The coating that takes the resistance with it

Ground that is settling hangs on a pile instead of holding it up, and the worst force in the pile is at a depth where nothing is applied.

The hero is the problem stated at working scale: a 750 mm pile, 30 m through consolidating ground under 10 m of fill, carrying 1,200 kN at its head. The neutral plane sits at 18.86 m, the drag accumulated above it is 1,131 kN, and the worst force in the pile is 2,331 kN — 1.94 times what the structure applied.

This rung is about the one term in that calculation anybody can change.

Four quantities, one of them a decision #

The drag above the neutral plane is $\tfrac{1}{2} a z_n^2$ with $a = \pi d \beta \gamma'$, and every symbol in it belongs to somebody else.

$\gamma'$ is the ground. Its effective unit weight is a soil property, measured rather than chosen.

$z_n$ is not an input at all. The neutral plane is found from equilibrium of the whole pile, so it is an output that moves when anything else does.

$d$ is the pile, and it is chosen — but it is chosen for the capacity the pile has to deliver, so it is not free. It is the same bind a member sized by one property and governed by another is always in.

$\beta$ is the interface, and it is the only one that is a decision. It is the ratio of shaft friction to vertical effective stress, and it is a property of the contact between the pile and the soil rather than of either of them alone.

Which means it can be engineered. Coat the shaft in bitumen and $\beta$ falls by an order of magnitude, because the soil is now sliding against a layer that shears at a low stress and creeps under sustained load.

A third of the friction has bought a sixth of the drag, and the reason is the second-order effect that makes this rung worth writing. The drag is $\tfrac12 a z_n^2$: reducing $a$ reduces it linearly, but it also raises the neutral plane, and the drag then falls with the square of the new depth. The two effects multiply.

They multiply in a particular direction, too. The plane rises from the bottom, and the friction is largest at the bottom because it is proportional to depth. The metres shed first are the metres pulling hardest.

Which free body produced the neutral plane #

The plane is the only quantity here that is not put in, so it is worth deriving once — and the free body that gives it is the whole pile rather than any part of it.

Take the entire pile, head to toe. Down through the head comes $Q$. Up from the base comes $Q_b$. And along the shaft comes friction, downward over the length above the plane and upward below it. Vertical equilibrium of that one body is

$$Q + \tfrac12 a z_n^2 = Q_b + \tfrac12 a (L^2 - z_n^2)$$

which rearranges to $z_n^2 = \tfrac12\left(L^2 + 2(Q_b - Q)/a\right)$, and that is the whole derivation.

Two features of it decide everything else in this essay.

It is one equation in one unknown, and the unknown is a length. No stiffness appears, no settlement appears, and no time appears. The plane is located by statics alone, on the assumption that the friction is fully mobilised in whichever direction the relative movement points — which is why this is a rigid-plastic model of a problem that is fundamentally about deformation.

And the right-hand side goes negative before the coating does its job. $z_n^2 \le 0$ requires $Q \ge Q_b + \tfrac12 a L^2$, which is exactly the pile’s own capacity. So the equation’s failure to have a root is not a numerical accident — it is the equation reporting, in the only way it can, that the pile is overloaded.

That is a property worth having in any model: the case where the physics runs out and the case where the arithmetic runs out should be the same case. Here they are, and a version of this calculation that returned zero drag instead of refusing would have said the smoothest pile was the best one.

Where the arithmetic stops #

Take $\beta$ to 0.05 and the calculation refuses to return an answer, and the refusal is the point of this section.

At $\beta = 0.05$ the whole shaft, top to toe, is worth 477 kN. The base takes 600. Together that is 1,077 kN, and the head load is 1,200. There is no depth at which the friction reverses, because there is no equilibrium — the pile is not carrying its load at all.

That is the trap the whole essay exists for. The coating removes the drag and the shaft resistance over the same length, in the same proportion, through the same mechanism, because they are the same friction with two different signs. A calculation that reports only the drag reports a monotonic improvement all the way to a pile that cannot stand up.

And the failure mode changes character on the way. At $\beta = 0.3$ this is a structural problem — a pile overstressed at 19 m. At $\beta = 0.07$ it is a geotechnical one — a pile with almost no shaft resistance, relying on 600 kN of end bearing and 668 kN of shaft, with a factor of safety of 1.06 on a base capacity nobody measured. The cure has moved the question to a different discipline and made the margin thinner.

Which is why bitumen coatings are specified over a stated depth rather than over the whole pile: coat the settling zone and leave the founding stratum bare. And that is a detailing decision about where the neutral plane will be after the coating has moved it, which is the circularity this rung is really about.

The plane is an output, and it moves with everything #

The gradient of one half is exact and it is worth seeing why, because it is the same equilibrium argument in a third form.

Add $\delta Q$ at the head. The plane rises until vertical equilibrium is restored, which requires that the drag shed above it plus the resistance gained below it equal $\delta Q$. Those two are equal — it is the same friction on the same metres, counted with opposite signs — so each is $\delta Q / 2$, and the worst force $Q + \text{drag}$ rises by $\delta Q - \delta Q/2$.

Downdrag is therefore self-limiting in a way no other load is. Loading the pile more heavily reduces the drag on it, which is the opposite of everything else in structural engineering and is the reason a lightly loaded pile in settling ground is in more trouble than a heavily loaded one.

Nothing about the soil differs across that line. The same sand, the same friction angle, the same effective stress, the same $\beta$ — and the friction on one side is holding the pile up while the friction on the other is pulling it down. What differs is which of the two is settling faster, which is a statement about time rather than about material.

The drag is a settlement problem wearing a force’s clothes #

Every number on this page is a force, and none of them is what the pile is actually being asked for.

Downdrag arrives because the ground moves down relative to the pile. It stops arriving the moment that relative movement stops — and it stops when the pile has settled enough to keep up. So the drag is displacement-controlled, and displacement-controlled actions have a property that force-controlled ones do not: they are relieved by the deformation they cause.

That is the reason downdrag is treated so differently in the two limit states, and the difference is larger than it looks.

At the ultimate limit state the drag is very nearly not a load at all. If the pile is about to fail in bearing, it is about to move down a great deal, and the moment it moves down faster than the ground the friction along the whole shaft reverses and starts holding it up. A pile at its geotechnical capacity has no drag on it. Codes reflect this by not adding the drag to the load in the bearing check, which reads as an omission and is a mechanism.

At the serviceability limit state it is entirely a load, because there the question is how far the pile head goes down, and the drag is what pushes it there.

And structurally it is always a load, because the 2,331 kN in the hero is a real force in a real section at 19 m depth, whatever the soil is doing. The section has to carry it.

So the same quantity is ignored, counted, and counted again in three checks on one pile — which looks like inconsistency and is three different questions. The imposed strain that goes away if the structure yields is the same idea in a frame: an action that is a deformation is present exactly as long as the structure is stiff enough to resist it.

Why a bigger pile is worse #

The instinct on discovering a pile is overstressed at depth is to make it bigger, and it is the wrong instinct twice over.

The drag is collected on the perimeter, which grows linearly with the diameter, so a bigger pile collects more drag. And the load is carried by the area, which grows with the square — so a bigger pile does resist the drag better than it collects it, and the stress falls.

Both are true and they answer different questions. If the problem is a stress in the concrete, a bigger pile helps. If the problem is a force — a splice, a pile cap connection, a steel section’s capacity, the tension in a raked pile’s reinforcement — a smaller pile helps, and a larger one makes the number worse.

Which of the two is the problem is decided by what is at the depth in question, and on a bored pile with no joints in it the answer is usually the stress. On a driven precast pile with a mechanical splice at 15 m, it is emphatically the force.

The same shape, wherever a traction changes sign #

The picture this essay draws — a force in a member that grows along part of its length and falls along the rest, with a maximum in the interior — belongs to a family, and recognising it is most of the transferable content.

A shear connector line on a composite beam. The interface shear points one way over one half of the span and the other way over the other, and the axial force in the slab peaks at midspan where no connector is doing anything.

A post-tensioned tendon. Friction along the duct acts against the jacking direction, so the prestress falls away from the live end — and on a tendon stressed from both ends the two friction losses meet at an interior point which is the tendon’s own neutral plane.

A bar developing its bond length. The force that arrives along a length is the general statement, and where a bar may stop is the detailing question it produces: whenever a member gathers its force from a surface traction rather than from an end load, the maximum is inside the member and its position is found from an equilibrium rather than read off a support.

And a slab on ground. Friction under the slab restrains its shrinkage, acts inward from both edges, and puts the largest tension in the middle of the pour — which is where slabs on ground crack, at a location no end condition explains.

The design consequence is the same in all four: the section to check is not at either end, and the position of the worst section moves when the loading changes. A drawing that details the ends generously and the middle economically has got it exactly backwards, and it is a natural thing to draw because the ends are where the connections are.

What the site can and cannot see #

None of this is visible once the piles are in, and it is worth listing what evidence exists.

A static load test measures the wrong thing. It loads the pile head and reads the settlement, over hours, on a pile whose surrounding ground is not settling. What it returns is the pile’s resistance with the friction acting upward along the whole shaft — the state the pile is in during the test and never afterwards. A test pile that reaches 2,400 kN says nothing about a working pile at 1,200 kN with 1,131 kN of drag on it, because the two have different force diagrams over the same length.

Instrumented piles are the only direct evidence, and they are rare. Strain gauges down the shaft give the force diagram itself, which is what every figure on this page draws, and the Scandinavian and Canadian test programmes that established the neutral-plane method were exactly that: piles with gauges left in the ground for years while the fill consolidated around them.

And the coating’s condition cannot be checked at all. Bitumen is applied in the yard, the pile is then driven through the ground it will live in, and whatever the driving did to the coating — stripped it, dragged it down, plugged it with soil — has happened in the one place nobody can look. A property that must be delivered once and can never be verified is the same class of risk a preloaded joint’s faying surface carries, and it is the honest argument against relying on a coating for more than it has to deliver.

Which brings the practical position round to where the arithmetic pointed. Use the coating to remove most of the drag, size the pile for the drag that remains assuming the coating underperforms, and never let the coated case be the one that provides the capacity.

What to carry away #

$\beta$ is the only term anybody chooses, and it is a property of the contact rather than of the pile or the soil.

Reducing it works better than it should, because the neutral plane rises as well and sheds the deepest, hardest-pulling metres first — a third of the friction bought a sixth of the drag.

And it removes the resistance along with the drag. Past a certain smoothness the pile has no equilibrium, and a calculation that reports only the drag reports an improvement all the way to a pile that will not stand.

A bigger pile drags harder in force and less hard in stress, so which way to move depends on whether the problem is a section or a splice.

Where the model stops #

The coating is uniform over the whole shaft. Real practice coats the settling zone only, so the pile has two $\beta$ values and the neutral plane has to be found for a shaft whose friction is discontinuous — usually iteratively, since the discontinuity is placed by an answer that depends on it.

Bitumen creeps, and that is what it is for. Its shear resistance depends on rate and temperature, so a slip coating is very effective against the slow settlement it is designed for and much less effective against a fast one — which is the opposite of the usual direction and makes it useless against, say, a seismic settlement.

The neutral plane is found from force equilibrium and not from compatibility. The honest version finds it where the pile’s settlement curve crosses the ground’s, and the two answers differ — the compatibility solution puts it shallower when the pile is compressible.

And no group effect is modelled. A pile inside a group is shielded by its neighbours: the block of soil between them settles with them rather than past them, so the interior piles drag much less than the corners. The ground’s load depends on what is carrying it, and here it depends on what is carrying it next door.

Nothing here is a settlement calculation. The drag is a force computed from an assumed relative movement, and how much the ground actually settles — and over what years — is a question the structure’s own supports ask too and neither of them can answer from statics.

The ladder from here #

Later rungs on this anchor: the compatibility solution, where the plane is found from two settlement curves crossing. The coated pile with two interface factors, and where to stop the coating. Heave drag, the same picture upside down, where swelling ground puts a pile into tension at a depth chosen by the same arithmetic — and pulls apart a pile with no tension reinforcement. Downdrag on a raked pile, where the drag has a horizontal component. And group effects resolved properly, with the block weight as the upper bound and the corner piles governing.

The mechanism was understood before it was calculated. Piles driven through fill in the 1930s were found to be carrying far more than had been put on them, and the explanation — that the ground was hanging on rather than holding up — was available immediately to anyone who drew the free body. What took until the 1960s and the Scandinavian test programmes was the neutral plane: the observation that the drag is not a load to be added but a redistribution to be located, and that the depth where it is worst is decided by an equilibrium the structure above knows nothing about.