A steel wire with a cross-sectional area of 0.5 mm^2 is secured between two fixed supports. The wire is just taut at 20^ ^ ^ C . Calculate the tension in the wire when the temperature drops to 0^ ^ ^ C . The coefficient

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Quantrex Academy · Accepted Answer

Correct answer: A. 24 N

The thermal stress developed in the wire due to a change in temperature is given by

\dfrac{F}{A} = Y \alpha \Delta T

The tension in the wire is

F = Y A \alpha \Delta T

Given:

Y = 2.0 \times 10^{11} \text{ N m}^{-2}

A = 0.5 \text{ mm}^2 = 0.5 \times 10^{-6} \text{ m}^2

\alpha = 1.2 \times 10^{-5} ^{}^{{}^{\circ}} \textC^{-1}

\Delta T = 20^{}^{{}^{\circ}} \text{C} - 0^{}^{{}^{\circ}} \text{C} = 20^{}^{{}^{\circ}} \text{C}

Substituting the values:

F = (2.0 \times 10^{11}) \times (0.5 \times 10^{-6}) \times (1.2 \times 10^{-5}) \times 20

F = 1.0 \times 10^5 \times 1.2 \times 10^{-5} \times 20

F = 1.2 \times 20 = 24 \text{ N}

A steel wire with a cross-sectional area of 0.5 mm^2 is secured...

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