Normal stress and strain

Strength of Materials CIE 102
[3 1 0 4]
for
First Year B.E. Degree Students
1

COURSE CONTENT IN BRIEF
1. Simple stress and strain
2. Statically indeterminate problems and thermal stresses
3. Shearing force and bending moment
4. Stresses due to bending
5. Stresses due to shearing
6. Slope and deflection of beams
7. Stresses due to Torsion in circular shaft
8. Variation of stress at a point
9. Stresses due to fluid pressure in thick and thin cylinder
10. Stability of columns
2

Books for Reference
1. Machanics of Materials, by E.P.Popov
2. Machanics of Materials, by E J Hearn
3. Strength of materials, by Beer and Johnston
4. Strength of materials, by F L Singer & Andrew Pytel
5. Strength of Materials, by B.S. Basavarajaiah & P. Mahadevappa
6. Strength of Materials, by Ramamruthum
7. Strength of Materials, by S S Bhavikatti
3

•Normal stress and strain
•Hooke’s law
•Modulus of elasticity
•Tension test on ductile and brittle materials
• Factor of safety and allowable stress
•Poisson's ratio
•Shear stress and shear strain
•Modulus of rigidity
•Tapering bar and stepped bar subjected to axial load
CHAPTER – I
Simple stress & strain
4

The subject strength of materials deals with the relations between
externally applied loads and their internal effects on bodies. The
bodies are no longer assumed to be rigid and the deformations,
however small, are of major interest
Alternatively the subject may be called the mechanics of solids.
CHAPTER – I Introduction
The subject, strength of materials or mechanics of materials involves
analytical methods for determining the strength , stiffness
(deformation characteristics), and stability of various load carrying
members.
5

Engineering Mechanics
Mechanics of Solids Mechanics of Fluids
Rigid Bodies Deformable
Bodies
Statics Dynamics
Strength of
Materials
Theory of
Elasticity
Theory of
Plasticity
Ideal
Fluids
Viscous
Fluids
Compressible
Fluids
Branches of Mechanics
6

GENERAL CONCEPTS
STRESS
No engineering material is perfectly rigid and hence, when
a material is subjected to external load, it undergoes
deformation.
While undergoing deformation, the particles of the material
offer a resisting force (internal force). When this resisting
force equals applied load the equilibrium condition exists
and hence the deformation stops.
These internal forces maintain the externally applied forces
in equilibrium.
7

Stress = internal resisting force / resisting cross sectional area
The internal force resisting the deformation per unit area is
called as stress or intensity of stress.
STRESS
8
A
R
=

gigapascal, 1GPa = 1×109
N/m2
= 1×103
MPa
= 1×103
N/mm2
SI unit for stress
N/m2
also designated as a pascal (Pa)
Pa = N/m2
kilopascal, 1kPa = 1000 N/m2
megapascal, 1 MPa = 1×106
N/m2
= 1×106
N/(106
mm2
) = 1N/mm2
1 MPa = 1 N/mm2
STRESS
9

AXIAL LOADING – NORMAL STRESS
Consider a uniform bar of cross
sectional area A, subjected to a tensile
force P.
Consider a section AB normal to the
direction of force P
Let R is the total resisting force acting
on the cross section AB.
Then for equilibrium condition,
R = P
Then from the definition of stress,
normal stress = σ = R/A = P/A
P
P
P
R
BA
R
P
STRESS
10
σ = Normal StressSymbol:

Direct or Normal Stress:
AXIAL LOADING – NORMAL STRESS
Intensity of resisting force perpendicular to or normal to the
section is called the normal stress.
Normal stress may be tensile or compressive
Tensile stress: stresses that cause pulling on the surface of the
section, (particles of the materials tend to
pull apart causing extension in the direction of
force)
Compressive stress: stresses that cause pushing on the surface of
the section, (particles of the materials tend to
push together causing shortening in the
direction of force)
STRESS
11

• The resultant of the internal forces for an
axially loaded member is normal to a
section cut perpendicular to the member
axis.
A
P
A
F
ave
A
=
∆
∆
=
→∆
σσ
0
lim
• The force intensity on that section is
defined as the normal stress.
STRESS
12

Illustrative Problem
A composite bar consists of an aluminum section rigidly fastened
between a bronze section and a steel section as shown in figure.
Axial loads are applied at the positions indicated. Determine the
stress in each section.
Bronze
A= 120 mm2
4kN
Steel
A= 160 mm2
Aluminum
A= 180 mm2
7kN2kN13kN
300mm 500mm400mm
Example 1
13

To calculate the stresses, first determine the forces in each
section.
For equilibrium condition algebraic sum of forces on LHS
of the section must be equal to that of RHS
4kN 7kN2kN13kN
To find the Force in bronze section,
consider a section bb1
as shown in the figure
Bronze
14
b
b1

Force acting on Bronze section is 4kN, tensile
Stress in Bronze section =
Force in Bronze section
Resisting c/s area of the Bronze section
=
2
22
/33.33
120
10004
120
4
mmN
mm
N
mm
kN
=
×
= = 33.33MPa
Tensile stress
15
Bronze
4kN 7kN2kN13kN
13kN 2kN 7kN
Bronze
4kN 4kN
(= )
b1
b

Illustrative ProblemForce in Aluminum section
Force acting on Aluminum section is 9kN, Compressive
16
4kN 7kN2kN13kN
2kN 7kN
Aluminum
9kN4kN 13kN
Aluminum
(= )

Illustrative ProblemForce in steel section
Force acting on Steel section is 7kN, Compressive
17
4kN 7kN2kN13kN
7kN
steel
7kN
4kN 2kN13kN
steel

Stress in Steel section =
Force in Steel section
Resisting cross sectional area of the Steel section
=
2
22
/75.43
160
10007
160
7
mmN
mm
N
mm
kN
=
×
=
Stress in Aluminum
section =
Force in Al section
Resisting cross sectional area of the Al section
=
2
22
/50
180
10009
180
9
mmN
mm
N
mm
kN
=
×
=
= 43.75MPa
Compressive stress
= 50MPa
Compressive stress
18

STRAIN
STRAIN :
When a load acts on the material it will undergo deformation.
Strain is a measure of deformation produced by the application of
external forces.
If a bar is subjected to a direct load, and hence a stress, the bar will
change in length. If the bar has an original length L and change in
length by an amount δL, the linear strain produced is defined as,
Strain is a dimension less quantity.
19
L
Lδ
ε =
Original length
Change in length
=Linear strain,

Linear Strain
strainnormal
stress
==
==
L
A
P
δ
ε
σ
L
A
P
A
P
δ
ε
σ
=
==
2
2
LL
A
P
δδ
ε
σ
==
=
2
2
20

STRESS-STRAIN DIAGRAM
In order to compare the strength of various materials it is necessary
to carry out some standard form of test to establish their relative
properties.
One such test is the standard tensile test in which a circular bar of
uniform cross section is subjected to a gradually increasing tensile
load until failure occurs.
Measurement of change in length over a selected gauge length of the
bar are recorded throughout the loading operation by means of
extensometers.
A graph of load verses extension or stress against strain is drawn as
shown in figure.
21

Typical tensile test curve for mild steel
22
Proportionality limit

Typical tensile test curve for mild steel showing upper yield point and
lower yield point and also the elastic range and plastic range
23

Limit of Proportionality :
From the origin O to a point called proportionality limit the stress
strain diagram is a straight line. That is stress is proportional to
strain. Hence proportional limit is the maximum stress up to which
the stress – strain relationship is a straight line and material behaves
elastically.
From this we deduce the well known relation, first postulated by
Robert Hooke in 1678, that stress is proportional to strain.
Beyond this point, the stress is no longer proportional to strain
A
PP
P =σ Load at proportionality limit
Original cross sectional area
=
24
Stress-strain Diagram

Elastic limit:
It is the stress beyond which the material will not return to its
original shape when unloaded but will retain a permanent
deformation called permanent set. For most practical purposes it can
often be assumed that points corresponding proportional limit and
elastic limit coincide.
Beyond the elastic limit plastic deformation occurs and strains are
not totally recoverable. There will be thus some permanent
deformation when load is removed.
A
PE
E =σ Load at elastic limit
=
25

Yield point:
It is the point at which there is an appreciable elongation or yielding
of the material without any corresponding increase of load.
A
PY
Y =σ Load at yield point
=
26

Ultimate strength:
It is the stress corresponding to maximum load recorded during the
test. It is stress corresponding to maximum ordinate in the stress-
strain graph.
A
PU
U =σ Maximum load taken by the material
=
27

It is the stress at failure. For most ductile material including
structural steel breaking stress is somewhat lower than ultimate
strength because the rupture strength is computed by dividing the
rupture load (Breaking load) by the original cross sectional area.
A
PB
B =σ load at breaking (failure)
=
28
Rupture strength (Nominal Breaking stress):
True breaking stress,
load at breaking (failure)
Actual cross sectional area=

The capacity of a material to allow these large plastic deformations
is a measure of ductility of the material.
After yield point the graph becomes much more shallow and covers
a much greater portion of the strain axis than the elastic range.
Ductile Materials:
The capacity of a material to allow large extension i.e. the ability to
be drawn out plastically is termed as its ductility. Material with high
ductility are termed ductile material.
Example: Low carbon steel, mild steel, gold, silver, aluminum.
29

A measure of ductility is obtained by measurements of the
percentage elongation or percentage reduction in area, defined as,
Percentage elongation
increase in gauge length (up to
fracture)
original gauge length
×100=
Percentage reduction in
area original area
×100=
Reduction in cross sectional area
of necked portion (at fracture)
30
Brittle Materials :
A brittle material is one which exhibits relatively small extensions
before fracture so that plastic region of the tensile test graph is
much reduced.
Example: steel with higher carbon content, cast iron, concrete, brick

STRESS-STRAIN : DUCTILE MATERIAL
31

Stress-Strain Diagram: Ductile Materials
2
2
1000
)(
)(
in
lb
ksi
insquareinch
lbpounds
psi
×
=
=
32

Stress-Strain Diagram: Brittle Materials
Stress-strain diagram for a typical brittle material
33

Stress-Strain Test
Machine used to test tensile test specimen
Test specimen with tensile load
L = gauge length
34

HOOKE”S LAW
For all practical purposes, up to certain limit the relationship
between normal stress and linear strain may be said to be linear for
all materials
Thomas Young in 1807 introduced a constant of proportionality
that came to be known as Young’s modulus.
stress (σ) α strain (ε)
stress (σ)
strain (ε) = constant
stress (σ)
strain (ε) = E
35
Modulus of Elasticity
Young’s Modulus
= or

HOOKE”S LAW
Young’s Modulus is defined as the ratio of normal stress to linear
strain within the proportionality limit.
From the experiments, it is known that strain is always a very small
quantity, hence E must be large.
For Mild steel, E = 200GPa = 2×105
MPa = 2×105
N/mm2
stress (σ)
strain (ε)=E =
LA
PL
L
L
A
P
δ
δ
=÷
The value of the Young’s modulus is a definite property of a
material
36

Material
Density
(kg/m3
)
Young's Modulus
109
N/m2
Ultimate Strength Su
106
N/m2
Yield Strength Sy
106
N/m2
Steel 7860 200 400 250
Aluminum 2710 70 110 95
Glass 2190 65 50 ...
Concrete 2320 30 40 ...
Wood 525 13 50 ...
Bone 1900 9 170 ...
Polystyrene 1050 3 48 ...
37

Hooke’s Law: Modulus of Elasticity
Below the yield stress
ElasticityofModulus
orModulusYoungs=
=
E
Eεσ
Strength is affected by alloying, heat
treating, and manufacturing process but
stiffness (Modulus of Elasticity) is not.
Stress-strain diagram for Iron and
different grades of steel
38

Stress-strain diagram
Hard drawn wire materials Various types of nylon
and polycarbonate
39

Elastic vs. Plastic Behavior
• If the strain disappears
when the stress is removed,
the material is said to
behave elastically.
• When the strain does not return to zero after the stress is removed,
the material is said to behave plastically.
• The largest stress for
which this occurs is called
the elastic limit.
40

Elastic vs. Plastic Behavior
For certain materials, for example, high carbon steel and non-ferrous
metals, it is not possible to detect any difference between the upper
and lower yield points and in some cases no yield point exists at all.
In such cases a proof stress is used to indicate beginning of plastic
strain.
Proof stress is the stress corresponding to a fixed permanent strain in
stress-strain diagram.
For example: 0.1% proof stress indicates that stress which, when
removed, produces a permanent strain or “set” of 0.1% of the original
gauge length.
41

Proof stress
Determination of 0.1%
proof stress
Permanent deformation or “set”
after straining beyond yield point
42

Deformations Under Axial Loading
AE
P
E
E ===
σ
εεσ
• From Hooke’s Law:
• From the definition of strain:
L
δ
ε =
• Equating and solving for the
deformation,
AE
PL
=δ
• With variations in loading, cross-
section or material properties,
∑=
i ii
ii
EA
LP
δ
43

A specimen of steel 20mm diameter with a gauge length of
200mm was tested to failure. It undergoes an extension of
0.20mm under a load of 60kN. Load at elastic limit is 120kN.
The maximum load is 180kN. The breaking load is 160kN.
Total extension is 50mm and the diameter at fracture is 16mm.
Find:
a) Stress at elastic limit
b) Young’s modulus
c) % elongation
d) % reduction in area
e) Ultimate strength
f) Nominal breaking stress
g) True breaking stress
Example 2
44

Solution:
a) Stress at elastic limit,
Example 2
σE =
Load at elastic limit
Original c/s area
MPa
mm
N
mm
kN
A
PE
97.38197.381
16.314
120
22
====
b) Young’s Modulus,
GPa
MPa
mm
N
mm
mm
mm
kN
L
L
A
P
E
98.190
190980
190980
101
98.190
200
20.0
16.314
60
23
2
=
=
=
×
==== −
δε
σ
(consider a load which is within the elastic limit)
45

c) % elongation,
Example 2
% elongation =
Final length at fracture – original length
Original length
%25100
200
50
=×=
%36100
16.314
4
1616.314
2
=×
×−
=
π
d) % reduction in area =
Original c/s area -Final c/s area at fracture
Original c/s area
46

e) Ultimate strength,
Example 2
Ultimate strength =
Maximum load
Original c/s area
)(
/96.572
16.314
180 2
2
MPa
mmN
mm
kN
==
f) Nominal breaking Strength
MPa
kN
29.509
16.314
160
==
g) True breaking Strength
MPa
mm
kN
38.795
06.201
160
2
==
47
Breaking load
Original c/s area
=
Breaking load
c/s area at fracture
=

A composite bar consists of an aluminum section rigidly fastened
between a bronze section and a steel section as shown in figure.
Axial loads are applied at the positions indicated. Determine the
change in each section and the change in total length. Given
Ebr = 100GPa, Eal = 70GPa, Est = 200GPa
Bronze
A= 120 mm2
4kN
Steel
A= 160 mm2
Aluminum
A= 180 mm2
7kN2kN13kN
300mm 500mm400mm
Example 3
48

From the Example 1, we know that,
Pbr = +4kN (Tension)
Pal = -9kN (Compression)
Pst = -7kN (Compression)
Example 3
stress (σ)
strain (ε)=E =
LA
PL
δ
=
AE
PL
L =δChange in length =
Change in length of
bronze = )/(10100120
3004000
232
mmNmm
mmN
Lbr
××
×
=δ
= 0.1mm
49
Deformation due to
compressive force is
shortening in length, and
is considered as -ve

Illustrative ProblemExample 3
=++ stalbr LLL δδδ
Change in total
length =
= -0.109mm
Change in length of
aluminum section = )/(1070180
4009000
232
mmNmm
mmN
Lal
××
×−
=δ
Change in length of
steel section = )/(10200160
5007000
232
mmNmm
mmN
Lst
××
×−
=δ
= -0.286mm
+0.1 – 0.286 - 0.109
= -0.295mm
50

2P Steel
Aluminum 2P4P
2.8m0.8m
Example 4
An aluminum rod is fastened to a steel rod as shown. Axial
loads are applied at the positions shown. The area of cross
section of aluminum and steel rods are 600mm2
and 300mm2
respectively. Find maximum value of P that will satisfy the
following conditions.
a)σst ≤ 140 MPa
b)σal ≤ 80 MPa
c)Total elongation ≤ 1mm,
Take Eal = 70GPa, Est = 200GPa
51

To find P, based on the condition, σst ≤ 140 MPa
solution
Stress in steel must be less than or equal to 140MPa.
Hence, σst =
= 140MPa
st
st
A
P
= 2
/140
2
mmN
A
P
st
==
kNN
A
P st
2121000
2
140
==
×
=
2P Steel
Aluminum 2P4P
4P2P 2P
2P2P Tensile
52

To find P, based on the condition, σal ≤ 80 MPa
solution
Stress in aluminum must be less than or equal to 80MPa.
Hence, σal =
= 80MPa
al
al
A
P
= 2
/80
2
mmN
A
P
al
==
kNN
A
P al
2424000
2
80
==
×
=
2P Steel
Aluminum 2P4P
4P2P 2P
2P 2P Compressive
53

To find P, based on the condition, total elongation ≤ 1mm
Solution
Total elongation = elongation in aluminum + elongation in steel.
stal AE
PL
AE
PL






+





=1mm





 +
+




 −
=
stst
st
alal
al
EA
PL
EA
PL 22
1mm






××
×+
+





××
×−
= 33
10200300
28002
1070600
8002 PP
1mm
P = 18.1kN
Ans: P = 18.1kN (minimum of the three values)
54

Derive an expression for the total extension of the tapered bar of
circular cross section shown in the figure, when subjected to an axial
tensile load W
WW
A B
L
Diameter
d1
Diameter
d2
55

Consider an element of length, δx at a distance x from A
B
WW
A
xd1 d2
dx
Diameter at x,
( ) x
L
dd
d ×
−
+= 12
1 c/s area at x, ( )2
1
2
44
kxd
dx
+==
ππ
xkd ×+= 1
Change in length over a
length dx is ( ) 











×+
=





=
Ekxd
Wdx
AE
PL
dx
2
1
4
π
length L is ( )
∫












×+
=
L
Ekxd
Wdx
0 2
1
4
π
56

Put d1+kx = t,
Then k dx = dt
length L is ( )
∫












×+
=
L
Ekxd
Wdx
0 2
1
4
π
( )
∫












×
=
L
Et
k
dt
W
0 2
4
π
LLL
kxdEk
W
tEk
Wt
Ek
W
0100
12
)(
1414
1
4






+
−
=




−
=





−
=
+−
πππ
E
dd
WL
dEd
WL
×
==
4
4
2121
ππ
57

A two meter long steel bar is having uniform diameter of 40mm for a
length of 1m, in the next 0.5m its diameter gradually reduces to
20mm and for remaining 0.5m length diameter remains 20mm
uniform as shown in the figure. If a load of 150kN is applied at the
ends, find the stresses in each section of the bar and total extension of
the bar. Take E = 200GPa.
500mm
Ф = 40mm
Ф = 20mm
150kN
150kN
500mm1000mm
58

Example 6
59
If we take a section any where along the length of the bar, it is
subjected to a load of 150kN.
500mm
Ф = 40mm
Ф = 20mm
150kN
150kN
500mm1000mm
2
1
3
MPa
kN
MPa
kN
MPa
kN
d
kN
MPa
kN
46.477
4
20
150
46.477
4
20
150
37.119
4
40
150
4
150
37.119
4
40
150
23
2min.2,
2.max,222
21
==
==
==⇒=
==
π
σ
π
σ
π
σ
π
σ
π
σ

60
If we take a section any where along the length of the bar, it is
subjected to a load of 150kN.
500mm
Ф = 40mm
Ф = 20mm
150kN
150kN
500mm1000mm
2
1
3
( )
( ) mm
E
kN
l
mm
E
kN
dEd
PL
l
mm
E
kN
l
194.1
4
20
500150
597.0
2040
50015044
597.0
4
40
1000150
23
21
2
21
=
×
×
=
=
×××
××
==
=
×
×
=
π
δ
ππ
δ
π
δ
mml 388.2total, =δ

Derive an expression for the total extension of the tapered bar AB of
rectangular cross section and uniform thickness, as shown in the
figure, when subjected to an axial tensile load W.
WW
A
B
L
d1
d2
b
b
61

W W
A B
x
d1
d2
b
b
dx
depth at x,
( ) x
L
dd
d ×
−
+= 12
1
c/s area at x, ( )bkxd += 1
xkd ×+= 1
length dx is ( ) 





×+
=





=
Ebkxd
Wdx
AE
PL
dx 1
62

length L is ( )∫ 





×+
=
L
Ebkxd
Wdx
0
1
( )12 loglog dd
kEb
P
ee −
××
=
( )
( )12
12
loglog
302.2
dd
ddEb
LP
−
−××
××
=
63

Derive an expression for the total extension produced by self weight
of a uniform bar, when the bar is suspended vertically.
L
Diameter
d
64

P1 = weight of the bar below
the section,
= volume × specific weight
= (π d2
/4)× x × γ
= A× x ×γ
Extension of the
element due to
weight of the bar
below that,
AE
dxxA
AE
dxP
AE
PL
dx
)(1 ρ××
==





=
P1x
Diameter
d
dx dx
element
65

The above
expression can
also be written as
Hence the total extension entire
bar E
L
E
x
AE
dxxA
L
L
22
)( 2
0
2
0
γγγ
=





=
××
= ∫
AE
PL
AE
LAL
A
A
E
L
×=
×
=×=
2
1
2
)(
2
2
γγ
Where, P = (AL)×γ
= total weight of the bar
66

SHEAR STRESS
Consider a block or portion of a material shown in Fig.(a) subjected to
a set of equal and opposite forces P. Then there is a tendency for one
layer of the material to slide over another to produce the form of
failure as shown in Fig.(b)
The resisting force developed by any plane ( or section) of the block
will be parallel to the surface as shown in Fig.(c).
P
P
Fig. a Fig. c
P
P
R
R
Fig. b
The resisting forces acting parallel to the surface per unit area is called
as shear stress.
67

Shear stress (τ) =
If block ABCD is subjected to shearing stress as shown in Fig.(d),
then it undergoes deformation. The shape will not remain
rectangular, it changes into the form shown in Fig.(e), as AB'
C'D.
Shear strain
Shear resistance
Area resisting shear A
P
=
This shear stress will always be tangential to the area on which it acts
τ
B
Fig. d
τ
D
C
A
τB'
D
C'
A
τ
B C
Fig. e
68

The angle of deformation is measured in radians and hence is
non-dimensional.
φφ ≈=
′
= tanstrainshear
AB
BB
D
τ
B' C'
A
τ
Fig. e
B
C
φ
The angle of deformation is then termed as shear strainφ
69
Shear strain is defined as the
change in angle between two
line element which are
originally right angles to one
another.

SHEAR MODULUS
For materials within the proportionality limit the shear strain is
proportional to the shear stress. Hence the ratio of shear stress to shear
strain is a constant within the proportionality limit.
For Mild steel, G= 80GPa = 80,000MPa = 80,000N/mm2
Shear stress (τ)
Shear strain (φ) = constant
The value of the modulus of rigidity is a definite property of a
material
= G
Shear Modulus
or
Modulus of Rigidity
=
70

Example: Shearing Stress
• Forces P and P‘ are applied transversely
to the member AB.
A
P
=aveτ
• The corresponding average shear stress is,
• The resultant of the internal shear force
distribution is defined as the shear of the
section and is equal to the load P.
• Corresponding internal forces act in the
plane of section C and are called shearing
forces.
• The shear stress distribution cannot be
assumed to be uniform.
71

Double shear
Consider the simple riveted lap joint shown in the Fig.(a). When
load is applied to the plates as shown in the figure the rivet is subjected
to shear forces tending to shear it on one plane as indicated.
Shear stress τ (in double shear) = P/2A
But the joint with two cover plates, shown in Fig.(b), the rivet is
subjected to possible shearing on two faces, which is called as double
shear. In such cases twice the area of the rivet is resisting the applied
forces so that the shear stress set up is given by
72
Fig. a
P
P
Fig. b
P
P

A
F
A
P
==aveτ
Single Shear
A
F
A
P
2
ave ==τ
Double Shear
Examples
73

Pin Shearing Stresses
• The cross-sectional area for
pins at D
26
2
2
m10491
2
mm25 −
×=





== ππ rA
MPa102
m10491
N1050
26
3
, =
×
×
== −
A
P
aveCτ
• The force on the pin at C is
equal to the force exerted by
the rod BC,
example: Shearing Stress
To find the shearing stress in pin.
74
Rod BC

• The cross-sectional area for pins at
D, & E
26
2
2
m10491
2
mm25 −
×=





== ππ rA
• The pin at A is in double shear with a
total force equal to the force exerted
by the boom AB,
MPa7.40
)m10491(2
kN40
2 26, =
×
== −
A
P
aveAτ
Example: Double shear
75
Rod AB

State of simple shear
Force on the face AB = P = τ × AB × t
Consider an element ABCD in a strained material subjected to shear
stress, τ as shown in the figure
Where, t is the thickness of the element.
τ
τ
A B
CD
Force on the face DC is also equal to P
76

The element is subjected to a clockwise moment
Now consider the equilibrium of the element.
(i.e., ΣFX = 0, ΣFY = 0, ΣM = 0.)
P
P
A B
CD
But, as the element is actually in equilibrium, there must be another
pair of forces say P' acting on faces AD and BC, such that they
produce a anticlockwise moment equal to ( P × AD )
For the force diagram shown,
ΣFX = 0, & ΣFY = 0,
But ΣM ≠ 0
77
P × AD = (τ × AB × t) × AD
force

Eq.(1) can be written as
If τ ' is the intensity of the shear stress on the
faces AD and BC, then P ' can be written as,
P ' = τ ' × AD × t
P ' × AB = P × AD
= (τ × AB × t)× AD ----- (1)
P
P
A B
CD
P ' P '
(τ ' × AD× t ) × AB = (τ × AB × t) × AD ----- (1)
τ ' = τ
78
τ
τ
A B
CD
τ ' τ '

Thus in a strained material a shear stress is always accompanied by
a balancing shear of same intensity at right angles to itself. This
balancing shear is called “complementary shear”.
The shear and the complementary
shear together constitute a state of
simple shear
A B
CD
τ'= τ
τ
τ
τ'= τ
79

Direct stress due to pure shear
Consider a square element of side ‘a’ subjected to shear stress as shown
in the Fig.(a). Let the thickness of the square be unity.
Fig.(b) shows the deformed shape of the element. The length of diagonal
DB increases, indicating that it is subjected to tensile stress. Similarly
the length of diagonal AC decreases indicating that compressive stress.
a
A B
CD
τ
τ
τ
τ
a
Fig.(a).
A
B
CD
τ
τ
τ
τa
a
Fig.(b).
80

Now consider the section, ADC of the element, Fig.(c).
Resolving the forces in σn direction, i.e., in the X-direction shown
a
Fig.(c).
a
A
CD
( )a2
a
A σn
CD
τ
τ
a
X
For equilibrium
( ) ( )
τσ
τσ
=
××−×××=
=∑
n
n aa
Fx
45cos212
0
81

Therefore the intensity of normal tensile stress developed on plane
BD is numerically equal to the intensity of shear stress.
Similarly it can be proved that the intensity of compressive stress
developed on plane AC is numerically equal to the intensity of
shear stress.
82

Consider the rectangular bar shown in Fig.(a) subjected to a tensile
load. Under the action of this load the bar will increase in length by
an amount δL giving a longitudinal strain in the bar of
POISSON’S RATIO
l
l
l
δ
ε =
Fig.(a)
83

The associated lateral strains will be equal and are of opposite sense to
the longitudinal strain.
POISSON’S RATIO
The bar will also exhibit, reduction in dimension laterally, i.e. its
breadth and depth will both reduce. These change in lateral dimension
is measured as strains in the lateral direction as given below.
d
d
b
b
lat
δδ
ε −=−=
Provided the load on the material is retained within the elastic range the
ratio of the lateral and longitudinal strains will always be constant. This
ratio is termed Poisson’s ratio (µ)
POISSON’S RATIO
Lateral strain
Longitudinal strain
=
l
l
d
d
δ
δ )(−
=
l
l
b
b
δ
δ )(−
OR
84

Poisson’s Ratio = µ
For most engineering metals the value of µ lies between 0.25 and 0.33
In general
x
x
y
y
l
l
l
l
δ
δ−
= OR
x
x
z
z
l
l
l
l
δ
δ−
=
z
y
x
PP
Lx
Ly
Lz
85
Poisson’s Ratio
Lateral strain
Strain in the direction of
load applied
=

Poisson’s Ratio = µ
In general
z
y
x
PxPx
Lx
LyLz
Strain in X-direction = εx
x
x
l
lδ
=
Strain in Y-direction = εy
x
x
y
y
l
l
l
l δ
µ
δ
==
Strain in Z-direction = εz
x
x
z
z
l
l
l
l δ
µ
δ
==
86

Poisson’s RatioLoad applied in Y-direction
Poisson’s Ratio
Lateral strain
load applied
=
y
y
x
x
l
l
l
l
δ
δ−
= OR
y
y
z
z
l
l
l
l
δ
δ−
=
z
y
x
Py
Lx
Ly
Lz
Py
y
y
x
x
l
l
l
l δ
µ
δ
==
87

Poisson’s RatioLoad applied in Z-direction
Poisson’s Ratio
Lateral strain
load applied
=
z
z
x
x
l
l
l
l
δ
δ−
= OR
z
z
y
y
l
l
l
l
δ
δ−
=
y
z
x
Pz
Lx
Ly
Lz
Pz
z
z
x
x
l
l
l
l δ
µ
δ
==
88

Load applied in X & Y direction
z
y
x
PxPx
Lx
LyLz
Py
Py
EE
xy σ
µ
σ
−=
EE
xy σ
µ
σ
µ −−=
EE
yx
σ
µ
σ
−=
89

General case:
z
y
xPxPx
Py
Py
Pz
Pz
90
EEE
zyx
x
σ
µ
σ
µ
σ
ε −−=
EEE
zxy
y
σ
µ
σ
µ
σ
ε −−=
EEE
xyz
z
σ
µ
σ
µ
σ
ε −−=
σx
σz
σy
σx
σz σy

Bulk Modulus
A body subjected to three mutually perpendicular equal direct stresses
undergoes volumetric change without distortion of shape.
If V is the original volume and dV is the change in volume, then
dV/V is called volumetric strain.
A body subjected to three mutually perpendicular equal direct stresses
then the ratio of stress to volumetric strain is called Bulk Modulus.
Bulk modulus, K






=
V
dV
σ
91

Relationship between volumetric strain and linear strain
Relative to the unstressed state, the change in
volume per unit volume is
( )( )( )[ ] [ ]
eunit volumperin volumechange
111111
1
=
++=
−+++=−+++=
zyx
zyxzyx
dV
εεε
εεεεεε
Consider a cube of side 1unit, subjected to
three mutually perpendicular direct stresses as
shown in the figure.
92

Relationship between volumetric strain and linear strain






−−=
EEE
zyx σ
µ
σ
µ
σ






−−+
EEE
zxy σ
µ
σ
µ
σ






−−+
EEE
xyz σ
µ
σ
µ
σ
zyx
V
dV
εεε ++=
Volumetric strain
93
( )zyx
E
σσσ
µ
++
−
=
21

For element subjected to uniform hydrostatic pressure,
( )
( )µ
µ
2-13KE
or
modulusbulk
213
=
=
−
=
E
K
σσσσ === zyx
( )
( )3
21
21
σ
µ
σσσ
µ
EV
dV
EV
dV
zyx
−
=
++
−
=






=
V
dV
K
σ
94

Relationship between young’s modulus of elasticity (E) and modulus
of rigidity (G) :-
τ
τ
A1
φφ
B1
A
D
B
a
a
45˚
C
H
95
Consider a square element ABCD of side ‘a’
subjected to pure shear ‘τ’. DA'B'C is the
deformed shape due to shear τ.
Drop a perpendicular AH to diagonal A'C.
σAC μσDB
E E
-Strain in the diagonal AC =
τ μ (- τ)
E E
= -
τ [ 1 + μ ]
E
= -------(1)
(A'C–AC) (A'C–CH) A'H
AC AC AC
Strain along diagonal AC = = =

E = 2G(1+ μ)
96
In Δle AA'H, Cos 45˚ = A'H/AA' => A'H= AA' × 1/√2
AC = √2 × AD ( AC = √ AD2
+AD2
)
Strain along the diagonal AC = =
φ
2
----(2)
A'H
AC
AA'
(√2 × √2 × AD)
=
Modulus of rigidity = G =
τ
φ
τ
G
=> φ =
Equating (1) & (3)
τ τ
2G E[1+μ]
=
Substituting in (2) Strain along the diagonal AC = -----------(3)τ

Substituting in (1)
E = 2G[ 1+(3K – 2G)/ (2G+6K)]
E = 18GK/( 2G+6K)
E = 9GK/(G+3K)
Relationship between E, G, and K:-
We have
E = 2G( 1+ μ) -----------(1)
E = 3K( 1- 2μ) -----------(2)
Equating (1) & (2)
2G( 1+ μ) =3K( 1- 2μ)
2G + 2Gμ=3K- 6Kμ
μ= (3K- 2G) /(2G +6K)
97

(1) A bar of certain material 50 mm square is subjected to an axial pull of
150KN. The extension over a length of 100mm is 0.05mm and decrease in
each side is 0.0065mm. Calculate (i) E (ii) μ (iii) G (iv) K
Solution:
(i) E = Stress/ Strain = (P/A)/ (dL/L) = (150×103
× 100)/(50 × 50 × 0.05)
E = 1.2 x 105
N/mm2
(ii) µ = Lateral strain/ Longitudinal strain = (0.0065/50)/(0.05/100) = 0.26
(iii) E = 2G(1+ μ)
G= E/(2 × (1+ μ)) = (1.2 × 105
)/ (2 × (1+ 0.26)) = 0.47 ×105
N/mm2
(iv) E = 3K(1-2μ)
K= E/(1-2μ) = (1.2 × 105
)/ (3 × (1- 2 × 0.26)) = 8.3 × 104
N/mm2
98

(2) A tension test is subjected on a mild steel tube of external diameter
18mm and internal diameter 12mm acted upon by an axial load of
2KN produces an extension of 3.36 x 10-3
mm on a length of 50mm
and a lateral contraction of 3.62 x 10-4
mm of outer diameter.
Determine E, μ,G and K.
(i) E = Stress/Strain = (2 ×103
× 50)/ (π/4(182
– 122
)× 3.36× 10-3
)
= 2.11× 105
N/mm2
ii) μ=lateral strain/longitudinal strain
= [(3.62 ×10-4
)/18]/[(3.36 × 10-3
)/50] = 0.3
iii) E = 2G (1 + μ)
G = E / 2(1+ μ) = (2.11 × 105
)/(2 × 1.3) = 81.15 × 103
N/mm2
iv) E = 3K(1 -2 μ) => K = E/ [3×(1-2 μ)]
5 3 2
99

Working stress: It is obvious that one cannot take risk of loading a
member to its ultimate strength, in practice. The maximum stress to
which the material of a member is subjected to in practice is called
working stress.
This value should be well within the elastic limit in elastic design
method.
Factor of safety: Because of uncertainty of loading conditions,
design procedure, production methods, etc., designers generally
introduce a factor of safety into their design, defined as follows
Factor of safety =
Allowable working
stress
Maximum stress
Allowable working
stress
Yield stress (or proof stress)
or
10
0

Homogeneous: A material which has a uniform structure throughout
without any flaws or discontinuities.
Malleability: A property closely related to ductility, which defines
a material’s ability to be hammered out into thin sheets.
10
1
Isotropic: If a material exhibits uniform properties throughout in all
directions it is said to be isotropic.
Anisotropic: If a material does not exhibits uniform properties
throughout in all directions it is said to be anisotropic or
nonisotropic.

Saint-Venant’s Principle
• Loads transmitted through
rigid plates result in uniform
distribution of stress and
strain.
• Concentrated loads result in
large stresses in the vicinity of
the load application point.
10
2

Saint-Venant’s Principle
• Saint-Venant’s Principle:
Stress distribution may be
assumed independent of the mode
of load application except in the
immediate vicinity of load
application points.
• Stress and strain distributions
become uniform at a relatively
short distance from the load
application points.
10
3
b

The normal stress at a particular point may
not be equal to the average stress but the
resultant of the stress distribution must
satisfy
∫∫ ===
A
ave dAdFAP σσ
10
4

A metallic bar 250mm×100mm×50mm is loaded as shown in the
figure. Find the change in each dimension and total volume. Take
E = 200GPa, Poisson's ratio, µ = 0.25
10
5
250
400kN50
100
2000kN
4000kN
4000kN
400kN
2000kN

Stresses in different
directions100
250
400kN50
2000kN
4000kN
100
250
50
MPa
mm
N
x 80
50100
1000400
2
=
×
×
=σ
MPa
mm
N
y 160
100250
10004000
2
=
×
×
=σ
MPa
mm
N
z 160
50250
10002000
2
=
×
×
=σ
10
6

Stresses in different direction
MPa80
MPa160
MPa160
EEE
zyx
x
σ
µ
σ
µ
σ
ε −−=
4
104
16016080 −
×=
+
−
−
−
+
=
EEE
x µµε
mml
l
l
l
x
x
x
x
1.0
104
250
4
=
×== −
δ
δδ
10
7

EEE
zxy
y
σ
µ
σ
µ
σ
ε −−=
( )3
101.1
16080160 −
×−=
+
−
+
−
−
=
EEE
y µµε
( )
mml
l
l
l
y
y
y
y
005.0
101.1
50
3
−=
×−== −
δ
δδ
10
8
MPa80
MPa160
MPa160

( )
mml
l
l
l
z
z
z
z
09.0
109
250
4
+=
×+== −
δ
δδ
EEE
xyz
z
σ
µ
σ
µ
σ
ε −−=
( )4
109
80160160 −
×+=
+
−
−
−
+
=
EEE
z µµε
10
9
MPa80
MPa160
MPa160

( )
( ) ( )
3
44
44
250
50100250102102
102109114
mmdV
VdV
V
dV
+=
××××=××=
×=×+−=
−−
−−
zyx
V
dV
εεε ++=
To find change in volume
( )
( )
( ) 4
10280
E
2-1
16016080
21
21
−
×==
+−+
−
=
++
−
=
µ
µ
σσσ
µ
EV
dV
EV
dV
zyx
Alternatively,
11
0
MPa80
MPa160
MPa160

A metallic bar 250mm×100mm×50mm is loaded as shown in the Fig.
shown below. Find the change in value that should be made in
4000kN load, in order that there should be no change in the volume
of the bar. Take E = 200GPa, Poisson's ratio, µ = 0.25
11
1
250
400kN
50
100
2000kN
4000kN

We know that
( )zyx
EV
dV
σσσ
υ
++
−
=
21
In order that change in volume to be zero
( )
( ) 0
21
0
=++
++
−
=
zyx
zyx
E
σσσ
σσσ
υ
( )
kNP
P
MPa
y
y
y
y
6000
100250
240
240
016080
−=
×
=−
−=
=+++
σ
σ
11
2
MPa80
MPa160
MPa160
The change in value should be an
addition of 2000kN compressive
force in Y-direction

Exercise Problems
1 An aluminum tube is rigidly fastened between a brass rod and
steel rod. Axial loads are applied as indicated in the figure.
Determine the stresses in each material and total deformation.
Take Ea=70GPa, Eb=100GPa, Es=200GPa
500mm 700mm600mm
steel
aluminum
brass
20kN 15kN 15kN 10kN
Ab=700mm2
Aa=1000mm2
As=800mm2
Ans: σb=28.57MPa, σa=5MPa, σs=12.5MPa, δl = - 0.142mm
11
3

Example 7
2. A 2.4m long steel bar has uniform diameter of 40mm for a
length of 1.2m and in the next 0.6m of its length its diameter
gradually reduces to ‘D’ mm and for remaining 0.6m of its
length diameter remains the same as shown in the figure. When
a load of 200kN is applied to this bar extension observed is
equal to 2.59mm. Determine the diameter ‘D’ of the bar. Take E
=200GPa
Ф = 40mm
Ф = D mm
200kN
200kN
500mm500mm1000mm
11
5

Exercise Problems
3 The diameter of a specimen is found to reduce by 0.004mm when
it is subjected to a tensile force of 19kN. The initial diameter of
the specimen was 20mm. Taking modulus of rigidity as 40GPa
determine the value of E and µ
Ans: E=110GPa, µ=0.36
4 A circular bar of brass is to be loaded by a shear load of 30kN.
Determine the necessary diameter of the bars (a) in single shear
(b) in double shear, if the shear stress in material must not exceed
50MPa.
Ans: 27.6, 19.5mm
11
6

Exercise Problems
5 Determine the largest weight W that can be supported by the two
wires shown. Stresses in wires AB and AC are not to exceed
100MPa and 150MPa respectively. The cross sectional areas of
the two wires are 400mm2
for AB and 200mm2
for AC.
Ans: 33.4kN
W
A
CB
300
450
11
7

Exercise Problems
6 A homogeneous rigid bar of weight 1500N carries a 2000N load
as shown. The bar is supported by a pin at B and a 10mm
diameter cable CD. Determine the stress in the cable
Ans: 87.53MPa
3m
A C
B
2000 N
3m
D
11
8

7. A stepped bar with three different cross-sectional areas, is
fixed at one end and loaded as shown in the figure. Determine
the stress and deformation in each portions. Also find the net
change in the length of the bar. Take E = 200GPa
250mm 270mm320mm
300mm2
450mm2
250mm2
10kN
40kN
20kN
Ans: -33.33, -120, 22.2MPa, -0.042, -0.192, 0.03mm, -0.204mm
11
9

8 The coupling shown in figure is constructed from steel of
rectangular cross-section and is designed to transmit a tensile
force of 50kN. If the bolt is of 15mm diameter calculate:
a) The shear stress in the bolt;
b) The direct stress in the plate;
c) The direct stress in the forked end of the coupling.
Ans: a)141.5MPa, b)166.7MPa, c)83.3MPa
12
0

Exercise Problems
9 The maximum safe compressive stress in a hardened steel punch
is limited to 1000MPa, and the punch is used to pierce circular
holes in mild steel plate 20mm thick. If the ultimate shearing
stress is 312.5MPa, calculate the smallest diameter of hole that
can be pierced.
Ans: 25mm
12
1
10. A rectangular bar of 250mm long is 75mm wide and 25mm
thick. It is loaded with an axial tensile load of 200kN, together
with a normal compressive force of 2000kN on face
75mm×250mm and a tensile force 400kN on face
25mm×250mm. Calculate the change in length, breadth,
thickness and volume. Take E = 200GPa & µ=0.3
Ans: 0.15,0.024,0.0197mm, 60mm3

Exercise Problems
11 A piece of 180mm long by 30mm square is in compression under
a load of 90kN as shown in the figure. If the modulus of elasticity
of the material is 120GPa and Poisson’s ratio is 0.25, find the
change in the length if all lateral strain is prevented by the
application of uniform lateral external pressure of suitable
intensity.
180
90kN30
30
Ans: 0.125mm
12
2

12. Define the terms: stress, strain, elastic limit, proportionality
limit, yield stress, ultimate stress, proof stress, true stress, factor
of safety, Young’s modulus, modulus of rigidity, bulk modulus,
Poisson's ratio,
13. Draw a typical stress-strain diagram for mild steel rod under
tension and mark the salient points.
14. Diameter of a bar of length ‘L’ varies from D1 at one end to D2
at the other end. Find the extension of the bar under the axial
load P
15. Derive the relationship between Young’s modulus and modulus
of rigidity.
12
3

17. A flat plate of thickness ‘t’ tapers uniformly from a width b1at
one end to b2 at the other end, in a length of L units. Determine
the extension of the plate due to a pull P.
18. Find the extension of a conical rod due to its own weight when
suspended vertically with its base at the top.
19. Prove that a material subjected to pure shear in two
perpendicular planes has a diagonal tension and compression of
same magnitude at 45o
to the planes of shear.
16. Derive the relationship between Young’s modulus and Bulk
modulus.
12
4

12
5
20 For a given materials E=1.1×105
N/mm2
&
G=0.43×105
N/mm2
.Find bulk modulus & lateral contraction
of round bar 40mm diameter & 2.5m length when stretched by
2.5mm. ANS: K=83.33Gpa, Lateral
contraction=0.011mm
21. The modulus of rigidity of a material is 0.8×105
N/mm2
, when
6mm×6mm bar of this material subjected to an axial pull of
3600N.It was found that the lateral dimension of the bar is
changed to 5.9991mm×5.9991mm. Find µ & E. ANS: µ=0.31,
E= 210Gpa.

Normal stress and strain

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Normal stress and strain