The stress in the steel rod is given by: $$\sigma_s=E_s(\epsilon_s-\alpha_s \Delta T)\tag{key}$$
where ##\epsilon_s## is the strain in the steel rod, Es is the Young's modulus of the steel rod, and ##\alpha_s## is the coefficient of linear expansion of the steel rod. This equation says that, if the steel rod expands with no constraint (i.e., with ##\sigma_s = 0##), the strain in the rod is ##\epsilon_s = \alpha_s \Delta T##, but, if the stress in the rod is greater than 0, the strain will be greater than ##\alpha_s \Delta T##. If ##\delta_s## is the downward displacement of the end of the steel rod, then ##\epsilon_s=\delta_s/L_s##, and $$\sigma_s=E_s\left(\frac{\delta_s}{L_s}-\alpha_s \Delta T\right)$$
Geometrically, the downward displacement of the end of the steel rod is related to the downward displacement of the end of the aluminum rod by:$$\frac{\delta_s}{0.6}=-\frac{\delta_A}{1.2}\tag{1}$$
The tensile stress in the aluminum rod is given by$$\sigma_A=E_A\frac{\delta _A}{L_A}$$
The tension in the steel rod is given by$$P_s=A_s\sigma_s=A_sE_s\left(\frac{\delta_s}{L_s}-\alpha_s \Delta T\right)\tag{2}$$
Similarly, the tension in the aluminum rod is given by $$P_A=A_AE_A\frac{\delta _A}{L_A}\tag{3}$$
If we combine Eqns. 1-3 with the moment balance on the bar ABC, we can solve for all the displacements, strains, stresses, and tensions. What is the moment balance on the bar ABC in terms of the tensions ##P_s## and ##P_A##?
The key to this whole analysis is the equation labeled "key." This equation takes into account the thermal expansion strain experienced by the rod plus whatever extra strain experienced by the rod to give the overall stress.
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