• # 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

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 $\theta =45{}^\circ$ ${{\sigma }_{n}}=\frac{1}{2}({{\sigma }_{x}}+{{\sigma }_{y}})$