• # question_answer According to the maximum shear stress theory failure, permissible twisting moment in a circular shaft is T. The permissible twisting moment in the same shaft as per the maximum principal stress 6 theory of failure will be                        A) T/2                              B) T          C) $\sqrt{2T}$                  D) 2T

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