Stresses and strains (Part 1)

MECHANICS OF SOLIDS
STRESSES & STRAINS (PART I)
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AMIE(I) STUDY CIRCLE(REGD.)
A FOCUSSED APPROACH
Stresses And Strains (Part I)
UNITS
Force
1 kilo-Newton (kN) = 103
N
1 mega-Newton (MN) = 106
N
1 giga- Newton (GN) = 109
N
Stress
The stress is expressed in units of kN/m2
, MN/m2
, N/mm2
, Pascal(N/m2
).
Various relationships are
1 kgf = 9.8 N
1 tonne = 10 kN
1 kgf/cm2
= 105
N/m2
= 0.1 N/mm2
1 N/mm2
= 106
N/m2
= 106
Pa = 1 MPa (Mega Pascal)
1 GPa = 103
MPa
EQUILIBRIUM OF BODIES
Free Body Diagram (FBD)
The free body diagram of an element of a member in equilibrium is the diagram of only that
member of element, as if made free from the rest, with all the external forces acting on it.
In many problems, it is essential to isolate the body under consideration from the other bodies
in contact and draw all the forces acting on the body. For this, first the body is drawn and
then applied forces, self weight and the reactions at the points of contact with other bodies are
drawn. Such a diagram of the body in which the body under consideration is freed from all
the contact surfaces and shows all the forces acting on it (including reactions at contact
surfaces), is called a Free Body Diagram (FBD).
Following are a few FBD.

MECHANICS OF SOLIDS
A FOCUSSED APPROACH
Body in Equilibrium
A body is said to be in equilibrium when it is at rest or has uniform motion. According to
Newton’s law of motion, it means the resultant of all the forces acting on a body in
equilibrium is zero. The resultant of coplanar system of forces acting on a body is zero when
 The algebraic sum of the component of forces along each of the two mutually
perpendicular directions is zero (translatory motion is zero).  x = 0 and y = 0
 The algebraic sum of moment of all the forces about any point in the plane is zero
(rotational moment is zero). M = 0
Example
A system of forces are acting at the corners of a rectangular block as shown in following
figure.

MECHANICS OF SOLIDS
A FOCUSSED APPROACH
Solution
Given : System of forces
Magnitude of the resultant force
Resolving forces horizontally,
ΣH = 25 – 20 = 5 kN
and now resolving the forces vertically
ΣV = (–50) + (–35) = – 85 kN
∴ Magnitude of the resultant force
R = (ΣH)2
+ (ΣV )2
= (5)2
+ (–85)2
= 85.15 kN
Direction of the resultant force
Let θ = Angle which the resultant force makes with the horizontal.
We know that
85
tan 17
5
V
H

 
   

or  = 86.60
Since ΣH is positive and ΣV is negative, therefore resultant lies between 270° and 360°. Thus
actual angle of the resultant force
= 360° – 86.6° = 273.4°
Problem
The forces 20 N, 30 N, 40 N, 50 N and 60 N are acting at one of the angular points of a
regular hexagon, towards the other five angular points, taken in order. Find the magnitude
and direction of the resultant force.
Answer: 155.8 N;  = 76.60

MECHANICS OF SOLIDS
A FOCUSSED APPROACH
Lami's Theorem
If three forces acting at a point are in equilibrium, the ratio of any of the forces to the sine of
the angle between the remaining two forces is the same.
Let three forces F1,F2, and F3 acting at a point be in equilibrium as shown in fig. above.
Then, 31 2
sin sin sin
FF F
  
 
Example (AMIE W2007, 10 marks)
An Electric light fixture weighing 15N hangs from a point C, by two strings AC and BC. AC
is inclined at 600 to the horizontal and BC at 450 to the vertical as shown in fig (6.54),
Determine the forces in the strings AC and BC.
Solution
First draw the F.B.D. of the electric light fixture,
Apply Lami's theorem at point ‘C’
T1/sin 150° = T2/sin 135° = 15/sin75°
T1 = 15.sin150°/ sin75°
T1 = 7.76N

MECHANICS OF SOLIDS
A FOCUSSED APPROACH
T2 = 15.sin135°/ sin75°
T2 = 10.98N
Example
A uniform wheel of 600 mm diameter, weighing 5KN rests against a rigid rectangular block
of 150mm height as shown in figure. Find the least pull, through the center of the wheel,
required just to turn the wheel over the corner A of the block. Also find the reaction of the
block. Take the entire surface to be smooth.
Solution
Let P = least pull required just to turn the wheel
Least pull must be applied normal to AO. F.B.D of wheel is shown in fig., from the fig.,
sin = 150/300, = 30º
AB = {(300)2
– (150)2
}1/2
= 260 mm
Now taking moment about point A, considering body is in equilibrium
P X 300 –5 X 260 = 0
P = 4.33 KN
Calculation for reaction of the block
Let R = Reaction of the block
Since body is in equilibrium, resolving all the force in horizontal direction and equate to zero,
R cos 30º – P sin 30º = 0
R = 2.5KN

MECHANICS OF SOLIDS
A FOCUSSED APPROACH
Example
A uniform wheel of 60 cm diameter weighing 1000 N rests against rectangular obstacle 15
cm high. Find the least force required which when acting through center of the wheel will
just turn the wheel over the corner of the block. Find the angle of force with horizontal.
Solution
Let, Pmin = Least force applied as shown in fig.
= Angle of the least force
From triangle OBC,
BC = BOsin
BC = 30sin
In Triangle BOD, BD = {(BO)2
– (OD)2
}1/2
BD = (302
– 152
)1/2
= 25.98
Taking moment of all forces about point B, We get
Pmin X BC – W X BD = 0
Pmin – W X BD/BC
Pmin = 1000 X 25.98 /30sin
We get minimum value of P when is maximum and maximum value of is at 90º i.e. 1,
putting sin=1
Pmin = 866.02N
Example
Three sphere A, B, C are placed in a groove shown in figure. The diameter of each sphere
is 100mm. Sketch the free body diagram of B. Assume the weight of spheres A, B, C as 1KN,
2KN and 1KN respectively.

MECHANICS OF SOLIDS
A FOCUSSED APPROACH
Solution
For ,
cos= 50/100, cos= .5 , = 60°
FBD of block B is given in fig.
Example
Two cylindrical identical rollers A and B, each of weight W are supported by an inclined
plane and vertical wall as shown in fig.
Assuming all surfaces to be smooth, draw free body
diagrams of
(i) roller A,
(ii) roller B
(iii) Roller A and B taken together.
Solution
Let us assumed

MECHANICS OF SOLIDS
A FOCUSSED APPROACH
W = Weight of each roller
R = Radius of each roller
RA = Reaction at point A
RB = Reaction at point B
RC = Reaction at point C
RD = Reaction at point D
FBD of Roller ‘B’
FBD of Roller ‘A’
FBD of Roller ‘B’ & ‘A’ taken together
SIMPLE STRESSES & STRAINS
Axially Loaded Bar
The simplest case to consider at the start is that of an initially straight metal bar of constant
cross section, loaded at its ends by a pair of oppositely directed collinear forces coinciding

MECHANICS OF SOLIDS
A FOCUSSED APPROACH
with the longitudinal axis of the bar and acting through the centroid of each cross section. For
static equilibrium the magnitudes of the forces must be equal. If the forces are directed away
from the bar, the bar is said to be in tension; if they are directed toward the bar, a state of
compression exists. These two conditions are shown below.
Bar in tension Bar in compression
Normal Stress
The intensity of normal force per unit area is termed as normal stress and is expressed in units
of force per unit area, e.g. N/m2
(Pascal).If the forces applied to the ends of the bar are such
that the bar is in tension, then resultant tensile stresses are set up in the bar; if the bar is in
compression we have compressive stresses.
Normal Strain
Suppose a test piece is placed in tension. The elongation per unit length, which is termed as
normal strain and denoted by a, may be found by dividing the total elongation  by the gauge
length L, i.e.  = /L. This is dimensionless.
Stress Strain Curve
Normal Stress () = P/A
where P is axial load in Newton and A is cross sectional
area.
Having obtained various values of  and , the
experimental data may be plotted as below.
For alloy steel stress strain diagram is given below.
Hooke’s Law
For stress strain curve above, it is evident that the relation between stress and strain is linear
for comparatively small values of the strain. This linear relation between elongation and the
axial force causing it was first noticed by Sir Robert Hooke and is called Hooke’s law.
 = E, where E denotes the slope of the straight-line portion OP in the curve.
Generalised Hook’s law to relate strain-stress for two dimensional state of stress is
 x x y
1
v
E
    
 y y x
1
v
E
    

P


MECHANICS OF SOLIDS
A FOCUSSED APPROACH
where v is Poisson ratio.
For three dimensional state of stress, it is
  x x y z
1
v
E
        
  y y x z
1
v
E
        
  z z x y
1
v
E
        
Gauge Length
Gauge length is the failure length of the parallel portion of the test piece over which
extensions are measured. For cylindrical test pieces, the gauge length used for calculation of
percentage elongation is a function of the cross-sectional area of the diameter of the test bar.
Yield Point
Yield point is the point at which there is an appreciable elongation or yielding of material
without any corresponding increase of load.
Yield Strength
The yield strength, closely associated with the yield point, is defined as the lowest stress at
which extension of the test piece increases without increase in load.
Ultimate Strength
Ultimate stress or ultimate strength corresponds to the highest point of the stress strain curve.
Ultimate strength is maximum load/original cross sectional area.
Percentage Elongation
The percentage elongation is the percentage increase in length of gauge length. If L0 is the
original gauge length and Lf is the final gauge length (measured after fracture) then
Percentage elongation = f 0
0
L L
x100
L

Example
A steel bar is subjected to loads as shown in fig.. Determine the change in length of the bar
ABCD of 18 cm diameter. E = 180 kN/mm2
.

MECHANICS OF SOLIDS
A FOCUSSED APPROACH
Solution
Ref given figure.
Since d = 180mm; E = 180 × 103
N/mm2
LAB = 300mm
LBC = 310mm
LCD = 310mm
From following figure
Load on portion AB = PAB = 50 × 103
N
Load on portion BC = PBC = 20 × 103
N
Load on portion CD = PCD = 60 × 103
N
Area of portion AB = Area of portion BC
= Area of portion CD = A = d2
/4
= (180)2
/4 = 25446.9 mm2
Using the relation l = Pl/AE
Lab = (50 × 103
× 300)/( 25446.9 × 180 × 103
) = 0.0033mm -----Compression
Lbc = (20 × 103
× 310)/( 25446.9 × 180 × 103
Lcd = (60 × 103
× 310)/( 25446.9 × 180 × 103
Since net change in length = – Lab – Lbc – Lcd
= – 0.0033 – 0.0012 – 0.0041
= – 0.00856 mm
Decrease in length = 0.00856mm

MECHANICS OF SOLIDS
A FOCUSSED APPROACH
Example
For the bar shown in Fig., calculate the reaction produced by the lower support on the bar.
Take E = 200 GN/m2. Find also the stresses in the bars.
Solution
Let R1 = reaction at the upper support;
R2 = reaction at the lower support when the bar touches it.
If the bar MN finally rests on the lower support,
we have
R1 + R2 = 55kN = 55000
N For bar LM, the total force = R1 = 55000 – R2 (tensile)
For bar MN, the total force =R2 (compressive)
L1 =extension of LM = [(55000 – R2) × 1.2]/[(110 × 10 – 6) × 200 × 109]
L2 = contraction of MN = [R2 × 2.4]/[(220 × 10–6) × 200 × 109]
In order that N rests on the lower support, we have from compatibility equation
L1 – L2 = 1.2/1000 = 0.0012 m
Or, [(55000 – R2) × 1.2]/[(110 × 10–6) × 200 × 109] – [R2 × 2.4]/[(220 × 10–6) ×
200 × 109] = 0.0012
on solving;
R2 = 16500N or, 16.5 KN
R1 = 55-16.5 = 38.5 KN
Stress in LM = R1/A1 = 38.5/110 × 10–6 = 0.350 × 106 kN/m2
= 350 MN/m2
Stress in MN = R2/A2 = 16.5/220 × 10–6 = 0.075 × 106 kN/m2
= 75 MN/m2

MECHANICS OF SOLIDS
A FOCUSSED APPROACH
Example
A member ABCD is subjected to point loads P1, P2, P3 andP4 as shown in figure.
Calculate the force P3 ,necessary for equilibrium if P1 = 120 kN, P2 = 220 kN and P4 = 160
kN. Determine also the net change in length of the member. Take E = 200 GN/m2
.
Solution
Modulus of elasticity E = 200 GN/ m2
= 2 × 105 N/mm2
.
Considering equilibrium of forces along the axis of the member.
P1 + P3 = P2 + P4;
120 + P3 = 220 + 160
Force P3 = 220 + 160 – 120 = 260 kN
The forces acting on each segment of the member are shown in the free body diagrams shown
below:
Let L1, L2 and L3, be the extensions in the parts 1, 2 and 3 of the steel bar respectively.
Then,
Extension of segment AB
= [(120 × 103
) × (0.75 × 103
)]/[1600 × (2 × 105
)] = 0.28125 mm
Compression of segment BC
= [(100 ×103
) × (1 × 103
)]/[625 × (2 × 105
)] = 0.8 mm
Extension of segment CD
= [(160 ×103
) × (1.2 × 103
)]/[900 × (2 × 105
)] = 1.0667 mm
Net change in length of the member
= l =0.28125 – 0.8 + 1.0667 = 0.54795 mm (increase)

MECHANICS OF SOLIDS
A FOCUSSED APPROACH
Problem
Find the total elongation of a steel bar as shown in following figure subjected to an axial
load of 1000 kN. E = 200 GPa.
Answer : 2.196 mm
Example
The bar shown in figure is subjected to an axial pull of 150 kN. Determine diameter of the
middle portion if stress there is limited to 125N/mm2
. Proceed to determine the length of this
middle portion if total extension of the bar is specified as 0.15 mm. Take modulus of elasticity
of bar material E = 2 × 105
N/mm2
.
Solution
Each of the segment of this composite bar is subjected to axial pull P =150 kN.
Axial Stress in the middle portion 2 = Axial pull/Area = 150 × 103
/[(/4).(d2
2
)]
 d2 = 39.1 mm
Since stress is limited to 125 N/mm2
, in the middle portion
125 = 150 × 103
/[(/4).(d2
2
)]
Diameter of middle portion
d2 = 39.1mm
(ii) Stress in the end portions, 1 = 3
= 150 × 103
/[(/4).(502
)] = 76.43 N/m2
Total change in length of the bar,
= change in length of end portions + change in length of mid portion
L = L1 + L2 + L3
= L1/E + L2/E + L3/E; Since E is same for all portions
= 1(L1 + L3)/E + L2/E
L1 + L3 = 300 – L2
Now putting all the values;

MECHANICS OF SOLIDS
A FOCUSSED APPROACH
0.15 = [76.43(300 – L2)]/2 × 105
+ 125L2/2 × 105
 L2 = 145.58 mm
Example (AMIE Winter 2012, 8 marks)
A bar of uniform cross sectional area “A” and length “L” hangs vertically from a rigid
support of the density of a material as  kg/m3
. Derive the expression for maximum stress
induced and elongation produced in the bar due to its own weight.
Solution
See following figure
Consider a small section of length x, at a distance x from the free end. The deformation
is given by
.
sW x
A E

 
where Ws = weight of portion below the section = A.x.

( )
.
Ax x x x
A E E
  
  
 Total deformation of the rod
2
0 2
L x gL
dx
E E
 
  
Maximum stress induced in the strip
Ax
x
A

  

MECHANICS OF SOLIDS
A FOCUSSED APPROACH
Example (AMIE S2006, 2012, Winter 2015, 10 marks)
Find the change in length of circular bar of uniform taper.
Solution
The stress at any cross section can be found by dividing the load by the area of cross section
and extension can be found by integrating extensions of a small length over whole of the
length of the bar. We shall consider the following cases of variable cross section:
Consider a circular bar that tapers uniformly from diameter d1 at the bigger end to diameter
d2 at the smaller end, and subjected to axial tensile load P as shown in figure.
Let us consider a small strip of length dx at a distance x from the bigger end.
Diameter of the elementary strip:
1 1 2[( ) ]/dx d d d x L  
= 1d kx ; where 1 2( ) /k d d L 
Cross sectional area of strip
2 2
1( )
4 4
s xA d d kx
 
  
Stress in the strip
2
2 1
1
4
( )( )
4
x
x
P P P
A d kxc kx

 
  

Strain in strip
2
1
4
( )
x
x
P
E d kx E



 

Elongation of the strip
2
1
4
( )
x x
Pdx
l dx
d kx E
 

 

The total elongation of this tapering bar can be worked out by integrating the above
expression between limits x = 0 to x = L.

MECHANICS OF SOLIDS
A FOCUSSED APPROACH
2 20 0
1 1
4 4
( ) ( )
L LPdx P dx
l
d kx E E d kx

 
 
  
=
1
1
10 0
( )4 4 1
( 1) ( )
LL
d kxP P
E x k EK d kx 

  
   
     
Putting the value of 1 2( ) /k d d l  in the above expression, we obtain
1 21 2 1
1
4 1 1
( )( )
PL
l
d d lE d d dd
l


 
 
    
 
=
1 2 1 1
4 1 1
( )
PL
E d d d d
 
 
  
=
1 2
4PL
Ed d
Example
A steel rod, circular in cross section, tapers from 2.5 cm diameter to 1.25 cm diameter in a
length of 50 cm. Find how much of this length will increase under a pull of 25 kN if E = 210
GPa.
Solution
Given: d1 = 1.25 cm, L = 50 cm, d2 = 2.5 cm, and P = 25 kN
Extension of a tapering circular bar
 =
1 2
4PL
Ed d
=
3
-4 9
4 x 25 x x 0.510
x 1.25 x 2.5x x 210 x10 10
= 0.2425 mm.
Example (AMIE Summer 2000, 2009, Winter 2006, 2012, 10 marks)
A tension bar is found to taper uniformly from (D - a) cm diameter to (D + a) cm. Prove that
the error involved in using the mean diameter to calculate Young’s Modulus is
2
10a
D
 
 
 
per
cent.
Solution
Given: d1 = D-a, d2 = D + a
Let L = Length of the bar and P = Load applied.
Mean diameter
1 2+d d
D
2


MECHANICS OF SOLIDS
A FOCUSSED APPROACH
Mean stress σ =
4P
D
If u = Extension of the bar
then strain ε = u/L
Young modulus
E = 2
4Pl
4 uD
Now for a tapering round bar
u =
1 2
4PL
Ed d
E =
1 2
4PL
ud d
 E =
4PL
(D - a)(D+a)u
= 22
4PL
( - )uaD
 Error in Young’s Modulus will be
4PL
u 22 2
1 1
-
- aD D
 
 
 
=
4PL
u
2
22 2
a
( - aD D
 
 
 
 Percentage error will be
4PL
u
2
22 2
a
( - )aD D
 
 
 
x
22
u( - )aD
4PL

x 100 =
2
2
a
D
x 100 =
2
10a
D
 
 
 
Hence Proved.
Example (AMIE S16, 8 marks)
A bar of steel of length "l" and is of uniform thickness ‘"t" The width of the bar varies
uniformly from a at one end to b (a > b) at the other end. Find the extension of the rod when
it carries an axial pull p.
Solution

MECHANICS OF SOLIDS
A FOCUSSED APPROACH
Figure
Consider any section X-X distance x from the bigger end.
Area of section
Width of section
=
a b
a x
L


=
2
a b
a kx wherek

 
Thickness of section = t
Area of section = t(a – kx)
Stress
Stress on the section
( )
P
x
t a kx
 

Extension
Extension of an elemental length
( )
Pdx
x
t a kx E
 

Total extension of the rod
0
[log ( ) log
( )
L
e e
P x P
L a kL a
tE a kx tEk

     

= log
P a
tEk a kL
 
  

MECHANICS OF SOLIDS
A FOCUSSED APPROACH
A steel bar AB of uniform thickness 2 cm, tapers uniformly from 100 mm to 50 mm in a length
of 400 mm. From first principles determine the elongation of plate; if an, axial tensile force
of 50 kN is applied on it. [E = 2 x 105
N/mm2
]
Solution
Consider a small element of length dx of the plate, at a distance x from the larger end. Then
at this section,
Width Wx = 100 (100 50)( / 400) 100
8
x
x mm
 
    
 
Cross section area Ax = thickness x width = 2
10 100
8
x
mm
 
 
 
Stress
3
250 10
/
10(100 / 8)
s
x
P x
N mm
A x
   

Strain
3
5
5 10 1
(100 / 8)(2 10 ) 40(100 / 8)
x
x
x
E x x x

   
 
 Elongation of the elementary length
40(100 / 8)
x x
dx
L xdx
x
  

The total change in length of the plate can be worked out by integrating the above identity
between the limits x = 0 and x = 400 mm.
i.e.
400 400
0 0
1
0.1386
40(100 / 8) 40 (100 / 8)
dx dx
L
x x
   
   mm
Problem
A flat steel plate is of trapezoidal form. The thickness of the plate is 15 mm and it tapers
uniformly from a width of 60 mm to 10 mm in a length of 30 cm. If an axial force of 100 kN
is applied at each end, determine the elongation of the plate. Take E = 204 kN/mm2
.
Answer : 0.351 mm

MECHANICS OF SOLIDS
A FOCUSSED APPROACH
Problem
A steel bar AB of uniform thickness 15 mm, tapers uniformly from 60 mm to 10 mm in a
length of 300 mm. From first principles determine the elongation of plate; if an, axial tensile
force of 120 kN is applied on it. [E = 2.04 x 105
N/mm2
]
Answer: = 0.422 mm
Example
A metal bar 5 cm x 5 cm section is subjected to an axial compressive load of 500 kN. The
contraction on a 20 cm gauge length is found to be 0.5 mm and the increase in thickness
0.045 mm. Find the value of Young’s modulus and Poisson’s ratio.
Solution
Given P = 500 kN, l=20 cm, Δl = 0.05 cm, Δt = 0.0045 cm.
Area of cross section A = 5 x 5 = 25 cm2
Longitudinal strain
ε = Δl/l = 0.05/20 = 0.0025 ( compressive)
stress σ = P/A = 500 x 103
/25 x 10-4
= 200 MPa (compressive)
Young’s Modulus
E = σ/ε = 200 x 106
/0.0025 = 80 GPa
Lateral strain
Δt/t = 0.0045/5 = 0.0009 (tensile)
Poisson’s Ratio
v = Lateral strain/Longitudinal = 0.0009/0.0025 = 0.36
Example
A 70 cm length of aluminium alloy bar is suspended from the ceiling so as to provide a
clearance of 0.03 cm between it and a 25 cm length of steel as shown in following figure.
Aal = 12.5 cm2
, Eal = 70 GN/m2
; As = 25 cm2
Es = 210 GN/m2
Determine the stress in the aluminium and in the steel due to a 300 kN load applied 50 cm
from the ceiling.
Solution
Elongation of AB under 300 kN load will be
3 -2
-4 9
300 x x 50 x10 10
12.5 x x 70 x10 10
= 1.71428 x 10-3
m = 0.171428 cm.

MECHANICS OF SOLIDS
A FOCUSSED APPROACH
This elongation is more than the clearance 0.03 cm between the bars. Hence bars DE and BC
will be subjected to compression whereas AB will remain in tension. Let P be compressive
force is BC and DE then the tensile force in AB will be (300-P) kN. Then,
3 -2
-4 9
(300 - P) x x 50 x10 10
12.5 x x 70 x10 10
-
3 -2
-4 9
P x x 20 x10 10
12.5 x x 70 x10 10
-
3 -2
-4 9
P x x 25 x10 10
25 x x 210 x10 10
= 0.03 x 10-2
 P = 166.854 kN
σAB =
3
-4
(300 - 166.854) x 10
12.5 x 10
= 106.52 MN/m2
(tensile)
σBC =
3
-4
166.854 x 10
12.5 x 10
= 133.48 MN/m2
(compressive)
σDE =
3
-4
166.854 x 10
25 x 10
= 66.74 Mn/m2
(compressive)
Example
A rigid bar AB is hinged at A and supported by a bronze rod 2 m long and a steel rod 1 m
long. A load of 80 tonnes is applied at the end B as shown in following figure. Using the data
in the following table, calculate the stress in each rod and reaction at A.
Area(mm2
) E(N/mm2
)
Steel 900 2.05 x 105
Bronze 600 0.82 x 105
Solution
There are three unknowns, i.e. reaction RA at A, force P1 in the steel rod and force P2 in the
bronze rod. To solve the problem, the two equations of static equilibrium(i.e. the force

MECHANICS OF SOLIDS
A FOCUSSED APPROACH
equation and moment equation) must be used in combination with the additional equations
obtained from the geometric relations between the elastic deformations.
Let suffix s stand for the steel rod and b for the bronze rod.
Taking moments about A, we get
 MA = 0 = (P1 x 1) + (P2 x 3) - (80000 x 4)
or P1 +3P2 = 320000 (1)
If Δ1 is the deformation of the steel rod and Δ2 that of bronze rod we have from similar
triangles,
s
1
= b
3
 1 1
1 1
P L
A E
=
1
3
2 2
2 2
P L
A E
or 1
5
x 1000P
900 x 2.05 x 10
=
1
3
2
5
x 2000P
600 x 0.82 x 10
from which P1 = 2.5 P2 (2)
Substituting the value of P1 in (1), we get
2.5P2 + 3P2 = 320000
From which, P2 = 58182 N
P1 = 145455 N
For finding the reactions at A, use the force equation(i.e.  P = 0). Assuming RA to act
downward,
P1 + P2 - RA - 80000 = 0
 RA = 145455 + 58182 - 80000 = 123637 N Answer
(The positive sign to RA shows that the assumed direction of RA is correct.)

MECHANICS OF SOLIDS
A FOCUSSED APPROACH
Example
A square bar of 25 mm side is held between two rigid plates and loaded by an axial pull
equal to 300 kN as shown in figure. Determine the reactions at end A and C and elongation
of the portion AB. Take E = 2 × 105
N/mm2
.
Solution
Cross section area of the bar A = 25 × 25 mm2
Since the bar is held between rigid support at the ends, the following observations need to be
made:
 Portion AB will be subjected to tension and portion BC will be under compression
 Since each ends are fixed and rigid and therefore total Elongation;
Lab – Lbc = 0; Elongation in portion AB equals shortening in portion BC. i.e.,
Lab = Lbc
 Sum of reactions equals the applied axial pull i.e., P = Ra + Rc
Apply second condition, we get
[Pab × Lab]/Aab.E = [Pbc × Lbc]/Abc.E
(Ra × 400)/(625 × 2 × 105
) = (Rc × 250)/(625 × 2 × 105
)
Rc = 1.6Ra (i)
Now apply third condition i.e., P = Ra + Rc
300 × 103
= Ra + 1.6 Ra
Ra = 1.154 × 105
N; Rc = 1.846 × 105
N
ELASTIC CONSTANTS
Modulus of Elasticity (E)
The quantity E, i.e. ratio of the unit stress to the unit strain, is the modulus of elasticity of the
material in tension, or, as it is often called Young’s modulus.
Bulk modulus (K)
It is defined as the ratio of uniform stress intensity to volumetric strain, within the elastic
limits and is denoted by K.

MECHANICS OF SOLIDS
A FOCUSSED APPROACH
Poisson’s Ratio
When a bar is subject to a simple tensile loading there is an increase in length of the bar in the
direction of the load, but a decrease in the lateral dimensions perpendicular to the load. The
ratio of the strain in the lateral direction to that in the axial direction is defined as Poisson’s
ratio. It is generally denoted by  or 1/m. Its values lies between 0.25 and 0.35.
Modulus of rigidity (G)1
It is defined as the ratio of shearing stress to shearing strain.
Volumetric Strain
When a member is subjected to stresses, it undergoes deformation in all directions. Hence,
there will be change in volume. The ratio of the change in volume to original volume is called
volumetric strain.
Thus v
V
e
V


where ev = Volumetric strain
δV = Change in volume
V = Original volume
It can be shown that volumetric strain is sum of strains in three mutually perpendicular
directions.
i.e., ev = ex + ey + ez
Let us consider a circular rod of length L and diameter ‘d’ as shown in figure.
Now 2
4
V d L


2
2
4 4
V d dL d L
 
   

2
2
4
V d L
d Ld L
  

 
v x z
d
e e e
d

  
1
In some books it is represented by N.

MECHANICS OF SOLIDS
A FOCUSSED APPROACH
In general for any shape volumetric strain may be taken as sum of strains in three mutually
perpendicular directions.
Relation between E and K
2
E 3K 1
m
 
  
 
Proof: Let a cube of side L be subjected to three mutually perpendicular like stresses of equal
intensity q. By the definition of bulk density
v
p
K
e

or v
V p
e
V K

  (i)
The total linear strain of each side
p p p
e
E mE mE
  

L p 2
e 1
L E m
  
   
 
(ii)
Now V = L3
 2
V 3L L  
or v
V 3 L 3p 2
e 3e 1
V L E m
   
     
 
(iii)
Equating (i) and (iii)
p 3p 2
1
K E m
 
  
 
or
2
E 3K 1
m
 
  
 
Relation between E and G
9KG
E
G 3K


Proof. Let us consider a square block ABCD, subjected to a state of simple shear q. There
will be some distortion.
Linear strain e of the diagonal AC = q/2G. (i)
Now, linear strain e of the diagonal AC, due to these two mutually perpendicular direct
stresses, is given by

MECHANICS OF SOLIDS
A FOCUSSED APPROACH
p p p 1
e 1
E mE E m
   
       
   
(ii)
Equating (i) and (ii)
q q 1
1
2G E m
 
  
 
or
1
E 2G 1
m
 
  
 
which gives relation between E and G.
We have already proved that
2
E 3K 1
m
 
  
 
Equating the two, we get
1 2
E 2G 1 3K 1
m m
   
         
Eliminating E, we get
1 3K 2N
m 6K 2N



Eliminating m, we get
9KN
E
N 3K


There is a 2% error in the determination of G. If E is assumed to be correctly determined,
what will be the error in the calculation of Poisson’s ratio when its correct value is 0.2?
Solution
We know 2 (1 )E G m 
Applying the principle of differential calculus, we can differentiate above equation keeping in
mind that E = 0, because E is constant.
 0 2 (1 ) 2G m G v   
or (1 )
G
v m
G

   
% error will be
(1 ) 2 1 0.2
100 100 100 12%
100 0.2
m G m
x x x
m m
    
      
 

MECHANICS OF SOLIDS
A FOCUSSED APPROACH
Example
A steel bar of rectangular cross section 20 mm x 10 mm is subjected to a pull of 20 kN in the
direction of its length. Taking E = 0.2 x 106
N/mm2
and m = 10/3, find the length of the sides
of the cross section and percentage decrease of area of cross section.
Solution
The strain in the direction of pull is
e1 =
p
m
=
P
AE
= 6
20000
20 x 10 x 0.2 x 10
= 5 x 10-4
Lateral strain = - 1e
m
= -
3
10
x 5 x 10-4
Hence 20 mm side is decreased by 20 x 1.5 x 10-4
= 0.0030 mm
10 mm side is decreased by 10 x 1.5 x 10-4
= 0.0015 mm
New area of cross section = (20 - 0.003)(10 - 0.0015) = ( 200 - 0.06)
 % decrease of area of cross section =
0.06
200
x 100 = 0.03 Answer
Example
A piece of steel 200 mm long and 20 mm x 20 mm cross section is subjected to a tensile force
of 40 kN in the direction of its length. Calculate the change in volume. Take 1/m = 0.3. E =
2.05 x 105
N/mm2
.
Solution
e1 = 5
40000
(20 x 20)(2.05 x )10
= 4.88 x 10-4
e2 = e3 = - 1e
m
= - 4.88 x 0.3 x 10-4
= -1.464 x 10-4
δV/V = e1 + e2 + e3 = [4.88 - (1.464 x 2)]10-4
= 1.952 x 10-4
V = 200 x 20 x 20 = 80000 mm3
 δV = 1.952 x 80000 x 10-4
= 15.62 mm3
Answer
Problem
A piece of steel 1 metre long and 3 cm x 3 cm cross section is subjected to a tensile force of
120 kN in the direction of its length. Calculate the change in volume. Take 1/m = 0.29 and E
= 205 kN/mm2
.
Answer: 246 mm3

MECHANICS OF SOLIDS
A FOCUSSED APPROACH
Problem
A metal specimen is compressed in the direction of its axis and means are employed to reduce
lateral expansion to one third of what it would be if free to expand. Calculate the modified
value of elastic constant. Prove that its value will be (9/8)E if m = 4.
Answer: E'  2
3m(m-1)
3 - 3m -4m
Example
A bar 30 mm in diameter was subjected to tensile load of 54 kN and the measured extension
on 300 mm gauge length was 0.112 mm and change in diameter was 0.00366 mm. Calculate
Poisson’s ratio and the values of three moduli.
Solution
Stress = 76.4 N/mm2
Linear strain = 0.112/300 = 3.73 x 10-4
E = stress/strain = 204.6 kN/mm2
Lateral strain = δd/d = 0.00366/30 = 1.22 x 10-4
But lateral strain = (1/m)e = (1/m) (3.73 x 10-4
)
1
m
(3.73 x 10-4
) = 1.22 x 10-4
 1/m = 0.326.
Again G =
E
1
2 1 +
m
 
 
 
=
204.6
2(1 + 0.326)
= 77.2 kN/mm2
K =
E
2
3 1 -
m
 
 
 
= 196 kN/mm2
where K = bulk modulus. G = Modulus of rigidity. Answer
Example
A 2m long rectangular bar of 7.5 cm × 5 cm is subjected to an axial tensile load of 1000kN.
Bar gets elongated by 2mm in length and decreases in width by 10 × 10–6 m. Determine the
modulus of elasticity E and Poisson's ratio of the material of bar.

MECHANICS OF SOLIDS
A FOCUSSED APPROACH
Solution
Given:
L = 2m;
B = 7.5cm = 0.075m;
D = 5cm = 0.05m
P = 1000kN
L = 2mm = 0.002m
b = 10 × 10-6
m.
Longitudinal strain
eL = et = L/L = 0.002/2 = 0.001
Lateral strain
b/b = 10 × 10-6
/0.075 = 0.000133
Tensile stress (along the length)
t = P/A = (1000 × 1000)/(0.075 × 0.05) = 0.267 × 109
N/m2
Modulus of elasticity,
E = t /et = 0.267 × 109
/0.001 = 267 × 109
N/m2
Poisson’s ratio = Lateral strain/ Longitudinal strain
= (b/b)/(L/L) = 0.000133/0.001
Problem
A bar of elastic material is subjected to direct stress in a longitudinal direction, and its strain
in the two directions at right angles are reduced to one half and one third respectively to
those which normally occur in an ordinary tension member. If E = 200 kN/mm2
and m = 4,
what is the value of the elastic constant?
Answer: 221 kN/mm2
Example (AMIE Summer 2015, 10 marks)
A 500 mm long bar has rectangular cross-section 20 mm x 40 mm. 1 lie bar is subjected to:
(i) 40 kN tensile force on 20 mm x 40 mm faces
(ii) 200 kN compressive force on 20 mm x 500 mm faces
(iii) 300 kN tensile force on 40 mm x 500 mm faces
Find the changes in dimensions and volume, if E = 2 x 105 N/mm2
and Poisson's ratio m =
0.3.

MECHANICS OF SOLIDS
A FOCUSSED APPROACH
Solution
Figure
The dimensions of the bar and forces are shown in given figure.
Stresses
The stresses on various planes are:
3
240 10
50 /
20 40
x
x
N mm
x
  
3
2300 10
15 /
500 40
y
x
N mm
x
  
3
2200 10
20 /
20 500
z
x
N m
x


  
Strains
  4
5
1 1
( ) 50 0.3(15 20) 2.575 10
2 10
x x y ze m x
E x
   
        
  5
5
1 1
( ) 15 0.3(50 20) 3 10
2 10
y y x ze m x
E x
   
        
  4
5
1 1
( ) 20 0.3(50 15) 1.975 10
2 10
x z x ye m x
E x
   
          
5
9 10x y ze e e e x 
   
Changes in dimension
x = 500 x ex = 500 x 2.575 x 10-4
= 0.1288 mm
y = 20 x ey = 20 x 3 x 10-5
= 6 x 10-4
mm
z = 40 x ez = -40 x 1.975 x 10-4
= -7.9 x 10-3

MECHANICS OF SOLIDS
A FOCUSSED APPROACH
Change in volume
V = Ve = 4 x 105
x 9 x 10-5
= 36 mm3
COMPOSITE SECTION
When a bar consists of two different
materials, it is said to be composite.
Since there ate two unknowns, two
equations will be required. The
conditions of equilibrium will provide
one equation for the stresses in the individual sections. The other equation can be obtained
from consideration of the deformation of the whole structure.
Let us take the case of solid bar enclosed in the hollow tube and subjected to a compressive
force P through rigid collars as shown in figure. Using suffix 1 for the bar and 2 for the tube,
we get from conditions of equilibrium
1 2P P P  (1)
Since the whole assembly in composite, the deformation of bar is equal to the deformation of
the tube. Thus
1 2  
or 1 2
1 1 2 2
P L P L
A E A E
 (2)
Solving
1
2 2
1 1
P
P
A E
1
A E


and 2
1 1
2 2
P
P
A E
1
A E


Example
A copper rod 25 mm in diameter is encased in steel tube 30 mm internal diameter and 35 mm
external diameter. The ends are rigidly attached. The composite bar is 500 mm long and is
subjected to an axial pull of 30 kN. Find the stresses induced in the rod and the tube. Take E
for steel = 2 x 105
N/mm2
and E for copper as 1 x 105
N/mm2
.

MECHANICS OF SOLIDS
A FOCUSSED APPROACH
Solution
Let us use suffix 1 for rod (copper) and 2 for tube (steel)
2 2
1A (25) 490.9mm
4

 
2 2 2
2A (35) (30) 255.3mm
4

    
Now 1 1 2 2p A p A P 30000N   (1)
also 1 2
1 2
p L p L
E E

or
5
1
1 2 2 25
2
E 2x10
p p p 2p
E 1x10
   (2)
From (1) 2 22p (490.9) p (255.3) 30000 
Solving p2 = 24.25 N/mm2
and p1 = 2p2 = 48.5 N/mm2
Problem
A copper rod 25 mm in diameter is encased in steel tube 30 mm internal diameter and 35 mm
external diameter. The ends are rigidly attached. The composite bar is 400 mm long and is
subjected to an axial pull of 40 kN. Find the stresses induced in the rod and the tube. Take E
for steel = 2 x 105
N/mm2
and E for copper as 1 x 105
N/mm2
.
Answer: steel: 64.68 N/mm2
; copper: 32.34 N/mm2
Example
Two steel plugs fit freely into the ends of steel tubular distance piece 400 mm long and are
drawn together by a steel bolt (500 mm long) and nut, the nut being tight fit in the beginning.
The nut is further tightened by 1/4 turn to draw the pieces together, the pitch of the bolt
thread being 2 mm. The pieces A and B are then subjected to forces of 50 kN tending to pull

MECHANICS OF SOLIDS
A FOCUSSED APPROACH
them apart. Calculate the stresses in the bolt and the tube. The area of cross section of the
bolt is 700 sq. mm and that of the tube is 500 sq. mm. E for steel = 200 GN/m2
.
Solution
Let p1 (tensile) and p2 (compressive) be the final stresses in the bolt and tube respectively.
For equilibrium,
(tensile force in 1) - (compressive force in 2) = external tensile force
 1 1 2 2p A p A 50000 
or 1 2700p 500p 50000 
or 1 2p 0.714p 71.43  (1)
From compatibility,
(total extension of bolt) + (total compression of tube) = movement of nut
or 1 2 movement of nut   
 1 1 2 2
1 2
p l p l 1
x2
E E 4
 
or 1 25 5
400 500
p p 0.5
2.05x10 2.05x10
 
or
5
1 2
2 0.5x2.05x10
p p 227.78
3 450
   (2)
Solving (1) and (2)
p1 = 138.62 N/mm2
(tensile) and p2 = 94.1 N/mm2
(compressive)
Problem
Two steel plugs fit freely into the ends of steel tubular distance piece 300 mm long and are
drawn together by a steel bolt (450 mm long) and nut, the nut being tight fit in the beginning.
The nut is further tightened by 1/4 turn to draw the pieces together, the pitch of the bolt
thread being 2 mm. The pieces A and B are then subjected to forces of 60 kN tending to pull
them apart. Calculate the stresses in the bolt and the tube. The area of cross section of the
bolt is 600 sq. mm and that of the tube is 400 sq. mm. E for steel = 205 GN/m2
.
Answer: p1 = 163.9 N/mm2
(tensile) and p2 = 95.8 N/mm2
(compressive)

MECHANICS OF SOLIDS
A FOCUSSED APPROACH
Example
Two vertical wires are suspended at a distance apart of 500 mm as shown in figure. Their
upper ends are firmly secured and their lower ends support a rigid horizontal bar which
carries load W. The left hand wire has a diameter of 1.6 mm and is made of copper and right
hand wire has an area of 0.8 mm and is made of steel. Both wires initially are exactly 4 m
long.
(a) Determine the position of the line of action of W, if due to W both wires extend by the
same amount, and
(b) Determine the slope of wire if a load of 500 N is hung at the centre of the bar. Neglect
the weight of bar and take E for steel = 200 kN/mm2
and E for copper = 120 kN/mm2
.
Solution
(a) Let suffix s stand for steel bar and c for copper bar.
Let Pc and Ps be the forces in copper and steel wire
respectively.
From statics c sP P W  (1)
Taking moment about steel wire
1
W(500 x)
P
500

 (2)
Taking moment about copper wire
2
Wx
P
500
 (3)
Since AB remains horizontal
c s  
or c s
c c s s
P P
A E A E
 (4)
 c c c
s s s
P A E 2.011 120
. x 2.399
P A E 0.503 200
  
Also c
s
P 500 x
P x

 , from (2) and (3)

500 x
2.399
x


i.e. x = 147.11 mm.
(b) Slope of wire if W is hung in the middle of bar
From (2)

MECHANICS OF SOLIDS
A FOCUSSED APPROACH
250
2
s c
W
P P N  
Now 5
250 4000
4.144
2.011 1.2 10
c
c
c c
Pl x
A E x x
    mm
5
250 4000
9.940
0.503 2 10
s
s
s s
Pl x
A E x x
    mm
If the  is the slope of the bar, we get
9.940 4.144
tan 0.011592
500


 
  = 0.66410
= 00
39.85' (clockwise)
Example
A beam weighing 50 N is held in horizontal position by three wires. The outer wires are of
brass of 1.8 mm dia and attached to each end of the beam. The central wire is of steel of 0.9
mm diameter and attached to the middle of the beam. The beam is rigid and the wires are of
the same length and unstressed before the beam is attached. Determine the stress induced in
each of the wire. Take Young's modulus for brass as 80 GN/m2 and for steel as 200 GN/m2.
Solution
If Pb denotes load taken by each brass wire and Ps denotes load taken by steel wire, then
Total load P = 2Pb + Ps = 2bAb + sAs (i)
As the beam is horizontal, all the wire extend by the same amount. Further since each wire is
of same length, the wires would experience the same amount of strain, thus
es = eb
s/Es = b/Eb
s = (Es × b)/Eb = (200 × b)/80 = 2.5b (ii)
Putting the value of equation (ii) in equation (i)
50 = 2b(/4)(1.8)2
+ 2.5b (/4)(0.9)2

MECHANICS OF SOLIDS
A FOCUSSED APPROACH
50 = 6.678 b
b = 7.49 N/mm2
s = 2.5b = 18.71 N/mm2
Problem
A load of 20 kN is jointly supported by the vertical rods equal in length, each 15 mm
diameter, and equidistant in a vertical plane. Initially, the rods are so adjusted as to share
the load equally. The outer rods are of steel and the middle one of copper. Calculate the total
stresses if a further load of 15 kN is added. Take Ec = 110 kN/mm2
; Es = 205 kN/mm2
.
Answer: 71.2 N/mm2
, 55.75 N/mm2
A solid steel cylinder 500 mm long and 70 mm diameter is placed inside an aluminium
cylinder having 75 mm inside diameter and 100 mm outside diameter. The aluminium
cylinder is 0.16 mm. longer than the steel cylinder. An axial load of 500kN is applied to the
bar and cylinder through rigid cover plates as shown in Fig.. Find the stresses developed in
the steel cylinder and aluminium tube. Assume for steel, E = 220 GN/m2
and for Al E = 70
GN/m2
Solution
Since the aluminium cylinder is 0.16 mm longer than the steel cylinder, the load required to
compress this cylinder by 0.16 mm will be found as follows :
E = stress/ strain = P.L/A.L
Or P = E.A.L/L
= 70 × 109
× /4 (0.12
- 0.0752
) × 0.00016/0.50016 = 76944 N
When the aluminium cylinder is compressed by its extra length 0.16 mm, the load then
shared by both aluminium as well as steel cylinder will be,
500000 – 76944 = 423056 N

MECHANICS OF SOLIDS
A FOCUSSED APPROACH
Let es = strain in steel cylinder
ea = strain in aluminium cylinder
s = stress produced in steel cylinder
a = stress produced in aluminium cylinder
Es = 220 GN/m2
Ea = 70 GN/m2
As both the cylinders are of the same length and are compressed by the same amount
es = ea
s /Es = a /Ea
or; s = Es/Ea.a= (220 × 109
/70 × 109
).
a = (22/7). a
Also Ps + Pa = P
or; s As + a Aa = 423056
(22/7). a As + a Aa = 423056 ...(i)
As = /4 (0.072
) = 0.002199 m2
Aa = /4 (0.12 – 0.0752
) = 0.003436 m2
Putting the value of As and Aa in equation (i) we get
a = 27.24 × 106
N/m2
= 27.24 MN/m2
and S = 22/7 × 27.24 = 85.61 MN/m2
Stress in the aluminium cylinder due to load 76944 N
= 76944/ /4 (0.12 - 0.0752)
= 23.39 × 109
N/m2
= 22.39 MN/m2
Total stress in aluminium cylinder
= 27.24 + 22.39 = 49.63 MN/m2
and stress in steel cylinder = 85.61 MN/m2
Example
A concrete column 300mm x 300mm in section, is reinforced by 10 longitudinal 20mm
diameter round steel bars. The column carries a compressive load of 450KN. Find load
carried and compressive stress produced in the steel bars and concrete. Take Es = 200GN/m2
and Ec = 15GN/m2
.

MECHANICS OF SOLIDS
A FOCUSSED APPROACH
Solution
Cross sectional Area of column = 300 × 300 = 90000mm2
Area of steel bars As = 10 × (/4)(20)2
= 3141.59mm2
Area of concrete = 90000 – 3141.59 = 86858.4 mm2
Each component (concrete and steel bars) shorten by the same amount under the compressive
load, and therefore
Strain in concrete = strain in steel
c/Ec = s/Es
Where c, s are stress induced in concrete and steel respectively
s = (Es c)/Ec
= (200 × 109
/15 × 109
) c = 13.33 c
Further,
Total load on column = Load carried by steel + load carried by concrete
P = s.As + c.Ac
450 × 103
= 13.33c × 3141.59 + c. 86858.4
= 128735.8 c
c = 3.59 N/mm2
s = 13.33 × 3.59 = 46.95 N/mm2
Load carried by concrete Pc = cAc = 3.59 × 86858.4 = 311821.656 N = 311.82KN
Load carried by steel Ps = sAs = 46.95 × 3141.59 = 147497.65 N = 147.49 KN
Problem
A reinforced concrete column 500 mm diameter has four steel rods of 30 mm diameter
embedded in it and carries a load of 680 kN. Find the stresses in steel and concrete. Take E
for steel = 2.04 x 105
N/mm2
and E for concrete = 0.136 x 105
N/mm2
. Find also the adhesive
force between steel and concrete.
Answer: ps = 43.24 N/mm2
, pc = 2.88 N/mm2
, 112.475 kN

MECHANICS OF SOLIDS
A FOCUSSED APPROACH
ASSIGNMENT
EQUILIBRIUM OF BODIES
Q.1. (AMIE S12, 4 marks): State what do you mean by equilibrium equation and compatibility condition.
Q.2. (AMIE S15, 18, W16, 4 marks): What is free body diagram? Write the purpose of free body diagrams.
Discuss steps to construct free body diagrams with examples.
Q.3. (AMIE W17, 2 marks): State Lami's theorem.
Q.4. (AMIE S18, 4 marks): Write the statement of Varignon's theorem or principle of moments.
Answer: The theorem states that the moment of a resultant of two concurrent forces about any point is equal to
the algebraic sum of the moments of its components about the same point. In other words, "If many coplanar
forces are acting on a body, then the algebraic sum of moments of all the forces about a point in the plane of the
forces is equal to the moment of their resultant about the same point."
Q.5. (AMIE W17, 2 marks): Define coefficient of friction.
Answer: For qualitative definition Coefficient of friction is defined as the level of roughness or smoothness of a
surface. Higher the coefficient of friction higher the roughness. The coefficient of friction value 0 indicates fully
smooth surface.
For quantitative definition, its defined as the ratio of frictional force acting on between two surface to the normal
reaction between two surface.
Q.6. (AMIE W17, 8 marks): Sum of magnitudes of forces in 16 N. The line of action of their resultant is
perpendicular to the smaller force. Determine the magnitudes of these two forces if the magnitude of the
resultant is 8 N and the angle of indication between the two forces.
Answer: 6 N and 10 N
Hint: See figure.
Q.7. (AMIE W17, 8 marks): A thin form rod AB is hinged at lower end A and the upper end B is attached by a
horizontal string of length 4 m from a point C, 1.5 m vertically above B. If the weight of the rod is 5 N find the
tension in the horizontal string.
Answer: 13.33 N
Q.8. (AMIE S18, 10 marks): A rectangular plate is held in equilibrium by application of forces shown in Fig.
Determine the magnitude of unknown force P and reaction at the hinge. Neglect the weight of plate.

MECHANICS OF SOLIDS
A FOCUSSED APPROACH
Answer: 19.4365 N
Q.9. (AMIE W05, 10 marks): Two equal cylinders, each weighing 900 N are placed in a box as shown in
figure. Neglecting friction between the cylinders and the box, estimate the reactions at A, B and C.
Answer: 900 N, 1800 N, 900 N
Q.10. (AMIE S07, 11, 14 marks): Three cylinders A, B, C of weights 100N, 200N, 300 N and radii 100 mm,
200 mm, 150 mm are placed in a rectangular ditch. Neglecting friction, determine reactions at various contact
points.
Answer: RG = -1.35 N, RD = 89.69 N, RF = 139.48 N, RH = 242.9 N, RI = -91 N
Q.11. (AMIE S12, 10 marks): Cylindrical roller A, weighing 1500 kg having diameter 50 cm, and another
cylinder roller By weighing 1200 kg having diameter 42 cm, are kept in a horizontal channel of width 72 cm.
Curved surfaced touches the base of the channel and one vertical face of the channel, and curved surface of B
also touches the other vertical face of the channel. Assuming no friction, find the reaction at all the contact
points.
Answer: 822.35 Kg, 822.35 Kg and 2700 Kg
Q.12. (AMIE W06, 10 marks): The figure below shows a smooth cylinder of radius 0.2 m supporting a rod AB
0.6 m long weighing 100 N. The end A of the rod is hinged to the horizontal surface AD. The cylinder is also
attached to the hinge by a string of length 0.2 m. Find the tension in the string.
Answer: 113.394 N
SIMPLE STRESSES & STRAINS
Q.13. (AMIE W15, S11, 13, 17, 3 marks): Write the expressions for generalised Hook’s law.

MECHANICS OF SOLIDS
A FOCUSSED APPROACH
Q.14. (AMIE W17, 6 marks): What do you mean by plain stress and plain strain condition?
Q.15. (AMIE W11, 5 marks): Express stresses x, y and z in terms of strains x, y and z using generalised
Hook’s law.
Q.16. (AMIE S05, 10 marks): The aluminium bar with a cross section area of 160 mm2
carries the axial loads
at the positions shown in following figure.
Given that E = 70 GPa, compute total change in length of the bar.
Answer: 4.227 mm
Q.17. (AMIE S08, 13, 8 marks): A mild steel bar, 6 m long, is 5 cm in diameter for 3 m of its length and 2.5
cm in diameter for the remaining length. The bar is in tension and the stress on the smallest section is 112 MPa.
Find the total elongation of the bar and the change in diameter at the smallest section. Given: E = 200 GPa and
Poisson's ratio = 0.15.
Answer: 0.21 cm, - 0.021 cm
Q.18. (AMIE S08, 13, 8 marks): A rod of square section of side D at one end tapers to a square section at the
other end. If its length is l, find the increase in length if it is subjected to an axial pull P.
Answer: Pl/EDd
Q.1. (AMIE S15, 6 marks): A bar of length 50 cm has varying cross-section. It carries a load of 25 kN. Find
the extension, if the cross- section is given by [5 + (x2
/100)] cm2
, where x is the distance from one end in cm.
Neglect weight of the bar. Take E = 2 x 1011
N/m2
.
Answer: 0.004 cm
Hint:
50 50
20 0
7
5 (2 10 )
100
Pdx
d
x
x
   
 
 
 
 
Q.19. (AMIE S17, 8 marks): Find an expression for elongation of a bar of rectangular section and a conical
section due to self weight.
Answer: WL/2AE
ELASTIC CONSTANTS
Q.1. (AMIE S11, 17, W10, 16, 17, 5 marks): Define Poisson's ratio. Write the expressions for modulus of
rigidity (G) and bulk modulus (K) in terms of Young’s modulus of elasticity (E) and Poisson’s ratio ().
Establish the relation between E, K and G.
Q.2. (AMIE S17, 6 marks): Define bulk modulus. Deduce the relation
E = 3K (1 - 2)
Q.20. (AMIE W06, 10 marks): Define (i) Modulus of elasticity (ii) Modulus of rigidity (iii) Bulk modulus and
establish the relationship among them.
Q.1. (AMIE W11, 16, 6 marks): What is volumetric stress and volumetric strain? How is the volumetric strain
determined when a circular rod is loaded axially?
Q.2. (AMIE S06, 8 marks): A specimen of 15 mm diameter and 200 mm long is subjected to tensile test and
data at proportional and elastic limits were recorded as below:

MECHANICS OF SOLIDS
A FOCUSSED APPROACH
Limit
Stress
(MPa)
Increase in length
(mm)
Reduction in diameter
(mm)
Proportional 340 00.90 ?
Elastic 350 01.00 0.0225
Find modulus of elasticity, Poisson’s ratio and reduction in diameter.
Answer: 70 GPa, 0.3, 0.02025 mm
Hint: E = stress/strain = 350 x 106
/0.005 = 70 GPa
Q.21. (AMIE W05, 06, 16, S06, 10, 6 marks): What is generalised Hook’s law? Write the governing equations
for two dimensional stet of stress and three dimensional state of stress.
Q.22. (AMIE S06, 4 marks): For two dimensional stress system, show that
 1 1 22
E
v
1 v
    

and  2 1 22
E
v
1 v
    

where v is Poisson ratio.
Q.23. (AMIE W08, 10 marks): A bar of 20 mm diameter is subjected to an axial tensile load of 120 kN, under
which 200 mm gauge length of this bar elongates by an amount of 3.5 x 10-4
m. Determine the modulus of
elasticity of the bar material. if  = 0.3, determine its change in diameter.
Answer: 0.0105 mm
Q.24. (AMIE S09, 8 marks): A bar 30 mm x 30 mm x 250 mm long is subjected to a pull of 90 kN in the
direction of its length. The extension of the bar was found to be 0.125 mm, while the decrease in each lateral
dimension is found to be 0.00375 mm. Find the Young's modulus, Poisson's ratio, modulus of rigidity and bulk
modulus for the material of the bar.
Answer: E = 20 x 104
N/mm2
, Poisson's ratio = 0.25, G = 8 x 104
N/mm2
, K = 13.34 x 104
N/mm2
Q.1. (AMIE S16, 6 marks): A cube of material is under the action of a uniform compressive stress on its faces.
If the volume of the cube decreases by 0.1%, find the uniform stress on the cube. Also, find the percentage
reduction in length of a side of the cube. Take E = 2 x 105
N/mm2
and v = 0.3.
Answer:  = 167 N/mm2
; 0.033%
Q.2. (AMIE S17, 8 marks): A bar of 24 mm diameter and 400 mm length is acted upon by an axial load of 38
kN. The elongation of the bar and the change in diameter are 0.165 mm and 0.0031 mm respectively. Determine
(i) Poisson's ratio (ii) values of the three moduli.
Answer: m = 0.313; E = 203636 MPa; G = 77546 MPa; K = 181494 mPa

MECHANICS OF SOLIDS
A FOCUSSED APPROACH
COMPOSITE SECTION
Q.25. (AMIE S06, 10 marks): A structural member 5 m long is made up of two materials. The first 1.7 m of its
length is of brass and is 7.5 cm2
in cross section and the remainder of its length is of steel and is 6.0 cm2
in cross
section. The bar is in tension under load P Newton and the total elongation of the bar is 0.1 cm. Determine (i)
magnitude of load (ii) work done in elongation of the bar. Take Es = 210 GPa and Eb = 84 GPa.
Answer: 18.8 kN, 940 N-cm
Q.26. (AMIE W12, 10 marks): A copper sleeve, 21 mm internal and 27 mm external diameter, surrounds a 20
mm steel bolt, one end of the sleeve being in contact with the shoulder of the bolt. The sleeve is 60 mm long.
After putting a rigid washer on the other end of the sleeve, a nut is screwed on the bolt through 10°. If the pitch
of the threads is 2.5 mm, find the stresses induced in the copper sleeve and steel bolt. Take E = 200 GN/m2
and
E = 90 GN/m2
.
Answer: c = 0.078 N/mm2
, s = 0.056 N/mm2
Hint: Contraction induced due to turning nut through 100
= (10/360) x 2.5 = 2.5/36 mm
Contraction of copper sleeve + extension of the steel bolt = 2.5/36 mm
Q.27. (AMIE W11, 8 marks): Two rods A and B of equal free length hang vertically 60 cm apart and support a
rigid bar horizontally. The bar remains horizontal when carrying a load of 5000 kg at 20 cm from rod A . If the
stress in B is 50 N/mm2
, find the stress in rod A and the areas of A and B. Take EA - 200,000 N/mm2
and EB =
90,000 N/mm2
.
Answer: 111.11 N/mm2
, AB = 327 mm2
, AA = 294.3 mm2
Q.28. (AMIE W09, 10 marks): An aluminium solid cylinder of 7.5 cm diameter fits loosely inside a steel tube
having 10 cm external diameter and 8 cm internal diameter. The steel tube is 0.02 cm longer than aluminium
cylinder and is 250 cm long before the load is applied. Calculate the safe load which can be placed on a rigid flat
plate on the lop of the steel tube. Safe stress for steel is 95 MPa and for aluminium 65 MPa, Es = 210 GPa, and
Eal = 70GPa.
Answer: 408.83 kN
Q.29. (AMIE W11, 8 marks): A 75 mm diameter compound bar is constructed by shrinking a circular brass
bush on to the outside of a 50 mm diameter solid steel rod. If the compound bar is then subjected to an axial
compressive load of 160 kN, determine the load carried by the steel rod and brass bush and the compressive
stresses set up in each material. Take Esteel =210,000 N/mm2
and Ebrass = 90,000 N/mm2
.
Answer: b = 34.82 N/mm2
, s = 81.48 N/mm2
Q.30. (AMIE W09, 10 marks): A load of 300 kN is applied on a shoo concrete column 250 mm x 250 mm.
The column is reinforced by steel bars of total area 5600mm2
If modulus of elasticity of steel is 15 limes that of
concrete, find the stresses in concrete and steel. If this stress in concrete should not exceed 4N/mm2
. find the
area of steel required so dial the column may support a load of 600kN.
Answer: s = 60 N/mm2
, c = 31.95 N/mm2
, As = 6250 mm2
Q.31. (AMIE W15, 10 marks): A copper rod of 40 nun diameter is surrounded tightly by a cast iron tube of 80
mm external diameter, the ends being firmly fastened together. When put to a compressive load of 30.000 N,
what load will be shared by each? Also determine the amount by which the compound bar shortens if it is 2000
mm long?
Answer: 3750 N, 26.250 N, 0.0796 mm
(For online support such as eBooks, video lectures, audio lectures, unsolved papers, online
objective questions, test series and course updates, visit www.amiestudycircle.com)

Stresses and strains (Part 1)

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Stresses and strains (Part 1)