Question

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

• A. $0.04$%
• B. $0.08$%
• C. $0.16$%
• D. $0.24$%

Answer 1

The correct option is C $0.16$%

Given:

force, $F=100KN$

radius, $L=10mm$

length, $L=1m$

Area $A=\pi r^2$

Stress $\sigma =\frac{F}{A}=\frac{{10}^5}{\pi ({10}^{-2}{)}^2}$

Elongation

$\Delta L=\frac{\left(F/A\right)L}{Y}=\frac{\left(\frac{{10}^5}{\pi ({10}^{-2}{)}^2}\right)\left(1m\right)}{2\times {10}^{11}N{m}^{-2}}=1.59\times {10}^{-3}m=1.59mm$

Strain produced in the rod is

$Strain=\frac{\Delta L}{L}=\frac{1.59\times {10}^{-3}m}{1m}=1.59\times {10}^{-3}=0.16$