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Slope stability analysis

Advanced and Design Soil Mechanics Reviewed 24 Oct 2026 2 min read

Natural hillsides, cuttings, embankments and dams all stand only because the soil's shear strength along any possible slip surface exceeds the shear stress driving movement. Slope stability analysis finds the weakest such surface and expresses safety as a factor of safety: available strength divided by the strength needed to just hold the slope.

For an infinite slope with seepage parallel to the surface, a closed-form expression shows that a dry cohesionless slope is stable up to its angle of friction, while full seepage roughly halves the safe slope. For finite slopes, circular slip surfaces are analysed with the method of slices: the soil above the trial circle is cut into vertical slices, and forces or moments on each slice are summed. The ordinary (Fellenius) method is simple but conservative; Bishop's simplified method includes inter-slice normal forces and is the usual hand method; Janbu, Spencer and Morgenstern-Price handle non-circular surfaces and satisfy more equilibrium conditions. Software searches thousands of surfaces to find the minimum factor.

Diagram: Indiacon original, not to scale.

Water is the main trigger of slope failures in India, especially in monsoon: rising pore pressure cuts effective stress and strength. Remedies include drainage (surface drains, horizontal drains), flattening or benching, toe buttresses, retaining structures, soil nails, anchors and vegetation. Monitoring with inclinometers and piezometers checks behaviour over time.

Key formulas
  • Infinite slope, dry sand: F = tan φ′ / tan β
  • With seepage parallel to slope: F ≈ (γ′/γsat) tan φ′ / tan β
  • Ordinary slices: F = Σ[c′l + (W cos α − ul) tan φ′] / Σ W sin α
  • Bishop: F = Σ{[c′b + (W − ub) tan φ′] / mα} / Σ W sin α
Solved example

Problem. A long natural slope in sand at 22° has φ′ = 34° and γsat = 20 kN/m³. Find the factor of safety when dry and when water flows parallel to the surface.

  1. Dry: F = tan 34° / tan 22° = 0.675 / 0.404 = 1.67.
  2. Seepage: γ′ = 20 − 9.81 = 10.19 kN/m³; ratio 10.19/20 = 0.51.
  3. F = 0.51 × 1.67 = 0.85.
  4. The slope is stable dry but fails with full seepage, which is why monsoon slides are common.

Answer. 1.67 dry; about 0.85 with parallel seepage

Quick quiz

1What is the factor of safety of a slope?

Available shear strength divided by the strength needed for equilibrium.

2Which method of slices is the usual hand method?

Bishop's simplified method.

3Why do slopes fail in heavy rain?

Pore water pressure rises, reducing effective stress and shear strength.

Code note
  • IS 2720 (series) (India): Methods of test for soils: water content, limits, compaction, shear and others, each in its own part.

Clauses and documents to read

  • IS 7894:1975 (stability analysis of earth dams)
  • IRC:SP:48 (hill road manual, cited by number)

Cited by number and clause only. We do not reproduce clause text, tables or figures (BIS Act 2016, s.10(5) and s.11). Check the current edition and amendments before use.

References and further reading

Standard texts, listed in no order of merit. We do not copy text, figures, tables or solved problems from any book; our explanations and examples are original.

  • Gopal Ranjan, A. S. R. Rao, Basic and Applied Soil Mechanics, New Age International, latest edition.
  • B. M. Das, Principles of Geotechnical Engineering, Cengage, latest edition.
  • K. Terzaghi, R. B. Peck, G. Mesri, Soil Mechanics in Engineering Practice, Wiley, 3rd edition.
  • J. A. Knappett, R. F. Craig, Craig's Soil Mechanics, CRC Press, latest edition.

Free and official sources

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