Mechanics

Module V: Concept of Stress and Strain

Lesson 21: Introduction to Stress - I

When a body, rigid or flexible, is subjected to external loading, it resists

change in shape and volume. This is effected by the internal force system

set-up inside the body, in general, at every point. If the body is sectioned

by an arbitrary plane, the exposed surface is subjected to an equilibrating

force distribution (force per unit area) comprising both normal and tangential

distributions.

Figure 1:

Partitioning the surface into small area elements, as shown in Fig. 1, one

can express the infinitesimal forces transmitted/resisted by an area element

in terms of this areal force distribution. The distribution is usually referred

to as the stress field, and define the normal and shear stresses on the surface.

The stress field is denoted by xx representing normal stress, and xy and

xz representing shear stress components along y and z axes, respectively.

The first subscript indicates the surface normal, while the second subscript

indicates the coordinate direction of the stress/force. One may, therefore,

define

xx = limA0

PxxA

, xy = limA0

PxyA

, xz = limA0

PxzA

.

It is customary to represent tensile normal stresses as positive. The unit

of stress in SI system is N/m2 or Pa (Pascal). Stress like force, cannot be

measured directly.

1 Stress Tensor and State of Stress

It we section an externally loaded body at an arbitrary point by three mutu-

ally perpendicular planes with normals along x, y and z axes, one can obtain

nine stress components which can be arranged as a matrix

[] =

xx xy xzyx yy yz

zx zy zz

This matrix is known as the stress tensor.

These stresses may be visualized through an infinitesimal element of the

body, as shown in Fig. 2, where the stresses on the three opposite surfaces are

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Figure 2:

not shown. From force equilibrium, these will be equal and opposite. Note

that no body forces have been considered as yet. Moment equilibrium about

the x-axis, assuming no body couples, reads

(zydxdy)dz + (yzdxdz)dy = 0

zy = yz.

Similarly, moment equilibrium about the y and z axes yields xz = zx, and

xy = yx. Thus, the stress tensor [] is symmetric with only six independent

stress components, namely xx, yy, zz, xy, xz, yz.

The stress tensor has been constructed using the stresses on the three

mutually orthogonal and specially oriented planes. However, given [] at a

point, one can determine the normal and shear stresses on any other plane

through that point. The stress tensor is, therefore, said to represent the (tri-

3

axial) state of stress at a point. Let ndA represent an oriented infinitesimal

area element on an arbitrary plane with unit normal n passing through the

point at which [] is given. We define the infinitesimal force vector

dP = []ndA = tdA

where t is known as the traction vector. The infinitesimal normal force is

then given by dPn = ndP = nTdP, where nT represents the transpose of the

column vector n. Hence, the normal stress on the arbitrary plane is obtained

nn =dPndA

= n dPdA

= n t = nT []n.

If s1 and s2 are two unit tangent (surface) vectors on the arbitrary plane, the

shear stresses are similarly obtained as

n1 = s1 t = sT1 []n, n2 = s2 t = sT2 []n.

If the nonzero components of the stress tensor occur only in one plane (say

the x-y plane), the stress tensor may be represented in a reduced form as

[] =

[xx xyyx yy

]

Such a stress distribution is known as a plane state of stress, or a plane stress

distribution.

2 Axially Loaded Bars

Consider an axially loaded long and slender bar, as shown in Fig. 3. The

corresponding state of stress at A is also shown, and can be represented in

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Figure 3:

Figure 4:

the x-y coordinate system as

[] =

[ 0

0 0

]

where = P/A. In the order determine the stresses on a plane with unit

normal n = cos i + sin j and unit tangent vector s1 = sin i+ cos j (as

shown), we determine the traction vector t = []n = cos i. Therefore, the

normal and shear stresses on the inclined surface are given by, respectively,

nn = n t = cos2

n1 = s1 t = sin cos

Note that the shear stress n1 is negative, implying that it is in a direction

opposite to the chosen direction of the surface tangent vector s1. The stresses

on the inclined surface are visualized in Fig. 4.

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Figure 5:

Figure 6:

Problem 1

Draw the axial force and axial stress diagrams for the axially-loaded bar

shown in Fig. 5.

Solution

We start by drawing the FBD of the bar, as shown in Fig. 6. The axial force

diagram is drawn by taking sections (perpendicular to the axis of the bar)

at different locations, and using the resulting FBD of the segment of the bar

to calculate the axial force at that section. The sign convention followed for

the axial force and stress is also shown in Fig. 6.

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Figure 7:

Problem 2

Determine the normal and shear stress on the rivet of cross-sectional area A

connecting the two plates, as shown in Fig. 7.

Solution

Representing the force P in - coordinates, we have P = P (sin 60 +

cos 60). Normal stress on the rivet

n =P sin 60

A=

3P

2A.

Shear stress on the rivet

s =P cos 60

A=

P

2A.

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