When the soil near the surface cannot carry the load, stop trying to spread it and start trying to reach past it. A pile carries load two ways at once — on its tip and along its sides — and knowing which one dominates changes everything about how you design it.
Piles are expensive. You use them when a spread footing genuinely cannot do the job — and there are five distinct reasons that happens.
The upper strata cannot carry the load at any practical footing size, but a competent layer exists deeper. Reach it.
Bearing capacity might be adequate but a shallow foundation would settle far too much. Piles bypass the compressible layer.
Transmission towers, tall buildings under wind, and offshore structures need tension and lateral capacity a footing simply cannot supply.
Bridge piers must be founded below the depth that flood flow can wash away, which can be many metres.
Piles anchored below the active zone escape seasonal swelling and shrinking that would jack a footing up and down.
Piling can be an order of magnitude more expensive per unit load. It is the answer only after a spread or raft foundation has been shown to fail.
End bearing: the tip rests on rock or dense stratum and carries nearly everything. Friction: no competent layer within reach, so load is shed along the shaft. Most real piles are a combination.
Driven piles displace soil sideways, densifying sand and increasing capacity — but they cause noise and vibration. Bored (cast-in-situ) piles remove soil, so there is no densification and less shaft friction, but they are quiet and can be very large.
Precast/prestressed concrete is the workhorse. Steel H-piles and pipe piles drive well through hard layers and splice easily. Timber is cheap and durable if kept permanently below the water table.
Point (end bearing) resistance plus skin (shaft) friction. They mobilise at very different displacements: shaft friction is fully developed after only about 5–10 mm of movement, while end bearing needs 10–25% of the pile diameter. At working loads, friction is doing almost all the work even in a nominally end-bearing pile.
Axial load in the pile decreases with depth as friction sheds it into the surrounding soil. Whatever is left at the bottom is the end bearing. The shape of that curve tells you immediately which mechanism dominates.
Nq* is Meyerhof's bearing capacity factor for piles — far larger than the shallow-foundation Nq (81 vs 23 at φ' = 32°) because the failure zone is confined by soil on all sides. The limiting value on the right almost always governs.
K ≈ 1.4 for driven displacement piles, ≈ 1−sinφ' for bored. δ' ≈ 0.5–0.8φ'. p is the pile perimeter.
Both q' and σ'v stop increasing below a critical depth of roughly 15–20 pile diameters. Arching in the sand means the vertical stress acting on the pile does not keep growing indefinitely with depth. Ignore this and a long pile's capacity is wildly overestimated — the calculator below shows the effect directly.
Nc* = 9 for a deep foundation, against 5.14 for a shallow one — again because the failure surface is fully confined.
α ≈ 1.0 for soft clay (cu < 25 kPa), falling to about 0.5 at cu = 100 kPa and 0.4–0.45 for stiff clay beyond that.
In clay, friction dominates overwhelmingly. Worked Example 2 gets 634 kN of shaft friction against just 86 kN of end bearing — the tip contributes 12%. Shortening a friction pile is far more damaging than in sand, where the tip matters more.
Switch soil type and watch the balance between tip and shaft flip. Try pushing the length well past the critical depth in sand to see friction accumulate linearly rather than quadratically.
Piles are almost never used singly — they come in groups under a common cap. And a group of nine piles does not carry nine times a single pile.
θ = arctan(D/s) in degrees, m × n piles at spacing s. Group capacity Qg = η · (mn) · Qu.
In clay a closely spaced group can fail as one enormous block rather than as individual piles. Compute the capacity of the group perimeter acting as a single deep footing, and take the smaller of the two answers.
Skin friction is normally your friend: the soil grips the pile and holds it up. But if the soil settles more than the pile, that grip reverses direction and starts dragging the pile down.
A pile driven through a soft clay layer that is still consolidating — under a new fill, a lowered water table, or its own weight. The clay settles past the pile, and friction along that length becomes a downward load instead of an upward resistance.
You lose the friction you were counting on over that length, and you gain an extra load the pile must carry. A layer that should have contributed +200 kN might instead impose −200 kN — a 400 kN swing. Mitigation: bitumen coating on the shaft through the settling layer, or a sleeve.
Negative skin friction is a load, not a reduction in resistance. Add it to the applied structural load before comparing against allowable capacity — do not simply subtract it from Qu.
Capacity answers "will it hold?" — and just as with shallow foundations, that is only half the design. A pile group can be nowhere near failure and still settle enough to distress the structure above it.
The pile as a compression member. ξ ≈ 0.5 for uniform or parabolic friction distribution, 0.67 for triangular. Small for concrete, larger for long steel piles.
Soil compressing beneath the point, with qwp = Qwp/Ap and Iwp ≈ 0.85. Usually the largest of the three.
Soil dragged down along the shaft, with \( I_{ws} = 2 + 0.35\sqrt{L/D} \). Typically the smallest term.
\( S_e = S_{e1} + S_{e2} + S_{e3} \). Worked Example 4 gets 1.1 + 8.7 + 0.7 mm — the tip term is 82% of the total. That is the opposite of the capacity picture, where the shaft carried 69%. A pile can get most of its strength from friction while getting most of its settlement from the point.
Bg = width of the pile group. A group settles considerably more than any single pile in it, because their stress bulbs overlap and the group loads a much deeper volume of soil.
Every equation in this chapter is an empirical correlation with real scatter — capacity predictions are commonly off by 30% or more. The load test is the only way to actually find out, and on any significant project it pays for itself.
Apply load in steps of roughly 25% of the estimated allowable load, holding each until settlement stabilises, and record the load-settlement curve.
Sometimes there is a clear plunging failure. More often there is not, so a criterion is used — for example net settlement reaching 10% of the pile diameter, or Davisson's offset line.
A verified capacity permits a lower factor of safety (often 2 instead of 3), which frequently saves more in pile lengths across the site than the test cost.
Problem: A 0.4 m square precast concrete pile is driven 12 m into sand with γ = 18 kN/m³, φ' = 32°. Take K = 1.4, δ' = 0.7φ', and a critical depth of 15D. Find the allowable capacity with FS = 3.
Problem: The same 0.4 m × 12 m pile is instead driven into clay with cu = 60 kPa. Take α = 0.55. Find the allowable capacity with FS = 3.
Problem: Nine of the clay piles from Example 2 are arranged in a 3 × 3 group at 1.2 m centres (s = 3D). Find the group capacity using Converse–Labarre.
Problem: The pile of Example 1 (0.4 m square, L = 12 m, Qall = 434 kN) carries its working load. Ep = 21 × 10⁶ kPa, Es = 30,000 kPa, μs = 0.3, ξ = 0.6. Find the settlement of a single pile, then of a 3 × 3 group at 1.2 m centres.
1. A pile in sand is driven to 30 m instead of 12 m. Why doesn't capacity increase as much as expected?
2. Why does α decrease as clay becomes stiffer?
3. A pile passes through soft clay that is still consolidating under a new fill. What happens to the friction along that length?
4. A 3×3 group at 3D spacing has η = 0.73. What does that mean?
5. In Example 2 the tip supplied only 12% of capacity. What does that imply for design?