The columns that lean

A tall building’s perimeter is the best place to put its lateral system, because it is the furthest from the centre and a moment of resistance is a force times a distance. The framed tube takes that idea to its conclusion: closely spaced columns all round the plan, joined by deep spandrel beams, so that the whole perimeter acts as one hollow cantilever.

It does not act as one hollow cantilever, and the reason is worth reading carefully because the fix follows straight from it.

Which free body produced the number #

Take one flange face of the tube and ask how axial force gets into a column in the middle of it.

The overturning moment arrives at the base as a couple between the two faces perpendicular to the wind. The corner columns are part of both faces and receive their force directly. A column in the middle of a flange face receives nothing directly: the only route to it is in-plane shear in the frame between it and the corner, transmitted bay by bay by the spandrel beams bending in the plane of the face.

Every bay of that transmission is a shear-flexible link, so the force runs out before it gets there. The middle column carries 13.4 N/mm² against the corner’s 78.5 — the corner columns take more than their share — and the tube behaves as though only 45% of the flange width existed. Its effective second moment is 64% of the gross.

That is shear lag, and it is a property of the spandrels, not of the plan. The distinction is the whole essay: a defect that looks geometric turns out to be a stiffness, and a stiffness can be changed.

One more feature of the mechanism is worth having before the fix, because it decides where the fix has to work. The governing equation for the flange’s warping is driven by the rate of change of the moment rather than by the moment itself — which is the shear. So the lag is not a constant property of the section. It is largest where the shear has accumulated and it vanishes where the moment does.

Two demands are being carried at once here, and the profile is the record of their changing ratio: near the top the accumulated moment is small and the storey shear is what there is, and near the base the moment has been collecting for the whole height. Every conclusion later in this essay about the angle of a diagonal is the same fact read a second time.

The number that decides it #

Hold the plan, the height and the column areas fixed and vary only the in-plane racking stiffness of one bay of perimeter frame. The overstress and the efficiency move a long way, and the ten-fold increase drawn on that axis is exactly what putting a diagonal across the face buys, because a diagonal replaces bending in the spandrel with axial force in a strut.

So the argument for the braced tube — and for the diagrid, which is the same argument taken further — lives on that axis rather than in the plan. It is not about making the building wider or the columns bigger. It is about the mechanism by which one part of the perimeter tells another part what it is carrying.

How far the spandrel alone can go #

The obvious lever is the spandrel, because it is the member the whole mechanism is named after. It has a ceiling, and the ceiling is visible in the formula rather than in any picture.

The racking stiffness of one bay is a storey of column bending and a bay of spandrel bending in series:

$$Gt = \frac{12E}{p\,h_s\left(\dfrac{h_s}{I_c} + \dfrac{p}{I_b}\right)}$$

For the frame drawn — columns at 3.0 m, a 3.8 m storey, $I_c = 1.4\times10^9$ and $I_b = 7.0\times10^8$ mm⁴ — the two terms in that bracket are $2.71\times10^{-6}$ and $4.29\times10^{-6}$. The spandrel is the larger of the two, so there is real ground to be had. But the column term does not move when the beam gets deeper, so the most a spandrel of unlimited depth could ever buy is

$$\frac{2.71 + 4.29}{2.71} = 2.58$$

— a factor of two and a half, against the factor of ten the axis in the previous figure was drawn to.

Ten times the beam stiffness has bought 2.81 where the ceiling is 2.58 in racking terms, which is most of what deepening a beam can do and is the wrong order of magnitude. No amount of beam depth changes that, because the column is in series with it. Moving both terms at once is not a deeper beam. It is a different mechanism, and there is only one available.

What a triangle does that a rectangle cannot #

The underlying fact is the one every truss rests on, and it is worth restating in the form that makes the tube case obvious.

A quadrilateral panel of pin-jointed bars has zero shear stiffness: it is a mechanism, not a soft structure. Making the nodes rigid gives it some, by bending the members, and the amount is 0.35% of what a continuous sheet of the same stretching stiffness would give. Adding one diagonal gives it 65%.

A perimeter frame is that quadrilateral with rigid nodes, which is the triangle that cannot fold with its triangle taken out. The spandrels are stiff, the columns are stiff, and the mechanism is still bending — three orders of magnitude worse than axial action at the same material cost. That is the whole of the case, and it is the same argument a gridshell makes about a curved surface.

The same face with the bending taken out #

Put the diagonal in and both terms of that series move together, because the bay is no longer carrying its racking by bending two members. It is carrying it by stretching and shortening one. Ten times is the honest figure for what that is worth, and it is the point already marked on the design axis two sections above. Drawn across the face, it looks like this.

The middle columns have gone from 13.4 N/mm² to 36.4 and the corner has come down from 78.5 to 55.5. Nothing about the plan has moved: the same 36 m face, the same thirteen columns, the same column area, the same height. What changed is the route the axial force takes to reach the middle of the face, and the whole of the improvement is in that route.

The defect also stops being a whole-building one, which is the part the single number for efficiency conceals.

Read the two height profiles together and the finding is not that the braced perimeter is a third better. It is that the framed tube is losing stiffness over its whole height while the braced one loses it only low down — which is where a stiffening measure is cheapest to make and where a designer would be adding material anyway.

Two demands, two angles #

A diagrid’s diagonals carry both the storey shear and the overturning moment, and the two want different geometry.

The shear at a level is carried by the horizontal components of the diagonal forces, so a shallow diagonal is efficient for it — at 35° to the horizontal a diagonal delivers most of its force horizontally.

The moment is carried by the vertical components acting as a couple across the plan, so a steep diagonal is efficient for it — at 90° a member is a column and delivers all of its force vertically.

The ratio of the two demands changes up the height. Near the top, the accumulated moment is small and the shear is what there is; near the base, the moment has been accumulating for the whole height and dominates. So the optimum angle is shallow at the top and steep at the base, and the classical answer of somewhere around 65 to 70 degrees is a compromise across the height rather than an optimum anywhere.

The same split appears in the drift, and the number that decides it is $\alpha H = H\sqrt{GA/EI}$ — below one the building behaves as a cantilever and above about six as a frame, with everything interesting in between. A diagrid raises $GA$ by an order of magnitude without touching $EI$, which moves the building down that axis: more of a cantilever, less of a frame.

Why the pair is better than either #

The reason a diagrid is worth the trouble is not simply that it is stiffer. It is that its two mechanisms are stiff in opposite places.

A wall alone bends and a frame alone shears, and tied together at every floor they reach less than either does on its own — less, in fact, than two springs in parallel would give. The extra comes from the systems restraining each other, and it works because each is stiff exactly where the other is not; how a tall building stands still is the essay about that interaction. A diagrid does the same thing inside a single system, because a triangulated perimeter is simultaneously a cantilever with a large $EI$ and a truss with a large $GA$.

The depth it does that at is the plan dimension of the building, which is the largest depth any structure in a building ever gets, and a moment of resistance is linear in it. That is why the perimeter is worth the trouble of using at all, and why the only question left is the one this essay has been asking: by what mechanism does one side of it talk to the other.

The arithmetic of a diagonal’s angle #

The two demands can be written down, and the optimum falls out of one differentiation.

Let a diagonal stand at $\theta$ to the horizontal. Its force resolves into $N\cos\theta$ horizontally and $N\sin\theta$ vertically, so a module of the perimeter carrying a storey shear $V$ and a moment $M$ needs

$$N_V = \frac{V}{n\cos\theta}, \qquad N_M = \frac{M}{c\, d\, \sin\theta}$$

with $n$ the number of diagonals sharing the shear on a face, $d$ the plan depth and $c$ a constant of the arrangement. The material required is proportional to the force times the member’s length, and the length goes as $1/\sin\theta$ for a module of fixed height. Multiplying through, the shear term’s cost goes as $1/(\sin\theta\cos\theta)$ and the moment term’s as $1/\sin^2\theta$.

The first is minimised at exactly 45° and the second falls monotonically toward 90°. So the optimum for a module is a weighted mean of those two, pulled toward the vertical in proportion to how much of the demand is moment — which is the height-dependence described above, now with a formula behind it. A building where the moment term is three times the shear term optimises at about 67°, and the flatness of the sum near its minimum is why almost every diagrid ever built sits between 60 and 70 degrees whatever its proportions.

The flatness is the useful part. It means the angle can be chosen for the façade module, the node fabrication or the floor-to-floor height, and cost a per cent or two rather than a factor.

What it costs in robustness #

The efficiency has a price, and it is the sharpest one in this essay.

In a framed tube the columns carry gravity and the spandrels carry shear, so losing a spandrel costs stiffness and losing a column costs a gravity path, and the two failures are separate. In a diagrid the same member does both jobs. Lose a diagonal and the gravity load above it has no route down and the shear across it has no route out, at once.

Worse, a triangulated panel that loses a member is not a weaker structure — it is a mechanism, and nothing about the strength of the survivors enters the answer at all. Real diagrids are therefore detailed with continuous nodes rather than pins and with enough member continuity that the loss of one diagonal is carried by bending in its neighbours, which is a redundancy bought back by exactly the mechanism the form was chosen to avoid.

The plan, which is not off the hook #

Making the perimeter efficient in bending does nothing for the other thing a perimeter has to do.

A triangulated perimeter is an excellent torsional structure, because a closed tube resists twist by shear flow round the perimeter and the diagonals carry that shear axially too. That is a genuine second advantage of the form and it is rarely stated: the diagrid’s torsional stiffness improves by the same order of magnitude as its lateral stiffness, which matters because plan eccentricity has to be assumed even where none is designed, and because the corner that moves most is where the drift limit is actually checked.

Where the model stops #

A diagrid node is not a pin and not a rigid joint. Every quantity here comes from a model that is one or the other. A real node is a fabricated casting or a welded assembly with six members meeting at it, and its stiffness, its tolerance and its cost are the dominant practical facts about the form.

Nothing here is second-order. A diagrid’s columns are inclined, so the gravity load in them has been amplified by the same P-delta softening as anything else, and a form whose lateral stiffness comes from members that also carry the gravity load has the two coupled more tightly than a separated system does.

Gravity in an inclined member has a horizontal component. A column carries its load straight down; a diagonal does not, and the horizontal components have to close round the plan. They do, on a symmetric building under symmetric load, and they do not during construction, under a pattern load, or where the plan changes shape up the height.

The floor plate has more to do. Every level of a diagrid has to deliver the diagonals’ out-of-balance horizontal forces back into the ring, so the floor is a working member rather than a passenger. That is a duty a diaphragm has in any building and a much larger one here.

How the load actually gets into it #

A diagrid receives its gravity load from floor plates and its wind load from the façade, and both arrive at levels rather than at nodes.

That mismatch is the form’s least glamorous problem. A diagrid module is typically six to eight storeys tall, so five or seven of every eight floors frame into the middle of a diagonal rather than into a node. The diagonal therefore carries local bending from the floor reactions on top of the axial force the whole system was designed to give it, and the bending is about the member’s weak axis as often as its strong one.

The usual resolution is a horizontal ring beam at each level, spanning between the diagonals and picking the floor up, which turns the floor reactions into a set of point loads at the nodes and leaves the diagonals purely axial. That ring is also what closes the horizontal components of the inclined members, so it is doing two jobs and is the member most likely to be under-drawn at scheme stage.

The consequence for a designer is that a diagrid is not simply a perimeter frame with different members. It is a system with an extra hierarchy in it — floors onto rings, rings onto nodes, nodes onto diagonals — and each level of that hierarchy is a place the mechanism can go back to bending without anybody deciding that it should.

The ring beam is what makes the diagonal possible #

The previous section gave the ring beam two jobs. It has a third, and without it the whole form does not stand up.

A module six storeys tall at 68° has a diagonal

$$\ell = \frac{6 \times 3.9}{\sin 68°} = 25.2\ \text{m}$$

long, and half of the diagonals in any diagrid are in compression. A 25 m strut is not a member anybody wants: it is what the perimeter would be if the floors did nothing, and its buckling load is $\pi^2EI/25.2^2$.

The floors intersect it every 3.9 m of height, which along the member is

$$\frac{3.9}{\sin 68°} = 4.21\ \text{m}$$

so if the ring beam at each level holds the diagonal laterally, the buckling length is 4.21 m rather than 25.2, and

$$\frac{P_{cr,\text{braced}}}{P_{cr,\text{free}}} = \left(\frac{25.2}{4.21}\right)^2 = 36$$

A factor of thirty-six, from a beam that is drawn to pick up a floor. The diagonal’s section is decided by that ratio and by nothing else: at 4.21 m it is a stocky member working near its squash load, and at 25.2 m it would be a slender one at a fraction of it, needing several times the steel to carry the same force.

Two consequences follow and both are the kind that get lost between disciplines.

The ring beam is a bracing member with a bracing member’s duties. A brace has to be stiff enough as well as strong enough, and it has to carry the restraint force the diagonal’s own out-of-straightness generates — a couple of per cent of 25 MN of diagonal force is not a nominal number. A ring sized only for the floor reaction is not a brace.

And a level without one is a level where the buckling length doubles. A double-height lobby, a plant floor with the ring omitted for ductwork, an atrium cutting the ring on one face: each takes the diagonal from 4.21 m to 8.4 m and quarters its critical load, locally, in a member that is also carrying the gravity of everything above it. This is the same hazard an interrupted load path always is, arriving through an element nobody thinks of as structural.

The count of joints, which is the argument nobody draws #

The diagrid’s economics are usually stated as a saving in steel. The larger part of them is a saving in joints, and the two are not the same thing at all.

Take the 36 m square plan at 40 storeys. As a framed tube with columns at 3 m centres, each face has twelve columns and each storey has a spandrel connection at every one of them:

$$12 \times 4 \times 40 = 1{,}920 \ \text{beam-to-column joints}$$

As a diagrid with six-storey modules and diagonals at 9.45 m across the plan, there are about sixteen nodes on each module ring and seven rings up the height:

$$16 \times 7 \approx 112\ \text{nodes}$$

A factor of seventeen. That is the number the form is actually sold on, because a connection is where the fabrication hours, the inspection, the tolerance and most of the site risk live, and a joint costs roughly what it costs whether it is carrying 200 kN or 2,000.

The trade is not free, and the two things it costs are worth naming because they are what makes a diagrid a specialist structure rather than a default one.

Each node is a much harder object. Six members meet at it, at compound angles, carrying tens of meganewtons, and it is a casting or a shop-welded assembly rather than a plate and some bolts. The saving is in count and the cost is in kind.

And the fabrication unit has grown by a factor of six. A framed tube’s column is a 4 m piece; a diagrid’s diagonal is 25 m, which is past what most road transport takes and past what most cranes lift in one at height. So the member is spliced — and a splice in the middle of a 25 m strut is a joint the count above did not include, placed at the point of maximum buckling moment.

That last point is where the two sections of this essay meet. The splice wants to be at a node, where the force changes and the member is held; the transport wants it at the middle, where the member is longest and least restrained. Which one wins is decided by a road.

What the picture cannot show #

The stress distribution across a face is drawn at the base, and it changes all the way up. Shear lag is worst where the moment is largest and disappears near the top, so the effective width is a function of height and the single number quoted for it is an average of a varying thing.

Nor does any figure show the reason the form is used as often as it is, which is that it is visible. A diagrid is a structure whose mechanism is legible from the street, and that has been a large part of the argument for building them — a consideration outside this collection’s subject and not outside the decision.

The generalisation #

The habit worth carrying is the question this essay turns on: by what mechanism does one part of a structure tell another part what it is carrying?

The framed tube’s answer is bending in a spandrel, and it is a poor answer — three orders of magnitude below what axial action gives at the same material cost. The diagrid’s answer is a diagonal. A gridshell’s is the same. A plate girder’s web is the same question answered by shear in a continuum, and a Vierendeel is the same question answered badly on purpose because somebody needed the opening.

The general form is that shear transfer is where structures lose their efficiency, because bending is what a structure does when it has no straight line available. Wherever a load path contains a member bending in its own plane to pass a force along, there is an order of magnitude to be had by putting a diagonal there instead — and the reason not to is almost always that something has to pass through.