GRAPHIC STATICS IN ARCHES AND CURVED BEAMS

Transcription

1 TU DELFT - FACULTY OF ARCHITECTURE GRAPHIC STATICS IN ARCHES AND CURVED BEAMS FINDING FORCE EQUILIBRIUM THROUGH TOTAL MINIMUM COMPLEMENTARY ENERGY Niels van Dijk

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3 GRAPHIC STATICS IN ARCHES AND CURVED BEAMS Date Name Niels van Dijk Student number Address Juniusstraat 161 Postal code 2625 WZ Place of residence Delft Phone number University Faculty Department Graduation track Graduation lab Main mentor Second mentor External examiner Technical University of Delft Architecture Architectural Engineering + Technology Building technology Computation and technology Ir. A. Borgart Chair of structural mechanics A.Borgart@tudelft.nl Ir. T.R. Welman Chair of design informatics T.R.Welman@tudelft.nl Ir. E.J. van der Zaag E.J.vanderZaag@tudelft.nl i P age

4 PART I INTRODUCTION BACKGROUND... 3 RESEARCH QUESTION AND OBJECTIVES... 5 THESIS OUTLINE... 6 PART II THEORY GRAPHIC STATICS... 9 STATIC-GEOMETRIC ANALOGY MINIMUM COMPLEMENTARY TOTAL ENERGY PART III METHOD DEVELOPMENT EXCEL PROBLEM AND HYPOTHESES LINE OF THRUST INSIDE THE MATERIAL LINE OF THRUST OUTSIDE THE MATERIAL OWN WEIGHT PART IV METHOD DEVELOPMENT GRASSHOPPER EXCEL TO GRASSHOPPER GRASSHOPPER EXPLORATORY METHODS EXPLORATION FOR OTHER SOLUTIONS FINAL GRASSHOPPER METHOD DEFINITION DESCRIPTIONS GRASSHOPPER PART V RESULTS CONCLUSIONS LIST OF FIGURES LITERATURE LIST APPENDIX ii P age

5 GRAPHIC STATICS IN ARCHES AND CURVED BEAMS PART I INTRODUCTION BACKGROUND... 3 RESEARCH QUESTION AND OBJECTIVES... 5 THESIS OUTLINE... 6 PART II THEORY GRAPHIC STATICS Head-tails method Cable analysis Arch analysis Practice Thrust network analysis STATIC-GEOMETRIC ANALOGY Plate calculations Shell calculations MINIMUM COMPLEMENTARY TOTAL ENERGY Complementary energy Normal forces Bending moments Rewriting Application of minimum complementary energy PART III METHOD DEVELOPMENT EXCEL PROBLEM AND HYPOTHESES Problem Hypothesis LINE OF THRUST INSIDE THE MATERIAL LINE OF THRUST OUTSIDE THE MATERIAL Test compressive complementary energy Test compressive and bending complementary energy Relating forces Forces construction Unknown horizontal support Review OWN WEIGHT Formula catenary Discretization Conclusions PART IV METHOD DEVELOPMENT GRASSHOPPER EXCEL TO GRASSHOPPER Grasshopper over Excel Differences Excel to Grasshopper Line of thrust Calculation of supports Goal of the script GRASSHOPPER EXPLORATORY METHODS Validation Limitations EXPLORATION FOR OTHER SOLUTIONS iii P age

6 13.1 Mathematical approach N l Combining calculations Graphical relations between construction, force polygon and line of thrust Relation force polygon line of thrust Relation total area line of thrust force polygon Changing parameters of force polygon Changing vertical support Changing horizontal support Changing divisions Application Proof of equal areas Area construction Area line of thrust Minimum complementary energy Conclusions FINAL GRASSHOPPER METHOD Closing line Finding the closing line Application closing line Validation Using the script Stresses Conclusions DEFINITION DESCRIPTIONS GRASSHOPPER Calculation description Graphical representation Visual check Display component Differences different scripts PART V RESULTS CONCLUSIONS Process Calculation method and component Recommendations LIST OF FIGURES LITERATURE LIST APPENDIX iv P age

7 GRAPHIC STATICS IN ARCHES AND CURVED BEAMS Part I Introduction 1 P age

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9 GRAPHIC STATICS IN ARCHES AND CURVED BEAMS 1 Background Throughout history people created and built the most fascinating structures. Mankind developed a good sense for how to construct certain structures, first by trial and error, later by learning how to analyse whether structures would collapse or not. With the development of computer technology more complex building shapes could be modelled, not only in design but also in structural analysis. Simple constructions, like beams and columns, are easily calculated analytically by hand. More complex to calculate are plates and beam networks, but it is still possible to do this by hand. For shell structures, it is nearly impossible to make analytical hand calculations. This is only possible for a small basic set of shell shapes. For more complicated structures, Finite Element Methods are needed to gain insight in the stresses and displacements caused by certain load cases. These FEM calculations are numerical calculations for complicated mechanical problems. The most efficient way of transferring loads is trough axial (stretching and compression) forces. A structurally good shell uses this efficiency and acts mostly in compression. When architects like Gaudi started designing structures with mainly compression, an analogy with chains was used. Chains only transfer loads trough stretching forces, inverting the shape of the hanging chain, gives a structure with only compression (Huerta 2005, Block, DeJong et al. 2006). When it is not possible to build the exact inverse shape of a hanging chain, bending moments are introduced. With bending in a material, large deformations can be expected, and bending becomes the most important factor dimensioning your structure. Figure 1 Left; hanging chain models by Gaudi (Daniel 2007) Right; Sagrada Familia, Barcelona, by Gaudi (Mak 2013) Due to the minimum presence of bending forces, compression structures can be designed with little structural height, compared to ordinary structures. This slenderness is part of the elegance, and probably the allure, of shell structures. Famous shell builders are architect Felix Candela and engineer Heinz Isler. Candela designed his shells trough mathematical models, using only basic shapes like hyperbolic paraboloids, making sure he could perform hand calculations on his designs. Isler was more of a mystery man, never revealing his real way of designing. What is known is that he used a lot of physical models to test different shells and shapes to create his final design. 3 P age

10 Figure 2 Left; Deitingen service station, by Isler (BauNetz 2009) Right; the Manantiales Restaurant, Xochilmico, by Candela (Lukas 2013) Until present days, there is still no way to get analytical information about freeform shells. If a designer wants to see the structural performance of a shell design, the Finite Element Method is essential. The way of calculating in this software is numerical, so the relation between geometry and structural performance is lost. In other words, the direct relation between geometrical en structural properties of a shell is unknown. This in itself is not a problem, since FEM is mostly used in the final stages of a design, when the shape is already determined. Here it is used to check material thicknesses and strengths, to see if the construction will fail or not. With the development of computer technology, a great new range of software became available. Tools like Grasshopper are used by designers to make parametric relations in 3D modelling software, giving more flexibility in designing complex shapes. The strength of this parametric way of modelling is that a designer can create a large set of different designs in a matter of seconds. The downside here is that the designer has to choose one of these designs. Designers can use different goals in a design and assess if a certain option, generated by a parametric model, fulfils these goals. If these goals are related to the field of structural mechanics, or other properties that can be expressed in numbers for that matter, these goals can be optimized for chosen values. This means that a certain parameter, or set of parameters, can be varied until the chosen output reaches a chosen value. This value, for example deflections in a structure, then has to be calculated by a structural mechanics component. If a FEM calculation is used, the designer can see if the design gives the desired results in numbers, but if this is not the case, it is not directly clear how changing the design can improve the performance. In this thesis the power of this parametric software is used to find ways to analyse geometrical properties of shells and give insight in the relations between geometry and structural properties, with the goal that designers whom are using parametric models, can implement this insight in their models. So when a designer does not retrieve the desired results, the parametric software can give graphical and geometrical output to give new leads for the designer to improve his or her design. This direct feedback is lacking in FEM calculations. 4 P age

11 GRAPHIC STATICS IN ARCHES AND CURVED BEAMS 2 Research question and objectives Summarizing, with a growing use of more complex shapes in buildings and structures, there is still little understanding of the relation between geometrical and structural properties of shell structures. There is no analytical method to calculate shell structures in a quick manner. Numerical calculations with FEM programs give no clear insight in the relations between geometry and structural properties and require expertise and a detailed model of the design to acquire accurate results. When these analytical relations are known, this can be modelled in a real time calculation component. Ideally, with a calculation component for both the architect and the engineer, the flexibility to make early design decisions for shape and structural functionality increases. In reality, the architect would want more feedback in the beginning of the design process on general flow of forces and to get an idea on where the critical points in the shell are situated. Since this is the time the general shape is determined, it is useful to get direct feedback and see the consequences of this shape. The structural engineer would want more detailed feedback like stresses, strains and deformation to dimension the reinforcements in concrete or the connection details to the foundation. This is generally found in FEM calculations, but for the engineer it is also good to see how the shape of a structure can improve the structural performances, instead of seeing only numbers in the output. This brings forward the following research question for this thesis; How can analytical relations and structural analogies be used to create a parametric calculation tool, to be used to calculate and design shells and thin plate structures? In recent years, a lot of research in this field is already done by Borgart and his students at the TU Delft; this makes it possible to give a more detailed problem description. In Part II 5 the staticgeometric theory is explained, introducing a conceptual split to divide the calculations in two sets. The analytical solution for these sets is already modelled in two different Grasshopper tools (Liang 2012, Oosterhuis 2012). The main objective of this thesis is to describe a better relation between structural and geometrical properties of shell structures using simple analytical methods. Second objective is to use this understanding to create a calculation component to make this knowledge accessible. To achieve these objectives, first research in the field of structural mechanics is done to get a better understanding of the problem. The starting point where the theses by Liem, van den Dool, Oosterhuis and Liang. Combining the calculations prepared in the latter two theses should be done step by step, so for the scope of this thesis, a research on beam and arch like structures is started. The thesis gives recommendations and a direct link on how to start with the calculations for shell structures. The secondary objective is to use the acquired insights on these types of structures and built them in a calculation component in Grasshopper. The component can be implemented in parametric design models and give a designer feedback on different generated variants of a design, and help him or her choose which variant to develop further. 5 P age

12 3 Thesis outline This thesis proposes a method of finding the normal forces and bending moments in arches and curved beams, as a stepping stone to finding these forces in shells and curved structures. The thesis has five main parts, each consisting of different chapters. Part I In part I the introduction to the subject is made with a problem background, description and definition. Also an indication is made of the work done before this thesis to place the problem in a broader perspective. Part II In part II different theories are described, that are used in the development of the. A wide literature research led to a more defined problem definition and a few theories and concepts that are used in the calculation method. Part III In different steps, all described in their own chapter, the development of the calculation method is documented. Each step uses a more complicated model and is tested, before the next step is made. Also the reasons for different research directions are explained and the consequences of these choices. In part III the focus is on the development in Excel. Part IV In this part, the calculation method is continued in Grasshopper. The difference with the Excel calculations is explained and in different chapters the method is expanded and tested to be more flexible. Part V In the final part the conclusions and recommendations are made. Also the literature list and appendices are included. 6 P age

13 GRAPHIC STATICS IN ARCHES AND CURVED BEAMS Part II Theory In this part different analysis methods for shell and plate structures are described. The work done by some of Borgarts former students is analysed and used to continue the search for an analytical solution to the shell problem. The theories used before are: Graphic statics Minimum complementary normal energy To continue the research, two methods are added: Static-geometric analogy Minimum complementary total energy The graphic statics theory is an important part for the calculation method. It is an intuitive and graphical way to get a better understanding of the equilibrium in cables and funicular structures. The theory on minimum complementary energy uses the idea that each mechanical system contains a minimum level of energy. This knowledge can be used to solve statically indeterminate structures. A combination of these theories is used by both Liem (Liem 2011) and van den Dool (Dool 2012) to create compressive only structures by generating structures using graphic statics and then calculating the energy of the system, optimizing the structures for minimum energy. The theses by Oosterhuis (Oosterhuis 2012) and Liang (Liang 2012) focus on the calculations for shell and plate structures. These calculations are based on the static-geometric analogy, splitting the shell calculations in two sets of calculations. The first step in finding the right preconditions for this split is made in this thesis where this split is solved for beam and arch elements. Since in these structures not only normal forces occur, like in the structures by Liem and van den Dool, the total complementary energy is necessary to finalize the calculations. 7 P age

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15 GRAPHIC STATICS IN ARCHES AND CURVED BEAMS 4 Graphic statics The graphic static method is an intuitive and graphical way to describe equilibrium in systems using force directions and (closed) force polygons. 4.1 Head-tails method If three forces act on a point in equilibrium, the magnitudes of the forces can be graphically determined when at least one magnitude and all three directions are known. Using the head-tails method, a closed force polygon can be created by plotting the known force with a chosen scale. The other forces can be measured in this scaled drawing, giving the magnitudes of the forces in the system (Dool 2012). If the polygon is not closed, which is possible, the point being analysed is not in equilibrium, and the distance needed to close the polygon is the resulting force and the direction of the movement of the point. Figure 3 Left; Force diagram and force polygon of a point in equilibrium Right; force diagram and force polygon of a point not in equilibrium with resulting displacement indicated in red 4.2 Cable analysis This method of analysing points in equilibrium can also be used to analyse cable structures. A cable structure only acts in tension. This means that the slopes of the cables and the directions of the applied loads, for example gravity, give all the directions of the forces acting in the system. One can make a force polygon for each point on the cable. Since each cable segment length is connected to two points, all these polygons can be connected, eliminating all the intermediate forces. The only forces remaining are the applied loads and the outer forces in the force polygon, representing the support reactions of the system. Also in this case, there can only be equilibrium if the polygon is closed. If one inverts this funicular or catenary shape, an arch with only compression is formed. Figure 4 Left; cable with two forces applied Centre; force polygons of point B and C in equilibrium Right; total force polygon of the system in equilibrium with F AB and F CD representing the support reactions 9 P age

16 4.3 Arch analysis This graphical trick can be used to analyse the equilibrium and the magnitudes of the forces in an arch. Huerta and Block, both working on the analysis of masonry arches, used chain models to study their structural performances. In a masonry arch, different blocks are stacked on each other, and act solely in compression, since no tension can be transferred trough the friction of the blocks. Using a drawing of a random Block illustrates the relation between the arch, the inverted hanging chain and the force polygons (see Figure 5). Figure 5 Thrust line in a random arched structure (Block, DeJong et al. 2006) In (a) the different blocks are portrayed by point loads corresponding to their weight, also their lines of influence are indicated. Below the arch, a hanging chain is drawn, loaded with the weights from the arch. In (b) the internal forces of one block are shown, and closed in a force polygon in (c) (Block, DeJong et al. 2006). This can be done for all the separate blocks, making the total force polygon of the arch in (d). The inverted hanging chain in the arch is called the line of thrust. This is an imaginary set of points where all the forces in the arch make equilibrium using only compressive actions (O'Dwyer 1996, Huerta 2005). 10 P age

17 GRAPHIC STATICS IN ARCHES AND CURVED BEAMS The difference between the chain and the arch is that the chain has two points as supports, so only one possible way for the forces to flow through the material is possible. In the arch, the complete thickness of the blocks can be used, making different paths for the line of thrust possible (see Figure 6) (Huerta 2005). Figure 6 Different positions of a chain through the thickness of the material (Huerta 2005) For the cable, the shape of the cable is the exact same shape as the forces from the force polygon. In the arch the path the forces follow can vary, as long as it stays inside the material. This path or line visualizes the points in the arch making equilibrium with the weight of the blocks using only compressive actions; this line is called the line of thrust and can be seen as an inverted catenary. The force polygon and the construction are each other s reciprocal figures, meaning they are directly related to each other, and inverting one, gives the shape of the other. Changing the arch, directly changes the line of thrust and the force polygon (Maxwell 1868). Figure 7, by Block, illustrates this relation of the line of thrust and the force polygon. Figure 7 Relation between the line of thrust and the force polygon in an arch (Block, DeJong et al. 2006) 4.4 Practice In engineering practice, graphic statics was very popular due to its intuitive use. In the beginning of the 20 th century this method was mostly replaced by analytical solutions that did not require any drawing skills. One can imagine that this method is easy for a cable or an arch, but much more complicated for structures like trusses and statically indeterminate systems. Italian mathematician Luigi Cremona ( ) developed a method to analyse trusses with graphic statics using the so-called Cremona diagrams. With a few steps it is possible to draw the 11 P age

18 reciprocal figure for a truss consisting of only triangles. In Figure 8 can be seen that the drawings become more complicated for bigger trusses and systems (Block, Mele et al. 2012). Figure 8 Cremora diagram of statically determinate truss (Wikipedia 2008) Drawing the correct reciprocal figure for a statically indeterminate system is somewhat problematic. One can imagine that, when using the head-tails method on the cable system in Figure 9, the problem starts with where to place the last force to make equilibrium. In theory, all the lines are possible, since all of the lines give a closed force polygon. In practise, this is not the case. If one would enter this system in a FEM program, only one correct answer will be returned. How this problem can be solved is described in Part II P age

19 GRAPHIC STATICS IN ARCHES AND CURVED BEAMS Figure 9 Statically indeterminate cable system, the dotted lines indicate different possibilities to close the force polygon with force F 3 Now, with the development of computer technologies, new possibilities emerge for graphic statics to be used, since the polygons are constructed through parametric rules, the drawing can be automated or scripted. 4.5 Thrust network analysis Before mentioned Philippe Block conducted a research for his PhD on the application of graphic statics in three-dimensional structures like shells and vaults. He describes a method to find the spatial equivalent of the line of thrust, the thrust network (Block 2009). In short, the method uses the same steps as in two-dimensional structures, but it needs an extra step in between. A compressive structure like a dome is discretized in a network of bars (G). The network is used to make a planar projection of the system (Γ). This planar projection is used to make a reciprocal diagram (Γ*). Figure 10 Relation between discrete network (G), planar projection or primal grid (Γ) and the reciprocal diagram or dual grid (Γ ) (Block 2009) 13 P age

20 If the construction contains only three valiant nodes, the reciprocal grid is not complicated to construct. When a higher valency than three occurs, the system becomes statically indeterminate and more shapes in the reciprocal diagram are possible. The outer lines in the reciprocal grids determine the horizontal supports of the structures. So scaling the grid, changes the height of the thrust network, just like the two-dimensional force polygon. Figure 11 Left; Discrete construction (G) and primal grid (Γ) Right; Different possible dual grids (Γ ) (Liem 2011) This is an extreme simplification of the method, since the application of two dimensional reciprocal figures suffices in this thesis. The description is solely to illustrate the power of graphical methods in vault design and to give an idea of how a step to shell structures can be made. To find more about the thrust network analysis, read the PhD thesis by P. Block (Block 2009). 14 P age

21 GRAPHIC STATICS IN ARCHES AND CURVED BEAMS 5 Static-geometric analogy A benefit of shell structures is the fact that shells can carry loads by a combination of bending and stretching actions. The downside to these properties is the analysis of the structure. The applied load is divided over the structure trough bending and stretching. To simplify this problem, a split can be made in an elementary piece of shell to two distinct surfaces, one surface to take the bending actions (B-surface), and one surface to take the stretching actions (S-surface). This theory only holds only if an appropriate distribution of loads is introduced over the two surfaces, p B and p S for bending and stretching. The stretching surface can be seen as a shell analysed with the thin plate theory. The bending surface is closely related to a normal plate (Calladine 1977, Calladine 1983). 5.1 Plate calculations The stretching surface is analysed using thin plate equations. A thin plate is a flat constructive element loaded in-plane. In such calculations, the bending stiffness is assumed to be 0, so only normal and shear forces in-plane occur. The set of calculations to describe the relation between stress, strain and deformations can be found in most books on mechanics of plates, for example (Beranek 1977). Since these calculations only deal with loads in-plane, an extra step is needed to transfer out of plane loads. To solve this, a thin plate surface has to be curved, like a membrane (Beranek 1976). Membranes have a bending stiffness of 0, just like plates, so a link between these calculations is made and used by Calladine to calculate the S-surface (Calladine 1983). The total set of formulas is shown in the next paragraph. For plates with a thickness the assumption that the bending stiffness is 0 will not hold. For these structures a different set of calculations is needed. In contrast to the thin plate, in these calculations, loads perpendicular to the surface are used and transferred to the supports to normal actions. Calladine uses this set of calculations on the B-surface. How to derive these formulas can be found in for example (Beranek 1977) and (Calladine 1983). The total set of formulas is shown in the next paragraph. 5.2 Shell calculations With the split of the initial surface element, the equilibrium equations also split. In a surface with only stretching, there are only normal forces (N x, N y, N xy ). In the B-surface there are shear forces (Q x, Q y ) and bending moments (M x, M y, M xy ). The resulting surfaces and forces can be seen in Figure 1 a b and c. Figure 12 Surface element split in two distinct surfaces (Calladine 1977) 15 P age

22 Now this conceptual split is only possible if two requirements are met. In Figure 12 the applied load is indicated with p, and is divided over the S and B-surface. So first to ensure a good force distribution p = p S + p B (5.1) The second requirement is found in the shape of the shell. Since the split is only conceptual, the two separate surfaces should still keep the same geometry. The geometry is defined by the Gaussian curvature (k) and the change in curvature (g). Both surfaces have their own equations for (g) thus giving a (g S, g B ). Since these should coincide g S = g B (5.2) to make sure both surfaces still fit the original combined surface. The equations for both surfaces can be written down in two columns; S-surface N x R 1 + N y R 2 = +p S B-surface k y R 1 + k x R 2 = +g B N x x + N yx y = 0 k y x + k xy y = 0 N y y + N xy x = 0 k y y + k yx x = 0 N xy = N yx N x = 2 φ y 2 N y = 2 φ x 2 N xy = N yx = 2 φ x y ε x = 1 Et (N x vn y ) ε y = 1 Et (N y vn x ) 1 2 γ xy = 1 2 γ yx = (1 + v) N Et xy (1 + v) N Et yx 2 ε y x 2 2 γ xy x y + 2 ε x y 2 = g S 2 M x x 2 k yx = k xy k y = 2 w y 2 k x = 2 w x 2 k xy = k yx = 2 w x y M y = D(k y + vk x ) M x = D(k x + vk y ) M yx = D(1 v)k yx M xy = D(1 v)k xy 2 M xy x y + 2 M y y 2 = p B 1 2 γ xy = 1 2 γ yx M yx = M xy 16 P age

23 GRAPHIC STATICS IN ARCHES AND CURVED BEAMS In 2010 M. Oosterhuis modelled a tool to solve the left row of calculations, so the stretching calculations. In 2012, D. Liang expanded this tool and developed a second tool to solve the right row, so the bending calculations (Liang 2012, Oosterhuis 2012). With these calculations the internal forces and moments, strains, curvatures, rotations and displacements of a surface can be calculated. Figure 13 Screenshots of the two tools modelled by Liang With this theory and these equations all shell structures can by mathematically analysed, if the split for p is known. A step in finding the split to fulfil this requirement is a computational scheme by Pavlovic, a former student of Calladine. With his set up and some time and patience the correct distribution between p S and p B can be found (Pavlovic 1984). This value for p is the value missing to link the tools from Oosterhuis and Liang together. Figure 14 Computational diagram to find the force distribution (Pavlovic 1984) 17 P age

24 6 Minimum complementary total energy For simple structures, one can solve all the calculations and equations with direct methods. For more complicated calculations, sometimes indirect methods are used. Nowadays FEM calculations are based on indirect methods, and the method of minimum complementary energy is also one of these methods. In this chapter a description of complementary energy is given, and in the last paragraph of the chapter an example is sketched to show the calculation possibilities with this energy. In the developed calculation method in this thesis, the same principle of calculation with energy is used. 6.1 Complementary energy Consider an elastic material under a known uniaxial stress σ. The material, with length, height and width 1, deforms with strain ε. The work needed to perform this deformation is called the deformation energy, E s, and can be seen in Figure 15 as the dark grey area. Deformation energy is potential energy, stored in the material. Figure 15 Stress-strain diagram for elastic material On the area above the curve is the complementary energy, E c. It is difficult to make a physical definition of complementary energy. The above shown figure applies to elastic materials, where the elasticity of the material changes under higher strains. When we consider a linear-elastic material, the relation between ε and σ is simplified and considered to be linear. Through Hooke s law the relation can be drawn as Figure 16 shows. Figure 16 Stress-strain diagram for linear-elastic material The slope of the line is defined by the material property E, the Young s modulus. The relation is defined by σ = Eε (6.1) and thus ε = σ E (6.2) 18 P age

25 GRAPHIC STATICS IN ARCHES AND CURVED BEAMS In Figure 16 can be seen that the complementary energy can be calculated by using the area of a triangle E c = 1 εσ (6.3) 2 and combined with formula (6.2) E c = 1 2 σ 2 E (6.4) This is the general formula for calculating the complementary energy in a cube of material. Next is described how this theory can be applied on bars in normal and bending loading. 6.2 Normal forces A bar loaded in normal force N gives a deformation ε, extension or compression, depending on the direction of the force. Applying the general rules for complementary energy for a bar with length 1 and force N = Aσ (6.5) Substituting (6.1) in (6.5) gives N = AEε (6.6) this gives a new graph, where the slope is defined by both the area of the section of the bar A and the Young s modulus E. Figure 17 Normal force-strain diagram Using formula (6.6) and Figure 17, the formula of the normal complementary energy, E c,n can be defined E c,n = 1 2 N 2 EA (per unit of length) (6.7) Since this is per unit of length, and it can be useful to know the energy in a bar or part of a construction, the length is added to the calculation giving E c,n = 1 2 N 2 EA l (6.8) 6.3 Bending moments Now for a bar in bending moment M. Bending moments cause bending deformations, expressed in curvature, κ, defined by κ = 1. For a bar with length 1 and force R 19 P age

26 M = EIκ (6.9) the graph can be plotted again. Here I is the second moment of area or second moment of inertia. Figure 18 Bending moments-curvature diagram From Figure 18 and formula (6.9), the complementary energy E c,m can be calculated. E c,m = 1 2 M 2 EI (per unit of length) (6.10) Just as the normal energy, here the length of a bar is also important and added to the formula E c,m = 1 2 M 2 EI l (6.11) Now the energy of a system can be calculated. The same principle can be used to calculate shear and torsion energies, bus this is not relevant for this thesis. 6.4 Rewriting The formulas as shown before can be simplified. In both formulas 1 can be removed, since these 2E values do not change over the different bars, meaning the value of the outcome will change, but not the position of the minimum. Also A and I can be rewritten A = dt (6.12) I = 1 12 dt3 (6.13) By rewriting the values for A and I in the formulas (6.8) and (6.11) the value for d (depth of the beam) can be eliminated from the equations, ending with the summation E c = l(n M2 ) (6.14) t 2 Here the depth d of the material is irrelevant, and the only relevant value is the thickness of the structure, t. One can imagine that sweeping the structure in one direction, creating a shell like shape, will not change the flow of forces compared to a beam. 6.5 Application of minimum complementary energy In nature, the theory of least work applies. Pierre-Louis Moreau de Maupertuis describes this principle, known as Maupertuis principle, in a French text from 1744, ACCORD DE DIFFÉRENTES LOIX DE LA NATURE QUI AVOIENT JUSQU'ICI PARU INCOMPATIBLES. This text holds the formula for the path a load follows through a physical system, meaning that nature finds equilibrium using least energy as 20 P age

27 GRAPHIC STATICS IN ARCHES AND CURVED BEAMS possible. This is can be summarized in a translated quote by Maupertuis in nature is thrifty in all its actions (Wikipedia 2013, Wikipedia 2013). Figure 19 Statically indeterminate cable system, the dotted lines indicate different possibilities to close the force polygon with force F 3, using minimum energy can solve this problem In a practical application this means, if one has a statically indeterminate system, one can make an infinite amount of force polygons (see Part II 4.2). Finding the correct polygon, the one that holds the forces in the construction, can be done using the principle of least complementary energy. For complicated structures it is possible to let a computer calculate the energy for different versions of the force polygon and return the option with the least energy. For more simple structures it is possible to express the total energy in a mathematical formula and find the minimum through differentiating. For more elaborate examples see (Blaauwendraad 2002, Borgart and Liem 2011, Liem 2011) 21 P age

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29 GRAPHIC STATICS IN ARCHES AND CURVED BEAMS Part III Method development Excel In this part the calculation method is developed and step by step described. In the first chapter the problem that needs to be solved is described and a hypothesis is presented. In the following chapters, the hypothesis is tested, starting with simple constructions calculated by hand and with (small) Excel sheets. The calculations expand over the chapters to a wider range of constructions and they are compared to calculations made with software used in building design practice (GSA Suite). 23 P age

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31 GRAPHIC STATICS IN ARCHES AND CURVED BEAMS 7 Problem and hypotheses The theory from Part II is used to define a more detailed problem description and a hypothesis is given on how to solve the stated problem. 7.1 Problem The problem to solve is to find the division or split in bending moments and normal forces in shells and curved structures as described in the theory by Calladine. To narrow down the research, the first structures used to solve this split are arches and curved beams. These structures also use a combination of normal forces and bending moments to transfer loads, but only in one plane. This makes the first steps in the research smaller and better to understand. To start the research, only hinged supports are used, so the bending moments in the supports are zero. When searching for the division in these forces, it is a good starting point to know some points where the bending moments are zero. In the previous chapters different theories are described. From these theories two important principles are used through the development of the calculation method; - Each construction contains minimum complementary energy - Each system has a line of thrust 7.2 Hypothesis With the theory on reciprocal figures, force polygons and minimum energy, a theory or hypothesis is developed. The hypothesis holds that, when the line of thrust is not inside the material, the forces in a construction can be related to the forces in the line of thrust, and vice versa, through their reciprocal relation. With these forces, the complementary energy of a construction can be calculated (see Part II 6). Of all the lines of thrust, the one returning the lowest complementary energy should be the correct line Part II 6.5. In different steps, constructions and calculations, this hypothesis is tested to find the forces in constructions. 25 P age

32 8 Line of thrust inside the material For a given span and (projected) load, endless compressive only structures can be generated (Liem 2009). Each structure in its own is a line of thrust, but only one line of thrust contains the least compressive complementary energy (see Figure 20). Figure 20 Different lines of thrust and their force polygons, one line contains the least energy with a given load case and span With a projected load and the help of a polar figure, a parabolic shape can be generated as explained in Part II 4.3 (Heyman 1998). This shape is the line of thrust and is mathematically described by y(x) = a(x h) 2 + k (8.1) with h and k defined as shown in Figure 21 (Ravenstein 2002). a can be expressed in h and k by substituting the points from the parabola in the formula. This gives a value of a = k h 2. Figure 21 Values needed to define the parabolic shape of the line of thrust 26 P age

33 GRAPHIC STATICS IN ARCHES AND CURVED BEAMS In the line of thrust, the top of the structure, which is only part of the parabola, is called f instead of k. So for the same span and load, different parabolic shapes can be generated by changing the height of the parabola, f. By changing the height f the forces in the construction change, and so do the support reactions. The horizontal support reaction H for parabolic shaped structures is calculated with H = ql2 8f (8.2) Now, with two simple equilibrium equations, at each random chosen point the forces in the line of thrust, and thus the construction, can be calculated. When a section is made through the structure at the random picked point, the equilibrium state of this part can be reviewed. The q and x are known for each point to calculate. The vertical support reaction, V sup, is half the total load, since only symmetrical structures are used. This leads to the value N V. F V = 0 F H = 0 Figure 22 In blue a part of a structure and in red the active forces F V + (q x) = V sup and F H = H sup. With this, the compressive energy can be calculated for each parabolic shape. First the normal forces in a set of points from the discretized line are calculated. With the mathematical description of the parabola, the lengths of the bars between the points can be found. To calculate the energy the average normal forces over the bar length are used. Now the formula described in Part II 6.4 for E c,n can be filled out to find the compressive complementary energy. If an arch has a given thickness of material and only acts in compression, this is the way to find the correct line of thrust in the material. This solves the problem described by (Block, Mele et al. 2012), where only a set of correct possible answers is found. 27 P age

34 9 Line of thrust outside the material In Part III 8 the assumption was made that the material was not able to handle tensile forces, and this assumption is used in a lot of research on arches and vaults (O'Dwyer 1996, Heyman 1998, Huerta 2005, Block, DeJong et al. 2006, Block 2009, Block, Mele et al. 2012). In building practice, reinforced concrete is sometimes used. This material can handle tensile forces as well as compressive forces, making it possible for the line of thrust to reach equilibrium outside the material. This paragraph explores the relation between the material and the line of thrust. 9.1 Test compressive complementary energy To see the relation between the line of thrust and the construction, two constructions with known horizontal supports H are tested. This way the height of the thrust line can be calculated with formula (8.2). For a portal frame and a semi-circle as drawn in Figure 23, the formulas for H are respectively H = 3l 2 ql 2 2 3l 2 +2l 1 12l 1 (Beranek 1977) (9.1) H = 4 qr (Dicke and Eisma 1978) (9.2) 3π Figure 23 Left; Portal frame Right; semi-circle Calculating H and using formula (8.2) leads to two values for f, one for each structure. This means that under influence of the shape of the structure, the line of thrust changes. When calculating the compressive energy of the relating lines of thrust, it becomes clear that the parabola calculated through the formula for H does not contain the minimum energy. In Figure 24 the Excel with the calculations for the portal frame is shown, here can be seen that the f calculated trough H holds more energy than the f in the upper calculations. For both the frame and the semi-circle the values are listed in Table 1. Table 1 Frame Semi- circle Hand calculated H 2,45 N 2,54 N Hand calculated f 3,67 m 3,53 m Calculated E c,n for f 228,9 N 227,4 N f with minimum energy 3,18 m 3,18 m Calculated E c,n minimum 225,7 N 225,7 N 28 P age

35 GRAPHIC STATICS IN ARCHES AND CURVED BEAMS Figure 24 Excel calculations for portal frame for two different values of f returning different values for the complementary energy With this test can be concluded that the line of thrust with the least compressive complementary energy is not the same as the line of thrust found by the formula (8.2). Also, the shape of the construction influences the height of the line of thrust. 9.2 Test compressive and bending complementary energy With the idea in mind that the minimum energy should be found in the construction, and not the line of thrust, the next step in the development is made. Calculating the energy in the construction instead of the line of thrust means that the forces in the line of thrust need to be linked to the forces in the construction Relating forces To get a better understanding of the rules on how to translate forces, Figure 25 is used. In this figure, a force (F) acting outside a material is shown. To find the resulting forces (N) and (M) inside the material, this force must be translated. A force outside a material with distance (e) perpendicular from the centre line of the material can be replaced by the same force in the material, plus a bending moment of M = N e. 29 P age

36 Figure 25 Translating and decomposing forces The force F is not in the same direction as the material, so this force needs to be decomposed in two forces, one acting in the direction of the centre line of the material (N) and one acting perpendicular to this centre line (Q) Forces construction With the relation between the forces of the line of thrust and the construction described, at different points in the construction the normal forces and bending moments can be calculated. To show this, the semi-circle is used. To determine the forces and bending moments, the exact dimensions of the arc are needed. The arc can be described by; g(x) = x(2r x) with Domain x = 0,2r (9.3) This formula gives a half circle with radius r. Combining (8.1), (8.2) and (9.3) can generate a set of points related to either the arc or the parabola of the thrust line. A(0; 0) B(2r; 0) R(r; 0) F(r; f) 30 P age

37 GRAPHIC STATICS IN ARCHES AND CURVED BEAMS Figure 26 In blue the construction of a semi-circular arch, in red a line of thrust with height f and in green the line j(x) to determine the eccentricity e To describe the shape of the thrust line, the common formula for a parabola and a known vertex is used (8.1). With the vertex point coordinates (h; k). Taking point F for the vertex, the formula for the thrust line can be derived gg(x) = f f(x r)2 r 2 (9.4) Here r is the radius of the semi-circle and f the height of the vertex point of the line of thrust. Line j(x) is used to calculate the bending moments in point P. The bending moment is calculated by multiplying the normal force in point Q (found in Part III 9.2.1), times the distance to the arch, the eccentricity. This distance should be calculated perpendicular from the parabola to the arch. At a random point Q on the thrust line the tangent can be found by differentiating formula gg(x) to gg (x) = 2f(x r) r 2 (9.5) To get the tangent normal to point Q, 1 gg (x) can be used, giving gg (x) = r2 2f(r x) (9.6) With the common formula for a straight line y(x), chosen point Q and formula (9.6) for the slope dy y(x) = dy x + c (9.7) dx Q (x; f f(x r)2 ) r 2 31 P age dx

38 line j(x) can be constructed r 2 j(x) = x + (f f x gg r 2f r x gg 2 r 2 + x ggr 2 2f r x gg ) (9.8) where x gg is the x value of point Q, so in this formula it is not a variable but a constant. Next is to calculate the location of point P by solving j(x) = g(x). With points P and Q known, distance PQ and e (see Part III for eccentricity) can be determined with PQ = e = (x Q x P ) 2 + (y Q y P ) 2 (9.9) This makes it possible to calculate the bending moments M at any given point P on the arc, just as the normal forces in the line of thrust described in Part III To calculate the normal force in this point, the force used for the bending moment from the line of thrust needs to be decomposed in a part parallel to the arch, and a part perpendicular. First step in doing so is determining the angle between the two vectors in de relating points. This can be done with formula (9.10) and see Figure 27. tan θ = m 1 m 2 1+m 1 m 2 (9.10) Figure 27 Two lines, their slopes and their angle Here (m 1, m 2 ) are the slopes in the two points, and those can be calculated with the derivative of formula (9.3) and with formula (9.5). With the angle known, the component parallel to the arch can be calculated with N x = F cos θ (9.11) solving the last step in finding the forces in a point in a construction. All these steps are modelled in Excel, so for a large amount of points in a mathematically described structure, the forces can be calculated. The average value over a bar is used to calculate the energy using the formulas for E c,n and E c,m. Combined, the total complementary energy can be calculated. For both the frame and the half circle, where the horizontal support reactions are known, this method is applied in Excel. For the frame the calculations are easier, but for the semi-circle a new Excel with some auto solvers was used. Also new dimensions for the arch are used, so the results are not directly comparable to Part III 9.1. Substituting the formula for H for the semi-circular arch in the 32 P age

39 GRAPHIC STATICS IN ARCHES AND CURVED BEAMS formula for f of the line of thrust, gives a result of f = 6,48 m. Trying different heights in the Excel sheet to calculate the bending moments, the normal forces and the energy gives the results as seen in Table 2 (and see appendix for Excel calculations). Table 2 f E c 6,6 113,43 6,5 111,08 6,4 110,25 6,3 110,99 As can be seen, changing the height of the line of thrust increases the energy in both directions, making the theory on the thrust line and the energy more plausible. The results are reason enough to further explore this direction. 9.3 Unknown horizontal support When designing more freeform shapes and arches, the horizontal support reaction is usually unknown. The calculation as performed in the last paragraph uses H to define the shape of the line of thrust, and returns the construction with the least complementary energy. So if H is not known, different values for the height of the parabola that is the line of thrust, f, can be tried and one finds the correct line when the returned complementary energy is minimum. First, in an Excel script, the calculation was performed for the semi-circular arch. By trying different values for f the construction with the least complementary energy can be found. The returned values for N and M are compared to the same shape and loading calculated in GSA Suite. A comparison of the values for M can be found in Figure 28 and for the support reactions in Table 3. In the values for bending moment M, maximum deviations of 4,1% are found and an average deviation of 2,6%. 3000,0 Bending moments Bending moments (Nm) 2000,0 1000,0 0,0-1000,0-2000,0 0,00 0,75 1,50 2,25 3,00 3,75 4,50 5,25 6,00 6,75 7,50 8,25 9,00 9,75 10,50 Calculated GSA Difference -3000,0 Horizontal distance (m) Figure 28 Bending moments comparison, the arch spans 11 meters and has a loading of 1000 N/m Table P age

40 GSA [N] Calculated [N] Deviation [%] Horizontal sup ,5 Vertical sup ,7 Resulting sup ,2 In Table 4 the different f-values are indicated, including their returned energy values. The f value with the lowest E c should be the correct line of thrust. Calculated with formula (8.2) a thrust line with a height of 6,48 m was found, applying this height in the Excel calculation returns a value for E c of 9,71E + 08 N 2. In the table the lowest energy value is for a height of 6,4 m, so there is a small deviation. Table 4. f [m] E c,m [N 2 ] E c,n [N 2 ] E c,n + E c,m [N 2 ] 6,2 6,77E+08 3,13E+08 9,91E+08 6,3 6,63E+08 3,12E+08 9,74E+08 6,4 6,58E+08 3,10E+08 9,69E+08 6,5 6,63E+08 3,09E+08 9,72E+08 6,6 6,77E+08 3,08E+08 9,85E+08 A possible explanation for this deviation can be the inaccuracy of the Excel calculations. The step j(x) = g(x) is solved using goal seek, making one cell variable, Excel changes this value until a given, other, cell is set to a value 0. This changing is done with fourteen digits, causing small deviations in all the answers due to rounding errors. Since the calculation for this half circular arch was already made with a known H, a second arch is also calculated with different f-values. The flat arch spans 11,0 m and is 5,5 2 1 m high Figure 29. Figure 29 Flat arch, part of a circle 34 P age

41 GRAPHIC STATICS IN ARCHES AND CURVED BEAMS To make the calculations in Excel a mathematical description of the arch is needed, and its differentiated formula. g(x) = 60,5 (x 5,5) 2 5,5 (9.12) g (x) = x 5,5 (9.13) 60,5 (x 5,5) 2 With these formulas the thrust line with the least complementary energy was found by trying different values for f, the height with the least complementary energy for this case is f = 2,42 m. This height does not have a meaning directly, other than that it defines the shape of the line of thrust. To see if the values for the normal and bending forces in this case are representative, a calculation in GSA was used for comparison. In Figure 30 a comparison between the two calculations values for the bending moments M is plotted. Expressed in percentages an average deviation of 2,7% is found and one extreme point with a deviation of 10,6%. This deviation can be explained by the low value of that point, so a slight difference in bending moment returns a big error in percentages. 400,0 Bending moments Bending moments (Nm) 200,0 0,0-200,0-400,0-600,0-800,0 0,00 0,50 1,00 1,50 2,00 2,50 3,00 3,50 4,00 4,50 5,00 5,50 6,00 6,50 7,00 7,50 8,00 8,50 9,00 9,50 10,00 10,50 11,00 Calculated GSA -1000,0 Horizontal distance (m) Figure 30 Bending moments comparison flat arch 9.4 Review After testing different structures in the final Excel script and comparing the results with GSA, this method gives results that encourage further research in this direction. There are some limitations to the calculations: - Only projected loads can be applied - Only structures with a mathematically described shape can be calculated - The calculated structures are symmetrical - The supports of the structure are on one horizontal line - The amount of calculated points is not easily changed in Excel In the following chapters these limitations are one by one tackled, to test the method on more general shapes and structures. 35 P age

42 10 Own weight The next step in testing the hypothesis of the relation between the line of thrust and the forces in the construction, is to use the own weight of the structure. When designing shells, the structures own weight is the most important load to calculate, since this is the prevailing case (Heyman 1977). Also, when for example a snow load is applied, this can be seen as own weight, and not as a projected load Formula catenary For the projected loads the shape of the thrust line is a parabola, when the load is discretised, the only relevant factors in the point loads are the span, the accuracy of the discretization and the magnitude of the original projected load. This means that the shape of the line of thrust does not change, only the height f. To find the correct line of thrust in a masonry arch, the link to a hanging chain is used. A hanging chain forms an all tension structure under influence of its own weight. Robert Hooke (1675) was first to discover this relation and formulated it in Latin. ut pendet continuum flexile, sic stabit contiguum rigidum inversum (Hooke 1676) Translated in English it says, as hangs the flexible line, so but inverted will stand the rigid arch. So if one builds an arch in the exact same shape as a hanging cable, the tension in the cable will be compression in the arch. The mathematical formula of this catenary shape was later described. This formula is used to calculate the support reactions in catenary systems. gg catenary (x) = H gx (cosh 1) (10.1) g H Here the shape of the catenary is expressed by the distance in the x, direction under influence of a given horizontal support H and self weight of g (Heyman 1998). This theory and formula are used to design compression only masonry arches. Different lines of thrust are possible within the thickness of the material (Huerta 2005, Block, DeJong et al. 2006), so an exact answer is difficult to find, just like with the projected load. While testing this formula in the Excel calculations for the projected loads, a problem occurred. Using this formula for the shape of the line of thrust instead of the parabolic formula, inconsistent results were returned from the sheet. Comparing formula (9.4) from the projected load to formula (10.1) for the catenary, there is a big difference in the variables for both formulas. In the parabola, the only variables are the span of the structure and the height of the vertex point of the parabola. In the catenary the horizontal support and the own weight are the two variables. So to make the parabola or catenary higher or lower to test different lines of thrust, the length of the line of thrust changes. In the parabola, the total load is separated from the length, so the total load stays the same, where in the catenary, the length of the line of thrust is directly related to the load, so of the length of the line of thrust changes, this influences the total support reactions. If the total load of the catenary changes due to different lengths of the line, the total vertical and horizontal support reactions change. This makes the forces in the line of thrust generated by the catenary formula not directly applicable in the structure, since the weight of the structure is not matching the weight of the line of thrust. Trying to solve this problem by unitizing the weight of the 36 P age

43 GRAPHIC STATICS IN ARCHES AND CURVED BEAMS structure and applying this on the formula, or fixing the variable H, since it is equal of the whole structure, did not return consistent values, so a new approach is needed to find the line of thrust in the structures loaded in own weight Discretization A big difference between projected loads and own weight can be seen when one tries to discretize the loads in to point loads and compares two different structures with both load cases. This discretizing is also done in the Excel calculation, where for a lot of different points the forces are calculated. When a projected load is discretized in to point loads with equal magnitude, the lines of influence are spaced equally (Figure 31 blue). When the same is done for constructions loaded with own weight, the lines of influence are not equally spaced. The distance between these lines changes when the slope of the construction changes (Figure 31 red). In the projected load, the horizontal distance x is always equal, in the own weight situation, the length along the line l is always equal. Figure 31 Two structures in projected load (blue) and own weight (red) If the load is discretized in point loads, an arch with a set of point loads is obviously the result. How to construct the line of thrust for this situation is explained in the theory of Part II 4. By using the discretized loads and creating a force polygon, the slopes for all the separate parts for the line of trust are known. So instead of using a mathematical solution to draw the line of thrust, now a graphical method is used Conclusions Mathematical descriptions for the line of thrust in own weight situations are hard to define due to the relation between the weight and the length of the structure and the line of thrust. Discretizing the structure in a force polygon and using that to create the line of thrust is a better and more intuitive approach. This method can be used for both the projected loads and for the own weight, the only difference is the horizontal distance between the lines of influence for each discrete load. This method uses mainly graphical information. This change in approach also needs a different method in modelling. This and the fact that the Excel calculations had gotten inaccurate already due to the use of solvers, motivated the switch to using Grasshopper. 37 P age

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45 GRAPHIC STATICS IN ARCHES AND CURVED BEAMS Part IV Method development Grasshopper In the following part the step from Excel calculations to Grasshopper scripting is explained. In the first chapter the advantage of Grasshopper over Excel and other software is described, also the differences between the Excel and the Grasshopper calculations are shown, while the method stays the same. Finally the requirements and functionality of the script is described. In the chapters following the general description, the different models are described and more research for the development of the method is explained. The technical description of the script is separated from the description of the method, to keep the explanations clear for readers not interested in any of the scripted model. 39 P age

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47 GRAPHIC STATICS IN ARCHES AND CURVED BEAMS 11 Excel to Grasshopper The conclusion from the Excel part of the development was that these calculations provide a good result for the equilibrium in arches and beams in projected loads, but the capabilities of the program were too limited. To use a more graphical approach, the step to use Grasshopper was made Grasshopper over Excel Grasshopper is a plug-in for Rhinoceros, a 3D modelling program. In Grasshopper parametric relations can be scripted for the modelling and transforming of geometry in Rhinoceros. This means that graphical information from modelled geometry can be analysed in Grasshopper, and after analysing graphical information can be returned, without having to deal with the exact mathematical formulas for the shapes. Since the use of force polygons and constructing the line of thrust is a set of reparative steps applicable on all structures, it is just a matter of modelling the correct steps in Grasshopper, and for each shape drawn in Rhinoceros by a designer, Grasshopper can analyse the structural performances. When modelling the calculations steps in a different program, more focused on calculations, like MatLab, a designer would have to make a translation from his or her designing software to the MatLab program for an analysis. An extra reason to use Grasshopper and not any other software is that the groundwork by Oosterhuis and Liang is also done in Grasshopper. It is logical to keep using Grasshopper to connect their two separate scripts in order to minimize errors in data streams Differences Excel to Grasshopper There are some important differences in the approach in the Excel calculations and the Grasshopper script. Apart from the general calculations, like when calculating a line length in Excel Pythagoras can be used, in Grasshopper this information can be found with a single component, two big differences are explained Line of thrust Finding the line of thrust in a graphical way uses the force polygon. By generating a force polygon, the slopes of the line of thrust are found, but this is a purely graphical approach. In Excel the formula for a parabola is used to predict the shape of the line of thrust, and the value for f is changed to test different heights and calculate the energy levels. In Grasshopper the shape of the line of thrust is changed by changing the shape of the force polygon, and then the lowest energy value is found Calculation of supports In the Excel sheet symmetrical structures are tested. In this case the vertical support reactions are always known, since this is half the total load. This knowledge is used in the calculations of the normal forces in the line of thrust. In the Grasshopper script, the goal is to calculate more random shapes, so a new method for asymmetry and the vertical supports V is found in two steps, both explained in the development of the script Goal of the script The script in Grasshopper was developed in three steps, an exploratory script to test the method more and develop the calculation method, and a final script, which is more organised and is possibly used by other people. In between these two scripts, a transitional script was made. The next step for this organised script is to turn it in to a component, so it can be used in parametric design models. 41 P age

48 The main goal of the exploratory script is to test the developed calculation method for asymmetrical structures and structures in own weight. The script can be messy and unwieldy, once the idea is scripted, easier or quicker modelling solutions can be found for the final model. The final model has more requirements and functionalities. A list of (functional) requirements is made for the model; The script calculates the values for M, N and the support reactions for arches and curved beams, and returns them in a clear and visual manner The script uses force polygons, lines of thrust and the theory of least complementary energy to find before stated values Both projected loads and own weight can be calculated The script can be used by people with little structural knowledge, to show the relation between the line of thrust and the forces in a structure The script has a clear set up for different calculation steps, so it can be used and expanded by others interested in graphic statics in arches and curved beams The script can be used in parametric models as a calculation or analytical component 42 P age

49 GRAPHIC STATICS IN ARCHES AND CURVED BEAMS 12 Grasshopper exploratory methods The main goal of the use of the new approach in Grasshopper is to find a way to calculate own weight situations and to solve the equations for asymmetrical structures. The method for calculating all the relevant information of the construction can by simplified in a flowchart, see Figure 32. First the relation with the Excel method will be explained and the method will be explained through the different blocks of the flowchart. Figure 32 Flowchart for finding minimum complementary energy through different values for H First the input needs to be defined. In Excel a mathematical formula is needed for a construction, but in Grasshopper one can pick a randomly drawn curve, choose a load and choose the refinement of the calculation steps. The structure and its load need to be cut in pieces, to discretize the load in point loads to make force polygons 43 P age

50 With the discretized load, the vertical supports can be calculated. To find the vertical support reactions of a structure with the supports on the same horizontal line, the equilibrium around one of the support points can be analysed. The summation of the bending moments around this point should be zero, ( M = 0) so calculating the bending moments around this point can give the vertical support reaction, see Figure 33. Since the grey support reactions cross the chosen (red) point, the bending moments due to these forces are zero. One support reaction (V b ) remains and can be calculated. The total vertical force is known, so simple subtraction gives the other support reaction. Figure 33 Bending moments on one of the supports Now the force polygon can be created, as described in Part II 4. The position of the connecting point is still unknown and is determined by the variable H. Figure 34 Constructing the force polygon for different values of H 44 P age

51 GRAPHIC STATICS IN ARCHES AND CURVED BEAMS With the force polygon the line of thrust can be drawn. The line of thrust changes slope between two lines of influence, and the angles of the slopes correspond to the angles in the force polygon. The height of the force polygon can be changed by changing variable H. Figure 35 Different values of H give different heights in lines of thrust With the calculation power of Grasshopper, the forces in the structure are easily calculated when a line of thrust is drawn, this happens in the last four blocks, with the complementary energy as a result The iteration found in the flow chart is the variable H, different values can be tried to find the structure with the least complementary energy, just like the Excel calculations Validation The results of the Grasshopper script are tested and reviewed in this paragraph. Four different shapes are used to test the results of the Grasshopper interface, a half circle (1.), a flat arch (2.), an asymmetric arch (3.) and a more free form asymmetric curve (4.) with two peaks instead of one, see Figure 36. Figure 36 Four tested structures with their lines of thrust and force polygons 45 P age

52 Table 5. Values for different shapes. Shape 4 has two lines of values from the script. Due to the decimals in Grasshopper all H values between 13392N and 13410N return the same value for the energy Shape # H [N] V l [N] V r [N] M max M min Compl. E tot (x10^9) [Nm] [Nm] 1 Script ,037 GSA Dev. 1% 0% 0% -3% -7% 2 Script ,7506 GSA Dev. -1% 0% 0% -2% 0% 3 Script ,3930 GSA Dev. -1% 0% 0% 4% 4% 4 Script , ,09 GSA Dev. 0% 0% 0% 0% -1% 0% 0% 0% 1% -1% To check if the results are accurate, the shapes tested in Grasshopper, are also calculated in GSA 1 and compared for support reactions (both horizontal and vertical), maximum bending moments and their positions in the construction and the course of the normal forces. Since it is difficult to compare the results on the exact same two points in both programs, the course of normal forces and the locations of the maximum bending moments are checked visually. The support reactions and maximum bending moments can be checked and deviations expressed in percentages (see Table 5). The calculation method in Grasshopper returns good results for the check in percentages. Some minor deviations are found. These can be explained by three reasons; Because of the discretization, average forces and moments over each bar length are used, giving an approximation of the answer, not the exact answer Due to rounding errors in Grasshopper values may have small deviations In the discretization, the centre of each line segment is used, but for a block with a thickness, the centre of gravity is not exactly in the middle of the centre line (see Figure 37). This error 1 Oasys GSA Suite is a software package developed in collaboration with Arup and is used by this company to perform structural calculations. The program uses the Finite Element Method to calculate the modelled structures. 46 P age

53 GRAPHIC STATICS IN ARCHES AND CURVED BEAMS might be small, but it is important to know the impact of the choice to discretize the loads. Figure 37 In grey the actual centre of gravity, in black the centre of the line, a small difference is found in the discretization. The drawing is an exaggeration to illustrate the difference 12.2 Limitations The main focus of this model is to proof the calculation method that was developed works. To proof this, only curves with supports on a horizontal line are tested. With these curves the method described in this chapter can be applied. If the supports are not on the same horizontal line, the calculations become more complicated. This is the next step to solve in the research. 47 P age

54 13 Exploration for other solutions The solution given in the exploratory model from Part IV 12.1 has the disadvantage that the user needs to find the correct answer manually and iterative. Finding a formula or expression that solves the problem in one calculation would be a more graceful solution. In this chapter two options for finding this solution are discussed, a mathematical approach and a graphical approach. The downside when a direct solution is that the user does not learn from the process. By finding the correct answer by hand, the user is forces to get an understanding of the mechanism and sees directly how the force polygon and line of thrust are influenced by the support reactions. The first approach focuses on the formula for the energy. As shown in Figure 38, the energy can be plotted in relation to the horizontal support reaction and this gives a parabolic shape. Differentiating the formula for this parabola and solving this for zero, gives the horizontal support with the least energy. Figure 38 The complementary energy of the structure as a function of H, the horizontal support reaction The other option is based on the idea that pictures as shown below present the only correct answer. So in the relation between the construction, the line of thrust and the force polygon should hold the information needed to find the correct equilibrium solution. Figure 39 Structure, line of thrust and force polygon, only one combination is the actual equilibrium solution 48 P age

55 GRAPHIC STATICS IN ARCHES AND CURVED BEAMS 13.1 Mathematical approach To find the formula for the least complementary energy, the first step was to solve this for compressive only structures, so the line of thrust is inside the material. This means that only normal complementary energy is included in the calculations. The formula presented in Part II 6.2 for normal complementary energy is E c,n = 1 2 N 2 EA l (13.1) This can be simplified for calculation purposes if the assumption is made that the section over the whole construction is equal and made of one material. This gives the formula E c,n = N 2 l (13.2) This calculates the energy for one bar, if the total energy is needed, a summation of all the bars must be made E c,n = N 2 il i (13.3) If the formula for the total energy can be expressed with H as the only variable, a differentiated formula can be used to find the least complementary energy. The formula for the energy has only two variables, N and l, so both variables are expressed in H N The value for the normal force N in the bar i of the line of thrust is equal to the force F i in the relating force polygon. In the force polygon shown in Figure 40, the load on the structure is discretised in point loads, g. V can be calculated in the method described in Part IV 12. In the formulas shown below, the forces 1 to 5 are expressed in a known V and the unknown H through Pythagoras. F 1 = (H 2 + V 1 2 ) F 2 = (H 2 + (V 1 g) 2 ) F 3 = (H 2 + (V 1 2 g) 2 ) F 4 = (H 2 + (V 1 3 g) 2 ) F 5 = (H 2 + (V 1 4 g) 2 ) F i = (H 2 + (V 1 (i 1) g) 2 ) (13.4) Figure 40 Force polygon with four discrete loads and a known V and unknown H 49 P age

56 l The length of a bar in the line of thrust can be expressed in the slope of the bar (angle θ i ) and the horizontal displacement ( x i ). The horizontal displacement is defined by the lines of influence of the discrete loads (Figure 41) and is a value given by the construction. The angle θ i can be found in the force polygon and is defined by the calculations as shown in the formulas below. The numbers can be found in Figure 40. tan θ 1 = V 1 H tan θ 2 = V 1 g H tan θ 3 = V 1 (2 g) H tan θ 4 = V 1 (3 g) H tan θ 5 = V 1 (4 g) H tan θ i = V 1 ((i 1) g) H (13.5) Figure 41 Discrete loads and their relating distances Combining calculations The formulas (13.4) and (13.5) as deduced in the last two paragraphs can be rewritten and combined with E c,n = N 2 il i In to E c,n = δx i ( V 1 ((i 1) g) ) (H 2 + (V H 1 (i 1) g ) 2 ) (13.6) In this formula, a summation of all the normal complementary energies is found with the only variable H. Now this formula can be differentiated and set to zero. Since the aim was to find a more insightful and quick-to-use-calculation, this direction became too complicated. The realization that this is only working for systems where the line of thrust coincides with the structure, made an end to this path in the research Graphical relations between construction, force polygon and line of thrust To find a more intuitive, flexible and easy to understand method, the next paragraph describes the relations between the force polygon, the line of thrust and the construction. This research direction started with two things, first the idea that, if the three images combined give the only correct answer, there should be some relation between the images that can help find the answer. The second is that the area under the line of thrust is equal to the area under the construction. This was found by coincidence while modelling the exploratory Grasshopper script. 50 P age

57 GRAPHIC STATICS IN ARCHES AND CURVED BEAMS Figure 42 In black the area of the line of thrust, in grey the area of the construction Using this second principle, illustrated in Figure 43, the flowchart to illustrate the script gets shorter. No longer is it necessary to know all the forces to calculate the energy, but this is still an iterative process. To illustrate the accuracy of the method using the equal areas, the horizontal supports and total complementary energy levels are indicated in Table 6. Table 6. Comparison with GSA Shape # H [N] Compl. E tot (x10^9) 1 Script ,037 eq A ,046 GSA Script ,7506 eq A ,7510 GSA Script ,3930 eq A ,3979 GSA Script , eq A ,09 GSA Figure 43 Shortened flowchart Relation force polygon line of thrust To find out why the area of the thrust line is equal to the area of the structure, research is done for the different geometrical relations. The ideal starting point for finding the correct equilibrium state, is if all the information can be found in one of the drawings. First the relation between the line of thrust and the force polygon is analysed. The line of thrust is considered as if it were a structure in the calculation method. So the relation is between a structure and a force polygon. The construction in the example is a four barred system, as shown in Figure 44. The bars have a length, l i, and in the bar a force acts, F i. The angles between the bars (α ij ) relate to 51 P age

58 the angles in the force polygon, as indicated in the drawing. The force polygon contains only forces, and have no length, only the ration between the forces is relevant. Figure 44 Four barred force diagram and a possible force polygon or reciprocal figure, highlighted the areas that are related The area highlighted in the construction, between bar l 1 and bar l 2, has a relation to the area highlighted in the force polygon, between force F 1 and F 2. The area in the construction will be called A 1,2 and the area in the force polygon will be called A f,1,2. Using the formula for the area of a triangle (Wikipedia 2013) A = 1 2 l il j sin(α ij ) (13.7) Both areas can be expressed. A 1,2 = 1 2 l 1l 2 sin(α 12 ) A f,1,2 = 1 2 F 1F 2 sin(180 α 12 ) and rewritten in to sin α 12 = 2A 1,2 l 1 l 2 sin(180 α 12 ) = 2A f,1,2 F 1 F 2 And since sin a = sin(180 a) (Wikipedia 2013) A f,1,2 F 1 F 2 = A 1,2 l 1 l 2 (13.8) Now the relation between areas in the force polygon and the area in the line of thrust can be found. Applying these formulas on the line of thrust instead of a four bar system gives the relation as shown in Figure 45. This does not give any information on the total area under the line of thrust, but only on the areas of two force directly linked to each other. Also these areas overlap, so the area of the force polygon is not directly related to the area under the line of thrust. 52 P age

59 GRAPHIC STATICS IN ARCHES AND CURVED BEAMS Figure 45 Line of thrust and force polygon, highlighted the areas that are mathematically related Relation total area line of thrust force polygon To find the total area under the line of thrust, Lopshits formula is introduced. Lopshits developed a formula to find the area of any given polygon oriented in a 2D space, using the lengths of the members of the figure, and the angles between the connecting members (Lopshits 1956). A = 1 2 (a 1[a 2 sin(θ 1 ) + a 3 sin(θ 1 + θ 2 ) + + a n 1 sin(θ 1 + θ θ n 2 )] +a 2 [a 3 sin(θ 2 ) + a 4 sin(θ 2 + θ 3 ) + + a n 1 sin(θ θ n 2 )] + + a n 2 [a n 1 sin(θ n 2 )]) (13.9) 53 P age

60 The formula and the proof for the method is somewhat unwieldy to use, so to indicate its relevance a small example of a line of thrust is used. The formula can be rewritten in separate area calculations for triangles, and the polygon can be cut in different triangles. Figure 46 Members and angles numbered n 2 = 1 2 a 1a 2 sin(θ 1 ) Figure 47 Step one n 3 = 1 2 a 1a 3 sin(θ 1 + θ 2 ) a 2a 3 sin(θ 2 ) Figure 48 Step two n 4 = 1 2 a 1a 4 sin(θ 1 + θ 2 + θ 3 ) a 2a 4 sin(θ 2 + θ 3 ) a 3a 4 sin(θ 3 ) Figure 49 Step three And then the summation A = n 2 + n 3 + n 4 So when the outer angles of the line of thrust and the element lengths are known, Lopshits can be applied. To relate this total area to an area in the force polygon, the element lengths and angles should be related to the lengths and angles in the force polygon. 54 P age

61 GRAPHIC STATICS IN ARCHES AND CURVED BEAMS Figure 50 Calculating length a 2 in the force polygon In Figure 50 the relation between length a 2 in the line of thrust is linked to the force polygon by projecting the distance between two discrete loads in the force polygon (δx 2 ). Since the slope of F 2 in the force polygon is equal to the slope of the element in the line of thrust, the length can be calculated using the slope and horizontal distance in het force polygon. The relation can be mathematically written as F i = H a i δx i (13.10) Also the angles θ 1 and θ 2 in both figures are related, similar to the relation described in Part IV and Part IV When this way of calculating is displayed visually, the total area of the line of thrust is represented by a part of the area in the force polygon, see Figure 51. Figure 51 Area of the construction related to the force polygon. The different dotted lines indicate the different lengths between the discrete loads 55 P age

62 13.3 Changing parameters of force polygon To get a better understanding of the influences of the different parameters that define the force polygon (horizontal support, vertical support and the divisions in the discretization), an inventory was made of the consequences of changing each parameter. All the relevant information in the force polygon for the test are the area; - of the total force polygon A f,total - of one of the force triangles in the polygon A f,i - of the thrust line projected in the force polygon A total - of one of the projected triangles of the thrust line projected in the force polygon A i - under the line of thrust A thr - the force density (F/l) In the figures used to support this paragraph, A i and A total are indicated with grey areas. The areas A f,i and A f,total are not indicated separately, to keep the pictures clear. The dimensions of δx i found in the construction are set equal for all the separate force polygons for explanation purposes, and are indicated with a red dotted line. The energy changes under influence of the changing support is also checked to see if the energy changes linear or in a (parabolic) curve with a minimum or maximum Changing vertical support By changing the vertical support reaction in the force polygon, the coordinate where the forces meet moves up and down (see Figure 52). In the table and figure can be seen that the areas of all the relevant geometry stays equal. The area of a triangle can be described by A = 1 wh, where w is the 2 width of the base and h is the height of the triangle. Changing V does not change the height or width of the figures, so evidently A thr does not change either. Areas A f,total A f,i A total A i A thr Energy F.D. Changes Equal Equal Equal Equal Equal Parabolic Equal Figure 52 Changing vertical support reaction Changing horizontal support By changing the horizontal support reaction in the force polygon, the coordinate where the forces meet moves left and right (see Figure 53). When H is 0, the point is on the same line as the vertical projected forces. Increasing H moves the point further away. This is used as reference to see if the areas increase, decrease or stay the same. 56 P age

63 GRAPHIC STATICS IN ARCHES AND CURVED BEAMS Areas A f,total A f,i A total A i A thr Energy F.D. Changes Higher Higher Smaller Smaller Smaller Parabolic Higher Figure 53 Changing horizontal support reaction When H is increased, the areas change as indicated in the figure and table. Since δx i moves along with H, the height of the triangle for A total stays the same, while the width of the base decreases. This means the area of the line of thrust, which is linked to A total through Lopshits, decreases Changing divisions By changing the divisions in the discretization, the shape of the force polygon outline does not change, only the internal arrangement changes (see Figure 54). Checking this influence is done for completeness, but it will not change the outcome of the calculations that much. The influences are displayed in the table. Areas A f,total A f,i A total A i A thr Energy F.D. Changes Equal Smaller Smaller Smaller Equal Equal Smaller Figure 54 Changing divisions in discretization The general shape of the force polygon and thus the area A f,totaal does not change. The areas A f,i and A i become smaller with higher divisions. The divisions are made in the structure, meaning that the δx i value changes when the divisions change. Changing this distance changes the height of the triangle for A total, while the base also changes resulting in a smaller area for higher divisions. Here can be seen that Lopshits formula is needed to relate the area of the force polygon to the area of the line of thrust. With higher divisions, more triangles for A i are generated. 57 P age

64 Application Combining the findings in the influences of the horizontal and vertical supports on the force polygon leads to a new method to finding the forces in the construction. Because the area under the line of thrust should be equal to the area under the construction, first the H has to be varied until this requirement is fulfilled. Since changing V does not change the areas in the force polygon, this value can be varied until the least complementary energy is fulfilled. This means that there is no more need to calculate the vertical supports through equilibrium in bending moments around one of the supports. Since this was only possible for constructions with the supports on one horizontal line, now also non even supports can be analysed. The new flowchart for the calculation method can be seen in Figure 55. Figure 55 Flowchart with two variables This method is modelled and tested in a transition model looking a lot like the exploratory model. The results on a curve with the supports on different heights are shown in table 7. The results give a close match to the GSA calculations. Table 7. M H [N] V l [N] V r [N] max M min [Nm] [Nm] GSA Grasshopper % 0% 0% -3% 1% 58 P age

65 GRAPHIC STATICS IN ARCHES AND CURVED BEAMS Figure 56 Tested curve with supports at different heights 13.4 Proof of equal areas The final application of the equality of both areas under the construction and the line of thrust needs to be proven. To do so, the mathematical descriptions of the area under the construction, the area under the line of thrust and the total complementary energy are needed. To find the areas, Lopshits can be used. To find the complementary energy, the start made in Part IV 13.1 can be used, but the bending moments need to be introduced. To proof that when these areas are equal, the correct horizontal support is used, the complementary energy is used. In this way can be shown that when the areas are equal, the system contains the least complementary energy, meaning that situation gives the correct equilibrium solution. The mathematical description of these three different calculations is made with vector calculations. With vector calculations the use of x- and y-coordinates helps defining all the necessary values Area construction The construction that needs to be analysed is expressed in an area with the use of Lopshits formula and the coordinates of the points where the discrete loads act. With these coordinates, the relevant information about the construction (element lengths and outer angles) can be described and added in Lopshits formula. Each coordinate of the construction is given by two numbers, for example point A; A x A. The vector y between two points, for example between point A and point B, is written by subtracting one vector from the other; AB = B x A x B y A y And then this can be rewritten in the actual length of the vector by; AB = (B x A x ) 2 + (B y A y ) 2 The angle θ between two vectors, for example AB and BC, can be written as; 59 P age

66 cos θ = AB BC AB BC Since the formula Lopshits developed uses sin θ and sin(cos 1 (x)) = 1 x 2, the formula for calculating the area of a triangle with discrete points A, B and C becomes; Area ABC = 1 2 AB BC 1 ( AB BC AB BC )2 With this expression, the total area of the construction can be found using the coordinate points and Lopshits formula Area line of thrust The same calculation steps can be made for the area under the line of thrust. The problem here is to find the correct y-coordinates for the points in the line of thrust. The x-coordinates are equal to the coordinates of the construction, since they both use the line of influence of the discrete loads. The relation between the line of thrust and the force polygon is already described in Part IV In the definition of the area, the force polygon was used and not the line of thrust is used. This is because there are less unknown values are used in the force polygon. The only two variables in the calculation of the energy and the areas should be the horizontal and one of the vertical supports. The length of the force can be found by expressing the vectors in the force polygon in variables, for H example ; F 2 where g is the value for the discrete load. This can be expressed in a V (1 g) length; F i = (H 2 + (V 1 (i 1) g) 2 ) This is similar to formula (13.4) derived in Part IV The vectors of the forces and the lengths of these forces can be used to calculate the angle between two relating forces. Because the angle in the force polygon is equal to the outer angles in the line of thrust, only the forces are used to calculate the angle θ 1 for the area under the line of thrust. cos θ = F 1 F 2 F 1 F 2 The lengths of the forces do no good in the calculation of the area, the length of the elements of the line of thrust are needed. As found in the exploration of Lopshits formula, the length of the element can be related to the force in this element, the distance between two discrete loads and the horizontal support; a i = F iδx i H Now this can be combined in the calculations for the triangle between element a 1 and a 2 of the line of thrust; A 12 = 1 2 a 1a 2 1 ( F 1 F 2 F 1 F 2 )2 With this expression the total area under the line of thrust can be found using the forces in the force polygon and the distances between the discrete loads from the construction. 60 P age

67 GRAPHIC STATICS IN ARCHES AND CURVED BEAMS Minimum complementary energy To calculate the minimum complementary energy, the normal forces and bending moments in the construction are needed. This energy is the link between the (area of the) construction and the (area of the) line of thrust, using information from both the construction and the line of thrust. Approximately the same steps as in the script can be used to find the complementary energy, only the steps need to be described with mathematical formulas. The relations shown in Figure 25, page 30, need to be expressed in vectors. Bending moments To choose a point for the forces to relate, the same is done as in the script, the middle of the part in the line of thrust is used. The middle of a vector can be found using; AB mıddle = 1 A x + B x 2 A y + B = y A x +B x 2 A y +B y 2 where A and B are the beginning and end points of the vector. From this middle point the distance perpendicular to the structure is needed. So the vector of the line needs to be rewritten in a perpendicular vector. For AB = B x A x B y A this means AB = (B y A y ) y B x A x Then the distance between the line of thrust and the construction can be calculated by finding the intersection point of these two lines, displayed in vectors. For this the formulas for the lines need to be described. The general formula for the intersection of two vector lines is A x A y + t m x m y = B x B y + s n x n y Here A and B are the starting points of the lines, represented in vectors. m and n indicate the slope of the line. t and s are the variables, at the intersection of the two lines t = s. The values for the construction for the starting point and the slope are easily found. The values for the slopes perpendicular to the line of thrust can be found in the force polygon, only the starting points get problematic. In the first part of the chapter the lengths of the elements in the line of thrust where found in the force polygons, but the information about the height of each member relative to the construction is still unknown. If this is expressed, the coordinates of point s on the construction can be found and finally the distance between the middle point on the line of thrust to point s can be derived. If this value is multiplied by the force F from the force polygon, the bending moment in point s is known. Normal forces 61 P age

68 To find the normal forces in point s, the force F from the force polygon needs to be decomposed in a force N in the same direction as the construction. This can be done by using the slopes of the two lines to calculate the angle between the slopes, just as the calculation used in the area. cos θ = AB F 1 AB F 1 The normal force can then be found with N i = F i cos θ. Energy calculation To find the total energy of the construction, the lengths between the calculated forces and bending moments has to be known. This can be done by calculating the vector lengths between all the points s calculated in the step before. Now with the formulas for the energy from Part II 6.4, the total complementary energy of the system can be calculated. The formulas in this chapter on calculating with vectors are found in (Craats 2000, Linden 2002) Conclusions When all three of these steps are completely expressed mathematically, a way to proof the areas of the construction and the line of thrust are equal when the energy is minimum should be found. Expected is that some variables in the equations can be omitted, for example the V, since in Part IV is found that this does not influence the area under the line of thrust. For the scope of this thesis, the mathematical proof is not directly relevant. A description on a direction for an answer is indicated. In five different examples in Grasshopper the areas are equal when the energy is minimum. 62 P age

69 GRAPHIC STATICS IN ARCHES AND CURVED BEAMS 14 Final Grasshopper method The final Grasshopper method is described and illustrated. Combining all the information and theory explained before, a method is developed to find the division in normal forces and bending moments in arches and curved beams, using graphic statics and the theory of least complementary energy. The method can be applied on both projected and own weight cases, for shapes that are symmetrical or asymmetrical and have even or uneven supports. The flowchart for the exploratory script is already given and explained in Part IV 12, the flowchart for the transition script is shown in Part IV For the final script some new theory and a new flowchart is needed Closing line In order to reduce the variables in the method and increase the understanding of the relation between the force polygon, line of thrust and the construction, it is important to find the exact division of the load over the vertical supports. The vertical support reactions are directly related to the vertical position of the chosen pole. First idea was to find the centre of the construction and use the x-coordinate to find the height of the pole. Testing this idea in the Grasshopper script did not return a good division over the supports, compared to the old method using bending moments around the support or the method with two variables. Figure 57 Centre point of the construction in relation to the vertical supports in the force polygon The point O can be more precisely placed with the use of the closing line. In the examples shown in Part III 9 the line of thrust changes height if the point of the force polygon is moved in the direction of the horizontal support H. This is because the supports of the construction are on the same horizontal line. If the constructions supports are not one a horizontal line this is different. Moving the point only in the H direction will create thrust lines that are not representing an equilibrium state. 63 P age

70 Figure 58 Left; construction and a force polygon and thrust line with H is 2244 N Right; construction and a force polygon and thrust line with H is 2937, the later does not present equilibrium Here is where the closing line gets important. The closing line has the same direction as the imaginary line that can be drawn between the two supports. If one equilibrium state is found, the closing line can be drawn trough the point O. If one moves the point over this line instead of only in the H direction, only equilibrium situations are found. Since the slope of the movement is known, only one variable is left to find the equilibrium state with the least complementary energy Finding the closing line This closing line is only helping if the location is known relative to the force polygon. To find the position of the closing line the following method can be applied. Figure 59 a. With the discrete loads from a construction, a force polygon can be generated using a randomly chosen point O. With this force polygon, a line of thrust can be generated, starting in one of the supports of the construction. Most likely the line will not end in the second support. An imaginary line can be drawn trough the start of the construction and the end of the line of thrust. This line can be moved to point O, and a point can be marked where the line intersects the vertical loads of the force polygon. Figure 59 b. A second imaginary line can be drawn between the start and end of the construction. This line can then be moved to the intersection in the loads. The moved line is then the closing line. Moving point O over this line generates only equilibrium solutions. Figure 59 Finding the closing line (Beranek 1980) 64 P age

71 GRAPHIC STATICS IN ARCHES AND CURVED BEAMS Application closing line With the closing line known, and the amount of possible lines of thrust greatly decreased, a new calculation method can be scripted. As shown in the previous paragraph, one iteration in making a random force polygon can determine the position of the closing line, so the flowchart of the calculation can be adjusted to incorporate this first force polygon and first line of thrust. Then the closing line can be found and this can be used to create only lines of thrust in equilibrium A new flowchart is drawn to illustrate the new procedure. The value to optimize in the loop is now the energy, the line of thrust with the least complementary energy is the line giving the correct normal forces and bending moments in the beam or arch. Figure 60 Flowchart calculation using closing line Shape Script Method H 1 Transition 2750 Final Energy 2779 Areas Transition 6814 Final Energy 6843 Areas Transition 4354 Final Energy 4314 Areas Transition Final Energy Areas Transition 3208 Final Energy 3274 Areas Validation In order to test the new method with the energy, the results of the transitional script are compared to the results of the new method. Only the horizontal support reactions are checked, since the way to calculate the forces on the constructions are equal to both scripts. In the table can be seen that there are rather big differences in the results. In the script the complementary energy was minimized to find the horizontal supports. If the areas are set equal, instead of the energy to a minimum, more accurate results are found. This can be explained by the calculation steps. There are more steps needed to calculate the forces and energy in the system, with more chance of rounding errors and inaccuracies. The different ways of finding the support reactions and the relating horizontal supports are listed in the table. 65 P age

72 14.2 Using the script To analyse a curved beam or arch with the Grasshopper script, these are the steps one need to take (with the script opened on a computer this part is easier to understand, otherwise continue at Part IV 14.3 Stresses); Set your Rhino curve in Grasshopper in the Input box, note that the curve should be drawn in the negative x direction (from right to left) in Rhino, in the xz-plane. If you want to calculate projected load, set the Boolean to True and set the Loading in Input. If you want to calculate own weight, set the Boolean to False and select the loading in the Beam section. Here you can choose between a load in N/m or to set a weight for the material. In the beginning of a design process the designer has no dimensions for the beam, so you can set the density to zero, and use the additional loading just to see the influences of the shape. When more is known about the material and the section, the actual weight can be used. Now in the Solver box, you can use the slider to generate different lines of thrust. Between the supports of the curve a red dot is shown. When the correct line of thrust is selected, the dot turns green. To help the used, the area is plotted next to the slider; this value should be (close) to zero. When the dot turns green, the correct equilibrium is found. In the Display component the user can choose to display different outputs, like the line of thrust, force polygon, forces, support reactions and stresses. The output is explained more in the definition description in Part IV Stresses To check if structures do not fail in certain load cases, stress calculations are needed. As an output option for the script, the stresses in the beam can be viewed. In order to calculate the stresses, the normal forces, bending moments and beam section need to be known. The stresses due to normal forces are calculated with σ N = N A The stresses due to bending are calculated with σ M = M w Here w is the stiffness for bending, defined by w = z, with z the distance from the centre of the I beam to the outer fibres of the material and I the second moment of area. This gives the bending stresses in positive and negative in the outer fibres of the beam. Combining the two stresses in the total stresses in the top and bottom of the beam gives σ = N A ± Mz I With this stress known, the designer or engineer can check if the material with the chosen section and load will break or not. In the script the yield strength of the material can be used as input, and the script will show in orange and green at what points the stresses become too high. 66 P age

73 GRAPHIC STATICS IN ARCHES AND CURVED BEAMS 14.4 Conclusions The result of the calculation method that is developed is a script in Grasshopper that can analyse curved beams and arches in own weight and projected loads. The output of the script can be both graphical and numerical, and it can be used in both the beginning of the design stage to optimize shapes for flow of forces, and in the end to check for stresses. The graphical output can tell the designer in a quick way how the shape of the construction performs. In general, the larger the distance between the line of thrust and the construction, the higher the stresses. This is directly clear when looking at Figure 61. The line of thrust and force polygon can be displayed. This does nog directly give quantitative information about the construction, but can be used in the concept phase of a design or for educational purposes. Students can increase their knowledge and understanding of mechanics in beams and arches by trying different shapes and see how these shapes change the force polygon and line of thrust, and how the line of thrust influences the forces in the structure. Figure 61 Output generated by Grasshopper, in orange the stresses higher than the yield strength of the material, the line of thrust if far outside the material of the construction. The larger numbers indicate the support reactions. 67 P age

74 15 Definition descriptions Grasshopper In Part IV 12, Part IV 13 and Part IV 14 all three Grasshopper calculation methods are shown in flowcharts. In this chapter the final script will be elaborated on in more detail. It is recommended to have the actual script opened in Grasshopper while reading this, so the reader can see what is being described. The description is split in three parts, first part explains the final calculation method in more detail, the second part explains the graphical representation of the solver and the output and the final and third part describes the differences between the exploratory script, the transitional script and the final script Calculation description The calculation is described per block. In the overview (Figure 62) different coloured blocks can be found, each block has its own function and can be related to the colours in the flowchart (Figure 63). Figure 62 Grasshopper script overview, with different coloured blocks Input In the input box the parameters for the calculation are set. First the Rhino curve that needs to be analysed has to be set in the curve input. Then one can choose in a Boolean to calculate projected load (True) or own weight (False). The two other parameters are the divisions of the discretization (a number slider) and the load of the structure. By changing the divisions more calculation points are introduced, slowing the script in linear steps. When a division number higher than around 60 is chosen, a good accuracy of the results is achieved. Discretization The discretization is done in two different paths, one for the projected load and one for the own weight calculation. After these two paths an extra block is used to select the right data to use in the rest of the calculation. In the own weight is done in two blocks. In the first block the beam material and section can be selected with some sliders. These combined give the weight of the beam. If the material and section of the construction is still Figure 63 Flowchart of method, colours match with Figure P age

75 GRAPHIC STATICS IN ARCHES AND CURVED BEAMS unknown, the additional loading can be selected. With the additional loading the section of the beam is not relevant. The weight is used in combination with the division slider and the length of each segment to generate the discrete point loads. In the projected load the horizontal distance between the supports is measured and divided by the slider. The x coordinates of these division points determine the position of the discrete load. The extra block uses the Boolean from the input to select what data to use. 1 st force polygon In this block the force polygon is constructed using a random chosen point. The discrete loads are placed in a straight vertical line to make the total load of the structure. With the horizontal and vertical supports collected from the Solver box (explained in Part IV 14.2) the origin point of the force polygon is created. With all the points known, the lines of the force polygon can be drawn. 1 st line of thrust Here the line of thrust is plotted, using mathematical calculations. The process as described in Part IV is iterative, making it slow if the division count increases. The component in Grasshopper is built mathematically. Because the slopes of the lines are known from the force polygon, and the x-coordinates are known from the discrete loads, the z-coordinates of each point can be calculated. Combining these coordinates returns the shape of the line of thrust. This line of thrust does not end in the second support of the construction, so this is not an equilibrium solution. Closing line In the closing line component, the vector between the start of the construction and the end of the 1 st line of thrust is used to find the intersection point with the total vertical load. From this intersection, the vector from the beginning to the end of the construction is plotted, which is the closing line (see Part IV ). 2 nd force polygon The information from the 1 st force polygon is used to create the 2 nd force polygon, with the closing coordinate at the closing line. 2 nd line of thrust With the information from the 2 nd force polygon the actual line of thrust is generated. All the possible lines of thrust created in this component give equilibrium. Eccentricity The eccentricity is calculated by analysing the separate segments of the polyline that is the line of thrust. The centre of each segment is picked, and there a line is plotted perpendicular to the slope of the segment, this is done by rotating the vector of the line 69 P age

76 segment. The intersection point of each of these lines with the curve is found, returning the distance from the centre of the segment on the line of thrust to the curve, which is the eccentricity. Forces calculation With a few mathematical steps the normal forces and bending moments are calculated, each in its own block. By shifting lists, the average values over the bar lengths are calculated giving the values for the complementary energy. Complementary energy In the final box, the values for the forces and the bar lengths of the structure are collected and the complementary energy is calculated Graphical representation All the blocks not covered by the coloured areas in Figure 62 are used for graphical representation and output. Two things are important for the representation. A visual check was built in the Grasshopper definition to help the user find the correct equilibrium state for the line of thrust. Also, the equilibrium state and the forces in the structure need to be displayed Visual check The visual check uses the Solver, the area subtraction and the visual check blocks. In the Solver the user uses one slider to change the force polygon over the closing line to find the equilibrium state of the structure. To help the user, a red dot or circle is plotted between the supports of the structure. The circle turns green when the area of the construction is equal to the area of the line of thrust by using a Similarity component. The accuracy of the component brings forth a range of values for the horizontal support returning a green circle. If a more precise approach is needed, a number representing the precise difference in areas is given in the first panel in the Solver, this value should be set to 0. The second panel in the solver box shows the value for total complementary energy. Since the original approach was to minimize this energy, this value is still displayed, but not directly used to find equilibrium. See Part IV for the explanation Display component When the correct equilibrium state for the beam or arch is found, one can choose different results to display in the Rhino view. Arrows portraying the support reactions are always shown. One can choose to ad one or a combination of the following options; Support reactions These are shown next to the arrows for the support, the values are in Newton. The values for the support reactions are found in the force polygon. The coordinate of point O relative to the total load in the force polygon is used to find the values for the horizontal and vertical support reactions. They are then displayed by previewing their vectors at the supports (Figure 64). 70 P age

77 GRAPHIC STATICS IN ARCHES AND CURVED BEAMS Figure 64 Support reactions Normal forces or bending moments This indicates the normal forces or bending moments (in Newton or Newton meter) at the division points. The values are projected at the points in the construction using Tags. In order to prevent the numbers from becoming unclear, the division points are all moved up by a small step (0.01 in the z direction) Figure 65 Bending moments 71 P age

78 Line of thrust Force polygon The line of thrust and force polygon can be displayed, the geometry for these parts is collected in their respective blocks and added in the display component. When trying to find the equilibrium solution it is advised to have the line of thrust displayed (True), since it can help with the insight of the forces and on which way to move the solver slider. Figure 66 Line of thrust and force polygon Beam geometry Shows the chosen geometry / section of the arch. This does not add any information to the calculation. This option is only relevant for beams loaded in own weight, since a projected load is not related to the section of the beam. A rectangle is plotted perpendicular at the beginning of the curve and swiped along the curve to create a beam. Figure 67 Beam geometry 72 P age

79 GRAPHIC STATICS IN ARCHES AND CURVED BEAMS Stresses (normal stress, bending stress or a combination ±) The stresses in the beam are calculated as described in Part IV With a if then component scripted in VB a colour value is selected for the text tags to be displayed in either green or orange. One output gives the stresses in the top of the beam, the other in the bottom. Figure 68 Combined stresses 15.3 Differences different scripts The first big difference between the exploratory model, the transitional model and the final model is the component calculating the vertical support reactions. In the exploratory model, the vertical supports are calculated through a set of mathematical steps with the use of the bending moments around one of the supports. In the transitional model, the vertical support is not calculated, but set as a variable. In the final model this is completely absent; the closing line is used to generate horizontal and vertical support reactions with a slider. The second difference is in the overall workings of the script. The exploratory model was not intended to be used for any other purpose than research, so the script has no real structure or overview between the different blocks. In the transitional model this was rebuild in a more structured way with clear in- and output. This has the advantage that no data is leaving the block half way, but all the data can be found at the beginning and end of the block. This way the model is easier accessible for further research and development. The third big difference is the presence of the display component. In the transitional script a first test was made in finding clear ways of displaying the results of the calculations, where in the exploratory model the forces could only be found within the different components. In the final script one clear way to show the output is scripted, so users do not need to find their own results. 73 P age

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83 GRAPHIC STATICS IN ARCHES AND CURVED BEAMS 16 Conclusions The goal of this chapter is to reflect on the process, the developed method and the developed component. Also recommendations are made on how to improve and/or expand the method and the component Process During the process a lot of theories were tested and tried. Some of them where so far off track that they are not documented in this thesis to keep the description short and clear. Because a lot of theory is already known in the field of structural mechanics, it was simple to verify intermediate results during the research with hand calculations or calculations in different software suites. This prevented the research from running too far off track at any point. At some points, when new directions where found, that where not directly related to the main goal, it is tempting to do more research in that field. This happened for example with the area relations in the force polygon and the line of thrust. At the end, this research also paid off, so overall no big side tracks delayed the research of the main goal. The initial goal of the research was to find the distribution in normal and bending forces in shell structures. Because researching new theories can take a lot of time, and one never knows when a good solution is found, a research like this can take much more time than expected. Half way it became clear that there is so much to find and learn in arches and beam structures, that the goal was adjusted to cover only these structures and not make the step to shells and vaults. Recommendations are made on how to continue this line of research in shells Calculation method and component The developed method holds a quick manner to find the distribution of normal forces and bending moments in arches and curved beam structures. The method is incorporated in the calculation component, so these results are discussed simultaneously. In order of the chapters, all (sub) conclusions and other findings are stated; Using the line of thrust and the complementary normal energy, the correct equilibrium solution for arches without bending moments and a hinged support can be found. The line of thrust returning the least complementary normal energy gives the right equilibrium. This only holds for structures in projected loads. Constructions with a known (simple) mathematical description, hinged supports, a projected load and the supports on one horizontal line, can be analysed using an Excel calculation sheet to find the construction with the least total complementary energy. This can be done by relating the normal forces in the line of thrust to the normal forces and bending moments in the construction. By trying lines of thrust with different heights, the construction with the least complementary energy can be found. For calculating structures with their own weight, the formula for catenary structures is used. This can only be done if the length of the catenary is equal to the length of the structure. The support reactions of a catenary structure change when the length of the catenary changes. So trying different heights will not 77 P age

84 return the structure with the least energy. There is no other mathematical description to find the line of thrust, so a step to a more graphical approach is made. The first graphical approach modelled in Grasshopper returns values close to the calculations made in GSA. In this calculation the own weight of the structure is used. Due to rounding errors and the discretization of the structure small deviations between the GSA calculations and the Grasshopper calculations are found. An extra calculation step is introduced to make the component also applicable to asymmetrical structures. The method still only works for hinged supports and supports on one horizontal line. The area under the line of thrust seems to be equal to the area under the construction. Applying this idea in the component gives the opportunity to calculate structures where the supports are not on a horizontal line. For this both the vertical and horizontal supports are variables to test different lines of thrust. Not all lines of thrust generated present an equilibrium solution. Using the closing line in the calculation method eliminates all the solutions that do not present equilibrium. With his extra information the calculation works with one variable. Searching for equal areas presents more accurate results than searching for the minimum energy. This is caused by the different calculations steps in the script. The script gives visual feedback that can be used in the beginning of the design process to give easy to understand feedback on the relation between the forces in the structure and the shape of the structure. The script gives numerical output to check if the stresses in the structure become too high, and shows the support reactions, bending moments and normal forces in the structure. The script is modelled in a clear manner, so others interested in this theory can use the script and develop the script further Recommendations Some recommendations and improvements are listed for further research and on how to improve the calculation method and script. Finding the exact mathematical method to find the equilibrium that returns least complementary energy is still a big challenge. If this method is found, the process of analysing arches and beams can be linearized and a script running with mathematical calculations is faster than a script using graphical analytics. Downside here is the chance that a black box is created around the calculation. 78 P age

85 GRAPHIC STATICS IN ARCHES AND CURVED BEAMS Increasing the flexibility of the component to make it applicable on more structures and load cases is a good direction for expansion. For example the possibility to calculate point loads on the structure, or loads in different directions. This addition is a pure scripting challenge, but the results should be validated. The method of calculating and using graphic statics should be valid in more complex load cases as well. Another example is the addition of more support cases, like clamped supports. In this case the line of thrust does not start and end in the support points, so more lines are possible. How to solve this and find the leas energy is a calculation challenge. No direct leads on how to solve this problem are found. Yet another addition can be a component where you can select section for the beam and select certain materials, so the user does not have to script this manually. Adding deformation calculations to the component improves the use for designing purposes. Especially when the script is linearized, the component can be implemented in a generative algorithm component enabling structures and beams to be optimized for bending moments, normal forces, deflections or stresses. To apply the method on vaults and shells, some things should be taken into consideration. First important step is that, just like beams and arches, a valid discretization needs to be generated. This force network, representing the structure, needs to be transformed in to a thrust network as described by P. Block. After that the different parameters defining the height of the thrust network need to be defined, and their influences on how they change the shape of the thrust network. The forces in the different bars can be related in a similar fashion as in the beams and arches. Important to take in consideration is the fact that in shells, one can find hoop forces. How these forces can be included in the calculations needs its own research. 79 P age

86 17 List of figures Figure 1 Left; hanging chain models by Gaudi (Daniel 2007) Right; Sagrada Familia, Barcelona, by Gaudi (Mak 2013)... 3 Figure 2 Left; Deitingen service station, by Isler (BauNetz 2009) Right; the Manantiales Restaurant, Xochilmico, by Candela (Lukas 2013)... 4 Figure 3 Left; Force diagram and force polygon of a point in equilibrium Right; force diagram and force polygon of a point not in equilibrium with resulting displacement indicated in red... 9 Figure 4 Left; cable with two forces applied Centre; force polygons of point B and C in equilibrium Right; total force polygon of the system in equilibrium with FAB and FCD representing the support reactions... 9 Figure 5 Thrust line in a random arched structure (Block, DeJong et al. 2006) Figure 6 Different positions of a chain through the thickness of the material (Huerta 2005) Figure 7 Relation between the line of thrust and the force polygon in an arch (Block, DeJong et al. 2006) Figure 8 Cremora diagram of statically determinate truss (Wikipedia 2008) Figure 9 Statically indeterminate cable system, the dotted lines indicate different possibilities to close the force polygon with force F Figure 10 Relation between discrete network (G), planar projection or primal grid (Γ) and the reciprocal diagram or dual grid (Γ ) (Block 2009) Figure 11 Left; Discrete construction (G) and primal grid (Γ) Right; Different possible dual grids (Γ ) (Liem 2011) Figure 12 Surface element split in two distinct surfaces (Calladine 1977) Figure 13 Screenshots of the two tools modelled by Liang Figure 14 Computational diagram to find the force distribution (Pavlovic 1984) Figure 15 Stress-strain diagram for elastic material Figure 16 Stress-strain diagram for linear-elastic material Figure 17 Normal force-strain diagram Figure 18 Bending moments-curvature diagram Figure 19 Statically indeterminate cable system, the dotted lines indicate different possibilities to close the force polygon with force F3, using minimum energy can solve this problem Figure 20 Different lines of thrust and their force polygons, one line contains the least energy with a given load case and span Figure 21 Values needed to define the parabolic shape of the line of thrust Figure 22 In blue a part of a structure and in red the active forces Figure 23 Left; Portal frame Right; semi-circle Figure 24 Excel calculations for portal frame for two different values of f returning different values for the complementary energy Figure 25 Translating and decomposing forces Figure 26 In blue the construction of a semi-circular arch, in red a line of thrust with height f and in green the line j(x) to determine the eccentricity e Figure 27 Two lines, their slopes and their angle Figure 28 Bending moments comparison, the arch spans 11 meters and has a loading of 1000 N/m Figure 29 Flat arch, part of a circle Figure 30 Bending moments comparison flat arch Figure 31 Two structures in projected load (blue) and own weight (red) Figure 32 Flowchart for finding minimum complementary energy through different values for H Figure 33 Bending moments on one of the supports Figure 34 Constructing the force polygon for different values of H Figure 35 Different values of H give different heights in lines of thrust Figure 36 Four tested structures with their lines of thrust and force polygons P age

87 GRAPHIC STATICS IN ARCHES AND CURVED BEAMS Figure 37 In grey the actual centre of gravity, in black the centre of the line, a small difference is found in the discretization. The drawing is an exaggeration to illustrate the difference Figure 38 The complementary energy of the structure as a function of H, the horizontal support reaction Figure 39 Structure, line of thrust and force polygon, only one combination is the actual equilibrium solution Figure 40 Force polygon with four discrete loads and a known V and unknown H Figure 41 Discrete loads and their relating distances Figure 42 In black the area of the line of thrust, in grey the area of the construction Figure 43 Shortened flowchart Figure 44 Four barred force diagram and a possible force polygon or reciprocal figure, highlighted the areas that are related Figure 45 Line of thrust and force polygon, highlighted the areas that are mathematically related Figure 46 Members and angles numbered Figure 47 Step one Figure 48 Step two Figure 49 Step three Figure 50 Calculating length a2 in the force polygon Figure 51 Area of the construction related to the force polygon. The different dotted lines indicate the different lengths between the discrete loads Figure 52 Changing vertical support reaction Figure 53 Changing horizontal support reaction Figure 54 Changing divisions in discretization Figure 55 Flowchart with two variables Figure 56 Tested curve with supports at different heights Figure 57 Centre point of the construction in relation to the vertical supports in the force polygon Figure 58 Left; construction and a force polygon and thrust line with H is 2244 N Right; construction and a force polygon and thrust line with H is 2937, the later does not present equilibrium Figure 59 Finding the closing line (Beranek 1980) Figure 60 Flowchart calculation using closing line Figure 61 Output generated by Grasshopper, in orange the stresses higher than the yield strength of the material, the line of thrust if far outside the material of the construction. The larger numbers indicate the support reactions Figure 62 Grasshopper script overview, with different coloured blocks Figure 63 Flowchart of method, colours match with Figure Figure 64 Support reactions Figure 65 Bending moments Figure 66 Line of thrust and force polygon Figure 67 Beam geometry Figure 68 Combined stresses P age

88 18 Literature list BauNetz. (2009). "Pionier der Schale." from Zum_Tod_von_Heinz_Isler_ html. Beranek, W. J. (1976). Berekeningen van platen deel 1 theoretische grondslagen. Delft, TU Delft. Beranek, W. J. (1977). Toegepaste mechanica K-3 vlakke constructiedelen elasticiteitstheorie. Delft, TU Delft. Beranek, W. J. (1980). Krachtswerking in liggers. Delft, TU Delft. Blaauwendraad, J. (2002). Theory of elasticity energy principles and variational methods. Delft, TU Delft. Block, P. (2009). Thrust network analysis ecploring three-dimensional equilibrium. PhD, Massachusetts institure of technology. Block, P., M. DeJong and J. Ochsendorf (2006). "As hangs the flexible line: equilibrium of masonry arches." Nexus network journal 8(2). Block, P., T. v. Mele, L. Lachauer and M. Rippmann (2012). "Geometry-based understanding of structures." Journal of the international association for shell and spatial structures 53(4). Borgart, A. and Y. A. B. F. Liem (2011). Force network analysis using complementary energy. Delft. Calladine, C. R. (1977). "The static-geometric analogy in the equations of thin shell structures." Mathematical proceedings of the Cambridge Philosophical Society 82. Calladine, C. R. (1983). Theory of shell structures. Cambridge, Cambridge University Press. Craats, J. v. d. (2000). Vectoren en matrices een inleiding in de lineaire algebra. Utrecht, Epsilon uitgaven. Daniel. (2007). "Picasaweb." from Dicke, D. and W. A. Eisma (1978). Statisch onbepaalde spanten. Delft, TU Delft. Dool, M. v. d. (2012). Optimizing shell structures. Delft. Heyman, J. (1977). Equilibrium of shell structures. Oxford, Clarendon. Heyman, J. (1998). Hooke's cubico-parabolical conoid. The Royal society. London. Hooke, R. (1676). A Description of Helioscopes, and Some Other Instruments. Oostenrijk, T. R. Huerta, S. (2005). The use of simple models in the teaching of the essentials of masonry arch behaviour. Theory and practice of construction: knowledge, means and models. G. Monchi. Ravenna. Liang, D. (2012). A parametric structural design tool (Grasshopper interface) for plate structures. Delft. Liem, Y. A. B. F. (2009). Geometric properties of adaptable arches. Delft. Liem, Y. A. B. F. (2011). Graphic statics in funicular design. Delft. Linden, C. v. d. (2002). Vademecum van de wiskunde. Ter Roye, Het Spectrum B.V. Lopshits, A. M. (1956). Computation of areas of oriented figures. Boston, D.C. Heath and company. Lukas, P. (2013). "What's in a name." from Mak, B. (2013). "Amanecemetropolis.net Sagrada Familia." from Maxwell, J. C. (1868). On reciprocal diagrams in space, and their relation to Airy's function of stress. Cambridge. O'Dwyer, D. (1996). "Funicular analysis of masonry vaults." Computers and structures 73: Oosterhuis, M. (2012). A parametric structural design tool for plate structures. Delft. Pavlovic, M. N. (1984). "A statically determinate truss model for thin shells: two-surface analysis (bending theory)." International journal for numerical methods in engineering P age

89 GRAPHIC STATICS IN ARCHES AND CURVED BEAMS Ravenstein, W. v. (2002). "Functievoorschrift opstellen van een parabool." from Wikipedia. (2008). "Cremora diagram." from Wikipedia. (2013). "Formule van Heron." Retrieved :06, from Wikipedia. (2013). "Maupertuis' principle." Retrieved :56, from Wikipedia. (2013). "Principle of least action." Retrieved :17, from Wikipedia. (2013). "Sine." Retrieved :22, from 83 P age

90 19 Appendix Calculation sheet in Excel 84 P age

91 GRAPHIC STATICS IN ARCHES AND CURVED BEAMS All the GSA calculations are made with the following properties; Material Section Loading In the different cases, first the support reactions are shown, then the maximum normal forces, shear forces and bending moments. The last image shows a contour picture of the structure with the bending moments illustrated. On the top right corner is the legend for the colours. The units of the numbers are indicated in the screenshots. 85 P age

92 Case 1 86 P age

93 GRAPHIC STATICS IN ARCHES AND CURVED BEAMS Case 2 87 P age

94 Case 3 88 P age

95 GRAPHIC STATICS IN ARCHES AND CURVED BEAMS Case 4 89 P age

96 Case 5 90 P age