The pipe decides what the soil weighs

Ask what load a buried pipe carries and the obvious answer is the weight of the soil above it: the unit weight, times the width, times the depth of cover. It is a tidy answer, it takes one line, and it can be out by a factor of nearly three in either direction.

The reason is that soil is not a fluid. It has shear strength, and shear strength means that a column of it can hand part of its weight to the column beside it — if the two columns move relative to one another. Which one moves down relative to which is decided by the conduit, and the sign of the answer follows from it.

Which free body produced the number #

A horizontal slice of the prism of backfill directly over the conduit, of width $B$ and thickness $dh$.

Four things act on it. Its own weight, $\gamma B\,dh$. The vertical stress from the slice above, $\sigma B$. The vertical stress it delivers to the slice below, $(\sigma + d\sigma)B$. And, on its two vertical faces, a shear — because the sides of the prism are rubbing against the ground beside them.

The shear is friction, so its magnitude is $\mu$ times the horizontal stress on the face, which is $K\sigma$. Two faces gives $2K\mu\sigma\,dh$, and vertical equilibrium is

$$B\,d\sigma = \gamma B\,dh - 2K\mu\sigma\,dh$$

if the prism is settling relative to the sides, so that the shear acts upward and takes load away. Integrating,

$$\sigma = \frac{\gamma B}{2K\mu}\left(1 - e^{-2K\mu h/B}\right)$$

which is exactly the pressure that stops growing — Janssen’s silo equation, with a trench for a silo and backfill for grain. The same equation, the same free body, and a completely different question.

The load on the conduit is that stress times its width, conventionally written $W_c = C_d\gamma B^2$ with $C_d = (1 - e^{-2K\mu H/B})/2K\mu$, and $C_d$ is always less than $H/B$. A rigid pipe in a narrow trench carries less than the weight of the soil over it, and the deeper it is buried the smaller the fraction.

The sign that flips #

Now lay the same pipe on the ground and build an embankment over it.

There is no trench and no undisturbed side; there is an interior prism directly over the pipe and an exterior prism either side of it, both of the same fill. The pipe is stiffer than the fill, so it settles less than the fill does. The interior prism therefore settles less than the exterior one, the shear on the interface acts downward on it, and the sign in the differential equation reverses:

$$C_c = \frac{e^{2K\mu H/B_c} - 1}{2K\mu} > \frac{H}{B_c}$$

The exponential goes the other way and the coefficient grows without limit. At three widths of cover, $C_c$ is 1.7 times $H/B_c$; the pipe carries 172% of the prism above it, and the excess is soil that belongs over somebody else.

Two identical pipes, identical fill, identical depth: 64% in one case and 172% in the other. And nothing physical distinguishes the two situations except which direction the ground moved relative to the conduit, which is a consequence of the construction method and not of the design.

The third case, which is what is actually built #

There is a way to have the trench answer without the trench, and it is why the pipes going into the ground today are mostly plastic.

Make the pipe softer than the fill beside it. It then deflects under load, the interior prism settles more than the exterior one, the shear acts upward, and the pipe receives less than the prism weight — the trench answer, in an embankment, achieved by the pipe rather than by the excavation.

What the pipe then does with that load is the second half of the argument and is at least as counter-intuitive. A flexible pipe does not carry its load in bending. It deflects into an ellipse, pushes outward at its springings into the soil either side, and mobilises a horizontal soil pressure that holds its shape. The standard expression for the deflection — the Iowa formula — has the two stiffnesses in series in its denominator:

$$\Delta x = \frac{D_L K W r^3}{EI + 0.061E'r^3}$$

and for a plastic pipe in ordinary compacted fill, $0.061E'r^3$ is about 99% of that denominator. The pipe contributes one per cent of its own stiffness against ovalling; the trench contributes the rest.

Which turns pipe specification into trench specification. A pipe of a given stiffness in well-compacted granular fill deflects 2% of its diameter and in poor fill deflects 18% — a factor of nine, from the material nobody is buying.

Where the plane of equal settlement is #

The embankment case has a limit the equation above does not show, and it is worth knowing because the exponential otherwise grows without bound.

The interior and exterior prisms settle differently only while the shear between them is being mobilised. Higher up the embankment the two have equalised, because the difference in settlement introduced at the conduit has been absorbed by the compression of the fill — and above that level, the plane of equal settlement, there is no relative movement and no shear at all. Fill above the plane simply loads both prisms as a uniform surcharge.

So the coefficient grows exponentially only up to the plane and linearly above it, and where the plane sits depends on how much the conduit and the ground beneath it settle relative to the fill either side. Marston and Spangler parametrised that with a settlement ratio and a projection ratio, and the honest position is that neither is knowable to better than a factor of two on a real job. The complete-projection case — the plane above the top of the fill, which is the upper bound — is what the exponential above computes, and it is what a careful designer uses because the alternative is a calculation resting on two guessed numbers.

Which produces one of this collection’s cleaner examples of a defensible conservatism: use the bound, know that it is a bound, and know which parameter would relax it. That is a different intellectual position from applying a factor of safety, and a better one.

What the pipe is actually checked against #

It is worth separating the three checks a buried pipe gets, because the arching argument feeds each of them differently.

A rigid pipe is checked in bending, as a ring carrying its load with a moment at the crown, invert and springings — which is the force that is only a radius with the uniform part removed and the second harmonic left behind. The load is the whole of the answer, so all of the arching argument goes straight into it, and the difference between the trench and the embankment cases is the difference between one wall thickness and another.

A flexible pipe is checked on deflection, not strength, and the check is a serviceability one about the joints staying sealed and the ovalling staying below about 5%. The load matters, but the trench matters more, and a designer’s effort belongs in the compaction specification rather than in the pipe.

And both are checked for buckling, because a ring in compression under external pressure can collapse into a two-lobed shape. The ground restrains that mode strongly — the pressure that needs no direction works out how much — so it only governs where the ground is soft, saturated or has been washed out.

The interesting consequence is that the three checks want different things. Bending wants a small load, which wants a narrow trench. Deflection wants stiff side fill, which wants a wide trench and good compaction beside the pipe. Buckling wants the fill to stay put. A narrow trench, which minimises the load, is also the hardest place to compact fill beside a pipe — and a great many failures of buried pipe are compaction failures in trenches too narrow to get a rammer into.

Where the model stops #

Once the mechanism is understood, the movement becomes a design variable — and there is a technique that does exactly that.

The induced trench, or imperfect ditch, is used where a rigid culvert has to go under a high embankment and the embankment load would be intolerable. Build the fill up around and over the culvert; then dig a trench in the fill directly above it and backfill that trench with something soft — straw historically, compressible foam now — and continue the embankment over the top.

The compressible layer settles far more than the fill either side. The interior prism therefore settles more than the exterior, the shear reverses to upward, and the culvert receives the trench answer even though it is under twenty metres of embankment. Reductions to a third of the embankment load are routine.

That is a remarkable thing to be able to do: the load on a structure has been reduced by putting something weak above it, and no member has been strengthened. It belongs to the same family as the part that is meant to be weak and as a lining with joints in it — cases where the design decision is to make something softer so that the forces go somewhere else.

Why the load is an imposed deformation, again #

The thread running through this is one the other essays about imposed deformation keep meeting.

The pipe’s load is not applied. It is the consequence of a relative displacement between two columns of soil, and the amount of load that gets transferred is proportional to how much the two are prevented from moving relative to one another — which is proportional to the stiffness of whatever is preventing it. That is the arithmetic of a restraint, not of a load.

So the same three statements hold here as for a tunnel lining and for a member built to the wrong length: a stiffer element takes more; a softer element takes less; and the way to reduce the force is to remove the restraint rather than to strengthen the member carrying it.

What makes the buried pipe the clearest of the three is that the sign is visible. The shear on the prism’s sides is an arrow that points up in one case and down in the other, and the arrow’s direction is the whole answer.

Where the model stops #

The prism’s sides were assumed to slip fully. Full mobilisation of $\mu$ needs relative movement, and a rigid pipe in a shallow trench may not produce enough of it — in which case less arching occurs than the equation says and the load is higher. The equation is unconservative in exactly the shallow case where it is most likely to be applied without thought.

$K$ and $\mu$ were separate constants and they are not. The product $K\mu$ is what appears, it is usually taken between 0.11 and 0.19 depending on the fill, and it sits in an exponent — so the uncertainty in it is amplified. Two defensible values of $K\mu$ can give conduit loads differing by 40%.

The trench had vertical sides. A battered trench has a width that grows with height, which changes the geometry of the free body and reduces the arching; and a trench in rock with a wide bench at the top of the pipe is not the case at all.

Water was left out. A saturated trench has no arching worth the name: below the water table the effective stress on the prism’s sides collapses, the friction with it, and the pipe receives something much closer to the full prism weight plus the water pressure. A trench that fills after a storm is a different structural problem from the one designed, and it arrives without notice.

A live load at the surface was left out too. A wheel load on a shallow pipe spreads through the fill and arrives as a pressure that has nothing to do with the arching argument — and it is usually the governing case at less than about a metre of cover, which is exactly the depth range where the arching reduction is smallest. The two effects are largest at opposite ends of the depth axis, which is why the design curve for a buried pipe has a minimum in the middle.

And the load was static and permanent. Arching is mobilised by movement and can be lost. Vibration from traffic, wetting and drying of the fill, or a subsequent excavation beside the trench can all relax the shear on the prism’s sides and hand the pipe its full prism load years after it was built. Arching is not a reduction that is safe to rely on for a long time, which is why the trench formula is used with more caution than the embankment one.

The generalisation #

The idea to keep is that a load path is decided by relative movement, and relative movement is decided by relative stiffness — so “what is the load” is a question that cannot be answered without knowing what the structure is.

That is uncomfortable, because a load is supposed to be an input and a structure an output, and the design process is arranged that way on every drawing board. It works for gravity, which does not care. It fails for every load that arrives through a restraint: earth pressure, which depends on how much the wall has moved; a prop force, which depends on when it was installed; a thermal force, which depends on what is holding it; a settlement moment, which is proportional to the stiffness that resists it. In every one of those, the stiffest path takes the load is not a design tool but a description of who has volunteered.

The buried pipe is the sharpest example because the volunteering is done by soil that has no opinion and is not on any drawing. Between the trench answer and the embankment answer lies a factor of nearly three, and the whole of it is in an arrow whose direction nobody applies.

The practical form of the habit is a question to ask before accepting any load: what would have to move for this load to be different? For a floor’s own weight the answer is nothing, and the load is a fact. For a buried pipe, a propped wall, a restrained beam or a pile in a settling fill, the answer is something that has already moved or is about to — and the load is a result rather than an input. The second kind is where the surprises are, and the free body is a choice is the tool for finding them: draw the boundary somewhere the movement crosses it, and the mechanism becomes visible.