Question

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

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

Answer 1

Answer: (C)

Elongation,
$\Delta L$ $=\frac{\left(F/A\right)L}{Y}$ $=\frac{\left(3.18\times 10^{8}N{m}^{-2}\right)\left(1m\right)}{2\times 10^{11}N{m}^{-2}}$
$=1.59\times 10^{-3}m$ $=1.59mm$
Strain produced in the rod is
Strain $=\frac{\Delta L}{L}$ $=\frac{1.59\times 10^{-3}m}{1m}$ $=1.59\times 10^{-3}$ $=0.16\%$