Question

A steel rod is clamped at its two ends and rests on a fixed horizontal base. The rod is unstrained at 20°C.
Find the longitudinal strain developed in the rod if the temperature rises to 50°C. Coefficient of linear expansion of steel = 1.2 × 10–5 °C–1.

Answer 1

The longitudinal strain developed in the rod if the temperature rises to 50°C is  $\bold{3.6 \times 10^{-4}}$

Explanation:

Temperature at which rod rest without strain on a set horizontal base, $T_1$ = 20 °C. Then Heat the rod to temperature, $T_2$ = 50 °C

Therefore, the temperature changes,

$\Delta T=T_2-T_1=30 ^{\circ}C$

Linear expansion coefficient for steel, $\alpha = 1.2 \times 10^{-5}$ °C-1

Let L be the length of the heated rod, and let L ' be the length of the heated rod.

Let the longitudinal strain in the rod develop be S.

$L^{\prime}=L(1+\alpha \Delta T)$

$\Delta L=L\Alpha \Delta T$

$S=\frac{\Delta L}{L}=\frac{L \alpha \Delta T}{L}=\alpha \Delta T$

$S =1.2 \times 10^{-5}\times (50-20)$

$=1.2 \times 10^{-5} \times 30$

$=36 \times 10^{-5}$

$S=3.6 \times 10^{-4}$

The $3.6 \times 10^{-4}$ strain will be opposite to the expansion direction.

Answer 2

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$3.6 \times {10}^{ - 4}$