Question 3 of 4: Shear Zones, Faulting and Mohr–Coulomb Stress Analysis

Question C: Shear Zones, Faulting and Mohr–Coulomb Stress Analysis (40 marks)

(C1) Shear zone geometry (4 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Shear-zone width $w = 1\text{ m}$; dextral slip (relative end-to-end offset of an originally perpendicular marker) $= 10\text{ m}$.

Find. (a) angular shear $\psi$ and shear strain $\gamma$. (b) elongation $e$ of the offset marker.

Approach. The marker, the zone boundary and the offset form a right triangle: shear strain is the offset-to-width ratio, angular shear its arctangent, and the marker's new length is the triangle's hypotenuse.

1. Shear strain. $\gamma = \dfrac{\text{slip}}{w} = \dfrac{10}{1} = 10$.
2. Angular shear. $\psi = \arctan(\gamma) = \arctan(10) = \boxed{84.3^{\circ}}$.
3. Elongation. The marker's original length is the zone width, $L_0=w=1\text{ m}$; after shear it is the hypotenuse $L_1=\sqrt{w^2+\text{slip}^2}=\sqrt{1^2+10^2}=10.05\text{ m}$. Elongation $e=\dfrac{L_1-L_0}{L_0}=\dfrac{10.05-1}{1}=\boxed{9.05}$ (i.e. 905%).

(C2) Ductile shear zone with strain ellipses (4 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

A ductile shear zone accommodates displacement by distributed internal strain rather than by a discrete slip surface, so an offset marker (here a dike) is not cut cleanly — it is progressively dragged and thinned/rotated across the zone's width, producing a smooth sigmoidal deflection rather than a sharp break. Because shear strain is heterogeneous across the zone (maximum on the centre-line, decaying to zero at the undeformed wall rock), the foliation (fabric) trajectory is itself sigmoidal: it starts parallel to the zone boundary far outside the zone, curves progressively into closer parallelism with the shear direction toward the centre, and is most rotated (steepest angle to the boundary) exactly on the centre-line. The strain ellipses record the same gradient: circular (undeformed) far from the zone, becoming progressively more eccentric (elongate) and more rotated toward parallelism with the shear plane as the centre is approached, with the long axis of each ellipse tracking the local fabric trajectory.

(C3) Shear-sense indicators (4 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Image 1. Indicator: asymmetric (σ-type) porphyroclasts — the dark, elongate lozenge-shaped clasts have tails/wings that are asymmetrically developed on either side of the clast, wrapped by the finer anastomosing foliation. Reading the tail asymmetry (longer/more deflected tail trailing to the lower-right of each clast) against the sub-horizontal foliation gives a dextral sense of shear.

Image 2. Indicator: S–C fabric — the finer, wavy foliation (S-surfaces, the finite-strain fabric) is cut by a straighter, more continuous, lower-angle band (a C- or C′-surface, a discrete shear band). The acute angle between S and C opens in the transport direction; here it opens to the right, giving the same dextral sense of shear as Image 1.

(C4) Anderson's theory of faulting (12 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

(a) Three assumptions.

1. The Earth's surface is a free surface that cannot transmit shear stress, so it must be a principal stress plane — one principal stress axis is always exactly vertical, and the other two are horizontal, everywhere at/near the surface. 2. The crust is treated as a homogeneous, isotropic material that fails according to the Coulomb (Navier–Coulomb) brittle failure criterion, with a fixed cohesion and coefficient of internal friction. 3. Faults form as conjugate pairs at a constant angle to $\sigma_1$ ($\theta=45^{\circ}-\phi/2$ from $\sigma_1$) that does not vary with depth or confining pressure — because only three mutually perpendicular arrangements of ($\sigma_1,\sigma_2,\sigma_3$) are possible with one axis vertical, this restricts all faults to exactly three geometric classes.

(b) The three fault types. Using an illustrative friction angle $\phi\approx30^{\circ}$ (so the conjugate half-angle from $\sigma_1$ is $45^{\circ}-15^{\circ}=30^{\circ}$):

(C5) Mohr–Coulomb analysis of a pre-existing plane of weakness (16 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. $\sigma_1 = 100$ MPa (vertical), $\sigma_3 = 20$ MPa (horizontal), $\phi_f = 35^{\circ}$, pre-existing plane dips $30^{\circ}$ E (cohesionless, $C_0=0$).

Find. (a) cross-section + sense of shear; (b) $\sigma_n$; (c) $\tau$; (d) $\tau_{\text{failure}}$; (e) dry stability and, if needed, $P_p$ to cause failure.

Approach. Resolve $\sigma_1,\sigma_3$ onto the plane with the Mohr-circle stress-transformation equations, using $\theta=90^{\circ}-\text{dip}=60^{\circ}$ (angle measured from $\sigma_1$ to the plane); compare the resolved shear stress to the cohesionless Coulomb strength $\sigma_n\tan\phi_f$; if it already fails dry, back-solve the Coulomb law with pore pressure only to show the required $P_p\le 0$.

1. (a) Configuration and sense of shear. $\sigma_1$ vertical and $\sigma_3$ horizontal is the Andersonian normal-faulting stress state, so any plane that fails here does so with normal (dip-slip, hanging-wall-down) motion. The plane dips 30° E, so its hanging wall (the block lying structurally above the inclined surface) is the east block, which slides down-dip (down and to the east) relative to the west (footwall) block. Rotating the cross-section so the plane is horizontal (the standard way to read apparent lateral sense off an inclined surface) puts the hanging wall — the geometrically upper block — moving toward the right; by the same top-right/bottom-left convention the exam's own C1 figure uses for "dextral," this plane's sense of shear is $\boxed{\text{dextral}}$.
2. (b) Normal stress. With $\theta=90^{\circ}-30^{\circ}=60^{\circ}$ measured from $\sigma_1$ to the plane, the fundamental stress-transformation equation gives $$\sigma_n=\frac{\sigma_1+\sigma_3}{2}+\frac{\sigma_1-\sigma_3}{2}\cos2\theta=\frac{100+20}{2}+\frac{100-20}{2}\cos120^{\circ}=60-20=\boxed{40\text{ MPa}}$$
3. (c) Shear stress. $$\tau=\frac{\sigma_1-\sigma_3}{2}\sin2\theta=40\sin120^{\circ}=\boxed{34.6\text{ MPa}}$$
4. (d) Shear stress at failure. The plane is a pre-existing (cohesionless) plane of weakness, so the Coulomb Failure Law reduces to pure frictional sliding, $C_0=0$: $$\tau_{\text{failure}}=\sigma_n\tan\phi_f=40\tan35^{\circ}=\boxed{28.0\text{ MPa}}$$
5. (e)(i) Dry stability. Comparing the resolved shear stress to the frictional strength: $\tau=34.6\text{ MPa} > \tau_{\text{failure}}=28.0\text{ MPa}$. The resolved shear stress already exceeds the plane's frictional (Coulomb) strength with zero pore pressure, so $\boxed{\text{yes, it fails even under dry conditions}}$ — the applied stress state is not marginal, it is well past the sliding threshold for this orientation.
6. (e)(ii) Required pore pressure. Since it already fails dry, this part is answered for completeness by back-solving the Coulomb law with pore pressure, $\tau=(\sigma_n-P_p)\tan\phi_f$: $$P_p=\sigma_n-\frac{\tau}{\tan\phi_f}=40-\frac{34.6}{\tan35^{\circ}}=\boxed{-9.5\text{ MPa}}$$ A negative "required" pore pressure is the arithmetic signature of a plane that is already critically stressed at $P_p=0$: no pore pressure is required to cause failure (a suction, physically impossible in a fluid-saturated rock, would be needed to prevent it).