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} ]