# For long column the equivalent length for both end fixed is

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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 ${L_e} = \frac{L}{{\sqrt 2 }}$ ${P_b} = \frac{{{\pi ^2}E{I_{}}}}{{2L^2}}$