Question

What will be the stress at $-{20}^{o}C$, if a steel rod with a cross-sectional area of $150m{m}^2$ is stretched between two fixed point ? The tensile load at $-{20}^{o}C$ is $5000N$: (Assume $\alpha =11.7\times {10}^{-6}{/}^{o}C$ and $Y=2000\times {10}^{11}N/m^2$)

• A. $93.6\times {10}^{6}N/m^2$
• B. $9.36\times {10}^{6}N/m^2$
• C. $936\times {10}^{6}N/m^2$
• D. $0.936\times {10}^{6}N/m^2$

Answer 1

Answer: (C)

$936\times {10}^{6}N/m^2$

By stress strain relationship, we know that,

$\frac{F}{A}=Y\frac{\Delta l}{l}$

$\frac{F}{A}=Y\frac{l\alpha \Delta T}{l}=Y\alpha \Delta T$

$\frac{F}{A}=(200\times {10}^{11})(11.7\times {10}^{-6})(20-(-20\left)\right)=936\times {10}^{6}N/m^2$

Answer is option C.