Rigid bar ABC: aluminium wire, steel bar

Assumptions: the figure gives only the length of the steel bar (1 m), so the aluminium wire BD is also taken as 1 m long. $E_{al} = 72\ \text{GPa}$ and $E_{st} = 210\ \text{GPa}$ (the units in the question are a misprint). The bar ABC is rigid and hinged at A; both rods are in tension, holding the bar against the 30 kN load.

Let $F_B$ = force in the aluminium wire (2 m from A) and $F_C$ = force in the steel bar (3 m from A); the 30 kN load is 4 m from A.

Equilibrium

Taking moments about the hinge A:

$$2F_B + 3F_C = 30\times 4 = 120\ \text{kN·m}$$

Compatibility

The rigid bar rotates about A, so the deflection is proportional to the distance from A:

$$\frac{\delta_C}{\delta_B} = \frac{3}{2}$$ $$\delta_B = \frac{F_B L}{A_{al}E_{al}} = \frac{F_B(1000)}{4(72\,000)} = 0.003472\,F_B, \qquad \delta_C = \frac{F_C L}{A_{st}E_{st}} = \frac{F_C(1000)}{2(210\,000)} = 0.002381\,F_C$$

($F$ in N, $\delta$ in mm). Hence

$$0.002381\,F_C = 1.5(0.003472)F_B \Rightarrow F_C = 2.1875\,F_B$$

Substituting in the moment equation: $2F_B + 3(2.1875)F_B = 120$ kN, so

$$F_B = 14.01\ \text{kN}, \qquad F_C = 30.66\ \text{kN}$$

Stresses

$$\sigma_{al} = \frac{14015}{4} = 3503.6\ \text{N/mm}^2, \qquad \sigma_{st} = \frac{30657}{2} = 15328.5\ \text{N/mm}^2$$

These are very large because the given areas are very small (if the areas were in cm², the stresses would be 100 times smaller: 35.0 and 153.3 N/mm²).

Reaction at A

$\Sigma V = 0$, taking upward forces as positive: $R_A + F_B + F_C = 30$, so

$$R_A = 30 - 14.01 - 30.66 = -14.67\ \text{kN}$$

The negative sign means $R_A = 14.67\ \text{kN}$ acts downward.

Answer: $\sigma_{al} = 3503.6\ \text{N/mm}^2$, $\sigma_{st} = 15328.5\ \text{N/mm}^2$, $R_A = 14.67\ \text{kN}$ downward.