A pile has no length until the ground gives it one

Every member drawn on this site is handed a length. A beam spans from one support to another; a column runs from one floor to the next; a truss chord is as long as the panel it crosses. The length is in the drawing, and every formula that follows uses it.

A pile is not like that. It goes into the ground until the ground is good enough, and when it is pushed sideways at its head there is no support at the bottom to span to — only more ground, of a stiffness that grows as it goes down. Nothing in the geometry says how much of the pile is involved. The length has to come out of the calculation, and where it comes from is the whole of this page.

That is not unique to piles, and it is worth naming the company it keeps. A shell forgets a held edge over a distance the shell decides; a beam on a continuous elastic restraint forgets how long it is; a bar transfers its force into concrete over a length nobody drew. Each of those is a governing equation with a length built into it out of two stiffnesses, and the length is a property of the pair rather than of either. The pile is the same construction with the awkward feature that one of the two stiffnesses belongs to the ground.

The only length that can be built #

The governing equation is a beam on an elastic foundation whose modulus grows with depth, which is what a normally consolidated sand does:

$$EI\,\frac{d^4y}{dz^4} + n_h\, z\, y = 0.$$

There are two constants in it and they have different dimensions. $EI$ is a force times a length squared; $n_h$ is a force per unit length cubed. The only combination of the two that has the dimensions of a length is

$$T = \left(\frac{EI}{n_h}\right)^{1/5},$$

and dimensional analysis alone therefore guarantees that every answer this equation gives is a function of $z/T$ and nothing else. The characteristic length is not an approximation or a fitted parameter; it is the only length the problem contains.

For a 600 mm bored pile at a cracked stiffness of $1.2\times10^{14}$ N·mm² in ordinary medium-dense sand, $T$ is 1.89 m.

The fifth power is worth noticing before anything is done with it. Every other characteristic length in this collection is a square root or a fourth root — the shell’s is $\sqrt{rt}$, the Winkler beam’s is $(4EI/k)^{1/4}$ — because the equations they come from have a lower power of the coordinate in them. The extra power here is the modulus gradient, and it is the reason this problem is so much less sensitive to its inputs than any of its relatives.

What the numbers come out as #

Solving the equation with a lateral load $P$ at a free head and nothing at the toe gives the classical coefficients, and they are recovered here by finite differences rather than quoted:

$$y_0 = 2.43\,\frac{P T^3}{EI}, \qquad M_{max} = 0.772\, P T \ \text{at}\ z = 1.32\,T.$$

Matlock and Reese’s published values, obtained from a power-series solution in the nineteen-fifties, are 2.435 and 0.772. Getting them back from a different method is the check that the model is the one everybody else is using.

For the pile drawn, carrying 150 kN at its head: the head moves 20.5 mm, the worst bending moment is 219 kNm, and it occurs 2.5 m down — a quarter of the way to the depth most people would guess, and at a section nobody can see.

Those three numbers are the design, and none of them was assumed. The load and the section were chosen; everything else is what the equation returned.

That last point is the practical one. The design moment in a laterally loaded pile is not at the ground surface, which is where an intuition trained on cantilevers puts it, and it is not at the toe. It is about one and a third characteristic lengths down, and where the reinforcement is curtailed matters accordingly.

Four characteristic lengths is the whole pile #

Solve the same problem for shorter and shorter piles and the head deflection tells a very clear story.

Below about two $T$ the pile is a lever with almost nothing holding its foot, and the head deflection is several times the long-pile value — seven times at one $T$. Between two and about three and a half it falls quickly. Past about four $T$ it stops moving to within one per cent, and adding more pile changes nothing at all.

The reason is that the soil reaction $p = n_h z y$ requires both a stiffness and a deflection, and by four $T$ the deflection has died away to nothing. The ground down there is perfectly capable of pushing; it is never asked to.

For the pile drawn, four $T$ is 7.6 m of a 20 m pile. The remaining twelve metres are carrying axial load down to a bearing stratum and are irrelevant to the horizontal one. That is the answer to the commonest request in the subject: a pile that deflects too much sideways cannot be fixed by making it longer, and the reason is that the extra length is in a part of the ground the problem never reaches.

The fifth root, which is the practical content #

The exponent is the reason this subject behaves the way it does, and it is worth taking seriously rather than noting.

$T \propto n_h^{-1/5}$ means that a hundredfold change in the ground’s stiffness — from a very soft silt to a dense gravel, which is most of the range there is — changes the characteristic length by a factor of $100^{1/5} = 2.51$. The head deflection goes as $T^3/EI$ at fixed $EI$, so it changes by a factor of $100^{3/5} = 15.8$, which is large but not as large as the hundred that went in.

Two consequences follow and both are counter-intuitive.

The site investigation matters less than it feels like it should. Arguments about whether $n_h$ is 5 or 8 MN/m³ move the answer by six per cent. The order of magnitude has to be right; the second figure does not.

The section matters more than it feels like it should. $T \propto (EI)^{1/5}$ too, so doubling the pile’s stiffness buys 15 per cent of length. But the head deflection is $2.43\,PT^3/EI \propto EI^{3/5}/EI = EI^{-2/5}$: doubling the stiffness reduces the deflection by 24 per cent. A serviceability problem on a laterally loaded pile is solved with diameter, not with depth and not with better ground.

Which free body produced the number #

Cut the pile at depth $z$ and draw the part above the cut.

Crossing the cut are a shear and a moment. Applied to the free body from outside are the head load $P$ and — this is the part that distinguishes the problem — a distributed soil reaction over the whole embedded length above the cut, whose intensity at every depth is $n_h z y(z)$: proportional to the local deflection, which is what is being solved for.

That circularity is the whole difficulty and the reason there is no elementary closed form. Horizontal equilibrium says the shear at depth $z$ is $P$ minus the integral of the soil reaction above it, and the soil reaction depends on the shape the pile has taken, which depends on the shear. The equation solves them together.

It also explains where the maximum moment is. The moment is the integral of the shear; the shear falls from $P$ at the head as the soil reaction accumulates; and the moment peaks where the shear passes through zero — at the depth where the soil above the cut has taken exactly the applied load. The peak is at the depth where the ground has finished pushing back, which is a statement about the soil and not about the pile.

What a cap is actually worth #

Cast a cap across the heads of several piles and the head can no longer rotate. That is a different boundary condition, and it changes two things by different amounts.

The head deflection falls from 20.5 mm to 7.8 mm — a factor of 2.6, which is the reason caps exist.

The maximum moment moves. In the free-head pile it is 219 kNm at 2.5 m down; in the fixed-head one it is 263 kNm at the head itself. The cap has not reduced the moment at all. It has increased it by twenty per cent and moved it into the pile-to-cap connection.

That is the trade nobody states. A cap buys deflection with moment, and puts the moment in the one location that is congested with reinforcement from three directions, awkward to detail, and universally drawn on the general arrangement as a simple bearing. A great many pile caps are detailed as though the pile heads were pinned and analysed as though they were fixed.

The depth to fixity, and what it is for #

The structure above the ground does not want a distributed soil reaction. It wants a column with a base, so that a frame model can be built.

The depth to fixity is that substitution: the length of an equivalent cantilever, fixed at its base, which has the same head deflection under the same load. It is a result rather than a choice, and it comes out at $1.94\,T$ for the free-headed pile drawn and $2.23\,T$ for the fixed-headed one — 3.7 and 4.2 m.

That is what should be drawn in the frame model: a column extending below ground by that much, with a fixed base. Two things about it are worth knowing.

It is defined by deflection, so the equivalent cantilever gets the head movement right and the moment distribution wrong. The real pile’s moment peaks below ground; the equivalent cantilever’s peaks at its own base, further down. Using it for the pile’s own design is a mistake; using it for the structure above is exactly right.

And it depends on the head condition, so a pile in a cap and a pile with a pinned head are different columns in the frame model, even though they are the same pile in the same ground.

Where the linear model breaks, and what replaces it #

Everything above is linear: the soil reaction is proportional to the deflection at every depth, for ever.

Soil is not linear, and it stops being linear early. Near the surface — in the top two or three diameters, which is exactly where the deflection is largest — the ground can only push so hard before it flows around the pile or heaves in front of it. The limit is a passive resistance of a few times the overburden pressure, and a real pile at working load has usually reached it over the top metre.

That is the p–y method: replace the linear spring at each depth with a curve, one that is stiff at small movement and flat past a limiting resistance, and solve the same differential equation with a non-linear foundation. The peak soil reaction the linear model here computes — 77.6 N/mm on the pile drawn — is the number to check against that limit, and where it exceeds it the linear answer under-predicts both the deflection and the moment.

The reinforcement, which is the reason the depth matters #

A steel pile has the same section everywhere and the location of the maximum moment is a matter of curiosity. A reinforced concrete pile does not, and there the location is the design.

The cage in a bored pile is expensive, awkward to place and often shortened deliberately: a full-length cage in a twenty-metre pile is a great deal of steel doing nothing for the eighteen metres where there is no moment. So the cage is curtailed, and the question is where.

The moment diagram answers it, and it answers it in a way that catches people out. The peak is at 1.32 $T$ — 2.5 m — but the diagram does not stop there. It falls to zero at about 4 $T$ and reverses, weakly, below that. A cage curtailed at three metres because the peak is at two and a half stops in a region still carrying most of the peak moment, and the curtailment has to run past the point of contraflexure with the same shift and anchorage every curtailed bar needs.

The rule that follows is worth stating as a length: the cage runs to at least six characteristic lengths, or to wherever the moment has genuinely gone, whichever is deeper. On the pile drawn that is 11 m of cage in a 20 m pile — much more than the four $T$ the deflection cared about, and much less than full length.

The group, where the ground has already been used #

A single pile in a large volume of ground gets the whole of that ground’s stiffness. Several piles in a row do not, and the reason is the same one that makes a group settle more than one pile.

A pile in front of another is pushing into ground that the one behind it has already loaded and softened. The effect is expressed as a p-multiplier — a factor of perhaps 0.4 to 0.8 applied to the soil resistance depending on spacing and position in the group — and it is largest for the trailing rows.

The consequence is that a group of $n$ piles carries considerably less than $n$ times what one pile carries laterally, at the same deflection, whereas in axial load the group is closer to additive. Lateral group efficiency is much poorer than axial group efficiency, and it is the direction most people’s intuition gets wrong, because axial capacity is the quantity everybody has a feel for.

The asymmetry has a clean reason. Axial load is carried into the ground over the whole shaft and toe, so the volumes two neighbouring piles are stressing overlap only at depth. Lateral load is carried into the top few characteristic lengths, so two piles a few diameters apart in the direction of loading are working the same small wedge of ground, one behind the other. Proximity is a mild nuisance in one direction and the whole problem in the other, and the spacing that solves the second is measured in diameters along the line of the load rather than in any direction at all.

Where the model stops #

The modulus gradient is a model of a soil profile, not a soil profile. Real ground is layered, and a stiff crust over soft clay behaves nothing like $n_h z$ — the crust does almost all the work and the characteristic length is meaningless.

Cracked stiffness is a guess. A concrete pile’s $EI$ falls as it cracks, and the calculation above uses a single reduced value throughout. The real member has a stiffness that varies down its length with the moment it is carrying.

Cyclic loading softens the ground. A pile under repeated lateral load — wind on a mast, waves on a monopile, traffic on an abutment — accumulates deflection cycle by cycle, and the static solution is the first cycle only.

The axial and lateral problems are solved separately and are one member. A pile carrying a large axial load and bending is a beam–column, and the axial force takes stiffness away from it exactly as it does from a column above ground. The linear model here has no axial force in it at all.

And the linear spring is not a soil. It has no strength limit, no memory and no gap. A real pile leaves a gap behind itself in stiff clay after a few cycles, and thereafter it is a member with no support on one side at all.

Where the ladder goes #

Later rungs on this anchor: the p–y method, and the curves for sand and for clay. Layered profiles, where the characteristic length stops existing. Group effects and p-multipliers, and their asymmetry with axial group behaviour. Cyclic degradation and accumulated deflection. The monopile, where the pile is short in its own characteristic lengths and rotates rather than bends. Pile-head connection detailing, which is where the fixed-head moment goes. Lateral load tests, and what they can and cannot calibrate. And the general question this belongs to: what a structure does when the drawing does not contain a length.