A point in two dimensional stress state subjected to biaxial stre
A point in two dimensional stress state subjected to biaxial stress is shown in figure below. What is the normal stress acting on the plane AB inclined at an angle of θ with the vertical plane ?
A. σ cos2 θ
B. σ
C. σ sin θ cos θ
D. zero
Please scroll down to see the correct answer and solution guide.
Right Answer is: B
SOLUTION
Concept:
ii) If σx and σy are normal stress on vertical and horizontal plane respectively and this plane is accompanied by shear stress τxy then normal stress and shear stress on plane a-a, which is inclined at an angle θ from plane of σx.
\(\sigma _{1\left( {a - a} \right)}' = \frac{{{\sigma _x} + {\sigma _y}}}{2} + \left( {\frac{{{\sigma _x} - {\sigma _y}}}{2}} \right) \cdot \cos 2\theta + {\tau _{xy}}\sin 2\theta \)
\(\sigma _{2\left( {a - a} \right)}' = \left( {\frac{{{\sigma _x} + {\sigma _y}}}{2}} \right) - \left( {\frac{{{\sigma _x} - {\sigma _y}}}{2}} \right) \cdot \cos 2\theta - {\tau _{xy}}\sin 2\theta \)
\({\tau _{\left( {a - a} \right)}} = - \left( {\frac{{{\sigma _x} - {\sigma _y}}}{2}} \right)\sin 2\theta + {\tau _{xy}}\cos 2\theta \)
Calculation:
Here the plane is making an angle θ with vertical but angle is in clockwise direction, so take –θ.
\({\sigma _n} = \frac{{\sigma + \sigma }}{2} + \left( {\frac{{\sigma - \sigma }}{2}} \right)\cos - 2\theta \)
= σ
