= Solution
<Pore fluid pressure> $P$ reduces the compressive load carried across grain contacts or an accessible <fault>. In the usual elementary approximation,
$$
\boldsymbol\sigma'=\boldsymbol\sigma^c-P\mathbf I,\qquad \sigma_i'=\sigma_i-P,\quad\sigma_n'=\sigma_n-P,\quad\tau'=\tau.
$$
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 is
$$
P_{\mathrm{crit}}=\sigma_n-\frac{|\tau|-c}{\mu_f},
$$
provided 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 $\sigma_3-P\leq-T_0$, fluid pressure can instead open a tensile fracture. A porous-continuum <effective stress> law may use $\sigma_i'=\sigma_i-\alpha P$ 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.<Stress and strength lecture notes by MIT OpenCourseWare>[Lecture 10, frictional strength modified by pore-fluid pressure.]{ref}
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