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Engineer associated with Mohr's circle and strength criteria for stressed materials.

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.

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.

Use compression-positive principal stresses ${\sigma }_1\ge {\sigma }_2\ge {\sigma }_3$. The two-dimensional Mohr's circle shown is the ${\sigma }_1$–${\sigma }_3$ circle: its horizontal axis is compressive normal stress ${\sigma }_n$, its vertical axis signed shear stress $\tau$, its centre is $C=({\sigma }_1+{\sigma }_3)/2$, and its radius is $R=({\sigma }_1-{\sigma }_3)/2$. The intercepts $({\sigma }_1,0)$ and $({\sigma }_3,0)$ represent principal stress planes, on which shear stress vanishes. The top and bottom points have $|\tau |=R$, the maximum shear stress.

In the rock sketch, $\theta$ is the counterclockwise angle from the ${\sigma }_1$ direction to the plane's unit normal vector $\mathbf{n}=(\cos \theta ,\sin \theta )$. Choose the clockwise unit tangent $\mathbf{t}=(\sin \theta ,-\cos \theta )$ and define $\tau =\mathbf{t}\cdot {\mathbf{\sigma }}^c\mathbf{n}$, with ${\mathbf{\sigma }}^c=\mathrm{diag}({\sigma }_1,{\sigma }_3)$ compression positive. Thus a positive $\theta$ produces a positive $2\theta$ rotation on the plotted circle. The actual traction on the positive-normal face is $-{\mathbf{\sigma }}^c\mathbf{n}=-{\sigma }_n\mathbf{n}-\tau \mathbf{t}$. Reversing the tangent convention reverses the plotted shear sign, not the physical prediction.

In three dimensions there are circles for each pair of principal stresses. The largest is the ${\sigma }_1$–${\sigma }_3$ circle; a general three-dimensional plane requires all three principal stresses and its normal, rather than a single angle on this circle.

In the simplest strength-cutoff model, opening begins when the least compressive effective stress reaches minus the tensile strength: ${\sigma }_3^{\prime }=-T_0$ with compression positive. In Mohr's circle, this is contact with the vertical line ${\sigma }_n^{\prime }=-T_0$ at $\tau =0$. The criterion represents an opening mechanism and differs from frictional shear failure.