Euler’s Bucking load on a column of length ‘I’ with both hinged, is given as
(E = Young’s Modulus; I = Moment of Inertia of Column Cross-section)
1
\(\frac{2\pi EI}{L^2}\)
2
\(3\frac{\pi^2 EI}{L^2}\)
3
\(\frac{\pi^2 EI}{L^2}\)
4
\(\frac{2\pi^2 EI}{L^2}\)