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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