Three, and what three is a property of

Four essays have now used the column analogy, on four different structures: a straight member fixed at both ends, a portal frame, a fixed arch and a frame with two different column heights. Each time the construction survived a change that looked as though it should have broken it, and each time the reason given was local: an integral does not care whether its path is straight, a product of inertia is just a third second moment.

There is one reason underneath all of them and it is a count.

A closed ring cut once has three redundants. A plane section has three stress resultants. The analogy is that correspondence and nothing else, so it works exactly as far as the correspondence holds and not one structure further.

Every shape, and only one ring #

The same code solved all four and there is no branch in it that asks which. It takes the centreline as a polyline from one fixed foot to the other, a second moment for each segment and a load; it computes an area, a centroid and three second moments at a width of $1/EI$; and it solves

$$M_0 = -\frac{P}{A}, \qquad \begin{bmatrix} I_x & I_{xy} \\ I_{xy} & I_y \end{bmatrix}\begin{bmatrix} H \\ V \end{bmatrix} = -\begin{bmatrix} Q_x \\ Q_y \end{bmatrix} .$$

A frame fixed at both feet is a closed ring, and that is the fact the count rests on. The ring is not only the frame: it runs up one leg, along the top, down the other and back through the ground, which is rigid and closes the loop. That is why a frame on two fixed feet is three times redundant and a frame on two pins is one time redundant — pinning a foot puts a hinge in the ring.

The arch is the same ring with the corners taken out of it, and the polygonised arch in the gallery above and the analytic arch whose elastic centre hangs in the air are the same calculation twice.

Where the ring was cut, which turns out not to matter #

This is the property a hand method most needs and most rarely gets tested for. The release is free: cut the ring anywhere, compute a released diagram, integrate it over the section, and the three redundants that come out are different numbers each time. The moments they produce are identical.

That has to be true and it is worth saying why rather than only observing it. The released structure plus its three redundants is the real structure, whatever release was chosen; the redundants are the actions on the cut faces, and cutting somewhere else asks about the actions on different faces. The final internal forces belong to the structure and the redundants belong to the cut.

But one of the three is physical, and it is worth knowing which. The thrust $H$ comes out at 12.19 kN from all three cuts. It is the beam’s axial force, and a member has one axial force wherever it is cut, so $H$ is a reaction rather than a bookkeeping quantity. $M_0$ and $V$ are the moment and the shear at a particular section of the beam, and a beam’s moment and shear vary along it — so they change with the cut, exactly as they should.

An engineer who quotes “the thrust” from a column-analogy calculation is quoting something real. One who quotes “the redundant moment” is quoting a number that depends on a choice nobody recorded.

The frame the count runs out on #

The two-bay frame has three fixed feet and two beams, and if the analogy could be applied to it there would be a section somewhere with six stress resultants. There is not.

The count is arithmetic rather than opinion. A plane frame’s degree of static indeterminacy is $3c - r$, where $c$ is the number of independent closed cycles in the frame-plus-ground and $r$ the number of releases. A single-bay portal on fixed feet: one cycle, no releases, three. A two-bay portal: two cycles, no releases, six. A three-pinned arch: one cycle, three hinges, zero — determinate, which is why it needs no analogy at all and can be solved by drawing.

And the dashed curve is the part worth dwelling on, because it is the natural thing to try. Drop the middle column, apply the analogy to what is left, and treat the middle column as something to be added afterwards. The answer is −184.5 kN·m at the knees against the real −32.7.

That is not an approximation and it should not be read as one. It is the exact answer to a single-bay frame of 16 m span, which is a different structure carrying four times the free moment. The error is 190 per cent of one bay’s free moment, and it is large for a reason that has nothing to do with the analogy: a middle column halves the span, and halving a span quarters a moment.

The arch is the same ring #

An arch looks like a different problem from a portal and it is the same one. Two fixed springings and a curved member between them close one ring through the ground, exactly as two fixed feet and a beam do; the count is three either way; and the analogy is exact on both.

Polygonising it changes nothing, which is the test that makes the claim about topology rather than about curves. Cutting the arch into forty-eight straight members adds forty-seven nodes and forty-seven members together, and a cycle count that adds one of each is a cycle count that does not move. The stiffness solution in the figure is that polygon; the analogy is the continuum; the two agree to three hundredths of a per cent, and the residue is the lumping of the load at the polygon’s own nodes.

That equivalence is worth having explicitly, because the arch and the frame are treated as separate subjects everywhere else. What separates them is the shape of $M_s$ and the values of the integrals, not the method — and a stepped frame is further from a portal, in every way except the count, than an arch is.

Two bays, and the other way it can be got wrong #

The unequal-bay case says something the symmetric one hides. The real frame’s two outer knees carry −33.5 and −9.3 kN·m, a factor of three and a half apart, because the short bay attracts less moment and hands less to its column. The outer ring gives the same number at both, −118.1, because to the outer ring there is no middle column and nothing to be unequal about.

So the outer-ring answer is not a bound either. It is not conservative, it is not unconservative, it is not an envelope: it is the answer to a structure that was not asked about, and its errors have no sign. A calculation that is wrong by a factor is easier to catch than one that is wrong by a factor that varies, and this one varies with the bay ratio.

The honest alternatives for two bays are the ones that do not pretend: solve the six-by-six flexibility problem, or use moment distribution, or use a stiffness solver. Cross wrote moment distribution five years after the analogy, and the two methods’ domains are the complement of each other — the analogy is exact and closed-form on one ring, and the distribution iterates on any number of them.

What does extend, which is the idea rather than the method #

The correspondence stops at one ring. The idea that produced it does not, and it is worth separating them because the idea is the more useful of the two.

The idea is: apply the redundants at the point that makes the flexibility matrix diagonal. The force method writes $\mathbf f\,\mathbf X = -\boldsymbol\delta$, and $\mathbf f$ depends on where the redundants are applied as well as on what they are. For one ring, applying them at the elastic centre makes the three-by-three diagonal — and then the analogy is available, because a diagonal three-by-three is a section’s $P/A$ and two bending stresses.

For two rings the same move gives a block-diagonal six-by-six: each ring’s own elastic centre diagonalises its own three-by-three block, and the coupling between the two rings survives in the off-diagonal block. So the analogy does not extend and its habit does: a two-bay frame solved with each bay’s redundants at that bay’s elastic centre has a much better-conditioned matrix than one solved with the redundants at the feet, and that is worth something to anybody solving it by hand.

The same reasoning is why moment distribution converges. Passing a moment round a joint and back is an iteration on a matrix that is nearly diagonal already, and how nearly decides how fast. Cross published both methods within five years of each other and they are the same observation applied twice: pick the variables that make the coupling small, then either invert what is left or iterate on it.

The whole of it, once, as a count #

Take the frames in the gallery and count them.

A portal on fixed feet. Members: two columns and a beam. Nodes: two feet, two knees. With the ground as a member closing the loop, there is one independent cycle: foot–knee–knee–foot–ground–foot. Three redundants.

A pitched portal. One more member and one more node, which changes the cycle’s shape and not its count. Three.

A stepped frame. Five members, six nodes including the feet, and still one cycle. Three.

An arch. One member, curved, two feet. One cycle. Three. Polygonise it into forty-eight straight members and the count does not move, because adding a node in the middle of a member adds a member and a node together.

A two-bay portal. Three columns, two beams. Two independent cycles, because there are two ways round that are not the same way. Six.

A three-pinned arch. One cycle, three releases. Zero, and it is the shape the whole of graphic statics exists to solve.

The counting rule is topological — it depends on how the members are connected and not at all on where they are — which is why it is the right statement of the analogy’s domain. Every geometric change made to the frame so far was a change that could not affect a count.

The same count in the rest of the subject #

A count of independent cycles is not a fact about frames. It is the shape of every determinacy question here, and the four versions are worth putting side by side because they look unrelated and are not.

A pin-jointed truss is determinate when $m + r = 2j$, and indeterminate by $m + r - 2j$. That expression is a cycle count in disguise: adding a member to a triangulated truss closes a loop and adds one redundancy, exactly as adding a bay to a frame closes a loop and adds three. The difference is the three: a pin-jointed loop transmits one redundant force and a rigid-jointed loop transmits three, because a rigid joint carries a moment as well as two force components. Counting the unknowns is that arithmetic.

A grillage is a plane frame’s three-dimensional cousin with the loads out of plane, and a closed cell in one carries three redundants too — a moment about each of two axes and a shear — so the same 3c applies with a different three.

And the count can lie, which is the standing warning on all of it. A count that does not see it is the case where the arithmetic says determinate and the structure is a mechanism, because the members happen to be arranged so that the equations are not independent. The cycle count here has the same exposure: it says how many redundants there are, and it says nothing about whether the structure stands up.

What the count does do, reliably, is mark a method’s domain. The analogy needs three and gets three from one ring, and every change made to the frame so far — curving the member, tilting the beam, stepping the frame, splaying the legs, moving a foot — was a change that could not touch a count.

Which free body produced the number #

Every frame in the gallery is cut once, at the mid-point of its own path by arc length, leaving two bent cantilevers each fixed at its own foot. The released moment at a station is the moment of the loads lying between that station and the cut, with a vertical load’s lever arm horizontal whatever the member’s shape. The three redundants are a couple, a horizontal force and a vertical force applied on a rigid arm at the elastic centre, and the three conditions are that the cut does not open.

Every one of them is checked against a stiffness solution of the same frame, polygonised into short members with the load lumped at the nodes, which shares no step with the analogy. The worst disagreement across the four shapes is 0.24 per cent of the free moment, and it is the lumping.

The two-bay frame is solved only by stiffness, because the analogy has nothing to say about it, and the dashed curve beside it is the analogy’s answer to a different question honestly labelled.

What the picture cannot show #

Anything about axial or shear flexibility. Every integral is a bending flexibility. A frame whose columns are short and stout relative to its beam carries a larger share of its deformation in shear and in axial shortening, and the analogy will not see it — which is the same limitation the conjugate beam has for the same reason.

A ring that is not plane. A frame that leaves the plane has six redundants at one cut rather than three, and the section that would be its drawing would need six stress resultants. There is a three-dimensional version of the analogy with a six-by-six in it, and it is not a section any more.

Support movements between the two feet. The settlement case of the frame bent by a foot that moved is inside the domain, because an imposed displacement changes the right-hand side and not the count. Two supports that both move and a third that does not is a two-ring problem again.

Whether any of it is worth doing. A stiffness solver answers all of these frames in milliseconds and does not care about cycles. The analogy’s remaining value is that it gives the thrust as a formula in the frame’s own geometry, so what stiffening the beam does can be read rather than re-run.

The assumption every one of these constructions rests on #

That the structure is linear and elastic, so that a released structure plus its redundants is the real structure. Every step of every one of these constructions has used superposition — the released diagram plus the redundants’ diagrams equals the answer — and superposition is exactly what a frame that has yielded anywhere does not have.

That assumption is invisible in the arithmetic because it is in the setup: there is no line of the calculation at which it is invoked, and no term that goes to zero when it fails. The place it shows is in the settlement case, where a frame that cracks sheds the moments the settlement made and keeps the ones the load made — so the two cases that were added cannot be added any more, and neither of them is wrong on its own.

Still open: the analogy as a way of choosing variables #

The four essays above have treated the column analogy as a method: a construction that takes a frame and returns its redundants. The last section of this one suggests it is better read as a choice of variables that happens to have a picture attached, and that reading has a consequence none of these essays has followed up.

If the elastic centre is the point that diagonalises a flexibility matrix, then every redundant structure has one — not only the ones with three redundants, and not only the ones in a plane. A continuous beam has an elastic centre for each span, a grillage has one for each cell, a three-dimensional frame has a point and an orientation. What those points are, whether they can be found without building the matrix they diagonalise, and whether any of them has a picture as good as a section drawn at a width of $1/EI$, is the question left open.