Rigid bar ABCD with brass and steel rods

The load at D turns the bar clockwise about the pin A: B moves down and stretches the brass rod BE (tension, hung from the ceiling); C moves down and compresses the steel rod CF (standing on the ground). Let the forces be $F_b$ (tension) and $F_s$ (compression); the distances from A are B = 2 m, C = 4 m and D = 5 m.

 ceiling ==========E==================
                   |
                   | brass (2 m)
 A-------2m--------B-------2m-------C---1m---D
 (pin)                              |        | 15 kN
                                    |        v
                                    | steel (3 m)
 ground ============================F=========

Equilibrium (moments about A)

Both rods resist the load (brass pulls B up, steel pushes C up):

$$F_b(2) + F_s(4) = 15\times5 = 75\ \text{kN·m}$$

Compatibility

The bar is rigid, so the deflection is proportional to the distance from A: $\delta_C = 2\delta_B$.

$$\frac{F_s(3000)}{600\times200} = 2\,\frac{F_b(2000)}{1000\times100} \;\Rightarrow\; 25\,F_s = 40\,F_b \;\Rightarrow\; F_s = 1.60\,F_b$$

(forces in kN with E in kN/mm²; both sides multiplied by 1000).

Solution

$$\begin{aligned} 2F_b + 4(1.6F_b) &= 75 \Rightarrow F_b = 8.929\ \text{kN},\quad F_s = 14.286\ \text{kN}\\ \sigma_b &= \frac{8928.6}{1000} = 8.93\ \text{N/mm}^2\\ \sigma_s &= \frac{14285.7}{600} = 23.81\ \text{N/mm}^2 \end{aligned}$$

Answer: $\sigma_{brass} = 8.93$ N/mm² (tension) and $\sigma_{steel} = 23.81$ N/mm² (compression).