Temperature stresses in bars with a gap

Take bar A as the left bar and bar B as the right bar, and take the temperature rise as $\Delta T = 40^\circ\text{C}$. The bars first expand freely; if the free expansion exceeds the gap, the walls prevent the excess and a compressive force $P$ (equal in both bars) develops.

Free thermal expansion

$$\begin{aligned} \delta_A &= \alpha_A\,\Delta T\,L_A = 12\times10^{-6}\times40\times400 = 0.192\ \text{mm}\\ \delta_B &= \alpha_B\,\Delta T\,L_B = 23\times10^{-6}\times40\times300 = 0.276\ \text{mm}\\ \delta_{total} &= 0.468\ \text{mm} > \text{gap} = 0.26\ \text{mm} \end{aligned}$$

The expansion that is prevented is $\delta_{prevented} = 0.468 - 0.26 = 0.208$ mm, so the bars come into contact and stresses are induced.

Compatibility

The compressive force $P$ shortens the bars by exactly the prevented expansion:

$$\frac{P L_A}{A_A E_A} + \frac{P L_B}{A_B E_B} = 0.208$$ $$P\left(\frac{400}{1000\times2\times10^5} + \frac{300}{800\times1\times10^5}\right) = P\,(5.7500\times10^{-6}) = 0.208$$ $$P = 36173.9\ \text{N} = 36.17\ \text{kN}$$

Stresses

$$\sigma_A = \frac{P}{A_A} = \frac{36173.9}{1000} = 36.17\ \text{N/mm}^2,\qquad \sigma_B = \frac{P}{A_B} = \frac{36173.9}{800} = 45.22\ \text{N/mm}^2$$

Answer: both bars are in compression: $\sigma_A = 36.17$ N/mm² (bar A) and $\sigma_B = 45.22$ N/mm² (bar B); common force $P = 36.17$ kN.