Graphic statics using discontinuous Airy stress functions

Abstract

It is known that the equilibrium of two-dimensional trusses can be represented using Maxwell reciprocal diagrams and polyhedral Airy stress functions, with the change in slope of the stress function corresponding to a tension force. This article generalises the analysis to include two-dimensional frames, showing how a discontinuity in the value of the Airy stress function corresponds to a bending moment.

Keywords

Airy stress function frames graphic statics kirigami Maxwell reciprocal diagrams

Introduction

Graphic statics grew from the 18th- and 19th-century works of Varignon,¹ Culmann,² Cremona,³ Maxwell,^4,5 Rankine⁶ and others. The chronology is laid out in Kurrer.⁷ Although a highly intuitive method of both visualising and determining the forces in structures, it is now rarely taught outside architecture schools, where Allen and Zalewski⁸ is a popular text. In engineering schools, it has long been supplanted by the analytical and matrix methods of computational structural analysis, such as the finite element method. However, graphic statics is presently undergoing something of a resurgence. Fundamental work by Crapo, Whiteley and others from the 1970s onwards^9–11 laid foundations for structural topology and rigidity theory upon which much of the following work draws. Recent advances in applications, particularly to structural masonry, are exemplified by the works of Ochsendorf, Block, Van Mele and Akbarzadeh,^12–15 as well as the works of Micheletti,¹⁶ Fraternali and Carpentieri^17,18 and Angelillo et al.¹⁹ The current special issue^20–22 contains further examples which emphasise how graphic statics and reciprocal figures can be understood in terms of piecewise planar (‘polyhedral’) Airy stress functions. A stress function is a function whose various curvatures define an equilibrium stress field. For polyhedral stress functions, all curvature is concentrated at the edges of the plane faces, such that the stress field is zero everywhere except along a set of discrete lines, these being the force-carrying bars of the truss. This leads to a powerful framework which helps to analyse the equilibrium of axial forces in trusses.

The contribution of this article is to explain how this description may be extended to include the analysis of bending moments in frames via the use of discontinuous Airy stress functions. The resulting procedures are also highly visual and intuitive and are closely related to the Japanese art of kirigami (the folding of paper with cuts). This article begins with proofs that the discontinuity in the Airy stress function corresponds to a bending moment, and readers interested in the more visual aspects may choose to skip these. However, for those with an interest in applied mathematics, the work contained in this article could be considered a special case of the much more complicated problem of a three-dimensional (3D) Cosserat continuum containing a thin shell structure.^23,24

A simple proof

Consider a two-dimensional (2D) Airy stress function $ϕ (x, y)$ consisting of two planar regions connected by a curved surface over a narrow region, as shown in Figure 1(a). The narrow central region will be the bar of the structure, and for the initial proof, we consider the bar to be straight and of constant width, with the planar regions corresponding to the stress-free space outside the bar. From this continuous, smooth stress function, we will develop results in the discontinuous limit, as the bar width is allowed to shrink to zero.

Figure 1.

(a) An Airy stress function consisting of linear regions either side of a bar, (b) a cross section of the stress function transverse to the bar and (c) the normal stresses $σ_{y y}$ parallel to the bar.

We choose a coordinate system with $y$ along the bar and $x$ orthogonal to the bar, with the boundaries of the bar being at $x_{1}$ and $x_{2}$ . Over the two linear regions, the stress function has differing gradients $g_{1}$ and $g_{2}$ , and it is continuous in value and slope at the bar boundaries at $x_{1}$ and $x_{2}$ (see Figure 1(b)).

The stress $σ_{y y}$ in the $y$ direction is given by the second derivative

σ_{y y} = \frac{\partial^{2} ϕ}{\partial x^{2}}

(1)

(Note that ‘stress’ $σ_{y y}$ in such a 2D problem means that the force is in the $y$ direction per unit length in the $x$ direction and thus has units of kN/m.) The stress is thus non-zero only over the curved section between $x_{1}$ and $x_{2}$ (see Figure 1(c)). The total force $T_{y}$ in the $y$ direction is thus

T_{y} = \int_{x_{1}}^{x_{2}} σ_{y y} d x = \int_{x_{1}}^{x_{2}} \frac{\partial^{2} ϕ}{\partial x^{2}} d x = {[\frac{\partial ϕ}{\partial x}]}_{x_{1}}^{x_{2}} = g_{2 x} - g_{1 x}

(2)

Since this is independent of $x_{1}$ and $x_{2}$ , it remains true for $x_{2} - x_{1}$ arbitrarily small, and we thus have the known result that the tension force in the bar is equal to the change in slope of the stress function transverse to the bar.

The bending moment $M_{1 z}$ about the z-axis passing through the left-hand edge $x_{1}$ is given by

M_{1 z} = \int_{x_{1}}^{x_{2}} (x - x_{1}) σ_{y y} d x = \int_{x_{1}}^{x_{2}} (x - x_{1}) \frac{\partial^{2} ϕ}{\partial x^{2}} d x

(3)

= {[(x - x_{1}) \frac{\partial ϕ}{\partial x}]}_{x_{1}}^{x_{2}} - \int_{x_{1}}^{x_{2}} \frac{\partial ϕ}{\partial x} d x

(4)

= (x_{2} - x_{1}) g_{x 2} - [ϕ_{2} - ϕ_{1}]

(5)

The magnitude of the moment is thus equal to the vertical mismatch of the two planes at $x_{1}$ (see Figure 1). Again, this remains true as $x_{1}$ and $x_{2}$ become arbitrarily close, whereupon the mismatch is the stress function discontinuity. This is the main result: that for a piecewise linear stress function made of planar surfaces over polygonal regions, the bending moment in the bar that is the shared edge of two adjacent polygons is given by the change in stress function across the bar.

For the shear stress in the bar, we use the stress function definition

τ_{y x} = - \frac{\partial^{2} ϕ}{\partial x \partial y}

(6)

The shear force $S_{x y}$ in the bar is thus

\begin{array}{l} S_{y x} = \int_{x_{1}}^{x_{2}} τ_{y x} d x = - \int_{x_{1}}^{x_{2}} \frac{\partial^{2} ϕ}{\partial x \partial y} d x \\ = - {[\frac{\partial ϕ}{\partial y}]}_{x_{1}}^{x_{2}} = - [g_{y 2} - g_{y 1}] \end{array}

(7)

That is, the shear force in the bar is given by the difference in slopes parallel to the bar, and again, the result holds as $x_{2} - x_{1}$ shrinks to zero.

At the bar boundaries, the forces normal to the bar are given by

σ_{x x} = \frac{\partial^{2} ϕ}{\partial y^{2}} = \frac{\partial g_{y}}{\partial y}

(8)

and since at the bar boundaries, the stress function is linear in the $y$ direction, it has zero second derivative $\partial^{2} ϕ / \partial y^{2}$ . We thus have zero applied stress normal to the bar boundaries. The shear stress there is $\partial^{2} ϕ / \partial x \partial y = \partial g_{y} / \partial x$ , and since for the planar regions, the gradient $g_{y}$ is constant with $x$ , we conclude that there are zero shear stresses applied to the bar boundaries. The derivation above thus applies to bars which are not loaded at their surfaces.

A more general case

We now relax some of the assumptions of the simple derivation above to allow for curved bars and applied distributed loads transverse and parallel to the bar. This more general case is illustrated in Figure 2. The Airy stress function is now non-planar over the adjoining spaces, and the bar region is curved. The bar is defined by a curve $C$ , and at a point on $C$ , we define a local coordinate system $(x, y)$ with $y$ tangent to $C$ and $x$ orthogonal to $y$ . The bar thickness is thus given by $x_{2} - x_{1}$ , which may vary as one moves along the bar. As before, the Airy stress function is everywhere continuous in value and in its first derivatives, and we develop the discontinuous limit as the bar width tends to zero.

Figure 2.

(a) A more general Airy stress function, (b) a cross section transverse to the bar and (c) the normal stresses parallel to the bar are now non-zero over regions outside the bar.

For the axial force $T_{y}$ in the bar, the $y$ direction stresses $σ_{y y}$ are still given by equation (1), and although there now exist stresses in the regions outside the bar, the integral across the bar is still given by equation (2), the only technical difference being that the slopes $g_{2 x}$ and $g_{1 x}$ are the components of the local gradient at the edges of the bar. The tension along the bar is thus given by the change in transverse slope across the bar, as before.

The previous derivations for the bending moment and shear force are similarly unchanged, excepting only that slopes are defined locally at the bar boundaries. The difference now is that bars may be loaded at their surfaces via distributed loads.

The forces $σ_{x x}$ normal to the bar faces are given by equation (8). That is, the applied normal stress is given by the curvature of the stress function along the boundary. For example, to apply a uniformly distributed load (UDL) to a straight bar, we may adopt a stress function which is parabolic in the $y$ direction.

The surface tractions parallel to the bar are given by

τ_{x y} = - \frac{\partial^{2} ϕ}{\partial x \partial y} = - \frac{\partial g_{x}}{\partial y}

(9)

That is, the applied surface shear stress is given by the rate at which the transverse gradient $(g_{x})$ changes in the $y$ direction. Thus, to apply a uniformly distributed traction along the surface of a straight bar, we may choose a hyperbolic stress function of the form $ϕ = - c x y$ . The transverse gradient $g_{x} = \partial ϕ / \partial x = - c y$ and thus $τ_{x y} = - \partial g_{x} / \partial y = c$ .

To apply a combination of uniformly distributed normal and shear stresses simultaneously to a bar, the external stress function may be chosen to be some appropriate quadratic function.

To apply concentrated loads, one simply treats the applied force in the manner of a bar and has a continuous external stress function with a kink in the slope. To apply a point moment, one similarly has an external stress function possessing a discontinuity.

We shall illustrate these with some examples. However, before proceeding, we first illustrate the connection to kirigami.

Kirigami

It is known that the traditional continuous Airy stress function is related geometrically to the yield line collapse mechanisms of slabs and thus to small-displacement origami. This follows from the fact that requirements for a yield line collapse mechanism of a slab to be compatible are formally equivalent to the conditions for the state of self-stress of a pin-jointed truss to be in equilibrium.²⁵ This is illustrated in Figure 3, which shows a single graph interpreted as the yield lines of the plastic collapse mechanism of a slab (Figure 3(a)) and as the bars in a 2D truss (Figure 3(b)). Compatibility of the slab collapse mechanism requires that the vector sum of the hinge rotations at any joint is zero, and similarly, equilibrium of the truss requires the vector sum of the bar forces meeting at any joint to be zero. The two problems are thus formally equivalent. The only difference is in the way that structural engineers tend to read the two problems. In the former case, most engineers would immediately ‘read’ that the three longer hinge lines must meet at a point because the quad slabs must remain rigidly planar as the out-of-plane mechanism develops. Fewer engineers would read the truss problem in the same manner, and yet the geometric condition that the truss can maintain a state of self-stress is exactly the same – the three bars connecting the two triangles must meet at a point.

Figure 3.

The formal equivalence between the compatibility of a yield line collapse mechanism and the equilibrium state of self-stress in a truss: (a) yield line collapse mechanism and (b) truss.

The reason that the diagrams tend to be read differently is that in the truss, the air spaces between the bars are not usually read as being rigid planar polygons. However, Maxwell^4,5 observed that the condition for a 2D truss to be capable of maintaining a state of self-stress was that it be the 2D projection of a 3D plane-faced polyhedron.That is, the truss can carry a self-stress if and only if the air spaces between the bars are considered to be rigid polygons, and these polygonal regions can be tilted out of the page into three dimensions in the manner of a small-displacement yield line collapse mechanism.

Having tilted the truss diagram into such a 3D polyhedron, one has created the Airy stress function for the truss. The stress function is identical to the polyhedron. The mathematical derivations above showing that bar forces are given by the change in slope of the stress function as one moves from one air-space region to another are thus identical to the obvious statement that the hinge rotations of the yield line collapse mechanism are given by the change in gradient of the rigid slab regions as one moves from one region to another across a hinge line.

Clearly, compatible yield line collapse mechanisms are thus equivalent to small-displacement origami patterns. We emphasise that this is only true for small out-of-plane displacements. Large-displacement origami patterns are more restrictive than small-displacement (first-order) yield line collapse mechanisms. For example, the collapse mechanism of Figure 3(a) is compatible only for small displacements. Larger displacements would generate in-plane forces, and compatibility would require additional hinge lines to preserve the zero Gaussian curvature of the deflected slab.²⁶

We have thus established the equivalence of continuous Airy stress functions for 2D trusses with yield line collapse mechanisms of slabs, plane-faced polyhedra and small-displacement origami patterns. However, in the earlier section, we introduced the idea that discontinuous Airy stress functions could represent bending moments in 2D frames. For this case, the following notions are equivalent:

Discontinuous Airy stress functions;

Yield line collapse mechanisms of slabs with free edges;

Polyhedra, some of whose faces are vertical;

Small-displacement kirigami.

Kirigami is the Japanese art of paper folding, which is like origami, but which allows cuts to be made in the paper before folding. The presence of the cuts opens up a wealth of possible folded shapes that are not accessible by origami. We emphasise that the equivalence here is only with small-displacement kirigami.

In the next section, we shall give examples of how to represent the forces and moments in a variety of 2D frames using discontinuous Airy stress functions, and the easiest way to visualise these examples is often via kirigami. Indeed, it is usually much easier to read the allowable possibilities via the intuitive understanding of what is possible with folding a piece of cut paper than it is from our trained knowledge of the conditions of equilibria. The fact that engineering undergraduates often have difficulty in visualising bending moment diagrams (BMDs) is a long-standing complaint of many academics. It is hoped that the constructions that follow may provide some useful input into that debate. Indeed, the possibility of teaching BMDs using paper and scissors is an intriguing prospect.

In the kirigami view, the BMD manifests itself as the air gap that opens along any cut. If the original sheet of paper contains a drawing of the frame in $(x, y)$ coordinates, then the BMD appears along the cut lines as having a magnitude in the out-of-plane $z$ direction. The traditional convention for plane frames is to draw the BMD in the $(x, y)$ plane of the frame, with some sign convention such as the diagram being drawn on the tension side of the beam. In the kirigami interpretation, the bending moment $M_{z}$ is drawn out of the plane, in the $z$ direction, and although this is an unusual convention for 2D frames, it is nevertheless rational and accords with a possible 3D representation of biaxial bending moments and torsion at any cross section as a single vector in 3D.

Sign convention

The basic definitions of equations (1) and (8) show that positive normal stresses $σ_{x x}$ and $σ_{y y}$ correspond to positive second derivatives of the stress function. It follows that a ‘tension positive’ convention implies that a stress function having a parabolic valley profile (such as $ϕ = x^{2}$ ) will create a constant tension stress field. This translates to origami as ‘valley folds correspond to tensile forces’ and to kirigami as ‘the lower side of the stress function at the discontinuity corresponds to the tensile side of the bending moment’. We shall adopt this sign convention throughout.

Some examples

The rather elementary result that bending moments correspond to discontinuities in the Airy stress function now allows a wide class of 2D frame problems to be solved in an intuitive graphical manner.

Figure 4 shows a simply supported beam with a central point load $W$ . The stress function is defined over the full plane. It has three creases that extend to infinity, and a discontinuity. The changes of slope across the creases correspond to the applied load and the reactions. The discontinuity in stress function along the beam is triangular in shape and corresponds to the BMD. Although there is a discontinuity in the value of the stress function on crossing the beam in the $y$ direction, there is no discontinuity in the slope $\partial ϕ / \partial y$ ; thus, the beam carries no axial force.

Figure 4.

The discontinuous stress function for a simply supported beam carrying a central point load. (The diagram shows a single-valued stress function acting over the full plane, but this could equally have been drawn as a double-layer stress function (i.e. as a closed polyhedron) between the folds.)

Figure 5 shows a pin-footed portal frame. This system possesses a state of self-stress corresponding to the Airy stress function shown. The stress function is zero everywhere except within the portal bay, where it rises linearly from the crease between the pinned feet. The stress function thus has a triangular discontinuity on each column and a constant discontinuity along the beam. Rotating these vertical faces about their beam axes to lie back in the horizontal plane reveals the BMD. The tension force between the feet that is in equilibrium with these moments is given by the fold angle across the crease at the base.

Figure 5.

The state of self-stress in a pin-footed portal.

Figure 6 shows a Γ-shaped cantilever carrying a point load at its tip. The stress function and BMD are shown.

Figure 6.

The stress function and BMD for a tip-loaded Γ-shaped cantilever.

Figure 7 shows a circular beam loaded by diametrically opposite forces. The loads may be created by folding the stress function in the region outside the circle. Inside the circle, the stress function is a rigid disc. This has 3 degrees of freedom – it can be moved up or down, and it may be tilted about the $x$ or $y$ axes. These freedoms span all possible states of self-stress of the beam, which may be caused by differential temperatures or shrink-fit construction.The fact that there can only be 3 degrees of freedom in the self-stress is immediately evident from the kirigami construction. Of the many possible equilibrium solutions, two are illustrated. In the first example, there is little or no self-stress, and the bending moment is thus close to the elastic solution of the moment induced purely by the applied loads. In the second example, there is a large self-stress creating tensions or compressions on the inside or outside of the pipe. In both cases, the kirigami description with bending moments drawn perpendicular to the plane of the structure is arguably easier to understand than the traditional description with moments drawn in-plane.

Figure 7.

A circular beam subject to diametrically opposite forces. The rigid central disc has 3 degrees of freedom, and two of the possibilities are shown, one having a small and one a large self-stress.

Figure 8 shows a similar case but with the applied loads not diametrically opposed. The conceptual ease with which the possible bending moments can be visualised should by now be apparent.

Figure 8.

A circular beam with loads that are not diametrically opposed.

A fixed-fixed beam is statically indeterminate and admits multiple equilibrium solutions. Figure 9 shows two possible solutions when a concentrated moment is applied at midspan. The first case is shear free, while in the second case, there are shears in the beam, represented by the differing slopes of the stress function above and below the beam. The applied concentrated moment is applied via a discontinuity in the stress function which extends to a point at infinity and similarly the reactant moments.

Figure 9.

Two of the possible equilibrium solutions for a fixed-fixed beam carrying a concentrated moment at midspan.

Figure 10 shows a pin-footed portal loaded with a UDL. The UDL is applied via a region over which the stress function is parabolic. The general solution is any system of forces in equilibrium with the applied load plus some multiple of the single state of self-stress in the unloaded portal. This latter may be mobilised by arbitrary flexing of the hinge line between the portal feet.

Figure 10.

An Airy stress function for a pin-footed portal carrying a uniformly distributed load. Per Figure 5, there is a state of self-stress in the unloaded frame which corresponds to arbitrary tilting of the central plane about the hinge line between the column feet.

Figure 11 shows a Γ-shaped cantilever with a uniformly distributed surface shear along the cross-beam. The stress function representation has two parabolic functions, each creating a state of uniaxial stress. Where these overlap, they add to create a hyperbolic stress function over a triangular region of pure shear above the beam. As explained in section ‘A more general case’, a hyperbolic stress function adjacent to a beam creates a surface traction along the beam. The purpose of the parabolic regions is merely to allow the stress function to be continued smoothly to infinity without introducing any additional forces to the system.

Figure 11.

Loading via surface shear tractions. Parabolic stress functions create uniaxial stresses which overlap to create a hyperbolic stress function, thereby applying surface tractions along the cross-beam.

The use of curved stress functions – as in the previous two examples – will present a challenge to models predicated on projections of plane-faced polyhedra.

The curved stress functions in these examples have been extended to points at infinity for equilibration of the external loads. Although this would appear to present problems for more complicated cases – for example, when applying distributed loads to internal floors in a building frame – such cases could be treated as the superposition of a number of simpler problems.

Relation to Maxwell reciprocal diagrams

Central to graphic statics is the concept of the dual pair of diagrams representing form and force. The form diagram shows the structural bars and the lines of action of applied forces. Lengths in the force diagram give the magnitude of the equilibrium forces in the corresponding bars. There are two possible conventions for 2D trusses: in the Cremona convention, forces are drawn as vectors parallel to their corresponding bars, while in the Maxwell convention, forces are drawn perpendicular to their bars. In this article, we shall adopt the Maxwell convention because this accords with the construction of reciprocal diagrams via gradients of Airy stress functions. In that construction, normals to the planar regions of the stress function between bars are intersected with a reciprocal plane to create the reciprocal diagram. All such surface normals are represented as vectors based at the origin (0, 0, 0) in a space $(x, y, h)$ . The structural diagram is a graph on the $(x, y)$ plane, and the stress function $ϕ (x, y)$ gives the third coordinate $h$ . The surface normals are then intersected with some plane (such as $h = 1$ ) to give the reciprocal diagram.¹¹

Thus far, all diagrams have been form diagrams, showing the structure and the lines of action of the applied forces, together with a representation of the Airy stress function over the 2D form diagram. We now construct the reciprocal force diagrams for some cases of interest, by considering the normals to the stress function.

This construction utilises the fundamental duality of 3D projective geometry in that planar polygonal faces in the original diagram map to vertices in the reciprocal diagram.

Example

For the centrally loaded, simply supported beam in Figure 12, there are three planar regions $I, I I and I I I$ which define reciprocal points $1,2 and 3$ on the reciprocal plane $h = 1$ . All three regions are infinite in extent. The polyhedral stress function over the original also contains a fourth surface $I V$ , a vertical face at the cut discontinuity which represents the bending moment in the beam. The normal to this vertical face is horizontal, and thus, it intersects the $h = 1$ plane at a point at infinity, which is the fourth point, 4, of the reciprocal diagram. The reciprocal points 1–4 are then connected by lines representing the magnitude of the forces in the bars that separate the various regions of the original diagram. The resulting reciprocal diagram is shown, with lower and upper case labelling showing the correspondence between a bar and its force.

Figure 12.

A centrally loaded, simply supported beam and its reciprocal: (a) the beam and the lines of action of the applied forces and reactions, that is, the form diagram, (b) the Airy stress function over the form diagram, whose normals define the reciprocal figure, the force diagram, (c) the dual Airy stress function over the force diagram, whose normals match the points on the original figure and (d) a vertical slice through both stress functions.

The requirement that the regions of the stress function remain planar means that the stress function has only a single degree of freedom and can be parameterised by the height $M$ , the magnitude of the stress function discontinuity at the centre of the beam.

Because the bar $f$ separates the original regions $I I$ and $I I I$ , the reciprocal points 2 and 3 are connected by the reciprocal line $F$ . Elementary geometry shows that

\frac{M}{L / 2} = \tan θ = \frac{F}{1} whence F = \frac{2 M}{L}

(10)

The reciprocal line $D$ similarly has length $2 M / L$ . The reciprocal line $E$ , which represents the magnitude $W$ of the applied central force along line $e$ , is thus $W = D + F = 4 M / L$ . We thus have $W = 4 M / L$ , which may be written in its more familiar form as $M = W L / 4$ . This value of $W$ also matches the slope discontinuity of the central ridge of the stress function, this being $2 \tan θ = 4 M / L$ by elementary geometry.

We thus have a consistent picture. The units of the stress function are kilonewton metre, these being the units of bending moment, and discontinuities in the stress function are thus, as we have demonstrated, bending moments.

For this simple case, the reciprocal diagram is not only dual to the original (as it must be by definition), but it is also topologically equivalent. If the reciprocal diagram so created is now considered to be a form diagram, this can carry a dual Airy stress function whose normals define the original geometry. Let reciprocal node 3 move down out of the plane by a distance $Δ$ , then to re-create the original geometry at the correct scale, we require

\frac{Δ}{F} = \tan β = \frac{L / 2}{1} whence Δ = \frac{F L}{2} = \frac{2 M}{L} \frac{L}{2} = M

(11)

That is, the dual stress function should be mobilised by moving the reciprocal node 3 down by a distance which is also equal to $M$ . The units of the dual stress function are thus also kilonewton metre, and the topological equivalence shows that the dual to a centrally loaded, simply supported beam is another such centrally loaded, simply supported beam. This dual beam has length $E = 4 M / L = W$ and is loaded by a central force of magnitude $2 \tan β = 2 M / F = L$ . In summary, the reciprocal to a simply supported beam of length $L$ carrying a central point load $W$ is a simply supported beam of length $W$ carrying a central point load $L$ , illustrating how the duality between form and forces persists even as graphic statics extends to cover bending.

Finally, we emphasise that given a pair of form and force diagrams, with the form diagram containing a beam, then while the axial forces are given by lengths in the force diagram, the bending moments are given by the discontinuity of the stress function over the form diagram.

Funiculars and points at infinity

In the analysis of reciprocal diagrams for trusses presented in McRobie et al.,²¹ it was explained how the presence of force lines of action extending off to infinity presented problems when interpreting the resulting reciprocal diagram as a form diagram. Nodal equilibrium was then plagued by ‘zero-times-infinity = finite’ issues, which, although not incorrect, were certainly ungainly. These issues were resolved in that article by ‘coning’ the force polygon on the reciprocal plane. That is, an arbitrary origin (a ‘pole’) was chosen on the reciprocal plane, and this was connected by coordinate vectors to the nodes of the applied force polygon. Dual to this ‘coning’ was a funicular back on the original diagram, which removed all the troublesome infinities. Figure 13 illustrates the case for a simple truss bridge. The lines of the applied force and the two reactions extend to infinity on the original diagram. The reciprocal diagram is shown, and the problem occurs at the left- and right-hand ends of the force polygon. Interpreting this as a form diagram, we have the problem that the forces in the members $H$ , $J$ and $K$ are given by the lengths $h$ , $j$ and $k$ in the original diagram and are thus infinite. Nevertheless, nodal equilibrium at node 1 requires that the vertical components of these horizontal forces be in equilibrium with the (finite) vertical force $a$ in the member $A$ that connects there.

Figure 13.

A truss bridge and its reciprocal, illustrating the equilibrium problem at the reciprocal abutments. The infinities may be removed by coning the force perimeter, leading to a funicular below the original structure.

By coning the perimeter force polygon as shown in Figure 13, the resulting dual diagram is the original with a funicular below it, and all infinities are neatly removed. Nodal equilibrium at node 1 is readily achieved since the vertical force in member $A$ can be readily carried by the vertical component in the cone member $P$ .

With the new understanding that discontinuities in the Airy stress function can represent the bending moments in beams, the need to cone the applied force polygon is obviated. Instead, the applied force polygon can be considered to be a beam, and previously unresolved vertical components of forces in the reciprocal structure can then be equilibrated against shear forces in that beam. For the example shown, then, the reciprocal diagram for the centrally loaded truss is a beam that is stressed against a truss. The reciprocal structure, interpreted as a form diagram, carries no external loads. However, it can carry a state of self-stress, via the interaction of the beam with the truss above it.

Figure 14 shows the Airy stress function and its dual. Let the bridge span be $L$ and the central load $W$ , then the original stress function is raised by $δ$ , with geometry giving

\tan θ = \frac{δ}{L / 2} = \frac{W / 2}{1} where δ = \frac{W L}{4}

(12)

Figure 14.

By treating the lower chord of the reciprocal as a beam, the need for coning is removed.

Similarly, for the dual stress function to re-create the original, it is lowered by $Δ$ , and geometry requires

\tan β = \frac{Δ}{W / 2} = \frac{L / 2}{1} where Δ = \frac{W L}{4}

(13)

Considering the reciprocal diagram now to be the form diagram of a truss of span $W$ , the bar forces are given by the lengths in the original diagram. Bar A thus carries a force $a = L / 2$ , which we may consider to be compressive. The structure thus resembles half a bicycle wheel, with an arch $A D E B$ pulled into compression by the tension spokes $C, G, F$ . The forces in the arch and spokes are equilibrated via a beam $H J K$ along the bottom chord. The previously unresolved vertical component of force at node 1, carried by member $A$ , is $L / 2$ . This must equilibrate with the shear in the beam, which, from equation (7), is equal to the difference in slopes of the (dual) stress function parallel to the beam. On one side, the slope is $\tan β$ , and on the other side, it is zero; thus, the shear in the beam is $\tan β$ , which equals $L / 2$ as required. The arch, spokes and beam possess an equilibrium state of self-stress, and there is no requirement for any coning.

Funicular beams

The word ‘funicular’ derives from the Latin for ‘rope’, and in many cases, their use in graphic statics is as a system of ropes which are tensioned to apply the external forces. However, rope forces are purely axial and cannot sustain moments. Therefore, to apply moments or to represent support moments of frames, it is apparent that any ‘funicular’ may need to contain beams. Figure 15 gives an example of a cantilever with a base moment, and the corresponding ‘funicular’ requires a beam connecting the column base to the line of the applied force. More generally, then, loads may be applied via systems resembling the control paddles of marionettes, with both strings and beams.

Figure 15.

The use of a ‘funicular’ beam to carry base moments of a loaded cantilever. Note how this has essentially turned the Γ-shaped cantilever of Figure 6 into the pin-footed portal of Figure 5.

Varignon

Somewhat remarkably, this discontinuous stress function interpretation can be applied to the famous diagram of 1725 by Varignon¹ (see Figure 16), perhaps the first historical example of a graphic statics form-and-force pair. The original form diagram is a rope from which loads are suspended. The force diagram consists of a set of triangles, each triangle being the closed polygon of forces necessary for nodal equilibrium at each node of the original structure.

Figure 16.

Varignon’s 1725 drawing of a form diagram – a loaded rope – and its associated force diagram. Forces have been drawn parallel to the associated members (Cremona convention).

The original Varignon diagrams use the (later) Cremona convention, but they have been redrawn here in the Maxwell convention (Figure 17(a) and (b)) and the members and forces re-labelled in a consistent manner. The orientation of both is then such that each resembles a bridge. If Varignon’s force diagram is now interpreted as a form diagram, then to re-create the original requires a dual Airy stress function with a discontinuity, giving the bending moment in a beam along its lower chord (Figure 17(c) and (d)). We arrive at the conclusion that Varignon’s diagram shows a pair of dual structures, with Varignon’s original form-force pairing showing a loaded rope suspension bridge and its purely axial forces. Taking the alternative perspective, we find that the form diagram shows a cable-stayed beam carrying no external load, and the force diagram (Varignon’s original form diagram) gives the axial forces associated with a state of self-stress in that bridge, with the corresponding bending moment in the cable-stayed beam being given by the discontinuity in the dual stress function that re-creates Varignon’s original. This duality between a loaded suspension bridge and the state of self-stress in a cable-stayed bridge has been in front of us for almost three centuries.

Figure 17.

A reinterpretation of Varignon’s diagram. If Varignon’s force diagram is interpreted as a form diagram, the pair corresponds to the state of self-stress in a cable-stayed beam under the action of no externally applied loads. The bending moment diagram is the discontinuity in the stress function over the second diagram which is necessary to re-create the first: (a) and (b) Varignon’s form and force diagrams; (c) the dual stress function over the force diagram, this having a vertical face along one edge; (d) the vertical face, and below this, the diagram has been redrawn to have a constant baseline – a more familiar representation of a bending moment diagram; and (e) the final interpretation, reading Varignon’s original force diagram as a form diagram. There is a central column supporting four tension cables. These are stressed against a continuous beam, and the bending moment in the beam is shown out of the plane of the diagram.

Summary and conclusion

This article has demonstrated how the Airy stress function description of reciprocal figures of 2D trusses may be extended to deal with 2D frame structures, with bending moments being given by discontinuities in the stress function. Beams and frames, subject to both point and distributed loads, are thus now amenable to analysis by graphic statics.

The stress function and associated moment diagram can be readily visualised using kirigami, a feature which may be of educational value in the teaching of fundamental structural behaviour.

Footnotes

Acknowledgements

The authors are grateful for conversations with Chris Calladine (who suggested a number of the examples), Simon Guest, Bill Baker, Toby Mitchell, Arek Mazurek and Marina Konstantatou.

Declaration of conflicting interests

The author(s) declared no potential conflicts of interest with respect to the research, authorship and/or publication of this article.

Funding

The author(s) received no financial support for the research, authorship and/or publication of this article.

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