Shear strength
Soils fail in shear: grains slide and roll past each other along a surface. Foundations punching into the ground, slopes sliding and retaining walls tipping are all shear failures. Shear strength is therefore the property that decides bearing capacity, slope stability and earth pressure.
The Mohr–Coulomb criterion describes strength as τf = c′ + σ′ tan φ′, written in effective stresses: friction grows with the effective normal stress between grains, plus a small cohesion intercept for some soils. Effective stress matters because pore water carries part of the normal stress but cannot resist shear. Sands are usually free draining, so drained strength applies under static loading; under earthquake or other very rapid loading, loose saturated sands can behave undrained and may liquefy. Clays loaded quickly do not drain, so their short-term strength is the undrained shear strength cu, which does not change with total stress. Long-term stability of clay, after pore pressures settle, uses drained parameters and is often more critical for cuttings.
Strength is measured in the laboratory by the direct shear box, the triaxial test (unconsolidated undrained, consolidated undrained with pore pressure, or consolidated drained), the unconfined compression test for clays and the laboratory vane. On site, the vane shear test, cone penetration and standard penetration tests give strength indirectly. Dense sands and overconsolidated clays show a peak then soften to a lower residual strength, important for old landslides.
Decide first whether the loading is drained or undrained; that choice decides which strength parameters to use.
Mohr–Coulomb: τf = c′ + σ′ tan φ′Effective stress: σ′ = σ − uUndrained clay: τf = cu (φu = 0)Unconfined compression: cu = qu / 2
Using total-stress friction angles from a quick test for long-term stability of a clay cutting.
Problem. Two drained shear box tests on a sand give failure shear stresses of 58 kPa at σ′ = 100 kPa and 116 kPa at σ′ = 200 kPa. Find c′ and φ′, then the strength at σ′ = 150 kPa.
- Slope tan φ′ = (116 − 58) / (200 − 100) = 0.58, so φ′ = 30.1°.
- Intercept c′ = 58 − 100 × 0.58 = 0.
- At 150 kPa: τf = 150 × 0.58 = 87 kPa.
Answer. c′ = 0, φ′ ≈ 30°; strength 87 kPa at 150 kPa
Quick quiz
1Why use effective stress for soil strength?
Pore water carries part of the normal stress but cannot resist shear.
2Which strength applies to clay loaded quickly?
Undrained shear strength cu.
3Name two lab tests for shear strength.
Direct shear and triaxial (also unconfined compression, lab vane).
- IS 2720 (series) (India): Methods of test for soils: water content, limits, compaction, shear and others, each in its own part.
- IS 1498 (India): Classification and identification of soils for general engineering purposes.
Clauses and documents to read
- IS 2720 (Part 13):1986 (direct shear test)
- IS 2720 (Part 11):1993 (UU triaxial)
- IS 2720 (Part 12):1981 (CU and CD triaxial)
- IS 2720 (Part 10):1991 (unconfined compressive strength)
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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