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Physicist and engineer whose work includes friction and strength of materials. His name appears in the Mohr-Coulomb failure criterion.

One of a pair of fault orientations that can form under a common principal stress field with opposite senses of resolved shear stress. For newly formed, optimally oriented faults obeying the same Mohr-Coulomb failure criterion, their acute intersection angle is $90^{\circ }-\varphi$, where $\varphi$ is the friction angle. The maximum compressive principal stress bisects that acute angle. This permits an estimate $\mu =\tan \varphi$ only under the common-formation and optimal-orientation assumptions, rather than for arbitrary inherited faults.

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.

Use a rock strength envelope: at each depth plot the smaller differential stress needed for frictional failure and for sustained ductile material creep at a specified strain rate. Shallow rock generally strengthens with depth because effective confining pressure increases; sufficiently deep hot rock weakens because thermally activated material creep becomes efficient.

At shallow depth, Byerlee's law or a Mohr-Coulomb failure criterion gives a relation of the formwhere $C$ is cohesion, $\mu$ a coefficient of friction, ${\sigma }_n$ the compressive normal stress and $P_f$ the pore pressure. Increasing overburden generally increases effective stress, hence the frictional strength. Conversion from fault shear stress to differential stress depends on the stress regime and fault orientation; a sketch should not label a universal numerical slope. High pore pressure or weak fault gouge can lower the frictional strength substantially.

A simplified dislocation creep relation isHere $Q$ is activation energy, $V^{\ast }$ activation volume, $R$ the molar gas constant and $T$ absolute temperature; $A$ and $n$ depend on composition and deformation mechanism. At fixed strain rate, increasing temperature usually decreases the required stress strongly. The competing effects of pressure, water weakening of rocks, grain size and composition also matter. The intersection with frictional failure gives a brittle-ductile transition, not a change from solid to liquid.

For the Indian continental interior, a useful first sketch has an upper-crustal strength peak, a weaker lower continental crust, and renewed strength in lithospheric mantle beneath the Moho. Quartz-bearing rocks can undergo ductile material creep at conditions where colder, drier olivine-rich mantle retains appreciable strength. This creates a possible two-peak, or “jelly sandwich”, rock strength envelope. With further depth the mantle also weakens. A cold craton can retain strong lithospheric mantle to much greater depths than a tectonically heated continental region.

For the central Pacific, show thin mafic oceanic crust above a relatively thick layer of strong lithospheric mantle. There need not be a conspicuous weak lower-crustal layer. Much of the integrated strength can reside below the Moho, with a broad strength peak followed by ductile weakening towards the asthenosphere. Cooling and age control the thickness: young hot oceanic lithosphere would give a much shallower and weaker profile than an old central-Pacific plate.

The plotted Moho depths of about 38 km and 7 km, and all plotted stresses, are illustrative assumptions. “India” alone does not specify a unique geothermal gradient, water content or strain rate, and the Pacific includes plates of different ages. The robust comparison is the possible crust–mantle layering in a continent versus the thin oceanic crust and mantle-dominated strength of an old oceanic plate. The Moho is a compositional boundary; neither it nor the deepest small plotted strength is automatically the seismic lithosphere-asthenosphere boundary.

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.

Treat the two strands as newly formed conjugate faults obeying the same Mohr-Coulomb failure criterion, rather than merely inherited faults reactivated in a later stress field. A fault strike is an unoriented line, so $296^{\circ }$ is equivalent to $116^{\circ }$. The acute angle between the fault strikes is therefore $116^{\circ }-60^{\circ }=56^{\circ }$. The formal compressive acute bisector puts ${\sigma }_1$ at azimuth $88^{\circ }$ and ${\sigma }_3$ at $178^{\circ }$. This orientation must also be checked against the mapped senses of strike-slip faulting, as discussed below.

Let $\beta =28^{\circ }$ be the angle between either fault plane and ${\sigma }_1$. Tangency of Mohr's circle to $|\tau |=c+\mu {\sigma }_n^{\prime }$, where ${\sigma }_n^{\prime }$ is effective stress normal to the plane and $c$ is cohesion, gives $\beta =45^{\circ }-\varphi /2$ and $\mu =\tan \varphi$. ThusIn the Mohr circle, the fault planes are $112^{\circ }$ apart around the circle, because a physical plane rotation is doubled. The figure uses $c=0$ to display both tangencies; a common nonzero cohesion changes the intercept, not the inferred friction angle. The inference of $\mu$ requires optimally oriented conjugate faults at formation: the two fault strikes alone cannot measure a friction coefficient for arbitrarily reactivated faults.

Consistency limitation: the mapped westward motion of the Anatolian block corresponds to left-lateral strike-slip faulting on the East Anatolian strand and right-lateral strike-slip faulting on the North Anatolian strand. East–west maximum compression for the acute-bisector construction instead predicts the opposite pair. The stress orientation compatible with the mapped senses lies near the north–south, obtuse bisector; the planes are then about $62^{\circ }$ from ${\sigma }_1$, which would give $\varphi =90^{\circ }-2(62^{\circ })=-34^{\circ }$ in the optimal-formation formula. A negative friction angle is inadmissible for ordinary frictional strength. Thus $0.675$ is the formal magnitude inferred from unoriented angles, while the full map does not support a self-consistent positive-friction, optimally formed conjugate pair. The common-stress assumption can still describe inherited or anisotropically controlled faults, but then their orientations no longer determine a unique coefficient.

The offset layered sequence indicates reverse slip on the west-dipping fault: its western hanging wall moved up relative to the eastern footwall. With the fault strike perpendicular to the east–west image plane, an Andersonian faulting regime has ${\sigma }_1$ horizontal east–west, ${\sigma }_2$ horizontal north–south, and ${\sigma }_3$ vertical. Draw opposing compression arrows east–west and a smaller opposing pair vertically; the intermediate pair points into and out of the page.

For example, in east, north, up coordinates, a compression-positive effective stress tensor consistent with a $30^{\circ }$ reverse fault and zero cohesion isThese magnitudes illustrate one possible stress tensor, rather than a measurement from the photograph. With $\mu =1/\sqrt{3}$, the ${\sigma }_1$–${\sigma }_3$ Mohr circle is tangent to the Mohr-Coulomb failure criterion at a plane $30^{\circ }$ from horizontal. The original failure-mechanics figure above sketches these principal stress directions.

Use compression-positive effective stress normal to the plane ${\sigma }_n^{\prime }$ on the horizontal axis and shear stress $\tau$ on the vertical axis. For any chosen pair of principal stresses, Mohr's circle has centre $C=({\sigma }_1^{\prime }+{\sigma }_3^{\prime })/2$ and radius $R=({\sigma }_1^{\prime }-{\sigma }_3^{\prime })/2$.

For the reverse fault example, $C=60\mathrm{MPa}$ and $R=30\mathrm{MPa}$. Draw the two Mohr-Coulomb failure criterion lines $\tau =\pm {\sigma }_n^{\prime }/\sqrt{3}$. The tangencies areThese are shear failure points with nonzero shear stress. The two smaller three-dimensional Mohr circles use $({\sigma }_1^{\prime },{\sigma }_2^{\prime })$ and $({\sigma }_2^{\prime },{\sigma }_3^{\prime })$; they lie inside the outer circle and do not touch the envelope in this example.

For the opening geological joints, use the tensile failure criterion ${\sigma }_n^{\prime }=-T_0$. The example gives $C=3.5\mathrm{MPa}$, $R=6.5\mathrm{MPa}$ and $T_0=3\mathrm{MPa}$. The circle first touches this vertical cutoff at $({\sigma }_n^{\prime },\tau )=(-3,0)\mathrm{MPa}$. Thus the geological joints form on planes normal to ${\sigma }_3$ and have zero resolved shear stress. A complete brittle envelope can combine this tensile cutoff with compressive Mohr-Coulomb failure criterion, using sufficient cohesion that shear failure does not occur first. The failure-mechanics figure displays these two distinct constructions.

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]

Loss of load-bearing strength accompanied by relative displacement tangential to a surface, rather than primarily opening normal to it. The Mohr-Coulomb failure criterion models frictional shear failure by $|\tau |=c+\mu {\sigma }_n^{\prime }$, using cohesion, a friction coefficient and compressive effective stress normal to the potential sliding plane.