PART II
Mechanics of Deformable Bodies
COURSE CONTENT IN
BRIEF
6. Simple stresses and strains
7. Statically indeterminate problems and thermal stresses
8. Stresses on inclined planes
9. Stresses due to fluid pressure in thin cylinders
6. Simple stresses and strains
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
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.
Alternatively the subject may be called the mechanics of solids.
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.
STRESS
The internal force resisting the deformation per unit area is
called as stress or intensity of stress.
Stress = internal resisting force / resisting cross sectional
area
R

A
STRESS
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/(106mm2) = 1N/mm2
1 MPa = 1 N/mm2
gigapascal, 1GPa = 1×109 N/m2
= 1×103 MPa
= 1×103 N/mm2
AXIAL LOADING – NORMAL STRESS
P
Consider a uniform bar of cross
sectional area A, subjected to a
tensile force P.
P
R
Consider a section AB normal to
the direction of force P
Let R is the total resisting force
acting on the cross section AB.
B
A
STRESS
Then for equilibrium condition,
R
P
R=P
Then from the definition of stress,
normal stress = σ = R/A = P/A
P
Symbol:
σ = Normal Stress
AXIAL LOADING – NORMAL STRESS
STRESS
Direct or 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
• The resultant of the internal forces for
an axially loaded member is normal
to a section cut perpendicular to the
member axis.
• The force intensity on that section is
defined as the normal stress.
F
  lim
A0 A
P
 ave 
A
Illustrative Problems
Q 6.1
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.
4kN
Bronze
A= 120 mm2
300mm
Aluminum
A= 180 mm2
13kN
400mm
Steel
A= 160 mm2
2kN
7kN
500mm
To calculate the stresses, first determine the forces in
each section.
To find the Force in bronze section,
consider a section bb1 as shown in the figure
b
4kN
13kN
2kN
Bronze
b1
For equilibrium condition algebraic sum of forces on
LHS of the section must be equal to that of RHS
7kN
b
4kN
13kN
2kN
7kN
Bronze
b1
4kN
Bronze
4kN
(=
13kN
2kN
7kN
)
Force acting on Bronze section is 4kN, tensile
Stress in Bronze
Force in Bronze section
section =
Resisting cross sectional area of the Bronze section
4kN
4 1000 N
2


33
.
33
N
/
mm
= 33.33MPa
= 120mm2 120mm2
(Tensile stress)
Force in Aluminum section
4kN
13kN
2kN
7kN
Aluminum
4kN
13kN
9kN
(=
2kN
Aluminum
Force acting on Aluminum section is 9kN,
(Compressive)
7kN
)
Force in steel section
4kN
13kN
2kN
7kN
steel
7kN
4kN
13kN
2kN
steel
7kN
Force acting on Steel section is 7kN, ( Compressive)
Stress in Aluminum
section
=
Force in Al section
Resisting cross sectional area of the Al section
=
9kN
9 1000 N
2


50
N
/
mm
180mm2 180mm2
= 50MPa
Compressive stress
Stress in Steel section =
Force in Steel section
Resisting cross sectional area of the Steel section
7kN
7 1000 N
2


43
.
75
N
/
mm
= 43.75MPa
= 160mm2 160mm2
(Compressive stress)
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 changes 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,
L
Change in length
=
Linear strain,  
Original length
L
Strain is a dimensionless quantity.
Linear Strain
P
A
2 


2L L

P
   st ress
A
2P P


2A A



L
 normalst rain

L
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.
STRESS-STRAIN DIAGRAM
Proportionality limit
Typical tensile test curve for mild steel
STRESS-STRAIN DIAGRAM
Typical tensile test curve for mild steel showing upper yield point
and lower yield point and also the elastic range and plastic range
Stress-strain Diagram
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, that stress is proportional to strain.
Beyond this point, the stress is no longer proportional to strain
PP
P 
= Load at proportionality limit
Original cross sectional area
A
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.
PE Load at proportional limit
E 
= Original cross sectional area
A
Stress-strain Diagram
Yield point:
It is the point at which there is an appreciable elongation or
yielding of the material without any corresponding increase of
load.
PY
Y 
=
A
Load at yield point
Original cross sectional area
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.
PU
U 
=
A
Maximum load taken by the material
Original cross sectional area
Stress-strain Diagram
Rupture strength (Nominal Breaking stress):
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.
PB
B 
=
A
load at breaking (failure)
Original cross sectional area
True breaking stress =
load at breaking (failure)
Actual cross sectional
area
Stress-strain Diagram
After yield point the graph becomes much more shallow and
covers a much greater portion of the strain axis than the
elastic range.
The capacity of a material to allow these large plastic
deformations is a measure of ductility of the material
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
Stress-strain Diagram
A measure of ductility is obtained by measurements of the
percentage elongation or percentage reduction in area,
defined as,
increase in gauge length (up to fracture)
×100
=
original gauge length
Percentage elongation
Reduction in cross sectional area
of necked portion (at fracture)
Percentage reduction in =
area
original area
Cup and cone fracture for a Ductile
Material
×100
Stress-strain Diagram
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 diagram for a typical brittle material
HOOKE’S LAW
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
stress (σ) α strain (ε)
stress (σ)
constant
=
strain (ε)
Thomas Young introduced a constant of proportionality that
came to be known as Young’s modulus.
stress (σ)
E
=
strain (ε)
=
Young’s Modulus
or
Modulus of Elasticity
HOOKE’S LAW
Young’s Modulus is defined as the ratio of normal stress to
linear strain within the proportionality limit.
stress (σ)
E = strain (ε) =
P L PL


A L AL
The value of the Young’s modulus is a definite property of a
material
From the experiments, it is known that strain is always a very
small quantity, hence E must be large.
For Mild steel, E = 200GPa = 2×105MPa = 2×105N/mm2
Deformations Under Axial Loading
• From Hooke’s Law:
  E


E

P
AE
• From the definition of strain:


L
• Equating and solving for the
deformation,
PL

AE
• With variations in loading, crosssection or material properties,
Pi Li
 
i Ai Ei
Q.6.2
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
Solution:
a) Stress at elastic limit,
σE =
Load at elastic limit
Original c/s area
PE
120kN


 381.97 N
2  381.97 MPa
2
mm
A 314.16mm
(consider a load which is within the elastic
limit)
60kN
P
2

190.98
314
.
16
mm
A
N
E 



190980
0.20mm
mm2
 L
110 3
L
200mm
 190980 MPa
 190.98GPa
b) Young’s Modulus,
c) % elongation,
Final length at fracture – original length
% elongation =
Original length
50

100  25%
200
d) % reduction in area =

Original c/s area -Final c/s area at fracture
Original c/s area
2


16
314.16 
314.16
4 100  36%
e) Ultimate strength,
Maximum load
Ultimate strength =
Original c/s area
f) Nominal breaking
Breaking load
Strength = Original c/s area

180kN
2

572
.
96
N
/
mm
314.16mm2
( MPa)
160kN

 509.29MPa
314.16
g) True breaking
Breaking load
Strength =
c/s area at fracture
160kN

 795.38MPa
2
201.06mm
Q.6.3
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
4kN
Bronze
A= 120 mm2
300mm
Aluminum
A= 180 mm2
13kN
400mm
Steel
A= 160 mm2
2kN
7kN
500mm
From the Example 1, we know that,
Pbr = +4kN (Tension)
Deformation due to
compressive force is
shortening in length, and is
considered as -ve
Pal = -9kN (Compression)
Pst = -7kN (Compression)
PL
stress (σ)
= 
E =
strain (ε)
AL
Change in length
Change in length of
bronze
=
=
PL
L 
AE
4000 N  300mm
Lbr 
120mm2 100 103 ( N / mm2 )
= 0.1mm
Change in length of
aluminum section =
 9000 N  400mm
= -0.286mm
Lal 
2
3
2
180mm  70 10 ( N / mm )
Change in length of
steel section
=
 7000 N  500mm
= -0.109mm
Lst 
160mm2  200 103 ( N / mm2 )
Change in total
length
= Lbr  Lal  Lst  +0.1 – 0.286 - 0.109
= -0.295mm
Q.6.4
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.
Take Eal = 70GPa,
Est = 200GPa
a)σst ≤ 140 MPa
b)σal ≤ 80 MPa
c)Total elongation ≤ 1mm,
2P
Aluminum
0.8m
4P
2P
Steel
2.8m
To find P, based on the condition, σst ≤ 140 MPa
Stress in steel must be less than or equal to 140MPa.
Hence, σst =
= 140MPa 
Pst 2 P

 140 N / mm2
Ast Ast
140  Ast
P
 21000 N  21kN
2
2P
Aluminum
4P
2P
2P
Steel
4P
2P
2P 2P
Tensile
To find P, based on the condition, σal ≤ 80 MPa
Stress in aluminum must be less than or equal to
80MPa.
Hence, σal =
2P
Pal

 80 N / mm2

Aal
Aal
= 80MPa
80  Aal
P
 24000 N  24kN
2
2P
Aluminum
4P
Steel
2P
4P 2P
2P
2P Compressive
2P
To find P, based on the condition, total elongation ≤ 1mm
Total elongation = elongation in aluminum + elongation in
steel.
 PL   PL 
1mm

 

 AE  al  AE  st
1mm
  2 PLal    2 PLst 
  

 
 Aal Eal   Ast Est 
1mm
  2 P  800    2 P  2800 


3  
3 
 600  70 10   300  200 10 
P = 18.1kN
Ans: P = 18.1kN (minimum of the three values)
Q.6.5
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
W
W
B
A
Diameter
d1
L
Diameter
d2
Consider an element of length, δx at a distance x from A
W
W
A
d1
B
x
dx
Diameter at x,  d1

d 2  d1 

x
L
 d1  k  x
d2
c/s area at x,
2

d1





PL 
Wdx


Change in length over a     

 AE  dx  d  kx2  E 
length dx is
1
4





Change in length over a
L
Wdx

 
0
length L is

 d  kx2  E 


1
4

4


4
d1  kx2
Consider an element of length, δx at a distance x from A




L
Wdx

 
0
  d  kx2  E 


1
4

Change in length over a
length L is
dt 

W


L
k 
 
0
  t 2  E 


4

L
Put d1+kx = t,
Then k dx = dt

4W  t
4W   1
 4W



 


Ek   1  0 Ek  t  0 Ek
 2 1
4WL
WL


Ed1d 2 d1d 2  E
4
L
L
 1



(
d

kx
)
 1
0
Q.6.6
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.
150kN
150kN
1000mm
Ф = 40mm
500mm 500mm
Ф = 20mm
2
150kN
3
150kN
1
1000mm
Ф = 40mm
500mm 500mm
Ф = 20mm
If we take a section any where along the length of the bar, it is
subjected to a load of 150kN.
150kN
 119.37 MPa
2
 40
4
150kN
150kN
2 



 119.37 MPa
2 , max .
2
2
d
 40
4
4
150kN
 2,min. 
 477.46 MPa
2
 20
4
150kN
3 
 477.46 MPa
2
 20
4
1 
2
150kN
3
150kN
1
1000mm
Ф = 40mm
500mm 500mm
Ф = 20mm
If we take a section any where along the length of the bar, it is
subjected to a load of 150kN.
150kN 1000
l1 
 0.597 mm
2
 40  E
4
4 PL
4 150kN  500
l2 

 0.597 mm
Ed1d 2   E  40  20


150kN  500
l3 
 1.194mm
2
 20  E
4


total, l  2.388mm
Q.6.7
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.
W
W
d1
b
d2
B
A
L
b
W
d1
b
W
d2
B
A
x
b
dx
Consider an element of length, δx at a distance x from A
depth at x,
 d1

d 2  d1 

x
L
 d1  k  x
Change in length over a
length dx is
c/s area at x,  d1  kxb


Wdx
 PL 


  
 AE  dx  d1  kxb  E 
Change in length over a
length L is


Wdx

  
0
 d1  kxb  E 
L
P
log e d 2  log e d1 

b E k
2.302  P  L
log d 2  log d1 

b  E  d 2  d1 
Q.6.8
Derive an expression for the total extension produced by self
weight of a uniform bar, when the bar is suspended vertically.
L
Diameter
d
dx
x
Diameter
d
P1 = weight of the bar below
element
the section,
= volume × specific weight
dx
= (π d2/4)× x × 
P1
= A× x ×
Extension of
the element
due to weight
of the bar
below that,
P1dx ( A  x   )dx
 PL 




AE
AE
 AE  dx
Hence the total extension
entire bar

L
0
The above expression
can also be written as
L
( A  x   )dx  x 
L2
  
AE
 2E  0 2E
2
L2
A ( AL )  L 1 PL

 
 
2E A
2 AE
2 AE
Where, P = (AL)×
= total weight of the bar
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 failure as shown in Fig.(b)
P
P
R
R
P
Fig. a
Fig. b
P
Fig. c
The resisting force developed by any plane ( or section) of the
block will be parallel to the surface as shown in Fig.(c).
The resisting forces acting parallel to the surface per unit area is
called as shear stress.
Shear stress (τ)
Shear resistance
Area resisting shear
=

P
A
This shear stress will always be tangential to the area on which
it acts
Shear strain
If block ABCD 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.
τ
τ
B'
C'
C
C
B
B
A
τ
Fig. d
D
A
τ
Fig. e
D
τ
B'
B
C

A
Shear strain is defined as
C'
the change in angle
between two line element
which are originally right
angles to one another.
τ
D
Fig. e
BB
shear strain 
 tan   
AB
The angle of deformation

is then termed as shear strain
The angle of deformation is measured in radians and hence
is non-dimensional.
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.
Shear Modulus
Shear stress (τ)
or
=
G
=
constant
=
Shear strain (φ)
Modulus of Rigidity
The value of the modulus of rigidity is a definite property
of a material
For Mild steel, G= 80GPa = 80,000MPa = 80,000N/mm2
example: Shearing Stress
• Forces P and P‘ are applied
transversely to the member AB.
• Corresponding internal forces act in
the plane of section C and are
called shearing forces.
• The resultant of the internal shear
force distribution is defined as the
shear of the section and is equal to
the load P.
• The corresponding average shear stress is,
 ave 
P
A
• The shear stress distribution cannot be
assumed to be uniform.
State of simple shear
Consider an element ABCD in a strained material
subjected to shear stress, τ as shown in the figure
τ
A
B
D
C
τ
Force on the face AB = P = τ × AB × t
Where, t is the thickness of the
element.
Force on the face DC is also equal to
P
State of simple shear
Now consider the equilibrium of the element.
(i.e., ΣFx = 0, ΣFy = 0, ΣM = 0.)
For the force diagram shown,
P
A
B
D
C
ΣFx = 0, & ΣFy = 0,
But ΣM = 0
The element is subjected
force
to a clockwise moment
P × AD = (τ × AB × t) × AD
P
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 )
State of simple shear
P ' × AB = P × AD
= (τ × AB × t)× AD ----- (1)
A
If τ1 is the intensity of the shear
stress on the faces AD and BC,
then P ' can be written as,
Equn.(1) can be written as
A
P'
P
τ
τ'
C
B
τ'
D
(τ ' × AD× t ) × AB = (τ × AB × t) × AD ----- (1)
τ' =τ
B
P'
D
P ' = τ ' × AD × t
P
τ
C
State of simple shear
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”.
τ
B
A
The shear and the
complementary shear together
constitute a state of simple
shear
τ'= τ
D
τ'= τ
τ
C
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.
τ
τ
A
a
τ
τ
a
D
B
B
A
τ
Fig.(a).
C
a
τ
a
D
τ
τ
C
Fig.(b).
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.
Direct stress due to pure shear
Now consider the section, ADC of the element, Fig.(c).
X
σn
A
τ
A
a
a
a
a
D
 2 a
τ
D
C
C
Fig.(c).
Resolving the forces in σn direction, i.e., in the X-direction
shown
For equilibrium
 Fx  0
n 
n 


2  a 1  2  a  cos 45
Direct stress due to pure shear
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.
POISSON’S RATIO
Poisson’s Ratio:
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
l
l 
l
Fig.(a)
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.
 lat  
b
b

d
d
The associated lateral strains will be equal and are of
opposite sense to the longitudinal strain.
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)
( d )
d

POISSON’S RATIO =
l
Longitudinal strain
l
Lateral strain
( b )
b
OR l
l
Poisson’s Ratio = µ
For most engineering metals the value of µ lies between 0.25 and
0.33
In general
y
Lz
Ly
P
P
x
Lx
z
 l y
Poisson’s
Ratio
Lateral strain
=
Strain in the direction of
load applied

l x
ly
lx
 l z
OR  l
x
lz
lx
Poisson’s Ratio = µ
In general
y
Lz
Ly
Px
Px
x
Lx
z
Strain in X-direction = εx
Strain in Y-direction = εy
l x

lx
Strain in Z-direction = εz


l z
lz

l x
lx
l y
ly

l x
lx
Load applied in Y-direction
y
Py
Lz
Ly
x
Lx
z
Py
 l x
Poisson’s
Ratio
Lateral strain
=

Strain in the direction of
load applied
Strain in X-direction = εx

l x
lx

l y
lx
ly
l y
ly
 l z
OR  l
y
lz
ly
Load applied in Z-direction
y
Pz
Lz
Ly
x
Lx
Pz
z
Poisson’s
Ratio
 l x
Lateral strain
=

Strain in the direction of
load applied
Strain in X-direction = εx

l x
lx

l z
lz
l z
 l y
lx
lz
OR

l z
ly
lz
Load applied in X & Y direction
y
Py
Strain in X-direction = εx
Ly
Lz
Px
Px

x
Lx
z
Py
Strain in Y-direction = εy
Strain in Z-direction = εz

y
E

 
y
E
x
E

x
E
x
E

y
E
Py
General
case:
Pz
y
Strain in X-direction = εx
Px
Px
z
x
x 
Py
Pz
Strain in Y-direction = εy
y 
y
E

x
E
σy
z
E
Strain in Z-direction = εz
y
x
z
z   

E
E

x
E
E

y
E

σz
σx
σx
σz
σy
z
E
Bulk Modulus
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
 dV 


 V 
Relationship between volumetric strain and linear strain
Consider a cube of side 1unit, subjected to
three mutually perpendicular direct
stresses as shown in the figure.
Relative to the unstressed state, the change
in volume per unit volume is



dV
 1  1   x 1   y 1   z   1  1   x   y   z
1
 x y z
 change in volume per unit volum e

Relationship between volumetric strain and linear strain
Volumetric strain
dV
 x y z
V
y
x
z   y

 

   
    
  x   z     z   y    x 
E
E  E
E
E   E
E
E 
E
1  2
 x   y   z 

E
For element subjected to uniform hydrostatic pressure,
x y z 
dV 1  2 
 x   y   z 

V
E
dV 1  2 
3 

V
E
K
E
K
 bulk modulus
31  2 
or
E  3K 1 - 2 

 dV 


V


Relationship between young’s modulus of elasticity (E)
and modulus of rigidity (G) :A1 A
H 45˚
a
a
φτ
φ
D
B1 B
τ
C
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.
Strain in the diagonal AC = τ /E – μ (- τ /E)
[ σn = τ ]
= τ /E [ 1 + μ ] -----------(1)
Strain along the diagonal AC=(A'C–AC)/AC=(A'C–CH)/AC=A'H/AC
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 = AA'/ (√2 × √2 × AD)=φ/2 ----(2)
Modulus of rigidity = G = τ /φ
φ = τ /G
Substituting in (2)
Strain along the diagonal AC = τ /2G -----------(3)
Equating (1) & (3)
τ /2G = τ /E[1+μ]
E=2G(1+ μ)
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)
Substituting in (1)
E = 2G[ 1+(3K – 2G)/ (2G+6K)]
E = 18GK/( 2G+6K)
E = 9GK/(G+3K)
(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 105N/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 ×105N/mm2
(iv) E = 3K(1-2μ)
K= E/(1-2μ) = (1.2 × 105)/ (3 × (1- 2 × 0.26)) = 8.3 × 104N/mm2
(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 103mm on a length of 50mm and a lateral contraction of 3.62
x 10-4mm 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× 105N/mm2
ii) μ=lateral strain/longitudinal strain = [(3.62 ×10-4)/18]/[(3.36 × 10-3)/50]
iii) E = 2G (1 + μ)
= 0.3
G = E / 2(1+ μ) = (2.11 × 105)/(2 × 1.3) = 81.15 × 103N/mm2
iv) E = 3K(1 -2 μ)
K = E/ [3×(1-2 μ)] = (2.11×105)/{3×[1-(2 × 0.3)]} = 175.42 ×103N/mm2
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 = Maximum stress or Yield stress (or proof stress)
Allowable working
Allowable working
stress
stress
Malleability: A property closely related to ductility, which
defines a material’s ability to be hammered out in to thin
sheets
Homogeneous: A material which has a uniform structure
throughout without any flaws or discontinuities.
Isotropic: If a material exhibits uniform properties throughout
in all directions ,it is said to be isotropic.
Anisotropic: If a material does not exhibit uniform properties
throughout in all directions ,it is said to be anisotropic or
nonisotropic.
Q.6.9
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
2000kN
4000kN
400kN
400kN
50
100
2000kN
250
4000kN
4000kN
400kN
50
2000kN
250
Stresses in different
directions
100
y 
4000 1000 N
 160MPa
2
250 100mm
400 1000 N
x 
 80MPa
2
100  50mm
50
100
250
2000 1000 N
z 
 160MPa
2
250  50mm
Stresses in different direction
160MPa
80MPa
160MPa
x 
x
E

y
E

z
E
 80
 160
 160
x 


 4 10 4
E
E
E
l x
lx

l x
250
 4 10  4
l x  0.1mm
160MPa
80MPa
160MPa
y 
y
E

x
E

z
E
 160
 80
 160
3
y 


  1.110
E
E
E

l y
ly

l y
50

  1.110 3
l y  0.005mm


160MPa
80MPa
160MPa
z 
z
E

y
E

x
E
 160
 160
 80
z 


  9 10 4
E
E
E

l z
lz

l z
250

  9 10  4
l z  0.09mm


To find change in volume
dV
 x y z
V
160MPa
80MPa
dV
 4  11  910  4  2 10  4
V
dV  2 10  4  V  2 10 4  250 100  50
160MPa

Alternatively,

dV  250mm3
dV 1  2
 x   y   z 

V
E
dV 1  2
 80  160  160

V
E
1 - 2
80  2 104

E


Q.6.10
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
4000kN
400kN
50
100
2000kN
250
160MPa
We know that
80MPa
160MPa
dV 1  2
 x   y   z 

V
E
In order that change in volume to be
zero
1  2
 80  
y
 160  0
 y  240MPa
 240 
0

x
E

x
 y  z 
 y  z  0
Py
250 100
Py  6000kN
The change in value should be an
addition of 2000kN compressive force
in Y-direction
Exercise Problems
Q1.
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
20kN
Ab=700mm2
brass
Aa=1000mm2
15kN
15kN
As=800mm2
steel
10kN
aluminum
500mm
600mm
700mm
Ans: σb=28.57MPa, σa=5MPa, σs=12.5MPa, δl = - 0.142mm
Q2.
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
200kN
200kN
1000mm
500mm 500mm
Ф = 40mm
Ф = D mm
Q3. 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
Q4. 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
Q5. 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
C
B
300
450
A
W
Q6. 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
D
C
A
3m
2000 N
B
3m
Ans: 87.53MPa
Q7. 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
300mm2
20kN
250mm
450mm2
250mm2
320mm
Ans: -33.33, -120, 22.2MPa,
40kN
10kN
270mm
-0.042, -0.192, 0.03mm, -0.204mm
Q8.
a)
b)
c)
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:
The shear stress in the bolt;
The direct stress in the plate;
The direct stress in the forked end of the coupling.
Ans: a)141.5MPa, b)166.7MPa, c)83.3MPa
Q9. 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
Q10. 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
Q11. 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.
90kN
30
30
180
Ans: 0.125mm
Q12. 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,
Q13. Draw a typical stress-strain diagram for mild steel rod
under tension and mark the salient points.
Q14 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
Q15. Derive the relationship between Young’s modulus and
modulus of rigidity.
Q16 Derive the relationship between Young’s modulus and
Bulk modulus.
Q17 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.
Q18 Find the extension of a conical rod due to its own weight
when suspended vertically with its base at the top.
Q19. 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.
Q20.
For a given material E=1.1×105N/mm2&
G=0.43×105N/mm2 .Find bulk modulus & lateral
contraction of round bar of 40mm diameter & 2.5m
length when stretched by 2.5mm.
ANS:
K=83.33Gpa, Lateral contraction=0.011mm
Q21.
The modulus of rigidity of a material is 0.8×105N/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.