CHAPTER 8: PHYSICS OF MATTERS

CHAPTER 8

PHYSICS
OF MATTERS

KMM

CHAPTER 8.0 : PHYSICS OF MATTERS

8.1 Stress and strain
8.2 Young’s Modulus
8.3 Heat Conduction
8.4 Thermal Expansion

8.1 Stress and Strain

LEARNING OUTCOME:
At the end of this chapter, students should be able to:
a) Distinguish between stress,   F and strain,   L for tensile and compression force.

A Lo
b) Analyse the graph of stress-strain, σ − ε for a metal under tension.

c) Explain elastic and plastic deformations.
d) Analyse graph of force–elongation, F−Δ for brittle and ductile materials.

a) Distinguish between stress,   F and strain,   L for tensile and compression force.

A Lo

Tensile stress Compressive stress

• Consider a rod that initially • Consider a rod that initially

has uniform cross-sectional has uniform cross-sectional

area, A and length L0. area, A and length L0.
Stretch the rod by applying Compress the rod by applying

the forces of equal the forces of equal

magnitude F but opposite magnitude F but opposite
directions at the both ends directions at the both ends

and the length of the rod will and the length of the rod will

increase (stretch) by decrease (compress) by

amount L as shown in figure. amount L as shown in figure.

 Tensile stress is defined as the ratio of  Compressive stress is defined as the ratio
the perpendicular tensile force, F to of the perpendicular compression force,
the cross-sectional area, A. F to the cross-sectional area, A.

Tensile stress,  F Compressive stress,  F

A A

F : the force acts perpendicular to F : the force acts perpendicular to
the cross section area. the cross section area.

A: cross - sectional area A: cross - sectional area
 Unit: newton per square meter, (N m-2)
 Unit: newton per square meter, (N m-2)
or Pascal (Pa) or kg m-1 s-2 or Pascal (Pa) or kg m-1 s-2
 Type of quantity: Scalar quantity
 It occurs when equal and opposite forces  Type of quantity: Scalar quantity

are directed away from each other  It occurs when equal and opposite
(tension or stretching force)and causes forces are directed toward each other
the object to elongate. (squeezing force) and causes the object
to compress.

 Tensile strain is defined as the ratio of  Compressive strain is defined as the ratio

the increase in length, L to original of the decrease in length, L to original

length, Lo . length, Lo .

Tensile strain,  L  L  L0 Compressive strain,  L  L  L0

L0 L0 L0 L0

L : the increase in length (elongation) L : the decrease in length (contraction)
L : final length L : final length
L0 : original (initial) length L0 : original (initial) length

 L is larger than original length, Lo  L is smaller than original length, Lo
 ε will become positive value  ε will become negative value

 Unit: Unitless  Unit: Unitless
 Type of quantity: Scalar quantity  Type of quantity: Scalar quantity

*Strain also called the fractional change in length

BENDING is a combination of tension and compression.

b) Analyse the graph of stress-strain, σ − ε for a metal under tension.
E

A graph of stress against strain for metal under tension

OA

 The graph is a straight line, where stress is directly proportional to the strain
(extension is directly proportional to the force

 Along the region OA, wire obeys Hooke Law. (F = kL)

 When the stress is removed, the wire returns to its original length.
 Wire experiences elastic deformation.
 Wire shows elastic behaviour.
 Point A is called proportionality limit.
 Beyond point A, stress no longer directly proportional to the strain.

Point B

 The point is called an elastic limit.
 Along path OAB when the stress is removed, the wire returns to its original length
 Wire experiences elastic deformation.
 Wire shows elastic behaviour.
 Beyond point B, the wire is permanently stretched and will never regain its original

shape or length.

Point C
 The point is called the yield point.
 Plastic deformation starts.
 The planes of the atoms slide across each other resulting in a sudden increase in elongation

and the wire thins uniformly.
 The wire has a permanent extension OP when the force is removed.
 When the force is further increased, extension increases rapidly.

CD
 The wire experiences plastic deformation.
 When the force or stress is removed, the wire will not return to its original shape

and length.

Point D
 The point is a called maximum force (stress) or breaking stress.
 After point D the wire thins and “necks” are formed.

DE
 The diameter of the wire becomes extremely small and uneven.
 The wire experiences plastic deformation.

Point E
 This point is called breaking point.
 The wire brakes or fractures.

c) Explain elastic and plastic deformations.
 A deformation is a change in the size or shape of an object.

 An elastic deformation is a change in an object’s shape or size in which the
object will return to its original size when the stress/force is removed.

 A plastic deformation is a change in an object’s shape or size in which the object
will not return to its original size when the stress/force is removed.

 The comparison between elastic and plastic deformations.

Elastic Deformation Plastic Deformation
A temporary change in an object’s shape A permanent change in an object’s shape
and size. and size.
The object will return to its original size The object will not return to its original
when the stress/force is removed. size when the stress/force is removed.
It behaves within the elastic limit. Occurs when the object is deformed
beyond its elastic limit.
It does obey Hooke’ law It does not obey Hooke’s law
The energy stored during elastic The energy is converted into heat during
deformation is fully recovered when the plastic deformation.
force is removed.
Occurs in region OB from the stress-strain Occurs in region CE from the stress-strain
graph. graph.

d) Analyse graph of force–elongation, F−Δ for brittle and ductile materials.

Force, F
Force, F
Y : Yield point
U : Ultimate strength
F : Fracture point

Elongation, L Elongation, L

Figure A Figure B

A graph of force against elongation A graph of force against elongation
for brittle material for ductile material

Brittle material (Figure A)

Force, F - can withstand little or no plastic elongation.
- does not experience plastic deformation.
Elongation, L - only experiences elastic deformation.
- the material breaks just after the elastic limit is
Figure A reached (the yield point).
- the ultimate strength is coincident with the fracture point.
- no necking occurs.
- example material : Glass

Force, F Figure B

Ductile material (Figure B)

Elongation, L

- can withstand large elongation even after it has begun to yield (there is a long

region between the yield at point Y and the ultimate strength at point U)

- undergoes both elastic and plastic deformations.

- there is a long region between the ultimate strength at point U and the fracture

point F in which the cross sectional area of the material is decreasing rapidly

and necking is occurring.

- the material breaks after experiencing plastic deformation.

- example material : Steel, Copper, Aluminum.

Example 8.1.1

A steel bar of rectangular cross-section, 3 cm by 2 cm, carries an axial load of 30 kN. If the
length of the bar is initial 20 cm and stretched by 0.01 cm, calculate the tensile stress and
strain.

30 kN 3 cm 30 kN

2 cm

Solution 8.1.1 20 cm

Tensile stress,   F  30 103  5.0107 N m2
A
0.02 0.03

Tensile strain,  L  0.01  5.0104

L0 20

Example 8.1.2

A cylindrical block is 30 cm long and has a circular cross-section 10 cm in diameter. It carries

a total compressive load of 70 kN, and under this load it contracts by 0.02 cm. Calculate the

compressive stress and compressive strain. F = 70 kN
d =10 cm

Solution 8.1.1

Compressive stress, l = 30 cm

  F F  70 103  8.9 106 N m2
A
 d2  0.102 

4 4

Compressive strain,  L   0.02  6.7 104 F = 70 kN

L0 30

8.2 Young’s Modulus

LEARNING OUTCOME:

At the end of this chapter, students should be able to:

a) Define and use Young’s Modulus, Y   .


b) Apply strain energy, U  1 FL from force-elongation graph.
2

c) Apply strain energy per unit volume, U  1  from stress-strain graph.

V2

8.2 Young’s Modulus

Young’s Modulus is defined as the ratio of the tensile stress to the tensile strain if the

proportionality limit has not been exceeded.

Y  tensile stress Y 
tensile strain 

F

Y  A  Y  F Lo
L
AL
Lo

 Unit: newton per square meter, (N m-2) or Pascal (Pa) or kg m-1 s-2

 Type of quantity: Scalar
 Young’s Modulus is a property of a material (it measure the resistance of the material to

elongation or compression) and is independent of the object’s size or shape.

Strain Energy

 The work done in stretching the wire is Force Plastic
stored as potential energy (U) in the wire deformation
and is called as strain energy in the wire. Elastic
deformation D

 The strain energy can be obtained from F A
area under the force-elongation graph of
this wire until the proportionality limit proportionality limit
(Hooke’s law) as shown in figure.
L Elongation

Area under graph

Strain energy,U  area under force  elongation graph

 1 FL

2

Where
force constant, k = gradient of F-L graph

 Unit: joules (J)
 Type of quantity: Scalar quantity

Strain Energy Per Unit Volume

Strain energy per unit volume can be obtained stress Plastic
from the area under the stress-strain graph deformation

until the proportionality limit (straight line graph) Elastic D

as shown in figure. deformation

Strain energy  U  1   A

volume V 2 proportionality limit
 area under stress  strain graph

where
Young’s Modulus, Y = gradient of stress-strain graph

 Unit: joules per cubic meters (J m-3)  strain
 Type of quantity: Scalar quantity
Area under graph

Example 8.2.1

A concrete block weighing 4000 N is attached to one end of a steel bar. The other end of the
bar is fixed to the ceiling. The bar is circular, with a cross-sectional area equal to 10 mm2 and
initial length equal to 50 cm. If the bar elongated by 0.05 mm, calculate
a) the tensile stress
b) the tensile strain
c) the Young’s Modulus.

Solution 8.2.1

a) the tensile stress

F W 1000  1.0 108 N m2
A A 10 106
 Tensile stress,  

b) the tensile strain

L  Lo L 0.05 10 3  1.0 104
Lo Lo 50 102
 Tensile strain,   

c) the Young’s Modulus.

Young’s Modulus,Y   1.0 108  1.0 1012 N m2
 1.0 104

Example 8.2.2

A copper wire of length 3.0 m stretches by 1.5 mm when a load is suspended vertically at
its end. If the Young ' s Modulus of copper is 1.2 x 1011 N m-2. Determine the stress on
the wire.

Solution 8.2.2

From Y  


 Y  L  1.5 10 3   6107
Lo 3.0
 Stress,
 Y  1.2 1011  N m-2

Example 8.2.3
The graph shows the behavior of a wire when it is stretched until it starts to undergo
plastic deformation.

What is the strain energy stored the wire?

Solution 8.2.3

Strain energy,U  area under force  elongation graph

 1 FL

2

  1 45011.5103  2.59 J
2

Example 8.2.4
A metal wire of is stretched by forces of various magnitude. The stress against strain
graph of wire is as shown in figure below.

a) Calculate the strain energy per unit volume stored in the wire.
b) Determine the Young’s modulus of the metal wire.

Solution 8.2.4

a) Calculate the total strain energy per unit volume stored in the wire.

Strain energy per unit volume  area under stress  strain graph

U  1 

V2

   1 10.5107 1.5103  78.75103 J m-3
2

b) Determine the Young’s Modulus of the metal wire.

 Y
  gradient of the graph  10.5 107  7 1010 N m-2
 1.5 10 3

8.3 Heat conduction

LEARNING OUTCOME:

At the end of this chapter, students should be able to:

a) Define heat conduction. through a cross-sectional

b) Solve problems related to rate of heat transfer, Q  kA T

tL

area (remarks: maximum two insulated objects in series)

c) Analyse graphs of temperature-distance (T-L) for heat conduction through insulated and

non-insulated rods. (remarks: maximum two rods in series)

8.3 Heat conduction
 Heat is defined as the energy that is transferred from a body at a higher

temperature to one at a lower temperature , by conduction, convection or radiation.
 Heat always transferred from a hot region (higher temperature) to a cool region

(lower temperature) until thermal equilibrium is achieved.

 Heat is transferred by three mechanisms,
1) Conduction
2) Convection
3) Radiation

32

 Heat conduction is defined as the process whereby heat is transferred through a
substance from a region of high temperature to a region of lower temperature.

The mechanism of heat conduction through solid A B
material

 Suppose a rod is heated at one end (A).

 Before the rod being heated all the molecules vibrate about their equilibrium

position.

 As the rod is heated the molecules at the hot end (A) vibrate with increasing

amplitude, thus the kinetic energy increases.

 While vibrating the hot molecules collide with the neighboring colder molecules

result in transfer of kinetic energy to the colder molecules.

 This transfer of energy will continue until the cold end (B) of the rod become hot.

Rate of heat flow, Q
t

T1  T2

L

 Consider a uniform cylinder conductor of length L and cross sectional area A with
temperature T1 at one end and T2 at the other end as shown in figure above.

 The heat flows to the right because T1 is greater than T2.

34

 The rate of heat flow , Q through the conductor is given by: 35
t
Q  kA T
tL

Q : rate of heat flow
t
k : thermal conductivity
A : cross - sectional area
ΔT

: temperature gradient
L

: the ratio of the temperature difference between two points
to the distance between these points

: T f  Ti  T2  T1
LL

 The rate of heat flow through an object depends on :
1. Thermal conductivity, k.
2. Cross-sectional area through which the heat flow, A.
3. Thickness of the material, L.

4. Temperature difference between the two sides of the material, T.

 The negative sign because the temperature T decreases as the distance, L
increases.

 Unit: J s-1 or Watt (W).
 Type of quantity : Scalar quantity

36

 Thermal conductivity, k is the ability of the
material to conduct heat.

 Materials with large k are called conductors.
 Those with small k are called insulators.

 SI unit for k is Wm-1 oC 1

37

Heat conduction through insulated rod

T1 A T2 T1  T2

L

 Consider heat conduction through an insulated rod which has cross sectional area A
and length L as shown above.

 If the rod is completely covered with a good insulator, no heat loss from the sides
of the rod.

 By assuming no heat is lost to the surroundings, therefore heat can only flow

through the cross sectional area from higher temperature region, T1 to lower
temperature region, T2.

38

T1 insulator T2

Temperature, T insulator Q is constant along the rod as
t
T1 T1  T2 no heat escapes to the sides

T2
0 length, L

 The red lines (arrows) represent the direction of heat flow.
 When the rod is in steady state (the temperature falls at a constant rate) thus the

rate of heat flows is constant along the rod.
 This causes the temperature gradient will be constant along the rod as shown in

figure above. 39

Combination 2 insulated rods in series

lc T3 lD T1  T3  T2

insulator and

T Material C Material D T2 kC  kD
1

insulator
Temperature,T

T1

T3

length, x

T2

0 lC  lD  length,L

40

 When steady state is achieved , the rate of heat flow through both materials is same

(constant).

 Q    Q  AC  AD
 t C  t D

 From the equation of thermal conductivity, we get

dQ k 1
k   dt  dT 
 dx 
A dT
dx

But kC  kD

 T    T 
 L C  L D

Temperature gradient for rod C < Temperature gradient for rod D 41

Heat conduction through non-insulated rod

X Y

T1 T2

Temperature, T T1  T2

T1 X

T2 Y
0 length, L

 The metal is not covered with an insulator, thus heat is lost to the surroundings

from the sides of the rod.
 The lines of heat flow are divergent and the temperature fall faster near the hotter

end than that near the colder end. 42

 Less heat is transferred to Y.

 This causes the temperature gradient gradually decreases along the rod and result

a curve graph where the temperature gradient at X higher than that at Y as

shown in figure below. Temperature,T

 From dQ  kA dT T1 X
dt dx

 And from the graph T2 Y
T at X  T at Y 0
LL length, L

 Thus Q at X  Q at Y where A and k are the same along the rod.
tt

Temperature gradient , T at any point on the rod is given by the slope of the 43

L

tangent at that point.

Combination 2 non-insulated rods in series

lP T3 lQ T1  T3  T2 and kP  kQ

T1 T2

Material P Material Q T at M  T at N Q at M  Q at N
LL tt
Temperature,T Q at O  Q at S
N tt
T1 M
O T at O  T at S 44
T3 S LL

length, x

T2  lP  lQ length,L
0

Example 8.3.1

How much heat flows through a sheet of copper 1.2 m  0.8 m of thickness 3.0 mm
for 1 hour if the temperatures of the surfaces are 30 °C and 28 °C respectively?
(Thermal conductivity of copper = 401 W m-1 K-1)

Solution 8.3.1 Heat, Q  kA T t

From Q  kA T L
tL
28  30 
Q  4011.2  0.8 1 60  60
3103 

 9.24108 J

Example 8.3.2

Calculate the rate of heat transfer on a cold day through a rectangular window that is
1.2 m wide and 1.8 m high, has a thickness of 6.2 mm, a thermal conductivity value of
0.27 W m-1 °C-1. The temperature inside the home is 21°C and the temperature outside
the home is -4°C.

Solution 8.3.2

Q  kA T
tL

 0.27 1.2 1.8  4  21

6.2 103

 2352 J s-1

Example 8.3.3

An aluminum rod and a copper rod of equal length 2.0 m and cross-sectional area 2 cm2 are
welded together in series. One end is kept at a temperature of 10 °C and the other at 30 °C as
shown in figure below. Calculate the rate of heat flows through the rod. (Thermal
conductivity of aluminum is 200 W m-1 °C-1 and of copper is 390 W m-1 °C-1 ).

Solution 8.3.3

Aluminum Copper ACu= AAl = 2.0 cm2 = 210-4 m2

30 °C T=? 10 °C

lAl= 2.0 m lCu= 2.0 m

 Q    Q 
 t  Al  t Cu

Aluminum Copper ACu= AAl = 2.0 cm2= 210-4 m2

30 °C T=? 10 °C

lAl= 2.0 m lCu= 2.0 m

  kA T     kA T 
 L  Al  L Cu

  200 2 104  T  30     390 2 104  10  T  
2 2

T  30  1.9510  T 

T  16.78 C

Aluminum Copper ACu= AAl = 2.0 cm2= 210-4 m2

30 °C T=? 10 °C

lAl= 2.0 m lCu= 2.0 m

The rate of heat flows through the rod,

Q    kA T 
t L  Al

   kA T 
 L  Al

  200 2104 16.78  30 
  2

 0.2644 J s-1