The load that depends on what carries it

A tank holding six metres of liquid is a straightforward object. The pressure at any depth is the unit weight times the depth, the resultant sits at the third point, and nothing about the tank changes either. Fill the same space with sand and the load falls to a third of that — then rises back to a half if the wall is prevented from moving, and to three times the liquid value if the wall is pushed the other way.

Nothing about the sand changed. What changed was what the wall did.

That is the argument in one figure, and it has two halves. The five terms have different shapes, so they act at different heights, so the resultant is not where a triangle would put it. And the largest of the five is the water, which is not soil at all.

The load a wall is given is not a load #

Every other load here arrives as a number. Snow does not care how much the roof sags under it; a lorry does not weigh less because the bridge deflected; the load a beam is given is a decision made before the analysis and held fixed while it runs.

Retained soil is the exception, because soil has shear strength. A wedge of it holds part of its own weight up on friction between its grains and leans on the wall only for the part it cannot — and how much it manages depends on whether it has been allowed to slide at all, which is to say on whether the wall moved.

So the load is a function of the displacement of the thing resisting it, and statics has no machinery for that. Equilibrium equations contain forces and distances and no displacements whatever, which is why there are only ever three of them: nowhere in $\Sigma F = 0$ is there a place to put “and the wall moved six millimetres”.

Three coefficients, and a factor of nine between the ends #

Rankine’s answer is three states rather than one number:

$$K_a = \tan^2\!\left(45° - \frac{\phi}{2}\right), \qquad K_0 = 1 - \sin\phi, \qquad K_p = \tan^2\!\left(45° + \frac{\phi}{2}\right).$$

At a friction angle of 30° those are $0.333$, $0.500$ and $3.000$.

Nine hundred and seventy-two kilonewtons per metre against a hundred and eight: a factor of nine on the load, same soil, same wall, decided by which way it moved.

The reciprocal relation is no coincidence. Active and passive are the same Mohr circle touching the same failure envelope from the two sides — the horizontal stress is the minor principal stress in one and the major in the other — so the two are inverses by construction, and anything raising one lowers the other by as much.

How far the wall has to move, and it is not symmetric #

The three coefficients are not options to choose between. They are the two ends and the starting point of a curve whose abscissa is the movement of the wall.

Six millimetres is nothing — the movement a masonry wall gets from mortar creep in its first winter, or a cantilever from the elastic rotation of its own base. Almost every wall reaches the active state by accident, which is why designing for it is honest rather than optimistic.

The other end is the problem. A hundred and fifty millimetres at the top of a six-metre wall is not a serviceability event, it is a demolition notice for whatever stands on the retained ground. A designer who quotes $K_p$ is quoting a number that arrives only after a movement nobody would accept — and the figure measures how much arrives early: the slope at the origin is 26 962 one way and 16 177 the other, per metre of wall movement per metre of run. The curve has a kink there. The soil is stiffer when let go than when pushed.

That asymmetry is why every code discounts passive resistance to a half or a third of its theoretical value, and the solver behind these figures does the same: the sliding check below mobilises half the passive term.

The consequence runs the other way too. A wall that is held — propped by a basement slab, keyed into rock, restrained by a building sitting on it — never reaches the active state and carries $K_0$: $0.500$ against $0.333$, 1.50 times the load an active design was checked for, produced by making the wall better.

The third point belongs to the triangle and to nothing else #

A pressure growing linearly with depth is a triangle, its resultant is $K\gamma H^2/2$, and it acts at $H/3$ above the base. That third point is why a retaining wall is a wedge in section and why its overturning moment goes as $H^3$ rather than $H^2$.

A real wall never has a pure triangle behind it, and the figure at the top of the page is the reason. A surcharge is constant with depth, so its resultant sits at mid-height — twice the lever arm per unit of thrust, which is why a lorry parked near the top of a wall is so much worse than its weight suggests. Water is confined below the table, so its triangle is short and sits low.

Two hundredths of the height sounds like a rounding error and is not. The overturning moment, the eccentricity of the base resultant, whether the base lifts, what the bearing peaks at — all of them are $P \times \bar{y}$ and nothing else, so the lever ratio is not a detail of the load but the load’s whole contribution to the answer.

Which free body produced these numbers #

The thrust and the resultant above came from cutting the soil, not the wall. The free body is a vertical plane at the back of the heel, from the retained surface down to base level; the force on it is the integral of the pressure profile and the line of action is that integral’s first moment over the integral. The generator sums the five terms and checks the total against a numerical integration of the profile it drew, agreeing to seven decimal places.

The wall’s own checks come from a different body: the wall, the base slab and the column of soil standing on the heel, together, cut on that same vertical plane and on the underside of the base. $\Sigma M = 0$ about the front toe gives overturning; $\Sigma H = 0$ along the underside gives sliding; $\Sigma V$ with the moment about the base centre gives the eccentricity.

The soil on the heel is the part left out, and it is a third of the restoring weight. It is inside the free body because the cut was made behind it, and the cut goes there rather than at the back of the stem because Rankine’s pressure applies to a vertical plane in soil, not to a concrete surface. Choosing that plane is what makes the heel structural — the same column of soil pushing the wall over is also sitting on it, holding it down, and which body was cut decides which of the two roles it plays.

The water is the largest term #

Soil pushes with a fraction of its weight. Water pushes with all of it: no shear strength, so its coefficient is exactly one — and it acts on top of the soil’s effective stress rather than instead of it.

The water behind a wall pushes harder than the soil does. That one comparison reorganises the subject: a drain is not a durability accessory but the structural element that removes the largest load, and the friction angle — the number the calculation appears to be about — does not enter the water term at all.

A sliding factor of 0.91 means the wall is moving. Nothing was built wrong and no load was miscalculated: a pipe silted up. That is the commonest way a retaining wall is lost, and the arithmetic above is why — the failure that arrives first is not the one the section was sized for. It is also a case where friction is asked for more than it has, and friction, being an inequality rather than an equation, gives no warning until it is exceeded.

A rigid block is not a wall #

The temptation, having read that weight is the only thing resisting overturning, is to treat the wall as a heavy block under a uniform pressure. The substitution fails twice.

The load is the first failure. A block takes a uniform pressure at a lever arm of $H/2$; earth pressure acts at $H/3$, and no single uniform pressure reproduces both the force and the moment — the one matching the moment is $K\gamma H/3$, the one matching the force is $K\gamma H/2$, and for a pure triangle they always differ by exactly three halves.

The weight is the second, and the larger. A block’s restoring moment is $W \cdot B/2$, because a block’s weight acts at the middle of its base. The wall’s sits at 61.6% of the base, pushed back by the column of soil on the heel. That offset is what a heel is for, and it is invisible to any routine that centres the weight — which is why the wall solver borrows the block’s pressure calculation and none of its load or weight arithmetic.

The base has its own middle third #

The eccentricity out of that free body still has to be checked against the base, which is the middle third again, and the problem a base plate solves at a smaller scale.

The arithmetic under the base is the ordinary one. While the resultant stays inside the middle third the peak pressure rises linearly with the eccentricity and has exactly doubled by the edge of it: 324 kN on a 3.5 by 1 m base is 93 kPa spread uniformly and 185 kPa at $e = B/6$. Past that edge the base lifts, the contact shortens, and the peak rises much faster than the load, because the same force is being carried on less and less ground.

The drained wall sits inside that limit and the undrained one is just outside it. Being outside is not a collapse: the pressure under the toe now rises much faster than the load, and the wall starts to rotate about it. Rotation relieves the active pressure, which is stabilising, and raises the toe pressure, which is not — and which of those wins is a question no figure here can answer.

The number everyone argues about is not the number that decides it #

A site investigation report is read for its friction angle, and a designer given 30° rather than 35° will feel the difference is the whole calculation. It is worth measuring how much of the answer that parameter actually carries, because the water term above has already suggested the answer.

Take the drained wall, no surcharge, 6 m of soil at 18 kN/m³. At $\phi = 30°$ the thrust is 108.0 kN/m. At 35° it is $K_a = 0.271$ and 87.8; at 25° it is $K_a = 0.406$ and 131.5. So a five-degree uncertainty either way — which is a great deal in a soil report — moves the thrust by about a fifth.

Now the same wall with the drain blocked. Of the 185.7 kN/m in the first figure, 78.5 is water and does not contain $\phi$ at all; the surcharge and soil terms making up the remaining 107.2 all scale with $K_a$. Swinging the friction angle by the same five degrees now gives totals of 165.7 and 209.1 — a swing of about an eighth rather than a fifth, because nearly half the load has stopped listening.

Set that beside the drain itself, which took the same wall from 128.0 to 185.7 kN/m, and the ranking is unambiguous. Blocking one pipe is worth more than being wrong about the friction angle by five degrees in the unfavourable direction, and it is worth more again on the checks rather than the load, because the water arrives low down and drags the lever arm with it. The parameter that is measured, argued over and reported to a decimal place is the second most important thing in the calculation; the first is whether a 100 mm pipe is still open, and no soil report has ever contained it.

Where the model stops #

The wall back is smooth and vertical and the ground behind it is level. Rankine assumes all three. A rough back attracts a downward shear from the settling soil, which reduces the horizontal thrust and tilts its line of action. Coulomb’s wedge handles that; Rankine cannot, and everything here is Rankine.

The soil is at failure everywhere. Both limiting coefficients describe soil that has reached its shear strength throughout the wedge. A real backfill is at failure only where it has moved enough, and the mobilisation curve between the end points is fitted rather than derived — monotone, finite-sloped at the origin, chosen because measured wall tests look like that.

Compaction is not in it. Backfill rolled in layers is pushed against the wall by the roller, and the locked-in pressure near the top can exceed $K_0$ substantially.

The water table is a line. Drawn static and level. A real one rises after rain with a lag, and the transient state during a storm is worse than either steady condition.

Plane strain, per metre run. Corners, buttresses, the ends of a wall and any variation of retained height along it are outside the model entirely.

The company this load keeps #

A load that depends on the displacement of what resists it is rare in statics — the moving structures have several, but among the static ones this site has three. Ponding is one: the roof deflects, the deflection makes room for water, the water deepens the deflection. Second-order effects are the second: the frame leans, the weight on it acquires a lever arm, the lever arm makes it lean further. Both are positive feedback, both have a critical stiffness at which they run away, and both turn a serviceability calculation into a stability problem.

Earth pressure is the third, and it is the only one whose feedback has the other sign. The wall moves away and the load goes down. It is self-limiting: the more it succeeds the less there is of it, which is why walls that are visibly out of plumb are so often still standing. Push the wall the other way and the sign reverses, which is what makes the passive side stiff.

Its nearest relative here is an imposed movement rather than an applied force — a load whose size is set by stiffness rather than by statics, and which a redundant structure hands to whatever is stiffest. Earth pressure is that family with the sign flipped: a movement that relieves rather than one that loads.

What these pictures cannot show #

Every figure here is a free body in equilibrium at one instant, and the argument is about a sequence. The mobilisation curve is drawn as though the wall could be placed anywhere along it; a real wall arrives at its position over years, and the path matters, because soil that has been to the active state and back does not return along the same curve. Nothing on this page is drawn twice.

Nor do the drawings show what the wall stands on. The base pressures in the last figure are demands, and the ground has its own capacity, its own settlement and its own failure surface, which may pass beneath the whole wall and take the slope with it. That surface loses most of the walls that are lost, and it is off the edge of every picture here, where a thrust line that leaves the drawing goes on existing.

And the thrust is drawn as an arrow at a height when it is grains pushing on grains. What lets it become an arrow is that the wall is rigid and rotates about its base — the idealisation the collection makes whenever it draws a connection as a point. A flexible sheet-pile wall bows outward in the middle and sheds its own pressure toward the stiff ends, and the triangle is then simply wrong.

The ladder from here #

Later rungs on this anchor. Coulomb’s wedge, wall friction, and the sloping backfill Rankine cannot take. Compaction pressure, and why the top metre of a backfilled wall is not active. At-rest pressure in propped basements, where the wall never gets its six millimetres. Anchored walls, whose distribution is not triangular and whose prop force depends on the excavation sequence. Embedded cantilevers, which rotate about a point below dredge level with pressure on both sides. Seismic earth pressure, where the wedge is accelerated and the load rises exactly when the wall can least take it. Soil arching and Janssen’s silo, where the pressure stops growing with depth. Passive failure with wall friction, where the plane surface becomes a log spiral and the plane answer is unsafe by a third. And global stability, the surface below the wall, which is not a structural calculation at all.

Coulomb published the wedge in 1773, in a paper on maxima and minima that also carries the friction law and the first bending theory worth having; Rankine’s stress-field version followed in 1857 and gives the coefficients above their closed form. Neither had the mobilisation curve. That came from measurements on model walls in the 1930s, and it is why two answers that differ, and that both describe soil at failure, turned out to be the end points of one question rather than rivals.