The pressure that stops growing

Water in a tank presses on the wall with $\gamma h$. Nothing stops it: go deeper and the pressure grows, in exact proportion, forever.

Grain in a silo does not do that, and the difference is one property that water does not have. Grain has shear strength. It can transmit a shear stress to the wall it is resting against, and once it does, part of its own weight is being carried by the wall rather than by whatever is underneath.

That is not a small correction. At the base of a 30 m fill in an 8 m silo the vertical pressure is 85 kN/m² against a liquid’s 270, and the wall is carrying 69% of the stored weight.

Which free body produced the number #

A horizontal slice of the stored solid, of thickness $dz$, across the whole cross-section.

Four things act on it. Its own weight, $\gamma A\,dz$, downward. The vertical pressure from the slice above, $p_v A$, downward. The vertical pressure from the slice below, $(p_v + dp_v)A$, upward. And the friction on the wall around its perimeter, $\mu p_h U dz$, upward, where $p_h$ is the horizontal pressure pushing the solid against the wall.

Sum them:

$$A\,dp_v = \gamma A\,dz - \mu p_h U\,dz$$

Now the closure that makes it solvable. The horizontal pressure is taken as a fixed fraction of the vertical one, $p_h = K p_v$ — the same assumption every earth-pressure argument on this site makes, and just as much an assumption here. Substituting and writing $R = A/U$ for the hydraulic radius:

$$\frac{dp_v}{dz} = \gamma - \frac{\mu K}{R}p_v$$

which is a first-order linear equation whose solution is

$$p_v(z) = \frac{\gamma R}{\mu K}\left(1 - e^{-\mu K z/R}\right)$$

The derivative goes to zero. As $p_v$ grows, the friction it generates grows with it, and at $p_v = \gamma R/\mu K$ the friction is carrying the whole of each new slice’s weight and the pressure stops rising.

One term in that slice is not a force the material supplies willingly, and the solution has already decided what it is.

The three numbers in it, and what they do #

$R = A/U$, the hydraulic radius. For a circle it is $D/4$ — 2.0 m for the 8 m silo here. It is the only geometry in the answer, and it is the ratio of what is being carried to what is available to carry it by friction. A wide silo has a large $R$, a high ceiling and behaves nearly like a tank; a narrow one has a small $R$ and saturates almost at once.

$\mu$, the friction between the solid and the wall. Not the internal friction of the material — that is a different number and it is $K$’s business. This is a contact property, and it depends as much on the wall as on the contents: a smooth-lined steel silo has half the wall friction of a concrete one, which halves the saturation depth’s denominator and doubles the ceiling.

$K$, the ratio of horizontal to vertical pressure. Between about 0.4 and 0.6 for granular solids at rest, and the least well known of the three. It is the same $K$ that appears in earth pressure and it carries the same caveat: it depends on the state of the material rather than on the material.

All three appear only in the products $\gamma R/\mu K$ and $R/\mu K$. The problem has two parameters, not four.

The exponent is the capstan’s #

The equation $dp_v/dz = \gamma - (\mu K/R)p_v$ has a first cousin on this site, and the relationship is exact rather than poetic.

A rope wrapped round a bollard obeys $dT/d\theta = \mu T$, giving $T = T_0 e^{\mu\theta}$ — the capstan equation, and the reason a sailor holds a hawser under tons of load with one hand. The mechanism there is: friction is proportional to the normal force, the normal force is proportional to the tension, so the tension grows exponentially in the wrapped angle.

The mechanism here is: friction is proportional to the horizontal pressure, the horizontal pressure is proportional to the vertical pressure, so the vertical pressure approaches its ceiling exponentially in the depth.

Same statement, same equation, different variable. Feeding this silo’s $\mu K$ and its depth-to-radius ratio into the capstan routine gives $e^{3.038} = 20.85$, which is exactly the exponential factor in the silo solution.

The two problems even share their surprise. In both, the exponential means the answer is dominated by a ratio — wrapped angle for the rope, depth over hydraulic radius for the silo — and both saturate so fast that the practical question becomes “how many turns” or “how many diameters” rather than anything about magnitude.

How much the wall carries, and why slenderness decides #

The base takes $p_v(H) \times A$ and the wall takes the rest, so the wall’s share is

$$1 - \frac{p_v(H)}{\gamma H}$$

which depends on nothing but the ratio $H/D$ once the material and the wall are fixed:

Past about four diameters the base pressure has stopped changing altogether — it is at the ceiling — so everything added above that point goes into the wall and nothing goes into the floor.

That single row is the whole reason silos are structurally different from tanks. A tall silo’s wall is a vertical load-carrying element, carrying tens of thousands of kilonewtons of stored material by friction, on top of resisting the horizontal pressure. The failure that follows is a vertical buckle of the wall rather than a burst — which is not what anybody expects a silo to do.

That makes the wall a surface that carries by being curved in two directions at once: a thin cylinder in hoop tension from the horizontal pressure, and in vertical compression from the friction. The friction hands it 371 kN per metre of perimeter at the base here — a membrane force that nothing about the stored material’s weight suggests would be there, and the one that decides the wall’s thickness against buckling.

The same equation with a different name #

Silos are not the only place a saturating pressure appears, and the family is worth naming because the mechanism transfers exactly.

A trench. The soil in a narrow trench arches across it and hands part of its weight to the sides, so the vertical stress on a pipe at the bottom is far less than $\gamma H$ — which is the trench condition every buried pipe is designed for, and it is Janssen with the trench walls in place of the silo’s.

A tunnel. The ground above a tunnel arches over it for the same reason, and the load the lining is designed for is a fraction of the overburden. The fraction is set by a hydraulic radius that here is the tunnel’s own width.

A pile. A pile in sand carries load by shaft friction, and the shaft friction depends on the horizontal stress against the pile, which depends on the vertical stress in the ground — and the vertical stress stops growing with depth once the pile’s own zone of influence starts arching. The observation that shaft friction reaches a “critical depth” and then stops increasing is the same saturation in the same equation.

What the three share is the ingredient water does not have: a material that can transmit shear to a boundary, and a boundary close enough that the shear matters. The hydraulic radius is the measure of “close enough”, and everything else follows.

What the material is doing in every one of them is forming an arch nobody drew. Arching in a granular solid is a thrust line finding its way to two supports, the same object a three-pinned arch has its hinges put in on purpose to make determinate; the silo’s supports happen to be its own walls, and the thrust delivered to them is the friction this page is about.

The horizontal pressure, which is what bursts it #

$p_h = Kp_v$ saturates too, at $\gamma R/\mu$ — 38 kN/m² at the base of the silo drawn.

For a circular silo that produces a hoop tension of $p_h D/2 = 152$ kN per metre of height, which is the number the wall’s horizontal reinforcement or its plate thickness is sized on. It is constant over the lower part of a tall silo, because the pressure has saturated — so a tall silo’s hoop reinforcement is uniform below a few diameters and graded above, which is the opposite arrangement to a tank’s.

For a rectangular silo it is much worse. A flat wall has to span between corners in bending rather than carry the pressure as a hoop tension, so the same 38 kN/m² produces a moment rather than a membrane force, and rectangular silos are correspondingly heavier for the same capacity. The hydraulic radius does not care about the shape — a square silo of 8 m side has exactly the same $R = 2$ m and the same pressures — so the difference is entirely in what the wall does with them.

One consequence of the saturation is easy to get wrong on the way out. A load spread over a surface has a resultant, and where that resultant acts is decided by the shape of the distribution rather than by its total: a liquid’s triangle puts it at $H/3$, while the silo’s profile is very nearly uniform over the lower part of a tall wall and puts it much higher. Any overturning or foundation calculation on the wall needs the right one, and the two differ by a sixth of the height.

One length, and why a tabletop model predicted full-size silos #

Write the solution with the characteristic depth in it and something becomes visible that the four parameters conceal:

$$p_v(z) = \gamma z_0\left(1 - e^{-z/z_0}\right), \qquad z_0 = \frac{R}{\mu K}$$

There is one length in the problem, and the depth enters only as the ratio $z/z_0$. Everything else — the friction coefficient, the pressure ratio, the diameter — has been absorbed into it, and the unit weight sets the scale of the pressures without affecting their shape at all.

The consequence is that all silos are the same silo. Two of them holding the same material against the same wall, one 8 m across and one 16 m, have identical curves once depth is measured in units of $z_0$; the second simply has a $z_0$ twice as large, so it reaches the same fraction of its ceiling at twice the depth. A silo is “slender” or “squat” according to $H/z_0$, and that ratio is what the table two sections up is really tabulating — $H/D$ stands in for it only because the material and the wall were held fixed.

This is why Janssen’s tabletop experiment settled the question for structures a hundred times its size. He was not measuring silos; he was measuring the shape of a dimensionless curve, and the shape does not know how big the apparatus is. It is the same argument that makes any model test worth doing, and it is unusually clean here because the problem has exactly one dimensionless group in it.

Two design consequences fall straight out, and they are worth stating because they cut against instinct.

Every pressure scales with the diameter and none of them with the height. The ceiling is $\gamma z_0 = \gamma R/\mu K$, so doubling the diameter doubles the vertical pressure at the base, doubles the horizontal pressure, and doubles the hoop tension per metre — while quadrupling the stored volume. Capacity bought by widening is paid for at the full rate. Capacity bought by heightening a silo that is already several $z_0$ deep costs nothing in pressure at all, because the pressure stopped rising; it costs only more wall, carrying more accumulated friction.

And the ceiling contains no reference to how much is stored. A silo filled to four diameters and a silo filled to forty put the same horizontal pressure on their walls at the base and the same vertical pressure on their floors. That is the property this collection has met once before, in a substructure whose uplift does not grow when the building does — an action with a ceiling set by geometry rather than by quantity — and it is rare enough in structures to be worth recognising when it appears.

The number a designer actually needs #

Three quantities come out of the solution and each sizes a different part of the structure.

The horizontal pressure sizes the wall’s hoop reinforcement, and it saturates — so on a tall silo it is a constant below a few diameters, at $\gamma R/\mu$.

The wall friction sizes the wall’s vertical reinforcement or its plate thickness against buckling, and it accumulates — 371 kN per metre of perimeter at the base here, growing all the way down.

The base pressure sizes the floor and the foundation, and it saturates too, at a value that on a slender silo is a small fraction of the stored weight divided by the area.

The uncomfortable part is that the three are not conservative in the same direction. A low estimate of $\mu$ gives a high base pressure and a low wall friction; a high one gives the opposite. So there is no single conservative choice of the friction coefficient, and a silo has to be checked with a high value for the wall and a low value for the floor — two calculations with two different numbers for the same physical property.

That is a shape worth carrying past this page. A parameter that appears on both sides of a load path has no conservative value, and the honest response is two load cases rather than a cautious number.

The floor is the plainest instance of it. What it is designed for is a pressure whose whole uncertainty sits in the load and none of it in a resistance — the character it shares with the class of checks where weight is the only thing resisting, and the reason a silo floor is sized by a number nobody can measure to better than a factor of two.

Where the model stops #

Everything is at rest. The Janssen solution is the filling case, and silos fail while emptying. When the solid starts to flow, the material at the wall moves down relative to the material at the centre, the principal stress directions rotate, and $K$ jumps — sometimes by a factor of two or three, in a band that travels up the silo as a “switch”. The pressure spike that follows is the single commonest cause of silo failure and this equation does not contain it.

$K$ is a constant. It is not: it depends on the stress state, on whether the material is being loaded or unloaded, and on how the silo was filled. The whole solution is an envelope around a number nobody can measure well.

And the pressure is uniform across the section. It is not — the vertical stress is higher at the centre than at the wall — so $p_v$ here is a mean, and the friction is computed from a $p_h$ derived from that mean rather than from the pressure actually at the wall.

Nor is the filling symmetric. Eccentric filling or eccentric discharge produces a pressure distribution that is not axisymmetric, which puts bending into a wall designed for hoop tension, and is the second commonest cause of failure.

What the pictures cannot show #

The curves are drawn for a silo that has been filled and left. A real one is filled, emptied, and filled again, dozens of times a year, and the pressures are different every time depending on how fast and from where.

Nor can they show the material. Grain, cement, coal and wood chips have wall friction coefficients spanning a factor of three and $K$ values spanning a factor of two, so the same silo behaves quite differently depending on what is in it — and a silo designed for one product and later used for another is a structure being asked a question it was not designed for.

The assumption the figure rests on #

$p_h = Kp_v$, with $K$ constant and known. Everything downstream — the saturation, the depth, the wall’s share, the hoop tension, the exponent — is that one closure. It is the same closure earth-pressure theory makes, it is unavoidable if a one-dimensional equation is wanted, and it is the reason silo design codes wrap the elegant result above in a set of multipliers whose job is to cover the range $K$ actually takes.

There is a second assumption of the same kind sitting beside it, and it is the one the first figure of the free-body section drew. The friction is at its limit everywhere. Janssen’s solution assumes full mobilisation over the whole height, which requires the solid to be settling relative to the wall all the way down — true while a silo is being filled, and not true afterwards. A silo standing full and undisturbed has less friction on its walls than the equation credits it with, and correspondingly more pressure on its floor.

The history, and the year it was measured #

Janssen published this in 1895, and what makes the paper worth remembering is not the derivation — the differential equation is a page — but that he measured it first.

He built a model silo with a floor that could be weighed independently of the walls, filled it with grain, and read off how much of the fill the floor was carrying as the depth grew. The measurement showed the floor’s share falling toward an asymptote; the equation came afterwards, to explain a curve that was already on paper.

That order matters because the closure $p_h = Kp_v$ is not derivable from anything. It is a modelling choice, and the only reason to believe it is that the solution it produces has the shape the measurements have. A century and a quarter of silo design rests on a fitted constant in a one-dimensional model — which is a fair description of a great deal of this subject, stated plainly for once.

The ladder from here #

Later rungs on this anchor: the discharge case and the switch pressure, where $K$ changes and a travelling band of high pressure runs up the wall. Mass flow against funnel flow, and how the hopper geometry decides which one happens. Eccentric discharge, which is the load case that turns a hoop-tension problem into a bending one. The hopper itself, where the wall is inclined and the friction has a vertical and a horizontal component. Silo quaking, a self-excited oscillation driven by stick-slip at the wall. And the measurement problem underneath all of it: why $\mu$ and $K$ are quoted as ranges, and what a designer does with a load that has a factor of two of genuine uncertainty in it.