Question

A steel rod is clamped at its two ends and rests on a fixed horizontal base. The rod is unstrained at ${20}^{\circ }C$. Find the longitudinal strain developed in the rod if the temperature rises to ${50}^{\circ }C$. Coefficient of linear expansion of steel $=1.2\times {10}^{-5}{}^{\circ }{C}^{-1}$.

Answer 1

Given, ${\theta }_1={20}^{\circ }C,{\theta }_2={50}^{\circ }C$

${\alpha }_{steel}=1.2\times {10}^{-5}/{}^{\circ }C$

Longitudinal strain = ?

Strain $=\frac{\Delta L}{L}-\frac{L\alpha \Delta \theta }{L}=\alpha \Delta \theta$

$=1.2\times {10}^{-5}\times (50-20)$

$=1.2\times {10}^{-5}\times 30$

$=36\times {10}^{-5}=3.6\times {10}^{-4}$

The strain is opposite to the direction of expansion.