Structural Engineering & Statics
where the load goes
Internal forces

The moment that was moved on purpose

The elastic analysis of a continuous beam gives one set of moments. It is not the only set the beam is allowed to have, and taking a smaller one at the support is legal, cheaper, and paid for in a rotation that has to be delivered before the design exists.

Assumes The moment over the support, and what it buys, After the first yield, which is not the end and One support too many, and what it costs to know.

20 min read 11 figures One support too manyThe load must go somewhereWhich failure arrives first

Run a beam over two equal spans, put a uniform load on it, and solve. The stiffness method returns a hogging moment of wL2/8wL^2/8 over the middle support and a sagging peak of 9wL2/1289wL^2/128 in each span — for the beam drawn here, 240 kNm and 135 kNm. Those are the moments the beam has.

They are not the moments the beam must be designed for.

A designer may reduce the support moment by a stated fraction, take whatever span moments statics then requires, and provide sections for those instead. The support moment falls to 168 kNm, the span moment rises to 163, and the largest moment anywhere on the beam has gone from 240 to 168 — thirty per cent less section, for the same beam under the same load. Nothing was approximated and nothing was neglected.

The same load, two diagrams, both in equilibriumOne span of a pair of 8 m spans under 30 kN/m, drawn twice. The elastic solution puts 240 kNm over the support and 135 in the span. Reducing the support moment by 30% and taking what statics then gives leaves 168 and 163: the section the beam needs falls from 240 kNm to 168, a saving of 30%. Both curves are in equilibrium with the same load — the mid-span ordinate plus half the support moment is the free moment 240 kNm for either — and the second is legitimate for that reason alone. What it costs is 9.1 milliradians of rotation at the support, which the section has to be able to deliver.012345678-200-100100200distance along the span (m)bending moment (kNm, sagging up)elastic 240redistributed 168163 kNmβ = 30% · the section needed falls 240 → 168 kNm · the hinge turns 9.1 mrad
Fig. 1 The same span drawn twice. The dashed curve is the elastic solution; the solid one has had thirty per cent taken off its support moment and the span moment that statics then demands. Both are in equilibrium with the same load, and the check is written on the figure: the mid-span ordinate plus half the support moment is the free moment of 240 kNm for either of them.

Which free body produced the number

Cut a single span out of the pair and draw it with whatever moments the rest of the structure applies at its ends.

The outer end is a simple support and carries nothing. The inner end carries the support moment MsM_s, hogging. Everything between them is decided by statics alone: the moment at a distance xx is the free moment of a simply supported span plus the straight line joining the two end values,

M(x)=wx(L−x)2−MsxLM(x) = \frac{wx(L-x)}{2} - \frac{M_s x}{L}

and the peak is wherever the derivative vanishes, at x∗=L/2−Ms/wLx^* = L/2 - M_s/wL.

That is the whole of it. MsM_s appears as an input, not as an output. The elastic analysis supplies one value for it; any other value produces a perfectly good set of internal forces in equilibrium with the same load, in a beam with the same span and the same supports. What the elastic analysis added — and what redistribution discards — is the requirement that the beam’s two spans have the same slope where they meet.

2 continuous spans against 2 simple onesThe 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 240.0 to 135.0, and a hogging moment of 240.0 appears over the supports where there was none.moment135.0 sagging240.0 hogging240.0 if the spans were simplereactions 90.0 300.0 90.0 — the inner supports carry far more than a sharethe continuous case needed stiffness; the comparison did not
Fig. 2 Continuity puts the support moment there in the first place. Two simple spans have none; running the beam through drops the sagging peak by a third and creates a hogging moment where there was nothing at all. Redistribution is the same trade run backwards, part of the way, on purpose.

Why a different diagram is allowed

The permission comes from the lower-bound theorem, and it is worth stating in the form that makes the licence obvious.

If a set of internal forces can be found that is in equilibrium with the applied load and nowhere exceeds the strength of the material, the structure will not collapse under that load.

It does not say the structure will distribute its forces that way. It says only that it could, and that is enough, because a structure with somewhere to go does not collapse. The same theorem licences a strut-and-tie model in a region where sections do not exist, and a chosen crack angle in a cracked web. Moment redistribution is that argument applied to the tamest structure in the subject.

The condition it attaches is not a small one. The structure has to be able to reach the distribution it is being designed for, which means the sections that are being asked to shed moment have to rotate while holding it.

The collapse mechanism of a propped cantileverA collapse mechanism, with the hinge position found by searching rather than quoted. Every position gives an upper bound on the collapse load; the lowest is 30.60, at a hinge 58.6 per cent along, which is a coefficient of 11.657 times Mp over the square of the span.sagging hinge at 4.69hinge at the fixed endlowest upper bound: 30.60every hinge position gives an upper bound on the collapse loadassumed position of the sagging hingecoefficient 11.66 Mp ÷ L²
Fig. 3 The mechanism the redistribution is heading toward. A hinge over the support and one in the span turn the beam into a mechanism, and the load at which that happens is the collapse load. Every intermediate distribution between the elastic one and this is available, and the design is a choice of where to stop.

The redistribution that is the plastic design

Push the redistribution until the two moments are equal and something exact happens.

The search here is a bisection on the fraction β\beta, and it stops at 31.4 per cent, where the support and span moments are both wL2/11.657wL^2/11.657. That denominator is not a coincidence and it was not looked up: it is the collapse coefficient of a propped cantilever, computed elsewhere on this site by searching over the position of the sagging hinge, and it appears here from an entirely different question — where do two moments become equal?

The reason the two questions have one answer is that a beam of uniform capacity collapses when both hinge locations reach that capacity at once, and the balanced design is precisely the one that provides no more than either of them needs.

Its collapse load, computed from the mechanism rather than assumed, comes out at exactly the design load. Every less-redistributed design has a margin: the fully elastic one collapses at 1.46 times its design load, because the section it provides is set by a support moment the beam sheds long before it fails.

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. 4 Two limits crossing, which is the shape of this whole argument. An elastic design is governed by the support moment and a fully redistributed one by the span moment; between them is the point where neither governs and the section is smallest. The same picture appears wherever two capacities move in opposite directions as a parameter is turned.

What it costs, in radians

Reducing the support moment by ΔM\Delta M is releasing the redundant by that much, and the released structure opens up a relative rotation at that support of

θ=ΔM⋅2L3EI\theta = \Delta M \cdot \frac{2L}{3EI}

— each span contributing L/3EIL/3EI, the flexibility of a member with a moment at one end and a simple support at the other. For the beam drawn, thirty per cent of 240 kNm over a stiffness of 42,000 kNm² gives 9.1 milliradians.

Nothing about plasticity enters that calculation. It is the compatibility the elastic solution was enforcing, measured. The support section has to turn through that angle while continuing to hold 168 kNm, and whether it can is a question about the section rather than about the beam.

What it costs to reach the plastic moment, for two shapesMoment against curvature for two cross-sections of identical area (3000 mm²) and identical depth (200 mm), in mild steel, each divided by its own first-yield moment and its own first-yield curvature. The rectangle has a shape factor of 1.50 and reaches 98% of its plastic moment at 4.3 times the curvature at first yield; The I-section has a shape factor of 1.09 and reaches 98% of its plastic moment at 1.2 times the curvature at first yield. The dashed lines are the rigid-plastic moments, computed from the equal-area axis rather than read off the curves, and no curve reaches its own.02468101200.511.5curvature ÷ curvature at first yieldmoment ÷ moment at first yieldrectangle: 1.50× the yield moment, at 4.3× the yield curvatureI-section: 1.09× the yield moment, at 1.2× the yield curvature
Fig. 5 The section’s own answer: a moment-curvature curve whose plateau is what makes redistribution possible at all. A section with a long flat top can rotate at constant moment and shed what it cannot hold; one that loses capacity as it curves cannot.
Moment against rotation, for three real jointsThree connections on one plot, with the classification boundaries for a beam of EI/L = 5250 drawn as rays through the origin. web cleats is pinned, flush end plate is semi-rigid, extended end plate is rigid. The boundaries are multiples of EI/L, so the same joint is rigid on a short stiff beam and semi-rigid on a long slender one.00.0050.010.0150.020.0250.030.0350.04050100150200rotation, radiansmoment, kN·mrigid abovepinned belowweb cleats — pinnedflush end plate — semi-rigidextended end plate — rigid
Fig. 6 Rotation capacity measured on real connections rather than on an idealised section. The three curves have wildly different stiffnesses and wildly different ultimate rotations, and the second is what a redistributed design is spending. A joint that reaches its moment and stops is a joint that cannot redistribute anything.

Why the limit is a section classification

Codes cap redistribution at something like thirty per cent, and the cap is usually met as a number in a clause. It is a rotation requirement wearing a percentage.

The rotation a hinge can deliver depends on how much of the section can yield before something local goes wrong, and what goes wrong is the plate buckling before the fibre yields. A stocky section reaches its plastic moment and holds it through a large rotation. A slender one reaches a smaller moment and loses it immediately. So the permitted redistribution is tied to the section class, and the tie is not administrative: it is the same rotation, arrived at from the other end.

Where the class limits come fromThe width-to-thickness ratio at which two kinds of plate reaches its own elastic critical stress at the yield stress, for three steel grades. A flange outstand (buckling coefficient 0.43) derives to 18.6, 15.2, 13.3 at 235, 355, 460 N/mm², against quoted limits of 14.0, 11.4, 10.0; A web, in bending (buckling coefficient 4) derives to 56.8, 46.2, 40.6 at 235, 355, 460 N/mm², against quoted limits of 42.0, 34.2, 30.0. The derived number is the larger every time, and by the same factor at every grade — flange outstand 1.33, web, in bending 1.35 — because both the derivation and the quoted limit go as one over the root of the yield stress. A constant ratio is what a fixed knockdown looks like: the derivation is for a perfect plate and the quoted limit is for a rolled one, carrying residual stress and not quite flat.flange outstandk = 0.4318.6 at 23515.2 at 35513.3 at 460quoted: 14ε1.33× the quoted limit, at every gradeweb, in bendingk = 456.8 at 23546.2 at 35540.6 at 460quoted: 42ε1.35× the quoted limit, at every grade0102030405060width ÷ thickness
Fig. 7 Which class a section falls into, and what the grade of steel does to the answer. A higher yield stress makes every plate relatively more slender, because the limit is a ratio of the plate’s own dimensions to a length that scales with 235/fy\sqrt{235/f_y} — so the stronger steel is the one that redistributes less.

There is a second reason the cap exists and it is about serviceability rather than collapse. A beam designed for a redistributed diagram is not redistributing under working loads: at any load below first yield the beam has the elastic distribution, whatever the design assumed. So the support region is carrying more moment than it was designed for, and cracks there open wider than the calculation suggests. The redistribution is a promise about behaviour at the ultimate limit state and says nothing about the beam a person is standing on.

The two hundred years it took to be allowed

The elastic analysis of a continuous beam is older than the permission to ignore it by most of a century, and the order in which the two arrived is worth a paragraph.

Continuous-beam theory was complete by the middle of the nineteenth century, and it arrived as a triumph: a method that could say what a redundant structure carries, where statics alone had been able to say nothing. The moments it produced were treated as facts about the beam, because that is exactly what they had been produced as. A design that used any other set of moments was, straightforwardly, a design that had got the analysis wrong.

What changed was not the analysis but a proof about collapse. The plastic theorems of the nineteen-fifties said that a structure with somewhere to go does not fail at the first yield, and that any equilibrium set of internal forces the material can sustain is a safe design. That turns the elastic diagram from the answer into an answer, and it does so without contradicting a single line of the elastic derivation — the elastic moments remain exactly right for a beam whose every section is still elastic, which the beam is, right up to the load nobody designs for.

The residue of the older view is visible in how redistribution is written down. It is expressed as a reduction from the elastic diagram, capped as a percentage of it, rather than as a diagram in its own right. The elastic solution is still the reference even in the clause that permits departing from it.

The same freedom, without the plasticity

There is a version of this that needs no yielding at all, and it is worth separating because it is often confused with the one above.

A redundant structure’s internal forces depend on the relative stiffnesses of its members. Make the support region less stiff than the analysis assumed — because it has cracked, because the section is smaller there, because the joint is not rigid — and the moment moves away from it elastically, with nothing anywhere past yield. That is not redistribution in the technical sense; it is a different analysis of a different structure. But it produces the same picture, and it means the elastic diagram is sensitive to a stiffness nobody measured.

A joint is springs in seriesThe five components of an end-plate joint, with each bar the flexibility it contributes. The column flange in bending is 39.62% of the total on its own, and doubling its stiffness raises the joint's by a factor of 1.25 — while doubling the stiffest component buys 1.05. The joint's rotational stiffness is 30818.34 kN·m per radian.flexibility contributed by each componentthey add, so the softest dominates — Sj = 30818.34 kN·m/radwhat doubling it buyscolumn web in shear21.89%×1.12column web in compression11.55%×1.06column flange in bending39.62%×1.25end plate in bending18.09%×1.1bolts in tension8.85%×1.05
Fig. 8 A joint made of springs in series, which is where the assumed rigidity actually comes from. A connection with a real stiffness makes the frame’s moments a function of that stiffness, and the elastic distribution moves without anything having yielded.

The practical consequence is a reason to be sceptical of the elastic diagram in both directions. It is not a measurement. It is the output of an analysis whose inputs include a set of stiffnesses that are estimates, and a beam whose support region has cracked has already redistributed some of its moment without being asked.

The neutral axis is wherever the first moment vanishesA 300 by 500 section with 1200 mm² of steel at a depth of 450, carrying 150 kNm after the tension side has cracked. The neutral axis is no longer at mid-depth: it has risen to 137.0 mm from the top, which is where the first moment of the compression zone plus the transformed steel vanishes. The compression is 18.0 N/mm² at the top fibre and the steel carries 309 N/mm²; the resulting couple is 371 kN on a lever arm of 404 mm, which multiplies back to the 150 kNm applied. The uncracked section would have had 3422×10⁶ mm⁴ against the cracked 1139×10⁶ — a loss of 67% of the stiffness.x = 1371200 mm² of steel, n = 7.5b = 30018.0 N/mm²371 kN in the steelz = 404C = T = 371 kN · C·z = 150.0 kNm = the applied momentcracked I 1139×10⁶ mm⁴ against uncracked 3422×10⁶ — 67% of the stiffness gone
Fig. 9 The stiffness that has already changed. A cracked section is perhaps a third of the uncracked one’s stiffness, and the hogging region over a support cracks first because that is where the moment is largest — so the real structure begins shedding support moment at a fraction of the design load, with no plasticity anywhere.

The frame, where the shed moment has to land somewhere

A beam is the easy case, because the moment taken off the support goes into the same member. In a frame it does not.

A beam end that sheds 30 per cent of its hogging moment is a beam end whose joint is no longer in balance with what it was: the column above and below were sized for a share of the elastic beam moment, and the moment they receive falls with it. That is a saving, and it is also a hazard, because the beam is now being designed on the assumption that it will hinge at the joint — and if the columns are the weaker members, the hinge forms in them instead. A mechanism with hinges in the columns of a storey is not a redistribution; it is a storey collapse.

This is the whole of what capacity design is about, and it turns up here in miniature: choosing to redistribute is choosing where the structure will yield, and a choice about where a structure yields is only worth anything if the rest of it is strong enough to make the choice come true.

A portal frame swaying under 30A portal frame pushed sideways, solved by the stiffness method because statics cannot divide the load between two columns. The base shears come out at 15.0 and 15.0 and add to the applied 30; the peak moment is 34.3. The sway is drawn hugely exaggerated, and the moment diagram is plotted on each member's tension face.30H 15.0 M 34.3H 15.0 M 34.3the two base shears add to the applied 30 — the split came from stiffness, not staticsthe sway is exaggerated; a real frame at this load moves a fraction of a millimetre
Fig. 10 The frame the same argument moves into. Redistributing a beam’s support moment changes what the joint delivers to the columns, and the columns were sized from the diagram that has just been changed. In a beam the shed moment goes into the span; in a frame it goes into whichever member is least able to refuse it.

Where the model stops

The two spans are identical, and the arithmetic of the balanced case depends on it. A pair of unequal spans has a support moment that is not shared symmetrically, the balanced redistribution is different on each side, and there is no single β\beta that equalises everything.

The rotation is calculated on the released elastic structure. The real hinge is a finite length of beam whose curvature is distributed over it rather than concentrated at a point, and the rotation it delivers depends on that length — which depends in turn on the shape of the moment diagram near the support, on how much of the length is cracked, and on the bar spacing. The 9.1 milliradians is a demand computed exactly and compared with a capacity known roughly.

And the load is one load. A beam whose imposed load can sit on either span has several diagrams rather than one, and redistribution has to be applied to each of them separately and consistently — which is more constraining than it sounds, because the diagram that governs the support and the one that governs the span are different load cases.

What the picture cannot show

Both curves are drawn as though they existed at once. They do not. The elastic one is the beam’s state at low load; the redistributed one is a state it reaches only near collapse, if the sections deliver the rotation. In between there is a beam that is partly one and partly the other, and the figure has no way to draw a sequence.

Nor can it show what is happening in the support section while the moment is being shed. The picture is of bending moments, and the event is a rotation at constant moment. The one quantity the whole argument turns on is the one quantity a bending-moment diagram does not contain.

The assumption the figure rests on

The flexural rigidity is constant along the beam, and the rotation quoted depends on it directly. In a reinforced concrete beam it is not constant: the hogging region cracks and the sagging region may not, so the real structure’s stiffness distribution favours redistribution before anybody chooses any. Taking EIEI as uniform makes the computed rotation an overestimate of what has to be delivered plastically and an underestimate of what has already happened elastically, and the two errors do not cancel.

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. 11 The diagram as an integral, which is the reason redistribution is bounded rather than free. The moment curve’s curvature is the load, and no redistribution can change it: the parabola between the end values is fixed, so moving one end moves the whole curve rigidly and the free moment is invariant. That single fact is what makes the identity on the first figure hold at every β\beta.

The ladder from here

Later rungs on this anchor: redistribution in a frame, where the moment shed from a beam end arrives in a column that was not designed for it. The interaction with pattern loading, where each arrangement has to be redistributed separately and the envelope of redistributed diagrams is not the redistribution of the envelope. Redistribution in a prestressed member, where the secondary moments are themselves a self-equilibrating field and the question of what may be redistributed has an extra term. The rotation capacity of a real hinge, measured rather than classified. And the limit-state version of the same argument, in which the elastic analysis is abandoned entirely and the design is a mechanism from the start.

Where this goes next

moment19.6 sagging24.5 hogging30.6 if the spans were simplereactions 14.0 38.5 38.5 14.0 — the inner supports carry far more than a sharethe continuous case needed stiffness; the comparison did not 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. sagging hinge at 4.69hinge at the fixed endlowest upper bound: 7.29every hinge position gives an upper bound on the collapse loadassumed position of the sagging hingecoefficient 11.66 Mp ÷ L² After the first yield, which is not the end A steel beam whose extreme fibre has reached yield has not failed. It has started forming a hinge, and collapse waits until there are enough hinges to make a mechanism. 1200 kNtie 750 kNstrut 960 kN38.7°z = 1600strut 3.0 N/mm² over 812 mm · limit 15.8bursting across each strut 240 kN · tie steel 1724 mm² When there is no section to design Beam theory needs a section, and a section needs the strain to be linear across it. Within about a depth of a support, a load, a corner or a hole it is not — and those are the regions structures actually fail in. 00.5%1%2%2%050100150200250300350strainstress, N/mm²mild steelcast iron The property that appears in none of the equations Ductility is in no design formula on this site. Every method on this site depends on it — and a brittle structure does not merely fail early, it makes the analysis wrong.

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Bending momentCollapse mechanismCompatibilityContinuityDuctilityEquilibriumFree bodyIndeterminacyLower bound theoremMoment redistributionPlastic hingePlastic momentRotation capacitySection classificationStiffness