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A column of height h with rectangular cross-section of a × 2a has a buckling load of P. If the cross-section is changed to 0.5a × 3a and height changed to 1.5h, the buckling load of the redesigned column will be

For column, the buckling load is taken as,

\({\rm{P}} = \frac{{{\pi ^2}{\rm{E}}\:{{\rm{I}}_{{\rm{min}}}}}}{{{\rm{L}}_{\rm{e}}^2}}\)

Where

E = Modulus of Elasticity

Imin = Minimum moment of inertia

Le = effective length of column.

End Condition Both end hinged One end fixed other free Both end fixed One end fixed and other hinged
Effective length (Le) L 2L 1/2 L/√2

Calculation:

In first case: Height = h, and cross-sectional area = a × 2a

\({\rm{P}} = \frac{{{\pi ^2}{\rm{E}}\:\left( {\frac{{2{\rm{a}} \times {{\rm{a}}^3}}}{{12}}} \right)\:}}{{{{\rm{h}}^2}}} = \frac{{{\pi ^2}{\rm{E}}{{\rm{a}}^4}}}{{6{{\rm{h}}^2}}}\)

In second case: Height = 1.5 h, and cross-sectional area = 3a × 0.5a

\({\rm{P'}} = \frac{{{\pi ^2}{\rm{E}}\:\left[ {\frac{{3{\rm{a}} \times {{\left( {0.5{\rm{a}}} \right)}^3}}}{{12}}} \right]}}{{{{\left( {1.5\:{\rm{h}}} \right)}^2}}} = \frac{1}{{12}} \times \frac{{{\pi ^2}{\rm{E}}{{\rm{a}}^4}}}{{6{{\rm{h}}^2}}} = \frac{{\rm{P}}}{{12}}\)