A shell only if the grid takes shear

Curving a surface buys an enormous amount. A flat plate spends its thickness on a lever arm of a few millimetres; curve the same sheet and the load is carried in the surface itself, at a thirtieth of the material. Every gridshell ever built is an attempt to buy that saving with members instead of with a continuous surface — a lattice following the same shape, with air where the sheet was.

Whether it works comes down to a property that appears on no drawing and in no member schedule.

The third resultant #

A membrane state is three stress resultants, not two. There is a direct force along $x$, a direct force along $y$, and a shear $N_{xy}$ in the plane of the surface.

The first two are carried by members lying in those directions, which any grid has. The third is carried by neither, and there is nothing subtle about why: it is a shear across a quadrilateral, and four bars pinned at their ends resist it with nothing at all. The quadrilateral simply becomes a parallelogram.

That would be a curiosity if $N_{xy}$ were small. It is not, and it is not optional either: on a barrel vault it is the mechanism by which load travels along the barrel to whatever supports it. Under a symmetric load the shear is modest, which is why an unbraced grid can look perfectly satisfactory on a uniform-load check. Under snow on one side, or wind, the shear is the whole load path, and a grid without it is a row of arches that have to carry an asymmetric load in bending.

Which free body produced the number #

Take one cell, of side $s$, and impose a shear flow $q$ on its four edges. Ask what resists it.

Pinned quadrilateral. Nothing. There is no member on the diagonal and the four bars are two-force members. The shear stiffness is exactly zero — not small, zero — and the cell is a mechanism.

One diagonal per cell. The shear is resolved into the diagonal, which carries it axially. Smeared over the cell, this gives $G t = Et/4$: a quarter of the axial stiffness, because the diagonal is at 45° and the resolution costs a $\cos^2$ at each end.

Triangulated. A third member direction, so every panel is a triangle and every load path is axial. $Gt = \tfrac{3}{8}Et$ for an equilateral layout, and the grid is essentially as stiff in shear as it is in tension.

Rigid nodes. No third member. The shear is carried by the four members bending in double curvature over the cell, which is the Vierendeel mechanism. Per cell that is $12EI/s^2$; smeared onto a unit width, which divides the member’s own $EI$ by $s$ as well, it is

$$Gt = \frac{12 E I_{\text{member}}}{s^3}$$

For a grid of 2,400 mm² members at 1.5 m centres, those come to 126,000, 84,000 and 448 kN/m against a continuous sheet of the same stretching stiffness at 129,000. Rigid nodes supply 0.35% of a sheet’s shear stiffness.

The first of those four answers — the zero — is not a stiffness calculation at all. It is a count, and it is available before any member has a section, which is worth seeing on its own terms because it is the same count that decides whether a space frame stands up.

Whether an assembly of bars is a structure or a mechanism is settled by that arithmetic before anything is sized, and the quadrilateral cell is on the wrong side of the line. Everything that follows is about how expensive it is to get back across it.

What that costs, as a displacement #

A stiffness three orders down is easy to nod at and hard to feel, so it is worth converting.

Take a 30 m barrel rising 6 m, under 1.2 kN/m² of asymmetric load. The shear flow carrying that load along the barrel is of order $wS/2 = 18$ kN/m, the shear strain is $q/Gt$, and the racking accumulates over the span. An ordinary serviceability limit of span/250 is 120 mm.

A metre and a fifth, on a thirty-metre span. That is not a serviceability failure, it is a different structure — and the point is that every one of those four grids has the same members, the same weight, the same surface and the same drawing. The only difference is what happens at a crossing.

And the discrepancy widens with span, because the racking is $qS/Gt$ with $q = wS/2$ in it, so it carries the square of the span while the serviceability limit carries only the first power.

Half again the span has taken the rigid-jointed grid from ten times its limit to fifteen times it, on identical members. That is what a mechanism with a small stiffness standing in for it looks like as the structure gets bigger: the deficiency is not a fixed penalty to be absorbed, it grows faster than the thing being checked against.

Why a finer mesh does not rescue it #

The rigid-node stiffness carries $1/s^3$, so halving the spacing multiplies it by eight. That is better than the square instinct suggests, and it is still not enough.

Eight times 0.35% is 2.8%, the racking falls from 1,205 mm to 151, and the limit is 120. So a grid at 750 mm centres — twice as many members, four times as many rigid joints, every one of them a welded or bolted moment connection — is still outside a serviceability limit that a single diagonal per cell cleared by a factor of nineteen.

That comparison is the practical content of this essay. The cheap fix and the expensive fix are not on the same curve. Refining the mesh is a large intervention chasing a small number; adding a diagonal changes which term is in the denominator.

It is also why the historical gridshells that work are triangulated or cable-braced, and why the ones that are quadrilateral with rigid nodes are either very small, very deep in section, or stiffened by their glazing — which is a fourth mechanism, and one that is usually neither drawn nor relied upon in the calculation while being relied upon in practice — the load path nobody put in the model, arriving on the stiffness side rather than the strength side.

The equivalent continuum, and what it is fair to ask of it #

Everything above went through a smeared surface: a lattice replaced by a sheet with the same stiffnesses per unit width. That step is doing a great deal of work and it is worth saying exactly what it assumes.

A member of area $A$ at spacing $s$ contributes $EA/s$ of stretching stiffness per unit width, so the equivalent sheet has $Et = EA/s$ and an equivalent thickness of $A/s$ — here 1.6 mm, on a grid whose members are 80 mm deep. The sheet is two orders thinner than the grid it represents, which is the correct answer for stretching and a badly wrong one for anything else.

That mismatch is the whole reason the shear question has to be asked separately. If the smeared sheet were a real 1.6 mm sheet it would have $G = E/2(1+\nu)$ automatically and none of this essay would exist. It is not a sheet; it is a bookkeeping device that reproduces one stiffness by construction and reproduces the others only if a mechanism has been provided for them.

The same caution applies in the other direction. The grid’s bending stiffness per unit width is $EI_{\text{member}}/s$, which corresponds to an equivalent thickness of about 45 mm — thirty times the stretching one. An equivalent shell with two different thicknesses depending on which question is asked is not a shell; it is three independent numbers wearing a shell’s name, and they have to be tracked as three.

That mismatch also says what a bigger member buys, which is less than it looks. The rigid-node share of a sheet is $12EI/s^2$ over $EA/2.6s$, so it goes as $I/A$ — as the square of the member’s depth and nothing else. Scale the section up by half in each dimension and the area rises by 2.25, the second moment by 5.06, and the share by 2.25.

So more than twice the material, spent on every member of the roof, halves the shortfall and does not close it. The fix is not a bigger member; it is a member in the direction that has none. That is the sentence this essay exists to earn, and the equivalent continuum is where it comes from: a stiffness that is missing because no mechanism supplies it is not recovered by improving the mechanisms that were never the problem.

The rise, which decides whether there is anything to buy #

None of the above matters if there is no membrane state worth having, and whether there is comes down to one ratio.

A shallow surface carries load with large membrane forces and small curvature; a deep one with small forces. The membrane force goes as $wR$ and the radius as $S^2/8f$, so at a rise-to-span ratio of 1:5, as here, the arch force is moderate. Flatten it to 1:20 and the same load produces four times the force in the same members, in a surface whose stability is far worse.

That is the same relationship an air-supported roof has with its rise, and it is worth naming because it puts a floor under the shape. A gridshell has an economic band of rise-to-span — roughly 1:8 to 1:3 — outside which it is either a flat grid pretending to be a shell or a dome with an awkward footprint.

What the diagonal costs, which is the reason anyone hesitates #

If a diagonal is worth nineteen rigid nodes, the obvious question is why anyone builds a quadrilateral gridshell at all. There are three answers and none of them is structural, which is worth stating plainly.

Glazing. A quadrilateral cell takes a flat rectangular pane. A triangular one takes a triangular pane, of which no two on a doubly curved surface are the same, and every vertex is a point where six panes and six members meet. The fabrication cost of a triangulated envelope is not a small premium over a quadrilateral one.

Appearance. A quadrilateral grid reads as a lattice; a triangulated one reads as a solid. On a roof intended to be read as a membrane over a space, the difference is the whole design.

Node count. A triangulated grid has half again as many members and the same number of nodes, but every node now carries six members converging in three dimensions at angles that vary across the surface. A quadrilateral node has four members at ninety degrees, and can be a repeated part.

So the real design choice is rarely “diagonal or rigid node”. It is “diagonal, or rigid node, or a cable cross in the plane of the cell” — which is a diagonal that can be prestressed, weighs almost nothing, and disappears visually, at the cost of only working in tension so that two are needed per cell. That third option is what most of the elegant quadrilateral gridshells actually use, and it is the reason they look unbraced and are not. A cable is the member that solves the problem while looking like nothing: it carries tension only, so a cell needs a pair rather than one, and the prestress in them is what keeps the slack one from going out of action before the taut one has taken up. The next section but one prices that prestress, because it is not free either.

The failure that actually happens, which is not racking #

A gridshell that clears the racking check does not then fail by racking. It fails by buckling, and the shear stiffness is in that answer too.

A shell’s buckling load depends on both its membrane stiffness and its bending stiffness, and a grid smeared into an equivalent shell has both — but the equivalent shell is orthotropic and shear-weak, and a shear-weak shell buckles at a fraction of what the isotropic formula gives. So the same property that decides whether the grid can carry an asymmetric load also decides whether it can carry a symmetric one without going.

Two features of grid buckling deserve naming because they differ from a solid shell’s.

It is often local. A single node can snap through while the rest of the surface stays put, especially on a shallow grid where the members meeting at a node are nearly coplanar. That is the shallow-frame problem at the scale of one joint, and it has nothing to do with the global mode. It is also the counting argument of the third figure arriving in a form the count cannot see, because the members are still all there and the arithmetic is still satisfied.

A gridshell node approaches that state as the surface flattens, which is why the local mode belongs to shallow grids: the count says structure, the geometry says mechanism, and the disagreement is a rank rather than a number of members. The counting rule cannot detect it and neither can a member schedule.

It is savagely imperfection-sensitive, in the way any shell is. Node positions out by a few millimetres — which is what a fabricated grid delivers — remove a large fraction of the classical load, and the knockdown factor is the largest single number in the design. A grid’s imperfections are node positions rather than a surface’s waviness, and they are larger; the shape of the penalty is the same.

The prestress a cable cross needs is a compression the grid pays for #

The cable pair was offered as the elegant answer and its price was left as a sentence. It has a number, and the number lands in the check the essay ends on.

A cell of side $s$ under a shear flow $q$ delivers $qs$ along each edge, and a diagonal at 45° resolving it carries

$$N = \sqrt{2}\,q\,s = 1.414 \times 18 \times 1.5 = 38.2\ \text{kN}$$

One diagonal would take that in compression. A cable cannot, so both cables have to be prestressed to at least 38.2 kN — otherwise the one that should be shortening goes slack before the design load is reached, and the cell is back to a mechanism at exactly the moment it is needed.

Now follow the prestress. Two cables in tension $T_0$, pulling inward along both diagonals, put each edge member into compression of

$$T_0\cos 45° = 0.707 \times 38.2 = 27\ \text{kN}$$

with nothing applied to the structure at all. Against the barrel’s own membrane force — $wR = 1.2 \times 18.75 = 22.5$ kN/m, which at 1.5 m spacing is 33.8 kN per member — that is an 80 per cent increase in the compression every member carries.

And compression is what the grid fails by. The last section of this essay says the real failure is buckling, that a grid’s buckling is imperfection-sensitive and often local at a node — and the bracing that solved the shear problem has added four fifths again to the force driving it.

So the cable cross is not free and it is not nearly free. It is a trade of a shear mechanism for a compression, made at a rate the drawing does not show, and it is the reason a cable-braced quadrilateral gridshell often ends up with members no lighter than the triangulated one it was chosen instead of.

The edge disturbance is smaller than one cell #

The essay has said the smearing is an approximation “with a length in it” and left the length unstated. It is worth computing, because for a grid it comes out on the wrong side of the mesh.

A shell’s bending boundary layer decays over roughly $\sqrt{Rt}$. The barrel’s radius is $R = S^2/8f = 900/48 = 18.75$ m, and the grid’s equivalent thickness for bending is 45 mm, so

$$\ell = \sqrt{18.75 \times 0.045} = 0.92\ \text{m}$$

against a member spacing of 1.5 m. The disturbance dies out inside a single cell.

Which means the equivalent continuum cannot represent the edge at all. A smeared shell would predict a bending field decaying over 0.92 m and hand back a moment distribution across a region the grid has no members in — a continuum answer about a length shorter than the discretisation it stands for is not an approximation, it is a category error.

Two consequences follow, and both are ordinary practice arrived at from the mechanism rather than from habit.

The edge is analysed as members, always. However good the smeared model is in the interior, the last row or two of cells has to be a frame analysis, because that is where the surface’s actions turn into a beam’s and the length scale of the turn is below the mesh.

And the edge member is a real member. The membrane shear the whole essay is about has to be delivered into something at the boundary — an edge beam, a tie, an arch, a foundation — and whatever that is carries the accumulated shear along its own length as an axial force. On this barrel, 18 kN/m accumulated over 15 m of half-span is 270 kN in the edge, in a member drawn as a trim.

The same statement in the other direction is what the timber gridshells rely on. A boundary that cannot take the shear leaves the surface to carry it in bending over 0.92 m — which is why an unrestrained free edge on a gridshell curls, and why every one of them that stands has something substantial running round it.

Where this model stops #

The smearing is an approximation with a length in it. Replacing a lattice by an equivalent continuum is only defensible when the thing being carried varies slowly compared with the cell size. Near a support, a boundary, or a concentrated load, the grid has to be analysed as members — and those are exactly the places a shell has its edge disturbance anyway.

The cladding is ignored. Glass in a stiff gasket, a plywood skin, a taut membrane — all of these carry in-plane shear, and some real quadrilateral gridshells rely on it heavily. It is left out here because it is usually left out of the calculation while being present in the building, which is a gap worth naming rather than closing quietly.

The nodes are assumed to be either pins or rigid. Real ones are neither, and a gridshell node that is nominally rigid at a fraction of a member’s stiffness supplies a fraction of the already-small number above. Semi-rigid is the honest description of nearly every gridshell connection ever made.

What the picture cannot show #

The cell drawing shows a shear and a racking. It does not show that the shear is carried round the surface — a gridshell’s shear field circulates, and the racking of any one cell is compatible with its neighbours’ only because the whole field is. That compatibility is what makes the smeared continuum a fair model in the interior and what breaks at every free edge.

Nor does it show the timber gridshells, where the whole argument is turned inside out. A lath grid bent into shape on site starts as a flat pinned quadrilateral mat — deliberately, because that is the only state it can be laid out and pushed into shape in — and is braced afterwards, once curved, with a third layer or with cables. The mechanism is not a defect in that construction; it is the erection method, and the shell exists only from the moment it is taken away. That is an erection-stage argument in which the erection state is not an unfortunate intermediate but the reason the geometry is achievable at all.

The generalisation #

The habit worth carrying is to ask, of any assembly meant to replace a continuum: which of the continuum’s actions has a path, and which has none?

A grid replacing a plate carries bending in two directions and does not carry the twisting moment unless its members can be twisted. A truss replacing a beam carries the moment as a chord couple and carries the shear only through its diagonals — which is why a Vierendeel needs rigid joints and a Pratt does not. A grid replacing a shell carries two membrane forces and needs a mechanism invented for the third.

In every case the missing action is the one that has no member lying along it, and in every case the fix is a member in that direction rather than a bigger member in the ones already there. The question is cheap to ask and it is answered before anything is sized.