What is the change in Euler’s buckling load, if the diameter of t
| What is the change in Euler’s buckling load, if the diameter of the column is reduced by 10%?
A. 4
B. 6
C. 34
D. 59
Please scroll down to see the correct answer and solution guide.
Right Answer is: C
SOLUTION
Concept:
Crippling load for the column:
\(\rm{\begin{array}{l} P = \frac{{{\pi ^2}EI}}{{L_e^2}} \Rightarrow P \propto I \propto {d^4}\\\end{array}}\)
Calculation:
\(\rm{ \frac{{{P_2}}}{{{P_1}}} = \frac{{d_2^4}}{{d_1^4}} = {\left( {\frac{{{d_2}}}{{{d_1}}}} \right)^4} = {\left( {\frac{{0.9d}}{d}} \right)^4} =0.6561}\)
\({P_2} = 0.6561\;{P_1}\)
∴ Percentage reduction is (P1 - P2) /P1 = 34.39 %.