= Solution
Use compression-positive <effective stress> normal to the plane $\sigma'_n$ on the horizontal axis and <shear stress> $\tau$ on the vertical axis. For any chosen pair of <principal stresses>, <Mohr's circle> has centre $C=(\sigma'_1+\sigma'_3)/2$ and radius $R=(\sigma'_1-\sigma'_3)/2$.
For the <reverse fault> example, $C=60\ \mathrm{MPa}$ and $R=30\ \mathrm{MPa}$. Draw the two <Mohr-Coulomb failure criterion> lines $\tau=\pm\sigma'_n/\sqrt3$. The tangencies are
$$
(\sigma'_n,\tau)=(45,\pm15\sqrt3)\ \mathrm{MPa}.
$$
These are <shear failure> points with nonzero <shear stress>. The two smaller three-dimensional <Mohr circles> use $(\sigma'_1,\sigma'_2)$ and $(\sigma'_2,\sigma'_3)$; they lie inside the outer circle and do not touch the envelope in this example.
For the opening <geological joints>, use the <tensile failure criterion> $\sigma'_n=-T_0$. The example gives $C=3.5\ \mathrm{MPa}$, $R=6.5\ \mathrm{MPa}$ and $T_0=3\ \mathrm{MPa}$. The circle first touches this vertical cutoff at \b[$(\sigma'_n,\tau)=(-3,0)\ \mathrm{MPa}$]. Thus the <geological joints> form on planes normal to $\sigma_3$ and have zero resolved <shear stress>. A complete brittle envelope can combine this tensile cutoff with compressive <Mohr-Coulomb failure criterion>, using sufficient <cohesion> that shear failure does not occur first. The failure-mechanics figure displays these two distinct constructions.
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