Page 90 - CITS - WCS - Mechanical

WORKSHOP SCIENCE - CITS

F
Stress =
A
2000
=
r π 2
2000
=
3.14 x 0.8 x 0.8
2000
=
2.0096
= 995.2 kg/cm 2

Stress
Young’s modulus =
Strain
995.2
2 x 10 ==
6
Strain
995.2
Strain = 6
2 x 10
= 0.0005

7 A tensile load of 2000 kg is applied on a rectangular rod of 2 cm x 1 cm whose length is 2 metres.
Calculate the elongation in length as E = 2 x 10 Kg/cm .
2
6
F = 2000 Kg.
L = 2 m = 200 cm
1
E = 2 x 10 kg/cm 2
6
Rectangular rod length = 2 cm
Breadth = 1 cm

Force(F) Force
Stress(σ) = =
Area(A) l x b
2000
=
2 x 1
= 1000 kg/cm 2

Stress
E =
Strain
1000
2 x 10 6 =
Strain
1000
Strain = 6
2 x 10
= 0.0005

= Strain

= 0.0005

77

CITS : WCS - Mechanical - Exercise 7 CITS : WCS - Mechanical - Exercise 7