Beam deflection used for column buckling load

Given: simply supported beam, $L = 5$ m = 5000 mm, central load $W = 800$ N, maximum (central) deflection $\delta = 30$ mm.

Step 1: find $EI$

For a simply supported beam with a central point load:

$$\delta = \frac{WL^3}{48EI} \;\Rightarrow\; EI = \frac{WL^3}{48\delta} = \frac{800\times5000^3}{48\times30} = 6.944\times10^{10}\ \text{N mm}^2$$

Step 2: buckling load as a column

For one end fixed and the other hinged, the effective length is $L_e = 0.7L$, which corresponds to the standard textbook form $P_{cr} = 2\pi^2EI/L^2$:

$$P_{cr} = \frac{2\pi^2EI}{L^2} = \frac{2\pi^2\times6.944\times10^{10}}{5000^2} = 54{,}831\ \text{N}$$

(Using the exact root of $\tan kL = kL$, that is $P_{cr} = 20.19EI/L^2$, gives about 56 kN.)

Answer: buckling load $\approx 54.8$ kN.