An aluminium rod and two steel rods, all having an identical length...

Let the final length of the composite system be $l and the cross-sectional area of each rod be$A$.<br><br>The natural length of the aluminium rod at temperature$\theta$is$l_{0a} = l_0(1 + \alpha_a \theta)$.<br><br>The natural length of each steel rod at temperature$\theta$is$l_{0s} = l_0(1 + \alpha_s \theta)$.<br><br>The mechanical strain in the aluminium rod is$\epsilon_a = \dfrac{l - l_{0a}}{l_0}$.<br><br>The mechanical strain in each steel rod is$\epsilon_s = \dfrac{l - l_{0s}}{l_0}$.<br><br>The tension in the aluminium rod is$F_a = Y_a A \epsilon_a = Y_a A \left(\dfrac{l - l_{0a}}{l_0}\right).

The tension in each steel rod is $F_s = Y_s A \epsilon_s = Y_s A \left(\dfrac{l - l_{0s}}{l_0}\right)$.

Since the rods are rigidly joined and there are no external forces, the net internal force must be zero:

$F_a + 2F_s = 0$

$Y_a A \left(\dfrac{l - l_{0a}}{l_0}\right) + 2 Y_s A \left(\dfrac{l - l_{0s}}{l_0}\right) = 0$

$Y_a (l - l_{0a}) + 2 Y_s (l - l_{0s}) = 0$

$l(Y_a + 2Y_s) = Y_a l_{0a} + 2 Y_s l_{0s}$

Substituting the values of $l_{0a}$ and $l_{0s}$:

$l = \dfrac{Y_a l_0(1 + \alpha_a \theta) + 2 Y_s l_0(1 + \alpha_s \theta)}{Y_a + 2Y_s}$

$l = l_0 \dfrac{Y_a + 2Y_s + Y_a \alpha_a \theta + 2 Y_s \alpha_s \theta}{Y_a + 2Y_s}$

$l = l_0 \left[ 1 + \dfrac{\alpha_a Y_a + 2 \alpha_s Y_s}{Y_a + 2Y_s} \theta \right]$