Question

Two steel rods and an aluminium rod of equal length $l_0$ and equal cross-section are joined rigidly at their ends as shown in the figure below. All the rods are in a state of zero tension at 0°C. Find the length of the system when the temperature is rised to $\theta$. Coefficient of linear expansion of aluminium and steel are $\alpha_a$ and $\alpha_s$ respectively. Young's modulus of aluminium is $Y_a$ and of steel is $Y_s$.

Answer 1

The final length of the system is $L = l_0 \left( 1 + \theta \left( \frac{2\alpha_s Y_s + Y_a \alpha_a}{Y_a + 2Y_s} \right) \right)$.

Answer 2

When the temperature rises, each rod attempts to expand. Due to rigid joining, they must reach a common final length $L$. The total strain in each rod is the sum of thermal strain and strain due to internal stress. For steel rods, the strain is $\frac{L - l_0}{l_0} = \alpha_s \theta + \frac{\sigma_s}{Y_s}$, and for the aluminium rod, it's $\frac{L - l_0}{l_0} = \alpha_a \theta + \frac{\sigma_a}{Y_a}$. Since the final lengths are equal, the total strains are equal: $\alpha_s \theta + \frac{\sigma_s}{Y_s} = \alpha_a \theta + \frac{\sigma_a}{Y_a}$. Considering force equilibrium, with two steel rods and one aluminium rod, the total force from steel must balance the force from aluminium. Assuming $\alpha_a > \alpha_s$, aluminium is in compression ($\sigma_a < 0$) and steel is in tension ($\sigma_s > 0$). The equilibrium condition is $2 |\sigma_s| A = |\sigma_a| A$, which simplifies to $2 \sigma_s = -\sigma_a$. Substituting this into the strain equality and solving for $\sigma_s$ gives $\sigma_s = \frac{Y_s Y_a (\alpha_a - \alpha_s) \theta}{Y_a + 2Y_s}$. The change in length $L - l_0$ is then calculated using the strain equation for steel: $L - l_0 = l_0 \left( \alpha_s \theta + \frac{\sigma_s}{Y_s} \right)$. Substituting the expression for $\sigma_s$ and simplifying leads to the final length $L = l_0 + (L - l_0) = l_0 \left( 1 + \theta \left( \frac{2\alpha_s Y_s + Y_a \alpha_a}{Y_a + 2Y_s} \right) \right)$.