Question 26.3.47: If the diameter of a long column is reduced by 20%, the perc...

(d)

EXPLANATION

Euler’s buckling load,

P=\frac{\pi^2 E I}{L^2} \text { where } I=\frac{\pi}{64} d^4

For circular column when diameter is reduced by 20%, then dia. of new column = 0.8 d

∴ New moment of inertia,  I^*=\frac{\pi}{4} \times(0.8  d)^4=\frac{\pi}{4} \times d^4 \times(0.8)^4

Initial buckling load,      P=\frac{\pi^2 E}{L^2} \times \frac{\pi}{64} d^4

New buckling load,          P^*=\frac{\pi^2 E}{L^2} \times \frac{\pi}{64} \times d^4 \times 0.8^4

∴      \% \quad \text { reduction in load }=\frac{P-P^*}{P} \times 100 \\ \space \\  =\frac{\left(\frac{\pi^2 E}{L^2} \times \frac{\pi}{64} d^4-\frac{\pi^2 E}{L^2} \times \frac{\pi}{64} d^4 \times 0.8^4\right)}{\frac{\pi^2 E}{L^2} \times \frac{\pi}{4} d^4} \times 100 \\ \space \\ =\frac{1-0.8^4}{1}=(1-0.4096) \times 100= \pmb{5 9 \% .}