For the aluminum rod
$P_{al} = (\sigma A)_{al}$
$P_{al} = 120(40)$
$P_{al} = 4800 \, \text{ N}$
$\delta_{al} = \left( \dfrac{\sigma L}{E} \right)_{al}$
$\delta_{al} = \dfrac{120(5)(1000)}{70000}$
$\delta_{al} = \frac{60}{7} \, \text{ mm}$
For the steel beam
$t_{B/C} = \dfrac{1}{EI}(Area_{BC}) \cdot \bar{X}_B$
$t_{B/C} = \dfrac{1}{200\,000(50 \times 10^6)} \left[ \frac{1}{2}(2)(9600)(\frac{4}{3}) - \frac{1}{2}(2)(2P)(\frac{2}{3}) - \frac{1}{2}(2)(4P)(\frac{4}{3}) \right](1000^3)$
$t_{B/C} = \dfrac{1}{10\,000}(12\,800 - \frac{20}{3}P)$
$t_{B/C} = \frac{32}{25} - \frac{1}{1500}P$
From the figure
$\delta_{al} = -t_{B/C}$
$\frac{60}{7} = -(\frac{32}{25} - \frac{1}{1500}P)$
$\frac{1}{1500}P = \frac{1724}{175}$
$P = 14\,777.14 \, \text{ N}$
$P = 14.78 \, \text{ kN}$ answer
