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
. −