Structural form

The floor is a beam lying down

A floor plate spans horizontally between the walls that resist a lateral load, carries a distributed inertia load, and has chords, a web and a span-to-depth ratio like any other beam. Its stiffness decides whether the walls share the load by their stiffness or by the area of floor nearest them — and the familiar tributary answer turns out to be neither limit.

Assumes How a tall building stands still, One support too many, and what it costs to know and The deflection that is not bending.

Push a building sideways and the load arrives as an inertia force or a wind pressure spread over a floor. It has to reach the walls, cores and frames that resist it, and those are in a few places while the load is everywhere.

The thing that gets it there is the floor plate, working in its own plane. That is not a metaphor: a floor plate carrying a lateral load is a beam whose span is the distance between the resisting elements, whose depth is the width of the building, whose web is the slab and whose flanges are its two edges — a deep beam whose depth is the building. It has a moment diagram, a shear diagram, chord forces and a deflection — all of them lying flat.

Whether the floor shares the load out by stiffness or by areaThe share of a uniform storey force taken by each of three equally stiff walls, against the stiffness of the floor plate that spans between them. A plate far stiffer than the walls translates almost rigidly, every wall deflects the same and the share is the ratio of stiffnesses — 60% to the middle wall here, the same as everyone else. A plate far softer than the walls behaves as a continuous beam over them and the middle wall takes 33%. Neither end is the tributary-area answer of 50%, which assumes a plate that is both soft and discontinuous over the wall, and which nothing here ever reaches.0.020.1011025000.20.40.60.8floor-plate stiffness ÷ wall stiffnessshare of the storey forcethe middle wallan end walltributary arearigid plate← soft plate
Fig. 1 The share of a uniform storey force taken by each of three equally stiff walls, against the stiffness of the plate spanning between them. At the stiff end every wall takes a third, which is the ratio of the stiffnesses. At the soft end the middle wall takes 62.5%, which is the continuous-beam answer. The dashed line is the tributary-area answer, and the curve crosses it rather than approaching it.

Which free body produced the number

Take a strip of floor between two cuts perpendicular to its span. The inertia load on it acts sideways; the only things that can carry it sideways are the two cut faces of the slab. So the internal actions are an in-plane shear and an in-plane moment, and the model is a beam.

Model the walls as springs — each a stiffness kk resisting a displacement of the floor at its own position — and the plate as a beam of in-plane bending stiffness EIEI and shear stiffness GAGA. The problem is then a beam on elastic supports, solved by exactly the machinery a beam on discrete supports needs, and the answer contains both flexibilities in series.

How much of a deflection belongs to the beamThe share of the total deflection that is the beam's own bending, against the stiffness of what it sits on. A 24 m beam on three supports under a uniform load: on rigid supports every millimetre is the beam's, and the share falls away as the supports soften until almost none of it is. The beam drawn beside this figure sits at 68% — so 32% of what it does is happening somewhere a beam calculation never looks. The two flexibilities are in series, which means the softer one governs and stiffening the other buys nothing.4020080030002000000.20.40.60.81stiffness of each supportfraction of the deflection that is bendingrigid supportsare over here68%the beam drawn
Fig. 2 The other half of the same solve, read the other way round: how much of a beam’s deflection is its own bending rather than its supports’ movement. Here the beam is the floor and the supports are the walls, and the ratio between the two flexibilities is the only thing the answer depends on.

The dimensionless quantity is the ratio of the plate’s own flexibility to the walls’. Nothing else matters — not the absolute stiffness of either, not the load, not the number of storeys.

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.40 per unit lengthshear480.0moment2880.0 at x = 12.00the moment peaks exactly where the shear passes through zero
Fig. 3 The floor plate read as the beam it is: 24 m between end walls, 40 kN/m of storey force, and the diagrams that follow. The moment peaks at 2,880 kNm in the middle of the plan and the shear at 480 kN at each wall — numbers about a floor, obtained by the arithmetic of a beam.

The two limits, and the answer that is neither

Three equally stiff walls at 0, 12 and 24 m, carrying a uniform 40 kN/m.

Rigid plate. The floor translates almost as a rigid body, every wall deflects by the same amount, and each carries kuk u. Equal stiffnesses give 33.3% each. This is the assumption that plan torsion rests on entirely.

Flexible plate. The floor bends between the walls, and the walls act as rigid supports. A continuous beam over three supports puts 62.5% on the middle one and 18.75% on each end — the familiar 5wL/45wL/4 at an interior support of two equal spans.

Tributary area would give 25/50/25, on the reasoning that each wall takes the floor nearest it.

The tributary answer is not either limit. It sits between them and the curve passes through it at one particular stiffness ratio, on its way from one limit to the other.

That is worth stating plainly because the tributary rule is taught as the flexible-diaphragm answer. It is the answer for a plate that is soft and simply supported at every wall — a floor discontinuous over the wall, which timber and metal-deck diaphragms often nearly are and a concrete slab never is. A continuous soft plate gives the middle wall a quarter more than tributary area does.

Where a beam's load comes fromA 12 × 8 m panel carrying 5 kN/m², divided at 45° from the corners. The long beams take a trapezoid of 32.0 m² each and the short beams a triangle of 16.0 m²; the four areas sum to 96.0 m², which is the panel, so no load has been invented or lost. The line loads quoted are the uniform equivalents; the real distributions peak at 20.00 kN/m at midspan.32.0 m²13.33 kN/m32.0 m²13.33 kN/m16.0 m²10.00 kN/m16.0 m²10.00 kN/m12 m8 m96.0 m² divided, 96.0 m² of panel — the division closes
Fig. 4 The tributary rule where it is exact: a load divided between beams by geometry alone, because the members are simply supported and share nothing. Every additional continuity between the elements takes the answer away from this and toward stiffness.

Rigidity is a comparison

There is no such thing as a rigid diaphragm. There is a plate whose in-plane deflection is small compared with the deflection of the walls it sits on, and the comparison has two quantities in it.

The usual criterion is that the diaphragm’s own mid-span deflection under the storey force should be less than half the average storey drift of the walls. Both sides move with the building: a stiff core makes the walls’ side small and pushes the plate toward being flexible; a slender building makes the drift large and pushes it toward being rigid.

The shear part is not a curve at allA 300 × 600 rectangle spanning 2,400 mm, a span-to-depth ratio of 4, with its bending deflection and its total drawn together. Bending gives 0.0254 mm and shear a further 0.0049 mm, so 16.3% of the movement is the term beam theory drops. The shear part is drawn separately beneath: two straight lines meeting under the load, each limb straight to 2e-16 of the peak. It is the shear diagram integrated once where the moment diagram above it is integrated twice, so it carries one degree of curvature less than the bending shape — none at all here. The real movement is 1 in 79,116 of the span; both panels are drawn at about 11,932 times it.100 kNthe gap is the shear: 16.3% of the totalbending alone, and what the beam really doesthe shear part alone, magnified 3 times furthertwo straight lines meeting under the loadshearγ = V/GAs is a slope the section is racked through, not a curvature —so this diagram is integrated once, where the moment diagram above it is integrated twice
Fig. 5 And the plate’s own deflection is not all bending. A floor plate 24 m long and 12 m deep has a span-to-depth ratio of two, which is deep-beam territory — the shear deformation is 36% of the total at this proportion, and a bending-only calculation of diaphragm flexibility understates it by that much.

That last point is the one most often missed. Shear deflection is negligible for a slender beam and dominant for a stubby one, and a floor plate in plan is about as stubby as structural elements get. A 24 by 12 m plate is at L/d=2L/d = 2; a long thin plan at 60 by 12 is at 5 and behaves much more like a beam; a square plan is at 1 and barely deflects in plane at all.

So the plan’s proportions decide the answer, not the slab. A 200 mm slab in a square plan is rigid by an enormous margin. The same slab in a plan five times as long as it is deep, on stiff cores, may not be.

Two centres, and the distance between them is a torqueA storey 30 by 18 m with its walls drawn heavy, pushed in one direction by 1000 kN. The force acts through the centre of mass and the storey turns about the centre of rigidity — the stiffness-weighted centroid of the walls, at x = 10.8 m — and the distance between the two is an eccentricity of 4.20 m before the 5% that has to be assumed anyway. The table below the plan splits each wall's force into its direct share and its torsional one. Torsion relieves the walls near the centre of rigidity and loads the far ones, so the wall in trouble is not the wall carrying the most: east façade is asked for 98% more than its direct share, and the walls at right angles to the push carry 55 kN each with nothing applied along them at all.centre of masscentre of rigiditye = 4.20 m1000 kNcore west800 kN direct− 93 torsional= 707 kNeast façade200 kN direct+ 196 torsional= 396 kNsouth0 kN direct− 55 torsional= 55 kNnorth0 kN direct+ 55 torsional= 55 kNcentre of rigidity at x = 10.80 m, y = 9.00 mtorsional radius r = 10.56 m against a plan radius of 10.10 mtorsionally stiff by the usual criterion
Fig. 6 What a rigid diaphragm licenses: treating the storey as one body with a single translation and a single rotation, so that the walls share by stiffness and the eccentricity produces a torque. Every number on that page depends on this one being true.

Chords, and the flanges nobody drew

A beam in bending has flanges. A floor plate has two edges, and the in-plane moment is carried as a tension along one and a compression along the other:

T=C=MdT = C = \frac{M}{d}

with dd the depth of the plan. For the 24 m span above, M=2,880M = 2{,}880 kNm and d=12d = 12 m, so the chord force is 240 kN — a tie running along the whole length of one edge of the building and a strut along the other.

The chord is usually not a member anybody drew. It is the reinforcement in the slab edge, the perimeter beam, the spandrel, or the top flange of the edge beam — whatever is continuous along the edge and can be shown to carry 240 kN. Where the edge is interrupted, by a stair, a lift shaft or an atrium, the chord has to be traced round the interruption or the diaphragm has no flange there.

A Warren truss of 8 panelsA Warren truss under equal panel-point loads. The joint equilibrium equations were assembled and solved; 14 members came out in tension, 15 in compression and 2 carrying nothing.48.0120.0168.0192.0192.0168.0120.048.0-84.0-144.0-180.0-192.0-180.0-144.0-84.0-67.950.9-50.933.9-33.917.0-17.0-17.017.0-33.933.9-50.950.9-67.9tensioncompression2 carrying nothing
Fig. 7 The same structure made explicit. A braced roof — a horizontal truss between the eaves — is a diaphragm whose web has been drawn as members and whose chords are the eaves beams. Every steel-framed building with no slab has one of these, and it makes the chord force visible because it is a member with a number on it.

A collector, or drag strut, is the other member the diaphragm needs and the analysis does not produce. The shear the plate delivers is spread along the whole depth of the plan, and the wall receiving it is a few metres long — so something has to gather the shear from the full width and drag it into the wall. That member carries an axial force equal to the wall’s reaction less whatever arrives directly along its length, and it sits in a line of slab or beam that looks structurally uneventful.

Two beams tied together, and the deeper one takes 89% of the loadTwo simply supported beams of 6 m, one twice as deep as the other, tied together at midspan so that they have to move as one. A load of 100 kN stands on the tie. Point stiffness is 48EI/L³, so the deeper beam is 8 times as stiff — depth cubed, nothing else — and the load divides in that ratio: 11.1 kN into the shallow beam and 88.9 kN into the deep one, 11% against 89%. Both midspan points move 50000.00 mm, which is the whole of the argument: the geometry of the load never entered it. The deflection is drawn 0 times full size — the real sag is 50000.00 mm on a 6 m span, about 1 in 0.P = 100 kNthe shallow beam takes 11.1 kN11% of it — one part of the stiffness in 9the deep beam takes 88.9 kN89% of it — 8 times the stiffness of its neighbourone load, two beams, one deflection: 50000.0 mm each
Fig. 8 The stiffest path takes the load, which is the whole of the rigid limit. What makes a diaphragm interesting is that the path’s own stiffness is comparable with the stiffness of the elements it is choosing between, so the sharing is negotiated rather than decided.

The shear in the plate, and where it is worst

The moment gives the chords. The shear gives everything else, and it is distributed across the depth of the plan exactly as it is across the depth of a beam.

Shear stress across a sectionThe distribution of shear stress over a tall rectangle, computed as VQ/It by accumulating the first moment of the area above every height. The peak is 0.14 against a mean of 0.09 — a ratio of 1.50 — and it falls at the neutral axis, where the bending stress is zero.neutral axispeak 0.1stressflow, q = VQ ÷ Imean stress 0.09 — the value a shear divided by an area would givepeak 1.50× that, and in the place bending ignoresthe flow is continuous; the stress jumps wherever the width does
Fig. 9 Shear flow accumulated as VQ/It over the depth of a rectangular section, which is the shape a floor plate has in plan. The peak is at the middle and 1.5 times the average, and the two edges — where the chords are — carry no shear at all. Read at the plan’s own scale, 480 kN across a 12 m depth is an average shear flow of 40 kN per metre.

Two things follow, and neither is obvious from a plan.

The middle of the plan is the busiest part of the diaphragm. Not the edges, where the chord forces are, and not near the walls in the depth direction. That is where an atrium usually goes.

The connection between the slab and everything else is a shear connection. The plate delivers its shear to the wall through whatever joins them — starter bars, a shear stud, a nailed edge — and it delivers it along the wall’s length, at a shear flow of the wall reaction divided by that length. A 396 kN reaction into a 6 m wall is 66 kN per metre of interface, which is a real number for a detail nobody draws.

The plate has to exist before it can do this

A diaphragm is the one part of a lateral system that arrives last and is assumed throughout.

A steel frame is erected, and for some weeks it has no floor plate at all — only decking, or bare beams. During that period the walls and cores are not connected to one another in plan, the wind still blows, and whatever holds the frame in position is temporary bracing whose job is exactly the job the slab will later do. That is the most dangerous day in miniature: a structure whose completed load path is understood and whose incomplete one is not drawn anywhere.

Built as two beams, used as oneBending moments in a two-span beam erected as simple spans under 12 kN/m and made continuous before the remaining 18 kN/m arrived, against the same beam built continuous from the start. The support moment is 324 kNm rather than 540 — 60% of it — and the midspan moment is 378 rather than 270, which is 140%. Both diagrams are in equilibrium with the same total load; they differ only in when the joint was made, which appears nowhere on the drawing.540 kNm built continuous324 staged378270same beam, same load, different history
Fig. 10 The general form of the problem: what a structure carries depends on what was present when the load arrived. A diaphragm is the extreme case, because before the slab is cast it is not there at all and afterwards it is assumed rigid.

The same argument applies to precast floors. Hollow-core units with no structural topping are a set of parallel planks; they become a diaphragm only through the grout in their joints and the tie steel across them, and the shear flow computed above has to cross every one of those joints. A precast floor’s diaphragm capacity is a joint calculation, not a slab calculation.

Two walls, and one number that does not move

Whether the floor shares the load out by stiffness or by areaThe share of a uniform storey force taken by each of two equally stiff walls, against the stiffness of the floor plate that spans between them. A plate far stiffer than the walls translates almost rigidly, every wall deflects the same and the share is the ratio of stiffnesses — 50% to the middle wall here, the same as everyone else. A plate far softer than the walls behaves as a continuous beam over them and the middle wall takes 50%. Neither end is the tributary-area answer of 50%, which assumes a plate that is both soft and discontinuous over the wall, and which nothing here ever reaches.0.020.1011025000.20.40.60.8floor-plate stiffness ÷ wall stiffnessshare of the storey forcethe middle wallan end walltributary arearigid plate← soft plate
Fig. 11 The same plate on two walls instead of three. Now the sharing is 50–50 at every plate stiffness — a symmetric structure has nothing to negotiate, and the whole of the diaphragm question disappears.

That degenerate case is worth having, because it explains why the subject is invisible in so much of practice. A simple rectangular building with a wall at each end has no distribution problem at all: symmetry fixes the shares whatever the diaphragm does. The question only arises when there are three or more resisting elements, or two of unequal stiffness, or a plan that twists — and then it arrives with all of the above at once.

What the number is worth knowing for

Three consequences follow from the sharing curve, and each decides something a designer does.

The wall forces are wrong if the classification is wrong. For the three-wall plan above, the middle wall is designed for 320 kN under a rigid assumption and 600 kN under a flexible one — nearly double. Getting the classification wrong in one direction overloads a wall; getting it wrong in the other overloads its neighbours, since the shares must always sum to the applied load.

The check has to be made in both directions. A plan long in one direction and short in the other has a plate that is slender in plan one way and stubby the other, so the same floor can be flexible for a north–south push and rigid for an east–west one. There is no single answer for a building.

And it decides whether torsion exists at all. Plan torsion is a consequence of the storey rotating as one body. A flexible diaphragm does not rotate as one body, so the eccentricity that produces a torque in a rigid analysis produces something else — a distribution along the plate — and the two analyses do not merely differ in magnitude, they contain different mechanisms.

Which is why the diaphragm question is asked first and every other lateral calculation waits on its answer.

Where the model stops

The plate has been treated as elastic and uncracked. A concrete diaphragm carrying its design load has cracked in plane, its effective stiffness is perhaps half the gross value, and the cracking is not uniform. Since only the ratio of two flexibilities matters, the error partly cancels — but only if the walls have cracked in the same proportion, and they have not.

Openings have been ignored entirely. A floor with a large atrium is a beam with a hole in its web, and a hole in a web is a Vierendeel panel whose local flexibility can exceed the whole plate’s. A plan with a re-entrant corner is worse: the corner is a stress concentration in a member with no flange there.

The walls have been modelled as springs at a point. A shear wall has length, and the diaphragm delivers shear to it along that length rather than at its centre. For a long wall the difference is real and it changes the chord force distribution near the wall.

And nothing here is dynamic. A flexible diaphragm has its own mass and its own natural frequency in plan, and a floor whose in-plane period approaches the building’s can amplify the storey force it is meant to be distributing — which is a possibility that a static stiffness ratio cannot see.

What the pictures cannot show

Every figure here draws the floor plate as a line, because a beam is a line. The thing it stands for is a two-dimensional plate whose stress field is genuinely two-dimensional near every wall and every opening, and the beam model is a Saint-Venant approximation that is good in the middle and poor at the ends — which is where the walls are.

The sharing curve is drawn against a stiffness ratio running over four decades, and no real building moves along it. A given building is one point; the curve exists to show which side of the tributary line that point falls on, and how far.

And nothing in these drawings shows the deflection. A diaphragm at its serviceability limit deflects in plane by a few millimetres over tens of metres, which is a thousandth of the line used to draw it.

The ladder from here

Later rungs on this anchor: the semi-rigid diaphragm as codes define it, and why a binary classification survives when the underlying quantity is continuous. Collector and drag forces in full, including the overstrength factors they are designed with because they must not fail before the wall does. Diaphragms with openings and re-entrant corners, where the plate model gives way to a strut-and-tie one. Timber and metal-deck diaphragms, whose stiffness is dominated by fastener slip rather than by material. Transfer diaphragms at podium levels, where the whole of a tower’s shear changes plane. And the historical case: diaphragm action was relied on for decades before it was calculated, and the first time anybody measured one it turned out to be several times stiffer than assumed and to be carrying forces nobody had drawn.

Named alongside this one

Essays reaching for the same objects. Nobody chose these; they are what the concept index makes visible.

What links here

Every essay whose body links to this one.

The objects this essay names

Each one links to every other essay that touches it.

Accidental eccentricityCentre of rigidityChord forceCollectorDeep beamDiaphragmErection stabilityIn plane stiffnessLateral systemLoad sharingPlan torsionShear deflectionShear stiffnessShear wallSpan to depthSupport flexibilityTemporary worksTributary area