Chapter 3
STRESS
Dr. HQ Fan
Department of Building and Real Estate
Friday, 23 Sep 2022
Learning outcomes:
Upon completing the study of this chapter,
students are able to:
 evaluate the stress in a structural element
due to the applied loads; and
 evaluate the deformation of axially loaded
members.
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Outline
1.
2.
3.
4.
5.
6.
Introduction
Normal stress
Shear stress
Normal strain
Shear strain
Hooke’s Law

7.
Deformation of Members Under Axial Loading

8.
Example 1
Example 2
Poisson’s Ratio
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1. Introduction
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We have so far learnt how to solve for
forces in a truss and a beam.
However, our main purpose for
studying structural analysis is to be able
to eventually design the structure.
Hence, we need to study the concept of
stress.
Stress can be defined loosely as the
ratio between force and the area (that
the force is applied on) i.e.
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P
s
A

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where s is the stress, P is the applied
force, and A is the area.
In another words, stress is similar to
pressure.
We will examine further different kinds
of stresses and the relationship between
stress and deformation.
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2. Normal stress
P
s
A


The most basic kind of stress is the
normal (axial) stress, s.
It is defined as the ratio of the axial
force (as in the case of the force in a
truss member) to the area
perpendicular to the force.
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where s is the normal stress,
P is the applied force, and A
A is the area perpendicular
to the force. The unit for
stress is N/m2 (Pa). However
it is a common practice to
use the unit of MPa
(i.e. 1 x 106 Pa).
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s
P
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3. Shear stress

If the applied force is parallel to the
surface on which it acts, shear stress
will be produced.
V
t
A


Where t is the shear stress, V is the
applied shear force and A is the area
parallel to the force.
Again the common unit for shear stress
is MPa.
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V
V
t
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V
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4. Normal Strain
When a member is subject to an axial
load, it will be lengthened if the load is
tension.
On the other hand, it will be shortened
if the load is compression.
The elongation or the shortening (i.e.
the change in length) in a member is
usually defined as d (delta).
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However, if we would like to
compare the deformation
behaviour of different materials,
it is commonly to use the term
strain (e – epsilon).
 Strain is defined as the ratio of
the change in length to the
original length of the member.

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P
d
L
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where e is the normal strain, d is the
change in length and L is the original
length of the member.
Since d and L have the same unit of
length,
therefore, e has no units (or mm/mm).
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5. Shear Strain
If a member is subject to a shear force,
shear deformation will occur.
Consider an element subject to the
shear stress as shown in the figure
below. d
s
Important note:
g
t
Shear stress in an
L
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element must be drawn
as a set as shown in the
figure. Why?
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
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Again, the shear strain, g has no units.
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6. Hooke’s Law
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When a member is subject to applied
load, it will deform.
For example, consider a bar supporting
a weight (P) as shown in the figure.
By changing the magnitude of the
weight, we can measure different
elongations of the bar.
If we plot these data, we will obtain the
following
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P
L
d
Weight
d
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If we divide the weight, P, by the crosssectional area of the bar, A, and the
deformation d, by the length of the bar,
L, we will then obtain the so-called
stress-strain curve of a material.
The slope of the stress-strain curve is
known as the modulus of elasticity, E,
or the Young’s modulus.
It is a measure of how “stiff” the
material is.
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
The relationship between s and e is
called the Hooke’s law:
s=Ee
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That is, for an elastic body, stress is
proportional to strain.
It should be noted that E has a unit of
stress such as MPa, or kPa etc.
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Example 1
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A 1000 mm long, 12 mm diameter rod made of
steel is subjected to the loading shown in the
figure below. If the elongation of the rod is 1 mm,
determine the modulus of elasticity E for the steel.
22.5 kN
1000 mm
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7. Deformation of Members Under
Axial Loading

Consider a homogeneous
rod CD of length L and
uniform cross-sectional
area of A subjected to a
concentric axial load of P
as shown in the figures
below.
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D
D
L
C
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d
C
P
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
We have discussed the axial stress and
the resulting axial stress for rod CD is:
P
s
A
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If the normal stress, s, is within the
elastic range of the material, then Hooke’s
law applies.
s  Ee
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
Substitute Eq. [2] to Eq. [1], we can write
Recall that
e
d
P
Ee 
A

L

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E
d
L
P

A
PL
d
AE
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[3]
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The above equation may be used only if
the rod is homogeneous (constant E), has
a uniform cross-sectional area A, and is
loaded at its end.
If the rod is loaded at other points, or if it
consists of several material portions of
various cross sections and possibly of
different materials, we must divide it into
component parts which satisfy individually
the required conditions for the applications
of Eq. [3].
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
The following equation (Eq. [4]) may be used:
d 
i

Pi Li
Ai Ei
[4]
where Pi, Li, Ai, and Ei are the internal force,
length, cross-sectional area, and modulus of
elasticity corresponding to part i, and d is the
deformation of the entire rod.
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Example 2
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Determine the deformation of the steel rod shown
in the figure below. E = 200 000 MPa
A = 0.003 m2
A = 0.0018 m2
70 kN
100 kN
2m
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1.5 m
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8. Poisson’s Ratio
 When a member is subjected to an axial
load, elongation of the member will occur.
 We have defined in the previous section
the concept of axial strain, e. However, it
is quite obvious to see that when the
member elongates in one direction, there
will be a corresponding contraction
(shrinking) in the orthogonal direction.
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P
P
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
If we considered the two-dimensional
case:
y
P
x
P
Ly
dy/2
dx/2
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Lx
dx/2
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For an isotropic material (mechanical
properties independent of direction), the
contraction will be the same for both y
and z directions.
If we divide the contraction (dy) by its
original length Ly, we will then obtain a socalled lateral strain, ey, which is in the y
direction. The ratio of the lateral strain to
the axial strain (ex) is called the Poisson’s
ratio.
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lateral strain

axial strain
ey
ez


ex
ex
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In fact, Poisson’s ratio is a material property
similar to the modulus of elasticity.
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Example 3
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A 500 mm long 16 mm diameter rod made of a
homogeneous, isotropic material is observed to increase in
length by 300 mm, and to decrease in diameter by 2.4 mm
when subjected to an axial load of 12 kN. Determine the
modulus of elasticity and Poisson’s ratio of the material.
500
mm
300 mm
12 kN
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Example 4
The assembly consists of two posts AD and CF made of A-36 steel and having
a cross-sectional area of 1000 mm2, and a 2014-T6 aluminium post BE having
a cross sectional area of 1500 mm2. If a central load of 400 kN is applied to the
rigid cap, determine the normal stress in each post. There is a small gap of 0.1
mm between the post BE and the rigid member ABC.
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The assembly consists of two posts AD and CF made of A-36 steel and having
a cross-sectional area of 1000 mm2, and a 2014-T6 aluminium post BE having
a cross sectional area of 1500 mm2. If a central load of 400 kN is applied to the
rigid cap, determine the normal stress in each post. There is a small gap of 0.1
mm between the post BE and the rigid member ABC.
(Acknowledgement: the example is “Mechanics of
Materials” 8th Edition, By Ferdinand Beer and E.
Johnston and John DeWolf and David Mazurek,
McGraw-Hill Education, 2020, available from PolyU
library)
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Example 5
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The rigid bar BDE is supported by two links AB and CD. Link AB is made of
aluminium (E = 70 GPa) and has a cross-sectional area of 500 mm2, Link
CD is made of steel (E = 200 GPa) and has a cross-sectional area of 600
mm2, For the 30-kN force shown, determine the deflection (a) of B, (b) of D,
(c) of E.
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
The rigid bar BDE is supported by two links AB and CD. Link AB is made of
aluminium (E = 70 GPa) and has a cross-sectional area of 500 mm2, Link
CD is made of steel (E = 200 GPa) and has a cross-sectional area of 600
mm2, For the 30-kN force shown, determine the deflection (a) of B, (b) of D,
(c) of E.
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
The rigid bar BDE is supported by two links AB and CD. Link AB is made of
aluminium (E = 70 GPa) and has a cross-sectional area of 500 mm2, Link
CD is made of steel (E = 200 GPa) and has a cross-sectional area of 600
mm2, For the 30-kN force shown, determine the deflection (a) of B, (b) of D,
(c) of E.
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
The rigid bar BDE is supported by two links AB and CD. Link AB is made of
aluminium (E = 70 GPa) and has a cross-sectional area of 500 mm2, Link
CD is made of steel (E = 200 GPa) and has a cross-sectional area of 600
mm2, For the 30-kN force shown, determine the deflection (a) of B, (b) of D,
(c) of E.
(Acknowledgement: the example is from
“Mechanics of Materials” (10th Edition) by
Russell C. Hibbeler, Pearson, 2016,
available from PolyU library)
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Next week
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HQ Fan
Chapter 4. Beam, Part 1
Friday, 30 Sep 2022
N002
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