The beam that sits on the ground

Every beam so far in this collection has been held at points. Two supports, or three, or a wall at one end — and between them nothing, so the whole of the beam’s job is to carry the load across the gap to the places where the ground is waiting.

A footing does not work like that. Lay a concrete strip on soil, put a column on it, and there is no gap. The ground is under every millimetre of it, and it pushes back, and how hard it pushes at any point is decided by how far the strip has gone down at that point. There is no span to cross. There is nothing to cross to.

The picture answers a question nobody had thought to ask, which is where the disturbance ends. A beam on two supports is disturbed along its whole length by definition — the moment diagram runs from one support to the other and nothing in it stays at zero. This one goes quiet: six metres out the strip is neither pressed into the ground nor pulled out of it, and past eight metres nothing is happening at all, on a strip that could be a hundred metres long.

The equation with one more term #

The ordinary beam equation says that bending stiffness times the fourth derivative of the deflected shape equals the load intensity:

$$EI\,y'''' = q.$$

Winkler’s model adds one term. The ground is replaced by a bed of independent springs, each of stiffness $k$ per unit length of beam, and the spring under any point pushes up with $k y$:

$$EI\,y'''' + k\,y = q.$$

That is the whole modification, and it changes the character of the problem completely. The homogeneous equation $EI y'''' + ky = 0$ has four exponential roots arranged on a square in the complex plane, and every one of them carries the same length scale:

$$\beta = \left(\frac{k}{4EI}\right)^{1/4}, \qquad \frac{1}{\beta} = \text{the characteristic length}.$$

$1/\beta$ is 2.56 m for the strip above. Nothing in the problem chose it. Neither the strip’s length nor the column load appears in it; it is manufactured out of the beam’s stiffness and the ground’s, and the structure then has to live at that scale.

For a point load $P$ on a beam long enough that its ends do not matter, the solution is three functions of $\beta x$ and nothing else:

$$y = \frac{P\beta}{2k}e^{-\beta x}(\cos\beta x + \sin\beta x), \quad M = \frac{P}{4\beta}e^{-\beta x}(\cos\beta x - \sin\beta x), \quad V = -\frac{P}{2}e^{-\beta x}\cos\beta x.$$

The two peaks are worth reading slowly. The settlement under the load is $P\beta/2k$ and the moment there is $P/4\beta$, and neither contains the length of the beam. A footing is not longer or shorter in the arithmetic that sizes its reinforcement. It is only longer or shorter in metres.

Which free body produced 195 kN per metre #

The number in the caption comes from a body small enough to draw in the margin.

Cut a slice of the strip between $x$ and $x + dx$. Four things act on it: a shear and a moment on each cut face, whatever load $q\,dx$ is applied from above, and the ground’s push $k y\,dx$ from below. Vertical equilibrium gives $dV/dx = ky - q$; moments about one face give $dM/dx = V$; and $M = -EI y''$ from the usual bending assumption. Eliminating $M$ and $V$ returns the equation above.

The term worth naming is $ky\,dx$. The contact pressure is not applied. It is a reaction whose magnitude is set by the answer — it cannot be drawn on the free body until the free body has been solved, which is where a redundant structure puts an analyst and for the same reason: the ground and the beam must agree about a displacement, which is a compatibility condition rather than an equilibrium one. It is the same category of load as retained soil pushing on a wall.

The second free body settles whether the solution is a solution. Cut the strip at the column and take the half to the right. The shear on that face is $-P/2$, which is all statics can say; everything else acting on that body is contact pressure, and it has to add to $P/2$. Integrating $k y$ outwards gives

$$\int_0^{\infty} k\,\frac{P\beta}{2k}\,e^{-\beta x}(\cos\beta x + \sin\beta x)\,dx = \frac{P}{2}$$

exactly, in closed form. Taking moments of the same distribution about the cut gives $P/4\beta$ — the 641 kNm in the caption — because $\int_0^{\infty}x e^{-\beta x}(\cos\beta x + \sin\beta x)dx$ is exactly $1/2\beta^2$. The closed form and a finite-difference solve of the same differential equation, sharing no code with it, agree to two parts in ten thousand of the peak.

Everything happens in βx #

Since the solution is three functions of $\beta x$, everything interesting about it happens at a fixed value of $\beta x$ — the same value for every beam, every soil and every load there has ever been.

Four stations, and the last of them is the argument of the essay. $e^{-\pi} = 0.0432$: one $\pi/\beta$ from the column — 8.05 m for this strip — the beam has forgotten. Past two or three characteristic lengths, nothing knows the load happened. Concrete out there is not in the load path and would leave the answer unchanged by being absent.

The station at $3\pi/4$ deserves its own sentence. At 6.04 m the settlement is zero, and beyond it the solution says the strip is pulled up, reaching 4.3% of the peak at 8.05 m. That is a real prediction for a strip cast into the ground and a fiction for one merely resting on it — a point a later section returns to.

The length the structure did not choose #

The contrast with an ordinary beam is sharpest in what sets the scale of the movement.

A simply supported beam has one length in it and a designer picked it. Halve the span and the deflection falls by a factor of sixteen, because the deflection goes as the fourth power of a length that was chosen. The beam on the ground has a length nobody picked, and halving anything a designer controls does not halve it — $1/\beta$ moves as the fourth root of $EI/k$, so quadrupling the slab’s stiffness lengthens the disturbance by $\sqrt{2}$ and no more. The same exponent that is a punishment in the span-governed world arrives upside down here, and is a mercy.

There is one more thing an ordinary beam does that this one does not. On discrete supports, a support that moves is a defect, and an expensive one.

Ten millimetres of settlement adds forty per cent to the peak moment there, and the same movement at a real support is one of the most awkward load cases in structural engineering. On the elastic foundation the settlement is not a defect at all: it is the mechanism by which the ground carries the load, and 3.90 mm of it is what generates the 195 kN per metre holding the strip up. Movement that has to be designed against in one place is the load path in the other.

Three decades of soil, three-quarters of a decade of consequence #

The fourth root has a practical consequence large enough to change how a geotechnical report should be read.

Soft ground is neither obviously the worse case nor obviously the better one. It settles more and bends more and presses less hard at the peak — which is why a footing designed for one soil and built on another is usually still a footing rather than a mistake.

Three decades of soil, three-quarters of a decade of consequence. Worth saying plainly, because it inverts where the anxiety usually sits: the one number a site investigation is least sure of is the one the answer is least sensitive to. The load, known to a few per cent, appears to the first power; the soil, known to a factor of three, appears to the fourth root, and a factor of three in $k$ is 1.32 in the answer.

A raft is not a beam spanning between columns #

The commonest way to get a raft wrong is to treat it as an upside-down floor: assume a uniform contact pressure $P/L$, take the columns as supports and the pressure as the load, and compute $PL/8$. That calculation is not merely approximate. It is wrong in two directions at once, and the two errors are reciprocals.

The reciprocity is neither a coincidence nor an approximation. Both ratios are built out of the same $\beta L$, so their product is identically one, and there is exactly one raft length at which a rigid calculation is right about both things — $L = 2/\beta$, 5.13 m here. Every other length is wrong in both.

The directions matter more than the magnitudes. The rigid calculation overstates the bending moment, which shows up as reinforcement nobody needed, and understates the peak contact pressure, which shows up as settlement nobody predicted. One error is expensive and safe; the other is cheap and not.

A factor of 3.12 on a 16 m raft, growing linearly with every further metre, is the practical form of the claim that the peak moment contains no length. The raft is not the span. Whatever the concrete’s outline, the structure carrying each column is the two or three characteristic lengths around it; the rest is a floor slab that happens to be at the bottom of the building.

This is the stiffest path taking the load, applied along one member rather than between members. The split is decided by relative stiffness rather than by geometry — not by tributary width, not by the decisions that give a beam its load — and here the competitors are the strip and the ground beneath it.

The pressure that cannot pull #

The uplift beyond $3\pi/4$ is where the model has to be watched, and it connects to an older result on this site.

Soil takes compression and nothing else. The middle-third rule is that requirement written for a section, and a footing base is where it is usually met first: a footing under a column with any moment on it has a resultant that can wander out of the kern and leave part of the base wanting to lift.

The Winkler solution above produces negative contact pressure past 6.04 m and takes it entirely seriously, because the springs in the model can pull. Real ones cannot. Re-solving with springs that go slack in tension shortens the disturbance a little and raises the peak pressure a little, and the correction is small precisely because the uplift it removes is only 4.3% of the settlement. Under a footing already lifting off from an applied moment the correction is not small, and the linear solution is not a slight overestimate but the wrong problem.

The same fourth root, from a curved surface #

The most surprising place this arithmetic turns up is not a foundation at all.

A thin shell carries load in its own surface and needs almost no thickness to do it. At its edges, where the membrane solution cannot satisfy the boundary conditions, bending is forced back in — and that edge disturbance dies away over a length of $\pi/[3(1-\nu^2)]^{1/4}\sqrt{Rt}$. Same fourth root, same equation: a cylindrical shell’s radial equilibrium is $D w'''' + (Et/R^2) w = p$, in which the curvature term $Et/R^2$ does exactly the job the springs do here.

A shell is a beam on an elastic foundation, and the foundation is its own curvature. One argument arrives from a bed of springs and the other from a curved surface; they meet at a fourth root, and both then find the disturbance dead within a couple of characteristic lengths — so an edge stiffener on a shell, like the far end of a raft, is a local matter rather than a global one.

Where the model stops #

The springs do not know about each other. Real soil is a continuum, so pressing down at one point drags its neighbours down too. A rigid plate on a Winkler bed settles uniformly with uniform pressure; the same plate on a real elastic half-space settles uniformly with pressure peaking at its edges, theoretically without bound. Winkler misses that entirely, which is why raft edges crack in ways this analysis does not predict.

$k$ is not a property of the soil. It is a property of the soil and the size of the loaded area, because a wider footing stresses a deeper volume. A value measured with a 300 mm plate is not the value for a 16 m raft, and using it as though it were is a far larger error than any uncertainty in the table.

Everything here is elastic and instantaneous. A clay settles for years, and the long-term settlement is a consolidation problem with no $k$ in it at all. This calculation gives the immediate movement and the internal forces, not the settlement anybody argues about in a dispute.

The load is a point. A real column delivers through a base a fraction of a metre across, and the peak moment of that distributed load is slightly below $P/4\beta$ — a small correction while the base is small against $1/\beta$.

The strip is prismatic and the ground uniform. A raft thickened under a column changes $EI$, therefore $\beta$, therefore the length over which everything is happening — so the thickening changes the problem it was added to solve.

The figures cannot show the one thing a reader most wants to see, which is that the bowl is real: 3.90 mm deep over 16.4 m, a slope of about 1 in 4207, drawn here between 110 and 217 times full size. At true scale each of these pictures is a straight horizontal line with a row of arrows under it — and the arrows would still be right, because the contact pressure is a genuine internal quantity revealed by cutting rather than an artefact of the exaggeration. What is exaggerated is the deflection. What is not is the moment, the pressure, or the place where the strip stops touching.

The second thing no picture here shows is the assumption underneath all of it: that the settlement at a point depends on the pressure at that point and nowhere else. Every number in this essay rests on that; it is false, and it survives because the fourth root is forgiving enough that a badly wrong $k$ gives an only slightly wrong $1/\beta$.

History, briefly #

Emil Winkler published the spring bed in 1867, and it was applied almost at once to the thing it was invented for: railway sleepers, which are short beams on ballast carrying concentrated loads. Zimmermann worked out the consequences for track in 1888, and the arrival of a second wheel one characteristic length away — where the first wheel’s disturbance has fallen only to 51% — is why rail seat design was a superposition problem from the start.

The reference treatment is Hetényi’s Beams on Elastic Foundation of 1946, and the classification by $\beta L$ used in the raft figures is his: below $\pi/4$ a beam behaves rigidly, above $\pi$ it behaves as though infinitely long, and between the two it is genuinely finite and the algebra much worse. A 16 m raft on ordinary ground is comfortably in the second category, which is what makes the rigid calculation indefensible rather than merely rough.

The ladder from here #

Later rungs on this anchor: the four Hetényi functions and the finite-beam solution, where end conditions stop being negligible. Two loads a characteristic length apart, and the superposition that decides sleeper spacing. The moment under a wall load rather than a column, which is the plane-strain version and has a different $\beta$. Beams on a bed that cannot pull, solved by iteration on the contact length. The elastic half-space alternative and the edge singularity Winkler cannot produce. Coupled springs — Pasternak and Vlasov — and the second length scale they introduce. The subgrade modulus as a function of footing size, which is the honest version of the third section here. Pile groups, where the beam runs vertically and the springs run along it. And the shell edge disturbance derived properly, so that the two fourth roots can be set beside each other and seen to be one equation.

The argument that carries furthest is not about foundations at all. A structure supported continuously by something proportional to its own movement acquires a length nobody specified, and past a few of those lengths it stops responding. True of a raft, a rail, a shell edge, a pipeline in soil, a floating ice sheet under a vehicle and a beam on elastomeric bearings — and in each case the first useful question is not how long the structure is but how many characteristic lengths long it is.