A) \[\frac{1}{5}\]
B) \[\frac{3}{4}\]
C) \[\frac{9}{5}\]
D) \[\frac{11}{6}\]
Correct Answer: C
Solution :
\[{{M}_{e}}=\frac{1}{2}\left[ M+\sqrt{{{M}^{2}}+{{T}^{2}}} \right]\] \[=\frac{1}{2}\left[ 400+\sqrt{{{400}^{2}}+{{300}^{2}}} \right]\] =450 k.N.m \[{{\sigma }_{b}}=\frac{32M}{\pi {{d}^{3}}}=\frac{32}{\pi {{d}^{3}}}\times 450\] \[{{T}_{e}}=\sqrt{{{M}^{2}}+{{T}^{2}}}=500\,\,kN.m\] \[\tau =\frac{16T}{\pi {{d}^{3}}}=\frac{16}{\pi {{d}^{3}}}\times 500\] \[\frac{{{\sigma }_{b}}}{\tau }=\frac{2\times 450}{500}=\frac{9}{5}=1.8\]You need to login to perform this action.
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