Question
Download Solution PDFThe buckling load for a given material depends upon:
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFExplanation:
Column:
- If a beam element is under a compressive load and its length is an order of magnitude larger than either of its other dimensions such a beam is called a column.
- Due to its size, its axial displacement is going to be very small compared to its lateral deflection called buckling.
Euler's Buckling Load:
\( {P_{cr}} = \frac{{{{\rm{\pi }}^2}E{I_{\min }}}}{{l_e^2}}\)
where, Pcr = critical load for buckling; E = Young's Modulus (GPa); Imin = Area moment of inertia, le = effective length; kmin = minimum radius of gyration
Slenderness Ratio (S):
The ratio between the length and least radius of gyration.
\(S = \frac{{{l_e}}}{{{k_{\min }}}}\)
Hence, the buckling load depends on the Slenderness Ratio and modulus of elasticity.
Additional Information
The buckling load for various end conditions is given in the table below.
End conditions |
Le |
Buckling load |
Both ends hinged |
Le = L |
\({P_b} = \frac{{{\pi ^2}E{I_{}}}}{{L^2}}\) |
Both ends fixed |
\({L_e} = \frac{L}{2}\) |
\({P_b} = \frac{{{4\pi ^2}E{I_{}}}}{{L^2}}\) |
One end fixed and another end is free |
Le = 2L |
\({P_b} = \frac{{{\pi ^2}E{I_{}}}}{{4L^2}}\) |
One end fixed and another end is hinged |
\({L_e} = \frac{L}{{\sqrt 2 }}\) |
\({P_b} = \frac{{{2\pi ^2}E{I_{}}}}{{L^2}}\) |
Last updated on May 19, 2025
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