1. **Problem Statement:** A rod consists of two cylindrical portions AB (steel) and BC (brass), fixed at both ends. Given the temperature rise $\Delta T = 50 ^\circ C$, find the compressive force $P$ induced in the rod. 2. **Given Data:** - Steel (AB): $E_s = 200$ GPa, $\alpha_s = 11.7 \times 10^{-6} / ^\circ C$, $d_{AB} = 30$ mm, $L_{AB} = 0.250$ m - Brass (BC): $E_b = 105$ GPa, $\alpha_b = 20.9 \times 10^{-6} / ^\circ C$, $d_{BC} = 50$ mm, $L_{BC} = 0.300$ m 3. **Calculate cross-sectional areas:** $$A_{AB} = \frac{\pi}{4} d_{AB}^2 = \frac{\pi}{4} (30)^2 = 706.86\ \text{mm}^2 = 706.86 \times 10^{-6} \ \text{m}^2$$ $$A_{BC} = \frac{\pi}{4} d_{BC}^2 = \frac{\pi}{4} (50)^2 = 1963.5\ \text{mm}^2 = 1.9635 \times 10^{-3} \ \text{m}^2$$ 4. **Calculate free thermal expansion $\delta_t$ if unconstrained:** $$\delta_t = L_{AB} \alpha_s \Delta T + L_{BC} \alpha_b \Delta T$$ $$= (0.250)(11.7 \times 10^{-6})(50) + (0.300)(20.9 \times 10^{-6})(50)$$ $$= 459.75 \times 10^{-6} \ \text{m}$$ 5. **Calculate shortening $\delta_p$ due to compressive force $P$:** $$\delta_p = \frac{P L_{AB}}{E_s A_{AB}} + \frac{P L_{BC}}{E_b A_{BC}}$$ $$= \frac{0.250 P}{(200 \times 10^9)(706.86 \times 10^{-6})} + \frac{0.300 P}{(105 \times 10^9)(1.9635 \times 10^{-3})}$$ $$= 3.2235 \times 10^{-6} P$$ 6. **Set net elongation to zero (fixed ends):** $$\delta_p = \delta_t$$ $$3.2235 \times 10^{-6} P = 459.75 \times 10^{-6}$$ 7. **Solve for $P$:** $$P = \frac{459.75 \times 10^{-6}}{3.2235 \times 10^{-6}} = 142.62 \times 10^{3} \ \text{N} = 142.6 \ \text{kN}$$ **Final answer:** The compressive force induced in the rod is $\boxed{142.6\ \text{kN}}$.