Question

A steel wire of cross-sectional area 0.5 mm2 is held between two fixed supports. If the wire is just taut at 20°C, determine the tension when the temperature falls to 0°C. Coefficient of linear expansion of steel is 1.2 × 10–5 °C–1 and its Young's modulus is 2.0 × 10–11 Nm–2.

Answer 1

The tension when the temperature falls to 0°C is 24 N

Explanation:

The steel wire cross-sectional area, A = 0.5 $mm^2$ = 0.5 × 10–6$m^2$

The wire temperature, $T_1$= 20 °C,

Reduced temperature  $T_2$= 0 °C

Temperature change, Δθ = $T_1- T_2$  = 20 °C

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

Young’s modulus, $\gamma = 2 \times 10^{-11} Nm^{-2}$

Let L be the steel wire's initial length and let L be the steel wire's length when the temperature is reduced to 0 ° C.

Decrease in length due to compression, $\Delta L = L^{\prime}- L= L\alpha \Delta \Theta$                …(1)

Let the applied TENSION be F.

$\gamma=\frac{s t r e s s}{s t r a i n}=\left(\frac{F}{A}\right) \times\left(\frac{\Delta L}{L}\right)$

$\Delta L=\frac{F L}{A Y}$-----------------------(2)

The length change due to generated tension is given by (1) and (2).

Thus we get on equating (1) and (2):

$&L \alpha \Delta \theta=\frac{F L}{A Y}$

$&F=\alpha \Delta \theta A Y$

$=1.2 \times 10^{-5} \times(20-0) \times 0.5 \times 10^{-6} \times 2 \times 10^{11}$

=1.2×20

F=24 Nm

The tension created when the temperature drops to 0 ° C therefore is 24 N.

Answer 2

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24 N