OpenSpecEngineering

Finds the factor of safety of a long, uniform slope against sliding on a plane parallel to its surface. This is the infinite slope model [2], [3]. It suits shallow slides, where a thin layer of soil slips over firmer ground below: hillsides with weathered soil over rock, embankment faces and cut slopes in sand. Engineers use it to check a slope angle, or to see how much a rising water table weakens a slope.

Equations

τ = γ·z·sin β·cos β
s = c′ + γ′·z·cos²β·tan φ′
FS = s / τ

β is the slope angle and z the vertical depth from the surface to the slip plane. γ is the total unit weight of the soil: the moist weight for a dry slope, the saturated weight γ_sat with seepage. c′ and φ′ are the effective cohesion and friction angle. τ is the shear stress that gravity puts on the slip plane, and s the shear strength available there. For a dry slope γ′ = γ. With the water table at the surface and seepage parallel to the slope, the pore pressure lowers the effective normal stress, so γ′ = γ_sat − γ_w [1], [3]. That roughly halves the friction part of the strength.

For a clean sand (c′ = 0), FS = tan φ′ / tan β when dry, whatever the depth. The page rates FS of 1.5 or more as Stable, 1.0 to 1.5 as Marginal and below 1.0 as Unstable. Picking a soil group fills in a typical φ′ for compacted fill [4] and sets c′ to 0; the unit weight stays yours to enter.

Limits and common mistakes

• The model ignores the ends of the slide. It suits a slide much longer than it is deep; for a deep or circular failure use a method of slices.
• Use effective-stress strength (c′, φ′) from drained tests. Small values of c′ raise FS a lot on shallow planes; don’t count on cohesion that wetting or root decay can remove.
• The seepage case is the worst one: water at the surface, flowing parallel to it. A lower water table gives a result between the two cases.

Worked example

A 25° hillside has 2.5 m of silty soil over rock. Tests gave c′ = 4 kPa, φ′ = 32° and γ_sat = 20 kN/m³. Is it safe after heavy rain saturates it?

1. Choose Pore-Water Condition Seepage parallel to slope, water table at surface.
2. Enter Slope Angle 25, Effective Cohesion 4, Effective Friction Angle 32 and Soil Unit Weight 20.
3. Enter Depth to Failure Plane 2.5 m, keep Unit Weight of Water 9.81 and press Solve. The Solution shows: Driving shear stress τ: 19.1511 kPa
   Available shear strength: 17.0754 kPa
   Factor of safety (FS): 0.891614
   Assessment: Unstable

The same slope with Dry / no seepage gives FS 1.5489, so it is the water that brings it down. Check by hand: τ = 20 × 2.5 × sin 25° × cos 25° = 19.15 kPa, and s = 4 + 10.19 × 2.5 × 0.8214 × 0.6249 = 17.08 kPa, so FS = 0.892.