Structural Engineering & Statics
where the load goes
Concept

Distribution factor — where it appears

The share of a joint's unbalanced moment that one branch takes, equal to its stiffness over the sum of the stiffnesses meeting there. It sums to one at every joint, which is what makes moment distribution converge, and it is a ratio of stiffnesses rather than of strengths.

Named by 7 essays across 2 fields — each of them below, with the objects they name alongside 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.

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.

deflection · Moment distribution One of these two curves is a stiffness and the other is a statement of statics. The torque a spandrel beam carries, against how much of its torsional stiffness is left. The rising curve is compatibility torsion — a floor beam framing into the side of the spandrel, which shares its fixed-end moment of 197 kNm between the spandrel's torsional stiffness and its own flexural one. Uncracked, the spandrel takes 51% of it, or 100 kNm; at a quarter of that stiffness it takes 21%, or 41 kNm, and the floor beam picks up what was shed. The flat line is equilibrium torsion — a canopy cantilevering 2.2 m off the same spandrel, whose 116 kNm is fixed by statics and contains no stiffness at all. The first can be designed away by accepting a rotation. The second cannot be designed away by anything.

The torsion that goes away if you let it

A spandrel beam attracts a torque in proportion to its own torsional stiffness. Crack it and the stiffness falls by a factor of four, the torque falls with it, and nothing has failed — because the floor beam it was competing with picks up exactly what was shed. A canopy hung off the same spandrel is a different animal entirely.

internal-forces · Compatibility torsion 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 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.

deflection · Moment distribution 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 The perimeter the edge leaves, and where its middle is. A 400 × 400 mm column at the edge of a 260 mm slab (d = 225 mm) carrying 12 kN/m², on a 7.2 m end span. Heavy line: the control perimeter 2d out, cut by the slab's edge — 2,614 mm, against 4,427 for the same column inside the slab. Wide band beneath it: the reduced perimeter u1 of EN 1992-1-1, 2,214 mm, its legs cut back from the edge. Open circle: the perimeter's centroid, 363 mm in from the column's centre. Filled: the load's resultant, 631 mm in, set by the slab's end moment of 170 kN·m on a shear of 269 kN — 269 mm beyond it, into the slab. The general multiplier on the shear is 1.78; u1/u1 gives 1.18 and the shortcut 1.4.

The perimeter whose middle is not the column's

At an edge column the slab's own edge cuts the control perimeter, and what is left has its centroid 363 mm inside the column's centre. The slab's end moment moves the shear's resultant inward too, by a distance that grows with the end span. What twists the perimeter is the gap between the two: nothing at a 5 m end span, enough by 7.2 m to put the check at 1.23 of its resistance where the 1.4 shortcut reads 0.97. And a bigger column makes it worse.

internal-forces · Punching shear

Named alongside it

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

Fixed-end momentMoment distributionCarry-overContinuityIterationConvergenceSlope-deflectionDegrees of freedomFree bodyIndeterminacyJoint stiffnessRotational stiffness

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