For this system, the stress tensor is given in dyadic notation by: $$\vec{\ \sigma}=\sigma_{xx}\vec{i_x}\vec{i_x}+\sigma_{yy}\vec{i_y}\vec{i_y}$$A unit normal vector to a plane of arbitrary orientation is given by: $$\vec{\ n}=\cos{\theta}\vec{i_x}+\sin{\theta}\vec{i_y}$$where ##\theta## is the angle relative to the x axis. If we dot the stress tensor with the unit normal, we obtain the traction vector on the plane: $$\vec{\ t}=\vec{\ \sigma}\centerdot \vec{\ n}=\sigma_{xx}\cos{\theta}\vec{i_x}+\sigma_{yy}\sin{\theta}\vec{i_y}$$The shear component (shear stress) of this vector on the plane is obtained by dotting the traction vector with the unit tangent to the plane: $$\vec{\ t}\centerdot \vec{i_t}=\vec{\ t}\centerdot (-\sin{\theta}\vec{i_x}+\cos{\theta}\vec{i_y})=(\sigma_{yy}-\sigma_{xx})\sin{\theta}\cos{\theta}=\frac{(\sigma_{yy}-\sigma_{xx})}{2}\sin{2\theta}$$The normal stress component of the traction vector is obtained by dotting the traction vector with the unit normal: $$\vec{\ t}\centerdot \vec{\ n}=\vec{\ t}\centerdot (\cos{\theta}\vec{i_x}+\sin{\theta}\vec{i_y})=\sigma_{xx}\cos^2{\theta}+\sigma_{yy}\sin^2{\theta}=\frac{(\sigma_{xx}+\sigma_{yy})}{2}+\frac{(\sigma_{xx}-\sigma_{yy})}{2}\cos{2\theta}$$The shear stress has maximum magnitude at ##2\theta=\pi/2##. At this value of theta, the normal stress is equal to ##\frac{(\sigma_{xx}+\sigma_{yy})}{2}##
