Railways Technical Ability Engineering Mechanics and Strength of Materials Question Bank Strength of Materials

  • 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) 2.8

    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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