Take $E= 2x 10^5$ MPa. Also determine the Rankine crippling load for the same column. Take $\sigma_c= 350 $ MPa and $\alpha^{'}$= (1/7500)

$l = 2500\hspace{0.05cm}mm\\ A = \frac{\pi}{4}(D^2 - d^2) = 706.86\hspace{0.05cm}mm^2\\$ $I_{xx} = I_{yy} = \frac{\pi}{64}(D^4 - d^4) = 28.98\hspace{0.05cm}\times\hspace{0.05cm}10^5\hspace{0.05cm}mm^4\ K = \sqrt{\frac{I_{min}}{A}} = 16\hspace{0.05cm}mm\ \sigma_c = 350\hspace{0.05cm}N/mm^2\ \alpha = \frac{1}{7500}\ I = AK^2 = 706.86\hspace{0.05cm}\times\hspace{0.05cm}16^2 = 181\hspace{0.05cm}\times\hspace{0.05cm}10^3\hspace{0.05cm}mm^4$ Both ends are hinged, therefore; $L_c = L = 2500\hspace{0.05cm}mm$ $P_{Euler} = \frac{\pi ^2}{L_c^2}EI = \frac{\pi^2}{2500^2}\hspace{0.05cm}\times\hspace{0.05cm}2\hspace{0.05cm}\times\hspace{0.05cm}10^5\hspace{0.05cm}\times\hspace{0.05cm}181\hspace{0.05cm}\times10^3\ \hspace{0.25cm} = 57.20\hspace{0.05cm}\times\hspace{0.05cm}10^3\hspace{0.05cm}N = 57.2\hspace{0.05cm}KN$ $P_{Rankine} = \frac{\sigma_{cr}\hspace{0.05cm}\times\hspace{0.05cm}A}{1 + \alpha(\frac{L_c}{K})^2}\ \hspace{0.25cm} = \frac{350\hspace{0.05cm}\times\hspace{0.05cm}76.86}{1 + 7500(\frac{2500}{16})^2}\ \hspace{0.25cm} = 58.37\hspace{0.05cm}KN$