Structural Engineering & Statics
where the load goes
Deflection

One support too many, and what it costs to know

Add a redundant restraint and the load has two routes to the ground. Equilibrium cannot say how it splits, and the answer turns out to depend on stiffness — which is a different kind of question.
9 min read 6 figures One support too manyThe load must go somewhere

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.

One support too manyThe same uniformly loaded beam with three sets of restraints, and the bending moment in each. Adding restraint moves moment from mid-span to the supports and lowers the peak — but only the first case can be solved by statics.simply supportedstatics alonesag 32.0propped at one endneeds stiffnesssag 18.0hog 32.0built in at both endsneeds stiffnesssag 10.7hog 21.3the load never changes; only what is holding the endsthe built-in case peaks at two-thirds of the simple span's moment
Fig. 1 The same uniformly loaded beam with three sets of restraints, and the bending moment in each. Only the first can be solved by statics; the other two need to know the stiffness, and both have lower peaks as a result.

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 EIEI, which never appeared in statics 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.

Which limit arrives firstUtilisation of the strength limit and of the deflection limit, against span. Strength grows as the square of the span and deflection as the fourth power, so the two cross — and past the crossing a beam is sized by how far it moves rather than by what it can carry.0.60.811.21.41.61.8200.511.5span, relative to the firstthey cross heredeflection runs out at 1.40strength runs out at 1.54the limitstrengthdeflection
Fig. 2 Two limits growing at different rates. Stiffness decides which route load takes in a redundant structure, so anything that changes relative stiffness — cracking, creep, a repair — redistributes the forces.

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.

Counting unknowns against equationsThree frames differing by one member. Two equilibrium equations per joint, one unknown per member and one per restraint: fewer unknowns than equations is a mechanism, equal is solvable by statics, more needs stiffness.m 4 + r 3 − 2j 8 = -1a mechanismm 5 + r 3 − 2j 8 = 0statically determinatem 6 + r 3 − 2j 8 = +1one member too manystatics can answer only the middle case
Fig. 3 Three frames differing by one member. The third has a route to spare, which is what robustness means and what statics alone cannot resolve.

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 deflected shape is the moment, integrated twiceA loaded beam with its deflected shape above and its bending-moment diagram below. The shape was obtained by integrating the moment twice and fitting the constants to the supports; the vertical scale is exaggerated enormously.24the largest movement, at x = 8.00momentdrawn at roughly three hundred times the real deflection —a beam at its serviceability limit moves about a three-hundredth of its span
Fig. 4 A cantilever’s deflected shape. Propping the tip would remove most of this movement, and the prop’s share of the load would depend on how stiff the prop is relative to the beam.

The counting rule, revisited

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

A Pratt truss of 6 panelsA Pratt truss under equal panel-point loads. The joint equilibrium equations were assembled and solved; 10 members came out in tension, 9 in compression and 2 carrying nothing.tensioncompression2 carrying nothing
Fig. 5 A determinate truss. Its member forces came out of equilibrium alone, because the count of unknowns against equations happened to balance — and a single extra diagonal would have 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 EIEI, 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.

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.

Load, shear and moment — a simple spanThe applied load, the shear force it produces and the bending moment that follows, drawn one above another to the same horizontal scale. Shear is the integral of the load and moment is the integral of shear.4 per unit lengthshear16.0moment32.0 at x = 4.00the moment peaks exactly where the shear passes through zero
Fig. 6 The moment diagram for a determinate beam. Everything here came from statics; the equivalent diagram for a continuous beam depends on the stiffness of every span, and no amount of summing forces will produce it.

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 a compatibility calculation that no figure on this page shows. The picture presents three answers of which only one is derivable from anything else here.

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.

Where this goes next

m 4 + r 3 − 2j 8 = -1a mechanismm 5 + r 3 − 2j 8 = 0statically determinatem 6 + r 3 − 2j 8 = +1one member too manystatics can answer only the middle case Counting the unknowns, and finding out whether statics can answer Two equations per joint, one unknown per member, one per restraint. Subtract, and the sign of the answer says whether the structure is a mechanism, solvable, or beyond what equilibrium alone can settle. 00.050.10.150.20.250.30.3501020304050overhang, as a fraction of the spanbest at 21% — peak 8.8sagging, mid-spanhogging, over the supportmoving the supports in by a third of the way halves the worst moment Where to put the supports, which is not at the ends Moving the supports of a uniformly loaded beam inward by about a fifth of its length halves the worst bending moment. The load has not changed and nor has the beam. 0.60.811.21.41.61.8200.511.5span, relative to the firstthey cross heredeflection runs out at 1.40strength runs out at 1.54the limitstrengthdeflection Stiffness is not strength, and usually it is the one that governs A beam can be nowhere near failure and still be unusable, because it has moved too far. For most long-span members that limit arrives first, and stronger steel does not help at all.