A) 2
B) 4
C) 8
D) 16
Correct Answer: A
Solution :
\[{{P}_{x}}=\frac{{{\pi }^{2}}E{{I}_{x}}}{{{L}^{2}}}=\frac{{{\pi }^{2}}EB{{H}^{3}}}{12{{L}^{2}}}\] \[{{P}_{y}}=\frac{4{{\pi }^{2}}E{{I}_{y}}}{{{L}^{2}}}=\frac{4{{\pi }^{2}}EB{{H}^{3}}}{12{{L}^{2}}}\] For \[{{P}_{x}}={{P}_{y}}\] \[\frac{{{\pi }^{2}}EB{{H}^{3}}}{12{{L}^{2}}}=\frac{4{{\pi }^{2}}EB{{H}^{3}}}{12{{L}^{2}}}\] Or \[{{H}^{2}}=4{{B}^{2}}\] \[\frac{H}{B}=2\]You need to login to perform this action.
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