Shear Strength of Soil — Mohr-Coulomb Criterion

1. Introduction to Shear Strength

Shear strength is the maximum shear stress a soil can sustain before it fails (slides along a failure plane). It is the single most important mechanical property of soil — it governs the bearing capacity of foundations, the stability of slopes and embankments, the lateral earth pressure on retaining walls, and the load-carrying capacity of piles.

Soil shear strength comes from two sources: friction (resistance to sliding between particles, proportional to normal stress) and cohesion (interparticle bonding from electrostatic forces, cementation, and suction). Sands and gravels derive strength almost entirely from friction; clays have both cohesion and friction, though the relative contributions depend on drainage conditions and stress history.

A critical distinction: shear strength must always be expressed in terms of effective stresses for long-term (drained) analysis. For short-term (undrained) analysis of saturated clays, total stress parameters are used because pore pressures are not measured during rapid loading.

2. Mohr-Coulomb Failure Criterion

2.1 General Form

2.2 Shear Strength Parameters for Common Soils

2.3 Principal Stress Relationship at Failure

3. Mohr’s Circle of Stress

Mohr’s circle is a graphical tool that shows the normal and shear stresses acting on any plane through a soil element, given the principal stresses. At failure, the Mohr’s circle is tangent to the Mohr-Coulomb failure envelope.

3.1 Circle Geometry

3.2 Angle of Failure Plane

3.3 Graphical Determination of c and φ

Plot Mohr’s circles for at least two triaxial tests with different cell pressures. Draw the best-fit common tangent to the circles — its y-intercept gives c (or c′) and its inclination gives φ (or φ′). More circles give a more reliable failure envelope.

4. Direct Shear Test (IS 2720 Part 13)

4.1 Test Setup and Procedure

A soil specimen (60 mm × 60 mm × 25 mm or 100 mm square) is placed in a split shear box. A vertical normal load is applied. The lower box is pushed horizontally at a constant rate while the upper box is held fixed; the horizontal (shear) force and horizontal displacement are recorded until failure.

4.2 Results and Parameters

4.3 Advantages and Disadvantages

5. Triaxial Compression Test

The triaxial test is the most versatile and widely used shear strength test. A cylindrical soil specimen (38 mm × 76 mm or 50 mm × 100 mm standard) is enclosed in a rubber membrane inside a pressurised cell. All-round cell pressure σ₃ is applied first, then a vertical deviator stress (σ₁ − σ₃) is applied until failure.

5.1 Stress State in Triaxial Test

5.2 Failure Criterion from Triaxial Test

6. CD, CU, and UU Triaxial Tests

6.1 Consolidated Drained (CD) Test

6.2 Consolidated Undrained (CU) Test

6.3 Unconsolidated Undrained (UU) Test

6.4 Summary Comparison

7. Pore Pressure Parameters (Skempton, 1954)

Skempton defined pore pressure parameters A and B to quantify how applied total stresses generate pore water pressure in undrained loading.

A positive A_f means pore pressure increases during shear (contractive behaviour — typical of loose sands and NC clays). A negative A_f means pore pressure decreases during shear (dilative behaviour — dense sands and heavily OC clays).

8. Undrained Shear Strength of Saturated Clay

8.1 φ_u = 0 Analysis

For a fully saturated clay tested under undrained conditions (no drainage), applying an additional all-round pressure simply increases both total stresses equally — the effective stresses do not change and neither does the soil structure. Therefore the shear strength (which depends on effective stress) is unchanged, and the failure envelope in total stress space is horizontal (φ_u = 0):

8.2 Variation of c_u with Depth

For a NC clay deposit, c_u increases approximately linearly with depth because effective overburden pressure (and hence pre-consolidation pressure) increases with depth. A useful empirical relationship (Skempton, 1957):

8.3 Unconfined Compression Test (UCT)

8.4 Consistency from q_u

9. Vane Shear Test (IS 2720 Part 30)

The field vane shear test measures the in-situ undrained shear strength of soft to medium clays. A four-bladed cruciform vane is pushed into the undisturbed clay and slowly rotated. The torque required to shear a cylindrical surface is measured.

After measuring the peak undisturbed strength, the vane is rapidly rotated several times to fully remould the soil, then the torque is measured again to give the remoulded strength. The ratio of peak to remoulded gives the sensitivity S_t.

Bjerrum (1972) correction for anisotropy and strain rate: c_u,corrected = μ × c_u,vane, where μ = 1.7 − 0.54 log(PI).

10. Sensitivity and Thixotropy

10.1 Sensitivity (S_t)

High sensitivity is dangerous — disturbance during construction or earthquake shaking can cause a sudden large loss of strength, leading to catastrophic slope failure or foundation failure.

10.2 Thixotropy

Thixotropy is the partial or complete recovery of shear strength after remoulding, if the soil is left undisturbed. It is caused by gradual re-establishment of electrochemical bonds between clay particles over time (hours to days). The thixotropic ratio = regained strength / remoulded strength (should be ≤ 1 always).

11. Worked Examples

Example 1 — Direct Shear Test: Find c and φ

Problem: Three direct shear tests on a soil sample give the following results at failure:

Find cohesion c and angle of internal friction φ.

Solution — Linear Regression

Verify: at σ = 50: τ_f = 12 + 50(0.56) = 12 + 28 = 40 ✓; at σ = 200: τ_f = 12 + 200(0.56) = 12 + 112 = 124 ✓

Example 2 — Triaxial Test: Mohr’s Circle and Failure Envelope

Problem: Two CU triaxial tests on a saturated clay give the following total stress results at failure:

Find: (a) total stress parameters c and φ, (b) effective stress parameters c′ and φ′.

(a) Total Stress Parameters

(b) Effective Stress Parameters

Example 3 — GATE-Style: Stress at Failure on Failure Plane

Problem (GATE CE type): A soil has c′ = 20 kPa, φ′ = 30°. A triaxial test gives σ₃′ = 100 kPa at failure. Find: (a) σ₁′ at failure, (b) shear stress on the failure plane, (c) normal stress on the failure plane.

(a) Major Principal Stress at Failure

(b) & (c) Stresses on Failure Plane

12. Common Mistakes

Mistake 1: Using Total Stress Parameters for Long-Term Stability Analysis

What happens: c_u and φ_u = 0 (from UU test) are used to compute long-term bearing capacity or slope FOS. Because undrained parameters reflect the short-term strength before drainage, using them for long-term design can be either unconservative (for NC clays where drainage increases strength) or overly conservative (for OC clays where drainage may reduce strength).

Fix: Long-term (drained) analysis → use effective stress parameters c′ and φ′ with pore pressures from steady-state seepage. Short-term (end-of-construction) → use c_u and φ_u = 0 in total stress analysis.

Mistake 2: Assuming φ_u = 0 for All Clays

What happens: φ_u = 0 applies only to fully saturated clays in undrained conditions (B = 1). For partially saturated clays or for clayey soils with significant sand content, φ_u > 0 and the UU failure envelope is not horizontal.

Fix: φ_u = 0 only when S = 100 % and no drainage occurs during shearing. If saturation is less than about 85 %, φ_u will be measurably positive.

Mistake 3: Confusing the Angle of the Failure Plane with the Angle of Internal Friction

What happens: The failure plane is assumed to make an angle of φ′ with the horizontal (major principal plane), instead of the correct α_f = 45° + φ′/2. For φ′ = 30° this gives 30° instead of 60° — a major error in computing stresses on the failure plane.

Fix: The failure plane makes an angle of 45° + φ′/2 with the major principal plane (σ₁ plane), or equivalently 45° − φ′/2 with the minor principal plane (σ₃ plane).

Mistake 4: Forgetting to Subtract Pore Pressure When Plotting Effective Stress Mohr’s Circles

What happens: In CU tests, Mohr’s circles are plotted in total stress space and the resulting c and φ are reported as effective parameters. The correct effective parameters require the circles to be plotted with σ₁′ = σ₁ − u and σ₃′ = σ₃ − u.

Fix: For effective stress Mohr’s circles, subtract the pore pressure at failure from both principal stresses: σ₁′ = σ₁ − u_f; σ₃′ = σ₃ − u_f. Note that the radius (= (σ₁ − σ₃)/2) is the same in total and effective stress because u cancels in the difference.

Mistake 5: Using Shear Stress at 45° as the Maximum Shear Stress

What happens: The maximum shear stress in a Mohr’s circle is (σ₁−σ₃)/2, which occurs on the plane inclined at 45° to the principal planes. Students assume this is also the failure plane. But the failure plane is at 45° + φ′/2 (not 45°), where the ratio τ/σ′ is maximised relative to the Mohr-Coulomb envelope — not where τ alone is maximum.

Fix: Maximum shear stress plane = 45° to principal planes. Failure plane = 45° + φ′/2 to major principal plane. These coincide only when φ = 0 (undrained saturated clay).

13. Frequently Asked Questions

Q1. Why does a saturated clay have φ_u = 0 in undrained loading?

When a fully saturated clay is subjected to additional cell pressure in a UU test (undrained condition), the applied stress is carried entirely by pore water (B = 1). The effective stresses — which control shear strength — do not change. Therefore, regardless of the total confining pressure applied, the soil fails at the same deviator stress. In Mohr’s circle terms, all UU circles for a saturated clay have the same diameter (same deviator stress at failure), and their common tangent is a horizontal line at height c_u — giving φ_u = 0. This is not a material property; it is a consequence of the undrained loading condition for a saturated soil.

Q2. What is the difference between the effective friction angle φ′ and the angle of repose?

The angle of repose is the steepest angle at which a dry granular material can pile up without sliding — it is a macroscopic field observation, not a laboratory-measured property. For a dry cohesionless sand, the angle of repose ≈ φ′_peak (peak effective friction angle), because the conditions (zero pore pressure, negligible cohesion, slow drainage) approximate a drained shear test. However, φ′ measured in the triaxial or direct shear test is more precise and accounts for the confining stress, density, and stress path — the angle of repose is a rough field estimate of φ′ only for dry granular soils without fines. For wet soils, the angle of repose can exceed φ′ due to apparent cohesion from capillarity.

Q3. Why is the CD test rarely used for clay in practice?

The CD test requires shearing so slowly that no excess pore pressure develops — for a low-permeability clay this can take days or weeks per test specimen. The strain rate for a CD test is typically 0.001–0.01 mm/min for a standard 76 mm specimen, meaning a single test takes 5–15 days. In practice, the CU test with pore pressure measurement is used instead — it takes only a few hours for shearing (after consolidation), and the pore pressure measurements allow computation of effective stresses. The effective strength parameters c′ and φ′ obtained from CU(+u) tests are equivalent to those from CD tests for most practical purposes.

Q4. How does the Mohr-Coulomb criterion relate to the bearing capacity of a foundation?

The bearing capacity of a shallow foundation is fundamentally a shear failure problem — the soil beneath the foundation shears along a failure surface when the applied stress exceeds the soil’s shear strength. Terzaghi’s bearing capacity equation (q_u = cN_c + qN_q + 0.5γBN_γ) is derived by assuming the soil obeys the Mohr-Coulomb failure criterion on a composite failure surface consisting of a straight active wedge below the footing, a radial fan zone, and a passive wedge at the edge. The bearing capacity factors N_c, N_q, and N_γ are directly functions of φ — they are derived from the geometry of the Mohr-Coulomb failure envelope. Higher φ → larger N values → higher bearing capacity.