The floor is a beam lying down

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.

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 $k$ resisting a displacement of the floor at its own position — and the plate as a beam of in-plane bending stiffness $EI$ and shear stiffness $GA$. 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.

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.

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 $k u$. 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 $5wL/4$ 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.

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.

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 $L/d = 2$; 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.

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:

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

with $d$ the depth of the plan. For the 24 m span above, $M = 2{,}880$ kNm and $d = 12$ 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.

The same structure can be made explicit rather than assumed. 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, and it is checked as the beam it is rather than trusted as a plate.

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.

That is the stiffest path taking the load in its plan form, and the diaphragm is what decides whether the phrase means anything here at all.

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.

The plate carries that shear the way any beam does — accumulated as VQ/It over its depth, which is the shear nobody draws with a floor plate as its section — and the collectors at the walls are where the accumulated flow has to be gathered into something a connection can take.

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.

The general form of the problem is that what a structure carries depends on what was present when the load arrived, and a diaphragm is the extreme case: a floor that is a load on the day it is cast and a structural element some days later.

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 #

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.

The plate has a frequency of its own #

Everything above is static, and the load a diaphragm most often carries is not. A floor plate has mass — most of the building’s mass, in fact — and an in-plane stiffness, so it has an in-plane natural frequency, and whether that frequency matters is decided by the same proportions as everything else on this page.

Treat the plate as a shear beam spanning between the walls, since shear is what it deforms in:

$$f_1 = \frac{1}{2L}\sqrt{\frac{GA_v}{\bar m}}$$

with $GA_v$ the plate’s shear rigidity and $\bar m$ its mass per metre of span. Two cases, both ordinary:

A 200 mm concrete slab, 12 m deep, spanning 12 m between walls. $GA_v = 12.5\times10^6 \times 2.4 = 3.0\times10^7$ kN, mass 5 kN/m² over 12 m, so $\bar m = 6.1$ t/m:

$$f_1 = \frac{1}{24}\sqrt{\frac{3.0\times10^7}{6.1}} = 92\ \text{Hz}$$

A bare metal roof deck, 12 m deep, spanning 30 m, at a shear stiffness of 3,000 kN per metre of width and a mass of 1 kN/m²:

$$f_1 = \frac{1}{60}\sqrt{\frac{36{,}000}{1.22}} = 2.9\ \text{Hz}$$

Ninety-two hertz and 2.9. The first is so far above anything a building is excited at that the plate is a rigid mass and nothing else. The second sits inside the range a single-storey building with masonry or tilt-up walls occupies — three to five hertz — which means the diaphragm is not distributing the storey force so much as responding to it, with its own dynamic amplification on top of the building’s.

That is the mechanism behind a rule that looks arbitrary in every code that carries it: the out-of-plane anchorage force for a wall attached to a flexible roof diaphragm is specified several times larger than the force the same wall would be designed for on a rigid one. The wall is not heavier. It is hung off the middle of a plate that is amplifying, and the middle of the plate is where the amplification is largest — the same amplification a floor gives a machine sitting on it, with the roof as the floor and the wall as the machine.

It also says which buildings the effect belongs to, and the answer is the unglamorous ones. A tall building’s floors are concrete and its diaphragms are at 92 Hz. A warehouse, a supermarket, a hangar — long plans, light roofs, heavy perimeter walls — is where the plate’s own dynamics reach down into the range that matters, and those are the buildings that lose their walls in earthquakes.

The chord is a continuity problem, not a capacity one #

The 240 kN chord force is worth converting into what it actually asks for, because the answer explains why diaphragms fail by detailing rather than by strength.

At 435 N/mm² it needs

$$A_s = \frac{240{,}000}{435} = 552\ \text{mm}^2$$

— two 20 mm bars. Along the edge of a 24 m building, carrying the entire in-plane moment of a storey. The interface to the wall is the same story: 66 kN per metre of wall needs 152 mm² per metre across the joint, which is H12 bars at 300 centres, and every slab has more than that in it already for other reasons.

So the diaphragm asks for almost no material. What it asks for is that the material be continuous, and continuity is the thing a floor plan is full of reasons to interrupt: a stair, a lift shaft, a riser, a movement joint, a change of level, a precast unit that stops short of the edge. Two 20 mm bars are trivial to provide and easy to omit at the one place they are needed, because the place they are needed looks like every other place.

The consequences of that asymmetry are worth stating as a habit rather than as a calculation.

A chord lapped without thought is a chord that is not there. 552 mm² of bar delivers 240 kN only where its lap does, and a lap in the middle of the span is at the point of maximum moment.

A gap in the edge is a gap in the flange. The plate has no other flange to fall back on — unlike a beam, whose web can carry some moment, a diaphragm’s web is a slab whose in-plane bending stiffness about its own centreline is what the chords were computed to avoid using.

And nothing about it shows up in a check. A slab with no chord steel passes every gravity calculation, every deflection limit and every crack check, and its diaphragm capacity is whatever happens to be continuous along the edge. That is the shape of an omission rather than an error, and the only defence against an omission is to look for the thing rather than to check the thing that was found.

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.