The corner columns take more than their share

The framed tube was an idea about where to put the material. Instead of bracing a tall building somewhere in the middle and hanging floors off it, put every column on the perimeter, tie them together with deep spandrel beams at every floor, and the building becomes a hollow cantilever with the whole plan as its lever arm.

The arithmetic of that is unarguable. A tube 30 metres by 40 has a second moment about a hundred times a core’s, and the second moment is where the stiffness of a tall building comes from.

The arithmetic assumes the flange works. It mostly does not.

Which free body produced the number #

Cut the tube at a height and take everything above it. The overturning moment there has to be carried by axial forces in the perimeter columns, and there is no argument about the total: the forces have to add to zero and their moments have to add to $M$.

The argument is about the distribution, and it starts with a question about how a column in the middle of the windward face finds out that it is supposed to be carrying anything.

It is not being pushed down by the floors — the floors carry gravity, and the overturning force is over and above that. What pushes it down is the shear in the spandrel beams either side of it, delivered from the column next door, which got it from the column next to that, and so on back to the corner. The corner column is the only one connected to the web face, which is where the storey shear actually enters the tube.

So the axial force walks along the flange, bay by bay, through the in-plane shear of a frame — and a frame in racking is not a plate in shear. Its shear stiffness is one storey of column bending in series with one bay of spandrel bending, and for the frame here

$$G t = \frac{12E}{p\,h_s\left(\dfrac{h_s}{I_c} + \dfrac{p}{I_b}\right)} = 30{,}800\ \mathrm{N/mm}$$

which against the columns smeared into a plate 20 mm thick is an effective shear modulus of 1,540 N/mm² — two per cent of steel’s. The perimeter of a framed tube is, in shear, a very soft material.

The equation, and the term that drives it #

Reissner’s assumption is the simplest one that can be wrong in the right direction: let the longitudinal displacement across the flange be its corner value plus a parabolic deficit,

$$u(x,z) = u_b(z) + u_1(z)\left(1 - \frac{x^2}{b^2}\right)$$

and find $u_1$ by making the whole thing stationary. Two coupled equations come out; eliminating the curvature leaves

$$u_1'' - k^2 u_1 = S\,M'(z)$$

The driving term is the derivative of the moment, which is the shear. That is the single most useful thing on this page, because it says a tube carrying a moment with no shear — a pure couple applied at the roof — has no shear lag at all, whatever its plan. Lag is not a consequence of the flange being wide. It is a consequence of force having to travel along the flange, and force travels along the flange only when the moment is changing.

What it costs, twice #

Shear lag charges twice, and the two charges are usually quoted separately by people who do not realise they are the same defect.

In strength, the corner column carries 1.52 times what plane sections says. A designer sizing the perimeter from a section analysis is fifty per cent under on the one column that decides the frame, and comfortably over on the eight in the middle of each face.

In stiffness, the tube deflects 171 mm at the top against the 123 a plane-sections cantilever would give — so its effective second moment is 72 per cent of its gross. Every drift check, every period calculation, every second-order amplification inherits that factor.

The effective width is the same statement in a third currency: the face carries its resultant on 51 per cent of its own width. Half the flange of the tube is not in the tube.

The fix is on the shear axis, not the plan #

Because $k^2$ is proportional to $Gt$, everything about shear lag is a statement about the racking stiffness of the perimeter frame — and the perimeter frame’s racking stiffness is one storey of column bending in series with one bay of spandrel bending.

Ten times that stiffness takes the concentration from 1.52 to 1.15 and the efficiency from 72 per cent to 95. A tenth of it takes them to 2.13 and 48.

Ten times is not a spandrel. A spandrel that deep would fill the window. What gives ten times is a diagonal, which replaces bending with axial action and takes the racking stiffness up by an order of magnitude with a fraction of the steel — and that is the braced tube, whose entire justification is this curve and not any argument about the diagonal carrying load.

The other route is a bundled tube: interior frames running across the plan, which do not stiffen the flange so much as shorten it. Two tubes side by side have a shear-lag length of their own, and each of them starts again at its own corner. The reason a bundled tube can be far more efficient than a single one of the same footprint has nothing to do with the extra material and everything to do with the fact that lag is a length effect and the length has been halved.

What a diagonal actually does #

A braced tube — a diagonal running across several storeys of the face, intersecting the columns as it goes — is usually explained as a way of getting the perimeter to carry the storey shear axially instead of by bending. That is true and it is the smaller half.

The larger half is that the diagonal makes the flange work. A face with a diagonal on it distributes axial force from corner to middle at the stiffness of a truss rather than of a Vierendeel frame, which is the order-of-magnitude change the efficiency curve above needs, and the columns in the middle of the face start carrying what a section analysis said they would.

That is why the diagonals on a braced tube run across the flange faces as well as the web faces, which a shear argument alone would not require: on the web face the diagonal carries storey shear, and on the flange face it carries almost none and is there to remove the lag.

The floor plate is what makes it one section #

Everything above treats the four faces as parts of a single closed section, which is a statement about compatibility rather than about geometry: the windward face is shortening, the leeward face is lengthening, and the two web faces are shearing, all of them consistently with one set of section rotations.

What enforces that is the floor. A floor plate is a stiff diaphragm in its own plane, and at every level it holds the four faces in a rectangle, transfers the storey shear into the web faces, and stops the flange faces from simply bowing out on their own.

Remove it — at an atrium, at a plant floor, at a double-height entrance — and the tube stops being a tube over that height. The consequence is not a small loss of stiffness; it is a change of structural type over a few metres of building, with everything above it supported on whatever is left.

A rule that turns out to be about the wrong quantity #

The received summary of shear lag is that it gets worse as the flange gets wider. It does, and that is not the mechanism.

Widening the face from nine metres to fifty-seven takes the concentration from 1.07 to 2.23 on this building. But the same range of concentrations is available at fixed width by varying the spandrel, and the parameter both routes move is $kb$ — the flange half-width measured against the shear-lag length $1/k$, which contains $\sqrt{G/E}$ and the plan depth and nothing about how wide the face happens to be on its own.

So the right statement is that lag is bad when the flange is wide compared with the distance over which the frame can distribute force. That distance is 71 metres here, which is why a 30-metre face is badly lagged on a building whose columns are three metres apart, and why the same face with a diagonal on it is not lagged at all.

The corner column belongs to two faces at once #

The one column the analysis above never resolves is the one it is most about.

A corner column is part of the flange face and part of the web face, and it is where the shear flow turns. In the web it is the last column of a row being sheared vertically; in the flange it is the first of a row being fed axially. Both faces deliver their force into the same member, and the model here treats it as belonging to the flange because that is the face whose distribution is in question.

The practical consequence is a detail rather than a number. A corner column carries the highest axial force in the building, sees the largest change in that force from one storey to the next, and is connected to spandrels in two directions at right angles — so it is simultaneously the most heavily loaded member, the one with the most connections, and the one whose splices are hardest to detail. Every framed tube ever built has a corner column heavier than anything else on the plan, and the reason is not the plan; it is this curve.

Where the model stops #

The webs are assumed to stay plane. They very nearly do — they are loaded by shear along their own length rather than through their width — but “very nearly” is doing some work at a plan aspect ratio far from one, and a tube much deeper than it is wide has some warping in the web too.

The parabola is an assumption, not a solution. Reissner’s one-parameter shape gives the right mechanism and the right dependence on $Gt$; a finite element model of the same frame gives a flange profile that is flatter in the middle and steeper at the corner, so the numbers here understate the corner slightly and overstate the middle.

Nothing here has floors in it. A real tube’s floor plates are stiff diaphragms that tie the four faces together at every level, and they are what makes the plan behave as one section at all. Removing them does not make the lag worse so much as make the question meaningless.

And the columns are smeared into a plate. That is exact for the axial part and an approximation for the shear part, because a frame’s racking stiffness is a discrete property of a bay and the continuum treats it as a modulus. The approximation is good while there are many bays across the face, which is the case the tube exists for.

What the pictures cannot show #

The first figure draws eleven columns as tick marks under a smooth curve, which is a continuum’s picture of eleven discrete members. The real distribution is eleven numbers, and the corner one is the one that gets designed; the smoothness is the model’s, not the building’s.

Nor can any of these figures show the thing that decided the form in practice, which is the window. A framed tube’s spandrels are as deep as the architecture will allow and its columns are as close together as the architecture will allow, and both of those constraints are about daylight. The curve of efficiency against racking stiffness is the engineering; where a particular building sits on it was settled by a facade drawing.

The assumption the figure rests on #

The spandrel is treated as prismatic and rigidly connected to the column at each end, so that a bay of frame has the racking stiffness written above.

Real spandrels are not, in the direction that matters. They are connected to columns through joints with finite stiffness, they are often haunched or hunched at the ends, and in a concrete tube they are cast monolithically with the column so that the effective span is shorter than the bay. Each of those changes $Gt$, which appears under a square root in $k$ and therefore under a fourth root in most of the answers — a robustness that is worth noticing, because it means the qualitative picture survives a lot of uncertainty about a detail nobody models.

What it does not survive is a spandrel that has been changed. A tube whose ground-floor spandrels have been removed to make an entrance, or whose top storeys have shallower beams for a plant room, has a shear-lag length that varies up the building, and the disturbance that produces is exactly the kind a transfer structure is built to deal with — a discontinuity in a load path that the members either side of it were designed assuming was continuous.