Structural Engineering & Statics
where the load goes
Internal forces

The ground that hangs on instead of holding up

A pile is driven through fill that has not finished settling. The fill goes down past the shaft, the friction along that length turns round, and the pile is now carrying the soil rather than the other way about. The worst force is not at the head and not at the toe, and nothing at that depth can be seen.

Assumes The free body is a choice, and choosing it well is the whole skill, The force that arrives along a length and The beam that sits on the ground.

15 min read 9 figures The load must go somewhereScale changes everythingThe statics of things that do not move

A pile is a member that gets its load in at one end and gives it out along its length, and almost every drawing of one shows the giving-out happening in a single direction. The soil holds the shaft, the shaft holds the pile, the pile holds the building. The axial force is largest where the load is applied and smallest at the toe, and a section chosen for the head load is a section chosen for the worst place in the member.

That picture is right whenever the pile is settling faster than the ground around it, which is what happens when a pile is loaded and the ground is not. It is wrong, and wrong by a factor that matters, the moment the ground has business of its own.

The worst force in a pile is not at the top of it. A 0.6 m pile 24 m long through ground that is settling, carrying 800 kN at its head. Above the neutral plane the soil moves down past the shaft and the friction acts downward, so the axial force grows with depth; below it the friction acts upward in the ordinary way and the force falls again to the 300 kN the base takes. The maximum is 1282.87 kN at 13.78 m — 1.60 times the load applied, and it is at a depth where nothing is applied, nothing is connected and nothing can be inspected. A pile section chosen for the head load is under-sized by that factor over the middle third of its length.
Fig. 1 A 600 mm pile, 24 m long, carrying 800 kN at its head through ground that is still consolidating. Above 13.77 m the fill moves down past the shaft and the friction acts downward, so the axial force grows with depth instead of falling; below it the friction acts upward in the usual way and the force comes back down to the 300 kN the base takes. The maximum is 1,282.87 kN, 1.60 times the applied load, and it occurs at a depth where nothing is applied, nothing is connected, and nothing will ever be inspected. The dashed line is the diagram the ordinary calculation draws.

Nothing about the soil changes across that depth. The same fill, the same friction angle, the same effective stress, the same interface. What changes is a comparison: above it the ground is going down faster than the pile, and below it the pile is going down faster than the ground. Friction has no opinion about which body is loading which. It opposes relative motion, and relative motion is a difference of two settlements.

The free body, which is the whole pile and not a section of it

Take the entire pile as the free body — head to toe, with the soil cut away and replaced by what it does — and write ΣV=0\Sigma V = 0 down its axis. Which body is drawn is the decision that produces the answer, and a free body cut anywhere short of the toe cannot see this at all. Four things appear on it, and only one of them was designed.

The applied load QQ pushes down at the head. The base resistance QbQ_b pushes up at the toe. Along the shaft there is friction, and it is not one term but two: a downward drag over the length where the ground is settling faster, and an upward shaft resistance over the length where the pile is. The depth that separates them is the neutral plane, and it is the only unknown in the equation.

The unit friction is taken here by the effective-stress route: τ=βσv′=βγ′z\tau = \beta \sigma'_v = \beta \gamma' z, so it grows linearly with depth and the force it delivers over any interval is the area under a straight line. For the pile drawn, β=0.3\beta = 0.3 and γ′=9\gamma' = 9 kN/m³, so the friction at the toe is 64.8 kN/m² and at 1 m depth it is 2.7. With a perimeter of 1.885 m that makes the friction 5.089 kN per metre, per metre of depth. Equilibrium is then

Q+a2zn2  =  Qb+a2(L2−zn2),Q + \tfrac{a}{2} z_n^2 \;=\; Q_b + \tfrac{a}{2}\left(L^2 - z_n^2\right),

a quadratic whose one positive root is

zn=L2+2(Qb−Q)/a2.z_n = \sqrt{\frac{L^2 + 2(Q_b - Q)/a}{2}}.

That is the whole of the model, and every number on this page comes out of it.

The same friction, drawn with the sign it actually has. Unit shaft friction against depth, with its sense rather than its magnitude. It grows linearly with depth in both halves because it is a fraction of the vertical effective stress and that grows linearly — so the largest friction on the whole shaft, 64.8 kN/m², acts at the toe, and the second largest acts just above the neutral plane and points the wrong way. The plane sits at 13.78 m, 57.4 per cent of the way down. Nothing about the soil differs across it: the same sand, the same friction angle, the same effective stress. What differs is which of the two is settling faster.
Fig. 2 The same shaft friction drawn with its sign rather than its magnitude. It grows linearly with depth in both halves, because it is a fraction of the vertical effective stress and that grows linearly — so the largest friction anywhere on the shaft acts at the toe, and the second largest acts just above the neutral plane and points the wrong way. The plane sits at 13.77 m, 57.4 per cent of the way down.

The shape of that drawing is what makes the problem worse than intuition suggests. The drag is not distributed evenly over the length it acts on, in the way a uniform load’s resultant is not where a uniform load is. It is heaviest at the bottom of the length it acts on, immediately above the plane, which is the deepest and least accessible part of the whole arrangement.

The number with nothing in it

Set Q=0Q = 0 and Qb=0Q_b = 0 — a pile carrying nothing at all, resting on nothing, in ground that is settling — and the root collapses to

zn=L2=0.7071 L.z_n = \frac{L}{\sqrt 2} = 0.7071\,L.

There is no soil in that expression. No β\beta, no unit weight, no diameter, no length except as a fraction of itself. A pile doing no work whatever, in any consolidating ground, on any planet, has its neutral plane at 70.71 per cent of its length, and carries at that depth a drag of exactly half its own shaft capacity — 732.9 kN of the 1,465.7 kN the shaft could deliver in the ordinary direction.

That is worth sitting with, because it says the phenomenon is not a soil property being unlucky. It is a consequence of the friction growing with depth and the equilibrium having to close, and the only way to move the plane is to put something at one end or the other. Which is the next question.

Load it more, and the drag goes away

The neutral plane is where it is because of what the pile is carrying. Add load at the head, and the pile has to find more resistance to balance it; the only place that resistance can come from is shaft that used to be dragging and now has to be holding. The plane rises.

A kilonewton added at the head is half a kilonewton inside the pile. The worst axial force anywhere in the pile, and the drag that is part of it, against the load applied at the head. The applied load is a straight line at 45 degrees to the axes because it is itself; the drag falls as the load rises, because the neutral plane climbs to meet the extra load and sheds the friction above it; and the sum of the two rises at a gradient of exactly 0.500. Loading a pile more heavily reduces the drag on it, and the worst force inside it grows at half the rate the load does. At the 800 kN drawn the drag is 482.87 kN and the worst force is 1282.87 kN, 1.60 times the load anybody applied.
Fig. 3 The worst force anywhere in the pile, and the drag that is part of it, against the load applied at the head. The applied load is a straight line at 45° because it is itself; the drag falls as the load rises; and the sum climbs at a gradient of exactly 0.500 over the whole range drawn. Loading a pile more heavily reduces the drag on it.

Differentiate. With zn2=(L2+2(Qb−Q)/a)/2z_n^2 = \left(L^2 + 2(Q_b - Q)/a\right)/2 the drag is D=a2zn2D = \tfrac{a}{2}z_n^2, so dD/dQ=−12dD/dQ = -\tfrac12 exactly, and the maximum force Nmax⁡=Q+DN_{\max} = Q + D therefore satisfies

dNmax⁡dQ=12.\frac{dN_{\max}}{dQ} = \tfrac12.

Every kilonewton added at the head raises the worst force inside the pile by half a kilonewton. The other half is paid for out of drag that no longer happens. Raising the load from 800 kN to 1,600 kN takes the neutral plane from 13.77 m to 5.71 m, the drag from 482.87 kN to 82.87 kN, and the maximum force from 1,282.87 kN to 1,682.87 kN — 800 kN of load for 400 kN of force.

This is why the common statement of the problem, add the drag to the design load, is not merely conservative but structurally the wrong shape. It treats two quantities as independent when one is a function of the other. A pile designed by adding the drag computed at the working load, and then checked at a higher load with the same drag added, is being checked against a state that cannot occur.

The reason it cannot occur is that downdrag is an imposed displacement, not an applied force. It is the same distinction that separates a temperature change from a load and a shrinkage strain from a stress: the ground is not pushing with a force it has decided on, it is moving, and the force is whatever the interface can transmit before it slips. Give the pile more work to do and the interface transmits less.

What it does not excuse

The half-a-kilonewton finding is about the force in the pile, and it says nothing about two other checks.

The first is geotechnical capacity, and here the argument runs the other way. At ultimate load the pile is moving down through the ground everywhere along it, the whole shaft is resisting, and the neutral plane has gone to the head. There is no drag at collapse — so drag must not be added to the load in a bearing-capacity check, and adding it is the double count that this arithmetic exposes. The ground’s own collapse mechanism does not know about consolidation.

The second is settlement, and it is the check that actually governs. The pile has to go down far enough for the ground below the neutral plane to mobilise the resistance the equilibrium demands, and that movement is added to whatever the building was allowed. It is also, awkwardly, the movement that feeds back: the pile going down reduces the relative motion that caused the drag in the first place, so the honest model is a compatibility problem rather than the equilibrium one on this page.

Two beams, or one beam four times as stiff. Two 200 × 150 planks spanning 4 m under 6 per millimetre. Loose, they have 112.5×10⁶ mm⁴ between them and deflect 16.2 mm, with the two faces at the interface sliding past one another. Bonded, the pair has 450.0×10⁶ — exactly 4 times as much, because doubling a depth cubes — and deflects 4.0 mm at half the extreme-fibre stress. Nothing was added but a restraint on slip. With connectors of stiffness 200 the same beam deflects 5.4 mm, which is 89% of the way from one bound to the other.
Fig. 4 The same shape of question at a scale that fits on a bench. Two members connected along an interface that transmits shear only as far as the slip permits — and the whole behaviour of the pair depends on the relative movement across the joint rather than on either component alone. A pile in consolidating ground is that arrangement with one of the two components ten metres thick and unreinforced.

Length is the wrong variable to reach for

The instinct on being shown a pile in trouble is to make it longer, and inside the settling layer that is exactly backwards.

Drag grows as the square of the length, and nothing else does. The worst force in the pile against the length of pile passing through the settling ground. The unit friction grows linearly with depth, so the drag accumulated above the neutral plane is the area under a straight line and therefore a square — 482.87 kN at 24 m and 1408.13 kN at 36.1 m, a factor of 2.92 for half as much length again. Shorter than 14.4 m there is no drag at all, because the whole shaft and the base together cannot reach the applied load and every metre of the pile is holding it up. The head load and the base resistance are held throughout. A longer pile is not a safer one here: every extra metre through the settling layer is another metre of ground hanging on the shaft, and the deepest metres hang hardest.
Fig. 5 The worst force in the pile against the length of pile passing through settling ground, with the head load and base resistance held. Because the unit friction grows linearly with depth, the drag is the area under a straight line and therefore grows as the square: 482.87 kN at 24 m, 1,408.13 kN at 36.1 m. Shorter than 14.4 m there is no drag at all — the whole shaft and the base together cannot reach the applied load, so every metre of the pile is holding it up.

The left-hand end of that curve is worth as much as the right. Below about 14 m the drag is zero, and not because the ground has stopped settling. It is because the pile is working hard enough that it needs every metre of its shaft to stand up, so there is no length left over to be dragged by. A short heavily-loaded pile has no downdrag problem; a long lightly-loaded one has the worst possible version of it. That ordering is the opposite of almost every other check in this collection.

Diameter does something stranger again. Doubling the pile from 600 mm to 1.2 m doubles the perimeter, so the drag rises from 482.87 kN to 1,215.66 kN and the maximum force from 1,282.87 to 2,015.66. But the area went up by four, so the stress fell from 4.54 N/mm² to 1.78. A bigger pile attracts more drag and cares much less about it, which is a scaling argument of the kind that runs through the whole subject with the sign reversed.

The same physics, fifty times smaller

Nothing about the analysis is specific to piles. It is the transfer of an axial force into a member through friction distributed along its surface, and this site has already drawn that object at another scale.

The bond stress is crowded against the loaded end. A 20 mm bar embedded 806 mm, with the force in it and the bond stress on it plotted along the embedment. Uniform bond — the assumption behind every development length ever tabulated — is a flat stress and a straight line of force. An elastic bond of the same peak strength is neither: the slip is largest where the bar is pulled and dies away over 1/α = 471 mm, so the far end of the bar is doing almost nothing. At the code's own length of 40 diameters the elastic bond is 55 per cent used. The uniform answer is what the bond looks like after it has yielded along the whole length, which is a statement about ductility rather than about strength.
Fig. 6 A reinforcing bar developing its force through bond: the same free body, the same integral along a perimeter, the same accumulation of force with distance. What the bar case does not have is a sign change, because the concrete around a bar is not settling past it. Add that one feature and the diagram grows an interior maximum.

The comparison is more than decorative, because it names precisely what is unusual here. Every other member in this collection has a monotone axial force diagram between its load points. This one does not, and the reason is that the sign of the surface traction is decided by a kinematic comparison rather than by the statics — which is a thing no equilibrium calculation can be asked.

What the ground is actually doing

The friction depends on effective stress, so it depends on the water, and the water is the part of the problem most likely to move.

Five loads behind one wall, and the water is the biggest. The horizontal pressure on a 6 m wall retaining soil at 19 kN/m³ with a friction angle of 30°, a surcharge of 10 kPa and the water table 2 m down, drawn once as the profile the wall feels and then once per term. The terms are surcharge 20.0 kN/m at 3.00 m, soil above water 12.7 kN/m at 4.67 m, soil at the water table 50.7 kN/m at 2.00 m, submerged soil 27.2 kN/m at 1.33 m, water 78.5 kN/m at 1.33 m, and they sum to 189.0 kN/m — matched to 7e-8 by integrating the drawn profile numerically. The largest single term is the water, at 78.5 kN/m: water has no shear strength, so its coefficient is exactly one where the soil's is 0.333, and it acts on top of the soil's effective stress rather than instead of it. The combined resultant sits at 1.91 m above the base, 0.319 of the height rather than the third point at 2.00 m that a pure triangle would give.
Fig. 7 The five things behind a retaining wall, of which the water is the largest. The same accounting applies down a pile shaft: what multiplies β is the effective vertical stress, so lowering a water table raises γ′ over that depth and raises the drag with it. Dewatering a site to build on it is a common way of causing the settlement that then loads the piles.

And the parameter itself is not well known. β\beta for a driven pile in a soft clay fill might be quoted between 0.2 and 0.35, which is a range of 1.75 in the drag straight away — and unlike most soil parameters it enters the answer linearly rather than through a root.

A range of 1000 in the ground is a range of 5.6 in the answer. The characteristic length of the same 540 × 10³ kNm² strip on eight soils, each drawn as the band its subgrade modulus is quoted over rather than as a point. From 5 to 5000 × 10³ kN/m³ is a factor of 1000, and 1/β = (4EI/k)^¼ turns it into a factor of 5.62 — the fourth root, 5.62, exactly. So the softest ground here gives 4.56 m and the stiffest 0.81 m, and the design moment P/4β moves by the same 5.62 rather than by 1000. Eight soils span 3.0 decades of stiffness and 0.75 decades of length. Arguing about the subgrade modulus to two figures is not where the uncertainty is.
Fig. 8 The consolation, drawn for a different problem on the same ground. A thousandfold range in the soil’s stiffness compresses to a factor of 5.6 in what a beam on it does, because the answer depends on a fourth root. Downdrag has no such mercy: the drag is proportional to β and to γ′, so the uncertainty arrives at full size.

A group is not nine of these

Piles are rarely alone, and a group changes the question rather than multiplying it.

The group is not weaker; it is very much softer. A 3 × 3 pile cap on the left, with each pile's share of 4.0 MN and 4.5 MNm in meganewtons — N/n plus M·y/Σy², the same three terms in the same order as a bolt group under an eccentric load and a section under biaxial bending. The corner piles take 1.56 times the average and a pile added at the centroid would change that by nothing at all, because it adds to neither second moment. On the right is the effect a bolt group cannot have: the piles share ground, so the stress bulbs overlap and the group settles 3.9 times as much as a single pile at the same load per pile, rising to 14.2 for 144 of them. The capacity check everyone makes — block failure against the sum of the piles — comes out at 4.54 here and does not govern at all. The check nobody tabulates is the one that does.
Fig. 9 A nine-pile group, which behaves as a block rather than as nine members. For downdrag the block matters more than for anything else: the fill inside the group cannot settle past the piles independently of the piles, so the drag on an interior pile is limited by the weight of the fill in its own tributary prism rather than by the friction its perimeter could develop.

For a group at close spacing the sum of the individual drags computed pile by pile can exceed the entire weight of settling fill enclosed by the group, which is impossible: the soil cannot pull down harder than it weighs. The block limit is the correct answer there and it is often a small fraction of the pile-by-pile one. Meanwhile the corner piles of the group have soil on two open sides and get very nearly the full individual value, so a group has a drag distribution as well as a total — and it is the reverse of the load distribution, which is heaviest at the corners too. The two pile up.

Where the model stops

The neutral plane is located by equilibrium alone. The proper location is where the settlement of the pile equals the settlement of the soil, which is a compatibility condition and requires a settlement profile for the ground and a load–transfer curve for the shaft. The equilibrium root used here is the limiting case in which the interface is fully mobilised in both directions, which is a good approximation for a soft fill over a firm founding stratum and a poor one when the ground is stiff and the movements are small.

The friction is fully mobilised everywhere. Real interfaces need a few millimetres of relative movement to reach the values used, and near the neutral plane there is by definition almost none — so there is a transition zone of a metre or two where the friction is passing smoothly through zero rather than reversing at a point. The maximum force is real; the sharp corner in the diagram is not.

The pile is rigid relative to the ground. Its own elastic shortening under the force diagram drawn is 0.39 mm, which is small against the settlements that cause the problem and is not always so. A long slender pile shortens enough to change the relative movement it is being dragged by.

The load is constant. The imposed part of a building’s load comes and goes, and the neutral plane moves with it. A pile loaded to 800 kN under permanent load and 1,200 kN under full imposed load has two neutral planes and two force diagrams, and the worst force is not at the worst load — it is at the lowest sustained load, which is the case nobody draws.

Nothing here is time-dependent. Consolidation stops. The drag is a transient in the life of the structure, however long a transient, and a check that treats it as permanent is checking a state that will pass. Whether it passes before or after the pile has to survive it is a programme question rather than a structural one.

And the drawing cannot show the thing that caused it. Every figure on this page is of a pile. The object doing the work is a layer of fill several metres thick going down by a hundred millimetres over five years, and no figure of a structural member has anywhere to put that.

The ladder from here

Later rungs on this anchor: the compatibility solution, where the neutral plane is found from two settlement curves crossing rather than from a force balance, and where the answer moves. Bitumen slip coatings, which reduce β\beta by an order of magnitude over the drag length and are the standard cure — and the detailing question of where to stop the coating, which is a decision about where the neutral plane will be after the coating has moved it. Downdrag on a raked pile, where the drag has a horizontal component nobody expected. The same phenomenon in reverse as heave drag, where ground swells past a pile and puts it into tension, and where a pile with no tension reinforcement pulls apart at a depth chosen by the same arithmetic. Group effects properly resolved, with the block weight as an upper bound and the corner piles as the governing members. And the transferable half of all of it — a member whose surface traction changes sign along its length has an interior maximum in its internal force, and the shear connectors of a composite beam, the bond along a post-tensioned tendon and the friction under a slab on ground are all the same picture waiting to be drawn.

Where this goes next

The group is not weaker; it is very much softer. A 3 × 3 pile cap on the left, with each pile's share of 9.0 MN and 4.5 MNm in meganewtons — N/n plus M·y/Σy², the same three terms in the same order as a bolt group under an eccentric load and a section under biaxial bending. The corner piles take 1.25 times the average and a pile added at the centroid would change that by nothing at all, because it adds to neither second moment. On the right is the effect a bolt group cannot have: the piles share ground, so the stress bulbs overlap and the group settles 3.9 times as much as a single pile at the same load per pile, rising to 14.2 for 144 of them. The capacity check everyone makes — block failure against the sum of the piles — comes out at 4.54 here and does not govern at all. The check nobody tabulates is the one that does. Nine piles, and four times the settlement A pile cap divides its load between its piles by the same three terms a bolt group uses and a section under biaxial bending uses. What a bolt group does not have is neighbours it shares ground with — and the group effect that matters is not the strength check everybody makes, but a stiffness effect nobody tabulates. The bond stress is crowded against the loaded end. A 20 mm bar embedded 806 mm, with the force in it and the bond stress on it plotted along the embedment. Uniform bond — the assumption behind every development length ever tabulated — is a flat stress and a straight line of force. An elastic bond of the same peak strength is neither: the slip is largest where the bar is pulled and dies away over 1/α = 471 mm, so the far end of the bar is doing almost nothing. At the code's own length of 40 diameters the elastic bond is 55 per cent used. The uniform answer is what the bond looks like after it has yielded along the whole length, which is a statement about ductility rather than about strength. The force that arrives along a length A bolt takes its force at a hole and a weld along a line. A reinforcing bar has no such place — it is a smooth cylinder in a hole of its own shape, and the only thing stopping it sliding out is a stress smeared over its surface. So the force in it is not a number, it is a function of position. The ground pushes back hardest where the beam has gone down furthest. A strip 16.4 m long and 1 m wide on ground of subgrade modulus 50 × 10³ kN/m³, carrying 1000 kN at its centre. The beam settles 3.90 mm under the load and the ground pushes back in proportion — the arrows are k times the settlement above them, peaking at 195 kN per metre — so the pressure diagram is the settlement bowl and not an assumed distribution. The characteristic length 1/β is 2.56 m: the bowl crosses zero at 6.04 m, which is 3π/4 of it, and beyond that the arrows reverse because the beam has lifted off. By 8.05 m — one π/β — the disturbance is 4.3% of what it was, which is why the moment 641 kNm and the peak pressure 195 kN/m contain no length at all. The settlement is drawn 217 times full size — the real bowl is 3.90 mm deep over 16.4 m, about 1 in 4207 — and at true scale the beam would be a straight line. The beam that sits on the ground Every other beam in this collection is held at points. A footing is held everywhere, by something that pushes back in proportion to how far it is pushed — and that single change hands the structure a length it did not choose. Two or three of those lengths from the column, nothing knows the load happened. Two identical pipes, and one carries three times the other. Load per metre on a buried conduit against the depth of cover, in trench widths, with the weight of the prism of soil directly above it drawn between them. A conduit laid in a narrow trench is stiffer than nothing and softer than the sides: the backfill settles relative to the undisturbed ground, the friction on the trench walls acts upward, and the conduit gets 64% of the prism. Lay the same conduit on the ground and build an embankment over it and it is now stiffer than the fill beside it, the interior prism settles less, the friction acts downward, and it gets 172% — a factor of 2.71 between two pipes with nothing different but which way the ground moved. The equation is Janssen's, the same one a silo wall obeys, with a trench for a silo; both curves start on the prism line, because with no depth there is no shear to redistribute anything. This is why a flexible pipe is buried rather than a rigid one: making the conduit weaker moves it down the page. The pipe decides what the soil weighs A buried conduit is not loaded by the soil above it. It is loaded by whatever share of that soil the relative movement leaves it — and which way the shear on the sides of the prism acts depends on whether the conduit settles more or less than the ground beside it. Two identical pipes under identical fill, one carrying two thirds of the prism and one carrying nearly twice it.

Named alongside this one

Essays reaching for the same objects. Nobody chose these; they are what the concept index makes visible.

The objects this essay names

Each one links to every other essay that touches it.

Axial forceCompatibilityDowndragEffective stressLoad pathNeutral planeSettlementShaft friction