Question
Download Solution PDFWhat will be the effect on Euler’s crippling load for a column with both end hinged, if the length of the column is halved?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDF
Concept:
|
End Condition |
Both end hinged |
One end fixed other free |
Both end fixed |
One end fixed and other hinged |
|
Effective length (Le) |
L |
2L |
L/2 |
\(\frac{L}{{\sqrt 2 }}\) |
Crippling load:
\(P = \frac{{{\pi ^2}EI}}{{L_e^2}}\)
Calculation:
\({P_c} = \frac{{{\pi ^2}EI}}{{L_e^2}}\), where Le = effective length of column
\(\Rightarrow{P_c}\ \ \alpha \ \ \frac{1}{{L_e^2}}\)
Effective length for 1st case = L, effective length for 2nd case = L/2
Then, \(\frac{{{{\left( {{P_c}} \right)}_2}}}{{{{\left( {{P_c}} \right)}_1}}} = {\left( {\frac{{{L_e}_1}}{{{L_e}_2}}} \right)^2}={\left( {\frac{{L}}{{L/{2}}}} \right)^2}\)
\( \Rightarrow {\left( {{P_c}} \right)_2} = 4 \times {\left( {{P_c}} \right)_1}\)
Last updated on Jul 15, 2025
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