Deflection

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

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.

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.

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.

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 nullVector 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: 3.40° at mid-span with the ends restrained against warping, against 9.09° if they are not, a factor of 2.67. What that angle does is move the flange tip sideways by 11.9 mm — 82 per cent of the member's own vertical deflection, and 37 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, 80 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 further sideways than four 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. At the 10 per cent rise drawn that is 0.10 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 0.10 per cent that the rib shortening removes from the thrust leaves 28 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.

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.

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 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.

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.