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ISBN: 0073380288
Author: Beer, Johnston, Dewolf,
and Mazurek
Title: MECHANICS OF MATERIALS
Front endsheets
Color: 4
Pages: 2, 3
U.S. Customary Units and Their SI Equivalents
Quantity U.S. Customary Units SI Equivalent
Acceleration ft/s2
0.3048 m/s2
in./s2
0.0254 m/s2
Area ft2
0.0929 m2
in2
645.2 mm2
Energy ft ? lb 1.356 J
Force kip 4.448 kN
lb 4.448 N
oz 0.2780 N
Impulse lb ? s 4.448 N ? s
Length ft 0.3048 m
in. 25.40 mm
mi 1.609 km
Mass oz mass 28.35 g
lb mass 0.4536 kg
slug 14.59 kg
ton 907.2 kg
Moment of a force lb ? ft 1.356 N ? m
lb ? in. 0.1130 N ? m
Moment of inertia
Of an area in4
0.4162 3 106
mm4
Of a mass lb ? ft ? s2
1.356 kg ? m2
Power ft ? lb/s 1.356 W
hp 745.7 W
Pressure or stress lb/ft2
47.88 Pa
lb/in2
(psi) 6.895 kPa
Velocity ft/s 0.3048 m/s
in./s 0.0254 m/s
mi/h (mph) 0.4470 m/s
mi/h (mph) 1.609 km/h
Volume, solids ft3
0.02832 m3
in3
16.39 cm3
Liquids gal 3.785 L
qt 0.9464 L
Work ft ? lb 1.356 J
SI Prefixes
Multiplication Factor Prefix† Symbol
1 000 000 000 000 5 1012
tera T
1 000 000 000 5 109
giga G
1 000 000 5 106
mega M
1 000 5 103
kilo k
100 5 102
hecto‡ h
10 5 101
deka‡ da
0.1 5 1021
deci‡ d
0.01 5 1022
centi‡ c
0.001 5 1023
milli m
0.000 001 5 1026
micro m
0.000 000 001 5 1029
nano n
0.000 000 000 001 5 10212
pico p
0.000 000 000 000 001 5 10215
femto f
0.000 000 000 000 000 001 5 10218
atto a
† The first syllable of every prefix is accented so that the prefix will retain its identity.
Thus, the preferred pronunciation of kilometer places the accent on the first syllable, not
the second.
‡ The use of these prefixes should be avoided, except for the measurement of areas and vol-
umes and for the nontechnical use of centimeter, as for body and clothing measurements.
Principal SI Units Used in Mechanics
Quantity Unit Symbol Formula
Acceleration Meter per second squared p m/s2
Angle Radian rad †
Angular acceleration Radian per second squared p rad/s2
Angular velocity Radian per second p rad/s
Area Square meter p m2
Density Kilogram per cubic meter p kg/m3
Energy Joule J N ? m
Force Newton N kg ? m/s2
Frequency Hertz Hz s21
Impulse Newton-second p kg ? m/s
Length Meter m ‡
Mass Kilogram kg ‡
Moment of a force Newton-meter p N ? m
Power Watt W J/s
Pressure Pascal Pa N/m2
Stress Pascal Pa N/m2
Time Second s ‡
Velocity Meter per second p m/s
Volume, solids Cubic meter p m3
Liquids Liter L 1023
m3
Work Joule J N ? m
† Supplementary unit (1 revolution 5 2p rad 5 3608).
‡ Base unit.
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ISBN: 0073380288
Author: Beer, Johnston, Dewolf,
and Mazurek
Title: MECHANICS OF MATERIALS
Front endsheets
Color: 4
Pages: 2, 3
U.S. Customary Units and Their SI Equivalents
Quantity U.S. Customary Units SI Equivalent
Acceleration ft/s2
0.3048 m/s2
in./s2
0.0254 m/s2
Area ft2
0.0929 m2
in2
645.2 mm2
Energy ft ? lb 1.356 J
Force kip 4.448 kN
lb 4.448 N
oz 0.2780 N
Impulse lb ? s 4.448 N ? s
Length ft 0.3048 m
in. 25.40 mm
mi 1.609 km
Mass oz mass 28.35 g
lb mass 0.4536 kg
slug 14.59 kg
ton 907.2 kg
Moment of a force lb ? ft 1.356 N ? m
lb ? in. 0.1130 N ? m
Moment of inertia
Of an area in4
0.4162 3 106
mm4
Of a mass lb ? ft ? s2
1.356 kg ? m2
Power ft ? lb/s 1.356 W
hp 745.7 W
Pressure or stress lb/ft2
47.88 Pa
lb/in2
(psi) 6.895 kPa
Velocity ft/s 0.3048 m/s
in./s 0.0254 m/s
mi/h (mph) 0.4470 m/s
mi/h (mph) 1.609 km/h
Volume, solids ft3
0.02832 m3
in3
16.39 cm3
Liquids gal 3.785 L
qt 0.9464 L
Work ft ? lb 1.356 J
SI Prefixes
Multiplication Factor Prefix† Symbol
1 000 000 000 000 5 1012
tera T
1 000 000 000 5 109
giga G
1 000 000 5 106
mega M
1 000 5 103
kilo k
100 5 102
hecto‡ h
10 5 101
deka‡ da
0.1 5 1021
deci‡ d
0.01 5 1022
centi‡ c
0.001 5 1023
milli m
0.000 001 5 1026
micro m
0.000 000 001 5 1029
nano n
0.000 000 000 001 5 10212
pico p
0.000 000 000 000 001 5 10215
femto f
0.000 000 000 000 000 001 5 10218
atto a
† The first syllable of every prefix is accented so that the prefix will retain its identity.
Thus, the preferred pronunciation of kilometer places the accent on the first syllable, not
the second.
‡ The use of these prefixes should be avoided, except for the measurement of areas and vol-
umes and for the nontechnical use of centimeter, as for body and clothing measurements.
Principal SI Units Used in Mechanics
Quantity Unit Symbol Formula
Acceleration Meter per second squared p m/s2
Angle Radian rad †
Angular acceleration Radian per second squared p rad/s2
Angular velocity Radian per second p rad/s
Area Square meter p m2
Density Kilogram per cubic meter p kg/m3
Energy Joule J N ? m
Force Newton N kg ? m/s2
Frequency Hertz Hz s21
Impulse Newton-second p kg ? m/s
Length Meter m ‡
Mass Kilogram kg ‡
Moment of a force Newton-meter p N ? m
Power Watt W J/s
Pressure Pascal Pa N/m2
Stress Pascal Pa N/m2
Time Second s ‡
Velocity Meter per second p m/s
Volume, solids Cubic meter p m3
Liquids Liter L 1023
m3
Work Joule J N ? m
† Supplementary unit (1 revolution 5 2p rad 5 3608).
‡ Base unit.
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MECHANICS OF
MATERIALS
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SIXTH EDITION
MECHANICS OF
MATERIALS
Ferdinand P. Beer
Late of Lehigh University
E. Russell Johnston, Jr.
Late of University of Connecticut
John T. Dewolf
University of Connecticut
David F. Mazurek
United States Coast Guard Academy
TM
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About the Authors
As publishers of the books written by Ferd Beer and Russ John-
ston, we are often asked how did they happen to write the books
together, with one of them at Lehigh and the other at the University
of Connecticut.
The answer to this question is simple. Russ Johnston’s first teach-
ing appointment was in the Department of Civil Engineering and Me-
chanics at Lehigh University. There he met Ferd Beer, who had joined
that department two years earlier and was in charge of the courses in
mechanics. Born in France and educated in France and Switzerland
(he held an M.S. degree from the Sorbonne and an Sc.D. degree in the
field of theoretical mechanics from the University of Geneva), Ferd
had come to the United States after serving in the French army during
the early part of World War II and had taught for four years at Williams
College in the Williams-MIT joint arts and engineering program. Born
in Philadelphia, Russ had obtained a B.S. degree in civil engineering
from the University of Delaware and an Sc.D. degree in the field of
structural engineering from MIT.
Ferd was delighted to discover that the young man who had
been hired chiefly to teach graduate structural engineering courses
was not only willing but eager to help him reorganize the mechanics
courses. Both believed that these courses should be taught from a few
basic principles and that the various concepts involved would be best
understood and remembered by the students if they were presented
to them in a graphic way. Together they wrote lecture notes in statics
and dynamics, to which they later added problems they felt would
appeal to future engineers, and soon they produced the manuscript
of the first edition of Mechanics for Engineers. The second edition of
Mechanics for Engineers and the first edition of Vector Mechanics for
Engineers found Russ Johnston at Worcester Polytechnic Institute and
the next editions at the University of Connecticut. In the meantime,
both Ferd and Russ had assumed administrative responsibilities in
their departments, and both were involved in research, consulting,
and supervising graduate students—Ferd in the area of stochastic pro-
cesses and random vibrations, and Russ in the area of elastic stability
and structural analysis and design. However, their interest in improv-
ing the teaching of the basic mechanics courses had not subsided, and
they both taught sections of these courses as they kept revising their
texts and began writing together the manuscript of the first edition of
Mechanics of Materials.
Ferd and Russ’s contributions to engineering education earned
them a number of honors and awards. They were presented with the
Western Electric Fund Award for excellence in the instruction of en-
gineering students by their respective regional sections of the Ameri-
can Society for Engineering Education, and they both received the
Distinguished Educator Award from the Mechanics Division of the
v
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same society. In 1991 Russ received the Outstanding Civil Engineer
Award from the Connecticut Section of the American Society of Civil
Engineers, and in 1995 Ferd was awarded an honorary Doctor of En-
gineering degree by Lehigh University.
John T. DeWolf, Professor of Civil Engineering at the University
of Connecticut, joined the Beer and Johnston team as an author on
the second edition of Mechanics of Materials. John holds a B.S. de-
gree in civil engineering from the University of Hawaii and M.E. and
Ph.D. degrees in structural engineering from Cornell University. His
research interests are in the area of elastic stability, bridge monitor-
ing, and structural analysis and design. He is a registered Professional
Engineering and a member of the Connecticut Board of Professional
Engineers. He was selected as the University of Connecticut Teaching
Fellow in 2006.
David F. Mazurek, Professor of Civil Engineering at the United
States Coast Guard Academy, joined the team in the fourth edition.
David holds a B.S. degree in ocean engineering and an M.S. degree in
civil engineering from the Florida Institute of Technology, and a Ph.D.
degree in civil engineering from the University of Connecticut. He is
a registered Professional Engineer. He has served on the American
Railway Engineering & Maintenance of Way Association’s Commit-
tee 15—Steel Structures for the past seventeen years. Professional
interests include bridge engineering, structural forensics, and blast-
resistant design.
vi About the Authors
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Contents
Preface xii
List of Symbols xviii
1 Introduction—Concept of Stress 2
1.1 Introduction 4
1.2 A Short Review of the Methods of Statics 4
1.3 Stresses in the Members of a Structure 7
1.4 Analysis and Design 8
1.5 Axial Loading; Normal Stress 9
1.6 Shearing Stress 11
1.7 Bearing Stress in Connections 13
1.8 Application to the Analysis and Design of Simple
Structures 13
1.9 Method of Problem Solution 16
1.10 Numerical Accuracy 17
1.11 Stress on an Oblique Plane under Axial Loading 26
1.12 Stress under General Loading Conditions;
Components of Stress 27
1.13 Design Considerations 30
Review and Summary for Chapter 1 42
2 Stress and Strain—Axial
Loading 52
2.1 Introduction 54
2.2 Normal Strain under Axial Loading 55
2.3 Stress-Strain Diagram 57
*2.4 True Stress and True Strain 61
2.5 Hooke’s Law; Modulus of Elasticity 62
2.6 Elastic versus Plastic Behavior of a Material 64
2.7 Repeated Loadings; Fatigue 66
2.8 Deformations of Members under Axial Loading 67
2.9 Statically Indeterminate Problems 78
2.10 Problems Involving Temperature Changes 82
2.11 Poisson’s Ratio 93
2.12 Multiaxial Loading; Generalized Hooke’s Law 94
*2.13 Dilatation; Bulk Modulus 96
vii
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viii Contents 2.14 Shearing Strain 98
2.15 Further Discussion of Deformations under Axial Loading;
Relation among E, n, and G 101
*2.16 Stress-Strain Relationships for Fiber-Reinforced Composite
Materials 103
2.17 Stress and Strain Distribution under Axial Loading;
Saint-Venant’s Principle 113
2.18 Stress Concentrations 115
2.19 Plastic Deformations 117
*2.20 Residual Stresses 121
Review and Summary for Chapter 2 129
3 Torsion 140
3.1 Introduction 142
3.2 Preliminary Discussion of the Stresses in a Shaft 144
3.3 Deformations in a Circular Shaft 145
3.4 Stresses in the Elastic Range 148
3.5 Angle of Twist in the Elastic Range 159
3.6 Statically Indeterminate Shafts 163
3.7 Design of Transmission Shafts 176
3.8 Stress Concentrations in Circular Shafts 179
*3.9 Plastic Deformations in Circular Shafts 184
*3.10 Circular Shafts Made of an Elastoplastic Material 186
*3.11 Residual Stresses in Circular Shafts 189
*3.12 Torsion of Noncircular Members 197
*3.13 Thin-Walled Hollow Shafts 200
Review and Summary for Chapter 3 210
4 Pure Bending 220
4.1 Introduction 222
4.2 Symmetric Member in Pure Bending 224
4.3 Deformations in a Symmetric Member in Pure Bending 226
4.4 Stresses and Deformations in the Elastic Range 229
4.5 Deformations in a Transverse Cross Section 233
4.6 Bending of Members Made of Several Materials 242
4.7 Stress Concentrations 246
*4.8 Plastic Deformations 255
*4.9 Members Made of an Elastoplastic Material 256
*4.10 Plastic Deformations of Members with a Single Plane of
Symmetry 260
*4.11 Residual Stresses 261
4.12 Eccentric Axial Loading in a Plane of Symmetry 270
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ixContents4.13 Unsymmetric Bending 279
4.14 General Case of Eccentric Axial Loading 284
*4.15 Bending of Curved Members 294
Review and Summary for Chapter 4 305
5 Analysis and Design of Beams for
Bending 314
5.1 Introduction 316
5.2 Shear and Bending-Moment Diagrams 319
5.3 Relations among Load, Shear, and Bending Moment 329
5.4 Design of Prismatic Beams for Bending 339
*5.5 Using Singularity Functions to Determine Shear and Bending
Moment in a Beam 350
*5.6 Nonprismatic Beams 361
Review and Summary for Chapter 5 370
6 Shearing Stresses in Beams and
Thin-Walled Members 380
6.1 Introduction 382
6.2 Shear on the Horizontal Face of a Beam Element 384
6.3 Determination of the Shearing Stresses in a Beam 386
6.4 Shearing Stresses txy in Common Types of Beams 387
*6.5 Further Discussion of the Distribution of Stresses in a
Narrow Rectangular Beam 390
6.6 Longitudinal Shear on a Beam Element of Arbitrary
Shape 399
6.7 Shearing Stresses in Thin-Walled Members 401
*6.8 Plastic Deformations 404
*6.9 Unsymmetric Loading of Thin-Walled Members;
Shear Center 414
Review and Summary for Chapter 6 427
7 Transformations of Stress and
Strain 436
7.1 Introduction 438
7.2 Transformation of Plane Stress 440
7.3 Principal Stresses: Maximum Shearing Stress 443
7.4 Mohr’s Circle for Plane Stress 452
7.5 General State of Stress 462
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x Contents 7.6 Application of Mohr’s Circle to the Three-Dimensional
Analysis of Stress 464
*7.7 Yield Criteria for Ductile Materials under Plane Stress 467
*7.8 Fracture Criteria for Brittle Materials under Plane Stress 469
7.9 Stresses in Thin-Walled Pressure Vessels 478
*7.10 Transformation of Plane Strain 486
*7.11 Mohr’s Circle for Plane Strain 489
*7.12 Three-Dimensional Analysis of Strain 491
*7.13 Measurements of Strain; Strain Rosette 494
Review and Summary for Chapter 7 502
8 Principal Stresses under a Given
Loading 512
*8.1 Introduction 514
*8.2 Principal Stresses in a Beam 515
*8.3 Design of Transmission Shafts 518
*8.4 Stresses under Combined Loadings 527
Review and Summary for Chapter 8 540
9 Deflection of Beams 548
9.1 Introduction 550
9.2 Deformation of a Beam under Transverse Loading 552
9.3 Equation of the Elastic Curve 553
*9.4 Direct Determination of the Elastic Curve from the Load
Distribution 559
9.5 Statically Indeterminate Beams 561
*9.6 Using Singularity Functions to Determine the Slope and
Deflection of a Beam 571
9.7 Method of Superposition 580
9.8 Application of Superposition to Statically Indeterminate
Beams 582
*9.9 Moment-Area Theorems 592
*9.10 Application to Cantilever Beams and Beams with Symmetric
Loadings 595
*9.11 Bending-Moment Diagrams by Parts 597
*9.12 Application of Moment-Area Theorems to Beams with
Unsymmetric Loadings 605
*9.13 Maximum Deflection 607
*9.14 Use of Moment-Area Theorems with Statically Indeterminate
Beams 609
Review and Summary for Chapter 9 618
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xiContents
10 Columns 630
10.1 Introduction 632
10.2 Stability of Structures 632
10.3 Euler’s Formula for Pin-Ended Columns 635
10.4 Extension of Euler’s Formula to Columns with Other End
Conditions 638
*10.5 Eccentric Loading; the Secant Formula 649
10.6 Design of Columns under a Centric Load 660
10.7 Design of Columns under an Eccentric Load 675
Review and Summary for Chapter 10 684
11 Energy Methods 692
11.1 Introduction 694
11.2 Strain Energy 694
11.3 Strain-Energy Density 696
11.4 Elastic Strain Energy for Normal Stresses 698
11.5 Elastic Strain Energy for Shearing Stresses 701
11.6 Strain Energy for a General State of Stress 704
11.7 Impact Loading 716
11.8 Design for Impact Loads 718
11.9 Work and Energy under a Single Load 719
11.10 Deflection under a Single Load by the
Work-Energy Method 722
*11.11 Work and Energy under Several Loads 732
*11.12 Castigliano’s Theorem 734
*11.13 Deflections by Castigliano’s Theorem 736
*11.14 Statically Indeterminate Structures 740
Review and Summary for Chapter 11 750
Appendices A1
A Moments of Areas A2
B Typical Properties of Selected Materials Used in
Engineering A12
C Properties of Rolled-Steel Shapes A16
D Beam Deflections and Slopes A28
E Fundamentals of Engineering Examination A29
Photo Credits C1
Index I1
Answers to Problems An1
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Preface
OBJECTIVES
The main objective of a basic mechanics course should be to develop
in the engineering student the ability to analyze a given problem in
a simple and logical manner and to apply to its solution a few fun-
damental and well-understood principles. This text is designed for
the first course in mechanics of materials—or strength of materials—
offered to engineering students in the sophomore or junior year. The
authors hope that it will help instructors achieve this goal in that
particular course in the same way that their other texts may have
helped them in statics and dynamics.
GENERAL APPROACH
In this text the study of the mechanics of materials is based on the
understanding of a few basic concepts and on the use of simplified
models. This approach makes it possible to develop all the necessary
formulas in a rational and logical manner, and to clearly indicate the
conditions under which they can be safely applied to the analysis and
design of actual engineering structures and machine components.
Free-body Diagrams Are Used Extensively. Throughout the
text free-body diagrams are used to determine external or internal
forces. The use of “picture equations” will also help the students
understand the superposition of loadings and the resulting stresses
and deformations.
Design Concepts Are Discussed Throughout the Text When-
ever Appropriate. A discussion of the application of the factor
of safety to design can be found in Chap. 1, where the concepts of
both allowable stress design and load and resistance factor design are
presented.
A Careful Balance Between SI and U.S. Customary Units Is
Consistently Maintained. Because it is essential that students be
able to handle effectively both SI metric units and U.S. customary
units, half the examples, sample problems, and problems to be
assigned have been stated in SI units and half in U.S. customary
units. Since a large number of problems are available, instructors can
assign problems using each system of units in whatever proportion
they find most desirable for their class.
Optional Sections Offer Advanced or Specialty Topics. Topics
such as residual stresses, torsion of noncircular and thin-walled mem-
bers, bending of curved beams, shearing stresses in non-symmetrical
xii
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xiiimembers, and failure criteria, have been included in optional sec-
tions for use in courses of varying emphases. To preserve the integ-
rity of the subject, these topics are presented in the proper
sequence, wherever they logically belong. Thus, even when not
covered in the course, they are highly visible and can be easily
referred to by the students if needed in a later course or in engi-
neering practice. For convenience all optional sections have been
indicated by asterisks.
CHAPTER ORGANIZATION
It is expected that students using this text will have completed a
course in statics. However, Chap. 1 is designed to provide them with
an opportunity to review the concepts learned in that course, while
shear and bending-moment diagrams are covered in detail in Secs.
5.2 and 5.3. The properties of moments and centroids of areas are
described in Appendix A; this material can be used to reinforce the
discussion of the determination of normal and shearing stresses in
beams (Chaps. 4, 5, and 6).
The first four chapters of the text are devoted to the analysis
of the stresses and of the corresponding deformations in various
structural members, considering successively axial loading, torsion,
and pure bending. Each analysis is based on a few basic concepts,
namely, the conditions of equilibrium of the forces exerted on the
member, the relations existing between stress and strain in the mate-
rial, and the conditions imposed by the supports and loading of the
member. The study of each type of loading is complemented by a
large number of examples, sample problems, and problems to be
assigned, all designed to strengthen the students’ understanding of
the subject.
The concept of stress at a point is introduced in Chap. 1, where
it is shown that an axial load can produce shearing stresses as well
as normal stresses, depending upon the section considered. The fact
that stresses depend upon the orientation of the surface on which
they are computed is emphasized again in Chaps. 3 and 4 in the
cases of torsion and pure bending. However, the discussion of com-
putational techniques—such as Mohr’s circle—used for the transfor-
mation of stress at a point is delayed until Chap. 7, after students
have had the opportunity to solve problems involving a combination
of the basic loadings and have discovered for themselves the need
for such techniques.
The discussion in Chap. 2 of the relation between stress and
strain in various materials includes fiber-reinforced composite mate-
rials. Also, the study of beams under transverse loads is covered in
two separate chapters. Chapter 5 is devoted to the determination of
the normal stresses in a beam and to the design of beams based
on the allowable normal stress in the material used (Sec. 5.4). The
chapter begins with a discussion of the shear and bending-moment
diagrams (Secs. 5.2 and 5.3) and includes an optional section on the
use of singularity functions for the determination of the shear and
bending moment in a beam (Sec. 5.5). The chapter ends with an
optional section on nonprismatic beams (Sec. 5.6).
Preface
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Chapter 6 is devoted to the determination of shearing stresses
in beams and thin-walled members under transverse loadings. The
formula for the shear flow, q 5 VQyI, is derived in the traditional
way. More advanced aspects of the design of beams, such as the
determination of the principal stresses at the junction of the flange
and web of a W-beam, are in Chap. 8, an optional chapter that may
be covered after the transformations of stresses have been discussed
in Chap. 7. The design of transmission shafts is in that chapter for
the same reason, as well as the determination of stresses under com-
bined loadings that can now include the determination of the prin-
cipal stresses, principal planes, and maximum shearing stress at a
given point.
Statically indeterminate problems are first discussed in Chap. 2
and considered throughout the text for the various loading conditions
encountered. Thus, students are presented at an early stage with a
method of solution that combines the analysis of deformations with
the conventional analysis of forces used in statics. In this way, they
will have become thoroughly familiar with this fundamental method
by the end of the course. In addition, this approach helps the stu-
dents realize that stresses themselves are statically indeterminate and
can be computed only by considering the corresponding distribution
of strains.
The concept of plastic deformation is introduced in Chap. 2,
where it is applied to the analysis of members under axial loading.
Problems involving the plastic deformation of circular shafts and of
prismatic beams are also considered in optional sections of Chaps. 3,
4, and 6. While some of this material can be omitted at the choice
of the instructor, its inclusion in the body of the text will help stu-
dents realize the limitations of the assumption of a linear stress-strain
relation and serve to caution them against the inappropriate use of
the elastic torsion and flexure formulas.
The determination of the deflection of beams is discussed in
Chap. 9. The first part of the chapter is devoted to the integration
method and to the method of superposition, with an optional section
(Sec. 9.6) based on the use of singularity functions. (This section
should be used only if Sec. 5.5 was covered earlier.) The second part
of Chap. 9 is optional. It presents the moment-area method in two
lessons.
Chapter 10 is devoted to columns and contains material on the
design of steel, aluminum, and wood columns. Chapter 11 covers
energy methods, including Castigliano’s theorem.
PEDAGOGICAL FEATURES
Each chapter begins with an introductory section setting the purpose
and goals of the chapter and describing in simple terms the material
to be covered and its application to the solution of engineering
problems.
Chapter Lessons. The body of the text has been divided into
units, each consisting of one or several theory sections followed by
sample problems and a large number of problems to be assigned.
xiv Preface
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xvEach unit corresponds to a well-defined topic and generally can be
covered in one lesson.
Examples and Sample Problems. The theory sections include
many examples designed to illustrate the material being presented
and facilitate its understanding. The sample problems are intended
to show some of the applications of the theory to the solution of
engineering problems. Since they have been set up in much the same
form that students will use in solving the assigned problems, the
sample problems serve the double purpose of amplifying the text and
demonstrating the type of neat and orderly work that students should
cultivate in their own solutions.
Homework Problem Sets. Most of the problems are of a practi-
cal nature and should appeal to engineering students. They are pri-
marily designed, however, to illustrate the material presented in the
text and help the students understand the basic principles used in
mechanics of materials. The problems have been grouped according
to the portions of material they illustrate and have been arranged in
order of increasing difficulty. Problems requiring special attention
have been indicated by asterisks. Answers to problems are given at
the end of the book, except for those with a number set in italics.
Chapter Review and Summary. Each chapter ends with a
review and summary of the material covered in the chapter. Notes
in the margin have been included to help the students organize their
review work, and cross references provided to help them find the
portions of material requiring their special attention.
Review Problems. A set of review problems is included at the end
of each chapter. These problems provide students further opportunity
to apply the most important concepts introduced in the chapter.
Computer Problems. Computers make it possible for engineering
students to solve a great number of challenging problems. A group
of six or more problems designed to be solved with a computer can
be found at the end of each chapter. These problems can be solved
using any computer language that provides a basis for analytical cal-
culations. Developing the algorithm required to solve a given problem
will benefit the students in two different ways: (1) it will help them
gain a better understanding of the mechanics principles involved;
(2) it will provide them with an opportunity to apply the skills acquired
in their computer programming course to the solution of a meaning-
ful engineering problem. These problems can be solved using any
computer language that provide a basis for analytical calculations.
Fundamentals of Engineering Examination. Engineers who
seek to be licensed as Professional Engineers must take two exams.
The first exam, the Fundamentals of Engineering Examination,
includes subject material from Mechanics of Materials. Appendix E
lists the topics in Mechanics of Materials that are covered in this exam
along with problems that can be solved to review this material.
Preface
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SUPPLEMENTAL RESOURCES
Instructor’s Solutions Manual. The Instructor’s and Solutions
Manual that accompanies the sixth edition continues the tradition of
exceptional accuracy and keeping solutions contained to a single page
for easier reference. The manual also features tables designed to assist
instructors in creating a schedule of assignments for their courses.
The various topics covered in the text are listed in Table I, and a
suggested number of periods to be spent on each topic is indicated.
Table II provides a brief description of all groups of problems and a
classification of the problems in each group according to the units
used. Sample lesson schedules are also found within the manual.
MCGRAW-HILL CONNECT ENGINEERING
McGraw-Hill Connect EngineeringTM
is a web-based assignment and
assessment platform that gives students the means to better connect
with their coursework, with their instructors, and with the important
concepts that they will need to know for success now and in the
future. With Connect Engineering, instructors can deliver assign-
ments, quizzes, and tests easily online. Students can practice impor-
tant skills at their own pace and on their own schedule. With Connect
Engineering Plus, students also get 24/7 online access to an eBook—
an online edition of the text—to aid them in successfully completing
their work, wherever and whenever they choose.
Connect Engineering for Mechanics of Materials is available at
www.mcgrawhillconnect.com
McGRAW-HILL CREATE™
Craft your teaching resources to match the way you teach! With
McGraw-Hill CreateTM
, www.mcgrawhillcreate.com, you can easily
rearrange chapters, combine material from other content sources, and
quickly upload content you have written like your course syllabus or
teaching notes. Arrange your book to fit your teaching style. Create
even allows you to personalize your book’s appearance by selecting
the cover and adding your name, school, and course information.
Order a Create book and you’ll receive a complimentary print review
copy in 3–5 business days or a complimentary electronic review copy
(eComp) via email in minutes. Go to www.mcgrawhillcreate.com
today and register to experience how McGraw-Hill Create empowers
you to teach your students your way.
McGraw-Hill Higher Education and Blackboard®
have
teamed up.
Blackboard, the Web-based course-management system, has
partnered with McGraw-Hill to better allow students and faculty to
use online materials and activities to complement face-to-face teach-
ing. Blackboard features exciting social learning and teaching tools
that foster more logical, visually impactful and active learning oppor-
tunities for students. You’ll transform your closed-door classrooms
into communities where students remain connected to their educa-
tional experience 24 hours a day.
This partnership allows you and your students access to
McGraw-Hill’s Connect and Create right from within your Black-
board course—all with one single sign-on.
xvi Preface
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xviiNot only do you get single sign-on with Connect and Create, you
also get deep integration of McGraw-Hill content and content engines
right in Blackboard. Whether you’re choosing a book for your course
or building Connect assignments, all the tools you need are right
where you want them—inside of Blackboard.
Gradebooks are now seamless. When a student completes an
integrated Connect assignment, the grade for that assignment auto-
matically (and instantly) feeds your Blackboard grade center.
McGraw-Hill and Blackboard can now offer you easy access to
industry leading technology and content, whether your campus hosts
it, or we do. Be sure to ask your local McGraw-Hill representative
for details.
ADDITIONAL ONLINE RESOURCES
Mechanics of Materials 6e also features a companion website (www.
mhhe.com/beerjohnston) for instructors. Included on the website are
lecture PowerPoints, an image library, and animations. Via the website,
instructors can also request access to C.O.S.M.O.S., a complete online
solutions manual organization system that allows instructors to create
custom homework, quizzes, and tests using end-of-chapter problems
from the text. For access to this material, contact your sales representa-
tive for a user name and password.
Hands-On Mechanics. Hands-On Mechanics is a website
designed for instructors who are interested in incorporating three-
dimensional, hands-on teaching aids into their lectures. Developed
through a partnership between McGraw-Hill and the Department
of Civil and Mechanical Engineering at the United States Military
Academy at West Point, this website not only provides detailed
instructions for how to build 3-D teaching tools using materials
found in any lab or local hardware store but also provides a com-
munity where educators can share ideas, trade best practices, and
submit their own demonstrations for posting on the site. Visit www.
handsonmechanics.com to see how you can put this to use in your
classroom.
ACKNOWLEDGMENTS
The authors thank the many companies that provided photographs
for this edition. We also wish to recognize the determined efforts
and patience of our photo researcher Sabina Dowell.
Our special thanks go to Professor Dean Updike, of the Depart-
ment of Mechanical Engineering and Mechanics, Lehigh University
for his patience and cooperation as he checked the solutions and
answers of all the problems in this edition.
We also gratefully acknowledge the help, comments and sug-
gestions offered by the many reviewers and users of previous editions
of Mechanics of Materials.
John T. DeWolf
David F. Mazurek
Preface
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a Constant; distance
A, B, C, . . . Forces; reactions
A, B, C, . . . Points
A, A Area
b Distance; width
c Constant; distance; radius
C Centroid
C1, C2, . . . Constants of integration
CP Column stability factor
d Distance; diameter; depth
D Diameter
e Distance; eccentricity; dilatation
E Modulus of elasticity
f Frequency; function
F Force
F.S. Factor of safety
G Modulus of rigidity; shear modulus
h Distance; height
H Force
H, J, K Points
I, Ix, . . . Moment of inertia
Ixy, . . . Product of inertia
J Polar moment of inertia
k Spring constant; shape factor; bulk modulus;
constant
K Stress concentration factor; torsional spring constant
l Length; span
L Length; span
Le Effective length
m Mass
M Couple
M, Mx, . . . Bending moment
MD Bending moment, dead load (LRFD)
ML Bending moment, live load (LRFD)
MU Bending moment, ultimate load (LRFD)
n Number; ratio of moduli of elasticity; normal
direction
p Pressure
P Force; concentrated load
PD Dead load (LRFD)
PL Live load (LRFD)
PU Ultimate load (LRFD)
q Shearing force per unit length; shear flow
Q Force
Q First moment of area
List of Symbols
xviii
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xixr Radius; radius of gyration
R Force; reaction
R Radius; modulus of rupture
s Length
S Elastic section modulus
t Thickness; distance; tangential deviation
T Torque
T Temperature
u, v Rectangular coordinates
u Strain-energy density
U Strain energy; work
v Velocity
V Shearing force
V Volume; shear
w Width; distance; load per unit length
W, W Weight, load
x, y, z Rectangular coordinates; distance; displacements;
deflections
x, y, z Coordinates of centroid
Z Plastic section modulus
a, b, g Angles
a Coefficient of thermal expansion; influence
coefficient
g Shearing strain; specific weight
gD Load factor, dead load (LRFD)
gL Load factor, live load (LRFD)
d Deformation; displacement
e Normal strain
u Angle; slope
l Direction cosine
n Poisson’s ratio
r Radius of curvature; distance; density
s Normal stress
t Shearing stress
f Angle; angle of twist; resistance factor
v Angular velocity
List of Symbols
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This chapter is devoted to the study of
the stresses occurring in many of the
elements contained in these excavators,
such as two-force members, axles,
bolts, and pins.
2
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C H A P T E R
3
Introduction—Concept of Stress
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4
Chapter 1 Introduction—Concept
of Stress
1.1 Introduction
1.2 A Short Review of the Methods
of Statics
1.3 Stresses in the Members of a
Structure
1.4 Analysis and Design
1.5 Axial Loading; Normal Stress
1.6 Shearing Stress
1.7 Bearing Stress in Connections
1.8 Application to the Analysis and
Design of Simple Structures
1.9 Method of Problem Solution
1.10 Numerical Accuracy
1.11 Stress on an Oblique Plane
Under Axial Loading
1.12 Stress Under General Loading
Conditions; Components of Stress
1.13 Design Considerations
1.1 INTRODUCTION
The main objective of the study of the mechanics of materials is to
provide the future engineer with the means of analyzing and design-
ing various machines and load-bearing structures.
Both the analysis and the design of a given structure involve
the determination of stresses and deformations. This first chapter is
devoted to the concept of stress.
Section 1.2 is devoted to a short review of the basic methods of
statics and to their application to the determination of the forces in the
members of a simple structure consisting of pin-connected members.
Section 1.3 will introduce you to the concept of stress in a member of
a structure, and you will be shown how that stress can be determined
from the force in the member. After a short discussion of engineering
analysis and design (Sec. 1.4), you will consider successively the normal
stresses in a member under axial loading (Sec. 1.5), the shearing stresses
caused by the application of equal and opposite transverse forces
(Sec. 1.6), and the bearing stresses created by bolts and pins in the
members they connect (Sec. 1.7). These various concepts will be
applied in Sec. 1.8 to the determination of the stresses in the members
of the simple structure considered earlier in Sec. 1.2.
The first part of the chapter ends with a description of the
method you should use in the solution of an assigned problem (Sec.
1.9) and with a discussion of the numerical accuracy appropriate in
engineering calculations (Sec. 1.10).
In Sec. 1.11, where a two-force member under axial loading is
considered again, it will be observed that the stresses on an oblique
plane include both normal and shearing stresses, while in Sec. 1.12 you
will note that six components are required to describe the state of stress
at a point in a body under the most general loading conditions.
Finally, Sec. 1.13 will be devoted to the determination from
test specimens of the ultimate strength of a given material and to
the use of a factor of safety in the computation of the allowable load
for a structural component made of that material.
1.2 A SHORT REVIEW OF THE METHODS OF STATICS
In this section you will review the basic methods of statics while
determining the forces in the members of a simple structure.
Consider the structure shown in Fig. 1.1, which was designed
to support a 30-kN load. It consists of a boom AB with a 30 3 50-mm
rectangular cross section and of a rod BC with a 20-mm-diameter
circular cross section. The boom and the rod are connected by a pin
at B and are supported by pins and brackets at A and C, respectively.
Our first step should be to draw a free-body diagram of the structure
by detaching it from its supports at A and C, and showing the reac-
tions that these supports exert on the structure (Fig. 1.2). Note that
the sketch of the structure has been simplified by omitting all unnec-
essary details. Many of you may have recognized at this point that
AB and BC are two-force members. For those of you who have not,
we will pursue our analysis, ignoring that fact and assuming that the
directions of the reactions at A and C are unknown. Each of these
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5
reactions, therefore, will be represented by two components, Ax and
Ay at A, and Cx and Cy at C. We write the following three equilib-
rium equations:
1l o MC 5 0: Ax10.6 m2 2 130 kN210.8 m2 5 0
Ax 5 140 kN (1.1)
y1 o Fx 5 0: Ax 1 Cx 5 0
Cx 5 2Ax Cx 5 240 kN (1.2)
1x o Fy 5 0: Ay 1 Cy 2 30 kN 5 0
Ay 1 Cy 5 130 kN (1.3)
We have found two of the four unknowns, but cannot determine the
other two from these equations, and no additional independent
equation can be obtained from the free-body diagram of the struc-
ture. We must now dismember the structure. Considering the free-
body diagram of the boom AB (Fig. 1.3), we write the following
equilibrium equation:
1l o MB 5 0: 2Ay10.8 m2 5 0 Ay 5 0 (1.4)
Substituting for Ay from (1.4) into (1.3), we obtain Cy 5 130 kN.
Expressing the results obtained for the reactions at A and C in vector
form, we have
A 5 40 kN y Cx 5 40 kN z, Cy 5 30 kNx
We note that the reaction at A is directed along the axis of the boom
AB and causes compression in that member. Observing that the com-
ponents Cx and Cy of the reaction at C are, respectively, proportional
to the horizontal and vertical components of the distance from B to
C, we conclude that the reaction at C is equal to 50 kN, is directed
along the axis of the rod BC, and causes tension in that member.
800 mm
50 mm
30 kN
600 mm
d ϭ 20 mm
C
A
B
Fig. 1.1 Boom used to support a 30-kN load.
Fig. 1.2
30 kN
0.8 m
0.6 m
B
Cx
Cy
Ay
C
AAx
Fig. 1.3
30 kN
0.8 m
Ay By
A BAx Bz
1.2 A Short Review of the Methods of Statics
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These results could have been anticipated by recognizing that
AB and BC are two-force members, i.e., members that are sub-
jected to forces at only two points, these points being A and B for
member AB, and B and C for member BC. Indeed, for a two-force
member the lines of action of the resultants of the forces acting at
each of the two points are equal and opposite and pass through
both points. Using this property, we could have obtained a simpler
solution by considering the free-body diagram of pin B. The forces
on pin B are the forces FAB and FBC exerted, respectively, by mem-
bers AB and BC, and the 30-kN load (Fig. 1.4a). We can express
that pin B is in equilibrium by drawing the corresponding force
triangle (Fig. 1.4b).
Since the force FBC is directed along member BC, its slope is
the same as that of BC, namely, 3/4. We can, therefore, write the
proportion
FAB
4
5
FBC
5
5
30 kN
3
from which we obtain
FAB 5 40 kN FBC 5 50 kN
The forces F9AB and F9BC exerted by pin B, respectively, on boom AB
and rod BC are equal and opposite to FAB and FBC (Fig. 1.5).
Knowing the forces at the ends of each of the members, we
can now determine the internal forces in these members. Passing
a section at some arbitrary point D of rod BC, we obtain two por-
tions BD and CD (Fig. 1.6). Since 50-kN forces must be applied
at D to both portions of the rod to keep them in equilibrium, we
conclude that an internal force of 50 kN is produced in rod BC
when a 30-kN load is applied at B. We further check from the
directions of the forces FBC and F9BC in Fig. 1.6 that the rod is
in tension. A similar procedure would enable us to determine that
the internal force in boom AB is 40 kN and that the boom is in
compression.
Fig. 1.4
(a) (b)
FBC
FBC
FAB FAB
30 kN
30 kN
3
5
4
B
Fig. 1.5
FAB F'AB
FBC
F'BCB
A B
C
Fig. 1.6
C
D
B
D
FBC
FBC F'BC
F'BC
6 Introduction—Concept of Stress
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71.3 STRESSES IN THE MEMBERS OF A STRUCTURE
While the results obtained in the preceding section represent a first
and necessary step in the analysis of the given structure, they do not
tell us whether the given load can be safely supported. Whether rod
BC, for example, will break or not under this loading depends not
only upon the value found for the internal force FBC, but also upon
the cross-sectional area of the rod and the material of which the rod
is made. Indeed, the internal force FBC actually represents the resul-
tant of elementary forces distributed over the entire area A of the
cross section (Fig. 1.7) and the average intensity of these distributed
forces is equal to the force per unit area, FBCyA, in the section.
Whether or not the rod will break under the given loading clearly
depends upon the ability of the material to withstand the corre-
sponding value FBCyA of the intensity of the distributed internal
forces. It thus depends upon the force FBC, the cross-sectional area
A, and the material of the rod.
The force per unit area, or intensity of the forces distributed
over a given section, is called the stress on that section and is
denoted by the Greek letter s (sigma). The stress in a member of
cross-sectional area A subjected to an axial load P (Fig. 1.8) is
therefore obtained by dividing the magnitude P of the load by the
area A:
s 5
P
A
(1.5)
A positive sign will be used to indicate a tensile stress (member in
tension) and a negative sign to indicate a compressive stress (mem-
ber in compression).
Since SI metric units are used in this discussion, with P ex-
pressed in newtons (N) and A in square meters (m2
), the stress s
will be expressed in N/m2
. This unit is called a pascal (Pa). How-
ever, one finds that the pascal is an exceedingly small quantity and
that, in practice, multiples of this unit must be used, namely, the
kilopascal (kPa), the megapascal (MPa), and the gigapascal (GPa).
We have
1 kPa 5 103
Pa 5 103
N/m2
1 MPa 5 106
Pa 5 106
N/m2
1 GPa 5 109
Pa 5 109
N/m2
When U.S. customary units are used, the force P is usually
expressed in pounds (lb) or kilopounds (kip), and the cross-sectional
area A in square inches (in2
). The stress s will then be expressed in
pounds per square inch (psi) or kilopounds per square inch (ksi).†
†The principal SI and U.S. customary units used in mechanics are listed in tables inside
the front cover of this book. From the table on the right-hand side, we note that 1 psi is
approximately equal to 7 kPa, and 1 ksi approximately equal to 7 MPa.
Fig. 1.7
A
FBCFBC
A
ϭ
Fig. 1.8 Member with an axial load.
(a) (b)
A
P
A
P' P'
P
ϭ
1.3 Stresses in the Members of a Structure
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1.4 ANALYSIS AND DESIGN
Considering again the structure of Fig. 1.1, let us assume that rod BC
is made of a steel with a maximum allowable stress sall 5 165 MPa.
Can rod BC safely support the load to which it will be subjected? The
magnitude of the force FBC in the rod was found earlier to be 50 kN.
Recalling that the diameter of the rod is 20 mm, we use Eq. (1.5) to
determine the stress created in the rod by the given loading. We
have
P 5 FBC 5 150 kN 5 150 3 103
N
A 5 pr2
5 pa
20 mm
2
b
2
5 p110 3 1023
m22
5 314 3 1026
m2
s 5
P
A
5
150 3 103
N
314 3 1026
m2 5 1159 3 106
Pa 5 1159 MPa
Since the value obtained for s is smaller than the value sall of the
allowable stress in the steel used, we conclude that rod BC can safely
support the load to which it will be subjected. To be complete, our
analysis of the given structure should also include the determination
of the compressive stress in boom AB, as well as an investigation of
the stresses produced in the pins and their bearings. This will be
discussed later in this chapter. We should also determine whether
the deformations produced by the given loading are acceptable. The
study of deformations under axial loads will be the subject of Chap. 2.
An additional consideration required for members in compression
involves the stability of the member, i.e., its ability to support a given
load without experiencing a sudden change in configuration. This
will be discussed in Chap. 10.
The engineer’s role is not limited to the analysis of existing
structures and machines subjected to given loading conditions. Of
even greater importance to the engineer is the design of new struc-
tures and machines, that is, the selection of appropriate components
to perform a given task. As an example of design, let us return to
the structure of Fig. 1.1, and assume that aluminum with an allow-
able stress sall 5 100 MPa is to be used. Since the force in rod BC
will still be P 5 FBC 5 50 kN under the given loading, we must have,
from Eq. (1.5),
sall 5
P
A
A 5
P
sall
5
50 3 103
N
100 3 106
Pa
5 500 3 1026
m2
and, since A 5 pr2
,
r 5
B
A
p
5
B
500 3 1026
m2
p
5 12.62 3 1023
m 5 12.62 mm
d 5 2r 5 25.2 mm
We conclude that an aluminum rod 26 mm or more in diameter will
be adequate.
8 Introduction—Concept of Stress
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91.5 AXIAL LOADING; NORMAL STRESS
As we have already indicated, rod BC of the example considered in
the preceding section is a two-force member and, therefore, the
forces FBC and F9BC acting on its ends B and C (Fig. 1.5) are directed
along the axis of the rod. We say that the rod is under axial loading.
An actual example of structural members under axial loading is pro-
vided by the members of the bridge truss shown in Photo 1.1.
Returning to rod BC of Fig. 1.5, we recall that the section we
passed through the rod to determine the internal force in the rod
and the corresponding stress was perpendicular to the axis of the
rod; the internal force was therefore normal to the plane of the sec-
tion (Fig. 1.7) and the corresponding stress is described as a normal
stress. Thus, formula (1.5) gives us the normal stress in a member
under axial loading:
s 5
P
A
(1.5)
We should also note that, in formula (1.5), s is obtained by
dividing the magnitude P of the resultant of the internal forces dis-
tributed over the cross section by the area A of the cross section; it
represents, therefore, the average value of the stress over the cross
section, rather than the stress at a specific point of the cross section.
To define the stress at a given point Q of the cross section, we
should consider a small area DA (Fig. 1.9). Dividing the magnitude
of DF by DA, we obtain the average value of the stress over DA.
Letting DA approach zero, we obtain the stress at point Q:
s 5 lim
¢Ay0
¢F
¢A
(1.6)
Photo 1.1 This bridge truss consists of two-force members that may be in
tension or in compression.
Fig. 1.9
P'
Q
⌬A
⌬F
1.5 Axial Loading; Normal Stress
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10 Introduction—Concept of Stress In general, the value obtained for the stress s at a given point
Q of the section is different from the value of the average stress
given by formula (1.5), and s is found to vary across the section.
In a slender rod subjected to equal and opposite concentrated loads
P and P9 (Fig. 1.10a), this variation is small in a section away from
the points of application of the concentrated loads (Fig. 1.10c), but
it is quite noticeable in the neighborhood of these points (Fig.
1.10b and d).
It follows from Eq. (1.6) that the magnitude of the resultant of
the distributed internal forces is
#dF 5 #A
s dA
But the conditions of equilibrium of each of the portions of rod
shown in Fig. 1.10 require that this magnitude be equal to the mag-
nitude P of the concentrated loads. We have, therefore,
P 5 #dF 5 #A
s dA (1.7)
which means that the volume under each of the stress surfaces in
Fig. 1.10 must be equal to the magnitude P of the loads. This, how-
ever, is the only information that we can derive from our knowledge
of statics, regarding the distribution of normal stresses in the various
sections of the rod. The actual distribution of stresses in any given
section is statically indeterminate. To learn more about this distribu-
tion, it is necessary to consider the deformations resulting from the
particular mode of application of the loads at the ends of the rod.
This will be discussed further in Chap. 2.
In practice, it will be assumed that the distribution of normal
stresses in an axially loaded member is uniform, except in the imme-
diate vicinity of the points of application of the loads. The value s
of the stress is then equal to save and can be obtained from formula
(1.5). However, we should realize that, when we assume a uniform
distribution of stresses in the section, i.e., when we assume that the
internal forces are uniformly distributed across the section, it follows
from elementary statics† that the resultant P of the internal forces
must be applied at the centroid C of the section (Fig. 1.11). This
means that a uniform distribution of stress is possible only if the line
of action of the concentrated loads P and P9 passes through the cen-
troid of the section considered (Fig. 1.12). This type of loading is
called centric loading and will be assumed to take place in all straight
two-force members found in trusses and pin-connected structures,
such as the one considered in Fig. 1.1. However, if a two-force mem-
ber is loaded axially, but eccentrically as shown in Fig. 1.13a, we find
from the conditions of equilibrium of the portion of member shown
in Fig. 1.13b that the internal forces in a given section must be
†See Ferdinand P. Beer and E. Russell Johnston, Jr., Mechanics for Engineers, 5th ed.,
McGraw-Hill, New York, 2008, or Vector Mechanics for Engineers, 9th ed., McGraw-Hill,
New York, 2010, Secs. 5.2 and 5.3.
Fig. 1.10 Stress distributions at
different sections along axially loaded
member.
(a) (b) (c) (d)
P' P' P' P'
P
Fig. 1.11
C
P
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11
equivalent to a force P applied at the centroid of the section and a
couple M of moment M 5 Pd. The distribution of forces—and, thus,
the corresponding distribution of stresses—cannot be uniform. Nor
can the distribution of stresses be symmetric as shown in Fig. 1.10.
This point will be discussed in detail in Chap. 4.
1.6 SHEARING STRESS
The internal forces and the corresponding stresses discussed in Secs.
1.2 and 1.3 were normal to the section considered. A very different
type of stress is obtained when transverse forces P and P9 are applied
to a member AB (Fig. 1.14). Passing a section at C between the
points of application of the two forces (Fig. 1.15a), we obtain the
diagram of portion AC shown in Fig. 1.15b. We conclude that inter-
nal forces must exist in the plane of the section, and that their resul-
tant is equal to P. These elementary internal forces are called shearing
forces, and the magnitude P of their resultant is the shear in the
section. Dividing the shear P by the area A of the cross section, we
Fig. 1.12
C
P
P'
Fig. 1.13 Eccentric axial loading.
MC
d
d
(a) (b)
P'P'
P
P
Fig. 1.14 Member with
transverse loads.
A B
P'
P
Fig. 1.15
A C
A C
B
(a)
(b)
P
P
PЈ
P'
1.6 Shearing Stress
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12 Introduction—Concept of Stress obtain the average shearing stress in the section. Denoting the shear-
ing stress by the Greek letter t (tau), we write
tave 5
P
A
(1.8)
It should be emphasized that the value obtained is an average
value of the shearing stress over the entire section. Contrary to what
we said earlier for normal stresses, the distribution of shearing
stresses across the section cannot be assumed uniform. As you will
see in Chap. 6, the actual value t of the shearing stress varies from
zero at the surface of the member to a maximum value tmax that may
be much larger than the average value tave.
Shearing stresses are commonly found in bolts, pins, and rivets
used to connect various structural members and machine compo-
nents (Photo 1.2). Consider the two plates A and B, which are con-
nected by a bolt CD (Fig. 1.16). If the plates are subjected to tension
forces of magnitude F, stresses will develop in the section of bolt
corresponding to the plane EE9. Drawing the diagrams of the bolt
and of the portion located above the plane EE9 (Fig. 1.17), we con-
clude that the shear P in the section is equal to F. The average
shearing stress in the section is obtained, according to formula (1.8),
by dividing the shear P 5 F by the area A of the cross section:
tave 5
P
A
5
F
A
(1.9)
Photo 1.2 Cutaway view of a connection with a bolt in shear.
Fig. 1.16 Bolt subject to single shear.
C
D
A
F
E'B
E
F'
Fig. 1.17
C C
D
F
PEЈE
(a) (b)
F
F'
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13
The bolt we have just considered is said to be in single shear.
Different loading situations may arise, however. For example, if
splice plates C and D are used to connect plates A and B (Fig. 1.18),
shear will take place in bolt HJ in each of the two planes KK9 and
LL9 (and similarly in bolt EG). The bolts are said to be in double
shear. To determine the average shearing stress in each plane, we
draw free-body diagrams of bolt HJ and of the portion of bolt located
between the two planes (Fig. 1.19). Observing that the shear P in
each of the sections is P 5 Fy2, we conclude that the average shear-
ing stress is
tave 5
P
A
5
Fy2
A
5
F
2A
(1.10)
1.7 BEARING STRESS IN CONNECTIONS
Bolts, pins, and rivets create stresses in the members they connect,
along the bearing surface, or surface of contact. For example, con-
sider again the two plates A and B connected by a bolt CD that we
have discussed in the preceding section (Fig. 1.16). The bolt exerts
on plate A a force P equal and opposite to the force F exerted by
the plate on the bolt (Fig. 1.20). The force P represents the resultant
of elementary forces distributed on the inside surface of a half-
cylinder of diameter d and of length t equal to the thickness of the
plate. Since the distribution of these forces—and of the correspond-
ing stresses—is quite complicated, one uses in practice an average
nominal value sb of the stress, called the bearing stress, obtained by
dividing the load P by the area of the rectangle representing the
projection of the bolt on the plate section (Fig. 1.21). Since this area
is equal to td, where t is the plate thickness and d the diameter of
the bolt, we have
sb 5
P
A
5
P
td
(1.11)
1.8 APPLICATION TO THE ANALYSIS AND DESIGN
OF SIMPLE STRUCTURES
We are now in a position to determine the stresses in the members
and connections of various simple two-dimensional structures and,
thus, to design such structures.
Fig. 1.19
K
L
H
J
K'
L'
F
FC
FD
F
P
P
(a) (b)
Fig. 1.18 Bolts subject to double shear.
K
AB
L
E H
G J
C
D
K'
L'
FF'
Fig. 1.20
A
C
D
d
t
F
P
F'
Fig. 1.21
A d
t
1.8 Application to the Analysis and
Design of Simple Structures
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14 Introduction—Concept of Stress As an example, let us return to the structure of Fig. 1.1 that
we have already considered in Sec. 1.2 and let us specify the supports
and connections at A, B, and C. As shown in Fig. 1.22, the 20-mm-
diameter rod BC has flat ends of 20 3 40-mm rectangular cross
section, while boom AB has a 30 3 50-mm rectangular cross section
and is fitted with a clevis at end B. Both members are connected at
B by a pin from which the 30-kN load is suspended by means of a
U-shaped bracket. Boom AB is supported at A by a pin fitted into a
double bracket, while rod BC is connected at C to a single bracket.
All pins are 25 mm in diameter.
Fig. 1.22
800 mm
50 mm
Q ϭ 30 kN Q ϭ 30 kN
20 mm
20 mm
25 mm
30 mm
25 mm
d ϭ 25 mm
d ϭ 25 mm
d ϭ 20 mm
d ϭ 20 mm
d ϭ 25 mm
40 mm
20 mm
A
A
B
B
B
C
C
B
FRONT VIEW
TOP VIEW OF BOOM AB
END VIEW
TOP VIEW OF ROD BCFlat end
Flat end
600 mm
a. Determination of the Normal Stress in Boom AB and
Rod BC. As we found in Secs. 1.2 and 1.4, the force in rod BC is
FBC 5 50 kN (tension) and the area of its circular cross section is
A 5 314 3 1026
m2
; the corresponding average normal stress is
sBC 5 1159 MPa. However, the flat parts of the rod are also under
tension and at the narrowest section, where a hole is located, we
have
A 5 120 mm2140 mm 2 25 mm2 5 300 3 1026
m2
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15The corresponding average value of the stress, therefore, is
1sBC2end 5
P
A
5
50 3 103
N
300 3 1026
m2 5 167 MPa
Note that this is an average value; close to the hole, the stress will
actually reach a much larger value, as you will see in Sec. 2.18. It is
clear that, under an increasing load, the rod will fail near one of the
holes rather than in its cylindrical portion; its design, therefore, could
be improved by increasing the width or the thickness of the flat ends
of the rod.
Turning now our attention to boom AB, we recall from Sec. 1.2
that the force in the boom is FAB 5 40 kN (compression). Since the
area of the boom’s rectangular cross section is A 5 30 mm 3 50 mm 5
1.5 3 1023
m2
, the average value of the normal stress in the main
part of the rod, between pins A and B, is
sAB 5 2
40 3 103
N
1.5 3 1023
m2 5 226.7 3 106
Pa 5 226.7 MPa
Note that the sections of minimum area at A and B are not under
stress, since the boom is in compression, and, therefore, pushes on
the pins (instead of pulling on the pins as rod BC does).
b. Determination of the Shearing Stress in Various
Connections. To determine the shearing stress in a connection
such as a bolt, pin, or rivet, we first clearly show the forces exerted
by the various members it connects. Thus, in the case of pin C of
our example (Fig. 1.23a), we draw Fig. 1.23b, showing the 50-kN
force exerted by member BC on the pin, and the equal and opposite
force exerted by the bracket. Drawing now the diagram of the portion
of the pin located below the plane DD9 where shearing stresses occur
(Fig. 1.23c), we conclude that the shear in that plane is P 5 50 kN.
Since the cross-sectional area of the pin is
A 5 pr2
5 pa
25 mm
2
b
2
5 p112.5 3 1023
m22
5 491 3 1026
m2
we find that the average value of the shearing stress in the pin at
C is
tave 5
P
A
5
50 3 103
N
491 3 1026
m2 5 102 MPa
Considering now the pin at A (Fig. 1.24), we note that it is in
double shear. Drawing the free-body diagrams of the pin and of the
portion of pin located between the planes DD9 and EE9 where shear-
ing stresses occur, we conclude that P 5 20 kN and that
tave 5
P
A
5
20 kN
491 3 1026
m2 5 40.7 MPa
Fig. 1.23
50 kN
50 kN
(a)
C
(b)
Fb
D'
D
d ϭ 25 mm
50 kN
(c)
P
Fig. 1.24
(a)
(b)
40 kN
40 kN
Fb
Fb
A
D'
E'
D
E
d ϭ 25 mm
(c)
40 kN
P
P
1.8 Application to the Analysis and
Design of Simple Structures
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16 Introduction—Concept of Stress Considering the pin at B (Fig. 1.25a), we note that the pin
may be divided into five portions which are acted upon by forces
exerted by the boom, rod, and bracket. Considering successively
the portions DE (Fig. 1.25b) and DG (Fig. 1.25c), we conclude that
the shear in section E is PE 5 15 kN, while the shear in section G
is PG 5 25 kN. Since the loading of the pin is symmetric, we con-
clude that the maximum value of the shear in pin B is PG 5 25 kN,
and that the largest shearing stresses occur in sections G and H,
where
tave 5
PG
A
5
25 kN
491 3 1026
m2 5 50.9 MPa
c. Determination of the Bearing Stresses. To determine the
nominal bearing stress at A in member AB, we use formula (1.11)
of Sec. 1.7. From Fig. 1.22, we have t 5 30 mm and d 5 25 mm.
Recalling that P 5 FAB 5 40 kN, we have
sb 5
P
td
5
40 kN
130 mm2125 mm2
5 53.3 MPa
To obtain the bearing stress in the bracket at A, we use t 5 2(25 mm)
5 50 mm and d 5 25 mm:
sb 5
P
td
5
40 kN
150 mm2125 mm2
5 32.0 MPa
The bearing stresses at B in member AB, at B and C in mem-
ber BC, and in the bracket at C are found in a similar way.
1.9 METHOD OF PROBLEM SOLUTION
You should approach a problem in mechanics of materials as you
would approach an actual engineering situation. By drawing on your
own experience and intuition, you will find it easier to understand
and formulate the problem. Once the problem has been clearly
stated, however, there is no place in its solution for your particular
fancy. Your solution must be based on the fundamental principles of
statics and on the principles you will learn in this course. Every step
you take must be justified on that basis, leaving no room for your
“intuition.” After an answer has been obtained, it should be checked.
Here again, you may call upon your common sense and personal
experience. If not completely satisfied with the result obtained, you
should carefully check your formulation of the problem, the validity
of the methods used in its solution, and the accuracy of your
computations.
The statement of the problem should be clear and precise. It
should contain the given data and indicate what information is
required. A simplified drawing showing all essential quantities
involved should be included. The solution of most of the problems
you will encounter will necessitate that you first determine the reac-
tions at supports and internal forces and couples. This will require
Fig. 1.25
(a)
(b)
(c)
1
2 FAB ϭ 20 kN
FBC ϭ 50 kN
1
2 FAB ϭ 20 kN
1
2 FAB ϭ 20 kN
1
2 Q ϭ 15 kN
1
2 Q ϭ 15 kN
1
2 Q ϭ 15 kN
1
2 Q ϭ 15 kN
Pin B
D
D
D
E
E
G
G
PE
PG
H
J
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17the drawing of one or several free-body diagrams, as was done in
Sec. 1.2, from which you will write equilibrium equations. These
equations can be solved for the unknown forces, from which the
required stresses and deformations will be computed.
After the answer has been obtained, it should be carefully
checked. Mistakes in reasoning can often be detected by carrying
the units through your computations and checking the units
obtained for the answer. For example, in the design of the rod
discussed in Sec. 1.4, we found, after carrying the units through
our computations, that the required diameter of the rod was
expressed in millimeters, which is the correct unit for a dimension;
if another unit had been found, we would have known that some
mistake had been made.
Errors in computation will usually be found by substituting the
numerical values obtained into an equation which has not yet been
used and verifying that the equation is satisfied. The importance of
correct computations in engineering cannot be overemphasized.
1.10 NUMERICAL ACCURACY
The accuracy of the solution of a problem depends upon two items:
(1) the accuracy of the given data and (2) the accuracy of the com-
putations performed.
The solution cannot be more accurate than the less accurate of
these two items. For example, if the loading of a beam is known to
be 75,000 lb with a possible error of 100 lb either way, the relative
error which measures the degree of accuracy of the data is
100 lb
75,000 lb
5 0.0013 5 0.13%
In computing the reaction at one of the beam supports, it would
then be meaningless to record it as 14,322 lb. The accuracy of the
solution cannot be greater than 0.13%, no matter how accurate the
computations are, and the possible error in the answer may be as
large as (0.13y100)(14,322 lb) < 20 lb. The answer should be prop-
erly recorded as 14,320 6 20 lb.
In engineering problems, the data are seldom known with an
accuracy greater than 0.2%. It is therefore seldom justified to write
the answers to such problems with an accuracy greater than 0.2%.
A practical rule is to use 4 figures to record numbers beginning with
a “1” and 3 figures in all other cases. Unless otherwise indicated, the
data given in a problem should be assumed known with a comparable
degree of accuracy. A force of 40 lb, for example, should be read
40.0 lb, and a force of 15 lb should be read 15.00 lb.
Pocket calculators and computers are widely used by practicing
engineers and engineering students. The speed and accuracy of these
devices facilitate the numerical computations in the solution of many
problems. However, students should not record more significant fig-
ures than can be justified merely because they are easily obtained.
As noted above, an accuracy greater than 0.2% is seldom necessary
or meaningful in the solution of practical engineering problems.
1.10 Numerical Accuracy
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18
SOLUTION
Free Body: Entire Hanger. Since the link ABC is a two-force member,
the reaction at A is vertical; the reaction at D is represented by its compo-
nents Dx and Dy. We write
1l oMD 5 0: 1500 lb2115 in.2 2 FAC110 in.2 5 0
FAC 5 1750 lb FAC 5 750 lb tension
a. Shearing Stress in Pin A. Since this 3
8-in.-diameter pin is in single
shear, we write
tA 5
FAC
A
5
750 lb
1
4p10.375 in.22
tA 5 6790 psi ◀
b. Shearing Stress in Pin C. Since this 1
4-in.-diameter pin is in double
shear, we write
tC 5
1
2 FAC
A
5
375 lb
1
4 p10.25 in.22
tC 5 7640 psi ◀
c. Largest Normal Stress in Link ABC. The largest stress is found
where the area is smallest; this occurs at the cross section at A where the 3
8-in.
hole is located. We have
sA 5
FAC
Anet
5
750 lb
1 3
8 in.211.25 in. 2 0.375 in.2
5
750 lb
0.328 in2
sA 5 2290 psi ◀
d. Average Shearing Stress at B. We note that bonding exists on
both sides of the upper portion of the link and that the shear force on each
side is F1 5 (750 lb)y2 5 375 lb. The average shearing stress on each surface
is thus
tB 5
F1
A
5
375 lb
11.25 in.211.75 in.2
tB 5 171.4 psi ◀
e. Bearing Stress in Link at C. For each portion of the link, F1 5
375 lb and the nominal bearing area is (0.25 in.)(0.25 in.) 5 0.0625 in2
.
sb 5
F1
A
5
375 lb
0.0625 in2
sb 5 6000 psi ◀
5 in.
500 lb
10 in.
A D
Dx
FAC
Dy
E
C
-in. diameter
750 lb
FAC ϭ 750 lb FAC ϭ 750 lb
1
4
-in. diameter3
8
FAC ϭ 375 lb1
2
FAC ϭ 375 lb1
2
CA
F1 ϭ F2 ϭ FAC ϭ 375 lb1
2
FAC ϭ 750 lb
-in. diameter3
8
in.
1.25 in.
1.25 in.
1.75 in.
3
8
FAC
F2 F1
A
B
375 lb
F1 ϭ 375 lb
-in. diameter1
4
1
4
in.
SAMPLE PROBLEM 1.1
In the hanger shown, the upper portion of link ABC is 3
8 in. thick and the
lower portions are each 1
4 in. thick. Epoxy resin is used to bond the upper
and lower portions together at B. The pin at A is of 3
8-in. diameter while a
1
4-in.-diameter pin is used at C. Determine (a) the shearing stress in pin A,
(b) the shearing stress in pin C, (c) the largest normal stress in link ABC,
(d) the average shearing stress on the bonded surfaces at B, (e) the bearing
stress in the link at C.6 in.
7 in.
1.75 in.
5 in.
1.25 in.
10 in.
500 lb
A
B
C
D
E
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19
SAMPLE PROBLEM 1.2
The steel tie bar shown is to be designed to carry a tension force of magnitude
P 5 120 kN when bolted between double brackets at A and B. The bar will
be fabricated from 20-mm-thick plate stock. For the grade of steel to be used,
the maximum allowable stresses are: s 5 175 MPa, t 5 100 MPa, sb 5
350 MPa. Design the tie bar by determining the required values of (a) the
diameter d of the bolt, (b) the dimension b at each end of the bar, (c) the
dimension h of the bar.
A B
SOLUTION
a. Diameter of the Bolt. Since the bolt is in double shear, F1 5 1
2 P 5
60 kN.
t 5
F1
A
5
60 kN
1
4 p d2
100 MPa 5
60 kN
1
4 p d2
d 5 27.6 mm
We will use d 5 28 mm ◀
At this point we check the bearing stress between the 20-mm-thick plate
and the 28-mm-diameter bolt.
tb 5
P
td
5
120 kN
10.020 m210.028 m2
5 214 MPa , 350 MPa OK
b. Dimension b at Each End of the Bar. We consider one of the
end portions of the bar. Recalling that the thickness of the steel plate is
t 5 20 mm and that the average tensile stress must not exceed 175 MPa,
we write
s 5
1
2 P
ta
175 MPa 5
60 kN
10.02 m2a
a 5 17.14 mm
b 5 d 1 2a 5 28 mm 1 2(17.14 mm) b 5 62.3 mm ◀
c. Dimension h of the Bar. Recalling that the thickness of the steel
plate is t 5 20 mm, we have
s 5
P
th
175 MPa 5
120 kN
10.020 m2h
h 5 34.3 mm
We will use h 5 35 mm ◀
d
F1 ϭ P
P
F1
F1
1
2
b
d
h
t ϭ 20 mm
P
P' ϭ 120 kN
a
t
a
db
1
2
P1
2
P ϭ 120 kN
t ϭ 20 mm
h
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PROBLEMS
20
1.1 Two solid cylindrical rods AB and BC are welded together at B
and loaded as shown. Knowing that the average normal stress must
not exceed 175 MPa in rod AB and 150 MPa in rod BC, determine
the smallest allowable values of d1 and d2.
1.2 Two solid cylindrical rods AB and BC are welded together at B and
loaded as shown. Knowing that d1 5 50 mm and d2 5 30 mm,
find the average normal stress at the midsection of (a) rod AB,
(b) rod BC.
1.3 Two solid cylindrical rods AB and BC are welded together at B
and loaded as shown. Determine the magnitude of the force P for
which the tensile stress in rod AB has the same magnitude as the
compressive stress in rod BC.
d2
d1
40 kN
30 kN
B
C
250 mm
300 mm
A
Fig. P1.1 and P1.2
2 in.
3 in.
30 kips
30 kips
C
A
B
30 in. 40 in.
P
Fig. P1.3
1.4 In Prob. 1.3, knowing that P 5 40 kips, determine the average
normal stress at the midsection of (a) rod AB, (b) rod BC.
1.5 Two steel plates are to be held together by means of 16-mm-diameter
high-strength steel bolts fitting snugly inside cylindrical brass
spacers. Knowing that the average normal stress must not exceed
200 MPa in the bolts and 130 MPa in the spacers, determine the
outer diameter of the spacers that yields the most economical and
safe design.
Fig. P1.5
100 m
15 mm
10 mm
b
a
B
C
A
Fig. P1.6
1.6 Two brass rods AB and BC, each of uniform diameter, will be
brazed together at B to form a nonuniform rod of total length
100 m which will be suspended from a support at A as shown.
Knowing that the density of brass is 8470 kg/m3
, determine (a) the
length of rod AB for which the maximum normal stress in ABC is
minimum, (b) the corresponding value of the maximum normal
stress.
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21Problems1.7 Each of the four vertical links has an 8 3 36-mm uniform rectan-
gular cross section and each of the four pins has a 16-mm diameter.
Determine the maximum value of the average normal stress in the
links connecting (a) points B and D, (b) points C and E.
0.2 m
0.25 m
0.4 m
20 kN
C
B
A
D
E
Fig. P1.7
1.8 Knowing that link DE is 1
8 in. thick and 1 in. wide, determine the
normal stress in the central portion of that link when (a) u 5 0,
(b) u 5 908.
1.9 Link AC has a uniform rectangular cross section 1
16 in. thick and
1
4 in. wide. Determine the normal stress in the central portion of the
link.
60 lb
F
D
E
JC D
B
A
8 in.
2 in.
4 in.
12 in.
4 in.
6 in.
Fig. P1.8
B
C
A
3 in.
7 in.
30Њ
6 in.
240 lb
240 lb
Fig. P1.9
1.10 Three forces, each of magnitude P 5 4 kN, are applied to the
mechanism shown. Determine the cross-sectional area of the uni-
form portion of rod BE for which the normal stress in that portion
is 1100 MPa.
0.100 m
0.150 m 0.300 m 0.250 m
P P P
E
A B C
D
Fig. P1.10
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22 Introduction—Concept of Stress 1.11 The frame shown consists of four wooden members, ABC, DEF,
BE, and CF. Knowing that each member has a 2 3 4-in. rectan-
gular cross section and that each pin has a 1
2-in. diameter, deter-
mine the maximum value of the average normal stress (a) in
member BE, (b) in member CF.
1.12 For the Pratt bridge truss and loading shown, determine the aver-
age normal stress in member BE, knowing that the cross-sectional
area of that member is 5.87 in2
.
40 in.
45 in.
15 in.
4 in.
A
B C
D
E F
4 in.
30 in.
30 in.
480 lb
Fig. P1.11
9 ft
80 kips 80 kips 80 kips
9 ft 9 ft 9 ft
12 ft
B D F
H
GEC
A
Fig. P1.12
D
B
E
A
Dimensions in mm
100
450
250
850
1150
500 675 825
CG
F
Fig. P1.13
1.13 An aircraft tow bar is positioned by means of a single hydraulic
cylinder connected by a 25-mm-diameter steel rod to two identical
arm-and-wheel units DEF. The mass of the entire tow bar is 200 kg,
and its center of gravity is located at G. For the position shown,
determine the normal stress in the rod.
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23Problems1.14 A couple M of magnitude 1500 N ? m is applied to the crank of
an engine. For the position shown, determine (a) the force P
required to hold the engine system in equilibrium, (b) the average
normal stress in the connecting rod BC, which has a 450-mm2
uniform cross section.
1.15 When the force P reached 8 kN, the wooden specimen shown failed
in shear along the surface indicated by the dashed line. Determine
the average shearing stress along that surface at the time of failure.
200 mm
80 mm
M
60 mm
B
A
C
P
Fig. P1.141.16 The wooden members A and B are to be joined by plywood splice
plates that will be fully glued on the surfaces in contact. As part
of the design of the joint, and knowing that the clearance between
the ends of the members is to be 1
4 in., determine the smallest
allowable length L if the average shearing stress in the glue is not
to exceed 120 psi.
1.17 A load P is applied to a steel rod supported as shown by an alu-
minum plate into which a 0.6-in.-diameter hole has been drilled.
Knowing that the shearing stress must not exceed 18 ksi in the
steel rod and 10 ksi in the aluminum plate, determine the largest
load P that can be applied to the rod.
15 mm
90 mm
WoodSteel
PP'
Fig. P1.15
5.8 kips
A
L
B
4 in.
5.8 kips
in.1
4
Fig. P1.16
1.6 in.
0.25 in.
0.6 in.
P
0.4 in.
Fig. P1.17
1.18 Two wooden planks, each 22 mm thick and 160 mm wide, are
joined by the glued mortise joint shown. Knowing that the joint
will fail when the average shearing stress in the glue reaches
820 kPa, determine the smallest allowable length d of the cuts if
the joint is to withstand an axial load of magnitude P 5 7.6 kN.
P' P
20 mm20 mm20 mm20 mm20 mm20 mm20 mm20 mm20 mm20 mm20 mm20 mm20 mm20 mm20 mmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmm0 mm0 mm0 mm0 mm0 mm0 mm20 mm
20 mm20 mm20 mm20 mm20 mm20 mm20 mm20 mm20 mm20 mm20 mm20 mm20 mm20 mm20 mm22020202020202020202020
160 mm111111160 m160 m160 m160 m160 m160 m160160160160160160 mm
d
GlueueueueueGlueGlueGlueGlueGlueGlueGlueGlueGlueGlue
Fig. P1.18
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24 Introduction—Concept of Stress 1.19 The load P applied to a steel rod is distributed to a timber support
by an annular washer. The diameter of the rod is 22 mm and the
inner diameter of the washer is 25 mm, which is slightly larger
than the diameter of the hole. Determine the smallest allowable
outer diameter d of the washer, knowing that the axial normal
stress in the steel rod is 35 MPa and that the average bearing stress
between the washer and the timber must not exceed 5 MPa.
1.20 The axial force in the column supporting the timber beam shown is
P 5 20 kips. Determine the smallest allowable length L of the bearing
plate if the bearing stress in the timber is not to exceed 400 psi.
P
d
22 mm
Fig. P1.19
6 in.
L
P
Fig. P1.20
1.21 An axial load P is supported by a short W8 3 40 column of cross-
sectional area A 5 11.7 in2
and is distributed to a concrete founda-
tion by a square plate as shown. Knowing that the average normal
stress in the column must not exceed 30 ksi and that the bearing
stress on the concrete foundation must not exceed 3.0 ksi, deter-
mine the side a of the plate that will provide the most economical
and safe design.
1.22 A 40-kN axial load is applied to a short wooden post that is
supported by a concrete footing resting on undisturbed soil.
Determine (a) the maximum bearing stress on the concrete foot-
ing, (b) the size of the footing for which the average bearing stress
in the soil is 145 kPa.
1.23 A 5
8-in.-diameter steel rod AB is fitted to a round hole near end C of
the wooden member CD. For the loading shown, determine (a) the
maximum average normal stress in the wood, (b) the distance b for
which the average shearing stress is 100 psi on the surfaces indicated
by the dashed lines, (c) the average bearing stress on the wood.
a aP
Fig. P1.21
P ϭ 40 kN
b b
120 mm 100 mm
Fig. P1.22
D
A
C
B
b
1500 lb
750 lb
750 lb
4 in.
1 in.
Fig. P1.23
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25Problems1.24 Knowing that u 5 408 and P 5 9 kN, determine (a) the smallest
allowable diameter of the pin at B if the average shearing stress
in the pin is not to exceed 120 MPa, (b) the corresponding aver-
age bearing stress in member AB at B, (c) the corresponding
average bearing stress in each of the support brackets at B.
1.25 Determine the largest load P that can be applied at A when u 5 608,
knowing that the average shearing stress in the 10-mm-diameter pin
at B must not exceed 120 MPa and that the average bearing stress
in member AB and in the bracket at B must not exceed 90 MPa.
1.26 Link AB, of width b 5 50 mm and thickness t 5 6 mm, is used to
support the end of a horizontal beam. Knowing that the average
normal stress in the link is 2140 MPa, and that the average shearing
stress in each of the two pins is 80 MPa, determine (a) the diameter
d of the pins, (b) the average bearing stress in the link.
1.27 For the assembly and loading of Prob. 1.7, determine (a) the average
shearing stress in the pin at B, (b) the average bearing stress at B in
member BD, (c) the average bearing stress at B in member ABC,
knowing that this member has a 10 3 50-mm uniform rectangular
cross section.
1.28 The hydraulic cylinder CF, which partially controls the position of
rod DE, has been locked in the position shown. Member BD is
5
8 in. thick and is connected to the vertical rod by a 3
8-in.-diameter
bolt. Determine (a) the average shearing stress in the bolt, (b) the
bearing stress at C in member BD.
16 mm
750 mm
750 mm
12 mm
50 mm B
A
C
P
Fig. P1.24 and P1.25
b
d
t
B
A
d
Fig. P1.26
1.8 in.
8 in.
4 in. 7 in.
D
F
E
A
C
B
400 lb
20Њ
75Њ
Fig. P1.28
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26 Introduction—Concept of Stress
1.11 STRESS ON AN OBLIQUE PLANE
UNDER AXIAL LOADING
In the preceding sections, axial forces exerted on a two-force
member (Fig. 1.26a) were found to cause normal stresses in that
member (Fig. 1.26b), while transverse forces exerted on bolts and
pins (Fig. 1.27a) were found to cause shearing stresses in those
connections (Fig. 1.27b). The reason such a relation was observed
between axial forces and normal stresses on one hand, and trans-
verse forces and shearing stresses on the other, was because stresses
were being determined only on planes perpendicular to the axis
of the member or connection. As you will see in this section, axial
forces cause both normal and shearing stresses on planes which
are not perpendicular to the axis of the member. Similarly, trans-
verse forces exerted on a bolt or a pin cause both normal and
shearing stresses on planes which are not perpendicular to the axis
of the bolt or pin.
(a)
(b)
P
P
P'
P'
P'
Fig. 1.26 Axial forces.
Fig. 1.28
P'
P'
P'
P
A
A0
P
V
F
P'
(a)
(c)
(b)
(d)
P
P'
PP
P' P'
(a) (b)
Fig. 1.27 Transverse forces.
Consider the two-force member of Fig. 1.26, which is subjected
to axial forces P and P9. If we pass a section forming an angle u with
a normal plane (Fig. 1.28a) and draw the free-body diagram of the
portion of member located to the left of that section (Fig. 1.28b),
we find from the equilibrium conditions of the free body that the
distributed forces acting on the section must be equivalent to the
force P.
Resolving P into components F and V, respectively normal and
tangential to the section (Fig. 1.28c), we have
F 5 P cos u V 5 P sin u (1.12)
The force F represents the resultant of normal forces distributed
over the section, and the force V the resultant of shearing forces
(Fig. 1.28d). The average values of the corresponding normal and
shearing stresses are obtained by dividing, respectively, F and V by
the area Au of the section:
s 5
F
Au
t 5
V
Au
(1.13)
Substituting for F and V from (1.12) into (1.13), and observing from
Fig. 1.28c that A0 5 Au cos u, or Au 5 A0ycos u, where A0 denotes
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27the area of a section perpendicular to the axis of the member, we
obtain
s 5
P cos u
A0ycos u
t 5
P sin u
A0ycos u
or
s 5
P
A0
cos2
u t 5
P
A0
sin u cos u (1.14)
We note from the first of Eqs. (1.14) that the normal stress s
is maximum when u 5 0, i.e., when the plane of the section is per-
pendicular to the axis of the member, and that it approaches zero as
u approaches 908. We check that the value of s when u 5 0 is
sm 5
P
A0
(1.15)
as we found earlier in Sec. 1.3. The second of Eqs. (1.14) shows that
the shearing stress t is zero for u 5 0 and u 5 908, and that for
u 5 458 it reaches its maximum value
tm 5
P
A0
sin 45° cos 45° 5
P
2A0
(1.16)
The first of Eqs. (1.14) indicates that, when u 5 458, the normal
stress s9 is also equal to Py2A0:
s¿ 5
P
A0
cos2
45° 5
P
2A0
(1.17)
The results obtained in Eqs. (1.15), (1.16), and (1.17) are
shown graphically in Fig. 1.29. We note that the same loading may
produce either a normal stress sm 5 PyA0 and no shearing stress
(Fig. 1.29b), or a normal and a shearing stress of the same magni-
tude s9 5 tm 5 Py2A0 (Fig. 1.29 c and d), depending upon the
orientation of the section.
1.12 STRESS UNDER GENERAL LOADING
CONDITIONS; COMPONENTS OF STRESS
The examples of the previous sections were limited to members
under axial loading and connections under transverse loading. Most
structural members and machine components are under more
involved loading conditions.
Consider a body subjected to several loads P1, P2, etc. (Fig.
1.30). To understand the stress condition created by these loads at
some point Q within the body, we shall first pass a section through
Q, using a plane parallel to the yz plane. The portion of the body to
the left of the section is subjected to some of the original loads, and
to normal and shearing forces distributed over the section. We shall
denote by DFx
and DVx
, respectively, the normal and the shearing
P'
(a) Axial loading
(b) Stresses for ϭ 0
m ϭ P/A0
(c) Stresses for ϭ 45°
(d) Stresses for ϭ –45°
' ϭ P/2A0
'ϭ P/2A0
m ϭ P/2A0
m ϭ P/2A0
P
Fig. 1.29
Fig. 1.30
P1
P4
P3
P2y
z
x
1.12 Stress Under General Loading Conditions;
Components of Stress
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28 Introduction—Concept of Stress
forces acting on a small area DA surrounding point Q (Fig. 1.31a).
Note that the superscript x is used to indicate that the forces DFx
and DVx
act on a surface perpendicular to the x axis. While the nor-
mal force DFx
has a well-defined direction, the shearing force DVx
may have any direction in the plane of the section. We therefore
resolve DVx
into two component forces, DVx
y and DVx
z, in directions
parallel to the y and z axes, respectively (Fig. 1.31b). Dividing now
the magnitude of each force by the area DA, and letting DA approach
zero, we define the three stress components shown in Fig. 1.32:
sx 5 lim
¢AS0
¢Fx
¢A
txy 5 lim
¢AS0
¢Vy
x
¢A
txz 5 lim
¢AS0
¢Vz
x
¢A
(1.18)
We note that the first subscript in sx, txy, and txz is used to indicate
that the stresses under consideration are exerted on a surface per-
pendicular to the x axis. The second subscript in txy and txz identifies
the direction of the component. The normal stress sx is positive if
the corresponding arrow points in the positive x direction, i.e., if the
body is in tension, and negative otherwise. Similarly, the shearing
stress components txy and txz are positive if the corresponding arrows
point, respectively, in the positive y and z directions.
The above analysis may also be carried out by considering the
portion of body located to the right of the vertical plane through Q
(Fig. 1.33). The same magnitudes, but opposite directions, are
obtained for the normal and shearing forces DFx
, DVy
x
, and DVz
x
.
Therefore, the same values are also obtained for the corresponding
stress components, but since the section in Fig. 1.33 now faces the
negative x axis, a positive sign for sx will indicate that the corre-
sponding arrow points in the negative x direction. Similarly, positive
signs for txy and txz will indicate that the corresponding arrows
point, respectively, in the negative y and z directions, as shown in
Fig. 1.33.
Fx
P2 P2
P1
y
z
x
y
z
x
P1
A
Fx
⌬
⌬
⌬Vx
⌬
Vx
⌬
(a) (b)
Q Q
z
Vx
⌬ y
Fig. 1.31
y
z
x
x
xy
Q
xz
Fig. 1.32
y
z
x
x
xy
xz
Q
Fig. 1.33
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29Passing a section through Q parallel to the zx plane, we define
in the same manner the stress components, sy, tyz, and tyx. Finally,
a section through Q parallel to the xy plane yields the components
sz, tzx, and tzy.
To facilitate the visualization of the stress condition at point
Q, we shall consider a small cube of side a centered at Q and the
stresses exerted on each of the six faces of the cube (Fig. 1.34).
The stress components shown in the figure are sx, sy, and sz,
which represent the normal stress on faces respectively perpen-
dicular to the x, y, and z axes, and the six shearing stress compo-
nents txy, txz, etc. We recall that, according to the definition of the
shearing stress components, txy represents the y component of the
shearing stress exerted on the face perpendicular to the x axis,
while tyx represents the x component of the shearing stress exerted
on the face perpendicular to the y axis. Note that only three faces
of the cube are actually visible in Fig. 1.34, and that equal and
opposite stress components act on the hidden faces. While the
stresses acting on the faces of the cube differ slightly from the
stresses at Q, the error involved is small and vanishes as side a of
the cube approaches zero.
Important relations among the shearing stress components will
now be derived. Let us consider the free-body diagram of the small
cube centered at point Q (Fig. 1.35). The normal and shearing
forces acting on the various faces of the cube are obtained
by multiplying the corresponding stress components by the area DA
of each face. We first write the following three equilibrium
equations:
oFx 5 0 oFy 5 0 oFz 5 0 (1.19)
Since forces equal and opposite to the forces actually shown in Fig.
1.35 are acting on the hidden faces of the cube, it is clear that Eqs.
(1.19) are satisfied. Considering now the moments of the forces
about axes x9, y9, and z9 drawn from Q in directions respectively
parallel to the x, y, and z axes, we write the three additional
equations
oMx¿ 5 0 oMy¿ 5 0 oMz¿ 5 0 (1.20)
Using a projection on the x9y9 plane (Fig. 1.36), we note that the
only forces with moments about the z axis different from zero are
the shearing forces. These forces form two couples, one of counter-
clockwise (positive) moment (txy DA)a, the other of clockwise (nega-
tive) moment 2(tyx DA)a. The last of the three Eqs. (1.20) yields,
therefore,
1loMz 5 0: (txy DA)a 2 (tyx DA)a 5 0
from which we conclude that
txy 5 tyx (1.21)
The relation obtained shows that the y component of the shearing
stress exerted on a face perpendicular to the x axis is equal to the x
1.12 Stress Under General Loading Conditions;
Components of Stress
yz
yx
xy
xzzx
zy
y
z
x
a
Qa
a
z
y
x
Fig. 1.34
x⌬A
z⌬A
y⌬A
Q
z
y
x
zy⌬A
yx⌬A
yz⌬A
xy⌬A
zx⌬A
xz⌬A
Fig. 1.35
yx⌬A
yx⌬A
xy⌬A
xy⌬A x⌬A
x⌬A
y⌬A
y ⌬A
x'
a
z'
y'
Fig. 1.36
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30 Introduction—Concept of Stress component of the shearing stress exerted on a face perpendicular to
the y axis. From the remaining two equations (1.20), we derive in a
similar manner the relations
tyz 5 tzy tzx 5 txz (1.22)
We conclude from Eqs. (1.21) and (1.22) that only six stress
components are required to define the condition of stress at a given
point Q, instead of nine as originally assumed. These six components
are sx, sy, sz, txy, tyz, and tzx. We also note that, at a given point,
shear cannot take place in one plane only; an equal shearing stress
must be exerted on another plane perpendicular to the first one. For
example, considering again the bolt of Fig. 1.27 and a small cube at
the center Q of the bolt (Fig. 1.37a), we find that shearing stresses
of equal magnitude must be exerted on the two horizontal faces of
the cube and on the two faces that are perpendicular to the forces
P and P9 (Fig. 1.37b).
Before concluding our discussion of stress components, let us
consider again the case of a member under axial loading. If we con-
sider a small cube with faces respectively parallel to the faces of the
member and recall the results obtained in Sec. 1.11, we find that the
conditions of stress in the member may be described as shown in Fig.
1.38a; the only stresses are normal stresses sx exerted on the faces of
the cube which are perpendicular to the x axis. However, if the small
cube is rotated by 458 about the z axis so that its new orientation
matches the orientation of the sections considered in Fig. 1.29c and
d, we conclude that normal and shearing stresses of equal magnitude
are exerted on four faces of the cube (Fig. 1.38b). We thus observe
that the same loading condition may lead to different interpretations
of the stress situation at a given point, depending upon the orientation
of the element considered. More will be said about this in Chap 7.
1.13 DESIGN CONSIDERATIONS
In the preceding sections you learned to determine the stresses in
rods, bolts, and pins under simple loading conditions. In later chap-
ters you will learn to determine stresses in more complex situations.
In engineering applications, however, the determination of stresses
is seldom an end in itself. Rather, the knowledge of stresses is used
by engineers to assist in their most important task, namely, the design
of structures and machines that will safely and economically perform
a specified function.
a. Determination of the Ultimate Strength of a Material. An
important element to be considered by a designer is how the material
that has been selected will behave under a load. For a given material,
this is determined by performing specific tests on prepared samples
of the material. For example, a test specimen of steel may be pre-
pared and placed in a laboratory testing machine to be subjected to
a known centric axial tensile force, as described in Sec. 2.3. As the
magnitude of the force is increased, various changes in the specimen
are measured, for example, changes in its length and its diameter.
(a) (b)
P
P'
Q
Fig. 1.37
(b)
(a)
m m
P
P'
P'
P
P
2A
z
x
y
'
45Њ
x
x
P
A
P
2A
'
'
' ϭ
ϭ
ϭ
Fig. 1.38
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31Eventually the largest force which may be applied to the specimen is
reached, and the specimen either breaks or begins to carry less load.
This largest force is called the ultimate load for the test specimen and
is denoted by PU. Since the applied load is centric, we may divide the
ultimate load by the original cross-sectional area of the rod to obtain
the ultimate normal stress of the material used. This stress, also known
as the ultimate strength in tension of the material, is
sU 5
PU
A
(1.23)
Several test procedures are available to determine the ultimate
shearing stress, or ultimate strength in shear, of a material. The one
most commonly used involves the twisting of a circular tube (Sec.
3.5). A more direct, if less accurate, procedure consists in clamping
a rectangular or round bar in a shear tool (Fig. 1.39) and applying
an increasing load P until the ultimate load PU for single shear is
obtained. If the free end of the specimen rests on both of the hard-
ened dies (Fig. 1.40), the ultimate load for double shear is obtained.
In either case, the ultimate shearing stress tU is obtained by dividing
the ultimate load by the total area over which shear has taken place.
We recall that, in the case of single shear, this area is the cross-
sectional area A of the specimen, while in double shear it is equal
to twice the cross-sectional area.
b. Allowable Load and Allowable Stress; Factor of Safety. The
maximum load that a structural member or a machine component will
be allowed to carry under normal conditions of utilization is consider-
ably smaller than the ultimate load. This smaller load is referred to as
the allowable load and, sometimes, as the working load or design load.
Thus, only a fraction of the ultimate-load capacity of the member is
utilized when the allowable load is applied. The remaining portion of
the load-carrying capacity of the member is kept in reserve to assure
its safe performance. The ratio of the ultimate load to the allowable
load is used to define the factor of safety.† We have
Factor of safety 5 F.S. 5
ultimate load
allowable load
(1.24)
An alternative definition of the factor of safety is based on the use
of stresses:
Factor of safety 5 F.S. 5
ultimate stress
allowable stress
(1.25)
The two expressions given for the factor of safety in Eqs. (1.24) and
(1.25) are identical when a linear relationship exists between the load
and the stress. In most engineering applications, however, this rela-
tionship ceases to be linear as the load approaches its ultimate value,
and the factor of safety obtained from Eq. (1.25) does not provide a
1.13 Design Considerations
†In some fields of engineering, notably aeronautical engineering, the margin of safety is
used in place of the factor of safety. The margin of safety is defined as the factor of safety
minus one; that is, margin of safety 5 F.S. 2 1.00.
P
Fig. 1.39 Single shear test.
P
Fig. 1.40 Double shear test.
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32 Introduction—Concept of Stress true assessment of the safety of a given design. Nevertheless, the
allowable-stress method of design, based on the use of Eq. (1.25), is
widely used.
c. Selection of an Appropriate Factor of Safety. The selec-
tion of the factor of safety to be used for various applications is one
of the most important engineering tasks. On the one hand, if a factor
of safety is chosen too small, the possibility of failure becomes unac-
ceptably large; on the other hand, if a factor of safety is chosen
unnecessarily large, the result is an uneconomical or nonfunctional
design. The choice of the factor of safety that is appropriate for a
given design application requires engineering judgment based on
many considerations, such as the following:
1. Variations that may occur in the properties of the member
under consideration. The composition, strength, and dimen-
sions of the member are all subject to small variations during
manufacture. In addition, material properties may be altered
and residual stresses introduced through heating or deforma-
tion that may occur during manufacture, storage, transporta-
tion, or construction.
2. The number of loadings that may be expected during the life of
the structure or machine. For most materials the ultimate stress
decreases as the number of load applications is increased. This
phenomenon is known as fatigue and, if ignored, may result in
sudden failure (see Sec. 2.7).
3. The type of loadings that are planned for in the design, or that
may occur in the future. Very few loadings are known with
complete accuracy—most design loadings are engineering esti-
mates. In addition, future alterations or changes in usage may
introduce changes in the actual loading. Larger factors of safety
are also required for dynamic, cyclic, or impulsive loadings.
4. The type of failure that may occur. Brittle materials fail sud-
denly, usually with no prior indication that collapse is immi-
nent. On the other hand, ductile materials, such as structural
steel, normally undergo a substantial deformation called yield-
ing before failing, thus providing a warning that overloading
exists. However, most buckling or stability failures are sudden,
whether the material is brittle or not. When the possibility of
sudden failure exists, a larger factor of safety should be used
than when failure is preceded by obvious warning signs.
5. Uncertainty due to methods of analysis. All design methods are
based on certain simplifying assumptions which result in calcu-
lated stresses being approximations of actual stresses.
6. Deterioration that may occur in the future because of poor main-
tenance or because of unpreventable natural causes. A larger fac-
tor of safety is necessary in locations where conditions such as
corrosion and decay are difficult to control or even to discover.
7. The importance of a given member to the integrity of the whole
structure. Bracing and secondary members may in many cases
be designed with a factor of safety lower than that used for
primary members.
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33In addition to the these considerations, there is the additional
consideration concerning the risk to life and property that a failure
would produce. Where a failure would produce no risk to life and
only minimal risk to property, the use of a smaller factor of safety
can be considered. Finally, there is the practical consideration that,
unless a careful design with a nonexcessive factor of safety is used,
a structure or machine might not perform its design function. For
example, high factors of safety may have an unacceptable effect on
the weight of an aircraft.
For the majority of structural and machine applications, factors
of safety are specified by design specifications or building codes writ-
ten by committees of experienced engineers working with profes-
sional societies, with industries, or with federal, state, or city agencies.
Examples of such design specifications and building codes are
1. Steel: American Institute of Steel Construction, Specification
for Structural Steel Buildings
2. Concrete: American Concrete Institute, Building Code Require-
ment for Structural Concrete
3. Timber: American Forest and Paper Association, National
Design Specification for Wood Construction
4. Highway bridges: American Association of State Highway Offi-
cials, Standard Specifications for Highway Bridges
*d. Load and Resistance Factor Design. As we saw previously,
the allowable-stress method requires that all the uncertainties associ-
ated with the design of a structure or machine element be grouped
into a single factor of safety. An alternative method of design, which
is gaining acceptance chiefly among structural engineers, makes it pos-
sible through the use of three different factors to distinguish between
the uncertainties associated with the structure itself and those associ-
ated with the load it is designed to support. This method, referred
to as Load and Resistance Factor Design (LRFD), further allows the
designer to distinguish between uncertainties associated with the live
load, PL, that is, with the load to be supported by the structure, and
the dead load, PD, that is, with the weight of the portion of structure
contributing to the total load.
When this method of design is used, the ultimate load, PU, of
the structure, that is, the load at which the structure ceases to be
useful, should first be determined. The proposed design is then
acceptable if the following inequality is satisfied:
gDPD 1 gLPL # fPU (1.26)
The coefficient f is referred to as the resistance factor; it accounts
for the uncertainties associated with the structure itself and will
normally be less than 1. The coefficients gD and gL are referred to
as the load factors; they account for the uncertainties associated,
respectively, with the dead and live load and will normally be greater
than 1, with gL generally larger than gD. While a few examples or
assigned problems using LRFD are included in this chapter and in
Chaps. 5 and 10, the allowable-stress method of design will be used
in this text.
1.13 Design Considerations
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34
SOLUTION
Free Body: Entire Bracket. The reaction at C is represented by its com-
ponents Cx and Cy.
1 l oMC 5 0: P(0.6 m) 2 (50 kN)(0.3 m) 2 (15 kN)(0.6 m) 5 0 P 5 40 kN
oFx 5 0: Cx 5 40 k
C 5 2C2
x 1 C2
y 5 76.3 kNoFy 5 0: Cy 5 65 kN
a. Control Rod AB. Since the factor of safety is to be 3.3, the allow-
able stress is
sall 5
sU
F.S.
5
600 MPa
3.3
5 181.8 MPa
For P 5 40 kN the cross-sectional area required is
Areq 5
P
sall
5
40 kN
181.8 MPa
5 220 3 1026
m2
Areq 5
p
4
dAB
2
5 220 3 1026
m2
dAB 5 16.74 mm ◀
b. Shear in Pin C. For a factor of safety of 3.3, we have
tall 5
tU
F.S.
5
350 MPa
3.3
5 106.1 MPa
Since the pin is in double shear, we write
Areq 5
Cy2
tall
5
176.3 kN2y2
106.1 MPa
5 360 mm2
Areq 5
p
4
dC
2
5 360 mm2
dC 5 21.4 mm Use: dC 5 22 mm ◀
The next larger size pin available is of 22-mm diameter and should be used.
c. Bearing at C. Using d 5 22 mm, the nominal bearing area of each
bracket is 22t. Since the force carried by each bracket is Cy2 and the allow-
able bearing stress is 300 MPa, we write
Areq 5
Cy2
sall
5
176.3 kN2y2
300 MPa
5 127.2 mm2
Thus 22t 5 127.2 t 5 5.78 mm Use: t 5 6 mm ◀
SAMPLE PROBLEM 1.3
Two forces are applied to the bracket BCD as shown. (a) Knowing that the
control rod AB is to be made of a steel having an ultimate normal stress of
600 MPa, determine the diameter of the rod for which the factor of safety
with respect to failure will be 3.3. (b) The pin at C is to be made of a steel
having an ultimate shearing stress of 350 MPa. Determine the diameter of
the pin C for which the factor of safety with respect to shear will also be
3.3. (c) Determine the required thickness of the bracket supports at C
knowing that the allowable bearing stress of the steel used is 300 MPa.
t t
A
D
B
dAB
C
0.6 m
0.3 m 0.3 m
50 kN 15 kN
P
50 kN 15 kN0.6 m
0.3 m 0.3 m
D
B
C
P
Cx
Cy
C
C
dC
F2
F1
F1 ϭ F2 ϭ 1
2
d ϭ 22 mm
t C
1
2
C
1
2
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35
SOLUTION
The factor of safety with respect to failure must be 3.0 or more in each of
the three bolts and in the control rod. These four independent criteria will
be considered separately.
Free Body: Beam BCD. We first determine the force at C in terms
of the force at B and in terms of the force at D.
1l oMD 5 0: B114 in.2 2 C18 in.2 5 0 C 5 1.750B (1)
1l oMB 5 0: 2D114 in.2 1 C16 in.2 5 0 C 5 2.33D (2)
Control Rod. For a factor of safety of 3.0 we have
sall 5
sU
F.S.
5
60 ksi
3.0
5 20 ksi
The allowable force in the control rod is
B 5 sall1A2 5 120 ksi2 1
4p 1 7
16 in.22
5 3.01 kips
Using Eq. (1) we find the largest permitted value of C:
C 5 1.750B 5 1.75013.01 kips2 C 5 5.27 kips ◀
Bolt at B. tall 5 tUyF.S. 5 (40 ksi)y3 5 13.33 ksi. Since the bolt is in
double shear, the allowable magnitude of the force B exerted on the bolt is
B 5 2F1 5 21tall A2 5 2113.33 ksi211
4 p213
8 in.22
5 2.94 kips
From Eq. (1): C 5 1.750B 5 1.75012.94 kips2 C 5 5.15 kips ◀
Bolt at D. Since this bolt is the same as bolt B, the allowable force is
D 5 B 5 2.94 kips. From Eq. (2):
C 5 2.33D 5 2.3312.94 kips2 C 5 6.85 kips ◀
Bolt at C. We again have tall 5 13.33 ksi and write
C 5 2F2 5 21tall A2 5 2113.33 ksi211
4 p211
2 in.22
C 5 5.23 kips ◀
Summary. We have found separately four maximum allowable values
of the force C. In order to satisfy all these criteria we must choose the
smallest value, namely: C 5 5.15 kips ◀
SAMPLE PROBLEM 1.4
The rigid beam BCD is attached by bolts to a control rod at B, to a hydraulic
cylinder at C, and to a fixed support at D. The diameters of the bolts used
are: dB 5 dD 5 3
8 in., dC 5 1
2 in. Each bolt acts in double shear and is made
from a steel for which the ultimate shearing stress is tU 5 40 ksi. The con-
trol rod AB has a diameter dA 5 7
16 in. and is made of a steel for which the
ultimate tensile stress is sU 5 60 ksi. If the minimum factor of safety is to
be 3.0 for the entire unit, determine the largest upward force which may
be applied by the hydraulic cylinder at C.
DC
B
A
6 in.
8 in.
D
DB
C
B C
6 in. 8 in.
F1
F1
B
3
8
in.
B ϭ 2F1
C
F2
F2
1
2
in.
C ϭ 2F2
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PROBLEMS
36
1.29 The 1.4-kip load P is supported by two wooden members of uni-
form cross section that are joined by the simple glued scarf splice
shown. Determine the normal and shearing stresses in the glued
splice.
1.30 Two wooden members of uniform cross section are joined by the
simple scarf splice shown. Knowing that the maximum allowable
tensile stress in the glued splice is 75 psi, determine (a) the largest
load P that can be safely supported, (b) the corresponding shearing
stress in the splice.
1.31 Two wooden members of uniform rectangular cross section are
joined by the simple glued scarf splice shown. Knowing that
P 5 11 kN, determine the normal and shearing stresses in the
glued splice.
60Њ
5.0 in.
3.0 in.
P'
P
Fig. P1.29 and P1.30
75 mm
150 mm
45454545454545454545454545454545454545454545454545ЊЊЊЊЊЊЊЊЊЊЊЊЊ
P'
P
Fig. P1.31 and P1.32
1.32 Two wooden members of uniform rectangular cross section are
joined by the simple glued scarf splice shown. Knowing that
the maximum allowable shearing stress in the glued splice is
620 kPa, determine (a) the largest load P that can be safely applied,
(b) the corresponding tensile stress in the splice.
1.33 A steel pipe of 12-in. outer diameter is fabricated from 1
4-in.-thick
plate by welding along a helix that forms an angle of 258 with a
plane perpendicular to the axis of the pipe. Knowing that the maxi-
mum allowable normal and shearing stresses in the directions
respectively normal and tangential to the weld are s 5 12 ksi and
t 5 7.2 ksi, determine the magnitude P of the largest axial force
that can be applied to the pipe.
1.34 A steel pipe of 12-in. outer diameter is fabricated from 1
4-in.-thick
plate by welding along a helix that forms an angle of 258 with a
plane perpendicular to the axis of the pipe. Knowing that a 66 kip
axial force P is applied to the pipe, determine the normal and
shearing stresses in directions respectively normal and tangential
to the weld.
25Њ
P
Weld
in.1
4
Fig. P1.33 and P1.34
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37Problems1.35 A 1060-kN load P is applied to the granite block shown. Determine
the resulting maximum value of (a) the normal stress, (b) the shear-
ing stress. Specify the orientation of that plane on which each of
these maximum values occurs.
1.36 A centric load P is applied to the granite block shown. Knowing
that the resulting maximum value of the shearing stress in the
block is 18 MPa, determine (a) the magnitude of P, (b) the orienta-
tion of the surface on which the maximum shearing stress occurs,
(c) the normal stress exerted on that surface, (d) the maximum
value of the normal stress in the block.
1.37 Link BC is 6 mm thick, has a width w 5 25 mm, and is made of
a steel with a 480-MPa ultimate strength in tension. What is the
safety factor used if the structure shown was designed to support
a 16-kN load P?
1.38 Link BC is 6 mm thick and is made of a steel with a 450-MPa
ultimate strength in tension. What should be its width w if the
structure shown is being designed to support a 20-kN load P with
a factor of safety of 3?
1.39 A 3
4-in.-diameter rod made of the same material as rods AC and
AD in the truss shown was tested to failure and an ultimate load
of 29 kips was recorded. Using a factor of safety of 3.0, determine
the required diameter (a) of rod AC, (b) of rod AD.
140 mm
140 mm
P
Fig. P1.35 and P1.36
A B
C
D
480 mm
600 mm
90Њ
w
P
Fig. P1.37 and P1.38
1.40 In the truss shown, members AC and AD consist of rods made of
the same metal alloy. Knowing that AC is of 1-in. diameter and
that the ultimate load for that rod is 75 kips, determine (a) the
factor of safety for AC, (b) the required diameter of AD if it is
desired that both rods have the same factor of safety.
1.41 Link AB is to be made of a steel for which the ultimate normal
stress is 450 MPa. Determine the cross-sectional area of AB for
which the factor of safety will be 3.50. Assume that the link will
be adequately reinforced around the pins at A and B.
10 kips 10 kips
10 ft 10 ft
5 ft
A
B C
D
Fig. P1.39 and P1.40
0.4 m
35Њ
B
A
C D
E
0.4 m 0.4 m
8 kN/m
20 kN
Fig. P1.41
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38 Introduction—Concept of Stress 1.42 A steel loop ABCD of length 1.2 m and of 10-mm diameter is
placed as shown around a 24-mm-diameter aluminum rod AC.
Cables BE and DF, each of 12-mm diameter, are used to apply the
load Q. Knowing that the ultimate strength of the steel used for
the loop and the cables is 480 MPa and that the ultimate strength
of the aluminum used for the rod is 260 MPa, determine the larg-
est load Q that can be applied if an overall factor of safety of 3 is
desired.
1.43 Two wooden members shown, which support a 3.6-kip load, are
joined by plywood splices fully glued on the surfaces in contact. The
ultimate shearing stress in the glue is 360 psi and the clearance
between the members is 1
4 in. Determine the required length L of
each splice if a factor of safety of 2.75 is to be achieved.
240 mm
180 mm
24 mm
C
D
Q
A
10 mm
180 mm
240 mm
F
Q'
12 mm
B
E
Fig. P1.42
1.44 Two plates, each 1
8-in. thick, are used to splice a plastic strip as
shown. Knowing that the ultimate shearing stress of the bonding
between the surfaces is 130 psi, determine the factor of safety with
respect to shear when P 5 325 lb.
1.45 A load P is supported as shown by a steel pin that has been inserted
in a short wooden member hanging from the ceiling. The ultimate
strength of the wood used is 60 MPa in tension and 7.5 MPa in
shear, while the ultimate strength of the steel is 145 MPa in shear.
Knowing that b 5 40 mm, c 5 55 mm, and d 5 12 mm, determine
the load P if an overall factor of safety of 3.2 is desired.
Fig. P1.43
3.6 kips
3.6 kips
L
5.0 in.
in.1
4
P'
P
in.5
8 in.3
4
in.1
4
in.2 1
4
Fig. P1.44
1
2
40 mm
d
c
b
P
1
2 P
Fig. P1.45
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39Problems1.46 For the support of Prob. 1.45, knowing that the diameter of
the pin is d 5 16 mm and that the magnitude of the load is
P 5 20 kN, determine (a) the factor of safety for the pin, (b) the
required values of b and c if the factor of safety for the wooden
member is the same as that found in part a for the pin.
1.47 Three steel bolts are to be used to attach the steel plate shown to
a wooden beam. Knowing that the plate will support a 110-kN
load, that the ultimate shearing stress for the steel used is 360 MPa,
and that a factor of safety of 3.35 is desired, determine the required
diameter of the bolts.
1.48 Three 18-mm-diameter steel bolts are to be used to attach the steel
plate shown to a wooden beam. Knowing that the plate will support
a 110-kN load and that the ultimate shearing stress for the steel
used is 360 MPa, determine the factor of safety for this design.
1.49 A steel plate 5
16 in. thick is embedded in a horizontal concrete slab
and is used to anchor a high-strength vertical cable as shown. The
diameter of the hole in the plate is 3
4 in., the ultimate strength of
the steel used is 36 ksi, and the ultimate bonding stress between
plate and concrete is 300 psi. Knowing that a factor of safety of
3.60 is desired when P 5 2.5 kips, determine (a) the required
width a of the plate, (b) the minimum depth b to which a plate of
that width should be embedded in the concrete slab. (Neglect the
normal stresses between the concrete and the bottom edge of the
plate.)
110 kN
Fig. P1.47 and P1.48
a
b
P
3
4
in.
5
16
in.
Fig. P1.49
1.50 Determine the factor of safety for the cable anchor in Prob. 1.49
when P 5 3 kips, knowing that a 5 2 in. and b 5 7.5 in.
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40 Introduction—Concept of Stress 1.51 In the steel structure shown, a 6-mm-diameter pin is used at C
and 10-mm-diameter pins are used at B and D. The ultimate
shearing stress is 150 MPa at all connections, and the ultimate
normal stress is 400 MPa in link BD. Knowing that a factor
of safety of 3.0 is desired, determine the largest load P that can
be applied at A. Note that link BD is not reinforced around the
pin holes.
18 mm
Top view
Side view
Front view
160 mm 120 mm
6 mm
A
A
B
C
B
D
C
B
D
P
Fig. P1.51
1.52 Solve Prob. 1.51, assuming that the structure has been redesigned
to use 12-mm-diameter pins at B and D and no other change has
been made.
1.53 Each of the two vertical links CF connecting the two horizontal mem-
bers AD and EG has a uniform rectangular cross section 1
4 in. thick
and 1 in. wide, and is made of a steel with an ultimate strength in
tension of 60 ksi. The pins at C and F each have a 1
2-in. diameter and
are made of a steel with an ultimate strength in shear of 25 ksi.
Determine the overall factor of safety for the links CF and the pins
connecting them to the horizontal members.
2 kips
10 in.
10 in.
16 in.
C
A
B
E
D
F G
Fig. P1.53
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41Problems1.54 Solve Prob. 1.53, assuming that the pins at C and F have been
replaced by pins with a 3
4-in. diameter.
1.55 In the structure shown, an 8-mm-diameter pin is used at A, and
12-mm-diameter pins are used at B and D. Knowing that the ulti-
mate shearing stress is 100 MPa at all connections and that the
ultimate normal stress is 250 MPa in each of the two links joining
B and D, determine the allowable load P if an overall factor of
safety of 3.0 is desired.
180 mm200 mm
Top view
Side view
Front view
8 mm
20 mm
8 mm
8 mm
12 mm
12 mm
B
C
B
D D
A
B CA
P
Fig. P1.55
1.56 In an alternative design for the structure of Prob. 1.55, a pin of
10-mm diameter is to be used at A. Assuming that all other speci-
fications remain unchanged, determine the allowable load P if an
overall factor of safety of 3.0 is desired.
*1.57 The Load and Resistance Factor Design method is to be used to
select the two cables that will raise and lower a platform support-
ing two window washers. The platform weighs 160 lb and each
of the window washers is assumed to weigh 195 lb with equip-
ment. Since these workers are free to move on the platform, 75%
of their total weight and the weight of their equipment will be
used as the design live load of each cable. (a) Assuming a resis-
tance factor f 5 0.85 and load factors gD 5 1.2 and gL 5 1.5,
determine the required minimum ultimate load of one cable.
(b) What is the conventional factor of safety for the selected
cables?
*1.58 A 40-kg platform is attached to the end B of a 50-kg wooden
beam AB, which is supported as shown by a pin at A and by a
slender steel rod BC with a 12-kN ultimate load. (a) Using the
Load and Resistance Factor Design method with a resistance
factor f 5 0.90 and load factors gD 5 1.25 and gL 5 1.6, deter-
mine the largest load that can be safely placed on the platform.
(b) What is the corresponding conventional factor of safety for
rod BC?
P P
Fig. P1.57
1.8 m
2.4 m
A B
C
Fig. P1.58
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42
REVIEW AND SUMMARY
This chapter was devoted to the concept of stress and to an introduc-
tion to the methods used for the analysis and design of machines and
load-bearing structures.
Section 1.2 presented a short review of the methods of statics
and of their application to the determination of the reactions exerted
by its supports on a simple structure consisting of pin-connected
members. Emphasis was placed on the use of a free-body diagram
to obtain equilibrium equations which were solved for the unknown
reactions. Free-body diagrams were also used to find the internal
forces in the various members of the structure.
The concept of stress was first introduced in Sec. 1.3 by consider-
ing a two-force member under an axial loading. The normal stress
in that member was obtained by dividing the magnitude P of the
load by the cross-sectional area A of the member (Fig. 1.41). We
wrote
s 5
P
A
(1.5)
Section 1.4 was devoted to a short discussion of the two prin-
cipal tasks of an engineer, namely, the analysis and the design of
structures and machines.
As noted in Sec. 1.5, the value of s obtained from Eq. (1.5)
represents the average stress over the section rather than the stress
at a specific point Q of the section. Considering a small area DA
surrounding Q and the magnitude DF of the force exerted on DA,
we defined the stress at point Q as
s 5 lim
¢Ay0
¢F
¢A
(1.6)
In general, the value obtained for the stress s at point Q is
different from the value of the average stress given by formula
(1.5) and is found to vary across the section. However, this varia-
tion is small in any section away from the points of application of
the loads. In practice, therefore, the distribution of the normal
stresses in an axially loaded member is assumed to be uniform,
except in the immediate vicinity of the points of application of the
loads.
However, for the distribution of stresses to be uniform in a
given section, it is necessary that the line of action of the loads P
and P9 pass through the centroid C of the section. Such a loading is
called a centric axial loading. In the case of an eccentric axial loading,
the distribution of stresses is not uniform. Stresses in members sub-
jected to an eccentric axial loading will be discussed in Chap 4.
Axial loading. Normal stressAxial loading. Normal stress
A
P'
P
Fig. 1.41
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43When equal and opposite transverse forces P and P9 of magnitude
P are applied to a member AB (Fig. 1.42), shearing stresses t are
created over any section located between the points of application
of the two forces [Sec 1.6]. These stresses vary greatly across the
section and their distribution cannot be assumed uniform. However,
dividing the magnitude P—referred to as the shear in the section—
by the cross-sectional area A, we defined the average shearing stress
over the section:
tave 5
P
A
(1.8)
Shearing stresses are found in bolts, pins, or rivets connecting two
structural members or machine components. For example, in the
case of bolt CD (Fig. 1.43), which is in single shear, we wrote
tave 5
P
A
5
F
A
(1.9)
while, in the case of bolts EG and HJ (Fig. 1.44), which are both in
double shear, we had
tave 5
P
A
5
Fy2
A
5
F
2A
(1.10)
Bolts, pins, and rivets also create stresses in the members they con-
nect, along the bearing surface, or surface of contact [Sec. 1.7]. The
bolt CD of Fig. 1.43, for example, creates stresses on the semicylin-
drical surface of plate A with which it is in contact (Fig. 1.45). Since
the distribution of these stresses is quite complicated, one uses in
practice an average nominal value sb of the stress, called bearing
stress, obtained by dividing the load P by the area of the rectangle
representing the projection of the bolt on the plate section. Denoting
by t the thickness of the plate and by d the diameter of the bolt, we
wrote
sb 5
P
A
5
P
td
(1.11)
In Sec. 1.8, we applied the concept introduced in the previous
sections to the analysis of a simple structure consisting of two pin-
connected members supporting a given load. We determined succes-
sively the normal stresses in the two members, paying special
attention to their narrowest sections, the shearing stresses in the
various pins, and the bearing stress at each connection.
The method you should use in solving a problem in mechanics of
materials was described in Sec. 1.9. Your solution should begin with
a clear and precise statement of the problem. You will then draw
one or several free-body diagrams that you will use to write equi-
librium equations. These equations will be solved for unknown
forces, from which the required stresses and deformations can be
computed. Once the answer has been obtained, it should be care-
fully checked.
Transverse forces. Shearing stressTransverse forces. Shearing stress
Single and double shearSingle and double shear
Review and Summary
Bearing stress
Method of Solution
A C B
P
PЈ
Fig. 1.42
Fig. 1.43
C
D
A
F
E'B
E
F'
K
AB
L
E H
G J
C
D
K'
L'
FF'
Fig. 1.44
A
C
D
d
t
F
P
F'
Fig. 1.45
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44 Introduction—Concept of Stress The first part of the chapter ended with a discussion of numeri-
cal accuracy in engineering, which stressed the fact that the accuracy
of an answer can never be greater than the accuracy of the given
data [Sec. 1.10].
In Sec. 1.11, we considered the stresses created on an oblique section
in a two-force member under axial loading. We found that both nor-
mal and shearing stresses occurred in such a situation. Denoting by
u the angle formed by the section with a normal plane (Fig. 1.46)
and by A0 the area of a section perpendicular to the axis of the
member, we derived the following expressions for the normal stress
s and the shearing stress t on the oblique section:
s 5
P
A0
cos2
u t 5
P
A0
sin u cos u (1.14)
We observed from these formulas that the normal stress is maximum
and equal to sm 5 PyA0 for u 5 0, while the shearing stress is maxi-
mum and equal to tm 5 Py2A0 for u 5 458. We also noted that t 5 0
when u 5 0, while s 5 Py2A0 when u 5 458.
Next, we discussed the state of stress at a point Q in a body under
the most general loading condition [Sec. 1.12]. Considering a small
cube centered at Q (Fig. 1.47), we denoted by sx the normal stress
exerted on a face of the cube perpendicular to the x axis, and by txy
and txz, respectively, the y and z components of the shearing stress
exerted on the same face of the cube. Repeating this procedure for
the other two faces of the cube and observing that txy 5 tyx, tyz 5
tzy, and tzx 5 txz, we concluded that six stress components are
required to define the state of stress at a given point Q, namely, sx,
sy, sz, txy, tyz, and tzx.
Section 1.13 was devoted to a discussion of the various concepts
used in the design of engineering structures. The ultimate load of a
given structural member or machine component is the load at which
the member or component is expected to fail; it is computed from
the ultimate stress or ultimate strength of the material used, as deter-
mined by a laboratory test on a specimen of that material. The ulti-
mate load should be considerably larger than the allowable load, i.e.,
the load that the member or component will be allowed to carry
under normal conditions. The ratio of the ultimate load to the allow-
able load is defined as the factor of safety:
Factor of safety 5 F.S. 5
ultimate load
allowable load
(1.24)
The determination of the factor of safety that should be used in the
design of a given structure depends upon a number of consider-
ations, some of which were listed in this section.
Section 1.13 ended with the discussion of an alternative approach to
design, known as Load and Resistance Factor Design, which allows
the engineer to distinguish between the uncertainties associated with
the structure and those associated with the load.
Stresses on an oblique section
Stress under general loading
Factor of safety
Load and Resistance Factor Design
P'
P
Fig. 1.46
yz
yx
xy
xzzx
zy
y
z
x
a
Qa
a
z
y
x
Fig. 1.47
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45
REVIEW PROBLEMS
1.59 A strain gage located at C on the surface of bone AB indicates that
the average normal stress in the bone is 3.80 MPa when the bone
is subjected to two 1200-N forces as shown. Assuming the cross
section of the bone at C to be annular and knowing that its outer
diameter is 25 mm, determine the inner diameter of the bone’s
cross section at C.
1200 N
1200 N
C
A
B
Fig. P1.59
B
A
C
0.5 in.
0.5 in.
1.8 in.
1.8 in.
45Њ
60Њ
5 kips
5 kips
Fig. P1.60
1.60 Two horizontal 5-kip forces are applied to pin B of the assembly
shown. Knowing that a pin of 0.8-in. diameter is used at each
connection, determine the maximum value of the average normal
stress (a) in link AB, (b) in link BC.
1.61 For the assembly and loading of Prob. 1.60, determine (a) the average
shearing stress in the pin at C, (b) the average bearing stress at C in
member BC, (c) the average bearing stress at B in member BC.
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46 Introduction—Concept of Stress 1.62 In the marine crane shown, link CD is known to have a uniform
cross section of 50 3 150 mm. For the loading shown, determine
the normal stress in the central portion of that link.
A
D
C
B
3 m25 m15 m
35 m
80 Mg
15 m
Fig. P1.62
1.63 Two wooden planks, each 1
2 in. thick and 9 in. wide, are joined by
the dry mortise joint shown. Knowing that the wood used shears
off along its grain when the average shearing stress reaches 1.20 ksi,
determine the magnitude P of the axial load that will cause the
joint to fail.
P'
P
␣
a
b
Fig. P1.64
1.64 Two wooden members of uniform rectangular cross section of
sides a 5 100 mm and b 5 60 mm are joined by a simple glued
joint as shown. Knowing that the ultimate stresses for the joint
are sU 5 1.26 MPa in tension and tU 5 1.50 MPa in shear and
that P 5 6 kN, determine the factor of safety for the joint when
(a) a 5 208, (b) a 5 358, (c) a 5 458. For each of these values of
a, also determine whether the joint will fail in tension or in shear
if P is increased until rupture occurs.
2 in.
1 in.
P'
2 in.
1 in. 9 in.
P
in.5
8
in.5
8
Fig. P1.63
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47Review Problems1.65 Member ABC, which is supported by a pin and bracket at C and
a cable BD, was designed to support the 16-kN load P as shown.
Knowing that the ultimate load for cable BD is 100 kN, determine
the factor of safety with respect to cable failure.
A
D
B
C
0.4 m
30Њ
40Њ
0.8 m
0.6 m
P
Fig. P1.65
1.67 Knowing that a force P of magnitude 750 N is applied to the pedal
shown, determine (a) the diameter of the pin at C for which the
average shearing stress in the pin is 40 MPa, (b) the correspond-
ing bearing stress in the pedal at C, (c) the corresponding bearing
stress in each support bracket at C.
diameterdiameter
x
B
E F
D
CA
xE
xF
60 in.
-in.5
8
-in.1
2
2000 lb
Fig. P1.66
9 mm
125 mm
75 mm
300 mm
5 mm
A B
C
C
D
P
Fig. P1.67
1.66 The 2000-lb load may be moved along the beam BD to any posi-
tion between stops at E and F. Knowing that sall 5 6 ksi for the
steel used in rods AB and CD, determine where the stops should
be placed if the permitted motion of the load is to be as large as
possible.
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48 Introduction—Concept of Stress 1.68 A force P is applied as shown to a steel reinforcing bar that has
been embedded in a block of concrete. Determine the smallest
length L for which the full allowable normal stress in the bar can
be developed. Express the result in terms of the diameter d of the
bar, the allowable normal stress sall in the steel, and the average
allowable bond stress tall between the concrete and the cylindri-
cal surface of the bar. (Neglect the normal stresses between the
concrete and the end of the bar.)
A
1.25 in.
2.4 kips
2.0 in.
B
Fig. P1.69 and P1.70
1.69 The two portions of member AB are glued together along a plane
forming an angle u with the horizontal. Knowing that the ultimate
stress for the glued joint is 2.5 ksi in tension and 1.3 ksi in shear,
determine the range of values of u for which the factor of safety
of the members is at least 3.0.
P
L d
Fig. P1.68
1.70 The two portions of member AB are glued together along a plane
forming an angle u with the horizontal. Knowing that the ultimate
stress for the glued joint is 2.5 ksi in tension and 1.3 ksi in shear,
determine (a) the value of u for which the factor of safety of the
member is maximum, (b) the corresponding value of the factor of
safety. (Hint: Equate the expressions obtained for the factors of
safety with respect to normal stress and shear.)
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49
COMPUTER PROBLEMS
The following problems are designed to be solved with a computer.
1.C1 A solid steel rod consisting of n cylindrical elements welded
together is subjected to the loading shown. The diameter of element i is
denoted by di and the load applied to its lower end by Pi, with the magni-
tude Pi of this load being assumed positive if Pi is directed downward as
shown and negative otherwise. (a) Write a computer program that can used
with either SI or U.S. customary units to determine the average stress in
each element of the rod. (b) Use this program to solve Probs. 1.2 and 1.4.
1.C2 A 20-kN load is applied as shown to the horizontal member ABC.
Member ABC has a 10 3 50-mm uniform rectangular cross section and
is supported by four vertical links, each of 8 3 36-mm uniform rectangular
cross section. Each of the four pins at A, B, C, and D has the same diam-
eter d and is in double shear. (a) Write a computer program to calculate
for values of d from 10 to 30 mm, using 1-mm increments, (1) the maxi-
mum value of the average normal stress in the links connecting pins B
and D, (2) the average normal stress in the links connecting pins C and
E, (3) the average shearing stress in pin B, (4) the average shearing stress
in pin C, (5) the average bearing stress at B in member ABC, (6) the
average bearing stress at C in member ABC. (b) Check your program by
comparing the values obtained for d 5 16 mm with the answers given for
Probs. 1.7 and 1.27. (c) Use this program to find the permissible values
of the diameter d of the pins, knowing that the allowable values of the
normal, shearing, and bearing stresses for the steel used are, respectively,
150 MPa, 90 MPa, and 230 MPa. (d) Solve part c, assuming that the thick-
ness of member ABC has been reduced from 10 to 8 mm.
Element n
Element 1
Pn
P1
Fig. P1.C1
0.2 m
0.25 m
0.4 m
20 kN
C
B
A
D
E
Fig. P1.C2
1.C3 Two horizontal 5-kip forces are applied to pin B of the assembly
shown. Each of the three pins at A, B, and C has the same diameter d and
is in double shear. (a) Write a computer program to calculate for values of
d from 0.50 to 1.50 in., using 0.05-in. increments, (1) the maximum value
of the average normal stress in member AB, (2) the average normal stress
B
A
C
0.5 in.
0.5 in.
1.8 in.
1.8 in.
45Њ
60Њ
5 kips
5 kips
Fig. P1.C3
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50 Introduction—Concept of Stress in member BC, (3) the average shearing stress in pin A, (4) the average
shearing stress in pin C, (5) the average bearing stress at A in member AB,
(6) the average bearing stress at C in member BC, (7) the average bearing
stress at B in member BC. (b) Check your program by comparing the values
obtained for d 5 0.8 in. with the answers given for Probs. 1.60 and 1.61.
(c) Use this program to find the permissible values of the diameter d of the
pins, knowing that the allowable values of the normal, shearing, and bearing
stresses for the steel used are, respectively, 22 ksi, 13 ksi, and 36 ksi. (d) Solve
part c, assuming that a new design is being investigated in which the thick-
ness and width of the two members are changed, respectively, from 0.5 to
0.3 in. and from 1.8 to 2.4 in.
1.C4 A 4-kip force P forming an angle a with the vertical is applied
as shown to member ABC, which is supported by a pin and bracket at C
and by a cable BD forming an angle b with the horizontal. (a) Knowing
that the ultimate load of the cable is 25 kips, write a computer program
to construct a table of the values of the factor of safety of the cable for
values of a and b from 0 to 458, using increments in a and b correspond-
ing to 0.1 increments in tan a and tan b. (b) Check that for any given
value of a, the maximum value of the factor of safety is obtained for b 5
38.668 and explain why. (c) Determine the smallest possible value of the
factor of safety for b 5 38.668, as well as the corresponding value of a,
and explain the result obtained.
␣
A
D
B
C
12 in.18 in.
15 in.
P
Fig. P1.C4
P
a
b
P'
␣
Fig. P1.C5
1.C5 A load P is supported as shown by two wooden members of uni-
form rectangular cross section that are joined by a simple glued scarf splice.
(a) Denoting by sU and tU, respectively, the ultimate strength of the joint
in tension and in shear, write a computer program which, for given values
of a, b, P, sU and tU, expressed in either SI or U.S. customary units, and
for values of a from 5 to 858 at 58 intervals, can calculate (1) the normal
stress in the joint, (2) the shearing stress in the joint, (3) the factor of safety
relative to failure in tension, (4) the factor of safety relative to failure in
shear, (5) the overall factor of safety for the glued joint. (b) Apply this pro-
gram, using the dimensions and loading of the members of Probs. 1.29 and
1.31, knowing that sU 5 150 psi and tU 5 214 psi for the glue used in
Prob. 1.29, and that sU 5 1.26 MPa and tU 5 1.50 MPa for the glue used
in Prob. 1.31. (c) Verify in each of these two cases that the shearing stress
is maximum for a 5 458.
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51Computer Problems1.C6 Member ABC is supported by a pin and bracket at A, and by two
links that are pin-connected to the member at B and to a fixed support at
D. (a) Write a computer program to calculate the allowable load Pall for
any given values of (1) the diameter d1 of the pin at A, (2) the common
diameter d2 of the pins at B and D, (3) the ultimate normal stress sU in
each of the two links, (4) the ultimate shearing stress tU in each of the
three pins, (5) the desired overall factor of safety F.S. Your program should
also indicate which of the following three stresses is critical: the normal
stress in the links, the shearing stress in the pin at A, or the shearing stress
in the pins at B and D (b and c). Check your program by using the data
of Probs. 1.55 and 1.56, respectively, and comparing the answers obtained
for Pall with those given in the text. (d) Use your program to determine the
allowable load Pall, as well as which of the stresses is critical, when d1 5
d2 5 15 mm, sU 5 110 MPa for aluminum links, tU 5 100 MPa for steel
pins, and F.S. 5 3.2.
180 mm200 mm
Top view
Side view
Front view
8 mm
20 mm
8 mm
8 mm
12 mm
12 mm
B
C
B
D D
A
B CA
P
Fig. P1.C6
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This chapter is devoted to the study of
deformations occurring in structural
components subjected to axial loading.
The change in length of the diagonal
stays was carefully accounted for in the
design of this cable-stayed bridge.
52
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C H A P T E R
53
Stress and Strain—Axial
Loading
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54
Chapter 2 Stress and Strain—
Axial Loading
2.1 Introduction
2.2 Normal Strain Under Axial
Loading
2.3 Stress-Strain Diagram
*2.4 True Stress and True Strain
2.5 Hooke’s Law; Modulus of
Elasticity
2.6 Elastic versus Plastic Behavior of
a Material
2.7 Repeated Loadings; Fatigue
2.8 Deformations of Members Under
Axial Loading
2.9 Statically Indeterminate Problems
2.10 Problems Involving Temperature
Changes
2.11 Poisson’s Ratio
2.12 Multiaxial Loading; Generalized
Hooke’s Law
*2.13 Dilatation; Bulk Modulus
2.14 Shearing Strain
2.15 Further Discussions of
Deformations Under Axial
Loading; Relation Among E, n,
and G
*2.16 Stress-Strain Relationships for
Fiber-Reinforced Composite
Materials
2.17 Stress and Strain Distribution
Under Axial Loading; Saint-
Venant’s Principle
2.18 Stress Concentrations
2.19 Plastic Deformations
*2.20 Residual Stresses
2.1 INTRODUCTION
In Chap. 1 we analyzed the stresses created in various members and
connections by the loads applied to a structure or machine. We also
learned to design simple members and connections so that they
would not fail under specified loading conditions. Another important
aspect of the analysis and design of structures relates to the deforma-
tions caused by the loads applied to a structure. Clearly, it is impor-
tant to avoid deformations so large that they may prevent the structure
from fulfilling the purpose for which it was intended. But the analysis
of deformations may also help us in the determination of stresses.
Indeed, it is not always possible to determine the forces in the mem-
bers of a structure by applying only the principles of statics. This is
because statics is based on the assumption of undeformable, rigid
structures. By considering engineering structures as deformable and
analyzing the deformations in their various members, it will be pos-
sible for us to compute forces that are statically indeterminate, i.e.,
indeterminate within the framework of statics. Also, as we indicated
in Sec. 1.5, the distribution of stresses in a given member is statically
indeterminate, even when the force in that member is known. To
determine the actual distribution of stresses within a member, it is
thus necessary to analyze the deformations that take place in that
member. In this chapter, you will consider the deformations of a
structural member such as a rod, bar, or plate under axial loading.
First, the normal strain P in a member will be defined as the
deformation of the member per unit length. Plotting the stress s
versus the strain P as the load applied to the member is increased
will yield a stress-strain diagram for the material used. From such a
diagram we can determine some important properties of the mate-
rial, such as its modulus of elasticity, and whether the material is
ductile or brittle (Secs. 2.2 to 2.5). You will also see in Sec. 2.5 that,
while the behavior of most materials is independent of the direction
in which the load is applied, the response of fiber-reinforced com-
posite materials depends upon the direction of the load.
From the stress-strain diagram, we can also determine whether
the strains in the specimen will disappear after the load has been
removed—in which case the material is said to behave elastically—or
whether a permanent set or plastic deformation will result (Sec. 2.6).
Section 2.7 is devoted to the phenomenon of fatigue, which
causes structural or machine components to fail after a very large
number of repeated loadings, even though the stresses remain in the
elastic range.
The first part of the chapter ends with Sec. 2.8, which is devoted
to the determination of the deformation of various types of members
under various conditions of axial loading.
In Secs. 2.9 and 2.10, statically indeterminate problems will
be considered, i.e., problems in which the reactions and the inter-
nal forces cannot be determined from statics alone. The equilib-
rium equations derived from the free-body diagram of the member
under consideration must be complemented by relations involving
deformations; these relations will be obtained from the geometry
of the problem.
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55In Secs. 2.11 to 2.15, additional constants associated with isotropic
materials—i.e., materials with mechanical characteristics independent
of direction—will be introduced. They include Poisson’s ratio, which
relates lateral and axial strain, the bulk modulus, which characterizes
the change in volume of a material under hydrostatic pressure, and
the modulus of rigidity, which relates the components of the shearing
stress and shearing strain. Stress-strain relationships for an isotropic
material under a multiaxial loading will also be derived.
In Sec. 2.16, stress-strain relationships involving several distinct
values of the modulus of elasticity, Poisson’s ratio, and the modulus
of rigidity, will be developed for fiber-reinforced composite materials
under a multiaxial loading. While these materials are not isotropic,
they usually display special properties, known as orthotropic proper-
ties, which facilitate their study.
In the text material described so far, stresses are assumed uni-
formly distributed in any given cross section; they are also assumed
to remain within the elastic range. The validity of the first assump-
tion is discussed in Sec. 2.17, while stress concentrations near circu-
lar holes and fillets in flat bars are considered in Sec. 2.18. Sections
2.19 and 2.20 are devoted to the discussion of stresses and deforma-
tions in members made of a ductile material when the yield point of
the material is exceeded. As you will see, permanent plastic deforma-
tions and residual stresses result from such loading conditions.
2.2 NORMAL STRAIN UNDER AXIAL LOADING
Let us consider a rod BC, of length L and uniform cross-sectional
area A, which is suspended from B (Fig. 2.1a). If we apply a load P
to end C, the rod elongates (Fig. 2.1b). Plotting the magnitude P of
the load against the deformation d (Greek letter delta), we obtain a
certain load-deformation diagram (Fig. 2.2). While this diagram con-
tains information useful to the analysis of the rod under consider-
ation, it cannot be used directly to predict the deformation of a rod
of the same material but of different dimensions. Indeed, we observe
that, if a deformation d is produced in rod BC by a load P, a load
2P is required to cause the same deformation in a rod B9C9 of the
same length L, but of cross-sectional area 2A (Fig. 2.3). We note
that, in both cases, the value of the stress is the same: s 5 PyA.
On the other hand, a load P applied to a rod B0C0, of the same
B B
C
C
L
A
P
␦
(a) (b)
Fig. 2.1 Deformation
of axially-loaded rod.
P
␦
Fig. 2.2 Load-deformation diagram.
2.2 Normal Strain under Axial Loading
2P
B'B'
C'
C'
L
2A
␦
Fig. 2.3
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56 Stress and Strain—Axial Loading cross-sectional area A, but of length 2L, causes a deformation 2d in
that rod (Fig. 2.4), i.e., a deformation twice as large as the deforma-
tion d it produces in rod BC. But in both cases the ratio of the
deformation over the length of the rod is the same; it is equal to
dyL. This observation brings us to introduce the concept of strain:
We define the normal strain in a rod under axial loading as the
deformation per unit length of that rod. Denoting the normal strain
by P (Greek letter epsilon), we write
P 5
d
L
(2.1)
Plotting the stress s 5 PyA against the strain P 5 dyL, we
obtain a curve that is characteristic of the properties of the material
and does not depend upon the dimensions of the particular specimen
used. This curve is called a stress-strain diagram and will be dis-
cussed in detail in Sec. 2.3.
Since the rod BC considered in the preceding discussion had
a uniform cross section of area A, the normal stress s could be
assumed to have a constant value PyA throughout the rod. Thus,
it was appropriate to define the strain P as the ratio of the total
deformation d over the total length L of the rod. In the case of a
member of variable cross-sectional area A, however, the normal
stress s 5 PyA varies along the member, and it is necessary to
define the strain at a given point Q by considering a small element
of undeformed length Dx (Fig. 2.5). Denoting by Dd the deforma-
tion of the element under the given loading, we define the normal
strain at point Q as
P 5 lim
¢xy0
¢d
¢x
5
dd
dx
(2.2)
Since deformation and length are expressed in the same units,
the normal strain P obtained by dividing d by L (or dd by dx) is a
dimensionless quantity. Thus, the same numerical value is obtained
for the normal strain in a given member, whether SI metric units or
U.S. customary units are used. Consider, for instance, a bar of length
L 5 0.600 m and uniform cross section, which undergoes a deforma-
tion d 5 150 3 1026
m. The corresponding strain is
P 5
d
L
5
150 3 1026
m
0.600 m
5 250 3 1026
m/m 5 250 3 1026
Note that the deformation could have been expressed in microme-
ters: d 5 150 mm. We would then have written
P 5
d
L
5
150 mm
0.600 m
5 250 mm/m 5 250 m
and read the answer as “250 micros.” If U.S. customary units are
used, the length and deformation of the same bar are, respectively,
P
BЉ BЉ
CЉ
CЉ
2L
A
2␦
Fig. 2.4
␦ ␦⌬x+ x +
Q
Q
⌬xx
⌬
P
Fig. 2.5 Deformation of axially-
loaded member of variable cross-
sectional area.
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57L 5 23.6 in. and d 5 5.91 3 1023
in. The corresponding strain is
P 5
d
L
5
5.91 3 1023
in.
23.6 in.
5 250 3 1026
in./in.
which is the same value that we found using SI units. It is customary,
however, when lengths and deformations are expressed in inches or
microinches (min.), to keep the original units in the expression obtained
for the strain. Thus, in our example, the strain would be recorded as
P 5 250 3 1026
in./in. or, alternatively, as P 5 250 min./in.
2.3 STRESS-STRAIN DIAGRAM
We saw in Sec. 2.2 that the diagram representing the relation between
stress and strain in a given material is an important characteristic of
the material. To obtain the stress-strain diagram of a material, one
usually conducts a tensile test on a specimen of the material. One
type of specimen commonly used is shown in Photo 2.1. The cross-
sectional area of the cylindrical central portion of the specimen has
been accurately determined and two gage marks have been inscribed
on that portion at a distance L0 from each other. The distance L0 is
known as the gage length of the specimen.
The test specimen is then placed in a testing machine (Photo 2.2),
which is used to apply a centric load P. As the load P increases, the
2.3 Stress-Strain Diagram
Photo 2.1 Typical
tensile-test specimen.
Photo 2.2 This machine is used to test tensile test specimens,
such as those shown in this chapter.
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58 Stress and Strain—Axial Loading distance L between the two gage marks also increases (Photo 2.3). The
distance L is measured with a dial gage, and the elongation d 5 L 2
L0 is recorded for each value of P. A second dial gage is often used
simultaneously to measure and record the change in diameter of the
specimen. From each pair of readings P and d, the stress s is computed
by dividing P by the original cross-sectional area A0 of the specimen,
and the strain P by dividing the elongation d by the original distance
L0 between the two gage marks. The stress-strain diagram may then
be obtained by plotting P as an abscissa and s as an ordinate.
Stress-strain diagrams of various materials vary widely, and dif-
ferent tensile tests conducted on the same material may yield differ-
ent results, depending upon the temperature of the specimen and
the speed of loading. It is possible, however, to distinguish some
common characteristics among the stress-strain diagrams of various
groups of materials and to divide materials into two broad categories
on the basis of these characteristics, namely, the ductile materials
and the brittle materials.
Ductile materials, which comprise structural steel, as well as
many alloys of other metals, are characterized by their ability to yield
at normal temperatures. As the specimen is subjected to an increas-
ing load, its length first increases linearly with the load and at a very
slow rate. Thus, the initial portion of the stress-strain diagram is a
straight line with a steep slope (Fig. 2.6). However, after a critical
value sY of the stress has been reached, the specimen undergoes a
large deformation with a relatively small increase in the applied load.
This deformation is caused by slippage of the material along oblique
surfaces and is due, therefore, primarily to shearing stresses. As we
can note from the stress-strain diagrams of two typical ductile mate-
rials (Fig. 2.6), the elongation of the specimen after it has started to
yield can be 200 times as large as its deformation before yield. After
a certain maximum value of the load has been reached, the diameter
of a portion of the specimen begins to decrease, because of local
instability (Photo 2.4a). This phenomenon is known as necking. After
necking has begun, somewhat lower loads are sufficient to keep the
specimen elongating further, until it finally ruptures (Photo 2.4b).
We note that rupture occurs along a cone-shaped surface that forms
an angle of approximately 458 with the original surface of the speci-
men. This indicates that shear is primarily responsible for the failure
of ductile materials, and confirms the fact that, under an axial load,
P
PЈ
Photo 2.3 Test specimen with
tensile load.
Yield Strain-hardening
Rupture
0.02
(a) Low-carbon steel
0.0012
0.2 0.25
60
40
20
Necking
⑀
Y
(ksi)
U
B
Rupture
(b) Aluminum alloy
0.004
0.2
60
40
20
⑀
Y
(ksi)
U
B
Fig. 2.6 Stress-strain diagrams of two typical ductile materials.
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59
shearing stresses are largest on surfaces forming an angle of 458 with
the load (cf. Sec. 1.11). The stress sY at which yield is initiated is
called the yield strength of the material, the stress sU corresponding
to the maximum load applied to the specimen is known as the ulti-
mate strength, and the stress sB corresponding to rupture is called
the breaking strength.
Brittle materials, which comprise cast iron, glass, and stone, are
characterized by the fact that rupture occurs without any noticeable
prior change in the rate of elongation (Fig. 2.7). Thus, for brittle
materials, there is no difference between the ultimate strength and
the breaking strength. Also, the strain at the time of rupture is much
smaller for brittle than for ductile materials. From Photo 2.5, we
note the absence of any necking of the specimen in the case of a
brittle material, and observe that rupture occurs along a surface per-
pendicular to the load. We conclude from this observation that nor-
mal stresses are primarily responsible for the failure of brittle
materials.†
The stress-strain diagrams of Fig. 2.6 show that structural steel
and aluminum, while both ductile, have different yield characteris-
tics. In the case of structural steel (Fig. 2.6a), the stress remains
constant over a large range of values of the strain after the onset of
yield. Later the stress must be increased to keep elongating the
specimen, until the maximum value sU has been reached. This is
due to a property of the material known as strain-hardening. The
Photo 2.4 Tested specimen of a
ductile material.
†The tensile tests described in this section were assumed to be conducted at normal
temperatures. However, a material that is ductile at normal temperatures may display the
characteristics of a brittle material at very low temperatures, while a normally brittle mate-
rial may behave in a ductile fashion at very high temperatures. At temperatures other than
normal, therefore, one should refer to a material in a ductile state or to a material in a
brittle state, rather than to a ductile or brittle material.
Rupture
⑀
BU ϭ
Fig. 2.7 Stress-strain diagram for a
typical brittle material.
2.3 Stress-Strain Diagram
Photo 2.5 Tested
specimen of a
brittle material.
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60 Stress and Strain—Axial Loading yield strength of structural steel can be determined during the ten-
sile test by watching the load shown on the display of the testing
machine. After increasing steadily, the load is observed to suddenly
drop to a slightly lower value, which is maintained for a certain
period while the specimen keeps elongating. In a very carefully con-
ducted test, one may be able to distinguish between the upper yield
point, which corresponds to the load reached just before yield starts,
and the lower yield point, which corresponds to the load required to
maintain yield. Since the upper yield point is transient, the lower
yield point should be used to determine the yield strength of the
material.
In the case of aluminum (Fig. 2.6b) and of many other ductile
materials, the onset of yield is not characterized by a horizontal por-
tion of the stress-strain curve. Instead, the stress keeps increasing—
although not linearly—until the ultimate strength is reached. Necking
then begins, leading eventually to rupture. For such materials, the
yield strength sY can be defined by the offset method. The yield
strength at 0.2% offset, for example, is obtained by drawing through
the point of the horizontal axis of abscissa P 5 0.2% (or P 5 0.002),
a line parallel to the initial straight-line portion of the stress-strain
diagram (Fig. 2.8). The stress sY corresponding to the point Y
obtained in this fashion is defined as the yield strength at 0.2%
offset.
A standard measure of the ductility of a material is its percent
elongation, which is defined as
Percent elongation 5 100
LB 2 L0
L0
where L0 and LB denote, respectively, the initial length of the tensile
test specimen and its final length at rupture. The specified minimum
elongation for a 2-in. gage length for commonly used steels with yield
strengths up to 50 ksi is 21 percent. We note that this means that
the average strain at rupture should be at least 0.21 in./in.
Another measure of ductility which is sometimes used is the
percent reduction in area, defined as
Percent reduction in area 5 100
A0 2 AB
A0
where A0 and AB denote, respectively, the initial cross-sectional area
of the specimen and its minimum cross-sectional area at rupture. For
structural steel, percent reductions in area of 60 to 70 percent are
common.
Thus far, we have discussed only tensile tests. If a specimen
made of a ductile material were loaded in compression instead of
tension, the stress-strain curve obtained would be essentially the
same through its initial straight-line portion and through the begin-
ning of the portion corresponding to yield and strain-hardening. Par-
ticularly noteworthy is the fact that for a given steel, the yield strength
is the same in both tension and compression. For larger values of
the strain, the tension and compression stress-strain curves diverge,
and it should be noted that necking cannot occur in compression.
Rupture
0.2% offset
⑀
Y
Y
Fig. 2.8 Determination of yield
strength by offset method.
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61
For most brittle materials, one finds that the ultimate strength in
compression is much larger than the ultimate strength in tension.
This is due to the presence of flaws, such as microscopic cracks or
cavities, which tend to weaken the material in tension, while not
appreciably affecting its resistance to compressive failure.
An example of brittle material with different properties in ten-
sion and compression is provided by concrete, whose stress-strain
diagram is shown in Fig. 2.9. On the tension side of the diagram, we
first observe a linear elastic range in which the strain is proportional
to the stress. After the yield point has been reached, the strain
increases faster than the stress until rupture occurs. The behavior of
the material in compression is different. First, the linear elastic range
is significantly larger. Second, rupture does not occur as the stress
reaches its maximum value. Instead, the stress decreases in magni-
tude while the strain keeps increasing until rupture occurs. Note that
the modulus of elasticity, which is represented by the slope of the
stress-strain curve in its linear portion, is the same in tension and
compression. This is true of most brittle materials.
*2.4 TRUE STRESS AND TRUE STRAIN
We recall that the stress plotted in the diagrams of Figs. 2.6 and 2.7
was obtained by dividing the load P by the cross-sectional area
A0 of the specimen measured before any deformation had taken
place. Since the cross-sectional area of the specimen decreases as P
increases, the stress plotted in our diagrams does not represent the
actual stress in the specimen. The difference between the engineer-
ing stress s 5 PyA0 that we have computed and the true stress
st 5 PyA obtained by dividing P by the cross-sectional area A of
the deformed specimen becomes apparent in ductile materials after
yield has started. While the engineering stress s, which is directly
proportional to the load P, decreases with P during the necking
phase, the true stress st, which is proportional to P but also inversely
proportional to A, is observed to keep increasing until rupture of
the specimen occurs.
Linear elastic range
Rupture, compression
Rupture, tension
⑀
U, tension
U, compression
Fig. 2.9 Stress-strain diagram for concrete.
*2.4 True Stress and True Strain
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62 Stress and Strain—Axial Loading Many scientists also use a definition of strain different from
that of the engineering strain P 5 dyL0. Instead of using the total
elongation d and the original value L0 of the gage length, they use
all the successive values of L that they have recorded. Dividing each
increment DL of the distance between the gage marks, by the cor-
responding value of L, they obtain the elementary strain DP 5 DLyL.
Adding the successive values of DP, they define the true strain Pt:
Pt 5 o¢P 5 o1¢LyL2
With the summation replaced by an integral, they can also express
the true strain as follows:
Pt 5 #
L
L0
dL
L
5 ln
L
L0
(2.3)
The diagram obtained by plotting true stress versus true strain
(Fig. 2.10) reflects more accurately the behavior of the material. As we
have already noted, there is no decrease in true stress during the neck-
ing phase. Also, the results obtained from tensile and from compressive
tests will yield essentially the same plot when true stress and true strain
are used. This is not the case for large values of the strain when the
engineering stress is plotted versus the engineering strain. However,
engineers, whose responsibility is to determine whether a load P will
produce an acceptable stress and an acceptable deformation in a given
member, will want to use a diagram based on the engineering stress
s 5 PyA0 and the engineering strain P 5 dyL0, since these expressions
involve data that are available to them, namely the cross-sectional area
A0 and the length L0 of the member in its undeformed state.
2.5 HOOKE’S LAW; MODULUS OF ELASTICITY
Most engineering structures are designed to undergo relatively small
deformations, involving only the straight-line portion of the corre-
sponding stress-strain diagram. For that initial portion of the diagram
(Fig. 2.6), the stress s is directly proportional to the strain P, and
we can write
s 5 EP (2.4)
This relation is known as Hooke’s law, after Robert Hooke (1635–1703),
an English scientist and one of the early founders of applied mechan-
ics. The coefficient E is called the modulus of elasticity of the material
involved, or also Young’s modulus, after the English scientist Thomas
Young (1773–1829). Since the strain P is a dimensionless quantity, the
modulus E is expressed in the same units as the stress s, namely in
pascals or one of its multiples if SI units are used, and in psi or ksi if
U.S. customary units are used.
The largest value of the stress for which Hooke’s law can be used
for a given material is known as the proportional limit of that material.
In the case of ductile materials possessing a well-defined yield point,
as in Fig. 2.6a, the proportional limit almost coincides with the yield
point. For other materials, the proportional limit cannot be defined as
t
⑀t
Yield
Rupture
Fig. 2.10 True stress versus true
strain for a typical ductile material.
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63easily, since it is difficult to determine with accuracy the value of the
stress s for which the relation between s and P ceases to be linear.
But from this very difficulty we can conclude for such materials that
using Hooke’s law for values of the stress slightly larger than the actual
proportional limit will not result in any significant error.
Some of the physical properties of structural metals, such as
strength, ductility, and corrosion resistance, can be greatly affected by
alloying, heat treatment, and the manufacturing process used. For
example, we note from the stress-strain diagrams of pure iron and of
three different grades of steel (Fig. 2.11) that large variations in the
yield strength, ultimate strength, and final strain (ductility) exist among
these four metals. All of them, however, possess the same modulus of
elasticity; in other words, their “stiffness,” or ability to resist a deforma-
tion within the linear range, is the same. Therefore, if a high-strength
steel is substituted for a lower-strength steel in a given structure, and
if all dimensions are kept the same, the structure will have an increased
load-carrying capacity, but its stiffness will remain unchanged.
For each of the materials considered so far, the relation between
normal stress and normal strain, s 5 EP, is independent of the
direction of loading. This is because the mechanical properties of
each material, including its modulus of elasticity E, are independent
of the direction considered. Such materials are said to be isotropic.
Materials whose properties depend upon the direction considered
are said to be anisotropic.
An important class of anisotropic materials consists of fiber-
reinforced composite materials. These composite materials are obtained
by embedding fibers of a strong, stiff material into a weaker, softer
material, referred to as a matrix. Typical materials used as fibers are
graphite, glass, and polymers, while various types of resins are used as
a matrix. Figure 2.12 shows a layer, or lamina, of a composite material
consisting of a large number of parallel fibers embedded in a matrix.
An axial load applied to the lamina along the x axis, that is, in a direc-
tion parallel to the fibers, will create a normal stress sx in the lamina
and a corresponding normal strain Px which will satisfy Hooke’s law as
the load is increased and as long as the elastic limit of the lamina is
not exceeded. Similarly, an axial load applied along the y axis, that is,
in a direction perpendicular to the lamina, will create a normal stress
sy and a normal strain Py satisfying Hooke’s law, and an axial load
applied along the z axis will create a normal stress sz and a normal
strain Pz which again satisfy Hooke’s law. However, the moduli of elas-
ticity Ex, Ey, and Ez corresponding, respectively, to each of the above
loadings will be different. Because the fibers are parallel to the x axis,
the lamina will offer a much stronger resistance to a loading directed
along the x axis than to a loading directed along the y or z axis, and
Ex will be much larger than either Ey or Ez.
A flat laminate is obtained by superposing a number of layers
or laminas. If the laminate is to be subjected only to an axial load
causing tension, the fibers in all layers should have the same orienta-
tion as the load in order to obtain the greatest possible strength. But
if the laminate may be in compression, the matrix material may not
besufficientlystrongtopreventthefibersfromkinkingorbuckling. The
lateral stability of the laminate may then be increased by positioning
Quenched, tempered
alloy steel (A709)
High-strength, low-alloy
steel (A992)
Carbon steel (A36)
Pure iron
⑀
Fig. 2.11 Stress-strain diagrams for
iron and different grades of steel.
2.5 Hooke’s Law; Modulus of Elasticity
Layer of
material
Fibers
y