Rigid bar AB hinged at A: 100 kN

Assumptions (from the figure): measured from B the supports are at 1 m (copper rod, 2 m long), then 3 m (steel rod, 1 m long), and the hinge A is at 4 m. So the lever arms from A are: steel 1 m, copper 3 m, load 4 m. Both rods are in tension and the bar is rigid.

Data: $A_s = 1000\ \text{mm}^2$, $A_c = 800\ \text{mm}^2$, $E_s = 200\ \text{GPa}$, $E_c = 100\ \text{GPa}$, $L_s = 1$ m, $L_c = 2$ m.

Equilibrium (moments about A)

$$F_s(1) + F_c(3) = 100(4) = 400\ \text{kN·m}$$

Compatibility

The rigid bar rotates about A, so $\delta_c = 3\delta_s$:

$$\frac{F_c L_c}{A_cE_c} = 3\,\frac{F_s L_s}{A_sE_s} \Rightarrow \frac{F_c(2)}{800\times10^{-6}(100\times10^9)} = 3\,\frac{F_s(1)}{1000\times10^{-6}(200\times10^9)}$$ $$F_c = 0.6\,F_s$$

Forces and stresses

$F_s + 3(0.6F_s) = 400 \Rightarrow 2.8F_s = 400$

$$F_s = 142.86\ \text{kN}, \qquad F_c = 85.71\ \text{kN}$$ $$\sigma_s = \frac{142857}{1000} = 142.86\ \text{MPa}, \qquad \sigma_c = \frac{85714}{800} = 107.14\ \text{MPa}$$

Reaction at A

$$\Sigma V = 0: \quad R_A = F_s + F_c - 100 = 142.86 + 85.71 - 100 = 128.57\ \text{kN}$$

The rods pull the bar up with more than the 100 kN load, so $R_A$ acts downward.

Answer: $\sigma_s = 142.86\ \text{MPa}$, $\sigma_c = 107.14\ \text{MPa}$ (both tensile), $R_A = 128.57\ \text{kN}$ downward.