Page 89 - CITS - WCS - Mechanical

WORKSHOP SCIENCE - CITS

15000
=
10
= 1500 kg/cm 2
l
Strain =
L
1.36
=
100
= 0.0136

Stress
Young’s modulus =
Strain
1500
E =
0.0136
= 110300 kg/cm 2

5 What force is required to stretch a steel wire of 10 mm long and 10 mm dia. to double its length. E of
steel is 205 KN/cm .
2
d = 10 mm = 1 cm
r = 0.5 cm
L = 1 cm
1
L = 2 cm
2
= L - L = 2 - 1 = 1 cm
2 1
E = 205 KN/cm 2

Strain = = = 1
Stress
E =
Strain
Stress
205 =
1
Stress = 1 x 205 = 205 KN/cm 2
Force(F)
Stress =
Area(A)
Force
205 =
3.14 x 0.5 x 0.5
Force = 205 x 3.14 x 0.5 x 0.5
= 161 KN

6 A wire of 1.6 cm diameter is subjected to a tensile load of 2000 Kg. Find the stress and strain if young’s
modulus = 2 x 10 kg/cm .
2
6
F = 2000 kg
d = 1.6 cm
r = 0.8 cm

E = 2 x 10 Kg/cm 2
6

76

CITS : WCS - Mechanical - Exercise 7