$A$ brass rod of length $50\; cm$ and diameter $3.0\; mm$ is joined to a steel rod of the same length and diameter. What is the change in length of the combined rod at $250\; ^{\circ}C$,if the original lengths are at $40.0\; ^{\circ}C$? Is there a 'thermal stress' developed at the junction? The ends of the rod are free to expand. (Coefficient of linear expansion of brass $= 2.0 \times 10^{-5}\; K^{-1}$,steel $= 1.2 \times 10^{-5}\; K^{-1}$)

(N/A) Initial temperature,$T_{1} = 40.0\; ^{\circ}C$
Final temperature,$T_{2} = 250\; ^{\circ}C$
Change in temperature,$\Delta T = T_{2} - T_{1} = 210\; ^{\circ}C$
Length of each rod,$l = 50\; cm$
Coefficient of linear expansion of brass,$\alpha_{b} = 2.0 \times 10^{-5}\; K^{-1}$
Coefficient of linear expansion of steel,$\alpha_{s} = 1.2 \times 10^{-5}\; K^{-1}$
Change in length of brass rod,$\Delta l_{b} = l \alpha_{b} \Delta T = 50 \times (2.0 \times 10^{-5}) \times 210 = 0.21\; cm$
Change in length of steel rod,$\Delta l_{s} = l \alpha_{s} \Delta T = 50 \times (1.2 \times 10^{-5}) \times 210 = 0.126\; cm$
Total change in length of the combined rod,$\Delta l = \Delta l_{b} + \Delta l_{s} = 0.21 + 0.126 = 0.336\; cm$
Since the ends of the rod are free to expand,there is no constraint on the expansion,and therefore no 'thermal stress' is developed at the junction.