Mohr's circle for plane stress

Plane stress describes a thin region where the out-of-plane stress components are negligible. Take tensile normal stress as positive. Define positive $\tau_{xy}$ to act in the positive $y$ direction on the face whose outward normal points in the positive $x$ direction. The other shear arrows must balance the element's moment. Keep this sign convention throughout: changing it halfway through the calculation changes the predicted orientation. Mohr's circle finds the two in-plane principal stresses. In a full three-dimensional check, the additional principal stress is zero under the plane-stress assumption, so include it when ranking all three principal stresses.

Plot normal stress $\sigma$ horizontally and shear stress $\tau$ vertically, with positive $\tau$ upward. Mark $A=(\sigma_x,\tau_{xy})$ and $B=(\sigma_y,-\tau_{xy})$. The midpoint is $C=(\sigma_x+\sigma_y)/2$ and the radius is $R=\sqrt{[(\sigma_x-\sigma_y)/2]^2+\tau_{xy}^2}$. The circle meets the normal-stress axis at $C+R$ and $C-R$, which are the principal stresses. The largest magnitude of in-plane shear stress is $R$ and occurs where the circle reaches its top or bottom. Do not confuse that with a three-dimensional maximum shear check.

An element rotated physically through $\theta$ counterclockwise corresponds to moving around this particular Mohr plot through $2\theta$ clockwise from point $A$. To find the major principal direction without an ambiguous tangent quadrant, calculate $\theta_p=\tfrac12\operatorname{atan2}(2\tau_{xy},\sigma_x-\sigma_y)$, then check it gives the larger stress. For a rotated face, use $\sigma_{x'}=C+D\cos 2\theta+\tau_{xy}\sin 2\theta$ and $\tau_{x'y'}=-D\sin 2\theta+\tau_{xy}\cos 2\theta$, where $D=(\sigma_x-\sigma_y)/2$. Angles describe face-normal directions, not lines drawn along the face.