Structural Engineering & Statics
where the load goes
Equilibrium

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.

Assumes Weight is the only thing resisting it, The load that is spread out, and the force that replaces it and The middle third.

25 min read 10 figures The load that will not hold stillWhich failure arrives firstThe load must go somewhere

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.

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 2 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 above water 12.0 kN/m at 4.67 m, soil at the water table 48.0 kN/m at 2.00 m, submerged soil 27.2 kN/m at 1.33 m, water 78.5 kN/m at 1.33 m, and they sum to 185.7 kN/m — matched to 7e-8 by integrating the drawn profile numerically. The largest single term is the water, at 78.5 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 1.90 m above the base, 0.317 of the height rather than the third point at 2.00 m that a pure triangle would give.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
Fig. 1 The horizontal pressure on a 6 m wall retaining soil at 18 kN/m³ and 30° of friction, with 10 kPa of surcharge and the water table 2 m down, drawn once whole and then once per term. Surcharge 20.0 kN/m at 3.00 m, soil above the water 12.0 at 4.67 m, soil at the water table 48.0 at 2.00 m, submerged soil 27.2 at 1.33 m, water 78.5 at 1.33 m — summing to 185.7 kN/m, with the combined resultant at 1.90 m rather than at the third point.

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 ΣF=0\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:

Ka=tan⁡2 ⁣(45°−ϕ2),K0=1−sin⁡ϕ,Kp=tan⁡2 ⁣(45°+ϕ2).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 0.3330.333, 0.5000.500 and 3.0003.000.

One soil, one wall, and a factor of nineThe pressure on a 6 m wall retaining dry soil at 18 kN/m³ with a friction angle of 30°, in the three states Rankine's theory allows, drawn to one scale and with no surcharge so that the three thrusts stand in the ratio of the three coefficients exactly. Active is 0.333 and 108.0 kN/m; at rest is 0.500 and 162.0 kN/m; passive is 3.000 and 972.0 kN/m. That is a spread of 9.00 from end to end, decided entirely by which way the wall moved and by how far — and the active and passive pair are exact reciprocals, Ka·Kp = 1.000. All three resultants sit at the same third point, 2.00 m above the base, because all three profiles are the same triangle scaled.active — the wall moved awayK = 0.333108.0 kN/mat 2.00 m, the third pointat rest — the wall did not moveK = 0.500162.0 kN/mat 2.00 m, the third pointpassive — the wall pushed inK = 3.000972.0 kN/mat 2.00 m, the third pointKp/Ka = 9.00 and Ka·Kp = 1.000 — the same Mohr circle touching the same envelope from the two sides
Fig. 2 The same 6 m wall and the same dry soil in the three states, with no surcharge, so the three thrusts stand in the ratio of the coefficients exactly: active 108.0 kN/m, at rest 162.0 and passive 972.0. A spread of 9.00 end to end, and Ka·Kp = 1.000 — active and passive are exact reciprocals.

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.

Which coefficient applies is decided by how far the wall movesThe pressure coefficient on a 6 m wall against the wall's own movement, negative into the soil and positive away from it. A wall that has not moved carries K₀ = 0.500. Letting it retreat 6.0 mm — one thousandth of the height — lets the soil carry its own weight on shear and brings the coefficient down to 0.342, close to the active limit of 0.333. Pushing it the other way reaches 2.87 against a passive limit of 3.00, but only after 150 mm — 25 times as far. Both halves of the axis are at the same scale, which is the argument: the curve has a kink at the origin, with a stiffness of 26962 on the active side and 16177 on the passive, and mobilising passive resistance fully would move the wall further than anything standing on it can tolerate. A wall designed for the active state and then not allowed to move carries 1.50 times what it was checked for.-200-150-100-50000.511.522.53movement of the wall (mm) — negative into the soilpressure coefficientat rest: 0.50 at 0 mmactive: 0.34 at 6 mmpassive: 2.87 at -150 mmKp = 3.00Ka = 0.333passive costs25× the movement
Fig. 3 The pressure coefficient against the wall’s own movement, positive away from the soil and negative into it, both halves at the same scale. A wall that has not moved carries K₀ = 0.500. Letting it retreat 6.0 mm — a thousandth of its height — brings it to 0.342, near the active limit of 0.333. Pushing it the other way reaches 2.87 against a limit of 3.00, but only after 150 mm: twenty-five times as far.

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 KpK_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 K0K_0: 0.5000.500 against 0.3330.333, 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 KγH2/2K\gamma H^2/2, and it acts at H/3H/3 above the base. That third point is why a retaining wall is a wedge in section and why its overturning moment goes as H3H^3 rather than H2H^2.

A triangular load and the force that replaces itA triangular distributed load with its resultant computed by integration: an area of 108.0 acting at 4.00 from the left. The two moment diagrams below show what the substitution costs — the reactions are identical and the peak moment is not.resultant 108.0at x = 4.00, the centroid of the areamomentspread: 83.1replaced: 144.0reactions agree exactly (36.00 and 36.00); the peak moment does not
Fig. 4 The same triangle drawn as a span load rather than a wall pressure: 6 m long, peaking at 36 kPa, which is Ka·γ·H for this soil. Integration puts the resultant of 108.0 at 4.00 from the pointed end — the third point above the base of a wall. The moment diagrams below show what replacing the spread load by that resultant costs: the reactions agree exactly at 36.00 each and the peak moment does not, 83.1 against 144.0.

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.

The third point belongs to the triangle and to nothing elseWhere the resultant of the pressure on a 6 m wall acts, as a fraction of the height, against the surcharge on the retained ground. With no surcharge and no water the profile is a pure triangle and the answer is exactly one third: 0.3333. A surcharge is constant with depth, so its resultant sits at mid-height and drags the combined one upward — at 40 kPa it has reached 0.404. Water pulls the other way, because its own triangle is confined to the bottom 4 m and acts low: with the table at 2 m the curve starts at 0.295. The case the other views are drawn at, 10 kPa with water, sits at 0.317 — and every consequence downstream, the overturning moment above all, is that number times the thrust.05101520253035400.280.30.320.340.360.380.40.42surcharge on the retained ground (kPa)height of the resultant, over Hexactly ⅓ at 0.3333the drawn case: 0.317no water tablewater at 2 mone third
Fig. 5 Where the resultant acts, as a fraction of the retained height, against the surcharge behind the wall. With no surcharge and no water it is exactly 0.3333 and the curve is flat. A surcharge drags it upward — 0.404 at 40 kPa. Water pulls it down, its triangle being confined to the bottom 4 m: with the table 2 m down the curve starts at 0.295. The case the other figures are drawn at, 10 kPa with water, sits at 0.317.

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×yˉP \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. ΣM=0\Sigma M = 0 about the front toe gives overturning; ΣH=0\Sigma H = 0 along the underside gives sliding; ΣV\Sigma V with the moment about the base centre gives the eccentricity.

The wall stands, and its weight does not act at the middle of its baseA retaining wall 6 m high on a 3.5 m base, with a 1 m toe and a 2.10 m heel, holding soil at 18 kN/m³ under a surcharge of 10 kPa and drained. The thrust is 128.0 kN/m acting at 2.16 m above the base, so the overturning moment about the toe is 276 kNm per metre run. Against it the wall musters 324 kN of weight — stem, base and the column of soil standing on the heel — and that weight acts at 61.6% of the base rather than at its middle, which is what a heel is for. The restoring moment is 698 kNm, a factor of 2.53 against overturning and 1.26 against sliding. The base resultant lands 0.447 m off centre against a middle third of ±0.583 m, so the pressure under the base runs from 163 to 22 kPa and the whole width is in bearing. A rigid block carrying the same weight under the uniform pressure that matches this moment returns 2.05 instead of 2.53, because a block centres its weight and a wall does not.534220821middle thirdthrust 128.0 kN/m at 2.16 msurcharge 10 kPabearing 163 down to 22 kPa across the whole baseresultant 0.447 m off centre against a middle third of ±0.583 m — inside itoverturning 2.53sliding 1.26weight 324 kN at 61.6% of the base
Fig. 6 A 6 m wall on a 3.5 m base with a 1 m toe and a 2.10 m heel, drained, under 10 kPa of surcharge. The thrust of 128.0 kN/m at 2.16 m gives 276 kNm about the toe; against it the free body musters 324 kN at 61.6% of the base, restoring 698 kNm. Overturning 2.53, sliding 1.26, and the base resultant 0.447 m off centre against a middle third of ±0.583 m, so the whole width bears, at 163 down to 22 kPa.

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 doesRetained soil drawn as the fluid it is equivalent to — the density a real liquid would need in order to push on the wall as hard, which is Ka·γ — against the soil's friction angle, for unit weights of 18 and 20 kN/m³. Water is the horizontal line at 9.81, and the curves cross it at 17.1° and 20.0°: above those angles the soil is the lighter load and below them it is the heavier one. At the 30° the other views are drawn at, a soil of 18 kN/m³ pushes like a fluid of 6.00 kN/m³ — 0.61 of what the water in it pushes with. So a wall holding saturated ground is carrying more water than soil, the water term does not care about the friction angle at all, and a blocked drain is the commonest way a retaining wall is lost.10152025303540455002468101214friction angle of the soil (degrees)equivalent fluid density (kN/m³)soil at 18 kN/m³soil at 20 kN/m³6.00 at 30°water: 9.81
Fig. 7 Retained soil as the liquid it is equivalent to — the density a real fluid would need to push as hard, which is Ka·γ — against the friction angle. Water is the horizontal line at 9.81, and the curves for 18 and 20 kN/m³ cross it at 17.1° and 20.0°. At the 30° everything else here is drawn at, soil of 18 kN/m³ pushes like a fluid of 6.00: 0.61 of what the water in it pushes with.

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.

The resultant has left the middle third and the heel has liftedA retaining wall 6 m high on a 3.5 m base, with a 1 m toe and a 2.10 m heel, holding soil at 18 kN/m³ under a surcharge of 10 kPa with the water table 2 m down. The thrust is 185.7 kN/m acting at 1.90 m above the base, so the overturning moment about the toe is 353 kNm per metre run. Against it the wall musters 338 kN of weight — stem, base and the column of soil standing on the heel — and that weight acts at 61.9% of the base rather than at its middle, which is what a heel is for. The restoring moment is 734 kNm, a factor of 2.08 against overturning and 0.91 against sliding. The base resultant lands 0.625 m off centre against a middle third of ±0.583 m, so the pressure under the base runs from 200 to 0 kPa and 0.12 m of it has lifted off. A rigid block carrying the same weight under the uniform pressure that matches this moment returns 1.68 instead of 2.08, because a block centres its weight and a wall does not.53427614721middle thirdthrust 185.7 kN/m at 1.90 msurcharge 10 kPabearing peaks at 200 kPa over 3.38 m of contactresultant 0.625 m off centre against a middle third of ±0.583 m — 0.12 m of the base has liftedoverturning 2.08sliding 0.91weight 338 kN at 61.9% of the base
Fig. 8 The same wall, soil and surcharge, with the drain blocked and the water table 2 m down. The thrust rises from 128.0 to 185.7 kN/m and its lever falls from 2.16 to 1.90 m, so the moment about the toe goes from 276 to 353 kNm; the saturated heel takes the weight from 324 to 338 kN. Overturning falls from 2.53 to 2.08 and sliding from 1.26 to 0.91, and the base resultant leaves the middle third at 0.625 m against ±0.583, lifting 0.12 m of the heel and peaking at 200 kPa.

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.

Weight is the only thing holding it downA body 3.5 m wide and 6 m tall weighing 324 kN, under a wind pressure of 15.33 kN/m². The wind delivers 92 kN and an overturning moment of 276 kNm about the leeward toe; the weight restores 567 kNm, a factor of 2.05. The resultant lands 0.85 m from the centre against a middle third of ±0.58 m, so the base is lifting over 0.81 m of its width.92 kNW = 324 kNmiddle third: ±0.58 mresultant at 0.85 mrestoring 567 kNmoverturning 276 kNmfactor 2.05
Fig. 9 The drained wall above as a rigid block of the same base width carrying the same 324 kN, under the uniform pressure of 15.33 kPa that reproduces its overturning moment exactly. The moment is right at 276 kNm and the answer is not: the restoring moment is 567 kNm rather than 698, a factor of 2.05 against the wall’s true 2.53. The resultant lands 0.85 m off centre against a middle third of ±0.58, so the block reports a base lifting over 0.81 m where the real wall is wholly in bearing.

The load is the first failure. A block takes a uniform pressure at a lever arm of H/2H/2; earth pressure acts at H/3H/3, and no single uniform pressure reproduces both the force and the moment — the one matching the moment is KγH/3K\gamma H/3, the one matching the force is KγH/2K\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⋅B/2W \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 pressure runs away outside the middle thirdPeak bearing pressure under a 3.5 × 1 m base carrying 324 kN, against the eccentricity of the load. Inside the middle third the line is straight and the pressure has doubled by the time it reaches the edge of it: 93 kPa at the centre, 185 kPa at e = B/6. Beyond that the base lifts, the contact length shortens, and the curve turns upward without limit — at e = 1.35 m the peak is 535 kPa on 1.21 m of base.00.20.40.60.811.20100200300400500eccentricity of the resultant (m)peak pressure (kPa)B/6: the base is on the point of liftinguniform: 93 kPa
Fig. 10 Peak pressure under a 3.5 by 1 m base carrying 324 kN, against the eccentricity of the resultant. Inside the middle third the relation is straight and the pressure has doubled by the edge of it — 93 kPa uniform, 185 kPa at e = B/6. Past that the base lifts, the contact shortens, and the curve turns upward without limit: 535 kPa on 1.21 m of contact at e = 1.35 m.

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.

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 K0K_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.

Where this goes next

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. 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. rectangle±100 of 600 mm33.3% of the depth The middle third A material that cannot be pulled imposes a condition on where the load may land, and the condition is a region rather than a point. For a rectangle it is the famous middle third; for every other section it is a shape nobody quotes, and one ordinary section's is nearly twice as generous as the rule allows. 00.020.040.060.080.10.120.1400.20.40.60.81sag at the middle of the baywater depth ÷ critical depthλ = 1.00, approached and never reachedreversed construction camber: δ₀ = 0.01 The water that will not run off A flat roof deflects, the deflection makes room for water, the water deepens the deflection. It is the same equation as a buckling column, with rain in place of the axial load.

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Bearing pressureFrictionLateral pressureLever armMiddle thirdOverturningResultant