Question

A steel rod of length $6\text{m}$ and diameter $20\text{mm}$ is fixed between two rigid support. Determine the stress in the rod, when the temperature increases by $80{}^{\circ }\text{C}$ and the ends yield of $1\text{mm}$ is allowed.
Take Young's modulus, $Y=2\times {10}^6{\text{N/cm}}^2$ and thermal coefficient of linear expansion, $\alpha =12\times {10}^{-6}/{}^{\circ }\text{C}$.

• A. $1587{\text{N/cm}}^2$
• B. $1687{\text{N/cm}}^2$
• C. $1787{\text{N/cm}}^2$
• D. $1887{\text{N/cm}}^2$

Answer 1

The correct option is A $1587{\text{N/cm}}^2$

Given:
Length of rod, $l=6\text{m}$
Diameter of rod, $d=20\text{mm}$
Change in temperature, $\Delta T=80{}^{\circ }\text{C}$
Allowed ends yeild, $\delta l=1\text{mm}$
Young's modulus, $Y=2\times {10}^6{\text{N/cm}}^2$
Co-efficient of linear expansion, $\alpha =12\times {10}^{-6}/{}^{\circ }C$

Increase in length due to increase in temperature,
$\Delta l=l\alpha \Delta T$
$\Rightarrow \Delta l=6\times 12\times {10}^{-6}\times 80=5.76\times {10}^{-3}\text{m}$
$\Rightarrow \Delta l=5.76\text{mm}$
So, net expansion,
$\Delta L=\Delta l-\delta l=5.76-1=4.76\text{mm}$
Now, Young's modulus,
$Y=\frac{\text{stress}}{\text{strain}}$
$\Rightarrow \text{stress}=Y\times \text{strain}$
$\Rightarrow \text{stress}=Y\times \frac{\Delta L}{l}$
$\Rightarrow \text{stress}=2\times {10}^6\times \frac{4.76}{6\times {10}^3}\approx 1587{\text{N/cm}}^2$

Why this question? Caution: Stress is produced in the bar because of restriction in the expansion of length. If we allow bar to expand freely, then stress will be zero.