When both ends of a column are fixed, the crippling load is F. If one end of the column is made free, the value of crippling load will be changed to ______.

Concept:

Crippling load, \({{\rm{P}}_{{\rm{cr}}}} = \frac{{{{\rm{\pi }}^2}{\rm{EI}}}}{{{\rm{L}}_{\rm{e}}^2}}\)

Where,

L_e = Effective length, and I = Moment of Inertia about bending axis

For a given E and I:

\({{\rm{P}}_{{\rm{cr}}}} \propto \frac{1}{{{\rm{L}}_{\rm{e}}^2}}\)

OR

\({\left( {{{\rm{P}}_{{\rm{cr}}}}} \right)_1}{\rm{\;}}{\left( {{\rm{L}}_{\rm{e}}^2} \right)_1} = {\left( {{{\rm{P}}_{{\rm{cr}}}}} \right)_2}{\left( {{\rm{L}}_{\rm{e}}^2} \right)_2}\)       ---(1)

Now

When both ends are fixed: \({L_{e1}} = \frac{L}{2}\)

When one end is free and another end is fixed:

Le₂ = 2L and (P_cr)₂ = P₂

Using equation (1), we get

\(\left( {\rm{F}} \right) \times {\left( {\frac{{\rm{L}}}{2}} \right)^2} = \left( {{{\rm{P}}_2}} \right) \times {\left( {2{\rm{L}}} \right)^2}\)

\(\Rightarrow {{\rm{P}}_2} = \frac{{\rm{F}}}{{16}}\)