Mechanics of solids explaining power point show.ppt

Dr. Alice Mathai
Professor
Dept. of Civil Engg.

SYLLABUS
Module – 1
Review of statics,
Concept of stress and strain – types, Stress – strain relation
Hooke’s law, Young’s modulus of elasticity.
Stress-strain diagram of mild steel.
Factor of safety, working stress.
Axially loaded bars with uniform cross section–stress, strain and
deformation.
Deformation of axially loaded bars with varying cross section and bars
with varying axial loads.
Statically indeterminate systems (number of unknowns restricted to two).

1. H. J. Shah andS. B. Junnarkar, Mechanics of Structures Vol - I,
Charotar Publishing House.
2. R. K. Bansal, A Text book of Strength of Materials, Laxmi
Publications (P) Ltd, New Delhi.
3. B. C. Punmia, Ashok K. Jain, Arun Kumar Jain, Mechanics of
Materials, Laxmi Publications (P) Ltd, New Delhi.

REFERENCES
• 1. Egor P. Popov, Engineering Mechanics of Solids, Prentice
Hall International Series.
• 2. James M Gere, S.P. Timoshenko, Mechanics of Materials,
CBS Publishers and Distributors, New Delhi.
• 3. R.C. Hibbeler, Mechanics of Materials (edn.10), Pearson
• 4. S. Ramamrutham and R. Narayanan, Strength of Materials,
Dhanpat Rai Publishing Co (P) Ltd.
• 5. Rattan, Strength of Materials, McGraw Hill Education India.

Review of statics
Mechanics, science concerned with the motion of bodies under the
action of forces(Dynamics), including the special case in which
a body remains at rest.(Statics)
Mechanics Engineering Mechanics Statics deals with rigid
bodies
• What are the principles of statics?
Statics is a branch of mechanics which studies the effects and
distribution of forces of rigid bodies which are and remain at
rest. In this area of mechanics, the body in which forces are
acting is assumed to be rigid.
• The deformation of non-rigid bodies is treated in Strength of
Materials

Review of statics…
• A rigid body is one which does not deform, in
other words the distance between the individual
particles making up the rigid body remains
unchanged under the action of external forces.
• A deformable body because of the action of
loads there will be stresses and strains and
corresponding deformations

How do you solve for statics?
• Draw a free-body diagram of the entire system.
• Write equilibrium equations to compute as
many unknown support reactions &
internal forces

What are the conditions of equilibrium?
• An object is in equilibrium if ;
• The resultant force acting on the object is zero.
• The sum of the moments acting on an object
must be zero
In 2D - ∑H =0,∑V =0& ∑M = 0 3 nos.
In3D - ∑ X=∑Y= ∑Z= ∑MX= ∑ MY =∑MZ = 0 6 nos.

MECHANICS OF SOLIDS
• Introduction:-
• Mechanics of Solids, also called Strength of materials,
is a subject which deals with the behaviour of solid
objects under the action of loads.
• Expert in Solid Mechanics - Solid Mechanician
• The strength of a material is its ability to withstand an
applied load without failure.
• Predicts how geometric and physical properties of
structure will influence its behaviour under service
conditions.

Structures or Machines
• assembly of members connected together
•perform useful functions and/or withstand applied
loads

Torsion
S
A
N
M
Bending
M
Axial
tension
N
Axial
compression
S
Compression Machine
base
arms
screw
Cross
head
Hand wheel

Properties of Material:-
Strength
•Stress – Resistance to deformation
•Stresses can occur isolated or in combination.
• Is structure strong enough to withstand loads applied to it ?
The max. stress the structure can take is its strength
The resistance by which the material opposes deformation is known as
Strength of Material
Stiffness
•Is it stiff enough to avoid excessive deformations and deflections?
•All materials are deformable and mechanics of solids takes this into
account.
Strength and stiffness of structures is function of size and shape, certain
physical properties of material.

Properties of Material:-
• Elasticity
• Plasticity
• Ductility
• Malleability
• Brittleness
• Toughness
• Hardness
•Creep
•Shrinkage
•Homogeneous
•Isotropic

INTERNAL FORCE:- STRESS
• Axial Compression
• Shortens the bar
• Crushing
• Buckling
n
m
P P
P= A
• Axial tension
•Stretches the bars &
tends to pull it apart
• Rupture
m n
=P/A
P
P

• Resistance offered by the material per unit cross-
sectional area is called STRESS.
 = P/A
Unit of Stress:
Pascal = 1 N/m2
kN/m2 , MN/m2 , GN/m2
1 MPa = 1 N/mm2
Permissible stress or allowable stress or working
stress = yield stress or ultimate stress /factor of
safety.

Strain
•It is defined as deformation per unit length
Linear strain
• it is the ratio of change in length to original length
•Tensile strain = increase in length = 
(+ Ve) () Original length L
Compressive strain = decrease in length = 
(- Ve) () Original length L
P

L
Strain is dimensionless quantity.

Example : 1
A short hollow, cast iron cylinder with wall thickness of
10 mm is to carry compressive load of 100 kN.
Compute the required outside diameter `D’ , if the
working stress in compression is 80 N/mm2.
Solution:  = 80N/mm2;
P= 100 kN = 100*103 N
as  = P/A
A =(/4) *{D2 - (D-20)2} = (/4) *{D2 - D2
+40D-400}
substituting in above eq. and solving. D =
49.8 mm
D
d
10 mm

Example: 2
A Steel wire hangs vertically under its weight. What is
the greatest length it can have if the allowable tensile
stress t =200 MPa? Density of steel =80 kN/m3.
Solution:
t =200 MPa= 200*103 kN/m2 ;
=80 kN/m3.
Wt. of wire P=(/4)*D2*L* 
c/s area of wire A=(/4)*D2
t = P/A
solving above eq. L =2500m
L

A


O
stress
strain
.
E
Elasticity
• Elasticity is the tendency of solid materials to
return to their original shape after being forces
are applied on them.
• When the forces are removed, the object will
return to its initial shape and size if the material is
elastic.
• In other words, The deformation disappears
completely, after removal of external forces
Linear elastic Nonlinear elastic

stress
strain


Strain
Stress
Stress- Strain Curve for Mild Steel (Ductile Material)
Plastic state
Of material
Elastic State
Of material
Yield stress
Point
E = modulus of
elasticity
Ultimate stress point
Breaking stress point
A
B
C
D
E
F
A- Limit of proportionality
B- Elastic limit
C – Upper yield point
D – Lower yield point
E – Ultimate stress
B – Breaking stress
E1 -Actual ultimate stress
F1 – Actual breaking stress
F1
E1

•21. .
Limits of proportionality
•If a tensile force applied to a uniform bar of mild steel is gradually increased
and the corresponding extension of the bar is measured, then provided the
applied force is not too large, a graph depicting these results is likely to be as
shown in Figure.
•Since the graph is a straight line, extension is directly proportional to the
applied force. (Hooke’s Law) The point on the graph where extension is no
longer proportional to the applied force is known as the limit of proportionality.
Elastic Limit
• Is the maximum stress from which an elastic body will recover its original
state after the removal of the deforming force.
• It differs widely for different materials.
• It is very high for a substance like steel and low for a substance like lead

Yield limit
•When specimen is stressed beyond elastic limit, strain increases more rapidly
than the stress. Because, sudden elongation of the specimen takes place,
without appreciable increase in the stress. This phenomena is known as
yielding of material.
• The stress corresponding to point of upper yield point is called yield stress.
• The portion between upper yield point and lower yield point is called yield
stage.
Ultimate stress
•Because of the plastic deforms, the material strain hardens and further strain
beyond lower yield point requires an increase in stress.
•The maximum stress reached at point E is called ultimate stress.
•In other words, Stress corresponding to the maximum load taken by the
specimen is called ultimate stress.
Strain hardening
•The phenomenon of increase in stress from D to E is known as strain
hardening.
•During strain hardening, the extension of the specimen is quite large. Also if
the specimen has mill scale or rust, it will be flaked off.
Concept of actual stresses

Modulus of Elasticity:
• Stress required to produce a strain of unity.
• i.e. the stress under which the bar would be stretched to
twice its original length . If the material remains elastic
throughout , such excessive strain.
• Represents slope of stress-strain line OA.
A


O
stress
strain
Value of E is same
in Tension &
Compression.
 =E 
E

A


O
• Hooke’s Law:-
Up to elastic limit, Stress is proportional to strain
  
 =E ; where E=Young’s modulus
=P/A and  =  / L
P/A = E ( / L)
 =PL /AE
E

Example:4 An aluminium bar 1.8 meters long has a
25 mm square c/s over 0.6 meters of its length and
25 mm circular c/s over other 1.2 meters . How
much will the bar elongate under a tensile load
P=17500 N, if E = 75000 Mpa.
Solution :-  = ∑PL/AE
=17500*600 / (252*75000) +
17500*1200/(0.785*252*75000) =0.794 mm
0.6 m
1.2 m
25 mm sq.sect 25 mm cir..sect
17500 N

15 kN
1 m 1 m 2 m
20 kN 15 kN
Example: 5 A prismatic steel bar having cross sectional
area of A=300 mm2 is subjected to axial load as shown in
figure . Find the net increase  in the length of the bar.
Assume E = 2 x 10 5 MPa.( Ans  = -0.17mm)
 = 20000*1000/(300*2x10 5)-15000*2000/(300*2 x10 5)
= 0.33 - 0.5 = -0.17 mm (i.e.contraction)
C B A
20
20 C
0
0 B
15 15
A
Solution:

9 m
x
5 m
3m
A = 445 mm 2
E = 2 x 10 5
A = 1000 mm 2
E = 1 x 10 5
A B
Example: 6 A rigid bar AB, 9 m long, is supported by two
vertical rods at its end and in a horizontal position under a
load P as shown in figure. Find the position of the load P so
that the bar AB remains horizontal.
P

9 m
x
5 m
3m
A B
P
P(9-x)/9 P(x)/9


(9 - x)*3=x*5*1.1236
27-3x=5.618 x
8.618 x=27
x = 3.13 m
For the bar to be in horizontal position, Displacements
at A & B should be same,
A = B
(PL/AE)A =(PL/AE)B
=
{P(x)/9}*5
0.000445*2*105
{P(9-x)/9}*3
(0.001*1*105)

P
P
X
L
d1 d2
dx
x
Extension of Bar of Tapering cross Section
from diameter d1 to d2:-
Bar of Tapering Section:
dx = d1 + [(d2 - d1) / L] * X
 = Px / E[ /4{d1 + [(d2 - d1) / L] * X}2]

 = 4 P x /[E {d1+kx}2 ]
= - [4P/  E] x 1/k [ {1 /(d1+kx)}]
=- [4PL/  E(d2-d1)] {1/(d1+d2 -d1) - 1/d1}
 = 4PL/( E d1 d2)
Check :-
When d = d1=d2
 =PL/ [( /4)* d2E ] = PL /AE (refer -24)

L
0
L
0

Types of Stress…
The stress developed in a body depends upon
how the external forces are applied over it.
On this basis, there are two types of stress ,
i. Normal Stress
ii. Tangential Stress or shear stress

P
P/2 P/2
P
• Connection should withstand full load P transferred
through the pin to the fork .
• Pin is primarily in shear which tends to cut it across at
section m-n .
• Average shear Stress =>  =P/(2A) (where A is cross
sectional area of pin)
• Note: Shearing conditions are not as simple as that for
direct stresses.
Direct Shear:--
Pin Pin
m
n
Fork

Shear stress 
=Shear resistance/shearing area
=P/Lx1

SHEAR STRAIN
Up to the elastic limit,
shear stress ()  shearing strain()
 = G 
Expresses relation between shear stress
and shear strain. /=N;
where
Modulus of Rigidity =G=  / 
Shear strain  = angular deformation in radians
= transverse displacement/distance AD=DD1/AD= dl/AD

 = 4 P dx /[E {d1+kx}2 ]
= - [4P/  E] x 1/k [ {1 /(d1+kx)}] dx
=- [4PL/  E(d2-d1)] {1/(d1+d2 -d1) - 1/d1}
 = 4PL/( E d1 d2)
Check :-
When d = d1=d2
 =PL/ [( /4)* d2E ] = PL /AE

L
0
L
0

`` P
P
X
L
d1 d2
dx
x
Q. Find extension of tapering circular bar under axial pull for the
following data: d1 = 20mm, d2 = 40mm, L = 600mm, E = 200GPa. P
= 40kN
L = 4PL/( E d1 d2)
= 4*40,000*600/(π* 200,000*20*40)
= 0.38mm. Ans.

P
P
X
L
b1 b2
bx
x
Bar of Tapering Section:
bx = b1 + [(b2 - b1) / L] * X = b1 + k*x,
Extension for the elemental length x =
 = Px / [Et(b1 + k*X)], k = (b2 - b1) / L
Extension of Tapering bar of uniform
thickness t, width varies from b1 to b2:-

L = L = Px / [Et(b1 + k*X)],

L
0

L
0
= P/Et ∫ x / [ (b1 + k*X)],
= - P/Etk * loge [ (b1 + k*X)]
0
L
,
= PLloge(b1/b2) / [Et(b2 – b1)]

L
0

P
P
X
L
b2 b1
bx
x
Take b1 = 200mm, b2 = 100mm, L = 500mm
P = 40kN, and E = 200GPa, t = 20mm
δL= PLloge(b1/b2) / [Et(b1 – b2)]
= 40000*500loge(200/100)/[200000*20 *100]
= 0.03465mm
Q. Calculate extension of Tapering bar of
uniform thickness t, width varies from b1 to
b2:-
P/Et ∫ x / [ (b1 + k*X)],

Elongation of a Bar of circular tapering section
due to self weight:
=Wx*x/(AxE)
(from  =PL/AE )
now Wx=1/3* AxX 
where Wx=Wt.of the bar
so = X *x/(3E)
so now
L = X *x/(3E)
= /(3E) Xdx= [/3E ] [X2 /2]
= L2/(6E)

L
0

L
0
x
L
d
A B
X

Let W=total weight of bar = (1/3)*(/4*d2)L 
 =12W/ (*d2L)
so,
L = [12W/ (*d2L)]*(L2/6E)
=2WL/ (*d2E)
=WL/[2*(*d2/4)*E]
=WL /2*A*E

Calculate elongation of a Bar of circular tapering
section due to self weight:Take L =10m, d =
100mm,  = 7850kg/m3
L = L2/(6E)
7850*9.81*10000*10000*/
[6*200000*10003]
= 0.006417mm
x
L
d
A B
X

P + dP
P
dx
X
Extension of Uniform cross section bar subjected
to uniformly varying tension due to self weight
PX=  A x
d = PX dx / A E;
 = PX dx/AE=  A x dx/AE
 = ( /E)  x dx= ( L2/2E)
If total weight of bar W=  A L  = W/AL
=WL/2AE (compare this results with slide-26)
L
0
L
0
L
0
L
d

dx
X
Q. Calculate extension of Uniform cross section bar
subjected to uniformly varying tension due to self weight
L
d
Take L = 100m, A = 100mm2 , density =
7850kg/m3
 = ( L2/2E)
 = 850*9.81*100000*100000/
[2*200000*10003 ]
= 1.925mm

Bar of uniform strenght:(i.e.stress is constant at all
points of the bar.)
dx
L
x
Area = A2
Area = A1
Force = p*(A*dA)
Force = p*(A+dA)
dx
comparing force at BC level of strip
of thickness dx
A
B C
D
B C
P(A + dA) = Pa + w*A*dx,
where w is density of the material hence
dA/A = wdx/p, Integrating logeA = wx/p + C,
at x = 0, A = A2 and x = L, A = A1, C = A2
loge(A/A2) = wx/p OR A = ewx/p
Down ward force
of strip = w*A*dx,

dx
L
x
Area = A2
Area = A1
Force = p*(A*dA)
Force = p*(A+dA)
dx
A
B C
D
B C
A = ewx/p
(where A is cross section area at any
level x of bar of uniform strenght )
Down ward force of
strip = w*A*dx,

dx
L
x
Area = A2
Area = A1
A
B C
D
p = 700000/5000 = 140MPa
A1 =A2 ewx/p
A1 = 5000*e8000*9.81*20000/[140*10003]
= 5056.31mm2
Q. A bar of uniform strength has following data.
Calculate cross sectional area at top of the bar.
A2 = 5000mm2 , L = 20m, load
at lower end = 700kN, density
of the material = 8000kg/m3

L B
D
P
P
L+L
B-B
D-D
POISSONS RATIO:- = lateral contraction per Unit axial
elongation, (with in elastic limit)
L(1+)
B(1-)
D(1-
)
= (B/B)/(L/L);
= (B/B)/()
So B =  B;
New breadth =
B -B = B -  B
=B(1 -   )
Sim.,New depth=
D(1- )

for isotropic materials  = ¼ for steel  = 0.3
Volume of bar before deformation V= L * B*D
new length after deformation L1=L + L = L + L = L (1+ )
new breadth B1= B - B = B -  B = B(1 -  )
new depth D1= D - D = D -  D = D(1 -  )
new cross-sectional area = A1= B(1- )*D(1- )= A(1-   )2
new volume V1= V - V = L(1+  )* A(1-   )2
 AL(1+  - 2   )
Since  is small
change in volume = V =V1-V = AL  (1-2 )
and unit volume change = V/ V = {AL  (1-2 )}/AL
V/ V =  (1-2 )

In case of uniformly varying tension, the elongation ‘’
is just half what it would be if the tension were equal
throughout the length of the bar.

Example: 7 A steel bar having 40mm*40mm*3000mm
dimension is subjected to an axial force of 128 kN.
Taking E=2*105N/mm2 and  = 0.3,find out change in
dimensions.
Solution:
given b=40 mm,t=40mm,L=3000mm
P=128 kN=128*103 N, E=2*105 mm2,  =0.3
L=?, b=?, t=?
t = P/A = 128*103 /40*40= 80 N/mm2
128 kN
128 kN
3000 mm 40
40

now  = t/E=80/(2*105 )=4*10-4
 = L/L ==> L=  *L=4*10-4 *3000 = 1.2 mm
(increase)
b= - *( *b)= -0.3*4*10-4*40 = 4.8*10-3 mm
(decrease)
t = - *( *t)= -0.3*4*10-4*40 = 4.8*10-3 mm
(decrease)

Change in volume = [3000 + 1.2) * (40 – 0.0048) *
(40 – 0.0048)] – 3000*40*40
= 767.608 mm3
OR by using equation (derivation is in chapter
of volumetric stresses and strains)
dv = p*(1-2µ)v/E
= (128000/40*40)*0.4*3000*40*40/200000
= 768mm3

Example: 8 A strip of 20 mm*30 mm c/s and
1000mm length is subjected to an axial push of 6
kN. It is shorten by 0.05 mm. Calculate (1) the
stress induced in the bar. (2) strain and young's
modulus & new cross-section. Take  =0.3
Solution:given,
c/s =20 mm*30 mm, A =600mm2,L=1000 mm,
P=6 kN=6*103 N, L =0.05 mm,  = ?, =?,E =?.
1.  = P/A =6000/600 =10 N/mm2 -----(1)
2  = L /L=0.05/1000 =0.00005 -----(2)
 =E  ==>E = /  =10/0.00005 = 2*105 N/mm2

3 Now,
New breadth B1 =B(1- )
=20(1-0.3*0.00005)
=19.9997 mm
New Depth D1 = D(1- )
=30(1-0.3*0.00005)
= 29.9995mm

Example: 9 A iron bar having 200mm*10 mm
c/s,and 5000 mm long is subjected to an axial pull
of 240 kN.Find out change in dimensions of the
bar. Take E =2*105 N/mm2 and  = 0.25.
Solution: b =200 mm,t = 10mm,so A = 2000mm2
 = P/A=240*103 / 2000 =120N/mm2
now =E   = /E =120/2*105=0.0006
= L /L L =  *L=0.0006*5000=3 mm
b = -*( *b)= -0.25*6*10-4*200
= 0.03 mm(decrease)
t = -*( *t) = -0.25*6*10-4*10
= 1.5*10-3 mm (decrease)

Composite Sections:
• as both the materials deforms axially by
same value strain in both materials are same.
s = c = 
s /Es= c /E (=  = L /L) _____(1) & (2)
•Load is shared between the two materials.
Ps+Pc = P i.e. s *As + c *Ac = P ---(3)
(unknowns are s, c and L)
Concrete
Steel
bars

Example: 10 A Concrete column of C.S. area 400 x
400 mm reinforced by 4 longitudinal 50 mm
diameter round steel bars placed at each corner
of the column carries a compressive load of 300
kN. Calculate (i) loads carried by each material &
compressive stresses produced in each material.
Take Es = 15 Ec Also calculate change in length
of the column. Assume the column in 2m long.
400 mm
4-50 bar
400
mm
Take Es = 200GPa

Solution:-
Gross C.S. area of column =0.16 m2
C.S. area of steel = 4*π*0.0252 = 0.00785 m2
Area of concrete =0.16 - 0.00785=0.1521m2
Steel bar and concrete shorten by same amount. So,
s = c => s /Es = c /Ec = > s= cx (Es /Ec)
= 15c

load carried by steel +concrete=300000 N
Ws +Wc= 300000
s As + c Ac = 300000
15 c x 0.00785 + c x0.1521 = 300000
c = 1.11 x 10 6 N/ m2
s =15x c=15 x1.11x 10 6=16.65 x10 6 N/ m2
Ws =16.65x10 6 x0.00785 / 10 3 =130.7 kN
Wc = 1.11x 10 6 x 0.1521/103= 168.83 kN
(error in result is due to less no. of digits
considered in stress calculation.)

we know that,
s /Es= c /E (=  = L /L) _____(1) & (2)
c = 1.11 MPa
s =15x c=15 x1.11x 10 6=16.65 MPa
The length of the column is 2m
Change in length
dL = 1.11*2000/[13.333*1000] = 0.1665mm
OR
dL = 16.65*2000/[200000] = 0.1665mm

Example: 10 A Concrete column of C.S. area 400 x 400
mm reinforced by 4 longitudinal 50 mm diameter round
steel bars placed at each corner of the column. Calculate
(1) maximum axial compressive load the column can
support &(ii) loads carried by each material &
compressive stresses produced in each material. Take
Also calculate change in length of the column. Assume
the column in 2m long. Permissible stresses in steel and
concrete are 160 and 5MPa respectively. Take Es =
200GPa and Ec = 14GPa.
400 mm
4-50 bar
400
mm

Solution:-
s = c => s /Es = c /Ec = > s= cx (Es /Ec)
= 14.286 c

Solution:-
s = c => s /Es = c /Ec = > s= cx (Es /Ec) = cx ( 200/14)
= 14.286c
So s = 14.286c
s = 160 then c = 160/14.286 = 11.2MPa > 5MPa, Not valid
c = 5MPa then s = 14.286*5 = 71.43 MPa <120MPa,Valid

Permissible stresses in each material are
c = 5MPa & s = 71.43 MPa
We know that
s As + c Ac = W
[71.43 x 0.00785 + 5 x0.1521]*10002 / 1000 = 1321.22kN
Load in each materials are
Ws =71.43x0.00785 x1000 =560.7255 kN
Wc = 5x 0.1521x1000 = 760.5kN

we know that,
s /Es= c /E (=  = L /L) _____(1) & (2)
c = 5 MPa
s =71.43 MPa
The length of the column is 2m
Change in length
dL = 5*2000/[14000] = 0.7143mm
OR
dL = 71.43*2000/[200000] = 0.7143mm

Example: 11 A copper rod of 40 mm diameter is
surrounded tightly by a cast iron tube of 80 mm diameter, the
ends being firmly fastened together. When it is subjected to a
compressive load of 30 kN, what will be the load shared by
each? Also determine the amount by which a compound bar
shortens if it is 2 meter long. Eci=175 GN/m2,Ec= 75 GN/m2 .
copper
Cast iron
80 mm
Cast iron
40 mm
2 meter

Area of Copper Rod =Ac = (/4)* 0.042 = 0.0004 m2
Area of Cast Iron =Aci= (/4)* (0.082 - 0.042) = 0.0012 m2
ci /Eci = c /Ec or
175 x 10 9
75 x 10 9
= 2.33
ci = 2.33 c
ci / c = Eci/Ec =

Now,
W = Wci +Wc
30 = (2.33 c ) x 0.012  + c x 0.0004 
c = 2987.5 kN/m2
ci = 2.33 x c = 6960.8kN/m2
load shared by copper rod = Wc = c Ac
= 2987.5 x 0.0004 
= 3.75 kN
Wci = 30 -3.75 = 26.25 kN

Strain c=c / Ec = L /L
L = (c /Ec) x L = [2987.5/(75 x 10 9)] x 2
= 0.0000796 m
= 0.0796 mm
Decrease in length = 0.0796 mm

R1
A1 = 110
mm2
1.2 m
2.4 m
L
M
N
R2
1.2 mm
For the bar shown in figure,
calculate the reaction produced
by the lower support on the bar.
Take E= 2*108 kN/m2.Find also
stresses in the bars.
A2 = 220
mm2
55
kN
Example: 12

Solution:-
R1+R2 = 55
 L1 =(55-R2)*1.2 / (110*10-6)*2*108 (LM extension)
 L2 =R2*2.4 / (220*10-6)*2*108 (MN contraction)
( Given:  L1-  L2 =1.2 /1000=0.0012)
(55-R2)*1.2 / [(110*10-6)*2*108 ] -R2*2.4 /[ (220*10-6)*2*108 ]
=0.0012
so R2 = 16.5 kN Since R1+R2 = 55 kN,
R1=38.5 kN
Stress in LM = Force/area = 350000 kN/m2
Stress in MN =75000 kN/m2

•Dealing with machines and structures an engineer
encounters members subjected to tension,
compression and shear.
•The members should be proportioned in such a
manner that they can safely & economically withstand
loads they have to carry.

100 mm
30000 N
Example: 3 Three pieces of wood having 37.5 x 37.5 mm
square C.S. are glued together and to the foundation as
shown in figure. If the horizontal force P=30000 N is applied
to it, what is the average shear stress in each of the glued
joints.(ans=4 N/mm2)
Plan
37.5
37.5
30000 N
Solution:-
P=30000N;glued c.s area=37.5x100mm x2 surfaces
Shear stress  = P/c.s area = 4N/mm2

Temperature stresses:-
Material
Change in temp.
Expands/ Shortens
no constraint is
present
Material
Constrained
No Expansion/
contraction
Temperature
stresses
Induced in material

Bar
Constraint
L
Uniform temp. increased to tº
Expansion =L t
but =PL/AE=P/A *L/E = tp L/E
so tp = *E/L = L t *E / L =  tE
tp= compressive , if temp. increases
tp= tensile, if temp. decreases
Suppose the support yield by an amount 
tp=( - )*E/L =(L t - )*E/L

Composite Section:- (Temp. stresses .)
E of Copper > steel
Steel(S)
Copper(C)
s
t s

c

c
t
s
t =Free expansion of steel due to rise in temp.
c
t =Free expansion of copper due to rise in temp.
s
 =Additional extension in steel to behave as
composite section
c
 =contraction in copper to behave as
composite section
Extension in steel = Contraction in copper
L

S = C
s
t + s
 = c
t - c

s
+ c
 = c
t - s
t
PL(1/AsEs +1/AcEc)= Lt(c - s) ----(1)
P = t(c - s)/ (1/AsEs +1/AcEc)
Substituting in eq.(1)
s = P /As and c = P /Ac
s/Es +c/Ec = t(c - s)
s+ c= t (c - s) strain relation
Steel(S)
Copper(C)
s
t s

c

c
t

A railway is laid so that there is no
stress in rail at 10º C. If rails are 30 m long Calculate,
1. The stress in rails at 60 º C if there is no allowance
for expansion.
2. The stress in the rails at 60 º C if there is an
expansion allowance of 10 mm per rail.
3. The expansion allowance if the stress in the rail is
to be zero when temperature is 60 º C.
4. The maximum temp. to have no stress in the rails
if the expansion allowance is 13 mm/rail.
Take  = 12 x 10 -6 per 1ºC E= 2 x 10 5 N/mm 2
Example: 13

Solution:
1. Rise in temp. = 60 º - 10 º = 50 ºC
so stress =  t E =12 x 10 -6 x50x 2 x 10 5
= 120 MPa
2. tp x L/E =  = (L t -10)
= (30000 x 12 x 10 -6 x50-10)
= 18 -10 = 8 mm
tp =E /L =8x 2 x 10 5 /30000
= 53.3 MPa

3. If stresses are zero ,
Expansion allowed =(L t )
= (30000 x 12 x 10 -6 x50)
=18 mm
4. If stresses are zero
tp =E /L*(L t -13)=0
L t=13
so t=13/ (30000 x 12 x 10 -6 )=360 C
allowable temp.=10+36=460c.

Example: 14
A steel bolt of length L passes through a copper tube
of the same length, and the nut at the end is turned
up just snug at room temp. Subsequently the nut is
turned by 1/4 turn and the entire assembly is raised
by temp 550C. Calculate the stress in bolt if
L=500mm,pitch of nut is 2mm, area of copper tube
=500sq.mm,area of steel bolt=400sq.mm
Es=2 * 105 N/mm2 ;s =12*10-6 /0C
Ec=1 * 105 N/mm2 ;c= 17.5*10-6 /0C

Solution:-
Two effects
(i) tightening of nut
(ii)raising temp.
tensile stress in steel = compressive force in copper
[Total extension of bolt
+Total compression of tube] =Movement of Nut
[s+  c] = np ( where p = pitch of nut)

(PL/AsEs + s L t) +(PL/AcEc- c L t)=np
P (1/AsEs +1/AcEc) = t(c - s)+np/L
so P[1/(400*2*105) + 1/(500*1*105) ]
=(17.5-12)*10-6 +(1/4)*2/500
so P=40000N
so ps=40000/400 = 100 MPa(tensile)
and pc=40000/500=80 MPa(compressive)

Example: 15 A circular section tapered bar is rigidly
fixed as shown in figure. If the temperature is raised
by 300 C, calculate the maximum stress in the bar.
Take
E=2*105 N/mm2 ; =12*10-6 /0C
1.0 m
D2=200 mm
D1=100 mm
X dX
P P
A
B

With rise in temperature compressive force P is
induced which is same at all c/s.
Free expansion = L  t = 1000*12*10-6*30
=0.36 mm
Force P induced will prevent a expansion of 0.36 mm
 = 4PL/(E*d1*d2) = L  t
Or P = (/4)*d1*d2  t E=1130400 N
Now Maximum stress = P/(least c/s area)
=1130400/(.785*1002) = 144MPa

Example: 16 A composite bar made up of aluminum
and steel is held between two supports.The bars are
stress free at 400c. What will be the stresses in the
bars when the temp. drops to 200C, if
(a) the supports are unyielding
(b)the supports come nearer to each other by 0.1 mm.
Take E al =0.7*105 N/mm2 ;al =23.4*10-6 /0C
ES=2.1*105 N/mm2 s =11.7*10-6 /0C
Aal=3 cm2 As=2 cm2

Steel Aluminum
60cm 30cm
2 cm2
3 cm2

Free contraction =Ls s t+ LALAlt
=600*11.7*10-6*(40-20)+300*23.4*
10-6*(40-20)=0.2808 mm.
Since contraction is checked tensile stresses will be
set up. Force being same in both
As s=Aal al
2 s= 3 al ==> s= 1.5 al
Steel
Aluminum
60cm 30cm
2 cm2
3 cm2

contraction of steel bar s
= (s/Es)*Ls
=[600/(2.1*105)]* s
contra.of aluminum bar al
= (al/Eal)*Lal
=[300/(0.7*105)]* al
(a) When supports are unyielding
s
+ al
=  (free contraction)
=[600/(2.1*105)]* s +[300/(0.7*105)]* al
=0.2808 mm

=[600/(2.1*105)]* s +[300/(0.7*105)]* al
=0.2808; but
s=1.5 al
al =32.76 N/mm2(tensile)
s =49.14 N/mm2(tensile)
(b) Supports are yielding
s
+ al
= ( - 0.1mm)
al =21.09 N/mm2(tensile)
s =31.64 N/mm2(tensile)

Example: 17 A copper bar 30 mm dia. Is completely
enclosed in a steel tube 30mm internal dia. and 50
mm external dia. A pin 10 mm in dia.,is fitted
transversely to the axis of each bar near each end. To
secure the bar to the tube.Calculate the intensity of
shear stress induced in the pins when the temp of the
whole assembly is raised by 500K
Es=2 * 105 N/mm2 ;s =11*10-6 /0K
Ec=1 * 105 N/mm2 ;c= 17*10-6 /0K

Solution
Copper bar Ac =0.785*302=706.9 mm2
steel bar As =0.785*(502- 302)=1257.1 mm2
[s /Es] +[ c/Ec] = (c - s)*t
[s / 2 * 105] +[ c/ 1 * 105] =(17-11)*10-6*50
s +2 c=60-----(1)
copper
steel
steel
10
30
10
10Ø Pin

Since no external force is present
sAs= cAc
s= cAc/As=[706.9/1257.1]*c
=0.562 c---(2)
substituting in eq.(1)
c=23.42 N/mm2
Hence force in between copper bar &steel tube
=cAc=23.42*706.9=16550N

C.S. area of pin = 0.785*102 =78.54 mm2
pin is in double shear
so shear stress in pin
=16550/(2*78.54)=105.4N/mm2
pin

SHRINKING ON:
d<D
D=diameter of wheel
d = diameter of steel tyre
increase in temp = toC
dia increases from d--->D
•tyre slipped on to wheel, temp. allowed to fall
•Steel tyre tries to come back to its
original position
•hoop stresses will be set up.
D
d

Tensile strain
 = (D - d) / d =(D-d)/d
so hoop stress = = E
= E*(D - d)/d

Example: 18
A thin steel tyre is to be shrunk onto a rigid wheel of
1m dia. If the hoop stress is to be limited to
100N/mm2, calculate the internal dia. of tyre. Find
also the least temp. to which the tyre must be heated
above that of the wheel before it could be slipped on.
Take  for the tyre = 12*10-6/oC
E =2.04 *105N/mm2

Solution:
= E*(D - d)/d
100 = 2.04*106(D - d)/d
or
(D - d)/d =4.9*10-4
or D/d =(1+4.9*10-4)
so d =0.99951D=0.99951*1000=999.51 mm

Now
D = d(1 + t)
or
t =(D/d)-1 = (D-d)/d =4.9*10 - 4
t =(D-d)/d *1/ 
=4.9*10-4/12*-6
=40.85 0 C

ELASTIC CONSTANTS:
Any direct stress produces a strain in its own
direction and opposite strain in every direction
at right angles to it.
Lateral strain /Longitudinal strain
= Constant
= 1/m = = Poisson’s ratio
Lateral strain = Poisson’s ratio x
Longitudinal strain
y =  x -------------(1)

Single direct stress along longitudinal axis
L
d b
x
x
x
y
x= x/E (tensile)
y=  x =  [x/E] (compressive)
Volume = L b d
V=bd L - d Lb - L bd
V/ V = L/L - b/b - d/d
= x - y - z = x-  x- x= x- 2 x= x(1-2 )

d
L
b
x
x
x
y
= x - y - z = x-  x- x= x- 2 x= x(1-2 )
= [x/E] x (1-2 )
Volumetric strain= v =[x/E] x (1-2 ) –
-----(2)
or v =[x/E] x (1-2/m)
v =[x/E] x (1-2/m)

Stress x along the axis and y and z
perpendicular to it.
x
z
y
x= x/E - y/mE - z/mE-----(i) -------(3)
y= y/E - z/mE - x/mE-----(ii)
z= z/E - x/mE - y/mE-----(iii)
Note:- If some of the stresses have opposite
sign necessary changes in algebraic signs of
the above expressions will have to be made.

Upper limit of Poisson’s Ratio:
adding (i),(ii) and (iii)
x+ y+ z=(1 - 2/m)(x+ y + z)/ E- -------(4)
known as DILATATION
For small strains represents the change in
volume /unit volume.

x y z
x x/E - x/E - x/E
y
x
x
y
y - y/E - y/E
y/E
z
z/E
- z/E - z/E
z
z
Sum all

Example: 19
A steel bar of size 20 mm x 10mm is subjected to a
pull of 20 kN in direction of its length. Find the
length of sides of the C.S. and decrease in C.S.
area. Take E=2 x 10 5 N/mm2 and m=10/3.

x= x/E= (P/Ax) x (1/E)
= (20000/(20x10)) x1/( 2 x105)=5 x 10 -4(T)
Lateral Strain =y=- x=-x/m =-1.5x10 -4(C)
side decreased by 20x1.5x10 -4=0.0030mm
side decreased by 10x1.5x10 -4=0.0015mm
new C.S=(20-0.003)(10-.0015)=199.94mm2
% decrease of area=(200-199.94)/200 x100
=0.03%

Example: 20
A steel bar 200x20x20 mm C.S. is subjected to a
tensile force of 40000N in the direction of its length.
Calculate the change in volume.
Take 1/m =0.3 and E = 2.05 *105 MPa.
Solution:
x= x/E= (P/A) x (1/E) =40000/20*20*2.05*105=
4.88*10-4
y= z=-(1/m)* x= -0.3* 4.88*10-4
= -1.464 *10-4

Change in volume:
V/ V= x + y+ z=(4.88 - 2*1.464)*10-4
=1.952 *10-4
V=200*20*20=80000 mm3
V=1.952*10-4*80000=15.62 mm3

YOUNG’S MODULUS (E):--
Young’s Modulus (E) is defined as the Ratio of
Stress () to strain ().
E =  /  -------------(5)

BULK MODULUS (K):--
• When a body is subjected to the identical stress  in
three mutually perpendicular directions, the body undergoes
uniform changes in three directions without the distortion of
the shape.
• The ratio of change in volume to original volume has
been defined as volumetric strain(v )
•Then the bulk modulus, K is defined as K=  / v





 
K=  / v
BULK MODULUS (K):--
Where, v = V/V
Change in volume
=
Original volume
Volumetric Strain
=
-------------(6)

MODULUS OF RIGIDITY (N): OR
MODULUS OF TRANSVERSE ELASTICITY OR
SHEARING MODULUS
Up to the elastic limit,
shear stress ()  shearing strain()
 = N 
Expresses relation between shear stress and shear strain.
/=N;
where
Modulus of Rigidity = N =  /  -------------(7)

YOUNG’S MODULUS E =  / 
K =  / v
BULK MODULUS
MODULUS OF RIGIDITY N =  / 
ELASTIC CONSTANTS
-------------(5)
-------------(6)
-------------(7)

COMPLEMENTRY STRESSES:“A stress in a given
direction cannot exist without a balancing shear
stress of equal intensity in a direction at right angles
to it.”
C
A
B
D
Moment of given couple=Force *Lever arm
= (.AB)*AD
Moment of balancing couple= (’.AD)*AB
so (.AB)*AD=(’.AD)*AB => = ’
Where =shear stress & ’=Complementary shear
stress


 
’
’

State of simple shear: Here no other stress is acting
- only simple shear.
Let side of square = b
length of diagonal AC =2 .b
consider unit thickness perpendicular to block.
 
’
’
A
B C
D



Equilibrium of piece ABC
the resolved sum of  perpendicular to the diagonal =
2*(*b*1)cos 450= 2 .b
if  is the tensile stress so produced on the diagonal
(AC*1)=2 .b
(2 .b)=2 .b
so
= 
 
’
’
A
B C
D



Similarly the intensity of compressive stress on
plane BD is numerically equal to .
“Hence a state of simple shear produces pure
tensile and compressive stresses across planes
inclined at 45 0 to those of pure shear, and
intensities of these direct stresses are each equal
to pure shear stress.”
 
’
’
A
B C
D
 

SHEAR STRAIN:
 


A
B C
D

 A
B
C
D
B’
C’
D’
/2
/2
B
A
C
B” C’’
 
D
State of simple
Shear on Block
Total
change in
corner
angles +/-

Distortion with
side AD fixed
F

Since
 is extremely small,
we can assume
BB” = arc with A as centre ,
AB as radius.
So, =BB”/AB=CC”/CD
Elongation of diagonal AC can be nearly taken as FC”.
Linear strain of diagonal = FC”/AC
= CC”cos 45/CDsec45
B
A
C
B” C’’
 
D
F

 = CC”/2CD = (1/2) 
but = /N (we know N= / )
so
 =  /2N ------(8)
Linear strain ‘’is half the shear strain ‘’.
B
A
C
B” C’’
 
D
F

RELATION BETWEEN ELASTIC CONSTANTS
(A) RELATION BETWEEN E and K
Let a cube having a side L be subjected to three
mutually perpendicular stresses of intensity 
By definition of bulk modulus
K= / v
Now v = v /V = /K ---------------------------(i)
x
z
y

The total linear strain for each side
 =/E -  /(mE) -  /(mE)
so L / L =  =(/E) *(1-2 /m)-------------(ii)
now V=L3
V = 3 L2 L
V/V = 3 L2 L/ L3= 3 L/L
= 3 (/E) * (1-2 /m) ------------------(iii)

Equating (i) and (iii)
/K = 3( /E)(1-2 /m)
E = 3 K(1-2 /m) -----(9)

(B) Relation between E and N
D
B
A
C
B” C’’
 
Linear strain of diagonal AC,
 = /2 = /2N --------------------------(i)
F




A
B C
D



State of simple shear produces tensile and
compressive stresses along diagonal
planes and
 = 
Strain  of diagonal AC, due to these two
mutually perpendicular direct stresses
 = /E - (- /mE) = (/E)*(1+1/m) ---(ii)
But  = 
so  = ( /E)*(1+1/m) ------------------(iii)

From equation (i) and (iii)
 /2N = ( /E)(1+1/m)
OR
E =2N(1+1/m)-------(10)
But E = 3 K (1-2 /m)------(9)
Eliminating E from --(9) & --(10)
 = 1/m = (3K - 2N) / (6K +2N)-----(11)
Eliminating m from –(9) & --(10)
E = 9KN / (N+3K) ---------(12)

(C) Relation between E ,K and N:--
 =1/m=(3K-2N)/(6K+2N)------(11)
E = 3K (1-2 /m) --------(9)
E = 9KN / (N+3K) -------(12)
E = 2N(1+1/m) -------(10)
(D) Relation between ,K and N:--

Example: 21
(a) Determine the % change in volume of a
steel bar of size 50 x 50 mm and 1 m long,
when subjected to an axial compressive
load of 20 kN.
(b) What change in volume would a 100 mm
cube of steel suffer at a depth of 5 km in sea
water?
Take E=2.05 x 10 5N/mm2 and
N = 0.82 x 10 5N/mm2

Solution: (a)
V/V = v = (/E)(1-2 /m)
[ = P/A = 20000/50 x 50 =8 kN/cm2]
so now
V/V=- (8 / 2.05 x 10 5 )(1 - 2/m)
= -3.902 *10 -5(1 - 2/m)----------------------(i)
Also E = 2N(1+1/m) -----------------------(10)
(1 +1/m)=E/2N =2.05 x 10 5 /(2 * 0.82 x 10 5 )
so 1/m =0.25

Substituting in ----(i)
V/V = -3.902*10 -5(1-2(0.25))=-1.951* 10 -5
Change in volume=-1.951*10-5 *1000*50*50
V = 48.775 mm2
% Change in volume=(48.775/ 50*50*1000)*100
=0.001951 %

Solution:(b)
Pressure in water at any depth ‘h’ is given by
p=wh taking w= 10080N/m3 for sea water
and h = 5km=5000m
p=10080*5000=50.4 *106N/m2 = 50.4N/mm2
E = 3K(1-2/m)

We have 1/m =0.25
so E = 3K(1-0.5) or K=E/1.5 = 2/3(E)
K=2/3 * 2.05* 10 5 =1.365 * 10 5 =N/mm2
now by definition of bulk modulus
K= /v or v = /K
but v = V/V
V/V = /K
V= 50.4 /1.365 * 10 5 * 100 3 =369.23 mm3

Example: 22 A bar 30 mm in diameter was
subjected to tensile load of 54 kN and
measured extension of 300 mm gauge length
was 0.112 mm and change in diameter was
0.00366 mm. Calculate Poisson’s Ratio and
the value of three moduli.
Solution:
Stress = 54 *103/(/4*d2) = 76.43 N/mm2
=Linear strain = L/L=0.112/300
=3.733*10-4

E=stress/strain =76.43/3.733* 10-4
=204741 N/mm2=204.7 kN/mm2
Lateral strain= d/d = 0.00366/30=1.22*10-4
But lateral strain =1/m* 
so 1.22*10-4=1/m *3.733*10-4
so 1/m=0.326
E=2N(1+1/m) or N=E/[2*(1+1/m)]
so N=204.7/[2*(1+0.326)]=77.2 kN/mm2

E = 3 K *(1-2 /m)
so K=E/[3*(1-2/m)]=204.7/[3*(1-2*0.326)]
K=196kN/mm2

Example: 23 Tensile stresses f1 and f2 act at right
angles to one another on a element of isotropic
elastic material. The strain in the direction of f1
is twice the direction of f2. If E for the material is
120 kN/mm3, find the ratio of f1:f2. Take
1/m=0.3
f2
f2
f1
f1
1 = 2 2
So ,f1/E –f2/mE =
2(f2/E –f1/mE)
f1/E +2f1/mE = 2f2/E +f2/mE

So
(f1/E)(1+2/m) =(f2/E)(2+1/m)
f1(1+2*0.3) =f2(2+0.3)
1.6f1=2.3f2
So f1:f2 = 1:1.4375

Example: 24 A rectangular block 250 mmx100
mmx80mm is subjected to axial loads as
follows.
480 kN (tensile in direction of its length)
900 kN ( tensile on 250mm x 80 mm faces)
1000kN (comp. on 250mm x100mm faces)
taking E=200 GN/m2 and 1/m=0.25 find
(1) Change in volume of the block
(2) Values of N and K for material of the block.

x =480x103/(0.1*0.08)=60 *106N/m2 (tens.)
y=1000x103/(0.25*0.1)=40*106N/m2(comp)
z=900x103/(0.25*0.08)=45*106N/m2(tens.)
x= (60 *106/E)+(0.25* 40*106/E)
- (0.25* 45*106/E)=(58.75* 106/E)
 y= -(40 *106/E)-(0.25* 45*106/E)
- (0.25* 60*106/E)=(- 66.25* 106/E)
z= (45 *106/E)-(0.25* 60*106/E)
+ (0.25* 40*106/E)=(40* 106/E)

Volumetric strain = v = x + y + z
=(58.75* 106/E)- (66.25* 106/E)+ (40* 106/E)
=32.5*106/E
v = V/V
so V= v V
=32.5*106*[(0.25*0.10*0.08)/(200*109)]*109
=325 mm3(increase)

Modulus of Rigidity
E = 2N(1+1/m)
so N=E/[2*(1+1/m)]=200/[2(1+0.25)]=80GN/m2
Bulk Modulus:
E = 3K(1-2/m)
so K=E/[3(1-2/m)]=200/[3(1-2*0.25)=133.33
GN/m2

Example: 25 For a given material
E=110GN/m2 and N=42 GN/M2. Find the bulk
modulus and lateral contraction of a round bar
of 37.5 mm diameter and 2.4 m long when
stretched by 2.5 mm.
Solution:
E=2N(1+1/m)
110*109=2*42*109(1+1/m)
gives 1/m =0.32

Now E = 3K(1-2/m)
110 x 109=3K(1-2*0.31)
gives K=96.77 GN/m2
Longitudinal strain =
L/L=0.0025/2.4=0.00104
Lateral strain=.00104*1/m=0.00104*0.31
=0.000323
Lateral Contraction=0.000323*37.5=0.0121mm

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