Question

What will be the Euler buckling load of a column when it is fixed at one and free at other end?
1. \(\dfrac{\pi^2 EI}{l^2}\)
2. \(\dfrac{\pi^2 EI}{4l^2}\)
3. \(\dfrac{2\pi^2 EI}{l^2}\)
4. \(\dfrac{4\pi^2 EI}{l^2}\)
5. \(\dfrac{4\pi^2 EI}{2l^2}\)

Answer 1

Correct Answer - Option 2 : \(\dfrac{\pi^2 EI}{4l^2}\)

Concept:

Euler buckling load for a column is given by,

\({{\rm{P}}_{\rm{E}}}{\rm{\;}} = {\rm{\;}}\frac{{{{\rm{\pi }}^2}{\rm{EI}}}}{{{\rm{L}}_{\rm{e}}^2}}\)

Where, Le is the effective length of the column that depends on the end support conditions, and EI is the flexural rigidity of the column.

Now, effective length (Le) for different end conditions in terms of actual length (L) are listed in the following table:

Support Conditions	Effective length (Le)
Both ends hinged/pinned	Le = L
One end hinged other end fixed	Le = L/√2
Both ends fixed	Le = L/2
One end fixed and the other end free	Le = 2L

 As Effective length L_e = 2L

 ∴  Euler buckling load of a column when it is fixed at one and free at other end is \({{\rm{P}}_{\rm{E}}}{\rm{\;}} = {\rm{\;}}\frac{{{{\rm{\pi }}^2}{\rm{EI}}}}{{{\rm{4L}}_{\rm}^2}}\)