Plane Failure Slope Stability with Tension Crack: 24-MMP-B1 Applied Rock Mechanics, December 2015, Q4

Question 4: Plane Failure Slope Stability with Tension Crack (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.

4.1–4.5 — Wedge geometry and factor of safety

Given. Slope height H = 30 m; face angle 62°; plane dip ψ_p = 35°; tension crack depth z = 9.9 m, located b = 10 m behind the crest on a horizontal upper surface; c = 100 kPa, φ = 34°; γ_rock = 27 kN/m³, γ_water = 9.81 kN/m³.

Find. The factor of safety for each of the four water/strength scenarios, and the sensitivity of the slope to them.

Check: with BOTH the tension-crack depth (9.9 m) and its horizontal offset behind the crest (10 m) given, the usual textbook shortcut of assuming the sliding plane passes through the toe is over-determined and must be checked. Projecting the 35° plane from the crack base toward the face shows it actually daylights on the FACE at 3.07 m above the toe, not at the toe itself — so the wedge geometry below uses that true daylight point, not the toe, as the lower vertex of the sliding mass.

Plane-failure wedge geometry: the 35° sliding plane, projected from the tension-crack base, daylights on the excavated face 3.07 m above the toe (point D), not at the toe — found by intersecting the plane and face lines rather than assumed.

Approach. Locate the crest, tension-crack top/base and the plane's true daylight point on the face by coordinate geometry; compute the wedge cross-sectional area (shoelace formula) and weight; then apply the standard plane-failure factor of safety with tension-crack water forces, $FS=\dfrac{cA+(W\cos\psi_p-U-V\sin\psi_p)\tan\phi}{W\sin\psi_p+V\cos\psi_p}$, where U is the uplift force on the sliding plane and V is the horizontal thrust from water in the tension crack, both triangular pressure distributions referenced to the water column height Z_w in the crack.

Locate the crest and the true daylight point. With the toe at the origin, the crest is at $(H/\tan62^\circ,\,H)=(15.95,\,30.0)$ m; the tension-crack base is at $(15.95+10,\,30-9.9)=(25.95,\,20.1)$ m. Intersecting the face line ($y=x\tan62^\circ$) with the plane through the crack base ($y=20.1-\tan35^\circ\,(25.95-x)$) gives the daylight point $D=(1.63,\,3.07)$ m — above the toe, confirming the plane exits on the face rather than below grade.
Wedge area and weight. The sliding mass is bounded by the face (D to crest), the horizontal ground (crest to crack top), the tension crack (crack top to crack base) and the sliding plane (crack base back to D). Shoelace formula on these four vertices gives area $A_{xs}=\boxed{255.0\ \text{m}^2}$ (per metre of slope length), so weight $W=\gamma_{rock}A_{xs}=27\times255.0=\boxed{6885\ \text{kN/m}}$.
Sliding-plane length. Distance from the crack base to D: $L=\sqrt{(25.95-1.63)^2+(20.1-3.07)^2}=\boxed{29.69\ \text{m}}$.
4.3 — Fully drained (U=V=0). $FS=\dfrac{cL+W\cos\psi_p\tan\phi}{W\sin\psi_p}=\dfrac{100(29.686)+6885(0.81915)(0.67451)}{6885(0.57358)}=\dfrac{2968.6+3804.1}{3949.1}=1.715=\boxed{1.72}$.
4.1 — Crack filled to Z_w=4.1 m. $V=\tfrac12\gamma_wZ_w^2=\tfrac12(9.81)(4.1)^2=82.5\ \text{kN/m}$; $U=\tfrac12\gamma_wZ_wL=\tfrac12(9.81)(4.1)(29.686)=597.0\ \text{kN/m}$. $FS=\dfrac{cL+(W\cos\psi_p-U-V\sin\psi_p)\tan\phi}{W\sin\psi_p+V\cos\psi_p}=\boxed{1.58}$.
4.2 — Crack completely full, Z_w=z=9.9 m. $V=\tfrac12(9.81)(9.9)^2=480.7\ \text{kN/m}$; $U=\tfrac12(9.81)(9.9)(29.686)=1441.5\ \text{kN/m}$. $FS=\boxed{1.29}$.
4.4 — Drained, cohesion lost (c=0). $FS=\dfrac{W\cos\psi_p\tan\phi}{W\sin\psi_p}=\dfrac{3804.1}{3949.1}=\boxed{0.96}$ — below 1.0, i.e. the slope is predicted to fail.

Question 4 — factor of safety by scenario
Scenario	Factor of safety
4.3 – fully drained	1.72
4.1 – crack filled to 4.1 m	1.58
4.2 – crack completely full (9.9 m)	1.29
4.4 – drained, cohesion = 0	0.96 (fails)

4.5 — Sensitivity. Rising water level in the tension crack steadily erodes the factor of safety — from 1.72 (drained) to 1.58 at partial fill and 1.29 when the crack is completely full, a 25% reduction — because the water simultaneously reduces the effective normal force on the plane (uplift U) and adds a destabilising horizontal thrust (V) at the back of the wedge. Even fully saturated, however, the slope stays just above FS=1 as long as cohesion is intact. Losing cohesion entirely (case 4.4, drained) is far more damaging: FS drops to 0.96, below the fully-saturated case, because cohesion alone supplies roughly 44% of the resisting force at this geometry (2968.6 kN of the 6772.7 kN numerator in the drained case). The slope is therefore MORE sensitive to loss of cohesion — e.g. from blast-vibration damage to the discontinuity — than to water infiltration on its own, and any design or monitoring programme should treat blast-vibration control and discontinuity-integrity monitoring as at least as important as drainage measures.

Topic. Plane failure analysis with a tension crack and water pressure (Hoek & Bray method).

Key relations. $FS=\dfrac{cA+(W\cos\psi_p-U-V\sin\psi_p)\tan\phi}{W\sin\psi_p+V\cos\psi_p}$; triangular water pressure distributions $U=\tfrac12\gamma_wZ_wL$ (uplift on the plane) and $V=\tfrac12\gamma_wZ_w^2$ (thrust in the tension crack).

Why this works. The tension crack is assumed water-filled to some height Z_w, with the water free to drain along the plane to the (atmospheric) daylight point on the face, giving a triangular pressure distribution that peaks at the crack and falls to zero at daylight; integrating that pressure over the crack face gives V, and over the plane length gives U. Both forces reduce the frictional resisting term and V also adds directly to the driving (sliding) term.

Common pitfall. Assuming the sliding plane passes through the toe by default when the question independently specifies both the tension-crack depth and its offset behind the crest — the two values together over-determine the geometry, and the plane's true daylight point must be solved for rather than assumed, as done here.

Source. Wyllie & Mah, Rock Slope Engineering (after Hoek & Bray), 4th ed., ch. 6 (plane failure, tension-crack water pressure models).

Question 4 of 6: Plane Failure Slope Stability with Tension Crack

Question 4: Plane Failure Slope Stability with Tension Crack (20 marks)

4.1–4.5 — Wedge geometry and factor of safety