AAnalytical graphic statics
Tamás BaranyaiFebruary 25, 2019
Abstract
Graphic statics is undergoing a renaissance, with computerized visual representation becomingboth easier and more spectacular as time passes. While methods of the past are revived and tweaked,little emphasis has been placed on studying the details of these methods. Due to the considerableadvances of our mathematical understanding since the birth of graphic statics, we can learn manyinteresting and beautiful things by examining these old methods from a more modern viewpoint.As such, this work shows the mathematical fabric joining different aspects of graphic statics, likedualities, reciprocal diagrams and discontinuous stress functions.
Graphic statics was born with the works of Culmann [1] and Bow [2], and it contained two diagrams,namely the force diagram and the geometrical or form diagram. The force diagram representedmagnitudes of forces with lengths of the corresponding line segments, while each line in the formdiagram gave the line of action of the corresponding force. In the early days of graphic statics,these two diagrams were connected by algorithmic methods. As structures became more and morecomplex, this algorithmic approach became too cumbersome, and graphic methods gave way toalgebraic ones. With the emergence of computers and their huge visualising power, these algorithmicmethods became feasible once again and are proposed as optimizing tools [3, 4, 5].The next step in the evolution of graphic statics came with Maxwell [6], who gave a 3 dimensionalgraphical construction to solve a two dimensional truss. His method introduced two new aspectsto graphic statics: One was the use of projective geometrical dualities (a symmetric polarity inparticular) to establish connection between the form and force diagrams, the other was the notion ofa discontinuous stress function corresponding to the discontinuous structure the truss is. These twonotions went on to live somewhat distinct evolutionary paths as more research followed.The set of usable projective geometrical dualites was quickly expanded with null-polarites (sym-plectic polarites) by Cremona [7]. Polarites were considered to be special in this regard by researchers[8] until very recently, when it has been shown [9] that any projective duality is usable for this purposeand in infinitely many ways.The idea of using a stress function for analysing discontinuous structures was quickly supportedby Klein and Wieghardt [10] with some rigorous mathematics, but the next paper came much later.It was in the 1980-s when the topic emerged again, interestingly for spatial trusses first [11], thenfor planar ones [12]. With the renewed interest a number of works [13, 14] described the use ofthis tool, essentially coining it Airy stress function (similarly to the continuum case). Apart fromtrusses, a recent pair of papers [15, 16] tackled the case of planar and spatial frames respectively.Yet, even these fairly recent works, while noting that after defining polyhedral stress functions oneought to have moments as function values and mentioning the possible use of projective dualities,did not explain the connection between the proposed elements.What appears to be missing is the explanation of the mathematical fabric tying these conceptstogether. This paper attempts to do that, with the simplest mathematical tool possible. The tool ofchoice is the notion of dual spaces associated to vector-spaces, motivated by the modern mathematicalapproach to projective geometry [17]. a r X i v : . [ phy s i c s . c l a ss - ph ] F e b The dual nature of forces
The main point of this paper is that forces can be considered both as vectors and as linearfunctionals, and graphic statics used both approaches in its diagrams. In order to see this, we willoccasionally need projective homogeneous coordinates. Although they are becoming fairly standard,the reader is provided with a brief introduction to them in the Appendix, if necessary.Let us consider a planar body acted on by a planar force system comprised of forces f i ( i ∈{ , . . . n } ) and let us have an orthogonal x, y coordinate-frame in the plane. We can describe anyforce f i acting in the plane with the triplet f i = ( F x,i , F y,i , M i ) , where F x,i and F y,i are the projectionsof the force to the coordinate axes and M i is the moment of the force with respect to an axisperpendicular to the plane passing through the origin (with respect to the origin for short). Let v ∈ R be an arbitrary vector, and let v i be defined as v i := v i − + f i . (1)The equilibrium of forces (cid:80) f i = 0 ( i ∈ { , . . . n } ) holds precisely if v n = v , and this can begraphically represented with a closed, arrow-continuous chain of vectors f i . Although this diagramhas not been used often in this 3 dimensional form, the projection of this to the M = 0 plane isnothing else then the well known "force diagram".Now let s i be defined as s i := ( − F y,i , F x,i , M i ) (2)and let us note, that the the scalar product (cid:104) ( q x , q y , , s i (cid:105) = − q x F y,i + q y F x,i + M i is the momentof the force with respect to the point ( q x , q y ) . We have embedded our 2D problem into 3D with z = 1 and we can think of s i as a linear functional, which can be evaluated at points of thevector-space the planar body is in. Naturally s i uniquely corresponds to (represents) force f i , andthe line of action of force f i is the intersection of the zero-space of s i with the plane z = 1 . Letus also note, how the triplets ( q x , q y , can be thought of as carefully chosen representants fromthe equivalence class ( q x , q y , ∼ , which is the projective point coordinate form of a point in the x, y plane. In this context the line of action of f i is precisely the equivalence class s i ∼ . Note,how given two line representants s i and s j , the sum s i + s j represents a line passing through theintersection point of s i and s j . The corresponding mechanical property is that the sum of twoforces has to pass through the intersection of the two original forces. After this has been pointedout, it is easy to see how this functional formulation lies under the "form diagram" of graphic statics.Since this paper is about graphic statics, we will want to draw this functional. One way to doit, is to measure up the value (cid:104) ( q x , q y , , s i (cid:105) over each point ( q x , q y , resulting in plane − F y,i x + F x,i y, − z, + M i = 0 , or with homogeneous coordinates ( − F y,i , F x,i , − , M i ) ∼ (3)(alternatives to this will be addressed later, in Section 3). Since a linear combination of functionals is also a functional and a linear combination of plane-representants also represents a plane, we are able to apply a projective duality to the 3 dimensionalforce diagram introduced earlier and extract the lines of action of the forces involved, as well ascreate the moment diagram of the structure. Any projective duality is applicable for this purpose,with the appropriate representation of moments. This representation will be addressed later, first afew examples are provided that will serve as a basis the general case can be reduced to.Let us number the forces according to how they follow each other on the body, and create the 3dimensional force diagram as defined above. The vertices of this diagram are then mapped to planesby a duality, which is describable with an invertible matrix equivalence class D ∼ as: p i ∼ := ( v i , ∼ D ∼ . (4)Due to the way the force diagram is defined, p i ∼ − p i − ∼ represents force f i and p i ∼ − p ∼ representsthe resultant of the force system (cid:80) i f i . Here i may also run on a subset of i ∈ { , . . . n } , depending n the forces we wish to consider the resultant of. The problem that we do not see these planesdirectly on the dual figure can be solved by observing the relation of the parts forming these linearcombinations. The orthogonal projection to the z = 0 image plane of the intersection of p i ∼ and p i − ∼ will be the line of action of f i , while the projection of the intersection of p i ∼ and p ∼ willbe the line of action of (cid:80) i f i . Furthermore, the moment diagram of the structure appears betweenplane p ∼ and p i ∼ , in the planes the vertical projecting rays running through the structure graze; asthe moment the structure is subjected to at any point is caused by the resultant (cid:80) i f i at that point. A basic example is presented in Figure 1a. It is in equilibrium with forces f =(1 , − , − (5) f =(0 , , (6) f =( − , , (7) f =(1 , − , − . (8)Let us pick v = (0 , , and create the force diagram. It is shown in Figure 2a. We can now use anull-polarity to map v i ∼ to p i ∼ as ( F x , F y , M, ∼ (cid:55)→ ( − F y , F x , − , M ) ∼ (9)which is the drawing of the functionals introduced in (3). Let us note how p ∼ is the x, y plane,meaning the intersection of p i ∼ with it is already the line of action of the resultant without anyprojection. Furthermore, if the resultant of the forces at a given point of the structure is (cid:80) i f i , thebending moment the structure is subjected to at that point can be measured vertically over thepoint, between p i ∼ and the x, y plane. Correspondingly, the projection required to get the lines ofaction of the forces is an orthogonal projection along the vertical direction (see Figure 2b).(a) Geometry of the first example (b) Geometry of the second example Figure 1
The reference plane for the moment diagram is not necessarily the z = 0 plane. Illus-trating this, consider the second example presented in Figure 1b. The statical equilibrium is satisfiedwith forces f =(0 , − , (10) f =(0 , , (11) f =(0 , − / , − / (12) f =(1 , − , − (13) f =( − , / , / . (14)The corresponding force diagram is shown in Figure 3a. The starting point was again v = (0 , , .However, instead of the null polarity defined above, the duality used to get the form diagram was ( F x , F y , M, ∼ (cid:55)→ ( − F y + 1 , F x , − , M ) ∼ (15) a) Force diagram of the structure shown in Figure 1a. (b) Moment functionals evaluated above the structuregiven in Figure 1a. The moment diagram is shown inred. the lines of actions of the forces are denoted with s i while the lines of action of the resultant with r i . Figure 2 which is no longer a polarity. It also maps v = (0 , , to plane z = x (plane p ∼ ). This means themoment diagram appears between planes p ∼ and p i ∼ , and the intersections of p ∼ and p i ∼ wouldhave to be projected down as well to get the lines of action of the resultants. The required projectionis again an orthogonal vertical one, as it can be seen in Figure 3b. In the case of multiple bodies we might not want to have the same starting point for eachof the loops in the force diagram. To see such an example, consider the structure presented in Figure4a, with the corresponding free-body diagram in Figure 4b. The indexing convention adopted wasthat f i,j is the j − th force acting on body i . The statical equilibrium is satisfied with forces f , =(1 / , , − (16) f , =(0 , − ,
3) = − f , (17) f , =( − / , ,
1) = − f , (18) f , =(0 , − , −
3) = − f , (19) f , =( − / , , (20) f , =(0 , − , . (21)Since all bodies are subjected to precisely 3 forces, the statical equilibrium requires these forcesto meet at a single point, implying the planarity of each corresponding loop in the force diagram.We can arrange these loops similarly to Cremona’s force plan, but in 3 dimensions, resulting in apolyhedron as presented in Figure 5a. Each face of this polyhedron represents the equilibrium ofa (sub-) body. We still have 3 degrees of freedom in deciding where to place the diagram (in thiscase v , = 0 was chosen), but the relative starting positions of the loops are determined by thisconstruction. (In this case, we have v , = v , = v , − f , .) The duality used to get p i from v i wasthe same as in (9), resulting in p , = (0 , , − , ∼ and p , = p , = (0 , − / , − , ∼ . (22)This means that while the moment diagram of the third body can be measured above or belowthe z = 0 plane, the moment diagrams of the first and second bodies appear between the plane y/ z = 1 and planes p i,j ( i, j ∈ { , } ). a) Force diagram of the structure shown in Figure 1b (b) Moment functionals evaluated above the structuregiven in Figure 1b. The moment diagram is shown inred. the lines of actions of the forces are denoted with s i . Figure 3
An earlier work [9] generalized Maxwell’s duality-based method to include all projective dualities.The proposed method consisted of two polyhedra such that one is a dual of the other under someprojective duality. The projected image (with some projection) of the edges of one of the polyhedragave the 2D form diagram of the structure under consideration (which was a truss, but we see nowhow it need not necessarily be a truss). Similarly, a possibly different projection of the edges of thedual polyhedron provided the 2D force diagram (Cremona force-plan) of the structure. If we wantto generalize this to our 3D force and 3D form diagrams, we can decompose arbitrary duality as theduality given in (9) and a pair of projective transformations T g and T f , acting on the 3D form and3D force diagrams respectively. The correctness of this approach was proven in the aforementionedpaper. There the problem is reduced to multiple possible dualities instead of the one given in (9),but with the help of affine transformations. Here the strict choice of (9) requires that T g and T f aremore general, projective transformations.Consider point q ∼ of the structure, and let m ∞∼ denote the point at infinity in the verticaldirection, which was the projection centre in the examples provided. Let us denote the intersectionof line ←−−−−→ q ∼ , m ∞∼ with planes p ∼ and p i ∼ with m ∼ and m i ∼ respectively. As the examples have shown,the slice of the moment diagram corresponding to q ∼ is the line segment m ∼ m i ∼ . By treating themoment functionals in a metric way, we implicitly imposed a scale on this line segment, which can berepresented by adding point m ∼ to this line, at one unit distance from m ∼ in the positive direction.Since m ∼ m ∼ = 1 holds, we can express the value of the moment diagram as a cross ratio m ∼ m i ∼ = m ∼ m i ∼ m ∼ m ∼ = ( m ∞∼ , m ∼ , m ∼ , m i ∼ ) (23)using that m ∞∼ is an ideal point. The importance of this is that the cross ratio is a projective in-variant. Using the notation that T g ( m ∼ ) = m (cid:48)∼ , the value of the moment diagram can be calculatedas the cross ratio ( m (cid:48)∞∼ , m (cid:48) ∼ , m (cid:48) ∼ , m (cid:48) i ∼ ) , which is the same as m ∼ m i ∼ since projective transfor-mations preserve cross ratios.As for the 3D force diagrams, we will also measure moments as cross ratios on the projecting rays.Starting again with the examples provided, let l ∼ be an arbitrary projecting ray (vertical line inthis case). The fact that a force system (cid:80) f i has moment (cid:80) M i with respect to the origin can be a) Geometry of the third example (b) Free-body diagram of the third example Figure 4 expressed with the cross ratio ( l ∞∼ , l ∼ , l ∼ , l i ∼ ) = (cid:88) M i (24)where l ∞∼ is the ideal point of the projecting ray and l ∼ , l ∼ and l i ∼ are the horizontal projectionsof v , v + (0 , , and v i to the projecting ray l ∼ . Since we are defining this value with a horizontalprojection, it is easy to see how the choice of l ∼ does not affect the value of the cross ratio. If wewish to use a general duality, we will have to construct, or we will get as a result a 3D force diagramthat is a projective transform of the example diagram, under transformation T f . The figure can beinterpreted/created with the help of the cross ratio ( T f ( l ∞∼ ) , T f ( l ∼ ) , T f ( l ∼ ) , T f ( l i ∼ )) = (cid:88) M i . (25)While at first this representation might look strange, in terms of actually representing a valuein a planar image it is better then the area-based representational idea of the so-called Rankine-reciprocals. Since the cross ratio is projective-invariant, it is preserved during the operation in whichwe project the 3D figure to our 2D image plane (paper or computer screen). As such, given a rulerand a calculator it is possible to accurately reconstruct the values of the moment diagram from asingle image. Contrast this with the theoretically impossible problem of determining and accurately comparing the areas of polygons which were originally skew to the image plane. In light of all this, we can understand Maxwell’s [6] and Cremona’s [7] graphical truss solvingmethods as a special case of this phenomenon. Special in the sense that the bars of the structurecoincide with the lines of action of the forces, and special in the sense that the force diagrams ofthe forces acting on a joint form planar polygons. (Since all forces f i acting on a joint have to passthrough the location of the joint, say ( j x , j y ) , we have (cid:104) s i , ( j x , j y , (cid:105) = 0 for all i implying this.)These planar polygons formed the faces of the polyhedra Maxwell related to what is now calledAiry stress function. This brings us to the works of Hegedüs [12] and Williams and McRobie [15],who found moment diagrams while creating discontinuous stress functions to analyse discontinuousstructures. We have shown where these diagrams come from and how these stress-functions can beconsidered as the evaluation of the appropriate moment functionals. Also, the choice of v or p ,carries the same degrees of freedom as the stress function has: due to differentiating twice, stressfunctions F ( x, y ) and F ( x, y ) + ax + by + c ( a, b, c ∈ R ) produce the same stresses. With this, wehave seen the underlying connection between different past approaches and fully understood thereason why they work.Furthermore, it was also shown how to interpret the moment values in the generalized version[9] of Maxwell’s reciprocal construction. The approach provided here can also be useful for moderngraphical analysis of structures, with the newly proposed method to construct the line of action of a) Force diagram of the structure shown in Figure 4a.Note the different loops corresponding to the differentbodies. Each loop is planar, since each set of forces meetat a single point. (b) Moment functionals evaluated above the structuregiven in Figure 4a. The moment diagram is shown inred. the lines of actions of the forces are denoted with s i,j . Note the different reference planes p , = p , and p , . Figure 5 the resultant from the (3 dimensional) force diagram.We have also seen how the need for a 3 dimensional duality to pass from vector to functionalform arises from the 3 dimensionality of the forces involved. Correspondingly, the generalizationof this method to 3 dimensional objects loaded with arbitrary forces remains an interesting openquestion, as one might attempt a 6 dimensional approach. There is a proposition to use 4 dimensionalpolyhedra[18] present in current literature, which is tied to Rankine’s representational idea. Whilethe higher dimensional approach can be simpler (the Dandelin spheres, usually also presented inprojective geometry classes are a classic example of this), the representation of these 6 componentsraises a few, interesting questions.
Appendix
Consider a vector-space V over the real numbers, and the following equivalence relation x ∼ y ⇐⇒ y ∈ { λx | λ ∈ R \ { }} ( x, y ∈ V ) . (26)The set of all such y is called the equivalence class of x (denoted with x ∼ ), while the vectorsthemselves are called representants of the equivalence class. If V is n dimensional, we can considereach equivalence class as a point in an n − dimensional projective space. How one embeds theaffine space into the projective one carries certain freedom in terms of coordinates. In this paperpoint x ∈ R n − is mapped to the equivalence class ( x , ∼ . In keeping with the duality principleof projective geometry, hyperplanes of the projective space can also be thought of as equivalenceclasses, such that hyperplane h ∼ contains point x ∼ , if and only if (cid:104) h, x (cid:105) = 0 holds. References [1] K. Culmann.
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