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

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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 Pb = 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

\({L_e} = \frac{L}{{\sqrt 2 }}\)

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

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