

, fl

t

/1/If 40" /

Y

x

600

Gage c

I Gage b Gage b

MU

y

Gage c 120°

~._s0°

I

so".

Gage a

B _

Gage a

Figum P346

Figure P3-67

SUMMARY When a system of loads is applied to a machine component or structural element, individual points of the body generally move. A study of displacements in which neither the size nor the shape of the body is changed is the concern of courses in rigid-body mechanics. When displacements induced by applied loads cause the size and/or shape of a body to be altered, individual points of the body move relative to one another. Inthe design of structural elements or machine components, the deformations experienced by the body as a result of the applied loads often represent as important a design consideration as the stresses. Strain (deformation per l.1l'li'[ length) is the quantity used to measure the intensity of deformation, just as stress (force per unit area) is used to measure the intensity of an internal force. Vtrhen the deformation 8 is in the same direction as the gage length L, the strain is called normal strain

est = 6..3

(3-1)

When the deformation 8 is perpendicular to the gage length L, the strain is called shear strain 5

yavg = 3’ = ran¢

t12) 5 “C =—=—=-0.006003 EBCABC 30,000(4) 1“

A. “S

The negative sign indicates that segment BC of the bar decreases in length. (d) For segment CD,

P,;,,LC,, __ +45(4)(12) _ _ aw_ _ ECDACD 30,000“) -+0.0l8001n. The positive sign indicates that segment CD of the bar increases in length. The deformations of the individual segments 5,15, 53¢;-, and 5(1) are then added algebraically to give the change in length of the complete bar. Thus, 540 =

PL

=5AB+5ac+3co

= -1-0.06560 — 0.00600 -1- 0.01800 = -1-0.07760 ill. '5 -1-0.0776 ill.

Alli.

The positive sign indicates that the complete bar increases in length.

I

L

Example Pl‘Obl€I]l 5-3 A homogeneous bar of uniform cross section A hangs vertically while suspended fi'om one end, as shown in Fig. 5-5a. Determine

(a) The elongation ofthe bar due to its own weight W in terms of W, L, A, and E. (b) The elongation of the bar if the bar is also subjected to an axial tensile force P at its lower end.

Figure 5-5(a)

193

194

CHAPTER 5 AXIAL LOADING APPLICATIONS AND PRESSURE vrsslus

P.=»;

T L

"T. = "rP;=vAx

SOLUTION (a) A free-body diagram of a segment of the bar, Fig. 5-5b, shows that the axial force is a function ofx, the distance from the free end ofthe bar. Thus, Eq. 5-4 is applicable. The weight of the segment of the bar shown in Fig. 5-Sb is W, = y V, = 1/Ax, where y is the specific weight of the material of which the bar is made and P} is the volume of the bar segment. From Eq. 5-4,

(b)

Figure 5-5(b)

L

5=f

P

1

L

Y

L

—xdx=—f yAxdx=—[ xdx

0 EA:

EA 4—oc

2° l 0) and a temperature increase (AT > 0) both cause a stretch of the bar (8 > 0). The stress at a discontinuity in a structural or machine element may be considerably greater than the nominal or average stress on the section. The ratio of the maximum stress at a discontinuity to the nominal stress on the section is called the stress concentration factor. Thus, the maximum normal stress at a discontinuity in a centrically loaded member is

0 = Kg

(5-5)

Graphs or tables for K, the stress concentration factor, can be found in numerous design handbooks, and they may be based on either the gross area or the net area (area at the reduced section). It is important when using such stress concentration graphs or tables to ascertain whether the factors are based on the gross or net section. Stress concentration is a very localized effect. For example, the stress on the botmdary of a hole in a large plate under uniform unidirectional tension is 3 times the nominal stress—the stress in regions far removed lrom the hole. However, at a distance of one hole diameter fi"om the edge of the hole, the stress is only about 7 percent greater than the nominal stress. Stress concentration is not significant in the case of static loading ofa ductile material because the material will yield inelastically in the region of high stress. As a result of the accompanying redistribution of stress, equilibrium may be established and no harm done. However, if the load is an impact or repeated load, the material may fiucture. Also, if the material is brittle, even a static load may cause fiacture.

271

CHAPTER 5

AXIAL LOADING APPLICATIONS AND PRESSURE VESSEIS

A pressure vessel is described as thin-walled when the ratio of the wall thickness to the radius of the vessel is so small that the distribution of normal stress on a plane perpendicular to the surface of the vessel is essentially uniform throughout the thickness of the vessel. Ifthe ratio of the wall thickness to the irmer radius of the vessel is less than about 0.1, the maximum normal stress is less than 5 percent greater than the average. In a spherical, thin-walled pressure vessel, the normal stress on a section that passes through the center of the sphere is called a meridional or axial stress and is given by

an = g

(5-1)

where r and t are the radius and wall thickness of the pressure vessel, and p is the internal pressure. There are no shearing stresses on any of these planes since there are no loads to induce them. In a cylindrical, thin-walled pressure vessel, the stress on a longitudinal plane is called a hoop or circumferential stress and is given by

0,, = 1%

(5-811)

The normal stress on a transverse plane is called an axial stress and has the sanre value as in a spherical pressure vessel



Figure P5-135

-I so kip 40 kip Figure P5-133

5-134* A tension member consists of a 50-mm-diameter brass (E = 100 GPa) bar connected to a 32-mm-diameter stainlesssteel (E = 190 GPa) bar, as shown in Fig. P5-134. Determine the maximum load P that can be applied to the bar if the normal stress in the brass must be limited to 200 MPa, the normal stress in the steel must be limited to 500 MPa, and the total elongation of the bar must be limited to 5.60 mm.

5-136* A rigid bar ABC is supported by two links, as shown in Fig. P5-136. Link BD is made of an aluminum alloy (E = 73 GPa) and has a cross-sectional area of 1250 mm’. Link CE is made of structural steel (E = 200 GPa) and has a crosssectional area of750 mm}. Determine the normal stress in each of the links and the deflection ofpoint A as the 50-kN load P is applied.

D

mi200mm E

Brass

400\n1m

B ' _

P

Steel

"

J

1.5m#l.0m

A

B

c

¥}e00nuui-l-—300isuia| P

Flam P5-134

Figure P5-136

5-135 Two rigid bars (AB and BC) and a 172-in.-diameter structural steel (E = 30,000 ksi) tie rod AC are used to support a 3000-lb load P, as shown in Fig. P5-135. Determine the normal stress in the tie rod and the change in length of the 30-in. reduced section of the tie rod as the load P is applied.

5-137 Bar A of Fig. P5-137 is a steel (E = 30,000 ksi) rod that has a cross-sectional area of 1.24 in? Member B is a brass (E = 15,000 ksi) post that has a cross-sectional area of 4.00 in? Determine the maximum permissible value for the load P if the allowable normal stresses are 30 ksi for the steel and 20 ksi for the brass.

CHAPTER S

AXIAL IDADING APPLICATIONS AND PRESSURE VKQSEIS

Iii 50 in. ii

|_mg.

-

rd _

-

I

R131

.

s in.

L

A

. I

5' .

C

"1

Q

B i

16in.

10in. Unleaded 0.009 in.

. l5in.

s

5t——g—___

10.000 '

I

4in.-‘ll-4in.

. E

em» “gm P5‘137

___

_

-—».—' .

am.

Figure P5-139

5-138 Bar C of Fig. P5-138 is an aluminum alloy (E = 73 GPa) rod that has a cross-sectional area of 625 mmz. Member D is a wood (E = 12 GPa) post that has a cross-sectional area of 2500 mmz. Determine the maximum permissible value for the load P if the allowable normal stresses are 100 MPa for the aluminum and 30 MPa for the wood.

50mm 0.09mrn unloladed

T

/l0Omm I 200mm E -

150 mm :.;_.'.=.=_,=

5-140 A 90'Imn‘diameter bmss (E = 100 GPa) bar is securely fastened to a 50-mm-diameter steel (E = 200 GPa) bar. The ends of the composite bar are then attached to rigid supports, as shown in Fig. P5-140. Determine the stresses in the brass and the steel alter a temperature drop of 70*C occurs. The thermal coeflicients of expansion for the brass and steel are l7.6(l0"")/°C and l1.9(10—°)/"C, respectively.

P

Tl

=—-

E__T___A

C

300 mm 800 mm

='@'-

I

Figure P5-138

5 139* The pin-connected structure shown in Fig. P5-139 occupies the position shown when unloaded. When the loads D = 16 kip and E = 8 kip are applied to the structure, the rigid bar C must become horizontal. Bar A is made of an aluminum alloy (E = 10,600 ksi), and bar B is made of bronze (E = 15,000 ksi). If the normal stresses in the bars must be limited to 20 ksi in the aluminum alloy and 15 ksi in the bronze, determine a. The minimum cross-sectional areas that will be satisfactory for the bars. b. The changes in length of rods A and B.

480 mm

Figure P5-140

5-141 Three bars, each 50 mm wide >< 25 mm thick x 4 m long, are connected and loaded as shown in Fig. P5-141. Bar A is made of Monel, which has a pmportional limit of 400 MPa and a modulus of elasticity of 180 GPa. Bar B is made of a magnesium alloy that has a proportional limit of 100 MPa and a modulus of elasticity of 40 GPa. Bar C is made of structural steel (elastoplastic) that has a proportional limit and yield point of 240 MPa and a modulus of elasticity of 200 GPa. Determine a. The normal stress in each of the bars after a 650-kN load P is applied. b. The vertical displacement (deflection) of pin D produced by the 650-kN load.

tuzvtuw PROBLEMS Z75 5-143* The conical water tank shown in Fig. P5-143 was fabricated from 1/8-in.-thick steel plate. When the tank is completely full ofwater (specific weight y = 62.4 lb/ft‘), determine the axial and hoop stresses 0,, and cn, at a point in the wall 8 fl below the apex of the cone.

B A

c \

12

4 5

3

D Smooth

.1""

“I

P

Figure P5-141

_-

'

0]7¢'_

---

S-142* The rigid barAB of Fig. P5-142 is horizontal and bar C and post D are unstressed before the load P is applied. Bar C has a cross-sectional area of 600 mm’ and is made of a lowcarbon steel (elastoplastic) that has a proportional limit and yield point of240 MPa and a modulus ofelasticity of200 GPa. Post D has a cross-sectional area of 2000 mmz and is made of cold-rolled brass, which has a proportional limit of 410 MPa and a modulus of elasticity of 100 GPa. Determine a. The normal stresses in bar C and post D after a 100-kN load P is applied. h. The vertical displacement (deflection) of point A produced by the 100-kN load.

50mm

100mm 200mm

0-we BE

P

A

““‘°”“°“ I:.'!J,=e%=I Rigid 15° my

C

300 mm

.

-1-

' ' an ,-

'

10ft

\5fi

Figtme P5-143

5-144 A thin-walled cylindrical pressure vessel has an outside diameter of 2 m and a wall thickness of 10 mm. The vessel is made of steel with a modulus of elasticity of 200 GPa and a Poisson’s ratio of 0.30. During proof testing of the vessel, an axial strain of 300 um/m is recorded. Determine a. h. c. d.

The internal pressure applied to the vessel. The axial and hoop stresses in the vessel. The maximtun shearing stress in the vessel. The hoop strain present when the axial strain was measured.

5-145 A gun barrel with an inside diameter of 3.00 in. and an outside diameter of 7.00 in. is made of steel having a yield strength of 50 ksi. Determine the maximum internal pressure that may be applied to the gun barrel before yielding occurs. 5-146* A hydraulic cylinder with an inside diameter of 200 mm and an outside diameter of 450 mm is made of steel (E = 210 GPa and v = 0.30]. For an internal pressure of 125 MPa, determine

Figure P5-142

a. The maximum tensile stress in the cylinder. h. The change in internal diameter of the cylinder.

Chapter 6 Torsional Loading of Shafts

6-1 INTRODUC'l'ION The problem oftransmitting a torque (a couple) from one plane to a parallel plane is fi"equently encountered in the design of machinery. The simplest device for accomplishing this function is a circular shaft such as that connecting an electric motor with a pump, compressor, or other machine. A modified fi"ee-body diagram (the weight and bearing reactions are not shown because they contribute no usefirl information to the torsion problem) of a shaft used to transmit a torque from a driving motorA to a coupling B is shown in Fig. 6-1. The resultant of the electromagnetic forces applied to armature A of the motor is a couple resisted by the resultant of the bolt forces (another couple) acting on the flange coupling B. The circular shaft transmits the torque fi"om the armature to the coupling. Typical torsion problems involve determinations of significant stresses in and deformations of shafts. A segment of the shaft between transverse planes a—a and b—b of Fig. 6-1 will be studied. The complicated stress distributions at the locations of the torqueapplying devices are beyond the scope of this elementary treatment of the torsion problem. A free-body diagram of the segment of the shafi between sections a—a and b—b is shown in Fig. 6-2 with the torque applied by the armature indicated on the left end as T The resisting torque T, at the right end of the segment is the resultant of the moment of the differential forces dF acting on the transverse plane b—b. The force dF is equal to 1: pdA where 1: P is the shearing stress on the transverse plane at a distance p from the center of the shaft and dA is a differential

Electromagnetic -

f°r°f5 "' q

“'1 . .. . ._ 1- in

Q

.--An-nature a

T

a

| b

Figure 6-1

._______

_

/

--- Bolt

forces

6-2 TORSIONAL SI-[BAKING srruuu a b

Tr’ I

T

-r (IF = T9112!

1f’,T\.;1 _. -

I 1 /

J4*\‘ dp i\ é 2' 1

{J

b

Figure 6-2

area. For circular sections, the shearing stress on any transverse plane is always perpendicular to the radius to the point. The resisting torque is statically equivalent to the sum of the torques produced by dF

r.=f pdF=f Prpdfl 3163

BIB3

1 \ \

normal stresses occur in collinear but oppositely directed pairs, and thus they have

0',=0

/

\

Eq. 2-11).

WI

295

(bl z\""

6:0!

Tn r fl

on /

Equations 2-12 and 2-13 then give 0,, = 0,, co52 6 +0). sin? 9 +21:,_,. sin6 oosfl

' or

rxy

(a)

=0+0+2r,ysinacosa =21:,,ysino'coso¢

I

1:), ‘The torque T is generated by a shear stress r,,. = Tc/J’. For the applied torque of Fig. 6-13a, the shear stress will act in the direction shown on Fig. 6-13b (which is in the positive sense as defined in Section 2-8). The double subscript naming convention for the shear stress was described in Section 2-7.

(C) Figure 6-13

T

296 CHAPTER 6 TORSIONAL LOADING or srurrs

:0

WI-\

riri

Slrcss

QQ

4s LA 0__

\O

ca '

135°

l ED @ D

é

___0

—-:9, - - — — — — - — - — — - — - — - — — - — -

— -————-—-———— -————-—

Figure 6-14 and rm = -(0, — a_,.) sin 9 cos 9 + r,_,,(cos2 6 — sinz 6)

(b)

= 0 + r,y(cos2 a — sinz or) = r,y(cos2 or — sinz or) Expressing Eqs. (a) and (b) in terms of the double angle 20¢ yields 0,, = rxy sin 2a

(6-9)

r,,, = rxy cos 20:

(6-10)

In Eqs. 6-9 and 6-10, 0,, is the normal stress on the inclined plane and r,,, is the shearing stress on the same plane. The shearing stress 1:X, is found using the elastic torsion formula (Eq. 6-6). At a given point of the circular shafi tn. is constant, and thus Eqs. 6-9 and 6-10 show that the stresses 0,, and r,,, are functions of the angle of the inclined plane ct. The results obtained from Eqs. 6-9 and 6-10 are shown on the graph of Fig. 6-14, from which it is apparent that the maximum shearing stress occurs on both transverse (0: = 0) and longitudinal (a = 90“) planes. The graph also shows that maximum normal stresses occur on planes oriented at 45° with the axis of the bar and perpendicular to the surface of the bar. On one of these planes (oz = 45‘ on Fig. 6-14), the normal stress is tension, and on the other (oz = 135“), the normal stress is compression. Furthermore, all of these maximum stresses have the same magnitude; hence, the elastic torsion formula gives the magnitude of both the maximum normal stress and the maximum shearing stress at a point in a circular shaft subjected to pure torsion (the only loading is a torque). Any of the stresses discussed previously may be significant in a given problem. Compare, for example, the failures shown in Fig. 6-15. In Fig. 6-15a, the steel rear axle of a truck split longitudinally. One would also expect this type of failure to occur in a shaft of wood with the grain running longitudinally. In Fig. 6-15b, the compressive stress caused the thin-walled aluminum alloy tube to buckle along one 45° plane, while the tensile stress caused tearing on the other 45° plane. Buckling of thin-walled tubes subjected to torsional loading is a matter of pararnoimt concern to the designer. In Fig. 6- 1 Sc, the tensile stresses caused the gray cast iron barto fail in ter1sion—typical of any brittle material subjected to torsion. In Fig. 6-15d, the low-carbon steel bar failed in shear on a plane that is almost transverse—a typical failure for a ductile material. The reason the fracture in Fig. 6-15d did not occur on a transverse plane is that under the large plastic twisting deformation before

as srruzssss or ontrqurr PLANES

297

(11)

(5) (C)

(___

L, —_._ B

0

‘

_%¥_11-—?____g

~- 1.5 rrt

4,--

0

tc

0 Tg

Figure rs-34 6-35 The hollow circular steel (G = 12,000 ksi) shaft of Fig. P6-35 is in equilibrirun under the torques indicated. Determine a. The maximum compressive stress in the shaft at a point near the outside stuface of the shaft. b. The maximum compressive stress in the shalt at a point near the inside surface of the shaft. c. The rotation of end D with respect to end A.

s kip-it /I \

“—

.-~":—-3fi__A__./

10 1tip'it \

1 I 11

- -I1-._f'-_'__:- T“-ad T“ \"' ~ I ‘ " _l 1

-r------5a__

../ 7—---4 it __f-I -.$“‘|lL—;’

/""'—-_

_

:"

4-0'6

3011!) N

100 N

600 N

500 N

Figure P6-36

Challenging Problem 6-37 When the two torques shown in Fig. P6-37 are applied to the steel (G = 12,000 ksi) shalt, point A moves 0.172 in. in the direction indicated by torque T1. Determine a. The torque T1. b. The maximum tensile stress in section BC of the shaft. c. The maximum compressive stress in section CD ofthe shaft.

Q‘ l\-7

25$] N

500 N

_»21n.

‘TX5?;

am

D

D

Figure P6-35

4 in. ../

--:. \-~ \.

6-36 Five 600-mm-diameter pulleys are keyed to a 40-mindiameter solid steel (G = 76 GPa) shaft, as shownin Fig. P6-36. The pulleys carry belts that are used to drive machinery in a factory. Belt tensions for normal operating conditions are indicated on the figure. Each segment of the shaft is 1.5 m long. Determine a. The maximtun shearing stress in each segment of the shaft. b. The maximum tensile and compressive stresses in the shaft. c. The rotation ofend E with respect to end A.

l“*-.,,

9000 lb-ft ‘\

48in.

, '“"~\

a,

"T

‘-..__ "~>

,

Zin.

" R24 in.aw / ~"‘~1/-'4

/_,

-1

,4,--' em. BI

T1

Figure P6-37

6-6 POWER TRANSMISSION One of the most common uses for the circular shaft is the transmission of power;

69

therefore, no discussion oftorsion would be adequate without including this topic. Power is the time rate of doing work, and the basic relationship for work done by a constant torque is Wt = T¢ where Wt is work and ¢ is the angular displacement of the shaft in radians. The time derivative of this expression gives

aw

a’

Power: -07* = TI? = Tm

(6-11)

where a'W;,1/dt is power in lb - ft per nrirrute (or similar units), T is a constant torque

6'10

in lb - it, and cu is the angular velocity of the shaft (assumed constant) in radlmin.

6-6 rowan TRANSMISSION All units, of course, may be changed to any other consistent set of units. Since the angular velocity is usually given in revolutions per minute (rpm), the conversion of revolutions to radians will often be necessary. Also, in the English system of units, power is usually given in units of horsepower, and the relation 1 hp = 33,000 lb - fi/min will be found useful. In the International (SI) system of units, power is given in watts [N ~ mfs). Solution of a power transmission problem is illustrated in the following example problem.

301

‘Pm 6.11

1 Example Problem 6-7 A diesel engine for 8. small commercial boat operates at 200 rpm and delivers 800 hp through a gearbox with a ratio of 4 to 1 to the propeller as shown in Fig. 6-17. Both the shaft from the engine to the gearbox and the propeller shalt are to be solid and made of heat-treated alloy steel. Determine the minimum permissible diameters for the two shafis if the allowable shearing stress is 20 ksi and the angle of twist in a 10-fi length of the propeller shaft is not to exceed 4°. Neglect power loss in the gearbox and assume (incorrectly because ofthrust stresses) that the propeller shaft is subjected to pure torsion. SOLUTION The first step is the determination of the torques to which the shafts are to be subjected. By means of the expression, power = Tco, the torques are obtained as follows:

ll!at Motor

5/’

I-‘I-151 _n= F

Gear Propeller -i_ .

Figure 6-17

P The angular velocity at in Eq. 6-11 must be in either radians per second or radians per minute; 1 revolution equals 2 J’! radians.

800(33,000) = T1(200)[2:r) from which T, = 2l,010lb- it which is the torque at the crank shaft of the engine. Because the propeller shafi speed is four times that of the crank shaft and power loss in the gearbox is to be neglected, the torque on the propeller shalt is one-fourth that on the crank shafi and is equal to 5252 lb - ft. The torsion formula can be used to determine the shaft sizes necessary to satisfy the stress specification. For the main shafi,

.1 _ 5 _ 21,010(12) _ (1/2)¢§ ¢

I

20003)

cf = 8.024

and

cl C] = 2.002 in.

or the shaft fi'0m the engine to the gearbox should be d = 2c| = 2(2.002) = 4.004 in. 1' 4.00 in.

Ans.

The torque on the propeller shaft is one-fourth that on the main shaft, and this is the only change ir1 the expression for cf; therefore,

cg = 8.024/4

C2 = 1.2612 in.

P Note that specifying the power that the shaft must transmit is just another way of specifying the torque that the shalt must withstand. After determining the torque from the given power, the application of the shearing stress equation (Eq. 6-6) and the angle of twist equation (Eq. 6-7) is the same as in previous examples.

302

crmrrsn 6 ronsronu LOADING or srurrs The size of the propeller shaft needed to satisfy the distortion specification is found using Eq. 6-7b. ILL 6=— JG

.7!

s2s2(12)(10)(12)

E01) = i (M2/2)(12)(19 ) cg = 5.747

Cg = 1.5483 > 1.2612

Therefore, the propeller shaft must be d = 20; = 2(l.5483) = 3.0966 in. '5 3.l0in.

Ans.

1 PROBLEMS MecM0vie Activities and Problems MM6.6 Gear trains: power transmission (two shafts). Concept checkpoints. Basic calculations involving power transmission in two shafts connected by gears. N[M6.7 Gear trains: power transmission (three shafts). Concept checkpoints. Basic calculations involving power transmission in three shafts connected by gears.

Introductory Problems 6-38* The shaft of a diesel engine is being designed to transmit 240 kW at l 80 rpm. Determine the minimum diameter required if the maximum shearing stress in the shaft is not to exceed 80 MPa. 6-39 A steel (G = 12,000 ksi) shaft with a 4-in. diameter must not twist more than 0.06 rad in a 20-ft length. Determine the maximum power that the shaft can transmit at 270 rpm.

Intermediate Problems 6-40* A 3-m long hollow steel (G = 80 GPa) shaft has an outside diameter of 100 mm and an inside diameter of 60 mm. The maximtnn shearing stress in the shaft is 80 MPa, and the angular velocity is 200 rpm. Determine a. The power being transmitted by the shaft. b. The magnitude of the angle of twist in the shaft. 6-41* The hydraulic turbines in a water-power plant rotate at 60 rpm and are rated at 20,000 hp. The 30-in.-diameter shaft between the turbine and the generator is made of steel (G = 12,000 ksi) and is 20 ft long. Determine a. The maximum shearing stress in the shaft at rated load. b. The angle of twist in the 20-ft length at rated load.

6-42 A solid circular steel (G = 80 GPa) shaft 1.5 m long transmits 200 kW at a speed of 400 rpm. If the allowable shearing stress is 70 MPa and the allowable angle of twist is 0.045 rad, determine a. The minimum permissible diameter for the shaft. b. The speed at which this power can be delivered if the shearing stress is not to exceed 50 MPa in a shaft with a diameter of 75 mm. 6-43 The engine of an automobile supplies 162 hp at 3800 rpm to the drive shaft. If the maximum shearing stress in the drive shaft must be limited to 5 ksi, determine a. The minimum diameter required for a solid drive shaft. b. The maximum inside diameter permitted for a hollow drive shaft if the outside diameter is 3 in. c. The percentage savings in weight realized ifthe hollow shaft is used instead of the solid shaft 6-44 A hollow shaft of aluminum alloy (G = 28 GPa) is to |:ransmit l200 kW at 1800 rpm. The shearing stress is not to exceed 100 MPa, mid the angle of twist is not to exceed 0.20 rad in a 3-m length. Determine the minimtun permissible outside diameter if the inside diameter is to be three-fourths of the outside diameter. 6-45* A motor delivers 350 hp at 1800 rpm to a gearbox, which reduces the speed to 200 rpm to drive a ball mill. If the maximum shearing stress in the steel shafts (G = 12,000 ksi) is not to exceed 15 ksi and the angle of twist in a 10-ft length is not to exceed 0.10 rad, determine the minimum permissible diameter for each of the shafts.

Challenging Problems 6-46* A motor supplies 200 kW at 250 rpm to gearA of the factory drive shaft shown in Fig. P6-46. Gears B and C transfer

6-1 STATICALLY INDE‘l'ERltflNATE MEMBERS 6 ft

125 kW, and 75 kW, respectively, to operating machinery in the factory. For an allowable shearing stress of 75 MPa, determine a. The minimum permissible diameter d1 for shaft AB. h. The minimum permissible diameter dg for shaft BC. c. The rotation of gear C with respect to gear A if both shafts are made of steel (G = 80 GPa) and have diameters of 75 mm.

us teeth

s rt st 4 it 1

303 ,

|I'l-I.‘ til‘ —_Uil—:[-1 It -__ ‘ ~ :T ._

Motor

GearA

"

GearB

96 teeth

Figure P6-47

A

B

Bearing -~

d1

dz

l

.l

C

l 1

lm

-

2m

aim

6-48 A motor provides 180 kW of power at 400 rpm to the drive shafts shown in Fig. P6-48. The maximum shearing stress in the three solid steel (G = 80 GPa) shafts must not exceed 70 MPa. Gears A, B, and C supply 40 kW, 60 kW, and 80 kW, respectively, to operating units in the plant. Determine a. The minimum satisfactory diameter for shaft D. b. The minimum satisfactory diameter for shaft E. c. The minimum satisfactory diameter for shaft E

Figure P6-46

2.50 m T 2 m 24 teeth --

2.25 rn Tl

-gt?‘

6-47 The motor shown in Fig. P6-47 develops 100 hp at a speed of 360 rpm. Gears A and B deliver 40 hp and 60 hp, respectively, to operating units in a factory. Ifthe maximum shearing stress in the shafts must be limited to 12 ksi, determine a. The minimum satisfactory diameter for the motor shaft. b. The minimum satisfactory diameter for the power shaft.

'\

-

|

6-7 S'l'A'l'lCALLY INDETERMINATE MEMBERS All problems discussed in the preceding sections of this chapter were statically detemrinate; therefore, only the equations of equilibrium were required to determine

the resisting torque at any section. Occasionally, torsionally loaded members are

be determined. A simplified angle of twist diagram will often be of assistance in obtaining the correct equations. The following examples illustrate the procedures to be followed in solving statically indeterminate torsion problems.

Example Problem 6-8 The circular Shafi AC ofFig. 6-18a is fixed to rigid walls atA and C. The solid sectionAB is made ofannealed bronze (GAB =

45 GPa), and the hollow section BC is made ofaluminum alloy (G5-C = 28 GPa). There is no stress in the shaft before the 30-kN - m torque is applied. Determine

the maximum shearing stresses in both the bronze and aluminum portions ofthe shaft after the torque is applied.

|

'g|

‘.

‘ Geanl

48 teeth -" i F _ '5 __-1| LI 5 __ !~ 24 teeth - —" O ' O Gear C

Figure P6-48

constructed and loaded such that the member is statically indeterminate (the number of independent equilibrium equations is less than the number of unknowns). When this occurs, distortion equations, which involve angles of twist, must be written until the total number of equations agrees with the number ofunknowns to

'

D

6-12

304 crrxrrsn 6 TORSIONAL LOADING or sn.u1'rs 4

'

1°°e.m

. ../'

,Ls , .-60rm'n

r~

(>

t

T

ac = +70

30 kN-m

2m

_

,

J1,/1 O“ B “*__-:_;__‘_v\ ‘~:

.4

s

TC

TA

1.4rrt

|

30 1,14-m

C

T,,B=_TA

(=0

(C)

11/ 6

30 kN-m

,

\

A

B

., >5-\ '\

C

7I

08.-‘A Qaxc

(4)

(11) Figure 6-18

SOLUTION A free-body diagram of the shaft is shown in Fig. 6-18b. The torques TA and T¢;- at the supports are unknown. A summation of moments about the axis of the

shaft gives

1",, +TC =30(lO3)N~m

P The torque diagram (Fig. 6-18c) is calculated by drawing free-body diagrams and solving the equations of equilibrium for sections of the shaft to the left of B and to the right of B. Alternatively, the torque diagram can be drawn by observing that it jumps TA at the left end ofthe shaft; it jumps 30 kN - m (in the opposite direction) at section B; and it jumps TC (in the same direction as TA) at section C.

This is the only independent equation of equilibrium relating the two rmlcnown torques TA and TC; therefore, the problem is statically indeterminate. A second equation can be obtained from the deformation of the shaft since the rotation of the two ends of the shaft are not independent.

The torque diagram for the shaft (shown in Fig. 6-13c) represents the results of applying the equilibrium equations to fi'ee-body diagrams of portions of the shafi. The torque in every section ofthe shalt betweenA and B is TAB = —T,;,

and the torque in every section of the shaft between B and C is TBC = +T¢. Where the torque is negative the shaft rotates in a negative sense, and where the torque is positive the shaft rotates in a positive sense as shown in the rotation

diagram of Fig. 6-18d. However, because the two ends of the shaft are attached to the walls, the total rotation of the shaft must be zero 9mm = 631.4 + 96/B = 0

P An alternative interpretation of‘ Eq. (b) is that the rotations of the two segments must have equal magnitude of opposite sense. That is, 95,4 = ~6(-,5.

(0)

(bl

Since Eq. (a) is expressed in terms of T,1 and Tc, the convenient form of Eq. (b) for use here is Eq. 6-7b. Substituting Eq. 6-7b into Eq. (b) and using TAB = -1), and TBC = +T¢ gives TALAB = m TCLBC m Gxs JAB Gsc Jsc

( C)

6-1 STATICALLY INDETERMINATE MEMBERS

305

For the two segments of the shafi, the polar second moments of area are JAB = (zr/2)[504) = 9.8l7(l0“) mm" = 9.8l7(l0‘6)m4

> It is the torques in me mrque diagmm

JBC = (77/2)(504 _ 304) _ 8-545[106)mm4 _ 8-545(l0_6) m4

(Fig. 6-18c) that are used in the angle of twist equation. That is, the torque in the lefl segment of the shafl is LB = -TA and the torque in the right segment of the shaft is

Therefore,

T/1(2)

_

Tc(1-4)

Tar: = +T¢~-

45(l09)(9.8l7)(l0“°) _ 28(109)(8.545)(l0“) from which TC = 0.7737 TA

(d)

Solving Eqs. (a) and (d) simultaneously yields

r,, = l6,9l4N~m = l6.9l4kN~m T¢=13,086N-m=l3.086kN-m

,86-IMPQ

The stresses in the two portions of the shaft can then be obtained by using Eq. 6-6. Thus, L

t _ T,,c,,,, _ 16.91-¢(103)(50)(10"3) ”” _ 1,“; _ 9.81"/(10-6)

= 86.l4(l06)N/m2 2 se.1 MPa

i

"m to

Ans.

"i_‘-J J

-’,\.

-

Xx

->‘0

Figure P6-49 6-50* A steel (G = 80 GPa) tube with an inside diameter of 100 mm and an outside diameter of 125 mm is encased in a Monel (G = 65 GPa) tube with an inside diameter of 125 mm and an outside diameter of 175 mm, as shown in Fig. P6-50. The tubes are connected at the ends to form a composite shaft. The shaft is subjected to a torque of 10 kN - m. Determine

6-51 A 3-in.-diameter cold-rolled steel (G = 11,600 ksi) shaft, for which the maximum allowable shearing stress is 15 ksi, exhibited severe corrosion in a certain installation. It is proposed to replace the shaft with one in which an aluminum alloy (G = 4000 ksi) tube 1/4 in thick is bonded to the outer surface of the cold-rolled steel shaft to produce a composite shaft. If the maximum allowable shearing stress in the aluminum alloy shell is 12 ksi, determine a. The maximum torque that the original shaft can transmit. b. The maximum torque that the replacement shaft can transmit. 6-52* A composite shalt consists ofa bronze (G = 45 GPa) sleeve with an outside diameter of 80 mm and an inside diameter of 60 mm over a solid aluminum (G = 28 GPa) rod with an outside diameter of 60 mm. Ifthe allowable shearing stress in the bronze is 150 MPa, determine a. The maximum torque T that can be transmitted by the composite shaft. b. The maximum shearing stress in the aluminum rod when the maximum torque is being transmitted. 6-53 Two 3-in.-diameter solid circular steel (G = 11,600 ksi) and bronze (G _ 6500 ksi) shafts are rigidly connected and supported as shown in Fig. P6-53. A torque T is applied at the

61 STATICALLY Il\lDB‘l'ERM1NATE MEMBERS junction of the two shafts as indicated. The allowable shearing stresses are 18 ksi for the steel and 6 ksi for the bronze. Determine the maximum torque T that can be applied.

if

a. The rotation of a cross section at C. b. The rotation of a cross section at C if the steel shell is assumed to be rigid. c. The percent error introduced by assuming the steel shell to be rigid.

__..---~ Steel T

I ,-

311

‘

H

5“ 5 n__

Annealed bronze

-

, |

V"

¢ ’

, .

2 (4l‘1n. "

>5»

sn

--._

' >

.

12in.

:-

-'

__

'

"as"!-"3 lfl.

Figure P6-53

4'in.x

‘

C

Figure P6-55

Intermediate Problems 6-54* A composite shaft, as shown in Fig. P6-54, consists of a solid brass (G = 39 GPa) core with an outside diameter of" 40 mm covered by a steel (G = 80 GPa) tube with an inside diameter of 40 mm and a wall thickness of 20 mm, which is in turn covered by an aluminum alloy (G = 28 GPa) sleeve with an inside diameter of 80 mm and a wall thickness of 10 mm. The three materials are bonded so that they act as a unit. Detemiine a. The maximum shearing stress in each material when the assembly is transmitting a torque of 15 kN - m. b. The angle of twist in a 3-m length when the assembly is transmitting a torque of‘ 10 kN - m.

\“-\

\ ____ 5‘;-. \ 100 nun ‘T\i_\\~,\\.:‘\\ ,80mm ~__\\ ‘~,_-$’___ 40mm Aluminum——’ “'-‘“-.. ‘*-\‘"* \ 1.,

/

j¥~\\ 1

\\

\

‘ Q /'

’.\\\

53,51 -~'

Brass =

Figure P6-54

6-55* The composite shaft shown in Fig. P6-55 is used as a torsional spring. The solid circular polymer (G = 150 ksi) portion of the shaft is encased in and firmly attached to a steel (G = 12,000 ksi) sleeve for part of its length. If a torque T of 1000 lb - in. is being transmitted by the composite shaft, determine

6-56 A hollow steel (G = 80 GPa) tube with an outside diameter of 100 mm and an inside diameter of S0 mm is covered with a Monel (G = 65 GPa) tube that has an outside diameter of 125 mm and an inside diameter of 102 nun. The tubes are connected at the ends to form a composite shaft. If the allowable shearing stress in the steel is 70 MPa and the allowable shearing stress in the Monel is 85 MPa, determine a. The maximtun torque T that the composite shalt can transmit. b. The angle of twist in a 2.5-m length when the composite shaft is transmitting the maximum torque. 6-57* A solid aluminum alloy (G = 4000 ksi) rod with an outside diameter of 2 in. is used as a shaft. A hollow steel (G = 12,000 ksi) tube with an inside diameter of 2 in. is placed over the rod to increase the torque-transmitting capacity ofthe shaft. The tube and the rod are attached at the ends to form a composite shaft. Determine the minimum tube thickness required to permit the torque-transmitting capacity of the shaft to be increased by 50 percent. 6-58 A composite shaft consists of a bronze (G = 45 GPa) shell that has an outside diameter of 100 mm bonded to a solid steel (G = 80 GPa) core. Determine the diameter of the steel core when the torque resisted by the steel core is equal to the torque resisted by the bronze shell. 6-59 The steel (G = 12,000 ksi) shaft shown in Fig. P6-59 is

attached to rigid walls at both ends. The right 10 ft of the shaft is hollow, having an inside diameter of 2 in. Determine a. The maximum shearing stress in the shaft. b. The angle of rotation of the section where the torque is applied with respect to its no-load position.

312

cmwren 6 TORSIONAL uoxumc or sums

_::_.;.

:’

4 Ks it

25 ki

nfl

1

2 i-1-

/ J _ I‘ \‘ /,

\*,./" B ~ __,- V _\~\\

.(Xx

4 in.

N‘ :.\‘\

19 fixx

7'. Q z

A

If

3 fix;\~§

--.I

I

I

/

1

.~

;____

'rdA

\

\

\

/

~.

'

\

-.

\

\

~.

\

b

~.

\

\

\‘~

\-\ \ ‘*-

/

1/ \,\’;I' T I\

/ / /

1

\ Q; \‘I

__.1 »--

I I I

L_ \

\

\l____

\

\

(p=f

,‘M.

“'-t

,/

\T’i"/%‘in. Figure 6-34

SOLUTION Once or and 5 are lorown, Eqs. 6-13 and 6-14 can be used to calculate the maximum shearing stress and the angle of twist. Since bla = 31/(1/2) = 6, the values of or and ,8 from Fig. 6-30 are

6=6=am and thus, T Tm“ : uazb =

=11,l11psi§l1,110psi O.30(l/2) (3)

Ans.

and

TL

6=—i fla3bG

2500 12 =L = 0.06661 rad 2 0.0661666 0.30(1/2) (3)(4000)(10 )

Ans.

336 CHAPTER 6 TORSIONAL LOADING or srurrs

I

EXHIIIPIO 6-17 A rectangular box section of aluminum alloy has outside dimensions 100 X 50 mm, as shown in Fig. 6-35. The plate thickness is 2 mm for the 50-rrmi sides and 3 mm for the 100-mm sides. If the maximum shearing stress must be limited to 95 MPa, determine the maximum torque that can be applied to the section. Neglect stress concentrations.

3 mm

l. 2mm

i.

50mm

100mm

6

il

Figure 6-as

SOLUTION The maximum stress will occur in the thinnest plate; therefore,

P Recall that the shear flow qr = rt is constant. Therefore, as the section thickness decreases, the shear stress must increase, and the maximum shear stress will occur in the thirmest portion of the section.

q = n = 9s(10‘)(2)(1o-3) = l9O(lO3)N/III The torque that can be transmitted by the section is given by Eq. 6-15 as

P The areaA is the area enclosed by the median line of the section. In this case, A is the area ofthe rectangle (100-l-1) mm wide and (50-1.5-1.5) mm tall.

T

2qA 2(l90)(103)(100 — 2)(50 — 3)(10_6) 1750N - m

Ans.

1 PROBLEMS Introductory Problems 6-119* The allowable shearing stress for the aluminum alloy (G = 4000 ksi) bar shown in Fig. P6-119 is 12 ksi. Determine the maximum permissible torque that may be applied to the bar.

Qiqil in.

Figure P6-119

in.

6-120* Thetwobars shown in Fig. P6-120aremade ofaluminum (G = 28 GPa). The cross sectional areas and lengths ofthe two bars are identical. Ifthe maximum shearing stress must be limited to 25 MPa., determine a. The maximum permissible torque that may be applied to each bar. b. The angle of twist for each bar when the torque of part a is being applied.

6-12 roasros ormnr-wxrranruans-snrmr now 337 6-123 A torque of 125 kip - in. will be applied to the hollow, thinwalled, aluminum alloy section shown in Fig P6-123. If the maximum shearing stress must be limited to 8 ksi, determine the minimum thickness required for the section.

f'_'\

T1.-J 400 mm

4-0mm

L-Smm

T1-J CW

‘

l31'.n.iI

Sin.

'l'3in.

Figure P6-123

6

Figure P6-121] 6-121 Two aluminum alloy bars {G = 4000 ksi) ofidentical length are rigidly attached to fixed supports at one end. One bar (see Fig. P6-121a) is square. The second bar was machined from a square bar of the same dimensions as the first bar, as shown in Fig. P6-l2lb. If the maximum shearing stress must be limited to 12 ksi, determine a. The maximum permissible torque thatmay be applied to the free end of each bar. h. The angle of twist for each bar when the torque of part a is being applied if the bars are 3 ft long.

Eiiiiiiiiiif

1.5111.

_________ __ L J

(H)

(5)

Figure P6-121

6-124 A 500-mm-wide x 3-mm-thick x 2-m-long aluminum sheet is to be formed into a hollow section by bending through 360° and welding (butt-weld) the long edges together. Assume a median length of 500 mm (no stretching of the sheet due to bending). If the maximum shearing stress must be limited to '75 lV[Pa, determine the maximum torque that can be carried by the hollow section if a. h. c. d.

The shape of the section is a circle. The shape of the section is an equilateral triangle. The shape of the section is a square. The shape of the section is a 150 x 100-mm rectange.

Intermediate Problems 6-125* A solid rectangular bar of aluminum alloy (G = 4000 ksi) is subjected to the torques shown in Fig. P6-125. IfT| = 10,000 lb - in., T; = 30,000 lb a = 2in., b = 3 in., and L1 = L; = 30 in., determine a. The maximum shearing stress in the bar. h. The rotation of end C with respect to the support atA.

6-122* A torque of 2.0 kN - m will be applied to the hollow, thin-walled, aluminum section shown in Fig. P6-122. If the maximum shearing stress must be limited to 40 MPa, determine the minimum thickness required for the section.

ir

L,

T 12 zl

L

C 2-—-_-

Figure P6-125

l— 100 mm —=l

Figure P6-122

6-126* The torsion member shown in Fig. P6-125 has the cross section shown in Fig. P6-126. If T; = T1 = 2T, a = 75 mm,

338 cruumarr 6 ronsroxu mxnrsc or sums b = 100 mm, and L1: L2 = 800mm, determine the maximum torque T that may be applied to the bar if the maximum shearing stress in the bar must be limited to 80 MPa.

Challenging Problems 6-129* A torque box from an airplane wing is shown in Fig. P6-129. Curves AB and CD are 40.2 in. long. The mean depth of the box is 12.5 in. The box is made of an aluminum alloy with an allowable shearing stress of 8 ksi. Determine the maximum torque that can be applied to the box.

10mm 75 mm l

5 mm

1

l001'rnni1l

F’

in

l.

Figure P6-126

-“-3

.

13 tn. i ., D

5.

0.05 in.

.

_ B

0.04111. 0.0-1 in.

.1"

10 rn. L C

+40 in. A

Figure P6-129 6-127 Two torsion members (see Fig. P6-127) made of the same material have the same length and the same weight. For the same allowable shear stress, determine a. The ratio of the torque that can be carried by the circular bar to the torque that can be carried by the square bar. b. The ratio of the angles of twist when the torques of part a are being carried by the two sections.

6-130 A cross section of an airplane fuselage made of aluminum alloy is shown in Fig. P6- 130. For an applied torque of 200 kN - m and an allowable shearing stress of 50 MPa, determine the minimum thickness of sheet (constant for the entire periphery) required to resist the torque.

5.. . \

lfl O

IDID __

B

_

15°

6*o

|"T‘|

Figure P6-127

Figure P6-130 6-128 The 50 >< 50-mm square torsion member shown in Fig. P6-128 is made of an aluminum alloy (G = 28 GPa). If the maximum shearing stress must not exceed 80 MPa and the rotation ofendD with respect to end/I must not exceed 0.035 rad, determine the maximum torque T that can be carried by the bar.

6-131 The 1.5 x 1.5-in. square bar of aluminum alloy (G = 4000 ksi) shown in Fig. P6-131 is rigidly attached to supports at A and D. Determine the reactions at the supports if T1 = 8000lb-in., T; =0,andL1=L; =L; = 1.5 it.

e

400Trnm T

400 mm 400 rnrn

Figure P6-128

37

B

l..l,,lL§~,.

,1 Figure P6-131

6-15 nssrss PROBLEMS 359 6-15 DESIGN PROBLEMS Design, withina limited context, has beendiscussed previously in Chapter5. In that chapter, design was linrited to axially loaded members and to pins. This chapter will extend design to solid or hollow circular bars subjected to static torsional loading. Design of these bars will be limited to proportioning a torsionally loaded member to perform a specified frmction without failure. In this chapter, failure will refer to failure by yielding or failure by fi'acture. Furthermore, design will be limited to circular bars made ofductile materials for which the significant failure stress is the shearing stress at yield or fracture. This book will not address the important issue of fatigue loading. That topic is covered in later courses. For many materials the yield strength in shear is not given in tables such as Tables B-17 and B-18. It will be shown in Chapter 10 that, for ductile materials, the yield strength in shear is either 0.5 or 0.577 times the tensile yield strength, depending upon the criterion selected for failure. For the present, the more conservative value of 0.5 will be used. Design for combined loading will be discussed in Chapter 7.

1 Example Problem 6-18 A 56161 circular shalt 4 ft long made 6f 2014-T4 wrought aluminum is subjected to a torsional load of 10,000 lb ~ in. If failure is by yielding and a factor of safety (FS) of 2 is specified, select a suitable diameter for the shafi if 2014-T4 wrought aluminum bars are available with diameters in increments of l/8 in. SOLUTION The failure criterion is (refer to Chapter 5) Strength 3 (Factor of safety)(Stress) For a torsionally loaded circular shaft, stress refers to the magnitude ofthe shearing stress r,, = T,0/J . For a failure mode ofyielding, strength is the yield strength in shear 1:,-, which for 2014-T4 wrought aluminum is listed in Table B-17 as 24 ksi. Thus, T, T 16 S T 1,, > F5(7c) =1=5(%/2) =fl

T

rrr

rrd3

Solving for the diameter,

d > [l6(FS)(T):|'/3 _ [16(2)(10,000)]"’ 1:1,. :rr(24)(l03) d 3 1.619 in. Since bars are available in 1/'8-in. increments, the smallest permissible bar is

5 dmh, = lg in.

Ans.

340 CHAPTER 6 rorrsronm LOADING or srnrrs

1 EXHIIIPIC Pl'Ol.')l6lI1 6- 1 9

A solid circular shaft 2 m long is to transmit

1000 kW at 600 rpm. Failure is by yielding, and the factor of safety is 1.75. If the shaft is made of structural steel, select a suitable diameter for the shalt if bars are available with diameters in increments of 10 mm. SOLUTION Since failure is by yielding, the strength is the yield strength in shear. Table B-

18 lists the yield strength in tension as 250 MPa, but does not give a value for the yield strength in shear. According to the discussion in Section 6-13, 17,. will be taken as ayr'2. Thus r_,, = 250f2 = 125 MPa. Using the results of Example Problem 6-18,

l6(FS)(T) ‘/3

_

d > l

H

Tv

i

(11)

The torque may be formd using Eq. 6-ll Power = Tm = T(2:rrN) or Power T= 7 2rrN

b ( )

Combining Eqs. (a) and (b) gives

a, > 8(FS)(Power) “'3 _|:

d,"

rr2Nr_,,

:|

M600/60)(125)(10‘)

d 3 0.1043m = l04.3mm Since bars are available in 10-mm increments, the smallest permissible bar is dm-,,, = 110 mm

Ans.

i EXHIIIPIC P1‘OlJlBlIl 6-20 A stepped shaft is subjected to the torques shown in Fig. 6-36. Both segments of the shaft are made of 6061-T6 wrought aluminum. For a factor of safety of 2.0 against failure by yielding, determine the required diameters for the two segments of the shaft. SOLUTION For failure by yielding, the significant strength in the shaft is the shearing yield strength. From Table B-17, the shearing yield strength for 6061-T6 wrought

615 DESIGN PROBLEMS 341 12,000 lb-in. _

1000 lb-in.

I

A-.

A 13,000 lb-in. B

C

Figure 6-36 aluminum is 26 ksi. Using the results of Example Problem 6-18,

16(FS)(T) ‘/3

dam J1’ Ty l i

For segment AB, I13

dxa Z |¥6(2)(100:]):|

= 0.7317 in.

:1-(26)(10 )

For segment BC,

P Of course, a fillet will be necessary at the change of diameter to reduce the stress concentration. If the fillet radius r is about 0.7(0.732 in.) 2 0.5 the stress concentration factor would be about 2, and the actual slzress at the change of diameter would be 2 (13 ksi) = 26 ksi = r,,-,|d.

162 12,000 ‘/3 a‘,,¢> L =1.675in. _l I(Z5)(l03) l

The minimum diameters are then d,;B = 0.732 in.

Ans.

dgc = 1.675 in.

Ans.

1 Example Ptfllllfllll 6-21 A steel pipe will be used as a shaft to transmit 100 kW at 120 rpm. Failure is by yielding (cry = 250 MPa), and the factor ofsafety is 1 .5. Determine the lightest-weight standard steel pipe that can be used for the shaft. SOLUTION The failure criterion Ty 2

can be solved for (J/c) to yield J FS(T) FS(Power) _2_r=m c

ry

ry (2rr N)

Substituting the given numerical values

5>

1.s(100)(103)

6 - 125(106)(2n')(120/60)

342

CHAPTER 6 TORSIONAL LOADING or srurrs O1‘

1 3 95.4900-°) 1113 = 95.4-9(lU3)11111'13 C

where

5 = Ti > 95 49(103) 1111113 c

Zrg

_

I

(0)

Equation (a) can be satisfied for an infinite number of hollow pipes having different radius ratios. Table B-14 can be used to select a pipe that satisfies the requirement that J/c 3 95 .49(103) n1m3. However, Table B-14 lists the properties I and S, where I is the rectangular second moment of area with respect to a diameter of the pipe, and S is the section modulus, S = I/c. Since J = 21 for circular sections (either solid or hollow), the required section modulus for the PIIJC 1S

P Recall that the polar second moment of area of a cross section is equal to the sum of the second moments relative to the x-and y-a.xes,J = I, +1_,. Since I, = I, = U4 :rrr4 for a circular section, J = 21 = 21} = 21,. = l/2 rrr4.

3

s = 1 = ll 2 i95'49(1°) = 47.75(lO3)m1I13 c

2 c

2

(b)

The lightest-weight standard pipe in Table B-14, with s 3 47.7500’) 111m3, is a 102-mm-nolninal-diameter pipe. Any pipe in Table B-l4 that has a section modulus greater than S = 47.75(103) 1111113 would satisfy the stress requirement, but the 102-mm pipe is the lightest, since it has the smallest cross-sectional area.

1 PROBLEMS Introductory Problems

b. If solid structural steel shaits are available with diameters in increments of 10 mm, determine the minimum diameter that can be used. c. Compare the weights of the two shafts.

6-132* A motor is to transmit 150 kW to a piece of mechanical equipment. The power is transmitted through a solid structrual steel shafi. Failure isby yielding, and the factorofsafety is 1.25. The designer has the freedom to operate the motor at 60 rpm or at 6000 rpm. For each case, determine the minimum shaft diameter. Shafls are available with diameters in increments of 5 mm. If weight is important, which speed would be used?

6-135 A shafi is to transmit 100 hp at 200 rpm. The designer has a variety of solid bars and standard steel pipes to select from. Both the bars and the pipes are made of structural steel, the failure mode is yielding, and the factor of safety is 2.

6 133* A 3-ft-long steel pipe is subjected to a torque of 1200 lb-ft at each end. The pipe is made of 0.2% C hardened steel, failure is by yielding, and the factor of safety is 1.5. Determine the nominal diameter of the lightest standard-weight steel pipe that can be used for the shafl.

a. Select the lightest standard-weight steel pipe that can be used. h. Select a suitable solid shaft if they are available with diameters in increments of U8 in. c. lfweight is critical, which shalt should be used?

6 134 A standard-weight structural steel pipe must transmit 150 kW at 60 rpm. The failure mode is yielding and the factor of safety is 1.5. a. Select the lightest standard-weight steel pipe that can be used.

Intermediate Problems 6-136* The motor shown in Fig. P6-136 supplies a torque of 1000 N - m to shaft ABCDE. The torques removed at C,

6-15 DESIGN PROBLEMS 343 f— zso mm —»

D,andEare500N-m,300N-m,and200N-m,respectively. The shaft is the same diameter throughout and is made of 0.4% C hot-rolled steel. For a factor of safety of 3 and failure by yielding, select a suitable diameter for the shaft if shafls are available with diameters in increments of 10 mm.

’\ ._

-1°" ‘Ts. |=IE

s- e

.A

Him

B

C

600 mm

50 mm

Ii

P

Figure P6-138

Figure P6-136

6-137 A torque of 30,000 lb - in is supplied to the factory drive shaft of Fig. P6-137 by a belt that drives pulley A. A torque of 10,000 lb - in. is removed by pulley B and 20,000 lb - in. by pulley C. The shaft is made of structural steel and has a constant diameter over its length. Segment AB of the shafl is 3 ft long, and segment BC is 4 it long. Failure is by yielding, and the factor of safety is 2.25. Select a suitable diameter for the shaft if shafts are available with diameters in increments of 1/8 in.

6-139 The motor shown in Fig. P6-139 supplies a torque of 380 lb - it to shaft BCD. The torques removed at gears C and D are 220 lb - ft and 160 lb - ft, respectively. The shaft BCD has a constant diameter and is made of 0.4% C hotrolled steel, failure is by yielding, and the factor of safety is 2. Determine a. The minimum allowable diameter of the shaft if shafts are available with diameters in increments of 1/8 in. b. The minimum allowable diameter of the bolts used in the coupling, if eight bolts are used, the material is structural steel, the mode of failure is yielding, and the factor of safety is 1.5. The diameter of the bolt circle is d 1 = 3.5 in., and the bolts are available with diameters in increments of 1/16 in.

A C

--

15‘

/Bearing

l!i|g’|'!l!i|%i|l.—

'

-

o " 0

FigureP6-137

= -

Q9

— =

4

4,

o __ 0

it

Challenging Problems 6-138* The band brake shown in Fig. P6-138 is part of a hoist. . . . . mg machine. The coeflicient of friction between the 500-1nmdiameter drum and the fiat belt is 0.20. The maximum actuating force P that can be applied to the brake arm is 490 N. Rotation of the drurn is clockwise. What miniminn-size shalt should be used to transmit the resisting torque developed by the brake to the machine if the shaft is to be made of 0.4% C hot-rolled steel? The factor of safety is 3 for failure by yielding. Circular steel bars are available with diameters in increments of 5 mm.

p

A

‘

,Z\/

l '0 T

C

\

t

.-.

§.r|\.l5i.r||.. _

fi—_ =' .i_

Figure P6-139

_.

344 CHAPTER 6 TORSIONAL LOADING or SHAFPS 6 140 A shaft used to transmit power is constructed by joining two solid segments of shaft with a collar, as shown in Fig. P6-140 The collar has an mside diameter equal to the diameter of the shaft and both the collar and the shaft are made of the same material The collar is securely bonded to the shaft segments Determine the ratio of the diameters of the collar and shaft such that the splice can transmit the same power as the shaft and at the same maximum shearing stress level. Is the solution dependent on the material selected‘?

dc 1' 1'17 “ ‘- » ‘ _ __ “'/- - ..'./ii I T _ -. T T “‘ i

d ,’ ‘I .'

Figure P6440

SUMMARY The problem of transmitting a torque (a couple) fi'om one plane to a parallel plane is frequently encountered in the design of machinery. The simplest device for accomplishing this fiinction is a circular shaft. The resisting torque is statically equivalent to the sum of the torques produced by the shear stresses T}=\/i

pdF=/i

BIBB

p1.'pdA

(6-1)

flffii-1

where p is the distance from the axis ofthe shaft to the element of area dA. The law of variation of the shearing st1'ess on the transverse plane (1: as a fimction of radial position p) must be known before the integral of Eq. 6-1 can be evaluated. If the assumption is made that a plane transverse cross section before twisting remains plane after twisting and a diameter of the section remains straight, the distortion of the shaft can be expressed as c9 ye = I

p9 and

yp = T

(6'2)

O1‘

1* a=ip C

we

The angle 8 is called the angle of twist. Equation 6-3 indicates that the shearing strain is zero at the center of the shaft and increases linearly with respect to the distance p from the axis of the shalt. This equation can be combined with Eq. 6-1 once the relationship between the shearing stress r and the shearing strain y is known. Since no assumptions have been made about the relationship between the stress and the strain or about the type ofmaterial of which the shaft is made, Eq. 6-3 is valid for elastic or inelastic action and for homogeneous or heterogeneous materials, provided the strains are not too large (tan y E 7/). If the assumption is made that Hooke’s law (r = Gy) applies (stresses must be below the proportional liriiit of the material), Eq. 6-3 can be written r

e=fp

we

When Eq. 6-4 is substituted into Eq. 6-1, the result is

Ti‘ "D

rp = 7

and

17, =

-:;*

(6-6)

SUMMARY where J is the polar second moment of the cross-sectional area of the shaft. Equation 6-6 indicates that the shearing stress rp, like the shearing strain y,-,, is zero at the center ofthe shaft and increases linearly with respect to the distance p from the axis of the shaft. Both the shearing strain y and the shearing stress 1: are maximum when p = c. Equation 6-6 is known as the elastic torison formula and is valid for both solid and hollow circular shafts. Frequently, the amount of twist in a shaft is important. Equations 6-2, 6-6, and Hooke’s law (tr = Gy) can be combined to give

L L e=yL=l p

or

pG

TL e=# GJ

(6-711,11)

The angle of twist determined from the above expressions is for a length of shaft of constant diameter (J = constant), constant material properties (G = constant), and carrying a torque T,. Ideally, the length of shaft should not include sections too near to (within about one-half shaft diameter of) places where mechanical devices (gears, pulleys, or couplings) are attached. Forpiactical purposes, however, it is customary to neglect distortions at connections and to compute angles as if there were no discontinuities. If T,, G, orJ is not constant along the length of the shaft, Eq. 6-7b takes the form n Tril-1 0=2 _ GM,

(6-7C)

where each term in the summation is for a length L where T,, G, and J are constant. If T,., G, 01.)’ is a fimction ofx (the distance along the length of the shaft), the angle of twist is found using

9=

LT,dx

fa i G1

= \t

‘t

\

Fig. 7-5 with the distortion greatly exaggerated. When Fig. 7-5 was drawn, the assumption was made that a plane section before bending remains a plane after bending. For this to be strictly true, it is necessary that the beam be bent only with couples (no shear on transverse planes). Also, the beam must be so proportioned that it will not buckle and the loads applied so that no twisting occurs (this last liniitation will be satisfied ifthe loads are applied in a plane of symmetry—a sufficient though not a necessary condition). When a beam is bent only with couples, the deformed shape ofall longitudinal elements (also referred to as fibers) is an arc ofa circle. Precise experimental measurements indicate that at some distance c above the bottom of the beam, longitudinal elements undergo no change in length. The curved surface formed by these elements (at radius p in Fig. '7-5) is referred to as the neutral surface of the beam, and the intersection of this surface with any cross section is called the neutral axis ofthe section. All elements (fibers) on one side of the neutral surface are compressed, and those on the opposite side are elongated. As shown in Fig. 7-5, the fibers above the neutral surface of the beam of Fig. 7-4-a (on the same side as the center of curvature) are compressed and the fibers below the neutral surface (on the side opposite the center of curvature) are elongated. The xy-axes of Fig. 7-4c lie in the plane of symmetry; the origin of the coordinate system lies on the neutral surface. Finally, the assumption is made that all longitudinal elements have the same initial length. This assumption imposes the restriction that the beam be initially straight and of constant cross section; however, inpractice, considerable deviation fi'om these last restrictions is ofien tolerated. The longitudinal strain 6, experienced by a longitudinal element that is located a distance y fi'om the neutral surface of the beam is determined by using the definition of normal strain as expressed by Eq. 3-1. Thus,

€_§_q-n "_r_

L,-

7-2 rrrrxnrrxr. s'r'rtu1ts 4 T

\

.

Lo)

\

@

21

I Top of beam

lfliinae

X

7.5 kip

7.5 kip

@\

1

1

\

7.0 in.

\

\

"\\ t-r

' 2n

211

2n

'

\

l

‘

\ \ O \

\

web

G \ \

\ it >-r

1

\

\

fiom beam, Discenttanceroin.f

\

\

\ 0 \

2l 1

\ \

\ \ 0

‘

\

3 l

t

\

p

\\

~.

\|

T" Bottom of beam

4 comp. rs

10

5

0 5 Strain, iniin. (10)-4

10

Figure 7-6 where L; is the final length of the fiber after the beam is loaded and L,- is the initial length of the fiber before the beam is loaded. From the geometry of the beam

segment shown in Fig. 7-5,

_ Ax‘ — Ax _ (p — J/)(A9) — pt/18) _ 1 éx —

AI

—

W39)

— “fly

(7-2)

Equation 7-2 indicates that the strain developed in a fiber is directly proportional to the distance of the fiber from the neutral surface of the beam (recall that y is measured fi"om the neutral surface). This variation can be demonstrated experimentally by means of strain gages attached to an I-bearn, as shown in Fig. 7-1. The strains, as measured by gages on two different sections, are plotted against the vertical position of the gages on the beam in Fig. 7-6. Curve I represents strains on a section at the center of the beam where pure bending occurs (no transverse shear), and curve 2 shows strains at a section near one end of the beam where both flexural (nornral) stresses and transverse shearing stresses exist. These curves are both straight lines within the limits of the accuracy of the measuring equipment.‘ IA more exact analysis using principles developed in the theory of elasticity indicates that curve 2 should be curved slightly. Note: Other experiments indicate that a plane section of an initially curved beam will also remain plane after bending and that deformations will still be proportional to the distance ofthe fiber from the neutral surface. The strain, however, will not be proportional to this distance, since each deformation must be divided by a different original length.

15

Ten.

flange

553

354 crntrrrrt 1 rtsxtrrmr. r.ornrsc= srrtsssss IN starts Note that Eq. 7-2 is valid for elastic or inelastic action so long as the beam does not twist or buckle and the transverse shearing stresses are small. Problems in this book will be assumed to satisfy these restrictions.

7-5 FLEXURAL STRESSES With the acceptance of the premise that the longitudinal strain ex is proportional to the distance ofthe fiber from the neutral surface ofthe beam, the law ofvariation of the nonnal stress 0, on the transverse plane can be determined by using a tensilecompressive stress-strain diagram for the material used in fabricating the beam. For many real materials, the tension and compression stress-strain diagrams are identical in the linearly elastic range. Although the diagrams may differ somewhat in the inelastic range, the differences can be neglected for most real problems. For beam problems in this book, the compressive stress-strain diagram will be assumed to be identical to the tensile diagram unless otherwise noted. For the special case of linearly elastic action, the relationship between stress tr, and strain 5, is given by Hooke ‘s law, Eq. 4-10 (since the state of stress is uniaxial) as

ax : Esx

('1)

Substituting Eq. 7-2 into Eq. (a) yields E Us : E51‘ : _';y

J’

Fv. @£— x

-2, l R

I

fa

(7'3)

Equation 7-3 shows that the normal stress 0, on the transverse cross section of the beam varies linearly with distance y fi'om the neutral surface. Also, since plane cross sections remain plane, the normal stress 0,, is uniformly distributed in the z-direction (see Fig. 7-4c). With the law of variation of flexural stress known, Fig. 7-4 can now be redrawn as shown in Fig. 7-7. The forces FL: and Fr are the resultants of the compressive and tensile flextnal stresses, respectively. Since the sum of the forces in the x-direction must be zero, FC is equal to FT; hence, they forrrr a couple of magnitude M,. The resisting moment M, developed by the normal stresses in a typical beam with loading in a plane of symmetry but of arbitrary cross section, such as the one shown in Fig. 7-8, is given by Eq. 7-l as

Figure 7-7 ll-f,=—LydF=—j;y01d/1

(bl

Since y is measured from the neutral surface, it is first necessary to locate this surface by means of the equilibritmr equation SF, = 0, which gives

2fl=LH=LqM=O

@

7-3 rtsxttrutsrrtsssss 355 Jiz

J’

M —i, Neutral surface

-

dy

x

Z

N

|i"-1

ea

Ai

_|_

_

_

Figure 7-8

Substituting Eq. 7-3 into Eq. (c) yields

/W-/ta)“ A

A P Ef

E

P Ay dA=——pyc A=0

=———

J

(7-4

whereyg is the distance from the neutral axis to the centroidal axis c—c ofthe cross section that is perpendicular to the plane of bending. Since neither (E/p) norA is zero, yc must equal zero. Thus,forflexural loading and linearly elastic action, the neutral axis passes through the centroid of the cross section. Since the nornral stress 0, varies linearly with distance y from the neutral stnface, the maximum normal stress crmx on the cross section can be written as E am, = — —c

(7-5)

P

where c is the distance to the surface of the beam (top or bottom) farthest from

the neutral surface. Ifthe quantity (E/p) is eliminated between Eqs. 7-3 and 7-5, a useful relationship between the maximum stress am, on a transverse cross section and the stress crx at an arbitrary distance y from the neutral surface is obtained. Thus,

@=3%m=3Q c

we

c

Substitution of Eq. 7-6 into Eq. (b) yields

M=-fyQm=-fiffim A

6‘

A

an

in which am, = 0,, (equals 0,, evaluated at y = c). The integral fyz ¢iA is called the second moment ofarea. Second moments of area of several common shapes are given in Table A-1, Appendix A. Second moments ofmore complex areas can usually be derived (without integration) from combinations ofthese simple shapes, as shown in the next section. A discussion ofsecond moments of area is presented in Appendix A.

356 crntrrsrt 1 rtsxtrrmr. rmnrsc srrtsssss IN starts 7-4 THE ELASTIC FLEXURE FORMULA When the integral [A yz dA in Eq. 7-7 is replaced by the symbol I, the elasticflexure fimnula is obtained as M at = -77’

< 2-in. flange plates to a 24 x 1-in. web plate. The beam is loaded in the plane of symmetry parallel to the web. On a section where the resisting moment M. = 1000 kip - ft, determine the maximum flexural stress. 7-12* Determine the maximum tensile and compressive flexural stresses on a section where the resisting moment M = -3 l< 3f4-in. plates to the flanges of the beam, as shown in Fig. P7-9. The maximum flexural stress in both the original and modified beams must be limited to 18 ksi. Determine

r-—100mm—*| 25 mm

a. The maximum resisting moment that the original beam can support. b. The maximum resisting moment that the modified beam can

200mm 25mm

support.

25 mm

iii

ii200nmiFigure P7-12

Figure P7-9

7-13 Determine the maximum tensile and compressive flexural stresses on a section where the resisting moment M = —20,000 lb - in. if the beam has the cross section shown in Fig. P7-13. The beam is loaded in the vertical plane of symmetry.

564 cnarrstt 7 FLEXURAL 1.oxn11ve= sntassrs [N saws 3 in.

produces a flexural strain of + 1200 ,u.m/m at a point on the top surface of the beam. Determine a. The maximum flexural stress at the point. b. The resisting moment M developed in the beam on a transverse cross section through the point.

Figure P7-13 7-14 Determine the maximum resisting moment M that can be supported by a beam having the cross section shown in Fig. P7-14 if the maximum flexural stress must be limited to 110 MPa.

7-17 A steel (E = 29,000 ksi) bar with a rectangular cross section is bent over a rigid mandrel (R = 12 in.) as shown in Fig. P7-17. If the maximum flexural stress in the bar is not to exceed the yield strength (0,, = 36 ksi) of the steel, determine the maximum allowable thickness h for the bar.

rh

T

TS Him

MU

I

so

ii!-20mm-ti

R

QM

_

\

Figure P7-17

25?“

i@—25rrun—i'—25 mm—*i

7-18* An aluminum alloy (E = 73 GPa) bar with a rectangular cross section is bent over a rigid mandrel as shown in Fig. P7-17. The thiclcness h of the bar is 25 rmn. If the maximum flexural stress in the bar must be limited to 100 MPa, determine the minimum allowable radius R for the mandrel.

Figure P7-14

Challenging Problems 7-15* A beam has the cross section shown in Fig. P7-15. On a section where the resisting moment is -30 kip - ft, determine a. The maximum tensile flexural stress. b. The maximum compressive flexural stress.

7-19 The cantilever beam shown in Fig. P7-19 is subjected to a moment M = 15,000 lb - in at its free end. The beam has a 2 x 2-in. square cross section. Determine the maximum tensile flexural stress in the beam.

Ti 1111.

M = 15,000 lb-in.

4in. 15 in.

tin. -isin.i=|—;

Ii

Figure P7-19

Figure P7-15

7-16* A hardened steel (E = 210 GPa) bar with a 50-mm square cross section is subjected to a flexural form of loading that

7-20 Determine the maximum flexural stress on a section where the resisting moment M = +100 kN - m if the beam has the cross section shown in Fig. P7-20. The beam is loaded in the vertical plane of symmetry.

7-4 THE ELASTIC rmxutts rottiuuut

365

Computer Problems

rm" 250 mm 25 mm |v

7-23 A beam with a hollow circular cross section (see Fig. P7-23) is being designed to support a maximum moment of 100 kip in. If the wall thickness is fixed (r = r, ~ r,- = 0.25 in.),

25mm

a. Calculate and plot the maximum flexural stress in the beam as a function of the outside diameter do (2.5 ir1. 5 do 5 6 in.). b. What is the smallest outside diameter that can be used if the maximum flexural stress in the beam must not exceed 40 ksi?

150mm mm

100mm

Figure P7-20 7-21* A steel pipe with an outside diameter of 4 in. and an inside diameter of 3 in. is simply supported at the ends and carries two concentrated loads, as shown in Fig. P7-21. On section A-/t, which is 5 ft from the right support, determine

Figure P7-23

a. The flexural stress at point A on the cross section. b. The flexural stress at point B on the cross section.

1000 lb

7-24 A beam with a solid rectangular cross section (see Fig. P7-24) is being designed to support amaximum moment of 6 kN - m. Ifthe height of the beam is to be twice the width of the beam (h = 217),

2000 lb

c """"""" “

i

ll

ll ll

ll

ll

.|_L

.0

H

@>-

% Section A—A

:x_:-_:_:x

a. Calculate and plot the maximum flexural stress in the beam as a function of the height (50 mm 5 h 5 400 mm). b. What is the smallest height that canbe used ifthe maximum flexural stress in the beam must not exceed 20 ksi?

2 ii

Figure P7-21

7-22 The cantilever beam shown in Fig. P7-22a is subjected to a moment M at its free end. The cross section of the beam is shown in Fig. P7-22b. Ifthe allowable stresses are 90 MPa (T) and 140 MPa (C), determine the maximum moment that can be applied to the beam.

J1

Figure P7-24 |*120 rmni

2&1 120 mm

K 1.2 m

(a) Figure P7-22

i

—’ i*— 40 rrnn

(bl

7-25 An S 4 x 9.5 structural steel beam is carrying a moment M that produces a maximum flexural stress of 20 ksi. The moment M must be increased by 75 percent, but the maximum flexural stress must not be increased. In order to strengthen the beam, rectangular steel plates are to be attached to the top and bottom flanges as shown in Fig. P7-25. If the thickness t of the plates must not exceed 1/2 in., prepare a design curve that shows the acceptable values of plate width b as a fimction of plate thickness t(0 in. 5 t 5 1/2 in.).

366 crnrrrx 7 rtaxuiuu. LOADING: smsssss nv BEAMS

I

b

15

and plot the percent increase in load-carrying capability (MM) of the beam as a function of the thickness (t) of the flanges.

5

i-‘Z150 111111?‘ 30mm S4>< 25 structural steel T-section is loaded and supported as shown in Fig. P7-61. Determine the maximum tensile and compressive flexural stresses in the beam.

Challenging Problems 7-64* Draw complete shear and bending moment diagrams for the beam ABCD shown in Fig. P7-64.

6000 lb

100016111 A

B

| I~

6fl

l

BRN/m

an I

T

6ft

__ Rlgldbar

1.5m

‘I

3

Figure P7-61

,,

is C

Ri ‘db

gl W

I .

D

— ui-—-I*~—-1m—-- -—1m-—--"

7-62* A C254 x 30 structural steel channel is loaded and supported as shown in Fig. P7-62. Determine the maximum tensile and compressive flexural stresses in the beam.

lrn

5 kN

Figure P7'64

Jr‘

3 "N 4 kN-m

D x

A

pendix B) is loaded and supported as shown in Fig. P7-65. The

I

I

lm

'

B

lm

1-65* An s 15 >< 50 American standard steel beam (see Ap-

C

I

l

total length ofthe beam is 1s 11. Ifthe allowable flexural stress

I

is 15,000 psi, determine the maximum permissible value for

the distributed load w.

Figure P7-62

—T 7-63 An 8 x 8-in. nominal size structural timber (see Appendix B) is supported by two brick colurrms, as shown in Fig. P7-63. Asstune that the brick colunms transmit only vertical forces to the timber beam. The beam supports the roof of a building through three timber columns. Columns A and C transmit forces of 1.8 kip to the beam; column B transmits a force of 23 kip_ a. Draw complete shear and bending moment diagrams for the beam. b. Determine the maximum tensile and compressive flexural stresses in the beam

A

B

W lb/3 A

'1'

C $JFB

7

Lfi Figure P7-65

J

21‘ fi

7_65 Draw complete shear and bending moment diagmms for segments AB and CD of the structure shown in Fig. P7-66.

C

3 ‘(N

A

B

1.5 kN.~’n1

1.5 11] Figure P7-63

Figure P7-66

0.75 m

I

1.5 m

C

390 cmwrsn 1 ELEXURAL LOADING: STRESSES [N muss 7-67* MemberAB supports a 55-lb sign, as shown in Fig. P7-67. Determine a. The maximum tensile flexural stress in the 1/2-in.-nominaldiameter standard steel pipe AB (see Appendix B). b. The normal stress in the 3/16-in.-diameter wire BC. c. The shearing stress in the 1/4-in.-diameter pin at A, which is in double shear.

7-69 A tractor is moving slowly over a bridge, as shown in Fig. P7-69. The forces exerted on one beam of the bridge by the tractor are 4050 lb by the rear wheels and 1010 lb by the front wheels. Determine the position x of the tractor for which the bending moment in the beam is maximum.

C 4050 lb

35 in.

V

16in.

..

‘

32111.

-1

..

I

;.I

.;-

16111.

.

A

I

~q B

first

e tr1;

1010 lb

'7 I

1

40 it

Figure P7-69

7-70* A body with a mass of 1500 kg is supported by a roller on an I-beam, as shown in Fig. P7-70. The roller moves slowly along the beam, thereby causing the shear force K and the bending moment M, to be functions ofx. a. Draw complete shear and bending moment diagrams when the roller is at position x. b. Determine the position of the roller when the bending moment is maximum.

Figure P7-67

7-68 An S457 x 81 American standard steel beam (see Appendix B) is loaded and supported as shown in Fig. P7~68. The segments of the beam are connected with a smooth pin at D. a. Draw complete shear and bending moment diagrams for the beam. b. Determine the maximum tensile and compressive flexural stresses in the beam.

In

10m

A

B

1500 kg so kN/m

1.5m

/I Sm°°‘h pi“

40l(.N/tl'l

1! T

Figure P7-68

2m

1'

C1 1 B .;,:)t._ 1.5m (E 1.5m T

T

T

40 k.N1‘m E‘

Figure P7-70

7-71 Two beams AD and EH are spliced, as shown in Fig. P7-71. Draw complete shear and bending moment diagrams for beam CF.

1-1 SHEARING SIRESSES [N BEAMS

10 kip B A 'r= iii D C 11— L

10ft

>5

5fi—-ii

rs 7.511

c -1| _ P V 5ft

i

=1 H —* lOfi

Figur'eP7-71

7-7 SHEARING STRESSES IN BEAMS The discussion of shearing stresses in beams was delayed while flex1n'al stresses were studied in Section 7-3. This procedure seems to be in keeping with the historical record on the study of beam stresses. From the time of Coulomb’s paper, which contained the correct theory of the distribution of flexural stresses, approximately seventy years elapsed before the Russian engineer D. J. Jourawski (1821-1891), while designing timber railroad bridges in 1844-1850, developed the elementary shear stress theory used today. In 1856, Saint-Venant developed a rigorous solution for shearing stresses in beams; however, the elementary solution of Jourawski is the one in general use today by engineers and architects because it yields adequate results and is much easier to apply. The method requires use of the elastic flexure formula in its development; therefore, the formula developed is limited to elastic action. The shearing stress evaluation discussed ir1 this section is used as follows: 1. For timber beams, because of their longitudinal plane of low shear resistance. 2. In design codes for shear stresses in the webs of I-beams. 3. For the evaluation of principal stresses at the junction between the flange and the web in certain wide-flange beams. If one constructs a beam by stacking flat slabs one on top of another without fastening them together, and then loads this beam in a direction normal to the surface of the slabs, the resulting deformation will appear somewhat like that in Fig. 7-19a. This sanre type of deformation can be observed by ta.king a pack of cards and bending them, and noting the relative motion of the ends of the cards with respect to each other. The fact that a solid beam does not exhibit this relative movement of longitudinal elements (see Fig. 7-19b, in which the beam is identical to that of Fig. 7-19a except that the layers are glued together) indicates the presence of shearing stresses on longitudinal planes. Evaluation of these shearing stresses will be determined by means of equilibrium and the fi"ee-body diagram of the short portion of a beam with a rectangular cross section shown in Fig. 7-20a. The normal force dF acting on a differential area dA = t dy on a cross section of the beam is equal to a,dA. The resultant of these differential forces is F = faxdzl integrated over the area of the cross section, where 0,, is the flexural stress at a distance y from the neutral surface and is given by the expression 0, = —M,y/I. Therefore, the resultant normal force F1 on the left end of the segment

391

392

ctr.-urrss 7 FLEXIJRAL LOADLYG: sntsssrs mt BEAMS

Figure 7-19 from y] to the top of the beam is“ M

M

C

F1=——fydA=——f y(rdy) I

dy

Mcafella Al

A

I

V1

Similarly, the resultant force F1 on the right side of the element is

F2 = -L +IAM)fCy(rdy)

.l’i.l’ C

V—AV

Ax

.

V,

-

(H)

VR

I

1

F,‘_I+:}l_’i-, i

lj,=rrAx

(5) Figure 7-20

These forces are shown on the free-body diagram of Fig. 7—20b. Also shown on the free-body diagram of Fig. 7—20b are the resultants of the vertical shear stresses on the left VL and right VR sides ofthe element and the resultant ofthe horizontal shear stress VH on the bottom of the element. A summation of forces in the horizontal direction yields AM

‘

V5=F1—F1=—Tf yfldy) HV

The average shearing stress rm.-g is the horizontal shear force V” divided by the horizontal shear area A, = I Ax between sections A and B. Thus, V Tavg:

AM C :_fij;|y(tdy)

‘if M is positive, then the normal stress above the neutral axis (where y is positive) will be negative (eotnpression) and the force 1"; will be a compressive forcer lf.-14 is negative, then the normal stress above the neutral axis will be positive (tension) and the force I-'1 will be a tensile force as drawn.

1-1 srrrtuuuc srrusssss [N BEAMS

393

l.nthelimitasAx—>O

_

AM

1

C

a‘M

1

'°

Y = £E»1oE(_i7r)_£, ’>"’>’-niwilf, ’Y"Y

(“J

The shear K at the beam section where the stress is to be evaluated is given by Eq. 7-1 1c as K = dM/dx. The integral of Eq. (a) is the first moment of the portion of the cross-sectional area between the transverse litre where the stress is to be evaluated and the extreme fiber of the beam. This integral is designated Q, and when values of V} and Q are substituted into Eq. (a), the formula for the horizontal (or longitudinal) shearing stress becomes

m=—%§

cm

The minus sign in Eq. (b) is needed to satisfy Eq. 7-la and is consistent with the sign convention for shearing stresses (Fig. 7-4c). At each point in the beam, the horizontal (longitudinal) and vertical (transverse) shearing stresses have

0 6»

the same magnitude (1.3, = 13,); hence, Eq. (b) also gives the vertical shearing stress at a point in a beam (averaged across the width).5 For the balance of this chapter, magnitudes of V, and Q will be used to determine the magnitude of the shearing stress ‘r and Eq. (b) will be written as

KQ

1.’ = T

(7-12)

The sense of the stress 1: will be determined fi'om the sense of the shear V, on transverse planes and fi"om rxy = 'r_,,, on longitudinal planes. Because the flexure formula was used in the derivation of Eq. 7-12, it is subject to the same assumptions and limitations as the flexure formula. Although the stress given by Eq. 7-12 is associated with a particular point in a beam, it is averaged across the thickness t and hence is accurate only if t is not too great. The variation of shearing stress on a transverse cross section of a beam will be demonstrated by using the rectangular cross section shown in Fig. 7-21a. The transverse shearing stress at any point of the section at a distance yl from the neutral axis is from Fig. 7-21b and Eq. 7-12,

Y

KQ

Vf

It

n,,y

Eii

Neutral axis

rial

(H)

l—'—‘l dyg y

C

Yr

(bl

Vf‘ 1:,/‘y V '1/2 V =—IL yc=— 21

rt’

—-vi l

@

L l'!1.E.X

Equation (c) indicates that the transverse shearing stress on a rectangular cross section has a parabolic distribution, as shown in Fig. 7-21c. The shearing stress acts in the direction of the shear force V that produces the stress. The maximum 51f the shear force of V is positive (downward on section B), then the horizontal shear stress will be negative (to the right on the bottom of the element) and the vertical shear stress will also be negative (downward on section B—in the same direction as the shear force).

(C) Figure 7-21

394 CHAPTER 7 rrnxrnuu. LOADING: snrsssss [N snuirs shearing stress occurs when yl = O (at the neutral axis) and has a magnitude

Vhz

Vhz

3V

3V

81

8(lh /12)

2th

2A

¢m,=_=m=__=__ L

T1

The maximum shearing stress given by Eq. 7-13 (Fig. 7-22) exists on both the transverse plane and the longitudinalplane along the neutral surface. This equation is useful in the design oftimber beams with rectangular cross sections since timber

i

1;

1:»'=T.v=r=

_ ML»

(7-13)

ls:

has a low shearing strength parallel to the grain.

Figure 1-22

7.12

Equation 7-13 is valid for a rectangular section and should not be used for other sections. For a rectangular section the maximum shearing stress is 1.5 times the average shearing stress (ravg = V/A). For a rectangular section having a depth twice the width, the maximum stress as computed by Saint-Venant’s more rigorous method is about 3 percent greater than that given by Eq. 7-13. If the beam is square, the error is about 12 percent. Ifthe width is four times the depth, the error is almost 100 percent, from which one must conclude that, ifEq. 7-12 were applied to a point in the flange of an I-beam or T-section, the result would be worthless. Furthermore, if Eq. 7-12 is applied to sections where the sides of the beam are not parallel, such as a triangular section, the average transverse shearing stress is subject to additional error because the variation of transverse shearing stress is greater when the sides are not parallel. A second illustration ofthe variation of shearing stress on a transverse section of a beam will be demonstrated by using the inverted T-shaped beam shown in Fig. 7-23a. For this section,

yc : 2(10)(7) + 10(2)(1) Z 4m 2(l0) + 10(2) c1 = 8 in. and C2 = 4 in.

Note that in Eq. 7-12, V and I a.re constant for any section, and only Q and I vary for different points in the section. The transverse shearing stress at any point in the stem of the section a distance yl from the neutral axis is from Fig. 7-23a

2 in.

L C‘

1

.

y

10 Ill. Y1

_ V Neutral axis

C2

yc ii 10 in. ii

to Figure 7-23

___ 2 in

V :_.

(B)

§_§,E'§u

1-1 srrauunc srrutssss [N cams

395

and Eq. 7-12, r

V,Q

V!“

rt

Ir,,y'y

=?—-=-—-

V

d

V

= ,—,

An expression for the average shearing stress in the flange can be written in a similar manner and is

V V I = 2—I(¢% —yi) = Z-,(42 -14?)

. (-4 < yi < -2 H1-J

These are parabolic equations for the theoretical stress distribution, and the results are shown in Fig. 7-23b. The diagram has a discontinuity at thejunction ofthe flange and stem because the thickness of the section changes abruptly. The distribution in the flange is fictitious because the stress at the top of the flange must be zero (a free surface). From Fig. 7-23b and Eq. 7- 12, one may conclude that, in general, the maximurnfi longitudinal and transverse shearing stress occrus at the neutral surface at a section where the transverse shear V, is maximum. There may be exceptions such as a beam with a cross section in the form of a Greek cross with

‘St 7.13

is ‘Pm “T

Q/t at the neutral surface less than the value some distance from the neutral surface. Another example of importance is the determination of the shearing stress in an I-beam. Consider the W203 x 22 section [I = 20.0(10") mm“] shown in Fig. 7-24a, and let the shear Ff, on the section be 37.5 kN. Equation 7-12 may be used to calculate the shearing stress at various distances yi fi'om the neutral axis of the beam. For example, at the neutral axis (yl = O) QM = 102(8)(99) + 95(6.2)(47.5) = 108.76(103) rnm3 = l08.76(10_°)m3

TNA = V'Q”” 1:, = 37'5(103)(l08'76)(l0_6) 20.0(10-‘=)(s.2)(10-3) = 32.89(10")N/m2 2 32.9 MPa

206 mm 6.2 mm

_.l _i)l8n'nn 102 mm W 203 X 22 section

(a)

,1 = 1.485 MPa

1 = 24.4 MPa

Similarly, at the junction between the web and the flange with 1‘ = 6.2 mm (in the web), rm“ = 32.9 MPa

Q; = 102(8)(99) = 80.78(103) mm3 = 80.78(10_") m3

1 = I/‘Q’ = 37'5(w3)(80'78)(l0‘6) = 24 43(1O6)N/m2 2 24 4MPa "' 11,. 20.0(1o-@)(6.2)(10-3) ' ' The shearing stress distribution for the complete cross section ofthe beam is shown in Fig. 7-24b. Equation 7-12 gives r = 1.485 MPa at the junction of the web and flange with t = 102 mm (in the flange). However, this result is incorrect because the bottom surface of the top flange (or the top surface of the bottom flange) is a free surface and thus 1.’ = 0. A similar result was obtained for the inverted T-section, shown in Fig. 7-23. More advanced methods of the theory of elasticity must be used to derive a correct solution. “In this book the term maximum, as applied to a longitudinal and transverse shearing stress, will mean the average stress across the thickness I at a point where such average has the maximum value.

Ll, ll’

‘tug = 1-;:b= 31.8 MP3

Shearing stress distribution

(bl Figure 7-24

396 cinirrrir 7 rrnxrnuu. LOADING: snrsssss [N nnuns J’

Neutral axis

The variation of shearing stress over the depth of the web is small, and the shearing stresses in the flanges are small compared to those in the web. As a result, the majority of the shear force V, is carried by the web. In the design of I-beams, the maximum shearing stress calculated by using Eq. 7-12 is approximated by dividing the shear force V, by the area of the web, that is,

V1 rm = T

(7-14)

web Z

‘

For the example being considered,

( (=1)

10 his

aw’ /‘Tate " _

__N

= 2.13 pCI'CCI1t

(a) The average shearing stress on a horizontal plane 4 in. above the bottom of the beam and 6 ft from the lefi support. (b) The maximum transverse shearing stress in the beam. (c) The average shearing stress in the joint between the flange and the stem at a section 6 ft from the left support. (d) The force transmitted from the flange to the stem by the glue in a 12-in. length of the joint centered 6 fi fi'om the lefi support. (e) The maximum tensile flexural stress in the beam.

_‘__A___ \1.~;4=%l1'14

8 in.

SOLUTION The second moment of the cross-sectional area about the neutral axis is

10 in.

IN, = ammo)’ + 2(10)(3)2 + é(10)(2)3 +10(2)(3)= = 533.3 in.“ -42 in. (5) Figure 7-27(a-b)

(a) The shear force V, on a cross section 6 ft fi'om the left support is +900 lb, as shown on Fig. 7-27c. The first moment Q4 for the bottom 4 in. of the stem (see Fig. 7-27e) is

Q4 = yc4A4 = 5(2)(4) = 43iI1-3

1-1 srmuusc srrtsssras [N arms $99 The average shearing stress on a horizontal plane 4 in. above the bottom of the beam is then given by Eq. 7- 12 as

1:4 = E = = 40.50 psi E 40.5 psi I.\,-At.‘ 533.3(2)

+ V,=+900lb V, 0 V,=-900 lb

Ans.

(b) The maxi.n1um transverse shearing stress in the beam will occur at the neu-

tral axis on the cross section supporting the largest shear force V}. The first

(C) +

M, 0

moment QM for the pa.rt of the stem below the neutral axis is

Mm = srso 111-11 (1

ma,‘

>1 L. 21.11.

L5. -1 lin.

Figure P7-95

(K1)

(11)

7-96 The simple beam shown in Fig. P7-96a has the cross section shown in Fig. P7-96b. Ifthe allowable stresses are 75 MPa shear and 120 MPa tension at point A (just above the flange), determine the maximum allowable load P.

250 mm

P

ml-'nml*—80nu'rt—i' 0.6 rn

'

_ 0

0

2m

(*1)

120 nun L

'

i

5 mtn '44

; 10mm

1

T

1

350mm

It-I I

B 7 1 rn

(.*—.~

-_._____

57"“.

,1

I \\

I \ H ?'~\ I \

‘I

'€—\

( 3.

._A=g r

\»s=1 fStacrctoencossentraK tion E“0 ,

fa Slconronccssntral E‘-1 Q \

,-' Probable shape of curve for \ ‘I /’ )|< 200 mm deep has 25-mmdiameter hole drilled from top to bottom of the beam on the centerline of a cross section. Determine the percent reduction in strength produced by the presence of the hole.

P

_

10 hits

12 ih.—~i

Figure P7-121

Challenging Problems 7-122* A 0.4% C hot-rolled steel (see Appendix B) bar with a rectangular cross section will be loaded as a cantilever beam. The bar has a depth h of 200 mm and has a 25-mm-diameter hole drilled from top to bottom of the beam on the centerline of a cross section where a bending moment of 50 kN - m must be supported. If a factor of safety of 4 with respect to failure by yielding is specified, determine the minimum acceptable width b for the bar.

422

CHAPTER 7 FLEXURAL LOADING: sntsssss [N BEAMS Computer Problems

7-123 A load of 5000 lb is supported by the beams shown in Fig. P7-123. The beams are 2 in. wide >< 4 in. deep. The holes for the threaded rods have a 5E8-in. diameter. If the maximum tensile flexural stresses in the beams must not exceed 20 ksi, determine the maximum permissible span for the top beam.

I

Li

E.

us

Ins 1./cl

Lo

.‘-1

7-124 A steel bar with a rectangular cross section will be loaded as a cantilever beam. The bar has a 10-mm-diameter hole drilled from top to bottom of the beam on the centerline ofa cross section where a bending moment of 10 kN - m must be supported without exceeding a stress of 200 MPa. Prepare a curve showing the acceptable combinations of beam width b and beam depth h. Limit the minimum depth of the beam to 30 mm so that the ratio of the beam depth h to the hole diameter d is greater than 3.

.17.

‘T

tr-'1—

.

z

7-125 Two beams with rectangular cross sections, similar to the ones shown in Fig. P7-123, support a concentrated load P of 4.5 kip. The top beam has a span of 42 in. If the maximum tensile flexural stress in the top beam must not exceed 15 ksi, prepare a curve showing the acceptable combinations of beam width b and beam depth I1 ifthe hole diameters are 314 in. Limit the minimum depth of the beam to 2.5 in. so that the ratio of the beam depth h to the hole diameter d is greater than 3.

:

P

Figure P7-123

7-1 1 INELASTIC BEHAVIOR OF FLEXURAL MEMBERS A large proportion of structural and machine designs are based on elastic analysis,

for which the flexure formula is applicable. However, for some designs, particularly when the weight of the structure is important (aircraft design, for example), the limitation requiring stresses to remain below the proportional limit of the material results in uneconomical or ineflicient designs. Therefore, this limitation is sometimes discarded and higher stress levels are tolerated in the design. This section is concemed with the analysis of stresses in instances in which the proportional limit of the material is exceeded or the material does not exhibit a linear stress-strain relationship. In both of these cases, Hooke’s law does not apply. The basic approach to problems involving inelastic action is the same as that outlined in Sections 7-2 and 7-3 for linearly elastic action. A plane section is assumed to remain plane; therefore, a linear distribution of strain exists. If a stressstrain diagram is available for the material, it can be used with the strain distribution to obtain a stress distribution forthe beam. The stress and strain distributions shown in Fig. 7-36a and b are for a beam of symmetrical cross section made of a material

6 = Mo’! G

mu Iii7 0

y

6°

C

N.S. l

I

1

/

I

/

I

I

I

E

\Q

E

I

‘—“‘.1—‘ BE

(0) Figure 7-36

7

OT

(-5)

(C)

we

7-11 INELASTIC BEHAVIOR or FLEXl]RAL MEMBERS

Jr’

J’

J’

J“

Neutral surface

(11) Figure 7-37

(b)

(C)

having the stress-strain diagram shown in Fig. 7-36c. If an equation can be written for the stress distribution, the resisting moment can be obtained by using Eq. 7-1. If it is not practical to obtain an equation for the stress distribution, the beam depth can be divided into layers of finite thickness and the stress associated with each layer determined. The method of Section 7-3 can then be used to evaluate the resisting moment. As a simplifying technique, the stress-strain diagram can be approximated by a series ofstraight lines resulting in a simplified stress distribution diagram and a reduction in the amount of work involved. If the material is elastoplastic (see Section 5-7), the large strains associated with yielding will permit plastic action to take place as indicated in Fig. 7-37. The stress distribution for fiilly elastic action is shown in Fig. 7-37a, for partially plastic action in Fig. 7-3 7b, and foressentially complete plastic action in Fig. 7-370. There must always be a slight amount of elastic action if the strain at the neutral surface is zero. The plastic action shown in Fig. 7-37¢: is idealized in Fig. 7-3 7d; this is the assumed distribution used for plastic analysis of structural steel beams. In members subjected to inelastic action, the neutral axis does not necessarily coincide with a centroidal axis as it does when the action is elastic. If the section is symmetrical and the stress-strain diagrams for tension and compression are identical, the neutral axis for plastic action coincides with the centroidal axis. However, if the section is unsymmetrical (for example, a T-section) or if the stressstrain relations for tension and compression differ appreciably (for example, cast iron), the neutral axis shifts away from the fibers that first experience inelastic action, and it is necessary to locate the axis before the resisting moment can be evaluated. The neutral axis is located by using the equation

):F,=f 0,4/i=0

(11)

3183

and solving for the location ofthe axis where 0, is zero. II1 Eq. (a), x is perpendicular to the cross section of the beam. The following Example Problems illustrate the concepts of inelastic analysis as applied to the pure bending of beams that are loaded in a plane of summetry.

EXHIDPIB P1‘Ol]lCIIl 7-15 A beam having the T cross section of Fig. 7-38 is made of elastoplastic steel (E = 200 GPa) with a proportional limit (equal to the yield point) of 240 MPa. Determine

__

423

424 CHAPTER 7 rtsxmuu. LOADING: smsssrs [N sntnts (a) The bending moment (applied in the vertical plane of symmetry) that will produce a longitudinal strain of —0.00l2 mfm at point B on the lower face of

f—l00mm*' 1 25rn.m

the flange.

B

(b) The bending moment required to produce completely plastic action in the beam. 200 mm

SOLUTION (a) The strain distribution, stress-strain, and stress distribution diagrams are shown in Figs. 7-39:1, b, and c, respectively. Because of the inelastic action and tmsyrnmetrical section, the location ofthe neutral axis must first be determined by using Eq. (a), the stress distribution diagram shown in Fig. 7-390,

25 mm

Figure 7-38

o,MPa

J’

y

25......1"

_::lzs mm I

a

NS. ' ___

240 — —

-e

Em

+240

240

‘m

—0.00l2

H -

e Lm G y

--- Q = zoo ooo E

_Q

0, MPa 2:0 y

II IIII 1,”,i

200-2a 0.0012

(H)

s, 5 m

L______

(b)

(< l6 mm thick, and has a web that is 14 mm thick. Determine r

50mm

T

a. The width b of the plates if the shear center of the section must be located at the center of the web. b. The maximum vertical shearing stress at the section when the vertical shear at the section is 40 kN.

25mm > '

r—2_§mm

l

25mm

0 it

”

l

25mm > 25 mm

.l

Figure P7-140

liiil 7-141 An eccentric H-section is made by welding two l >< 10-in. steel flanges to a 1 x 12-in. steel web, as shown in Fig. P7-141. The section is used as a 5-it cantilever beam that carries a concentrated load of 100 kip at the free end. Asstu-ne that all of the section is effective in resisting flexural stresses and that only the web resists vertical shearing stresses. a. Locate the shear center of the section with respect to the center of the web. b. Prepare a sketch showing the distribution of shearing stress throughout the cross section.

Figure P7-142

Intermediate Problems 7-143" All metal in the cross section shown in Fig. P7-143 has a thickness of lf4 in. The dimensions shown are centerline dimensions for the flanges and the web. a. Locate the shear center of the section with respect to the center of the web. b. Determine the shearing stress at point O when the vertical shear at the section is 1500 lb.

440 CHAPTER 7 PLEXURAL LOADING: sritsssas [N sntnrs Challenging Problems 7-146* A tl1in-walled cylindrical tube cut longitudinally to make a semicylinder is used as a cantilever beam. The load acts parallel to the cut section as shown in Fig. P7-146.

u11

Oc-

a. Locate the shear center of the section with respect to the center of the tube (dimension e). b. Determine the shearing stress at point A if the radius R of the section is 25 mm, the thickness t is 2.5 mm, and the load P is 440 N.

Q-ric12*

t 4 in. —-l

Figure P7-143 P 7-144* Locate the shear center for the cross section shown in Fig. P7-144 and determine the maximum shearing stress produced on the cross section by a vertical shear of 6 kN.

‘II

Shear " center

T

T2 __ _

A

2.

90mm

l 60mm

6n1rn

8

Figure P7-146

so rnrn l

90 mm

7-147* A thin-walled slotted tube is used as a cantilever beam. The beam is loaded as shown in Fig. P7-147. ii

90mm

-i

90mm

Figure P7-144 7-145 A thin-walled box section (see Fig. P7-145) is used as a cantilever beam. Locate the shear center of the section with respect to the center of the web.

a. Locate the shear center of the section with respect to the center of the tube (dimension e). b. Determine the shearing stress at point A if the radius R of the section is 2 in., the thickness I is 0.10 in., and the load P is 110 lb.

P dB _-

A/

I

A

L 1.

A

Small

l*u—-F Figure P7-145

B Figure P7-147

\ \.

\.A

\ “En

ii

or Small

7-13 I~‘l.l1XURAL srnassas uv BEAMS orrwo MA'l'liRIAl.S 441 7-148 A cantilever beam with the cross section shown in Fig. P7-148 is subjected to a vertical concentrated load of 4 kN at the free end. The section has a constant thickness of 4 mm.

7-149 Locate the shear center for the cross section shown in Fig. P7-149 and determine the maximum shearing stress produced on the cross section by a vertical shear force of 300 lb.

a. Locate the shear center of the section with respect to the center of the web. h. Prepare a sketch showing the distribution of shearing stress throughout the cross section. The dimensions shown in Fig. P7-148 are centerline dimensions for the flanges, web, and extensions. Assume that all of the section is effective in resisting flexural stresses but that the vertical shearing force resisted by the flanges is negligible. The vertical shearing force in the 20-mm extensions is not negligible.

%in \

2 in

.r '< s in. deep has a 3.5-in.-wide x 1/2-in.-deep aluminum alloy plate securely fastened to its bottom face, as shown in Fig. 7-48:1. The moduli of elasticity for the

timber and 8.ll.1II1l]111II1 alloy are 1250 ksi and 10,000 ksi, respectively. Determine the maximum flexural stress in each material if the applied moment is 75 kip-in.

lf—4in.—'1

|*—4in.—\

R

%

a = 5.293 in.

c—~—-—

-—»~ 7:: b = 3.20‘? in.

5"*2,51BlL15 in.*|

‘IIe.

li~w=s(s.s)=2sin.g.~|

(11)

(1')

Figure 1-4s SOLUTION The ratio of moduli for the aluminum and timber is

n_EB_E,,_l0,000_8 _E,,_E,_1250 ' The actual cross section (timber A and aluminum B) and the transformed timber cross section are shown in Figs. 7-4 8a and b, respectively. The neutral axis ofthe transformed section is located by the principle of moments as

yc

=

EM EA

=

2s(1/2)(1/4) + 4(s)(4.s) 2B(l/2) + 4(8)

=3.2U7' .

"1

above the bottom of the section. The second moment of area of the transformed timber cross section with respect to the neutral axis is

I = l13(2s)(1/2)’ + 2s(1/2)(2.9s7)’ + 113(4)(s)’ + 4(s)(1.293)2 = 34-6.9in.4 The maximum flexural stress in the timber is

Mc _ 7s(10 3 )(s.293) _1 1444 10,

Urmax r\:

T-fir

1 144 psi

‘ )1“

Ans.

P Ifthe aluminum alloy plate were replaced with a piece of wood 8 x 3.5 in. = 28 in. wide, the force and moment on the wooden flange would be the same as the force and moment on the aluminum plate even though the stresses at points on the wooden flange would be only one-eighth the stress at an equivalent point on the aluminum plate.

444 cnsmzrr 7 rrrxuruu. LOADING: srrrsssrs [N sruuus and for the aluminum is

11 12,,

3.201

cam,‘ = ;(E)a,mx = 5293 (8)(ll44.4) "5 5550 psi

Ans.

I PROBLEMS MecMovie Activities and Problems MM7.18 Introducing the transformed area method. Example; Try one. Determine bending stresses in a composite beam using the transformed area method. .\l[M7.19 Aluminum and brass composite beam. Example; Try one. Given allowable stresses for two materials, determine the largest allowable moment that can be applied to the beam cross section.

of the materials if the composite beam is simply supported and carries a concentrated load 30 kN in the center of the beam. 7-153 A 4-in-wide x 6-in-deep timber cantilever beam 6 it long is reinforced by bolting two 112 x 6-in. structural steel plates to the sides of the timber beam, as shown in Fig. P7-153. The moduli of elasticity for the timber and steel are 1600 ksi and 29,000 ksi, respectively. Determine the maximum tensile flexural stress in each ofthe materials whenacoupleM = -10 kip - ft is applied to the free end of the beam.

Introductory Problems 7-150* A timber beam 150 mm wide x 350 mm deep has a 150mm-wide x 15-mm-thick steel plate fastened securely to its top face. The moduli of elasticity for the timber and steel are l0 GPa and 200 GPa, respectively. Determine the maximum flexural stress in the timber when the maximum flexural stress in the steel is 75 MPa (T). 7-151* A timber beam 6 in. wide x 12 in. deep has a 6-in.-wide >< l/2-in.-thick steel plate fastened securely to its bottom face. The moduli of elasticity for the timber and steel are 1500 ksi and 30,000 ksi, respectively. Determine the maximum flexural stress in the steel when the maximum flexural stress in the timber is 1250 psi (C). 7-152 A composite beam 225 mm wide x 300 mm deep x 4 m long is made by bolting two 100-mm-wide x 300-mm-deep timber planks to the sides of a 25 x 300-mm structural aluminum plate, as shown in Fig. P7-152. The moduli of elasticity for the timber and aluminum are 8 GPa and 73 GPa, respectively. Determine the maximum tensile flexural stress in each W‘ k 25 mm

300 n1.rn

l 100 mml L100 mrrrl Figure P7-152

6 in.

§m._r| |r—4rn._-l i_§m. Figure P7-153

Intermediate Problems 7-154* A 50-mm-wide >< 80-mm-deep wood (E = 10 GPa) beam will be reinforced with 3-mm-thick structural aluminum (E = 70 GPa) plates on its top and bottom faces. A maximum bending moment of3 kN-m must he resistedby the composite beam. If the allowable flexural stresses are 15 MPa in the wood and 135 MPa in the aluminum, determine the minimum width required for the aluminum plates. 7-155 A cantilever beam 6 ti long carries a concentrated load of 4000 lb at the free end. The beam consists of a 4-in.-wide x 10-in.-deep timber section reinforced with 4-in.-wide x 3/4in.-thick steel plates on the top and bottom surfaces. The moduli ofelasticity for the wood and steel are 1600 ksi and 30,000 ksi, respectively. Determine the maximum flexural stresses in the wood and in the steel. 7-156 A 50-mm-wide x 125-mm-deep polymer (E = 1.40 GPa) beam will be reinforced with a 6-mm-thick brass (E = 100 GPa) plate on its bottom face. If the allowable flexural stresses

1-14 rraxrrmri srruzssrrs IN ruunrroxcrrn coscrurnz sums 445 are 6 MPa in the polymer and 60 MPa in the brass, determine the width ofbrass plate required to have the allowable stresses in the two materials occur simultaneously.

are 12 GPa and 200 GPa, respectively. If the allowable flexural stresses are 10 MPa in the wood and 75 MPa in the steel, determine the maximum load P that can be applied at the center of a simply supported beam having a span of 4 m.

Challenging Problems 7-157* A timber beam 8 in. wide x 15 in. deep has an 8-in.wide x 1/2-in-thick steel plate securely fastened to its bottom face. The beam will be simply supported, will have a span of 16 ft, and will carry a uniformly distributed load over its entire length. The moduli of elasticity for the wood and steel are 1600 ksi and 30,000 ksi, respectively. If the allowable flexural stresses are 1600 psi in the wood and 18,000 psi in the steel, determine the rnaximmn allowable magnitude for the distributed load. 7-158* A 150-mm wide x 300-mm-deep timber beam 5 m long is reinforced with 150-mm-wide >< 15-mm-thick steel plates on the top and bottom faces. The beam is simply supported and carries a uniformly distributed load of 20 kN/m over its entire length. The moduli of elasticity for the timber and steel are 13 GPa and 200 GPa, respectively. Determine the maximum tensile flexural stresses in the timber and in the steel. 7-159 A 6-in-wide >< 12-in.-deep timber (E = 1500 ksi) beam will be reinforced with steel (E = 30,000 ksi) plates on its top and bottom faces. The beam will be simply supported, will have a span of 20 ft, and will carry a concentrated load of 5000 lb at the center ofthe span. Ifthe allowable flexural stresses are 1 ksi in the wood and 10 ksi in the steel, prepare a curve showing the acceptable combinations ofplate width and plate thickness. Limit the plate thicknesses to values less than 3r'4 in. 7-160 A timber beam 200 mm wide x 350 mm deep has a 200mm-wide x 16-mm-thick steel plate securely fastened to its bottom face. The moduli of elasticity for the wood and steel

Computer Problems 7-161 A timber beam 8 in. wide >< 15 in. deep is to be strengthened by adding 8-in.-wide x t-in.-thick steel plates to its top and bottom faces. The moduli of elasticity for the wood and steel are 1600 ksi and 30,000 ksi, respectively. If the allowable flexural stresses are 2.4 ksi in the wood and 18 ksi in the steel, a. Determine the maximum moment that can be carried by the beam without the steel plates. h. Compute and plot the percent increase in moment-carrying capacity of the beam gained by adding the steel plates, for 0 5 t 5 2 in. c. Compute and plot the maximum stresses in the wood and in the steel when the beam is loaded to capacity for 0 5 t 5 2 in. 7-162 A timber beam 150 mm wide x 300 mm deep is to be strengthened by fastening 50-mm-thick x w-mm-wide aluminum alloy plates to its top and bottom faces. The moduli of elasticity for the wood and aluminum alloy are 13 GPa and 73 GPa, respectively. a. If a maximum bending moment of 75 kN - m must be resisted by the composite beam, compute and plot the maximum flexural stresses in the wood and in the aluminum for 0 5 w 5 150 mm. b. If the allowable flexural stresses are 15 MPa in the wood and 135 MPa in the aluminum alloy, determine the minimum width w for the aluminum alloy plates.

7-14 FLEXURAL STRESSES IN REINFORCED CONCRETE BEAMS Concrete is widely used ir1 beam construction because it is economical, readily available, fireproof, and exhibits a reasonable compressive strength. However, concrete has relatively little tensile strength; therefore, concrete beams must be reinforced with another material, usually steel, that can resist the tensile forces. The design of reinforced concrete beams is beyond the scope of this book. According to MacGregor,g reinforced concrete beams are designed when the structure reaches a limit state (when the structure or an element ofthe structure becomes unfit for its intended use). The limit state is divided into two groups: one leads to collapse of the structure; the second does not cause collapse of the structure. The second limit state is referred to as the serviceability limit state and is based on a “Reinforced Concrete Mechanics and Design, 3rd ed., J. G. McG1-egor, Upper Saddle River, N. J. Prentice Hall, 1997.

446

CHAPTER 7

FLEXURAL LOADING STRESSES IN BEAMS

linear distribution of strain, as in Section 7-13. Thus, at the service load the beam acts elastically. The transformed section method of Section 7-13 provides a satisfactory procedure for analyzing reinforced concrete beam problems for the serviceability state. The transformed section used for these problems consists of the actual concrete on the compression side of the neutral axis plus the equivalent amount of hypothetical concrete (which is able to develop tensile stresses) on the tension side of the neutral axis required to replace the steel reinforcing rods. The actual concrete on the tension side of the beam is assumed to crack to the neutral surface; therefore, it has no tensile load-carrying ability and is neglected. The solution for maximum stresses in a given beam or for the maximum bending moment with given allowable stresses consists of three steps. 1. Locate the neutral axis for the transformed section. 2. Determine the second moment of area of the transformed section with respect to the neutral axis. 3. Use the flexure formula to determine the required stresses or moment. In the design of a reinforced concrete beam to carry a specified moment with a balanced design (a balanced design means that the allowable stresses in the two materials are reached simultaneously), the following four equations can be written in terms of four unknown properties of the cross section and solved simultaneously. 1. 2. 3. 4.

The flexure formula for the allowable stress in the concrete. The flexure formula for the allowable stress ir1 the steel. The moment equation for the location of the neutral axis. The equation for the second moment of area of the cross section with respect to the neutral axis.

The design ofa reinforced concrete beam also requires consideration ofother factors such as the bond (shearing) stresses between the concrete and reinforcing steel, the diagonal tensile stresses that may be developed, and the amount of concrete that is needed beyond the reinforcing bars. Discussion of such topics can be found in textbooks devoted to reinforced concrete design. The procedure for analyzing or designing reinforced concrete beams using the transformed section method is illustrated in the following examples.

i EXEIIIIPIC P1‘0b1€1Il 7-2 1 A simple reinforced concrete beam carries a uniformly distributed load of 1500 lb/ft on a span of 16 ft. The beam has a rectangular cross section 12 in. wide x 21 in. deep, and 2 in.2 of steel reinforcing rods are placed with their centers 3 in. from the bottom of the beam, as shown in Fig. 7-49a. The moduli of elasticity for the concrete and steel are 2500 ksi and 30,000 ksi, respectively. Determine the maximum flexural stress in the concrete and the average normal stress in the steel. SOLUTION The ratio of the moduli of elasticity is

n

Z

E. E,

i

Z

30,000 2500

Z

12

7-14 '*—l2m.H

*—l2in.—&~_§i 0

v ~

_

~

v

r~1_@f~§~—1

0

-

1 1. , n |5' .91 ‘ |-1 @ 5. u ‘

_

er .

a a >

a -Q 0 ‘-1. s '-u "‘ 1 0 0 1.

Q

0_o._i'4 s Figure P7-163

.

'-

_ ‘ Z’ -‘I :2.s in.

Figure P7-165

7 164* A reinforced concrete beam (see Fig. P7-164} has a 200-mrn-wide x 350-mm-deep cross section with four 15mm-diameter steel bars placed 75 mm from the bottom of the beam. The maximum moment supported by the beam is 15 kN-m. The moduli of elasticity of the concrete and steel are 15 GPa and 200 GPa, respectively. Determine the maximum average tensile stress in the steel and the maximum compressive stress in the concrete at the section of maximum moment.

¢

>. u

s

'.s

-> \

1. ".\ -' ' 1 no ' ..., ‘ s, . n

i

'

7-166" A simply supported reinforced concrete beam 200 mm wide with a depth to the center of steel of 300 mm has a span of 4 m. The tension reinforcement consists of three 16-mindiameter steel bars. The ratio of the moduli of elasticity is 12. If the allowable stresses are 6.5 MPa in the concrete and 120 MPa in the steel, determine the maximum load per meter of length that can be uniformly distributed over the middle half of the beam. 7-167 A simply supported reinforced concrete beam is 8 in. wide and has a span of 12 it. The tension reinforcement, which is located 16 in. below the top surface of the beam, consists of three 7/8-in.-diameter steel bars. The moduli ofelasticity of the concrete and steel are 2400 ksi and 30,000 ksi, respectively. If the allowable stresses are 1000 psi in the concrete and 16,000 psi in the steel, determine the maximum load per foot of length that can be uniformly distributed over the full length of the beam.

i2°°“"“l \ ‘

Intermediate Problems

-

. ' . . —r 75 mm Figure P7-164

Challenging Problems 7-168* A steel reinforced concrete beam of balanced design has a width of 300 mm and a depth to center of steel of 500 mm. Use 16.5 GPa and 198 GPa for the moduli of elasticity of the concrete and steel, respectively. If the allowable stresses are 7 MPa in the concrete and 125 MPa in the steel, determine a. The required cross-sectional area for the steel rods. b. The maximum moment that can be resisted by the beam.

7 165 A simply supported reinforced concrete beam carries a uniformly distributed load of 820 lblfl on a span of 13 ft. The beam has a rectangular cross section 10 in. wide >< 18 in. deep. The tension reinforcement consists of three 31'4in.-diameter steel bars placed 2.5 in. from the bottom of the beam, as shown in Fig. P7-165. The moduli of elasticity of the concrete and steel are 2200 ksi and 30,000 ksi, respectively. Determine the average tensile stress in the steel and the maximum compressive stress in the concrete at the section of maximum moment.

7-1 69 A steel reinforced concrete beam of balanced design is

needed to support a uniformly distributed load of 1000 lblfi on a simply supported span of 16 ft. The width of the beam must be 10 in., and the allowable stresses are 800 psi in the concrete and 16,000 psi in the steel. The moduli ofelasticity of the concrete and steel are 2400 ksi and 30,000 ksi, respectively. Determine a. The required cross-sectional area for the steel rods. b. The depth from the top surface of the beam to the center of the steel rods.

450 CHAPTER '1 rrrxtrrmr. LOADING snrsssss IN BEAMS 7-15 FLEXURAL STRESSES IN CURVED BEAMS One of the assumptions made in the development of the flexural stress theory in Section 7-2 was that all longitudinal elements ofa beam have the same length, thus restricting the theory to initially straight beams of constant cross section. Although considerable deviation from this restriction can be tolerated ir1 real problems, when the initial curvature of the beam becomes significant, the linear variation of strain over the cross section is no longer valid, even though the assumption of the plane cross section remaining plane is valid. A theory will now be developed for a beam, subjected to pure bending, having a constant cross section and a constant or slowly varying initial radius of curvature in the plane of bending. The development will be limited to linearly elastic action. Figure 7-51a is the elevation of part of such a beam. The xy-plane is the plane ofbending and a plane of symmetry. The radius of curvature (distance to the center of curvature) of the neutral surface is R and the radius of curvature to some other surface in the beam (located at a distance y from the neutral surface) is p. Since a plane section before bending remains plane aflter bending, the longitudinal deformation of any element will be proportional to the distance of the element fi"om the neutral surface, as indicated in Fig. 7-51a, from which 5;

5 = 5}’

(11)

The tmstrained length ofany longitudinal element is p9; therefore, Eq. (a) in terms of strain becomes I";6€|'

"961 = T1’ _./'

_,->-\‘\‘ I,}\.. _/

r" .,‘

.\

\

R.

'0 '1

__\

1-. \M-\

/

.|'4‘\

1\-

/.-’

“\ i._.

1;‘ ._

"

-.

T‘-._ \

\

/, ..\

,.-

J

;\

" _,¢

/1

t»-- _ \Q1‘ \

L-an___

\

J?

‘Q: __

2"

I

II I

II

IIIII

I I__rn.I

\'\

Q‘?!

'-=-|

A

-

‘fl 61-?

X - — — - — -—

1 '1

‘

I

I

1M,

a

I I — _\ _ — _ — —

— — — — — _ _ — _

_ -.

50

(H) Figure 7-51

an

(5)

7-15 FLEXURAL srttsssas or CURVED BEAMS fi"om which the expression for the longitudinal strain distribution becomes P','€;y

G’

T16;

y

=—-——-—=———--i—

b p

b R—y

b

U

which shows that the strain does not vary linearly with y (Fig. 7-51b), as was the case for the initially straight beam. Since the action is elastic, Hooke’s law applies; therefore, 6, = cr,,fE (for cry = 0, = 0), which when substituted into Eq. (b) yields

,_m2_'g Y x—

b p—

b R—y

(C,

which indicates that the flexural stress distribution as well as the strain distribution is not linear with y. The location of the neutral surface is obtained fi"om the equation SF, = 0; thus,

[a,d.4='@f JldA=0 A

P

AP

Since y=R— p, whereRis the radius of the neutral surface,

R-PdA=0 5

A

P

Since r,», 0;, and b are not zero, it follows that

I EM = 0 A

(7-23)

P

Equation 7-23 may be written R—p dA f_a1=Rf__fd.1=o A P A P A OI"

R = fii

(7-24)

A P l.n general, Eq. 7-23 or 7-24 should be solved for R for each specific problem; however, the general solution for a rectangular cross section of width I is easily obtained as follows: r,, R _

f rt

rd

r,

Jtdp=f IE4 tdp=0 P rt P rr

Since R is a constant, R ln(r.,/r,-) — (ro — r,-) = O

451

452

CHAPTER 7 rirzxtrrmr. LOADING snrsssss IN BEAMS fi'om which the radius of the neutral surface is To — I’;

R:

l11(n»/rt)

(d)

d/1 The term f — ir1 Eq. 7-24 is tabulated and shown in Table B-20 for several cross P

sections. A The resisting moment in terms of the flexural stress is obtained from the equilibrium equation EM = 0; thus, 1' 2 11 R— 2 M,=—L(vldA)y=—}%_L%dA=_’%_L%dA

re)

The value of R for a given problem, obtained from Eq. 7-24-, can be substituted ir1 Eq. (e) and a solution for M thus obtained. However, it will, in general, be found more convenient to write Eq. (e) in the following form:

M,.=%[j;I(R—p)dA—RL$dA:| From Eq. 7-23, the second integral inthe brackets is zero, and when R—p is replaced by y in the first integral, the resisting moment is given by the expression

M,=mf,,d,,=u,.,,C b

,1

b

0.,

where yg is the y-coordinate ofthe centroid of the cross-sectional area A measured fi"om the neutral axis NA. For a positive moment, yg must always be negative, indicating that the neutral axis of the cross section is always displaced from the centroid toward the center of curvature. Replacing (r,-0,-)/b in Eq. (f) by (p0,)/y from Eq. (c) and solving for 0,, gives U

X

My Myc

My (R — y)Ay-:;-

= L = L

(7- 25)

which is the expression for the elastic flexural stress at any point in an initially ctnved beam. The preceding development is for pure bending and neglects radial compressive stresses that occur within the material. These compressive stresses are usually very small. If the beam is loaded with forces (instead of couples), additional stresses will occur on the radial planes. Because the action is elastic, the principle of superposition applies and the additional normal stresses can be added to the flexural stresses obtained from Eq. 7-25. The application ofEqs. 7-23 and 7-25 is illustrated in the following examples.

Example Problem 7-23 A segment of a curved beam (see Fig. 7-52a) of high-strength steel which has a proportional limit of 95 ksi has the trapezoidal cross section shown i.n Fig. 7-52b. The beam is subjected to a moment M of —l00 kip ~ in.

7-15 rmxtrrutr. sntsssrs [N currvsn nrmrs 453 Jr‘

1- 1 n.—"— 1 111.41

f7i\\ \

/.

_ 3 - ’/fll

3 III.

/

R

/./'

|

z \ \

’.’

III Is

.i| T

|

\I M ‘

\

\\

1 \\

\

I.‘

\\

‘M

p Qldfi;/’ -:[__..~

\ \“__

r,’

In 1

-5!-I5..

('1)

(5)

Figure 7-52

(a) Determine the flexural stresses at the top and bottom surfaces. (b) Sketch the flexural stress distribution in the beam. (c) Determine the percentages of error if the flexure formula for a straight beam

(Eq. 7-8) were used for part (a). SOLUTION The first step will be to locate the neutral surface by using Table B-20 of Ap-

pendix B. The cross-sectional area is

2 1

A = %(s-3) =4.5in.2 and

p 6-3 3 ' ' f. @=@);@1n§-2+1=107941n The value of R is found using Eq. 7-25: A

A

4.5

.

.9

The radial distance from the center of ctnvature to the centroid of the cross-

sectional area is

3(4+1)+s(2+2)

,

s(2+1)

m

~= is =4-3333

r‘

Therefore,yg = R — rc = 4.1683 — 4.3333 = —O.1645 in.

_

yL

3

in.

454 CHAPTER 7 rrrxrrruu. LOADING: snrsssss 11v BEAIHS (a) The required stresses from Eq. 7-25 are At the bottom ofthe beam (y = —l.83l2 in. and p = r,, = 6 in.): M,y —100(—1.8312) 0,5 = — = L = -41.23 ksi E 41.2 ksi(C) pAyC 6(4.5)(—0.1645)

Ans.

At the top ofthe beam (y = +1.688 in. and p = r,- = 3 ir1.):

my -100(+1.16ss) W , = — = 4 = -t-52.631-ts = 52.6ks G” p/iyg 3(4.5)(-0.1645) ' 1 (T)

+52.6 1.169 in. N.A.

1.83 1 in.

41.2

Figure 7-53

or ksi

Ans.

These stresses are well below the proportional limit of the material. (b) Plotting the two stresses from part (a) and zero stress at the neutral surface will indicate that the curve must be shaped as in Fig. 7-53. (c) The cross-sectional second moment of area with respect to the centroidal axis parallel to the neutral axis is

1 = %(1)(s)3 + 1(s)(0.1ss7)2 + %(1)(s)’ + %(1)(3)(0.3333)1 = 3.25 tn.‘ At the bottom of the beam (12 = —1.6667 in.):

M,y

—100(—l.6667)

U)-3 = —T = —

_~

,

= —51.28I(S1 = 51.3 I(51(C)

At the top ofthe beam (y = +1.3333 in.):

M,y

—100(+1.3333)

G)-7' = —T = —

_~

,

= -I-4I.02I(Sl = 4-1.0I(S1(T)

Therefore, the errors are

51.2341-2341.23 (100) = +2-4.37 2 24.4%high atbottom 41.02 — 52.63 (100) 52.63

= -22.05 2 22.1%low at top

Ans.

i EXHIIIPIB PFOITIIEIII 7-24 The elevation and cross section of a segment of a punch press frame are shown ir1 Fig. 7-54. The fiame is made of a gray cast iron that has a proportional limit in tension of 100 MPa. Determine the maximum tensile and compressive flexural stresses produced by a moment M of —40 kN~m. SOLUTION Assume that the neutral axis of the cross section is to the right of the flange and apply Eq. 7-24; thus, using Table B-20 results in

A = Z [00, - 7,-)1 = 240(s0) + 40(240) = 24,000 mm:

7-15 risxtrrutr. smsssrs [N currvsn rams 455 NA. _ -71 my_;‘_ii?

i 90m.ttl—i

—300mm —-—

?-O I, \

A /\

\

120mm

\

_ /Oentroid

=300 ‘ t

I\ ,._‘?' 11

y

~

more 40mm -

*7

- 4<

120mm

\,

l_

M . . . Dimensions in mm

7

till 240 rmn 60 n1.m

Figure 7-54

and

IA

dA _= p

ii

r, =2401n—+401 360 600 b1r1— _ . 7,]

300

"360

= 64.190mm Therefore,

A

24,000

A P Therefore, the assumption about the location of the neutral axis is correct.

The location of the centroid of the cross section with respect to the center of curvature is found using the principle of moments for areas. Thus,

_ s0(240)(3s0) +40(240)(4s0) _ "' _

e0(240) + 40(240)

_ 390 mm

The distance fi'om the neutral axis to the centroid is yc = R —rC = 373.89 -390 = —l6.l1rnrn From Eq. 7-25 the stresses are: At the outside ofthe stem ()2 = —226.11 mm and p = 600 mm):

U _ M,y _ -40(10’)(-226.11)(10"‘) "° — pziyg _ 600(10-‘)(24,000)(10-‘=')(-16.11)(10-3)

= -3s.99(10='=) N/H12 '5 39.0 MIPa (c)

Ans.

456 CHAPTER 7 rrrxtrrutr. LOADING: snrsssrs 11v BEAMS At the inside ofthe flange ()1 = +73.89 mm and p = 300 mm):

U__ M,y _

-40(10’)(+73.s9)(10-1)

’“ _ ,0AyC _ 300(10-3)(24,000)(10-‘=)(-16.11)(10-3) = 25.48(106)N/m2 E 25.5 MPa (T)

Ans.

The low stresses indicate elastic action.

1 PROBLEMS Introductory Problems

7-170 A curved rectangularbeam with a width 7, a depth d (in the

All dimensions in millimeters

m@

radial direction), and an inside radius r,- of 10d is subjected to a moment M in the plane of curvature. Show that the error in computing the maximum flexural stress by Eq. 7-8 (the flexure formula) instead of Eq. 7-25 (the curved beam fonnula) is approximately 3.7 percent low.

7-171* A curved beam having an inside radius of 6 ir1. has the cross section shown in Fig. P7-171. The beam is subjected to a momentM in the plane of curvature. Determine the dimension b needed to make the flexural stress at the inside curved surface equal in magnitude to the flexural stress at the outside cruved surface of the beam

'

i% El l

‘

M

M

Cross section

Figure P7472

_ _ _ _ 7-173* Determine the maximum tensile and compressive flexural stresses in the curved beam of Fig. P7-173 if the magnitude of the moment M shown in the figure is 30 kip - ft.

8 in.—>;

M f

'

2 in.

M / "‘

am-A 1;,,_

8 in2in.4m2in. T

b

M

Cross section

2 in.

ts-7

8 in.

l in.

\

Figure P7-171

C1-055 section

M

Figure P7-173

Challenging Problems Intermediate Problems 7-172 Determine the maximum tensile and compressive flexural stresses in the curved beam of Fig. P7-172 if the magnitude of the moment M shown in the figure is 20 kN - m.

7-174* The curved beam ofFig. P7-174 is subjected to a bending moment M in the plane of curvature, as shown in the figure. Determine the maximum permissible magnitude for M if the flexural stress is not to exceed 35 MPa (T) or 140 MPa (C).

7-16 |\

1

/

\

\

f¢———~

w

300 mm

-‘

L‘

,,_ I I

'

5\ /

,-\/

,tw/\8

»a‘_

\_-3' 751-am M

_-

\-/ M

-hm

I’,

_

L

_1m11n._2m‘

"‘~.

\

[ I

COMBINED LOADING: AXIAL, PRESSURE, FLEXURAL, AND TORSIONAL

L .5= 24m

Y iii.

P

R

S

50mm I-—25

50mm Cross section

Figure P7-174 '7-175 The cross section of a segment of a crane hook is shown in Fig. P7-175. The curvature and loading are in the xy-plane. Determine the flexural stresses at the inside curved surface and at the outside curved surface if the magnitude ‘ y _of the _ resisting moment M. shown on the cross section IS 70 kip - in.

43*-.

1 in.Z

N. A.

if 2 in.

Centroid '

Figure P7-175

7-16 COMBINED LOADING: AXIAL, PRESSURE, FLEXURAL, AND TORSIONAL The stresses and strains produced bythe fundamental types of loads (axial, pressure of thin-walled vessels, torsional, and flexural) have now been analyzed. Many machine and structural elements are subjected to a combination of any two or all four of these types of loads, and a procedure for calculating the nomial and shearing stresses resulting from such loads at a point on a given plane is required. The procedure used to solve such problems is the same as that previously developed for axial and torsional loads. First, the member is sectioned to expose a cross section normal to the axis of the member, and a fi"ee-body diagram of the part of the member to one side of the section is drawn. Then, the internal forces at the section are found using the equations of equilibrium. The internal forces may be a combination of an axial force, a shear force, a torque, or a bending moment. Using the formulas previously developed, the normal and shear stresses are then calculated separately at the point of interest on the cross section for each of the internal forces. Normal stresses due to the axial force, the internal pressure, and the bending moment are then added (or subtracted) to obtain the normal stress at the point. Similarly, the shearing stresses at the point due to the shear force and torque are added (or subtracted) to obtain the shearing stress at the point. Once the normal and shearing stresses at a point are known, the stress transformation equations (Eqs. 2-12 and 2-13) can be used to determine normal and shearing stresses on other planes through the point. Principal stresses and maximum shearing stresses at the point and the planes on which they act can be determined by using Mohr’s circle or Eqs. 2-14 through 2-18. The combined normal and shearing stresses at a point on a cross section can be determined using the above procedure (method of superposition), provided the combined stresses do not exceed the proportional limit of the material. The following example problems illustrate the procedure.

EXHIIIPIB Pfflblfllll 7-25 The solid 100-mm-diameter shafi shown in Fig. 7-55a is subjected to an axial compressive force P = 200 kN and a vertical force V = 100 kN. For point/1 on the outside surface of the shaft, determine

Mr mi

0 4» 7.21

458 cnsrrsx 7 rrsxtnuu. LOADING: STRESSES [N snms

J‘

Z

it

V

t

,

All

Vr

‘ 500mm

P

an

\.

(-11) 100 l-{N

MI’

2001-:N

Jr

(0)

in

4i

25.5 MPa a

(d)

Figure 7-55

16.98 MPa

(e)

(a) The x- and y-components of stress. (b) The principal stresses and the maximum shearing stress at the point. SOLUTION (a) Since point A is on the outside surface (a free surface) of the shafi, a state of plane stress exists at the point. The coordinate system is selected as shown in Fig. 7-55b. This is the same coordinate system for which the equations in this chapter were developed. Passing a transverse plane through point A

and isolating the segment of the shafi to the right of point A results in the free—body diagram shown in Fig. 7-55c. The intemal forces on the transverse cross section are an axial compressive force F = 200 kN, a shear force I/Z = 100 kN, and a moment M, = 50 kN ~ m. The directions of the stresses on the lefi face of the element are in accordance with the directions of the forces that produce the stresses, that is, in the directions of the intemal forces.

These stresses are shown in Fig. 7-55d. The magnitudes of the stresses are determined as follows.

F

200(10*)

6

2

0, = Z = ii, = 25.46(10 )N/m = 25.46 MPa (rr/4)(0. 100)

_ I/,Q _ Vvbm-A) _ r—

It

—

It

—

V ‘L E 2

(J'rr4)

— 2 4 tr)

_ 4V» — 3 JT?'2

100 10 3 = 16.97"/[10°)N/m2 = 16.917 MPa = 4L2) 3:rr(0.050) There is no normal stress due to M since pointA is on the neutral axis.

7-16 cosmnnm tomato: AXIAL, Passscns,r|.sxuiui1., mo rossiotw. 459 The x- and y-components of stress on the element at A are 0,, = 25.46 MPa "5 25.5 MPa (C)

ABS-

r,,. = -16.97? MPa E -16.98 MPa

ADS-

There is no normal stress ir1 the y-direction (the circumferential direction),

since there is no force to cause such a stress. The shear stresses on the four sides of the element have the same magnitude. The shear stresses that meet at a comer either both point toward the comer or both point away from the corner. These stresses are shown on the element of Fig. 7-55e. (b) Equation (2-15) is used to calculate the principal stresses. The stresses for use in this equation are 0, = —25.461\/[Pa

0,, = 0

1:“, = -16.97? MPa

Substituting these stress components into Eq. 2-15 yields _a,,+o,,i

ox—a,, 2

Upl.p2 — T

T

2 +1-'xy

-25.46 0 1 I -25.46-0 2 +(-16.97102 = pf = -12.73 :t 21.22 Thus, the principal stresses are GP] = -12.73 + 21.22 = +8.490 MPa '5 8.49 MPa (T)

Alls-

crp; = -12.73 — 21.22 = -33.95 MPa '5 34.0 MPa (C)

Alls-

gp3 = 6, = 0

Ans.

Since the two in-plane principal stresses are of opposite sign and op; = oz = O, amax _ Umin

rm“ = 5 _ s.490 - (-33.95) _ 2 = 21.2 l\/[Pa

Ans.

1 EXHIIIPIB Pl‘0b1€Ill 7-26 The cast iron frame of a small press is shaped as shown in Fig. 7-56.-:1. The cross section a—a of the flame is shown in

Fig. 7-56b. Axis c—c passes through the centroid of the cross section. For a load Q of 16 kip, and assuming linearly elastic action, determine

460 cnsrrss 7 rmxmm. LOADING: snrsssrs [N BEAMS J’ M

-E.

r—1Ii8in.

Q Q

15 in.

la

A

G

B

‘

18in.

_

u|4lt4L_u| 5'5"55'

16kip

a—4_—

\ My

if/—a—x I

i

to 5.

.4 Q

F

(11)

(Q)

(C)

Figure 7-56(a-e)

(a) The normal stress distribution on section a—a. (b) The location of the neutral axis (line of zero stress). (c) The principal stresses for the critical points on section a—a.

SOLUTION (a) The frame is sectioned at a—a, and a free-body diagram for the part of the frame above the section is shown in Fig. 7-56c. Since the load acts in a plane of

symmetry, there are three independent equations ofequilibrium. The intemal forces at the section are fotmd using the equations of equilibrium as follows:

+_>2F,,—0; +1“EF,,=0: + 2M.-,=0=

V,=0 l6—F=0 M,— 1s(1s)=0

V,-=0 F=l6kip M, =288kip-in.

Thus, there are two intemal forces on the section, an axial force F which produces a constant normal stress 0,1 = F/A over the section, and a bending moment M,, which produces a linear variation of normal stress 0,2 = —M,.x/I over the section. The cross-sectional area A and the second moment I of the

cross-sectional area with respect to the centroidal axis c—c are

A = 12(2) + 8(2) = 40 in.2 1 = 1lE(12)(2)’ + 12(2)(2)’ + %(2)(s)’ + 2(s)(3)1 = 333.3 in.4 The stresses o',,| and 0,2 due to the internal forces are F 16,000 _ , o,.| = I = T =+400ps1= 4-00ps1 (T)

which is constant over the cross-sectional area. The maximum tensile flexural stress occurs at the left edge of section a—a (point A) and is

M 283 101 -3 , , 0,2,, = -% = -% = +2592 pS1 = 2592 p51 (T)

7-16

COMBINED LOADING: AXIAL, PRESSURE, FLEXURAL, AND TORSIONAL

The maximum compressive flexural stress occurs at the right edge of section a—a (point B) and is

M1

2ss(10’)(+7)

,

,

0y2;,- = —'TB = T = —6049ps1 = 6049ps1 (C) The distributions of stresses 0,4 and 0,; are shown in Figs. 7-56d and e, respectively. Superirnposing the normal stresses at points A and B gives 0,,,, = 0,4,, + 0,2,, = 400 + 2592 = +2992 psi E 2990 psi (T) 0,3 = 0,13 + 0,23 = 400 — 6049 = -5649 psi '5 5650 psi (C) The distribution of normal stress on section a—a is shown in Fig. 7-56f. (b) The location of the neutral axis (line of zero stress) can be determined as the place where the compressive flexural stress is 400 psi because this stress will just balance the axial tensile stress of 400 psi. Thus,

M,x

288(lO3)(x)

.

“» = -T = -W = -“°°P=“ from which

I = 3i33'3(400) = 0.4629 in. 2 0.463 in. 2ss(103)

Ans.

P The normal stress due to bending varies I-tom zero at the centroidal axis (c-c on Fig. 7-56b) to a maxiinurn at the edges of the “beam.” The neutral axis (the line of zero stress) lies 0.463 in. to the right of the certtroidal axis.

to the right of the centroidal axis of the cross section. (c) With no shearing stresses on section 0-0, the normal stresses are principal stresses, and the critical points, as observed from the stress distribution of Fig. 7-56f, are the left and right edges of the section. The principal stresses for the right edge are 0,, = 5650 psi (C)

and

0

Ans.

and

0

Ans.

The principal stresses for the lefi edge are

0,, = 2990 psi (T)

The principal stresses are shown on elements A and B of Fig. 7-56g.

_ 2-0453 in. |"3in.*'[>

_.

-

__

_.

_P

_

“P7 (4)

Figure 7-S6(d-g)

__

|

x

2990psi

I

5650 psi

A

400p§i

i

E

Mx

"2"-T (=1)

2990 psi

5650 psi

(D

(3)

462

CHAPTER 7 rrsxurur. LOADING snrsssrzs [N BEAMS

‘I EXHIIIPIB PTOIJIBIII 7-27 A gray cast iron compression member is subjected to a vertical load Q = 100 kN, as shown in Fig. 7-570. Determine the normal stresses at each comer of section ABCD. SOLUTION Since stresses are required at points A, B, C, and D, the member is sectioned through these points, and a fi'ee-body diagram of the portion of the member

above the section is drawn, as shown in Fig. 7-57b. In general, the internal forces acting on a section of a member consist of shear forces P} and I/Q, an axial force F, a torque I} = 114,, and bending moments M, and 114,. The six equations of equilibrium used to determine these internal forces are

zF,=. BF, = zF,— 99¢

=0 -100=0 .N'~q>F=0

EM, = EMy = Q9Q

=0 =—100kN=100kN(C) 51$ =0

M; — lU0(0.06U) = 0 =0

My = 6kN-II1 T, =0

M, — l00(0.075) = 0

M, = 7.5 kN - m

Thus, there are three internal forces on the section: (1) an axial force F which produces a constant normal stress 01 = F/A over the section; and (2) two bending moments M, and AL each of which produces a linear variation of normal stress 0 = Mc/I over the section. The cross-sectional area A and the second

moments L, and L, of the area with respect to the centroidal x and z axes are

A = 1s0(120) = 18.O0(l03)mm2 = l8.0O(1O_3)1'n2 1 1, = E(1s0)(120)3 =21.60(10‘)mm“ = 21.60(10 _ ‘)m“ 1

I, = E(l20)(150)3 = 33.75(10‘)mm“ = 33.7'5(1O_6)m4

5 “=1

Q

y

100104

Q.

Bs>5Q

375rmn XE‘ \F_______

/

A

1

“--___

:-T.‘-‘q

' '

I‘

C

1,11

_"'~

A

\

I

I

K“-___ I

FT“-" B .

.

._

_

-

2/ .1/V: M

F

B

~,_“_""‘ _\i ‘_‘_‘_‘_-_-_?.1"@

Figure 7-57(a-b)

C

V

.

N \

(8

v=—— ——

E1, 2

Lx + — L2

2

At the left end of the beam where x = 0, the deflection is PL3

SPL3

ZEIL

Ebhz

Ans.

v=——=—,L

It is also interesting to note how the maximum flexural stress varies along the length of the beam. From Eq. 7-9,

Mc

h P45)

referred to as a constant stress beam.

6PL

“W = T = W = W E

P Equation (a) indicates that the maximum flexural stress does not depend on position x but is constant along the entire length of the beam. This type of beam is fiequently

(4)

I

1 PROBLEMS MecMuvie Activities and Problems

Y

I

MM8.l Beam bormdary condition game. Determine appropriate botmdary conditions necessary to determine beam deflections using the double integration method.

w

A

Introductory Problems

B L

8-1* A beam is loaded and supported as shown in Fig. P8-1. Determine a. The equation of the elastic curve. Use the designated axes. b. The deflection at the lett end of the beam. c. The slope at the lefl end of the beam.

Figure P8-2 8-3 A beam is loaded and supported as shown in Fig. P8-3. Determine a. The equation of the elastic curve. Use the designated axes. b. The deflection midway between the supports. c. The slope at the lefl end of the beam. J’

l

..

ix Cb

.=

L

A

B L

Figure P8-l

Figure P8-3

8-2* A beam is loaded and supported as shown in Fig. P8-2. Determine

8-4* A beam is loaded and supported as shown in Fig. P8-4. Determine

a. The equation of the elastic curve. Use the designated axes. b. The deflection at the right end of the beam. c. The slope at the right end of the beam.

a. The equation of the elastic curve. Use the designated axes. b. The deflection at the left end of the beam. c. The slope at the left end of the beam

498 crntvrss s FLEXURAL r.o1tn11~1c= stun nsrrscnons 24 i(N‘t1'l ( A

B

iflefl

30mm A

B

L 2 mil

ii L

Figure P8-4

to

(b)

Figure P8-8 8-5 A beam is loaded and supported as shown in Fig. P8-5. Determine a . The equation of the elastic curve . Use the designated axes. b. The deflection midway between the supports. c. The slope at the right end of the beam. J?

l

8-9 The beam shown in Fig. P8-9 is a W10 >< 30 structural steel (E = 29,000 ksi) wide-flange section (see Appendix B). Determine a. The equation of the elastic c1n"ve. Use the designated axes. b. The deflection at the lefl end of the beam if w = 2000 lb/ft and L = 10 ft.

w L

A

’|'

B L

L12

Figure rs-5

l|||ii|||l1 n___s it

8-6 For the steel beam [E = 200 GPa and I = 32.0(10°) mm“] shown in Fig. P8-6, determine the deflection at a section midway between the supports. Figure P8-9

1 A B 1 at M» “ta 1

10ltN

10kN

8-10* A cantilever beam is fixed at the lefl end and carries a uniformly distributed load w over the full length of the beam. In addition, the right end is subjected to a moment of 3wL2/8, as shown in Fig. P8-10. Determine a. The equation of the elastic curve. Use the designated axes. b. The maximum deflection in the beam if] = 2.5 (106) mm‘, E=210 GPa,L =31n,andw= l§OON:"m.

8-7* For the steel beam (E = 30,000 ksi and I = 32.1 in‘) shown in Fig. P8-7, determine the deflection at a section midway between the supports.

Y I w

7500 lb< 120-mm aluminum (E = 70 GPa) bars, as shown in Fig. P8-Sb. Determine the deflection at the left end of the beam.

8-11* A beam is loaded and supported as shown in Fig. P8-l 1. Determine a. The equation of the elastic curve. Use the designated axes. b. The deflection midway between the supports.

es nmnscnonnrnnmmmnou 499 11¢:

J’ wL2

|

T

“’

A

A

I L

I

T

i

20m

B

~

smmm 1

B

c

I

.

ljm

1

Figure P8-14

Figure P8-l 1 8-15 Determine the maximum deflection for the beam shown in Fig. P8-15. 8-12 A beam is loaded and supported as shown in Fig. P8-12. Determine a. The equation of the elastic curve for the interval between the supports. Use the designated axes. b. The deflection midway between the supports.

453- .

L DM

~

L

.B

L

Figure P8-15 )’

I A

W ___

B 5

L

c mi ' )1a

' .5‘ U3 i

8-16* A boy with a mass of60 kg is standing on a 40 x 300-mm wood (E = 10 GPa) diving board, as shown in Fig. P8-16. If the lengthAB is 0.6 m and the length BC is 1.5 m, determine the maximum deflection in the diving board when the boy is standing on the end of the board.

Figure P8-12

Intermediate Problems 8-13* A beam is loaded and supported as shown in Fig. P8-13. Determine

A

a. The equation of the elastic curve. Use the designated axes. b. The slope at the lefl and right ends of the beam. c. The deflection midway between the supports.

Y

A ‘+1./'2

P

QL

B

_ C

‘

u

1¢

n a 1

u

' u

—*A_

Figure P8-l6

C M27

8-17 Determine the maximum deflection for the beam shown in Fig. P8-17.

Figure P8-13

T 8-14* A 100 x 300-mm timber having a modulus of elasticity of 8 GPa is loaded and supported as shown in Fig. P8-14. Determine a. The deflection at the 7-kN load. b. The deflection at the flee end of the beam.

A

w

I

B

L Figure P8-17

l

L

S00 CHAPTER s FLEXIIRAL LOADING: BEAM niiruicnons W

8-18 A timber beam 150 mm wide x 300 mm deep is loaded and supported as shown in Fig. P8-18. The modulus of elasticity of the timber is 10 GPa. Apointer is attached to the right end ofthe beam The load acts at the midpoint of the span. Determine

A

B

C

I

L

a. The deflection of the right end of the pointer. b. The maximum deflection of the beam.

U2

=

Figure P8-2 1

P = 8900 N

_ 6.

8-22* A beam is loaded and supported as shown in Fig. P8-22 Determine a. The equation of the elastic curve. Use the designated axes. b. The deflection at the lefl end of the beam.

.=

5.5m

- 1.5m

u‘

Figure rs-1s w(x)=Icx2-

8-19* The cantilever beam shown in Fig. P8-19 has a second moment of area of I in the interval AB and a second moment of area of 21 in the interval BC. Determine the deflection at end A of the beam.

AL

L

B

Figure P8-22

Y

P

I

‘I’ L.

I

8-23* Determine the deflection midway between the supports for beam AB of Fig. P8-23. Segment BC of the beam is rigid.

A-

B‘

1

L

L

L

2

C

W

Figure P8-19 8-20 A beam is loaded and supported as shown in Fig. P8-20. Determine a. The equation of the elastic curve of member AB. Use the designated axes. b. The slope at the right end of the beam AB. c. The deflection midway between the supports. y

L1-

C U2

IIIIIIIIII é B

AI

ee Figure P8-23 8-24 The beam shown in Fig. P8-24 is a WT 203 >< 37 structural steel (E = 200 GPa) T-section (see Appendix B). Determine

P

a. The equation of the elastic curve for the region of the beam between the supports. Use the designated axes. b. The deflection midway between the supports if w = 5.5 kNim andL = 3.5 m.

A Rigid A ‘I é

B.

J‘

2.Le"3

I

Figure rs-20 8-21 A beam is loaded and supported as shown in Fig. P8-21. Determine a. The maximum deflection between the supports. b. The deflection at the right end of the beam.

A

W B

—e L

Figure P8-24

E

WLZ

IC 1:: LI3

I

s-3 DEFLEC'1‘l0NBY]1\I’l‘EGRATION 8-25 The beam shown in Fig. P8-25 is a W8 x 40 structural steel (E = 29,000 ksi) wide-flange section (see Appendix B). Determine a. The equation of the elastic cunre for the region of the beam between the supports. Use the designated axes. b. The deflection midway between the supports if w = 240 lb,/ft and L = 16 fl.

S01

8-28 The simply supported beam ABCD shown in Fig. P8-28 has a second moment of area of 21 in the center section BC and a second moment of area of I in the other two sections near the supports. Determine a. The deflection at section B. b. The maximum deflection in the beam.

P

wL=r2 X (

f

U

A

D

e.m.L2.Ie..Lt

0

U2

L

U2

,

I

B-

C

1

Figure P8-25

Figure P8-28

Challenging Problems

8-29* Determine the maximum deflection for the beam shown in Fig. P8-29.

8-26* The cantilever beamABC shown in Fig. P8-26 has a second moment of area of 21 in the interval AB and a second moment of area ofI in the interval BC. Determine

J’

a. The deflection at section B. b. The deflection at section C.

I

P

w

P

i

c

‘LIBLI A

L

‘B

AC

Figure P8-26 8-27* Abeam/IB is loaded and supported as shown in Fig. P8-27. The load P is applied through a collar that can be positioned on the load bar DE at any location in the interval L/4 < a < 31.14. Determine a. The equation of the elastic Cl..l.1'VB for beam AB. b. The location of the load P for maximum deflection at end B. c. The location of load P for zero deflection at end B.

Figure P8-29

8-30 A beam is loaded and supported as shown in Fig. P8-30. Determine a. The deflection at the lefl end of the beam. b. The deflection midway between the supports.

)"

__W _‘

Ali:

y

I

':

A ix

'

D

‘TlB C

P

Figure P8-27

.1.

c 2L

2

Figure P8-30

E \_

a

L

F

\

“~-\- Rigid

8-31 The cantilever beam ABC shown in Fig. P8-31 has a second moment of area of 41 in the interval AB and a second moment of area of I in the interval BC. Determine a. The deflection at section B. b. The deflection at section C.

S02

CHAPTER s FLEXURAL LOADING: arm nrruzcnons 2w

‘"5

8-34 A beam is loaded and supported as shown in Fig. P8-34. Determine the deflection at the lefl end of the distributed load.

A L¥B LAC

Figure P8-31 8-32* A beam is loaded and supported as shown in Fig. P8-32. Determine the deflection midway between the supports.

A; —= '

I_ITITl 0

'

: I

B

L

:

L

'

I

Figure P8-34

Q

8-35 The cantilever beam ABC shown in Fig. P8-35 has a second moment of area of 41 in the interval AB and a second moment of area of I in the interval BC. Determine

Figure rs-32 8-33* A beam is loaded and supported as shown in Fig. P8-33. Determine

a. The deflection at section B. b. The deflection at section C.

a. The slope at the left end of the beam. b. The maximum deflection between the supports.

I

A

i

%. . L

Figure rs-as

xii ;

L

l

Zw

A

B

L+. .

W

I1: c Li~riLi-

Figure P8-35

8-4 DEFLECTIONS BY INTEGRATION OF SHEAR FORCE OR LOAD EQUATIONS In Section 8-3, the equation ofthe elastic curve was obtained by integrating Eq. 8-1 and applying the appropriate boundary conditions to evaluate the two constants of integration. In a similar manner, the equation of the elastic c1u've can be obtained

from load and shear force equations. The differential equations that relate deflection v to load w(x) or deflection v to shear force V(x) are obtained by substituting Eq. 7-1 lc or Eq. 7-1 ld, respectively, into Eq. 8- l. Thus, dzv El F = M(x)

[8-1)

d3v 4

El g = w(x)

(s-3)

When Eq. 8-2 or 8-3 is used to obtain the equation ofthe elastic curve, either three or four integrations will be required instead of the two integrations required with

s-4 nnrnzcrtons BY mrscmrtos or snssn roan: on mm squtmoss 503 Eq. 8-1. These additional integrations will introduce additional constants of integration. The botmdaty conditions, however, now include conditions on the shear forces and bending moments, in addition to the conditions on slopes and deflections. Use of a particular differential equation is usually based on mathematical convenience or personal preference. In those instances when the expression for the load is easier to write than the expression for the moment, Eq. 8-3 would be preferred over Eq. 3-l. The following examples illusnate the use of Eq. 8-3 for calculating beam deflections.

1 Example Problem 8-5 A beam is loaded and supported as shown

Y

in Fig. 8-9. Determine

i

(a) The equation of the elastic curve. (b) The maximum deflection of the beam.

SOLUTION

W

,4 |’Fi*

3

-

'

I.

-

:-it

(a) Since the equation for the load distribution [w(x) = w = constant] is given, Eq. S-3 will be used to determine the equation of the elastic curve. In Section 7-6 (see Fig. 7-16), the upward direction was considered positive for a distributed load w; therefore, Eq. 8-3 is written as

Figure 8-9

dfv E1 F = w(x) = —w Successive integration gives d3

Elé = V(x)= —wx+C1 d2

2

EIé=M(x)=-‘%+c,x+c2 dv

3

2

E15:-%+c1%+c2x+c, wx4 x3 x2 EI =__ ccc c V 24+‘6+’2+’x+“ The four constants of integration are determined by applying the boundary conditions. Thus, At x = 0, v = 0;

therefore, C4 = 0

At x = O, M = 0;

therefore, C2 = 0

At x = L, M = 0;

therefore, C1 = %

Atx=L,v=0;

3 therefore,C3=—& 24

Thus,

__ 24EI[x W 4_ 2Lx 3 +Lx] 3 v_

Ans.

P The constant C1 could also have been determined from a boundary condition involving the shear force V. For example, the shear force jumps upward by wL/2 across the left support. Therefore, at x = 0 the shear force equation gives V = w(0) + C, = wL/2 and the first constantofintegration is C t = wLf2.

S04 CHAPTER s rmxmm. LOADING: mam nnruzcnonts (b) The maximum deflection occurs at x = L./2, which when substituted into the equation of the elastic curve gives

v

y

SWL4

mu

SWL4 L

Ans

= —— = — 384-E1 384-E1

.

i Example Problem 8-6 A beam is loaded and supported as shown in Fig. 8-10. Determine

I -— w(x) = w cos (1m'2L) w

.1

A

I

B

I

(a) The equation of the elastic curve. (b) The deflection at the right end of the beam. (c) The support reactions IQ and Eli; at the left end of the beam.

La;

Figure s-10

SOLUTION (a) Since the equation for the load distribution is given and the moment equation is not easy to write, Eq. 8-3 will be used to determine the deflections. l.n Section 7-6 (see Fig. 7-16), the upward direction was considered positive for

a distributed load w; therefore, Eq. 8-3 is written as d4v mc EYE =w(x) = —w cos E Successive integration gives a‘3v

2wL , rrx

E1‘? =V(x)=—?s1nE+C|

dzv

4-WL2

Jrx

HE = M(I)= TCOSE -l-C1I+C2

dv

8wL3 _ rrx

x2

?SlI1iI +C1?+Cg.t+C3

EIv=—

16wL4 rrx x3 x2 H4 cos 2L +C16 +C;2 +C3x+C4

The four constants of integration are deterrriined by applying the boundary conditions. Thus,

At x = 0, v = 0; dv

therefore, C4 =

l6wL4 4 rt

Atx=O,a=O;

the1-efore,C3=0

Atx=L,V=0;

therefore,C|= w

Atx=L,M=0;

2

L

rr 2wL2 therefore,Cg=—T

s-4 onrtscrrons mt lNI'EGRA'l'ION or SHEAR FORCE on roan squmous SOS Thus,

v = -L 4SL“cosE - 1r3Lx3 + 3rr3L2x2 - 4&1,‘

31451

Ans.

21:

(b) The deflection at the right end of the beam is

v3 = v,:;_ = -3n"%H(-1:31.“ + 31%“ - 4815“) (219 - 48)wL4 = -0.04795 L 4 E1 = _______ (311451)

W /

An. 5

(c) The shear force V(x) and bending moment M(x) at any distance x from the support are

V(x) = %[1-sin 2wL

Jrx

M(x) = ?[2L cos E + rrx - er] Thus, the support reactions at the lefi end of the beam are

2 L K1=V,=n=l— rr 2[Jr —2)wL2

MA = Mx=0 = -g

Ans.

Ans.

1 PROBLEMS Introductory Problems 8-36* For the beam and loading shown in Fig. P8-36, determine a. The equation of the elastic curve. b. The maximum deflection for the beam.

w -L.

A

B L

Figure P8-3'7

J’

I

A

W

ti.

8-38 For the beam and loading shown in Fig. P8-38, determine

B

a. The equation of the elastic curve. b. The deflection midway between the supports. J’

Figure P8-36

‘

‘me W

s-31* Forthebeam and loading showninFig. P8-37, determine a. The equation of the elastic curve. b. The maximum deflection for the beam.

f‘

I L

Figure P8-38

B

S06 cnavran s rurxtrruu. LOADING: nun nsruzcnons Intermediate Problems

19(3) = w.r3fL3

8-39* A beam is loaded and supported as shown in Fig. P8-39 Determine a. The equation of the elastic curve. b. The deflection at the left end of the beam. c. The support reactions Kt; and MI-3.

A .

_ _

i

L

B _

.

W L

Figure P8-41

Challenging Problems 8-42* A beam is loaded and supported as shown in Fig. P8-42. Determine

w(x) = wx31'L3 -\

W L

| i .1: A

a. The equation of the elastic curve. b. The deflection at the left end of the beam. c. The support reactions VI; and MB.

B L

Figure P8-39

.1"

i

-- w(.r) = w cos (1txf2L)

W

_L A

8-40* A beam is loaded and supported as shown in Fig. P8-40. Determine

ti.

B

Figure P8-42

a. The equation of the elastic cune. b. The maximum deflection of the beam.

8-43* A beam is loaded and supported as shown in Fig. P8-43 Determine a. b. c. d. e.

y

I

wo=)=e1-

The equation of the elastic curve. The deflection midway between the supports. The maximum deflection of the beam. The slope at the lefi end of the beam. The support reactions RA and RB.

if J.

A .

(— w(x) = w sin (rtx."2L}

l_

B L

.0

Figure rs-40

A

mt B

I

is

Figure rs-43 8-41 A beam is loaded and supported as shown in Fig. P8-41 Determine a. b. c. d.

The equation of the elastic curve. The deflection midway between the supports. The maximum deflection of the beam. The support reactions RA and RB.

8-44 A beam is loaded and supported as shown in Fig. P8-44 Determine a. b. c. d.

The equation of the elastic curve. The deflection midway between the supports. The slope at the lefl end of the beam. The support reactions RA and R5.

s-5 srucumnnrrtmcrross

ta we=Wa

Figure P8-183 8-184 An S178 x 30 American standard section is used for a beam that is loaded and supported, as shown in Fig. P8-184. Determine a. The reactions at the supports A, B, and C. b. The maximum flexural and transverse shearing stresses in the beamifw= l0kN/mandl. =2m.

A g l

B L 2w

w Figure P8-180

8-181* Draw complete shear force and bending moment diagrams for the beam shown in Fig. P8-181.

Figure P8-184

s-10 DESIGN PROBLEMS 567 8-185 An S6 x 17.25 American standard section is used for a beam that is loaded and supported, as shown in Fig. P8-185. Determine

W

C

a. The reac1'ions atth e supportA,B, s an ac . b. The maximum flexural and transverse shearing stresses in

%—

i

the beam ifw = 500lb1fi and L = 6 ft.

Figure P8485

%—

L

8-10 DESIGN PROBLEMS in Chapter 7 when the design of beams was discussed, either flexural strength or shear strength was the controlling parameter. In this chapter an additional parameter, deflection, will be introduced. Thus, the design of a beam may be based on flexural stress, shearing stress, or deflection. The design procedure is similar to that presented in Chapter 7. The beam is first designed based on flexural stress, and then checked for shearing stress and deflection. Ifthe shearing stress and deflection

are within allowable limits, the design is satisfactory. If either the shearing stress or the deflection is greater than the allowable value, the beam must be redesigned until all of the allowable limits are satisfied. Clearly, this is a trial-and-error process. The following examples illustrate the procedures for designing beams where allowable limits are given for flextnal stress, shearing stress, and deflection.

i Example Problem 8-27 An air-dried Douglas fir timber (E = 13 GPa) beam is loaded as shown in Fig. 8-39a. If the allowable flexural stress is 8 MPa, the allowable shearing stress is 0.7 It/[Pa, and the allowable deflection is 14 mm, determine the lightest-weight standard structural timber that can be used

for the beam. SOLUTION Load (free-body), shear force, and bending moment diagrams for the beam of Fig. 8-39a are shown in Figs. 8-39b, c, and d, respectively. Since the load is uniformly distributed over the full length of the beam,

1 1a,, = R, = 51350) (5) = 2l25N rm, = 2l25N Mm = %(2125)(2.5) = 2656N-111 w = 850 Nfm

A

L B

(8 l Figure 8-39(2)

+

2L

i

S68 CHAPTER s Frsxuruu. LOADING: error nsrrrcnoss w = s50 Nfm

1 =2l25N 2.5m L 3*:Z

w

RB=2l25N 2.5 m

125

l\JAx

0

-

A4” N . m

-2125

- - -. ->

§e_ _ _ _ _ _656 _ _ _'i

+

+o

§ E - -_ - - _-_- r-p

Figure 8-39(b—d) The minimum section modulus needed to satisfy the allowable value of the

flexural stress is given by Eq. 7-9 as

s 2 E = fl = 332.0(10-6) I113 = 332.0(l03)IlI]1I13

e,.1

s(10 )

The lightest-weight standard structural timber listed in Table B-16 (Appendix B) withS 3 332.0(lO3) mm3 is a timber with nominal dimensions of 5 1 x 254 mm. Some properties of this timber that will be needed later are

Mass/unit length = 6.38 kg/m

Area = 9.88(lO3)l'I1lIl2 = 9.88(l0_3)m2 I = 4s.3(10°)riu;ri* = 4-8.3(l0_6)m“ s = 4-00(l03)nm13 = 400(10-“mi For a rectangular section, the area is A = bk, the second moment of area is I =

bhii12, and the maximum shearing stress occurs at the neutral axis. Then,

Q .( 2) 8 h

h

=—b—

bhz

=7

s-10 nssrerv PROBLEMS and the maximtnn shearing stress is Vr Q Vr(bh2/8) Vr Vr 2125 --=1.5-I = -It = (bh3/l2)b be =1.sA =1.s-i9.8s(10-3) = 0.3226(l06)Nfm2 = 0.32215 MPa < 0.7 It/[Pa Thus, the shearing stress requirement is satisfied. For a simply supported beam with a uniformly distributed load, case 7 of Table B- 19 (Appendix B), gives the maximinn deflection as

I

|_ 5wL“ _

s(8s0)(s)‘

v‘“‘“‘ ' 38451 _ 384(13)(10"‘)(48.3)(10-°) =1l.0l7(l0_3)m= l1.0l7mm < 14mm Therefore, the deflection requirement is satisfied. The 51 X 254-mm standard structural timber satisfies the requirements for flexural stress, shearing stress, and deflection; however, the analysis thus far neglected the weight of the beam. The timber beam weighs (6.38)(9.8 1) = 62.59 N/m. Adding this uniformly distributed load to the applied load gives a imiformly distributed load w = 850 + 62.6 = 912.6 Nfm. For this loading, the maximum shear force is 2282 N and the maximum bending moment is 2852 N - m. The section modulus needed to satisfy the flexrnal stress requirement is

s 3 % = g = 356.5110-‘=) 1113 = 35851103) mm3 < 4OO(lO3)II1.I1'13 den

8(lO )

where S = 400 (103) mm3 is the actual section modulus of the cross section originally selected (51 x 254 mm nominal). Thus, the 51 x 254-mm timber satisfies the flexural stress requirement. The maxirnimi shearing stress and the maximum deflection with the beam weight included are ‘Ema; =

V

=

2282

= 0.3465(l06)Nfm2 — 0.3465 MPa < 0.7 we Ivmi = 5wL4 = s(912.6)(s)“ 384151

384(13)(10°)(48.3)(10-6)

= ll.828(l0'3)m = 11.828 rr1n1 0.5 in. Since |vm,d| is greater than van, a new section must be selected with sufficient I to satisfy the deflection requirement, Thus,

Iv HI > swrx‘ + Pb(3L2—4b2) ‘"‘ — 38451 4851 Solving for! yields

swrxt

Pb(3L2 - 4112)

I3 L+i 334E|vmi1i| 435 lvruiril

_ 5(500/12)(192)‘ + 1000(72)[3(192)1 -402?] _ 60 Min, — 384(29)(10°)(0.5)

48(29)(10°)(0.5)

—

'

'

The lightest S-shape in Table B-3 (Appendix B) with I Z 60.14 in.“ is an S8 x 23 beam withI = 64.9 in.‘ and S = 16.2 in.3. The S8 x 23 beam satisfies the requirements for flexural stress, shearing stress, and deflection. Consider now the effect of the beam’s weight on the deflection. The S8 x 23 beam weighs 23 lb./ft. Adding the weight of the beam to the 500 lb/ft applied distributed load gives a uniformly distributed load w = 500 + 23 = 523 lb/It, resulting in maximum values of shear force and bending moment of 4809 lb and 19,870 lb - ft, respectively. The value of I required as a result of this increase in load is

SWL4 + 5 Pb(3L2 - 41,1) 12 7 384Elvmidl 48Ell"midl _ 5(s23/12)(192)“ + 1000(72)[3(192)“-4(12)2] _ 62 49, ,, T 384(29)(10°)(0.5)

48(29)(10°)(0.5)

T

' m‘

For the S8 x 23 beam, I = 64.9 in.“ > 62.49 in.4; therefore, the addition of the weight of the beam does not change the selection of the beam. An S8 x 23 beam satisfies all requirements of flexural stress, shearing stress, and deflection.

PROBLEMS Introductory Problems 8 186* A 3 m-long simply supported beam is loaded with a uniformly distributed load of 2 6 kN/m over its entire length. The

beam is made of air-dried Douglas fir (E = 13 GPa) with an allowable flexural stress of 8 MPa and an allowable shearing stress of 0.7 MPa. The maximum deflection at the center of

s-10 nssrsn PROBLEMS 573 the span must not exceed 10 mm. Select the lightest standard structural timber that cart be used for the beam. 8-187* An air-dried Douglas fir (E = 1900 ksi) beam is simply supported and has a span of 16 ft. 'I‘he beam is subjected to a uniformly distributed load of 800 lb/it over its entire length. If the allowable flexural stress is 1200 psi, the allowable shearing stress is 90 psi, and the allowable deflection at the middle of the span is 1/2 in., select the lightest standard structural timber that can be used to support the load. 8-188 A standard structural steel (E = 200 GPa) pipe is to support the load shown in Fig. P8-188. The allowable flexural stress and deflection are 150 MPa and 5 mm, respectively. Select the lightest permissible standard steel pipe that can be used to support the load. Neglect the effects of shear.

Intermediate Problems 8-190* A solid circular shaft made of ASTM A36 steel (E = 200 GPa) is supported by bearings spaced 1.5 m apart. The shaft is to support a 4-kN load perpendicular to the shaft; the load may be placed at any point between the bearings. The allowable flexural stress is 152 MPa, the allowable shearing stress is 100 MPa, and the allowable deflection is 5 mm. If shafts are available with diameters in increments of 5 mm, determine the smallest-diameter shaft that can be used to support the load. Neglect the weight of the shaft. 8-191 A simply supported structural steel (E = 29,000 ksi) beam has a span of 24 ft and carries a uniformly distributed load of 1200 lb/it. The bearn has an allowable flexural stress of 24 ksi, an allowable shearing stress of 14 ksi, and an allowable deflection of' 1 / 360 ofthe span. Select the lightest American standard (S-shape) beam that can be used to support the loading.

Challenging Problems

< 3-in. steel (E = 30,000 ksi) bars, as shown in Fig. P8-l95b. Determine a. The radius of curvature of the beam. b. The deflection at the right end of the beam. c. The deflection 3 ft from the support.

T _zT_.

.

w _t_

x

;A

B‘

_

L

1 in _>

.3‘

I

=1-:

~_-,'l..-"

-< '.'-.116/'-..

w.< 5/32 in. thick. Determine a. The slendemess ratio. b. The Euler buckling load. Use E = 1900 ksi for the wood. 9-6 Determine the allowable compressive load that an 89-minnominal-diameter standard structural steel pipe can support if it is 5 m long and a factor of safety of 2 is specified.

9-7 Determine the allowable compressive load that a 5 >< 5-in air-dried red oak timber can support if it is 18 it long and a factor of safety of 3 is specified.

Intermediate Problems 9-8* A 2.5-m-long column with the cross section shown in Fig. P9-8 is constructed from two pieces of air-dried Douglas fir timber. The timbers are nailed together so that they act as a unit. Determine a. The slendemess ratio. b. The smallest slendemess ratio for which the Euler buckling load equation is valid. c. The Euler buckling load. 50mm 50mm 50mm .=

-‘if.-.3535: ‘ ‘

.:

-4"/' . l-.

150 -

1'1

-

-mm

is?" 2-33-51 ff; - 1-3": 1+?-I 22:11::-isi

v ';.- .--,r

“

’

sum

9-9* A WT6 >< 36 structural steel section (see Appendix B for cross-sectional properties) is used for a 15-it-long column Determine a. The slendemess ratio. b. The smallest slendemess ratio for which the Euler buckling load equation is valid. c. The Euler buckling load. 9-10 A WTl52 x 89 structural steel section (see Appendix B for cross-sectional properties) is used for a 6-m-long column Determine a. The slendemess ratio. b. The Euler buckling load. Use E = 200 GPa for the steel. c. The axial stress in the column when the Euler load is applied. 9-ll Determine the maximum allowable compressive load for a l0-ft-long aluminum (E = 10,000 ksi) column having the

S86 CHAPIER9 COLUMNS cross section shown in Fig. P9-ll if a factor of safety of 2.25 is specified.

l‘‘

250 mm i 25

if.

}>2 in.

25 mm

2 in.—*l G P

100mm

: .,._5'g!-5'_i:.'

ml.

ml;

Figure P9-ll

Figure P9-14

9-12 Two L 127 x 76 x 12.7-mm structural steel angles (see Appendix B for cross-sectional properties) are used for a column that is 4.5 m long. Determine the total compressive load required to buckle the two members if

Challenging Problems

a. They act independently of each other. b. They are riveted together as shown in Fig. P9-12.

l

Figure P9-12

150 mm

9-15* The 2-in.-diameter solid circular column shown in Fig. P9-15 is made of an aluminum alloy [E = 10,000 ksi and or = l2.5(10'°)!° F]. If the column is initially stress free, determine the temperature increase that will cause the column to buckle.

i

10 it

‘i

Figure P9-15

9-13* Determine the maximum allowable compressive load for a 10-it-long aluminum alloy (E = 10,000 ksi) column having the cross section shown in Fig. P9-13 if a factor of safety of 2.50 is specified.

9-16* A 60-kN load is supported by a tie rod AB and a pipe strut BC, as shown in Fig. P9-16. The tie rod has a diameter of 30 mm and is made of steel with a modulus of elasticity of 210 GPa and a yield strength of3 60 MPa. The pipe strut has an inside diameter of 50 mm and a wall thickness of 15 mm and is made of an aluminum alloy with a modulus of elasticity of 73 GPa and a yield strength of 280 Ml-"a. Determine the factor of safety with respect to failure by yielding or buckling for the structure.

3 in_

. 4 in.

<

1.5 m kw .4

a

4in.

_l Z0 m

lie

60 kN

Figure P9-13 9-14 Determine the maximum allowable compressive load for a 6.5-m-long steel (E = 200 GPa) coltnnn having the cross section shown in Fig. P9-14 if a factor of safety of 1.92 is specified.

C‘

Figure P9-16

smzcrs or DIFFERENT IDBALIZED arm commons 9-17 A column 20 ft long is made by riveting three S10 x 25.4 structural steel sections (see Appendix B for cross-sectional properties] together as shown in Fig. P9-17. Determine the maximum compressive load that this column can support. Use E = 29,000 ksi.

The channels are made of structural steel with a modulus of elasticity of 29,000 ksi and a yield strength of 36 ksi. Determine the maximum load P that can be applied to the truss if a factor of safety of 1.75 with respect to failure by yielding and a factor of safety of 4 with respect to failure by buckling are specified.

2P J‘,

587

ix

A

P

1211

B

l8fi

1 c

Figure P9-17 9-18* Two C229 x 30 structural steel channels (see Appendix B for cross-sectional properties) are used for a column that is 12 I11 long. Determine the total compressive load required to buckle the two members if they are laced together back to back 150 mm apart, as shown in Fig. P9-18.

Figure P9-18

l ’ 9 P‘

' ' "

’ 18 ft T ’

Figure P9-19

y

9-20 A simple pin-connected truss is loaded and supported, as shown in Fig. P9-20. All members of the truss are WTl02 x 43 sections (see Appendix B for cross-sectional properties) made of structural steel with a modulus of elasticity of 200

150 mm

a. The factor of safety with respect to failure by yielding. b. The factor of safety with respect to failure by buckling.

i 5-" Liming be"

X

L

GPa and a yield strength of250 MPa. Determine

I

aa

9-19 A simple pin-connected truss is loaded and supported as shown in Fig. P9-19. The members ofthe truss were fabricated by bolting two C10 x 30 channel sections (see Appendix B for cross-sectional properties) back to back to form an H-section.

1. E

4m

15 kN

Figure P9-20

9-5 EFFECTS 0F DIFFERENT IDEALIZED END CONDITIONS The Euler buckling fonnula, as expressed by either Eq. 9-1 or Eq. 9-2, was derived for a column with pinned ends. The Euler equation changes for columns with difierent end conditions such as the four common ones shown in Fig. 9-4. While it is possible to set up the differential equation with the appropriate boundary conditions to determine the Euler equation for each new case, a more

conunon approach makes use ofthe concept ofan effective length. The pin-ended column, by definition, has zero bending moments at each end. The length L in

4m

30 kN

i

'

S88 cnarvrsns commvs i

P

P

P

I

we I

I

‘

\

I

I

I

I

I

.I

I

LE5-=L'=L

I

L’ E 0.71.

I

L’ =-LIZ

I

I

L=L'f2

I

I

f

II Pinned or round ends

Fixed or built-in ends

One free end, one fixed end

One pinned end, one fixed end

(H)

(bl

(C)

(d)

Figure 9-4

the Euler equation, therefore, is the distance between successive points of zero bending moment. All that is needed to modify the Euler column formula for use with other end conditions is to replace L by L’, where L’ is defined as the efléctive length of the colurrm (the distance between two successive inflection points or points of zero moment). The ends of the column in Fig. 9-4b are built in or fixed. Since the deflection curve is symmetrical, the distance between successive points of zero moment (inflection points) is half the length of the column. Thus, the effective length L’ of a fixed-end column for use in the Euler column formula is half the true length

(L’ = 0.5L). The colurrm in Fig. 9-4c, being fixed at one end and free at the other end, has zero moment only at the fiee end. Ifa mirror image of this column is visualizedbelow the fixed end, however, the effective length between points ofzero moment is seen to be twice the actual length of the column (L’ = ZL). The column in Fig. 9-4d is fixed at one end and pinned at the other end. The effective length of this column cannot be determined by inspection, as could be done in the previous two cases; therefore, it is necessary to solve the differential equation to determine the effective length. This procedure yields L’ — 0.7L. Apin-ended column is usually loaded through apin that, as a result offriction, is not completely free to rotate; hence, there will always be an indeterminate moment at the ends ofa pin-connected column that will reduce the distance between the infiectionpoints to a value less than L. Also, it is impossible to support a column

so that all rotation is eliminated; therefore, the effective length of the column in Fig. 9-4b will be somewhat greater than LIZ. As a result, it is usually necessary

9-5 srrlzcrs or DIFFERENT IDBALIZED END CONDITIONS to modify the effective colurrm lengths indicated by the ideal end conditions. The amount of the corrections will depend on the individual application. In summary, the term L/r in all column formulas in this book is interpreted to mean the effective slendemess ratio L’/r. In the problems in this book, the length given for a member is assumed to be the effective length unless otherwise noted.

-I Example Problem 9-4 A structural steel (E = 29,000 ksi) colunm 10 ft long must support an axial compressive load P, as shown in Fig. 9-5. The column has a l x 2-in. rectangular cross section. The left end ofthe colurrm is fixed; the pin and bracket arrangement at the right end allows rotation about the pin but prevents rotation about a vertical axis. Determine the maximum safe load for the column if a factor of safety of 2 with respect to failure by buckling is specified. J-'

Z

.

I It

. ___

p__,.

_—‘_—‘_—‘_——‘l'

Figure 9-5

SOLUTION The relationship between factor of safety and loads is

FS

P“

_ Pall

(a)

For failure by buckling, the ultimate load Pu is the Euler buckling load P”, which

is

e. =

251

(b)

Substituting Eq. (b) into Eq. (a) and rewriting yields 2

Pall =

(L) (F-5')

(5')

The second moment of area I and the effective length L’ depend on the plane in which the column buckles; either the xy- or xz-plane. If buckling occurs in the xy-plane, the left end of the column is fixed and the right end is pinned. Thus, the eflbctive length L’ is L’ = 0.7L = 0.7(l0)(l2) = 84 in.

589

S90 CHAPTER9 COLUMNS and the second moment of area I is

1 1 1 23=0.666"/' .4 1=_6a3=_ 12 ,2()() H1 Therefore, for buckling in the xy-plane,

P,“ = rr2EI , = n1(29)(10°)(0.6667) 2 = 13,522 lb (17) (F3)

(34) (2)

For buckling in the xz-plane, both the lefi end of the colunm and the right end of the column are fixed. Thus, the effective length L’ is L’ = 0.5L = 0.5(l0)(l2) = 60 in. and the second moment of area I is 1 3 = —(2)(l) 1 3 = 0.16667 m. - 4 I = —hb 12 12 Therefore, for buckling in the xz-plane: Pan =

JT2EI 2

(H) (Fm

=

n1(29)(1o°)(o.16667)

(601%)

= 5626 lb

For a colunm with this cross section and end conditions, buckling will occur in the xz-plane and the safe load that may be applied is

P,“ = 6626 lb z 6.63 kip

Ans.

1 PROBLEMS Introductory Problems 9-21* A flnished2 x 4 timber (actual size 1 5/8 in. x 3 112 in. x 10 ft long) is used as a fixed-end, fixed-end column. Ifthe modulus ofelasticity for the timber is 1600 ksi and a factor ofsafety of 3 with respect to failure by buckling is specified, determine the maximum safe load for the colunm. 9-22* An L102 >< 76 x 6.4-mm aluminum alloy (E = 70 GPa) angle is used for a fixed-end, pirmed-end column having an actual length of 3 m. Determine the maximum safe load for the column if a factor of safety of 1.75 with respect to failure by buckling is specified. See Appendix B for cross-sectional properties; they are the same as those for a steel angle of the same size. 9-23 A W8 x 15 structural steel section (see Appendix B for cross-sectional properties) is used for a fixed-end, free-end column having an actual length of 10 ft. Determine the maximum

safe load for the column if a factor of safety of 2 with respect to failure by buckling is specified. Use E = 29,000 ksi. 9-24 Determine the maximum load that a 50-mm x 75-mm x 2.5-m-long aluminum alloy (E = 73 GPa) bar earl support with a factor of safety of 3 with respect to failure by buckling if it is used as a fixed-end, pinned-end column. 9-25* A 6-in. >< 6-in. >< 20-ft-long timber (E = 1900 ksi) is used as a fixed-end, pinned-end colunm to support a 40,000-lb load. Determine the factor of safety based on the Euler buckling load. 9-26* A W254 x 33 structural steel (E = 200 GPa) section is used for a column with an actual length of 6 in The colunm can be considered fixed at both ends for bending about the axis of the cross section with the smallest second moment of area and pinned at both ends for bending about the axis with the largest second moment of area. Determine the maximum axial

smzcrs or DIFFERENT IDEALIZBD arm commons compressive load P that can be supported by the column if a factor of safety of 1.9 with respect to failure by buckling is specified. 9-27 A W10 x 22 structural steel (E = 29,000 ksi) section is used for a colunm with an actual length of 20 Pr. The column can be considered pinned at one end and fixed at the other end for bending about the axis of the cross section with the largest second moment of area and fixed at both ends for bending about the axis with the smallest second moment of area. Determine the maximum axial compressive load P that can be supported by the column if a factor of safety of 3 with respect to failure by buckling is specified.

of the bar permits free vertical movement but no lateral movement or rotation. The bottom of the bar is free to move laterally but cannot rotate. Determine the maximum load P that can be applied if a factor of safety of 2.5 with respect to failure by buckling is specified.

P

1

9-28 A S127 x l5 structural steel (E = 200 GPa) section (see Appendix B for cross-sectional properties) will be used for a 12-m-long pinned-end, pinned-end colunm to support a 60-kN load. Equally spaced lateral braces will be installed to prevent buckling about the weak axis. If the braces offer no restraint to bending of the column and no restraint to buckling about the strong axis, determine

611

a. The spacing required for the lateral braces. b. The maximum load that the column can support once the lateral braces are installed.

Intermediate Problems

is Figure P9-31

9-29* A WT’? x 24 structural steel section (see Appendix B for cross-sectional properties) is used for a column with an actual length of 20 ft. If the modulus of elasticity for the steel is 29,000 ksi and a factor of safety of 2 with respect to failure by buckling is specified, determine the maximum safe load for the coltmtn under the following support conditions. a. b. c. d.

591

Pinned-pinned. Fixed-free. Pinned-fixed. Fixed-fixed.

9-30* A solid circular rod with diameter D, length L, and modulus ofelasticity E will be used to support an axial compressive load P. The support system for the rod is shown in Fig. P9-30. Determine the critical buckling load in terms ofD, L, and E if buckling occurs in the plane of the page.

9-32* A structural steel (E = 200 GPa) bar has a diameter of 50 mm, is 5 m long, and supports an axial compressive load P, as shown in Fig. P9-32. End A is fixed. The support at end B permits free movement in the x- and z-directions but no rotation about the z-axis. Determine the maximum load P that can be applied if a factor of safety of 2 with respect to failure by buckling is specified.

—\= _—n W

PP L3

A

P

P

L

_—'

5m

Figure P9-32 L12

L12

Figure P9-3|]

9-31 The structural steel (E = 29,000 ksi) bar shown in Fig. P9-31 has a 1.0-in. diameter and is 6 ft long. The support at the top

9-33 A 2-in.-diameter by 24-it-long solid, circular alumimun alloy (E 10,000 ksi) bar is used to transmit a 4000-lb force, as shown in Fig. P9-33. If the supports permit buckling only in the plane of the page, determine the factor of safety with respect to failure by buckling.

S92

cmtvrsrts cowuss

“I i

4000 lb

L/3

Figure P9-33

Te

ca0 F

it

9-34 Two 25-mm-diameter structural steel (200 GPa) columns support a 400-kg mass, as shown in Fig. P9-34. The two short struts at the sides of the mass prevent horizontal movement of the mass but offer no constraint to vertical motion. Determine the factor of safety with respect to failure by buckling.

9-37 A free-body diagram for the fixed-end, pinned-end column shown in Fig. 9-4d is shown in Fig. P9-37. Use the freebody diagram to develop the differential equation ofthe elastic curve. Verify that the effective length for the fixed-end, pinnedend colunm is L’ = 0.7L by solving the differential equation of the elastic curve and applying the appropriate boundary conditions.

P

; x

r_

4001;;

-5

I41;

t

F

F

Figure P9-37

3m

9-38 A rigid block is supponed by two fixed-end, fixed-end columns, as shown in Fig. P9-38a. Determine the effective lengths of the columns by solving the differential equation of the elastic curve and applying the appropriate boundary conditions. Assume that buckling occurs as shown in Fig. P9-38b.

I

Figure P9-34

2P

Challenging Problems 9-35 Verify the effective length for the fixed-end, fixed-end column shown in Fig. 9-4b by solving the differential equation of the elastic curve and applying the appropriate boundary conditions. Place the origin of the xy-coordinate system at the lower end of the colunm with the x-axis along the axis of the tmdeformed column. 9-36 Verify the effective length for the fixed-end, free-end column shown in Fig. 9-4c by solving the differential equation of the elastic curve and applying the appropriate boundary conditions. Place the origin of the xy-coordinate system at the lower end of the column with the x-axis along the axis of the undeformed column.

I

I

L ta)

I < 10’

= 5.833l\/[Pa

responding Euler buckling load for this 2E

colunm would be 11, = a = 49s ><

Thus, the allowable load is

P = q,,,(A) = 5.333110°)(150)(200)(10"°) = l74.99(l03)N 2 175.0 kN

(L/1')

Ans-

103 N, which is 2.8 times larger than the allowable load permitted by Code 4 of Table 9-1.

1 PROBLEMS Introductory Problems 9-39* An air-dried red oak (E = 1800 ksi and E = 4.6 ksi) timber colunm with an effective length of 5 R has a 3 >< 3-in. rectangular cross section. Determine the maximum compressive load permitted by Code 4. 9-40* Douglas fir(E = 1 1 GPa and E = 7.6 MPa) timber columns with 200 >< 300-mm rectangular cross sections will be used to support axial compressive loads. Determine the maximum compressive load permitted by Code 4 if a. The effective length of the column is 2 m. b. The effective length of the column is 4 m. c. The effective length of the column is 6 m. 9-41 A 2.5-in.-diameter standard-weight steel pipe colunm is 8 ft long, is pinned at both ends, and supports an axial com-

pressive load P. If E = 29,000 ksi and 0,. = 36 ksi, determine the maximum load permitted by Code 1. 9-42* A W254 x 89 structural steel (E = 200 GPa and 11,. = 250 MPa) column is pinned at both ends, is 3 m long, and supports an axial compressive load P, Detemiine the maximum load permitted by Code 1. 9-43 A 3.0-in.-diameter solid circular 6061-T6 aluminum alloy bar is to be used as a colunm with an effective length of 30 in. Determine the maximum axial compressive load P permitted by Code 3. 9-44 A 2014-T6 aluminum alloy tube with an outside diameter of 100 mm and an inside diameter of80 mm is used for a column with an effective length of 1.0 m. Determine the maximum axial compressive load P permitted by Code 2.

S98 crnn'rax9 commas Intermediate Problems 9-45* Three structural steel bars with a 1 x 4-in. rectangular cross section will be used for an 8-ft-long fixed-ended column. Determine the maximum compressive load permitted by Code 1 if a. The three bars act as independent axially loaded members. b. The three bars are welded together to form an H-colunm. 9-46* Three hollow circular structural steel tubes with inside diameters of50 mm and outside diameters of80 mm will be used for a 3 .5-m-long pin-ended column. Determine the maximum compressive load permitted by Code 1 if a. The three tubes act as independent axially loaded members. b. The three tubes are welded together as shown in Fig. P9-46.

Figure P9-48

9-49* Four L4 x 3 >< 3/8-in. structural steel angles 11 it long are used as a pin-ended colunm. Determine the maximtun load permitted by Code 1 if the angles are welded together to form a 6 x 8-in. box section as shown in Fig. P9-49.

-—

Figure P9-46

l-—6 1n.—l Figure P9-49 9-47 Two C10 x 15.3 structural steel channels 12 ft long are used as a fixed-ended, pin-ended colunm. Determine the maximum load permitted by Code 1 if the channels are welded together to form a 10 x 5.2-in. box section, as shown in Fig. P9-47.

9-50 Two L102 x 76 x 9.5-mm structural steel angles 7 m long are used as a pin-ended colunm. Determine the maximum load permitted by Code 1 if the angles are fastened together to form a 102 >< 76-mm box section as shown in Fig. P9-50.

loinl

5.2in.

'

Figure P9-47

9-48 Four L76 x 76 >< 12.7-mm structural steel angles 5 m long are used as a fixed-ended colunm. Determine the maximum load permitted by Code 1 if the angles are fastened together as shown in Fig. P9-48.

l—r6 mm—l Figure P9-50

9-51 An L5 x 5 x 3/4-in. aluminum alloy 2014-T6 angle will be used as a fixed-ended, pin-ended column to support a load of 120 kip. The cross-sectional properties of steel and aluminum angles are the same (see Appendix B). Determine the maximum permissible length permitted by Code 2.

9-4

9-52* A ccltunn of aluminum alloy 2014-T6 is composed of two L127 x 127 x 19.1-mm angles riveted together as shown in Fig. P9-52. The length between end cotmections is 3 n1, and the end connections are such that there is no restraint to bending about the y-axis; but restraint to bending about the x-axis reduces the effective length to 2.1 m. Determine the maximum axial compressive load permitted by Code 2. The crosssectional properties of aluminmn and steel angles are the same (see Appendix B).

EMPIRICAL COLUMN FORM C LOADING

5

101.6 rmn J’

i

r1_1=48.3 mm I'2_2 =24.l |'l'.|.|'l'|.

f 152.4mm

A=29l0rum2

I

x

- -1

T

i2L3mm —1

l*—20.8 mm

2

Figure P9-54

J?

I

x

Figure P9-52

9-55 Three 2 x 4-in. timber studs are nailed together to form an 8-ft-long column with the cross section shown in Fig. P9-55 The colunm is fixed at the bottom and pinned at the top If E = 1600 ksi and E = 1100 psi, use Code 4 to determine the maximum permissible axial compressive load that may be applied.

|-—4.—-|

9-53* A strut of aluminum alloy 6061-T6 having the cross section shown in Fig. P9-53 is to carry an axial compressive load of 165 kip. The strut is 4 ft long and is fixed at the bottom and pinned at the top. Determine the dimension d of the cross section using Code 3.

*.-'r;f. i:iI.- :; -1" :",l"1'l.l"-A-51

2m’ N)

l15"+5'nL Figure P9-55

.1

—2.z Figure P9-53

9-54 Two L152 x 89 x 12.7-mm angles of aluminum alloy 2014T6 are welded together as shown in Fig. P9-54 to form a 4.75m-long pin-ended column. The pins provide no restraint to bending about the x-axis but reduce the effective length to 3.25 m for bending about the y-axis. The cross-sectional properties of one angle are given on the figure, where C is the centroid ofone angle. Determine the maximum axial compressive load permitted by Code 2.

9-56 A sand bin is supported by four 200 x 250-mm fixed-ended rectangular timber columns 4.5 m long. Assume that the load is equally divided among the four columns and that the column loads are axial. If the columns are made of timber with E = 13 GPa and F, = 9 MPa, determine the maximum load permitted by Code 4.

Challenging Problems 9-57* A 25-ft-long plate and angle colunm consists of four L 5 x 3 1/2 x 1/2-in. structural steel angles riveted to a 10 x 1/2 IIL structural steel plate as shown in Fig. P9-57. Determine the maximum safe load permitted by Code 1 if a. The column is pinned at both ends. b. The colunm is fixed at the base and pinned at the top.

600 CHAPTER9 comruvs '

the effective length to 36 in.. Determine the maximum load permitted by Code 1.

. 10m.

60 in?-1'

'- 3 in. +4 lin

A

r

l

Ii

gm

2 in. 1 in. —x_

O

ii

‘\

_14,- in

L inj l; s1u.—LII

Figure P9-57

Sectionl-1 _A

A

\--_. Pins -

~

Figure P9-59

9-58* Four C178 x 22 structural steel channels 12 m long are used to fabricate a column with the cross section shown in Fig. P9-58. The column is fixed at the base and pinned at the top. The pin at the top offers no restraint to bending about the yaxis; but for bending about the x-axis, the pin provides restraint sufficient to reduce the effective length to 7.5 m. Determine the maximum axial compressive load permitted by Code 1.

9-60 A connecting rod made of SAE 4340 heat-treated steel has the cross section shown in Fig. P9-60. The pins at the ends of the rod are parallel to the x-axis and are 1250 mm apart. Assume that the pins offer no restraint to bending about the xaxis but provide essentially complete fixity for bending about the y-axis. Determine the maximum axial compressive load permitted by Code 1.

J’ J’ l

T‘ i1‘T1

I

Figm_eP9_58

20111111

x 50mm 10mm

10mm

9-59 The machine pa-1 shown in Fig. P9-59 is made ofSAE 4340 heat-treated steel and carries an axial compressive load. The pins at the ends offer no restraint to bending about the axis of the pin, but restraint about the perpendicular axis reduces

30 mm

l

I“

2° mm

II‘

A 50 mm

Figure P9-60

9-5 ECCENTRICALLY LOADED COLUMNS Although a given column will support the maximum load when the load is applied

centrically, it is sometimes necessary to apply an eccentric load to a column. For example, a beam supporting a floor load in a building may in turn be supported by an angle riveted or welded to the side ofa colurrm, as shown in Fig. 9-9. Frequently, the floorbearn is framed into the colunm with a stiff connection, and the colunm

is then subjected to a bending moment due to continuity of the floorbearn and column. The eccentricity of the load, or the bending moment, will increase the stress in the colunm or reduce its load-carrying capacity. Two methods will be

presented for computing the allowable load on a colunm subjected to an eccentric load (or an axial load combined with a bending moment).

9-5 ECCENTRICALLY 1.0110110 cotumrs 601 <

P

x

12;:-_ ii»

y

Figure 9-9

9-5-1 Allowable Stress Method This method is based on the specification that the sum of the direct and bending stresses (P/A + Mc/T) shall not exceed the allowable stress prescribed for a centric load by the appropriate colunm formula, P+Mc<

A

I_aa1l

93

(-)

l.n Eq. 9-3, 0,11 is the allowable stress for a centric load and is calculated using the equations of Table 9-l and using the largest value of the slendemess ratio for the cross section irrespective of the axis about which bending occurs. Values for c and I used in calculating the bending stress, however, depend on the axis about which the bending occurs. Equation 9-3, which is prescribed by most modem codes, usually produces a conservative design. 9-5-2 I1lUB1'€l.Cti0I1 Mfithfld One of the modern expressions for treating combined loads is known as the interaction formula, of which several types are in use. The analysis for compression members subjected to bending and direct stress may be derived as follows. It is assumed that the stress in the colunm can be written as P+Mc< — — cr A I-211 and, when this expression is divided by 0,“, it becomes

PA

M 1

flan

0611

L + if 5 1

(9-4)

When considering eccentrically loaded colimrns, the value of 0,,“ will, in general, be difi'erent for the two terms. In the first term P/A represents an axial stress on a colunm; therefore, the value of 0,“ should be the average allowable stress on an axially loaded column as obtained by an empirical formula such as those presented ir1 Table 9- 1. In the second term, Mcfl represents the flexural stress induced in the member as a result of the eccentricity of the load or an applied bending moment;

602

CHAITER9 cowmrs therefore, the corresponding value of can should be the allowable flexural stress. Since the two values of can are different, one recommended form of Eq. 9-4 is the following interaction formula

PA M 1 L + —c/— 5 1 an

(9-5)

ab

in which P/A is the average axial stress on the eccentrically loaded colunm, 0,, is the allowable average axial stress for an axially loaded column (note that the greatest value of L/r should be used to calculate an), Mc/I is the flexural stress in the column, and ob is the allowable flexural stress. When Eq. 9-5 is used to select the most economical section, it will usually not be possible to obtain a section that will exactly satisfy the equation. Any section that makes the sum of the terms on the lefi side of the equation less than unity is considered safe, and the safe section giving the largest sum (less than unity) is the most efficient section.

i EXHIIIPIC Pfflblfilll 9-7 A W457 x 144 wide-flange section is used for the colunm shown ir1 Fig. 9-10. The column is made of steel (E = 200 GPa, 0}. = 290 MPa, ob = 190 MPa) and has an efiective length of 6 m. An eccentric load P (e = 125 mm) is applied on the centerline of the web as shown in the figure. Determine the maximum safe load according to (a) The allowable stress method. (b) The interaction method.

SOLUTION The cross-sectional properties for a W457 X 14-4 wide-flange section are

A = 18.365 mmz rm = 67.3 m.m

c = 472/2 = 236mm rm = 33."/(10°) m1n4

rum = 199mm

Inm = 728(l0°)rnn14

l, B x

Figure 9-10

Y

9-5

Since the column is made of steel, Code 1 of Table 9-1 will be used to calculate aa|1. The equation for determining can depends on the slendemess range.

Q: /2115 = 12n1(200)(10“) =l16_68

290001)

< 18.75 structural steel sections 25 ft long are laced back to back, as shown in Fig. P9-67, to form a pinended column. Determine the maximum load permitted by the interaction method if the load is applied at point A of the cross section. LetE = 29,000 ksi, 0_.. = 36 ksi, and 0;, = 24 ksi.

9-5 ECCENTRICALLY roman commts 605 3’

75 mm —v

/-- Lacing bars

hT2S mm .25 rnm

4-L

I

O

A

1.5 in. | x

x

II

Figure P9-70 3’

9-71 Two L6 x 3 l/2 x 1/2-in, structural steel (E = 29,000 ksi and 0y = 36 ksi) angles are welded together as shown in Fig. P9-71 to form a column with an effective length of 12 ii. Use the allowable stress method to determine

Figure P9-67

9-68 A hollow square steel member with outside and inside dimensions of 100 mm and 70 mm (the walls are 15 mm thick) functions as a pin-ended colm-nn 4 m long. Determine the maximum load that the colunm can carry if the load is applied with a known eccentricity of 15 mm along a diagonal of the square as shown in Fig. P9-68. Use the allowable stress method and let E = 200 GPa and 0,. = 250 MPa.

a. The maximum axial compressive load P that can be supported by the coltmtn. b. The maximum and minimum values for the distance d when a 50-kip load is applied.

Load

3%in.

applied hcrc \_

+

J’ 3/%ir1.

d

I 15 mm Figure P9-68

Challenging Problems 9-69* A WT8 x 25 structural steel T-section is to be used as a compression member to transmit an eccentric load of 100 kip. The effective length of the member is 10 it. Determine the maximum eccentricity e permitted ifthe point of load application is on the centerline of the stem between the outer edge of the flange and the centroidal axis of the column. Use the interaction method with E = 29,000 ksi, 0, = 36 ksi, and 0,, = 24 ksi. 9-70* Two 50 x 150-mm structural steel bars 5 m long will be welded together, as shown in Fig. P9-70, and used for a pin-ended column. Determine the maximum load permitted if the load is applied at the point on the cross section indicated in Fig. P9-70. Use the allowable stress method and let E = 200 GPa and 0, = 250 MPa.

‘--7 6111.-ei Figure P9-71 9-72 A 2014-T6 aluminum alloy compression member with an effective length of 1.25 m has the T cross section shown in Fig. P9-72. a. Determine the maximum axial compressive load P permitted by Code 2. b. Use the allowable stress method to determine the maximum bending moment M in the yz-plane that can be applied as shown when the column is supporting a 175-kN axial load. P M

_,.__28.2 mm

y

I

|

1.

l

1', = 30.5 mm ry = 20.8 mm

A = 2050 r|:|.m2 Figure P9-72

606 CHAPTER9 commts 9-6 DESIGN PROBLEMS In previous chapters design usually involved strength as a controlling parameter. Since buckling is an elastic phenomenon, the modulus of elasticity (stifliness) is a more significant parameter than is strength (such as yield strength if failure is by yielding) if the column length is in the slender range. If the column is in the intermediate range, both yield strength and stifihess may be important parameters. “Then designing columns using the representative codes listed in Table 9-1, a designer must be aware of several factors. The codes are for specific materials, that is, materials with a specific value of yield strength and modulus of elasticity. In addition, some codes include a factor of safety, while others require that a factor of safety be introduced. Each of the codes has a range of applicability for the slendemess ratio. Finally, all codes in Table 9-1 are for axially loaded members. For example, Code 2 is limited to a specific material, 2014-T6 aluminum alloy. The factor of safety (FS) is included in the code. If the slendemess ratio lies between 12 and 55, the colunm is in the intermediate range and the empirical column formula is valid. If the slendemess ratio exceeds 55, a form of the Euler formula is used with a factor of safety of approximately 1.94 included. For a slendemess ratio less than 12, the axially loaded member is in the compressionblock range where buckling does not occur. In this text, design will be limited to columns subjected to axial loads. Factors such as residual stress, out-of-straightness, and local buckling may be found in design codes established by professional organizations. The following examples illustrate the use of the codes in Table 9-1.

1 EXHIIIPIE Pfflblfllll 9-8 Select the lightest structural steel widefiange section listed in Appendix B to support an axial compressive load of 150 kip as a 15-ft column. Use Code 1. SOLUTION When a rolled section is to be selected to support a specified load, it is usually necessary to make several trial solutions since there is no direct relationship between areas and radii of gyration for different structural shapes. The best section is usually the section with the least area (smallest mass) that will support the load. A minimum area can be obtained by assuming L/r = 0. The loadcarrying capacity of various sections with areas larger than this minimum can then be calculated, using the proper colunm formula, to determine the lightest one that will carry the specified load. If L/r is small,

F8; g = 1.661 140 for steel columns). Short compression members (Lfr < 40 for steel columns) can be treated as compression blocks where yielding occurs before buck-

ling. Columns that lie between these extremes are analyzed by using empirical formulas (column design codes).

1 REVIEW PROBLEMS 9-85* A 20-it-long timber (E = 1200 ksi and 0, = 2.4 ksi) column has the cross section shown in Fig. P9-85. The timbers are nailed together so that they act as a unit. Determine a. The slendemess ratio. b. The smallest slendemess ratio for which the Euler buckling load equation is valid. c. The Euler buckling load. d. The axial stress in the column when the Euler load is applied.

rzu.

2in. T2in.> T

2a-1.

9-87 Determine the maximum compressive load that a W36 x 160 structural steel column (see Appendix B for cross-sectional properties) can support if it is 30 ii long and a factor of safety of 2.24 is specified. 9-88* A 3-in-long column with the cross section shown in Fig. P9-88 is fabricated ii-om three pieces of timber (E = l3 GPa and 0, = 35 MPa). The timbers are nailed together so that they act as a unit. Determine a. The slendemess ratio. b. The smallest slendemess ratio for which the Euler buckling load equation is valid. c. The Euler buckling load. d. The axial stress in the colunm when the Euler load is applied.

F

ii T

T ZOT

6ir1.

T

Z...

ml

I

50mm

-

50mm

A

20 rmn -l

Figure P9-85

Figure P9-88

9-86* Determine themaximum compressive load thataW'I‘l78 X 51 structural steel colunm (see Appendix B for cross-sectional properties] can support if it is 8 m long and a factor of safety of 1.92 is specified.

9-89 Determine the maximum allowable compressive load for a 12-ft-long aluminum alloy (E = 10,600 ksi) column having the cross section shown in Fig. P9-89 if a factor of safety of 2.25 is specified.

612

caxvrarw commas l

4in.

1*;

”

lin.

ix 8 in. 1 in.

Figure P9-91 lin.

lrisail T Figure 1:949

9-92* A structural steel W356>< 122 wide-flange section will be used for a 9-m-long pin-ended colunm. Determine the maximum axial compressive load permitted by Code 1 if

9 90 A 25-mm-diameter tie rod and a pipe strut with an inside diameter of 100 mm and a wall thickness of 25 mm are used to support a 100-kN load as shown in Fig. P9-90. Both the tie rod and the pipe strut are made of structural steel with a modulus of elasticity of 200 GPa and a yield strength of 250 MPa. Determine a. The factor of safety with respect to failure by yielding. b. The factor of safety with respect to failure by buckling.

A

2.5

a. The column is unbraced throughout is total length. b. The column is braced such that the effective length for bending about the y-axis is reduced to 6 m. 9-93 A cold-rolled steel tension bar AB and a structural steel (E = 29,000 ksi and Uy = 36 ksi) compression strut BC are used to support a load P = 100 kip, as shown in Fig. P9-93. Assume that the pins at B and C offer no restraint to bending about the x-axis but provide end conditions which are essentially fixed at C and free atB forbending about they-axis. Select the lightest structural steel tee section listed in Appendix B that canbe used for the strut BC. Use Code 1.

at tn

4 \ B

\-

4.5 rn

100 rm

"

3

ta

fimill

4

6

Figure P9-90

.\5 I00 kip \..§.\

'

\

A

Figure P9-93 9 91* Three S10 x 35 structural steel sections 30 ft long are used to fabricate a column with the cross section shown in Fig. P9-91. The column is fixed at the base and pinned at the top. The pin at the top offers no restraint to bending about the yaxis; but for bending about the x-axis, the pin provides restraint sufficient to reduce the effective length to 20 ft. Determine the maximum axial compressive load permitted by Code 1.

9-94 The compression member AB of the truss shown in Fig. P994 is a structural steel W254 x 67 wide-flange section with the xx-axis lying in the plane of the truss. The member is continuous from A to B. Consider all connections to be the equivalent of pin ends. If Code l applies, determine

asviawrsomalus 613 a. The maximum safe load for member AB. h. The maximum safe load for member/IB ifthe bracing member CD is removed.

y tr‘?--t_ /r \~__~ F

,.

I

/ 6m 1"‘

~__~_

./

UP2 > Up3)- When Mohrs circle (dashed) forany state ofstress is tangent to the envelope of Fig. 10-29b, failure occurs. The principal stresses Up] and 0P3, the ultimate tensile strength a,,,, and the ultimate compressive strength am, are related by the equation

E - E =1 Um‘

Um‘

ttczs)

10-7 FAILURE rusonnzs Fort nnrrrts MA'l'liRIAl.S where 01,, is the magnitude of the compressive strength, and a,,| and op; carry algebraic signs. A state of stress rs safe when

E-2

01>“ DI

Comparison between theory and experimental data (for gray cast iron) is shown in Fig. 10-30, which is adapted from Blake.5 In the first quadrant the p1'incipal stresses have the same sign, and the maxirrunn-normal-stress theory and the Coulomb-Mohr theory are identical. Note that the failure envelope for the maximum-normal-stress theory is not square as was the case for ductile materials, where the tensile and compressive yield strengths were equal (see Fig. 10-19b). The two theories give different predictions of failure in the fourth quadrant, where the principal stresses have opposite sig;ns. In the fourth quadrant, the maximumnormal-stress theory is unsafe because the data lie inside the failure envelope. The Coulomb-Mohr theory is conservative because the data lie outside the failure envelope. The line for pine torsion, op, = —0,,3, is also shown in Fig. 10-30. This line gives two predictions for the ultimate shear strength, that is, where the line intersects the envelopes for the two failure theories. As previously mentioned, one ofthe characteristics of a brittle material is that the ultimate shear strength is approximately equal to the ultimate tensile strength. This is predicted by the n1axirnurn-norrnalstress theory, whereas the Coulomb-Mohr theory predicts a value somewhat less.

om 3 , , -6“ ______ _ _ ‘ ' ‘ A‘ ‘A '/$5

“pr

' C°m‘_1 *3 glgilssltggi C°m°I

6111/“P3 Z *1

A

I

A

%\ AA

' — — — Coulomb—Mohr theory

um:

Figure 10-so 1 EXHIIIPIB Pl‘0blC1I1 1 0- 12 At the critical point in a machine component, the principal stresses (plane stress) are as shown in Fig. 10-31. The failure strengths for the material are am = 2.5 ksi and om = 16 ksi. Use the Coulomb-Mohr theory to determine if this state of stress is safe. SOLUTION The state of stress is safe if the stresses satisfy Eq. 10-27.

4 ksi

UP1 _ Upg < I Um‘

Gar — 2l(Sl

Substituting the principal stresses from Fig. 10-31 (after noting that op; is compressive and that out is the magnitude of the ultimate compressive strength) into Eq. 10-26 yields Figure 10-31

2

(— 4)

25 — — 16 = 1. 05 > 1

Thus, the state of stress shown in Fig. 10-31 is unsafe.

651

Ans.

5Handbook of Mechanics, Materials, and Stmcnmes, Alexander Blake, Ed, Wiley-Interscience, New York, 1985.

652

cnsrmzn 10 runner surmons mo nnzoruss or rmuns

2°? ’““‘

1 Example Problem 10- 13 A 200-mm-diameter solid circular shaft T

T §\ i

§

is subjected torque tensile T, as strength shown inof620 Fig. MPa 10-32a. shaft iscompressive made of a material withto an aultimate and The an ultimate strength of 820 l\/[Pa. Determine the maximum permissible value for the torque T according to the Coulomb-Mohr theory of failtne.

(Q) m t

I

I

SOLUTION The state of stress for a shaft subjected to a torsional load T is shown in Fig. 10-32b. The shearing stress r is given by Eq. 6-6 as

i _i

r=E ?T[0'100) = 636.61" J = 140.200)“/32

(5) Figure 19_3g

The principal stresses a.re op, = —crp3 = 1: = 636.61"

op; = 0

The maximum pemrissible value of T according to the Coulomb-Mohr theory is given by Eq. 10-26 as

an in-, Um‘

URL‘ —

01'

636.6T

(—fi36.6T)

it-i@=1 620(l0 ) 820(10 ) Solving for T yields T = 0.5546(106)N - m E 555kN - m

Ans.

I PROBLEMS Introductory Problems

5 ksi

10-59* The state of plane stress at the critical point in a machine component is shown in Fig. P10-59. The failure strengths for the material are 26 ksi in tension and 97 ksi in compression. Use the Coulomb-Mohr theory to determine if this state of stress is safe.

4 ki _ l I

I L _

Figure P10-59

13 ksi

10-7 FAIIJJIIB 'l'HBORlltS won tt1u'm.s MATIIRIALS l0-60* Two states of plane stress are shown in Fig. P10-60. The failure strengths for the material are 152 MPa in tension and 572 MPa in compression. Use the Coulomb-Mohr theory to determine if these states of stress are safe.

ep, = I00 MPa

653

10-63 The solid circular cast iron shaft shown in Fig. P10-63 is subjected to a torque T = 240 kip - in. and a bending moment M = 110 kip - in. The failure strengths for this material are 43 ksi in tension and 140 ksi in compression Use the CoulombMohr theory to determine the minimum permissible diameter for this shaft.

Gm = 200 MPa M T

6p]=10owa

oP1=75 MP3

0"‘,

M

(-1 Flgune P10-60

(b)

Figure P10-63

l0-61 The state of plane stress at the critical point in a machine component rnade cfcast iron is shown in Fig. P10-61. The failure strengths for this material are 30 ksi in tension and 108 ksi in compression. Use the Coulomb-Mohr theory to determine if this state of stress is safe.

10-65* A solid circular gray cast iron shaft is loaded as shown in Fig. P10-65. The failure strengths for this type of cast iron are 36.5 ksi in tension and 124 ksi in compression. Use the Coulomb-Mohr theory to determine the minimum permissible diameter for the shaft. Neglect the effects of transverse shear.

_9_ksi

I

10-64 A thin-walled cylindrical ]J1‘6SSl.l1’6 vessel has an inside diameter of 300 mm and a wall thickness of 5 mm. The vessel is made of a material with am = 276 MPa and on, = 340 MPa. Use the Coulomb-Mohr theory to determine the maximum internal pressure that the vessel can safely support.

Challenging Problems

lflkfi

i

I

T

I 22 ksi 54ltip

‘r

._,

Steel

Figure P10-68

009-> 45 kip

I8 kip

Figure P10-69

l

'" " so m " 27 kN-m Aluminnm~/

4-00

10 mm ll]-70 A solid circular steel shah ofdiameterd and lengthL is subjected to a constant torque T. Compare the total elastic strain

tuzvurw PROBLEMS 657 energy stored in this shaft with the total elastic strain energy stored in an axially loaded bar of the same size and at the same maximum tensile stress level. ll]-71* A weight of 40 lb is dropped from a height of 3 ft onto the center of a small rigid platform as shown in Fig. P10-71. The two steel (E = 30,000 ksi) rods supporting the platform each have cross-sectional areas of 2.5 in? and are 8 ft long. Determine

10-74* The bronze (E = 80 GPa) beam of Fig. P10-7'4 is 75 mm wide X 25 mm deep. Each of the supporting coil springs has a modulus of 10 kN/m. From what height should the block W, with a mass of 5 kg, be dropped in order to produce a total deflection at the center of the beam equal to four times the deflection produced by the same mass when it is slowly applied to the beam?

a. The impact factor. b. The maximum tensile stress developed in the rods. c. The maximum deflection of the platform.

I

i

zslnm

1.5 m

Wliijl

I;

i

I

1.5 m

Figure P10-‘T4

Lo)

10-75* A vertical energy load is applied to the coil spring (modulus = 200 lb/in.) of the beam-spring system shown in Fig. P10-75. If the spring itself absorbs 720 lb - in. of energy, determine the energy absorbed by the steel (E = 30,000 ksi) beam.

‘(T29

Figure P10-71

ll]-72 An S 305 x 74 structural steel section (see Appendix B) is used as a simply supported beam 3 m long. What weight falling on the center of the span fiom a height of 0.75 rn will produce a maximum flexural stress of 120 MPa in the beam? The web of the beam is vertical.

U

ll]-73 When the 90-lb block B of Fig. P10-73 was dropped from a point 2 in. above the top of the coil spring C (modulus = 1700 lb/ft), point D on the steel (E = 29,000 ksi) cantilever beam A was observed to deflect 2.4 in. downward. Determine the impact factor.

11%

It-J to

s- . . ;..ii‘ —Ii

Figure P10-75

Ii Bj2in. 3 in.

\

C

i_l 1 mini: D

ll

Figure P10-73

5fl

10-'76 The frame shown in Fig. P10-76 is subjected to a load P of 100 kN. The material is ductile and has a proportional limit of 220 MPa in tension and compression. Determine the factor of safety with respect to failure by yielding according to

>

a. The maxirnum-shearing-stress theory of failure. b. The rnaxirnum-distortion-energy theory of failtue.

658

cnarrax 10 manor snrrnons AND rnsonrss or I~'AlI.IJ]lB

50mm

P

350mm

50"“

P

_50mm A—-

-—A

= l

150mm

50 mm Section A-A

a. The principal stress 0,; = 12 ksi tension. b. The principal stress op; = 12 ksi compression 1|]-78 A state ofplane stress at the critical point in a machine component is shown in Fig. P10-78. The material is brittle with om = 68 MPa and 0,, = 206 MPa. Use the Coulomb-Mohr theory of failure to determine if the state of stress is safe.

Figure P10-76

10-77* A material with a proportional limit of 36 ksi in tension and compression is subjected to a biaxial state of stress. Two of the principal stresses are cpl = 20 ksi (T) and op, = 0, Determine the factor of safety with respect to failure by yielding according to each of the theories of failure if

i~40Nl.P8

50 MPa Qi

Figure P10-78

Appendix A Second Moments ofArea

A-1 INTRODUCTION The centroid of an area is located by considering the first moment of the area about an axis. This computation requires evaluation of an integral of the form L x dA. in the analysis of stresses and deflections in beams and shafts, an expression of the form ff! x2 dA is frequently encotmtered in which dA represents an element ofarea and x represents the distance from the element to some axis in, or perpendicular to, the plane of the area. An expression of the form L1 x2 dA is known as the second moment of the area. In the analysis of the angular motion of rigid bodies, an expression of the form fm r2 dm is encountered, ir1 which dm represents an element of mass and r represents the distance from the element to some axis. Euler‘ gave the name “moment of inertia" to expressions of the form fm r2 dm. Because of the similarity between the two types of integrals, both have become widely known as moments of inertia. In this text, the integrals involving areas will be referred to as “second moments of area.” Methods used to determine second moments of area are discussed in this appendix.

A-2 SECOND MOMENT OF PLANE AREAS

z

The second moment of an area with respect to an axis will be denoted by the symbol I for an axis in the plane of the area and by the symbol J for an axis perpendicular to the plane of the a.rea. The particular axis about which the second moment is taken will be denoted by subscripts. Thus, the second moments of the area A shown in Fig. A-1 with respect to x- and y-axes in the plane of the area are

L,=fy2a‘A A

and

Iy=fx2dA

(A-1)

O _ __r__\ \

J’

x

ax

2:35’ dA

as .4

Figure A-l

A

The quantities I, and I}. are sometimes referred to as the rectangular second moments of the area A. Similarly, the second moment of the area A shown in Fig. A-1 with respect to a z-axis, which is perpendicular to the plane of the area at the origin O of the lleeonhard Euler (1707-1783), a noted Swiss mathematician and physicist.

659

APPENDIX A

SECOND MOMENTS OF AREA

xy-coordinate system, is

J, = I flax = I (11 + y2) as

(A-2)

J, = f x2dA + I y2a'A = 1,. + L.

(A-3)

A

A

Thus,

A

A

The quantity J, is known as the polar second moment of the area A. The second moment of an area can be visualized as the sum of a number of terms, each consisting of an area multiplied by a distance squared. Thus, the dimensions of a second moment are a length raised to the fourth power (L“). Common imits are IIHI14 and in.‘ Also, the sign of each term summed to obtain the second moment is positive, because either a positive or negative distance squared is positive. Therefore, the second moment of an area is always positive.

A-2-1 Parallel Axis Theorem for Second Moments of Area

y:

Vifhen the second moment of an area has been determined with respect to a given axis, the second moment with respect to a parallel axis can be obtained by means of the parallel axis theorem. ifone of the axes (say the x-axis) passes through the centroid of the area, as shown in Fig. A-2, the second moment of the area about a parallel x’-axis located a distance d, from the x-axis is

-\< ——Arca A

. or-+- -" 5; g.=~ M |I

I

I

I,»=f(y-l-d,,)2dA=fy2dA+2d,fydA+dff dA

IH

A

A

A

A

\\TT wk.

>‘\

since d, is the same for every element of area dxl. The first integral is the second moment L, of the area with respect to the x-axis and the last integral is the total area A. Therefore,

Figure A-2 I,»=1}+2d,[ydA+dfA A

(a)

The integral ff, y dA is the first moment ofthe area with respect to the x-axis. Since the x-axis passes through the centroid C of the area, the first moment is zero and Eq. (a) becomes

1,, = 1,6 + dfx

(A-4)

where Ixc is the second moment of the area with respect to the x-axis through the centroid and ti, is the distance between the x- and Jr’-axes. l.r1 a similar manner it can be shown that

J, = 1,6 + (df + dj) A = 1,6 + d§A

(A-5)

where J,3 is the polar second moment of the area with respect to the z-axis through the centroid and d, is the distance between the 2- and 2'-axes. The parallel axis theorem states that the second moment of an area with respect to any axis ir1 the plane of the area is equal to the second moment of the area with respect to a parallel axis through the centroid of the area added to the

A-2

SECOND MOMENT or PLANE AREAS

product of the area and the square of the distance between the two axes (refer to Eqs. A-4 and A-5). The theorem also indicates that the second moment of an area with respect to an axis through the centroid of the area is less than that for any parallel axis because

1,6 = 1,,» - dfx

F X

(A-6)

*1 t—”’“ dA

As a point of caution, note that the parallel axis theorem (Eq. A-4) is valid only for transfers to or fi'om a centroidal axis. That is, if x’ and x” are two parallel axes (neither of which passes through the centroid of an a.rea), then

r,- = 1,6 + df,A = (1,, - df,/1) + d§,A = 1,, + (43, - d},)A as 1,» + (dfl - d,1)2A

661

“Areal! I

O

(H) P‘

A-2-2 Second Moments of Areas by Integration Rectangular and polar second moments of area were defined in Section A-2. When the second moment of a plane area with respect to a line is determined by using Eq. A-1 or A-2, it is possible to select the element of area dA in various ways and to express the area of the element i11 terms of either polar or Cartesian coordinates. In some cases, an element of area with dimensions a'A = dy dx, as shown in Fig. A- l, may be required. This type of element has the slight advantage that it can be used for calculating both 1,, and iv, but it has the greater disadvantage of requiring double integration. Most problems can be solved with less work by choosing elements of the type shown in Figs. A-3a and A-3b. The following should be considered when selecting an element of area dA for a specific problem. 1. If all parts of the element of area are the same distance from the axis, the second moment can be determined directly by using Eq. A-1 or A-2. Thus, the element shown in Fig. A-2 can be used to determine either L or Iy directly, but a double integration is required. The element shown in Fig. A-3a can be used to determine directly because the dimension x is constant for the element. The element shown in Fig. A-3a is not suitable for determining I, directly because the y-dimension is not constant for the element. Similarly, the element shown in Fig. A-Sb is suitable for determining L, directly but not Iy. A single integration would be required with elements of the type shown in Fig. A-3. 2. If the second moment of the element ofarea with respect to the axis about which the second moment of the area is to be found is known, the second moment ofthe area can be found by summing the second moments of the individual elements that make up the area. For example, ifthe second moment til, for the rectangular area d/1 in Fig. A-4 is known, the second moment L, for the complete area/1 is simply L, = fl, dlx. 3. If both the location of the centroid of the element and the second moment of the element about its centroidal axis parallel to the axis of interest for the complete area are known, the parallel axis theorem can ofien be used to simplify the solution of a problem. For example, consider the area shown in Fig. A-5. If both the distance d, and the second moment dlxc forthe rectangular element dA are lcnown, then by the parallel axis theorem d1, = dlxc + dfdA. The second moment for the complete area A is then simply L = [A dlx.

K Area A

s

Ls ._e_-J

0

I

(5) Figure A-3 J’

.7? /AreaA dA =ydx I

0

Figure A-4

J’ /—Al‘6fi A

dz!

if x‘. 41 O

Figure A-5

l .

APPENDIX A

SECOND MOMENTS OF AREA

From the previous discussion it is evident that either single or double integration may be required for the determination of second moments of area, depending on the element of a.rea dA selected. Vlfhen double integration is used, all parts of the element will be the same distance from the moment axis, and the second moment of the element can be written directly. Special care must be taken ir1 establishing the limits for the two integrations to see that the correct area is included. If a strip element is selected, the second moment can usually be obtained by a single integration, but the element must be properly selected in order for its second moment about the reference axis to be either known or readily calculated by using the parallel axis theorem. The following exarnples illustrate the procedure for determining the second moments of areas by integration.

J’

dA=bdy

dy

i Example Pfflbifilll A-1 Determine the second moment for the rectangle shown ir1 Fig. A-6a with respect to (a) The base of the rectangle. (b) An axis through the centroid parallel to the base. (c) An axis through the centroid normal to the area.

:a~

~._l;e

@e__b___l

‘

(H) Figure A-6(a) P The second moment I, can be computed with a single integration if a thin strip element parallel to the x-axis is used. J’

h

C oi

Ix,

l

(a) An element of area dA = bdy, as shown in Fig. A-6a,will be used. Since all parts of the element are located a distance y fi"om the x-axis, Eq. A-l can be used directly for the determination of the second moment L, about the base of the rectangle. Thus,

:1

3 iv

I, =/ly2dA=f y2bdy=

A

0

3

0

3 = &

3

Ans.

This result will be used fi'equently in later examples, when elements of the type shown ir1 Fig. A-4 are used to determine second moments about the x-axis. (b) The parallel axis theorem (Eq. A-6) will be used to determine the second moment [,5 about an axis that passes through the centroid of the rectangle (see Fig. A-6b) and is parallel to the base. Thus,

L 2

O Le 1» el

SOLUTION

1’°_"_"_3_2()_12 —r a'2A—E ’121>h—@ x

(b) Figure A-6(b) P If an area has an axis of symmetry, the centroid is located on that axis; if the area has two axes of symmetry, the centroid is located at the point of intersection of the two axes.

A "5'

This result will be used frequently in later examples when elements of the type shown in Fig. A-5 are used to determine second moments about the x-axis. (c) The second moment 13.; for the rectangle can be deterrnined in an identical manner. It can also be obtained fiom the preceding solution by interchanging b and h; that is,

hb3

@@=rE

A-2

SECOND MOMENT OF PLANE AREAS

The polar second moment J2;-; about the z-axis through the centroid of the

rectangle is given by Eq. A-3 as

bhi’

hbi’

an

LC = LC +I_t¢ = E + E = E012 +112)

Ans. .3’

P

/Kxde

d

Example PTOOICIII A-2 Detemiine the second moment of area for the circle shown in Fig. A-7 with respect to a diameter of the circle.

1' r’

SOLUTION Polar coordinates are convenient for this problem. An element of area dA = pd9 dp, as shown in Fig. A-7, will be used. If the x-axis is selected as the diameter about which the second moment of area is to be determined, then y = p sin 9. Application of Eq. A-1 yields 211

=

O

\\\'°

y

= P S111' e X

R

R

t=ffiM=f f A

6 \\\\

Figure A-7

U

2" R 3 '29d d9=—[———:l R“ 0 sin20 2” L _£'° S1“ ‘J 4 2 4 U

ma‘ 4

=i

A115.

P Polar coordinates are usually more efficient when circularboundaries are involved.

1

EXHIIIPIB PPOIJIBIII A-3 Determine the polar second moment of area for the circle shown in Fig. A-8 with respect to

Y

(a) An axis through the center of the circle and normal to the plane of the area. (b) An axis through the edge of the circle and normal to the plane of the area.

SOLUTION (a) Polar coordinates are convenient for this problem. An element of area dA =

X

2:rp dp, as shown ir1 Fig. A-B, will be used. Since all parts of the element are located a constant distance p from the center of the circle, Eq. A-2 can be used directly for the determination of the polar second moment J, about an axis through the center of the circle and normal to the plane of the area. Thus, Figure A-8 R

R

NR4

J, = I r2 dA = f p2(22rp dp) =1 211,03 dp = — A 0 0 2

Ans.

This result could have been obtained from the solution of Example Problem A-2 and use of Eq. A-3. Thus, rrR“

rrR“

rrR4

#=4+@=:r+1r=‘? (b) The parallel axis theorem (Eq. A-5) will be used to determine the polar second moment J, about an axis that passes through the edge of the circle and is

P The second moment J, can be computed with a single integration if a thin armular element at a constant distance p from the z-axis is used.

664 APPENDIX A SECOND MOMENTS or AREA no1'rr1alto the plane of the area. Thus,

R4

3 R4

J, = 1,,-+ df/I = % + R2(rrR2) = “T

Y

Ans.

i EXHIIIPIB PIOOICIII A - 4 Determine the second moment of area for the shaded region of Fig. A-9 with respect to

l

ll’

lin.

fir

y2=2x

-V

1 in.

l

l ,

(a) The x-axis. (b) An axis through the origin of the xy-coordinate system and normal to the plane of the area. SOLUTION (a) An element of area dA = x dy = (yz/2) dy, as shown in Fig. A-3b, will be used. Since all parts of the element are located a distance y from the x-axis, Eq. A-l can be used directly for the determination of the second moment I,

about the x-axis. Thus,

|*i2in.;

Figure A-9

yz

A

P The second moment I, can be computed with a single integration if a thin strip element parallel to the x-axis is used.

2 y4

ys 2

2

10 1

1,=f y2dA=fy2(—)dy=f —dy= [-1 =3.10tn:‘ A

2

1

Ans. (b) The same element of area can be used to obtain the second moment I, if the result of Example Problem A-1 is used as the known value for dly. Thus,

bk’ dye)’ dye’-‘/2)’ y“

dI’:3:

3

:

3

:24”

Sunmiing all such elements yields

2 ye ly =L dly =fi

y7 =

Z

127

.

= fi =0.756ll1.4

Once I} and 13, are known for the area, the polar second moment for an axis through the origin of the my-coordinate system and normal to the plane of the

area is obtained by using Eq. A-3. Thus, J, = I, + I), = 3.10 + 0.756 = 3.856 E 3.86 in.4

Ans.

A-2-3 Radius of Gyration of Areas Since the second moment ofan area has the dimensions of length to the fourth power, it can be expressed as the

A-2

area A multiplied by a length k squared. Thus, from Eqs. A-l and A-2,

1,:

y2dA=Akf

k,= —

x2a‘A=Akj

r,.= — k\[451 ,,[45( y>] 3 dy fi"omwhich +30

1

3

[M =LdI_vC =l15 E|:E(30—y)] dy

--l _4s E 50

300- y )4 +3U—1171ss 45 _15— ‘ mm“

From the definition of a radius of gyration,

11,6 —

Ans.

1. ~ 117,183 N t,..¢- = ,1 fit = ti i = l0.206mm = 10.21 rr1m

Ans.

A

(bl

/126,563

= l0.607mm E 10.61 nun

kxc =

=

1125

The radii of gyration k,- and k,,- can be determined by using the parallel axis theorem for radii of gyration. Thus,

it,» =

+ er; = ,/(10.s07)2 + (40? = 4-l.4m.n1

Ans.

12,,» = ,/kfc + er; = ,/(10.20s)’ + (4012 = 41.3mm

Ans.

A-2

SECOND MOMENT OF PLANE AREAS

A-2-4 Second Moments of Composite Areas The second moments IL, I,., and J, of an areaA with respect to any set ofx-, y-, and z-coordirrate axes were defined as

L,=fy2dA

1,=fs’aA

J,=[1-"ax

A

A

A

Frequently in engineering practice, an irregular area A will be encountered which can be broken up into a series of simple areas A1, A2, A3, . . ., A, for which the integrals have been evaluated and tabulated. The second moment of the irregular area, the composite area, with respect to any axis is equal to the sum of the second moments of the separate parts of the area with respect to the specified axis. For example,

L=fy2dA A

=fy1e1A,+f y2dA2+f y2dA3+~-+f y2dA,, A,

/1

.1.

A,

=Ixl+Ix2+Ix3+"'+Ixn

When an area such as a hole is removed from a larger area, its second moment must be subtracted fi'om the second moment of the larger area to obtain the resulting second moment. Thus, for the case of a square plate with a hole,

r =1 +1 I El Therefore,

1

l

=1_ I

Table A-l contains a listing ofthe values ofthe integrals for frequently encotmtered shapes such as rectangles, triangles, circles, and semicircles. Tables listing second Table A-1

y‘

T

Second Moments of Plane Areas

b

.v

bk,

2

Ix

'2

I X

bhi 3

A = an

E _x'

Y

IJI13

1»

I?

,1, 3

‘ kt" x.

L--34-;-_

1‘

ZR

1,113 I-=— * 12

_1 A_2bh

APPENDIX A

Table A-1

SECOND MOMENTS OF AREA

Second Moments of Plane Areas (continued)

J"

1,, = 5rrR“ -4-

rrR"

il

A = rrR2 I

R It

J"

J-rR“

SR4

‘X = T ‘ w :rrR4

x

I

4" = Y

R4

;.=”_

X

s 1 .4=—231'R2

E 311;

xi

y‘

‘R F

J’

R4

1 _;-[R4 " 16

I

4R‘ 9n‘

1,=L x

I6

1 R2 A=—471'

.1

E

3“-

mm‘

‘-:

,'i\

I,=T(6——s in

“~,=e ‘l

Le

H

1,. = as-=1 e + —sin /""\

2Rsin6

“=5? A =6R2

IQ Ix.) »—~~—~

moments of area and other properties for the cross sections of common structural shapes are found in engineering handbooks and in data books prepared by industrial

organizations such as the American Institute of Steel Construction. An abbreviated listing is also included in Appendix B. In some instances the second moments IXC, I,.C, and I,3 of a composite shape with respect to centroidalx-, y, and z-axes of the composite may be required These quantities can be determined by first evaluating the second moments IX», Iy» , and I2» of the composite with respect to any convenient set of parallel x’-, y I -

and z’-axes and then transferring these second moments to the centroidal axes by using the parallel axis theorem.

A-2

SECOND MOMENT OF PLANE AREAS

Y 1 EXHIIIPIB PIOIJICIII A-7 A beam having the cross section shown in Fig. A-12a is constructed by gluing a 2 X 6-in. wooden plank 10 it long to a second 2 x 6-in. wooden plank also 10 It long. Determine the second moment of the cross-sectional area with respect to

5' E

(a) The x-axis.

"

(b) The y-axis. (c) The 3113- axis, which passes through the centroid of the area and is parallel to the x-axis.

in ‘I ..

fir . 6 m‘

lis

—'*—r

(11) SOLUTION (a) As shown in Fig. A-12b, the cross-sectional area can be divided into two simple rectangles. Since both rectangles have an edge along the x-axis, their second moments ofarea are just bh3/3. Therefore, the second moment ofarea of the entire area with respect to the x-axis is

1

1

A = 1,1 + 1.2 = 5(4)(2)‘* + 5(2)(8)’ = 352 in.‘

y

Ans.

C

_T d

(b) Using the sanre division of areas as in part (a), the first area has an edge along the y-axis and its second moment of area is given by bh3/3. However, the parallel axis theorem is needed for the second rectangle, since neither the centroid of the rectangle nor either edge is along the y-axis. Therefore, the second moment of area of the entire area with respect to the y-axis is Iy = I,,| +I_,,2

1 3 1 3 2 _ 4 = 5(2)“) + E(8)(2) + 16(5) = 448 111-

A115-

(c) The centroid will be located using the same division of areas as above: d(8 + 16) = 1(8) + 4(16)

d = 3 in.

Then, using the parallel axis theorem for both rectangles, the second moment of area of the entire area with respect to the xc-axis is

IxC = IxCl + IxC2 =

1

3

+ 8(3 — 1)

2

+

1

3

+ l6(4 — 3)

2

- 4

= 136.0 111.

Ans. Note that, since the If-HXIS passes through the centroid of the entire area, the second moments of area I, and IXC are related by the parallel axis theorem L = Lg? -l- Adz

= 136.0 + 24(3)2 = 352 in.“

(b)

X

Figure A42

P The second rnornent of a composite area with respect to a given axis equals the sum of the second moments of the separate parts of the area with respect to the same axis.

APPENDIX A

SECOND MOMENTS OF AREA

i EXHIIIPIB PPOIJICIII A-8 Determine the second moment ofthe shaded area shown in Fig. A-13a with respect to (a) The x-axis.

(b) The y-axis. (c) An axis through the origin O of the xy-coordinate system and normal to the plane of the area.

y I

1001m'n

a no 0“ 1"

lso 100mm J

___________ _ m 60mm ii

' —,_— so rmn L

P150 mm

Figure A-13

c

.4

I

50 mm

(=1)

(b)

SOLUTION As shown in Fig. A-13b, the shaded area can be divided into a 100 x 200-mm rectangle (A), with a 60-mm-diameter circle (B) and a 100-mm-diarneter halfcircle (C) removed, and a 100 x 200-mm triangle (D). The second moments for these areas, with respect to the x- andy-axes, can be obtained by using information

fi'om Table A-1, as follows: (a) For the rectangle (shape A),

1003 = 66.66"/(10“)mn1“ 1,1 = bh3 T = 200 —%—2 For the circle (shape B), R4

1,2 = Iec + 4521 = 51- + d§(eR2) = @ +(s01)(n)(301) = "/.70s(10‘=)nnn“ For the half-circle (shape C), R4 R2 1,3 = 1,43 —i-dg/1 = % +d_3(%)

= L204) +(s0)1{L;°)2] = 12.212(1o°) nnn‘

A-2 SECOND MOMENT or PLANE AREAS

673

For the triangle (shape D), Ixli = I.rC + at?!‘

bh3

bh

... + -( . )

=_

41 _

200 1003

= Q+ 36

100 2 200 100

100 + — 3

Q =183.333(l0°)mm“ 2

For the composite area, 1:1" : [xi _ I12 _ Ix3 + Ix-1

= 66.s61(10°) - 7.70s(1o“)- 12.2"/2(1o6) + 1s3.333(10°')

= 230.o23(106) 2 23o(1o@) mm‘

Ans.

(b) For the rectangle (shape A),

I b3h 2003000) , , ,1 = T = f =266.s61(10 )n1m For the circle (shape B),

1,2 = on + djzl R4

= % + siren’) = @ + (1502)(n)(302) = s4.2s3(10°)nnn** For the half-circle (shape C),

R4

50“

1,, = % = $ = 2.4s4(10°)nnn‘* For the triangle (shape D),

bh3

100 2003

1,, = E = % = 66.6s1(10‘)nnn" For the composite area, I2 = Iyl _ I21 _ 1&3 + I24

= 2es.6s7(106) - 64.253(1o'=) - 2.4s4(10°)+ 6s.61s7(10@) = 26s.e27(1o6) 2 2s7(1o@) IIIIH4

Ans

P When a hole is present in a larger area, its second moment must be subtracted fiorn the second moment ofthe larger area in order to obtain the required second moment.

674 APPENDIX A snconrn MOMENTS or AREA (c) For the composite area,

J2 = Ix + Iy

= 230.o23(10°) + 266.621(10‘) = 496.ss0(10“) '5 497(10‘)mm“

Ans-

“ Example A-9 A column with the cross section shown in Fig. A-14a is constructed from a W24 x 84 wide-flange section and a C12 x 30 channel section. Determine the second moments and radii of gyration of the cross-sectional area with respect to horizontal and vertical axes through the centroid of the cross section. Cl2x3O

- W24 X s4 (11) 0‘m T3F‘:

eoolsou

_,__\.=

,' 1 1-'8_8°i

l

0.510 in.

J 'IIII

I I FEI I I5'I I |—-Q . I_ | I I

5'; In

" C ‘ I | I I I

I

I

In“H

—>I

3.12sm. H

I_.—-_.-

i-—S—M—3—*|

A = 24.7 in.F‘-'PII\-I 1,, = 2310 in.‘ 1}, = 94.4 in.‘ (*5)

A = s.s2 in? 1,. = 162 in.‘ 1,.= 5.14 mi‘ (C)

(J)

Figure A-14 SOLUTION Properties and dimensions for the structural shapes can be obtained from Appendix B. The properties and dimensions for the wide-flange section are shown in Fig. A-14b. The properties and dimensions for the channel section are shown in Fig. A-14c. In Figs. A-14-b and c, the x-axis passes through the centroid of the wide-flange section and a parallel x’-axis passes through the centroid of the channel. A centroidal xg-axis for the composite section (see Fig. A-14d) can be located by using the principle of moments as applied to areas. The total area A 1' for the composite section is Ar = Awp + AC1; = 24.7 + 8.82 = 33.52 in.2

A-2

The moment of the composite area about the x-axis is A7"(_)I(_')7' = Aw;-{yc)w.r + Ac;-1(y¢-)¢-H = 24.7(O) + 8.82(11.3S6) = 104.835 in.3

snconn MOMENI‘ or PLANE mass 675 P The principle of moments can be used to locate the centroid of any composite body if first moments of the individual parts are known or can be determined.

The distance (y¢)¢ from the x-axis to the centroid of the composite section is

104.235

,

U’c);- = g = 3.12s m. The second moment [I,,C)W,- for the wide-flange section about the centroida1xCaxis of the composite section is determined by using the parallel axis theorem. Thus,

P The parallel axis theorem for second moments is valid only for transfers to or from a parallel axis through the centroid of the area.

(I,C)w;.- = (I,,)W;- + (_VQ')%’,-I,-AW]-' = 2370 +(3.128)2(24.7) = 2611.7 in.‘ Similarly, the second moment (I,C)¢;.; for the channel about the centroidal 2:5axis of the composite section is (IxC)CH = Ux')c;-1 + l(yC)r;'1-1 _ (yC)T]2ACH = 5.14 +(11.886 — 3.128)2(8.82) = 681.7 in.4 For the composite area, (1xC')r = (IIC)W1-‘+[[XC)(.‘11' = 2611.7 -1- 681.7 = 3293.4 '5 3290111.‘

ADS.

The y-axis passes through the centroid of both areas; therefore, the second moment (IN); for the composite section is

(*ry6')1' = (INCJWF + Uy6')c11 = 94.4 + 162 = 256.4 E 256in.4

Ans.

The radius of gyration about the xg-axis for the composite section is

(1 -)- “Z = 3293.4 ~ , (k,C)T=[% 33-52 ‘/2 =9.912=9.91 m.

Ans.

The radius of gyration about the yc-flXlS for the composite section is

(1 -) "2 256.4 ‘/2 N _ (k,.¢).»-=[% = 33.52 =2.766=2.771n.

}’

Ans‘

x

"""""7

A-2-5 Mixed Second Moments ofAreas The mixed second moment (commonly called the area product of inertia) dlxy of the element ofarea dA shown in Fig. A-15 with respect to the x- and y-axes is defined as the product of the two coordinates of the element multiplied by the area of the element; thus dlxy = xydA

as Ir I

O

Figure A-l5

x

APPENDIX A

SECOND MOMENTS OF AREA

The mixed second moment (area product of inertia) of the total area A about the x- and y-axes is the sum of the mixed second moments of the elements of the area; thus,

1,_.= I d1,.,= I xydA A

dA'

The dimensions of the mixed second moment are the same as the dimensions for the rectangular or polar second moments, but since the product xy can be either positive or negative, the mixed second moment can be positive, negative, or zero. Recall that rectangular or polar second moments are always positive. The mixed second moment of an area with respect to any two orthogonal axes is zero when either of the axes is an axis of symmetry. This statement can be demonstrated by means of Fig. A-16, which is symmetrical with respect to the y-axis. For every element of area dz! on one side of the axis of symmetry, there is a corresponding element of area d A‘. on the opposite side of the axis such that the mixed second moments of dA and dA' will be equal in magnitude but opposite in sign. Thus, they will cancel each other ir1 the summation and the resulting mixed second moment for the total area will be zero. The parallel axis theorem for mixed second moments can be derived from Fig. A-17 in which the x- and y-axes pass through the centroid C of the area and are parallel to the x’- and y’-axes. The mixed second moment with respect to the x’- and y’-axes is

dd .1

Figure A-16

d)‘

X

(A-12)

A

i

I,-,.- =1’; x'y'dA = L (dy +x)(d, +y)dA

I dd i

. 1

C

W

y

x

=d,,d,.f dA+d_,f ydA+d,f xdA+fxydA A

_ _ _ ._~< >< C15 x x x C12 X x >< C10 x x x x C9 >< x >< C8 x x x C7 x >< x C6 x x x C5 >< x C4 >< x C3 >< X x

17.1 15.3 13.5 12.6 14.7 1 1.8 9.96 8.82 7.35 6.09 8.82 7.35 5.88 4.49 5.88 4.41 3.94 5.51 4.04 3.38 4.33 3.60 2.87 3.83 3.09 2.40 2.64 1.97 2.13 1.59 1.76 1.47 1.21

Area

58 51.9 45.8 42.7 50 40 33.9 30 25 20.7 30 25 20 15.3 20 15 13.4 18.75 13.75 11.5 14.75 12.25 9.8 13 10.5 8.2 9 6.7 7.25 5.4 6 5 4.1

18.00 18.00 18.00 18.00 15.00 15.00 15.00 12.00 12.00 12.00 10.00 10.00 10.00 10.00 9.00 9.00 9.00 8.00 8.00 8.00 7.00 7.00 7.00 6.00 6.00 6.00 5.00 5.00 4.00 4.00 3.00 3.00 3.00

Flange

web

Width Thickness

Thickness

(in.]

4.200 4.100 4.000 3.950 3.716 3.520 3.400 3.170 3.047 2.942 3.033 2.886 2.739 2.600 2.648 2.485 2.433 2.527 2.343 2.260 2.299 2.194 2.090 2.157 2.034 1.920 1.885 1.750 1.721 1.584 1.596 1.498 1.410

(in.)

0.625 0.625 0.625 0.625 0.650 0.650 0.650 0.501 0.501 0.501 0.436 0.436 0.436 0.436 0.41 3 0.413 0.413 0.390 0.390 0.390 0.366 0.366 0.366 0.343 0.343 0.343 0.320 0.320 0.296 0.296 0.273 0.273 0.273

0.700 0.600 0.500 0.450 0.716 0.520 0.400 0.510 0.387 0.282 0.673 0.526 0.379 0.240 0.448 0.285 0.233 0.487 0.303 0.220 0.419 0.314 0.210 0.437 0.314 0.200 0.325 0.190 0.321 0.184 0.356 0.258 0.170

Axis X_X I (iIt.4)

676 627 578 554 404 349 3 15 162 144 129 103 91.2 78.9 67.4 60.9 51.0 47.9 44.0 36.1 32.6 27.2 24.2 21.3 17.4 15 .2 13 .1 8.90 7.49 4.59 3 .85 2.07 1.85 1.66

S

75.1 69.7 64.3 61.6 53.8 46.5 42.0 27.0 24.1 21.5 20.7 18.2 15.8 13.5 13.5 11.3 10.6 11.0 9.03 8.14 7.78 6.93 6.08 5.80 5.06 4.38 3.56 3.00 2.29 1.93 1.38 1.24 1.10

Axis Y r (it1.)

6.29 6.41 6.56 6.64 5.24 5.44 5.62 4.29 4.43 4.61 3.42 3.52 3.66 3.87 3.22 3.40 3.48 2.82 2.99 3.11 2.51 2.60 2.72 2.13 2.22 2.34 1.83 1.95 1.47 1.56 1.08 1.12 1.17

Courtesy of The American Institute of Steel Construction. ‘C means channel, followed by the nominal depth in inches, then the weight in pounds per foot of length. 1Not part of the American Standard Series.

I

S

17.8 16.4 15.1 14.4 11.0 9.23 8.13 5.14 4.47 3.88 3.94 3.36 2.81 2.28 2.42 1.93 1.76 1.98 1.53 1.32 1.38 1.17 0.968 1.05 0.866 0.693 0.632 0.479 0.433 0.319 0.305 0.247 0.197

5.32 5.07 4.82 4.69 3.78 3.37 3.1 1 2.06 1 .88 1 .73 1 .65 1 .48 1 .32 1 .16 1 .17 1 .01 0.962 1 .01 0.854 0.781 0.779 0.703 0.625 0.642 0.564 0.492 0.450 0.378 0.343 0.283 0.268 0.233 0.202

-Y r (in.)

If (in.)

1.02 1.04 1.06 1.07 0.867 0.886 0.904 0.763 0.780 0.799 0.669 0.676 0.692 0.713 0.642 0.661 0.669 0.599 0.615 0.625 0.564 0.571 0.581 0.525 0.529 0.537 0.489 0.493 0.450 0.449 0.416 0.410 0.404

0.862 0.858 0.866 0.877 0.798 0.777 0.787 0.674 0.674 0.698 0.649 0.617 0.606 0.634 0.583 0.586 0.60 1 0.565 0.553 0.57 1 0.532 0.525 0.540 0.514 0.499 0.51 1 0.478 0.484 0.459 0.457 0.455 0.438 0.436

APPENDIX B TABLES 011 mtomzlmas 689 1’

[

X—

-—X < 8 x 1 >< 718 x 3/4 x 578 x 112 L6 >< 6 x 1 x 7/8 x 3/4 x 518 x 112 x 318 L5 >< 5 x 7/8 x 3/4 x 5:'8 x l/2 x 318 L4 >< 4 x 314 x 5/8 x 1/2 x 3/8 >< 114 L3§ x 3% x l/2 x 378 x 114 L3 x 3 x 1/2 x 3/8 x 1/4

L2§ >< 25 >< 1:2 x x L2 x 2 x x x

3:'8 l/4 318 114 l/8

51.0 45.0 38.9 32.7 26.4 37.4 33.1 28.7 24.2 19.6 14.9 27.2 23.6 20.0 16.2 12.3 18.5 15.7 12.8 9.8 6.6 1 1.1 8.5 5.8 9.4 7.2 4.9 7.7 5.9 4.1 4.7 3.19 1.65

Axis Z—Z

Area

I

s

P"

X5; 01' Jig

P"

(in-2)

(in-4)

(ill-3)

(ill-)

(in.)

(in-)

15.0 13.2 11.4 9.61 7.75 11.0 9.73 8.44 7.11 5.75 4.36 7.98 6.94 5.86 4.75 3.61 5.44 4.61 3.75 2.86 1.94 3.25 2.48 1.69 2.75 2.11 1.44 2.25 1.73 1.19 1.36 0.938 0.484

89.0 79.6 69.7 59.4 48.6 35.5 31.9 28.2 24.2 19.9 15.4 17.8 15.7 13.6 11.3 8.74 7.67 6.66 5.56 4.36 3.04 3.64 2.87 2.01 2.22 1.76 1.24 1.23 0.984 0.703 0.479 0.348 0.190

15.8 14.0 12.2 10.3 8.36 8.57 7.63 6.66 5.66 4.61 3.53 5.17 4.53 3.86 3.16 2.42 2.81 2.40 1.97 1.52 1.05 1.49 1.15 0.794 1.07 0.833 0.577 0.724 0.566 0.394 0.351 0.247 0.131

2.44 2.45 2.47 2.49 2.50 1 .80 1 .81 1 .83 1 .84 1 .86 1 .88 1 .49 1.5 1 1 .52 1 .54 1 .56 1 .19 1 .20 1 .22 1 .23 1 .25 1 .06 1 .07 1 .09 0.898 0.9 1 3 0.930 0.739 0.753 0.769 0.594 0.609 0.626

2.37 2.32 2.28 2.23 2.19 1.86 1.82 1.78 1.73 1.68 1.64 1.57 1.52 1.48 1.43 1.39 1.27 1.23 1.18 1.14 1.09 1.06 1.01 0.968 0.932 0.888 0.84 0.806 0.762 0.717 0.636 0.592 0.546

1.56 1.57 1.58 1.58 1.59 1 . 17 1.17 1.17 1.18 1.18 1.19 0.973 0.975 0.978 0.983 0.990 0.778 0.779 0.782 0.788 0.795 0.683 0.687 0.694 0.584 0.58 .592 0.487 0.487 0.491 0.389 0.391 0.398

Courtesy of The American Institute of Steel Construction.

.u1m:1\mx B TABLES 011 Pnomzlmss 691 1'

+1-P‘

Z\

L

X

\1\ Y

X 1/

—’|?~."—

N

Table B-8 Equal Leg Angles [SI Units) Axis X-X or Y—Y Size and

Thickness

(111111) L203 >< 203 >< x x x x L152 x 152 >< x x x x x L127 x 127 x x x x x L102 x 102 x x x >< x L89 x 89 >< x x L76 x 76 x x >< L64 x 64 >< >< >< L51 x 51 >< x x

Mass

Area

(1121111) (1111112) 25.4 222 191 159 123 25.4 222 191 159 12] 95 22.2 191 153 121 95 191 159 123 9.5 64 12.7 95 64 12.7 95 6.4 12.7 9.5 6.4 9.5 6A 32

75.9 67.0 57.9 48.7 39.3 55.7 49.3 42.7 36.0 29.2 22.2 40.5 35.1 29.8 24.1 18.3 27.5 23.4 19.0 14.6 9.8 16.5 12.6 8.6 14.0 10.7 7.3 11.5 8.8 6.1 7.0 4.75 2.46

9675 8515 7355 6200 5000 7095 6275 5445 4585 3710 2815 5150 4475 3780 3065 2330 3510 2975 2420 1845 1250 2095 1600 1090 1775 1360 929 1450 1115 768 877 605 312

Axis Z—Z

I

S

(10°

(103

r

XC O1‘ yg

r

111111‘)

1111113)

(111111)

(111111)

(111111)

62.0 62.2 62.7 63.2 63.5 45.7 46.0 46.5 46.7 47.2 47.8 37.8 38.4 38.6 39.1 39.6 30.2 30.5 31.0 31.2 31.8 26.9 27.2 27.7 22.8 23.2 23.6 18.8 19.1 19.5 15.1 15.5 15.9

60.2 58.9 57.9 56.6 55.6 47.2 46.2 45.2 43.9 42.7 41.7 39.9 38.6 37.6 36.3 35.3 32.3 31.2 30.0 29.0 27.7 26.9 25.7 24.6 23.7 22.6 21.4 20.5 19.4 18.2 16.2 15.0 13.9

39.6 39.9 40.1 40.1 40.4 29.7 29.7 29.7 30.0 30.0 30.2 24.7 24.8 24.8 25.0 25.1 19.8 19.8 19.9 20.0 20.2 17.3 17.4 17.6 14.8 14.9 15.0 12.4 12.4 12.5 9.88 9.93 10.1

37.0 33.1 29.0 24.7 20.2 14.8 13.3 11.7 10.1 8.28 6.61 7.41 6.53 5.66 4.70 3.64 3.19 2.77 2.31 1.81 1.27 1.52 1.19 0.837 0.924 0.732 0.516 0.512 0.410 0.293 0.199 0.145 0.079

259 229 200 169 137 140 125 109 92.8 75.5 57.8 84.7 74.2 63.3 51.8 39.7 46.0 39.3 32.3 24.9 17.2 24.4 18.8 13.0 17.5 13.7 9.46 11.9 9.28 6.46 5.75 4.05 2.15

692

APPENDIX B 'r.m1.ss or Pnomrrms Y

-1. 1-xc

Z

X L

1 ‘\~-—-J-—X .._ '

J’c

Ira

Table B-9 Unequal Leg Angles (U.S. Cusmmary Units) Size and

Axis X—X

Thickness (in.)

Weight Area (lb/11) (in?)

L9 x 4 x 5/8 x 1/2 L8 x 6 x 1 x 3/4 x 1/2 L8 x 4 x 1 x 3/4 x 1/2 L7 x 4 x 3/4 x 1/2 x 3/8 L6 x 4 x 3/4 x 1/2 x 3/8 L6 x 3§x1/2 x 3/8 L5 x 3§x 3/4 x 1/2 x 3/8 x 1/4 L5 x 3 x l/2 x 3/8 x 1/4 L4 x 3§x 1/2 x 3/8 x 1/4 L4 x 3 x 1/2 x 3/8 x 1/4 L3§ x 3 x l/2 x 3/8 x 1/4

7.73 26.3 6.25 21.3 44.2 13.0 9.94 33.8 6.75 23.0 37.4 11.0 8.44 28.7 5.75 19.6 7.69 26.2 5 .25 17.9 3.98 13.6 6.94 23.6 4.75 16.2 3.61 12.3 4.50 15.3 11.7 3.42 5.81 19.8 4.00 13.6 3.05 10.4 2.06 7.0 3.75 12.8 2.86 9.8 1.94 6.6 3.50 11.9 2.67 9.1 6.2 1.81 3.25 11.1 2.48 8.5 1.69 5.8 10.2 3.00 2.30 7.9 1.56 5.4 2.75 9.4 2.11 7.2 1.44 4.9 2.50 8.5 1.92 6.6 1.31 4.5 2.25 7.7 1.73 5.9 1.19 4.1 1.55 5.3 3.62 1.06

L3§ X2; >< 3/s >< 1/4 L3 >< 2; >< 1,12 x 3/8

>< 1.3 >< 2 >< >< >< L2; >< 2 >< ><

1/4 1:2 3/s 1/4 3/3 1,14

1

Axis Y—Y

Axis Z—Z

S

P’

If

?'

(111-1

I (in.“)

(111-3)

(111-)

(in.)

(in.)

Tana

3.36 3.31 2.65 2.56 2.47 3.05 2.95 2.86 2.5 1 2.42 2.37 2.08 1.99 1.94 2.08 2.04 1.75 1.66 1.61 1.56 1.75 1.70 1.66 1.25 1.21 1.16 1.33 1.28 1.24 1.13 1.08 1.04 1.20 1.16 1.1 1 1.00 0.956 0.911 1.08 1.04 0.993 0.813 0.787

8.32 6.92 38.8 30.7 21.7 11.6 9.36 6.74 9.05 6.53 5.10 8.68 6.27 4.90 4.25 3.34 5.55 4.05 3.18 2.23 2.58 2.04 1.44 3.79 2.95 2.09 2.42 1.92 1.36 2.33 1.85 1.30 1.36 1.09 0.777 1.30 1.04 0.743 0.672 0.543 0.392 0.514 0.372

2.65 2.17 8.92 6.92 4.79 3.94 3.07 2.15 3.03 2.12 1.63 2.97 2.08 1.60 1.59 1.23 2.22 1.56 1.21 0.830 1.15 0.888 0.614 1.52 1.17 0.808 1.12 0.866 0.599 1.10 0.851 0.589 0.760 0.592 0.412 0.744 0.581 0.404 0.474 0.371 0.260 0.363 0.254

1.04 1.05 1.73 1.76 1.79 1.03 1.05 1.08 1.09 1.1 1 1.13 1.12 1.15 1.17 0.972 0.988 0.977 1.01 1.02 1.04 0.829 0.845 0.861 1.04 1.06 1.07 0.864 0.879 0.896 0.881 0.897 0.914 0.704 0.719 0.735 0.722 0.736 0.753 0.546 0.559 0.574 0.577 0.592

0.858 0.810 1.65 1.56 1.47 1.05 0.953 0.859 1.01 0.917 0.870 1.08 0.987 0.941 0.833 0.787 0.996 0.906 0.861 0.814 0.750 0.704 0.657 1.00 0.955 0.909 0.827 0.782 0.736 0.875 0.830 0.785 0.705 0.660 0.614 0.750 0.706 0.661 0.583 0.539 0.493 0.581 0.537

0.847 0.854 1.28 1.29 1.30 0.846 0.852 0.865 0.860 0.872 0.880 0.860 0.870 0.877 0.759 0.767 0.748 0.755 0.762 0.770 0.648 0.654 0.663 0.722 0.727 0.734 0.639 0.644 0.651 0.621 0.625 0.631 0.534 0.537 0.544 0.520 0.522 0.528 0.428 0.430 0.435 0.420 0.424

0.216 0.220 0.543 0.551 0.558 0.247 0.258 0.267 0.324 0.335 0.340 0.428 0.440 0.446 0.344 0.350 0.464 0.479 0.486 0.492 0.357 0.364 0.371 0.750 0.755 0.759 0.543 0.551 0.558 0.714 0.721 0.727 0.486 0.496 0.506 0.667 0.676 0.684 0.414 0.428 0.440 0.614 0.626

I’

ya"

(111-4)

S (in?)

(in.)

64.9 53.2 80.8 63.4 44.3 69.6 54.9 38.5 37.8 26.7 20.6 24.5 17.4 13.5 16.6 12.9 13.9 9.99 7.78 5.39 9.45 7.37 5.11 5.32 4.18 2.91 5.05 3.96 2.77 3.45 2.72 1.91 3.24 2.56 1.80 2.08 1.66 1.17 1.92 1.53 1.09 0.912 0.654

11.5 9.34 15.1 11.7 8.02 14.1 10.9 7.49 8.42 5.81 4.44 6.25 4.33 3.32 4.24 3.24 4.28 2.99 2.29 1.57 2.91 2.24 1.53 1.94 1.49 1.03 1.89 1.46 1.00 1.45 1.13 0.776 1.41 1.09 0.755 1.04 0.810 0.561 1.00 0.781 0.542 0.547 0.381

2.90 2.92 2.49 2.53 2.56 2.52 2.55 2.59 2.22 2.25 2.27 1.88 1.91 1.93 1.92 1.94 1.55 1.58 1.60 1.62 1.59 1.61 1.62 1.23 1.25 1.27 1.25 1.26 1.28 1.07 1.09 1.11 1.09 1.10 1.12 0.913 0.928 0.945 0.924 0.940 0.957 0.768 0.784

.u1m:1\mx B TABLES 011 Pnomzlmns 693 1' Z

I

xi “\.—~—-4—x " E ya -\ T 11 z

P ___

Table B-10 Unequal Leg Angles (SI Units)

1’

Axis X—X

Size and

Axis Y—Y

Thickness

Mass

Area

1 (101

s (101

r

yc

1 (101

s (101

r

If

r

(111111)

(ks/111)

(1111112)

mm‘)

mm’)

(111111)

(111111)

mm‘)

mml)

(111111)

(111111)

(111111)

Tan Q!

39.1 31.7 65.8 50.3 34.2 55.7 42.7 29.2 39.0 26.6 20.2 35.1 24.1 18.3 22.8 17.4 29.5 20.2 15.5 10.4 19.0 14.6 9.82 17.7 13.5 9.22 16.5 12.6 8.63 15.2 11.8 8.04 14.0 10.7 7.29 12.6 9.82 6.70 11.5 8.78 6.10 7.89 5.39

4985 4030 8385 6415 4355 7095 5445 3710 4960 3385 2570 4475 3065 3230 2905 2205 3750 2580 1970 1330 2420 1845 1250 2260 1725 1170 2095 1600 1090 1935 1485 1005 1775 1360 929 1615 1240 845 1450 1115 768 1000 684

27.0 22.1 33.6 26.4 18.4 29.0 22.9 16.0 15.7 11.1 8.57 10.2 7.24 5.62 6.91 5.37 5.79 4.16 3.24 2.24 3.93 3.07 2.13 2.21 1.74 1.21 2.10 1.65 1.15 1.44 1.13 0.795 1.35 1.07 0.749 0.866 0.691 0.487 0.799 0.637 0.454 0.380 0.272

188 153 247 192 131 231 179 123 138 95.2 72.8 102 71.0 54.4 69.5 53.1 70.1 49.0 37.5 25.7 47.7 36.7 25.1 31.8 24.4 16.9 31.0 23.9 16.4 23.8 18.5 12.7 23.1 17.9 12.4 17.0 13.3 9.19 16.4 12.8 8.88 8.96 6.24

73.7 74.2 63 .2 64.3 65.0 64.0 64.8 65.8 56.4 57.2 57.7 47.8 48.5 49.0 48.8 49.3 39.4 40.1 40.6 41.1 40.4 40.9 41.1 31.2 31.8 32.3 31.8 32.0 32.5 27.2 27.7 28.2 27.7 27.9 28.4 23.2 23 .6 24.0 23.5 23.9 24.3 19.5 19.9

85.3 84.1 67.3 65.0 62.7 77.5 74.9 72.6 63.8 61.5 60.2 52.8 50.5 49.3 52.8 51.8 44.5 42.2 40.9 39.6 445 43.2 42.2 31.8 30.7 29.5 33.8 32.5 31.5 28.7 27.4 26.4 30.5 29.5 28.2 25.4 24.3 23.1 27.4 26.4 25.2 20.7 20.0

3.46 2.88 16.1 12.8 9.03 4.83 3.90 2.81 3.77 2.72 2.12 3.61 2.61 2.04 1.77 1.39 2.31 1.69 1.32 0.928 1.07 0.849 0.599 1.58 1.23 0.870 1.01 0.799 0.566 0.970 0.770 0.541 0.566 0.454 0.323 0.541 0.433 0.309 0.280 0.226 0.163 0.214 0.155

43.4 35.6 146 113 78.5 64.6 50.3 35.2 49.7 34.7 26.7 48.7 34.1 26.2 26.1 20.2 36.4 25.6 19.8 13.6 18.8 14.6 10.1 24.9 19.2 13.2 18.4 14.2 9.82 18.0 13.9 9.65 12.5 8.70 6.75 12.2 9.52 6.62 7.77 6.08 4.26 5.95 4.16

26.4 26.7 43.9 44.7 45.5 26.2 26.7 27.4 27.7 28.2 28.7 28.4 29.2 29.7 24.7 25.1 24.8 25.7 25.9 26.4 21.1 21.5 21.9 26.4 26.9 27.2 21.9 22.3 22.8 22.4 22.8 23.2 17.9 18.3 18.7 18.3 18.7 19.1 13.9 14.2 14.6 14.7 15.0

21.8 20.6 41.9 39.6 37.3 26.7 24.2 21.8 25.7 23.3 22.1 27.4 25.1 23.9 21.2 20.0 25.3 23.0 21.9 20.7 19.1 17.9 16.7 25.4 24.3 23.1 21.0 19.9 18.7 22.2 21.1 19.9 17.9 16.8 15.6 19.1 17.9 16.8 14.8 13.7 12.5 14.8 13.6

21.5 21.7 32.5 32.8 33.0 21.5 21.6 22.0 21.8 22.1 22.4 21.8 22.1 22.3 19.3 19.5 19.0 19.2 19.4 19.6 16.5 16.6 16.8 18.3 18.5 18.6 16.2 16.4 16.5 15.8 15.9 16.0 13.6 13.6 13.8 13.2 13.3 13.4 10.9 10.9 11.0 10.7 10.8

0.216 0.220 0.543 0.551 0.558 0.247 0.258 0.267 0.324 0.335 0.340 0.428 0.440 0.446 0.344 0.350 0.464 0.479 0.486 0.492 0.357 0.364 0.371 0.750 0.755 0.759 0.543 0.551 0.558 0.714 0.721 0.727 0.486 0.496 0.506 0.667 0.676 0.684 0.414 0.428 0.440 0.614 0.626

L229 x 102 x x L203 x 152 x x x L203 x 102 x x x L178 x 102 x x x L152 x 102 x x x L152 x 89 x x L127 x 89 x x x x L127 x 76 x x x L102 x 89 x x x L102 >< 76 x x x L89 x 76 x x x L89 x 64 x x x L76 x 64 x x x L76 x 51 x x x L64 x 51 x x

15.9 123 25.4 191 123 25.4 191 123 191 123 9.5 191 123 95 12.7 95 191 123 95 6.4 12.7 95 6A 12.7 9.5 6A 12.7 95 6A 12.7 95 6A 12.7 95 6.4 12.7 95 6A 12.7 9.5 6A 95 6A

“is Z—Z

694 APPENDIX B TABLES 0B PROPERTIES Y -1‘J"C X_-L

T Table B-1 1

1’

Structural Tees (U.S. Customary Units)

Designation‘

Area (in?)

W'I‘l8x115 x80 W'I‘l5x66 x54 W'I‘l2x52 x47 x42 x31 WT9x38 x30 x25 x20 WT8x50 x25 x20 x13 WT7x60 x41 x34 x24 x15 x11 WT6x60 x48 x36 x25 x15 x8 WT5x56 x44 x30 x15 x6 WT4x29 x20 x12 x9 x5 WT3x10 x6 WT2x6.5

33.8 23.5 19.4 15.9 15.3 13.8 12.4 9.11 11.2 8.82 7.33 5.88 14.7 7.37 5.89 3.84 17.7 12.0 9.99 7.07 4.42 3.25 17.6 14.1 10.6 7.34 4.40 2.36 16.5 12.9 8.82 4.42 1.77 8.55 5.87 3.54 2.63 1.48 2.94 1.78 1.91

Depth of Tee

Flange

X

Stem

Axis X—X

Axis Y—Y

Thickness (in.)

Thickness (in.)

I

S

T

yc

(111-1

Width (in-)

(in.4)

(111.3)

(in.)

17.950 18.005 15.155 14.915 12.030 12.155 12.050 11.870 9.105 9.120 8.995 8.950 8.485 8.130 8.005 7.845 7.240 7.155 7.020 6.985 6.920 6.870 6.560 6.355 6.125 6.095 6.170 5.995 5.680 5.420 5.110 5.235 4.935 4.375 4.125 3.965 4.070 3.945 3.100 3.015 2.080

16.470 12.000 10.545 10.475 12.750 9.065 9.020 7.040 11.035 7.555 7.495 6.015 10.425 7.070 6.995 5.500 14.670 10.130 10.035 8.030 6.730 5.000 12.320 12.160 12.040 8.080 6.520 3.990 10.415 10.265 10.080 5.810 3.960 8.220 8.070 6.495 5.250 3.940 6.020 4.000 4.060

1 .260 1 .020 1 .000 0.760 0.750 0.875 0.770 0.590 0.680 0.695 0.570 0.525 0.985 0.630 0.505 0.345 0.940 0.85 5 0.720 0.595 0.385 0.33 5 1 .105 0.900 0.670 0.640 0.440 0.265 1 .250 0.990 0.680 0.510 0.210 0.810 0.560 0.400 0.330 0.205 0.365 0.280 0.345

0.760 0.650 0.615 0.545 0.500 0.515 0.470 0.430 0.425 0.415 0.355 0.315 0.585 0.380 0.305 0.250 0.590 0.510 0.415 0.340 0.270 0.230 0.710 0.550 0.430 0.370 0.260 0.220 0.755 0.605 0.420 0.300 0.190 0.510 0.360 0.245 0.230 0.170 0.260 0.230 0.280

934 740 421 349 189 186 166 131 71.8 64.7 53.5 44.8 76.8 42.3 33.1 23.5 51.7 41.2 32.6 24.9 19.0 14.8 43.4 32.0 23.2 18.7 13.5 8.70 28.6 20.8 12.9 9.28 4.35 9.12 5.73 3.53 3.41 2.15 1.76 1.32 0.526

67.0 55.8 37.4 32.0 20.0 20.3 18.3 15.6 9.83 9.29 7.79 6.73 11.4 6.78 5.35 4.09 8.61 7.14 5.69 4.48 3.55 2.91 8.22 6.12 4.54 3.79 2.75 2.04 6.40 4.77 3.04 2.24 1.22 2.61 1.69 1.08 1.05 0.717 0.693 0.564 0.321

5.25 5.61 4.66 4.69 3.51 3.67 3.67 3.79 2.54 2.71 2.70 2.76 2.28 2.40 2.37 2.47 1.71 1.85 1.81 1.87 2.07 2.14 1.57 1.51 1.48 1.60 1.75 1.92 1.32 1.27 1.21 1.45 1.57 1.03 0.988 0.999 1.14 1.20 0.774 0.861 0.524

(111)

I (in.“)

S (in?)

r (in.)

4.01 4.74 3.90 4.01 2.59 2.99 2.97 3.46 1.80 2.16 2.12 2.29 1.76 1.89 1.81 2.09 1.24 1.39 1.29 1.35 1.58 1.76 1.28 1.13 1.02 1.17 1.27 1.74 1.21 1.06 0.884 1.10 1.36 0.874 0.735 0.695 0.834 0.953 0.560 0.677 0.440

470 147 98.0 73.0 130 54.5 47.2 17.2 76.2 25.0 20.0 9.55 93.1 18.6 14.4 4.80 247 74.2 60.7 25.7 9.79 3.50 172 135 97.5 28.2 10.2 1.41 118 89.3 58.1 8.35 1.09 37.5 24.5 9.14 3.98 1.05 6.64 1.50 1.93

57.1 24.6 18.6 13.9 20.3 12.0 10.5 4.90 13.8 6.63 5.35 3.17 17.9 5.26 4.12 1.74 33.7 14.6 12.1 6.40 2.91 1.40 28.0 22.2 16.2 6.97 3.12 0.706 22.6 17.4 11.5 2.87 0.551 9.13 6.08 2.81 1.52 0.532 2.21 0.748 0.950

3.73 2.50 2.25 2.15 2.91 1.98 1.95 1.38 2.61 1.69 1.65 1.27 2.51 1.59 1.57 1.12 3.74 2.48 2.46 1.91 1.49 1.04 3.13 3.09 3.04 1.96 1.52 0.773 2.68 2.63 2.57 1.37 0.785 2.10 2.04 1.61 1.23 0.841 1.50 0.918 1.00

Courtesy of The American Institute of Steel Construction. ‘WT means structural T-section (cut from a W-section), followed by the nominal depth in inches, then the weight in pounds per foot of length.

APPENDIX B TABLES 011 Pnomzlrrms 695

T

_|_

—x

_

W

Table B-12 Structural Tees (SI Units) Flange

Stem

Axis X—X

Axis Y—Y

Area

Depth of Tee

Thickness

1 (101

s (101

r

J"c

1 (101

s (101

r

(1111112)

(111111)

Width (111111)

Thickness

Designation‘

(111111)

(mm)

mm‘)

mm3)

(111111)

(111111)

mm‘)

mm‘)

(111111)

WT457 >< x WT381 x x WT305 x x x >< WT229 x x x x WT203 x x >< x WTl78 x x x x x x WTl52 x x >< x x x WT127 x >< x x x WT102 x X x >< x WT76 x X WT51 x

21805 15160 12515 10260 9870 8905 8000 5875 7225 5690 4730 3795 9485 4755 3800 2475 11420 7740 6445 4560 2850 2095 11355 9095 6840 4735 2840 1525 10645 8325 5690 2850 1140 5515 3785 2285 1695 955 1895 1150 1230

455.9 457.3 384.9 378.8 305.6 308.7 306.1 301.5 231.3 231.6 228.5 227.3 215.5 206.5 203.3 199.3 183.9 181.7 178.3 177.4 175.8 174.5 166.6 161.4 155.6 154.8 156.7 152.3 144.3 137.7 129.8 133.0 125.3 111.1 104.8 100.7 103.4 100.2 78.7 76.6 52.8

418.3 304.8 267.8 266.1 323.9 230.3 229.1 178.8 280.3 191.9 190.4 152.8 264.8 179.6 177.7 139.7 372.6 257.3 254.9 204.0 170.9 127.0 312.9 308.9 305.8 205.2 165.6 101.3 264.5 260.7 256.0 147.6 100.6 208.8 205.0 165.0 133.4 100.1 152.9 101.6 103.1

32.0 25.9 25.4 19.3 19.1 22.2 19.6 15.0 17.3 17.7 14.5 13.3 25.0 16.0 12.8 8.8 23.9 21.7 18.3 15.1 9.8 8.5 28.1 22.9 17.0 16.2 11.2 6.7 31.8 25.1 17.3 13.0 5.3 20.6 14.2 10.2 8.4 5.2 9.3 7.1 8.8

19.3 16.5 15.6 13.8 12.7 13.1 11.9 10.9 10.8 10.5 9.0 8.0 14.9 9.7 7.7 6.4 15.0 13.0 10.5 8.6 6.9 5.8 18.0 14.0 10.9 9.4 6.6 5.6 19.2 15.4 10.7 7.6 4.8 13.0 9.1 6.2 5.8 4.3 6.6 5.8 7.1

389 308 175 145 78.7 77.4 69.1 54.5 29.9 26.9 22.3 18.6 32.0 17.6 13.8 9.78 21.5 17.1 13.6 10.4 7.91 6.16 18.1 13.3 9.66 7.78 5.62 3.62 11.9 8.66 5.37 3.86 1.81 3.80 2.39 1.47 1.42 0.895 0.733 0.549 0.219

1098 914 613 S24 328 333 300 256 161 152 128 110 187 111 87.7 67.0 141 117 93.2 73.4 58.2 47.7 135 100 74.4 62.1 45.1 33.4 105 78.2 49.8 36.7 20.0 42.8 27.7 17.7 17.2 11.7 11.4 9.24 5.26

133 142 118 119 89.2 93.2 93.2 96.3 64.5 68.8 68.6 70.1 57.9 61.0 60.2 62.7 43.4 47.0 46.0 47.5 52.6 54.4 39.9 38.4 37.6 40.6 44.5 48.8 33.5 32.3 30.7 36.8 39.9 26.2 25.1 25.4 29.0 30.5 19.7 21.9 13.3

102 120 99.1 102 65.8 75.9 75.4 87.9 45.7 54.9 53.8 58.2 44.7 48.0 46.0 53.1 31.5 35.3 32.8 34.3 40.1 44.7 32.5 28.7 25.9 29.7 32.3 44.2 30.7 26.9 22.5 27.9 34.5 22.2 18.7 17.7 21.2 24.2 14.2 17.2 11.2

196 61.2 40.8 30.4 54.1 22.7 19.6 7.16 31.7 10.4 8.32 3.98 38.8 7.74 5.99 2.00 103 30.9 25.3 10.7 4.07 1.46 71.6 56.2 40.6 11.7 4.25 0.587 49.1 37.2 24.2 3.48 0.454 15.6 10.2 3.80 1.66 0.437 2.76 0.624 0.803

936 403 305 228 333 197 172 80.3 226 109 87.7 51.9 293 86.2 67.5 28.5 552 239 198 105 47.7 22.9 459 364 265 114 51.1 11.6 370 285 188 47.0 9.03 150 99.6 46.0 24.9 8.72 36.2 12.3 15.6

94.7 63.5 57.2 54.6 73.9 50.3 49.5 35.1 66.3 42.9 41.9 32.3 63.8 40.4 39.9 28.4 95.0 63.0 62.5 48.5 37.8 26.4 79.5 78.5 77.2 49.8 38.6 19.6 68.1 66.8 65.3 34.8 19.9 53.3 51.8 40.9 31.2 21.4 38.1 23.3 25.4

171 119 98 80 77 70 63 46 57 45 37 30 74 37 30 19 89 61 51 36 22 16 89 71 54 37 22 12 83 65 45 22 9 43 30 18 13 7 15 9 10

‘WT means structural T-section (cut from a W-section), follcrwed by the nominal depth in rnrn, than the mass in kg per meter of length.

696 .u111|:1~rn1x B TABLES or Psomrrms Table B-13 Properties of Standard Steel Pipe (U.S. Customary Units) Dimensions Nominal Outside Diam. d (in.)

Inside Diam. di

Diam. do (in.]

wan

Properties

Thickness Weight I

S

F’

(in.)

w (lb/11)

A

(111-)

(111-2)

(111-4)

(111-3)

(111)

0.622 0.824 1 .049 1 .380 1 .6 10 2.067 2.469 3.068 3.548 4.026 5.047 6.065 7.98 1 10.020 12.000

0.109 0.113 0.133 0.140 0.145 0.154 0.203 0.216 0.226 0.237 0.258 0.280 0.322 0.365 0.375

0.85 0.13 1.68 2.27 2.72 3.65 5.79 7.58 9.11 10.79 14.62 18.97 28.55 40.48 49.56

0.250 0.333 0.494 0.669 0.799 1.075 1.704 2.228 2.680 3.174 4.300 5.581 8.399 11.91 14.58

0.017 0.037 0.087 0.195 0.310 0.666 1.530 3.017 4.787 7.233 15.16 28.14 72.49 160.7 279.3

0.04 1 0.07 1 0.13 3 0.23 5 0.326 0.56 1 1 .064 1 .724 2.39 3.21 5.45 8.50 16.81 29.9 43.8

0.26 0.33 0.42 0.54 0.62 0.79 0.95 1.16 1.34 1.51 1.88 2.25 2.94 3.67 4.38

1.500 1.939 2.323 2.900 3.826 5.761

0.200 0.218 0.276 0.300 0.337 0.432

3.63 5.02 7.66 10.25 14.98 28.57

1.068 1.477 2.254 3.016 4.407 8.405

0.39 1 0.868 1 .924 3.894 9.610 40.49

0.412 0.73 1 1 .33 8 2.23 4.27 12.22

0.61 0.77 0.92 1.14 1.48 2.20

1.100 1.503 1.771 2.300 3.152 4.897

0.400 0.436 0.552 0.600 0.674 0.864

6.41 9.03 13.69 18.58 27.54 53.16

1.885 2.656 4.028 5.466 8.101 15.64

0.568 1.311 2.871 5.993 15.28 66.33

0.564 1 .104 1 .997 3.42 6.79 20.0

0.55 0.70 0.84 1.05 1.37 2.06

I

Standard Weight 1 vu

— —

v~..1z:->1.6:._-m- .L»-

>-:- MO0 O\Ln-l=nu-I rMl\J-

0 840 1.050 1.315 1.660 1.900 2.375 2.875 3.500 4.000 4.500 5.563 6.625 8.625 10.750 12.750

Extra Strong 1.900 2 2.375

11

2;

2.375

3 4 6

3.500 4.500 6.625

Double Extra Strung

11

2

2% 3 O\-in

1.900 2.315 2.275 3.500 4.500 6.625

.u1m:1\mx B TABLES 011 Pnomzlrrms 697 Table B-111 Properties of Standard Steel Pipe (SI Units) Dimensions Nominal Outside Inside Wall Diam. Diam. Diam. Thickness d do di I

(111111)

(111111) (111111)

Properties Mass TH

A

(111111)

(kg/111)

(1111112)

I

S

( 10° )

( 103 )

(111111‘)

(111111’)

7'

(111111)

Standard Weight 21.3 26.7 33.4 42.2 48.1 60.3 73.0 88.9 101.6 114.3 141.3 168.3 219.1 273.1 323.9

15.8 20.9 26.6 35.1 40.9 52.5 62.7 77.9 90.1 102.3 128.2 154.1 202.7 254.5 304.8

2.77 2.87 3.38 3.56 3.68 3.91 5.16 5.49 5.74 6.02 6.55 7.11 8.18 9.27 9.53

1.264 1.681 2.499 3.376 4.045 5.428 8.611 11.27 13.55 16.05 21.74 28.21 42.46 60.20 73.71

161.3 214.8 318.7 431.6 515.5 693.5 1099 1437 1729 2048 2774 3600 5419 7684 9406

Extra Strong 38 48.3 51 60.3 64 70.0 76 88.9 102 114.3 152 168.3

38.1 49.3 59.0 73.7 97.2 146.3

5.08 5.54 7.01 7.62 8.56 10.97

5.399 7.466 1 1.39 1 5.24 22.28 42.49

689 953 1454 1946 2843 5423

0.163 0.361 0.801 1.621 4.000 16.85

10.16 11.07 14.02 15.24 17.12 21.95

9.53 13.43 20.36 27.63 40.96 79.06

1216 1714 2600 3526 5226 10090

0.236 0.546 1.195 2.494 6.360 27.61

13 19 25 32 38 51 64 76 89 102 127 152 203 254 305

0.007 0.015 0.036 0.081 0.129 0.277 0.637 1.256 1.992 3.011 6.310 11.71 30.2 66.9 116.3

0.672 6.6 1.163 8.5 2.179 10.7 3.851 13.7 5.342 15.8 9.193 20.0 17.44 24.1 28.25 29.5 39.17 34.0 52.60 38.4 89.31 47.8 139.3 57.2 275.5 74.7 490 93.2 718 11 1.3

6.75 1 1.98 21 .93 36.54 69.67 200

15.4 19.5 23.5 29.0 37.6 55.9

Double Extra Stmng 38 51 64 76 102 152

48.3 60.3 70.0 88.9 114.3 168.3

27.9 38.2 45.0 58.4 80.1 124.4

0.564 1.104 1.997 3.42 6.79 20.0

13.9 17.9 21.4 26.7 34.8 52.3

698 APPENDIX B 'r.m1.ss or Psomrrms Table B-15 Properties of Standard Structural Timber (U.S. Customary Units) Dimensions‘

Properties

W

Area A

Second Moment I

Section Modulus S

(lblfi)

(in?)

(inf)

(in?)

1.64 2.54 3.39 4.29 5.19 3.65 5.66 7.55 9.57 11.6 8.40 11.4 14.5 17.5 20.6 15.6 19.8 23.9 28.0 32.0 25.0 30.3 35.6 40.9 46.1 36.7 43.1 49.5 55.9 62.3

5.89 9.14 12.2 15.4 18.7 13.1 20.4 27.2 34.4 41.7 30.3 41.3 52.3 63.3 74.3 56.3 71.3 86.3 101 116 90.3 109 128 147 166 132 155 178 201 224

6.45 241 571 116 206 14A 538 127 259 459 763 193 393 697 1128 264 536 951 1538 2327 679 1204 1948 2948 4243 1458 2358 3569 5136 7106

3.56 8.57 15 .3 24.4 35.8 7.94 19.1 34.0 54.5 79.9 27.7 51.6 82.7 121 167 70.3 113 165 228 300 143 209 289 380 485 253 349 460 587 729

Nominal

Size bxh (in.)

2x4 6 8 10 12 4x4 6 8 10 12 6x6 8 10 12 14 8x8 10 12 14 16 10 x 10 12 14 16 18 12 >< 12 14 16 18 20

Dressed

Weight

Size (in)

1§x3§

1% 11 1%

111

31x35 8 8

1:. 11 11 11;

sgxsg

11 11 11;

13; 7% >

ZEI

_ Pb(L2—b2)3"2 9,/51.1-:1

atx = ./(L1 - b=);s

I/92 Z-=|r-".L—

Mxz v = i

atx=L

6:_Pb(L2—b2) ‘ 61.21

I

n9'|

“‘“_—+iML2 2151

PL3 vmu = -Z

atx = 0

.5” Ci

mm

4351 mx=Lfl

6 _ PL:

v = —%(3L2 — 4x2) L

OE-TEE

2 — +1551 atx=L

7

Y

W

6| : _ WL3 24E! at x = O

N

5wL4

334.91 mx=Lfl

Vmnx = _ i

Fm

:9 HiQ

vmu

6 _

WL

“+2451

v = - 2—:2_I(;’ - 213;’ + L’)

atx =L ML} V1-:mx=_i

9~/521 atx = L/J5

X

‘ to Q_

€:7

1»-m

/92

atx=0

ML

"1 “"5 atx=L

ML!

v

==1==w not max

= —i

1651

v = - ,2 —X2)

APPENDIX B TABLES or Pnornmms Table B-20 Properties of Areas for llurved Bums

fig

A = bu, - r.-)

L

1' dA 1? 1,.

L} r,,

A

1"c = '—" J’2 "’

r,,

P

7:

I

b

=~—=l

2,-+0

A=i(7',,—I';')

lei:

5%

A P

!'(j_%

bra

I-n:!_b

"0-"I

Ti

foo

A = pm, -r.-)

YT i L

Ln,

r I = ?i(2-b1+ bz) +?'a(b! +252)

‘

Ffl.

3(b1 + b2)

/‘Q= 1n'l_b|_|_b2 AP To-T‘; T3"

A=:rrb2

T_

/);d%i=21r(r¢—‘/rgjbz)

L :-~-

A=:rrbh dz!

Zarb

f.7=T(’"~"5"") ‘T'i"rT A = (c — r;)b| + (d — c)tw + (r,, — d]b;

I

r, = §[1>1(¢= -,3) +W= -¢11+b2(r3 -d=) dA =b,m5+¢,,1n-+b,1n'Z d j-

§'——% ~‘*"

A

Kr

I‘?

c_

mop-no

P

I‘;

C

d

703

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nmax 705 INDEX Allowable load, 594-595 Allowable stress, 594-595, 601-602 Allowable stress design, 264-265 Allowable stress method, 601 American standard beams (S-shapes), 686-687 American standard channels (C-shapes), 688-689 Angle of twist, 278, 281-282, 287-289 Angle sections (L-shapes): equal legs (table), 690-691 unequal legs (table), 692-693 Applied force, 3 Area: mixed second moment of, 356, 412, 675-676 polar second moment of, 280, 618, 660 second moment of, 280, 355, 659-682 Assumptions: in centr'ic loading formula, 189-190 in elastic curve equation, 487-488 in flexure formula, 352-354 in shearing stress formula, 391-394 in torsion formula, 277-279 Axial: deformation, 189-190 force, 48 loading, 48, 189-190 strain, 123, 159, 177 stress, 247 Axial-force diagram, 53, 192-193 Axially loaded members: inelastic behavior, 239-240 statically detenuinate, 189-190, 201-203 statically indeterminate, 209-212 Balanced design, 446 Beam deflections: due to shearing stress, 530-531 statically detenninate: by energy methods, 532-536 by integration methods, 489-490, 502-503 with singularity flmctions, 507-509 by superposition methods, 520 statically indeterminate: by energy methods, 560 by integration methods, 542 by superposition methods, 550 Beams: bending moment in, 366-368 bending moment diagram for, 367, 379-380 curved, 430-432 definition of, 349 deflection of, 489, 502, 507, 520, 530, 532 elastic curves for, 487-490

flexural strains in, 352-354 flexural stress formula for, 356 flexural stresses in, 354-356 load-shear relation for, 376-3 79 neutral axis in, 352, 355 normal stresses in, 354-356 of two materials, 441-442 reinforced concrete, 445-446 section modulus of, 356 shear diagram for, 379-380 shear-moment relation for, 376-379 shearing force in, 351, 366-368 shearing stress formula for, 393 shearing stresses in, 391-396 statically determinate, 351 statically iI1dC1ICHl'1lI1fi1§C, 351 types of, 349-351 Bearing stress, 51, 55-56 Bending moment: defined, 35 diagrams, 367, 379-380 equations, 366-368 sign convention for, 367 Boundary conditions: for beams, 489, 491 for columns, 580 Buckling, defined, 578 Cantilever beam, defined, 350 Castigliano, Alberto, 532 Castigliano’s theorem, 532-535, 560 Center: of flexure, 431 of twist, 431 Channel sections (C-shapes), 688-689 Circular shafis: deformations in, 277-279 elastic torsion formula for, 279-281 normal stresses in, 295-297 shearing stresses in, 279-281, 295-297 statically indeterminate, 303 strains in, 277-279 Clapeyron, B.P.E., 532 Clapeyron’s theorem, 532-533 Clebsch, A., 507 Coefficient: of thermal expansion, 176 of selected materials (table), 700-701 Column codes, 594-595 Column formulas: empirical, 592-595 Euler, 581

706 nmsx Colunms: centric loading of, 579-581 critical load for, 581 critical stress, 581 defined, 578 eccentric loading of, 600-602 effect of end conditions on, 587-5 89 effective length of, 588 empirical formulas for, 592-595 Euler’s formula for, 581 slendemess ratio for, 581 Combined loadings, 254-255, 315-316, 457 Compatibility equation, 209 Components: of a stress, 66, 73 ofa strain, 129-130 Compressive strain, 124 Compressive stress, 49 Concentrated force, 2 Concept: of strain, 2, 122 of stress, 2, 48 Continuous beam, defined, 350 Coulomb, C.A., 277, 391 Coulomb’s theory, 638 Couple: bending, 35 twisting, 35 Creep: defined, 157 limit, 159 Critical buckling load, 581 Critical stress, 581 Curvature, 488 Curved beams, 430-432 Cylindrical pressure vessels,: thick-walled, 257-262 thin-walled, 247-249 Deflection of beams: by energy methods, 532-536, 560 by integration method, 489-490, 502-503, 542 with singularity filnctions, 507-509 by superposition method, 520, 550 Deformation: axial, 123, 189-190 defined, 121-122 diagram, 212, 214 flexural, 352-354 plastic, 157 torsional, 277-279, 281-283 Design: balanced, 446 for axial loads, 264-265 for bending loads, 475-478, 567

for columns, 606 for torsional loads, 339 Diagram: angle of twist, 281-283, 287-289 axial-force, 53, 192-193 bending-moment, 367, 379-380 deformation, 212, 214 displacement, 209 polar, 308 shear force, 367, 379-380 stress-strain, 153

for selected materials, 156, 15s, 16c»1s1 239440

torque, 284 Differential equation of the elastic curve: for beams, 487-489 for columns, 580 Displacement, 121, 202-203, 281-283 Distortion, defined, 129 Double shear, defined, 49 Ductile material, defined, 158 Ductility, defined, 158 Duleau, A., 277, 331 Eccentrically loaded columns, 600-602 allowable stress method, 601 interaction method, 601-602 Effective column length, 588 for various end conditions, 588-5 89 Elastic: failure, 264 flexure formula, 356 limit, defined, 157 torsion formula, 279-281 Elastic constants, related, 159, 168 Elastic curve, differential equation of: for beams, 487-489 for columns, 580 Elasticity, modulus of: defined, 155-156 of selected materials (table), 700-701 Elastoplastic material, 158, 239, 325, 423 Elongation, percent: defined, 159 of selected materials (table), 700-701 Empirical column formulas, 592-595 End conditions: for beams, 350 for columns, 587-589 Equations of equilibrium, 4 Equation of elastic curve, 491, 493, 496, 4 97 Equilibrium: equations, 4 of a deformable body, 30 of a rigid body, 4-5 states of, 578

nroax 707 Euler, Leonard, 579 Euler buckling loaci 581 Extensometer, 155 External forces, 3

Hoop stress, 247 Horsepower, 301 Huber, M.T. 639 Huber-I-Ienke-von Mises theory, 639

Factor of safety, 265 Failure: defined, 264 elastic, 264 fracture, 264 mode, 264 slip, 264 yielding, 264 Flexural: loading, 351 strains, 352-3 54 stresses, 354-356, 450-452 Flexure fonnula, 356 Force: applied, 3 axial, 48 concentrated, 2 contact, 2 defined, 2 distributed, 2 dynamic, 624 external, 3 gravitational, 2 internal, 3, 34-35 normal, 35, 49, 66, 73 resultant, 66, 72-73, 109 shear, 35, 50, 66 surface, 2 transverse shear, 393-395 work of, 532, 615 Free-body diagram, 5

Impact factor, 626 Inelastic behavior: axial loading, 239-240 flexural loading, 422-423 torsional loading, 325-326 Integration method: for beam deflections, 489-490, 502, 503, 542 with singularity functions, 507-509 interaction method, 601-602 Internal forces, 3, 34-35 Isotropic materials: defined, 164 generalized Hooke’s law for, 164-168

Galileo, 1 General state of stress, 108-1 13 Generalized Hooke’s law, 164-168 Gravitational force, 2 Guest, J.J., 638 Guest’s theory, 638 Gyration, radius of, 664-665 Heaviside, O. 507 Homogeneous material, 177 Hooke’s law: for a biaxial state of stress, 155 for a general state of stress, 164-168 for a plane state of stress, 164-166 for shearing stresses, 166-168 for a triaxial state of stress, 164-168 for a uniaxial state of stress, 155 Hooke, Robert, 155

Joints, method of, 10 Jourawski, D.J., 391 Lame, G., 257 Limit state, 445 Load: allowable, 594-595 axial, 48, 189-190 combined, 254-255, 315-316, 457 concentrated, 2 critical buckling, 581 dead, 3 distributed, 2 dynamic, 624 energy, 624 Euler buckling, 581 flexural, 351 impact, 4, 624 repeated, 4 shear: double, 49 punching, 49 single, 49 static, 3 sustained, 3 torsional, 276 transverse, 366-367, 393, 431 Load resistance factor design (LRFD), 264 Load-shear relation for beams, 376-379 Longitudinal strain, 159 Matching conditions, 490 Materials: selected stress-strain diagrams for, 156, 158, 160-161, 239-240 table of selected properties, 700-701

708 nmsx Maximmn: in-plane shearing strain, 136 in-plane shearing stress, 88 normal strain, 136 normal stress, 66-67, 87, 100, 110-111, 296 shearing strain, 136 shearing stress, 66-67, 88, 100-101, 111-112, 296 Mechanics of materials, defined, 1, 3 Meridional stress, 247 Method: ofjoints, 10 of sections,10 of superposition: for beam deflections, 520, 550 for combined stresses, 254-255, 315-316, 457 for thermal strains, 177 Mixed second moments, 356, 412, 675-676 Modulus: of elasticity: defined, 155-156 for selected materials (table), 700-701 of resilience, defined, 616 of rigidity: defined, 157 for selected materials (table), 700-701 of toughness, defined, 616 secant, 157 shear, 157 tangent, 157 Young’s, 156 Mohr, Otto, 98 Mohrh circle: for strains, 140-141 for stresses, 98-102 Moment diagram, 367, 379-380 Neutral: axis, 352, 355 surface, 352, 355 Newton (unit of force), 52 Normal: force, 35, 49, 66, 73 strain, 122, 130-131 axial, 123 flexural, 352-354 maximum and minimum, 136 sip convention for, 124, 133 stress, axial, 247 flexural, 354-356, 450-452 maximum and minimum, 66-67, 87, 100, 110-111 sign convention for, 66, 74-77 Orthotropic materials, 181 Overhanging beam, defined, 349

Parallel-axis theorem, 660-661, 676 Pascal (unit of stress), 52 Percent: elongation, 159 of selected materials (table), 700-701 reduction in area, 159 Permanent set, 157 Pin-ended columns, defined, 579 Plane strain, 130-133 transformation of, 130-133 Plane stress, 74-77 transformation of, 75-77 Plastic deformation, defined, 157 Poisson, Simeon, 159 Poisson’s ratio, 145, 159 Polar second moment, 280, 618, 660 Power transmission, 300-301 Pressure vessels: cylindrical, 247-249 spherical, 246-247 thick-walled, 257-262 thin-walled, 246-250 Principal: planes, 87, 110 strains, 135-136 stresses, 85-90, 100, 110-111 defined, 87 in flexural members, 405-407 by Mohr’s circle, 100-102 Principle of superposition, 164 Proportional limit, 157 Punching shear, 49 Radius of gyration, 664-665 Rankine, W.J.M., 637 Reactions, 3 Recovery, defined, 157 Reinforced beams, 441 Reinforced concrete beams, 445-446 Resilience, modulus of, 616 Resistance strain gage, 143-144 Resisting: moment, 351, 367 shear, 351, 366-367 torque, 276 Restrained beam, defined, 350 Resultant stress, 72, 109 Rigid body, defined, 4 Rigidity, modulus of, 157 for selected materials (table), 700-701 Rosette analysis, 144 Safety, factor of, 265 Saint-Venant, Barre de, 237, 332, 391 Saint-Venant’s principle, 237, 406

nrosx 709 Secant modulus, defined, 157 Second moments: ofarea, 280, 355, 659-682 of composite shapes, 669-670 dimensions of, 660 mixed, 356, 412, 675-676 parallel-axis theorem, 660-661, 676 polar, 280, 618, 660 principal, 679-681 radius of gyration for, 664-665 rectangular, 659 tables of, for conunon shapes, 669-670, 677 transfer of axes for, 660-661 Section modulus, 356 Sections, method of, 10-11 Service load, 446 Shear, defined, 49-50, 66 double, 49 horizontal, 392-395 longitudinal, 393-395 punching, 49 resisting, 351 single, 49 transverse, 393-395 vertical, 393-395 Shear center, 431-434 Shear flow, 333-334, 433 Shear force, 35, 50, 66 Shear force diagram, 367, 379-380 Shear modulus, defined, 157 Shear-moment relation for beams, 376-379 Shearing strain, 122, 123, 131-132 maximum, formula for, 136 sign convention for, 124, 133 Shearing stress, 49, 66 maximum, formula for, 66-67, 88-89, 101, 296 sign convention for, 66, 74, 76-77 Shrink fitting, 257 Sign convention: for beam deflections, 487-488 for bending moments, 367 for normal strains, 124, 133 for normal stresses, 66, 74, 76-77 for shearing strains, 124, 133 for shearing stresses, 66, 74, 76-77 for transverse shear in beams, 367 for torsion, 282 Simple beam, defined, 349 Single shear, defined, 49 Singularity fimctions, 508-509 Slenderness ratio, 581 Slip, defined, 157 Slip failure, 264 Spherical pressure vessels, 246-247

Statically indetenuinate members: with axial loads, 209-212 with flexural loads: by energy methods, 560 by integration methods, 542 by superposition methods, 550 with torsional loads, 303 Stifi' (design), 30 Strain, defined, 2, 122 axial, 123, 159, 177 compressive, 124 flexural, 352-354 lateral, 177 longitudinal, 159 maximum normal, 136 maximum shearing, 136 in-plane, 136 measurement, 142-145, 155 nominal, 155 normal, 122, 130-131 plane, 1 30-1 33 principal, 135-136 shearing, 122, 123,131-132 sign convention for, 124, 133 tensile, 124 thermal, 176-177 torsional shearing, 277-279 total 177 transformation equations for 130-133 transverse 177 true, 155 two-dimensional, 130-133 units of, 124 Strain energy, 532-536, 615-619 for axial loads, 617 for bending loads, 619 for torsional loads, 618 for transverse shear, 619 Strain energy intensity, 533, 616 Strain energy method, 625-627 Strain gage, 143-144, 155 Strain hardening, 157, 239 Strain rosette, 144 Strength: ultimate, 158 of selected materials (table), 700-701 yield, 158 of selected materials (table), 700-701 Stress, defined, 2, 48 allowable, 53 axial, 247 average bearing, 51, 55-56 average normal, 49 average shearing, 49 average total, 66

710 nmsx Stress, defined, (Contal) bearing, 51, 55-S6 circumferential, 247 compressive, 49 critical, 581 flexural, 354-356, 450-452 hoop, 247 maximum normal, 65-67, 85-90, 100, 110-111, 296 by Mohr’s circle, 100-102 maximum shearing, 66-67, 85-90, 100-101, 111-1 12, 296 in-plane, 88 by Mohr’s circle, 100-102 meridional, 247 normal, 49, 66 compressive, 49 on inclined plane, 65-68 tensile, 49 plane, 74-77 principal, 85-90, 405-407 by Mohr’s circle, 100-102 punching shear, 49 resultant, 72, 109 shearing, 49, 66, 279-281, 391-396 sign convention, 66, 74, 76-77 tangential, 66, 247 tensile, 49 thermal, 225 torsional shearing, 279-281 lII'HIlSf0l'IT1fl1Il0I1 equations, 75-77 true, 155 two-dirnensional, 74-77 units of, 51-52 Stress Concentrations: axial loading, 234-237 flexural loading, 418-420 torsional loading, 322-323 Stress concentration factor: axial loading, 235-236 flexural loading, 419-420 torsional loading, 322-323 Stress ellipsoid, 1 10-111 Stress-strain diagrams, 153 for selected materials, 156, 158, 160-161, 239-240 Stress-strain equations: for isotropic materials, 164-168 for orthotropic materials, 180-183 Stress-strain-temperature equations, 176-177 Stress trajectory, 234, 406 Stress vector, 72-74, 109 Structural tees (T-shapes), 694-695 Superposition, principle of, 164

Superposition method: for beam deflections, 520, 550 for combined loading, 254-255, 315-316, 457 for 11116111131 strains, 177 Surface force, 2 Table: of column codes, 594 of deflections and slopes of beams, 702 of properties of areas, 669-670, 677 for curved beams, 703 of properties of materials, 700-701 of properties of rolled-steel shapes, 684-695 of properties of standard steel pipe, 696-697 of properties of structural timber, 698-699 Tangent modulus, defined, 157 Temperature effects, 176-177, 225 Tensile: strain, 124 stress, 49 test, 153-155 Theories of failure: Coulomb-Mohr, 650-651 Coulomb’s Theory, 638 distortion-energy, 639-642 Guest’s theory, 638 Huber-Henke-von Mises theory, 639 maximum-normal-stress, 637-638, 650-651 maximum-shear-stress, 638-639 Rankine’s theory, 637 Thermal expansion, coefficient of, 176 for selected materials (table), 700-701 Thermal: strain, 176-177 stress, 225 Thick-walled pressure vessels, 257-262 Thin-walled pressure vessels: cylindrical, 247-249 spherical, 246-247 Torque, 35, 276 Torque diagram, 284, 286 Torsion: of circular shafis, 276-283 of noncircular shafts, 331-333 of thin-walled tubes, 333-334 Torsional displacement, 281-283 Torsional loading, 276 Torsional shearing strain, 277-2 79 Torsional shearing stresses, 279-281 Transformation equations: for plane strain, 130-133 for plane stress, 75-77 Transformed cross section, 442 Transverse shearing force, 393-395

nrosx 711 True: strain, 155 stress, 155 Truss analysis: method ofjoints, 10 method of sections, 10-11 Truss assumptions, 9 Twist, angle of, 278, 281-282 Twisting couple, 35 Twisting moment, 35 Ultimate: elongation, defined, 159 strength, 146 defined, 158 for selected materials (table), 700-701 Units of measurement: of power, 300-301 of second moments of area, 660 of strain, 124 of stress, 51-52

Unsymmetrical bending, 410-413 Von Mises, R. 639 Warping of cross sections, 277-278 Watt (unit of power) 301 Weight, defined, 2 Wide-flange sections (W-shapes), 684-685 Work: defined, 300 of a couple, 300 ofa force, 532, 615 Work hardening, 157 Work-kinetic energy method, 627-629 Yield: point, defined, 158 for selected materials (table), 700-701 strength, defined, 158 for selected materials (table), 700-701 Young’s modulus, 156 Young, Thomas, 156

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Answers

Chapter 1

1-2

T= (153.33 +3.667W) 115 1~"= (133.33 +2.667W) 115 n=nwN T, = 317N 73 = 1717 N

1-4

N,-,~ = 424N 51> 45*

1-1

1-7

1-9

1-10

1-12 1-15 1-17 1-18 1-20 1-23 1-25

1-26

1-29

A = 300 N -> 1>=10.46115 *1; 15* N, = 24.1115 11> 753 Np = 14.94115 is 75> co; 33.3 kip (c) cs; 23.1 kip (T) FG: 17.32 kip (T) P = 9.131 kN 4! 30= Np - 3.93 kN is 603 N, = s.06 kN its 60~= A = 2 kN ¢ M, = llkN-m '1 A = 61.5115 51> 77.59 B = 13.33 115 -1 A = 200.3115 as 56.43 ND = 130.0115 F = 342 N = 603 N = 439 N = 925 N >2*F§*?é.-"I13-I:4'1= 224 115