10.3 Lateral Load Behavior of Deep Foundations

10.3 Lateral Load Behavior of Deep Foundations

Deep foundations supporting bridge piers, high-rise buildings, offshore platforms, and retaining walls are frequently subjected to substantial horizontal loads (H) and overturning moments (M). The design of laterally loaded piles requires evaluating both the geotechnical limit state (soil ultimate lateral bearing failure) and the structural limit state (excessive pile deflection or flexural plastic hinge formation).

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Pile Stiffness & Relative Embedment Regimes

The fundamental response of a laterally loaded pile depends on whether it behaves as a short (rigid) pile or a long (flexible) pile.

• Short (Rigid) Pile: The pile flexural rigidity (EI) is high relative to soil stiffness. Under lateral load, the pile rotates or translates as a rigid body without significant bending, causing soil capacity breakdown along its entire embedment depth.
• Long (Flexible) Pile: The pile flexures significantly under load. The lower portion of the pile remains virtually unaffected by lateral forces, and structural failure occurs when the internal bending moment reaches the yield moment capacity $M_y$ of the pile cross-section, forming a plastic hinge.

               RIGID VS FLEXIBLE LATERALLY LOADED PILE
   FREE-HEAD SHORT (RIGID) PILE          FREE-HEAD LONG (FLEXIBLE) PILE
       H                                     H
   ────┬────>                                ────┬────>
       │                                         │
       │   Deflection y                          │   Deflection y
       │  \                                     │  \n       │   \                                    │   \  Bending
       │    \  Rigid Rotation                   │    │  Curve
       │     \                                  │   ╱
       │      \                                 │  │   Plastic Hinge (My)
       │       \                                │ ╱
       │        \ Rotation Center               ││  Zero Deflection
       ▼         \                              ▼└────── at depth > 4T

Stiffness Length Parameters

1. Cohesionless Soil / Granular Strata (Linearly Increasing Subgrade Modulus $k_h = n_h \cdot z$): The relative stiffness factor (T) is:

   $$T = \sqrt[5]{\frac{E I}{n_h}}$$

   Where (E I) is pile flexural rigidity (kN·m² or lb·in²), and $n_h$ is the modulus of subgrade reaction rate of increase (kN/m³ or lb/in³).

   • Rigid Pile Criteria: Embedded length $L \le 2 T$
   • Flexible (Long) Pile Criteria: Embedded length $L \ge 4 T$

2. Cohesive Soil / Overconsolidated Clay (Constant Subgrade Modulus $K = k_h \cdot B$): The elastic length parameter $L_e$ is:

   $$L_e = \sqrt[4]{\frac{E I}{K}} = \sqrt[4]{\frac{E I}{k_h B}}$$
   • Rigid Pile Criteria: Embedded length $L \le 2.0 L_e$
   • Flexible (Long) Pile Criteria: Embedded length $L \ge 3.5 L_e$

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Broms' Ultimate Lateral Load Method

Broms (1964) developed widely used analytical solutions to determine ultimate lateral load capacity $H_u$. The method relies on simplified earth pressure profiles at ultimate soil failure.

1. Broms' Cohesionless Soil (Sand) Assumptions

• Ultimate soil resistance per unit length $p_u$ increases linearly with depth: $$p_u = 3 \cdot \gamma' \cdot z \cdot B \cdot K_p$$ Where $K_p = \tan^2(45^\circ + \phi'/2) = \frac{1 + \sin\phi'}{1 - \sin\phi'}$.

Ultimate Capacity Formulas for Long Flexible Pile in Sand:

• Free-Head Condition (pile top free to rotate, load applied at height (e) above ground): $$M_{max} = H_u \left( e + 0.54 \sqrt{\frac{H_u}{\gamma' B K_p}} \right) = M_y$$ Solving for $H_u$ yields the ultimate lateral load required to induce structural yield moment $M_y$.
• Fixed-Head Condition (pile top restrained against rotation by rigid pile cap): $$M_{max} = 0.50 H_u \left( e + 0.54 \sqrt{\frac{H_u}{\gamma' B K_p}} \right) = M_y$$ Key Insight: A fixed-head pile in sand achieves approximately 2.0 times higher ultimate lateral capacity than an equivalent free-head pile because fixing the pile head distributes bending moments between the pile head and the underground plastic hinge.

2. Broms' Cohesive Soil (Clay) Assumptions

• Soil pressure $p_u = 0$ from ground surface (z = 0) to depth (z = 1.5 B) (due to surface soil breakaway/tension cracks).
• Soil pressure $p_u = 9 c_u B$ constant at depths below (z = 1.5 B).

Ultimate Capacity Formulas for Long Flexible Pile in Clay:

• Free-Head Condition: $$M_{max} = H_u \left( e + 1.5 B + 0.5 f \right) = M_y \quad \text{where } f = \frac{H_u}{9 c_u B}$$
• Fixed-Head Condition: $$M_{max} = H_u \left( 1.5 B + 0.5 f \right) = 2 M_y$$

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Non-Linear p-y Curve Method

While Broms' method checks ultimate load capacity, serviceability design (predicting working deflection (y(z)) and bending moment (M(z))) requires non-linear soil-structure modeling.

The laterally loaded pile is modeled as a beam on a non-linear Winkler foundation governed by the 4th-order differential equation:

$$E I \frac{d^4 y}{dz^4} + P_x \frac{d^2 y}{dz^2} + p(z, y) = 0$$

Where:

• (y) = lateral pile deflection at depth (z)
• $P_x$ = axial structural load on the pile (incorporating P-Delta destabilizing effects)
• (p(z, y)) = non-linear soil reaction force per unit length at deflection (y)

                   NON-LINEAR p-y CURVE CONCEPT
  Pile Deflection y(z)                      Soil Reaction Curve p(y)
    │    ┌───┐                                Soil Reaction p
    │    │ █ │                                  ▲      p_ult (Ultimate Resistance)
    ├────┼───┼────> Load H                      │     ┌─────────────────
    │    │ █ │                                  │   ╱
    │   ╱  █  \                                 │  ╱   Initial Modulus Es = Epy
    │  │   █   │                                │ ╱
    │ ╱    █    \                               │╱
    ││     █     │                              └──────────────────────> Deflection y
    ▼      █      ▼                             (p-y non-linear spring at depth z)

Matlock (1970) Soft Clay p-y Curve Formulation

For soft saturated clays under short-term static loading, the non-linear p-y relationship is:

$$\frac{p}{p_u} = 0.5 \left( \frac{y}{y_{50}} \right)^{1/3} \quad \text{for } y \le 8 y_{50}$$

Where:

• $y_{50} = 2.5 \cdot \epsilon_{50} \cdot B$ (deflection at 50% of ultimate soil reaction)
• $\epsilon_{50}$ = strain corresponding to 50% maximum stress in an undrained triaxial test (typically 0.005 for stiff clay to 0.020 for soft clay)
• $p_u = 3 c_u B + \gamma' z B + J c_u z$ (increasing from surface to $9 c_u B$ at depth $z_r$)

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PE-Style Worked Example: Broms' Method in Sand

Problem: A steel pipe pile with outer diameter $D = 610\text{ mm}$ (0.61 m), wall thickness $t = 12.7\text{ mm}$, yield moment capacity $M_y = 680\text{ kN}\cdot\text{m}$, and flexural rigidity $EI = 1.15 \times 10^5\text{ kN}\cdot\text{m}^2$ is driven to a depth $L = 16.0\text{ m}$ into loose-to-medium sand. Soil properties:

• Effective unit weight $\gamma' = 10.0\text{ kN/m}^3$
• Friction angle $\phi' = 34^\circ$
• Modulus of subgrade reaction rate $n_h = 12,000\text{ kN/m}^3$
• The pile has a free-head boundary condition, and lateral load (H) acts at height $e = 1.2\text{ m}$ above the ground surface.

Calculate:

1. Stiffness factor (T) and verify if the pile is flexible (long).
2. Ultimate lateral load $H_u$ using Broms' method.
3. Allowable working lateral load $H_{all}$ using a factor of safety (FS = 2.5).

Solution Step-by-Step:

1. Verify Long (Flexible) Pile Behavior:

$$T = \sqrt[5]{\frac{EI}{n_h}} = \sqrt[5]{\frac{1.15 \times 10^5\text{ kN}\cdot\text{m}^2}{12,000\text{ kN/m}^3}} = \sqrt[5]{9.5833} = 1.571\text{ m}$$

• Critical embedded length for flexible pile: $L_{crit} = 4 T = 4 \cdot (1.571\text{ m}) = 6.28\text{ m}$.
• Since embedded pile length $L = 16.0\text{ m} > 6.28\text{ m}$, the pile behaves as a long flexible pile.

2. Broms' Passive Pressure Coefficient $K_p$:

$$K_p = \tan^2(45^\circ + 34^\circ/2) = \tan^2(62^\circ) = (1.8807)^2 = 3.537$$

• Earth pressure factor denominator: $$\gamma' \cdot B \cdot K_p = 10.0\text{ kN/m}^3 \cdot 0.61\text{ m} \cdot 3.537 = 21.576\text{ kN/m}^2$$

3. Broms' Moment Equation for Free-Head Long Pile in Sand:

$$M_{max} = H_u \left( e + 0.54 \sqrt{\frac{H_u}{\gamma' B K_p}} \right) = M_y$$ $$680 = H_u \left( 1.2 + 0.54 \sqrt{\frac{H_u}{21.576}} \right)$$ $$680 = 1.2 H_u + 0.54 \cdot \frac{H_u^{1.5}}{\sqrt{21.576}} = 1.2 H_u + 0.11626 \cdot H_u^{1.5}$$

Let us solve this non-linear equation for $H_u$ via iteration:

• Try $H_u = 180\text{ kN}$: $1.2(180) + 0.11626(180)^{1.5} = 216 + 0.11626(2414.9) = 216 + 280.76 = 496.76\text{ kN}\cdot\text{m} < 680$
• Try $H_u = 220\text{ kN}$: $1.2(220) + 0.11626(220)^{1.5} = 264 + 0.11626(3263.1) = 264 + 379.37 = 643.37\text{ kN}\cdot\text{m} < 680$
• Try $H_u = 229\text{ kN}$: $1.2(229) + 0.11626(229)^{1.5} = 274.8 + 0.11626(3465.7) = 274.8 + 402.92 = 677.72\text{ kN}\cdot\text{m}$
• Try $H_u = 229.6\text{ kN}$: $1.2(229.6) + 0.11626(229.6)^{1.5} = 275.52 + 0.11626(3479.3) = 275.52 + 404.50 = 680.02\text{ kN}\cdot\text{m}$ (Matches $M_y = 680$)

Thus, ultimate lateral load $H_u = 229.6\text{ kN}$.

4. Depth to Maximum Bending Moment $z_m$:

$$z_m = 0.54 \sqrt{\frac{H_u}{\gamma' B K_p}} = 0.54 \sqrt{\frac{229.6}{21.576}} = 0.54 \cdot 3.262 = 1.76\text{ m below ground}$$

5. Allowable Lateral Design Load:

$$H_{all} = \frac{H_u}{FS} = \frac{229.6\text{ kN}}{2.5} = 91.84\text{ kN} \approx 91.8\text{ kN}$$