Page 87 - CITS - WCS - Mechanical

WORKSHOP SCIENCE - CITS

2 A bar of steel 2.5 cm diameter was subjected to compressive load of 4500 kg. The compression in a
length of 20 cm was found to be 0.008 cm. Find the Young’s modulus of elasticity of bar.
Solution

Diameter of bar d = 2.5 cm
Force applied i.e. compressive load = 4500 kg
Original length of bar = 20 cm
Change in length = 0.008 cm
π
\ Area of original cross-section = d 2
π 2 4
= = × 2.5
4

π × 6.25
4
= cm 2
Force applied
\ Stress ==
Area of original cross section

4500
π × 6.25
=
4
4500× 4
=
π × 6.25
2880
=
π
2880
\ Stress = π Kg/cm 2

Change in length
\ Strain =
Original length
0.008 8 1000
= =
20 20
8 4
= =
20× 1000 10000
4
\ Strain =
10000
Stress
\ Young’s modulus =
Strain
2880 4
= ÷
π 10000

2880 10000
= ×
π 4
7200000
=
π
= 2292000 Kg/cm 2

= 2.292 x 10 Kg/cm 2
6

74

CITS : WCS - Mechanical - Exercise 7