Structural Engineering & Statics
where the load goes
Field

Equilibrium

Nothing moves, so everything adds to nothing — and the free body decides what everything means.
A beam, its loads and its reactions. A 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.

Everything adds to nothing, and that is the whole of statics

A structure that stays put obeys two statements — the forces on it sum to zero, and so do the moments. Every number in the subject comes out of those two sentences.

 The same beam, cut at x = 5. A beam separated at one station. On the exposed face a shear force and a bending moment appear, equal and opposite on the two pieces, with values obtained by summing the forces on whichever piece is easier.

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.

 Three forces must meet at a point. A body held by two supports and one load. The reaction at the roller is vertical and the load's direction is given, so their lines of action fix a meeting point — and the pin reaction has to point at it.

Three forces must meet at a point, and a drawing can find it

A body held by exactly three forces has their lines of action concurrent. That is a theorem, it is enough to solve for direction and magnitude, and for a century it was done with a straightedge.

 Counting unknowns against equations. Three frames differing by one member. Two equilibrium equations per joint, one unknown per member and one per restraint: fewer unknowns than equations is a mechanism, equal is solvable by statics, more needs stiffness.

Counting the unknowns, and finding out whether statics can answer

Two equations per joint, one unknown per member, one per restraint. Subtract, and the sign of the answer says whether the structure is a mechanism, solvable, or beyond what equilibrium alone can settle.

 A truss cut through panel 3. The truss severed through one panel, with everything to the right removed and the three cut member forces drawn on the exposed faces. Taking moments about the marked joint removes two of the three unknowns, so one equation gives the third: -52.94.

Answering one question without solving the rest

A truss of fifty members can be interrogated about one of them. Cut through three, take moments about the point where two of them meet, and the third falls out in a single line.

 A triangular load and the force that replaces it. A triangular distributed load with its resultant computed by integration: an area of 24.0 acting at 5.33 from the left. The two moment diagrams below show what the substitution costs — the reactions are identical and the peak moment is not.

The load that is spread out, and the force that replaces it

A distributed load can be swapped for a single force at its centroid. The reactions come out identical and the bending moment does not, and knowing which side of the cut the swap is legitimate on is most of the skill.

 A beam, its loads and its reactions. A 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. The moment of everything on the body is summed about three marked points, and every sum comes to zero. Nothing here is a new equation: with only vertical forces on the body, the moment about a point does not depend on that point's height, so every centre anywhere on the plane returns the same equation — and the third equilibrium equation, the horizontal sum, reads nothing equals nothing.

The equation that is not new, and the three that are

A plane free body yields exactly three independent equations. Most attempts at a fourth are one of the first three wearing different clothes — and on a beam under vertical load, one of the three is already saying nothing.

 The count is necessary and not sufficient. Two pin-jointed frames, each satisfying m + r = 2j exactly. One of them folds anyway, because the equations are not independent; the ghosted outline is the motion that costs no member any change of length, drawn at an exaggeration of 0.55 of the span.

The count that does not see it

A frame can have exactly as many unknowns as equations and fold up anyway. The count asks whether there are enough equations; it never asks whether they are different from one another.

 Weight is the only thing holding it down. A body 2.5 m wide and 6 m tall weighing 120 kN, under a wind pressure of 1 kN/m². The wind delivers 48 kN and an overturning moment of 144 kNm about the leeward toe; the weight restores 150 kNm, a factor of 1.04. The resultant lands 1.20 m from the centre against a middle third of ±0.42 m, so the base is lifting over 2.35 m of its width.

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.

 Where a beam's load comes from. A 8 × 6 m panel carrying 5 kN/m², divided at 45° from the corners. The long beams take a trapezoid of 15.0 m² each and the short beams a triangle of 9.0 m²; the four areas sum to 48.0 m², which is the panel, so no load has been invented or lost. The line loads quoted are the uniform equivalents; the real distributions peak at 15.00 kN/m at midspan.

The load a beam is given is a decision

Every beam calculation so far has started with a load per metre, handed over as though it were a property of the beam. It is not. It is the answer to a prior question nobody draws, and two defensible answers to it differ by sixty per cent on the same floor.

 Three legs, three equations, one answer. A rigid top on three legs carrying 120 kN at (0.3, 0.2) m. The three equilibrium equations available — one vertical and two moments — leave three unknowns, so the system is exactly determinate and the reactions are 56.8, 13.6, 49.6 kN. Move the load anywhere and the answer moves with it; nothing about the legs' stiffness enters.

Six equations, and the drawing shows three

Every essay so far has taken place on a piece of paper, where equilibrium is three equations and a structure is a diagram. The real object has six, the extra three are the ones nobody writes down, and the difference between three legs and four is not a matter of degree.

 The reaction lies inside the cone, so the block stands. A 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.

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.

 Five loads behind one wall, and the water is the biggest. The 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 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.

 A general force system is a screw, not a force. Two forces that are neither parallel, nor concurrent, nor coplanar. Their resultant is (30, 0, -60) and no point in space reduces them to it alone: the moment left on the central axis is (14.40, 0.00, -28.80), which is parallel to the resultant and has a pitch of 0.4800. The simplest object equivalent to the system is therefore a force and a couple about the same line — a wrench — and a planar drawing can never produce one, because a coplanar system has its moment perpendicular to its resultant by construction.

Moving a force, and what it costs

Every free body begins by putting a force somewhere convenient. That move is free along the force's own line, costs a couple across it, and in three dimensions leaves behind something no choice of point can remove.

 A basement is a boat. A 20 by 30 m substructure dug 6 m into ground whose water table stands 2 m down. The head on the underside of the base slab is 4.0 m, so the pressure there is 39.2 kN/m² over the whole plan — 23.5 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 0.80. The structure floats if the water reaches 2.82 m below the ground, and a base slab alone would have to be 1.64 m thick to hold it down.

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.

 The force runs along the chord whatever route the member takes. A member pinned at two points 10 m apart, loaded only at those two points, and bowed 1.2 m off the line between them. Equilibrium leaves the two end forces no choice: equal, opposite, and along the chord. So at every section the axial force is P cos α, the shear is P sin α, and the bending moment is the force times the perpendicular offset — 240 kNm at the crown for 200 kN at 1.2 m, which is a multiplication rather than an analysis. Straighten the member and the moment diagram is identically zero; that is the case a truss member is in, and the reason it carries one number.

The member with only one direction

If a body is in equilibrium under forces applied at exactly two points, those forces are equal, opposite and along the line joining the points. It is three conclusions from two equations, it is the shortest real theorem in statics, and nearly everything that follows depends on it without saying so.

 A tank grows without limit; a silo stops. Vertical pressure against depth in a 8 m silo of a solid weighing 9 kN/m³, beside the straight line a liquid of the same weight would have produced. The free body is a slice: its own weight in, the wall friction out, and the friction is μK times the vertical pressure that generates it. The result saturates at γR/μK = 89 kN/m² and reaches 63% of it at one characteristic depth, R/μK = 9.9 m. At the base the pressure is 85 kN/m² against a liquid's 270 — 69% less — and the wall has taken 69% of the stored weight down with it. The exponent is the capstan's, and for the same reason.

The pressure that stops growing

A tank of liquid presses harder the deeper it gets, without limit. A silo of grain does not. Wall friction carries part of the weight, the pressure that generates the friction is proportional to the pressure being carried, and the equation that follows is the one that describes a rope round a bollard.

 Balanced, and four times as heavy on the bearing. A bascule leaf of 900 kN whose centroid is 9 m from the trunnion, balanced by 2700 kN at 3 m on the other side. What balancing achieves is exactly one thing: the moment about the pivot is zero at every opening angle, because both terms carry the same cosine. What it costs is two things that are not zero. The reaction on the trunnion becomes 4.0 times the leaf's own weight, since both weights are still there. And the rotational inertia rises by 33%, so the balanced leaf is the hardest one to start and to stop — which is why the counterweight is put as close to the pivot as it will fit, at the price of being heavy: the same balance at twice the radius weighs 1350 kN and carries 1.25 times the inertia.

Balanced, and four times as heavy

A counterweight cancels a moment about a pivot, and that is the only thing it cancels. The bearing beneath carries both weights, the inertia rises as the square of the radius, and a load that moves cannot be balanced at more than one position at all.

 The envelope is not a state of the structure. Every arrangement of the imposed load on three spans — 8 of them, since each span is loaded or not — drawn faintly, with the greatest sagging and greatest hogging at each station drawn over them. Each faint curve is a real state of equilibrium and satisfies the free-moment identity exactly: mid-span ordinate minus the mean of the end moments is wL²/8, to 3e-16 of it. The envelope satisfies it nowhere, missing by up to 23% — because it is assembled from different load cases at different stations and no arrangement of load produces it. seven of the 8 arrangements are needed to build it; the rest never govern anywhere.

The envelope is not a structure

A continuous beam whose imposed load may sit on any span has eight load cases, and every one of them is a genuine state of equilibrium. The curve the design is made against is not one of them — it is assembled from different cases at different stations, and it fails the identity all eight satisfy exactly.

 The two theorems close on the answer from opposite sides. A pinned-base portal frame under 200 kN at mid-span and 80 kN at the eaves, with a plastic moment of 200 kNm. The three horizontal lines are mechanisms — work equations, each an upper bound: the beam mechanism gives 1.000, sway 1.250 and the combined one 0.714. The curve is the lower bound: for each value of the one redundant, the largest load factor whose moment field stays inside the plastic moment everywhere. Its peak is 0.714, and it touches the lowest mechanism exactly — so the collapse load is known rather than bracketed. Using the beam mechanism instead would have claimed 1.40 times the real capacity, and the sway one 1.75 times, both of them on the wrong side.

Two ways of being wrong

Plastic analysis has two theorems and they point in opposite directions. Any equilibrium field that nowhere exceeds the plastic moment gives a load at or below the collapse load; any mechanism gives one at or above it. Only one of those errors is safe.

 The point the rafter turns about, which is off the frame. A pitched portal of 8 m span and 4.0 m to the eaves, with a 1.5 m rise, collapsing. Each rigid part of the mechanism rotates about some point: the left column about its base hinge, the right about its own. The rafter between them does neither, and its centre is found by one rule — two bodies joined at a hinge share that hinge, so the second body's centre lies on the line through the first body's centre and the hinge, extended. Two hinges give two lines and they cross at (8.0, 11.0) metres, which is 5.5 m above the ridge and outside any drawing of the frame itself. From there the whole collapse is two ratios of lengths and no trigonometry: the load factor is 1.339. Flatten the roof and the centre descends; make the two lines parallel and it goes to infinity, which is the statement that the rafter translates instead of turning.

The point the mechanism turns about

A collapsing frame is a chain of rigid pieces, and every piece is rotating about some point. Find those points and the whole collapse load reads off two ratios of lengths, with no trigonometry anywhere — and for a pitched roof the point in question is well above the top of the drawing.

 At thirty degrees each leg carries the whole load. A 100 kN lift on two legs at 60 degrees to the horizontal. Each leg is a two-force member, so its force is the vertical share it carries divided by the sine of its angle: T = W/(2 sin β) = 57.7 kN, which is 0.58 times the whole load in each. The curve on the right is that division, and it is the reason the rule about sling angles exists rather than a convention: at sixty degrees a leg carries 0.58 W, at forty-five 0.71, at thirty exactly 1.00, and at fifteen 1.93. And the horizontal components do not disappear — they run through the thing being lifted, which here carries 57.7 kN of compression between the two pick points. That is what a spreader beam is for: it takes the compression as a designed strut so the legs above it can stand up.

The angle that doubles the force

A crane hook takes the weight and nothing else. What holds the load is a set of legs running down to it at an angle, and each leg is a two-force member — so at thirty degrees from the horizontal each of two legs carries the entire load, and the difference goes into the thing being lifted as compression.

 The wind pushes on one face and pulls on three. A 30 × 20 m building in plan, with the measured pressure coefficient on each face and the arrows drawn in the direction the pressure acts. Only the windward face is pushed; the other three are sucked, and the side faces are sucked hardest of all at c_p = -0.7. The horizontal resultant is 842 kN at a velocity pressure of 0.9 kPa, and the arithmetic of it is the whole point: the leeward suction pulls the building downwind, so it ADDS, supplying 38% of the answer, while the two side faces cancel each other exactly and supply none of it. The coefficients are wind tunnel data; what is computed is the free body they are applied to.

Most of it is suction

A wind load is drawn as arrows pressing on the windward face, which is where about three fifths of it comes from. The rest is a pull on the back. The two side faces carry the largest suctions on the building and contribute nothing at all to the answer — and the inside of the building, which nobody draws, decides whether the roof stays on.

 The snow that left the roof is standing against the wall. A roof in section with a 1.0 m obstruction at its downwind end, drawn with the vertical scale exaggerated 4 times because a metre of snow on twenty metres of roof is thinner than the line it would be drawn with. The balanced layer is 0.48 kN/m² everywhere; the wedge against the wall holds the snow that 30% of the 20 m upwind gave up, so its area is fixed by conservation of mass rather than chosen. That makes it 0.85 m deep and 3.4 m long, with a peak load of 2.60 kN/m² — 5.4 times the balanced value, and still only 27% of the snow on the roof. An intensity several times the design load, made of a quantity nobody would notice had moved.

The load that arrives where the wind stops

Snow is the one load a structure is given rather than subjected to. What falls is spread evenly over a whole region; what a member carries is whatever the wind left above it, and the wind piles it against whatever gets in the way. The heaviest patch on a roof is usually a quarter of the snow on it.

 The load that makes itself worse. The amplification of a deflection against the ratio of applied load to buckling load. A structure at half its buckling load deflects twice as far as first-order analysis predicts, and the curve runs away well before the load is reached.

The load that is really a lean

No frame is ever plumb. The columns are out of upright by something like a three-hundredth, and every tonne of gravity load standing on that lean has a horizontal component. The force that represents it is not a safety allowance — it is an exact statics substitution for a geometry nobody drew.

 Deflection goes as the fourth power of the span. Deflection against span for a constant load intensity and section, with two slower relationships drawn faintly behind it for comparison: the load itself, which grows in proportion to the span, and the bending moment, which grows as its square. Doubling the span multiplies the deflection by sixteen, while the moment only quadruples.

The weight that has to be known before it can be found

Every other load arrives from outside and can be looked up. A structure's own weight depends on how big it is, and how big it is depends on the load — so the first calculation on any project is a fixed point, and the fraction of a member spent carrying itself turns out to be the square of its span as a fraction of a span it can never reach.

 A bearing capacity is a mechanism, and here it is. Prandtl's collapse mechanism under a 3.0 m footing in a soil of 32° friction. A rigid wedge is driven down with the footing at 61° to the horizontal; a fan of radial shear turns the stress through exactly ninety degrees on a logarithmic spiral whose growth rate is tanφ; and a passive wedge at 29° has to be pushed up and out of the way. Nothing here is empirical — every angle is a function of φ alone — and the mechanism reaches 15.9 m from the centre, which is 10.6 times the footing's half width. That is why two footings closer together than about four widths do not have separate bearing capacities.

The ground is a mechanism

Bearing capacity is met as a formula with three terms and a table of coefficients, and that presentation hides what it is. Underneath is a plastic collapse mechanism — a rigid wedge, a fan of radial shear on a logarithmic spiral, and a passive wedge that has to be pushed up and out of the way — and every coefficient in the table is a property of that one drawing.

 The check that everything adds up, and the error it cannot see. Four versions of the same 3-bay, 4-storey frame, with the global equilibrium residual each one produces — the sum of the reactions against the sum of the applied loads, as a fraction of the applied total. It is the first thing every analysis prints and it is worth having: a lost restraint and a load entered in the wrong unit both show up immediately, at 8% and 32%, because both change what the structure is carrying. The fourth bar is the point. A member whose stiffness is wrong by a factor of ten redistributes the internal forces completely — the second bar shows the change in the member forces, 24% — and the global residual is exactly zero, because the wrong answer is still in equilibrium with the same loads. Equilibrium is one equation per degree of freedom of the whole body, and a stiffness error lives entirely in the many equations underneath it. A model can satisfy every equilibrium check ever devised and be a model of a different structure.

The check that cannot see the error

Every analysis prints a global equilibrium residual, and it is the first thing anybody looks at. It catches a lost restraint and a load entered in the wrong unit immediately. It is structurally incapable of catching a member whose stiffness is wrong by a factor of ten, because the wrong answer is still in equilibrium with the same loads.

 The force nobody applied, and the speed it wins at. Lateral force per unit weight for a vehicle on a 400 m curve, against speed. The rising curve is what the free body demands — v²/gR, which is the body's own acceleration written on the other side of the equation — and the flat line is what 6.0° of cant supplies from the weight. They cross at 73 km/h, which is the speed the curve was set out for; below it the deficiency has the other sign and the rail is pushed the other way. The upper line is overturning, at b/2h = 0.399 — and there is no mass in that number, so a loaded vehicle and an empty one go over at the same 160 km/h and only the height of the load decides. At the 108 km/h drawn the deficiency is 0.124 of the weight, which is 49 kN on this 40 tonne vehicle.

The force that is really an acceleration

Every other load in this collection is applied by something. This one is applied by nothing at all — it is the body's own acceleration, written on the other side of the equation so that statics can be used on a problem statics has no business with. The move is legitimate, it is a hundred and eighty years old, and it is exactly half done more often than it is done.

 Two identical pipes, and one carries three times the other. Load per metre on a buried conduit against the depth of cover, in trench widths, with the weight of the prism of soil directly above it drawn between them. A conduit laid in a narrow trench is stiffer than nothing and softer than the sides: the backfill settles relative to the undisturbed ground, the friction on the trench walls acts upward, and the conduit gets 64% of the prism. Lay the same conduit on the ground and build an embankment over it and it is now stiffer than the fill beside it, the interior prism settles less, the friction acts downward, and it gets 172% — a factor of 2.71 between two pipes with nothing different but which way the ground moved. The equation is Janssen's, the same one a silo wall obeys, with a trench for a silo; both curves start on the prism line, because with no depth there is no shear to redistribute anything. This is why a flexible pipe is buried rather than a rigid one: making the conduit weaker moves it down the page.

The pipe decides what the soil weighs

A buried conduit is not loaded by the soil above it. It is loaded by whatever share of that soil the relative movement leaves it — and which way the shear on the sides of the prism acts depends on whether the conduit settles more or less than the ground beside it. Two identical pipes under identical fill, one carrying two thirds of the prism and one carrying nearly twice it.

 The circle is searched for, and the first guess is 39 per cent optimistic. The same slope with 81 trial circles evaluated, each one through the toe and each one giving its own factor of safety. There is no equation whose solution is the answer: the slip surface is a shape the ground chooses, so the calculation is a search over shapes and the answer is the smallest number found — 1.191 against 1.650 for the circle a first guess puts through the toe from above the middle of the slope, which is 39 per cent optimistic. A slope analysis that reports one circle has reported nothing.

The surface that has to be searched for

Every other check in this collection is made at a section somebody drew. A slope has no section — the failure surface is a shape the ground chooses, so the calculation is a search over shapes, and the answer is the smallest number found rather than the solution of anything.

 Every pressure points at the pin, so the water lifts nothing. A radial gate of radius 8 m holding 6 m of water, with its pivot 6 m above the sill. The pressure on a curved surface cannot be obtained by multiplying anything by anything, so it is integrated round the arc: the horizontal component comes to 176.6 kN/m and the vertical to 110.5. Both are recoverable without any integral at all — the horizontal is the pressure force on the surface's own vertical projection, γH²/2 = 176.6, and the vertical is the weight of the water standing above it, 110.5. They agree to 0.000 per cent. And because every pressure is normal to a circle, every one of them passes through the centre: the moment of the whole 208 kN/m about the pivot is -3.4e-15 kNm, against 353 for a flat gate on the same hinge.

Every pressure points at the pin

Pressure acts normal to a surface, so on a curved one every element pushes in a different direction and no multiplication gives the resultant. Two free bodies recover it without an integral — and on a circular surface a third observation makes the whole force disappear from the equation a hoist has to satisfy.

 Independence, and the floor it never goes below. The reduction factor a column is allowed, two ways. The Eurocode storey rule falls to 0.70 and stops. The independence argument — n bays each with a mean and a standard deviation, summed — gives (1 + zv/√n)/(1 + zv), which falls faster and stops at 0.503: a floor with no n in it at all, decided only by how variable the load is and how far out the fractile is drawn. The mean is never reduced away, because every bay really does carry its mean. At ten storeys the two differ by 10.0 points.

The load that is never all there at once

A column at the bottom of twenty storeys is designed for the imposed load of twenty floors added up, and twenty floors do not reach their own worst day together. The reduction that follows is not a discount on the safety margin. It is the central limit theorem, and it has a floor it never goes below.

 The reaction lies inside the cone, so the block stands. A block of 48 on a plane at 22°, against a coefficient of friction of 0.6. Resolving across and along the plane gives a normal force of 44.5 and a friction demand of 18.0, against a capacity of μN = 26.7 — a ratio of 0.67. Added together the two make one contact reaction leaning 22.0° from the normal, and the admissible reactions fill a cone of half-angle arctan μ = 31.0°. 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 22.0°, which is why the angle of repose is a material property and the size of a heap of sand is not.

The area that is not in the equation

Friction is proportional to the force pressing two surfaces together and independent of how large they are, which sounds like an approximation and is not. The area is absent because the contact that carries the load is a tiny fraction of the contact that is drawn, and that fraction grows in exact proportion to the load.

 Weight is the only thing holding it down. A body 1.6 m wide and 4.5 m tall weighing 22 kN, under a wind pressure of 1 kN/m². The wind delivers 2 kN and an overturning moment of 4 kNm about the leeward toe; the weight restores 18 kNm, a factor of 4.35. The resultant lands 0.18 m from the centre against a middle third of ±0.27 m, so the base is still wholly in bearing.

Whether it tips or slides

A free body pushed sideways has two ways of leaving, and which one it takes is decided before any load is known. The condition is a width divided by a height set against a coefficient of friction, and the weight, the wind pressure and the depth of the body all cancel out of it.

 A preloaded joint, before and after it slips. Eight preloaded bolts at 100 kN each, on two friction faces at μ = 0.35. The joint carries 560 kN by friction with the bolts in tension and not in shear at all; past that it slips into bearing and carries 900 kN with the bolts now in shear. Two different mechanisms, one joint.

The force that is capped on purpose

Everywhere else in this collection friction is a nuisance whose value nobody controls, checked with a coefficient known to one figure. In a friction damper the inequality is the design intent — the device is specified so that a member behind it can never be asked for more than a stated force.

 Two load sets with the same resultant and different work. The two ways of putting a uniform load of 10 kN/m onto a beam element 6.00 m long. Both put 30.0 kN at each node, so both have the same resultant and the same moment about any point — they are equivalent for a rigid body. The consistent set adds a couple of 30.0 kN·m at each end, in opposite senses, which is what makes it do the same virtual work over the element's shape functions as the real load does. The couples cancel in the resultant, which is exactly why the resultant cannot see them, and they are the whole difference between an exact answer and one that is a third out.

Equivalent in work, not in resultant

Two force systems with the same resultant and the same moment about every point are interchangeable — for a rigid body. A finite element is not a rigid body, and substituting one for the other on a beam element leaves the tip of a cantilever a third too low with no warning of any kind.

 Both checks pass, and the contact lets go. A contact pressed together by 1000, with μ = 0.4. Every tangential force the contact can supply lies inside a disc of radius μN = 400.0, because the friction law bounds the length of the force and not its components. The contact is asked for 300.0 one way and 300.0 the other way — 75% and 75% of the radius taken one at a time — and 424.3 together, 106% of it. Each one-direction check passes and the force does not fit: the square those checks describe reaches √2 times further at its corners than the contact can.

Seventy-five per cent each way

A contact asked for friction in two directions at once can supply a force of a certain length pointing any way it likes, so its limit is a disc and not a square. Two checks made one direction at a time, each passing at seventy-five per cent, describe a contact that has already let go.

 The pier grips, gives, and grips again. A sliding bearing carrying 3000 kN on a pier head of 20 kN/mm, dragged by a deck expanding at 1.7 mm an hour, with a static coefficient of 0.05 and a kinetic one of 0.03. The force in the pier climbs while the bearing grips, reaches 150 kN, and falls in a fraction of a second to 30 kN: the pier springs back under only the kinetic friction, overshoots the 90 kN that friction would hold it at, and grips again. The swing is 120 kN — 2.00 times the 60 kN between the two coefficients — and the pier head jumps 6.00 mm each time, three times in 12 hours.

The pier that moves in jumps

A sliding bearing whose static friction is larger than its kinetic friction does not release a slow thermal movement as a drift. It grips, gives and grips again, and each time the force in the pier swings by twice the difference between the two coefficients — whatever the pier is made of.

 Same deck, same load, and two pier forces. A deck bearing on a pier of 20 kN/mm with μ = 0.03, taken to the same final state two ways: the deck moves 4 mm over the pier, and the bearing's load rises from 2000 to 4000 kN. Moved first, while the bearing carries 2000 kN, the pier force reaches the limit of 60 kN and the bearing slides for the rest of the movement; the load arriving afterwards raises the limit and changes nothing, and the pier is left carrying 60 kN. Loaded first, the limit is 120 kN before the deck moves, the bearing grips throughout, and the pier carries 80 kN. Both states are at the same displacement under the same load, and both satisfy equilibrium and the friction bound; the order is the only difference, and it appears in neither.

The order the loads arrived in

Statics allows a contact with friction a whole range of forces and has no way to choose between them. A real structure does choose, and what it chooses by is the order in which things happened to it — so the force in a pier under a sliding bearing is a record of its history, not a function of its loads.

 The resultant has left the base, and it tips. A body on three supports weighing 60 kN, pushed sideways by 16 kN at a height of 3 m in the plan direction 270°. The push moves the resultant of weight and push 0.80 m from under the weight, and the base — the convex hull of the supports, shaded — lets it go 0.75 m that way before the edge drawn heavy becomes a tipping line: a factor of 0.94, found both along the ray and by moments about that edge. The dashed rosette is the same reach in every direction, from 0.75 m toward the middle of the nearest edge to 1.50 m toward the furthest support. To hold it the support opposite the tipping edge would have to pull 1.3 kN, which a support standing on the ground cannot do, so it lifts and the body turns about that edge.

Half as far between the legs

A body standing on feet, legs or pads has for its base the polygon its supports enclose, and how far its weight can be pushed before it tips depends on which way it is pushed. A three-legged stand pushed toward the gap between two legs has exactly half the reach it has pushed toward one of them.

 Four forces pair off, and the line joining the pairs carries both resultants. A beam 8 m long held by the vertical link at the left end, the strut at 6 m, the horizontal link at the right end, under a load of 10 kN at 3 m inclined at −60.0°. Pairing the load with line A: the two cross at P, (0.00, 5.20) m, and lines B and C cross at Q, (6.00, 0.00). The resultant of the first pair passes through P and that of the second through Q, and since they balance each other both lie on PQ, dashed. The force polygon on the right is the load, then A, B and C, closing where it began, with PQ's direction as the diagonal that splits it into two triangles. The forces are A 4.33 kN along its line, B 6.12 kN along its line, C 0.67 kN against its line, as the three equations of equilibrium give them.

The line that pairs four forces

Three forces in equilibrium meet at a point; four need not. But they pair off. The resultant of two passes through the point where their lines cross, the resultant of the other two through theirs, and the two resultants must share the line joining those points. Culmann's line turns a four-force body into two triangles — and the method of sections into a drawing.

 The pole, the strings, and the resultant of any number of forces. Five downward loads on a span of 10 m — 30 kN at 1.5 m, 20 kN at 3.5 m, 45 kN at 5.0 m, 25 kN at 7.0 m, 35 kN at 8.5 m — adding to 155.0 kN. On the right, the loads laid end to end down one line, with a pole 60.0 kN to the left of it and a ray drawn to every division between them. On the left, the funicular polygon: each segment parallel to the ray of the loads it has passed, so the shape is the one a string carrying these loads would hang in. The first and last strings are extended until they cross, at 5.24 m, and that crossing is where the 155.0 kN resultant acts — the same station the moment sum Σ P x / Σ P gives, 5.24 m, reached with no pole in it at all.

The pole decides the drawing, not the answer

Five forces will not pair off the way four do. They need a point that is nowhere on the structure — chosen freely, by whoever is holding the pencil — and the string of lines it generates. Every choice draws a different polygon and finds the same resultant, and the shape it draws turns out to be the beam's bending moment diagram.

 A pin's force does not pass through its centre. Left, a pin of radius 0.15 m in its hole, with a coefficient of friction of 0.15. The reaction at the contact is inclined by φ = 8.5° to the radius through it, because the friction it can develop is that fraction of the force pressing the surfaces together, and the perpendicular distance from the pin's centre to that inclined line is R sin φ = 0.022 m. Every position the contact can take gives a line tangent to the same circle, shaded. Right, a link 0.48 m long pinned at both ends, at the same scale — 3.2 pin radii, which is a stubby linkage rather than a structural tie, drawn that way because at the thirty radii an ordinary tie has, the two circles are smaller than the pencil: its force is a common tangent to the two circles — the two solid lines for the two senses of rotation, the two dashed ones for the senses in which its ends turn oppositely — and the dashed centre line every construction in this collection draws is none of them. A pin of this size carrying 5.0 MN delivers a couple of 111.3 kN·m to whatever it is pinned to, which is the same offset read as a moment rather than as a distance.

The pin that is not a point

Every line of action drawn so far passes exactly through a pin's centre, which is true of a frictionless pin and of nothing else. A real one carries its force tangent to a small circle instead, so a link's line is a band, a construction's answer is a range, and a support drawn as a hinge hands a couple of a hundred kilonewton-metres to whatever it is pinned to.

 The column under the first interior support is given a quarter more. Two equal spans of 8.0 m carrying 5.0 kN/m, continuous over three supports. The tributary rule gives each interior support one span's worth of load, 40.0 kN, and each end support half of that. The continuous beam gives 15.0 kN, 50.0 kN, 15.0 kN — ratios of 0.75, 1.25, 0.75 to what the areas say. The reactions still add to the whole load, because they must; what has moved is which support gets it. The end supports are relieved because the span next to them hogs over the first interior support, lifting their end of it.

The column given more than its rectangle

The tributary rule draws a rectangle round each column and hands it whatever stands inside. A floor is continuous over its columns, and a continuous beam does not give each support the load above it — the first interior one takes a quarter more and the end ones a quarter less. On a grid the two directions multiply, and two columns on the same floor differ by a factor of nearly three.

 The push that tips it, in every direction. The tipping push in each plan direction, drawn as a distance from the centre, for a 6000 kN body with its weight 15 m up and the push at 18 m. On rigid ground the curve is the hull's: weakest along the axes at 1333 kN. On four equal pads it shrinks, and more toward the corners. With the pads as built — stiffnesses 20000, 20000, 8000, 20000 kN/m — it is no longer symmetric: the weakest direction is 15°, on the soft pad's side, at 1039 kN.

It tips inside its own hull

On rigid ground a body tips when its resultant reaches the edge of its base, and how stiff its supports are has nothing to do with it. On pads that settle, the body leans as it is pushed, the lean moves its weight, and the push that tips it falls by one number a site engineer already has: settlement times the height of the weight, over the square of the half-width. Toward a corner the loss doubles, and one soft pad makes the weakest direction one nobody checks.

 The section, once the joint has supplied the fourth equation. Everything above a cut through panel 3. Four severed members; three equations. Moments about the mid-joint of the horizontal remove both diagonals and leave both legs. Moments about the point where the legs' lines meet, 36.0 m up, remove both legs and leave both diagonals. With the joint's result that the diagonals are equal and opposite, that second equation has one unknown: 20.83 kN in each diagonal. The legs follow: 26.1 T and 86.3 C.

The cut that needs a joint first

The method of sections works because a cut through three members leaves three unknowns and a point about which two of them have no moment. A K-braced tower has no such cut anywhere: every section severs two legs and two diagonals. One joint in the middle of a horizontal supplies the missing equation, and only in that order does each step have one unknown — after which the diagonals turn out to be carrying not the shear but the moment about the point where the legs would meet.

 A roof truss and its reciprocal figure. Left, a pitched roof truss of 8 m under three loads of 10.0 kN, its 13 members drawn in the colour of their force — 5 in tension, 6 in compression and 2 carrying nothing — with a letter on every region outside it between one external force and the next and a number on every cell inside it. Right, the force diagram: every lettered or numbered space is a point, every member is the line between the two spaces it separates, drawn parallel to the member and as long as its force, and every joint of the frame is a closed polygon. The eight joints and eleven spaces of the frame have become eight polygons and eight distinct points — fewer points than spaces, because some spaces land on one point, as the two either side of a member carrying nothing always do. Force times length adds to 190.0 kN·m over the tension members and 250.0 over the compression members, and the difference, −60.0, is fixed by the loads and where they act, whatever frame carries them. Each point was placed by crossing one member, and the 8 crossings not used to place anything all close to within 5e-15 kN.

Every space a point, every joint a polygon

A truss's force diagram is a second drawing of the truss in which the joints have become polygons and the spaces between members have become points. Maxwell showed in 1864 that the exchange runs both ways, so a designer can draw the forces first and ask what shape carries them. His theorem also says which frames have such a diagram at all, and the answer is a surprise, because it is about polyhedra.

 Two lines and a triangle: a three-hinged arch drawn. A three-hinged arch of 20.0 m span, springings at (0.0, 0.0) and (20.0, 0.0) m and the crown hinge at (10.0, 5.0), under 100.0 kN at 5.0 m. The right half carries no load, so it is a two-force member and its reaction lies along the line from its springing through the crown hinge. That line meets the load's line at K, 7.50 m up, and the left reaction must pass through K too. The triangle of the load and the two reaction directions gives the reactions as 90.1 kN at A and 55.9 kN at B, with a horizontal thrust of 50.0 kN — the values four equilibrium equations return, to 7e-15 kN. The shape of the rib entered nowhere.

The arch that is only its three hinges

A three-hinged arch's reactions come from three points and nothing else, so a parabola, a circle and a portal frame on the same hinges push on their abutments identically. Move a load across and the point where the reactions cross runs along two straight lines through the crown. That is the arch's influence line, drawn with a straightedge — and friction in the hinges it was built around blurs it.

 Putting a funicular through three points. Four loads — 40.0 kN at 3.0 m, 60.0 kN at 7.0 m, 30.0 kN at 12.0 m, 50.0 kN at 16.0 m — and three points the polygon must pass through: A and B at the springings and C, 5.0 m above their chord at 10.0 m. A trial pole, dashed, draws a polygon from A that ends 6.39 m below B. The ray through the trial pole parallel to its own closing line cuts the load line at Q, 95.0 kN from the top, which is the left reaction of a simple beam on A and B and does not depend on the pole at all. Every pole whose polygon passes through A and B lies on the line through Q parallel to A B; the one whose polygon also reaches C is 98.0 kN from the load line, which is the moment at C of that simple beam, 490.0 kN·m, divided by C's height above the chord. That is the three-hinged arch's thrust — the four equilibrium equations give 98.0 kN.

The polygon that runs out of freedom

A funicular polygon for given loads has exactly three freedoms, so it can be made to pass through three chosen points and no more. Three points is a three-hinged arch, and the drawing solves it. Take the crown hinge away and one freedom is left over. The drawing then offers a whole family of thrust lines and cannot say which one the arch uses — the rib's stiffness decides, and stiffness is not on the paper.

 Two pencil lines, and where they cross. Two lines drawn with a pencil 0.2 mm wide are two bands, drawn here much wider than a pencil so the shape can be seen, and they cross not at a point but in a parallelogram. At 60° apart the parallelogram's long diagonal is 2.0 pencil widths — 0.40 mm; at 12° apart the parallelogram's long diagonal is 9.6 pencil widths — 1.91 mm. The crossing's uncertainty along the bisector is the width divided by twice the sine of half the angle, so it grows without limit as the lines turn parallel, and a construction that finds a point by crossing two lines inherits it.

How wrong a drawing is

A pencil line is a band, and two bands cross in a parallelogram that grows as they turn parallel. The accuracy of a graphical construction is therefore a property of the angles it makes, not of the hand that made it — and the worst case is the shallow arch, the structure the method was most used on. Measured properly, the drawing's error there is the size of the builder's, and the check draughtsmen relied on cannot see it.

 Four ways of being nearly a mechanism. Four measures of the two-bar frame against the angle its bars make with the line between the supports, on logarithmic axes: the bar force as a multiple of the load, the condition number of the equilibrium matrix, the largest relative change of force from a one-millimetre error in any joint's position, and the joint's sag under the load as a share of the rise, for bars of axial stiffness 200 MN over a span of 8.0 m, loaded with 50 kN. The first three grow as one over the angle: at 2° the force is 14.6 times the load, the condition number 54, and a millimetre of error changes the force by 0.7 per cent. The sag grows as the cube: 4.3 per cent of the rise at 8°, 41 per cent at 4°, 3.1 times it at 2°. The measure that ends the analysis is not the arithmetic's.

Rigid by every test, and still folding

Counting the unknowns says whether a frame can be solved and the rank of its equations says whether it can stand, and both answers are yes or no. A frame a few degrees from a critical form passes both and is still nearly a mechanism — and of the ways that shows, the last to arrive is the one the rank test measures. Its own sag under load eats a tenth of its geometry at six degrees; sixteen-figure arithmetic would not notice anything until well below half a degree.

 The toe that runs out of ground. A block 6.0 m square on ground that bears 300 kPa at most — 10800 kN over the whole base — pushed 8.0 m up, weighing 3000 kN, drawn at three stages of the push with the lean exaggerated eight times and the contact pressure under the base; the dashed line is the ground's bearing limit. At a push of 629 kN it touches the ground over 3.9 m of its 6.0 and has not yet brought the ground to its limit anywhere. At a push of 770 kN it touches the ground over 2.5 m of its 6.0 and has brought the ground to its limit under 0.8 m. At a push of 776 kN it touches the ground over 2.3 m of its 6.0 and has brought the ground to its limit under 1.1 m. The last is the tipping push: the heel has lifted, the toe is pressing the ground at its limit over a widening block, and the lever arm of the weight about that block is shrinking as the block grows.

The ballast that helps it over

On rigid ground a heavier block is harder to tip, in exact proportion to its weight — that is the whole of the overturning check. On ground with a bearing capacity it is not. The toe presses the ground to its limit, the weight's lever arm shrinks as the yielded block under the toe grows, and the tipping push peaks when the block weighs half of what the ground under it can bear. Past that, every tonne of ballast added to steady it brings it closer to going over.

 The forces the girder can hold with nothing on it. The one self-stress state of a K-braced girder of eight 3 m panels, 3 m deep: member forces in equilibrium at every joint with no load and no reaction. It lives entirely in the two middle panels — twelve members: the four diagonals meeting at the middle joint of the central vertical in tension, and the four central chord members and the half-verticals at the outer side of each middle panel in compression, at 0.89 and 0.45 of the diagonal force. It is symmetric about mid-span, so a symmetric load can call on it as freely as any other: symmetry does not supply the missing equation. Every other member of the girder is idle in it, drawn faint.

The redundancy only the sun can find

A K-braced girder that is symmetric about mid-span has one member more than statics can resolve, and it is at the centre, where two K's meet at one joint. Symmetry does not remove it, because the forces it allows are symmetric themselves. Loads barely touch it — a section calculation that ignores it is exact for a load at almost any joint. What finds it is a millimetre of misfit or a sunlit top chord, which put forces into the middle of the girder that no load calculation contains.

 A downward pressure that pulls the corners up. The nodal forces that do the same work as a uniform pressure on one eight-node serendipity quadrilateral, as fractions of the whole load, each node's share being the integral of its shape function over the element. The shares are −1/12, −1/12, −1/12, −1/12, 1/3, 1/3, 1/3, 1/3 at its eight nodes, adding up to one. The four corner forces are negative: the pressure pushes down and the work-equivalent forces at the corners pull up.

The corners the pressure pulls up

Replace a uniform pressure on a beam element by its work-equivalent nodal loads and the ends acquire couples the resultant cannot see. Do the same on an eight-node plate element and the corners acquire forces pointing the wrong way: a pressure pushing down is represented by the corners being pulled up, a twelfth of the load each. It is correct, it is the vector that makes the solution the best one available, and it is the reason a printout of nodal forces is not a picture of where a load goes.

 Weakest the day it is finished. The factor of safety against tipping of a 15,000 kN tower on a 10 m square base over soft clay (undrained strength 35 kPa, gaining 0.22 of the effective stress the tower puts into it, consolidating with a 90 per cent time of 7.1 years), through twelve years, for construction times of three months, one year and four years; its wind push grows as it rises. Built in three months, it bottoms out at 1.14 on the day it is finished; built in one year, it bottoms out at 1.33 on the day it is finished; built in four years, it bottoms out at 1.60 on the day it is finished. After that every curve climbs toward 1.93, the factor once the clay has consolidated. Had the finished weight arrived in an instant the factor would be 0.92; on ground that could not fail it would be 3.00.

The ground that arrives after the weight

A block's resistance to tipping on ground that can yield is a parabola in its weight, with its peak at half of what the ground can bear. On soft clay the ground's capacity is not fixed: it grows as the clay drains under the weight, and it grows years after the weight has arrived. A tower on such ground is at its weakest on the day it is finished — 1.33 against the 1.93 it will have once the clay has drained — and a silo filled in a week walks over the top of the parabola and down the far side, which is the other way a structure can be built.

 Three grounds, three tipping pushes. The push against the lean it produces, in thousandths of a radian, for the 6 m square, 3,000 kN block on ground that bears 300 kPa at most, half of it mobilised at 7.5 mm of settlement, pushed 8 m up, its weight 5.0 m up, on the three grounds matched at their limit and at half of it; dashed, the rigid-plastic limit, 813 kN. Linear, capped: 776 kN at a lean of 13.2 thousandths; hyperbolic: 701 kN at a lean of 26.2 thousandths; S-shaped: 770 kN at a lean of 15.2 thousandths. The S-shaped ground behaves almost exactly like the capped one; the hyperbola, which never quite reaches its limit, tips 10 per cent sooner at twice the lean.

The ground that never quite gives way

On a bed of springs capped at the ground's bearing limit, a block pushed sideways tips at a definite push, below the limit the rigid-plastic parabola gives. Real ground is curved: it softens all the way to its limit and never quite reaches it. On such ground the block has no tipping push of its own. Its resistance creeps toward the limit, and the peak it does have is made by its own weight leaning with it — 14 per cent below the limit with the weight 5 m up, nearly at it with the weight at the base, where the lean simply runs away.

 Every one of these soils failed that slope. For a 10 m slope at 35° in soil of 19 kN/m³ with a pore pressure ratio of 0.25, over a firm stratum 5 m below its toe, which has slipped: the pairs of effective cohesion and friction angle that put its critical circle at a factor of safety of exactly one, by Bishop's method. At no friction the cohesion is 32.1 kPa; at 20° it is 12.4; at 40°, 2.0; and the line reaches no cohesion at 48.1°, where a skin of soil slides parallel to the face. Labelled, the depth of each pair's slip below the original ground: 13.6 m at 0°, 8.8 m at 5°, 7.4 m at 10°, 5.2 m at 20°, 3.6 m at 30°, 2.3 m at 40°.

A failed slope gives a line, not a soil

A slope that has slipped is the one full-scale strength test geotechnics ever gets, and the standard way to read it is to set its factor of safety to one and solve for the soil. One equation cannot give two strengths. What it gives is a line of cohesions and friction angles, every one of which fails the slope exactly, and the repair designed on one point of that line can be worth a factor of 1.07 or of 1.51. The slip itself says which point: each soil on the line fails along a different surface, and the depth of the scar is the second equation.

 The roller's pressure, not the soil's. Horizontal pressure down the back of a wall, for backfill of 20 kN/m³ and φ′ = 30° behind a wall 3.0 m high, compacted by a roller of 60 kN per metre of its width. Dashed, the soil's own active pressure, Kₐ·γz (Kₐ = 0.33), and at-rest pressure, K₀·γz (K₀ = 0.50). Solid, Ingold's compaction envelope: rising as the passive limit γz/Kₐ to 27.6 kPa at 0.46 m, then constant at √(2pγ/π) = 27.6 kPa all the way down, since the active pressure would not reach it until 4.15 m. Its thrust is 76.6 kN per metre of wall against the active 30.0, and its moment about the base 106.3 kN·m against 30.0.

The pressure the roller leaves behind

A retaining wall is designed for the pressure of the soil it holds, a triangle growing with depth. The soil behind most walls was not tipped there. It was placed in layers and rolled, and each pass of the roller pushed the soil sideways against the wall harder than the soil's own weight ever could, and left most of that push locked in when it moved on. Behind a 3 m wall compacted by an ordinary vibrating roller the locked-in pressure is a constant 28 kPa from half a metre down, the thrust two and a half times the active triangle's and the moment at the base three and a half.

 The floor under the reduction rises with the correlation. The least factor a column carrying ever more floors may take, against the correlation between floors, for an office imposed load with a coefficient of variation of 0.6, its characteristic value the 95th percentile, loads normal: (1 + zv√ρ)/(1 + zv). With independent floors it is 0.503, the floor the independence argument found; at a correlation of 0.1 it is 0.660, at 0.3 0.775, at 0.5 0.855. EN 1991-1-1's storey rule falls to ψ₀ = 0.70 (dashed), and the two cross at a correlation of 0.157: floors that move together any more than that make the rule's tall-building reduction larger than the averaging can pay for.

The floors that fill up together

A column may be designed for less than the sum of its floors' imposed loads because the floors do not reach their worst day together — an argument that assumes each floor's load is independent of every other's. Let any two floors be correlated, and the reduction the averaging can pay for shrinks, and stops shrinking at a higher floor. At a correlation of 0.16 that floor reaches the 0.7 the storey rule falls to. One tenant on five floors halves what twenty floors can average away.

 Closer piles add pull-out the block cannot pay for. For 600 mm piles 15 m long under a 20 × 30 m raft, in ground whose shaft friction is 0.30 of the effective vertical stress and whose submerged unit weight is 10 kN/m³: the summed pull-out of every pile on a square grid (thin, rising steeply as the spacing closes), the submerged weight of the block of ground the group would lift (dashed, 90 MN whatever the spacing), and the group's capacity, the lesser (solid). The two meet at 2.06 m — 3.4 pile diameters — where s² = π·d·β·L/2. Closer than that the group lifts the block and the extra piles add nothing; the most piles worth having is about 142.

The piles that would lift the ground

A basement too light to stay down can be pinned down with tension piles, each holding by the friction on its shaft. Add enough of them, close enough together, and the ground between them stops being something they grip and becomes something they carry: the group lifts out as a block, and the block's weight is all it can hold. For 600 mm piles 15 m long that happens closer than 2.06 m apart, a spacing with no raft, no water and no load in it — and every pile past that point is paid for twice.

 Each slip is a shear test at one normal stress. The soil's strength, τ = 8 + σ′·tan 24° (line), and the mean effective normal stress and shear stress along the critical slip of each of five slopes that fail in it: 55° and 3.8 m high, 8 kPa; 45° and 5.2 m high, 13 kPa; 35° and 8.2 m high, 23 kPa; 28° and 14.1 m high, 42 kPa; 22° and 41.5 m high, 109 kPa. Every slip is one point on the line. Two slips fix the line as two shear-box tests do, and fix it well only if their normal stresses are far apart: the steep, short slopes test the soil where cohesion is most of its strength, the flat, tall ones where friction is.

Two slips are a shear test

One failed slope gives a line of soils, every one of which fits the failure. Two failed slopes give two lines, and where they cross is a soil that fits both. Whether the crossing means anything depends on how differently the two slopes loaded the ground: two slips at one angle cross only at zero cohesion, two at nearby angles cross anywhere in a twenty-degree band, and a flat, tall slope beside a steep, short one pins the soil to within two degrees — because each slip is a shear test at one normal stress, and a strength line needs two tests far apart.

 Where a bending stem sheds its compaction pressure. The share of the rigid build-up's pressure shed at each depth by a cantilever stem 3 m high, bending stiffness 67,500 kN·m² a metre (a concrete stem about 300 mm thick), backfilled in 250 mm lifts of soil at 20 kN/m³ and 30°, each compacted by a roller of 60 kN a metre: nothing at the base, which does not move; little in the top lift, which nothing is placed after; most, 20 per cent, at 0.9 m down, where the pressure was locked early and the stem went on moving away from it.

The stem that bends away from the roller

A roller compacting backfill against a wall leaves a lateral pressure behind it that the soil's weight alone would never make, and for a wall that cannot move that residual is Ingold's envelope. A cantilever stem moves as the fill goes up, and a residual locked in against it relaxes as the stem bends away. But not everywhere: the base never moves, so it keeps its pressure, and the last lifts have nothing placed after them, so they keep theirs. A 300 mm stem sheds an eighth of the rigid wall's base moment and the softest stem a third — never more, because the two ends of the stem are where the relief cannot reach.

 How long the pumps must run is set by how fast the water comes back. The floors that must be cast before the pumps stop under a 20 × 30 m basement dug 6 m below a water table at the surface, its 0.9 m slab and walls weighing 18.7 MN against 35.3 MN of uplift when the water is back, so that the weight stays ahead of the water at every instant, against the water's recovery time on a logarithmic scale, for a floor every 5, 7 and 10 days. If the water returns at once, five floors — the building has to be nearly heavy enough on its own. With a floor every 5 days: 3 at a 3-day recovery, 2 at 7, 0 at 14; a floor every 7 days: 4 at a 3-day recovery, 3 at 7, 1 at 14; a floor every 10 days: 4 at a 3-day recovery, 3 at 7, 2 at 14. Dashed, the closed form for weight added continuously at a floor a week, which runs about one floor below the steps: a floor arrives at the end of its cycle, not through it.

The water that comes back while the floors go up

A basement below the water table is built dry, inside an excavation kept pumped, and the building that will one day hold it down is not there yet. When the pumps stop, the water comes back over days and the frame goes up a floor a week, and the basement is safe only if the weight stays ahead of the water at every instant. How long the pumps must run is not set by how heavy the building is. It is set by one ratio — how much weight the frame adds while the ground refills — and for a basement that needs five floors on it to hold itself down, a seven-day recovery asks for three and a fourteen-day one for one.

 A counterweight that follows the jib leaves only the payload, halved. The moment in the mast of a crane whose 50 m jib weighs 300 kN and luffs between 15° and 85°, lifting 60 kN at its tip, against the luffing angle, with a counterweight linked to the luff so that it balances the jib and half the payload at every angle: 1,449 kN·m loaded and −1,449 empty at 15°, 131 and −131 at 85°. What is left is the payload's moment less the half the counterweight was set for, plus or minus 1,449 kN·m at most, against 4,744 with a fixed counterweight.

The counterweight that follows the jib

A counterweight that stays put balances one position and leaves the mast a moment at every other. Make it move and it can balance anything that moves slowly enough to be followed. On a luffing crane the thing that moves most is not the payload but the jib itself, whose own moment about the mast changes more than twice as much through its luff as the payload's ever is — and a counterweight linked to the luff cancels that change exactly. What it cannot cancel is the payload, which leaves the ground in a second. The best any counterweight can do with that is wait halfway, and the mast is left with half the payload's moment at the longest radius, whatever else the counterweight does.

 The water pushes on the chord, and the abutments push back along the arch. One horizontal ring of an arch dam, one metre high, 50 m below the water, spanning a 100 m gap with a central angle of 133.6°, drawn in plan with the water on the convex side. The water presses 491 kPa normal to the face everywhere (arrows); by the projection rule its resultant is that pressure times the chord, 49,050 kN per metre of height, pointing straight downstream. Each abutment pushes back along the ring's tangent, at 66.8° to the chord, so the two thrusts are N = pw/2sin(θ/2) = 26,686 kN — the pressure times the ring's radius, 54.4 m, which is the hoop formula of a pipe, arrived at without a pipe.

The angle that saves the most concrete

Cut an arch dam into horizontal rings and each ring is a curved surface with water on one side. The projection rule gives the water's whole push on it without an integral — the pressure times the chord — and the two abutments, pushing back along the ring, turn that into a thrust equal to the pressure times the radius, the hoop formula of a pipe arrived at without a pipe. From there one line of calculus says that the ring using least concrete has a central angle of 133.6°, whatever the canyon, the depth or the concrete. That number explains why arch dams in narrowing canyons are drawn at a constant angle rather than a constant radius, and why the angle that suits the concrete is not the one that suits the rock.

 In a V-shaped canyon the rings carry the water, and the top rings more than theirs. The water pressure down the crown of a dam 100 m high in a V-shaped canyon 200 m wide at the crest, 5 m thick at the crest and 25 m at the base, with its rings at 133.6° (dashed), divided between the horizontal ring at each depth (solid) and the vertical cantilever at the crown (thin) so that the two deflect alike. At 4 m below the crest the ring carries 195% of the water there — the cantilever, loaded from below and free at the top, leans on it; at 49 m the cantilever carries 11%; at 89 m, 6%. Over the whole reservoir the cantilever carries 10% of the water and the rings the rest.

The water the valley floor takes

An arch dam is not a stack of rings. At its crown it is also a vertical cantilever fixed into the valley floor, and the water at each depth is shared between the ring there and the cantilever in whatever proportion makes them bend alike. In a V-shaped canyon the short lower rings are stiff and carry nearly everything, and the top rings carry up to twice their own water because the cantilever leans on them. In a U-shaped canyon the cantilever takes half the reservoir and bends the heel into tension that the ring formula never sees.

 Propped before it is filled, the stem keeps the roller's pressure. For a 3 m stem of stiffness 67,500 kN·m² a metre, fixed at its base, propped at its top by a slab cast first, backfilled in 250 mm lifts with a 60 kN/m roller: the lateral pressure on the stem after the last lift, in kPa, against depth — propped at its top before the fill (solid), the same stem as a free cantilever (dashed), a stem too stiff to move (dotted) and the active pressure. The propped stem moves at most 0.15 mm, in its middle, where the free cantilever's top moved 2.61 mm; it relaxes almost nothing, and its pressure lies on the stiff stem's along the whole height — the roller's residual, 27 kPa from half a metre down, four times the active pressure at mid-height.

The prop that holds what the roller left

A basement wall whose ground-floor slab is cast before the backfill spans between its base and the slab, and as the fill is compacted against it the wall bows by a sixth of a millimetre. A free cantilever stem leaned sixteen times as far and shed an eighth of the roller's pressure; the propped wall sheds one per cent. Its base moment falls to a third, which looks like a gain, but the pressure has not gone: the slab now carries 22.6 kN a metre against the 6 an active-pressure design gives it and the 9 of soil at rest.

 The water that rises after the basement is finished pushes on its slab first. Section through a line of columns of a 20 × 30 m basement dug 9 m, its 0.9 m slab held down at six columns and by 8 storeys above, designed for a water table 3 m below the ground. The water table as designed is 3 m below the ground, a 6 m head under the slab; dashed, the same water risen 1.0 m. The slab bows up between its columns and walls under the uplift less its own weight (exaggerated). As designed, the building holds itself down at 0.65 of its weight's capacity, the slab is at 0.37 of its bending strength and its cracks are 0.140 mm against the 0.200 mm long set for every crack in concrete that must keep water out. With the water 1.0 m higher: 0.76, 0.45, and 0.176 mm against the same 0.200.

The basement that leaks before it floats

A water table that rises after a basement is finished is a load that grows through the building's life, and the building's flotation check is the last thing it troubles. A metre of rise adds a sixth to the uplift on the whole building but over a quarter to the net load on the slab, which carries the water less its own weight. The slab's cracks, which a watertight basement is designed to, are the first thing to run out of room — and the limit for the cracks that pass right through it tightens as the water rises.

 When the slab lets go, the stem is a cantilever again. For a 3 m stem of stiffness 67,500 kN·m² a metre, fixed at its base and propped at its top by a slab cast before the fill, backfilled in 250 mm lifts with a 60 kN/m roller: the bending moment down the stem in kN·m a metre, the base's moment positive — propped as the fill left it; after the slab has shrunk 2.08 mm away and gone slack; and with the slab tied to the stem and shrunk 4.5 mm, pulling it; dashed, the stem as a cantilever from the start. Propped, the base carries 26.6. Slack, 69.6, 2.6 times as much and close to the cantilever's 82.9. Tied, 119.9, more than the cantilever ever carried, because the slab that propped the top now pulls it.

The slab that shrinks away from the wall

A basement wall propped by its ground-floor slab before it is backfilled keeps the roller's pressure, and its slab takes it. Then the slab dries. Two millimetres of shrinkage at each end — about a year for a 30 m slab indoors — and a slab that only bears on the wall has let go of it, leaving a stem designed as propped to carry 2.6 times the base moment it was given. Tying the slab into the wall to stop that makes it worse: the slab goes on shrinking and pulls, and the base moment passes the free cantilever's.

 Balancing the payload is a hedge between lifting it and losing it. The mast's design moment over the payload's own moment, against the share of the payload a counterweight driven from the load cell balances, for a crane whose 50 m jib weighs 300 kN, lifting 60 kN at 48.3 m, with a hoisting factor of 1.20 and a rebound factor of 0.50 for a payload let go at once. Lifting off, the mast carries 1.20 less the share balanced; let go, the share balanced plus 0.50, backwards. The larger of the two is least at a share of 0.35, where it is 0.85. The linked counterweight's half gives 1.00; balancing the whole payload gives 1.50, the worst of the three, because a crane balanced for its load is thrown backwards by all of it when the load goes.

The counterweight that knows the load

A counterweight driven from the hoist's load cell can follow the payload as the rope takes it up, so the step at pick-up that defeated every linked counterweight becomes a ramp a motor can keep pace with. Balancing the whole payload is then the worst thing to do with it. A crane balanced for its load is thrown backwards by all of it when the load is let go — 1.5 times the payload's moment with a rebound factor of 0.5, against 1.0 for the linked crane's half. The best share is a hedge between lifting the payload and losing it: 0.35 of it, for 0.85, which buys 18% more at the tip from the same mast.

 Split into what stays and what visits, the column needs far less. The factor a column may take on n floors' summed characteristic imposed loads, keeping the 50-year load at the 95th percentile, for office imposed load as a sustained part of mean 0.50 kN/m², renewed with each tenancy every five years on average, and intermittent loads of mean 0.2 kN/m² arriving 0.3 times a year on each floor for three days: floors let one by one, in tenancies of 5, and to one tenant; the one-number model with the same tenancies (dotted) and the storey rule (dashed). At 20 floors: 0.27, 0.37 and 0.60, against 0.75 for the one-number model and 0.73 for the rule. One floor's characteristic load is 3.24 kN/m².

The load that stays and the load that visits

An office floor's imposed load is two loads: a sustained one that arrives with a tenant and stays until the next, and intermittent ones — a party, a move, a crowd — that visit one floor for a few days. Simulated over fifty years, a floor's characteristic load is set mostly by its worst visit, and visits on different floors almost never coincide. So twenty floors let one by one need only 0.27 of their summed loads, in tenancies of five 0.37 and to one tenant 0.60, against the storey rule's 0.73. The rule is safe for offices by a wide margin. It stops being safe only where the load stays — an archive or a library let to one occupier — and there its column load is exceeded in a quarter of lives.

 The mechanism outcarries the frame that is nearly one. The load on the joint of two bars of 200 MN axial stiffness across 8.0 m against how far it moves down, followed to large displacement: the flat pair with its bars tensioned to 0, 50 and 200 kN with nothing applied, and the same bars raised 3 degrees with no tension, a rise of 210 mm. The raised frame is stiffer at first and snaps through at 11.1 kN, 89 mm down. The flat pair at 200 kN starts at 100 N/mm, carries 11.1 kN at 89 mm and 50 kN at 210 mm, and goes on stiffening; at 50 kN of tension, 50 kN at 242 mm; with none, at 252 mm.

The mechanism that stands up anyway

Two bars in a straight line between two pins are a mechanism by the count, by the rank and by every measure of how nearly a frame folds. Tension them and they carry load: 200 kN of tension gives the joint the stiffness of the same bars raised 1.8 degrees, and where the bars raised 3 degrees snap through at 11 kN, the tensioned flat pair carries 50 kN and goes on stiffening. The stiffness is a lack of fit of four millimetres, which 83 degrees of warmth takes away; compressed, the same pair buckles to rest crooked on one side or the other; and no tension can reach a mechanism that swings rather than shivers.

 A quarter of the visits shared uses up the whole reduction. The factor needed by 20 floors carrying office imposed load — a sustained part of mean 0.50 kN/m² renewed every five years on average, and visits of mean 0.2 kN/m² arriving 0.3 times a year on each floor for three days, against the share of the visits made by the whole building at once: floors let singly (solid) and to one tenant (dashed), and the storey rule's 0.73 (dotted). With no share, 0.26 and 0.60; with a tenth, 0.51 and 0.67; with a quarter, 0.68 and 0.75; with all, 0.96 and 0.97. The rule's factor is reached at a share of 0.34 let singly and 0.21 to one tenant.

The visits that come to every floor at once

A floor's characteristic imposed load is set mostly by its worst visit — a party, a move, a crowd — and visits to different floors almost never coincide, which is why a column carrying twenty office floors needs only about a quarter of their summed loads. Let some of the visits be made by the whole building at once and nothing changes on any one floor; one floor cannot tell. The column can. A tenth of the visits shared takes the twenty-floor factor from 0.26 to 0.51, a quarter to 0.68, and a third uses up the whole of the storey rule's reduction. The rule fails in exactly the buildings whose use synchronises the crowd, and it fails most where the load mostly visits.

 The pressure on the wall jumps where the bin meets the hopper. The normal pressure on the wall of a bin 8 m across holding 30 m of a solid weighing 9 kN/m³, wall friction 0.45 and K = 0.45, over a conical hopper of half-angle 20° with wall friction 0.40, the solid's internal friction 35°, down the bin and then down the hopper: in the bin, K times Janssen's vertical stress (solid) and that times the 1.35 discharge factor (dotted); in the hopper, filling (light) and emptying (dark). At the foot of the bin the wall carries 38.1 kPa, 51.4 with the discharge factor. On the hopper's top edge it carries 75.8 kPa filling and 95.7 emptying — 1.99 and 2.51 times the bin wall's. Half way down the hopper, 46.3 and 36.7 kPa.

The pressure that switches at the hopper

Below a silo's bin a conical hopper funnels the grain to its outlet, and the wall pressure there does not keep growing with depth as a liquid's would. It peaks at the hopper's top edge. Filling, the edge carries twice the bin wall's pressure just above it; emptying, the grain converges on the outlet, its largest stress turns from vertical towards horizontal, and the edge carries two and a half times. Half a metre above the apex the stress is 2.6 kPa whether the bin holds 5 m of grain or 60 — which is why a silo empties at the same rate full or nearly empty.

 The corridor's share of a column decides its reduction. The factor on the summed characteristic loads that a column carrying n floors needs, against n, for office imposed load as the earlier essays drew it — a sustained part of mean 0.50 kN/m² and visits of mean 0.2 kN/m² a third of a year apart on each floor — and on the corridor a crowd of mean 1.00 kN/m² at one evacuation a year, on every floor at once, with 0%, 10%, 25%, 50%, 100% of the column's tributary area corridor; dashed, the storey rule. At 20 floors: 0.26, 0.30, 0.49, 0.90, 0.98, against the rule's 0.73. The lines are simulated, 4,000 lives at each point, and carry a few hundredths of sampling noise.

The column under the escape corridor

A column's imposed-load reduction is allowed by the number of floors it carries, on the reasoning that floors are not crowded at the same moment. A column whose tributary area includes escape corridor carries a part of every floor that is crowded at exactly the same moment, every time the building is evacuated. Its reduction follows how much corridor its area holds, not how many floors it carries — a tenth costs little, two-fifths uses up the whole of the storey rule's allowance, and a stair-core column needs almost none. Against the nominal loads a designer writes down, the margin that saves the columns at the corridor's edge has run out by the time the area is mostly corridor.

 One alarm in a few hundred that empties the building sets the column's load. The factor on the summed characteristic loads that a column carrying 20 floors needs, for office load and a corridor crowd at one alarm a year, evacuated three floors at a time, against the share of alarms that escalate to every floor at once (logarithmic), with 25% and 50% of the tributary area corridor; dotted, each with every alarm a full evacuation; dashed, the storey rule's 0.73. 25% corridor: 0.26 with none escalating, 0.27 at one in five hundred, 0.33 at one in a hundred, 0.38 at one in twenty, against 0.49 in full every time; 50% corridor: 0.29 with none escalating, 0.39 at one in five hundred, 0.59 at one in a hundred, 0.75 at one in twenty, against 0.90 in full every time.

The stair that empties three floors at a time

A column beside an escape stair carries the corridor crowd of every evacuation, and when every floor empties at once its floors are loaded together and earn almost no reduction. Tall buildings are usually evacuated in phases instead, three floors at a time, so that the stair is never asked to carry everyone. Under such a stair the column earns its reduction back and more — 0.29 of the summed floor loads, where the storey rule allows 0.73. But a phased strategy is a plan for the usual alarm, and the column's characteristic load belongs to the unusual one. If one alarm in a hundred escalates to a full evacuation, the factor climbs back to 0.59; one in twenty, and it passes the rule.

 The seasons ride on the shrinkage. How far the slab's end has moved away from the fill since the fill was finished, over six years, for slabs 10, 20, 30 m between walls at 60 per cent humidity, swinging 3.0 K either way each year, propping a 3 m stem fixed at its base, its top held by the slab cast before its fill went up in 250 mm lifts rolled at 60 kN/m. The slab lets go of the stem at 2.08 mm (dashed). 10 m: a seasonal swing of ±0.15 mm on shrinkage reaching 1.28 mm by year six; it still touches the stem in the sixth year. 20 m: a seasonal swing of ±0.30 mm on shrinkage reaching 2.55 mm by year six; it last touches the stem 3.4 years on. 30 m: a seasonal swing of ±0.45 mm on shrinkage reaching 3.83 mm by year six; it last touches the stem 1.3 years on.

The seasons that stop reaching the wall

A ground-floor slab that has shrunk away from the basement wall it propped moves with the seasons as well, a few tenths of a millimetre at each end every year. It looks like a ratchet waiting to happen: each winter the slab presses the stem back into its fill, and each push might lock a little more pressure in. On a 30 m slab it never gets the chance — the seasons ride on a shrinkage ten times larger, and the slab last touches the wall sixteen months after the fill is finished. On a 10 m slab it touches every year for good, and the stem still loses most of its prop.

 Under a light building the first thing the pump was holding up is the building. The hours after the pump stops at which each check on the basement is used up — the slab's 0.2 mm cracks and the building's flotation — against the number of storeys over it, with the water table risen 3.0 m and ground recovering in 12 hours; drawn at 48 hours where a check is never used up. The cracks' deadline is 9.5 h whatever the building. Flotation's is at once under 2 storeys, 1.0 h under 4 storeys, 4.7 h under 5 storeys, 10.1 h under 6 storeys, never under 8 storeys; with 5 storeys or fewer it comes first.

The building held down by its pump

A basement whose water table has risen above its design level is often given back its design head by a pump under the slab, and from then on it is safe while the pump runs. When the pump stops the water comes back on the ground's own clock, and the checks the pump was protecting are lost in an order the building's weight decides. Under eight storeys only the slab's cracks were ever at risk; under five, the building's flotation check goes first, four and three-quarter hours after the pump stops, and the slab's strength is never in question at all. A structure that was safe by its weight has become one that is safe by a machine, and the machine's reliability is now one of its loads.

 A slow release moves the hedge up, and never past half the surge. The best share of the payload for the counterweight to balance, and the mast's design moment over the payload's at that share, against the release time the design can rely on, for the crane lifting 60 kN at 48.3 m, its mast and jib rocking backwards with a period of 2.0 s and a damping ratio of 0.02, with a hoisting factor of 1.20. Let go at once, 0.13 and 1.07; over 1.2 s, 0.36 and 0.84; over 6 s, 0.56 and 0.64. However slow the release, the share cannot pass 0.60, half the hoisting factor, because the payload's own moment is still let go even when its rebound is not.

The release the rope cannot slow

A crane whose counterweight knows the payload balances only a third of it, because a payload can be let go and the crane is then thrown backwards by whatever the counterweight was holding, and half as much again in the rebound. Slow the release and the rebound goes, so the hope was that a device on the rope could slow it. Nothing on the rope can: the rope holds only what the hook holds. How fast a load goes belongs to the payload, and what the crane owns is how fast it answers — its period and its damping. And with no rebound at all, the counterweight still should not balance more than 0.6 of the payload, because the payload's own moment is still let go.

 The silt hands the head back over decades, a season at a time. The basement is under five storeys, its water table 2.0 m above the design head and swinging ±0.5 m over the year, highest in April; pumped drains under its slab start 30 times as conductive as the ground. The head under the slab above the design head, against the years since the drains were new, for drains silting with time constants of 4, 8 and 16 years; dots, a standpipe read every October on the 8-year drains. With those drains the head reaches flotation in the wet season of year 24.1; the reading first exceeds 0.25 m in year 14.8, 9.4 years before.

The drain that fails a season at a time

A basement held at its design head by drains and a pump can fail suddenly, when the pump stops, or slowly, when fines settle in the drains and the head under the slab creeps up while the pump keeps running. The slow failure gives a standpipe read once a year a decade of warning, in its own time constants — if it is read in the wet season. And long before the silt floats the building on its own, it has made every stop of the pump more likely to.

All essays