• # question_answer From a tension test, the yield strength of steel is found to be $200\,\text{N/m}{{\text{m}}^{\text{2}}}\text{.}$ Using a factor of safety of 2 and applying maximum principal stress theory of failure, the permissible stress in the steel shaft subjected to torque will be: A) $50\,\text{N/m}{{\text{m}}^{\text{2}}}$                    B) $57.7\,\text{N/m}{{\text{m}}^{\text{2}}}$C) $86.6\,\text{N/m}{{\text{m}}^{\text{2}}}$                 D) $100\,\text{N/m}{{\text{m}}^{\text{2}}}$

${{\sigma }_{y}}=200\,\text{N/m}{{\text{m}}^{\text{2}}}\text{,}$ FOS = 2 Allowable stress, ${{\sigma }_{y}}=\frac{{{\sigma }_{y}}}{FOS}=\frac{200}{2}=100\,\text{N/m}{{\text{m}}^{\text{2}}}$ According to maximum principal stress theory,             ${{\sigma }_{1}}={{\sigma }_{y}}$ Under torsion, maximum principal stress,             ${{\sigma }_{1}}=\tau$ Permissible stress $=100\,\text{N/m}{{\text{m}}^{\text{2}}}$