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The diameter of long column is D buckling load is P. If the diameter is reduced by 50% new buckling load will be:
1. 0.4 P
2. 0.25 P
3. 0.0825 P
4. 0.0625 P

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Correct Answer - Option 4 : 0.0625 P


Design of columns using Euler’s theory of buckling:

\({{\bf{P}}_{\bf{b}}} = \;\frac{{{{\bf{\pi }}^2}{\bf{EAK}}_{{\bf{min}}}^2}}{{{\bf{L}}_{\bf{e}}^2}}\)     


E = Young’s modulus, A = area of cross-section, K = minimum radius of gyration, Pb = buckling load

The radius of gyration is given as:

\({\bf{K}} = \sqrt {\frac{{{{\bf{I}}_{{\bf{min}}}}}}{{\bf{A}}}} \)  

where I = moment of inertia, A = area of cross-section



Initial diameter = D, Initial buckling load =\(​​\;{{\rm{P}}_{{{\rm{b}}_1}}}\)   

Final diameter after 50% decreased = \(\frac{D}{2}\), final buckling load = \({{\rm{P}}_{{{\rm{b}}_2}}}\)

Calculation of radius of gyration:

\({{\rm{K}}_{{\rm{mi}}{{\rm{n}}_1}}} = \sqrt {\frac{{\frac{{\rm{\pi }}}{{64}}{\rm{\;}}{{\rm{D}}^4}}}{{\frac{{\rm{\pi }}}{4}{\rm{\;}}{{\rm{D}}^2}}}{\rm{\;\;}}} {\rm{\;}} = {\rm{\;}}\frac{{\rm{D}}}{4}\)

\({{\rm{K}}_{{\rm{mi}}{{\rm{n}}_2}}} = \sqrt {\frac{{\frac{{\rm{\pi }}}{{64}}{\rm{\;}}{{\left( {\frac{{\rm{D}}}{4}} \right)}^4}}}{{\frac{{\rm{\pi }}}{4}{\rm{\;}}{{\left( {\frac{{\rm{D}}}{2}} \right)}^2}}}{\rm{\;\;}}} {\rm{\;}} = {\rm{\;}}\frac{{\rm{D}}}{8}\)

Buckling load is given by:

\(\frac{{{{\rm{p}}_{{{\rm{b}}_1}}}}}{{{{\rm{P}}_{{{\rm{b}}_2}}}}} = \;\frac{{\frac{{{{\rm{\pi }}^2}{\rm{E}}\left( {\frac{{{{\rm{D}}^2}}}{4}} \right){{\left( {\frac{{\rm{D}}}{4}} \right)}^2}}}{{{{\rm{L}}^2}}}}}{{\frac{{{{\rm{\pi }}^2}{\rm{E}}\left( {\frac{{{{\rm{D}}^2}}}{{16}}} \right){{\left( {\frac{{\rm{D}}}{8}} \right)}^2}}}{{{{\rm{L}}^2}{\rm{\;}}}}}}=16\) 

\(\frac{{{{\rm{P}}_{{{\rm{b}}_1}}}}}{{{{\rm{P}}_{{{\rm{b}}_2}}}}}\) = 16

\({{\rm{P}}_{{{\rm{b}}_2}}} = \frac{{{{\rm{P}}_{{{\rm{b}}_1}}}}}{{16}}\)

\({{\rm{P}}_{{{\rm{b}}_2}}} = 0.0625{\rm{\;P}}\)

Euler's theory is applicable only for long columns

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