Rock Slope Stability

Open-pit walls fail along joints and faults. Learn failure modes (planar, wedge, toppling, circular), kinematic analysis, the planar-failure factor-of-safety equation with cohesion, friction and water pressure, acceptance criteria, and slope monitoring and depressurisation.

A large pit wall failure can move millions of tonnes (Bingham Canyon 2013: ≈ 145 Mt, predicted by monitoring so no one was hurt). Slope engineers balance steep walls for economics against the risk of failure.

An analogy

A rock block on a joint plane is like a book on a tilted desk: friction and 'glue' (cohesion) hold it, gravity pulls it down, and water underneath lifts it and makes it slide.

Failure modes

Planar: sliding on one joint daylighting in the face. Wedge: sliding on two intersecting joints. Toppling: steeply dipping joints into the wall causing columns to rotate. Circular: in weak or heavily fractured rock and soil-like material.

Kinematic analysis

Stereonets test whether joints can daylight: the plane must dip less steeply than the face but more steeply than the friction angle (Markland test). Only kinematically possible failures need detailed analysis.

Water

Water pressure in joints and tension cracks reduces effective normal stress and adds driving force. Horizontal drains, dewatering wells and depressurisation can raise FoS significantly and allow steeper walls.

Acceptance and monitoring

Criteria (Read & Stacey): FoS 1.1–1.3 and PoF 5–25 % at bench scale, FoS 1.2–1.5 and PoF ≤ 5–10 % for overall slopes, depending on consequence. Slope stability radar, prisms, extensometers and InSAR detect accelerating movement for timely evacuation.

Formulas

• Planar FoS: FoS = [cA + (W cos ψ − U) tan φ] / (W sin ψ)
• Dry cohesionless: FoS = tan φ / tan ψ
• Kinematic (planar): φ < ψ < slope face angle, strike within ±20° of face

Worked example: Planar block — dry

Given: ψ = 35°, φ = 30°, c = 25 kPa; A = 20 m²/m, W = 2000 kN/m

Find: FoS.

• Resisting = 25 × 20 + 2000 cos35° tan30° = 500 + 946 = 1446
• Driving = 2000 sin35° = 1147
• FoS = 1.26

Answer: 1.26 — acceptable for a bench, marginal overall

Worked example: Cohesionless joint

Given: c = 0, φ = 30°, ψ = 35°

Find: FoS.

• FoS = tan30°/tan35° = 0.577/0.700 = 0.82

Answer: 0.82 — fails without cohesion

Common mistakes

• Wrong: Ignoring water pressure. — Right: Water can reduce FoS by 20–40 %.
• Wrong: Relying on cohesion of joints long-term. — Right: Cohesion degrades — check c = 0 case.

Applications

• Pit wall design: Bench and inter-ramp angles by structural domain.
• Depressurisation: Drains and wells to steepen slopes.
• Monitoring & TARPs: Trigger action response plans from radar data.

Key takeaways

• Planar, wedge, toppling, circular modes.
• FoS = [cA + (W cosψ − U) tanφ]/(W sinψ).
• Water lowers FoS — drain slopes.
• Monitor with radar and prisms.

Going further

• Limit equilibrium software: Slide, SWedge, RocPlane for 2D/3D analysis.
• Numerical models: FLAC/3DEC for complex failure mechanisms.
• Probabilistic analysis: Probability of failure from parameter distributions.