PHYSICS
CHAPTER 12
Matter is made up of
many particles called
atoms or molecules.
 Matter can exist in one
of the following states i.e.
solid, liquid and gas.

Introduction
CHAPTER 12:
Mechanical properties of matter
(2 Hours)
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PHYSICS
CHAPTER 12
Learning Outcome:
12.1 Stress and strain (1 hour)
At the end of this chapter, students should be able to:




Define stress and strain for a stretched wire.
Sketch and explain the graph of stress-strain.
Distinguish between elastic and plastic deformation.
Sketch and distinguish F-e graph for elastic and ductile
materials.
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PHYSICS
CHAPTER 12
12.1 Stress and Strain
12.1.1 Elasticity of solids





is defined as the property of a material that enables them to
return to their original dimensions (shape and size) after an
applied force (stress) has been removed.
Deformation occurs when external forces act to stretch,
compress or shear a solid.
An object or a material which returns to its original length or
size after being distorted suffers elastic deformation, i.e. it
behaves within the elastic limit.
Plastic deformation occurs when a material is deformed
beyond its elastic limit.
The behaviour of materials under deformation can be described
by the following mechanical properties :
 Strength – ability of a material to withstand a force
without breaking.
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PHYSICS
CHAPTER 12
Stiffness – resistance of a material to changes in
shape and size.
 Elasticity
 Ductility – tendency of a material to change its size
and shape considerably before breaking.
 Brittleness – tendency of a material to break without
deforming.
Table 12.1 shows the examples of elastic and inelastic
materials.


Elastic material
Inelastic material
Sponge
Paper
Spring
Metal
Rubber
Plasticine
Table 12.1
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PHYSICS
12.1.2

CHAPTER 12
Stress,  and strain, 
Consider a rod that initially has uniform cross-sectional area, A
and length l0.Stretch the rod by applying the forces of equal
magnitude F but opposite directions at the both ends and the
rod will extent by amount e as shown in Figure 12.8.
l0
A
F
F
Figure 12.8

Simulation 12.1
e
Stress is defined as the ratio of the perpendicular force, F to
the cross-sectional area, A.
OR
F
stress,  

A
Where
F : the force act perpendicu lar to the cross section
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A : cross - sectional area
PHYSICS
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


CHAPTER 12
This type of stress is called tensile stress.
Stress is a scalar quantity.
The unit for stress is kg m1 s2 or N m2 or pascal (Pa).
Strain ( ) is defined as the ratio of extension (elongation), e
to original length, l0 .
Where
OR
Strain,  
e
l0



l  l0 e : extension (elongation)
l0
l : final length
l0 : original (initial) length
This type of strain is called tensile strain.
Strain is a scalar quantity and dimensionless (no unit).
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PHYSICS
CHAPTER 12
12.1.3 Force-extension and stress-strain graphs
Graphs for metal (ductile material)
Plastic
deformation
Force, F
Elastic
deformation
B
Extension, e
D
E
E
D
Plastic
deformation
C
A
A
T
OT
Extension, e O
Figure 12.8a
C
B
Elastic
deformation
Force, F
Figure 12.8b
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PHYSICS
CHAPTER 12
Stress , 
Elastic
deformation
B
Plastic
deformation
D
E
C
A
OT
Description
Figure 12.8c Strain, 
A : proportionality limit
B : elastic limit
C : yield point
D : point of maximum force (stress)
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E : fracture (breaking) point
PHYSICS

CHAPTER 12
Explanation for Figures 12.8a, 12.8b and 12.8c
OA - The force (stress) increases linearly with the extension
(strain) until point A. Point A is the proportionality limit.
- The straight line graph (OA) obeys Hooke’s law which
states “Below the proportionality limit, the restoring
force, Fs is directly proportional to the extension, e.”
Fs  ke where k : force (Hooke) constant
B
The negative sign indicates that the restoring force is the
opposite direction to increasing extension.
- This is the elastic limit of the material.
- Beyond this point, the material is permanently stretched
and will never regain its original shape and length. If the
force (stress) is removed, the material has a permanent
extension of OT.
- The area between the two parallel line (AO and CT)
represents the work done to produce the permanent
extension OT.
- OB region is known as elastic deformation.
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PHYSICS
C
CHAPTER 12
- The yield point marked a change in the internal structure
of the material.
- The plane (layer) of the atoms slide across each other
resulting in a sudden increase in extension and the
material thins uniformly.
oThis is because of the molecular structure is distorted.
oThe plane (layer) of the atoms slide across each other resulting in a
sudden increase in elongation and the material thins uniformly.
oYielding is usually accompanied by ‘thinning’ of the material.
before yield point
at yield point
Thickness of wire
decreased
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PHYSICS
CDE
D
E
CHAPTER 12
- This region is known as plastic deformation.
- When the force (stress) increases, the extension
(strain) increases rapidly.
- The force (stress) on the material is maximum and is
known as the breaking force (stress). This is
sometimes called the Ultimate Tensile Strength (UTS).
- This is the point where the material breaks or fractures.
Ductile materials
Brittle materials
- undergo plastic deformation before breaking.
- such as steel, copper, aluminium.
- do not show plastic behaviour (deformation).
- such as glass.
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PHYSICS

CHAPTER 12
Figure 12.9 shows the stress-strain graphs for various materials.
Stress , 
Steel
Glass
Copper
Aluminium
O
Strain, ε
Figure 12.9
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PHYSICS
CHAPTER 12
Graph for rubber (elastic material)
Stress , 
X
1
increasing stress
W
decreasing stress
Y
O
Figure 12.10
ε1 ε 2 Strain, ε
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PHYSICS

CHAPTER 12
Explanation for Figure 12.10
 Rubber undergoes elastic deformation.
 It is able to regain its original shape and length when the
stress is removed but does not obey Hooke’s law.

The strain produced when decreasing the stress (XY) is
greater than the strain produced when increasing the stress
(WX) as shown in Figure 12.10.

The shaded area is called the hysteresis loop and it
represents the energy loss per unit volume.
This energy lost in the form of heat dissipation.

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PHYSICS
CHAPTER 12
Learning Outcome:
12.2 Young modulus (1 hour)
At the end of this chapter, students should be able to:



Define and use Young’s modulus formulae.
Explain relationship between Young’s modulus and
Hooke’s law.
Derive and use strain energy,
1
U 
Fe
2

Deduce strain energy from the Fe graph and the
stressstrain graph.
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PHYSICS
CHAPTER 12
12.2 Young’s Modulus

is defined as the ratio of the tensile stress to the tensile
strain if the proportionality limit has not been exceeded.
OR
Tensile stress
Y 
Tensile strain
Y 
 F 


 A 
e

l
 0




Y 
F l0
Ae

It is a scalar quantity.

The unit of Young modulus is kg m1 s2 or N m2 or Pa.
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PHYSICS


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CHAPTER 12
Young’s modulus does not depend to the length of the wire
but it depend to the material made the wire.
Young’s modulus does not change if the length of the wire is
increase or decrease.
Table 12.2 shows the value of Young modulus for various
material.
Material
Y (GPa)
Aluminium
69
Copper
110
Steel
200
Nylon
3.7
Glass
70
Table 12.2
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PHYSICS
CHAPTER 12
Relationship between force constant, k and Young modulus,
Y for a wire

From the statement of Hooke’s law and definition of Young
modulus, thus
and
F  ke
YAe
F
l0
YAe
ke 
l0
k 
YA
l0
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PHYSICS
CHAPTER 12
12.2.5 Strain energy


When a wire is stretched by a load (force), work is done on the
wire and strain (elastic potential) energy is stored within.
Consider the force-extension graph of this wire until the
proportionality limit ( Hooke’s law) as shown in Figure 12.11.
Force
Proportionality limit
F
Strain energy
0
Figure 12.11
e
extension
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PHYSICS

CHAPTER 12
The total work done, W in stretching a wire from 0 to e is given
e
by
W   Fde  Shaded Area
0
W  strain energy 
1
Fe
2

From the definitions of tensile stress and tensile strain, thus
F
stress 
F  (stress ) A
A
e
e  (strain )l0
strain 
l0
strain energy 
strain energy
volume

1
2
1
(stress )(strain ) Al0
Volume
(stress )(strain )
2
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PHYSICS

CHAPTER 12
This strain energy per unit volume is the area under the
stress-strain graph until the proportionality limit (straight line
graph) as shown in Figure 12.12.
Stress

Proportionality limit
Strain energy per
unit volume
0

Strain
Figure 12.12
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PHYSICS
CHAPTER 12
Example 12.3 :
A thin steel wire initially 1.5 m long and of diameter 0.50 mm is
suspended from a rigid support. A mass of 3 kg is attached to the
lower end of the wire. Calculate
a. the extension of the wire,
b. the energy stored in the wire.
(Young’s modulus for steel = 2.0  1011 N m2)
3
Solution : m  3 kg; l0  1.5 m; d  0.5 10 m
Y  2.0 10 N m
11
2
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PHYSICS
CHAPTER 12
Solution : m  3 kg; l0  1.5 m; d  0.5 10
Y  2.0 10 N m
11
2
3
m
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PHYSICS
CHAPTER 12
Example 12.4 :
A copper wire LM is fused at one end, M to an iron wire MN as
shown in figure below.
L
M
N

F
The copper wire has length 0.90 m and cross-section area
0.90  106 m2. The iron has length 1.40 m and cross-section
area 1.30  106 m2. The compound wire is stretched and the total
length increases by 0.01 m. Determine
a. the ratio of the extension of copper wire to the extension of iron
wire,
b. the extension of each wire,
c. the applied force to the compound wire.
(Given Y iron = 2.10  1011 Pa ,Y copper = 1.30  1011 Pa )
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PHYSICS
CHAPTER 12
Solution :l0C  0.90 m; AC  0.90 10 6 m 2 ; l0I  1.40 m;
AI  1.30 10
6
m ; e total  0.01 m;
2
YC  1.30 10 Pa; YI  2.10 10 Pa;
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PHYSICS
Solution :
CHAPTER 12
PHYSICS
CHAPTER 12
Exercise 12.1 :
1. A support cable on a bridge has an area of cross-section of
0.0085 m2 and a length of 35 m. It is made of high tensile steel
whose Young’s modulus is 2.8 1011 Pa. The tension in the
cable is 720 kN.
Calculate
a. the extension of the cable.
b. the strain energy stored in the cable.
ANS. : 0.0106 m; 3.81 kJ
2. A wire of length 0.50 m is fixed horizontally between two
supports separated by 0.50 m. When a mass of 8.0 kg hangs
from the middle of the wire, the mid-point sags by 1.00 cm. The
diameter of the wire is 2.8 mm. Calculate the Young’s modulus
of the wire.
(Given g = 9.81 m s2)
ANS. : 1.99  1011 Pa
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PHYSICS
CHAPTER 12
THE END…
Next Chapter…
CHAPTER 13 :
Heat
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