Question

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

• A. 4
• B. 36
• C. 49
• D. 59

Answer 1

Answer:

D. 59

It is given by the formula $P_{cr} = \frac{\pi^ 2 EI_{min}}{ L_e ^ 2}$ where $I = \frac{\pi d ^ 4}{64}$ and then $P_{cr} =(\frac{ \pi^ 2 E}{ L_e ^ 2 })(\frac{\pi d^ 4}{ 64} )$ Hence, For the circular section, $P_ {cr} \propto d^ 4$ ; So the Reduction in Euler's Load is =$\frac{ (P_{cr})_i -(P_{cr})_f}{ (P_{cr} )_i}$= $\frac{d_{i}^ 4 - d_{f} ^ 4}{d_{i} ^ 4}$ ; Let the initial diameter of the column is d. Given that diameter of the column is reduced by 20%. Hence, $d_{f} =0.8d$ \Rightarrow$Reduction in load =$\frac {d ^ 4 - (0.8d) ^ 4}{d ^ 4} = 0.5904$ .. The Percentage Reduction in Load is 59.04%