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Railways NTPC (Technical Ability) Engineering Mechanics and Strength of Materials Question Bank Strength of Materials

  • 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) \[100\,\text{N/m}{{\text{m}}^{\text{2}}}\]

    Correct Answer: D

    Solution :

    \[{{\sigma }_{y}}=200\,\text{N/m}{{\text{m}}^{\text{2}}}\text{,}\] 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 \[=100\,\text{N/m}{{\text{m}}^{\text{2}}}\]


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