Choose what to take away

A redundant structure has more restraints than equilibrium can settle, so its internal forces depend on stiffness as well as on statics. There are two ways to get at them, and they are duals.

The stiffness method writes one equation per freedom and solves for displacements. The force method — older by a century, and the way every hand calculation before about 1960 was done — writes one equation per redundant and solves for forces.

Its procedure is worth stating in full because the whole essay is about one step of it:

1. Remove enough restraints to leave something statically determinate.
2. Compute the gap each release opens under the applied load.
3. Compute the gap a unit value of each redundant closes.
4. Require the gaps to vanish, and solve for the redundants.

$$\mathbf{f}\,\mathbf{X} = -\boldsymbol{\delta}_0$$

Step 1 is the one that has no rule attached to it. It is a choice, made by a person, and it is the reason this method is history rather than practice.

Which free body produced the number #

Two of them, and that is the point.

Release A: hinges over the interior supports. The released structure is a row of simple spans. Each hinge opens by the sum of the end rotations of the two spans meeting there, which for a uniform load is $wL_1^3/24EI + wL_2^3/24EI$. A unit pair of moments at that hinge closes it by $(L_1 + L_2)/3EI$, and affects its neighbour by $L/6EI$.

Release B: remove the interior supports. The released structure is one simple span of the whole 26 m. Each removed support deflects by the ordinary simple-beam deflection at that point, and a unit force at any of them lifts every other one — Maxwell’s influence coefficients, which are famously symmetric and famously not zero.

For the 8, 10, 8 metre beam under 12 kN/m these give:

Different unknowns, different units, different numbers throughout. And the bending moment diagram each produces agrees with the other to nine parts in $10^{15}$ — machine precision, everywhere along the beam.

Why the answer cannot depend on the choice #

That agreement is not luck and it is worth being able to see why, because it is what makes the method trustworthy.

A released structure plus its redundants, with the redundants at their correct values, is the original structure. It satisfies equilibrium — the released structure was in equilibrium and the redundants are self-equilibrating pairs. It satisfies compatibility — that is what the equations enforced. And for a linear elastic structure the state satisfying both is unique.

So any release that produces a determinate structure gets there, because they are all describing the same final state by a different route. The released structure is scaffolding, and it is gone by the end.

This is superposition doing something more than adding load cases. The load case and the redundant case are being superposed even though the redundant case has no external load in it at all: it is a self-stressed state, in equilibrium with nothing, and adding it changes the deflections without disturbing the equilibrium. Those states are exactly the self-stress states a redundant structure has, and there is one per redundancy.

Where the two releases differ, which is the whole practical content #

Look at the shape of the two matrices rather than their entries.

Release A is banded. A moment released at one support is felt only in the two spans either side of it, so $f_{ij} = 0$ whenever the supports are more than one apart. On six equal spans the matrix is tridiagonal: 52% of its entries are exactly zero.

Release B is full. A unit reaction anywhere on a single 48 m simply-supported span deflects every other point on it. Nothing is zero, and on six spans the matrix is 100% full.

That difference has a name. The tridiagonal equations from release A, written out, are

$$M_{i-1}L_i + 2M_i(L_i + L_{i+1}) + M_{i+1}L_{i+1} = -\frac{w L_i^3}{4} - \frac{w L_{i+1}^3}{4}$$

which is Clapeyron’s three-moment equation of 1857. Each equation involves three unknowns; the system can be solved by forward elimination in a single pass down the beam; and the arithmetic fits in a column of figures. Continuous beams of any number of spans were routine hand calculations for a century because of this, and they were routine for exactly this reason.

Release B on the same beam is a full system of simultaneous equations with no structure to exploit, and it would have been solved by nobody.

What “a good release” means, generally #

The rule that produced the three-moment equation generalises, and it is worth stating in a form that applies past continuous beams.

Choose releases whose effects are local. A redundant whose unit value disturbs only its own neighbourhood produces zeros in the matrix; a redundant whose unit value disturbs the whole structure produces none.

For a continuous beam that means moments rather than reactions, because a moment release is a hinge and a hinge does not transmit anything past the adjacent spans. For a rigid frame it means cutting members rather than removing supports. For a closed ring or a box it means cutting the section, which introduces three redundants at one place rather than three redundants spread over the structure.

The same instinct appears again in the stiffness method, where it becomes the node numbering problem — number the nodes so that connected ones are close together and the assembled matrix is banded. The physics is the same in both: a matrix’s sparsity is a picture of what is connected to what.

What the two matrices are actually made of #

It is worth looking at the entries rather than only at the pattern, because they say something about what a flexibility coefficient is.

$f_{ij}$ is the displacement at release $i$ caused by a unit redundant at release $j$. Maxwell’s reciprocal theorem — push here and measure there — says $f_{ij} = f_{ji}$ for every elastic structure, so the matrix is symmetric before anything has been computed. That is a free check on every coefficient, and on a hand calculation it was the check.

The diagonal entries are always positive: a unit force at a release moves that release in its own direction. The off-diagonals can be either sign, and their sign is the interesting part. On release A the off-diagonals are positive — hogging one support hogs its neighbour — while on a frame the same coefficients can easily come out negative, and the sign is a statement about how the structure passes a disturbance along.

The ratio of off-diagonal to diagonal is what decides how coupled the redundants are. On release A it is $L/6$ over $(L_1+L_2)/3$, which is about 0.28 for equal spans — weakly coupled, so a change at one support barely reaches the next. On release B it is 224.8 over 265.9, which is 0.85 — almost completely coupled, so nothing can be solved without everything.

That single ratio is the whole difference between a calculation a person can do and one they cannot, and it was fixed the moment the release was chosen.

The one thing the force method is still better at #

The stiffness method won, and it won completely: every analysis program in existence is a stiffness program. But there is a class of problem where the force method’s unknowns are the natural ones, and it is worth knowing because those problems recur.

When there are very few redundancies in a large structure, the force method has a very small system to solve while the stiffness method has a very large one. A continuous truss over five supports has hundreds of members and three redundants. A tied arch has one. A propped cantilever has one.

More importantly, when the question is “what does this one restraint carry?”, the force method answers it directly and the stiffness method has to compute a whole displacement field and then differentiate back out of it. The tie force in a tied arch, the prop reaction in a propped beam, the thrust in a two-hinged arch — these are the redundants, and they come out of the force method as the answer rather than as a by-product.

The same is true of temperature, settlement and shrinkage. A support that moves is not a load in the stiffness sense at all; it is a prescribed displacement, and it enters the force method as a non-zero right-hand side in exactly the place the load’s gaps went — which makes the whole class of imposed-deformation problems a one-line modification rather than a special case. A thermal restraint is the same shape of problem: a movement the structure wants, a restraint that refuses it, and a force that is whatever closing the gap requires.

The four-span case, where the arithmetic stops being symmetric #

Three equal-ish spans is a case with an answer that reads as obvious once seen: both interior supports carry $-98.61$ kNm, because the beam is symmetric.

Take four spans — 8, 10, 10, 8 — and the symmetry survives while the numbers stop being equal. The two outer interior supports carry $-97.94$ kNm and the middle one $-101.03$: a spread of 3%, in a beam whose two long spans are identical and adjacent.

The reason is that the middle support has two ten-metre spans meeting at it while the outer ones have a ten and an eight. A hinge at the middle opens more under the load, and closes less per unit moment, so it attracts more moment. That is the stiffest-path argument arriving in a place where nothing was made stiffer — the difference is in the lengths, and a length is what a flexibility is made of.

The practical content is small and worth having: a continuous beam’s worst support moment is not at its longest span but where its longest spans meet, and on an irregular beam the two are usually different supports. Neither is visible on a span-by-span check, and the fill of the flexibility matrix is what carries the information between them.

Where it stops #

It needs a released structure that is stable. Not every set of releases leaves something determinate — release the wrong ones and what is left is a mechanism, the flexibility matrix is singular, and the method fails with no useful diagnosis. On a beam this is easy to avoid. On a frame it is not, and it is another judgement the method needs and the stiffness method does not.

It is linear. Superposing a load case and a redundant case assumes the two do not interact, which fails as soon as second-order effects matter or a section yields. The stiffness method is no better in principle but is far easier to iterate.

The flexibilities have to be computed, and computing a deflection in a released structure is a virtual-work integral per coefficient. On $n$ redundants that is $n(n+1)/2$ integrals, all of them by hand. The three-moment equation is precious partly because it has done those integrals once, in general, for a case that recurs.

What the picture cannot show #

The two diagrams in the first figure lie on top of each other, which makes the result look inevitable. It is worth remembering how different the intermediate states are: release A’s beam sags in three separate spans with kinks over the supports, and release B’s beam is one long sagging curve. Neither of those shapes exists anywhere in the real structure, and neither is a stage the beam passes through.

That is the thing about the method most easily misread. The released structure is not a simplification of the beam and it is not the beam before the supports were built. It is a fiction chosen for arithmetic convenience, and its only property that matters is that it is determinate. A student who tries to attach physical meaning to it is looking for something that is not there.

Why it lost, which was not about mathematics #

The force method was the profession’s main tool for a hundred years and it is now taught for a fortnight and never used. Three things ended it, and only one of them is technical.

The unknowns cannot be chosen by a machine. A stiffness program is handed a geometry and derives its unknowns from it — three per node in a plane, always, with no decisions. A force-method program has to be handed a set of releases, and there is no general algorithm for choosing good ones. Automatic release selection was attempted and it does not work reliably: the choice that keeps the matrix sparse depends on the topology in a way that resists being written down.

Boundary conditions are awkward. A fixed support in the stiffness method is a row and column struck out. In the force method it is one more redundancy, one more release to choose, and one more column of flexibility coefficients to integrate.

And the sparsity advantage reversed. The force method’s whole appeal was a smaller system. That mattered enormously when the system was solved by hand and stopped mattering when it was not — and by then the stiffness method’s matrix, though larger, was structured: banded, symmetric, positive definite, and assembled by a scatter operation that needs no judgement whatever.

So the method that survived is the one whose arithmetic is worse and whose bookkeeping is mechanical, which is the ordinary outcome once the arithmetic is not being done by a person. It is worth noticing that the same trade decided moment distribution’s fate for the same reason and at the same time.

The generalisation #

The habit worth carrying is the separation between what a method computes and what it costs to compute it.

Every route through this problem gives the same numbers. The force method with hinges, the force method with reactions, moment distribution, the stiffness method, and a fine-element model all produce $-98.61$ kNm over the supports of that beam, because the structure has one answer and they are all looking for it.

What separates them is entirely the arithmetic: how many unknowns, how sparse, how automatable, how much judgement. Those are engineering questions about the calculation and not about the structure, and confusing the two produces both of the errors this essay is about — believing that a different release gives a different answer, and believing that two methods giving the same answer are therefore equally useful.

Which is also why the three-moment equation is worth knowing in an age that will never use it. It is not a way of getting an answer; it is a demonstration that the choice of unknowns is a design decision, made by a person, with an order-of-magnitude consequence — and that decision has not gone away, it has only moved into the node numbering.