Note Chap 11 SP015 (Deformation of solids) Ng Soon Lai

Chapter 11 Physics

CHAPTER 11: 1
Deformation of solids

(F2F:4.5 Hours)

Chapter 11 Physics

Overview:

Deformation of solids

Stress Strain Stress-strain Young’s
graph modulus

Elastic Ductile
material material

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Chapter 11 Physics

Learning Outcome:

11.1 Stress and strain (2 hours)

At the end of this chapter, students should be able to:
 Distinguish between stress and strain for tensile and

compression force.

 Discuss the graph of stress-strain,− for a metal under

tension.
 Discuss elastic and plastic deformations.

 Discuss graph of force-elongation, F−e for brittle and

ductile materials.

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Chapter 11 Physics

11.1 Stress and strain

11.1.1 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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Chapter 11 Physics

⚫ 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 11.1 shows the examples of elastic and inelastic

materials.

Elastic material Inelastic material

Sponge Paper

Spring Metal

Rubber Plasticine

Table 11.1

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Chapter 11 Physics
11.1.2 Stress,  and strain, 

Tensile 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 11.1a.

l0 A

F⊥ F⊥

eFigure 11.1a

 Stress is defined as the ratio of the perpendicular force, F⊥ to the

cross-sectional area, A.

OR Tensile stress,  = F⊥

A

Where

F⊥ : the force act perpendicu lar to the cross section
A : cross - sectional area
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Chapter 11 Physics

 Stress is a scalar quantity and its unit is given by

=F unit of  = kg m s−2

A = kg m 2 −2

m −1s

OR N m−2 OR pascal (Pa)

 Strain ( ) is defined as the ratio of extension (elongation), e to

original length, l0 .

OR Where
Tensile strain,  = e = l − l0
e : extension (elongatio n)
l0 l0
l : final length

l0 : original (initial) length

 Strain is a scalar quantity without unit.

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Chapter 11 Physics

Compressive stress and strain

 Consider a rod that initially has uniform cross-sectional area, A and length

l0. Compress the rod by applying the forces of equal magnitude F⊥ but

opposite directions at the both ends and the length of the rod will decrease

by amount e as shown in Figure 11.1b.

l0 A

F⊥ F⊥

Figure 11.1b e

Compressive stress,  = F⊥

A

Compressive strain,  = e = l0 − l 8

l0 l0
Where e : compression

Chapter 11 Physics

11.1.3 Force-extension and stress-strain graphs

Graphs for metal (ductile material)

Force, F Plastic Extension, e

Elastic deformation E
deformation
D D
E
Plastic
BC deformation

A

T C

OT Extension, e O B

Figure 11.2a A Elastic

deformation

Force, F

Figure 11.2b

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Chapter 11 Physics

S tress ,  Plastic
deformation
Elastic
deformation D
E

BC

A

OT Figure 11.2c Strain,  Description

A : proportionality limit

B : elastic limit

C : yield point

D : point of maximum force (stress)

E : fracture (breaking) point 10

Chapter 11 Physics

 Explanation for Figures 11.2a, 11.2b and 11.2c

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

The negative sign indicates that the restoring force is the
opposite direction to increasing extension.

B - 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. 11

Chapter 11 Physics

C - 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.

CDE - This region is known as plastic deformation.

- When the force (stress) increases, the extension
(strain) increases rapidly.

D - 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).

E - This is the point where the material breaks or fractures.

 Ductile materials - undergo plastic deformation before
 Brittle materials breaking.

- such as steel, copper, aluminium.

- do not show plastic behaviour
(deformation).

- such as glass. 12

Chapter 11 Physics

 Figure 11.3 shows the stress-strain graphs for various materials.

S tress , 

Glass Steel

Copper
Aluminium

O Strain, ε

Figure 11.3

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Chapter 11 Physics

Learning Outcome:

11.2 Young’s modulus (2.5 hours)

At the end of this chapter, students should be able to:
 Define Young’s modulus.

 Discuss strain energy from the force− elongation, F−e

graph.
 Discuss strain energy per unit volume from stress−strain,

− graph.

 Solve problems related to the Young’s modulus.

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Chapter 11 Physics

11.2 Young modulus

11.2.1 Young modulus, Y

 is defined as the ratio of the tensile stress to the tensile
strain if the proportionality limit has not been exceeded.

OR Y = Tensile stress
Tensile strain

 F⊥  Y = F⊥l0
Y= A Ae
 e 
l0

 It is a scalar quantity.

 Its unit is kg m−1 s−2 or N m−2 or Pa.

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Chapter 11 Physics

 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 11.2 shows the value of Young modulus for various
material.

Material Y (GPa)

Aluminium 69

Copper 110

Steel 200

Nylon 3.7

Glass 70

Table 11.2

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Chapter 11 Physics

11.2.2 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 F = k e
and F = YAe
l0
ke = YAe
l0

k = YA
l0

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Chapter 11 Physics

11.2.3 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 11.5.
Force

F Proportionality limit

Strain energy

e0 extension
Figure 11.5
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Chapter 11 Physics

 The total work done, W in stretching a wire from 0 to e is given

by
e

W = Fde = Shaded Area
0

W = strain energy,U = 1 Fe
2

 From the definitions of tensile stress and tensile strain, thus

stress = F F = (stress ) A
= eA
strain l0 = e = (strain )l0 ) Al0 Volume

strain energy 1 (stress )(strain
2

strain energy = 1 (stress )(strain )
volume 2

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Chapter 11 Physics

 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 11.6.
Stress

 Proportionality limit

Strain energy per
unit volume

0 Strain

Figure 11.6

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Chapter 11 Physics

Example 11.1 :

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)

Solution : m = 3 kg; l0 = 1.5 m; d = 0.5 10−3 m
Y = 2 .0  1011 N m −2

a. The cross sectional area of the wire is

( )A = d 2 A =  0.5 10−3 = 1.96 10−7 m2
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The applied force to the wire is given by

F =W = mg

= 3(9.81)

F = 29.4 N 21

Chapter 11 Physics

Solution : m = 3 kg; l0 = 1.5 m; d = 0.5 10−3 m
Y = 2 .0  1011 N m −2
a. By applying the Young’s modulus formulae, hence

Y = Fl0
Ae
(29.4)(1.5)

1.96 10−7 e
( )2.0 1011 =

e = 1.1310−3 m

b. By applying the equation of the strain energy, thus

Energy stored = 1 Fe
2

( )= 1 (29.4) 1.1310−3
2

Energy stored = 0.0166 J

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Chapter 11 Physics

Example 11.2 :

A copper wire LM is fused at one end, M to an iron wire MN as

shown in figure below. 
LM NF

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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Chapter 11 Physics

Solution :l0C = 0.90 m; AC = 0.90 10−6 m 2 ; l0I = 1.40 m;
AI = 1.30 10−6 m 2 ; e total = 0.01 m;

YYC==F1l.030thu1s0e11 P aF; lY0I = 2.10 1011 Pa;
=
a. Apply : Ae YA

Hence the ratio of extensions is given by

FC l0C

eC = YC AC and FC = FI = F
eI FI l0I

YI AI ( )( )eC (0.90)
eC = YI AI l0C ( )( )eeCI (1.40)
eI YC ACl0I = 2.10 1011 1.30 10−6
1.30 1011 0.90 10−6

= 1.5
eI 24

Chapter 11 Physics

Solution :
b. The total extension of the compound wire is

etotal = eC + eI and eC = 1.5eI

Hence the extension of the iron wire is

0.01 = 1.5eI + eI
eI = 4.0 10−3 m
and the entension of the copper wire is
( )eC

eC
= 1.5 4.0 10−3
= 6.0 10−3 m

c. The applied force is given by

Fl0C F (0.90)
ACeC
0.90 10−6 6.0 10−3
YC = ( ) ( )( )1.301011=

F = 780 N

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Chapter 11 Physics

Exercise 11.1 :

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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Chapter 11 Physics

THE END.

Next Chapter…

CHAPTER 12 :
Heat conduction and

thermal expansion

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