10.2 Shear Strength of Soils and Mohr-Coulomb Failure Criteria

10.2 Shear Strength of Soils and Mohr-Coulomb Failure Criteria

The shear strength of a soil mass represents its internal resistance per unit area to sliding or slippage along an internal plane. Foundation bearing capacity failure, retaining wall collapse, and slope instability are fundamentally shear failures—soil rarely fails under direct compression, but rather shears when internal shear stresses exceed available shear resistance.

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The Mohr-Coulomb Failure Criterion

In 1900, Christian Otto Mohr combined Coulomb's frictional model with stress transformation theory to formulate the classical Mohr-Coulomb failure criterion.

Total vs. Effective Stress Envelopes

Because soil shearing resistance is transmitted through interparticle mineral contact points, shear strength must be analyzed in terms of effective stresses:

• Effective Stress Formulation (Drained / Long-Term): $\tau_f = c' + \sigma' \tan \phi' = c' + (\sigma - u) \tan \phi'$ Where:

  • $\tau_f$ = shear strength on the failure plane
  • $c'$ = effective cohesion (apparent interparticle bonding or cementation)
  • $\sigma'$ = effective normal stress on the failure plane ($= \sigma - u$)
  • $\phi'$ = effective angle of internal friction (shearing resistance from particle interlock and friction)
  • $u$ = pore water pressure

• Total Stress Formulation (Undrained / Short-Term): $\tau_f = c_u + \sigma \tan \phi_u$ For fully saturated cohesive soils subjected to rapid undrained loading, $\phi_u = 0$, giving: $\tau_f = c_u = s_u$ Where $c_u$ (or $s_u$) is the undrained shear strength.

   Shear Stress (τ)
       ^
       |                          / Mohr-Coulomb Failure Envelope
       |                         /  τ = c' + σ' tan(φ')
       |                     ---*---
       |                 .-'    |    '-.
       |              .'        |        '.
       |             /          |          \
       |            ;           |           ;
       |            |     Mohr's Circle     |
       |            ;       at Failure      ;
       |             \                     /
       |              '.                 .'
       |                '-.           .-'
       |    c'             '---------'
       |----+-------------------|-----------|-----------------> Effective Normal
       0                       σ'3         σ'1                  Stress (σ')

Principal Stresses at Failure

From the geometric relationship between Mohr's circle of stress and its tangent failure envelope, a fundamental relationship connects major principal stress $\sigma'_1$ and minor principal stress $\sigma'_3$ at failure: $\sigma'_1 = \sigma'_3 \tan^2\left(45^\circ + \frac{\phi'}{2}\right) + 2 c' \tan\left(45^\circ + \frac{\phi'}{2}\right)$

Defining the flow value (passive pressure factor) $N_\phi$: $N_\phi = \tan^2\left(45^\circ + \frac{\phi'}{2}\right) = \frac{1 + \sin \phi'}{1 - \sin \phi'}$ $\sigma'_1 = \sigma'_3 N_\phi + 2 c' \sqrt{N_\phi}$

For clean, cohesionless soils ($c' = 0$, such as sands and gravels): $\sigma'_1 = \sigma'_3 \tan^2\left(45^\circ + \frac{\phi'}{2}\right) \implies \frac{\sigma'_1}{\sigma'_3} = \frac{1 + \sin \phi'}{1 - \sin \phi'}$

Orientation of the Failure Plane

Mohr's circle transformation demonstrates that the failure plane does not align with the plane of maximum shear stress (which occurs at $45^\circ$). Instead, because shear resistance increases with normal stress, failure occurs on a critical plane where the ratio of shear stress to shear strength is maximized.

The failure plane is inclined at an angle $\theta$ to the major principal plane (the plane upon which $\sigma'_1$ acts, typically the horizontal plane in standard compression tests): $\theta = 45^\circ + \frac{\phi'}{2}$

The angle made by the failure plane with the direction of major principal stress $\sigma'_1$ is: $\alpha = 90^\circ - \theta = 45^\circ - \frac{\phi'}{2}$

The normal and shear stresses acting on this failure plane are calculated directly: $\sigma'_f = \frac{\sigma'_1 + \sigma'_3}{2} + \frac{\sigma'_1 - \sigma'_3}{2} \cos(2\theta)$ $\tau_f = \frac{\sigma'_1 - \sigma'_3}{2} \sin(2\theta)$

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Laboratory Shear Testing Methods

Four standard laboratory test configurations determine soil shear strength parameters:

1. Direct Shear Test (ASTM D3080)

Soil is placed inside a horizontally split metal shear box. A constant normal force $N$ is applied vertically, while a horizontal shear force $T$ is applied to displace the lower box half until the soil shears.

• Advantages: Rapid, inexpensive, simple setup; excellent for clean cohesionless sands and gravels.
• Disadvantages: Failure plane is artificially forced along a pre-determined horizontal plane; non-uniform stress distribution at edges; drainage cannot be controlled for fine clays; pore water pressure cannot be measured.
• Failure criterion: $\tau_f = T_{\max} / A$ plotted against $\sigma' = N / A$.

2. Unconfined Compression Test (UCT, ASTM D2166)

A special case of the triaxial test where the specimen has no lateral confinement (confining cell pressure $\sigma_3 = 0$). Only applicable to cohesive, saturated fine-grained soils.

• Axial load is increased rapidly until the cylindrical sample fails at axial stress $q_u$ (the unconfined compressive strength).
• Because $\sigma_3 = 0$ and $\phi = 0$, the diameter of Mohr's circle equals $q_u$: $c_u = s_u = \frac{q_u}{2}$

3. Triaxial Shear Tests (ASTM D2850 / D4767)

The gold standard of geotechnical testing. A cylindrical specimen encased in a thin rubber membrane is placed inside a pressurized fluid chamber. Testing occurs in two stages: Stage 1 (Confining Stage) where cell fluid pressure $\sigma_3$ is applied; and Stage 2 (Shearing Stage) where an axial deviator stress $\Delta \sigma_d = \sigma_1 - \sigma_3$ is applied to cause shear failure.

Skempton's Pore Pressure Parameters

In undrained triaxial shearing, pore water pressure response is quantified by Skempton's coefficients $A$ and $B$: $\Delta u = B [ \Delta \sigma_3 + A (\Delta \sigma_1 - \Delta \sigma_3) ]$

• Parameter $B$: Governs the confining stage. For fully saturated soils ($S = 100\%$), water is virtually incompressible relative to the soil skeleton, yielding $B = 1.0$. For dry soils, $B = 0$.
• Parameter $A$: Governs the deviator shearing stage. At failure ($A_f$):
  • Normally consolidated soft clays: $A_f = +0.5 \text{ to } +1.0$ (contractive tendency, positive pore pressure buildup).
  • Heavily overconsolidated clays and dense sands: $A_f = -0.5 \text{ to } 0$ (dilatant tendency, negative pore pressure / suction).

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Field In-Situ Shear Testing

Because laboratory specimens suffer from disturbance during drilling, extraction, transport, and trimming, in-situ field testing is essential for sensitive deposits.

1. Field Vane Shear Test (VST, ASTM D2573)

A four-bladed cruciform metal vane is pressed into undisturbed cohesive soil at the bottom of a borehole and rotated at a standard rate ($0.1^\circ/\text{s}$) until a cylindrical shear surface fails.

            | |  Drive Rod
            | |
         +--+-+--+
         |  | |  |
         |  | |  |
       H |  | |  |  Vane Height (H = 2D standard)
         |  | |  |
         +--+-+--+
           <-D->    Vane Diameter (D)

The maximum torque $T$ mobilizes shear resistance along the cylindrical perimeter (area $\pi D H$) and the two circular flat ends (top and bottom): $T = T_{\text{side}} + 2 T_{\text{end}} = (c_u \pi D H) \left(\frac{D}{2}\right) + 2 \int_0^{D/2} c_u (2\pi r\,dr) r = c_u \pi D^2 \left( \frac{H}{2} + \frac{D}{6} \right)$

Solving for undrained shear strength $c_u$: $c_u = \frac{T}{\pi D^2 \left( \frac{H}{2} + \frac{D}{6} \right)}$

For a standard rectangular vane where $H = 2D$: $c_u = \frac{T}{\pi D^2 \left( D + \frac{D}{6} \right)} = \frac{6 T}{7 \pi D^3}$

2. Standard Penetration Test (SPT, ASTM D1586)

A standard split-spoon sampler ($50.8\text{ mm}$ OD, $34.9\text{ mm}$ ID) is driven into the bottom of a borehole using a $140\text{ lb}$ ($63.5\text{ kg}$) safety or donut hammer dropping freely from a height of $30\text{ in}$ ($760\text{ mm}$).

• The sampler is driven through three consecutive $6\text{-in}$ ($150\text{ mm}$) intervals.
• The first 6 inches is discarded as the seating drive.
• The SPT $N$-value is the sum of blows required for the second and third 6-inch intervals (blows per foot of penetration).

To account for equipment variations and depth, the raw field $N$-value is normalized: $N_{60} = \frac{N \cdot \eta_H \cdot \eta_B \cdot \eta_S \cdot \eta_R}{60}$ Where $\eta_H$ is hammer efficiency (safety hammer $\approx 60\%$, automatic trip hammer $\approx 80\%$), $\eta_B$ is borehole diameter correction, $\eta_S$ is sampler lining correction, and $\eta_R$ is rod length correction.

Correcting for effective overburden stress $\sigma'_0$ yields $(N_1)_{60}$: $(N_1)_{60} = C_N N_{60} = \sqrt{\frac{p_a}{\sigma'_0}} N_{60} \le 2.0 N_{60}$ Where $p_a = 100\text{ kPa} \approx 1\text{ atm}$ is reference atmospheric pressure (Liao & Whitman formula).

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Step-by-Step Worked Problem Examples

Worked Example: Consolidated-Undrained (CU) Triaxial Test Analysis

Problem: A consolidated-undrained (CU) triaxial compression test was performed on a saturated normally consolidated clay specimen ($c' = 0$). The confining cell pressure was maintained at $\sigma_3 = 150\text{ kPa}$. At failure, the deviator stress reached $\Delta \sigma_d = 180\text{ kPa}$, and the induced pore water pressure was measured to be $u_f = 60\text{ kPa}$.

1. Calculate the effective principal stresses at failure ($\sigma'_3$ and $\sigma'_1$).
2. Determine the effective angle of internal friction $\phi'$.
3. Determine the theoretical failure plane angle $\theta$ with respect to the major principal plane.
4. Compute the effective normal stress $\sigma'_f$ and shear stress $\tau_f$ acting on the failure plane.

Solution:

Step 1: Compute Total and Effective Principal Stresses $\sigma_3 = 150\text{ kPa}$ $\sigma_1 = \sigma_3 + \Delta \sigma_d = 150 + 180 = 330\text{ kPa}$ Apply Terzaghi's effective stress principle ($u_f = 60\text{ kPa}$): $\sigma'_3 = \sigma_3 - u_f = 150 - 60 = 90\text{ kPa}$ $\sigma'_1 = \sigma_1 - u_f = 330 - 60 = 270\text{ kPa}$

Step 2: Calculate Effective Friction Angle ($\phi'$) For a normally consolidated clay, effective cohesion $c' = 0$. Using the principal stress ratio relation: $\sin \phi' = \frac{\sigma'_1 - \sigma'_3}{\sigma'_1 + \sigma'_3} = \frac{270 - 90}{270 + 90} = \frac{180}{360} = 0.500$ $\phi' = \arcsin(0.500) = 30.0^\circ$

Step 3: Determine Failure Plane Orientation ($\theta$) The failure plane angle with the major principal (horizontal) plane is: $\theta = 45^\circ + \frac{\phi'}{2} = 45^\circ + \frac{30.0^\circ}{2} = 45^\circ + 15^\circ = 60.0^\circ$

Step 4: Compute Stresses on the Failure Plane $\sigma'_f = \frac{\sigma'_1 + \sigma'_3}{2} + \frac{\sigma'_1 - \sigma'_3}{2} \cos(2\theta)$ $\sigma'_f = \frac{270 + 90}{2} + \frac{270 - 90}{2} \cos(120^\circ) = 180 + 90(-0.5) = 180 - 45 = 135.0\text{ kPa}$ $\tau_f = \frac{\sigma'_1 - \sigma'_3}{2} \sin(2\theta) = \frac{270 - 90}{2} \sin(120^\circ) = 90 \times \frac{\sqrt{3}}{2} = 77.94\text{ kPa}$ Check with Mohr-Coulomb equation: $\tau_f = \sigma'_f \tan \phi' = 135.0 \tan(30^\circ) = 135.0 \times 0.57735 = 77.94\text{ kPa} \quad (\text{Matches perfectly!})$

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