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

  • question_answer
      The state of plane stress at a point is described by \[{{\sigma }_{x}}={{\sigma }_{y}}\] and \[{{\tau }_{xy}}=0.\] The normal stress on the plane inclined at \[45{}^\circ \] to the x-plane will be       

    A) \[\sigma \]                                 

    B) \[\sqrt{2\sigma }\]

    C) \[\sqrt{3\sigma }\]                      

    D) \[2\sigma \]  

    Correct Answer: A

    Solution :

    \[{{\sigma }_{1}}=\frac{1}{2}({{\sigma }_{x}}+{{\sigma }_{y}})+\frac{1}{2}({{\sigma }_{x}}-{{\sigma }_{y}})\,cos2\theta \] At \[\theta =45{}^\circ \] \[{{\sigma }_{1}}=\frac{1}{2}(\sigma +\sigma )+\frac{1}{2}(\sigma -\sigma )=\sigma \]


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