Question
Download Solution PDFWhen the both ends of a column are fixed, the crippling load is P. If one end of the column is made free, the value of crippling load will be changed to
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
Crippling load, \({{\rm{P}}_{{\rm{cr}}}} = \frac{{{{\rm{\pi }}^2}{\rm{EI}}}}{{{\rm{L}}_{\rm{e}}^2}}\)
Where,
Le = 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 (Pcr)2 = P2
Using equation (1), we get
\(\left( {\rm{P}} \right) \times {\left( {\frac{{\rm{L}}}{2}} \right)^2} = \left( {{{\rm{P}}_2}} \right) \times {\left( {2{\rm{L}}} \right)^2}\)
\(\Rightarrow {{\rm{P}}_2} = \frac{{\rm{P}}}{{16}}\)
∴ Crippling load becomes \(\frac{P}{{16}}\)Last updated on Jun 14, 2025
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