Steel rod with two brass rods: stresses

The rigid plate rests on all three rods, so each rod shortens by the same amount: $\delta_s = \delta_b$. The problem states 50 kN in the text (the figure shows 60 kN); the solution is worked for 50 kN, and the 60 kN values follow in proportion.

Compatibility

$$\frac{\sigma_s L_s}{E_s} = \frac{\sigma_b L_b}{E_b} \Rightarrow \frac{\sigma_s(200)}{200\times10^3} = \frac{\sigma_b(250)}{100\times10^3} \Rightarrow \sigma_b = 0.40\,\sigma_s$$

Equilibrium

$$P = \sigma_sA_s + 2\sigma_bA_b = \sigma_s\left[1000 + 2(800)(0.40)\right] = \sigma_s\,(1640)$$ $$50\times10^3 = 1640\,\sigma_s \Rightarrow \sigma_s = 30.488\ \text{N/mm}^2,\quad \sigma_b = 12.195\ \text{N/mm}^2$$

Load carried: steel 30.49 kN, each brass rod 9.76 kN (total 50.00 kN).

Answer: for 50 kN, $\sigma_{steel} = 30.49$ N/mm² and $\sigma_{brass} = 12.20$ N/mm² (compressive). If the load is 60 kN, the stresses become 36.59 N/mm² and 14.63 N/mm².