Correct Answer - Option 3 : 80 kN
Concepts:
Buckling load for the column,
\({P_e} = \frac{{{\pi ^2}E{I_{min}}}}{{L_e^2}}\)
Where,
Pe = Buckling load, Imin = Moment of inertia about centroids axis, Le = Effective length.
Effective length (Le),
When both ends of the column hinged (Le) = L
When both ends of the column fixed (Le) = \(\frac{L}{2}\)
Calculation:
Given:
Buckling load for both ends hinged = 20kN and Le = L
\({P_e} = \frac{{{\pi ^2}E{I_{min}}}}{{L^2}}\) = 20 kN.
Then, buckling load for the column when both ends fixed and Le = \(\frac{L}{2}\)
\({P_e} = \frac{{{4\pi ^2}E{I_{min}}}}{{L^2}}\)
Pe= 4 × 20 = 80 kN.