10.1 Driven Pile Axial Capacity & Settlement

10.1 Driven Pile Axial Capacity & Settlement

Driven deep foundations are structural members (steel H-piles, pipe piles, precast prestressed concrete piles, or timber piles) forced into the ground using impact or vibratory hammers. Driven piles displace soil (full-displacement or partial-displacement) or cut through it (non-displacement open-ended pipes or H-piles), significantly altering the surrounding soil stress state.

Static Ultimate Axial Capacity Formulation

The ultimate compressive load capacity $Q_u$ of a single driven pile is computed as the sum of ultimate skin friction (shaft resistance) $Q_s$ and ultimate end bearing (tip resistance) $Q_b$, minus the effective pile weight $W_p$ (which is typically neglected in standard geotechnical calculations unless extraordinary pile self-weight exists):

$$Q_u = Q_s + Q_b - W_p = \sum_{i=1}^{n} f_{s,i} A_{s,i} + q_b A_b$$

Where:

• $f_{s,i}$ = unit skin friction (side resistance) along pile segment (i) (kPa or psf)
• $A_{s,i}$ = perimeter surface area of pile segment (i) = $P \cdot \Delta z_i$ (m² or ft²)
• (P) = pile perimeter (m or ft)
• $q_b$ = unit end bearing capacity at the pile tip (kPa or psf)
• $A_b$ = gross cross-sectional area of the pile tip (m² or ft²)

To determine the allowable design load $Q_{all}$ under Working Stress Design (WSD) or Allowable Stress Design (ASD):

$$Q_{all} = \frac{Q_u}{FS} \quad \text{or} \quad Q_{all} = \frac{Q_s}{FS_s} + \frac{Q_b}{FS_b}$$

Standard factors of safety range between (FS = 2.0) (when verified by field static load testing or dynamic wave equation analysis) and (FS = 3.0) (for empirical static design without dynamic load testing).

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Axial Capacity in Cohesionless Soils (Sands & Gravels)

In granular soils, skin friction and end bearing depend primarily on the effective vertical stress $\sigma'_v$, the internal friction angle $\phi'$, soil-pile interface friction angle $\delta'$, and lateral earth pressure coefficient (K).

1. Unit Side Resistance (Skin Friction)

Unit side resistance at depth (z) is calculated as:

$$f_s = K \cdot \sigma'_v \cdot \tan(\delta') \le f_{s,max}$$

• Earth Pressure Coefficient ((K)): For driven displacement piles (e.g., closed-end pipe, precast concrete), driving compacts the sand, elevating (K) above $K_o$ (typically $K = 1.0\text{ to } 2.0$). For non-displacement piles (e.g., H-piles), $K \approx K_o = 1 - \sin\phi'$.
• Interface Friction Angle ($\delta'$): Usually taken as $\delta' = 0.60\phi'\text{ to } 0.80\phi'$ depending on pile material (steel: $\delta' \approx 0.7\phi'$; smooth concrete: $\delta' \approx 0.8\phi'$; rough concrete: $\delta' \approx 1.0\phi'$).
• Limiting Skin Friction ($f_{s,max}$): Due to soil arching and particle crushing along the pile shaft, vertical effective stress reaches a pseudo-critical threshold at depths of (15D) to (20D). Design standards cap $f_s$ at maximum thresholds (typically 100 kPa to 150 kPa / 2,000 psf to 3,000 psf).

2. Unit End Bearing Capacity

Unit end bearing in sand is derived from bearing capacity theory:

$$q_b = \sigma'_v \cdot N_q \le q_l$$

• $\sigma'_v$ = effective vertical stress at the pile tip (kPa or psf)
• $N_q$ = bearing capacity factor for deep foundations (from Meyerhof, Berezantsev, or Nordlund charts, which are significantly higher than shallow foundation $N_q$ values due to deep embedment containment).
• $q_l$ = limiting unit end bearing capacity (Meyerhof recommendation: $q_l = 0.5 \cdot N_q \cdot \tan\phi'$ in MPa, or $q_l = 50 N_q \tan\phi'$ in kPa).

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Axial Capacity in Cohesive Soils (Clays & Silts)

Design in clay strata utilizes total stress (alpha method) or effective stress (beta and lambda methods).

1. Total Stress Analysis: Alpha ($\alpha$) Method

The $\alpha$-method relates unit side resistance directly to undrained shear strength $c_u$:

$$f_s = \alpha \cdot c_u$$

Where $\alpha$ is an empirical adhesion factor reflecting clay remolding, stress relief, and pile insertion disturbance. API RP 2A guidelines express $\alpha$ as a function of the normalized strength ratio $\psi = c_u / \sigma'_v$:

$$\alpha = \begin{cases} 1.0 & \text{for } \psi \le 0.5 \ 0.5 \cdot \psi^{-0.5} & \text{for } \psi > 0.5 \quad (\text{with } \alpha \le 1.0) \end{cases}$$

For soft to medium clays ($c_u < 50$ kPa), $\alpha \approx 0.8 - 1.0$. For stiff to hard overconsolidated clays ($c_u > 150$ kPa), $\alpha$ drops to (0.35 - 0.50).

Unit end bearing in clay is evaluated at deep failure conditions using Skempton's limit:

$$q_b = N_c \cdot c_u = 9 \cdot c_u$$

Where $N_c = 9.0$ for deep embedded circular or square foundations ($D/B \ge 5$).

2. Effective Stress Analysis: Beta ($\beta$) Method

Applicable for long-term drained conditions or fully reconsolidated clays:

$$f_s = \beta \cdot \sigma'_v = (K \cdot \tan\delta') \cdot \sigma'_v$$

For normally consolidated clays, $\beta = (1 - \sin\phi') \tan\phi' \approx 0.25 - 0.35$. For overconsolidated clays, $\beta = (1 - \sin\phi') \sqrt{OCR} \tan\phi'$.

3. Lambda ($\lambda$) Method

The $\lambda$-method incorporates mean effective vertical stress $\bar{\sigma}'_v$ and mean undrained shear strength $\bar{c}_u$ over the total pile penetration length (L):

$$f_{avg} = \lambda \cdot \left( \bar{\sigma}'_v + 2 \bar{c}_u \right)$$

Where $\lambda$ decreases nonlinearly from (0.32) at pile length $L = 10\text{ m}$ to (0.12) at $L = 70\text{ m}$.

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Pile Elastic Settlement (Vesic's Method)

The elastic settlement (S) of a single driven pile under working load $Q_w$ consists of three structural and geotechnical components:

$$S = S_1 + S_2 + S_3$$

1. Elastic Compression of Pile Stem ($S_1$):

   $$S_1 = \frac{(Q_{wp} + \xi Q_{ws}) L}{A_p E_p}$$

   Where $Q_{wp}$ is the load carried at the pile tip under working load, $Q_{ws}$ is load carried by skin friction, $E_p$ is pile elastic modulus, $A_p$ is pile cross-sectional area, and $\xi$ is skin friction distribution factor ($\xi = 0.5$ for uniform or parabolic distribution, $\xi = 0.67$ for triangular distribution increasing with depth).

2. Settlement Caused by Load at Pile Tip ($S_2$):

   $$S_2 = \frac{Q_{wp} C_p}{D \cdot q_b}$$

   Where $C_p$ is an empirical coefficient (0.02 for sand/gravel, 0.04 for clay), and (D) is pile diameter/width.

3. Settlement Caused by Load Transmitted Along Shaft ($S_3$):

   $$S_3 = \frac{Q_{ws} C_s}{L \cdot q_b}$$

   Where $C_s = (0.93 + 0.16 \sqrt{L/D}) C_p$.

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PE-Style Worked Example: Driven Pile Capacity

Problem: A closed-end steel pipe pile with an outer diameter $D = 406\text{ mm}$ (0.406 m) is driven to a depth $L = 18\text{ m}$ into a layered soil profile. The soil profile consists of:

• Layer 1 (0 to 6 m): Soft to medium clay, $\gamma_1 = 17.5\text{ kN/m}^3$, undrained shear strength $c_u = 40\text{ kPa}$.
• Layer 2 (6 to 18 m): Dense sand, $\gamma_{sat} = 19.5\text{ kN/m}^3$, effective friction angle $\phi' = 36^\circ$, (K = 1.4), $\delta' = 28^\circ$.
• The groundwater table is located at $z = 3.0\text{ m}$ depth. Below GWT, unit weight of water $\gamma_w = 9.81\text{ kN/m}^3$.

Calculate:

1. Ultimate shaft resistance $Q_s$
2. Ultimate tip resistance $Q_b$ (using $N_q = 55$, capped at $q_l = 10\text{ MPa}$)
3. Allowable design load $Q_{all}$ using a factor of safety (FS = 2.5).

Solution Step-by-Step:

1. Pile Geometry Parameters:

• Perimeter $P = \pi \cdot D = \pi \cdot 0.406\text{ m} = 1.2755\text{ m}$
• Tip Area $A_b = \frac{\pi}{4} D^2 = \frac{\pi}{4} (0.406)^2 = 0.1295\text{ m}^2$

2. Layer 1 (Soft Clay, 0 to 6 m):

• Shaft Area $A_{s1} = P \cdot 6.0\text{ m} = 1.2755 \cdot 6.0 = 7.653\text{ m}^2$
• Adhesion Factor $\alpha$: For $c_u = 40\text{ kPa}$, $c_u < 50\text{ kPa}$, so $\alpha = 1.0$.
• Unit Skin Friction $f_{s1} = \alpha \cdot c_u = 1.0 \cdot 40 = 40\text{ kPa}$
• Shaft Capacity Layer 1: $Q_{s1} = f_{s1} \cdot A_{s1} = 40 \cdot 7.653 = 306.1\text{ kN}$

3. Layer 2 (Dense Sand, 6 to 18 m):

• Thickness $\Delta z_2 = 12.0\text{ m}$

• Shaft Area $A_{s2} = P \cdot 12.0\text{ m} = 1.2755 \cdot 12.0 = 15.306\text{ m}^2$

• Compute effective stress profile $\sigma'_v(z)$:

  • $\sigma'_v(z=0) = 0$
  • $\sigma'_v(z=3\text{ m}) = 3.0 \cdot 17.5 = 52.5\text{ kPa}$
  • $\sigma'_v(z=6\text{ m}) = 52.5 + 3.0 \cdot (17.5 - 9.81) = 52.5 + 23.07 = 75.57\text{ kPa}$
  • $\sigma'_v(z=18\text{ m}) = 75.57 + 12.0 \cdot (19.5 - 9.81) = 75.57 + 116.28 = 191.85\text{ kPa}$

• Average effective stress in Layer 2 (6 to 18 m):

  $$\bar{\sigma}'_{v2} = \frac{75.57 + 191.85}{2} = 133.71\text{ kPa}$$

• Unit Skin Friction Layer 2:

  $$f_{s2} = K \cdot \bar{\sigma}'_{v2} \cdot \tan(\delta') = 1.4 \cdot 133.71 \cdot \tan(28^\circ) = 1.4 \cdot 133.71 \cdot 0.5317 = 99.53\text{ kPa}$$

  (This is below typical limit $f_{s,max} = 120\text{ kPa}$, so $f_{s2} = 99.53\text{ kPa}$).

• Shaft Capacity Layer 2: $Q_{s2} = f_{s2} \cdot A_{s2} = 99.53 \cdot 15.306 = 1523.4\text{ kN}$

• Total Ultimate Shaft Resistance:

  $$Q_s = Q_{s1} + Q_{s2} = 306.1 + 1523.4 = 1829.5\text{ kN}$$

4. Ultimate End Bearing $Q_b$ at $z = 18\text{ m}$:

• Effective vertical stress at pile tip: $\sigma'_v(z=18) = 191.85\text{ kPa}$
• Calculated unit end bearing: $q_b = \sigma'_v \cdot N_q = 191.85 \text{ kPa} \cdot 55 = 10,551.75\text{ kPa} = 10.55\text{ MPa}$
• Check limiting unit end bearing $q_l = 10.0\text{ MPa} = 10,000\text{ kPa}$.
• Since $10.55\text{ MPa} > 10.0\text{ MPa}$, use $q_b = 10,000\text{ kPa}$.
• Total Ultimate Base Capacity: $$Q_b = q_b \cdot A_b = 10,000 \text{ kPa} \cdot 0.1295\text{ m}^2 = 1295.0\text{ kN}$$

5. Total Ultimate Capacity and Allowable Load:

• Ultimate Capacity $Q_u = Q_s + Q_b = 1829.5 + 1295.0 = 3124.5\text{ kN}$
• Allowable Design Load: $$Q_{all} = \frac{Q_u}{FS} = \frac{3124.5\text{ kN}}{2.5} = 1249.8\text{ kN} \approx 1250\text{ kN}$$