Structural Engineering & Statics
where the load goes
Generator

The answer arrives in instalments

Rendered here at the parameters it defaults to, with every essay that calls it — which is the same list as the blast radius of changing it.
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.

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.

9 essays call moment-distribution. The drawing above is what it returns with no arguments at all; every call below passes it something, because a placement that passes nothing draws whichever member of the family the generator happens to default to rather than the one its essay argues about.

Where it is called

Changing this generator changes every one of these figures.

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. Deflection

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.

 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. Deflection

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. Deflection

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.

 A force may be moved anywhere, at the price of a couple. A 80 kN force applied 250 mm off the centreline of a body, and the same force applied ON the centreline together with a couple of 20 kNm. The two systems are equivalent: they have the same resultant force and the same moment about every point in space, so no equilibrium equation written about the body can tell them apart. What they are not is the same loading — the stresses inside the body differ, and they differ over a distance of about the body's own depth. The offset is drawn to a scale that keeps the arrow on the body; the number beside it is the real one. Internal forces

The moment the beam left behind

A beam reaction is drawn arriving on a column's centreline. It arrives on a cleat a hundred millimetres out from the face, and the difference is a couple that goes into the column and has to be shared between the lengths above and below it. Nothing about it appears in a frame model whose members meet at nodes.

 The same load, two diagrams, both in equilibrium. One span of a pair of 9 m spans under 30 kN/m, drawn twice. The elastic solution puts 304 kNm over the support and 171 in the span. Reducing the support moment by 30% and taking what statics then gives leaves 213 and 207: the section the beam needs falls from 304 kNm to 213, a saving of 30%. Both curves are in equilibrium with the same load — the mid-span ordinate plus half the support moment is the free moment 304 kNm for either — and the second is legitimate for that reason alone. What it costs is 13.0 milliradians of rotation at the support, which the section has to be able to deliver. Internal forces

The moment that was shed has to land

Redistribution takes a moment off a beam's support and pays for it with rotation. On a beam that is the whole story. In a frame the support is a column, the shed moment does not vanish, and it arrives at a member whose section was chosen from the diagram it has just left.

 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. Deflection

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.

 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. Deflection

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. Deflection

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. Deflection

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 library, page 3 of 7 — where moment-distribution sits