Structural Engineering & Statics
where the load goes
Field

Deflection

Stiffness is not strength. What moves, how far, and why the answer goes as the fourth power of the span.
Which limit arrives first. Utilisation of the strength limit and of the deflection limit, against span. Strength grows as the square of the span and deflection as the fourth power, so the two cross — and past the crossing a beam is sized by how far it moves rather than by what it can carry.

Stiffness is not strength, and usually it is the one that governs

A beam can be nowhere near failure and still be unusable, because it has moved too far. For most long-span members that limit arrives first, and stronger steel does not help at all.

 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.

Span to the fourth, which is why spans are short

Doubling a span multiplies its deflection by sixteen. No other relationship in ordinary structural work is that steep, and it is the reason long spans are always a different kind of structure.

 Where the peak moment goes when a restraint is added. The same uniformly loaded beam with three sets of restraints, and the bending moment in each. Adding restraint moves moment from mid-span to the supports and lowers the peak — but only the first case can be solved by statics.

One support too many, and what it costs to know

Add a redundant restraint and the load has two routes to the ground. Equilibrium cannot say how it splits, and the answer turns out to depend on stiffness — which is a different kind of question.

 The deflection at x = 4, by virtual work. Three diagrams: the moment from the real load, the moment from a unit load placed where the answer is wanted, and their product. The area under the third, divided by EI, is the deflection — 213.33 here. No standard case was consulted, so the method works for any load pattern at all.

One deflection, without solving everything

To find how far one point of a structure moves, put an imaginary force of one unit there, multiply two moment diagrams together, and integrate. The answer arrives without ever solving for the deflected shape.

 Maxwell's reciprocal theorem. A load at one point and the deflection it causes at another, against the same load moved to the second point and the deflection read at the first. Both integrals return 93.3335, and neither calculation was told about the other. The two deflected shapes are entirely different; the two readings are identical.

The theorem that swaps the question round

Push here and measure there; push there and measure here. The two readings are identical, for every elastic structure, whatever its shape — and that fact turns an influence line into something a model can be asked for directly.

 3 continuous spans against 3 simple ones. The bending moment in a continuous beam whose support 1 has settled by 0.01. Three curves: the moment the load causes, the moment the settlement causes on its own — dashed, peaking at 73.5, and in equilibrium with no applied load at all — and their sum, which is what the beam carries, peaking at 73.5 against 24.5 without the settlement. The settlement field is proportional to EI: a stiffer beam is punished harder for the same movement, which is the opposite of every intuition load-carrying gives.

The support that moved

A redundant structure knows things statics cannot see. Settle one support by ten millimetres and a complete set of bending moments appears — in equilibrium with no load at all, and larger for a stiffer beam.

 The water that will not run off. Water depth against sag at the middle of the bay, for a roof bay starting with a reversed construction camber of 0.01. There is no critical value to reach: the deflection grows from the first increment, slowly at first and then without bound as the ratio approaches 1.00 — which is the depth at which the bay fills faster than it stiffens, and which the roof bay therefore never attains. The perfect roof bay, drawn for comparison, sits on the vertical axis until it arrives there and then has no answer at all.

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.

 It moves, or it pushes. Never both, and never neither. A 30 m steel member 30 °C warmer than it was built, in three conditions. Free, it grows 10.8 mm and carries nothing. Held, it moves nothing and carries 75.6 MPa in compression — which is E·α·ΔT and contains neither the length nor the area of the member, so the identical stress arises in a two-metre strut. Held by a spring it does some of each: 3.2 mm of movement and 53.2 MPa, and the split is decided by the spring rather than by the member.

The movement nobody applied

A temperature change is the only load in this collection that a structure can decline. Let it move and it produces a movement with no stress; hold it and it produces a stress with no movement — and that stress contains no length, no area and no second moment, so a bracket and a bridge girder carry exactly the same one.

 The answer arrives in instalments. The hogging moment at support 1 of a three-span beam, cycle by cycle. It starts at the fixed-end moment of 53.3 kNm — the value with every joint clamped — and settles at 64.0 kNm against an exact 64.0. The error falls by about a factor of four per cycle: 14.00, 3.50, 1.95, 0.59 kNm after one, two, three and four. Two cycles is an engineering answer and nobody had to invert anything.

Solved by passing it around

An indeterminate structure needs simultaneous equations, and for thirty years engineers solved them without writing any down. Clamp every joint, release one, share out what is left over, pass half of it along, and repeat — and the answer walks in, three figures correct after four cycles.

 The shear part is not a curve at all. A 300 × 600 rectangle spanning 2,400 mm, a span-to-depth ratio of 4, with its bending deflection and its total drawn together. Bending gives 0.0254 mm and shear a further 0.0049 mm, so 16.3% of the movement is the term beam theory drops. The shear part is drawn separately beneath: two straight lines meeting under the load, each limb straight to 2e-16 of the peak. It is the shear diagram integrated once where the moment diagram above it is integrated twice, so it carries one degree of curvature less than the bending shape — none at all here. The real movement is 1 in 79,116 of the span; both panels are drawn at about 11,932 times it.

The deflection that is not bending

Engineer's beam theory computes a deflection as the second integral of a moment, and that calculation silently drops a term. The beam also shears, and the shear deflection is not a correction to the curve — it is a different shape, and for a deep member it is most of the answer.

 Every member's share of the movement, and they are not the members expected. A Pratt truss of six panels at a depth of 0.85, carrying 10 kN at each top node, with the movement of the bottom chord at mid-span attributed member by member. The unit-load sum δ = ΣF·f·L/EA gives 631.06 at EA = 1: 47.2% from four top chords, 28.5% from six bottom chords, 22.3% from six diagonals, 2.0% from five verticals. The single worst member is a top chord at mid-span at 14.8% of the whole. Each member is drawn at the width of its own share. The same deflection from a stiffness solution that shares none of this arithmetic is 631.06, a relative residual of 3.6e-15.

Which member moved the roof

A beam sags because it curves. A truss has no curvature anywhere — it comes down because every one of its members changes length, and the sum of those changes, weighted member by member, is a ranking that names which ones are worth stiffening. Usually not the ones a designer worries about.

 The area is the rotation, and its first moment is the movement. A 6 m cantilever under a tip load of 10, with the M/EI diagram beneath it. The shaded area is 180.00, which by the first theorem is the change of slope along the whole member. Its centroid is at 2.000 m, and the first moment about the tip is 720.00 — which by the second theorem is the deviation from the tangent, and for a cantilever that tangent is horizontal, so it is the deflection itself. Integrating the curvature twice instead gives 720.00.

The area of a diagram is a rotation

A deflection is the double integral of a bending moment, and the two constants of integration are the whole difficulty. Mohr's theorems replace them with two pictures — an area, and where that area's centre of gravity sits.

 Four camber rules, and what each leaves on the finished beam. The same 12 m composite beam, cambered against four different things, followed through its own load history. Positive is a sag and negative a hog, and the point at the left of each line is the shape it was fabricated to. Cambering against the wet concrete leaves 12.7 mm of sag at the end and a flat beam on the day the slab is poured; cambering against the total load leaves the beam dead flat when fully loaded and hogged 37.9 mm — one part in 316 of the span — before anything is on it at all.

Built to the wrong shape on purpose

A cambered beam is fabricated curved upward so that load bends it down to something like straight. Nothing in the analysis changes, no stress anywhere is altered, and almost every mistake made with it is a bookkeeping mistake about which loads count.

 Two differences up the same building, peaking in different places. Differential shortening between a perimeter column and the core of a 40-storey building, plotted up the height. The part driven by load peaks at level 20 — exactly half way up, because a floor near the top has almost nothing built above it to shorten what is beneath, and a floor near the bottom has almost nothing beneath it to shorten. The part driven by shrinkage does not care what is above it at all and accumulates all the way to the roof. Their sum is worst at level 40, at 43 mm, which across a 9 m bay is a floor out of level by one in 208.

The columns are shorter than the core

Every column in a tall building gets shorter as the building is built on top of it, and the core beside it gets shorter by a different amount. The floors between them tilt by the difference — and the difference is largest exactly half way up, because a floor near the top has almost nothing built above it and a floor near the bottom has almost nothing beneath it.

 How much of a deflection belongs to the beam. The share of the total deflection that is the beam's own bending, against the stiffness of what it sits on. A 8 m beam on two supports under a uniform load: on rigid supports every millimetre is the beam's, and the share falls away as the supports soften until almost none of it is. The beam drawn beside this figure sits at 51% — so 49% of what it does is happening somewhere a beam calculation never looks. The two flexibilities are in series, which means the softer one governs and stiffening the other buys nothing.

The deflection that belongs to the support

A beam calculation answers a question about a beam sitting on things that do not move. Real ones sit on bearings, on other beams and on columns that shorten, and every one of those is a spring in series with the member — so a deflection is the sum of two things and only one of them is a property of the beam.

 A derivative taken with a ruler, and the step that makes it worst. Castigliano's theorem says the deflection is ∂U/∂P, and the derivative here is taken numerically — two solves at ±dQ and a central difference. Against the unit-load answer of 1.720635e-2 it agrees to 1.6e-13, which for a linear structure it must: ∂N/∂P is exactly the force a unit load produces, so the two expressions are the same sum written twice. The error against step size is the classic pair of straight lines — truncation falling as the step shrinks, round-off rising as the difference of two nearly equal energies loses its digits — meeting near dQ = 1.2e+1. For a linear structure the truncation term is exactly zero, so what is drawn here is round-off alone.

The deflection that is a derivative

A structure's strain energy is one number. Differentiate it with respect to a load and out comes the displacement under that load — and the trick that makes it a method rather than an identity is that the load does not have to be there.

 The building does not care how far it went down; it cares how much it tilted. Five footings on soil that is 35% as stiff under one of them, carrying 60 kN/m. They settle between 12 and 54 mm, and the number that matters is neither of those: it is the angular distortion between neighbours, 4.32 per thousand, or one in 231 — against a limit of one in 500 for cracking in finishes, which this does not. A building that went down half a metre uniformly would be undamaged and would need a new front step; this one has moved a twentieth as far and has cracked.

The settlement that matters is the difference

A building that goes down half a metre uniformly is undamaged and needs a new front step. One that goes down a twentieth as far, unevenly, has cracked. The superstructure can even the difference out — and the only way it can do so is by carrying the difference itself, as a force.

 One coefficient, and nothing else in it. The deflected shapes of one beam under four load cases, each scaled so that its mid-span deflection is the same, with the tangent at the left-hand support drawn on each. The end rotation is that deflection times a coefficient that depends only on the shape of the load: 3.20 for a uniform load, 3.00 for a load at mid-span, 2.99 for a triangular load, 3.60 for a load on half the span. Every material property, every second moment and the span itself cancel out of the ratio θL/δ, so a beam at any deflection limit has an end rotation that is known before anything about it is: at L/360 it is 8.89 milliradians, or 0.51 of a degree.

The angle nobody limits

Every serviceability rule in this collection limits a displacement. What a bearing, a joint and a cladding gap actually have to accommodate is an angle — and the angle is locked to the displacement by a coefficient that contains no material, no section and no span.

 One drift, two motions, opposite curvatures. The sideways movement of a 120 m building under a uniform wind, drawn as the sum of the two mechanisms that produce it. The bending curve is a cantilever's: flat at the base, steepening upward, concave one way. The racking curve is a stack of parallelograms: steepest at the base and flattening, concave the other. They add to 366 mm at the roof, of which 61% is bending. The one group that decides the split is αH = H√(GA/EI) = 2.48: below one the racking dominates and the building behaves as a frame, above about six the bending does and it behaves as a cantilever, and everything interesting is in between.

Two motions with one name

A tall building's sway is two movements added. A frame racks like a stack of parallelograms, worst at the bottom; a cantilever bends about its base, worst at the top. The total at roof level says nothing about which storey is worst, and on this building it is neither.

 The beam is stiffer than its cracked section and softer than its gross one. Moment against mid-span deflection for a 300 × 550 mm beam spanning 8.0 m, with the two bounds it lies between. The uncracked line is what the gross transformed section gives; the cracked line is what the section at a crack gives; and the curve between them is the member, because between the cracks the concrete is still carrying tension and the average curvature is not either section's. At the service load the deflection is 23.2 mm — span over 345 — against 8.3 uncracked and 25.2 fully cracked, a factor of 3.03 between the bounds. The interpolation ζ = 1 − β(M_cr/M)² sits it 88 per cent of the way across, and β falls from one to a half under sustained or repeated load because the bond that does the dragging deteriorates.

Stiffer than its cracked section says

At a crack the concrete below the neutral axis has gone and the steel carries the tension alone. Between the cracks it has not gone — bond drags it back into tension, the steel strain drops, and the curvature averaged over a length of beam is neither section's.

 Half the beam does nearly all of the deflecting. The virtual-work integrand M·m/EI along the member, normalised to its own peak, with the running share of the answer beside it. The integrand is a density: it says how much of the deflection each millimetre of the beam produced. For this case the half nearest the root supplies 87.5 per cent of it, and the rest of the member supplies the remainder. Stiffening the busy 50 per cent by 1.5 times takes the deflection down by 29.2 per cent; the same material spent on the quiet end takes it down by 4.2 — a factor of 7.0 for the same steel. The map of what is contributing is not the map of where the moment is largest, and the second is the one that gets drawn.

Where a deflection comes from

The unit-load method gives a deflection as an integral, and this collection has treated that integral as a number to evaluate. It is not a number. It is a density, and it says which millimetres of the member produced the answer — which is not the same map as where the moment is largest.

 Almost all of it is exactly zero. The stiffness matrix of a 2-bay, 3-storey plane frame: 36 freedoms, of which 16.2 per cent of the 36² entries are non-zero. The zeros are not small numbers; they are absences. A member reaches only the two nodes at its ends, so it can contribute nothing to any row belonging to a node it does not touch, and every such entry is zero exactly rather than nearly. The non-zeros therefore sit in a band of width 13 about the diagonal. Before the supports are applied the matrix is singular, and its null space has exactly three dimensions — the three rigid-body motions a plane frame has with respect to the ground, which is the same statement the null vector of the equilibrium matrix makes about a truss that is a mechanism, arrived at from the other end.

The matrix that replaced the hand methods

Moment distribution passes moments round a frame until they stop moving. Virtual work computes one deflection at a time. Both are exact and both stop scaling in the low tens of members. What replaced them adds no physics at all — the whole of the invention is the bookkeeping.

 Two different structures released, and one bending moment diagram. The bending moment in a continuous beam of 8, 10, 8 m under 12 kN/m, solved twice by the force method with different redundants. The first release puts a hinge over each interior support, so the released structure is a row of simple spans and the redundants are moments. The second removes each interior support, so the released structure is one simple span of the whole length and the redundants are reactions. The two released structures have nothing in common — different shapes, different deflections, different everything — and the diagrams they produce lie on top of each other to 9e-15 of the peak moment. Which restraints are released is a choice about the arithmetic and not about the structure, which is a fact worth trusting: it means a hand calculation can pick whichever release makes the sums easiest and be sure of the answer.

Choose what to take away

The other machine for a redundant structure works by removing restraints until what is left can be solved by statics, then putting back exactly enough force to close the gaps that opened. Which restraints are removed does not change the answer at all, and changes the arithmetic completely — one choice gives a tridiagonal matrix a person can solve on paper, and another gives a full one.

 The corner is where the plate twists, and it has to be held down. The twisting moment M_xy over a 6 × 6 m simply-supported panel under 10 kN/m², from Navier's double series. It is zero along both centrelines and largest at the corners, which is the opposite of the bending moments and is why no strip reading contains it: 31% of the load crosses this panel in twist, and a strip can only bend. At a free corner the twisting moment is statically equivalent to a downward point force of 2M_xy — 26.7 kN here, 7.4% of the whole load on the panel, applied at a point — so the corner has to be held down. A panel whose corners are free lifts there, deflects more, and cracks diagonally across them. The deflection at the centre is 2.53 mm, against 4.05 mm for the strip that ignores all this — 60% more, which is the size of what the twist is carrying.

A third of the load crosses sideways

A slab spanning both ways is usually explained as two beams sharing a load by a fourth power, and the explanation is not merely approximate — it is missing a mechanism. A real plate carries load three ways, and the third one has no beam strip in it: it is twisting, it accounts for a third of the load on a square panel, and it is why the corners lift.

 The least reliable number in the material decides the answer, briefly. What a twenty per cent error in the concrete's tensile strength does to a computed deflection, against how far past cracking the beam is. Well past the cracking moment it does almost nothing — at 1.9 times M_cr the spread is 56 per cent — because the section is nearly fully cracked and the interpolation has run out. Just above cracking it does everything: at 1.19 times M_cr the same twenty per cent moves the deflection by 3658 per cent. Tensile strength is the property with the widest scatter and the least direct test, and a beam designed to sit near its cracking moment has put the answer on it.

The curvature nobody applied

Concrete shrinks as it dries, by about half a millimetre in every metre. In a symmetrically reinforced member that is a shortening and nothing else. In a member with more steel in one face than the other — which is every beam and every slab — the steel holds one side back and the section bends, with no load on it at all.

 One member's stiffness, scattered into the freedoms it touches. A member's own six-by-six stiffness matrix relates the forces at its two ends to the displacements there, and it is written in the member's own axes. Assembly is two operations and no physics: rotate it into the structure's axes, then add each of its thirty-six entries into the row and column of the global freedom that entry belongs to. Every member does the same, and the sum is the structure. The shaded rows and columns are the six freedoms this one member reaches; every other entry it contributes is exactly zero, and that is the whole reason a global stiffness matrix is sparse. Nothing here is an approximation — the result is the same equilibrium and the same compatibility a hand method writes, in an order a machine can follow.

The answer that depends on how it was divided

Every computed answer in this collection came out of a structure chopped into pieces — elements, strips, stations, trial positions. The chopping is invisible in the result and it is not neutral: some divisions give the exact answer, some give one that is always too stiff, and one of them changes nothing but the cost of getting there.

 The gap is a sum of five things and only one of them is computed. What a 30 mm movement joint is asked to accommodate, by three combination rules. The top bar is every term at its extreme, added: 33.2 mm, which assumes the hottest day, the fullest floor, the whole of the shrinkage and the worst-placed wall arrive together. The chance of that is about 1.5%. The bottom bar treats them as independent and asks for 16.0 mm. The middle bar is the rule used for actions and almost never for movements — one term at its full value and the rest at their coincidence factors — and gives 25.5 mm. The segments across the top bar are the terms themselves, and the ordering is the finding: the largest is tolerance at 10.0 mm, which is not a structural quantity at all, and the smallest is deflection at 3.2 mm — the only one anybody computes carefully, and 10% of the total.

The gap nobody computed

A movement joint is sized by adding up everything the structure will do to it, and the deflection calculation — the only term anybody computes carefully — is usually the smallest one in the list. The largest is a construction tolerance, which is not a structural quantity at all, and the sum of the extremes is nearly twice what treating them as independent would ask for.

 Every reason a building is stiffer than its model, added up. The computed natural frequency of a floor, and the same frequency after each source of stiffness that was deliberately left out is put back. Not one of them is a modelling error. Cladding and partitions are stiffness nobody is allowed to rely on for strength; a nominally pinned connection is never pinned; a slab acts with its beam whether or not shear connectors were provided; and concrete between the cracks is stiffer than a cracked section assumes. Together they multiply the stiffness by 1.83 and the frequency by 1.35, because a frequency is the square root of a stiffness, and every factor is halved on the way through. The asymmetry is the finding: leaving stiffness out makes a deflection conservative and a vibration check unconservative in the direction that matters, since a stiffer floor has a higher frequency and sits further from the footfall range. The model here reads 4.40 Hz against a 5.2 Hz criterion and fails it; the floor reads 5.95 Hz and passes. The correction that would have got it right is exactly the stiffness nobody is willing to count on.

Stiffer than the model said

Measured natural frequencies of finished buildings come out between ten and sixty per cent above the values computed for them, consistently and in one direction only. Nothing on the list of reasons is a modelling error: every one is a real source of stiffness deliberately left out — and leaving stiffness out is conservative for deflection and unconservative for vibration.

 The same curve, computed twice and from opposite ends. A cantilever of 6 m, with its deflected shape drawn from a double integration of M/EI and the bending moment of its conjugate beam drawn on top of it. The conjugate is the same span carrying M/EI as a load, with its supports transformed — a real fixed end becomes a free end and a real free end becomes a fixed one, because a fixed end has no slope and no deflection and the conjugate therefore needs no shear and no moment there. The two curves agree to 4.7e-6 of the largest deflection, which is the trapezium rule and not the method. The largest deflection is 36.00 mm against the closed form's 36.00. The reaction of the conjugate beam is -9.000 milliradians, which is the real beam's rotation at that support — so the whole of a slope calculation is one reaction.

The beam whose moment is a deflection

A beam's bending moment is the second integral of its load. Its deflection is the second integral of M/EI. They are the same problem, so a deflection can be found by loading a fictitious beam with M/EI and asking a statics question — and the only thing to remember is the supports, which are not remembered but derived, one boundary condition at a time.

 The check that depends on a date. Total deflection and the deflection occurring after the brittle finishes are built, for one 12 m beam, against the day those finishes go up. The total barely moves — the beam ends up where it ends up. The increment falls from 32 mm at a week to 14 mm at a year, because creep is fast at first and slow later and a partition built early inherits nearly all of it: 44% of the final creep has already happened by day 28. The span/500 limit is 24 mm and the span/250 limit is 48; this beam passes the first only after day 25. Camber subtracts from both terms of the difference and therefore changes the upper curve and not the lower one, which is the reason a cambered beam can satisfy every total-deflection check and still crack the wall.

The limit that depends on a date

Total deflection can nearly always be met, and on a long span it is met with camber. The limit that actually decides the member is the other one — the deflection occurring after the brittle finishes are built — and camber does nothing for it whatever, because it is subtracted from both terms of a difference. The same beam passes or fails on the day the partitions went up.

 Two answers added, and the answer to the two together, drawn on top of each other. A 8 m beam under a 60 kN point load at mid-span (152.38 mm), under 12 kN/m of uniform load (152.38 mm), and under both at once (304.76 mm). The sum of the first two is 304.76 mm, and the residual between it and the third is zero — not small, zero, to the last bit of the arithmetic. That exactness is not a numerical accident: the governing equation is linear in the load, so the response is a linear operator applied to it, and a linear operator distributes over addition by definition. Every calculation that adds one load case to another is standing on that one line.

The addition everything else rests on

Influence lines add, the unit-load method adds, moment distribution adds, load combinations add, and a stiffness matrix is linear by construction. All of it stands on one sentence with three hypotheses in it — and when they fail, two of the failures point in opposite directions.

 The deformation with no limit against it. A 8 m open section carrying 12 kN/m at an eccentricity of 75 mm from its shear centre. The torque is small — 900 Nmm per mm — and the twist is not: 2.55° at mid-span with the ends restrained against warping, against 9.09° if they are not, a factor of 3.57. What that angle does is move the flange tip sideways by 8.9 mm — 62 per cent of the member's own vertical deflection, and 28 per cent of the span/250 that vertical deflection is checked against. The same load on a closed section of the same depth moves it 0.15 mm, 60 times less. No code gives a limit for this quantity, so it is the one movement in the collection that is computed only after somebody has complained about it.

The movement with no limit against it

Every code in the world gives a deflection limit. None gives a twist limit — and a beam loaded off its shear centre twists. On an open section a modest eccentricity moves the flange tip sideways by three fifths of the sag that does get checked, and nothing anywhere says whether that is acceptable.

 A tenth of a per cent of the thrust is all of the moment. The thrust a two-hinged arch loses to its own axial shortening, against rise-to-span. The flexibility equation's denominator has two terms — ∫y²ds/EI for bending and ∫cos²θ ds/EA for shortening — and their ratio is about (15/8)(i/f)², the square of the radius of gyration over the RISE. The loss is a little more than that ratio, because the released rib also shortens under its own shear, and at the 10 per cent rise drawn it is 0.11 per cent of the thrust, which sounds like a rounding error and is not: a parabolic arch under a uniform load is funicular, so the rigid solution has NO crown moment at all, and the 0.11 per cent that the rib shortening removes from the thrust leaves 29 kNm behind. The correction that is a tenth of a per cent of the thrust is a hundred per cent of the bending. At a two per cent rise the loss is 2.6 per cent, because a shallow arch's thrust is enormous and its lever arm is not.

The arch that gets shorter

A parabolic arch under a uniform load is funicular, so the perfect solution gives it no bending at all. Then the rib shortens under its own thrust by a tenth of a per cent, and every kilonewton-metre of moment the arch will ever carry comes from that.

 Maxwell's reciprocal theorem. A load at one point and the deflection it causes at another, against the same load moved to the second point and the deflection read at the first. Both integrals return 307.5006, and neither calculation was told about the other. The two deflected shapes are entirely different; the two readings are identical.

An influence line is a deflected shape

Finding where a load has to stand to be worst means solving the structure once for every position it could stand in. Reciprocity says the answer is a single deflected shape — release the quantity being asked about, move it by a unit, and the shape the structure takes is the influence line.

 Every member's share of the movement, and they are not the members expected. A Pratt truss of eight panels at a depth of 1, carrying 15 kN at each top node, with the movement of the bottom chord at mid-span attributed member by member. The unit-load sum δ = ΣF·f·L/EA gives 2019.41 at EA = 1: 49.4% from six top chords, 30.8% from eight bottom chords, 16.8% from eight diagonals, 3.0% from seven verticals. The single worst member is a top chord at mid-span at 11.9% of the whole. Each member is drawn at the width of its own share. The same deflection from a stiffness solution that shares none of this arithmetic is 2019.41, a relative residual of 1.4e-14.

The member that is not worth stiffening

A truss's deflection is a sum of one term per member, and a term is zero whenever either force in its product is. A vertical carrying the whole of a panel load can contribute nothing at all to the movement — which a total can never show and a per-member sum shows nothing else.

 The shear deflection is not a correction. The share of a sandwich panel's deflection that is shear rather than bending, against how slender the panel is. A solid beam at a span-to-depth ratio of 20 spends about a per cent of its deflection on shear; this panel spends 25% at the same ratio, because its core is 2800 times softer in shear than its faces are in tension. At the 37 of the panel drawn it is 9%. The curve falls as the square of the span because bending grows as the fourth power and shear as the second, so the term that is negligible for a long panel is the whole answer for a short one.

The stiffness that belongs to the span

A section's flexural rigidity is a property of the section, and a member's is not. Once shear deformation is counted the effective stiffness contains the span, so the same panel is a different member at three metres and at six — and for a sandwich the correction is not a correction.

 The answer arrives in instalments. The hogging moment at support 1 of a three-span beam, cycle by cycle. It starts at the fixed-end moment of 98.0 kNm — the value with every joint clamped — and settles at 156.9 kNm against an exact 156.9. The error falls by about a factor of four per cycle: 21.03, 5.92, 1.54, 0.60 kNm after one, two, three and four. Two cycles is an engineering answer and nobody had to invert anything.

Why it converges, and how fast

Moment distribution is an iteration, and iterations do not always converge. This one always does, at a rate the beam's own proportions fix — about a factor of four per cycle on a regular beam and considerably worse on an irregular one, which is where the method's reputation for two cycles being enough comes from and where it stops being true.

 The deflection a building can take, and which way it is bending. The limiting deflection ratio — the sag or hog of a building divided by its length — at a critical tensile strain of 0.075 per cent, against how long the building is compared with its height. Two curves, and the gap between them is the whole finding: hogging is worse than sagging by exactly 2.0 times, because the neutral axis of a hogging building sits near the bottom and the tension face is the full height of the wall. The minimum is at L/H = 1.6, where the two mechanisms cross: a squat building cracks diagonally in shear and a long low one cracks in bending at the extreme fibre.

The crack is a strain, not a slope

Angular distortion is what every table of settlement limits is written in, and it is a proxy. What cracks a building is a tensile strain in it, and reading the building as a deep beam says which of two mechanisms supplies it, why hogging is exactly twice as damaging as sagging, and why a horizontal stretch of a few millimetres takes half the capacity away before the settlement starts.

 A portal on a stepped base — the two passes added. Bending moments on a single-bay portal with columns of 5 m and 3.5 m under one horizontal beam of 9 m, carrying 10 kN/m down and 60 kN across, with fixed bases. This frame is the two passes added. The corner moments are 5.8 and 99.1 kNm, and the short column's top carries 17.01 times what the tall one does. The diagram is drawn on the tension side of each member.

Every joint balanced, and the frame still leaning

Moment distribution enforces one equation per joint, and a frame free to translate has one more equation than it has joints. So a table that balances perfectly can describe a structure held up by a prop nobody built — and finding the prop, then removing it, is a second pass whose unknown is a distance rather than a rotation.

 Four things a far end can be doing, and what each is worth. The rotational stiffness of a member at one end, for four conditions at the other, each drawn as the shape the member takes when the near end is rotated through one unit. They are the same expression evaluated four times — M = (2EI/L)(2θ_a + θ_b) — and the only thing that changes is what the far end is known to be doing. A held far end gives 4EI/L and carries over a half; a free one gives 3EI/L and carries over nothing; a far end rotating equally and oppositely gives 2EI/L and carries over minus one, which is what a symmetric structure does to a member crossing its axis; and a far end rotating equally and in the same sense gives 6EI/L and carries over one. None of the four is an approximation. Each removes a freedom that was going to be discovered by iteration.

Told what the far end is doing

Moment distribution discovers, cycle by cycle, that the pinned end of a beam carries no moment — a fact known before any arithmetic started. Telling it instead changes one stiffness from 4EI/L to 3EI/L and the work from thirty numbers to eight, for the identical answer. Cutting the beam on its own axis of symmetry gets it in two.

 Every number in the table is a rotation, and none of them is a moment. The rotation contributions of a three-span beam of 8, 10, 8 m under 24 kN/m, sweep by sweep. There are six of them, one per member end, and not one is a bending moment: the moment is assembled at the end from M = FEM + 2m′ + m′ of the far end, and until that is done the table holds quantities that mean nothing on their own. That is the trade. A moment distribution stopped after two cycles hands over moments that are wrong by a known amount and can be used; this table stopped after two sweeps hands over nothing that can be read at all — and it gets there in four sweeps against Cross's own count on the same beam, writing six numbers a sweep rather than one per distributed member end plus a carry-over.

The table that cannot be read halfway

Kani's method converges at exactly the rate moment distribution does, sweep for sweep and digit for digit, because it is the same iteration. What it changes is what is written in the boxes — rotations rather than moments — and that buys a shorter table that repairs its own mistakes and cannot be stopped early.

 The whole deflected shape, found from the members' changes of length. A Pratt truss of eight panels at a depth of 1, carrying 15 kN at each top joint, at EA = 1, drawn as built and in its deflected shape, with every movement magnified the same number of times. The shape comes from Williot's construction and Mohr's correction: every joint's movement from the members' extensions alone, less the rigid rotation the supports forbid. The mid-span joint L4 moves down 2019.41 units, and the dots at every joint are a stiffness solution sharing none of that arithmetic, agreeing to 1e-14 of the largest movement. One construction gives all sixteen joints at once.

The drawing that is right except for a rotation

Williot's construction finds every joint of a truss from its members' changes of length alone, in one drawing — and puts the roller six thousand units off its support. The error is one rigid rotation, Mohr's diagram takes it away, and a drawing started from the member symmetry holds still never makes it.

 Soft in shear, the wall gives up moment to the span. The bending moment along a 300 × 600 rectangle fixed at its left end and propped at its right, under a uniform load, per unit load and span, at span-to-depth ratios of 4, 2, 1, and for the same beam treated as rigid in shear, dashed. Rigid in shear the wall carries 0.125 wL² and the prop 0.375 of the load. Counting the shear it deforms by, the wall moment falls to 0.119 wL² at 4, 0.105 wL² at 2, 0.070 wL² at 1, and the prop's share rises to 0.381, 0.395, 0.430. The load has not changed and the beam is no weaker: its forces have moved, because in a redundant beam they come from how it deforms.

The shear that moves the moments

In a statically determinate beam, shear deformation adds movement and changes no force. In a redundant one the forces come from how the beam deforms, so a stubby beam soft in shear carries less moment at its wall and more in its span — and the carry-over factor, the one half every hand method passes along, falls to nothing at φ = 2 and then changes sign.

 A fixed-ended member's end moments are the edge stresses of a column. A prismatic member, fixed at both ends under a point load 0.333 of the span from its left end, drawn three ways. At the top, the member. In the middle, the column the analogy puts in its place: a strip as long as the member and as wide at each point as 1/EI there, with its centroid, the elastic centre, 0.500 of the span from the left. At the bottom, the simply supported moment diagram, dashed, which is the load on that column; the straight line, which is the stress that load produces, P/A + M(x − x̄)/I; and the member's own moment diagram, which is the difference. The end moments are the column's edge stresses: 0.148 PL at the left and 0.074 at the right, which are Pab²/L² and Pa²b/L², and the largest sagging moment is 0.099 PL.

The fixed-end moment is a column stress

A fixed end forbids a change of slope and a deviation, so on a member fixed at both ends Mohr's two theorems both have the answer nothing. Written with 1/EI as a width, those two conditions are the P/A + My/I of a short column under an eccentric load — the end moments are its edge stresses, and a haunched member's are read from where its elastic centre has moved.

 A portal frame, and the section it is a drawing of. A single-bay portal of 6.0 m span and 5.0 m height with fixed feet, carrying 10.0 kN/m on its beam, with the beam's second moment equal to the columns'. On the right, the analogous column: the frame's own centreline drawn as a section as wide as 1/EI at every point, so it is narrow where the frame is stiff. Its area is 16.000 and its elastic centre sits 1.56 m below the beam — inside the frame, on no member at all. The released moments loaded onto that section give a direct stress of 33.8 kN·m and a bending stress whose gradient is the horizontal thrust, 6.4 kN. Together they give 10.6 kN·m at the feet, −21.2 at the knees and 23.8 at the crown, against a free moment of 45.0. The same frame solved by stiffness gives 10.6, −21.2 and 23.8.

The elastic centre is not on the frame

Cross's analogy turns a member fixed at both ends into a short column and its end moments into edge stresses. A closed frame's analogous column is the frame's own outline, its elastic centre is a point hanging in mid-air inside it, and the horizontal thrust a gravity load produces is that section's bending stress about the axis through that point.

 A unit load carried by the prop taken away. A beam fixed at its left end and propped at its right, under 4 kN/m over 8 m, asked how far it moves at 4 m. The real moment is the propped cantilever's own, with -32.0 kNm at the wall. The unit load is carried by the prop taken away, whose moment diagram peaks at 4.00. Their product has 98.92 of area on one side and -13.33 on the other, and the net, divided by EI, is 85.33 — the propped cantilever's closed-form deflection, 85.33, which no part of this calculation was given.

Any structure will carry the unit load

Virtual work has two readings and each is free exactly where the other is bound. A unit load needs only something to stand on in equilibrium, so the deflection of a beam statics cannot solve comes out of a cantilever statics can. A virtual displacement needs only to fit together, so a reaction comes out of pushing a mechanism — and on a redundant beam the unknown cancels out of the equation and nothing is found at all.

 Three routes to one deflection, and the one that is wrong. Two 2500 mm bars of 250 MPa proof stress meeting at a loaded apex, the drop of the apex against the load. The line is geometry: each bar's extension from its own stress, divided by the sine of its slope. The dots are the derivative of the total complementary energy with respect to the load, and lie on it. The dashed line is the derivative of the strain energy — Castigliano's theorem applied to a material that is not linear — which leaves the truth by ten per cent at 99 kN and at 170 kN gives 308 mm for a deflection of 46.0.

The other area under the curve

Castigliano's theorem says a deflection is the derivative of the strain energy with respect to the load, and it is true only while the material is linear. Past that, the right energy is the area on the other side of the stress–strain curve. On two aluminium bars at their proof stress the strain energy gives a deflection four times too large, and on a redundant truss minimising it picks a set of forces in perfect equilibrium that no deformed shape can produce.

 A fixed arch, and the section it is a drawing of. A parabolic arch of 30.0 m span and 6.0 m rise, fixed at both springings, carrying 20.0 kN/m over the span. On the right the analogous column: the arch's own centreline drawn as a section of width ds/EI, so it is narrow where the arch is stiff. Its area is 32.99 and its elastic centre sits 3.86 m above the springings — 2.14 m below the crown and on no part of the arch at all, which is the point of it. The released moments loaded onto that section give a direct stress of 3054.0 kN·m and a bending stress whose gradient is the horizontal thrust, 375.0 kN.

The centre that hangs in the air

Cross's analogy turns a closed frame into a short column and reads its redundant moments off as that column's stresses. Nothing in it requires the frame to be straight. Run round a fixed arch, the analogous column's centroid is a point two thirds of the way up the rise, on no part of the arch at all, and the bending stress about it is the arch's horizontal thrust. For a parabola under a uniform load the answer it returns is a diagram of nothing, which is exactly right.

 An unsymmetric frame, and a section with a product of inertia. A portal of 8.0 m span whose columns are 5.0 m and 3.0 m, carrying 10.0 kN/m on its beam, and beside it the analogous column — the frame's own centreline at a width of 1/EI. The section is no longer symmetric about a vertical line, so it has a product of inertia of −18.4 against Ix = 31.6 and Iy = 168.0, and its principal axes are tilted 7.54° from the horizontal — the dashed pair through the elastic centre. A section with a product of inertia does not bend about the axis it is loaded about, so the thrust and the redundant shear come out of a two-by-two rather than out of two divisions.

The axes that have to be turned first

Make one column of a portal shorter than the other and the analogous column stops being symmetric about a vertical line. It acquires a product of inertia, its principal axes tilt seven and a half degrees, and the thrust and the redundant shear stop being two separate divisions. Using the elastic centre and nothing else — which is what the symmetric construction looks like from outside — reports 17.0 kN·m at the left foot where the frame carries 27.8.

 A frame bent by nothing at all. A portal of 8.0 m span and 5.0 m columns at EI = 20000.0 kN·m², with its right foot settled 10.0 mm and no load on it anywhere. The moment diagram is drawn on the members: −3.95 kN·m at the left foot, −3.95 at the left knee, −0.00 at the crown, 3.95 and 3.95 on the right. The settlement is drawn hugely magnified; at true scale it is 10.0 mm on an 8.0 m frame. The diagram is antisymmetric, the vertical force the settlement develops is 0.99 kN, and every one of those numbers is proportional to EI.

The load that is not a load

Settle one foot of a portal frame by ten millimetres and the frame develops moments with nothing applied to it anywhere. In the analogous column the case is simpler than a load case, not harder — the section carries no direct stress at all, and the whole answer is one bending stress. And it scales the wrong way: the moments are proportional to EI, so the stiffer the frame, the more a settlement costs it.

 Every shape the analogy reaches, which is every single ring. Four frames of 10.0 kN/m on an 8 m span, each solved by the column analogy and each checked against a stiffness solution of the same frame: a portal, largest moment 40.6 kN·m, agreeing to 0.24 per cent; a pitched portal, largest moment 35.6 kN·m, agreeing to 0.15 per cent; a stepped frame, largest moment 46.6 kN·m, agreeing to 0.16 per cent; splayed legs, largest moment 16.9 kN·m, agreeing to 0.09 per cent. The diagrams are drawn normal to each member. Nothing in the method asks what shape the frame is: it needs an area, a centroid and three second moments of the centreline, and a polyline has all five however it bends.

Three, and what three is a property of

The column analogy works because a closed ring cut once has three redundants and a plane section has three stress resultants. That match is the whole method, and it is topological rather than geometric — a portal, a pitch, a step, a splay and a polygonised arch are all one ring and all exact. Add a second bay and the frame has six redundants with nothing to be a drawing of, and the outer ring on its own is out by 190 per cent.

 One volume of steel, divided three ways. A Pratt truss of eight panels at a depth of 1, carrying 10 kN at each top joint, drawn three times with every member as wide as its area, the total volume the same in each. With equal areas the mid-span deflection is 43503; fully stressed, with area in proportion to force, 32702; with area in proportion to the square root of the product of each member's real and virtual forces — the division that makes mid-span as stiff as this steel can make it — 31318. The fully stressed truss is 4 per cent short of the stiffest; the equal-area truss is 39 per cent short. No member is allowed less than 10 per cent of the equal area; the deflections are in units of load × length / (E × volume).

The truss that is stiff by accident

Give a truss a fixed volume of steel and ask how to divide it among the members. Sized for strength — every member at the same stress — its mid-span deflection comes within four per cent of the stiffest that steel can make, although stiffness was never asked about. The reason is an inequality, and the same inequality says where the accident stops: at the quarter point the strength design is sixty-nine per cent short of the best.

 The strength design, and the least steel that is stiffer. A Pratt truss of eight panels at a depth of 1, carrying 10 kN at each top joint, each member drawn as wide as its section from a catalogue whose sections step by 25 per cent in area, the smallest 10 per cent of the largest member's need. Above, every member at the smallest section that carries its force. Below, the least steel that makes mid-span 1.50 times as stiff with every member still strong enough — the discrete optimum — with the 24 members it made larger drawn in the second colour: six of six top chord members, eight of eight bottom chord members, six of eight diagonals, four of seven verticals. The optimum uses 44 per cent more steel than the strength design, and it puts it where a member's real and virtual forces are both large; members either force leaves small are not touched.

The calculus answer, rounded, is the worst one

A real truss is built from a catalogue, and every member is rounded up to the next section. That rounding costs its stiffness almost nothing: the few per cent that separate the strength design from the stiffest survive it. What does cost is the next step. When a deflection limit governs, the obvious move — take the continuous optimum and round it up — needs more steel than any other way of stiffening the truss, and the exact discrete answer is within one per cent of a bound no catalogue can beat.

 The rotation a bearing sees is made before it arrives. The end rotation of a 12 m pretensioned beam, 300 mm wide and 700 mm deep, with a 160 mm in-situ slab 1.2 m wide, in 60 per cent humidity, from the prestress's release at three days to thirty years, with sagging positive. Release turns each end 4.8 milliradians upward; creep takes it to 7.9 by day 28, when the beam is set on its bearings. The slab brings it back by 0.9, the surfacing by 0.1, and thirty years of creep take it to 8.1 upward. The band after erection is the imposed load (0.9 down) and a night with the top cooler (0.6 down) above the line, and a sunny day with the top warmer (1.0 up) below it. Every value in the band is below zero: under every load it will carry, the beam's ends still point up, and a level bearing is turned the same way for its whole life.

The angle made in the casting yard

A bridge bearing is designed for the rotation of the beam it carries, and the rotation is listed as a sum of load, temperature, creep and a tolerance. Followed through the life of a pretensioned beam, the largest term is none of those. It is the upward turn the prestress gives the beam's ends in the casting yard, before any bearing exists, and under every load the beam will ever carry its ends still point up.

 The strain-energy route halves the sag. The midpoint deflection of a cable 10.0 m long, pretensioned to 60 kN, with an axial stiffness EA of 64 MN, loaded across its span at midpoint, against the load, found three ways. Solid: the deflection itself, from the cable's geometry; the dots, the derivative of the complementary energy with respect to the load, lie on it. Dashed: the derivative of the strain energy. Dotted: the pretension alone, as if the cable's tension did not grow. At 10.0 kN the cable sags 212 mm; the strain-energy route gives 107 mm, 0.51 of it, and the pretension alone 417 mm. At 20.0 kN: 294, 128 and 833 mm. The strain-energy route understates the deflection of a member that stiffens by the same mechanism that made it overstate one that softens.

The sag the strain energy halves

The derivative of the strain energy overstates the deflection of a member that softens. Turn the curve the other way up — a cable that grows stiffer as it sags, a hanger that takes up its play — and the same derivative understates it: by half for a pretensioned cable carrying a modest load, by two thirds for one with no pretension, and by the whole of the slack for a member that has any. The error is one ratio in both directions, and the least-work shortcut built on it hands the load to the member that stiffens.

 Fully stressed, with a choice of diagonals. Two trusses sized so that every member is at the allowable stress under the full load, each member drawn as wide as its area, tension and compression in two colours. Above, an eight-panel Pratt truss as deep as a panel is long, with a counter-diagonal in each of its six interior panels, loaded by 10 at every bottom joint, found by resizing and reanalysing until nothing changes; below, the same truss without its counters. The counters in the two middle panels have shrunk to nothing (dotted) and the four nearer the supports have stayed, working in compression beside the diagonals in tension. The truss that kept them needs 1,200 units of steel against the Pratt's 1,220, and deflects 24.0 at mid-span against 26.0.

The truss whose forces follow its sections

In a determinate truss each member's force is fixed before its section is chosen, so sizing for strength and sizing for stiffness can be done in either order. Put a counter-diagonal in every panel and they cannot. The fully stressed design becomes an iteration that starves some counters to nothing and keeps others, lands on a different truss from every start, and — whichever it lands on — weighs the same and deflects the same. Then enlarge one group of members to stiffen it, and a vertical nobody touched is overloaded by 58 per cent.

 A deck holds the middle of the span. The twist along an 8 m open section on forks, free to warp, carrying 12 kN/m at 75 mm from its shear centre — 900 Nmm of torque per mm of span. With nothing fastened to it the beam twists 6.25° at mid-span; a deck resisting the top flange's rotation at 5.0 kNm per metre per radian holds it to 4.20°, and one four times as stiff to 2.10°. The deck works where the beam is weakest, in the middle; near the supports the beam's own torsional stiffness does most of the work whatever is fastened to it.

A deck is a spring, not a wall

An open-section beam loaded off its shear centre twists, and the usual reassurance is that the deck fastened to its top flange will stop it. The deck resists the flange's rotation with a stiffness per metre of span, and that stiffness has to be compared with the beam's own. On an 8 m beam a screwed deck of ordinary stiffness removes a third of the twist; on a 16 m beam the same deck removes three quarters, because the beam's torsional stiffness falls with the square of its span and the deck's does not.

 What a millimetre on each member does at mid-span. A Pratt truss of eight panels, 24.0 m long and 2.4 m deep, carrying 60 kN of dead load at each top node, drawn with its depth exaggerated. Each member is drawn as thick as, and labelled with, the distance mid-span moves for a millimetre's change in that member's length: its force under a unit load at mid-span. The middle top-chord members move it 2.50 mm per mm — the span over four depths — and the chord members fall off towards the supports: 1.25, 1.88, 2.50, 2.50, 1.88, 1.25 along the top chord. The diagonals move it 0.80 per mm and most verticals 0.50; three verticals — the one at mid-span and the two beside the supports — carry nothing under a load at mid-span and move it not at all (dashed). A truss's shape is its members' lengths, weighted like this.

The camber that lives in the member lengths

A beam is cambered by bending it; a truss cannot be, because it has no curvature to bend — its shape is nothing but the lengths of its members. So a truss's camber is a cutting list, and each millimetre on that list arrives at mid-span multiplied by the member's force under a unit load there: two and a half for the middle chords of a truss ten times as long as it is deep, nothing at all for some verticals. Build only the chords to their dead-load strain and a quarter of the sag stays, the webs' share. Cut every member to a millimetre of scatter and the mid-span misses by six, an eighth of the camber.

 The same rib, fixed, bends at both ends as well as the crown. The bending moment that axial shortening leaves in a 60 m parabolic concrete rib at a 6 m rise (EI 420,000 kN·m², EA 20,400,000 kN) under 60 kN/m, sagging upward, from a frame solve of the rib. Dashed: two-hinged, sagging throughout, 29.4 kN·m at the crown. Solid: fixed at the springings, 57.9 kN·m sagging at the crown and 112.4 kN·m hogging at each springing. The funicular load itself leaves no moment in either rib; all of this is the 0.11 and 0.63 per cent of the thrust the two ribs lose by getting shorter.

The springings that make shortening worse

Fixing an arch at its springings is the stiffer, cheaper and usual way to build one in concrete, and it makes the arch six times as sensitive to its own shortening. The thrust it loses acts at the elastic centre, two thirds of the way up, so the moment lands at the springings as well as the crown, twice as large and the other way round — at the section the fixed arch is designed at, not away from it.

 How stiff an abutment has to be to count as fixed. The shortening moments in a 60 m parabolic concrete rib at a 6 m rise (EI 420,000 kN·m²) under 60 kN/m as shares of the fixed rib's, against the rotational stiffness of each springing as a multiple of EI/L on a logarithmic scale: at the springings (solid, −112 kN·m when fixed) and the crown (dashed, 58 kN·m fixed, 29 two-hinged). The springing moment is half the fixed value at kθ = 11.3 EI/L and nine tenths at 102 EI/L; at EI/L itself it is 8 per cent. The crown moment never falls below the two-hinged rib's 51 per cent of the fixed value.

The abutment that spreads before it turns

A fixed arch is fixed only if its abutments hold still, and they have two ways not to. Turning releases the springing moment, but slowly: an abutment has to be a hundred times the rib's EI/L to hold nine tenths of it, and a footing on stiff clay under a slender concrete rib already is. Spreading does the opposite. The rib's own shortening is worth 13.6 mm of spread, so an abutment that gives 7 mm a side under the thrust doubles every secondary moment in the arch — and a footing on dense sand gives 8.

 Fit the counters last, and the tolerance locks in nothing. The largest force random length errors lock into an eight-panel Pratt truss with a counter-diagonal in each of its six interior panels, 24.0 m long and 3.0 m deep, carrying 60 kN of dead load at each interior bottom joint, against the errors' standard deviation: the median draw (solid) and the 95th percentile (dashed) with every member fitted in the shop, and the same with the counters left loose until the truss carries its dead load and then cut to the gap (on the axis). With everything fitted, 174 and 288 kN at 1 mm, 348 and 577 at 2 — straight lines, because the force is proportional to the misfit. With the counters fitted last, nothing at any tolerance: the Pratt truss without its counters is determinate, and a determinate truss takes any length error by moving.

The counter fitted after the load

A truss with a counter-diagonal in every panel cannot take a change of length for free, so a camber cut into its members might seem to lock force into it. One cutting list locks in nothing — every member shortened by its own dead-load stretch — and the simpler lists lock in little. What locks in force is the shop: a millimetre of error per member puts 174 kN into some member of a truss whose largest dead-load force is 471. Leave the counters loose until the dead load is on and cut them to the gap, and the same errors lock in nothing at all.

 Five ways to share one deflection among four members. The sway at the top of the left column of a portal 12 m wide and 6 m high with fixed bases, an IPE 500 beam on HEB 300 columns, carrying 25 kN/m on the beam and 40 kN sideways at the top of the left column, 10.38 mm, found five times by the unit load: on the frame itself and on four releases of it. Each bar is split into what each member's integral contributes — left column, left half of the beam, right half, right column — and the red line is the total, which every bar reaches. In millimetres, member by member: the frame itself −3.72, 2.16, 0.66, 11.28; left column alone 10.38, 0.00, 0.00, 0.00; right column alone 0.00, 0.17, 0.17, 10.04; three hinges −15.78, 4.90, 1.41, 19.84; pin and roller −31.43, 27.10, 14.55, 0.15. A member's share can be negative, and on one structure it is the whole answer.

One column is enough to find the sway

The unit load that finds a deflection may be carried on any structure that can carry it, so in a fixed-base portal the sway can be found from one column's two end moments and nothing else. Carry it on a three-hinged frame instead and the same sway comes out of all four members, one of them contributing a negative amount. Every structure gives the same total and a different account of where it came from — which means "how much of the sway is axial shortening" has no answer until a structure is named, and only one structure's answer is worth having.

 A point at mid-span is worth twice its total spread out. The mid-span twist of an open section 8 m long on forks, carrying 12 kN/m at 75 mm from its shear centre, against the stiffness of one rotational spring at mid-span, logarithmic; dashed, the twist with the uniform deck of 5 kN·m per metre per radian, whose whole restraint is 40 kN·m per radian. A mid-span spring of 19.5 kN·m per radian — 49 per cent of the deck's total — twists the beam as little as the deck does; the deck's whole total at mid-span leaves 3.13°.

The restraint that works where the twist is

A deck fastened along a beam's top flange resists its twist a little everywhere. A secondary beam framing in at mid-span resists it a lot at one point. Given the same total stiffness, the point does better — half the deck's total at mid-span holds the beam as well as the whole deck — because it works where the beam twists most. But a point restraint has a price the deck never charged: it takes the torque through one connection, up to three fifths of the whole applied torque, and if it is stiff it moves the worst twist out to the quarter-points rather than removing it.

 A secondary holds more than it twists once its connection passes a few tens of kilonewton-metres per radian. The mid-span twist of the 8 m primary beam of 12 kN/m at 75 mm off its shear centre, with a 6 m 356 × 171 × 51 UB framing in at mid-span whose 60 kN reaction acts 60 mm off the shear centre, against the rotational stiffness of the secondary's connection on a logarithmic scale, in series with the secondary's own 3EI/L of 14,453 kN·m/rad. Dashed, the 6.25° with no secondary. On the deck's side the secondary is worse than none below 33 kN·m/rad and better above it: 9.12° at 10, 3.28 at 100, 0.47 at 1,000. On the far side its torque opposes the deck's (0.89° at 10); as a pair, the torques cancel (1.05° at 100). EN 1993-1-8 calls this secondary's connection nominally pinned below 2,409 kN·m/rad (dotted).

The secondary beam that twists what it holds

A secondary beam framing into a twisting primary at mid-span is the best restraint a primary can have — concentrated exactly where the twist is. It is also a load: its reaction arrives on the primary's web, off the shear centre, as a torque at exactly the same place. Which wins is decided by the connection, and the threshold is low: a fin plate stiffer than about 33 kN·m per radian, a seventieth of what still counts as a pin, makes the secondary hold far more than it twists. The bolts' free play is another matter, because the torque acts from the first millimetre and the restraint only once the play is taken up.

 A cambered girder is an arch on its slings. A 40 m precast girder of 11 kN/m, lateral stiffness 2.55 × 10⁵ kN·m², hung 0.90 m below its roll axis, picked 2 per cent of its length in from each end, with 100 mm of camber at mid-length, drawn in elevation with the camber exaggerated. The girder's centre of gravity is two-thirds of the camber above the line through its ends, 67 mm; its lifting points, on the arc 2 per cent in from each end, are 8 mm above it. It rolls about the line through the picks (dashed), so the camber has lifted its centre of gravity 59 mm towards the roll axis: the girder hangs as though its hook were that much lower.

The camber that lowers the hook

A camber is built into a girder so that it ends up level in the finished structure, and nothing about it matters until then — except on the day it is lifted. A cambered girder hanging from its ends is a shallow arch, its centre of gravity two-thirds of the camber above the line through its lifting points, and it rolls about that line as though its hook were that much lower. On a long precast girder already close to its lifting limit, a hundred millimetres of camber takes the factor of safety against roll from 2.3 to 1.4. The camber grows while the girder waits in the yard, so the same girder becomes harder to lift every week it is stored. Pick it a fifth of its length in from each end and the camber drops out of the problem altogether.

 Two members each half-way to ponding are unstable together. The girder's ponding parameter Cp against the joist's Cs. Either member alone is stable up to 1; together the roof is stable only below the solid curve, (1 − Cp)(1 − Cs) = (8/π²)·Cp·Cs, which meets the diagonal at 0.526 — two members each about half-way to ponding on their own are critical together. The dashed line is the simplified rule Cp + 0.9Cs ≤ 0.25. The marked roof is a 12 by 9 m bay whose girders and joists are each sized to deflect L/240 under 1.5 kN/m²: Cp = 0.26 and Cs = 0.19, stable, with its joists' sag amplified 1.79 times where the joist alone would say 1.24 and the girder alone 1.35.

The roof that ponds on its girders

A flat roof's joists sit on girders, and the water fills the sum of both their sags. Two members each half-way to ponding on their own are unstable together, and long before that the growth of the water is far larger than either member's own amplification says. AISC's simplified rule turns out to be a line of constant growth, and a member sized to a deflection limit has a ponding parameter proportional to its span — so a bay sized to L/240 under 1.5 kN/m² meets the rule up to 6 m, doubles its water at 12 and ponds at 24.5.

 A stiff building bridges the trough it stands in. The settlement of a building 30 m long centred on the trough, over a trough 30 mm deep whose inflection points are 15 m from its centre (dashed: the ground with no building), measured from the building's settlement under its own weight, for relative stiffnesses EI/(kₛ·B⁴) of 10⁻⁴, 10⁻³ and 10⁻². At 10⁻⁴ its deflection ratio is 95% of the ground's; at 10⁻³ its deflection ratio is 66% of the ground's; at 10⁻² its deflection ratio is 17% of the ground's. The stiff building settles almost as a rigid body and leaves the trough's curvature to the ground beneath it.

The building the trough cannot bend

A tunnel leaves a settlement trough, and damage criteria are applied to the trough's deflection ratio over the length of each building above it. But a building is not the ground: it has a bending stiffness, and it takes only as much of the ground's curvature as that stiffness lets it. One number decides how much — the building's stiffness over the soil's modulus times its length to the fourth — and over two decades of it a building goes from following the trough to bridging it. Because the length enters to the fourth power, the same building is stiff when short and soft when long.

 The gaps that open after the arch is closed. For the 60 m fixed concrete rib rising 6 m under 60 kN/m, closed at 28 days with its abutments jacked apart to take up the day's gap: the gap between the rib and its springings that each movement opens after closure, in millimetres, against the years since, on a log scale — the rib's creep under its thrust, its shrinkage, and the ground consolidating under the thrust by 1.0 times its 8 mm immediate spread with 90% done at 10 years. After a hundred years they are 24.8, 16.2 and 8.0 mm, 48.9 in all, against the 13.6 mm the thrust shortened the rib on the day. The rib resists them while it creeps itself, so the gap it acts as if fixed against is 20.9 mm (fine dashes).

The jacks the arch forgets

Jack a fixed concrete arch's abutments apart before the crown is closed, by exactly the gap its thrust and its footings open on the day, and the springings start with no moment at all. A century later they carry 173 kN·m — half as much again as the 112 the jacks were set to remove. The thrust's own moment never relaxes; the jacks' correction does, as the rib creeps; and shrinkage and the ground's consolidation arrive after the jacks have gone. No single setting cancels movements that run on different clocks.

 A building that crosses the trough at an angle stands on a warped surface. Plan of a 30 × 12 m building 10 m high, its walls 300 mm thick, over a trough 30 mm deep with its inflection 15 m from the tunnel's axis, its long axis 45° from square across the tunnel. The lines are the greenfield trough's settlement contours, 5, 10, 15, 20, 25 mm, parallel to the tunnel's axis (dashed). The corners settle A 27.4, B 18.4, C 27.4, D 18.4 mm, so they do not lie in a plane: A and C together settle 18.1 mm more than B and D. A building spanning square across the trough has no such warp, and one lying along it has none either; between the two, the trough twists the building as well as bending it.

The building the trough twists

A building that crosses a tunnel's settlement trough at an angle stands on ground that is warped as well as sagged. Its four corners do not settle into a plane, and beyond about twenty degrees of skew the warp is larger than the sag. Whether the building follows that warp or resists it depends on a property its bending never needed — whether its walls and floors close into a tube — and the same walls, closed or open, differ a thousandfold in their stiffness against it.

 The joint's moment takes the sign of the day it was cast. For two 12 m pretensioned beams, 300 mm wide and 700 mm deep, with a 160 mm in-situ slab 1.2 m wide, in 60 per cent humidity, made continuous over their pier by a joint cast with the slab: the moment the joint is asked to carry to hold their ends together, sagging positive, over thirty years, with the joint cast when the beams are 28, 90 and 365 days old; dashed, the joint's cracking moment either way, 140 kN·m. Cast at 28 days the joint is sagged to 211 kN·m; at 90 days it ends at 40; at a year it is hogged to −214.

The joint cast over the pier

Two pretensioned beams made continuous over a pier by a joint cast with the slab have ends that are still turning when the joint goes in, and the joint stops them. What it is then asked to carry depends on the day it is cast. At four weeks the beams' creep is still lifting their ends and the joint is sagged to half as much again as its cracking moment. At a year the beams have nearly stopped and the slab's shrinkage hogs the joint by as much the other way. Cast at about sixteen weeks, it is asked for almost nothing.

 Where the outlet sits decides whether the roof ponds at all. The load the water puts in the more heavily loaded member of a square bay, over what the same head of water would put in it on a rigid roof, against the bay's side, with girders and joists each sized to deflect L/240 under 1.5 kN/m². With the level fixed by a scupper it is 2.01 at 12 m and 3.92 at 18 m, and the bay ponds at 24.5 m. With a drain on the girder line it is 1.24 and 1.41; with a drain at the bay's middle, 0.82 and 0.75, falling as the bay grows, because the water's surface goes down with the roof.

The roof that drains towards its sag

The depth of water a ponding check takes as given is set by the drainage: an overflow's height above the roof, plus the head a storm needs to leave through it. Which head a roof can carry depends less on how stiff it is than on where the overflow sits. With the level fixed by a scupper in the parapet, a 12 by 9 m bay allows 85 mm of water and an 18 m square bay 39. With a drain at the middle of the bay, whose inlet goes down as the roof sags, the same bays allow 182 and 205 mm — more than a rigid roof would — because the water's surface follows the sag down instead of filling it.

 The shear term cuts the beam's stiffness, not its share. For 20 storeys of two 6.0 m walls 0.35 m thick either side of a 2.4 m opening, coupling beams 0.35 m wide, 60 kN at every floor, against the coupling beams' depth: each beam's stiffness with its shear deformation counted over its stiffness in bending alone, the degree of coupling likewise, and the largest beam's shear likewise. At a depth of 0.6 m (span over depth 4.0) the stiffness is 0.85 of the bending figure, the coupling 0.977 and the shear 0.961; at 2 m (1.2), 0.33, 0.973 and 0.937. Where the shear term is large the beam is stiff enough that its stiffness has stopped deciding anything.

The coupling beam asked for shear

The coupling beams of a pair of shear walls are the stubbiest members in a building, and the shear term in their deflection is largest exactly there: a beam two metres deep over a 2.4 m opening is a third as stiff as bending alone says. The walls hardly notice. The pair's degree of coupling moves by under 3%, the beams' shears by under 7%, because a beam that stiff has stopped deciding how much of the overturning the walls carry as a couple. What decides it is the shear the beams can give. A 600 mm beam limited to 2 N/mm² starts to yield at 1.9 times the storey force; past the last yield the pair is two cantilevers again, and the beams supply rotation instead.

 What it costs depends on the load it must still carry. The steel in an eight-panel Pratt truss with a counter-diagonal in each interior panel, loaded by 10 at every bottom joint, sized to survive the loss of any one interior diagonal, against the share of the full load it must carry with the diagonal gone; the intact fully stressed design's 1,200 and the Pratt truss's 1,220 as level lines. At a third of the full load the robust truss needs 1,221, at a half 1,252, at two thirds 1,304, at the full load 1,555, and at four thirds — two thirds with the loss counted as sudden, doubled — 2,074. From about a half the interior diagonals grow in proportion to the load, each sized by its partner's loss; past the full load the chords grow too, because the damaged truss is then carrying more than the intact one was sized for.

The truss sized to lose a diagonal

A Pratt truss with a counter-diagonal in every interior panel, sized fully stressed, starves its middle counters to nothing and saves a little steel over the plain Pratt. Ask it to survive the loss of any one diagonal at the accidental two thirds of its load and the family of equal designs collapses to one: every counter comes back, sized by its partner's loss, and the truss needs 1,304 units of steel against 1,200 — 9 per cent for an alternative path the Pratt cannot have at any price. A lost chord can be survived too, by the panel's diagonals turned into a chord, for twice the steel. Four members in the end triangles cannot be survived at all.

 Allowed to yield, the damaged truss is hardly lighter. The steel in an eight-panel Pratt truss with a counter-diagonal in each interior panel, loaded by 10 at every bottom joint, sized to survive the loss of any one interior diagonal, against the share of the full load it must carry with the diagonal gone: checked elastically (solid) and by the lower-bound theorem, every member allowed to yield (dashed); the intact design's 1,200 as a level line. At a third, 1,221 elastic and 1,206.7 plastic; at two thirds, 1,304 elastic and 1,296.7 plastic; at the full load, 1,555 elastic and 1,545.0 plastic; at four thirds, 2,074 elastic and 2,060.0 plastic. The plastic design saves 7.4 units at two thirds of the load out of the 104 that robustness costs.

The section that yielding cannot reach

A truss that loses a member in an accident is allowed to yield while it finds another way to stand, so a counter-braced truss sized by the lower-bound theorem ought to need far less steel than one sized elastically. It needs 1,296.7 units against 1,304 — seven of the 104 that robustness costs. One diagonal gone, the damaged panel is determinate at its own section, and no redistribution can change a force a cut decides; yielding can reach only the verticals. What does move the price is a counter that buckles: good for half its tension in compression, it adds 120 units, more than robustness did.

 The best compensation is less than the prediction. For every floor of a 40-storey building of 3.6 m storeys, its perimeter columns at 12 N/mm² and its core at 6 under the finished load, cast a floor every seven days, with 25 per cent of the load arriving as finishes 90 days after the frame is complete and creep coefficient 2.2, from its fit-out to thirty years, with the creep anywhere within ±30 per cent of the design value: the largest amount any floor stands high (dashed) and stands low (light), and the worse of the two (solid), against the share of its predicted thirty-year movement each floor is set high by. Uncompensated, a floor ends 31.9 mm low; compensated in full, a floor stands 16.0 mm high at its fit-out; the worst is smallest, 11.0 mm, at 0.79 of the prediction, where level 40 standing high 0 days after its fit-out under creep ×0.70 balances level 33 standing low after thirty years under creep ×1.30.

The floor that is level on one date

Each floor of a tall concrete building is set high by the amount its columns are predicted to drop against the core, so that it arrives level. Counted in time, the drop does not peak at mid-height and stop at the roof: the worst floor of a forty-storey building is level 31, the roof moves 24 mm, and half the movement is still to come when the floors are fitted out. Set high by all of the prediction, a floor is level on one date thirty years away and 16 mm high at handover. With the creep known to ±30 per cent, setting it high by about four fifths of the prediction makes the worst floor's error 11 mm instead of 16. And none of it changes what the partitions see.

 A joint poured twice ends near where the slab's day puts it. For two 12 m pretensioned beams, 300 mm wide and 700 mm deep, with a 160 mm in-situ slab 1.2 m wide, in 60 per cent humidity, made continuous over their pier: the moment in the joint, sagging positive, over thirty years, with its first stage — the diaphragm and the beams' bottom connection — cast at 28 days and the slab at 42, 90 and 365 (solid), against the joint cast once, with each slab (dashed). With the slab at 42 days it ends at 139 kN·m against 160 poured once; with the slab at 90 days it ends at 41 kN·m against 40 poured once; with the slab at 365 days it ends at −165 kN·m against −214 poured once. The first stage starts each history early and the slab's weight kicks it into hog on the day it is cast; after that each solid line runs beside its dashed one.

The joint poured twice over the pier

A continuity joint over a pier can be cast in two stages — the diaphragm and the beams' bottom connection first, the slab weeks later — in the hope that the early pour catches the beams' creep and the late one leaves the slab's shrinkage little to do. It does neither. From the slab's day on the composite section holds exactly what a single pour on that day would have held, and the first stage adds only its own small account, of either sign. It trims a year-old slab's hog by a fifth, and it asks a joint with half the cracking moment to carry the beams through the months between.

 Closing late beats jacking again on the same date. For the 60 m concrete rib rising 6 m under 60 kN/m, struck at 28 days, on ground that spreads 8 mm at once and as much again as it consolidates, 90 per cent in 10 years: the springing moment after a century, against a date in the arch's first years — left three-hinged and closed on that date (solid), or fixed at striking and jacked back to no moment on that date (dashed); dotted, fixed and jacked at striking only, −173. At three months, −124 against −140; at a year, −84 against −117; at ten years, −15 against −76. The difference is the memory of the years before: the jacked arch's early restraint goes on relaxing after the second jacking, and the open arch had none.

The arch that waits with its hinges open

A concrete arch fixed on the day it is struck has to restrain every millimetre its rib creeps and shrinks and its ground settles for the rest of its life, and jacking it again a year later cannot make it forget the year before. Struck as a three-hinged arch and left that way, it takes the fastest part of every movement for nothing, because a determinate arch has no moment to spend on a change of length. Closed at a year it ends the century with half the springing moment of the arch fixed at striking, and a third less than the arch jacked again on the same date. The price is a crown that has dropped 78 mm while it waited, and a rib that is no longer the shape of its own thrust.

 How much of the drain has to work. The load in the more heavily loaded member, with the roof's own 0.5 kN/m², over its 2.0 kN/m² capacity, for a 12 by 9 m bay in a 150 mm/h storm, with a drain at its middle that passes the storm at 30 mm when clear and a 400 mm scupper at its edge whose sill is 50 mm above the supports, against the share of a clear drain's flow the drain still passes. Members at L/180, L/240, L/360 and rigid: with the drain fully blocked, 1.25, 0.99, 0.84, 0.66; a tenth open, 0.80, 0.77, 0.71, 0.62; half open, 0.44, 0.45, 0.46, 0.48; clear, 0.37, 0.37, 0.38, 0.40. A softer roof is the worse with the drain blocked and very slightly the better once about a fifth of it works.

The drain that is mostly blocked

A roof's ponding check is made with its primary drains blocked, and a drain is seldom blocked outright: leaves and gravel choke it to a fraction of its flow. A fraction turns out to be nearly enough. On a 12 by 9 m bay sized to L/240, a drain passing a seventh of its clear flow still takes the whole of a 150 mm/h storm, because the water standing over it at the bottom of the sag is the bay's own deflection plus the level. A softer roof's drain works harder still. Only near complete blockage does the scupper take over, and only there does stiffness help.

 The middle reaction is a dimension, read off in millimetres. The middle reaction under dead load of a Pratt truss continuous over two 24.0 m spans, 3.0 m deep, carrying 60 kN of dead load at each interior bottom joint, against the height of its middle bearing above the level drawn, cut to the continuous list (solid) and as two simple spans (dashed). At the drawn level the continuous list lands with 572 kN, the analysis's value, and the simple-span list with 480 kN, the two spans' own share. Every millimetre the bearing stands high adds 2.6 kN: 10 mm high, 598 kN; 10 mm low, 545; the simple-span list's 92 kN shortfall is the bearing 35 mm low.

The pier's share is cut into the members

A truss continuous over a middle support has one reaction that statics cannot find, and the lengths its members are cut to decide it. Cambered to land level under its dead load on three bearings, a two-span truss puts 572 kN on its pier; cambered span by span, as two simple trusses, the same members on the same bearings put 480 kN there and carry the dead load with no moment over the pier at all. The list chooses the moment diagram. And a millimetre of shop tolerance on every member moves the pier's share as much as a bearing set fourteen millimetres out of level, so a reaction the design depends on has to be set by a jack rather than by a drawing.

 The column line's sum agrees with every analysis, because it cannot see the beams. The roof sway of a 4-bay, 10-storey concrete frame, 8 m bays and 3.5 m storeys, 600 mm columns and 400 × 700 mm beams, 40 kN/m on every beam and 40 kN of wind at every floor, against the beams' stiffness as a share of their gross value, from the whole frame (line) and from the windward column line's integral of the moments that analysis gives it (dots). At the gross stiffness, 15.7 mm; at 0.5, 25.4; at 0.35, the usual allowance for cracked beams, 33.3; at 0.2, 52.0. The windward column's base moment moves from 126 to 235 kN·m, and the line's integral follows the sway exactly at every one.

The check that cannot see the beams

The unit load that finds a frame's sway may be carried on any structure that can carry it, and in a frame of four bays and ten storeys it can be carried down one line of columns alone. Then the roof sway is that line's real moments against the unit load's, a few dozen products by hand — exact for the frame as drawn, for its beams cracked to a third of their stiffness, for its joints made springs and for its bases made springs, the last only once the bases' rotation is given its own term. It is exact because the unit load never enters a beam. Which makes it a check on the analysis's arithmetic that cannot say anything about the analysis's beams.

 The survey knows the creep once most of the floors are set. The doubt about the creep of a 40-storey building cast a floor every seven days, its creep known to ±30 per cent, as the half-width of its 95% band, after a survey of every floor set by then — each read to a standard error of 1.0 mm — against the floor it is made after: from the survey alone (drawn at 100% where it is larger), and combined with the design's ±30%; beside them, the share of the building's floors still to be cast, which are all a survey can correct. The survey alone beats the design's doubt only after floor 25, with 38% of the floors still to cast; the combined doubt halves after floor 31, with 23% still to cast.

The survey that arrives after the floors

A tall building's lower storeys start shortening months before its upper floors are cast, so a survey every ten floors should tell the creep the building really has and let the floors still to come be set for it. It does tell it, to a millimetre's precision, after the twenty-fifth floor. By then most of the floors whose height it was meant to correct are already set, and the floor that ends furthest from level is the one no survey can help.

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