Question

The diameter of long column is D buckling load is P. If the diameter is reduced by 50% new buckling load will be:

• A. 0.4 P
• B. 0.25 P
• C. 0.0825 P
• D. 0.0625 P

Answer 1

Answer:

D. 0.0625 P

Given: Initial diameter = D , Initial buckling load = $P_{b_1}$ ; Final diameter after 50% decreased = D/2, final buckling load = $P_{b_2} </p><p style="color: rgb(0,0,0); margin-top: 2px; margin-bottom: 2px" data-pxy="true">Design of columns using Euler's theory of buckling:$P_b =\frac{\pi ^ 2 EA K_{min} ^ 2}{ L_e ^ 2}$radius of gyration:$K_{{min}_1}=\sqrt{\frac{\frac{\pi}{64} D^4}{\frac{\pi}{4}D^2}}$=$\frac{D}{4}$and$K_{{min}_2}=\sqrt{\frac{\frac{\pi}{64}(\frac{D}{4})^4}{\frac{\pi}{4}(\frac{D}{2})^2}}$=$\frac{D}{8}$therefore Buckling load is given by:$\frac{P_{b_1}}{P_{b_2}}=\frac{\frac{\pi^2E(\frac{D^2}{4})(\frac{D}{4})^2}{L^2}}{\frac{\pi^2E(\frac{D^2}{16})(\frac{D}{8})^2}{L^2}}$=$16$$\Rightarrow \frac{P_{b_1}}{P_{b_2}}=16 \Rightarrow P_{b_2}=\frac{P_{b_1}}{16} \Rightarrow P_{b_2}=.0625 P$