1 What is a shell structure and how do they work? Classification of structural shapes Structures can be classified in many ways according to their shape, their function and the materials from which they are made. A structure or structural element may be a fully threedimensional solid object like a ball bearing, or it might have some dimensions notable smaller than others. If it is largely straight and one dimensional then we might call it a beam, a girder, a column, or possibly a strut or tie if it is designed to work particularly in compression or tension. We may be very concerned with the cross-sectional shape of these elements, solid rectangle, I section, circular tube and so on, but the size of the cross-section is small compared to the length. There is no absolute rule as how small is small, except that some mathematical formulae only work well if the span to depth ratio is big enough. If an element is curved and one-dimensional then it might be an arch or the parabolic cable of a suspension bridge. Arches and suspension cables rely on curvature for the thrust or tension in the element to resist lateral loads. Both straight and curved structural elements are much more efficient at carrying axial forces than shear forces and bending moments. But compressive axial forces may cause the object to element deflect sideways and buckle. It is possible to build up beams or arches from assemblies of smaller one-dimensional objects, to produce a truss or a truss arch. A flat two-dimensional object might be described as a plate or slab or wall, depending upon the job it is called to do. Again there is no absolute rule as to how thin the object has to be before you think of it as primarily two-dimensional. Plates are very efficient at carrying inplane loading, for example when acting as a wall, but less efficient when carrying lateral loading as a floor slab. A curved two-dimensional structure is a shell. If a beam is defined by a straight line, an arch by a curved line and a plate by a plane, then a shell is defined by a curved surface. Again there is no absolute rule as to how thin a structure has to be before it is called a shell. The relationship between a shell and a plate is similar to that between an arch and a beam. In each case the curved object can use its curvature to resist a load perpendicular to its axis or surface by tensions or compressions parallel to its axis or surface. Shell structures include birds eggs, concrete shells, pressure vessels, ships hulls, sails, balloons, inflatable boats, car tyres, monocoque car bodies, aircraft fuselages, masonry vaults and the skull. It may seem odd to describe a balloon as a shell, but it certainly corresponds to a structure formed from a thin curved surface. The balloon fabric is so thin that it will immediately buckle in compression and can therefore only carry tension forces. Masonry structures can only work in compression
2 with the line of thrust within the cross-section. Reinforced concrete shells and steel car bodies can work in both tension and compression, and may also have appreciable bending stiffness. Just as a beam can be made from an assemblage of straight elements in the form of a truss, a shell can be made from linear elements joined together. Examples include grid, lattice and reticulated (meaning net-like ) shells, geodesic domes, cable nets, fishing nets and spiders webs. The words grid, lattice and reticulated all mean very much the same thing, although some authors like to use one or other for particular types of shell. Usually the elements that make up the shell are all in one layer, but sometimes they are in two or more layers. There is no clear distinction between structural forms. How wide does a beam have to be before it becomes a slab? A wide arch is a cylindrical shell. A cylinder is a surface formed by moving a curve along a straight line and the most common cylinder is the circular cylinder (often just called a cylinder ), but a cylinder can have any shape. This relationship between arches and shells makes it worth thinking as much as possible about arches before moving onto shells with more complicated geometries. There is some mathematics in the following section, but it does not have to be followed in detail to get the gist of the argument. Funicular arches The word funicular comes from the Latin for a slender rope. If a rope carries a uniform vertical load per unit horizontal length, then it will automatically hang in the shape of a parabola. The parabola is the funicular shape corresponding to a uniform load per unit horizontal length and the uniform load per unit horizontal length is the funicular load corresponding to a parabolic cable. A suspension bridge deck is roughly horizontal and therefore, in the absence of traffic, it applies a uniform load per unit horizontal length to the suspension cable via the hangers. The own weight of the cable itself is uniform per unit arc length along the cable, rather than per unit horizontal length. This is usually neglected in the case of a suspension bridge because the deck is so much heavier than the cable, but if a cable is hanging under its own weight it will hang in a catenary rather than a parabola. The work catenary comes from the Latin for a chain. Some people use Figure 1. Parabola and catenary (upper the words funicular and catenary curve) interchangeably, but others prefer to reserve the word catenary for the particular case of the rope or cable hanging under its own weight. If a cable is only carrying vertical loads then the horizontal component of tension in the cable, H = T cosλ = constant.
3 Here T is the tension in the cable and λ is the slope. The vertical component of tension, V = T sinλ = Htanλ = H dy dx. Thus in the case of a uniform load per unit horizontal length, w, vertical equilibrium gives This can be integrated twice to give w = dv dx = H d 2 y dx 2. y = x2 2c. where c = w H which has the units of length. Thus we have a parabola. We don t need to worry about the constants of integration, they just move the curve around. If the loading, w, is constant per unit arc length, s, then w = dv ds = dx dv ds dx This can be integrated to give dv = cosλ dx = 1 dv 1+ tan 2 λ dx = 1 1 + dy dx dy dx = sinh x c. 2 H d 2 y dx 2. Again c = w H and we have left out the constant of integration because it just moves the curve sideways. Integrating again, y c = cosh x c 1 in which the constant of integration is chosen so that the curve goes through the origin. This is the upper curve in figure 1, while the lower curve is the parabola It can be seen that the two curves are identical when their slope is low and they only peal apart when the load per unit horizontal length on the catenary increases with slope. The function cosh is the hyperbolic cosine, coshθ = eθ + e θ, in which e = 2.718... is 2 Euler s number.
4 The parabola and catenary have a number of different and sometimes interesting properties. The catenary is one of the few curves where there is a simple relationship between x and y and the arc length along the curve, s, starting from the bottom, s c = sinh x c = 2 y c + 1 1. The tension in the catenary is T = H cosh x c It is relatively easy to find the funicular load for a given shape of cable or funicular shape for a given load, either by doing a simple physical experiment or mathematically. Having found the shape it can be inverted, or turned upside down, to find the best shape for the equivalent compression structure or arch. If an arch is carrying a funicular load, there will be no bending moment in it, which is equivalent to saying that the line of thrust is along the axis of the arch. If a non-funicular load is added, it will produce bending moments and cause a deviation of the line of thrust, possibly causing the collapse of a masonry arch.. The concept of funicular load applies particularly to structures that have to carry one dominant load case, perhaps their own weight or some permanent load due to water or soil. The arch at Torcello (figure 2) has to carry the extra of the masonry supporting the steps. This means that more curvature is required towards the supports than mid-span. Figure 2. Torcello, Henri Cartier-Bresson 1956 Even if the arch is only supporting its own weight, it may vary along its length. Close examination of the photograph of the Taq-i Kisra (figure 3) shows that the arch (or cylindrical shell) is thinner at the Figure 3. Taq-i Kisra or the Great arch of Ctesiphon, Iraq
5 top. This means that the funicular shape is no longer the catenary and the curvature at the top needs to be reduced to take into account the reduced loading. Let us suppose that we want to make a circular cylindrical shell of varying thickness so its own weight is funicular. If t is the thickness of the arch, R is its radius and is its weight per unit volume, so that t = t = 1 R H Rcos 2 λ dv dλ = H R d dλ tanλ ( ) = H R sec2 λ. in which H R is a constant with the units of length. Note that H is now a compression, and also has unit of force per unit width. Figure 4 shows how the shell gets thicker as it approaches the vertical. The stress in the arch, H σ = t cosλ = Rcosλ, Figure 4. Funicular circular arch which reduces away from the top, the reverse of what happens with a catenary. It seems arbitrary to choose a cylindrical shell of uniform thickness (which has the catenary as the funicular shape) or to choose a circular shape. Instead one might say that the compressive stress, σ, should be constant. If that is the case, Figure 5. Constant stress arch t = dv ds = H d ( ds tanλ) = H sec2 λ dλ ds. and H = σt cosλ.
6 Thus These can be integrated to give σ s = log cosλ e 1 sinλ σ x = λ σ y = log e ( cosλ) t = t 0 cosλ where t 0 is the thickness at the top of the arch. Figure 5 shows a constant stress arch or cylindrical shell. Let us now think about scale. If the arch is made from a weak masonry we might have = 25 10 3 N/m 3 ds σ dλ = 1 cosλ = cosλ 1 sin λ sinλ cosλ dx σ dλ = 1. ds σ dλ = tan λ and σ = 0.5MPa = 0.5 10 6 N/m 2. Therefore σ = 20m which means that if we decided to use the part of the arch in figure 5 between -1.25 and +1.25 on the horizontal axis, the span would be 50m. If we used concrete at a stress of 25MPa, the corresponding span will be 1250m, or 1.25kilometres. Figure 6 shows a comparison of the catenary, circular and constant stress arches. They all have the same thickness and curvature (and therefore stress) and at the top. However the catenary will have stress increased by a factor of 2.5 at the supports. So the catenary is not a particularly good shape for an arch or cylindrical shell, unless practical considerations mean that it has to have a constant thickness. Figure 6. Three arches, circular (top), constant stress (middle) and catenary (bottom)
7 Membrane stresses in shells What does this mean for shells? The answer is, possibly, nothing, because shells are potentially so much more efficient than arches or they can be if everything is as it should be, which it rarely is. Consider a cable hanging under some combination of loads (possibly horizontal as well as vertical). If it lies in a plane, we have two equations of equilibrium, resolving horizontally and vertically. There is only one force unknown, the tension in the cable. Thus we have effectively one equation left over to find the shape, and this is what we did in the previous section. The same applies to a cable in three dimensions, now we have three equations of equilibrium and therefore two equations to determine the shape. The equivalent to axial tension or compression in an arch are membrane stresses in a shell. Membrane stresses correspond to forces that are tangent to the surface defining the shell. They are usually specified as a force per unit width, so that to find the stress in the material in MPa (that is N/mm 2 ) one has to divide the membrane stress by the thickness, as we did for the constant stress cylindrical shell. For shells with ribs, grid shells or fabrics more complicated calculations are necessary to convert membrane stress to material stress. The ribs may only work in one direction and for a fabric the tension per yarn is the membrane stress divided by the number of yarns per unit width. Shells may also experience bending and twisting moments and corresponding shear forces perpendicular to the surface. However shells (including grid shells) are much stronger and stiffer under membrane action than under bending and therefore shells should always be designed to work as much as possible using membrane action. Chris Calladine s book on shell structures 1 discusses these two modes, bending and membrane action, in detail. A string bag (figure 7) is a shell structure. Like fishing nets, string bags only have strings running in two directions. This is to allow the squares on the surface to change to diamond shapes, thus enabling the surface to take up double curvature. If there were a third set of strings, the surface would be Figure 7 String bag 1 Calladine, C. R., Theory of Shell Structures, Cambridge University Press 1988
8 fully triangulated, and would not be able to take up double curvature without wrinkling. Similarly it is not possible to flatten the skin of half a grapefruit, because the skin is effectively triangulated. It is possible to roll it up (figure 8) and we will return to this later. Figure 8. Two grapefruit half skins, one rolled-up The point of this discussion is that membrane stress potentially involves stresses in three directions, or two orthogonal principal stresses and their orientation, again three values. In three dimensions we have three equations of equilibrium, one in each of the x, y and z directions, or alternatively, normal to the surface and two directions tangent to the surface. Thus we have the same number of equations as unknowns and this suggests that shell structures are statically determinate. If this is the case, then all loads can potentially be taken by membrane action and are therefore funicular. A structure is statically determinate if the internal forces and support reactions can be found using only the equations of equilibrium. However it NOT enough to just have the right number of equations, they also have to have a solution. In the case of shell structures the equations are differential equations, or the equivalent algebraic equations in a computer analysis. They are partial differential equations in which the coefficients are not constant; they vary with the curvature of the shell. It is not a question of whether one is clever enough to solve the equations; it is a question of whether a solution exists at all, particularly one that satisfies the boundary conditions at the edge of the shell. The equations of shell structures, particularly membrane action, show a very deep level of interaction between stress and curvature. Curvature has units of 1 length and it is the reciprocal of the radius of curvature, the higher the curvature, the lower the radius of curvature. Membrane stress and surface curvature are both the same sort of mathematical objects, symmetric second order tensors. Just as there are principal stresses and a Mohr s circle for stress, there are principal curvatures and a Mohr s circle for curvature. Green and
9 Zerna 2 is a wonderful book on the subject, but expect to spend a number of months doing not much else to understand it. The normal component of load is the product of the membrane stresses and the curvature. Here the product has to take into account the directions of stresses and curvatures. The Gaussian curvature is the product of the two principal curvatures. It is positive for a dome-like or synclastic surface and negative for a saddle-shaped or anti-clastic surface (horse s saddle, not bicycle). Cylindrical shells have zero Gaussian curvature since they only curve in one direction. Gauss s Theorema Egregium Gauss s Theorema Egregium (illustrious theorem in Latin), states that the Gaussian curvature of a surface can be calculated by measuring lengths on the surface. In the case of the string bag, this would correspond to measuring the diagonals of the diamonds. When the grapefruit is rolled up one principal curvature is increased, but the other is reduced such that their product remains constant. Membrane stresses try and stop lengths on a surface changing. Inextensible deformation of a surface is deformation in which lengths on the surface remain constant, and therefore the Gaussian curvature also remains constant. A developable surface is a surface with zero Gaussian curvature and can be made by bending a sheet of paper without changing lengths on the paper. Developable surfaces include cylinders and cones. Thus designing a surface to work by membrane action is equivalent to choosing a surface which cannot be bent inextensibly. The Cohn-Vossen theorem tells us that a closed surface with positive Gaussian curvature cannot be deformed inextensibly. This is why eggs are so stiff and strong. On the other hand half an egg, like half a grapefruit, can be deformed inextensibly. One might think that a cooling tower (a hyperboloid of one sheet) with its big hole at the top can be deformed inextensibly, but it cannot. The difference between the grapefruit and the cooling tower is that one has positive Gaussian curvature and the other has negative. Thus the Gaussian curvature and the boundary conditions determine whether a structure can work by membrane action alone, in other words, be incapable of inextensional deformation. This is a very difficult question, with no general answer. To complicate matters a shell which works by membrane action alone still has to have sufficient bending stiffness to resist buckling, unless it is a tension structure like a tent or sail. If a shell can deform inextensibly, then it cannot carry all loads by membrane action, but it will be able to carry some loads funicular loads. 2 A. E. Green and W. Zerna, 1969, Theoretical Elasticity, 2 nd edition, Oxford University Press
10 Thus the mathematics is very difficult which is why people rely on computer analysis, particularly the finite element method. The computer results will show how much the shell relies on bending, and therefore to what extent membrane action is incapable of carrying the load. On the other hand the computer analysis won t tell you why the shell has to rely on bending moments, and what to do to make things better. Hyperbolic paraboloids The Calgary Saddledome (figure 9) is a hyperbolic paraboloid (hypar) shell and is approximately elliptic in plan. Let us write the surface as z c = x2 a y2 2 b 2 and the plan of the boundary as x 2 A 2 + y2 B 2 = 1. Figure 9 The Calgary Saddledome The shell is attached to a ring beam and if we assume that the ring beam is only supported in the vertical direction by the stands, then the horizontal component of membrane tension in the shell is resisted by compression in the ring beam, a bit like a warped tennis racquet. If F is the horizontal component of compressive thrust in the ring beam, consideration of equilibrium of the shell and ring beam in the horizontal plane tells us that F = constant and that the horizontal components of membrane stress per unit horizontal length are F B in the x direction and F A in the y direction. Vertical equilibrium now produces a vertical load per unit plan area, w = 2Fc AB A a B 2 b 2 When the engineers were designing the roof they found that their computer printouts (this was before graphic displays) showed very high bending stresses, but they didn t know why. It was only when they realised that they had chosen. A a = B 2 that it became clear what the 2 b
11 problem was. So they lowered the mid-point of the shell (decreased a and increased b ) and found the stresses dropped dramatically. Figure 10 shows a shell by Milo Ketchum made up of a number of hyperbolic paraboloids. Each hyperbolic paraboloid has straight line boundaries and the stresses in the shells are tensions in the sagging direction and compression in the arching direction exactly what one wants. The tensions and compressions cancel out in the direction perpendicular to the boundaries, but add in the tangential direction, producing an axial force in the boundary, which goes straight down to the ground. Shells of revolution The simplest shell forms to analyse are shells of revolution under a radially symmetric load. For simplicity we will consider only a load due to self-weight, t per unit surface area. Symmetry means that we only have two, not three components of membrane stress. The hoop stress, σ H, acts around a circle in a horizontal plane. The longitudinal stress, σ L, acts up the slope perpendicular to the hoop stress. Membrane stresses are usually taken as tensile positive, even in a structure we know will be predominately in compression. The shape of the shell is described by the curve that is rotated around the z axis: in which u is a parameter. Figure 10 Hyperbolic paraboloid shell by Milo Ketchum ( ) ( ) z = z u r = r u tan λ = dz dr = ds du = dr du 2 dz du dr du + dz du 2
12 The principal curvature in the longitudinal direction is dλ ds and the principal curvature in the hoop direction is sinλ r. The sinλ appears in this expression since the radius of curvature is the perpendicular distance from the curve to the z axis. The equilibrium equation in the vertical direction is 2πrt = d ( ds 2πrσ sinλ) L or tr = d ( ds rσ sin λ) L. This comes from consideration of vertical equilibrium of a horizontal ring. In the normal direction the product of the membrane stresses and the curvatures is equal to the normal component of load, t cosλ = σ L dλ ds + σ H sin λ. r Here the principal stresses and principal curvatures are parallel so multiplying stress and curvature is easy. The shell is funicular, regardless of how the thickness varies, because the hoop stress can vary to maintain equilibrium. For a spherical shell λ = u r = Rsin λ z = Rcosλ s = Ru dλ ds = 1 R sin λ = 1 r R Here both the curvatures are negative. This is just a matter of sign convention, but their product is positive, corresponding to the positive Gaussian curvature. Thus trsin λ = R d ( ds σ L sin 2 λ) = d ( dλ σ L sin 2 λ) t cosλ = σ L + σ H R so that, if the thickness is constant,
13 σ L sin 2 λ = Rt cosλ + constant of integration. σ L and σ H must be equal when λ 0, and therefore, σ L = This is negative, corresponding to compression. Thus σ H = Rt cosλ Rt ( 1 cosλ ). sin 2 λ ( 1 cosλ ) sin 2 λ = Rt cosλ 1 1 + cosλ 1 The hoop stress is compressive until cosλ 1 + cosλ becomes negative at λ = ±51.8º, or 38º above the horizontal. This causes problems for masonry domes. Things can be improved by reducing the thickness towards the top, but the best thing to do is to change both the thickness and the shape. Let us suppose that the ideal shell has the membrane stresses (force per unit width) σ L = σ H = tσ, where σ is a constant stress with units force over area. So we now have constant stress in both directions. Then in the vertical direction and in the normal direction. The second equation can be written σ rt = d ds dλ cosλ = cosλ σ dr + sin λ r which can be substituted into the first, so that σ rt = sinλ d dz ( rt sin λ ) σ t cosλ = t dλ ds + sin λ r = 1 r d ( dr r sinλ) = tan λ r d dz r sin λ. ( ) ( rt sin λ ) = r sin2 λ dt dz + t sinλ d dz r sin λ = r sin 2 λ dt dz + σ rt cos2 λ ( )
14 dt dz = σ t t = t 0 e. ( ) σ z z 0 Thus the thickness increases exponentially as z decreases (σ is negative). Figure 11 constant stress shells shell of revolution has larger The shape of the middle surface of the shell is determined solely by equilibrium in the normal direction. However, this equation probably cannot be solved analytically and figure 11 shows a numerical solution. The smaller scale shell in figure 11 is the two dimensional cylindrical shell from figures 5 and 6. It can be seen that the shell of revolution will span roughly twice as far for the same stress, 100m for the weak masonry and 2.5kilometres for concrete this is if the shell is only carrying its own weight. The thickness doesn t come into the expression for maximum span, but if the shell is too thin other loads will dominate the stresses and the shell may buckle. Buckling span than cylindrical shell Shell buckling is particularly nasty because shell structures are so efficient, almost no deflection occurs and then suddenly there is total collapse. Paradoxically, the less efficient the shell, in terms of shape, triangulation of the surface and boundary support, the better it behaves in buckling. This is because bending action of shells requires much more deflection than membrane action and therefore small irregularities in shell geometry and other initial imperfections have little effect. For a properly supported shell working primarily by membrane action, experiments show that the theoretical eigenvalue buckling load can never be reached, even when the utmost care is taken to eliminate initial imperfections. The analysis of shell buckling by hand calculations is effectively impossible even eignvalue analysis of a spherical shell is very difficult, and gives wildly optimistic answers. This means that there is no option but to use computer analysis, but this is quite an esoteric area, and even though many programs offer shell buckling, the results should be treated with a great deal of circumspection. There is still a place for physical model tests for shell buckling.