Module_SP015

z Learning outcomes

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

11.1 Stress and Strain 11.2 Young’s modulus

(a) Distinguish between stress and strain for (a) Define Young’s Modulus.
tensile and compression force.
(Lecture : C1&C2,CLO1, PLO1, MQF LOD1)
(Lecture : C1&C2,CLO1, PLO1, MQF LOD1)
(Tutorial : C3&C4, CLO3,PLO4, CTPS3, (b) Discuss strain energy from force-
MQF LOD6) elongation graph.

(b) Discuss the graph of stress-strain for a (Lecture : C1&C2,CLO1, PLO1, MQF LOD1)
metal under tension.
(c) Discuss strain energy per unit
(Lecture : C1&C2,CLO1, PLO1, MQF LOD1) volume from stress-strain graph.
(c) Discuss elastic and plastic deformation.
(Lecture : C1&C2,CLO1, PLO1, MQF LOD1) (Lecture : C1&C2,CLO1, PLO1, MQF LOD1)
(d) Discuss graph of force-elongation, f-e for
(d) Solve problems related to the
brittle and ductile materials. Young’s Modulus.
(Lecture : C1&C2,CLO1, PLO1, MQF LOD1)
(Tutorial : C3&C4, CLO3,PLO4, CTPS3,
MQF LOD6)

z 11.0 Introduction

When looking at a force problem, you're probably used to
being concerned only about how the object moves after
being affected by the forces acting on it. What you
probably haven't thought too much about is how the
structure of the object might be affected by that same
force. A common example of this happens with bridges.
When people or vehicles move over a bridge, their weight
creates a downwards force. The bridge doesn't change
position due to this weight, but it can bend.

Although solid may be thought as having a definite shape and volume, it’s possible to change
its shape and volume by applying external forces. A sufficiently large force will
permanently deform or break an object, but otherwise, when the external forces are
removed, the object tends to return to its original shape and size. This is called elastic
properties.
The elastic properties of a material is an important criteria to determine the strength of the
material. The elastic properties of a material are discussed in terms of stress and strain.
To help ensure that a bridge can hold the amount of weight that it was designed to withstand
without breaking under heavy traffic, it undergoes something called a stress test.

z 11.1 Stress and strain

Stress ( ) is defined as force acting per unit cross sectional area.

Mathematically, we can write this as: A long bar clamped at one end is
stretched by the amount of e

under the action of a force ⃗ .

: Force perpendicular to the cross

section. Figure 1
A: Cross sectional area

SI unit for stress is kg m‒1s‒2 or N m‒2 or Pa (Pascal)

The amount of stress applied to an object will determine the level of strain the object
experiences, where strain is the amount of deformation an object undergoes due to
stress.

Strain ( ) is defined as ratio of the change in length to the

original length. .

Mathematically, we can write this as: Figure 2

e : amount of elongation or compression
: original length Tensile Compression
No unit for strain.

Note: or
Amount of elongation,
Amount of compression,
where is the final length

Depending on how force is applied to the object, it can undergo different types of
stress and strain.

Two of the most common types are tensile and compressive stress and strain.

When an object is under tension it is experiencing an increase in length. A

rubber band being stretched out is a common example of an object experiencing

tensile stress and strain.

The opposite of tension is compression, where an object is undergoing a
decrease in length. If you've ever squeezed a rubber ball or a pet's squeak toy
in your hands, you were creating a compressive stress and strain in the

object.

Any object is capable of experiencing tensile and compressive stress and strain, but
not all react to that stress to the same degree.

Graph of stress‒strain for a metal under tension

Stress, (Pa) Plastic
deformation
Elastic
deformation

Once stress is Permanent deformed by stress
removed returns
to original D
size/shape

E A : proportionality limit
B : elastic limit
AB CElastic but no longer linear C : yield point
D : point of maximum
Elastic limit is
reached when stress (force)
the graph line E : fracture (breaking)
start to curve.
point
OP Strain, OP : permanent strain when

stress is removed.

Elastic deformation

1) Elastic deformation occurs when an object can return to its original length or size
after being distorted.

2) Point O to point B is known as elastic deformation.

Point Explanation
OA
o Stress increase linearly with strain until point A. Point A is the
proportionality limit.

o The straight line graph (OA) obeys Hooke’s Law which states “Below the
proportionality limit, the restoring force, is directly proportional to the
extension ”

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

B o This is the elastic limit of the material.
o Beyond this point, the material is permanently stretched and will never
regain its original shape and length.
o If the stress is removed, the material has a permanent extension OT.

Plastic Deformation

1) Occur when an object has been deformed beyond its elastic limit.
2) At this stage, the object can be extended by a large amount with a small addition

of force.
3) After the force (stress) is removed, the object cannot regain its original length or

shape.
4) Point CDE is known as plastic deformation.

Point Explanation
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
D
sudden increase in extension and the material thins uniformly.
E  When stress increases, strain increases rapidly.
 Stress on the material is maximum and is known as breaking stress

(force).
 This is sometimes called the Ultimate Tensile Strength (UTS).
 After this point, the material will thin and “necks” are formed.
 This is the point where the material breaks or fractures.



Graph of stress‒strain for various materials

Stress, Steel Ductile materials

Glass Copper o undergo plastic
Aluminium deformation before
Bone breaking.
Strain, ε
O o Example: steel, copper,
aluminium.

Brittle materials

o do not undergo or show
very little plastic
deformation before
breaking.

o Example : glass, bone,
concrete.

Graph of force‒elongation for brittle and ductile material

A brittle material A ductile material

Force, (N) Force, (N)

Elastic Elastic Plastic
deformation deformation deformation

Maximum fracture D
force
The fracture of a brittle
material is sudden with Maximum E
little or no plastic force
deformation. fracture

Brittle AB C Ductile
material material
(glass, bone, (steel,
concrete) aluminium)

O elongation, (m) OP elongation, (m)

z 11.2 Young’s Modulus

Measures resistance of material to change its shape when a force is
applied to it.

Young’s modulus, is defined as the ratio of the tensile stress

to the tensile strain if the proportionality limit has not been
exceeded.
Mathematically, we can write this as:







Scalar quantity ; SI unit is kg m‒1s‒2 or N m‒2 or Pa (Pascal)

Young’s modulus, is the gradient of the stress-strain graph in the
elastic region, where Hooke’s law is obeyed.

Value of Young’s

modulus is large for a

Limit of proportionality stiff material ‒ slope

(gradient) of graph is

steep, difficult to

deform.

Gradient = Young’s modulus Small slope– Large slope–
Young’s modulus is Young’s
Stress, small, material is modulus is
Stress, flexible, easy tolarge, material
stretch, large strain is stiff, hard to
for little stress. stretch, little
strain for large
Strain, stress.

Strain,

Young’s modulus is a property of material, does not depend to the
length, weight or shape of the material but it depend to the type of
material.

Table below shows the approximate values of Young’s modulus for
various materials.

Material
Aluminium
Bone
Brass
Copper
Steel
Glass
Lead
Concrete
Tungsten
Diamond

Relationship between force constant, and
Young’s modulus,

From the statement of Hooke’s law and definition of Young modulus,
thus

Hooke’s law:
Young’s modulus :



into :



where : Young’s modulus
: force constant : original length
: cross sectional area

Strain energy,

o When a wire is stretched by a load (force), work is done on the wire and
strain (elastic potential) energy is stored within it.

o Consider the force-extension graph of this wire until the proportionality
limit ( Hooke’s law) as shown in Figure below.

Force (N) Limit of proportionality The total work done, in stretching
the wire from 0 to is given by
F

0 e elongation (m)

Work done is stored as strain energy thus

where
: strain energy (unit : J)
: Force (unit : N)
: elongation or compression (unit : m)



strain energy can also calculated from

graph force‒elongation :

Strain energy, = Area under graph

Strain energy per unit volume

From the definition: Tensile strain
Tensile stress




Therefore:



Strain energy per unit volume,

where
: strain energy per volume (unit : J m‒3)
: Stress (unit : N m‒2 @ Pa)
: strain ( no unit)

Stress, (Pa) From stress-strain graph, Strain energy
per unit volume is given by


SI unit for strain energy per volume

0 Strain, is J m−3

Example 1

For safety in climbing, a mountaineer uses a nylon rope that is 50 m long
and 1.0 cm in diameter. When supporting a 90-kg climber, the rope
elongates 1.6 m. Find its Young’s modulus.

Solution

Given : m; 1 cm ; ;m
Find :

Use :

Pa

Example 2

Bone has a Young’s modulus of about 18 Pa. Under

compression, it can withstand a stress of about 160 Pa

before breaking. Assume that a femur (thighbone) is 0.50 m

long, and calculate the amount of compression this bone can

withstand before breaking.

Solution Pa ; 160 Pa ; m

Given : 18
Find :

Example 3

A 80 N block is suspended at the end of a wire of length
80 cm and diameter 1.0 mm. The extension of the wire is
0.97 mm. Calculate the strain energy per unit volume in the
wire.

Solution

Given : 80 N ; 1 m ; m ; 0.97 m
Find :



0.97 J m‒3

1×



Topic 12
Heat conduction and

thermal expansion

Subtopics : 12.1 Heat conduction

12.2 Thermal expansion

MODE Face to Non
face SLT Face to
Lecture face SLT
Tutorial 1
1
3
3

Learning outcomes

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

12.1 Heat conduction 12.2 Thermal expansion

a) Define heat conduction a) Define coefficient of linear, area, and volume
thermal expansion.
(Lecture : C1&C2,CLO1, PLO1, MQF LOD1)
(Lecture : C1&C2,CLO1, PLO1, MQF LOD1)
b) Solve problem related to rate of heat transfer,
b) Solve problems related to thermal expansion of
linear, area and volume (include expansion of
= − liquid in a container).
through a cross-sectional area (maximum two
insulated object in series). (Tutorial : C3&C4, CLO3,PLO4, CTPS3, MQF LOD6)

(Tutorial : C3&C4, CLO3,PLO4, CTPS3, MQF LOD6)

c) Discuss graph of temperature – distance , T-x
for heat conduction through insulated and
non-insulated rods (maximum two insulated
object in series).

(Tutorial : C3&C4, CLO3,PLO4, CTPS3, MQF LOD6)

12.0 Introduction

Heat vs Temperature

Heat Definition Temperature

A form of energy which transferred A relative measurement of hotness
from one object to another due to or coldness of an object.
difference in temperatures
between the two objects. Unit of SI unit : Kelvin (K)
measurement Other unit : ;
SI unit : Joule (J)
Property - Scalar quantity that is same for any
- Heat flows from region of higher 2 systems in thermal equilibrium.
temperature to lower temperature.
- Increases when heated
- Heat can be transfer through - Decreases when cooled
conduction, convection and
radiation.

12.1 Heat conduction

Definition

Heat conduction is defined as a process whereby heat is transferred through a solid from
a region of high temperature to a region of lower temperature by direct contact of matter.

Transmission of heat

Rate of heat transfer, where


: rate of heat transfer
1) Consider a properly insulated slab of material of
uniform cross sectional area of , and thickness .
One face of the slab is at a temperature and the
other face is at temperature

∶ cross − sectional area

∶ thermal conductivity


: temperature gradient

∶ difference in temperature
between the cold and hot faces
= −

2) The heat flows to the right from hotter face to the ∶ thickness of the material
colder one. conducting heat (distance
between hot and cold faces)
3) It is found experimentally that the rate of heat
transfer is given by the relation: 4) Negative sign in equation above
indicates that heat flow is always in the
direction of decreasing temperature

5) The rate of heat transfer is a scalar 6) Thermal conductivity, is a
quantity. Its SI unit is or Watt (W)
proportionality constant that depends

Thermal conductivity, on the material. Substances that are

good conductors have large thermal

1) It is the characteristic of heat conducting conductivities, whereas good

ability of a material. insulators have low thermal

2) It is the indicator of how fast a material able to conductivities.

conduct heat.

3) Thermal conductivity, is defined as a rate of Substance Thermal conductivity,
(W m-1 K-1)
heat flows perpendicularly through the unit cross Silver
Copper 420
sectional area of a solid, per unit temperature Steel 380
Glass 40
gradient along the direction of heat flow. Good 0.84
Brick 0.70
thermal Wood 0.15
Wool 0.04
conductor Styrofoam 0.01

(metals)

4) SI unit for : W–1 m–1 K–1 @ W m–1 °C–1 poor thermal
5) Scalar quantity conductor
(insulators)

Heat conduction through jointed rods Because 1 2 heat flows through the two
rods from the hotter end 1 towards the cooler end
Consider 2 different properly insulated rods, X at 2 .
and Y of same uniform cross-sectional area,
but different lengths and thermal conductivities When the rods are in steady state, rate of
jointed in series as shown in figure below. heat flows along both rods are the same.



Rate of heat flow through X = Rate of heat flow through Y

Material X Material Y

kX Heat flow kY

lx ly (Solution key for jointed rods)
Thus we have:






Example 1
=
One wall of a house consists of 0.019-m-thick
plywood backed by 0.076-m-thick insulation, as
Figure below shows. − − = − −

The temperature at the inside surface is 25.0
°C, while the temperature at the outside surface . − = . ( ) −
is 4.0 °C, both being constant. The thermal . .
conductivities of the insulation and the plywood
are, respectively, 0.030 and 0.080 J/(s m °C), . − = . ( − )
and the area of the wall is 35 m2. Find = . ℃
(a) the temperature at the insulation–plywood
interface (b) heat conducted through the wall in one hour.

Rate of heat flows along both materials are the same.
Consider the insulation :

. −
= − .

( × ) = − . ( )

= . × J

Graph of temperature-distance ( )

(1) Heat conduction through insulated rod o For properly insulated rod in
steady condition, there will be no
X Insulator Y heat escapes from sides to the
surrounding.

o In steady condition, the rate of the
T (°C) heat flow, through the rod is



constant.

o Temperature gradient ( temperature
difference per unit length) is also

constant (same everywhere) along
the length of the rod.

o Temperature varies linearly with
distance along the rod. Decreasing
0 x (m) of temperature is uniform.

x

(2) Heat conduction through non-insulated rod

X Y o For non insulated conductor, heat can
escape from the sides of it. (Heat is lost

to the surrounding)

o The quantity of heat flowing from hot end

T (°C) x through a cross section per second gets

lesser and lesser as we move along the rod

towards the cold end.

o This means that becomes smaller and

smaller as we move towards cold end.

o The same is true for

o Hence we get a curve in temperature-
distance ( ) graph whose gradient
0 x (m) decreases as we move towards the cold
x end.

(3) Heat conduction through insulated combination of rods in series

and o When steady-state has been achieved,
the rate of heat flow through both

Material  Material  materials is the same.

(°C) o From the equation of thermal conductivity,

Smaller , steeper temperature gradient

(i) o If then Line
(ii)
Temperature gradient in material 1 is smaller
than temperature gradient in material 2.

o If then Line

Temperature gradient in material 1 is greater

(m) than temperature gradient in material 2.



(4) Heat conduction through non-insulated combination of rods in series

Material 

Material  o Heat loss to surrounding.
o Temperature decreases non
(°C)
uniformly with distance (curve line
graph).
o Values of are different, thus the
curves are different for both rods.


(m)



12.2 Thermal expansion

Definition
Thermal expansion is the change in the dimensions of a body due to a change
in its temperature.

o Most of the materials expand on heating and contract on cooling.

o The expansion in length is called linear expansion.

o The expansion in the area is called area expansion.

o The expansion in volume is called volume expansion

o For solids, thermal expansion occurs through linear (length) , area and volume

depending on the geometrical shape of the material.

o Liquid and gas will only experience volume expansion.
o How ? Rising of temperature will cause the molecules to be active in

motion and push the adjacent molecules to the side which causes the
material to expand.

Linear Expansion Initial
temperature
Consider a rod of initial length .
Initial length
If the rod is subjected to a rise in temperature
the length of the rod increases by and

the new final length is .

Experiments show that change in length : Final length final
and temperature

expressed in equation: Coefficient of linear expansion,

Is defined as fractional change in length per degree
∆ : change in length = − change in temperature.
∆ : change in temperature = −
: coefficient of linear expansion * ∆ : fractional change in length

The new final length is given by

Its SI unit is K−1. It is generally in the order of 10−5 .

Area Expansion Relation between and

Consider a plate of initial area .

If the plate is subjected to a rise in temperature
the area of the plate increases by and

the new final area is .

Experiments show that change in area :

and

Thus the change in area can be calculated as: * Limit for isotropic expansion
(uniform expansion in all
dimensions )

: change in area = Coefficient of area expansion,
: change in temperature =
Is defined as fractional change in area per degree
: coefficient of area expansion change in temperature.

The new final area is given by: * ∆ : fractional change in area



Its SI unit is K−1.

Volume Expansion Relation between and

Consider a given solid of initial volume .

If the solid is subjected to a rise in temperature
its volume increases by and the new

final volume is .

Experiments show that change in volume :

and

Thus the change in volume can be calculated * Limit for isotropic expansion
as: (uniform expansion in all
dimensions )


: change in volume = Coefficient of volume expansion,
: change in temperature =
Is defined as fractional change in volume per degree
: coefficient of volume expansion change in temperature.

The new final volume is given * ∆ : fractional change in volume

by:

Its SI unit is K−1.

Prove : = 2 Prove : = 3 Extra note for better understanding

Lo L L

Lo Ao L

L A Lo L

∆ = − Lo
∆ = ( + ∆ ) − Lo
∆ = ( (1 + ∆ )) −
∆ = −
= (1 + 2α∆ + ∆ ) − ∆ = ( + ∆ ) −
= + 2α∆ + ∆ − ∆ = ( (1 + ∆ )) −
Since α is generally in the order of 10−5 , ∆ ∆ = (1 + ∆ ) −
can be negligible.
∆ = 2 ∆ = (1 + 3 ∆ + 3 ∆ + ∆ ) −
∆ = ∆
Comparing with : ∆ = ∆ Since α is generally in the order of 10−5 , α and
→ = 2 can be negligible.
∆ = (1 + 3 ∆ ) −

∆ = + 3 ∆ −

∆ = 3 ∆

∆ = ∆
Comparing with : ∆ = ∆

→ = 3

As Table below indicates, each substance or material has its own characteristic coefficient
of expansion. For a solid, the coefficient of volume expansion is approximately three times
the linear expansion coefficient = 3 (This assumes that the coefficient of linear
expansion of the solid is the same in all directions.)

change in
temperature on
the Celsius
scale equals the
change on the
Kelvin scale.

Example:

Expansion of liquid in a container Extra volume
of liquid
Consider liquid that fills a container to the brim. When we heat it, not only created by
the liquid but also the container will expand. thermal
expansion has
The liquid will overflow or not when temperature rises depends on the change in to go
volume of both the liquid and the container. somewhere
Ultimately we would need to compare the volume expansions of both materials

If the liquid expands less than the container's expansion, the level of liquid would
be lower.

If the liquid and the container both expand the same amount for the same increase in
temperature (meaning that their volume expansion coefficients are equal), then the
liquid will still fill the container to the brim., there would be no overflow.

But, generally coefficient of volume expansion for liquid
much more greater than that of solid ( liquid > solid), thus
liquid expands greater than the container's expansion
and liquid would overflow from the container.

We can write a relation in apparent expansion as:



Example 2 We can conceptualize the situation by imagining that the ends
of both bolts expand and move toward each other as the
An electronic device has been poorly designed so temperature rises.
that two bolts attached to different parts of the
device almost touch each other in its interior, as in The sum of the changes in length of the two bolts must equal
Figure below. The steel and brass bolts are at the length of the initial gap between the ends.
different electric potentials and if they touch, a
short circuit will develop, damaging the device. If ∆ + ∆ = 5.0 × 10 m
the initial gap between the ends of the bolts is
5.0 m at 27 °C, at what temperature will the bolts ∆ + ∆ = 5.0 × 10 m
touch?
Given: = 19 × 10 ℃ and ∆ ( + ) = 5.0 × 10 m

= 11 × 10 ℃ ∆ = 19 × 10 5.0 × 10
0.030 + (11 × 10 )(0.010)

∆ = 7.4℃

Temperature at which the bolts touch is

= + ∆
= 27 + 7.4

Example 3 (b) How significant is the change in volume of the flask?

A volumetric flask made of Pyrex is calibrated mL
at 20.0°C. It is filled to the 100-mL mark with
35.0°C acetone. (a) What is the volume of the contract as temperature drops
acetone when it cools to 20.0°C?

Given: coefficient of linear expansion for Pyrex,
; coefficient of

volume expansion for acetone,

mL

contract as temperature drops

The significant volume change of flask is

mL ∆ −0.0144
= ∆ = −0.225 × 100% = 6.4%

about 6.4% of the change in the acetone’s
volume

Topic 13
Gas laws and kinetic theory

13.1 Ideal gas equation MODE Face to Non
13.2 Kinetic Theory of Gases face SLT Face to
13.3 Molecular kinetic energy and internal energy Lecture face SLT
Tutorial 1.5
1.5
4
4

LEARNING OUTCOMES

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

13.1 Ideal ggas equations 13.3 Molecular Kinetic Energy and Internal
a) Solve problems related to ideal gas equation Energy

(Tutorial : C3&C4, CLO3,PLO4, CTPS3, MQF LOD6) a) Discuss translational kinetic energy of a

b) Discuss the following graphs of an ideal gas: molecule
I. p-V graph at constant temperature = =
II. V-T graph at constant pressure
III. p-T graph at constant volume b) Discuss degree of freedom, f for

(Lecture : C1&C2,CLO1, PLO1, MQF LOD1) monoatomic, diatomic and polyatomic

13.2 Kinetic Theory of Gases gas molecules.
a) State the assumptions of kinetic theory of gases.
b) Discuss root mean square (rms) speed of gas c) State the principle of equipartition of

molecules. energy.

(Lecture : C1&C2,CLO1, PLO1, MQF LOD1) d) Discuss internal energy of gas.

c) Solve problems related to root mean square (rms) (Lecture : C1&C2,CLO1, PLO1, MQF LOD1)
speed of gas molecules.
d) Solve problems related to the equations, e) Solve problems related to internal
and pressure energy
= =

(Tutorial : C3&C4, CLO3,PLO4, CTPS3, MQF LOD6) =

(Tutorial : C3&C4, CLO3,PLO4, CTPS3, MQF LOD6)

4 variables used to
describe gases

Amount of gas, , Molar mass can be obtained from
= , periodic table but need to change to
unit kg/mol

Pressure, Force per unit area gas particles exert on the wall of their container
1 atm = . × Pa

Volume, Volume gas is the volume of the container
1L = × m3

Temperature, Usually measured in ℃ but need to change to Kelvin ( )
= ℃ + .

13.1 Ideal gas equation

Boyle’s law

The pressure, of a fixed mass of gas at constant
temperature is inversely proportional to its volume,

− graph at constant
∝
( constant) temperature



constant (m3)

Note : and represent variables initially, and and
represent the variables after the change is made

Charles’s law

The volume, of a fixed mass of gas at constant pressure is
directly proportional to its absolute temperature,

∝ − graph at
( constant) constant
pressure



constant

Note : and represent variables initially, and and
represent the variables after the change is made

Pressure’s law (Guy Lussac)

The pressure, of a fixed mass of gas at constant volume
is directly proportional to its absolute temperature,

∝ − graph at constant
( constant) volume







constant

Note : and represent variables initially, and and
represent the variables after the change is made

o These laws can be combined into single general relation between, , , and of
a fixed quantity of gas:
write in mathematic form:

o One mole of any gas at Standard Temperature
and Pressure STP ( = 273.15 K, =101.3 kPa)
occupies volume of = 22.4 liters = 22.4 dm3 :

o The value 8.31 J K─1 mol ─ 1 is found experimentally same for all gases.
o The value 8.31 J K─1 mol ─ 1 is known as molar gas constant ,