The roof that carries everything as shear

The surface that carries by being curved got shell action from a dome: a curved surface has no lever arm, so it needs no bending, and thin shells become possible. A dome has two curvatures of the same sign and it has the classic set of forces: compression along the meridians, and hoop forces that change sign at a particular latitude. Two curvatures of opposite sign took a surface of the other kind, a cable net, where the curvature in one family hangs and the other holds it down, and the whole thing works only because it has been pulled into shape by tension.

A surface with opposite curvatures made of a material that can take compression as well as tension is something else again. The simplest of those is the hyperbolic paraboloid, $z = cxy$. Félix Candela built a series of them in concrete in Mexico through the 1950s, down to 42 mm thick across more than 40 metres in the restaurant at Xochimilco. They seem at once exotic and simple, and the reason is that they can be drawn with a ruler. The whole of the surface is made of straight lines, in two families, so it can be cast on straight planks. It was natural to assume that the straight lines had something to do with how it stands up.

They have nothing to do with it. What the surface actually does is one of the plainest results in shell theory, and it can be had from one equation.

A membrane equation with a single term left in it #

A membrane under a vertical load $q$ per unit of plan has three forces per unit length: $N_x$, $N_y$ and $N_{xy}$. For a shallow surface $z(x, y)$, vertical equilibrium says

$$N_x \frac{\partial^2 z}{\partial x^2} + 2N_{xy}\frac{\partial^2 z}{\partial x\,\partial y} + N_y\frac{\partial^2 z}{\partial y^2} = -q.$$

That is the membrane equation, and it just says that a force on a curved line has a component normal to the line. For $z = cxy$, the second derivatives along $x$ and along $y$ are both zero, as the surface is straight in those directions, and the mixed derivative is $c$. The first and third terms disappear. What remains is

$$2c\,N_{xy} = -q, \qquad\text{so}\qquad N_{xy} = -\frac{q}{2c} = -\frac{q\,ab}{2f},$$

where $f$ is the rise at a corner of a plan with half-sides $a$ and $b$, and $c = f/ab$.

And the horizontal equilibrium is satisfied with $N_x = N_y = 0$, along with a constant $N_{xy}$. So on a 20 m square with corners 3 m above and below the centre, $c = 0.03$ per metre, under 2 kPa, the shear is $2/0.06 = 33.3$ kN/m. It is the same at every point of the surface, with no reference to the position, and it requires no calculation beyond dividing two numbers. For an 80 mm sheet that is 0.42 MPa.

The picture is where this becomes intuitive. A shear on a square element is a tension along one diagonal and an equal compression along the other, as anyone who has drawn Mohr’s circle knows. And on this surface, the diagonals are not straight. In coordinates turned 45 degrees, $z = c(u^2 - v^2)/2$, and it is a parabola hanging along one diagonal and a parabola arching along the other. So at every point of the surface, a cable and an arch cross at right angles, carrying equal force. Vertical equilibrium for each is force times curvature, $N_T c$ from the cable and $N_C c$ from the arch, and they sum to $q$ when each takes half.

The surface is two sets of parabolas, half hanging and half arching, with equal forces in the two. The straight lines come into it only because a saddle has this particular property, and they carry no force whatsoever in the membrane state. Their job is to make the formwork cheap. Candela himself was entirely clear about this, and he designed with that one number and a slide rule. The geometry has done the rest of the work, as it did for the shape that carries itself.

One line of algebra against a real structure #

Membrane theory has the reputation of being approximate, so it’s worth asking how far to trust a single line. Here the check is a lattice: nodes on the surface, twelve cells a side, with bars along the two straight directions and across both diagonals in the proportions that match a thin sheet of Poisson’s ratio one third, and a bending stiffness across each set of three nodes in line, so the sheet can bend if it wishes. Edge members run around the boundary and the four corners are held.

The agreement at the centre is within 2 per cent across a range of rises from 1 m to 8 m. That is a strong result, as the lattice has been told nothing about shear, about diagonals or about hyperbolas. The second lesson from the figure is the reciprocal: halve the rise and the shear doubles. A shallow hypar is elegant and it pays in stress, and in this regard it behaves exactly as a shallow arch does: the hinge put in on purpose has an arch’s thrust going as one over its rise for the same reason, a curvature that has been divided into a load.

How large a force is this? Take the same 80 mm of sheet and lay it flat on the same 20 m square, simply supported.

A factor of 86. The plate’s bending stress goes as one over the square of its thickness and the shell’s as one over the thickness, so the ratio is larger the thinner the sheet. An 80 mm concrete flat plate would crack under its own weight at that span, with a stress of 36 MPa against a tensile strength of three or four. A twisted one is at less than half an MPa, and it doesn’t care about the tensile strength as long as the concrete has a modest mesh along the tension diagonal. That is what a metre or so of corner displacement does. The sheet itself is not any stronger, and the entire benefit is in how the load finds the supports.

Uniform, except where it has to arrive #

A uniform shear on a finite sheet has to go somewhere when it reaches the edge. Each edge’s shear flow pushes along it, and that’s the only thing the sheet can give the edge, since a membrane has no force normal to itself. So an edge member receives a constant force per unit length along its axis, and its axial force builds up linearly. Which way that happens depends on where it is held.

The dashed lines are the membrane theory’s prediction of these forces: $N_{xy}$ times the distance along the edge. With two low corners held and the two high corners free in plan, which is the classic gabled umbrella, the force starts at nothing at the high corner and collects to $N_{xy} \times 2a = 667$ kN at the low corner, where the whole of it has to be delivered. With all four corners held, the same shear divides at the mid-point, so the edge is in tension along one half and compression along the other, with nothing at its centre. Those are two entirely different sets of edge members for exactly the same sheet and the same stress. The lattice’s values at its end cells are close to the line, a little lower. Its end cell’s centre is 0.83 m in from the corner, where membrane theory reads 306 kN and 611 kN, and the lattice is about 8 per cent below that. Along the two-corner edge it falls further under the line, and that is something the one-line theory has no way of saying. Just past mid-length the theory wants 361 kN, and the edge member carries 243. The rest, 92 kN, is in the three strips of sheet beside it, which are being stretched along with it and act as its flange. Make the edge member eight times heavier and it takes 342 of the total, and sixty times heavier and it takes all of it. Membrane theory has the sheet carrying shear and nothing else, which is right over most of the surface and wrong in the strip that lies against a light edge.

An edge member in compression with a force of 667 kN over 20 m is a strut with a length to match. It is what limits the roof, as the sheet is hardly stressed. A good part of the structure’s real engineering is in those edge beams, and they need to be either very heavy or held along their length. It’s also why a pair of these, back to back, leaning on each other, or four as an umbrella on a central column, are so favoured: the edge forces of adjoining units balance one another, and only their difference remains. A shell only if the grid takes shear made the corresponding point about gridshells, where shear in the surface is a thing that has to be supplied. Here, the shear isn’t a side effect but the whole of the load path, so a hypar built as a lattice of straight timbers along the straight lines, with no diagonal boarding, would not work at all. The timber hypars that have been built all have layers of boarding laid across the diagonals.

And the sheet’s uniform shear isn’t quite uniform. Over most of the surface it is within a few per cent of the membrane figure, and it changes rapidly in the outer cells, where it has to deliver its force at a point. It climbs to more than one and a half times the membrane value next to the corners, at which the whole of the structure’s load gets concentrated into a support, and real shells are usually thickened there. The whole price is at the edge, the dome essay said, and in this case too the bending and the disturbance are concentrated there.

What the one equation does not contain #

Everything so far has been stress. Consider how to find a deflection. Membrane strains from the stresses are small. To work out the shape’s movement, the strains would have to be integrated with the geometry, and here the surface brings a surprise.

Suppose the deflection $w$ of the surface is any function of the form $g(x) + h(y)$, that is, the sheet’s straight lines are moved each to a new height while remaining straight. The strains that this causes in a membrane are in shear, and the shear strain can be cancelled by in-plane displacement, $u(y)$ and $v(x)$, along those lines’ own directions, with no strain along any of them. So a family of possible movements exists that strains the membrane not at all. Mathematically, the boundary of this surface lies along the characteristic lines of the membrane equation, which is exactly where boundary conditions don’t determine a solution. When a lattice with no bending was tried for this roof, the solver reported it as a mechanism, and that’s an accurate report. A hyperbolic paraboloid held by its edges is a statically sufficient structure with a shape that is only weakly determined. The shape is held by the small bending stiffness of the sheet and by the edge members, in both axial and bending stiffness.

This figure is the unexpected finding about these roofs. The shell stress is settled by the rise, and the deflection is settled mainly by the edge members. With a tenfold change in deflection across the range of edge areas, the sheet’s shear hardly moves. A designer who has verified the stress with the one-line calculation and looked no further has learned nothing about the roof’s stiffness. That is compounded by a lower rise, which increases both the shear and the edge force and magnifies the sag: a third of the rise gives between four and ten times the deflection.

In the early days of building hypars, this was underestimated. Several shallow concrete roofs on large spans deflected far more than expected over the years, as the creep of the edge members worked on that weak membrane stiffness, and the problem has been written about extensively, by Schnobrich among others, in reviews of how far membrane analysis can be trusted for shallow forms. Rules of thumb grew up that set a minimum rise against the span and asked for edge members to be propped or tied. Those limits read like caution, and this figure is what they’re really saying, even if the theory they were born from couldn’t show it. On this roof, a rise-to-half-span ratio of 3 to 10 gives edges with stiffness to spare, and a 1 in 10 version has a deflection five times as large, which then creeps, as the deflection that arrives three years late would predict.

There’s a further limit at low rise. The compression diagonal is an arch made of an 80 mm sheet, and it can buckle. The critical force, read as a cylinder of radius $1/c$, is

$$N_{cr} = \frac{E\,t^2\,c}{\sqrt{3(1 - \nu^2)}},$$

Reissner’s form, with a value of 2,830 kN/m for this roof against 33 kN/m applied, a safety factor of 85. A factor of 85 looks far beyond any need, but the rise enters twice, once in the applied shear and once in the critical force, so the factor goes with the square of the rise: it’s 9 at 1 m of rise. Then a classical critical load for a thin shell needs a large knockdown, and creep in concrete takes the modulus down by a further factor of three. Over a thin shallow roof the margin can vanish. That’s another argument for a rise and a thickness that the stress calculation alone never asks for.

How the sheet is reinforced follows from all this #

The steel in such a sheet has an unusual brief. Tension runs along one diagonal only, at 33 kN/m, and with a concrete sheet that is 0.42 MPa, well under what the concrete itself could carry if it were not for shrinkage and creep. So the mesh has less to do with strength than with keeping the tension diagonal from opening, and Candela’s thin shells had very little in them. Mesh laid along the straight lines, which is much easier on the planks, is at 45 degrees to the tension and carries it inefficiently, by the same geometry as the angle in a beam’s links. Each of the two directions then has to carry the whole of what a single layer along the diagonal would, so it takes twice the steel, and that is usually accepted for the sake of the planks.

Where this lattice is not a shell #

The analysis is linear. At large deflection, the shape has changed, with the shear going up as the curvature falls. A shallow shell loaded close to its limit has a softening response, in the manner of the arch that turns aside before it snaps. The figures here are all first-order.

The lattice stands in for a sheet. Bars and diagonals in that pattern match a sheet at one Poisson’s ratio, and the bending stiffness across a line of nodes ignores a plate’s twisting term. It is good enough to recover the shear to 2 per cent and to show that the edges control the sag. It isn’t good enough to give the bending moments near a corner, which matter for reinforcement, and a shell-element analysis would be needed.

The material is uniform and uncracked. Concrete in tension along a diagonal is cracked, or will be. Its mesh then controls the stiffness across one diagonal only, and the arch and cable’s agreement of equal force no longer implies equal stiffness. In reality, a cracked hypar carries more through its compression diagonal than through the tension one.

A uniform load on plan. Snow does approximate this, but self-weight is on the surface and its projected value differs slightly from point to point. Wind and asymmetric snow are not uniform, and unlike the dome, this surface has to carry an unbalanced load by bending or by a much less even membrane state. A roof that can’t resist such a load in membrane action needs bending strength it may not have.

The edges are straight bars on a corner support. In a real roof the edge member has an eccentricity from the middle of the sheet, and a real umbrella has a column at the low corner, where a thrust has to go somewhere. That thrust goes through a tie or into the ground, and the thrust that never reaches the ground is the reminder that a tie is the last part of the load path, and a part that has to be there.

Still open: four units and one column #

An inverted umbrella is four hypars sharing a central column, whose interior edges are all compression members leaning against each other in balance, and whose outer edges have a much smaller force. Its balance is exact under uniform load and it disappears as soon as one quadrant has snow and its neighbour doesn’t. An unbalanced umbrella puts a moment on its column and on the valley members. It’s how these roofs have mostly failed, and how much the sheet’s shear can redistribute around the four units, and how much is then thrust on a single column’s bending, is what a single square with its four corners held can’t say. That will take the same lattice with a re-entrant boundary, and a model of the column’s stiffness to go with it.