Question

A column with one end fixed and one end free has a critical buckling load of 100 N. For the same column, if the free end is replaced with a pinned end then the critical buckling load will be ____________ N (round off to the nearest integer).

Answer 1

Concept:

Euler's buckling load \(P_e = \frac{{{\pi ^2}EI_{min}}}{{{L_e^2}}}\)

where L_e = effective length of a column

1. If one end of the column is fixed and the other end free L_e = 2L
2. If one end fixed and the other end is pinned or hinged L_e = \(\frac{L}{\sqrt2}\)
3. If both ends of the column are fixed Le = \(\frac{L}{2}\)
4. If both ends of a column are hinged Le = L

Calculation:

Given:

For the column whose one end fixed and another end free:

Buckling load P₁ = 100 N, L₁ = 2L

If the free end is replaced by pinned joint L₂ = \(\frac{L}{\sqrt2}\)

Since the column is the same \(P_e \propto \frac{1}{L_e^2}\)

\(\frac{P_2}{P_1} = \frac{L_1^2}{L_2^2}\)

P₂ = \(100 \times\frac{4L^2}{\frac{L^2}{2}}\)  = 800 N