Lecture no.1

Introduction
External loads
Surface forces
Body Forces
Equilibrium
Internal loading / Internal Forces
Sign Conventions
Axial forces, SF, BM
General State of stress
General condition of loading
Components of stress

Analysis and design of any structure or machine involve two major
questions: (a) Is the structure strong enough to withstand the loads
applied to it and (b) is it stiff enough to avoid excessive deformations
and deflections? In Statics, the members of a structure were treated as
rigid bodies; but actually all materials are deformable and this property
will henceforth be taken into account. Thus Strength of Materials may
be regarded as the statics of deformable or elastic bodies.

BRANCHES OF ENGINEERING MECHANICS

Both the strength and stiffness of a structural member are functions of
its size and shape and also of certain physical properties of the material
from which it is made. These physical properties of materials are
largely determined from experimental studies of their behavior in a
testing machine. The study of Strength of Materials is aimed at
predicting just how these geometric and physical properties of a
structure will influence its behavior under service conditions. The
applications of the subject are broad in scope and will be found in all
branches of engineering.

The primary objective of the Strength of material / Mechanic of
Material is the development of relationships between the loads applied
to a nonrigid body and the internal forces and deformations induced in
the body.
Mechanics of materials is a branch of mechanics that develops
relationships between the external loads applied to a deformable body
and the intensity of internal forces acting within the body. This subject
is also concerned with computing the deformations of the body, and it
provides a study of the body's stability when the body is subjected to
external forces.

The forces that act on a structure include the applied
loads and the resulting reaction forces.
The applied loads are the known loads that act on a
structure. They can be the result of the structure’s own
weight, occupancy loads, environmental loads, and so on.
The reactions are the forces that the supports exert on a
structure. They are considered to be part of the external
forces applied.

Forces on a structure can arise from many sources, such as the
structure's own weight, any objects placed on it, wind pressure and so
forth. Force is a vector quantity, that is, it has both magnitude and
direction. The SI unit of force is the Newton (N), which is defined as
the force required to impart an acceleration of one metre per second per
second to a mass of one kilogram (that is, 1 N = 1 kg m/s2). An object
placed on a structure will thus impart a vertical force equal to its mass
multiplied by the acceleration due to gravity (g = 9.81 m/s2).

Surface Forces. As the name implies, surface forces are caused by the
dir contact of one body with the surface of another. In all cases these
forces distributed over the area of contact between the bodies, Fig. 1
1a. In particular, if this area is small in comparison with the total
surface area of the body then the surface force may be idealized as a
single concentrated force, which is applied to a point on the body, Fig.
11a. For example, this might be done to represent the effect of the
ground on the wheels of a bicycle when studying the loading on the
bicycle. If the surface loading is applied along a narrow area, the
loading may be idealized as a linear distributed load, w(s). Here the
loading is measured as having an intensity of force/length along the
area and is represented graphically by a series of arrows along the line s

Concentrated force
idealization

Linear distributed
load idealization

Fig. 1-1 (a)

s, Fig. 11a. The loading along the length of a beam is a typical
example of where this idealization is often applied, Fig. 11b. The
resultant force FR of w(s) is equivalent to the area under the distributed
loading curve, and this resultant acts through the centroid C or
geometric center of this area.
Body Force. A body force occurs when one body exerts a force on
another body without direct physical contact between the bodies.
Examples include the effects caused by the earth's gravitation or its
electromagnetic field. Although body forces affect each of the particles
composing the body, these forces are normally represented by a single
concentrated force acting on the body. In the case of gravitation, this
force is called the weight of the body and acts through the body's center
of gravity.

The forces on a body can also give rise to moments, which tend to cause
the body to rotate about an axis. The moment of a force about an axis is
simply equal to the magnitude of the force multiplied by the
perpendicular distance from the axis to the line of action of the force.
Consider, for example, the lever AB shown in Figure 2.1 (a). The effect
of the force P acting at B is to impart both a direct force P and a
moment M = Pa on the hinge at A, as shown in Figure 2.1 (b).
For a general case of forces and moments in space each force in
the resultant of three forces Fx, Fy and Fz and similarly each moment is
the resultant of three couples Mx, My and Mz.

SUMMARY
Type of force system possible resultants
Collinear …………………………………………… Force
concurrent , coplanar ……………………………. Force
parallel , coplanar…………………………………. Force or a couple
Nonconcurrent , nonparallel ,coplanar ………… Force or a couple
concurrent , noncoplanar………………………… Force
parallel , noncoplanar ……………………………. Force or a couple
Nonconcurrent , nonparallel , noncoplanar…….. Force or a couple , or a force and a couple

The surface forces that develop at the supports or points of
support between bodies are called reactions. For twodimensional
problems, i.e., bodies subjected to coplanar force systems, the supports
most commonly encountered are shown in Table 11. Note the symbol
used to represent each support and the type of reactions it exerts on its
contacting member. One possible way to determine a type of support
reaction is to imagine the attached member as being translated or
rotated in a particular direction. If the support prevents translation in a
given direction, then a force must be developed on the member in that
direction. Likewise, if rotation is prevented, a couple moment must be

exerted on the member. For example, a roller support only prevents
translation in the contact direction, perpendicular or normal to the
surface. Hence, the roller exerts a normal force F on the member at the
point of contact. Since the member is free to rotate about the roller, a
couple moment cannot be developed by the roller on the member at the
point of contact.
Remember that the concentrated forces and couple moment
shown in Table 11 actually represent the resultants of distributed
surface forces that exist between each support and its contacting
member. Although it is these resultants that are determined in practice,
it is generally not important to determine the actual surface load
distribution, since the area over which it acts is considerably smaller

than the total surface area of the contacting member.

2.8 SUPPORTS AND EXTERNAL REACTIONS

In structural analysis terminology the consequence of subjecting a
structure to an action is termed the "response" of the structure. A
structure subjected to an action (whether static or dynamic) will move
(translate and/or rotate) indefinitely unless the action is resisted in some
way. "Supports" are provided as the means of preventing free
movement of the structure. A statically loaded structure will deform
into a new shape but remain attached to its supports and at rest in its
new position. A dynamically loaded structure will remain attached to

supports but its deformed shape will change as a function of time.
"Damping" will eventually bring it to rest but a finite length of time is
subjected to simultaneously applied loads and come to rest
instantaneously. Supports prevent free motion of the structure as a
whole by developing forces to counteract the load actions. The
counteracting forces are commonly termed "external reactions" or
simply "reactions."

Equilibrium of a body requires both a balance of forces, to prevent the
body from translating or moving along a straight or curved path, and a
balance of moments, to prevent the body from rotating. These
conditions can be expressed mathematically by the two vector
equations
ΣF =0
(1–1)
Σ MO = 0

Here, Σ F represents the sum of all the forces acting on the body, and Σ
MO is the sum of the moments of all the forces about any point O either
on or off the body. If an x, y, z coordinate system is established with the
origin at point O, the force and moment vectors can be resolved into
components along the coordinate axes and the above two equations can
be written in scalar form as six equations, namely,

Σ Fx = 0 Σ Fy = 0 Σ Fz = 0
(1–2)
ΣMx = 0 ΣMy = 0 ΣMz = 0

Often in engineering practice the loading on a body can be
represented as a system of coplanar forces. If this is the case, and the
forces lie in the x–y plane, then the conditions for equilibrium of the
body can be specified by only three scalar equilibrium equations; that
is,

Σ Fx = 0

Σ Fy = 0 (1–3)

Σ MO = 0

In this case, if point O is the origin of coordinates, the moments will be
directed along the z axis, which is perpendicular to the plane that
contains the forces.
Successful application of the equations of equilibrium requires
complete specification of all the known and unknown forces that act on
the body. The best way to account for these forces is to draw the body's
free-body diagram. Obviously, if the freebody diagram is drawn
correctly, the effects of all the applied forces and couple moments can
be accounted for when the equations of equilibrium are written.

Consider a body of arbitrary shape acted upon by the forces shown in
Fig. 12. In statics, we would start by determining the resultant of the
applied forces to determine whether or not the body remains at rest. If
the resultant is zero, we have static equilibrium—a condition generally
prevailing in structures. If the resultant is not zero, may apply inertia
forces to bring about dynamic equilibrium. Such cases are discussed
later under dynamic loading. For the present, we consider only cases
involving static equilibrium.
In strength of materials, we make an additional investigation of
the internal distribution of the forces. This is done by passing an

Figure 12 Exploratory section a-a through loaded member.

exploratory section aa through the body and exposing the internal
forces acting on the exploratory section that are necessary to maintain
the equilibrium of either segment. In general, the internal forces reduce
to a force and a couple that, for convenience, are resolved into
components that are normal and tangent to the section, as shown in Fig.
13.
The origin of the reference axes is always taken at the centroid
which is the key reference point of the section. Although we are not yet
ready to show why this is so, we shall prove it as we progress; in
particular, we shall prove it for normal forces in the next article. If the x
axis is normal to the section, the section is known as the x surface or,
more briefly, the x face.

Figure 1-3
Components of internal effects on exploratory section a-a.

The notation used in Fig. 13 identifies both the exploratory section and
the direction of the force or moment component. The first subscript
denotes the face on which the component acts; the second subscript
indicates the direction of the particular component. Thus Pxy is the force
on the x face acting in the y direction.

Each component reflects a different effect of the applied loads
on the member and is given a special name, as follows:

Pxx Axial force. This component measures the pulling (or pushing)
action perpendicular to the section. A pull represents a tensile
force that tends to elongate the member, whereas a push is a
compressive force that tends to shorten it. It is often denoted by P.

Pxy, Pxz Shear forces. These are components of the total resistance
to sliding the portion to one side of the exploratory section
past the other. The resultant shear force is usually
designated by V, and its components by Vy and Vz to
identify their directions.
Mxx Torque. This component measures the resistance· to
twisting the member and is commonly given the symbol T.
Mxy, Mxz Bending moments. These components measure the
resistance to bending the member about the y or z axes and
are often denoted merely by My or Mz.

2.3 Internal forces in structures
So far, we have concentrated on the external forces applied to structures
the applied loads and the support reactions. In order for the structure to
transmit the external loads to the ground, internal forces must be
developed within the individual members. The aim of the design
process is to produce a structure that is capable of carrying all these
internal forces, which may take the form of axial forces, shear forces,
bending moments or torques.
Consider first a twodimensional beam where the applied forces
and reactions all lie in a single plane (Figure 2.14(a)). The internal
forces at a point in the structure can be found by splitting it at that point

and drawing free body diagrams for the two sides (Figure 2.14(b)). The
requirements of equilibrium state that not only must the resultant force
on the entire structure be zero, but the resultant on any segment of it
must also be zero. It is therefore clear that there must be forces acting at
the cut point, as shown. These are drawn on the free body diagrams of
the segmented structure as though they were external loads, but they are
in fact the internal forces in the beam. The forces can be thought of as
the external forces that would have to be applied to the cut beam in
order to produce the same deformations as in the original uncut beam.
The forces shown are an axial force T, a transverse force S, known as a
shear force, and a bending moment M.

For equilibrium at the cut point, the forces acting on the faces
either side of the cut must be equal and opposite; this means that, when
the two segments are put together to form the complete structure, there
is no resultant external load at that point.
For a member in threedimensional space, a total of six internal
forces must be considered, as shown in Figure 2.15. Here, there is again
an axial force T in the x direction, and the resultant shear force has been
resolved into components Sy and Sz parallel to the y and z axes

respectively. There are also moments about each of the three axes: My
tends to cause the structure to bend in the horizontal (x z) plane; Mz
causes bending in the vertical (x y) plane; Mx causes the member to

being, however, we will restrict ourselves to twodimensional systems.

2.3.1 Sign convention for internal forces

Before defining the internal forces more fully, we need to extend the
sign convention introduced earlier. For a twodimensional system,
positive forces act in the positive x and y directions, and a positive
moment about the z-axis acts from the positive x towards the positive y-
axis, that is anticlockwise. We can also define a positive face of a
member as one whose outward normal is in a positive axis direction.
Thus, for the beam segment in Figure 2.16(a), the righthand face is a
positive x-face, the top surface is a positive y-face and the other two
faces are negative.

We can now define a positive internal force as one which acts
either in a positive direction on a positive face or in a negative
direction on a negative face. Conversely, a negative force either acts in
a negative direction on a positive face or vice versa.

EQUILIBRIUM AND STRESS
We will not attempt a thorough review of equilibrium methods of statics, since
presumably the reader is familiar with analytical concepts that lead to computation of
equivalentforce systems and reactions for loadcarrying members. For now the
discussion will be conceptual. Later, certain topics of statics will be reviewed as
considered pertinent.
The beamtype structure of Fig. 1.1 is considered to be in equilibrium with
six actions occurring at each end of the beam. There are six and only six possible
actions that can occur: three forces and three couples, which are evaluated using the
equilibrium equations of statics. A section passed through the beam anywhere along
its axis can be viewed as shown in Fig. 1.2. Internal forces and couples act on the cut
section as illustrated and have magnitudes necessary to produce equilibrium for the
free body. Specifically, we classify the forces and couples as in Fig. 1.3. The force

z

x

FIGURE 1.1
Beamtype structure showing the six actions of statics.

vector acting along the beam axis is shown in Fig. 1.3a and causes either tension or
compression on the section, depending upon its direction. The remaining two force
vectors (Fig. 1.3b) produce shear loading on the cut section that is characterized by
the forces acting tangent to the cut section as opposed to acting normal to the cut
section as in the case of the axial force. The three couples of Fig. 1.2 are illustrated in
Figs. 1.3ce as vectors. The axialcouple vector of Fig. 1.3c represents a twisting
couple whose direction is determined using the righthandscrew rule. The twisting
couple, referred to as torque, causes a shear action to occur on the cut cross section.
The couples of Figs. 1.3d and e are referred to as bending moments, and the vector
representation is interpreted as illustrated. It turns out that these couples cause a
combination of tension and compression on the cut section.
The early chapters of this text are devoted to analytical methods for
computing the magnitude and direction of these six actions and then the computation
of the corresponding stresses. In particular, Chapter 2 deals with the action of Fig.

1.3a. Chapter 4 is devoted to the twisting couple of Fig. 1.3c applied to members of
circular cross section. Analytical methods for computing shear forces and bending
moments of Figs. 1.3b, d, and e are discussed in Chapter 5. Chapter 6 deals with the
methods for computing stress caused by the bending moments of Figs. 1.3d and e, and
Chapter 7 is devoted to the derivation of computational methods for the shear stress
produced by the shear forces of Fig. 1.3b. Loadcarrying members subject to various
combinations of the forces and couples are analyzed in Chapter 8. This analysis is the
culmination of the study of equivalent force systems and their associated stresses.

INTERNAL STRESSES AND STRESS
RESULTANTS (INTERNAL FORCES)

Framed members subjected to load must develop internal
stresses to resist the loads and prevent a material failure. The integrated
effects of stresses are force quantities of a particular magnitude and
direction. A frequent terminology used to refer to these force quantities
is stress resultants. “Internal forces” is an alternative terminology.
Internal force, that can be created in framed members are categorized
into four types.

1.Axial force
2.shear force
3.flexural moment
4.torsional moment

The determination of these force quantities is one of the basic aims of
structural analysis. The particular force quantities created in a given
structure depend upon its behavior. For a planar frame member, the
significant internal forces are the hear and axial direct forces and the
flexural moment. Figure 2.11 depicts the manner in which these
generalized forces are developed by the internal stresses. The resultants
of the shear stress distribution τ, and the axial stress distribution σ1,

also a major cause of deflections, and this subject is also discussed as is
deflection due to shear. Internal forces developed in each type of
framed structure are presented in Table 2.1.

Table 2.1. Internal Forces (Stress Resultants) for the
Basic Structural Types

Schematic of Assumed
Structural Type Description
Member Forces

Plane truss and Only axial Force, F 1, is
space truss significant.
F1

Only inplane shear force F1 ,
flexural moment F2 , and axial
Beam and plane
frame force F3, are significant. For
beams, axial force generally
F3 does not exist.
F2

Schematic of Assumed
Structural Type Description
Member Forces

F1 Only shear force F1 and
flexural moment F2 (both
Grid transverse to the plane of the
grid) and the torsional moment
F2 F3 F3 are significant.

F2 F5
All stress resultants are signif
icant: axial force F1, shear
Space Frame forces F2 and F3 torsional
F3 moments F4, and flexural
F1
moments F5 and F6·
F6 F4

PROBLEM: ANALYZE THE TRUSS BY STIFFNESS
METHOD

ANALYSIS OF GRIDS

Definition:

A grid is a structure that has loads applied perpendicular to its plane.
The members are assumed to be rigidly connected at the joints.
A very basic example of a grid structure is floor system as shown

y-axis

Wzzj,δ zzj
Wzzi,δ zzi

Wzj,δ zj
Wzi,δ zi

x-axis
Wxxi,δ xxi Wxi,δ xi Wxxj,δ xxj Wxj,δ xj

Wyi,δ yi
is

Wyj,δ yj
x
z-a

Wyyi,δ yyi Wyyj,δ yyj

FORCES AND DISPLACEMENT IN
LOCAL COORDINATES

when integrated over the member cross section, are the transverse shear
force V and the longitudinal (or axial) force P, respectively. Flexural
stresses are the source of two internal forces. Tensile and compressive
region of this stress distribution,σ2 produce the compressive force C and
tensile force T, respectively. Unlike the resultant P. which acts at the
centroid. the resultants T and C are separated by a distance, and neither
of their lines of action coincides with the centroid. Consequently, the
net effect of T and C is the creation of an internal resisting moment, M.
Shear and moment resultants that develop in an individual
framed member exhibit a relationship to each other and to the
transverse loads applied to the member. These fundamental
interrelationships will be formulated in Chapter 3. Flexural moment is

It is common to refer to loads as “forces that are applied to a structure.”
"Applied" is taken to mean “having an identifiable location or point of
application.” Each of the six types of framed structures were illustrated
in Fig. 1.4. Inspection of that figure indicates the applied load types that
are included in each of the six idealized models.

1. Planar truss. Concentrated loads applied in two orthogonal
directions at the joints.
2. Space truss. Concentrated loads applied in three ol1hogonal
directions at the joints.
3. Beams, a. Transverse loads and flexural moments that are applied
along the member and in the plane of the beam. These can be
concentrated or distributed in nature, b. concentrated moments

applied at the joints and acting in the plane of the beam.
• Grid. a. Concentrated or distributed loads (transverse to the plane of
the grid) and flexural moments applied along the member length. b,
Concentrated or distributed torsional moments acting along the
member length. c, Concentrated loads (transverse to the plane of the
frame) and outofplane moments applied at the joints.
• Planar frame. a, Transverse loads and flexural moments that are
applied along the member and in the plane of the frame. These can
be concentrated or distributed in nature. b, Concentrated loads in two
orthogonal directions and inplane moments applied at the joints
and in the plane of the frame.
• Space frame. a. Transverse loads and flexural moments that are

applied along the member length. These may be concentrated or
distributed in nature and act in any plane passing through the entire
member. b, Concentrated or distributed torsional moments acting
along the member length. c, Concentrated loads and moments
applied at the joints and in any of the three orthogonal planes.

Various load categories were described in Section 1.6. The
weight of objects (either dead or live load) and the hydrostatic pressure
of water are two examples of applied loads. In each case there is
contact between the load source and the loaded structure.
In a strict technical sense. loads arc not always applied to the
structure. Frequently. structures are subjected to phenomena not com
monly referred to as loads. Temperature change and shrinkage of

material are two examples. Each of these phenomena cause a structure
to experience strain and stress and consequently, to deform. These are
the same kinds of effects as caused by applied loads. When such effects
are included, it is conventional to refer to the general category of loads
as "actions." In this text the terms "load" and "action" are treated as
synonomous, both meaning any effect that causes stress and/or strain in
a structure.
any action must satisfy Newton’s first law, i.e., it must cause a
reaction. The fundamental concepts of how a structure an action are
developed in the balance of this chapter.

General State of stress
As we have stated, the six actions can combine to
produce a combination of stresses. The question of exactly
what is stress should be answered in a very simplistic manner
that is also quite exact in a theoretical sense. We idealize a
loadcarrying body as shown in Fig. 1.4 and say that the body
is a continuum. Essentially, a continuum is a

y

z x

FIGURE 1.4
Idealized body.

collection of material particles, and its exact size is not important for this discussion.
Materials are made up of clusters of molecules, and every material has a definite
molecular structure. On a microscopic scale a material is composed of a space with
atoms at specific locations. The continuum model is a collection of many molecules
and is large enough that the individual molecular interactions for the material can be
ignored and the total of all molecular interactions can be averaged and the continuum
can be assigned some overall gross property to describe its behavior. The continuum
is considered to be quite large compared to an atomistic model; however, it can still
be imagined to be small enough to be of differential size. In other words, we can
effect the mathematical concept of the limit of some quantity with respect to a length
dimension. We assume that as we find the limiting value of a quantity as a length
parameter approaches zero that we do not violate the material assumption of a
continuum; the continuum still exists even though it may be of differential size.

Consider the continuum of Fig. 1.4 to be in statical equilibrium and imagine
a slice taken through the continuum, as shown in Fig. 1.5. The continuum is still in
equilibrium since internal forces and couples at the slice can be viewed as external
balancing forces and couples. We are concerned with forces acting on the smooth, cut
surface that are illustrated as acting on individual, small surface areas. On a small
scale these forces are considered to be acting in a direction either normal or tangent to
their respective areas. The forces originate from the molecular interactions at the
microscopic level; however, as we have stated, we assume that microscopic forces are
averaged and their resultants act on the individual areas. We do not consider small
couples acting on each element even though they may exist. It has been demonstrated,
both experimentally and analytically, that smallbody couples can be ignored for the
general theory of mechanics of materials.
Every force depicted in Fig. 1.5 can be resolved into one normal component
and two shear components. The definition of normal and shear arises naturally;
normal

forces act normally to their area and shear forces act tangentially to their area.
Visualize one small area ΔA as shown in Fig. 1.5b. The force ΔF is shown in
components ΔFx, ΔFy, and ΔFz. Normal stress is denoted by σ (sigma) and is defined
as the normal force per unit area; hence, the terminology normal stress. Normal stress
is given as

∆Fx
σx = (1.1)
∆A
The Greek letter τ (tau) is usually used for shear stress and is the shear force divided
by the area, or

∆Fy
τy = (1.2)
∆A

and

∆Fz
τz = (1.3)
∆A
A more complete description is given in Chapter 9 concerning shear and normal
stresses that act at a point such as the small area of Fig. 1.5b.
Previously, it was noted that each action illustrated in Fig. 1.3 produced
either a normal stress or shear stress. In every case we begin with a definition of
stress, either normal or shear, as given by Eqs. (1.1), (1.2), and (1.3), and establish a
theory that describes how a particular action causes a stress. The distribution of stress
that occurs on the cross section of the member because of the action of forces and
couples is dependent upon the way the loadcarrying member deforms. In the case of
the axial force, Fig. 1.3a, the member either elongates or shortens in the axial
direction. If we assume a uniform deformation, meaning that every point on the cross

section deforms an equal amount parallel to the beam axis, then we must investigate
the limitation of the theory subject to that assumption.
It turns out that each of the six actions produces some corresponding
deformation of the member, and prior to developing a theory for stress distribution we
must establish the deformation characteristics of the member when subjected to a
particular load. Chapters 29 are devoted to the study of concepts that have been
briefly introduced in this section.

Internal Loadings. One of the most important applications of statics in
the analysis of problems involving mechanics of materials is to be able
to determine the resultant force and moment acting within a body,
which are necessary to hold the body together when the body is
subjected to external loads. For example, consider the body shown in
Fig. 1-2a, which is held in equilibrium by the four external forces.* In
order to obtain the internal Loadings acting on a specific region within
the body, it is necessary to use the method of sections. This requires
that an imaginary section or "cut" be made through the region where
the internal loadings are to be determined. The two parts of the body
are
* The body's weight is not shown, since it is assumed to be quite small, and therefore
negligible compared with the other loads.

then separated, and a freebody diagram of one of the parts is drawn. If
we consider the section shown in Fig. 1-2a, then the resulting freebody
diagram of the bottom part of the body is shown in Fig. 12b. Here it
can be seen that there is actually a distribution of internal force acting
on the "exposed" area of the section. These forces represent the effects
of the material of the top part of the body acting on the adjacent
material of the bottom part. Although this exact distribution may be
unknown, we can use statics to determine the resultant internal force
and moment, FR and MRo this distribution exerts at a specific point O on
the sectioned area, Fig. 1-2c.
Since the entire body is in equilibrium, then each part of the body
is also in equilibrium. Consequently, FR and MRo can be determined by

applying Eqs. 11 to anyone of the two parts of the sectioned body.
When doing so, note that FR acts through point 0, although its computed
value will not depend on the location of this point. On the other hand,
MRo does depend on this location, since the moment arms must extend
from O to the line of action of each force on the freebody diagram. It
will be shown in later portions of the text that point O is most often
chosen at the centroid of the sectioned area, and so we will always
choose this location for O, unless otherwise stated. Also, if a member is
long and slender, as in the case of a rod or beam, the section to be
considered is generally taken perpendicular to the longitudinal axis of
the member. This section is referred to as the cross section.

Later in this text we will show how to relate the resultant
internal force and moment to the distribution of force on the sectioned
area, Fig. 12b, and thereby develop equations that can be used for
analysis and design of the body. To do this, however, the components
of FR and MRo acting both normal or perpendicular to the sectioned area
and within the plane of the area, must be considered. If we establish x,
y, z axes with origin at point O, as shown in Fig; 12d, then FR and MRo
can each be resolved into three components. Four different types of
loadings can then be defined as follows:
Nz is called the normal force, since it acts perpendicular to the area.
This force is developed when the external loads tend to push or pull on

V is called the shear force, and it can be determined from its two
components using vector addition, V = Vx + Vy. The shear force lies in
the plane of the area and is developed when the external loads tend to
cause the two segments of the body to slide over one another.
Tz is called the torsional moment or torque. It is developed when the
external loads tend to twist one segment of the body with respect to the
other.
M is called the bending moment. It is determined from the vector
addition of its two components, M = Mx + My. The bending moment is
caused by the external loads that tend to bend the body about an axis
lying within the plane of the area.

In this text, note that representation of a moment or torque is
shown in three dimensions as a vector with an associated curl, Fig. I2d.
By the right-hand rule, the thumb gives the arrowhead sense of the
vector and the fingers or curl indicate the tendency for rotation (twist or
bending).

Each of the six unknown x, y, z components of force and
moment shows in Fig. 12d can be determined directly from the six
equations of equilibrium, that is, Eqs. 12, applied to either segment of
the body. If, however, the body is subjected to a coplanar system of
forces, then only normalforce, shear, and bendingmoment components
will exist at the section. To show this, consider the body in Fig. 13a. If
it is in equilibrium, then the internal resultant components, acting at the
indicated section, can be determined by first "cutting" the body into two
parts, as shown in Fig. 13b, and then applying the equations of
equilibrium to one of the parts. Here the internal resultants consist of
the normal force N, shear force V, and bending moment Mo. These
loadings must be equal in magnitude and opposite in direction on each

of the sectioned parts (Newton's third law). Furthermore, the magnitude
of each unknown is determined by applying the three equations of
equilibrium to either one of these parts, Eqs. 13. If we use the x, y, z
coordinate axes, with origin established at point O, as shown on the left
segment, then a direct solution for N can be obtained by applying ΣFx =

0, and V can be obtained directly from ΣFy = 0. Finally, the bending

moment Mo can be determined directly by summing moments about

point O (the z axis), Σ Mo = 0, in order to eliminate the moments caused
by the unknowns N and V.

The method of sections is used to determine the internal loadings at a
point located on the section of a body. These resultants are statically
equivalent to the forces that are distributed over the material on the
sectioned area. If the body is static, that is, at rest or moving with
constant velocity, the resultants must be in equilibrium with the
external loadings acting on either one of the sectioned segments of the
body.
We will now present a procedure that can be used for applying
the method of sections to determine the internal resultant normal force,
shear force, bending moment, and torsional moment at a specific

Support Reactions. First decide which segment of the body is to be
considered. Then before the body is sectioned, it will be necessary to
determine the support reactions or the reactions at the connections only
on the chosen segment of the body. This is done by drawing the free
body diagram for the entire body, establishing a coordinate system, and
then applying the equations of equilibrium to the body.

Free-Body Diagram. Keep all external distributed loadings, couple
moments, torques, and forces acting on the body in their exact
locations, then pass an imaginary section through the body at the point
where the internal resultant loadings are to be determined. If the body
represents a member of a structure or mechanical device, this section is

often taken perpendicular to the longitudinal axis of the member. Draw
a freebody diagram of one of the' 'cut" segments and indicate the
unknown resultants N, V, M, and T at the section. In most cases, these
resultants are placed at the point representing the geometric center or
centroid of the sectioned area. In particular, if the member is subjected
to a coplanar system of forces, only N, V, and M act at the centroid.
Establish the x, y, z coordinate axes at the centroid and show the
resultant components acting along the axes.

Equations of Equilibrium. Apply the equations of equilibrium to
obtain the unknown resultants. Moments should be summed at the
section, about the axes where the resultants act. Doing this eliminates

unknown forces N and V and allows a direct solution for M (and T). If
the solution of the equilibrium equations yields a negative value for a
resultant, the assumed directional sense of the resultant is opposite to
that shown on the freebody diagram.
The following examples illustrate this procedure numerically and also
provide a review of some of the important principles of statics. Since
statics plays such an important role in the analysis of problems in
mechanics of materials, it is highly recommended that one solves as
many problems as possible of those that follow these examples.

In Sec. 1.2 we showed how to determine the internal resultant force and
moment acting at a specified point on the sectioned area of a body, Fig.
19a. It was stated that these two loadings represent the resultant effects
of the actual distribution of force acting over the sectioned area, Fig. 1
9b. Obtaining this distribution of internal loading, however, is one of
the major problems in mechanics of materials.

Later it will be shown that to solve this problem it will be
necessary to study how the body deforms under load, since each
internal force distribution will deform the material in a unique way.
Before this can be done, however, it is first necessary to develop a
means for describing the internal force distribution at each point on the
sectioned area. To do this we will establish the concept of stress.
Consider the sectioned area to be subdivided into small areas,
such as the one ΔA shown shaded in Fig. 19b. As we reduce ΔA to a
smaller and smaller size, we must make two assumptions regarding the
properties of the material. We will consider the material to be
continuous, that is, to consist of a continuum or uniform distribution of
matter having no voids, rather than being composed of a finite number

of distinct atoms or molecules. Furthermore, the material must be
cohesive, meaning that all portions of it are connected together, rather
than having breaks, cracks, or separations. Now, as the subdivided area
ΔA of this continuouscohesive material is reduced to one of
infinitesimal size, the distribution of force acting over the entire
sectioned area will consist of an infinite number of forces, each acting
on an element ΔA located at a specific point on the sectioned area. A
typical finite yet very small force ΔF, acting on its associated area ΔA,
is shown in Fig. 19c. This force, like all the others, will have a unique
direction, but for further discussion we will replace it by two of its
components, namely, ΔFn and ΔFt, which are taken normal and tangent
to the area, respectively. As the area Δ A approaches zero, so do the

force ΔF and its components; however, the quotient of the force and
area will, in general, approach a finite limit. This quotient is called
stress, and as noted, it describes the intensity of the internal force on a
specific plane (area) passing through a point.

Normal Stress. The intensity of force, or force per unit area, acting
normal to ΔA is defined as the normal stress, σ(sigma). Mathematically
it can be expressed as

∆Fn
σ = lim (1–4)
∆A→0 ∆A

If the normal force or stress "pulls" on the area element ΔA as shown in
Fig. I9c, it is referred to as tensile stress, whereas if it "pushes" on ΔA
it is called compressive stress.
Shear Stress. Likewise, the intensity of force, or force per unit area,
acting tangent to ΔA is called the shear stress, τ(tau). This component
is expressed mathematically as

∆Ft
τ = lim (1–5)
∆A→0 ∆A

In Fig. I9c, note that the orientation of the area ΔA completely
specifies the direction of ΔFn, which is always perpendicular to the area.

On the other hand, each shear force ΔFt can act in an infinite number of
directions within the plane of the area. Provided, however, the direction
of ΔF is known, then the direction of ΔFt can be established by
resolution of ΔF as shown in the figure.
Cartesian Stress Components. To specify further the direction of the
shear stress, we will resolve it into rectangular components, and to do
this we will make reference to x, y, z coordinate axes, oriented as shown

in Fig. 1–10a. Here the element of area ΔA = Δx Δy and the three
Cartesian components of ΔF are shown in Fig. 110b. We can now
express the normalstress component as

∆Fz
σ z = lim
∆A→0 ∆A

z

F1

(b)

Fig. 1.11

z

(d)

(c) Fig. 1.11

and the two shearstress components as

∆Fx
τ zx = lim
∆A→0 ∆A

∆Fy
τ zy = lim
∆A→0 ∆A

The subscript notation z in σz is used to reference the direction of the
outward normal line, which specifies the orientation of the area ~A.
Two subscripts are used for the shearstress components, τzx and τzy.
The z specifies the orientation of the area, and x and y refer to the
direction lines for the shear stresses.

To summarize these concepts, the intensity of the internal force
at a point in a body must be described on an area having a specified
orientation. This intensity can then be measured using three
components of stress acting on the area. The normal component acts
normal or perpendicular to the area, and the shear components act
within the plane of the area. These three stress components are shown
graphically in Fig. 1l0c.
Now consider passing another imaginary section through the
body parallel to the x–z plane and intersecting the front side of the
element shown in Fig.110a. The resulting freebody diagram is shown
in Fig. 111a. Resolving the force acting on the area ΔA = Δx Δz into its
rectangular components, and then determining the intensity of these

force components, leads to the normal stress and shearstress
components shown in Fig. 111b. Using the same notation as before, the
subscript y in σy, τyx, and τyz refers to the direction of the normal line

associated with the orientation of the area, and x and z in τyx and τyz
refer to the corresponding direction lines for the shear stress. Lastly,
one more section of the body parallel to the y–z plane, as shown in Fig.
111c, gives rise to normal stress σx and shear stresses τxy and τxz, Fig. 1
11d. If we continue in this manner, using corresponding parallel x
planes, we can "cut out" a cubic volume element of material that
represents the state of stress acting around the chosen point in the body,
Fig. 112.

Equilibrium Requirements. Although each of the six faces of the
element in Fig. 112 will have three components of stress acting on it, if
the stress around the point is constant, some of these stress components
can be related by satisfying both force and moment equilibrium for the
element. To show the relationships between the components we will
consider a freebody diagram of the element, Fig. 113a. This element
has a volume of ΔV = Δx Δy Δz, and in accordance with Eqs. 14 and 1
5, the forces acting on each face are determined from the product of the
average stress times the area of the face. For simplicity, we have not
labeled the "dashed" forces acting on the "hidden" sides of the element.
Instead, to view, and thereby label, some of these forces, the element is
shown from a front view in Fig. 113b. Here it should be noted that the

z

x
Fig. 1.13 (a) Element free-body diagram

z

Fig. 1.13 (b) Element free-body diagram

force components on the "hidden" sides of the element are designated
with stresses having primes, and these forces are shown in the opposite
direction to their counterparts acting on the opposite faces of the
element.
If we now consider force equilibrium in the y direction, we have

CONCEPT OF STRESS AT A GENERAL
POINT IN AN ARBITRARILY LOADED
MEMBER

In Arts. 13 and 14 the concept of stress was introduced by considering
the internal force distribution required to satisfy equilibrium in a
portion of a bar under centric load. The nature of the force distribution
led to uniformly distributed normal and shearing stresses on transverse
planes through the bar. In more complicated structural members or
machine components the stress distributions will not be uniform on
arbitrary internal planes; therefore. a more general concept of the state
of stress at a point is needed.

Consider a body of arbitrary shape that is in equilibrium under
the action of a system of applied forces. The nature of the internal force
distribution at an arbitrary interior point O can be studied by exposing
an interior plane through O as shown in Fig. 113a. The force
distribution required on such an interior plane to maintain equilibrium
of the isolated part of the body, in general, will not be uniform;
however, any distributed force acting on the small area ΔA surrounding
the point of interest O can be replaced by a statically equivalent
resultant force ΔFn through O and a couple ΔMn. The subscript n
indicates that the resultant force and couple are associated with a
particular plane through Onamely, the one having an outward normal
in the n direction at O. For any other plane through O the values of ΔF

and ΔM could be different. Note that the line of action of ΔFn or ΔMn

may not coincide with the direction of n. If the resultant force ΔFn is
divided by the area ΔA, an average force per unit area (average
resultant stress) is obtained. As the area ΔA is made smaller and
smaller, the couple ΔMn vanishes as the force distribution becomes
more and more uniform. In the limit a quantity known as the stress
vector4 or resultant stress is obtained. Thus,

4
The component of a tensor on a plane is a vector; therefore, on
a particular plane, the stresses can be treated as vectors.

∆Fn
S n = lim
∆A→0 ∆A

In Art. 13 it was pointed out that materials respond to components of
the stress vector rather than the stress vector itself. In particular, the
components normal and tangent to the internal plane were important.
As shown in Fig. 1-13b the resultant force ΔFn can be resolved into the

components ΔFnn and ΔFnt. A normal stress σn and a shearing stress τn
are then defined as
∆Fnn
σ n = lim
∆A→0 ∆A

∆Fnt
τ n = lim
∆A→0 ∆A

For purposes of analysis it is convenient to reference stresses to some
coordinate system. For example, in a Cartesian coordinate system the
stresses on planes having outward normals in the x, y. and z directions
are usually chosen. Consider the plane having an outward normal in the
x direction. In this case the normal and shear stresses on the plane will
be σx and τx, respectively. Since τx, in general, will not coincide with

the y or z axes, it must be resolved into the components τxy and τxz, as
shown in Fig. 113c. Unfortunately the state of stress at a point in a

stress vector since the stress vector itself depends on the orientation of
the plane with which it is associated. An infinite number of planes can
be passed through the point, resulting in an infinite number of stress
vectors being associated with the point. Fortunately it can be shown
(see Art. 19) that the specification of stresses on three mutually
perpendicular planes is sufficient to describe completely the state of
stress at the point. The rectangular components of stress vectors on
planes having outward normals in the coordinate directions are shown
in Fig. 114. The six faces of the small element are denoted by the
directions of their outward normals so that the positive x face is the one
whose outward normal is in the direction of the positive x axis. The
coordinate axes x, y, and z are arranged as a righthand system. The

sign convention for stresses is as follows. Normal stresses (indicated by
the symbol σ and a single subscript to indicate the plane on which the
stress acts) are positive if they point in the direction of the outward
normal. Thus normal stresses are positive if tensile. Shearing stresses
are denoted by the symbol τ followed by two subscripts; the first
subscript designates the plane on which the shearing stress acts and the
second the coordinate axis to which it is parallel. Thus, τxy is the
shearing stress on an x plane parallel to the z axis. A positive shearing
stress points in the positive direction of the coordinate axis of t he
second subscript if it acts on a surface with an outward normal in the
positive direction. Conversely, if the outward normal of the surface is in
the negative direction, then the positive shearing stress points in the

negative direction of the coordinate axis of the second subscript. The
stresses shown on the element in Fig. 114 are all positive.

STRESS UNDER GENERAL LOADING
CONDITIONS; COMPONENTS OF STRESS

The examples of the previous sections were limited to members under axial
loading and connections under transverse loading. Most structural members and
machine components are under more involved loading conditions.
Consider a body subjected to several loads P 1, P2, etc. (Fig. 1.32). To
understand the stress condition created by these loads at some point Q within the
body, we shall first pass a section through Q, using a plane parallel to the yz plane.
The portion of the body to the left of the section is subjected to some of the original
loads, and to normal and shearing forces distributed over the section. We shall denote
by ΔFx and ΔVx, respectively, the normal and the shearing forces acting on a small

area ΔA surrounding point Q (Fig. 1.33a). Note that the superscript x is used to
indicate that the forces ΔFx and ΔVx act on a surface perpendicular to the x axis.
While the normal force ΔFx has a welldefined direction, the shearing force ΔVx may
have any direction in the plane of the section. We therefore resolve ΔVx into two
component forces, ΔVxy and ΔVxz in directions parallel to the y and z axes,
respectively (Fig. 1.33 b). Dividing now the magnitude of each force by the area ΔA,
and letting ΔA approach zero, we define the three stress components shown in Fig.
1.34:
x
∆F
σ x = lim
∆A→0 ∆A

∆V yx ∆Vzx
τ xy = lim τ xz = lim (1.18)
∆A→0 ∆A ∆A→0 ∆A

y τxy

x
z
Fig. 1.34

We note that the first subscript in σx, τxy and τxz is used to indicate that the stresses
under consideration are exerted on a surface perpendicular to the x axis. The second
subscript in τxy and τxz identifies the direction of the component. The normal stress σx
is positive if the corresponding arrow points in the positive x direction, i.e., if the
body is in tension, and negative otherwise. Similarly, the shearing stress components
τxy and τxz are positive if the corresponding arrows point, respectively, in the positive y
and z directions.
The above analysis may also be carried out by considering the portion of
body located to the right of the vertical plane through Q (Fig. 1.35). The same
magnitudes, but opposite directions, are obtained for the normal and shearing forces
ΔFx, ΔVxy and ΔVxz Therefore, the same values are also obtained for the
corresponding stress components, but since the section in Fig. 1.35 now faces the
negative x axis, a positive sign for σx will indicate that the corresponding arrow points

corresponding arrows point, respectively, in the negative y and z directions, as shown
in Fig. 1.35.
y

σx

Fig. 1.35
z

Passing a section through Q parallel to the zx plane, we define in the same
manner the stress components, σy, τyz, and τyx Finally, a section through Q parallel to

the xy plane yields the components σz, τzx and τzy.
To facilitate the visualization of the stress condition at point Q, we shall
consider a small cube of side a centered at Q and the stresses exerted on each of the
six faces of the cube (Fig. 1.36). The stress components shown in the figure are σx, σy

and σz which represent the normal stress on faces respectively perpendicular to the x,

y, and z axes, and the six shearing stress components τxy τxz etc. We recall that, ac

cording to the definition of the shearing stress components, τxy represents the y
component of the shearing stress exerted on the face perpendicular to the x axis, while
τyx represents the x component of the shearing stress exerted on the face perpendicular
to the y axis. Note that only three faces of the cube are actually visible in Fig. 1.36,
and that equal and opposite stress components act on the hidden faces. While the

involved is small and vanishes as side a of the cube approaches zero.
Important relations among the shearing stress components will now be
derived. Let us consider the freebody diagram of the small cube centered at point Q
(Fig. 1.37). The normal and shearing forces acting on the various faces of the cube are
obtained by multiplying the corresponding stress components by the area ΔA of each
face. We first write the following three equilibrium equations:

∑F x =0 ∑F y =0 ∑F z =0 (1.19)

Since forces equal and opposite to the forces actually shown in Fig. 1.37 are acting on
the hidden faces of the cube, it is clear that Eqs. (1.19) are satisfied. Considering now
the moments of the forces about axes Qx', Qy', and Qz' drawn from Q in directions
respectively parallel to the x, y, and z axes, we write the three additional equations

∑M x =0 ∑M y =0 ∑M z =0 (1.20)

Using a projection on the x'y' plane (Fig. 1.38), we note that the only forces with
moments about the z axis different from zero are the shearing forces. These forces
form two couples, one of counterclockwise (positive) moment (τxy ΔA)a, the other of

clockwise (negative) moment –(τxy ΔA)a. The last of the three Eqs. (1.20) yields,
therefore,
+ ↑ ∑Mz = 0: (τ xy ∆A) a − (τ yx ∆A) a = 0 (1.21)
from which we conclude that
τ xy = τ yx

The relation obtained shows that the y component of the shearing stress exerted on a
face perpendicular to the x axis is equal to the x component of the shearing stress
exerted on a face perpendicular to the y axis. From the remaining two equations
(1.20), we derive in a similar manner the relations

similarly, τ yz = τ zy and τ yz = τ zy (1.22)

We conclude from Eqs. (1.21) and (1.22) that only six stress components are
required to define the condition of stress at a given point Q, instead of nine as
originally assumed. These six components are ux, uy, uz, T XY' Tyz, and T ZX. We
also note that, at a given point, shear cannot take place in one plane only; an equal
shearing stress must be exerted on another plane perpendicular to the first one. For
example, considering again the bolt of Fig. 1.29 and a small cube at the center Q of
the bolt (Fig. 1.39a), we find that shearing stresses of equal magnitude must be
exerted on the two horizontal faces of the cube and on the two faces that are
perpendicular to the forces P and P' (Fig. 1.39b). Before concluding our discussion of
stress components, let us consider again the case of a member under axial loading. If
we consider a small cube with faces respectively parallel to the faces of the member
and recall the results obtained in Sec. 1.11, we find that the conditions of stress in the
member may be described as shown in Fig. lAOa; the only stresses are normal
stresses U x exerted on the faces of the cube which are perpendicular to the x axis.

is rotated by 45° about the z axis so that its new orientation matches the orientation
of the sections considered in Fig. 1.31c and d, we conclude that normal and shearing
stresses of equal magnitude are exerted on four faces of the cube (Fig. 1.40b). We
thus observe that the same loading condition may lead to different interpretations of
the stress situation at a given point, depending upon the orientation of the element
considered. More will be said about this in Chap 7.

Stress, Strain, and Their Relationships

2.1 Introduction

The concepts of stress and strain are two of the most important concepts within the
subject of mechanics of materials or mechanics of deformable bodies. They are
discussed in detail in this chapter, particularly as they relate to twodimensional
situations.
In the case of stress as well as in the case of strain, emphasis is placed on the
use of the semigraphical procedure known as the Mohr’s circle solution. The
underlying mathematical concepts leading to Mohr's circle are developed and
discussed. Examples are solved to illustrate the use of this powerful semigraphical
method of solution, which will be used throughout this book whenever problems are
encountered dealing with stress and strain analysis.

The treatment given to the concepts of stress and strain in this book differs
from that in other books in several important respects, the most significant of which is
the fact that the sign convention adopted for strain is compatible with the sign
convention for stress as they relate to the construction of the corresponding Mohr's
circles. This approach is advantageous in that it makes the construction of the stress
and strain Mohr's circles identical.
The discussion relating stress to strain in this chapter is limited to the range
of material behavior within which the strain varies linearly with stress. This procedure
frees the students from information which, although very important, is extraneous for
the time being. A more complete discussion of material behavior is provided in
Chapters 3 and 4.

2.2 Concept of Stress at a Point

If a body is subjected to external forces, a system of internal forces is
developed. These internal forces tend to separate or bring closer together the material
particles that make up the body. Consider, for example, the body shown in Figure
2.1(a), which is subjected to the external forces Fl, F2, ... , Fi. Consider an imaginary
plane that cuts the body into two parts, as shown. Internal forces are transmitted from
one part of the body to the other through this imaginary plane. Let the freebody
diagram of the lower part of the body be constructed as shown in Figure 2,1 (b). The
forces F1, F2, and F3 are held in equilibrium by the action of an internal system of
forces distributed in some manner through the surface area of the imaginary plane.
This system of internal forces may be represented by a single resultant force R and/or
by a couple. For the sake of simplicity in introducing the concept of stress, only the
force R is assumed to exist. In general, the force R may be decomposed into a

component Fn perpendicular to the plane and known as the normal force, and a

component Ft, parallel to the plane and known as the shear force.

If the area of the imaginary plane is to be A, then Fn / A and Ft / A represent,
respectively, average values of normal and shear forces per unit area called stresses.
These stresses, however, are not, in general, uniformly distributed throughout the area
under consideration, and it is therefore desirable to be able to determine the
magnitude of both the normal and shear stresses at any point within the area. If the
normal and shear forces acting over a differential element of area ΔA in the
neighborhood of point O are ΔFn and ΔFt respectively, as shown in Figure 2.1 (b),
then the normal stress σ and the shearing stress, are given by the following
expressions: ∆Fn
σ = lim
∆A→0 ∆A
(2.1)
∆Ft
τ = lim
∆A → 0 ∆A

In the special case where the components Fn and Fr are uniformly distributed over
the entire area A, then σ = Fn/A and τ = Ft/A.
Note that a normal stress acts in a direction perpendicular to the plane on
which it acts and it can be either tensile or compressive. A tensile normal stress is one
that tends to pull the material particles away from each other, while a compressive
normal stress is one that tends to push them closer together. A shear stress, on the
other hand, acts parallel to the plane on which it acts and tends to slide (shear)
adjacent planes with respect to each other. Also note that the units of stress (σ or τ)
consist of units of force divided by units of area. Thus, in the British gravitational
system of measure, such units as pounds per square inch (psi) and kilopounds per
square inch (ksi) are common. In the metric (SI) system of measure, the unit that has
been proposed for stress is the Newton per square meter (N/m 2), which is called the
pascal and denoted by the symbol Pa. Because the Pascal is a very small quantity,
another SI unit that is widely used is the mega Pascal (106 pascals) and is denoted by
the symbol MPa. This unit may also be written as MN/m2.

Components of Stress

In the most general case, normal and shear stresses at a point in a body may
be considered to act on three mutually perpendicular planes. This most general state
of stress is usually referred to as triaxial. It is convenient to select planes that are
normal to the three coordinates axes x, y, and z and designate them as the X, Y, and Z
planes, respectively. Consider these planes as enclosing a differential volume of
material in the neighborhood of a given point in a stressed body. Such a volume of
material is depicted in Figure 2.2 and is referred to as a three-dimensional stress
element. On each of the three mutually perpendicular planes of the stress element,
there acts a normal stress, and a shear stress which is represented by its two
perpendicular components.
The notation for stresses used in this text consists of affixing one subscript to
a normal stress, indicating the plane on which it is acting, and two subscripts to a

shear stress, the first of which designates the plane on which it is acting and the
second its direction. For example, σx is a normal stress acting on the X plane, τxy is a

shear stress acting on the X plane and pointed in the positive y direction, and τxz is a
shear stress acting in the X plane and pointed in the positive z direction.
It is observed from Figure 2.2 that three stress components exist on each of
the three mutually perpendicular planes that define the stress element. Thus there
exists a total of nine stress components that must be specified in order to define
completely the state of stress at any point in the body. By considerations of the
equilibrium of the stress element, it can easily be shown that, τxy = τyx, τxz = τzx, and τyz

= τzy so that the number of stress components required to completely define the state
of stress at a point is reduced to six.

By convention, a normal stress is positive if it points in the direction of the outward
normal to the plane. Thus a positive normal stress produces tension and a negative
normal stress produces compression. A component of shear stress is positive if it is
pointed along the positive direction of the coordinate axis and if the outward normal
to its plane is also in the positive direction of the corresponding axis. If, however, the
outward normal is in the negative direction of the coordinate axis, a positive shear
stress will also be in the negative direction of the corresponding axis. The stress
components shown in Figure 2.2 are all positive. It should be noted, however, that
such a sign convention for shear stress is rather cumbersome. It is only used in the
analysis of triaxial stress problems that are usually dealt with in advanced courses
such as the theory of elasticity.
A complete study of the triaxial or threedimensional state of stress is beyond the
scope of this chapter, and the analysis that follows is limited to the special case in
which the stress components in one direction are all zero. For example, if all the stress

components in the z direction are zero (i.e., τxz = τyz = τz = 0), the stress condition
reduces to a biaxial or twodimensional state of stress in the xy plane. This state of
stress is referred to as plane stress. Fortunately, many of the problems encountered in
practice are such that they can be considered plane stress problems.

2.4 Analysis of Plane Stress
As mentioned previously the state of stress known as plane stress is one in
which all the stress components in one direction vanish. Thus, if it is assumed that all
the components in the z direction shown in Figure 2.2 are zero (i.e., τxz = τyz = τz = 0),
the stress element shown in Figure 2.3(a) is obtained and it is the most general plane
stress condition that can exist. It should be observed that a stress element is in reality
a schematic representation of two sets of perpendicular planes passing through a point
and that the element degenerates into a point in the limit when both dx and dy
approach zero.

From considerations of the equilibrium of forces on the stress element in Figure
2.3(a), it can be shown that τxy = τyx. Thus assume that the depth of the stress element
into the paper is a constant equal to h. Since, by definition, force is the product of
stress and the area over which it acts, then a summation of moments of all forces
about a z axis through point O leads to the following equation:

τ xy = (h dy )dx − τ yx = (h dx)dy = 0
from which

τ xy = τ yx

STRESS AT A POINT
Stress at a point is terminology that means exactly what it says. Refer to Fig.
8.2 of Example 8.1. Stress at a point C would imply that the stresses at point C
are to be computed. The point must be drawn large enough for it to be
visualized. Therefore the concept of a stress block or material element is
necessary. The stress block is the point enlarged for the practical purpose of
drawing it and is referred to as a stress element since its actual size is
elemental. Point C of Fig. 8.2 is shown enlarged in Fig. 8.la. Physically, the
element can be visualized as a small square near the outside surface of the
beam located at point C.
The stresses are identified as acting on the edge of the elemental
square and is the standard concept of stress at a point in two dimensions.

(a)

FIGURE 8.1
(a)Element viewed along the z
axis;
(b) positive stresses acting on an x
element; (b)

(c)

FIGURE 8.1
(c) element viewed along the x axis;
(d) element viewed along the y axis. (d)

Actually, the elemental square is an elemental cube located at point C.
The cube is shown in Fig. 8.1b. The subscript notation identifies the direction
of the stress at the face, or area, of the cube on which it acts. For instance, the
stress σxx is the normal stress acting on the face perpendicular to the x axis.
The first subscript identifies the face, or area, of the cube. The second
subscript identifies the direction of the stress. A shear stress on the face of the
cube normal to the y axis and acting in the x direction would be τyx as shown
in Fig. 8.1b. The complete threedimensional description of stress at a point is
illustrated in the figure.
The discussion in the previous chapters concerning shear stresses on
mutually perpendicular planes would imply that τxy = τyx τyz = τzy and τxz = τzx.
The subscript notation lends itself toward identifying this equivalence. The
nine components shown in Fig. 8.lb are conveniently presented using the
following format.

(8.5)

The use of the boldface σ will represent the stress at a point. Equation
(8.5) represents the stress, σ, as a ninecomponent quantity and is referred to
as a stress tensor, which has certain mathematical properties that are useful in
advanced studies.
Stress at a point will be viewed two dimensionally in this chapter.
Figure 8.la is actually the cube as it is viewed along the z axis;
therefore, even though the stresses appear to be applied along the edge of the
element, they are actually applied to a surface that is perpendicular to the
plane of the page. The stress components of Fig. 8.la will be written as

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