Structural Engineering & Statics
where the load goes
Equilibrium

A basement is a boat

Every load in this collection presses down and is resisted by strength. Hydrostatic uplift presses up, is resisted by weight, and does not care what is built on it — so the check contains no material property at all. It is a ratio of two weights, and one of them is water.

Assumes The load that depends on what carries it, Weight is the only thing resisting it and The free body is a choice, and choosing it well is the whole skill.

19 min read 11 figures The load must go somewhereThe statics of things that do not moveWhich failure arrives first

Every load so far in this collection presses down or sideways, arrives from something built or from the weather, and is resisted by a strength. Hydrostatic uplift is none of those things. It presses up, it arrives from a water table nobody controls, it is resisted by weight, and the check that governs it contains no material property of any kind.

A substructure below the water table displaces water, and the water pushes back. That is Archimedes rather than statics, and the arithmetic is the same either way: the pressure on the underside of the base slab is the depth of water above that slab times the unit weight of water, applied over the whole plan area.

A basement is a boatA 20 by 30 m substructure dug 6 m into ground whose water table stands 0 m down. The head on the underside of the base slab is 6.0 m, so the pressure there is 58.9 kN/m² over the whole plan — 35.3 MN of it, pushing upward. Nothing about the structure changes that number. What resists it is weight: 18.7 MN of concrete and whatever is built above, giving a factor of 0.53. The structure floats if the water reaches 2.82 m below the ground, and a base slab alone would have to be 2.45 m thick to hold it down.water table58.9 kN/m² upward, everywhereweight 18.7 MN6 muplift 35.32 MN · weight 18.72 MNfactor against flotation 0.53no strength appears anywhere in that ratio
Fig. 1 A 20 × 30 m substructure dug 6 m into ground whose water table stands at the surface. The head on the underside of the base slab is the full 6 m, so the pressure there is 58.9 kN/m² over 600 m² — 35.3 MN pushing upward, against 18.7 MN of concrete. The factor against flotation is 0.53, and the structure floats.

The check with no strength in it

Write it out:

factor=weight+friction on the walls+tie-downγw h A\text{factor} = \frac{\text{weight} + \text{friction on the walls} + \text{tie-down}}{\gamma_w\,h\,A}

There is no fyf_y, no fckf_{ck}, no EE, no section modulus and no member anywhere in that expression. It is a weight divided by a weight of water, and the only design variables available are:

  • add weight — thicken the slab, add ballast, count the building above;
  • add friction — mobilise shear on the outside of the retaining walls;
  • tie it down — anchors or tension piles into the ground beneath;
  • or stop the water arriving — drainage, which changes hh rather than resisting it.

That is the whole list. Nothing about making the structure stronger appears on it.

The only load case that gets better as the water goes awayFactor against flotation for a 20 by 30 m substructure 6 m deep, against how far down the water table stands. With the water at the surface the head under the slab is the full 6 m and the factor is 0.53; it reaches one at about 2.82 m and rises steeply after that, because the head falls linearly while the weight resisting it does not move at all. The curve has no material property anywhere in it: it is a weight divided by a weight of water, and the only design variables are ballast, tie-down and drainage.012345601234depth of the water table below ground (m)weight ÷ upliftit floats below thisno strengthin this ratio
Fig. 2 Factor against flotation against the depth of the water table below ground. With the water at the surface it is 0.53; it reaches one at 2.82 m and rises steeply after that, because the head falls linearly while the weight resisting it does not move at all.

The curve has a shape worth noting: it is a reciprocal, not a straight line, because the resistance is fixed and the action is falling. So a small change in the assumed water table near the critical depth changes the factor a great deal, and a large change well above it changes almost nothing.

Which free body produced the number

The free body is the whole substructure, cut on a plane just beneath its base slab and on planes just outside its walls.

Across the bottom face acts the water pressure, γwh\gamma_w h upward, uniform over the whole plan. It is uniform because the base is horizontal and hydrostatic pressure depends only on depth.

Across the outside of the walls acts water pressure horizontally, which balances itself, and shear on the soil interface if the wall is rough and the soil is willing to supply it.

And downward acts the weight of everything inside the cut: the slab, the walls, the floors, the building above.

A uniform load and the force that replaces itA uniform distributed load with its resultant computed by integration: an area of 1180.0 acting at 10.00 from the left. The two moment diagrams below show what the substitution costs — the reactions are identical and the peak moment is not.resultant 1180.0at x = 10.00, the centroid of the areamomentspread: 2950.0replaced: 5900.0reactions agree exactly (590.00 and 590.00); the peak moment does not
Fig. 3 The uplift as a distributed load and as the single force it is equivalent to. A uniform pressure over a flat base has its resultant at the centroid of the plan, which is why flotation has no lever arm in it and why the check is a force balance rather than a moment one.

There is nothing else, and in particular there is no reaction from the ground. The structure is not resting on anything — that is precisely the condition being investigated. A free body with no bearing reaction is unusual in this subject and it is the reason the check feels unfamiliar.

A beam, its loads and its reactionsA free body: the beam cut away from everything it touches, with the forces that were doing the touching drawn on it. The reactions are computed from the loads, so their arrows are to scale relative to each other.12 per unit length48.048.0ΣM about one support gives the other reaction; ΣF then gives the first
Fig. 4 Choosing the free body is the whole of the method, and here the choice is unusually stark: the cut has to go under the base slab, because a cut anywhere higher leaves the uplift on the wrong side of it and produces a structure standing comfortably on ground that is trying to lift it.

The load that does not scale with the building

A basement is a boatA 20 by 30 m substructure dug 6 m into ground whose water table stands 3 m down. The head on the underside of the base slab is 3.0 m, so the pressure there is 29.4 kN/m² over the whole plan — 17.7 MN of it, pushing upward. Nothing about the structure changes that number. What resists it is weight: 18.7 MN of concrete and whatever is built above, giving a factor of 1.06. The structure floats if the water reaches 2.82 m below the ground, and a base slab alone would have to be 1.23 m thick to hold it down.water table29.4 kN/m² upward, everywhereweight 18.7 MN6 muplift 17.66 MN · weight 18.72 MNfactor against flotation 1.06no strength appears anywhere in that ratio
Fig. 5 The same substructure with the water table at 3 m. The head has halved, the pressure with it, and the factor is 1.06 — the structure is stable by a margin that depends entirely on a number measured in a borehole some years earlier.

Uplift is the one action in this collection with the following property: it does not grow when the building does.

A taller building has more wind, more seismic mass, more gravity load, more foundation, more of everything. Its uplift is unchanged, because uplift depends on the plan area of the substructure and the depth of the water — neither of which the storeys above alter. So the resistance grows and the action does not.

storeys above weight factor
0 18.7 MN 0.53
1 23.5 0.67
2 28.3 0.80
3 33.1 0.94
4 37.9 1.07

Three and a half storeys of ordinary building are enough to hold this substructure down. Which means the governing case is not the finished building at all — it is the substructure standing empty, which is a real condition that exists for months during construction, and it is the case the design has to satisfy.

That is the same structure of argument as the most dangerous day and it arrives from the opposite direction: there, the capacity was missing; here, the load that provides the capacity is.

Ballast, friction, and tying it down

Ballast is the direct answer and it is expensive in the currency that matters underground: to take this substructure from 0.53 to a factor of 1.10, an extra 1.40 m of concrete is needed over the whole plan. That is 840 m³ of concrete occupying 1.4 m of a basement that was excavated in order to have space in it.

Wall friction is the cheap answer and the one hardest to justify. The outside of a 6 m deep retaining wall has 600 m² of contact with the soil, and mobilising even 20 kN/m² of shear on it is worth 12.0 MN — which takes the factor from 0.53 to 0.87 on its own.

The reaction lies inside the cone, so the block standsA block of 100 on a plane at 15°, against a coefficient of friction of 0.35. Resolving across and along the plane gives a normal force of 96.6 and a friction demand of 25.9, against a capacity of μN = 33.8 — a ratio of 0.77. Added together the two make one contact reaction leaning 15.0° from the normal, and the admissible reactions fill a cone of half-angle arctan μ = 19.3°. Equilibrium is possible exactly when the demanded reaction lies inside that cone, which here it does. The weight enters neither the cone nor the lean: a block of any weight on this slope leans its reaction by the same 15.0°, which is why the angle of repose is a material property and the size of a heap of sand is not.15°reaction, leaning 15.0° from the normalthe cone: half-angle arctan μ = 19.3°W = 100demand 25.9 against a capacity of 33.8 — F/μN = 0.77the weight appears nowhere in the cone — only the direction of the reaction is asked about
Fig. 6 Friction as a force that is whatever it needs to be, up to a limit. Wall friction against uplift is exactly that: it is not a force that exists until the structure tries to rise, it develops only as the structure moves, and the movement needed to mobilise it fully is more than a basement can tolerate.

That last point is why wall friction is usually taken conservatively or ignored. It needs relative movement to develop, the movement it needs is millimetres, and a basement that has risen by millimetres has already broken its own waterproofing.

Tension piles or ground anchors are the reliable answer. They convert a weight problem into a strength problem, which is the one thing everything else on this page cannot do — and they reintroduce a material property to a check that had none.

Five loads behind one wall, and the water is the biggestThe horizontal pressure on a 6 m wall retaining soil at 18 kN/m³ with a friction angle of 30°, a surcharge of 10 kPa and the water table 0 m down, drawn once as the profile the wall feels and then once per term. The terms are surcharge 20.0 kN/m at 3.00 m, soil at the water table 0.0 kN/m at 3.00 m, submerged soil 61.1 kN/m at 2.00 m, water 176.6 kN/m at 2.00 m, and they sum to 257.7 kN/m — matched to 6e-13 by integrating the drawn profile numerically. The largest single term is the water, at 176.6 kN/m: water has no shear strength, so its coefficient is exactly one where the soil's is 0.333, and it acts on top of the soil's effective stress rather than instead of it. The combined resultant sits at 2.08 m above the base, 0.346 of the height rather than the third point at 2.00 m that a pure triangle would give.the whole profile257.7 kN/mat 2.08 m0246surcharge20.0 kN/mat 3.00 msoil at the water table0.0 kN/mat 3.00 msubmerged soil61.1 kN/mat 2.00 m0246water176.6 kN/mat 2.00 mresultant at 2.08 m, which is 0.346 of the height — one third only for a pure trianglethe water term is the largest single one, at 176.6 kN/m of 257.7 kN/m
Fig. 7 The load that depends on what carries it: the same wall, from the side, with the water separated from the soil. The water pressure on the wall is the largest single component and it is the same water that is pushing up under the slab — one water table producing a horizontal load, a vertical load and a buoyant reduction of the soil’s own weight, all at once.

Drainage is a structural decision

The fourth option changes hh rather than resisting γwhA\gamma_w h A, and it is the only one that removes the load rather than balancing it.

An underdrain beneath the slab, connected to a pumped sump, holds the water table at the level of the drain. That reduces the head to nearly zero and the uplift with it, and it turns a structural problem into a mechanical one — with the whole of the structure’s stability now depending on a pump.

Which is why it is usually refused. The design life of a basement is a century; the design life of a pump is fifteen years; and the failure mode of the arrangement is that the pump stops and the structure floats. It is the load that will not hold still in a form no analysis can see: a load whose magnitude depends on a maintenance regime. Where drainage is used, it is used with a relief valve — a one-way valve in the slab that admits water and floods the basement rather than letting it lift, which is a decision to accept a flooded basement in preference to a broken one — an alternative load path whose alternative is not structural at all.

The general form is worth keeping. Uplift is one of very few structural actions that can be designed away rather than designed for, and every scheme that does so trades a structural certainty for a mechanical one.

Weight against weight, twice

Weight is the only thing holding it downA body 4 m wide and 12 m tall weighing 900 kN, under a wind pressure of 1 kN/m². The wind delivers 36 kN and an overturning moment of 216 kNm about the leeward toe; the weight restores 1800 kNm, a factor of 8.33. The resultant lands 0.24 m from the centre against a middle third of ±0.67 m, so the base is still wholly in bearing.36 kNW = 900 kNmiddle third: ±0.67 mresultant at 0.24 mrestoring 1800 kNmoverturning 216 kNmfactor 8.33
Fig. 8 Weight is the only thing resisting it — the overturning check, which shares the whole of this page’s character. Both are ratios of one weight to another action, both contain no strength, and both are the checks a structure passes or fails as a rigid body rather than as an assembly of members.
Both failures are decided by the same two numbersFactors of safety against overturning and against sliding, for a body 4 m wide weighing 900 kN under a wind pressure of 1 kN/m², as its height grows. Overturning falls as the square of the height and sliding as the first power, so they cross: below 34.6 m the body overturns at a factor of one, and uplift at one edge has already begun at 20.0 m — a ratio of exactly √3, whatever the numbers are.051015200123456height of the body (m)factor of safetyuplift starts at 20.0 moverturningsliding
Fig. 9 And the two failures that come with it. Overturning falls as the square of the height and sliding as the first power, so they cross — a shape flotation does not share, because it has only one mode and no lever arm anywhere in it.

The comparison is worth making because it reveals what is unusual about flotation. Overturning has a lever arm, so the geometry of the body matters. Sliding has a friction coefficient, so a material property sneaks back in. Flotation has neither: it is the only check in this collection whose answer is unchanged by every dimension of the structure except the two that set its plan area and its depth.

The slab itself

A two-way slab is a one-way slab as soon as it is not squareThe share of the load carried by the strips spanning the short way, against the ratio of the sides. The two families of strips cross at the centre and must deflect equally there, and a strip's deflection goes as the fourth power of its span — so at a ratio of 1.33 the short strips already take 76% and at 2 they take 94%. The panel drawn here is 8 × 6 m, a ratio of 1.33, and its short strips take 76.0%. Two-way action is worth having at a ratio of one and worth almost nothing by two.11.522.530.40.50.60.70.80.91long span ÷ short spanshare taken by the short strips8 × 6 m: 76.0%by 2 : 1 it is a one-way slab
Fig. 10 And a second calculation that follows the first. Once the substructure is held down — by piles, by anchors, or by the columns it carries — the base slab spans between those hold-downs with the uplift pressure applied to its underside. A 58.9 kN/m² upward load on a slab spanning 8 by 6 m is a substantial slab, and it is designed upside down.

This is where the thickness argument returns, correctly. The slab is not thick because thickness resists pressure; it is thick because it has to span between the piles or columns that hold it down, under a uniform pressure applied from below, with the reinforcement in the top face where a designer’s hand is not used to putting it.

A check made on a perimeter, not on a sectionOne bay of a flat slab, 7.2 m square, on a 600 × 600 mm column. The heavy closed line is the control perimeter, 2d from the column face with its corners rounded at that radius — 11196 mm long against 2400 mm round the column itself. The shaded area inside it delivers no shear across it and is subtracted from the load; everything outside arrives through the perimeter. At 59 kN/m² that is 2476 kN across 11196 × 700 mm, a shear stress of 0.316 N/mm² against a resistance of 0.520.2d = 1400column7.2 m bayperimeter u₁ = 11196 mmshear to carry V = 2476 kNv = 0.316 against 0.520 N/mm²61% of the resistance used
Fig. 11 Including a punching check upside down: the column pushes down through the slab while the water pushes up around it, which is the same geometry as a flat slab with the sign of everything reversed. The load inside the control perimeter is subtracted here for a different reason — it is genuinely acting on the column rather than crossing the perimeter.

The two ways of counting the same weight

There is a bookkeeping trap in this subject that is worth spelling out, because both ways of doing the arithmetic are correct and mixing them is not.

The buoyancy view. The structure displaces a volume of water; the upthrust is the weight of that volume; the structure floats if it weighs less. This is Archimedes and it needs the whole submerged volume, walls and voids included.

The pressure view. The water pushes up on the underside of the base slab at γwh\gamma_w h and horizontally on the walls, and the horizontal components cancel. This is the free-body calculation made above, and it needs the plan area and the head, and nothing else.

The two give the same answer for a closed box, and they do not look as though they should. The reconciliation is that the pressure view’s uplift, γwhA\gamma_w h A, is exactly the weight of a prism of water of plan area AA and depth hh — which is the volume the structure displaces below the water table, provided its walls are vertical and its base flat.

Where they diverge is instructive. A structure with a stepped base, a projecting toe, or a sloping wall has a displaced volume that is not hAhA, and the buoyancy view is then the reliable one; a structure with an open top below the water table has no buoyancy argument at all and the pressure view still works. Pick one and stay in it, and never add a buoyant upthrust to a hydrostatic pressure on the same surface.

Where the model stops

The water table is a design assumption, not a measurement. It is measured in boreholes at one time of year, in one season of one decade, and the value used for design is a conservative envelope of what it might become. The whole calculation is exact and its principal input is not.

The soil’s own weight has been left out. Soil above any part of the structure — a heel projecting beyond the wall, a backfill over a lower basement — counts, and it counts at its submerged weight below the water table, which is roughly half its bulk weight. Getting that halving wrong is the commonest arithmetic error in the subject.

Nothing here is a seepage calculation. In a low-permeability soil the full hydrostatic head may take a long time to develop, and in a fissured rock it may develop instantly. The uplift used in design is the long-term steady state, which is correct and is not what the structure sees on the first day.

And the factor of safety is applied differently from every other one on this site. Uplift is a permanent action favourable to nothing, so it is factored up while the weight resisting it is factored down — and the two partial factors are applied to quantities that are both simply weights. There is no characteristic strength to take a fraction of, which puts the check in the same family as overturning and outside every other one on this site.

What the pictures cannot show

The uplift arrows in the hero figure are drawn as a row of discrete arrows with a pressure band beneath them. The real action is a continuous pressure on a surface, and the drawing has to make it look like a set of forces in order to show that it is there at all.

Nor can any figure show that the water is on the other side of the concrete. Every pressure drawn here acts on a surface that is also a waterproofing problem, and the two disciplines meet at exactly the plane this calculation is made on.

And the factor of 0.53 is a number about one substructure with one water table. It is drawn near enough to failure to make the arithmetic legible, which is a choice about a figure rather than a statement about basements.

The ladder from here

Later rungs on this anchor: tension piles and ground anchors, and the pull-out mechanisms that decide their capacity. The submerged unit weight of soil, and why the buoyancy of the ground is the term most often dropped. Uplift during construction, where the dewatering is switched off in stages and the governing case is a partly built structure with the pumps already stopped. Seepage and the flow net, which is where the pressure distribution comes from when the ground is not uniform. Relief valves and the decision to flood. And the historical case: several large basements and dry docks have floated, and in every recorded instance the structure was intact, the calculation was right about the structure, and the water table was not where the calculation said.

Where this goes next

the whole profile185.7 kN/mat 1.90 m0246surcharge20.0 kN/mat 3.00 msoil above water12.0 kN/mat 4.67 msoil at the water table48.0 kN/mat 2.00 m0246submerged soil27.2 kN/mat 1.33 mwater78.5 kN/mat 1.33 mresultant at 1.90 m, which is 0.317 of the height — one third only for a pure trianglethe water term is the largest single one, at 78.5 kN/m of 185.7 kN/m The load that depends on what carries it Every other load in this collection is a number the structure is given. Retained soil is not — it pushes with a fraction of its own weight, and the fraction is decided by how far the wall moves. Six millimetres of retreat on a six-metre wall takes a third off the load, and being held still puts it back. 48 kNW = 120 kNmiddle third: ±0.42 mresultant at 1.20 mrestoring 150 kNmoverturning 144 kNmfactor 1.04 Weight is the only thing resisting it A structure that is strong enough everywhere can still be blown over, and nothing in its material properties has any part in whether it is. The whole answer is a weight and a width — and the failure begins long before anything tips, at the moment one edge stops pressing down. 2012.57.5shear 7.5moment 22.5the cut, at x = 5nothing was applied here — the internal forces are what the left-hand piece needs The free body is a choice, and choosing it well is the whole skill Cutting a structure open is not a step in the method. It is the method — and where the cut is made decides whether the answer takes one line or twenty. 15°reaction, leaning 15.0° from the normalthe cone: half-angle arctan μ = 19.3°W = 100demand 25.9 against a capacity of 33.8 — F/μN = 0.77the weight appears nowhere in the cone — only the direction of the reaction is asked about The force that is whatever it needs to be Every other force in statics has a value the equations produce. Friction has an inequality instead, so it takes whatever value equilibrium demands and the bound only ever says no — which means a problem with friction in it has a range of answers rather than one.

The objects this essay names

Each one links to every other essay that touches it.

BallastBuoyancyControl perimeterDrainageEffective depthEquilibriumFactor of safetyFlotationFree bodyHydrostatic pressureOverturningPunching shearTension pileUpliftWater table