A) T/2
B) T
C) \[\sqrt{2T}\]
D) 2T
Correct Answer: D
Solution :
According to the maximum shear stress theory, \[{{\tau }_{\max }}=\frac{{{\sigma }_{1}}-{{\sigma }_{2}}}{2}={{\tau }_{y}}=\frac{{{\sigma }_{y}}}{2}\] \[\frac{16T}{\pi {{d}^{3}}}=\frac{{{\sigma }_{y}}}{2}\] \[\therefore \,\,\,\,\,\,\,\,{{\sigma }_{y}}=\frac{32T}{r{{d}^{3}}}\] According to maximum principal stress theory, \[{{\sigma }_{1}}={{\sigma }_{y}}\frac{32T}{r{{d}^{3}}}=2\left( \frac{16T}{\pi {{d}^{3}}} \right)\] Hence, permissible torque = 2T
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