Hong Life CalculatorsUnderstanding the infinite slope stability calculator — concept and principle
Long natural slopes often fail along a shallow slip surface parallel to the ground, such as a weathered soil layer sliding over rock. The infinite slope method treats such a slope as one long uniform block and compares the shear strength available on the slip plane with the shear stress driving it downhill. This calculator gives the factor of safety and a stability label, including slope-parallel groundwater.
How it is calculated
FS = [c + (γ − m γ_w) z cos²β tan φ] / (γ z sin β cos β)
- c: cohesion (kPa); φ: friction angle; γ: unit weight (kN/m³).
- z: vertical depth to the slip plane (m); β: slope angle.
- m: fraction of z that is saturated, from 0 (dry) to 1 (water at the surface); γ_w = 9.81 kN/m³.
The label reads "FS ≥ 1.5" at 1.5 or more, "marginal" from 1 to 1.5 and "unstable" below 1.
The principle
Take a column of soil of depth z on the slip plane. Its weight per unit plan area, γz, splits into a component along the slope, γz sin β cos β, which drives sliding, and a component normal to it, γz cos²β, which presses the plane together. Pore water pressure from parallel seepage, m γ_w z cos²β, reduces that normal stress. The Mohr–Coulomb strength c + σ′ tan φ divided by the driving stress is the factor of safety. End effects are ignored. Dividing through shows two parts: c/(γz sin β cos β) plus (1 − mγ_w/γ) tan φ / tan β.
Worked example
- Inputs: c = 5 kPa, φ = 30°, γ = 19 kN/m³, z = 2 m, β = 25°, m = 0.
- Normal stress = 19 × 2 × cos²25° = 38 × 0.8214 = 31.21 kPa; × tan 30° = 18.02 kPa.
- Resisting stress = 5 + 18.02 = 23.02 kPa.
- Driving stress = 19 × 2 × sin 25° × cos 25° = 14.55 kPa.
- FS = 23.02 / 14.55 = 1.582, which is labelled FS ≥ 1.5.
Check: cohesion part 0.344 plus friction part tan 30°/tan 25° = 1.238 gives the same 1.582.
| Saturated fraction m (0–1) | Factor of safety |
|---|---|
| 0 | 1.581659882 |
| 1 | 0.9423929626 |
| 2 | 0.3031260434 |
| 3 | -0.3361408758 |
| 4 | -0.975407795 |
| 5 | -1.614674714 |
| 6 | -2.253941633 |
How to read the result
FS below 1 means failure is expected; values just above 1 offer little margin for uncertain strength. Many guidelines ask for about 1.3 to 1.5 for long-term static slopes. Groundwater is decisive: with m = 1 the same slope drops to about 0.94. For a dry slope with no cohesion, FS = tan φ / tan β, so a sand slope is stable only if it is flatter than its friction angle.
Common mistakes
- Using perpendicular depth instead of vertical depth z.
- Relying on cohesion in shallow, weathered soils where c may be close to zero.
- Using m = 0 for slopes that become saturated in heavy rain.
- Applying the method to short slopes or deep rotational failures.
Frequently asked questions
Why does rain trigger landslides?
Rising water increases m, which cuts the normal effective stress and the friction resistance while the driving stress stays the same.
What depth should I use?
The depth of the weak layer or soil–rock contact, found by investigation.
Does vegetation help?
Roots can add some cohesion near the surface, but it is hard to quantify and often ignored.
Sources: Skempton, A. W. and DeLory, F. A. (1957), Stability of natural slopes in London Clay, Proc. 4th ICSMFE, London · Duncan, J. M., Wright, S. G. and Brandon, T. L., Soil Strength and Slope Stability, Wiley