Told what the far end is doing

Balance the joints of a three-span beam and watch what happens at its two ends. Each is a simple support, so it carries no moment. The first cycle finds a fixed-end moment sitting there and cancels it. The second cycle finds a carry-over has arrived from the interior joint and cancels that. The third cancels the next one, and so on, in a sequence that converges to the fact that a pin carries nothing — which was true of the drawing before any arithmetic began.

The method is spending its cycles discovering something it was told.

That figure is the whole content of what follows, and the important word in it is known. Every one of the four numbers is exact. None is a simplification, an average, or a value fitted to make something come out. Each is what the near end of a member resists when the far end is doing a particular thing, and the modification consists of noticing that the far end’s behaviour was on the drawing all along.

What 4EI/L was assuming #

The stiffness a distribution normally uses is $4EI/L$, and it is not a general property of a member. It is the stiffness of a member whose far end cannot rotate at all.

Where that comes from is the shape in the first panel of the figure above. Rotate one end of a beam through a unit angle while holding the other rigidly, and the moment needed is $4EI/L$; the far end, being held, develops $2EI/L$ in the process, which is where the carry-over of one half comes from — it is a ratio of those two numbers and nothing else.

Now ask what the method is doing at a real pinned support. It is applying $4EI/L$ to a member whose far end is emphatically not held, then discovering the error, cancelling it, carrying half of the cancellation back, and repeating. The iteration converges because the beam is diagonally dominant, and what it converges to is the answer that would have been available immediately from $3EI/L$.

The three modifications, and what each one knows #

3EI/L, for a far end free to rotate. Set $M_{ba} = 0$ in the slope-deflection expression for the far end and it fixes $\theta_b = -\theta_a/2$. Substitute back and $M_{ab} = 3EI\theta_a/L$. The carry-over is zero — the far end has no moment for anything to be carried to.

2EI/L, for a far end rotating equally and oppositely. This is what a symmetric structure does to a member crossing its axis of symmetry: the two halves are mirror images, so the two ends of the central member rotate by the same amount in opposite senses. Put $\theta_b = -\theta_a$ in and the stiffness is $2EI/L$. The carry-over is $-1$: whatever the near end gets, the far end gets the negative of it, exactly, and there is nothing iterative about it.

6EI/L, for a far end rotating equally and in the same sense. This is antisymmetry, which is what a symmetric structure under an antisymmetric load does — a continuous beam with the load reversed on one half, or the sway of a symmetric frame. The carry-over is $+1$.

The four together are one formula asked four questions, and it is worth being explicit about which formula. Slope-deflection says $M_{ab} = (2EI/L)(2\theta_a + \theta_b)$ for a member with no chord rotation. Set $\theta_a = 1$ and the four values of $\theta_b$ — nought, $-\tfrac12$, $-1$, $+1$ — give $4$, $3$, $2$, $6$. The whole family is one line of algebra, and the four cases in every textbook table are four substitutions into it.

Measured in numbers written down #

The comparison that matters is not accuracy, because there is no accuracy to compare. It is arithmetic.

Thirty against two is a factor of fifteen, and it is worth saying what the factor is made of, because it is not one effect.

Releasing the end spans removes two joints from the sweep entirely. Those joints were being balanced every cycle and were contributing nothing but the cancellation of their own carry-overs, so the work per cycle falls from six numbers to four — and, more importantly, the starting point improves so much that two cycles are enough where five were needed.

Cutting on the axis of symmetry removes half the beam. One joint is distributed rather than two, the carry-over across the middle span is exact rather than iterative, and one cycle finishes it.

The gap at cycle zero is the thing to look at. The full method starts 128 kNm from the answer and the modified ones start 5.22 kNm from it, before any distribution has happened at all. Almost the whole of the saving is in the starting point rather than in the rate of convergence, and the starting point is a fixed-end moment.

The factors move too, and that is the third thing to change #

Changing a member’s stiffness changes its share of every out-of-balance at the joint it meets, and this is the part that is easiest to carry out of a table by habit.

At the first interior support of the 8-10-8 beam, the full method has the 8 m span offering $4EI/8$ and the 10 m span offering $4EI/10$. The shares are 0.556 and 0.444, and the shorter span takes more, which is the ordinary result.

Release the end and the 8 m span offers $3EI/8$ instead. It is now the weaker of the two, at 0.484 against 0.516, and the middle span takes the larger share. The two members have swapped which of them is stiffer, without either of them changing. What changed is what is known about the far end of one of them.

Cut on the axis of symmetry and the middle span offers $2EI/10$ — half what it did — and the shares become 0.652 and 0.348. The end span now takes nearly two thirds.

Three sets of distribution factors, one beam, all three correct, all three converging to the same moments. It is the clearest available demonstration that a distribution factor is not a property of a member: it is a property of a member and what the analysis knows about the rest of the structure. The same warning applies wherever stiffness decides a share, and it is why a factor copied from a previous job is a hazard rather than a saving.

Two cycles, written out #

The modified run is short enough to follow completely, which is worth doing once.

Start: the released end spans carry $wL^2/8 = 24 \times 64/8 = 192$ kNm at the interior joints and nothing at the pins. The middle span carries $wL^2/12 = 24 \times 100/12 = 200$ at each of its ends. Only joints 1 and 2 are distributed, and by symmetry they do the same thing.

Cycle one. At joint 1 the out-of-balance is $192 - 200 = -8$ kNm. Share it: $-8 \times 0.484 = -3.87$ to the end span and $-8 \times 0.516 = -4.13$ to the middle. The end span carries over nothing; the middle span carries half of its share to joint 2, which is doing the mirror image of the same thing. The support moment is now 195.87.

Cycle two. Each joint finds the carry-over the other one sent it, shares that, and the support moment reaches 197.13. Cycle three gives 197.21 and cycle four 197.22, at which point the numbers have stopped moving in the second decimal place.

Eight numbers to three-figure accuracy, on a beam that took thirty by the ordinary route. The full run’s first cycle, by comparison, produces a support moment of 196.44 and its second produces 205.06 — an overshoot of eight kilonewton-metres in the wrong direction, on the way to the same place.

The half of the modification that gets left out #

This is the error the shortcut invites, and it is worse than not using the shortcut, because it converges.

A span released at one end does not start from $wL^2/12$ at both ends. It starts from $wL^2/8$ at the held end and nothing at the release, because a propped cantilever under a uniform load has exactly that.

Take the stiffness without the moment and the arithmetic is perfectly well behaved. Every joint balances, the iteration converges as fast as it ever did, and the answer is wrong — because the structure being described has a released end that was nevertheless given a fixed end’s starting moment, and no such structure exists.

The failure has no symptom. There is no out-of-balance left over, no residual, and no disagreement between the numbers on the sheet. That is the shape of defect this collection keeps meeting: an assumption made in one place and not carried through to another, invisible to every check that reads what is written down.

The safest way to hold it is that a modified stiffness and its fixed-end moment are one object. Neither is a modification on its own, and a table that has one without the other is not a faster version of the method — it is a different problem.

What the full method reaches, and how #

Set the two side by side. The full method is not wrong, it is uninformed, and watching it work is the best argument for informing it.

The overshoot in that curve is worth a sentence, because it is the visible cost of not knowing. The first cycle cancels the fixed-end moments at the two pinned ends, which sends a large carry-over inward; the second cycle receives it and overcorrects; the third receives the reply. The method is having a conversation with itself about a fact that was on the drawing, and the oscillation is the sound of it.

The modified run has no such feature. Its curve approaches from one side and stops, because there is nothing at the ends to send anything.

The beam where it matters most #

The saving above is real and it is on a well-behaved beam. Make the beam awkward and the modification is worth more rather than less.

The reason is worth stating because it inverts the usual expectation about shortcuts. A convergence rate depends on the beam’s proportions; a starting point does not. The full method’s arithmetic grows as the beam gets more irregular, and the symmetric run’s does not grow at all, because it is not iterating over anything the irregularity affects. The worse the beam, the more the modification is worth, which is the opposite of how an approximation behaves.

A joint where only some of the members know #

The clean cases above have every member at a joint modified the same way. Real structures do not oblige.

Consider the first interior support of a beam whose left span runs out to a pinned end and whose right span continues into three more spans. The left member is a candidate for $3EI/L$; the right member is not, because its far end is an interior joint that will rotate by an amount nobody knows yet. So the joint has one modified member and one ordinary one, and the distribution factors are computed from $3EI/L_1$ and $4EI/L_2$ — a mixture, which is correct and which looks like an inconsistency on the sheet.

The carry-overs are mixed in the same way and for the same reason: nothing is carried to the released pin, and a half is carried along the continuing span. A table with a column of zeros in one place and halves in another is not evidence that somebody lost their place.

The rule that resolves it is a rule about knowledge rather than about members. A modification is available exactly when the far end’s rotation is determined by something outside the iteration — a pin that fixes the moment there at zero, or a symmetry that fixes the rotation as the negative of the near end’s. An interior joint’s rotation is determined by the iteration itself, which is the thing being computed, so there is nothing to substitute and $4EI/L$ stands.

That is also why the modifications cannot be chained. Releasing the end span does not then make the next span releasable; its far end is still an ordinary joint, and the fact that its neighbour got a modification tells it nothing.

Symmetry needs an axis, and the axis has to sit inside a span #

Symmetry needs an axis, and the axis has to be inside a span. Everything above cuts the beam at the middle of its central span, which requires an odd number of spans arranged symmetrically. A four-span beam is symmetric about a support, and the member crossing that axis is not one member but two — which needs the antisymmetric stiffness as well and is a different construction. The machinery here refuses that case rather than reporting it as a slow one.

A load has to be symmetric too. The 2EI/L stiffness says the two halves of the structure do mirror-image things, which is true only when the loading is a mirror image as well. The pattern loading a continuous beam is actually checked for is deliberately not symmetric, so the halving applies to one of the load cases and not to the governing one. What is done in practice is to split an unsymmetric load into a symmetric part and an antisymmetric part, solve each on half the structure with 2EI/L and 6EI/L, and add — which is superposition doing the work, and doubles the number of runs while quartering each.

The modifications assume nothing else is released. A member with a pin at both ends has no rotational stiffness at all, and a joint where every member is like that is a mechanism rather than a slow calculation. The distribution factor at such a joint is a division by zero and the method has nothing to say.

And none of it applies to the sway pass. A frame free to translate needs a second distribution whose unknown is a displacement, and the stiffnesses above are all derived for members whose ends do not move relative to one another. A modified stiffness saves work inside each pass and does nothing about the fact that there are two of them.

Nothing bends through a radian, and no figure holds a table #

The four shapes in the first figure are drawn at a unit rotation, which is an angle of one radian and is fifty-seven degrees. Nothing bends like that. The shapes are correct as shapes — the stiffness is a linear property, so the picture at one radian and the picture at a thousandth are the same picture at different scales — but a reader should not take the curvature seriously, and the far end of the antisymmetric case is genuinely rotating as much as the near end, which no beam in service does.

The other absence is the sheet itself. The count of numbers written is the whole finding of this essay and it is not visible in any figure, because a table is not a picture. Thirty numbers against two is an argument about a person with a pencil, and it stopped being an argument about anything else in about 1965.

The assumption the whole comparison rests on #

Every count above assumes the arithmetic is done by hand and that a number written is a unit of cost. On a computer none of this matters: a solver assembles the stiffness matrix with $4EI/L$ everywhere, applies the boundary conditions as constraints, and the pinned end costs a struck-out row rather than five cycles.

So the honest description of the modified stiffnesses is that they are a hand method’s response to a hand method’s problem, and they are obsolete in the same way the whole technique is.

What survives is the idea underneath them, and it is not obsolete at all. A modified stiffness is a boundary condition applied early. The far end’s behaviour was known, the method was going to discover it by iteration, and telling it instead removed a freedom. That is the same move a solver makes when it eliminates a constrained freedom before factorising rather than penalising it afterwards, and it is the same move an effective length makes about a column’s ends — where the four end conditions produce factors of 0.5, 0.7, 1.0 and 2.0 by exactly the analogous argument about what the far end is doing.

Two tables of four numbers, in two different fields, from one question asked of one member.

There is a historical detail that makes the point sharper than any of the arithmetic. The modified stiffnesses are not a later refinement bolted onto Cross’s method by people trying to speed it up; they are older than it, because they are slope-deflection results and slope-deflection predates moment distribution by about twenty years. The shortcut was available before the method it shortens existed. What Cross contributed was a way of solving the equations without writing them, and what the modifications do is put a few of the equations back — the ones whose answers were already known — so that the iteration has less to find.

That is a general shape and it is worth carrying out of this. An iterative method’s cost is set by how much it has to discover, and every boundary condition applied in advance is something it does not. The reason a hand computer cared and a solver does not is only that a solver’s discovery is cheap; the accounting is identical, and it is the same accounting that decides how many modes an eigenvalue routine has to sweep before it can stop.

What survives when nothing is done by hand #

Later rungs on this anchor: the antisymmetric half-structure in full, with the pattern-load decomposition that makes it useful. Haunched members, where the stiffness and the carry-over are both properties of a taper and neither is a round number. Members with a modified stiffness meeting at a joint whose other members do not have one, which is the ordinary case and where the distribution factors stop being obvious. And the same iteration written in rotations rather than moments, which shortens the table again and takes away the ability to read it halfway.