Past exam of the year 2 of the earth sciences (geology) course of the University of Oxford / 2025 / Paper 3 / 3 / b

For a plane whose normal lies in the ${\sigma }_1$–${\sigma }_3$ plane, rotate through $2\theta$ from $({\sigma }_1,0)$ in the stated sign convention. The circle point givesThe normal-to-plane angle is essential: if the angle supplied is between the plane itself and ${\sigma }_1$, use its complement for $\theta$. The principal stress planes have zero shear stress, and planes at $45^{\circ }$ to those principal stress planes have maximum $|\tau |$.

For an arbitrary plane in a fully three-dimensional stress field, let $n_i$ be its direction cosines in the principal stress basis. ThenThese follow by resolving the traction into its normal and tangential components; the compression-positive tensor is the negative of the usual tension-positive Cauchy stress tensor. The family of Mohr circles visualizes the possible combinations, but the two-dimensional $2\theta$ rule alone does not specify an arbitrary three-dimensional orientation.^[ref]

Draw a failure envelope in the $({\sigma }_n^{\prime },\tau )$ plane. A useful brittle shear-failure approximation is the Mohr-Coulomb failure criterionwhere $c$ is cohesion, $\phi$ the internal friction angle, and ${\sigma }_n^{\prime }$ 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 impliesWith the angle convention above, the upper tangent point is at $2\theta =90^{\circ }+\phi$. Its plane therefore makes $45^{\circ }-\phi /2$ with the maximum-compression direction; there is a conjugate plane on the other side. For $\phi =30^{\circ }$, 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 ${\sigma }_3^{\prime }\le -T_0$, 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.^[ref]

Pore fluid pressure $P$ reduces the compressive load carried across grain contacts or an accessible fault. In the usual elementary approximation,At fixed total stress, increasing $P$ translates the entire Mohr circle left by $P$ without changing its radius or the differential stress. The circle therefore approaches the frictional failure envelope even though the applied shear has not increased. Equivalently, in total-stress coordinates the strength on an existing plane is $c+{\mu }_f({\sigma }_n-P)$, reduced by ${\mu }_{f}P$. For a known orientation and constant frictional parameters, the ideal critical pressure isprovided the plane is in contact and sliding failure is reached before tensile opening.

For the diagram's illustrative ${\sigma }_1=80$, ${\sigma }_3=20$, $c=10\mathrm{MPa}$ and ${\mu }_f=0.6$, the dry circle is below failure. The freely oriented tangency threshold is about $P=8.4\mathrm{MPa}$; the $15\mathrm{MPa}$ circle is already beyond that threshold and shows the loading tendency, not a stable post-failure state. If $P$ grows enough that ${\sigma }_3-P\le -T_0$, fluid pressure can instead open a tensile fracture. A porous-continuum effective stress law may use ${\sigma }_i^{\prime }={\sigma }_i-\alpha P$ with a coefficient $\alpha$ different from one. Also, changing fluid pressure can change total stress through constrained deformation; pure horizontal translation assumes those total stresses remain fixed.^[ref]