Structural Engineering & Statics
where the load goes
Concept

Convergence — where it appears

The approach of an iterative calculation to its exact answer as the iteration proceeds, and the rate at which the remaining error falls. It decides how many cycles a hand method needs and, for a numerical model, whether refining the mesh is improving the answer or merely changing it.

Named by 7 essays across 3 fields — each of them below, with the objects they name alongside it.

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.

deflection · Discretisation 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.

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

Equivalent in work, not in resultant

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

equilibrium · Force couple 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.

deflection · Moment distribution 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.

deflection · Moment distribution Three quantities converging at three different rates. How much of the exact answer a truncated modal analysis of a ten-storey building reaches, against how many modes it keeps. The mass count is the rule — ninety per cent, reached at two modes. The base shear is ahead of it: one mode carries 85 per cent of the mass and 98 per cent of the base shear. The force in the top storey is behind it, at 80 per cent on one mode and 94 on two. The rule is written in the quantity that converges fastest, and it is checked against a quantity nobody plots.

The modes that were left out

Nobody runs every mode a model has, and the rule for how many is a mass count — enough to account for ninety per cent of the structure. The rule is written in the one quantity that converges fastest. On a twenty-storey frame two modes give the base shear to within one per cent and the force in the top storey to within twenty.

dynamics · Mode shapes Two capacity curves that agree, from two answers that do not. Base shear against roof drift for the same eight-storey frame with a soft ground storey, pushed with a fixed triangular pattern and with one recomputed from the tangent stiffness at every step. At a roof displacement of 426 mm they differ by 8.7 per cent — which is the number a capacity-spectrum procedure reads, and it is the number that is nearly the same. The storey drifts underneath these two curves differ by a factor of 3.8 at the fourth floor, and the drift is what the analysis was run for.

The pattern that stopped describing the building

A pushover analysis pushes with a load pattern chosen to resemble the first mode, and by the time the structure has done anything worth analysing it no longer has that mode. Recomputing the pattern as the frame softens changes the base shear by nine per cent and the drift at the fourth floor by a factor of four.

dynamics · Pushover

Named alongside it

The objects these essays reach for when they reach for this one.

Degrees of freedomFixed-end momentCarry-overContinuityDistribution factorIterationMoment distributionStiffness matrixStiffness methodDiscretisationEigenvalueFlexibility

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