Fault Stress and Pore-Pressure-Induced Failure: 18-Geol-A4 Structural Geology, May 2017, Q4

Question D: Fault Stress and Pore-Pressure-Induced Failure (13 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.

σ₁ (vertical)	82 MPa
σ₃ (horizontal, E–W)	30 MPa
Fault strike / dip	N–S / 40° W
Cohesion, C	10 MPa
Friction coefficient, μ	0.5 (φ=tan⁻¹0.5≈26.6°)

Find. The normal stress σ_n and shear stress τ resolved on the fault plane; whether Coulomb failure occurs at the given (dry, P_p=0) stress state; and, if not, the pore pressure P_p required to induce failure.

Fig. D — Mohr-circle construction: the dry circle (blue, centre 56 MPa, radius 26 MPa) lies entirely below the Coulomb envelope, so the fault point (σn=60.5, τ=25.6) is stable; a pore pressure of 29.3 MPa shifts the circle left (orange, dashed) until it just touches the envelope at the same τ, inducing failure.

Approach. Because the fault strikes N–S (parallel to σ₂, whose magnitude does not affect stresses on this plane) and σ₁/σ₃ are vertical/horizontal E–W, the problem reduces to a 2-D Mohr-circle construction in the vertical E–W plane: locate the fault’s pole angle from σ₁, read (or compute) σ_n and τ from the circle, compare the resulting point to the Coulomb envelope, and if it lies below the envelope, translate the circle left by the pore pressure needed to bring it tangent to the envelope.

Locate the fault’s pole angle θ from σ₁. A plane striking N–S and dipping 40°W has a pole that plunges (90−40)=50° to the east (standard stereographic pole-to-plane relation: pole trend = dip azimuth+180°, pole plunge = 90°−dip). Since σ₁ is vertical, the angle between the pole and σ₁ is θ=90°−50°=40°.
Resolve σ_n and τ on the fault (2θ=80° on the Mohr circle). Using $\sigma_n=\dfrac{\sigma_1+\sigma_3}{2}+\dfrac{\sigma_1-\sigma_3}{2}\cos2\theta$ and $\tau=\dfrac{\sigma_1-\sigma_3}{2}\sin2\theta$: centre $=\frac{82+30}{2}=56$ MPa, radius $=\frac{82-30}{2}=26$ MPa. $$\sigma_n = 56+26\cos80^\circ = 56+26(0.1736) = \boxed{60.5\ \text{MPa}}$$ $$\tau = 26\sin80^\circ = 26(0.9848) = \boxed{25.6\ \text{MPa}}$$ (Cross-checked directly via the Cauchy traction formula $\mathbf t=\boldsymbol\sigma\cdot\mathbf n$ on the pole unit vector — identical result.)
Test Coulomb failure at P_p=0. The shear strength available on this plane at the current (dry) normal stress is $$\tau_{\text{available}} = C+\mu\sigma_n = 10+0.5(60.5) = \boxed{40.3\ \text{MPa}}$$ Since the actual resolved shear stress τ=25.6 MPa is well below the 40.3 MPa needed for slip, the fault does NOT fail at the given, dry stress state — the Mohr circle plots entirely below the failure envelope (Fig. D).
Find the pore pressure required for failure. Raising P_p shifts the circle left (reduces σ_n→σ_n−P_p) without changing τ, so failure occurs once $$\tau = C+\mu(\sigma_n-P_p) \ \Rightarrow\ P_p = \sigma_n-\frac{\tau-C}{\mu} = 60.5-\frac{25.6-10}{0.5} = 60.5-31.2 = \boxed{29.3\ \text{MPa}}$$ (At this P_p, the effective stresses are σ₃′=30−29.3≈0.7 MPa and σ₁′≈52.7 MPa — both still compressive, confirming the failure mode is frictional slip on the fault, not tensile fracturing.)

Question D — Final results
Part	Quantity	Result
A	Normal stress on fault	σ_n = 60.5 MPa
A	Shear stress on fault	τ = 25.6 MPa
B	Shear strength available (dry)	40.3 MPa > 25.6 MPa — no failure
C	Pore pressure required for failure	P_p ≈ 29.3 MPa

Topic. Mohr–Coulomb analysis of a pre-existing fault and pore-pressure-induced (effective-stress) reactivation.

Key relations. Plane-resolution: $\sigma_n=\tfrac{\sigma_1+\sigma_3}{2}+\tfrac{\sigma_1-\sigma_3}{2}\cos2\theta$, $\tau=\tfrac{\sigma_1-\sigma_3}{2}\sin2\theta$ (θ = angle from σ₁ to the plane’s pole); Coulomb failure $\tau=C+\mu\sigma_n'$; effective stress (Terzaghi) $\sigma'=\sigma-P_p$.

Why this works. Pore pressure is isotropic — it subtracts equally from every normal stress and therefore shifts the whole Mohr circle rigidly along the σ_n axis without altering its radius (the differential stress, and hence τ on every plane, is unchanged). A fault that is stable under the current total stresses can therefore be brought to failure purely by raising fluid pressure, with zero change in the tectonic (total) stress field — the mechanism behind injection-induced seismicity and fault reactivation in hydrocarbon/geothermal/CO₂ operations.

Common pitfall. Using the fault’s dip angle directly as θ in the Mohr formula (here 40°) instead of the angle from σ₁ to the plane’s pole — for a plane dipping δ, the pole-based θ measured from a vertical σ₁ is (90°−δ), not δ itself; getting this angle backwards silently swaps the sin/cos terms and gives the wrong (σ_n,τ) pair.

Source. Davis & Reynolds, Structural Geology of Rocks and Regions, 3rd ed., Ch.6 (Mohr circle construction, Coulomb failure, effective stress); Fossen, Structural Geology, 2nd ed., Ch.7–8.

Question 4 of 5: Fault Stress and Pore-Pressure-Induced Failure

Question D: Fault Stress and Pore-Pressure-Induced Failure (13 marks)