2.3 Shear Strength Principles & Laboratory Testing

2.3 Shear Strength Principles & Laboratory Testing

1. Mohr-Coulomb Failure Criterion

The shear strength of soil ($\tau_f$) is the maximum resistance per unit area that a soil mass can offer against sliding along an internal failure surface under applied stress.

Effective Stress Form of Mohr-Coulomb Criterion:

$$\tau_f = c' + \sigma' \tan\phi' = c' + (\sigma - u) \tan\phi'$$

Where:

• $c'$: Effective cohesion intercept (interparticle chemical bonding or cementation).
• $\phi'$: Effective angle of internal friction (interlocking and mineral friction).
• $\sigma'$: Effective normal stress acting on the failure plane.

Principal Stresses at Failure

In Mohr's circle of stress, failure occurs when the circle becomes tangent to the Mohr-Coulomb envelope. The major ($\sigma'_1$) and minor ($\sigma'_3$) effective principal stresses at failure satisfy:

$$\sigma'_1 = \sigma'_3 \tan^2\left(45^\circ + \frac{\phi'}{2}\right) + 2 c' \tan\left(45^\circ + \frac{\phi'}{2}\right) = \sigma'_3 N_\phi + 2 c' \sqrt{N_\phi}$$

Where $N_\phi = \tan^2\left(45^\circ + \frac{\phi'}{2}\right) = \frac{1 + \sin\phi'}{1 - \sin\phi'}$.

Failure Plane Orientation

The failure plane is inclined at an angle $\theta$ relative to the major principal stress plane (the horizontal plane when vertical stress is major):

$$\theta = 45^\circ + \frac{\phi'}{2}$$

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2. Laboratory Shear Strength Testing Methods

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3. Detailed Triaxial Testing Framework

A cylindrical soil specimen ($H:D = 2:1$) is encased in a rubber membrane, placed in a pressurized chamber ($\sigma_3$), and loaded axially by a piston ($\Delta\sigma = \sigma_1 - \sigma_3$, termed the deviator stress).

Unconsolidated-Undrained (UU / Total Stress) Test

• Cell pressure $\sigma_3$ applied with drainage closed.
• Axial deviator stress $\Delta\sigma$ increased to failure with drainage closed.
• For fully saturated soils ($S = 100\%$), increasing $\sigma_3$ causes an equal pore pressure increase ($\Delta u = \Delta\sigma_3$). Consequently, effective principal stress remains unchanged.
• Mohr circles of total stress at failure for various $\sigma_3$ have identical diameters ($\Delta\sigma_f$).
• Total stress envelope is horizontal: $$\phi_u = 0 \quad \text{and} \quad \tau_f = c_u = S_u = \frac{\sigma_{1,f} - \sigma_{3,f}}{2}$$

Consolidated-Undrained (CU) Test with Pore Pressure Measurement

• Specimen consolidated under cell pressure $\sigma_3$ with drainage open ($u \to 0$).
• Drainage valves closed, axial load applied to failure while measuring excess pore pressure $u_f$.
• Effective principal stresses at failure: $$\sigma'_3 = \sigma_3 - u_f \quad \text{and} \quad \sigma'_1 = \sigma_1 - u_f = \sigma_3 + \Delta\sigma_f - u_f$$
• Allows determination of both total parameters ($c, \phi$) and effective parameters ($c', \phi'$).

Skempton's Pore Pressure Parameters:

$$\Delta u = B \left[ \Delta\sigma_3 + A (\Delta\sigma_1 - \Delta\sigma_3) \right]$$

• For saturated soils, $B = 1.0$.
• Parameter $A$ at failure ($A_f$): $A_f < 0$ (heavily overconsolidated/dilative), $A_f = 0.5 - 1.0$ (normally consolidated clay), $A_f > 1.0$ (sensitive/collapsible clay).

Consolidated-Drained (CD) Test

• Specimen fully consolidated under $\sigma_3$.
• Axial shearing performed at an extremely slow strain rate to ensure $u \approx 0$ throughout.
• Total stresses equal effective stresses ($\sigma_1 = \sigma'_1, \sigma_3 = \sigma'_3$).
• For normally consolidated clays and clean sands, effective cohesion is zero ($c' = 0$).

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4. Drained vs. Undrained Soil Response & Critical State Concept

Granular Soil Behavior (Sands)

• Loose Sand: Contractive during shear (volume decreases), generating positive pore pressure in undrained loading. Reaches peak strength smoothly.
• Dense Sand: Dilative during shear (volume expands due to particle interlocking), generating negative pore pressure in undrained loading. Exhibits sharp peak strength followed by strain softening to critical state.
• Critical State: At large shear strains, all specimens reach the critical state void ratio ($e_{crit}$) where continuous shearing occurs at constant volume and constant shear stress without further volume change.

Fine-Grained Soil Behavior (Clays)

• Normally Consolidated Clay ($OCR = 1$): Contractive behavior, positive $A_f$, $c' \approx 0$.
• Overconsolidated Clay ($OCR > 4-8$): Dilative behavior, negative $A_f$, exhibits true/apparent effective cohesion $c' > 0$.

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5. Stress Paths in $p'-q$ Space

To track continuous stress states during testing without drawing multiple Mohr circles, stress paths plot invariant stress parameters:

$$p' = \frac{\sigma'_1 + \sigma'_3}{2} \quad \text{and} \quad q = \frac{\sigma'_1 - \sigma'_3}{2} = \frac{\Delta\sigma}{2}$$

The failure line in $p'-q$ space is defined by:

$$q = a' + p' \tan\alpha'$$

Converting $a'$ and $\alpha'$ to Mohr-Coulomb parameters $c'$ and $\phi'$:

$$\sin\phi' = \tan\alpha' \quad \Longleftrightarrow \quad \phi' = \arcsin(\tan\alpha')$$ $$c' = \frac{a'}{\cos\phi'}$$

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6. Comprehensive Worked Example

Problem Statement: A Consolidated-Undrained (CU) triaxial compression test with pore pressure measurement was conducted on a saturated normally consolidated clay ($c' = 0$). The test results at failure are:

• Chamber cell pressure: $\sigma_3 = 200 \text{ kPa}$
• Deviator stress at failure: $\Delta\sigma_f = 160 \text{ kPa}$
• Measured pore pressure at failure: $u_f = 100 \text{ kPa}$

Calculate:

1. Total principal stresses ($\sigma_1, \sigma_3$) and total principal stress ratio at failure.
2. Effective principal stresses ($\sigma'_1, \sigma'_3$) at failure.
3. Effective angle of internal friction ($\phi'$).
4. Orientation of the failure plane ($\theta$) relative to the horizontal major principal plane.
5. Effective normal stress ($\sigma'$) and shear stress ($\tau_f$) on the failure plane.

Solution:

1. Total Principal Stresses:

   $$\sigma_3 = 200 \text{ kPa}$$ $$\sigma_1 = \sigma_3 + \Delta\sigma_f = 200 + 160 = 360 \text{ kPa}$$ $$\text{Total Ratio } \frac{\sigma_1}{\sigma_3} = \frac{360}{200} = 1.80$$

2. Effective Principal Stresses:

   $$\sigma'_3 = \sigma_3 - u_f = 200 - 100 = 100 \text{ kPa}$$ $$\sigma'_1 = \sigma_1 - u_f = 360 - 100 = 260 \text{ kPa}$$

3. Effective Angle of Internal Friction ($\phi'$): For normally consolidated clay ($c' = 0$), the principal stress relationship is:

   $$\sigma'_1 = \sigma'_3 \tan^2\left(45^\circ + \frac{\phi'}{2}\right) = \sigma'_3 \left(\frac{1 + \sin\phi'}{1 - \sin\phi'}\right)$$ $$\frac{\sigma'_1}{\sigma'_3} = \frac{260}{100} = 2.60$$ $$\frac{1 + \sin\phi'}{1 - \sin\phi'} = 2.60 \implies 1 + \sin\phi' = 2.60 (1 - \sin\phi') = 2.60 - 2.60 \sin\phi'$$ $$3.60 \sin\phi' = 1.60 \implies \sin\phi' = \frac{1.60}{3.60} = 0.4444$$ $$\phi' = \arcsin(0.4444) = 26.38^\circ$$

4. Failure Plane Orientation ($\theta$):

   $$\theta = 45^\circ + \frac{\phi'}{2} = 45^\circ + \frac{26.38^\circ}{2} = 45^\circ + 13.19^\circ = 58.19^\circ$$

5. Stresses on Failure Plane:

   $$\sigma' = \frac{\sigma'_1 + \sigma'_3}{2} + \frac{\sigma'_1 - \sigma'_3}{2} \cos(2\theta) = \frac{260 + 100}{2} + \frac{260 - 100}{2} \cos(116.38^\circ)$$ $$\sigma' = 180 + 80 (-0.4444) = 180 - 35.56 = 144.44 \text{ kPa}$$ $$\tau_f = \sigma' \tan\phi' = 144.44 \times \tan(26.38^\circ) = 144.44 \times 0.4960 = 71.64 \text{ kPa}$$

   Alternatively, using $\tau_f = \frac{\sigma'_1 - \sigma'_3}{2} \sin(2\theta) = 80 \times \sin(116.38^\circ) = 80 \times 0.8958 = 71.66 \text{ kPa} \quad \checkmark$