Rigid bars PQ and RS: maximum force W

Assumed arrangement: both bars are 6 m long (3 m + 3 m) and pinned at P and R. The aluminium rod hangs from the ceiling to Q (upper bar), the steel rod joins Q to S (lower bar), and $W$ acts at the middle of RS.

Equilibrium

• Lower bar RS (moments about R): the steel rod pulls S up with $F_{st}$, so $F_{st}(6) = W(3)$ and $F_{st} = W/2$.
• Upper bar PQ (moments about P): the steel rod pulls Q down with $F_{st}$ and the aluminium rod pulls Q up with $F_{al}$, so $F_{al}(6) = F_{st}(6)$ and $F_{al} = F_{st} = W/2$.

Let $F = W/2$. Both rods are in tension.

Deformations

$$\begin{aligned} \delta_{al} &= \frac{F L}{AE} = \frac{F(2000)}{500\times70\times10^3} = 5.714\times10^{-5}\,F\ \text{mm}\\ \delta_{st} &= \frac{F(2000)}{300\times200\times10^3} = 3.333\times10^{-5}\,F\ \text{mm} \end{aligned}$$

Geometry of movement

• Q moves down by $\delta_{al}$ (bar PQ rotates about P).
• S moves down by $\delta_Q + \delta_{st} = \delta_{al} + \delta_{st}$ (bar RS rotates about R).
• The load point is at the middle of RS, so it moves $\tfrac12(\delta_{al}+\delta_{st})$.

Limiting the movement of W to 5 mm:

$$\tfrac12(\delta_{al}+\delta_{st}) = 5 \Rightarrow (5.714\times10^{-5}+3.333\times10^{-5})F = 10$$ $$F = 110526.3\ \text{N},\qquad W = 2F = 221052.6\ \text{N} = 221.05\ \text{kN}$$

Answer: $W_{max} = 221.05$ kN (force of 110.53 kN in each rod).