10.3 Lateral Earth Pressures (Rankine & Coulomb) and Retaining Walls

10.3 Lateral Earth Pressures (Rankine & Coulomb) and Retaining Walls

Earth retaining structures—such as gravity walls, reinforced concrete cantilever walls, bridge abutments, and sheet pile bulkheads—must support lateral thrust from retained soil masses and surface surcharges. Board exam problems in this domain require analyzing earth pressure distributions and checking the overall external stability of the retaining structure against overturning, sliding, and foundation bearing failure.

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The Three Lateral Earth Pressure States

The magnitude of lateral earth pressure exerted by a soil mass against a retaining wall depends strictly on the magnitude and direction of lateral wall displacement ($\Delta x / H$):

   Lateral Earth Pressure (σ_h)
       ^
       |                                      Passive State (K_p)
       |                                    .--------
       |                                  .'
       |                                .'
       |            At-Rest (K_0)     .'
       |                  *---------'
       |                .'
       |              .'
       |  .----------' Active State (K_a)
       |  
       +------------------|-------------------|-------------------> Wall Movement
       <-- Away from Soil (Active)      Into Soil (Passive) -->

1. At-Rest Earth Pressure ($K_0$)

Occurs when the wall is completely rigid and unyielding (zero lateral strain, $\epsilon_x = 0$), such as massive bridge abutments keyed into solid bedrock or basement retaining walls restrained by floor diaphragms.

• For normally consolidated cohesionless soils, Jaky's empirical equation applies: $K_0 \approx 1 - \sin \phi'$
• For overconsolidated soils, the coefficient increases with overconsolidation ratio ($OCR$): $K_{0,OC} = K_{0,NC} \sqrt{OCR} = (1 - \sin \phi') \sqrt{OCR}$

2. Active Earth Pressure ($K_a$)

Occurs when the wall tilts or translates away from the backfill. As the wall moves outward, the soil expands laterally, mobilizing internal shear strength along an active failure wedge. Lateral stress decreases until reaching a lower-bound minimum at active shear failure.

• Displacement Required: Very small outward movement, typically $\Delta x / H \approx 0.001 \text{ to } 0.004$ (roughly $1\text{ to } 4\text{ mm}$ per meter of wall height for sand).

3. Passive Earth Pressure ($K_p$)

Occurs when the wall is forced into the backfill (such as anchor slabs or the soil wedge resisting the toe of a retaining wall). Lateral stress increases as the soil is compressed, reaching an upper-bound maximum at passive shear failure.

• Displacement Required: Significantly larger inward displacement, typically $\Delta x / H \approx 0.01 \text{ to } 0.05$ (roughly 10 times the movement required for the active state).

$\text{Magnitude Relationship: } K_a < K_0 < K_p$

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Rankine's Lateral Earth Pressure Theory

Formulated by William John Macquorn Rankine in 1857, this theory assumes:

1. The backfill soil is homogeneous and isotropic.
2. The back of the retaining wall is perfectly vertical ($\beta = 90^\circ$) and frictionless (wall friction angle $\delta = 0$).
3. The backfill surface is planar (horizontal or uniformly inclined at angle $\alpha$).
4. The failure surface in the soil is a planar slip boundary.

Cohesionless Backfill ($c = 0$, Horizontal Surface)

For a dry, homogeneous sand of unit weight $\gamma$ and internal friction angle $\phi$:

• Active Pressure Coefficient ($K_a$): $K_a = \tan^2\left(45^\circ - \frac{\phi}{2}\right) = \frac{1 - \sin \phi}{1 + \sin \phi}$ Active lateral stress at depth $z$: $\sigma_a = \gamma z K_a$ Total active resultant thrust per unit length of wall: $P_a = \frac{1}{2} \gamma H^2 K_a \quad (\text{acting at } y = H/3 \text{ from the base})$

• Passive Pressure Coefficient ($K_p$): $K_p = \tan^2\left(45^\circ + \frac{\phi}{2}\right) = \frac{1 + \sin \phi}{1 - \sin \phi} = \frac{1}{K_a}$ Total passive resultant resistance: $P_p = \frac{1}{2} \gamma H^2 K_p \quad (\text{acting at } y = H/3 \text{ from the base})$

Surcharge and Groundwater Stratification

When external conditions act on the backfill, their lateral pressure contributions superimpose:

1. Uniform Surcharge ($q$): A surface surcharge load $q$ (in $\text{kPa}$) exerts a constant uniform lateral pressure along the entire wall height: $\sigma_{a,q} = q K_a \implies P_q = q K_a H \quad (\text{acting at } y = H/2)$
2. Water Table Effects: Water is an isotropic fluid with zero shear resistance ($K_w = 1.0$). If the groundwater table is located at depth $h_1$ below the crest:
   • Above the water table ($z \le h_1$): $\sigma'_v = \gamma z$, and lateral stress is $\sigma'_a = \gamma z K_a$.
   • Below the water table ($z > h_1$): effective vertical stress is $\sigma'_v = \gamma h_1 + \gamma' (z - h_1)$, where buoyant unit weight is $\gamma' = \gamma_{\text{sat}} - \gamma_w$. The lateral effective stress is $\sigma'_a = \sigma'_v K_a$.
   • Hydrostatic Water Thrust: The water pressure acts independently and must be added directly: $u = \gamma_w (z - h_1) \implies P_w = \frac{1}{2} \gamma_w (H - h_1)^2 \quad (\text{acting at } (H - h_1)/3)$

Cohesive Backfill ($c - \phi$ Soils) & Tension Cracks

For a cohesive soil possessing both cohesion $c$ and friction angle $\phi$, Rankine active pressure is: $\sigma_a = \gamma z K_a - 2 c \sqrt{K_a}$

At the ground surface ($z = 0$), the lateral stress is tensile: $\sigma_a = -2 c \sqrt{K_a}$. Because soil possesses negligible tensile strength, tension cracks develop down to a critical depth $z_c$ where $\sigma_a = 0$: $\gamma z_c K_a - 2 c \sqrt{K_a} = 0 \implies z_c = \frac{2 c}{\gamma \sqrt{K_a}}$

   Depth (z)
       |
       0  -2c√Ka (Tension Zone)
       |  /|
       | / |
       |/  |  zc = 2c / (γ√Ka)
      zc---|----------------- Tension Crack Depth
       |\  |
       | \ |  Active Compression Zone
       |  \|
       H   +----------------- σ_a = γ H Ka - 2c√Ka

Total active thrust calculations depend on crack condition:

• Condition 1 (No Water in Crack, Tension Ignored): $P_a = \frac{1}{2} (\gamma H K_a - 2 c \sqrt{K_a})(H - z_c) = \frac{1}{2} \gamma K_a (H - z_c)^2$ Acting at $y = (H - z_c) / 3$ above the base.
• Condition 2 (Crack Completely Filled with Rainwater): Rainwater fills the open fissure, exerting full hydrostatic thrust $P_w = \frac{1}{2} \gamma_w z_c^2$ on the crack walls: $P_{\text{total}} = P_a + P_w = \frac{1}{2} \gamma K_a (H - z_c)^2 + \frac{1}{2} \gamma_w z_c^2$

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Coulomb's Earth Pressure Theory

Formulated by Charles-Augustin de Coulomb in 1776, this wedge equilibrium theory addresses practical wall geometries:

• Accounts for wall friction angle $\delta$ (typically $\delta \approx \frac{1}{2}\phi' \text{ to } \frac{2}{3}\phi'$ between concrete and soil).
• Accounts for sloping back of wall (batter $\beta$) and sloping backfill (inclination $\alpha$).
• The active thrust $P_a$ acts at an angle $\delta$ to the normal drawn to the back face of the wall.

$\text{Active Coefficient } K_a = \frac{\sin^2(\beta + \phi)}{\sin^2\beta \sin(\beta - \delta) \left[ 1 + \sqrt{\frac{\sin(\phi + \delta) \sin(\phi - \alpha)}{\sin(\beta - \delta) \sin(\beta + \alpha)}} \right]^2}$

When the back of the wall is vertical ($\beta = 90^\circ$), the backfill is horizontal ($\alpha = 0$), and wall friction is zero ($\delta = 0$), Coulomb's equation reduces exactly to Rankine's $K_a$.

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Retaining Wall External Stability Checks

A rigid retaining wall must satisfy three fundamental stability criteria under service loads:

                 |   | Stem
                 |   |
                 |   |        Active Thrust P_a
                 |   |       <-----------------
                 |   |
         +-------+---+-----------+
         |  Toe  |   |    Heel   | Base Slab (Width B)
         +-------+---+-----------+
         |<--x-->|     Resultant R
                 ^----/ 

1. Stability Against Overturning

Evaluated by summing moments about the toe of the wall base slab: $FS_{ot} = \frac{\sum M_R}{\sum M_O} \ge 2.0 \quad (\text{or } 1.5 \text{ under earthquake load})$ Where $\sum M_R$ is the sum of resisting moments (from concrete weight and soil weight above the heel) and $\sum M_O$ is the sum of overturning moments (from horizontal earth and surcharge thrusts).

2. Stability Against Sliding

Evaluated along the base-soil interface: $FS_{sl} = \frac{\sum F_R}{\sum F_d} = \frac{(\sum V) \tan \delta_b + c_a B + P_p}{P_h} \ge 1.5$ Where $\sum V$ is total vertical force, $\delta_b \approx \frac{2}{3}\phi'$ is base friction angle, $c_a \approx \frac{2}{3}c'$ is base adhesion, and $P_h$ is the horizontal driving thrust. Passive resistance $P_p$ in front of the toe is frequently discounted by $50\%$ or neglected entirely due to potential future utility trench excavation.

3. Base Bearing Pressure & Eccentricity Check

The resultant normal force $R = \sum V$ acts at location $\bar{x}$ from the toe: $\bar{x} = \frac{\sum M_R - \sum M_O}{\sum V}$ $\text{Eccentricity } e = \frac{B}{2} - \bar{x}$

To prevent tensile detachment at the heel of the foundation, the resultant must fall within the middle third of the base slab: $e \le \frac{B}{6}$

If $e \le B/6$, the contact pressure distribution is trapezoidal: $q = \frac{\sum V}{B} \left( 1 \pm \frac{6e}{B} \right)$ $q_{\max} = \frac{\sum V}{B} \left( 1 + \frac{6e}{B} \right) \le q_{\text{allowable}}$ $q_{\min} = \frac{\sum V}{B} \left( 1 - \frac{6e}{B} \right) \ge 0$

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Step-by-Step Worked Problem Examples

Worked Example: Cantilever Retaining Wall Stability Analysis

Problem: A reinforced concrete cantilever retaining wall supports a horizontal granular backfill. The total wall height is $H = 6.0\text{ m}$, base width is $B = 3.5\text{ m}$, base slab thickness is $0.6\text{ m}$, and vertical stem thickness is $0.4\text{ m}$. The stem is positioned such that the toe length is $0.8\text{ m}$ and the heel length is $2.3\text{ m}$ ($0.8 + 0.4 + 2.3 = 3.5\text{ m}$). Backfill properties: $\gamma = 18.0\text{ kN/m}^3$, $\phi = 30^\circ$, $c = 0$. A uniform surcharge $q = 20.0\text{ kPa}$ acts on the backfill surface. Unit weight of reinforced concrete is $\gamma_c = 24.0\text{ kN/m}^3$.

1. Calculate the total horizontal driving active thrust $P_h$ and overturning moment $M_O$ about the toe.
2. Compute the factor of safety against overturning ($FS_{ot}$).
3. Determine the base resultant eccentricity $e$ and maximum soil contact pressure $q_{\max}$.

Solution:

Step 1: Compute Earth Pressure and Overturning Thrust $K_a = \tan^2(45^\circ - 30^\circ/2) = \frac{1 - \sin 30^\circ}{1 + \sin 30^\circ} = \frac{0.50}{1.50} = \frac{1}{3} \approx 0.3333$ Driving forces:

• Soil active thrust: $P_{a1} = \frac{1}{2} \gamma H^2 K_a = \frac{1}{2} (18.0) (6.0)^2 (1/3) = 108.0\text{ kN/m}$, acting at $y_1 = 6.0/3 = 2.0\text{ m}$.
• Surcharge thrust: $P_{a2} = q K_a H = (20.0) (1/3) (6.0) = 40.0\text{ kN/m}$, acting at $y_2 = 6.0/2 = 3.0\text{ m}$. $P_h = 108.0 + 40.0 = 148.0\text{ kN/m}$ $M_O = (108.0 \times 2.0) + (40.0 \times 3.0) = 216.0 + 120.0 = 336.0\text{ kN}\cdot\text{m/m}$

Step 2: Compute Vertical Weights and Resisting Moments about Toe Stem height = $6.0 - 0.6 = 5.4\text{ m}$.

Step 3: Check Overturning Factor of Safety $FS_{ot} = \frac{\sum M_R}{M_O} = \frac{773.51}{336.0} = 2.30 \ge 2.0 \quad (\mathbf{SAFE})$

Step 4: Check Resultant Location and Base Contact Pressure $\bar{x} = \frac{\sum M_R - M_O}{\sum V} = \frac{773.51 - 336.0}{371.80} = \frac{437.51}{371.80} = 1.1767\text{ m from toe}$ $e = \frac{B}{2} - \bar{x} = \frac{3.5}{2} - 1.1767 = 1.75 - 1.1767 = 0.5733\text{ m}$ Check middle-third kern limit: $\frac{B}{6} = \frac{3.5}{6} = 0.5833\text{ m}$ Since $e = 0.5733\text{ m} \le 0.5833\text{ m}$, the resultant lies within the middle third (no tensile detachment at heel). $q_{\max} = \frac{\sum V}{B} \left( 1 + \frac{6e}{B} \right) = \frac{371.80}{3.5} \left( 1 + \frac{6(0.5733)}{3.5} \right) = 106.23 (1 + 0.9828) = 210.63\text{ kPa} \approx 210.6\text{ kPa}$ $q_{\min} = 106.23 (1 - 0.9828) = 1.83\text{ kPa} \ge 0$

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