 Solution

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]