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A frictional shear-strength approximation $|\tau |=c+{\sigma }_n^{\prime }\tan \phi$, with cohesion $c$ and internal friction angle $\phi$. Effective compression increases shear resistance. A separate tensile cutoff is generally needed at low confinement. The names refer to Christian Otto Mohr and Charles-Augustin de Coulomb.

For a cohesionless, optimally oriented normal fault with vertical maximum compression and horizontal minimum compression,where $\delta$ is fault dip and $\phi$ the internal friction angle. This is a geometric inference under the model, not a measurement of friction from dip alone.

For a compression-positive Mohr circle, put $C=({\sigma }_1^{\prime }+{\sigma }_3^{\prime })/2$ and $R=({\sigma }_1^{\prime }-{\sigma }_3^{\prime })/2$. Tangency to the cohesionless Mohr-Coulomb failure criterion $\tau ={\mu }_f{\sigma }_n^{\prime }$ gives $R/C=\sin \phi$. The contact point is $({\sigma }_n^{\prime },\tau )=(C{\cos }^2\phi ,C\sin \phi \cos \phi )$. The physical plane-normal angle to ${\sigma }_1^{\prime }$ is half the corresponding central angle, giving the stated dip. Equivalently the plane itself makes $45^{\circ }-\phi /2$ with ${\sigma }_1^{\prime }$.

The associated stress ratio is ${\sigma }_1^{\prime }/{\sigma }_3^{\prime }=(1+\sin \phi )/(1-\sin \phi )$. A reactivated nonoptimal plane, nonzero cohesion or unknown principal stress directions defeats the simple dip-only inversion.

A numerical answer needs an assumption about fault orientation. For a cohesionless fault optimally oriented for Mohr-Coulomb failure criterion, with vertical ${\sigma }_1$, write $\phi$ for the internal friction angle and $\delta$ for the fault dip. The angle between the plane and ${\sigma }_1$ is $90^{\circ }-\delta$, so the optimal normal-fault dip relation givesHence ${\mu }_f=\tan 40^{\circ }=0.839\simeq 0.84$ under that model.

For the Mohr circle, plot effective compression ${\sigma }_n^{\prime }$ horizontally and shear stress $\tau$ vertically. Draw a circle with diameter $[{\sigma }_3^{\prime },{\sigma }_1^{\prime }]$ tangent to $\tau ={\mu }_f{\sigma }_n^{\prime }$. Its centre and radius obey $R/C=\sin \phi$. One illustrative normalization is ${\sigma }_3^{\prime }=1$, ${\sigma }_1^{\prime }=4.59891$, with tangency at the resolved stresses on the $65^{\circ }$ plane. Pore fluid pressure shifts the total-stress circle horizontally; the friction law uses effective stress.

The dip does not determine the actual sliding coefficient of friction without the optimal-orientation assumption. A reactivated inherited plane may be nonoptimal, cohesion may matter, and the stress ratio and pore fluid pressure are not measured. For an arbitrary compression-positive stress state,at cohesionless sliding. Thus the requested single value is a conditional inference, not a friction measurement.

Assume an initially unfractured homogeneous isotropic material described by a linear Mohr-Coulomb failure criterion, with the given coefficient of friction ${\mu }_f=0.5$ representing the relevant internal friction angle. Use spatially uniform pore fluid pressure, effective stresses ${\sigma }_i^{\prime }={\sigma }_i-P_f$, and compression-positive signs. ThenThe Mohr circle in the ${\sigma }_1^{\prime },{\sigma }_3^{\prime }$ plane has centre $C=({\sigma }_1^{\prime }+{\sigma }_3^{\prime })/2$ and radius $R=({\sigma }_1^{\prime }-{\sigma }_3^{\prime })/2$. At first failure it is tangent to the failure envelope. The radius to the positive-shear tangent point is perpendicular to the line of slope $\tan \phi$, hence makes angle $90^{\circ }+\phi$ with the positive normal-stress axis.

If $\theta$ is the angle between the fault normal and ${\sigma }_1$, the stress transformation isThus $2\theta =90^{\circ }+\phi$, $\theta =58.2825^{\circ }$, and the angle of the fault itself to horizontal ${\sigma }_1$ isThe two conjugate thrust faults therefore strike approximately east–west and dip about $32^{\circ }$ north and $32^{\circ }$ south. This is the fault dip, not the $58.3^{\circ }$ fault-normal angle. Both sediments and basement have the same optimum orientation under these assumptions because their stipulated friction slopes are equal; their cohesion and absolute failure stresses could still differ.

The first two panels in solution give the labeled construction. Constant cohesion shifts the straight failure envelope without changing its slope, and uniform pore fluid pressure translates the Mohr circle in normal-stress coordinates; neither changes this tangency angle. In the illustrative cohesionless case, $R/C=\sin \phi$, giving ${\sigma }_1^{\prime }/{\sigma }_3^{\prime }=(1+\sin \phi )/(1-\sin \phi )\simeq 2.618$. Pre-existing weaknesses, a nonuniform stress field, anisotropic strength, or a nonlinear failure envelope invalidate the simple optimum-orientation prediction.^[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]