SOM Topic 1.2

DJJ30103

1.2 STRENGTH OF MATERIALS
APPLY THE FORCES ON
1.2.1MATERIALS
State the Hooke’s

Prepared By: Azuan Kamaruddin

HOOKE’S LAW

The material remains elastic and returns to its original shape
and size after removal of the force. The elastic properties of

materials only in the range in which the force and
displacement are proportional

Strain α Stress

= Constant = E


*E is Young’s modulus

σ P PL
ϵ A∆L
E = = A =
∆L

L

SO,

= OR ∆ =

∆

DJJ30103

1.1 STRENGTH OF MATERIALS

1.2.2APPLY THE FORCES ON
MATERIALS Sketch the general shape of stress versus strain to

common engineering materials which is elastic and

brittle when subjected to load in the tensile test. Label

the following points;

- Elasticity limit

- Plasticity limit

- Yield point

- Maximum tensile strength

Prepared By: Azuan Kamaruddin

STRESS STRAIN GRAPH

MECHANICAL BEHAVIOUR CAN OBTAIN FROM STRESS STRAIN GRAPH

LOAD AT POINT C
YIELD STRESS = ORIGINAL CROSS SECTIONAL AREA

MAXIMUM LOAD
MAXIMUM TENSILE STRENGTH = ORIGINAL CROSS SECTIONAL AREA

EXTENSION
PERCENTAGE ELONGATION = ORIGINAL LENGTH X 100%

PERCENTAGE REDUCTION IN AREA = ORIGINAL AREA − NECKING AREA X 100%

ORIGINAL AREA

EXAMPLE

Load, kN 0 20 40 60 80
Extension, mm 0 0.030 0.062 0.091 0.121

In a tensile test on mild steel specimen of 14.28mm diameter and gauge length 50mm, the following reading
had recorded as per shown in table.

Final gauge length after fracture = 64.45mm
Diameter of gauge at the neck after fracture = 11.51mm

Calculate:

- Young modulus
- Percentage of elongation
- Percentage reduction in area

SOLUTION

Load, kN 0 20 40 60 80
Extension, mm 0 0.030 0.062 0.091 0.121

In a tensile test on mild steel specimen of 14.28mm diameter and gauge length 50mm, the following reading

had recorded as per shown in table. = 64.45mm
Final gauge length after fracture E = ∆

Diameter of gauge at the neck after fracture = 11.51mm (80 − 20)(50)
Calculate: E = ( 4 14.282)(0.121 − 0.030)

- Young modulus = 60 50 205.84 / 2
- Percentage of elongation 160.16 0.091 =
- Percentage reduction in area −
2 − 2 =
4 4
=
4 2

14.282 − 11.512 64.45 − 50
= 14.282 = 50 100%

= 28.9%

= 35.03%

DJJ30103

1.1 STRENGTH OF MATERIALS
APPLY THE FORCES ON
1.2.3MATERIALS
Define and calculate problems related to the following:
- Stress and strain
- Young’s modulus
- Safety factor
- Poisson’s ratio
- Strain energy

Prepared By: Azuan Kamaruddin

STRESS =


σ =


EXAMPLE: 30kN

Find the tensile stress for steel bar

25mm 30kN

EXAMPLE: 30kN

Find the tensile stress for steel bar

25mm 30kN

σ =


σ =
−

= 60 x

STRAIN =


ε = δ


EXAMPLE: 50mm

Find the tensile strain for steel bar if final
elongantion is 50.03mm

EXAMPLE: 50mm

Find the tensile strain for steel bar if final
elongantion is 50.03mm

ε = δ


ε = . −



= 60 x −

YOUNG MODULUS =


E = σ
ε

EXAMPLE:

Refer stress and strain had calculate in
previous, please find young modulus

E =
−

= 1 x

WORKING STRESS =


* Working stress maximum allowable stress for the component
SAFETY FACTOR =



EXAMPLE:
Calculate safety factor if ultimate tensile stress is 230MN/m2 and σ = 60MN/m2

SF =


= 3.833

STRAIN ENERGY, U = . ∆


EXAMPLE:

Calculate strain energy absorbed by a steel bar of diameter 30mm, length 600mm when
subjected to a tensile load of 150kN. Given E = 200GN/m2

∆ =



U = . ∆

U = .

. .

U = ( ) ( − )

( − ) ( )


= 47.74J

POISSON’ RATIO, V = = ∆



∆
=

EXAMPLE: = − . . −
=
Original diameter = 12 mm
Diameter of neck = 11.997 mm = . − = . −
Original length = 50mm
Final length = 50.044 mm

V = . − = 0.284
. −