Page 88 - CITS - WCS - Mechanical

WORKSHOP SCIENCE - CITS

3 A force of 10 tonnes is applied axially on a rod of 12 cm dia. the original length is 100 mm.If modulus
of elasticity is 2 x 10 kg/cm . Calculate stress and strain developed in the rod.
12
2
Solution
Force applied = 10 tones = 10 x 1000 kg
= 10 kg
4
Diameter (d) = 12 mm = 1.2 cm
Young’s modulus (E) = 2 x 10 kg/cm 2
12
Force applied
Stress = Area of original cross section

10 4
= π 12 12
4 × 10 × 10

10 4 × 4 ×10 ×10
=
π ×12 ×12

10 6
=
36 π
= 8841 kg/cm 2
\ Stress = 8841 kg/cm 2

We know 4 10 4
Stress 1.2 1.2
= Young’s modulus
Strain 10 4
4
Strain x Young’s modulus = Stress
1.2 1.2 3.142
Stress
Strain == 40000 s Modulus
Young'
4.52448
= 8841
2 10 12
4420.5
= 4420.5
12
10
12
10
= 4420.5 x 10 -12
Stress = 8841 kg/cm 2
Strain = 4420.5 x 10
-12
4 A bar of 100 cm elongates to 101.36 cm when a load of 15000 kg is applied to it. Take the area of cross
section of bar as 10 cm . Find the stress, strain and young’s modulus.
2
L = 100 cm
1
L 2 = 101.36 cm

= L - L
2 1
= 101.36 - 100 = 1.36 cm
F = 15000 kg
A = 10 cm 2

Force(F)
Stress =
Area(A)

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