Take $E= 8x10^4$ MPa. Compare Euler critical load by taking $\sigma_c$= 550 MPa and $\alpha^{'}$ = (1/1600). For what length of the column the critical load by the Euler’s and Rankine formula will be equal to each other.

Since both ends are hinged, $L_c = L = 6\hspace{0.05cm}m$

$A = \frac{\pi}{4}(D^2 - d^2) = 13744\hspace{0.05cm}mm^2 = 13.74\hspace{0.05cm}\times\hspace{0.05cm}10^3\hspace{0.05cm}mm^2\\ L = 6\hspace{0.05cm}m = 6000\hspace{0.05cm}mm\\ E = 8\hspace{0.05cm}\times\hspace{0.05cm}10^4\\ \alpha = \frac{1}{1600}$

$I_{xx} = I_{yy} = \frac{\pi}{64}(D^4 - d^4)\\ \hspace{0.5cm} = 53.68\hspace{0.05cm}\times\hspace{0.05cm}10^3\hspace{0.05cm}mm^4\\ K = \sqrt{\frac{I_{min}}{A}} = \sqrt{\frac{53.68\hspace{0.05cm}\times\hspace{0.05cm}10^3}{13744}}\\ \hspace{0.05cm} = 62.50$

$P_{Euler} = …