Cohesion of the weakness plane (pre-existing, hence taken frictional-only)
cf = 0
Find. (a) The real-world geometry and sense of shear on the plane; (b) the normal stress σn; (c) the shear stress τ; (d) the shear strength τf at failure, whether the plane fails dry, and the pore pressure Pp needed if not; (e) confirmation of (b)–(d) on a plotted Mohr circle.
Approach. Resolve σ1 and σ3 onto the plane using the fundamental stress-transformation equations with θ measured from the σ1 direction to the plane's own normal (θ=dip, since a plane's normal is inclined from vertical by exactly the plane's dip — check: a horizontal, 0°-dip bed has a vertical normal, θ=0), compare the resolved shear stress to the Mohr–Coulomb (cohesionless) failure line, and back-solve for the pore pressure that would bring the plane to failure via the effective-stress principle.
Fig. C1 – Real-world configuration (dip assumed to the east — see check note): σ1 vertical, σ3 horizontal, weakness plane dipping 30°, with the resolved normal stress σn and down-dip shear traction τ shown on the plane.
Set up the angle from σ1 to the plane's normal. The plane dips 30° from horizontal, so its normal is inclined θ=30° from the vertical σ1 direction (a 0°-dip plane has a vertical normal, θ=0; a 90°-dip plane has a horizontal normal, θ=90° — θ tracks the dip directly).
Normal stress on the plane (b).
$$\sigma_n=\frac{\sigma_1+\sigma_3}{2}+\frac{\sigma_1-\sigma_3}{2}\cos2\theta=\frac{120+40}{2}+\frac{120-40}{2}\cos60^{\circ}=80+40(0.5)$$
$$\boxed{\sigma_n = 100.0\ \text{MPa}}$$
Shear stress on the plane (c).
$$\tau=\frac{\sigma_1-\sigma_3}{2}\sin2\theta=40\sin60^{\circ}$$
$$\boxed{\tau = 34.6\ \text{MPa}}$$
As a check, \((\sigma_n-80)^2+\tau^2=(20)^2+34.6^2\approx40^2\), confirming the point lies exactly on the Mohr circle of radius 40 MPa centred at 80 MPa.
Shear strength at failure, dry (d-i, d-ii). The pre-existing plane is treated as cohesionless (\(c_f=0\), friction-only, since only φf is given):
$$\tau_f=\sigma_n\tan\phi_f=100.0\tan40^{\circ}$$
$$\boxed{\tau_f = 83.9\ \text{MPa}}$$
Since the resolved shear stress \(\tau=34.6\) MPa is well BELOW the dry shear strength \(\tau_f=83.9\) MPa, the point (100.0, 34.6) plots comfortably below the Mohr–Coulomb envelope — the plane does NOT fail under dry conditions.
Pore pressure required to trigger failure (d-iii). Effective stress reduces the normal stress acting across the plane, \(\sigma_n'=\sigma_n-P_p\), while the resolved \(\tau\) itself is unaffected by pore pressure (pore pressure is isotropic). Failure occurs when \(\tau\) equals the reduced strength \(\sigma_n'\tan\phi_f\):
$$\tau=(\sigma_n-P_p)\tan\phi_f\ \Rightarrow\ P_p=\sigma_n-\frac{\tau}{\tan\phi_f}=100.0-\frac{34.6}{\tan40^{\circ}}$$
$$\boxed{P_p \approx 58.7\ \text{MPa}}$$
This is a large overpressure (comfortably below lithostatic \(\sigma_1=120\) MPa, but well above \(\sigma_3=40\) MPa), consistent with the plane being far from failure under dry conditions in step 4.
Fig. C2 – Mohr circle (centre 80 MPa, radius 40 MPa) with the plotted point (100.0, 34.6) at 2θ=60°, plus the dry Mohr–Coulomb envelope τ=σntan40° — the plotted point sits well inside (below) the envelope, confirming no failure at Pp=0 (d-ii). The dashed orange circle is the same circle shifted left by Pp=58.7 MPa (effective centre 21.3 MPa, σ3′=−18.7, σ1′=61.3 MPa): the plane’s point moves to (41.3, 34.6), which lies exactly on the envelope (41.3 tan40°=34.6), confirming d-iii.
Check: the source states only that the plane "dips 30°," not which compass direction — dip-to-the-east is adopted here as the stated assumption (per the exam's own Note 1, "submit a clear statement of any assumptions made"). The magnitudes of σn, τ, τf and Pp are unaffected by this choice, but the right-lateral/left-lateral call in (a) would flip to left-lateral if the plane instead dipped west. Since σ1 is vertical and exceeds σ3, this is an Andersonian extensional (normal-faulting) stress regime regardless of the assumed dip azimuth — the plane, if it slipped, would always show hanging-wall-down, normal-sense shear.
