 Solution

Use compression-positive principal stresses ${\sigma }_1\ge {\sigma }_2\ge {\sigma }_3$. The two-dimensional Mohr's circle shown is the ${\sigma }_1$–${\sigma }_3$ circle: its horizontal axis is compressive normal stress ${\sigma }_n$, its vertical axis signed shear stress $\tau$, its centre is $C=({\sigma }_1+{\sigma }_3)/2$, and its radius is $R=({\sigma }_1-{\sigma }_3)/2$. The intercepts $({\sigma }_1,0)$ and $({\sigma }_3,0)$ represent principal stress planes, on which shear stress vanishes. The top and bottom points have $|\tau |=R$, the maximum shear stress.

In the rock sketch, $\theta$ is the counterclockwise angle from the ${\sigma }_1$ direction to the plane's unit normal vector $\mathbf{n}=(\cos \theta ,\sin \theta )$. Choose the clockwise unit tangent $\mathbf{t}=(\sin \theta ,-\cos \theta )$ and define $\tau =\mathbf{t}\cdot {\mathbf{\sigma }}^c\mathbf{n}$, with ${\mathbf{\sigma }}^c=\mathrm{diag}({\sigma }_1,{\sigma }_3)$ compression positive. Thus a positive $\theta$ produces a positive $2\theta$ rotation on the plotted circle. The actual traction on the positive-normal face is $-{\mathbf{\sigma }}^c\mathbf{n}=-{\sigma }_n\mathbf{n}-\tau \mathbf{t}$. Reversing the tangent convention reverses the plotted shear sign, not the physical prediction.

In three dimensions there are circles for each pair of principal stresses. The largest is the ${\sigma }_1$–${\sigma }_3$ circle; a general three-dimensional plane requires all three principal stresses and its normal, rather than a single angle on this circle.