If the diameter of a long column is reduced by 20 %, the percentage of reduction in Euler's buckling load is _______.

Question

Question

This question was previously asked in

JKSSB JE CE 2021 Official Paper Shift 1 (Held on 28 Oct 2021)

Answer 1

Concept:

Buckling load given by Euler is:

\(P = \frac{{{\pi ^2}EI}}{{L_{eff}^2}}\)

where,

E = Modulus of elasticity of the material

I = Moment of Inertia

\(I = \frac{\pi }{{64}}{D^4}\)

Leff = Effective length of the column

Calculation:

Let P1 be the buckling load of the column when the diameter of the column is "D"

\({P_1} \propto D^4\)

Let P2 be the buckling load of the column when the diameter is reduced by 20%, it becomes 0.8D.

\({P_2} \propto {\left( {0.80D} \right)^4}\)

\(\% \;reduction = \;\frac{{{P_1} - {P_2}}}{{{P_1}}} \times 100\)

\(\% \;reduction = \;\frac{{{D^4} - {{\left( {0.80D} \right)}^4}}}{{{D^4}}} \times 100\)

% reduction = 59.04 %