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

  • question_answer In the case of bi-axial state of normal stress, the normal stress on \[45{}^\circ \] plane is equal to:     

    A) The sum of the normal stresses       

    B) Difference of the normal stresses       

    C) Half the sum of the normal stresses    

    D) Half the difference of the normal stresses

    Correct Answer: C

    Solution :

    Normal stress on oblique plane under biaxial loading, \[{{\sigma }_{n}}=\frac{1}{2}({{\sigma }_{x}}+{{\sigma }_{y}})+\frac{1}{2}({{\sigma }_{x}}-{{\sigma }_{y}})\,cos2\theta \] For \[\theta =45{}^\circ \] \[{{\sigma }_{n}}=\frac{1}{2}({{\sigma }_{x}}+{{\sigma }_{y}})\]


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