Question

What will be the stress at $-{20}^{\circ }C,$ if a steel rod with a cross-sectional area of $150{mm}^2$ is stretched between two fixed point ? The tensile load at ${20}^{\circ }C$ is $5000$
$N:$ (Assume $\alpha =11.7\times {10}^{-6}{/}^{\circ }C$ and $Y=200\times {10}^{11}N/m^2)$

• A. $12.7\times {10}^{6}N/m^2$
• B. $1.27\times {10}^{6}N/m^2$
• C. $127\times {10}^{6}N/m^2$
• D. $0.127\times {10}^{6}N/m^2$

Answer 1

The correct option is C $127\times {10}^{6}N/m^2$

$\begin{array}{c}\text{Let length of rod at}{0}^{\circ }f,{20}^{\prime }c,-{20}^{\circ }\text{c are}\\ \text{L,}{L}_1,L_2\text{rerpectively.}\end{array}$

$\begin{array}{c}\Delta L_2=\Delta L_1+2L_1\Delta T\\ \text{where}\Delta L_2\text{and}\Delta L_1\text{are elongation of rods.}\\ \text{by Hookes law}\end{array}$

$\Delta L_2=\Delta L_1+2L\alpha \Delta T$

$\begin{array}{cc}\frac{F_{2}L}{AY}& =\frac{F_{1}L}{AY}+2L\alpha \Delta T\\ \frac{F_2}{A}& =\frac{F_1}{A}+2Y\alpha \Delta T\end{array}$

$\begin{array}{cc}\text{Stress at}{20}^{\circ }C& =\frac{5000}{150\times {10}^{-6}}+2x\text{2x}{10}^{11}x\\ & =33.3\times {10}^6+93.6\times {10}^6\\ & =126.9\times {10}^6\\ & =127\times {10}^{6}N/m^2\end{array}$