The buckling load for a given material depends upon: 

This question was previously asked in
JKSSB JE Mechanical 02 Nov 2021 Shift 1 Official Paper
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  1. slenderness ratio and modulus of elasticity
  2. Poisson's ratio and modulus of elasticity
  3. Poisson's ratio and slenderness ratio
  4. slenderness ratio and cross sectional area

Answer (Detailed Solution Below)

Option 1 : slenderness ratio and modulus of elasticity
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Detailed Solution

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Explanation:

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, l= 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

\({L_e} = \frac{L}{{\sqrt 2 }}\)

\({P_b} = \frac{{{2\pi ^2}E{I_{}}}}{{L^2}}\)
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