Internal forces

The moment over the support, and what it buys

Run a beam over its supports instead of stopping at each one, and the mid-span moment falls by a third while a new moment appears where there was none. Nothing was added but continuity.

Assumes The diagram is an integral, and that is why it can be drawn by eye and One support too many, and what it costs to know.

Three beams of seven metres, each sitting on its own pair of supports, have a mid-span moment of wL2/8wL^2/8 apiece. Replace them with one beam of twenty-one metres running over four supports, and the largest sagging moment drops by more than a third.

Nothing was added. The same steel spans the same distances under the same load. What changed is that the beam is now continuous over its supports, so it can hog there — and hogging over a support relieves sagging between them.

3 continuous spans against 3 simple ones. The bending moment in a continuous beam, solved by the stiffness method, drawn over the moment in the same spans made simply supported. The peak sagging moment falls from 30.6 to 19.6, and a hogging moment of 24.5 appears over the supports where there was none.
Fig. 1 Three continuous spans against the same three spans made simple. The peak sagging moment falls substantially, and a hogging moment appears over the interior supports where there was none before.

The mechanism is a rotation that does not happen

A simply supported beam’s ends rotate freely under load. Two adjacent simple spans rotate toward each other over the support between them, each independently.

Make them one member and that is no longer allowed. The beam has one slope at that station, not two, so the two spans have to agree — and the moment that forces them to agree is the hogging moment over the support. It exists because a rotation was prevented.

That is a compatibility condition, not an equilibrium one, which is why continuity puts the problem beyond statics. Equilibrium is satisfied by any set of reactions summing correctly; picking the real one requires knowing that the beam is continuous and how stiff each span is. The counting rule says how many extra conditions are needed: one per redundant support, which for four supports on a single beam is two.

Once the hogging moment exists, its effect on the span is arithmetic. The sagging moment at mid-span becomes the simply supported value minus the average of the end moments:

Mmid=wL28Mleft+Mright2.M_{\text{mid}} = \frac{wL^2}{8} - \frac{M_{\text{left}} + M_{\text{right}}}{2}.

Every bit of hogging bought at the supports is subtracted from the sagging in the span. The moment has not been reduced; it has been moved to somewhere the beam can be thicker.

The numbers, and the support that carries too much

For equal spans under uniform load the results are worth memorising, because they recur everywhere.

Two equal spans: hogging over the middle support is wL2/8wL^2/8, sagging in each span is 0.070wL20.070wL^2 — against 0.125wL20.125wL^2 simply supported. The peak has fallen by nearly half.

2 continuous spans against 2 simple ones. The bending moment in a continuous beam, solved by the stiffness method, drawn over the moment in the same spans made simply supported. The peak sagging moment falls from 30.6 to 17.2, and a hogging moment of 30.6 appears over the supports where there was none.
Fig. 2 The two-span case at 7 m each under 5 kN/m, solved by the stiffness method and drawn over the same two spans made simple. The peak sagging moment falls from 30.6 kNm to 17.2, and a hogging moment of 30.6 appears over the middle support where there was none. The hogging is exactly the simple span’s own peak, which is the coincidence the coefficients above record: for two equal spans the support moment is wL2/8wL^2/8 and so is the free moment.

That coincidence is worth not reading as a rule. It holds for two equal spans and for nothing else, and the three-span case in the opening figure already breaks it — the hogging there is 0.100wL20.100wL^2 rather than 0.1250.125, because the third span is pulling the other way.

Three equal spans: hogging over the interior supports is 0.100wL20.100wL^2, sagging in the outer spans is 0.080wL20.080wL^2 and in the middle span 0.025wL20.025wL^2.

The reactions are the part that surprises people. For two equal spans the middle support carries

Rmid=108wL=1.25wL,R_{\text{mid}} = \frac{10}{8}wL = 1.25wL,

against the wLwL it would carry if the two spans were simple — twenty-five per cent more — while each end support carries only 0.375wL0.375wL instead of 0.5wL0.5wL.

A column under the middle of a continuous beam therefore takes considerably more load than a tributary-area calculation suggests, and the end columns take considerably less. That is not an academic point: apportioning a spread load by tributary area is how loads are estimated at the start of every project, and it is wrong in a direction that under-designs the interior column and over-designs the outer one.

3 continuous spans against 3 simple ones. The bending moment in a continuous beam, solved by the stiffness method, drawn over the moment in the same spans made simply supported. The peak sagging moment falls from 50.6 to 20.3, and a hogging moment of 30.3 appears over the supports where there was none.
Fig. 3 Unequal spans, which is the ordinary case. The long middle span dominates the hogging over both interior supports, and the short outer spans are carrying moment generated by a span they do not touch.

The unequal-span figure shows something the equal-span case hides: a span’s moments depend on its neighbours. A short span next to a long one is hogged by its neighbour’s load, and can end up with more hogging over its supports than sagging in its middle — a beam that is bent the wrong way along most of its length, sized by a load it does not carry.

Loading it in patterns

The moment diagrams above assume every span is loaded. The worst case for any given quantity almost never is.

Consider the hogging over one interior support. It is produced by the loads on the two spans either side of it, and relieved by loads on the spans beyond those. So the worst hogging comes from loading the two adjacent spans and leaving the next ones empty — an arrangement nobody would draw by instinct and which is enforced in every code.

For the sagging in one span, the reverse: load that span and leave its neighbours empty, so that no hogging arrives to relieve it. The pattern that maximises one quantity minimises another, and a beam of five spans has a considerable number of arrangements to check.

The result is that a continuous beam is not designed against a moment diagram. It is designed against a moment envelope — the outer boundary of every diagram from every load pattern, which has more hogging at the supports and more sagging in the spans than any single arrangement produces. Finding the worst position of a load is exactly what influence lines are for, and pattern loading is that question asked span by span.

The envelope is not a state of the structure. Every arrangement of the imposed load on three spans — 8 of them, since each span is loaded or not — drawn faintly, with the greatest sagging and greatest hogging at each station drawn over them. Each faint curve is a real state of equilibrium and satisfies the free-moment identity exactly: mid-span ordinate minus the mean of the end moments is wL²/8, to 3e-16 of it. The envelope satisfies it nowhere, missing by up to 23% — because it is assembled from different load cases at different stations and no arrangement of load produces it. seven of the 8 arrangements are needed to build it; the rest never govern anywhere.
Fig. 4 Every arrangement of the imposed load on three spans — eight of them, since each span is either loaded or not — drawn faintly, with the greatest sagging and greatest hogging at each station drawn over the top. Seven of the eight are needed to build the outline; the eighth never governs anywhere.

The envelope has a property that is easy to state and easy to forget: it is not a state of the structure. Each faint curve behind it is a real equilibrium and satisfies the free-moment identity exactly — the mid-span ordinate minus the mean of the end moments is wL2/8wL^2/8, to three parts in 101610^{16}, which is the arithmetic of the solver rather than a property of the beam. The envelope satisfies it nowhere, missing by as much as 23 per cent, because it is assembled from different load cases at different stations and no arrangement of load produces it.

That matters as soon as anything is computed from the envelope rather than checked against it. A deflection, a crack width, a reaction, a shear diagram: none of them may be read off an outline that no loading produces. The envelope is a list of demands to be met one section at a time, and the arrangement that governs one section is not the arrangement that governs the next.

There is a practical shortcut worth knowing. Dead load is present in every pattern, so only the imposed load is patterned, and for a floor where the imposed load is a modest fraction of the total the envelope is not much wider than the all-spans-loaded diagram. For a bridge, where the moving load dominates, it is very much wider.

Half of it goes to the far end

The reason a span’s moments depend on its neighbours has a precise form, and it is the mechanism behind the hand method that made continuous beams tractable for thirty years.

Take a single span, held against rotation at its far end, and apply a moment at the near end. The near end rotates; the far end cannot; and a moment appears at the far end of exactly half the applied one. That ratio is the carry-over factor, and for a prismatic member it is a half regardless of length, stiffness or material.

The stiffness solver behind these figures returns 50.0050.00 for an applied 100100, which is worth noting because nothing in the matrix knows it should. The factor is a consequence of the cubic shape a beam bends into, arriving as a number rather than as an assumption.

The companion quantity is the distribution factor. Where several members meet at a joint, an out-of-balance moment divides among them in proportion to their stiffnesses, 4EI/L4EI/L for a member fixed at its far end and 3EI/L3EI/L for one pinned there. Load goes where the stiffness is, applied to a moment at a joint.

Those two numbers are the whole of moment distribution. Clamp every joint, compute the fixed-end moments, then release the joints one at a time: at each release, distribute the out-of-balance moment in proportion to stiffness, carry half of each distributed amount to the far end, and move on. The out-of-balance moments shrink with each pass and the process converges, usually within three or four cycles, to the exact elastic answer.

What makes it remarkable as a method is that every step is physically meaningful. It is not an algebraic manipulation that happens to converge; it is a sequence of releases, each of which is a structure that could exist. An engineer running it can watch the moment move from support to span and stop when the changes stop mattering — which is a property no matrix solution has, and the reason it was still being taught long after computers made it unnecessary.

Continuity that nobody designed

The most common continuity problem is not a continuous beam analysed badly. It is a beam that was analysed as simply supported and built continuous by accident.

A precast floor unit sitting on a beam, with a structural topping poured over both, is continuous whether or not the calculation said so. A “pinned” steel connection with two rows of bolts through a substantial end plate has real rotational stiffness. A concrete slab cast in one pour over a supporting beam has full continuity, whatever the design assumed about it.

The consequence is hogging where none was expected, and hogging puts tension on the top, where — in a member designed as simply supported — there is no reinforcement at all. The result is a crack over the support: usually not dangerous, since the section simply rotates until the moment it cannot carry has been shed to the span, and always unwelcome, since a crack in a floor is a defect that has to be explained.

The profession’s response is a small, standing insurance policy: nominal top steel over the supports of nominally simply supported slabs, sized not from any analysis but as a fraction of the span reinforcement. It exists to control a moment the design says is zero.

The deflected shape is the moment, integrated twice. A 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.
Fig. 5 The deflected shape of a beam free to rotate at its ends. A real end connection with any stiffness at all prevents part of that rotation, and the moment that prevents it appears whether the calculation asked for it or not.

The general lesson is worth taking beyond floors. A structural model states which restraints exist; the building supplies whatever restraints its details actually provide; and where the two differ, the building wins. Assuming a connection is pinned when it is not puts moment where the analysis says there is none, and that is exactly where nothing has been provided to resist it.

What continuity costs

The benefits are large and the costs are the reason simply supported spans are still built deliberately.

Sensitivity to settlement. A continuous beam whose middle support settles by a few millimetres redistributes its moments substantially, because the settlement is a change to a compatibility condition. The same movement under a simply supported beam changes nothing at all. Bridges on soft ground are frequently made determinate for exactly this reason, and the redundancy that provides the efficiency provides the sensitivity.

Tension on the top. Hogging puts the top fibres in tension. In reinforced concrete that means the main steel must be at the top over every support and at the bottom in every span, with laps arranged past the points of contraflexure — a considerably more elaborate detail than the straight bottom bars of a simple span, and one where a mistake is invisible after the concrete is poured.

The compression flange is unrestrained. Over a support, the bottom flange is in compression, and the floor slab that restrains the top flange does nothing for it. A continuous steel beam has its stability problem exactly where its moment is largest, which is a considerably worse arrangement than the simply supported case where the two coincide favourably, and it is a stability problem rather than a strength one.

The peak is sharper than it looks. A sagging moment diagram is a parabola with a flat top, so a span designed for its maximum is very nearly right for ten per cent of its length either side. A hogging peak over an internal support is a cusp — two curves meeting at an angle, because the support reaction is a step in the shear — so the moment falls away steeply on both sides. The reinforcement or the flange sized for it is fully used at one section and over-provided a metre away, and the curtailment has to be got right in a region where the moment is changing fast.

Continuity has to be built. A splice capable of transmitting the full hogging moment is a moment connection with all the cost that implies, and a beam that is continuous on the drawing and pinned in the fabrication is a beam designed for one structure and built as another.

Load, shear and moment — a cantilever. The 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.
Fig. 6 A cantilever’s diagram, hogging throughout. The region of a continuous beam over its support behaves like two cantilevers back to back, with the same consequence: tension on top, compression on the unrestrained bottom flange.

Two ways to the same benefit

Continuity is not the only way to put hogging into a beam, and comparing the alternatives is instructive because they differ in what they demand rather than in what they deliver.

An overhang does it determinately. Moving the supports inboard makes the ends cantilever and hog, relieving mid-span exactly as continuity does, and the arrangement is solvable by statics and insensitive to settlement. What it costs is a beam that sticks out past its supports, which is available at the end of a structure and not in the middle of one.

A haunch or a deeper section over the support does not reduce the moment; it accommodates it. And in a continuous beam it does something more, because stiffening a region attracts more moment to it — so haunching a support increases the hogging there, which further relieves the span. The structure obliges by sending load to whatever is stiffest, which is convenient here and hazardous in general.

A Gerber hinge gives up the redundancy while keeping most of the benefit: real hinges inserted at the points of contraflexure make the beam determinate again, with the moment diagram it had while continuous. Nineteenth-century railway bridges used the arrangement constantly, because a determinate structure on settling piers is worth a great deal.

Two of these move and the third cannot. The first span of a 3-span beam under 5 kN/m, with the stiffness of the middle span swept over a factor of 25. The support moment and the mid-span moment both move — that is what redundancy does, and it is the whole reason a continuous beam has to be analysed rather than read off. Their combination does not: the mid-span ordinate plus the average of the two end moments is 30.6 kNm at every point on this axis, which is wL²/8 for this span and this load and contains nothing else. The largest departure anywhere on the sweep is 2.3e-16 of the value, which is the arithmetic of the stiffness solution rather than a property of the beam. Continuity buys a distribution and not a capacity, and this is the line that says so.
Fig. 7 The first span of a three-span beam under 5 kN/m, with the stiffness of the middle span swept over a factor of twenty-five. The support moment and the mid-span moment both move, which is what redundancy does and the whole reason a continuous beam has to be analysed rather than read off a table. Their combination does not move at all: the mid-span ordinate plus the average of the two end moments is 30.6 kNm at every point on the axis, and the largest departure anywhere on the sweep is two parts in 101610^{16}.

That flat line is the statics underneath every arrangement on this page. Haunching the support, stiffening a neighbour, making the middle span twenty-five times stiffer than the others — each redistributes, and none of them changes the sum. Continuity buys a distribution and not a capacity. The free moment wL2/8wL^2/8 is fixed by the load and the span, and everything a designer can do is decide which end of the beam has to carry which part of it.

The reaction that goes negative

There is a load case a continuous beam has that a simple one does not, and it catches people because it produces a reaction pointing the wrong way.

Load one span of a two-span beam and leave the other empty. The loaded span deflects; continuity forces the empty span to rotate with it at the shared support; and because the empty span has nothing pushing it down, its far end lifts. If the loaded span is long enough relative to the empty one, the reaction at that far support becomes negative — the beam is trying to pull the support upward.

A support that can only push then does nothing, the beam lifts off it, and the structure that is being analysed is not the structure that exists. For a bridge deck this appears as uplift at an abutment bearing; for a floor beam it appears as a support that unloads rather than reverses; and in either case the standard answers are to hold the beam down, to make the arrangement heavy enough that the dead load never lets uplift occur, or to accept the lift-off and analyse the structure as though the support were not there.

The case is a good reminder that pattern loading is not only about maximising a moment. It also changes which supports are participating, and a load arrangement that reduces every moment in the beam can still be the one that governs a bearing detail.

The moment in 2 continuous spans. The bending moment in a continuous beam, solved by the stiffness method. The peak sagging moment is 33.4, and over the interior support a hogging moment of 38.1 appears — a moment no simply supported span carries at all, because continuity is what puts it there.
Fig. 8 Two unequal spans. The long span dominates everything: it sets the hogging over the shared support, it bends the short span backwards along most of its length, and under pattern loading it is what lifts the short span’s far end.

Where the model stops

Elastic behaviour and known stiffness. The distribution depends on the relative EIEI of the spans, and for a concrete beam that cracks in the hogging regions first, the stiffness ratio changes during loading. The analysis has to assume a stiffness distribution before it can compute the one that would justify it.

No support settlement. As above, and it is the largest single caveat.

Ductility, if redistribution is claimed. Codes permit the elastic hogging moment to be reduced by a stated percentage, with the sagging increased to compensate, on the argument that the section will redistribute plastically. The permission is conditional on rotation capacity, and the limits differ by section class for that reason.

Prismatic members. The standard coefficients assume a beam of constant section, and the second moment of area is what varies when it is not. A haunched one has a different distribution, and the haunch’s effect on the stiffness has to be included rather than treated as a local thickening.

Simple supports. Every figure here assumes the beam sits on knife edges. A real beam framing into a column has partial restraint at the end supports as well, which puts hogging into the end spans that none of the standard coefficients contain.

The figures carry one honest simplification. The continuous diagram and the simply supported one are drawn together to make the comparison legible, and they are diagrams of two different structures — the second is not a stage the first passes through, and no amount of loading turns one into the other. The comparison is between two design decisions taken before anything was built.

The ladder from here

Later rungs on this anchor: the three-moment equation. Moment distribution and the carry-over factor. Fixed-end moments as a table and as a derivation. Pattern loading and moment envelopes. Support settlement as a load case. Moment redistribution and the code limits on it. Continuity in composite construction, where the hogging region is cracked concrete over a steel beam. The Gerber girder and determinacy by inserted hinge. And continuous frames, where the same argument runs round corners as well as along a line.

Clapeyron published the three-moment equation in 1857, and it made continuous beams calculable for the first time — three unknown support moments at a time, in a chain that could be solved span by span. It held the field for seventy years, until Hardy Cross found a way to do the same job by iteration that could be taught in an afternoon.

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ContinuityHogging and saggingMoment envelopeMoment redistributionPattern loadingPoint of contraflexureReaction distributionStiffness