Resting on a fixed horizontal base, a steel rod is clamped at both...

Given:

Initial temperature, T_1 = 20^{}^{{}^{\circ}} C

Final temperature, T_2 = 50^{}^{{}^{\circ}} C

Change in temperature, \Delta T = 50^{}^{{}^{\circ}} C - 20^{}^{{}^{\circ}} C = 30^{}^{{}^{\circ}} C

Coefficient of linear expansion, \alpha = 1.2 \times 10^{-5} ^{}^{{}^{\circ}} C^{-1}

If the rod were free to expand, the increase in length would be $\Delta L = L \alpha \Delta T$.

Since the rod is clamped at both ends, it cannot expand. The rigid supports exert a compressive force to maintain the original length.

The longitudinal (mechanical) strain developed in the rod is given by:

$\text{Strain} = -\dfrac{\Delta L}{L} = -\alpha \Delta T$

Substituting the given values:

$\text{Strain} = -(1.2 \times 10^{-5}) \times 30$

$\text{Strain} = -3.6 \times 10^{-4}$

The negative sign indicates that the strain developed is compressive.