Pile Foundations

Negative skin friction (NSF), also called down-drag, is the downward drag force on a pile shaft when the surrounding soil settles more than the pile. The soil then moves down relative to the pile and acts as an extra load instead of supporting the pile.

Typical situation

A pile is driven through a soft compressible clay layer to a firm stratum (rock, dense sand). If a fill is placed over the clay after driving, the clay consolidates under the fill and settles, dragging the pile shaft down. Down-drag acts down to the neutral point, below which the pile moves more than the soil and normal (positive) skin friction acts.

  Fill  ////////////////
  GL ---------------------   soil settles
  Soft |  | <- soil moves down
  clay |P |   NSF acts down
  - - -|  |- - neutral point
  Firm |  |   positive friction
  layer|__|   end bearing

Causes

• Fill placed on soft clay after driving the piles.
• Lowering of the water table (increase in effective stress, consolidation).
• Consolidation of remoulded clay disturbed by pile driving.
• Heavy surcharge or loads near the pile, and settlement of loose sand under vibration.

Effects

• Adds a downward load $Q_{nsf}$ on the pile, so the net capacity is reduced: $Q_{all} = \dfrac{Q_{ult} - Q_{nsf}}{FS}$ (or the working load plus $Q_{nsf}$ must be less than the capacity).
• Excessive settlement or structural overstress of the pile, especially for end-bearing piles.
• Tilting or cracking of the structure above, and settlement of the piles relative to the surrounding ground.

Remedial and preventive measures

• Coat the pile shaft over the compressible layer with bitumen (or use a sleeve) to reduce friction.
• Use a casing or a larger-diameter driven hole filled with bentonite slurry.
• Preload or drive the piles after the consolidation of the fill is complete.
• Use lightweight fill, or avoid surcharging, and keep the water table from being lowered.
• Increase the pile capacity (longer or larger piles) to carry the drag load.

A pile foundation transfers loads to deeper, stronger strata or by shaft friction. It is preferred over a shallow foundation in the following conditions.

1. Weak surface soil: the top layers (soft clay, loose sand, fill, peat) are too weak or compressible, and a firm layer, rock or dense sand lies at a reasonable depth below.
2. Heavy loads: large column loads, such as tall buildings, bridges and silos, that need more area than shallow footings can give (footings would overlap).
3. Uplift and lateral loads: structures with tension (transmission towers, offshore platforms) or large horizontal loads, e.g. from wind, waves or earthquakes, such as retaining structures and quays.
4. High water table or scour: where excavation for a shallow footing is difficult, or river-bed scour may remove the soil near the surface (bridge piers).
5. Expansive or collapsible soil: piles pass through the active zone to stable soil.
6. Excessive settlement of shallow foundations: where settlement or differential settlement would be unacceptable.
7. Neighbouring excavation or future excavation: when the soil near the structure may be removed.
8. Compaction of loose granular soil by driven piles, or where shallow foundations in deep soft soil are uneconomical.

Single pile

The negative skin friction acts over the length $L_n$ of the pile that lies in the settling clay (down to the neutral point).

Unit negative skin friction: $f_n = \alpha c_u$ (total stress method, $\alpha \approx 1$ for soft clay), or $f_n = \beta\sigma'_v$ (effective stress method, Bjerrum and Burland, $\beta = 0.2$ to $0.25$ for clay).

$$Q_{nsf} = P \times L_n \times f_n = \pi D\,L_n\,c_u \quad (\alpha = 1)$$

where $P$ is the perimeter of the pile and $D$ the diameter.

Group of piles

A group in clay moves with the soil mass between the piles, so the drag is the lesser of:

1. The sum of the drag on all piles, $Q_{ng} = n \times Q_{nsf}$.
2. The weight of the settling soil enclosed by the group, together with the shear along the perimeter of the block:

$$Q_{ng} = \gamma B_g L_g H_f + c_u\,P_g\,H_f$$

where $B_g \times L_g$ is the plan of the group, $P_g = 2(B_g + L_g)$ its perimeter, $H_f$ the thickness of the compressible layer (or fill) that settles, and $\gamma$ its unit weight.

The calculated drag is added to the structural load when checking the pile capacity, since it acts in the same direction.

Given: 16 piles (4 × 4), $D = 0.8$ m, $L = 10$ m, spacing $s = 2.4$ m, clay; $c_u = 40$ kN/m² for the top 5 m and $60$ kN/m² for the lower 5 m; adhesion factor $\alpha = 0.6$.

The group capacity is the smaller of (1) the sum of the capacities of individual piles and (2) the capacity of the block of soil enclosing the group.

1. Single pile

$$\begin{aligned} A_p &= \frac{\pi}{4}(0.8)^2 = 0.5027\ \text{m}^2,\quad \text{perimeter } = \pi(0.8) = 2.513\ \text{m} \\ Q_s &= \alpha c_{u1}P L_1 + \alpha c_{u2}PL_2 \\ &= 0.6(40)(2.513)(5) + 0.6(60)(2.513)(5) = 301.6 + 452.4 = 754.0\ \text{kN} \\ Q_p &= 9c_uA_p = 9 \times 60 \times 0.5027 = 271.4\ \text{kN} \\ Q_u &= 754.0 + 271.4 = 1025.4\ \text{kN} \end{aligned}$$

Sum for 16 piles: $16 \times 1025.4 = 16\,406.7$ kN.

2. Block failure

$$\begin{aligned} B_g = L_g &= 3(2.4) + 0.8 = 8.0\ \text{m},\quad P_g = 4 \times 8 = 32\ \text{m} \\ Q_{u,block} &= P_g\sum(c_uL) + 9c_uB_gL_g \\ &= 32\,(40 \times 5 + 60 \times 5) + 9 \times 60 \times 8 \times 8 \\ &= 16\,000 + 34\,560 = 50\,560\ \text{kN} \end{aligned}$$

3. Group capacity

Block capacity (50 560 kN) is greater than the sum of individual piles (16 407 kN), so the piles act individually and the efficiency is 1.

Answer: ultimate load capacity of the pile group $= 16\,407$ kN (about 16.4 MN).

Given: 16 square piles in a 4 × 4 group, side $b = 0.30$ m, $L = 12$ m, soft clay, $\alpha = 0.8$. The value of $c_u$ is not legible, but it cancels out, as both the individual and block capacities are proportional to $c_u$. Point bearing and skin friction are both considered.

For 100% efficiency, block capacity = sum of individual pile capacities.

Individual pile

$$\begin{aligned} Q_u &= \alpha c_u(4bL) + 9c_ub^2 \\ &= c_u\left[0.8(4)(0.3)(12) + 9(0.3)^2\right] \\ &= c_u(11.52 + 0.81) = 12.33\,c_u \end{aligned}$$

Sum for 16 piles $= 16 \times 12.33\,c_u = 197.28\,c_u$.

Block of side $B_g$ (square)

$$Q_{block} = c_u(4B_g)L + 9c_uB_g^2 = c_u\left(48B_g + 9B_g^2\right)$$

Equate

$$\begin{aligned} 9B_g^2 + 48B_g - 197.28 &= 0 \\ B_g &= \frac{-48 + \sqrt{48^2 + 4(9)(197.28)}}{2(9)} = 2.721\ \text{m} \end{aligned}$$

For a 4 × 4 group $B_g = 3s + b$:

$$s = \frac{2.721 - 0.30}{3} = 0.807\ \text{m}$$

Answer: centre-to-centre spacing of piles $\approx 0.81$ m ($\approx 2.7b$) for 100% efficiency. A larger spacing also gives an efficiency of 100%.

Pile foundation: a pile is a long, slender structural member of timber, concrete or steel, driven into or formed in the ground to transfer the structure's load to a deeper, stronger stratum or to the surrounding soil by end bearing and/or shaft friction. A pile foundation is a group of piles connected by a pile cap.

(i) Based on material

• Timber piles: economical and light, used for light loads and temporary works; need protection from rot and borers above the water table.
• Concrete piles: precast (reinforced or prestressed) or cast-in-situ.
• Steel piles: H-piles, pipe piles (open or closed end); high load capacity and easy to splice; prone to corrosion.
• Composite piles: two materials, e.g. timber below the water table with concrete above.

(ii) Based on method of installation

• Driven piles: installed by hammering or vibration (precast concrete, steel, timber); displace soil.
• Bored (drilled) piles: a hole is bored, then filled with concrete (bored cast-in-situ).
• Driven cast-in-situ piles: a casing or shell driven, and then filled with concrete (e.g. Franki, Raymond).
• Jacked and screw piles: pressed or screwed into ground where vibration must be avoided.

(iii) Based on load transfer

• End-bearing (point-bearing) piles: pass through weak soil and rest on rock or dense soil; load is carried mainly by the tip.
• Friction (floating) piles: carry load by skin friction along the shaft in deep soft soil.
• Combined (friction and end bearing): most piles in practice.

Pile foundations are classified in several ways.

1. According to material

2. According to load transfer

• End-bearing piles: tip rests on a hard stratum or rock; shaft passes through weak soil. Load is carried by the point.
• Friction piles: the load is transferred by skin friction along the shaft, in thick soft soil where no hard layer is reachable.
• Combined piles: both end bearing and friction.

3. According to method of forming

• Precast piles: cast and cured in a yard, then driven.
• Cast-in-situ piles: concreted in a hole in the ground, with or without casing (bored, driven cast-in-situ).

4. According to method of installation

• Driven: by hammer (drop, diesel, hydraulic) or vibratory driver.
• Bored: soil removed by auger or rotary rig and concrete placed.
• Jacked (pressed) and screwed piles for low-vibration sites.

Other classes: by displacement (large, small, non-displacement), by use (bearing, tension, compaction, batter, anchor) and by size (small-diameter micropiles).

Types of pile by use (function)

1. Bearing (end-bearing) piles: carry vertical loads to a firm stratum.
2. Friction piles: carry load by skin friction in deep soft soil.
3. Tension (uplift) piles: resist upward loads by skin friction.
4. Compaction piles: densify loose granular soil, without carrying much load.
5. Batter (inclined) piles: resist lateral and inclined loads.
6. Laterally loaded piles: resist horizontal forces, as in retaining walls and jetties.
7. Anchor piles: hold down sheet piles or bulkheads against pull.
8. Fender and dolphin piles: absorb impact of ships at harbours.
9. Sheet piles: form walls to retain soil and water.

Use of tension piles

Tension piles resist uplift in: transmission towers and tall chimneys (wind overturning), basements and underground tanks below the water table (buoyancy), offshore platforms and jetties, bridge piers subject to overturning, and foundations of water tanks and dams that need hold-down. They are anchored in the cap and may have an under-reamed bulb to increase resistance.

A bored cast-in-situ pile is formed by drilling a hole in the ground and filling it with reinforced concrete. Because soil is removed, there is little displacement and vibration.

Procedure

1. Set out the pile position and set up the boring rig (auger or rotary).
2. Boring: drill the hole to the required depth. In loose or collapsing soil use a temporary casing or bentonite slurry to keep the hole stable.
3. Cleaning: remove loose soil and cuttings from the bottom (using a cleaning bucket) and check depth and verticality.
4. Reinforcement: lower the prefabricated steel cage with spacers.
5. Concreting: place concrete through a tremie pipe from the bottom up (so that it does not segregate), keeping the tremie embedded in the concrete. The slurry is displaced upwards.
6. Casing withdrawal: pull out the casing as the concrete rises, keeping enough concrete head inside.
7. Curing and cut-off: after setting, cut the pile head to the cut-off level and connect to the pile cap.

Piles are grouped by the amount of soil they displace when installed.

The choice of pile type depends on the following factors.

1. Soil and ground conditions: type and depth of firm strata, soft or loose layers, boulders, groundwater and aggressiveness of soil or water (corrosion, sulphates).
2. Magnitude and type of load: vertical, lateral or uplift; heavy loads need large concrete or steel piles.
3. Length of pile required: timber for short lengths, steel for very long piles, precast concrete for medium.
4. Installation method and equipment available, noise and vibration limits, and overhead clearance.
5. Effect on adjacent structures: displacement piles may heave soil and damage nearby buildings.
6. Durability: exposure to marine, sulphate or varying water levels (timber rots above water table).
7. Availability of material and local skill, transport and handling (precast piles are heavy).
8. Cost and time: total cost including driving, cap and testing.
9. Group action and settlement requirements, and the need for load testing.

The vertical load capacity of a pile can be estimated by:

1. Static (analytical) methods: from soil properties using the formula $Q_u = Q_p + Q_s$ ($\alpha$-, $\beta$- and $\lambda$-methods, Meyerhof, Terzaghi).
2. Dynamic formulae: from the pile driving data (Engineering News, Hiley, Danish formulae) and the wave equation.
3. In-situ penetration tests: SPT and CPT correlations (Meyerhof).
4. Pile load test: direct test on a trial or working pile, the most reliable.

Bearing capacity of a pile in sand

The ultimate capacity is the sum of end bearing and shaft friction, $Q_u = Q_p + Q_s$.

Static formula

$$Q_p = A_p\,q'N_q^{*} \le A_p\,q_l,\qquad q_l = 50N_q^{*}\tan\phi\ (\text{kN/m}^2)$$ $$Q_s = \sum K\sigma'_v\tan\delta\,(\pi D)\,\Delta L$$

where $q'$ is the effective overburden pressure at the tip (limited to the critical depth, $15D$ to $20D$), $K$ the earth pressure coefficient (about 1 for loose, 2 for dense sand), $\delta \approx 0.75\phi$ the pile–soil friction angle, and $\sigma'_v$ the average effective stress over each layer. The allowable load is $Q_u/FS$ with $FS = 2.5$ to $3$.

SPT (Meyerhof) method: $Q_p = 40N\dfrac{L}{D}A_p \le 400NA_p$ and $f_s = 2\bar N$ (kN/m², driven piles), $Q_s = f_sA_s$.

Group effect

When piles are close together, the stress zones overlap, so the group behaves differently from the sum of single piles. The efficiency is $\eta = \dfrac{\text{capacity of group}}{n \times \text{capacity of one pile}}$.

  Single pile        Group (overlapping zones)
     |  |             | || || |
    /    \            /  \/ \/  \   deeper, wider
   / bulb \          /  stress   \  stressed zone

• In loose sand a driven group compacts the soil, so $\eta \ge 1$ (taken as 1 for design); in dense sand or bored piles in sand, $\eta \le 1$.
• Group capacity: $Q_g = \eta\,nQ_{u}$, with $\eta$ found from the Converse–Labarre formula $\eta = 1 - \theta\dfrac{(n-1)m + (m-1)n}{90mn}$, where $\theta = \tan^{-1}(D/s)$, or taken as 1 for driven piles at $s \ge 2.5D$.
• Settlement of a group is larger than that of a single pile at the same load per pile, because a larger zone of soil is stressed.

Pile load test: application and limitations

Application: carried out by applying load in increments (usually through a hydraulic jack against a kentledge or anchor piles) and recording the settlement (IS 2911 Part 4).

• It gives the actual ultimate load (the load at which settlement reaches about 10% of the pile diameter, or the point of a break in the curve) and the safe load, i.e. the smaller of 50% of the load at 10% $D$ settlement or two-thirds of the load at 12 mm settlement.
• It checks the design assumptions, and verifies the quality of the installed pile.
• Used to determine the settlement under working load, lateral load capacity (horizontal test) and uplift capacity (pullout test).

Limitations:

• It is costly and time-consuming (the test takes days).
• It shows the behaviour of a single pile; a group has a different capacity (deeper stress zone), and group settlement is not found.
• Time effects (set-up, consolidation in clay) may not be reflected unless the test is carried out after a rest period.
• The test depends on the correctness of the test set-up; it covers only the test location.

Dynamic pile formulae (Engineering News, Hiley, Danish) estimate the capacity from the set per blow, but they are inaccurate for the following reasons.

1. Energy losses are uncertain. The energy lost in the hammer, cap, cushion, pile compression and soil quake cannot be assessed correctly. Efficiency is only assumed.
2. Dynamic resistance differs from static resistance. Pore water pressure, soil damping and strain rate during driving give a resistance that differs from the static capacity, especially in clays and saturated fine sands.
3. Impact is treated as a rigid-body collision. The formulae use a coefficient of restitution, which ignores wave propagation in the pile (the stress wave takes time to travel the length).
4. Time effects. In clays the strength after driving is regained (set-up) with time, whereas in dense saturated sands there may be relaxation, which the set at the end of driving does not show.
5. Soil variation along the pile (layers) and the shaft–tip load distribution are not accounted for.
6. Only suitable for granular soil and not for clay or long piles; the results may differ by as much as 50% or more. For these reasons a factor of safety of 2 to 6 is used, and the pile load test is preferred for important structures.

A laterally loaded pile is a pile that has to resist horizontal forces or moments in addition to (or instead of) vertical load. The load is transferred to the soil by lateral soil reaction along the shaft, and the pile bends like a beam on an elastic foundation.

Examples

• Piles under retaining walls, bridge abutments and quay walls (earth and water pressure).
• Jetties, wharves and offshore platforms subject to waves, berthing and mooring forces.
• Transmission towers, tall chimneys and high-rise buildings due to wind and earthquake.
• Bridge piers subject to braking forces and flowing water.

Behaviour and analysis

• Short (rigid) piles fail by rotation of the pile or by failure of the soil. Long (flexible) piles fail by yielding of the pile material, with a plastic hinge.
• Analysis: Broms' method (ultimate lateral load for short and long piles in cohesive and cohesionless soil), subgrade reaction method (Winkler spring model with $k_h$), and p–y curves for non-linear soil.
• The lateral capacity is increased by larger diameter, by fixity to the cap, or by using batter (inclined) piles to carry the horizontal component.

The settlement of a pile group in sand is greater than that of a single pile carrying the same load per pile, as a larger volume of soil is stressed. The main methods are:

1. Vesic's method (empirical):

$$S_g = S\sqrt{\frac{B_g}{D}}$$

where $S$ is the settlement of a single pile at the same working load, $B_g$ is the width of the group and $D$ the pile diameter.

2. Meyerhof's SPT method:

$$S_g\ (\text{mm}) = \frac{2q\sqrt{B_g}\,I}{\bar N},\qquad I = 1 - \frac{D}{8B_g} \ge 0.5$$

where $q$ is the net pressure on the group (kN/m²) and $\bar N$ the average SPT value within a depth $B_g$ below the tip (for CPT: $S_g = \dfrac{qB_gI}{2q_c}$).

3. Equivalent raft method: the pile load is assumed to act on an imaginary raft of the group size at a depth $\tfrac{2}{3}L$ from the top (for friction piles) with a 2:1 load spread, and the settlement of the raft is computed from elasticity or from SPT/plate test data.

4. Elastic (interaction) methods (Poulos): settlement from interaction factors between piles.

Uplift (tension) piles resist upward or pull-out forces in a structure, where the load tries to lift the foundation.

• Uses: transmission towers and chimneys (wind overturning), basement floors and tanks under the water table (buoyancy), anchoring of docks and offshore structures, and bridge piers resisting overturning.
• Resistance: only the skin friction along the shaft acts (there is no end bearing), plus the weight of the pile. It is about 70 to 80% of the friction in compression. Ultimate uplift: $Q_{up} = \alpha c_u\pi DL + W_p$ (clay) or $\sum K\sigma'_v\tan\delta\,\pi D\Delta L + W_p$ (sand).
• Under-reamed (belled) piles have a larger resistance from the bulb. The pile is also anchored to the cap by reinforcement for full tension.
• Group: the uplift capacity of a group is the lesser of the sum of individual piles, and the weight of the block of soil plus pile cap and the shear around the block.
• A factor of safety of 2 to 3 is used.

The behaviour of a pile in a group differs from that of an isolated single pile because the stress zones of neighbouring piles overlap.

Factors: spacing (generally 2.5 to 3D), number and arrangement, type of soil, method of installation and the length of pile. Group capacity is the smaller of the sum of the capacities of individual piles and the block capacity. Settlement must be checked for the group, not for the single pile.

An inclined (batter) pile is driven at an angle to the vertical, usually at a slope (batter) of 1 horizontal to 4 to 6 vertical (up to about 1 in 3).

  Load  |      \  /  
   <--- |       \/   horizontal component
     \   \     /\ 
      \   \   /  \
       pile   pile

• The axial load $Q$ in the inclined pile has a horizontal component $Q\sin\theta$ that resists lateral loads, so the pile resists horizontal loads mainly by axial thrust (more efficient than bending of a vertical pile).
• Used where large lateral forces act, such as retaining walls, quay walls, bridge abutments, jetties, dams, and towers.
• They are installed in groups with vertical piles, or in opposing pairs.
• Limitations: higher cost and more difficult driving. Bending may arise when the ground settles around them (NSF/downdrag), and the capacity is reduced if the soil settles.

Given: 16 piles (4 × 4), $D = 0.40$ m, $L = 12$ m, $c_u = 35$ kPa, spacing $s = 0.80$ m (c/c, $= 2D$), $\alpha = 0.8$, FoS $= 2.5$.

1. Individual pile

$$\begin{aligned} A_p &= \frac{\pi}{4}(0.4)^2 = 0.1257\ \text{m}^2,\quad P = \pi(0.4) = 1.257\ \text{m} \\ Q_s &= \alpha c_uPL = 0.8 \times 35 \times 1.257 \times 12 = 422.2\ \text{kN} \\ Q_p &= 9c_uA_p = 9 \times 35 \times 0.1257 = 39.6\ \text{kN} \\ Q_u &= 461.8\ \text{kN} \end{aligned}$$

Sum of 16 piles $= 16 \times 461.8 = 7389.0$ kN.

2. Block failure

$$\begin{aligned} B_g = L_g &= 3(0.8) + 0.4 = 2.8\ \text{m},\quad P_g = 4(2.8) = 11.2\ \text{m} \\ Q_{block} &= P_gLc_u + 9c_uB_gL_g \\ &= 11.2 \times 12 \times 35 + 9 \times 35 \times 2.8^2 \\ &= 4704.0 + 2469.6 = 7173.6\ \text{kN} \end{aligned}$$

3. Group capacity

The block capacity (7173.6 kN) is smaller than the sum of individual piles (7389.0 kN), so the block failure governs. Efficiency $= 7173.6/7389.0 = 0.97$.

$$Q_{all} = \frac{7173.6}{2.5} = 2869.4\ \text{kN}$$

Answer: allowable load of the pile group $\approx 2869$ kN.

Given: 16 driven concrete piles (4 × 4), $B = 0.40$ m, spacing $2.5B = 1.0$ m, FOS $= 2.5$, critical depth $D_{cr} = 12B = 4.8$ m, water table at ground level. Layers: loose sand 0–3 m ($\gamma = 16$, $\phi = 30^\circ$); medium dense sand 3–9 m ($\gamma_{sat} = 19$, $\phi = 35^\circ$); dense sand 9–14 m ($\gamma_{sat} = 20$, $\phi = 40^\circ$). The pile length is 14 m.

Assumptions: the given $\gamma$ values are saturated unit weights, $\gamma_w = 9.81$ kN/m³; concrete pile $\delta = 0.75\phi$; $K = 1$ (loose), $1.5$ (medium dense, interpolated) and $2$ (dense); $N_q$ for $\phi = 40^\circ$ is read from the Berezantsev curve as about 130 (an approximate chart reading). Effective stress is constant below $D_{cr}$.

Effective stresses

$\gamma' = 6.19,\ 9.19,\ 10.19$ kN/m³ for the three layers.

$$\sigma'_v(3\text{ m}) = 6.19 \times 3 = 18.57,\qquad \sigma'_v(D_{cr}) = 18.57 + 9.19 \times 1.8 = 35.11\ \text{kN/m}^2$$

Shaft friction ($Q_s = K\sigma'_{avg}\tan\delta\,P\,\Delta L$, $P = \pi \times 0.4 = 1.257$ m)

$Q_s = 451.2$ kN.

End bearing

$$\begin{aligned} q' &= 35.11\ \text{kN/m}^2 \ (\text{at } D_{cr}) \\ q_p &= q'N_q = 35.11 \times 130 = 4564.6\ \text{kN/m}^2 \le 50N_q\tan\phi = 5454\ \text{kN/m}^2 \\ Q_p &= 4564.6 \times 0.1257 = 573.6\ \text{kN} \end{aligned}$$

Capacity

$$\begin{aligned} Q_u &= 451.2 + 573.6 = 1024.8\ \text{kN} \\ Q_{all} &= \frac{1024.8}{2.5} = 409.9\ \text{kN per pile} \end{aligned}$$

For driven piles in sand at 2.5B spacing the group efficiency is taken as $\eta = 1$ (soil is compacted), so

$$Q_{g,all} = 16 \times 409.9 = 6559\ \text{kN}$$

Answer: allowable load of the group $\approx 6560$ kN (410 kN per pile). If the conservative Converse–Labarre efficiency ($\theta = 21.8^\circ$, $\eta = 0.637$) were used, the group value would be about 4176 kN.

Given: consolidated clay 16 m deep, $c_u = 30$ kN/m², $\gamma = 22$ kN/m³, load $Q = 2000$ kN, $\alpha = 0.6$, FS $= 3$.

Trial pile: diameter $D = 0.40$ m, length $L = 12$ m (the pile stays in the clay, with 4 m of clay below the tip).

Capacity of one pile

$$\begin{aligned} Q_s &= \alpha c_u\pi DL = 0.6 \times 30 \times \pi(0.4)(12) = 271.4\ \text{kN} \\ Q_p &= 9c_uA_p = 9 \times 30 \times 0.1257 = 33.9\ \text{kN} \\ Q_u &= 305.4\ \text{kN} \end{aligned}$$

Number of piles

$$n = \frac{FS \times Q}{Q_u} = \frac{3 \times 2000}{305.4} = 19.6 \ \Rightarrow\ 20\ \text{piles}$$

Arrange them in a 5 × 4 pattern at spacing $s = 3D = 1.2$ m.

Group check

• Sum of individual piles: $20 \times 305.4 = 6107$ kN.
• Block: $B_g = 3(1.2) + 0.4 = 4.0$ m, $L_g = 4(1.2) + 0.4 = 5.2$ m.

$$Q_{block} = 2(B_g + L_g)Lc_u + 9c_uB_gL_g = 2(9.2)(12)(30) + 9(30)(4.0)(5.2) = 6624 + 5616 = 12\,240\ \text{kN}$$

• Group capacity $=$ the smaller $= 6107$ kN, so $\eta = 1$.
• Allowable group load $= 6107/3 = 2036$ kN $> 2000$ kN. O.K.

Answer: provide 20 piles of 0.40 m diameter and 12 m length in a 5 × 4 group at 1.2 m c/c (allowable load 2036 kN > 2000 kN), joined by a pile cap of about 5.2 m × 4.0 m plus edge distance. Check the settlement of the group.

Given: 12 circular piles in a rectangular (4 × 3) pattern, $s/d = 3.0$, $q_u = 80$ kN/m² so $c_u = 40$ kN/m², $L = 15$ m, $\alpha = 0.85$. Efficiency $= 100\%$ means block capacity $\ge$ sum of individual pile capacities.

Individual pile (diameter $d$)

$$Q_u = \alpha c_u\pi dL + 9c_u\frac{\pi d^2}{4} = 40\left(40.06\,d + 7.07\,d^2\right)$$

Sum for 12 piles $= 40(480.7\,d + 84.8\,d^2)$.

Block (4 piles by 3 piles, $s = 3d$)

$$\begin{aligned} B_g &= 2s + d = 7d,\quad L_g = 3s + d = 10d \\ Q_{block} &= 2(B_g + L_g)Lc_u + 9c_uB_gL_g = 40\left(510\,d + 630\,d^2\right) \end{aligned}$$

The block capacity ($510d + 630d^2$) is larger than the sum of individual piles ($480.7d + 84.8d^2$) for every diameter, so the group fails as individual piles and the efficiency is 100% for any $d$ when $s = 3d$. The ratio $s/d$ governs, not the diameter.

Minimum spacing

Setting block $=$ sum for a general $s/d$ (checked in Python) gives $s/d = 2.11$ as the smallest ratio for 100% efficiency. Thus $s/d = 3$ is satisfactory.

The diameter is then fixed by the load to be carried or practical size. For example, with $d = 0.40$ m:

$$s = 3 \times 0.40 = 1.2\ \text{m},\quad Q_u(\text{one pile}) = 686.1\ \text{kN},\quad Q_{u,group} = 8233\ \text{kN}$$

Answer: spacing $= 3d$ (for example $s = 1.2$ m with a pile diameter of 0.40 m, centre to centre), which gives an efficiency of 100%; the minimum possible spacing for 100% efficiency is about $2.1d$. The data given do not fix the diameter, so a practical value of 0.3 to 0.5 m is adopted.

Given: $D = 0.35$ m, $q_u = 80$ kPa so $c_u = 40$ kPa, $\alpha = 0.80$, ultimate load $Q_u = 1200$ kN, end bearing is 30% of the total load.

$$\begin{aligned} Q_p &= 0.30 \times 1200 = 360\ \text{kN} \\ Q_s &= 1200 - 360 = 840\ \text{kN} \end{aligned}$$

Shaft friction: $Q_s = \alpha c_u\pi DL$

$$\begin{aligned} 840 &= 0.80 \times 40 \times \pi \times 0.35 \times L = 35.19\,L \\ L &= \frac{840}{35.19} = 23.87\ \text{m} \end{aligned}$$

Answer: length of pile required $\approx 24$ m.

Given: $D = 0.30$ m, unit skin friction $f_s = 60$ kN/m² (ultimate), allowable load $250$ kN, tip carries 20% of the total load.

Assumption: a factor of safety of 2.5 is applied to the allowable load to obtain the ultimate load.

$$\begin{aligned} Q_u &= 2.5 \times 250 = 625\ \text{kN} \\ Q_s &= 0.80 \times 625 = 500\ \text{kN} \quad (\text{shaft share}) \end{aligned}$$ $$\begin{aligned} Q_s &= f_s\,\pi D\,L = 60 \times \pi \times 0.30 \times L = 56.55\,L \\ L &= \frac{500}{56.55} = 8.84\ \text{m} \end{aligned}$$

Answer: length of pile $\approx 8.9$ m, say 9 m. (If no factor of safety is applied, i.e. the shaft carries 200 kN, $L = 3.54$ m.)

Given: driven concrete pile $D = 0.40$ m, length 14 m, water table at 3 m. Layers: loose sand 0–3 m ($\gamma = 16$, $\phi = 30^\circ$); soft clay 3–9 m ($\gamma = 16$, $c = 15$ kN/m²); dense sand 9–14 m ($\gamma = 20$, $\phi = 40^\circ$). For concrete: $\delta = 0.75\phi$, $K = 1$ (loose) and $2$ (dense).

Assumptions: $\gamma_w = 9.81$ kN/m³; below the water table the given $\gamma$ values are saturated values; effective stress is limited at the critical depth $L' = 15D = 6$ m (Das/Meyerhof) for both shaft friction and the tip; $N_q$ for $\phi = 40^\circ$ is read from the Berezantsev curve as about 130 (approximate chart reading); $\alpha = 1.0$ for soft clay ($c < 25$ kPa); factor of safety $= 3$ (not given). Pile area $A_p = 0.1257$ m², perimeter $P = 1.257$ m.

Effective stress

$$\sigma'_v(3\ \text{m}) = 16 \times 3 = 48\ \text{kN/m}^2,\qquad \sigma'_v(6\ \text{m}) = 48 + (16 - 9.81)(3) = 66.57\ \text{kN/m}^2 \ (\text{constant below})$$

Shaft resistance

$Q_s = 37.5 + 113.1 + 483.0 = 633.6$ kN.

End bearing

$$\begin{aligned} q_pN_q &= 66.57 \times 130 = 8654\ \text{kN/m}^2 \\ q_l &= 50N_q\tan\phi = 50(130)\tan 40^\circ = 5454\ \text{kN/m}^2\ (\text{limiting value governs}) \\ Q_p &= 5454 \times 0.1257 = 685.4\ \text{kN} \end{aligned}$$

Capacity

$$Q_u = 633.6 + 685.4 = 1318.9\ \text{kN},\qquad Q_{all} = \frac{1318.9}{3} = 439.6\ \text{kN}$$

Answer: allowable pile load capacity $\approx 440$ kN (ultimate $\approx 1319$ kN). The result depends on the chart value of $N_q$.

Given: closed-end steel pipe piles, 3 × 4 group (12 piles), $D = 0.30$ m, $L = 15$ m, spacing $0.9$ m. Layers: 0–6 m ($c_u = 50$, $\alpha = 0.74$); 6–12 m ($c_u = 65$, $\alpha = 0.62$); 12–15 m ($c_u = 90$, $\alpha = 0.50$). FS $= 2.5$.

1. Single pile

$$\begin{aligned} P &= \pi(0.3) = 0.9425\ \text{m},\quad A_p = 0.07069\ \text{m}^2 \\ \sum\alpha c_uL &= 0.74(50)(6) + 0.62(65)(6) + 0.50(90)(3) = 222.0 + 241.8 + 135.0 = 598.8\ \text{kN/m} \\ Q_s &= 0.9425 \times 598.8 = 564.4\ \text{kN} \\ Q_p &= 9c_uA_p = 9 \times 90 \times 0.07069 = 57.3\ \text{kN} \\ Q_u &= 621.6\ \text{kN} \end{aligned}$$

Sum for 12 piles $= 12 \times 621.6 = 7459.3$ kN.

2. Block failure

$$\begin{aligned} B_g &= 2(0.9) + 0.3 = 2.1\ \text{m},\quad L_g = 3(0.9) + 0.3 = 3.0\ \text{m},\quad P_g = 2(2.1 + 3.0) = 10.2\ \text{m} \\ Q_{block} &= P_g\sum c_uL + 9c_uB_gL_g \\ &= 10.2(50 \times 6 + 65 \times 6 + 90 \times 3) + 9(90)(2.1)(3.0) \\ &= 9792 + 5103 = 14\,895\ \text{kN} \end{aligned}$$

3. Group capacity

Group capacity $=$ smaller of $7459.3$ kN and $14\,895$ kN $= 7459.3$ kN (efficiency 1).

$$Q_{all} = \frac{7459.3}{2.5} = 2983.7\ \text{kN}$$

Answer: allowable load of the pile group $\approx 2984$ kN.

Given: 20 piles in a 4 × 5 pattern, $d = 0.3$ m, $L = 20$ m, $q_u = 80$ kN/m² so $c_u = 40$ kN/m², $\alpha = 0.5$. "The same possibility of individual and block failure" means the block capacity equals the sum of the individual pile capacities (efficiency $= 1$). Point bearing is included.

Individual piles

$$\begin{aligned} Q_s &= \alpha c_u\pi dL = 0.5 \times 40 \times \pi \times 0.3 \times 20 = 377.0\ \text{kN} \\ Q_p &= 9c_uA_p = 9 \times 40 \times 0.0707 = 25.4\ \text{kN} \\ Q_u &= 402.4\ \text{kN} \\ \sum Q_u &= 20 \times 402.4 = 8048.8\ \text{kN} \end{aligned}$$

Block (4 piles in one direction, 5 in the other, spacing $s$)

$$\begin{aligned} B_g &= 3s + 0.3,\quad L_g = 4s + 0.3,\quad P_g = 2(B_g + L_g) = 14s + 1.2 \\ Q_{block} &= P_g Lc_u + 9c_uB_gL_g \\ &= (14s + 1.2)(20)(40) + 9(40)(3s + 0.3)(4s + 0.3) \end{aligned}$$

Equate

$$\begin{aligned} 800(14s + 1.2) + 360(12s^2 + 2.1s + 0.09) &= 8048.8 \\ 4320\,s^2 + 11\,956\,s + 992.4 &= 8048.8 \\ s &= 0.50\ \text{m} \end{aligned}$$

Answer: centre-to-centre spacing $\approx 0.50$ m ($\approx 1.7d$) for equal chances of individual and block failure. This is smaller than the practical minimum of $2.5d$ to $3d$ (0.75 to 0.9 m) needed for driving, so at a practical spacing the block capacity is higher and the group fails as individual piles (efficiency 1).

Given: 16 piles (taken as a 4 × 4 group), $d = 0.60$ m, through $H = 3.5$ m of unconsolidated (settling) soil, $q_u = 60$ kN/m² so $c_u = 30$ kN/m², $\gamma = 16$ kN/m³, $\alpha = 1$.

The printed spacing of "800 cm" is taken as 800 mm = 0.80 m; a spacing of 8 m is not reasonable for a 16-pile group.

The negative skin friction (NSF) on the group is the smaller of (1) the sum of the NSF on individual piles and (2) the NSF on the block of soil.

1. Individual piles

$$\begin{aligned} Q_{nsf} &= \alpha c_u\pi dH = 1 \times 30 \times \pi \times 0.6 \times 3.5 = 197.9\ \text{kN} \\ 16Q_{nsf} &= 3166.7\ \text{kN} \end{aligned}$$

2. Block

$$\begin{aligned} B_g &= 3(0.8) + 0.6 = 3.0\ \text{m},\quad P_g = 4(3.0) = 12\ \text{m} \\ Q_{ng} &= \gamma B_g^2H + P_gHc_u \\ &= 16 \times 3.0^2 \times 3.5 + 12 \times 3.5 \times 30 \\ &= 504.0 + 1260.0 = 1764.0\ \text{kN} \end{aligned}$$

Answer: negative skin friction on the group $\approx 1764$ kN (the block value governs). If the spacing were really 8 m, the piles would act separately and the NSF would be $16Q_{nsf} \approx 3167$ kN.

Given: 9 piles (3 × 3, square pattern), $d = 0.30$ m, $L = 12$ m, $q_u = 75$ kN/m² so $c_u = 37.5$ kN/m². Point bearing is neglected, and the efficiency is to be 1. The adhesion factor is not given, so $\alpha = 1.0$ is assumed.

For efficiency $= 1$, block capacity = sum of individual pile capacities (skin friction only).

Individual piles

$$\sum Q_u = 9\,\alpha c_u\pi dL = 9 \times 1.0 \times 37.5 \times \pi \times 0.3 \times 12 = 3817.0\ \text{kN}$$

Block

$$Q_{block} = 4B_g\,L\,c_u = 4 \times B_g \times 12 \times 37.5 = 1800\,B_g$$

Equate

$$\begin{aligned} 1800\,B_g &= 3817.0 \Rightarrow B_g = 2.12\ \text{m} \\ B_g &= 2s + d \Rightarrow s = \frac{2.12 - 0.30}{2} = 0.91\ \text{m} \end{aligned}$$

Thus $s/d = 3.0$. In general $B_g = \dfrac{9\alpha\pi d}{4}$.

Answer: centre-to-centre spacing $\approx 0.91$ m ($\approx 3d$) for an efficiency of 1.

Given: 16 piles (4 × 4), $d = 0.50$ m, spacing $s = 1.0$ m, soft clay $c_u = 30$ kN/m², $\alpha = 0.6$, point bearing neglected. The length is printed as 90 m; 9 m is taken as the intended value (the efficiency does not depend on $L$, and for 90 m the capacities are 10 times larger).

Sum of individual piles

$$Q_u = \alpha c_u\pi dL = 0.6 \times 30 \times \pi \times 0.5 \times 9 = 254.5\ \text{kN},\quad 16Q_u = 4071.5\ \text{kN}$$

Block

$$\begin{aligned} B_g &= L_g = 3(1.0) + 0.5 = 3.5\ \text{m},\quad P_g = 14\ \text{m} \\ Q_{block} &= P_gLc_u = 14 \times 9 \times 30 = 3780\ \text{kN} \end{aligned}$$

Group capacity and efficiency

The smaller value governs: $Q_{g} = 3780$ kN.

$$\eta = \frac{3780}{4071.5} = 0.928\ (92.8\%)$$

Answer: ultimate load capacity of the group $= 3780$ kN and efficiency $= 0.93$ (block failure governs). For $L = 90$ m as printed: $Q_g = 37\,800$ kN, $\eta = 0.93$.