• # question_answer A shaft is subjected to a maximum bending stress of $80\,N/m{{m}^{2}}$ and maximum shearing stress equal to $30\,N/m{{m}^{2}}$ at a particular section. If the yield point in tension of the material is $280\,N/m{{m}^{2}},$ and maximum shear stress theory of failure is used, then the factor of safety obtained will be: A) 2.5                               B) 2.8C) 3.0                               D) 3.5

$\sigma =80\,\text{N/m}{{\text{m}}^{\text{2}}}\text{,}$ $\tau =30\,\text{N/m}{{\text{m}}^{\text{2}}}\text{,}$$\sigma =280\,\text{N/m}{{\text{m}}^{\text{2}}},$ ${{\sigma }_{y}}=280\,\text{N/m}{{\text{m}}^{\text{2}}}$ According to maximum principal stress theory, yield stress in shear,             ${{\tau }_{y}}=\frac{{{\sigma }_{y}}}{2}=\frac{280}{2}=140\,\text{N/m}{{\text{m}}^{\text{2}}}$ Maximum shear stress induced, ${{\tau }_{y}}=\frac{1}{2}\sqrt{{{\sigma }^{2}}+4{{\tau }^{2}}}=50\,\text{N/m}{{\text{m}}^{\text{2}}}$ $=\frac{1}{2}\sqrt{{{\sigma }^{2}}+4{{\tau }^{2}}}$             $=\frac{1}{2}\sqrt{{{80}^{2}}+4+{{30}^{2}}}$ Factor of safety $=\frac{{{\tau }_{y}}}{{{\tau }_{\max }}}=\frac{140}{50}=2.8$