Soil Mechanics Laboratory · Geotechnical Engineering

Direct Shear Calculator — Cohesion c and Friction Angle φ

The direct shear test in a shear box (ASTM D3080) provides the Mohr-Coulomb shear strength parameters: cohesion c and friction angle φ. This calculator fits the linear envelope τ = c + σ·tan φ by least squares regression to the (σn, τ) pairs measured in the laboratory, typically 3-4 points with different normal stresses. It also calculates residual (post-peak) parameters useful for slope stability analysis with prior movements and large deformation diagnostics.

What is it and when to apply it?

Direct shear is the most widely used test to characterize shear strength in granular and cohesive-frictional soils: sands, silts, and clays without fines. It is fast (1-3 days) and economical, but it imposes the failure plane and does not measure pore pressures, which limits its accuracy in saturated clays (where triaxial CU with u measurements is preferred). It is used for foundation design, retaining walls, slopes, and fills. In rocks, it is applied to contacts and discontinuities (barrera-barrera Roger Hoek 1977, ISRM).

Applied Formulas

Mohr-Coulomb Criterion:

τ = c + σn · tan φ

Linear Regression of n points:

tan φ = [n·Σ(σn·τ) − Σσn·Στ] / [n·Σσn² − (Σσn)²]

c = (Στ − tan φ · Σσn) / n

Coefficient of Determination R²:

R² = 1 − SSres/SStot, where SSres = Σ(τobs − τpred)²

Residual Parameters: cr ≈ 0 (cohesion is lost after failure); φr is obtained from the descending branch or test with large displacement (reverse direct shear)

ASTM D3080 Acceptance Criteria: 3-4 points with σn covering the project range; shear rate 0.1-1 mm/min (drained); R² > 0.95

Calculate online

Enter 3 pairs (σn, τ) and obtain c and φ by Mohr-Coulomb linear regression.

Least squares regression with 3 points. If c is slightly negative, it is reported as 0 (sand with no real cohesion).

Calculation Example

Input data — direct shear on silty sand SM, Los Angeles embankment
σn (kPa)τpeak (kPa)τresidual (kPa)
504632
1007661
200138121
300198181

Calculation for peak envelope with 4 points: Σσn = 650, Στ = 458, Σσn² = 140,000, Σ(σn·τ) = 100,600. tan φ = (4·100,600 − 650·458)/(4·140,000 − 650²) = (402,400 − 297,700)/(560,000 − 422,500) = 104,700/137,500 = 0.7614. φ = arctan 0.7614 = 37.3°. c = (458 − 0.7614·650)/4 = (458 − 495)/4 = −9.25 kPa. Since c is negative (small), it is interpreted as sand with no real cohesion and reported as c = 0, φ = 37.3° (ASTM standard practice). Residual envelope: Σσn = 650, Στr = 395, Σσn·τr = 87,350. tan φr = (4·87,350 − 650·395)/137,500 = (349,400 − 256,750)/137,500 = 92,650/137,500 = 0.6738. φr = 33.9°, cr ≈ 0. Peak R² = 0.999 (very good). Loss due to failure: Δφ = 37.3 − 33.9 = 3.4°, typical behavior of dense silty sand.

Result: Peak: c = 0, φ = 37.3° · Residual: cr = 0, φr = 33.9° · R² = 0.999.

Interpretation of Results

The peak parameters (c = 0, φ = 37.3°) correspond to a dense silty sand in an undisturbed state and are suitable for designing new foundations on virgin ground. The residual parameters (φ = 33.9°) apply to slopes with a history of movement, compacted fills undergoing reconsolidation, or post-failure analysis. The peak-residual difference of 3° is moderate; in sensitive clays or loessial soils, the difference can exceed 20°, and long-term stability requires residual parameters mandatorily.

Reference Standards

Frequently Asked Questions

When do I use peak and when residual?

Peak: design of new structures, foundations, slopes excavated in virgin ground. Residual: analysis of slopes with prior sliding (reactivation), dams with detected stability problems, back-analysis of failures. If in doubt, use residual (conservative).

Why direct shear vs. triaxial?

Direct shear is fast and cheap but imposes the failure plane and does not measure u. Triaxial CU in saturated clay measures u and allows obtaining effective c' and φ' separated from total ones. In clean sands, both give similar results; in clays, triaxial is more accurate.

What shear rate should I use?

For sands 0.5-1.0 mm/min. For silts 0.1-0.3 mm/min. For clays 0.005-0.05 mm/min (fully drained). ASTM D3080 specifies t50 from the oedometer × 20-50 to estimate the minimum rate. Very high rate underestimates c in clays due to u generation without dissipation.

Square or circular box?

Square is the ASTM standard (60×60 mm or 100×100 mm). Circular (63.5 mm) used in older equipment; the same area is assumed. For samples with coarse particles, a 100×100 mm box is used to minimize size effects. Never particles > 1/10 of the box side.

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