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

  • question_answer Principal stresses at a point in plane stressed element are \[{{\sigma }_{x}}={{\sigma }_{y}}=500k\text{g/c}{{\text{m}}^{\text{2}}}.\] Normal stress on the plane inclined at \[45{}^\circ \] to x-axis will be:

    A) 0                                 

    B) \[500\,k\text{g/c}{{\text{m}}^{\text{2}}}\]

    C) \[707\,k\text{g/c}{{\text{m}}^{\text{2}}}\]                   

    D) \[1000\,k\text{g/c}{{\text{m}}^{\text{2}}}\]

    Correct Answer: B

    Solution :

    \[{{\sigma }_{n}}=\frac{1}{2}({{\sigma }_{x}}+{{\sigma }_{y}})+\frac{1}{2}({{\sigma }_{x}}-{{\sigma }_{y}})\,cos\theta \] For \[\theta =45{}^\circ \] \[{{\sigma }_{n}}=\frac{1}{2}({{\sigma }_{x}}+{{\sigma }_{y}})+\frac{1}{2}(500+500)=500\,\text{kg/c}{{\text{m}}^{\text{2}}}\]

You need to login to perform this action.
You will be redirected in 3 sec spinner