First slide

Stress and strain

Question

A steel rod of length 1 m and radius 10 mm is stretched by a force 100 kN along its length. The percentage strain produced in the rod is $Y_{steel} = 2 \times 10^{11} {Nm}^{- 2}$

Moderate

A

0.04%
B

0.08%
C

0.16%
D

0.24%

Solution

Elongation,
$ΔL = \frac{(F / A) L}{Y} = \frac{(3 .18 \times 10^{8} {Nm}^{- 2}) (1 m)}{2 \times 10^{11} {Nm}^{- 2}}$
$= 1.59 \times 10^{- 3} m = 1.59 mm$
Strain produced in the rod is
Strain $= \frac{ΔL}{L} = \frac{1 .59 \times 10^{- 3} m}{1 m} = 1 .59 \times 10^{- 3} = 0 .16 %$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App