Rigid bar AB: vertical displacement of F

Assumptions: the dimensions marked "m" in the figure are read as millimetres (rod length 300 mm, AB = 400 mm, AF = 200 mm). The rods AC (steel, 20 mm dia.) and BD (aluminium, 40 mm dia.) are vertical posts under the rigid bar, so both are in compression.

Rod properties

$$A_{st} = \frac{\pi}{4}(20)^2 = 314.16\ \text{mm}^2, \qquad A_{al} = \frac{\pi}{4}(40)^2 = 1256.64\ \text{mm}^2$$ $$A_{st}E_{st} = 314.16\times200\,000 = 6.283\times10^{7}\ \text{N}, \qquad A_{al}E_{al} = 1256.64\times70\,000 = 8.796\times10^{7}\ \text{N}$$

Bar remains horizontal (as stated in the question)

If AB stays horizontal, both rods shorten by the same amount $\delta$, and this is also the vertical displacement of F. Each rod force is $P = AE\delta/L$, and together they carry the 90 kN load:

$$\frac{\delta}{L}\left(A_{st}E_{st} + A_{al}E_{al}\right) = 90\,000 \Rightarrow \delta = \frac{90\,000\times300}{6.283\times10^{7}+8.796\times10^{7}} = 0.179\ \text{mm}$$

Loads in the rods: $P_{st} = \dfrac{6.283\times10^{7}\times0.179}{300} = 37.5$ kN and $P_{al} = 90 - 37.5 = 52.5$ kN.

For the bar to stay horizontal, the load must act where these rod forces balance it. Taking moments about A: $90\,x = 52.5\times400$, so $x = 233.3$ mm from A.

If the load is exactly at the mid-point (AF = 200 mm)

Then $P_A = P_B = 45$ kN, and

$$\delta_A = \frac{45\,000\times 300}{314.16\times 200\,000} = 0.2149\ \text{mm}, \qquad \delta_B = \frac{45\,000\times 300}{1256.64\times 70\,000} = 0.1535\ \text{mm}$$ $$\delta_F = \frac{\delta_A + \delta_B}{2} = \frac{0.2149+0.1535}{2} = 0.1842\ \text{mm}$$

In this case the bar tilts slightly (by $0.0614/400$ rad).

Answer: with AB remaining horizontal, the vertical displacement of F is $\delta_F = 0.179\ \text{mm}$ downward (steel 37.5 kN, aluminium 52.5 kN, load 233.3 mm from A). If the load is taken exactly at the mid-point, $\delta_F \approx 0.184$ mm and the bar tilts slightly.