 Solution

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]