Slope Stability

The balance that keeps a slope standing

Every slope is a quiet contest between two things. Gravity pulls the soil mass downhill, generating a driving shear stress along any surface the mass might slide on. The soil's own shear strength resists that motion, gripping the grains together with friction and cohesion. As long as strength wins, the hillside stands. The instant the driving stress equals the strength, the slope is at the point of failure — and a landslide follows.

Slope stability analysis turns this contest into a single number, the factor of safety (FS):

      available shear strength
FS = ───────────────────────────
      shear stress required for equilibrium

FS > 1   stable (strength to spare)
FS = 1   limiting equilibrium — failure imminent
FS < 1   the slope is already moving

The trick is that we do not know in advance where the soil will slide. So the analysis assumes a trial slip surface, computes the FS for it, then repeats for thousands of other candidate surfaces. The surface that produces the lowest FS is the critical one — the slope's weakest link — and that value is reported as the FS of the slope.

Where shear strength comes from: Mohr–Coulomb

The resisting strength of soil at any point on the slip surface is given by the Mohr–Coulomb failure criterion, written in effective stresses:

τ_f = c′ + σ′ · tan φ′

where:
  τ_f = shear strength at failure        (kPa)
  c′  = effective cohesion               (kPa)
  σ′  = effective normal stress on plane (kPa)
  φ′  = effective friction angle         (degrees)
  σ′  = σ − u   (total stress minus pore water pressure u)

Two mechanisms supply strength. Cohesion c′ is a constant grip — the electrochemical bonding of clay particles — present even when there is no confining pressure. Friction grows with the effective normal stress σ′ pressing the grains together. The split matters: a clean sand has c′ ≈ 0 and lives entirely on friction, while a stiff clay carries much of its strength as cohesion.

The dependence on effective stress is the single most important fact in geotechnical engineering. Water pressure u in the pores pushes the grains apart, so only σ′ = σ − u clamps them. Raise the water table and σ′ falls, friction falls, and the slope loses strength without anyone touching it. This is why rain, not earthquakes, triggers most landslides.

The circular slip surface and the method of slices

For a roughly homogeneous clay slope, the critical slip surface is very nearly a circular arc. The sliding mass rotates about the circle's center O. We can therefore take moments about O: the driving moment is the weight of the mass times its horizontal offset from O, and the resisting moment is the shear strength integrated along the arc times the radius R.

But the normal stress, water pressure, and even the soil type vary along the arc, so the strength is not uniform. The method of slices handles this by cutting the sliding wedge into vertical strips and treating each slice's base separately:

Ordinary (Fellenius / Swedish) method of slices:

       Σ [ c′·ℓ + (W·cos α − u·ℓ)·tan φ′ ]
FS = ──────────────────────────────────────
                 Σ [ W·sin α ]

per slice:
  W = γ · b · h        slice weight   (γ = unit weight)
  α = base inclination angle
  ℓ = b / cos α        base length of slice
  u = pore pressure at the slice base
  W·sin α  = driving component (down the arc)
  W·cos α  = normal component (clamps the base)

Bishop's simplified method improves on this by accounting for the horizontal forces between slices; it satisfies vertical equilibrium and is solved iteratively (FS appears on both sides). It typically gives an FS a few percent higher than the conservative Ordinary method and is the workhorse of routine design. Janbu's, Spencer's and Morgenstern–Price methods relax further assumptions and satisfy full force and moment equilibrium, at the price of more computation.

The infinite-slope shortcut

When the failure surface is a plane parallel to the ground — typical of a thin soil mantle sliding over bedrock — the analysis collapses to a closed form that needs no slices. For a dry cohesionless slope at angle β:

Dry, cohesionless infinite slope:
  FS = tan φ′ / tan β

→ a sand slope is stable as long as β < φ′ (the angle of repose)

With seepage parallel to the slope (fully saturated):
  FS = (γ′ / γ_sat) · (tan φ′ / tan β)   ≈  0.5 · tan φ′ / tan β

→ saturating the slope roughly halves the factor of safety

That second result is sobering: a dry sand slope safely standing at β = 30° with φ′ = 35° (FS ≈ 1.21) drops to FS ≈ 0.61 when fully saturated with seepage — it fails. The angle of repose you see in a dry sandpile is exactly the angle at which FS = 1 for dry sand.

Comparing analysis methods

Method	Slip surface	Equilibrium satisfied	Inter-slice forces	Typical use
Infinite slope	Planar, surface-parallel	Force (1 element)	None	Shallow colluvium over rock
Ordinary / Fellenius	Circular arc	Moment only	Ignored	Hand check; conservative (low FS)
Bishop's simplified	Circular arc	Moment + vertical force	Horizontal only	Routine design workhorse
Janbu's simplified	Any shape	Horizontal + vertical force	Horizontal only	Non-circular / wedge surfaces
Spencer's	Any shape	Full force + moment	Constant inclination	Critical or back-analysis cases
Morgenstern–Price	Any shape	Full force + moment	Variable function f(x)	Complex layered geometry

Worked example: a simple cut slope

A 6 m high cutting is excavated at 1V:2H (β ≈ 26.6°) in a stiff clay with effective cohesion c′ = 10 kPa, friction angle φ′ = 25°, and unit weight γ = 19 kN/m³. A trial circular slip surface bounds a sliding mass of weight W = 1,150 kN per metre run, acting at a horizontal offset of 3.0 m from the circle center; the arc length is 14 m at radius R = 12 m, with an average normal effective stress of 45 kPa on the base. Estimate the factor of safety.

Driving moment   M_d = W · x = 1,150 × 3.0 = 3,450 kN·m

Resisting strength per metre of arc:
  τ_f = c′ + σ′·tan φ′ = 10 + 45 × tan 25°
      = 10 + 45 × 0.466 = 30.97 kPa

Total resisting force along the arc:
  S = τ_f × arc length = 30.97 × 14 = 433.6 kN

Resisting moment  M_r = S · R = 433.6 × 12 = 5,203 kN·m

         M_r     5,203
FS  =  ─────── = ─────── = 1.51   ✓ acceptable for a permanent cut
         M_d     3,450

This single circle gives FS = 1.51. A full analysis would search hundreds more circles — flatter, deeper, shallower — to confirm none gives a lower value. If a deeper circle that passes through a softer clay layer returned, say, FS = 1.18, that would govern the design.

Water: the usual culprit

Because strength rides on effective stress σ′ = σ − u, the position of the water table is often the difference between a safe slope and a failure. Three water-related triggers dominate the landslide record:

• Rainfall infiltration. Prolonged or intense rain raises the water table and pore pressures, cutting σ′ and friction. It also adds weight. A slope safe in summer can fail in a wet spring.
• Rapid drawdown. When a reservoir or river level drops faster than the slope can drain, high internal pore pressures remain while the external water support vanishes — a classic earth-dam and riverbank failure mode.
• Perched water and seepage. Water trapped above a low-permeability seam produces locally high u and seepage forces that drive shallow translational slides.

The standard defences attack u directly: horizontal drains, drainage blankets, toe drains and relief wells all lower the water table and restore effective stress without moving a single grain of slope soil.

Failure modes and how engineers fight them

• Rotational (circular) slide. A deep arc rotates the mass about a center; typical of homogeneous clay. Fix: flatten the slope, build a toe berm, or unload the crest.
• Translational slide. Movement along a planar weak layer parallel to the surface — colluvium over rock, or a clay seam. Fix: soil nails, anchors, or drainage of the seam.
• Wedge / block failure. Two intersecting discontinuities release a block, common in rock cuts. Fix: rock bolts and shotcrete.
• Flow (debris flow, earthflow). Saturated soil loses almost all strength and runs like a fluid. Fix: debris basins, check dams, slope drainage.
• Progressive failure. Strain-softening clays shed strength as they deform; the slope can creep then suddenly collapse once the peak strength is exceeded along enough of the arc. Fix: design to the residual (not peak) strength.
• Liquefaction-induced flow slide. Loose saturated sand under seismic shaking loses strength to near zero. Fix: densify the soil, install drains, or avoid loose saturated fills.

Seismic loading and the pseudo-static method

An earthquake adds a horizontal inertial force to the sliding mass. The simplest design tool, the pseudo-static method, applies a constant horizontal force k_h·W (k_h is the seismic coefficient, often 0.10–0.20) and recomputes the FS. Because earthquakes are brief and some permanent slope movement is usually tolerable, the acceptable seismic FS is lower — often 1.0 to 1.1. More refined design uses Newmark sliding-block analysis, which integrates the acceleration record above the yield acceleration to predict centimetres of permanent displacement rather than a pass/fail number.