Question 1 of 5: Fault reactivation and critical pore pressure

Question 1: Fault reactivation and critical pore pressure (20 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. A fault plane already carries a known normal and shear stress; its strength is described by a Mohr-Coulomb (cohesion + friction) criterion.

Find. Whether the fault is currently at the point of slip; if not, the pore pressure build-up Δu required to trigger slip.

Approach. Compare the acting shear stress against the Mohr-Coulomb shear strength at the given (total) normal stress; if stable, reduce the effective normal stress with an unknown pore pressure u until the effective-stress strength criterion is exactly met by the acting shear stress, and solve for u.

1. Shear strength available at the current (dry) stress state. With no pore pressure yet acting, the fault's frictional strength is $$\tau_{max}=c+\sigma_n\tan\phi=10+40\tan45^\circ=10+40(1)=50\ \text{MPa}$$ $$\boxed{\tau_{max}=50\ \text{MPa}}$$
2. Compare against the acting shear stress. The acting shear stress τ = 40 MPa is less than the available strength of 50 MPa, so the fault is not currently at the point of slipping — it has a 10 MPa shear-strength reserve.
3. Introduce pore pressure and re-apply the effective-stress criterion. Fluid pressure u lowers the effective normal stress clamping the fault (Terzaghi's principle, σ'_n = σ_n − u) without changing the acting shear stress, so the criterion for renewed slip becomes $$\tau=c+(\sigma_n-u)\tan\phi$$ Substituting the known τ = 40 MPa and solving for u: $$40=10+(40-u)\tan45^\circ\ \Rightarrow\ 30=40-u\ \Rightarrow\ u=10\ \text{MPa}$$ $$\boxed{\Delta u_{crit}=10\ \text{MPa}}$$