Round bar fixed-free (effective length 2 m): buckling load

Given: the bar is clamped at the bottom and free at the top (cantilever column). The effective length is $L_e = 2$ m, so the actual length is $L = L_e/2 = 1$ m. A horizontal force $H = 300$ N at the top produces a deflection $\delta = 20$ mm.

Step 1: find $EI$ from the lateral load

Acting as a cantilever:

$$\delta = \frac{HL^3}{3EI} \;\Rightarrow\; EI = \frac{300\times(1000)^3}{3\times20} = 5\times10^{9}\ \text{N mm}^2$$

Step 2: buckling load

$$P_{cr} = \frac{\pi^2EI}{L_e^2} = \frac{\pi^2\times5\times10^9}{(2000)^2} = 12{,}337\ \text{N}$$

Answer: buckling load $\approx 12.34$ kN.

If the 2 m were taken as the actual length (so $L_e = 4$ m), $EI = 4\times10^{10}\ \text{N mm}^2$ and $P_{cr} = 24.67$ kN; the usual reading of "effective length" gives 12.34 kN.