Hydrofracture and Fault Reactivation: 18-Geol-A4 Structural Geology, December 2017, Q4

Question D: Hydrofracture and Fault Reactivation (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.

Given. Greatest principal stress σ₁ = 341 MPa; least principal stress σ₃ = 95 MPa; cohesion c = 75 MPa; angle of internal friction φ = 28°.

Find. (A) pore fluid pressure P_f to hydrofracture the intact rock; (B) normal and shear stress on the resulting shear fractures; (C) pore fluid pressure to reactivate those (now cohesionless) faults.

Approach. Raising pore fluid pressure lowers the effective stress σ′=σ−P_f uniformly, which shifts the Mohr circle to the LEFT on the σ_n–τ diagram without changing its radius; find the P_f at which the circle first becomes tangent to the Mohr–Coulomb envelope (A), read the tangent point for the stress state on the new fracture (B), then reapply the tangency condition with zero cohesion (Byerlee frictional sliding on the now-discrete fault surface) to find the reactivation pressure (C).

Total-stress Mohr circle. $$C = \frac{\sigma_1+\sigma_3}{2} = \frac{341+95}{2} = 218 \text{ MPa}, \qquad R = \frac{\sigma_1-\sigma_3}{2} = \frac{341-95}{2} = 123 \text{ MPa}$$ Pore pressure shifts the circle's centre to C−P_f while R stays fixed at 123 MPa.
(A) Pore pressure to hydrofracture. Tangency of the effective-stress circle to τ=c+σ_ntanφ requires the perpendicular distance from the shifted centre to the envelope to equal R: $$(C-P_f)\sin\phi + c\cos\phi = R$$ Solving for the effective centre at failure, $$C-P_f = \frac{R - c\cos\phi}{\sin\phi} = \frac{123 - 75(0.883)}{0.4695} = \frac{123-66.2}{0.4695} = 120.9 \text{ MPa}$$ so $$P_f = 218 - 120.9 = \boxed{97.1 \text{ MPa}}$$
(B) Stress on the new shear fractures. The tangent point of a Mohr circle (centre C′, radius R) on the envelope τ=c+σ_ntanφ is at $$\sigma_n' = C' - R\sin\phi = 120.9 - 123(0.4695) = 63.2 \text{ MPa}, \qquad \tau = R\cos\phi = 123(0.883) = \boxed{108.6 \text{ MPa}}$$ Check on the envelope: c+σ_n′tanφ = 75+63.2(0.5317) = 108.6 MPa ✓. In total-stress terms, σ_n = σ_n′+P_f = 63.2+97.1 = 160.3 MPa (this total normal stress is unaffected by pore pressure, since P_f is isotropic and does not change the geometry of the plane relative to σ₁/σ₃).
(C) Pore pressure to reactivate the fault. The new fracture is now a discrete, cohesionless surface, so it reactivates by pure frictional sliding, τ=μσ_n′, with μ=tanφ=0.5317. Because the plane's orientation relative to σ₁/σ₃ is fixed, its total normal stress (160.3 MPa) and shear stress (108.6 MPa) do not change with P_f — only the effective normal stress does: $$\tau = \mu(\sigma_{n,\text{total}} - P_{f2}) \ \Rightarrow\ P_{f2} = \sigma_{n,\text{total}} - \frac{\tau}{\mu} = 160.3 - \frac{108.6}{0.5317} = 160.3 - 204.3 = \boxed{-44.0 \text{ MPa}}$$ Independent cross-check via Sibson's critical pore-pressure-ratio formula for an optimally-oriented cohesionless fault, R_f=(√(μ²+1)+μ)²=2.77, solving (σ₁−P_f2)/(σ₃−P_f2)=R_f gives the same P_f2=−44.0 MPa.

Check: the negative result in (C) is not an error — it means that once the fracture exists as a cohesionless surface, the SAME remote stress state (σ₁=341, σ₃=95) already exceeds its frictional sliding strength even at zero excess pore pressure (dry, P_f=0: τ=108.6 MPa > μσ_n=0.5317×160.3=85.2 MPa). Physically, a real fault cannot sustain a negative fluid pressure, so the result should be read as "no additional pore pressure is needed — the fault is already critically stressed and would reactivate spontaneously," which is the expected fault-weakening outcome: a cohesionless discontinuity is always easier to reactivate than it was to create as a new fracture through intact, cohesive rock (P_f2=−44.0 MPa < P_f=97.1 MPa). The friction coefficient for reactivation is assumed equal to tanφ of the intact rock (i.e. residual ≈ peak friction), the standard simplifying assumption absent separate residual-strength data.

Fig. D — Mohr diagram: raising P_f shifts the effective-stress circle left (grey→blue) until tangent to the Coulomb envelope, giving the hydrofracture pressure, tangent-point stress state, and (by the same geometry with c=0) the reactivation pressure.

Question D — final results
Quantity	Value
(A) Pore pressure to hydrofracture, P_f	97.1 MPa
(B) Normal stress on new fracture, σ_n′ (effective) / total	63.2 MPa / 160.3 MPa
(B) Shear stress on new fracture, τ	108.6 MPa
(C) Pore pressure to reactivate the fault, P_f2	−44.0 MPa (already unstable at P_f=0)

Topic. Effective-stress (Terzaghi) Mohr–Coulomb analysis of hydraulic fracturing and fault reactivation.

Key relations. σ′=σ−P_f; Mohr circle centre C=(σ₁+σ₃)/2, radius R=(σ₁−σ₃)/2 (P_f shifts C, never changes R); tangency condition (C−P_f)sinφ+c cosφ=R; tangent-point stress σ_n′=C′−R sinφ, τ=R cosφ; cohesionless (Byerlee) frictional reactivation τ=μσ_n′, μ=tanφ.

Why this works. Fluid pressure enters the Coulomb criterion only through the effective normal stress, so it can only translate the Mohr circle horizontally — it can never change the differential stress (σ₁−σ₃), which is why the SAME tangency geometry (just with c=0 instead of c=75) answers both the intact-rock fracture pressure (A) and the fault reactivation pressure (C) from one consistent diagram.

Common pitfall. Forgetting that a discrete fracture, once formed, has essentially zero cohesion and must be re-solved with c=0 rather than reusing the intact-rock envelope for part (C); and reporting the tangent-point normal stress as if it were the total stress rather than the EFFECTIVE stress that the Coulomb criterion is actually evaluated against.

Source. Davis & Reynolds, Structural Geology of Rocks and Regions, 3rd ed., Ch.5 (Mohr–Coulomb failure, effective stress, hydraulic fracturing, fault reactivation).

Question 4 of 5: Hydrofracture and Fault Reactivation

Question D: Hydrofracture and Fault Reactivation (12 marks)