Transcript

Chapter 2: Stress & Strain MM211: Solid Mechanics

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Chapter 2 Stress and Strain

2.0 INTRODUCTION

In the first Chapter we discussed the equations of statics, and how to determine the
ground reactions for any structure. The method can also be used to determine the
internal loads carried by the members or parts of a body. We now need to define how
these internal loads are distributed and carried by the material and the deformation they
create.

2.1 STRESS (4
th
SI p. 5-14, p. 23-27, p. 47-61)

Consider an element of continuous (no voids) and cohesive (no cracks, breaks and
defects) material subjected to a number of externally applied loads as shown in Fig.
2.1(a). It is understood that the member is in equilibrium



Fig. 2.1 External and internal forces in a structural member

If we now cut this body, the applied forces can be thought of as being distributed over
the cut area A as in Fig. 2.1b). Now if we look at infinitesimal regions AA, we assume
the resultant force in this infinitesimal area is AF. In fact, AF is also a distributed force.
When AA is extremely small, we can say that the distributed force AF is uniform. In
other words, if we look at the whole sectioned area, we can say that the entire area A is
subject to an infinite number of forces, where each one (of magnitude AF) acts over a
small area of size AA. Now, we can define stress.

Definition: Stress is the intensity of the internal force on a specific plane passing
through a point.
Mathematically, stress can be expressed as

A
F
A
A
A
=
÷ A 0
lim o (2.1)

Dividing the magnitude of the internal force AF by the acting area AA, we obtain the
stress. If we let AA approach zero, we obtain the stress at a point. In general, the stress
could vary in the body, which depends on the position that we are concerned with. The
stress is one of most important concepts that we introduce here in mechanics of solids.

Chapter 2: Stress & Strain MM211: Solid Mechanics

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Normal and Shear Stress
As we know; force is a vector that has bot h magnitude and direction. But in the
definition of stress, we only consider the magnitude of the force. Obviously, this may
easily confuse us. Let’s still take patch AA as an example. As we can see, force AF is not
perpendicular to the sectioned infinitesimal area AA. If we only take the magnitude of
the force into account, apparently, the stress may not reflect the real mechanical status
at this point. In other words, we need to consider both magnitude and direction of the
force.

Now let’s resolve the force AF in the normal and tangential directions of the acting area
as Fig. 2.1(b). The intensity of the force or force per unit area acting normally to
section A is called Normal Stress, o (sigma), and it is expressed as:

A
F
n
A
A
A
=
÷ A 0
lim o (2.2)

If this stress “pulls” on the area it is referred as Tensile Stress and defined as Positive. If
it “pushes” on the area it is called Compressive Stress and defined as Negative.
The intensity or force per unit area acting tangentially to A is called Shear Stress, t
(tau), and it is expressed as:


A
F
t
A
A
A
=
÷ A 0
lim t (2.3)

Average Normal Stress (4
th
SI p. 7-8)
To begin, we only look at beams that carry tensile or compressive loads and which are
long and slender. Such beams can then be assumed to carry a constant stress, and Eq.
(2.2) can be simplified to:


A
F
= o (2.4)

We call this either Average Normal Stress or Uniform Uniaxial Stress.

Units of Stress
The units in the SI system is the Newton per square meter or
Pascal, i.e.: Pa = N/m
2
.
In engineering, Pa seems too small, so we usually use:
Kilo Pascal KPa (=Pa× 10
3
) e.g.
20,000Pa= 20kPa
Mega Pascal MPa (=Pa× 10
6
) e.g.
20,000,000Pa= 20MPa
Giga Pascal GPa (=Pa× 10
9
) e.g.
20,000,000,000Pa = 20GPa

Example 2.1: An 80 kg lamp is supported by a single
electrical copper cable of diameter d = 3.15 mm. What is the stress carried by the
cable.

Chapter 2: Stress & Strain MM211: Solid Mechanics

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To determine the stress in the wire/ cable as Eq. (2.4), we need the cross sectional area
A of the cable and the applied internal force F:

MPa
A
F
N mg F
m
d
A
6 . 100
10 793 . 7
784
784 81 . 9 80
10 793 . 7
3
00315 . 0
4
6
2 6
2 2
=
×
= =
= × = =
× =
×
= =
÷
÷
o
t t


Allowable Stress (4
th
SI p. 28-29)
From Example 2.1, we may be concerned whether or not 80kg would be too heavy, or
say 100.6MPa stress would be too high for the wire/ cable, from a safety point of view.
Indeed, stress is one of most important indicators of structural strength. When the stress
(intensity of force) of an element exceeds some level, the structure will fail. For
convenience, we usually adopt allowable force or allowable stress to measure the
threshold of safety in engineering.

Moreover, there are the following several reasons that we must take into account in
engineering:
- The load for design may be different from the actual load.
- The size of the structural member may not be very precise due to manufacturing
and/ or assembly.
- Various defects in the material due to manufacturing processing or material
handling.

One simple method to consider such uncertainties is to use a number called the Factor
of
Safety, F.S., which is a ratio of failure load F
fail
(found from experimental testing)
divided by the allowable one F
allow


allow
fail
F
F
S F = . . (2.5)

If the applied load is linearly related to the stress developed in the member, as in the
case of using o = F/A, then we can define the factor of safety as a ratio of the failure
stress o
fail
to the allowable stress o
allow



allow
fail
S F
o
o
= . . (2.6)

Usually, the factor of safety is chosen to be greater than 1 in order to avoid the potential
failure. This is dependent on the specific design case. For nuclear power plant, the
factor of safety for some of its components may be as high as 3. For an aircraft design,
the higher the F.S. (safer), the heavier the structure, therefore, the higher the
operational cost. So we need to balance the safety and cost.

The value of F.S. can be found in design codes/ standards and engineering handbooks.
More often, we use Eq. (2.6) to compute the allowable stress:


Chapter 2: Stress & Strain MM211: Solid Mechanics

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. .S F
fail
allow
o
o = (2.7)

Example 2.2: In Example 2.1, if the maximum allowable stress for copper is o
Cu,allow
=
50MPa, please determine the minimum size of the wire/ cable from the material
strength point of view.

Mathematically,
allow Cu
d
mg
A
F
,
2
4
o
t
o s = =

Therefore: mm
mg
d
allow Cu
469 . 4 10 469 . 4
4
3
,
= × = >
÷
to


Obviously, the lower the allowable stress, the bigger the cable size. Stress is an
indication of structural strength and elemental size.

In engineering, there are two significant problems associated with stress as follows.

Problem (1) Stress Analysis: for a specific structure, we can determine the stress
level. With the stress level, we then justify the safety and reliability of a
structural member, i.e. known size A and load F, to determine stress
level: o = F / A
Problem (2) Engineering Design: Inversely, we can design a structural member based
on the allowable stress so that it can satisfy the safety requirements, i.e.
known material’s allowable stress o
allow
and load F, to design the element
size: A ≥ F / o
allow

2.2 DEFORMATION(4
th
SI p. 61-75)

Whenever a force is applied to a body, its shape and size will change. These changes are
referred as deformations. These deformations can be thought of being either positive
(elongation) or negative (contraction) in sign as shown in Fig. 2.2.



Fig. 2.2 Deformation due to applied axial forces

It is however very hard to make a relative comparison between bodies or structures of
different size and length as their individual deformations will be different. This requires
the development of the concept of Strain, which relates the body’s deformation to its
initial length.




Chapter 2: Stress & Strain MM211: Solid Mechanics

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2.3 STRAIN(4
th
SI p.48-50; &4th p. 67-69; p. 105)

Normal Strain
The elongation (+ ve) or contraction (−ve) of a body per unit length is termed Strain.



Fig. 2.3 Generalized deformation due to applied forces

Let’s take the arbitrarily shaped body in Fig. 2.3 as an example. Consider the
infinitesimal line segment AB that is contained within the undeformed body as shown in
Fig. 2.3(a). The line AB lies along the n-axis and has an original length of AS. After
deformation, points A and B are displaced to A’ and B’ and in general the line becomes
a curve having a length AS’ The change in length of the line is therefore AS-AS’. We
consequently define the generalized strain mathematically as

( ) n along A B
S
S S
A B
÷
A
A ÷ ' A
=
÷
lim c (2.8)

Average Normal Strain
If the stress in the body is everywhere constant, in other words, the deformation is
uniform in the material (e.g. uniform uniaxial tension or compression) as shown in Fig.
2.2, the strain can be computed by


L
L
L
L L
Original
Original Deformed
A
=
÷
= c (2.9)

i.e. the change in length of the body over its original length,


Unit of Strain
From Eqns. (2.8) and (2.9), we notice that t he normal strain is a dimensionless quantity
since it is a ratio of two lengths. Although this is the case, it is common in practice to
state it in terms of a ratio of length units. i.e. meters per meter ( m/m)

Usually, for most engineering applications ε is very small, so measurements of strain are
in micrometers per metre (µm/ m) or (µ/ m).

Sometimes for experiment work, strain is expressed as a percent, e.g. 0.001 m/ m =
0.1%.
A normal strain of 480µm for a one-meter length is said:


Chapter 2: Stress & Strain MM211: Solid Mechanics

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c = 480× 10
-6
= 480(µm/ m) = 0.0480% = 480µ (micros) = 480µs (micro strain)

Example 2.3: In Example 2.1, if it is measured that the cable was elongated by 1.35 mm
due to the weight of the light, what would its strain be?

s µ c 900 10 900
5 . 1
00135 . 0
6
= × = =
÷



2.4 STRESS-STRAIN RELATIONSHIP, HOOKE'S LAW(4
th
SI p.56)

Material Test and Stress-Strain Diagram
Material strength depends on its ability to sustain a load without undue deformation or
failure. This property is inherent to the material itself and must be determined by
experiment. One of the most important tests to perform in this regard is the tension or
compression test. To do so, a bunch of standard specimens are made. The test is
performed in a tensile test machine but in most cases in a universal testing machine
(UTM). Shown in Fig. 2.4 is the specimen and test result displayed in a Stress-Strain
Diagram.



Fig. 2.4 Material test and Stress-Strain Diagram

The Stress-Strain diagram consists of 4 stages during the whole process, elastic, yielding,
hardening and necking stages respectively. From the yielding stage, some permanent
plastic deformation occurs. About 90% of engineering problems are only concerned
with the elastic deformation in structural members and mechanical components. Only
10% of engineering work concerns itself with plasticity and other nonlinear stages (e.g.
metal forming). In this subject, we are only involved in the linear elastic region, in which
the relationship between the strain and stress is linear.

Hooke’s Law
The stress-strain linear relationship was discovered by Robert Hook in 1676 and is
known as Hooke's law. It is mathematically represented by Eq. (2.10),

c o E = (2.10)


Chapter 2: Stress & Strain MM211: Solid Mechanics

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where E is termed the Modulus of Elasticity or Young's Modulus with units of N/ m
2
or
Pa. For most of engineering metal materials, GPa is used as the appropriate units, e.g.
mild steel is about 200GPa~ 210GPa.

2.5 POISSON'S RATIO(4
th
SI Ed p. 84)

Definition
When a deformable body is stretched by a tensile force, not only does it elongate but it
also contracts laterally, i.e. it would contract in the other two dimensions as shown in
Fig 2.5(a). Likewise, a compressive force acting on a deformable body will cause it to
contract in the direction of the force but its sides will expand laterally as in Fig. 2.5(b).
When the load P is applied to the bar it changes the bar’s length by o and its radius by
o
r
. The strain in the axial direction and in lateral/ radial directions are respectively:


L
axial
o
c =
r
r
lateral
o
c =

In the early 1800s, French scientist Poisson realized that within the elastic range the ratio of
these two strains is a constant. We named the constant Poisson’s ratio or v (Nu)


Axial
Lateral
Strain Axial
Strain Lateral
c
c
v ÷ = ÷ =
(2.11)

Fig. 2.5 Relationship of the axial strain with the lateral strain

The negative sign is used here since longitudinal elongation (positive strain) causes lateral
contraction (negative strain), and vice versa. So Poisson’s ration is positive, i.e. v≥0.

Remarks
- The lateral strain is caused only by the axial force. No force or stress acts in
lateral direction;
- Lateral strain is the same in all lateral directions;
- Usually 0 ≤ v ≤ 0.5. For most linearly elastic materials v= 0.3;
- Poisson’s ratio is a constant.

Strain in Lateral Direction
For bars subjected to a tensile stress o
x
, the strains in the y and z planes are:


Chapter 2: Stress & Strain MM211: Solid Mechanics

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¦
¦
¦
¹
¦
¦
¦
´
¦
÷ = ÷ =
÷ = ÷ =
=
E
E
E
x
x z
x
x y
x
x
vo
vc c
vo
vc c
o
c
(2.12)

2.6 THERMALSTRAIN(4
th
SI Ed p. 74)

Thermal Deformation
When the temperature of a body is changed, its overall size will also change. In other
words, temperature change may cause the dimension or shape change in the material.
More specially, if the temperature increases, generally a material expands. Whereas if
the temperature decreases, the material will contract. It is supposed that this is a
common sense for anyone (aha, but for some special materials such as NiTi alloys or
shape memory alloys, when you heat them they contract) .



Fig. 2.6 Thermal and mechanical deformation

For the majority of engineering materials this relationship is linear. If we assume that the
material is homogeneous and isotropic, from experiment, we can find a linear relation
between thermal deformation and temperature change as:

o
T
= o ⋅AT ⋅ L (2.13)

Where: o : Coefficient of thermal expansion, units are strain per ºC
AT : algebraic change in temperature (ºC ) (increase + ; decrease −)
oT : algebraic change in length (“+ ” = elongation; “−” = contraction)

Thermal Strain

T
L
T
Thermal
A · = = o
o
c (2.14)





Chapter 2: Stress & Strain MM211: Solid Mechanics

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Coupled Strain Status
If we consider both mechanical strain c
o
and thermal strain c
T
in the structure as shown
in Fig. 2.6(b), by referring to Eq. (2.12), the total strains in all directions would be
computed as:


¦
¦
¦
¹
¦
¦
¦
´
¦
÷ A = + =
÷ A = + =
+ A = + =
E
T
E
T
E
T
z
z T z
y
y T y
x
x T x
o
o c c c
o
o c c c
o
o c c c
o
o
o
(2.15)

2.7 ELASTICDEFORMATION OF AXIALLYLOADEDMEMBER(4
th
SI Ed., p.61)

Now we are going to find the elastic deformation of a member subjected to axial loads.
Let’s consider a generalized bar shown in Fig. 2.7, which has a gradually varying cross-
sectional area along its length L. For a more general case, the bar is subjected to
concentrated loads at its right end and also a variable external load distributed along its
length (such as a distributed load could be for example, to represent the weight of a
vertical bar or friction forces acting on bar surface). Here we wish to find the relative
displacement o of one end with respect to the other.


Fig. 2.7 Thermal and mechanical deformation

We pick a differential element of length dx and cross-sectional area A(x). A FBD can be
drawn for the element here shown in Fig. 2.7. Assume that resultant internal axial force
is represented as P(x). The load P(x) will deform the element into the shape indicated
by the dashed outline.

The average stress in the cross-sectional area would be ( )
( )
( ) x A
x P
x = o

The average strain in the cross-sectional area would be ( )
dx
d
x
o
c =

Provided these quantities do not exceed the proportional limit, we can relate them using
Hook’s law, i.e. o = Ec

Therefore

Chapter 2: Stress & Strain MM211: Solid Mechanics

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( )
( )
( )
|
.
|

\
|
=
dx
d
x E
x A
x P o


Re-organize the equation, we have


( )
( ) ( )
dx
x A x E
x P
d = o

For the entire length L of the bar, we must integrate this expression to find the required
end displacement


( )
( ) ( )
}
=
L
dx
x A x E
x P
0
o (2.16)

Where: o = displacement between two points
L = distance between the points
P(x) = Internal axial force distribution
A(x) = Cross-sectional area
E(x) = Young’s modulus

Constant Load and Cross-Sectional Area
In many engineering cases, the structural members experience a constant load and have
a constant cross-sectional area and are made of one homogenous material, i.e.
P(x) = P = constant (no axially distributed load)
A(x) = A = constant (uniform cross-sectional area)
E(x) = E = constant (homogeneous material)
From Eq. (2.16), we have


EA
PL
= o (2.17)

Multi-Segment Bar
If the bar is subjected to several different axial forces or cross-sectional areas or Young’s
moduli, the above equation can be used for each segment. The total displacement can
be computed from algebraic addition as


¿
=
i i i
i i
E A
L P
o (2.18)

Example 2.4: The composite bar shown in the figure is made of two segments, AB and
BC, having cross-sectional areas of A
AB
= 200mm
2
and A
BC
= 100mm
2
. Their Young’s
moduli are E
AB
= 100GPa and E
BC
= 210GPa respectively. Find the total displacement
at the right end.


Chapter 2: Stress & Strain MM211: Solid Mechanics

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Step 1 FBDs for Segments AB
and BC. Assume the internal
forces are in tension.

Step 2 Equilibriums
Internal force in AB

kN P
F F P F
AB
AB x
30
0
1 2
÷ =
+ ÷ ÷ = = ÷
¿
+


(Opposite to our assumption of
tension, so Segment AB is in
compression)

Internal force in BC

( ) tension in kN F P
F P F
BC
BC x
10
0 0
1
1
= =
= + ÷ = = ÷
¿
+


Step 3 Compute the total deformation by using Eq. (2.18)

( ) left towards mm m
A E
L P
A E
L P
AC
BC BC
BC BC
AB AB
AB AB
BC AB AC
4 004 . 0 002 . 0 006 . 0
10 100 10 210
2 . 4 10 10
10 200 10 100
4 10 30
6 9
3
6 9
3
÷ = ÷ = + ÷ =
× × ×
× ×
+
× × ×
× × ÷
= + = + =
÷ ÷
o
o o o

2.8 STATICALLYINDETERMINATE MEMBERS LOADEDAXIALLY(4
th
SI Ed p. 47)

Statically Determinate and Indeterminate
When a bar is supported at one end and subjected to an axial force P at the other end
as shown in Fig. 2.8(a), there is only one unknown reaction force F
A
. By using the
equations of statics, the unknown reaction can easily be determined. Such a system with
the same number of unknown reactions as equations of statics is called statically
determinate. – i.e. known reactions can be determined strictly from equilibrium equations.


Chapter 2: Stress & Strain MM211: Solid Mechanics

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Fig. 2.8 Statically determinate and indeterminate structures

If the bar is also restricted at the free end as shown in Fig. 2.8(b), it has 2 unknown
reactions F
A
and F
B
, one known force P and one equation of statics:


P F F
P F F F
B A
B A y
= +
= ÷ + = = | +
¿
0 0
(2.19)

It cannot be solved unless we introduce one more condition. If the system has more
unknown forces than equations of statics it is called statically indeterminate.

Compatibility Conditions
What we need is an additional equation that specifies how the structure is displaced due
to the applied loading. Such an equation is usually termed the compatibility equation.

Since there are 2 unknown and only 1 equation of statics herein, what we need is an
additional equation that specifies how the structure is displaced due to the applied loading.
Such an equation is usually termed the compatibility equation or kinematic condition(s).

In order to determine the compatibility for this example we need to determine how
point C is going to move, and how much point B moves in relation to point A. Now,
since both ends of the bar are fully fixed, then the total change in length between A and
B must be zero.

Basically the amount that length AC elongates CB contracts as shown in Fig. 2.9, so the
equation can be written as:

0 = +
CB AC
o o (2.20)


Chapter 2: Stress & Strain MM211: Solid Mechanics

13


Fig. 2.9 Compatibility condition

Let’s look at the free body diagram for segment AC and CB as in Fig. 2.9. ( Now a FBD
can be at any level of the structural system or structural members). Therefore, for
segment AC,


( )
( ) + = =
+ = = ÷ = = | +
¿
elongation
E A
L F
E A
L P
Tension F P P F F
AC AC
AC A
AC AC
AC AC
AC
A AC AC A y
o
0 0
(2.21)

For segment CB,


( )
( ) ÷ ÷ = ÷ =
÷ ÷ = = + = = | +
¿
n Contractio
E A
L F
E A
L P
n Compressio F P P F F
CB CB
CB B
CB CB
CB CB
CB
B CB CB B y
o
0 0
(2.22)

Compatibility condition:

0 =
|
|
.
|

\
|
÷ + = + =
CB CB
CB B
AC AC
AC A
CB AC
E A
L F
E A
L F
o o o (2.23)

Combining Compatibility equation (2.23) with the equation of statics (2.19), we now
can solve for the two unknowns F
A
and F
B
as,


¦
¦
¹
¦
¦
´
¦
= +
= ÷
P F F
E A
L F
E A
L F
B A
CB CB
CB B
AC AC
AC A
0
(2.24)


Chapter 2: Stress & Strain MM211: Solid Mechanics

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P
E A
L
E A
L
E A
L
F P
E A
L
E A
L
E A
L
F e i
CB CB
CB
AC AC
AC
CB CB
CB
A
CB CB
CB
AC AC
AC
AC AC
AC
B
+
=
+
= . .

If A
AC
E
AC
= A
CB
E
CB
= Const, we have

P
L
L
F and P
L
L
F
CB
A
AC
B
= = (2.25)

Example 2.5: Two bars made of Copper and Aluminium are fixed to the rigid abutments.
Originally, there is a gap of 5mm between the ends as shown in the figure. Determine the
average normal stress in both bars if we increase the temperature from 10°C to 210°C.


Let’s firstly look at the copper bar. When the bar system is heated up from 10°C to 210°C, the
copper bar expands towards the right by o
T ,Cu
. After the copper bar touches the aluminium
bar, a mechanical force F will develop, which will prevent the copper bar from expanding
further. We assume that due to such a mechanical force, the copper bar is pressed back by δ
F
,Cu
. The real total deformation of the copper bar will be computed as

Cu F Cu T Cu , ,
o o o ÷ = (elongation + , Contraction −)

Similarly, we have

Al F Al T Al , ,
o o o ÷ = (elongation + , Contraction −)


Chapter 2: Stress & Strain MM211: Solid Mechanics

15
Because these two expanding bars should fill the gap, we prescribe a compatibility
condition as

005 . 0 = +
Cu Al
o o (2.26)

From these two equations, we have

( ) ( ) 005 . 0
, , , ,
= ÷ + ÷
Cu F Cu T Al F Al T
o o o o
i.e 005 . 0 =
|
|
.
|

\
|
÷ A +
|
|
.
|

\
|
÷ A
Al Al
Al
Al Al
Cu Cu
Cu
Cu Cu
A E
FL
TL
A E
FL
TL o o

N
A E
L
A E
L
TL TL
F
Al Al
Al
Cu Cu
Cu
Al Al Cu Cu
2 . 206
10 85 . 7 10 69
8 . 0
10 85 . 7 10 110
4 . 0
005 . 0 8 . 0 200 23 4 . 0 200 10 17 005 . 0
5 9 5 9
6
=
× × ×
+
× × ×
÷ × × + × × ×
=
+
÷ A + A
=
÷ ÷
÷
o o

The average normal stress can be computed as


MPa A
F
63 . 2 10 85 . 7
2 . 206
5
= ×
= =
÷
o

The deflections can then be calculated as

÷ = × = × ÷ × =
× × ×
×
÷ × × × =
|
|
.
|

\
|
÷ A = ÷ =
÷ ÷ ÷
÷
÷
mm m
A E
FL
TL
Cu Cu
Cu
Cu Cu Cu F Cu T Cu
35 . 1 10 35 . 1 10 55 . 9 10 36 . 1
10 85 . 7 10 110
4 . 0 2 . 206
4 . 0 200 10 17
3 6 5
5 9
6
, ,
o o o o

÷ = × = × ÷ × =
× × ×
×
÷ × × × =
|
|
.
|

\
|
÷ A = ÷ =
÷ ÷ ÷
÷
÷
mm m
A E
FL
TL
Al Al
Al
Al Al Al F Al T Al
65 . 3 10 65 . 3 10 05 . 3 10 68 . 3
10 85 . 7 10 69
8 . 0 2 . 206
8 . 0 200 10 23
3 6 3
5 9
6
, ,
o o o o


2.9 AVERAGE SHEAR STRESS (4
th
SI Ed p. 10)

The intensity or force per unit area acting tangentially to A is called Shear Stress, (tau),
and it is expressed as in Eq. (2.27 ) as:


A
F
t
A
A
A
=
÷ A 0
lim t (2.27)

In order to show how the shear stress can develop in a structural member, let’s take a
block as an example. The block is supported by two rigid bodies and an external force F
is applied vertically as shown in Fig. 2.10. If the force is large enough, it will cause the
material of the block to deform and fail along the vertical planes as shown.

Chapter 2: Stress & Strain MM211: Solid Mechanics

16

A FBD of the unsupported center segment indicat e that shear force V=F/2 must be
applied at each section to hold the segment in equilibrium.



Fig. 2.10 Average shear stress

¿
= ÷ = = | + 0 2 0 F V F
y

∴V = F / 2

The average shear stress distributed over each sectioned area that develops the shear
force is defined by


A
V
avg
= t (2.28)

tavg = assume to be the same at each point over the section
V = Internal shear force
A = Area at the section


2.10 STRESS CONCENTRATIONS (4
th
SI Ed p. 107)
For a uniform cross-sectional bar that has applied an axial force, both experiment and
theory of elasticity finds that the normal stress will be uniformly distributed over the
cross-section

Stress Concentration
However, if we drill a hole for some reason in the component, the typical example is to
build a connection with other structural elements. For such a case, if we cut at the
hole’s center plane, we find that the stress distribution is no longer uniform, as in Fig.
2.11(a). It may distribute over such a small area in a highly non-uniform pattern. We
call this phenomenon Stress Concentration.

Stress Concentration Factor
In engineering practice, though, the actual stress distribution does not have to be
determined. Instead, only the maximum stress at these sections must be known, and the
member is then designed to resist this highest stress when the axial load is applied. The

Chapter 2: Stress & Strain MM211: Solid Mechanics

17
specific values of the maximum normal stress at the critical section can be determined
by experimental methods or by advanced mathematical techniques using the theory of
elasticity. The results of these investigations are usually reported in graphical form (as in
Fig. 2.11(c)) in terms of Stress Concentration Factor K.


avg
K
o
o
max
= (2.29)

in which o
avg
=P/A’ is the assumed average stress as in Fig. 2.11(b). Provided K has been
known from the figures or tables (as in Fig. 2.11(c)), and the average normal stress has
been calculated from o
avg
=P/A’, where A’ is the smallest cross-sectional area. Then from
the above equation the maximum stress at the cross section can be computed as:


A
P
K K
avg
'
= = o o
max
(2.30)

Fig. 2.11 Stress concentration

Stress concentration occurs in the case that there is a sudden change in cross-sectional
area. By observing Fig. 2.11(c), it is interesting to note that the bigger the ratio of
change in the sectional area, the higher the stress concentration.

2 . If we only take the magnitude of the force into account. O bviously. and it is expressed as:   lim Ft A0 A (2. apparently. force F is not perpendicular to the sectioned infinitesimal area A. The intensity or force per unit area acting tangentially to A is called Shear Stress.2) If this stress “pulls” on the area it is referred as Tensile Stress and defined as Positive.0 0 0 Pa= 2 0 kPa M Pascal ega M Pa (=Pa× 106) e. we need to consider both magnitude and direction of the force. 2 0 .2 ) can be simplified to:  F A (2 .4 ) We call this either Average Normal Stress or Uniform Uniaxial Stress. If it “pushes” on the area it is called Compressive Stress and defined as Negative. so we usually use: K Pascal ilo K Pa (=Pa× 103) e. Such beams can then be assumed to carry a constant stress. As we can see. Chapter 2: Stress & Strain 2 MM211: Solid Mechanics .0 0 0 .  (tau).e.Normal and Shear Stress As we know. (2 .1 5 mm. U of Stress nits The units in the SI system is the Newton per square meter or Pascal.0 0 0 . we only consider the magnitude of the force. we only look at beams that carry tensile or compressive loads and which are long and slender. Pa seems too small. Let’s still take patch A as an example. Now let’s resolve the force F in the normal and tangential directions of the acting area as Fig.: Pa = N/m2. force is a vector that has both magnitude and direction.1 (b). i. 2 0 . The intensity of the force or force per unit area acting normally to section A is called Normal Stress. this may easily confuse us.3) A verage Normal Stress (4 th SI p.1: An 8 0 kg lamp is supported by a single electrical copper cable of diameter d = 3 . and Eq.  (sigma).0 0 0 . 2 0 . In engineering. the stress may not reflect the real mechanical status at this point. But in the definition of stress.g.0 0 0 Pa= 2 0 MPa G Pascal iga G Pa (=Pa× 109) e. 7 -8 ) To begin. and it is expressed as:   lim Fn A0 A (2. In other words. What is the stress carried by the cable.g.g.0 0 0 Pa = 2 0 GPa Example 2.

from a safety point of view. Indeed.5) If the applied load is linearly related to the stress developed in the member.S. there are the following several reasons that we must take into account in engineering:  The load for design may be different from the actual load.81  784 N A  d 2   0.6 ) to compute the allowable stress: Chapter 2: Stress & Strain 3 MM211: Solid Mechanics . For convenience. 2 8 -2 9 ) From Example 2 .  F fail Fallow (2. More often.6 MPa stress would be too high for the wire/ cable. the factor of safety is chosen to be greater than 1 in order to avoid the potential failure. (2 . F.S . The value of F. we may be concerned whether or not 8 0 kg would be too heavy. stress is one of most important indicators of structural strength.6MPa A 7. the higher the F. as in the case of using  = F/A. we need the cross sectional area A of the cable and the applied internal force F:  7.S. (safer). So we need to balance the safety and cost. the higher the operational cost. This is dependent on the specific design case. the structure will fail..793  10 6 A llowable Stress (4 th SI p.To determine the stress in the wire/ cable as Eq. then we can define the factor of safety as a ratio of the failure stress fail to the allowable stress allow F . For an aircraft design.4 ). we use Eq.6) Usually.   fail  allow (2. which is a ratio of failure load Ffail (found from experimental testing) divided by the allowable one Fallow F . Moreover. or say 1 0 0 . When the stress (intensity of force) of an element exceeds some level. the heavier the structure. we usually adopt allowable force or allowable stress to measure the threshold of safety in engineering.  Various defects in the material due to manufacturing processing or material handling. (2 .00315 2  F 784   100.S .793  10 6 m 2 4 3 F  mg  80  9. therefore. For nuclear power plant. O ne simple method to consider such uncertainties is to use a num ber called the Factor of Safety.1 . the factor of safety for some of its components may be as high as 3 .S.  The size of the structural member may not be very precise due to manufacturing and/ or assembly. can be found in design codes/ standards and engineering handbooks.

469  10 3  4. In engineering. These changes are referred as deformations. i. With the stress level. please determine the minimum size of the wire/ cable from the material strength point of view. known material’s allowable stress allow and load F.allow = 5 0 MPa. 2.2: In Example 2 . i. Stress is an indication of structural strength and elemental size. its shape and size will change. Chapter 2: Stress & Strain 4 MM211: Solid Mechanics .  F mg  2   Cu. there are two significant problems associated with stress as follows. Problem (1 ) Stress Analysis: for a specific structure. 2 . 6 1 -7 5 ) EFO A N Whenever a force is applied to a body.469mm O bviously. to design the element size: A ≥ F / allow Problem (2 ) 2.allow  4. to determine stress level:  = F / A Engineering Design: Inversely.2 D RM TIO (4 th SI p. Mathematically. Fig.e. we then justify the safety and reliability of a structural member. known size A and load F.2 Deformation due to applied axial forces It is however very hard to make a relative comparison between bodies or structures of different size and length as their individual deformations will be different.S .allow A d 4 Therefore: d 4mg  Cu. (2 . we can design a structural member based on the allowable stress so that it can satisfy the safety requirements. the lower the allowable stress. the bigger the cable size.7 ) Example 2. if the maximum allowable stress for copper is Cu. This requires the development of the concept of Strain. allow   fail F .2 . These deformations can be thought of being either positive (elongation) or negative (contraction) in sign as shown in Fig.e. we can determine the stress level.1 . which relates the body’s deformation to its initial length.

U of Strain nit From Eqns. After deformation.2 . Consider the infinitesimal line segment AB that is contained within the undeformed body as shown in Fig.e.g.9 ).3 as an example. p. in other words.2. the strain can be computed by  LDeformed  LOriginal LOriginal  L L (2 . i.3 (a). meters per meter ( m/m) Usually. (2 .e. Fig. The line AB lies along the n-axis and has an original length of S. A normal strain of 4 8 0 m for a one-meter length is said: Chapter 2: Stress & Strain 5 MM211: Solid Mechanics . uniform uniaxial tension or compression) as shown in Fig. 2.3 Generalized deformation due to applied forces Let’s take the arbitrarily shaped body in Fig. it is common in practice to state it in terms of a ratio of length units.8 ) and (2 . 2 . 6 7 -6 9 .4 8 -5 0 .9 ) i. for most engineering applications ε is very small. points A and B are displaced to A’ and B’ and in general the line becomes a curve having a length S’ The change in length of the line is therefore S-S’. 0 . We consequently define the generalized strain mathematically as   lim S   S B A S B  A along n (2. strain is expressed as a percent. so measurements of strain are in micrometers per metre (m/ m) or (/ m).3 STRA (4 th SI p. Although this is the case. the deformation is uniform in the material (e.8) A verage Normal Strain If the stress in the body is everywhere constant. we notice that the normal strain is a dimensionless quantity since it is a ratio of two lengths.g. 2 .1 % . Sometimes for experiment work. 2 . 1 0 5 ) IN Normal Strain The elongation (+ ve) or contraction (−ve) of a body per unit length is termed Strain. e.0 0 1 m/ m = 0 . &4 th p. the change in length of the body over its original length.

(2 . This property is inherent to the material itself and must be determined by experiment. hardening and necking stages respectively. elastic. 2. H ooke’s L aw The stress-strain linear relationship was discovered by Robert Hook in 1 6 7 6 and is known as Hooke's law. From the yielding stage.5 2. a bunch of standard specimens are made.5 6 ) IN A NSH OK A M aterial Test and Stress-Strain D iagram Material strength depends on its ability to sustain a load without undue deformation or failure.3 5 mm due to the weight of the light.4 STRESS-STRA REL TIO IP.1 0 ) Chapter 2: Stress & Strain 6 MM211: Solid Mechanics .1 0 ). in which the relationship between the strain and stress is linear. O nly 1 0 % of engineering work concerns itself with plasticity and other nonlinear stages (e. O ne of the most important tests to perform in this regard is the tension or compression test.4 is the specimen and test result displayed in a Stress-Strain Diagram. we are only involved in the linear elastic region. About 9 0 % of engineering problems are only concerned with the elastic deformation in structural members and mechanical components. what would its strain be?  0.3: In Example 2 . It is mathematically represented by Eq. To do so. = 4 8 0 × 1 0 -6 = 4 8 0 (m/ m) = 0 . H O E'S L W(4 th SI p.00135  900  10 6  900s 1. metal forming). The test is performed in a tensile test machine but in most cases in a universal testing machine (UTM). yielding. Fig. In this subject. 2 . some permanent plastic deformation occurs.g. if it is measured that the cable was elongated by 1 .0 4 8 0 % = 4 8 0  (micros) = 4 8 0 s (micro strain) Example 2.4 Material test and Stress-Strain Diagram The Stress-Strain diagram consists of 4 stages during the whole process.1 . Shown in Fig.   E (2 .

For most of engineering metal materials.3 .e. When the load P is applied to the bar it changes the bar’s length by  and its radius by r. Usually 0 ≤ v ≤ 0 . Remarks     The lateral strain is caused only by the axial force. GPa is used as the appropriate units. not only does it elongate but it also contracts laterally. the strains in the y and z planes are: Chapter 2: Stress & Strain 7 MM211: Solid Mechanics . 2. 84) ISSO TIO D efinition When a deformable body is stretched by a tensile force.5 (a). We named the constant Poisson’s ratio or v (Nu)    Lateral Strain   Lateral Axial Strain  Axial (2 . i.5 PO N'S RA (4th SI Ed p. Poisson’s ratio is a constant. i. The strain in the axial direction and in lateral/ radial directions are respectively:  axial   L  lateral  r r In the early 1800s. 2 . No force or stress acts in lateral direction.where E is termed the Modulus of Elasticity or Young's Modulus with units of N/ m2 or Pa. Lateral strain is the same in all lateral directions. So Poisson’s ration is positive. 2. e.5 . For most linearly elastic materials v= 0 . it would contract in the other two dimensions as shown in Fig 2 .5 (b). v≥0 .e.1 1 ) Fig. and vice versa. mild steel is about 2 0 0 GPa~ 2 1 0 GPa. Strain in L ateral D irection For bars subjected to a tensile stress x. a compressive force acting on a deformable body will cause it to contract in the direction of the force but its sides will expand laterally as in Fig. French scientist Poisson realized that within the elastic range the ratio of these two strains is a constant. Likewise.g.5 Relationship of the axial strain with the lateral strain The negative sign is used here since longitudinal elongation (positive strain) causes lateral contraction (negative strain).

units are strain per ºC (2 .1 2 ) 2. its overall size will also change. 7 4 ) A IN Thermal D eformation When the temperature of a body is changed.  x  x  E   x   y   x   E   x      x  z E  (2 . the material will contract.1 3 ) : algebraic change in temperature (ºC ) (increase + .1 4 ) Chapter 2: Stress & Strain 8 MM211: Solid Mechanics . if the temperature increases. If we assume that the material is homogeneous and isotropic. It is supposed that this is a common sense for anyone (aha. but for some special materials such as NiTi alloys or shape memory alloys. generally a material expands. when you heat them they contract) . from experiment. Whereas if the temperature decreases. temperature change may cause the dimension or shape change in the material. 2.6 TH ERM LSTRA (4 th SI Ed p. decrease −) : algebraic change in length (“+ ” = elongation. In other words.6 Thermal and mechanical deformation For the majority of engineering materials this relationship is linear. we can find a linear relation between thermal deformation and temperature change as: T =  ⋅T ⋅ L Where:  T : Coefficient of thermal expansion. “−” = contraction) T Thermal Strain  Thermal  T L    T (2 . Fig. More specially.

to represent the weight of a vertical bar or friction forces acting on bar surface).1 5 ) 2.1 2 ). The load P(x) will deform the element into the shape indicated by the dashed outline. p. 2 .7 .6 1 ) A EFO A N F X L Y O D EM Now we are going to find the elastic deformation of a member subjected to axial loads.7 EL STICD RM TIO O A IA L L A EDM BER(4 th SI Ed. the bar is subjected to concentrated loads at its right end and also a variable external load distributed along its length (such as a distributed load could be for example. A FBD can be drawn for the element here shown in Fig.e. (2 . the total strains in all directions would be computed as: x   x   T   x  T  E  y   y   T   y  T  E  z       T  T z  z E  (2 . 2.7 Thermal and mechanical deformation We pick a differential element of length dx and cross-sectional area A(x).6 (b). 2 .  = E Therefore Chapter 2: Stress & Strain 9 MM211: Solid Mechanics . Assume that resultant internal axial force is represented as P(x). 2 . by referring to Eq.C oupled Strain Status If we consider both mechanical strain  and thermal strain T in the structure as shown in Fig.. For a more general case. Fig. Let’s consider a generalized bar shown in Fig. The average stress in the cross-sectional area would be  x   The average strain in the cross-sectional area would be  x   P x  Ax  d dx Provided these quantities do not exceed the proportional limit. i. we can relate them using Hook’s law. which has a gradually varying crosssectional area along its length L. Here we wish to find the relative displacement  of one end with respect to the other.7 .

The total displacement can be computed from algebraic addition as   i Pi Li Ai Ei (2 . having cross-sectional areas of AAB = 2 0 0 mm2 and ABC = 1 0 0 mm2 . P(x) = P = constant (no axially distributed load) A(x) = A = constant (uniform cross-sectional area) E(x) = E = constant (homogeneous material) From Eq. the above equation can be used for each segment. Their Young’s moduli are EAB = 1 0 0 GPa and EBC = 2 1 0 GPa respectively.1 6 ).4: The composite bar shown in the figure is made of two segments. we have d  P x  dx E x Ax  For the entire length L of the bar. AB and BC. we must integrate this expression to find the required end displacement   0 L P x  dx E x Ax   L P(x) A(x) E(x) = = = = = displacement between two points distance between the points Internal axial force distribution Cross-sectional area Young’s modulus (2 . Chapter 2: Stress & Strain 10 MM211: Solid Mechanics . Find the total displacement at the right end. we have  PL EA (2 .e.1 8 ) Example 2. (2 .1 7 ) M ulti-Segment Bar If the bar is subjected to several different axial forces or cross-sectional areas or Young’s moduli. i.1 6 ) Where: C onstant L and C oad ross-Sectional A rea In many engineering cases. the structural members experience a constant load and have a constant cross-sectional area and are made of one homogenous material.P x   d   E x   Ax   dx  Re-organize the equation.

004m  4mm towards left  P L PAB L AB  30  10 3  4 10  10 3  4. known reactions can be determined strictly from equilibrium equations. the unknown reaction can easily be determined. 4 7 ) INA EM OD X LY Statically D eterminate and Indeterminate When a bar is supported at one end and subjected to an axial force P at the other end as shown in Fig. there is only one unknown reaction force FA. Step 2 Equilibriums Internal force in AB   Fx  0   PAB  F2  F1  PAB  30kN (O pposite to our assumption of tension. By using the equations of statics.Step 1 FBDs for Segments AB and BC. (2 . 2 .006  0. Assume the internal forces are in tension.2  BC BC   E AB AAB E BC ABC 100  10 9  200  10 6 210  10 9  100  10 6 2.1 8 )  AC   AB   BC   AC  0. – i.e. so Segment AB is in compression) Internal force in BC    Fx  0   PBC  F1  0   PBC  F1  10kN in tension  Step 3 Compute the total deformation by using Eq. Such a system with the same number of unknown reactions as equations of statics is called statically determinate.8 (a). Chapter 2: Stress & Strain 11 MM211: Solid Mechanics .8 STA A L IND TIC L Y ETERM TE M BERS L A EDA IA L (4 th SI Ed p.002  0.

Such an equation is usually termed the compatibility equation or kinematic condition(s). If the system has more unknown forces than equations of statics it is called statically indeterminate. and how much point B moves in relation to point A.8 Statically determinate and indeterminate structures If the bar is also restricted at the free end as shown in Fig. Basically the amount that length AC elongates CB contracts as shown in Fig. C ompatibility C onditions What we need is an additional equation that specifies how the structure is displaced due to the applied loading.2 0 ) Chapter 2: Stress & Strain 12 MM211: Solid Mechanics . it has 2 unknown reactions FA and FB. 2 . one known force P and one equation of statics:    Fy  0  FA  FB  P  0  FA  FB  P (2 . Now. what we need is an additional equation that specifies how the structure is displaced due to the applied loading. In order to determine the compatibility for this example we need to determine how point C is going to move. 2 . Since there are 2 unknown and only 1 equation of statics herein.1 9 ) It cannot be solved unless we introduce one more condition. Such an equation is usually termed the compatibility equation. 2.8 (b).9 . since both ends of the bar are fully fixed. so the equation can be written as:  AC   CB  0 (2 .Fig. then the total change in length between A and B must be zero.

   Fy  0  FA  PAC  0  PAC  FA Tension   (2 .9 .2 3 ) Combining Compatibility equation (2 .2 2 )  CB   PCB LCB F L   B CB ACB ECB ACB ECB Contractio n   Compatibility condition:    AC   CB  FA L AC  FB LCB  0   AAC E AC  ACB ECB    (2 . ( Now a FBD can be at any level of the structural system or structural members).2 1 )  AC  PAC L AC F L  A AC AAC E AC AAC E AC elongation   For segment CB.2 4 ) Chapter 2: Stress & Strain 13 MM211: Solid Mechanics .1 9 ). we now can solve for the two unknowns FA and FB as. 2 .    Fy  0 FB  PCB  0  PCB   FB Compression   (2 . 2. FB LCB  FA L AC A E  A E  0  AC AC CB CB    FA  FB  P  (2 .9 Compatibility condition Let’s look at the free body diagram for segment AC and CB as in Fig. for segment AC.Fig. Therefore.2 3 ) with the equation of statics (2 .

the copper bar expands towards the right by T .e. we have  Al  T . we have FB  L AC P L and FA  LCB P L (2 .Cu   F .i.5: Two bars made of Copper and Aluminium are fixed to the rigid abutments. which will prevent the copper bar from expanding further. Contraction −) Similarly. The real total deformation of the copper bar will be computed as  Cu   T . the copper bar is pressed back by δF . Originally. FB  L AC AAC E AC L AC LCB  AAC E AC ACB ECB P FA  LCB ACB ECB L AC LCB  AAC E AC ACB ECB P If AACEAC = ACBECB = Const.Cu.Cu .Cu (elongation + . Let’s firstly look at the copper bar. Al (elongation + . there is a gap of 5mm between the ends as shown in the figure. Determine the average normal stress in both bars if we increase the temperature from 10°C to 210°C. Al   F . Contraction −) Chapter 2: Stress & Strain 14 MM211: Solid Mechanics .2 5 ) Example 2. After the copper bar touches the aluminium bar. a mechanical force F will develop. When the bar system is heated up from 10°C to 210°C. We assume that due to such a mechanical force.

and it is expressed as in Eq. Al   F .Cu   Cu TLCu   ECu ACu  110  10 9  7. Chapter 2: Stress & Strain 15 MM211: Solid Mechanics .8   Al   T . (tau).2  0.1 0 .8  0. we have (2 .2 6 )  i.005    Cu TLCu   Al TLAl  0.55  10 6  1.4  23  200  0.2 7 ) as:   lim Ft A0 A (2 .2 7 ) In order to show how the shear stress can develop in a structural member. we prescribe a compatibility condition as  Al   Cu  0. Al    T .4   Cu   T .Cu   F . Al    Al TLAl   E Al AAl  69  10 9  7.65mm  2.9 A VERA E SH R STRESS (4 th SI Ed p.35  10 3 m  1.85  10 5    FL  206.68  10 3  3.8  9 5 9 110  10  7.2  A 7.36  10 5  9.005   FL Al    Al TLAl    E Al AAl     FLCu  Cu TLCu   ECu ACu  F    0.05  10 6  3. let’s take a block as an example.2 N The average normal stress can be computed as  F 206.35mm  Al   23  10 6  200  0.85  10 5  206.8  3.85  10 5    FL  206. Al   F .2  0.Because these two expanding bars should fill the gap.005 LCu L Al  ECu ACu E Al AAl 17  10 6  200  0. The block is supported by two rigid bodies and an external force F is applied vertically as shown in Fig. (2 .4 0.Cu   0. it will cause the material of the block to deform and fail along the vertical planes as shown.e T .005 From these two equations. If the force is large enough.85  10 69  10  7.63MPa The deflections can then be calculated as Cu   17  10 6  200  0.85  10 5  2. 1 0 ) G EA The intensity or force per unit area acting tangentially to A is called Shear Stress.Cu   F .005 0.4  1.65  10 3 m  3. 2 .

the typical example is to build a connection with other structural elements. Stress C oncentration Factor In engineering practice. The Chapter 2: Stress & Strain 16 MM211: Solid Mechanics . the actual stress distribution does not have to be determined. We call this phenomenon Stress Concentration.10 Average shear stress    Fy  0  2V  F  0 ∴V = F / 2 The average shear stress distributed over each sectioned area that develops the shear force is defined by  avg   avg V A V A (2 .1 1 (a). though. only the maximum stress at these sections must be known. both experiment and theory of elasticity finds that the normal stress will be uniformly distributed over the cross-section Stress C oncentration However. Fig.2 8 ) = assume to be the same at each point over the section = Internal shear force = Area at the section 2. if we drill a hole for some reason in the component. 1 0 7 ) TIO For a uniform cross-sectional bar that has applied an axial force. we find that the stress distribution is no longer uniform.A FBD of the unsupported center segment indicat e that shear force V=F/2 must be applied at each section to hold the segment in equilibrium. and the member is then designed to resist this highest stress when the axial load is applied. For such a case. 2 . as in Fig. if we cut at the hole’s center plane. Instead. 2.10 STRESS C NC O ENTRA NS (4 th SI Ed p. It may distribute over such a small area in a highly non-uniform pattern.

1 1 (b).3 0 ) Fig. K  max  avg (2 . Provided K has been known from the figures or tables (as in Fig.1 1 (c). 2 . where A’ is the smallest cross-sectional area.1 1 (c)) in terms of Stress Concentration Factor K. Chapter 2: Stress & Strain 17 MM211: Solid Mechanics . 2 . Then from the above equation the maximum stress at the cross section can be computed as:  max  K avg  K P A (2 . By observing Fig. and the average normal stress has been calculated from avg=P/A’.specific values of the maximum normal stress at the critical sectio n can be determined by experimental methods or by advanced mathematical techniques using the theory of elasticity.1 1 (c)). the higher the stress concentration. 2 .11 Stress concentration Stress concentration occurs in the case that there is a sudden change in cross-sectional area. 2 . it is interesting to note that the bigger the ratio of change in the sectional area. 2.2 9 ) in which avg=P/A’ is the assumed average stress as in Fig. The results of these investigations are usually reported in graphical form (as in Fig.