Structural form

The columns that lean

A framed tube carries its wind load by bending the spandrel beams between its columns, and it does it badly — the corner columns take nearly six times what the middle ones do. Tilt the columns instead, so the perimeter is triangulated, and the same shear is carried axially. The concentration falls to 1.23 and the tube recovers most of the stiffness the plan said it had.

Assumes The corner columns take more than their share, How a tall building stands still and The angle that uses half of itself.

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.

The corner columns take what the middle ones did notAxial stress in the columns across one flange face of a 36 by 36 m framed tube, at the base. Plane sections says the flat line: every column on the face at the same distance from the neutral axis, therefore at the same stress. The solved distribution is the curve — 78.5 N/mm² at the corner against 13.4 in the middle, a ratio of 5.88. The middle columns lag because the only route the axial force has into them is the in-plane shear of the spandrel frame, bay by bay from the corner. The face is carrying its resultant on an effective width of 45 per cent, and the tube deflects as though its second moment were 64 per cent of the gross.-15-10-551015020406080across the face (m from the centre)column stress (N/mm²)13 columns at 3.0 mcorner 78.513.4 in the middleplane sectionseffective width 45% · stiffness 64% of gross
Fig. 1 Axial stress in the columns across one face of a framed tube at its base. Plane sections says the flat line; the solved distribution is the curve, at 78.5 N/mm² in the corner against 13.4 in the middle.

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.

The number that decides it

Shear lag is a property of the spandrels, not of the planThe corner column's overstress and the tube's stiffness, against the racking stiffness of one bay of the perimeter frame. At the frame drawn — 3.0 m bays, a 3.8 m storey — the corner carries 1.78 times what plane sections predicts, the middle of the face carries 0.18 of the corner, and the tube deflects as though its second moment were 63 per cent of the gross. Ten times the racking stiffness — which is what a diagonal across the face buys, replacing bending with axial action — takes the concentration to 1.23 and the efficiency to 92 per cent. That is the braced tube, and the argument for it is on this axis rather than in the plan.0.1×0.3×10×00.511.522.5racking stiffness of the perimeter framecorner ÷ plane sectionscorner overstressstiffness kept1.78 as builtspandrel I = 0.70 × 10⁹ mm⁴ · shear-lag length 88 m against a 220 m height
Fig. 2 The corner overstress and the tube’s efficiency, against the racking stiffness of one bay of the perimeter frame. Ten times the racking stiffness takes the concentration from 1.78 to 1.23 and the efficiency from 63% to 92%.

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.

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.

Four ways to make a cell resist being racked, and one that is not oneOne cell of a grid shell under the membrane shear it has to carry, by the four mechanisms available for carrying it, with the racking each produces over a 36 m span under 1.2 kN/m² of asymmetric load. A serviceability limit of span/250 is 144 mm. Four pin-jointed bars in a quadrilateral have **no** in-plane shear stiffness whatever — the cell folds, and the answer is not a large deflection but a mechanism. Rigid nodes carry the shear by bending the members over a cell, which smears to 12EI/s³ and comes to 0.35% of what a continuous sheet of the same stretching stiffness gives: 1736 mm, ten times the limit. One diagonal per cell, or a third member direction, carries it axially instead and lands within a factor of two of the sheet. That is the whole difference between a grid shell and a row of arches.pinneda mechanismno shear stiffness at allrigid1736 mmGt = 0.35% of a sheet'sbraced9.3 mmGt = 65% of a sheet'striangulated6.2 mmGt = 98% of a sheet's
Fig. 3 Four ways to make a cell resist being racked, and one that is not one. Four pin-jointed bars in a quadrilateral have no in-plane shear stiffness whatever — the cell folds — while rigid nodes supply 0.35% of a sheet’s and one diagonal lands within a factor of two of it.

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.

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.

One drift, two motions, opposite curvaturesThe sideways movement of a 160 m building under a uniform wind, drawn as the sum of the two mechanisms that produce it. The bending curve is a cantilever's: flat at the base, steepening upward, concave one way. The racking curve is a stack of parallelograms: steepest at the base and flattening, concave the other. They add to 1278 mm at the roof, of which 73% is bending. The one group that decides the split is αH = H√(GA/EI) = 3.31: below one the building is a cantilever and above about six it is a frame, and everything interesting is in between.0200400600800100012001400020406080100120140160sideways movement (mm)height (m)bendingracking1278 mmroof drift 1 in 125 · worst storey 1 in 111
Fig. 4 One drift split into the two motions that produce it. The bending part is flat at the base and steepens upward; the racking part is steepest at the base. Their sum is 1278 mm at the roof, of which 73% is bending.

The same split appears in the drift, and the number that decides it is αH=HGA/EI\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 GAGA by an order of magnitude without touching EIEI, 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.

Two shapes that are the wrong way up for each otherDeflected shapes of a 40-storey building under a uniform wind, drawn to the same scale. The wall alone bends: its shape is flattest at the base and steepest at the top, reaching 3139 mm. The frame alone shears: it is steepest at the base where the storey shear is largest, reaching 547 mm. Tied together at every floor they reach 367 mm — less than a quarter of either, and less than the 466 mm two springs in parallel would give, because each is stiff exactly where the other is not.wall 3139 mmframe 547 mmtogether 367 mm40 storeys at 3.9 m · 40 kN per floor
Fig. 5 Deflected shapes of a tall building under a uniform wind. The wall alone bends and reaches 3139 mm; the frame alone shears and reaches 547; tied together at every floor they reach 367 — less than either and less than two springs in parallel would give.

Two springs in parallel would give 466 mm; the real pair gives 367. 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 the interaction. A diagrid does the same thing inside a single system, because a triangulated perimeter is simultaneously a cantilever with a large EIEI and a truss with a large GAGA.

Chord force against truss depthThe force in a truss chord for a fixed bending moment, against the depth of the truss. The relationship is a reciprocal: the chords form a couple whose lever arm is the depth, so a shallow truss pays for it steeply.0.511.5205001000150020002500depth of the truss1333800533381the same moment, resisted by a longer lever arm
Fig. 6 The reciprocal that makes the whole perimeter worth using. The depth that matters here is the plan dimension of the building, and it is the largest depth any structure in a building ever gets.

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 θ\theta to the horizontal. Its force resolves into NcosθN\cos\theta horizontally and NsinθN\sin\theta vertically, so a module of the perimeter carrying a storey shear VV and a moment MM needs

NV=Vncosθ,NM=McdsinθN_V = \frac{V}{n\cos\theta}, \qquad N_M = \frac{M}{c\, d\, \sin\theta}

with nn the number of diagonals sharing the shear on a face, dd the plan depth and cc a constant of the arrangement. The material required is proportional to the force times the member’s length, and the length goes as 1/sinθ1/\sin\theta for a module of fixed height. Multiplying through, the shear term’s cost goes as 1/(sinθcosθ)1/(\sin\theta\cos\theta) and the moment term’s as 1/sin2θ1/\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.

Take that one away and there is no structureA 10-panel warren truss under 20 kN at each top node, before and after member 4 is removed. What is left is a mechanism: the assembled stiffness matrix is singular, and no set of member forces holds the load in any position. Nothing about the strength of the remaining members enters the answer.intactmember 4 removeda mechanismrank deficient: the frame folds
Fig. 7 A truss before and after one member is removed. What is left is a mechanism — the stiffness matrix is singular, and no set of member forces holds the load in any position.

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.

Two null spaces of one matrix, and the count is their differenceTwo pin-jointed frames, each with the forces it can carry with nothing applied to it drawn on its bars — tension one colour, compression the other, thickness in proportion. That force set is the null space of the equilibrium matrix; a mechanism is the null space of its transpose; and Maxwell's count b + r − 2j is the difference of their dimensions and knows neither of them separately. A square with both diagonals has s = 1 and m = 0. Two bars in a straight line has s = 1 and m = 1 with a count of 0, so the count is satisfied by a frame that both folds and can be prestressed — and the prestress stiffness is positive, which is why a tensioned pair of collinear bars is stiff at all.a square with both diagonalss = 1 · m = 0Maxwell's count: +1nothing to stiffentwo bars in a straight lines = 1 · m = 1Maxwell's count: +0prestress stiffness +1.000
Fig. 8 The two null spaces of one equilibrium matrix. A frame can have a set of forces it carries with nothing applied, a motion that costs no member any length, or both — and the member count is only the difference between the two.

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.

Two centres, and the distance between them is a torqueA storey 36 by 22 m with its walls drawn heavy, pushed in one direction by 1600 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 = 15.0 m — and the distance between the two is an eccentricity of 3.00 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 wall is asked for 25% more than its direct share, and the walls at right angles to the push carry 95 kN each with nothing applied along them at all.centre of masscentre of rigiditye = 3.00 m1600 kNwest wall800 kN direct− 50 torsional= 750 kNeast wall800 kN direct+ 199 torsional= 999 kNsouth wall0 kN direct− 95 torsional= 95 kNnorth wall0 kN direct+ 95 torsional= 95 kNcentre of rigidity at x = 15.00 m, y = 9.00 mtorsional radius r = 17.02 m against a plan radius of 12.18 mtorsionally stiff by the usual criterion
Fig. 9 A storey pushed in one direction, with the centre of mass and the centre of rigidity marked. The distance between them is a torque, and the walls at right angles to the push carry 95 kN each with nothing applied along them.

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.

Every path to the ground goes through the linkA braced bay 8 m by 4 m whose two diagonals stop 800 mm apart instead of meeting. The storey shear reaches the ground through the diagonals, and the vertical components they deliver to the beam have to pass through the segment between them: the link carries 50% of the applied shear as a shear force, at a lever arm short enough that its ends reach 0 kNm while the rest of the beam carries 0. The deflected shape drawn is the solved one, magnified — the real drift under this load is 0.009 mm. Everything outside the link is designed to stay elastic while the link is yielding, which is what makes the mechanism a choice rather than a hope.storey shearthe link — 800 mm0.71 kN in each diagonal per kN of shearlink shear 0.50 kN · storey drift 0.009 mm · deflection magnified 5067×
Fig. 10 Where a diagonal’s force goes at the end of it. The path to the ground runs through the segment between the diagonals, and it carries half the applied shear at a very short lever arm.

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.

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.

Named alongside this one

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

The objects this essay names

Each one links to every other essay that touches it.

Axial forceBracingDiagridDriftEffective widthFramed tubeLateral systemLoad pathMechanismPlan torsionRackingRobustnessShear lagStiffnessTruss