8.1 Lateral Earth Pressure Theories

Introduction to Lateral Earth Pressure

Lateral earth pressure is the horizontal stress exerted by soil onto a retaining structure. In geotechnical engineering design, determining the magnitude, direction, and point of application of lateral earth thrust is essential for evaluating the stability of retaining walls, sheet pile bulkheads, braced cuts, and basement walls. The magnitude of lateral stress depend on the intrinsic strength properties of the soil (friction angle $\phi'$, cohesion $c'$), groundwater conditions, surface surcharges, wall geometry, and—most critically—the magnitude and direction of lateral wall displacement.

The relationship between horizontal effective stress ($\sigma_h'$) and vertical effective stress ($\sigma_v'$) is defined by the lateral earth pressure coefficient ($K$):

$$K = \frac{\sigma_h'}{\sigma_v'}$$

Three distinct lateral earth pressure states exist depending on structural boundary movement:

1. At-Rest State ($K_0$): The wall experiences zero horizontal displacement ($\Delta x = 0$). The soil remains in an elastic, non-yielded state.
2. Active State ($K_a$): The retaining wall moves outward away from the backfill soil mass. The backfill expands horizontally, reducing horizontal stress until shear failure occurs along internal slip planes. This represents the minimum lateral pressure exerted on a wall.
3. Passive State ($K_p$): The retaining wall is forced inward toward the soil mass, compressing the soil horizontally until passive plastic failure is induced throughout the soil block. This represents the maximum lateral resistance the soil mass can mobilize.

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Wall Displacement Requirements

To transition from the at-rest condition to fully mobilized limit equilibrium states, the retaining structure must undergo finite lateral deformation. The displacement required to reach the active or passive state is typically expressed as a ratio of wall top displacement ($\Delta$) to total wall height ($H$).

Key Principle: Mobilizing passive resistance requires substantially larger wall displacements (roughly 5 to 10 times greater) than mobilizing active pressure. In structural design, relying on full passive resistance is valid only if the structural movement necessary to mobilize it can be tolerated.

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At-Rest Earth Pressure ($K_0$)

For rigid basement walls, bridge abutments restrained at the deck level, or rock-socketed retaining structures where wall yield is prevented, lateral stresses must be calculated using the at-rest coefficient $K_0$.

From isotropic linear elasticity theory, assuming zero lateral strain ($\epsilon_h = 0$):

$$K_0 = \frac{\nu}{1 - \nu}$$

where $\nu$ is Poisson's ratio of the soil.

For coarse-grained, normally consolidated soils, Jaky (1944) proposed the widely accepted empirical relationship:

$$K_0 = 1 - \sin\phi'$$

For overconsolidated soils, mechanical preloading locks in horizontal effective stress. Mayne and Kulhawy (1982) modified Jaky's equation to account for Overconsolidation Ratio ($\text{OCR}$):

$$K_0 = (1 - \sin\phi') \cdot \text{OCR}^{\sin\phi'}$$

The total at-rest force per unit width ($P_0$) acting on a wall of height $H$ with dry soil of unit weight $\gamma$ is:

$$P_0 = \frac{1}{2} K_0 \gamma H^2$$

The point of application of $P_0$ acts at a height of $\bar{y} = H/3$ above the base of the wall.

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Rankine's Earth Pressure Theory (1857)

Rankine's theory evaluates stress states in a soil mass undergoing plastic equilibrium. It assumes:

1. The soil backfill is homogeneous, isotropic, and cohesionless or cohesive-frictional.
2. The back-face of the retaining wall is vertically smooth (zero wall friction, $\delta = 0$).
3. The interface between the wall back and backfill is vertical ($\theta = 90^\circ$).
4. The shear failure plane is planar.

1. Rankine Active State

As a smooth vertical wall yields outward, horizontal principal stress $\sigma_h'$ decreases while vertical stress $\sigma_v'$ remains constant ($\sigma_v' = \sigma_1'$, $\sigma_h' = \sigma_3'$). Failure occurs when the Mohr stress circle expands to touch the Mohr-Coulomb failure envelope.

The Rankine active lateral pressure coefficient ($K_a$) for horizontal ground is:

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

The failure plane in the soil mass makes an angle of $\alpha_a$ with the horizontal:

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

For cohesive-frictional soils ($c' > 0, \phi' > 0$), effective active lateral stress at depth $z$ is:

$$\sigma_a' = \sigma_v' K_a - 2 c' \sqrt{K_a} = \gamma z K_a - 2 c' \sqrt{K_a}$$

At shallow depths, $\sigma_a'$ is negative, indicating tensile stress. Soil cannot sustain tension; consequently, tension cracks develop down to a critical depth ($z_c$):

$$z_c = \frac{2 c'}{\gamma \sqrt{K_a}}$$

The depth of an unsupported vertical cut ($H_c$) in cohesive soil occurs where total active thrust equals zero:

$$H_c = 2 z_c = \frac{4 c'}{\gamma \sqrt{K_a}}$$

Caution for PE Exam: When calculating total active thrust ($P_a$) on a wall with cohesive backfill, if tension cracks exist, tensile stresses are neglected over depth $z_c$. If water fills the tension crack, a hydrostatic force $P_w = \frac{1}{2} \gamma_w z_c^2$ must be added horizontally at the upper boundary.

2. Rankine Passive State

When a smooth wall is pushed into the soil mass, horizontal stress increases until it becomes the major principal stress ($\sigma_h' = \sigma_1'$, $\sigma_v' = \sigma_3'$). The Rankine passive coefficient ($K_p$) is:

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

The passive shear failure planes are inclined at an angle $\alpha_p$ relative to the horizontal:

$$\alpha_p = 45^\circ - \frac{\phi'}{2}$$

Passive lateral stress including cohesion is:

$$\sigma_p' = \sigma_v' K_p + 2 c' \sqrt{K_p} = \gamma z K_p + 2 c' \sqrt{K_p}$$

3. Rankine Theory with Sloping Backfill ($\beta$)

If the backfill is inclined at an angle $\beta$ with the horizontal, Rankine active pressure acts parallel to the inclined ground surface at a height of $H/3$ above the base. The active coefficient $K_a(\beta)$ is:

$$K_a(\beta) = \cos\beta \frac{\cos\beta - \sqrt{\cos^2\beta - \cos^2\phi'}}{\cos\beta + \sqrt{\cos^2\beta - \cos^2\phi'}}$$

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Coulomb's Earth Pressure Theory (1776)

Coulomb's limit equilibrium method considers a rigid soil wedge bounded by the wall back face and an inclined planar failure surface inside the soil mass. Coulomb's theory explicitly includes:

• Wall friction angle ($\delta$), typically assumed to be $\frac{1}{2}\phi'$ to $\frac{2}{3}\phi'$.
• Wall batter/inclination angle ($\theta$), measured from the horizontal.
• Sloping backfill angle ($\beta$).

Coulomb Active Pressure Coefficient ($K_a$)

By considering force equilibrium of the sliding wedge under wedge self-weight ($W$), wall reaction ($P_a$), and failure plane resultant ($R$), the active coefficient is expressed as:

$$K_a = \frac{\sin^2(\theta + \phi')}{\sin^2\theta \sin(\theta - \delta) \left[ 1 + \sqrt{\frac{\sin(\phi' + \delta) \sin(\phi' - \beta)}{\sin(\theta - \delta) \sin(\theta + \beta)}} \right]^2}$$

The active resultant thrust $P_a$ acts at an angle $\delta$ relative to the normal to the back face of the wall.

Coulomb Passive Pressure Coefficient ($K_p$)

Similarly, the Coulomb passive pressure coefficient is:

$$K_p = \frac{\sin^2(\theta - \phi')}{\sin^2\theta \sin(\theta + \delta) \left[ 1 - \sqrt{\frac{\sin(\phi' + \delta) \sin(\phi' + \beta)}{\sin(\theta + \delta) \sin(\theta + \beta)}} \right]^2}$$

Important Engineering Insight: Coulomb's passive theory overestimates passive resistance when $\delta > \phi'/3$ because real failure surfaces are curved (log-spiral) rather than strictly planar. For wall friction $\delta > \phi'/3$, engineers must use log-spiral earth pressure charts (e.g., Caquot-Kerisel or NAVFAC DM-7) to avoid unsafe passive estimates.

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Water Table and Surcharge Effects

1. Influence of Groundwater

When a water table is present behind a wall at depth $z_w$ below the top of the wall:

1. Above the water table ($z \le z_w$), vertical stress is calculated using dry or moist unit weight: $\sigma_v' = \gamma z$.
2. Below the water table ($z > z_w$), vertical effective stress is computed using submerged unit weight ($\gamma' = \gamma_{sat} - \gamma_w$): $\sigma_v' = \gamma z_w + \gamma' (z - z_w)$.
3. Pore water pressure ($u = \gamma_w (z - z_w)$) acts equally in all directions and is added directly to effective horizontal stress: $\sigma_h = K \sigma_v' + u$.

2. Uniform Surcharge Loads

A uniform surcharge load ($q$) applied at the ground surface increases vertical stress uniformly throughout the soil column by $\Delta\sigma_v = q$. This creates a constant lateral pressure increment:

$$\Delta\sigma_h = K q$$

The resultant surcharge force is $P_q = K q H$, acting at mid-height ($\bar{y} = H/2$).

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Worked Numerical Example: Multi-Layer Pressure Calculation

Problem Statement

A $6.0\text{ m}$ high vertical retaining wall supports a backfill consisting of two horizontal soil layers with a uniform surface surcharge of $q = 15.0\text{ kPa}$. The soil properties are:

• Layer 1 ($0\text{ to } 3.0\text{ m}$): Moist unit weight $\gamma_1 = 18.0\text{ kN/m}^3$, $\phi_1' = 30^\circ$, $c_1' = 0$.
• Layer 2 ($3.0\text{ to } 6.0\text{ m}$): Saturated unit weight $\gamma_{sat2} = 20.0\text{ kN/m}^3$, $\phi_2' = 34^\circ$, $c_2' = 0$.
• Groundwater table is located at $z = 3.0\text{ m}$ (top of Layer 2). Unit weight of water $\gamma_w = 9.81\text{ kN/m}^3$.
• Use Rankine's earth pressure theory.

Calculate the total active thrust ($P_a$) per meter length of wall and its point of application ($\bar{y}$) above the base.

Step-by-Step Solution

1. Calculate Active Earth Pressure Coefficients

• Layer 1: $K_{a1} = \tan^2(45^\circ - 30^\circ/2) = \tan^2(30^\circ) = \frac{1}{3} \approx 0.3333$.
• Layer 2: $K_{a2} = \tan^2(45^\circ - 34^\circ/2) = \tan^2(28^\circ) = 0.2827$.

2. Compute Effective Vertical and Horizontal Stresses at Key Depths

At $z = 0\text{ m}$ (Top of Layer 1):

• $\sigma_v' = 15.0\text{ kPa}$ (Surcharge)
• $\sigma_{h1}' = K_{a1} \cdot \sigma_v' = 0.3333 \times 15.0 = 5.00\text{ kPa}$
• $u = 0 \implies \sigma_{h1} = 5.00\text{ kPa}$

At $z = 3.0\text{ m}$ (Bottom of Layer 1):

• $\sigma_v' = 15.0 + (18.0 \times 3.0) = 69.0\text{ kPa}$
• $\sigma_{h1}' = 0.3333 \times 69.0 = 23.00\text{ kPa}$
• $u = 0 \implies \sigma_{h1} = 23.00\text{ kPa}$

At $z = 3.0\text{ m}$ (Top of Layer 2):

• Vertical effective stress is continuous: $\sigma_v' = 69.0\text{ kPa}$
• Switch to Layer 2 coefficient $K_{a2}$:
• $\sigma_{h2}' = K_{a2} \cdot \sigma_v' = 0.2827 \times 69.0 = 19.51\text{ kPa}$
• $u = 0 \implies \sigma_{h2} = 19.51\text{ kPa}$

At $z = 6.0\text{ m}$ (Base of Wall):

• $\gamma' = \gamma_{sat2} - \gamma_w = 20.0 - 9.81 = 10.19\text{ kN/m}^3$
• $\sigma_v' = 69.0 + (10.19 \times 3.0) = 69.0 + 30.57 = 99.57\text{ kPa}$
• $\sigma_{h2}' = 0.2827 \times 99.57 = 28.15\text{ kPa}$
• $u = \gamma_w \times 3.0 = 9.81 \times 3.0 = 29.43\text{ kPa}$
• Total horizontal stress: $\sigma_{h2} = 28.15 + 29.43 = 57.58\text{ kPa}$

3. Break Soil Pressure Diagram into Geometric Components

• Component 1 (Layer 1 Surcharge Rectangle, $z=0$ to $3\text{ m}$):

  • $F_1 = 5.00 \times 3.0 = 15.00\text{ kN/m}$
  • Moment arm about base: $y_1 = 3.0 + 1.5 = 4.50\text{ m}$

• Component 2 (Layer 1 Soil Triangle, $z=0$ to $3\text{ m}$):

  • $F_2 = \frac{1}{2} \times (23.00 - 5.00) \times 3.0 = 27.00\text{ kN/m}$
  • Moment arm about base: $y_2 = 3.0 + \frac{3.0}{3} = 4.00\text{ m}$

• Component 3 (Layer 2 Effective Stress Rectangle, $z=3$ to $6\text{ m}$):

  • $F_3 = 19.51 \times 3.0 = 58.53\text{ kN/m}$
  • Moment arm about base: $y_3 = 1.50\text{ m}$

• Component 4 (Layer 2 Soil Effective Stress Triangle, $z=3$ to $6\text{ m}$):

  • $F_4 = \frac{1}{2} \times (28.15 - 19.51) \times 3.0 = 12.96\text{ kN/m}$
  • Moment arm about base: $y_4 = \frac{3.0}{3} = 1.00\text{ m}$

• Component 5 (Hydrostatic Water Triangle, $z=3$ to $6\text{ m}$):

  • $F_5 = \frac{1}{2} \times 29.43 \times 3.0 = 44.15\text{ kN/m}$
  • Moment arm about base: $y_5 = \frac{3.0}{3} = 1.00\text{ m}$

4. Total Active Thrust ($P_a$) and Resultant Point of Application ($\bar{y}$)

$$\sum P_a = F_1 + F_2 + F_3 + F_4 + F_5 = 15.00 + 27.00 + 58.53 + 12.96 + 44.15 = 157.64\text{ kN/m}$$ $$\sum M_{base} = (15.00 \times 4.5) + (27.00 \times 4.0) + (58.53 \times 1.5) + (12.96 \times 1.0) + (44.15 \times 1.0)$$ $$\sum M_{base} = 67.50 + 108.00 + 87.80 + 12.96 + 44.15 = 320.41\text{ kN}\cdot\text{m/m}$$ $$\bar{y} = \frac{\sum M_{base}}{P_a} = \frac{320.41}{157.64} = 2.03\text{ m above the base}$$