On the duality of space trusses and plate structures of rigid plates and elastic edges

Abstract

Dualities have been known to map space trusses and plate structures to each other since the 1980s. Yet the computational similarity of the two has not been used to solve the unfamiliar plate structure with the methods of the well-known truss. This article gives a method to find the forces and displacements of a plate structure with rigid plates and elastic edges, using a dual truss. The plates are assumed to be rigid in their respective planes only and deformable otherwise. The method provided is applicable for both statically determinate and indeterminate structures, subjected to both statical and kinematical loads.

Keywords

computational similarity duality plate structure truss

Introduction

Dualities as an articulated projective geometrical concept have emerged in the field of statics with Maxwell¹ and Cremona and Beare² and were expanded at that time by Klein and Wieghardt.³ While these methods were able to give one the forces of a (planar) truss, with the spread of algebraic methods they were almost forgotten in engineering circles. It was in the 1980s, when projective geometry experienced a renaissance, that the rigidity of a truss being a projective invariant was discovered.^4,5 Investigation of the duality of engineering structures followed, both in two-dimensional (2D) between plane trusses and grillages^6–8 and in three-dimensional (3D) space between spatial trusses and plate (sheet) structures.⁹ Parallel to this geometric renaissance, engineer Baracs¹⁰ and architect Wester^11,12 began researching and popularising plate structures for their efficiency and clarity. While even at this point plate structures themselves were not new and their load bearing has been studied since the 1950s,^13,14 till date very few homogeneous, cast concrete plate structures have been built. At present though, with the spread of automation into construction, the use of smaller, prefabricated plates forming a spatial structure is getting more obvious and economical. This motivates researchers to study, for instance, glass¹⁵ and wooden¹⁶ plate structures. New ways are being tested to connect the prefabricated plates on-site. In case of glass this is usually done by glueing,^17–19 while wooden elements easily allow the use of traditional finger joints and mechanical elements as well.^20,21 In any case the on-site joints between the prefabricated elements are generally softer and they are generally responsible for the majority of the displacements occurring in the structure. This motivated the choice in the presented model to describe such prefabricated structures, with plates rigid in their respective planes (deformable perpendicular to these planes) connected by elastic edges. For monolithic (reinforced concrete) structures – although classical theory suggests maximal stresses and thus maximal displacements at the edges – this assumption is less natural. The general question of how stiffness distribution influences the applicability of the presented model remains to be properly investigated in a follow-up work.

The continuation of the research on projective geometry in mechanics turned back to the path set by Maxwell, with its sight set on geometric representation of forces of spatial structures^22,23 and the extension of such diagrams to kinematics of the structures.^24–26 While La Magna et al.²⁷ mention the possibility of using a dual truss structure to compute the edge forces of a plate structure, a method to do this in the general case of a possibly statically indeterminate (hyperstatic) structure subjected to both statical and kinematical loads is hitherto undeveloped.

Notation and modelling

Trusses and plate structures

Here, a quick review is given on the statical and compatibility equations of the two models for trusses and plate structures, presenting the similarity of the two models and to provide the objects of comparison.

The well-known model of a truss comprises pin- or ball-joints and elastic bars connecting them. Loads are restricted to concentrated forces acting on the joints. The system of statical and compatibility equations takes the form^28,29 of

[\begin{matrix} 0 & G^{'} \\ {G^{'}}^{T} & F^{'} \end{matrix}] [\begin{matrix} ϵ \\ ϕ^{'} \end{matrix}] + [\begin{matrix} π^{'} \\ 0 \end{matrix}] = [\begin{matrix} 0 \\ 0 \end{matrix}]

(1)

The upper block contains the equilibrium equations of the joints in global frames, while the lower block contains the compatibility equations of the bars in their local (one-dimensional) frames. In the case of a space-truss with n joints and l bars, $G^{'} \in ℝ^{3 n \times l}$ is the geometrical matrix while matrix $F^{'} \in ℝ^{l \times l}$ is in fact diagonal and contains the stiffnesses of the bars on its main diagonal. The vector $ϵ \in ℝ^{3 n}$ contains the displacements of the joints such that

ϵ = (\begin{matrix} ϵ_{1} \\ ⋮ \\ ϵ_{n} \end{matrix}) and ϵ_{i} = (\begin{matrix} ϵ_{i, 1} \\ ϵ_{i, 2} \\ ϵ_{i, 3} \end{matrix})

(2)

Here $ϵ_{i}$ is the 3D displacement of joint $i .$ Vector $π^{'}$ contains the loads of the vertices similarly. Vector

ϕ^{'} = (\begin{matrix} {ϕ^{'}}_{1} \\ ⋮ \\ {ϕ^{'}}_{l} \end{matrix})

(3)

contains the bar forces, ${ϕ^{'}}_{i}$ being the force in bar $i .$

A model of a plate structure comprises plates, subjected only to loads in their planes. In the following, the plates are considered to be rigid against forces acting in their planes, while their joins with other plates are considered to be elastic. The plates are considered to be so soft perpendicular to their planes that deformations in this direction happen without significant resistance. Hence, corresponding to each plate only the displacements happening in the plane of the plate are relevant (cause forces), and any contact force between two plates acts such that its line of action is the line of intersection of the planes of the plates.

Proposition I

With the proper choice of equations and parameters, the system of equilibrium and compatibility equations of a plate structure can be cast in the form

[\begin{matrix} 0 & G \\ G^{T} & F \end{matrix}] [\begin{matrix} δ \\ ϕ \end{matrix}] + [\begin{matrix} π \\ 0 \end{matrix}] = [\begin{matrix} 0 \\ 0 \end{matrix}]

(4)

The top block contains the moment equations of the plates (in a coordinate system specified later) and the bottom block contains the compatibility equations of the edges in their local one-dimensional coordinate systems. Here, G is the geometrical matrix, F is the stiffness matrix of the elastic joins, $ϕ$ contains the edge forces with which the plates support each other, $δ$ contains the small displacements (rotations) of the plates and $π$ contains the loads. While $G$ is more intuitive, F might need to be clarified. If along edge k two plates are joined with relative deformation t_k (measured in the direction of the edge as translation) then the arising force $ϕ_{k}$ acting on both plates is such that $t_{k} = F_{k, k} | ϕ_{k} |$ holds. The comparison of the compatibility conditions in case of a truss and the plate structure can be seen in Figure 1. Most of the modelling assumptions are already present in plate structure literature,^14,15,30 while the concentration of the deformations to the edges appears to be new. At the moment the exact conditions (of realistic structures) under which this idea is applicable are not known.

Figure 1.

Comparison of compatibility conditions. Left: Displacements of joints $P'_{u}$ and $P'_{v}$ of a truss. Middle: The change of barlength $({t^{'}}_{k})$ assuming small displacements. Right: Rotational displacements of plates $P_{u}$ and $P_{v}$ of a plate structure, as well as the change of length of a fictitious spring joining them (assuming small displacements).

Restrictions on the loads are present in both models. In case of the truss loads acting on the bars are transferred to the joints through bending of the bars. Yet, as the dominant load bearing is done through normal forces in the bars, the conceptional model handles the loads as they acted directly at the joints. The plate model behaves similarly: In reality, the loads acting on the plates are transferred to the edges through bending. In the model, the loads are assumed to act at the edges, decomposed into components lying in the planes of the plates as required by the reduced dimensionality of the static equations. For plate structures built from ‘short’ elements (where the bending moments in the plates are not dominant) load reductions like this^14,15 and even more drastic ones³⁰ are common in engineering theory. For plate structures from ‘long’ elements (where the bending moments in the plates are dominant), the method presented here is not applicable.

It can also be noted that modern trusses are built with rigid connections between the bars, yet a pin-jointed theoretical model is used to capture the behaviour of the structure. One reason behind this is the possibility of plastic deformations in case of steel trusses. Such material-specific investigations are beyond the scope of this article.

Duals of points, lines, and planes are naturally given by projective geometry. This geometric skeleton can be clad with components of the two mechanical models. The correspondence of the primal-dual components is presented in Table 1. In the following, the relationships of the primal and dual elements will be shown through the connectivity graphs of the structures. In case of a truss the vertices of the connectivity graph represent the joints, while edges represent the bars. Two vertices are connected with an edge if there is an appropriate bar in the structure.

Table 1.

The corresponding components of the models.

	Plate structure	Truss
Geometry	Plane	Point
	Line	Line
	Point	Plane
Model parts	Rigid plate	Joint
Model parts	Elastic edge	Elastic bar
Mechanical quantities	Edge forces	Bar forces
	Planar loads of plates	Loads on joints
	Displacements of plates	Displacements of joints
	Stiffness of edges	Stiffness of bars
Equations	Equilibrium of coplanar forces	Equilibrium of concurrent forces
Equations	Edge directional compatibility	Bar directional compatibility

In case of a plate structure, the graph vertices represent the plates, and there is a graph edge joining two vertices if there is a corresponding elastic edge joining the two appropriate plates. As the location of the elements is determined by the projective duality which interchanges incidence relations (of spatial points, lines, and planes), it can be seen that the connectivity graphs of a plate structure and the dual truss are isomorphic. This is also required for our plan to give a correspondence between equations of the two structures. We can write a 3D equilibrium equation corresponding to each graph vertex, containing unknowns corresponding to the graph edges incident with the graph vertex. In the case of a plate structure, these are three moment equations (in general), in the case of a truss three projections of the vectorial force equilibrium. We can also have compatibility equations corresponding to each graph edge. In the case of a truss this is the compatibility of the bar force (through the elongation of the bar) with the bar directional relative displacement of the two joints. In the case of a plate structure, the edge force (through deformation of the elastic edge) has to be compatible with the edge directional relative displacement of the two appropriate plates. These systems of equations are presented in equations (1) and (4). While there are multiple ways we could make such dual transformations, we will restrict ourselves to the canonical duality of projective geometry. With its help, transformation matrices $Δ,$ $Φ$ and $Π$ will be presented satisfying

δ = Δ ϵ

(5)

ϕ = Φ ϕ^{'}

(6)

π = Π π^{'}

(7)

F = Φ^{- 1} F^{'} Φ^{- 1}

(8)

Rigidity of the structures is not only influenced by the topology, but the possibly degenerate geometry as well. Fortunately for this article, such degeneracy is an incidence condition, meaning rigidity is a projective invariant.⁹ As such, if the dual truss is not rigid with ball-joints only, then the primal plate structure cannot carry (all of) its loads with plate action only, and the model presented here is not applicable.

Review of projective geometry and screw theory

We will make use of concepts from projective geometry as well as screw theory. For the sake of clarity, they are provided here. Let us start by defining the following equivalence relation on a vector space $V$

x \sim y \Leftrightarrow y \in {λ x | λ \in ℝ \ {0}} (x, y \in V)

(9)

Let us denote the arising equivalence class of x with $x_{\sim} .$ This notion enables us to assign (homogeneous) coordinates to points, planes, and lines of the projective space as follows.

Homogeneous coordinates of planes and points can be attained by setting the dimension of the vectors in equation (9) to be 4. Each equivalence class uniquely represents a point, and all points are represented this way. Although the projective space does not discriminate between its points, in human thought we often think of the 3D projective space $(P G (3))$ as a union of Euclidean space $(E^{3})$ and the ideal plane at infinity (which is $P G (2)$ essentially). We will use the convention that if $p \in E^{3},$ we will use the representant of form $(1, p),$ while in case of ideal points all representants are of form $(0, λ p) .$ If no distinction is to be made, we will use the form ${(p_{0}, p)}_{\sim} .$

Similarly, each plane is represented with a single equivalence class and each equivalence class represents a unique plane. Denoting the standard inner product with $〈, 〉,$ the incidence relations take the following form: point $P$ with coordinates $p_{\sim}$ lies in plane $S$ with coordinates $s_{\sim}$ if and only if $〈 p_{\sim}, s_{\sim} 〉 = 0$ holds. As it can be seen, different choices of representant from the same equivalence class do not affect this relation.

Homogeneous coordinates of lines are attained in a slightly different way. Let us have $l = (l_{1}, l_{2}, l_{3}),$ $\bar{l} = (l_{4}, l_{5}, l_{6})$ such that the tuple $(l, \bar{l}) \in ℝ^{6}$ is not the 0 vector. The equivalence class ${(l, \bar{l})}_{\sim}$ represents a line of $P G (3)$ if and only if

〈 l, \bar{l} 〉 = l_{1} l_{4} + l_{2} l_{5} + l_{3} l_{6} = 0

(10)

holds, and all lines of $P G (3)$ are represented this way. This way of representation is called the Plücker coordinates of a line and equation (10) the Plücker identity.

Here, it is convenient to present a pair of additional identities, see Pottmann and Wallner³¹ for more details:

Coordinates of the line passing through points ${(p_{0}, p)}_{\sim}$ and ${(q_{0}, q)}_{\sim}$ are given by

{(l, \bar{l})}_{\sim} = (p_{0} q - q_{0} p, p \times q)_{\sim}

(11)

This returns $0$ , iff the two points coincide. Similarly, the common line of planes ${(u_{0}, u)}_{\sim}$ and ${(v_{0}, v)}_{\sim}$ are given by

{(l^{*}, {\bar{l}}^{*})}_{\sim} = (u \times v, u_{0} v - v_{0} u)_{\sim}

(12)

which also gives $0$ iff the two planes are identical.

Observing equations (11) and (12) shows that any line passing through a given point arise from the linear combination of three non-coplanar lines passing through the point, and any line lying in a plane arises as a linear combination of three non-concurrent lines lying in the plane.

Dualities are incidence preserving one-to one maps between points and planes of $P G (3) .$ With the interchange of incidence relations (line joining points–line in which planes intersect), they induce a corresponding one to one mapping on lines. Dualities with period 2 are called polarities and have often been used in graphical statics before.^1,2 The effect of any duality on homogeneous coordinates can be represented with an invertible matrix equivalence class $M_{\sim},$ such that the dual of point ${(p_{0}, p)}_{\sim}$ is plane $(p_{0}, p) M_{\sim},$ and the dual of plane ${(s_{q}, s)}_{\sim}$ is point $(p_{0}, p) M_{\sim}^{- T} .$ A corresponding linear map can be constructed acting on the Plücker coordinates of lines.

The ’canonical’ duality is the one represented with the four-dimensional identity matrix (its equivalence class), mapping point ${(p_{0}, p)}_{\sim}$ into plane ${(p_{0}, p)}_{\sim}$ and plane ${(p_{0}, p)}_{\sim}$ into point ${(p_{0}, p)}_{\sim} .$ As it can be seen, it has period 2 thus it is a polarity as well. (In a more geometric approach it can be thought of as being represented by the purely complex unit sphere in the complex projective space.) The induced line-to line mapping in this case is given as ${(l, \bar{l})}_{\sim} \mapsto {(\bar{l}, l)}_{\sim} .$

So far we have homogeneous coordinates, but we would like to do mechanics with metric quantities. In order to have some, let us introduce the notion of the oriented line segment (an idea is dating back to at least Klein).³²

Definition

An oriented line segment $(l, \bar{l})$ is a sliding-vector bound to line ${(l, \bar{l})}_{\sim},$ with length $| | (l, \bar{l}) | |$ and a direction, which are given by:

If $| | l | | > 0,$ then $| | (l, \bar{l}) | | = | | l | |$ and the direction is given by $l$ (finite line).

If $| | l | | = 0,$ then $| | (l, \bar{l}) | | = | | \bar{l} | |$ and the direction is given by $\bar{l}$ (ideal line).

We can now give the effect of the duality on this metric quantity similar to the line, as

(l, \bar{l}) \mapsto (\bar{l}, l)

(13)

Screw theory was originally proposed by Sir R Ball,³³ providing a connection between Plücker coordinates of lines and force/velocity/displacement systems. Here, a quick excerpt is presented, the reader may find the modern (engineering) interpretation in Davidson and Hunt³⁴ or Gallardo-Alvarado.³⁵ For the mathematical minded, there is Pottmann and Wallner³¹ and even Felix Klein.³²

The effect of system of forces and moments in $E^{3}$ can be given (after a choice of coordinate system) as the vector couple $(R, T),$ where $R$ is a force acting at the chosen origin and $T$ is a moment both containing the original moments and accounting for the translation of the elements of the force system into the chosen origin. Generally this vector couple is called a dyname or wrench. In the special case of $〈 R, T 〉 = 0$ (the Plücker identity, see equation (10)) the dyname can be reduced into a single force (if $R \neq 0)$ or a single moment (if $R = 0$ ). In the first case, $(R, T)$ is a representant from the ${(R, T)}_{\sim}$ equivalence class, corresponding to a regular oriented line segment in $P G (3),$ which is the line of action of the resulting force. In the second case it represents a line segment lying in the ideal plane (see equation (11)). In the following, we will consider only dynames satisfying equation (10) and think of moments as forces lying in the ideal plane, acting along the corresponding ideal lines.

By using oriented line segments, a force F acting along line ${(l, \bar{l})}_{\sim}$ can be represented as

\frac{‖ F ‖}{‖ (l, \bar{l}) ‖} (l, \bar{l})

(14)

assuming the force and the line segment point to the same direction.

Although in screw theory mostly forces and instantaneous velocities are represented as screws, small displacements can be handled this way as well, since they behave similarly to instantaneous velocities. We only need the formulation with the oriented line segment, as follows: One can describe a small rotation with angle $α$ around axis ${(l, \bar{l})}_{\sim}$ as

\frac{α}{‖ (l, \bar{l}) ‖} (l, \bar{l})

(15)

The sign of $α$ is according to the handedness of the coordinate system.

The following operation will be used to describe the relation of forces and the displacements they cause

(l, \bar{l}) \circ (m, \bar{m}) : = 〈 l, \bar{m} 〉 + 〈 \bar{l}, m 〉

(16)

From the mathematical point of view, this is an indefinite inner product, satisfying linearity, symmetry but not positive definiteness. Also, (different) lines ${(l, \bar{l})}_{\sim}$ and ${(m, \bar{m})}_{\sim}$ intersect if and only if ${(l, \bar{l})}_{\sim} \circ {(m, \bar{m})}_{\sim} = 0 .$ From the physical point of view, this operation can represent many things. The relevant ones are as follows:

The quantity $(| | F | | / (| | (l, \bar{l}) | |)) (1 / | | (m, \bar{m}) | |) (l, \bar{l}) \circ (m, \bar{m})$ is the moment of force $(| | F | | / (| | (l, \bar{l}) | |)) (l, \bar{l})$ to the axis ${(m, \bar{m})}_{\sim}$

Given a rigid body rotated with $(α / | | (m, \bar{m}) | |) (m, \bar{m}),$ the quantity $(α / | | (m, \bar{m}) | |) (1 / | | (l, \bar{l}) | |) (m, \bar{m}) \circ (l, \bar{l})$ is the $(l, \bar{l})$ directional component of the displacement of all points of the body lying on ${(l, \bar{l})}_{\sim}$

Note how the canonical duality satisfies the following: Given two line segments $(l, \bar{l})$ and $(m, \bar{m}),$ as well as their duals $(l, \bar{l})^{'}$ and $(m, \bar{m})^{'},$ we have

(l, \bar{l}) \circ (m, \bar{m}) = (l, \bar{l})^{'} \circ (m, \bar{m})^{'}

(17)

which will greatly simplify our analysis.

Bases and dual bases

Due to the linear nature of the problem, the effect of the duality can be described with its effect on appropriately chosen bases. First, a plate in plane $P$ and joint at dual point $P'$ will be considered. Let us have an orthonormal base in the finite part of $P G (3),$ given by $b_{1},$ $b_{2}$ and $b_{3},$ such that the origin is neither $P'$ nor lying in $P$ . With this restriction both point and plane can be represented with the same vector: $(1, n)$ . We will also consider the additional condition of $b_{1} ∥ n,$ and we will take note of what it changes. Now two pairs of bases can be created, bound to $P$ and $P'$ , respectively.

The first one will be spanned by $d_{1},$ $d_{2}$ and $d_{3},$ where

d_{j} : = (〈 n, b_{j} 〉 n, n \times b_{j}) j = 1 \dots 3

(18)

All line segments perpendicular to $P$ can be given as a linear combination of these, since they all pass through $(0, n)$ and are linearly independent. Thus, any rotation (or translation) of $P$ can be given as

\sum_{j = 1 \dots 3} \frac{μ_{j}}{‖ d_{j} ‖} d_{j} | μ_{j} \in ℝ

(19)

If the additional condition of $b_{1} ∥ n$ is satisfied, then $d_{2}$ and $d_{3}$ intersect $P$ in ideal points, thus pure translations of the plate in the plane are represented with linear combinations where $μ_{1} = 0 .$

The next base will be spanned by $f_{1},$ $f_{2}$ and $f_{3},$ where

f_{j} : = (n \times b_{j}, b_{j}) j = 1 \dots 3

(20)

Any oriented line segment lying in plane $P$ can be given as a linear combination of these (see equation (12)). Thus, any force acting in $P$ can be given as

\sum_{j = 1 \dots 3} \frac{μ_{j}}{‖ f_{j} ‖} f_{j} | μ_{j} \in ℝ

(21)

The satisfaction of the additional condition of $b_{1} ∥ n$ means $f_{1}$ is the ideal line of the plane and a moment acting in the plane is given by $μ_{2} = μ_{3} = 0,$ while $μ_{1} = 0$ means the force is passing through the point where $d_{1}$ intersects $P$ .

The corresponding dual base for forces passing through $P'$ will be ${{f^{'}}_{j}}$ where

{f^{'}}_{j} : = (b_{j}, n \times b_{j}) j = 1 \dots 3

(22)

It is easy to see (from equation (11)) that any force passing through $P^{'}$ can be given as a linear combination of these. The dual base for the displacements of the joint at $P^{'}$ will be ${{d^{'}}_{j}}$ where

{d^{'}}_{j} : = (n \times b_{j}, 〈 n, b_{j} 〉 n) j = 1 \dots 3

(23)

They all lie in a plane passing through the origin, perpendicular to $n .$ Any displacement of $P$ can be expressed with the help of these, such that

\sum_{j = 1 \dots 3} \frac{μ_{j}}{‖ {d^{'}}_{j} ‖} {d^{'}}_{j} | μ_{j} \in ℝ

(24)

To see that this indeed spans $E^{3}$ and how the displacements scale one can express the translation caused at $P'$ as

\sum_{k = 1 \dots 3} b_{k} \sum_{j = 1 \dots 3} \frac{μ_{j}}{‖ {d^{'}}_{j} ‖ ‖ {f^{'}}_{k} ‖} {d^{'}}_{j} \circ {f^{'}}_{k}

(25)

where

{d^{'}}_{j} \circ {f^{'}}_{k} = 〈 (n \times b_{j}), (n \times b_{k}) 〉 + 〈 n, b_{j} 〉 〈 n, b_{k} 〉

(26)

{d^{'}}_{j} \circ {f^{'}}_{k} = 〈 n, n 〉 〈 b_{j}, b_{k} 〉 - 〈 n, b_{j} 〉 〈 n, b_{k} 〉 + 〈 n, b_{j} 〉 〈 n, b_{k} 〉

(27)

{d^{'}}_{j} \circ {f^{'}}_{k} = 〈 n, n 〉 〈 b_{j}, b_{k} 〉

(28)

This, the orthonormality of ${b_{j}}$ and the fact that $\forall j | | {f^{'}}_{j} | | = 1$ shows that the translation of $P$ from the rotation described in equation (24) is

{‖ n ‖}^{2} \sum_{j = 1 \dots 3} \frac{μ_{j}}{‖ {d^{'}}_{j} ‖} b_{j}

(29)

With this and equation (17), we have also discovered which base vector (line segment) intersects which one. Here, the satisfaction of the additional condition of $b_{1} ∥ n$ implies that ${d^{'}}_{1}$ lies at infinity. A drawing of the bases is presented in Figure 2 in the general case and in Figure 3 if $b_{1} ∥ n$ is satisfied.

Figure 2.

Left: Bases associated with plane $(1, n) .$ Elements of ${f_{j}}$ are lying in it while elements of ${d_{j}}$ are perpendicular to it. Right: Dual bases, associated with point $(1, n) .$ Elements of ${{f^{'}}_{j}}$ are passing through it, while elements of ${{d^{'}}_{j}}$ lie on a plane passing through the origin of ${b_{j}} .$

Figure 3.

Bases an dual bases if $b_{1} ∥ n$ is satisfied. Some line segments are drawn twice, their apparently opposite direction is attributed to the fact that all lines are loops in $P G (3) .$ Left: Bases associated with plane $(1, n) .$ Line segment $f_{1}$ is along the ideal line of the plane, accounting for the moments in the plane. Line segments $d_{2}$ and $d_{3}$ also lie in infinity, representing pure translations of the plane. Right: Dual bases, associated with point $(1, n) .$ Line segment ${d^{'}}_{1}$ lies along an ideal line.

Remark

The restriction of the origin of ${b_{j}}$ not being a vertex of the truss or in a plane of a plate is a necessary condition of having two finite dual structures.

Extension to multiple plates and joints

Trusses and plate structures consist of multiple plates/joints and edges/bars. The vector representing plane $P_{i}$ and point $P'_{i}$ will be denoted with $(1, n_{i}) .$ Consequently there will be multiple bases attached to points and planes, for instance, the jth element of the base for forces attached to $P_{i}$ will be denoted with $f_{i, j} .$ The extension of the additional condition on base directions takes the form of $b_{i, 1} ∥ n_{i}$ for all $i .$ Moreover, a line segment along edge k and dual bar k will be denoted with $e_{k}$ and ${e^{'}}_{k},$ respectively.

Now the validity of Proposition I can be shown. If edge force $ϕ_{k}$ acts along edge $e_{k}$ on plate $P_{i},$ its moment to axis $d_{i, j}$ is given by

ϕ_{k} \frac{e_{k} \circ d_{i, j}}{‖ e_{k} ‖ ‖ d_{i, j} ‖}

(30)

Similarly, the $e_{k}$ directional component from the small rotation $δ_{i, j}$ around axis $d_{i, j}$ is

δ_{i, j} \frac{e_{k} \circ d_{i, j}}{‖ e_{k} ‖ ‖ d_{i, j} ‖}

(31)

It is apparent that if we want to use the same matrix (although transposed) in both the equilibrium and the compatibility equations, the required equilibrium equations are the moment equations around axes $d_{i, j} .$ By introducing the notation $r = 3 (i - 1) + j,$ the elements of $G$ can be given as

G_{r, k} = {\begin{cases} \frac{e_{k} \circ d_{i, j}}{‖ e_{k} ‖ ‖ d_{i, j} ‖} if ϕ_{k} acts on the plate P_{i} \\ \frac{- e_{k} \circ d_{i, j}}{‖ e_{k} ‖ ‖ d_{i, j} ‖} if - ϕ_{k} acts on the plate P_{i} \\ 0 otherwise . \end{cases}

(32)

Please note, how considering plate $P_{u}$ and $P_{v}$ joined by edge $e_{k},$ if force $ϕ_{k}$ acts on $P_{u},$ then force $- ϕ_{k}$ acts on plate $P_{v} .$ Details on a practical sign convention will be provided later, when the transformation of forces is discussed.

In order to see how $G^{T} δ$ gives the relative displacements of the plates along edges, let us expand the kth row of $G^{T} δ$ as

\pm (\sum_{j = 1 \dots 3} δ_{u, j} \frac{e_{k} \circ d_{u, j}}{‖ e_{k} ‖ ‖ d_{u, j} ‖} - \sum_{j = 1 \dots 3} δ_{v, j} \frac{e_{k} \circ d_{v, j}}{‖ e_{k} ‖ ‖ d_{v, j} ‖})

(33)

The minus sign is required since plates $P_{u}$ and $P_{v}$ move relative to each other, its appearance can be seen through the force–reaction force reasoning shown above. The first part of the sum

d_{u}^{k} : = \sum_{j = 1 \dots 3} δ_{u, j} \frac{e_{k} \circ d_{u, j}}{‖ e_{k} ‖ ‖ d_{u, j} ‖}

(34)

is the $e_{k}$ directional component of the displacements of all the points of $P_{u}$ lying on edge $e_{k}$ (see Figure 4). The second part of the relative displacement, $d_{v}^{k},$ can be defined similarly. These are absolute displacements, with respect to a stationary coordinate system lying on $e_{k} .$ The choice of its direction influences weather the ± sign is + or –.

Figure 4.

Relative displacement of plates $P_{u}$ and $P_{v},$ calculated as $d_{u}^{k} - d_{v}^{k} .$ The edge is doubled to conveniently show the absolute displacements $d_{u}^{k}$ and $d_{v}^{k},$ as well as their derivation. While the vectorial displacements ( $d^{a}$ and $d^{b}$ ) vary depending on which point ( $a$ or $b$ ) of the plate we pick along edge $e_{k},$ the $e_{k}$ directional component is always the same. This can be seen through a planar reasoning in the plane of the plate.

In the following, we will describe how statics and kinematics of a plate structure translate to statics and kinematics of the dual truss.

Finding the transformation matrices

Transformation of forces

Consider plate $P_{i}$ acted upon by force

\sum_{j = 1 \dots 3} \frac{π_{i, j}}{‖ f_{i, j} ‖} f_{i, j}

(35)

and supported along its $e_{k}$ edges with $ϕ_{k}$ edge forces ( $k \in ℐ$ some index-set). The equilibrium of forces is captured with the equation

\sum_{j = 1 \dots 3} \frac{π_{i, j}}{‖ f_{i, j} ‖} f_{i, j} + \sum_{k \in ℐ} \frac{ϕ_{k}}{‖ e_{k} ‖} e_{k} = 0

(36)

In case of the dual vertex at point $P'$ , the equilibrium of the dual forces and the dual load gives the equation

\sum_{j = 1 \dots 3} \frac{{π^{'}}_{i, j}}{‖ {f^{'}}_{i, j} ‖} {f^{'}}_{i, j} + \sum_{k \in ℐ} \frac{{ϕ^{'}}_{k}}{‖ {e^{'}}_{k} ‖} {e^{'}}_{k} = 0

(37)

Both equations (36) and (37) are linear combinations where the corresponding vectors ( $f_{i, j}$ to ${f^{'}}_{i, j}$ and $e_{k}$ to ${e^{'}}_{k})$ differ only in a common permutation. This implies that given the additional constraints

\frac{π_{i, j}}{‖ f_{i, j} ‖} = \frac{{π^{'}}_{i, j}}{‖ {f^{'}}_{i, j} ‖} and \frac{ϕ_{k}}{‖ e_{k} ‖} = \frac{{ϕ^{'}}_{k}}{‖ {e^{'}}_{k} ‖}

(38)

the two equilibrium equations are equivalent. Since this is precisely what we want, we determine $Π$ and $Φ$ such that equation (38) is satisfied. Both $Φ$ and $Π$ are diagonal, the $k, k th$ element of $Φ$ is

Φ_{k, k} = \frac{‖ e_{k} ‖}{‖ {e^{'}}_{k} ‖}

(39)

while $Π$ is more conveniently given in block-diagonal form as

Π = [\begin{matrix} Π_{1} \\ ⋱ \\ Π_{n} \end{matrix}]

(40)

where

Π_{i} = [\begin{matrix} ‖ f_{i, 1} ‖ \\ ‖ f_{i, 2} ‖ \\ ‖ f_{i, 3} ‖ \end{matrix}]

(41)

since $| | {f^{'}}_{i, j} | | = 1$ for all $i$ and $j .$

Solving equation (1) tells us the magnitude of bar forces, and whether they represent tension or compression. In order to use the dual structure effectively, one must see how the sign convention appears in the plate structure. If bar k joining vertices $n_{i}$ and $n_{j}$ is under tension, the effect of its force on vertex $i$ can be given as

\frac{{ϕ^{'}}_{k}}{‖ n_{j} - n_{i} ‖} (n_{j} - n_{i}, n_{i} \times n_{i})

(42)

such that ${ϕ^{'}}_{k} > 0 .$ This means the dual force is of form

\frac{ϕ_{k}}{‖ n_{i} \times n_{j} ‖} (n_{i} \times n_{j}, n_{j} - n_{i})

(43)

such that $ϕ_{k} > 0 .$ This segment is pointing towards $n_{i} \times n_{j} .$ We may observe then that if bar k is under tension, then the dual edge force acting on plate i is such that $n_{i},$ $n_{j}$ and the edge force $ϕ_{k} (n_{i} \times n_{j})$ are ordered according to the handedness of the coordinate system. Interchanging $n_{i}$ and $n_{j}$ and remembering the anti-commutativity of the cross product shows, this satisfies Newton’s third law of forces and reactions. While $n_{i} \neq n_{j}$ is a trivial requirement, $n_{i} ∥ n_{j} \Leftrightarrow | | n_{i} \times n_{j} | | = 0$ would cause problems. In this case however, the intersection of the planes would lie in the ideal plane. Since we consider only finite structures, this never happens and the provided explanation is general enough.

Transformation of displacements

The displacement $d_{i}$ of $P_{i}$ can be given as

d_{i} = \sum_{j = 1 \dots 3} \frac{δ_{i, j}}{‖ d_{i, j} ‖} d_{i, j}

(44)

while the dual displacement ${d^{'}}_{i}$ is

{d^{'}}_{i} = \sum_{j = 1 \dots 3} \frac{{δ^{'}}_{i, j}}{‖ {d^{'}}_{i, j} ‖} {d^{'}}_{i, j}

(45)

Similar to the case of the forces, these are linear combinations in which the vectors differ only in a common permutation. Furthermore, the method we want to create is good if it maps unit displacement to unit displacement as

\frac{1}{‖ d_{i, j} ‖} \mapsto \frac{1}{‖ {d^{'}}_{i, j} ‖}

(46)

implying

\frac{δ_{i, j}}{‖ d_{i, j} ‖} = \frac{{δ^{'}}_{i, j}}{‖ {d^{'}}_{i, j} ‖}

(47)

However, what we usually prefer in case of a truss is the translational displacement of $P_{i}^{'},$ in the form of ∑ $j = 1 \dots 3 ϵ_{i, j} b_{i, j} .$ We can use equation (29) to get precisely that and have

δ_{i, j} = \frac{‖ d_{i, j} ‖}{{‖ n_{i} ‖}^{2}} ϵ_{i, j}

(48)

From this we see that the matrix $Δ$ can also be given in block-diagonal form, as

Δ = [\begin{matrix} Δ_{1} \\ ⋱ \\ Δ_{n} \end{matrix}]

(49)

where

Δ_{i} = \frac{1}{{‖ n_{i} ‖}^{2}} [\begin{matrix} ‖ d_{i, 1} ‖ \\ ‖ d_{i, 2} ‖ \\ ‖ d_{i, 3} ‖ \end{matrix}]

(50)

While we see that $Φ$ is determined only by the geometry of the structure and the centre of the duality (the origin of the coordinate system in this case), $Π$ and $Δ$ are sensitive to directions of $b_{i, j} .$ More precisely, if the additional condition of $b_{i, 1} ∥ n_{i}$ holds, then $Π^{- 1} = Δ$ is true.

Transformation of the system of equations

Now we can examine how and when the systems of equations can be transformed into each other. In case of the equilibrium equations, we can write

G ϕ + π = 0

(51)

G Φ ϕ^{'} + Π π^{'} = 0

(52)

Π^{- 1} G Φ ϕ^{'} + π^{'} = 0

(53)

implying

G^{'} = Π^{- 1} G Φ

(54)

In case of the compatibility equations, this is

G^{T} δ + F ϕ = 0

(55)

G^{T} Δ ϵ + F Φ ϕ^{'} = 0

(56)

Φ G^{T} Δ ϵ + Φ F Φ ϕ^{'} = 0

(57)

which would imply

G^{'} = Δ G Φ

(58)

F^{'} = Φ F Φ

(59)

Two things are noteworthy. First, the connection between F and $F^{'}$ is determined only by the geometry of the structure and the centre of the duality, not the choice of bases. Second, if the bases are chosen such that $b_{i, 1} ∥ n_{i}$ (or $b_{i, j} ∥ n_{i}$ for all $i$ and any j) is satisfied, then there is no contradiction between equations (54) and (58) and the whole system of equations can be transformed.

This condition however is not necessary to use this method. In fact, it might be useful to use bases given by assuming $b_{i, j} = b_{k, j} \forall i, k \in {1 \dots n},$ resulting in the frame at each joint having the same orientation. The matrices $Π, Δ$ and $Φ$ can be constructed as given, and they can be used to transform the loads and the stiffness matrix. The dual truss can be solved with whatever equations one likes, and the resulting forces and displacements can be transformed back.

A numerical example

Consider the plate structure seen in Figure 5. The vertical plates are fixed in their planes supporting plate 5 in a statically indeterminate way. The force $π_{5}$ is of magnitude 1 kN, and there is an additional support displacement: plate 1 is rotated around the x-axis with 1/10 degrees (0.0017 radians).

Figure 5.

Plate structure arising from five plates (the vertical ones are fixed, supporting ones) and four internal edges.

The geometry of the plates is given by vectors

(1, n_{1}) = (1, - 1, 0, 0)

(60)

(1, n_{2}) = (1, - 1, - 1, 0)

(61)

(1, n_{3}) = (1, 1, - 1, 0)

(62)

(1, n_{4}) = (1, 1, 0, 0)

(63)

(1, n_{5}) = (1, 0, - 0.2, - 0.4) .

(64)

The plates are assumed to be glued together with polyurethane glue in 1 mm thickness and in 2 cm width. By assuming the glue has shear modulus of 1 N/mm² and after working out the edge lengths the stiffness matrix takes the form of

F = [\begin{matrix} 0.4472 & 0 & 0 & 0 \\ 0 & 0.3333 & 0 & 0 \\ 0 & 0 & 0.3333 & 0 \\ 0 & 0 & 0 & 0.4472 \end{matrix}] 10^{- 7} m / kN

(65)

The internal edges will be numbered such that edge k joins plate k with plate 5, this way the edges can be represented with

e_{1} = (0, - 0.4, 0.2, 1, - 0.2, - 0.4)

(66)

e_{2} = (0.4, - 0.4, 0.2, 1, 0.8, - 0.4)

(67)

e_{3} = (0.4, 0.4, - 0.2, - 1, 0.8, - 0.4)

(68)

e_{4} = (0, 0.4, - 0.2, - 1, - 0.2, - 0.4)

(69)

and from this the matrix relating bar and edge forces is

Φ = [\begin{matrix} 0.4082 & 0 & 0 & 0 \\ 0 & 0.4472 & 0 & 0 \\ 0 & 0 & 0.4472 & 0 \\ 0 & 0 & 0 & 0.4082 \end{matrix}]

(70)

From this one can have the stiffness matrix of the dual truss as

F^{'} = [\begin{matrix} 0.7454 & 0 & 0 & 0 \\ 0 & 0.6667 & 0 & 0 \\ 0 & 0 & 0.6667 & 0 \\ 0 & 0 & 0 & 0.7454 \end{matrix}] 10^{- 8} m / kN

(71)

We will go with the strategy of $b_{i, 1} = (1, 0, 0),$ $b_{i, 2} = (0, 1, 0)$ and $b_{i, 3} = (0, 0, 1)$ for all i. This has the benefit of having the local bases of the truss corresponding to the directions of the global coordinate system. Thus, the relevant bases to this problem are ${d_{1, j}},$ ${d_{5, j}}$ and ${f_{5, j}}$ . They are given by

d_{1, 1} = (1, 0, 0, 0, 0, 0)

(72)

d_{1, 2} = (0, 0, 0, 0, 0, - 1)

(73)

d_{1, 3} = (0, 0, 0, 0, 1, 0)

(74)

d_{5, 1} = (0, 0, 0, 0, - 0.4, 0.2)

(75)

d_{5, 2} = (0, 0.04, 0.08, 0.4, 0, 0)

(76)

d_{5, 3} = (0, 0.08, 0.16, - 0.2, 0, 0)

(77)

f_{5, 1} = (0, - 0.4, 0.2, 1, 0, 0)

(78)

f_{5, 2} = (0.4, 0, 0, 0, 1, 0)

(79)

f_{5, 3} = (- 0.2, 0, 0, 0, 0, 1)

(80)

From this one can work out the relevant parts of $Δ$ and $Π$ . It turns out that $Δ_{1}$ is the 3D identity matrix and

Δ_{5} = [\begin{matrix} 5 & 0 & 0 \\ 0 & 1.1180 & 0 \\ 0 & 0 & 4.4721 \end{matrix}] \frac{1}{m}

(81)

Π_{5} = [\begin{matrix} 0.4472 & 0 & 0 \\ 0 & 0.4 & 0 \\ 0 & 0 & 0.2 \end{matrix}]

(82)

The decomposition of the loads into the appropriate bases gives

δ_{1} = [\begin{matrix} 0.0017 \\ 0 \\ 0 \end{matrix}] rad and π_{5} = [\begin{matrix} - 1 \\ 0 \\ 0 \end{matrix}] kN

(83)

which can be transformed to truss loads as

{π^{'}}_{5} = Π_{5}^{- 1} π_{5} = [\begin{matrix} - 2.2361 \\ 0 \\ 0 \end{matrix}] kN

(84)

and

ϵ_{1} = Δ_{1}^{- 1} δ_{1} = [\begin{matrix} 0.0017 \\ 0 \\ 0 \end{matrix}] m

(85)

Now everything is given in the problem of the dual truss, which can be seen in Figure 6, along with the directions of the arising bar forces as well. Importantly, as our choice of local bases does not satisfy $b_{i, j} ∥ n_{i}$ (for all $i$ and any j), we cannot transform G to G′ or vice versa. However, this step is not necessary to solve the plate structure as we can handle the truss with the known methods, obtaining

ϵ_{5} = [\begin{matrix} 0.0005 \\ - 0.0009 \\ - 0.0017 \end{matrix}] m and ϕ^{'} = [\begin{matrix} - 4.5643 \\ 5.5899 \\ - 5.5899 \\ 4.5643 \end{matrix}] 10^{4} kN

(86)

Figure 6.

Dual truss with five vertices and four internal bars. The supporting vertices are represented with ◦, while the internal one with •.

This can be transformed back to the plate structure giving

δ_{5} = [\begin{matrix} 0.0025 \\ - 0.0010 \\ - 0.0078 \end{matrix}] and ϕ = [\begin{matrix} - 1.8634 \\ 2.4999 \\ - 2.4999 \\ 1.8634 \end{matrix}] 10^{4} kN

(87)

Since the bases and dual bases are not normed, the actual rotation of $P_{5}$ is not $| | δ_{5} | |,$ but one has to compute

\begin{array}{l} d_{5} = 0.0175 d_{5, 1} - 0.0195 d_{5, 2} - 0.1561 d_{5, 3} \\ = (0, - 0.0007, - 0.0013, 0.0012, - 0.0010, 0.0005) \end{array}

(88)

The norm of this can be taken as defined in case of a line segment, returning the magnitude of the rotation of plate 5 to be $0.0015$ radians or 0.0850 degrees. The axis and direction of this are given by the line segment $d_{5} .$

Summary

A metric correspondence was provided between forces and small displacements of a plate structure (made of plates rigid in their planes, soft perpendicular to their planes, joined by elastic edges) and those of its dual: a truss with elastic bars. Since the computational methods of trusses are well developed and widely known, this ability to turn unfamiliar problems into familiar ones may give a useful tool in the hands of structural engineers or researchers. Especially now, when automation propels the spread of prefabricated plate structures, where the joining edges are usually softer then the plates itself.

As the provided method is based on a conceptional model, it comes with the corresponding advantages and disadvantages. By focusing on one type of behaviour, it sacrifices some of its applicability to general shell structures. In exchange, the behaviour focused on may be easier to grasp and to be used as a design principle. Furthermore, the arising system of linear equations is relatively small compared to finite-element approaches and its generation can be easily automated. As such it can be used in real-time form finding and computational design. In comparison, the finite-element approach is typically time consuming both in its modelling and computational aspect. The use of the presented model may eventually be also analogous to the use of the pin-joint model in modern truss design, where engineers start with the pin-joint model as a concept and then model the possibly rigid connections in detail if necessary. As the material and geometric limitations of the presented model are as of yet unknown, further investigations of these aspects are still required.

Footnotes

Acknowledgements

The author wishes to thank professors P.L. Várkonyi, T. Tarnai and I. Sajtos for useful discussions on the subject.

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.

ORCID iD

Tamás Baranyai

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