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 with stand 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 non-rigid 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.

1-1a. 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

(a)Fig. 1-1

(b)Fig. 1-1

s, Fig. 1-1a. The loading along the length of a beam is a typical

example of where this idealization is often applied, Fig. 1-1b. 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.

Type of force systemCollinear ……………………………………………

concurrent , coplanar …………………………….

parallel , coplanar………………………………….

Nonconcurrent , nonparallel ,coplanar …………

concurrent , noncoplanar…………………………

parallel , noncoplanar …………………………….

Nonconcurrent , nonparallel , noncoplanar……..

possible resultantsForce

Force

Force or a couple

Force or a couple

Force

Force or a couple

Force or a couple , or a force and a couple

SUMMARY

The surface forces that develop at the supports or points of

support between bodies are called reactions. For two-dimensional

problems, i.e., bodies subjected to coplanar force systems, the supports

most commonly encountered are shown in Table 1-1. 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 1-1 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

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

its

than the total surface area of the contacting member.

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

0

0

OM

F(1–1)

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,

000

000

zyx

zyx

MMM

FFF(1–2)

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,

0

0

0

O

y

x

M

F

F

(1–3)

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 free-body 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. 1-2. 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 1-2 Exploratory section a-a through loaded member.

exploratory section a-a 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.

1-3.

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

The notation used in Fig. 1-3 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:

Axial force. This component measures the pulling (or pushing)

action perpendicular to the section. A pull rep resents 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.

Pxx

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.

Torque. This component measures the resistance· to

twisting the member and is commonly given the symbol T.

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.

Pxy, Pxz

Mxy, Mxz

Mxx

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 two-dimensional 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 three-dimensional 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

twist about its longitudinal axis, and is called a torque. For the time

zx

y

Figure 2.15

being, however, we will restrict ourselves to two-dimensional 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 two-dimensional 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 anti-clockwise. 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 right-hand 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

equivalent-force systems and reactions for load-carrying members. For now the

discussion will be conceptual. Later, certain topics of statics will be reviewed as

considered pertinent.

The beam-type structure of Fig. 1.1 is considered to be in equilib rium 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 equi librium for the

free body. Specifically, we classify the forces and couples as in Fig. 1.3. The force

z

x

FIGURE 1.1 Beam-type structure showing the six actions of statics.

y

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. 1.3c-e as vectors. The axial-couple vector of Fig. 1.3c represents a twisting

couple whose direction is determined using the right-hand-screw 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 com pression 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. Load-carrying members subject to various

combina tions 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 , 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.

Structural TypeSchematic of Assumed

Member ForcesDescription

Plane truss and space truss

Only axial Force, F1, is significant.

Beam and plane frame

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

F2

F3

F1

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

Structural TypeSchematic of Assumed

Member ForcesDescription

Grid

Only shear force F1 and flexural moment F2 (both transverse to the plane of the grid) and the torsional moment

F3 are significant.

Space Frame

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

F1

F2 F3

F3

F2 F5

F1

F4F6

PROBLEM: ANALYZE THE TRUSS BY STIFFNESS METHOD

ANALYSIS OF GRIDSANALYSIS 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

x-axisWxj,xjWxxj,xxjWxxi,xxi Wxi,xi

Wyi,yi

Wyyi,yyi

Wyj,yj

Wyyj,yyj

Wzj,zj

Wzzj,zzj

Wzi,zi

Wzzi,zzi

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 rela tionship 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 out of-plane 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 in-plane mo ments 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 al ways 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

load-carrying body as shown in Fig. 1.4 and say that the body

is a continuum. Essentially, a continuum is a

FIGURE 1.4

Idealized body.

xz

y

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 molecu lar 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 small-body 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

y

FIGURE 1.5

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

A

Fxx

(1.1)

The Greek letter τ (tau) is usually used for shear stress and is the shear force divided

by the area, or

A

Fyy

(1.2)

and

A

Fzz

(1.3)

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 load-carrying 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 2-9 are devoted to the study of concepts that have been

briefly introduced in this section.

(a)Fig. 1-1

(b)Fig. 1-2

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 free-body diagram of one of the parts is drawn. If

we consider the section shown in Fig. 1-2a, then the resulting free-body

diagram of the bottom part of the body is shown in Fig. 1-2b. 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. 1-1 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 free-body 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.

(c)Fig. 1-2

(d)Fig. 1-2

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. 1-2b, 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; 1-2d, 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

the two segments of the body.

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. I-2d.

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).

Fig. 1-3

Fig. 1-3

F3

Each of the six unknown x, y, z components of force and

moment shows in Fig. 1-2d can be determined directly from the six

equations of equilibrium, that is, Eqs. 1-2, applied to either segment of

the body. If, however, the body is subjected to a coplanar system of

forces, then only normal-force, shear, and bending-moment

components will exist at the section. To show this, consider the body in

Fig. 1-3a. 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. 1-3b, 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. 1-3. 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

location in a body.

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 free-body 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

the

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 free-body 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.

1-9a. 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. 1-9b. 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. Further more, 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 continuous-cohesive 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. 1-9c. 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.

Fig. 1.9 (a)

Fig. 1.9

(b) (c)

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

If the normal force or stress "pulls" on the area element ΔA as shown in

Fig. I-9c, it is referred to as tensile stress, whereas if it "pushes" on ΔA

it is called compressive stress.

A

Fn

A

lim0

Shear Stress. Likewise, the intensity of force, or force per unit area,

acting tangent to ΔA is called the shear stress, (tau). This component

is expressed mathematically as

(1–4)

Fig. 1.9

A

Ft

A

lim0

(1–5)

In Fig. I-9c, 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. 1-10b. We can now

express the normal-stress component as

A

Fz

Az

lim

0

Fig. 1.10 (a)

Fig. 1.10

(c)(b)

F1

z

Fig. 1.11

(b)

z

Fig. 1.11(c)

(d)

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 shear-stress components, zx and zy.

The z specifies the orientation of the area, and x and y refer to the

direction lines for the shear stresses.

and the two shear-stress components as

A

F

A

F

y

Azy

x

Azx

lim

lim

0

0

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. 1-l0c.

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.1-10a. The resulting free-body diagram is shown

in Fig. 1-11a. 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 shear-stress

components shown in Fig. 1-11b. Using the same notation as before,

the subscript y in σy, yx, and yz refers to the direction of the normal line

associated with the orientation of the area, and x and z in yx and yz

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.

1-11c, gives rise to normal stress σx and shear stresses xy and xz, 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. 1-12.

Equilibrium Requirements. Although each of the six faces of the

element in Fig. 1-12 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 free-body diagram of the element, Fig. 1-13a. This element

has a volume of ΔV = Δx Δy Δz, and in accordance with Eqs. 1-4 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. 1-13b. Here it should be noted that the

x

z

Fig. 1.13 (a) Element free-body diagram

Fig. 1.13 (b) Element free-body diagram

z

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. 1-3 and 1-4 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; there fore. 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

distribu tion at an arbitrary interior point O can be studied by exposing

an interior plane through O as shown in Fig. 1-13a. 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 O-namely, 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.

FIG. 1—13

FIG. 1—13

FIG. 1—13

In Art. 1-3 it was pointed out that materials respond to components of

the stress vector rather than the stress vector itself. In particular, the

com ponents 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 n

are then defined as

A

FS n

An

lim

0

A

Fnn

An

lim

0

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 direc tion. In this case the normal and shear stresses on the plane will

be σx and x, respectively. Since x, in general, will not coincide with

the y or z axes, it must be resolved into the components xy and xz, as

shown in Fig. 1-13c. Unfortunately the state of stress at a point in a

material is not completely defined by these three components of the

A

Fnt

An

lim

0

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. 1-9) that the specification of stresses on three mutually

perpendicular planes is sufficient to describe com pletely 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. 1-14. 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

coordi nate axes x, y, and z are arranged as a right-hand system. The

FIG. 1—14

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 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, xy 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. 1-14 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 P1, 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

Fig. 1.32

Fig. 1.33

(a) (b)

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 per pendicular to the x axis.

While the normal force ΔFx has a well-defined 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 ΔVx

z 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:

A

V

A

V

A

F

xz

Axz

xy

Axy

x

Ax

limlim

lim

00

0

(1.18)

y τxy

x

zFig. 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 ΔVx

z 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

in the negative x direction. Similarly, positive signs for τxy and τxz will indicate that the

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

in Fig. 1.35.

σx

y

zFig. 1.35

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 ex erted on each of the

six faces of the cube (Fig. 1.36). The stress com ponents shown in the figure are σx, σy

and σz which represent the nor mal 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

stresses acting on the faces of the cube differ slightly from the stresses at Q the error

Fig. 1.36

z

y

x

Fig. 1.37

involved is small and vanishes as side a of the cube approaches zero.

Important relations among the shearing stress components will now be

derived. Let us consider the free-body diagram of the small cube centered at point Q

(Fig. 1.37). 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:

000 zyx FFF (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

000 zyx MMM (1.20)

Fig. 1.38

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 shear ing 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,

0:0 aAaAM yxxyz

zyyzzyyz andsimilarly,

from which we conclude that

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 man ner the relations

yxxy

(1.22)

(1.21)

We conclude from Eqs. (1.21) and (1.22) that only six stress com ponents 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 mag nitude 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 con sider 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. However, if the small cube

Fig. 1.39

(a) (b)

Fig. 1.40

is ro tated 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 two-dimensional

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 free-body

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

Figure 2.1

Figure 2.1

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:

A

F

A

F

t

A

n

A

lim

lim

0

0

(2.1)

In the special case where the components Fn and Fr are uniformly distributed over

the entire area A, then σ = Fn/A and = 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 )

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/m2), 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, xy is a

shear stress acting on the X plane and pointed in the positive y direction, and xz 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, xy = yx, xz = zx, and yz

= zy 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 three-dimensional 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., xz = yz = z = 0), the stress condition

reduces to a biaxial or two-dimensional 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., xz = yz = z = 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 xy = yx. 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:

0)()( dydxhdxdyh yxxy

from which

yxxy

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.

x

FIGURE 8.1 (a)Element viewed along the z

axis;(b) positive stresses acting on an

element;

(a)

(b)

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

(c)

(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 yx as shown

in Fig. 8.1b. The complete three-dimensional 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 xy = yx yz = zy and xz = zx.

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 nine-component 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