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 points? The tensile load at $20^{\circ} C$ is $5000\, N$. (Assume $\alpha=11.7 \times 10^{-6 \circ} C^{-1}$ and $Y=200 \times 10^{11} N m^{-2}$ )

• A. $12.7 \, \times 10^{6}Nm^{- 2}$
• 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.
$\frac{F_{2} L}{A Y}=\frac{F_{1} L}{A Y}+2\alpha L\Delta T$
$\left(\sigma \right)_{2}=\frac{F_{2}}{A}=\frac{F_{1}}{A}+2\alpha \Delta TY=\frac{5000}{150 \, \times \left(10\right)^{- 6}}+\left(2\right) \, \left(11.7 \, \times \left(10\right)^{- 6}\right) \, \left(20\right) \, \left(2 \, \times \left(10\right)^{11}\right)$
$=127 \, \times 10^{6}Nm^{- 2}$