Question

For long column the equivalent length for both end fixed is
1. L_e = L
2. \(L_e ~=~\frac{L}{\sqrt2}\)
3. \(L_e ~=~\frac{L}{2}\)
4. L_e = 2L

Answer 1

Correct Answer - Option 3 : \(L_e ~=~\frac{L}{2}\)

Explanation:

According to Euler buckling load is determined by:

\({P_b} = \frac{{{\pi ^2}E{I_{}}}}{{L_e^2}}\)

where P_b = Buckling load, E = Young’s modulus of elasticity, Imin = Minimum moment of inertia, Le = Effective/Equivalent length

Buckling load for various end conditions is given in the table below.

End conditions	Le	Buckling load
Both ends hinged	Le = L	\({P_b} = \frac{{{\pi ^2}E{I_{}}}}{{L^2}}\)
Both ends fixed	\({L_e} = \frac{L}{2}\)	\({P_b} = \frac{{{4\pi ^2}E{I_{}}}}{{L^2}}\)
One end fixed and another end is free	Le = 2L	\({P_b} = \frac{{{\pi ^2}E{I_{}}}}{{4L^2}}\)
One end fixed and another end is hinged	\({L_e} = \frac{L}{{\sqrt 2 }}\)	\({P_b} = \frac{{{\pi ^2}E{I_{}}}}{{2L^2}}\)