Question

What will be the stress at $-20^{\circ }C$, if a steel rod with a cross-sectional area of $150m{m}^2$ is stretched between two fixed points? The tensile load at $20^{\circ }C$ is $5000N$. (Assume $\alpha =11.7\times 10^{-6\circ }{C}^{-1}$ and $Y=200\times 10^{11}N{m}^{-2}$ )

• A. $12.7\times 10^{6}N{m}^{-2}$
• B. $84.2\times 10^{6}N{m}^{-2}$
• C. $127\times 10^{6}N{m}^{-2}$
• D. $0.842\times 10^{6}N{m}^{-2}$

Answer 1

Answer: (C)

Let L = free length at $0^{\circ }C$
$L_0=$ final stretched length in each case.
$L_1$ and $L_2$ are free length at $+20^{\circ }C$ and $-20^{\circ }C$ respectively.


We know $s_2=s_1+2\alpha L\Delta T$
Here, $s_1$ and $s_2$ are load deformation.
$\frac{F_{2}L}{AY}=\frac{F_{1}L}{AY}+2\alpha L\Delta T$
${\left(\sigma \right)}_2=\frac{F_2}{A}=\frac{F_1}{A}+2\alpha \Delta TY=\frac{5000}{150\times {\left(10\right)}^{-6}}+\left(2\right)\left(11.7\times {\left(10\right)}^{-6}\right)\left(20\right)\left(2\times {\left(10\right)}^{11}\right)$
$=127\times 10^{6}N{m}^{-2}$