1.1.

Introduction
There are two major parts to this c hapter. Review of some important fundamentals of static s and
mec hanics of solids, the concept of stress, modes of load transmission, general sign c onvention for
stress and forc e resultants that will be used throughout the book, and analysis and design principles
are provided first. This is followed with treatment for c hanging the components of the state of stress
given in one set of c oordinate axes to any other set of rotated axes, as well as variation of stress
within and on the boundaries of a load-carrying member. Plane stress and its transformation are of
basic importanc e, since these c onditions are most c ommon in engineering practic e. The chapter is
thus also a brief guide and introduc tion to the remainder of the text.

Mechanics of Materials and Theory of Elasticity

The basic struc ture of matter is characterized by nonuniformity and discontinuity attributable to its
various subdivisions: molecules, atoms, and subatomic partic les. Our c oncern in this text is not with
the particulate struc ture, however, and it will be assumed that the matter with whic h we are
c oncerned is homogeneous and c ontinuously distributed over its volume. There is the c lear
implic ation in suc h an approach that the smallest element cut from the body possesses the same
properties as the body. Random fluctuations in the properties of the material are thus of no
c onsequenc e. This approach is that of c ontinuum mechanics, in which solid elastic materials are
treated as though they are continuous media rather than composed of discrete molecules. Of the
states of matter, we are here c onc erned only with the solid, with its ability to maintain its shape
without the need of a container and to resist continuous shear, tension, and c ompression.

In contrast with rigid-body statics and dynamics, which treat the external behavior of bodies (that
is, the equilibrium and motion of bodies without regard to small deformations associated with the
application of load), the mechanic s of solids is concerned with the relationships of external effec t
(forces and moments) to internal stresses and strains. Two different approac hes used in solid
mec hanics are the mec hanics of materials or elementary theory (also c alled the technical theory)
and the theory of elasticity. The mechanics of materials focuses mainly on the more or less
approximate solutions of prac tical problems. The theory of elastic ity c oncerns itself largely with more
mathematical analysis to determine the exac t stress and strain distributions in a loaded body. The
differenc e between these approaches is primarily in the nature of the simplifying assumptions used,
described inSection 3.1.

External forces acting on a body may be classified as surfac e forc es and body forc es. A surfac e
force is of the c onc entratedtype when it ac ts at a point; a surface forc e may also be
distributed uniformly or nonuniformly over a finite area. Body forc es are associated with the mass
rather than the surfaces of a body, and are distributed throughout the volume of a body.
Gravitational, magnetic , and inertia forces are all body forc es. They are specified in terms of forc e
per unit volume. All forces acting on a body, including the reactive forc es c aused by supports and
body forc es, are c onsidered to be external forces. Internal forc es are the forc es that hold together
the particles forming the body. Unless otherwise stated, we assume in this text that body forces c an
be neglec ted and that forces are applied steadily and slowly. The latter is referred to as static
loading.

In the International System of Units (SI), forc e is measured in newtons (N). Because the newton is a
small quantity, the kilonewton (kN) is often used in prac tice. In the U.S. Customary System, forc e is
expressed in pounds (lb) or kilopounds (kips). We define all important quantities in both systems of
units. However, in numerical examples and problems, SI units are used throughout the text
c onsistent with international c onvention. (Table D.2 compares the two systems.)

Historical Development
The study of the behavior of members in tension, compression, and bending began with Leonardo da
Vinci (14521519) and Galileo Galilei (15641642). For a proper understanding, however, it was
nec essary to establish accurate experimental desc ription of a materials properties. Robert Hooke
(16151703) was the first to point out that a body is deformed subject to the ac tion of a force. Sir
Isaac Newton (16421727) developed the concepts of Newtonian mechanics that bec ame key
elements of the strength of materials.

Leonard Euler (17071783) presented the mathematic al theory of columns in 1744. The renowned
mathematician Joseph-Louis Lagrange (17361813) received credit in developing a partial differential
equation to describe plate vibrations. Thomas Young (17731829) established a coefficient of
elasticity, Youngs modulus. The advent of railroads in the late 1800s provided the impetus for muc h
of the basic work in this area. Many famous scientists and engineers, including Coulomb, Poisson,
Navier, St. Venant, Kirchhoff, and Cauc hy, were responsible for advanc es in mechanic s of materials
during the eighteenth and nineteenth centuries. The British physic ist William Thomas Kelvin (1824
1907), better known by his knighted name, Sir Lord Kelvin, first demonstrated that torsional moments
acting at the edges of plates could be dec omposed into shearing forces. The prominent English
mathematician Augustus Edward Hough Love (18631940) introduced simple analysis of shells, known
as Loves approximate theory.

Over the years, most basic problems of solid mec hanics had been solved. Stephan P. Timoshenko
(18781972) made numerous original contributions to the field of applied mechanic s and wrote
pioneering textbooks on the mec hanics of materials, theory of elasticity, and theory of elastic
stability. The theoretic al base for modern strength of materials had been developed by the end of
the nineteenth century. Following this, problems assoc iated with the design of airc raft, spac e
vehicles, and nuc lear reac tors have led to many studies of the more advanced phases of the
subject. Consequently, the mechanic s of materials is being expanded into the theories of elastic ity
and plasticity.

In 1956, Turner, Clough, Martin, and Topp introduc ed the finite element method, which permits the
numeric al solution of c omplex problems in solid mec hanics in an economical way. Many contributions
in this area are owing to Argyris and Zienkiewicz. The rec ent trend in the development is
c harac terized by heavy reliance on high-speed c omputers and by the introduc tion of more rigorous
theories. Numerical methods presented in Chapter 7 and applied in the chapters following have clear
application to c omputation by means of electronic digital computers. Research in the foregoing areas
is ongoing, not only to meet demands for treating c omplex problems but to justify further use and
limitations on whic h the theory of solid mechanic s is based. Although a widespread body of
knowledge exists at present, mec hanics of materials and elasticity remain fascinating subjects as
their areas of application are continuously expanded.[*] The literature dealing with various aspects of
solid mec hanics is voluminous. For those seeking more thorough treatment, selected references are
identified in brac kets and c ompiled at the end of each chapter.
[*]
Historica l re views of m e cha nics of m a teria ls and the theory of ela sticity are give n in
R efs. 1.1 through 1.5.
1.2. Scope of Treatment
As stated in the preface, this book is intended for advanc ed undergraduate and graduate engineering
students as well as engineering professionals. To make the text as c lear as possible, attention is
given to the fundamentals of solid mechanic s and c hapter objectives. A special effort has been made
to illustrate important principles and applications with numerical examples. Emphasis is placed on a
thorough presentation of several classic al topic s in advanced mechanic s of materials and applied
elasticity and of selec ted advanced topic s. Understanding is based on the explanation of the physic al
behavior of members and then modeling this behavior to develop the theory.

The usual objective of mec hanics of material and theory of elasticity is the examination of the load-
c arrying capacity of a body from three standpoints: strength, stiffness, and stability. Recall that
these quantities relate, respectively, to the ability of a member to resist permanent deformation or
fracture, to resist deflection, and to retain its equilibrium configuration. For instance, when loading
produc es an abrupt shape c hange of a member, instability oc curs; similarly, an inelastic deformation
or an excessive magnitude of deflection in a member will cause malfunc tion in normal service. The
foregoing matters, by using the fundamental principles (Sec. 1.3), are discussed in later c hapters for
various types of struc tural members. Failure by yielding and frac ture of the materials under c ombined
loading is taken up in detail in Chapter 4.

Our main concern is the analysis of stress and deformation within a loaded body, whic h is
acc omplished by applic ation of one of the methods described in the next sec tion. For this purpose,
the analysis of loads is essential. A struc ture or mac hine c annot be satisfac tory unless its design is
based on realistic operating loads. The princ ipal topics under the heading of mechanic s of solidsmay
be summarized as follows:

1. Analysis of the stresses and deformations within a body subjec t to a prescribed system of
forces. This is acc omplished by solving the governing equations that describe the stress and
strain fields (theoretical stress analysis). It is often advantageous, where the shape of the
structure or conditions of loading prec lude a theoretical solution or where verification is
required, to apply the laboratory techniques of experimental stress analysis.

2. Determination by theoretic al analysis or by experiment of the limiting values of load that a

structural element c an sustain without suffering damage, failure, or compromise of func tion.

3. Determination of the body shape and selection of the materials that are most efficient for
resisting a presc ribed system of forces under specified conditions of operation such as
temperature, humidity, vibration, and ambient pressure. This is the design function.

The design func tion, item 3, clearly relies on the performanc e of the theoretical analyses under items
1 and 2, and it is to these that this text is directed. Particularly, emphasis is plac ed on the
development of the equations and methods by which detailed analysis can be acc omplished.

The ever-inc reasing industrial demand for more sophisticated struc tures and mac hines calls for a
good grasp of the c onc epts of stress and strain and the behavior of materialsand a considerable
degree of ingenuity. This text, at the very least, provides the student with the ideas and information
nec essary for an understanding of the advanced mec hanics of solids and enc ourages the c reative
process on the basis of that understanding. Complete, carefully drawn free-body diagrams facilitate
visualization, and these we have provided, all the while knowing that the subjec t matter can be
learned best only by solving problems of prac tical importanc e. A thorough grasp of fundamentals will
prove of great value in attac king new and unfamiliar problems.
1.3. Analysis and Design
Throughout this text, a fundamental proc edure for analysis in solving mec hanics of solids problems is
used repeatedly. The complete analysis of load-carrying structural members by the method of
equilibrium requires consideration of three conditions relating to c ertain laws of forces, laws of
material deformation, and geometric compatibility. These essential relationships, c alled the basic
princ iples of analysis, are:

1. Equilibrium Conditions. The equations of equilibrium of forces must be satisfied throughout

the member.

2. Material Behavior. The stressstrain or forc e-deformation relations (for example, Hookes
law) must apply to the material behavior of which the member is c onstructed.

3. Geometry of Deformation. The c ompatibility conditions of deformations must be satisfied:

that is, each deformed portion of the member must fit together with adjacent portions. (Matter
of compatibility is not always broac hed in mechanic s of materials analysis.)

The stress and deformation obtained through the use of the three princ iples must conform to the
c onditions of loading imposed at the boundaries of a member. This is known as satisfying
the boundary conditions. Applications of the prec eding procedure are illustrated in the problems
presented as the subject unfolds. Note, however, that it is not always nec essary to exec ute an
analysis in the exac t order of steps listed previously.

As an alternative to the equilibrium methods, the analysis of stress and deformation c an be

acc omplished by employing energy methods (Chap. 10), which are based on the c oncept of strain
energy. The aspect of both the equilibrium and the energy approac hes is twofold. These methods
c an provide solutions of acceptable ac curac y where configurations of loading and member shape are
regular, and they can be used as the basis of numerical methods (Chap. 7) in the solution of more
realistic problems.

Engineering design is the proc ess of applying sc ience and engineering techniques to define a
structure or system in detail to allow its realization. The objective of a mechanic al design proc edure
includes finding of proper materials, dimensions, and shapes of the members of a structure or
mac hine so that they will support presc ribed loads and perform without failure. Machine design is
c reating new or improved machines to accomplish specific purposes. Usually, structural design deals
with any engineering disc ipline that requires a struc tural member or system.

Design is the essence, art, and intent of engineering. A good design satisfies performance, c ost, and
safety requirements. Anoptimum design is the best solution to a design problem within given
restric tions. Effic ienc y of the optimization may be gaged by such criteria as minimum weight or
volume, optimum cost, and/or any other standard deemed appropriate. For a design problem with
many c hoices, a designer may often make decisions on the basis of experience, to reduce the
problem to a single variable. A solution to determine the optimum result becomes straightforward in
such a situation.

A plan for satisfying a need usually includes preparation of individual preliminary design.
Eac h preliminary design involves a thorough c onsideration of the loads and ac tions that the
structure or machine has to support. For eac h situation, an analysis is necessary. Design dec isions,
or choosing reasonable values of the safety factors and material properties, are significant in the
preliminary design process.

The role of analysis in design may be observed best in examining the phases of a design process.
This text provides an elementary treatment of the concept of design to meet strength
requirements as those requirements relate to individual machine or structural components. That is,
the geometrical c onfiguration and material of a c omponent are preselec ted and the applied loads are
specified. Then, the basic formulas for stress are employed to select members of adequate size in
eac h c ase. The following is rational procedure in the design of a load-carrying member:

1. Evaluate the most likely modes of failure of the member. Failure criteria that predict the
various modes of failure under anticipated conditions of service are disc ussed in Chapter 4.

2. Determine the expressions relating applied loading to suc h effects as stress, strain, and
deformation. Often, the member under consideration and conditions of loading are so
signific ant or so amenable to solution as to have been the subjec t of prior analysis. For these
situations, textbooks, handbooks, journal articles, and technical papers are good sourc es of
information. Where the situation is unique, a mathematical derivation specific to the c ase at
hand is required.

3. Determine the maximum usable value of stress, strain, or energy. This value is obtained either
by referenc e to c ompilations of material properties or by experimental means such as simple
tension test and is used in c onnec tion with the relationship derived in step 2.

4. Select a design factor of safety. This is to acc ount for unc ertainties in a number of aspec ts of
the design, including those related to the ac tual service loads, material properties, or
environmental factors. An important area of uncertainty is connec ted with the assumptions
made in the analysis of stress and deformation. Also, we are not likely to have a sec ure
knowledge of the stresses that may be introduced during machining, assembly, and shipment of
the element.

The design factor of safety also reflects the consequenc es of failure; for example, the
possibility that failure will result in loss of human life or injury or in costly repairs or danger to
other c omponents of the overall system. For these reasons, the design fac tor of safety is also
sometimes c alled the fac tor of ignorance. The uncertainties enc ountered during the design
phase may be of such magnitude as to lead to a design c arrying extreme weight, volume, or
c ost penalties. It may then be advantageous to perform thorough tests or more exacting
analysis rather to rely on overly large design fac tors of safety.

The true factor of safety, usually referred to simply as the factor of safety, c an be determined
only after the member is constructed and tested. This factor is the ratio of the maximum load
the member c an sustain under severe testing without failure to the maximum
load ac tually c arried under normal service conditions, the working load. When a linear
relationship exists between the load and the stress produced by the load, the factor of safety
n may be expressed as

Equation 1.1

Maximum usable stress represents either the yield stress or the ultimate stress. The allowable
stress is the working stress. The factor of safety must be greater than 1.0 if failure is to be
avoided. Values for factor of safety, selec ted by the designer on the basis of experience and
judgment, are about 1.5 or greater. For most applic ations, appropriate factors of safety are
found in various construc tion and manufacturing codes.

The foregoing proc edure is not always conducted in as formal a fashion as may be implied. In some
design proc edures, one or more steps may be regarded as unnec essary or obvious on the basis of
previous experience. Suffice it to say that c omplete design solutions are not unique, involve a
c onsideration of many factors, and often require a trial-and-error proc ess [Ref. 1.6]. Stress is only
one consideration in design. Other phases of the design of c omponents are the prediction of the
deformation of a given c omponent under given loading and the c onsideration of buckling (Chap. 11).
The methods of determining deformation are discussed in later chapters. Note that there is a very
c lose relationship between analysis and design, and the examples and problems that appear
throughout this book illustrate that connection.

We conc lude this section with an appeal for the reader to exercise a degree of skepticism with
regard to the applic ation of formulas for whic h there is unc ertainty as to the limitations of use or the
areas of applic ability. The relatively simple form of many formulas usually results from rather severe
restric tions in its derivation. These relate to simplified boundary conditions and shapes, limitations on
stress and strain, and the neglec t of c ertain complic ating factors. Designers and stress analysts
must be aware of such restric tions lest their work be of no value or, worse, lead to dangerous
inadequac ies.

In this chapter, we are concerned with the state of stress at a point and the variation of
stress throughout an elastic body. The latter is dealt with in Sections 1.8 and 1.16 and the former
in the balance of the c hapter.
1.4. Conditions of Equilibrium
A structure is a unit consisting of interconnected members supported in such a way that it is
c apable of c arrying loads in static equilibrium. Structures are of four general types: frames, trusses,
mac hines, and thin-walled (plate and shell) structures. Framesand mac hines are struc tures
c ontaining multiforce members. The former support loads and are usually stationary, fully restrained
structures. The latter transmit and modify forces (or power) and always c ontain moving parts.
The truss provides both a practic al and ec onomic al solution, particularly in the design of bridges and
buildings. When the truss is loaded at its joints, the only forc e in each member is an axial forc e,
either tensile or compressive.

The analysis and design of structural and mac hine c omponents require a knowledge of the
distribution of forc es within such members. Fundamental concepts and conditions of static equilibrium
provide the nec essary background for the determination of internal as well as external forces.
In Section 1.6, we shall see that components of internal-forc es resultants have spec ial meaning in
terms of the type of deformations they cause, as applied, for example, to slender members. We note
that surface forc es that develop at support points of a structure are called reactions. They
equilibrate the effec ts of the applied loads on the structures.

The equilibrium of forces is the state in which the forces applied on a body are in balance.
Newtons first law states that if the resultant forc e acting on a particle (the simplest body) is zero,
the particle will remain at rest or will move with constant veloc ity. Statics is c onc erned essentially
with the case where the particle or body remains at rest. A complete free-body diagram is essential
in the solution of problems concerning the equilibrium.

Let us c onsider the equilibrium of a body in spac e. In this three-dimensional case, the conditions of
equilibrium require the satisfaction of the following equations of statics:

Equation 1.2

The foregoing state that the sum of all forc es acting on a body in any direction must be zero; the
sum of all moments about any axis must be zero.

In a planar problem, where all forces act in a single (xy) plane, there are only three independent
equations of static s:

Equation 1.3

That is, the sum of all forc es in any (x, y) directions must be zero, and the resultant moment about
axis z or any point A in the plane must be zero. By replacing a force summation with an equivalent
moment summation in Eqs. (1.3), the following alternativesets of conditions are obtained:

Equation 1.4a
provided that the line c onnec ting the points A and B is not perpendic ular to the x axis, or

Equation 1.4b

Here points A, B, and C are not c ollinear. Clearly, the judic ious selec tion of points for taking moments
c an often simplify the algebraic computations.

A struc ture is statically determinate when all forces on its members c an be found by using only the
c onditions of equilibrium. If there are more unknowns than available equations of statics, the problem
is called statically indeterminate. The degree of static indeterminacy is equal to the difference
between the number of unknown forces and the number of relevant equilibrium c onditions. Any
reaction that is in exc ess of those that can be obtained by statics alone is termed redundant. The
number of redundants is therefore the same as the degree of indeterminac y.
1.5. Definition and Components of Stress
Stress and strain are most important c onc epts for a comprehension of the mec hanics of solids. They
permit the mec hanical behavior of load-c arrying c omponents to be desc ribed in terms fundamental to
the engineer. Both the analysis and design of a given mac hine or structural element involve the
determination of stress and material stressstrain relationships. The latter is taken up in Chapter 2.

Consider a body in equilibrium subject to a system of external forces, as shown in Fig. 1.1a. Under
the action of these forc es, internal forces are developed within the body. To examine the latter at
some interior point Q, we use an imaginary plane to c ut the body at a sec tion aa through Q, dividing
the body into two parts. As the forc es ac ting on the entire body are in equilibrium, the forc es acting
on one part alone must be in equilibrium: this requires the presence of forces on plane aa. These
internal forces, applied to both parts, are distributed continuously over the cut surfac e. This
process, referred to as the method of sections (Fig. 1.1), is relied on as a first step in solving all
problems involving the investigation of internal forc es.

Figure 1.1. Method of sections: (a) Sectioning of a loaded body; (b) free body with external
and internal forces; (c) enlarged area A with components of the force F.

[View full size im age ]

A free-body diagram is simply a sketch of a body with all the appropriate forc es, both known and
unknown, acting on it. Figure 1.1b shows such a plot of the isolated left part of the body. An
element of area A, loc ated at point Q on the c ut surface, is acted on by force F. Let the origin of
c oordinates be placed at point Q, with x normal and y, z tangent to A. In general, Fdoes not lie
along x, y, or z.

Decomposing F into c omponents parallel to x, y, and z (Fig. 1.1c), we define the normal
stress x and the shearing stresses xy and xz:

Equation 1.5

These definitions provide the stress components at a point Q to which the area A is reduced in the
limit. Clearly, the expressionA 0 depends on the idealization discussed in Section 1.1. Our
c onsideration is with the average stress on areas, which, while small as compared with the size of
the body, is large compared with interatomic distances in the solid. Stress is thus defined adequately
for engineering purposes. As shown in Eq. (1.5), the intensity of force perpendicular, or normal, to
the surface is termed the normal stress at a point, while the intensity of force parallel to the surfac e
is the shearing stress at a point.

The values obtained in the limiting process of Eq. (1.5) differ from point to point on the surface
as F varies. The stress components depend not only on F, however, but also on the orientation of
the plane on which it acts at point Q. Even at a given point, therefore, the stresses will differ as
different planes are considered. The complete description of stress at a point thus requires the
specification of the stress on all planes passing through the point.

Bec ause the stress ( or ) is obtained by dividing the force by area, it has units of force per unit
area. In SI units, stress is measured in newtons per square meter (N/m2), or pascals (Pa). As the
pascal is a very small quantity, the megapascal (MPa) is commonly used. When U.S. Customary
System units are used, stress is expressed in pounds per square inch (psi) or kips per square inch
(ksi).

It is verified in Section 1.12 that in order to enable the determination of the stresses on an infinite
number of planes passing through a point Q, thus defining the stresses at that point, we need only
specify the stress components on three mutually perpendicular planes passing through the point.
These three planes, perpendic ular to the c oordinate axes, contain three hidden sides of an
infinitesimal cube (Fig. 1.2). We emphasize that when we move from point Q to point Q the values
of stress will, in general, change. Also, body forces can exist. However, these cases are not
discussed here (see Sec. 1.8), as we are now merely interested in establishing the terminology
nec essary to spec ify a stress component.

Figure 1.2. Element subjected to three-dimensional stress. All stresses have positive sense.

The general c ase of a three-dimensional state of stress is shown in Fig. 1.2. Consider the stresses
to be identical at points Qand Q and uniformly distributed on each face, represented by a single
vec tor ac ting at the center of each face. In acc ordance with the foregoing, a total of nine
sc alar stress components defines the state of stress at a point. The stress c omponents can be
assembled in the following matrix form, wherein each row represents the group of stresses ac ting on
a plane passing throughQ(x, y, z):

Equation 1.6
We note that in indic ial notation (refer to Sec. 1.17), a stress c omponent is written as ij, where the
subscripts i and j each assume the values of x, y, and z as required by the foregoing equation.
The double subsc ript notation is interpreted as follows: The first subscript indicates the direc tion of
a normal to the plane or fac e on which the stress c omponent ac ts; the second subsc ript relates to
the direction of the stress itself. Repetitive subsc ripts are avoided in this text, so the normal
stresses xx , yy , and zz are designated x , y , and z, as indic ated in Eq. (1.6). A fac e or plane is
usually identified by the axis normal to it; for example, the x fac es are perpendic ular to the x axis.

Sign Convention
Referring again to Fig. 1.2, we observe that both stresses labeled yx tend to twist the element in a
c loc kwise direc tion. It would be c onvenient, therefore, if a sign c onvention were adopted under
whic h these stresses c arried the same sign. Applying a convention relying solely on the c oordinate
direction of the stresses would clearly not produc e the desired result, inasmuch as the yx stress
acting on the upper surfac e is directed in the positive x direc tion, while yx acting on the lower
surface is directed in the negative x direction. The following sign convention, whic h applies to both
normal and shear stresses, is related to the deformational influence of a stress and is based on the
relationship between the direc tion of an outward normal drawn to a particular surface and the
directions of the stress components on the same surface.

When both the outer normal and the stress component fac e in a positive direction relative to the
c oordinate axes, the stress is positive. When both the outer normal and the stress component fac e
in a negative direction relative to the coordinate axes, the stress is positive. When the normal points
in a positive direction while the stress points in a negative direc tion (or vice versa), the stress is
negative. In acc ordance with this sign convention, tensile stresses are always positive and
c ompressive stresses always negative. Figure 1.2 depic ts a system of positive normal and shear
stresses.

Equality of Shearing Stresses

We now examine properties of shearing stress by studying the equilibrium of forc es (see Sec. 1.4)
acting on the cubic element shown in Fig. 1.2. As the stresses acting on opposite faces (whic h are
of equal area) are equal in magnitude but opposite in direc tion, translational equilibrium in all
directions is assured; that is, F x = 0, F y = 0, and F z = 0. Rotational equilibrium is established by
taking moments of the x-, y-, and z-direc ted forces about point Q, for example. From Mz = 0,

( xy dy dz)dx + ( yx dx dz)dy = 0

Simplifying,

Equation 1.7a
Likewise, from My = 0 and Mx = 0, we have

Equation 1.7b

Henc e, the subsc ripts for the shearing stresses are commutative, and the stress tensor is symmetric .
This means that shearing stresses on mutually perpendic ular planes of the element are equal.
Therefore, no distinction will hereafter be made between the stress
c omponents xy and yx , xz and zx , or yz and zy . In Section 1.8, it is shown rigorously that the
foregoing is valid even when stress components vary from one point to another.

Some Special Cases of Stress

Under particular circ umstances, the general state of stress (Fig. 1.2) reduces to simpler stress
states, as briefly described here. These stresses, whic h are commonly encountered in prac tice, are
given detailed consideration throughout the text.

a. Triaxial Stress. We shall observe in Section 1.13 that an element subjec ted to only
stresses 1, 2, and 3 acting in mutually perpendic ular directions is said to be in a state of
triaxial stress. Such a state of stress c an be written as

Equation a

The absence of shearing stresses indicates that the preceding stresses are the principal
stresses for the element. A special case of triaxial stress, known as spherical or dilatational
stress, oc curs if all principal stresses are equal (see Sec. 1.14). Equal triaxial tension is
sometimes c alled hydrostatic tension. An example of equal triaxial compression is found in a
small element of liquid under static pressure.

b. Two-Dimensional or Plane Stress. In this case, only the x and y faces of the element are
subjected to stress, and all the stresses ac t parallel to the x and y axes, as shown in Fig.
1.3a. The plane stress matrix is written

Equation 1.8

Figure 1.3. (a) Element in plane stress; (b) two-dimensional presentation of

plane stress; (c) element in pure shear.

[View full size image]

 Although the three-dimensional nature of the element under stress should not be forgotten, for
the sake of c onvenienc e we usually draw only a two-dimensional view of the plane stress
element (Fig. 1.3b). When only two normal stresses are present, the state of stress is
c alled biaxial. These stresses oc cur in thin plates stressed in two mutually perpendicular
directions.

c. Pure Shear. In this c ase, the element is subjec ted to plane shearing stresses only, for
example, xy and yx (Fig. 1.3c). Typical pure shear occ urs over the cross sections and on
longitudinal planes of a c ircular shaft subjected to torsion.

d. Uniaxial Stress. When normal stresses ac t along one direction only, the one-dimensional state
of stress is referred to as a uniaxial tension or compression.
1.6. Internal Force-Resultant and Stress Relations
Distributed forces within a load-carrying member can be represented by a static ally equivalent
system c onsisting of a forc e and a moment vector acting at any arbitrary point (usually the centroid)
of a sec tion. These internal force resultants, also called stress resultants, exposed by an imaginary
c utting plane c ontaining the point through the member, are usually resolved into c omponents normal
and tangent to the c ut section (Fig. 1.4). The sense of moments follows the right-hand sc rew rule,
often represented by double-headed vec tors, as shown in the figure. Each component c an be
associated with one of four modes of forc e transmission:

1. The axial forc e P or N tends to lengthen or shorten the member.

2. The shear forces Vy and Vz tend to shear one part of the member relative to the adjac ent part
and are often designated by the letter V.

3. The torque or twisting moment T is responsible for twisting the member.

4. The bending moments My and Mz cause the member to bend and are often identified by the
letter M.

Figure 1.4. Positive forces and moments on a cut section of a body and components of the
force dF on an infinitesimal area dA.

A member may be subject to any or all of the modes simultaneously. Note that the same sign
c onvention is used for the forc e and moment c omponents that is used for stress; a positive force (or
moment) component acts on the positive fac e in the positive coordinate direction or on a negative
fac e in the negative coordinate direction.

A typical infinitesimal area dA of the cut sec tion shown in Fig. 1.4 is ac ted on by the components of
an arbitrarily directed forcedF, expressed using Eq. (1.5) as dF x = x dA, dF y = xy dA,
and dF z = xz dA. Clearly, the stress components on the c ut section c ause the internal force
resultants on that section. Thus, the incremental forces are summed in the x, y, and z directions to
give

Equation 1.9a
In a like manner, the sums of the moments of the same forces about the x, y, and z axes lead to

Equation 1.9b

where the integrations proceed over area A of the cut sec tion. Equations (1.9) represent
the relations between the internal forc e resultants and the stresses. In the next paragraph, we
illustrate the fundamental c oncept of stress and observe how Eqs. (1.9) c onnec t internal force
resultants and the state of stress in a spec ific case.

Consider a homogeneous prismatic bar loaded by axial forces P at the ends (Fig. 1.5a).
A prismatic bar is a straight member having c onstant c ross-sectional area throughout its length. To
obtain an expression for the normal stress, we make an imaginary cut (sec tion aa) through the
member at right angles to its axis. A free-body diagram of the isolated part is shown in Fig. 1.5b,
wherein the stress is substituted on the cut sec tion as a replacement for the effec t of the removed
part. Equilibrium of axial forces requires that P = x dA or P = A x . The normal stress is therefore

Equation 1.10

Figure 1.5. (a) Prismatic bar in tension; (b) Stress distribution across cross section.

[View full size im age ]

where A is the cross-sec tional area of the bar. Bec ause Vy , Vz, and T all are equal to zero, the
second and third of Eqs. (1.9a) and the first of Eqs. (1.9b) are satisfied by xy = xz = 0.
Also, My = Mz = 0 in Eqs. (1.9b) requires only that x be symmetrically distributed about
the y and z axes, as depicted in Fig. 1.5b. When the member is being extended as in the figure, the
resulting stress is a uniaxial tensile stress; if the direction of forces were reversed, the bar would be
in c ompression under uniaxial c ompressive stress. In the latter c ase, Eq. (1.10) is applicable only to
c hunky or short members owing to other effects that take place in longer members.[*]
[*]
Further discussion of unia x ia l com pre ssion stress is found in Section 11.6, where we ta k e up the
classification of colum ns.
Similarly, application of Eqs. (1.9) to torsion members, beams, plates, and shells is presented as the
subject unfolds, following the derivation of stressstrain relations and examination of the geometric
behavior of a particular member. Applying the method of mechanic s of materials, we develop
other elementary formulas for stress and deformation. These, also called the basic formulas of
mechanic s of materials, are often used and extended for application to more c omplex problems in
advanc ed mechanic s of materials and the theory of elasticity. For reference purposes to preliminary
discussions, Table 1.1 lists some commonly encountered cases. Note that in thin-walled vessels
(r/t 10) there is often no distinc tion made between the inner and outer radii because they are
nearly equal. In mechanics of materials, r denotes the inner radius. However, the more accurate shell
theory (Sec . 13.11) is based on the average radius, whic h we use throughout this text. Eac h
equation presented in the table desc ribes a state of stress assoc iated with a single force, torque,
moment component, or pressure at a section of a typical homogeneous and elastic structural member
[Ref. 1.7]. When a member is acted on simultaneously by two or more load types, c ausing various
internal force resultants on a section, it is assumed that eac h load produces the stress as if it were
the only load ac ting on the member. The final or combined stress is then determined by superposition
of the several states of stress, as discussed in Section 2.2.

Table 1.1. Commonly Used Elementary Formulas for Stress[a]

1. Prismatic Bars of Linearly Elastic Material

[View full size image]

where

x = normal axial stress I = moment of inertia about neutral

axis (N.A.)
= shearing stress due to torque

xy = shearing stress due to vertical J = polar moment of inertia of circular

shear forc e c ross sec tion

P = axial force b = width of bar at which xy is

c alculated
T = torque

V = vertical shear force r = radius

M = bending moment about z axis Q = first moment about N.A. of the

area beyond the point at
A = c ross-sectional area which xy is calculated
 y, z = c entroidal principal axes of the
area

2. Thin-Walled Pressure Vessels

[View full size image]

where

= tangential stress in c ylinder wall p = internal pressure

a = axial stress in c ylinder wall t = wall thickness

= membrane stress in sphere wall r = mean radius

[a]De taile d de rivations a nd lim ita tions of the use of these form ula s a re discusse d in Sections
1.6, 5.7, 6.2, a nd 13.13.

The mec hanics of materials theory is based on the simplifying assumptions related to the pattern of
deformation so that the strain distributions for a cross sec tion of the member can be determined. It
is a basic assumption that plane sec tions before loading remain plane after loading. The assumption
c an be shown to be exac t for axially loaded prismatic bars, for prismatic circ ular torsion members,
and for prismatic beams subjected to pure bending. The assumption is approximate for other beam
situations. However, it is emphasized that there is an extraordinarily large variety of c ases in which
applications of the basic formulas of mec hanics of materials lead to useful results. In this text we
hope to provide greater insight into the meaning and limitations of stress analysis by solving problems
using both the elementary and exact methods of analysis.
1.7. Stresses on Inclined Sections
The stresses in bars, shafts, beams, and other structural members c an be obtained by using the
basic formulas, such as those listed in Table 1.1. The values found by these equations are for
stresses that oc cur on cross sections of the members. Recall that all of the formulas for stress are
limited to isotropic , homogeneous, and elastic materials that behave linearly. This sec tion deals with
the states of stress at points loc ated on inclined sections or planes under axial loading. As before,
we use stress elements to represent the state of stress at a point in a member. However, we now
wish to find normal and shear stresses acting on the sides of an element in any direction.

The directional nature of more general states of stress and finding maximum and minimum values of
stress are disc ussed inSections 1.10 and 1.13. Usually, the failure of a member may be brought
about by a c ertain magnitude of stress in a certain direc tion. For proper design, it is necessary to
determine where and in what direction the largest stress oc curs. The equations derived and the
graphic al technique introduced here and in the sections to follow are helpful in analyzing the stress
at a point under various types of loading. Note that the transformation equations for stress are
developed on the basis of equilibrium c onditions only and do not depend on material properties or on
the geometry of deformation.

Axially Loaded Members

We now consider the stresses on an inclined plane aa of the bar in uniaxial tension shown in Fig.
1.6a, where the normal x to the plane forms an angle with the axial direction. On an isolated part
of the bar to the left of section aa, the resultant P may be resolved into two c omponents: the
normal force Px = P cos and the shear forc e Py = P sin , as indicated in Fig. 1.6b. Thus, the
normal and shearing stresses, uniformly distributed over the area Ax = A/c os of the inclined plane
(Fig. 1.6c), are given by

Equation 1.11a

Equation 1.11b

Figure 1.6. (a) Prismatic bar in tension; (b, c) side views of a part cut from the bar.
where x = P/A. The negative sign in Eq. (1.11b) agrees with the sign convention for shearing
stresses desc ribed in Section 1.5. The foregoing process of determining the stress in proc eeding
from one set of coordinate axes to another is c alled stress transformation.

Equations (1.11) indicate how the stresses vary as the inclined plane is cut at various angles. As
expec ted, x is a maximum ( max ) when is 0 or 180, and xy is maximum ( max ) when is 45
or 135. Also, . The maximum stresses are thus

Equation 1.12

Observe that the normal stress is either maximum or a minimum on planes for which the shearing
stress is zero.

Figure 1.7 shows the manner in which the stresses vary as the section is cut at angles varying
from = 0 to 180. Clearly, when > 90, the sign of xy in Eq. (1.11b) changes; the shearing
stress changes sense. However, the magnitude of the shearing stress for any angle determined
from Eq. (1.11b) is equal to that for + 90. This agrees with the general conclusion reached in the
preceding section: shearing stresses on mutually perpendicular planes must be equal.

Figure 1.7. Example 1.1. Variation of stress at a point with the inclined section in the bar
shown in Fig. 1.6a.
We note that Eqs. (1.11) can also be used for uniaxial compression by assigning to P a negative
value. The sense of each stress direction is then reversed in Fig. 1.6c.

Example 1.1. State of Stress in a Tensile Bar

Compute the stresses on the inclined plane with = 35 for a prismatic bar of a cross-
sec tional area 800 mm2, subjected to a tensile load of 60 kN (Fig. 1.6a). Then
determine the state of stress for = 35 by calc ulating the stresses on an adjoining
face of a stress element. Sketc h the stress configuration.

Solution

The normal stress on a cross section is

Introduc ing this value in Eqs. (1.11) and using = 35, we have

x = x cos2 = 75 (cos 35)2 = 50.33 MPa

xy = x sin c os = 75(sin 35)(cos 35) = 35.24 MPa

The normal and shearing stresses acting on the adjoining y face are, respectively, 24.67
MPa and 35.24 MPa, as calculated from Eqs. (1.11) by substituting the angle + 90 =
125. The values of x and xy are the same on opposite sides of the element. On the
basis of the established sign c onvention for stress, the required sketch is shown in Fig.
1.8.

Figure 1.8. Example 1.1. Stress element for = 35.

1.8. Variation of Stress within a Body
As pointed out in Section 1.5, the components of stress generally vary from point to point
in a stressed body. These variations are governed by the conditions of equilibrium of statics.
Fulfillment of these conditions establishes certain relationships, known as the differential
equations of equilibrium, which involve the derivatives of the stress components.

Consider a thin element of sides dx and dy (Fig. 1.9), and assume that x , y , xy ,
and yx are func tions of x, y but do not vary throughout the thickness (are independent
of z) and that the other stress c omponents are zero. Also assume that
the x and ycomponents of the body forces per unit volume, F x and F y , are independent
of z and that the z component of the body force F z= 0. This combination of stresses,
satisfying the conditions described, is the plane stress. Note that because the element is
very small, for the sake of simplicity, the stress components may be considered to be
distributed uniformly over each face. In the figure they are shown by a single vector
representing the mean values applied at the c enter of each face.

Figure 1.9. Element with stresses and body forces.

As we move from one point to another, for example, from the lower-left c orner to the
upper-right corner of the element, one stress component, say x , ac ting on the
negative x fac e, changes in value on the positive x face. The stresses y , xy ,
and yxsimilarly change. The variation of stress with position may be expressed by a
truncated Taylors expansion:

Equation a

The partial derivative is used bec ause x is a function of x and y. Treating all the
components similarly, the state of stress shown in Fig. 1.9 is obtained.

We c onsider now the equilibrium of an element of unit thickness, taking moments of force
about the lower-left c orner. Thus, Mz= 0 yields
Neglecting the triple produc ts involving dx and dy, this reduc es to xy = yx . In a like
manner, it may be shown that yz = zy and xz = zx , as already obtained in Section 1.5.
From the equilibrium of x forc es, F x = 0, we have

Equation b

Upon simplific ation, Eq. (b) bec omes

Equation c

Inasmuc h as dx dy is nonzero, the quantity in the parentheses must vanish. A similar

expression is written to describe the equilibrium of y forc es. The x and y equations yield the
following differential equations of equilibrium for two-dimensional stress:

Equation 1.13

The differential equations of equilibrium for the case of three-dimensional stress may be
generalized from the prec eding expressions as follows:

Equation 1.14
A suc cinc t representation of these expressions, on the basis of the range and summation
conventions (Sec. 1.17), may be written as

Equation 1.15a

where xx = x, xy = y, and xz = z. The repeated subsc ript is j, indicating summation. The

unrepeated subscript is i. Here i is termed the free index, and j, the dummy index.

If in the foregoing expression the symbol /x is replaced by a comma, we have

Equation 1.15b

where the subsc ript after the comma denotes the coordinate with respect to whic h
differentiation is performed. If no body forces exist, Eq. (1.15b) reduc es to ij,j = 0,
indic ating that the sum of the three stress derivatives is zero. As the two equilibrium
relations of Eqs. (1.13) c ontain three unknowns ( x , y , xy ) and the three expressions of
Eqs. (1.14) involve the six unknown stress components, problems in stress analysis
are internally statically indeterminate.

In a number of practical applic ations, the weight of the member is the only body force. If
we take the y axis as upward and designate by the mass density per unit volume of the
member and by g, the gravitational ac celeration, then F x = F z = 0 and F y = g in Eqs.
(1.13) and (1.14). The resultant of this force over the volume of the member is usually so
small compared with the surface forces that it can be ignored, as stated in Section 1.1.
However, in dynamic systems, the stresses caused by body forc es may far exc eed those
associated with surfac e forces so as to be the principal influenc e on the stress field.[*]
[*]In this ca se , the body is not in sta tic e quilibrium , a nd the ine rtia force term s ax, ay, a nd
az (where ax , ay , a nd az a re the com ponents of acce lera tion) m ust be include d in the body
force com pone nts Fx, Fy, and F z, re spectively, in Eqs. (1.14).

Application of Eqs. (1.13) and (1.14) to a variety of loaded members is presented in

sections employing the approach of the theory of elastic ity, beginning with Chapter 3. The
following sample problem shows the pattern of the body force distribution for an arbitrary
state of stress in equilibrium.

Example 1.2. The Body Forces in a Structure

The stress field within an elastic struc tural member is expressed as follows:

Equation d
Determine the body forc e distribution required for equilibrium.

Solution

Substitution of the given stresses into Eq. (1.14) yields

The body force distribution, as obtained from these expressions, is therefore

Equation e

The state of stress and body force at any specific point within the member may
be obtained by substituting the specific values of x, y, and z into Eqs. (d) and
(e), respectively.
1.9. Plane-Stress Transformation
A two-dimensional state of stress exists when the stresses and body forc es are
independent of one of the coordinates, here taken as z. Such a state is desc ribed by
stresses x , y , and xy and the x and y body forc es. Two-dimensional problems are of two
classes: plane stress and plane strain. In the case of plane stress, as described in the
previous sec tion, the stresses z, xz, and yz, and the z-directed body forces are assumed
to be zero. The c ondition that occ urs in a thin plate subjec ted to loading uniformly
distributed over the thickness and parallel to the plane of the plate typifies the state of
plane stress (Fig. 1.10). In the case of plane strain, the stresses xz and yz and the body
force F z are likewise taken to be zero, but z does not vanish[*] and can be determined
from stresses x and y .

[*] More deta ils and illustrations of these a ssum ptions a re give n in Chapter 3.

Figure 1.10. Thin Plate in-plane loads.

We shall now determine the equations for transformation of the stress components x , y ,
and xy at any point of a body represented by an infinitesimal element, isolated from the
plate illustrated in Fig. 1.10. The z-direc ted normal stress z, even if it is nonzero, need
not be considered here. In the following derivations, the angle loc ating the x axis is
assumed positive when measured from the x axis in a counterclockwise direction. Note
that, according to our sign convention (see Sec. 1.5), the stresses are indic ated as
positive values.

Consider an infinitesimal wedge cut from the loaded body shown in Fig. 1.11a, b. It is
required to determine the stresses x and xy, whic h refer to axes x, y making an
angle with axes x, y, as shown in the figure. Let side AB be normal to the x axis. Note
that in acc ordance with the sign c onvention, x and xy are positive stresses, as shown
in the figure. If the area of sideAB is taken as unity, then sides QA and QB have area
cos and sin , respectively.

Figure 1.11. Elements in plane stress.

[View full size im age ]

Equilibrium of forces in the x and y directions requires that

Equation 1.16

where px and py are the c omponents of stress resultant ac ting on AB in

the x and y direc tions, respec tively. The normal and shear stresses on the x plane
(AB plane) are obtained by projecting px and py in the x and y directions:

Equation a

From the foregoing it is clear that . Upon substitution of the stress

resultants from Eq. (1.16), Eqs. (a) become

Equation 1.17a

Equation 1.17b

Note that the normal stress y ac ting on the y face of an inclined element (Fig. 1.11c)
may readily be obtained by substituting + /2 for in the expression for x. In so doing,
we have

Equation 1.17c
Equations (1.17) c an be converted to a useful form by introducing the following
trigonometric identities:

The transformation equations for plane stress now become

Equation 1.18a

Equation 1.18b

Equation 1.18c

The foregoing expressions permit the c omputation of stresses acting on all possible
planes AB (the state of stress at a point) provided that three stress components on a set
of orthogonal faces are known.

Stress tensor. It is important to note that addition of Eqs. (1.17a) and (1.17c) gives the
relationships

x + y = x + y = c onstant

In words then, the sum of the normal stresses on two perpendicular planes is invariant
that is, independent of . This c onc lusion is also valid in the case of a three-dimensional
state of stress, as shown in Section 1.13. In mathematic al terms, the stresswhose
components transform in the preceding way by rotation of axes is termed tensor. Some
examples of other quantities arestrain and moment of inertia. The similarities between the
transformation equations for these quantities are observed in Sections 2.5 and C.4. Mohrs
circ le (Sec. 1.11) is a graphical representation of a stress tensor transformation.

Polar Representations of State of Plane Stress

Consider, for example, the possible states of stress corresponding to x = 14 MPa, y = 4
MPa, and xy = 10 MPa. Substituting these values into Eq. (1.18) and permitting to vary
from 0 to 360 yields the data upon which the c urves shown in Fig. 1.12are based. The
plots shown, called stress trajec tories, are polar representations: x versus (Fig. 1.12a)
and xy versus (Fig. 1.12b). It is observed that the direc tion of each maximum shear
stress bisec ts the angle between the maximum and minimum normal stresses. Note that the
normal stress is either a maximum or a minimum on planes at = 31.66 and = 31.66 +
90, respec tively, for which the shearing stress is zero. The conclusions drawn from this
example are valid for any two-dimensional (or three-dimensional) state of stress and are
observed in the sections to follow.

Figure 1.12. Polar representations of x and xy (in megapascals) versus .

[View full size im age ]

Cartesian Representation of State of Plane Stress

Now let us examine a two-dimensional condition of stress at a point in a loaded machine
component on an element illustrated inFig. 1.13a. Introducing the given values into the first
two of Eqs. (1.18), gives

x = 4.5 + 2.5 c os 2 + 5 sin 2

xy = 2.5 sin 2 + 5 cos 2

Figure 1.13. Graph of normal stress x and shearing stress xy with angle (for
180).

[View full size im age ]

In the foregoing, permitting to vary from 0 to 180 in increments of 15 leads to the data
from whic h the graphs illustrated inFig. 1.13b are obtained [Ref. 1.7]. This Cartesian
representation demonstrates the variation of the normal and shearing stresses versus
180. Observe that the direction of maximum (and minimum) shear stress bisects the angle
between the maximum and minimum normal stresses. Moreover, the normal stress is either a
maximum or a minimum on planes = 31.7 and = 31.7 + 90, respectively, for which the
shear stress is zero. Note as a check that x + y = max + min = 9 MPa, as expected.

The conclusions drawn from the foregoing polar and Cartesian representations are valid
for any state of stress, as will be seen in the next sec tion. A more convenient approac h to
the graphical transformation for stress is considered in Sections 1.11 and1.15. The manner
in whic h the three-dimensional normal and shearing stresses vary is discussed in Sections
1.12 through 1.14.
1.10. Principal Stresses and Maximum In-Plane Shear Stress
The transformation equations for two-dimensional stress indicate that the normal stress x
and shearing stress xy vary continuously as the axes are rotated through the angle .
To ascertain the orientation of xy corresponding to maximum or minimum x, the
nec essary c ondition d x/d = 0 is applied to Eq. (1.18a). In so doing, we have

Equation a

This yields

Equation 1.19

Inasmuc h as tan 2 = tan( + 2), two directions, mutually perpendicular, are found to
satisfy Eq. (1.19). These are the princ ipal directions, along whic h the principal or maximum
and minimum normal stresses act. Two values of p, corresponding to the 1 and 2 planes,
are represented by and , respectively.

When Eq. (1.18b) is compared with Eq. (a), it becomes clear that xy = 0 on a principal
plane. A principal plane is thus a plane of zero shear. The princ ipal stresses are determined
by substituting Eq. (1.19) into Eq. (1.18a):

Equation 1.20

Note that the algebraic ally larger stress given here is the maximum principal stress, denoted
by 1. The minimum princ ipal stress is represented by 2. It is necessary to substitute one
of the values p into Eq. (1.18a) to determine which of the two c orresponds to 1.

Similarly, employing the preceding approac h and Eq. (1.18b), we determine the planes of
maximum shearing stress. Thus, settingd xy/d = 0, we now have ( x y )c os 2 +
2 xy sin 2 = 0 or

Equation 1.21
The foregoing expression defines two values of s that are 90 apart. These directions may
again be denoted by attaching a prime or a double prime notation to s. Comparing Eqs.
(1.19) and (1.21), we also observe that the planes of maximum shearing stress are inc lined
at 45 with respect to the planes of principal stress. Now, from Eqs. (1.21) and (1.18b),
we obtain the extreme values of shearing stress as follows:

Equation 1.22

Here the largest shearing stress, regardless of sign, is referred to as the maximum shearing
stress, designated max. Normal stresses acting on the planes of maximum shearing stress
can be determined by substituting the values of 2s from Eq. (1.21) into Eqs. (1.18a) and
(1.18c):

Equation 1.23

The results are illustrated in Fig. 1.14. Note that the diagonal of a stress element toward
whic h the shearing stresses act is called the shear diagonal. The shear diagonal of the
element on which the maximum shearing stresses act lies in the direc tion of the algebraically
larger princ ipal stress as shown in the figure. This assists in predicting the proper
direc tion of the maximum shearing stress.

Figure 1.14. Planes of principal and maximum shearing stresses.

1.11. Mohrs Circle for Two-Dimensional Stress
A graphical technique, predic ated on Eq. (1.18), permits the rapid transformation of stress
from one plane to another and leads also to the determination of the maximum normal and
shear stresses. In this approach, Eqs. (1.18) are depicted by a stress c irc le, called Mohrs
circ le.[*] In the Mohr representation, the normal stresses obey the sign convention
of Section 1.5. However, for the purposes only of construc ting and reading values of
stress from Mohrs circ le, the sign convention for shear stress is as follows: If the shearing
stresses on opposite faces of an element would produce shearing forc es that result in
a c loc kwise c ouple, as shown in Fig. 1.15c, these stresses are regarded as positive.
Accordingly, the shearing stresses on the y faces of the element in Fig. 1.15a are taken as
positive (as before), but those on the x faces are now negative.
[*]Afte r O tto Mohr (18351918), profe ssor a t Dresde n P olyte chnic. For furthe r de tails, se e
Re f. 1.7, for e x am ple .

Figure 1.15. (a) Stress element; (b) Mohrs circle of stress; (c) interpretation of
positive shearing stresses.

[View full size im age ]

Given x , y , and xy with algebraic sign in acc ordance with the foregoing sign c onvention,
the procedure for obtaining Mohrs circ le (Fig. 1.15b) is as follows:

1. Establish a rec tangular c oordinate system, indicating + and +. Both stress sc ales
must be identical.

2.
Loc ate the center C of the c ircle on the horizontal axis a distance from
the origin.

3. Loc ate point A by c oordinates x and xy . These stresses may correspond to any
face of an element such as in Fig. 1.15a. It is usual to specify the stresses on the
positive x face, however.

4. Draw a circ le with center at C and of radius equal to CA.

5. Draw line AB through C.

The angles on the circle are measured in the same direction as is measured in Fig. 1.15a.
An angle of 2 on the circle corresponds to an angle of on the element. The state of
stress assoc iated with the original x and y planes corresponds to points A and B on the
circ le, respec tively. Points lying on diameters other than AB, suc h as A and B, define
states of stress with respec t to any other set of x and y planes rotated relative to the
original set through an angle .

It is clear that points A1 and B1 on the circle loc ate the princ ipal stresses and provide their
magnitudes as defined by Eqs. (1.19) and (1.20), while D and E represent the maximum
shearing stresses, defined by Eqs. (1.21) and (1.22). The radius of the c ircle is

Equation a

where

Thus, the radius equals the magnitude of the maximum shearing stress. Mohrs circle shows
that the planes of maximum shear are always loc ated at 45 from planes of principal stress,
as already indicated in Fig. 1.14. The use of Mohrs c irc le is illustrated in the first two of
the following examples.

Example 1.3. Principal Stresses in a Member

At a point in the structural member, the stresses are represented as in Fig.

1.16a. Employ Mohrs circle to determine (a) the magnitude and orientation of
the principal stresses and (b) the magnitude and orientation of the maximum
shearing stresses and associated normal stresses. In each c ase, show the
results on a properly oriented element; represent the stress tensor in matrix
form.

Figure 1.16. Example 1.3. (a) Element in plane stress; (b) Mohrs circle of
stress; (c) principal stresses; (d) maximum shear stress.
 Solution

Mohrs circle, c onstructed in acc ordance with the proc edure outlined, is shown
in Fig. 1.16b. The center of the c ircle is at (40 + 80)/2 = 60 MPa on the axis.

a. The principal stresses are represented by points A1 and B1 Hence, the

maximum and minimum principal stresses, referring to the c irc le, are

or

1 = 96.05 MPa and 2 = 23.95 MPa

The planes on whic h the principal stresses act are given by

Henc e
 Mohrs circle clearly indic ates that locates the 1 plane. The results
may readily be c hecked by substituting the two values of p into Eq.
(1.18a). The state of princ ipal stress is shown in Fig. 1.16c.

b. The maximum shearing stresses are given by points D and E. Thus,

It is seen that ( 1 2)/2 yields the same result. The planes on which
these stresses ac t are represented by

As Mohrs c ircle indicates, the positive maximum shearing stress acts on a

plane whose normal x makes an angle with the normal to the original
plane (x plane). Thus, + max on two opposite x faces of the element will
be directed so that a clockwise couple results. The normal stresses acting
on maximum shear planes are represented by OC, = 60 MPa on each
face. The state of maximum shearing stress is shown inFig. 1.16d. The
direc tion of the max s may also be readily predicted by recalling that they
act toward the shear diagonal. We note that, acc ording to the general
sign c onvention (Sec. 1.5), the shearing stress acting on the x plane
in Fig. 1.16d is negative. As a c hec k, if and the given
initial data are substituted into Eq. (1.18b), we obtain xy = 36.05 MPa,
as already found.

We may now desc ribe the state of stress at the point in the following
matrix forms:

These three representations, associated with the = 0, = 28.15,

and = 73.15 planes passing through the point, are equivalent.

Note that if we assume z = 0 in this example, a muc h higher shearing stress is

obtained in the planes bisec ting the x and z planes (Problem 1.56). Thus,
three-dimensional analysis, Section 1.15, should be considered for determining
the true maximum shearing stress at a point.

Example 1.4. Stresses in a Frame

The stresses ac ting on an element of a loaded frame are shown in Fig. 1.17a.
Apply Mohrs circ le to determine the normal and shear stresses ac ting on a plane
defined by = 30.

Figure 1.17. Example 1.4. (a) Element in biaxial stresses; (b) Mohrs circle
of stress; (c) stress element for = 30.

[View full size im a ge ]

 Solution

Mohrs circle of Fig. 1.17b describes the state of stress given in Fig. 1.17a.
Points A1 and B1 represent the stress components on the x and y fac es,
respectively. The radius of the c ircle is (14 + 28)/2 = 21. Corresponding to the
30 plane within the element, it is nec essary to rotate through 60
c ounterclockwise on the circle to locate pointA. A 240 counterc loc kwise
rotation loc ates point B. Referring to the circle,

x = 7 + 21 c os 60 = 17.5 MPa

y = 3.5 MPa

and xy = 21 sin 60 = 18.19 MPa

Figure 1.17c indic ates the orientation of the stresses. The results c an be
c hecked by applying Eq. (1.18), using the initial data.

Example 1.5. Cylindrical Vessel Under Combined Loads

A thin-walled c ylindric al pressure vessel of 250-mm diameter and 5-mm wall

thickness is rigidly attac hed to a wall, forming a cantilever (Fig. 1.18a).
Determine the maximum shearing stresses and the assoc iated normal stresses at
point A of the cylindrical wall. The following loads are applied: internal
pressure p = 1.2 MPa, torque T = 3 kN m, and direct force P = 20 kN. Show the
results on a properly oriented element.

Figure 1.18. Example 1.5. Combined stresses in a thin-walled cylindrical

pressure vessel: (a) side view; (b) free body of a segment; (c) and (d)
element A (viewed from top).

Solution

The internal force resultants on a transverse section through point A are found
from the equilibrium conditions of the free-body diagram of Fig. 1.18b. They
are V = 20 kN, M = 8 kN m, and T = 3 kN m. In Fig. 1.18c, the combined
axial, tangential, and shearing stresses are shown acting on a small element at
point A. These stresses are (Tables 1.1 and C.1)

We thus have x = 47.6 MPa, y = 30 MPa, and xy = 6.112 MPa. Note that for
element A, Q = 0; henc e, the direc t shearing stress d = xz = VQ/Ib = 0.

The maximum shearing stresses are from Eq. (1.22):

Equation (1.23) yields

To locate the maximum shear planes, we use Eq. (1.21):

Applying Eq. (1.18b) with the given data and 2s = 55.2, xy = 10.71 MPa.
Henc e, , and the stresses are shown in their proper directions in Fig.
1.18d.
1.12. Three-Dimensional Stress Transformation
The physical elements studied are always three dimensional, and hence it is desirable to
consider three planes and their associated stresses, as illustrated in Fig. 1.2. We note that
equations governing the transformation of stress in the three-dimensional case may be
obtained by the use of a similar approac h to that used for the two-dimensional state of
stress.

Consider a small tetrahedron isolated from a continuous medium (Fig. 1.19a), subject to a
general state of stress. The body forc es are taken to be negligible. In the figure, px, py ,
and pz are the Cartesian components of stress resultant p acting on oblique plane ABC. It is
required to relate the stresses on the perpendic ular planes intersec ting at the origin to the
normal and shear stresses on ABC.

Figure 1.19. Stress components on a tetrahedron.

The orientation of plane ABC may be defined in terms of the angles between a unit
normal n to the plane and the x, y, and zdirec tions (Fig. 1.19b). The direc tion cosines
associated with these angles are

Equation 1.24

The three direction cosines for the n direc tion are related by

Equation 1.25
The area of the perpendicular plane QAB, QAC, QBC may now be expressed in terms of A,
the area of ABC, and the direc tion cosines:

AQAB = Ax = A i = A(li + mj + nk) i = Al

The other two areas are similarly obtained. In so doing, we have altogether

Equation a

Here i, j, and k are unit vectors in the x, y, and z direc tions, respec tively.

Next, from the equilibrium of x, y, z-direc ted forces together with Eq. (a), we obtain, after
canceling A,

Equation 1.26

The stress resultant on A is thus determined on the basis of known

stresses x , y , z, xy , xz, and yz and a knowledge of the orientation of A. In the limit as
the sides of the tetrahedron approach zero, plane A c ontains point Q. It is thus
demonstrated that the stress resultant at a point is specified. This in turn gives the stress
components ac ting on any three mutually perpendicular planes passing through Q as shown
next. Although perpendic ular planes have been used there for c onvenienc e, these planes
need not be perpendicular to define the stress at a point.

Consider now a Cartesian c oordinate system x, y, z, wherein x c oinc ides with n and y, z
lie on an oblique plane. The x y zand xyz systems are related by the direction
cosines: l1 = cos (x, x), m1 = c os(x, y), and so on. The notation corresponding to a
complete set of direc tion cosines is shown in Table 1.2. The normal stress x is found by
projecting px , py , and pz in the xdirection and adding

Equation 1.27

Table 1.2. Notation for Direction Cosines

x y z

x l1 m1 n1

y l2 m2 n2
z l3 m3 n3

Equations (1.26) and (1.27) are c ombined to yield

Equation 1.28a

Similarly, by projec ting px , py , and pz in the y and z direc tions, we obtain, respectively,

Equation 1.28b

Equation 1.28c

Rec alling that the stresses on three mutually perpendicular planes are required to specify
the stress at a point (one of these planes being the oblique plane in question), the
remaining components are found by considering those planes perpendicular to the oblique
plane. For one suc h plane, n would now coincide with the y direction, and expressions for
the stresses y, yx, and yz would be derived. In a similar manner, the stresses z, zx
, and zy are determined when n c oincides with the z direction. Owing to the symmetry of
the stress tensor, only six of the nine stress components thus developed are unique. The
remaining stress c omponents are as follows:

Equation 1.28d

Equation 1.28e

Equation 1.28f
Equations (1.28) represent expressions transforming the quantities x , y , z, xy , xz,
and yz which, as we have noted, c ompletely define the state of stress. Quantities suc h as
stress (and moment of inertia, Appendix C), whic h are subject to suc h transformations, are
tensors of sec ond rank (see Sec. 1.9).

The equations of transformation of the components of a stress tensor, in indicial notation,

are represented by

Equation 1.29a

Alternatively,

Equation 1.29b

The repeated subsc ripts i and j imply the double summation in Eq. (1.29a), which, upon
expansion, yields

Equation 1.29c

By assigning r, s = x, y, z and noting that rs = sr, the foregoing leads to the six
expressions of Eq. (1.28).

It is interesting to note that, because x, y, and z are orthogonal, the nine direc tion
cosines must satisfy trigonometric relations of the following form:

Equation 1.30a

and

Equation 1.30b
From Table 1.2, observe that Eqs. (1.30a) are the sums of the squares of the c osines in
eac h row, and Eqs. (1.30b) are the sums of the products of the adjacent c osines in any
two rows.
1.13. Principal Stresses in Three Dimensions
For the three-dimensional case, it is now demonstrated that three planes of zero shear
stress exist, that these planes are mutually perpendic ular, and that on these planes the
normal stresses have maximum or minimum values. As has been disc ussed, these normal
stresses are referred to as princ ipal stresses, usually denoted 1, 2, and 3.
The algebraically largest stress is represented by 1, and the smallest by 3: 1 > 2 > 3.

We begin by again c onsidering an oblique x plane. The normal stress acting on this plane is
given by Eq. (1.28a):

Equation a

The problem at hand is the determination of extreme or stationary values of x. To

acc omplish this, we examine the variation of x relative to the direction c osines. Inasmuch
as l, m, and n are not independent, but connected by l2 + m2 + n2 = 1, only l andm may be
regarded as independent variables. Thus,

Equation b

Differentiating Eq. (a) as indicated by Eqs. (b) in terms of the quantities in Eq. (1.26), we
obtain

Equation c

From n2 = 1 l2 m2, we have n/l = l/n and n/m = m/n. Introduc ing these into Eq.
(c), the following relationships between the c omponents of p and n are determined:

Equation d

These proportionalities indicate that the stress resultant must be parallel to the unit normal
and therefore contains no shear component. It is c oncluded that, on a plane for which x
has an extreme or princ ipal value, a principal plane, the shearing stress vanishes.

It is now shown that three princ ipal stresses and three principal planes exist. Denoting the
principal stresses by p, Eq. (d) may be written as
Equation e

These expressions, together with Eq. (1.26), lead to

Equation 1.31

A nontrivial solution for the direc tion cosines requires that the c harac teristic determinant
vanish:

Equation 1.32

Expanding Eq. (1.32) leads to

Equation 1.33

where

Equation 1.34a

Equation 1.34b

Equation 1.34c
The three roots of the stress c ubic equation (1.33) are the principal stresses, c orresponding
to whic h are three sets of direc tion cosines, which establish the relationship of the principal
planes to the origin of the nonprincipal axes. The principal stresses are the characteristic
values or eigenvalues of the stress tensor ij. Since the stress tensor is a symmetric tensor
whose elements are all real, it has real eigenvalues. That is, the three princ ipal stresses
are real [Refs. 1.8 and 1.9]. The direc tion cosines l, m, and n are the eigenvectors of ij.

It is clear that the principal stresses are independent of the orientation of the original
coordinate system. It follows from Eq. (1.33) that the coefficients I1, I2, and I3 must
likewise be independent of x, y, and z, since otherwise the principal stresses would change.
For example, we can demonstrate that adding the expressions for x, y, and z given by
Eq. (1.28) and making use of Eq. (1.30a) leads to I1 = x + y + z = x + y + z.
Thus, the c oefficients I1, I2, and I3 represent three invariants of the stress tensor in three
dimensions or, briefly, the stress invariants. For plane stress, it is a simple matter to show
that the following quantities are invariant (Prob. 1.27):

Equation 1.35

Equations (1.34) and (1.35) are partic ularly helpful in checking the results of a stress
transformation, as illustrated in Example 1.7.

If now one of the princ ipal stresses, say 1 obtained from Eq. (1.33), is substituted into Eq.
(1.31), the resulting expressions, together with l2 + m2 + n2 = 1, provide enough
information to solve for the direc tion cosines, thus specifying the orientation of 1 relative to
the xyz system. The direction c osines of 2 and 3 are similarly obtained. A convenient way
of determining the roots of the stress c ubic equation and solving for the direc tion cosines is
presented in Appendix B, where a related computer program is also inc luded (see Table
B.1).

Example 1.6. Three-Dimensional Stress in a Hub

A steel shaft is to be forc e fitted into a fixed-ended cast-iron hub. The shaft is
subjected to a bending moment M, a torque T, and a vertical forc e P, Fig.
1.20a. Suppose that at a point Q in the hub, the stress field is as shown inFig.
1.20b, represented by the matrix
 Figure 1.20. Example 1.6. (a) Hub-shaft assembly. (b) Element in three-
dimensional stress.

Determine the principal stresses and their orientation with respec t to the original
c oordinate system.

Solution

Substituting the given stresses into Eq. (1.33) we obtain from Eqs. (B.2)

1 = 11.618 MPa, 2 = 9.001 MPa, 3 = 25.316 MPa

Succ essive introduction of these values into Eq. (1.31), together with Eq.
(1.30a), or applic ation of Eqs. (B.6) yields the direction c osines that define the
orientation of the planes on which 1, 2, and 3 act:

l1 = 0.0266, l2 = 0.6209, l3 = 0.7834

m1 = 0.8638, m2 = 0.3802, m3 = 0.3306

n1 = 0.5031, n2 = 0.6855, n3 = 0.5262

Note that the direc tions of the principal stresses are seldom required for
purposes of predic ting the behavior of structural members.

Example 1.7. Three-Dimensional Stress in a Machine Component

The stress tensor at a point in a machine element with respect to a Cartesian

c oordinate system is given by the following array:

Equation f
Determine the state of stress and I1, I2, and I3 for an x, y, z c oordinate
system defined by rotating x, y through an angle of = 45 counterc lockwise
about the z axis (Fig. 1.21a).

Figure 1.21. Example 1.7. Direction cosines for = 45.

Solution

The direction c osines corresponding to the prescribed rotation of axes are given
in Fig. 1.21b. Thus, through the use of Eq. (1.28) we obtain

Equation g

It is seen that the arrays (f) and (g), when substituted into Eq. (1.34), both
yield I1 = 100 MPa, I2 = 1400 (MPa)2, and I3 = 53,000 (MPa)3, and the
invariance of I1, I2, and I3 under the orthogonal transformation is confirmed.
1.14. Normal and Shear Stresses on an Oblique Plane
A cubic element subjected to principal stresses 1, 2, and 3 acting on mutually
perpendic ular princ ipal planes is called in a state of triaxial stress (Fig. 1.22a). In the
figure, the x, y, and z axes are parallel to the princ ipal axes. Clearly, this stress c ondition is
not the general case of three-dimensional stress, which was taken up in the last two
sections. It is sometimes required to determine the shearing and normal stresses acting on
an arbitrary oblique plane of a tetrahedron, as in Fig. 1.22b,given the principal stresses or
triaxial stresses acting on perpendic ular planes. In the figure, the x, y, and z axes are
parallel to the princ ipal axes. Denoting the direction c osines of plane ABC by l, m, and n,
Eqs. (1.26) with x = 1, xy = xz = 0, and so on, reduce to

Equation a

Figure 1.22. Elements in triaxial stress.

Referring to Fig. 1.22a and definitions (a), the stress resultant p is related to the principal
stresses and the stress c omponents on the oblique plane by the expression

Equation 1.36

The normal stress on this plane, from Eq. (1.28a), is found as

Equation 1.37

Substitution of this expression into Eq. (1.36) leads to

Equation 1.38a

or

Equation 1.38b

Expanding and using the expressions 1 l2 = m2 + n2, 1 n2 = l2 + m2, and so on, the
following result is obtained for the shearing stress on the oblique plane:

Equation 1.39

This c learly indic ates that if the principal stresses are all equal, the shear stress vanishes,
regardless of the c hoices of the direction cosines.

For situations in whic h shear as well as normal stresses act on perpendic ular planes (Fig.
1.22b), we have px, py , and pz defined by Eqs. (1.26). Then, Eq. (1.37) becomes

Equation 1.40

Henc e,

Equation 1.41

where is given by Eq. (1.40). Formulas (1.37) through (1.41) represent the simplified
transformation expressions for the three-dimensional stress.

It is interesting to note that substitution of the direc tion cosines from Eqs. (a) into Eq.
(1.25) leads to

Equation 1.42
whic h is a stress ellipsoid having its three semiaxes as the principal stresses (Fig. 1.23).
This geometrical interpretation helps to explain the earlier conc lusion that the principal
stresses are the extreme values of the normal stress. In the event that 1 = 2 = 3, a
state of hydrostatic stress exists, and the stress ellipsoid becomes a sphere. In this case,
note again that any three mutually perpendicular axes can be taken as the princ ipal axes.

Figure 1.23. Stress ellipsoid.

Octahedral Stresses
The stresses acting on an octahedral plane is represented by face ABC in Fig.
1.22b with QA = QB = QC. The normal to this oblique fac e thus has equal direc tion cosines
relative to the princ ipal axes. Sinc e l2 + m2 + n2 = 1, we have

Equation b

Plane ABC is clearly one of eight such faces of a regular oc tahedron (Fig. 1.24). Equations
(1.39) and (b) are now applied to provide an expression for the oc tahedral shearing stress,
whic h may be rearranged to the form

Equation 1.43

Figure 1.24. Stresses on an octahedron.

Through the use of Eqs. (1.37) and (b), we obtain the oc tahedral normal stress:

Equation 1.44

The normal stress acting on an oc tahedral plane is thus the average of the princ ipal
stresses, the mean stress. The orientations of oct and oct are indicated in Fig. 1.24. That
the normal and shear stresses are the same for the eight planes is a powerful tool for failure
analysis of ductile materials (see Sec. 4.8). Another useful form of Eq. (1.43) is developed
in Section 2.15.
1.15. Mohrs Circles in Three Dimensions
Consider a wedge shown in Fig. 1.25a, cut from the cubic element subjected to triaxial
stresses (Fig. 1.22a). The only stresses on the inc lined x face (parallel to the z axis) are
the normal stress x and the shear stress xy ac ting in the xy plane. Inasmuc h as the
foregoing stresses are determined from force equilibrium equations in the xy plane, they
are independent of the stress 3. Thus, the transformation equations of plane stress (Sec.
1.9) and Mohrs circle can be employed to obtain the stresses x and xy. The foregoing
conclusion is also valid for normal and shear stresses acting on inclined faces cut through
the element parallel to the x and y axes.

Figure 1.25. Triaxial state of stress: (a) wedge; (b) planes of maximum shear stress.

The stresses acting on elements oriented at various angles to the principal axes c an be
visualized with the aid of Mohrs circle. The c ubic element (Fig. 1.22a) viewed from three
different directions is sketc hed in Figs. 1.26a to c. A Mohrs c ircle is drawn corresponding
to eac h projection of an element. The cluster of three circles represents Mohrs c ircles for
triaxial stress (Fig. 1.26d). The radii of the circles are equal to the maximum shear
stresses, as indicated in the figure. The normal stresses acting on the planes of maximum
shear stresses have the magnitudes given by the abscissa as of the centers of the c irc les.

Figure 1.26. (ac) Views of elements in triaxial stresses on different principal axes;
(d) Mohrs circles for three-dimensional stress.
The largest shear stresses occ ur on planes oriented at 45 to the princ ipal planes. The
shear stress is a maximum located as the highest point on the outer c ircle. The value of
the absolute maximum shearing stress is therefore

Equation 1.45

acting on the planes that bisec t the planes of the maximum and minimum principal stresses,
as shown in Fig. 1.25b. It is noted that the planes of maximum shear stress may also be
ascertained by substituting n2 = 1 l2 m2 into Eq. (1.38b), differentiating with respec t
to l and m, and equating the resulting expressions to zero (Prob. 1.80).

Determining the absolute value of maximum shear stress is signific ant when designing
members made of ductile materials, since the strength of the material depends on its ability
to resist shear stress (Sec. 4.6). Obviously, as far as the stress magnitudes are concerned,
the largest circ le is the most significant one. However, all stresses in their various
transformations may play a role in causing failure, and it is usually instructive to plot all
three principal c irc les of stress, as depicted in the figure. An example of this type oc curs in
thin-walled pressurized c ylinders, where = 1, a = 2, and r = 3 = 0 at the outer
surface (Table 1.1). It is also interesting to note that, in special c ases, where two or all
principal stresses are equal, a Mohrs c ircle becomes a point.

Equations of Three Mohrs Circles for Stress

It has been demonstrated that, given the values of the principal stresses and of the
direction c osines for any oblique plane (Fig. 1.22b), the normal and shear stresses on the
plane may be asc ertained through the application of Eqs. (1.37) and (1.38). This may also
be ac complished by means of a graphic al technique due to Mohr [Refs. 1.10 through 1.12].
The latter procedure was used in the early history of stress analysis, but today it is
employed only as a heuristic device.

In the following discussion, we demonstrate that the aforementioned equations together

with the relation l2 + m2 + n2 = 1 are represented by three c ircles of stress, and the
coordinates (, ) loc ate a point in the shaded area of Fig. 1.26d [Ref. 1.13]. These
simultaneous equations are

Equation a

where l2 0, m2 0, and n2 0. Solving for the direction c osines, results in

Equation 1.46

Inasmuc h as 1 > 2 > 3, the numerators of Eqs. (1.46) satisfy

Equation b

as the denominators of Eqs. (1.46) are ( 1 2) > 0 and ( 1 3) > 0, ( 2 3) > 0 and
( 2 1) < 0, ( 3 1) < 0 and ( 3 2) < 0, respectively.

Finally, the prec eding inequalities may be expressed as follows

Equation 1.47

Equations (1.47) represent the formulas of the three Mohrs circ les for stress, shown in Fig.
1.26d. Stress points (, ) satisfying the equations for circles centered at C1 and C2 lie on
or outside c irc les, but for the c irc le c entered at C3 lie on or inside circle. We conclude
therefore that an admissible state of stress must lie on Mohrs c ircles or within the shaded
area enclosed by these c ircles.

Example 1.8. Analysis of Three-Dimensional Stresses in a Member

The state of stress on an element of a structure is illustrated in Fig. 1.27a.

Using Mohrs circle, determine (a) the principal stresses and (b) the maximum
shearing stresses. Show results on a properly oriented element. Also, (c) apply
the equations developed in Section 1.14 to c alculate the octahedral stresses.

Figure 1.27. Example 1.8. (a) Element in three-dimensional stress; (b)

Mohrs circles of stress; (c) stress element for .

[View full size im a ge ]

Solution

a. First, Mohrs c ircle for the transformation of stress in the xy plane is

 sketc hed in the usual manner as shown, centered at C2 with
diameter A2A3 (Fig. 1.27b). Next, we c omplete the three-dimensional
Mohrs circle by drawing two additional circles of
diameters A1A2 and A1A3 in the figure. Referring to the circ le, the princ ipal
stresses are 1 = 100 MPa, 2 = 40 MPa, and 3 = 60 MPa. Angle
, as tan . The results are sketc hed on a properly
oriented element in Fig. 1.27c.

b. The absolute maximum shearing stress, point B3, equals the radius of the
c ircle c entered at C3 of diameterA1 A3. Thus,

The maximum shearing stress occ urs on the planes 45 from the y
and z fac es of the element of Fig. 1.27c.

c. The octahedral normal stress, from Eq. (1.44), is

The octahedral shearing stress, using Eq. (1.43), is

Comments

A comparison of the results (see Fig. 1.27b) shows that

oct < 1 and oct < ( max )a

That is, the maximum principal stress and absolute maximum shear stress are
greater than their octahedral counterparts.
1.16. Boundary Conditions in Terms of Surface Forces
We now consider the relationship between the stress components and the given surface
forces acting on the boundary of a body. The equations of equilibrium that must be satisfied
within a body are derived in Section 1.8. The distribution of stress in a body must also be
such as to accommodate the conditions of equilibrium with respect to externally applied
forces. The external forc es may thus be regarded as a continuation of the internal stress
distribution.

Consider the equilibrium of the forces acting on the tetrahedron shown in Fig. 1.19b, and
assume that oblique fac e ABC is coincident with the surface of the body. The components
of the stress resultant p are thus now the surface forces per unit area, or surface
trac tions, px , py , and pz. The equations of equilibrium for this element,
representing boundary c onditions, are, from Eqs. (1.26),

Equation 1.48

For example, if the boundary is a plane with an x-directed surface normal, Eqs. (1.48)
give px = x , py = xy , and pz = xz; under these circ umstanc es, the applied surfac e force
components px , py , and pz are balanced by x , xy , and xz, respec tively.

It is of interest to note that, instead of prescribing the distribution of surface forces on the
boundary, the boundary c onditions of a body may also be given in terms of displac ement
components. Furthermore, we may be given boundary c onditions that prescribe surface
forces on one part of the boundary and displac ements on another. When displacement
boundary c onditions are given, the equations of equilibrium express the situation in terms of
strain, through the use of Hookes law and subsequently in terms of the displac ements by
means of straindisplacement relations (Sec. 2.3). It is usual in engineering problems,
however, to spec ify the boundary conditions in terms of surface forces, as in Eq. (1.48),
rather than surface displacements. This practice is adhered to in this text.
1.17. Indicial Notation
A system of symbols, called indicial notation, index notation, also known as tensor notation,
to represent c omponents of force, stress, displacement, and strain is used throughout this
text. Note that a particular class of tensor, a vec tor, requires only a single subsc ript to
desc ribe each of its c omponents. Often the c omponents of a tensor require more than a
single subsc ript for definition. For example, sec ond-order or second-rank tensors, such as
those of stress or inertia, require double subsc ripting: ij,Iij. Quantities such as temperature
and mass are scalars, classified as tensors of zero rank.

Tensor or indic ial notation, here briefly explored, offers the advantage of suc cinc t
representation of lengthy equations through the minimization of symbols. In addition,
physical laws expressed in tensor form are independent of the choice of c oordinate system,
and therefore similarities in seemingly different physic al systems are often made more
apparent. That is, indicial notation generally provides insight and understanding not readily
apparent to the relative newc omer to the field. It results in a saving of spac e and serves as
an aid in nonnumerical computation.

The displacement c omponents u, v, and w, for instance, are

written u1, u2, u3 (or ux , uy , uz) and c ollec tively as ui, with the understanding that the
subscript i can be 1, 2, and 3 (or x, y, z). Similarly, the coordinates themselves are
represented by x1, x2,x3, or simply xi(i = 1, 2, 3), and xx , xy , xz, or xi (i = x, y, z). Many
equations of elasticity become unwieldy when written in full, unabbreviated term; see, for
example, Eqs. (1.28). As the complexity of the situation desc ribed increases, so does that
of the formulations, tending to obsc ure the fundamentals in a mass of symbols. For this
reason, the more c ompac t indicial notation is sometimes found in publications.

Two simple c onventions enable us to write most equations developed in this text in indic ial
notation. These conventions, relative to range and summation, are as follows:

Range convention: When a lowerc ase alphabetic subscript is unrepeated, it

takes on all values indicated.

Summation convention: When a lowercase alphabetic subsc ript is repeated in a

term, then summation over the range of that subscript is indic ated, making
unnec essary the use of the summation symbol.

The introduction of the summation c onvention is attributable to A. Einstein (18791955).

This notation, in conjunc tion with the tensor c onc ept, has far-reac hing consequences not
restricted to its notational convenience [Refs. 1.14 and 1.15].