This document discusses key concepts in strength of materials including Young's modulus of elasticity, tensile and longitudinal forces, stress-strain graphs, proportional limit, elastic limit, yield point, plastic behavior, energy stored in strained materials, types of stress and strain, and moduli of elasticity including Young's modulus, bulk modulus, and shear modulus. It provides examples of calculating stresses, strains, energy storage, and the velocity of sound using these mechanical properties.

1. STRENGTH OF MATERIALS
Young’s Modules of Elasticity Y=
Y = =
A tensile or longitudinal force is a force which stretches or increases the length of a material
Fig 1: The graph of extension against weight or Force.
Key of the graphs
A = Proportionality limit
B = Elastic limit
C = Yield point
D = The material becomes thin and breaks breaking point

2. é = Permanent extension
Fig 2: The graph of stress against strain
S = Permanent strain.
0A = Hooke’s law obeyed.
0B = Elastic deformation region.
0C = Region of plastic deformation.
Features of the graphs
· The proportionality limit is the maximum stress beyond which Hooke’s law is not obeyed or
the ratio of stress/strain is no longer constant
· 0A: Hooke’s law is obeyed i.e. F = KX.
Stress α Strain i.e. stress/strain = constant.
· Elastic limit is the maximum stress before which the material returns to its original
shape and size.
· The yield point is the point beyond which the material changes from elastic behavior to
plastic behavior.
· Plastic behavior is a situation where a force may cause a large increase in length
unexpectedly and it will not return to its original shape or size.

3. · Energy stored in a strained wire suppose a wire of length is stretched by a force F.
Work done = Average force x distance
= Fe in Joules as SI unit
This is the elastic energy stored in the wire
Energy = Fe
But, Y = = Y =
F =
Energy = Fe = . e
Energy =
The energy per unit volume of a wire
=
= = Y
= ( . ). = .
= x Stress x strain
Energy per unit volume = x Stress x Strain

4. Fig 3: Graph of F Vs ℯ and Energy if elastic limit is not exceed
Energy stored = work done = F x x
Total work done = Area under the curve
= x 0A x AB
= x e x F
Energy ; If Hooke’s law is obeyed
This agrees with the earlier derived equation.
Plastic behavior is exhibited when the material is permanently strained after the force is
removed. The energy is converted to heat.
Example 1: Two vertical wires X and Y suspend at the same horizontal level are converted by a
light rod XY at the lower ends as shown. The wires have the same length l and cross-sectional
area A. A weight of 30N is placed at O on the rod, Where . Both wires are
stretched and the rod XY then remains horizontal

5. If wire X has young’s modulus 1.0 x 10N/M2 calculate young’s modulus for wire Y assuming
that the elastic limit is not exceed for both wires (N&P).
Solution
Since XY is horizontal = = e
Let and be the forces in X and Y
Taking moments
About point X: 30 x 1 = x 3
= = 10N
About point Y: 30 x 2 = x 3
= = 20N
For wire X, = =
=

6. For wire Y: = =
= = =
20 = 1.0 10
=
= 5 x NM-2
Example 2:
A rubber cord of a catapult has a cross-sectional area of 2m and an initial length of 0.20m
and is stretched to 0.24m to fire a small stone of 10g.Calculate the initial velocity of the object
when it just leaves the catapult(Assume Young’s modulus for rubber is b x pa and the
elastic limit is not exceeded).
Solution
K.E of the fire stone =Energy stored in stretched rubber
m = Fe
V =
Where F is given by Y= F is given by Y = .
F =
F =
F = = 240N

7. V = =
V = = 31.0m
Applications of elasticity
1. In the design of structures the engineer has to ensure that the stresses to be applied to the
structure do not exceed the elastic limit of the structure
2. Steel has very high strength (high modulus of elasticity) so it is used in construction of girders
spring etc. It can withstand very high stresses
3. In construction I and H cross section beams are made using steel bars in such a way that the
beams can withstand large stresses
Example 3:
A steel wire of radius 5 is fixed at the ceiling.Find the maximum weight which can be
hung from it, if it is allowed strain is and young’s modulus is 2.0 x pa
From Y=
Where Y = 2.0 x pa, = and
A = π = π (5 )2 = 2.5π
F = Y x x A

8. F=2 x x
F = 15.71 x 10
F = 157.1N
This force corresponds to the maximum stress of
σ = =
= 2 x pa or N/m-2
Example 4:
The breaking stress of steel is 8.0 x Nm-2. Find the greatest length of the steel wire that can
hang vertically without breaking if its density 9.9 x kg
The wire can break under its weight
Breaking weight of the wire
W = A
= A
= A 9.9 9.8
= 9.702 x A
Breaking force of wire:
Stress =
F = Stress x A
F = 8 x x A
Now, W = F
9.702 X A = 8 x A

9. =
= 0.8245723 x
= 8.246 x 104m
Energy stored in a strained wire = to the area under the graph of force F against the extension X
Area of F the element
A =
= =
= k = Fx
Now from Y = Y =
= k

10. TYPES OF STRESS
Stress =
SI unit of stress is Nm-2
Dimensions of stress
There are two main types of stress
i. Normal stress is the one is which force (F) is applied normally to the area
(A) and is defined as the normal force applied per unit area.
i.e. Normal Stress = =
Normal stress is of two kinds, namely
a) Normal tensile stress =
This stress causes an increase in length
DRAWING
b) Normal compressive stress =
This stress causes a decrease in length
Drawing

11. ii. The Tangential stress is one due to the force applied parallel to the surface on a
body which is equal to tangential force/parallel to force.
TYPES OF STRAIN
Strain = is unit less and dimensionless
There are three types of strain, namely Longitudinal, Volumetric and shearing strain
i) LONGITUDINAL STRAIN = =
ii) BULK STRAIN is the ratio of the change in volume per unit original volume i.e
Bulk Strain =
iii) SHEARING STRAIN is change in the shape of the body caused by a tangential force

12. Definition:
Shearing strain = = = =
= Angle of shear is the angle through which a vertical line is turned by a tangential force
TYPES OF MODULI OF ELASTICITY
There are three types of modulus of elasticity corresponding the three types of strain namely
young’s modules and shear modulus or modulus of rigidity
I. Young’s Modulus of Elasticity
Y =
Y = = Y =
ii. The Bulk modulus of Elasticity K is defined as

13. K =
=
Now, = change in pressure and Volume decreases
K = - in N
Or K = -V in N
Examples of moduli of elasticity
For steel K = 160 x N
For water K = 2.2 x N
For air at normal pressure K = N
High value of K indicates the difficulty to change the volume.
COMPRESSIBILITY is a measure of how hard or how easy it is to compress the material.
Compressibility is the reciprocal of the modulus of elasticity.
Example 1:
A spherical ball is subjected to a normal uniform pressure of N/ . If the bulk modulus of
the material of the ball is N , find the volume strain it suffers.
Solution:

14. = N , K = N
Using K = -v K =
=
= =
Example 2
The density of ocean water at the surface is 1030 kg/ . The bulk modulus water is 2.0 x N
. If the atmospheric pressure is N ,what is the density of the ocean water at a depth
where the pressure is 500 Atmospheres.By what % age is the water compressed?
Solution
Density =
=
V’ = final volume
The mass of water is the same 1030kg
From K= -V
V = 1 at the surface
2.0 x = - . (500 – 1) x
2.0 x =
= -
= -2.495 x - 0.025

15. V’ = V - V = (1-0.02495)
= 0.97505
Density at 500 Atm;
=
ρ= 1056 kg
% compression: x 100% = x 100%
= 2.5%
iii. SHEAR MODULUS OR MODULUS OF RIGIDITY
It is defined as the ratio of the tangential stress applied to the shear strain
i.e. Modulus of Rigidity =
η=
or η= where is in Rads
The SI unit of η is N or pa
Young’s modulus for steel = 20 x N
Modulus of rigidity for steel = 8 x N
Generally the modulus of rigidity for solids is much smaller than young’s modulus of elasticity.

16. The bulk modulus of elasticity of a gas if the change in the gas is isothermal.
PV= constant
Differentiating with respect to V gives,
P + V = 0
-V = + p
But -V = k, the bulk modulus
k =
The isothermal Bulk modulus of a gas is equal to the pressure of the gas.
The Bulk modulus of elasticity of a gas if the change is adiabatic
= Constant, k
Differentiating with respect to V
. + = 0
+ . = 0
-
The adiabatic modulus of elasticity
THE VELOCITY OF SOUND

17. The velocity of sound depends on the density and the modulus of elasticity. By dimensional
analysis
V = k
=
Equating indices gives
For T -1 = -2b b =
For M 0 = a + b a =-b = a = -
For L 1 = -3a – b RHS = -3 x (- ) - = 1 - = 1 = LHS
V = k
Mathematical analysis shows that k=1
V =
Now in gases sound propagates under adiabatic conditions where E = , the bulk modulus =
P
Velocity of sound in air is given by
V =