Rigid bar pinned at B with two vertical rods

The load $P$ at C rotates the rigid bar clockwise about the pin B: A moves up (stretching the aluminium rod fixed to the ground below) and D moves down (stretching the steel rod hung from the ceiling). Both rods are in tension; let the forces be $F_{al}$ and $F_{st}$.

 F_al                    F_st
  |                        ^
  v                        |
  A---1.5m---B---0.75---C---0.75---D
   (Al, 1.5 m)  pin     P=500 N  (steel, 0.9 m)

Equilibrium (moments about B)

Aluminium pulls A down and steel pulls D up; both give anticlockwise moments about B, balancing the clockwise moment of $P$:

$$F_{al}(1.5) + F_{st}(1.5) = P(0.75) = 500\times0.75$$ $$F_{al} + F_{st} = 250\ \text{N}$$

Compatibility

The bar is rigid, and A and D are both 1.5 m from the pin, so the upward movement of A equals the downward movement of D: $\delta_{al} = \delta_{st}$.

$$\frac{F_{al}(1500)}{300\times100\times10^3} = \frac{F_{st}(900)}{200\times200\times10^3} \;\Rightarrow\; 5\times10^{-5}F_{al} = 2.25\times10^{-5}F_{st} \;\Rightarrow\; F_{al} = 0.450\,F_{st}$$

Solution

$$\begin{aligned} F_{st}(1+0.450) &= 250 \Rightarrow F_{st} = 172.41\ \text{N}\\ F_{al} &= 77.59\ \text{N}\\ \sigma_{st} &= \frac{172.41}{200} = 0.862\ \text{N/mm}^2\\ \sigma_{al} &= \frac{77.59}{300} = 0.259\ \text{N/mm}^2 \end{aligned}$$

Answer: $\sigma_{steel} = 0.862$ N/mm² (tension) and $\sigma_{aluminium} = 0.259$ N/mm² (tension).