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.

Assumes Solved by passing it around, The matrix that replaced the hand methods and One support too many, and what it costs to know.

Moment distribution solves a frame by passing an out-of-balance round its joints, and the reason it was a great method is that it converges quickly enough to stop early. That is a claim about a rate, and a rate is a thing that can be measured.

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.
Fig. 1 The hogging moment at an interior support of three spans of 7, 9 and 7 m under 24 kN/m, cycle by cycle. It starts at the fixed-end moment of 98.0 kN·m — the value with every joint clamped — and settles at 156.9 against an exact 156.9. The error falls by about a factor of four per cycle: 21.03, 5.92, 1.54 and 0.60 kN·m after one, two, three and four.

What is fixed before anything happens

The iteration’s ingredients are all decided by geometry, and none of them changes during it.

What each member takes is decided before anything is distributed. Distribution factors at every joint of a three-span beam of 7, 9, 7 m. At each joint the out-of-balance moment is shared between the members meeting there in proportion to 4EI/L, so a short span takes more of it than a long one — the factors are properties of the geometry and are written down once, before any arithmetic happens. The fixed-end moments the method starts from are wL²/12: 98.0, 162.0, 98.0 kNm.
Fig. 2 The distribution factors at every joint. At each joint the out-of-balance is shared between the members meeting there in proportion to 4EI/L, so a short span takes more of it than a long one: 0.563 and 0.438 here. The fixed-end moments the method starts from are wL²/12 — 98.0, 162.0 and 98.0 kN·m.

Three numbers per joint and one per member, all written down once:

The fixed-end moment, which is the moment the member delivers to a joint that is clamped. It is the starting state and the whole of the load’s influence.

The distribution factor, 4EI/L4EI/L\frac{4EI/L}{\sum 4EI/L}, which shares an out-of-balance between the members at a joint. The factors at a joint sum to one, so nothing is created or lost.

And the carry-over factor, which is exactly 12\tfrac12 for a prismatic member with its far end fixed. Release one end of a member by a unit rotation and the far end picks up half the moment.

The iteration is nothing but repeated application of those three, and the interesting question is why applying them repeatedly should approach the right answer at all.

Which free body produced the number

The free body is one joint, and the operation is releasing it.

Clamp every joint. Each member then delivers its fixed-end moment to the joint it meets, and the joint is generally out of balance — the sum of the moments on it is not zero, and the clamp is holding the difference.

Now unclamp one joint, and only that one. It rotates until it is in balance, and the rotation puts a moment into each member meeting it in proportion to the member’s stiffness — which is the distribution factor. Half of each of those moments arrives at the member’s far end, which is still clamped, and puts that joint further out of balance.

Clamp the released joint again and move to the next. Every step satisfies rotational equilibrium at exactly one joint while disturbing its neighbours, and the disturbance is halved on the way there.

That is a precise description of a numerical method: Gauss–Seidel iteration on the stiffness equations, with one equation per joint rotation and the equations solved one at a time in sequence, each using the most recent values of the others.

Why it always converges

Gauss–Seidel does not converge for a general system. It converges for this one for a reason that is structural rather than numerical.

The stiffness matrix of a frame is symmetric and positive definite — symmetric because reciprocity says so, positive definite because the strain energy of a restrained elastic structure is positive for any non-zero displacement. Gauss–Seidel converges for every symmetric positive definite system, unconditionally, from any starting point.

So moment distribution cannot diverge on an elastic frame, ever. That is a theorem rather than an observation, and it is why a method with no error estimate and no stopping rule could be trusted by people who had no way of checking it.

The matrix is also diagonally dominant for an ordinary beam — a joint’s own stiffness exceeds the sum of the carry-overs reaching it — which is a stronger condition and is what makes the convergence fast rather than merely certain.

How fast, and what decides it

The rate is a property of the beam, and the factor of four is not universal.

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 72.0 kNm — the value with every joint clamped — and settles at 328.9 kNm against an exact 328.9. The error falls by about a factor of four per cycle: 99.40, 36.43, 8.64, 2.10 kNm after one, two, three and four. Two cycles is an engineering answer and nobody had to invert anything.
Fig. 3 Three spans of 6, 14 and 6 m under the same load. The starting fixed-end moment is 72.0 kN·m and the answer is 328.9 — four and a half times as far to travel — and the errors after one to four cycles are 99.40, 36.43, 8.64 and 2.10. Two cycles leaves 11 per cent of the answer outstanding against under four per cent on the regular beam.
What each member takes is decided before anything is distributed. Distribution factors at every joint of a three-span beam of 6, 14, 6 m. At each joint the out-of-balance moment is shared between the members meeting there in proportion to 4EI/L, so a short span takes more of it than a long one — the factors are properties of the geometry and are written down once, before any arithmetic happens. The fixed-end moments the method starts from are wL²/12: 72.0, 392.0, 72.0 kNm.
Fig. 4 The factors for the same beam: 0.700 and 0.300 rather than 0.563 and 0.438. A long span next to a short one is soft next to something stiff, so the short span takes 70 per cent of every out-of-balance — and the fixed-end moments it starts from are 72.0 and 392.0 kN·m, a ratio of 5.4.

Two things make the irregular beam slower and they compound.

The starting point is further away. The clamped state is a poor guess when the spans are very unequal, because the fixed-end moments differ by a factor of 5.4 and the balanced answer is nowhere near either.

And the distribution factors are further from a half. A factor of 0.7 sends most of the out-of-balance one way, so each cycle corrects one side well and the other badly. The convergence rate of Gauss–Seidel is governed by the spectral radius of the iteration matrix, and an unequal beam’s is larger.

The practical rule survives, with a caveat. Two cycles is enough for a regular beam and four for an irregular one, and the way to tell which is which is to look at the fixed-end moments before starting: if they are within a factor of two, the method will be quick.

Watching the error, which is the method’s own instrument

The cycle-by-cycle plots are the closest thing the method has to an error estimate, and reading them is a skill worth stating.

The increments fall geometrically. On the regular beam they are 21.03, 5.92, 1.54 and 0.60 — ratios of 3.55, 3.84 and 2.57, hovering around four. A sequence whose increments fall by a constant factor is converging geometrically, and the remaining error is roughly the last increment divided by (ratio − 1).

So the last increment is an error estimate. After the third cycle the increment was 1.54 and the ratio about 3.8, so the outstanding error is about 0.55 — which is very nearly the 0.60 the fourth cycle actually delivered. A designer who plots the increments has a stopping rule, and the plot costs nothing.

And a ratio that is not constant is a signal. If the increments stop falling geometrically, either an arithmetic error has been made or the frame has something in it — a very soft member, a nearly-mechanism — that makes the iteration matrix’s spectral radius approach one. Both are worth knowing about, and neither shows up in a single final number.

That is the property the matrix solvers gave away. An elimination is either right or wrong and shows nothing in between; an iteration shows its own health at every step, and the health is information about the structure rather than about the arithmetic.

It is the same answer as everything else

The method’s output is worth checking against a route sharing none of its arithmetic.

The same diagram, from two entirely different arithmetics. Bending moments on a three-span beam under 24 kN/m. The curve is the stiffness solution; the marked points are what moment distribution reached after eight cycles of hand arithmetic — -0.0, 156.9, 156.9, 0.0 kNm at the supports against an exact 0.0, 156.9, 156.9, -0.0. The two share no code and no equations, which is the only arrangement in which agreement is evidence of anything.
Fig. 5 The bending moments on the beam, with the curve from a stiffness solution and the marked points from eight cycles of hand arithmetic: −0.0, 156.9, 156.9 and 0.0 kN·m at the supports against an exact 0.0, 156.9, 156.9 and −0.0. The largest disagreement anywhere is 2.7 × 10⁻³ kN·m.
3 continuous spans against 3 simple ones. The bending moment in a continuous beam, solved by the stiffness method, drawn over the moment in the same spans made simply supported. The peak sagging moment falls from 243.0 to 86.1, and a hogging moment of 156.9 appears over the supports where there was none.
Fig. 6 The same diagram against the same spans made simply supported: 86.1 kN·m of sagging against 243.0, and 156.9 of hogging where there was none. The reactions are 61.6, 214.4, 214.4 and 61.6 — the inner supports carrying far more than a share.

The moment over the support is what continuity buys and what moment distribution computes, and the two figures are the same numbers reached by an iteration and by an elimination.

Where the carry-over of a half comes from

The one number in the method that looks arbitrary is not, and deriving it takes two lines.

A prismatic member with its far end fixed, rotated by θ at its near end, has a near-end moment of 4EIθ/L4EI\theta/L and a far-end moment of 2EIθ/L2EI\theta/L. Those are the slope-deflection coefficients, and the ratio between them is a half.

The 4 and the 2 come from integrating the curvature along a member whose ends are held: the near end’s rotation produces a moment diagram that is linear along the member, and the far end’s share of it is exactly half. Nothing about the load enters, because the carry-over is a property of a member’s response to an imposed rotation rather than to anything applied.

Two consequences follow that a designer meets constantly.

A member pinned at its far end has a carry-over of zero and a stiffness of 3EI/L3EI/L rather than 4EI/L4EI/L. There is nothing at the far end to receive a carry-over, and the near-end stiffness is three-quarters of the built-in value. That is the “modified stiffness” every textbook mentions, and it is a considerable saving in hand work — an end span with a pinned support needs no cycles at all beyond its first balance.

And a haunched member has neither. Its stiffness and its carry-over are both properties of the taper, they are not 4EI/L and a half, and they have to come from a table. That is why haunched portal frames were among the last structures to be done by hand, and among the first to go to a computer.

The other hand method, and why this one won

Before moment distribution the standard route was the force method, and the comparison says what Cross actually contributed.

The same problem, and one of these can be solved by hand. The flexibility matrices for a 5-span continuous beam, one per choice of redundant, with the magnitude of each entry shaded and the exact zeros left empty. Releasing a moment at a support is felt only in the two spans either side of it, so the matrix is tridiagonal and each equation involves three unknowns — which is the three-moment equation, and is the whole reason continuous beams could be solved on paper for a century. Releasing a support instead is felt everywhere: a unit reaction at any interior support deflects every other point on the beam, so the matrix is full at 100% against 63%. Both give the same bending moment everywhere. The structure did not change; the bookkeeping did, and the bookkeeping was the invention.
Fig. 7 The flexibility matrices for a five-span beam, one per choice of redundant. Releasing a moment at each support is felt only in the two spans either side, so the matrix is tridiagonal at 63 per cent non-zero; releasing a support instead is felt everywhere and the matrix is full at 100 per cent. Both give the same bending moment.
Two different structures released, and one bending moment diagram. The bending moment in a continuous beam of 7, 9, 7 m under 24 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 7e-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.
Fig. 8 The same beam solved twice by the force method with different redundants — hinges over the supports, and the supports themselves removed. The two released structures have nothing in common, and the diagrams they produce lie on top of each other to 7 × 10⁻¹⁵ of the peak moment.

The tridiagonal matrix is the three-moment equation, and it is why continuous beams could be solved on paper for a century. What it could not do was frames: choosing what to take away from a multi-storey frame produces a released structure nobody can visualise and a flexibility matrix that is full.

Moment distribution needs no released structure and no matrix at all. Its unknowns are joint rotations, which every joint has whether the frame is a beam or a building, and its bookkeeping is one column per member on a sheet of paper. That is what made continuous frames routine, and it is the whole of the invention.

What the starting point is, and why it matters

Every iteration begins somewhere, and the clamped state is a deliberate choice rather than an obvious one.

Clamping every joint gives a state that is in equilibrium with the load — each member is a fixed-ended beam and carries its share correctly — and violates compatibility at every joint, since the rotations that should be shared are all zero. The iteration then restores compatibility one joint at a time while never leaving equilibrium.

That is the opposite of the force method, which starts from a state that satisfies compatibility trivially — a released structure, free to deform — and violates the redundant conditions, then restores them by solving simultaneously.

Both are ways of arriving at a state satisfying both, and the choice decides everything about the arithmetic. Cross’s route keeps equilibrium exactly at every step, so a calculation stopped early is still a state the structure could be in — a legitimate, if not the elastic, distribution of moments. The force method’s route keeps compatibility, and a calculation stopped early is not in equilibrium and is nothing at all.

That is more than an aesthetic difference. A moment distribution stopped after two cycles is a lower-bound solution: it is in equilibrium with the load, and if the members can carry what it says, the structure is safe. Nothing about the force method’s intermediate states has that property, and it is the reason a partially converged hand calculation was ever acceptable.

What it became

The method was obsolete within forty years, and the way it went is worth knowing because it did not go away.

The matrix replaced the hand methods by solving the same equations directly rather than iteratively, and for a small frame direct elimination is faster and exact. But Gauss–Seidel and its descendants did not disappear; they went into the solvers for problems too large to eliminate, where an iterative method’s cost per step and its ability to stop early are decisive again.

So a modern finite-element solver on a very large model is running an algorithm whose behaviour Hardy Cross’s readers would have recognised: a sweep, a residual, a rate of convergence and a judgement about when to stop.

What was lost is the visibility. A designer doing moment distribution watched the answer arrive and could see which joints were still moving; the same designer with a matrix solver sees a number. That is a fair trade for the size of problems now solved, and it is the reason the method is still taught after being useless for fifty years.

The frame, which is what it was for

Everything above is a beam because a beam makes the rate visible, and the method was not invented for beams.

A continuous beam has one rotation per joint and two members meeting at each. A frame has one rotation per joint and three or four members meeting at each, and the arithmetic is identical: the same fixed-end moments, the same factors in proportion to 4EI/L, the same carry-over of a half.

That invariance is the whole point. Nothing about the procedure changes when a column arrives at a joint, so a designer who could balance a continuous beam could balance a multi-storey frame with no new technique — only more columns on the sheet.

The comparison with the force method is where the size of that gift shows. For a beam the two are comparable; for a frame of nn storeys and mm bays the force method needs a released structure with 3nm3nm redundants chosen so that the released frame is stable and visualisable, and the flexibility matrix is full. Cross’s method needs a sheet of paper with a column per member.

A method whose difficulty does not grow with the structure’s connectivity is a rare thing, and it is why the technique spread from a ten-page paper to every design office in the world inside a decade.

What to carry away

The factors and the fixed-end moments are geometry, written down once. Nothing about them changes during the iteration.

Convergence is guaranteed by symmetry and positive definiteness, not by good behaviour. An elastic frame cannot make it diverge.

The rate is a property of the beam. A regular beam converges by about four per cycle; an unequal one much more slowly, and the fixed-end moments say which before starting.

And it is Gauss–Seidel. That is not a modern reinterpretation — it is what the method is, and it explains both its certainty and its speed.

Where the model stops

Sway is not in it. Everything above assumes the joints rotate and do not translate. A frame free to sway needs a second pass with a sway correction, and the two are coupled — which is where the hand method stops being pleasant.

The members are prismatic. The carry-over factor is exactly one half only for a member of constant section with its far end fixed; a haunched member has its own carry-over and its own stiffness, and both have to be looked up.

Everything is elastic and first-order. A frame that has yielded has different stiffnesses, so the distribution factors are wrong, and one that is swaying has second-order moments the iteration does not contain.

Nothing here is an error bound. The observed rate says the method is converging and not how close it is. The honest stopping rule is to watch the increments, which is what the figures plot and what a hand calculation shows.

And the joints are released one at a time. Releasing them all at once is Jacobi rather than Gauss–Seidel, converges more slowly, and is what a person does by accident when balancing a whole row of joints before carrying anything over.

The lower-bound property also explains a habit that otherwise looks like laziness. A designer who stopped after two cycles and detailed for the moments obtained was not accepting an approximate answer; they were detailing for an equilibrium state, which the lower-bound theorem says is safe. The elastic state was never sacred, and a moment shed on purpose is the same move made deliberately.

The method converges to the same answer three other methods reach by different routes. Choosing what to take away gets there by releasing redundants rather than by balancing joints, and the matrix that replaced the hand methods gets there in one step by solving the whole thing at once — which is faster and says nothing about which joint was doing the work.

The ladder from here

Later rungs on this anchor: the sway correction, and the two-pass procedure for a frame free to translate. Haunched members, where the stiffness and the carry-over are both properties of the taper. Modified stiffnesses for members with a pinned far end, where 4EI/L becomes 3EI/L and the carry-over becomes zero. Symmetry and antisymmetry, which halve the work by halving the structure. Kani’s method, which is the same iteration written in terms of rotations rather than moments. And the convergence rate computed rather than observed, from the spectral radius of the iteration matrix — which turns the rule of thumb above into a number.

Hardy Cross published the method in a ten-page paper in 1930. It is one of the very few genuinely great pieces of engineering method, it made a generation of continuous structures possible, and it was obsolete within forty years — replaced by a computer doing the same iteration without needing it to be elegant. What is worth noticing is that the replacement was not a different idea; it was the same idea, run by something that does not get tired at the tenth joint.

Named alongside this one

Essays reaching for the same objects. Nobody chose these; they are what the concept index makes visible.

The objects this essay names

Each one links to every other essay that touches it.

Carry-overContinuityConvergenceDistribution factorFixed-end momentFlexibilityForce methodFree bodyIterationMoment distributionRedundancyStiffness method