0 (not stated — a pre-existing plane of weakness/joint is assumed cohesionless, frictional-sliding only)
Find. (a) The real-world geometry and sense of shear on the plane; (b) the normal stress \(\sigma_n\); (c) the shear stress \(\tau\); (d) the shear strength \(\tau_f\) at failure, whether the plane fails dry, and the pore pressure \(P_p\) needed if not; (e) confirmation of (b)–(d) on a plotted Mohr circle.
Approach. Resolve \(\sigma_1\) and \(\sigma_3\) onto the plane using the fundamental stress-transformation equations with \(\theta\) measured from the \(\sigma_1\) direction to the plane's normal, 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.
(a) Assumed configuration: \(\sigma_1\) vertical, \(\sigma_3\) horizontal, plane of weakness dipping 30° to the east (the source states only the dip magnitude, so the dip direction is stated here as an assumption per the exam's own Note 1). With the plane's up-dip side to the west, the block above the plane is driven down-and-east (down-dip) relative to the block below — a top-to-the-east sense that reads as right-lateral (dextral) in this 2-D cross-section.
Set up the angle from σ₁ to the plane's normal. The plane dips 30° from horizontal, so its normal is inclined \(\theta = 90^{\circ}-30^{\circ}=60^{\circ}\) from the vertical \(\sigma_1\) direction.
Normal stress on the plane (b).
$$\sigma_n=\frac{\sigma_1+\sigma_3}{2}+\frac{\sigma_1-\sigma_3}{2}\cos2\theta=\frac{100+30}{2}+\frac{100-30}{2}\cos120^{\circ}=65-17.5$$
$$\boxed{\sigma_n = 47.5\ \text{MPa}}$$
Shear stress on the plane (c).
$$\tau=\frac{\sigma_1-\sigma_3}{2}\sin2\theta=35\sin120^{\circ}$$
$$\boxed{\tau = 30.3\ \text{MPa}}$$
As a check, \((\sigma_n-65)^2+\tau^2=(-17.5)^2+30.3^2\approx35^2\), confirming the point lies exactly on the Mohr circle of radius 35 MPa.
Shear strength at failure, dry (d-i, d-ii). Treating the pre-existing plane as cohesionless (\(c_f=0\), friction only):
$$\tau_f=\sigma_n\tan\phi_f=47.5\tan40^{\circ}$$
$$\boxed{\tau_f = 39.9\ \text{MPa}}$$
Since the resolved shear stress \(\tau=30.3\) MPa is LESS than the dry shear strength \(\tau_f=39.9\) MPa, the point (47.5, 30.3) plots 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 \(\tau\) is unaffected by pore pressure. Failure occurs when the resolved \(\tau\) equals the reduced strength \(\sigma_n'\tan\phi_f\):
$$\tau=(\sigma_n-P_p)\tan\phi_f\ \Rightarrow\ P_p=\sigma_n-\frac{\tau}{\tan\phi_f}=47.5-\frac{30.3}{\tan40^{\circ}}$$
$$\boxed{P_p \approx 11.4\ \text{MPa}}$$
This is comfortably below \(\sigma_3=30\) MPa, so a physically realistic pore-fluid overpressure could indeed trigger slip on this plane.
(e) Mohr circle check: centre 65 MPa, radius 35 MPa. The plotted point (47.5, 30.3) sits below the \(\tau_f=\sigma_n\tan40^{\circ}\) envelope, confirming no dry failure; the dashed green circle shows the effective-stress circle shifted left by \(P_p=11.4\) MPa, now tangent to the envelope at the same \(\tau\).
Final results – Question C
Quantity
Value
Sense of shear (assumed dip to the east)
Right-lateral (dextral)
Normal stress, σn
47.5 MPa
Shear stress, τ
30.3 MPa
Shear stress required at failure, τf
39.9 MPa
Fails dry?
No (τ < τf)
Pore pressure Pp needed to trigger failure
11.4 MPa
Check: the source states only that the plane "dips 30°," not which compass direction it dips. 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 \(\sigma_n\), \(\tau\), \(\tau_f\) and \(P_p\) are unaffected by this choice, but the right-lateral/left-lateral call in (a) would flip to left-lateral if the plane instead dipped to the west.
