= Solution
Draw a <failure envelope> in the $(\sigma_n',\tau)$ plane. A useful brittle shear-failure approximation is the <Mohr-Coulomb failure criterion>
$$
|\tau|=c+\mu_f\sigma_n',\qquad\mu_f=\tan\varphi,
$$
where $c$ is <cohesion>, $\varphi$ the <internal friction angle>, and $\sigma_n'$ the effective compressive <normal stress>. A circle wholly inside the envelope is below this ideal failure threshold. Increasing the <differential stress> expands it until it first touches the envelope: the tangent points identify the failing plane orientations. For a linear envelope, tangency implies
$$
R=C'\sin\varphi+c\cos\varphi,\qquad \sigma_1'=\sigma_3'\frac{1+\sin\varphi}{1-\sin\varphi}+\frac{2c\cos\varphi}{1-\sin\varphi}.
$$
With the angle convention above, the upper tangent point is at $2\theta=90^\circ+\varphi$. Its plane therefore makes $45^\circ-\varphi/2$ with the maximum-compression direction; there is a conjugate plane on the other side. For $\varphi=30^\circ$, this is $30^\circ$, rather than the $45^\circ$ plane of maximum <shear stress>. An existing weak <fault> can reactivate first, with its own lower <cohesion> and friction, and need not have this optimum new-fracture orientation.
For tensile <brittle fracture>, add a <tensile strength> cutoff: failure occurs when $\sigma_3'\leq-T_0$, where $T_0>0$. The opening fracture is approximately perpendicular to the minimum-compression direction. Real <rock> envelopes curve, especially at small confinement; a straight shear envelope must not be extrapolated as an exact tensile law. High confinement can instead favor distributed or ductile deformation, so merely drawing a larger circle does not establish that brittle failure is the active mechanism.<Stress and strength lecture notes by MIT OpenCourseWare>[Lecture 10, frictional strength and Mohr construction; the tensile cutoff is a separate condition.]{ref}
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