Rigid bar AB hinged at C: 40 kN

Assumptions: the bar is rigid and hinged at C; A is 4 m and B is 2 m from C, on opposite sides. The load at B pulls B down, so the copper rod at B is in tension. The bar lifts at A, so the steel rod at A is in compression (it pushes the bar down at A). Both rods are 1 m long.

Data: $A_s = 400\ \text{mm}^2$, $E_s = 200\ \text{kN/mm}^2$, $A_c = 600\ \text{mm}^2$, $E_c = 110\ \text{kN/mm}^2$.

Equilibrium (moments about hinge C)

$$F_c(2) + F_s(4) = 40(2) \Rightarrow F_c + 2F_s = 40\ \text{kN}$$

Compatibility

If the bar rotates through a small angle $\phi$, then $\delta_B = 2\phi$ (elongation of the copper rod) and $\delta_A = 4\phi$ (shortening of the steel rod), so $\delta_A = 2\delta_B$:

$$\frac{F_sL}{A_sE_s} = 2\,\frac{F_cL}{A_cE_c} \Rightarrow \frac{F_s}{400(200)} = \frac{2F_c}{600(110)}$$ $$F_s = \frac{2\times 400\times200}{600\times110}F_c = 2.4242\,F_c$$

Substituting: $F_c + 2(2.4242F_c) = 40$, which gives

$$F_c = 6.839\ \text{kN}, \qquad F_s = 16.580\ \text{kN}$$

Stresses

$$\sigma_s = \frac{16580}{400} = 41.45\ \text{N/mm}^2\ \text{(compressive)}, \qquad \sigma_c = \frac{6839}{600} = 11.40\ \text{N/mm}^2\ \text{(tensile)}$$

Answer: steel $\approx 41.45\ \text{N/mm}^2$, copper $\approx 11.40\ \text{N/mm}^2$.