A) 2.5
B) 2.8
C) 3.0
D) 3.5
Correct Answer: B
Solution :
\[\sigma =80\,\text{N/m}{{\text{m}}^{\text{2}}}\text{,}\] \[\tau =30\,\text{N/m}{{\text{m}}^{\text{2}}}\text{,}\]\[\sigma =280\,\text{N/m}{{\text{m}}^{\text{2}}},\] \[{{\sigma }_{y}}=280\,\text{N/m}{{\text{m}}^{\text{2}}}\] 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 \[=\frac{{{\tau }_{y}}}{{{\tau }_{\max }}}=\frac{140}{50}=2.8\]
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