Past exam of the year 2 of the earth sciences (geology) course of the University of Oxford / 2023 / Paper 1 / 2 / c / Solution

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.