The document is an instructor's manual for the seventh edition of "Mechanics of Materials" by Ferdinand P. Beer and co-authors, designed for engineering students taking their first or an advanced course in this subject. It provides an overview of course structure, core topics, problem sets, and instructional resources, emphasizing the analytical methods used in mechanics, including stress analysis and deformation considerations. The manual includes detailed descriptions of chapters and sections, covering key concepts such as stress, strain, torsion, and bending.

2. Instructors and Solutions Manual
to accompany
Mechanics of Materials
Seventh Edition
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
Prepared by
Amy Mazurek
PROPRIETARY AND CONFIDENTIAL
This Manual is the proprietary property of McGraw-Hill Education and protected by copyright and other state and federal laws.
This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This
document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.
Mecánica de Materiales - Ferdinand P. Beer 7ma. Edición.

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4. iii
TO THE INSTRUCTOR
As indicated in its preface, Mechanics of Materials is designed for the first course in mechanics
of materials—or strength of materials—offered to engineering students in the sophomore or
junior year. However, because of the large number of optional sections that have been included
and the maturity of approach that has been achieved, this text can also be used to teach a more
advanced course.
The text has been divided into units, each corresponding to a well-defined topic and consisting of
one or several theory sections followed by sample problems and a large number of problems to
be assigned. In order to accommodate courses of varying emphases, considerably more material
has been ‘included than can be covered effectively in a single three-credit-hour course. To assist
the instructors in making up a schedule of assignments that best fits their classes, the various
topics presented in the text have been listed in Table I and both a minimum and a maximum
number of periods to be spent on each topic have been suggested. Topics have been divided into
three categories: core topics that will probably be covered in every course; additional topics that
can be selected to complement this core to form courses of various emphases; and finally topics
that can be used with more advanced students.
The problems have been grouped according to the portions of material they illustrate and have
been arranged in order of increasing difficulty, with problems requiring special attention
indicated by asterisks. The instructor’s attention is called to the fact that problems have been
arranged in groups of six or more, all problems of the same group being closely related. This
means that the instructor will easily find additional problems to amplify a particular point that
has been brought up in the discussion of a problem assigned for homework. Since half of the
problems are stated in SI units and half in U.S. customary units, it also means that the instructor
has the choice of assigning problems using SI units and problems using U.S. customary units in
whatever proportion is found to be most desirable for a given class. To assist in the preparation
of homework assignments 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. It should also be noted
that answers to all problems with a number set in roman type are given at the end of the text,
while problems with a number set in italic are not.
In Table III six alternative lists of possible assignments have been suggested. Four of these lists
consist of problems whose answers are given at the end of the text, and two of problems whose
answers are not. Half of the problems in each list are stated in SI units and half in U.S. customary
units. For those instructors who wish to emphasize the use of SI units, four additional lists of
problems have been given in Table IV, in which 75% of the problems use SI units. Since the lists
in Tables III and IV cover the entire text, instructors will want to select those groups of sections
that are best suited to the course they wish to teach. For a typical one-semester course consisting
of 42 class meetings and including four quizzes, as many as 38 of the 46 available groups can be
selected.
Since the approach used in this text differs in a number of respects from the approach used in
other books, the instructor is advised to read the preface to Mechanics of Materials, in which the
authors have outlined their general philosophy. Attention is particularly called to the fact that
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5. iv
statically indeterminate problems are first discussed in Chapter 2 and are 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, and will have become thoroughly familiar with it by the end of
the course. The concept of plastic deformation is also introduced in Chap. 2, where it is applied
to the analysis of members under axial loading, while problems involving the plastic deformation
of circular shafts and of prismatic beams are considered in optional sections of Chaps. 3 and 4,
respectively. On the other hand, while the concept of stress at a point is introduced in Chap. 1,
the discussion of the transformation of stresses is delayed until Chap. 7, after students have
discovered for themselves the need for special techniques, such as Mohr’s circle. In this edition,
shear and bending-moment diagrams are introduced at the beginning of Chap. 5 and applied
immediately to the design of beams in that chapter and in the next.
A brief description, chapter by chapter, of the topics included in the text will be found in the
following pages. It is hoped that this material will help instructors organize their courses to best
fit the needs of their students.
The authors of Mechanics of Materials, 7/e, wish to thank Professor Dean P. Updike of the
Department of Mechanical Engineering and Mechanics at Lehigh University and Amy Mazurek
for having written the problem solutions contained in this Manual.
John T. DeWolf
David F. Mazurek
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DESCRIPTION OF THE MATERIAL CONTAINED IN
MECHANICS OF MATERIALS, Seventh Edition
Chapter 1
Introduction–Concept of Stress
The main purpose of this chapter is to introduce the concept of stress. After a short review of
Statics in Section 1.1 emphasizing the use of free-body diagrams, Sections 1.2 through 1.2
discuss normal stresses under an axial loading, shearing stresses—with applications to pins and
bolts in single and double shear—and bearing stresses. This section also introduces the student
to the concepts of analysis and design. Section 1.2A emphasizes the fact that stresses are
inherently statically indeterminate and that, at this point, normal stresses under an axial loading
can only be assumed to be uniformly distributed. Moreover, such an assumption requires that the
axial loading be centric.
Section 1.2D is devoted to the application of these concepts to the analysis of a simple structure.
Section 1.2E describes how students should approach the solution of a problem in mechanics of
materials using the SMART methodology: Strategy, Modeling, Analysis and Reflect & Think.
Section 1.2E also discusses the numerical accuracy to be expected in such a solution. Problems
included in the first lesson also serve as a review of the methods of analysis of trusses, frames,
and mechanisms learned in statics.
Section 1.3 discusses the determination of normal and shearing stresses on oblique planes under
an axial loading, while Section 1.4 introduces the components of stress under general loading
conditions. This section emphasizes the fact that the components of the shearing stresses exerted
on perpendicular planes, such as τxy and τyx, must be equal. It also introduces the students to the
concept of transformation of stress. However, the study of the computational techniques
associated with the transformation of stress at a point is delayed until Chapter 7, after students
have discovered for themselves the need for such techniques.
Section 1.5 is devoted to design considerations. It introduces the concepts of ultimate load,
ultimate stress, and factor of safety. It also discusses the reasons for the use of factors of safety in
engineering practice. The section ends with an optional presentation of an alternative method of
design, Load and Resistance Factor Design.
Chapter 2
Stress and Strain–Axial Loading
This chapter is devoted to the analysis and design of members under a centric axial loading.
Section 2.1A introduces the concept of normal strain, while Section 2.1B describes the general
properties of the stress-strain diagrams of ductile and brittle materials and defines the yield
strength, ultimate strength, and breaking strength of a material. Section 21C, which is optional,
defines true stress and true strain. Section 2.1D introduces Hooke’s law, the modulus of
elasticity, and the proportional limit of a material. It defines as isotropic those materials whose
mechanical properties are independent of the direction considered and as anisotropic those
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whose mechanical properties depend upon that direction. Among the latter are fiber-reinforced
composite materials, that are described in this section.
Section 2.1E discusses the elastic and the plastic behavior of a material and defines its elastic
limit, while Section 2.1F is devoted to fatigue and the behavior of materials under repeated
loadings. The first lesson of Chapter 2 ends with Section 2.1G, which shows how Hooke’s law
can be used to determine the deformation of a rod of uniform or variable cross section under one
or several loads, and introduces the concept of relative displacement. Section 2.2 discusses
statically indeterminate problems involving members under an axial load. As indicated in the
preface of the text and in the introduction to this manual, the authors believe it is important to
introduce the students at an early stage to the concept of statical indeterminacy and to show them
how the analysis of deformations can be used in the solution of problems that cannot be solved
by the methods of statics alone. It will also help them realize that stresses, being statically
indeterminate, can be computed only by considering the corresponding distribution of strains.
Section 2.3 discusses the thermal expansion of rods and shows how to determine stresses in
statically indeterminate members subjected to temperature changes.
Section 2.4 introduces the concept of lateral strain for an isotropic material and defines
Poisson’s ratio. Section 2.5 discusses the multiaxial loading of a structural element and derives
the generalized Hooke’s law for such a loading. Since this derivation is based on the application
of the principle of superposition, this principle is also introduced in Section 2.5, and the
conditions under which it can be used are clearly stated. Section 2.6 is optional. It discusses the
change in volume of a material under a multiaxial loading and defines the dilatation and the bulk
modulus or modulus of compression of a given material.
Section 2.7 introduces the concept of shearing strain. It should be noted that the authors define
the shearing strain as the change in the angle formed by the faces of the element of material
considered, and not as the angle through which one of these faces rotates. Hooke’s law for
shearing stress and strain and the modulus of rigidity are also introduced in this section, as well as
the generalized Hooke’s law for a homogeneous, isotropic material under the most general stress
conditions. Section 2.8 points out that strains, just as stresses, depend upon the orientation of the
planes considered. It also establishes the fact that the constants E, v, and G are not independent
from each other and derives Eq. (2.35), that expresses the relation among these three constants.
Section 2.9, which is optional, extends the stress-strain relationships to fiber-reinforced
composite materials. The relations obtained are expressed by Eqs. (2.37) and (2.39) and involve
three different values of the modulus of elasticity and six different values of Poisson’s ratio.
Section 2.10 discusses the distribution of the normal stresses under a centric axial loading and
shows that this distribution depends upon the manner in which the loads are applied. However,
except in the immediate vicinity of the points of application of the loads, the distribution of
stresses can be assumed uniform. This result verifies Saint-Venant’s principle. Section 2.11
discusses stress concentrations near circular holes and fillets in flat bars under axial loading.
Section 2.12 is devoted to the plastic deformation of members under centric axial loads and
introduces the concept of an elastoplastic material. As stated in the preface of the text, the
authors believe that students should be exposed to the concept of plastic deformation in the first
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8. vii
course in mechanics of materials, if only to let them realize the limitations of the assumption of a
linear stress-strain relation in engineering applications. By introducing this concept early in the
course in connection with axial loading, rather than later with torsion or bending, one makes it
easier for the students to understand and accept it. For the same reason, residual stresses are
discussed in Section 2.13 in connection with axial loading. However, since some instructors may
not want to include the concept of residual stresses in an elementary course, this section is
optional and can be omitted without any prejudice to the understanding of the rest of the text.
Chapter 3
Torsion
The Introduction introduces this type of loading, while Section 3.1 establishes the relation that
must be satisfied, on the basis of statics, by the shearing stresses in a given section of a shaft
subjected to a torque. This condition, however, does not suffice to determine the stresses, and
one must analyze the deformations that occur in the shaft. This is done in Section 3.1A, where it
is proved that the distribution of shearing strains in a circular shaft is linear. It should be noted
that the discussion presented in See. 3.1B is based solely on the assumption of rigid end plates,
rather than on arbitrary and gratuitous assumptions regarding the deformations of a shaft. The
results obtained in this and the following sections clearly depend upon the validity of this
assumption, but can be extended to other loading conditions through the application of Saint-
Venant’s principle.
Section 3.1C is devoted to the analysis of the shearing stresses in the elastic range and presents
the derivation of the elastic torsion formulas for circular shafts. The section ends with remarks
on the transformation of stresses in torsion and the comparison between the failures of ductile
and brittle materials in torsion.
The formula for the angle of twist of a shaft in the elastic range is derived in Section 3.2. This
section also contains various applications involving the twisting of single and gear-connected
shafts. Section 3.3 deals with the solution of problems involving statically indeterminate shafts.
Section 3.4 is devoted to the design of transmission shafts and begins with the determination of
the torque required to transmit a given power at a given speed, both in SI and U.S. customary
units. Note that the effect of bending on the design of transmission shafts will be discussed in
Section 8.2, which is optional. Section 3.5 discusses stress concentrations at fillets in circular
shafts.
Sections 3.6 through 3.7 deal with the plastic deformations and residual stresses in circular
shafts and are optional. Since a similar presentation of the plastic deformations and residual
stresses of members in pure bending is given in Chapter 4, the instructor may decide to include
only one of these presentations in the course. Section 3.6 describes the general method for the
determination of the torque corresponding to a given maximum shearing stress in a shaft made of
a material with a nonlinear stress-strain diagram, while Sections 3.7 and 3.8 deal, respectively,
with the deformations and the residual stresses in shafts made of an elastoplastic material.
Sections 3.9 and 3.10 are also optional. They are devoted, respectively, to the torsion of solid
members and thin-walled hollow shafts of noncircular section.
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Chapter 4
Pure Bending
The Introduction defines this type of loading and shows how the results obtained in the following
sections can be applied to the analysis of other types of loadings as well, namely, eccentric axial
loadings and transverse loadings. Sections 4.1 and 4.1A establish the relation that must be
satisfied, on the basis of statics, by the normal stresses in a given section of a member subjected
to pure bending. This condition, however, does not suffice to determine the stresses, and one
must analyze the deformations that occur in the member. This is done in Section 4.1B, where it is
proved that the distribution of normal stresses in a symmetric member in pure bending is linear.
It should be noted that no assumption is made in this discussion regarding the deformations of
the member, except that the couples should be applied in such a way that the ends of the member
remain plane. Whether this can actually be accomplished is discussed at the end of Section 4.3.
Section 4.2 is devoted to the analysis of the normal stresses in the elastic range and presents the
derivation of the elastic flexure formulas. It also defines the elastic section modulus and ends
with the derivation of the formula for the curvature of an elastic beam. Section 4.3 discusses the
anticlastic curvature of members in pure bending and also states the loading conditions required
for the ends of the member to remain plane.
Section 4.4 discusses the determination of stresses in members made of several materials and
defines the transformed section of such members. It also shows how the transformed section can
be used to determine the radius of curvature of the member. The section ends with a discussion
of the stresses in reinforced-concrete beams. Section 4.5 deals with the stress concentrations at
fillets and grooves in flat bars under pure bending.
Section 4.6 is optional. This section discusses the plastic deformations and residual stresses in
members subjected to pure bending in much the same way that these were discussed in Sections 3.6
through 3.8 in the case of members in torsion. Section 4.6 describes the general method for the
determination of the bending moment corresponding to a given maximum normal stress in a
member possessing two planes of symmetry and made of a material with a nonlinear stress-strain
diagram. Section 4.6A deals with members made of an elastoplastic material and derives
formulas relating the thickness of the elastic core and the radius of curvature with the applied
bending moment in the case of members with a rectangular cross section. It also defines the
shape factor and the plastic section modulus of members with a nonrectangular section. Section
4.6B deals with the determination of the plastic moment of members made of an elastoplastic
material and possessing a single plane of symmetry, while Section 4.6C discusses residual
stresses.
Section 4.7 shows how the stresses due to a two-dimensional eccentric axial loading can be
obtained by replacing the given eccentric load by a centric load and a couple, and superposing
the corresponding stresses. Attention is called to the fact that the neutral axis does not pass
through the centroid of the section.
Section 4.8 deals with the unsymmetric bending of elastic members. It is first shown that the
neutral axis of a cross section will coincide with the axis of the bending couple if, and only if, the
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axis of the couple is directed along one of the principal centroidal axes of the cross section. It is
then shown that stresses due to unsymmetric bending can always be determined by resolving the
given bending couple into two component couples directed along the principal axes of the
section and superposing the corresponding stresses. Sample Problem 4.10 has been included in
this section to provide an opportunity for students to use material on Mohr’s Circle to determine
stresses in non-symmetrical sections. The material is based on what students will have learned in
a study of Statics where the transformation of moments of inertia is covered for problems where
it is necessary to look at the rotation of coordinate axes. Problems 4.141 through 4.143 are also
asterisked and are best solved with the use of Mohr’s Circle.
This method of analysis is extended in Section 4.9 to the determination of the stresses due to an
eccentric axial loading in three-dimensional space. The eccentric load is replaced by an equivalent
system consisting of a centric load and two bending couples, and the corresponding stresses are
superposed.
Section 4.10 is optional; it deals with the bending of curved members.
Chapter 5
Analysis and Design of Beams
for Bending
In the Introduction beams are defined as slender prismatic members subjected to transverse loads
and are classified according to the way in which they are supported. It is shown that the internal
forces in any given cross section are equivalent to a shear force V and a bending couple M. The
bending couple M creates normal stresses in the section, while the shear force V creates shearing
stresses. The former is determined in this chapter, using the flexure formula (5.1), while the latter
will be discussed in Chapter 6.
Since the dominant criterion in the design of beams for strength is usually the bending stresses in
the beam, the determination of the maximum value of the bending moment in the beam is the
most important factor to be considered. To facilitate the determination of the bending moment in
any given section of the beam, the concept of shear and bending-moment diagrams will be
introduced in Section 5.1, using free-body diagrams of various portions of the beam.
An alternative method for the determination of shear and bending-moment diagrams, based on
relations among load, shear, and bending moment, is presented in Section 5.2. To maintain the
interest of the students, most of the problems to be assigned are focused on the engineering
applications of these methods and call for the determination, not only of the shear and bending
moment, but also of the normal stresses in the beam.
Section 5.3 is devoted to the design of prismatic beams based on the allowable normal stress for
the material used. Sample Problems and problems to be assigned include wooden beams of
rectangular cross section, as well as rolled-steel W and S beams. An optional paragraph on page 372
describes the application of Load and Resistance Factor Design to beams under transverse loading.
Section 5.4 introduces the concept of singularity functions and shows how these functions can
provide an alternative and effective method for the determination of the shear and bending
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11. x
moment at any point of a beam under the most general loading condition. While this section is
optional, it should be included in the lesson schedule if singularity functions are to be used later
for the determination of the slope and deflection of a beam (Section 9.3). It is pointed out on
page 387 that singularity functions are particularly well suited to the use of computers, and
several optional problems requiring the use of a computer (Probs. 5.118 through 5.125) have
been included in this assignment.
Section 5.5, which is optional, is devoted to nonprismatic beams, such as forged or cast beams
designed to be of constant strength, and rolled-steel beams reinforced with cover plates.
Chapter 6
Shearing Stresses in Beams and Thin-Walled Members
It is shown in the Introduction that a transverse load creates shearing stresses as well as normal
stresses in a beam. Considering first the horizontal face of a beam element, it is shown in
Sections 61A that the horizontal shear per unit length q, or shear flow, is equal to VQ/I. This
result is applied in Concept Application 6.1 to the determination of the shear force in the nails
connecting three planks forming a wooden beam, as well as in Probs. 6.1 through 6.4. Probs. 6.5
through 6.8 apply the same concepts to steel beams made of sections bolted together
In Section 6.1B the average shearing stress τave exerted on the horizontal face of the beam
element is obtained by dividing the shear flow q by the width t of the beam:
ave
VQ
It
  (6.6)
Note that since the shearing stresses τxy and τyx exerted at a given point are equal, the expression
obtained also represents the average shearing stress exerted at a given height on a vertical section
of the beam. This formula is used to determine shearing stresses in a beam made of glued planks
in Sample Prob. 6.1 and to design a timber beam in Sample Prob. 6.2. Problems 6.9 through 6.12
and 6.21 and 6.22 call for the determination of shearing stresses in various types of beams.
Section 6.1C explores shearing stresses in common beam types In Concept Applications 6.2 and
6.3 the designs obtained on the basis of normal stresses, respectively, for a timber beam in
Sample Prob. 5.7 and for a rolled-steel beam in Sample Prob. 5.8 are checked and found to be
acceptable from the point of view of shearing stresses.
Section 6.2 is optional and discusses the distribution of stresses in a narrow rectangular beam.
In Section 6.3 the expression q = VQ/I obtained on Section 6.1A for the shear flow on the
horizontal face of a beam element is shown to remain valid for the curved surface of a beam
element of arbitrary shape. It is then applied in Concept Application 6.4 and in Probs. 6.29
through 6.33 to the determination of the shearing forces and shearing stresses in nailed and glued
vertical surfaces.
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Section 6.4 deals with the determination of shearing stresses in thin-walled members and shows
that Eq. (6.6) can be applied to the determination of the average shearing stress in a section of
arbitrary orientation.
Section 6.5, which is optional, describes the formation of plastic zones in beams subjected to
transverse loads.
Section 6.6, which is also optional, deals with the unsymmetric loading of thin-walled members,
the determination of the shear center, and the computation of the shearing stresses caused by a
shearing force exerted at the shear center.
Chapter 7
Transformations of Stress and Strain
After a short introduction, formulas for the transformation of plane stress under a rotation of
axes are derived in Section 7.1A, while the principal planes of stress, principal stresses, and
maximum shearing stress are determined in Section 7.1B.
Section 7.2 is devoted to the use of Mohr’s circle. It should be noted that the convention used in
the text provides for a rotation on Mohr’s circle in the same sense as the corresponding rotation
of the element; in other words, this convention is the same as that used in statics for the
transformation of moments and products of inertia. Attention is called to the statement at the
bottom of page 493 of the text and the accompanying Fig. 7.15.
Section 7.3 discusses the general (three-dimensional) state of stress and establishes the fact that
three principal axes of stress and three principal stresses exist. Section 7.4 shows how three
different Mohr’s circles can be used to represent the transformations of stress associated with
rotations of the element about the principal axes. The results obtained are used to show that in a
state of plane stress, the maximum shearing stress does not necessarily occur in the plane of
stress.
Section 7.5 is optional. Section 7.5A presents the two criteria most commonly used to predict
whether a ductile material will yield under a given state of plane stress, while Section 7.5B
discusses the two criteria used to predict the fracture of brittle materials.
Section 7.6 deals with stresses in thin-walled pressure vessels; it is limited to the analysis of
cylindrical and spherical pressure vessels.
The second part of the chapter (Sections 7.7 through 7.9) deals with transformations of strain
and is optional. Section 7.7A presents the derivation of the formulas for the transformation of strain
under a rotation of axes. It should be noted that this derivation is based on the consideration of an
oblique triangle (Fig. 7.51) and the use of the law of cosines, and that the determination of the
shearing strain is facilitated by the use of Eq. (7.43), which relates it to the normal strain along
the coordinate axes and their bisector.
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Section 7.7B introduces Mohr’s circle for plane strain, and Section 7.8 discusses the three-
dimensional analysis of strain and its application to the determination of the maximum shearing
strain in states of plane strain and of plane stress. Section 7.9 deals with the use of strain rosettes
for the determination of states of plane strain.
Chapter 8
Principal Stresses under a Given Loading
This chapter is devoted to the determination of the principal stresses and maximum shearing
stress in beams, transmission shafts subjected to transverse loads as well as to torques, and
bodies of arbitrary shape under combined loadings.
In the Introduction it is shown that, while only normal stresses occur on a square element with
horizontal and vertical faces located at the surface of a beam, shearing stresses will occur if the
element is rotated through 45o
(Fig. 8.1). The reverse situation is observed for an element with
horizontal and vertical faces subjected only to shearing stresses (Fig. 8.2). The analysis of beams,
therefore, should include the determination of the principal stresses and maximum shearing
stress at various points.
This is done in Section 8.1 for cantilever beams of various rectangular sections subjected to a
single concentrated load at their free end. It is found that the principal stress σmax does not exceed
the maximum normal stress σm determined by the method of Chapter 5 except very close to the
load. While this result holds for most beams of nonrectangular section, it may not be valid for
rolled-steel W and S beams, and the analysis and design of such beams should include the
determination of the principal stress σmax at the junction of the web with the flanges of the beam.
(See Sample Probs. 8.1 and 8.2, and Probs. 8.1 through 8.14).
Section 8.2 is devoted to the analysis and design of transmission shafts using gears or sprocket
wheels to transmit power to and from the shaft. These shafts are subjected to transverse loads as
well as to torques. The design of such shafts is the subject of Sample Prob. 8.3 and Probs. 8.15
through 8.30.
The determination of the stresses at a given point K of a body due to a combined loading is the
subject of Section 8.3. First, the loading is reduced to an equivalent system of forces and couples
in a section of the body containing K. Next, the normal and shearing stresses are determined at K.
Finally, using one of the methods of transformation of stresses presented in Chapter 7, the
principal planes, principal stresses, and maximum shearing stress may be determined at K. This
procedure is illustrated in Concept Application 8.1 and Sample Probs. 8.4 and 8.5.
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Chapter 9
Deflection of Beams
The relation derived in Chapter 4 between the curvature of a beam and the bending moment is
recalled in Section 9.1 and used to predict the variation of the curvature along the beam. In
Section 9.1A, the equation of the elastic curve for a beam is obtained through two successive
integrations, after the bending moment has been expressed as a function of the coordinate x.
Concept Applications 9.1 and 9.2 show how the boundary conditions can be used to determine
the two constants of integration in the cases of a cantilever beam and of a simply supported
beam. Concept Application 9.3 indicates how to proceed when the bending moment must be
represented by two different functions of x.
Section 9.1B is optional; it shows in the case of a beam supporting a distributed load, how the
equation of the elastic curve can be obtained directly from the function representing the load
distribution through the use of four successive integrations.
Section 9.2 is devoted to the analysis of statically indeterminate beams and to the determination
of the reactions at their supports. It is suggested that a minimum of two lessons be spent on
Sections 9.1 through 9.2 if neither the use of singularity functions (Section 9.3) nor the
moment-area method (Sections 9.5 through 9.6) are to be covered in the course.
Section 9.3 is devoted to the use of singularity functions for the determination of beam
deflections and slopes. It is optional and assumes that Section 5.4 has been covered previously. It
is recommended that both Sections 5.4 and 9.3 be included in the course, since singularity
functions provide the students with an effective and versatile method for the determination of
deflections and slopes under the most diverse loading conditions. In addition, and as indicated
earlier, singularity functions are well suited to the use of computers.
Section 9.4A discusses the method of superposition for the determination of beam deflections
and slopes. It shows how the expressions given in Appendix D for various simple loadings can
be used to obtain the deflection and slope of a beam supporting a more complex loading. In
Section 9.4B, the method of superposition is applied to the determination of the reactions at the
supports of statically indeterminate beams.
Sections 9.5 through 9.6 are optional. They deal with the application of the moment-area
methods to the determination of the deflection of beams and may be omitted in courses that place
a greater emphasis on analytical methods and make use of singularity functions. It should be
noted, however, that these methods provide a very practical means for the determination of the
deflection and slope of beams of variable cross section.
The two moment-area theorems are derived in Section 9.5A and are immediately applied in
Section 9.5B to the computation of the slope and deflection of cantilever beams and beams with
symmetric loadings (simply supported or overhanging beams). Section 9.5C shows how to draw
a bending-moment diagram by parts. This approach greatly facilitates the determination of
moment areas in all but the simplest loading situations.
Mecánica de Materiales - Ferdinand P. Beer 7ma. Edición.

15. xiv
Section 9.6 deals with simply supported and overhanging beams with unsymmetric loadings.
The analysis of such beams hinges on the use of a reference tangent drawn through one of the
supports after the tangential deviation of the second support has been computed from the
bending-moment diagram. Section 9.6B describes how to locate the point of maximum deflection
and how to compute that deflection.
Section 9.6C deals with the analysis of statically indeterminate beams and the determination of
the reactions at their supports.
Chapter 10
Columns
Section 10.1 introduces the concept of stability of a structure. The example considered in this
section consists of a block supported by two spring-connected rigid rods. It is shown that the
position of equilibrium in which both rods are aligned is stable if this position is the only
possible position of equilibrium of the system. The same criterion is applied to an elastic
pin-ended column in Section 10.1A in order to derive Euler’s formula. Section 10.1B shows how
Euler’s formula for pin-ended columns can be used to determine the critical load of columns
with other end conditions.
Section 10.2 is optional; it deals with the eccentric loading of a column and gives the derivation
of the secant formula.
Section 10.3 discusses the design of columns under a centric load. Empirical formulas developed
by various engineering associations for the design of steel columns, aluminum columns, and wood
columns are presented in Section 10.3A. Section 10.3B is devoted to an optional discussion of
the application of Load and Resistance Factor Design to steel columns. As noted at the end of
this section, the design formulas presented in this section are intended to provide introductory
examples of different design approaches. These formulas do not provide all the requirements that
are needed for more comprehensive designs often encountered in engineering practice.
Section 10.4 discusses the design of columns under an eccentric load and presents two of the
most frequently used methods: the allowable-stress method and the interaction method.
Chapter 11
Energy Methods
Section 11.1A introduces the concept of strain energy by considering the work required to
stretch a rod of uniform cross section. This work, which is equal to the area under the load-
deformation curve, represents the strain energy of the rod. The strain-energy density is defined in
Section. 11.1B, as well as the modulus of toughness and the modulus of resilience of a given
material. The formula for the elastic strain energy associated with normal stresses is derived in
Section 11.2A, as well as the expressions for the strain energy corresponding to an axial loading
and to pure bending. The formula for the strain energy associated with shearing stresses is
derived in Section 11.2B, as well as the expressions corresponding to torsion and transverse
loading.
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16. xv
Section 11.3, which is optional, covers the strain energy for a general state of stress and derives
an expression for the distortion energy per unit volume, both in the general case of three-
dimensional stress and in the particular case of plane stress.
Section 11.4A discusses impact loadings and Section 11.4B the design of a structure for an
impact load. To facilitate the solution of impact-loading problems, it is shown in Section 11.5A
that the strain energy of a structure subjected to a single concentrated load P can be obtained by
equating the strain energy to the work of P. (Appendix D is used to express the deflection in
terms of P ). As shown in Section 11.5B, the reverse procedure can be used to determine the
deflection of a structure at the point of application of a single load P or a single couple M; the
strain energy of the structure is computed from one of the formulas derived in Section 11.2, and
the work of P or M is equated to the expression obtained for the strain energy.
Sections 11.6 through 11.9 are optional. In Section 11.6 an expression for the strain energy of a
structure subjected to several loads is obtained by computing the work of the loads as they are
successively applied. Reversing the order in which the loads are applied, one proves Maxwell’s
reciprocal theorem. The expression obtained for the strain energy of the structure is used in
Section 11.7 to prove Castigliano’s theorem. Section 11.8 is devoted to the application of
Castigliano’s theorem to the determination of the deflection and slope of a beam and to the
deflection of a point in a truss. Finally, Section 11.9 deals with the application of Castigliano’s
theorem to the determination of the reactions at the supports of statically indeterminate structures
such as beams and trusses.
Mecánica de Materiales - Ferdinand P. Beer 7ma. Edición.

17. Mecánica de Materiales - Ferdinand P. Beer 7ma. Edición.

18. TABLE I: LIST OF TOPICS COVERED IN MECHANICS OF MATERIALS, Seventh Edition
Suggested Number of Periods
Core Additional Advanced
Sections Topics Topics Topics Topics
Chapter 1: Introduction – Concept of Stress
1.1-2 Stress Under Axial Loading 1-2
1.3-5 Components of Stress; Factor of Safety 1
Chapter 2: Stress and Strain – Axial Loading
2.1 Stress-Strain Diagrams; Deformations Under 1-2
Axial Loading
2.2-3 Statically Indeterminate Problems 1
2.4-5 Poisson’s Ratio; Generalized Hooke’s Law 1
*2.6 Dilatation; Bulk Modulus 0.25-0.5
2.7-8 Shearing Strain 0.5
*2.9 Stress-Strain Relationships for Fiber-Reinforced 0.5-1
Composite Materials
2.10-12 Stress Concentrations; Plastic Deformations 0.5-1
*2.13 Residual Stresses 0.5
Chapter 3: Torsion
3.1 Stresses in Elastic Range 1
3.2-3 Angle of Twist; Statically Indeterminate Shafts 1-2
3.4-5 Design of Transmission Shafts; Stress 1
Concentrations
*3.6-8 Plastic Deformations; Residual Stresses 1-2
*3.9-10 Noncircular Members;
Thin-Walled Hollow Shafts 1-2
Chapter 4: Pure Bending
4.1-3 Stresses in Elastic Range 1-2
4.4-5 Members Made of Several Materials; Stress 1-2
Concentrations
*4.6 Plastic Deformations; Residual Stresses 1-2
4.7 Eccentric Axial Loading 1-2
4.8-9 Unsymmetric Bending; General Eccentric 1-2
Axial Loading
*4.10 Bending of Curved Members 1-2
Chapter 5: Analysis and Design of Beams for Bending
5.1 Shear and Bending-Moment Diagrams 1-1.5
5.2 Using Relations Between w, V, and M 1-1.5
5.3 Design of Prismatic Beams in Bending 1-2
*5.4 Use of Singularity Functions to Determine
V and M 1-2
*5.5 Nonprismatic Beams 1-2
Chapter 6: Shearing Stresses in Beams and Thin-Walled Members
6.1 Shearing Stresses in Beams 1-2
*6.2 Shearing Stresses in Narrow Rectangular Beam 0.25
6.3-4 Shearing Stresses in Thin-Walled Members 1-2
*6.5 Plastic Deformations 0.25
*6.6 Unsymmetric Loading; Shear Center 1-2
Mecánica de Materiales - Ferdinand P. Beer 7ma. Edición.

19. TABLE I: LIST OF TOPICS COVERED IN MECHANICS OF MATERIALS,
Seventh Edition (CONTINUED)
Suggested Number of Periods
Core Additional Advanced
Sections Topics Topics Topics Topics
Chapter 7: Transformation of Stress and Strain
7.1 Transformation of Plane Stress 1-2
7.2 Mohr’s Circle for Plane Stress 1-2
7.3-4 Three-Dimensional Analysis of Stress 0.5-1
*7.5 Yield and Fracture Criteria 0.5-1
7.6 Thin-Walled Pressure Vessels 0.5-1
*7.7-8 Analysis of Strain; Mohr’s Circle 1-1.5
*7.9 Strain Rosette 0.5
Chapter 8: Principal Stresses under a Given Loading
8.1 Principal Stresses in a Beam 0.5-1
8.2 Design of Transmission Shafts 0.5-1
8.3 Stresses under Combined Loadings 1-3
Chapter 9: Deflection of Beams
9.1-1A Equation of Elastic Curve 0.5-1
*9.1B Direct Determination of Elastic Curve from 0.5
Load Distribution
9.2 Statically Indeterminate Beams 0.5 - 1
*9.3 Use of Singularity Functions 1-2
9.4 Method of Superposition 1-2
Application of Moment-Area Theorems to:
*9.5 Cantilever Beams and Beams with 1-2
Symmetric Loadings
*9.6A-6B Beams with Unsymmetric Loadings; Maximum 1-1.5
Deflection
*9.6C Statically Indeterminate Beams 0.5
Chapter 10: Columns
10.1 Euler’s Column Formula 1-2
*10.2 Eccentric Loading; Secant Formula 1
10.3 Design of Columns under a Centric Load 1-2
10.4 Design of Columns under an Eccentric Load 1-2
Chapter 11: Energy Methods
11.1-2 Strain Energy 1-2
11.3 Strain Energy for General State of Stress 0.5
11.4 Impact Loading 0.5-1
11.5 Deflections by Work-Energy Method 0.5-1
*11.6-8 Castigliano’s Theorem 1-2
*11.9 Statically Indeterminate Structures 1-2
_______ _______ _______
Total Number of Periods 24-41½ 21-38½ 3-6
Mecánica de Materiales - Ferdinand P. Beer 7ma. Edición.

20. TABLE II: CLASSIFICATION AND DESCRIPTION OF PROBLEMS
Mechanics of Materials Seventh Edition
Problem Number*
SI Units U.S. Units Problem description
CHAPTER 1: INTRODUCTION–CONCEPT OF STRESS
Normal stress under axial loading:
1.1,2 1.3,4 in bars
1.5,6
1.7,9 1.8,10 in pin-connected structures
1.13,14 1.11,12 in trusses and mechanisms
1.15,18 1.16,17 Shearing stress
1.20,21 1.19,22 Bearing stress between flat surfaces
1.24,25 1.23,28 Shearing and bearing stresses at pin-connected joints
1.26,27
1.29,30 1.31,32 Stresses on an oblique plane
1.35,36 1.33,34
Factor of safety:
1.38,39 1.37,42 in tension
1.40,41
1.43,44 1.45,46 in shear
1.47,48 1.49,50
1.53,54 1.51,52 in structures involving links and pins
1.55,56
*1.57 *1.58 Load and Resistance Factor Design
1.59,62 1.60,61 Review problems
1.63,66 1.64,65
1.67,68 1.69,70
1.C2,C4,C6 1.C1,C3, C5 Computer problems
CHAPTER 2: STRESS AND STRAIN - AXIAL LOADING
Stresses and deformations in statically determinate structures:
2.1,3 2.2,5 simple rods and wires
2.4,6 2.7,8
2.9,10 2.11,13 multiple-criteria problems
2.12,14
2.16,18 2.15,17 composite rods and members
2.19,20
2.21,24 2.22,23 members of trusses and simple frames
2.25,26 2.27,28
2.29 2.30 computation of deformations by integration
2.31 2.32 true strain
* Problems that do not involve any specific system of units have been indicated by underlining their number.
Answers are not given to problems with a number set in italic type.
Mecánica de Materiales - Ferdinand P. Beer 7ma. Edición.

21. TABLE II: CLASSIFICATION AND DESCRIPTION OF PROBLEMS
Mechanics of Materials Seventh Edition (CONTINUED)
Problem Number*
SI Units U.S. Units Problem description
Statically indeterminate structures (constant temperature):
2.33,34 2.35,36 with members undergoing equal deformations
2.37,38
2.41,42 2.39,40 composite rods with both ends restrained
2.43,45 2.44,46 with members undergoing unequal deformations
Statically indeterminate structures (with temperature changes):
2.47,48 2.49,50 with members undergoing equal deformations
2.51,54 2.52,53 composite rods with both ends restrained
2.55,56 with unequal deformations
2.57
2.60 2.58,59 rods with gaps
Poisson’s ratio and generalized Hooke’s law:
2.62,63 2.61,66 uniaxial loading
2.64,65
2.67,68 2.69,70 biaxial loading
2.73,74 2.71,72 derivation of formulas
2.77,78 2.75,76 Hooke’s law for shearing stress and strain
2.81,82 2.79,80
*2.84,*86 *2.83,*85 dilatation
*2.87,*88 problems involving cylindrical coordinates
*2.89 *2.90 Theory problems
*2.91,*92 Problems involving composites
2.93,94 2.95,96 Stress concentrations in flat bars
2.97,98 2.99,100
Plastic deformations under axial loading:
2.103,104 2.101,102 simple problems
2.105,106
2.107,108 2.111,112 more complex problems
2.109,110
2.113,114
*2.115
2.117,*118 2.116 problems involving temperature change
*2.122,*123 *2.119,*120 problems involving residual stresses
*2.121
2.124,127 2.125,126 Review problems
2.129,131 2.128,130
2.133,134 2.132,135
2.C1,C3,C6 2.C2,C4,C5 Computer problems
* Problems that do not involve any specific system of units have been indicated by underlining their number.
Answers are not given to problems with a number set in italic type.
Mecánica de Materiales - Ferdinand P. Beer 7ma. Edición.

22. TABLE II: CLASSIFICATION AND DESCRIPTION OF PROBLEMS
Mechanics of Materials Seventh Edition (CONTINUED)
Problem Number*
SI Units U.S. Units Problem description
CHAPTER 3: TORSION
Shearing stresses:
3.1,2 3.4,5 in simple shafts
3.3,6 3.7,8
3.11,12 3.9,10 in shafts subjected to several torques
3.13,14
3.17,18 3.15,16 in composite shafts
3.19,20
3.21,22 3.23,24 in gear-connected shafts
3.27,28 3.25,26
3.29 3.30 special problems
Angle of twist:
3.31,34 3.32,33 in simple shafts
3.35,36 in shafts subjected to several torques
3.37,38 3.39,40 in composite shafts
3.42,43 3.41,44 in gear-connected shafts
3.45,48 3.46,47 Design of shafts based on allowable stress and allowable angle of twist
3.49,50
Statically indeterminate shafts:
3.55,56 3.51,52 with inner core and outer shell of different materials
3.57,58 3.53,54
3.59
3.60,62 3.61,63 Special problems
Design of shafts:
3.65,67 3.64,66 easy problems with solid shafts
3.68,69 3.70,72 hollow shafts
3.71,73
3.74,75 3.76,77 gear-connected shafts
3.80,81 3.78,79 multiple-criteria problems
3.82,83
3.84,87 3.85,86 Stress concentrations in stepped shafts
3.88,90 3.89,91
* Problems that do not involve any specific system of units have been indicated by underlining their number.
Answers are not given to problems with a number set in italic type.
Mecánica de Materiales - Ferdinand P. Beer 7ma. Edición.

23. TABLE II: CLASSIFICATION AND DESCRIPTION OF PROBLEMS
Mechanics of Materials Seventh Edition (CONTINUED)
Problem Number*
SI Units U.S. Units Problem description
Plastic deformations of shafts:
Shafts made of elastoplastic material:
3.92,95 3.93,94 stresses in solid shafts
3.96,97
3.98,100 3.99,101 angle of twist for solid shafts
3.102,104 3.103,105
3.106,107 3.108,109 hollow and tapered shafts
3.112,113 3.110,111 Shafts made of a material with a nonlinear stress-strain diagram
3.116,117 3.114,115 Residual stresses and permanent angle of twist in shafts made of an
3.118,119 3.120 elastoplastic material
Bars with rectangular cross section:
3.123,124 3.121,122 easy problems
3.127,128 3.125,126
3.129,130 3.131,132 comparing circular and rectangular shafts
3.134 3.133
3.135,137 3.136,138 application to structural shapes
Thin-walled hollow shafts:
3.140,141 3.139,145 determine shearing stresses
3.142,143 3.146
3.144
3.147,149 3.148,150 Special problems
3.152,153 3.151,154 Review problems
3.156,157 3.155,159
3.158,161 3.160,162
3.C1,C2,C3 3.C4,C5,C6 Computer problems
* Problems that do not involve any specific system of units have been indicated by underlining their number.
Answers are not given to problems with a number set in italic type.
Mecánica de Materiales - Ferdinand P. Beer 7ma. Edición.

24. TABLE II: CLASSIFICATION AND DESCRIPTION OF PROBLEMS
Mechanics of Materials Seventh Edition (CONTINUED)
Problem Number*
SI Units U.S. Units Problem description
CHAPTER 4: PURE BENDING
Normal stresses:
4.1,3 4.2,5 in beams with horizontal plane of symmetry
4.4,6 4.7,8
4.11 4.9,10 in unsymmetrical beams (first locate centroid)
4.12,13 Resultant force on portion of cross section
4.14
4.16,17 4.15,18 Beams with different allowable stresses in tension and compression
4.19,20
4.21,22 4.23 Maximum stress and radius of curvature
4.24
4.25,29 4.26,27 Maximization of beam strength
4.28
4.31 4.30 Anticlastic curvature
4.32 Special problem on theory
Stresses in composite beams:
4.33,34 4.37,38 symmetric beams of two materials
4.35,36
4.39,40 4.41,42 unsymmetric beams of two materials
4.43,44 4.45,46 Radius of curvature in composite beams
4.49,50 4.47,48 Stresses in reinforced concrete beams
4.53,54 4.51,52
4.55,56 Beams of three materials
4.58 4.57 Composite beams with circular cross section
4.59 *4.60 Beams with different moduli of elasticity in tension and compression
4.61,62 4.63,64 Stress concentrations in flat bars in pure bending
4.65,66
Plastic deformation in pure bending (elastoplastic material):
4.67,68 4.69,70 plastic zone in rectangular beams
4.71,72
4.73,74 4.75,76 plastic zone in symmetric beams
4.77,78 4.79,80 plastic moment and shape factor of symmetric beams
4.81,82 4.84,86 plastic moment of unsymmetric beams
4.83,85
Residual stresses in symmetric beams:
4.87,88 4.89,90 after Mp has been applied and removed
4.91 4.92 after a given plastic zone has been developed
Special problems:
4.94,95 4.93 residual radius of curvature
4.96 4.97,98 Plastic deformation of beams with a nonlinear stress-strain diagram
* Problems that do not involve any specific system of units have been indicated by underlining their number.
Answers are not given to problems with a number set in italic type.
Mecánica de Materiales - Ferdinand P. Beer 7ma. Edición.

25. TABLE II: CLASSIFICATION AND DESCRIPTION OF PROBLEMS
Mechanics of Materials Seventh Edition (CONTINUED)
Problem Number*
SI Units U.S. Units Problem description
Eccentric loading in plane of symmetry of member:
4.99,102 4.100,101 find stress in symmetric section
4.103,104
4.106,107 4.105,109 design of symmetric section
4.108,110 4.111,112
4.115,116 4.113,114 find stress in unsymmetric section
4.117,120 4.118,119
4.121,122 4.124,125 computation of loads from strain measurements
4.123,126
Unsymmetric bending with one or two planes of symmetry:
4.127,129 4.128,130 bending moment at an angle with horizontal
4.133,134 4.131,132
4.136,137 4.135,138 section at an angle with horizontal
4.139,140
*4.141 *4.142,*143 Bending of unsymmetric section (principal axes must be determined)
General eccentric bending:
4.144,145 4.146,147 symmetric beam; find stresses
4.148,149
4.152,153 4.150,151 Bending of unsymmetric beams; determine largest bending moment
4.155 4.154,156
4.157,158 4.159,160 Special problems and problems on theory
Curved beams:
4.161,162 4.163,164 with rectangular cross section
4.165,166
4.169,170 4.167,168 under eccentric loading
4.173,174 4.171,172 with unsymmetric cross section
4.177,178 4.175,176 with circular cross section
4.179
4.180 4.181,182 with triangular cross section
4.185,186 4.183,184 with trapezoidal cross section
*4.187,188 4.189,*191 special problems and derivations of formulas
4.190
* Problems that do not involve any specific system of units have been indicated by underlining their number.
Answers are not given to problems with a number set in italic type.
Mecánica de Materiales - Ferdinand P. Beer 7ma. Edición.

26. TABLE II: CLASSIFICATION AND DESCRIPTION OF PROBLEMS
Mechanics of Materials Seventh Edition (CONTINUED)
Problem Number*
SI Units U.S. Units Problem description
4.192,195 4.193,194 Review problems
4.196,197 4.198,200
4.199,201 4.202,203
4.C1,C3 4.C2,C4 Computer problems
4.C5,C7 4.C6
CHAPTER 5: ANALYSIS AND DESIGN OF BEAMS FOR BENDING
Using the free-body diagram of a portion of a beam:
5.1,2 5.3,4 draw V and M diagrams (easy problems)
5.5,6
5.7,9 5.8,10 draw V and M diagrams and determine maximum values of |V| and |M|
5.11,13 5.12,14
5.15,17 5.16,20 find maximum normal stress in given beam section
5.18,19
5.22,23 5.21,25 draw V and M diagrams and find maximum normal stress in beam
5.24,26
5.27,29 5.28,31 determine given parameter to minimize normal stress in beam
5.30,33 5.32
Using relations among w, V and M whenever appropriate:
5.34,35 5.36,37 draw V and M diagrams (easy problems)
5.38,39
5.41,42 5.40,43 draw V and M diagrams and determine maximum values of |V| and |M|
5.44,45
5.46,48 5.47,49 find maximum normal stress in a given beam section
5.52,53 5.50,51 write equations for V and M and find maximum value of |M|
5.55,56 5.54,57 draw V and M diagrams and find maximum normal stress in beam
5.59,60 5.58,61
*5.62,*64 *5.63 Special problems
* Problems that do not involve any specific system of units have been indicated by underlining their number.
Answers are not given to problems with a number set in italic type.
Mecánica de Materiales - Ferdinand P. Beer 7ma. Edición.

27. TABLE II: CLASSIFICATION AND DESCRIPTION OF PROBLEMS
Mechanics of Materials Seventh Edition (CONTINUED)
Problem Number*
SI Units U.S. Units Problem description
5.65,66 5.67,68 Design of timber beams
5.69,70
5.73,74 5.71,72 Design of steel beams, W shapes
5.77,78 5.75,76 Design of steel beams, S shapes
5.79,80 5.81,82 Design of steel beams, miscellaneous shapes
5.83 5.84 Design of beams resting on ground
5.85,86 5.87,88 Find allowable load for beam of unsymmetric cross section with
5.89,90 allowable stresses in tension and compression
5.93 5.91,92 Design of steel beams, W shapes
*5.94,*95 *5.96,*97 Design of beams using LRFD
Using singularity functions write equations for V(x) and M(x) and
5.98,100 5.99,101 find M at given point in beam
5.102,105 5.103,104
5.106,108 5.107,109 find |M |max in beam
5.110,111 find σmax in beam
5.112,113 find |M |max and σmax in beam
5.116,117 5.114,115 design beam, knowing allowable stress
Using a computer and step functions,
5.118,119 5.120,121 calculate V and M along the beam
5.122,123 5.124,125 calculate V and M along the beam , and determine σmax in the beam
Nonuniform beams
Beams of constant strength:
5.128,129 5.126,127 beams of uniform width and variable depth
5.130,131
5.134,135 5.132,133 built-up timber beams
5.136 5.137 beams of circular cross section
5.138 5.139 beams of uniform depth and variable width
5.140,143 5.141,142 rolled-steel beams with cover plates
5.144,145 5.146,147
5.148,149 5.150,151 tapered beams
5.152,153 5.154,155 Review problems
5.156,157 5.158,159
5.161,162 5.160,163
5.C1,C2 5.C3,C4 Computer problems
5.C6 5.C5
* Problems that do not involve any specific system of units have been indicated by underlining their number.
Answers are not given to problems with a number set in italic type.
Mecánica de Materiales - Ferdinand P. Beer 7ma. Edición.

28. TABLE II: CLASSIFICATION AND DESCRIPTION OF PROBLEMS
Mechanics of Materials Seventh Edition (CONTINUED)
Problem Number*
SI Units U.S. Units Problem description
CHAPTER 6: SHEARING STRESSES IN BEAMS AND THIN-WALLED MEMBERS
6.1,2 6.3,6 Shearing forces in nails and bolts, using horizontal cuts
6.4,5 6.7,8
6.9,10 6.11,12 Shearing stresses in beams
6.13,14
6.16,18 6.15,17 Designs of beams for normal and shearing stresses
6.19,20
6.22,24 6.21,23 Beams with singly-symmetric sections
6.26,28 6.25,27 Beams with various geometric sections
6.30,32 6.29,31 Shearing forces and shearing stresses on arbitrary cuts due to vertical shear
6.33,34
6.35,36 6.38,39 Shearing stresses in extruded beams
6.37,42 6.40,41
6.44,46 6.43,45 Shearing stresses in bolts
6.48,49 6.47,50 Shearing stresses and shear flow in thin-walled members
6.53,54 6.51,52 Special problems
6.55,56 6.58,59 Shearing stresses in composite beams
6.57
6.60 Plastic behavior
6.61,62 6.63,64 Shear center in thin-walled beams with horizontal and vertical portions
6.65,68 6.66,67 Shear center and shearing stresses in extruded beams
6.70,72 6.69,71 Shear center in thin-walled beams with oblique portions
6.73,74 Shear center in thin-walled beams with circular portions
6.76,77 6.75,78 Problems involving location of shear center
6.79,80 Special problems
*6.81,*82 *6.85,*86 Shearing stresses in semicircular shapes, angle shapes and Z shapes
*6.83,*84 *6.87,*88
6.89,90 6.91,92 Review problems
6.94,95 6.93,96
6.97,99 6.98,100
6.C1 6.C2 Computer problems
6.C3,C4 6.C5,C6
* Problems that do not involve any specific system of units have been indicated by underlining their number.
Answers are not given to problems with a number set in italic type.
Mecánica de Materiales - Ferdinand P. Beer 7ma. Edición.

29. TABLE II: CLASSIFICATION AND DESCRIPTION OF PROBLEMS
Mechanics of Materials Seventh Edition (CONTINUED)
Problem Number*
SI Units U.S. Units Problem description
CHAPTER 7: TRANSFORMATION OF STRESS AND STRAIN
7.2,4 7.1,3 Find stresses on oblique plane from equilibrium of wedge
7.5,7 7.6,8 Find principal planes and stresses
7.9,11 7.10,12 Find planes of maximum shearing stress and corresponding stresses
7.14,16 7.13,15 Find stresses on a given plane
7.18,19 7.17,21 Stresses on oblique planes - simple applications
7.20,22
7.23,26 7.24,25 Find principal stresses and/or maximum shearing stress in loaded shaft
7.27,29 7.28,30 Special problems involving determination of a stress to satisfy a given
requirement
Using Mohr’s circle, determine:
7.31,33 7.32,34 principal planes and stresses, and maximum shearing stress
7.36,38 7.35,37 stresses on oblique plane
7.40,41 7.39,43 stresses on oblique plane - simple applications
7.42,44
7.45,48 7.46,47 principal stresses and/or maximum shearing stress in loaded shaft
7.49,51 7.50,52
7.53
7.55,56 7.54,57 find principal planes and stresses resulting from superposition of two
states of stress
7.58,59 7.60,63 find range of values of a parameter for which a certain stress will not be
7.61,62 exceeded
7.64 7.65 Derivation of a formula involving Mohr’s circle
7.68,69 7.66,67 In-plane and out-of-plane maximum shearing stress
7.70,71 7.72,73 Maximum shearing stress in a three dimensional state of stress
7.75,76 7.74,77 Maximum shearing stress (more advanced problems)
7.78,79
*7.80
7.81,82 7.83,84 Determine if material will yield under given state of stress. If not, find
7.85,86 7.87,88 the factor of safety
7.89,90 7.91,92 Will rupture occur under a given state of stress?
7.94,95 7.93,97 Find stress or load for which rupture will occur
7.96
* Problems that do not involve any specific system of units have been indicated by underlining their number.
Answers are not given to problems with a number set in italic type.
Mecánica de Materiales - Ferdinand P. Beer 7ma. Edición.

30. TABLE II: CLASSIFICATION AND DESCRIPTION OF PROBLEMS
Mechanics of Materials Seventh Edition (CONTINUED)
Problem Number*
SI Units U.S. Units Problem description
7.98,99 7.100,102 Spherical pressure vessels (easy problems)
7.101,103
7.104,105 7.107,109 Cylindrical pressure vessels (easy problems)
7.106,108 7.110,111
7.112,113 7.116,117 Stresses in welds in cylindrical pressure vessels
7.114,115 7.118,119
7.122,123 7.120,121 Pressure vessels subjected to external loadings
7.124,125
7.126,127 Shrunk fit rings
Find state of strain associated with given rotation:
7.129,131 7.128,130 using formulas of Sec. 7.10
7.133,135 7.132,134 using Mohr’s circle
Find principal strains and maximum shearing strain (in plane and out of plane):
7.138,139 7.136,137 for a state of plane stress
7.142,143 7.140,141 for a state of plane strain
7.144,145 7.146,149 Problems involving strain rosettes
7.147,148
7.154,155 7.150,151 Problems involving use of Mohr’s circle and Hooke’s law
7.156,157 7.152,153
7.159,160 7.158,162 Review problems
7.161,163 7.164,167
7.165,166 7.168,169
7.C1,C2 7.C3,C4 Computer problems
7.C5,C6 7.C7,C8
CHAPTER 8: PRINCIPAL STRESSES UNDER GIVEN LOADING CONDITIONS
Principal stresses in rolled-steel beams:
8.3,4 8.1,2 find σm in beam and σmax at junction of flanges and web
8.5,6 8.7,8 design beam, taking σm, τm, and σmax into account
8.9,10 8.11,12 check earlier design for σmax at junction of flanges and web
8.13,14
Design of transmission shafts:
8.15,16 8.19,20 loading represented by forces and couples
8.17,18 8.21,22
8.23,24 8.25,26 loading represented by input and output power
8.27,28 8.29,30
* Problems that do not involve any specific system of units have been indicated by underlining their number.
Answers are not given to problems with a number set in italic type.
Mecánica de Materiales - Ferdinand P. Beer 7ma. Edición.

31. TABLE II: CLASSIFICATION AND DESCRIPTION OF PROBLEMS
Mechanics of Materials Seventh Edition (CONTINUED)
Problem Number*
SI Units U.S. Units Problem description
For beam of rectangular cross section under axial, bending and transverse loading
(easy problems), find:
8.33,34 8.31,32 normal and shearing stresses
8.35,36 For beams of circular cross section under axial, bending, and transverse
loading, find:
8.38,40 8.37,39 normal and shearing stresses
8.42,43 8.41,44 principal stresses and maximum shearing stress
For beams of rectangular cross section under axial, bending and transverse
loading, find:
8.47,48 8.45,46 normal and shearing stresses
8.51,52 8.49,50 principal stresses and maximum shearing stress
For steel beams and structural tubes under axial, bending and transverse
loading, find:
8.53,54 normal and shearing stresses
8.55,56 8.57,58 principal stresses, principal planes, and maximum shearing stress
8.60 8.59 Special problems
*8.61 *8.62,*63 Problems involving torsion of rectangular sections or structural shapes
*8.64
8.66,67 8.65,69 Review problems
8.68,70 8.71,72
8.74,76 8.73,75
8.C1,C2 8.C4,C5 Computer problems
8.C3,C6 8.C7
* Problems that do not involve any specific system of units have been indicated by underlining their number.
Answers are not given to problems with a number set in italic type.
Mecánica de Materiales - Ferdinand P. Beer 7ma. Edición.

32. TABLE II: CLASSIFICATION AND DESCRIPTION OF PROBLEMS
Mechanics of Materials Seventh Edition (CONTINUED)
Problem Number*
SI Units U.S. Units Problem description
CHAPTER 9: DEFLECTION OF BEAMS
Using the integration method, determine the equation of the elastic curve
and the deflection and/or slope at specific points for:
9.3,4 9.1,2 cantilever beams
9.5 9.6
9.7 9.8 overhanging beams
simply supported beams
9.10 9.9 symmetrical loading
9.12 9.11 unsymmetrical loading
9.14,15 9.13 beams and loadings requiring the use of 2 equations and 4 constants
9.16 of integration
9.17 9.18 direct determination of the elastic curve from an analytic function of w(x)
For a statically indeterminate beam (first degree), determine:
9.19,21 9.20,23 reaction at the roller support
9.22,24
9.25,28 9.26,27 reaction at the roller support and draw the M diagram (use of 2 equations
and 4 constants of integration required)
9.30,32 9.29,31 reaction at the roller support and the deflection at a given point (use of 2 equations
and 4 constants of integration required)
9.33 9.34 For a statically indeterminate beam (second degree), determine the reaction at one end
and draw the M diagram
Using singularity functions, determine the equation of the elastic curve and
the deflection and/or slope at specified points:
9.35,36 9.37,38 easy problems
9.39,40 problems involving overhanging beams
9.41,44 9.42,43 problems with distributed loads
9.45,48 9.46,47 problems with numerical data
9.49,51 9.50,52 For a statically indeterminate beam (first degree), determine the reaction at
9.53,54 9.55,56 the roller support and the deflection at a specified point.
9.57 9.58 For a statically indeterminate beam (second degree), determine the reaction
at the roller support and the deflection at a specified point.
9.59,62 9.60,61 determine the maximum deflection
9.63,64 Problems involving the reduction of a load to a force-couple system.
* Problems that do not involve any specific system of units have been indicated by underlining their number.
Answers are not given to problems with a number set in italic type.
Mecánica de Materiales - Ferdinand P. Beer 7ma. Edición.

33. TABLE II: CLASSIFICATION AND DESCRIPTION OF PROBLEMS
Mechanics of Materials Seventh Edition (CONTINUED)
Problem Number*
SI Units U.S. Units Problem description
Using method of superposition, determine the deflection and slope at
specified points in:
9.65,67 9.66,68 cantilever beams
9.71,72 9.69,70 simply supported beams
9.73,74 9.75,76 cantilever beams (with numerical data)
9.77,78 simply supported beams (with numerical data)
9.79,80 9.82 statically indeterminate beams (first degree)
9.81
9.83 9.84 statically indeterminate beams (second degree)
9.85,88 9.86,87 combined beams, determinate (with numerical data)
9.90,91 9.89,92 statically indeterminate beams (with numerical data)
9.94 9.93 Combined bending and torsion of rods
Using the moment-area method, determine the slope and/or deflection
at specified points in
cantilever beams
9.95,96 9.97,98 with simple loadings
9.99,100
9.102,104 9.101,103 with numerical data
9.105,107 9.106,108 with variable EI
simply supported beams
9.109,110 9.111,112 with symmetric loadings
9.114 9.113
9.116 9.115 with variable EI
9.118,119 9.117,121 with numerical data
9.120,122
*9.123 *9.124 special problems
Simply supported and overhanging beams with unsymmetric loadings
9.125,126 9.127,128 simply supported beams (easy problems)
9.129,130 9.131,132 simply supported beams with numerical data
9.133,134 overhanging beams
9.136,138 9.135,137 overhanging beams (with numerical data)
9.139 9.140 simply supported beams with variable EI
Find maximum deflection for:
9.141,142 simply supported beams
9.143 9.144 with numerical data
9.145 9.146 overhanging beams
Statically indeterminate beams (first degree)
9.147,150 9.148,149 single span
9.152 9.151 two span beams, find all reactions
9.153 9.154 single span beams (with numerical data)
9.155 9.156 simply supported beams with additional elastic support at midspan
* Problems that do not involve any specific system of units have been indicated by underlining their number.
Answers are not given to problems with a number set in italic type.
Mecánica de Materiales - Ferdinand P. Beer 7ma. Edición.

34. TABLE II: CLASSIFICATION AND DESCRIPTION OF PROBLEMS
Mechanics of Materials Seventh Edition (CONTINUED)
Problem Number*
SI Units U.S. Units Problem description
9.158,161 9.157,159 Review problems
9.164,165 9.160,162
9.167,168 9.163,166
9.C1,C3 9.C2,C5 Computer problems
9.C4,C8 9.C6,C7
CHAPTER 10: COLUMNS
Stability of rigid-rod-and-spring systems:
10.1,2 10.4,5 single spring
10.3
10.7 10.6,8 systems with two or more springs
Application of Euler’s formula to the critical loading or pin-ended columns:
10.9 10.10 short struts
10.12,13 10.11 comparison of critical loads for various cross sections
10.14
Allowable loading for pin-ended columns:
10.16,18 10.15,17 rolled-steel shapes
10.19,20 multiple-member structures
10.24,25 10.21,22 columns with various end conditions
10.27,28 10.23,26
Application of the secant formula to the eccentric loading of columns:
10.29,30 10.32,34 find σmax and either deflection or e for a given load
10.31,33
10.36,37 10.35,38 find σmax and load for a given deflection an eccentricity
10.39,40
10.41,42 temperature induced loading
10.43,44 10.45,46 find Pall for given e, σmax and F.S. (using Fig. 10.24)
Design problems
10.47,50 10.48,49 find column length
10.51,52 10.53,54 find cross section of column
10.55,56 find factor of safety of column
* Problems that do not involve any specific system of units have been indicated by underlining their number.
Answers are not given to problems with a number set in italic type.
Mecánica de Materiales - Ferdinand P. Beer 7ma. Edición.

35. TABLE II: CLASSIFICATION AND DESCRIPTION OF PROBLEMS
Mechanics of Materials Seventh Edition (CONTINUED)
Problem Number*
SI Units U.S. Units Problem description
Analysis of columns under centric load:
columns with simple cross section:
10.57,59 10.58 steel columns
10.60 10.61 wood columns
10.62 10.63,64 aluminum columns
columns with built-up cross sections:
10.66,67 10.65,68 steel columns
10.69 wood columns
10.70 aluminum column
Design of columns under a centric load:
10.71,73 10.72 wood columns
10.75,76 10.74 aluminum columns
10.77,80 10.78,79 steel columns
10.83,84 10.81,82
Application of LRFD formulas:
*10.86 *10.85 analysis of columns under a centric loading
*10.88 *10.87 design of columns under a centric loading
Analysis of columns under an eccentric load:
10.89,90 steel columns, find allowable load
10.91,92 wood columns, find allowable load
10.93,94 aluminum columns, find allowable load
Design of columns under an eccentric load:
find maximum allowable length or allowable eccentricity:
10.95,96 steel columns
10.97,98
10.99,100 wood columns
10.101,102 aluminum columns
design cross section:
10.103,104 steel columns, rectangular cross section
10.105,106 steel columns, tube
10.107,108 wood columns, rectangular or circular cross sections
10.109,110 aluminum columns
10.111,112
10.115,116 10.113,114 steel column, wide-flange shape
10.118,119 10.117,120 Review problems
10.121,124 10.122,123
10.125,126 10.127,128
10.C1,C3,C6 10.C2,C4,C5 Computer problems
* Problems that do not involve any specific system of units have been indicated by underlining their number.
Answers are not given to problems with a number set in italic type.
Mecánica de Materiales - Ferdinand P. Beer 7ma. Edición.

36. TABLE II: CLASSIFICATION AND DESCRIPTION OF PROBLEMS
Mechanics of Materials Seventh Edition (CONTINUED)
Problem Number*
SI Units U.S. Units Problem description
CHAPTER 11: ENERGY METHODS
11.2,3 11.1,4 Modulus of resilience
Modulus of resilience and modulus of toughness:
11.6 11.5 from stress-strain diagram
11.7 11.8 from load-deflection diagram of a tensile test
Strain energy under axial loads:
11.10 11.9,11 strain energy under a given load
11.12,13 maximum allowable strain energy
11.14,15 factor of safety
11.16 11.17 strain energy by integration of approximate methods
11.19,20 11.18 strain energy of trusses
11.21
11.22 11.23 with numerical data
11.24,27 11.25,26 Strain energy in bending:
11.30,31 11.28,29 with numerical data
11.32 derivation of formulas
11.33 11.34 Strain energy in torsion:
11.35 by integration
11.36,37 11.38,39 Maximum-distortion-energy criterion for 3-dimensional state of stress
11.40 11.*41 Special problems
Impact loading:
11.42 11.43,44 of rods (horizontal impact)
11.45,46 of rods (vertical impact)
11.47
11.50,51 11.48,49 of beams (horizontal impact)
11.52,53 11.54,55 of beams (vertical impact)
11.56,57 problems on theory
Use of work-energy method to determine deflection or slope of:
11.58,61 11.59,60 prismatic beams
11.62,65 11.63,64 nonprismatic beams
11.66,69 11.67, 68 angle of twist of shafts
11.70 angle of twist of a thin-walled hollow shaft
11.71,72 11.73,75 deflection of a joint of a truss
11.74,76
* Problems that do not involve any specific system of units have been indicated by underlining their number.
Answers are not given to problems with a number set in italic type.
Mecánica de Materiales - Ferdinand P. Beer 7ma. Edición.

37. TABLE II: CLASSIFICATION AND DESCRIPTION OF PROBLEMS
Mechanics of Materials Seventh Edition (CONTINUED)
Problem Number*
SI Units U.S. Units Problem description
11.77,78 11.80,82 Work of several loads applied to a beam
11.79,81
Determinate structures. Use Castigliano’s theorem to determine:
11.83,86 11.84,85 deflection and/or slope of beams
11.88,89 11.87,90
11.93,94 11.91,92 deflection and/of slope of beams (with numerical data)
11.95,96 11.97,98
11.99,100 11.101,102 deflection of a given joint in a truss
11.103,104
11.105,106 11.107,109 deflection and/or slope at a given point in a bent or curved rod
11.108,110
Indeterminate structures (first degree). Use Castigliano’s theorem to determine:
11.113,114 11.111,112 reaction at roller support and draw bending-moment diagram
11.115,116
11.117,118 force in member(s) of a truss
11.119,121 11.120,122
11.123,126 11.124,125 Review problems
11.127,129 11.128,131
11.130,133 11.132,134
11.C3,C4,C5 11.C1,C2,C6 Computer problems
* Problems that do not involve any specific system of units have been indicated by underlining their number.
Answers are not given to problems with a number set in italic type.
Mecánica de Materiales - Ferdinand P. Beer 7ma. Edición.

38. TABLE
III:
SAMPLE
ASSIGNMENT
SCHEDULE
FOR
A
COURSE
IN
MECHANICS
OF
MATERIALS
Mechanics
of
Materials
7th
Edition
The
six
lists
of
suggested
assignments
contained
in
this
table
cover
the
entire
text.
The
instructor
can
select
those
assignments
that
are
covered
in
the
course.
50%
OF
THE
PROBLEMS
IN
EACH
OF
THESE
LISTS
USE
SI
UNITS
AND
50%
U.S.
CUSTOMARY
UNITS
|
<
-----For
these
lists
answers
to
all
of
the
problems
are
given
in
the
back
of
the
book------------>|
|<----
For
Lists
5
and
6
no
answers
are
given->|
Group
Sections
Topics
List
1
List
2
List
3
List
4
List
5
List
6
1
1.1
Introduction
2.
1.2
Stresses
in
the
Members
of
a
Structure
1.3,7,17,25
1.4,13,16,24
1.1,10,15,23
1.2,8,18,28
1.5,12,19,27
1.6,11,22,26
3
1.3-5
Components
of
Stress;
Factor
of
Safety
1.29,37,48,52
1.30,42,47,50
1.31,39,46,56
1.32,41,45,55
1.34,38,44,51
1.33,39,43,49
4
2.1
Stress-Strain
Diagram
2.1,13,19,27
2.4,11,20,23
2.2,14,15,24
2.7,9,17,21
2.5,12,16,28
2.8,10,18,22
5
2.2-3
Statically
Indeterminate
Problems
2.35,42,52,55
2.36,41,50,56
2.33,44,48,58
2.34,39,47,59
2.38,40,53,57
2.37,46,49,60
6
2.4-9
Generalized
Hooke’s
Law;
Poisson’s
Ratio
2.66,68,79,84
2.61,67,80,88
2.63,70,77,85
2.64,69,78,83
2.62,72,76,87
2.65,71,75,86
7
2.10-13
Stress
Concentration;
Plastic
Behavior
2.93,102,108,116
2.94,101,107,121
2.99,105,112,117
2.100,106,111,118
2.96,104,110,119
2.95,103,109,120
8
3.1
Stresses
in
Torsion
3.1,9,18,25
3.2,10,20,23
3.4,11,15,22
3.7,13,16,21
3.5,12,17,26
3.8,14,19,24
9
3.2-3
Angle
of
Twist,
Indeterminate
Shafts
3.32,38,46,57
3.33,41,47,56
3.31,40,45,53
3.35,44,48,54
3.34,39,51,58
3.36,43,52,55
10
3.4-5
Transmission
Shafts:
Stress
Concentrations
3.67,77,81,89
3.65,76,80,86
3.66,73,79,90
3.64,71,78,84
3.72,75,83,91
3.70,74,82,85
11
3.6-8
Plastic
Deformations;
Residual
Stresses
3.94,100,111,119
3.93,98,110,118
3.95,101,113,115
3.92,99,112,114
3.97,105,109,117
3.96,103,108,116
12
3.9-10
Noncircular
and
Thin-Walled
members
3.124,133,135,139
3.123,129,137,146
3.122,132,136,142
3.121,131,138,143
3.126,134,141,145
3.125,130,140,148
13
4.1-3
Stresses
and
Deformations
in
the
Elastic
Range
4.3,10,19,26
4.1,9,16,23
4.4,12,18,24
4.2,11,15,21
4.8,14,20,28
4.7,13,17,27
14
4.4-5
Members
Made
of
Composite
Materials
4.37,40,47,61
4.38,44,48,65
4.33,41,49,63
4.34,45,50,64
4.35,42,51,62
4.36,46,52,66
15
4.6
Plastic
Deformations;
Residual
Stresses
4.68,79,82,92
4.67,80,81,92
4.75,77,86,91
4.69,78,84,91
4.73,76,83,90
4.70,74,85,89
16
4.7
Eccentric
Axial
Loading
in
a
Plane
of
Symmetry
4.101,108,114,121
4.100,106,113,122
4.103,109,115,124
4.99,105,116,125
4.104,112,118,123
4.102,111,119,126
17
4.8-9
Unsymmetric
Bending
4.127,135,144,151
4.129,138,145,150
4.128,137,146,152
4.130,139,147,153
4.131,136,148,154
4.132,140,149,156
18
4.10
Bending
of
Curved
Members
4.162,169,175,185
4.161,173,181,185
4.164,167,177,183
4.163,171,179,184
4.166,168,176,186
4.165,172,182,186
19
5.1
Shear
and
Bending-Moment
Diagrams
5.7,16,26,31
5.9,20,23,32
5.8,15,25,29
5.12,18,21,33
5.14,19,24,28
5.10,17,22,28
20
5.2
Relations
Between
w,
V,
and
M
5.41,46,61,64
5.43,48,54,64
5.40,47,55,63
5.42,49,56,63
5.44,50,58,62
5.45,51,57,62
21
5.3
Design
of
Prismatic
Beams
in
Bending
5.69,71,79,92
5.65,76,80,91
5.68,73,81,89
5.70,72,82,89
5.67,78,88,93
5.66,75,87,90
22
5.4
Use
of
Singularity
Functions
to
Determine
V
and
M
5.103,108,115,118
5.104,106,114,119
5.100,109,117,120
5.98,107,110,124
5.99,111,113,125
5.101,112,116,121
23
5.5
Nonprismatic
Beams
5.128,132,143,151
5.130,139,140,150
5.126,134,147,149
5.127,135,141,149
5.131,137,142,148
5.129,133,146,148
24
6.1-2
Shearing
Stresses
in
a
Beam
6.2,12,19,23
6.1,11,18,21
6.7,13,17,24
6.3,9,15,22
6.8,14,20,27
6.6,10,16,25
25
6.3-5
Shearing
Stresses
in
Thin-Walled
Members
6.31,37,43,56
6.29,35,45,57
6.32,40,44,51
6.30,38,46,59
6.34,41,50,53
6.33,39,47,54
26
6.6
Shear
Center
6.61,69,76,87
6.62,71,77,88
6.63,70,75,81
6.64,72,78,82
6.66,73,79,85
6.67,74,80,86
27
7.1
Transformation
of
Plane
Stress
7.2,12,18,24
7.4,10,19,25
7.1,9,21,26
7.3,11,17,23
7.6,14,20,28
7.8,16,22,30
28
7.2
Mohr’s
Circle
for
Plane
Stress
7.33,39,41,61
7.34,40,43,55
7.32,35,47,49
7.31,37,46,51
7.38,44,50,60
7.36,42,52,54
29
7.3-5
Three-Dimensional
Stress
Analysis;
Fracture
Criteria
7.69,77,82,91
7.68,74,81,92
7.73,75,84,89
7.66,79,83,90
7.72,76,86,93
7.67,78,85,97
30
7.6
Stresses
in
Thin-Walled
Pressure
Vessels
7.100,108,117,124
7.102,106,118,122
7.98,111,112,120
7.103,109,113,121
7.99,110,116,123
7.101,107,119,125
31
7.7-9
Analysis
and
Measurement
of
Strain
7.131,140,147,150
7.133,136,148,151
7.128,143,146,157
7.132,139,149,156
7.130,138,145,152
7.134,142,144,153
32
8.1-2
Principal
Stresses
in
a
Beam;
Design
of
Shafts
8.4,12,17,20
8.3,11,15,19
8.2,13,21,24
8.1,9,22,23
8.7,10,16,25
8.8,14,18,26
33
8.3
Stresses
under
Combined
Loadings
8.31,40,49,53
8.32,38,46,55
8.35,39,48,58
8.36,37,47,57
8.34,44,50,54
8.33,41,45,56
34
9.1-2
Deflection
of
Beams
by
Integration
9.3,11,24,27
9.4,10,19,26
9.1,12,23,25
9.2,9,20,28
9.8,14,22,29
9.6,15,21,31
35
9.3
Use
of
Singularity
Functions
9.37,45,50,62
9.38,41,56,59
9.35,46,49,61
9.36,43,53,60
9.39,47,55,63
9.40,42,52,64
36
9.4
Method
of
Superposition;
Statically
Indet.
Beams
9.65,76,79,87
9.67,75,80,86
9.66,77,82,88
9.68,73,84,85
9.70,74,83,92
9.69,78,81,89
37
9.5
Moment-Area
Method
9.97,104,111,119
9.98,102,112,118
9.96,103,109,113
9.95,101,110,115
9.100,106,114,117
9.99,108,116,121
38
9.6
Moment-Area
Method;
Max
Deflection;
Indet.
Beams
9.129,137,146,148
9.125,135,142,154
9.132,133,144,153
9.128,136,145,147
9.131,134,143,149
9.127,138,141,151
39
10.1
Columns:
Euler’s
Formula
10.2,11,19,22
10.1,10,16,21
10.6,13,17,27
10.4,9,15,24
10.8,14,20,26
10.5,12,18,23
40
10.2
Eccentric
Loading
and
Secant
Formula
10.34,40,49,56
10.32,39,45,51
10.30,41,47,54
10.29,35,43,53
10.33,42,48,55
10.31,38,46,52
41
10.3
Design
of
Columns
under
a
Centric
Load
10.62,65,71,79
10.60,68,75,78
10.64,66,72,80
10.58,69,74,77
10.63,67,73,81
10.61,70,76,82
42
10.4
Design
of
Columns
under
an
Eccentric
Load
10.92,101,107,114
10.91,102,105,113
10.93,97,109,116
10.89,95,103,111
10.94,98,112,115
10.90,96,108,110
43
11.1-3
Elastic
Strain
Energy
11.6,11,20,28
11.2,9,21,25
11.5,12,18,30
11.1,10,23,24
11.8,14,19,29
11.4,13,22,26
44
11.4-5
Design
for
Impact
Loading;
Deflections
11.44,52,63,76
11.43,50,59,66
11.45,54,62,75
11.42,48,58,68
11.47,55,64,74
11.46,49,60,69
45
11.6-8
Castigliano’s
Theorem
11.80,83,91,99
11.82,88,97,103
11.77,85,93,101
11.78,90,95,102
11.79,84,92,100
11.81,87,98,104
46
11.9
Statically
Indeterminate
Structures
11.109,113,117
11.107,114,118
11.106,111,117
11.105,112,119
11.110,116,120
11.108,115,122
Mecánica de Materiales - Ferdinand P. Beer 7ma. Edición.

39. TABLE
IV:
SAMPLE
ASSIGNMENT
SCHEDULE
FOR
A
COURSE
IN
MECHANICS
OF
MATERIALS
Mechanics
of
Materials
7th
Edition
The
four
lists
of
suggested
assignments
contained
in
this
table
cover
the
entire
text.
The
instructor
can
select
those
assignments
that
are
covered
in
the
course.
75%
OF
THE
PROBLEMS
IN
EACH
OF
THESE
LISTS
USE
SI
UNITS
AND
25%
U.S.
CUSTOMARY
UNITS
ANSWERS
TO
ALL
OF
THESE
PROBLEMS
ARE
GIVEN
IN
THE
BACK
OF
THE
BOOK
Group
Sections
Topics
List
1a
List
2a
List
3a
List
4a
1
1.1
Introduction
2
1.2
Stresses
in
the
Members
of
a
Structure
1.3,7,20,25
1.4,13,21,24
1.1,9,15,23
1.2,14,18,28
3
1.3-5
Components
of
Stress;
Factor
of
Safety
1.29,37,48,54
1.30,42,47,53
1.36,40,46,56
1.35,41,45,55
4
2.1
Stress-Strain
Diagram
2.1,13,19,26
2.4,11,20,25
2.6,14,15,24
2.3,9,17,21
5
2.2-3
Statically
Indeterminate
Problems
2.35,42,54,55
2.36,41,51,56
2.33,45,48,58
2.34,43,47,59
6
2.4-9
Generalized
Hooke’s
Law;
Poisson’s
Ratio
2.66,68,81,84
2.61,67,82,88
2.63,74,77,85
2.64,73,78,83
7
2.10-13
Stress
Concentration;
Plastic
Behavior
2.93,102,108,122
2.94,101,107,123
2.98,105,112,117
2.97,106,111,118
8
3.1
Stresses
in
Torsion
3.1,9,18,28
3.2,10,20,27
3.3,11,15,22
3.6,13,16,21
9
3.2-3
Angle
of
Twist,
Indeterminate
Shafts
3.32,38,49,57
3.33,42,50,56
3.31,37,45,53
3.35,42,48,54
10
3.4-5
Transmission
Shafts:
Stress
Concentrations
3.67,77,81,88
3.65,76,80,87
3.69,73,79,90
3.68,71,78,84
11
3.6-8
Plastic
Deformations;
Residual
Stresses
3.94,100,107,119
3.93,98,106,118
3.95,104,113,115
3.92,102,112,114
12
3.9-10
Noncircular
and
Thin-Walled
members
3.124,133,135,144
3.123,129,137,147
3.128,132,136,142
3.127,131,138,143
13
4.1-3
Stresses
and
Deformations
in
the
Elastic
Range
4.3,10,19,25
4.1,9,16,22
4.6,12,18,24
4.4,11,15,21
14
4.4-5
Members
Made
of
Composite
Materials
4.37,40,53,61
4.38,44,54,65
4.33,39,49,63
4.34,43,50,64
15
4.6
Plastic
Deformations;
Residual
Stresses
4.68,79,82,88
4.67,80,81,87
4.72,77,86,91
4.71,78,84,91
16
4.7
Eccentric
Axial
Loading
in
a
Plane
of
Symmetry
4.101,108,117,121
4.100,106,120,122
4.103,110,115,124
4.99,107,116,125
17
4.8-9
Unsymmetric
Bending
4.127,135,144,153
4.129,138,145,155
4.133,137,146,152
4.134,139,147,153
18
4.10
Bending
of
Curved
Members
4.162,169,178,185
4.161,173,180,185
4.164,170,177,183
4.163,174,179,184
19
5.1
Shear
and
Bending-Moment
Diagrams
5.7,16,26,30
5.9,20,23,27
5.13,15,25,29
5.11,18,21,33
20
5.2
Relations
Between
w,
V,
and
M
5.41,46,60,64
5.43,48,59,64
5.40,53,55,63
5.42,52,56,63
21
5.3
Design
of
Prismatic
Beams
in
Bending
5.69,71,79,85
5.65,76,80,86
5.73,77,81,89
5.70,74,82,89
22
5.4
Use
of
Singularity
Functions
to
Determine
V
and
M
5.103,108,118,123
5.104,106,119,122
5.100,105,117,120
5.98,102,110,124
23
5.5
Nonprismatic
Beams
5.128,132,143,145
5.130,139,140,144
5.134,136,147,149
5.135,138,141,149
24
6.1-2
Shearing
Stresses
in
a
Beam
6.2,12,19,28
6.1,11,18,26
6.5,13,17,24
6.4,9,15,22
25
6.3-5
Shearing
Stresses
in
Thin-Walled
Members
6.31,37,49,56
6.29,35,48,57
6.32,42,44,51
6.30,36,46,59
26
6.6
Shear
center
6.61,69,76,83
6.62,71,77,84
6.65,70,75,81
6.68,72,78,82
27
7.1
Transformation
of
Plane
Stress
7.2,12,18,27
7.4,10,19,29
7.7,9,21,26
7.5,11,17,23
28
7.2
Mohr’s
Circle
for
Plane
Stress
7.33,41,58,61
7.34,40,48,55
7.32,47,49,59
7.31,45,46,51
29
7.3-5
Three-Dimensional
Stress
Analysis;
Fracture
Criteria
7.69,77,82,94
7.68,74,81,95
7.70,75,84,89
7.71,79,83,90
30
7.6
Stresses
in
Thin-Walled
Pressure
Vessels
7.100,108,114,124
7.102,106,115,122
7.98,105,112,120
7.103,104,113,121
31
7.7-9
Analysis
and
Measurement
of
Strain
7.131,140,147,155
7.133,136,148,154
7.129,143,146,157
7.135,139,149,156
32
8.1-2
Principal
Stresses
in
a
Beam;
Design
of
Shafts
8.4,12,17,27
8.3,11,15,28
8.5,13,21,24
8.6,9,22,23
33
8.3
Stresses
under
Combined
Loadings
8.31,40,51,53
8.32,38,52,55
8.35,43,48,58
8.36,42,47,57
34
9.1-2
Deflection
of
Beams
by
Integration
9.3,11,24,30
9.4,10,19,32
9.7,12,23,25
9.5,9,20,28
35
9.3
Use
of
Singularity
Functions
9.37,45,54,62
9.38,41,51,59
9.35,48,49,61
9.36,44,53,60
36
9.4
Method
of
Superposition;
Statically
Indet.
Beams
9.65,76,79,91
9.67,75,80,90
9.72,77,82,88
9.71,73,84,85
37
9.5
Moment-Area
Method
9.97,104,119,122
9.98,102,118,120
9.96,105,109,113
9.95,107,110,115
38
9.6
Moment-Area
Method;
Max
Deflection;
Indet.
Beams
9.129,137,146,152
9.125,135,142,150
9.130,133,144,153
9.126,136,145,147
39
10.1
Columns:
Euler’s
Formula
10.2,11,19,28
10.1,10,16,25
10.3,13,17,27
10.7,9,15,24
40
10.2
Eccentric
Loading;
Secant
Formula
10.34,40,50,56
10.32,39,44,51
10.30,37,47,54
10.29,36,43,53
41
10.3
Design
of
Columns
under
a
Centric
Load
10.62,65,71,83
10.60,68,75,84
10.59,66,72,80
10.57,69,74,77
42
10.4
Design
of
Columns
under
an
Eccentric
Load
10.92,101,106,107
10.91,102,104,105
10.93,99,109,116
10.89,100,103,111
43
11.1-3
Elastic
Strain
Energy
11.6,11,20,31
11.2,9,21,27
11.7,12,18,30
11.3,10,23,24
44
11.4-5
Design
for
Impact
Loading;
Deflections
11.44,52,65,76
11.43,50,61,66
11.45,53,62,75
11.42,51,58,68
45
11.6-8
Castigliano’s
Theorem
11.80,83,94,99
11.82,88,96,103
11.77,86,93,101
11.78,89,95,102
46
11.9
Statically
Indeterminate
Structures
11.109,113,121
11.107,114,121
11.106,113,117
11.105,112,119
Mecánica de Materiales - Ferdinand P. Beer 7ma. Edición.

40. C
CH
HA
AP
PT
TE
ER
R 1
1
Mecánica de Materiales - Ferdinand P. Beer 7ma. Edición.

41. Mecánica de Materiales - Ferdinand P. Beer 7ma. Edición.

42. PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
3
d1
d2
125 kN
125 kN
60 kN
C
A
B
0.9 m 1.2 m
PROBLEM 1.1
Two solid cylindrical rods AB and BC are
welded together at B and loaded as shown.
Knowing that 1 30 mm
d  and 2 50 mm,
d 
find the average normal stress at the
midsection of (a) rod AB, (b) rod BC.
SOLUTION
(a) Rod AB:
Force: 3
60 10 N tension
P  
Area: 2 3 2 6 2
1 (30 10 ) 706.86 10 m
4 4
A d
   
    
Normal stress:
3
6
6
60 10
84.882 10 Pa
706.86 10
AB
P
A
 

   

84.9 MPa
AB
  
(b) Rod BC:
Force: 3 3 3
60 10 (2)(125 10 ) 190 10 N
P       
Area: 2 3 2 3 2
2 (50 10 ) 1.96350 10 m
4 4
A d
   
    
Normal stress:
3
6
3
190 10
96.766 10 Pa
1.96350 10
BC
P
A
 
 
    

96.8 MPa
BC
   
Mecánica de Materiales - Ferdinand P. Beer 7ma. Edición.

43. PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
4
d1
d2
125 kN
125 kN
60 kN
C
A
B
0.9 m 1.2 m
PROBLEM 1.2
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 150 MPa in either rod, determine
the smallest allowable values of the diameters
d1 and d2.
SOLUTION
(a) Rod AB:
Force: 3
60 10 N
P  
Stress: 6
150 10 Pa
AB
  
Area: 2
1
4
A d


2
1
3
2 6 2
1 6
4
4 (4)(60 10 )
509.30 10 m
(150 10 )
AB
AB
AB
AB
P P
A
A
P
d
P
d




 

  


   

3
1 22.568 10 m
d 
  1 22.6 mm

d 
(b) Rod BC:
Force: 3 3 3
60 10 (2)(125 10 ) 190 10 N
P       
Stress: 6
150 10 Pa
BC
   
Area: 2
2
4
A d


2
2
3
2 3 2
2 6
4
4 (4)( 190 10 )
1.61277 10 m
( 150 10 )


 

 
 
   
 
BC
BC
P P
A d
P
d
3
2 40.159 10 m
d 
  2 40.2 mm
d  
Mecánica de Materiales - Ferdinand P. Beer 7ma. Edición.

44. PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
5
0.75 in.
1.25 in.
12 kips
P
B
C
25 in.
30 in.
A
PROBLEM 1.3
Two solid cylindrical rods AB and BC are welded together at B and loaded as
shown. Knowing that P = 10 kips, find the average normal stress at the
midsection of (a) rod AB, (b) rod BC.
SOLUTION
(a) Rod AB:
2 2 2
1
12 10 22 kips
(1.25) 1.22718 in
4 4
22
17.927 ksi
1.22718
AB
P
A d
P
A
 

  
  
   17.93 ksi
AB
  
(b) Rod BC:
2 2 2
2
10 kips
(0.75) 0.44179 in
4 4
10
22.635 ksi
0.44179
AB
P
A d
P
A
 


  
   22.6 ksi
AB
  
Mecánica de Materiales - Ferdinand P. Beer 7ma. Edición.

45. PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
6
0.75 in.
1.25 in.
12 kips
P
B
C
25 in.
30 in.
A
PROBLEM 1.4
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 stresses in
rods AB and BC are equal.
SOLUTION
(a) Rod AB:
2
2
2
2
12 kips
(1.25 in.)
4 4
1.22718 in
12 kips
1.22718 in
AB
P P
d
A
A
P
 

 
 



(b) Rod BC:
2 2
2
2
(0.75 in.)
4 4
0.44179 in
0.44179 in
BC
P P
A d
A
P
 


 


2 2
12 kips
1.22718 in 0.44179 in
5.3015 0.78539
AB BC
P P
P
 



 6.75 kips
P  
Mecánica de Materiales - Ferdinand P. Beer 7ma. Edición.

46. PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
7
1200 N
1200 N
C
A
B
PROBLEM 1.5
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.
SOLUTION


  
P P
A
A
Geometry: 2 2
1 2
( )
4

 
A d d
2 2 2
2 1 1
2 3 2
2 6
6 2
3
2
4 4
(4)(1200)
(25 10 )
(3.80 10 )
222.92 10 m
14.93 10 m
 




   
  

 
 
A P
d d d
d
d 2 14.93 mm

d 
Mecánica de Materiales - Ferdinand P. Beer 7ma. Edición.

47. PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
8
100 m
15 mm
10 mm
b
a
B
C
A PROBLEM 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.
SOLUTION
Areas: 2 2 6 2
2 2 6 2
(15 mm) 176.715 mm 176.715 10 m
4
(10 mm) 78.54 mm 78.54 10 m
4
AB
BC
A
A




   
   
From geometry, b  100  a
Weights: 6
6
(8470)(9.81)(176.715 10 ) 14.683
(8470)(9.81)(78.54 10 )(100 ) 652.59 6.526
AB AB AB
BC BC BC
W g A a a
W g A a a




   
     


Normal stresses:
At A,
6 3
652.59 8.157
3.6930 10 46.160 10

   
    
A AB BC
A
A
AB
P W W a
P
a
A
(1)
At B,
6 3
652.59 6.526
8.3090 10 83.090 10

  
    
B BC
B
B
BC
P W a
P
a
A
(2)
(a) Length of rod AB. The maximum stress in ABC is minimum when  

A B or
6 3
4.6160 10 129.25 10 0
   
a
35.71m

a 35.7 m
 
AB a 
(b) Maximum normal stress.
6 3
6 3
6
3.6930 10 (46.160 10 )(35.71)
8.3090 10 (83.090 10 )(35.71)
5.34 10 Pa


 
   
   
  
A
B
A B 5.34 MPa
  
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9
0.2 m
0.25 m
0.4 m
20 kN
C
B
A
D
E
PROBLEM 1.7
Each of the four vertical links has an 8 36-mm
 uniform rectangular
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.
SOLUTION
Use bar ABC as a free body.
3
3
3
3
0 : (0.040) (0.025 0.040)(20 10 ) 0
32.5 10 N Link is in tension.
0 : (0.040) (0.025)(20 10 ) 0
12.5 10 N Link is in compression.
     
 
     
  
C BD
BD
B CE
CE
M F
F BD
M F
F CE
Net area of one link for tension (0.008)(0.036 0.016)
  6 2
160 10 m

 
For two parallel links, 6 2
net 320 10 m

 
A
(a)
3
6
6
net
32.5 10
101.563 10
320 10
BD
BD
F
A
 

   

101.6 MPa
 
BD 
Area for one link in compression (0.008)(0.036)
 6 2
288 10 m

 
For two parallel links, 6 2
576 10 m

 
A
(b)
3
6
6
12.5 10
21.701 10
576 10
CE
CE
F
A
 

 
    

21.7 MPa
  
CE 
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10
10 in. 8 in.
2 in.
12 in.
4 in.
30⬚
120 lb
120 lb
C
A
B
PROBLEM 1.8
Link AC has a uniform rectangular cross section 1
8
in. thick and 1 in. wide.
Determine the normal stress in the central portion of the link.
SOLUTION
Use the plate together with two pulleys as a free body. Note that the cable tension causes at 1200 lb-in.
clockwise couple to act on the body.
0: (12 4)( cos30 ) (10)( sin30 ) 1200 lb 0
1200 lb
135.500 lb
16 cos30 10 sin30
        
   
  
B AC AC
AC
M F F
F
Area of link AC: 2
1
1 in. in. 0.125 in
8
  
A
Stress in link AC:
135.50
1084 psi 1.084 ksi
0.125
AC
AC
F
A
      
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11
0.100 m
0.150 m 0.300 m 0.250 m
P P P
E
A B C
D
PROBLEM 1.9
Three forces, each of magnitude P  4 kN, are applied to the mechanism
shown. Determine the cross-sectional area of the uniform portion of rod
BE for which the normal stress in that portion is 100 MPa.
SOLUTION
Draw free body diagrams of AC and CD.
Free Body CD: 0: 0.150 0.250 0
0.6
   

D
M P C
C P
Free Body AC: 0: 0.150 0.350 0.450 0.450 0
1.07
7.1333 (7.133)(4 kN) 28.533 kN
0.150
    
   
A BE
BE
M F P P C
F P P
Required area of BE:
3
6 2
6
28.533 10
285.33 10 m
100 10





   

BE
BE
BE
BE
BE
BE
F
A
F
A
2
285 mm

BE
A 
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12
4 kips
308
u
6 in.
12 in.
D
C
B
A
PROBLEM 1.10
Link BD consists of a single bar 1 in. wide and
1
2
in. thick. Knowing that each pin has a 3
8
-in.
diameter, determine the maximum value of the
average normal stress in link BD if (a)  = 0,
(b)  = 90.
SOLUTION
Use bar ABC as a free body.
(a) 0.
 
0: (18 sin30 )(4) (12 cos30 ) 0
3.4641 kips (tension)
     

A BD
BD
M F
F
Area for tension loading: 2
3 1
( ) 1 0.31250 in
8 2
  
    
  
  
A b d t
Stress: 2
3.4641 kips
0.31250 in
BD
F
A
   11.09 ksi
  
(b) 90 .
  
0: (18 cos30 )(4) (12 cos30 ) 0
6 kips i.e. compression.
      
 
A BD
BD
M F
F
Area for compression loading: 2
1
(1) 0.5 in
2
A bt
 
  
 
 
Stress: 2
6 kips
0.5 in
BD
F
A


  12.00 ksi
  
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13
9 ft
80 kips 80 kips 80 kips
9 ft 9 ft 9 ft
12 ft
B D F
H
G
E
C
A
PROBLEM 1.11
For the Pratt bridge truss and loading shown, determine the
average normal stress in member BE, knowing that the cross-
sectional area of that member is 5.87 in2
.
SOLUTION




Use entire truss as free body.
0: (9)(80) (18)(80) (27)(80) 36 0
120 kips
H y
y
M A
A
     

Use portion of truss to the left of a section cutting members
BD, BE, and CE.
12
0: 120 80 0 50 kips
15
y BE BE
F F F
       

2
50 kips
5.87 in
  
BE
BE
F
A
8.52 ksi
 
BE 
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14
40 in.
45 in.
15 in.
4 in.
A
B C
D
E F
4 in.
30 in.
30 in.
480 lb
PROBLEM 1.12
The frame shown consists of four wooden members, ABC,
DEF, BE, and CF. Knowing that each member has a 2  4-in.
rectangular cross section and that each pin has a 1
2
-in.
diameter, determine the maximum value of the average
normal stress (a) in member BE, (b) in member CF.
SOLUTION



Stress in tension member CF:
Add support reactions to figure as shown.
Using entire frame as free body,
0: 40 (45 30)(480) 0
900 lb
    

A x
x
M D
D
Use member DEF as free body.
Reaction at D must be parallel to BE
F and .
CF
F
4
1200 lb
3
 
y x
D D
4
0: (30) (30 15) 0
5
2250 lb
4
0: (30) (15) 0
5
750 lb
 
     
 
 
 
 
   
 
 

F BE Y
BE
E CE Y
CE
M F D
F
M F D
F
Stress in compression member BE:
Area: 2
2 in. 4 in. 8 in
  
A
(a)
2250
8


 
BE
BE
F
A
281psi
  
BE 
Minimum section area occurs at pin.
2
min (2)(4.0 0.5) 7.0 in
  
A
(b)
min
750
7.0
  
CF
CF
F
A
107.1psi
 
CF 
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15
D
B
E
A
Dimensions in mm
100
450
250
850
1150
500 675 825
C
G
F
PROBLEM 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.
SOLUTION




FREE BODY – ENTIRE TOW BAR:
2
(200 kg)(9.81 m/s ) 1962.00 N
0: 850 1150(1962.00 N) 0
2654.5 N
 
   

A
W
M R
R
FREE BODY – BOTH ARM & WHEEL UNITS:
100
tan 8.4270
675
 
  
2
0: ( cos )(550) (500) 0
500
(2654.5 N)
550 cos 8.4270
2439.5 N (comp.)
2439.5 N
(0.0125 m)



   



   
E CD
CD
CD
CD
CD
M F R
F
F
A
6
4.9697 10 Pa
   4.97 MPa
  
CD 
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16
C
A
B
E F G
D
200 mm
150 mm
150 mm
300 mm
400 mm
600 mm
800 N
PROBLEM 1.14
Two hydraulic cylinders are used to control the position
of the robotic arm ABC. Knowing that the control rods
attached at A and D each have a 20-mm diameter and
happen to be parallel in the position shown, determine the
average normal stress in (a) member AE, (b) member DG.
SOLUTION
Use member ABC as free body.
3
4
0: (0.150) (0.600)(800) 0
5
4 10 N
   
 
B AE
AE
M F
F
Area of rod in member AE is 2 3 2 6 2
(20 10 ) 314.16 10 m
4 4
A d
   
    
Stress in rod AE:
3
6
6
4 10
12.7324 10 Pa
314.16 10
AE
AE
F
A
 

   

(a) 12.73 MPa
AE
  
Use combined members ABC and BFD as free body.
4 4
0: (0.150) (0.200) (1.050 0.350)(800) 0
5 5
1500 N
   
     
   
   
 
F AE DG
DG
M F F
F
Area of rod DG: 2 3 2 6 2
(20 10 ) 314.16 10 m
4 4
A d
   
    
Stress in rod DG: 6
6
1500
4.7746 10 Pa
3.1416 10
DG
DG
F
A
 

    

(b) 4.77 MPa
DG
   
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17
PROBLEM 1.15
Determine the diameter of the largest circular hole that can be punched into a sheet of polystyrene 6 mm
thick, knowing that the force exerted by the punch is 45 kN and that a 55-MPa average shearing stress is
required to cause the material to fail.
SOLUTION
For cylindrical failure surface: A dt


Shearing stress: or
P P
A
A


 
Therefore,
P
dt



Finally,
3
6
3
45 10 N
(0.006 m)(55 10 Pa)
43.406 10 m
P
d
t
 






 
43.4 mm
d  
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18
2 in.
1 in.
P'
2 in.
1 in. 9 in.
P
in.
5
8
in.
5
8
PROBLEM 1.16
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.
SOLUTION
Six areas must be sheared off when the joint fails. Each of these areas has dimensions 5 1
8 2
in. in.,
 its area
being
2 2
5 1 5
in 0.3125 in
8 2 16
   
A
At failure, the force carried by each area is
2
(1.20 ksi)(0.3125 in ) 0.375 kips

  
F A
Since there are six failure areas,
6 (6)(0.375)
 
P F 2.25 kips

P 
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19
0.6 in.
3 in. Wood
Steel
P
P'
PROBLEM 1.17
When the force P reached 1600 lb, 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.
SOLUTION
Area being sheared: 2
3 in. 0.6 in. 1.8 in
  
A
Force: 1600 lb
P 
Shearing stress: 2
2
1600 lb
8.8889 10 psi
1.8 in
    
P
A
889 psi
  
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20
40 mm
8 mm
12 mm
P
10 mm
PROBLEM 1.18
A load P is applied to a steel rod supported as shown by an aluminum
plate into which a 12-mm-diameter hole has been drilled. Knowing that
the shearing stress must not exceed 180 MPa in the steel rod and 70 MPa
in the aluminum plate, determine the largest load P that can be applied to
the rod.
SOLUTION
For steel: 1
6 2
(0.012 m)(0.010 m)
376.99 10 m
A dt
 

 
 
6 2 6
1 1 1
3
(376.99 10 m )(180 10 Pa)
67.858 10 N
  
     
 
P
P A
A
For aluminum: 3 2
2 (0.040 m)(0.008 m) 1.00531 10 m
A dt
  
   
3 2 6 3
2 2 2
2
(1.00531 10 m )(70 10 Pa) 70.372 10 N
P
P A
A
  
       
Limiting value of P is the smaller value, so 67.9 kN
P  
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21
6 in.
L
P
PROBLEM 1.19
The axial force in the column supporting the timber beam shown is
P 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.
SOLUTION
Bearing area: 
b
A Lw
3
20 10 lb
8.33 in.
(400 psi)(6 in.)


 

  
b
b
b
P P
A Lw
P
L
w
8.33 in.

L 
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22
d
12 mm
PROBLEM 1.20
Three wooden planks are fastened together by a series of bolts to form
a column. The diameter of each bolt is 12 mm and the inner diameter
of each washer is 16 mm, which is slightly larger than the diameter of
the holes in the planks. Determine the smallest allowable outer
diameter d of the washers, knowing that the average normal stress in
the bolts is 36 MPa and that the bearing stress between the washers
and the planks must not exceed 8.5 MPa.
SOLUTION
Bolt:
2 2
4 2
Bolt
(0.012 m)
1.13097 10 m
4 4
d
A
  
   
Tensile force in bolt:
P
P A
A
 
  
6 4 2
3
(36 10 Pa)(1.13097 10 m )
4.0715 10 N

  
 
Bearing area for washer:  
2 2
4

 
w o i
A d d
and w
BRG
P
A


Therefore, equating the two expressions for Aw gives
 
2 2
2 2
3
2 2
6
2 4 2
3
4
4
4 (4.0715 10 N)
(0.016 m)
(8.5 10 Pa)
8.6588 10 m
29.426 10 m






 
 

 

 
 
o i
BRG
o i
BRG
o
o
o
P
d d
P
d d
d
d
d
29.4 mm

o
d 
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23
P 5 40 kN
b b
120 mm 100 mm
PROBLEM 1.21
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
footing, (b) the size of the footing for which the average bearing
stress in the soil is 145 kPa.
SOLUTION
(a) Bearing stress on concrete footing.
3
3 2 3 2
3
6
3
40 kN 40 10 N
(100)(120) 12 10 mm 12 10 m
40 10
3.3333 10 Pa
12 10



  
    

   

P
A
P
A
3.33 MPa 
(b) Footing area. 3 3
40 10 N 145 kPa 45 10 Pa

    
P
3
2
3
40 10
0.27586 m
145 10
P P
A
A



   

Since the area is square, 2

A b
0.27586 0.525 m
  
b A 525 mm

b 
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24
a a
P
PROBLEM 1.22
An axial load P is supported by a short W8 40
 column of cross-
sectional area 2
11.7 in

A and is distributed to a concrete foundation
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, determine the side a of
the plate that will provide the most economical and safe design.
SOLUTION
For the column,  
P
A
or
(30)(11.7) 351 kips

  
P A
For the 
a a plate, 3.0 ksi
 
2
351
117 in
3.0

  
P
A
Since the plate is square, 2

A a
117
 
a A 10.82 in.

a 
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25
b
d
t
B
A
d
PROBLEM 1.23
Link AB, of width b = 2 in. and thickness t = 1
4
in., is used to support the end of a
horizontal beam. Knowing that the average normal stress in the link is 20 ksi and
that the average shearing stress in each of the two pins is 12 ksi, determine (a) the
diameter d of the pins, (b) the average bearing stress in the link.
SOLUTION
Rod AB is in compression.
1
where 2 in. and in.
4
1
( 20)(2) 10 kips
4

  
 
     
 
 
A bt b t
P A
Pin: P
P
P
A
 
and 2
4
P
A d


(a)
4 4 (4)(10)
1.03006 in.
(12)
P
P
A P
d
  
   
1.030 in.
d  
(b)
10
38.833 ksi
(1.03006)(0.25)
b
P
dt
   
38.8 ksi
b
  
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26
16 mm
750 mm
750 mm
12 mm
50 mm B
A
C
P
␪
PROBLEM 1.24
Determine the largest load P which may be applied at A when
  60°, 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.
SOLUTION
Geometry: Triangle ABC is an isoseles triangle with angles shown here.
Use joint A as a free body.
Law of sines applied to force triangle:
sin 30 sin 120 sin 30
sin30
0.57735
sin 120
sin 30
sin 30
 
  

 


 

AB AC
AB
AB
AC
AC
P F F
F
P F
F
P F
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27
PROBLEM 1.24 (Continued)
If shearing stress in pin at B is critical,
2 2 6 2
6 6 3
(0.010) 78.54 10 m
4 4
2 (2)(78.54 10 )(120 10 ) 18.850 10 N
 



   
     
AB
A d
F A
If bearing stress in member AB at bracket at A is critical,
6 2
6 6 3
(0.016)(0.010) 160 10 m
(160 10 )(90 10 ) 14.40 10 N



   
     
b
AB b b
A td
F A
If bearing stress in the bracket at B is critical,
6 2
6 6 3
2 (2)(0.012)(0.010) 240 10 m
(240 10 )(90 10 ) 21.6 10 N



   
     
b
AB b b
A td
F A
Allowable FAB is the smallest, i.e., 14.40  103
N
Then from statics, 3
allow (0.57735)(14.40 10 )
 
P
3
8.31 10 N
  8.31 kN 
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28
16 mm
750 mm
750 mm
12 mm
50 mm B
A
C
P
␪
PROBLEM 1.25
Knowing that   40° and P  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 average
bearing stress in member AB at B, (c) the corresponding average
bearing stress in each of the support brackets at B.
SOLUTION
Geometry: Triangle ABC is an isoseles triangle with angles shown here.
Use joint A as a free body.
Law of sines applied to force triangle:
sin 20 sin110 sin50
sin110
sin 20
(9)sin110
24.727 kN
sin 20
AB AC
AB
P F F
P
F
 
  




 

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29
PROBLEM 1.25 (Continued)
(a) Allowable pin diameter.
2 2
4
2
2 2 


  
AB AB AB
P
F F F
A d d
where 3
24.727 10 N
 
AB
F
3
2 6 2
6
2 (2)(24.727 10 )
131.181 10 m
(120 10 )
 


   

AB
F
d
3
11.4534 10 m

 
d 11.45 mm 
(b) Bearing stress in AB at A.
3 6 2
3
6
6
(0.016)(11.4534 10 ) 183.254 10 m
24.727 10
134.933 10 Pa
183.254 10

 

    

   

b
AB
b
b
A td
F
A
134.9 MPa 
(c) Bearing stress in support brackets at B.
3 6 2
1 3
6
2
6
(0.012)(11.4534 10 ) 137.441 10 m
(0.5)(24.727 10 )
89.955 10 Pa
137.441 10

 

    

   

AB
b
A td
F
A
90.0 MPa 
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30
45 mm
200 mm
100 mm 175 mm
D
F
E
A
C
B
P
208
u
PROBLEM 1.26
The hydraulic cylinder CF, which partially controls the position of rod
DE, has been locked in the position shown. Member BD is 15 mm
thick and is connected at C to the vertical rod by a 9-mm-diameter
bolt. Knowing that P  2 kN and 75 ,
   determine (a) the average
shearing stress in the bolt, (b) the bearing stress at C in member BD.
SOLUTION
Free Body: Member BD.
40 9
0: (100 cos20 ) (100 sin 20 )
41 4
c AB AB
M F F
    
(2 kN)cos75 (175sin 20 ) (2 kN)sin75 (175cos20 ) 0
      
100
(40cos20 9sin 20 ) (2 kN)(175)sin(75 20 )
41
4.1424 kN
AB
AB
F
F
      

9
0: (4.1424 kN) (2 kN)cos75 0
41
0.39167 kN
     

x x
x
F C
C
40
0: (4.1424 kN) (2 kN)sin 75 0
41
5.9732 kN
     

y y
y
F C
C
5.9860 kN

C 86.2°
(a)
3
6
ave 2
5.9860 10 N
94.1 10 Pa 94.1 MPa
(0.0045 m)



    
C
A

(b)
3
6
5.9860 10 N
44.3 10 Pa 44.3 MPa
(0.015 m)(0.009 m)
b
C
td


     
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31
0.2 m
0.25 m
0.4 m
20 kN
C
B
A
D
E
PROBLEM 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  50-mm uniform rectangular cross
section.
PROBLEM 1.7 Each of the four vertical links has an 8  36-mm
uniform rectangular 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.
SOLUTION
Use bar ABC as a free body.
3
3
0 : (0.040) (0.025 0.040)(20 10 ) 0
32.5 10 N
     
 
C BD
BD
M F
F
(a) Shear pin at B.
2
  BD
F
A
for double shear
where 2 2 6 2
(0.016) 201.06 10 m
4 4
  
   
A d
3
6
6
32.5 10
80.822 10 Pa
(2)(201.06 10 )
 

  

80.8 MPa
  
(b) Bearing: link BD. 6 2
(0.016)(0.008) 128 10 m

   
A dt
1 3
6
2
6
(0.5)(32.5 10 )
126.95 10 Pa
128 10
BD
b
F
A
 

   

127.0 MPa
 
b 
(c) Bearing in ABC at B. 6 2
(0.016)(0.010) 160 10 m

   
A dt
3
6
6
32.5 10
203.12 10 Pa
160 10
BD
b
F
A
 

   

203 MPa
 
b 
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32
A
C
D E
B
12 in.
12 in.
15 in.
16 in. 16 in. 20 in.
1500 lb
PROBLEM 1.28
Two identical linkage-and-hydraulic-cylinder systems control the
position of the forks of a fork-lift truck. The load supported by the one
system shown is 1500 lb. Knowing that the thickness of member BD is
5
8
in., determine (a) the average shearing stress in the 1
2
-in.-diameter
pin at B, (b) the bearing stress at B in member BD.
SOLUTION
Use one fork as a free body.
0: 24 (20)(1500) 0
B
M E
   
1250 lb
E 
0: 0
x x
x
F E B
B E
   
 
1250 lb
x
B 
2 2 2 2
0: 1500 0 1500 lb
1250 1500 1952.56 lb
y y y
x y
F B B
B B B
    
    
(a) Shearing stress in pin at B.
2
2 2
pin pin
1
0.196350 in
4 4 2
A d
   
  
 
 
3
pin
1952.56
9.94 10 psi
0.196350
B
A
     9.94 ksi
  
(b) Bearing stress at B.
  
3
5
1
2 8
1952.56
6.25 10 psi
B
dt
     6.25 ksi
  
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33
75 mm
150 mm
45
45
45
45
45
45
45
45
45
45
45
45
45
45
45
45
45
45
45
45
45
45
45
45
45⬚
⬚
⬚
⬚
⬚
⬚
⬚
⬚
⬚
⬚
⬚
⬚
⬚
P'
P
PROBLEM 1.29
Two wooden members of uniform rectangular cross section are joined
by the simple glued scarf splice shown. Knowing that P  11 kN,
determine the normal and shearing stresses in the glued splice.
SOLUTION
3
3 2 3 2
0
2 3 2
3
3
0
90 45 45
11 kN 11 10 N
(150)(75) 11.25 10 mm 11.25 10 m
cos (11 10 )cos 45
489 10 Pa
11.25 10





     
  
    
 
   

P
A
P
A
489 kPa
  
3
3
3
0
sin 2 (11 10 )(sin90 )
489 10 Pa
2 (2)(11.25 10 )

 
 
   

P
A
489 kPa
  
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34
75 mm
150 mm
45
45
45
45
45
45
45
45
45
45
45
45
45
45
45
45
45
45
45
45
45
45
45
45
45⬚
⬚
⬚
⬚
⬚
⬚
⬚
⬚
⬚
⬚
⬚
⬚
⬚
P'
P
PROBLEM 1.30
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.
SOLUTION
3 2 3 2
0
3
0
90 45 45
(150)(75) 11.25 10 mm 11.25 10 m
620 kPa 620 10 Pa
sin 2
2
A
P
A





     
    
  

(a)
3 3
0
2 (2)(11.25 10 )(620 10 )
sin2 sin90



 
 

A
P
3
13.95 10 N
  13.95 kN

P 
(b)
2 3 2
3
0
cos (13.95 10 )(cos45 )
11.25 10

 
 
 

P
A
3
620 10 Pa
  620 kPa
  
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35
608
5.0 in.
3.0 in.
P'
P PROBLEM 1.31
The 1.4-kip load P is supported by two wooden members of uniform cross section
that are joined by the simple glued scarf splice shown. Determine the normal and
shearing stresses in the glued splice.
SOLUTION
2
0
2 2
0
1400 lb 90 60 30
(5.0)(3.0) 15 in
cos (1400)(cos30 )
15
P
A
P
A



      
 

  70.0 psi
  
0
sin 2 (1400)sin60
2 (2)(15)



 
P
A
40.4 psi
  
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36
608
5.0
3.0 in.
P'
P PROBLEM 1.32
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.
SOLUTION
2
0
2
0
(5.0)(3.0) 15 in
90 60 30
cos



 
     

A
P
A
(a) 0
2 2
(75)(15)
1500 lb
cos cos 30


  

A
P 1.500 kips

P 
(b)
0
sin 2 (1500)sin60
2 (2)(15)



 
P
A
43.3 psi
  
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37
6 in.
6 in.
P PROBLEM 1.33
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 2.5 ksi, determine
(a) the magnitude of P, (b) the orientation 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.
SOLUTION
2
0
max
max
(6)(6) 36 in
2.5 ksi
45 for plane of
A

 
 

 
(a) max 0 max
0
| |
| | 2 (2)(36)(2.5)
2
P
P A
A
 
    180.0 kips
P  
(b) sin 2 1 2 90
 
   45.0
   
(c) 2
45
0 0
180
cos 45
2 (2)(36)
P P
A A
      45 2.50 ksi
   
(d) max
0
180
36
P
A


  max 5.00 ksi
   
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38
6 in.
6 in.
P PROBLEM 1.34
A 240-kip load P is applied to the granite block shown. Determine the resulting
maximum value of (a) the normal stress, (b) the shearing stress. Specify the
orientation of the plane on which each of these maximum values occurs.
SOLUTION
2
0
2 2 2
0
(6)(6) 36 in
240
cos cos 6.67cos
36
A
P
A
   
 

   
(a) max tensile stress  0 at 90.0
  
max. compressive stress  6.67 ksi at 0
   
(b) max
0
240
2 (2)(36)
P
A
  
max 3.33 ksi
  
at 45
  
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39
208
P
Weld
10 mm
PROBLEM 1.35
A steel pipe of 400-mm outer diameter is fabricated from 10-mm thick
plate by welding along a helix that forms an angle of 20 with a plane
perpendicular to the axis of the pipe. Knowing that a 300-kN axial
force P is applied to the pipe, determine the normal and shearing
stresses in directions respectively normal and tangential to the weld.
SOLUTION
2 2 2 2
3 2
3 2
2 6
3
0.400 m
1
0.200 m
2
0.200 0.010 0.190 m
( ) (0.200 0.190 )
12.2522 10 m
20
300 10 cos 20
cos 21.621 10 Pa
12.2522 10
 

 



 
    
   
 
 
  
    

o
o o
i o
o o i
o
d
r d
r r t
A r r
P
A
21.6 MPa
   
3
6
3
0
300 10 sin 40
sin 2 7.8695 10 Pa
2 (2)(12.2522 10 )
  
  
    

P
A
7.87 MPa
  
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40
208
P
Weld
10 mm
PROBLEM 1.36
A steel pipe of 400-mm outer diameter is fabricated from 10-mm
thick plate by welding along a helix that forms an angle of 20° with a
plane perpendicular to the axis of the pipe. Knowing that the
maximum allowable normal and shearing stresses in the directions
respectively normal and tangential to the weld are 60 MPa
  and
36 MPa,
  determine the magnitude P of the largest axial force that
can be applied to the pipe.
SOLUTION
2 2 2 2
3 2
0.400 m
1
0.200 m
2
0.200 0.010 0.190 m
( ) (0.200 0.190 )
12.2522 10 m
20
 



 
    
   
 
 
o
o o
i o
o o i
d
r d
r r t
A r r
Based on 2
0
| | 60 MPa: cos
P
A
  
 
3 6
3
2 2
(12.2522 10 )(60 10 )
832.52 10 N
cos cos 20



 
   

o
A
P
Based on | | 30 MPa: sin 2
2
  
 
o
P
A
3 6
3
2 (2)(12.2522 10 )(36 10 )
1372.39 10 N
sin 2 sin 40



 
   

o
A
P
Smaller value is the allowable value of P. 833 kN
P  
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41
12 in.
9 in. 1 in.
C
D
Q
A
9 in.
12 in.
F
Q'
B
E
in.
1
2
in.
3
8
PROBLEM 1.37
A steel loop ABCD of length 5 ft and of 3
8
-in. diameter is placed as
shown around a 1-in.-diameter aluminum rod AC. Cables BE and DF,
each of 1
2
-in. diameter, are used to apply the load Q. Knowing that the
ultimate strength of the steel used for the loop and the cables is 70 ksi,
and that the ultimate strength of the aluminum used for the rod is 38 ksi,
determine the largest load Q that can be applied if an overall factor of
safety of 3 is desired.
SOLUTION
Using joint B as a free body and considering symmetry,
3 6
2 0
5 5
   
AB AB
F Q Q F
Using joint A as a free body and considering symmetry,
4
2 0
5
8 5 3
0
5 6 4
  
    
AB AC
AC AC
F F
Q F Q F
Based on strength of cable BE,
2
2 1
(70) 13.7445 kips
4 4 2
U U U
Q A d
 
 
 
   
 
 
Based on strength of steel loop,
2
,
2
6 6 6
5 5 5 4
6 3
(70) 9.2775 kips
5 4 8
U AB U U U
Q F A d

 

  
 
 
 
 
Based on strength of rod AC,
2 2
,
3 3 3 3
(38) (1.0) 22.384 kips
4 4 4 4 4 4
U AC U U U
Q F A d
 
 
    
Actual ultimate load QU is the smallest, 9.2775 kips
U
Q
 
Allowable load:
9.2775
3.0925 kips
. . 3
U
Q
Q
F S
   3.09 kips
Q  
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42
A B
C
D
480 mm
908
w
P
PROBLEM 1.38
Link BC is 6 mm thick, has a width w  25 mm, and is made of a steel with
a 480-MPa ultimate strength in tension. What was the safety factor used if the
structure shown was designed to support a 16-kN load P?
SOLUTION
Use bar ACD as a free body and note that member BC is a two-force member.
3
3
0:
(480) (600) 0
600 (600)(16 10 )
20 10 N
480 480
 
 

   
A
BC
BC
M
F P
F P
Ultimate load for member BC: U U
F A


6 3
(480 10 )(0.006)(0.025) 72 10 N
   
U
F
Factor of safety:
3
3
72 10
F.S.
20 10

 

U
BC
F
F
F.S. 3.60
 
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43
A B
C
D
480 mm
908
w
P
PROBLEM 1.39
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?
SOLUTION
Use bar ACD as a free body and note that member BC is a two-force member.
3
3
0:
(480) 600 0
600 (600)(20 10 )
25 10 N
480 480
 
 

   
A
BC
BC
M
F P
P
F
For a factor of safety F.S.  3, the ultimate load of member BC is
3 3
(F.S.)( ) (3)(25 10 ) 75 10 N
    
U BC
F F
But 

U U
F A
3
6 2
6
75 10
166.667 10 m
450 10



    

U
U
F
A
For a rectangular section, A  wt or
6
3
166.667 10
27.778 10 m
0.006
A
w
t



   
27.8 mm
w  
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44
1.4 m
0.75 m
0.4 m
B
A
C
PROBLEM 1.40
Members AB and BC of the truss shown are made of the same alloy. It is known
that a 20-mm-square bar of the same alloy was tested to failure and that an
ultimate load of 120 kN was recorded. If a factor of safety of 3.2 is to be
achieved for both bars, determine the required cross-sectional area of (a) bar
AB, (b) bar AC.
SOLUTION
Length of member AB:
2 2
0.75 0.4 0.85 m
  
AB
Use entire truss as a free body.
0: 1.4 (0.75)(28) 0
15 kN
   

c x
x
M A
A
0: 28 0
28 kN
   

y y
y
F A
A
Use Joint A as free body.
0.75
0: 0
0.85
(0.85)(15)
17 kN
0.75
x AB x
AB
F F A
F
   
 
0.4
0: 0
0.85
(0.4)(17)
28 20 kN
0.85
y y AC AB
AC
F A F F
F
    
  
For the test bar, 2 6 2 3
(0.020) 400 10 m 120 10 N
U
A P

    
For the material,
3
6
6
120 10
300 10 Pa
400 10
U
U
P
A
 

   

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45
PROBLEM 1.40 (Continued)
(a) For member AB: F.S. U U AB
AB AB
P A
F F

 
3
6 2
6
(F.S.) (3.2)(17 10 )
181.333 10 m
300 10
AB
AB
U
F
A



   

2
181.3 mm
AB
A  
(b) For member AC: F.S. U U AC
AC AC
P A
F F

 
3
6 2
6
(F.S.) (3.2)(20 10 )
213.33 10 m
300 10
AC
AC
U
F
A



   

2
213 mm
AC
A  
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46
1.4 m
0.75 m
0.4 m
B
A
C
PROBLEM 1.41
Members AB and BC of the truss shown are made of the same alloy. It is known
that a 20-mm-square bar of the same alloy was tested to failure and that an
ultimate load of 120 kN was recorded. If bar AB has a cross-sectional area of
225 mm2
, determine (a) the factor of safety for bar AB and (b) the cross-
sectional area of bar AC if it is to have the same factor of safety as bar AB.
SOLUTION
Length of member AB:
2 2
0.75 0.4 0.85 m
  
AB
Use entire truss as a free body.
0: 1.4 (0.75)(28) 0
15 kN
   

c x
x
M A
A
0: 28 0
28 kN
   

y y
y
F A
A
Use Joint A as free body.
0.75
0: 0
0.85
(0.85)(15)
17 kN
0.75
x AB x
AB
F F A
F
   
 
0.4
0: 0
0.85
(0.4)(17)
28 20 kN
0.85
y y AC AB
AC
F A F F
F
    
  
For the test bar, 2 6 2 3
(0.020) 400 10 m 120 10 N
U
A P

    
For the material,
3
6
6
120 10
300 10 Pa
400 10
U
U
P
A
 

   

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47
PROBLEM 1.41 (Continued)
(a) For bar AB:
6 6
3
(300 10 )(225 10 )
F.S.
17 10
U U AB
AB AB
F A
F F
 
 
  

F.S. 3.97
 
(b) For bar AC: F.S. U U AC
AC AC
F A
F F

 
3
6 2
6
(F.S.) (3.97)(20 10 )
264.67 10 m
300 10
AC
AC
U
F
A



   

2
265 mm
AC
A  
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48
1.4 ft
35⬚
B
A
C D
E
1.4 ft 1.4 ft
600 lb/ft
5 kips
PROBLEM 1.42
Link AB is to be made of a steel for which the ultimate normal stress is
65 ksi. Determine the cross-sectional area of AB for which the factor
of safety will be 3.20. Assume that the link will be adequately
reinforced around the pins at A and B.
SOLUTION
(4.2)(0.6) 2.52 kips
P  
0 : (2.8)( sin35 )
(0.7)(2.52) (1.4)(5) 0
D AB
M F
   
  
ult
ult
2
5.4570 kips
. .
( . .) (3.20)(5.4570 kips)
65 ksi
0.26854 in
AB
AB
AB
AB
AB
AB
F
F
A F S
F S F
A




 
 
 2
0.268 in
AB
A  
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49
16 kN
L
125 mm
6 mm
16 kN PROBLEM 1.43
Two wooden members are joined by plywood splice plates that are fully glued on
the contact surfaces. Knowing that the clearance between the ends of the members
is 6 mm and that the ultimate shearing stress in the glued joint is 2.5 MPa,
determine the length L for which the factor of safety is 2.75 for the loading shown.
SOLUTION
all
2.5 MPa
0.90909 MPa
2.75
  
On one face of the upper contact surface,
0.006 m
(0.125 m)
2
L
A


Since there are 2 contact surfaces,
all
3
6
2
16 10
0.90909 10
( 0.006)(0.125)
0.14680 m
P
A
L
L
 

 

 146.8 mm 
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50
16 kN
L
125 mm
6 mm
16 kN PROBLEM 1.44
For the joint and loading of Prob. 1.43, determine the factor of safety when
L = 180 mm.
PROBLEM 1.43 Two wooden members are joined by plywood splice plates that
are fully glued on the contact surfaces. Knowing that the clearance between the
ends of the members is 6 mm and that the ultimate shearing stress in the glued
joint is 2.5 MPa, determine the length L for which the factor of safety is 2.75 for
the loading shown.
SOLUTION
Area of one face of upper contact surface:
3 2
0.180 m 0.006 m
(0.125 m)
2
10.8750 10 m
A
A 


 
Since there are two surfaces,
3
all 3 2
all
all
16 10 N
2 2(10.8750 10 m )
0.73563 MPa
2.5 MPa
F.S. 3.40
0.73563 MPa
u
P
A






 


   
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51
P
PROBLEM 1.45
Three 3
4
-in.-diameter steel bolts are to be used to attach the steel plate shown to
a wooden beam. Knowing that the plate will support a load P = 24 kips and that
the ultimate shearing stress for the steel used is 52 ksi, determine the factor of
safety for this design.
SOLUTION
For each bolt,
2
2 2
3
0.44179 in
4 4 4
A d
   
  
 
 
(0.44179)(52)
22.973 kips
U U
P A
 

For the three bolts, (3)(22.973) 68.919 kips
U
P  
Factor of safety:
68.919
. .
24
U
P
F S
P
  . . 2.87
F S  
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52
P
PROBLEM 1.46
Three steel bolts are to be used to attach the steel plate shown to a wooden beam.
Knowing that the plate will support a load P = 28 kips, that the ultimate shearing
stress for the steel used is 52 ksi, and that a factor of safety of 3.25 is desired,
determine the required diameter of the bolts.
SOLUTION
For each bolt,
24
8 kips
3
P  
Required: ( . .) (3.25)(8.0) 26.0 kips
U
P F S P
  
2 2
4
4
4 (4)(26.0)
0.79789 in.
(52)
U U U
U
U
U
P P P
A d d
P
d



 
  
   0.798 in.
d  
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53
1
2
40 mm
d
c
b
P
1
2 P
PROBLEM 1.47
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
40 mm,
b  55 mm,
c  and 12 mm,
d  determine the load P if an
overall factor of safety of 3.2 is desired.
SOLUTION
Based on double shear in pin,
2
2 6 3
2 2
4
(2)(0.012) (145 10 ) 32.80 10 N
4

 

 
   
U U U
P A d
Based on tension in wood,
6
3
( )
(0.040)(0.040 0.012)(60 10 )
67.2 10 N
 
  
  
 
U U U
P A w b d
Based on double shear in the wood,
6
3
2 2 (2)(0.040)(0.055)(7.5 10 )
33.0 10 N
 
   
 
U U U
P A wc
Use smallest 3
32.8 10 N
 
U
P
Allowable:
3
3
32.8 10
10.25 10 N
. . 3.2

   
U
P
P
F S
10.25 kN 
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54
1
2
40 mm
d
c
b
P
1
2 P
PROBLEM 1.48
For the support of Prob. 1.47, knowing that the diameter of the pin is
16 mm

d and that the magnitude of the load is 20 kN,

P 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 members is the same as that found in part a
for the pin.
PROBLEM 1.47 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 40 mm,

b 55 mm,
c  and 12 mm,
d  determine the
load P if an overall factor of safety of 3.2 is desired.
SOLUTION
3
20 kN 20 10 N
  
P
(a) Pin: 2 2 6 2
(0.016) 2.01.06 10 m
4 4
  
   
A d
Double shear:
2 2
U
U
P P
A A
 
 
6 6 3
2 (2)(201.16 10 )(145 10 ) 58.336 10 N
 
     
U U
P A
3
3
58.336 10
. .
20 10

 

U
P
F S
P
. . 2.92

F S 
(b) Tension in wood: 3
58.336 10 N for same F.S.
 
U
P
where 40 mm 0.040 m
( )
U U
U
P P
w
A w b d
    

3
3
6
58.336 10
0.016 40.3 10 m
(0.040)(60 10 )



     

U
U
P
b d
w
40.3 mm
b  
Shear in wood: 3
58.336 10 N for same F.S.
 
U
P
Double shear: each area is A wc

2 2
  
U U
U
P P
A wc
3
3
6
58.336 10
97.2 10 m
2 (2)(0.040)(7.5 10 )



   

U
U
P
c
w
97.2 mm
c  
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55
a
b
P
3
4
in.
1
4
in.
PROBLEM 1.49
A steel plate 1
4
in. thick is embedded in a
concrete wall to anchor a high-strength 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 = 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 end of the plate.)
SOLUTION
Based on tension in plate,
( )
( )
. .


 


 
U U
U U
A a d t
P A
P a d t
F S
P P
Solving for a,
 
1
4
( . .) 3 (3.60)(2.5)
4 (36)
U
F S P
a d
t

   
(a) 1.750 in.
a  
Based on shear between plate and concrete slab,
perimeter depth 2( ) 0.300 ksi
U
A a t b 
    
2 ( ) . . U
U U U
P
P A a t b F S
P
 
   
Solving for b,
 
1
4
( . .) (3.6)(2.5)
2( ) (2) 1.75 (0.300)
U
F S P
b
a t 
 
 
(b) 7.50 in.
b  
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56
a
b
P
3
4
in.
1
4
in.
PROBLEM 1.50
Determine the factor of safety for the cable
anchor in Prob. 1.49 when P  2.5 kips, knowing
that a  2 in. and b  6 in.
PROBLEM 1.49 A steel plate 1
4
in. thick is
embedded in a concrete wall to anchor a high-
strength 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 = 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 end of the plate.)
SOLUTION
Based on tension in plate,
2
( )
3 1
2 0.31250 in
4 4
A a d t
 
  
  
  
  
(36)(0.31250) 11.2500 kips
U U
P A


 
11.2500
. . 4.50
3.5
U
P
F S
P
  
Based on shear between plate and concrete slab,
1
perimeter depth 2( ) 2 2 (6.0)
4
A a t b
 
     
 
 
2
27.0 in 0.300 ksi
U
A 
 
(0.300)(27.0) 8.10 kips
U U
P A

  
8.10
. . 3.240
2.5
U
P
F S
P
  
Actual factor of safety is the smaller value. . . 3.24
F S  
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57
P
6 in.
8 in.
4 in.
1
2
in.
A
B C D
PROBLEM 1.51
Link AC is made of a steel with a 65-ksi ultimate normal stress and has
a 1 1
4 2
 -in. uniform rectangular cross section. It is connected to a
support at A and to member BCD at C by 3
4
-in.-diameter pins, while
member BCD is connected to its support at B by a 5
16
-in.-diameter pin.
All of the pins are made of a steel with a 25-ksi ultimate shearing stress
and are in single shear. Knowing that a factor of safety of 3.25 is
desired, determine the largest load P that can be applied at D. Note that
link AC is not reinforced around the pin holes.
SOLUTION
Use free body BCD.
8
0 : (6) 10 0
10
B AC
M F P
 
  
 
 
0.48 AC
P F
 (1)
6
0 : 0
10
x x AC
F B F
   
6
1.25
10
 
x AC
B F P
0 : 6 4 0
   
C y
M B P
2 2
i.e.
3 3
y y
B P B P
  
2
2 2 2 2
1.25 1.41667 0.70588
3
 
     
 
 
x y
B B B P P P B (2)
Shear in pins at A and C.
2
2
pin
25 3
0.84959 kips
. . 4 3.25 4 8
U
AC
F A d
F S
  

   
   
   
   
Tension on net section of A and C.
net net
65 1 1 3
0.625 kips
. . 3.25 4 2 8
U
AC
F A A
F S


   
    
   
   
Smaller value of FAC is 0.625 kips.
From (1), (0.48)(0.625) 0.300 kips
P  
Shear in pin at B.
2
2
pin
25 5
0.58999 kips
. . 4 3.25 4 16
U
B A d
F S
  

   
   
   
   
From (2), (0.70588)(0.58999) 0.416 kips
P  
Allowable value of P is the smaller value. 0.300 kips
P  or 300 lb
P  
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58
P
6 in.
8 in.
4 in.
1
2
in.
A
B C D
PROBLEM 1.52
Solve Prob. 1.51, assuming that the structure has been redesigned to use
5
16
-in-diameter pins at A and C as well as at B and that no other changes
have been made.
PROBLEM 1.51 Link AC is made of a steel with a 65-ksi ultimate
normal stress and has a 1 1
4 2
 -in. uniform rectangular cross section. It is
connected to a support at A and to member BCD at C by 3
4
-in.-diameter
pins, while member BCD is connected to its support at B by a 5
16
-in.-
diameter pin. All of the pins are made of a steel with a 25-ksi ultimate
shearing stress and are in single shear. Knowing that a factor of safety of
3.25 is desired, determine the largest load P that can be applied at D.
Note that link AC is not reinforced around the pin holes.
SOLUTION
Use free body BCD.
8
0 : (6) 10 0
10
B AC
M F P
 
  
 
 
0.48 AC
P F
 (1)
6
0 : 0
10
y x AC
F B F
   
6
1.25
10
 
x AC
B F P
0 : 6 4 0
C y
M B P
   
2 2
i.e.
3 3
y y
B P B P
  
2
2 2 2 2
1.25 1.41667 0.70583
3
x y
B B B P P P B
 
     
 
 
(2)
Shear in pins at A and C.
2
2
pin
25 5
0.58999 kips
. . 4 3.25 4 16
U
AC
F A d
F S
  

   
   
   
   
Tension on net section of A and C.
net net
65 1 1 5
0.9375 kips
. . 3.25 4 2 16
U
AC
F A A
F S


   
    
   
   
Smaller value of FAC is 0.58999 kips.
From (1), (0.48)(0.58999) 0.283 kips
P  
Shear in pin at B.
2
2
pin
25 5
0.58999 kips
. . 4 3.25 4 16
U
B A d
F S
  

   
   
   
   
From (2), (0.70588)(0.58999) 0.416 kips
P  
Allowable value of P is the smaller value. 0.283 kips
P  or 283 lb
P  
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59
24 kN
C
A
B
E
D
F G
250 mm
400 mm
250 mm
PROBLEM 1.53
Each of the two vertical links CF connecting the two horizontal
members AD and EG has a 10  40-mm uniform rectangular cross
section and is made of a steel with an ultimate strength in tension of
400 MPa, while each of the pins at C and F has a 20-mm diameter and
are made of a steel with an ultimate strength in shear of 150 MPa.
Determine the overall factor of safety for the links CF and the pins
connecting them to the horizontal members.
SOLUTION
3
3
0 : 0.40 (0.65)(24 10 ) 0
39 10 N
E CF
CF
M F
F
    
 
Based on tension in links CF,
6 2
6 6 3
( ) (0.040 0.02)(0.010) 200 10 m (one link)
2 (2)(400 10 )(200 10 ) 160.0 10 N
U U
A b d t
F A



     
     
Based on double shear in pins,
2 2 6 2
6 6 3
(0.020) 314.16 10 m
4 4
2 (2)(150 10 )(314.16 10 ) 94.248 10 N
U U
A d
F A
 



   
     
Actual FU is smaller value, i.e. 3
94.248 10 N
U
F  
Factor of safety:
3
3
94.248 10
. .
39 10
U
CF
F
F S
F

 

. . 2.42
F S  
Mecánica de Materiales - Ferdinand P. Beer 7ma. Edición.

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60
24 kN
C
A
B
E
D
F G
250 mm
400 mm
250 mm
PROBLEM 1.54
Solve Prob. 1.53, assuming that the pins at C and F have been replaced
by pins with a 30-mm diameter.
PROBLEM 1.53 Each of the two vertical links CF connecting the two
horizontal members AD and EG has a 10  40-mm uniform rectangular
cross section and is made of a steel with an ultimate strength in tension of
400 MPa, while each of the pins at C and F has a 20-mm diameter and
are made of a steel with an ultimate strength in shear of 150 MPa.
Determine the overall factor of safety for the links CF and the pins
connecting them to the horizontal members.
SOLUTION
Use member EFG as free body.
3
3
0 : 0.40 (0.65)(24 10 ) 0
39 10 N
E CF
CF
M F
F
    
 
Based on tension in links CF,
6 2
6 6 3
( ) (0.040 0.030)(0.010) 100 10 m (one link)
2 (2)(400 10 )(100 10 ) 80.0 10 N
U U
A b d t
F A



     
     
Based on double shear in pins,
2 2 6 2
6 6 3
(0.030) 706.86 10 m
4 4
2 (2)(150 10 )(706.86 10 ) 212.06 10 N
U U
A d
F A
 



   
     
Actual FU is smaller value, i.e. 3
80.0 10 N
U
F  
Factor of safety:
3
3
80.0 10
. .
39 10
U
CF
F
F S
F

 

. . 2.05
F S  
Mecánica de Materiales - Ferdinand P. Beer 7ma. Edición.

100. PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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on a website, in whole or part.
61
180 mm
200 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 C
A
P
PROBLEM 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 ultimate
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.
SOLUTION
Statics: Use ABC as free body.
10
0 : 0.20 0.18 0
9
    
B A A
M F P P F
10
0 : 0.20 0.38 0
19
    
A BD BD
M F P P F
Based on double shear in pin A, 2 2 6 2
(0.008) 50.266 10 m
4 4
  
   
A d
6 6
3
3
2 (2)(100 10 )(50.266 10 )
3.351 10 N
. . 3.0
10
3.72 10 N
9
 
 
   
  
U
A
A
A
F
F S
P F
Based on double shear in pins at B and D, 2 2 6 2
(0.012) 113.10 10 m
4 4
  
   
A d
6 6
3
3
2 (2)(100 10 )(113.10 10 )
7.54 10 N
. . 3.0
10
3.97 10 N
19
 
 
   
  
U
BD
BD
A
F
F S
P F
Based on compression in links BD, for one link, 6 2
(0.020)(0.008) 160 10 m

  
A
6 6
3
3
2 (2)(250 10 )(160 10 )
26.7 10 N
. . 3.0
10
14.04 10 N
19
 
 
   
  
U
BD
BD
A
F
F S
P F
Allowable value of P is smallest,  3
3.72 10 N
 
P 3.72 kN

P 
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101. PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
62
180 mm
200 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 C
A
P
PROBLEM 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
specifications remain unchanged, determine
the allowable load P if an overall factor of
safety of 3.0 is desired.
PROBLEM 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 ultimate 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.
SOLUTION
Statics: Use ABC as free body.
10
0: 0.20 0.18 0
9
    
B A A
M F P P F
10
0: 0.20 0.38 0
19
    
A BD BD
M F P P F
Based on double shear in pin A, 2 2 6 2
(0.010) 78.54 10 m
4 4
  
   
A d
6 6
3
3
2 (2)(100 10 )(78.54 10 )
5.236 10 N
. . 3.0
10
5.82 10 N
9
 
 
   
  
U
A
A
A
F
F S
P F
Based on double shear in pins at B and D, 2 2 6 2
(0.012) 113.10 10 m
4 4
  
   
A d
6 6
3
3
2 (2)(100 10 )(113.10 10 )
7.54 10 N
. . 3.0
10
3.97 10 N
19
 
 
   
  
U
BD
BD
A
F
F S
P F
Based on compression in links BD, for one link, 6 2
(0.020)(0.008) 160 10 m

  
A
6 6
3
3
2 (2)(250 10 )(160 10 )
26.7 10 N
. . 3.0
10
14.04 10 N
19
 
 
   
  
U
BD
BD
A
F
F S
P F
Allowable value of P is smallest,  3
3.97 10 N
 
P 3.97 kN

P 
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63
1.8 m
2.4 m
A B
C PROBLEM 1.57
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 0.90
  and load factors
1.25
 
D and L
  1.6, determine the largest load that can be safely
placed on the platform. (b) What is the corresponding conventional
factor of safety for rod BC?
SOLUTION
1 2
1 2
3
0 : (2.4) 2.4 1.2
5
5 5
3 6
   
  
A
M P W W
P W W
For dead loading, 1 2
3
(40)(9.81) 392.4 N, (50)(9.81) 490.5 N
5 5
(392.4) (490.5) 1.0628 10 N
3 6
   
   
   
   
   
D
W W
P
For live loading, 1 2 0
W mg W
 
5
3

L
P mg
From which
3
5
 L
P
m
g
Design criterion:   
 
D D L L U
P P P
3 3
3
(0.90)(12 10 ) (1.25)(1.0628 10 )
1.6
5.920 10 N
 


   
 
 
U D D
L
L
P P
P
(a) Allowable load.
3
3 5.92 10
5 9.81


m 362 kg

m 
Conventional factor of safety:
3 3 3
1.0628 10 5.920 10 6.983 10 N
       
D L
P P P
(b)
3
3
12 10
. .
6.983 10

 

U
P
F S
P
. . 1.718
F S  
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64
P P PROBLEM 1.58
The Load and Resistance Factor Design method is to be used to select
the two cables that will raise and lower a platform supporting two
window washers. The platform weighs 160 lb and each of the window
washers is assumed to weigh 195 lb with equipment. 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 resistance factor 0.85
  and load factors
D
  1.2 and 1.5,
L
  determine the required minimum ultimate load of
one cable. (b) What is the corresponding conventional factor of safety
for the selected cables?
SOLUTION
  
 
D D L L U
P P P
(a)
 


 D D L L
U
P P
P
1 3
(1.2) 160 (1.5) 2 195
2 4
0.85
   
   
   
   
 629 lb

U
P 
Conventional factor of safety:
1
160 0.75 2 195 372.5 lb
2
       
D L
P P P
(b)
629
. .
372.5
 
U
P
F S
P
. . 1.689
F S  
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65
A
D
C
B
3 m
25 m
15 m
35 m
80 Mg
15 m
PROBLEM 1.59
In the marine crane shown, link CD is known to have a
uniform cross section of 50  150 mm. For the loading
shown, determine`` the normal stress in the central portion
of that link.
SOLUTION
Weight of loading: 2
(80 Mg)(9.81 m/s ) 784.8 kN
 
W
Free Body: Portion ABC.
0: (15 m) (28 m) 0
28 28
(784.8 kN)
15 15
1465 kN
  
 
 
 A CD
CD
CD
M F W
F W
F
3
6
1465 10 N
195.3 10 Pa
(0.050 m)(0.150 m)

 
    
CD
CD
F
A
195.3 MPa
CD
   
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66
B
A
C
0.5 in.
0.5 in.
1.8 in.
1.8 in.
45⬚
60⬚
5 kips
5 kips
PROBLEM 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.
SOLUTION
Use joint B as free body.
Law of Sines:
10
sin 45 sin60 sin95
 
  
AB BC
F F
7.3205 kips

AB
F
8.9658 kips

BC
F
Link AB is a tension member.
Minimum section at pin: 2
net (1.8 0.8)(0.5) 0.5 in
A   
(a) Stress in
net
7.3205
:
0.5
AB
AB
F
AB
A
   14.64 ksi
 
AB 
Link BC is a compression member.
Cross sectional area is 2
(1.8)(0.5) 0.9 in
A  
(b) Stress in BC:
8.9658
0.9

 
 
BC
BC
F
A
9.96 ksi
  
BC 
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67
B
A
C
0.5 in.
0.5 in.
1.8 in.
1.8 in.
45⬚
60⬚
5 kips
5 kips
PROBLEM 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.
PROBLEM 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.
SOLUTION
Use joint B as free body.
Law of Sines:
10
8.9658 kips
sin 45 sin60 sin95
  
  
AB BC
BC
F F
F
(a) Shearing stress in pin at C.
2
  BC
P
F
A
2 2 2
(0.8) 0.5026 in
4 4
8.9658
8.92
(2)(0.5026)
P
A d
 

  
  8.92 ksi
  
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68
PROBLEM 1.61 (Continued)
(b) Bearing stress at C in member BC.   BC
b
F
A
2
(0.5)(0.8) 0.4 in
8.9658
22.4
0.4

  
 
b
A td
22.4 ksi
 
b 
(c) Bearing stress at B in member BC.   BC
b
F
A
2
2 2(0.5)(0.8) 0.8 in
8.9658
11.21
0.8

  
 
b
A td
11.21 ksi
 
b 
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69
PROBLEM 1.62
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.
SOLUTION
At each bolt location the upper plate is pulled down by the tensile force Pb of the bolt. At the same time, the
spacer pushes that plate upward with a compressive force Ps in order to maintain equilibrium.

b s
P P
For the bolt, 2
4


 
b b
b
b b
F P
A d
or 2
4



b b b
P d
For the spacer, 2 2
4
( )


 

s s
s
s s b
P P
A d d
or 2 2
( )
4


 
s s s b
P d d
Equating b
P and ,
s
P
2 2 2
( )
4 4
200
1 1 (16)
130
b b s s b
b
s b
s
d d d
d d
 
 


 
   
   
   
 
 
25.2 mm

s
d 
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70
200 mm
80 mm
M
60 mm
B
A
C
P PROBLEM 1.63
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.
SOLUTION


Use piston, rod, and crank together as free body. Add wall reaction H
and bearing reactions Ax and Ay.
3
0 : (0.280 m) 1500 N m 0
5.3571 10 N
    
 
A
M H
H
Use piston alone as free body. Note that rod is a two-force member;
hence the direction of force FBC is known. Draw the force triangle
and solve for P and FBE by proportions.
2 2
3
200 60 208.81 mm
200
17.86 10 N
60
  
   
l
P
P
H
(a) 17.86 kN

P 
3
208.81
18.6436 10 N
60
BC
BC
F
F
H
   
Rod BC is a compression member. Its area is
2 6 2
450 mm 450 10 m

 
Stress:
3
6
6
18.6436 10
41.430 10 Pa
450 10
 
  
    

BC
BC
F
A
(b) 41.4 MPa
  
BC 
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71
60 lb
F
D
E
J
C D
B
A
8 in.
2 in.
4 in.
12 in.
4 in.
6 in.
␪
PROBLEM 1.64
Knowing that the link DE is 1
8
in. thick and 1 in. wide, determine
the normal stress in the central portion of that link when
(a) 0 ,
   (b) 90 .
  
SOLUTION
Use member CEF as a free body.
0 : 12 (8)(60 sin ) (16)(60 cos ) 0
40 sin 80 cos lb
 
 
     
  
C DE
DE
M F
F
2
1
(1) 0.125 in
8

 
 
 
 

DE
DE
DE
DE
A
F
A
(a) 0: 80 lb
   
DE
F
80
0.125



DE 640 psi
  
DE 
(b) 90 : 40 lb
    
DE
F
40
0.125



DE 320 psi
  
DE 
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72
D
A
C
B
b
1500 lb
750 lb
750 lb
4 in.
1 in. PROBLEM 1.65
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.
SOLUTION
(a) Maximum normal stress in the wood.
2
net
net
5
(1) 4 3.375 in
8
1500
444 psi
3.375

 
  
 
 
  
A
P
A
444 psi
  
(b) Distance b for  = 100 psi.
For sheared area see dotted lines.
2
1500
7.50 in.
2 (2)(1)(100)


 
  
P P
A bt
P
b
t
7.50 in.

b 
(c) Average bearing stress on the wood.
1500
2400 psi
5
(1)
8
    
 
 
 
b
b
P P
A dt
2400 psi
 
b 
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73
18 mm
Top view
Side view
Front view
160 mm 120 mm
6 mm
A
A
B
C
B
D
C
B
D
P
PROBLEM 1.66
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.
SOLUTION
Use free body ABC.
0 : 0.280 0.120 0
   
C BD
M P F
3
7
BD
P F
 (1)
0 : 0.160 0.120 0
B
M P C
   
3
4
P C
 (2)
Tension on net section of link BD:
6
3 3 3
net net
400 10
(6 10 )(18 10)(10 ) 6.40 10 N
. . 3

  
 

      
 
 
 
U
BD
F A A
F S
Shear in pins at B and D:
6
2 3 2 3
pin
150 10
(10 10 ) 3.9270 10 N
. . 4 3 4
  
 
 
  
     
  
  
 
U
BD
F A d
F S
Smaller value of FBD is 3
3.9270 10 N.

From (1), 3 3
3
(3.9270 10 ) 1.683 10 N
7
 
   
 
 
P
Shear in pin at C:
6
2 3 2 3
pin
150 10
2 2 (2) (6 10 ) 2.8274 10 N
. . 4 3 4
  
 
 
  
     
  
  
 
U
C A d
F S
From (2), 3 3
3
(2.8274 10 ) 2.12 10 N
4
 
   
 
 
P
Smaller value of P is allowable value. 3
1.683 10 N
 
P 1.683 kN

P 
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74
A
D
B
C
0.4 m
30⬚
40⬚
0.8 m
0.6 m
P
PROBLEM 1.67
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.
SOLUTION
Use member ABC as a free body, and note that member BD is a two-force member.
0 : ( cos40 )(1.2) ( sin 40 )(0.6)
( cos30 )(0.6)
( sin30 )(0.4) 0
1.30493 0.71962 0
c
BD
BD
BD
M P P
F
F
P F
    
 
  
 
3 3
3
3
3
1.81335 (1.81335)(16 10 ) 29.014 10 N
100 10 N
100 10
. .
29.014 10
    
 

 

BD
U
U
BD
F P
F
F
F S
F
. . 3.45

F S 
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75
P
L d
PROBLEM 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 all
 in the steel, and the average
allowable bond stress all
 between the concrete and the cylindrical
surface of the bar. (Neglect the normal stresses between the
concrete and the end of the bar.)
SOLUTION
For shear,
all all

  

 
A dL
P A dL
For tension, 2
2
all all
4
4


 

 
   
 
A d
P A d
Equating, 2
all all
4

  

dL d
Solving for L, min all all
/4
 

L d 
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76
A
1.25 in.
2.4 kips
2.0 in.
B
␪
PROBLEM 1.69
The two portions of member AB are glued together along a plane
forming an angle  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  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 the
normal and shearing stresses.)
SOLUTION
2
0 (2.0)(1.25) 2.50 in
A  
At the optimum angle, ( . .) ( . .)
 

F S F S
Normal stress:
0
2
, 2
0
cos
cos


 

  
U
U
A
P
P
A
0
,
2
( . .)
cos




 
U
U A
P
F S
P P
Shearing stress:
0
,
0
0
,
sin cos
sin cos
( . .)
sin cos
U
U
U
U
A
P
P
A
A
P
F S
P P




  
 

 
  
 
Equating,
0 0
2
sin cos
cos
 
 


U U
A A
P
P
Solving,
sin 1.3
tan 0.520
cos 2.5
 

 
   
U
U
(a) opt 27.5
   
(b) 0
2 2
(12.5)(2.50)
7.94 kips
cos cos 27.5
U
U
A
P


  

7.94
. .
2.4
 
U
P
F S
P
. . 3.31

F S 
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77
A
1.25 in.
2.4 kips
2.0 in.
B
␪
PROBLEM 1.70
The two portions of member AB are glued together along a plane
forming an angle  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  for which the factor of safety of the members is
at least 3.0.
SOLUTION
2
0 (2.0)(1.25) 2.50 in.
2.4 kips
( . .) 7.2 kips
U
A
P
P F S P
 

 
Based on tensile stress,
2
0
0
2
cos
(2.5)(2.50)
cos 0.86806
7.2
 



  
U
U
U
U
P
A
A
P
cos 0.93169
  21.3
   21.3
  
Based on shearing stress,
0 0
sin cos sin 2
2
   
 
U U
U
P P
A A
0
2 (2)(2.50)(1.3)
sin 2 0.90278
7.2

   
U
U
A
P
2 64.52
   32.3
   32.3
  
Hence, 21.3 32.3

    
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78
Element n
Element 1
Pn
P1
PROBLEM 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 magnitude Pi of this load being assumed positive if Pi is
directed downward as shown and negative otherwise. (a) Write a computer program
that can be 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 Problems 1.1 and 1.3.
SOLUTION
Force in element i:
It is the sum of the forces applied to that element and all lower ones:
1
i
i k
k
F P

 
Average stress in element i:
2
1
Area
4
Ave. stress

 

i i
i
i
A d
F
A
Program outputs:
Problem 1.1 Problem 1.3
Element Stress (MPa) Element Stress (ksi)
1 84.883 1 22.635
2 96.766 2 17.927

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79
0.2 m
0.25 m
0.4 m
20 kN
C
B
A
D
E
PROBLEM 1.C2
A 20-kN load is applied as shown to the horizontal member ABC.
Member ABC has a 10 50-mm
 uniform rectangular cross section
and is supported by four vertical links, each of 8 36-mm
 uniform
rectangular cross section. Each of the four pins at A, B, C, and D has the
same diameter 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,
(i) the maximum value of the average normal stress in the links
connecting pins B and D, (ii) the average normal stress in the links
connecting pins C and E, (iii) the average shearing stress in pin B,
(iv) the average shearing stress in pin C, (v) the average bearing stress
at B in member ABC, and (vi) the average bearing stress at C in member
ABC. (b) Check your program by comparing the values obtained for
d  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 thickness of member ABC
has been reduced from 10 to 8 mm.
SOLUTION
Forces in links.
F.B. diagram of ABC:
0: 2 ( ) ( ) 0
( )/2( ) (tension)
C BD
BD
M F BC P AC
F P AC BC
   

0: 2 ( ) ( ) 0
( )/2( ) (comp.)
B CE
CE
M F BC P AB
F P AB BC
   

(i) Link BD.
Thickness  L
t
( )
/
BD L L
BD BD BD
A t w d
F A

 
 
(iii) Pin B.
2
/( /4)
B BD
F d
 

(v) Bearing stress at B.
Thickness of member AC
AC t

Sig Bear /( )
BD AC
B F dt

(vi) Bearing stress at C.
Sig Bear /( )
CE AC
C F dt

(ii) Link CE.
Thickness  L
t
/
CE L L
CE CE CE
A t w
F A


 
(iv) Pin C.
2
/( /4)
C CE
F d
 

Shearing stress in ABC under Pin B.
( /2)
0: 2 2
2
B AC AC AC
y B BD
BD
AC
AC AC
F t w
F F F
F
w




  

P = 20 kN
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80
PROBLEM 1.C2 (Continued)
Program Outputs
Input data for Parts (a), (b), (c):
P  20 kN, AB  0.25 m, BC  0.40 m, AC  0.65 m,
TL  8 mm, WL  36 mm, TAC  10 mm, WAC  50 mm
(c) Answer: 16 mm 22 mm
d
  (c)
Check: For d  22 mm, Tau AC = 65 MPa < 90 MPa O.K.     
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81
PROBLEM 1.C2 (Continued)
Input data for Part (d): P  20 kN,
AB = 0.25 m, BC = 0.40 m,
AC = 0.65 m, TL = 8 mm, WL = 36 mm,
TAC  8 mm, WAC  50 mm
(d) Answer: 18 mm 22 mm
d
  (d)
Check: For d = 22 mm, Tau AC = 81.25 MPa < 90 MPa O.K.
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82
B
A
C
0.5 in.
0.5 in.
1.8 in.
1.8 in.
45⬚
60⬚
5 kips
5 kips
PROBLEM 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 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, (i) the
maximum value of the average normal stress in member AB, (ii) the
average normal stress in member BC, (iii) the average shearing stress
in pin A, (iv) the average shearing stress in pin C, (v) the average
bearing stress at A in member AB, (vi) the average bearing stress at C
in member BC, and (vii) the average bearing stress at B in member BC.
(b) Check your program by comparing the values obtained for
d  0.8 in. with the answers given for Problems 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 thickness and width of the two members are
changed, respectively, from 0.5 to 0.3 in. and from 1.8 to 2.4 in.
SOLUTION
Forces in members AB and BC.
Free body: Pin B.
From force triangle:
2
sin 45 sin60 sin75
2 (sin 45 /sin75 )
2 (sin60 /sin75 )
BC
AB
AB
BC
F
F P
F P
F P
 
  
  
  
(i) Max. ave. stress in AB.
Width  w
Thickness  t
( )
/
AB
AB AB AB
A w d t
F A

 

(iii) Pin A.
2
( /2)/( /4)
A AB
F d
 

(v) Bearing stress at A.
Sig Bear /
AB
A F dt

(vii) Bearing stress at B in member BC.
Sig Bear /2
BC
B F dt

(ii) Ave. stress in BC.
/
BC
BC BC BC
A wt
F A



(iv) Pin C.
2
( /2)/( /4)
C BC
F d
 

(vi) Bearing stress at C.
Sig Bear /
BC
C F dt

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83
PROBLEM 1.C3 (Continued)
Program Outputs
Input data for Parts (a), (b), (c):
P = 5 kips, w = 1.8 in., t = 0.5 in.
(c) Answer: 0.70 in. 1.10 in.
d
  (c)

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84
PROBLEM 1.C3 (Continued)
Input data for Part (d),
P = 5 kips, w  2.4 in., t  0.3 in.
(d) Answer: 0.85 in. 1.25 in.
d
  (d)
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85
a
b
A
D
B
C
12 in.
18 in.
15 in.
P
PROBLEM 1.C4
A 4-kip force P forming an angle  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  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  and  from 0 to 45, using increments in  and 
corresponding to 0.1 increments in tan  and tan  . (b) Check that
for any given value of , the maximum value of the factor of safety is
obtained for 38.66
   and explain why. (c) Determine the smallest
possible value of the factor of safety for 38.66 ,
   as well as the
corresponding value of , and explain the result obtained.
SOLUTION
(a) Draw F.B. diagram of ABC:
ult
0: ( sin )(1.5 in.) ( cos )(30 in.)
( cos )(15 in.) ( sin )(12 in.) 0
15 sin 30 cos
15 cos 12 sin
. . /
 
 
 
 
  
  




C
M P P
F F
F P
F S F F
ult
Output for 4 kips and 20 kips:
 
P F
(b) When 38.66°, tan 0.8
 
  and cable BD is perpendicular to the lever arm BC.
(c) . . 3.579
F S  for 26.6 ;
   P is perpendicular to the lever arm AC.
Note: The value . . 3.579
F S  is the smallest of the values of F.S. corresponding to 38.66
   and the
largest of those corresponding to 26.6 .
   The point 26.6 , 38.66
 
    is a “saddle point,” or
“minimax” of the function . . ( , ).
F S  
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86
P
a
b
P'
a
PROBLEM 1.C5
A load P is supported as shown by two wooden members of uniform rectangular cross
section that are joined by a simple glued scarf splice. (a) Denoting by U
 and ,
U

respectively, the ultimate strength of the joint in tension and in shear, write a
computer program which, for given values of a, b, P, U
 and ,
U
 expressed in either
SI or U.S. customary units, and for values of  from 5 to 85 at 5 intervals, can be
used to calculate (i) the normal stress in the joint, (ii) the shearing stress in the joint,
(iii) the factor of safety relative to failure in tension, (iv) the factor of safety relative
to failure in shear, and (v) the overall factor of safety for the glued joint. (b) Apply
this program, using the dimensions and loading of the members of Probs. 1.29 and
1.31, knowing that 150 psi
 
U and 214 psi
 
U for the glue used in Prob. 1.29, and
that 1.26 MPa
 
U and 1.50 MPa
 
U for the glue used in Prob. 1.31. (c) Verify in
each of these two cases that the shearing stress is maximum for 45 .
a  
SOLUTION
(i) and (ii) Draw the F.B. diagram of lower member:
0: cos 0 cos
 
     
x
F V P V P
0: sin 0

   
y
F F P sin
F P 

Area /sin
ab 

Normal stress: 2
( / ) sin
Area
F
P ab
 
 
Shearing stress: ( / ) sin cos
Area
V
P ab
  
 
(iii) F.S. for tension (normal stresses):
/
U
FSN  

(iv) F.S. for shear:
/
U
FSS  

(v) Overall F.S.:
F.S.  The smaller of FSN and FSS.
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87
PROBLEM 1.C5 (Continued)
Program Outputs
Problem 1.29
150 mm
75 mm
11 kN
1.26 MPa
1.50 MPa







U
U
a
b
P
ALPHA SIG (MPa) TAU (MPa) FSN FSS FS
5 0.007 0.085 169.644 17.669 17.669
10 0.029 0.167 42.736 8.971 8.971
15 0.065 0.244 19.237 6.136 6.136
20 0.114 0.314 11.016 4.773 4.773
25 0.175 0.375 7.215 4.005 4.005
30 0.244 0.423 5.155 3.543 3.543
35 0.322 0.459 3.917 3.265 3.265
40 0.404 0.481 3.119 3.116 3.116
45 0.489 0.489 2.577 3.068 2.577  (b), (c)
50 0.574 0.481 2.196 3.116 2.196 
55 0.656 0.459 1.920 3.265 1.920 
60 0.733 0.423 1.718 3.543 1.718 
65 0.803 0.375 1.569 4.005 1.569 
70 0.863 0.314 1.459 4.773 1.459 
75 0.912 0.244 1.381 6.136 1.381 
80 0.948 0.167 1.329 8.971 1.329 
85 0.970 0.085 1.298 17.669 1.298 
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88
PROBLEM 1.C5 (Continued)
Problem 1.31
5 in.
3 in.
1400 lb
150 psi
214 psi
U
U
a
b
P







ALPHA SIG (psi) TAU (psi) FSN FSS FS
5 0.709 8.104 211.574 26.408 26.408
10 2.814 15.961 53.298 13.408 13.408
15 6.252 23.333 23.992 9.171 9.171
20 10.918 29.997 13.739 7.134 7.134
25 16.670 35.749 8.998 5.986 5.986
30 23.333 40.415 6.429 5.295 5.295
35 30.706 43.852 4.885 4.880 4.880
40 38.563 45.958 3.890 4.656 3.890
45 46.667 46.667 3.214 4.586 3.214 
 (c)
50 54.770 45.958 2.739 4.656 2.739 
55 62.628 43.852 2.395 4.880 2.395 
60 70.000 40.415 2.143 5.295 2.143  (b)
65 76.663 35.749 1.957 5.986 1.957 
70 82.415 29.997 1.820 7.134 1.820 
75 87.081 23.333 1.723 9.171 1.723 
80 90.519 15.961 1.657 13.408 1.657 
85 92.624 8.104 1.619 26.408 1.619 


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89
180 mm
200 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 C
A
P
PROBLEM 1.C6
Member ABC is supported by a pin and
bracket at A and by two links, which 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 all
P for any given
values of (i) the diameter 1
d of the pin at A,
(ii) the common diameter d2 of the pins at B
and D, (iii) the ultimate normal stress U in
each of the two links, (iv) the ultimate shearing
stress U in each of the three pins, and (v) the
desired overall factor of safety F.S. (b) 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. (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 1 
d
2 15 mm,

d 110 MPa
 
U for aluminum links,
100 MPa
 
U for steel pins, and F.S.  3.2.
SOLUTION
(a) F.B. diagram of ABC:
200
0:
380
200
0:
180
  
  
A BD
B A
M P F
M P F
(i) 1
For given of Pin :
d A 2
1
2( / )( /4),
A U
F FS d
 
 1
200
180
A
P F

(ii) 2
For given of Pins and :
d B D 2
2
2( / )( /4),
BD U
F FS d
 
 2
200
380
BD
P F

(iii) For ultimate stress in links BD: 3
200
2( / )(0.02)(0.008),
380

 
BD U BD
F FS P F
(iv) For ultimate shearing stress in pins: 4
P is the smaller of 1
P and 2.
P
(v) For desired overall F.S.: 5
P is the smaller of 3
P and 4.
P
If 3 4
< ,
P P stress is critical in links.
If 4 3
<
P P and 1 2
< ,
P P stress is critical in Pin A.
If 4 3
P P
 and 2 1,
P P
 stress is critical in Pins B and D.
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90
PROBLEM 1.C6 (Continued)
Program Outputs
(b) Problem 1.55. Data: 1 2
8 mm, 12 mm, , 250 MPa, 100 MPa, . . 3.0
U U
d d F S
 
    
all 3.72 kN.
P  Stress in Pin A is critical. 
(c) Problem 1.56. Data: 1 2
10 mm, 12 mm, 250 MPa, 100 MPa, . . 3.0
 
    
U U
d d F S
all 3.97 kN.
P  Stress in Pins B and D is critical. 
(d) Data: 1 2 15mm, 110 MPa, 100 MPa, . . 3.2
U U
d d F S
 
    
all 5.79 kN.
P  Stress in links is critical. 
Mecánica de Materiales - Ferdinand P. Beer 7ma. Edición.

130. C
CH
HA
AP
PT
TE
ER
R 2
2
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93
PROBLEM 2.1
A nylon thread is subjected to a 8.5-N tension force. Knowing that 3.3 GPa

E and that the length of the
thread increases by 1.1%, determine (a) the diameter of the thread, (b) the stress in the thread.
SOLUTION
(a) Strain:
1.1
0.011
100
L

   
Stress: 9 6
(3.3 10 )(0.011) 36.3 10 Pa
 
    
E
P
A
 
Area: 9 2
6
8.5
234.16 10 m
36.3 10


   

P
A
Diameter:
9
6
4 (4)(234.16 10 )
546 10 m
 



   
A
d 0.546 mm
d  
(b) Stress: 36.3 MPa
  
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94
PROBLEM 2.2
A 4.8-ft-long steel wire of 1
4
-in.-diameter is subjected to a 750-lb tensile load. Knowing that E = 29 × 106
psi,
determine (a) the elongation of the wire, (b) the corresponding normal stress.
SOLUTION
(a) Deformation:
2
;
4
PL d
A
AE

  
Area:
2
2 2
(0.25 in.)
4.9087 10 in
4
A
 
  
2 2 6
(750 lb)(4.8 ft 12 in./ft)
(4.9087 10 in )(29 10 psi)
 


 
2
3.0347 10 in.
 
  0.0303 in.
  
(b) Stress:
P
A
 
Area: 2 2
(750lb)
(4.9087 10 in )
 


4
1.52790 10 psi
   15.28 ksi
  
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95
PROBLEM 2.3
An 18-m-long steel wire of 5-mm diameter is to be used in the manufacture of a prestressed concrete beam.
It is observed that the wire stretches 45 mm when a tensile force P is applied. Knowing that 200 GPa,
E 
determine (a) the magnitude of the force P, (b) the corresponding normal stress in the wire.
SOLUTION
(a) , or
PL AE
P
AE L

  
with 2 2 6 2
1 1
(0.005) 19.6350 10 m
4 4
  
   
A d
6 2 9 2
(0.045 m)(19.6350 10 m )(200 10 N/m )
9817.5 N
18 m

 
 
P
9.82 kN
P  
(b)
6
6 2
9817.5 N
500 10 Pa
19.6350 10 m
 
   

P
A
500 MPa
  
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96
PROBLEM 2.4
Two gage marks are placed exactly 250 mm apart on a 12-mm-diameter aluminum rod with E = 73 GPa and
an ultimate strength of 140 MPa. Knowing that the distance between the gage marks is 250.28 mm after a load
is applied, determine (a) the stress in the rod, (b) the factor of safety.
SOLUTION
(a) 0
L L
  
250.28 mm 250 mm
 
0.28 mm

0

 
L
0.28 mm
250 mm

4
1.11643 10
 
 
 E
9 4
(73 10 Pa)(1.11643 10 )

  
7
8.1760 10 Pa
 
81.8 MPa
  
(b) F.S. u



140 MPa
81.760 MPa

1.71233
 F.S. 1.712
 
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97
PROBLEM 2.5
An aluminum pipe must not stretch more than 0.05 in. when it is subjected to a tensile load. Knowing that
6
10.1 10
E   psi and that the maximum allowable normal stress is 14 ksi, determine (a) the maximum
allowable length of the pipe, (b) the required area of the pipe if the tensile load is 127.5 kips.
SOLUTION
(a)  
PL
AE
Thus,
6
3
(10.1 10 )(0.05)
14 10
EA E
L
P
 


  

36.1 in.

L 
(b)  
P
A
Thus,
3
3
127.5 10
14 10
P
A


 

2
9.11 in

A 
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98
PROBLEM 2.6
A control rod made of yellow brass must not stretch more than 3 mm when the tension in the wire is 4 kN.
Knowing that E = 105 GPa and that the maximum allowable normal stress is 180 MPa, determine (a) the
smallest diameter rod that should be used, (b) the corresponding maximum length of the rod.
SOLUTION
(a)
2
;
4
 
P d
A
A


Substituting, we have
2
3
6
3
4
4
4(4 10 N)
(180 10 Pa)
5.3192 10 m





  
 
 
 
 



 
P P
d
d
d
d
5.32 mm
d  
(b) ;

  
 
E
L
Substituting, we have




  
E
E L
L
9 3
6
(105 10 Pa)(3 10 m)
(180 10 Pa)

 


L 1.750m
L  
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99
PROBLEM 2.7
A steel control rod is 5.5 ft long and must not stretch more than 0.04 in. when a 2-kip tensile load is applied to it.
Knowing that 6
29 10 psi,
E   determine (a) the smallest diameter rod that should be used, (b) the
corresponding normal stress caused by the load.
SOLUTION
(a) 6
(2000 lb)(5.5 12 in.)
0.04 in.
(29 10 psi)
:
PL
AE A


 

2 2
1
0.113793 in
4
A d

 
0.38063 in.
d  0.381 in.
d  
(b) 2
2000 lb
17575.8 psi
0.113793 in
P
A
    17.58 ksi
  
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100
PROBLEM 2.8
A cast-iron tube is used to support a compressive load. Knowing that 6
10 10
 
E psi and that the maximum
allowable change in length is 0.025%, determine (a) the maximum normal stress in the tube, (b) the minimum
wall thickness for a load of 1600 lb if the outside diameter of the tube is 2.0 in.
SOLUTION
(a) 0.00025
100
L
 
 
;

  
 
E
L


  E
L
6
(10 10 psi)(0.00025)
  
3
2.50 10 psi
  
2.50 ksi
  
(b) 2
3
1600 lb
; 0.64 in
2.50 10 psi


    

P P
A
A
 
2 2
4

 
o i
A d d
2 2 4

 
i o
A
d d
2
2 2 2
4(0.64 in )
(2.0 in.) 3.1851 in

  
i
d
1.78469 in.
i
d
 
1 1
( ) (2.0 in. 1.78469 in.)
2 2
   
o i
t d d
0.107655 in.
t 
0.1077

t 
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101
PROBLEM 2.9
A 4-m-long steel rod must not stretch more than 3 mm and the normal stress must not exceed 150 MPa when
the rod is subjected to a 10-kN axial load. Knowing that 200 GPa,
E  determine the required diameter of
the rod.
SOLUTION
4 m
L 
3 6
9 3
3 10 m, 150 10 Pa
200 10 Pa, 10 10 N
E P
 

   
   
Stress:
3
6 2 2
6
10 10 N
66.667 10 m 66.667 mm
150 10 Pa



    

P
A
P
A


Deformation:
3
6 2 2
9 3
(10 10 )(4)
66.667 10 m 66.667 mm
(200 10 )(3 10 )






    
 
PL
AE
PL
A
E
The larger value of A governs: 2
66.667 mm
A 
2 4 4(66.667)
4
A
A d d

 
   9.21 mm
d  
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102
PROBLEM 2.10
A nylon thread is to be subjected to a 10-N tension. Knowing that 3.2
E  GPa, that the maximum allowable
normal stress is 40 MPa, and that the length of the thread must not increase by more than 1%, determine the
required diameter of the thread.
SOLUTION
Stress criterion:
6
9 2
6
9
2 6
40 MPa 40 10 Pa 10 N
10 N
: 250 10 m
40 10 Pa
250 10
: 2 2 564.19 10 m
4
P
P P
A
A
A
A d d




 



   
    


    
0.564 mm
d 
Elongation criterion:
1% 0.01
:
L
PL
AE


 

9
9 2
9
6 2
/ 10 N/3.2 10 Pa
312.5 10 m
/ 0.01
312.5 10
2 2 630.78 10 m
P E
A
L
A
d

 




   

   
0.631 mm
d 
The required diameter is the larger value: 0.631 mm
d  
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103
PROBLEM 2.11
A block of 10-in. length and 1.8 × 1.6-in. cross section is to support a centric compressive load P. The
material to be used is a bronze for which E  14 × 106
psi. Determine the largest load that can be applied,
knowing that the normal stress must not exceed 18 ksi and that the decrease in length of the block should be at
most 0.12% of its original length.
SOLUTION
Considering allowable stress, 3
18 ksi or 18 10 psi
 

Cross-sectional area: 2
(1.8 in.)(1.6 in.) 2.880 in
 
A
3 2
4
(18 10 psi)(2.880 in )
5.1840 10 lb
or 51.840 kips
  
 
 
P
P A
A
 
Considering allowable deformation, 0.12% or 0.0012 in.

L

2 6
4
(2.880 in )(14 10 psi)(0.0012 in.)
4.8384 10 lb
or 48.384 kips
 
    
 
 
 
PL
P AE
AE L
P


The smaller value for P resulting from the required deformation criteria governs.
48.4 kips 
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104
PROBLEM 2.12
A square yellow-brass bar must not stretch more than 2.5 mm when it is subjected to a tensile load. Knowing
that 105
E  GPa and that the allowable tensile strength is 180 MPa, determine (a) the maximum allowable
length of the bar, (b) the required dimensions of the cross section if the tensile load is 40 kN.
SOLUTION
6 3
9 3
180 10 Pa 40 10 N
105 10 Pa 2.5 10 m

 
   
   
P
E
(a)
9 3
6
(105 10 )(2.5 10 )
1.45833 m
180 10
PL L
AE E
E
L





 
 
  

1.458 m
L  
(b)
3
6 2 2
6
40 10
222.22 10 m 222.22 mm
180 10





    

P
A
P
A
2
222.22
A a a A
   14.91 mm
a  
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105
72 in.
54 in.
72 in.
B
A
C
D
P ⫽ 130 kips PROBLEM 2.13
Rod BD is made of steel 6
( 29 10 psi)
E   and is used to brace the axially
compressed member ABC. The maximum force that can be developed in member
BD is 0.02P. If the stress must not exceed 18 ksi and the maximum change in
length of BD must not exceed 0.001 times the length of ABC, determine the
smallest-diameter rod that can be used for member BD.
SOLUTION
3
0.02 (0.02)(130) 2.6 kips 2.6 10 lb
    
BD
F P
Considering stress, 3
18 ksi 18 10 psi
   
2
2.6
0.14444 in
18
    
BD BD
F F
A
A


Considering deformation, (0.001)(144) 0.144 in.
  
3
2
6
(2.6 10 )(54)
0.03362 in
(29 10 )(0.144)



    

BD BD BD BD
F L F L
A
AE E
Larger area governs. 2
0.14444 in

A
2 4 (4)(0.14444)
4

 
   
A
A d d 0.429 in.

d 
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106
3.5 m
4.0 m
2.5 m
B
A C
P
PROBLEM 2.14
The 4-mm-diameter cable BC is made of a steel with 200 GPa.

E Knowing
that the maximum stress in the cable must not exceed 190 MPa and that the
elongation of the cable must not exceed 6 mm, find the maximum load P that
can be applied as shown.
SOLUTION
2 2
6 4 7.2111 m
  
BC
L
Use bar AB as a free body.
4
0: 3.5 (6) 0
7.2111
0.9509
 
   
 
 

A BC
BC
M P F
P F
Considering allowable stress, 6
190 10 Pa
  
2 2 6 2
6 6 3
(0.004) 12.566 10 m
4 4
(190 10 )(12.566 10 ) 2.388 10 N
 
 


   
       
BC
BC
A d
F
F A
A
Considering allowable elongation, 3
6 10 m
 
 
6 9 3
3
(12.566 10 )(200 10 )(6 10 )
2.091 10 N
7.2111
BC BC
BC
BC
F L AE
F
AE L


 
  
     
Smaller value governs. 3
2.091 10 N
BC
F  
3 3
0.9509 (0.9509)(2.091 10 ) 1.988 10 N
BC
P F
     1.988 kN
P  
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107
P
1.25-in. diameter
4 ft
3 ft
d
A
B
C
PROBLEM 2.15
A single axial load of magnitude P = 15 kips is applied at end C of the steel
rod ABC. Knowing that E = 30 × 106
psi, determine the diameter d of portion
BC for which the deflection of point C will be 0.05 in.
SOLUTION
i
C
i i AB BC
PL PL PL
A E AE AE

   
  
   
   

4 ft 48 in.; 3 ft 36 in.
AB BC
L L
   
2 2
2
(1.25 in.)
1.22718 in
4 4
  
AB
d
A
 
Substituting, we have
3
6 2
15 10 lb 48 in. 36 in.
0.05 in.
30 10 psi 1.22718 in
  

 
  
 
  
  BC
A
2
0.59127 in
BC
A 
2
4
BC
d
A


4
 BC
A
or d

2
4(0.59127 in )
d


0.86766 in.
d 
0.868 in.
d  
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108
36 mm 28 mm
25 mm
250 mm
PROBLEM 2.16
A 250-mm-long aluminum tube ( 70 GPa)
E  of 36-mm outer
diameter and 28-mm inner diameter can be closed at both ends by
means of single-threaded screw-on covers of 1.5-mm pitch. With
one cover screwed on tight, a solid brass rod ( 105 GPa)
E  of
25-mm diameter is placed inside the tube and the second cover is
screwed on. Since the rod is slightly longer than the tube, it is
observed that the cover must be forced against the rod by rotating
it one-quarter of a turn before it can be tightly closed. Determine
(a) the average normal stress in the tube and in the rod, (b) the
deformations of the tube and of the rod.
SOLUTION
2 2 2 2 2 6 2
tube
2 2 2 6 2
rod
9
tube 9 6
tube tube
rod 6
rod rod
( ) (36 28 ) 402.12 mm 402.12 10 m
4 4
(25) 490.87 mm 490.87 10 m
4 4
(0.250)
8.8815 10
(70 10 )(402.12 10 )
(0.250)
(105 10 )(490.87 1
 
 






      
    
   
 
  
 
o i
A d d
A d
PL P
P
E A
PL P
E A
9
6
* 6
* *
tube rod tube rod
9 9 6
3
3
9
4.8505 10
0 )
1
turn 1.5 mm 0.375 mm 375 10 m
4
or
8.8815 10 4.8505 10 375 10
0.375 10
27.308 10 N
(8.8815 4.8505)(10 )

     



  


  
 
    
 
 
   
    

  

P
P P
P
(a)
3
6
tube 6
tube
27.308 10
67.9 10 Pa
402.12 10
 

   

P
A
tube 67.9 MPa
  
3
6
rod 6
rod
27.308 10
55.6 10 Pa
490.87 10
 

      

P
A
rod 55.6 MPa
   
(b) 9 3 6
tube (8.8815 10 )(27.308 10 ) 242.5 10 m
  
     tube 0.243 mm
  
9 3 6
rod (4.8505 10 )(27.308 10 ) 132.5 10 m
  
       rod 0.1325 mm
   
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109
P 5 350 lb
A B C D
1 in. 1 in.
1.6 in. 2 in.
0.4 in.
1.6 in.
P 5 350 lb
PROBLEM 2.17
The specimen shown has been cut from a 1
4
-in.-thick sheet
of vinyl (E = 0.45 × 106
psi) and is subjected to a 350-lb
tensile load. Determine (a) the total deformation of the
specimen, (b) the deformation of its central portion BC.
SOLUTION
3
6
(350 lb)(1.6 in.)
4.9778 10 in.
(0.45 10 psi)(1in.)(0.25 in.)
 
   

AB
AB
AB
PL
EA
3
6
(350 lb)(2 in.)
15.5556 10 in.
(0.45 10 psi)(0.4 in.)(0.25 in.)
 
   

BC
BC
BC
PL
EA
3
4.9778 10 in.
  
  
CD AB
(a) Total deformation:
  
AB BC CD
   
3
25.511 10 in.
 
 
3
25.5 10 in.
 
  
(b) Deformation of portion BC :
3
15.56 10 in.
BC
 
  
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110
375 mm
1 mm
C
D A
B
P PROBLEM 2.18
The brass tube ( 105 GPa)

AB E has a cross-sectional area of
140 mm2
and is fitted with a plug at A. The tube is attached at B to a
rigid plate that is itself attached at C to the bottom of an aluminum
cylinder ( 72 GPa)
E  with a cross-sectional area of 250 mm2
. The
cylinder is then hung from a support at D. In order to close the
cylinder, the plug must move down through 1 mm. Determine the
force P that must be applied to the cylinder.
SOLUTION
Shortening of brass tube AB:
2 6 2
9
9
9 6
375 1 376 mm 0.376 m 140 mm 140 10 m
105 10 Pa
(0.376)
25.578 10
(105 10 )(140 10 )




      
 
   
 
AB AB
AB
AB
AB
AB AB
L A
E
PL P
P
E A
Lengthening of aluminum cylinder CD:
2 6 2 9
9
9 6
0.375 m 250 mm 250 10 m 72 10 Pa
(0.375)
20.833 10
(72 10 )(250 10 )




     
   
 
CD CD CD
CD
CD
CD CD
L A E
PL P
P
E A
Total deflection: A AB CD
  
  where 0.001 m
A
 
9 9
0.001 (25.578 10 20.833 10 )
 
    P
3
21.547 10 N
P   21.5 kN
P  
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111
0.4 m
0.5 m
P
Q
20-mm diameter
60-mm diameter
A
B
C
PROBLEM 2.19
Both portions of the rod ABC are made of an aluminum for which 70 GPa.
E 
Knowing that the magnitude of P is 4 kN, determine (a) the value of Q so that
the deflection at A is zero, (b) the corresponding deflection of B.
SOLUTION
(a) 2 2 6 2
(0.020) 314.16 10 m
4 4
AB AB
A d
  
   
2 2 3 2
(0.060) 2.8274 10 m
4 4
BC BC
A d
  
   
Force in member AB is P tension.
Elongation:
3
6
9 6
(4 10 )(0.4)
72.756 10 m
(70 10 )(314.16 10 )
 


   
 
AB
AB
AB
PL
EA
Force in member BC is Q  P compression.
Shortening:
9
9 3
( ) ( )(0.5)
2.5263 10 ( )
(70 10 )(2.8274 10 )
 

 
    
 
BC
BC
BC
Q P L Q P
Q P
EA
For zero deflection at A, BC AB
 

9 6 3
2.5263 10 ( ) 72.756 10 28.8 10 N
 
       
Q P Q P
3 3 3
28.3 10 4 10 32.8 10 N
Q       32.8 kN

Q 
(b) 6
72.756 10 m
AB BC B
   
    0.0728 mm
  
AB 
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112
0.4 m
0.5 m
P
Q
20-mm diameter
60-mm diameter
A
B
C
PROBLEM 2.20
The rod ABC is made of an aluminum for which 70 GPa.
E  Knowing that
6 kN

P and 42 kN,

Q determine the deflection of (a) point A, (b) point B.
SOLUTION
2 2 6 2
2 2 3 2
(0.020) 314.16 10 m
4 4
(0.060) 2.8274 10 m
4 4
AB AB
BC BC
A d
A d
 
 


   
   
3
3 3 3
6 10 N
6 10 42 10 36 10 N
0.4 m 0.5 m
  
        
 
AB
BC
AB BC
P P
P P Q
L L
3
6
6 9
3
6
3 9
(6 10 )(0.4)
109.135 10 m
(314.16 10 )(70 10 )
( 36 10 )(0.5)
90.947 10 m
(2.8274 10 )(70 10 )
AB AB
AB
AB A
BC BC
BC
BC
P L
A E
P L
A E







   
 
 
    
 
(a) 6 6 6
109.135 10 90.947 10 m 18.19 10 m
A AB BC
     
        0.01819 mm
A
   
(b) 6
90.9 10 m 0.0909 mm
B BC
  
      or 0.0909 mm
  
B 
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113
4.0 m 4.0 m
2.5 m
D C
A
B
228 kN PROBLEM 2.21
For the steel truss ( 200 GPa)
E  and loading shown, determine
the deformations of the members AB and AD, knowing that their
cross-sectional areas are 2400 mm2
and 1800 mm2
, respectively.
SOLUTION
Statics: Reactions are 114 kN upward at A and C.
Member BD is a zero force member.
2 2
4.0 2.5 4.717 m
  
AB
L
Use joint A as a free body.
2.5
0: 114 0
4.717
y AB
F F
   
215.10 kN
AB
F  
4
0: 0
4.717
x AD AB
F F F
   
(4)( 215.10)
182.4 kN
4.717
AD
F

  
Member AB:
3
9 6
( 215.10 10 )(4.717)
(200 10 )(2400 10 )
AB AB
AB
AB
F L
EA
 
 
 
 
3
2.11 10 m

   2.11 mm
  
AB 
Member AD:
3
9 6
(182.4 10 )(4.0)
(200 10 )(1800 10 )
AD AD
AD
AD
F L
EA
 

 
 
3
2.03 10 m

  2.03 mm
 
AD 
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114
15 ft
8 ft
8 ft
8 ft
D
C
F
E
G
A
B
30 kips
30 kips
30 kips
PROBLEM 2.22
For the steel truss 6
( 29 10 psi)
E   and loading shown, determine the
deformations of the members BD and DE, knowing that their cross-
sectional areas are 2 in2
and 3 in2
, respectively.
SOLUTION
Free body: Portion ABC of truss
0: (15 ft) (30 kips)(8 ft) (30 kips)(16 ft) 0
48.0 kips
E BD
BD
M F
F
    
 
Free body: Portion ABEC of truss
0: 30 kips 30 kips 0
60.0 kips
    
 
x DE
DE
F F
F

3
2 6
( 48.0 10 lb)(8 12 in.)
(2 in )(29 10 psi)

  
 

BD
PL
AE
3
79.4 10 in.
 
  
BD 

3
2 6
( 60.0 10 lb)(15 12 in.)
(3 in )(29 10 psi)

  
 

DE
PL
AE
3
124.1 10 in.
 
  
DE 
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115
6 ft 6 ft
5 ft
C
D E
A
B
28 kips 54 kips
PROBLEM 2.23
Members AB and BC are made of steel 6
( 29 10 psi)
E   with cross-
sectional areas of 0.80 in2
and 0.64 in2
, respectively. For the loading
shown, determine the elongation of (a) member AB, (b) member BC.
SOLUTION
(a) 2 2
6 5 7.810 ft 93.72 in.
   
AB
L
Use joint A as a free body.
3
5
0: 28 0
7.810
43.74 kip 43.74 10 lb
   
  
y AB
AB
F F
F
3
6
(43.74 10 )(93.72)
(29 10 )(0.80)


 

AB AB
AB
AB
F L
EA
0.1767 in.
 
AB 
(b) Use joint B as a free body.
3
6
0: 0
7.810
(6)(43.74)
33.60 kip 33.60 10 lb
7.810
   
   
x BC AB
BC
F F F
F
3
6
(33.60 10 )(72)
(29 10 )(0.64)


 

BC BC
BC
BC
F L
EA
0.1304 in.
BC
  
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116
6 m
5 m
C
D
A
B
P PROBLEM 2.24
The steel frame ( 200 GPa)
E  shown has a diagonal brace BD with
an area of 1920 mm2
. Determine the largest allowable load P if the
change in length of member BD is not to exceed 1.6 mm.
SOLUTION
3 2 6 2
2 2 9
9 6 3
3
1.6 10 m, 1920 mm 1920 10 m
5 6 7.810 m, 200 10 Pa
(200 10 )(1920 10 )(1.6 10 )
7.81
78.67 10 N



 
 
    
    

  
 
 
BD BD
BD BD
BD BD
BD
BD BD
BD BD BD
BD
BD
A
L E
F L
E A
E A
F
L
Use joint B as a free body. 0:
x
F
 
5
0
7.810
 
BD
F P
3
3
5 (5)(78.67 10 )
7.810 7.810
50.4 10 N

 
 
BD
P F
50.4 kN

P 
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117
P
125 mm
225 mm
225 mm
150 mm
E
D
A B
C
PROBLEM 2.25
Link BD is made of brass ( 105 GPa)
E  and has a cross-sectional area of
240 mm2
. Link CE is made of aluminum ( 72 GPa)
E  and has a cross-
sectional area of 300 mm2
. Knowing that they support rigid member ABC,
determine the maximum force P that can be applied vertically at point A if
the deflection of A is not to exceed 0.35 mm.
SOLUTION
Free body member AC:
0: 0.350 0.225 0
1.55556
   

C BD
BD
M P F
F P
0: 0.125 0.225 0
0.55556
   

B CE
CE
M P F
F P
9
9 6
9
9 6
(1.55556 )(0.225)
13.8889 10
(105 10 )(240 10 )
(0.55556 )(0.150)
3.8581 10
(72 10 )(300 10 )
 
 




    
 
    
 
BD BD
B BD
BD BD
CE CE
C CE
CE CE
F L P
P
E A
F L P
P
E A
Deformation Diagram:
From the deformation diagram,
Slope:
9
9
17.7470 10
78.876 10
0.225
 



 
   
B C
BC
P
P
L
9 9
9
13.8889 10 (0.125)(78.876 10 )
23.748 10
  
 

 
   
 
A B AB
L
P P
P
Apply displacement limit. 3 9
0.35 10 m 23.748 10
  
   
A P
3
14.7381 10 N
P   14.74 kN
P  
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118
260 mm
18 kN 18 kN
240 mm
180 mm
C
D
E
F
A
B
PROBLEM 2.26
Members ABC and DEF are joined with steel links (E  200 GPa).
Each of the links is made of a pair of 25 × 35-mm plates. Determine
the change in length of (a) member BE, (b) member CF.
SOLUTION
Free body diagram of Member ABC:
0:
 
B
M
(0.26 m)(18 kN) (0.18 m) 0
 
CF
F
26.0 kN
CF
F 
0:
 
x
F
18 kN 26.0 kN 0
  
BE
F
44.0 kN
 
BE
F
Area for link made of two plates:
3 2
2(0.025 m)(0.035 m) 1.750 10 m

  
A
(a)
3
9 3 2
6
( 44.0 10 N)(0.240 m)
(200 10 Pa)(1.75 10 m )
30.171 10 m
 

 
 
 
  
BE
BE
F L
EA
0.0302 mm
BE
   
(b)
3
9 3 2
6
(26.0 10 N)(0.240 m)
(200 10 Pa)(1.75 10 m )
17.8286 10 m
 


 
 
 
BF
CF
F L
EA
0.01783 mm
CF
  
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119
P = 1 kip
10 in.
22 in.
18 in.
A
E
D
B C
PROBLEM 2.27
Each of the links AB and CD is made of aluminum 6
( 10.9 10 psi)
 
E
and has a cross-sectional area of 0.2 in2
. Knowing that they support the
rigid member BC, determine the deflection of point E.
SOLUTION
Free body BC:
3
0: (32) (22)(1 10 ) 0
687.5 lb
     

C AB
AB
M F
F
3
0: 687.5 1 10 0
312.5 lb
     

y CD
CD
F F
F
3
6
3
6
(687.5)(18)
5.6766 10 in.
(10.9 10 )(0.2)
(312.5)(18)
2.5803 10 in.
(10.9 10 )(0.2)
 
 


    

    

AB AB
AB B
CD CD
CD C
F L
EA
F L
EA
Deformation diagram:
3
6
3.0963 10
Slope
32
96.759 10 rad
 



 
 
 
B C
BC
L
3 6
3
2.5803 10 (22)(96.759 10 )
4.7090 10 in.
  
 

 
   
 
E C EC
L
3
4.71 10 in.
E
 
   
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120
12.5 in.
D
C
A
x
B
50 lb
16 in.
4 in.
E
1
16 in.
PROBLEM 2.28
The length of the 3
32
-in.-diameter steel wire CD has been adjusted so that
with no load applied, a gap of 1
16
in. exists between the end B of the rigid
beam ACB and a contact point E. Knowing that 6
29 10 psi,
E  
determine where a 50-lb block should be placed on the beam in order to
cause contact between B and E.
SOLUTION
Rigid beam ACB rotates through angle  to close gap.
3
1/16
3.125 10 rad
20
 
  
Point C moves downward.
3 3
3
2
2 3 2
4 4(3.125 10 ) 12.5 10 in.
12.5 10 in.
3
6.9029 10 in
4 32
C
CD C
CD
CD CD
CD
CD
A d
d
F L
EA
 
 
 

 


    
  
 
   
 
 

6 3 3
(29 10 )(6.9029 10 )(12.5 10 )
12.5
200.18 lb
  
  
 

CD CD
CD
CD
EA
F
L
Free body ACB:
0: 4 (50)(20 ) 0
    
A CD
M F x
(4)(200.18)
20 16.0144
50
3.9856 in.
  

x
x
For contact, 3.99 in.
x  
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121
PROBLEM 2.29
A homogeneous cable of length L and uniform cross section is suspended from one end. (a) Denoting by  the
density (mass per unit volume) of the cable and by E its modulus of elasticity, determine the elongation of the
cable due to its own weight. (b) Show that the same elongation would be obtained if the cable were horizontal
and if a force equal to half of its weight were applied at each end.
SOLUTION
(a) For element at point identified by coordinate y,
2
0
0
weight of portion below the point
( )
( ) ( )
( ) 1
2

 

 


 
 
  
  
  
 
 

L
L
P
gA L y
Pdy gA L y dy g L y
d dy
EA EA E
g L y g
dy Ly y
E E
2
2
2
  
 
 
 
 
g L
L
E
2
1
2

 
gL
E

(b) Total weight: 

W gAL
2
1 1
2 2
 

   
EA EA gL
F gAL
L L E
1
2
F W
 
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122
h
A a
b
P PROBLEM 2.30
A vertical load P is applied at the center A of the upper section of a homogeneous
frustum of a circular cone of height h, minimum radius a, and maximum radius b.
Denoting by E the modulus of elasticity of the material and neglecting the effect
of its weight, determine the deflection of point A.
SOLUTION
Extend the slant sides of the cone to meet at a point O and place the origin of the coordinate system there.
From geometry, tan
b a
h



1 1
, , tan
tan tan

 
  
a b
a b r y
At coordinate point y, 2
A r


Deformation of element of height dy:  
Pdy
d
AE
2 2 2
tan

  
 
P dy P dy
d
E r E y
Total deformation:
1
1
1
1
2 2 2 2
1 1
1 1 1
tan tan tan
b
b
A
a
a
P dy P P
y a b
E y E E

     
 
 
    
 
 
   

1 1 1 1
2
1 1
( )
tan
b a P b a
P
a b Eab
E 
 
 
  A
Ph
Eab


  
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123
PROBLEM 2.31
Denoting by  the “engineering strain” in a tensile specimen, show that the true strain is ln(1 ).
 
 
t
SOLUTION
0
0 0 0
ln ln ln 1 ln(1 )
 
 
 

     
 
 
t
L
L
L L L
Thus, ln(1 )
 
 
t 
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124
PROBLEM 2.32
The volume of a tensile specimen is essentially constant while plastic deformation occurs. If the initial
diameter of the specimen is d1, show that when the diameter is d, the true strain is 1
2 ln( / ).
 
t d d
SOLUTION
If the volume is constant, 2 2
1 0
4 4
d L d L
 

2
2
1 1
2
0
2
1
0
ln ln

 
   
 
 
   
 
t
d d
L
L d
d
d
L
L d
1
2ln
 
t
d
d

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125
40 mm
60 mm
Aluminum plates
(E = 70 GPa)
300 mm
Brass core
(E = 105 GPa) Rigid
end plate
P
h
h
PROBLEM 2.33
An axial centric force of magnitude P  450 kN is applied
to the composite block shown by means of a rigid end
plate. Knowing that h  10 mm, determine the normal
stress in (a) the brass core, (b) the aluminum plates.
SOLUTION
;
A B A B
P P P
  
   
and
 
A B
A A B B
P L P L
E A E A
 
Therefore, ( ) ; ( )
   
 
   
   
A A A B B B
P E A P E A
L L
 
Substituting,  
A A A B B
P E A E A
L

 
   
 
 
A A B B
P
L E A E A

 

3
9 9
3
(450 10 N)
(70 10 Pa)(2)(0.06 m)(0.01m) (105 10 Pa)(0.06 m)(0.04 m)
1.33929 10


  
 
Now,   
E
(a) Brass-core: 9 3
8
(105 10 Pa)(1.33929 10 )
1.40625 10 Pa

  
 
B

140.6 MPa
B
  
(b) Aluminum: 9 3
7
(70 10 Pa)(1.33929 10 )
9.3750 10 Pa
A
 
  
 
93.8 MPa
A
  
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126
40 mm
60 mm
Aluminum plates
(E = 70 GPa)
300 mm
Brass core
(E = 105 GPa) Rigid
end plate
P
h
h
PROBLEM 2.34
For the composite block shown in Prob. 2.33, determine
(a) the value of h if the portion of the load carried by the
aluminum plates is half the portion of the load carried by
the brass core, (b) the total load if the stress in the brass is
80 MPa.
PROBLEM 2.33. An axial centric force of magnitude
P  450 kN is applied to the composite block shown by
means of a rigid end plate. Knowing that h  10 mm,
determine the normal stress in (a) the brass core, (b) the
aluminum plates.
SOLUTION
;
a b a b
P P P
  
   
and
a b
a a b b
P L P L
E A E A
 
 
Therefore,
( ) ; ( )
a a a b b b
P E A P E A
L L
 
 
   
 
(a)
1
2
a b
P P

1
( ) ( )
2
a a b b
E A E A
L L
 
   

   
   
1
2
b
a b
a
E
A A
E
 
  
 
1 105 GPa
(40 mm)(60 mm)
2 70 GPa
 
  
 
a
A
2
1800 mm
a
A 
2
1800 mm 2(60 mm)( )
 h
15.00 mm
h  
(b)
1
and
2
   
b
b b b b a b
b
P
P A P P
A
 
a b
P P P
 
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127
PROBLEM 2.34 (Continued)
1
( )
2
b b b b
P A A
 
 
( )1.5
b b
P A


6
(80 10 Pa)(0.04 m)(0.06 m)(1.5)
 
P
5
2.880 10 N
 
P
288 kN
P  
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128
4.5 ft
18 in.
P PROBLEM 2.35
The 4.5-ft concrete post is reinforced with six steel bars, each with a 1
8
1 -in. diameter.
Knowing that Es  29 × 106
psi and Ec = 4.2 × 106
psi, determine the normal stresses
in the steel and in the concrete when a 350-kip axial centric force P is applied to the
post.
SOLUTION
Let portion of axial force carried by concrete.
portion carried by the six steel rods.


c
s
P
P
( )
c c c
c
c c
s s s
s
s s
c s c c s s
c c s s
P L E A
P
E A L
P L E A
P
E A L
P P P E A E A
L
P
L E A E A







 
 
   

 

2 2 2
2 2 2
2
3
4
6 2 6 2
6
6 (1.125 in.) 5.9641in
4 4
(18 in.) 5.9641 in
4 4
248.51in
4.5 ft 54 in.
350 10 lb
2.8767 10
(4.2 10 psi)(248.51 in ) (29 10 psi)(5.9641in )

  
   

 
 
   
  
s s
c c s
A d
A d A
L
 
 

6 4
(29 10 psi)( 2.8767 10 ) 8.3424 10 psi
s s
E
   
        s  8.34 ksi 
6 4 3
(4.2 10 psi)( 2.8767 10 ) 1.20821 10 psi

      
c c
E
  c  1.208 ksi 
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129
4.5 ft
18 in.
P
PROBLEM 2.36
For the post of Prob. 2.35, determine the maximum centric force that can be applied if the
allowable normal stress is 20 ksi in the steel and 2.4 ksi in the concrete.
PROBLEM 2.35 The 4.5-ft concrete post is reinforced with six steel bars, each with
a 1
8
1 -in. diameter. Knowing that Es  29 × 106
psi and Ec  4.2 × 106
psi, determine
the normal stresses in the steel and in the concrete when a 350-kip axial centric force P
is applied to the post.
SOLUTION
Allowable strain in each material:
Steel:
3
4
6
20 10 psi
6.8966 10
29 10 psi

 

   

s
s
s
E
Concrete:
3
4
6
2.4 10 psi
5.7143 10
4.2 10 psi

 

   

c
c
c
E
Smaller value governs. 4
5.7143 10

 
  
L
Let c
P = Portion of load carried by concrete.
s
P = Portion of load carried by 6 steel rods.

 
 
   
 
 
c
c c c c c
c c
P L
P E A E A
E A L

 
 
   
 
 
s
s s s s s
s s
P L
P E A E A
E A L
2 2 2
6
6 (1.125 in.) 5.9641in
4 4
 
 
  
 
 
s s
A d
2 2 2 2 2
(18 in.) 5.9641in 2.4851 10 in
4 4
 
     
 
 
c c s
A d A
 
 
   
c s c c s s
P P P E A E A
6 2 2 6 2 4
[(4.2 10 psi)(2.4851 10 in ) (29 10 psi)(5.9641in )](5.7143 10 )

     
P
5
6.9526 10 lb
 
P
695 kips
P  
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130
300 mm
60 mm
Aluminium shell
E 70 GPa
Brass core
E 105 GPa
25 mm PROBLEM 2.37
An axial force of 200 kN is applied to the assembly shown by means of
rigid end plates. Determine (a) the normal stress in the aluminum shell,
(b) the corresponding deformation of the assembly.
SOLUTION
Let Pa = Portion of axial force carried by shell.
Pb = Portion of axial force carried by core.
, or
a a a
a
a a
P L E A
P
E A L
 
 
, or
b b b
b
b b
P L E A
P
E A L
 
 
Thus, ( )
a b a a b b
P P P E A E A
L

   
with 2 2 3 2
2 3 2
[(0.060) (0.025) ] 2.3366 10 m
4
(0.025) 0.49087 10 m
4
a
b
A
A




   
  
9 3 9 3
6
[(70 10 )(2.3366 10 ) (105 10 )(0.49087 10 )]
215.10 10


 
     
 
P
L
P
L
Strain:
3
3
6 6
200 10
0.92980 10
215.10 10 215.10 10

 

    
 
P
L
(a) 9 3 6
(70 10 )(0.92980 10 ) 65.1 10 Pa
  
     
a a
E 65.1 MPa
a
  
(b) 3
(0.92980 10 )(300 mm)
L
  
   0.279 mm
  
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131
300 mm
60 mm
Aluminium shell
E 70 GPa
Brass core
E 105 GPa
25 mm PROBLEM 2.38
The length of the assembly shown decreases by 0.40 mm when an axial
force is applied by means of rigid end plates. Determine (a) the magnitude
of the applied force, (b) the corresponding stress in the brass core.
SOLUTION
Let Pa = Portion of axial force carried by shell and Pb = Portion of axial force carried by core.
, or
a a a
a
a a
P L E A
P
E A L
 
 
, or
b b b
b
b b
P L E A
P
E A L
 
 
Thus, ( )
a b a a b b
P P P E A E A
L

   
with 2 2 3 2
2 3 2
[(0.060) (0.025) ] 2.3366 10 m
4
(0.025) 0.49087 10 m
4
a
b
A
A




   
  
9 3 9 3 6
[(70 10 )(2.3366 10 ) (105 10 )(0.49087 10 )] 215.10 10
 
 
       
P
L L
with 0.40 mm, 300 mm
L
  
(a) 6 3
0.40
(215.10 10 ) 286.8 10 N
300
P     287 kN
P  
(b)
9 3
6
3
(105 10 )(0.40 10 )
140 10 Pa
300 10




 
    

b b
b
b
P E
A L
140.0 MPa

b
 
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132
B
C
15 in.
25 in.
1.25 in.
A
6 kips
6 kips
2 in.
PROBLEM 2.39
A polystyrene rod consisting of two cylindrical portions AB and BC is restrained at
both ends and supports two 6-kip loads as shown. Knowing that 6
0.45 10 psi,
 
E
determine (a) the reactions at A and C, (b) the normal stress in each portion of
the rod.
SOLUTION
(a) We express that the elongation of the rod is zero.
2 2
4 4
0
BC BC
AB AB
AB BC
P L
P L
d E d E
 
   
But AB A BC C
P R P R
   
Substituting and simplifying,
2 2
0
C BC
A AB
AB BC
R L
R L
d d
 
2 2
25 2
15 1.25
   
 
   
 
 
BC
AB
C A A
BC AB
d
L
R R R
L d
4.2667
C A
R R
 (1)
From the free body diagram, 12 kips
A C
R R
  (2)
Substituting (1) into (2), 5.2667 12

A
R
2.2785 kips
A
R  2.28 kips
A
R   
From (1), 4.2667(2.2785) 9.7217 kips
C
R  
9.72 kips
C
R   
(b) 2
4
2.2785
(1.25)



  
AB A
AB
AB AB
P R
A A
1.857 ksi
AB
   
2
4
9.7217
(2)
BC C
BC
BC BC
P R
A A 

 
   3.09 ksi
BC
   
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133
A
B
C
D
E
F
20 in.
16 in.
P
PROBLEM 2.40
Three steel rods (E = 29 × 106
psi) support an 8.5-kip load P. Each of the
rods AB and CD has a 0.32-in2
cross-sectional area and rod EF has a 1-in2
cross-sectional area. Neglecting the deformation of bar BED, determine
(a) the change in length of rod EF, (b) the stress in each rod.
SOLUTION
Use member BED as a free body.
By symmetry, or by 0:
 
E
M
CD AB
P P

0: 0
     
y AB CD EF
F P P P P
2 AB EF
P P P
 
  
CD CD
AB AB EF EF
AB CD EF
AB CD EF
P L
P L P L
EA EA EA
  
Since and ,
AB CD AB CD AB CD
L L A A  
  
Since points A, C, and F are fixed, , ,
B AB D CD E EF
     
  
Since member BED is rigid, E B C
  
 
0.32 16
0.256
1 20
      
AB AB EF EF AB EF
AB EF EF EF
AB EF EF AB
P L P L A L
P P P P
EA EA A L
2 2(0.256 ) 1.512
8.5
5.6217 kips
1.512 1.512
0.256(5.6217) 1.43916 kips
    
  
  
AB EF EF EF EF
EF
AB CD
P P P P P P
P
P
P P
(a) 3
(5.6217)(16)
0.0031016 in.
(29 10 )(1)
   

EF EF
EF
EF
P L
EA
0.00310 in.
EF
  
(b)
1.43916
4.4974 ksi
0.32
 
   
AB
AB CD
AB
P
A
  4.50 ksi
AB CD
 
  
5.6217
5.6217 ksi
1
EF
EF
EF
P
A
    5.62 ksi
EF
  
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134
180
40-mm diam. 30-mm diam.
120
100
Dimensions in mm
100
A C
D E
60 kN 40 kN
Brass
Steel B
PROBLEM 2.41
Two cylindrical rods, one of steel and the other of brass, are joined at
C and restrained by rigid supports at A and E. For the loading shown
and knowing that 200 GPa
s
E  and 105 GPa,
b
E  determine
(a) the reactions at A and E, (b) the deflection of point C.
SOLUTION
A to C: 9
2 3 2 3 2
6
200 10 Pa
(40) 1.25664 10 mm 1.25664 10 m
4
251.327 10 N
E
A
EA
 
 
    
 
C to E: 9
2 2 6 2
6
105 10 Pa
(30) 706.86 mm 706.86 10 m
4
74.220 10 N
E
A
EA
 
 
   
 
A to B:
6
12
180 mm 0.180 m
(0.180)
251.327 10
716.20 10



 
 

 
A
A
AB
A
P R
L
R
PL
EA
R
B to C: 3
3
6
12 6
60 10
120 mm 0.120 m
( 60 10 )(0.120)
251.327 10
447.47 10 26.848 10

 
  
 
 
 

   
A
A
BC
A
P R
L
R
PL
EA
R
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135
PROBLEM 2.41 (Continued)
C to D: 3
3
6
9 6
60 10
100 mm 0.100 m
( 60 10 )(0.100)
74.220 10
1.34735 10 80.841 10

 
  
 
 
 

   
A
A
BC
A
P R
L
R
PL
EA
R
D to E: 3
3
6
9 6
100 10
100 mm 0.100 m
( 100 10 )(0.100)
74.220 10
1.34735 10 134.735 10

 
  
 
 
 

   
A
A
DE
A
P R
L
R
PL
EA
R
A to E:
9 6
3.85837 10 242.424 10
    
 
   
   
AE AB BC CD DE
A
R
Since point E cannot move relative to A, 0
AE
 
(a) 9 6 3
3.85837 10 242.424 10 0 62.831 10 N
 
     
A A
R R 62.8 kN
 
A
R 
3 3 3 3
100 10 62.8 10 100 10 37.2 10 N
         
E A
R R 37.2 kN
 
E
R 
(b) 9 6
1.16367 10 26.848 10
    
     
C AB BC A
R
9 3 6
6
(1.16369 10 )(62.831 10 ) 26.848 10
46.3 10 m
 

    
  46.3 m
C
 
  
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136
180
40-mm diam. 30-mm diam.
120
100
Dimensions in mm
100
A C
D E
60 kN 40 kN
Brass
Steel B
PROBLEM 2.42
Solve Prob. 2.41, assuming that rod AC is made of brass and
rod CE is made of steel.
PROBLEM 2.41 Two cylindrical rods, one of steel and the
other of brass, are joined at C and restrained by rigid supports at A
and E. For the loading shown and knowing that 200
s
E  GPa
and 105 GPa,
b
E  determine (a) the reactions at A and E, (b) the
deflection of point C.
SOLUTION
A to C: 9
2 3 2 3 2
6
105 10 Pa
(40) 1.25664 10 mm 1.25664 10 m
4
131.947 10 N
E
A
EA
 
 
    
 
C to E: 9
2 2 6 2
6
200 10 Pa
(30) 706.86 mm 706.86 10 m
4
141.372 10 N
E
A
EA
 
 
   
 
A to B:
6
9
180 mm 0.180 m
(0.180)
131.947 10
1.36418 10



 
 

 
A
A
AB
A
P R
L
R
PL
EA
R
B to C: 3
3
6
12 6
60 10
120 mm 0.120 m
( 60 10 )(0.120)
131.947 10
909.456 10 54.567 10

 
  
 
 
 

   
A
A
BC
A
P R
L
R
PL
EA
R
C to D: 3
3
6
12 6
60 10
100 mm 0.100 m
( 60 10 )(0.100)
141.372 10
707.354 10 42.441 10

 
  
 
 
 

   
A
A
CD
A
P R
L
R
PL
EA
R
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137
PROBLEM 2.42 (Continued)
D to E: 3
3
6
12 6
100 10
100 mm 0.100 m
( 100 10 )(0.100)
141.372 10
707.354 10 70.735 10

 
  
 
 
 

   
A
A
DE
A
P R
L
R
PL
EA
R
A to E:
9 6
3.68834 10 167.743 10
    
 
   
   
AE AB BC CD DE
A
R
Since point E cannot move relative to A, 0
AE
 
(a) 9 6 3
3.68834 10 167.743 10 0 45.479 10 N
 
     
A A
R R 45.5 kN
 
A
R 
3 3 3 3
100 10 45.479 10 100 10 54.521 10
E A
R R
          54.5 kN
 
E
R 
(b) 9 6
2.27364 10 54.567 10
    
     
C AB BC A
R
9 3 6
6
(2.27364 10 )(45.479 10 ) 54.567 10
48.8 10 m
 

    
  48.8 m
C
 
  
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138
A
B
D E
F
C
550 mm
75 mm 100 mm
225 mm
2 kN
PROBLEM 2.43
Each of the rods BD and CE is made of brass (E  105 GPa)
and has a cross-sectional area of 200 mm2
. Determine the
deflection of end A of the rigid member ABC caused by the
2-kN load.
SOLUTION
Let  be the rotation of member ABC as shown.
Then 1 1
0.625 0.075 0.1
  
A B C
     
But BD BD
B
P L
AE
 
9 6
(105 10 )(200 10 )(0.075 )
0.225
B
BD
BD
EA
P
L
 

 
 
6
7 10 
 
Free body ABC:
 CE CE
C
P L
AE

9 6
(105 10 )(200 10 )(0.1 )
0.225
C
CE
CE
EA
P
L
 

 
 
6
9.3333 10 
 
From free body of member ABC:
0: (0.625)(2000) 0.075 0.1 0
    
F BD CE
P P
M
or 6 6
(0.625)(2000) 0.075(7 10 ) 0.1(9.3333 10 ) 0
 
    
3
0.85714 10 rad
 
 
and 3 3
0.625 0.625(0.85714 10 ) 0.53571 10 m
 
    
A
 
0.536 mm
  
A 
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139
P
F
C
D
B
A
E
12 in.
12 in.
12 in.
8 in.
10 in.
PROBLEM 2.44
The rigid bar AD is supported by two steel wires of 1
16
-in. diameter
(E  29 × 106
psi) and a pin and bracket at A. Knowing that the
wires were initially taut, determine (a) the additional tension in each
wire when a 220-lb load P is applied at D, (b) the corresponding
deflection of point D.
SOLUTION
Let  be the notation of bar ABCD.
Then 12
B
 

24

C
 
BE BE
B
P L
AE
 
2
6 1
(29 10 ) (12 )
4 6
10



 
  
 
 
BE
BE
BE
EA
P
L
3
106.765 10 
 
 CF CF
C
P L
EA

2
6 1
(29 10 ) (24 )
4 16
18



 
  
 
 
CE
CF
CF
EA
P
L
3
118.628 10 
 
Using free body ABCD,
0: 12 24 36 0
    
A BE CF
P P P
M
3 6
(12)(106.765 10 ) (24)(118.628 10 ) (36)(220) 0
    
 
6
3
4.1283 10 (36)(220)
1.91847 10 rad

 
 
 
(a) 3 3
(106.765 10 )(1.91847 10 ) 204.83 lb

   
BE
P 205 lb
BE
P  
3 3
(118.628 10 )(1.91847 10 ) 227.58 lb
CF
P 
    228 lb
CF
P  
(b) 3 3
36 (36)(1.91847 10 ) 69.1 10 in.
   
    
D
0.0691 in. 
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140
P
A
D B
L L
C
L
3
4
PROBLEM 2.45
The rigid bar ABC is suspended from three wires of the same material. The cross-
sectional area of the wire at B is equal to half of the cross-sectional area of the
wires at A and C. Determine the tension in each wire caused by the load P shown.
SOLUTION
3
0: 2 0
4
    
A C B
LP LP LP
M
3 1
8 2
C B
P P P
 
5
0: 2 0
4
     
C A B
LP LP LP
M
5 1
8 2
A B
P P P
 
Let l be the length of the wires.
5 1
8 2
 
  
 
 
A
A B
P l l
P P
EA EA

2
( /2)
 
B
B B
P l l
P
E A EA

3 1
8 2
 
  
 
 
C
C B
P l l
P P
EA EA

From the deformation diagram,
A B B C
   
  
or
1
( )
2
B A c
  
 
1 5 1 3 1
( / 2) 2 8 2 8 2
B B B
l l
P P P P P
E A EA
 
   
 
 
5 1 1
;
2 2 5
B B
P P P P
  0.200
B
P P
 

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141
PROBLEM 2.45 (Continued)
5 1 21
8 2 5 40
A
P
P P P
 
  
 
 
0.525
A
P P
 
3 1 11
8 2 5 40
C
P
P P P
 
  
 
 
0.275

C
P P 
Check:
1.000 Ok
  
A B C
P P P P 
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142
D
P
B C
E
15 in.
8 in.
8 in.
8 in.
F
A
8 in.
PROBLEM 2.46
The rigid bar AD is supported by two steel wires of 1
16
-in. diameter
6
( 29 10 psi)
 
E and a pin and bracket at D. Knowing that the wires
were initially taut, determine (a) the additional tension in each wire when
a 120-lb load P is applied at B, (b) the corresponding deflection of point B.
SOLUTION
Let  be the rotation of bar ABCD.
Then 24 8
A C
   
  
AE AE
A
P L
AE
 
6 2
1
4 16
3
(29 10 ) ( ) (24 )
15
142.353 10
A
AE
AE
EA
P
L
 



 
 
CF CF
C
P L
AE
 
 
2
6 1
4 16
3
(29 10 ) (8 )
8
88.971 10
C
CF
CF
EA
P
L
 



 
 
Using free body ABCD,
0:
D
M
 
3 3
3
24 16 8 0
24(142.353 10 ) 16(120) 8(88.971 10 ) 0
0.46510 10 rad
 
 
   
     
 
AE CF
P P P
哷
(a) 3 3
(142.353 10 )(0.46510 10 )

  
AE
P 66.2 lb
AE
P  
3 3
(88.971 10 )(0.46510 10 )

  
CF
P 41.4 lb
CF
P  
(b) 3
16 16(0.46510 10 )
  
  
B
3
7.44 10 in.
B
 
   
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143
Brass core
E 105 GPa
20.9 10–6
/C
Aluminum shell
E 70 GPa
23.6 10–6
/C
25 mm
60 mm
PROBLEM 2.47
The aluminum shell is fully bonded to the brass core and the
assembly is unstressed at a temperature of 15 C.
 Considering only
axial deformations, determine the stress in the aluminum when the
temperature reaches195 C.

SOLUTION
Brass core:
6
105 GPa
20.9 10 / C
E
 

  
Aluminum shell:
6
70 GPa
23.6 10 / C
E
 

  
Let L be the length of the assembly.
Free thermal expansion:
195 15 180 C
T
    
Brass core: ( ) ( )
T b b
L T
 
 
Aluminum shell: ( ) ( )
 
T a a
L T
 
Net expansion of shell with respect to the core: ( )( )
a b
L T
  
  
Let P be the tensile force in the core and the compressive force in the shell.
Brass core: 9
2 2
6 2
105 10 Pa
(25) 490.87 mm
4
490.87 10 m
( )
b
b
P b
b b
E
A
PL
E A



 
 
 

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144
PROBLEM 2.47 (Continued)
Aluminum shell: ( )
p a
a a
PL
E A
 
9
2 2
3 2
3 2
70 10 Pa
(60 25 )
4
2.3366 10 mm
2.3366 10 m
( ) ( )
( )( )
a
a
P b P a
b a
b b a a
E
A
PL PL
L T KPL
E A E A

  
 

 
 
 
 
 
    
where
9 6 9 3
9 1
1 1
1 1
(105 10 )(490.87 10 ) (70 10 )(2.3366 10 )
25.516 10 N
b b a a
K
E A E A
 
 
 
 
   
 
Then
6 6
9
3
( )( )
(23.6 10 20.9 10 )(180)
25.516 10
19.047 10 N
b a T
P
K
 
 

 

  


 
Stress in aluminum:
3
6
3
19.047 10
8.15 10 Pa
2.3366 10
 

      

a
a
P
A
8.15 MPa
a
   
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145
Brass core
E 105 GPa
20.9 10–6
/C
Aluminum shell
E 70 GPa
23.6 10–6
/C
25 mm
60 mm
PROBLEM 2.48
Solve Prob. 2.47, assuming that the core is made of steel ( 200
s
E  GPa,
6
11.7 10 / C)
s
 
   instead of brass.
PROBLEM 2.47 The aluminum shell is fully bonded to the brass core
and the assembly is unstressed at a temperature of 15 C.
 Considering
only axial deformations, determine the stress in the aluminum when the
temperature reaches195 C.

SOLUTION
Aluminum shell: 6
70 GPa 23.6 10 / C
 
   
E
Let L be the length of the assembly.
Free thermal expansion: 195 15 180 C
T
    
Steel core: ( ) ( )
T s s
L T
 
 
Aluminum shell: ( ) ( )
T a a
L T
 
 
Net expansion of shell with respect to the core: ( )( )
a s
L T
  
  
Let P be the tensile force in the core and the compressive force in the shell.
Steel core: 9 2 2 6 2
200 10 Pa, (25) 490.87 mm 490.87 10 m
4
( )



     

s s
P s
s s
E A
PL
E A
Aluminum shell: 9
2 2 3 2 3 2
70 10 Pa
( )
(60 25) 2.3366 10 mm 2.3366 10 m
4
( ) ( )


  

 

     
 
a
P a
a a
a
P s P a
E
PL
E A
A
( )( )
 
    
a s
s s a a
PL PL
L T KPL
E A E A
where
9 6 9 3
9 1
1 1
1 1
(200 10 )(490.87 10 ) (70 10 )(2.3366 10 )
16.2999 10 N
 
 
 
 
   
 
s s a a
K
E A E A
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146
PROBLEM 2.48 (Continued)
Then
6 6
3
9
( )( ) (23.6 10 11.7 10 )(180)
131.412 10 N
16.2999 10
a s T
P
K
   

    
   

Stress in aluminum:
3
6
3
131.412 10
56.241 10 Pa
2.3366 10

      

a
a
P
A
 56.2 MPa
a
   
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147
12 in.
1 in.
1 in.
Steel core
E 29 106 psi
Brass shell
E 15 106 psi
in.
1
4
in.
1
4
in.
1
4
in.
1
4
PROBLEM 2.49
The brass shell 6
( 11.6 10 / F)
 
  
b is fully bonded to the steel core
6
( 6.5 10 / F).
 
  
s Determine the largest allowable increase in
temperature if the stress in the steel core is not to exceed 8 ksi.
SOLUTION
Let axial force developed in the steel core.
s
P 
For equilibrium with zero total force, the compressive force in the brass shell is .
s
P
Strains: ( )
( )
s
s s
s s
s
b b
b b
P
T
E A
P
T
E A
 
 
  
   
Matching: s b
 

( ) ( )
s s
s b
s s b b
P P
T T
E A E A
 
     
1 1
( )( )
s b s
s s b b
P T
E A E A
 
 
   
 
 
(1)
2
2
(1.5)(1.5) (1.0)(1.0) 1.25 in
(1.0)(1.0) 1.0 in
  
 
b
s
A
A
6
3 3
5.1 10 / F
(8 10 )(1.0) 8 10 lb
 


   
    
b s
s s s
P A
9 1
6 6
1 1 1 1
87.816 10 lb
(29 10 )(1.0) (15 10 )(1.25)
 
    
 
s s b b
E A E A
From (1), 9 3 6
(87.816 10 )(8 10 ) (5.1 10 )( )
 
    T
137.8 F
  
T 
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148
6 ft
10 in.
10 in.
PROBLEM 2.50
The concrete post 6
( 3.6 10
c
E   psi and 6
5.5 10 / F)
 
  
c is reinforced with six steel
bars, each of 7
8
-in. diameter 6
( 29 10
s
E   psi and 6
6.5 10 / F).
 
  
s Determine the
normal stresses induced in the steel and in the concrete by a temperature rise of 65°F.
SOLUTION
2
2 2
7
6 6 3.6079 in
4 4 8
   
  
 
 
s
A d
2 2 2
10 10 3.6079 96.392 in
    
c s
A A
Let c
P  tensile force developed in the concrete.
For equilibrium with zero total force, the compressive force in the six steel rods equals .
c
P
Strains: ( ) ( )
c c
s s c c
s s c c
P P
T T
E A E A
   
      
Matching: c s
 
 ( ) ( )
c c
c s
c c s s
P P
T T
E A E A
 
     
6
6 6
3
1 1
( )( )
1 1
(1.0 10 )(65)
(3.6 10 )(96.392) (29 10 )(3.6079)
5.2254 10 lb
 

 
   
 
 
 
  
 
 
 
 
c s c
c c s s
c
c
P T
E A E A
P
P
3
5.2254 10
54.210 psi
96.392
c
c
c
P
A


   54.2 psi
c
 

3
5.2254 10
1448.32 psi
3.6079
c
s
s
P
A


       1.448 ksi
s
   
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149
B
C
250 mm
300 mm
A
50-mm diameter
30-mm diameter
PROBLEM 2.51
A rod consisting of two cylindrical portions AB and BC is restrained at both
ends. Portion AB is made of steel ( 200 GPa,
s
E  6
11.7 10 / C)
s
 
   and
portion BC is made of brass ( 105 GPa,
b
E  6
20.9 10 / C).
b
 
   Knowing
that the rod is initially unstressed, determine the compressive force induced in
ABC when there is a temperature rise of 50 C.

SOLUTION
2 2 2 6 2
2 2 3 2 3 2
(30) 706.86 mm 706.86 10 m
4 4
(50) 1.9635 10 mm 1.9635 10 m
4 4
AB AB
BC BC
A d
A d
 
 


    
     
Free thermal expansion:
6 6
6
( ) ( )
(0.250)(11.7 10 )(50) (0.300)(20.9 10 )(50)
459.75 10 m
T AB s BC b
L T L T
  
 

   
   
 
Shortening due to induced compressive force P:
9 6 9 3
9
0.250 0.300
(200 10 )(706.86 10 ) (105 10 )(1.9635 10 )
3.2235 10

 

 
 
   
 
P
s AB b BC
PL PL
E A E A
P P
P
For zero net deflection, P T
 

9 6
3
3.2235 10 459.75 10
142.624 10 N
 
  
 
P
P 142.6 kN
P  
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150
A B C
1 -in. diameter
1
2
24 in. 32 in.
2 -in. diameter
1
4
PROBLEM 2.52
A rod consisting of two cylindrical portions AB and BC is restrained at both
ends. Portion AB is made of steel 6
( 29 10 psi,
 
s
E 6
6.5 10 / F)
 
  
s and
portion BC is made of aluminum 6
( 10.4 10 psi,
 
a
E 6
13.3 10 /°F).
 
 
a
Knowing that the rod is initially unstressed, determine (a) the normal stresses
induced in portions AB and BC by a temperature rise of 70°F, (b) the
corresponding deflection of point B.
SOLUTION
2 2 2 2
(2.25) 3.9761 in (1.5) 1.76715 in
4 4
 
   
AB BC
A A
Free thermal expansion. 70 F
T
  
Total:
6 3
6 3
3
( ) ( ) (24)(6.5 10 )(70) 10.92 10 in.
( ) ( ) (32)(13.3 10 )(70) 29.792 10 in.
( ) ( ) 40.712 10 in.
 
 
  
 
 

     
     
   
T AB AB s
T BC BC a
T T AB T BC
L T
L T
Shortening due to induced compressive force P.
9
6
9
6
24
( ) 208.14 10
(29 10 )(3.9761)
32
( ) 1741.18 10
(10.4 10 )(1.76715)




   

   

AB
P AB
s AB
BC
P BC
a BC
PL P
P
E A
PL P
P
E A
Total: 9
( ) ( ) 1949.32 10
   
   
P P AB P BC P
For zero net deflection, P T
 
 9 3
1949.32 10 40.712 10
 
  
P 3
20.885 10 lb
 
P
(a)
3
3
20.885 10
5.25 10 psi
3.9761
AB
AB
P
A


       5.25 ksi
AB
   
3
3
20.885 10
11.82 10 psi
1.76715
BC
BC
P
A


       11.82 ksi
BC
   
(b) 9 3 3
( ) (208.14 10 )(20.885 10 ) 4.3470 10 in.
P AB
  
    
3 3
( ) ( ) 10.92 10 4.3470 10
B T AB P AB
    
          3
6.57 10 in.
B
 
   
or
9 3 3
( ) (1741.18 10 )(20.885 10 ) 36.365 10 in.
P BC
  
    
3 3 3
( ) ( ) 29.792 10 36.365 10 6.57 10 in.
B T BC P BC
     
             (checks)
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151
A B C
1 -in. diameter
1
2
24 in. 32 in.
2 -in. diameter
1
4
PROBLEM 2.53
Solve Prob. 2.52, assuming that portion AB of the composite rod is made of
aluminum and portion BC is made of steel.
PROBLEM 2.52 A rod consisting of two cylindrical portions AB and BC is
restrained at both ends. Portion AB is made of steel 6
( 29 10 psi,
 
s
E
6
6.5 10 / F)
 
  
s and portion BC is made of aluminum 6
( 10.4 10 psi,
 
a
E
6
13.3 10 /°F).
 
 
a Knowing that the rod is initially unstressed, determine
(a) the normal stresses induced in portions AB and BC by a temperature rise of
70°F, (b) the corresponding deflection of point B.
SOLUTION
2 2
(2.25) 3.9761 in
4

 
AB
A 2 2
(1.5) 1.76715 in
4

 
BC
A
Free thermal expansion. 70 F
T
  
6 3
6 3
( ) ( ) (24)(13.3 10 )(70) 22.344 10 in.
( ) ( ) (32)(6.5 10 )(70) 14.56 10 in.
T AB AB a
T BC BC s
L T
L T
 
 
 
 
     
     
Total: 3
( ) ( ) 36.904 10 in.
T T AB T BC
   
   
Shortening due to induced compressive force P.
9
6
9
6
24
( ) 580.39 10
(10.4 10 )(3.9761)
32
( ) 624.42 10
(29 10 )(1.76715)




   

   

AB
P AB
a AB
BC
P BC
s BC
PL P
P
E A
PL P
P
E A
Total: 9
( ) ( ) 1204.81 10
   
   
P P AB P BC P
For zero net deflection, P T
 
 9 3
1204.81 10 36.904 10
 
  
P 3
30.631 10 lb
P  
(a)
3
3
30.631 10
7.70 10 psi
3.9761
AB
AB
P
A


       7.70 ksi
AB
   
3
3
30.631 10
17.33 10 psi
1.76715
BC
BC
P
A


       17.33 ksi
BC
   
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152
PROBLEM 2.53 (Continued)
(b) 9 3 3
( ) (580.39 10 )(30.631 10 ) 17.7779 10 in.
  
    
P AB
3 3
( ) ( ) 22.344 10 17.7779 10
B T AB P AB
    
          3
4.57 10 in.
 
  
B 
 or 9 3 3
( ) (624.42 10 )(30.631 10 ) 19.1266 10 in.
  
    
P BC
3 3 3
( ) ( ) 14.56 10 19.1266 10 4.57 10 in. (checks)
B T BC P BC
     
           
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153
PROBLEM 2.54
The steel rails of a railroad track (Es  200 GPa, αs  11.7 × 102–6
/C) were laid at a temperature of 6C.
Determine the normal stress in the rails when the temperature reaches 48C, assuming that the rails (a) are
welded to form a continuous track, (b) are 10 m long with 3-mm gaps between them.
SOLUTION
(a) 6 3
( ) (11.7 10 )(48 6)(10) 4.914 10 m
T T L
   
      
12
9
(10)
50 10
200 10
P
PL L
AE E
 
 

    

3 12
4.914 10 50 10 0
T P
   
 
      
6
98.3 10 Pa
    98.3 MPa
  
(b) 3 12 3
4.914 10 50 10 3 10
T P
   
  
       
3 3
12
6
3 10 4.914 10
50 10
38.3 10 Pa

 

  


   38.3 MPa
   
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154
15 mm
40 mm
2 m
5 mm
Steel
Brass
Steel
P⬘
P
PROBLEM 2.55
Two steel bars 6
( 200 GPa and 11.7 10 / C)
 
   
s s
E are used to
reinforce a brass bar 6
( 105 GPa, 20.9 10 / C)
 
   
b b
E that is subjected
to a load 25 kN.
P  When the steel bars were fabricated, the distance
between the centers of the holes that were to fit on the pins was made
0.5 mm smaller than the 2 m needed. The steel bars were then placed in
an oven to increase their length so that they would just fit on the pins.
Following fabrication, the temperature in the steel bars dropped back to
room temperature. Determine (a) the increase in temperature that was
required to fit the steel bars on the pins, (b) the stress in the brass bar
after the load is applied to it.
SOLUTION
(a) Required temperature change for fabrication:
3
0.5 mm 0.5 10 m
T
 
  
Temperature change required to expand steel bar by this amount:
3 6
3 6
, 0.5 10 (2.00)(11.7 10 )( ),
0.5 10 (2)(11.7 10 )( )
21.368 C
T s
L T T
T T
T
   
 
     
     
   21.4 C
 
(b) Once assembled, a tensile force P*
develops in the steel, and a compressive force P*
develops in the
brass, in order to elongate the steel and contract the brass.
Elongation of steel: 2 6 2
(2)(5)(40) 400 mm 400 10 m
s
A 
   
* *
9 *
6 9
(2.00)
( ) 25 10
(400 10 )(200 10 )
 

   
 
P s
s s
F L P
P
A E
Contraction of brass: 2 6 2
(40)(15) 600 mm 600 10 m
b
A 
   
* *
9 *
6 9
(2.00)
( ) 31.746 10
(600 10 )(105 10 )
 

   
 
P b
b b
P L P
P
A E
But ( ) ( )
P s P b
 
 is equal to the initial amount of misfit:
3 9 * 3
* 3
( ) ( ) 0.5 10 , 56.746 10 0.5 10
8.8112 10 N
    
     
 
P s P b P
P
Stresses due to fabrication:
Steel:
* 3
* 6
6
8.8112 10
22.028 10 Pa 22.028 MPa
400 10
s
s
P
A
 

    

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155
PROBLEM 2.55 (Continued)
Brass:
* 3
* 6
6
8.8112 10
14.6853 10 Pa 14.685 MPa
600 10
b
b
P
A
 

        

To these stresses must be added the stresses due to the 25-kN load.
For the added load, the additional deformation is the same for both the steel and the brass. Let  be the
additional displacement. Also, let Ps and Pb be the additional forces developed in the steel and brass,
respectively.
6 9
6
6 9
6
(400 10 )(200 10 )
40 10
2.00
(600 10 )(105 10 )
31.5 10
2.00
s b
s s b b
s s
s
b b
b
P L P L
A E A E
A E
P
L
A E
P
L

  
  


  
 
  
   
 
  
   
Total: 3
25 10 N
s b
P P P
   
6 6 3 6
40 10 31.5 10 25 10 349.65 10 m
   
  
      
6 6 3
6 6 3
(40 10 )(349.65 10 ) 13.9860 10 N
(31.5 10 )(349.65 10 ) 11.0140 10 N
s
b
P
P


    
    
3
6
6
3
6
6
13.9860 10
34.965 10 Pa
400 10
11.0140 10
18.3566 10 Pa
600 10
s
s
s
b
b
b
P
A
P
A





   


   

Add stress due to fabrication.
Total stresses:
6 6 6
34.965 10 22.028 10 56.991 10 Pa
s
       57.0 MPa
 
s
6 6 6
18.3566 10 14.6853 10 3.6713 10 Pa
      
b 3.67 MPa
b
  
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156
15 mm
40 mm
2 m
5 mm
Steel
Brass
Steel
P⬘
P
PROBLEM 2.56
Determine the maximum load P that may be applied to the brass bar of
Prob. 2.55 if the allowable stress in the steel bars is 30 MPa and the
allowable stress in the brass bar is 25 MPa.
PROBLEM 2.55 Two steel bars ( 200 GPa and 11.7
s s
E 
   10–6
/C)
are used to reinforce a brass bar ( 105 GPa, 20.9
b b
E 
   10–6
/C)
that is subjected to a load 25 kN.
P  When the steel bars were fabricated,
the distance between the centers of the holes that were to fit on the pins
was made 0.5 mm smaller than the 2 m needed. The steel bars were then
placed in an oven to increase their length so that they would just fit on the
pins. Following fabrication, the temperature in the steel bars dropped back
to room temperature. Determine (a) the increase in temperature that was
required to fit the steel bars on the pins, (b) the stress in the brass bar after
the load is applied to it.
SOLUTION
See solution to Problem 2.55 to obtain the fabrication stresses.
*
*
22.028 MPa
14.6853 MPa
s
b




Allowable stresses: ,all ,all
30 MPa, 25 MPa
s b
 
 
Available stress increase from load.
30 22.028 7.9720 MPa
25 14.6853 39.685MPa
  
  
s
b


Corresponding available strains.
6
6
9
6
6
9
7.9720 10
39.860 10
200 10
39.685 10
377.95 10
105 10
s
s
s
b
b
b
E
E







   


   

6
Smaller value governs 39.860 10
 
  
Areas: 2 6 2
(2)(5)(40) 400 mm 400 10 m
s
A 
   
2 6 2
(15)(40) 600 mm 600 10 m
b
A 
   
Forces 9 6 6 3
9 6 6 3
(200 10 )(400 10 )(39.860 10 ) 3.1888 10 N
(105 10 )(600 10 )(39.860 10 ) 2.5112 10 N
s s s
b b b
P E A
P E A


 
  
      
      
Total allowable additional force:
3 3 3
3.1888 10 2.5112 10 5.70 10 N
       
s b
P P P 5.70 kN

P 
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157
20
20 20
200
0.15
Dimensions in mm
30
A A
Section A-A
PROBLEM 2.57
An aluminum rod (Ea  70 GPa, αa  23.6 × 106
/C) and a steel link
(Es × 200 GPa, αa  11.7 × 106
/C) have the dimensions shown at a
temperature of 20C. The steel link is heated until the aluminum rod
can be fitted freely into the link. The temperature of the whole
assembly is then raised to 150C. Determine the final normal stress
(a) in the rod, (b) in the link.
SOLUTION
150 C 20 C 130 C
f i
T T T
        
Unrestrained thermal expansion of each part:
Aluminum rod: ( ) ( )
T a a
L T
 
 
6
( ) (0.200 m)(23.6 10 / C)(130 C)
 
   
T a
4
6.1360 10 m

 
Steel link: ( ) ( )
T s s
L T
 
 
6
( ) (0.200 m)(11.7 10 / C)(130 C)
 
   
T s
4
3.0420 10 m

 
Let P be the compressive force developed in the aluminum rod. It is also the tensile force in the steel link.
Aluminum rod: ( )
P a
a a
PL
E A
 
9 2
(0.200 m)
(70 10 Pa)( /4)(0.03 m)
P



9
4.0420 10
  P
Steel link: ( )
P s
s s
PL
E A
 
9 2
(0.200)
(200 10 Pa)(2)(0.02 m)
P


9
1.250 10
  P
Setting the total deformed lengths in the link and rod equal gives
3
(0.200) ( ) ( ) (0.200) (0.15 10 ) ( ) ( )

      
T s P s T a P a
   
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158
PROBLEM 2.57 (Continued)
3
( ) ( ) 0.15 10 ( ) ( )
P s P a T a T s
   

    
9 9 3 4 4
1.25 10 4.0420 10 0.15 10 6.1360 10 3.0420 10
    
        
P P
4
8.6810 10 N
 
P
(a) Stress in rod: 
P
A

4
8
2
8.6810 10 N
1.22811 10 Pa
( /4)(0.030 m)


 
   
R
122.8 MPa
R
   
(b) Stress in link:
4
8
2
8.6810 10 N
1.08513 10 Pa
(2)(0.020 m)


  
L
108.5 MPa
 
L 
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159
Bronze
A 2.4 in2
E 15 106 psi
12 10–6
/F
0.02 in.
14 in. 18 in.
Aluminum
A 2.8 in2
E 10.6 106 psi
12.9 10–6
/F
PROBLEM 2.58
Knowing that a 0.02-in. gap exists when the temperature is 75 F,

determine (a) the temperature at which the normal stress in the
aluminum bar will be equal to 11 ksi, (b) the corresponding exact
length of the aluminum bar.
SOLUTION
3
3 3
11 ksi 11 10 psi
(11 10 )(2.8) 30.8 10 lb


    
     
a
a a
P A
Shortening due to P:
3 3
6 6
3
(30.8 10 )(14) (30.8 10 )(18)
(15 10 )(2.4) (10.6 10 )(2.8)
30.657 10 in.
b a
P
b b a a
PL PL
E A E A


 
 
 
 
 
Available elongation for thermal expansion:
3 3
0.02 30.657 10 50.657 10 in.
  
    
T
But ( ) ( )
T b b a a
L T L T
  
   
6 6 6
(14)(12 10 )( ) (18)(12.9 10 )( ) (400.2 10 )
  
        
T T T
Equating, 6 3
(400.2 10 ) 50.657 10 126.6 F
 
      
T T
(a) hot cold 75 126.6 201.6 F
      
T T T hot 201.6 F
 
T 
(b) ( ) a
a a a
a a
PL
L T
E A
 
  
3
6 3
6
(30.8 10 )(18)
(18)(12.9 10 )(26.6) 10.712 10 in.
(10.6 10 )(2.8)
 

    

3
exact 18 10.712 10 18.0107 in.

   
L 18.0107 in.

L 
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160
Bronze
A 2.4 in2
E 15 106 psi
12 10–6
/F
0.02 in.
14 in. 18 in.
Aluminum
A 2.8 in2
E 10.6 106 psi
12.9 10–6
/F
PROBLEM 2.59
Determine (a) the compressive force in the bars shown after a
temperature rise of 180 F,
 (b) the corresponding change in length of
the bronze bar.
SOLUTION
Thermal expansion if free of constraint:
6 6
3
( ) ( )
(14)(12 10 )(180) (18)(12.9 10 )(180)
72.036 10 in.
  
 

   
   
 
T b b a a
L T L T
Constrained expansion: 0.02 in.
 
Shortening due to induced compressive force P:
3 3
72.036 10 0.02 52.036 10 in.
P
  
    
But b a b a
P
b b a a b b a a
PL PL L L
P
E A E A E A E A

 
   
 
 
9
6 6
14 18
995.36 10
(15 10 )(2.4) (10.6 10 )(2.8)

 
   
 
 
 
P P
Equating, 9 3
3
995.36 10 52.036 10
52.279 10 lb
 
  
 
P
P
(a) 52.3 kips
P  
(b) ( ) b
b b b
b b
PL
L T
E A
 
  
3
6 3
6
(52.279 10 )(14)
(14)(12 10 )(180) 9.91 10 in.
(15 10 )(2.4)
 

    

3
9.91 10 in.
 
 
b 
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161
Aluminum
A 5 2000 mm2
E 5 75 GPa
5 23 3 10–6
/8C
A B
300 mm 250 mm
0.5 mm
a
Stainless steel
A 5 800 mm2
E 5 190 GPa
5 17.3 3 10–6
/8C
a
PROBLEM 2.60
At room temperature (20 C)
 a 0.5-mm gap exists between the ends of
the rods shown. At a later time when the temperature has reached
140C, determine (a) the normal stress in the aluminum rod, (b) the
change in length of the aluminum rod.
SOLUTION
140 20 120 C
    
T
Free thermal expansion:
6 6
3
( ) ( )
(0.300)(23 10 )(120) (0.250)(17.3 10 )(120)
1.347 10 m
T a a s s
L T L T
  
 

   
   
 
Shortening due to P to meet constraint:
3 3 3
1.347 10 0.5 10 0.847 10 m
P
   
     
9 6 9 6
9
0.300 0.250
(75 10 )(2000 10 ) (190 10 )(800 10 )
3.6447 10

 

 
   
 
 
 
 
 
   
 
 
a s a s
P
a a s s a a s s
PL PL L L
P
E A E A E A E A
P
P
Equating, 9 3
3
3.6447 10 0.847 10
232.39 10 N
 
  
 
P
P
(a)
3
6
6
232.39 10
116.2 10 Pa
2000 10
a
a
P
A
 

      

116.2 MPa
  
a 
(b) ( ) a
a a a
a a
PL
L T
E A
 
  
3
6 6
9 6
(232.39 10 )(0.300)
(0.300)(23 10 )(120) 363 10 m
(75 10 )(2000 10 )
 


    
 
0.363 mm
 
a 
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162
in. diameter
5.0 in.
P'
P
5
8
PROBLEM 2.61
A standard tension test is used to determine the properties of an experimental plastic.
The test specimen is a 5
8
-in.-diameter rod and it is subjected to an 800-lb tensile
force. Knowing that an elongation of 0.45 in. and a decrease in diameter of 0.025 in.
are observed in a 5-in. gage length, determine the modulus of elasticity, the modulus
of rigidity, and Poisson’s ratio for the material.
SOLUTION
2
2 2
5
0.306796 in
4 4 8
   
  
 
 
A d
800 lb
P 
3
800
2.6076 10 psi
0.306796
y
P
A
    
0.45
0.090
5.0
  
y
y
L


0.025
0.040
0.625

   
x
x
d


3
3
2.6076 10
28.973 10 psi
0.090

   
y
y
E


3
29.0 10 psi
E   
0.040
0.44444
0.090

   
x
y
v


0.444

v 
3
3
28.973 10
10.0291 10 psi
2(1 ) (2)(1 0.44444)

   
 
E
v
 3
10.03 10 psi
   
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163
640 kN
2 m
PROBLEM 2.62
A 2-m length of an aluminum pipe of 240-mm outer diameter and 10-mm
wall thickness is used as a short column to carry a 640-kN centric axial load.
Knowing that E  73 GPa and v  0.33, determine (a) the change in length of
the pipe, (b) the change in its outer diameter, (c) the change in its wall
thickness.
SOLUTION
0.240 0.010 2.0
  
o
d t L
3
2 0.240 2(0.010) 0.220 m 640 10 N
      
i o
d d t P
 
2 2 3 2
(0.240 0.220) 7.2257 10 m
4 4
  
     
o i
A d d
(a)
3
9 3
(640 10 )(2.0)
(73 10 )(7.2257 10 )
 

   
 
PL
EA
3
2.4267 10 m

   2.43 mm
 
 
3
2.4267
1.21335 10
2.0


    
L


3
(0.33)( 1.21335 10 )
  
     
LAT v
4
4.0041 10
 
(b) 4 2
(240 mm)(4.0041 10 ) 9.6098 10 mm
 
     
o o LAT
d d 
0.0961 mm
o
d
  
4 3
(10 mm)(4.0041 10 ) 4.0041 10 mm
 
     
LAT
t t
0.00400 mm
t
  
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164
10
200 mm
150 mm
4
200 kN 200 kN
PROBLEM 2.63
A line of slope 4:10 has been scribed on a cold-rolled yellow-brass
plate, 150 mm wide and 6 mm thick. Knowing that E  105 GPa and
v  0.34, determine the slope of the line when the plate is subjected to
a 200-kN centric axial load as shown.
SOLUTION
3 2
(0.150)(0.006) 0.9 10 m
A 
  
3
6
3
200 10
222.22 10 Pa
0.9 10
x
P
A
 

   

6
3
9
222.22 10
2.1164 10
105 10
x
x
E

 

   

3
(0.34)(2.1164 10 )
y x
  
    
3
0.71958 10
   
3
3
4(1 )
tan
10(1 )
4(1 0.71958 10 )
10(1 2.1164 10 )
0.39887





 

 

y
x



tan 0.399
  
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165
2.75 kN
2.75 kN
50 mm
A B
12 mm
PROBLEM 2.64
A 2.75-kN tensile load is
applied to a test coupon
made from 1.6-mm flat
steel plate (E  200 GPa,
v  0.30). Determine the
resulting change (a) in the
50-mm gage length, (b) in
the width of portion AB
of the test coupon, (c) in
the thickness of portion
AB, (d) in the cross-
sectional area of portion
AB.
SOLUTION
2
(1.6)(12) 19.20 mm
A  
6 2
19.20 10 m

 
3
2.75 10 N
 
P
3
6
2.75 10
19.20 10
x
P
A
 

 

6
143.229 10 Pa
 
6
6
9
143.229 10
716.15 10
200 10
x
x
E

 

   

6 6
(0.30)(716.15 10 ) 214.84 10
y z x
    
        
(a) 6 6
0.050 m (0.50)(716.15 10 ) 35.808 10 m
   
     
x x
L L
0.0358 mm 
(b) 6 6
0.012 m (0.012)( 214.84 10 ) 2.5781 10 m
   
       
y y
w w
0.00258 mm
 
(c) 6 9
0.0016 m (0.0016)( 214.84 10 ) 343.74 10 m
   
       
z z
t t
0.000344 mm
 
(d) 0 0 0 0 0 0 0
(1 ) (1 ) (1 )
     
       
y z y z y z
A w t w t A w t
0 0 0 0 0
( negligible term) 2
  
      
y z y
A A A w t w t
6 9 2
(2)(0.012)(0.0016)( 214.84 10 ) 8.25 10 m
 
      2
0.00825 mm
 
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166
200 mm
22-mm diameter
75 kN 75 kN
PROBLEM 2.65
In a standard tensile test, a steel rod of 22-mm diameter is subjected to a
tension force of 75 kN. Knowing that 0.3

v and 200 GPa,

E
determine (a) the elongation of the rod in a 200-mm gage length, (b) the
change in diameter of the rod.
SOLUTION
3 2 2 6 2
75 kN 75 10 N (0.022) 380.13 10 m
4 4
P A d
  
      
3
6
6
6
6
9
6
75 10
197.301 10 Pa
380.13 10
197.301 10
986.51 10
200 10
(200 mm)(986.51 10 )
x
x x
P
A
E
L



 




   


   

  
(a) 0.1973 mm
 
x 
 6 6
(0.3)(986.51 10 ) 295.95 10
y x
v
   
        
 6
(22 mm)( 295.95 10 )
  
   
y y
d 
 (b) 0.00651 mm
  
y 
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167
2.5 in. PROBLEM 2.66
The change in diameter of a large steel bolt is carefully measured as the
nut is tightened. Knowing that 6
29 10 psi
 
E and 0.30,

v determine
the internal force in the bolt if the diameter is observed to decrease by
3
0.5 10 in.


SOLUTION
3
3
3
0.5 10 in. 2.5 in.
0.5 10
0.2 10
2.5
y
y
y
d
d






   

     
3
3
0.2 10
: 0.66667 10
0.3
y y
x
x
v
v
 




 
     
6 3 3
(29 10 )(0.66667 10 ) 19.3334 10 psi
x x
E
  
     
2 2 2
(2.5) 4.9087 in
4 4
 
  
A d
3 3
(19.3334 10 )(4.9087) 94.902 10 lb
x
F A

    
94.9 kips
F  
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168
240 mm
600 mm
C
D
A
B
50 mm
PROBLEM 2.67
The brass rod AD is fitted with a jacket that is used to apply a
hydrostatic pressure of 48 MPa to the 240-mm portion BC of the rod.
Knowing that E  105 GPa and v  0.33, determine (a) the change in
the total length AD, (b) the change in diameter at the middle of the rod.
SOLUTION
 
 
6
6 6
9
6
6 6
9
6
48 10 Pa, 0
1
( )
1
48 10 (0.33)(0) (0.33)( 48 10 )
105 10
306.29 10
1
( )
1
(0.33)( 48 10 ) 0 (0.33)( 48 10 )
105 10
301.71 10
x z y
x x y z
y x y z
p
E
E
  
   
   


      
  
      

 
   
       

 
(a) Change in length: only portion BC is strained. L  240 mm
6
(240)( 301.71 10 ) 0.0724 mm
y y
L
  
      
(b) Change in diameter: d  50 mm
6
(50)( 306.29 10 ) 0.01531mm
   
      
x z x
d 
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169
x
z
3 in.
4 in.
z
y
x
A
B
C
D
PROBLEM 2.68
A fabric used in air-inflated structures is subjected to a biaxial
loading that results in normal stresses 18 ksi
 
x and 24 ksi
 
z .
Knowing that the properties of the fabric can be approximated as
E  12.6 × 106
psi and v  0.34, determine the change in length of
(a) side AB, (b) side BC, (c) diagonal AC.
SOLUTION
 
 
6
6
3
6
18 ksi 0 24 ksi
1 1
( ) 18,000 (0.34)(24,000) 780.95 10
12.6 10
1 1
( ) (0.34)(18,000) 24,000 1.41905 10
12.6 10
  
   
   


  
      

        

x y z
x x y z
z x y z
E
E
(a) 6
( ) (4 in.)(780.95 10 ) 0.0031238 in.
AB x
AB
  
   
0.00312 in. 
(b) 3
( ) (3 in.)(1.41905 10 ) 0.0042572 in.

   
BC z
BC
 
0.00426 in. 
Label sides of right triangle ABC as a, b, c.
Then 2 2 2
c a b
 
Obtain differentials by calculus.
2 2 2
cdc ada bdb
 
a b
dc da db
c c
 
But 2 2
4 in. 3 in. 4 3 5 in.
    
a b c
0.0031238 in. 0.0042572 in.
   
AB BC
da db
 
(c)
4 3
(0.0031238) (0.0042572)
5 5
AC dc
   
0.00505 in. 
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170
y 6 ksi
x 12 ksi
1 in.
A B
C
D
1 in.
PROBLEM 2.69
A 1-in. square was scribed on the side of a large steel pressure vessel.
After pressurization the biaxial stress condition at the square is as
shown. Knowing that E  29 × 106
psi and v  0.30, determine the
change in length of (a) side AB, (b) side BC, (c) diagonal AC.
SOLUTION
3 3
6
6
3 3
6
6
1 1
( ) 12 10 (0.30)(6 10 )
29 10
351.72 10
1 1
( ) 6 10 (0.30)(12 10 )
29 10
82.759 10
x x y
y y x
E
E
  
  


 
     
 

 
 
     
 

 
(a) 6 6
0
( ) (1.00)(351.72 10 ) 352 10 in.
   
    
AB x
AB 
(b) 6 6
0
( ) (1.00)(82.759 10 ) 82.8 10 in.
   
    
BC y
BC 
(c) 2 2 2 2
0 0
6 2 6 2
6
0 0
( ) ( ) ( ) ( ) ( )
(1 351.72 10 ) (1 82.759 10 )
1.41452
( ) 2 ( ) 307 10
 
 

     
     

   
x y
AC AB BC AB BC
AC AC AC 
or use calculus as follows:
Label sides using a, b, and c as shown.
2 2 2
c a b
 
Obtain differentials. 2 2 2
 
cdc ada bdc
from which
a b
dc da dc
c c
 
6 6
6 6
6
But 100 in., 1.00 in., 2 in.
351.72 10 in., 82.8 10 in.
1.00 1.00
(351.7 10 ) (82.8 10 )
2 2
307 10 in.
 

 
 

  
     
    
 
AB BC
AC
a b c
da db
dc
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171
PROBLEM 2.70
The block shown is made of a magnesium alloy, for which
45
E  GPa and 0.35.
v  Knowing that 180
x
   MPa,
determine (a) the magnitude of y
 for which the change in
the height of the block will be zero, (b) the corresponding
change in the area of the face ABCD, (c) the corresponding
change in the volume of the block.
SOLUTION
(a)
6
0 0 0
1
( )
(0.35)( 180 10 )
y y z
y x y z
y x
v v
E
v
  
   
 
  
  
   
6
63 10 Pa
   63.0 MPa
y
   
6
3
9
6
3
9
1 (0.35)( 243 10 )
( ) ( ) 1.890 10
45 10
1 157.95 10
( ) 3.510 10
45 10
     
 
   


 
          

 
        

z z x y x y
x y
x x y Z
v
v v
E E
v
v v
E E
(b) 0
0
(1 ) (1 ) (1 )
( ) ( )
x z
x x z z x z x z x z
x z x z x z x z x z
A L L
A L L L L
A A A L L L L
     
     

      
       
3 3
(100 mm)(25 mm)( 3.510 10 1.890 10 )
 
     
A 2
4.05 mm
  
A 
(c) 0
0
(1 ) (1 ) (1 )
(1 )
( small terms)
  
           
  

   
       
      
x y z
x x y y z z
x y z x y z x y y z z x x y z
x y z x y z
V L L L
V L L L
L L L
V V V L L L
3 3
(100)(40)(25)( 3.510 10 0 1.890 10 )
 
      
V 3
162.0 mm
  
V 
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172
x
␴
z
␴
z
y
x
A
B
C
D
PROBLEM 2.71
The homogeneous plate ABCD is subjected to a biaxial loading as
shown. It is known that 0
z
 
 and that the change in length of
the plate in the x direction must be zero, that is, 0.
x
  Denoting
by E the modulus of elasticity and by v Poisson’s ratio, determine
(a) the required magnitude of ,
x
 (b) the ratio 0 / .
z
 
SOLUTION
0
0
, 0, 0
1 1
( ) ( )
z y x
x x y z x
v v v
E E
   
     
  
    
(a) 0
x v
 
 
(b)
2
2
0 0 0
1 1 1
( ) ( 0 )
z x y z
v
v v v
E E E
      

         0
2
1
z
E
v





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173
(b)
(a)
␶m ␶m
P
P'
P'
P
P
2A
z
x
y
'
45⬚
␴x
␴x
P
A
P
2A
␴ '
␴
'
␴
'
␴ ⫽
⫽
⫽
PROBLEM 2.72
For a member under axial loading, express the normal strain  in a
direction forming an angle of 45 with the axis of the load in terms of the
axial strain x by (a) comparing the hypotenuses of the triangles shown in
Fig. 2.43, which represent, respectively, an element before and after
deformation, (b) using the values of the corresponding stresses of  and
x shown in Fig. 1.38, and the generalized Hooke’s law.
SOLUTION
Figure 2.49
(a) 2 2 2
2 2 2 2
2 2 2 2
[ 2(1 )] (1 ) (1 )
2(1 2 ) 1 2 1 2
4 2 2 2
x x
x x x x
x x x x
v
v v
v v
  
     
     

    
 
       
 
    
Neglect squares as small. 4 2 2
x x
v
  
  
1
2
x
v
 

  
(A) (B)
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174
PROBLEM 2.72 (Continued)
(b)
1
2
1
2
x
v
E E
v P
E A
v
E
 


 
  

 


1
2
x
v


 
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175
x
␴
y
␴
PROBLEM 2.73
In many situations, it is known that the normal stress in a given direction is zero.
For example, 0
z
  in the case of the thin plate shown. For this case, which is
known as plane stress, show that if the strains x and y have been determined
experimentally, we can express , ,
 
x y and z
 as follows:
2 2
( )
1
1 1
x y y x
x y z x y
v v v
E E
v
v v
   
    
 
    

 
SOLUTION
0
1
( )
z
x x y
v
E

  

  (1)
1
( )
y x y
v
E
  
   (2)
Multiplying (2) by v and adding to (1),
2
2
1
or ( )
1
x y x x x y
v E
v v
E v
     

   


Multiplying (1) by v and adding to (2),
2
2
1
or ( )
1
y x y y y x
v E
v v
E v
     

   


1
( )
z x y
v
v v
E E
  
    
E
 2
2
( )
1
(1 )
( ) ( )
1
1
x y y x
x y x y
v v
v
v v v
v
v
   
   
  


     



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176
x
x
␴
z
z
␴
y
y
␴
y
x
z (a) (b)
␴
PROBLEM 2.74
In many situations, physical constraints
prevent strain from occurring in a given
direction. For example, 0
z
  in the case
shown, where longitudinal movement of
the long prism is prevented at every point.
Plane sections perpendicular to the
longitudinal axis remain plane and the
same distance apart. Show that for this
situation, which is known as plane strain,
we can express ,
z ,
x and y as follows:
2
2
( )
1
[(1 ) (1 ) ]
1
[(1 ) (1 ) ]
z x y
x x y
y y x
v
v v v
E
v v v
E
  
  
  
 
   
   
SOLUTION
1
0 ( ) or ( )
z x y z z x y
v v v
E
      
       
2
2
1
( )
1
[ ( )]
1
[(1 ) (1 ) ]
x x y z
x y x y
x y
v v
E
v v
E
v v v
E
   
   
 
  
   
    
2
2
1
( )
1
[ ( )]
1
[(1 ) (1 ) ]
y x y z
x y x y
y x
v v
E
v v
E
v v v
E
   
   
 
   
    
    
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177
4.8 in.
3.2 in.
2 in.
P
PROBLEM 2.75
The plastic block shown is bonded to a rigid support and to a vertical plate to
which a 55-kip load P is applied. Knowing that for the plastic used 150
G  ksi,
determine the deflection of the plate.
SOLUTION


2
3
3
3
3
3
(3.2)(4.8) 15.36 in
55 10 lb
55 10
3580.7 psi
15.36
150 10 psi
3580.7
23.871 10
150 10
2 in.


 
 
 

  
 
   


A
P
P
A
G
G
h
3
3
(2)(23.871 10 )
47.7 10 in.
h
  

  
 
0.0477 in.
   
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178
4.8 in.
3.2 in.
2 in.
P
PROBLEM 2.76
What load P should be applied to the plate of Prob. 2.75 to produce a 1
16
-in.
deflection?
PROBLEM 2.75 The plastic block shown is bonded to a rigid support and to a
vertical plate to which a 55-kip load P is applied. Knowing that for the plastic
used 150
G  ksi, determine the deflection of the plate.
SOLUTION
3
3
1
in. 0.0625 in.
16
2 in.
0.0625
0.03125
2
150 10 psi
(150 10 )(0.03125)
4687.5 psi



 
 

  
 
  

h
h
G
G
2
3
(3.2)(4.8) 15.36 in
(4687.5)(15.36)
72.0 10 lb
A
P A

 
 
 
72.0 kips 
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179
a a
c
b
A
B
P
PROBLEM 2.77
Two blocks of rubber with a modulus of rigidity 12 MPa
G  are
bonded to rigid supports and to a plate AB. Knowing that 100 mm
c 
and 45 kN,
P  determine the smallest allowable dimensions a and b of
the blocks if the shearing stress in the rubber is not to exceed 1.4 MPa
and the deflection of the plate is to be at least 5 mm.
SOLUTION
Shearing strain:
a G
 
  
6
6
(12 10 Pa)(0.005 m)
0.0429 m
1.4 10 Pa
G
a



  

42.9 mm
a  
Shearing stress:
1
2
2
P P
A bc
  
3
6
45 10 N
0.1607 m
2 2(0.1 m)(1.4 10 Pa)


  

P
b
c
160.7 mm
b  
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180
a a
c
b
A
B
P
PROBLEM 2.78
Two blocks of rubber with a modulus of rigidity 10
G  MPa are bonded
to rigid supports and to a plate AB. Knowing that 200
b  mm and
125
c  mm, determine the largest allowable load P and the smallest
allowable thickness a of the blocks if the shearing stress in the rubber
is not to exceed 1.5 MPa and the deflection of the plate is to be at least
6 mm.
SOLUTION
Shearing stress:
1
2
2
P P
A bc
  
3
2 2(0.2 m)(0.125 m)(1.5 10 kPa)
P bc
   75.0 kN

P 
Shearing strain:
a G
 
  
6
6
(10 10 Pa)(0.006 m)
0.04 m
1.5 10 Pa
G
a



  

40.0 mm
a  
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181
8 in.
b
a
P
PROBLEM 2.79
An elastomeric bearing (G  130 psi) is used to support a bridge girder
as shown to provide flexibility during earthquakes. The beam must not
displace more than 3
8
in. when a 5-kip lateral load is applied as shown.
Knowing that the maximum allowable shearing stress is 60 psi,
determine (a) the smallest allowable dimension b, (b) the smallest
required thickness a.
SOLUTION
Shearing force: 5 kips 5000 lb
P  
Shearing stress: 60 psi
 
2
5000
, or 83.333 in
60
   
P P
A
A


and (8 in.)( )

A b
(a)
83.333
10.4166 in.
8 8
A
b    10.42 in.
b  
 3
60
461.54 10 rad
130




    
(b) 3
0.375 in.
But , or
461.54 10
 

 
  

a
a
0.813 in.
a  
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182
8 in.
b
a
P
PROBLEM 2.80
For the elastomeric bearing in Prob. 2.79 with b  10 in. and a  1 in.,
determine the shearing modulus G and the shear stress  for a
maximum lateral load P  5 kips and a maximum displacement
0.4 in.


SOLUTION

Shearing force: 5 kips 5000 lb
P  
Area: 2
(8 in.)(10 in.) 80 in
A  
Shearing stress:
5000
80
 
P
A
 62.5 psi

 
Shearing strain:
0.4 in.
0.400 rad
1in.

   
a

Shearing modulus:
62.5
0.400


 
G 156.3 psi

G 
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183
150 mm
100 mm
30 mm
B
A
30 mm
P PROBLEM 2.81
A vibration isolation unit consists of two blocks of hard rubber bonded
to a plate AB and to rigid supports as shown. Knowing that a force of
magnitude 25 kN
P  causes a deflection 1.5 mm
  of plate AB,
determine the modulus of rigidity of the rubber used.
SOLUTION
3 3
3
3
1 1
(25 10 N) 12.5 10 N
2 2
(12.5 10 N)
833.33 10 Pa
(0.15 m)(0.1m)
    

   
F P
F
A

3
3
1.5 10 m 0.03 m
1.5 10
0.05
0.03





  

  
h
h
3
6
833.33 10
16.67 10 Pa
0.05
G



   
16.67 MPa
G  
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184
150 mm
100 mm
30 mm
B
A
30 mm
P PROBLEM 2.82
A vibration isolation unit consists of two blocks of hard rubber with a
modulus of rigidity 19 MPa
G  bonded to a plate AB and to rigid
supports as shown. Denoting by P the magnitude of the force applied
to the plate and by  the corresponding deflection, determine the
effective spring constant, / ,
k P 
 of the system.
SOLUTION
Shearing strain:
h

 
Shearing stress:
G
G
h

 
 
Force:
1 2
or
2
GA GA
P A P
h h
 

  
Effective spring constant:
2
P GA
k
h

 
with 2
(0.15)(0.1) 0.015 m 0.03 m
A h
  
6 2
6
2(19 10 Pa)(0.015 m )
19.00 10 N/m
0.03 m
k

  
3
19.00 10 kN/m
k   
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185
PROBLEM 2.83
A 6-in.-diameter solid steel sphere is lowered into the ocean to a point where the pressure is 7.1 ksi (about
3 miles below the surface). Knowing that 6
29 10
E   psi and 0.30,
v  determine (a) the decrease in
diameter of the sphere, (b) the decrease in volume of the sphere, (c) the percent increase in the density of the
sphere.
SOLUTION
For a solid sphere, 3
0 0
3
3
6
(6.00)
6
113.097 in


  



   
x y z
V d
p
3
3
6
6
7.1 10 psi
1
( )
(1 2 ) (0.4)(7.1 10 )
29 10
97.93 10
   

  
  
 
   

  
x x y z
v v
E
v p
E
Likewise, 6
97.93 10
y z
  
   
6
293.79 10
x y z
e    
     
(a) 6 6
0 (6.00)( 97.93 10 ) 588 10 in.
  
        
x
d d 6
588 10 in.

  
d 
(b) 6 3 3
0 (113.097)( 293.79 10 ) 33.2 10 in
 
        
V V e 3 3
33.2 10 in

  
V 
(c) Let mass of sphere. constant.
m m
 
0 0 0 (1 )
m V V V e
  
   
0 0
0 0 0
2 3 2 3
6
6
0
0
1
1 1 1
(1 ) 1
(1 ) 1
293.79 10
100% (293.79 10 )(100%)
  
 
 




      
 
          
   

  
 
V
m
V e m e
e e e e e e
e
0.0294% 
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186
s
E5 105 GPa
y 5 258 MPa
n 5 0.33
135 mm
85 mm PROBLEM 2.84
(a) For the axial loading shown, determine the change in height and the
change in volume of the brass cylinder shown. (b) Solve part a,
assuming that the loading is hydrostatic with 70
x y z
  
    MPa.
SOLUTION
0
2 2 3 2 3 2
0 0
3 3 6 3
0 0 0
135 mm 0.135 m
(85) 5.6745 10 mm 5.6745 10 m
4 4
766.06 10 mm 766.06 10 m
  

 
     
    
h
A d
V A h
(a) 6
6
6
9
0, 58 10 Pa, 0
1 58 10
( ) 552.38 10
105 10
  

    
    

         

x y z
y
y x y z
v v
E E
6
0 (135 mm)( 552.38 10 )
 
    
y
h h 0.0746 mm
  
h 
6
6
9
(1 2 )
1 2 (0.34)( 58 10 )
( ) 187.81 10
105 10

   

  
       

y
x y z
v
v
e
E E
3 3 6
0 (766.06 10 mm )( 187.81 10 )

     
V V e 3
143.9 mm
  
V 
(b) 6 6
6
6
9
70 10 Pa 210 10 Pa
1 1 2 (0.34)( 70 10 )
( ) 226.67 10
105 10
     
     
         
  
        

x y z x y z
y x y z y
v
v v
E E
6
0 (135 mm)( 226.67 10 )
 
    
y
h h 0.0306 mm
  
h 
6
6
9
1 2 (0.34)( 210 10 )
( ) 680 10
105 10
   
  
      

x y z
v
e
E
3 3 6
0 (766.06 10 mm )( 680 10 )

     
V V e 3
521mm
  
V 
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187
11 kips
11 kips
8 in.
1 in. diameter PROBLEM 2.85*
Determine the dilatation e and the change in volume of the 8-in. length
of the rod shown if (a) the rod is made of steel with E  29 × 106
psi
and v  0.30, (b) the rod is made of aluminum with E  10.6 × 106
psi
and v  0.35.
SOLUTION
2 2 2
3
(1) 0.78540 in
4 4
11 10 lb
A d
P
 
  
 
Stresses : 3
3
11 10
14.0056 10 psi
0.78540
0

   
 
x
y z
P
A

 
(a) Steel. 6
29 10 psi 0.30
  
E v
3
6
6
1 14.0056 10
( ) 482.95 10
29 10


      

x
x x y z
v v
E E

   
6
6
1
( ) (0.30)(482.95 10 )
144.885 10

     

          
  
x
y x y z x
v
v v v
E E
6
1
( ) 144.885 10
         
x
z x y z y
v
v v
E E

    
6
193.2 10
    
x y z
e    
6 3 3
(0.78540)(8)(193.2 10 ) 1.214 10 in
 
       
v ve Le 

(b) Aluminum. 6
10.6 10 psi 0.35
  
E v
3
3
6
14.0056 10
1.32128 10
10.6 10
x
x
E

 

   

3 6
(0.35)(1.32128 10 ) 462.45 10
   
       
y x
v
6
462.45 10
z y
  
   
6
396 10
x y z
e    
     
6 3 3
(0.78540)(8)(396 10 ) 2.49 10 in
 
       
v ve Le 
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188
PROBLEM 2.86
Determine the change in volume of the 50-mm gage length segment AB in Prob. 2.64 (a) by computing the
dilatation of the material, (b) by subtracting the original volume of portion AB from its final volume.
PROBLEM 2.64 A 2.75-kN tensile load is applied to a test coupon made from 1.6-mm flat steel plate
(E  200 GPa, v  0.30). Determine the resulting change (a) in the 50-mm gage length, (b) in the width of
portion AB of the test coupon, (c) in the thickness of portion AB, (d) in the cross-sectional area of portion AB.
2.75 kN
2.75 kN
50 mm
A B
12 mm
SOLUTION
(a) 2 6 2
0 (12)(1.6) 19.2 mm 19.2 10 m

   
A
3
0 0 0
Volume (50)(19.2) 960 mm
  
V L A
3
6
6
0
2.75 10
143.229 10 Pa 0
19.2 10
  


     

x y z
P
A
6
6
9
1 143.229 10
( ) 716.15 10
200 10
x
x x y z
E E

    

      

3 6
(0.30)(716.15 10 ) 214.84 10
    
        
y z x
6
286.46 10
    
x y z
e    
6 3
0 (960)(286.46 10 ) 0.275 mm

    
v v e 
(b) From the solution to problem 2.64,
0.035808 mm 0.0025781 0.00034374 mm
    
x y z
  
The dimensions when under the 2.75-kN load are
Length: 0 50 0.035808 50.035808 mm

    
x
L L
Width: 0 12 0.0025781 11.997422 mm

    
y
w w
Thickness: 0 1.6 0.00034374 1.599656 mm

    
z
t t
Volume: 3
(50.03581)(11.997422)(1.599656) 960.275 mm
  
V Lwt 
3
0 960.275 960 0.275 mm
     
V V V 
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189
A
B
R1
80 mm
R2
P PROBLEM 2.87
A vibration isolation support consists of a rod A of radius 1 10
R  mm and
a tube B of inner radius 2 25
R  mm bonded to an 80-mm-long hollow
rubber cylinder with a modulus of rigidity G  12 MPa. Determine the
largest allowable force P that can be applied to rod A if its deflection is not
to exceed 2.50 mm.
SOLUTION
Let r be a radial coordinate. Over the hollow rubber cylinder, 1 2.
R r R
 
Shearing stress  acting on a cylindrical surface of radius r is
2
P P
A rh


 
The shearing strain is
2
P
G Ghr



 
Shearing deformation over radial length dr:
2
d
dr
P dr
d dr
Gh r


 


 
Total deformation.
2 2
1 1
2
1
2 1
2
1 2 1
2
ln (ln ln )
2 2
2
ln or
2 ln( / )
R R
R R
R
R
P dr
d
Gh r
P P
r R R
Gh Gh
R
P Gh
P
Gh R R R
 

 
 

 
  
 
 
1 2
Data: 10 mm 0.010 m, 25 mm 0.025 m, 80 mm 0.080 m
R R h
     
6 3
6 3
3
12 10 Pa 2.50 10 m
(2 )(12 10 )(0.080)(2.50 10 )
16.46 10 N
ln (0.025/0.010)
G
P




   
 
   16.46 kN 
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190
A
B
R1
80 mm
R2
P PROBLEM 2.88
A vibration isolation support consists of a rod A of radius R1 and a tube B of
inner radius R2 bonded to a 80-mm-long hollow rubber cylinder with a
modulus of rigidity G  10.93 MPa. Determine the required value of the ratio
R2/R1 if a 10-kN force P is to cause a 2-mm deflection of rod A.
SOLUTION
Let r be a radial coordinate. Over the hollow rubber cylinder, 1 2.
R r R
 
Shearing stress  acting on a cylindrical surface of radius r is
2
P P
A rh


 
The shearing strain is
2
P
G Ghr



 
Shearing deformation over radial length dr:
2


 





d
dr
d dr
P dr
dr
Gh r
Total deformation.
2 2
1 1
2
1
2 1
2
1
2
ln (ln ln )
2 2
ln
2
R R
R R
R
R
P dr
d
Gh r
P P
r R R
Gh Gh
R
P
Gh R
 

 

 
  

 
6
2
3
1
2
1
2 (2 )(10.93 10 )(0.080)(0.002)
ln 1.0988
10.10
exp(1.0988) 3.00
R Gh
R P
R
R
   
  
  2 1
/ 3.00

R R 
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191
PROBLEM 2.89
The material constants E, G, k, and v are related by Eqs. (2.33) and (2.43). Show that any one of these constants
may be expressed in terms of any other two constants. For example, show that (a) k  GE/(9G  3E) and
(b) v  (3k  2G)/(6k  2G).
SOLUTION
and
3(1 2 ) 2(1 )
E E
k G
v v
 
 
(a) 1 or 1
2 2
E E
v v
G G
   
2 2
3[2 2 4 ] 18 6
3 1 2 1
2
  
  
 
 
 
 
 
 
 
E EG EG
k
G E G G E
E
G
9 6


EG
k
G E

(b)
2(1 )
3(1 2 )
k v
G v



3 6 2 2
3 2 2 6
  
  
k kv G Gv
k G G k
3 2
6 2



k G
v
k G

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192
PROBLEM 2.90
Show that for any given material, the ratio G/E of the modulus of rigidity over the modulus of elasticity is
always less than 1
2
but more than 1
3
. [Hint: Refer to Eq. (2.43) and to Sec. 2.13.]
SOLUTION
or 2(1 )
2(1 )
E E
G v
v G
  

Assume 0
v  for almost all materials, and 1
2
v for a positive bulk modulus.
Applying the bounds,
1
2 2 1 3
2
E
G
 
  
 
 
Taking the reciprocals,
1 1
2 3
G
E
 
or
1 1
3 2
G
E
  
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193
Ex 50 GPa
Ey 15.2 GPa
Ez 15.2 GPa
xz 0.254
xy 0.254
zy 0.428
y
z
x
PROBLEM 2.91
A composite cube with 40-mm sides and the properties shown is
made with glass polymer fibers aligned in the x direction. The cube
is constrained against deformations in the y and z directions and is
subjected to a tensile load of 65 kN in the x direction. Determine
(a) the change in the length of the cube in the x direction, (b) the
stresses , ,
x y
  and .
z

50 GPa 0.254
15.2 GPa 0.254
15.2 GPa 0.428
 
 
 
x xz
y xy
z zy
E v
E v
E v
SOLUTION
Stress-to-strain equations are
yx y
x zx z
x
x y z
v v
E E E

 
    (1)
xy x y zy z
y
x y z
v v
E E E
  
     (2)
yz y
xz x z
z
x y z
v
v
E E E

 
     (3)
xy yx
x y
v v
E E
 (4)
yz zy
y z
v v
E E
 (5)
zx xz
z x
v v
E E
 (6)
The constraint conditions are 0 and 0.
y z
 
 
Using (2) and (3) with the constraint conditions gives
1 zy xy
y z x
y z x
v v
E E E
  
  (7)
1
yz xz
y z x
y z x
v V
E E E
  
   (8)
1 0.428 0.254
or 0.428 0.077216
15.2 15.2 50
0.428 1 0.254
or 0.428 0.077216
15.2 15.2 50
     
     
   
     
y z x y z x
y z x y z x
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194
PROBLEM 2.91
(Continued)
Solving simultaneously, 0.134993
  
 
y z x
Using (4) and (5) in (1),
1 xy xz
x x y z
x x
v v
E E E
   
  
2 6 2
3
6
6
3
6
9
1
[1 (0.254)(0.134993) (0.254)(0.134993)]
0.93142
(40)(40) 1600 mm 1600 10 m
65 10
40.625 10 Pa
1600 10
(0.93142)(40.625 10 )
756.78 10
50 10







  

   

   


  

x x
x
x
x
x
x
E
E
E
A
P
A
(a) 6
(40 mm)(756.78 10 )
  
  
x x x
L 0.0303 mm
 
x 
(b) 6
40.625 10 Pa
  
x 40.6 MPa
 
x 
 6 6
(0.134993)(40.625 10 ) 5.48 10 Pa
 
    
y z  5.48 MPa
 
 
y z 
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195
Ex 50 GPa
Ey 15.2 GPa
Ez 15.2 GPa
xz 0.254
xy 0.254
zy 0.428
y
z
x
PROBLEM 2.92
The composite cube of Prob. 2.91 is constrained against deformation in
the z direction and elongated in the x direction by 0.035 mm due to a
tensile load in the x direction. Determine (a) the stresses , ,
x y
  and z
and (b) the change in the dimension in the y direction.
50 GPa 0.254
15.2 GPa 0.254
15.2 GPa 0.428
x xz
y xy
z zy
E v
E v
E v
 
 
 
SOLUTION
yx y
x zx z
x
x y z
v v
E E E

 
    (1)
xy x y zy z
y
x y z
v v
E E E
  
     (2)
yz y
xz x z
z
x y z
v
v
E E E

 
     (3)
xy yx
x y
v v
E E
 (4)
yz zy
y z
v v
E E
 (5)
zx xz
z x
v v
E E
 (6)
Constraint condition: 0
Load condition : 0




z
y
From Equation (3),
1
0 xz
x z
x z
v
E E
 
  
(0.254)(15.2)
0.077216
50
  
  
xz z
z x x
x
v E
E
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196
PROBLEM 2.92
(Continued)
From Equation (1) with 0,
y
 
1 1
1 1
[ 0.254 ] [1 (0.254)(0.077216)]
0.98039
0.98039
    
  



   
   


zx xz
x x z x z
x z x x
x z x
x x
x
x
x x
x
v v
E E E E
E E
E
E
But 6
0.035 mm
875 10
40 mm
x
x
x
L

 
   
(a)
9 6
3
(50 10 )(875 10 )
44.625 10 Pa
0.98039


 
  
x 44.6 MPa
 
x 
0
y
  
6 6
(0.077216)(44.625 10 ) 3.446 10 Pa
    
z 3.45 MPa
 
z 
From (2),
6 6
9 9
6
1
(0.254)(44.625 10 ) (0.428)(3.446 10 )
0
50 10 15.2 10
323.73 10
   

  
 
   
 
  
xy zy
y x y z
x y z
v v
E E E
(b) 6
(40 mm)( 323.73 10 )
  
   
y y y
L 0.0129 mm
  
y 
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197
120 mm
60 mm r
P
15 mm
PROBLEM 2.93
Knowing that, for the plate shown, the allowable stress is 125 MPa, determine
the maximum allowable value of P when (a) r  12 mm, (b) r  18 mm.
SOLUTION
2 6 2
(60)(15) 900 mm 900 10 m
120 mm
2.00
60 mm
A
D
d

   
 
(a)
12 mm
12 mm 0.2
60 mm
  
r
r
d
max
From Fig. 2.60b, 1.92 
 
P
K K
A
6 6
3
max (900 10 )(125 10 )
58.6 10 N
1.92
A
P
K
 
 
   
58.3 kN
 
(b)
18 mm
18 mm, 0.30 From Fig 2.60b, 1.75
60 mm
   
r
r K
d
6 6
3
max (900 10 )(125 10 )
64.3 10 N
1.75
 
 
   
A
P
K
64.3 kN
 
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198
120 mm
60 mm r
P
15 mm
PROBLEM 2.94
Knowing that P  38 kN, determine the maximum stress when (a) r  10 mm,
(b) r  16 mm, (c) r  18 mm.
SOLUTION
2 6 2
(60)(15) 900 mm 900 10 m
10 mm
2.00
60 mm

   
 
A
D
d
(a)
10 mm
10 mm 0.1667
60 mm
  
r
r
d
max
From Fig. 2.60b, 2.06 
 
KP
K
A
3
6
max 6
(2.06)(38 10 )
87.0 10 Pa
900 10

  

 87.0 MPa
 
(b)
16 mm
16 mm 0.2667
60 mm
  
r
r
d
From Fig. 2.60b, 1.78

K
3
6
max 6
(1.78)(38 10 )
75.2 10 Pa
900 10

  

 75.2 MPa
 
(c)
18 mm
18 mm, 0.30
60 mm
From Fig 2.60b, 1.75
  

r
r
d
K
3
6
max 6
(1.75)(38 10 )
73.9 10 Pa
900 10
 

  

73.9 MPa
 
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199
A
d rf
P
1
2
in.
1
8
3 in.
3
8
in.
11
16
4 in.
PROBLEM 2.95
A hole is to be drilled in the plate at A. The diameters of the
bits available to drill the hole range from 1
2
to 11
/2 in. in
1
4
-in. increments. If the allowable stress in the plate is 21 ksi,
determine (a) the diameter d of the largest bit that can be
used if the allowable load P at the hole is to exceed that at
the fillets, (b) the corresponding allowable load P.
SOLUTION
At the fillets:
4.6875 0.375
1.5 0.12
3.125 3.125
D r
d d
   
From Fig. 2.60b, 2.10
K 
2
min
all
max all
min
min all
all
(3.125)(0.5) 1.5625 in
(1.5625)(21)
15.625 kips
2.10
 

 
 
  
A
P
K
A
A
P
K
At the hole: net ( 2 ) , from Fig. 2.60a
 
A D r t K
net all
max all all
net

 
   
P A
K P
A K
with all
4.6875 in. 0.5 in. 21 ksi
D t 
  
Hole diam. r 2
d D r
  2 /
r D K net
A all
P
0.5 in. 0.25 in. 4.1875 in. 0.107 2.68 2.0938 in2
16.41 kips
0.75 in. 0.375 in. 3.9375 in. 0.16 2.58 1.96875 in2
16.02 kips
1 in. 0.5 in. 3.6875 in. 0.213 2.49 1.84375 in2
15.55 kips
1.25 in. 0.625 in. 3.4375 in. 0.267 2.41 1.71875 in2
14.98 kips
1.5 in. 0.75 in. 3.1875 in. 0.32 2.34 1.59375 in2
14.30 kips
(a) Largest hole with all 15.625 kips
P  is the 3
4
-in.-diameter hole. 
(b) Allowable load all 15.63 kips
P  
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200
A
d rf
P
1
2
in.
1
8
3 in.
3
8
in.
11
16
4 in.
PROBLEM 2.96
(a) For 13 kips
P  and 1
2
in.,
d  determine the maximum
stress in the plate shown. (b) Solve part a, assuming that the
hole at A is not drilled.
SOLUTION
Maximum stress at hole:
Use Fig. 2.60a for values of K.
2
net
max
net
2 0.5
0.017, 2.68
4.6875
(0.5)(4.6875 0.5) 2.0938 in
(2.68)(13)
16.64 ksi
2.0938

  
  
  
r
K
D
A
P
K
A
Maximum stress at fillets:
Use Fig. 2.60b for values of K.
0.375 4.6875
0.12 1.5
3.125 3.125
r D
d d
    2.10
K 
2
min
max
min
(0.5)(3.125) 1.5625 in
(2.10)(13)
17.47 ksi
1.5625

 
  
A
P
K
A
(a) With hole and fillets: 17.47 ksi 
(b) Without hole: 17.47 ksi 
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201
P
9 mm
9 mm
9 mm
96 mm
60 mm
A
rf
PROBLEM 2.97
Knowing that the hole has a diameter of 9 mm, determine (a) the radius
rf of the fillets for which the same maximum stress occurs at the hole A
and at the fillets, (b) the corresponding maximum allowable load P if
the allowable stress is 100 MPa.
SOLUTION
For the circular hole,
1
(9) 4.5 mm
2
r
 
 
 
 
2 2(4.5)
96 9 87 mm 0.09375
96
r
d
D
    
6 2
net (0.087 m)(0.009 m) 783 10 m
A dt 
   
From Fig. 2.60a, hole 2.72
K 
hole
max
net
K P
A
 
6 6
3
net max
hole
(783 10 )(100 10 )
28.787 10 N
2.72
A
P
K
 
 
   
(a) For fillet, 96 mm, 60 mm
D d
 
96
1.60
60
D
d
 
6 2
min (0.060 m)(0.009 m) 540 10 m
A dt 
   
6 6
fillet min max
max fillet 3
min
(5.40 10 )(100 10 )
28.787 10
1.876
K P A
A P
K



 
  



From Fig. 2.60b, 0.19 0.19 0.19(60)
   
f
f
r
r d
d
11.4 mm
f
r  
(b) 28.8 kN
P  
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202
rA 5 20 mm
rB 5 15 mm
B
A
64 mm
88 mm
P
t
PROBLEM 2.98
For 100 kN,
P  determine the minimum plate thickness t required if the
allowable stress is 125 MPa.
SOLUTION
At the hole: 20 mm 88 40 48 mm
A A
r d
   
2 2(20)
0.455
88
A
A
r
D
 
From Fig. 2.60a, 2.20
K 
max
net max
3
3
6
(2.20)(100 10 N)
36.7 10 m 36.7 mm
(0.048 m)(125 10 Pa)
A A
KP KP KP
t
A d t d
t



   

   

At the fillet:
88
88 mm, 64 mm 1.375
64
B
B
D
D d
d
   
15
15 mm 0.2344
64
B
B
B
r
r
d
  
From Fig. 2.60b, 1.70
K 
max
min B
KP KP
A d t
  
3
3
6
max
(1.70)(100 10 N)
21.25 10 m 21.25 mm
(0.064 m)(125 10 Pa)
B
KP
t
d 


    

The larger value is the required minimum plate thickness.
 36.7 mm
t  
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203
P
P
t 5
2 in.
3 in.
5
8
in.
r 5 1
4
in.
PROBLEM 2.99
(a) Knowing that the allowable stress is 20 ksi, determine the maximum
allowable magnitude of the centric load P. (b) Determine the percent
change in the maximum allowable magnitude of P if the raised portions
are removed at the ends of the specimen.
SOLUTION
3 0.250
1.50 0.125
2 2
D r
d d
   
From Fig. 2.60b, K  2.08
2
min (0.625)(2) 1.25 in
  
A td
(a) min max
max
min
(1.25)(20)
12.0192 kips
2.08

    
A
KP
P
A K
12.02 kips
P  
(b) Without raised section, K  1.00
min max (1.25)(20) 25 kips
P A 
  
25 12.02
% change 100%
12.02

 
 
 
 
108.0%
 
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204
3
4
in.
1
2
in.
1
2
in.
5 in.
1 in.
6
rf 5
P
PROBLEM 2.100
A centric axial force is applied to the steel bar shown. Knowing that
all 20 ksi
  , determine the maximum allowable load P.
SOLUTION
At the hole: 0.5 in. d 5 1 4 in.
   
r
2 2(0.5)
0.2 From Fig. 2.60a, 2.51
5
  
r
K
d
2
net (0.75)(4) 3 in
  
A td
max
net

KP
A

net max (3)(20)
23.9 kips
2.51
  
A
P
K

At the fillet :
6.5
6.5 in., d 5 in., 1.3
5
D
D
d
   
0.5
0.5 in. 0.1
5
r
r
d
  
From Fig. 2.60b, K  2.04
2
min (0.75)(5) 3.75 in
  
A td
max
min

KP
A

min max (3.75)(20)
36.8 kips
2.04
  
A
P
K

Smaller value for P controls. 23.9 kips
P  
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205
L
B
A
P
PROBLEM 2.101
The cylindrical rod AB has a length L  5 ft and a 0.75-in. diameter; it is made of a mild steel
that is assumed to be elastoplastic with E  29 × 106
psi and 36 ksi

Y
 . A force P is applied
to the bar and then removed to give it a permanent set .
P Determine the maximum value of
the force P and the maximum amount m
 by which the bar should be stretched if the desired
value of P is (a) 0.1 in., (b) 0.2 in.
SOLUTION
2 2 2
(0.75) 0.44179 in 5 ft 60 in.
4 4
    
A d L
 
3
3
(60)(36 10 )
0.074483 in.
29 10

 

   

Y
y Y
L
L
E
When m
 exceeds ,
Y thus causing permanent stretch ,
p the maximum force is
3 3
(0.44179)(36 10 ) 15.9043 10 lb

    
m Y
P A
15.90 kips
P  
'
so that
       
     
p m m Y m p Y
(a) 0.1in. 0.1 0.074483 0.1745 in.
 
   
p m 
(b) 0.2 in. 0.2 0.074483 0.274 in.
 
   
p m 
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206
L
B
A
P
PROBLEM 2.102
The cylindrical rod AB has a length L  6 ft and a 1.25-in. diameter; it is made of a mild steel
that is assumed to be elastoplastic with E  29 × 106
psi and 36 ksi

Y
 . A force P is
applied to the bar until end A has moved down by an amount m
 . Determine the maximum
value of the force P and the permanent set of the bar after the force has been removed,
knowing (a) 0.125 in.,

m
 (b) 0.250 in.

m

SOLUTION
2 2 2
(1.25) 1.22718 in 6 ft 72 in.
4 4
 
    
A d L
3
3
(72)(36 10 )
0.089379 in.
29 10

 

   

Y
Y Y
L
L
E
If 3
, (1.22718)(36 10 )
  
   
m Y m Y
P A
3
44.179 10 lb 44.2 kips
  
(a) 0.125 in. so that 44.2 kips
 
 
m Y m
P 
 0.089379

 
    
m Y
Y
P L L
AE E
0.125 0.089379 0.356 in.
  
    
p m 
(b) 0.250 in. so that 44.2 kips
 
 
m Y m
P 
  
  Y
0.250 0.089379 0.1606 in.
  
    
p m 
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207
6 mm
9-mm diameter
0.4 m
0.7 m
1.25 m
C B
D
A
C⬘
1
d
PROBLEM 2.103
Rod AB is made of a mild steel that is assumed to be elastoplastic
with E  200 GPa and 345 MPa

Y
 . After the rod has been
attached to the rigid lever CD, it is found that end C is 6 mm too
high. A vertical force Q is then applied at C until this point has
moved to position 
C . Determine the required magnitude of Q and
the deflection 1
 if the lever is to snap back to a horizontal position
after Q is removed.
SOLUTION
2 2 6 2
(9) 63.617 mm 63.617 10 m
4
AB
A
 
   
Since rod AB is to be stretched permanently,
6 6
max
( ) (63.617 10 )(345 10 )
AB AB Y
F A  
   
3
21.948 10 N
 
0: 1.1 0.7 0
   
D AB
M Q F
3 3
max
0.7
(21.948 10 ) 13.9669 10 N
1.1

   
Q 13.97 kN 
3
3
max
9 6
( ) (21.948 10 )(1.25)
2.15625 10 m
(200 10 )(63.617 10 )
 


    
 
AB AB
AB
AB
F L
EA
3
3.0804 10 rad
0.7

 

   
AB
3
1 1.1 3.39 10 m
  

   3.39 mm 
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208
6 mm
9-mm diameter
0.4 m
0.7 m
1.25 m
C B
D
A
C⬘
1
d
PROBLEM 2.104
Solve Prob. 2.103, assuming that the yield point of the mild steel is
250 MPa.
PROBLEM 2.103 Rod AB is made of a mild steel that is assumed to
be elastoplastic with E  200 GPa and 345 MPa

Y
 . After the rod
has been attached to the rigid lever CD, it is found that end C is 6 mm
too high. A vertical force Q is then applied at C until this point has
moved to position 
C . Determine the required magnitude of Q and
the deflection 1
 if the lever is to snap back to a horizontal position
after Q is removed.
SOLUTION
2 2 6 2
(9) 63.617 mm 63.617 10 m
4
AB
A
 
   
Since rod AB is to be stretched permanently,
6 6
max
( ) (63.617 10 )(250 10 )
AB AB Y
F A  
   
3
15.9043 10 N
 
0: 1.1 0.7 0
   
D AB
M Q F
3 3
max
0.7
(15.9043 10 ) 10.12 10 N
1.1
   
Q 10.12 kN 
3
3
max
9 6
( ) (15.9043 10 )(1.25)
1.5625 10 m
(200 10 )(63.617 10 )



    
 
AB AB
AB
AE
F L
EA

3
2.2321 10 rad
0.7


   
AB


3
1 1.1 2.46 10 m

   
  2.46 mm 
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209
P
40-mm
diameter
30-mm
diameter
1.2 m
0.8 m
C
B
A
PROBLEM 2.105
Rod ABC consists of two cylindrical portions AB and BC; it is made of a mild
steel that is assumed to be elastoplastic with 200 GPa
E  and 250 MPa.
Y
 
A force P is applied to the rod and then removed to give it a permanent set
2
p
  mm. Determine the maximum value of the force P and the maximum
amount m
 by which the rod should be stretched to give it the desired permanent
set.
SOLUTION
2 2 6 2
2 3 2 3 2
6 6 3
max min
(30) 706.86 mm 706.86 10 m
4
(40) 1.25664 10 mm 1.25664 10 m
4
(706.86 10 )(250 10 ) 176.715 10 N
AB
BC
Y
A
A
P A






   
    
     
max 176.7 kN
P  
3 3
9 6 9 3
3
(176.715 10 )(0.8) (176.715 10 )(1.2)
(200 10 )(706.86 10 ) (200 10 )(1.25664 10 )
1.84375 10 m 1.84375 mm
BC
AB
AB BC
P L
P L
EA EA
  


  
    
   
  
or 2 1.84375
p m m p
     
 
      3.84 mm
m
  
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210
P
40-mm
diameter
30-mm
diameter
1.2 m
0.8 m
C
B
A
PROBLEM 2.106
Rod ABC consists of two cylindrical portions AB and BC; it is made of a mild
steel that is assumed to be elastoplastic with 200 GPa
E  and 250 MPa.
Y
 
A force P is applied to the rod until its end A has moved down by an amount
5 mm.
m
  Determine the maximum value of the force P and the permanent
set of the rod after the force has been removed.
SOLUTION
2 2 6 2
2 3 2 3 2
6 6 3
max min
(30) 706.86 mm 706.86 10 m
4
(40) 1.25664 10 mm 1.25644 10 m
4
(706.86 10 )(250 10 ) 176.715 10 N
AB
BC
Y
A
A
P A






   
    
     
max 176.7 kN
P  
3 3
9 6 9 3
3
(176.715 10 )(0.8) (176.715 10 )(1.2)
(200 10 )(706.68 10 ) (200 10 )(1.25664 10 )
1.84375 10 m 1.84375 mm
BC
AB
AB BC
P L
P L
EA EA
  


  
    
   
  
5 1.84375 3.16 mm
p m
  
     3.16 mm
p
  
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211
190 mm
190 mm
C
B
A
P
PROBLEM 2.107
Rod AB consists of two cylindrical portions AC and BC, each with a cross-sectional
area of 1750 mm2
. Portion AC is made of a mild steel with 200 GPa
E  and
250 MPa,
Y
  and portion CB is made of a high-strength steel with 200 GPa
E 
and 345 MPa.
Y
  A load P is applied at C as shown. Assuming both steels to be
elastoplastic, determine (a) the maximum deflection of C if P is gradually increased
from zero to 975 kN and then reduced back to zero, (b) the maximum stress in each
portion of the rod, (c) the permanent deflection of C.
SOLUTION
Displacement at C to cause yielding of AC.
6
, 3
, , 9
(0.190)(250 10 )
0.2375 10 m
200 10

  

    

AC Y AC
C Y AC Y AC
L
L
E
Corresponding force. 6 6 3
, (1750 10 )(250 10 ) 437.5 10 N
 
     
AC Y AC
F A
9 6 3
3
(200 10 )(1750 10 )(0.2375 10 )
437.5 10 N
0.190
C
CB
CB
EA
F
L
  
  
      
For equilibrium of element at C,
3
( ) 0 875 10 N
AC CB Y Y AC CB
F F P P F F
      
Since applied load 3 3
975 10 N 875 10 N,
P     portion AC yields.
3 3 3
437.5 10 975 10 N 537.5 10 N
CB AC
F F P
        
(a)
3
3
9 6
(537.5 10 )(0.190)
0.29179 10 m
(200 10 )(1750 10 )
CB CD
C
F L
EA
 


    
 
0.292 mm 
(b) Maximum stresses: , 250 MPa
 
 
AC Y AC 250 MPa 
3
6
6
537.5 10
307.14 10 Pa 307 MPa
1750 10
 

       

BC
BC
F
A
307 MPa
 
(c) Deflection and forces for unloading.
 3 3
3
3
9 6
975 10 2 487.5 10 N
(487.5 10 )(0.190)
0.26464 10 m
(200 10 )(1750 10 )


 
   
       
    
      

   
 
AC AC CB CB AC
CB AC AC
AB
AC CB AC AC
P L P L L
P P P
EA EA L
P P P P P


3 3
3
0.29179 10 0.26464 10
0.02715 10 m
    


     
 
p m
0.0272 mm 
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212
190 mm
190 mm
C
B
A
P
PROBLEM 2.108
For the composite rod of Prob. 2.107, if P is gradually increased from zero until the
deflection of point C reaches a maximum value of 0.3 mm
m
  and then decreased
back to zero, determine (a) the maximum value of P, (b) the maximum stress in each
portion of the rod, (c) the permanent deflection of C after the load is removed.
PROBLEM 2.107 Rod AB consists of two cylindrical portions AC and BC, each with a
cross-sectional area of 1750 mm2
. Portion AC is made of a mild steel with 200
E  GPa
and 250 MPa,
Y
  and portion CB is made of a high-strength steel with 200

E GPa
and 345 MPa.
Y
  A load P is applied at C as shown. Assuming both steels to be
elastoplastic, determine (a) the maximum deflection of C if P is gradually increased
from zero to 975 kN and then reduced back to zero, (b) the maximum stress in each
portion of the rod, (c) the permanent deflection of C.
SOLUTION
Displacement at C is 0.30 mm.
m
  The corresponding strains are
3
3
0.30 mm
1.5789 10
190 mm
0.30 mm
1.5789 10
190 mm
m
AC
AC
m
CB
CB
L
L






   
      
Strains at initial yielding:
6
, 3
, 9
6
, 3
, 9
250 10
1.25 10 (yielding)
200 10
345 10
1.725 10 (elastic)
200 10







   


     

Y AC
Y AC
Y BC
Y CB
E
E
(a) Forces: 6 6 3
(1750 10 )(250 10 ) 437.5 10 N
AC Y
F A  
     
9 6 3 3
(200 10 )(1750 10 )( 1.5789 10 ) 552.6 10 N
CB CB
F EA   
        
For equilibrium of element at C, 0
AC CB
F F P
  
3 3 3
437.5 10 552.6 10 990.1 10 N
       
AC CD
P F F 990 kN 
(b) Stresses: ,
:  

AC Y AC
AC 250 MPa 
3
6
6
552.6 10
: 316 10 Pa
1750 10
 

     

CB
CB
F
CB
A
316 MPa
 
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213
PROBLEM 2.108 (Continued)
(c) Deflection and forces for unloading.
3 3
3
3
9 6
2 990.1 10 N 495.05 10 N
(495.05 10 )(0.190)
0.26874 10 m 0.26874 mm
(200 10 )(1750 10 )
AC AC CB CB AC
CB AC AC
AB
AC CB AC AC
P L P L L
P P P
EA EA L
P P P P P

 

 
  
       
    
       

    
 
0.30 mm 0.26874 mm
  
   
p m 0.0313mm 
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214
1 m
A
B C
Q
D E
1 m
2 m
PROBLEM 2.109
Each cable has a cross-sectional area of 100 mm2
and is made of an
elastoplastic material for which 345 MPa
Y
  and 200 GPa.
E  A force
Q is applied at C to the rigid bar ABC and is gradually increased from 0 to
50 kN and then reduced to zero. Knowing that the cables were initially taut,
determine (a) the maximum stress that occurs in cable BD, (b) the
maximum deflection of point C, (c) the final displacement of point C.
(Hint: In part c, cable CE is not taut.)
SOLUTION
Elongation constraints for taut cables.
Let rotation angle of rigid bar .
  ABC
CE
BD
AB AC
L L


  
1
2
  
 
AB
BD CE CE
AC
L
L
(1)
Equilibrium of bar ABC.
0: 0
A AB BD AC CE AC
M L F L F L Q
   
1
2
AB
CE BD CE BD
AC
L
Q F F F F
L
    (2)
Assume cable CE is yielded. 6 6 3
(100 10 )(345 10 ) 34.5 10 N
CE Y
F A 
     
From (2), 3 3 3
2( ) (2)(50 10 34.5 10 ) 31.0 10 N
BD CE
F Q F
       
Since 3
34.5 10 N,
BD Y
F A   cable BD is elastic when 50 kN.
Q 
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215
PROBLEM 2.109 (Continued)
(a) Maximum stresses. 345 MPa
CE Y
 
 
3
6
6
31.0 10
310 10 Pa
100 10
BD
BD
F
A
 

   

310 MPa
BD
  
(b) Maximum of deflection of point C.
3
3
9 6
(31.0 10 )(2)
3.1 10 m
(200 10 )(100 10 )
BD BD
BD
F L
EA
 


   
 
From (1), 3
2 6.2 10 m
C CE BD
   
    6.20 mm  
Permanent elongation of cable CE: ( ) ( )

 
  Y CE
CE p CE
L
E
max
6
3 3
9
( ) ( )
(345 10 )(2)
6.20 10 2.75 10 m
200 10

 
 
 

    

Y CE
CE P CE
L
E
(c) Unloading. Cable CE is slack ( 0)
CE
F  at 0.
Q 
From (2), 2( ) 2(0 0) 0
BD CE
F Q F
    
Since cable BD remained elastic, 0.
BD BD
BD
F L
EA
   0 
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216
1 m
A
B C
Q
D E
1 m
2 m
PROBLEM 2.110
Solve Prob. 2.109, assuming that the cables are replaced by rods of the
same cross-sectional area and material. Further assume that the rods are
braced so that they can carry compressive forces.
PROBLEM 2.109 Each cable has a cross-sectional area of 100 mm2
and is made of an elastoplastic material for which 345 MPa
Y
  and
200 GPa.
E  A force Q is applied at C to the rigid bar ABC and is
gradually increased from 0 to 50 kN and then reduced to zero. Knowing
that the cables were initially taut, determine (a) the maximum stress
that occurs in cable BD, (b) the maximum deflection of point C, (c) the
final displacement of point C. (Hint: In part c, cable CE is not taut.)
SOLUTION
Elongation constraints.
Let rotation angle of rigid bar .
ABC
 
 
  
BC CE
AB AC
L L
1
2
  
 
AB
BD CE CE
AC
L
L
(1)
Equilibrium of bar ABC.
0: 0
   
A AB BD AC CE AC
M L F L F L Q
1
2
AB
CE BD CE BD
AC
L
Q F F F F
L
    (2)
Assume cable CE is yielded. 6 6 3
(100 10 )(345 10 ) 34.5 10 N
 
     
CE Y
F A
From (2), 3 3 3
2( ) (2)(50 10 34.5 10 ) 31.0 10 N
       
BD CE
F Q F
Since 3
34.5 10 N,

  
BD Y
F A cable BD is elastic when 50 kN.
Q 
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217
PROBLEM 2.110 (Continued)
(a) Maximum stresses. 345 MPa
CE Y
 
 
3
6
6
31.0 10
310 10 Pa
100 10
BD
BD
F
A
 

   

310 MPa
BD
  
(b) Maximum of deflection of point C.
3
3
9 6
(31.0 10 )(2)
3.1 10 m
(200 10 )(100 10 )
BD BD
BD
F L
EA
 


   
 
From (1), 3
2 6.2 10 m
C CE BD
   
    6.20 mm  
Unloading. 3
50 10 N, CE C
Q  
  
  
From (1), 1
2
BD C
 
 

Elastic
9 6 1
6
2
(200 10 )(100 10 )( )
5 10
2





 

 
   
C
BD
BD C
BD
EA
F
L
9 6
6
(200 10 )(100 10 )( )
10 10
2
 


 
 
 
   
CE C
CE C
CE
EA
F
L
From (2), 6
1
2
12.5 10 
   
   
CE BD C
Q F F
Equating expressions for ,

Q 6 3
12.5 10 50 10

  
C
3
4 10 m
C
 
  
(c) Final displacement. 3 3 3
( ) 6.2 10 4 10 2.2 10 m
C C m C
     

        2.20 mm  
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218
P
14 in.
2.0 in.
P'
in.
1
2
in.
3
16 3
16
in.
PROBLEM 2.111
Two tempered-steel bars, each 3
16
in. thick, are bonded to a 1
2
-in. mild-steel
bar. This composite bar is subjected as shown to a centric axial load of
magnitude P. Both steels are elastoplastic with 6
29 10 psi
 
E and with
yield strengths equal to 100 ksi and 50 ksi, respectively, for the tempered
and mild steel. The load P is gradually increased from zero until the
deformation of the bar reaches a maximum value 0.04 in.
m
  and then
decreased back to zero. Determine (a) the maximum value of P, (b) the
maximum stress in the tempered-steel bars, (c) the permanent set after the
load is removed.
SOLUTION
For the mild steel, 2
1
1
(2) 1.00 in
2
 
 
 
 
A
3
1
1 6
(14)(50 10 )
0.024138 in.
29 10



  

Y
Y
L
E
For the tempered steel, 2
2
3
2 (2) 0.75 in
16
 
 
 
 
A
3
2
2 3
(14)(100 10 )
0.048276 in.
29 10



  

Y
Y
L
E
Total area: 2
1 2 1.75 in
  
A A A
1 2.
Y m Y
  
  The mild steel yields. Tempered steel is elastic.
(a) Forces: 3 3
1 1 1 (1.00)(50 10 ) 50 10 lb

    
Y
P A
3
3
2
2
(29 10 )(0.75)(0.04)
62.14 10 lb
14
 
   
m
EA
P
L
3
1 2 112.14 10 lb 112.1 kips
    
P P P 112.1 kips

P 
(b) Stresses: 3
1
1 1
1
50 10 psi 50 ksi
Y
P
A
 
    
3
3
2
2
2
62.14 10
82.86 10 psi 82.86 ksi
0.75


    
P
A
82.86 ksi 
Unloading:
3
6
(112.14 10 )(14)
0.03094 in.
(29 10 )(1.75)


   

PL
EA
(c) Permanent set: 0.04 0.03094 0.00906 in.
p m
  
     0.00906 in. 
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219
P
14 in.
2.0 in.
P'
in.
1
2
in.
3
16 3
16
in.
PROBLEM 2.112
For the composite bar of Prob. 2.111, if P is gradually increased from zero
to 98 kips and then decreased back to zero, determine (a) the maximum
deformation of the bar, (b) the maximum stress in the tempered-steel bars,
(c) the permanent set after the load is removed.
PROBLEM 2.111 Two tempered-steel bars, each 3
16
in. thick, are bonded to
a 1
2
-in. mild-steel bar. This composite bar is subjected as shown to a centric
axial load of magnitude P. Both steels are elastoplastic with 6
29 10
E   psi
and with yield strengths equal to 100 ksi and 50 ksi, respectively, for the
tempered and mild steel. The load P is gradually increased from zero until
the deformation of the bar reaches a maximum value 0.04 in.
m
  and
then decreased back to zero. Determine (a) the maximum value of P, (b) the
maximum stress in the tempered-steel bars, (c) the permanent set after the
load is removed.
SOLUTION
Areas: Mild steel: 2
1
1
(2) 1.00 in
2
 
 
 
 
A
Tempered steel: 2
2
3
2 (2) 0.75 in
16
 
 
 
 
A
Total: 2
1 2 1.75 in
  
A A A
Total force to yield the mild steel:
3 3
1 1 (1.75)(50 10 ) 87.50 10 lb
Y
Y Y Y
P
P A
A
 
      
,
Y
P P
 therefore, mild steel yields.
Let 1 force carried by mild steel.
P 
2 force carried by tempered steel.
P 
3 3
1 1 1 (1.00)(50 10 ) 50 10 lb

    
P A
3 3 3
1 2 2 1
, 98 10 50 10 48 10 lb
P P P P P P
         
(a)
3
2
6
2
(48 10 )(14)
(29 10 )(0.75)


 

m
P L
EA
0.0309 in. 
(b)
3
3
2
2
2
48 10
64 10 psi
0.75


   
P
A
64.0 ksi 
Unloading:
3
6
(98 10 )(14)
0.02703 in.
(29 10 )(1.75)


   

PL
EA
(c) 0.03090 0.02703 0.003870 in.
P m
  
    
 0.00387 in. 
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220
1.7 m
1 m
2.64 m
C
B
E
D
A
Q
a
PROBLEM 2.113
The rigid bar ABC is supported by two links, AD and BE, of uniform
37.5 6-mm
 rectangular cross section and made of a mild steel that is
assumed to be elastoplastic with 200 GPa
E  and 250 MPa.
Y
 
The magnitude of the force Q applied at B is gradually increased
from zero to 260 kN. Knowing that 0.640 m,
a  determine (a) the
value of the normal stress in each link, (b) the maximum deflection
of point B.
SOLUTION
Statics: 0: 0.640( ) 2.64 0
C BE AD
M Q P P
    
Deformation: 2.64 , 0.640
    
  
A B a
Elastic analysis:
2 6 2
9 6
6
6 6
9
9 6
6
6 6
(37.5)(6) 225 mm 225 10 m
(200 10 )(225 10 )
26.47 10
1.7
(26.47 10 )(2.64 ) 69.88 10
310.6 10
(200 10 )(225 10 )
45 10
1.0
(45 10 )(0.640 ) 28.80 10
  
 
 
  
 




   
 
   
   
  
 
   
   
AD A A A
AD
AD
AD
BE B B B
BE
A
EA
P
L
P
A
EA
P
L
9
128 10 
  
BE
BE
P
A
From statics,
2.64
4.125
0.640
BE AD BE AD
Q P P P P
   
6 6 6
[28.80 10 (4.125)(69.88 10 )] 317.06 10
 
     
Y
 at yielding of link AD: 6 9
250 10 310.6 10
  
    
AD Y
6
6 6 3
804.89 10
(317.06 10 )(804.89 10 ) 255.2 10 N
 

 
    
Y
Y
Q
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221
PROBLEM 2.113 (Continued)
(a) Since 3
260 10 ,
Y
Q Q
   link AD yields. 250 MPa
AD
  
6 6 3
(225 10 )(250 10 ) 56.25 10 N
  
     
AD Y
P A
From statics, 3 3
4.125 260 10 (4.125)(56.25 10 )
BE AD
P Q P
     
3
3
6
6
27.97 10 N
27.97 10
124.3 10 Pa
225 10
 
 

   

BE
BE
BE
P
P
A
124.3 MPa
 
BE 
(b)
3
6
9 6
(27.97 10 )(1.0)
621.53 10 m
(200 10 )(225 10 )
BE BE
B
P L
EA
 


   
 
0.622 mm
B
   
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222
1.7 m
1 m
2.64 m
C
B
E
D
A
Q
a
PROBLEM 2.114
Solve Prob. 2.113, knowing that a  1.76 m and that the magnitude of
the force Q applied at B is gradually increased from zero to 135 kN.
PROBLEM 2.113 The rigid bar ABC is supported by two links, AD
and BE, of uniform 37.5 6-mm
 rectangular cross section and made
of a mild steel that is assumed to be elastoplastic with 200 GPa
E 
and 250 MPa.
Y
  The magnitude of the force Q applied at B is
gradually increased from zero to 260 kN. Knowing that 0.640 m,

a
determine (a) the value of the normal stress in each link, (b) the
maximum deflection of point B.
SOLUTION
Statics: 0: 1.76( ) 2.64 0
C BE AD
M Q P P
    
Deformation: 2.64 , 1.76
   
 
A B
Elastic Analysis:
2 6 2
9 6
6
6 6
9
9 6
6
6 6
(37.5)(6) 225 mm 225 10 m
(200 10 )(225 10 )
26.47 10
1.7
(26.47 10 )(2.64 ) 69.88 10
310.6 10
(200 10 )(225 10 )
45 10
1.0
(45 10 )(1.76 ) 79.2 10
  
 
 
  
 




   
 
   
   
  
 
   
   
AD A A A
AD
AD
AD
BE B B B
BE
BE
A
EA
P
L
P
A
EA
P
L
9
352 10 
  
BE
P
A
From statics,
2.64
1.500
1.76
   
BE AD BE AD
Q P P P P
6 6 6
[73.8 10 (1.500)(69.88 10 )] 178.62 10
 
     
Y
 at yielding of link BE: 6 9
250 10 352 10
  
    
BE Y Y
6
6 6 3
710.23 10
(178.62 10 )(710.23 10 ) 126.86 10 N
Y
Y
Q
 

 
    
(a) Since 3
135 10 N ,
Y
Q Q
   link BE yields. 250 MPa
BE Y
 
  
6 6 3
(225 10 )(250 10 ) 56.25 10 N
 
     
BE Y
P A


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223
PROBLEM 2.114 (Continued)
From statics, 3
1
( ) 52.5 10 N
1.500
   
AD BE
P Q P
3
6
6
52.5 10
233.3 10
225 10
 

   

AD
AD
P
A
233 MPa
 
AD 
From elastic analysis of AD, 3
6
751.29 10 rad
69.88 10
 
  

AD
P
(b) 3
1.76 1.322 10 m
B
  
   1.322 mm
B
   
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224
1.7 m
1 m
2.64 m
C
B
E
D
A
Q
a
PROBLEM 2.115
Solve Prob. 2.113, assuming that the magnitude of the force Q
applied at B is gradually increased from zero to 260 kN and then
decreased back to zero. Knowing that 0.640 m,
a  determine (a) the
residual stress in each link, (b) the final deflection of point B.
Assume that the links are braced so that they can carry compressive
forces without buckling.
PROBLEM 2.113 The rigid bar ABC is supported by two links, AD
and BE, of uniform 37.5 6-mm
 rectangular cross section and made
of a mild steel that is assumed to be elastoplastic with 200 GPa
E 
and 250 MPa.
Y
  The magnitude of the force Q applied at B is
gradually increased from zero to 260 kN. Knowing that
0.640 m,
a  determine (a) the value of the normal stress in each
link, (b) the maximum deflection of point B.
SOLUTION
See solution to Problem 2.113 for the normal stresses in each link and the deflection of Point B after loading.
6
6
6
250 10 Pa
124.3 10 Pa
621.53 10 m
AD
BE
B


 
 
 
 
The elastic analysis given in the solution to Problem 2.113 applies to the unloading.
6
3
6
6 6
317.06 10
260 10
820.03 10
317.06 10 317.06 10


 
 

    
 
Q
Q
Q
9 9 6 6
9 9 6 6
6
310.6 10 (310.6 10 )(820.03 10 ) 254.70 10 Pa
128 10 (128 10 )(820.03 10 ) 104.96 10 Pa
0.640 524.82 10 m
 
 
 



       
       
 
  
AD
BE
B
(a) Residual stresses.
6 6 6
, res 250 10 254.70 10 4.70 10 Pa
   

        
AD AD AD 4.70MPa
  
6 6 6
, res 124.3 10 104.96 10 19.34 10 Pa
  
       
BE BE BE 19.34 MPa
 
(b) 6 6 6
, 621.53 10 524.82 10 96.71 10 m
     

       
B P B B 0.0967 mm
  
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225
L
B
A
PROBLEM 2.116
A uniform steel rod of cross-sectional area A is attached to rigid supports and is
unstressed at a temperature of 45 F.
 The steel is assumed to be elastoplastic
with 36
Y
  ksi and 6
29 10 psi.
E   Knowing that 6
6.5 10 / F,
 
  
determine the stress in the bar (a) when the temperature is raised to 320 F,

(b) after the temperature has returned to 45 F
 .
SOLUTION
Let P be the compressive force in the rod.
Determine temperature change to cause yielding.
3
6 6
( ) ( ) 0
36 10
( ) 190.98 F
(29 10 )(6.5 10 )

  

 
        

    
 
Y
Y
Y
Y
L
PL
L T L T
AE E
T
E
But 320 45 275 F ( )
       Y
T T
(a) Yielding occurs. 36.0 ksi
Y
 
    
Cooling:
6 6 3
( T) 275 F
( ) 0
( )
(29 10 )(6.5 10 )(275) 51.8375 10 psi
   
 


  

   
      

 
   
      
P T
P L
L T
AE
P
E T
A
(b) Residual stress:
3 3
res 36 10 51.8375 10 15.84 10 psi

         
Y
   15.84 ksi 
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226
B
A C
A 500 mm2
A 300 mm2
150 mm 250 mm
PROBLEM 2.117
The steel rod ABC is attached to rigid supports and is unstressed
at a temperature of 25 C.
 The steel is assumed elastoplastic, with
E  200 GPa and 250 MPa.
y
  The temperature of both
portions of the rod is then raised to 150 C
 . Knowing that
6
11.7 10 / C,
 
   determine (a) the stress in both portions of the
rod, (b) the deflection of point C.
SOLUTION
6 2
6 2
500 10 m 0.150 m
300 10 m 0.250 m


  
  
AC AC
CB CB
A L
A L
Constraint: 0
P T
 
 
Determine T to cause yielding in portion CB.
( )
AC CB
AB
AC CB
AC CB
AB AC CB
PL PL
L T
EA EA
L L
P
T
L E A A


   
 
  
 
 
At yielding, 6 6 3
(300 10 )(2.50 10 ) 75 10 N
Y CB Y
P P A  
      
3
9 6 6 6
( )
75 10 0.150 0.250
90.812 C
(0.400)(200 10 )(11.7 10 ) 500 10 300 10

  
 
  
 
 
  
   
 
   
 
AC CB
Y
Y
AB AC CB
L L
P
T
L E A A
Actual T: 150 C 25 C 125 C ( )
       Y
T
Yielding occurs. For 3
( ) , 75 10 N
Y Y
T T P P
     
(a)
3
6
6
75 10
150 10 Pa
500 10
Y
AC
AC
P
A
 


      

150.0 MPa
AC
   
 
   
Y
CB Y
CB
P
A
250 MPa
CB
   
(b) For ( ) , portion remains elastic.
Y
T T AC
  
/
3
6 6
9 6
( )
(75 10 )(0.150)
(0.150)(11.7 10 )(125) 106.9 10 m
(200 10 )(500 10 )
Y AC
C A AC
AC
P L
L T
EA
 
 

   

     
 
Since Point A is stationary, 6
/ 106.9 10 m
C C A
  
   0.1069 mm
C
   
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227
B
A C
A 500 mm2
A 300 mm2
150 mm 250 mm
PROBLEM 2.118
Solve Prob. 2.117, assuming that the temperature of the rod is raised to
150C and then returned to 25 C.

PROBLEM 2.117 The steel rod ABC is attached to rigid supports and is
unstressed at a temperature of 25 C.
 The steel is assumed elastoplastic,
with 200

E GPa and 250
 
Y MPa. The temperature of both portions
of the rod is then raised to 150 C.
 Knowing that 6
11.7 10 / C,
 
  
determine (a) the stress in both portions of the rod, (b) the deflection
of point C.
SOLUTION
6 2 6 2
500 10 m 0.150 m 300 10 m 0.250 m
 
     
AC AC CB CB
A L A L
Constraint: 0
P T
 
 
Determine T to cause yielding in portion CB.
( )


   
 
  
 
 
AC CB
AB
AC CB
AC CB
AB AC CB
PL PL
L T
EA EA
L L
P
T
L E A A
At yielding, 6 6 3
(300 10 )(250 10 ) 75 10 N
Y CB Y
P P A  
      
3
9 6 6 6
75 10 0.150 0.250
( )
(0.400)(200 10 )(11.7 10 ) 500 10 300 10
90.812 C
   
    
    
   
   
 
 
 
AC CB
Y
Y
AB AC CB
L L
P
T
L E A A
Actual : 150 C 25 C 125 C ( )
        Y
T T
Yielding occurs. For 3
( ) , 75 10 N
     
Y Y
T T P P
Cooling:
 
( )
( ) 125 C
 

 
   

AC CB
AC CB
AB
L L
A A
EL T
T P
6 6
9 6
3
0.150 0.250
500 10 300 10
(200 10 )(0.400)(11.7 10 )(125)
103.235 10 N
 

 
 
  

Residual force: 3 3 3
res 103.235 10 75 10 28.235 10 N (tension)

       
Y
P P P
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228
PROBLEM 2.118
(Continued)
(a) Residual stresses.
3
res
6
28.235 10
500 10
AC
AC
P
A
 

 

56.5 MPa
AC
  
3
res
6
28.235 10
300 10
CB
CB
P
A
 

 

9.41 MPa
 
CB 
(b) Permanent deflection of point C. res
  AC
C
AC
P L
EA
0.0424 mm
C
   
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229
P
14 in.
2.0 in.
P'
in.
1
2
in.
3
16 3
16
in.
PROBLEM 2.119
For the composite bar of Prob. 2.111, determine the residual stresses in
the tempered-steel bars if P is gradually increased from zero to
98 kips and then decreased back to zero.
PROBLEM 2.111 Two tempered-steel bars, each 3
16
in. thick, are
bonded to a 1
2
-in. mild-steel bar. This composite bar is subjected as
shown to a centric axial load of magnitude P. Both steels are
elastoplastic with 6
29 10
E   psi and with yield strengths equal to
100 ksi and 50 ksi, respectively, for the tempered and mild steel. The
load P is gradually increased from zero until the deformation of the bar
reaches a maximum value 0.04
m
  in. and then decreased back to
zero. Determine (a) the maximum value of P, (b) the maximum stress
in the tempered-steel bars, (c) the permanent set after the load is
removed.
SOLUTION
Areas. Mild steel: 2
1
1
(2) 1.00 in
2
 
 
 
 
A
Tempered steel: 2
2
3
(2) (2) 0.75 in
16
 
 
 
 
A
Total: 2
1 2 1.75 in
  
A A A
Total force to yield the mild steel: 3 3
1 1 (1.75)(50 10 ) 87.50 10 lb
 
      
Y
Y Y Y
P
P A
A
P PY; therefore, mild steel yields.
Let 1
2
3 3
1 1 1
3 3 3
1 2 2 1
3
3
2
2
2
force carried by mild steel
force carried by tempered steel
(1.00)(50 10 ) 50 10 lb
, 98 10 50 10 48 10 lb
48 10
64 10 psi
0.75




    
         

   
Y
P
P
P A
P P P P P P
P
A
Unloading.
3
3
98 10
56 10 psi
1.75
P
A


    
Residual stresses.
Mild steel: 3 3 3
1,res 1 50 10 56 10 6 10 psi
   

         6 ksi
 
Tempered steel: 3 3 3
2,res 2 1 64 10 56 10 8 10 psi
  
        8.00 ksi 
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230
P
14 in.
2.0 in.
P'
in.
1
2
in.
3
16 3
16
in.
PROBLEM 2.120
For the composite bar in Prob. 2.111, determine the residual stresses in
the tempered-steel bars if P is gradually increased from zero until the
deformation of the bar reaches a maximum value 0.04
m
  in. and is
then decreased back to zero.
PROBLEM 2.111 Two tempered-steel bars, each 3
16
in. thick, are bonded
to a 1
2
-in. mild-steel bar. This composite bar is subjected as shown to a
centric axial load of magnitude P. Both steels are elastoplastic
with 6
29 10
E   psi and with yield strengths equal to 100 ksi and 50 ksi,
respectively, for the tempered and mild steel. The load P is gradually
increased from zero until the deformation of the bar reaches a maximum
value 0.04
m
  in. and then decreased back to zero. Determine (a) the
maximum value of P, (b) the maximum stress in the tempered-steel bars,
(c) the permanent set after the load is removed.
SOLUTION
For the mild steel,
3
2 1
1 1 6
1 (14)(50 10 )
(2) 1.00 in 0.024138 in.
2 29 10



 
    
 

 
Y
Y
L
A
E
For the tempered steel,
3
2 2
2 2 6
3 (14)(100 10 )
2 (2) 0.75 in 0.048276 in.
16 29 10



 
    
 

 
Y
Y
L
A
E
Total area: 2
1 2 1.75 in
  
A A A
1 2
  
 
Y m Y
The mild steel yields. Tempered steel is elastic.
Forces: 3 3
1 1 1
6
3
2
2
(1.00)(50 10 ) 50 10 lb
(29 10 )(0.75)(0.04)
62.14 10 lb
14


    

   
Y
m
P A
EA
P
L
Stresses:
3
3 3
1 2
1 1 2
1 2
62.14 10
50 10 psi 82.86 10 psi
0.75
  

       
Y
P P
A A
Unloading: 3
112.14
64.08 10 psi
1.75
P
A
     
Residual stresses. 3 3 3
1,res 1 50 10 64.08 10 14.08 10 psi 14.08 ksi
  
          
3 3 3
2,res 2 82.86 10 64.08 10 18.78 10 psi 18.78 ksi
   
         
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231
PROBLEM 2.121
Narrow bars of aluminum are bonded to the two sides of a thick steel plate as
shown. Initially, at 1 70 F,
T   all stresses are zero. Knowing that the
temperature will be slowly raised to T2 and then reduced to T1, determine
(a) the highest temperature T2 that does not result in residual stresses, (b) the
temperature T2 that will result in a residual stress in the aluminum equal to
58 ksi. Assume 6
12.8 10 / F
 
  
a for the aluminum and 6
6.5 10 / F
 
  
s
for the steel. Further assume that the aluminum is elastoplastic, with
E 6
10.9 10
  psi and 58
Y
  ksi. (Hint: Neglect the small stresses in the
plate.)
SOLUTION
Determine temperature change to cause yielding.
3
6 6
( ) ( )
( )( )
58 10
( ) 844.62 F
( ) (10.9 10 )(12.8 6.5)(10 )
  
   

  
    
      

    
  
a Y s Y
a s Y Y
Y
Y
a s
PL
L T L T
EA
P
E T
A
T
E
(a) 2 1 ( ) 70 844.62 915 F
      
Y Y
T T T 915 F
 
After yielding,
( ) ( )
Y
a s
L
L T L T
E

  
    
Cooling:
( ) ( )
  

  
    
a s
P L
L T L T
AE
The residual stress is
res ( )( )
Y Y a s
P
E T
A
    

     
Set res
( )( )
Y
Y Y a s
E T
 
   
 
    
3
6 6
2 (2)(58 10 )
1689 F
( ) (10.9 10 )(12.8 6.5)(10 )

  

    
  
Y
a s
T
E
(b) 2 1 70 1689 1759 F
      
T T T 1759 F
 
If 2 1759 F,
 
T the aluminum bar will most likely yield in compression.
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232
440 mm
a ⫽ 120 mm
F
C B
A
PROBLEM 2.122
Bar AB has a cross-sectional area of 1200 mm2
and is made of a steel
that is assumed to be elastoplastic with 200 GPa
E  and 250 MPa.
Y
 
Knowing that the force F increases from 0 to 520 kN and then
decreases to zero, determine (a) the permanent deflection of point C,
(b) the residual stress in the bar.
SOLUTION
2 6 2
1200 mm 1200 10 m
A 
  
Force to yield portion AC: 6 6
3
(1200 10 )(250 10 )
300 10 N
AC Y
P A 
   
 
For equilibrium, 0.
  
CB AC
F P P
3 3
3
300 10 520 10
220 10 N
     
  
CB AC
P P F
3
9 6
3
3
6
6
(220 10 )(0.440 0.120)
(200 10 )(1200 10 )
0.29333 10 m
220 10
1200 10
183.333 10 Pa



 
  
 
 

 

  
CB CB
C
CB
CB
P L
EA
P
A


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233
PROBLEM 2.122
(Continued)
Unloading:
( )

  

    
 
  
 
 
AC AC CB CB AC CB
C
AC BC CB
AC
P L P L F P L
EA EA EA
L L FL
P
EA EA EA
3
3
3 3 3
3
6
6
3
6
6
(520 10 )(0.440 0.120)
378.18 10 N
0.440
378.18 10 520 10 141.820 10 N
378.18 10
315.150 10 Pa
1200 10
141.820 10
118.183 10 Pa
1200 10
(378.


 
    

 
        
 
    

 
      

 
CB
AC
AC CB
CB AC
AC
AC
BC
BC
C
FL
P
L L
P P F
P
A
P
A



3
3
9 6
18 10 )(0.120)
0.189090 10 m
(200 10 )(1200 10 )



 
 
(a) 3 3 3
, 0.29333 10 0.189090 10 0.104240 10 m
C p C C
     
       
 0.1042 mm
 
(b) 6 6 6
, res 250 10 315.150 10 65.150 10 Pa
AC Y AC
  
        
 65.2 MPa
  
6 6 6
, res 183.333 10 118.183 10 65.150 10 Pa
CB CB CB
  
         
 65.2 MPa
  
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234
440 mm
a ⫽ 120 mm
F
C B
A
PROBLEM 2.123
Solve Prob. 2.122, assuming that 180
a  mm.
PROBLEM 2.122 Bar AB has a cross-sectional area of 1200 mm2
and
is made of a steel that is assumed to be elastoplastic with 200
E  GPa
and 250
Y
  MPa. Knowing that the force F increases from 0 to
520 kN and then decreases to zero, determine (a) the permanent
deflection of point C, (b) the residual stress in the bar.
SOLUTION
2 6 2
1200 mm 1200 10 m
A 
  
Force to yield portion AC: 6 6
3
(1200 10 )(250 10 )
300 10 N
AC Y
P A 
   
 
For equilibrium, 0.
  
CB AC
F P P
3 3
3
300 10 520 10
220 10 N
CB AC
P P F
     
  
3
9 6
3
3
6
6
(220 10 )(0.440 0.180)
(200 10 )(1200 10 )
0.23833 10 m
220 10
1200 10
183.333 10 Pa



 
  
 
 

  

  
CB CB
C
CB
CB
P L
EA
P
A


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235
PROBLEM 2.123
(Continued)
Unloading:
( )
AC AC CB CB AC CB
C
AC BC CB
AC
P L P L F P L
EA EA EA
L L FL
P
EA EA EA

  

    
 

  
 
 
3
3
3 3 3
3
3
9 6
3
6
6
(520 10 )(0.440 0.180)
307.27 10 N
0.440
307.27 10 520 10 212.73 10 N
(307.27 10 )(0.180)
0.23045 10 m
(200 10 )(1200 10 )
307.27 10
256.058 10 Pa
1200 10





 
    

 
        

   
 
 
    

CB
AC
AC CB
CB AC
C
AC
AC
FL
P
L L
P P F
P
A
3
6
6
212.73 10
177.275 10 Pa
1200 10
 
  
     

CB
CB
P
A
(a) 3 3 3
, 0.23833 10 0.23045 10 0.00788 10 m
C p C C
     
       
 0.00788 mm
 
(b) 6 6 6
,res 250 10 256.058 10 6.0580 10 Pa
AC AC AC
  
        
 6.06 MPa
  
6 6 6
,res 183.333 10 177.275 10 6.0580 10 Pa
CB CB CB
  
         
 6.06 MPa
  

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236
P
l l
C
B
A
␦
PROBLEM 2.124
The uniform wire ABC, of unstretched length 2l, is attached to the
supports shown and a vertical load P is applied at the midpoint B.
Denoting by A the cross-sectional area of the wire and by E the
modulus of elasticity, show that, for ,
l
  the deflection at the
midpoint B is
3
 
P
l
AE
SOLUTION
Use approximation.
sin tan
l

 
 
Statics: 0: 2 sin 0

   
Y AB
F P P
2sin 2
 
 
AB
P Pl
P
Elongation:
2
2
AB
AB
P l Pl
AE AE


 
Deflection:
From the right triangle,
2 2 2
2 2
( )
 

  

AB
l l
l 2 2
2  
  
AB AB
l l
3
1
2 1 2
2

 

 
  
 
 

AB
AB AB
l l
l
Pl
AE
3
3 3
Pl P
l
AE AE
 
   
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237
B
d
C
A
12 in.
18 in.
1.5 in.
2.25 in.
28 kips
E
D
28 kips PROBLEM 2.125
The aluminum rod ABC ( 6
10.1 10 psi),
E   which consists of two
cylindrical portions AB and BC, is to be replaced with a cylindrical steel
rod DE ( 6
29 10 psi)
E   of the same overall length. Determine the
minimum required diameter d of the steel rod if its vertical deformation is
not to exceed the deformation of the aluminum rod under the same load
and if the allowable stress in the steel rod is not to exceed 24 ksi.
SOLUTION
Deformation of aluminum rod.
3
6 2 2
4 4
28 10 12 18
10.1 10 (1.5) (2.25)
0.031376 in.
BC
AB
A
AB BC
BC
AB
AB BC
PL
PL
A E A E
L
L
P
E A A
 
  
 
 
 
 
 

 
 
 
  

Steel rod. 0.031376 in.
 
3
2
6
3
2
3
(28 10 )(30)
0.92317 in
(29 10 )(0.031376)
28 10
1.16667 in
24 10
PL PL
A
EA E
P P
A
A





    


    

Required area is the larger value. 2
1.16667 in
A 
Diameter:
4 (4)(1.16667)
A
d
 
  1.219 in.

d 
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238
C
B
A
3 in.
2 in.
30 kips 30 kips
P ⫽ 40 kips
40 in.
30 in.
PROBLEM 2.126
Two solid cylindrical rods are joined at B and loaded as shown. Rod AB is
made of steel 6
( 29 10
 
E psi), and rod BC of brass 6
( 15 10 psi).
E  
Determine (a) the total deformation of the composite rod ABC, (b) the
deflection of point B.
SOLUTION
Portion AB: 3
2 2 2
6
3
3
6
40 10 lb
40 in.
2 in.
(2) 3.1416 in
4 4
29 10 psi
(40 10 )(40)
17.5619 10 in.
(29 10 )(3.1416)
AB
AB
AB
AB
AB AB
AB
AB AB
P
L
d
A d
E
P L
E A
 
 
 


  
 

   

Portion BC: 3
2 2 2
6
3
3
6
20 10 lb
30 in.
3 in.
(3) 7.0686 in
4 4
15 10 psi
( 20 10 )(30)
5.6588 10 in.
(15 10 )(7.0686)
BC
BC
BC
BC
BC BC
BC
BC BC
P
L
d
A d
E
P L
E A
 
 
  


  
 
 
    

(a) 6 6
17.5619 10 5.6588 10
AB BC
    
      3
11.90 10 in.
 
   
(b) B BC
 
  3
5.66 10 in.
B
 
   
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239
3 mm
A
B
40 mm
100 kg
20 mm
Brass strip:
E 5 105 GPa
a 5 20 3 1026
/8C
PROBLEM 2.127
The brass strip AB has been attached to a fixed
support at A and rests on a rough support at B.
Knowing that the coefficient of friction is 0.60
between the strip and the support at B, determine
the decrease in temperature for which slipping
will impend.
SOLUTION
Brass strip:
6
105 GPa
20 10 / C
E
 

  
0: 0
    
y
F N W N W
0: 0
( ) 0
  

 
 
     
       
x
F P N P W mg
PL P mg
L T T
EA EA EA
Data:
2 6 2
2
9
0.60
(20)(3) 60 mm 60 10 m
100 kg
9.81 m/s
105 10 Pa
A
m
g
E



   


 
9 6 6
(0.60)(100)(9.81)
(105 10 )(60 10 )(20 10 )
T 
 
  
4.67 C
  
T 
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240
2 in.
2 in.
3 in.
C
D
A
B
P'
P
1 -in. diameter
1-in. diameter
1
2
1 -in. diameter
1
2
PROBLEM 2.128
The specimen shown is made from a 1-in.-diameter cylindrical steel rod
with two 1.5-in.-outer-diameter sleeves bonded to the rod as shown.
Knowing that 6
29 10 psi,
 
E determine (a) the load P so that the
total deformation is 0.002 in., (b) the corresponding deformation of the
central portion BC.
SOLUTION
(a) i i i
i i i
PL L
P
A E E A
    
1
2
4
i
i i
i
L
P E A d
A



 
  
 
 
L, in. d, in. A, in2
L/A, in1
AB 2 1.5 1.7671 1.1318
BC 3 1.0 0.7854 3.8197
CD 2 1.5 1.7671 1.1318
6.083  sum
6 1 3
(29 10 )(0.002)(6.083) 9.353 10 lb
P 
    9.53 kips
P  
(b)
3
6
9.535 10
(3.8197)
29 10
BC BC
BC
BC BC
PL L
P
A E E A


  

3
1.254 10 in.
 
  
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241
24 kN
F
E
A
B
C
D
300 mm
250 mm
400 mm
250 mm
40 mm
G
PROBLEM 2.129
Each of the four vertical links connecting the
two rigid horizontal members is made of
aluminum ( 70 GPa)

E and has a uniform
rectangular cross section of 10  40 mm. For
the loading shown, determine the deflection
of (a) point E, (b) point F, (c) point G.
SOLUTION
Statics. Free body EFG:
3
0: (400)(2 ) (250)(24) 0
7.5 kN 7.5 10 N
F BE
BE
M F
F
    
    
3
0: (400)(2 ) (650)(24) 0
19.5 kN 19.5 10 N
E CF
CF
M F
F
   
  
Area of one link:
2
6 2
(10)(40) 400 mm
400 10 m

 
 
A
Length: 300 mm 0.300 m
 
L
Deformations.
3
6
9 6
3
6
9 6
( 7.5 10 )(0.300)
80.357 10 m
(70 10 )(400 10 )
(19.5 10 )(0.300)
208.93 10 m
(70 10 )(400 10 )
BE
BE
CF
CF
F L
EA
F L
EA






 
    
 

   
 
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242
PROBLEM 2.129 (Continued)
(a) Deflection of Point E. | |
 

E BF 80.4 m
 
 
E 
(b) Deflection of Point F. F CF
 
 209 m
F
 
  
Geometry change.
Let  be the small change in slope angle.
6 6
6
80.357 10 208.93 10
723.22 10 radians
0.400
 

 

   
   
E F
EF
L
(c) Deflection of Point G. G F FG
L
  
 
6 6
6
208.93 10 (0.250)(723.22 10 )
389.73 10 m
    

     
 
G F FG
L
390 m
 
 
G 
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243
4 ft
8 in.
8 in.
P
PROBLEM 2.130
A 4-ft concrete post is reinforced with four steel bars, each with a 3
4
-in. diameter.
Knowing that 6
29 10
s
E   psi and 6
3.6 10
c
E   psi, determine the normal
stresses in the steel and in the concrete when a 150-kip axial centric force P is
applied to the post.
SOLUTION
2
2
3
4 1.76715 in
4 4

 
 
 
 
 
 
 
 
s
A
2 2
8 62.233 in
  
c s
A A
6
6
(48)
0.93663 10
(1.76715)(29 10 )
 
   

s s
s s
s s
P L P
P
A E
6
6
(48)
0.21425 10
(62.233)(3.6 10 )
 
   

c c
c c
c c
P L P
P
A E
But :
s c
 
 6 6
0.93663 10 0.21425 10
 
  
s c
P P
0.22875
s c
P P
 (1)
Also, 150 kips
s c
P P P
   (2)
Substituting (1) into (2), 1.22875 150 kips
c
P 
122.075 kips
c
P 
From (1), 0.22875(122.075) 27.925 kips
s
P  
27.925
1.76715
s
s
s
P
A
     15.80 ksi
s
   
122.075
62.233
c
c
c
P
A
     1.962 ksi
c
   
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244
100 mm
2 m
A
C
D
B E
3 m
150 mm
PROBLEM 2.131
The steel rods BE and CD each have a 16-mm diameter
( 200 GPa);
E  the ends of the rods are single-threaded with a
pitch of 2.5 mm. Knowing that after being snugly fitted, the nut at
C is tightened one full turn, determine (a) the tension in rod CD,
(b) the deflection of point C of the rigid member ABC.
SOLUTION
Let  be the rotation of bar ABC as shown.
Then 0.15 0.25
B C
   
 
But turn
CD CD
C
CD CD
P L
E A
 
 
turn
9 2
4
3 6
( )
(200 10 Pa) (0.016 m)
(0.0025m 0.25 )
2m
50.265 10 5.0265 10

 


 

 
   
CD CD
CD C
CD
E A
P
L
or
BE BE BE BE
B BE B
BE BE BE
P L E A
P
E A L
 
 
9 2
4
6
(200 10 Pa) (0.016 m)
(0.15 )
3 m
2.0106 10





 
BE
P
From free body of member ABC:
0: 0.15 0.25 0
A BE CD
M P P
   
6 3 6
0.15(2.0106 10 ) 0.25(50.265 10 5.0265 10 ) 0
 
     
3
8.0645 10 rad
 
 
(a) 3 6 3
50.265 10 5.0265 10 (8.0645 10 )
CD
P 
    
3
9.7288 10 N
  9.73 kN
CD
P  
(b) 3
0.25 0.25(8.0645 10 )
C
  
  
3
2.0161 10 m

  2.02 mm
C
   
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245
8 in.
Aluminum shell
1.25 in.
Steel
core
0.75 in.
PROBLEM 2.132
The assembly shown consists of an aluminum shell 6
( 10.6 10 psi,
 
a
E
a  12.9  106
/°F) fully bonded to a steel core 6
( 29 10 psi,
 
s
E
s  6.5  106
/°F) and is unstressed. Determine (a) the largest allowable
change in temperature if the stress in the aluminum shell is not to exceed
6 ksi, (b) the corresponding change in length of the assembly.
SOLUTION
Since ,
a s
 
 the shell is in compression for a positive temperature rise.
Let 3
6 ksi 6 10 psi
a
     
 
2 2 2 2 2
(1.25 0.75 ) 0.78540 in
4 4
 
    
a o i
A d d
2 2 2
(0.75) 0.44179 in
4 4
 
  
s
A d
a a s s
P A A
 
  
where P is the tensile force in the steel core.
3
3
(6 10 )(0.78540)
10.667 10 psi
0.44179
a a
s
s
A
A



    
3 3
6 3
6 6
( ) ( )
( )( )
10.667 10 6 10
(6.4 10 )( ) 0.93385 10
29 10 10.6 10
s a
s a
s a
s a
a s
s a
T T
E E
T
E E
T
 
  
 
 
 
     
   
 
     
 
(a) 145.91 F
T
   145.9 F
  
T 
(b)
3
6 3
6
10.667 10
(6.5 10 )(145.91) 1.3163 10
29 10
  

    

or
3
6 3
6
6 10
(12.9 10 )(145.91) 1.3163 10
10.6 10
  
 
    

 3
(8.0)(1.3163 10 ) 0.01053 in.
  
   
L  0.01053 in.
  
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246
3.5 in.
5.5 in. 2.2 in.
P
PROBLEM 2.133
The plastic block shown is bonded to a fixed base and to a horizontal
rigid plate to which a force P is applied. Knowing that for the
plastic used 55
G  ksi, determine the deflection of the plate
when 9
P  kips.
SOLUTION
Consider the plastic block. The shearing force carried is 3
9 10 lb
 
P
The area is 2
(3.5)(5.5) 19.25 in
 
A
Shearing stress:
3
9 10
467.52 psi
19.25
P
A


  
Shearing strain: 3
467.52
0.0085006
55 10

   

G
But (2.2)(0.0085006)

  
   
h
h
0.01870 in.

 
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247
150
300
75
150
P⬘
75
Dimensions in mm
P
15
r 5 6
60
PROBLEM 2.134
The aluminum test specimen shown is subjected to two equal and opposite
centric axial forces of magnitude P. (a) Knowing that E  70 GPa
and all
  200 MPa, determine the maximum allowable value of P and the
corresponding total elongation of the specimen. (b) Solve part a, assuming that
the specimen has been replaced by an aluminum bar of the same length and a
uniform 60  15-mm rectangular cross section.
SOLUTION
6 9
all
2 6 2
min
200 10 Pa 70 10 Pa
(60 mm)(15 mm) 900 mm 900 10 m


   
   
E
A
(a) Test specimen. 75 mm, 60 mm, 6 mm
D d r
  
75 6
1.25 0.10
60 60
D r
d d
   
From Fig. 2.60b, max
1.95
P
K K
A

 
6 6
3
max (900 10 )(200 10 )
92.308 10 N
1.95
A
P
K
 
 
    92.3 kN
P  
Wide area * 2 3 2
(75 mm)(15 mm) 1125 mm 1.125 10 m

   
A
3
9 3 6 3
6
92.308 10 0.150 0.300 0.150
70 10 1.125 10 900 10 1.125 10
7.91 10 m
i i i
i i i
PL L
P
A E E A
   

  
      
 
   
 
  0.791 mm
  
(b) Uniform bar.
6 6 3
all (900 10 )(200 10 ) 180 10 N
P A 
      180.0 kN
P  

3
3
6 9
(180 10 )(0.600)
1.714 10 m
(900 10 )(70 10 )
PL
AE
 


   
 
 1.714 mm
  
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248
L
C
P
P
k
m
B
B'
C'
PROBLEM 2.135
The uniform rod BC has a cross-sectional area A and is made of a
mild steel that can be assumed to be elastoplastic with a modulus of
elasticity E and a yield strength .
Y Using the block-and-spring
system shown, it is desired to simulate the deflection of end C of the
rod as the axial force P is gradually applied and removed, that is, the
deflection of points C and C should be the same for all values of P.
Denoting by  the coefficient of friction between the block and the
horizontal surface, derive an expression for (a) the required mass m
of the block, (b) the required constant k of the spring.
SOLUTION
Force-deflection diagram for Point C or rod BC.
For
max

 

 
 
 
Y Y
C C
Y Y
P P A
PL EA
P
EA L
P P A
Force-deflection diagram for Point C of block-and-spring system.
0: 0
    
y
F N mg N mg
0: 0
    
x f f
F P F P F
If block does not move, i.e., or ,
  
  
f
F N mg P mg
then or
c c
P
P k
K
 
 
 
If P  mg, then slip at occurs.

 
m
P F mg
If the force P is the removed, the spring returns to its initial length.
(a) Equating PY and Fmax,

 

  Y
Y
A
A mg m
g

(b) Equating slopes,
EA
k
L
 
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249
Element n Element 1
Pn P1
PROBLEM 2.C1
A rod consisting of n elements, each of which is homogeneous and
of uniform cross section, is subjected to the loading shown. The length
of element i is denoted by ,
i
L its cross-sectional area by ,
i
A modulus of
elasticity by ,
i
E and the load applied to its right end by i
P , the magnitude
Pi of this load being assumed to be positive if i
P is directed to the right
and negative otherwise. (a) Write a computer program that can be used
to determine the average normal stress in each element, the deformation
of each element, and the total deformation of the rod. (b) Use this
program to solve Probs. 2.20 and 2.126.
SOLUTION
For each element, enter
, ,
i i i
L A E
Compute deformation
Update axial load i
P P P
 
Compute for each element
/
/
i i
i i i i
P A
PL A E




Total deformation:
Update through n elements
i
  
 
Program Outputs
Problem 2.20
Element Stress (MPa) Deformation (mm)
1 19.0986 0.1091
2 12.7324 0.0909
Total Deformation  0.0182 mm
Problem 2.126
Element Stress (ksi) Deformation (in.)
1 12.7324 0.0176
2 2.8294 0.0057
Total Deformation  0.01190 in.

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250
Element n Element 1
Pn
P2
A
B
PROBLEM 2.C2
Rod AB is horizontal with both ends fixed; it consists of n elements, each
of which is homogeneous and of uniform cross section, and is subjected to
the loading shown. The length of element i is denoted by ,
i
L its cross-
sectional area by ,
i
A its modulus of elasticity by ,
i
E and the load applied
to its right end by ,
i
P the magnitude i
P of this load being assumed to be
positive if i
P is directed to the right and negative otherwise. (Note that
P1 0.)
 (a) Write a computer program which can be used to determine the
reactions at A and B, the average normal stress in each element, and the
deformation of each element. (b) Use this program to solve Probs. 2.41
and 2.42.
SOLUTION
We Consider the reaction at B redundant and release the rod at B
Compute with 0
B B
R
 
For each element, enter
, ,
i i i
L A E
Update axial load
i
P P P
 
Compute for each element
/
/
i i
i i i i
P A
PL A E




Update total deformation
B B i
  
 
Compute due to unit load at
B B

Unit 1/
Unit /
i i
i i i i
A
L A E




Update total unit deformation
Unit Unit Unit
B B i
  
 
Superposition
For total displacement at 0
B 
Unit 0
B B B
R
 
 
Solving:
/Unit
B B B
R  
 
Then: A i B
R P R
  
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251
PROBLEM 2.C2 (Continued)
For each element
Unit
Unit
i B i
i B i
R
R
  
  
 
 
Program Outputs
Problem 2.41
RA 62.809 kN
RB 37.191 kN
 
 
Element Stress (MPa) Deformation (mm)
1 52.615
 0.05011

2 3.974 0.00378
3 2.235 0.00134
4 49.982 0.04498
Problem 2.42
RA 45.479 kN
RB 54.521 kN
 
 
Element Stress (MPa) Deformation (mm)
1 77.131
 0.03857

2 20.542
 0.01027

3 11.555
 0.01321

4 36.191 0.06204

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252
Element n Element 1
A
B
0
␦
PROBLEM 2.C3
Rod AB consists of n elements, each of which is homogeneous and of
uniform cross section. End A is fixed, while initially there is a gap 0
between end B and the fixed vertical surface on the right. The length of
element i is denoted by ,
i
L its cross-sectional area by ,
i
A its modulus of
elasticity by ,
i
E and its coefficient of thermal expansion by .
i
 After the
temperature of the rod has been increased by ,
T
 the gap at B is closed
and the vertical surfaces exert equal and opposite forces on the rod.
(a) Write a computer program which can be used to determine the
magnitude of the reactions at A and B, the normal stress in each element,
and the deformation of each element. (b) Use this program to solve
Probs. 2.59 and 2.60.
SOLUTION
We compute the displacements at B.
Assuming there is no support at B,
enter , , ,
i i i i
L A E 
Enter temperature change T. Compute for each element.
i i i
LT
 

Update total deformation.
B B i
  
 
Compute due to unit load at .
B B
Unit /
 
i i i i
L A E
Update total unit deformation.
Unit Unit Unit
  
 
B B i
Compute reactions.
From superposition,
0
( )/Unit
B B B
R   
 
Then
A B
R R
 
For each element,
/
/

 
 
 
i B i
i i i B i i i
R A
LT R L A E
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253
PROBLEM 2.C3 (Continued)
Program Outputs
Problem 2.59.
52.279 kips
R 
Element Stress (ksi) Deformation (10* 3 in.)

1 21.783
 9.909
2 18.671
 10.091
Problem 2.60.
232.390 kN
R 
Element Stress (MPa) Deformation (microm)
1 116.195
 363.220
2 290.487
 136.780

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254
Plate
␴
A1, E1, ( Y)1
L
␴
A2, E2, ( Y)2
P
PROBLEM 2.C4
Bar AB has a length L and is made of two different materials of given
cross-sectional area, modulus of elasticity, and yield strength. The bar is
subjected as shown to a load P that is gradually increased from zero until
the deformation of the bar has reached a maximum value m
 and then
decreased back to zero. (a) Write a computer program that, for each of 25
values of m
 equally spaced over a range extending from 0 to a value
equal to 120% of the deformation causing both materials to yield, can be
used to determine the maximum value m
P of the load, the maximum
normal stress in each material, the permanent deformation p of the bar,
and the residual stress in each material. (b) Use this program to solve
Probs. 2.111 and 2.112.
SOLUTION
Note: The following assumes 1 2
( ) ( )
Y Y
 
Displacement increment
2 2
0.05( ) /
m Y L E
 

Displacements at yielding
1 1 2 2
( ) / ( ) /
A Y B Y
L E L E
   
 
For each displacement If
1 1
2 2
1 1 2 2
:
/
/
( / )( )
m A
m
m
m m
E L
E L
P L A E A E
 
 
 



 
If
1 1
2 2
1 1 2 2
:
( )
/
( / )
A m B
Y
m
m m
E L
P A L A E
  
 
 
 


 
If
1 1
2 2
1 1 2 2
:
( )
( )
m B
Y
Y
m
P A A
 
 
 
 


 
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255
PROBLEM 2.C4 (Continued)
Permanent deformations, residual stresses
Slope of first (elastic) segment
1 1 2 2
Slope ( )/
( /Slope)
P m m
A E A E L
P
 
 
 
1 res 1 1
2 res 2 2
( ) ( /( Slope))
( ) ( /( Slope))
m
m
E P L
E P L
 
 
 
 
Program Outputs
Problems 2.111 and 2.112
DM
10**  3 in.
PM
kips
SIGM (1)
ksi
SIGM (2)
ksi
DP
10**  3 in.
SIGR (1)
ksi
SIG (2)
ksi
0.000 0.000 0.000 0.000 0.000 0.000 0.000
2.414 8.750 5.000 5.000 0.000 0.000 0.000
4.828 17.500 10.000 10.000 0.000 0.000 0.000
7.241 26.250 15.000 15.000 0.000 0.000 0.000
9.655 35.000 20.000 20.000 0.000 0.000 0.000
12.069 43.750 25.000 25.000 0.000 0.000 0.000
14.483 52.500 30.000 30.000 0.000 0.000 0.000
16.897 61.250 35.000 35.000 0.000 0.000 0.000
19.310 70.000 40.000 40.000 0.000 0.000 0.000
21.724 78.750 45.000 45.000 0.000 0.000 0.000
24.138 87.500 50.000 50.000 0.000 0.000 0.000
26.552 91.250 50.000 55.000 1.379 2.143
 2.857
28.966 95.000 50.000 60.000 2.759 4.286
 5.714
31.379 98.750 50.000 65.000 4.138 6.429
 8.571


2.112 
33.793 102.500 50.000 70.000 5.517 8.571
 11.429
36.207 106.250 50.000 75.000 6.897 10.714
 14.286
38.621 110.000 50.000 80.000 8.276 12.857
 17.143
41.034 113.750 50.000 85.000 9.655 15.000
 20.000 2.111 
43.448 117.500 50.000 90.000 11.034 17.143
 22.857
45.862 121.250 50.000 95.000 12.414 19.286
 25.714
48.276 125.000 50.000 100.000 13.793 21.429
 28.571
50.690 125.000 50.000 100.000 16.207 21.429
 28.571
53.103 125.000 50.000 100.000 18.621 21.429
 28.571
55.517 125.000 50.000 100.000 21.034 21.429
 28.571
57.931 125.000 50.000 100.000 23.448 21.429
 28.571
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256
1
2
d
1
2
d
D
r P
P9
PROBLEM 2.C5
The plate has a hole centered across the width. The stress concentration factor
for a flat bar under axial loading with a centric hole is
2 3
2 2 2
3.00 3.13 3.66 1.53
r r r
K
D D D
     
   
     
     
where r is the radius of the hole and D is the width of the bar. Write a computer
program to determine the allowable load P for the given values of r, D, the
thickness t of the bar, and the allowable stress all
 of the material. Knowing
that t 1
4
in., 3.0 in.,
D
  and 16 ksi,
 
all determine the allowable load P for
values of r from 0.125 in. to 0.75 in., using 0.125 in. increments.
SOLUTION
Enter
all
, , ,
r D t 
Compute K
2.0 /
RD r D

2 3
3.00 3.13 3.66 1.53
K RD RD RD
   
Compute average stress
ave all/K
 

Allowable load
all ave ( 2.0 )
P D r t

 
Program Output
Radius
(in.)
Allowable Load
(kips)
0.1250 3.9802
0.2500 3.8866
0.3750 3.7154
0.5000 3.4682
0.6250 3.1523
0.7500 2.7794

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257
L
A
B
c
P
2c
PROBLEM 2.C6
A solid truncated cone is subjected to an axial force P as shown. The exact
elongation is 2
( )/(2 ).
PL c E
 By replacing the cone by n circular cylinders
of equal thickness, write a computer program that can be used to calculate
the elongation of the truncated cone. What is the percentage error in the
answer obtained from the program using (a) 6,
n  (b) 12,
n  (c) 60?

n
SOLUTION
For 1 to :
( 0.5)( / )
2 ( / )
i
i i
i n
L i L n
r c c L L

 
 
Area:
2
i
A r


Displacement:
( / )/( )
P L n AE
 
 
Exact displacement:
2
exact /(2.0 )
PL c E
 

Percentage error:
exact exact
Percent = 100( )/
  

Program Output
n Approximate Exact Percent
6 0.15852 0.15915 0.40083

12 0.15899 0.15915 0.10100

60 0.15915 0.15915 0.00405


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297. Mecánica de Materiales - Ferdinand P. Beer 7ma. Edición.

298. C
CH
HA
AP
PT
TE
ER
R 3
3
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261
T
18 mm PROBLEM 3.1
Determine the torque T that causes a maximum shearing stress of 70 MPa in the steel
cylindrical shaft shown.
SOLUTION
4
max
3
max
3 6
;
2
2
(0.018 m) (70 10 Pa)
2
641.26 N m





 

 
 
Tc
J c
J
T c
641 N m
 
T 
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262
T
18 mm PROBLEM 3.2
For the cylindrical shaft shown, determine the maximum shearing stress caused by
a torque of magnitude T  800 N  m.
SOLUTION
4
max
max 3
3
6
;
2
2
2(800 N m)
(0.018 m)
87.328 10 Pa





 



 
Tc
J c
J
T
c
max 87.3 MPa
  
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263
2.4 m
30 mm
45 mm
T
PROBLEM 3.3
(a) Determine the torque T that causes a maximum shearing stress of
45 MPa in the hollow cylindrical steel shaft shown. (b) Determine the
maximum shearing stress caused by the same torque T in a solid cylindrical
shaft of the same cross-sectional area.
SOLUTION
(a) Given shaft:  
4 4
2 1
4 4 6 4 6 4
6 6
3
3
2
(45 30 ) 5.1689 10 mm 5.1689 10 m
2
(5.1689 10 )(45 10 )
5.1689 10 N m
45 10
J c c
J
Tc J
T
J c
T







 
     
 
 
   

5.17 kN m
T   
(b) Solid shaft of same area:
 
2 2 2 2 3 2
2 1
2
4
3
3
6
3
(45 30 ) 3.5343 10 mm
or 33.541 mm
2
,
2
(2)(5.1689 10 )
87.2 10 Pa
(0.033541)
A c c
A
c A c
Tc T
J c
J c
 







     
  
  

  
87.2 MPa
  
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264
3 in.
4 ft
T
PROBLEM 3.4
(a) Determine the maximum shearing stress caused by a 40-kip  in. torque T in
the 3-in.-diameter solid aluminum shaft shown. (b) Solve part a, assuming that
the solid shaft has been replaced by a hollow shaft of the same outer diameter
and of 1-in. inner diameter.
SOLUTION
(a)
4
(40 kip in.)(1.5 in.)
(1.5 in.)
2
7.5451 ksi





 
 
 

Tc
J
7.55 ksi
  
(b)
4 4
(40 kip in.)(1.5 in.)
[(1.5 in.) (0.5 in.) ]
2
7.6394 ksi







Tc
J
7.64 ksi
  
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265
T = 40 kip · in.
T'
3 in.
T'
T = 40 kip · in.
T
4 in.
(b)
(a)
PROBLEM 3.5
(a) For the 3-in.-diameter solid cylinder and loading shown, determine the
maximum shearing stress. (b) Determine the inner diameter of the 4-in.-
diameter hollow cylinder shown, for which the maximum stress is the
same as in part a.
SOLUTION
(a) Solid shaft:
4
1 1
(3.0 in.) 1.5 in.
2 2
2

  

c d
J c
max
3
3
2
2(40 kip in.)
(1.5 in.)
7.5451 ksi








Tc
J
T
c
max 7.55 ksi
  
(b) Hollow shaft:
1 1
(4.0 in.) 2.0 in.
2 2
  
o
c d
 
4 4
2
max
4 4
max
4
4
2
2(40 kip in.)(2.0 in.)
(2.0 in.)
(7.5451 ksi)
9.2500 in 1.74395 in.
and 2 3.4879 in.





 
 

 
  
 
o i
o o
o
i o
i
i i
c c
J T
c c
Tc
c c
c
d c
3.49 in.
i
d  
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266
60 mm
30 mm
D
200 mm
T ⫽ 3 kN · m
PROBLEM 3.6
A torque 3 kN m
T   is applied to the solid bronze cylinder shown.
Determine (a) the maximum shearing stress, (b) the shearing stress at point D
which lies on a 15-mm-radius circle drawn on the end of the cylinder, (c) the
percent of the torque carried by the portion of the cylinder within the 15-mm
radius.
SOLUTION
(a) 3
4 3 4 6 4
3
3 3
6
6
1
30 mm 30 10 m
2
(30 10 ) 1.27235 10 m
2 2
3 kN 3 10 N
(3 10 )(30 10 )
70.736 10 Pa
1.27235 10
m
c d
J c
T
Tc
J
 


 


   
    
  
 
   

70.7 MPa
m
  
(b) 3
15 mm 15 10 m
D
 
  
3 6
3
(15 10 )(70.736 10 )
(30 10 )
D
D
c

 
 

 
 

35.4 MPa
D
  
(c) 3
2
D D D D
D D D D
D D
T J
T
J
  
  

  
3 3 6
3
(15 10 ) (35.368 10 ) 187.5 N m
2
187.5
100% (100%) 6.25%
3 10
D
D
T
T
T
 
    
  

6.25% 
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267
4 in.
8 in.
ds
t 5 in.
1
4
3 in.
D
C
A
B
T
PROBLEM 3.7
The solid spindle AB is made of a steel with an allowable shearing stress of
12 ksi, and sleeve CD is made of a brass with an allowable shearing stress
of 7 ksi. Determine (a) the largest torque T that can be applied at A if the
allowable shearing stress is not to be exceeded in sleeve CD, (b) the
corresponding required value of the diameter s
d of spindle AB.
SOLUTION
(a) Analysis of sleeve CD:
 
2
1 2
4 4 4 4 4
2 1
3
3
2
1 1
(3) 1.5 in.
2 2
1.5 0.25 1.25 in.
= (1.5 1.25 ) 4.1172 in
2 2
(4.1172)(7 10 )
= 19.21 10 lb in.
1.5
o
c d
c c t
J c c
J
T
c
 

  
    
   

   
19.21 kip in.
T   
(b) Analysis of solid spindle AB:
3
3 3
3
3
=
19.21 10
1.601in
2 12 10
(2)(1.601)
1.006 in. 2
s
Tc
J
J T
c
c
c d c





   

  
2.01 in.

d 
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268
4 in.
8 in.
ds
t 5 in.
1
4
3 in.
D
C
A
B
T
PROBLEM 3.8
The solid spindle AB has a diameter ds  1.5 in. and is made of a steel with an
allowable shearing stress of 12 ksi, while sleeve CD is made of a brass with
an allowable shearing stress of 7 ksi. Determine the largest torque T that can
be applied at A.
SOLUTION
Analysis of solid spindle AB:
1
0.75 in.
2
s
c d
 
3
=
2
Tc J
T c
J c
 
 
 
3 3 3
(12 10 )(0.75) 7.95 10 lb in.
2
T

    
Analysis of sleeve CD: 2
1 1
(3) 1.5 in.
2 2
  
o
c d
 
1 2
4 4 4 4 4
2 1
3
3
2
1.5 0.25 1.25 in.
(1.5 1.25 ) 4.1172 in
2 2
(4.1172)(7 10 )
19.21 10 lb in.
1.5
 

    
    

    
c c t
J c c
J
T
c
The smaller torque governs. 3
7.95 10 lb in.
T   
 7.95 kip in.
 
T 
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269
6.8 kip · in.
72 in.
C
10.4 kip · in.
3.6 kip · in.
B
48 in.
A
PROBLEM 3.9
The torques shown are exerted on pulleys A, B, and C.
Knowing that both shafts are solid, determine the
maximum shearing stress in (a) shaft AB, (b) shaft BC.
SOLUTION
(a) Shaft AB: 3
3.6 10 lb in.
AB
T   
4
max 3
3
3
max 3
1 1
(1.3) 0.65 in.
2 2
2
2
(2)(3.6 10 )
8.35 10 psi
(0.65)





  

 

  
c d
J c
Tc T
J c
max 8.35 ksi
  
(b) Shaft BC: 3
4
3
3
max 3 3
6.8 10 lb in.
1 1
(1.8) 0.9 in.
2 2
2
2 (2)(6.8 10 )
5.94 10 psi
(0.9)


 
  
  


   
BC
BC
T
c d
J c
T
c
max 5.94 ksi
  
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270
6.8 kip · in.
72 in.
C
10.4 kip · in.
3.6 kip · in.
B
48 in.
A
PROBLEM 3.10
The shafts of the pulley assembly shown are to be
redesigned. Knowing that the allowable shearing
stress in each shaft is 8.5 ksi, determine the smallest
allowable diameter of (a) shaft AB, (b) shaft BC.
SOLUTION
(a) Shaft AB: 3
3.6 10 lb in.
AB
T   
3
max
4
max 3
3
3
3
3
max
8.5 ksi 8.5 10 psi
2
2
2 (2)(3.6 10 )
0.646 in.
(8.5 10 )
AB
Tc T
J c
J c
T
c




 
  
  

  

2 1.292 in.
AB
d c
  
(b) Shaft BC: 3
3
max
3
3
3
3
max
6.8 10 lb in.
8.5 10 psi
2 (2)(6.8 10 )
0.7985 in.
(8.5 10 )
BC
BC
T
T
c

 
  
 

  

2 1.597 in.
BC
d c
  
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271
D
dCD ⫽ 21 mm
B
dBC ⫽ 18 mm
C
60 N · m
48 N · m
A
dAB ⫽ 15 mm
144 N · m
PROBLEM 3.11
Knowing that each portion of the shafts AB, BC,
and CD consist of a solid circular rod, determine
(a) the shaft in which the maximum shearing
stress occurs, (b) the magnitude of that stress.
SOLUTION
Shaft AB:
max 3
max 3
48 N m
1
7.5 mm 0.0075 m
2
2
(2)(48)
72.433 MPa
(0.0075)
T
c d
Tc T
J c
 
  
 
 




Shaft BC: 48 144 96 N m
T     
max 3 3
1 2 (2)(96)
9 mm 83.835 MPa
2 (0.009)

 
     
Tc T
c d
J c
Shaft CD: 48 144 60 156 N m
T      
max 3 3
1 2 (2 156)
10.5 mm 85.79 MPa
2 (0.0105)
Tc T
c d
J c

 

     
Answers: (a) Shaft CD (b) 85.8 MPa 
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272
D
dCD ⫽ 21 mm
B
dBC ⫽ 18 mm
C
60 N · m
48 N · m
A
dAB ⫽ 15 mm
144 N · m
PROBLEM 3.12
Knowing that an 8-mm-diameter hole has been
drilled through each of the shafts AB, BC, and CD,
determine (a) the shaft in which the maximum
shearing stress occurs, (b) the magnitude of that
stress.
SOLUTION
Hole: 1 1
1
4 mm
2
c d
 
Shaft AB:
 
2 2
4 4 4 4 9 4
2 1
48 N m
1
7.5 mm
2
(0.0075 0.004 ) 4.5679 10 m
2 2
T
c d
J c c
  
 
 
     
2
max 9
(48)(0.0075)
78.810 MPa
4.5679 10
Tc
J
 
  

Shaft BC: 48 144 96 N m
T      2 2
1
9 mm
2
c d
 
 
4 4 4 4 9 4
2 1
2
max 9
(0.009 0.004 ) 9.904 10 m
2 2
(96)(0.009)
87.239 MPa
9.904 10
J c c
Tc
J
 



     
  

Shaft CD: 48 144 60 156 N m
T       2 2
1
10.5 mm
2
c d
 
 
4 4 4 4 9 4
2 1 (0.0105 0.004 ) 18.691 10 m
2 2
J c c
  
     
2
max 9
(156)(0.0105)
87.636 MPa
18.691 10
Tc
J
 
  

Answers: (a) Shaft CD (b) 87.6 MPa 
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273
54 mm
46 mm
46 mm
40 mm
A
B
C
D
E
TB = 1.2 kN · m
TC = 0.8 kN · m
TD = 0.4 kN · m
PROBLEM 3.13
Under normal operating conditions,
the electric motor exerts a torque of
2.4 kN  m on shaft AB. Knowing
that each shaft is solid, determine the
maximum shearing stress in (a) shaft
AB, (b) shaft BC, (c) shaft CD.
SOLUTION
(a) Shaft AB: 3 1
2.4 10 N m, 0.027 m
2
AB
T c d
    
3
6
3 3
2 2(2.4 10 )
77.625 10 Pa
(0.027)
AB
Tc T
J c

 

     77.6 MPa 
(b) Shaft BC:
1
2.4 kN m 1.2 kN m 1.2 kN m, 0.023 m
2
BC
T c d
       
3
6
3 3
2 (2)(1.2 10 )
62.788 10 Pa
(0.023)
BC
Tc T
J c

 

     62.8 MPa 
(c) Shaft CD: 3 1
0.4 10 N m 0.023 m
2
CD
T c d
    
3
6
3 3
2 (2)(0.4 10 )
20.929 10 Pa
(0.023)
CD
Tc T
J c

 

     20.9 MPa 
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274
54 mm
46 mm
46 mm
40 mm
A
B
C
D
E
TB = 1.2 kN · m
TC = 0.8 kN · m
TD = 0.4 kN · m
PROBLEM 3.14
In order to reduce the total mass of
the assembly of Prob. 3.13, a new
design is being considered in which
the diameter of shaft BC will be
smaller. Determine the smallest
diameter of shaft BC for which the
maximum value of the shearing
stress in the assembly will not be
increased.
PROBLEM 3.13 Under normal
operating conditions, the electric
motor exerts a torque of 2.4 kN  m
on shaft AB. Knowing that each shaft
is solid, determine the maximum
shearing stress in (a) shaft AB,
(b) shaft BC, (c) shaft CD.
SOLUTION
See solution to Problem 3.13 for maximum shearing stresses in portions AB, BC, and CD of the shaft. The
largest maximum shearing value is 6
max 77.625 10 Pa
   occurring in AB.
Adjust diameter of BC to obtain the same value of stress.
3
2
Tc T
J c


 
3
3 6 3
6
2 (2)(1.2 10 )
9.8415 10 m
(77.625 10 )
T
c
 


   

3 3
21.43 10 m 2 42.8 10 m
c d c
 
     42.8 mm 
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275
B
C
Brass
T
A
Steel
PROBLEM 3.15
The allowable shearing stress is 15 ksi in the 1.5-in.-diameter steel rod AB and 8 ksi in
the 1.8-in.-diameter brass rod BC. Neglecting the effect of stress concentrations,
determine the largest torque that can be applied at A.
SOLUTION
4 3
max max
, ,
2 2
Tc
J c T c
J
 
 
  
Rod AB: max
3
1
15 ksi 0.75 in.
2
(0.75) (15) 9.94 kip in.
2
c d
T


  
  
Rod BC: max
3
1
8 ksi 0.90 in.
2
(0.90) (8) 9.16 kip in.
2
c d
T


  
  
The allowable torque is the smaller value. 9.16 kip in.
T   
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276
B
C
Brass
T
A
Steel
PROBLEM 3.16
The allowable shearing stress is 15 ksi in the steel rod AB and 8 ksi in the brass rod BC.
Knowing that a torque of magnitude 10 kip in.
T   is applied at A, determine the
required diameter of (a) rod AB, (b) rod BC.
SOLUTION
3
max
max
2
, ,
2
Tc T
J c
J



  
(a) Rod AB: max
10 kip in. 15 ksi

  
T
3 3
(2)(10)
0.4244 in
(15)
0.7515 in.
c
c

 
 2 1.503 in.
d c
  
(b) Rod BC: max
10 kip in. 8 ksi
T 
  
3 2
(2)(10)
0.79577 in
(8)
0.9267 in.
c
c

 
 2 1.853 in.
d c
  
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277
750 mm
600 mm
TB ⫽ 1200 N · m
TC ⫽ 400 N · m
dAB
B
A
C
dBC
PROBLEM 3.17
The solid shaft shown is formed of a brass for which the
allowable shearing stress is 55 MPa. Neglecting the effect of
stress concentrations, determine the smallest diameters dAB and
dBC for which the allowable shearing stress is not exceeded.
SOLUTION
6
max
3
max 3
max
55 MPa 55 10 Pa
2 2




  
  
Tc T T
c
J c
Shaft AB:
3
3
6
1200 400 800 N m
(2)(800)
21.00 10 m 21.0 m
(55 10 )
AB
T
c


   
   

minimum 2 42.0 mm
AB
d c
  
Shaft BC:
3
3
6
400 N m
(2)(400)
16.667 10 m 16.67 mm
(55 10 )
BC
T
c


 
   

minimum 2 33.3 mm
BC
d c
  
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278
750 mm
600 mm
TB ⫽ 1200 N · m
TC ⫽ 400 N · m
dAB
B
A
C
dBC
PROBLEM 3.18
Solve Prob. 3.17, assuming that the direction of TC is reversed.
PROBLEM 3.17 The solid shaft shown is formed of a brass for
which the allowable shearing stress is 55 MPa. Neglecting the effect
of stress concentrations, determine the smallest diameters dAB and dBC
for which the allowable shearing stress is not exceeded.
SOLUTION
Note that the direction of TC has been reversed in the figure.
6
max
3
max 3
max
55 MPa 55 10 Pa
2 2




  
  
Tc T T
c
J c
Shaft AB:
3
3
6
1200 400 1600 N m
(2)(1600)
26.46 10 m 26.46 mm
(55 10 )
AB
T
c


   
   

minimum 2 52.9 mm
AB
d c
  
Shaft BC:
3
3
6
400 N m
(2)(400)
16.667 10 m 16.67 mm
(55 10 )
BC
T
c


 
   

minimum 2 33.3 mm
BC
d c
  
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279
D
A
B
90 mm
dAB
C
T
PROBLEM 3.19
The solid rod AB has a diameter dAB  60 mm and is made of a steel for
which the allowable shearing stress is 85 MPa. The pipe CD, which has
an outer diameter of 90 mm and a wall thickness of 6 mm, is made of an
aluminum for which the allowable shearing stress is 54 MPa. Determine
the largest torque T that can be applied at A.
SOLUTION
Rod AB: 6
all
1
85 10 Pa 0.030 m
2
c d
    
3
all
all all
3 6 3
2
(0.030) (85 10 ) 3.605 10 N m
2
J
T c
c
 


 
    
Pipe CD: 6
all 2 2
1
54 10 Pa 0.045 m
2
c d
    
 
1 2
4 4 4 4 6 4
2 1
6 6
3
all
all
2
0.045 0.006 0.039 m
(0.045 0.039 ) 2.8073 10 m
2 2
(2.8073 10 )(54 10 )
3.369 10 N m
0.045
c c t
J c c
J
T
c
 



    
     
 
    
Allowable torque is the smaller value. 3
all 3.369 10 N m
T   
3.37 kN m
 
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280
D
A
B
90 mm
dAB
C
T
PROBLEM 3.20
The solid rod AB has a diameter dAB  60 mm. The pipe CD has an outer
diameter of 90 mm and a wall thickness of 6 mm. Knowing that both the
rod and the pipe are made of a steel for which the allowable shearing
stress is 75 MPa, determine the largest torque T that can be applied at A.
SOLUTION
6 all
all all
75 10 Pa
J
T
c

   
Rod AB: 4
3 3 6
all all
3
1
0.030 m
2 2
(0.030) (75 10 )
2 2
3.181 10 N m
c d J c
T c

 

  
  
  
Pipe CD:
 
2 2 1 2
4 4 4 4 6 4
2 1
6 6
3
all
1
0.045 m 0.045 0.006 0.039 m
2
(0.045 0.039 ) 2.8073 10 m
2 2
(2.8073 10 )(75 10 )
4.679 10 N m
0.045
c d c c t
J c c
T
  

      
     
 
   
Allowable torque is the smaller value. 3
all 3.18 10 N m
T   
3.18 kN m
 
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281
A
100 mm
40 mm
C
B
D
T ⫽ 1000 N · m
PROBLEM 3.21
A torque of magnitude 1000 N m
T   is applied
at D as shown. Knowing that the allowable
shearing stress is 60 MPa in each shaft, determine
the required diameter of (a) shaft AB, (b) shaft CD.
SOLUTION
1000 N m
100
(1000) 2500 N m
40
CD
B
AB CD
C
T
r
T T
r
 
   
(a) Shaft AB: 6
all
3 6 3
3 6
60 10 Pa
2 2 (2)(2500)
26.526 10 m
(60 10 )



 

 
     

Tc T T
c
J c
3
29.82 10 29.82 mm
c 
   2 = 59.6 mm
d c
 
(b) Shaft CD: 6
all
3 6 3
3 6
60 10 Pa
2 2 (2)(1000)
10.610 10 m
(60 10 )



 

 
     

Tc T T
c
J c
3
21.97 10 m 21.97 mm
c 
   2 = 43.9 mm
d c
 
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282
A
100 mm
40 mm
C
B
D
T ⫽ 1000 N · m
PROBLEM 3.22
A torque of magnitude 1000 N m
T   is applied
at D as shown. Knowing that the diameter of shaft
AB is 56 mm and that the diameter of shaft CD is
42 mm, determine the maximum shearing stress
in (a) shaft AB, (b) shaft CD.
SOLUTION
1000 N m
100
(1000) 2500 N m
40
CD
B
AB CD
C
T
r
T T
r
 
   
(a) Shaft AB:
1
0.028 m
2
c d
 
6
3 3
2 (2)(2500)
72.50 10
(0.028)

 
    
Tc T
J c
72.5 MPa 
(b) Shaft CD:
1
= 0.020 m
2
c d

6
3 3
2 (2)(1000)
68.7 10
(0.020)

 
    
Tc T
J c
68.7 MPa 
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283
F
TE
H
E
A
B
D
C
G
rG
rD
TF
PROBLEM 3.23
Under normal operating conditions, a motor
exerts a torque of magnitude TF at F. The
shafts are made of a steel for which the
allowable shearing stress is 12 ksi and have
diameters dCDE  0.900 in. and dFGH  0.800 in.
Knowing that rD  6.5 in. and rG  4.5 in.,
determine the largest allowable value of TF.
SOLUTION
all 12 ksi
 
Shaft FG:
3
all
,all all
3
1
0.400 in.
2
2
(0.400) (12) 1.206 kip in.
2
 


 
 
  
F
c d
J
T c
c
Shaft DE:
3
,all
3
, all
1
0.450 in.
2
2
(0.450) (12) 1.7177 kip in.
2
4.5
(1.7177) 1.189 kip in.
6.5



 

  
   
E all
G
F E F
D
c d
T c
r
T T T
r
Allowable value of F
T is the smaller. 1.189 kip in.
F
T   
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284
F
TE
H
E
A
B
D
C
G
rG
rD
TF
PROBLEM 3.24
Under normal operating conditions, a motor
exerts a torque of magnitude 1200 lb in.
 
F
T
at F. Knowing that 8 in.,

D
r 3 in.,

G
r and
the allowable shearing stress is 10.5 ksi in
each shaft, determine the required diameter of
(a) shaft CDE, (b) shaft FGH.
SOLUTION
all
3
3
1200 lb in.
8
(1200) 3200 lb in.
3
10.5 ksi =10,500 psi
2 2
or




 
   

  
F
D
E F
G
T
r
T T
r
Tc T T
c
J c
(a) Shaft CDE:
3 3
(2)(3200)
0.194012 in
(10,500)

 
c
0.5789 in. 2
DE
c d c
  1.158 in.
DE
d  
(b) Shaft FGH:
3 3
(2)(1200)
0.012757 in
(10,500)

 
c
0.4174 in. 2
FG
c d c
  0.835 in.
FG
d  
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285
B
4 in.
2.5 in.
E
G
H
A
D
F
C
TC
TF
PROBLEM 3.25
The two solid shafts are connected by gears as shown and are made
of a steel for which the allowable shearing stress is 7000 psi.
Knowing the diameters of the two shafts are, respectively,
1.6 in.
BC
d  and 1.25 in.,
EF
d  determine the largest torque TC
that can be applied at C.
SOLUTION
max 7000 psi 7.0 ksi
  
Shaft BC:
3
max
max
3
1.6 in.
1
0.8 in.
2
2
(7.0)(0.8) 5.63 kip in.
2
BC
C
d
c d
J
T c
c
 



 
 
  
Shaft EF:
3
max
max
3
1.25 in.
1
0.625 in.
2
2
(7.0)(0.625) 2.684 kip in.
2
EF
F
d
c d
J
T c
c
 



 
 
  
By statics,
4
(2.684) 4.30 kip in.
2.5
A
C F
D
r
T T
r
   
Allowable value of C
T is the smaller.
4.30 kip in.
C
T   
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286
B
4 in.
2.5 in.
E
G
H
A
D
F
C
TC
TF
PROBLEM 3.26
The two solid shafts are connected by gears as shown and are made of a
steel for which the allowable shearing stress is 8500 psi. Knowing that a
torque of magnitude 5 kip in.
C
T   is applied at C and that the assembly
is in equilibrium, determine the required diameter of (a) shaft BC,
(b) shaft EF.
SOLUTION
max 8500 psi 8.5 ksi
  
(a) Shaft BC:
3
max 3
max
3
5 kip in.
2 2
(2)(5)
0.7208 in.
(8.5)




 
  
 
C
T
Tc T T
c
J c
c
2 1.442 in.
BC
d c
  
(b) Shaft EF:
3
3
max
2.5
(5) 3.125 kip in.
4
2 (2)(3.125)
0.6163 in.
(8.5)
 
   
  
D
F C
A
r
T T
r
T
c
2 1.233 in.
EF
d c
  
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287
B
C
75 mm
A
D
E
F
30 mm
90 mm
T
30 mm
PROBLEM 3.27
For the gear train shown, the diameters of the three
solid shafts are:
20mm 25mm 40mm
AB CD EF
d d d
  
Knowing that for each shaft the allowable shearing
stress is 60 MPa, determine the largest torque T that
can be applied.
SOLUTION
Statics: AB
T T

75
2.5
30
90
(2.5 ) 7.5
30
CD AB C
CD AB
C B B
EF CD F
EF CD
F D D
T T r
T T T T
r r r
T T r
T T T T
r r r
   
   
Determine the magnitude of T so that the stress is 6
60 MPa 60 10 Pa.
 
3
shaft
2
Tc J
T c
J c
 
 
  
Shaft AB:
1
10 mm 0.010 m
2
AB
c d
  
6 3
(60 10 )(0.010) 94.2 N m
2
AB
T T T

    
Shaft CD:
1
12.5 mm 0.0125 m
2
CD
c d
  
6 3
2.5 (60 10 )(0.0125) 73.6 N m
2
CD
T T T

    
Shaft EF:
1
20 mm 0.020 m
2
  
EF
c d
6 3
7.5 (60 10 )(0.020) 100.5 N m
2
EF
T T T

    
The smallest value of T is the largest torque that can be applied. 73.6 N m
T   
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288
B
C
75 mm
A
D
E
F
30 mm
90 mm
T
30 mm
PROBLEM 3.28
A torque T  900 N  m is applied to shaft AB of the
gear train shown. Knowing that the allowable shearing
stress is 80 MPa, determine the required diameter of
(a) shaft AB, (b) shaft CD, (c) shaft EF.
SOLUTION
From statics,
75 mm
2.5
30 mm
90 mm
(2.5 ) 7.5
30 mm
AB
C
CD AB
B
E
EF BC
D
T T
r
T T T T
r
r
T T T T
r

  
  
(a) Shaft AB:
all 4
2
3
all
3
all
900 N m
2
2 2 900 N m
80 MPa


 
  
  
 


 
AB
AB
T T
Tc Tc
J c
T
c
T
c
19.2757 mm 38.6 mm
AB
c d
   
(b) Shaft CD: 2.5 2.5(900 N m)
CD
T T
  
3
all
2 2 2250 N m
80 MPa
CD
T
c
  

 
26.161 mm 52.3 mm
CD
c d
   
(c) Shaft EF: 7.5 7.5(900 N m)
  
EF
T T
3
all
2 2 6750 N m
80 MPa
EF
T
c
  

 
37.731 mm 75.5 mm
  
EF
c d 
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289
O O
c1
max
r
m
c2
0
(a) (b)
PROBLEM 3.29
While the exact distribution of the shearing stresses in a
hollow-cylindrical shaft is as shown in Fig. a, an approximate
value can be obtained for max by assuming that the stresses
are uniformly distributed over the area A of the cross section,
as shown in Fig. b, and then further assuming that all of the
elementary shearing forces act at a distance from O equal to
the mean radius 1 2
( )
½ c c
 of the cross section. This
approximate value 0 / ,
m
T Ar
  where T is the applied torque.
Determine the ratio max 0
/
  of the true value of the maximum
shearing stress and its approximate value 0
 for values
of 1 2
/ ,
c c respectively, equal to 1.00, 0.95, 0.75, 0.50, and 0.
SOLUTION
For a hollow shaft,
      
2 2 2 2
max 4 4 2 2 2 2 2 2
2 1 2 1 2 1 2 1
2 2 2
Tc Tc Tc Tc
J c c c c c c A c c

 
   
   
By definition, 0
2 1
2
( )
m
T T
Ar A c c
  

Dividing, max 2 2 1 1 2
2 2 2
0 2 1 1 2
( ) 1 ( / )
1 ( / )
c c c c c
c c c c


 
 
 

1 2
/
c c 1.0 0.95 0.75 0.5 0.0
max 0
/
  1.0 1.025 1.120 1.200 1.0

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290
c2
c1
PROBLEM 3.30
(a) For a given allowable shearing stress, determine the ratio T/w of the
maximum allowable torque T and the weight per unit length w for the
hollow shaft shown. (b) Denoting by 0
( / )
T w the value of this ratio for a
solid shaft of the same radius 2,
c express the ratio T/w for the hollow shaft
in terms of 0
( / )
T w and 1 2
/ .
c c
SOLUTION
 
2 2
2 1
weight per unit length
specific weight
totalweight
length


  




    
w
g
W
L
W gLA
w gA g c c
L L
  
2 2 2 2
4 4
2 1 2 1
all 2 1
all all all
2 2 2
2 2
c c c c
J c c
T
c c c
  
 
 

  
(a)  
2 2
1 2 all
T
c c
W

 
 
2 2
1 2 all
2
2
c c
T
w gc



 (hollow shaft) 
 1 0
c  for solid shaft 2 all
0 2


 

 
 
T c
w g
(solid shaft)
(b)
2
1
2
0 2
( / )
1
( / )
h
T w c
T w c
 
2
1
2
0 2
1
T T c
w w c
 
   
 
 
     
     

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291
PROBLEM 3.31
Determine the largest allowable diameter of a 3-m-long steel rod (G  77.2 GPa) if the rod is to be twisted
through 30° without exceeding a shearing stress of 80 MPa.
SOLUTION
3 6
6
3
9 3
30
3 m, 523.6 10 rad, 80 10 Pa
180
, , ,
(80 10 )(3.0)
5.9374 10 m 5.9374 mm
(77.2 10 )(523.6 10 )

 
   
 




     
     

   
 
L
TL GJ Tc GJ c G c L
T c
GJ L J JL L G
c
2 11.87
d c
  
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292
5000 ft
A
B
PROBLEM 3.32
The ship at A has just started to drill for oil on the ocean floor at a depth of 5000 ft.
Knowing that the top of the 8-in.-diameter steel drill pipe 6
( 11.2 10 psi)
G  
rotates through two complete revolutions before the drill bit at B starts to operate,
determine the maximum shearing stress caused in the pipe by torsion.
SOLUTION
TL GJ
T
GJ L
Tc GJ c G c
J JL L


 

 
  
1
2rev (2)(2 ) 12.566 rad, 4.0 in.
2
5000 ft 60,000 in.
 
    
 
c d
L
6
3
(11.2 10 )(12.566)(4.0)
9.3826 10 psi
60,000


   9.38 ksi
  
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293
A
3 ft
1.5 in.
T = 60 kip · in.
B
PROBLEM 3.33
(a) For the solid steel shaft shown, determine the angle of twist
at A. Use 6
11.5 10 psi.
 
G (b) Solve part a, assuming that the
steel shaft is hollow with a 1.5-in. outer radius and a 0.75-in.
inner radius.
SOLUTION
(a) 4 4
2 4
1.5 in. (1.5 in.) 7.9522 in
2
(60 kip in.)(36 in.)
(11,200 kips/in )(7.9522 in )
0.024252 radians



  

 

c J
TL
GJ
1.390
   
(b) 4 4 4
2 4
[(1.5 in.) (0.75 in.) ] 7.4552 in
2
(60 kip in.)(36 in.)
(11,200 kips/in )(7.4552 in )
0.025869 radians
J



  



1.482
   
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294
2.5 m
40 mm
50 mm
A
B
T0
PROBLEM 3.34
(a) For the aluminum pipe shown (G  27 GPa), determine the torque T0
causing an angle of twist of 2°. (b) Determine the angle of twist if the
same torque T0 is applied to a solid cylindrical shaft of the same length
and cross-sectional area.
SOLUTION
(a)
4 4 4 4
6 4
3
9
50 mm 0.050 m, 40 mm 0.040 m
( ) (0.050 0.040 )
2 2
5.7962 10 m
2 34.907 10 rad 2.5 m
27 10 Pa
 



   
   
 
    
 
o i
o i
c c
J c c
L
G
TL
GJ
 
9 6 3
0
3
(27 10 )(5.7962 10 )(34.907 10 )
2.5
2.1851 10 N m
  
  
 
  
GJ
T
L
0 2.19 kN m
T   
Area of pipe: 2 2 2 2 2
( ) (0.050 0.040 ) 2.8274 m
 
    
o i
A c c
(b) Radius of solid of same area:
4 4 6 2
3
0
9 6
0.030 m
(0.030) 1.27235 10 m
2 2
(2.1851 10 )(2.5)
0.15902 rad
(27 10 )(1.27235 10 )

 



 
   

  
 
s
s s
s
A
c
J c
T L
GJ
9.11
s
   
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295
300 N · m
A
200 N · m
1 m
1.2 m
0.9 m
44 mm
40 mm
B
C
48 mm
D
PROBLEM 3.35
The electric motor exerts a 500-N  m torque on
the aluminum shaft ABCD when it is rotating at a
constant speed. Knowing that 27 GPa
G  and that
the torques exerted on pulleys B and C are as shown,
determine the angle of twist between (a) B and C,
(b) B and D.
SOLUTION
(a) Angle of twist between B and C.
9
4 9
200 N m, 1.2 m
1
0.022 m, 27 10 Pa
2
367.97 10 m
2
BC BC
BC
T L
c d G
J c
 
  
   
  
3
/ 9 9
(200)(1.2)
24.157 10 rad
(27 10 )(367.97 10 )
B C
TL
GJ
 
   
 
/ 1.384
B C
   
(b) Angle of twist between B and D.
9
4 4 9 4
3
/ 9 9
1
500 N m, 0.9 m, 0.024 m, 27 10 Pa
2
(0.024) 521.153 10 m
2 2
(500)(0.9)
31.980 10 rad
(27 10 )(521.153 10 )
 



      
   
  
 
CD CD
CD
C D
T L c d G
J c
3 3 3
/ / / 24.157 10 31.980 10 56.137 10 rad
B D B C C D
     
        / 3.22
B D
   
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296
30 mm
46 mm
C
A
B
TA ⫽ 300 N · m
TB ⫽ 400 N · m
0.9 m
0.75 m
PROBLEM 3.36
The torques shown are exerted on pulleys A and B. Knowing
that the shafts are solid and made of steel (G  77.2 GPa),
determine the angle of twist between (a) A and B, (b) A and C.
SOLUTION
(a)
4 9 4
3
9 9
1
300 N m, 0.9 m, 0.015 m
2
(0.015) 79.522 10 m
2
(300)(0.9)
43.980 10 rad
(77.2 10 )(79.522 10 )





    
  
   
 
AB AB AB
AB
AB AB
AB
T L C d
J
T L
GJ
2.52
AB
   
(b)
4 9 4
3
9 9
1
300 400 700 N m, 0.75 m, 0.023 m
2
(0.023) 439.57 10 m
2
(700)(0.75)
15.4708 10 rad
(77.2 10 )(439.57 10 )
0.89
2.52 0.89



  



      
  
   
 
 
     
BC BC BC
BC
BC BC
BC
BC
BC
AC AB BC
T L C d
J
T L
GJ
3.41
AC
   
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297
Brass
200 mm
300 mm
A
B
C
Aluminum
100 N · m
PROBLEM 3.37
The aluminum rod BC ( 26 GPa)

G is bonded to the brass rod AB
( 39 GPa).

G Knowing that each rod is solid and has a diameter of 12 mm,
determine the angle of twist (a) at B, (b) at C.
SOLUTION
Both portions:
3
4 3 4 9 4
1
6 mm = 6 10 m
2
(6 10 ) 2.03575 10 m
2 2
100 N m
c d
J c
T
 

 
  
    
 
Rod AB: 9
39 10 Pa, 0.200 m
AB AB
G L
  
(a) 9 9
(100)(0.200)
0.25191 rad
(39 10 )(2.03575 10 )
AB
B AB
AB
TL
G J
  
   
 
14.43
B
   
Rod BC: 9
9 9
26 10 Pa, 0.300 m
(100)(0.300)
0.56679 rad
(26 10 )(2.03575 10 )
BC BC
BC
BC
BC
G L
TL
G J
 
  
  
 
(b) 0.25191 0.56679 0.81870 rad
C B BC
  
     46.9
C
   
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298
400 mm
375 mm
250 mm
D
60 mm
36 mm
TA ⫽ 800 N · m
TB ⫽ 1600 N · m
C
B
A
PROBLEM 3.38
The aluminum rod AB ( 27 GPa)

G is bonded to the brass
rod BD ( 39 GPa).

G Knowing that portion CD of the
brass rod is hollow and has an inner diameter of 40 mm,
determine the angle of twist at A.
SOLUTION
Rod AB: 9
4 4 9
3
/ 9 9
27 10 Pa, 0.400 m
1
800N m 0.018 m
2
(0.018) 164.896 10 m
2 2
(800)(0.400)
71.875 10 rad
(27 10 )(164.896 10 )
A B
G L
T c d
J c
TL
GJ
 




  
   
   
   
 
Part BC: 9
4 4 6 4
3
/ 9 6
1
39 10 Pa 0.375 m, 0.030 m
2
800 1600 2400 N m, (0.030) 1.27234 10 m
2 2
(2400)(0.375)
18.137 10 rad
(39 10 )(1.27234 10 )
B C
G L c d
T J c
TL
GJ
 




    
       
   
 
Part CD:
 
1 1
2 2
4 4 4 4 6 4
2 1
3
/ 9 6
1
0.020 m
2
1
0.030 m, 0.250 m
2
(0.030 0.020 ) 1.02102 10 m
2 2
(2400)(0.250)
15.068 10 rad
(39 10 )(1.02102 10 )
C D
c d
c d L
J c c
TL
GJ
 




 
  
     
   
 
Angle of twist at A. / / /
3
105.080 10 rad
A A B B C C D
   

  
 
6.02
A
   
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299
4 in.
8 in.
ds
t 5 in.
1
4
3 in.
D
C
A
B
T
PROBLEM 3.39
The solid spindle AB has a diameter 1.75 in.

s
d and is made of a steel with
6
11.2 10 psi
G   and all 12 ksi,
  while sleeve CD is made of a brass
with 6
5.6 10 psi
G   and all 7 ksi.
  Determine (a) the largest torque T
that can be applied at A if the given allowable stresses are not to be exceeded
and if the angle of twist of sleeve CD is not to exceed 0.375, (b) the
corresponding angle through which end A rotates.
SOLUTION
Spindle AB: 6
all
4 4 4
1
(1.75 in.) 0.875 in. 12 in., 12 ksi, 11.2 10 psi
2
(0.875) 0.92077 in
2 2

 
     
  
c L G
J c
Sleeve CD:
 
1 2 all
4 4 4 6
2 1
1.25 in., 1.5 in., 8 in., 7ksi
4.1172 in , 5.6 10 psi
2
c c L
J c c G


   
    
(a) Largest allowable torque T.
Ciriterion: Stress in spindle AB.
Tc J
T
J c

  
(0.92077)(12)
12.63 kip in.
0.875
T   
Critrion: Stress in sleeve CD.
4
2
4.1172 in
(7 ksi)
1.5 in.
J
T
c

  19.21kip in.
T  
Criterion: Angle of twist of sleeve CD. 3
0.375 6.545 10 rad
 
   
6
3
(4.1172)(5.6 10 )
(6.545 10 )
8
TL JG
T
JG L
  

   
18.86 kip in.
T  
The largest allowable torque is 12.63 kip in.
T   
(b) Angle of rotation of end A. / / /
3
6 6
12 8
(12.63 10 )
(0.92077)(11.2 10 ) (4.1172)(5.6 10 )
   
    
 
  
 
 
 
 
i i i
A A D A B C D
i i i i
T L L
T
J G J G
0.01908 radians
 1.093
A
   
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300
4 in.
8 in.
ds
t 5 in.
1
4
3 in.
D
C
A
B
T
PROBLEM 3.40
The solid spindle AB has a diameter 1.5 in.

s
d and is made of a steel with
6
11.2 10 psi
 
G and all 12 ksi,
  while sleeve CD is made of a brass with
6
5.6 10 psi
 
G and all 7 ksi.
  Determine the angle through which end A
can be rotated.
SOLUTION
Stress analysis of solid spindle AB: s
1
0.75 in.
2
c d
 
3
3 3 3
2
(12 10 )(0.75) 7.95 10 lb in.
2
Tc J
T c
J c
T
 
 

  
    
Stress analysis of sleeve CD: 2
1 1
(3) 1.5 in.
2 2
  
o
c d
 
1 2
4 4 4 4 4
2 1
3
3
2
1.5 0.25 1.25 in.
(1.5 1.25 ) 4.1172 in
2 2
(4.1172)(7 10 )
19.21 10 lb in.
1.5
 
 
    
    

    
c c t
J c c
J
T
c
The smaller torque governs. 3
7.95 10 lb in.
T   
Deformation of spindle AB: 0.75 in.
c 
4 4 6
3
6
0.49701in , 12 in., 11.2 10 psi
2
(7.95 10 )(12)
0.017138 radians
(11.2 10 )(0.49701)
AB
J c L G
TL
GJ
    

  



Deformation of sleeve CD: 4 6
3
6
4.1172 in , 8 in., 5.6 10 psi
(7.95 10 )(8)
0.002758 radians
(5.6 10 )(4.1172)

   

  

CD
J L G
TL
GJ
Total angle of twist: 0.019896 radians
  
  
AD AB CD 1.140
AD
   
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301
T
E
F
B
A
4.5 in.
6 in.
12 in.
8 in.
6 in.
D
C
PROBLEM 3.41
Two shafts, each of 7
8
-in. diameter, are
connected by the gears shown. Knowing that
6
11.2 10 psi
 
G and that the shaft at F is
fixed, determine the angle through which end
A rotates when a 1.2-kip  in. torque is applied
at A.
SOLUTION
Calculation of torques.
Circumferential contact force between gears B and E: AB EF E
EF AB
B E B
T T r
F T T
r r r
  
1.2 kip in. 1200 lb in.
6
(1200) 1600 lb in.
4.5
AB
EF
T
T
   
  
Twist in shaft FE.
6
4
4 3 4
3
/ 6 3
1 7
12 in., in., 11.2 10 psi
2 16
7
57.548 10 in
2 2 16
(1600)(12)
29.789 10 rad
(11.2 10 )(57.548 10 )
E F
L c d G
J c
TL
GJ
 




    
 
   
 
 
   
 
Rotation at E. 3
/ 29.789 10 rad
E F
E
  
  
Tangential displacement at gear circle: E E B B
r r
  
 
Rotation at B. 3 3
6
(29.789 10 ) 39.718 10 rad
4.5
E
B E
B
r
r
   
    
Twist in shaft BA. 3 4
3
/ 6 3
8 6 14 in. 57.548 10 in
(1200)(14)
26.065 10 rad
(11.2 10 )(57.548 10 )
A B
L J
TL
GJ




    
   
 
Rotation at A. 3
/ 65.783 10 rad
A B A B
   
    3.77
A
   
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302
30 mm
E
60 mm
30 mm
90 mm
0.5 m
0.1 m
0.2 m
0.4 m
0.2 m
B
D
C
A
T PROBLEM 3.42
Two solid steel shafts, each of 30-mm diameter, are
connected by the gears shown. Knowing that G  77.2 GPa,
determine the angle through which end A rotates when a
torque of magnitude T  200 N  m is applied at A.
SOLUTION
Calculation of torques.
Circumferential contact force between gears B and D: AB DE
B D
T T
F
r r
 
90
(200) 300 N m
60
D
DE AB
B
r
T T
r
   
Twist in shaft DE. 4 4 9 4
(0.015) 79.522 10 m
2 2
DE
J c
  
   
3
9 9
0.5 m
(300)(0.5)
24.434 10 rad
(77.2 10 )(79.522 10 )
 


   
 
DE
DE DE
DE
DE
L
T L
GJ
Rotation at D. 3
24.434 10 rad
D DE
  
  
Circumferential displacement at gear circles: D D B B
r r
  
 
Rotation at B. 3 3
90
(24.434 10 ) 36.651 10 rad
60
D
B D
B
r
r
   
    
Twist in shaft AB. 9 4
3
9 9
0.1 0.2 0.4 0.2 0.9 m, 79.522 10 m
(200)(0.9)
29.320 10 rad
(77.2 10 )(79.522 10 )
AB AB
AB AB
AB
AB
L J
T L
GJ




      
   
 
Rotation at A.
3 3
36.651 10 29.320 10 rad
  
 
 
   
A B AB
3
65.971 10 rad

  3.78 
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303
F
E
D
nr r
C
l
TA
B
A
nr
l
l
r
PROBLEM 3.43
A coder F, used to record in digital form the
rotation of shaft A, is connected to the shaft by
means of the gear train shown, which consists of
four gears and three solid steel shafts each of
diameter d. Two of the gears have a radius r and the
other two a radius nr. If the rotation of the coder F
is prevented, determine in terms of T, l, G, J, and n
the angle through which end A rotates.
SOLUTION
2
AB A
C AB A
CD AB
B
E CD A
EF CD
D
T T
r T T
T T
r n n
r T T
T T
r n n

  
  
2
3
3 3
4 2
1 1
1 1
EF EF A
E EF
E E A
D E
D
CD CD A
CD
A A A
C D CD
C C
B C
B
AB AB A
AB
T l T l
GJ n GJ
r T l
r n n GJ
T l T l
GJ nGJ
T l T l T l
nGJ GJ n
n GJ n
r Tl
r n GJ n n
T l T l
GJ GJ
 

 

  

 

  
  
 
 
     
 
 
 
   
 
 
 
4 2
1 1
1
A
A B AB
T l
GJ n n
  
 
    
 
 
  
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304
F
E
D
nr r
C
l
TA
B
A
nr
l
l
r
PROBLEM 3.44
For the gear train described in Prob. 3.43,
determine the angle through which end A rotates
when 5 lb in.,
 
T 2.4 in.,

l 1/16 in.,

d
6
11.2 10 psi,
 
G and 2.
n 
PROBLEM 3.43 A coder F, used to record in
digital form the rotation of shaft A, is connected to
the shaft by means of the gear train shown, which
consists of four gears and three solid steel shafts
each of diameter d. Two of the gears have a radius r
and the other two a radius nr. If the rotation of the
coder F is prevented, determine in terms of T, l, G,
J, and n the angle through which end A rotates.
SOLUTION
See solution to Prob. 3.43 for development of equation for .
A

2 4
1 1
1
A
Tl
GJ n n

 
  
 
 
Data: 6
4
4 6 4
1 1
5 lb in., 2.4 in., in., 11.2 10 psi
2 32
1
2, 1.49803 10 in
2 2 32
  
      
 
    
 
 
T l c d G
n J c
3
6 6
(5)(2.4) 1 1
1 938.73 10 rad
4 16
(11.2 10 )(1.49803 10 )
A
 

 
    
 
   
 53.8
A
   
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305
PROBLEM 3.45
The design specifications of a 1.2-m-long solid circular transmission shaft require that the angle of twist of
the shaft not exceed 4° when a torque of 750 N  m is applied. Determine the required diameter of the shaft,
knowing that the shaft is made of a steel with an allowable shearing stress of 90 MPa and a modulus of
rigidity of 77.2 GPa.
SOLUTION
3 4
750 N m, 4 69.813 10 rad, 1.2 m,
2

 
       
T L J c
6
9
90 MPa 90 10 Pa
77.2 GPa 77.2 10 Pa
G
   
  
Based on angle of twist. 4
3
4 4
9 3
2
2 (2)(750)(1.2)
18.0569 10 m
(77.2 10 )(69.813 10 )


  


 
   
 
TL TL
GJ Gc
TL
c
G
Based on shearing stress. 3
3
3 3
6
2
2 (2)(750)
17.441 10 m
(90 10 )
Tc T
J c
T
c


 

 
   

Use larger value. 3
18.0569 10 m 18.0569 mm
c 
   2 36.1 mm
d c
  
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306
A
L
a
P
B
C
PROBLEM 3.46
The solid cylindrical rod BC of length L  24 in. is attached to the rigid
lever AB of length a  15 in. and to the support at C. Design specifications
require that the displacement of A not exceed 1 in. when a 100-lb force P is
applied at A. For the material indicated, determine the required diameter of
the rod.
Steel: all  15 ksi, G  11.2  106
psi.
SOLUTION
At the allowable twist angle,
1
sin 0.06667
15
3.8226 0.066716 rad.
cos (100)(15)cos 3.8226 1496.7 lb in.




  
  
    
a
T Pa
Based on twist. 4
4
2 2

 

   
TL TL TL
c
GJ G
Gc
4
6
3 4
(2)(1496.7)(24)
(11.2 10 )(0.066716)
30.603 10 in 0.418 in.




  
c
c
Based on stress. 3
3
3 3 3
2 2
( 15,000 psi)
2(1496.7)
63.522 10 in 0.399 in.
(15,000)
 




    
   
Tc T T
c
J c
c c
Use larger value for design. 0.399 in.

c 2 0.837 in.
d c
  
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307
A
L
a
P
B
C
PROBLEM 3.47
The solid cylindrical rod BC of length L  24 in. is attached to the rigid
lever AB of length a  15 in. and to the support at C. Design specifications
require that the displacement of A not exceed 1 in. when a 100-lb force P is
applied at A. For the material indicated, determine the required diameter of
the rod.
Aluminum: all  10 ksi, G  3.9  106
psi.
SOLUTION
At the allowable twist angle,
1
sin 0.06667
15
3.8226 0.066716 rad
cos (100)(15)cos3.8226 1496.7 lb in.




  
  
    
a
T Pa
Based on twist. 4
4
2 2

 

   
TL TL TL
c
GJ G
Gc
4
6
3 4
(2)(1496.7)(24)
(3.9 10 )(0.066716)
87.888 10 in 0.544 in.




  
c
c
Based on stress. 3
3
3 3 3
2 2
( 10,000 psi)
2(1496.7)
95.283 10 in 0.457 in.
(10,000)
 




    
   
Tc T T
c
J c
c c
Use larger value for design. 0.544 in.

c 2 1.089 in.
 
d c 
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308
A
100 mm
40 mm
C
B
D
T 5 1000 N · m
400 mm
600 mm
PROBLEM 3.48
The design of the gear-and-shaft system shown
requires that steel shafts of the same diameter be
used for both AB and CD. It is further required
that max 60 MPa,
  and that the angle D
 through
which end D of shaft CD rotates not exceed 1.5.
Knowing that G  77.2 GPa, determine the
required diameter of the shafts.
SOLUTION
100
1000 N m (1000) 2500N m
40
B
CD D AB CD
C
r
T T T T
r
      
For design based on stress, use larger torque. 2500 N m
AB
T  
3
3 6 3
6
3
2
2 (2)(2500)
26.526 10 m
(60 10 )
29.82 10 m 29.82 mm, 2 59.6 mm
Tc T
J c
T
c
c d c


 


 
   

    
Design based on rotation angle. 3
1.5 26.18 10 rad
D
 
   
Shaft AB: 2500 N m, 0.4 m
AB
T L
  
(2500)(0.4) 1000
1000
100 1000 2500
40
AB
B AB
B
C B
C
TL
GJ GJ GJ
GJ
Gears
r
r GJ GJ

 
 
  

 



  
   
  
   

Shaft CD: 1000 N m, 0.6 m
CD
T L
  
4
2
4 9 4
9 3
3
(1000)(0.6) 600
2500 600 3100 3100
(2)(3100) (2)(3100)
976.46 10 m
(77.2 10 )(26.18 10 )
31.435 10 m 31.435 mm, 2 62.9 mm


  
  



  
     
   
 
    
CD
D C CD
D
TL
GJ GJ GJ
GJ GJ GJ G c
c
G
c d c
Design must use larger value for d. 62.9 mm
d  
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309
A
0.3 m
0.6 m
0.4 m C
B
500 N · m
300 N · m
D
PROBLEM 3.49
The electric motor exerts a
torque of 800 N m
 on the
steel shaft ABCD when it is
rotating at constant speed.
Design specifications require
that the diameter of the shaft
be uniform from A to D and
that the angle of twist between
A to D not exceed 1.5°.
Knowing that max 60 MPa
 
and G  77.2 GPa, determine
the minimum diameter shaft
that can be used.
SOLUTION
Torques:
300 500 800 N m
500 N m, 0
AB
BC CD
T
T T
   
  
Design based on stress. 6
60 10 Pa
  
3 6 3
3 6
3
2 2 (2)(800)
8.488 10 m
(60 10 )
20.40 10 m 20.40 mm, 2 40.8 mm
Tc T T
c
J c
c d c


 


     

    
Design based on deformation. 3
/ 1.5 26.18 10 rad
D A
 
   
/
/
/
/ / / / 4 4
2
4 9 4
9 3
/
3
0
(500)(0.6) 300
(800)(0.4) 320
620 620 (2)(620)
(2)(620) (2)(620)
195.292 10 m
(77.2 10 )(26.18 10 )
21.022 10 m 21.0




   

  




  
  
     
   
 
  
D C
BC BC
C B
AB AB
B A
D A D C C B B A
D A
T L
GJ GJ GJ
T L
GJ GJ GJ
GJ G c Gc
c
G
c 22 mm, 2 42.0 mm
 
d c
Design must use larger value of d. 42.0 mm
d  
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310
500 mm
300 mm
C
D
B
P
A
PROBLEM 3.50
A hole is punched at A in a plastic sheet by applying a 600-N
force P to end D of lever CD, which is rigidly attached to the
solid cylindrical shaft BC. Design specifications require that
the displacement of D should not exceed 15 mm from the
time the punch first touches the plastic sheet to the time it
actually penetrates it. Determine the required diameter of
shaft BC if the shaft is made of a steel with 77.2 GPa

G
and all 80 MPa.
 
SOLUTION
Torque: (0.300 m)(600 N) 180 N m
T rP
   
Shaft diameter based on displacement limit.
4
15 mm
0.005 rad
300 mm
2
r
TL TL
GJ Gc




  
 
4 9 4
9
3
2 (2)(180)(0.500)
14.843 10 m
(77.2 10 )(0.05)
11.038 10 m 11.038 m 2 22.1 mm
TL
c
G
c d c


   

    
  
Shaft diameter based on stress.
6
3
3 6 3
6
3
2
80 10 Pa
2 (2)(180)
1.43239 10 m
(80 10 )
11.273 10 m 11.273 mm 2 22.5 mm
Tc T
J c
T
c
c d c
 

 


   
   

    
Use the larger value to meet both limits. 22.5 mm
d  
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311
18 in.
12 in.
1.5 in.
2.0 in.
A
C
B
T 12.5 kip · in.
Aluminum
Brass
PROBLEM 3.51
The solid cylinders AB and BC are bonded together at B and are
attached to fixed supports at A and C. Knowing that the modulus of
rigidity is 3.7  106
psi for aluminum and 5.6 106
psi for brass,
determine the maximum shearing stress (a) in cylinder AB, (b) in
cylinder BC.
SOLUTION
The torques in cylinders AB and BC are statically indeterminate. Match the rotation B
 for each cylinder.
Cylinder AB:
4 4
6
6
1
0.75 in. 12 in.
2
0.49701 in
2
(12)
6.5255 10
(3.7 10 )(0.49701)
AB AB
B AB
c d L
J c
T L T
T
GJ

 
  
 
   

Cylinder BC:
4 4 4
6
6
1
1.0 in. 18 in.
2
(1.0) 1.5708 in
2 2
(18)
2.0463 10
(5.6 10 )(1.5708)
BC BC
B BC
c d L
J c
T L T
T
GJ
 
 
  
  
   

Matching expressions for :
B
6 6
6.5255 10 2.0463 10
AB BC
T T
 
  
3.1889
BC AB
T T
 (1)
Equilibrium of connection at :
B 3
0 12.5 10 lb in.
AB BC
T T T T
     
3
12.5 10
AB BC
T T
   (2)
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312
PROBLEM 3.51 (Continued)
Substituting (1) into (2), 3
4.1889 12.5 10
AB
T  
3 3
2.9841 10 lb in. 9.5159 10 lb in.
AB BC
T T
     
(a) Maximum stress in cylinder AB.
3
3
(2.9841 10 )(0.75)
4.50 10 psi
0.49701


   
AB
AB
T c
J
4.50 ksi
AB
  
(b) Maximum stress in cylinder BC.
3
3
(9.5159 10 )(1.0)
6.06 10 psi
1.5708
BC
BC
T c
J


    6.06 ksi
BC
  
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313
18 in.
12 in.
1.5 in.
2.0 in.
A
C
B
T 12.5 kip · in.
Aluminum
Brass
PROBLEM 3.52
Solve Prob. 3.51, assuming that cylinder AB is made of steel, for
which G  11.2  106
psi.
PROBLEM 3.51 The solid cylinders AB and BC are bonded together
at B and are attached to fixed supports at A and C. Knowing that the
modulus of rigidity is 3.7  106
psi for aluminum and 5.6  106
psi for
brass, determine the maximum shearing stress (a) in cylinder AB,
(b) in cylinder BC.
SOLUTION
The torques in cylinders AB and BC are statically indeterminate. Match the rotation B
 for each cylinder.
Cylinder AB: 4 4
6
6
1
0.75 in. 12 in. 0.49701 in
2 2
(12)
2.1557 10
(11.2 10 )(0.49701)
AB AB
B AB
c d L J c
T L T
T
GJ

 
    
   

Cylinder BC: 4 4 4
6
6
1
1.0 in. 18 in. (1.0) 1.5708 in
2 2 2
(18)
2.0463 10
(5.6 10 )(1.5708)
BC BC
B BC
c d L J c
T L T
T
GJ
 
 
     
   

Matching expressions for :
B
6 6
2.1557 10 2.0463 10 1.0535
 
   
AB BC BC AB
T T T T (1)
Equilibrium of connection at :
B 3
0 12.5 10
AB BC AB BC
T T T T T
      (2)
Substituting (1) into (2), 3
2.0535 12.5 10
 
AB
T
3 3
6.0872 10 lb in. 6.4128 10 lb in.
     
AB BC
T T
(a) Maximum stress in cylinder AB.
3
3
(6.0872 10 )(0.75)
9.19 10 psi
0.49701


   
AB
AB
T c
J
9.19 ksi
AB
  
(b) Maximum stress in cylinder BC.
3
3
(6.4128 10 )(1.0)
4.08 10 psi
1.5708
BC
BC
T c
J


    4.08 ksi
BC
  
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314
0.2 in.
Steel core
Brass jacket
6 ft
1.2 in.
T
T
A
B
PROBLEM 3.53
The composite shaft shown consists of a 0.2-in.-thick brass
jacket (G  5.6  106
psi) bonded to a 1.2-in.-diameter steel core
(Gsteel  11.2  106
psi). Knowing that the shaft is subjected to
5-kip  in. torques, determine (a) the maximum shearing stress in
the brass jacket, (b) the maximum shearing stress in the steel
core, (c) the angle of twist of end B relative to end A.
SOLUTION
Steel core: 1
4 4 4
1 1
6 6 2
1 1
1
0.6 in.
2
(0.6) 0.20358 in
2 2
(11.2 10 )(0.20358) 2.2801 10 lb in
c d
J c
G J
 
 
  
    
Torque carried by steel core. 1 1 1
T G J
L


Brass jacket:
 
2 1
4 4 4 4 4
2 2 1
6 6 2
2 2
0.6 0.2 0.8 in.
(0.8 0.6 ) 0.43982 in
2 2
(5.6 10 )(0.43982) 2.4630 10 lb in
 
    
    
    
c c t
J C C
G J
Torque carried by brass jacket. 2 2 2
T G J
L


Total torque: 1 2 1 1 2 2
3
6 6
1 1 2 2
3
( )
5 10
2.2801 10 2.4630 10
1.05416 10 rad/in.



   

 
   
 
T T T G J G J
L
T
L G J G J
(a) Maximum shearing stress in brass jacket.
6 3
max 2 max 2 2 (5.6 10 )(0.8)(1.05416 10 )

  
    
G G c
L
3
4.72 10 psi
  4.72 ksi 
(b) Maximum shearing stress in steel core.
6 3
max 1 max 1 1 (11.2 10 )(0.6)(1.05416 10 )

  
    
G G c
L
3
7.08 10 psi
  7.08 ksi 
(c) Angle of twist. ( 6 ft 72 in.)
L  
3 3
(72)(1.0542 10 ) 75.9 10 rad
L
L

  
    
4.35
  
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315
0.2 in.
Steel core
Brass jacket
6 ft
1.2 in.
T
T
A
B
PROBLEM 3.54
The composite shaft shown consists of a 0.2-in.-thick brass
jacket (G  5.6  106
psi) bonded to a 1.2-in.-diameter steel
core (Gsteel  11.2  106
psi). Knowing that the shaft is being
subjected to the torques shown, determine the largest angle
through which it can be twisted if the following allowable
stresses are not to be exceeded: steel  15 ksi and brass  8 ksi.
SOLUTION
max max max
all all
max
for each material.

 
 
 

G Gc
L
L Gc
Steel core: all max
3
all
6
1
15 ksi 15,000 psi, 0.6 in.
2
15,000
2.2321 10 rad/in.
(11.2 10 )(0.6)

 
   
  

c d
L
Brass jacket: all max
3
all
6
8 ksi 8000 psi, 0.6 0.2 0.8 in.
8000
1.78571 10 rad/in.
(5.6 10 )(0.8)

 
    
  

c
L
Smaller value governs. 3
all
1.78571 10 rad/in.
L
 
 
Allowable angle of twist: 6 ft 72 in.
L  
3 3
all
all (72)(1.78571 10 ) 128.571 10 rad
L
L

  
    
7.37
  
Mecánica de Materiales - Ferdinand P. Beer 7ma. Edición.

355. PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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316
250 mm
38 mm
1.4 kN · m
50 mm
C
200 mm
B
A
PROBLEM 3.55
Two solid steel shafts ( 77.2 GPa)

G are connected to a coupling disk B and
to fixed supports at A and C. For the loading shown, determine (a) the
reaction at each support, (b) the maximum shearing stress in shaft AB, (c) the
maximum shearing stress in shaft BC.
SOLUTION
Shaft AB:
4 4 9 4
9 9
3
1
, 0.200 m, 25 mm 0.025 m
2
(0.025) 613.59 10 m
2 2
(77.2 10 )(613.59 10 )
236.847 10
0.200
AB AB
AB AB
AB B
AB
AB
AB B B B
AB
T T L c d
T L
J c
GJ
GJ
T
L
 

  


    
    
 
   
Shaft BC:
4 4 9 4
9 9
3
1
, 0.250 m, 19 mm 0.019 m
2
(0.019) 204.71 10 m
2 2
77.2 10 )(204.71 10 )
63.214 10
0.250
BC BC
BC BC
BC B
BC
BC
BC B B
BC
T T L c d
T L
J c
GJ
GJ
T
L
 

 


    
    
  
   
Equilibrium of coupling disk. AB BC
T T T
 
3 3 3 3
3 3 3
3 3
1.4 10 236.847 10 63.214 10 4.6657 10 rad
(236.847 10 )(4.6657 10 ) 1.10506 10 N m
(63.214 10 )(4.6657 10 ) 294.94 N m
   


      
     
    
B B B
AB
BC
T
T
(a) Reactions at supports. 1105 N m
A AB
T T
   
295 N m
C BC
T T
   
(b) Maximum shearing stress in AB.
3
6
9
(1.10506 10 )(0.025)
45.0 10 Pa
613.59 10
AB
AB
AB
T c
J
 

   

45.0 MPa
AB
  
(c) Maximum shearing stress in BC.
6
9
(294.94)(0.019)
27.4 10 Pa
204.71 10
BC
BC
BC
T c
J
 
   

27.4 MPa
BC
  
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on a website, in whole or part.
317
250 mm
38 mm
1.4 kN · m
50 mm
C
200 mm
B
A
PROBLEM 3.56
Solve Prob. 3.55, assuming that the shaft AB is replaced by a hollow shaft
of the same outer diameter and 25-mm inner diameter.
PROBLEM 3.55 Two solid steel shafts ( 77.2 GPa)

G are connected to
a coupling disk B and to fixed supports at A and C. For the loading
shown, determine (a) the reaction at each support, (b) the maximum
shearing stress in shaft AB, (c) the maximum shearing stress in shaft BC.
SOLUTION
Shaft AB:
 
2
1
4 4 4 4 9 4
2 1
9 9
3
, 0.200 m, 25 mm 0.025 m
12.5 mm 0.0125 m
(0.025 0.0125 ) 575.24 10 m
2 2
(77.2 10 )(575.24 10 )
222.04 10
0.200
AB AB
AB
AB
AB B B B
AB
T T L c
c
J c c
GJ
T
L
 
  


   
 
     
 
   
Shaft BC:
4 4 9
9 9
3
1
, 0.250 m, 19 mm 0.019 m
2
(0.019) 204.71 10 m
2 2
77.2 10 )(204.71 10 )
63.214 10
0.250
BC BC
BC
BC
BC B B B
BC
T T L c d
J c
GJ
T
L
 
  


    
   
  
   
Equilibrium of coupling disk: AB BC
T T T
 
3 3 3
3
3 3 3
3 3
1.4 10 222.04 10 63.214 10
4.9079 10 rad
(222.04 10 )(4.9079 10 ) 1.08975 10 N m
(63.214 10 )(4.9079 10 ) 310.25 N m
 
 


    
 
     
    
B B
B
AB
BC
T
T
(a) Reactions at supports: 1090 N m
A AB
T T
   
310 N m
C BC
T T
   
(b) Maximum shearing stress in AB:
3
6
2
9
(1.08975 10 )(0.025)
47.4 10 Pa
575.243 10
AB
AB
AB
T c
J
 

   

47.4 MPa
AB
  
(c) Maximum shearing stress in BC:
6
9
(310.25)(0.019)
28.8 10 Pa
204.71 10
BC
BC
BC
T c
J
 
   

28.8 MPa
BC
  
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318
36 mm
30 mm
900 mm
600 mm
C
B
D
A
T ⫽ 500 N · m
PROBLEM 3.57
Two solid steel shafts are fitted with flanges that are then
connected by bolts as shown. The bolts are slightly
undersized and permit a 1.5° rotation of one flange with
respect to the other before the flanges begin to rotate as a
single unit. Knowing that G  77.2 GPa, determine the
maximum shearing stress in each shaft when a torque of T
of magnitude 500 N  m is applied to the flange indicated.
The torque T is applied to flange B.
SOLUTION
Shaft AB:
4 4 9 4
9 9
3
1
, 0.6 m, 0.015 m
2
(0.015) 79.522 10 m
2 2
(77.2 10 )(79.522 10 )
0.6
10.2318 10
 

 



   
   

 
 
 
AB
AB
AB AB
B
AB AB
AB AB
AB B B
B
T T L c d
J c
T L
G J
G J
T
L
Shaft CD: 4 4
9 4
9 9
3
1
, 0.9 m, 0.018 m, (0.018)
2 2 2
164.896 10 m
77.2 10 )(164.896 10 )
14.1444 10
0.9
 
  


     
 
  
   
CD CD CD
CD
CD CD
CD C C C
CD
T T L c d J c
J
G J
T
L
Clearance rotation for flange B: 3
1.5 26.178 10 rad
B
 
    
Torque to remove clearance: 3 3
(10.2318 10 )(26.178 10 ) 267.85 N m
AB
T  
     
Torque T to cause additional rotation : 500 267.85 232.15 N m
T
     
AB CD
T T T
  
 
3 3 3
232.15 (10.2318 10 14.1444 10 ) 9.5236 10 rad
  
 
      
3 3
3 3
(10.2318 10 )(9.5236 10 ) 97.444 N m
(14.1444 10 )(9.5236 10 ) 134.706 N m
AB
CD
T
T


     
     
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358. PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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319
PROBLEM 3.57 (Continued)
Maximum shearing stress in AB:
6
9
(267.85 97.444)(0.015)
68.9 10 Pa
79.522 10
AB
AB
AB
T c
J
 

   

68.9 MPa 
Maximum shearing stress in CD:
6
9
(134.706)(0.018)
14.70 10 Pa
164.896 10
CD
CD
CD
T c
J
 
   

14.70 MPa 
Mecánica de Materiales - Ferdinand P. Beer 7ma. Edición.

359. PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
320
36 mm
30 mm
900 mm
600 mm
C
B
D
A
T ⫽ 500 N · m
PROBLEM 3.58
Two solid steel shafts are fitted with flanges that are then
connected by bolts as shown. The bolts are slightly
undersized and permit a 1.5° rotation of one flange with
respect to the other before the flanges begin to rotate as a
single unit. Knowing that G  77.2 GPa, determine the
maximum shearing stress in each shaft when a torque of T
of magnitude 500 N  m is applied to the flange indicated.
The torque T is applied to flange C.
SOLUTION
Shaft AB:
4 4 9 4
9 9
3
1
, 0.6 m, 0.015 m
2
(0.015) 79.522 10 m
2 2
(77.2 10 )(79.522 10 )
0.6
10.2318 10
 

 



   
   

 
 
 
AB AB
AB
AB AB
B
AB AB
AB AB
AB B B
AB
B
T T L c d
J c
T L
G J
G J
T
L
Shaft CD: 4 4
9 4
9 9
3
1
, 0.9 m, 0.018 m (0.018)
2 2 2
164.896 10 m
77.2 10 )(164.896 10 )
14.1444 10
0.9
CD CD CD
CD
CD CD
CD C C C
CD
T T L c d J c
J
G J
T
L
 
  


     
 
  
   
Clearance rotation for flange C: 3
1.5 26.178 10 rad
C
 
    
Torque to remove clearance: 3 3
(14.1444 10 )(26.178 10 ) 370.27 N m
CD
T 
     
Torque T to cause additional rotation : 500 370.27 129.730 N m
T
     
AB CD
T T T
  
 
3 3 3
129.730 (10.2318 10 14.1444 10 ) 5.3220 10 rad
  
 
     
3 3
3 3
(10.2318 10 )(5.3220 10 ) 54.454 N m
(14.1444 10 )(5.3220 10 ) 75.277 N m
AB
CD
T
T


     
     
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321
PROBLEM 3.58 (Continued)
Maximum shearing stress in AB.
6
9
(54.454)(0.015)
10.27 10 Pa
79.522 10
AB
AB
AB
T c
J
 
   

10.27 MPa 
Maximum shearing stress in CD.
6
9
(370.27 75.277)(0.018)
48.6 10 Pa 48.6 MPa
164.896 10
CD
CD
CD
T c
J
 

    


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322
B
C
D
E
A
T
T′
PROBLEM 3.59
The steel jacket CD has been attached to the 40-mm-diameter steel
shaft AE by means of rigid flanges welded to the jacket and to the
rod. The outer diameter of the jacket is 80 mm and its wall thickness
is 4 mm. If 500-N  m torques are applied as shown, determine the
maximum shearing stress in the jacket.
SOLUTION
Solid shaft:
1
0.020 m
2
c d
 
4 4 9 4
(0.020) 251.33 10 m
2 2
S
J c
  
   
Jacket:
 
2
1 2
4 4 4 4
2 1
6 4
1
0.040 m
2
0.040 0.004 0.036 m
(0.040 0.036 )
2 2
1.3829 10 m
J
c d
c c t
J c c
 

 
    
   
 
Torque carried by shaft. /
S S
T GJ L


Torque carried by jacket. /
J J
T GJ L


Total torque.
6
6 6
( ) /
(1.3829 10 )(500)
423.1 N m
1.3829 10 251.33 10



 
     


   
   
S J S J
S J
J
J
S J
G T
T T T J J G L
L J J
J
T T
J J
Maximum shearing stress in jacket.
6
2
6
(423.1)(0.040)
12.24 10 Pa
1.3829 10
J
J
T c
J 
   

 12.24 MPa 
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323
B
L
A
T
2
c
2c
PROBLEM 3.60
A torque T is applied as shown to a solid tapered shaft AB. Show by
integration that the angle of twist at A is
4
7
12
TL
Gc



SOLUTION
Introduce coordinate y as shown.
cy
r
L

Twist in length dy:
4
4 4
4
4
2 2
4 4 4 4
2
4 4
4 3 4 3 3
4
4 3 4
2
2
2 2
2 1 2 1 1
3 24 3
2 7 7
24 12
L L
L L
L
L
Tdy Tdy TL dy
d
GJ Gc y
G r
TL dy TL dy
Gc y Gc y
TL TL
Gc y Gc L L
TL TL
Gc L Gc

 

 
 
 
  
 
   
    
   
 
 
 
 
 
 
 

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324
PROBLEM 3.61
The mass moment of inertia of a gear is to be determined experimentally by
using a torsional pendulum consisting of a 6-ft steel wire. Knowing that
G  11.2  106
psi, determine the diameter of the wire for which the torsional
spring constant will be 4.27 lb  ft/rad.
SOLUTION
Torsion spring constant 4.27 lb ft/rad 51.24 lb in./rad
   
K
4
4 6 4
6
/ 2
2 (2)(72)(51.24)
209.7 10 in
(11.2 10 )
0.1203 in.


 

   
   


T T GJ Gc
K
TL GJ L L
LK
c
G
c 2 0.241 in.
d c
  
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325
PROBLEM 3.62
A solid shaft and a hollow shaft are made of the same material
and are of the same weight and length. Denoting by n the ratio
1 2
/ ,
c c show that the ratio /
s h
T T of the torque Ts in the solid shaft
to the torque Th in the hollow shaft is (a) 2 2
(1 )/(1 )
n n
 
if the maximum shearing stress is the same in each shaft,
(b) 2
(1 )/(1 )
 
n n if the angle of twist is the same for each shaft.
SOLUTION
For equal weight and length, the areas are equal.
 
 
2 2 2 2 2 2 1/2
0 2 1 2 0 2
4 4 2 2 4 4 4 4
0 2 2 1 2
(1 ) (1 )
(1 ) (1 )
2 2 2 2
s h
c c c c n c c n
J c c n J c c c n
  
   
      
      
(a) For equal stresses,
0 2
4 2 2 2 2 1/2
2 2
2 2
4 4 2 1/2 2 2 1/2 2
0 2 2
2
(1 ) 1 (1 )
(1 ) (1 ) (1 )(1 ) 1


  
  
   
    
s h
s h
s s
h h
T c T c
J J
c n c
T J c n n
T J c c n c n n n n

(b) For equal angles of twist,
4 2 2 2 2 2
2
2
4 4 4 2
2
2
(1 ) (1 ) 1
(1 ) 1 1
s h
s h
s s
h h
T L T L
GJ GJ
c n
T J n n
T J c n n n


  
  
   
  

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326
C
t
A
L2
L1
B
D
r1
r2
T
PROBLEM 3.63
An annular plate of thickness t and modulus G is used to
connect shaft AB of radius r1 to tube CD of radius r2. Knowing
that a torque T is applied to end A of shaft AB and that end D
of tube CD is fixed, (a) determine the magnitude and location
of the maximum shearing stress in the annular plate, (b) show
that the angle through which end B of the shaft rotates with
respect to end C of the tube is
2 2
1 2
1 1
4
BC
T
Gt r r


 
 
 
 
 
SOLUTION
Use a free body consisting of shaft AB and an inner portion of the plate BC, the outer radius of this portion
being .

The force per unit length of circumference is .
t
2
0
(2 ) 0
2
M
t T
T
t
  

 
 
 

(a) Maximum shearing stress occurs at 1 max 2
1
2
T
r
tr
 

  
Shearing strain: 2
2
T
G GT
 


 
The relative circumferential displacement in radial length d is
d d d
d
d
    

 

 

2 3
2 2
T d Td
d
Gt GT
 
 



  
(b)
2
2 2
1 1
1
/ 3 3 2
1
2 2
2 2
r
r r
B C
r r
Td T d T
Gt Gt
Gt r
 
   
 
 
 



 
 
 
2 2 2 2
2 1 1 2
1 1 1 1
2 4
2 2
T T
Gt Gt
r r r r
 
   
   
    
   
   
   

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327
PROBLEM 3.64
Determine the maximum shearing stress in a solid shaft of 1.5-in. diameter as it transmits 75 hp at a speed of
(a) 750 rpm, (b) 1500 rpm.
SOLUTION
3
1
0.75 in. 75 hp (75)(6600) 495 10 lb in./s
2
      
c d P
(a)
3
3
750
12.5 Hz
60
495 10
6.3025 10 lb in.
2 2 (12.5)
 
 

    
f
P
T
f
3
3
3 3
2 (2)(6.3025 10 )
9.51 10 psi
(0.75)

 

    
Tc T
J c
9.51 ksi
  
(b)
3
3
1500
25 Hz
60
495 10
3.1513 10 lb in.
2 (25)

 

   
f
T
3
3
3
(2)(3.1513 10 )
4.76 10 psi
(0.75)



   4.76 ksi
  
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328
PROBLEM 3.65
Determine the maximum shearing stress in a solid shaft of 12-mm diameter as it transmits 2.5 kW at a
frequency of (a) 25 Hz, (b) 50 Hz.
SOLUTION
1
6 mm 0.006 m 2.5 kW 2500 W
2
c d P
    
(a)
2500
25 Hz. 15.9155 N m
2 2 (25)
P
f T
f
    
 
6
3 3
2 2(15.9155)
46.9 10 Pa
(0.006)

 
    
Tc T
J c
46.9 MPa
  
(b) 50 Hz.
f 
2500
7.9577 N m
2 (50)
T

  
6
3
2(7.9577)
23.5 10 Pa
(0.006)


   23.5 MPa
  
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329
PROBLEM 3.66
Using an allowable shearing stress of 4.5 ksi, design a solid steel shaft to transmit 12 hp at a speed of
(a) 1200 rpm, (b) 2400 rpm.
SOLUTION
3
all 4.5 ksi 4500 psi 12 hp (12)(6600) 79.2 10 lb in./s
       
P
(a)
3
3
3
1200
20 Hz
60
79.2 10
630.25 lb in.
2 2 (20)
2
2 (2)(630.25)
0.44675 in.
(4500)
 


 
 

   
 
  
f
P
T
f
Tc T
J c
T
c
2 0.893 in.
d c
  
(b)
3
3
2400
40 Hz
60
79.2 10
315.13 lb in.
2 2 (40)
(2)(315.13)
0.35458 in.
(4500)
 

 

   
 
f
P
T
f
c
2 0.709 in.
d c
  
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330
PROBLEM 3.67
Using an allowable shearing stress of 50 MPa, design a solid steel shaft to transmit 15 kW at a frequency of
(a) 30 Hz, (b) 60 Hz.
SOLUTION
6 3
all 50 MPa 50 10 Pa 15 kW 15 10 W
      
P
(a)
3
3
3
3 3
6
30 Hz:
15 10
79.577 N m
2 2 (30)
2
2 (2)(79.577)
10.0438 10 m 10.0438 mm
(50 10 )
 


 



   
 
    

f
P
T
f
Tc T
J c
T
c
2 20.1 mm
d c
  
(b)
3
3
3
6
60 Hz:
15 10
39.789 N m
2 2 (60)
(2)(39.789)
7.9718 10 m 7.9718 mm
(50 10 )
 




   
   

f
P
T
f
c
2 15.94 mm
d c
  
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331
75 mm
30 mm
PROBLEM 3.68
While a steel shaft of the cross section shown rotates at 120 rpm,
a stroboscopic measurement indicates that the angle of twist is
2° in a 4-m length. Using G  77.2 GPa, determine the power
being transmitted.
SOLUTION
Twist angle:
 
3
1 1
2 2
4 4 4 4
2 1
6 4
2 34.907 10 rad
1
0.015 m
2
1
0.0375 m
2
(0.0375 0.015 )
2 2
3.0268 10 m , 4 m

 


   
 
 
   
  
c d
c d
J c c
J L
9 6 3
3
(77.2 10 )(3.0268 10 )(34.907 10 )
4
2.0392 10 N m
120
120 rpm Hz 2 Hz
60
TL GJ
T
GJ L
T
f
 
  
  
  
  


3 3
(2 ) 2 (2)(2.0392 10 ) 25.6 10 W 25.6 kW
 
     
P f T 
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332
PROBLEM 3.69
Determine the required thickness of the 50-mm tubular shaft of Concept Application 3.7 if it is to transmit the
same power while rotating at a frequency of 30 Hz.
SOLUTION
From Example 3.07, 3
100kW 100 10 W
P   
6
all 2
1
60MPa 60 10 Pa 25 mm 0.025 m
2
c d
      
30 Hz

f
530.52 N m
2
  
P
T
f
 
 
4 4 2 2
2 1 4 4
2 1
2
2
Tc Tc
J c c
J c c



   

4 4 4 9 4
2
1 2 6
2 (2)(530.52)(0.025)
0.025 249.90 10 m
(60 10 )
 

     

Tc
c c
3
1 22.358 10 m = 22.358 mm
c 
 
2 1 25 mm 22.358 mm 2.642 mm
    
t c c 2.64 mm
t  
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333
PROBLEM 3.70
A steel drive shaft is 6 ft long and its outer and inner diameters are respectively equal to 2.25 in. and 1.75 in.
Knowing that the shaft transmits 240 hp while rotating at 1800 rpm, determine (a) the maximum shearing
stress, (b) the angle of twist of the shaft (G  11.2  106
psi).
SOLUTION
6
all
6
6
6 ft 72 in.
11 10 psi
7500 psi
6600 lb in./s
240 hp 1.584 10 lb in./s
hp
1 Hz
1800 rpm 30 Hz
60 rpm
1.584 10 lb in./s
(2 ): 8403.4 lb in.
2 2 (30 Hz)


 
 
 


   
 
 
    
L
G
P
f
P
P T f T
f
(a) 4 4
2
(8403.4 lb in.)(1.125 in.)
[(1.125 in.) (0.875 in.) ]
m
Tc
J 


 

5930 psi
m
  
(b) 6 4 4
2
(8403.4 lb in.)(72 in.)
(11 10 psi) [(1.125 in.) (0.875 in )]
TL
GJ 


 
 
3 180
34.48 10 radians
radians

 
 
   
 
1.98
   
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334
5 m
60 mm
25 mm
T
T′
PROBLEM 3.71
The hollow steel shaft shown all
( 77.2 GPa, 50 MPa)

 
G
rotates at 240 rpm. Determine (a) the maximum power that can
be transmitted, (b) the corresponding angle of twist of the shaft.
SOLUTION
 
2 2
1 1
4 4 4 4
2 1
6 4 6 4
6
6 6
3
1
30 mm
2
1
12.5 mm
2
[(30) (12.5) ]
2 2
1.234 10 mm 1.234 10 m
50 10 Pa
(50 10 )(1.234 10 )
2056.7 N m
30 10
m
m
m
c d
c d
J c c
Tc J
T
J c
 






 
 
   
   
 
 
    

Angular speed: 240 rpm 4 rev/sec 4 Hz
f   
(a) Power being transmitted. 3
2 2 (4)(2056.7) 51.7 10 W
P f T
 
   
51.7 kW
P  
(b) Angle of twist. 9 6
(2056.7)(5)
0.1078 rad
(77.2 10 )(1.234 10 )
TL
GJ
 
  
 
6.17
   
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335
3.5 in.
t
PROBLEM 3.72
A steel pipe of 3.5-in. outer diameter is to be used to transmit a torque of
3000 lb  ft without exceeding an allowable shearing stress of 8 ksi. A series
of 3.5-in.-outer-diameter pipes is available for use. Knowing that the wall
thickness of the available pipes varies from 0.25 in. to 0.50 in. in 0.0625-in.
increments, choose the lightest pipe that can be used.
SOLUTION
 
3
2
2 2
4 4
2 1
3
4 4 4 4
2
1 2 3
1
3000 lb ft 36 10 lb in.
1
1.75 in.
2
2T
π
2 (2)(36 10 )(1.75)
1.75 4.3655 in
π π(8 10 )
1.4455 in.


    
 
 


    


o
T
c d
Tc c
J c c
Tc
c c
c
Required minimum thickness: 2 1
t c c
 
1.75 1.4455 0.3045 in.
t   
Available thicknesses: 0.25 in., 0.3125 in., 0.375 in., etc.
Use 0.3125 in.
t  
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336
d2
40 mm
(a) (b)
PROBLEM 3.73
The design of a machine element calls for a 40-mm-outer-diameter
shaft to transmit 45 kW. (a) If the speed of rotation is 720 rpm,
determine the maximum shearing stress in shaft a. (b) If the speed of
rotation can be increased 50% to 1080 rpm, determine the largest inner
diameter of shaft b for which the maximum shearing stress will be the
same in each shaft.
SOLUTION
(a)
3
3
720
12 Hz
60
45 kW 45 10 W
45 10
596.83 N m
2 2 (12)
1
20 mm 0.020 m
2
 
 
  

   
  
f
P
P
T
f
c d
6
3 3
2 (2)(596.83)
47.494 10 Pa
(0.020)
Tc T
J c

 
     max 47.5 MPa
  
(b)
 
3
2 2
4 4
2 1
1080
18 Hz
60
45 10
397.89 N m
2 (18)
2
f
T
Tc Tc
J c c



 

  
 

4 4 2
1 2
4 4 9
1 6
2
(2)(397.89)(0.020)
0.020 53.333 10
(47.494 10 )
Tc
c c
c



 
   

3
1 15.20 10 m 15.20 mm
c 
   2 1
2 30.4 mm
d c
  
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337
C
150 mm
60 mm
B
A
F
60 mm
D
150 mm
E
PROBLEM 3.74
Three shafts and four gears are used to form a gear train that will
transmit power from the motor at A to a machine tool at F. (Bearings for
the shafts are omitted in the sketch.) The diameter of each shaft is as
follows: 16 mm,
AB
d  20 mm,
CD
d  28mm.
EF
d  Knowing that the
frequency of the motor is 24 Hz and that the allowable shearing stress
for each shaft is 75 MPa, determine the maximum power that can be
transmitted.
SOLUTION
6
all 75 MPa =75 10 Pa
  
Shaft AB: 3
3 3 6
all all
3
all all
1 2
0.008 m
2
(0.008) (75 10 ) 60.319 N m
2 2
24 Hz 2 2 (24)(60.319) 9.10 10 W
AB
AB AB
AB AB
AB
AB AB
Tc T
c d
J c
T c
f P f T
   
    
    


 

 
Shaft CD:
3 3 6
all all
3
3
all all
1
0.010 m
2
2
(0.010) (75 10 ) 117.81 N m
2 2
60
(24) 9.6 Hz 2 2 (9.6)(117.81) 7.11 10 W
150
CD CD
CD
CD
CD CD
B
CD AB CD
C
c d
Tc T
T c
J c
r
f f P f T
r
 
       
      
 
 

 
Shaft EF:
1
0.014 m
2
EF EF
c d
 
3 3 6
all all
3
all all
(0.014) (75 10 ) 323.27 N m
2 2
60
(9.6) 3.84 Hz
150
2 2 (3.84)(323.27) 7.80 10 W
EF
D
EF CD
E
EF
T c
r
f f
r
P f T
    
  
   
 

 
Maximum allowable power is the smallest value. 3
all 7.11 10 W 7.11kW
P    
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338
C
150 mm
60 mm
B
A
F
60 mm
D
150 mm
E
PROBLEM 3.75
Three shafts and four gears are used to form a gear train that will
transmit 7.5 kW from the motor at A to a machine tool at F. (Bearings
for the shafts are omitted in the sketch.) Knowing that the frequency of
the motor is 30 Hz and that the allowable stress for each shaft is 60
MPa, determine the required diameter of each shaft.
SOLUTION
3 6
all
7.5 kW 7.5 10 W 60 MPa = 60 10 Pa
P 
    
Shaft AB:
3
3
3
3 9 3
6
7.5 10
= 30 Hz 39.789 N m
2π 2π(30)
2 2
π
π
(2)(39.789)
422.17 10 m
π(60 10 )




   
   
  

AB AB
AB
AB
AB
AB AB
AB
P
f T
f
Tc T T
c
J c
c
3
7.50 10 m 7.50 mm
AB
c 
   2 15.00 mm
AB AB
d c
  
Shaft CD:
3
60 7.5 10
(30) 12 Hz 99.472 N m
150 2π 2π(12)

      
B
CD AB CD
C CD
r P
f f T
r f
3 6 3
3 6
2 2 2(99.472)
1.05543 10 m
π
π π(60 10 )
CD CD
CD CD
CD CD
Tc T T
c
J c
 
      
 
3
10.18 10 m 10.18 mm
CD
c 
   2 20.4 mm
CD CD
d c
  
Shaft EF:
3
3 6 3
3 6
60 7.5 10
(12) 4.8 Hz 248.68 N m
150 2π 2π(4.8)
2 (2)(248.68)
2.6886 10 m
π π(60 10 )
 

      
     

D
EF CD EF
E EF
EF
EF
EF EF
r P
f f T
r f
Tc T
c
J c
3
13.82 10 m 13.82 mm
EF
c 
   2 27.6 mm
EF EF
d c
  
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339
C
5 in.
3 in.
D
A
B
PROBLEM 3.76
The two solid shafts and gears shown are used to
transmit 16 hp from the motor at A operating at a speed
of 1260 rpm to a machine tool at D. Knowing that each
shaft has a diameter of 1 in., determine the maximum
shearing stress (a) in shaft AB, (b) in shaft CD.
SOLUTION
(a) Shaft AB: 3
16 hp (16)(6600) 105.6 10 lb in./sec
    
P
3
3
3
3
1260
21 Hz
60
105.6 10
800.32 lb in.
2 2 (21)
1
0.5 in.
2
2
(2)(800.32)
4.08 10 psi
(0.5)
 



 

   
 
 
  
AB
f
P
T
f
c d
Tc T
J c
4.08 ksi
AB
  
(b) Shaft CD: 3
5
(800.32) 1.33387 10 lb in.
3
    
C
CD AB
B
r
T T
r
3
3
3 3
2 (2)(1.33387 10 )
6.79 10 psi
(0.5)

 

   
T
c
6.79 ksi
CD
  
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340
C
5 in.
3 in.
D
A
B
PROBLEM 3.77
The two solid shafts and gears shown are used to
transmit 16 hp from the motor at A, operating at a speed
of 1260 rpm, to a machine tool at D. Knowing that the
maximum allowable shearing stress is 8 ksi, determine
the required diameter (a) of shaft AB, (b) of shaft CD.
SOLUTION
(a) Shaft AB: 3
16 hp (16)(6600) 105.6 10 lb in./sec
    
P
3
3
3
3
3
3
1260
21 Hz
60
8 ksi 8 10 psi
105.6 10
800.32 lb in.
2 2 (21)
2 2
(2)(800.32)
0.399 in.
(8 10 )
2 0.799 in.

 




 
  

   
  
 

 
AB
AB
f
P
T
f
Tc T T
c
J c
c
d c 0.799 in.
AB
d  
(b) Shaft CD: 3
5
(800.32) 1.33387 10 lb in.
3
    
C
CD AB
B
r
T T
r
3
3 3
3
2 (2)(1.33387 10 )
0.473 in.
(8 10 )
2 0.947 in.
CD
T
c
d c
 

  

  0.947 in.
CD
d  
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341
B
C
D
A
3
4
in.
5
8
in.
1
2
r ⫽ 4 in.
1
8
r 1 in.
⫽ PROBLEM 3.78
The shaft-disk-belt arrangement shown is used to transmit 3 hp from point A to
point D. (a) Using an allowable shearing stress of 9500 psi, determine the required
speed of shaft AB. (b) Solve part a, assuming that the diameters of shafts AB and
CD are, respectively, 0.75 in. and 0.625 in.
SOLUTION
3
3
9500 psi 3 hp (3)(6600) 19,800 lb in./s
2
2


 

    
  
P
Tc T
T c
J c
Allowable torques.
5
8
in.
 diameter shaft:
3
all
5 5
in., (9500) 455.4 lb in.
16 2 16
c T
 
   
 
 

3
4
in.
 diameter shaft: 3
8
3
all
3
in., (9500) 786.9 lb in.
2 8
  
   
 
 
c T
Statics: 1 2 1 2
( ) ( )
1.125
0.25
4.5
B B C C
B
B C C C
C
T r F F T r F F
r
T T T T
r
   
  
(a) Allowable torques. ,all ,all
455.4 lb in. 786.9 lb in.
B C
T T
   
Assume 786.9 lb in.
C
T  
Then (0.25)(786.9) 196.73 lb in. 455.4 lb in. (okay)
B
T     
19,800
2
2 2 (196.73)
AB
B
P
P f T f
T
  

 
16.02 Hz
AB
f  
(b) Allowable torques. ,all ,all
786.9 lb in. 455.4 lb in.
B C
T T
   
Assume 455.4 lb in.
C
T  
Then (0.25)(455.4) 113.85 lb in. 786.9 lb in.
B
T     
19,800
2
2 2 (113.85)
AB
B
P
P f T f
T
  

 
27.2 Hz
AB
f  
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342
PROBLEM 3.79
A 5-ft-long solid steel shaft of 0.875-in. diameter is to transmit 18 hp between a motor and a machine tool.
Determine the lowest speed at which the shaft can rotate, knowing that G  11.2  106
psi, that the maximum
shearing stress must not exceed 4.5 ksi, and the angle of twist must not exceed 3.5°.
SOLUTION
3
1
5 ft 60 in. 0.4375 in. 18 hp 118.8 10 lb in./s
2
L c d P
       
Stress requirement. 3
3
2
4.5 ksi 4.5 10 psi
 

    
Tc T
J c
3 3 3
(4.5 10 )(0.4375) 591.92 lb in.
2 2
T c
 

    
Twist angle requirement. 3
3.5 61.087 10 rad
TL
GJ
 

    
4 6 4 3
(11.2 10 )(0.4375) (61.087 10 )
2 (2)(60)
656.21lb in.
    
 
  
 
GJ Gc
T
L L
Maximum allowable torque is the smaller value. 591.92 lb in.
T  
3
118.8 10
2 31.943 Hz
2 2 (591.92)
P
P f T f
T

   

 
1917 rpm
f  
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343
PROBLEM 3.80
A 2.5-m-long steel shaft of 30-mm diameter rotates at a frequency of 30 Hz. Determine the maximum power
that the shaft can transmit, knowing that 77.2 GPa,
G  that the allowable shearing stress is 50 MPa, and that
the angle of twist must not exceed 7.5.
SOLUTION
1
15 mm 0.015 m 2.5 m
2
c d L
   
Stress requirement. 6
3 6 3
50 10 Pa
(50 10 )(0.015) 265.07 N m
2 2
Tc
J
J
T c
c
 
  

  
     
Twist angle requirement. 3 9
4
4 9 4 3
7.5 130.90 10 rad 77.2 10 Pa
2
(77.2 10 )(0.015) (130.90 10 ) 803.60 N m
2 2
G
TL TL
GJ Gc
T Gc



 



     
 
     
Smaller value of T is the maximum allowable torque.
265.07 N m
T  
Power transmitted at f  30 Hz.
3
2 2 (30)(265.07) 49.96 10 W
P f T
   
  50.0 kW
P  
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344
PROBLEM 3.81
The design specifications of a 1.2-m-long solid transmission shaft require that the angle of twist of the shaft
not exceed 4 when a torque of 750 N  m is applied. Determine the required diameter of the shaft, knowing
that the shaft is made of a steel with an allowable shearing stress of 90 MPa and a modulus of rigidity of
77.2 GPa.
SOLUTION
3
4
6 9
750 N m, 4 69.813 10 rad,
1.2 m,
2
90 MPa 90 10 Pa 77.2 GPa 77.2 10 Pa




     
 
     
T
L J c
G
Based on angle of twist. 4
3
4 4
9 3
2
2 (2)(750)(1.2)
18.06 10 m
(77.2 10 )(69.813 10 )
TL TL
GJ Gc
TL
c
G


  


 
   
 
Based on shearing stress. 3
3
3 3
6
2
2 (2)(750)
17.44 10 m
(90 10 )
Tc T
J c
T
c


 

 
   

Use larger value. 3
18.06 10 m 18.06 mm
c 
   2 36.1mm
d c
  
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345
d1 ⫽ 38 mm d2
PROBLEM 3.82
A 1.5-m-long tubular steel shaft (G  77.2 GPa) of 38-mm outer
diameter d1 and 30-mm inner diameter d2 is to transmit 100 kW
between a turbine and a generator. Knowing that the allowable shearing
stress is 60 MPa and that the angle of twist must not exceed 3°,
determine the minimum frequency at which the shaft can rotate.
SOLUTION
 
3
2 1
4 4 4 4 9 4
2 1
1.5 m, 3 52.360 10 rad
1 1
19 mm 0.019 m, 15 mm 0.015 m
2 2
(0.019 0.015 ) 125.186 10 m
2 2

 


    
     
     
o i
L
c d c d
J c c
Stress requirement. 6 2
60 10 Pa
Tc
J
 
  
9 6
2
(125.186 10 )(60 10 )
395.32 N m
0.019
J
T
c
 
 
   
Twist angle requirement.
TL
GJ
 
9 9 3
(77.2 10 )(125.186 10 )(52.360 10 )
337.35 N m
1.5
GJ
T
L
  
  
   
Maximum allowable torque is the smaller value. 337.35 N m
T  
Power transmitted: 3
100 kW 100 10 W 2
P P f T
    
Frequency:
3
100 10
47.2 Hz
2 2 (337.35)
P
f
T
 

   47.2 Hz
f  
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346
d1 ⫽ 38 mm d2
PROBLEM 3.83
A 1.5-m-long tubular steel shaft of 38-mm outer diameter d1 is to be
made of a steel for which all  65 MPa and G  77.2 GPa. Knowing
that the angle of twist must not exceed 4° when the shaft is subjected
to a torque of 600 N  m, determine the largest inner diameter d2 that
can be specified in the design.
SOLUTION
 
2
6 3
4 4
2 1
1
1.5 m 19 mm 0.019 m
2
65 10 Pa 4 69.813 10 rad
2
 


   
     
 
o
L c d
J c c
Stress requirement.
 
2 2
4 4
2 1
2
Tc Tc
J c c


 

4 4
2
4 4
1 2 6
3
2 (2)(600)(0.019)
0.019
(65 10 )
11.6889 10 m 11.6889 mm
Tc
c c
 

   

  
Twist angle requirement.
 
4 4
2 1
4 4
4 4
1 2 9 3
3
1
2
2 (2)(600)(1.5)
0.019
(77.2 10 )(69.813 10 )
12.448 10 m 12.4482 mm
TL TL
GJ G c c
TL
c c
G
c


   
 

   
 
  
Use smaller value of c1. 1 11.6889 mm
c  1
2 23.4 mm
i
d c
  
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347
90 mm
45 mm
r
PROBLEM 3.84
The stepped shaft shown must transmit 40 kW at a speed of 720 rpm.
Determine the minimum radius r of the fillet if an allowable stress of
36 MPa is not to be exceeded.
SOLUTION
Angular speed:
1 Hz
(720 rpm) 12 Hz
60 rpm
f
 
 
 
 
Power: 3
40 10 W
P  
Torque:
3
40 10
530.52 N m
2 2 (12)
 

   
P
T
f
In the smaller shaft, 45 mm, 22.5 mm 0.0225 m
d c
  
6
3 3
2 (2)(530.52)
29.65 10 Pa
(0.0225)

 
    
Tc T
J c
Using 6
max 36 MPa 36 10 Pa
    results in
6
max
6
36 10
1.214
29.65 10
K



  

From Fig 3.32 with
90 mm
2, 0.24
45 mm
D r
d d
  
0.24 (0.24)(45 mm)
r d
  10.8 mm
r  
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348
6 in.
5 in.
r
PROBLEM 3.85
The stepped shaft shown rotates at 450 rpm. Knowing that 0.5 in.,

r
determine the maximum power that can be transmitted without
exceeding an allowable shearing stress of 7500 psi.
SOLUTION
5 in.
6 in.
0.5 in.
6
1.20
5
0.5
0.10
5
d
D
r
D
d
r
d



 
 
From Fig. 3.32, 1.33
K 
For smaller side,
3
3 3
3
1
2.5 in.
2
2
(2.5) (7500)
138.404 10 lb in.
2 (2)(1.33)
450 rpm 7.5 Hz


  
 
 
    
 
c d
KTc KT
J c
c
T
K
f
Power. 3 6
2 2 (7.5)(138.404 10 ) 6.52 10 in. lb/s
P f T
     
 
Recalling that1hp 6600 in. lb/s,
  988 hp
P  
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349
2 in.
1.5 in.
r
T
T'
PROBLEM 3.86
Knowing that the stepped shaft shown transmits a torque of magnitude
2.50 kip in.,
 
T determine the maximum shearing stress in the shaft when
the radius of the fillet is (a) 1
8
in.,
r  (b) 3
16
in.
r 
SOLUTION
3 3
2
2 in. 1.5 in. 1.33
1.5
1
0.75 in. 2.5 kip in.
2
2 (2)(2.5)
3.773 ksi
(0.75)
D
D d
d
c d T
Tc T
J c
 
   
   
  
(a)
1
in. 0.125 in.
8
 
r r
0.125
0.0833
1.5
r
d
 
From Fig. 3.32, 1.42
K 
max (1.42)(3.773)
Tc
K
J
   max 5.36 ksi
  
(b)
3
in.
16

r 0.1875 in.
r 
0.1875
0.125
1.5
r
d
 
From Fig. 3.32, 1.33
K 
max (1.33)(3.773)
Tc
K
J
   max 5.02 ksi
  
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350
60 mm
30 mm
T
T' PROBLEM 3.87
The stepped shaft shown must rotate at a frequency of 50 Hz. Knowing
that the radius of the fillet is 8
r  mm and the allowable shearing stress
is 45 MPa, determine the maximum power that can be transmitted.
SOLUTION
3
3
3
2
2
1
30 mm 15 mm 15 10 m
2
60 mm, 8 mm
60 8
2, 0.26667
30 30
KTc KT c
T
J K
c
d c d
D r
D r
d d
 



  
    
 
   
From Fig. 3.32, 1.18
K 
Allowable torque.
3 3 6
(15 10 ) (45 10 )
202.17 N m
(2)(1.18)
T
 
 
  
Maximum power. 3
2 (2 )(50)(202.17) 63.5 10 W
P f T
   
  63.5 kW
P  
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351
60 mm
30 mm
T
T' PROBLEM 3.88
The stepped shaft shown must transmit 45 kW. Knowing that the allowable
shearing stress in the shaft is 40 MPa and that the radius of the fillet is
6
r  mm, determine the smallest permissible speed of the shaft.
SOLUTION
6
0.2
30
60
2
30
r
d
D
d
 
 
From Fig. 3.32, 1.26
K 
For smaller side,
1
15 mm 0.015 m
2
c d
  
3
3 3 6
3
3
3
2
(0.015) (40 10 )
168.30 N m
2 (2)(1.26)
45 kW 45 10 2
45 10
42.6 Hz
2 2 (168.30 10 )
KTc KT
J c
c
T
K
P P f T
P
f
T
 

   
   

  



  

 
42.6 Hz
f  
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352
r ⫽ ⫺
D
(D d)
1
2
d
Full quarter-circular fillet
extends to edge of larger shaft.
PROBLEM 3.89
A torque of magnitude 200 lb in.
T   is applied to the stepped shaft
shown, which has a full quarter-circular fillet. Knowing that 1 in.,
D 
determine the maximum shearing stress in the shaft when (a) 0.8 in.,
d 
(b) 0.9 in.
d 
SOLUTION
(a)
1.0
1.25
0.8
1
( ) 0.1 in.
2
0.1
0.125
0.8
D
d
r D d
r
d
 
  
 
From Fig. 3.32, 1.31
K 
For smaller side,
1
0.4 in.
2
c d
 
3
3
3
2
(2)(1.31)(200)
2.61 10 psi
(0.4)
KTc KT
J c



 
   2.61 ksi
  
(b)
1.0
1.111
0.9
1
( ) 0.05
2
0.05
0.05
1.0
D
d
r D d
r
d
 
  
 
From Fig. 3.32, 1.44
K 
For smaller side,
1
0.45 in.
2
c d
 
3
3 3
2 (2)(1.44)(200)
2.01 10 psi
(0.45)
KT
c

 
    2.01 ksi
  
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353
r ⫽ ⫺
D
(D d)
1
2
d
Full quarter-circular fillet
extends to edge of larger shaft.
PROBLEM 3.90
In the stepped shaft shown, which has a full quarter-circular fillet, the
allowable shearing stress is 80 MPa. Knowing that 30 mm,
D  determine
the largest allowable torque that can be applied to the shaft if
(a) 26 mm,
d  (b) 24 mm.
d 
SOLUTION
6
80 10 Pa
  
(a)
30 1 2
1.154 ( ) 2 mm 0.0768
26 2 26
D r
r D d
d d
      
From Fig. 3.32, 1.36
K 
Smaller side,
1
13 mm 0.013 m
2
c d
  
3
3 3 6
2
(0.013) (80 10 )
203 N m
2 (2)(1.36)
KTc KT
J c
c
T
K


  
 

    203 N m
T   
(b)
30 1 3
1.25 ( ) 3 mm 0.125
24 2 24
D r
r D d
d d
      
From Fig. 3.32, 1.31
K 
3 3 6
1
12 mm 0.012 m
2
(0.012) (80 10 )
165.8 N m
2 (2)(1.31)
c d
c
T
K
  
  

    165.8 N m
T   
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354
r ⫽ ⫺
D
(D d)
1
2
d
Full quarter-circular fillet
extends to edge of larger shaft.
PROBLEM 3.91
In the stepped shaft shown, which has a full quarter-circular fillet, 1.25 in.
D 
and 1 in.
d  Knowing that the speed of the shaft is 2400 rpm and that the
allowable shearing stress is 7500 psi, determine the maximum power that can
be transmitted by the shaft.
SOLUTION
1.25
1.25
1.0
1
( ) 0.15 in.
2
0.15
0.15
1.0
D
d
r D d
r
d
 
  
 
From Fig. 3.32, 1.31
K 
For smaller side,
1
0.5 in.
2
c d
 
3
3
3
2
(0.5) (7500)
1.1241 10 lb in.
(2)(1.31)
2400 rpm 40 Hz
KTc J c
T
J Kc K
T
f
  


  
   
 
3
3
2 2 (40)(1.1241 10 )
282.5 10 lb in./s
P f T
  
  
 
42.8 hp
P  
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355
c ⫽ 32 mm
T
T'
PROBLEM 3.92
The solid circular shaft shown is made of a steel that is assumed to be
elastoplastic with Y  145 MPa. Determine the magnitude T of the
applied torques when the plastic zone is (a) 16 mm deep, (b) 24 mm
deep.
SOLUTION
6
3 3 6
3
32 mm 0.32 m 145 10 Pa
(0.032) (145 10 )
2 2
7.4634 10 N m
Y
Y
Y Y
c
J
T c
c

  

   
   
  
(a) 16 mm 0.016 m
p
t   0.032 0.016 0.016 m
Y p
c t
     
3 3
3
3 3
3
4 1 4 1 0.016
1 (7.4634 10 ) 1
3 4 3 4 0.032
9.6402 10 N m

   
    
   
   
   
  
Y
Y
T T
c
9.64 kN m
T   
(b) 24 mm 0.024 m
p
t   0.032 0.024 0.008 m
Y p
c t
     
3 3
3
3 3
3
4 1 4 1 0.008
1 (7.4634 10 ) 1
3 4 3 4 0.032
9.9123 10 N m

   
    
   
   
   
  
Y
Y
T T
c
9.91kN m
T   
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356
PROBLEM 3.93
A 1.25-in.-diameter solid rod is made of an elastoplastic material with 5 ksi.
Y 
 Knowing that the elastic
core of the rod is 1 in. in diameter, determine the magnitude of the applied torque T.
SOLUTION
3
3 3 3
3
3 3
3
3 3
3
1
0.625 in.
2
5 10 psi
1
0.5 in.
2
(0.625) (5 10 )
2 2
1.91747 10 lb in.
4 1 4 1 0.5
1 (1.91747 10 ) 1
3 4 3 4 0.625
2.23 10 lb in.


  


 
 
 
   
  
   
    
   
   
   
  
Y
Y Y
Y
Y Y
Y
Y
c d
d
J
T c
c
T T
c
2230 lb in.
T   
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357
3 in. T
4 ft
PROBLEM 3.94
The solid shaft shown is made of a mild steel that is assumed to be
elastoplastic with 6
11.2 10 psi
 
G and 21ksi.
 
Y Determine the
maximum shearing stress and the radius of the elastic core caused by the
application of torque of magnitude (a) 100 kip in.,
 
T (b) 140 kip in.
T  
SOLUTION
4 4
1.5 in., 7.9522 in , 21ksi
2
Y
c J c


   
(a) 100 kip in.
T  
4
(100 kip in.)(1.5 in.)
7.9522 in


 
m
Tc
J
18.86 ksi
m
  
Since ,
m Y
 
 shaft remains elastic.
Radius of elastic core: 1.500 in.
c  
(b) 140 kip in.
T  
(140)(1.5)
26.4 ksi.
7.9522
  
m Impossible: 21.0 ksi
m Y
 
  
Plastic zone has developed. Torque at onset of yield is
7.9522
(21ksi) 111.33 kip in.
1.5
Y Y
J
T
c

   
Eq. (3.32):
3
3
4 1
1
3 4
Y
Y
T T
c

 
 
 
 
 
3
140
4 3 4 3 0.22743 0.6104
111.33
Y Y
Y
T
c T c
 
 
     
 
 
0.6104 0.6104(1.5 in.)
Y c
   0.916 in.
Y
  
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358
T
30 mm
1.2 m
PROBLEM 3.95
The solid shaft shown is made of a mild steel that is assumed to be
elastoplastic with G  77.2 GPa and Y  145 MPa. Determine the
maximum shearing stress and the radius of the elastic core caused by the
application of a torque of magnitude (a) T  600 N  m, (b) T  1000 N  m.
SOLUTION
(a) 600 N m:
 
T Since ,
Y
T T
 the shaft is elastic
Maximum shearing stress: m
Y Y
T
T



6
6
(145 10 )(600)
113.3 10 Pa
768.11
Y
m
Y
T
T



    113.3 MPa
m
  
Radius of elastic core: c
  15.00 mm
E
  
(b) 1000 N m:
 
T Since ,
Y
T T
 there is a plastic zone.
Maximum shearing stress: m Y
 
 145.0 MPa
m
  
Radius of elastic core:
3
3
3
4
1
3
(3)(1000)
4 3 4 0.46003
768.71
0.46003 (0.46003)(15 mm)




 
 
 
   
 
 
 
    
 
Y
Y
Y
Y
Y
T T
T
c T
c
6.90 mm
Y
  
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359
T
30 mm
1.2 m
PROBLEM 3.96
The solid shaft shown is made of a mild steel that is assumed to be
elastoplastic with 145 MPa.
 
Y Determine the radius of the elastic core
caused by the application of a torque equal to 1.1 TY, where TY is the
magnitude of the torque at the onset of yield.
SOLUTION
3
3
1 4
15 mm 1
2 3
4 3 4 (3)(1.1) 0.88790
Y
Y
Y
Y
c d T T
c
T
c T


 
 
   
 
 
 
 
 
    
0.88790 (0.88790)(15 mm)
Y c
   13.32 mm
Y
  
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360
PROBLEM 3.97
It is observed that a straightened paper clip can be twisted through several revolutions by the application of a
torque of approximately 60 mN  m. Knowing that the diameter of the wire in the paper clip is 0.9 mm,
determine the approximate value of the yield stress of the steel.
SOLUTION
3
3
1
0.45 mm 0.45 10 m
2
60 mN m 60 10 N m
P
c d
T


   
    
3 3
3
6
3 3 3
4 4 4 2
3 3 3 2 3
3 (3)(60 10 )
314 10 Pa
2 2 (0.45 10 )
  


 


    

   

Y
P Y Y Y
P
Y
J
T T c T c
c
T
c
314 MPa
Y
  
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361
1.2 m
15 mm
B
T
A
PROBLEM 3.98
The solid shaft shown is made of a mild steel that is assumed to be
elastoplastic with 77.2 GPa

G and 145 MPa.
 
Y Determine the angle of
twist caused by the application of a torque of magnitude (a) 600 N m,
 
T
(b) 1000 N m.
T  
SOLUTION
3
1
15 mm 15 10 m
2

   
c d
Torque at onset of yielding: 3
3 3 3 6
2
(15 10 ) (145 10 )
768.71 N m
2 2
Y
Y
Tc T
J c
c
T


   
 
 
   
(a) 600 N m.
T   Since ,
Y
T T
 the shaft is elastic.
4 3 4 9
2 (2)(600)(1.2)
0.11728 rad
(15 10 ) (77.2 10 )
TL TL
GJ c G

  
   
 
6.72
   
(b) 1000 N m.
T   Y
T T
 A plastic zone has developed.
3
3
4 3 4 9
3
4
1 4 3
3
2 (2)(768.71)(2.1)
0.15026 rad
(15 10 ) (77.2 10 )
(3)(1000)
4 0.46003
768.71
0.15026
0.32663 rad
0.46003 0.46003
Y Y
Y
Y
Y Y
Y
Y
Y
T
T T
T
T L T L
GJ c G
 
 

 





   
 
 
     
 
 
   
 
   
 
  
   18.71
   
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362
3 in. T
4 ft
PROBLEM 3.99
For the solid circular shaft of Prob. 3.94, determine the angle of twist
caused by the application of a torque of magnitude (a) 80 kip in.,
T  
(b) 130 kip in.
T  
SOLUTION
3
4 4 4
1 1
(3) 1.5 in. 21 10 psi
2 2
4 ft 48 in. (1.5) 7.9522 in
2 2
Y
c d
L J c

 
    
    
Torque at onset of yielding:
Tc J
T
J c

  
3
3
(21 10 )(7.9522)
111.330 10 lb in.
1.5
 
    
Y
Y
J
T
c
(a) 3
80 kip in. 80 10 lb in.
    
T
Since ,
Y
T T
 the shaft is fully elastic.
TL
GJ
 
3
3
6
(80 10 )(48)
43.115 10 rad
(11.2 10 )(7.9522)
 

  

2.47
   
(b)
3
3 4
130 kip in. 130 10 lb in. 1
3


 
 
 
         
 
 
 
Y
Y Y
T T T T T
3
3
6
3
3
3
3
3
3
(111.330 10 )(48)
60.000 10 rad
(11.2 10 )(7.9522)
(3)(130 10 )
4 3 4 0.79205
111.330 10
60.000 10
75.75 10 rad
0.79205 0.79205
Y
Y
Y
Y
Y
T L
GJ
T
T









   


    


    4.34
   
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363
1.2 m
15 mm
B
T
A
PROBLEM 3.100
For the solid shaft of Prob. 3.98, determine (a) the magnitude of the torque
T required to twist the shaft through an angle of 15, (b) the radius of the
corresponding elastic core.
PROBLEM 3.98 The solid shaft shown is made of a mild steel that is
assumed to be elastoplastic with 77.2 GPa
G  and 145 MPa.
Y
 
Determine the angle of twist caused by the application of a torque of
magnitude (a) 600 N m,
T   (b) 1000 N m.
T  
SOLUTION
3
6
3 9
1
15mm 15 10 m
2
15 0.2618 rad
(1.2)(145 10 )
0.15026 rad
(15 10 )(77.2 10 )
Y Y
Y
c d
L L
c cG

 



   
  

   
 
(a) Since ,
Y
 
 there is a plastic zone.
3
3 3 3 6
3 3
2
(15 10 ) (145 10 )
768.71 N m
2 2
4 1 4 1 0.15026
1 (768.71) 1
3 4 3 4 0.2618
976.5 N m
Y
Y
Y
Y
Y
Y
T c T
J c
c
T
T T


  



 
 
   
   
   
 
   
 
   
   
 
   
 
  977 N m
T   
(b) Y Y Y
L c
   
 
3
3
(15 10 )(0.15026)
8.61 10 m
0.2618






   
Y
Y
c
8.61 mm
Y
  
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364
PROBLEM 3.101
A 3-ft-long solid shaft has a diameter of 2.5 in. and is made of a mild steel that is assumed to be elastoplastic
with 21 ksi
Y
  and 6
11.2 10 psi.
G   Determine the torque required to twist the shaft through an angle of
(a) 2.5, (b) 5.
SOLUTION
3
4 4 4
3
3
3
3
6
1
3 ft 36 in., 1.25 in., 21 10 psi
2
(1.25) 3.835 in
2 2
(3.835)(21 10 )
64.427 10 lb in.
1.25
(64.427 10 )(36)
53.999 10 rad 3.0939
(11.2 10 )(3.835)
Y
Y Y
Y Y
Y
Y
L c d
J c
T c J
T
J c
T L
GJ

 


 
     
  

     

     

(a) 3
2.5 43.633 10 rad
 
    Y
  The shaft remains elastic.
TL
GJ
 
6 3
3
(11.2 10 )(3.835)(43.633 10 )
36
52.059 10 lb in.
 
 
 
  
GJ
T
L
52.1kip in.
T   
(b) 3
5 87.266 10 rad
 
    Y
  A plastic zone occurs.
3
3
3
3
3
3
4 1
1
3 4
4 1 53.999 10
(64.427 10 ) 1
3 4 87.266 10
80.814 10 lb in.
Y
Y
T T




 
 
 
   
 
 
 
 
 

 
    
 
 

 
 
   80.8 kip in.
T   
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365
PROBLEM 3.102
A 18-mm-diameter solid circular shaft is made of a material that is assumed to be elastoplastic with
145 MPa
 
Y and 77.2 GPa.
G  For a 1.2-m length of the shaft, determine the maximum shearing stress and
the angle of twist caused by a 200-N  m torque.
SOLUTION
6
3 3 6
1
145 10 Pa, 0.009 m, 1.2 m, 200 N m
2
(0.009) (145 10 ) 166.04 N m
2 2
Y
Y
Y Y
c d L T
J
T c
c

  

      
     
Y
T T
 (plastic region with elastic core) max 145.0 MPa
Y
 
  
3
4 4 9
3
3
3
2 (2)(166.04)(1.2)
250.43 10 radians
(0.009) (77.2 10 )
4 1
1
3 4
3 (3)(200)
4 4 0.38641 0.72837
166.04
Y Y
Y
Y
Y
Y Y
Y
T L T L
GJ c G
T T
T
T

 


 
 

    

 
 
 
 
 
 
     
 
 
3
3
250.43 10
343.82 10 radians
0.72837 0.72837





    
19.70
   
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366
PROBLEM 3.103
A 0.75-in.-diameter solid circular shaft is made of a material that is assumed to be elastoplastic with
20 ksi
Y 
 and G  11.2  106
psi. For a 4-ft length of the shaft, determine the maximum shearing stress and
the angle of twist caused by a 1800-lb  in. torque.
SOLUTION
3 1
20 ksi 20 10 psi, 0.375 in.,
2
4 ft 48 in., 1800 lb in.
     
   
Y c d
L T
3 3 6
(0.375) (20 10 )
2 2
1656.70 lb in.
Y
Y Y
J
T c
c
  

   
 
Y
T T
 (Plastic region with elastic core) max 20.0 ksi
Y
 
  
4 4 6
3
3
3
2 (2)(1656.70)(48)
0.22857 rad
(0.375) (11.2 10 )
4 1
1
3 4
3 (3)(1800)
4 4 0.7451 0.90471
1656.70
Y Y
Y
Y
Y
Y Y
Y
T L T L
GJ c G
T T
T
T

 


 
 
   

 
 
 
 
 
 
     
 
 
0.22857
0.25264 rad
0.90471 0.9047
Y

   
14.48
Y
   
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367
2 m
B
T
A
⫽ 300 N · m
12 mm
PROBLEM 3.104
The shaft AB is made of a material that is elastoplastic with
Y  90 MPa and G  30 GPa. For the loading shown,
determine (a) the radius of the elastic core of the shaft,
(b) the angle of twist at end B.
SOLUTION
(a)
6
3
3 6
12 mm 0.012 m
90 10 Pa
2
(0.012) (90 10 ) 244.29 N m
2
Y
Y
Y Y
c
J
T c
c

 


 
 
 
   
300 N m Y
T T
   Plastic region with elastic core
3 3
3 3
3
4 1 3 (3)(300)
1 1 1 0.31585
3 4 244.29
0.68102 (0.68102)(0.012) 8.17 10 m
 

 
 
      
 
 
 
   
Y Y
Y
Y
Y
Y
T
T T
T
c c
c
8.17 mm
Y
  
(b)
9
4 4 9
2 m
30 10 Pa
2 (2)(244.29)(2)
0.5000 rad
(0.012) (30 10 )
0.5000
0.734 rad
/ 0.68120
Y Y
Y
Y Y Y
Y
L
G
T L T L
JG c G
c c

 
  

 

 
   

   
42.1
   
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368
PROBLEM 3.105
A solid circular rod is made of a material that is assumed to be elastoplastic. Denoting by Y
T and φY,
respectively, the torque and the angle of twist at the onset of yield, determine the angle of twist if the torque is
increased to (a) 1.1 ,
Y
T T
 (b) 1.25 ,
Y
T T
 (c) 1.3 .
Y
T T

SOLUTION
3
3
3
3
4 1
1
3 4
3 1
4 or
3
4
Y
Y
Y
Y Y
Y
T T
T
T T
T


 
 
 
 
 
 
 
  

(a)
3
1
1.10 1.126
4 (3)(1.10)
Y Y
T
T


  

1.126
 
 Y 
(b)
3
1
1.25 1.587
4 (3)(1.25)
Y Y
T
T


  

1.587
 
 Y 
(c)
3
1
1.3 2.15
4 (3)(1.3)
Y Y
T
T


  

2.15
 
 Y 
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369
70 mm
30 mm
PROBLEM 3.106
A hollow shaft is 0.9 m long and has the cross section shown. The steel
is assumed to be elastoplastic with Y  180 MPa and G  77.2 GPa.
Determine (a) the angle of twist at which the section first becomes
fully plastic, (b) the corresponding magnitude of the applied torque.
SOLUTION
1 1 2 2
1 1
0.015 m 0.035 m
2 2
   
c d c d
(a) For onset of fully plastic yielding, 1
Y c
 
1
6
3
9
1
(0.9)(180 10 )
139.90 10 rad
(0.015)(77.2 10 )
Y Y
Y
Y
c
G L L
L
c G
   
  

 
    

   

8.02
   
(b)  
2
2
1
1
3
2 3 3
2 1
6 3 3 3
2
2 2
3 3
2
(180 10 )(0.035 0.015 ) 14.89 10 N m
3
 
    

   
     

c
c
p Y Y Y
c
c
T dp c c
 14.89 kN m
   14.89 kN m
 
p
T 
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370
70 mm
30 mm
PROBLEM 3.107
A hollow shaft is 0.9 m long and has the cross section shown. The steel
is assumed to be elastoplastic with Y  180 MPa and G  77.2 GPa.
Determine the applied torque and the corresponding angle of twist
(a) at the onset of yield, (b) when the plastic zone is 10 mm deep.
SOLUTION
(a) At the onset of yield, the stress distribution is the elastic distribution with max Y
 

 
2 2 1 1
4 4 4 4 6
2 1
6 6
3
2
max
2
1 1
0.035 m, 0.015 m
2 2
(0.035 0.015 ) 2.2777 10 m
2 2
(2.2777 10 )(180 10 )
11.7139 10 N m
0.035
 

 


   
     
 
       
Y Y
Y Y
c d c d
J c c
T c J
T
J c
11.71 kN m
 
Y
T 
3
3
9 6
(11.7139 10 )(0.9)
59.956 10 rad
(77.2 10 )(2.2777 10 )
Y
Y
T L
GJ
 


   
 
3.44
Y
   
(b) 2
0.010 m 0.035 0.010 0.025 m
Y
Y Y
Y
t c t
L L G

  

 
     
   
6
3
9
(180 10 )(0.9)
83.938 10 rad
(77.2 10 )(0.025)
Y
Y
L
G





   

4.81
   
Torque T1 carried by elastic portion: 1 Y
c  
 
Y
 
 at .
Y
 
 1
1
Y
Y
T
J

  where  
4 4
1 1
2


 
Y
J c
4 4 9 4
1
9 6
3
1
1
(0.025 0.015 ) 534.07 10 m
2
(534.07 10 )(180 10 )
3.8453 10 N m
0.025





   
 
    
Y
Y
J
J
T
Torque T2 carried by plastic portion:
 
2
2
3
2 3 3
2 2
6 3 3 3
2
2 2
3 3
2
(180 10 )(0.035 0.025 ) 10.2730 10 N m
3


 
     

   
     
 Y
Y
c
c
Y Y Y Y
T dp c
Total torque:
3 3
1 2
3
3.8453 10 10.2730 10
14.1183 10 N m
T T T
     
   14.12 kN m
 
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371
2.5 in.
3 in.
A
B
C
D
E
x
5 in.
T
T′
PROBLEM 3.108
A steel rod is machined to the shape shown to form a tapered solid shaft to which
a torque of magnitude 75 kip in.
 
T is applied. Assuming the steel to be elastoplastic
with 21 ksi
Y
  and 6
11.2 10 psi,
G   determine (a) the radius of the elastic core
in portion AB of the shaft, (b) the length of portion CD that remains fully elastic.
SOLUTION
(a) In portion AB.
1
1.25 in.
2
c d
 
3 3 3 3
3
3
3
3
3
3
(1.25) (21 10 ) 64.427 10 lb in.
2 2
4
1
3
3 (3)(75 10 )
4 4 0.79775
64.427 10
0.79775 (0.79775)(1.25) 0.99718 in.
AB Y
Y Y
Y
Y
Y
Y
Y
J
T c
c
T T
c
T
c T
c
  




      
 
 
 
 
 

    

   0.997 in.
Y
  
(b) For yielding at point C. 3
, , 75 10 lb in.
Y x
c c T
 
    
3
3
3
3
3
2
2 (2)(75 10 )
1.31494 in.
(21 10 )
C Y
x Y
x
x
Y
J
T c
c
T
c
 

 
 

  

Using proportions from the sketch,
1.50 1.31494
1.50 1.25 5
x



3.70 in.
x  
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372
2.5 in.
3 in.
A
B
C
D
E
x
5 in.
T
T′
PROBLEM 3.109
If the torque applied to the tapered shaft of Prob. 3.108 is slowly increased, determine
(a) the magnitude T of the largest torques that can be applied to the shaft, (b) the
length of the portion CD that remains fully elastic.
PROBLEM 3.108 A steel rod is machined to the shape shown to form a tapered solid
shaft to which a torque of magnitude 75 kip in.
T   is applied. Assuming the steel
to be elastoplastic with 21 ksi
Y
  and 6
11.2 10 psi,
 
G determine (a) the radius
of the elastic core in portion AB of the shaft, (b) the length of portion CD that remains
fully elastic.
SOLUTION
(a) The largest torque that may be applied is that which makes portion AB fully plastic.
In portion AB,
1
1.25 in.
2
c d
 
3 3 3 3
(1.25) (21 10 ) 64.427 10 lb in.
2 2
Y
Y Y
J
T c
c
  

      
For fully plastic shaft, 0
Y
 
3
3
4 1 4
1
3 4 3
Y
Y
T T T
c

 
  
 
 
 
3 3
4
(64.427 10 ) 85.903 10 lb in.
3
    
T 85.9 kip in.
T   
(b) For yielding at point C, 3
, , 85.903 10 lb in.
 
    
Y x
c c T
3
3
3
3
3
2
2 (2)(85.903 10 )
1.37580 in.
(21 10 )
x
Y
x x
x
Y
Tc T
J c
T
c


 
 

  

Using proportions from the sketch,
1.50 1.37580
1.50 1.25 5
x



2.48 in.
x  
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373
13.5
12
9
6
3
0 0.001 0.002 0.003
(ksi)
PROBLEM 3.110
A solid brass rod of 1.2-in. diameter is subjected to a torque that causes
a maximum shearing stress of 13.5 ksi in the rod. Using the - diagram
shown for the brass rod used, determine (a) the magnitude of the torque,
(b) the angle of twist in a 24-in. length of the rod.
SOLUTION
(a) max
1
13.5 ksi 0.600 in.
2
c d
   
From the stress-strain diagram, max 0.003
  
Let
max
z
c
 

 
1 1
2 3 2 3 2
0 0 0
2 2 2 where the integral is given by
       
   
  
c
T d c z dz c I I I z dz
Evaluate I using a method of numerical integration. If Simpson’s rule is used, the integration formula is
2
3


 
z
I wz
where w is a weighting factor. Using 0.25,
 
z we get the values given in the table below:
z  , ksi
 2
, ksi
z  w 2
, ksi
wz 
0 0 0 0.000 1 0
0.25 0.00075 4.5 0.281 4 1.125
0.5 0.0015 8.6 2.15 2 4.30
0.75 0.00225 12.2 6.86 4 27.45
1.0 0.003 13.5 13.5 1 13.5
46.375
2
wz 
 
3
(0.25)(46.375)
3.865 10 ksi
3
I 
  
(a) 3 3
2 2 (0.600) (3.865) 5.24 kip in.
 
   
T c I 5.24 kip in.
 
T 
(b) max
c
L

 
3
(24)(0.003)
120 10 rad
0.800
m
L
c

 
    6.88
   
Note: Answer may differ slightly due to reading of graph and choice of numerical integration formula.
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374
13.5
12
9
6
3
0 0.001 0.002 0.003
(ksi)
PROBLEM 3.111
A solid brass rod of 0.8-in. diameter and 30-in. length is twisted
through an angle of 10°. Using the - diagram shown for the brass
rod used, determine (a) the magnitude of the torque applied to the
rod, (b) the maximum shearing stress in the rod.
SOLUTION
(a) 3
3
max
10 174.53 10 rad
1
0.400 in. 30 in.
2
(0.4)(174.53 10 )
0.00233
30
c d L
c
L





   
  

  
Let
max 2
 

 
z
c
1 1
2 3 2 3 2
2 2
0 0 1/3
2 2 2 where the integral is given by
       
   
  
c
T d c z dz c I I I z dz
Evaluate I using a method of numerical integration. If Simpson’s rule is used, the integration formula is
2
3


 
z
I wz
where w is a weighting factor. Using
1
,
6
 
z we get the values given in the table below:
z  , ksi
 2
, ksi
z  w 2
, ksi
wz 
0 0 0 0 1 0
0.25 0.000583 3.5 0.219 4 0.88
0.5 0.001165 7.0 1.75 2 3.50
0.75 0.001748 10.0 5.625 4 22.50
1.0 0.00233 12.2 12.6 1 12.60
39.48
2
wz 
 
(0.25)(39.48)
3.29 ksi
3
I  
3 3
2
2 2 (0.400) (3.29)
T c I
 
  1.322 kip in.
T   
Note: Answer may differ slightly due to differences of opinion in reading the stress-strain curve.
(b) From the graph, max 12.60 ksi
  
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375
0
20
40
60
80
100
0.001 0.002 0.003
(MPa)
725 mm
d 5 50 mm
T'
T
PROBLEM 3.112
A 50-mm-diameter cylinder is made of a
brass for which the stress-strain diagram
is as shown. Knowing that the angle of
twist is 5° in a 725-mm length, determine
by approximate means the magnitude T
of torque applied to the shaft
SOLUTION
(a) 3
3
max
5 87.266 10 rad
1
0.025 m 0.725 m
2
(0.025)(87.266 10 )
0.00301
0.725
c d L
c
L


   
  

  



Let
max
z
c
 

 
1 1
2 3 2 3 2
2 2
0 0 1/3
2 2 2 where the integral is given by
c
T d c z dz c I I I z dz
   
  
       
Evaluate I using a method of numerical integration. If Simpson’s rule is used, the integration formula is
2
3
z
I wz 

 
where w is a weighting factor. Using 0.25,
z
  we get the values given in the table below.
z  , MPa
 2
, MPa
z  w 2
, MPa
wz 
0 0 0 0 1 0
0.25 0.00075 30 1.875 4 7.5
0.5 0.0015 55 13.75 2 27.5
0.75 0.00226 75 42.19 4 168.75
1.0 0.00301 80 80 1 80
283.75
2
6
283.75 10 Pa
wz 
 
 
6
6
(0.25)(283.75 10 )
23.65 10 Pa
3
I

  
3 3 6 3
2 2 (0.025) (23.65 10 ) 2.32 10 N m
 
     
T c I 2.32 kN m
T   
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376
PROBLEM 3.113
Three points on the nonlinear stress-strain diagram used in Prob. 3.112 are (0, 0), (0.0015, 55 MPa), and
(0.003, 80 MPa). By fitting the polynomial 2
 
  
T A B C through these points, the following approximate
relation has been obtained.
9 12 2
46.7 10 6.67 10
 
   
T
Solve Prob. 3.113 using this relation, Eq. (3.2), and Eq. (3.26).
PROBLEM 3.112 A 50-mm diameter cylinder is made of a brass for which the stress-strain diagram is as
shown. Knowing that the angle of twist is 5 in a 725-mm length, determine by approximate means the
magnitude T of torque applied to the shaft.
SOLUTION
1
rad, 0.025m, 0.725m
2
c d L
 
        
3
3
max
(0.025)(87.266 10 )
3.009 10
0.725
c
L





   
Let
max
1
2
0 0
2 2
c
z
c
T d c z dz
 
 
 
 
 

     
The given stress-strain curve is
 
2 2 2
max max
1
3 2 2 2
max max
0
1 1 1
3 2 3 2 4
max max
0 0 0
2 2
max max
2
2
1 1 1
2
3 4 5
A B C A B z C z
T c z A B z C z dz
c A z dz B z dz C z dz
c A B C
     
  
 
  
 
 
 
  
 
 

  
    
  
  
  
Data: 9 12
0, 46.7 10 , 6.67 10
A B C
     
9 3 3
max
2 12 3 2 3
max
1 1 1
0, (46.7 10 )(3.009 10 ) 35.13 10
3 4 4
1 1
(6.67 10 )(3.009 10 ) 12.08 10
5 5
A B
C




     
      
3 3 3
2 (0 (35.13 10 12.08 10 )) 2.26 10 N m
 
        
T 2.26 kN m
T   
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377
1.2 in.
35 ft
B
A
T
PROBLEM 3.114
The solid circular drill rod AB is made of a steel that is assumed to be elastoplastic
with 22 ksi
Y
  and 6
11.2 10 psi.
G   Knowing that a torque 75 kip in.
T   is
applied to the rod and then removed, determine the maximum residual shearing stress
in the rod.
SOLUTION
4 4 4
1.2 in. 35 ft 420 in.
(1.2) 3.2572 in
2 2
(3.2572)(22)
59.715 kip in.
1.2
Y
Y
c L
J c
J
T
c
 

  
  
   
Loading:
3
3
75 kip in.
4 1
1
3 4

 
 
 
 
 
 
Y
Y
T
T T
c
3
3
3 (3)(75)
4 4 0.23213
59.715
0.61458, 0.61458 0.73749 in.
Y
Y
Y
Y
T
T
c
c
c



    
  
Unloading: where 75 kip in.

   
T
T
J
At ,
  c
(75)(1.2)
27.63 ksi
3.2572
  
At ,
 
 Y
(75)(0.73749)
16.98 ksi
3.2572
  
Residual: res load
  
 
At ,
  c res 22 27.63 5.63 ksi
    
At ,
 
 Y res 22 16.98 5.02 ksi
   
res
maximum 5.63 ksi
  
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on a website, in whole or part.
378
1.2 in.
35 ft
B
A
T
PROBLEM 3.115
In Prob. 3.114, determine the permanent angle of twist of the rod.
PROBLEM 3.114 The solid circular drill rod AB is made of steel that is assumed to
be elastoplastic with 22 ksi
Y
  and 6
11.2 10 psi.
G   Knowing that a torque
75 kip in.
 
T is applied to the rod and then removed, determine the maximum
residual shearing stress in the rod.
SOLUTION
From the solution to Prob. 3.114,
4
1.2 in.
3.2572 in
0.61458
0.73749 in.
Y
Y
c
J
c






After loading, 35 ft 420 in.
Y Y
Y Y
L L L
L
L G
   
 
  
      
3
load 6
(420)(22 10 )
1.11865 rad 64.09
(0.73749)(11.2 10 )


   

During unloading, (elastic) 75 kip in.
TL
T
GJ
   
3
6
(75 10 )(420)
0.86347 rad 49.47
(11.2 10 )(3.2572)


    

Permanent angle of twist.
perm load 1.11865 0.86347 0.25518
  
     14.62
   
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on a website, in whole or part.
379
16 mm
0.6 m
B
T
A
PROBLEM 3.116
The solid shaft shown is made of a steel that is assumed to be
elastoplastic with 145 MPa
Y
  and 77.2 GPa.
G  The torque is
increased in magnitude until the shaft has been twisted through 6; the
torque is then removed. Determine (a) the magnitude and location of the
maximum residual shearing stress, (b) the permanent angle of twist.***
SOLUTION
3
3
max
6
9
max
4 4 9 4
3 3
0.016 m 6 104.712 10 rad
(0.016)(104.72 10 )
0.0027925
0.6
145 10
0.0018782
77.2 10
0.0018
0.67260
0.0027925
(0.016) 102.944 10 m
2 2
(0.016) (145 1
2 2
Y
Y
Y Y
Y
Y Y
c
c
L
G
c
J c
J
T c
c





 

 
  




    

  

  

  
   
    6
0 ) 932.93 N m
 
At end of loading.
3
3 3
load 3
4 1 4 1
1 (932.93) 1 (0.67433) 1.14855 10 N m
3 4 3 4
Y
Y
T T
c

   
      
   
   
 
Unloading: elastic 3
1.14855 10 N m
T   
At ,
  c
3
6
9
(1.14855 10 )(0.016)
178.513 10 Pa
102.944 10
T c
J
 
 
    

At ,
 
 Y
6 6
3
3
9 9
(178.513 10 )(0.67433) 120.376 10 Pa
(1.14855 10 )(0.6)
86.713 10 rad 4.97
(77.2 10 )(102.944 10 )
Y
T c
J c
T L
GJ


 


     
 
      
 
Residual: res load perm load
     
 
   
(a) At ,
  c 6 6 6
res 145 10 178.513 10 33.513 10 Pa
        res 33.5 MPa
  
At ,
 
 Y
6 6 6
res 145 10 120.376 10 24.624 10 Pa
       res 24.6 MPa
 
Maximum residual stress: 33.5 MPa at 16.00 mm
  
(b) 3 3 3
perm 104.712 10 86.713 10 17.9990 10 rad
   
      perm 1.03
   
Mecánica de Materiales - Ferdinand P. Beer 7ma. Edición.

419. PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
380
16 mm
0.6 m
B
T
A
PROBLEM 3.117
After the solid shaft of Prob. 3.116 has been loaded and unloaded as
described in that problem, a torque T1 of sense opposite to the original
torque T is applied to the shaft. Assuming no change in the value of ,
Y

determine the angle of twist 1
 for which yield is initiated in this second
loading and compare it with the angle Y
 for which the shaft started to
yield in the original loading.
PROBLEM 3.116 The solid shaft shown is made of a steel that is
assumed to be elastoplastic with 145 MPa
Y
  and 77.2 GPa.
G 
The torque is increased in magnitude until the shaft has been twisted
through 6; the torque is then removed. Determine (a) the magnitude
and location of the maximum residual shearing stress, (b) the permanent
angle of twist.
SOLUTION
From the solution to Prob. 3.116,
6
9 4
0.016 m, 0.6 m
145 10 Pa,
102.944 10 m
Y
c L
J


 
 
 
The residual stress at c
  is res 33.5 MPa
 
For loading in the opposite sense, the change in stress to produce reversed yielding is
6 6 6
1 res
9 6
1 1
1 1
145 10 33.5 10 111.5 10 Pa
(102.944 10 )(111.5 10 )
0.016
717 N m
Y
T c J
T
J c
  



       
 
   
 
Angle of twist at yielding under reversed torque.
3
3
1
1 9 9
(717 10 )(0.6)
54.16 10 rad
(77.2 10 )(102.944 10 )
T L
GJ
 


   
 
1 3.10
   
Angle of twist for yielding in original loading.
6
3
9
(0.6)(145 10 )
70.434 10 rad
(0.016)(77.2 10 )
 


 
 

   

Y Y
Y
Y
c
G L
L
cG
4.04
Y
   
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on a website, in whole or part.
381
5 m
25 mm
60 mm
T
T'
PROBLEM 3.118
The hollow shaft shown is made of a steel that is assumed to
be elastoplastic with 145 MPa
 
Y and 77.2 GPa.

G The
magnitude T of the torques is slowly increased until the
plastic zone first reaches the inner surface of the shaft; the
torques are then removed. Determine the magnitude and
location of the maximum residual shearing stress in the rod.
SOLUTION
1 1
2 2
1
12.5 mm
2
1
30 mm
2
c d
c d
 
 
When the plastic zone reaches the inner surface, the stress is equal to .
Y
 The corresponding torque is
calculated by integration.
 
2
1
2
2 3 3
2 1
6 3 3 3 3 3
(2 ) 2
2
2
3
2
(145 10 )[(30 10 ) (12.5 10 ) ] 7.6064 10 N m
3
      

   
  
  
  
       

C
C
Y Y
Y Y
dT dA d d
T d c c
Unloading. 3
7.6064 10 N m
T   
 
4 4 4 4 6 4 6 4
2 1
3 3
6
1
1 6
3 3
6
2
2 6
[(30) (12.5) ] 1.234 10 mm 1.234 10 m
2 2
(7.6064 10 )(12.5 10 )
77.050 10 Pa 77.05 MPa
1.234 10
(7.6064 10 )(30 10 )
192.63 10 Pa 192.63 MPa
1.234 10
 







       
  
     

  
     

J c c
T c
J
T c
J
Residual stress.
Inner surface: res 1 145 77.05 67.95 MPa
Y
  
    
Outer surface: res 2 145 192.63 47.63 MPa
Y
  
     
Maximum residual stress: 68.0 MPa at inner surface 
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382
5 m
25 mm
60 mm
T
T'
PROBLEM 3.119
In Prob. 3.118, determine the permanent angle of twist of the rod.
PROBLEM 3.118 The hollow shaft shown is made of a steel that
is assumed to be elastoplastic with 145 MPa
 
Y and
77.2 GPa.

G The magnitude T of the torques is slowly increased
until the plastic zone first reaches the inner surface of the shaft;
the torques are then removed. Determine the magnitude and
location of the maximum residual shearing stress in the rod.
SOLUTION
1 1
2 2
1
12.5 mm
2
1
30 mm
2
c d
c d
 
 
When the plastic zone reaches the inner surface, the stress is equal to .
Y
 The corresponding torque is
calculated by integration.
 
2
1
2
2 3 3
2 1
6 3 3 3 3 3
(2 ) 2
2
2
3
2
(145 10 )[(30 10 ) (12.5 10 ) ] 7.6064 10 N m
3
Y Y
c
Y Y
c
dT dA d d
T d c c
 
  
  
       

      

   

Rotation angle at maximum torque.
1 max
6
max 9 3
1
(145 10 )(5)
0.75130 rad
(77.2 10 )(12.5 10 )
Y
Y
Y
c
L G
L
Gc
 


 
 

  
 
Unloading. 3
7.6064 10 N m
T   
 
4 4 4 4 6 4 6 4
2 1
3
9 6
[(30) (12.5) ] 1.234 10 mm 1.234 10 m
2 2
(7.6064 10 )(5)
0.39922 rad
(77.2 10 )(1.234 10 )
 



       
 
   
 
J c c
T L
GJ
Permanent angle of twist.
perm max 0.75130 0.39922 0.35208 rad
  
     perm 20.2
   
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383
Y
Y
c
c0
␶
␶
1
3
PROBLEM 3.120
A torque T applied to a solid rod made of an elastoplastic material is increased until
the rod is fully plastic and them removed. (a) Show that the distribution of residual
shearing stresses is as represented in the figure. (b) Determine the magnitude of the
torque due to the stresses acting on the portion of the rod located within a circle of
radius c0.
SOLUTION
(a)
After loading: 3 3
load
4 4 2
0,
3 3 2 3
Y Y Y Y
T T c c
 
  
   
Unloading: load
3 3
2 2( ) 4
at
3
4
3
Y
Y
Tc T T
c
J c c
c
  
 

 
     
 
Residual: res
4 4
1
3 3
Y Y Y
c c
 
   
 
   
 
 
To find c0 set, res 0
0 and c
 
 
0
0
4 3
0 1
3 4
c
c c
c
    0 0.150
c c
 
(b)
0 (3/4)
2 2
0
0 0
3 4
(3/4)
3 4
3
0
4
2 2 1
3
4 1 3 4 1 3
2 2
3 3 4 3 4 3 4 4

       
 
 
 
  
 
 
 
   
     
   
   
     
       
 
   
 
c c
Y
c
Y Y
T d d
c
c
c
3 3 3
9 27 9
2 0.2209
64 256 128
Y Y Y
c c c

  
 
   
 
 
3
0 0.221 Y
T c

 
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Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
384
PROBLEM 3.121
Determine the largest allowable square cross section of a steel shaft of length 20 ft if the maximum shearing
stress is not to exceed 10 ksi when the shaft is twisted through one complete revolution. Use 6
11.2 10 psi.
 
G
SOLUTION
3
max
20 ft 240 in.
10 ksi 10 10 psi
1 rev 2 radians
L

 
 
  
 
max 2
1
T
c ab
  (1)
3
2
TL
c ab G
  (2)
Divide (2) by (1) to eliminate T.
2
1 1
3
max 2
2
c ab L c L
c bG
c ab G


 
Solve for b. 1 max
2
c L
b
c G



For a square section, 1.0
a
b

From Table 3.1,
1
2
3
6
0.208,
0.1406
(0.208)(240)(10 10 )
(0.1406)(11.2 10 )(2 )
c
c
b






0.0505 in.
b  
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385
PROBLEM 3.122
Determine the largest allowable length of a stainless steel shaft of 3 3
8 4
-in.
 cross section if the shearing stress
is not to exceed 15 ksi when the shaft is twisted through 15. Use 6
11.2 10 psi.
G  
SOLUTION
3
max
3
in. 0.75 in.
4
3
in. 0.375 in.
8
15 ksi 15 10 psi
15
15 rad 0.26180 rad
180



 
 
  
   
a
b
max 2
1
T
c ab
  (1)
3
2
TL
c ab G
  (2)
Divide (2) by (1) to eliminate T.
2
1 1
3
max 2
2
c ab L c L
c bG
c ab G


 
Solve for L. 2
1 max
c bG
L
c



0.75
2
0.375
a
b
 
Table 3.1 gives 1 2
0.246, 0.229
c c
 
6
3
(0.229)(0.375)(11.2 10 )(0.26180)
68.2 in.
(0.246)(15 10 )
L

 

68.2 in.
L  
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386
900 mm
25 mm
25 mm
15 mm
45 mm
A
A
B
B
(a)
(b)
T
T
PROBLEM 3.123
Using all 70 MPa

 and G  27 GPa, determine for each of the aluminum
bars shown the largest torque T that can be applied and the corresponding
angle of twist at end B.
SOLUTION
6 9
all 70 10 Pa 27 10 Pa 0.900 m
G L
     
(a) 1 2
45 mm 15 mm 3.0 From Table 3.1, 0.267, 0.263
a
a b c c
b
    
2
max 1 max
2
1
T
T c ab
c ab
 
 
2 6
(0.267)(0.045)(0.015) (70 10 ) 189.236 N m
T     189.2 N m
T   
3
3 3 9
2
(189.236)(0.900)
157.921 10 rad
(0.263)(0.045)(0.015) (27 10 )
TL
c ab G
 
   

9.05
   
(b) 1 2
25 mm 25 mm, 1.0 From Table 3.1, 0.208, 0.1406
    
a
a b c c
b
2
max 1 max
2
2 6
(0.208)(0.025)(0.025) (70 10 )
227.5 N m
 
 
 
 
T
T c ab
ab
228 N m
T   
3
3 3 9
2
(227.5)(0.900)
138.075 10 rad
(0.1406)(0.025)(0.025) (27 10 )
TL
c ab G
 
   

7.91
   
Mecánica de Materiales - Ferdinand P. Beer 7ma. Edición.

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387
900 mm
25 mm
25 mm
15 mm
45 mm
A
A
B
B
(a)
(b)
T
T
PROBLEM 3.124
Knowing that the magnitude of the torque T is 200 N  m and that
G  27 GPa, determine for each of the aluminum bars shown the
maximum shearing stress and the angle of twist at end B.
SOLUTION
9
200 N m 0.900 m 27 10 Pa
T L G
    
(a) 1 2
45 mm, 15 mm, 3.0 From Table 3.1, 0.267 0.263
    
a
a b c c
b
max 2 2
1
6
200
(0.267)(0.045)(0.015)
74.0 10 Pa
  
 
T
c ab
max 74.0 MPa
  
3 3 9
2
3
(200)(0.900)
(0.263)(0.045)(0.015) (27 10 )
166.904 10 rad


 

 
TL
c ab G
9.56
   
(b) 1 2
25 mm, 25 mm, 1.0 From Table 3.1, 0.208, 0.1406
    
a
a b c c
b
6
max 2 2
1
200
61.539 10 Pa
(0.208)(0.025)(0.025)
T
c ab
     max 61.5 MPa
  
3 3 9
2
3
(200)(0.900)
(0.1406)(0.025)(0.025) (27 10 )
121.385 10 rad


 

 
TL
c ab G
6.95
   
Mecánica de Materiales - Ferdinand P. Beer 7ma. Edición.

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388
25 in.
2.4 in.
1.6 in.
1 in.
4 in.
B
B
A
A
T
T
(b)
(a)
PROBLEM 3.125
Determine the largest torque T that can be applied to each of the two
brass bars shown and the corresponding angle of twist at B, knowing that
all 12 ksi
  and 6
5.6 10 psi.
G  
SOLUTION
6 3
all
25 in., 5.6 10 psi, 12 10 psi

    
L G
max 2
1
T
c ab
  or 2
1 max
T c ab 
 (1)
3
2
TL
c ab G
  or 1 max
2

 
c L
c bG
(2)
(a) 1 2
4 in., 1in., 4.0 From Table 3.1, 0.282, 0.281
    
a
a b c c
b
From (1), 2 3 3
(0.282)(4)(1) (12 10 ) 13.54 10
T     13.54 kip in.
T   
From (2),
3
6
(0.282)(25)(12 10 )
0.05376 radians
(0.281)(1)(5.6 10 )


 

3.08
   
(b) 1 2
2.4 in., 1.6 in., 1.5 From Table 3.1, 0.231, 0.1958
    
a
a b c c
b
From (1), 2 3 3
(0.231)(2.4)(1.6) (12 10 ) 17.03 10
T     17.03 kip in.
T   
From (2),
3
6
(0.231)(25)(12 10 )
0.0395 radians
(0.1958)(1.6)(5.6 10 )


 

2.26
   
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389
25 in.
2.4 in.
1.6 in.
1 in.
4 in.
B
B
A
A
T
T
(b)
(a)
PROBLEM 3.126
Each of the two brass bars shown is subjected to a torque of magnitude
12.5 kip · in.
T  Knowing that 6
5.6 10 psi,
G   determine for each
bar the maximum shearing stress and the angle of twist at B.
SOLUTION
25 in.,
L  6
5.6 10 psi,
G   3
12.5 10 lb in.
T   
(a) 4 in.,

a 1in.,

b 4.0
a
b

From Table 3.1, 1 2
0.282, 0.281
c c
 
3
3
max 2 2
1
12.5 10
11.08 10 psi
(0.282)(4)(1)
T
c ab


    max 11.08 ksi
  
3
3 3 6
2
(12.5 10 )(25)
0.04965 radians
(0.282)(4)(1) (5.6 10 )
TL
c ab G


  

2.84
   
(b) 2.4 in.,
a  1.6 in.,
b  1.5
a
b

From Table 3.1, 1 2
0.231, 0.1958
c c
 
3
3
max 2 2
1
12.5 10
8.81 10 psi
(0.231)(2.4)(1.6)
T
c ab


    max 8.81 ksi
  
6
3 3 6
2
(12.5 10 )(25)
0.02899 radians
(0.1958)(2.4)(1.6) (5.6 10 )
TL
c ab G


  

1.661
   
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390
30 mm
750 mm
B
b
A
T
PROBLEM 3.127
The torque T causes a rotation of 0.6 at end B of the aluminum bar
shown. Knowing that 15 mm

b and 26 GPa,

G determine the
maximum shearing stress in the bar.
SOLUTION
3
3
2
3
1
2
3
2 2
max 2 2
1
1 1
30 mm 0.030 m
15 mm 0.015 m
0.6 10.472 10 rad
30
2.0
15
a
b
c ab G
TL
T
c L
c ab G
c ab G c bG
T
c L
c ab c ab L
a
b



 


 
 
   
  
  
 
From Table 3.1,
1
2
3 9 3
max 3
6
0.246
0.229
(0.229)(15 10 )(26 10 )(10.472 10 )
(0.246)(750 10 )
5.07 10 Pa
c
c

 



  


  max 5.07 MPa
  
Mecánica de Materiales - Ferdinand P. Beer 7ma. Edición.

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391
30 mm
750 mm
B
b
A
T
PROBLEM 3.128
The torque T causes a rotation of 2 at end B of the stainless steel
bar shown. Knowing that 20 mm

b and 75 GPa,

G determine the
maximum shearing stress in the bar.
SOLUTION
3
3
2
3
2
3
2 2
max 2 2
1
1 1
30 mm 0.030 m
20 mm 0.020 m
2 34.907 10 rad
30
1.5
20



 


 
 
   
  
  
 
a
b
TL c ab G
T
L
c ab G
T c ab G c bG
c L
c ab c ab L
a
b
From Table 3.1,
1
2
3 9 3
6
max 3
0.231
0.1958
(0.1958)(20 10 )(75 10 )(34.907 10 )
59.2 10 Pa
(0.231)(750 10 )
c
c

 



  
  

max 59.2 MPa
  
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392
b
b
b
A B
PROBLEM 3.129
Two shafts are made of the same material. The cross section of shaft A is a
square of side b and that of shaft B is a circle of diameter b. Knowing that the
shafts are subjected to the same torque, determine the ratio /
A B
  of maximum
shearing stresses occurring in the shafts.
SOLUTION
A. Square: 1
2 3
1
1, 0.208 (Table 3.1)
0.208
A
a
c
b
T T
c ab b

 
 
B. Circle: 3 3
1 2 16
2
B
Tc T T
c b
J c b

 
   
Ratio:
1
0.3005
0.208 16
A
B
 


   0.944
A
B


 
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393
TA
TB
A
B
PROBLEM 3.130
Shafts A and B are made of the same material and have the same cross-
sectional area, but A has a circular cross section and B has a square
cross section. Determine the ratio of the maximum torques TA and TB
when the two shafts are subjected to the same maximum shearing stress
( ).
A B

  Assume both deformations to be elastic.
SOLUTION
Let c  radius of circular section A and b  side of square section B.
For equal areas 2 2
,
c b
 
b
c


Circle: 3
3
2
2

 

   
A A
A A A
T c T
T c
J c
Square:
From Table 3.1, 1 0.208
c 
3
1
2 3
1 1
A B
B B B
T T
T c b
c ab c b
 
   
Ratio:
3
3
3/2
3 3
1 1 1
1
2
2
2
B
B
A A
B B
B B
b
c
T
T c b c b c

 
 


  

  
For the same stresses,
1
(2)(0.208)
 

  
A
B A
B
T
T
1.356
A
B
T
T
 
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394
TA
TB
A
B
PROBLEM 3.131
Shafts A and B are made of the same material and have the same
length and cross-sectional area, but A has a circular cross section and B
has a square cross section. Determine the ratio of the maximum values
of the angles A
 and B
 when the two shafts are subjected to the
same maximum shearing stress ( ).
 

A B Assume both deformations
to be elastic.
SOLUTION
Let c  radius of circular section A and b  side of square section B.
For equal areas, 2 2
c b b c
 
  
Circle: max
A A A
A
c L
G L cG
  
 
   
Square: From Table 3.1, 1 2
0.208, 0.1406
c c
 
3
2 3
1
3
3 4
2
0.208
0.208
0.208 1.4794
0.1406
 
 

   
  
B B
B B B
B B B
B
T T
T b
c ab b
T L b L L
bG
c ab G b G
Ratio: 0.676 0.676
1.4794
A A A A
B B B B
L bG b
cG L c
   

   
   
For equal stresses, 0.676
B
A B
A

  

  1.198
B
A


 
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395
TA
TB
A
B
PROBLEM 3.132
Shafts A and B are made of the same material and have the same cross-
sectional area, but A has a circular cross section and B has a square cross
section. Determine the ratio of the angles A and B through which shafts
A and B are respectively twisted when the two shafts are subjected to the
same torque (TA  TB). Assume both deformations to be elastic.
SOLUTION
Let c  radius of circle section A and b  side of square section B.
For equal areas, 2 2
c b b c
 
  
Circle:
4
2
A
A
TL T L
JG c G


 
Square: 3 4
2 2
B
B
TL T L
c ab G c b G
  
From Table 3.1, 2 0.1406
c 
Ratio:
4
4
(0.1406)
2
A A
B B
T L b G
T L
c G



 
For ,

A B
T T
4 2
4
(0.1406)
0.883
2
 


 
A
B
c
c

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396
b
b b
b
2b
T
T
T
(a)
(b)
(c)
PROBLEM 3.133
A torque of magnitude T  2 kip  in. is applied to each of the steel
bars shown. Knowing that all 6 ksi,

 determine the required
dimension b for each bar.
SOLUTION
max
2 kip in. 6 ksi

  
T
(a) Circle:
1
2
c b

max 3 3
3 3
max
2 16
16 (16)(2)
1.698 in
(6)
Tc T T
J c b
T
b

 
 
  
   1.193 in.
b  
(b) Square: 1
, 1.0 From Table 3.1, 0.208
  
a
a b c
b
3 3
max 2 3
1 max
1 1
2
1.603 in
(0.208)(6)
T T T
b
c
c ab c b


      1.170 in.
b  
(c) Rectangle: 1
2 2.0, 0.246
a
a b c
b
  
3 3
max 2 3
1 max
1 1
2
0.668 in
2 (2)(0.246)(6)
2
T T T
b
c
c ab c b


      0.878 in.
b  
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397
b
b b
b
2b
T
T
T
(a)
(b)
(c)
PROBLEM 3.134
A torque of magnitude T  300 N  m is applied to each of the
aluminum bars shown. Knowing that all  60 MPa, determine
the required dimension b for each bar.
SOLUTION
6
max
300 N m 60 10 Pa

   
T
(a) Circle:
1
2
c b

max 3 3
3 6 3
6
max
2 16
16 (16)(300)
25.46 10 m
(60 10 )
Tc T T
J c b
T
b

 
 

  
   

3
29.4 10 m 29.4 mm
b 
   
(b) Square: 1
, 1.0 From Table 3.1, 0.208
  
a
a b c
b
3
max 2 3 6
1 max
1 1
6 3
300
(0.208)(60 10 )
24.04 10 m
T T T
b
c
c ab c b



    

 
3
28.9 10 m 28.9 mm
b 
   
(c) Rectangle: 1
2 , 2.0, 0.246
  
a
a b c
b
3
max 2 3
1 max
1 1
6
3 3
2
2
300
(2)(0.246)(60 10 )
10.16 10 m
T T T
b
c
c ab c b



   


 
3
21.7 10 m 21.7 mm
b 
   
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398
1.25 m
T
PROBLEM 3.135
A 1.25-m-long steel angle has an L127  76  6.4 cross section. From
Appendix C we find that the thickness of the section is 6.4 mm and that
its area is 1250 mm2
. Knowing that all  60 MPa and that G  77.2 GPa,
and ignoring the effect of stress concentrations, determine (a) the largest
torque T that can be applied, (b) the corresponding angle of twist.
SOLUTION
2
1 2
1250 mm 6.4 mm 0.0064 m
1250
195.313 mm 0.195313 m
6.4
195.313
30.518, 0.032768
6.4
1
1 0.630 0.32645
3
  
   
  
 
   
 
 
A b
A
a
b
a b
b a
b
c c
a
2
max 1 max
2
1
T
T c ab
c ab
 
  
(a) 2 6 3
(0.32645)(0.195313)(0.0064) (60 10 ) 156.696 10 N m
    
T
157.0 kN m
T   
(b)
2
1 max 1 max max
3 3
2
2 2
  
    
TL c ab L c L L
c bG bG
c ab G c ab G
6
3
9
(60 10 )(1.25)
151.797 10 rad
(0.0064)(77.2 10 )
 

  

8.70
   
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399
1 in.
a
a
L8 8 1
8 in.
8 in.
PROBLEM 3.136
A 36-kip  in. torque is applied to a 10-ft-long steel angle with an L8 8 1
  cross
section. From Appendix C, we find that the thickness of the section is 1 in. and that its
area is 15 in2
. Knowing that 6
11.2 10 psi,
G   determine (a) the maximum shearing
stress along line a-a, (b) the angle of twist.
SOLUTION
2
15 in
15 in., 1in., 15
1in.
A a
a b
t b
    
Since 1 2
1
5, 1 0.630
3
a b
c c
b a
 
   
 
 
or 1 2
1 0.630
1 0.3193
3 15
c c
 
   
 
 
3 6
36 10 lb in.; 120 in.; 11.2 10 psi
     
T L G
(a) Maximum shearing stress: max 2
1
T
c ab
 
3
3
max 2
36 10
7.52 10 psi
(0.3193)(15)(1)


   max 7.52 ksi
  
(b) Angle of twist: 2
2
TL
c ab G
 
3
3 6
(36 10 )(120)
0.08052 radians
(0.3193)(15)(1) (11.2 10 )


 

4.61
   
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400
T
W310 60
PROBLEM 3.137
A 4-m-long steel member has a W310  60 cross section. Knowing that
77.2 GPa
G  and that the allowable shearing stress is 40 MPa, determine
(a) the largest torque T that can be applied, (b) the corresponding angle of
twist. Refer to Appendix C for the dimensions of the cross section and
neglect the effect of stress concentrations. (Hint: consider the web and
flanges separately and obtain a relation between the torques exerted on the
web and a flange, respectively, by expressing that the resulting angles of
twist are equal.)
SOLUTION
all
W310 60, 4 m, 77.2 GPa, 40 MPa

   
L G
For one flange: From App. C, 203 mm, 13.1mm, / 15.50
a b a b
  
Eq. (3.45): 1 2
1 0.630
1 0.31979
3 15.50
c c
 
   
 
 
Eq. (3.44): 3 3 9
2
6
(4)
0.31979(0.203)(0.0131) (77.2 10 )
355.04 10
f f
f
f f
T L T
c ab G
T

 
 

 
(1)
For web: From App. C, 302 2(13.1) 275.8 mm, 7.49 mm, / 36.822
a b a b
    
Eq. (3.45): 1 2
1 0.630
1 0.32763
3 36.822
c c
 
   
 
 
Eq. (3.44): 3 9
(4)
0.32763(0.2758)(0.00749) (77.2 10 )
w
w
T
 

6
1364.64 10
w w
T
 
  (2)
Since angle of twist is the same for flanges and web,
6 6
: 355.04 10 1364.64 10
3.8436
f w f w
f w
T T
T T
   
   

(3)
But the sum of the torques exerted on the two flanges and on the web is equal to the torque T applied to the
member:
2 f w
T T T
  (4)
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401
PROBLEM 3.137 (Continued)
Substituting for f
T from (3) into (4),
2(3.8436 ) 0.115112
w w w
T T T T T
   (5)
From (3), 3.8436(0.115112 ) 0.44244
f f
T T T T
  (6)
For one flange:
From Eq. (3.43), 2 2 6
1 max 0.31979(0.203)(0.0131) (40 10 )
445.62 N m
f
T c ab 
  
 
Eq. (6): 445.62 0.44244T
 1007 N m
T   
For web: 2 2 6
1 max 0.32763(0.2758)(0.00749) (40 10 )
202.77 N m
w
T c ab 
  
 
Eq. (5): 202.77 0.115112T
 1762 N m
T   
(a) Largest allowable torque: Use the smaller value. 1007 N m
T   
(b) Angle of twist: Use ,
f
T which is critical.
Eq. (1): 6
(355.04 10 )(445.62) 0.161763 rad
f
  
    9.27
   
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402
b b
a
a
W8 31
PROBLEM 3.138
An 8-ft-long steel member with a W8  31 cross section is subjected to a 5-kip  in. torque.
The properties of the rolled-steel section are given in Appendix C. Knowing that
6
11.2 10 psi,
G   determine (a) the maximum shearing stress along line a-a, (b) the
maximum shearing stress along line b-b, (c) the angle of twist. (See hint of Prob. 3.137.)
SOLUTION
Flange:
 
1 2 3
2
3 3
2 2
3 3
8.00
8.00 in., 0.435 in., 18.3908
0.435
1
1 0.630 0.32192
3
where
(0.32192)(8.00)(0.435) 0.21199 in

 
   
    
  
 
f
f
f
f f f
f
a
a b
b
T L
b
c c
a c ab G
G G
T c ab K K c ab
L L
K
Web:
7.13
8.0 (2)(0.435) 7.13 in., 0.285 in., 25.018
0.285
a
a b
b
     
1 2 3
2
3 3
2 2
3 4
1 1 0.630 0.32494
3
where
(0.32494)(7.13)(0.285) 0.053632 in

 
 
    
 
 
  
 
w
w
w
w w w
w
b T L
c c
a c ab G
G G
T c ab K K c ab
L L
K
For matching twist angles, f w
  
 
Total torque. 2 (2 )
f w f w
G
T T T K K
L

   
, ,
2 2 2
(0.21199)(5000) (0.053632)(5000)
2219.3 lb in.; 561.46 lb in.
(2)(0.21199) 0.053632 (2)(0.21199) 0.053632

  
  
     
 
f w
f w
p w f w f w
f w
K T
G T K T
T T
L K K K K K K
T T
(a) 2 2
1
2219.3
4550 psi
(0.32192)(8.00)(0.435)
f
f
T
c ab
    4.55 ksi
f
  
(b) 2 2
1
561.46
2980 psi
(0.32494)(7.13)(0.285)
w
w
T
c ab
    2.98 ksi
w
  
(c)
2 (2 )
f w f w
G T TL
L K K G K K


  
 
where 8 ft 96 in.
L  
3
6
(5000)(96)
44.629 10 rad
(11.2 10 )[(2)(0.21199) 0.53632]
 
  
 
2.56
   
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403
a
6 in.
4 in.
in.
b
1
4
in.
1
4
in.
1
2
in.
1
2
PROBLEM 3.139
A 5-kip  ft torque is applied to a hollow aluminum shaft having the cross
section shown. Neglecting the effect of stress concentrations, determine the
shearing stress at points a and b.
SOLUTION
3 3
(5)(10 )(12) 60 10 lb in.
T    
Area bounded by center line:
2
(5.5)(3.75) 20.625 in
a bh
  
At Point a,
3
0.25 in.
2
60 10
(2)(0.25)(20.625)





t
T
ta
3
5.82 10 psi
  5.82 ksi
  
At Point b, 0.50 in.
t 
3
3
2
60 10
(2)(0.50)(20.625)
2.91 10 psi
T
ta
 


  2.91ksi
  
Area bounded by center line.
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404
90 mm
60⬚
a
b
PROBLEM 3.140
A torque T  750 kN  m is applied to a hollow shaft shown that has a uniform 8-mm
wall thickness. Neglecting the effect of stress concentrations, determine the shearing
stress at points a and b.
SOLUTION
Detail of corner.
1
tan 30
2
2 tan 30
8
6.928 mm
2 tan 30
90 2 76.144 mm
t e
t
e
b e
 


 

  
Area bounded by center line.
2 2
2 6 2
1 3 3 3
(76.144)
2 2 4 4
2510.6 mm 2510.6 10 m
0.008 m
a b b b
t

  
  

6
6
750
18.67 10 Pa
2 (2)(0.008)(2510 10 )
 
   

T
ta
18.67 MPa
  
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405
30 mm
60 mm
30 mm
a
b
PROBLEM 3.141
A 750-N  m torque is applied to a hollow shaft having the cross section shown
and a uniform 6-mm wall thickness. Neglecting the effect of stress concentrations,
determine the shearing stress at points a and b.
SOLUTION
Area bounded by center line.
2 2
6 2
2 (33) (60)(66) 7381 mm
2
7381 10 m
0.006 mat both and


  
 

a
t a b
Then at points a and b,
6
6
750
8.47 10 Pa
2 (2)(0.006)(7381 10 )
 
   

T
ta
8.47 MPa
  
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406
20 mm
20 mm
50 mm
50 mm PROBLEM 3.142
A hollow member having the cross section shown is formed from sheet
metal of 2-mm thickness. Knowing that the shearing stress must not
exceed 3 MPa, determine the largest torque that can be applied to the
member.
SOLUTION
Area bounded by center line.
2 6 2
6 6
(48)(18) (30)(18)
1404 mm 1404 10 m
0.002m
or 2 (2)(0.002)(1404 10 )(3 10 )
2
a
t
T
T ta
ta
 


 
  

    
16.85 N m
T   
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407
10 mm
10 mm
50 mm
50 mm PROBLEM 3.143
A hollow member having the cross section shown is formed from sheet
metal of 2-mm thickness. Knowing that the shearing stress must not
exceed 3 MPa, determine the largest torque that can be applied to the
member.
SOLUTION
Area bounded by center line.
2 6 2
(48)(8) (40)(8)
704 mm 704 10 m
0.002 m
a
t

 
  

6 6
2 (2)(0.002)(704 10 )(3 10 )
2
T
T ta
ta
  
     
8.45 N m
  
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408
b
40 mm
2 mm
4 mm
a
4 mm
55 mm
55 mm
PROBLEM 3.144
A 90-N  m torque is applied to a hollow shaft having the cross section
shown. Neglecting the effect of stress concentrations, determine the shearing
stress at points a and b.
SOLUTION
Area bounded by center line.
2 2 3 2
52 52 39 39 (39) 2377.6 mm 2.3776 10 m
4
90 N m
a
T
 
       
 
3 3 2
90 N m
2 2(4 10 m)(2.3776 10 m )
a
T
ta
  

 
 
4.73 MPa
a
  
 3 3 2
90 N m
2 2(2 10 m)(2.3776 10 m )
  

 
 
b
T
ta
9.46 MPa
b
  
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409
2 in.
2 in.
2 in.
d
3 in.
PROBLEM 3.145
A hollow member having the cross section shown is to be formed from sheet metal of
0.06-in. thickness. Knowing that a 1250-lb  in. torque will be applied to the member,
determine the smallest dimension d that can be used if the shearing stress is not to
exceed 750 psi.
SOLUTION
Area bounded by center line.
(5.94)(2.94) 2.06 17.4636 2.06
0.06 in., 750 psi, 1250 lb in.
2


   
   

a d d
t T
T
ta
2
1250
17.4636 2.06 13.8889
(2)(0.06)(750)


  
T
a
t
d
3.5747
1.735 in.
2.06
d   1.735 in.
d  
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410
2 in. d
2 in.
2 in.
3 in.
PROBLEM 3.146
A hollow member having the cross section shown is to be formed from sheet metal of
0.06-in. thickness. Knowing that a 1250-lb  in. torque will be applied to the member,
determine the smallest dimension d that can be used if the shearing stress is not to
exceed 750 psi.
SOLUTION
Area bounded by center line.
(5.94)(2.94 ) 1.94 17.4636 4.00
0.06 in., 750 psi, 1250 lb in.
2


    
   

a d d d
t T
T
ta
2
1250
17.4636 4.00 13.8889
(2)(0.06)(750)


  
T
a
t
d
3.5747
0.894 in.
4.00
d   0.894 in.
d  
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411
c1
O
c2
PROBLEM 3.147
A cooling tube having the cross section shown is formed from a sheet of stainless
steel of 3-mm thickness. The radii c1  150 mm and c2  100 mm are measured to
the center line of the sheet metal. Knowing that a torque of magnitude T  3 kN  m
is applied to the tube, determine (a) the maximum shearing stress in the tube, (b) the
magnitude of the torque carried by the outer circular shell. Neglect the dimension of
the small opening where the outer and inner shells are connected.
SOLUTION
Area bounded by center line.
 
2 2 2 2 3 2
1 2
3 2
(150 100 ) 39.27 10 mm
39.27 10 m
0.003 m
a c c
t
 

     
 

(a)
3
6
3
3 10
12.73 10 Pa
2 (2)(0.003)(39.27 10 )
T
ta
 

   

12.76 MPa
  
(b) 2
1 1 1 1
(2 ) 2
   
 
T c t c c t
2 6 3
2 (0.150) (0.003)(12.73 10 ) 5.40 10 N m

     1 5.40 kN m
T   
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412
1.1 in.
0.12 in.
0.08 in.
2.4 in.
a
b
PROBLEM 3.148
A hollow cylindrical shaft was designed to have a uniform wall thickness of
0.1 in. Defective fabrication, however, resulted in the shaft having the cross
section shown. Knowing that a 15-kip  in. torque is applied to the shaft,
determine the shearing stresses at points a and b.
SOLUTION
Radiusof outer circle 1.2 in.
Radiusof inner circle 1.1in.


Mean radius 1.15 in.

Area bounded by center line.
2 2 2
(1.15) 4.155 in
m
a r
 
  
At point a, 0.08 in.
t 
15
2 (2)(0.08)(4.155)
T
ta
   22.6 ksi
  
At point b, 0.12 in.
t 
15
2 (2)(0.12)(4.155)
T
ta
   15.04 ksi
  
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413
T
T
T'
T'
(a) (b)
PROBLEM 3.149
Equal torques are applied to thin-walled tubes of the same length L,
same thickness t, and same radius c. One of the tubes has been slit
lengthwise as shown. Determine (a) the ratio /
b a
  of the maximum
shearing stresses in the tubes, (b) the ratio /
b a
  of the angles of twist
of the shafts.
SOLUTION
Without slit:
Area bounded by center line. 2
a c


2
3
3
2 2
2
2


 

 
  
a
a
T T
ta c t
TL TL
J c t
GJ c tG
With slit:
2
2 , , 1
a c
a c b t
b t


   
1 2
2 2
1
3 3
2
1
3
3
2
3
2
b
b
c c
T T
c ab ct
T TL
c ab G ct G




 
 
 
(a) Stress ratio:
2
2
3 2 3
2
b
a
T c t c
T t
ct
 
 
  
3
b
a
c
t


 
(b) Twist ratio:
3 2
3 2
3 2 3
2
 
 
  
b
a
TL c tG c
TL
ct G t
2
2
3
b
a
c
t


 
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414
L
t
cm
T
T'
PROBLEM 3.150
A hollow cylindrical shaft of length L, mean radius ,
m
c and uniform
thickness t is subjected to a torque of magnitude T. Consider, on the
one hand, the values of the average shearing stress ave
 and the angle
of twist  obtained from the elastic torsion formulas developed in
Secs. 3.1C and 3.2 and, on the other hand, the corresponding values
obtained from the formulas developed in Sec. 3.10 for thin-walled
shafts. (a) Show that the relative error introduced by using the thin-
walled-shaft formulas rather than the elastic torsion formulas is the
same for ave
 and  and that the relative error is positive and
proportional to the ratio / .
m
tc (b) Compare the percent error
corresponding to values of the ratio / m
t c of 0.1, 0.2, and 0.4.
SOLUTION
Let 1
2 2
outer radius
  
m
c c t and 1
1 2
inner radius m
c c t
  
   
4 4 2 2
2 1 2 1 2 1 2 1
2 2 2 2
2 2
2 2
1
2 2
( )( )
2 2
1 1
(2 )
2 4 4
1
2
4
1
2
4
1
2
4
m m m m m
m m
m
m
m
m m
J c c c c c c c c
c c t t c c t t c t
c t c t
Tc T
J
c t t
TL TL
JG
c t c t G
 






     
 
     
 
 
 
 
 
 
 
 

 
 
 
 

 
 
Area bounded by center line. 2
ave 2
2 2




 
m
m
a c
T T
ta c t
2 2 2 2 3
(2 / )
4 4( ) 2


 
  


m
m m
TL ds TL c t TL
t
a G c G c tG
(a) Ratios:
 
2 2
2
ave
2 2
1
2 1
4 1
4
2
m
m m m
c t t
T t
T
c t c


 

    

 
2 2
2
2
3
1
1
2 1
4 1
4
2
m m
m
m
c t c tG
TL t
TL c
c tG


 

     
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415
PROBLEM 3.150 (Continued)
(b)
2
ave 2
2
1
1
1 1
4
m m
t
c
 
 
    

m
t
c
0.1 0.2 0.4
2
2
1
4 m
t
c
0.0025 0.01 0.04
% 0.25% 1.000% 4.00%
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416
12 in.
in.
1
4
45⬚
T
T' PROBLEM 3.151
A steel pipe of 12-in. outer diameter is fabricated from
1
4 -in. -thick plate by welding along a helix which forms an angle
of 45 with a plane perpendicular to the axis of the pipe.
Knowing that the maximum allowable tensile stress in the weld
is 12 ksi, determine the largest torque that can be applied to the
pipe.
SOLUTION
From Eq. (3.14) of the textbook,
45 max
 

hence,
 
3
max
2
1 2
4 4 4 4
2 1
12 ksi 12 10 psi
1 1
(12) 6.00 in.
2 2
6.00 0.25 5.75 in.
[(6.00) (5.75) ] 318.67 in.
2 2

 
  
  
    
    
o
c d
c c t
J c c
max
max
3
3
(12 10 )(318.67)
637 10 lb in.
6.00

  

   
Tc J
T
J c
T
637 kip in.
T   
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417
C
B
F
D
A
30 mm
25 mm
60 mm
75 mm
E
T
PROBLEM 3.152
A torque of magnitude 120 N m
 
T is applied to shaft AB of the gear
train shown. Knowing that the allowable shearing stress is 75 MPa in
each of the three solid shafts, determine the required diameter of
(a) shaft AB, (b) shaft CD, (c) shaft EF.
SOLUTION
Statics:
Shaft AB: AB A B
T T T T
  
Gears B and C: 25 mm, 60 mm
B C
r r
 
Force on gear circles.
60
2.4
25
B C
BC
B C
C
C B
B
T T
F
r r
r
T T T T
r
 
  
Shaft CD: 2.4
CD C D
T T T T
  
Gears D and E: 30 mm, 75 mm
D E
r r
 
Force on gear circles.
75
(2.4 ) 6
30
D E
DE
D E
E
E D
D
T T
F
r r
r
T T T T
r
 
  
Shaft EF: 6
EF E F
T T T T
  
Required diameters.
max 3
3
3
max
6
max
2
2
2
2 2
75 10 Pa
Tc T
J c
T
c
T
d c





 

 
 
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418
PROBLEM 3.152 (Continued)
(a) Shaft AB: 120 N m
AB
T T
  
3
3
6
2(120)
2 20.1 10 m
(75 10 )
AB
d


  

20.1 mm
AB
d  
(b) Shaft CD: (2.4)(120) 288 N m
CD
T   
3
3
6
(2)(288)
2 26.9 10 m
(75 10 )
CD
d


  

26.9 mm
CD
d  
(c) Shaft EF: (6)(120) 720 N m
EF
T   
3
3
3
(2)(720)
2 36.6 10 m
(75 10 )
EF
d


  

36.6 mm
EF
d  
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419
1.2 m
80 mm
1.6 m
42 mm
D
C
B
A
TA
240 mm 60 mm
PROBLEM 3.153
Two solid shafts are connected by gears as shown. Knowing
that 77.2 GPa
G  for each shaft, determine the angle
through which end A rotates when 1200 N m.
 
A
T
SOLUTION
Calculation of torques:
Circumferential contact force between gears B and C: AB CD C
CD AB
B C B
T T r
F T T
r r r
  
240
1200 N m (1200) 3600 N m
80
AB CD
T T
    
Twist in shaft CD: 9
1
0.030 m, 1.2 m, 77.2 10 Pa
2
c d L G
    
4 4 6 4
3
/ 9 9
(0.030) 1.27234 10 m
2 2
(3600)(1.2)
43.981 10 rad
(77.2 10 )(1.27234 10 )
C D
J c
TL
GJ
 




   
   
 
Rotation angle at C. 3
/ 43.981 10 rad
C C D
  
  
Circumferential displacement at contact points of gears B and C: C C B B
r r
  
 
Rotation angle at B. 3 3
240
(43.981 10 ) 131.942 10 rad
80
C
B C
B
r
r
   
    
Twist in shaft AB: 9
4 4 9 4
3
/ 9 9
1
0.021m, 1.6 m, 77.2 10 Pa
2
(0.021) 305.49 10 m
2 2
(1200)(1.6)
81.412 10 rad
(77.2 10 )(305.49 10 )
 
 

    
   
   
 
A B
c d L G
J c
TL
GJ
Rotation angle at A. 3
/ 213.354 10 rad
A B A B
   
    12.22
A
   
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420
B
C
A
TB
TA
␣
␣
0.625 in.
0.5 in.
PROBLEM 3.154
In the bevel-gear system shown, 18.43 .
   Knowing that the
allowable shearing stress is 8 ksi in each shaft and that the system is in
equilibrium, determine the largest torque A
T that can be applied at A.
SOLUTION
Using stress limit for shaft A,
3 3
1
8 ksi, 0.25 in.
2
(8)(0.25) 0.196350 kip in.
2 2
A
c d
J
T c
c

  

  
    
Using stress limit for shaft B,
3 3
1
8 ksi, 0.3125 in.
2
(8)(0.3125) 0.3835 kip in.
2 2

  

  
    
B
c d
J
T c
c
From statics, (tan )
(tan18.43 )(0.3835) 0.1278 kip in.
A
A B B
B
A
r
T T T
r
T

 
   
The allowable value of A
T is the smaller.
0.1278 kip in.
A
T   127.8 lb in.
A
T   
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421
A
8 in.
6 in.
5 in.
16 in.
2 in.
C
B
D
TA
TD
PROBLEM 3.155
The design specifications for the gear-and-shaft system
shown require that the same diameter be used for both
shafts, and that the angle through which pulley A will
rotate when subjected to a 2-kip  in. torque A
T while
pulley D is held fixed will not exceed 7.5 .
 Determine
the required diameter of the shafts if both shafts are made
of a steel with 6
11.2 10 psi
G   and all 12 ksi.
 
SOLUTION
Statics:
Gear B.
0:
B
M
 
0 /
B A B B
r F T F T r
  
Gear C. 0:
C
M
 
0
C D
C
D C A B
B
r F T
r
T r F T nT
r
 
  
5
2.5
2
C
B
r
n
r
  
Torques in shafts. AB A B CD C B A
T T T T T nT nT
    
Deformations: /
CD
C D
T L
GJ
  A
nT L
GJ

/
AB
A B
T L
GJ
  = A
T L
GJ
Kinematics: /
0 0 A
D C D C D
nT L
GJ
   
    
2
C A
B B C B B C C B
B
r n T L
r r n
r GJ
     
     
2 2
/
( 1)
A A A
A C B C
n T L T L n T L
GJ GJ GJ
  

    
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422
PROBLEM 3.155 (Continued)
Diameter based on stress.
Largest torque: m CD A
T T nT
 
3 3
all
3
3
3
3
3
2
12 10 psi, 2 10 lb in.
2 (2)(2.5)(2 10 )
0.6425 in., 2 1.285 in.
(12 10 )
  

 
       

    

m A
m m A
A
m
T c nT
T
J c
nT
c d c
Diameter based on rotation limit.
2
4
3
4
4
6
7.5 0.1309 rad
( 1) (2)(7.25)
8 16 24 in.
(2)(7.25) (2)(7.25)(2 10 )(24)
0.62348 in., 2 1.247 in.
(11.2 10 )(0.1309)
A A
A
n T L T L
L
GJ c G
T L
c d c
G



  
  

    

    

Choose the larger diameter. 1.285 in.
d  
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423
Steel core
Aluminum jacket
72 mm
54 mm
A
B
2.5 m
T
PROBLEM 3.156
A torque of magnitude 4 kN m
 
T is applied at end A of the
composite shaft shown. Knowing that the modulus of rigidity
is 77.2 GPa for the steel and 27 GPa for the aluminum,
determine (a) the maximum shearing stress in the steel core,
(b) the maximum shearing stress in the aluminum jacket,
(c) the angle of twist at A.
SOLUTION
Steel core: 4 4 9
1 1 1 1
9 9 3 2
1 1
1
0.027 m (0.027) 834.79 10
2 2 2
(77.2 10 )(834.79 10 ) 64.446 10 N m
c d J c
G J
  

     
     
Torque carried by steel core. 1 1 1 /
T G J L


Aluminum jacket:
 
1 1 2 2
4 4 4 4 6 4
2 2 1
9 6 3 2
2 2
1 1
0.027 m, 0.036 m
2 2
(0.036 0.027 ) 1.80355 10 m
2 2
(27 10 )(1.80355 10 ) 48.70 10 N m
c d c d
J c c
G J
  

   
     
     
Torque carried by aluminum jacket. 2 2 2 /
T G J L


Total torque: 1 2 1 1 2 2
( ) /
T T T G J G J L

   
3
3
3 3
1 1 2 2
4 10
35.353 10 rad/m
64.446 10 48.70 10
T
L G J G J
 

   
   
(a) Maximum shearing stress in steel core.
9 3
1 1 1 (77.2 10 )(0.027)(35.353 10 )
G G c
L

  
     6
73.7 10 Pa
  73.7 MPa 
(b) Maximum shearing stress in aluminum jacket.
9 3
2 2 2 (27 10 )(0.036)(35.353 10 )
G G c
L

  
     6
34.4 10 Pa
  34.4 MPa 
(c) Angle of twist. 3 3
(2.5)(35.353 10 ) 88.383 10 rad
L
L

  
     5.06
   
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424
100 mm
60 mm
500 mm
300 mm
A
B
45 mm
40 mm
C
D
T
PROBLEM 3.157
Ends A and D of the two solid steel shafts AB and CD are
fixed, while ends B and C are connected to gears as shown.
Knowing that the allowable shearing stress is 50 MPa in
each shaft, determine the largest torque T that may be
applied to gear B.
SOLUTION
Gears B and C:
40
100
C
B C C
B
r
r
  
  0.4
B C
 
 (1)
0:
C CD C
M T r F
   (2)
0:
B AB B
M T T r F
    (3)
Solve (2) for F and substitute into (3).
100
40
2.5
   
 
B
AB CD AB CD
C
AB CD
r
T T T T T T
r
T T T
(4)
Shaft AB: 0.3 m, 0.030 m
L c
 
3
/
4
(0.3)
235.77 10
(0.030)
2
 

    
AB AB AB
B B A
T L T T
JG G
G
(5)
Shaft CD: 0.5 m, 0.0225 m
L c
 
3
/
4
(0.5)
1242 10
(0.0225)
2
 

    
CD CD CD
C C D
T L T T
JG G
G
(6)
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425
PROBLEM 3.157 (Continued)
Substitute from (5) and (6) into (1).
3 3
0.4 : 235.79 10 0.4 1242 10
 
    
AB CD
B C
T T
G G
0.47462

CD AB
T T (7)
Substitute for CD
T from (7) into (4).
2.5 (0.47462 )
 
AB AB
T T T 2.1865
 AB
T T (8)
Solving (7) for AB
T and substituting into (8),
2.1865
0.47462
CD
T
T
 
  
 
4.6068
 CD
T T (9)
Stress criterion for shaft AB:
all 50 MPa
 
 
AB
3
3 6
2
(0.030 m) (50 10 Pa) 2120.6 N m
2
AB
AB AB AB AB
T c J
T c
J c

  

  
   
From (8), 2.1865(2120.6 N m) 4.64 kN m
T     
Stress criterion for shaft CD:
all 50 MPa:
CD
 
 
3 3 6
(0.0225 m) (50 10 Pa)
2 2
894.62 N m
 
 
   
 
CD
CD CD CD
T c
T c
J
From (7), 4.6068(894.62 N m) 4.12 kN m
T    
The smaller value for T governs.
4.12 kN m
T   
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426
5 m
25 mm
60 mm
T
T'
PROBLEM 3.158
As the hollow steel shaft shown rotates at 180 rpm, a stroboscopic
measurement indicates that the angle of twist of the shaft is 3.
Knowing that 77.2 GPa,
G  determine (a) the power being
transmitted, (b) the maximum shearing stress in the shaft.
SOLUTION
 
2 2
1 1
4 4 4 4
2 1
6 4 6 4
9 6
1
30 mm
2
1
12.5 mm
2
[(30) (12.5) ]
2 2
1.234 10 mm 1.234 10 m
3 0.05236 rad
(77.2 10 )(1.234 10 )(0.0536)
997.61 N m
5
 





 
 
   
   
  

 
   
c d
c d
J c c
TL
GJ
GJ
T
L
Angular speed: 180 rpm 3 rev/sec 3 Hz
f   
(a) Power being transmitted. 3
2 2 (3)(997.61) 18.80 10 W
P f T
   
 
18.80 kW
P  
(b) Maximum shearing stress.
3
2
6
(997.61)(30 10 )
1.234 10
m
Tc
J




 

6
24.3 10 Pa
  24.3 MPa
m
  
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427
2 in.
1.5 in.
r
T
T'
PROBLEM 3.159
Knowing that the allowable shearing stress is 8 ksi for the stepped shaft
shown, determine the magnitude T of the largest torque that can be
transmitted by the shaft when the radius of the fillet is (a) 3
16
in.,

r
(b) 1
4
in.

r
SOLUTION
2 in. 1.5 in. 1.33
D
D d
d
  
max
1
0.75 in. 8 ksi
2
c d 
  
max
Tc
K
J
  or
3
max max
2
J c
T
Kc K
 
 
(a)
3
in. 0.1875 in.
16
 
r r
0.1875
0.125
1.5
r
d
 
From Fig. 3.32, 1.33
K 
3
(8)(0.75)
(2)(1.33)
T

 3.99 kip in.
T   
(b)
1
in.
4
r  0.25 in.
r 
0.25
0.1667
1.5
r
d
 
From Fig. 3.32, 1.27
K 
3
(8)(0.75)
(2)(1.27)
T

 4.17 kip in.
T   
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428
0.5 in.
5 in.
0.2 in.
0.2 in.
0.2 in.
0.2 in.
0.5 in.
6 in.
1.5 in.
1.5 in.
PROBLEM 3.160
A hollow brass shaft has the cross section shown. Knowing that the
shearing stress must not exceed 12 ksi and neglecting the effect of stress
concentrations, determine the largest torque that can be applied to the
shaft.
SOLUTION
Calculate the area bounded by the center line of the wall cross section. The area is a rectangle with two semi-
circular cutouts.
5 0.2 4.8 in.
6 0.5 5.5 in.
1.5 0.1 1.6 in.
b
h
r
  
  
  
2 2 2
3
max max min
min
3 3
min max
2 (4.8)(5.5) (1.6) 18.3575 in
2
12 10 psi 0.2 in.
2
2 (2)(18.3575)(0.2)(12 10 ) 88.116 10 lb in.


 

 
    
 
 
   
     
a bh r
T
t
at
T at
88.1 kip in. 7.34 kip ft
   
T 
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429
C
B
F
E
D
A
d2
d1
T
T'
PROBLEM 3.161
Two solid brass rods AB and CD are brazed to a brass sleeve EF.
Determine the ratio 2 1
/
d d for which the same maximum shearing stress
occurs in the rods and in the sleeve.
SOLUTION
Let 1 1 2 2
1 1
and
2 2
c d c d
 
Shaft AB: 1
1 3
1 1
2
Tc T
J c


 
Sleeve EF:
 
2 2
2 4 4
2 2 1
2
Tc Tc
J c c


 

For equal stresses,
 
2
3 4 4
1 2 1
4 4 3
2 1 1 2
2 2
T Tc
c c c
c c c c
 


 
Let 2
1
c
x
c
 4 4
1 or 1
x x x x
   
Solve by successive approximations starting with 0 1.0.
x 
4 4 4
1 2 3
4 4
4 5
2
1
2 1.189, 2.189 1.216, 2.216 1.220
2.220 1.221, 2.221 1.221 (converged).
1.221 1.221
x x x
x x
c
x
c
     
   
  2
1
1.221
d
d
 
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430
6 ft
B
T
A
5 3 kip · in.
0.5 in.
PROBLEM 3.162
The shaft AB is made of a material that is elastoplastic with
12.5 ksi
Y 
 and G  4  106
psi. For the loading shown, determine
(a) the radius of the elastic core of the shaft, (b) the angle of twist of
the shaft.
SOLUTION
3 3
0.5 in.
(0.5) (1.25) 2.454 kip in.
2 2
  


    
Y
Y
c
J
T c
c
3 kip in. Y
T T
   plastic region with elastic core
3 3
3 3
4 1 3 (3)(2.454)
1 4 4 0.33307
3 4 3
Y Y
Y
Y
T
T T
T
c c
 
 
       
 
 
 
0.69318
Y
c

 (0.69318)(0.5) 0.347 in.
Y
   
6 3
4 9 3
6 ft 72 in. 4 10 psi 4 10 ksi
2 (2)(2.454)(72)
0.4499 rad
(0.5) (4 10 )

 
     
   

Y Y
Y
L G
T L T L
JG c G
0.4499
0.64904 rad 37.2
/ 0.69318
Y Y Y
Y
c c
  

 
       
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431
Element 1
Element n
B
A
Tn
T1
PROBLEM 3.C1
Shaft AB consists of n homogeneous cylindrical elements, which
can be solid or hollow. Its end A is fixed, while its end B is free,
and it is subjected to the loading shown. The length of element i
is denoted by ,
i
L its outer diameter by ,
i
OD its inner diameter
by ,
i
ID its modulus of rigidity by ,
i
G and the torque applied to
its right end by ,
i
T the magnitude i
T of this torque being
assumed to be positive if i
T is counterclockwise from end B and
negative otherwise. (Note that 0
i
ID  if the element is solid.)
(a) Write a computer program that can be used to determine the
maximum shearing stress in each element, the angle of twist of
each element, and the angle of twist of the entire shaft. (b) Use
this program to solve Probs. 3.35, 3.36, and 3.38.
SOLUTION
For each cylindrical element, enter
, , , ,
i i i i i
L OD ID G T
and compute
 
4 4
( /32)

 
i i i
J OD ID
Outline of program.
Update torque i
T T T
 
and compute
( /2)/
/




i i i
i i i i
T OD J
TL G J
Angle of twist of entire shaft, starting with 0,
  update through nth element
i
  
 
Program Outputs
Problem 3.35
Element
Maximum Stress
(MPa)
Angle of Twist
(degrees)
1.0000 11.9575 1.3841
2.0000 23.0259 1.8323
Angle of twist for entire shaft 3.2164
 
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432
PROBLEM 3.C1 (Continued)
Program Outputs (Continued)
Problem 3.36
Element
Maximum Stress
(MPa)
Angle of Twist
(degrees)
1.0000 56.5884 2.5199
2.0000 36.6264 0.8864
Angle of twist for entire shaft 3.4063
 
Problem 3.38
Element
Maximum Stress
(MPa)
Angle of Twist
(degrees)
1.0000 87.3278 4.1181
2.0000 56.5884 1.0392
3.0000 70.5179 0.8633
Angle of twist for entire shaft 6.0206
  
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433
A1
b1
A2
a2
B2
B1
An
an
Bn
bn –1
T0
PROBLEM 3.C2
The assembly shown consists of n cylindrical shafts, which can be solid
or hollow, connected by gears and supported by brackets (not shown).
End 1
A of the first shaft is free and is subjected to a torque 0 ,
T while
end n
B of the last shaft is fixed. The length of shaft i i
A B is ,
i
L its outer
diameter ,
i
OD its inner diameter ,
i
ID and its modulus of rigidity .
i
G
(Note that IDi 0
 if the element is solid.) The radius of gear i
A is ,
i
a
and the radius of gear i
B is .
i
b (a) Write a computer program that can
be used to determine the maximum shearing stress in each shaft, the
angle of twist of each shaft, and the angle through which end i
A rotates.
(b) Use this program to solve Probs. 3.41 and 3.44.
SOLUTION
Torque in shafts. Enter 0
1 1
( / )
i
i i i i
T T
T T A B
 


For each shaft, enter
i i i i
L OD ID G
Compute:  
4 4
( /32)
( /2)
i i i
i i i i
i i i i i
J OD ID
T OD J
T L G J



 


Angle of rotation at end 1:
A
Compute rotation at the “A” end of each shaft.
Start with angle n

 and update from n to 1, and add .
i
1 1
Angle Angle( )/
i i i
A B 
 
 
Program Output
Problem 3.41
Shaft No.
Max Stress
(ksi)
Angle of Twist
(degrees)
1 9.29 1.493
2 12.16 1.707
Angle through which A1 rotates 3.769
 
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434
PROBLEM 3.C2 (Continued)
Program Output (Continued)
Problem 3.44
Shaft No.
Max Stress
(ksi)
Angle of Twist
(degrees)
1 104.31 40.979
2 52.15 20.490
3 26.08 10.245
Angle through which A1 rotates 53.785
  
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435
Element 1
Element n
A
B
T2
Tn
PROBLEM 3.C3
Shaft AB consists of n homogeneous cylindrical elements, which
can be solid or hollow. Both of its ends are fixed, and it is
subjected to the loading shown. The length of element i is
denoted by ,
i
L its outer diameter by ,
i
OD its inner diameter by
,
i
ID its modulus of rigidity by ,
i
G and the torque applied to its
right end by ,
i
T the magnitude i
T of this torque being assumed
to be positive if i
T is observed as counterclockwise from end B
and negative otherwise. Note that 0
i
ID  if the element is solid
and also that 1 0.
T  Write a computer program that can be used
to determine the reactions at A and B, the maximum shearing
stress in each element, and the angle of twist of each element.
Use this program (a) to solve Prob. 3.55 and (b) to determine the
maximum shearing stress in the shaft of Sample Problem 3.7.
SOLUTION
We consider the reaction at B as redundant and release the shaft at B.
Compute with 0:
B B
T
 
For each element, enter
, , , ,
i i i i i
L OD ID G T  
1
Note: 0
 
B
T T
Compute
4 4
( /32)( )
i i i
J OD ID

 
Update torque
i
T T T
 
And compute for each element
( /2)
/
i i i
i i i i
T OD J
TL G J




Compute :
B Starting with 0
  and updating through n elements,
:
i i i B n
    
  
Compute due to unit torque at .
B B
Unit /2
Unit /
i i i
i i i i
OD J
L G J




For n elements,
Unit ( ) Unit ( ) Unit
B B i
i i
 
  
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436
PROBLEM 3.C3 (Continued)
Superposition:
For total angle at B to be zero, (Unit ( )) 0
B B B
T n
 
 
/(Unit ( ))
B B B
T n
 
 
Then ( )
A B
T T i T
  
For each element: Max stress: Total (Unit )
  
 
i i B i
T
Angle of twist: Total (Unit )
  
 
i i B i
T
Program Outputs Problem 3.55 0.295 kN m
1.105 kN m
A
B
T
T
  
  
Element max
 (MPa)
Angle of Twist
(degrees)
1 45.024
 –0.267
2 27.375 –0.267
Problem 3.05 51.733 lb ft
38.267 lb ft
A
B
T
T
  
  

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437
B
L
A
T
PROBLEM 3.C4
The homogeneous, solid cylindrical shaft AB has a length L, a diameter d, a
modulus of rigidity G, and a yield strength .
Y
 It is subjected to a torque T that
is gradually increased from zero until the angle of twist of the shaft has reached
a maximum value m
 and then decreased back to zero. (a) Write a computer
program that, for each of 16 values of m
 equally spaced over a range
extending from 0 to a value 3 times as large as the angle of twist at the onset of
yield, can be used to determine the maximum value m
T of the torque, the
radius of the elastic core, the maximum shearing stress, the permanent twist,
and the residual shearing stress both at the surface of the shaft and at the
interface of the elastic core and the plastic region. (b) Use this program to
obtain approximate answers to Probs. 3.114, 3.115, 3.116.
SOLUTION
At onset of yield:
3
2
Y Y Y
Y Y Y
Y
J
T c
c
T L T J L
L
GJ c GJ cG

 


 
 
  
 
 
Loading: .
m Y
T T
3
4 1
1
3 4
Y
m m
m
T T


 
 
 
   
 
 
 
Eq. (1)
Y
Y
m
c



 Eq. (2)
Unloading (elastic):
1 1
2 1 2
Angle of twist for unloading
at
at
 
   

     
 
  
  
m
u u
m
Y
Y
T L
GJ
c
T c
J
c
Superpose loading and unloading for 0 to 3
  
  Y using 0.2 Y
 increments.
1
When :
2
Y m Y Y m Y
Y Y
T T d
 
    
 
  
When : , use Eq. (1). , use Eq. (2).
  
Y m Y
T
1 1 2 2
Residual: m u R Y R Y

        
     
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438
PROBLEM 3.C4 (Continued)
max max
Interpolate between values at the values of or indicated,

T
Problems 3.114 and 3.115
PHIM
deg
TM
kip  in.
RY
in.
TAUM
ksi
PHIP
deg
TAUR1
ksi
TAUR2
ksi
0.000 0.000 1.200 0.000 0.000 0.000 0.000
7.878 11.943 1.200 4.400 0.000 0.000 0.000
15.756 23.886 1.200 8.800 0.000 0.000 0.000
23.635 35.829 1.200 13.200 0.000 0.000 0.000
31.513 47.772 1.200 17.600 0.000 0.000 0.000
39.391 59.715 1.200 22.000 0.000 0.000 0.000
47.269 68.101 1.000 22.000 2.346 1.092 –3.090
55.147 72.366 0.857 22.000 7.411 2.957 –4.661
63.025 74.761 0.750 22.000 13.710 4.786 –5.543
70.904 76.207 0.667 22.000 20.634 6.402 –6.076
max 75 kip in.
T
  
78.782 77.132 0.600 22.000 27.902 7.792 –6.417
86.660 77.751 0.545 22.000 35.372 8.980 –6.645
94.538 78.181 0.500 22.000 42.967 9.999 –6.803
102.416 78.488 0.462 22.000 50.642 10.878 –6.916
110.294 78.714 0.429 22.000 58.371 11.643 –6.999
118.173 78.883 0.400 22.000 66.138 12.313 –7.062
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439
PROBLEM 3.C4 (Continued)
Problem 3.116
PHIM
deg
TM
kN  m
RY
mm
TAUM
MPa
PHIP
deg
TAUR1
MPa
TAUR2
MPa
0.000 0.000 16.000 0.000 0.000 0.000 0.000
0.807 0.187 16.000 29.000 0.000 0.000 0.000
1.614 0.373 16.000 58.000 0.000 0.000 0.000
2.421 0.560 16.000 87.000 0.000 0.000 0.000
3.228 0.746 16.000 116.000 0.000 0.000 0.000
4.036 0.933 16.000 145.000 0.000 0.000 0.000
4.843 1.064 13.333 145.000 0.240 7.198 –20.363
5.650 1.131 11.429 145.000 0.759 19.486 –30.719
6.457 1.168 10.000 145.000 1.405 31.542 –36.533 max 6

  
7.264 1.191 8.889 145.000 2.114 42.197 –40.046
8.071 1.205 8.000 145.000 2.859 51.354 –42.292
8.878 1.215 7.273 145.000 3.624 59.184 –43.794
9.685 1.221 6.667 145.000 4.402 65.901 –44.837
10.492 1.226 6.154 145.000 5.188 71.699 –45.583
11.300 1.230 5.714 145.000 5.980 76.739 –46.132
12.107 1.232 5.333 145.000 6.776 81.152 –46.543

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440
L
A
2c
B
c
B
L
A
2c
r1
L/n
ri
rn
T
A
T
c
PROBLEM 3.C5
The exact expression is given in Prob. 3.158 for the angle
of twist of the solid tapered shaft AB when a torque T is
applied as shown. Derive an approximate expression
for the angle of twist by replacing the tapered shaft by
n cylindrical shafts of equal length and of radius
1
2
( )( / ),
  
i
r n i c n where 1, 2, . . . , .

i n Using for T, L,
G, and c values of your choice, determine the percentage
error in the approximate expression when (a) 4,
n 
(b) 8,
n  (c) n  20, (d) 100.
n 
SOLUTION
From Problem 3.158, exact expression:
4
7
12
TL
Gc



or 4 4
7
0.18568
12
TL TL
Gc Gc


  
 
  
  
Consider typical ith shaft:
Enter unit values of T, L, G, and c.
(Note: Specific values can be entered).
Enter initial value of zero for .

Enter number cylindrical shafts.

n
For 1 to ,
i n
 update .

  
  
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441
PROBLEM 3.C5 (Continued)
Program Output
Coefficient of 4
/ :
TL Gc
Exact coefficient from Problem 3.158 is 0.18568.
Number of elemental disks .
 n
n Approximate Exact Percent Error
4 0.17959 0.18568 –3.28185
8 0.18410 0.18568 –0.85311
20 0.18542 0.18568 –0.13810
100 0.18567 0.18568 –0.00554

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442
t
L
A
c
2c
B
T
PROBLEM 3.C6
A torque T is applied as shown to the long, hollow, tapered shaft AB of
uniform thickness t. Derive an approximate expression for the angle of twist
by replacing the tapered shaft by n cylindrical rings of equal length and of
radius 1
2
( )( / ),
i
r n i c n
   where 1, 2, . . . , .
i n
 Using for T, L, G, c and t
values of your choice, determine the percentage error in the approximate
expression when (a) 4,
n  (b) 8,
n  (c) 20,
n  (d) 100.
n 
SOLUTION
Since the shaft is long, c L, the angle  is small, and we can use t as the thickness of the n cylindrical
rings.
For ,
tan
c L
zc c c
L L

  
 
1
2
i
c
r n i
n
  
  
  
  
2 2 3
(Area) (2 ) 2
( / )
i i i i i
i
J r rt r tr
T L n
GJ
 

  
 
Enter unit values for T, L, G, t, and c.
(Note: Specific values can be entered is desired).
Enter initial value of zero for .

Enter number of cylindrical rings.

n
For 1 to ,
i n
 update .

  
  
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443
PROBLEM 3.C6 (Continued)
Program Output
Coefficient of 3
/ :
TL Gtc
Exact coefficient from Problem 3.153 is 0.05968.
Number of elemental disks .
 n
n Approximate Exact Percent Error
4 0.058559 0.059683 –1.883078
8 0.059394 0.059683 –0.483688
20 0.059637 0.059683 –0.078022
100 0.059681 0.059683 –0.003127

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483. Mecánica de Materiales - Ferdinand P. Beer 7ma. Edición.

484. C
CH
HA
AP
PT
TE
ER
R 4
4
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447
M 5 15 kN · m
Dimensions in mm
B
20 40 20
20
20
80
A
PROBLEM 4.1
Knowing that the couple shown acts in a vertical plane, determine
the stress at (a) point A, (b) point B.
SOLUTION
For rectangle: 3
1
12
I bh
Outside rectangle: 3
1
1
(80)(120)
12
I
6 4 6 4
1 11.52 10 mm 11.52 10 m
I
u u
Cutout: 3
2
1
(40)(80)
12
I
6 4 6 4
2 1.70667 10 mm 1.70667 10 m
I
u u
Section: 6 4
1 2 9.81333 10 m
I I I
u
(a) 40 mm 0.040 m
A
y
3
6
6
(15 10 )(0.040)
61.6 10 Pa
9.81333 10
A
A
My
I
V
u
u
u
61.6 MPa
A
V W
(b) 60 mm 0.060 m
B
y
3
6
6
(15 10 )( 0.060)
91.7 10 Pa
9.81333 10
B
B
My
I
V
u
u
u
91.7 MPa
B
V W
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448
2 in.
2 in.
1.5 in.
2 in.
2 in.
2 in.
A
B
M ! 25 kip · in.
PROBLEM 4.2
Knowing that the couple shown acts in a vertical plane, determine
the stress at (a) point A, (b) point B.
SOLUTION
For rectangle: 3
1
12
I bh
For cross sectional area:
3 3 3 4
1 2 3
1 1 1
(2)(1.5) (2)(5.5) (2)(1.5) 28.854 in
12 12 12
I I I I
(a) 2.75 in.
A
y
(25)(2.75)
28.854
A
A
My
I
V 2.38 ksi
A
V W
(b) 0.75 in.
B
y
(25)(0.75)
28.854
B
B
My
I
V 0.650 ksi
B
V W
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449
200 mm
220 mm
12 mm
12 mm
8 mm
C x
y
M
PROBLEM 4.3
Using an allowable stress of 155 MPa, determine the largest bending
moment M that can be applied to the wide-flange beam shown. Neglect the
effect of fillets.
SOLUTION
Moment of inertia about x-axis:
3 2
1
6 4
3 6 4
2
6 4
3 1
1
(200)(12) (200)(12)(104)
12
25.9872 10 mm
1
(8)(196) 5.0197 10 mm
12
25.9872 10 mm
u
u
u
I
I
I I
6 4 6 4
1 2 3
6
6 6
3
56.944 10 mm 56.944 10 m
1
with (220) 110 mm 0.110 m
2
with 155 10 Pa
(56.944 10 )(155 10 )
80.2 10 N m
0.110
u u
u
u u
u ˜
x
I I I I
Mc
c
I
I
M
c
M
V
V
V
80.2 kN m
˜
x
M W
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450
200 mm
220 mm
12 mm
12 mm
8 mm
C x
y
M
PROBLEM 4.4
Solve Prob. 4.3, assuming that the wide-flange beam is bent about the y axis
by a couple of moment My.
PROBLEM 4.3. Using an allowable stress of 155 MPa, determine the
largest bending moment M that can be applied to the wide-flange beam
shown. Neglect the effect of fillets.
SOLUTION
Moment of inertia about y axis:
3 6 4
1
3 3 4
2
6 4
3 1
1
(12)(200) 8 10 mm
12
1
(196)(8) 8.3627 10 mm
12
8 10 mm
u
u
u
I
I
I I
6 4 6 4
1 2 3
6
6 6
3
16.0084 10 mm 16.0084 10 m
1
with (200) 100 mm 0.100 m
2
with 155 10 Pa
(16.0084 10 )(155 10 )
24.8 10 N m
0.100
u u
u
u u
u ˜
y
y
I I I I
Mc
c
I
I
M
c
M
V
V
V
24.8 kN m
˜
y
M W
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451
M2
M1
0.1 in.
0.2 in.
0.5 in.
0.5 in.
(a)
(b)
PROBLEM 4.5
Using an allowable stress of 16 ksi, determine the largest couple that can be
applied to each pipe.
SOLUTION
(a) 4 4 4 4 3 4
3
(0.6 0.5 ) 52.7 10 in
4 4
0.6 in.
(16)(52.7 10 )
:
0.6
o i
I r r
c
Mc I
M
I c
S S
V
V
u
u
1.405 kip in.
˜
M W
(b) 4 4 3 4
3
(0.7 0.5 ) 139.49 10 in
4
0.7 in.
(16)(139.49 10 )
:
0.7
I
c
Mc I
M
I c
S
V
V
u
u
3.19 kip in.
˜
M W
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452
120 mm
30 mm
30 mm
M = 2.8 kN · m
r 5 20 mm
A
B
PROBLEM 4.6
Knowing that the couple shown acts in a vertical plane,
determine the stress at (a) point A, (b) point B.
SOLUTION
3 4
6 4
1 1
(0.120 m)(0.06 m) 2 (0.02 m)
12 12 4
2.1391 10 mm
ª º
˜
« »
¬ ¼
u
I
S
(a)
3
6 4
(2.8 10 N m)(0.03 m)
2.1391 10 mm
u ˜
u
A
A
M y
I
V
39.3 MPa
A
V W
(b)
3
6 4
(2.8 10 N m)(0.02 m)
2.1391 10 m
u ˜
u
B
B
M y
I
V
26.2 MPa
B
V W
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453
y
z C
PROBLEM 4.7
Two W4 13
u rolled sections are welded together as shown. Knowing that for the steel
alloy used 36 ksi
Y
V and 58 ksi
U
V and using a factor of safety of 3.0, determine the
largest couple that can be applied when the assembly is bent about the z axis.
SOLUTION
Properties of W4 u 13 rolled section.
(See Appendix C.)
2
4
Area 3.83 in
Width 4.060 in.
3.86 in
y
I
For one rolled section, moment of inertia about axis b-b is
2 2 4
3.86 (3.83)(2.030) 19.643 in
b y
I I Ad
For both sections, 4
2 39.286 in
width 4.060 in.
z b
I I
c
all
all
all
58
19.333 ksi
. . 3.0
(19.333)(39.286)
4.060
V
V V
V
U Mc
F S I
I
M
c
all 187.1 kip in.
˜
M W
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454
y
z
C
PROBLEM 4.8
Two W4 13
u rolled sections are welded together as shown. Knowing that for the steel
alloy used 58 ksi
U
V and using a factor of safety of 3.0, determine the largest couple
that can be applied when the assembly is bent about the z axis.
SOLUTION
Properties of W4 u 13 rolled section.
(See Appendix C.)
2
4
Area 3.83 in
Depth 4.16 in.
11.3 in
x
I
For one rolled section, moment of inertia about axis a-a is
2 2 4
11.3 (3.83)(2.08) 27.87 in
a x
I I Ad
For both sections, 4
2 55.74 in
depth 4.16 in.
z a
I I
c
all
all
all
58
19.333 ksi
. . 3.0
(19.333)(55.74)
4.16
V
V V
V
U Mc
F S I
I
M
c
all 259 kip in.
˜
M W
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455
D
C
B
A
6 in.
2 in.
3 in.
3 in.
15 kips 15 kips
3 in.
40 in. 40 in.
60 in.
PROBLEM 4.9
Two vertical forces are applied to a beam of the cross section shown. Determine
the maximum tensile and compressive stresses in portion BC of the beam.
SOLUTION
A 0
y 0
A y
c 18 5 90
d 18 1 18
6 36 108
0
108
3 in.
36
Y
Neutral axis lies 3 in. above the base.
3 2 3 2 4
1 1 1 1 1
3 2 3 2 4
2 2 2 2 2
4
1 2
1 1
(3)(6) (18)(2) 126 in
12 12
1 1
(9)(2) (18)(2) 78 in
12 12
126 78 204 in
I b h A d
I b h A d
I I I
top bot
5 in. 3 in.
y y
0
(15)(40) 600 kip in.
M Pa
M Pa
˜
top
top
(600)(5)
204
M y
I
V top 14.71 ksi (compression)
V W
bot
bot
(600)( 3)
204
M y
I
V
bot 8.82 ksi (tension)
V W
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456
D
C
B
A
25 kips 25 kips
20 in. 20 in.
60 in.
4 in.
1 in.
1 in.
1 in.
6 in.
8 in. PROBLEM 4.10
Two vertical forces are applied to a beam of the cross section shown.
Determine the maximum tensile and compressive stresses in portion
BC of the beam.
SOLUTION
A 0
y 0
A y
c 8 7.5 60
d 6 4 24
e 4 0.5 2
6 18 86
86
4.778 in.
18
o
Y
Neutral axis lies 4.778 in. above the base.
3 2 3 2 4
1 1 1 1 1
3 2 3 2 4
2 2 2 2 2
3 2 3 2 4
3 3 3 3 3
4
1 2 3
top b
1 1
(8)(1) (8)(2.772) 59.94 in
12 12
1 1
(1)(6) (6)(0.778) 21.63 in
12 12
1 1
(4)(1) (4)(4.278) 73.54 in
12 12
59.94 21.63 73.57 155.16 in
3.222 in.
I b h A d
I b h A d
I b h A d
I I I I
y y
ot 4.778 in.
0
(25)(20) 500 kip in.
˜
M Pa
M Pa
top
top
(500)(3.222)
155.16
My
I
V top 10.38 ksi (compression)
V W
bot
bot
(500)( 4.778)
155.16
My
I
V
bot 15.40 ksi (tension)
V W
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457
10 mm 10 mm
50 mm
10 mm
150 mm 150 mm
A D
B C
10 kN 10 kN
250 mm
50 mm
PROBLEM 4.11
Two vertical forces are applied to a beam of
the cross section shown. Determine the
maximum tensile and compressive stresses
in portion BC of the beam.
SOLUTION
2
, mm
A 0 , mm
y 3
0 , mm
Ay
c 600 30 3
18 10
u
d 600 30 3
18 10
u
e 300 5 3
1.5 10
u
1500 3
37.5 10
u
3
0
37.5 10
25mm
1500
Y
u
Neutral axis lies 25 mm above the base.
3 2 3 4 4
1 2 1
3 2 3 4
3
3 4 9 4
1 2 3
1
(10)(60) (600)(5) 195 10 mm 195 mm
12
1
(30)(10) (300)(20) 122.5 10 mm
12
512.5 10 mm 512.5 10 m
I I I
I
I I I I
u
u
u u
top bot
3
3 3
35 mm 0.035 m 25 mm 0.025 m
150 mm 0.150 m 10 10 N
(10 10 )(0.150) 1.5 10 N m
u
u u ˜
y y
a P
M Pa
3
top 6
top 9
(1.5 10 )(0.035)
102.4 10 Pa
512.5 10
My
I
V
u
u
u
top 102.4 MPa (compression)
V W
3
6
bot
bot 9
(1.5 10 )( 0.025)
73.2 10 Pa
512.5 10
M y
I
V
u
u
u
bot 73.2 MPa(tension)
V W
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458
72 mm
216 mm
36 mm
54 mm
108 mm
y
z C
PROBLEM 4.12
Knowing that a beam of the cross section shown is bent about a horizontal
axis and that the bending moment is 6 kN ˜ m, determine the total force
acting on the shaded portion of the web.
SOLUTION
The stress distribution over the entire cross section is given by the bending stress formula:
x
My
I
V
where y is a coordinate with its origin on the neutral axis and I is the moment of inertia of the entire cross
sectional area. The force on the shaded portion is calculated from this stress distribution. Over an area
element dA, the force is
x
My
dF dA dA
I
V
The total force on the shaded area is then
* *
My M M
F dF dA y dA y A
I I I
³ ³ ³
where *
y is the centroidal coordinate of the shaded portion and A*
is its area.
1
2
54 18 36 mm
54 36 54 36 mm
d
d
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459
PROBLEM 4.12 (Continued)
Moment of inertia of entire cross section:
3 2 3 2 6 4
1 1 1 1 1
3 2 3 2 6 4
2 2 2 2 2
6 4 6 4
1 2
1 1
(216)(36) (216)(36)(36) 10.9175 10 mm
12 12
1 1
(72)(108) (72)(108)(36) 17.6360 10 mm
12 12
28.5535 10 mm 28.5535 10 m
I b h A d
I b h A d
I I I
u
u
u u
For the shaded area,
* 2
*
* * 3 3 6
* * 3 6
6
3
(72)(90) 6480 mm
45 mm
291.6 10 mm 291.6 10 m
(6 10 )(291.6 10 )
28.5535 10
61.3 10 N
A
y
A y
MA y
F
I
u u
u u
u
u 61.3 kN
F W
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460
24 mm
12 mm 12 mm
20 mm 20 mm
20 mm
20 mm
24 mm
z
y
C
PROBLEM 4.13
Knowing that a beam of the cross section shown is bent about a horizontal axis
and that the bending moment is 4 kN m,
˜ determine the total force acting on
the shaded portion of the beam.
SOLUTION
Dimensions in mm:
3 3
6 6
6 4 6 4
1 1
(12 12)(88) (40)(40)
12 12
1.3629 10 0.213 10
1.5763 10 mm 1.5763 10 m
u u
u u
z
I
For use in Prob. 4.14,
3 3
6 6
6 4 6 4
1 1
(88)(64) (24 24)(40)
12 12
1.9224 10 0.256 10
1.6664 10 mm 1.6664 10 m
u u
u u
y
I
Bending about horizontal axis. 4 kN m
z
M ˜
6 4
6 4
(4 kN m)(0.044 m)
111.654 MPa
1.5763 10 m
(4 kN m)(0.020 m)
50.752 MPa
1.5763 10 m
z
A
z
z
B
z
M c
I
M c
I
V
V
˜
u
˜
u
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461
PROBLEM 4.13 (Continued)
Portion (1): 2 6 2
(44)(12) 528 mm 528 10 m
A
u
avg
1 1
(111.654) 55.83 MPa
2 2
A
V V
6 2
1 avg
Force (55.83 MPa)(528 10 m ) 29.477 kN
u
A
V
Portion (2): 2 6 2
(20)(20) 400 mm 400 10 m
A
u
avg
1 1
(50.752) 25.376 MPa
2 2
B
V V
6 2
2 avg
Force (25.376 MPa)(400 10 m ) 10.150 kN
u
A
V
Total force on shaded area 29.477 10.150 39.6 kN
W
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462
24 mm
12 mm 12 mm
20 mm 20 mm
20 mm
20 mm
24 mm
z
y
C
PROBLEM 4.14
Solve Prob. 4.13, assuming that the beam is bent about a vertical axis by a
couple of moment 4 kN m.
˜
PROBLEM 4.13. Knowing that a beam of the cross section shown is bent
about a horizontal axis and that the bending moment is 4 kN m,
˜ determine the
total force acting on the shaded portion of the beam.
SOLUTION
Bending about vertical axis. 4 kN m
˜
y
M
See Prob. 4.13 for sketch and 6 4
1.6664 10 m
u
y
I
6 4
6 4
(4 kN m)(0.032 m)
76.81MPa
1.6664 10 m
(4 kN m)(0.020 m)
48.01MPa
1.6664 10 m
V
V
˜
u
˜
u
D
y
E
y
Mc
I
Mc
I
Portion (1): 2 6 2
(44)(12) 528 mm 528 10 m
A
u
avg
1 1
( ) (76.81 48.01) 62.41MPa
2 2
D E
V V V
6 2
1 avg
Force (62.41MPa)(528 10 m ) 32.952 kN
u
A
V
Portion (2): 2 6 2
(20)(20) 400 mm 400 10 m
u
A
avg
1 1
(48.01) 24.005 MPa
2 2
E
V V
6 2
2 avg
Force (24.005 MPa)(400 10 m ) 9.602 kN
u
A
V
Total force on shaded area 32.952 9.602 42.6 kN
W
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463
M
1.5 in.
0.5 in.
1.5 in. 1.5 in.
0.5 in. 0.5 in.
0.5 in.
PROBLEM 4.15
Knowing that for the extruded beam shown the allowable stress is 12 ksi
in tension and 16 ksi in compression, determine the largest couple M that
can be applied.
SOLUTION
A 0
y 0
Ay
c 2.25 1.25 2.8125
d 2.25 0.25 0.5625
4.50 3.375
3.375
0.75 in.
4.50
Y
The neutral axis lies 0.75 in. above bottom.
top bot
3 2 3 2 4
1 1 1 1 1
2 2 3 2 4
2 2 2 2 2
4
1 2
2.0 0.75 1.25 in., 0.75 in.
1 1
(1.5)(1.5) (2.25)(0.5) 0.984375 in
12 12
1 1
(4.5)(0.5) (2.25)(0.5) 0.609375 in
12 12
1.59375 in
y y
I b h A d
I b h A d
I I I
My I
M
I y
V
V
Top: (compression)
(16)(1.59375)
20.4 kip in.
1.25
M ˜
Bottom: (tension)
(12)(1.59375)
25.5 kip in.
0.75
M ˜
Choose the smaller as Mall. all 20.4 kip in.
M ˜ W
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464
M
15 mm
d ! 30 mm
20 mm
40 mm PROBLEM 4.16
The beam shown is made of a nylon for which the allowable stress is
24 MPa in tension and 30 MPa in compression. Determine the largest
couple M that can be applied to the beam.
SOLUTION
2
, mm
A 0 , mm
y 3
0 , mm
Ay
c 600 22.5 3
13.5 10
u
d 300 7.5 3
2.25 10
u
Σ 900 3
15.75 10
u
3
0
15.5 10
17.5 mm The neutral axis lies 17.5 mm above the bottom.
900
Y
u
top
bot
3 2 3 2 3 4
1 1 1 1 1
3 2 3 2 3 4
2 2 2 2 2
3 4 9 4
1 2
30 17.5 12.5 mm 0.0125 m
17.5 mm 0.0175 m
1 1
(40)(15) (600)(5) 26.25 10 mm
12 12
1 1
(20)(15) (300)(10) 35.625 10 mm
12 12
61.875 10 mm 61.875 10 m
y
y
I b h A d
I b h A d
I I I
u
u
u u
| |
My I
M
I y
V
V
Top: (tension side)
6 9
(24 10 )(61.875 10 )
118.8 N m
0.0125
M
u u
˜
Bottom: (compression)
6 9
(30 10 )(61.875 10 )
106.1 N m
0.0175
M
u u
˜
Choose smaller value. 106.1 N m
M ˜ W
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465
M
15 mm
d ! 30 mm
20 mm
40 mm PROBLEM 4.17
Solve Prob. 4.16, assuming that 40 mm.
d
PROBLEM 4.16 The beam shown is made of a nylon for which the
allowable stress is 24 MPa in tension and 30 MPa in compression.
Determine the largest couple M that can be applied to the beam.
SOLUTION
2
, mm
A 0 , mm
y 3
0 , mm
Ay
c 600 32.5 3
19.5 10
u
d 500 12.5 3
6.25 10
u
Σ 1100 3
25.75 10
u
3
0
25.75 10
23.41 mm The neutral axis lies 23.41 mm above the bottom.
1100
Y
u
top
bot
3 2 3 2 3 4
1 1 1 1 1
2 2 3 2 3 4
2 2 2 2 2
3 4
1 2
40 23.41 16.59 mm 0.01659 m
23.41 mm 0.02341 m
1 1
(40)(15) (600)(9.09) 60.827 10 mm
12 12
1 1
(20)(25) (500)(10.91) 85.556 10 mm
12 12
146.383 10 mm 146.3
y
y
I b h A d
I b h A d
I I I
u
u
u 9 4
83 10 m
u
| |
My I
M
I y
V
V
Top: (tension side)
6 9
(24 10 )(146.383 10 )
212 N m
0.01659
M
u u
˜
Bottom: (compression)
6 9
(30 10 )(146.383 10 )
187.6 N m
0.02341
M
u u
˜
Choose smaller value. 187.6 N m
M ˜ W
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466
1.2 in.
0.75 in.
2.4 in.
M
PROBLEM 4.18
Knowing that for the beam shown the allowable stress is 12 ksi in tension and
16 ksi in compression, determine the largest couple M that can be applied.
SOLUTION
c rectangle d semi-circular cutout
2
1
2 2
2
2
1
2
(2.4)(1.2) 2.88 in
(0.75) 0.8836 in
2
2.88 0.8836 1.9964 in
0.6 in.
4 (4)(0.75)
0.3183 in.
3 3
(2.88)(0.6) (0.8836)(0.3183)
0.7247 in.
1.9964
A
A
A
y
r
y
Ay
Y
A
S
S S
6
6
Neutral axis lies 0.7247 in. above the bottom.
Moment of inertia about the base:
3 4 3 4 4
1 1
(2.4)(1.2) (0.75) 1.25815 in
3 8 3 8
b
I bh r
S S
Centroidal moment of inertia:
2 2 4
top
bot
1.25815 (1.9964)(0.7247) 0.2097 in
1.2 0.7247 0.4753 in.,
0.7247 in.
b
I I AY
y
y
| |
V
V
M y I
M
I y
Top: (tension side)
(12)(0.2097)
5.29 kip in.
0.4753
M ˜
Bottom: (compression)
(16)(0.2097)
4.63 kip in.
0.7247
M ˜
Choose the smaller value. 4.63 kip in.
M ˜ W
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467
54 mm
40 mm
80 mm
M
PROBLEM 4.19
Knowing that for the extruded beam shown the allowable stress is
120 MPa in tension and 150 MPa in compression, determine the largest
couple M that can be applied.
SOLUTION
2
, mm
A 0 , mm
y 3
0 , mm
Ay d, mm
c 2160 27 58,320 3
d 1080 36 38,880 3
Σ 3240 97,200
97,200
30 mm The neutral axis lies 30 mm above the bottom.
3240
Y
top bot
3 2 3 2 3 4
1 1 1 1 1
2 2 3 2 3 4
2 2 2 2 2
3 4 9 4
1 2
54 30 24 mm 0.024 m 30 mm 0.030 m
1 1
(40)(54) (40)(54)(3) 544.32 10 mm
12 12
1 1 1
(40)(54) (40)(54)(6) 213.84 10 mm
36 36 2
758.16 10 mm 758.16 10 m
y y
I b h A d
I b h A d
I I I
u
u
u u
| | | |
V
V
M y I
M
I y
Top: (tension side)
6 9
3
(120 10 )(758.16 10 )
3.7908 10 N m
0.024
M
u u
u ˜
Bottom: (compression)
6 9
3
(150 10 )(758.16 10 )
3.7908 10 N m
0.030
M
u u
u ˜
Choose the smaller as Mall. 3
all 3.7908 10 N m
M u ˜ all 3.79 kN m
M ˜ W
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468
M
48 mm
48 mm
48 mm
36 mm
36 mm
PROBLEM 4.20
Knowing that for the extruded beam shown the allowable stress is
120 MPa in tension and 150 MPa in compression, determine the largest
couple M that can be applied.
SOLUTION
2
, mm
A 0 , mm
y 3
0 , mm
Ay
c Solid rectangle 4608 48 221,184
d Square cutout –1296 30 –38,880
Σ 3312 182,304
182,304
55.04 mm
3312
Y Neutral axis lies 55.04 mm above bottom.
top
bot
3 2 3 2 6 4
1 1 1 1 1
3 2 3 2 6 4
2 2 2 2 2
6 4
1 2
96 55.04 40.96 mm 0.04096 m
55.04 mm 0.05504 m
1 1
(48)(96) (48)(96)(7.04) 3.7673 10 mm
12 12
1 1
(36)(36) (36)(36)(25.04) 0.9526 10 mm
12 12
2.8147 10 mm
u
u
u
y
y
I b h A d
I b h A d
I I I 6 4
2.8147 10 m
u
| | ?
M y I
M
I y
V
V
Top: (tension side)
6 6
3
(120 10 )(2.8147 10 )
8.25 10 N m
0.04096
u u
u ˜
M
Bottom: (compression)
6 6
3
(150 10 )(2.8147 10 )
7.67 10 N m
0.05504
u u
u ˜
M
Mall is the smaller value. 3
7.67 10 N m
M u ˜ 7.67 kN m
˜ W
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469
PROBLEM 4.21
Straight rods of 6-mm diameter and 30-m length are stored by coiling the
rods inside a drum of 1.25-m inside diameter. Assuming that the yield
strength is not exceeded, determine (a) the maximum stress in a coiled rod,
(b) the corresponding bending moment in the rod. Use 200 GPa.
E
SOLUTION
Let inside diameter of the drum,
1
diameter of rod, ,
2
radius of curvature of center line of rods when bent.
U
D
d c d
3
4 4 12 4
1 1 1 1
(1.25) (6 10 ) 0.622 m
2 2 2 2
(0.003) 63.617 10 m
4 4
D d
I c
U
S S
u
u
(a)
9
6
max
(200 10 )(0.003)
965 10 Pa
0.622
Ec
V
U
u
u 965 MPa
V W
(b)
9 12
(200 10 )(63.617 10 )
20.5 N m
0.622
EI
M
U
u u
˜ 20.5 N m
M ˜ W
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470
900 mm
8 mm
t
r
M
M' PROBLEM 4.22
A 900-mm strip of steel is bent into a full circle by two couples applied
as shown. Determine (a) the maximum thickness t of the strip if the
allowable stress of the steel is 420 MPa, (b) the corresponding moment
M of the couples. Use 200 GPa.
E
SOLUTION
When the rod is bent into a full circle, the circumference is 900 mm. Since the circumference is equal to
2S times U, the radius of curvature, we get
900 mm
143.24 mm 0.14324 m
2
U
S
Stress: or
Ec
E c
E
UV
V H
U
For 420 MPa
V and 200 GPa,
E
6
3
9
(0.14324)(420 10 )
0.3008 10 m
200 10
c
u
u
u
(a) Maximum thickness: 3
2 0.6016 10 m
t c
u
0.602 mm
t W
Moment of inertia for a rectangular section.
3 3 3 3
15 4
(8 10 )(0.6016 10 )
145.16 10 m
12 12
bt
I
u u
u
(b) Bending moment:
EI
M
U
9 15
(200 10 )(145.16 10 )
0.203 N m
0.14324
M
u u
˜ 0.203 N m
M ˜ W
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471
5 ft
PROBLEM 4.23
Straight rods of 0.30-in. diameter and 200-ft length are sometimes
used to clear underground conduits of obstructions or to thread wires
through a new conduit. The rods are made of high-strength steel and,
for storage and transportation, are wrapped on spools of 5-ft
diameter. Assuming that the yield strength is not exceeded,
determine (a) the maximum stress in a rod, when the rod, which is
initially straight, is wrapped on a spool, (b) the corresponding
bending moment in the rod. Use 6
29 10 psi
E u .
SOLUTION
Radius of cross section:
1 1
(0.30) 0.15 in.
2 2
r d
Moment of inertia: 4 4 6 4
(0.15) 397.61 10 in
4 4
I r
S S
u
1
5 ft 60 in. 30 in.
2
0.15 in.
D D
c r
U
(a)
6
3
max
(29 10 )(0.15)
145.0 10 psi
30
Ec
V
U
u
u max 145.0 ksi
V W
(b)
6 6
(29 10 )(397.61 10 )
30
EI
M
U
u u
384 lb in.
M ˜ W
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472
20 mm
12 mm
60 N · m
z
y
PROBLEM 4.24
A 60-N ˜ m couple is applied to the steel bar shown. (a) Assuming that
the couple is applied about the z axis as shown, determine the maximum
stress and the radius of curvature of the bar. (b) Solve part a, assuming
that the couple is applied about the y axis. Use 200 GPa.
E
SOLUTION
(a) Bending about z-axis.
3 3 3 4 9 4
1 1
(12)(20) 8 10 mm 8 10 m
12 12
20
10 mm 0.010 m
2
u u
I bh
c
6
9
(60)(0.010)
75.0 10 Pa
8 10
Mc
I
V
u
u
75.0 MPa
V W
3 1
9 9
1 60
37.5 10 m
(200 10 )(8 10 )
M
EI
U
u
u u
26.7 m
U W
(b) Bending about y-axis.
3 3 3 4 9 4
6
9
1 1
(20)(12) 2.88 10 mm 2.88 10 m
12 12
12
6 mm 0.006 m
2
(60)(0.006)
125.0 10 Pa
2.88 10
I bh
c
Mc
I
V
u u
u
u
125.0 MPa
V W
3 1
9 9
1 60
104.17 10 m
(200 10 )(2.88 10 )
M
EI
U
u
u u
9.60 m
U W
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473
C
80 mm
5 mm 5 mm
10 mm
10 mm
80 mm
M
PROBLEM 4.25
(a) Using an allowable stress of 120 MPa, determine the largest couple
M that can be applied to a beam of the cross section shown. (b) Solve
part a, assuming that the cross section of the beam is an 80-mm square.
SOLUTION
(a) 1 2
4
I I I , where I1 is the moment of inertia of an 80-mm square and I2 is the moment of inertia of
one of the four protruding ears.
3 3 6 4
1
1 1
(80)(80) 3.4133 10 mm
12 12
I bh u
3 2 3 2 3 4
2
6 4 6 4
1 2
6 6
1 1
(5)(10) (5)(10)(45) 101.667 10 mm
12 12
4 3.82 10 mm 3.82 10 mm , 50 mm 0.050 m
(120 10 )(3.82 10 )
0.050
u
u u
u u
?
I bh Ad
I I I c
Mc I
M
I c
V
V
3
9.168 10 N m
u ˜
9.17 kN m
˜ W
(b) Without the ears:
6 2
1 3.4133 10 m , 40 mm 0.040 m
u
I I c
6 6
3
(120 10 )(3.4133 10 )
10.24 10 N m
0.040
V
u u
u ˜
I
M
c
10.24 kN m
˜ W
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474
0.75 in.
0.2 in.
1.5 in.
0.1 in.
M
PROBLEM 4.26
A thick-walled pipe is bent about a horizontal axis by a couple M. The
pipe may be designed with or without four fins. (a) Using an allowable
stress of 20 ksi, determine the largest couple that may be applied if the
pipe is designed with four fins as shown. (b) Solve part a, assuming that
the pipe is designed with no fins.
SOLUTION
x
I of hollow pipe: 4 4 4
(1.5 in.) (0.75 in.) 3.7276 in
4
ª º
¬ ¼
x
I
S
x
I of fins: 3 2 3
1 1
2 (0.1)(0.2) (0.1 0.2)(1.6) 2 (0.2)(0.1)
12 12
ª º ª º
u
« » « »
¬ ¼ ¬ ¼
x
I
4
0.1026 in
(a) Pipe as designed, with fins:
4
3.8302 in , 1.7 in.
x
I c
4
all all
3.8302 in
20 ksi, (20 ksi)
1.7 in.
V V x
I
M
c
45.1 kip in.
˜
M W
(b) Pipe with no fins:
4
all 20 ksi, 3.7276 in , 1.5 in.
x
I c
V
4
all
3.7276 in
(20 ksi)
1.5 in.
V x
I
M
c
49.7 kip in.
˜
M W
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475
b
d
M'
M
PROBLEM 4.27
A couple M will be applied to a beam of rectangular cross section that
is to be sawed from a log of circular cross section. Determine the ratio
d/b for which (a) the maximum stress Vm will be as small as possible,
(b) the radius of curvature of the beam will be maximum.
SOLUTION
Let D be the diameter of the log.
2 2 2 2 2 2
3 2
1 1 1
12 2 6
D b d d D b
I
I bd c d bd
c
(a) Vm is the minimum when
I
c
is maximum.
2 2 2 3
2 2
1 1 1
( )
6 6 6
1 3 1
0
6 6 3
I
b D b D b b
c
d I
D b b D
db c
§ ·
¨ ¸
© ¹
2 2
1 2
3 3
d D D D
2
d
b
W
(b)
EI
M
U
U is maximum when I is maximum, 3
1
12
bd is maximum, or 2 6
b d is maximum.
2 2 6
( )
D d d
is maximum.
2 5 7
6 8 0
D d d
3
2
d D
2 2
3 1
4 2
b D D D
3
d
b
W
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476
h
h
C
h0
h0
M
PROBLEM 4.28
A portion of a square bar is removed by milling, so that its cross section
is as shown. The bar is then bent about its horizontal axis by a couple M.
Considering the case where h 0.9h0, express the maximum stress in the
bar in the form 0 ,
m k
V V where 0
V is the maximum stress that would
have occurred if the original square bar had been bent by the same
couple M, and determine the value of k.
SOLUTION
1 2
3 3
0
4 3 3 3 4
0 0
2
3 4
0
0
4 2
1 1
(4) (2) (2 2 )( )
12 3
1 4 4 4
3 3 3 3
3
4 (4 3 )
3
I I I
hh h h h
h h h h h h h h
c h
Mc Mh M
I h h h
h h h
§ · § ·
¨ ¸ ¨ ¸
© ¹ © ¹
V
For the original square, 0 0
0 2 3
0 0 0 0
, .
3 3
(4 3 )
V
h h c h
M M
h h h h
3 3
0 0
2 2
0 0 0 0 0
0
0.950
(4 3 ) (4 (3)(0.9) )(0.9 )
0.950
V
V
V V
h h
h h h h h h
0.950
k W
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477
h
h
C
h0
h0
M
PROBLEM 4.29
In Prob. 4.28, determine (a) the value of h for which the maximum stress
m
V is as small as possible, (b) the corresponding value of k.
PROBLEM 4.28 A portion of a square bar is removed by milling, so that
its cross section is as shown. The bar is then bent about its horizontal
axis by a couple M. Considering the case where h 0.9h0, express the
maximum stress in the bar in the form 0 ,
m k
V V where 0
V is the
maximum stress that would have occurred if the original square bar had
been bent by the same couple M, and determine the value of k.
SOLUTION
1 2
3 3
0
4 3 3 3 4
0 0
2 3
0
4 2
1 1
(4) (2) (2 2 )
12 3
1 4 4 4
3 3 3 3
4
3
§ · § ·
¨ ¸ ¨ ¸
© ¹ © ¹
I I I
hh h h h
h h h h h h h
I
c h h h h
c
I
c
is maximum at 2 3
0
4
0.
3
d
h h h
dh
ª º
« »
¬ ¼
2
0
8
3 0
3
h h h
0
8
9
h h W
2 3
3
0 0 0 0 3
0
4 8 8 256 729
3 9 9 729 256
I Mc M
h h h h
c I h
V
§ · § ·
¨ ¸ ¨ ¸
© ¹ © ¹
For the original square, 3
0
0 0 0
0
1
3
I
h h c h h
c
0
0 2
0 0
3
Mc M
I h
V
0
729 1 729
0.949
256 3 768
V
V
˜ 0.949
k W
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478
PROBLEM 4.30
For the bar and loading of Concept Application 4.1, determine (a) the radius of curvature ,
U (b) the radius of
curvature Uc of a transverse cross section, (c) the angle between the sides of the bar that were originally
vertical. Use 6
29 10 psi
u
E and 0.29.
v
SOLUTION
From Example 4.01, 4
30 kip in. 1.042 in
M I
˜
(a)
3
6 1
6
1 (30 10 )
993 10 in.
(29 10 )(1.042)
u
u
u
M
EI
U
1007 in.
U W
(b) 1 vc c
v v
H H
U Uc
6 1 6 1
1 1
(0.29)(993 10 )in. 288 10 in.
v
U U
u u
c
3470 in.
Uc W
(c) 6
length of arc 0.8
230 10 rad
radius 3470
b
T
U
u
c
0.01320
T q W
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479
z
x
y
C
A
M
PROBLEM 4.31
A W200 31.3
u rolled-steel beam is subjected to a couple M of moment
45 kN ˜ m. Knowing that 200
E GPa and 0.29,
v determine (a) the
radius of curvature ,
U (b) the radius of curvature Uc of a transverse cross
section.
SOLUTION
For W 200 31.3
u rolled steel section,
6 4
6 4
31.3 10 mm
31.3 10 m
I
u
u
(a)
3
3 1
9 6
1 45 10
7.1885 10 m
(200 10 )(31.3 10 )
U
u
u
u u
M
EI
139.1 m
U W
(b) 3 3 1
1 1
(0.29)(7.1885 10 ) 2.0847 10 m
U U
u u
c
v 480 m
Uc W
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480
2
!
2
!
2
!
2
!
y
y # $c
y # %c
y
x
x
y
PROBLEM 4.32
It was assumed in Sec. 4.1B that the normal stresses y
V in a
member in pure bending are negligible. For an initially straight
elastic member of rectangular cross section, (a) derive an
approximate expression for y
V as a function of y, (b) show that
( max max
) ( /2 )( )
V U V
y x
c and, thus, that y
V can be neglected in
all practical situations. (Hint: Consider the free-body diagram of
the portion of beam located below the surface of ordinate y and
assume that the distribution of the stress x
V is still linear.)
SOLUTION
Denote the width of the beam by b and the length by L.
T
U
L
Using the free body diagram above, with cos 1
2
T
|
0: 2 sin 0
2
2 1
sin
2
T
V V
T T
V V V V
U
6
|
³
³ ³ ³
y
y y x
c
y y y
y x x x
c c c
F bL bdy
dy dy dy
L L
But, max
( )
x x
y
c
V V
(a)
2
max max
( ) ( )
2
V V
V
U U
³
y
y
x x
y
c
c
y
y dy
c c
2 2
max
( )
( )
2
x
y y c
c
V
V
U
W
The maximum value occurs at 0
y y
V .
(b)
2
max max
max
( ) ( )
( )
2 2
x x
y
c c
c
V V
V
U U
W
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481
30 mm
6 mm
6 mm
30 mm
Aluminum
Brass PROBLEM 4.33
A bar having the cross section shown has been formed by securely
bonding brass and aluminum stock. Using the data given below,
determine the largest permissible bending moment when the
composite bar is bent about a horizontal axis.
Aluminum Brass
Modulus of elasticity 70 GPa 105 GPa
Allowable stress 100 MPa 160 MPa
SOLUTION
Use aluminum as the reference material.
1.0 in aluminum
/ 105/70 1.5 in brass
b a
n
n E E
For the transformed section,
1
3 2
1
1 1 1 1 1
3 3 3 4
12
1.5
(30)(6) (1.5)(30)(6)(18) 88.29 10 mm
12
u
n
I b h n A d
3 3 3 4
2
2 2 2
3 4
3 1
3 4
1 2 3
1.0
(30)(30) 67.5 10 mm
12 12
88.29 10 mm
244.08 10 mm
u
u
u
n
I b h
I I
I I I I
9 4
244.08 10 m
u
nMy I
M
I ny
V
V
Aluminum: 6
1.0, 15 mm 0.015 m, 100 10 Pa
u
n y V
6 9
3
(100 10 )(244.08 10 )
1.627 10 N m
(1.0)(0.015)
M
u u
u ˜
Brass: 6
1.5, 21 mm 0.021 m, 160 10 Pa
u
n y V
6 9
3
(160 10 )(244.08 10 )
1.240 10 N m
(1.5)(0.021)
M
u u
u ˜
Choose the smaller value 3
1.240 10 N m
u ˜
M 1.240 kN m
˜ W
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Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
482
32 mm
32 mm
8 mm 8 mm
8 mm
8 mm
Aluminum
Brass
PROBLEM 4.34
A bar having the cross section shown has been formed by securely
bonding brass and aluminum stock. Using the data given below,
determine the largest permissible bending moment when the
composite bar is bent about a horizontal axis.
Aluminum Brass
Modulus of elasticity 70 GPa 105 GPa
Allowable stress 100 MPa 160 MPa
SOLUTION
Use aluminum as the reference material.
For aluminum, 1.0
n
For brass, / 105/70 1.5
b a
n E E
Values of n are shown on the sketch.
For the transformed section,
3 3 3 4
1
1 1 1
3 3 3 3 3 4
2
2 2 2 2
3 4
3 1
3 4 9 4
1 2 3
1.5
(8)(32) 32.768 10 mm
12 12
1.0
(32)(32 16 ) 76.459 10 mm
12 12
32.768 10 mm
141.995 10 mm 141.995 10 m
n
I b h
n
I b H h
I I
I I I I
u
u
u
u u
| |
V
V
nMy I
M
I ny
Aluminum: 6
6 9
1.0, | | 16 mm 0.016 m, 100 10 Pa
(100 10 )(141.995 10 )
887.47 N m
(1.0)(0.016)
V
u
u u
˜
n y
M
Brass: 6
6 9
1.5, | | 16 mm 0.016 m, 160 10 Pa
(160 10 )(141.995 10 )
946.63 N m
(1.5)(0.016)
V
u
u u
˜
n y
M
Choose the smaller value. 887 N m
M ˜ W
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on a website, in whole or part.
483
30 mm
6 mm
6 mm
30 mm
Aluminum
Brass
PROBLEM 4.35
For the composite bar indicated, determine the largest permissible
bending moment when the bar is bent about a vertical axis.
PROBLEM 4.35. Bar of Prob. 4.33.
SOLUTION
Use aluminum as reference material.
1.0 in aluminum
/ 105/70 1.5 in brass
b a
n
n E E
For transformed section,
3
1
1 1 1
3 3 4
12
1.5
(6)(30) 20.25 10 mm
12
u
n
I b h
3
2
2 2 2
3 3 4
12
1.0
(30)(30) 67.5 10 mm
12
u
n
I b h
3 4
3 1 20.25 10 mm
u
I I
3 4 9 4
1 2 3 108 10 mm 108 10 m
I I I I
u u
V
V ?
nMy I
M
I ny
Aluminum: 6
1.0, 15 mm 0.015 m, 100 10 Pa
u
n y V
6 9
(100 10 )(108 10 )
720 N m
(1.0)(0.015)
M
u u
˜
Brass: 6
1.5, 15 mm 0.015 m, 160 10 Pa
u
n y V
6 9
(160 10 )(108 10 )
768 N m
(1.5)(0.015)
u u
˜
M
Choose the smaller value. 720 N m
M ˜ W
Aluminum Brass
Modulus of elasticity 70 GPa 105 GPa
Allowable stress 100 MPa 160 MPa
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484
32 mm
32 mm
8 mm 8 mm
8 mm
8 mm
Aluminum
Brass
PROBLEM 4.36
For the composite bar indicated, determine the largest permissible
bending moment when the bar is bent about a vertical axis.
PROBLEM 4.36 Bar of Prob. 4.34.
Aluminum Brass
Modulus of elasticity 70 GPa 105 GPa
Allowable stress 100 MPa 160 MPa
SOLUTION
Use aluminum as the reference material.
For aluminum, 1.0
n
For brass, / 105/70 1.5
b a
n E E
Values of n are shown on the sketch.
For the transformed section,
3 3 3 3 3 4
1
1 1 1 1
3 3 3 4
2
2 2 2
3 4
3 2
3 4 9 4
1 2 3
1.5
(32)(48 32 ) 311.296 10 mm
12 12
1.0
(8)(32) 21.8453 10 mm
12 12
21.8453 10 mm
354.99 10 mm 354.99 10 m
n
I h B b
n
I h b
I I
I I I I
u
u
u
u u
| |
V
V
nMy I
M
I ny
Aluminum: 6
6 9
3
1.0, | | 16 mm 0.016 m, 100 10 Pa
(100 10 )(354.99 10 )
2.2187 10 N m
(1.0)(0.016)
V
u
u u
u ˜
n y
M
Brass: 6
6 9
3
1.5 | | 24 mm 0.024 m 160 10 Pa
(160 10 )(354.99 10 )
1.57773 10 N m
(1.5)(0.024)
V
u
u u
u ˜
n y
M
Choose the smaller value.
3
1.57773 10 N m
M u ˜ 1.578 kN m
M ˜ W
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524. PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
485
10 in.
6 in.
in.
1
2
5 3
in.
1
2
5 3
PROBLEM 4.37
Wooden beams and steel plates are securely bolted together to form the composite
member shown. Using the data given below, determine the largest permissible bending
moment when the member is bent about a horizontal axis.
Wood Steel
Modulus of elasticity: 6
2 10 psi
u 6
29 10 psi
u
Allowable stress: 2000 psi 22 ksi
SOLUTION
Use wood as the reference material.
1.0 in wood
/ 29/2 14.5 in steel
s w
n
n E E
For the transformed section,
3 2
1
1 1 1 1 1 1
3
2 4
2 3 4
2
2 2 2
4
3 1
4
1 2 3
12
14.5 1 1
(5) (14.5)(5) (5.25) 999.36 in
12 2 2
1.0
(6)(10) 500 in
12 12
999.36 in
2498.7 in
n
I b h n A d
n
I b h
I I
I I I I
§ · § ·
¨ ¸ ¨ ¸
© ¹ © ¹
nMy I
M
I ny
V
V ?
Wood:
3
1.0, 5 in., 2000 psi
(2000)(2499)
999.5 10 lb in.
(1.0)(5)
u ˜
n y
M
V
Steel: 3
3
3
14.5, 5.5 in., 22 ksi 22 10 psi
(22 10 )(2499)
689.3 10 lb in.
(14.5)(5.5)
u
u
u ˜
n y
M
V
Choose the smaller value. 3
689 10 lb in.
u ˜
M 689 kip in.
˜
M W
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525. PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
486
10 in.
3 in.
in.
3 in.
1
2
PROBLEM 4.38
Wooden beams and steel plates are securely bolted together to form the composite
member shown. Using the data given below, determine the largest permissible bending
moment when the member is bent about a horizontal axis.
Wood Steel
Modulus of elasticity: 6
2 10 psi
u 6
29 10 psi
u
Allowable stress: 2000 psi 22 ksi
SOLUTION
Use wood as the reference material.
1.0 in wood
/ 29/2 14.5 in steel
s w
n
n E E
For the transformed section,
3 3 4
1
1 1 1
3 3 4
2
2 2 2
4
3 1
4
1 2 3
1.0
(3)(10) 250 in
12 12
14.5 1
(10) 604.17 in
12 12 2
250 in
1104.2 in
n
I b h
n
I b h
I I
I I I I
§ ·
¨ ¸
© ¹
nMy I
M
I ny
V
V ?
Wood:
3
1.0, 5 in., 2000 psi
(2000)(1104.2)
441.7 10 lb in.
(1.0)(5)
u ˜
n y
M
V
Steel: 3
3
3
14.5, 5 in., 22 ksi 22 10 psi
(22 10 )(1104.2)
335.1 10 lb in.
(14.5)(5)
u
u
u ˜
n y
M
V
Choose the smaller value. 3
335 10 lb in.
u ˜
M 335 kip in.
˜
M W
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526. PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
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487
PROBLEM 4.39
A copper strip (Ec = 105 GPa) and an aluminum strip (Ea = 75 GPa)
are bonded together to form the composite beam shown. Knowing
that the beam is bent about a horizontal axis by a couple of moment
M = 35 N ˜ m, determine the maximum stress in (a) the aluminum
strip, (b) the copper strip.
SOLUTION
Use aluminum as the reference material.
1.0 in aluminum
/ 105/75 1.4 in copper
c a
n
n E E
Transformed section:
2
, mm
A 2
, mm
nA 0, mm
y 3
0, mm
nAy
c 144 144 9 1296
d 144 201.6 3 604.8
Σ 345.6 1900.8
0
1900.8
5.50 mm
345.6
Y
The neutral axis lies 5.50 mm above the bottom.
3 2 3 2 4
1
1 1 1 1 1 1
3 2 3 2 4
2
2 2 2 2 2 2
4 9 4
1 2
1.0
(24)(6) (1.0)(24)(6)(3.5) 2196 mm
12 12
1.4
(24)(6) (1.4)(24)(6)(2.5) 1864.8 mm
12 12
4060.8 mm 4.0608 10 m
n
I b h n A d
n
I b h n A d
I I I
u
(a) Aluminum:
6
9
1.0, 12 5.5 6.5 mm 0.0065 m
(1.0)(35)(0.0065)
56.0 10 Pa 56.0 MPa
4.0608 10
n y
nMy
I
u
u
V
56.0 MPa
V W
(b) Copper:
6
9
1.4, 5.5 mm 0.0055 m
(1.4)(35)( 0.0055)
66.4 10 Pa 66.4 MPa
4.0608 10
n y
nMy
I
u
u
V
66.4 MPa
V W
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on a website, in whole or part.
488
PROBLEM 4.40
A copper strip (Ec = 105 GPa) and an aluminum strip (Ea = 75 GPa)
are bonded together to form the composite beam shown. Knowing
that the beam is bent about a horizontal axis by a couple of moment
M = 35 N ˜ m, determine the maximum stress in (a) the aluminum
strip, (b) the copper strip.
SOLUTION
Use aluminum as the reference material.
1.0 in aluminum
/ 105/75 1.4 in copper
c a
n
n E E
Transformed section:
2
, mm
A 2
, mm
nA 0, mm
Ay 3
0, mm
nAy
c 216 216 7.5 1620
d 72 100.8 1.5 151.8
Σ 316.8 1771.2
0
1771.2
5.5909 mm
316.8
Y
The neutral axis lies 5.5909 mm above the bottom.
3 2 3 2 4
1
1 1 1 1 1 1
3 2 3 2 4
2
2 2 2 2 2 2
4 9 4
1 2
1.0
(24)(9) (1.0)(24)(9)(1.9091) 2245.2 mm
12 12
1.4
(24)(3) (1.4)(24)(3)(4.0909) 1762.5 mm
12 12
4839 mm 4.008 10 m
u
n
I b h n A d
n
I b h n A d
I I I
(a) Aluminum:
6
9
1.0, 12 5.5909 6.4091mm 0.0064091
(1.0)(35)(0.0064091)
56.0 10 Pa
4.008 10
n y
nMy
I
u
u
V
56.0 MPa
56.0 MPa
V W
(b) Copper:
6
9
1.4, 5.5909 mm 0.0055909 m
(1.4)(35)( 0.0055909)
68.4 10 Pa
4.008 10
n y
nMy
I
u
u
V
68.4 MPa
68.4 MPa
V W
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on a website, in whole or part.
489
in.
5 3 1
2
6 in.
12 in.
M
PROBLEM 4.41
The 6 12-in.
u timber beam has been strengthened by bolting to it the steel
reinforcement shown. The modulus of elasticity for wood is 6
1.8 10 psi
u and for
steel, 6
29 10 psi.
u Knowing that the beam is bent about a horizontal axis by a
couple of moment 450 kip in.,
M ˜ determine the maximum stress in (a) the
wood, (b) the steel.
SOLUTION
Use wood as the reference material.
For wood, 1
For steel, / 29 /1.8 16.1111
s w
n
n E E
Transformed section: c wood d steel
421.931
112.278
3.758 in.
o
Y
The neutral axis lies 3.758 in. above the wood-steel interface.
3 2 3 2 4
1
1 1 1 1 1 1
3 2 3 2 4
2
2 2 2 2 2 2
4
1 2
1
(6)(12) (72)(6 3.758) 1225.91 in
12 12
16.1111
(5)(0.5) (40.278)(3.578 0.25) 647.87 in
12 12
1873.77 in
450 kip in. V
˜
n
I b h n A d
n
I b h n A d
I I I
nMy
M
I
(a) Wood: 1, 12 3.758 8.242 in.
n y
(1)(450)(8.242)
1.979 ksi
1873.77
V
w 1.979 ksi
w
V W
(b) Steel: 16.1111, 3.758 0.5 4.258 in.
n y
(16.1111)(450)( 4.258)
16.48 ksi
1873.77
V
s 16.48 ksi
s
V W
2
, in
A 2
, in
nA 0
y 3
0, in
nAy
c 72 72 6 432
d 2.5 40.278 0.25 10.069
112.278 421.931
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490
6 in.
12 in.
C8 11.5
M
PROBLEM 4.42
The 6 12-in.
u timber beam has been strengthened by bolting to it the steel
reinforcement shown. The modulus of elasticity for wood is 6
1.8 10 psi
u and for
steel, 6
29 10 psi.
u Knowing that the beam is bent about a horizontal axis by a
couple of moment 450 kip in.,
M ˜ determine the maximum stress in (a) the
wood, (b) the steel.
SOLUTION
Use wood as the reference material.
6
6
For wood, 1
29 10
For steel, 16.1111
1.8 10
s
w
n
E
n
E
u
u
For C8 11.5
u channel section,
2 4
3.38 in , 0.220 in., 0.571 in., 1.32 in
w y
A t x I
For the composite section, the centroid of the channel (part 1) lies 0.571 in. above the bottom of the section.
The centroid of the wood (part 2) lies 0.220 6.00 6.22 in.
above the bottom.
Transformed section:
3
0 2
478.93 in
3.787 in.
126.456 in
Y 0 0
d y Y
The neutral axis lies 3.787 in. above the bottom of the section.
2 2 4
1 1 1 1 1 1
3 2 3 2 4
2
2 2 2 2 2 2
4
1 2
(16.1111)(1.32) (54.456)(3.216) 584.49 in
1
(6)(12) (72)(2.433) 1290.20 in
12 12
1874.69 in
450 kip in
I n I n A d
n
I b h n A d
I I I
nMy
M
I
V
˜
(a) Wood: 1, 12 0.220 3.787 8.433 in.
n y
(1)(450)(8.433)
2.02 ksi
1874.69
w
V 2.02 ksi
w
V W
(b) Steel: 16.1111, 3.787 in.
n y
(16.1111)(450)( 3.787)
14.65 ksi
1874.67
V
s 14.65 ksi
s
V W
Part A, in2
nA, in2
, in.
y 3
, in
nAy d, in.
1 3.38 54.456 0.571 31.091 3.216
2 72 72 6.22 447.84 2.433
6 126.456 478.93
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on a website, in whole or part.
491
24 mm
6 mm
6 mm
Aluminum
Copper
PROBLEM 4.43
For the composite beam indicated, determine the radius of curvature
caused by the couple of moment 35 N ˜ m.
Beam of Prob. 4.39.
SOLUTION
See solution to Prob. 4.39 for the calculation of I.
1
9 9
1 35
0.1149 m
(75 10 )(4.0608 10 )
u u
a
M
E I
U
8.70 m
U W
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531. PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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on a website, in whole or part.
492
24 mm
9 mm
3 mm
Aluminum
Copper
PROBLEM 4.44
For the composite beam indicated, determine the radius of curvature
caused by the couple of moment 35 N m.
˜
Beam of Prob. 4.40.
SOLUTION
See solution to Prob. 4.40 for the calculation of I.
1
9 9
1 35
0.1164 m
(75 10 )(4.008 10 )
u u
a
M
E I
U
8.59 m
U W
Mecánica de Materiales - Ferdinand P. Beer 7ma. Edición.

532. PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
493
in.
5 3 1
2
6 in.
12 in.
M
PROBLEM 4.45
For the composite beam indicated, determine the radius of curvature caused by the
couple of moment 450 kip ˜ in.
Beam of Prob. 4.41.
SOLUTION
See solution to Prob. 4.41 for calculation of I.
4 6
3
3
6 1
6
1873.77 in 1.8 10 psi
450 kip in 450 10 lb in.
1 450 10
133.421 10 in.
(1.8 10 )(1873.77)
u
˜ u ˜
u
u
u
w
I E
M
M
EI
U
7495 in. 625 ft
U W
Mecánica de Materiales - Ferdinand P. Beer 7ma. Edición.

533. PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
494
6 in.
12 in.
C8 11.5
M
PROBLEM 4.46
For the composite beam indicated, determine the radius of curvature caused by the
couple of moment 450 kip ˜ in.
Beam of Prob. 4.42.
SOLUTION
See solution to Prob. 4.42 for calculation of I.
4 6
3
3
6 1
6
1874.69 in 1.8 10 psi
450 kip in. 450 10 lb in.
1 450 10
133.355 10 in.
(1.8 10 )(1874.69)
U
u
˜ u ˜
u
u
u
w
I E
M
M
EI
7499 in. 625 ft
U W
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534. PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
495
5.5 in.
6 in.
5.5 in.
4 in.
5.5 in.
5.5 in.
-in. diameter
5
8
PROBLEM 4.47
A concrete slab is reinforced by 5
8
-in.-diameter steel rods
placed on 5.5-in. centers as shown. The modulus of elasticity is
3 u 106
psi for the concrete and 29 u 106
psi for the steel.
Using an allowable stress of 1400 psi for the concrete and
20 ksi for the steel, determine the largest bending moment in a
portion of slab 1 ft wide.
SOLUTION
6
6
29 10
9.6667
3 10
s
c
E
n
E
u
u
Consider a section 5.5 in. wide.
2
2 2
5
0.3068 in
4 4 8
§ ·
¨ ¸
© ¹
s s
A d
S S
2
2.9657 in
s
nA
Locate the natural axis.
5.5 (4 )(2.9657) 0
2
x
x x
2
2.75 2.9657 11.8628 0
x x
Solve for x.
1.6066 in. 4 2.3934 in.
x x
3 2
3 2 4
1
(5.5) (2.9657)(4 )
3
1
(5.5)(1.6066) (2.9657)(2.3934) 24.591in
3
I x x
nMy I
M
I ny
V
V
Concrete: 1, 1.6066 in., 1400 psi
n y V
3
(24.591)(1400)
21.429 10 lb in.
(1.0)(1.6066)
M u ˜
Mecánica de Materiales - Ferdinand P. Beer 7ma. Edición.

535. PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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496
PROBLEM 4.47 (Continued)
Steel: 3
9.6667, 2.3934 in., 20 ksi=20 10 psi
u
n y V
3
3
(24.591)(20 10 )
21.258 10 lb in.
(9.6667)(2.3934)
M
u
u ˜
Choose the smaller value as the allowable moment for a 5.5 in. width.
3
21.258 10 lb in.
M u ˜
For a 1 ft = 12 in. width,
3 3
12
(21.258 10 ) 46.38 10 lb in.
5.5
M u u ˜
46.38 kip in.
˜
M 3.87 kip ft
˜ W
Mecánica de Materiales - Ferdinand P. Beer 7ma. Edición.

536. PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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on a website, in whole or part.
497
5.5 in.
6 in.
5.5 in.
4 in.
5.5 in.
5.5 in.
-in. diameter
5
8
PROBLEM 4.48
Solve Prob. 4.47, assuming that the spacing of the 5
8
-in.-diameter
steel rods is increased to 7.5 in.
PROBLEM 4.47 A concrete slab is reinforced by 5
8
-in.-diameter
steel rods placed on 5.5-in. centers as shown. The modulus of
elasticity is 3 × 106
psi for the concrete and 29 u 106
psi for the
steel. Using an allowable stress of 1400 psi for the concrete and
20 ksi for the steel, determine the largest bending moment in a
portion of slab 1 ft wide.
SOLUTION
6
6
29 10 psi
9.667
3 10 psi
u
u
s
c
E
n
E
Number of rails per foot:
12 in.
1.6
7.5 in.
Area of
5
-in.-
8
diameter bars per foot:
2
2
5
1.6 0.4909 in
4 8
S § ·
¨ ¸
© ¹
s
A
Transformed section, all concrete.
First moment of area:
12 4.745(4 ) 0
2
§ ·
¨ ¸
© ¹
x
x x
1.4266 in.
x
2
9.667(0.4909) 4.745 in
s
nA
3 2 4
1
(12)(1.4266) 4.745(4 1.4266) 43.037 in
3
NA
I
For concrete: all 1400 psi 1.4266 in.
c x
V
4
43.037 in
(1400 psi)
1.4266 in.
I
M
c
V 42.24 kip.in.
M W
For steel: all 20 ksi 4 4 1.4266 2.5734 in.
V
c x
4
steel 20 ksi 43.042 in
9.667 2.5734 in.
˜ ˜
I
M
n c
V
34.60 kip in.
˜
M W
We choose the smaller M. 34.60 kip in.
M ˜
Steel controls. 2.88 kip ft
˜
M W
Mecánica de Materiales - Ferdinand P. Beer 7ma. Edición.

537. PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
498
300 mm
540 mm
60 mm
25-mm
diameter
PROBLEM 4.49
The reinforced concrete beam shown is subjected to a positive bending
moment of 175 kN ˜ m. Knowing that the modulus of elasticity is 25 GPa
for the concrete and 200 GPa for the steel, determine (a) the stress in the
steel, (b) the maximum stress in the concrete.
SOLUTION
2 2 3 2
3 2
200 GPa
8.0
25 GPa
4 (4) (25) 1.9635 10 mm
4 4
15.708 10 mm
s
c
s
s
E
n
E
A d
nA
S S
§ ·
˜ u
¨ ¸
© ¹
u
Locate the neutral axis.
3
2 3 6
300 (15.708 10 )(480 ) 0
2
150 15.708 10 7.5398 10 0
u
u u
x
x x
x x
Solve for x.
3 3 2 6
15.708 10 (15.708 10 ) (4)(150)(7.5398 10 )
(2)(150)
177.87 mm, 480 302.13 mm
x
x x
u u u
3 3 2
3 3 2
9 4 3 4
1
(300) (15.708 10 )(480 )
3
1
(300)(177.87) (15.708 10 )(302.13)
3
1.9966 10 mm 1.9966 10 m
I x x
nMy
I
V
u
u
u u
(a) Steel: 302.45 mm 0.30245 m
y
3
6
3
(8.0)(175 10 )( 0.30245)
212 10 Pa
1.9966 10
V
u
u
u
212 MPa
V W
(b) Concrete: 177.87 mm 0.17787 m
y
3
6
3
(1.0)(175 10 )(0.17787)
15.59 10 Pa
1.9966 10
V
u
u
u
15.59 MPa
V W
Mecánica de Materiales - Ferdinand P. Beer 7ma. Edición.

538. PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
499
300 mm
540 mm
60 mm
25-mm
diameter
PROBLEM 4.50
Solve Prob. 4.49, assuming that the 300-mm width is increased to 350 mm.
PROBLEM 4.49 The reinforced concrete beam shown is subjected to a
positive bending moment of 175 kN ˜ m. Knowing that the modulus of
elasticity is 25 GPa for the concrete and 200 GPa for the steel, determine
(a) the stress in the steel, (b) the maximum stress in the concrete.
SOLUTION
2 2
3 2
3 2
200 GPa
8.0
25 GPa
4 (4) (25)
4 4
1.9635 10 mm
15.708 10 mm
s
c
s
s
E
n
E
A d
nA
S S
§ ·
¨ ¸
© ¹
u
u
Locate the neutral axis.
3
2 3 6
350 (15.708 10 )(480 ) 0
2
175 15.708 10 7.5398 10 0
u
u u
x
x x
x x
Solve for x.
3 3 2 6
15.708 10 (15.708 10 ) (4)(175)(7.5398 10 )
(2)(175)
167.48 mm, 480 312.52 mm
x
x x
u u u
3 3 2
3 3 2
9 4 3 4
1
(350) (15.708 10 )(480 )
3
1
(350)(167.48) (15.708 10 )(312.52)
3
2.0823 10 mm 2.0823 10 m
V
u
u
u u
I x x
nMy
I
(a) Steel: 312.52 mm 0.31252 m
y
3
6
3
(8.0)(175 10 )( 0.31252)
210 10 Pa
2.0823 10
V
u
u
u
210 MPa
V W
(b) Concrete: 167.48 mm 0.16748 m
y
3
6
3
(1.0)(175 10 )(0.16748)
14.08 10 Pa
2.0823 10
V
u
u
u
14.08 MPa
V W
Mecánica de Materiales - Ferdinand P. Beer 7ma. Edición.

539. PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
500
12 in.
2.5 in.
20 in.
4 in.
24 in.
1-in.
diameter
PROBLEM 4.51
Knowing that the bending moment in the reinforced concrete beam is
100 kip ˜ ft and that the modulus of elasticity is 6
3.625 10 psi
u for the
concrete and 6
29 10 psi
u for the steel, determine (a) the stress in the
steel, (b) the maximum stress in the concrete.
SOLUTION
6
6
2 2 2
29 10
8.0
3.625 10
(4) (1) 3.1416 in 25.133 in
4
s
c
s s
E
n
E
A nA
S
u
u
§ ·
¨ ¸
© ¹
Locate the neutral axis.
(24)(4)( 2) (12 ) (25.133)(17.5 4 ) 0
2
x
x x x
§ ·
¨ ¸
© ¹
2 2
96 192 6 339.3 25.133 0 or 6 121.133 147.3 0
x x x x x
Solve for x.
2
121.133 (121.133) (4)(6)(147.3)
1.150 in.
(2)(6)
x
3 17.5 4 12.350 in.
d x
3 2 3 2 4
1 1 1 1 1
3 3 4
2 2
2 2 4
3 3 3
4
1 2 3
1 1
(24)(4) (24)(4)(3.150) 1080.6 in
12 12
1 1
(12)(1.150) 6.1 in
3 3
(25.133)(12.350) 3833.3 in
4920 in
I b h A d
I b x
I nA d
I I I I
nMy
I
V where 100 kip ft 1200 kip in.
M ˜ ˜
(a) Steel: 8.0
n 12.350 in.
y
(8.0)(1200)( 12.350)
4920
s
V
24.1 ksi
s
V W
(b) Concrete: 1.0, 4 1.150 5.150 in.
n y
(1.0)(1200)(5.150)
4920
c
V 1.256 ksi
c
V W
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540. PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
501
8 in.
2 in.
16 in. -in. diameter
7
8
PROBLEM 4.52
A concrete beam is reinforced by three steel rods placed as shown. The
modulus of elasticity is 6
3 10 psi
u for the concrete and 6
29 10 psi
u for the
steel. Using an allowable stress of 1350 psi for the concrete and 20 ksi for the
steel, determine the largest allowable positive bending moment in the beam.
SOLUTION
6
6
2
2 2 2
29 10
9.67
3 10
7
3 (3) 1.8040 in 17.438 in
4 4 8
s
c
s s
E
n
E
A d nA
S S
u
u
§ ·§ ·
¨ ¸¨ ¸
© ¹© ¹
Locate the neutral axis:
2
8 (17.438)(14 ) 0
2
4 17.438 244.14 0
x
x x
x x
Solve for x.
2
17.438 17.438 (4)(4)(244.14)
5.6326 in.
(2)(4)
x
14 8.3674 in.
x
3 2 3 2 4
1 1
8 (14 ) (8)(5.6326) (17.438)(8.3674) 1697.45 in
3 3
s
I x nA x
V
V ?
nMy I
M
I ny
Concrete: 1.0, 5.6326 in., 1350 psi
n y V
3
(1350)(1697.45)
406.835 10 lb in. 407 kip in.
(1.0)(5.6326)
u ˜ ˜
M
Steel: 3
9.67, 8.3674 in., 20 10 psi
n y V u
3
(20 10 )(1697.45)
419.72 lb in. 420 kip in.
(9.67)(8.3674)
u
˜ ˜
M
Choose the smaller value. 407 kip in.
˜
M 33.9 kip ft
M ˜ W
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541. PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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502
b
d
PROBLEM 4.53
The design of a reinforced concrete beam is said to be balanced if the maximum stresses
in the steel and concrete are equal, respectively, to the allowable stresses s
V and .
c
V
Show that to achieve a balanced design the distance x from the top of the beam to the
neutral axis must be
1
V
V
s c
c s
d
x
E
E
where c
E and s
E are the moduli of elasticity of concrete and steel, respectively, and d is
the distance from the top of the beam to the reinforcing steel.
SOLUTION
( )
( )
1
1 1
1
V V
V
V
V V
V V
V
V
s c
s
c
s c s
c s c
c s
s c
nM d x Mx
I I
n d x d
n n
x x
E
d
x n E
d
x
E
E
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503
b
d
PROBLEM 4.54
For the concrete beam shown, the modulus of elasticity is 25 GPa for the concrete and
200 GPa for the steel. Knowing that b = 200 mm and d = 450 mm, and using an
allowable stress of 12.5 MPa for the concrete and 140 MPa for the steel, determine
(a) the required area As of the steel reinforcement if the beam is to be balanced, (b) the
largest allowable bending moment. (See Prob. 4.53 for definition of a balanced beam.)
SOLUTION
9
9
6
6
200 10
8.0
25 10
( )
( )
1 1 140 10
1 1 2.40
8.0 12.5 10
0.41667 (0.41667)(450) 187.5 mm
V V
V
V
V
V
u
u
u
˜
u
s
c
s c
s
c
s
c
E
n
E
nM d x Mx
I I
n d x d
n n
x x
d
x n
x d
Locate neutral axis. ( )
2
s
x
bx nA d x
(a)
2 2
2
(200)(187.5)
1674 mm
2 ( ) (2)(8.0)(262.5)
s
bx
A
n d x
2
1674 mm
s
A W
3 2 3 2
9 4 3 4
1 1
( ) (200)(187.5) (8.0)(1674)(262.5)
3 3
1.3623 10 mm 1.3623 10 m
V
V
u u
s
I bx nA d x
nMy I
M
I ny
(b) Concrete: 6
1.0 187.5 mm 0.1875 m 12.5 10 Pa
u
n y V
3 6
3
(1.3623 10 )(12.5 10 )
90.8 10 N m
(1.0)(0.1875)
M
u u
u ˜
Steel: 6
8.0 262.5 mm 0.2625 m 140 10 Pa
u
n y V
3 6
3
(1.3623 10 )(140 10 )
90.8 10 N m
(8.0)(0.2625)
M
u u
u ˜
Note that both values are the same for balanced design.
90.8 kN m
˜
M W
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504
Aluminum
Brass
Steel
Brass
Aluminum
1.5 in.
0.5 in.
0.5 in.
0.5 in.
0.5 in.
0.5 in.
PROBLEM 4.55
Five metal strips, each a 0.5 × 1.5-in. cross section, are bonded together to
form the composite beam shown. The modulus of elasticity is 30 × 106
psi
for the steel, 15 × 106
psi for the brass, and 10 × 106
psi for the aluminum.
Knowing that the beam is bent about a horizontal axis by a couple of
moment 12 kip in.,
˜ determine (a) the maximum stress in each of the
three metals, (b) the radius of curvature of the composite beam.
SOLUTION
Use aluminum as the reference material.
6
6
6
6
30 10
3.0 in steel
10 10
15 10
1.5 in brass
10 10
1.0 in aluminum
u
u
u
u
s
a
b
a
E
n
E
E
n
E
n
For the transformed section,
3 2 3 2
1
1 1 1 1 1 1
4
1
(1.5)(0.5) (0.75)(1.0)
12 12
0.7656 in
n
I b h n A d
3 2 3 2 4
2
2 2 2 2 2 2
3 3 4
3
3 3 3
4 4
4 2 5 1
5
4
1
1.5
(1.5)(0.5) (1.5)(0.75)(0.5) 0.3047 in
12 12
3.0
(1.5)(0.5) 0.0469 in
12 12
0.3047 in 0.7656 in
2.1875 in
¦ i
n
I b h n A d
n
I b h
I I I I
I I
(a) Aluminum:
(1.0)(12)(1.25)
6.86 ksi
2.1875
nMy
I
V W
Brass:
(1.5)(12)(0.75)
6.17 ksi
2.1875
nMy
I
V W
Steel:
(3.0)(12)(0.25)
4.11ksi
2.1875
nMy
I
V W
(b)
3
6 1
6
1 12 10
548.57 10 in.
(10 10 )(2.1875)
U
u
u
u
a
M
E I
1823 in. =151.9 ft
U W
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544. PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
505
Steel
Aluminum
Brass
Aluminum
Steel
1.5 in.
0.5 in.
0.5 in.
0.5 in.
0.5 in.
0.5 in.
PROBLEM 4.56
Five metal strips, each a 0.5 × 1.5-in. cross section, are bonded together to
form the composite beam shown. The modulus of elasticity is 30 × 106
psi
for the steel, 15 × 106
psi for the brass, and 10 × 106
psi for the aluminum.
Knowing that the beam is bent about a horizontal axis by a couple of
moment 12 kip · in., determine (a) the maximum stress in each of the
three metals, (b) the radius of curvature of the composite beam.
SOLUTION
Use aluminum as the reference material.
6
6
6
6
30 10
3.0 in steel
10 10
15 10
1.5 in brass
10 10
1.0 in aluminum
u
u
u
u
s
a
b
a
E
n
E
E
n
E
n
For the transformed section,
3 2
1
1 1 1 1 1 1
3 2
4
12
3.0
(1.5)(0.5) (3.0)(0.75)(1.0)
12
2.2969 in
n
I b h n A d
3 2 3 2 4
2
2 2 2 2 2 2
3 3 4
3
3 3 3
4 4
4 2 5 1
5
4
1
1.0
(1.5)(0.5) (1.0)(0.75)(0.5) 0.2031in
12 12
1.5
(1.5)(0.5) 0.0234 in
12 12
0.2031in 2.2969 in
5.0234 in
¦ i
n
I b h n A d
n
I b h
I I I I
I I
(a) Steel:
(3.0)(12)(1.25)
8.96 ksi
5.0234
nMy
I
V W
Aluminum:
(1.0)(12)(0.75)
1.792 ksi
5.0234
nMy
I
V W
Brass:
(1.5)(12)(0.25)
0.896 ksi
5.0234
nMy
I
V W
(b)
3
6 1
6
1 12 10
238.89 10 in.
(10 10 )(5.0234)
U
u
u
u
a
M
E I
4186 in. 349 ft
U W
Mecánica de Materiales - Ferdinand P. Beer 7ma. Edición.

545. PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
506
Brass
Aluminum
0.8 in.
PROBLEM 4.57
The composite beam shown is formed by bonding together a brass rod and an
aluminum rod of semicircular cross sections. The modulus of elasticity is
6
15 10 psi
u for the brass and 6
10 10 psi
u for the aluminum. Knowing that the
composite beam is bent about a horizontal axis by couples of moment
8 kip in.,
˜ determine the maximum stress (a) in the brass, (b) in the aluminum.
SOLUTION
For each semicircle, 2 2
0.8 in. 1.00531 in
2
r A r
S
,
4 4
0 base
4 (4)(0.8)
0.33953 in. 0.160850 in
3 3 8
S
S S
r
y I r
2 2 4
base 0 0.160850 (1.00531)(0.33953) 0.044953 in
I I Ay
Use aluminum as the reference material.
6
6
1.0 in aluminum
15 10
1.5 in brass
10 10
b
a
n
E
n
E
u
u
Locate the neutral axis.
0
0.17067
0.06791 in.
2.51327
Y
The neutral axis lies 0.06791 in.
above the material interface.
1 2
2 2 4
1 1 1 1
2 2 4
2 2 2 2
1 2
0.33953 0.06791 0.27162 in., 0.33953 0.06791 0.40744 in.
(1.5)(0.044957) (1.5)(1.00531)(0.27162) 0.17869 in
(1.0)(0.044957) (1.0)(1.00531)(0.40744) 0.21185 in
d d
I n I n Ad
I n I n Ad
I I I
4
0.39054 in
(a) Brass: 1.5, 0.8 0.06791 0.73209 in.
n y
(1.5)(8)(0.73209)
0.39054
nMy
I
V 22.5 ksi
V W
(b) Aluminium: 1.0, 0.8 0.06791 0.86791 in.
n y
(1.0)(8)( 0.86791)
0.39054
nMy
I
V
17.78 ksi
V W
A, in2
nA, in2
0, in.
y 3
0, in
nAy
c 1.00531 1.50796 0.33953 0.51200
d 1.00531 1.00531 0.33953 0.34133
6 2.51327 0.17067
Mecánica de Materiales - Ferdinand P. Beer 7ma. Edición.

546. PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
507
Steel
38 mm
10 mm
z
y
3 mm
6 mm
Aluminum
PROBLEM 4.58
A steel pipe and an aluminum pipe are securely bonded together to form
the composite beam shown. The modulus of elasticity is 200 GPa for the
steel and 70 GPa for the aluminum. Knowing that the composite beam is
bent by a couple of moment 500 N ˜ m, determine the maximum stress
(a) in the aluminum, (b) in the steel.
SOLUTION
Use aluminum as the reference material.
1.0 in aluminum
/ 200/70 2.857 in steel
s a
n
n E E
For the transformed section,
Steel: 4 4 4 4 3 4
(2.857) (16 10 ) 124.62 10 mm
4 4
s s o i
I n r r
S S
§ ·
u
¨ ¸
© ¹
Aluminium: 4 4 4 4 3 4
(1.0) (19 16 ) 50.88 10 mm
4 4
a a o i
I n r r
S S
§ ·
u
¨ ¸
© ¹
3 4 9 4
175.50 10 mm 175.5 10 m
s a
I I I
u u
(a) Aluminum: 19 mm 0.019 m
c
6
9
(1.0)(500)(0.019)
54.1 10 Pa
175.5 10
a
a
n Mc
I
V
u
u
54.1 MPa
a
V W
(b) Steel: 16 mm 0.016 m
c
6
9
(2.857)(500)(0.016)
130.2 10 Pa
175.5 10
s
s
n Mc
I
V
u
u
130.2 MPa
s
V W
Mecánica de Materiales - Ferdinand P. Beer 7ma. Edición.

547. PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
508
50 mm
100 mm
%
'
%
Et # Ec
1
2
Ec
M
PROBLEM 4.59
The rectangular beam shown is made of a plastic for which the
value of the modulus of elasticity in tension is one-half of its
value in compression. For a bending moment 600 N m,
M ˜
determine the maximum (a) tensile stress, (b) compressive stress.
SOLUTION
1
2
n on the tension side of neutral axis
1
n on the compression side
Locate neutral axis.
1 2
2 2
( ) 0
2 2
1 1
( ) 0
2 4
x h x
n bx n b h x
bx b h x
2 2
3 3 6 4
1 1
3 3 6 4
2 2
6 4 6
1 2
1 1
( ) ( )
2 2
1
0.41421 41.421 mm
2 1
58.579 mm
1 1
(1) (50)(41.421) 1.1844 10 mm
3 3
1 1 1
( ) (50)(58.579) 1.6751 10 mm
3 2 3
2.8595 10 mm 2.8595 10 m
§ ·
u
¨ ¸
© ¹
§ ·§ ·
u
¨ ¸¨ ¸
© ¹© ¹
u u
x h x x h x
x h h
h x
I n bx
I n b h x
I I I 4
(a) Tensile stress:
1
, 58.579 mm 0.058579 m
2
n y
6
6
(0.5)(600)( 0.058579)
6.15 10 Pa
2.8595 10
nMy
I
V
u
u
6.15 MPa
t
V W
(b) Compressive stress: 1, 41.421 mm 0.041421 m
n y
6
6
(1.0)(600)(0.041421)
8.69 10 Pa
2.8595 10
nMy
I
V
u
u
8.69 MPa
c
V W
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548. PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
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509
PROBLEM 4.60*
A rectangular beam is made of material for which the modulus of elasticity is t
E in tension and c
E in
compression. Show that the curvature of the beam in pure bending is
1
U r
M
E I
where
2
4
( )
t c
r
t c
E E
E
E E
SOLUTION
Use t
E as the reference modulus.
Then .
c t
E nE
Locate neutral axis.
2 2
( ) 0
2 2
( ) 0 ( )
1 1
x h x
nbx b h x
nx h x nx h x
h nh
x h x
n n
3
3
3 3 3
trans
3/ 2
3 3 3
3 3 2
1 1
( )
3 3 3 1 1
1
1 1 1
3 3 3
1 1 1
n h n
I bx b h x bh
n n
n n
n n n
bh bh bh
n n n
ª º
§ ·
§ ·
« »
¨ ¸
¨ ¸ ¨ ¸
« »
© ¹ © ¹
¬ ¼
u
3
trans
trans
3
trans
3 2
2 2
1 1
where
12
12
3 1
4 / 4
/ 1
t r
r t
t
r t
t c t t c
c t c t
M M
I bh
E I E I
E I E I
E I n
E E bh
I bh n
E E E E E
E E E E
U
u u
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549. PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
510
r
80 mm
40 mm
8 mm
M
PROBLEM 4.61
Knowing that 250 N m,
M ˜ determine the maximum stress in the beam
shown when the radius r of the fillets is (a) 4 mm, (b) 8 mm.
SOLUTION
3 3 3 4 9 4
1 1
(8)(40) 42.667 10 mm 42.667 10 m
12 12
20 mm 0.020 m
80 mm
2.00
40 mm
u u
I bh
c
D
d
(a)
4 mm
0.10
40 mm
r
d
From Fig. 4.27, 1.87
K
6
max 9
(1.87)(250)(0.020)
219 10 Pa
42.667 10
Mc
K
I
V
u
u
max 219 MPa
V W
(b)
8 mm
0.20
40 mm
r
d
From Fig. 4.27, 1.50
K
6
max 9
(1.50)(250)(0.020)
176.0 10 Pa
42.667 10
Mc
K
I
V
u
u
max 176.0 MPa
V W
Mecánica de Materiales - Ferdinand P. Beer 7ma. Edición.

550. PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
511
r
80 mm
40 mm
8 mm
M
PROBLEM 4.62
Knowing that the allowable stress for the beam shown is 90 MPa, determine
the allowable bending moment M when the radius r of the fillets is (a) 8 mm,
(b) 12 mm.
SOLUTION
3 3 3 4 9 4
1 1
(8)(40) 42.667 10 mm 42.667 10 m
12 12
20 mm 0.020 m
80 mm
2.00
40 mm
u u
I bh
c
D
d
(a)
8 mm
0.2
40 mm
r
d
From Fig. 4.27, 1.50
K
max
Mc
K
I
V
6 9
max (90 10 )(42.667 10 )
(1.50)(0.020)
I
M
Kc
V
u u
128.0 N m
M ˜ W
(b)
12 mm
0.3
40 mm
r
d
From Fig. 4.27, 1.35
K
6 9
(90 10 )(42.667 10 )
(1.35)(0.020)
M
u u
142.0 N m
M ˜ W
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551. PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
512
r
M
4.5 in.
in.
3
4
PROBLEM 4.63
Semicircular grooves of radius r must be milled as shown in the sides of
a steel member. Using an allowable stress of 8 ksi, determine the largest
bending moment that can be applied to the member when (a) r = 3
8
in.,
(b) r = 3
4
in.
SOLUTION
(a)
3
2 4.5 (2) 3.75 in.
8
4.5 0.375
1.20 0.10
3.75 3.75
§ ·
¨ ¸
© ¹
d D r
D r
d d
From Fig. 4.28, 2.07
K
3 3 4
1 1 3 1
(3.75) 3.296 in 1.875 in.
12 12 4 2
(8)(3.296)
6.79 kip in.
(2.07)(1.875)
§ ·
¨ ¸
© ¹
? ˜
I bh c
Mc I
K M
I Kc
V
V
6.79 kip in.
˜
V W
(b)
3 4.5 0.75
2 4.5 (2) 3.0 1.5 0.25
4 3.0 3.0
§ ·
¨ ¸
© ¹
D r
d D r
d d
From Fig. 4.28, 3 3 4
1 1 3
1.61 (3.0) 1.6875 in
12 12 4
§ ·
¨ ¸
© ¹
K I bh
1 (8)(1.6875)
1.5 in. 5.59 kip in.
2 (1.61)(1.5)
V
˜
I
c d M
Kc
W
Mecánica de Materiales - Ferdinand P. Beer 7ma. Edición.

552. PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
513
r
M
4.5 in.
in.
3
4
PROBLEM 4.64
Semicircular grooves of radius r must be milled as shown in the sides
of a steel member. Knowing that M = 4 kip · in., determine the
maximum stress in the member when the radius r of the semicircular
grooves is (a) r = 3
8
in., (b) r = 3
4
in.
SOLUTION
(a)
3
2 4.5 (2) 3.75 in.
8
4.5 0.375
1.20 0.10
3.75 3.75
§ ·
¨ ¸
© ¹
d D r
D r
d d
From Fig. 4.28, 2.07
K
3 3 4
1 1 3 1
(3.75) 3.296 in 1.875 in.
12 12 4 2
(2.07)(4)(1.875)
4.71ksi
3.296
§ ·
¨ ¸
© ¹
I bh c
Mc
K
I
V
4.71ksi
V W
(b)
3 4.5 0.75
2 4.5 (2) 3.00 in. 1.50 0.25
4 3.00 3.0
§ ·
¨ ¸
© ¹
D r
d D r
d d
From Fig. 4.28,
3 3 4
1 1 3 1
1.61 (3.00) 1.6875 in 1.5 in.
12 12 4 2
§ ·
¨ ¸
© ¹
K I bh c d
(1.61)(4)(1.5)
5.72 ksi
1.6875
Mc
K
I
V
5.72 ksi
V W
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553. PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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514
(a) (b)
100 mm
150 mm
18 mm
100 mm
150 mm
18 mm
M M PROBLEM 4.65
A couple of moment M = 2 kN · m is to be
applied to the end of a steel bar. Determine the
maximum stress in the bar (a) if the bar is
designed with grooves having semicircular
portions of radius r = 10 mm, as shown in Fig. a,
(b) if the bar is redesigned by removing the
material to the left and right of the dashed lines
as shown in Fig. b.
SOLUTION
For both configurations,
150 mm
100 mm
10 mm
150
1.50
100
10
0.10
100
D
d
r
D
d
r
d
For configuration (a),
Fig. 4.28 gives 2.21.
a
K
For configuration (b), Fig. 4.27 gives K 1.79.
b
3 3 6 4 6 4
1 1
(18)(100) 1.5 10 mm 1.5 10 m
12 12
1
50 mm 0.05 m
2
u u
I bh
c d
(a)
3
6
6
(2.21)(2 10 )(0.05)
147.0 10 Pa 147.0 MPa
1.5 10
u
u
u
KMc
I
V
147.0 MPa
V W
(b)
3
6
6
(1.79)(2 10 )(0.05)
119.0 10 Pa 119.0 MPa
1.5 10
u
u
u
KMc
I
V
119.0 MPa
V W
Mecánica de Materiales - Ferdinand P. Beer 7ma. Edición.

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515
(a) (b)
100 mm
150 mm
18 mm
100 mm
150 mm
18 mm
M M PROBLEM 4.66
The allowable stress used in the design of a
steel bar is 80 MPa. Determine the largest
couple M that can be applied to the bar (a) if
the bar is designed with grooves having
semicircular portions of radius r 15 mm, as
shown in Fig. a, (b) if the bar is redesigned by
removing the material to the left and right of
the dashed lines as shown in Fig. b.
SOLUTION
For both configurations,
150 mm 100 mm
15 mm
150
1.50
100
15
0.15
100
D d
r
D
d
r
d
For configuration (a), Fig. 4.28 gives 1.92.
a
K
For configuration (b), Fig. 4.27 gives 1.57.
b
K
3 3 6 4 6 4
1 1
(18)(100) 1.5 10 mm 1.5 10 m
12 12
1
50 mm 0.050 m
2
I bh
c d
u u
(a)
6 6
3
(80 10 )(1.5 10 )
1.250 10 N m
(1.92) (0.05)
u u
? u ˜
u
KMc I
M
I Kc
V
V
1.250 kN m
˜ W
(a)
6 6
3
(80 10 )(1.5 10 )
1.530 10 N m 1.530 kN m
(1.57)(0.050)
u u
u ˜ ˜
I
M
Kc
V
W
Mecánica de Materiales - Ferdinand P. Beer 7ma. Edición.

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on a website, in whole or part.
516
z
x
8 mm
12 mm
M PROBLEM 4.67
The prismatic bar shown is made of a steel that is assumed to be
elastoplastic with 300 MPa
Y
V and is subjected to a couple M
parallel to the x axis. Determine the moment M of the couple for which
(a) yield first occurs, (b) the elastic core of the bar is 4 mm thick.
SOLUTION
(a) 3 3 4
12 4
1 1
(12)(8) 512 mm
12 12
512 10 m
I bh
u
1
4 mm 0.004 m
2
c h
6 12
(300 10 )(512 10 )
0.004
38.4 N m
Y
Y
I
M
c
V
u u
˜ 38.4 N m
Y
M ˜ W
(b)
1 2
(4) 2 mm 0.5
2 4
Y
Y
y
y
c
2
2
3 1
1
2 3
3 1
(38.4) 1 (0.5)
2 3
52.8 N m
Y
Y
y
M M
c
ª º
§ ·
« »
¨ ¸
© ¹
« »
¬ ¼
ª º
« »
¬ ¼
˜ 52.8 N m
M ˜ W
Mecánica de Materiales - Ferdinand P. Beer 7ma. Edición.

556. PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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on a website, in whole or part.
517
z
x
8 mm
12 mm
M PROBLEM 4.68
Solve Prob. 4.67, assuming that the couple M is parallel to the z axis.
PROBLEM 4.67 The prismatic bar shown is made of a steel that is
assumed to be elastoplastic with 300 MPa
Y
V and is subjected to a
couple M parallel to the x axis. Determine the moment M of the couple
for which (a) yield first occurs, (b) the elastic core of the bar is 4 mm
thick.
SOLUTION
(a)
3 3 3 4
9 4
1 1
(8)(12) 1.152 10 mm
12 12
1.152 10 m
I bh
u
u
1
6 mm 0.006 m
2
c h
6 9
(300 10 )(1.152 10 )
0.006
57.6 N m
Y
Y
I
M
c
V
u u
˜ 57.6 N m
Y
M ˜ W
(b)
1 2 1
(4) 2 mm
2 6 3
Y
Y
y
y
c
2
2
3 1
1
2 3
3 1 1
(57.6) 1
2 3 3
83.2 N m
Y
Y
y
M M
c
ª º
§ ·
« »
¨ ¸
© ¹
« »
¬ ¼
ª º
§ ·
« »
¨ ¸
© ¹
« »
¬ ¼
˜ 83.2 N m
M ˜ W
Mecánica de Materiales - Ferdinand P. Beer 7ma. Edición.

557. PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
518
PROBLEM 4.69
A solid square rod of side 0.6 in. is made of a steel that is assumed to be elastoplastic with E = 29 × 106
psi
and 48 ksi.
Y
V Knowing that a couple M is applied and maintained about an axis parallel to a side of the
cross section, determine the moment M of the couple for which the radius of curvature is 6 ft.
SOLUTION
3 3 4
3 3
3
3 3
max 6
3
3
1 1
(0.6)(0.6) 10.8 10 in 0.3 in.
12 2
(10.8 10 )(48 10 )
1728 lb in. 6 ft 72 in.
0.3
0.3 48 10
4.16667 10 1.65517 10
72 29 10
1.65517 10
0.39724
4.16667 10
3
1
2
u
u u
˜
u
u u
u
u
u
Y
Y
Y
Y
Y Y
M
Y
I c h
I
M
c
c
E
y
c
M M
V
U
V
H H
U
H
H
2
2
1 3 1
(1728) 1 (0.39724)
3 2 3
ª º
§ · ª º
« »
¨ ¸ « »
© ¹ ¬ ¼
¬ ¼
Y
Y
c
2460 lb in.
M ˜ W
Mecánica de Materiales - Ferdinand P. Beer 7ma. Edición.

558. PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
519
PROBLEM 4.70
For the solid square rod of Prob. 4.69, determine the moment M for which the radius of curvature is 3 ft.
PROBLEM 4.69. A solid square rod of side 0.6 in. is made of a steel that is assumed to be elastoplastic with
E = 29 × 106
psi and 48 ksi.
Y
V Knowing that a couple M is applied and maintained about an axis parallel
to a side of the cross section, determine the moment M of the couple for which the radius of curvature is 6 ft.
SOLUTION
3 3 4
3 3
3
3 3
max 6
3
3
max
1 1
(0.6)(0.6) 10.8 10 in 0.3 in.
12 2
(10.8 10 )(48 10 )
1728 lb in. 3 ft 36 in.
0.3
0.3 48 10
8.3333 10 1.65517 10
36 29 10
1.65517 10
0.19862
8.3333 10
3
1
2
u
u u
˜
u
u u
u
u
u
Y
Y
Y
Y
Y Y
Y
I c h
I
M
c
c
E
y
c
M M
V
U
V
H H
U
H
H
2
2
1 3 1
(1728) 1 (0.19862)
3 2 3
ª º
§ · ª º
« »
¨ ¸ « »
© ¹ ¬ ¼
¬ ¼
Y
Y
c
2560 lb in.
M ˜ W
Mecánica de Materiales - Ferdinand P. Beer 7ma. Edición.

559. PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
520
18 mm
24 mm
x
y
M!
M
PROBLEM 4.71
The prismatic rod shown is made of a steel that is assumed to be elastoplastic with
E 200 GPa and 280 MPa
Y
V . Knowing that couples M and Mc of moment
525 N · m are applied and maintained about axes parallel to the y axis, determine
(a) the thickness of the elastic core, (b) the radius of curvature of the bar.
SOLUTION
3 3 3 4 3 4
6 9
2
2
1 1
(24)(18) 11.664 10 mm 11.664 10 m
12 12
1
9 mm 0.009 m
2
(280 10 )(11.664 10 )
362.88 N m
0.009
3 1
1 or 3 2
2 3
V
u u
u u
˜
§ ·
¨ ¸
© ¹
Y
Y
Y Y
Y
Y
I bh
c h
I
M
c
y y M
M M
c M
c
(2)(525)
3 0.32632 0.32632 2.9368 mm
362.88
Y
Y
y
y c
c
(a) core 2 5.87 mm
Y
t y W
(b)
9 3
6
(200 10 )(2.9368 10 )
2.09 m
280 10
u u
?
u
Y Y Y
Y
Y
y Ey
E
V
H U
U V
W
Mecánica de Materiales - Ferdinand P. Beer 7ma. Edición.

560. PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
521
18 mm
24 mm
x
y
M!
M
PROBLEM 4.72
Solve Prob. 4.71, assuming that the couples M and Mc are applied and
maintained about axes parallel to the x axis.
PROBLEM 4.71 The prismatic rod shown is made of a steel that is assumed to
be elastoplastic with E 200 GPa and 280 MPa
Y
V . Knowing that couples
M and Mc of moment 525 N · m are applied and maintained about axes parallel
to the y axis, determine (a) the thickness of the elastic core, (b) the radius of
curvature of the bar.
SOLUTION
3 3 3 4 9 4
6 9
2
2
1 1
(18)(24) 20.736 10 mm 20.736 10 m
12 12
1
12 mm 0.012 m
2
(280 10 )(20.736 10 )
483.84 N m
0.012
3 1
1 or 3 2
2 3
V
u u
u u
˜
§ ·
¨ ¸
© ¹
Y
Y
Y Y
Y
Y
I bh
c h
I
M
c
y y M
M M
c M
c
(2)(525)
3 0.91097 0.91097 10.932 mm
483.84
Y
Y
y
y c
c
(a) core 2 21.9 mm
Y
t y W
(b)
9 3
6
(200 10 )(10.932 10 )
7.81m
280 10
u u
?
u
Y Y Y
Y
Y
y Ey
E
V
H U
U V
W
Mecánica de Materiales - Ferdinand P. Beer 7ma. Edición.

561. PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
522
z
y
90 mm
60 mm
C
PROBLEM 4.73
A beam of the cross section shown is made of a steel that is assumed to be
elastoplastic with 200 GPa
E and 240 MPa.
VY For bending about the z axis,
determine the bending moment at which (a) yield first occurs, (b) the plastic
zones at the top and bottom of the bar are 30 mm thick.
SOLUTION
(a) 3 3 6 4 6 4
6 6
1 1
(60)(90) 3.645 10 mm 3.645 10 m
12 12
1
45 mm 0.045 m
2
(240 10 )(3.645 10 )
19.44 10N m
0.045
Y
Y
I bh
c h
I
M
c
V
u u
u u
u ˜
19.44 kN m
Y
M ˜ W
6
1 1
3
1
6
2 2
3
2
(240 10 )(0.060)(0.030)
432 10 N
15 mm 15 mm 0.030 m
1 1
(240 10 )(0.060)(0.015)
2 2
108 10 N
2
(15 mm) 10 mm 0.010 m
3
V
V
u
u
§ ·
u
¨ ¸
© ¹
u
Y
Y
R A
y
R A
y
(b) 3 3
1 1 2 2
2( ) 2[(432 10 )(0.030) (108 10 )(0.010)]
M R y R y
u u
3
28.08 10 N m
u ˜ 28.1 kN m
M ˜ W
Mecánica de Materiales - Ferdinand P. Beer 7ma. Edición.

562. PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
523
30 mm
30 mm
30 mm
30 mm
15 mm
15 mm
z
y
C
PROBLEM 4.74
A beam of the cross section shown is made of a steel that is assumed to be
elastoplastic with 200 GPa
E and 240 MPa.
Y
V For bending about the
z axis, determine the bending moment at which (a) yield first occurs,
(b) the plastic zones at the top and bottom of the bar are 30 mm thick.
SOLUTION
(a) 3 3 6 4
rect
3 3 3 4
cutout
6 3 6 4
6 4
1 1
(60)(90) 3.645 10 mm
12 12
1 1
(30)(30) 67.5 10 mm
12 12
3.645 10 67.5 10 3.5775 10 mm
3.5775 10 mm
I bh
I bh
I
u
u
u u u
u
6 6
3
1
45 mm 0.045 m
2
(240 10 )(3.5775 10 )
0.045
19.08 10 N m
Y
Y
c h
I
M
c
V
u u
u ˜ 19.08 kN m
Y
M ˜ W
6 3
1 1
1
6 3
2 2
2
(240 10 )(0.060)(0.030) 432 10 N
15 mm 15 mm 30 mm 0.030 m
1 1
(240 10 )(0.030)(0.015) 54 10 N
2 2
2
(15 mm) 10 mm 0.010 m
3
Y
Y
R A
y
R A
y
V
V
u u
u u
(b) 1 1 2 2
2( )
M R y R y
3 3
3
2[(432 10 )(0.030) (54 10 )(0.010)]
27.00 10 N m
u u
u ˜ 27.0 kN m
M ˜ W
Mecánica de Materiales - Ferdinand P. Beer 7ma. Edición.

563. PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
524
3 in.
3 in.
3 in.
3 in.
1.5 in. 1.5 in.
z
y
C
PROBLEM 4.75
A beam of the cross section shown is made of a steel that is assumed to
be elastoplastic with 6
29 10 psi
E u and 42 ksi.
Y
V For bending
about the z axis, determine the bending moment at which (a) yield first
occurs, (b) the plastic zones at the top and bottom of the bar are 3 in.
thick.
SOLUTION
(a) 3 2 3 2 4
1 1 1 1 1
3 3 4
2 2 2
4
3 1
4
1 2 3
1 1
(3)(3) (3)(3)(3) 87.75 in
12 12
1 1
(6)(3) 13.5 in
12 12
87.75 in
188.5 in
I b h A d
I b h
I I
I I I I
4.5 in.
(42)(188.5)
4.5
Y
Y
c
I
M
c
V
1759 kip in.
˜
Y
M W
1 1
1
2 2
2
(42)(3)(3) 378 kip
1.5 1.5 3.0 in.
1 1
(42)(6)(1.5)
2 2
189 kip
2
(1.5) 1.0 in.
3
Y
Y
R A
y
R A
y
V
V
(b) 1 1 2 2
2( ) 2[(378)(3.0) (189)(1.0)] 2646 kip in.
M R y R y
˜ 2650 kip in.
˜
M W
Mecánica de Materiales - Ferdinand P. Beer 7ma. Edición.

564. PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
525
3 in.
3 in.
3 in.
3 in.
1.5 in. 1.5 in.
z
y
C
PROBLEM 4.76
A beam of the cross section shown is made of a steel that is assumed to
be elastoplastic with 6
29 10 psi
E u and 42 ksi.
Y
V For bending
about the z axis, determine the bending moment at which (a) yield first
occurs, (b) the plastic zones at the top and bottom of the bar are 3 in.
thick.
SOLUTION
(a) 3 2 3 2 4
1 1 1 1 1
3 3 4
2 2 2
4
3 1
4
1 2 3
1 1
(6)(3) (6)(3)(3) 175.5 in
12 12
1 1
(3)(3) 6.75 in
12 12
175.5 in
357.75 in
4.5 in.
I b h A d
I b h
I I
I I I I
c
(42)(357.75)
3339 kip in.
4.5
˜
Y
Y
I
M
c
V
3340 kip in.
˜
Y
M W
1 1
1
2 2
2
(42)(6)(3) 756 kip
1.5 1.5 3 in.
1 1
(42)(3)(1.5)
2 2
94.5 kip
2
(1.5) 1.0 in.
3
V
V
Y
Y
R A
y
R A
y
(b) 1 1 2 2
2( ) 2[(756)(3) (94.5)(1.0)] 4725 kip in.
M R y R y
˜ 4730 kip in.
˜
M W
Mecánica de Materiales - Ferdinand P. Beer 7ma. Edición.

565. PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
526
z
y
90 mm
60 mm
C
PROBLEM 4.77
For the beam indicated (of Prob. 4.73), determine (a) the fully plastic moment
,
p
M (b) the shape factor of the cross section.
SOLUTION
From Problem 4.73, 200 GPa and 240 MPa
VY
E
2
1
6 2
1
6 6
3
(60)(45) 2700 mm
2700 10 m
(240 10 )(2700 10 )
648 10 N
45 mm 0.045 m
Y
A
R A
d
V
u
u u
u
(a) 3 3
(648 10 )(0.045) 29.16 10 N m
u u ˜
p
M Rd 29.2 kN m
p
M ˜ W
(b) 3 3 6 4 6 4
6 6
3
1 1
(60)(90) 3.645 10 mm 3.645 10 m
12 12
45 mm 0.045 m
(240 10 )(3.645 10 )
19.44 10 N m
0.045
Y
Y
I bh
c
I
M
c
V
u u
u u
u ˜
29.16
19.44
p
Y
M
k
M
1.500
k W
Mecánica de Materiales - Ferdinand P. Beer 7ma. Edición.

566. PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
527
30 mm
30 mm
30 mm
30 mm
15 mm
15 mm
z
y
C
PROBLEM 4.78
For the beam indicated (of Prob. 4.74), determine (a) the fully plastic moment
,
p
M (b) the shape factor of the cross section.
SOLUTION
From Problem 4.74, 200 GPa
E and 240 MPa
VY
(a) 1 1
6
3
1
(240 10 )(0.060)(0.030)
432 10 N
15 mm 15 mm 30 mm
0.030 m
Y
R A
y
V
u
u
2 2
6
3
2
(240 10 )(0.030)(0.015)
108 10 N
1
(15) 7.5 mm 0.0075 m
2
Y
R A
y
V
u
u
3 3
1 1 2 2
3
2( ) 2[(432 10 )(0.030) (108 10 )(0.0075)]
27.54 10 N m
p
M R y R y
u u
u ˜ 27.5 kN m
p
M ˜ W
(b) 3 3 6 4
rect
3 3 3 4
cutout
6 3 3 4
rect cutout
1 1
(60)(90) 3.645 10 mm
12 12
1 1
(30)(30) 67.5 10 mm
12 12
3.645 10 67.5 10 3.5775 10 mm
I bh
I bh
I I I
u
u
u u u
9 4
3.5775 10 m
u
1
45 mm 0.045 m
2
c h
6 9
3
(240 10 )(3.5775 10 )
19.08 10 N m
0.045
Y
Y
I
M
c
V
u u
u ˜
27.54
19.08
p
Y
M
k
M
1.443
k W
Mecánica de Materiales - Ferdinand P. Beer 7ma. Edición.

567. PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
528
3 in.
3 in.
3 in.
3 in.
1.5 in. 1.5 in.
z
y
C
PROBLEM 4.79
For the beam indicated (of Prob. 4.75), determine (a) the fully plastic
moment ,
p
M (b) the shape factor of the cross section.
SOLUTION
From Problem 4.75, 6
29 10 psi and 42 ksi.
Y
E V
u
(a) 1 1
1
2 2
2
(42)(3)(3) 378 kip
1.5 1.5 3.0 in.
(42)(6)(1.5) 378 kip
1
(1.5) 0.75 in.
2
Y
Y
R A
y
R A
y
V
V
1 1 2 2
2( ) 2[(378)(3.0) (378)(0.75)]
p
M R y R y
2840 kip in.
p
M ˜ W
(b) 3 2 3 2 4
1 1 1 1 1
3 3 4
2 2 2
4
3 1
4
1 2 3
1 1
(3)(3) (3)(3)(3) 87.75 in
12 12
1 1
(6)(3) 13.5 in
12 12
87.75 in
188.5 in
4.5 in.
I b h A d
I b h
I I
I I I I
c
(42)(188.5)
1759.3 kip in
4.5
2835
1759.3
Y
Y
p
Y
I
M
c
M
k
M
V
˜
1.611
k W
Mecánica de Materiales - Ferdinand P. Beer 7ma. Edición.

568. PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
529
3 in.
3 in.
3 in.
3 in.
1.5 in. 1.5 in.
z
y
C
PROBLEM 4.80
For the beam indicated (of Prob. 4.76), determine (a) the fully plastic
moment ,
p
M (b) the shape factor of the cross section.
SOLUTION
From Problem 4.76, 6
29 10
E u and 42 ksi
VY
(a) 1 1
1
2 2
2
(42)(6)(3) 756 kip
1.5 1.5 3.0 in.
(42)(3)(1.5) 189 kip
1
(1.5) 0.75 in.
2
Y
Y
R A
y
R A
y
V
V
1 1 2 2
2( ) 2[(756)(3.0) (189)(0.75)]
p
M R y R y
4820 kip in.
p
M ˜ W
(b) 3 2 3 2 4
1 1 1 1 1
3 3 4
2 2 2
4
3 1
4
1 2 3
1 1
(6)(3) (6)(3)(3) 175.5 in
12 12
1 1
(3)(3) 6.75 in
12 12
175.5 in
357.75 in
4.5 in.
I b h A d
I b h
I I
I I I I
c
(42)(357.75)
3339 kip in.
4.5
4819.5
3339
V
˜
Y
Y
p
Y
I
M
c
M
k
M
1.443
k W
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530
r ! 18 mm
PROBLEM 4.81
Determine the plastic moment p
M of a steel beam of the cross section shown,
assuming the steel to be elastoplastic with a yield strength of 240 MPa.
SOLUTION
For a semicircle, 2 4
;
2 3
r
A r r
S
S
Resultant force on semicircular section: Y
R A
V
Resultant moment on entire cross section:
3
4
2
3
p Y
M Rr r
V
Data: 6
240 MPa 240 10 Pa, 18 mm 0.018 m
Y r
V u
6 3
4
(240 10 )(0.018) 1866 N m
3
p
M u ˜
1.866 kN m
p
M ˜ W
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570. PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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531
50 mm
30 mm
10 mm
30 mm
10 mm
10 mm
PROBLEM 4.82
Determine the plastic moment p
M of a steel beam of the cross section shown,
assuming the steel to be elastoplastic with a yield strength of 240 MPa.
SOLUTION
Total area: 2
(50)(90) (30)(30) 3600 mm
A
2
1
2
1
1800 mm
2
1800
36 mm
50
A
A
x
b
2 3 3
1 1 1 1
2 3 3
2 2 2 2
2 3 3
3 3 3 3
2 3 3
4 4 4 4
(50)(36) 1800 mm , 18 mm, 32.4 10 mm
(50)(14) 700 mm , 7 mm, 4.9 10 mm
(20)(30) 600 mm , 29 mm, 17.4 10 mm
(50)(10) 500 mm , 49 mm, 24.5 10 mm
A y A y
A y A y
A y A y
A y A y
u
u
u
u
3 3 6 3
1 1 2 2 3 3 4 4
6 6 3
79.2 10 mm 79.2 10 m
(240 10 )(79.2 10 ) 19.008 10 N m
p Y i i
A y A y A y A y
M A y
V
u u
6 u u u ˜
19.01 kN m
p
M ˜ W
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532
36 mm
30 mm
PROBLEM 4.83
Determine the plastic moment p
M of a steel beam of the cross section shown,
assuming the steel to be elastoplastic with a yield strength of 240 MPa.
SOLUTION
Total area: 2
1
(30)(36) 540 mm
2
A Half area: 2
1
1
270 mm
2
A A
By similar triangles,
30 5
36 6
b
b y
y
Since 2 2
1 1
1 5 12
,
2 12 5
12
(270) 25.4558 mm
5
21.2132 mm
A by y y A
y
b
2 6 2
1
2 6 2
2
2 6 2
3 1 2
6
3 3 3
1 2 3
1
1
(21.2132)(25.4558) 270 mm 270 10 m
2
(21.2132)(36 25.4558) 223.676 mm 223.676 10 m
46.324 mm 46.324 10 m
240 10
64.8 10 N, 53.6822 10 N, 11.1178 10 N
1
8.485
3
i Y i i
A
A
A A A A
R A A
R R R
y y
V
u
u
u
u
u u u
3
3
2
3
3
3 mm 8.4853 10 m
1
(36 25.4558) 5.2721 mm 5.2721 10 m
2
2
(36 25.4558) 7.0295 mm 7.0295 10 m
3
y
y
u
u
u
1 1 2 2 3 3 911 N m
p
M R y R y R y
˜ 911 N m
p
M ˜ W
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Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
533
0.4 in. 0.4 in.
0.4 in.
1.0 in.
1.0 in.
PROBLEM 4.84
Determine the plastic moment Mp of a steel beam of the cross section shown,
assuming the steel to be elastoplastic with a yield strength of 42 ksi.
SOLUTION
2
1
2
2
1
1
2
2
2
2
3
3
(1.0 in.)(0.4 in.) 2(1.4 in.)(0.4 in.) 1.52 in
(1.52)
0.95 in.
2(0.4)
(1.0)(0.4) 0.4 in
1.2 0.95 0.25 in.
2(0.4)(1.4 0.95) 0.36 in
1
(1.4 0.95) 0.225 in.
2
2(0.4)(0.95) 0.760 in
1
(0.95) 0.475 i
2
A
x
R
y
R
y
R
y n.
1 2 3
1 2 3
( )( )
[(0.4)(0.25) (0.36)(0.225) (0.760)(0.475)](42)
p y
M R y R y R y V
22.8 kip in.
˜
p
M W
Mecánica de Materiales - Ferdinand P. Beer 7ma. Edición.

573. PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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534
5 mm
80 mm
5 mm
120 mm
t = 5 mm
PROBLEM 4.85
Determine the plastic moment Mp of the cross section shown when the beam is
bent about a horizontal axis. Assume the material to be elastoplastic with a
yield strength of 175 MPa.
SOLUTION
For ,
p
M the neutral axis divides the area into two equal parts.
Total area (100 100 120) 320
t t
1
Shaded area 2 (320)
2
80 mm
80
(80 mm) 64 mm
100
at t
a
b
6 2
1
6 2
2
6 2
3
2 2(0.08 m)(0.005 m) 800 10 m
2(100 ) 2(0.02 m)(0.005 m) 200 10 m
(120 mm) (0.120 m)(0.005 m) 600 10 m
u
u
u
A at
A a t
A t
6 2
1 1
6 2
2 2
6 2
3 3
(175 MPa)(800 10 m ) 140 kN
(175 MPa)(200 10 m ) 35 kN
(175 MPa)(600 10 m ) 105 kN
u
u
u
Y
Y
Y
R A
R A
R A
V
V
V
1 2 3
: (32 mm) (8 mm) (16 mm)
(140 kN)(0.032 m) (35 kN)(0.008 m) (105 kN)(0.016 m)
¦
z p
M M R R R
= 6.44 kN m
˜
p
M W
Mecánica de Materiales - Ferdinand P. Beer 7ma. Edición.

574. PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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on a website, in whole or part.
535
2 in.
4 in.
3 in.
in.
1
2
in.
1
2
in.
1
2
PROBLEM 4.86
Determine the plastic moment p
M of a steel beam of the cross section shown, assuming
the steel to be elastoplastic with a yield strength of 36 ksi.
SOLUTION
Total area: 2
2
1 1 1
(4) (3) (2) 4.5 in
2 2 2
1
2.25 in
2
A
A
§ · § · § ·
¨ ¸ ¨ ¸ ¨ ¸
© ¹ © ¹ © ¹
2 3
1 1 1 1
2 3
2 2 2 2
2 3
3 3 3 3
2 3
4 4 4 4
2.00 in , 0.75, 1.50 in
0.25 in , 0.25, 0.0625 in
1.25 in , 1.25, 1.5625 in
1.00 in , 2.75, 2.75 in
A y A y
A y A y
A y A y
A y A y
1 1 2 2 3 3 4 4
( )
(36)(1.50 0.0625 1.5625 2.75)
p Y
M A y A y A y A y
V
212 kip in.
˜
p
M W
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575. PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
536
z
y
90 mm
60 mm
C
PROBLEM 4.87
For the beam indicated (of Prob. 4.73), a couple of moment equal to the full
plastic moment p
M is applied and then removed. Using a yield strength of
240 MPa, determine the residual stress at 45mm
y .
SOLUTION
3
29.16 10 N m
p
M u ˜
See solutions to Problems 4.73 and 4.77.
6 4
max
3.645 10 m
0.045 m
p
I
c
M c
M y
I I
V
u
c at 45 mm
y c
LOADING UNLOADING RESIDUAL STRESSES
3
6
6
6 6
res
6
(29.16 10 )(0.045)
360 10 Pa
3.645 10
360 10 240 10
120 10 Pa
Y
V
V V V
u
c u
u
c u u
u res 120.0 MPa
V W
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576. PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
537
30 mm
30 mm
30 mm
30 mm
15 mm
15 mm
z
y
C
PROBLEM 4.88
For the beam indicated (of Prob. 4.74), a couple of moment equal to the
full plastic moment p
M is applied and then removed. Using a yield
strength of 240 MPa, determine the residual stress at 45mm
y .
SOLUTION
3
27.54 10 N m
p
M u ˜ (See solutions to Problems 4.74 and 4.78.)
6 4
max
3
6
6
3.5775 10 m , 0.045 m
at
(27.54 10 )(0.045)
346.4 10 Pa
3.5775 10
p
I c
M c
M y
y c
I I
V
V
u
c
u
c u
u
LOADING UNLOADING RESIDUAL STRESSES
6 6 6
res 346.4 10 240 10 106.4 10 Pa
Y
V V V
c u u u
res 106.4 MPa
V W
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577. PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
538
3 in.
3 in.
3 in.
3 in.
1.5 in. 1.5 in.
z
y
C
PROBLEM 4.89
A bending couple is applied to the bar indicated, causing plastic zones
3 in. thick to develop at the top and bottom of the bar. After the couple
has been removed, determine (a) the residual stress at 4.5 in.,
y
(b) the points where the residual stress is zero, (c) the radius of
curvature corresponding to the permanent deformation of the bar.
Beam of Prob. 4.75.
SOLUTION
See solution to Problem 4.75 for bending couple and stress distribution during loading.
6 3
4
2646 kip in. 1.5 in. 29 10 psi 29 10 ksi
42 ksi 188.5 in 4.5 in.
V
˜ u u
Y
Y
M y E
I c
(a)
(2646)(4.5)
63.167 ksi
188.5
(2646)(1.5)
21.056 ksi
188.5
c
cc Y
Mc
I
My
I
V
V
res
res
At , 63.167 42 21.167 ksi
At , 21.056 42 20.944 ksi
c
cc
Y
Y Y
y c
y y
V V V
V V V res 20.9 ksi
V W
LOADING UNLOADING RESIDUAL STRESSES
(b) 0
res 0 Y
My
I
V V
?
0
(188.5)(42)
2.99 in.
2646
Y
I
y
M
V
Answer: 0 2.99 in., 0, 2.99 in.
y W
(c) res
At , 20.944 ksi
Y
y y V
3
(29 10 )(1.5)
2077 in.
20.944
V U
U V
u
?
Ey Ey
173.1ft
U W
Mecánica de Materiales - Ferdinand P. Beer 7ma. Edición.

578. PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
539
3 in.
3 in.
3 in.
3 in.
1.5 in. 1.5 in.
z
y
C
PROBLEM 4.90
A bending couple is applied to the bar indicated, causing plastic zones
3 in. thick to develop at the top and bottom of the bar. After the couple
has been removed, determine (a) the residual stress at 4.5 in.
y ,
(b) the points where the residual stress is zero, (c) the radius of
curvature corresponding to the permanent deformation of the bar.
Beam of Prob. 4.76.
SOLUTION
See solution to Problem 4.76 for bending couple and stress distribution.
6 3
4725 kip in. 1.5 in. 29 10 psi 29 10 ksi
˜ u u
Y
M y E
4
42 ksi 357.75 in 4.5 in.
Y I c
V
(a)
res
res
(4725)(4.5)
59.434 ksi
357.75
(4725)(1.5)
19.8113 ksi
357.75
At , 59.434 42 17.4340 ksi
At , 19.8113 42 22.189 ksi
c
cc
c
cc
Y
Y
Y Y
Mc
I
My
I
y c
y y
V
V
V V V
V V V
59.4 ksi
c
V W
LOADING UNLOADING RESIDUAL STRESSES
(b) 0
res 0 0
Y
My
I
V V
?
0
(357.75)(42)
3.18 in.
4725
Y
I
y
M
V
Answer: 0 3.18 in., 0, 3.18 in.
y W
(c) res
At , 22.189ksi
Y
y y V
3
(29 10 )(1.5)
1960.43 in.
22.189
u
?
Ey Ey
V U
U V
163.4 ft.
U W
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579. PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
540
z
y
90 mm
60 mm
C
PROBLEM 4.91
A bending couple is applied to the beam of Prob. 4.73, causing plastic zones
30 mm thick to develop at the top and bottom of the beam. After the couple has
been removed, determine (a) the residual stress at 45 mm,
y (b) the points
where the residual stress is zero, (c) the radius of curvature corresponding to
the permanent deformation of the beam.
SOLUTION
See solution to Problem 4.73 for bending couple and stress distribution during loading.
3
6 4
28.08 10 N m 15 mm 0.015 m 200 GPa
240 MPa 3.645 10 m 0.045 m
Y
Y
M y E
I c
V
u ˜
u
(a)
3
6
6
3
6
6
(28.08 10 )(0.045)
346.67 10 Pa 346.67 MPa
3.645 10
(28.08 10 )(0.015)
115.556 10 Pa 115.556 MPa
3.645 10
V
V
u
c u
u
u
cc u
u
Y
Mc
I
M y
I
At res
, 346.67 240 106.670 MPa
Y
y c V V V
c res 106.7 MPa
V W
At res
, 115.556 240 124.444 MPa
Y Y
y y V V V
cc res 124.4 MPa
V
LOADING UNLOADING RESIDUAL STRESSES
(b) 0
res 0 0
V V
? Y
My
I
6 6
3
0 3
(3.645 10 )(240 10 )
31.15 10 m 31.15 mm
28.08 10
Y
I
y
M
V
u u
u
u
0 31.2 mm, 0, 31.2 mm
y
Answer: W
(c) 6
res
At , 124.444 10 Pa
Y
y y V u
9
6
(200 10 )(0.015)
124.444 10
Ey Ey
V U
U V
u
?
u
24.1 m
U W
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580. PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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541
C
z
y
1 in.
1 in.
1 in.
1 in.
1 in.
2 in.
PROBLEM 4.92
A beam of the cross section shown is made of a steel that is assumed to be
elastoplastic with E 29 × 106
psi and = 42 ksi.
Y
V A bending couple is
applied to the beam about the z axis, causing plastic zones 2 in. thick to
develop at the top and bottom of the beam. After the couple has been removed,
determine (a) the residual stress at y 2 in., (b) the points where the residual
stress is zero, (c) the radius of curvature corresponding to the permanent
deformation of the beam.
SOLUTION
Flange: 3 2 3 2 4
1 1 1 1 1
1 1
(3)(1) (3)(1)(1.5) 7.0 in
12 12
I b h A d
Web:
3 3 4
2 2 2
4
3 1
4
1 2 3
1 1
(1)(2) 0.6667 in
12 12
7.0 in
14.6667 in
2 in.
(42)(14.6667)
2
Y
Y
I b h
I I
I I I I
c
I
M
c
V
308 kip in.
˜
Y
M W
1 1
1
2 2
2
(42)(3)(1) 126 kips
1.0 0.5 1.5 in.
1 1
(42)(1)(1)
2 2
21 kips
2
(1.0) 0.6667 in.
3
Y
Y
R A
y
R A
y
V
V
1 2
1 2
2( ) 2[(126)(1.5) (21)(0.6667)]
M R R
y y 406 kip in.
˜
M W
6 3
4
406 kip in. 1.0 in. 29 10 29 10 ksi
42 ksi 14.6667 in 2 in.
˜ u u
Y
Y
M y E
I c
V
(a)
(406)(2)
55.364
14.6667
Mc
I
V c
(406)(1.0)
27.682
14.662
Y
My
I
V cc
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542
PROBLEM 4.92 (Continued)
res
At , 55.364 42 13.3640 ksi
Y
y c V V V
c
res 13.36 ksi
V W
res
At , 27.682 42 14.3180 ksi
c
Y Y
y y V V V
res 14.32 ksi
V W
(b) 0
res
0
0 0
(14.667)(42)
1.517 in
406
Y
Y
My
I
I
y
M
?
V V
V
0
: 1.517 in., 0,1.517 in.
y
Answer W
(c) res
3
At , 14.3180 ksi
(29 10 )(1.0)
2025.42 in.
14.3180
Y
y y
Ey Ey
u
?
V
V U
U V
168.8 ft
U W
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582. PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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543
PROBLEM 4.93*
A rectangular bar that is straight and unstressed is bent into an arc of circle of radius by two couples of
moment M. After the couples are removed, it is observed that the radius of curvature of the bar is R
U .
Denoting by Y
U the radius of curvature of the bar at the onset of yield, show that the radii of curvature satisfy
the following relation:
2
1 1 3 1
1 1
2 3
R Y Y
U U
U U U U
­ ½
ª º
§ ·
° °
« »
® ¾
¨ ¸
« »
© ¹
° °
¬ ¼
¯ ¿
SOLUTION
2
2
1 3 1
, 1 ,
2 3
Y
Y
Y Y
M
M M
EI
U
U U
§ ·
¨ ¸
¨ ¸
© ¹
Let m denote .
Y
M
M
2 2
2 2
2
2
3
1 3 2
2
1 1 1 1
1 1 3 1
1 1 1
2 3
Y Y Y
Y
R Y
Y Y Y
M
m m
M
M mM m
EI EI
m
U U
U U
U U U U U
U U U
U U U U U
§ ·
?
¨ ¸
¨ ¸
© ¹
­ ½
§ ·
­ ½ ° °
¨ ¸
® ¾ ® ¾
¨ ¸
¯ ¿ ° °
© ¹
¯ ¿
W
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583. PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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544
PROBLEM 4.94
A solid bar of rectangular cross section is made of a material that is assumed to be elastoplastic. Denoting by
Y
M and ,
Y
U respectively, the bending moment and radius of curvature at the onset of yield, determine (a) the
radius of curvature when a couple of moment 1.25 Y
M M is applied to the bar, (b) the radius of curvature
after the couple is removed. Check the results obtained by using the relation derived in Prob. 4.93.
SOLUTION
(a)
2
2
1 3 1
, 1
2 3
Y
Y
Y Y
M
M M
EI
U
U U
§ ·
¨ ¸
¨ ¸
© ¹
Let 1.25
Y
M
m
M
2
2
3 1
1 3 2 0.70711
2 3
Y Y
Y
M
m m
M
U U
U
U
§ ·
¨ ¸
¨ ¸
© ¹
0.707 Y
U U W
(b)
1 1 1 1 1 1.25
0.70711
Y
R Y Y Y
M mM m
EI EI
U U U U U U U
0.16421
Y
U
6.09
R Y
U U W
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545
B
A
16 mm 20 mm
M PROBLEM 4.95
The prismatic bar AB is made of a steel that is assumed to be elastoplastic and
for which 200 GPa
E . Knowing that the radius of curvature of the bar is
2.4 m when a couple of moment 350 N m
M ˜ is applied as shown,
determine (a) the yield strength of the steel, (b) the thickness of the elastic
core of the bar.
SOLUTION
2
2
2 2
2 2
3 2 2
2 2
2 2
2
2 2
3 1
1
2 3
3 1
1
2 3
3 (2 ) 1
1
2 12 3
1
1
3
Y
Y
Y Y
Y Y
Y
Y
M M
I
c E c
b c
c E c
bc
E c
U
U
V U V
V U V
U V
V
§ ·
¨ ¸
¨ ¸
© ¹
§ ·
¨ ¸
¨ ¸
© ¹
§ ·
¨ ¸
¨ ¸
© ¹
§ ·
¨ ¸
¨ ¸
© ¹
(a)
2 2
2
2 2
1
3
Y
Y
bc M
E c
§ ·
¨ ¸
¨ ¸
© ¹
U V
V Cubic equation for Y
V
Data: 9
200 10 Pa
420 N m
2.4 m
20 mm 0.020 m
1
8 mm 0.008 m
2
E
M
b
c h
U
u
˜
6 21 2
21 2 6
(1.28 10 ) 1 750 10 350
1 750 10 273.44 10
Y Y
Y Y
V V
V V
ª º
u u
¬ ¼
ª º
u u
¬ ¼
Solving by trial, 6
292 10 Pa
Y
V u 292 MPa
Y
V W
(b)
6
3
9
(292 10 )(2.4)
3.504 10 m 3.504 mm
200 10
Y
Y
y
E
V U
u
u
u
thickness of elastic core 2 7.01 mm
Y
y W
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546
M
60 mm
40 mm
A
M'
B
#
(MPa)
300
200
100
0 0.005 0.010
PROBLEM 4.96
The prismatic bar AB is made of an aluminum
alloy for which the tensile stress-strain diagram
is as shown. Assuming that the -
V H diagram is
the same in compression as in tension,
determine (a) the radius of curvature of the bar
when the maximum stress is 250 MPa, (b) the
corresponding value of the bending moment.
(Hint: For part b, plot V versus y and use an
approximate method of integration.)
SOLUTION
(a) 6
250 MPa 250 10 Pa
0.0064 from curve
1
h 30 mm 0.030 m
2
40 mm 0.040 m
m
m
c
b
V
H
u
1
1 0.0064
0.21333 m
0.030
m
c
H
U
4.69 m
U W
(b) Strain distribution: m m
y
u
c
H H H
where
y
u
H
Bending couple:
1
2 2
0 0
2 | | 2 | | 2
c c
c
M y bdy b y dy bc u du bc J
³ ³ ³
V V V
where the integral J is given by
1
0
| |
u du
V
³
Evaluate J using a method of numerial integration. If Simpson’s rule is used, the integration formula is
| |
3
u
J wu V
'
6
where w is a weighting factor.
Using 0.25,
u
' we get the values given in the table below:
u | |
H | |, (MPa)
V | |, (MPa)
u V w | |, (MPa)
wu V
0 0 0 0 1 0
0.25 0.0016 110 27.5 4 110
0.5 0.0032 180 90 2 180
0.75 0.0048 225 168.75 4 675
1.00 0.0064 250 250 1 250
1215 | |
wu V
m 6
6
(0.25)(1215)
101.25 MPa 101.25 10 Pa
3
J u
2 6 3
(2)(0.040)(0.030) (101.25 10 ) 7.29 10 N m
M u u ˜ 7.29 kN m
M ˜ W
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547
1.2 in.
0.8 in.
A
B
#
(ksi)
50
30
40
20
10
0
0.004 0.008
M
PROBLEM 4.97
The prismatic bar AB is made of a bronze alloy for which the tensile
stress-strain diagram is as shown. Assuming that the -
V H diagram is
the same in compression as in tension, determine (a) the maximum
stress in the bar when the radius of curvature of the bar is 100 in.,
(b) the corresponding value of the bending moment. (See hint given
in Prob. 4.96.)
SOLUTION
(a) 100 in., 0.8 in., 0.6 in.
b c
U
0.6
0.006
100
m
c
H
U
From the curve, 43.0 ksi
Vm W
(b) Strain distribution: where
m m
y y
u u
c
H H H
H
Bending couple:
1
2 2
0 0
2 | | 2 | | 2
c c
c
M y bdy b y dy bc u du bc J
³ ³ ³
V V V
where the integral J is given by
1
0
| |
u du
V
³
Evaluate J using a method of numerial integration. If Simpson’s rule is used, the integration formula is
| |
3
u
J wu V
'
6
where w is a weighting factor.
Using 0.25,
u
' we get the values given the table below:
u | |
H | |, ksi
V | |, ksi
u V w | |, ksi
wu V
0 0 0 0 1 0
0.25 0.0015 25 6.25 4 25
0.5 0.003 36 18 2 36
0.75 0.0045 40 30 4 120
1.00 0.006 43 43 1 43
224 | |
wu V
m 6
(0.25)(224)
18.67 ksi
3
J
2
(2)(0.8)(0.6) (18.67)
M 10.75 kip in.
˜
M W
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548
#
M
PROBLEM 4.98
A prismatic bar of rectangular cross section is made of an alloy for
which the stress-strain diagram can be represented by the relation
n
k
H V for 0
V ! and n
k
H V
for 0
V . If a couple M is
applied to the bar, show that the maximum stress is
1 2
3
m
n Mc
n I
V
SOLUTION
Strain distribution: where
m m
y y
u u
c c
H H H
Bending couple:
2
0 0
1
2
0
2 | | 2 | |
2 | |
c c c
c
y dy
M y bdy b y dy bc
c c
bc u du
³ ³ ³
³
V V V
V
For
1
,
| | n
n
m m
n
m
m m
K K
u u
H V H V
H V
V V
H V
§ ·
?
¨ ¸
© ¹
Then
1 1
1
1
1 1
2 2
0 0
2
1
2 2
0
1
2
2 2
2
2
2 2 1
2 1
2
n n
n
m m
m m
n
m
M bc u u du bc u du
u n
bc bc
n
n M
bc
³ ³
³
V V
V V
V
Recall that
3
2
2
1 (2 ) 2 1 2
12 3 3
I b c c
bc
c c I
bc
?
Then
2 1
3
m
n Mc
n I
V
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549
45 mm
30 mm
24 mm
15 mm
A
D
B
P
PROBLEM 4.99
Knowing that the magnitude of the horizontal force P is 8 kN,
determine the stress at (a) point A, (b) point B.
SOLUTION
2 6 2
3 3 3 4 9 4
3
3
(30)(24) 720 mm 720 10 m
45 12 33 mm 0.033 m
1 1
(30)(24) 34.56 10 mm 34.56 10 m
12 12
1
(24 mm) 12 mm 0.012 m 8 10 N
2
(8 10 )(0.033) 264 N m
A
e
I bh
c P
M Pe
u
u u
u
u ˜
(a)
3
6
6 9
8 10 (264)(0.012)
102.8 10 Pa
720 10 34.56 10
A
P Mc
A I
u
u
u u
V
102.8 MPa
A
V W
(b)
3
6
6 9
8 10 (264)(0.012)
80.6 10 Pa
720 10 34.56 10
B
P Mc
A I
V
u
u
u u
80.6 MPa
B
V W
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550
y
z x
6 kips
3 in.
A
C
b
PROBLEM 4.100
A short wooden post supports a 6-kip axial load as shown. Determine the
stress at point A when (a) 0,
b (b) 1.5 in.,
b (c) 3 in.
b
SOLUTION
2 2 2
4 4 4
3
(3) 28.27 in
(3) 63.62 in
4 4
63.62
21.206 in
3
6 kips
A r
I r
I
S
c
P M Pb
S S
S S
(a) 0 0
6
0.212 ksi
28.27
b M
P
A
V
212 psi
V W
(b) 1.5 in. (6)(1.5) 9 kip in.
b M ˜
6 9
0.637 ksi
28.27 21.206
P M
A S
V
637 psi
V W
(c) 3 in. (6)(3) 18 kip in.
b M ˜
6 18
1.061 ksi
28.27 21.206
P M
A S
V
1061 psi
V W
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551
P
P
r r
PROBLEM 4.101
Two forces P can be applied separately or at the same time to a plate that is welded
to a solid circular bar of radius r. Determine the largest compressive stress in the
circular bar, (a) when both forces are applied, (b) when only one of the forces is
applied.
SOLUTION
For a solid section, 2 4
, ,
4
A r I r c r
S
S
Compressive stress
2 3
4
F Mc
A I
F M
r r
V
S S
(a) Both forces applied. 2 , 0
F P M
2
2
P
r
V
S
W
(b) One force applied. ,
F P M Pr
2 2
4
r
F P
r r
V
S S
2
5
P
r
V
S
W
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552
x
z
y
30 kN
20 kN
100 kN
30 mm
60 mm
120 mm
90 mm
90 mm
A
B
D
C
PROBLEM 4.102
A short 120 × 180-mm column supports the three axial loads
shown. Knowing that section ABD is sufficiently far from the
loads to remain plane, determine the stress at (a) corner A,
(b) corner B.
SOLUTION
3 2
(0.120 m)(0.180 m) 21.6 10 m
A
u
2 4 2
1
(0.120 m)(0.180 m) 6.48 10 m
6
S
u
(30 kN)(0.03 m) (100 kN)(0.06 m) 5.10 kN m
M ˜
(a)
3 3
3 2 4 3
150 10 N 5.10 10 N m
21.6 10 m 6.48 10 m
A
P M
A S
u u ˜
u u
V
0.926 MPa
A
V W
(b)
3 3
3 2 4 3
150 10 N 5.10 10 N m
21.6 10 m 6.48 10 m
B
P M
A S
u u ˜
u u
V
14.81MPa
B
V W
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553
80 mm
80 mm
2
3
1
C
A
P
P
P
PROBLEM 4.103
As many as three axial loads, each of magnitude P 50 kN, can be applied to
the end of a W200 × 31.1 rolled-steel shape. Determine the stress at point A
(a) for the loading shown, (b) if loads are applied at points 1 and 2 only.
SOLUTION
For W200 31.3
u rolled-steel shape.
2 3 2
6 4 6 4
3970 mm 4.000 10 m
1 1
(210) 105 mm 0.105 m
2 2
31.3 10 mm 31.3 10 m
u
u u
A
c d
I
(a) Centric load: 3
3 50 50 50 150 kN 150 10 N
P u
3
6
3
3 150 10
37.783 10 Pa 37.8 MPa
3.970 10
P
A
u
u
u
V W
(b) Ececentric loading: 80 mm 0.080 m
e
3
3 3
3 3
6
3 6
2 50 50 100 kN 100 10 N
(50 10 )(0.080) 4.0 10 N m
2 100 10 (4.0 10 )(0.105)
38.607 10 Pa 38.6 MPa
3.970 10 31.3 10
A
P
M Pe
P Mc
A I
u
u u ˜
u u
u
u u
V W
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on a website, in whole or part.
554
z
x
y
A
C
b
25 mm
10 mm
30 mm
30 mm
10 mm
10 kN
10 kN
PROBLEM 4.104
Two 10-kN forces are applied to a 20 × 60-mm rectangular bar as
shown. Determine the stress at point A when (a) b 0, (b) b 15 mm,
(c) b 25 mm.
SOLUTION
3 2
2 6 3
(0.060 m)(0.020 m) 1.2 10 m
1
(0.020 m)(0.060 m) 12 10 m
6
A
S
u
u
(a)
3 2 6 3
0, (10 kN)(0.025 m) 250 N m
(20 kN)/(1.2 10 m ) (250 N m)/(12 10 m )
16.667 MPa 20.833 MPa
A
b M
˜
u ˜ u
V
4.17 MPa
A
V W
(b)
3 2 6 3
15 mm, (10 kN)(0.025 m 0.015 m) 100 N m
(20 kN)/(1.2 10 m ) (100 N m)/(12 10 m )
16.667 MPa 8.333 MPa
A
b M
˜
u ˜ u
V
8.33 MPa
A
V W
(c) 3 2
25 mm, 0: (20 kN)/(1.2 10 m )
A
b M
u
V
16.67 MPa
A
V W
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555
1 in.
(a) (b)
P'
P P'
P PROBLEM 4.105
Portions of a 1 1
-
2 2
in.
u square bar have been bent
to form the two machine components shown.
Knowing that the allowable stress is 15 ksi,
determine the maximum load that can be applied
to each component.
SOLUTION
The maximum stress occurs at point B.
3
15 ksi 15 10 psi
B
B
P Mc P Pec
KP
A I A I
V
V
u
where
2
3 3 4
1
1.0 in.
(0.5)(0.5) 0.25 in
1
(0.5)(0.5) 5.2083 10 in for all centroidal axes.
12
ec
K e
A I
A
I
u
(a) (b)
(a)
2
3
0.25 in.
1 (1.0)(0.25)
52 in
0.25 5.2083 10
c
K
u
3
( 15 10 )
52
B
P
K
V u
288 lb
P W
(b)
0.5
0.35355 in.
2
c
2
3
3
1 (1.0)(0.35355)
71.882 in
0.25 5.2083 10
( 15 10 )
71.882
B
K
P
K
V
u
u
209 lb
P W
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556
A D
18 mm
40 mm
12 mm
12 mm
P
B
PROBLEM 4.106
Knowing that the allowable stress in section ABD is 80 MPa,
determine the largest force P that can be applied to the bracket shown.
SOLUTION
2 6 2
3 3 4 9 4
(24)(18) 432 mm 432 10 m
1
(24)(18) 11.664 10 mm 11.664 10 m
12
1
(18) 9 mm 0.009 m
2
1
(18) 40 49 mm 0.049 m
2
A
I
c
e
u
u u
On line BD, both axial and bending stresses are compressive.
Therefore max
P Mc
A I
V
P Pec
A I
Solving for P gives max
1
P
ec
A I
V
§ ·
¨ ¸
© ¹
6
6 2 9 4
80 10 Pa
1 (0.049 m)(0.009 m)
432 10 m 11.664 10 m
P
u
§ ·
¨ ¸
u u
© ¹
1.994 kN
P W
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557
a
a
d
P'
P
PROBLEM 4.107
A milling operation was used to remove a portion of a solid bar
of square cross section. Knowing that a 30 mm, d 20 mm,
and all 60 MPa,
V determine the magnitude P of the largest
forces that can be safely applied at the centers of the ends of the
bar.
SOLUTION
3
3
2 2
1 1
, ,
12 2
2 2
6
3 ( ) 1 3( )
where
A ad I ad c d
a d
e
P Mc P Ped
A I ad ad
P P a d a d
KP K
ad ad
ad ad
V
V
Data:
3 2
2
30 mm 0.030 m 20 mm 0.020 m
1 (3)(0.010)
4.1667 10 m
(0.030)(0.020) (0.030)(0.020)
a d
K
u
6
3
3
60 10
14.40 10 N
4.1667 10
P
K
V u
u
u
14.40 kN
P W
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558
a
a
d
P'
P
PROBLEM 4.108
A milling operation was used to remove a portion of a solid bar of
square cross section. Forces of magnitude P 18 kN are applied at
the centers of the ends of the bar. Knowing that a 30 mm and
all 135 MPa,
V determine the smallest allowable depth d of the
milled portion of the bar.
SOLUTION
3
2
3
2
2
1 1
, ,
12 2
2 2
1 1
( ) 3 ( )
2 2
1
12
3 2 2
or 3 0
A ad I ad c d
a d
e
P a d d
P Mc P Pec P P P a d
A I ad I ad ad ad
ad
P P P
d d P
ad a
d
V
V V
Solving for d,
2
1 2 2
12
2
P P
d P
a a
­ ½
° °
§ ·
® ¾
¨ ¸
© ¹
° °
¯ ¿
V
V
Data: 3 6
0.030 m, 18 10 N, 135 10 Pa
a P V
u u
2
3 3
3 6
6
1 (2)(18 10 ) (2)(18 10 )
12(18 10 )(135 10 )
0.030 0.030
(2)(135 10 )
d
­ ½
ª º
u u
° °
u u
® ¾
« »
u ¬ ¼
° °
¯ ¿
3
16.04 10 m
u 16.04 mm
d W
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559
P
5 in.
12 kips PROBLEM 4.109
The two forces shown are applied to a rigid plate supported by a steel pipe of 8-in.
outer diameter and 7-in. inner diameter. Determine the value of P for which the
maximum compressive stress in the pipe is 15 ksi.
SOLUTION
4 4 4
all 15 ksi I (4 in.) (3.5 in.) 83.2 in
4 4
NA
S S
V
2 2 2
(4 in.) (3.5 in.) 11.78 in
A S S
Max. compressive stress is at point B.
2 4
12 (5 )(4.0 in.)
11.78 in 83.2 in
15 ksi 1.019 0.085 0.240
13.981 0.325
B
Q Mc P P
A I
P P
P
V
43.0 kips
P W
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560
P'
P'
P
P
d
d
h
PROBLEM 4.110
An offset h must be introduced into a solid circular rod of
diameter d. Knowing that the maximum stress after the offset
is introduced must not exceed 5 times the stress in the rod
when it is straight, determine the largest offset that can be
used.
SOLUTION
For centric loading, c
P
A
V
For eccentric loading, e
P Phc
A I
V
Given 5
e c
V V
4
2
5
(4)
4 1
64
4
2
2 4
P Phc P
A I A
d
Phc P I
h d
d
I A cA
d
S
S
§ ·
¨ ¸
© ¹
?
§ ·§ ·
¨ ¸¨ ¸
© ¹© ¹
0.500
h d W
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561
P'
P'
P
P
d
d
h
PROBLEM 4.111
An offset h must be introduced into a metal tube of 0.75-in. outer
diameter and 0.08-in. wall thickness. Knowing that the maximum
stress after the offset is introduced must not exceed 4 times the
stress in the tube when it is straight, determine the largest offset
that can be used.
SOLUTION
1
2 2 2 2
1
2
4 4 4 4
1
3 4
1
0.375 in.
2
0.375 0.08 0.295 in.
(0.375 0.295 )
0.168389 in
(0.375 0.295 )
4 4
9.5835 10 in
c d
c c t
A c c
I c c
S S
S S
u
For centric loading, cen
P
A
V
For eccentric loading, ecc
P Phc
A I
V
ecc cen
3
4 or 4
3 3 (3)(9.5835 10 )
(0.168389)(0.375)
P Phc P
A I A
hc I
h
I A Ac
V V
u
0.455 in.
h W
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562
1
1
2
4
3
16 kips PROBLEM 4.112
A short column is made by nailing four 1 × 4-in. planks to a 4 × 4-in. timber. Using
an allowable stress of 600 psi, determine the largest compressive load P that can be
applied at the center of the top section of the timber column as shown if (a) the
column is as described, (b) plank 1 is removed, (c) planks 1 and 2 are removed,
(d) planks 1, 2, and 3 are removed, (e) all planks are removed.
SOLUTION
(a) Centric loading: 0
P
M
A
V
2
2
(4 4) 4(1)(4) 32 in
(0.600 ksi)(32 in )
A
P
P
A
u
?
V
19.20 kips W
(b) Eccentric loading:
P Pec
M Pe
A I
V
2
(4)(4) (3)(1)(3) 28 in
A e y
(1)(4)(2.5)
0.35714 in.
28
Ay
y
A
¦
2
3 2 3 2 4
( )
1 1
(6)(4) (6)(4)(0.35714) (4)(1) (4)(1)(2.14286) 53.762 in
12 12
¦
I I Ad
1 (0.600 ksi)
11.68 kips
1 (0.35714)(2.35714)
28 53.762
§ ·
?
¨ ¸ ª º
© ¹
« »
¬ ¼
ec
P P
A I
V W
(c) Centric loading: 0
P
M
A
V
2
(6)(4) 24 in
(0.600 ksi)(24) 14.40 kips
A
P W
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563
PROBLEM 4.112 (Continued)
(d) Eccentric loading:
P Pec
M Pe
A I
V
2
(4)(4) (1)(4)(1) 20 in
A e x
2.5 2 0.5 in.
x
3 4
1
(4)(5) 41.667 in
12
I
(0.600 ksi)
7.50 kips
1 (0.5)(2.5)
20 41.667
P
ª º
« »
¬ ¼
W
(e) Centric loading: 0
P
M
A
V
2
2
(4)(4) 16 in
(0.600 ksi)(16.0 in ) 9.60 kips
A
P W
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564
0.75 in.
3 in.
3 in.
1 in.
1.5 in. 1.5 in.
a
a
B
A
0.75 in.
Section a–a
PROBLEM 4.113
A vertical rod is attached at point A to the cast
iron hanger shown. Knowing that the allowable
stresses in the hanger are all 5 ksi
V and
all 12 ksi
V , determine the largest downward
force and the largest upward force that can be
exerted by the rod.
SOLUTION
3
2
2
all all
3 2
3 2 3 2
(1 3)(0.5) 2(3 0.25)(2.5)
(1 3) 2(3 0.75)
12.75 in
1.700 in.
7.5 in
7.5 in
5 ksi 12 ksi
1
12
1 1
(3)(1) (3 1)(1.70 0.5) (1.5)(3) (1.5 3)(2.5 1.70)
12 12
10.825 in
c
c
Ay
X
A
X
A
I bh Ad
I
V V
u u
u u
§ ·
¨ ¸
© ¹
u u
¦
¦
¦
4
Downward force.
(1.5 in.+1.70 in.) (3.20 in.)
M P P
At D: D
P Mc
A I
V
(3.20) (1.70)
5 ksi
7.5 10.825
5 ( 0.6359)
P P
P
7.86 kips p
P
At E: E
P Mc
A I
V
(3.20) (2.30)
12 ksi
7.5 10.825
12 ( 0.5466)
P P
P
21.95 kips p
P
We choose the smaller value. 7.86 kips p
P W
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565
PROBLEM 4.113 (Continued)
Upward force.
(1.5 in. 1.70 in.) (3.20 in.)
M P P
At D: D
P Mc
A I
V
(3.20) (1.70)
12 ksi
7.5 10.825
12 ( 0.6359)
P P
P
18.87 kips n
P
At E: E
P Mc
A I
V
(3.20) (2.30)
5 ksi
7.5 10.825
5 ( 0.5466)
P P
P
9.15 kips n
P
We choose the smaller value. 9.15 kips n
P W
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566
0.75 in.
3 in.
3 in.
1 in.
1.5 in. 1.5 in.
a
a
B
A
0.75 in.
Section a–a
PROBLEM 4.114
Solve Prob. 4.113, assuming that the vertical
rod is attached at point B instead of point A.
PROBLEM 4.113 A vertical rod is attached at
point A to the cast iron hanger shown. Knowing
that the allowable stresses in the hanger are
all 5 ksi
V and all 12 ksi,
V determine the
largest downward force and the largest upward
force that can be exerted by the rod.
SOLUTION
3
2
2
all all
3 2
3 2 3 2
(1 3)(0.5) 2(3 0.25)(2.5)
(1 3) 2(3 0.75)
12.75 in
1.700 in.
7.5 in
7.5 in
5 ksi 12 ksi
1
12
1 1
(3)(1) (3 1)(1.70 0.5) (1.5)(3) (1.5 3)(2.5 1.70)
12 12
10.825 in
c
c
Ay
X
A
X
A
I bh Ad
I
V V
u u
u u
§ ·
¨ ¸
© ¹
u u
¦
¦
¦
4
Downward force.
all all
5 ksi 12 ksi
V V
(2.30 in. 1.5 in.) (3.80 in.)
M P
At D: D
P Mc
A I
V
(3.80) (1.70)
12 ksi
7.5 10.825
12 ( 0.4634)
P P
P
25.9 kips p
P
At E: E
P Mc
A I
V
(3.80) (2.30)
5 ksi
7.5 10.825
5 ( 0.9407)
P P
P
5.32 kips p
P
We choose the smaller value. 5.32 kips p
P W
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567
PROBLEM 4.114 (Continued)
Upward force.
all all
5 ksi 12 ksi
V V
(2.30 in. 1.5 in.) (3.80 in.)
M P P
At D: D
P Mc
A I
V
(3.80) (1.70)
5 ksi
7.5 10.825
5 ( 0.4634)
P P
P
10.79 kips n
P
At E: E
P Mc
A I
V
(3.80) (2.30)
12 ksi
7.5 10.825
12 ( 0.9407)
P P
P
12.76 kips n
P
We choose the smaller value. 10.79 kips n
P W
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568
32 mm
P'
P
a
a
B
A
4 mm
2 mm radius
20 mm
Section a–a
PROBLEM 4.115
Knowing that the clamp shown has been tightened until
400 N,
P determine (a) the stress at point A,
(b) the stress at point B, (c) the location of the neutral
axis of section a a.
SOLUTION
Cross section: Rectangle c Circle d
2
1
1
2 2
2
2
(20 mm)(4 mm) 80 mm
1
(20 mm) 10 mm
2
(2 mm) 4 mm
20 2 18 mm
A
y
A
y
S S
1
2
(80)(10) (4 )(18)
11.086 mm
80 4
20 8.914 mm
11.086 10 1.086 mm
18 11.086 6.914 mm
B
A
Ay
c y
A
c y
d
d
S
S
¦
¦
2 3 2 3 4
1 1 1 1
2 4 2 3 4
2 2 2 2
3 4 9 4
1 2
2 6 2
1 2
1
(4)(20) (80)(1.086) 2.761 10 mm
12
(2) (4 )(6.914) 0.613 10 mm
4
3.374 10 mm 3.374 10 m
92.566 mm 92.566 10 m
I I A d
I I A d
I I I
A A A
S
S
u
u
u u
u
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569
PROBLEM 4.115 (Continued)
32 8.914 40.914 mm 0.040914 m
(400 N)(0.040914 m) 16.3656 N m
e
M Pe
˜
(a) Point A:
3
6 9
6 6 6
400 (16.3656)(8.914 10 )
92.566 10 3.374 10
4.321 10 43.23 10 47.55 10 Pa
A
P Mc
A I
V
u
u u
u u u 47.6 MPa
A
V W
(b) Point B:
6 9
6 6 6
400 (16.3656)(11.086)
92.566 10 3.374 10
4.321 10 53.72 10 49.45 10 Pa
B
P Mc
A I
V
u u
u u u 49.4 MPa
B
V W
(c) Neutral axis: By proportions,
20
47.55 47.55 49.45
9.80 mm
a
a
9.80 mm below top of section W
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570
P'
P
A
B
C
a a
t
90$
PROBLEM 4.116
The shape shown was formed by bending a thin steel plate. Assuming that the
thickness t is small compared to the length a of a side of the shape, determine the
stress (a) at A, (b) at B, (c) at C.
SOLUTION
Moment of inertia about centroid:
3
3
1
2 2
12 2
1
12
a
I t
ta
§ ·
¨ ¸
© ¹
Area: 2 2 2 ,
2 2 2
a a
A t at c
§ ·
¨ ¸
© ¹
(a)
2 2 2 2
3
1
12
2
a a
A
P
P Pec P
A I at ta
V
2
A
P
at
V W
(b)
2 2
3
1
12
2
a a
B
P
P Pec P
A I at ta
V
2
B
P
at
V W
(c) C A
V V
2
C
P
at
V W
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571
50 mm
50 mm
P PROBLEM 4.117
Three steel plates, each of 25 150-mm
u cross section, are welded
together to form a short H-shaped column. Later, for architectural
reasons, a 25-mm strip is removed from each side of one of the
flanges. Knowing that the load remains centric with respect to the
original cross section, and that the allowable stress is 100 MPa,
determine the largest force P (a) that could be applied to the original
column, (b) that can be applied to the modified column.
SOLUTION
(a) Centric loading:
P
A
V
3 2 3 2
6 3
6
(3)(150)(25) 11.25 10 mm 11.25 10 m
( 100 10 )(11.25 10 )
1.125 10 N
A
P A
u u
u u
u
V
1125 kN
P W
(b) Eccentric loading (reduced cross section):
3 2
, 10 mm
A , mm
y 3 3
(10 mm )
A y , mm
d
M 3.75 87.5 328.125 76.5625
N 3.75 0 0 10.9375
O 2.50 87.5 218.75 98.4375
6 10.00 109.375
3
3
109.375 10
10.9375 mm
10.00 10
Ay
Y
A
6 u
6 u
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572
PROBLEM 4.117 (Continued)
The centroid lies 10.9375 mm from the midpoint of the web.
3 2 3 3 2 6 4
1 1 1 1 1
3 2 3 3 2 6 4
2 2 2 2 2
3 2 3 3 2 6 4
3 3 3 3 3
1 1
(150)(25) (3.75 10 )(76.5625) 22.177 10 mm
12 12
1 1
(25)(150) (3.75 10 )(10.9375) 7.480 10 mm
12 12
1 1
(100)(25) (2.50 10 )(98.4375) 24.355 10 mm
12 12
I b h A d
I b h A d
I b h A d
I I
u u
u u
u u
6 4 6 4
1 2 3
3
3 2
3
3
54.012 10 mm 54.012 10 m
10.9375 75 25 110.9375 mm 0.1109375 m
where 10.4375 mm 10.4375 10 m
10.00 10 m
1 1 (101.9375 10 )(0.1109375)
10.00 10 54.012
I I
c
M Pe e
P Mc P Pec
KP A
A I A I
ec
K
A I
u u
u
u
u
u u
V
2
6
6
3
122.465 m
10
( 100 10 )
817 10 N
122.465
P
K
u
u
V
817 kN
P W
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573
(a) (b)
y
y
y x
x
A
A
B
B
C
3 in.
3 in.
4 in.
2 in.
2 in. 2 in.
1 in.
P
PROBLEM 4.118
A vertical force P of magnitude 20 kips is applied
at point C located on the axis of symmetry of the
cross section of a short column. Knowing that
5 in.,
y determine (a) the stress at point A,
(b) the stress at point B, (c) the location of the
neutral axis.
SOLUTION
Locate centroid.
Part 2
, in
A , in.
y 3
, in
Ay
M 12 5 60
N 8 2 16
6 20 76
76
3.8 in.
20
Ay
y
A
6
6
Eccentricity of load: 5 3.8 1.2 in.
e
3 2 4 3 2 4
1 2
4
1 2
1 1
(6)(2) (12)(1.2) 21.28 in (2)(4) (8)(1.8) 36.587 in
12 12
57.867 in
I I
I I I
(a) Stress at A: 3.8 in.
A
c
20 20(1.2)(3.8)
20 57.867
A
A
P Pec
A I
V 0.576 ksi
A
V W
(b) Stress at B: 6 3.8 2.2 in.
B
c
20 20(1.2)(2.2)
20 57.867
B
B
P Pec
A I
V 1.912 ksi
B
V W
(c) Location of neutral axis:
1
0 0
57.867
2.411 in.
(20)(1.2)
P Pea ea
A I I A
I
a
Ae
V V ?
Neutral axis lies 2.411 in. below centroid or 3.8 2.411 1.389
in. above point A.
Answer:1.389 in. from point A W
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574
(a) (b)
y
y
y x
x
A
A
B
B
C
3 in.
3 in.
4 in.
2 in.
2 in. 2 in.
1 in.
P
PROBLEM 4.119
A vertical force P is applied at point C located
on the axis of symmetry of the cross section of
a short column. Determine the range of values
of y for which tensile stresses do not occur in
the column.
SOLUTION
Locate centroid.
2
, in
A , in.
y 3
, in
Ay
M 12 5 60
N 8 2 16
6 20 76
76
3.8 in.
20
i i
i
A y
y
A
6
6
Eccentricity of load:
3 2 4 3 2 4
1 2
4
1 2
3.8 in. 3.8 in.
1 1
(6)(2) (12)(1.2) 21.28 in (2)(4) (8)(1.8) 36.587 in
12 12
57.867 in
e y y e
I I
I I I
If stress at A equals zero, 3.8 in.
A
c
1
0
A A
A
P Pec ec
A I I A
V ?
57.867
0.761 in. 0.761 3.8 4.561 in.
(20)(3.8)
A
I
e y
Ac
If stress at B equals zero, 6 3.8 2.2 in.
B
c
1
0
57.867
1.315 in.
(20)(2.2)
1.315 3.8 2.485 in.
B B
B
B
P Pec ec
A I I A
I
e
Ac
y
V ?
Answer: 2.49 in. 4.56 in.
y W
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575
P
P
P
P
PROBLEM 4.120
The four bars shown have the same cross-sectional area. For the given
loadings, show that (a) the maximum compressive stresses are in the
ratio 4:5:7:9, (b) the maximum tensile stresses are in the ratio 2:3:5:3.
(Note: the cross section of the triangular bar is an equilateral triangle.)
SOLUTION
Stresses:
At A, 1
A A
A
Pec Aec
P P
A I A I
V
§ ·
¨ ¸
© ¹
At B, 1
B B
B
Pec Aec
P P
A I A I
V
§ ·
¨ ¸
© ¹
1
4
A
P
A
V W
2 4
1 1
2
2
2
2
1 1 1
, , ,
12 2 2
1 1
( )
2 2
1
1
12
1 1
( )
2 2
1
1
12
A B
A
B
A a I a c c a e a
a a a
P
A a
a a a
P
A a
V
V
­
°
°
§ ·
° § ·§ ·
¨ ¸
¨ ¸¨ ¸
°
© ¹© ¹
¨ ¸
°
¨ ¸
°
¨ ¸
®
© ¹
°
° § ·
§ ·§ ·
° ¨ ¸
¨ ¸¨ ¸
© ¹© ¹
° ¨ ¸
° ¨ ¸
° ¨ ¸
© ¹
¯
1
2
B
P
A
V W
2
5
A
P
A
V W
2 2 4
2 2
2
4
2
2
4
2
, ,
4
( )( )( )
1
4
( )( )( )
1
4
A
B
a
A c a c I c e c
P c c c
A c
P c c c
A c
S
S
S
S
V
S
S
V
S
­
?
°
°
° § ·
° ¨ ¸
° ¨ ¸
° ¨ ¸
® ¨ ¸
© ¹
°
° § ·
° ¨ ¸
°
¨ ¸
° ¨ ¸
¨ ¸
° © ¹
¯
2
3
B
P
A
V W
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576
PROBLEM 4.120 (Continued)
3
7
A
P
A
V W
2 4
3 3
2
4
3
2
4
3
2 1
2 12
2 2
( )
2 2
1
1
12
2 2
( )
2 2
1
1
12
A
B
A a c a I a e c
a a a
P
A a
a a a
P
A a
V
V
­
°
°
° § ·
§ ·§ ·
° ¨ ¸
¨ ¸¨ ¸
¨ ¸¨ ¸
° ¨ ¸
© ¹© ¹
° ¨ ¸
° ¨ ¸
® ¨ ¸
© ¹
°
° § ·
§ ·§ ·
° ¨ ¸
¨ ¸¨ ¸
¨ ¸¨ ¸
° ¨ ¸
© ¹© ¹
° ¨ ¸
° ¨ ¸
° ¨ ¸
© ¹
¯
3
5
B
P
A
V W
2
4
3
4
4
1 3 3
( )
2 2 4
1 3 3
36 2 96
2 3
3 2 3
A B
A s s s
I s s s
s
c s e c s
§ ·
¨ ¸
¨ ¸
© ¹
§ ·
¨ ¸
¨ ¸
© ¹
2
4
4
3
4 3 3
1
3
96
A
s s
s
P
A
s
V
§ ·
§ ·§ ·§ ·
¨ ¸
¨ ¸¨ ¸¨ ¸
¨ ¸
¨ ¸
© ¹© ¹
© ¹
¨ ¸
¨ ¸
¨ ¸
© ¹
4
9
A
P
A
V W
2
4
4
3
4 3 2 3
1
3
96
B
s s
s
P
A
s
V
§ ·
§ ·§ ·§ ·
¨ ¸
¨ ¸¨ ¸¨ ¸
¨ ¸
¨ ¸
© ¹© ¹
© ¹
¨ ¸
¨ ¸
¨ ¸
© ¹
4
3
B
P
A
V W
Mecánica de Materiales - Ferdinand P. Beer 7ma. Edición.

616. PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
577
30 mm
45 mm
15 mm
90 mm
25 mm
d
A
B P
PROBLEM 4.121
An eccentric force P is applied as shown to a steel bar of
25 90-mm
u cross section. The strains at A and B have been
measured and found to be
350 70
A B
H P H P
Knowing that 200
E GPa, determine (a) the distance d, (b) the
magnitude of the force P.
SOLUTION
3 2 3 2
3 3 6 4 6 4
1
15 45 30 90 mm 25 mm 45 mm 0.045 m
2
(25)(90) 2.25 10 mm 2.25 10 m
1 1
(25)(90) 1.51875 10 mm 1.51875 10 m
12 12
60 45 15 mm 0.015 m 15 45 30 mm 0.030 m
A B
h b c h
A bh
I bh
y y
u u
u u
Stresses from strain gages at A and B: 9 6 6
9 6 6
(200 10 )(350 10 ) 70 10 Pa
(200 10 )( 70 10 ) 14 10 Pa
A A
B B
E
E
V H
V H
u u u
u u u
A
A
My
P
A I
V (1)
B
B
My
P
A I
V (2)
Subtracting,
( )
A B
A B
M y y
I
V V
6 6
( ) (1.51875 10 )(84 10 )
2835 N m
0.045
A B
A B
I
M
y y
V V
u u
˜
Multiplying (2) by A
y and (1) by B
y and subtracting, ( )
A B B A A B
P
y y y y
A
V V
3 6 6
3
( ) (2.25 10 )[(0.015)( 14 10 ) ( 0.030)(70 10 )]
94.5 10 N
0.045
A B B A
A B
A y y
P
y y
V V
u u u
u
(a) 3
2835
0.030 m
94.5 10
M
M Pd d
P
?
u
30.0 mm
d W
(b) 94.5 kN
P W
Mecánica de Materiales - Ferdinand P. Beer 7ma. Edición.

617. PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
578
30 mm
45 mm
15 mm
90 mm
25 mm
d
A
B P
PROBLEM 4.122
Solve Prob. 4.121, assuming that the measured strains are
600 420
A B
H P H P
PROBLEM 4.121 An eccentric force P is applied as shown to a
steel bar of 25 90-mm
u cross section. The strains at A and B have
been measured and found to be
350 70
A B
H P H P
Knowing that 200
E GPa, determine (a) the distance d, (b) the
magnitude of the force P.
SOLUTION
3 2 3 2
3 3 6 4 6 4
1
15 45 30 90 mm 25 mm 45 mm 0.045 m
2
(25)(90) 2.25 10 mm 2.25 10 m
1 1
(25)(90) 1.51875 10 mm 1.51875 10 m
12 12
60 45 15 mm 0.015 m 15 45 30 mm 0.030 m
A B
h b c h
A bh
I bh
y y
u u
u u
Stresses from strain gages at A and B: 9 6 6
9 6 6
(200 10 )(600 10 ) 120 10 Pa
(200 10 )(420 10 ) 84 10 Pa
A A
B B
E
E
V H
V H
u u u
u u u
A
A
My
P
A I
V (1)
B
B
My
P
A I
V (2)
Subtracting,
( )
A B
A B
M y y
I
V V
6 6
( ) (1.51875 10 )(36 10 )
1215 N m
0.045
A B
A B
I
M
y y
V V
u u
˜
Multiplying (2) by A
y and (1) by B
y and subtracting, ( )
A B B A A B
P
y y y y
A
V V
3 6 6
3
( ) (2.25 10 )[(0.015)(84 10 ) ( 0.030)(120 10 )]
243 10 N
0.045
A B B A
A B
A y y
P
y y
u u u
u
V V
(a) 3
3
1215
5 10 m
243 10
M
M Pd d
P
? u
u
5.00 mm
d W
(b) 243 kN
P W
Mecánica de Materiales - Ferdinand P. Beer 7ma. Edición.

618. PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
579
40 mm
80 mm
P'
P
PROBLEM 4.123
The C-shaped steel bar is used as a dynamometer to determine the magnitude P
of the forces shown. Knowing that the cross section of the bar is a square of side
40 mm and that strain on the inner edge was measured and found to be 450P,
determine the magnitude P of the forces. Use 200 GPa.
E
SOLUTION
At the strain gage location,
9 6 6
2 6 2
3 3 4 9 4
(200 10 )(450 10 ) 90 10 Pa
(40)(40) 1600 mm 1600 10 m
1
(40)(40) 213.33 10 mm 213.33 10 m
12
80 20 100 mm 0.100 m
20 mm 0.020 m
E
A
I
e
c
P Mc P Pec
KP
A I A I
V H
V
u u u
u
u u
3 2
6 9
6
3
3
1 1 (0.100)(0.020)
10.00 10 m
1600 10 213.33 10
90 10
9.00 10 N
10.00 10
ec
K
A I
P
K
V
u
u u
u
u
u
9.00 kN
P W
Mecánica de Materiales - Ferdinand P. Beer 7ma. Edición.

619. PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
580
z
x
B A
A = 10.0 in2
Iz = 273 in4
y
A
z
x
6 in.
6 in.
P
Q
10 in.
PROBLEM 4.124
A short length of a rolled-steel column supports a rigid plate
on which two loads P and Q are applied as shown. The strains
at two points A and B on the centerline of the outer faces of the
flanges have been measured and found to be
6 6
400 10 in./in. 300 10 in./in.
A B
u u
H H
Knowing that E 29 u 106
psi, determine the magnitude of
each load.
SOLUTION
Stresses at A and B from strain gages:
6 6 3
6 6 3
(29 10 )( 400 10 ) 11.6 10 psi
(29 10 )( 300 10 ) 8.7 10 psi
u u u
u u u
A A
B B
E
E
V H
V H
Centric force: F P Q
Bending couple: 6 6
M P Q
5 in.
c
(6 6 )(5)
10.0 273
A
F Mc P Q P Q
A I
V
3
11.6 10 0.00989 0.20989
u
P Q (1)
(6 6 )(5)
10.0 273
B
F Mc P Q P Q
A I
V
3
8.7 10 0.20989 0.00989
u
P Q (2)
Solving (1) and (2) simultaneously,
3
44.2 10 lb 44.2 kips
u
P W
3
57.3 10 lb 57.3 kips
u
Q W
Mecánica de Materiales - Ferdinand P. Beer 7ma. Edición.

620. PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
581
P
C
B
A
y
z x
3 in.
5 in.
PROBLEM 4.125
A single vertical force P is applied to a short steel post as shown.
Gages located at A, B, and C indicate the following strains:
500
A
H P 1000
B
H P 200
C
H P
Knowing that 6
29 10 psi,
E u determine (a) the magnitude of
P, (b) the line of action of P, (c) the corresponding strain at the
hidden edge of the post, where 2.5 in.
x and 1.5 in.
z
SOLUTION
3 4 3 4 2
6 6
1 1
(5)(3) 11.25 in (3)(5) 31.25 in (5)(3) 15 in
12 12
2.5 in., 2.5 in., 2.5 in., 2.5 in.
1.5 in., 1.5 in., 1.5 in., 1.5 in.
(29 10 )( 500 10 ) 14,500 psi 14.5 ksi
x z
x z
A B C D
A B C D
A A
B
I I A
M Pz M Px
x x x x
z z z z
E
V H
V
u u
6 6
6 6
(29 10 )( 1000 10 ) 29,000 psi 29 ksi
(29 10 )( 200 10 ) 5800 psi 5.8 ksi
B
C C
E
E
H
V H
u u
u u
0.06667 0.13333 0.08
x A z A
A x z
x z
P M z M x
P M M
A I I
V (1)
0.06667 0.13333 0.08
x B z B
B x z
x z
P M z M x
P M M
A I I
V (2)
0.06667 0.13333 0.08
x C z C
C x z
x z
P M z M x
P M M
A I I
V (3)
Substituting the values for A
V , B
V , and C
V into (1), (2), and (3) and solving the simultaneous equations
gives
87 kip in. 90.625 kip in.
x z
M M
˜ ˜ (a) 152.3 kips
P W
90.625
152.25
z
M
x
P
(b) 0.595 in.
x W
87
152.25
x
M
z
P
0.571in.
z W
0.06667 0.13333 0.08
(0.06667)(152.25) (0.13333)(87) (0.08)( 90.625) 8.70 ksi
x D z D
D x z
x z
P M z M x
P M M
A I I
V
(c) Strain at hidden edge:
3
6
8.70 10
29 10
D
E
V
H
u
u
300
H P W
Mecánica de Materiales - Ferdinand P. Beer 7ma. Edición.

621. PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
582
b ! 40 mm
a ! 25 mm
20 mm
A
D
C
B
d
P
PROBLEM 4.126
The eccentric axial force P acts at point D, which must be located
25 mm below the top surface of the steel bar shown. For 60 kN,
P
determine (a) the depth d of the bar for which the tensile stress at
point A is maximum, (b) the corresponding stress at point A.
SOLUTION
3
3 2
1
12
1 1
2 2
1 1
12
1 4 6
2 2
A
A
A bd I bd
c d e d a
P Pec
A I
d a d
P P a
b d b d
d d
V
V
­ ½
° ° ­ ½
® ¾ ® ¾
¯ ¿
° °
¯ ¿
(a) Depth d for maximum :
V A Differentiate with respect to d.
2 3
4 12
0
V ­ ½
® ¾
¯ ¿
A
d P a
dd b d d
3
d a 75 mm
d W
(b)
3 3
6
3 3 3 2
60 10 4 (6)(25 10 )
40 10 Pa
40 10 75 10 (75 10 )
A
V
­ ½
u u
° °
u
® ¾
u u u
° °
¯ ¿
40 MPa
A
V W
Mecánica de Materiales - Ferdinand P. Beer 7ma. Edición.

622. PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
583
A
! 60#
B
z
y
16 mm
16 mm
40 mm 40 mm
M ! 300 N · m
D
C
PROBLEM 4.127
The couple M is applied to a beam of the cross section shown in a plane
forming an angle E with the vertical. Determine the stress at (a) point A,
(b) point B, (c) point D.
SOLUTION
3 3 4 9 4
3 6 4 6 4
1
(80)(32) 218.45 10 mm 218.45 10 m
12
1
(32)(80) 1.36533 10 mm 1.36533 10 m
12
16 mm
40 mm
z
y
A B D
A B D
I
I
y y y
z z z
u u
u u
300cos30 259.81 N m 300sin30 150 N m
y z
M M
q ˜ q ˜
(a)
3 3
9 6
(150)(16 10 ) (259.81)(40 10 )
218.45 10 1.36533 10
y A
z A
A
z y
M z
M y
I I
u u
u u
V
6
3.37 10 Pa
u 3.37 MPa
A
V W
(b)
3 3
9 6
(150)(16 10 ) (259.81)( 40 10 )
218.45 10 1.36533 10
y B
z B
B
z y
M z
M y
I I
u u
u u
V
6
18.60 10 Pa
u 18.60 MPa
B
V W
(c)
3 3
9 6
(150)( 16 10 ) (259.81)( 40 10 )
218.45 10 1.36533 10
y D
z D
D
z y
M z
M y
I I
u u
u u
V
6
3.37 10 Pa
u 3.37 MPa
D
V W
Mecánica de Materiales - Ferdinand P. Beer 7ma. Edición.

623. PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
584
! 30#
z
y
0.6 in.
0.4 in.
0.6 in.
M 400 lb · m
A B
D
C
PROBLEM 4.128
The couple M is applied to a beam of the cross section shown in a plane
forming an angle E with the vertical. Determine the stress at (a) point A,
(b) point B, (c) point D.
SOLUTION
3 3 4
3 3 4
1
(0.4)(1.2) 57.6 10 in
12
1
(1.2)(0.4) 6.40 10 in
12
0.6 in.
1
z (0.4) 0.2 in.
2
u
u
§ ·
¨ ¸
© ¹
z
y
A B D
A B D
I
I
y y y
z z
400cos 60 200 lb in., 400 sin60 346.41lb in.
y z
M M
q ˜ q ˜
(a) 3 3
( 346.41)(0.6) (200)(0.2)
57.6 10 6.40 10
y A
z A
A
z y
M z
M y
I I
u u
V
3
9.86 10 psi 9.86 ksi
u W
(b) 6 3
( 346.41)(0.6) (200)( 0.2)
57.6 10 6.4 10
y B
z B
B
z y
M z
M y
I I
u u
V
3
2.64 10 psi 2.64 ksi
u W
(c) 3 3
( 346.41)( 0.6) (200)( 0.2)
57.6 10 6.40 10
y D
z D
D
z y
M z
M y
I I
u u
V
3
9.86 10 psi 9.86 ksi
u W
Mecánica de Materiales - Ferdinand P. Beer 7ma. Edición.

624. PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
585
M ! 25 kN · m
! 15#
C
80 mm
80 mm
30 mm
20 mm
z
y
A B
D
PROBLEM 4.129
The couple M is applied to a beam of the cross section shown in a
plane forming an angle E with the vertical. Determine the stress at
(a) point A, (b) point B, (c) point D.
SOLUTION
3 3 6 4
6 4
25sin 15 6.4705 kN m
25cos15 24.148 kN m
1 1
(80)(90) (80)(30) 5.04 10 mm
12 12
5.04 10 m
y
z
y
y
M
M
I
I
q ˜
q ˜
u
u
3 3 3 6 4 6 4
1 1 1
(90)(60) (60)(20) (30)(100) 16.64 10 mm 16.64 10 m
3 3 3
z
I
u u
Stress:
y z
y z
M z M y
I I
V
(a) 6 4 6 4
(6.4705 kN m)(0.045 m) (24.148 kN m)(0.060 m)
5.04 10 m 16.64 10 m
A
V
˜ ˜
u u
57.772 MPa 87.072 MPa
29.3 MPa
A
V W
(b) 6 4 6 4
(6.4705 kN m)( 0.045 m) (24.148 kN m)(0.060 m)
5.04 10 m 16.64 10 m
B
V
˜ ˜
u u
57.772 MPa 87.072 MPa
144.8 MPa
B
V W
(c) 6 4 6 4
(6.4705 kN m)( 0.015 m) (24.148 kN m)( 0.100 m)
5.04 10 m 16.64 10 m
D
V
˜ ˜
u u
19.257 MPa 145.12 MPa
125.9 MPa
D
V W
Mecánica de Materiales - Ferdinand P. Beer 7ma. Edición.

625. PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
586
A
y
z
B
3 in.
2 in.
2 in. 4 in.
3 in.
C
M ! 10 kip · in.
! 20#
D
PROBLEM 4.130
The couple M is applied to a beam of the cross section shown in a plane
forming an angle E with the vertical. Determine the stress at (a) point A,
(b) point B, (c) point D.
SOLUTION
Locate centroid.
2
, in
A , in.
z 3
, in
Az
M 16 1 16
N 8 2 16
Σ 24 0
The centroid lies at point C.
3 3 4
3 3 4
1 1
(2)(8) (4)(2) 88 in
12 12
1 1
(8)(2) (2)(4) 64 in
3 3
1 in., 4 in.
4 in., 0
10 cos20 9.3969 kip in.
10 sin 20 3.4202 kip in.
z
y
A B D
A B D
z
y
I
I
y y y
z z z
M
M
q ˜
q ˜
(a)
(9.3969)(1) (3.4202)( 4)
88 64
y A
z A
A
z y
M z
M y
I I
V
0.321 ksi
A
V W
(b)
(9.3969)( 1) (3.4202)( 4)
88 64
y B
z B
B
z y
M z
M y
I I
V
0.107 ksi
B
V W
(c)
(9.3969)( 4) (3.4202)(0)
88 64
y D
z D
D
z y
M z
M y
I I
V
0.427 ksi
D
V W
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626. PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
587
A
2.5 in.
5 in.
2.5 in.
3 in.
y
z
b 5 508
3 in.
1 in.
1 in.
B
C
D
5 in.
M 5 60 kip · in.
PROBLEM 4.131
The couple M is applied to a beam of the cross section shown in a
plane forming an angle E with the vertical. Determine the stress at
(a) point A, (b) point B, (c) point D.
SOLUTION
3 2 4
3 4 2 2 4
60 sin 40 38.567 kip in.
60 cos 40 45.963 kip in.
3 in.
5 in.
1
(10)(6) 2 (1) 178.429 in
12 4
1
(6)(10) 2 (1) (1) (2.5) 459.16 in
12 4
q ˜
q ˜
ª º
« »
¬ ¼
ª º
« »
¬ ¼
z
y
A B D
A B D
z
y
M
M
y y y
z z z
I
I
S
S
S
(a)
( 38.567)(3) (45.963)(5)
178.429 459.16
y A
z A
A
z y
M z
M y
I I
V
1.149 ksi 1.149 ksi
A
V W
(b)
( 38.567)(3) (45.963)( 5)
178.429 459.16
y B
z B
B
z y
M z
M y
I I
V
0.1479 0.1479 ksi
B
V W
(c)
( 38.567)( 3) (45.963)( 5)
178.429 459.16
y D
z D
D
z y
M z
M y
I I
V
1.149 ksi
1.149 ksi
D
V W
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627. PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
588
A B
4 in.
1.6 in.
2.4 in.
4.8 in.
C
M 5 75 kip · in.
b 5 758
D
y
z
PROBLEM 4.132
The couple M is applied to a beam of the cross section shown in a
plane forming an angle E with the vertical. Determine the stress at
(a) point A, (b) point B, (c) point D.
SOLUTION
3 3 4
3 3 4
1 1
(4.8)(2.4) (4)(1.6) 4.1643 in
12 12
1 1
(2.4)(4.8) (1.6)(4) 13.5851 in
12 12
1.2 in.
2.4 in.
75sin15 19.4114 kip in.
75cos15 72.444 kip in.
z
y
A B D
A B D
z
y
I
I
y y y
z z z
M
M
q ˜
q ˜
(a)
(19.4114)(1.2) (72.444)(2.4)
4.1643 13.5851
y A
z A
A
z y
M z
M y
I I
V 7.20 ksi
A
V W
(b)
(19.4114)(1.2) (72.444)( 2.4)
4.1643 13.5851
y B
z B
B
z y
M z
M y
I I
V
18.39 ksi
B
V W
(c)
(19.4114)( 1.2) (72.444)( 2.4)
4.1643 13.5851
y D
z D
D
z y
M z
M y
I I
V
7.20 ksi
D
V W
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628. PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
589
! 30#
y
z
M 100 N · m
A
B
r 20 mm
C
D
PROBLEM 4.133
The couple M is applied to a beam of the cross section shown in a
plane forming an angle E with the vertical. Determine the stress at
(a) point A, (b) point B, (c) point D.
SOLUTION
2
4 2 4
4 3 4 9 4
4
4 3 4 9 4
4 8
8 2 3 8 9
(0.109757)(20) 17.5611 10 mm 17.5611 10 m
(20)
62.832 10 mm 62.832 10 m
8 8
4 (4)(20)
8.4883 mm
3 3
20 8.4883 11.5117 mm
z
y
A D
B
A
r
I r r r
I r
r
y y
y
z z
S S S
S S
S S
S S
§ ·§ · § ·
¨ ¸¨ ¸ ¨ ¸
© ¹© ¹ © ¹
u u
u u
20 mm 0
100cos30 86.603 N m
100sin30 50 N m
D B
z
y
z
M
M
q ˜
q ˜
(a)
3 3
9 9
(86.603)( 8.4883 10 ) (50)(20 10 )
17.5611 10 62.832 10
y A
z A
A
z y
M z
M y
I I
V
u u
u u
6
57.8 10 Pa
u 57.8 MPa
A
V W
(b)
3
9 9
(86.603)(11.5117 10 ) (50)(0)
17.5611 10 62.832 10
y B
z B
B
z y
M z
M y
I I
V
u
u u
6
56.8 10 Pa
u 56.8 MPa
B
V W
(c)
3 3
9 9
(86.603)( 8.4883 10 ) (50)( 20 10 )
17.5611 10 62.832 10
y D
z D
D
z y
M z
M y
I I
V
u u
u u
6
25.9 10 Pa
u 25.9 MPa
D
V W
Mecánica de Materiales - Ferdinand P. Beer 7ma. Edición.

629. PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
590
165 mm
310 mm
15#
M 16 kN · m
W310 $ 38.7
A
B
C
D
E
PROBLEM 4.134
The couple M is applied to a beam of the cross section shown in a plane
forming an angle E with the vertical. Determine the stress at
(a) point A, (b) point B, (c) point D.
SOLUTION
For W310 38.7
u rolled steel shape,
6 4 6 4
6 4 6 4
84.9 10 mm 84.9 10 m
7.20 10 mm 7.20 10 m
1 1
(310) 155 mm
2 2
1
(165) 82.5 mm
2
u u
u u
§ ·
¨ ¸
© ¹
§ ·
¨ ¸
© ¹
z
y
A B D E
A E B D
I
I
y y y y
z z z z
3 3
3 3
(16 10 ) cos15 15.455 10 N m
(16 10 ) sin 15 4.1411 10 N m
u q u ˜
u q u ˜
z
y
M
M
(a)
6
6
84.9 10
tan tan tan 15 3.1596
7.20 10
z
I
I
u
q
u
I T
72.4q
I
72.4 15 57.4
D q W
(b) Maximum tensile stress occurs at point E.
3 3 3 3
6 6
(15.455 10 )( 155 10 ) (4.1411 10 )(82.5 10 )
84.9 10 7.20 10
y E
z E
E
z y
M z
M y
I I
u u u u
u u
V
6
75.7 10 Pa 75.7 MPa
u W
Mecánica de Materiales - Ferdinand P. Beer 7ma. Edición.

630. PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
591
6 in.
3.33 in.
208
C
B
D
A
E
M 5 15 kip · in.
S6 3 12.5 PROBLEM 4.135
The couple M acts in a vertical plane and is applied to a beam oriented
as shown. Determine (a) the angle that the neutral axis forms with the
horizontal, (b) the maximum tensile stress in the beam.
SOLUTION
For S6 12.5
u rolled steel shape,
4
4
22.0 in
1.80 in
1
(3.33) 1.665 in.
2
1
(6) 3 in.
2
z
y
E A B D
A B D E
I
I
z z z z
y y y y
15 sin 20 5.1303 kip in.
15 cos 20 14.095 kip in.
q ˜
q ˜
z
y
M
M
(a)
22.0
tan tan tan (90 20 ) 33.58
1.80
z
y
I
I
q q q
I T
88.29q
I
88.29 70 18.29
D q q q W
(b) Maximum tensile stress occurs at point D.
(5.1303)( 3) (14.095)(1.665)
13.74 ksi
22.0 1.80
y D
z D
D
z y
M z
M y
I I
V W
Mecánica de Materiales - Ferdinand P. Beer 7ma. Edición.

631. PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
592
152 mm
13 mm
48.8 mm
C150 ! 12.2
M 6 kN · m
E
A
B
D
C
5# PROBLEM 4.136
The couple M acts in a vertical plane and is applied to a beam oriented as
shown. Determine (a) the angle that the neutral axis forms with the
horizontal, (b) the maximum tensile stress in the beam.
SOLUTION
6 4
6 4
6 sin 5 0.52293 kN m
6 cos 5 5.9772 kN m
C150 12.2
0.286 10 m
5.45 10 m
c
c
c
c
q ˜
q ˜
u
u
u
y
z
y
z
M
M
I
I
(a) Neutral axis:
6 4
6 4
5.45 10 m
tan tan tan 5
0.286 10 m
c
c
u
q
u
z
y
I
I
I T
tan 1.66718 59.044
I I q
5 54.0
D I D
q q W
(b) Maximum tensile stress at : 76 mm, z 13 mm
c c
E E
E y
6 4 6 4
(0.52293 kN m)(0.013 m) (5.9772 kN m)( 0.076 m)
0.286 10 m 5.45 10 m
y E z E
E
y z
M z M y
I I
c c
c c
c c ˜ ˜
u u
V
107.1MPa
E
V W
Mecánica de Materiales - Ferdinand P. Beer 7ma. Edición.

632. PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
593
A
B
C
M ! 400 N · m
30
D
E
5 mm
5 mm
18.57 mm
50 mm
50 mm
5 mm
z'
y'
Iy' ! 281 # 103
mm4
Iz' ! 176.9 # 103 mm4
PROBLEM 4.137
The couple M acts in a vertical plane and is applied to a beam oriented
as shown. Determine (a) the angle that the neutral axis forms with the
horizontal, (b) the maximum tensile stress in the beam.
SOLUTION
3 4 9 4
3 4 9 4
176.9 10 mm 176.9 10 m
281 10 mm 281 10 m
18.57 mm, 25 mm
400cos30 346.41 N m
400sin30 200 N m
z
y
E E
z
y
I
I
y z
M
M
c
c
c
c
u u
u u
c
q ˜
q ˜
(a)
9
9
176.9 10
tan tan tan30 0.36346
281 10
19.97
z
y
I
I
M T
M
c
c
u
˜ q
u
q
30 19.97
D q q 10.03
D q W
(b) Maximum tensile stress occurs at point E.
3 3
9 9
6 6 6
(346.41)( 18.57 10 ) (200)(25 10 )
176.9 10 281 10
36.36 10 17.79 10 54.2 10 Pa
y E
z E
E
z y
M z
M y
I I
c
c
c c
c
c u u
u u
u u u
V
54.2 MPa
E
V W
Mecánica de Materiales - Ferdinand P. Beer 7ma. Edición.

633. PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
594
A
B
in.
4 in.
4 in.
4 in.
0.859 in.
458
C
M 5 15 kip · in.
D
1
2
y'
z'
Iy' 5 6.74 in4
Iz' 5 21.4 in4
PROBLEM 4.138
The couple M acts in a vertical plane and is applied to a beam oriented
as shown. Determine (a) the angle that the neutral axis forms with the
horizontal, (b) the maximum tensile stress in the beam.
SOLUTION
4 4
21.4 in 6.74 in
0.859 in. 4 0.859 in. 3.141 in.
4 in. 4 in. 0.25 in.
15 sin 45 10.6066 kip in.
15 cos45 10.6066 kip in.
c c
c c
c
c
c c c
q ˜
q ˜
z y
A B D
A B D
y
z
I I
z z z
y y y
M
M
(a) Angle of neutral axis:
21.4
tan tan tan ( 45 ) 3.1751
6.74
z
y
I
I
M T
c
c
q
72.5
72.5 45
M
D
q
q q 27.5
D q W
(b) The maximum tensile stress occurs at point D.
(10.6066)( 0.25) ( 10.6066)( 3.141)
21.4 6.74
y D
z D
D
z y
M z
M y
I I
V
0.12391 4.9429
5.07 ksi
D
V W
Mecánica de Materiales - Ferdinand P. Beer 7ma. Edición.

634. PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
595
A
M ! 120 N · m
20
D
B
E
10 mm
10 mm
10 mm
10 mm
6 mm
y'
z' 6 mm
C
Iy' ! 14.77 # 103
mm4
Iz' ! 53.6 # 103 mm4
PROBLEM 4.139
The couple M acts in a vertical plane and is applied to a beam
oriented as shown. Determine (a) the angle that the neutral
axis forms with the horizontal, (b) the maximum tensile stress
in the beam.
SOLUTION
3 4 9 4
3 4 9 4
53.6 10 mm 53.6 10 m
14.77 10 mm 14.77 10 m
120 sin 70 112.763 N m
120 cos 70 41.042 N m
z
y
z
y
I
I
M
M
c
c
c
c
u u
u u
q ˜
q ˜
(a) Angle of neutral axis: 20q
T
9
9
53.6 10
tan tan tan 20 1.32084
14.77 10
52.871
52.871 20
z
y
I
I
c
c
u
q
u
q
q q
M T
M
D 32.9
D q W
(b) The maximum tensile stress occurs at point E.
9 9
16 mm 0.016 m
10 mm 0.010 m
(112.763)( 0.016) (41.042)(0.010)
53.6 10 14.77 10
E
E
y E
z E
E
z y
y
z
M z
M y
I I
V
c
c
c c
c
c
c
u u
6
61.448 10 Pa
u 61.4 MPa
E
V W
Mecánica de Materiales - Ferdinand P. Beer 7ma. Edición.

635. PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
596
M 5 750 N · m
208
90 mm
25 mm
25 mm
30 mm
C
B
A
PROBLEM 4.140
The couple M acts in a vertical plane and is applied to a beam oriented as
shown. Determine (a) the angle that the neutral axis forms with the horizontal,
(b) the maximum tensile stress in the beam.
SOLUTION
3 6 4
6 4
750 sin 20
256.5 N m
750 cos20
704.8 N m
1
2 (90)(25) 0.2344 10 mm
12
0.2344 10 m
c
c
c
c
c
c
q
˜
q
˜
ª º
u
« »
¬ ¼
u
y
y
z
z
y
y
M
M
M
M
I
I
3 6 4 6 4
1
(50)(90) 1.0125 10 mm 1.0125 10 m
36
z
I
c u u
(a) Neutral axis:
6 4
6 4
1.0125 10 m
tan tan tan 20
0.2344 10 m
c
c
u
q
u
z
y
I
I
I T
tan 1.5724 57.5q
I I
20 57.5 30 37.5
q q q q
D I
37.5
D q W
(b) Maximum tensile stress, at D: 30 mm z 25 mm
c c
D
y
6 4 6 4
(256.5 N m)( 0.030 m) (704.8 N m)(0.025 m)
0.2344 10 m 1.0125 10 m
y D z D
D
y z
M z M y
I I
c c
c c
c c ˜ ˜
u u
V
32.83 MPa 17.40 MPa 50.23 MPa
50.2 MPa
D
V W
Mecánica de Materiales - Ferdinand P. Beer 7ma. Edición.

636. PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
597
A
40 mm
10 mm
40 mm
10 mm 10 mm
70 mm
C
M ! 1.2 kN · m
y
z
Iy ! 1.894 # 106 mm4
Iz ! 0.614 # 106 mm4
Iyz ! $0.800 # 106
mm4
PROBLEM 4.141
The couple M acts in a vertical plane and is applied to a beam oriented
as shown. Determine the stress at point A.
SOLUTION
Using Mohr’s circle, determine the principal axes and principal moments of inertia.
6 4
6 4
6 4
: (1.894, 0.800) 10 mm
: (0.614, 0.800) 10 mm
: (1.254, 0) 10 mm
u
u
u
Y
Z
E
2 2 2 2 6 6 4
6 4 6 4 6 4
6 4 6 4 6 4
6
6
0.640 0.800 10 1.0245 10 mm
(1.254 1.0245) 10 mm 0.2295 10 mm 0.2295 10 m
(1.254 1.0245) 10 mm 2.2785 10 mm 2.2785 10 m
0.800 10
tan 2 1.25 25.67
0.640 10
v
u
m m
R EF FZ
I
I
FZ
FE
T T
u u
u u u
u u u
u
q
u
3 3
3 3
cos (1.2 10 ) cos 25.67 1.0816 10 N m
sin (1.2 10 ) sin 25.67 0.5198 10 N m
cos sin 45 cos 25.67 45 sin 25.67 21.07 mm
cos sin 45 cos 25.67 45 sin 25.67 60.05 mm
v m
u m
A A m A m
A A m A m
v A
A
v
M M
M M
u y z
v z y
M u
I
T
T
T T
T T
V
u q u ˜
u q u ˜
q q
q q
3 3 3 3
6 6
(1.0816 10 )(21.07 10 ) ( 0.5198 10 )(60.05 10 )
0.2295 10 2.2785 10
u A
u
M v
I
u u u u
u u
6
113.0 10 Pa
u 113.0 MPa
A
V W
Mecánica de Materiales - Ferdinand P. Beer 7ma. Edición.

637. PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
598
C
A
z
y
2.4 in.
2.4 in. 2.4 in.
2.4 in.
2.4 in.
2.4 in.
M ! 125 kip · in.
PROBLEM 4.142
The couple M acts in a vertical plane and is applied to a beam oriented as
shown. Determine the stress at point A.
SOLUTION
^ `
3 4
3 2 4
4
1
2 (7.2)(2.4) 66.355 in
3
1
2 (2.4)(7.2) (2.4)(7.2)(1.2) 199.066 in
12
2 (2.4)(7.2)(1.2)(1.2) 49.766 in
y
z
yz
I
I
I
­ ½
® ¾
¯ ¿
­ ½
® ¾
¯ ¿
Using Mohr’s circle, determine the principal axes and principal moments of inertia.
4 4
4 4
4
:(66.355 in , 49.766 in )
:(199.066 in , 49.766 in )
:(132.710 in , 0)
Y
Z
E
Mecánica de Materiales - Ferdinand P. Beer 7ma. Edición.

638. PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
599
PROBLEM 4.142 (Continued)
49.766
tan 2
66.355
2 36.87 18.435
m
m m
DY
DE
T
T T
q q
2 2 4
4
4
82.944 in
132.710 82.944 49.766 in
132.710 82.944 215.654 in
u
v
R DE DY
I
I
125sin18.435 39.529 kip in.
125cos18.435 118.585 kip in.
q ˜
q ˜
u
v
M
M
4.8 cos18.435 2.4 sin 18.435 5.3126 in.
4.8 sin 18.435 2.4 cos18.435 0.7589 in.
(118.585)(5.3126) (39.529)(0.7589)
215.654 49.766
A
A
A u A
A
u
u
M u M
I I
Q
Q
Q
Q
V
q q
q q
2.32 ksi
2.32 ksi
A
V W
Mecánica de Materiales - Ferdinand P. Beer 7ma. Edición.

639. PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
600
A
6 in.
2.08 in.
1.08 in.
0.75 in.
0.75 in.
4 in.
C
M 5 60 kip · in.
y
z
Iy 5 8.7 in4
Iyz 5 18.3 in4
Iz 5 24.5 in4
PROBLEM 4.143
The couple M acts in a vertical plane and is applied to a beam oriented
as shown. Determine the stress at point A.
SOLUTION
Using Mohr’s circle, determine the principal axes and principal moments of inertia.
4
4
4
:(8.7, 8.3) in
:(24.5, 8.3) in
:(16.6, 0) in
Y
Z
E
4
4
7.9 in
8.3 in
EF
FZ
2 2 4
4 4
8.3
7.9 8.3 11.46 in tan 2 1.0506
7.9
23.2 16.6 11.46 5.14 in 16.6 11.46 28.06 in
m
m u v
FZ
R
EF
I I
T
T
q
sin (60) sin 23.2 23.64 kip in.
cos (60) cos 23.2 55.15 kip in.
cos sin 3.92cos23.2 1.08 sin 23.2 4.03 in.
cos sin 1.08cos23.2 3.92 sin 23.2 0.552 in.
q ˜
q ˜
q q
q q
u m
v m
A A m A m
A A m A m
v A u A
A
v u
M M
M M
u y z
v z y
M u M v
I I
T
T
T T
T T
V
(55.15)( 4.03) (23.64)(0.552)
28.06 5.14
10.46 ksi
A
V W
Mecánica de Materiales - Ferdinand P. Beer 7ma. Edición.

640. PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
601
75 mm
125 mm
28 kN
28 kN
14 kN
A
D
B G
H
E
F
PROBLEM 4.144
The tube shown has a uniform wall thickness of 12 mm. For the
loading given, determine (a) the stress at points A and B, (b) the point
where the neutral axis intersects line ABD.
SOLUTION
Add y- and z-axes as shown. Cross section is a 75 mm u 125-mm rectangle with a 51 mm u 101-mm rectangular
cutout.
3 3 6 4 6 4
3 3 3 4 6 4
3 2 3 2
1 1
(75)(125) (51)(101) 7.8283 10 mm 7.8283 10 m
12 12
1 1
(125)(75) (101)(51) 3.2781 10 mm 3.2781 10 m
12 12
(75)(125) (51)(101) 4.224 10 mm 4.224 10 m
z
y
I
I
A
u u
u u
u u
Resultant force and bending couples:
3
14 28 28 70 kN 70 10 N
(62.5 mm)(14 kN) (62.5 mm)(28kN) (62.5 mm)(28 kN) 2625 N m
(37.5 mm)(14 kN) (37.5 mm)(28 kN) (37.5 mm)(28 kN) 525 N m
z
y
P
M
M
u
˜
˜
(a)
3
3 6 6
70 10 (2625)( 0.0625) ( 525)(0.0375)
4.224 10 7.8283 10 3.2781 10
y A
z A
A
z y
M z
M y
P
A I I
V
u
u u u
6
31.524 10 Pa
u 31.5 MPa
A
V W
3
3 6 6
70 10 (2625)(0.0625) ( 525)(0.0375)
4.224 10 7.8283 10 3.2781 10
y B
z B
B
z y
M z
M y
P
A I I
V
u
u u u
6
10.39 10 Pa
u 10.39 MPa
B
V W
(b) Let point H be the point where the neutral axis intersects AB.
6 3
3 6
0.0375 m, ?, 0
0
7.8283 10 70 10 ( 525)(0.0375)
2625 4.224 10 3.2781 10
0.03151 m 31.51 mm
V
§ · ª º
u u
¨ ¸ « »
¨ ¸ u u
¬ ¼
© ¹
H H H
y H
z H
z y
z H
H
z y
z y
M z
M y
P
A I I
I Mz
P
y
M A I
31.51 62.5 94.0 mm
Answer: 94.0 mm above point A. W
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602
20 mm
20 mm
20 mm
20 mm
60 mm
P
z x
O
a
y
PROBLEM 4.145
A horizontal load P of magnitude 100 kN is applied to the beam
shown. Determine the largest distance a for which the maximum
tensile stress in the beam does not exceed 75 MPa.
SOLUTION
Locate the centroid.
A, mm2 , mm
y 3
, mm
Ay
M 2000 10 3
20 10
u
N 1200 –10 3
12 10
u
3200 3
8 10
u
3
Y
8 10
3200
2.5 mm
Ay
A
u
¦
¦
Move coordinate origin to the centroid.
Coordinates of load point: , 2.5 mm
P P
X a y
Bending couples:
x P y
M y P M aP
3 2 3 2 6 4
1 1
(100)(20) (2000)(7.5) (60)(20) (1200)(12.5) 0.40667 10 mm
12 12
x
I u
6 4
0.40667 10 m
u
3 3 6 4 6 4
6 3
1 1
(20)(100) (20)(60) 2.0267 10 mm 2.0267 10 m
12 12
75 10 Pa, 100 10 N
u u
u u
y
y
x
A
x y
I
M x
P M y
P
A I I
V V
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603
PROBLEM 4.145 (Continued)
6 3 3 3
6
3 6 6
6
6 3
3
For point , 50 mm, 2.5 mm
2.0267 10 100 10 ( 2.5)(100 10 )( 2.5 10 )
75 10
50 10 3200 10 0.40667 10
2.0267 10
{31.25 1.537 75} 10 1.7111 10 N m
50 10
y x
y
x
y
I P M y
M A x y
x A I
M
­ ½
® ¾
¯ ¿
­ ½
u u u u
° °
u
® ¾
u u u
° °
¯ ¿
u
u u ˜
u
V
3
3
3
(1.7111 10 )
17.11 10 m
100 10
y
M
a
P
u
u
u
17.11 mm
a W
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604
1 in.
1 in.
1 in.
4 in. 5 in.
2.5 in.
P
a
PROBLEM 4.146
Knowing that P 90 kips, determine the largest distance a for
which the maximum compressive stress dose not exceed 18 ksi.
SOLUTION
2
3 3 4
3 3 4
(5 in.)(6 in.) 2(2 in.)(4 in.) 14 in
1 1
(5 in.)(6 in.) 2 (2 in.)(4 in.) 68.67 in
12 12
1 1
2 (1in.)(5 in.) (4 in.)(1in.) 21.17 in
12 12
x
z
A
I
I
Force-couple system at C: (2.5 in.)
x z
P P M P M Pa
For P 90 kips: 90 kips (90 kips)(2.5 in.) 225 kip in. (90kips)
x z
P M M a
˜
Maximum compressive stress at B: 18 ksi
B
V
(3 in.) (2.5 in.)
x z
B
x z
P M M
A I I
V
2 4 4
90 kips (225 kip in.)(3 in.) (90 kips) (2.5 in.)
18 ksi
14 in 68.67 in 21.17 in
18 6.429 9.830 10.628
1.741 10.628
a
a
a
˜
0.1638 in.
a W
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605
1 in.
1 in.
1 in.
4 in. 5 in.
2.5 in.
P
a
PROBLEM 4.147
Knowing that 1.25 in.,
a determine the largest value of P that
can be applied without exceeding either of the following
allowable stresses:
ten 10 ksi
V comp 18 ksi
V
SOLUTION
2
3 3 4
3 3 4
(5 in.)(6 in.) (2)(2 in.)(4 in.) 14 in
1 1
(5 in.)(6 in.) 2 (2 in.)(4 in.) 68.67 in
12 12
1 1
2 (1in.)(5 in.) (4 in.)(1in.) 21.17 in
12 12
x
z
A
I
I
Force-couple system at C: For 1.25 in.,
a
(2.5 in.)
x
P P M P
(1.25 in.)
y
M Pa
Maximum compressive stress at B: 18 ksi
B
V
2 4 4
(3 in.) (2.5 in.)
(2.5 in.)(3 in.) (1.25 in.)(2.5 in.)
18 ksi
14 in 68.67 in 21.17 in
18 0.0714 0.1092 0.1476
18 0.3282 54.8 kips
x z
B
x z
P M M
A I I
P P P
P P P
P P
V
Maximum tensile stress at D: 10 ksi
D
V
(3 in.) (2.5 in.)
10 ksi 0.0714 0.1092 0.1476
10 0.1854 53.9 kips
x z
D
x z
P M M
A I I
P P P
P P
V
The smaller value of P is the largest allowable value.
53.9 kips
P W
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on a website, in whole or part.
606
%
y
A
B
E
D
C
z
x
R ! 125 mm
150 mm
200 mm
P ! 4 kN PROBLEM 4.148
A rigid circular plate of 125-mm radius is attached to a solid
150 u 200-mm rectangular post, with the center of the plate directly above
the center of the post. If a 4-kN force P is applied at E with 30 ,
T q
determine (a) the stress at point A, (b) the stress at point B, (c) the point
where the neutral axis intersects line ABD.
SOLUTION
3
3 3
3 3
4 10 N (compression)
sin30 (4 10 )(125 10 )sin30 250 N m
cos30 (4 10 )(125 10 )cos30 433 N m
x
z
P
M PR
M PR
u
q u u q ˜
q u u q ˜
3 6 4 6 4
3 6 4 6 4
3 2 3 2
1
(200)(150) 56.25 10 mm 56.25 10 m
12
1
(150)(200) 100 10 mm 100 10 m
12
100 mm 75 mm
(200)(150) 30 10 mm 30 10 m
x
z
A B A B
I
I
x x z z
A
u u
u u
u u
(a)
3 3 3
3 6 6
4 10 ( 250)(75 10 ) ( 433)( 100 10 )
30 10 56.25 10 100 10
x A z A
A
x z
P M z M x
A I I
V
u u u
u u u
3
633 10 Pa 633 kPa
A
V u W
(b)
3 3 3
3 6 6
4 10 ( 250)(75 10 ) ( 433)(100 10 )
30 10 56.25 10 100 10
x B z B
B
x z
P M z M x
A I I
V
u u u
u u u
3
233 10 Pa 233 kPa
B
V u W
(c) Let G be the point on AB where the neutral axis intersects.
6 3 3
3 6
0 75 mm ?
0
100 10 4 10 ( 250)(75 10 )
433 30 10 56.25 10
G G G
x G z G
G
x z
z x G
G
z x
z x
P M z M x
A I I
I P M z
x
M A I
­ ½
­ ½ u u u
° °
® ¾ ® ¾
u u
° °
¯ ¿ ¯ ¿
V
V
3
46.2 10 m 46.2 mm
u Point G lies 146.2 mm from point A. W
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Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
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607
%
y
A
B
E
D
C
z
x
R ! 125 mm
150 mm
200 mm
P ! 4 kN PROBLEM 4.149
In Prob. 4.148, determine (a) the value of θ for which the stress at D
reaches it largest value, (b) the corresponding values of the stress at A, B,
C, and D.
PROBLEM 4.148 A rigid circular plate of 125-mm radius is attached to a
solid 150 u 200-mm rectangular post, with the center of the plate directly
above the center of the post. If a 4-kN force P is applied at E with
30 ,
T q determine (a) the stress at point A, (b) the stress at point B,
(c) the point where the neutral axis intersects line ABD.
SOLUTION
(a) 3 3 3
4 10 N (4 10 )(125 10 ) 500 N m
sin 500sin cos 500cos
x x
P PR
M PR M PR
T T T T
u u u ˜
3 6 4 6 4
3 6 4 6 4
3 2 3 2
1
(200)(150) 56.25 10 mm 56.25 10 m
2
1
(150)(200) 100 10 mm 100 10 m
2
100 mm 75 mm
(200)(150) 30 10 mm 30 10 m
x
z
D D
I
I
x z
A
u u
u u
u u
1 sin cos
x z
x z x z
P M z M x Rz Rx
P
A I I A I I
T T
V
­ ½
® ¾
¯ ¿
For V to be a maximum, 0
d
d
V
T
with ,
D D
z z x x
cos sin
0 0
D D D
x Z
d Rz Rx
P
d I I
V T T
T
­ ½
® ¾
¯ ¿
6 3
6 3
sin (100 10 )( 75 10 ) 4
tan
cos 3
(56.25 10 )(100 10 )
z D
x D
I z
I x
T
T
T
u u
u u
sin 0.8, cos 0.6
T T 53.1
T q W
(b)
3 3 3
3 6 6
4 10 (500)(0.8)(75 10 ) (500)(0.6)( 100 10 )
30 10 56.25 10 100 10
x A z A
A
x z
P M z M x
A I I
V
u u u
u u u
6 6
( 0.13333 0.53333 0.300) 10 Pa 0.700 10 Pa
u u 700 kPa
A
V W
6 6
( 0.13333 0.53333 0.300) 10 Pa 0.100 10 Pa
B
V u u 100 kPa
B
V W
6
( 0.13333 0 0) 10 Pa
C
V u 133.3 kPa
C
V W
6 6
( 0.13333 0.53333 0.300) 10 Pa 0.967 10 Pa
D
V u u 967 kPa
D
V W
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647. PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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608
C
0.5 in.
5 in.
1.43 in.
1.43 in.
5 in.
0.5 in.
y
z
M0
PROBLEM 4.150
A beam having the cross section shown is subjected to a couple 0
M that
acts in a vertical plane. Determine the largest permissible value of the
moment 0
M of the couple if the maximum stress in the beam is
not to exceed 12 ksi. Given: 4
11.3 in ,
y z
I I 2
4.75 in ,
A
min 0.983 in.
k (Hint: By reason of symmetry, the principal axes form
an angle of 45q with the coordinate axes. Use the relations 2
min min
I Ak
and min max .)
y z
I I I I
SOLUTION
0 0
0 0
2 2 4
min min
4
max min
sin 45 0.70711
cos 45 0.70711
(4.75)(0.983) 4.59 in
11.3 11.3 4.59 18.01 in
cos 45 sin 45 3.57 cos 45 0.93 sin 45 1.866 in.
cos 45 sin 45 0.93 cos 4
u
v
y z
B B B
B B B
M M M
M M M
I Ak
I I I I
u y z
v z y
q
q
q q q q
q q
0
min max
0 0
5 ( 3.57) sin 45 3.182 in.
0.70711
( 1.866) 3.182
0.70711 0.4124
4.59 18.01
v B u B B B
B
M u M v u v
M
Iv Iu I I
M M
q q
ª º
« »
¬ ¼
ª º
« »
¬ ¼
V
0
12
0.4124 0.4124
B
M
V
0 29.1 kip in.
˜
M W
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on a website, in whole or part.
609
C
0.5 in.
5 in.
1.43 in.
1.43 in.
5 in.
0.5 in.
y
z
M0
PROBLEM 4.151
Solve Prob. 4.150, assuming that the couple 0
M acts in a horizontal
plane.
PROBLEM 4.150 A beam having the cross section shown is subjected
to a couple 0
M that acts in a vertical plane. Determine the largest
permissible value of the moment 0
M of the couple if the maximum
stress in the beam is not to exceed 12 ksi. Given: 4
11.3 in ,
y z
I I
2
4.75 in ,
A min 0.983 in.
k (Hint: By reason of symmetry, the
principal axes form an angle of 45q with the coordinate axes. Use the
relations 2
min min
I Ak and min max .
y z
I I I I
)
SOLUTION
0 0
0 0
2 2 4
min min
4
max min
cos 45 0.70711
sin 45 0.70711
(4.75)(0.983) 4.59 in
11.3 11.3 4.59 18.01 in
cos 45 sin 45 0.93 cos 45 ( 3.57 sin 45 ) 1.866 in.
cos 45 sin 45 ( 3
u
v
y z
D D D
D D D
M M M
M M M
I Ak
I I I I
u y z
v z y
q
q
q q q q
q q
0
min max
0 0
.57) cos 45 (0.93) sin 45 3.182 in.
0.70711
( 1.866) 3.182
0.70711 0.4124
4.59 18.01
v D u D D D
D
v u
M u M v u v
M
I I I I
M M
q q
ª º
« »
¬ ¼
ª º
« »
¬ ¼
V
0
12
0.4124 0.4124
D
M
V
0 29.1 kip in.
˜
M W
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610
C
40 mm
10 mm 10 mm
10 mm
70 mm
y
z
40 mm
M0
PROBLEM 4.152
The Z section shown is subjected to a couple 0
M acting in a vertical
plane. Determine the largest permissible value of the moment 0
M of the
couple if the maximum stress is not to exceed 80 MPa. Given:
6 4
max 2.28 10 mm ,
I
u 6 4
min 0.23 10 mm ,
I
u principal axes
25.7q and 64.3q .
SOLUTION
6 4 6 4
max
6 4 6 4
min
0
0
6
6
2.28 10 mm 2.28 10 m
0.23 10 mm 0.23 10 m
cos 64.3
sin 64.3
64.3
tan tan
2.28 10
tan 64.3
0.23 10
20.597
87.22
v
u
v
u
v
u
I I
I I
M M
M M
I
I
T
M T
M
u u
u u
q
q
q
u
q
u
q
Points A and B are farthest from the neutral axis.
3
6 0 0
6
cos 64.3 sin 64.3 ( 45) cos 64.3 ( 35) sin 64.3
51.05 mm
cos 64.3 sin 64.3 ( 35) cos 64.3 ( 45) sin 64.3
25.37 mm
( cos 64.3 )( 51.05 10 ) ( sin 64.3 )(25.
80 10
2.28 10
B B B
B B B
v B u B
B
v u
u y z
v z y
M u M v
I I
M M
V
q q q q
q q q q
q u q
u
u
3
6
3
0
37 10 )
0.23 10
109.1 10 M
u
u
u
6
0 3
80 10
109.1 10
M
u
u
0 733 N m
M ˜ W
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611
C
40 mm
10 mm 10 mm
10 mm
70 mm
y
z
40 mm
M0
PROBLEM 4.153
Solve Prob. 4.152 assuming that the couple 0
M acts in a horizontal
plane.
PROBLEM 4.152 The Z section shown is subjected to a couple 0
M
acting in a vertical plane. Determine the largest permissible value of the
moment 0
M of the couple if the maximum stress is not to exceed 80 MPa.
Given: 6 4
max 2.28 10 mm ,
I
u 6 4
min 0.23 10 mm ,
I
u principal axes
25.7q and 64.3q .
SOLUTION
6 4 6 4
min
6 4 6 4
max
0
0
6
6
0.23 10 mm 0.23 10 m
2.28 10 mm 2.28 10 m
cos 64.3
sin 64.3
64.3
tan tan
0.23 10
tan 64.3
2.28 10
0.20961
11.84
v
u
v
u
v
u
I I
I I
M M
M M
I
I
T
M T
M
u u
u u
q
q
q
u
q
u
q
Points D and E are farthest from the neutral axis.
3
6
3
0
cos 25.7 sin 25.7 ( 5) cos 25.7 45 sin 25.7
24.02 mm
cos 25.7 sin 25.7 45cos25.7 ( 5) sin 25.7
38.38 mm
( cos 64.3 )( 24.02 10 )
0.23 10
( sin 64.3 )(38.38 10 )
2.28 10
D D D
D D D
v D u D D
D
v u
u y z
v z y
M u M v M
I I
M
V
q q q q
q q q q
q u
u
q u
u 6
6 3
0
3
0
80 10 60.48 10
1.323 10 N m
M
M
u u
u ˜ 0 1.323 kN m
M ˜ W
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612
1.5 in.
0.3 in.
1.5 in.
0.6 in.
0.3 in.
0.6 in.
M0
y
z C
PROBLEM 4.154
An extruded aluminum member having the cross section shown
is subjected to a couple acting in a vertical plane. Determine the
largest permissible value of the moment 0
M of the couple if the
maximum stress is not to exceed 12 ksi. Given:
4
max 0.957 in ,
I 4
min 0.427 in ,
I principal axes 29.4q and
60.6q
SOLUTION
4
max
4
min
0 0
0.957 in
0.427 in
sin 29.4 , cos29.4
u
v
u v
I I
I I
M M M M
q q
29.4
T q
0.427
tan tan tan 29.4
0.957
0.2514 14.11
v
u
I
I
M T
M
q
q
Point A is farthest from the neutral axis.
0.75 in., 0.75 in.
A A
y z
cos29.4 sin 29.4 1.0216 in.
cos29.4 sin 29.4 0.2852 in.
A A A
A A A
u y z
v z y
q q
q q
0 0
0
( cos29.4 )( 1.0216) ( sin 29.4 )( 0.2852)
0.427 0.957
1.9381
v A u A
A
v u
M u M V M M
I I
M
V
q q
0
12
1.9381 1.9381
A
M
V
0 6.19 kip in.
M ˜ W
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613
20 mm
20 mm
b = 60 mm
b = 60 mm
M0
z
y
C
PROBLEM 4.155
A beam having the cross section shown is subjected to a couple M0
acting in a vertical plane. Determine the largest permissible value of
the moment M0 of the couple if the maximum stress is not to exceed
100 MPa. Given: 4 4
/36 and /72.
y z yz
I I b I b
SOLUTION
4 4
6 4
4 4
6 4
60
0.360 10 mm
36 36
60
0.180 10 mm
72 72
u
u
y z
yz
b
I I
b
I
Principal axes are symmetry axes.
Using Mohr’s circle, determine the principal moments of inertia.
6 4
6 4 6 4
6 4 6 4
0.180 10 mm
2
0.540 10 mm 0.540 10 m
2
0.180 10 mm 0.180 10 m
u
u u
u u
yz
y z
v
y z
u
R I
I I
I R
I I
I R
0 0 0 0
6
6
sin 45 0.70711 , cos45 0.70711
0.540 10
45 tan tan tan 45 3
0.180 10
u v
v
u
M M M M M M
I
I
q q
u
q q
u
T M T
71.56
M q
Point A: 0 20 2 mm
A A
u v
3
3
0
0
6
6
0 3 3
(0.70711 )( 20 2 10 )
0 11.11 10
0.180 10
100 10
900 N m
111.11 10 111.11 10
u
u
u
u
˜
u u
v A u A
A
v u
A
M u M v M
M
I I
M
V
V
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614
PROBLEM 4.155 (Continued)
Point B:
60 20
mm, mm
2 2
B B
u v
3 3
60 20
0 0
2 2
6 6
3
0
6
0 3 3
(0.70711 ) 10 (0.70711 ) 10
0.540 10 0.180 10
111.11 10
100 10
900 N m
111.11 10 111.11 10
v B u B
B
v u
B
M M
M u M v
I I
M
M
u u
u u
u
u
˜
u u
V
V
Choose the smaller value. 0 900 N m
˜
M W
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615
M
A
B
E
C
D
h
b PROBLEM 4.156
Show that, if a solid rectangular beam is bent by a couple applied in a plane containing
one diagonal of a rectangular cross section, the neutral axis will lie along the other
diagonal.
SOLUTION
3 3
tan
cos , sin
1 1
12 12
z z
z y
b
h
M M M M
I bh I hb
T
T T
3
3
1
12
tan tan
1
12
z
y
bh
I b h
I h b
hb
M T ˜
The neutral axis passes through corner A of the diagonal AD. W
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616
A
A
B
B
C
C
D
D
z
z
x
x
b
h
y
h
6
b
6
(a) (b)
P
PROBLEM 4.157
(a) Show that the stress at corner A of the prismatic
member shown in part a of the figure will be zero if
the vertical force P is applied at a point located on the
line
1
/6 /6
x z
b h
(b) Further show that, if no tensile stress is to occur in
the member, the force P must be applied at a point
located within the area bounded by the line found in
part a and three similar lines corresponding to the
condition of zero stress at B, C, and D, respectively.
This area, shown in part b of the figure, is known as
the kern of the cross section.
SOLUTION
3 3
1 1
12 12
2 2
z x
A A
I hb I bh A bh
h b
z x
Let P be the load point.
2 2
3 3
1 1
12 12
( ) ( )
1
/6 /6
z P x P
z A x A
A
z x
b h
P P
P P
M Px M Pz
P M x M z
A I I
Px Pz
P
bh hb bh
P x z
bh b h
V
ª º
« »
¬ ¼
(a) For 0, 1 0 1
/6 /6 /6 /6
A
x z x z
b h b h
V
(b) At point E: 0 /6
E
z x b
?
At point F: 0 /6
F
x z h
?
If the line of action ( , z )
P P
x lies within the portion marked ,
A
T a tensile stress will occur at corner A.
By considering 0, 0,
B C
V V and 0,
D
V the other portions producing tensile stresses are identified.
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617
A
C
y
y
z
z
x
PROBLEM 4.158
A beam of unsymmetric cross section is subjected to a couple 0
M acting in the
horizontal plane xz. Show that the stress at point A, of coordinates y and z, is
2
z yz
A y
y z yz
zI yI
M
I I I
V
where Iy, Iz, and Iyz denote the moments and product of inertia of the cross section
with respect to the coordinate axes, and My the moment of the couple.
SOLUTION
The stress A
V varies linearly with the coordinates y and z. Since the axial force is zero, the y- and z-axes are
centroidal axes:
1 2
A C y C z
V
where 1
C and 2
C are constants.
2
1 2
1 2
1 2
2
1 2
1 2
2 2
2
2
0
z A
z yz
yz
z
y A
yz y
yz
yz y
z
z y y z yz
M y dA C y dA C yz dA
I C I C
I
C C
I
M z dA C yz dA C z dA
I C I C
I
I C I C
I
I M I I I C
³ ³ ³
³ ³ ³
V
V
2 2
1 2
2
z y
y z yz
yz y
y z yz
z yz
A y
y z yz
I M
C
I I I
I M
C
I I I
I z I y
M
I I I
V
W
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618
A
C
y
y
z
z
x
PROBLEM 4.159
A beam of unsymmetric cross section is subjected to a couple 0
M acting in the
vertical plane xy. Show that the stress at point A, of coordinates y and z, is
2
y yz
A z
y z yz
yI zI
M
I I I
V
where , ,
y z
I I and yz
I denote the moments and product of inertia of the cross
section with respect to the coordinate axes, and z
M the moment of the couple.
SOLUTION
The stress A
V varies linearly with the coordinates y and z. Since the axial force is zero, the y- and z-axes are
centroidal axes:
1 2
A C y C z
V
where 1
C and 2
C are constants.
2
1 2
1 2
2 1
2
1 2
1 1
2
1
1 2
2 2
0
y A
yz y
yz
y
z A
yz
z yz
y
y z y z yz
y z
y z yz
yz z
y z yz
M z dA C yz dA C z dA
I C I C
I
C C
I
M y dz C y dA C yz dA
I
I C I C
I
I M I I I C
I M
C
I I I
I M
C
I I I
³ ³ ³
³ ³ ³
V
V
2
2
y yz
A z
y z yz
I y I
M
I I I
V
W
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658. PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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619
A
B
C
P
D
y
x
z
xA
zA
PROBLEM 4.160
(a) Show that, if a vertical force P is applied at point A of the section
shown, the equation of the neutral axis BD is
2 2
1
A A
z x
x z
x z
r r
§ ·
§ ·
¨ ¸
¨ ¸
¨ ¸ ¨ ¸
© ¹ © ¹
where z
r and x
r denote the radius of gyration of the cross section with
respect to the z axis and the x axis, respectively. (b) Further show that, if
a vertical force Q is applied at any point located on line BD, the stress at
point A will be zero.
SOLUTION
Definitions:
2 2
,
x z
x z
I I
r r
A A
(a)
2 2
2 2
1 0
x A z A
z E x E A E A E
E
z x z x
A A
E E
z x
M Pz M Px
P M x M z P Px x Pz z
A I I A Ar Ar
P x z
x z
A r r
V
ª º
§ ·
§ ·
« »
¨ ¸
¨ ¸
¨ ¸ ¨ ¸
« »
© ¹ © ¹
¬ ¼
if E lies on neutral axis.
2 2 2 2
1 0, 1
A A A A
z x z x
x z x z
x z x z
r r r r
§ · § ·
§ · § ·
¨ ¸ ¨ ¸
¨ ¸ ¨ ¸
¨ ¸ ¨ ¸
¨ ¸ ¨ ¸
© ¹ © ¹
© ¹ © ¹
W
(b)
2 2
x E z E
z A x A E A E A
A
z y z x
M Pz M Px
P M x M z P Px x Pz z
A I I A Ar Ar
V
0 by equation from part (a). W
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Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
620
24 mm
50 mm
B
A
h
B
A
C
600 N · m 600 N · m
PROBLEM 4.161
For the curved bar shown, determine the stress at point A when
(a) h 50 mm, (b) h 60 mm.
SOLUTION
(a) 1 2
3
100
2 50
1
1 2
50 mm, 50 mm, 100 mm
(24)(50) 1.200 10 mm
50
R 72.13475 mm
ln
ln
1
( ) 75 mm
2
2.8652 mm
u
h r r
A
h
r
r
r r r
e r R
3
6
3 3 3
72.13475 50 22.13475 mm 50 mm
(600)(22.13475 10 )
77.3 10 Pa
(1.200 10 )(2.8652 10 )(50 10 )
u
u
u u u
A A
A
A
A
y r
My
Aer
V
77.3 MPa
W
(b)
3 2
1 2
110
2 50
1
1 2
3
3 3
60 mm, 50 mm, 110 mm, (24)(60) 1.440 10 mm
60
76.09796 mm
ln
ln
1
( ) 80 mm 3.90204 mm
2
76.09796 50 26.09796 mm 50 mm
(600)(26.09796 10 )
(1.440 10 )(3.90204 10 )(
u
u
u u
A A
A
A
A
h r r A
h
R
r
r
r r r e r R
y r
M y
Aer
V 6
3
55.7 10 Pa
50 10 )
u
u
55.7 MPa
W
Mecánica de Materiales - Ferdinand P. Beer 7ma. Edición.

660. PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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on a website, in whole or part.
621
24 mm
50 mm
B
A
h
B
A
C
600 N · m 600 N · m
PROBLEM 4.162
For the curved bar shown, determine the stress at points A and B
when h 55 mm.
SOLUTION
1 2
3 2
2
1 2
55 mm, 50 mm, 105 mm
(24)(55) 1.320 10 mm
50
74.13025 mm
105
ln ln
1 50
1
( ) 77.5 mm
2
3.36975 mm
u
h r r
A
h
R
r
r
r r r
e r R
3
6
3 3 3
74.13025 50 24.13025 mm 50 mm
(600)(24.13025 10 )
65.1 10 Pa
(1.320 10 )(3.36975 10 )(50 10 )
u
u
u u u
A A
A
A
A
y r
M y
Aer
V
65.1MPa
W
3
6
3 3 3
74.13025 105 30.86975 mm 105 mm
(600)( 30.8697 10 )
39.7 10 Pa
(1.320 10 )(3.36975 10 )(105 10 )
u
u
u u u
B B
B
B
B
y r
My
Aer
V
39.7 MPa W
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622
C
B
A
0.75 in.
4 kip · in.
3 in.
h
4 kip · in.
PROBLEM 4.163
For the machine component and loading shown, determine the
stress at point A when (a) 2 in.,
h (b) 2.6 in.
h
SOLUTION
4 kip in.
M ˜
Rectangular cross section: 2 1 2
3 in.
A bh r r r h
1 2
2
1
1
( ), ,
2 ln
h
r r r R e r R
r
r
(a) 2
2 in. (0.75)(2) 1.5 in
h A
1 3 2 1 in.
r
1
(3 1) 2 in.
2
r
3
1
2
1.8205 in. 2 1.8205 0.1795 in.
ln
R e
At point A: 1 1 in.
r r
( ) ( 4)(1 1.8205)
12.19 ksi
(1.5)(0.1795)(1)
A
M r R
Aer
V
12.19 ksi
A
V W
(b) 2
2.6 in. (0.75)(2.6) 1.95 in
h A
1 3 2.6 0.4 in.
r
1
(3 0.4) 1.7 in.
2
r
3
0.4
2.6
1.2904 in. 1.7 1.2904 0.4906 in.
ln
R e
At point A: 1 0.4 in.
r r
( ) ( 4)(0.4 1.2904)
11.15 ksi
(1.95)(0.4096)(0.4)
A
M r R
Aer
V
11.15 ksi
A
V W
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662. PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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623
C
B
A
0.75 in.
4 kip · in.
3 in.
h
4 kip · in.
PROBLEM 4.164
For the machine component and loading shown, determine the
stress at points A and B when 2.5 in.
h
SOLUTION
4 kip in.
M ˜
Rectangular cross section: 2
2.5 in. 0.75 in. 1.875 in.
h b A
2 1 2
3 in. 0.5 in.
r r r h
2
1
1 2
3.0
0.5
1 1
( ) (0.5 3.0) 1.75 in.
2 2
2.5
1.3953 in.
ln
ln
1.75 1.3953 0.3547 in.
r
r
r r r
h
R
e r R
At point A: 1 0.5 in.
r r
( ) ( 4 kip in.)(0.5 in. 1.3953 in.)
(0.75 in.)(2.5 in.)(0.3547 in.)(0.5 in.)
A
M r R
Aer
˜
V 10.77 ksi
A
V W
At point B: 2 3 in.
r r
( ) ( 4 kip in.)(3 in. 1.3953 in.)
(0.75 in. 2.5 in.)(0.3547 in.)(3 in.)
B
M r R
Aer
˜
u
V 3.22 ksi
B
V W
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on a website, in whole or part.
624
40 mm
60 mm
120 N · m
r1
PROBLEM 4.165
The curved bar shown has a cross section of 40 60 mm
u and an inner
radius 1 15 mm.
r For the loading shown, determine the largest tensile
and compressive stresses.
SOLUTION
1 2
2 6 2
2
1
1 2
40 mm, 15 mm, 55 mm
(60)(40) 2400 mm 2400 10 m
40
30.786 mm
55
ln ln
40
1
( ) 35 mm
2
u
h r r
A
h
R
r
r
r r r
4.214 mm
e r R
My
Aer
V
At 15 mm, 30.786 15 15.756 mm
r y
3
6
6 3 3
(120)(15.786 10 )
12.49 10 Pa
(2400 10 )(4.214 10 )(15 10 )
V
u
u
u u u
12.49 MPa
V W
At 55 mm, 30.786 55 24.214 mm
r y
3
6
6 3 3
(120)( 24.214 10 )
5.22 10 Pa
(2400 10 )(4.214 10 )(55 10 )
V
u
u
u u u
5.22 MPa
V W
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664. PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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625
40 mm
60 mm
120 N · m
r1
PROBLEM 4.166
For the curved bar and loading shown, determine the percent
error introduced in the computation of the maximum stress by
assuming that the bar is straight. Consider the case when
(a) 1 20 mm,
r (b) 1 200 mm,
r (c) 1 2 m.
r
SOLUTION
2 6 2
3 3 6 4 6 4
40 mm, (60)(40) 2400 mm 2400 10 m , 120 N m
1 1 1
(60)(40) 0.32 10 mm 0.32 10 mm , 20 mm
12 12 2
h A M
I bh c h
u ˜
u u
Assuming that the bar is straight,
8
6
6
(120)(20 10 )
7.5 10 Pa 7.5 MPa
(0.32 10 )
s
Mc
I
V
u
u
u
(a) 1 2
20 mm 60 mm
r r
2
1
40
36.4096 mm
60
ln ln
20
h
R
r
r
1 16.4096 mm
r R
1 2
1
( ) 40 mm
2
r r r
3.5904 mm
e r R
3
6
1
6 3 3
( ) (120)( 16.4096 10 )
11.426 10 Pa 11.426 MPa
(2400 10 )(3.5904 10 )(20 10 )
a
M r R
Aer
u
u
u u u
V
11.426 ( 7.5)
% error 100% 34.4%
11.426
u
W
For parts (b) and (c), we get the values in the table below:
1, mm
r 2, mm
r , mm
R , mm
r , mm
e , MPa
V % error
(a) 20 60 36.4096 40 3.5904 11.426 34.4 %
(b) 200 240 219.3926 220 0.6074 7.982 6.0 % W
(c) 2000 2040 2019.9340 2020 0.0660 7.546 0.6 % W
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on a website, in whole or part.
626
0.4 in.
0.4 in.
0.3 in.
0.8 in.
0.8 in.
1.2 in.
A A
C C
B
P9 P
b
B
PROBLEM 4.167
Steel links having the cross section shown are available with
different central angles E. Knowing that the allowable stress is
12 ksi, determine the largest force P that can be applied to a
link for which 90 .
E q
SOLUTION
Reduce section force to a force-couple system at G, the centroid of the cross section AB.
1 cos
2
a r
E
§ ·
¨ ¸
© ¹
The bending couple is .
M Pa
For the rectangular section, the neutral axis for bending couple only lies at
2
1
.
ln
r
r
h
R
Also, e r R
At point A, the tensile stress is
1 1 1
1
A A A
A
P My P Pay P ay P
K
A Aer A Aer A er A
V
§ ·
¨ ¸
© ¹
where
1
1 A
ay
K
er
and 1
A
y R r
A
A
P
K
V
Data: 1 2
1.2 in., 0.8 in., 1.6 in., 0.8 in., 0.3 in.
r r r h b
2
1.6
0.8
0.8
(0.3)(0.8) 0.24 in 1.154156 in.
ln
1.2 1.154156 0.045844 in., 1.154156 0.8 0.35416 in.
1.2(1 cos 45 ) 0.35147 in.
(0.35147)(0.35416)
1 4.3940
(0.045844)(0.8)
(0.24)(12)
0.65544 kips
4.3940
A
A R
e y
a
K
P
q
655 lb
P W
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666. PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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on a website, in whole or part.
627
0.4 in.
0.4 in.
0.3 in.
0.8 in.
0.8 in.
1.2 in.
A A
C C
B
P9 P
b
B
PROBLEM 4.168
Solve Prob. 4.167, assuming that E 60q.
PROBLEM 4.167 Steel links having the cross section shown are
available with different central angles E. Knowing that the
allowable stress is 12 ksi, determine the largest force P that can be
applied to a link for which 90 .
E q
SOLUTION
Reduce section force to a force-couple system at G, the centroid of the cross section AB.
1 cos
2
a r
E
§ ·
¨ ¸
© ¹
The bending couple is .
M Pa
For the rectangular section, the neutral axis for bending couple only lies at
2
1
.
ln
r
r
h
R
Also, e r R
At point A, the tensile stress is
1 1 1
1
A A A
A
P My P Pay P ay P
K
A Aer A Aer A er A
V
§ ·
¨ ¸
© ¹
where
1
1 A
ay
K
er
and 1
A
y R r
A
A
P
K
V
Data: 1 2
1.2 in., 0.8 in., 1.6 in., 0.8 in., 0.3 in.
r r r h b
2
1.6
0.8
0.8
(0.3)(0.8) 0.24 in 1.154156 in.
ln
1.2 1.154156 0.045844 in. 1.154156 0.8 0.35416 in.
(1.2)(1 cos30 ) 0.160770 in.
(0.160770)(0.35416)
1 2.5525
(0.045844)(0.8)
(0.24)(12)
1.128 kips
2.5525
A
A R
e y
a
K
P
q
1128 lb
P W
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on a website, in whole or part.
628
20 mm
20 mm
30 mm
30 mm
B A
C
a
5 kN
5 kN
PROBLEM 4.169
The curved bar shown has a cross section of 30 30 mm.
u Knowing
that the allowable compressive stress is 175 MPa, determine the largest
allowable distance a.
SOLUTION
Reduce the internal forces transmitted across section AB to a force-couple system at the centroid of the
section. The bending couple is
( )
M P a r
For the rectangular section, the neutral axis for bending couple only lies at
2
1
.
ln
h
R
r
r
Also, e r R
The maximum compressive stress occurs at point A . It is given by
1 1
1
1
1
( )
with
( )( )
Thus, 1
A A
A
A
P My P P a r y
A Aer A Aer
P
K y R r
A
a r R r
K
er
V
(1)
Data: 1 2 50
20
1
6 3
6 6
3
30
30 mm, 20 mm, 50 mm, 35 mm, 32.7407 mm
ln
35 32.7407 2.2593 mm, 30 mm, 12.7407 mm, ?
175 MPa 175 10 Pa, 5 kN 5 10 N
(900 10 )( 175 10 )
31.5
5 10
A
A
h r r r R
e b R r a
P
A
K
P
V
V
u u
u u
u
Solving (1) for 1
1
( 1)
,
K er
a r a r
R r
(30.5)(2.2593)(20)
108.17 mm
12.7407
a r
108.17 mm 35 mm
a 73.2 mm
a W
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629
90 mm
40 mm
14 mm
2500 N
B
A
PROBLEM 4.170
For the split ring shown, determine the stress at (a) point A,
(b) point B.
SOLUTION
2
1
1 2 2 1
2
45
20
1 2
1 1
40 20 mm, (90) 45 mm 25 mm
2 2
25
(14)(25) 350 mm 30.8288 mm
ln
ln
1
( ) 32.5 mm 1.6712 mm
2
r
r
r r h r r
h
A R
r r r e r R
Reduce the internal forces transmitted across section AB to a force-couple system at the centroid of the cross
section. The bending couple is
3
(2500)(32.5 10 ) 81.25 N m
M Pa Pr
u ˜
(a) Point A: 20 mm 30.8288 20 10.8288 mm
A A
r y
3
6 6 3 3
6
2500 (81.25)(10.8288 10 )
350 10 (350 10 )(1.6712 10 )(20 10 )
82.4 10 Pa
A
A
My
P
A AeR
V
u
u u u u
u 82.4 MPa
A
V W
(b) Point B: 45 mm 30.8288 45 14.1712 mm
B B
r y
3
6 6 3 3
6
2500 (81.25)( 14.1712 10 )
350 10 (350 10 )(1.6712 10 )(45 10 )
36.6 10 Pa
B
B
B
My
P
A Aer
V
u
u u u u
u 36.6 MPa
B
V W
Mecánica de Materiales - Ferdinand P. Beer 7ma. Edición.

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630
A
C
B
M' M
2 in.
3 in.
0.5 in. 2 in.
3 in.
0.5 in.
0.5 in.
PROBLEM 4.171
Three plates are welded together to form the curved
beam shown. For 8 kip in.,
M ˜ determine the stress
at (a) point A, (b) point B, (c) the centroid of the cross
section.
SOLUTION
1
1 1
ln ln
i i
i i
r i i
i i
i
A b h A
R
r r
dA b b
r r
Ar
r
A
6 6 6
6 6 6
6
6
³
Part b h A b 1
ln i
i
r
r
r Ar
M 3 0.5 1.5 0.462452 3.25 4.875
N 0.5 2 1.0 0.225993 4.5 4.5
O 2 0.5 1.0 0.174023 5.75 5.75
r
3
3.5
5.5
6
6 3.5 0.862468 15.125
3.5 15.125
4.05812 in., 4.32143 in.
0.862468 3.5
0.26331 in. 8 kip in.
R r
e r R M
˜
(a) 1 4.05812 3 1.05812 in.
A
y R r
1
( 8)(1.05812)
(3.5)(0.26331)(3)
A
A
My
Aer
V
3.06 ksi
A
V W
(b) 2 4.05812 6 1.94188 in.
B
y R r
2
( 8)( 1.94188)
(3.5)(0.26331)(6)
B
B
My
Aer
V
2.81 ksi
B
V W
(c) C
y R r e
8
(3.5)(4.32143)
C
C
My Me M
Aer Aer Ar
V
0.529 ksi
C
V W
Mecánica de Materiales - Ferdinand P. Beer 7ma. Edición.

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631
A
C
B
M' M
2 in.
3 in.
0.5 in. 2 in.
3 in.
0.5 in.
0.5 in.
PROBLEM 4.172
Three plates are welded together to form the curved beam
shown. For the given loading, determine the distance e
between the neutral axis and the centroid of the cross
section.
SOLUTION
1
1 1
ln ln
i i
i i
r i i
i i
i
A b h A
R
r r
dA b b
r r
Ar
r
A
6 6 6
6 6 6
6
6
³
Part b h A b 1
ln i
i
r
r
r Ar
M 3 0.5 1.5 0.462452 3.25 4.875
N 0.5 2 1.0 0.225993 4.5 4.5
O 2 0.5 1.0 0.174023 5.75 5.75
r
3
3.5
5.5
6
6 3.5 0.862468 15.125
3.5 15.125
4.05812 in., 4.32143 in.
0.862468 3.5
0.26331 in.
R r
e r R
0.263 in.
e W
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632
B
C
A
Mʹ
M 150 mm
135 mm
36 mm
45 mm
B
A
PROBLEM 4.173
Knowing that the maximum allowable stress is 45 MPa, determine
the magnitude of the largest moment M that can be applied to the
components shown.
SOLUTION
1
1 1
2ln ln
i i i
i i
r i i
i i
i i
i
A b h A
R
r r
dA b b
r r
Ar
r
A
6 6 6
6 6 6
6
6
³
Part , mm
i
b , mm
h 2
, mm
A
1
ln , mm
i
i
i
r
b
r
, mm
r 3
, mm
Ar
M 108 45 4860 28.3353 172.5 3
838.35 10
u
N 36 135 4860 18.9394 262.5 3
1275.75 10
u
r, mm
150
195
330
9720 47.2747 3
2114.1 10
u
3
1
1
1
6 6 3 3
3
2
9720 2114.1 10
205.606 mm 217.5 mm
47.2747 9720
11.894 mm
205.606 150 55.606 mm
(45 10 )(9720 10 )(11.894 10 )(150 10 )
14.03 kN m
(55.606 10 )
205.606 330 124
A
A
A
A
A
B
R r
e r R
y R r
My
Aer
Aer
M
y
y R r
u
u u u u
˜
u
V
V
2
.394 mm
B
B
My
Aer
V
2
6 6 3 3
3
(45 10 )(9720 10 )(11.894 10 )(330 10 )
(124.394 10 )
13.80 kN m
B
B
Aer
M
y
V
u u u u
u
˜
13.80 kN m
M ˜ W
Mecánica de Materiales - Ferdinand P. Beer 7ma. Edición.

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Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
633
B
C
A
Mʹ
M 150 mm
135 mm
36 mm
45 mm
B
A
PROBLEM 4.174
Knowing that the maximum allowable stress is 45 MPa, determine
the magnitude of the largest moment M that can be applied to the
components shown.
SOLUTION
1
1 1
ln ln
i i i
i i
r i i
i i
i i
i
A b h A
R
r r
dA b b
r r
Ar
r
A
6 6 6
6 6 6
6
6
³
, mm
i
b , mm
h 2
, mm
A
1
2
ln , mm
i
i
i
r
b
r
, mm
r 3
, mm
Ar
M 36 135 4860 23.1067 217.5 6
1.05705 10
u
N 108 45 4860 15.8332 307.5 6
1.49445 10
u
r, mm
150
285
330
9720 38.9399 6
2.5515 10
u
6
3
1
1
1
6 6 3 3
3
2
9720 2.5515 10
249.615 mm, 262.5 mm
38.9399 9720
12.885 mm, 20 10 N m
249.615 150 99.615 mm
(45 10 )(9720 10 )(12.885 10 )(150 10 )
8.49 kN m
(99.615 10 )
249.
A
A
A
A
A
B
R r
e r R M
y R r
My
Aer
Aer
M
y
y R r
u
u ˜
u u u u
˜
u
V
V
2
615 330 80.385 mm
B
B
My
Aer
V
2
6 6 3 3
3
(45 10 )(9720 10 )(12.885 10 )(330 10 )
(80.385 10 )
23.1 kN m
B
B
Aer
M
y
V
u u u u
u
˜
8.49 kN m
M ˜ W
Mecánica de Materiales - Ferdinand P. Beer 7ma. Edición.

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on a website, in whole or part.
634
120 lb
120 lb
d
A
B
r1
PROBLEM 4.175
The split ring shown has an inner radius r1 0.8 in. and a circular cross
section of diameter d 0.6 in. Knowing that each of the 120-lb forces is
applied at the centroid of the cross section, determine the stress (a) at point A,
(b) at point B.
SOLUTION
1
2 2 2
2 2 2 2
0.8 in.
0.8 0.6 1.4 in.
1 1
( ) 1.1in. 0.3 in.
2 2
(0.3) 0.28274 in for solid circular section
1 1
1.1 (1.1) (0.3) 1.079150 in.
2 2
1.1 1.0791503 0.020850 in.
A
B A
A B
r r
r r d
r r r c d
A c
R r r c
e r R
ª º ª º
« » « »
¬ ¼ ¬ ¼
S S
120 lb 0: 0 120 lb
x
Q F P Q P
¦
G
0 0: 2 0 2 (2)(1.1)(120) 264 lb in.
M rQ M M rQ
˜
¦
(a)
3
0.8 in.
( ) 120 ( 264)(0.8 1.079150)
16.05 10 psi
0.28274 (0.28274)(0.020850)(0.8)
u
A
A
A
A
r r
P M r R
A Aer
V
16.05 ksi
A
V W
(b)
3
1.4 in.
( ) 120 ( 264)(1.4 1.079150)
0.28274 (0.28274)(0.020850)(1.4)
9.84 10 psi
B
B
B
B
r r
P M r R
A Aer
V
u
9.84 ksi
B
V W
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Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
635
120 lb
120 lb
d
A
B
r1
PROBLEM 4.176
Solve Prob. 4.175, assuming that the ring has an inner radius r1 = 0.6 in. and
a cross-sectional diameter d 0.8 in.
PROBLEM 4.175 The split ring shown has an inner radius r1 0.8 in. and a
circular cross section of diameter d 0.6 in. Knowing that each of the 120-lb
forces is applied at the centroid of the cross section, determine the stress
(a) at point A, (b) at point B.
SOLUTION
1
2 2 2
2 2 2 2
0.6 in.
0.6 0.8 1.4 in.
1 1
( ) 1.0 in. 0.4 in.
2 2
(0.4) 0.50265 in for solid circular section.
1 1
1.0 (1.0) (0.4) 0.958258 in.
2 2
1.0 0.958258 0.041742 in.
ª º ª º
« » « »
¬ ¼ ¬ ¼
A
B A
A B
r r
r r d
r r r c d
A c
R r r c
e r R
S S
120 lb 0 0 120 lb
¦
G x
Q F P Q P
0 0: 2 0 2 (2)(1.0)(120) 240 lb in.
M rQ M M rQ
¦ ˜
(a)
3
0.6 in.
( ) 120 ( 240)(0.6 0.958258)
0.50265 (0.50265)(0.041742)(0.6)
7.069 10 psi
A
A
A
A
r r
P M r R
A Aer
V
u 7.07 ksi
A
V W
(b)
3
1.4 in.
( ) 120 ( 240)(1.4 0.958258)
0.50265 (0.50265)(0.041742)(1.4)
3.37 10 psi
B
B
B
B
r r
P M r R
A Aer
V
u 3.37 ksi
B
V W
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Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
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636
220 N
220 N
12 mm
16 mm
a
B A C
PROBLEM 4.177
The bar shown has a circular cross section of 14-mm diameter. Knowing
that a 32 mm, determine the stress at (a) point A, (b) point B.
SOLUTION
2 2 2 2
2 2 2
1
2
1
8 mm 12 8 20 mm
2
1 1
20 20 8 19.1652 mm
2 2
20 19.1652 0.83485 mm
(8) 201.06 mm
220 N
( ) 220(0.032 0.020) 11.44 N m
19.1652 12 7.1652 mm
19.1652 28
ª º ª º
« » « »
¬ ¼ ¬ ¼
˜
A
b
c d r
R r r c
e r R
A r
P
M P a r
y R r
y R r
S S
8.8348 mm
(a)
1
3
6 6 3 3
220 ( 11.44)(7.1652 10 )
201.06 10 (201.06 10 )(0.83485 10 )(12 10 )
u
u u u u
A
A
P My
A Aer
V
41.8 MPa
A
V W
(b)
2
3
6 6 3 3
220 ( 11.44)(8.8348 10 )
201.06 10 (201.06 10 )(0.83485 10 )(28 10 )
u
u u u u
B
B
P My
A Aer
V
20.4 MPa
B
V W
Mecánica de Materiales - Ferdinand P. Beer 7ma. Edición.

676. PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
637
220 N
220 N
12 mm
16 mm
a
B A C
PROBLEM 4.178
The bar shown has a circular cross section of 14-mm diameter. Knowing
that the allowable stress is 38 MPa, determine the largest permissible
distance a from the line of action of the 220-N forces to the plane
containing the center of curvature of the bar.
SOLUTION
2 2 2 2
2 2 2
1
1
8 mm 12 8 20 mm
2
1 1
20 20 8 19.1652 mm
2 2
20 19.1652 0.83485 mm
(8) 201.06 mm
220 N
( )
19.1652 12 7.1652 mm
ª º ª º
« » « »
¬ ¼ ¬ ¼
A
c d r
R r r c
e r R
A r
P
M P a r
y R r
S S
1 1 1
1
( ) ( )
1
( )
where 1
A A A
A
A
P My P P a r y P a r y
A Aer A Aer A er
KP a r y
K
A er
V
ª º
« »
¬ ¼
6 6
1
3 3
3
(38 10 )(201.06 10 )
34.729
220
( 1)
(34.729 1)(0.83485 10 )(12 10 )
(7.1652 10 )
0.047158 m
0.047158 0.020 0.027158
u u
u u
u
A
A
A
K
P
K er
a r
y
a
V
27.2 mm
a W
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638
16 mm
12 mm
M
C
PROBLEM 4.179
The curved bar shown has a circular cross section of 32-mm
diameter. Determine the largest couple M that can be applied to
the bar about a horizontal axis if the maximum stress is not to
exceed 60 MPa.
SOLUTION
2 2
2 2
16 mm 12 16 28 mm
1
2
1
28 28 16 25.4891mm
2
c r
R r r c
ª º
« »
¬ ¼
ª º
« »
¬ ¼
28 25.4891 2.5109 mm
e r R
max
V occurs at A, which lies at the inner radius.
It is given by max
1
A
My
Aer
V from which 1 max
.
A
Aer
M
y
V
Also, 2 2 2
(16) 804.25 mm
A c
S S
Data: 1 25.4891 12 13.4891mm
A
y R r
6 3 3 6
3
(804.25 10 )(2.5109 10 )(12 10 )(60 10 )
13.4891 10
M
u u u u
u
107.8 N m
M ˜ W
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639
90 mm
80 mm
A
B
100 mm
P
PROBLEM 4.180
Knowing that 10 kN,
P determine the stress at (a) point A, (b) point B.
SOLUTION
Locate the centroid D of the cross section.
90 mm
100 mm 130 mm
3
r
Force-couple system at D.
10 kN
(10 kN)(130 mm) 1300 N m
P
M Pr ˜
Triangular cross section.
2 6 2
1 1
(90 mm)(80 mm)
2 2
3600 mm 3600 10 m
A bh
u
2 2
1
1 1
(90) 45 mm
2 2
190 190 0.355025
ln 1 ln 1
90 100
126.752 mm
130 mm 126.752 mm 3.248 mm
h
R
r r
h r
R
e r R
(a) Point A: 100 mm 0.100 m
A
r
6 2 6 2 3
( ) 10 kN (1300 N m)(0.100 m 0.126752 m)
3600 10 m (3600 10 m )(3248 10 m)(0.100 m)
A
A
A
P M r R
A Aer
V
˜
u u u
2.778 MPa 29.743 MPa
32.5 MPa
A
V W
(b) Point B: 190 mm 0.190 m
B
r
6 2 6 2 3
( ) 10 kN (1300 N m)(0.190 m 0.126752 m)
3600 10 m (3600 10 m )(3.248 10 m)(0.190 m)
B
B
B
P M r R
A Aer
V
˜
u u u
2.778 MPa 37.01MPa
34.2 MPa
B
V W
Mecánica de Materiales - Ferdinand P. Beer 7ma. Edición.

679. PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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on a website, in whole or part.
640
2.5 in.
3 in.
2 in.
2 in.
3 in.
B
C
M
A M
PROBLEM 4.181
Knowing that 5 kip in.,
M ˜ determine the stress at (a) point A,
(b) point B.
SOLUTION
2
1 1 2 2
1 1
(2.5)(3) 3.75 in
2 2
2 1 3.00000 in.
2.5 in., 2 in., 0, 5 in.
A bh
r
b r b r
Use formula for trapezoid with 2 0.
b
2
1 2
2
1 2 2 1 1 2
1
2
5
2
1 ( )
2
( )ln ( )
(0.5)(3) (2.5 0)
2.84548 in.
[(2.5)(5) (0)(2)] ln (3)(2.5 0)
0.15452 in. 5 kip in.
h b b
R
r
b r b r h b b
r
e r R M
˜
(a) 1 0.84548 in.
A
y R r
1
(5)(0.84548)
(3.75)(0.15452)(2)
A
A
My
Aer
V 3.65 ksi
A
V W
(b) 2 2.15452 in.
B
y R r
2
(5)( 2.15452)
(3.75)(0.15452)(5)
B
B
My
Aer
V
3.72 ksi
B
V W
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on a website, in whole or part.
641
3 in.
M
M
B
A
C
3 in.
2 in.
2 in.
2.5 in.
PROBLEM 4.182
Knowing that 5 kip in.,
M ˜ determine the stress at (a) point A,
(b) point B.
SOLUTION
2
1 1 2 2
1 (2.5)(3) 3.75 in
2
2 2 4.00000 in.
0, 2 in., 2.5 in., 5 in.
A
r
b r b r
Use formula for trapezoid with 1 0.
b
2
1 2
2
1 2 2 1 1 2
1
2
1 ( )
2
( ) ln ( )
(0.5)(3) (0 2.5)
3.85466 in.
5
[(0)(5) (2.5)(2)] ln (3)(0 2.5)
2
0.14534 in. 5 kip in.
h b b
R
r
b r b r h b b
r
e r R M
˜
(a) 1 1.85466 in.
A
y R r
1
(5)(1.85466)
(3.75)(0.14534)(2)
A
A
My
Aer
V 8.51 ksi
A
V W
(b) 2 1.14534 in.
B
y R r
2
(5)( 1.14534)
(3.75)(0.14534)(5)
B
B
My
Aer
V 2.10 ksi
B
V W
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681. PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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642
6 in. 4 in.
C
B
B
A
A
b a
80 kip · in.
PROBLEM 4.183
Knowing that the machine component shown has a trapezoidal cross
section with 3.5 in.
a and 2.5 in.,
b determine the stress at
(a) point A, (b) point B.
SOLUTION
Locate centroid.
2
, in
A , in.
r 3
, in
Ar
M 10.5 6 63
N 7.5 8 60
6 18 123
2
1
2
1
1 2
2
1 2 2 1 1 2
2
10
4
123
6.8333 in.
18
( )
( )ln ( )
(0.5)(6) (3.5 2.5)
6.3878 in.
[(3.5)(10) (2.5)(4)]ln (6)(3.5 2.5)
0.4452 in. 80 kip in.
r
r
r
h b b
R
b r b r h b b
e r R M
˜
(a) 1 6.3878 4 2.3878 in.
A
y R r
1
(80)(2.3878)
(18)(0.4452)(4)
A
A
My
Aer
V 5.96 ksi
A
V W
(b) 2 6.3878 10 3.6122 in.
B
y R r
2
(80)( 3.6122)
(18)(0.4452)(10)
B
B
My
Aer
V
3.61 ksi
B
V W
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643
6 in. 4 in.
C
B
B
A
A
b a
80 kip · in.
PROBLEM 4.184
Knowing that the machine component shown has a trapezoidal cross
section with 2.5 in.
a and 3.5 in.,
b determine the stress at
(a) point A, (b) point B.
SOLUTION
Locate centroid.
2
, in
A , in.
r 3
, in
Ar
M 7.5 6 45
N 10.5 8 84
6 18 129
2
1
2
1
1 2
2
1 2 2 1 1 2
2
10
4
129
7.1667 in.
18
( )
( )ln ( )
(0.5)(6) (2.5 3.5)
6.7168 in.
[(2.5)(10) (3.5)(4)]ln (6)(2.5 3.5)
0.4499 in. 80 kip in.
r
r
r
h b b
R
b r b r h b b
e r R M
˜
(a) 1 2.7168 in.
A
y R r
1
(80)(2.7168)
(18)(0.4499)(4)
A
A
My
Aer
V 6.71 ksi
A
V W
(b) 2 3.2832 in.
B
y R r
2
(80)( 3.2832)
(18)(0.4499)(10)
B
B
My
Aer
V
3.24 ksi
B
V W
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644
20 mm
30 mm
35 mm
40 mm
a
a
B
A
B
A
250 N · m
250 N · m
Section a–a
PROBLEM 4.185
For the curved beam and loading shown, determine the stress at
(a) point A, (b) point B.
SOLUTION
Locate centroid.
2
, mm
A , mm
r 3
, mm
Ar
M 600 45 3
27 10
u
N 300 55 3
16.5 10
u
6 900 3
43.5 10
u
2
1
3
2
1
1 2
2
1 2 2 1 2 1
2
65
35
43.5 10
48.333 mm
900
( )
( )ln ( )
(0.5)(30) (40 20)
46.8608 mm
[(40)(65) (20)(35)]ln (30)(40 20)
1.4725 mm 250 N m
r
r
r
h b b
R
b r b r h b b
e r R M
u
˜
(a) 1 11.8608 mm
A
y R r
3
6
6 3 3
1
( 250)(11.8608 10 )
63.9 10 Pa
(900 10 )(1.4725 10 )(35 10 )
u
u
u u u
A
A
My
Aer
V 63.9 MPa
A
V W
(b) 2 18.1392 mm
B
y R r
3
6
6 3 3
2
( 250)( 18.1392 10 )
52.6 10 Pa
(900 10 )(1.4725 10 )(65 10 )
u
u
u u u
B
B
My
Aer
V 52.6 MPa
B
V W
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645
35 mm
60 mm
25 mm
40 mm
60 mm
15 kN
a
a
Section a–a
PROBLEM 4.186
For the crane hook shown, determine the largest tensile stress in
section a-a.
SOLUTION
Locate centroid.
2
, mm
A , mm
r 3
, mm
Ar
M 1050 60 3
63 10
u
N 750 80 3
60 10
u
6 1800 3
103 10
u
3
103 10
63.333 mm
1800
r
u
Force-couple system at centroid: 3
15 10 N
P u
2
1
3 3 3
2
1
1 2
2
1 2 2 1 1 2
2
100
40
(15 10 )(68.333 10 ) 1.025 10 N m
( )
( )ln ( )
(0.5)(60) (35 25)
63.878 mm
[(35)(100) (25)(40)]ln (60)(35 25)
4.452 mm
r
r
M Pr
h b b
R
b r b r h b b
e r R
u u u ˜
Maximum tensile stress occurs at point A.
1
1
3 3 3
6 6 3 3
23.878 mm
15 10 (1.025 10 )(23.878 10 )
1800 10 (1800 10 )(4.452 10 )(40 10 )
A
A
A
y R r
My
P
A Aer
V
u u u
u u u u
6
84.7 10 Pa
u 84.7 MPa
A
V W
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646
PROBLEM 4.187
Using Eq. (4.66), derive the expression for R given in Fig. 4.61 for a circular cross section.
SOLUTION
Use polar coordinate E as shown. Let w be the width as a function of E
2 2
2 sin
cos
sin
2 sin
w c
r r c
dr c d
dA w dr c d
E
E
E E
E E
2
0
2 sin
cos
dA c
d
r r c
³ ³
S E
E
E
2 2
0
2 2 2 2 2
0
(1 cos )
cos
cos ( )
2
cos
S
S
E
E
E
E
E
E
³ ³
³
dA c
d
r r c
r c r c
d
r c
2 2
0 0
0 0
2 2
2 2 1
0
2 2
2 2
2 2
2 ( cos ) 2( )
cos
2 2 sin
1
tan
2 2
2( ) tan
2 ( 0) 2 (0 0) 4 0
2
2 2
dr
r c d r c
r c
r c
r c
r c
r c
r c
r c r c
r r c
§ ·
˜
¨ ¸
© ¹
³ ³
S S
S
S
S
E E
E
E E
E
S
S
S S
2
2 2 2 2
2 2 2 2 2 2
2 2 2 2 2 2
2 2
2 2 2 2
1
2
2 2
1 1 1
2 2 2
( )
A c
A c c r r c
R
dA r r c r r c r r c
r
c r r c c r r c
r r c
r r c c
S
S
S S
u
³
W
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647
PROBLEM 4.188
Using Eq. (4.66), derive the expression for R given in Fig. 4.73 for a trapezoidal cross section.
SOLUTION
The section width w varies linearly with r.
0 1
1 1 2 2
1 0 1 1
2 0 1 2
1 2 1 1 2 1
1 2
1
2 1 1 2 2 1 0 0
2 1 1 2
0
at and at
( )
( )
w c c r
w b r r w b r r
b c c r
b c c r
b b c r r c h
b b
c
h
r b rb r r c hc
r b rb
c
h
2 2
1 1
2 2
1 1
0 1
0 1
2
0 1 2 1
1
2 1 1 2 2 1 2
1
2 1 1 2 2
1 2
1
ln
ln ( )
ln
ln ( )
³ ³ ³
r r
r r
r r
r r
c c r
dA w
dr dr
r r r
c r c r
r
c c r r
r
r b rb r b b
h
h r h
r b rb r
b b
h r
1 2
1
( )
2
A b b h
2
1
2
1
1 2
2
2 1 1 2 1 2
( )
( )ln ( )
r
dA
r
r
h b b
A
R
r b rb h b b
³
W
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648
PROBLEM 4.189
Using Equation (4.66), derive the expression for R given in Fig. 4.73 for a triangular cross section.
SOLUTION
The section width w varies linearly with r.
0 1
1 2
0 1 1
0 1 2
1 1 2 1
2
1 0 1 2
at and 0 at
0
( )
and
w c c r
w b r r w r r
b c c r
c c r
b c r r c h
br
b
c c c r
h h
2 2
1 1
2 2
1 1
0 1
0 1
2
0 1 2 1
1
2 2
1
2 2 2 2
1 1
ln
ln ( )
ln
ln ln 1
r r
r r
r r
r r
c c r
dA w
dr dr
r r r
c r c r
r
c c r r
r
br r b
h
h r h
br r r r
b b
h r h r
§ ·
¨ ¸
© ¹
³ ³ ³
1
2
A bh
2 2
2 2
1
1
1 1
2 2
ln 1
ln 1
r r
r r
dA
h r
r h r
bh h
A
R
b
³
W
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649
r1
r2
b1
b2
b3
r3
r4
PROBLEM 4.190
Show that if the cross section of a curved beam consists of two or more
rectangles, the radius R of the neutral surface can be expressed as
3
1 2
3
2 4
1 2 3
ln
b
b b
A
R
r
r r
r r r
ª º
§ ·
§ · § ·
« »
¨ ¸
¨ ¸ ¨ ¸
« »
© ¹ © ¹ © ¹
¬ ¼
where A is the total area of the cross section.
SOLUTION
3
1 2
1
1 3
2 4
1 2 3
1
ln
ln ln
i
i
i
i
b b
b b
i
i
A A
R
r
dA b
r r
A A
r r
r r
r r r r
6
6 6
ª º
§ · § ·
§ · § ·
« »
6 ¨ ¸ ¨ ¸
¨ ¸ ¨ ¸
« »
© ¹ © ¹ © ¹ © ¹
¬ ¼
³
W
Note that for each rectangle,
1
1
2
1
1
ln
i
i
i
i
i
i i
i
r
r
r
r
dr
dA b
r r
r
dr
b b
r r
³ ³
³
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650
r
!
r
!
x
!
x
!
C
R
b
r1
2
2
PROBLEM 4.191
For a curved bar of rectangular cross section subjected to a
bending couple M, show that the radial stress at the neutral
surface is
1
1
1 ln
r
r
M R
Ae R r
V
§ ·
¨ ¸
© ¹
and compute the value of r
V for the curved bar of Concept
Applications 4.10 and 4.11. (Hint: consider the free-body
diagram of the portion of the beam located above the neutral
surface.)
SOLUTION
At radial distance r,
( )
V
r
M r R M MR
Aer Ae Aer
For portion above the neutral axis, the resultant force is
1
1 1
1
1
1 1
( ) ln 1 ln
R
r r
r
R R
r r
H dA bdr
Mb MRb dr
dr
Ae Ae r
r
Mb MRb R MbR R
R r
Ae Ae r Ae R r
§ ·
¨ ¸
© ¹
³ ³
³ ³
V V
Resultant of :
Vn cos
r r
F dA
V E
³
/2 /2
/2 /2
/2
/2
cos ( ) cos
sin 2 sin
2
r r
r r
bR d bR d
bR bR
³ ³
T T
T T
T
T
V E E V E E
T
V E V
For equilibrium:
1
1
2 sin 0
2
2 sin 2 1 ln sin 0
2 2
r
r
F H
r
MbR R
bR
Ae R r
T
T T
V
§ ·
¨ ¸
© ¹
1
1
1 ln
r
r
M R
Ae R r
V
§ ·
¨ ¸
© ¹
W
Using results of Examples 4.10 and 4.11 as data,
2
1
8 kip in., 3.75 in , 5.9686 in., 0.0314 in., 5.25 in.
8 5.25 5.9686
1 ln
(3.75)(0.0314) 5.9686 5.25
r
M A R e r
˜
ª º
« »
¬ ¼
V 0.536 ksi
r
V W
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651
C
B
A
300 mm 300 mm
25 mm
25 mm
4 kN
4 kN
PROBLEM 4.192
Two vertical forces are applied to a beam of the cross section shown. Determine
the maximum tensile and compressive stresses in portion BC of the beam.
SOLUTION
2 2 2
1 1
2
2 2
4 (4)(25)
(25) 981.7 mm 10.610 mm
2 2 3 3
25
(50)(25) 1250 mm 12.5 mm
2 2
r
A r y
h
A bh y
S S
S S
1 1 2 2
1 2
(981.7)(10.610) (1250)( 12.5)
2.334 mm
981.7 1250
A y A y
y
A A
1
2 4 2 4 2 6 4
1 1 1 1 1
1 1
2 3 2 3 4
1 1 1 1
3 3 3 4
2
2 2
(25) (981.7)(10.610) 42.886 10 mm
8 8
10.610 ( 2.334) 12.944 mm
42.866 10 (981.7)(12.944) 207.35 10 mm
1 1
(50)(25) 65.104 10 mm
12 12
12.5
x
I I A y r A y
d y y
I I A d
I bh
d y y
S S
u
u u
u
2 3 2 3 4
2 2 2 2
3 4 9 4
1 2
top
bot
( 2.334) 10.166 mm
65.104 10 (1250)(10.166) 194.288 10 mm
401.16 10 mm 401.16 10 m
25 2.334 27.334 mm 0.027334 m
25 2.334 22.666 mm 0.022666 m
I I A d
I I I
y
y
u u
u u
3 3
0: (4 10 )(300 10 ) 1200 N m
M Pa M Pa
u u ˜
top 6
top 9
(1200)(0.027334)
81.76 10 Pa
401.16 10
My
I
V
u
u
top 81.8 MPa
V W
6
bot
bot 9
(1200)( 0.022666)
67.80 10 Pa
401.16 10
My
I
V
u
u
bot 67.8 MPa
V W
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652
0.018 in.
PROBLEM 4.193
A steel band saw blade that was originally straight passes over 8-in.-diameter pulleys
when mounted on a band saw. Determine the maximum stress in the blade, knowing that
it is 0.018 in. thick and 0.625 in. wide. Use 6
29 10 psi.
E u
SOLUTION
Band blade thickness: 0.018 in.
t
Radius of pulley:
1
4.000 in.
2
r d
Radius of curvature of centerline of blade:
1
4.009 in.
2
1
0.009 in.
2
r t
c t
U
Maximum strain:
0.009
0.002245
4.009
m
c
H
U
Maximum stress: 6
(29 10 )(0.002245)
m m
E
V H u
3
65.1 10 psi
m
V u 65.1 ksi
m
V W
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653
(a) (b)
a
M M
PROBLEM 4.194
A couple of magnitude M is applied to a square bar of side a. For
each of the orientations shown, determine the maximum stress
and the curvature of the bar.
SOLUTION
4
3 3
1 1
12 12 12
2
a
I bh aa
a
c
max 4
2
12
a
M
Mc
I a
V max 3
6M
a
V W
4
1
12
M M
EI a
E
U 4
1 12M
Ea
U
W
For one triangle, the moment of inertia about its base is
3 4
3
1
4
2 1
4
1 2
1 1
2
12 12 24
2
24
12
a a
I bh a
a
I I
a
I I I
§ ·
¨ ¸
© ¹
max 4 3
/ 2 6 2
/12
2
a Mc Ma M
c
I a a
V max 3
8.49M
a
V W
4
1
12
U
M M
EI a
E
4
1 12M
Ea
U
W
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654
40 mm
60 mm
PROBLEM 4.195
Determine the plastic moment p
M of a steel beam of the cross section shown, assuming
the steel to be elastoplastic with a yield strength of 240 MPa.
SOLUTION
Let c1 be the outer radius and c2 the inner radius.
1 1
2 2
1 2
1 2
3
3
1 2
4 4
2 3 2 3
2
3
a a b b
A y A y A y
c c
c c
c c
S S
S S
§ ·§ · § ·§ ·
¨ ¸¨ ¸ ¨ ¸¨ ¸
© ¹© ¹ © ¹© ¹
2
2
3 2
2 2 1 1 1
3 3
1 1 2 2 1
2
3
4
( )
3
p Y Y
A y A y c c
M A y A y c c
V V
Data: 6
240 MPa 240 10 Pa
Y
V u
1
2
6 3 3
3
60 mm 0.060 m
40 mm 0.040 m
4
(240 10 )(0.060 0.040 )
3
48.64 10 N m
p
c
c
M u
u ˜ 48.6 kN m
p
M ˜ W
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655
46 mm
50 mm
M # 300 N · m
30 mm
26 mm
PROBLEM 4.196
In order to increase corrosion resistance, a 2-mm-thick cladding of
aluminum has been added to a steel bar as shown. The modulus
of elasticity is 200 GPa for steel and 70 GPa for aluminum. For a
bending moment of 300 N ˜ m, determine (a) the maximum stress
in the steel, (b) the maximum stress in the aluminum, (c) the radius
of curvature of the bar.
SOLUTION
Use aluminum as the reference material.
1
n in aluminum
200
2.857
70
s
a
E
n
E
in steel
Cross section geometry:
Steel: 2
(46mm)(26mm) 1196 mm
s
A 3 4
1
(46 mm)(26 mm) 67,375 mm
12
s
I
Aluminum: 2 2
(50 mm)(30 mm) 1196 mm 304 mm
a
A
3 4 4
1
(50 mm)(30 mm ) 67,375 mm 45,125 mm
12
a
I
Transformed section.
4 9 4
(1)(45,125) (2.857)(67,375) 237,615 mm 237.615 10 m
a a s s
I n I n I
u
Bending moment. 300 N m
M ˜
(a) Maximum stress in steel: 2.857
s
n 13 mm 0.013 m
s
y
6
9
(2.857)(300)(0.013)
46.9 10 Pa
237.615 10
s s
s
n My
I
V
u
u
46.9 MPa
s
V W
(b) Maximum stress in aluminum: 1,
a
n 15 mm 0.015 m
a
y
6
9
(1)(300)(0.015)
18.94 10 Pa
237.615 10
a a
a
n My
I
V
u
u
18.94 MPa
a
V W
(c) Radius of curvative:
EI
M
U
9 9
(70 10 )(237.615 10 )
300
U
u u
55.4 m
U W
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656
P'
P
a a
t
t
80 mm
60 mm
Section a–a
A B
200 mm
80 mm
PROBLEM 4.197
The vertical portion of the press shown consists of a rectangular
tube of wall thickness t 10 mm. Knowing that the press has
been tightened on wooden planks being glued together until P
20 kN, determine the stress at (a) point A, (b) point B.
SOLUTION
Rectangular cutout is 60 mm 40 mm.
u
3 2 3 2
3 3 6 4
6 4
3
3 3
(80)(60) (60)(40) 2.4 10 mm 2.4 10 m
1 1
(60)(80) (40)(60) 1.84 10 mm
12 12
1.84 10 m
40 mm 0.040 m 200 40 240 mm 0.240 m
20 10 N
(20 10 )(0.240) 4.8 10 N m
A
I
c e
P
M Pe
u u
u
u
u
u u ˜
(a)
3 3
6
3 6
20 10 (4.8 10 )(0.040)
112.7 10 Pa
2.4 10 1.84 10
A
P Mc
A I
V
u u
u
u u
112.7 MPa
A
V W
(b)
3 3
6
3 6
20 10 (4.8 10 )(0.040)
96.0 10 Pa
2.4 10 1.84 10
B
P Mc
A I
u u
u
u u
V 96.0 MPa
B
V W
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657
x
y
z
P
P
P P
A
C
B
D
a
PROBLEM 4.198
The four forces shown are applied to a rigid plate supported by a solid steel
post of radius a. Knowing that P 24 kips and a 1.6 in., determine the
maximum stress in the post when (a) the force at D is removed, (b) the forces
at C and D are removed.
SOLUTION
For a solid circular section of radius a,
2 4
4
A a I a
S
S
Centric force: 2
4
4 , 0
x z
F P
F P M M
A a
V
S
(a) Force at D is removed.
3 , , 0
x z
F P M Pa M
2 2 2
4
3 ( )( ) 7
x
F M z P Pa a P
A I a a a
S
V
S S
(b) Forces at C and D are removed.
2 , ,
x z
F P M Pa M Pa
Resultant bending couple: 2 2
2
x z
M M M Pa
2
2 2 2
4
2 2 2 4 2
2.437 /
F Mc P Paa P
P a
A I a a a
S
V
S
S
Numerical data: 24.0 kips, a 1.6 in.
P
Answers: (a) 2
(7)(24.0)
(1.6)
V
S
20.9 ksi
V W
(b) 2
(2.437)(24.0)
(1.6)
V 22.8 ksi
V W
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658
25 mm
25 mm
r # 20 mm P # 3 kN
a PROBLEM 4.199
The curved portion of the bar shown has an inner radius of 20 mm.
Knowing that the allowable stress in the bar is 150 MPa, determine the
largest permissible distance a from the line of action of the 3-kN force to
the vertical plane containing the center of curvature of the bar.
SOLUTION
Reduce the internal forces transmitted across section AB to a force-couple system at the centroid of the
section. The bending couple is
( )
M P a r
For the rectangular section, the neutral axis for bending couple only lies at
2
1
ln
r
r
h
R
Also, e r R
The maximum compressive stress occurs at point A. It is given by
1 1
( )
A A
A
P My P P a r y P
K
A Aer A Aer A
V
with 1
A
y R r
Thus, 1
1
( )( )
1
a r R r
K
er
Data: 1 2
45
20
2 6 2
1
3 6
25 mm, 20 mm, 45 mm, 32.5 mm
25
30.8288 mm, 32.5 30.8288 1.6712 mm
ln
25 mm, (25)(25) 625 mm 625 10 m 10.8288 mm
3 10 N m, 150 10 Pa
A
h r r r
R e
b A bh R r
P V
u
u ˜ u
6 6
3
1
1
( 150 10 )(625 10 )
31.25
3 10
( 1) (30.25)(1.6712)(20)
93.37 mm
10.8288
93.37 32.5
AA
K
P
K er
a r
R r
a
V
u u
u
60.9 mm
a W
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659
400 lb
400 lb
400 lb
400 lb
3
2.5
1.5
0.5
3
2.5
Dimensions in inches
1.5 0.5
0.5
0.5
r 5 0.3 r 5 0.3
PROBLEM 4.200
Determine the maximum stress in each of the two machine
elements shown.
SOLUTION
For each case, (400)(2.5) 1000 lb in.
˜
M
At the minimum section,
3 4
1
(0.5)(1.5) 0.140625 in
12
0.75 in.
I
c
(a) / 3/1.5 2
D d
/ 0.3/1.5 0.2
r d
From Fig 4.32, 1.75
K
3
max
(1.75)(1000)(0.75)
9.33 10 psi
0.140625
KMc
I
V u max 9.33 ksi
V W
(b) / 3/1.5 2 / 0.3/1.5 0.2
D d r d
From Fig. 4.31, 1.50
K
3
max
(1.50)(1000)(0.75)
8.00 10 psi
0.140625
KMc
I
V u max 8.00 ksi
V W
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660
10 mm
120 mm
10 mm
120 mm
10 mm
M
PROBLEM 4.201
Three 120 u 10-mm steel plates have been welded together to form the
beam shown. Assuming that the steel is elastoplastic with
200 GPa
E and 300 MPa,
Y
V determine (a) the bending moment
for which the plastic zones at the top and bottom of the beam are
40 mm thick, (b) the corresponding radius of curvature of the beam.
SOLUTION
2
1
6 6 3
1 1
2
2
6 6 3
2 2
2
3
6 6 3
3 2
(120)(10) 1200 mm
(300 10 )(1200 10 ) 360 10 N
(30)(10) 300 mm
(300 10 )(300 10 ) 90 10 N
(30)(10) 300 mm
1 1
(300 10 )(300 10 ) 45 10 N
2 2
Y
Y
Y
A
R A
A
R A
A
R A
V
V
V
u u u
u u u
u u u
3 3 3
1 2 3
65 mm 65 10 m 45 mm 45 10 m 20 mm 20 10 m
y y y
u u u
(a) 1 1 2 2 3 3
2( ) 2{(360)(65) (90)(45) (45)(20)}
M R y R y R y
3
56.7 10 N m
u ˜ 56.7 kN m
M ˜ W
(b)
9 3
6
(200 10 )(30 10 )
300 10
Y Y Y
Y
y Ey
E
V
U
U V
u u
u
20.0 mm
U W
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661
B
A
4.5 in.
P Q
4.5 in.
PROBLEM 4.202
A short length of a W8 u 31 rolled-steel shape supports a rigid plate on which two
loads P and Q are applied as shown. The strains at two points A and B on the
centerline of the outer faces of the flanges have been measured and found to be
6 6
550 10 in./in. 680 10 in./in.
A B
H H
u u
Knowing that 6
29 10 psi,
E u determine the magnitude of each load.
SOLUTION
Strains:
6
550 10 in./in.
A
H
u 6
680 10 in./in.
B
H
u
6 6
1 1
( ) ( 550 680)10 615 10 in./in.
2 2
C A B
H H H
u
Stresses: 6 6
6 6
(29 10 psi)( 550 10 in./in.) 15.95 ksi
(29 10 psi)( 615 10 in./in.) 17.835 ksi
u u
u u
A A
C C
E
E
V H
V H
W8 u 31: 2
3
9.12 in
27.5 in
A
S
(4.5 in.)( )
M P Q
At point C: 2
; 17.835 ksi
9.12 in
C
P Q P Q
A
V
162.655 kips
P Q (1)
At point A: A
P Q M
A S
V
3
(4.5 in.)( )
15.95 ksi 17.835 ksi ;
27.5 in
P Q
11.5194 kips
P Q (2)
Solve simultaneously. 75.6 kips 87.1 kips
P Q
75.6 kips p
P W
87.1 kips p
Q W
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662
M1
M1
M'1
M'1
A
C
B
D
1
!
1
!
1
!
1
!
PROBLEM 4.203
Two thin strips of the same material and same cross
section are bent by couples of the same magnitude
and glued together. After the two surfaces of
contact have been securely bonded, the couples are
removed. Denoting by V1 the maximum stress and
by 1
U the radius of curvature of each strip while
the couples were applied, determine (a) the final
stresses at points A, B, C, and D, (b) the final radius
of curvature.
SOLUTION
Let b width and t thickness of one strip.
Loading one strip, 1
M M
3
1
1 1
,
12 2
I bt c t
1 1
1 2
1 1
3
1 1
1 12
M c M
I bt
M M
EI Et
V
V
U
After M1 is applied to each of the strips, the stresses are those given in the sketch above. They are
1 1 1 1
, , ,
A B C D
V V V V V V V V
The total bending couple is 2M1.
After the strips are glued together, this couple is removed.
3 3
1
1 2
2 , (2 )
12 3
M M I b t bt c t
c c
The stresses removed are
1 1
3 2
2
3
2 3
M y M y M y
I bt bt
V
c
c
1 1
1 1
2 2
3 1 3 1
, 0,
2 2
A B C D
M M
bt bt
V V V V V V
c c c c
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663
PROBLEM 4.203 (Continued)
(a) Final stresses: 1 1
1
( )
2
A
V V V
1
1
2
A
V V
W
1
B
V V W
1
C
V V
W
1 1
1
2
D
V V V
1
1
2
D
V V W
1 1
3 3
2
3
1 2 3 1 1
4
M M M
EI E bt Et
c
c c c
U U
(b) Final radius:
1 1 1 1
1 1 1 1 1 1 3 1
4 4
U U U U U U
c
1
4
3
U U W
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664
b 60 mm
h 40 mm
a
a
Steel
Aluminum
PROBLEM 4.C1
Two aluminum strips and a steel strip are to be bonded together to form
a composite member of width 60 mm
b and depth 40 mm.
h The
modulus of elasticity is 200 GPa for the steel and 75 GPa for the
aluminum. Knowing that 1500 N m,
M ˜ write a computer program
to calculate the maximum stress in the aluminum and in the steel for
values of a from 0 to 20 mm using 2-mm increments. Using appropriate
smaller increments, determine (a) the largest stress that can occur in the
steel, (b) the corresponding value of a.
SOLUTION
Transformed section: (all steel) steel
alum
E
n
E
3 3
1 1 1
2 ( ) ( 2 )
12 12 2
I bh nb b h a
ª º
§ ·
« »
¨ ¸
© ¹
¬ ¼
At Point 1: 2
alum
h
M
I
V
At Point 2: 2
steel
h
M a
n
I
V
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665
PROBLEM 4.C1 (Continued)
For 0 to 20 mm
a using 2-mm intervals: compute: n, alum steel
, , .
I V V
60 mm 40 mm 1500 N m
b h M ˜
Moduli of elasticity: Steel 200 GPa Aluminum 75 GPa
Program Output
a
mm
I
4 6
m /10
Sigma
Aluminum
MPa
Sigma
Steel
MPa
0.000 0.8533 35.156 93.750
2.000 0.7088 42.325 101.580
4.000 0.5931 50.585 107.914
6.000 0.5029 59.650 111.347
8.000 0.4352 68.934 110.294
10.000 0.3867 77.586 103.448
12.000 0.3541 84.714 90.361
14.000 0.3344 89.713 71.770
16.000 0.3243 92.516 49.342
18.000 0.3205 93.594 24.958
20.000 0.3200 93.750 0.000
Find ‘a’ for max. steel stress and the corresponding aluminum stress.
6.600 0.4804 62.447 111.572083
6.610 0.4800 62.494 111.572159
6.620 0.4797 62.540 111.572113
Max. steel stress 111.6 MPa occurs when a 6.61 mm.
Corresponding aluminum stress 62.5 MPa
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666
d x
y
tf
tw
bf
PROBLEM 4.C2
A beam of the cross section shown, made of a steel that is assumed to be
elastoplastic with a yield strength Y
V and a modulus of elasticity E, is bent
about the x axis. (a) Denoting by Y
y the half thickness of the elastic core, write
a computer program to calculate the bending moment M and the radius of
curvature U for values of Y
y from 1
2
d to 1
6
d using decrements equal
to 1
2
.
f
t Neglect the effect of fillets. (b) Use this program to solve Prob. 4.201.
SOLUTION
Compute moment of inertia .
x
I
3 3
1 1
( )( 2 )
12 12
x f f w f
I b d b t d t
Maximum elastic moment:
( /2)
x
Y Y
I
M
d
V
For yielding in the flanges,
(Consider upper half of cross section.)
2
d
c
Stress at junction of web and flange:
( /2) f
A Y
Y
d t
y
V V
Detail of stress diagram: Resultant forces:
1
2
3
4
1
( )
2
1
( )
3
2
[ ( )]
3
2
( )
3
ª º
¬ ¼
Y
Y Y f
Y Y f
f
a c y
a y y c t
a y y c t
a c t
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667
PROBLEM 4.C2 (Continued)
Bending moment.
4
1
2 n n
n
M R a
¦
Radius of curvature.
;
Y Y
Y Y
Y
y E
y
E
V
H U U U
V
For yielding in the web, (Consider upper half of cross section.)
5
6
7
1
2
1
[ ( )]
2
2
3
f
Y f
Y
a c t
a y c t
a y
Bending moment.
7
5
2 n n
n
M R a
¦
Radius of curvature. Y Y
Y Y
Y
y E
y
E
V
H U U U
V
Program: Key in expressions for n
a and n
R for 1 to 7.
n
For to ( ) at /2
Y f f
y c c t t decrements, compute 2 n n
M R a
6 for 1 to 4
n and ,
Y
Y
y E
U
V
then print.
For ( ) to /3 at /2
Y w f
y c t c t decrements, compute 2 n n
M R a
6 for 5 to 7
n and ,
Y
Y
y E
U
V
then print.
Input numerical values and run program.
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668
PROBLEM 4.C2 (Continued)
Program Output
For a beam of Problem 4.201,
Depth 140.00 mm
d Width of flange 120.00 mm
f
b
Thickness of flange 10.00 mm
f
t Thickness of web 10.00 mm
w
t
0.000011600 m to the 4th
I
Yield strength of steel sigma 300 MPa
Y
Yield moment 49.71 kip in.
Y
M ˜
(mm)
Y
y (kN m)
M ˜ (m)
U
For yielding still in the flange,
70.000 49.71 46.67
65.000 52.59 43.33
60.000 54.00 40.00
For yielding in the web,
60.000 54.00 40.00
55.000 54.58 36.67
50.000 55.10 33.33
45.000 55.58 30.00
40.000 56.00 26.67
35.000 56.38 23.33
30.000 56.70 20.00
25.000 56.97 16.67
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669
z
0.4
0.4
1.6
1.2
0.4
1.2
0.4 0.8
Dimensions in inches
0.4
0.8
B
E
D
A
y
M
#
#
C
PROBLEM 4.C3
An 8 kip in.
˜ couple M is applied to a beam of the cross section shown
in a plane forming an angle E with the vertical. Noting that the
centroid of the cross section is located at C and that the y and z axes
are principal axes, write a computer program to calculate the stress at
A, B, C, and D for values of E from 0 to 180° using 10° increments.
(Given: 4
6.23 in
y
I and 4
1.481 in .
z
I )
SOLUTION
Input coordinates of A, B, C, D.
(1) 2 (1) 1.4
(2) 2 (2) 1.4
(3) 1 (3) 1.4
(4) 1 (4) 1.4
A A
B B
C C
D D
z z y y
z z y y
z z y y
z z y y
Components of M.
sin
cos
y
z
M M
M M
E
E
Equation 4.55, Page 305:
( )
( )
( )
y
z
z y
M z n
M y n
n
I I
V
Program: For 0 to 180
E q using 10° increments.
For 1 to 4
n using unit increments.
Evaluate Equation 4.55 and print stresses.
Return
Return
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670
PROBLEM 4.C3 (Continued)
Program Output
Moment of couple: 8.00 kip in.
M ˜
Moments of inertia: 4 4
6.23 in 1.481 in
y z
I I
Coordinates of Points A, B, D, and E:
Point : (1) 2: (1) 1.4
Point : (2) 2: (2) 1.4
Point : (3) 1: (3) 1.4
Point : (4) 1: (4) 1.4
---Stress at Points---
A z y
B z y
D z y
E z y
Beta
q
A
ksi
B
ksi
D
ksi
E
ksi
0 –7.565 –7.565 7.565 7.565
10 –7.896 –7.004 7.673 7.227
20 –7.987 –6.230 7.548 6.669
30 –7.836 –5.267 7.193 5.909
40 –7.446 –4.144 6.621 4.970
50 –6.830 –2.895 5.846 3.879
60 –6.007 –1.558 4.895 2.670
70 –5.001 –0.174 3.794 1.381
80 –3.843 1.216 2.578 0.049
90 –2.569 2.569 1.284 –1.284
100 –1.216 3.843 –0.049 –2.578
110 0.174 5.001 –1.381 –3.794
120 1.558 6.007 –2.670 –4.895
130 2.895 6.830 –3.879 –5.846
140 4.144 7.446 –4.970 –6.621
150 5.267 7.836 –5.909 –7.193
160 6.230 7.987 –6.669 –7.548
170 7.004 7.896 –7.227 –7.673
180 7.565 7.565 –7.565 –7.565
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671
B
b
r1
M'
M
A
A
B
C
h
PROBLEM 4.C4
Couples of moment 2 kN m
M ˜ are applied as shown to a curved bar
having a rectangular cross section with 100 mm
h and 25
b mm. Write
a computer program and use it to calculate the stresses at points A and B for
values of the ratio 1/
r h from 10 to 1 using decrements of 1, and from 1
to 0.1 using decrements of 0.1. Using appropriate smaller increments,
determine the ratio 1/
r h for which the maximum stress in the curved bar is
50% larger than the maximum stress in a straight bar of the same cross
section.
SOLUTION
Input: 100 mm,
25 mm,
2 kN m
h
b
M ˜
For straight bar, straight
2
6
48 MPa
M
S
M
h b
V
Following notation of Section 4.15, key in the following:
2 1 2 1 1 2
; /ln ( ); : ; 2500
r h r R h r r r r r e r R A bh (I)
Stresses: 1 1 1
( )( )
A M r R Aer
V V 2 2 2
( )/( )
B M r R Aer
V V (II)
Since 100 mm,
h for 1 1
/ 10, 1000 mm.
r h r Also, 1 1
/ 10, 100
r h r
Program: 1
For 1000 to 100
r at 100 decrements,
using equations of Lines I and II, evaluate 2 1
, , , , ,
r R r e V and 2
V
Also evaluate ratio 1 straight
/
V V
Return and repeat for 1 100
r to 10 at 10 decrements.
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672
PROBLEM 4.C4 (Continued)
Program Output
2
Bending moment 2 kN m 100.000 in. 2500.00 mm
M h A
˜
Stress in straight beam 48.00 MPa
1
r
mm
rbar
mm
R
mm
e
mm
1
V
MPa
2
V
MPa
1 /
r h
–
ratio
–
1000 1050 1049 0.794 –49.57 46.51 10.000 –1.033
900 950 949 0.878 –49.74 46.36 9.000 –1.036
800 850 849 0.981 –49.95 46.18 8.000 –1.041
700 750 749 1.112 –50.22 45.95 7.000 –1.046
600 650 649 1.284 –50.59 45.64 6.000 –1.054
500 550 548 1.518 –51.08 45.24 5.000 –1.064
400 450 448 1.858 –51.82 44.66 4.000 –1.080
300 350 348 2.394 –53.03 43.77 3.000 –1.105
200 250 247 3.370 –55.35 42.24 2.000 –1.153
100 150 144 5.730 –61.80 38.90 1.000 –1.288
=====================================================
100 150 144 5.730 –61.80 38.90 1.000 –1.288
90 140 134 6.170 –63.15 38.33 0.900 –1.316
80 130 123 6.685 –64.80 37.69 0.800 –1.350
70 120 113 7.299 –66.86 36.94 0.700 –1.393
60 110 102 8.045 –69.53 36.07 0.600 –1.449
50 100 91 8.976 –73.13 35.04 0.500 –1.523
40 90 80 10.176 –78.27 33.79 0.400 –1.631
30 80 68 11.803 –86.30 32.22 0.300 –1.798
20 70 56 14.189 –100.95 30.16 0.200 –2.103
10 60 42 18.297 –138.62 27.15 0.100 –2.888
===================================================================
Find 1/
r h for max straight
( )/( ) 1.5
V V
52.70 103 94 8.703 –72.036 35.34 0.527 –1.501
52.80 103 94 8.693 –71.998 35.35 0.528 –1.500
52.90 103 94 8.683 –71.959 35.36 0.529 –1.499
Ratio of stresses is 1.5 for 1 1
52.8 mm or / 0.529.
r r h
[Note: The desired ratio 1/
r h is valid for any beam having a rectangular cross section.]
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673
M
h1
h2
b1
hn
bn
b2
PROBLEM 4.C5
The couple M is applied to a beam of the cross
section shown. (a) Write a computer program that,
for loads expressed in either SI or U.S. customary
units, can be used to calculate the maximum tensile
and compressive stresses in the beam. (b) Use this
program to solve Probs. 4.9, 4.10, and 4.11.
SOLUTION
Input: Bending moment M.
For 1 to ,
n n Enter n
b and n
h
Area n n
b h
' (Print)
1 1
( )/2 /2
n n n n
a a h h
[Moment of rectangle about base]
( Area) n
m a
' '
[For whole cross section]
; Area Area Area
m m m
' '
Location of centroid above base.
/Area
y m (Print)
Moment of inertia about horizontal centroidal axis.
For 1 to ,
n n 1 1
( )/2 /2
n n n n
a a h h
3 2
/12 ( )( )
n n n n n
I b h b h y a
I I I
'
' (Print)
Computation of stresses.
Total height: For 1 to ,
n
n n
H H h
Stress at top:
top
H y
M M
I
(Print)
Stress at bottom:
bottom
y
M M
I
(Print)
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674
PROBLEM 4.C5 (Continued)
Problem 4.9
Summary of cross section dimensions:
Width (in.) Height (in.)
9.00 2.00
3.00 6.00
Bending moment 600.000 kip in.
˜
Centroid is 3.000 mm above lower edge.
Centroidal moment of inertia is 204.000 in4
.
Stress at top of beam 14.706 ksi
Stress at bottom of beam 8.824 ksi
Problem 4.10
Summary of cross section dimensions:
Width (in.) Height (in.)
4.00 1.00
1.00 6.00
8.00 1.00
Bending moment 500.000 kip in.
˜
Centroid is 4.778 in. above lower edge.
Centroidal moment of inertia is 155.111 in4
.
Stress at top of beam 10.387 ksi
Stress at bottom of beam 15.401 ksi
Problem 4.11
Summary of cross section dimensions:
Width (mm) Height (mm)
50 10
20 50
Bending moment 1500.0000 N m
˜
Centroid is 25.000 mm above lower edge.
Centroidal moment of inertia is 512,500 mm4
.
Stress at top of beam 102.439 MPa
Stress at bottom of beam 72.171 MPa
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675
c
z
y
y
M
Dy
PROBLEM 4.C6
A solid rod of radius 1.2 in.
c is made of a steel that is assumed to be
elastoplastic with 29,000 ksi
E and 42 ksi.
Y
V The rod is subjected to a
couple of moment M that increases from zero to the maximum elastic
moment Y
M and then to the plastic moment .
p
M Denoting by Y
y the half
thickness of the elastic core, write a computer program and use it to calculate
the bending moment M and the radius of curvature U for values of Y
y from
1.2 in. to 0 using 0.2-in. decrements. (Hint: Divide the cross section into 80
horizontal elements of 0.03-in. height.)
SOLUTION
3 3
3 3
(42 ksi) (1.2 in.) 57 kip in.
4 4
4 4
(42 ksi) (1.2 in.) 96.8 kip in.
3 3
Y Y
p Y
M c
M c
S S
V
V
˜
˜
Consider top half of rod.
Let Number of elements in top half.
i
Height of each element:
' '
c
h h
L
2 2
For 0 to 1, Step 1:
( )
{( 0.5) } at midheight of element
If go to 100
( 0.5)
Stress in elastic core
go to 200
100 Stress in plastic zone
200 Area 2 ( )
Force (
Y
E Y
y
E Y
E
n i
y n h
z c n h z
y y
n h
z h
'
ª º
' m
¬ ¼
t
'
m
m
' '
' '
V V
V
V V
V
Repeat for 1.2 in.
to 0
At 0.2-in. decrements
Area)
Moment Force ( 0.5)
Moment
/
Print , , and .
Next
Y
Y
Y Y
Y
y
y
n h
M M
P y E
y M
½
°
°
°
°
°
°
°
°
°
°
°
¾
°
°
°
°
' ' ' °
°
' °
°
°
°
°
¿
V
U
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676
PROBLEM 4.C6 (Continued)
Program Output
Radius of rod 1.2 in.
Yield point of steel 42 ksi
Yield moment 57.0 kip in.
Plastic moment 96.8 kip in.
˜
˜
Number of elements in half of the rod 40
For 1.20 in.,
Y
y 57.1 kip in.
M ˜ Radius of curvature 828.57 in.
For 1.00 in.,
Y
y 67.2 kip in.
M ˜ Radius of curvature 690.48 in.
For 0.80 in.,
Y
y 76.9 kip in.
M ˜ Radius of curvature 552.38 in.
For 0.60 in.,
Y
y 85.2 kip in.
M ˜ Radius of curvature 414.29 in.
For 0.40 in.,
Y
y 91.6 kip in.
M ˜ Radius of curvature 276.19 in.
For 0.20 in.,
Y
y 95.5 kip in.
M ˜ Radius of curvature 138.10 in.
For 0.00 in.,
Y
y infinite
M Radius of curvature zero
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677
3 in.
2 in.
2.5 in.
C
B
a
A
PROBLEM 4.C7
The machine element of Prob 4.178 is to be redesigned by removing part
of the triangular cross section. It is believed that the removal of a small
triangular area of width a will lower the maximum stress in the element.
In order to verify this design concept, write a computer program to
calculate the maximum stress in the element for values of a from 0 to
1 in. using 0.1-in. increments. Using appropriate smaller increments,
determine the distance a for which the maximum stress is as small as
possible and the corresponding value of the maximum stress.
SOLUTION
See Figure 4.79, Page 289.
2 2
5 kip in. 5 in. 2.5 in.
M r b
˜
For 0 to 1.0 at 0.1 intervals,
a
1
1 2
1 2
3
2
( /( ))
Area ( )( /2)
h a
r a
b b a h a
b b h
1 2
1 1
+ ( /3) 2 /3 Area
2 2
x a b h h b h h
ª º
« »
¬ ¼
2 ( )
r r h x
2
1
2
1
1 2
2
1 2 2 1 1 2
( )
( )ln ( )
r
r
h b b
R
b r b r h b b
e r R
1 1
2 2
( )/[Area ( )( )]
( )/[Area ( )( )]
D
B
M r R e r
M r R e r
V
V
Print and Return
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678
PROBLEM 4.C7 (Continued)
Program Output
a
in.
R
in.
D
V
ksi
B
V
ksi b1 r e
0.00 3.855 8.5071
2.1014 0.00 4.00 0.145
0.10 3.858 7.7736
2.1197 0.08 4.00 0.144
0.20 3.869 7.2700
2.1689 0.17 4.01 0.140
0.30 3.884 6.9260
2.2438 0.25 4.02 0.134
0.40 3.904 6.7004
2.3423 0.33 4.03 0.127
0.50 3.928 6.5683
2.4641 0.42 4.05 0.119
0.60 3.956 6.5143
2.6102 0.50 4.07 0.111
0.70 3.985 6.5296
2.7828 0.58 4.09 0.103
0.80 4.018 6.6098
2.9852 0.67 4.11 0.094
0.90 4.052 6.7541
3.2220 0.75 4.14 0.086
1.00 4.089 6.9647
3.4992 0.83 4.17 0.078
Determination of the maximum compressive stress that is as small as possible.
a
in.
R
in.
D
V
ksi
B
V
ksi b1 r e
0.620 3.961 6.51198
2.6425 0.52 4.07 0.109
0.625 3.963 6.51185
2.6507 0.52 4.07 0.109
0.630 3.964 6.51188
2.6591 0.52 4.07 0.109
Answer: When 625 in.,
a the compressive stress is 6.51 ksi.
Mecánica de Materiales - Ferdinand P. Beer 7ma. Edición.

718. C
CH
HA
AP
PT
TE
ER
R 5
5
Mecánica de Materiales - Ferdinand P. Beer 7ma. Edición.

719. Mecánica de Materiales - Ferdinand P. Beer 7ma. Edición.

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681
B
w
A
L
PROBLEM 5.1
For the beam and loading shown, (a) draw the shear and bending-moment
diagrams, (b) determine the equations of the shear and bending-moment
curves.
SOLUTION
Reactions:
0: 0
2 2
B
L wL
M AL wL A
      
0: 0
2 2
A
L wL
M BL wL B
     
Free body diagram for determining reactions:
Over whole beam, 0 x L
 
Place section at x.
Replace distributed load by equivalent concentrated load.
0: 0
2
y
wL
F wx V
    
2
L
V w x
 
 
 
 

0: 0
2 2
J
wL x
M x wx M
     
2
( )
2
w
M Lx x
 
( )
2
w
M x L x
  
Maximum bending moment occurs at .
2
L
x 
2
max
8
wL
M  
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682
B
P
C
A
L
b
a
PROBLEM 5.2
For the beam and loading shown, (a) draw the shear and bending-moment
diagrams, (b) determine the equations of the shear and bending-moment
curves.
SOLUTION


Reactions:
0: 0
C
Pb
M LA bP A
L
    
0: 0
A
Pa
M LC aP C
L
    
From A to B, 0 x a
 
0: 0
y
Pb
F V
L
   
Pb
V
L
 
0: 0
J
Pb
M M x
L
   
Pbx
M
L
 
From B to C, a x L
 
0: 0
y
Pa
F V
L
   
Pa
V
L
  
0: ( ) 0
K
Pa
M M L x
L
     
( )
Pa L x
M
L

 
At section B, 2
Pab
M
L
 
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683
B
w0
A
L
PROBLEM 5.3
For the beam and loading shown, (a) draw the shear and bending-
moment diagrams, (b) determine the equations of the shear and bending-
moment curves.
SOLUTION
Free body diagram for determining reactions.
Reactions:
0 0
0: 0
2 2
    
y A A
w L w L
F R R
0 2
0: 0
2 3
  
    
  
  
A A
w L L
M M
2 2
0 0
3 3
  
A
w L w L
M
Use portion to left of the section as the free body.
Replace distributed load with equivalent concentrated load.
0 0
1
0: 0
2 2
     
y
w L w x
F x V
L
2
0 0
2 2
 
w L w x
V
L

2
0 0 0
0:
1
( ) 0
3 2 2 3
 
    
    
    
    
J
M
w L w L w x x
x x M
L
2 3
0 0 0
3 2 6
   
w L w Lx w x
M
L

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684
B
w
L
A
PROBLEM 5.4
For the beam and loading shown, (a) draw the shear and bending-
moment diagrams, (b) determine the equations of the shear and bending-
moment curves.
SOLUTION
Free body diagram for determining reactions.
Reactions:
0: 0
   
y A
F R wL
A
R wL

0: ( ) 0
2
 
    
 
 
A A
L
M M wL
2
0
2

A
w L
M
Use portion to the right of the section as the free body.
Replace distributed load by equivalent concentrated load.
0: ( ) 0
    
y
F V w L x
( )
V w L x
  
0: ( ) 0
2

 
     
 
 
J
L x
M M w L x
2
( )
2
  
w
M L x 
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685
B
P
P
C
A
a a
PROBLEM 5.5
For the beam and loading shown, (a) draw the shear and bending-moment
diagrams, (b) determine the equations of the shear and bending-moment
curves.
SOLUTION


From A to B: 0 x a
 
0: 0
    
y
F P V
V P
  
0: 0
J
M Px M
   
M Px
  
From B to C: 2
a x a
 
0: 0
     
y
F P P V
2
V P
  
0: ( ) 0
     
J
M Px P x a M
2
M Px Pa
   
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686
D
A
B
a a
C
L
w w PROBLEM 5.6
For the beam and loading shown, (a) draw the shear and bending-moment
diagrams, (b) determine the equations of the shear and bending-moment
curves.
SOLUTION
Reactions: A D wa
 
From A to B, 0 x a
 
0:
y
F
  0
wa wx V
  
( )
V w a x
  
0:
J
M
  ( ) 0
2
x
wax wx M
   
2
2
x
M w ax
 
 
 
 
 

From B to C, a x L a
  
0:
y
F
  0
wa wa V
  
0
V  
0:
J
M
  0
2
a
wax wa x M
 
    
 
 
2
1
2
M wa
 
From C to D, L a x L
  
0: ( ) 0
y
F V w L x wa
     
( )
V w L x a
   
0: ( ) ( ) 0
2
J
L x
M M w L x wa L x

 
       
 
 
2
1
( ) ( )
2
 
   
 
 
M w a L x L x 
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687
B
A C D E
3 kN 2 kN 2 kN
5 kN
0.3 m 0.3 m
0.3 m 0.4 m
PROBLEM 5.7
Draw the shear and bending-moment diagrams for the beam and
loading shown, and determine the maximum absolute value (a) of
the shear, (b) of the bending moment.
SOLUTION
Origin at A:

Reaction at A:
0: 3 2 5 2 0 2 kN
y A A
F R R
       
0: (3 kN)(0.3 m) (2 kN)(0.6 m) (5 kN)(0.9 m) (2 kN)(1.3 m) 0
      
A A
M M
 0.2 kN m
 
A
M 
From A to C:
0: 2 kN
y
F V
  
1 0: 0.2 kN m (2 kN) 0
M x M
     
0.2 2
  
M x
From C to D:
0: 2 3 0
y
F V
    
1kN
V  
2 0: 0.2 kN m (2 kN) (3 kN)( 0.3) 0
M x x M
        
0.7
M x
 
From D to E:
0: 5 2 0 3 kN
y
F V V
      
3 0: 5(0.9 ) (2)(1.3 ) 0
M M x x
       
1.9 3
M x
  
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688
PROBLEM 5.7 (Continued)
From E to B:
0: 2 kN
y
F V
  
4 0: 2(1.3 ) 0
M M x
     
2.6 2
M x
  

 (a) max
3.00 kN
V  
(b) max
0.800 kN m
M   
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689
100 lb 100 lb
250 lb
10 in.
25 in.
20 in.
15 in.
A B
C D E
PROBLEM 5.8
Draw the shear and bending-moment diagrams for the beam
and loading shown, and determine the maximum absolute value
(a) of the shear, (b) of the bending moment.
SOLUTION
Reactions:
0: (45 in.) 100 lb(15 in.) 250 lb(20 in.) 100 lb(55 in.) 0
C E
M R
     
200 lb
E
R 
0: 200 lb 100 lb 250 lb 100 lb 0
y C
F R
      
250 lb
C
R 
At any point, V is the sum of the loads and reactions to the left (assuming + ) and M the sum of their moments
about that point (assuming ).
(a) max 150.0 lb

V 
(b) max 1500 lb in.
 
M 

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690
PROBLEM 5.8 (Continued)
Detailed computations of moments:
0
A
M 
(100 lb)(15 in.) 1500 lb in.
C
M     
(100 lb)(35 in.) (250 lb)(20 in.) 1500 lb in.
D
M      
(100 lb)(60 in.) (250 lb)(45 in.) (250 lb)(25 in.) 1000 lb in.
      
E
M
(100 lb)(70 in.) (250 lb)(55 in.) (250 lb)(35 in.) (200 lb)(10 in.) 0
B
M      
(Checks)
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691
B
A
C D
25 kN/m
40 kN 40 kN
0.6 m 0.6 m
1.8 m
PROBLEM 5.9
Draw the shear and bending-moment diagrams for the beam and loading
shown, and determine the maximum absolute value (a) of the shear,
(b) of the bending moment.
SOLUTION
The distributed load is replaced with an equivalent concentrated load of 45 kN to compute the reactions.
(25 kN/m)(1.8 m) 45 kN

0: (40 kN)(0.6 m) 45 kN(1.5 m) 40 kN(2.4 m) (3.0 m) 0
A B
M R
      
62.5 kN
B
R 
0: 62.5 kN 40 kN 45 kN 40 kN 0
y A
F R
      
62.5 kN
A
R 
At C:
0: 62.5 kN
y
F V
  
1 0: (62.5kN)(0.6m) 37.5kN m
M M
    
At centerline of the beam:
0: 62.5 kN 40 kN (25 kN/m)(0.9 m) 0
y
F V
     
0
V 
2 0:
M
 
(62.5 kN)(1.5 m) (40 kN)(0.9 m) (25 kN/m)(0.9 m)(0.45 m) 0
   
M
47.625 kN m
M  
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692
PROBLEM 5.9 (Continued)
Shear and bending-moment diagrams:
(a) max
62.5 kN

V 
(b) max
47.6 kN m
 
M 
From A to C and D to B, V is uniform; therefore M is linear.
From C to D, V is linear; therefore M is parabolic.
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on a website, in whole or part.
693
B
A
C D
2.5 kips/ft 15 kips
6 ft 6 ft
3 ft
PROBLEM 5.10
Draw the shear and bending-moment diagrams for the beam and loading
shown, and determine the maximum absolute value (a) of the shear,
(b) of the bending moment.
SOLUTION
0: 15 (12)(6)(2.5) (6)(15) 0
B A
M R
     
18 kips
A
R 
0: 15 (3)(6)(2.5) (9)(15) 0
A B
M R
    
12 kips

B
R
Shear:
18 kips
A
V 
18 (6)(2.5) 3 kips
C
V   
to : 3 kips
C D V 
to : 3 15 12 kips
D B V    
Areas under shear diagram:
1
to : (6)(18 3) 63 kip ft
2
A C V dx
 
   
 
 

to : (3)(3) 9 kip ft
C D V dx   

to : (6)( 12) 72 kip ft
D B V dx     

Bending moments: 0
A
M 
0 63 63 kip ft
   
C
M
63 9 72 kip ft
   
D
M
72 72 0
  
B
M
max
18.00 kips

V 
  max
72.0 kip ft
 
M 
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733. PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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on a website, in whole or part.
694
B
A
C D E
3 kN 3 kN
300 mm 300 mm
200 mm
450 N ? m
PROBLEM 5.11
Draw the shear and bending-moment diagrams for the beam and loading
shown, and determine the maximum absolute value (a) of the shear,
(b) of the bending moment.
SOLUTION 0: (700)(3) 450 (300)(3) 1000 0
B
M A
     
2.55 kN
A 
0: (300)(3) 450 (700)(3) 1000 0
A
M B
      
3.45 kN
B 
At A: 2.55 kN 0
V M
 
A to C: 2.55 kN
V 
At C: 0:
C
M
 
(300)(2.55) 0
M
  
765 N m
M  
C to E: 0.45 N m
V   
At D: 0:
D
M
 
(500)(2.55) (200)(3) 0
M
   
675 N m
M  
At D: 0:
D
M
 
(500)(2.55) (200)(3) 450 0
M
    
1125 N m
M  
E to B: 3.45 kN
V  
At E: 0:
E
M
 
(300)(3.45) 0
M
  
1035 N m
M  
At B: 3.45 kN, 0
 
V M
(a) max
3.45 kN

V 
(b) max
1125 N m
 
M 
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734. PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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695
400 lb 1600 lb 400 lb
12 in. 12 in. 12 in. 12 in.
8 in.
8 in.
C
A
D E F
G
B
PROBLEM 5.12
Draw the shear and bending-moment diagrams for the beam and
loading shown, and determine the maximum absolute value (a) of
the shear, (b) of the bending moment.
SOLUTION
0: 16 (36)(400) (12)(1600)
G
M C
    
(12)(400) 0
  1800 lb
C 
0: 0
x x
F C G
     1800 lb
x
G 
0: 400 1600 400 0
y y
F G
       2400 lb
y
G 
A to E: 400 lb
V  
E to F: 2000 lb
V  
F to B: 400 lb
V 
At A and B, 0
M 
At ,
D
0: (12)(400) 0
D
M M
    4800 lb in.
  
M
At +
,
D 0:
D
M
  (12)(400) (8)(1800) 0
M
   9600 lb in.
 
M
At E, 0: (24)(400) (8)(1800) 0
E
M M
     4800 lb in.
 
M
At ,
F
0: (8)(1800) (12)(400) 0
F
M M
      19,200 lb in.
M   
At +
,
F 0: (12)(400) 0
F
M M
     4800 lb in.
  
M
(a) Maximum | | 2000 lb
V  
(b) Maximum | | 19,200 lb in.
M   
Mecánica de Materiales - Ferdinand P. Beer 7ma. Edición.

735. PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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696
B
A
C D
1.5 kN
1.5 kN
0.9 m
0.3 m
0.3 m
PROBLEM 5.13
Assuming that the reaction of the ground is uniformly distributed, draw
the shear and bending-moment diagrams for the beam AB and determine
the maximum absolute value (a) of the shear, (b) of the bending
moment.
SOLUTION
Over the whole beam,
0: 1.5 1.5 1.5 0
y
F w
     2 kN/m
w 
A to C: 0 0.3 m
x
 
0: 2 0
y
F x V
    (2 ) kN
V x

0: (2 ) 0
2
J
x
M x M
 
    
 
 
2
( ) kN m
M x
 
At ,
C 
0.3 m
x 
0.6 kN, 0.090 kN m
90 N m
V M
  
 
C to D: 0.3 m 1.2 m
x
 
0: 2 1.5 0
y
F x V
     (2 1.5) kN
V x
 
0: (2 ) (1.5)( 0.3) 0
2
J
x
M x x M
 
      
 
 
2
( 1.5 0.45) kN m
M x x
   
At the center of the beam, 0.75 m
x 
0 0.1125 kN m
112.5 N m
V M
   
  
At +
,
C 0.3 m, 0.9 kN
x V
  
(a) Maximum | | 0.9 kN 900 N
V   
(b) Maximum | | 112.5 N m
M   
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736. PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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697
B
C D E
2 kips/ft
24 kips
A
3 ft 3 ft 3 ft 3 ft
2 kips/ft PROBLEM 5.14
Assuming that the reaction of the ground is uniformly distributed, draw
the shear and bending-moment diagrams for the beam AB and determine
the maximum absolute value (a) of the shear, (b) of the bending moment.
SOLUTION
Over the whole beam,
0: 12 (3)(2) 24 (3)(2) 0
y
F w
      3 kips/ft
w 
A to C: (0 3 ft)
x
 
0: 3 2 0
y
F x x V
     ( ) kips
V x

0: (3 ) (2 ) 0
2 2
J
x x
M x x M
      2
(0.5 ) kip ft
M x
 
At C, 3 ft
x 
3 kips, 4.5 kip ft
V M
  
C to D: (3 ft 6 ft)
x
 
0: 3 (2)(3) 0
y
F x V
     (3 6) kips
V x
 
3
0: (3 ) (2)(3) 0
2 2
x
MK x x M
   
      
   
   
2
(1.5 6 9) kip ft
M x x
   
At ,
D
6 ft
x 
12 kips, 27 kip ft
V M
  
D to B: Use symmetry to evaluate.
(a) max
| | 12.00 kips
V  
(b) max
| | 27.0 kip ft
M   
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698
B
A
C
3 kN/m
1.5 m 1.5 m 2.2 m
100 mm
200 mm
10 kN PROBLEM 5.15
For the beam and loading shown, determine the maximum
normal stress due to bending on a transverse section at C.
SOLUTION
Using CB as a free body,
3
0: (2.2)(3 10 )(1.1) 0
C
M M
     
3
7.26 10 N m
M   
Section modulus for rectangle:
2
1
6
S bh

2 3 3
6 3
1
(100)(200) 666.7 10 mm
6
666.7 10 m

  
 
Normal stress:
3
6
6
7.26 10
10.8895 10 Pa
666.7 10
M
S
 

   

10.89 MPa
  
Mecánica de Materiales - Ferdinand P. Beer 7ma. Edición.

738. PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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699
750 lb
B
A
C D
150 lb/ft
750 lb
3 in.
12 in.
4 ft
4 ft
4 ft
PROBLEM 5.16
For the beam and loading shown, determine the maximum
normal stress due to bending on a transverse section at C.
SOLUTION
Reactions: 
C A by symmetry.
0: (2)(750) (12)(150) 0
y
F A C
     
1650 lb
A C
 
Use left half of beam as free body.
0:
E
M
 
(1650)(6) (750)(2) (150)(6)(3) 0
M
    
3
5700 lb ft 68.4 10 lb in.
    
M
Section modulus: 2 2 3
1 1
(3)(12) 72 in
6 6
S bh
 
  
 
 
Normal stress:
3
68.4 10
950 psi
72
M
S


  
950 psi
  
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700
B
A
C D E
150 kN 150 kN
2.4 m
0.8 m
0.8 m
0.8 m
W460 ⫻ 113
90 kN/m
PROBLEM 5.17
For the beam and loading shown, determine the maximum
normal stress due to bending on a transverse section at C.
SOLUTION
Use entire beam as free body.
0:
B
M
 
4.8 (3.6)(216) (1.6)(150) (0.8)(150) 0
A
    
237 kN

A
Use portion AC as free body.
0:
C
M
 
(2.4)(237) (1.2)(216) 0
309.6 kN m
M
M
  
 
For 6 3
W460 113, 2390 10 mm
S
  
Normal stress:
3
6 3
6
309.6 10 N m
2390 10 m
129.5 10 Pa
M
S 
 
 

 

129.5
  MPa 
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Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
701
B
A
a
a
30 kN 50 kN 50 kN 30 kN
2 m
5 @ 0.8 m 5 4 m
W310 3 52
PROBLEM 5.18
For the beam and loading shown, determine the maximum normal
stress due to bending on section a-a.
SOLUTION
Reactions: By symmetry, A B

0 : 80 kN
   
A B
y
F
Using left half of beam as free body,
0:
J
M
 
(80)(2) (30)(1.2) (50)(0.4) 0
M
    
3
104 kN m 104 10 N m
M     
For 3 3
6 3
W310 52, 747 10 mm
747 10 m
S

  
 
Normal stress:
3
6
6
104 10
139.2 10 Pa
747 10
M
S
 

   

139.2 MPa
  
Mecánica de Materiales - Ferdinand P. Beer 7ma. Edición.

741. PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
702
B
A
C
8 kN
1.5 m 2.1 m
W310 ⫻ 60
3 kN/m
PROBLEM 5.19
For the beam and loading shown, determine the maximum
normal stress due to bending on a transverse section at C.
SOLUTION
Use portion CB as free body.
3
0: (3)(2.1)(1.05) (8)(2.1) 0
23.415 kN m 23.415 10 N m
C
M M
M
     
    
For 3 3
6 3
W310 60, 844 10 mm
844 10 m
S

  
 
Normal stress:
3
6
6
23.415 10
27.7 10 Pa
844 10
M
S
 

   

27.7 MPa
  
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742. PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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on a website, in whole or part.
703
B
A
C D E F G
5
kips
5
kips
2
kips
2
kips
2
kips
6 @ 15 in. 5 90 in.
S8 3 18.4
PROBLEM 5.20
For the beam and loading shown, determine the maximum normal
stress due to bending on a transverse section at C.
SOLUTION
Use entire beam as free body.
0:
B
M
 
90 (75)(5) (60)(5) (45)(2) (30)(2) (15)(2) 0
A
      
9.5 kips
A 
Use portion AC as free body.
0: (15)(9.5) 0
142.5 kip in.
C
M M
M
   
 
For 3
8 18.4, 14.4 in
S S
 
Normal stress:
142.5
14.4
M
S
  
9.90 ksi
  
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704
B
A
C D E
25 kips 25 kips 25 kips
2 ft
1 ft 2 ft
6 ft
S12 ⫻ 35
PROBLEM 5.21
Draw the shear and bending-moment diagrams for the beam
and loading shown and determine the maximum normal stress
due to bending.
SOLUTION
0:
B
M
 
(11)(25) 10 (8)(25) (2)(25) 0 52.5 kips
C
    
C
0:
C
M
 
(1)(25) (2)(25) (8)(25) 10 0 22.5 kips
B
    
B
Shear:
A to C
: 25 kips
V  
C
to D
: 27.5 kips
V 
D
to E
: 2.5 kips
V 
E
to B: 22.5 kips
V  
Bending moments:
At C, 0: (1)(25) 0
C
M M
   
25 kip ft
M   
At D, 0: (3)(25) (2)(52.5) 0
D
M M
    
30 kip ft
M  
At E, 0: (2)(22.5) 0 45 kip ft
E
M M M
      
max 45 kip ft 540 kip in.
   
M
For 12 35
S  rolled steel section, 3
38.1 in
S 
Normal stress:
540
14.17 ksi
38.1
M
S
    14.17 ksi
  
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705
Hinge
2.4 m
0.6 m
1.5 m 1.5 m
C
B
A E
D
80 kN/m 160 kN
W310 ⫻ 60
PROBLEM 5.22
Draw the shear and bending-moment diagrams for the beam
and loading shown and determine the maximum normal
stress due to bending.
SOLUTION
Statics: Consider portion AB and BE separately.
Portion BE:
0:
E
M
 
(96)(3.6) (48)(3.3) (3) (160)(1.5) 0
C
   
248kN
 
C
56kN
 
E
0
A B E
M M M
  
At midpoint of AB:
0: 0
0: (96)(1.2) (96)(0.6) 57.6 kN m
y
F V
M M
  
     
Just to the left of C:
0: 96 48 144 kN
y
F V
      
0: (96)(0.6) (48)(0.3) 72 kN
C
M M
      
Just to the left of D:
0: 160 56 104 kN
0: (56)(1.5) 84 kN m
y
D
F V
M M
     
     

Mecánica de Materiales - Ferdinand P. Beer 7ma. Edición.

745. PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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706
PROBLEM 5.22 (Continued)

From the diagram,

3
max
84 kN m 84 10 N m
M      
For W310 60
 rolled-steel shape,
3 3
6 3
844 10 mm
844 10 m
x
S

 
 
Stress: max
m
M
S
 
3
6
6
84 10
99.5 10 Pa
844 10
m
 

  

99.5 MPa
m
  
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on a website, in whole or part.
707
H
A
7 @ 200 mm ⫽ 1400 mm
Hinge
30 mm
20 mm
C
B D E F G
300 N 300 N 300 N
40 N PROBLEM 5.23
Draw the shear and bending-moment diagrams for the beam
and loading shown, and determine the maximum normal
stress due to bending.
SOLUTION









Free body EFGH. Note that 0
E
M  due to hinge.
0: 0.6 (0.2)(40) (0.40)(300) 0
213.33 N
E
M H
H
    

0: 40 300 213.33 0
126.67 N
y E
E
F V
V
     

Shear:
to : 126.67 N m
to : 86.67 N m
to : 213.33 N m
E F V
F G V
G H V
 
 
  
Bending moment at F:
0: (0.2)(126.67) 0
25.33 N m
F F
F
M M
M
   
 
Bending moment at G:
0: (0.2)(213.33) 0
42.67 N m
G G
G
M M
M
    
 
Free body ABCDE.
0: 0.6 (0.4)(300) (0.2)(300)
(0.2)(126.63) 0
257.78 N
B
M A
A
   
 

0: (0.2)(300) (0.4)(300) (0.8)(126.67) 0.6 0
468.89 N
A
M D
D
      

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708
PROBLEM 5.23 (Continued)
Bending moment at B.
0: (0.2)(257.78) 0
51.56 N m
B B
B
M M
M
    
 
Bending moment at C.
0: (0.4)(257.78) (0.2)(300)
0
43.11 N m
C
C
C
M
M
M
   
 
 
Bending moment at D.
0: (0.2)(213.33) 0
25.33 N m
    
  
D D
D
M M
M
max 51.56 N m
 
M 
2 2
3 3 6 3
1 1
(20)(30)
6 6
3 10 mm 3 10 m

 
   
S bh
Normal stress:
6
6
51.56
17.19 10 Pa
3 10
 
  

17.19 MPa
  
max
342 N

V 
max
516 N m
 
M 
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on a website, in whole or part.
709
24 kN/m
64 kN ? m
B
A
C D
2 m 2 m 2 m
S250 ⫻ 52
PROBLEM 5.24
Draw the shear and bending-moment diagrams for the beam
and loading shown and determine the maximum normal stress
due to bending.
SOLUTION
Reactions:
0: 4 64 (24)(2)(1) 0 28 kN
D
M A
     
A
0: 28 (24)(2) 0 76 kN
y
F D
      
D
A to C: 0 2m
x
 
0: 28 0
28 kN
y
F V
V
    
 
0: 28 0
( 28 ) kN m
J
M M x
M x
   
  
C to D: 2m 4m
x
 
0: 28 0
28 kN
y
F V
V
    
 
0: 28 64 0
( 28 64) kN m
J
M M x
M x
    
   
D to B: 4m 6m
x
 
0:
24(6 ) 0
( 24 144) kN
y
F
V x
V x
 
  
  
2
0:
6
24(6 ) 0
2
12(6 ) kN m
J
M
x
M x
M x
 

 
   
 
 
   
3
max 56 kN m 56 10 N m
M     
For S250 52
 section, 3 3
482 10 mm
S  
Normal stress:
3
6
6 3
56 10 N m
116.2 10 Pa
482 10 m
M
S
 
 
   

116.2 MPa
  
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on a website, in whole or part.
710
B
A
C D
5 ft 5 ft
8 ft
W14 ⫻ 22
10 kips
5 kips PROBLEM 5.25
Draw the shear and bending-moment diagrams for the beam
and loading shown, and determine the maximum normal stress
due to bending.
SOLUTION
Reaction at C: 0: (18)(5) 13 +(5)(10) 0
10.769 kips
B
M C
C
   

Reaction at B: 0: (5)(5) (8)(10) 13 0
4.231 kips
C
M B
B
   

Shear diagram:
to : 5 kips
to : 5 10.769 5.769 kips
to : 5.769 10 4.231 kips
A C V
C D V
D B V

 

 
   
   
At A and B, 0
M 
At C, 0: (5)(5) 0
25 kip ft
C C
C
M M
M
  
  
At D, 0: (5)(4.231)
21.155 kip ft
D D
D
M M
M
   
 
max
5.77 kips
V  
max
| |
M occurs at C. max
| | 25 kip ft 300 kip in.
   
M 
For W14 22
 rolled-steel section, 3
29.0 in
S 
Normal stress:
300
29.0
M
S
   10.34 ksi
  
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on a website, in whole or part.
711
B
C D E
A
8 kN 8 kN
W310 ⫻ 23.8
1 m 1 m 1 m 1 m
W
PROBLEM 5.26
Knowing that 12 kN
W  , draw the shear and bending-moment
diagrams for beam AB and determine the maximum normal
stress due to bending.
SOLUTION
By symmetry, A B

0: 8 12 8 0
2 kN
y
F A B
A B
      
 
Shear: to : 2 kN
A C V

 
 to : 6 kN
C D V
 
   
 to : 6 kN
D E V
 
  
 to : 2 kN
E B V

   
Bending moment: max
6.00 kN

V 
At C, 0: (1)(2) 0
C C
M M
   
2 kN m
 
C
M
At D, 0: (2)(2) (8)(1) 0
D D
M M
    
4 kN m
D
M    
By symmetry, 2 kN m at .
M D
  2 kN m
 
E
M
max| | 4.00 kN m
 
M occurs at E. 
For 3 3 6 3
W310 23.8, 280 10 mm 280 10 m
x
S 
    
Normal stress:
3
max
max 6
| | 4 10
280 10
x
M
S
 

 

6
14.29 10 Pa
  max 14.29 MPa
  
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712
B
C D E
A
8 kN 8 kN
W310 ⫻ 23.8
1 m 1 m 1 m 1 m
W
PROBLEM 5.27
Determine (a) the magnitude of the counterweight W for which
the maximum absolute value of the bending moment in the beam
is as small as possible, (b) the corresponding maximum normal
stress due to bending. (Hint: Draw the bending-moment diagram
and equate the absolute values of the largest positive and
negative bending moments obtained.)
SOLUTION
By symmetry, A  B
0: 8 8 0
8 0.5
y
F A W B
A B W
      
  
Bending moment at C: 0: (8 0.5 )(1) 0
(8 0.5 ) kN m
C C
C
M W M
M W
     
  
Bending moment at D:
0: (8 0.5 )(2) (8)(1) 0
(8 ) kN m
D D
D
M W M
M W
      
  
Equate: 8 8 0.5
D C
M M W W
    
10.67 kN
W  
(a)
3
max
10.6667 kN
2.6667 kN m
2.6667 kN m 2.6667.10 N m
| | 2.6667 kN m
C
D
W
M
M
M

  
   
 
For W310 23.8
 rolled-steel shape,
3 3 6 3
280 10 mm 280 10 m
x
S 
   
(b)
3
6
max
max 6
| | 2.6667 10
9.52 10 Pa
280 10
x
M
S
 

   

max 9.52 MPa
  
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on a website, in whole or part.
713
B
A
C D
a 5 ft
8 ft
W14 ⫻ 22
10 kips
5 kips PROBLEM 5.28
Determine (a) the distance a for which the maximum absolute
value of the bending moment in the beam is as small as
possible, (b) the corresponding maximum normal stress due to
bending. (See hint of Prob. 5.27.)
SOLUTION
Reaction at B: 0: 5 (8)(10) 13 0
1
(80 5 )
18
C B
B
M a R
R a
    
 
Bending moment at D:
0: 5 0
5
5 (80 5 )
13
D D B
D B
M M R
M R a
    
  
Bending moment at C:
0 5 0
5
C C
C
M a M
M a
  
 
Equate:
5
5 (80 5 )
13
C D
M M
a a
 
 
4.4444 ft
a  ( ) 4.44 ft
a a  
Then (5)(4.4444) 22.222 kip ft
C D
M M
    
max
| | 22.222 kip ft 266.67 kip in.
   
M
For W14 22
 rolled-steel section, 3
29.0 in
S 
Normal stress:
266.67
9.20 ksi
29.0
M
S
    ( ) 9.20 ksi
b 
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714
B
A
a
C D
P Q
12 mm
18 mm
500 mm
500 mm
PROBLEM 5.29
Knowing that 480 N,
P Q
  determine (a) the distance a for
which the absolute value of the bending moment in the beam is
as small as possible, (b) the corresponding maximum normal
stress due to bending. (See hint of Prob. 5.27.)
SOLUTION
480 N 480 N
P Q
 
Reaction at A: 0: 480( 0.5)
480(1 ) 0
720
960 N
D
M Aa a
a
A
a
    
  
 
 
 
 
Bending moment at C: 0: 0.5 0
360
0.5 480 N m
C C
C
M A M
M A
a
    
 
   
 
 
Bending moment at D: 0: 480(1 ) 0
480(1 ) N m
D D
D
M M a
M a
     
   
(a) Equate:
360
480(1 ) 480
D C
M M a
a
    
0.86603 m
a  866 mm
a  
128.62 N 64.31 N m 64.31 N m
C D
A M M
     
(b) For rectangular section, 2
1
6
S bh

2 3 9 3
1
(12)(13) 648 mm 648 10 m
6
S 
   
6
max
max 9
| | 64.31
99.2 10 Pa
6.48 10
M
S
 
   

max 99.2 MPa
  
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754. PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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715
B
A
a
C D
P Q
12 mm
18 mm
500 mm
500 mm
PROBLEM 5.30
Solve Prob. 5.29, assuming that 480 N
P  and 320 N.
Q 
PROBLEM 5.29 Knowing that 480 N,
P Q
  determine
(a) the distance a for which the absolute value of the bending
moment in the beam is as small as possible, (b) the
corresponding maximum normal stress due to bending. (See
hint of Prob. 5.27.)
SOLUTION
480 N 320 N
P Q
 
Reaction at A: 0: 480( 0.5) 320(1 ) 0
560
800 N
D
M Aa a a
A
a
      
 
 
 
 
Bending moment at C: 0: 0.5 0
280
0.5 400 N m
C C
C
M A M
M A
a
    
 
   
 
 
Bending moment at D: 0: 320(1 ) 0
( 320 320 ) N m
D D
D
M M a
M a
     
   
(a) Equate:
280
320 320 400
D C
M M a
a
    
2
320 80 280 0 0.81873 m, 1.06873 m
a a a
    
Reject negative root. 819 mm
a  
116.014 N 58.007 N m 58.006 N m
C D
A M M
     
(b) For rectangular section, 2
1
6
S bh

2 3 9 3
1
(12)(18) 648 mm 648 10 m
6
S 
   
6
max
max 9
| | 58.0065
89.5 10 Pa
648 10
M
S
 
   

max 89.5 MPa
  
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716
Hinge
18 ft
B
a
C
4 kips/ft
W14 ⫻ 68
A
PROBLEM 5.31
Determine (a) the distance a for which the absolute value
of the bending moment in the beam is as small as possible,
(b) the corresponding maximum normal stress due to bending.
(See hint of Prob. 5.27.)
SOLUTION
For 3
W14 68, 103 in
x
S
 
Let (18 ) ft
b a
 
Segment BC:
By symmetry, B
V C

0: 4 0
2
y B
B
F V C b
V b
    

2 2
0: (4 ) 0
2
2 2 2 lb ft
J B
B
x
M V x x M
M V x x bx x
 
     
 
 
    
2 2 2
max
1
2 0
2
1 1
2 2
m m
dM
b x x b
dx
M b b b
   
  
Segment AB:
2
( )
0: 4( )
2
( ) 0
2( ) 2 ( )
K
B
a x
M a x
V a x M
M a x b a x

   
   
    
max
2 2
max
| | occurs at 0.
| | 2 2 2 2 (18 ) 36
M x
M a ab a a a a

       
(a) Equate the two values of max
| |:
M
 
2 2 2
2 2 1
2
1 1 1
36 (18 ) 162 18
2 2 2
1
54 162 0 54 (54) (4) (162)
2
a b a a a
a a a
     
     
54 50.9118 3.0883 ft
a    3.09 ft
a  
(b) max
| | 36 111.179 kip ft 1334.15 kip in.
M a
    
2
max
| | 1334.15
12.95 kips/in
103
x
M
S
    12.95 ksi
m
  
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717
B
d
A
L ⫽ 10 ft
PROBLEM 5.32
A solid steel rod of diameter d is supported as shown. Knowing that for
steel 3
490 lb/ft ,
  determine the smallest diameter d that can be
used if the normal stress due to bending is not to exceed 4 ksi.
SOLUTION
Let W  total weight.
2
4
W AL d L
 

 
Reaction at A:
1
2
A W

Bending moment at center of beam:
2 2
0: 0
2 2 2 4
8 32
C
W L W L
M M
WL
M d L


     
     
     
     
 
For circular cross section,  
1
2
c d

4 3 3
,
4 4 32
I
I c S c d
c
  
   
Normal stress:
2 2 2
32
3
32
d L
M L
S d
d


 
   
Solving for d,
2
L
d



Data:
3 3
3
2
10 ft (12)(10) 120 in.
490
490 lb/ft 0.28356 lb/in
12
4 ksi 4000 lb/in
L   
  
 


2
(120) (0.28356)
4000
d  1.021 in.
d  
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718
B
b
b
A D
C
1.2 m 1.2 m 1.2 m
PROBLEM 5.33
A solid steel bar has a square cross section of side b and is
supported as shown. Knowing that for steel 3
7860 kg / m ,
 
determine the dimension b for which the maximum normal stress
due to bending is (a) 10 MPa, (b) 50 MPa.
SOLUTION
Weight density: g
 

Let L  total length of beam.
2
W AL g b L g
 
 
Reactions at C and D:
2
W
C D
 
Bending moment at C:
0: 0
6 3
18
C
L W
M M
WL
M
  
   
  
  
 
Bending moment at center of beam:
0: 0
4 2 6 2 24
E
L W L W WL
M M M
     
      
     
     
2 2
max| |
18 18
WL b L g
M

 
For a square section, 3
1
6
S b

Normal stress:
2 2 2
3
| | /18
3
/6
M b L g L g
S b
b
 
   
Solve for b:
2
3
L g
b



Data: 3 2
3.6 m 7860 kg/m 9.81 m/s
L g

   (a) 6
10 10 Pa
   (b) 6
50 10 Pa
  
(a)
2
3
6
(3.6) (7860)(9.81)
33.3 10 m
(3)(10 10 )
b 
  

33.3 mm
b  
(b)
2
3
6
(3.6) (7860)(9.81)
6.66 10 m
(3)(50 10 )
b 
  

6.66 mm
b  
Mecánica de Materiales - Ferdinand P. Beer 7ma. Edición.

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719
B
w
A
L
PROBLEM 5.34
Using the method of Sec. 5.2, solve Prob. 5.1a.
PROBLEM 5.1 For the beam and loading shown, (a) draw the shear and
bending-moment diagrams, (b) determine the equations of the shear and
bending-moment curves.
SOLUTION
0: 0
2 2
B
L wL
M AL wL A
      
0: 0
2 2
A
L wL
M BL wL B
     
dV
w
dx
 
0
x
A
V V wdx wx
    

A
V V wx A wx
   
2
wL
V wx
  
dM
V
dx

0 0
2
2
2 2
x x
A
wL
M M V dx wx dx
wLx wx
 
   
 
 
 
 
2
2 2
A
wLx wx
M M
   2
( )
2
w
M Lx x
  
Maximum M occurs at
1
, where
2
x 
0
dM
V
dx
 
2
max
| |
8
wL
M  
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720
B
P
C
A
L
b
a
PROBLEM 5.35
Using the method of Sec. 5.2, solve Prob. 5.2a.
PROBLEM 5.2 For the beam and loading shown, (a) draw the shear and
bending-moment diagrams, (b) determine the equations of the shear and
bending-moment curves.
SOLUTION
0: 0
C
Pb
M LA bP A
L
    
0: 0
A
Pa
M LC aP C
L
    
At ,
A
0
Pb
V A M
L
  
A to B
: 0 x a
 
0
0 0
x
w wdx
 

0
A
V V
 
Pb
V
L
 
0 0
a a
B A
Pb Pba
M M V dx dx
L L
   
  B
Pba
M
L
 
At +
,
B
Pb Pa
V A P P
L L
     
B
to C: a x L
 
0 0
x
a
w wdx
 

0
C B
V V
 
Pa
V
L
  
( )
0
L
C B a
C B
Pa Pab
M M V dx L a
L L
Pab Pba Pab
M M
L L L
      
    

max
| |
Pab
M
L
 
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721
B
w0
A
L
PROBLEM 5.36
Using the method of Sec. 5.2, solve Prob. 5.3a.
PROBLEM 5.3 For the beam and loading shown, (a) draw the shear
and bending-moment diagrams, (b) determine the equations of the shear
and bending-moment curves.
SOLUTION
Free body diagram for determining reactions.
Reactions:
0 0
0: 0
2 2
    
y A A
w L w L
F V V
0 2
0: 0
2 3
  
    
  
  
A A
w L L
M M
2
0
3
 
A
w L
M
2
0
0
0 , ,
2 3
   
A A
x w L w L
w w V M
L
0
   
dV w x
w
dx L
2
0 0
0
2
    

x
A
w x w x
V V dx
L L
2
0 0
2 2
 
w L w x
V
L

2
2 2
  
o o
dM w L w x
V
dx L
2
0 0
0 0
2 2
x x
A
w L w x
M M V dx dx
L
 
   
 
 
 
 
3
0 0
2 6
 
w L w x
x
L
2 3
0 0 0
3 2 6
   
w L w L w x
M x
L

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722
B
w
L
A
PROBLEM 5.37
Using the method of Sec. 5.2, solve Prob. 5.4a.
PROBLEM 5.4 For the beam and loading shown, (a) draw the shear
and bending-moment diagrams, (b) determine the equations of the shear
and bending-moment curves.
SOLUTION
0: 0
y A A
F V wL V wL
    
2
0: ( ) 0
2 2
A A
L wL
M M wL M
 
      
 
 
dV
w
dx
 
0
    

x
A
V V wdx wx
V wL wx
  
dM
V wL wx
dx
  
2
0
( )
2
     

x
A
wx
M M wL wx dx wLx
2 2
2 2
   
wL wx
M wLx 
max
2
max 2


V wL
wL
M
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723
B
P
P
C
A
a a
PROBLEM 5.38
Using the method of Sec. 5.2, solve Prob. 5.5a.
PROBLEM 5.5 For the beam and loading shown, (a) draw the shear and
bending-moment diagrams, (b) determine the equations of the shear and
bending-moment curves.
SOLUTION
At A+
: A
V P
 
Over AB: 0
dV
w
dx
  
A
dM
V V P
dx
    
M Px C
  
1
0 at 0 0
M x C
  
M Px
  
At point B:   
x a M Pa
At point B+
: 2
V P P P
    
Over BC: 0
dV
w
dx
  
2
dM
V P
dx
   
2
2
M Px C
  
At B:   
x a M Pa
2 2
2
    
Pa Pa C C Pa
2
M Px Pa
   
At C: 2 3
  
x a M Pa 
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724
D
A
B
a a
C
L
w w
PROBLEM 5.39
Using the method of Sec. 5.2, solve Prob. 5.6a.
PROBLEM 5.6 For the beam and loading shown, (a) draw the shear and
bending-moment diagrams, (b) determine the equations of the shear and
bending-moment curves.
SOLUTION
Reactions: A D wa
 
A to B: 0 x a
  w w

,
A
V A wa
  0
A
M 
0
    

x
A
V V wdx wx
( )
V w a x
  
dM
V wa wx
dx
  
0 0
( )
x x
A
M M V dx wa wx dx
   
 
2
1
2
M wax wx
  
2
1
0
2
B B
V M wa
 
B to C: a x L a
   0
V  
0
dM
V
dx
 
0
x
B a
M M V dx
  

B
M M
 2
1
2
M wa
 






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725
PROBLEM 5.39 (Continued)
L a x L
  
C to D: [ ( )]

      

x
C L a
V V w dx w x L a
[ )]
  
V w L x a 
[ ( )]
x x
C L a L a
M M V dx w x L a dx
 
     
 
2
2 2
2
2 2
( )
2
( )
( ) ( )
2 2
( )
( )
2 2
x
L a
x
w L a x
x L a
w L a x L a
x L a
w L a x

 
   
 
 
 

      
 
 
 

    
 
 
2 2
2
1 ( )
( )
2 2 2
x L a
M wa w L a x
 

    
 
 

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on a website, in whole or part.
726
B
A C D E
3 kN 2 kN 2 kN
5 kN
0.3 m 0.3 m
0.3 m 0.4 m
PROBLEM 5.40
Using the method of Sec. 5.2, solve Prob. 5.7.
PROBLEM 5.7 Draw the shear and bending-moment diagrams
for the beam and loading shown, and determine the maximum
absolute value (a) of the shear, (b) of the bending moment.
SOLUTION
Free body diagram for determining reactions.
Reactions:
0: 3 kN 2 kN 5 kN 2 kN 0
y A
F V
      
2 kN
A
V  
0: (3 kN)(0.3 m) (2 kN)(0.6 m) (5 kN)(0.9 m) (2 kN)(1.3 m) 0
      
A A
M M
0.2 kN m
A
M  
Between concentrated loads and the vertical reaction,
the scope of the shear diagram is ,
 i.e., the shear is
constant. Thus, the area under the shear diagram is equal
to the change in bending moment.
A to C:
2 kN 0.6 0.4 kN
C A C
V M M M
     
C to D:
1kN 0.3 0.1kN m
       
D C D
V M M M
D to E:
3 kN 0.9 0.8 kN m
       
E D E
V M M M
E to B:
2 kN 0.8 0 (Checks)
B E B
V M M M
      
Mecánica de Materiales - Ferdinand P. Beer 7ma. Edición.

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727
100 lb 100 lb
250 lb
10 in.
25 in.
20 in.
15 in.
A B
C D E
PROBLEM 5.41
Using the method of Sec. 5.2, solve Prob. 5.8.
PROBLEM 5.8 Draw the shear and bending-moment diagrams
for the beam and loading shown, and determine the maximum
absolute value (a) of the shear, (b) of the bending moment.
SOLUTION
Free body diagram for determining reactions.
Reactions:
0: 100 lb 250 lb 100 lb 0
      
Y C E
F V V
450 lb
 
C E
V V 
0: (45 in.) (100 lb)(15 in.) (250 lb)(20 in.) (100 lb)(55 in.) 0
     
C E
M V
200 lb

E
V
250 lb
 
C
V
Between concentrated loads and the vertical reaction, the
scope of the shear diagram is ,
 i.e., the shear is constant.
Thus, the area under the shear diagram is equal to he change
in bending moment.
A to C:
100 lb, 1500, 1500 lb in.
C A C
V M M M
       
C to D:
150 lb 3000, 1500 lb in.
      
D C D
V M M M
D to E:
100 lb, 2500, 1000 lb in.
       
E D E
V M M M
E to B:
100 lb, 1000, 0 (Checks)
    
B E B
V M M M 
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728
B
A
C D
25 kN/m
40 kN 40 kN
0.6 m 0.6 m
1.8 m
PROBLEM 5.42
Using the method of Sec. 5.2, solve Prob. 5.9.
PROBLEM 5.9 Draw the shear and bending-moment diagrams for the
beam and loading shown, and determine the maximum absolute value
(a) of the shear, (b) of the bending moment.
SOLUTION
Free body diagram to determine reactions:
0:
(3.0 m) 45 kN(1.5 m) (40 kN)(0.6 m) (40 kN)(2.4 m) 0
A
B
M
V
 
   
62.5 kN
B
V  
0: 40 kN 45 kN 40 kN 62.5 kN 0
y A
F V
      
62.5 kN
A
V 
Change in bending moment is equal to area under shear diagram.
A to C: (62.5 kN)(0.6 m) 37.5 kN m
 
C to E:
1
(0.9 m)(22.5 kN) 10.125 kN m
2
 
E to D:
1
(0.9 m)( 22.5 kN) 10.125 kN m
2
   
D to B: ( 62.5 kN)(0.6 m) 37.5 kN m
   



Mecánica de Materiales - Ferdinand P. Beer 7ma. Edición.

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729
B
A
C D
2.5 kips/ft 15 kips
6 ft 6 ft
3 ft
PROBLEM 5.43
Using the method of Sec. 5.2, solve Prob. 5.10.
PROBLEM 5.10 Draw the shear and bending-moment diagrams for the
beam and loading shown, and determine the maximum absolute value
(a) of the shear, (b) of the bending moment.
SOLUTION
Reactions at supports A and B:
0: 15( ) (12)(6)(2.5) (6)(15) 0
B A
M R
     
18 kips
A
R  
0: 15 (3)(6)(2.5) (9)(15) 0
A B
M R
    
12 kips
B
R 
Areas under shear diagram:
A to C:
1
(6)(3) (6)(15) 63 kip ft
2
  
C to D: (3)(3) 9 kip ft
 
D to B: (6)( 12) 72 kip ft
   
Bending moments:
0
A
M 
0 63 63 kip ft
C
M    
63 9 72 kip ft
D
M    
72 72 0
B
M    
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730
0.5 m
4 kN
1 m 1 m
0.5 m
4 kN
A
E
D
C
B
F
PROBLEM 5.44
Draw the shear and bending-moment diagrams for the beam and loading
shown, and determine the maximum absolute value (a) of the shear, (b) of
the bending moment.
SOLUTION
0:
3 (1)(4) (0.5)(4) 0
2 kN
B
M
A
 
   
 
A
0: 3 (2)(4) (2.5)(4) 0
6 kN
A
M B
    
 
B
Shear diagram:
A to C: 2 kN
V 
C to D: 2 4 2 kN
V    
D to B: 2 4 6 kN
V     
Areas of shear diagram:
A to C: (1)(2) 2 kN m
V dx   

C to D: (1)( 2) 2 kN m
V dx     

D to E: (1)( 6) 6 kN m
V dx     

Bending moments:
0
A
M 
0 2 2 kN m
2 4 6 kN m
6 2 4 kN m
4 2 6kN m
6 6 0
C
C
D
D
B
M
M
M
M
M




   
   
   
   
  
(a) max
6.00 kN
V  
(b) max
6.00 kN m
M   
Mecánica de Materiales - Ferdinand P. Beer 7ma. Edición.

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731
300 N 300 N
200 mm
75 mm
200 mm 200 mm
A C D
B
F
E PROBLEM 5.45
Draw the shear and bending-moment diagrams for the beam and
loading shown, and determine the maximum absolute value (a) of
the shear, (b) of the bending moment.
SOLUTION
3
0:
0.075 (0.2)(300) (0.6)(300) 0
3.2 10 N
A
EF
EF
M
F
F
 
  
 
3
0: 0 3.2 10 N
x x EF x
F A F A
     
0: 300 300 0
600 N
y y
y
F A
A
    

Couple at D: 3
(0.075)(3.2 10 )
240 N m
D
M  
 
Shear:
A to C: 600 N
V 
C to B: 600 300 300 N
  
V
Areas under shear diagram:
A to C: (0.2)(600) 120 N m
V dx   

C to D: (0.2)(300) 60 N m
V dx   

D to B: (0.2)(300) 60 N m
V dx   

Bending moments:
0
A
M 
0 120 120 N m
120 60 180 N m
180 240 60 N m
60 60 0
C
D
D
B
M
M
M
M


   
   
    
   
Maximum 600 N
V  
Maximum 180.0 N m
 
M 
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732
B
A
C
3 kN/m
1.5 m 1.5 m 2.2 m
100 mm
200 mm
10 kN PROBLEM 5.46
Using the method of Sec. 5.2, solve Prob. 5.15.
PROBLEM 5.15 For the beam and loading shown, determine
the maximum normal stress due to bending on a transverse
section at C.
SOLUTION
0:
3 (1.5)(10) (1.1)(2.2)(3) 0
2.58 kN
C
M
A
A
 
   

0:
(1.5)(10) 3 (4.1)(2.2)(3) 0
14.02 kN
A
M
C
C
 
   

Shear:
A to D
: 2.58 kN
V 
D
to C
: 2.58 10 7.42 kN
   
V
C
: 7.42 14.02 6.60 kN
V    
B: 6.60 (2.2)(3) 0
V   
Areas under shear diagram:
A to D: (1.5)(2.58) 3.87 kN m
V dx   

D to C: (1.5)( 7.42) 11.13 kN m
V dx     

C to B:
1
(2.2)(6.60) 7.26 kN m
2
V dx
 
  
 
 

Bending moments:
0
A
M 
0 3.87 3.87 kN m
3.87 11.13 7.26 kN m
7.26 7.26 0
D
C
B
M
M
M
   
    
  
3
7.26 kN m 7.26 10 N m
C
M     
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772. PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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733
PROBLEM 5.46 (Continued)
For rectangular cross section, 2 2
1 1
(100)(200)
6 6
S bh
 
   
 
3 3 6 2
666.67 10 mm 666.67 10 m

   
Normal stress:
3
6
6
7.26 10
10.89 10 Pa
666.67 10

   

C
M
S

10.89 MPa 
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on a website, in whole or part.
734
750 lb
B
A
C D
150 lb/ft
750 lb
3 in.
12 in.
4 ft
4 ft
4 ft
PROBLEM 5.47
Using the method of Sec. 5.2, solve Prob. 5.16.
PROBLEM 5.16 For the beam and loading shown, determine the
maximum normal stress due to bending on a transverse section at C.
SOLUTION
Reactions: by symmetry
C A

0: (2)(750) (12)(150) 0
1650 lb
y
F A C
A C
     
 
Shear:
1650 lb

A
V
1650 (4)(150) 1050 lb
   
C
V
1050 750 300 lb
C
V    
300 (2)(150) 0
E
V   
Areas under shear diagram:
A to C:
1
(1650 1050)(4)
2
V dx
 
 
 
 

5400 lb ft
 
C to E:
1
(300)(2) 300 lb ft
2
V dx
 
  
 
 

Bending moments:
0

A
M
0 5400 5400 lb ft
5400 300 5700 lb ft
   
   
C
E
M
M
3
5700 lb ft 68.4 10 lb in.
    
E
M
For rectangular cross section, 2 2 3
1 1
(3)(12) 72 in
6 6
S bh
 
  
 
 
Normal stress:
3
68.4 10
950 psi
72
M
S


   
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735
B
A
a
a
30 kN 50 kN 50 kN 30 kN
2 m
5 @ 0.8 m 5 4 m
W310 3 52
PROBLEM 5.48
Using the method of Sec. 5.2, solve Prob. 5.18.
PROBLEM 5.18 For the beam and loading shown, determine
the maximum normal stress due to bending on section a-a.
SOLUTION
Reactions: By symmetry, A  B.
0: 80 kN
y
F
    
A B
Shear diagram:
A to C: 80 kN
V 
C to D: 80 30 50 kN
V   
D to E: 50 50 0
V   
Areas of shear diagram:
A to C: (80)(0.8) 64 kN m
V dx
   
C to D: (50)(0.8) 40 kN m
V dx
   
D to E: 0
V dx
 
Bending moments:
0
A
M 
0 64 64 kN m
C
M    
64 40 104 kN m
D
M    
104 0 104 kN m
E
M    
3
max
104 kN m 104 10 N m
M     
For 3 3 6 3
W310 52, 747 10 mm 747 10 m
S 
    
Normal stress:
3
6
6
104 10
139.2 10 Pa
747 10
M
S
 

   

139.2 MPa

 
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736
B
A
C D E F G
5
kips
5
kips
2
kips
2
kips
2
kips
6 @ 15 in. 5 90 in.
S8 3 18.4
PROBLEM 5.49
Using the method of Sec. 5.2, solve Prob. 5.20.
PROBLEM 5.20 For the beam and loading shown, determine the
maximum normal stress due to bending on a transverse section at C.
SOLUTION
Use entire beam as free body.
0:
B
M
 
90 (75)(5) (60)(5) (45)(2) (30)(2) (15)(2) 0
9.5 kips
A
      
 
A
Shear A to C: 9.5 kips
V 
Area under shear curve A to C: (15)(9.5)
142.5 kip in.
V dx
 
 
0
A
M 
0 142.5 142.5 kip in.
   
C
M
For 3
S8 18.4, 14.4 in
S
 
Normal stress:
142.5
14.4
M
S
   9.90 ksi
  
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737
w
L
A
B
x
w 5 w0 [x/L]1/2
PROBLEM 5.50
Determine (a) the equations of the shear and bending-moment curves for
the beam and loading shown, (b) the maximum absolute value of the
bending moment in the beam.
SOLUTION
1
1/2
2 0
0 1/2
 
     
 
 
dV x w x
w w
dx L L
3/2
1
1/2
2
3
  
o
w x
V C
L
0 at
 
V x L
0 1 1 0
2 2
0
3 3
   
w L C C w L
3/2
0
0 1/2
2 2
3 3
 
   
 
w x
V w L
L

5/2
0
2 0 1/2
2 2 2
3 3 5
    
dM w x
V M C w Lx
dx L
2 2 2
0 0 2 0
2 4 2
0 at 0
3 15 5
      
M x L C w L w L C w L
5/2
2
0
0 0
1/2
2 4 2
3 15 5
  
w x
M w Lx w L
L

2
0
max max
2
occurs at 0
5
 
M x M w L 
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738
w
A
L
B
x
w ⫽ w0 cos␲ x
2L
PROBLEM 5.51
Determine (a) the equations of the shear and bending-moment curves for
the beam and loading shown, (b) the maximum absolute value of the
bending moment in the beam.
SOLUTION
0
0
1
2
0
1 2
2
1
2
0
2 2
cos
2
2
sin
2
4
cos
2
0 at 0. Hence, 0.
4
0 at 0. Hence, .
dV x
w w
dx L
Lw x dM
V C
L dx
L w x
M C x C
L
V x C
L w
M x C






   
   
  
  
   
(a) 0
(2 / sin( /2 )
V Lw x L
  
  
2
0
(4 / )[1 cos( /2 )]
M L w x L

  
  
(b) 2 2
0
max
4 /
M w L 
 
Mecánica de Materiales - Ferdinand P. Beer 7ma. Edición.

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on a website, in whole or part.
739
B
x
w w ⫽ w0 sin
A
L
␲ x
L
PROBLEM 5.52
Determine (a) the equations of the shear and bending-moment curves for
the beam and loading shown, (b) the maximum absolute value of the
bending moment in the beam.
SOLUTION
0
0
1
2
0
1 2
2
2
1
1
sin
cos
sin
0 at 0 0
0 at 0 0 0
0
dV x
w w
dx L
w L x dM
V C
L dx
w L x
M C x C
L
M x C
M x L C L
C





   
  
  
  
    

(a) 0
cos
w L x
V
L


 
2
0
2
sin
w L x
M
L


 
0 at
2
dM L
V x
dx
  
(b)
2
0
max 2
sin
2
w L
M



2
0
max 2
w L
M

 
Mecánica de Materiales - Ferdinand P. Beer 7ma. Edición.

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on a website, in whole or part.
740
B
x
w
w ⫽ w0
A
L
x
L
PROBLEM 5.53
Determine (a) the equations of the shear and bending-moment curves for
the beam and loading shown, (b) the maximum absolute value of the
bending moment in the beam.
SOLUTION
0
2
0 1
3
0 1 2
1
2
1
6
dV x
w w
dx L
x dM
V w C
L dx
x
M w C x C
L
   
   
   
2
0 at 0 0
M x C
  
2
0 1 1 0
1 1
0 at 0
6 6
M x L w L C L C w L
     
(a)
2
2
0 0
1 1
2 6
x
V w w L
L
   2 2
0
1
( 3 )/
6
V w L x L
  
3
0 0
1 1
6 6
x
M w w Lx
L
   3
0
1
( / )
6
M w Lx x L
  
(b) max
M occurs when 2 2
0. 3 0
m
dM
V L x
dx
   
2 2
max 0
1
6
3 3 3 3
m
L L L
x M w
 
  
 
 
 
2
max 0
0.0642
M w L
 
Mecánica de Materiales - Ferdinand P. Beer 7ma. Edición.

780. PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
741
A
C D
B
2 ft 10 ft 3 ft
3 kips/ft
S10 3 25.4
PROBLEM 5.54
Draw the shear and bending-moment diagrams for the beam and
loading shown and determine the maximum normal stress due to
bending.
SOLUTION
0: (10) (5.5)(15)(3) 0 24.75 kips
C D D
M R R
     
0: (4.5)(15)(3) (10) 0 20.25 kips
D C C
M R R
     
Shear:
0
A
V 
0 (2)(3) 6 kips
C
V    
6 20.50 14.25 kips
C
V     
14.25 (10)(3) 15.75 kips
D
V     
15.75 24.75 9 kips
D
V     
9 (3)(3) 0
B
V   
Locate point E where 0:
V 
10
4.75 ft 10 5.25 ft
14.25 15.75
e e
e e

   
Areas of shear diagram:
A to C:
1
(2)( 6) 6 kip ft
2
V dx
 
    
 
 

C to E:
1
(4.75)(14.25) 33.84375 kip ft
2
V dx
 
  
 
 

E to D:
1
(5.25)( 15.75) 41.34375 kip ft
2
V dx
 
    
 
 

D to B:
1
(3)(9) 13.5 kip ft
2
V dx
 
  
 
 

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781. PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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on a website, in whole or part.
742
PROBLEM 5.54 (Continued)
Bending moments: 0
A
M 
0 6 6 kip ft
    
C
M
6 33.84375 27.84375 kip ft
27.84375 41.34375 13.5 kip ft
13.5 13.5 0
    
    
   
E
D
B
M
M
M
Maximum 27.84375 kip ft 334.125 kip in.
M    
For 3 334.125
S10 25.4, 24.6 in Normal stress: 13.58 ksi
24.6
    
M
S
S
 
Mecánica de Materiales - Ferdinand P. Beer 7ma. Edición.

782. PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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on a website, in whole or part.
743
A B
C
16 kN/m
1 m
1.5 m
S150 ⫻ 18.6
PROBLEM 5.55
Draw the shear and bending-moment diagrams for the beam
and loading shown, and determine the maximum normal
stress due to bending.
SOLUTION
0: 2.5 (1.75)(1.5)(16) 0
16.8 kN
B
M A
A
    

0: (0.75) (1.5)(16) 2.5 0
7.2 kN
A
M B
B
     

Shear diagram:
16.8 kN
16.8 (1.5)(16) 7.2 kN
7.2 kN
A
C
B
V
V
V

   
 
Locate point D where 0.
V 
1.5
24 25.2
16.8 7.2
1.05 m 1.5 0.45 m
d d
d
d d

 
  
Areas of the shear diagram:
A to D:
1
(1.05)(16.8) 8.82 kN m
2
V dx
 
  
 
 

D to C:
1
(0.45)( 7.2) 1.62 kN m
2
V dx
 
    
 
 

C to B: (1)( 7.2) 7.2 kN m
V dx     

Bending moments:
0
0 8.82 8.82 kN m
8.82 1.62 7.2 kN m
7.2 7.2 0
A
D
C
B
M
M
M
M

   
   
  
Maximum 3
| | 8.82 kN m 8.82 10 N m
M     
For S150 18.6
 rolled-steel section, 3 3 6 3
120 10 mm 120 10 m
S 
   
Normal stress:
3
6
6
| | 8.82 10
73.5 10 Pa
120 10
M
S
 

   

73.5 MPa
  
Mecánica de Materiales - Ferdinand P. Beer 7ma. Edición.

783. PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
744
A B
C
0.9 m 3 m
12 kN/m
9 kN
W200 3 19.3
PROBLEM 5.56
Draw the shear and bending-moment diagrams for the beam
and loading shown and determine the maximum normal
stress due to bending.
SOLUTION
0: (0.9)(9) (1.5)(3)(12) 3 0
C
M B
    
15.3 kN
B 
0: (3.9)(9) 3 (1.5)(3)(12) 0
B
M C
    
29.7 kN
C 
Shear:
to : 9 kN
 
A C V
: 9 29.7 20.7 kN

   
C V
: 20.7 (3)(12) 15.3 kN
B V    
max
20.7 kN

V 
Locate point E where 0.

V
3
36 (20.7)(3)
20.7 15.3

 
e e
e
1.725 ft 3 1.275 ft
e e
  
Areas under shear diagram:
A to C: (0.9)(9) 8.1kN m
V dx   

C to E:
1
(1.725)(20.7) 17.8538 kN m
2
V dx
 
  
 
 

E to B:
1
( 1.275)(15.3) 9.7538 kN m
2
V dx
 
    
 
 

Mecánica de Materiales - Ferdinand P. Beer 7ma. Edición.

784. PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
745
PROBLEM 5.56 (Continued)
Bending moments:
0
A
M 
0 8.1 8.1kN m
C
M     
8.1 17.8538 9.7538 kN m
9.7538 9.7538 0
    
  
E
B
M
M
3
max
9.7538 10 N m at point
  
M E 
For W200 19.3
 rolled-steel section, 3 3 6 3
162 10 mm 162 10 m

   
S
Normal stress:
3
6
6
9.7538 10
60.2 10 Pa 60.2 MPa
162 10
M
S
 

    

60.2 MPa

 
Mecánica de Materiales - Ferdinand P. Beer 7ma. Edición.

785. PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
746
A B
80 lb/ft
1600 lb
1.5 ft
9 ft
11.5 in.
1.5 in.
PROBLEM 5.57
Draw the shear and bending-moment diagrams for the beam
and loading shown, and determine the maximum normal
stress due to bending.
SOLUTION
0: (1600 lb)(1.5 ft) [(80 lb/ft)(9 ft)](7.5 ft) 12 0
650 lb 650 lb
A
M B
B
    
   
B
0: 1600 lb [(80 lb/ft)(9 ft)] 650 lb 0
1670 lb 1670 lb
y
F A
A
     
   
A
max
9
0.875 ft 2641 lb ft
70 lb 650 lb
x x
x M

   
1
(11.5 in.) 5.75 in.
2
c  
3 4
1
(1.5 in.)(11.5 in.) 190.1in
12
I  

max 2641 lb ft 31,690 lb in.
   
M
max
4
(31,690 lb in.)(5.75 in.)
190.1in

 
m
M c
I
 959 psi
m
  
Mecánica de Materiales - Ferdinand P. Beer 7ma. Edición.

786. PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
747
A
B
C
16 in. 24 in.
25 lb/in.
500 lb
S4 ⫻ 7.7
PROBLEM 5.58
Draw the shear and bending-moment diagrams for the beam and
loading shown and determine the maximum normal stress due to
bending.
SOLUTION
0: (16) (4)(40)(25) (24)(500) 0 1000 lb
B A A
M R R
      
0: (16) (20)(40)(25) (40)(500) 0 2500 lb
A B B
M R R
      
Shear:
1000 lb
A
V  
1000 (16)(25) 1400 lb
B
V     
1400 2500 1100 lb
B
V     
1100 (24)(25) 500 lb
C
V   
Areas of shear diagram:
A to B:
1
( 1000 1400)(16) 19,200 lb in.
2
V dx      

B to C:
1
(1100 500)(24) 19,200 lb in.
2
V dx
 
   
 
 

Bending moments:
0
A
M 
0 19,200 19,200 lb in.
B
M     
19,200 19,200 0
C
M    
Maximum 19.2 kip in.
M  
For S4 7.7
 rolled-steel section, 3
3.03 in
S 
Normal stress:
19.2
6.34 ksi
3.03
M
S
    
Mecánica de Materiales - Ferdinand P. Beer 7ma. Edición.

787. PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
748
A B
D
C
80 kN/m
W250 ⫻ 80
1.2 m 1.2 m
1.6 m
60 kN · m 12 kN · m
PROBLEM 5.59
Draw the shear and bending-moment diagrams for the
beam and loading shown and determine the maximum
normal stress due to bending.
SOLUTION
Reaction:
0: 4 60 (80)(1.6)(2) 12 0
76 kN
B
M A
      
 
A
Shear: 76 kN
A
V 
A to C : 76.0 kN
V  
76 (80)(1.6) 52 kN
D
V    
D to C : 52 kN
V  
Locate point where V  0.
( ) 80 76 0 0.95 m
V x x x
    
Areas of shear diagram:
A to C: (1.2)(76) 91.2 kN m
V dx
   
C to E:
1
(0.95)(76) 36.1kN m
2
V dx
   
E to D:
1
(0.65)( 52) 16.9 kN m
2
V dx
     
D to B: (1.2)( 52) 62.4 kN m
V dx
     
Bending moments: 60 kN m
A
M   
60 91.2 31.2 kN m
C
M     
31.2 36.1 67.3 kN m
E
M     
67.3 16.9 50.4 kN m
D
M    
50.4 62.4 12 kN m
B
M     
For 3 3
W250 80, 983 10 mm
S
  
Normal stress:
3
6
max 6 3
67.3 10 N m
68.5 10 Pa
983 10 m
M
S
 
 
   

68.5 MPa
m
  
Mecánica de Materiales - Ferdinand P. Beer 7ma. Edición.

788. PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
749
A C B
D
400 kN/m
W200 3 22.5
w0
0.3 m 0.3 m
0.4 m
PROBLEM 5.60
Knowing that beam AB is in equilibrium under the loading
shown, draw the shear and bending-moment diagrams and
determine the maximum normal stress due to bending.
SOLUTION
0
0
0: (1)( ) (0.4)(400) 0
160 kN/m
y
F w
w
   

Shear diagram: 0
A
V 
0 (0.3)(160) 48 kN
48 (0.3)(400) (0.3)(160) 48 kN
48 (0.3)(160) 0
C
D
B
V
V
V
  
    
   
Locate point E where V  0.
By symmetry, E is the midpoint of CD.
Areas of shear diagram:
A to C:
1
(0.3)(48) 7.2 kN m
2
 
C to E:
1
(0.2)(48) 4.8 kN m
2
 
E to D:
1
(0.2)( 48) 4.8 kN m
2
   
D to B:
1
(0.3)( 48) 7.2 kN m
2
   
Bending moments: 0
A
M 
0 7.2 7.2 kN
C
M   
7.2 4.8 12.00 kN
12.0 4.8 7.2 kN
7.2 7.2 0
E
D
B
M
M
M
  
  
  
3
max
12.00 kN m 12.00 10 N m
M     
For W200 22.5
 rolled-steel shape, 3 3 6 3
193 10 mm 193 10 m
x
S 
   
Normal stress:
3
6
6
12.00 10
62.2 10 Pa
193 10
M
S
 

   

62.2 MPa
  
Mecánica de Materiales - Ferdinand P. Beer 7ma. Edición.

789. PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
750
B
A
1.2 ft 1.2 ft
C
w0 50 lb/ft
T
w0
3
4
in.
PROBLEM 5.61
Knowing that beam AB is in equilibrium under the loading shown,
draw the shear and bending-moment diagrams and determine the
maximum normal stress due to bending.
SOLUTION
A to C: 0 1.2 ft
x
 
0
2
0
2
0
3 2
50 1 50 41.667
1.2
41.667 50
(41.667 50)
0 20.833 50
0 (20.833 50 )
6.944 25
x
A
x
A
x
x
w x
dV
w x
dx
V V x dx
dM
x x
dx
M M V dx
x x dx
x x
 
   
 
 
   
  
   
 
  
 



At C to B, use symmetry conditions. 1.2 ft,
x  30.0 lb
24.0 lb in.
V
M
 
  
Maximum | | 24.0 lb ft 288 lb in.
M    
Cross section:
1
(0.75) 0.375 in.
2 2
d
c
 
  
 
 
4 3 4
(0.375) 15.532 10 in
4 4
I c
  
 
   
 
 
Normal stress: 3
3
| | (2.88)(0.375)
6.95 10 psi
15.532 10
M c
I
 
   

6.95 ksi
  
Mecánica de Materiales - Ferdinand P. Beer 7ma. Edición.

790. PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
751
0.4 m
P Q 24 mm
0.2 m
0.5 m 0.5 m
60 mm
A
C D E F
B
0.3 m
PROBLEM 5.62*
The beam AB supports two concentrated loads P and Q. The
normal stress due to bending on the bottom edge of the beam
is 55 MPa
 at D and 37.5 MPa
 at F. (a) Draw the shear and
bending-moment diagrams for the beam. (b) Determine the
maximum normal stress due to bending that occurs in the beam.
SOLUTION
(a) 3 3 4
3 3 6 3
1
(24)(60) 432 10 mm 30 mm
12
14.4 10 mm 14.4 10 m
I c
I
S M S
c


   
     
At D, 6 6
(14.4 10 )(55 10 ) 792 N m
D
M 
    
At F, 6 6
(14.4 10 )(37.5 10 ) 540 N m
F
M 
    
Using free body FB, 0: 540 0.3 0
540
1800 N
0.3
F
M B
B
    
 
Using free body DEFB, 0: 792 3 (0.8)(1800) 0
2160 N
D
M Q
Q
     

Using entire beam, 0: 0.2 (0.7)(2160) (1.2)(1800) 0
3240 N
A
M P
P
     

0: 3240 2160 1800 0
3600 N
y
F A
A
     

Shear diagram and its areas:
A to C
: 3600 N
V  (0.2)(3600) 720 N m
AC
A   
C
to E
: 3600 3240 360 N
V    (0.5)(360) 180 N m
CE
A   
E
to B: 360 2160 1800 N
V     (0.5)( 1800) 900 N m
EB
A     
Bending moments:
0
0 720 720 N m
720 180 900 N m
900 900 0
A
C
E
B
M
M
M
M

   
   
  
max
| | 900 N m
M  
(b) Normal stress: 6
max
max 6
| | 900
62.5 10 Pa
14.4 10
M
S
 
   

max 62.5 MPa
  
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752
A
480 lb/ft
1 ft 1 ft
1.5 ft 1.5 ft
W8 31
8 ft
P
B
C D E F
Q PROBLEM 5.63*
The beam AB supports a uniformly distributed load of 480 lb/ft and
two concentrated loads P and Q. The normal stress due to bending
on the bottom edge of the lower flange is 14.85 ksi at D and
10.65 ksi at E. (a) Draw the shear and bending-moment diagrams
for the beam. (b) Determine the maximum normal stress due to
bending that occurs in the beam.
SOLUTION
(a) For W8 31
 rolled-steel section, 3
27.5 in
S 
M S

At D, (27.5)(14.85) 408.375 kip in.
D
M   
At E, (27.5)(10.65) 292.875 kip in.
E
M   
34.03 kip ft 24.41kip ft
D E
M M
   
Use free body DE.
0: 34.03 24.41 (1.5)(3)(0.48) 3 0
2.487 kips
E D
D
M V
V
      
 
0: 34.03 24.41 (1.5)(3)(0.48) 3 0
3.927 kips
D E
E
M V
V
     
 
Use free body ACD.
0: 1.5 (1.25)(2.5)(0.48) (2.5)(2.487) 34.03 0
25.83 kips
A
M P
      
 
P
0: (2.5)(0.48) 2.487 25.83 0
24.54 kips
y
F A
     
 
A
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753
PROBLEM 5.63* (Continued)
Use free body EFB.
0: 1.5 (1.25)(2.5)(0.48) (2.5)(3.927) 24.41 0
8.728 kips
B
M Q
Q
     

0: 3.927 (2.5)(0.48) 8.7 0
13.855 kips
y
F B
B
     

Areas of load diagram:
A to C: (1.5)(0.48) 0.72 kip ft
 
C to F: (5)(0.48) 2.4 kip ft
 
F to B: (1.5)(0.48) 0.72 kip ft
 
Shear diagram: 24.54 kips
A
V 
24.54 0.72 23.82 kips
C
V 
  
23.82 25.83 2.01kips
C
V 
   
2.01 2.4 4.41kips
F
V 
   
4.41 8.728 13.14 kips
F
V 
    
13.14 0.72 13.86 kips
B
V     
Areas of shear diagram:
A to C:
1
(1.5)(24.52 23.82) 36.23 kip ft
2
  
C to F:
1
(5)( 2.01 4.41) 16.05 kip ft
2
    
F to B:
1
(1.5)( 13.14 13.86) 20.25 kip ft
2
   
Bending moments: 0
A
M 
0 36.26 36.26 kip ft
C
M    
36.26 16.05 20.21kip ft
F
M    
20.21 20.25 0
B
M   
Maximum M occurs at C: max
36.26 kip ft 435.1kip in.
M    
(b) Maximum stress: max 435.1
27.5
M
S
   15.82 ksi
  
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754
C D B
A
2 kN/m
P
0.1 m 0.1 m 0.125 m
36 mm
18 mm
Q
PROBLEM 5.64*
Beam AB supports a uniformly distributed load of 2kN/m and two
concentrated loads P and Q. It has been experimentally determined
that the normal stress due to bending in the bottom edge of the beam
is 56.9 MPa at A and 29.9 MPa at C. Draw the shear and bending-
moment diagrams for the beam and determine the magnitudes of the
loads P and Q.
SOLUTION







3 3 4
3 3 6 3
1
(18)(36) 69.984 10 mm
12
1
18 mm
2
3.888 10 mm 3.888 10 m
I
c d
I
S
c

  
 
    
At A, 6
(3.888 10 )( 56.9) 221.25 N m
A A
M S 
      
At C, 6
(3.888 10 )( 29.9) 116.25 N m
C C
M S 
      
0: 221.23 (0.1)(400) 0.2 0.325 0
A
M P Q
     
0.2 0.325 181.25
P Q
  (1)
0: 116.25 (0.05)(200) 0.1 0.225 0
C
M P Q
     
0.1 0.225 106.25
P Q
  (2)
Solving (1) and (2) simultaneously, 500 N
P  
250 N
Q  
Reaction force at A: 400 500 250 0 1150 N m
A A
R R
     
1150 N 250
221.25 N m 116.25 N m 31.25 N m
A D
A C D
V V
M M M
 
        
max
| | 1150 N
V  
max
| | 221 N m
M   
Mecánica de Materiales - Ferdinand P. Beer 7ma. Edición.

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Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
755
1.8 kN 3.6 kN
C
B
A D h
0.8 m 0.8 m 0.8 m
40 mm
PROBLEM 5.65
For the beam and loading shown, design the cross section of the
beam, knowing that the grade of timber used has an allowable
normal stress of 12 MPa.
SOLUTION
Reactions:
0: 2.4 (1.6)(1.8) (0.8)(3.6) 0
D
M A
      2.4 kN
A 
0: (0.8)(1.8) (1.6)(3.6) 2.4 0
A
M D
      3 kN
D 
Construct shear and bending moment diagrams:
3
max
| | 2.4 kN m 2.4 10 N m
M     
all
6
3
max
min 6
all
6 3
3 3
2 2
3
3
2
12 MPa
12 10 Pa
| | 2.4 10
12 10
200 10 m
200 10 mm
1 1
(40)
6 6
200 10
(6)(200 10 )
40
M
S
S bh h
h




 

 

 
 
 
 


3 2
30 10 mm
  173.2 mm
h  
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756
A B
120 mm
h
10 kN/m
5 m
PROBLEM 5.66
For the beam and loading shown, design the cross section of the
beam, knowing that the grade of timber used has an allowable normal
stress of 12 MPa.
SOLUTION
Reactions:
A = B by symmetry
0: (5)(10) 0
y
F A B
    
25 kN
A B
 
From bending moment diagram,
3
max
| | 31.25 kN m 31.25 10 N m
M     
6
all
3
max 3 3
min 6
all
6 3
2 2 6
6
2 3 2
12 MPa 12 10 Pa
31.25 10
2.604 10 m
12 10
2.604 10 mm
1 1
(120) 2.604 10
6 6
(6)(2.064 10 )
130.21 10 mm
120

  

   

 
 
   
 
 

  
M
S
S bh h
h


361mm
h  
Mecánica de Materiales - Ferdinand P. Beer 7ma. Edición.

796. PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
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757
1.2 kips/ft
6 ft
a
a
B
A
PROBLEM 5.67
For the beam and loading shown, design the cross section of the beam,
knowing that the grade of timber used has an allowable normal stress of
1750 psi.
SOLUTION
Equivalent concentrated load:
1
(6)(1.2) 3.6 kips
2
P
 
 
 
 
Bending moment at A:
(2)(3.6) 7.2 kip ft = 86.4 kip in.
   
A
M
3
max
min
all
86.4
49.37 in
1.75
M
S

  
For a square section, 3
1
6
S a

3
6
a S

3
min (6)(49.37)
a  min 6.67 in.
a  
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758
4.8 kips 4.8 kips
2 kips 2 kips
F
b
A
2 ft 2 ft 3 ft 2 ft 2 ft
9.5 in.
B C D E
PROBLEM 5.68
For the beam and loading shown, design the cross section of the
beam, knowing that the grade of timber used has an allowable
normal stress of 1750 psi.
SOLUTION
For equilibrium, 2.8 kips
B E
 
Shear diagram:
A to B
: 4.8 kips
V  
B
to C
: 4.8 2.8 2 kips
V     
C
to D
: 2 2 0
V    
D
to E
: 0 2 2 kips
V   
E
to F: 2 2.8 4.8 kips
V   
Areas of shear diagram:
A to B: (2)( 4.8) 9.6 kip ft
   
B to C: (2)( 2) 4 kip ft
   
C to D: (3)(0) 0

D to E: (2)(2) 4 kip ft
 
E to F: (2)(4.8) 9.6 kip ft
 
Bending moments: 0
0 9.6 9.6 kip ft
9.6 4 13.6 kip ft
13.6 0 13.6 kip ft
13.6 4 9.6 kip ft
9.6 9.6 0
A
B
C
D
E
F
M
M
M
M
M
M

    
     
     
     
   
3
max
| | 13.6 kip ft 162.3 kip in. 162.3 10 lb in.
M       
Required value for S:
3
3
max
all
| | 162.3 10
93.257 in
1750
M
S


  
For a rectangular section,
2 2
3
1 1 ( )(9.5)
, 15.0417
12 2 6 6
I bh b
I bh c h S b
c
     
Equating the two expressions for S, 15.0417 93.257
b  6.20 in.
b  
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on a website, in whole or part.
759
C
A
B
D h
0.6 m 0.6 m
3 m
100 mm
6 kN/m
2.5 kN
2.5 kN PROBLEM 5.69
For the beam and loading shown, design the cross section of the
beam, knowing that the grade of timber used has an allowable
normal stress of 12 MPa.
SOLUTION
By symmetry, B C

0: 2.5 2.5 (3)(6) 0
y
F B C
       6.5 kN
B C
 
Shear:
A to B: 2.5 kN
2.5 6.5 9 kN
9 (3)(6) 9 kN
B
C
V
V
V



  
   
C to D: 9 6.5 2.5 kN
V     
Areas of the shear diagram:
A to B: (0.6)(2.5) 1.5 kN m
V dx   

B to E:
1
(1.5)(9) 6.75 kN m
2
V dx
 
  
 
 

E to C: 6.75 kN m
V dx   

C to D: 1.5 kN m
V dx   

Bending moments: 0
0 1.5 1.5 kN m
1.5 6.75 8.25 kN m
8.25 6.75 1.5 kN m
1.5 1.5 0
A
B
E
C
D
M
M
M
M
M

   
   
   
  
Maximum 3
| | 8.25 kN m 8.25 10 N m
M     
6
all
3
6 3 3 3
max
min 6
all
12 MPa 12 10 Pa
| | 8.25 10
687.5 10 m 687.5 10 mm
12 10
M
S



  

     

For a rectangular section, 2
1
6
S bh

3 2
3
2 3 2
1
687.5 10 (100)
6
(6)(687.5 10 )
41.25 10 mm
100
h
h
 
   
 

   203 mm
h  
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799. PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
760
A
B
150 mm
b
3 kN/m
C
2.4 m 1.2 m
PROBLEM 5.70
For the beam and loading shown, design the cross section of
the beam, knowing that the grade of timber used has an
allowable normal stress of 12 MPa.
SOLUTION
0: 2.4 (0.6)(3.6)(3) 0
B
M A
    2.7 kN
A 
0: (1.8)(3.6)(3) 2.4 0
A
M B
    8.1 kN
B 
Shear: 2.7 kN
2.7 (2.4)(3) 4.5 kN
4.5 8.1 3.6 kN
3.6 (1.2)(3) 0
A
B
B
C
V
V
V
V



   
   
  
Locate point D where 0.
V 
2.4
7.2 6.48
2.7 4.5
0.9 m 2.4 1.5 m
d d
d
d d

 
  
Areas of the shear diagram:
A to D:
1
(0.9)(2.7) 1.215 kN m
2
V dx
 
  
 
 

D to B:
1
(1.5)( 4.5) 3.375 kN m
2
V dx
 
    
 
 

B to C:
1
(1.2)(3.6) 2.16 kN m
2
V dx
 
  
 
 

Bending moments: 0
0 1.215 1.215 kN m
1.215 3.375 2.16 kN m
2.16 2.16 0
A
D
B
C
M
M
M
M

   
    
   
Maximum 3
| | 2.16 kN m 2.16 10 N m
M      6
all 12 MPa 12 10 Pa
   
3
6 3 3 3
min 6
all
| | 2.16 10
180 10 m 180 10 mm
12 10
M
S



     

For rectangular section, 2 2 3
1 1
(150) 180 10
6 6
S bh b
   
3
2
(6)(180 10 )
150
b

 48.0 mm
b  
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761
D
C
A
B E
F
6 ft
2 ft 2 ft 2 ft 2 ft
11 kips/ft 20 kips
20 kips
PROBLEM 5.71
Knowing that the allowable normal stress for the steel used is 24 ksi,
select the most economical wide-flange beam to support the loading
shown.
SOLUTION
By symmetry, 
A F
R R
0: 20 (6)(11) 20 0
y A F
F R R
      
50 kips
 
A F
R R
Maximum bending moment occurs at center of beam.
0: (7)(53) (5)(20) (1.5)(3)(11) 0
J J
M M
      
221.5 kip.ft 2658 kip in.
J
M   
3
max
min
all
2658
110.75 in
24
  
M
S

Shape S (in2
)
W24 68
 154
W21 62
 127
W18 76
 146
W16 77
 134
W14 82
 123
W12 96
 131
Use W21 62.
 
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762
2.75 kips/ft
24 kips
B
A C
9 ft 15 ft
PROBLEM 5.72
Knowing that the allowable normal stress for the steel used is 24 ksi, select
the most economical wide-flange beam to support the loading shown.
SOLUTION
0: 24 (12)(24)(2.75) (15)(24) 0
C
M A
      48 kips
A 
0: 24 (12)(24)(2.75) (9)(24) 0
A
M C
     42 kips
C 
Shear: 48
48 (9)(2.75) 23.25 kips
23.25 24 0.75 kips
0.75 (15)(2.75) 42 kips
A
B
B
C
V
V
V
V



  
   
    
Areas of the shear diagram:
A to B:
1
(9)(48 23.25) 320.6 kip ft
2
V dx
 
   
 
 

B to C:
1
(15)( 0.75 42) 320.6 kip ft
2
V dx
 
     
 
 

Bending moments: 0
0 320.6 320.6 kip ft
320.6 320.6 0
A
B
C
M
M
M

   
  
Maximum | | 320.6 kip ft 3848 kip in.
M    
all
3
min
all
24 ksi
| | 3848
160.3 in
24
M
S

  


3
Shape , (in )
W30 99 269
W27 84 213
W24 104 258
W21 101 227
W18 106 204
S

 



Lightest wide flange beam: W27 84
 @ 84 lb/ft 
Mecánica de Materiales - Ferdinand P. Beer 7ma. Edición.

802. PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
763
5 kN/m
70 kN 70 kN
A D
C
B
3 m 3 m
5 m
PROBLEM 5.73
Knowing that the allowable normal stress for the steel used is 160 MPa,
select the most economical wide-flange beam to support the loading shown.
SOLUTION
all
Section modulus
160 Mpa
 
6 3
max
min
all
3 3
286 kN m
1787 10 m
160 MPa
1787 10 mm
M
S 

   
 

Use W530 92.
 
Shape S, ( 3 3
10 mm )
W610 101
 2520
W530 92
 2080  
W460 113
 2390
W410 114
 2200
W360 122
 2020
W310 143
 2150
Mecánica de Materiales - Ferdinand P. Beer 7ma. Edición.

803. PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
764
C
D
A
B
0.8 m 0.8 m
2.4 m
50 kN/m PROBLEM 5.74
Knowing that the allowable normal stress for the steel used is 160 MPa,
select the most economical wide-flange beam to support the loading
shown.
SOLUTION
0: 3.2 (24)(3.2)(50) 0
D
M B
     120 kN
B 
0: 3.2 (0.8)(3.2)(50) 0
B
M D
    40 kN
D 
Shear: 0
0 (0.8)(50) 40 kN
40 120 80 kN
80 (2.4)(50) 40 kN
40 0 40 kN
A
B
B
C
D
V
V
V
V
V



   
   
   
    
Locate point E where 0.
V 
2.4
120 192
80 40
1.6 m 2.4 0.8 m
e e
e
e e

 
  
Areas:
1
to : (0.8)( 40) 16 kN m
2
1
to : (1.6)(80) 64 kN m
2
1
to : (0.8)( 40) 16 kN m
2
to : (0.8)( 40) 32 kN m
A B V dx
B E V dx
E C V dx
C D V dx
 
    
 
 
 
  
 
 
 
    
 
 
    




Bending moments: 0
0 16 16 kN m
16 64 48 kN m
48 16 32 kN m
32 32 0
A
B
E
C
D
M
M
M
M
M

    
    
   
  
Mecánica de Materiales - Ferdinand P. Beer 7ma. Edición.

804. PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
765
PROBLEM 5.74 (Continued)
Maximum 3
| | 48 kN m 48 10 N m
M     
6
all
3
6 3 3 3
min 6
all
160 MPa 160 10 Pa
| | 48 10
300 10 m 300 10 mm
160 10
M
S



  

     

3 3
Shape (10 mm )
W310 32.7 415
W250 28.4 308
W200 35.9 342
S

 

Lightest wide flange beam: W250 28.4 @ 28.4 kg/m
 
Mecánica de Materiales - Ferdinand P. Beer 7ma. Edición.

805. PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
766
3 kips/ft
18 kips
A
D
C
B
6 ft 6 ft
3 ft
PROBLEM 5.75
Knowing that the allowable normal stress for the steel used is 24 ksi,
select the most economical S-shape beam to support the loading shown.
SOLUTION
0: 12 (9)(6)(3) (3)(18) 0 9 kips
C
M A A
      
0: 12 (3)(6)(3) (15)(18) 0 27 kips
A
M C C
     
Shear: 9 kips
A
V 
B to C: 9 (6)(3) 9 kips
V    
C to D: 9 27 18 kips
V    
Areas:
A to E: (0.5)(3)(9) 13.5 kip ft
 
E to B: (0.5)(3)( 9) 13.5 kip ft
   
B to C: (6)( 9) 54 kip ft
   
C to D: (3)(18) 54 kip ft
 
Bending moments: 0
A
M 
0 13.5 13.5 kip ft
E
M    
13.5 13.5 0
B
M   
0 54 54 kip ft
C
M    
54 54 0
D
M   
all
Maximum 54 kip ft 648 kip in. 24 ksi
M     

3
min
648
27 in
24
S  
Shape 3
(in )
S
S12 31.8
 36.2
S10 35
 29.4
Lightest S-shaped beam: S12 31.8
 

Mecánica de Materiales - Ferdinand P. Beer 7ma. Edición.

806. PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
767
48 kips 48 kips 48 kips
A
D
E
C
B
6 ft
2 ft
2 ft
2 ft
PROBLEM 5.76
Knowing that the allowable normal stress for the steel used is 24 ksi,
select the most economical S-shape beam to support the loading shown.
SOLUTION
0: (12)(48) 10 (8)(48) (2)(48) 0 105.6 kips
E
M B
       
B
0: (2)(48) (2)(48) (8)(48) 10 0 38.4 kips
B
M E
       
E
Shear: A to B: 48 kips
V  
B to C: 48 105.6 57.6 kips
V    
C to D: 57.6 48 9.6 kips
V   
D to E: 9.6 48 38.4 kips
V    
Areas: A to B: (2)( 48) 96 kip ft
   
B to C: (2)(57.6) 115.2 kip ft
 
C to D: (6)(9.6) 57.6 kip ft
 
D to E: (2)( 38.4) 76.8 kip ft
  
Bending moments: 0
A
M 
0 96 96 kip ft
B
M     
96 115.2 19.2 kip ft
C
M     
19.2 57.2 76.8 kip ft
D
M    
76.8 76.8 0
E
M   
Maximum 96 kip ft 1152 kip in.
M    
all 24 ksi
 
3
min
all
1152
48 in
24
M
S

  
Shape 3
(in )
S
S15 42.9
