It tips inside its own hull

A body on legs tips when the resultant of its weight and whatever pushes it reaches the edge of the polygon its feet enclose. That essay was careful about what the statement assumes — a rigid body, standing on rigid ground — and careful to say that under those assumptions the answer does not depend on how the load was shared between the feet before the edge was reached. Tipping is decided by the hull, and the hull does not care.

It closed on the question of what happens when the ground is not rigid, and the question has a precise answer that is less comfortable than the one before it. A body on supports that compress leans as it is pushed. Leaning moves its weight toward the side that went down. The weight’s own moment is then partly overturning rather than wholly restoring, and the push needed to finish the job is smaller than the hull says.

How much smaller turns out to be one ratio, and the ratio is made of quantities that are written on a geotechnical report and a general arrangement drawing.

A tank on four pads #

Take a water tank on a braced tower, weighing 6,000 kN full, with its centre of gravity 15 m above the ground. It stands on four pad footings at the corners of an 8 m square, so each pad is 4 m from the centre along each axis. Wind on the tank and tower adds up to a horizontal resultant acting 18 m up.

On rigid ground, the stability of such a body is a moment balance about an edge. Pushed along an axis, the body tips about the line joining two pads when the push’s moment, $P \times 18$, reaches the weight’s, $6{,}000 \times 4$: at 1,333 kN. Pushed toward a corner, it tips about the diagonal through the two side pads, 5.66 m from the weight’s line, at 1,886 kN — stronger, as the rigid analysis of four legs found. And along that diagonal the far pad loses all its load at exactly half the tipping push, 943 kN, which is the result that made the diagonal a trap rather than a strength.

Now make the pads real. Each pad settles into the ground in proportion to its load: say 20,000 kN/m, so that under a quarter of the tank’s weight each goes down 75 mm, which is an unremarkable settlement for a pad on firm clay. The body is still rigid and the four supports are springs that cannot pull.

Springs alone change which pad lifts, and nothing else #

Before the lean is admitted, it is worth seeing that springs by themselves do not disturb the hull’s answer.

A rigid body on four springs moves as a plane: down by some amount at its centre and tilted about two axes. Three equations of equilibrium fix those three quantities, and each pad’s reaction is its stiffness times how far the plane has pressed it down. If all four stiffnesses are equal the reactions are linear in the position of the resultant — the formula the rigid analysis of four legs had to assume — and if they are not, they are still linear, around a different centre.

Either way, as the push grows, a pad lifts when the plane rises above it, the springs that remain carry the body, and the body tips when no three remaining springs can hold a plane in balance. That last condition is geometric. It happens when the resultant reaches the edge of the hull of the pads still in contact, and so, in a first-order analysis, stiffness moves the lift-offs and leaves the tipping point exactly where the rigid hull put it. With one pad softer than the others, a pad lifts at 762 kN along an axis instead of at the tipping point, and the tank still tips at 1,333.

What breaks this is not the springs. It is that the plane tilts, and a tilted plane carries the weight with it.

The lean moves the weight #

Tilt the tank by an angle $\theta$ and its centre of gravity, 15 m up, moves sideways by $15\theta$. The weight’s line of action has moved toward the side that went down. About the edge it tips on, the weight’s restoring lever has shortened by the same amount.

The pads pushed away from unload, the pads pushed toward load up, and both pairs leave their original 1,500 kN in straight lines, as they would on rigid supports. The difference is where the upwind pair runs out. Ignoring the lean, at 1,333 kN. Including it, at 1,240.

The arithmetic is short enough to do by hand, and doing it gives the whole essay’s ratio.

With four equal pads of stiffness $k$ at $\pm a$, the tank settles $s = W/4k$ uniformly under its own weight. When the upwind pair is just about to lift, the plane passes through zero at $x = -a$, so its tilt is $\theta = s/a$. The moment balance about the centre, now including the weight’s shift $h\theta$, is

$$P_{\text{tip}}\,h_P + W h \theta = W a,$$

and substituting $\theta = s/a$,

$$P_{\text{tip}} = \frac{Wa}{h_P}\left(1 - \frac{s\,h}{a^2}\right).$$

The first factor is the rigid hull’s push. The second is the lean’s toll, and for this tank it is $0.075 \times 15/16 = 0.070$. The solver that drew the figure does not use the formula; it solves the tilted equilibrium with pads removed as they lift, and its answer agrees with the closed form to the precision of the bisection.

The lean is a narrower base #

The closed form can be read a second way, and the second reading is the one a designer can act on.

Multiply it out: $P_{\text{tip}} h_P = W(a - sh/a)$. That is exactly the rigid-ground moment balance for a body whose base is not $a$ from its centre but $a - sh/a$. The lean is equivalent to moving every edge of the base inward by the distance the weight travels, which for the tank is $0.075 \times 15/4 = 0.28$ m. A tank designed on rigid ground with a half-width of 4 m is, on its own pads, a tank with a half-width of 3.72 m.

That puts the toll in the currency the base was designed in. To give the tank back its rigid-ground 1,333 kN, the half-width has to satisfy $a - 1.125/a = 4$, which is $a = 4.26$ m: another 26 cm from the centre to each pad, if the pads keep their stiffness. Widening the base helps twice, because the half-width is squared in the toll — the lean is a displacement at the weight’s height, and a wider base both turns that displacement into a smaller angle and moves the edge further away.

The other levers are all in the same ratio and are worth ranking. Halving the settlement halves the toll, and costs a bigger footing or better ground. Lowering the weight reduces it in direct proportion; a tank emptied to half its depth drops its centre of gravity and its weight together, so its toll falls faster than its weight. Stiffening the tower does nothing for the foundation’s share and is a separate term altogether.

And the reading explains why a factor of safety computed on rigid ground is not simply reduced by the toll. The factor is a ratio of restoring to overturning moments, $Wa/Ph_P$. On soft ground the restoring side is $W(a - sh/a)$, but the settlement $s$ is itself proportional to $W$, so the toll grows with the weight that was supposed to be resisting. A fuller tank is heavier, more stable against a given wind on rigid ground, and leans further on its pads, so on soft ground each extra kilonewton of weight buys less than the last — and once $sh/a^2$ passes one half, adding weight makes the body less stable rather than more.

A number that is already on the drawings #

The ratio $s h/a^2$ is worth dwelling on, because it is unusual for a stability quantity to be made of numbers nobody has to derive.

The settlement under permanent load is the number a foundation designer reports and a site engineer measures. The height of the weight is on the general arrangement. The half-width of the base is too. The stiffness of the ground, which is the genuinely uncertain quantity, enters only through a settlement that somebody has already had to estimate for a different reason.

It also says immediately which bodies care. A squat body with a wide base and a firm foundation has a ratio in the thousandths. The tank, at 0.07, loses seven per cent of its overturning resistance to its own lean. A chimney with its weight 40 m up on a raft 10 m across that settles 50 mm has $0.05 \times 40/25 = 0.08$. A crane on outrigger pads over made ground, where each pad might sink 100 mm and the weight sits 10 m above a base 3 m from centre to pad, has $0.1 \times 10/9 = 0.11$ — before the jib is swung out.

And the line has an end. At $sh/a^2 = 1$ the tipping push along an axis is zero. A body there needs no push at all: the least tilt moves its weight far enough to tilt it further, the pads’ resistance to rotation, $\sum k x^2$, is exactly used up by the weight’s $Wh$, and the body falls over under its own weight on ground that has not failed. That condition is written $Wh = \sum k x^2$, and it is a buckling load in every sense but the name — the same competition between a stiffness and a load times a length that decides how far a frame leans under its own gravity.

Hambly called it the leaning instability, in a 1985 paper arguing that the tower at Pisa stood close to it: that its foundation’s rotational stiffness was near enough to its weight times the height of that weight for any lean the soil allowed to feed itself, which is a better account of a lean that kept growing for centuries than any single settlement. What the rigid-ground analysis cannot represent at all — a body tipping with its resultant well inside its base — is the ordinary behaviour of a tall body on soft ground, approached rather than reached.

Toward a corner the loss is twice as fast #

Along an axis the lean costs the fraction $sh/a^2$. Toward a corner it costs about twice that, and the reason is the rigid analysis’s own finding turned round.

Pushed toward a corner, the far pad lifts at half the tipping push. From then on the tank stands on three pads, and the only one of them that resists rotation about the diagonal it will tip about is the near pad: the two side pads sit on that diagonal and contribute nothing. The restoring stiffness about that diagonal falls to a third the moment the far pad leaves, while the weight’s lean moment does not fall at all. So over the second half of the push the lean takes a much larger bite than over the first.

On rigid ground the diagonal was the strongest direction against tipping and the weakest against lift-off. On soft ground it keeps the second property and loses the first faster than any other direction. At a ratio of 0.2, the axis keeps 80 per cent of its rigid push and the diagonal 60; past 0.34 the diagonal’s three-pad phase has so little stiffness left that it cannot carry the lean at all, and the corner pad’s lift-off is the collapse.

That last point bears on what the rigid analysis said a lifted leg meant. There, a lifted leg along the diagonal was a warning: a foot no longer available for friction or relied-on bearing, with the body still at half its tipping push. On a soft enough base the warning and the event coincide.

One pad on softer ground #

Real pads are not equal. The ground under one corner of a site is looser fill, or the footing there was cast on a soft spot, and nobody finds out until the tank is full. Make the tank’s north-east pad, at $(+4, +4)$, 8,000 kN/m instead of 20,000.

Nothing is pushing yet, and already the tank is not where it was. A rigid plane on four springs of unequal stiffness is a table on an uneven floor: it settles toward the soft pad, tilts by 8.1 milliradians, and its reactions are no longer equal — 1,927 kN on each of the two stiff pads beside the soft one, 1,138 on the soft pad itself and 1,009 on the stiff pad opposite it. The diagonal pair through the soft pad has shed load to the other pair, which is the four-legged table’s self-stress state, set by the ground rather than by a short leg.

Pushed toward the soft pad the opposite pad lifts at 444 kN, less than half its rigid-ground value, and the tank tips at 1,223 kN. A first-order analysis of the same springs gives 1,886, the hull’s number, because a first-order analysis cannot see the lean that the soft pad created before any push arrived.

Putting every direction on one drawing shows the part that a check along chosen directions misses. On rigid ground the curve of tipping push against direction is the hull’s square — weakest along the axes, strongest toward the corners. On equal soft pads it shrinks, and shrinks more at the corners. With the soft pad it loses its symmetry, and its weakest point is at 15 degrees from an axis, on the soft pad’s side, at 1,039 kN.

Neither the axis nor the diagonal is the weakest direction for the body as built, and the difference is not large — 1,039 against 1,066 along the axis. The important number is the other one. The weakest push in any direction is 78 per cent of the rigid analysis’s weakest, for a tank whose pads settle between 50 and 142 mm under its weight, and a check that treats the ground as rigid reports a factor of safety about 1.28 times too high.

What the lean does not change #

It is worth being exact about which conclusions from rigid ground survive.

Sliding is untouched. The horizontal push is resisted by friction at the pads, and a small lean does not change the total vertical force the friction depends on. Where a body chooses between tipping and sliding, the lean moves only the tipping side of the comparison, so on soft ground a body that would have slid on rigid ground can tip instead.

The hull still decides where a body tips about. The tank still tips about a line joining pads; the lean changes how hard it has to be pushed to get there, not which line it turns on.

A lifted pad is still a lost pad. Every consequence the rigid analysis drew from lift-off — friction gone from that foot, a holding-down bolt asked for tension it was not designed for, the body now determinate on three supports — arrives at a smaller push on soft ground, and toward the corners at a much smaller one.

And the ground has to hold. Everything here treats the pads as linear springs that never fail. As the tank tips, one pad ends up carrying the whole 6,000 kN, four times its design share, and a pad under that load is well past any stiffness a settlement estimate was made for. The real body tips earlier still, by the amount the most loaded pad’s stiffness falls off as it approaches its bearing capacity.

Where the model stops #

Linear springs that cannot pull. Ground under a pad stiffens and softens with load, creeps under the permanent part of it, and has a bearing capacity at which its stiffness is zero. A linear spring is the first term of that behaviour and every result here is the small-settlement limit.

A rigid body. The tank’s tower is assumed not to deform. A braced tower sways under the same push, which adds its own lean to the foundation’s and moves the weight further; the two add, and the frame’s share is an analysis of its own.

Small tilts. The weight’s shift is $h\theta$, which is the first term of $h\sin\theta$. At the tilts drawn here — a few tens of milliradians at most — the error is below a tenth of a per cent.

A static push. Wind on a tank is a fluctuating load, and a body whose rotational stiffness has been reduced by its own weight has a longer rocking period, which moves it on the wind’s spectrum. Nothing here is dynamic.

Independent pads. Each pad settles only under its own load. Pads 8 m apart on a clay stratum are not independent — loading one depresses the ground under its neighbours — and the interaction reduces the differential settlement that the lean is made of, so for equal pads the numbers here are on the unfavourable side.

Nothing about how long it takes. A soft pad on clay settles over months. The tilt at the moment of filling is not the tilt a year later, and a body checked on its immediate settlement can have lost more of its hull by the time the design wind arrives.

Still open: the body whose ground stiffens under the side that goes down #

The springs here are linear, so the side that goes down is exactly as stiff as the side that comes up. Real ground is stiffer the more it is compressed, up to a point, and then softer as it approaches failure. The first effect restores some of what the lean took away, the second takes more, and which of them dominates at the push that matters depends on how far below its bearing capacity the most heavily loaded pad is working when the resultant nears the edge. Whether a body on nonlinear ground has a well-defined tipping push at all — or only a load at which its lean stops converging — is the question after this one.