A glass window is to be installed in an aluminium frame. On the...

Let the dimensions of the glass at $40^{{}^{{}^{\circ}}}\text{C}$ be $L_{g1} = 20\text{ cm}$ and $L_{g2} = 30\text{ cm}$.

Let the dimensions of the aluminium frame at $40^{{}^{{}^{\circ}}}\text{C}$ be $L_{a1}$ and $L_{a2}$.

When the temperature falls to $0^{{}^{{}^{\circ}}}\text{C}$, the drop in temperature is $\Delta T = 40^{{}^{{}^{\circ}}}\text{C}$.

To ensure no stress is exerted on the glass, the dimensions of the aluminium frame at $0^{{}^{{}^{\circ}}}\text{C}$ must be equal to the dimensions of the glass at $0^{{}^{{}^{\circ}}}\text{C}$.

$L_{a1}(1 - \alpha_a \Delta T) = L_{g1}(1 - \alpha_g \Delta T)$

$L_{a1} = L_{g1} \dfrac{1 - \alpha_g \Delta T}{1 - \alpha_a \Delta T} \approx L_{g1} [1 + (\alpha_a - \alpha_g) \Delta T]$

Substituting the given values:

\alpha_a - \alpha_g = (24 - 9) \times 10^{-6} = 15 \times 10^{-6} ^{{}^{{}^{\circ}}}\textC^{-1}

$(\alpha_a - \alpha_g) \Delta T = 15 \times 10^{-6} \times 40 = 600 \times 10^{-6} = 0.0006$

$L_{a1} = 20 (1 + 0.0006) = 20.012\text{ cm}$

$L_{a2} = 30 (1 + 0.0006) = 30.018\text{ cm}$

The required size of the aluminium frame is $20.012\text{ cm} \times 30.018\text{ cm}$.