One support too many, and what it costs to know

Put a beam on two supports and the reactions follow from two equations. Put it on three and they do not — any set of three reactions that adds up correctly and balances the moments is in equilibrium, and there are infinitely many.

The structure nevertheless picks one. Finding out which requires information equilibrium does not contain, and the information is how stiff everything is.

The second principle #

Equilibrium says the forces balance. It does not say the structure stays in one piece, and that is the missing condition.

Compatibility is the requirement that the deformed structure fits together — that a beam continuous over a support has the same slope on both sides of it, that a member framing into a joint rotates with the joint, that a support which does not settle has zero deflection above it.

For a determinate structure compatibility is automatic: there is exactly one force system, and whatever deflection it produces is the answer. For a redundant one it is the extra equation, and there is one of them per redundancy.

The classical method makes this explicit. Remove the redundant support to leave a determinate structure, compute how much the beam would deflect at that point, then work out what force applied there would push it back to zero. That force is the redundant reaction, and the calculation needed a deflection — which needed $EI$, which never appeared in statics at all.

That is the whole of the extra machinery. Equilibrium is an equation between forces and compatibility is an equation between movements, and the second one drags a stiffness into a problem that had contained no material property at all.

Load follows stiffness #

The general principle that comes out is short and has enormous reach: load goes where the stiffness is.

Two members sharing a load in parallel take shares in proportion to their stiffnesses. A beam continuous over several supports puts more moment where it is stiffer. A frame with one very stiff bay and several flexible ones sends nearly all the lateral load to the stiff one.

That is why a stiff core in a steel-framed building attracts almost all the wind load even though the frame is nominally capable of resisting it. It is why a stiff cladding panel can become structural without anybody intending it. And it is why an attempt to strengthen part of a structure can make things worse: a stiffened region attracts more load, and it may attract more than the stiffening added.

What redundancy buys #

Three things, and they are the reason nearly every real structure has it.

Lower peak forces. Built-in ends drop the mid-span moment to a third of the simply supported value and put two-thirds of it over the supports, where the beam can be haunched. Moving supports inboard achieves the same effect determinately, and continuity achieves it without an overhang.

Less deflection. A fixed-ended beam deflects a fifth as far as a simply supported one under the same load, because the end restraint curves the ends back. For a deflection-governed member, that is the difference between two sections.

Survival. A determinate structure has exactly enough members; lose one and it becomes a mechanism. A redundant structure loses a member and redistributes, which is robustness — and since the Ronan Point collapse in 1968 it has been a design requirement rather than a bonus.

What it costs #

The costs are less obvious than the benefits and are the reason determinate structures are still built deliberately.

Sensitivity to settlement. If one support of a three-support beam settles by a few millimetres, the reactions redistribute substantially. The same settlement under a two-support beam changes nothing at all. Bridges on soft ground are often made determinate for exactly this reason.

Sensitivity to temperature. A redundant structure restrained against expansion develops forces when it warms. A determinate one moves and develops none. That is why long bridges have expansion joints and why a continuous structure without them cracks.

Sensitivity to fabrication. A member made slightly too short in a redundant frame is stressed before any load arrives, and the whole frame is stressed with it. In a determinate frame it just fits differently.

Locked-in forces. All three of the above are the same phenomenon as far as the free body is concerned: a redundant structure can have internal forces with no external load at all. Prestressing exploits it deliberately; everything else suffers it accidentally.

Analysis. Historically the largest cost and now the smallest. Moment distribution, published by Hardy Cross in 1930, made hand analysis of continuous frames possible for the first time and was the standard method for thirty years. A computer does it now without comment.

What ductility does to the argument #

There is an escape from some of the sensitivity, and it depends on the material behaving well past its elastic limit.

If a redundant steel structure is overloaded, the most highly stressed section yields and forms a plastic hinge. It does not break — it rotates while continuing to carry its moment, and further load is redistributed to sections that have capacity left. Collapse requires enough hinges to form a mechanism, which for a redundant structure means several.

That means the collapse load of a ductile redundant structure does not depend on the elastic distribution at all, and therefore does not depend on the settlement, temperature or fabrication effects that troubled the elastic analysis. Locked-in forces are relieved by a small amount of yielding and stop mattering.

This is the deep justification for plastic design, and it is why ductility is treated as a material requirement rather than a bonus. A brittle redundant structure has all of the sensitivity and none of the escape, which is why unreinforced concrete and masonry are analysed so much more carefully than steel.

The compatibility sum, done on named bodies #

The claim that a compatibility condition supplies the missing equation is easy to assert and worth doing once, because the calculation uses two free bodies that are both the same structure with a support removed, which is a strange enough manoeuvre to deserve seeing.

Take a beam built into a wall at one end, propped on a roller at the other, carrying a uniform load $w$ over a span $L$. Three reactions, two equations: one redundant.

Body one — the released structure under the load. Delete the prop. What remains is a cantilever, which is determinate, and its tip deflects

$$\delta_1 = \frac{wL^4}{8EI}$$

downward. Nothing here needed the prop to exist; the beam has simply been asked what it would do without one.

Body two — the released structure under the redundant force. Delete the load instead, and apply an unknown upward force $R$ at the tip of the same cantilever. It lifts by

$$\delta_2 = \frac{RL^3}{3EI}.$$

The condition. The prop does not move, so the two must cancel:

$$\frac{RL^3}{3EI} = \frac{wL^4}{8EI} \quad\Longrightarrow\quad R = \frac{3wL}{8}.$$

The prop takes three-eighths of the load and the wall five-eighths, which is not the half-and-half that symmetry of appearance suggests, and the asymmetry came from stiffness rather than from equilibrium.

Everything else follows from statics once $R$ is known. The moment at the wall is $wL^2/2 - (3wL/8)L = wL^2/8$, hogging. The shear vanishes three-eighths of the span from the prop, so the largest sagging moment is $9wL^2/128$ there — about 56 per cent of the $wL^2/8$ a simply supported beam of the same span would have suffered. One prop, correctly analysed, has taken nearly half the peak moment out of the beam.

Two features of that calculation are worth carrying away. First, $EI$ appears in both deflections and then cancels: for a uniform beam the answer is independent of stiffness, and it is only when the parts have different stiffnesses that the material properties survive into the result. Second, the whole method rests on superposition — two separately computed responses added — which is legitimate only while everything is linear, and which is exactly the assumption second-order behaviour destroys.

Redundancy is a continuum, not a count #

The counting rule returns an integer, and the propped cantilever above was solved as though the prop were infinitely stiff. Real props are not, and running the same compatibility sum with a spring instead of a rigid support shows that redundancy is the limit of a continuum rather than a property a structure either has or does not.

Give the prop a stiffness $k$. It now settles by $R/k$ under its own reaction, so the compatibility condition is not “the tip does not move” but “the tip moves as far as the prop lets it”:

$$\frac{wL^4}{8EI} - \frac{RL^3}{3EI} = \frac{R}{k}$$

which rearranges to

$$R = \frac{3wL}{8}\cdot\frac{\alpha}{1+\alpha}, \qquad \alpha = \frac{kL^3}{3EI}.$$

The rigid answer is recovered as $\alpha \to \infty$ and the determinate one as $\alpha \to 0$, and $\alpha$ has a clean reading: it is the prop’s stiffness divided by the released structure’s own stiffness at that point, since $3EI/L^3$ is exactly a cantilever’s tip stiffness.

The numbers are unforgiving in a way the integer count conceals. At $\alpha = 1$ — a prop exactly as stiff as the beam it is propping — the prop takes half of what a rigid one would, and the beam is halfway between determinate and continuous. Reaching 90 per cent of the benefit needs $\alpha = 9$: a prop nine times stiffer than the member it is helping.

That is the arithmetic behind a great many things this collection has treated separately. It is why a nominally pinned base delivers partial fixity rather than none, why a semi-rigid connection delivers a fraction of the fixed-end moment set by $S_j L/EI$, and why a slender column used as a prop is worth much less than the drawing suggests.

And it sharpens what the counting rule is for. The count says how many compatibility equations have to be written. It says nothing about how much continuity is actually delivered, and the answer to that is a ratio of stiffnesses that runs continuously from nothing to everything — with the halfway point at a restraint as stiff as the thing it restrains, which is stiffer than most restraints are.

The redundancy that is not there #

The count says how many extra restraints exist. It does not say whether there is an alternative path, and the difference between those two has killed people.

The Silver Bridge at Point Pleasant, on the Ohio River, carried its deck from two chains of steel eyebars — flat bars pinned end to end, two bars per link. On 15 December 1967 a single eyebar failed from a corrosion crack about three millimetres deep at one pinhole. Its partner in the same link, now carrying everything alone, failed immediately after. The link separated, the chain unloaded, and the entire bridge fell into the river in under a minute, killing forty-six people.

By any counting rule the bridge was heavily redundant: hundreds of members, many more restraints than equations. By load path it had none at all, because the eyebar chain was the only route from the deck to the towers, and each link was two members whose failure modes were not independent. Redundancy that shares a cause is not redundancy.

The classification the collapse produced — fracture-critical, meaning a member whose failure would collapse the structure — is now applied to bridge elements everywhere, and it is a statement about paths rather than about counts. Modern eyebar and hanger arrangements use three or more bars per link precisely so that one failure leaves a working structure and, just as important, leaves visible evidence.

The general lesson reaches well beyond bridges. A transfer beam carrying six columns is a single point of failure inside a frame that counts as many times redundant. So is a single tie holding a cantilevered floor, and so is a connection detail repeated identically at every node, since a systematic error in it is present everywhere at once. The counting rule and the load path are answering different questions, and only one of them is about survival.

What the answer depends on that nobody measured #

Every redundant analysis has to be told the relative stiffnesses before it can start, and this is the cost that survives after the computer has removed all the others.

For a steel frame the stiffnesses are known well: the sections are manufactured to tolerance, the modulus is a constant to within a per cent or two, and the answer is as good as the model. For reinforced concrete it is not. A cracked section may be a third as stiff as an uncracked one, cracking spreads during loading rather than arriving all at once, and creep goes on changing the ratio for years. Choosing an uncracked stiffness for a beam and a cracked one for a column, or the reverse, moves the computed moments substantially — and neither choice is wrong, because both states occur.

The honest response is to stop treating the analysis as a measurement. The forces in a redundant structure are not a fact about it in the way that the total reaction is a fact about it; they are a fact about a stiffness distribution that was assumed. The usual practical answer is to bound it: analyse with a plausible range of stiffness ratios, check that no member is unsafe anywhere in the range, and rely on ductility to sort out the difference. Which is, once more, the argument of the previous section — the collapse load is robust to exactly the assumptions the elastic answer is fragile to.

The counting rule, revisited #

Everything in this essay was predicted by an arithmetic done before any of it was calculated.

A determinate truss is the cleanest case of it: its member forces come out of equilibrium alone, because the count of unknowns against equations happens to balance — and a single extra diagonal would put it beyond reach.

The count says whether a compatibility calculation is needed and how many extra conditions it requires. It does not say what the answer will be, and it does not say whether the redundancy is worth having — but it tells a designer, in ten seconds, which body of theory the structure belongs to.

Where the model stops #

Linear elastic behaviour. The stiffness-proportional distribution assumes everything stays elastic. Concrete cracks, which changes its stiffness during loading and redistributes as it goes, and the analysis has to guess a stiffness before it knows the answer.

Known stiffnesses. The distribution depends on relative $EI$, and for a composite or cracked member that is uncertain by a factor. A redundant analysis is only as good as its stiffness assumptions.

No support movement. Every figure here assumes rigid supports at fixed positions. A real foundation settles, and for a redundant structure that is a load case.

Small deflections. As everywhere else on this site — and for a sway-sensitive frame, the amplification interacts with the redistribution in a way neither analysis alone captures.

Superposition. The force method above added two separately computed responses, which is only legitimate while the structure is linear in both material and geometry. Once either fails — a section cracks, an axial force starts amplifying a sway — the two cases can no longer be computed apart and recombined, and the elegant release-and-restore argument has to be replaced by an analysis that follows the loading history in order.

Ductility available. The plastic argument requires it, and codes place limits on how much redistribution may be assumed for exactly that reason — a limit on how far the moment diagram may be reshaped.

The figures have a limitation worth naming: the three cases in the first figure are drawn with their end moments taken from standard results, which is honest for the values and hides where they came from. The propped and built-in cases were not solved by the equilibrium the rest of this site uses — they required the compatibility calculation the middle of this page draws in three parts, and the opening picture presents its answers without any of that working. Only the first of its three cases is derivable from statics alone.

The ladder from here #

Later rungs: the force method and the compatibility equations. The displacement method, which is what every structural program uses. Moment distribution, and why it made continuous frames tractable by hand. Fixed-end moments. Support settlement as a load case. Temperature effects in restrained structures. Prestress as a deliberate locked-in force. Plastic hinges and collapse mechanisms. Moment redistribution and the limits codes place on it. And robustness, tying and progressive collapse, which is the whole argument for redundancy expressed as a requirement.

Hardy Cross published moment distribution in a ten-page paper in 1930. It is one of the 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.