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POSITION ANALYSIS OF PLANAR TENSEGRITY STRUCTURES By JAHAN B BAYAT A DISSERTATION PRESENTED TO THE GRADUATE SCHOOL OF THE UNIVERSITY OF FLORIDA IN PARTIAL FULFILLMENT OF THE REQUIREMENTS FOR THE DEGREE OF DOCTOR OF PHILOSOPHY UNIVERSITY OF FLORIDA 2006 Copyright 2006 by Jahan B Bayat To the scientific community. ACKNOWLEDGMENTS I express my gratitude to my supervisory committee chair, Prof. Carl D. Crane. I wish to thank all my committee members for their time and effort. I gratefully acknowledge support from the Air Force Office of Scientific Research, Grant Number F496200010021; and from the Department of Energy, Grant Number DEFG04 86NE37967. TABLE OF CONTENTS A CK N OW LED GM EN TS ..................................................... .............. iv LIST OF TABLES ............... ............. ...................... vii LIST OF FIGURES .............. .....................viii ABSTRAC T ............. ............ ................. ........ ........x CHAPTER 1 LITERATURE REVIEW ...................................................................................................1 2 TWO SPRING PLANAR TENSEGRITY SYSTEM.........................................19 2 .1 Introdu action ......... ..... .... ......... ........................ 19 2.2 Problem Statement................... ....... ......................... ....... 20 2.3 Approach 1: Determine (Li, L2) to Minimize Potential Energy .........................21 2.3.1 Development of Geometric Equations .................................................21 2.3.2 Sylvester's Solution M ethod ......................................... 27 2.3.3 Solution of Geometry and Energy Equations........... ...................29 2.3.4 N um erical Exam ples ........................................ ................. 30 2.3.5 Discussion of Results ........................................35 2.4 Approach 2: Determine (cos 04 and cos 61) to Minimize Potential Energy.........35 2.4.1 Obtain Expression for cos0i in terms of cos4 ................ ...... ....36 2.4.2: Write Li in terms of cos0i and L2 in terms of cos04 ..............................38 2.4.3: W rite the potential energy equation ................................ ............... ..38 2.4.4: Express the equation dU/dcos04 = 0 in terms of the variables c4 and ci ..38 2.4.5 Construct Sylvester Matrix for this approach............... ............41 2.4.6 Numerical Example for Approach 2.........................................42 2.4.7 Comparison of Results for Approaches 1 and 2....................................44 3 THREE SPRING PLANAR TENSEGRITY....................... ...............45 3.1 Introduction ............... ...... ..... .... . ... .... ..........45 3.2 First Approach Problem Statement (Descriptive Parameters Li, L2) ................46 3.2.1 Development of Geometric Constraint Equation ....................................47 3.2.2 Development of Potential Energy Equations .............. .............50 3.2.3 Create Solution M atrix ...................................................... 53 3.2.4 Solution of Three Simultaneous Equations in Three Unknowns  C ontinu ation M eth od .................................................................................. 57 3.2.5 N um erical Exam ple ......................... ... .....................58 3.3 Second Approach Problem Statement (calculate cos94, cosil)...........................60 3.3.1 Development of Geometric Equations ................................... ...............61 3.3.2 Development of Potential Energy Equations ..............................63 3.4 Conclusion .................. ................... ......... .69 4 FOUR SPRING PLANAR TENSEGRITY....................................71 4.1 Introduction...................................... .............................. ........ 71 4.2 Problem Statement.................... .................. .. ............ ....... .... ....72 4.3 Development of Geometric and Potential Energy Constraint Equations .............73 4.4 Solution of Geometry and Energy Equations...................... ...............77 4.5 N um erical C ase Study ................................................. ............... 78 4.6 Conclusions.............................................. 82 5 FUTURE WORK, SUMMARY, AND CONCLUSION ........................................83 5.1 Future W ork ................................................... 83 5.1.1 The 3D Tensegrity Platform ......... ............... ................ 83 5.1.2 The 2D Tensegrity structures ............. ........................ 85 5.2: Sum m ary and Conclusion.................................................. 85 APPENDIX: SAMPLE OUTPUT ................................................87 LIST OF REFEREN CES .............................................................................. ................. 94 BIOGRAPHICAL SKETCH .................................................. ............... 97 LIST OF TABLES Table page 21. Eight real solutions. ................................................ ..... ...30 22. Complex solutions. .......................................... ...... ...... .3 1 23. Tw entyfour real solutions.................................... ............... 32 24. Eight feasible real solutions................................................... 34 25.Eight feasible real solutions..... ............ ........ ..... ........42 26.Tw enty com plex solutions..............................................................................43 27. Four nonfeasible real solutions. ......................................................... 43 31: Seven real solutions for threespring planar tensegrity system(units are inches) ......60 32. Real value answers using cosine of 61, 04 in radians and L3 in inches....................69 41. C efficient G D E and F ................................................................................... 77 42. Eighteen real solutions. (Solutions that are geometrically impossible are shaded.)..79 43. Identification of minimum potential energy state. (Solutions that do not have all positive second derivatives are shaded.) ..........................................81 44. Calculated forces in struts, and elastic ties. ................................. ......82 LIST OF FIGURES Figure page 11. Parallel prisms. ............................ .................. ......... 1 12. Tensegrity prisms. ....................... .................. ..... .. .2 13. Plan view of tensegrity and corresponding reinforced prisms. .............. ...............3 14. Component of a tensegrity structures. ............................ ...............5 15. 4strut tensegrity, 3D, top view. ................ ............... ...................................6 16. Tensegrity platform ................................................. ........... .7 17. Zeeman's catastrophe machine. ........... ......... ......... .... ...............9 18. Planar twospring system.......... ..... .............. .. ........10 19. Special threespring system ........... ......... ... ............................... 11 110. General planar threespring compliant mechanism ..........................................11 111. A representation of a dome which utilities tensegrity solutions' technology ........13 21. Two dimensional tensegrity structure................................................... ........ 19 22. Twospring planar tensegrity system ................................................................. 20 23. Identification of angles 04, 084, and 0"4. ............................................................... 21 24. Triangle 413. ........................... .............. .........22 25. Triangles 412. .......................................... ....... ...............23 26. Triangles 413. .......................................... ........ ...............23 27. Possible configurations for numerical example............................... ........... ....26 28. Four equilibrium configurations.................... .....................31 29. Realizable configurations for case 2............................ ......... ......... 34 viii 210. Tensegrity system with two struts, two ties and two springs. ...............................36 31. Tensegrity structure with two struts, a noncompliant (tie), and three springs. ........46 32. Planar tensegrity structure. .............................................. ............... 47 33. Triangle 432. ........................... ...........................47 34. Triangle 412. ........................... ...........................48 35. Triangle 413. ........................... ......................49 36. C ase 1, equilibrium solution. .........................................................................................60 41. Two strut, four spring planar tensegrity device.............................. ............ ...71 42. Four equilibrium solutions.................................................. 81 51. Threedimensional parallel platform of 3 struts, 3 springs, and 6 ties. ......................83 Abstract of Dissertation Presented to the Graduate School of the University of Florida in Partial Fulfillment of the Requirements for the Degree of Doctor of Philosophy POSITION ANALYSIS OF PLANAR TENSEGRITY STRUCTURES By Jahan B. Bayat August 2006 Chair: Carl Crane Major Department: Mechanical and Aerospace Engineering. Tensegrity is an abbreviation of tension and integrity. Tensegrity structures are spatial structures formed by a combination of rigid elements in compression (struts) and connecting elements that are in tension (ties). In threedimensional tensegrity structures no pair of struts touches, and the end of each strut is connected to noncoplanar ties, which are in tension. In twodimensional tensegrity structures, struts still do not touch. A tensegrity structure stands by itself in its equilibrium position and maintains its form solely because of the arrangement of its struts and ties. The potential energy of the system stored in the springs is at a minimum in the equilibrium position when no external force or torque is applied. A closedform solution of a twospring, threespring, and four spring planar tensegrity mechanism was developed to determine all possible equilibrium configurations when no external force or moment is applied. Here "closed form" means that all solution equilibrium poses will be determined, although for each case a high degree polynomial will have to be solved numerically. CHAPTER 1 LITERATURE REVIEW We examined the literature related to tensegrity systems, selfdeployable tensegrity systems, and parallel mechanisms with compliant elements. Knight et al. [1] reported on the line geometries of a family of tensegrity structures called skew prisms (antiprisms, tensegrity prisms) with pairs of triangles, squares, pentagonals, hexagonals, and so on, located at tops and bases. They used the quality index with SPS connectors (S is a ballandsocket joint, and P is a sliding joint). The quality index measures the geometric stability of inparallel devices (which in three dimensions consist of a pair of rigid platforms connected by legs that are kinematical S PS connectors). Geometric stability depends on the geometry of the legs. Figures 11 A and 11 C show a sequence of parallel prisms with parallel ties of fixed length. (A) (C) Figure 11. Parallel prisms. A relative anticlockwise rotation can be achieved by rotating the top of each prism through its vertical axes as shown in Figures 12 B and 12 D to yield a corresponding righthanded tensegrity prism that is completed by inserting struts on the diagonals of the skew quadrilaterals. Lefthanded tensegrity prisms are mirror images of righthanded prisms and are obtained by rotating the tops in a clockwise direction and interchanging 180 the ties and struts. The angle a is unique for each tensegrity prism (a = 90 180) n where n is the total number of sides in the upper or lower polygons. The value of a (together with the size of the tops and bases and their distance apart) enables one to compute the length of the struts and ties that defines a unique configuration for the tensegrity prism. (B) (D) Figure 12. Tensegrity prisms. The quality index for each skew prism is zero, which means that it has instantaneous mobility. Further examination revealed that all the sets of connectorlines for each prism belong to a linear complex of lines within a 5system of screws. Each of the sets of lines is reciprocal to a single screw. Adding additional ties along the diagonals of the skew prism faces can easily form reinforced skew prisms. Unloaded tensegrity prisms that are stable in the sense that they are configurations of minimum potential energy, but they have instantaneous mobility simply because the connectorlines belong to a linear complex. Reinforced tensegrity prisms where additional ties are inserted are completely stable and shown in Figure 13. Tensegrity Prisms Reinforced TI; it% Prisms Figure 13. Plan view of tensegrity and corresponding reinforced prisms. Pellegrino [2] shows how deployable structures are able to change their shape from a packed configuration to operational form. Usually, energy is stored in the structure when packed and released during unpacking when the operational configuration is required. Some simple examples are a spring loaded umbrella, a retractable roof of a car, and a space radio telescopic antenna. Deployable structures are used for ease of transportation and storage. The essential requirement is that the transformation process should be autonomous and reliable, and without causing any damage to nearby structures. Examples of deployable structures that undergo large geometric transformations are coiled rods, flexible shells, lattice column, and membranes. A deployable structure using lattice columns with several longitudinal elements, called longerons, braced at regular intervals by short members perpendicular to the longerons and by diagonal members, has low dead weight but can support high load. This structure has a very small wind resistance in atmosphere and minimal particle meteoroid damage in space. Tamai [3] presents a brief analogy between compatibility and equilibrium of a finite linkage mechanism. Some linkages are known whose degree of freedom in some specific position is greater than the expected degree of freedom as calculated from the equilibrium equation. An example is a four bar linkage with equal length opposite bars. The linkage has two shapes: one associated with a parallelogram and the other with an anti parallelogram. Plotting these angles as a function of each other provides two curves with a common point. This point is when all four bars lie on a straight line. At this instant the mechanism may move from the parallelogram configuration to the anti parallelogram configuration. Duffy et al. [46] analyze a three dimensional tensegrity structure that is made up of elastic and rigid elements (Figure 14). In this assembly, the elastic elements are under tension and the rigid elements under compression. The papers present the static position analysis problem and determine the position assumed by the structure when external loads are applied, and when the system is presented by changing the free lengths of the compliant elements. Tlop tie Connection l IL f Bottom tie Figure 14. Component of a tensegrity structures. The mathematical formulation to find the equilibrium positions of the structure is based on the virtual work principle. The obtained equations are solved using numerical methods. Some assumptions are made to simplify the derivation of the mathematical model. These assumptions are the absence of internal dissipative forces and the manner by which the external forces are applied. The numerical method determines the coordinates of the strut end points in the equilibrium position. A force balance is then conducted to validate the results. Stern [7] presents the position analysis of a symmetric nstrut tensegrity system. A three dimensional nstruts system with two platforms, one at the top and the other at the bottom, is considered in the analysis. A static analysis of the internal forces is conducted on the top and bottom platforms. The relationship between the geometry of the structure and the internal forces are investigated. Tensegrity structures with different number of struts are analyzed. The results of each analysis are compared with the results of other analyses to obtain common patterns in all systems by relating the results to the number of struts in the systems. The patterns are formulated into equations based on the number of struts in the system. The formulation is done for 3, 4, 5, and 6 struts tensegrity systems with solid top and bottom platforms. For example, a plan view of 4 strut tensegrity system is shown in Figure 15. HH G6 DD CC AA a BB EE FF Figure 15. 4strut tensegrity, 3D, top view. Marshall [8] introduces a parallel platform device that incorporates tensegrity principles. The device, shown in Figure 15, replaces the struts, of a tensegrity structure with prismatic actuators, and each elastic member with a cablespring combination in series. The length of the three prismatic actuators and the length of the cables that are in series with the spring are adjustable and thus the device has six degrees of freedoms. This study shows that in order to achieve an arbitrary position and orientation of the top platform, an external wrench must be applied to maintaining equilibrium. This study also shows that the device's compliance characteristics can be varied while maintaining its position and orientation. A reverse analysis of the device is presented in which the desired pose and total potential energy are given and the lengths of the prismatic actuators and the cables are determined. The effect of a seventh leg, another prismatic actuator, is also analyzed and found to satisfactorily implement the needed external wrench (Figure 16). Figure 16. Tensegrity platform. Gantes and Konitopoulou [9] discusses bistable deployable structures. A bistable structure is self standing and stress free when fully closed or fully deployed. It exhibits incompatibilities between the member lengths at intermediate geometric configurations during the deployment process, which leads to the occurrence of secondorder strains and stresses resulting in a snapthrough phenomenon that "locks" the structures in their deployed configuration. Until now the geometric shapes that were possible in the deployed configuration were only flat or curved with constant curvature. This limitation is addressed in the paper by proposing a geometric design methodology for deployable arches of arbitrary curvature, accounting also for the discrete joint size, and applying it successfully for the geometric design of a semielliptical arch. The arch is then modeled with finite elements, and a geometrically nonlinear analysis is performed in order to verify the deploy ability feature. Further verification is provided by the construction of a smallscale physical model. A preliminary structural design indicates the overall feasibility of the arch for short to medium spans and light loads. Deployable structures are prefabricated space frames consisting of straight bars linked together in the factory as a compact bundle, which can then be unfolded into large span, load bearing structural shapes by simple articulation. Because of this feature they offer significant advantages in comparison to conventional, nondeployable structures for a wide spectrum of applications ranging from temporary structures to the aerospace industry, being mainly characterized by their feature of transforming and adapting to changing needs. From a structural point of view, deployable structures have to be designed for two completely different loading conditions, under service loads in the deployed configuration, and during deployment. The structural design process is very complicated and requires successive iterations to achieve some balance between desired flexibility during deployment and desired stiffness in the deployed configuration From a geometric point of view, the whole idea of this type of deployable structure is based on the socalled scissorlike elements, pairs of bars connected to each other at an intermediate point through a pivotal connection which allows them to rotate freely about an axis perpendicular to their common plane but restrains all other degrees of freedom, while, at the same time, their end points are hinged to the end points of other scissorlike elements. Yin et al. [10] present a special planar threespring mechanism that is designed to control contact forces. An energy function is defined to describe the behavior of this kind of mechanism and it can be used to perform the catastrophe analysis of this mechanism. The analysis result can be used as a design and control tool. By comparing the three spring system and a twospring system, it was found that the threespring mechanism has better stability than the twospring system. A threespring mechanism which can be used to control a general contact force in a plane is also analyzed. Intuitively, Yin et al. [10] showed a catastrophe occurs whenever a smooth change of parameters gives rise to a discontinuous change in behavior. A well known example that can easily be made to demonstrate a catastrophe is Zeeman's catastrophe machine (see Figurel7). Zeeman's machine can be constructed by attaching two linear springs to a single point C on a disk that can rotate about 0. One of the springs is attached to a fixed pivot at point A and the other spring is attached to point B which can be moved in the x, y plane. The position of point B is called the controlling parameter as it dictates the position of the disk which is defined by the angle between OA and OM. Y O< A Figure 17. Zeeman's catastrophe machine. Parallel compliant mechanisms are classically known for their role in mounting and suspension systems. Recently, however, they have been instrumental in the field of robotic force control. A new theory for the simultaneous control of displacement and force for a partially constrained endeffecter has been proposed in Yin et al. [10]. Figure 18 shows a planar compliant mechanism which is actually a planar two spring system. It consists of a pair of linear springs connected at one end to a movable base and at the other end to a common pivot which is the axis of the wheel contacting to a surface. The base is connected to a planar two freedom PP (P denotes prismatic pair) manipulator The contact force can be controlled by displacing the two prismatic joints of the manipulator. The required displacements can be calculated from the stiffness mapping. This kind of control was called kinestatic control by Griffis and Duffy [11]. In order to design the planar twospring system it is necessary to compute a spring stiffness which will generate a range of displacements of the movable base which can be produced by the prismatic joints over a required range of change in contact force. Clearly if the system is overdesigned and the spring stiffness is very high it is always possible to generate any necessary changes in contact force. However such a system will be too sensitive to errors because very small displacements of the platform will generate large changes in contact force. On the other hand if the springs are too soft there can be stability problems. Figure 18. Planar twospring system. Yin et al. [10] are also considered another compliant mechanism. That is shown in Figure 19. It consists of three linear springs joined to the triangular frame fixed points and connected at the axis of the wheel. The triangular frame is connected to the planar two freedom PP manipulator. This mechanism is a special threespring system. The catastrophe analysis of this system was presented. Comparing the results demonstrates that threespring system has better stability characteristics than the twospring system. [I Figure 19. Special threespring system. A more general planar threespring compliant mechanism is shown in Figure 110. This mechanism is connected to the planar three freedom RRR (R denotes revolute pair) manipulator. It can be used to control both force and moment. The catastrophe analysis of this mechanism was performed in Yin et al. [10]. Figure 110. General planar threespring compliant mechanism. Crane and Duffy [12] studied kinematics of robot manipulators and Crane and Bayat [13] presented cases of two dimensional tensegrity structures that proved to be mathematically challenging in spite of their simple configurations. Different methods were tried to solve these problems. It was expected the final closed form solution be a simple mathematical expression. In each case the intermediate mathematical expressions become complicated to solve and the relevant matrices become large to manipulate. Roth and Whiteley [14] propose a technology based on tensegrity for tough, rigid, largescale domes that are also economical to construct. The development of a structural technology to economically cover large areas would be useful for warehouses, permanent or temporary protection for archaeological and other vulnerable sites, largescale electrical or electromagnetic shielding and exclusion or containment of flying animals or other objects. Structures based on such a technology can serve as frameworks in which environmental control, energy transformation and food production facilities could be embedded. The space application is also possible by using selfdeployed structures. Summary Advantages are improved rigidity, ethereal, resilient, equallength struts, simple Joints. The resulting list represents an initial attempt to identify applications for which the technology would be suitable. As the technology develops and is tested against them, some or all of the applications may be winnowed out and other suitable applications not in the list may become apparent. The current list is as follows: Superstructures for embedded substructures allow the substructures to escape terrestrial confines where this is useful (e.g. in congested or dangerous areas, urban areas, flood plains or irregular, delicate or rugged terrains). 1. Economic largescale protection of storage, archaeological, agricultural, construction or other sites. 2. Refugee or hiking shelters. 3. Frames over cities for environmental control, energy transformation and food production. 4. Largescale electrical or electromagnetic shielding. 5. Exclusion or containment of flying animals or other objects. 6. Spherical superstructures for space stations. 7. Earthquakeresistant applications. These structures are extremely resilient and testing would very likely show they could withstand large structural shocks like earthquakes. Thus, they would likely be desirable in areas where earthquakes are a problem. A tensegrities, pneumatic, structure performed very well during recent earthquakes in Japan. * Lowenvironmentalimpact shells for musical performances. * Indoor/outdoor pavilions for trade shows etc. * Supports to hold sunscreen protection for vulnerable amphibians. * Watersheds to keep rain water from percolating through contaminated soils into groundwater, perhaps temporarily during insitu remediation. * Frames for hanging plants or other objects to dry. * Pergola, trellis, or topiary framework. * Micrometeorite protection, sunshielding for Martian colonies. Figure 111. A representation of a dome which utilities tensegrity solutions' technology Skelton et al. [15] present a solution so the tensegrity structures are reduced to linear algebra problems, after first characterizing the problem in a vector space where direction cosines are not needed. That is, we describe the components of all members by vectors as opposed to the usual practice of characterizing it as static problem in terms of the magnitude of tension and compression. While this approach transfer the problem into vector space to describe the problem, the advantage is that the vector space makes the mathematical structure of the problem amenable to linear algebra treatment for both two and three dimensional tensegrity structures. This paper characterizes the static equilibrium some tensegrity structures. Furthermore, it uses vectors to describe each element (springs and studs), eliminating the need to use direction cosines and the subsequent transcendental functions that followthe use of trigonometry equations. In this paper, the authors choose to represent a tensegrity structure as an oriented graph in real three dimensional space R3 defined in terms of nodes and directed branches which are all represented as vectors in R3. A loop is any closed path in the graph. The advantage of this approach is that the both the magnitude and the direction cosines of the forces are contained in vectors which can be solved using linear algebra. Thus linear algebra plays a larger role in this approach compared to the usual approach in mechanics and finite element methods using direction cosines. In this oriented graph, the nodes consist of the ends of the bars. Hence if there are n bars, then there are 2n nodes. There are two types of directed branches; the string branch (in vectors) and the bar branches (in vectors). Geometric connectivity of the structure maintained such that each directed branch can undergo a displacement in reaching its equilibrium state. String vectors can change both their length and orientation while bar vectors can only change their orientation. Node vectors can change both their length and orientation but subject to a Law of Geometric Connectivity is stated as follows: The vector sum of all branch vectors in any loop is zero. These loop equations are in the form of a set of linear algebraic equations in the branch vectors. In this study of tensegrity structures, force equilibrium are such that spring can only take tension and bars sustain compressive forces. We therefore choose to distinguish between the string (or tensile) forces and the bar (or compressive) forces which are defined in terms of the string and bar vectors respectively. This paper reduces the study of the tensegrity equilibrium to a series of linear algebra problems using directed graph theory. Of course the existence conditions for the linear algebra problems are nonlinear in the design variables. The presented procedure and formulation give some insight to solution of tensegrity structures and identifies the free parameters that may be used to achieve desired structural shapes. Tibert [16] presents adjustable tensegrity structure. A tensegrity structure is a lightweight consisting of compression members surrounded by a network of tension members. They can be easily dismantled providing possibilities for reusable and modular structures. Tensegrities adapt their shape by self adjusting their tension and compression in their members. They can adapt to changing environment when they are equipped with sensors and actuators. A fullscale prototype of an adjustable tensegrity is built and tested at Swiss Federal Institute of Technology. This paper begins with a description of important aspects of the design, assembly, and static testing. Tests show that the structure behaves linearly when subjected to vertical loads applied to a single joint. Nonlinearities are detected for small displacements, for loads applied to several joints and for adjusting combinations of telescoping compression members. To predict behavior, dynamic relaxationa nonlinear methodhas been found to be reliable. Appropriate strut adjustments found by a stochastic search algorithm are identified for the control goal of constant roof slope and for the load conditions studied. When adjusting struts, an excessive number of adjustable members does not necessarily lead to improved performance. Tensegrity is an abbreviation of tensileintegrity as termed by one R. Buckminster Fuller. He described tensigrity as "small islands of compression in a sea of tension." This is a description of a network of light tension members that provide rigidity to a limited number of discontinuous compression members. Some researchers have found that discontinuous compressions are not a necessity for creating a tensegrity and more efficient structures can be built if compression elements are allowed to join. Defining a tensegrity structure as any selfstressing structure composed of struts and cables would include structures such as a bicycle wheel. In order to add precision terminology, a definition has been proposed by Motro and Raducanu. They describe a tensegrity as a stable system that contains a discontinuous set of components in compression inside a network of components in tension. Several independent efforts have contributed to the invention and development of tensegrity structures. Le Ricolais designed many unusual lightweight cablestrut structures. The sculptor Kenneth Snelson started in 1948 with sculptures employing members in continuous tension and discontinuous compression. Applications of the tensegrity principle in nature have recently been found through studies that use a tensegrity model to describe observations of how cells respond to stress. Transforming tensegrity from sculptures into practical structural has been a challenge that began with several studies of their geometric, nonlinear behavior and states of self stress. Reasons for the lack of test data are related to the fabrication assembly process as well as to the lack of rigidity unless prestressed members are used. Joint design is the biggest challenge in constructing a fullscale tensegrity. Although it is the problem for all space structures, tensegrities present particular challenges. Joints need to be pin jointed, modular, and light in order to take advantage of tensegrities ease of dismantling and potential reuse. In addition to the complexities of constructing the structure itself, simulation of the behavior is not straightforward. Tensegrity structures exhibit geometric nonlinear behavior. In addition, as it is common with other pinjointed structures, nodal friction should be included in the simulation of a truss structure for precision control. Finally, for fullscale construction, assembly sequences, fabrication tolerances, and construction lack offit need to be considered. Fest et al. [17] present a study of adjustable tensegrity structures. A tensegrity structure is consisting of compression members surrounded by a network of tension members. A tensegrity structure is a lightweight structure. Tensegrity structures can be easily dismantled providing possibilities for reusable and modular structures. Tensegrities adapt their shape by self adjusting their tension and compression in their members. They can adapt to changing environment when they are equipped with sensors and actuators. A fullscale prototype of an adjustable tensegrity is built and tested at Swiss Federal Institute of Technology. This paper presents important aspects of the design, assembly, and static testing. Tests show that the structure behaves linearly when it is subjected to vertical loads applied to a single joint. Nonlinearities are observed when loads applied to several joints and for adjusting combinations of telescoping compression members. Appropriate strut adjustments found by a search algorithm for the load conditions studied. When adjusting struts, an excessive number of adjustable members do not necessarily lead to equilibrium. Active tensegrity structures have the potential to widen the scope for innovative, lightweight, and reusable structural systems. Lessons learn through the construction and testing of an adjustable fullscale prototype can be used in construction and design of practical tensegrity structures. This work expected to contribute to the development of tensigrity and improve in their design and performance during service. CHAPTER 2 TWO SPRING PLANAR TENSEGRITY SYSTEM 2.1 Introduction This Chapter presents two approaches to solve the forward position analysis of a twostrut tensegrity system with two compliant ties and two noncompliant ties (Figure 2 1). ties 2 3 struts 4 1 Figure 21. Two dimensional tensegrity structure. Here it is assumed that the lengths of the struts and noncompliant ties are known together with the spring constants and free lengths of the two compliant ties. The objective is to determine all possible equilibrium poses for the device when no external loads are applied. Gravity loads are neglected. The objective of this effort is to determine, in closedform, all possible equilibrium deployed positions in stable condition (minimum potential energy) of a planar tensegrity system wherein two of the ties are compliant. Figure 22 shows the system which is comprised of two struts (compression members a12 and a34), two noncompliant ties (tension members a41 and a23), and two elastic tensile members (springs), one connected between points 1 and 3 and one between points 2 and 4. It should be noted in Figure 22 that strut a34 passes through a slit cut in strut a12 and as such the two struts do not intersect or collide. 23 3 2 a34 4 14 Figure 22. Twospring planar tensegrity system. 2.2 Problem Statement The problem statement for the twospring plane tensegrity system can be explicitly written as given: al2, a34 lengths of struts a23, a41 lengths of noncompliant ties ki, Lol spring constant and free length of compliant tie between points 4 and 2 k2, Lo2 spring constant and free length of compliant tie between points 3 and 1 find: Li length of spring 1 at equilibrium position L2 length of spring 2 at equilibrium position It should be noted that the problem statement could be formulated in a variety of ways, that is, a different variable (such as the relative angle between strut a34 and tie a41) could have been selected as the generalized parameter for this problem. Two solution approaches are presented. 2.3 Approach 1: Determine (L1, L2) to Minimize Potential Energy 2.3.1 Development of Geometric Equations Figure 23 shows the three angles 64, 0'4, and 0"4 which must satisfy the relation 4 + 4' = + 4" (21) Figure 24 shows the triangle formed by side a34, a23, and L1. A cosine law for this triangle can be written as 2 2 2 1 +a34 +La34 coSe4 23 2 2 2 (22) 3 823 2 L2 L1 12 434 014 4 a41 Figure 23. Identification of angles 04, 0'4, and 0"4. 4A'4 Figure 24. Triangle 413. Solving for cos94' yields 2 2 2 cos4e, a23 _1 a34 (23) 2L, a34 Figure 25 shows the triangle formed with sides a41, a12, and L1. A cosine law for this triangle can be written as L2 2 2 1 41 +LLa41 Cos 4 a12 (24) 2 2 2 Solving for cos94" yields 2a2 2 cosO4 12 1 41 (25) 2L, a41 Figure 26 shows the triangle formed by sides a41, a34 and L2. A cosine law for this triangle can be written as 342 412 L 2 34 + + a34 a41 Cos04 2 2 2 (26) Solving for cos94 yields 2 2 2 L, a3 a COS 4 L2 a34 a41 2a34 a41 (27) 2 4 a41 1 11"4 Figure 25. Triangles 412. 3 a34 / L: a41 Figure 26. Triangles 413. Equating the cosine of the left and right sides of 21 yields cos (04 + 4') = COS (7T + 04") (28) Expanding this equation yields cos04 cos04' sin04 sin04' = cos04" (29) Rearranging 29 yields cos04 cos04' + cos04" = sin04 sin04' (210) Squaring both sides of 210 gives (cos04)2 (cos04')2 + 2 cos04 cos04' cos04" + (cos04")2 = (sin04)2 (sin4')2 (211) Substituting for (sin04)2 and (sin04')2 in terms of cos04 and cos04' gives (cos04)2 (cos04')2 + 2 cos04 cos04' cos04" + (cos04")2 (1cos284) (1cos204') (212) Equations 23, 25, and 27 are substituted into 212 to yield a single equation in the parameters L1 and L2 which can be written as A L24 + B L22 + C = 0 (213) where A =L2, B = L4 +B2 L12 + B, C = C2 L2 +C0 (214) and B2 = (a232 + a342 + a412 + a122), Bo = (a12 a41) (a12 + a41) (a23 a34) (a23 + a34), C2 = (a34 a41) (a34 + a41) (a23 a12) (a23 + a12), Co = (a41a23 + a34a12) (a41a23 a34a12) (a41 + a23 a122 a342) (215) Equation 213 expresses length L2 as a function of the length Li. The function is quadratic with respect to (L2)2 and (Li)2 and thus there will be two possible values for (L2)2 for each value of (L1)2. As a verification of Equation 213 a numerical example is presented to show how four values of L2 can occur for a given value of L1. For this example, assume a41 = 1.0 m, al2 = 2.0 m, a23 = 2.0 m, and a34 = 2.5 m. Further, assume the value of L1 is given as L1 = 2.25 m. Evaluating the coefficients in Equations 215 and 214 and substituting into Equation 213 gives (5.0625) L24 + (58.3242) L22 + (110.2500) = 0 (216) Solving this equation for L2 yields four answers; L2a = 3.0228, L2b = 3.0228, L2c = 1.5438, L2d = 1.5438 (217) Figure 27 shows the mechanism in four configurations where for each of these configurations L2 can have a positive or negative value, thereby giving eight possible states of the mechanism. However, due to the symmetry associated with the reflected solutions about a41, only four values of L2 (or two values of L22) exist. This example was presented to show that the degree of the polynomial which relates L1 and L2 is indeed fourth order. 2 2 4 1 4 1 1 4 1 4 3 2 2 Figure 27. Possible configurations for numerical example. The potential energy of the system can be evaluated as 1 1 U= k, (L L, )2 +k2 (L L02)2 2 2 (218) At equilibrium, the potential energy will be a minimum. This condition can be determined as the configuration of the mechanism whereby the derivative of the potential energy taken with respect to the length L1 equals zero, i.e. dU dL =k, (L L1)+k2 (L2 L02) d = 0 SdL (219) The derivative dL2/dL1 can be determined via implicit differentiation from Equation 213 as dL L, [L2 (L2 +2L2 a23 a4 2 34 a12) + (a122 a23)(a412 a342) dL, L[L (L2 +2L2 a23 2 a4 a34 a12) + (a122 a)(a23 a342))] (220) Substituting Equation 220 into Equation 219 and regrouping gives D L25 + E L24 + F L23 + G L22 + H L2 + J = 0 (221) where D = D1 L1, E = E1 L1, F = F3 Li3 + F2 L12 + Fi L1, G = G3 Li3 + Gi Li, H= H5 L1 + H4 L14 + H3 Li3 + H2 L12 + Hi L1 + Ho, J= Ji Li (222) and where, D1 = k2, El = k2 LO2, F3 = 2 (k2 ki), F2 = 2 ki Lol, F = k2 (a122 + a232 + a342 + a412), G3 = 2 k2 L02, Gi = k2 L2 (a122 + a232 + a342 + a412), H5 = ki, H4 = ki Lol, H3 = ki (a122 + a232 + a342 + a412), H2 = ki Lo, (a122 + a232 + a342 + a412), H1 = ki (a342 a232) (a412 a122) + k2 (a342 a412) (a232 a122), Ho = ki Lo, (a342 a232) (a412 a122), Ji = k2 Lo2 (a342 a412) (a122 a232) (223) Equations 213 and 221 represent two equations in the two variables L1 and L2. The simultaneous solution of these equations is a necessary condition that the system is in equilibrium. 2.3.2 Sylvester's Solution Method A system of m equations of order n in the variable Y may be written as (Gantes and Konitopoulou [9]) Al,n Y" + Al,n1 Y"1 + A1,n2 y"2 + ... + AI,o = 0 A2,n Y" + A2,n1 Y"1 + A2,n2 y"2 + ... + A2,0 = 0 , Am,n Yn + Am,n1 Yn1 + Am,n2Yn2 + ... + Am,o = 0 (224) The terms Un = Y" Un1 = Y1, U = Y, Uo = 1 may be substituted into Equation 224 to yield Al,n Un + Al,n1 Un1 + Al,n2 Un2 + ... + AI,o Uo = 0 A2,n Un + A2,n1 Un1 + A2,n2 Un2 + ... + A2,0 Uo = 0 , Am,n Un + Am,n1 Un1 + Am,n2 Un2 + ... + Am,o Uo = 0 (225) Equation 225 represents a system of m homogeneous equations with n+1 unknowns (the term Uo will be treated as an unknown for the time being) and can be written in matrix form as An, A,_,,i Ao V U, O A2,n A2n1 ... A U 0 Am,n Am,nI Am,O Uo 0 (226) If the number of equations is less than the number of unknowns, additional equations can be created by multiplying selected equations from set of Equation 224 by powers of Y to obtain a new system. For example, multiplying the first equation from the set of Equation 224 by Y will introduce one new equation and one new unknown, Y"+1 (which will be represented by Un+1). Now, multiplying the second equation of the set of Equation 224 by Y will yield another equation, but no new variables are introduced. Ultimately, the system will be formed where the new total number of homogeneous equations equals the new total number of unknowns. In other words, the coefficient matrix will be a square matrix. Now, one solution to the set of 'homogeneous' equations is the trivial solution where all the Ui terms equal zero. However, this cannot occur since Uo=l. Thus it must be the case that the set of equations are linearly dependent which means that the determinant of the coefficient matrix must equal zero. Now consider that the terms Aij of the coefficient matrix are polynomial functions of order q of another variable called X, that is Ai = ai X' + bi X'1 + ci Xq2 + ... + Zi = 0 (227) where ai, bi, ci, ... zi are constants. Evaluation of the determinant of the coefficient matrix will result in a polynomial in the variable X that must equal zero so that the set of equations in the variables Ui will be linearly dependent. This solution approach will now be applied to Equations 213 and 221 which represent two equations in the two unknowns L1 and L2. 2.3.3 Solution of Geometry and Energy Equations Equations 213 and 221 can be solved by using Sylvester's variable elimination procedure by multiplying Equation 213 by L2, L22, L23, and L24 and Equation 221 by L2, L22, L23 to yield a total of nine equations that can be written in matrix form as 0 0 0 D E F G H J L28 0 SO O O0 A 0 B O C L27 0 SO O0 A 0 B O C O L26 0 O O D E F G H J 0 L5 0 0 0 A 0 B 0 C 0 0 L24 0 O D E F G H J O O L23 0 O A O B O C O O O L2 0 DE F G H J 0 0 0 L2 0 A 0 B O C 1 0 (228) This set of equations can only be solved if the determinant of the 9x9 coefficient matrix is equal to zero. Expansion of this determinant yields a 30th degree polynomial in the variable L1. When the determinant was expanded symbolically, it was seen that the two lowest order coefficients were identically zero. Thus the polynomial can be divided throughout by Li2 to yield a 28th degree polynomial. The coefficients of the 28th degree polynomial were obtained symbolically in terms of the given quantities. They are not presented here due to their length and complexity. Values for L2 that correspond to each value of L1 can be determined by first solving Equation 213 for four possible values of L2. Only one of these four values also satisfies Equation 221. 2.3.4 Numerical Examples Example 1: The following parameters were selected for this numerical example: Strut lengths a12 = 3 in. a34 = 3.5 in. Noncompliant tie lengths a41 = 4 in. a23 = 2 in. Spring 1 free length & spring constant Lol = 0.5 in. ki = 4 lbf/in. Spring 2 free length & spring constant L02 = 1 in. k2 = 2.5 lbf/in. Eight real and twenty complex roots were obtained for the 28th degree polynomial in L1. Real values for L1 and the corresponding values of L2 are shown in Table 21. Table 21. Eight real solutions. Case 1 2 3 4 5 6 7 8 L1, in. 5.485 5.322 1.741 1.576 1.628 1.863 5.129 5.476 L2, in. 2.333 2.901 1.495 1.870 1.709 1.354 3.288 2.394 The values of L1 and L2 listed in Table 21 satisfy the geometric constraints defined by Equation 213 and the energy condition defined by Equation 221. Each of these eight cases was analyzed to determine whether it represented a minimum or maximum potential energy condition and cases 3, 4, 5, and 6 were found to be minimum states. A free body analysis of struts a12 and a34 was performed to show that these bodies were indeed in equilibrium for each of these four cases. Figure 28 shows the four static equilibrium configurations of the system. Note that there are several cases where the 2 3 2 3 4 1 4 1 Case3 Case 4 3 2 . 2 3 3 spring in compression with a negative spring length spring in tension 4 Case 5 1 4 Case 6 1 4 Figure 28. Four equilibrium configurations. spring lengths are negative, such as for example the spring between points 2 and 4 in case 4. In this case the spring is acting to pull point 2 towards point 4 which is counter intuitive for the normal spring that is in compression. There are twenty complex roots for L1 in this example. Unique corresponding values for L2 were obtained for each case (Table 22). In fact there are corresponding values for L2 such that the complex pair ofLi and L2 satisfies both Equations 213 and 221. This result means that no extraneous terms were introduced in the elimination procedure. These complex roots were not analyzed further. Table 22. Complex solutions. Case L1, inches L2, inches 1 7.0285 0.0648i 0.3225 + 2.5486i 2 7.0285 + 0.0648i 0.3224 2.5486i 3 2.0919 2.4639i 6.5995 + 0.3617i 4 2.0919 + 2.4639i 6.5995 0.3617i 5 1.7904 2.0044i 6.5232 + 0.0738i 6 1.7904 + 2.0044i 6.5232 0.0738i 7 0.7211 0.1827i 0.2411 1.2909i 8 0.7211 + 0.1827i 0.2411 + 1.2909i 9 0.005823 1.7651i 0.05803 0.01682i 10 0.005823 + 1.7651i 0.05803 + 0.01682i 11 0.3062 0.4101i 0.1220 0.9493i 12 0.3062 + 0.4101i 0.1220 + 0.9493i 13 0.8140 0.5553i 0.5055 +0.9541i Table 22. Continued. Case L1, inches L2, inches 14 0.8140 + 0.5553i 0.5055 +0.9541i 15 2.3264 1.9808i 6.1894 + 0.2416i 16 2.3264 + 1.9808i 6.1894 0.2416i 17 2.9910 2.5413i 6.2069 0.9310i 18 2.9910 + 2.5413i 6.2069 + 0.9310i 19 7.0351 0.1067i 0.4518 2.6442i 20 7.0351 + 0.1067i 0.4518 + 2.6442i Example 2: For this case, a C language program was written that would randomly select values for the parameters a12, a23, a34, a41, ki, Lol, k2, and L02 in attempt to find a case where the number of real roots for L1 was twenty eight. Such a case was not found, but a set of inputs resulting in twenty four real values for L1 is presented here. The input parameters for this case were selected as: Strut lengths a12 = 3.309848 in. a34 = 3.002692 in. Noncompliant tie lengths a41 = 7.335484 in. a23 = 7.978210 in. Spring 1 free length & spring constant Lol = 0.953703 in. ki = 0.193487 lbf/in. Spring 2 free length & spring constant L02 = 0.607318 in. k2 = 9.108249 lbf/in. Twenty four real and four complex roots were obtained for L1. The real values for Li and the corresponding values of L2 are shown in Table 23. The four complex values for L1 were 1.2818 18.2760i and 0.8019 18.2940i. Table 23. Twentyfour real solutions. Case L1, inches L2, inches 1 19.0226 2.5540 2 16.9651 2.8637 3 11.7045 0.6215 4 11.0691 4.3323 5 11.0649 4.3325 6 10.5742 4.6685 Table 23. Continued. Case L1, inches L2, inches 7 10.5716 4.6687 8 5.1348 10.3382 9 5.1343 10.3382 10 3.7712 11.2881 11 3.7704 11.2881 12 0.02315 11.7686 13 0.02315 11.7686 14 3.7733 11.2881 15 3.7737 11.2881 16 5.1330 10.3382 17 5.1334 10.3382 18 10.5735 4.6686 19 10.5755 4.6685 20 11.0631 4.3326 21 11.0664 4.3324 22 11.7046 0.6193 23 17.4121 2.7902 24 19.5242 2.4884 At first glance, it appears that a set of input parameters have been found whereby most of the roots of the resulting polynomial equation in L1 and the corresponding values of L2 are real. However, from Figure 22, it is apparent that the system will be realizable only if a41 a12 <L  < a41 + a12, a34 a23 <L  < a34 + a23 , a23 ai2 < L2 a23 + a12 , a41 a34 < L21 K a41 + a34 . (229) For this numerical case, the system will be realizable only if 4.9755 in. < L1 < 10.6453 in. , 4.6684 in. < L21 < 10.3382 in. (230) Table 24 lists the cases which satisfy these conditions. Table 24. Eight feasible real solutions. Case L1, inches L2, inches 6 10.5742 4.6685 7 10.5716 4.6687 8 5.1348 10.3382 9 5.1343 10.3382 16 5.1330 10.3382 17 5.1334 10.3382 18 10.5735 4.6686 19 10.5755 4.6685 The spring L2 is at an extreme limit value in every one of the cases listed in Table 24. When IL21 = 10.3382 in., strut a34, tie a41, and the spring L2 are collinear. When IL21 = 4.668, strut a12, tie a23, and spring L2 are collinear. Both configurations are shown in Figure 29. 2 a34 a41 3 4 L1 4 L2 L  L 2 4 2 Figure 29. Realizable configurations for case 2. 2.3.5 Discussion of Results It is apparent that obtaining values of L1 and L2 that simultaneously satisfy Equations 213 and 221 is a necessary condition for equilibrium. Satisfying these equations is not sufficient, however, to guarantee that the system is physically realizable. The real values for L1 and L2 that were calculated in the second example, but which violate the conditions of Equation 229 are an interesting case. Here some or all the angles 04, 04' and 04" will be complex, yet the condition of Equation 21 can still be satisfied. Presented above was the technique to obtain all possible equilibrium positions of a planar tensegrity system that incorporates two compliant members. The approach of satisfying geometric constraints while simultaneously finding positions where the derivative of the total potential energy with respect to the generalized coordinate equaled zero resulted in a 28th degree polynomial in a single variable. Although the resulting polynomial was of higher degree than anticipated, an analysis of the real and complex solutions indicates that no extraneous solutions were introduced during the variable elimination procedure. Complex solutions satisfy the geometry equation but they can not be used to construct the structure, hence they are not the solution. 2.4 Approach 2: Determine (cos 04 and cos 01) to Minimize Potential Energy The objective of this approach is to investigate, in closedform, possible equilibrium positions using the cosine of two angles, 04 and 01, as the descriptive parameters for the system. Figure 210 shows the tensegrity system which is comprised of two struts (compression members a12 and a34), two noncompliant ties (tension members a41 and a23), and two elastic tensile members (springs), one connected between points 1 and 3 and one between points 2 and 4. It should be noted in Figure 210 that strut a34 passes through a slit cut in strut a12 and as such the two struts do not intersect or collide. v2 a23 3 k1/x, k \ L0 Sa i34 a12 4 s o a41 Figure 210. Tensegrity system with two struts, two ties and two springs. The problem statement is written as given a41, a12, a23, a34 kl, k2, Lol, Lo2 find cose4 (and corresponding value of cosO1) when the system is in equilibrium The solution approach is outlined as follows: 1. Obtain expression for cos6O in terms of cos94 2. Write Li in terms of cosq 1 and L2 in terms of cose4 3. Write the potential energy equation 4. Obtain an equation in the variables cos04 and cos6l which corresponds to the condition dU/dcos94 = 0 5. Utilize Sylvester' method to obtain values for cos04 and cos6l that satisfy the equation from step 1 and step 4. 2.4.1 Obtain Expression for cos9i in terms of cos94 The cosine law for the quadrilateral 1234 is written as 2 Z41 a23 2 (231) where by definition 2 Z41 12 (X4S1 + Y4 1) + 4 12 2 (232) X4 = a34 S4 (233) Y4 = (a41 + a34 C4) (234) 2 2 a a Z4 =34 + 41 +a34414 2 2 (235) Substituting Equation 231 into 232 and rearranging gives 2 2 al2X4, = a12Y41 +Z4 12 a23 2 2 (236) Substituting for X4, Y4, and Z4 and then squaring this equation and substituting s42=1C42 and si2=c12 yields A c +Bc+D = (237) where A = A, C4 + A2 A, = 2 a122 a34 a41 2 2 2 2 A2 a12 a34 a12 a41 B = B1 C42 + B2 C4+B3 B1 = 2 al2 a41 a342 B2 = a2 a34 (a342 a12 3 a412 + a23 ) B3= a12 a4 (a412 a122 a342 + a232) D = Di c42 + D2 C4 + D3 2 2 2 2 Di = a34 a41 a12 a34 D2 = a34 a41 (a412 a122 a342 + a232) D3 = (a122 + a23 2 a12 a34 a34 a412) (a122 + a232 + 2 a12 a34 a342 a412) / 4 (2.38) Equation (237) provides a relationship between ci and c4. 2.4.2: Write L1 in terms of cos8i and L2 in terms of cos04 A cosine law for triangle 412 that expresses Li in terms of ci may be written as 2 2 2 al2 41 L, 2+ + al2 41 1 1 2 2 2 (239) A cosine law for the triangle 341 that expresses L2 in terms of c4 may be written as a342 42 L2 + + a34 41c 2( 2 2 2 (240) 2.4.3: Write the potential energy equation The energy stored in the springs is given by U = 1/2 ki (LiLoi)2 +1/2 k2 (L2LO2)2 (241) 2.4.4: Express the equation dU/dcos04 = 0 in terms of the variables c4 and ci The derivative of the potential energy U with respect to c4 may be written as d k,(L L )2 d k2(L2 L )2 dU 2( L ) dL dc, 2 2 02) dL2 dc4 dL, dc, dc4 dL2 dc4 (242) dU dL dc dL = k,(L, L01) d +k2(L2 L02) 2 dc4 dc1 dc4 dc4 (243) From Equations 239 and 240 dL al2 a41 dc, L (244) dL2 34 a41 dc4 L2 (245) From Equation 237 dc1, A, c,2 + 2B1 c, c4 + B2 C1 + 2D1, 4 + D2 dc4 2A, c, c4 + 2 A2 + B,42 + B2 c4 + B3 (246) Substituting Equations 244, 245, and 246 into 243 gives dU 1 a 41) A, c,2 +2B c, c4 +B2 1 +2D c4 +D2 134 a41 k Al Lo) + k 2L LO2L02) dc4 2A, c, c4 +2A2c1 +B c42 +B,2 4 +B3 2L (247) Equating Equation 247 to zero, dividing throughout by a41 and rearranging gives k,(L L1)L2 a12 F + k2(L2 L2)L a34G =0 (248) where F=A, c,2 +2B, c, c4 +B2 C +2D, c4 +D2 (249) G = 2A, c c4 + 2 A c + B C42 + B2 4 + B3 (250) Rearranging this Equation 248 gives L L L2[k, a12 F+k 2a34 G] = L, [k2 LO2 a34 G] +L2 [k, L a12F] (25 1) Squaring both sides and rearranging gives L,2 L22 [k, a12 F + k2 a34 G]2 L,2 [k2 L02 a34 G]2 L22 [k, Lo, a12 F]2 2L, L2 [k2 L02 a34 G][k, L,, a12 F] (252) Squaring both sides again gives L14 L24 [k, a12 F + k2 a34 G]4 + L4 [k2 L02 a34 G]4 + L24 [k, Lo0 a12 F]4 214 L,2 [k, a12 F]2 [k2 L02 a34 G]2 2L12 L24 [k, a12 F]2 [k, L01 a12 F]2 +2L,2 L2 [k2 L02 a34 G]2 [k, L01 a12 F]2 =4L12 L22 [k2 L02 a34 G]2 [k1 Lo0 a12 F]2 (253) Rearranging gives, L14 L24 [k a12 F + k2 a34 G]4 + L14 [k2 L02 a34 G]4 + L24 [k, Lo0 a12 F]4 214 L22 [k, a12 F]2 [k2 L02 a34 G]2 2L12 L24 [k, a12 F]2 [k, L,0 a12 F]2 2L,2 L22 [k2 L2 3a4 G]' [k Lo, a12 F]2 = 0 (254) Substituting for L12 and L22 using Equations 239 and 240 and factoring the polynomial in terms of c4 and Ci gives Cio c110 + C9 C19 + C8 ci8 + C7 c17 + C6 c16 + C5 c15 + C4 C14 + C3 C13 + C2 C12 + C1c1 + Co =0 (255) where Cl = C10,2 C42 + C10,1 C4 + C10,0 C9 =C9,3 C43 + C9,2 C42 + C9,l C4 + C9,0 Cs = C8,4 C44 + C8,3 C43 + C8,2 42 + C8,1 4 + C,0 C7 = C7,5 c45 + C7,4 C44 + C7,3 C43 + C7,2 C42 + C7,1 c4 + C7,0 C6 = C6,6 C46 + C6,5 C45 + C6,4 C44 + C6,3 C43 + C6,2 C42 + C6,1 c4 + C6,0 C5 =C5,7 C47 + C5,6 C46 + C5,5 C45 + C5,4 C44 + C5,3 C43 + C5,2 C42 + c5,1 4 + 5,0 C4= C4,8 C48 + C4,7 C47 + C4,6 C46 + C4,5 C45 + C4,4 C44 + C4,3 C43 + C4,2 C42 + C4,1 C4 + C4,0 C3 = C3,9 C49 + C3,8 C48 + C3,7 C47 + C3,6 C46 + C3,5 C45 + C3,4 C44 + C3,3 C43 + C3,2 C42 + C3,1 C4 + C3,0 C2 C2,10 C410 + C2,9 C49 + C2,8 C48 + C2,7 C47 + C2,6 C46 + c2,5 C45 + c2,4 C44 + c2,3 C43 + C2,2 C42 + C2,1 C4 + C2,0 Cl =Cl,o c410 + C1,9 C49 + C1,8 C48 + C1,7 C47 + C1,6 C46 + C1,5 C45 + C1,4 C44 + C1,3 C43 + Cl,2 C42 + C1,1 C4 + C1,0 Co = Co,l0 C410 C + C,9 C49 + Co,8 C48 + Co,7 C47 + Co,6 C46 + Co,5 C45 + Co,4 C44 + Co,3 C43 + Co,2 C4 + C,1 4 + CO (256) All the terms Cij have been obtained symbolically and are defined in terms of known mechanism parameters. For example, Clo,o = 8 ki4 a126 a414 a34 Ai4 + 4k14 a126 a416 A14+ 4k14 a126 a412 a344 Ai4 Cio,i= 16k1 a12 a41 a34 A1 + 16 k14 a12 a41 a34 A1 C10,2 = 16 ki4 a126 a414 a342 A14 C42 (257) The remaining terms are quite lengthy and are not listed here. 2.4.5 Construct Sylvester Matrix for this approach Multiplying Equation 237 by cl, c12, c 3, c 4, c 5, cl6, c 7, cl8, c19 and 255 by cl gives 12 equations that can be written as follows 0 O O O O O O O 0 A B D c" 0 0 o 0 0 O O O O A B D 0 c10o 0 O O O O O O O A B D 0 O c0 9 0 O O O O O O A B D 0 O 0 c 8 0 O O O O O A B D 0 O O O c 7 0 O O O O A B D O O O O O c 6 0 O O O A B D O O O O O O c 5 O O O A B D 0 O O O O O O c 4 0 0 A B D 0 O O O O O O O c3 0 A B D 0 O O O O O O O O c 2 0 O CO C9, C8 C7 C6 C5 C4 C3 C2 C CO c, 0 C, C9 C, C, C, C C C3 C2 C C 1 0 (258) L J .(258) Expansion of the 12 x 12 determinant yields a 32nd degree polynomial in the parameter c4. 2.4.6 Numerical Example for Approach 2 Parameters of Example 1 (section 2.3.4) are used for this numerical example: Strut lengths a12 = 3 in. a34 = 3.5 in. Noncompliant tie lengths a41 = 4 in. a23 = 2 in. Spring 1 free length & spring constant Lol = 0.5 in. ki = 4 lbf/in. Spring 2 free length & spring constant L02 = 1 in. k2 = 2.5 lbf/in. For this example, a Maple program was written that would solve the determinant of Sylvester matrix in Equation 258. The result was a polynomial of degree 32 with respect to the variable cosO4. The solution to the polynomial for this numerical example resulted in 12 real and 20 complex roots. Eight of the real roots were in acceptable range of 1 to +1. For each value of cos04 the corresponding value of cos6i was calculated such that Equations 237 and 255 are simultaneously satisfied. These values, as well as calculated values for L1 and L2, are presented in Table 25. Table 25.Eight feasible real solutions. Case c4 (radian) c1 (radian) L, (inches) L2 (inches) 1 0.8144902900 0.2120649698 5.4853950884 2.3332963547 2 0.7083900124 0.1385596475 5.3221641782 2.9008756697 3 0.9290901467 0.9154296793 1.7405998094 1.4951507917 4 0.8840460933 0.9381755562 1.5760033787 1.8699490332 5 0.9046343262 0.9312332602 1.6280054527 1.7088706403 6 0.9434161072 0.8970754537 1.8628443599 1.3543814070 7 0.6228159543 0.0544134519 5.1289299905 3.2880318246 8 0.8042767224 0.2077177765 5.4758767915 2.3937944296 Complex values for cl that correspond to each complex value of c4 were determined and are presented in Table 26. Although these solutions are not realizable, it is important to show that since the c4, c1 pairs satisfy the two equations, no extraneous solutions were introduced in the variable elimination method. Table 26.Twenty complex solutions. Case C4 C1 1 1.2513573 0.08533821 1.0200425+0.0625503 I1 2 1.2513573 + 0.0853382 I 1.0200425 0.0625503 I1 3 1.2371852 0.05868661 1.0164651 + 0.03791361 4 1.2371852+0.05868661 1.0164651 0.03791361 5 1.06627740.02229991 1.0214122+0.01103171 6 1.0662774 + 0.0222999 I1 1.0214122 0.0110317 I1 7 1.0406024 0.0082765 I1 1.0447401 + 0.0104582 I1 8 1.0406024 + 0.0082765 I 1.0447401 0.0104582 I 9 1.0322412 0.0344383 I 1.0269253 + 0.0377413 I 10 1.0322412 + 0.0344383 I 1.0269253 0.0377413 I 11 1.0087191 0.0028359 I 1.1658773 +0.0338462 I1 Table 26. Continued. Case C4 Ci 12 1.0087191 + 0.00283591 1.1658773 0.03384621 13 0.3360373 0.4127430 I1 0.9379955 + 0.6334257 I 14 0.3360373 + 0.4127430 I1 0.9379955 0.6334257 I 15 0.3571365 0.10681791 0.9796431 + 0.3840155 I 16 0.3571365 + 0.10681791 0.9796431 0.3840155 I 17 0.51056850.03439811 1.0755053 0.29906131 18 0.5105685+0.03439811 1.0755053 +0.2990613 I1 19 0.54185770.17049791 1.1122907+0.42951361 20 0.5418577+0.17049791 1.11229070.42951361 Lastly, there were four real values of cos04 that were not in the range 1<: cos4 C 1. These values are listed in Table 27. Corresponding values for cosOi could not be obtained that would satisfy both Equations 237 and 248. Thus it must be concluded that four extraneous roots were introduced in this second solution approach. Table 27. Four nonfeasible real solutions. Case c4 (radian) 1 0.7460150 101 2 1.0119084 3 1.0061580 4 0.7460150 101 2.4.7 Comparison of Results for Approaches 1 and 2 The numerical example from Section 2.3.4 was examined using this second approach. It was not possible to symbolically expand the determinant of the coefficient matrix in Equation 258. Rather the coefficients of the polynomials A, B, D, and Ci through C1o were obtained numerically and then the determinant was expanded to obtain the single polynomial equation in c4. The numerical example of the second approach resulted in a 32nd degree polynomial in the variable c4. Table 25 shows the resulting values of c4 that solved this polynomial as well as the corresponding values of ci such that Equations 237 and 255 are simultaneously satisfied. The real solutions are identified in Table 25. Values of Table 25 were tested in the differential of potential energy to identify equilibrium cases. Analysis showed that cases 3, 4, 5, and 6 of Table 25 represent equilibrium configurations. These cases are identically similar to cases presented in Table 21. Comparison of the results of the numerical example using the two approaches outlined in this chapter showed that both approaches agree on the correct numerical answer. Hence, both approaches are valid in finding equilibrium configuration were closed form solution for this structure can not be found. CHAPTER 3 THREE SPRING PLANAR TENSEGRITY 3.1 Introduction This chapter considers the case of a planar tensegrity system with one non compliant member, a41, three compliant members, springs L1, L2, L3, and two struts, a34 and a12 (see Figure 31). In Figure 31 the two struts do not intersect but one passes through a slit cut in the other one. The objective of this study is to find all possible equilibrium configurations in stable condition (minimum potential energy) when given the lengths of the struts and the noncompliant tie together with the free length and spring constants for the three compliant members. The device shown in Figure 31 is a two degree of freedom system. Two parameters must be specified, in addition to the constant mechanism parameters, in order to define the configuration of the device. These two parameters will be referred to as the descriptive parameters for the system. One obvious set of descriptive parameters are the angles 04 and 01. Considering the noncompliant member a41 as being fixed to ground, specification of 04 will define the location of point 3. Similarly, specification of 01 will define the location of point 2. Two approaches to solve this problem are presented in this chapter. Both aim to find a set of descriptive parameters that minimize the potential energy in the system. In the first approach, the lengths of the compliant members L1 and L2 are chosen as the descriptive parameters. Derivatives of the potential energy equation are obtained with respect to L1 and L2 and values for the descriptive parameters are obtained such that these derivatives are zero, corresponding to either a minimum or maximum potential energy state. In the second approach, the cosines of the angles 04 and 01 were chosen as the descriptive parameters. The cosines of the angles were chosen rather than the angles themselves in the hope that the resulting equations would be simpler in that, for example, a single value of cos04 accounts for the obvious symmetry in solutions that will occur with respect to the fixed member a41. L3 3 L, a3 L2 L0 4 L04a41 Figure 31. Tensegrity structure with two struts, a noncompliant (tie), and three springs. 3.2 First Approach Problem Statement (Descriptive Parameters L1, L2) The problem statement can be explicitly written as: Given: a12, a34 lengths of struts, a41 length of noncompliant tie ki, Lol spring constant and free length of compliant tie between points 4 and 2 k2, Lo2 spring constant and free length of compliant tie between points 3 and 1 k3, L03 spring constant and free length of compliant tie between points 2 and 3 Find: L1 length of spring 1 at equilibrium position, L2 length of spring 2 at equilibrium position, 47 L3 corresponding length of spring 3 at equilibrium position 3.2.1 Development of Geometric Constraint Equation Figure 32 shows the nomenclature that is used. L1, L2, and L3 are the extended lengths of the compliant ties between points 4 and 2, points 3 and 1, and points 2 and 3. L 3 2 L aL2 014 4 04 0 1141 Figure 32. Planar tensegrity structure. The analysis starts by considering different triangles in this structure. Figure 33 shows the triangle formed by side as4, L3, and Li. 3 Figure 33. Triangle 432. 48 A cosine law for this triangle can be written as L2 2 L,2 1 4 +L a34 COS41 2 2 2 (31) Solving for cos94' yields r L3 L1 a34 cos94 1 2Li a34 (32) Figure 34 shows the triangle formed by side a41, a12, and L1. A cosine law for this triangle can be written as 2 2 2 L1 41 +L a41 cos411 12 2 2 2 (33) Solving for cos94" yields 2 2 2 cos1 4 = a12 a41 2Li a41 (34) 2 L a12 a 1 Figure 34. Triangle 412. Figure 35 shows the triangle formed by a41, a34, and L2. A cosine law for this triangle can be written as a2 42 L 2 34 + + a34a41 Cos04 2 2 2 (35) Figure 35. Triangle 413. Solving for cos04 yields 2 2 2 cos04 L2 a34 a41 cos4 2a a 2a34 a41 From Figure 32 it is apparent that 04 + 04 = 7T+ 04" Equating the cosine of the left and right sides of Equation 37 yields cos (04 + 04') = COS (7 + 04") and expanding this Equation 38 yields cos04 cos4' sin04 sin04' = cos04" (36) (37) (38) (39) Rearranging (39) gives cos04 cos04' + cos04" = sin04 sin04' . (310) Squaring both sides yields (cos94)2 (cos94')2 + 2 cos04 cos04' cos04" + (cos94")2 = (sinO4)2 (sinO4')2 (311) Substituting for (sin94)2 and (sin94')2 in terms of cos04 and cos04' gives (cos94)2 (cos94')2 + 2 cos04 cos04' cos04" + (cos94")2 = (1Cos284) (1cos284') (312) Equations 32, 34, and 36 are substituted into Equation 312 to yield a single equation in the parameters L1, L2, and L3 which can be written as G1 L34 + (G2 L22 + G3) L32 + (G4 L24 + G5 L22 + G6) = 0 (313) where G1 = a412, G2 = G2a LI2 + G2b, G3 = G3a LI2 + G3b, G4 = G4a L1, G5 = G5a Li4 + G5b LI2 + G5c, G6 = G6a Li2 + G6b. (314) and where G2a = 1, G2b = a12 a412, G3a = (a34 a412), G3b = a42 (a41 a12 a34) a12 a34 , G4a = 1, Gsa = 1, G5b = (a122 a342 a41 2), G5c = a342 (a412 a122), G6a = a122 (a412 a342), G6b = a122 a342 (a122 + a342 a412). (315) 3.2.2 Development of Potential Energy Equations The potential energy of the system can be written as 1 1 1 U =k, (L, L01)2 +k2 (L2 L02)2 + k3 (L3 L03)2 (316) 2 2 2 At equilibrium, the potential energy will be a minimum. This condition can be determined as the configuration of the mechanism whereby the derivative of the potential energy taken with respect to the descriptive parameters L1 and L2 both equal zero. The geometric constraint equation, Equation 313, contains three unknown terms, L1, L2, and L3. From this equation, L3 can be considered as a dependent variable of L1 and L2. The following two expressions may be written: aU aL =k,j(L,L0o)+k3(L3LO3) 3 =0 (317) aL, aL, aU aL =k2(L2LO2)+k3(L3LO3) 3 =0 (318) OL2 OL2 The derivatives OL3/Li and OL3/OL2 can be determined via implicit differentiation from Equation 313 as aL3 L, [2L,2 L2 + G2a L22 L32 + G3a L32 + L2 + G5b L22 +G6a (319 BL, L3 [2G, L32 + G2a L,2 L22 + G2b L22 + G3a LI2 +G3b] aL3 L2 [2L,2 L,2 + G2a L12 L32 + G2b L32 + L4 + G5b L2 +G5J 3 8L2 L3[2GL32 + G2a L12 L22 + G2b L22 + G3a LI2 +G3b Substituting (319) into (317) and rearranging gives (Di L22+D2) L33 + (D3 L22+D4) L32 + (D5 L24+D6 L22+D7) L3 + (D8 L24+D9 L22+D1) = 0 (321) where D =DiaLi, D2 = D2a L1 + D2b, D3 = D3a L, D4 = D4a L, D= D5a L1, D6a L13 + D6b L12 + D6c LI + D6d , D7a Li3 + D7b Li2 + D7c Ll + D7d, Ds8a L1, D9a L13 + D9b LI =DioaLi (3.22) and Dia = G2a k3 , D2a = 2 Gi ki G3a k3, D2b = 2 Gi k Lo0 , D3a = G2a k3 L03 , D4a = G3a k3 L03 , D5a= k3, D6a = G2a ki 2 k3 D6b = G2a ki Lo0 D6c = G2b kl G5b k3 D6d = G2b kl Lo0 , D7a = G3a ki D7b = G3a ki Lo0 D7c = G3b ki G6a k3 D7d = G3b ki Lo0 , Dsa = k3 L03 , D9a = 2 k3 L03 D9b = G5b k3 L03 DlOa = G6a k3 L03 (323) Substituting (320) into (318) and rearranging gives (El L2 + E2) L33 + (E3 L2) L32 + (E4 L23 + E5 L22 E6 L2 + E7) L3 + (E8 L23 + E9 L2) = 0 (324) where Ela Ll2 + Elb ,  2 Gi k2 L02 , E3a LI2 + E3b , E4a LI2 + E4b , E5a Li2 + E5b , E6a Li4 + E6b Li2 + E6c , E7a Li + E7b , E8a L , E9a Li4 + E9b Li2 + E9c( G2a k3 , G2a k3 L03 , G2a k2 2 k3 , G2a k2 L02 , k3 , G3a k2 L02 , 2 k3 L03 , k3 L03 , G2b k3 + 2 G k2 , G2b k3 L03 , G2b k2 , G2b k2 L02 , G3a k2 G5b k3 E6c = G3b k2 G5c k3 , G3b k2 L02 , E9b = G5b k3 L03 , E9c = G5c k3 L03 . 3.2.3 Create Solution Matrix Equations 313, 321, and 324 are three equations in the three unknowns L1, L2, and L3. Sylvester's method, reference 11, is applied in order to obtain sets of values for these parameters that simultaneously satisfy all three equations. In this solution, the parameter L1 is embedded in the coefficients of the three equations to yield three equations in the apparent unknowns L2 and L3. Determining the condition that these new coefficients (which contain Li) must satisfy such that the three equations can have common roots for L2 and L3 will yield a single polynomial in Li. (326) (325) Equation 313 was multiplied by L2, L3, L2L3, L32, L22, L2L3 2, L22L3, L22L32, L33, L23, L2L33, L22L33, L23L3, L23L32, and L23L33. Equation 321 was multiplied by L3, L2, L32, L2L32, L33, L22, L2L33, L22L3, L2 L32, L23, L34, L2 L3, L2 L32, and L2L34. Equation 3 24 was multiplied by L2, L3, L2L3, L32, L22, L2L32, L22L3, L22L32, L33, L23, L2L33, L2L33, L23L3, L23L32, L24, L34, L24L3, L2 L32, L2L34, and L22L34. This resulted in a set of 52 equations that can be written in matrix for as M = 0. (327) The vector h is written as 73 5 5 3 7 7 2 6 3 5 4 4 5 3 6 2 7 7 = [L27L3 L25L35, L23L37, L27L32 L26L33, L25L34 L24L35, L23L36, L22L37, L27L3, L2 L3 L2 L3 L2 L3 L2 L3 L2 L3 L2L3 L2 L2 L3, L2 L3 L2 L3 L2 L3 , 2 5 6 7 6 5 42 3 3 2 4 5 6 5 4 L2 L3 L2L3 L37, L26, L2 5L3, L24L3 L2 L3 L2 L3 L2L3 L36, L2 L2 4L3, L2 L3 L2 L3 L2L3 L35, L24, L2 L3, L2 L3 L2L3 L34, L23, L2 L3, L2L32, L3 , L22, L2L3, L32, L2, L3, T (328) The coefficient matrix M is a 52x52 matrix. To show this 52x52 coefficient matrix, it is subdivided into the four 26x26 submatrices M11, M12, M21, M22 as Mu M12 M= (329) The four sub matrices are: 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 G4 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 D, 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 D, 0 D 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 D, O D 0 0 0 0 D O D O O O 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 E4 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 E, 0 E, 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 E, 0 E, 0 0 0 0 E8 0 0 0 0 0 0 0 0 0 0 0 E, 0 E, 0 0 0 0 E8 E, E, E, 0 0 0 0 0 0 0 0 0 0 0 0 0 0 O O O O O O O O O O O O O O O O O 0 0 0 0 0 0 O O O 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 G4 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 G, 0 G, 0 G, 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 G, 0 G, 0 G, 0 0 0 0 0 0 0 0 0 0 0 0 0 0 G, 0 G, 0 G, 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 D, O D, 0 0 0 0 D8 O D, 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 D, O D, 0 0 0 0 D 0 O 0 0 0 0 0 D, OD, 0 0 0 D 0 D, 0 00 0 0 O D6 0 D, 0 0 0 0 0 0 0 0 0 0 0 0 0 D, 0 D, 0 0 0 0 D O D O O O O O O0 0 0 0 0 D, 0 D, 0 0 0 D 0 D, 0 0 0 0 0 0 D, 0 D, 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 E, 0 E, 0 0 0 O O O O O O O O O O O O O O O O O E, 0 El 0 0 0 0 E E 0 0 0 0 0 0 0 0 0 0 0 0 E, 0 E, 0 0 0 0 E8 E, E, E, 0 0 0 0 0 0 0 0 E, 0 E, 0 0 0 E8 E, E, E, 0 0 0 0 0 E, 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 E, 0 E, 0 0 0 0 E E, E, E, 0 0 0 0 0 0 0 0 0 E4 0 E, 0 0 0 E E, E, E,2 0 0 0 0 0 E6 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 G, 0 G, 0 G, 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 G 0 G, 0 G, 0 O O O O0 0 0 0 0 0 0 0 0 0 0 0 G, 0 G, 0 G, 0 0 0 0 0 0 0 0 0 0 0 0 0 0 G, 0 G 0 G, 0 0 0 0 0 0 0 0 0 0 G, 0 G3, 0 0 0 0 0 0 0 0 0 0 0 0 0 G 0 G2 0 G, 0 0 0 0 0 0 0 0 0 0 0 G, 0 0 0 G, 0 G2 0 G, 0 0 0 0 0 0 0 0 0 0 G, 0 G, 0 0 0 0 0 G, G2 G, 0 0 0 0 0 0 0 0 G, 0 G 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 D, O D, 0 0 0 0 D8 O D, 0 0 0 0 0 0 D, 0 0 0 0 0 0 D, 0 D, 0 0 0 D8 0 D, 0 0 0 0 0 0 D6 0 D2 0 0 0 0 0 D, O D, 0 0 0 D 0 D3 0 0 0 0 0 0 D6 0 D2 O O O O D, D, D, 0 D 0 D, 0 0 0 0 0 D6 0 D2 0 0 0 0 D, 0 D, 0 0 0 0 0 0 D, D, 0 D8 0 D, 0 0 0 0 0 D6 0 D2 0 0 0 0 D, 0 D, 0 0 0 0 0 0 0 0 0 0 0 0 EO 0 El 0 0 0 0 E E, E E, O O O O O E, 0 0 0 0 0 0 0 0 0 0 0 0 0 E, 0 El 0 0 0 0 E0 E0 E0 E 0 0 0 0 0 E4 0 E, 0 0 0 E E, E, E2 0 0 0 0 0 E, 0 0 0 0 0 0 E, E4E, 0 E E, E 0 0 0 0 E 0 0 0 0 0 0 E,E, 0 0 0 0 0 0 0 0 0 0 0 0 E, 0 E, 0 0 0 E, E, E, E2 0 0 0 0 0 E, 0 0 0 0 0 E, E, O 0 E, E, E E2 0 0 0 0 E, 0 0 0 0 0 0 E, E O 0 0 0 56 0 0 0 0 0 0 0 0 0 0 0 G, 0 G, 0 G, 0 0 0 0 G, O G, 0 0 G, O 0 0 0 0 0 D, 0 D, 0 0 D, O D, 0 0 0 D, O D, D, 0 D, O D, D,, O 0 0 0 0 0 0 0 0 0 0 0 E, 0 E, 0 Eg E, E, E, 0 E, O E, E, 0 D, 0 D, 0 0 0 D, O D, 0 0 0 0 D, O D, 0 D, 0 D, 0 0 D, 0 D,, 0 0 0 0 0 0 0 E, 0 E, 0 0 E, E, E3 E2 0 0 E, 0 0 E, E, 0 0 0 0 0 0 0 0 0 0 0 E, 0 E, 0 0 E, E, E3 E2 0 0 E, O 0 E, E, 0 0 0 E, 0 E, 0 0 0 E, E, E3 E2 0 0 0 E, 0 0 0 E, E, 0 0 0 0 0 0 0 0 0 0 0 0 G, 0 G2 0 G, O 0 0 0 0 0 G, O G3 0 0 0 0 G, O 0 0 0 0 0 0 0 G, 0 G2 0 G, 0 0 0 0 0 0 G, 0 G3 0 0 0 0 G, 0 0 G2 0 G, 0 0 0 0 0 0 0 0 G, O G3 0 0 0 0 0 0 G, 0 0 0 0 0 D, 0 0 0 D, 0 D3 0 0 0 0 D, O D2 0 D, 0 D, 0 0 D, 0 D,0 0 0 D, O D3 0 0 0 0 0 D, O D2 0 0 D, 0 D, 0 0 0 D, 0 0 D,0 0 0 0 0 D, O D2 0 0 0 D, 0 D, 0 0 0 0 D, 0 0 0 D,0 0 0 0 0 0 0 0 O12 E, 0 E, 0 0 0 E, E, E3 E2 0 0 0 E, 0 0 0 E, E 0 0 0 0 0 0 0 E, 0 0 0 E, E, E3 E2 0 0 0 E, 0 0 0 E, E, 0 0 0 0 0 0 0 0 E, E, E3 E2 0 0 0 0 E, 0 0 0 0 E, E, 0 0 0 0 0 0 0 0 0 0 0 E, E3 E2 0 0 0 0 E, O 0 0 0 E, E, 0 0 0 0 0 0 0 0 0 0 0 0 O E 0 0 0 0 0 E E 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 G, 0 G2 0 G, 0 0 0 0 0 0 0 0 G, O G3 0 0 0 0 0 0 G, O 0 0 G2 0 G, 0 0 0 0 0 0 0 0 G O0 G3 O O O O O 0 G, 0 0 0 0 0 0 0 0 0 0 0 G, O G3 0 0 0 0 0 0 0 0 G, O 0 0 0 0 0 0 O 0 0 0 0 0 G 0 G3 O O O O O O 0 G 0 0 0 0 0 0 0 0 G 0 G3 O O O O O O O O 0 G 0 0 0 0 0 0 0 0 0 0 0 0 O 0 D O0 D2 0 0 O D, 0 D 0 0 0 0 D O O O0 D 0 0 0 0 0 0 D3 0 0 0 0 0 D, O D2 0 0 D, 0 D, 0 0 0 D, 0 0 D,0 0 0 0 0 0 O D, 0 D, O 0 D, 0 0 0 0 D 0 0 0 0 0 0 0 0 0 0 0 D6 0 D2 0 0 0 D, 0 D, 0 0 0 0 D, 0 0 0 D,, 0 0 0 0 0 0 0 0 D, 0 D, 0 0 0 0 0 D, 0 0 0 0 D,, 0 0 0 0 0 0 0 0 0 0 0 0 0 E, E, E3 E2 0 0 0 0 E, 0 0 0 0 E, E, 0 0 0 0 0 0 0 0 0 0 E3 E2 0 0 0 0 E, 0 0 0 0 E, E, 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 E O 0 0 0 0 E, E, 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 E, E 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 E 0 0 0 0 E E 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 E 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 G O0 G3 O O O O O O O 0 G, 0 0 0 0 0 0 0 0 0 0 G 0 G 0 0 0 0 0 0 0 0 G6 0 0 0 0 0 0 0 0 0 O G5 O G3 O 0 O 0 O O O O O O G6 0 0 0 0 0 0 0 0 0 0 0 0 000 000 0 G 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 G3 0 0 0 00 0 0 0 00 G 0 0 0 0 0 0 0 0 0 0 0 0 0 M = 0 0 00 000 G 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 O G 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 D2 0 0 0 D, 0 D, 0 0 0 0 D, 0 0 0 D,, 0 0 0 0 0 0 0 0 0 O O D, O D, O O O O O D, 0 0 O O D,, 0 O 0 0 0 0 0 0 0 0 0 D, 0 0 0 0 0 D, 0 0 00 D,, 0 0 0 0 0 0 0 0 0 0 0 0 0 O D, 0 0 0 0 0 D,, 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 D, 0 0 0 0 0 Do 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 E 0 O O O O 0 E, E7 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 E 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 E E 0 00 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 000 00 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 (330) In order for a solution for the vector ) to exist, it is necessary for the 52 resulting equations to be linearly dependent. This will occur if the determinant of M equals zero. It was not possible to symbolically expand the determinant of matrix M. A numerical case was analyzed and a polynomial of degree 158 in the variable L1 was obtained. It was not possible to solve this high degree polynomial for the values of L1, although several commercial and inhouse written algorithms were attempted. Because of this, a different method was attempted to solve the set of equations 313, 321, and 324. 3.2.4 Solution of Three Simultaneous Equations in Three Unknowns  Continuation Method The continuation method (Morgan [18], Mora [19], Cullen [20], Strang [21], Garcia and Li [22], Morgan [2324], Wampler et al., [25]) which is a numerical technique to solve a set of equations in multiple variables are used. This is as opposed to Sylvester's method which would lead to a symbolic solution of the problem. A concise description of the continuity method is presented by Tsai, 1999. Suppose one wishes to solve the set of equations F(x) which are defined by f I(xi, x2,***,x)= 0 F(x):f 2X X2,**.,Xn)= 0 (331) fnxtx,,2...,X)=0 F(x) is called the target system. The continuation method begins by first estimating the total number of possible solution sets (sets of values for L1, L2, and L3 for our case) that satisfy the given equations. For example, Bezout's theorem states that a polynomial of total degree n has at most n isolated solutions in the complex Euclidean space. Including solutions at infinity, the Bezout number of a polynomial system is equal to the total degree of the system. Next, an initial system, G(x) =0, is obtained, whose solution will be of the same degree as that of F(x), but whose solution set is known in closed form. In other words, G(x) maintains the same polynomial structure as F(x). Finally, a homotopy function H(x, t) is prepared such as H(x, t) =y (1t) G(x) + t F(x) (332) where y is a random complex constant. When t=0, the homotopy function equals the initial system, G(x). When t=l, the homotopy function equals the target system, F(x). Recall that the solutions to G(x) are known. As the parameter t is increased in small steps from 0 to 1, the solutions of H(x, t) can be tracked (referred to as path tracking) and when t =1, these solutions will be the solutions to the original target system. If the degree of the solution set was overestimated, some of the solutions will track to infinity and these can easily be discarded. 3.2.5 Numerical Example The following information is given: strut lengths: a12 = 14 in. a34 = 12 in. noncompliant tie lengths: a41 = 10 in. spring 1 free length & spring constant: Loi = 8 in. ki = 1 lbf/in. spring 2 free length & spring constant: Lo2 = 2 in. k2 = 2.687 Ibf/in. spring 3 free length & spring constant: L03 = 2.5 in. k3 = 3.465 Ibf/in. Based on these values, the coefficients in equations 313, 321, and 324 were evaluated to yield the three equations 100 L34 + [(Li2 + 96) L22 + 44 L12 52224] L32 + L12 L24 + (L14 440 L12 13824) L22 8624 L12 + 6773760 = 0 (333) (2.5 Li L22 + 90 L1 1600) L33 + (8.663 L1 L22 + 381.165 L1) L32 + [2.5 Li L24 + (6 L13 + 8 L12 + 1196 L1 768) L22 + 44 L13 352 L12 30664 L1 + 417792] L3 + 8.663 Li L24 + (17.326 L13 3811.651 L1) L22 74708.354 L1 = 0 (334) [(2.5 L12 + 160) L2 1074.637] L33 + (831.633 8.663 L12) L2 L32 + [(7 L12 + 192) L23 + (515.826 + 5.373 L12) L22 + (2.5 L14 + 1188 L12 69888) L2 236.420 L12 + 280609.161] L3 + 17.326 Li2 L23 + (119755.135 + 8.663 L4 3811.651 L12) L2 = 0 (335) The continuation method was run on this set of three equations in three unknowns to obtain all solution sets for the three spring lengths, L1, L2, and L3, for the particular numerical example. The software PHCpack (Verschelde [26]) was used to implement the method. The PHCpack software estimated the number of possible solutions to be 136. Seven real solutions were obtained and these are listed in Table 31. Table 31: Seven real solutions for threespring planar tensegrity system(units are inches) Case L1 L2 L3 1 13.000 8.000 7.017 2 11.376 10.371 5.333 3 7.585 9.097 10.106 4 11.029 12.557 3.044 5 13.969 5.800 9.164 6 14.248 9.373 4.774 7 13.181 11.599 2.488 An equilibrium analysis was conducted for the seven cases and only the first case was indeed in equilibrium. Case 1 is shown in Figure 36. 2 3L3 a34 L2 a12 a41 Figure 36. Case 1, equilibrium solution. 3.3 Second Approach Problem Statement (calculate cos04, cos01) In the second approach, the cosines of the angles 04 and 01 were chosen as the descriptive parameters. The cosines of the angles were chosen rather than the angles themselves in the hope that the resulting equations would be simpler in that, for example, a single value of cos04 accounts for the obvious symmetry in solutions that will occur with respect to the fixed member a41. The problem statement is presented as follows: Given: Strut lengths a12, a34, tie length a41, and the spring constants and free lengths ki, k2, k3, Lol, Lo2, L03. Find: cos04 and cosOi when the system is in equilibrium. The solution approach is as follows: 1. Obtain expressions for L1, L2, and L3 in terms of cos04 and cosO1 2. Write the potential energy equation 3. Determine values of cos04 and cosi1 such that dU/dcos94 = dU/d cosi1= 0 3.3.1 Development of Geometric Equations Figure 31 shows the nomenclature used. Springs L1, L2, and L3 are the extended lengths of the compliant ties between points 4 and 2, 2 and 3, and 1 and 3 respectively. The unloaded lengths of springs are given by Lol, L02, and L03. A cosine law for the quadrilateral 1234 can be written as: Z41 3 (336) where, 2 Z41 =a12(X4S + Y4C) + Z4 + 12 (337) and where X4= a34 S4 (338) Y4 (a41 + a34 04) (339) 2 2 a a Z4 34 + 41 +3441c4. (340) 2 2 Substituting (337) into (336) and rearranging gives a 2 2 a12X2s, = a,,Yc, + Z, + 2 (341) 2 2 Substituting (338), (339), and (340) into (341), then squaring it, and substituting for S42= 1 C42 and si2=C12 and multiplying the entire equation by 4 yields L34 +AL32 + B = 0 (342) where A = A, c4 + A2 c1 + A3 c1 C4 + A4 (343) B = B1 c12 + B2 c12 C4 + B3 c1 + B4 c1 c4 + B5 ci c42 + B6 C42 + B7 c4 + B8 (344) and where A, = 4 a34 a41 A2 = 4 al2 a41 A3 = 4 al2 a34 A4 = 2 (a122 + a342 + a412) B1 = 4 a122 (a342 + a412) B2 = 8 a12 a34 a41 B3 = 4 al2 a41 (a412 + a122 + a342) B4 = 4 al2 a34 (a342 + a122 + 3 a412) B5 = 8 al2 a41 a342 B6 = 4 a342 (a122 + a412) B7 = 4 a34 a41 (a412 + a342 + a122) Bs = (a412 + a342 2 al2 a34 + a122) (a412 + a342 + 2 al2 a34 + a12) (345) Equation 342 expresses L3 as a function of c4 and Ci. A cosine law for triangle 4 12 may be written as 2 2 2 al2 41 12 41 (346) A cosine law for the triangle 341 may be written as 4 + 1 +3441c4 2 (347) 2 2 2 Equations 342, 346, and 347 define the three spring lengths in terms of the variables cos04 and cosOi. 3.3.2 Development of Potential Energy Equations The total potential energy stored in all three springs is given by, U = 1/2 k (LiLoi)2 + 1/2k2 (L2LO2)2 + 1/2 k3 (L3LO3)2 (348) Differentiating the potential energy with respect to c4 and ci and then evaluating values for c4 and ci that cause the derivative of the potential energy to equal zero, will identify configurations of either minimum or maximum potential energy. These derivatives may be written as, dU/dc4 = ki (Li Lol) dLi/dc4 + k2 (L2 L02) dL2/dc4 + k3 (L3 L03) dL3/dc4 (349) dU/dci = ki (Li Lol) dLi/dci + k2 (L2 L02) dL2/dci + k3 (L3 L03) dL3/dci (350) Since from 346, Li is not a function of c4, dLi/dc4 = 0. Similarly, from 347, L2 is not a function of ci and thus dL2/dci = 0. Equations 349 and 350 reduce to dU/dc4 = k2 (L2 L02) dL2/dc4 + k3 (L3 L03) dL3/dc4 (351) dU/dci = ki (Li Lol) dLi/dci + k3 (L3 L03) dL3/dci (352) The term dL2/dc4 is evaluated from 347 as dL2 a3441 (353) dc4 L2 and the term dL1/dc, is evaluated from 346 as dL1 al2a41 (354) dc, Li Implicit differentiation of 342 for L3 with respect to c4 and Ci yields dL3 (A, + A3,c,)L32 + B2C12 + B4CI + 2B5Ic4 + 2B64 + B (355) dc4 2L,3 (2L,32 + AIc4 + A2c, + A3CC4 + A4) dL3 (A2 + A3C4)L32 + 2Bici + 2B2c1c4 + B3 + B4c4 + B5c42 (356) dc, 2L3(2L32 + AIc4 + A2c, + A3CC4 + A4) Substituting 353 and 355 into 351 and 354 and 356 into 352 and equating to zero yields (Mi L33 + M2 L32 + M3 L3 + M4) L2 + M5 L33 + M6 L3 = 0 (357) (Ni L33 + N2 L32 + N3 L3 + N4) L1 + N5 L33 + N6 L3 = 0 (358) Where the coefficients Mi and Ni, i=1..6, are functions of C4 and Ci as M1 = k3 (A3 c1 + A) + 4 k2 a34 a41 M2 = k3 L03 (A3 c1 + Ai) M3 = k3 B2 c12 + [2 (k2 a34 a41A3 k3 B5) c4 + 2 k2 a34 a41 A2 k3 B4) ci1 + 2 (k2 a34 a41 A, k3 B6) C4 k3 B7 + 2 k2 a34 a41 A4 M4 = k3 LO3 B2 C12 + k3 LO3 B4 c1 + 2 k3 Lo3 Bs cl c4 + 2 kL032C3 B6 C4 + k3 L03 B7 M5 = 4 k2 a34 a41 L02 M6 = 2 k2 (a34 a41LO2 A2 1i + a34 a41 Lo2 A3 ci 04 + a34 a41 L02 A1 c4 + a34 a41 L02 A4) (359) N1 = k3 (A3 C4 + A2)+ 4 ki ai2 a4i N2 = k3 Lo3 (A3 c4 + A2) N3 = 2 ki a12 a41 (A1 C4 + A2 C1 + A3 C1 04 + A4) k3 (B4 04 + B5 42 +2 Bi Ci + 2 B2 ci C4 + B3) N4 = k3 Lo3 (B4 C4 + B5 c42 + 2 Bl cl + B2 c1 c4 + B3) N5 = 4 ki a12 a41 Loi N6 = 2 ki a12 a41 Lo, (A4 + Al c4 + A2 Ci1 + A3 C1 C4) (360) Equations 357 and 358 may be written as (Mi L33 + M2 L32 + M3 L3 + M4) L2 = M5 L33 M6 L3, (361) (Ni L33 + N2 L32 + N3 L3 + N4) Li = N5 L33 N6 L3 (362) Squaring both sides of both equations gives (Mi L33 + M2 L32 + M3 L3 + M4)2 L22 = (M5 L33 + M6 L3)2 (363) (Ni L33 + N2 L32 + N3 L3 + N4)2 L12 Li (N5 L33 + N6 L3)2 (364) Using 346 and 347 to substitute for L12 and L22 will yield two equations in the parameters ci, C4, and L3. These two equations can be arranged as (pi C12 + p2 C1 + p3) C43 + (p4 C13 + p5 C12 + p6 C1 + P7) C42 + (P8 ci4 + p9 ci3 + pio C12 + pll C1 + p12) C4 + (p13 C14 + p14 C13 + 15 C12 + pl16 C1 + p17) = 0 (365) (qi ci + q2) C44 + (q3 C12 + q4 C1 + q5) C43 + (q6 C13 + q7 C12 + q8 c1 + q9) C42 + (q10 C13 + q11 C2 + q12 C1 + q13) 04 + (q4 C13 + q15 C12 + q16 1 + q17) = 0 (366) where Pi = pia L32 + lb L3+ Plc P2 = P2a L32 + P2b L3 + P2c p3 = P3a L32 + P3b L3 + P3c P4 = P4a L32 + P4b L3 + P4c 5 = p5a L34 + P5b L33 + P5c L32 + P5d L3 + P5e P6 = p6a L34 + P6b L33 + P6c L32 + P6d L3 + P6e P7 P7a L34 + P7b L33 + P7c L32 + P7d L3 + P7e p8 = p8a L32 + P8b L3 + P8c P9 = 9a L34 + P9b L33 + p9c L32 + P9d L3 + P9e P10 PlOa L36 + plOb L35 + plOc L34 + pld L33 + ploe L32 + plf L3 + plOg PH11 lla L36 + llb L35 + p L34 + lld L33 + pile L32 + pllf L3 + llg P12 = Pl2a L36 + Pl2b L35 + p12c L34 + Pl2d L33 + Pl2e L32 + Pl2f L3 + Pl2g P13 Pl3a L32 + Pl3b L3 + Pl3c P14 = pl4a L34 + l4b L33 + 14c L32 + l4d L3 + Pl4e P15 Pl5a L36 + Pl5b L35 + p15 L34 + Pl5d L33 + Pl5e L32 + Pl5f L3 + Pl5g P16 Pl6a L36 + Pl6b L35 + Pl6 L34 + Pl6d L33 + Pl6e L32 + Pl6f L3 + Pl6g P17 = Pl7a L36 + Pl7b L35 + 17 L34 + Pl7d L33 + Pl7e L32 + pl7f L3 + Pl7g (367) qu = qla L32 +qlb L3 +qlc q2 q= 2a L32 + q2b L3 + q2c q3 = 3a L32 + q3b L3 + q3c q4 = q4a L34 + q4b L33 + q4c L32 + q4d L3 + q4e q5 q5a L34 + q5b L33 + q5c L32 + q5d L3 + q5e q6 = q6a L32 + q6b L3 + q6c q7 = q7a L34 + q7b L33 + q7c L32 + q7d L3 + q7e q8 = q8a L36 + q8b L35 + q8c L34 + q8d L33 + q8e L32 + q8f L3 + q8g q9 = q9a L36 + q9b L35 + q9c L34 + q9d L33 + q9e L32 + q9f L3 + q9g qio = qloa L32 + qiob L3 + qioc qii = q1ua L34 + qb L33 + q11iic L32 + qld L3 + qiie q12 q2a L36 + q12b L35 + q12c L34 + q12d L33 + ql2e L32 + q12f L3 + ql2g q13 = q13a L36 + q13b L35 + q13c L34 + q13d L33 + ql3e L32 + q13f L3 + q13g q14 = q14a L32 + q14b L3 + q14c qi5 = qi5a L34 + q15b L33 + q15c L32 + q15d L3 + qi5e q16 q16a L36 + q16b L35 + q16c L34 + q16d L33 + q16e L32 + q16f L3 + q16g q17 = q17a L36 + q17b L35 + q17c L34 + q17d L33 + ql7e L32 + ql7f L3 + ql7g (368) The coefficients pia through ql7g are functions of the given constant parameters. These coefficients are defined in the Appendix. The coefficients pia through ql7g have been obtained symbolically. For example, the terms pia through plc are written as pla = 8 k22 a343 a413 A32 16 k2 a342 a412 A3 k3 B5 + 8 k32 B52 a34 a41 pib = 16 k32 B52 L03 a34 a41 + 16 k2 a342 a412 A3 k3 L03 B5 plc = 8 k32 L032 B52 a34 a41 The remaining terms are not listed here due to their complexity. Similarly, equation 342 is an equation in the same parameters, ci, c4, and L3. This equation can be factored as (ri c1 + r2) C42 + (r3 C12 + r4 c1 + r5) 04 + (r6 C12 + r7 1 + r8) = 0 (369) where ri = 8 a12 a41 a34 r2 = 4 a342 (a122 + a412) r3 = 8 a34 a41 a12 r4 = (4 a12 a34) L32 + 4 a12 a34 (a342 + a122 + 3 a412) r5 = (4 a34 a41) L32 + 4 a34 a41 (a342 + a122 + a412) r6 = 4 a122 (a342 + a412) r7 = (4 a12 a41) L32 + 4 a12 a41 (a412 + a342 + ai22) rs = L34 2 (a342 + a12 + a412) L32 (370) 3.3.3 Solution of Equation Set for c4, C1, and L3 Equations 365, 366, and 369 are factored such that the parameter L3 is embedded in the coefficients pi through r8. Sylvester's method was used in an attempt to solve these equations for all possible sets of values of L3, ci, and c4. Equation 365 was 2 2 3 4 2 multiplied by ci, ci2, c4, c42, c4 c14, c12c4, c1c42, and cic42 to obtain 10 equations 2 3 2 2 3 (including itself). Equation 366 was multiplied by ci, ci2, cl c4, c4 cic4, c1 c4, ci c4, cic42 and cl2C42 to obtain 11 equations (including itself). Equation 369 was multiplied by 2 3 4 2 3 4 2 3 4 2 22 32 42 3 ci, ci ci ci c4, C4 C4 C4 c1C4, c1 c4, c1 c4, c1 c4, c1C4 C1 c4 C1 c4 C1 c4 c1C4 , c12 43, c3ci14, c1i44 and C12C44 to obtain 22 equations including itself. This resulted in a set of 43 "homogeneous" equations in 43 unknowns. The condition for a solution to this set of equations was that they are linearly dependent, i.e. the determinant of the 43 x43 coefficient matrix must equal zero. This would yield a polynomial in the single variable L3. Due to the complexity of the problem, it was not possible to expand this determinant symbolically. Similarly, it was not possible to root the polynomial that resulted from a particular numerical example. Because of this, the Continuation Method was used again to determine solutions for the same numerical example presented in Section 3.2.5. Substituting the given numerical values into equations 365, 366, and 369 are yield three equations for this numerical example. For this case, the coefficients pia through ql7g and r, through r8 are evaluated and presented in Appendix b. The continuation method was run on this set of three equations in three unknowns to obtain all solution sets for the three variables, L3, ci, and c4, for the particular numerical example. The software PHCpack (Verschelde [26]) was again used to implement the method. The PHCpack software estimated the number of possible solutions to be 136. Seven real solutions were obtained and these are listed in Table 32. Table 32. Real value answers using cosine of 01, 04 in radians and L3 in inches. Case Cosine 01 Cosine 04 L3 1 0.453571523 0.749999925 7.01658755 2 0.594965126 0.568509675 5.33336481 3 0.851695156 0.671871253 10.1059910 4 0.622722694 0.359683532 3.04401046 5 0.360201938 0.876513972 9.16391281 6 0.332122619 0.650641695 4.77378239 7 .0436671807 0.456122910 2.48758125 The lengths of the springs L1 and L2 were calculated for each of the seven cases and it was determined that the seven real solutions here correspond directly with the seven real solutions that were determined from the first problem formulation and listed in Table 31. As before, case 1 was the only case that was indeed in equilibrium and the mechanism is shown in this configuration in Figure 36. 3.4 Conclusion Two approaches were presented to solve the threespring planar tensegrity equilibrium problem. Both approaches resulted in high degree polynomials. These polynomials could not be derived symbolically and were in fact difficult to derive numerically. A continuation approach was presented to solve the set of three equations for each of the two solution approaches. A force balance analysis was conducted to identify all realizable solutions; in this case one. It is apparent that further analysis of Sylvester's variable elimination procedure may be necessary to reduce the number of simultaneous equations, and thereby the size of the univariate polynomial that results. As a result of the work in this chapter it can be said that the degree of the solution has been bounded, although further work may yield a simpler result. CHAPTER 4 FOUR SPRING PLANAR TENSEGRITY 4.1 Introduction In this chapter planar tensegrity mechanisms with two struts and four elastic ties (see Figure 41) are analyzed to determine all possible equilibrium configurations for the device when no external forces or moments are applied. The stable equilibrium position is determined by identifying the configurations at which the potential energy stored in the four springs is a minimum. 3 L, a12 a34 4 L4 1 Figure 41. Two strut, four spring planar tensegrity device Four nonlinear equations are obtained in the variables L1, L2, L3, and L4 which correspond to the equilibrium lengths of the four compliant ties. One equation is associated with the geometry constraint that must be satisfied in order for the mechanism to be assembled. The other three equations correspond to the derivative of the potential energy equation taken with respect to three of the compliant tie lengths. Solution of these four equations in order to find sets of values for the parameters Li through L4 that simultaneously satisfy all four equations proved to be a very difficult process. For this case, the continuity method was used in an attempt to find all equilibrium solutions for two numerical cases. The results of the continuity method are then checked to first verify that the set of solutions simultaneously satisfy all four equations and then to determine whether each solution set represents a minimum potential energy and an equilibrium state. 4.2 Problem Statement The problem statement can be explicitly written as: Given a12, a34 lengths of struts, ki, Lo, spring constant and its free length between points 4 and 2, k2, Lo2 spring constant and its free length between points 3 and 1, k3, L03 spring constant and its free length between points 3 and 2, k4, L04 spring constant and its free length between points 4 and 1. Find at equilibrium position Li length of spring 1, L2 length of spring 2, L3 length of spring 3, L4 length of spring 4, It should be noted that the problem statement could be formulated in a variety of ways. The solution presented here uses L1, L2, and L3 as the three generalized parameters, knowledge of which completely describes the system (i.e. the value for L4, the last remaining parameter, can be deduced). The geometry equation that relates L4 in terms of the three generalized parameters is derived first. Following this, a derivative of the potential energy equation is taken with respect to each of the three generalized parameters to yield an additional three equations. 4.3 Development of Geometric and Potential Energy Constraint Equations The geometry equation is derived in a manner similar to that presented in Section 3.2.1. The parameter a41 in equations (3.15) can be replaced by L4 to yield an equation in the unknown terms L1, L2, L3, and L4. This equation can be rearranged as G1 (Li4 L24) + G2 (L14 L22) + G3 (L32 L12 L22) + G4 (L12 L22) + G5 (L32 L22) + G6 (L22) + G7( L32 L12 )+ G( L12 )+ G9( L34 ) + Go0( L32 )+ G1 = 0 (41) where G1I=l G2=l G3=1 G4 = L42 a122 a342 G= a122 L42 G6 = L4 a34 a342 a122 G7 = L4 LI2 2 2 2 2 Gs = a34 a12 a12 L4 G9 = L42 Glo = L4 a122 L4 a342 a122 a342 L42 Gl = L4 a34 a12 + a34 a12 a12 a34 (42) The potential energy of the system can be evaluated as 1 1 1 1 U= k, (L, L)2 +k2 (L2 L02)2 + k3 (L3 L03)2 + k4 (L4 L04)2 (43) 2 2 2 2 At equilibrium, the potential energy will be a minimum. This condition can be determined as the configuration of the mechanism whereby the derivative of the potential energy taken with respect to the lengths L1, L2, and L3 all equal zero, i.e. dU dL = k, (L L1) + k4(L4 L04) = (44) dL, dL, dU dL =k2(L2 L2)+k4(L4 L L04 = 0 (45) dL2 dL2 dU dL = k3(L3 L3) +k4(L4 L4) (46) dL3 dL3 The derivatives dL4/dL1, dL4/dL2, and dL4/dL3 are determined by implicit differentiation of Equation 41 as dL4 _L, [L24 L22a342 +2L22L2 L22L32 L22122 L22L42 + a122L42 L42L32 a342122 + L32a342)] (47) dL, L4 (L22 342 L22L2 L2 L32 + a122L,2 L32L,2 a342 a122 + 2L L32 +L34 L32 a342 L32 a122 dL4 _L [2L22L,2 +a342L2 L,4 +L3,2 L+a12 L42 342 +L42L2 +a342 a122 + L4L32 L3a 122)] (48) dL2 L4 (L22 a342 L22L2 L22L32 + a122L2 L32L,2 a342 a122 +2L L32 + L34 L32a342 L32a122) dL4 L3 [L,2L2 L2L42 L2 a12 L42L2 +L44 + 2L42L32 L42 342 a122L42 a342 12 a3422)] (49) dL3 L4 (L22 342 L22L2 L22L32 + a122,12 L32L2 a342 a122 +2L42L32 + L34 L32a342 L32a122) Substituting Equation 47 into Equation 44 and rearranging gives Di(Li L24) + D2(Li3 L22) + D3(L12 L22) + D4(L32 Li L22) + Ds(L1 L22) + D6(L32 L22) + D7(L22) + D8(L34) + D9(L32) + Dio + Dn1(L32 Li3) + D12(L3) + D13(L32 Li2) + D14(L12) + D15(L34 Li) + D16(L32 Li) + D17 (Li) = 0 (410) where D1 = k4L4 + k4LO4 , D2 = 2k4L4 + 2k4LO4 kiL4 , D3 = kiL4Lo1 , D4 = kiL4k4LO4 + k4L4 , D5 = k4Lo4L42 + k4L4 + kiL4342 + k4L4a122 k4L04a122 + k4a34 L4 k4LO4a342 D6 = kiL4Lo1 , D7 = kiL4Lola342 Ds = kiL4Lol , D9 = kiL4Lolal22kiL4 Loi+kiL4Loia34 , Dio = kiL4Loia34 a122 D1 = kiL4, D12 = kiL4a122 D13 = kiL4Lo1 , D14 = kiL4Loia2 , Di5 = kiL4, D16 = 2kiL4 + k4L4 + k4Lo4a342 kL4a122 k4a342L4 + kiL4a34 k4LO4L4 D17 = k4a122L43 kiL4a342al22 + k4L4a342a122 + k4LO4a122L42 k4L04a342a12. (411) Substituting Equation 48 into Equation 45 and rearranging gives E1(L12L23) + E2(L32L23) + E3(L23) + E4(12L22) + E5(L32L22) + E6(L22) + E7(L14L2) + E8(L32L2) + E9(L2L2) + E10(L34L2) + E(L32L12L2) + E12(L2) + E13(L32) + E14(L12) + E15(L34) + E16 = 0 (412) where E1 = k2L4 2k4L4 + 2k4LO4 , E2 = k2L4 , E3 = k2L4a342 E4 = k2L4LO2 , E5 = k2L4LO2 , E6 = k2L4L02a342 E7 = k4Lo4 k4L4 , Es = k4L4 k4LO4L42 + 2k2L4 k2L4a342 k2L4a122 k4L4a122 + k4L4ai122 E9 = k4L4 + k2L4122 + k4L4342 + k4L4122 k4LO4a342 k4L04122 k4L4Lo4L42 E0o = k2L4 , E1 = k2L4 + k4L4 k4LO4 , E12 = k4L43a34 + k4L4342al22 k4L04a342a122 k2L4342al22 + k4LO4L42a342 E13 = k2L4L02a12 2k2L43L02 + k2L4LO2a34 + k2L4LO2L2 E14 = k2L4LO2a122 El5 = k2L4LO2 , E16 = k2L4L02a342al22 (413) Lastly, substituting Equation 49 into Equation 46 and rearranging gives Fi(L2L22) + F2(L3L12L22) + F3(L33L22) + F4(L32L22) + F5(L3L22) + F6(L22) + F7(L33L12) + F8(L3 Li2) + F9(L12) + Fo1(L35) + Fll(L34) + F12(L33) + Fl3(L32) + F14(L3) + F15 = 0 (414) where FI = k3L4LO3 , F2 = k4L4 k4LO4 k3L4 , F3 = k3L4 , F4 = k3L4LO3 , F5 = k3L4a342 k4L4a122 + k4L4 k4L04L42 + k4Lo4a122 F6 = k3L4LO3a342 F7 = k3L4 + k3L4LO3 , Fs = k4L4 + k3L4a122 k4L4a342 k4L04L42 + k4LO4a34 F9 = k3L4L3a122 , Flo = k3L4, F11 = k3k4L03 , Fl2 = 2k3L43 k3L4a342 k3L4a122 + 2k4LO4L42 2k4L43 , F13 = 2k3L43L03 + k3L4LO3a342 + k3L4L03a122 F14 = k4L43a342 + k4a122L43 + L4Lo4L4 + k3L4a342a122 + k4L4a342 a122 k4LO4L4 a34 k4LO4a122L42 k4L04a342a12 k4L45 F15 = k3L4LO3a34 ai12 4.4 Solution of Geometry and Energy Equations The problem at hand is to determine sets of values for the parameters L1, L2, L3, and L4 that will simultaneously satisfy Equations (41), (410), (412), and (414). Sylvester's method is one approach for solving this set of equations for all possible solutions. The parameter L4 has been embedded in the coefficients of Equations (41), (410), (412), and (414) and Table 41 shows the 37 products of L1, L2, and L3 that exist in these four equations. The Table indicates which products that are multiplied by which coefficients for the four equations. Table 41. Coefficient G, D, E, and F. L1x L2Y L3z G's D's E's F's 1 L1TL24 Gi 2 L1i4L22 G2 3 L32 L12 L22 G3 4 LiL22 G4 D3 E4 Fi 5 L32 2 G D6 E5 F4 6 L22 G6 D7 E6 F6 7 L32 L2 G7 D13 8 L1 Gs D14 E14 F9 9 L34 G9 D8 Ei5 F11 10 L32 Gio D9 E13 F13 11 1 Gil Dio E16 Fi5 12 L1L24 D_ 13 L1 L22 D2 14 L32 L L22 D4_ 15 LiL22 D5 16 L32Ll3 Di_ 17 L13 D12 18 L34L1 Di5 19 L32 LL1 D16 20 L1 D17 21 L2 L23 E _ 22 L32 L23 E2 23 L2 3 E3 24 L14L2 E7 25 L32 L2 E _ 26 Li2 L2 E9 (415) Table 41. Continued. L1x L2Y L3 G's D's E's F's 27 L3 4 L2 E __ 28 L3 L1L L2 En _ 29 L2 E1_ 30 L3Li21 L22 F2 31 L33 L2 2F3 32 L3 L22 F5 33 L33 L1 F7 34 L3Li2 F8 35 L35 Fio 36 L33 F12 37 L3 F14 It is apparent from the complexity of the four equations that attempting to multiply the four equations by different products of L1, L2, and L3 in order to obtain a set of "homogeneous" equations where the number of equations equals the number of unknowns is impractical. For this reason, a numerical case will be considered and this case will be solved using the continuation approach. 4.5 Numerical Case Study The following parameters were selected as a numerical example: Strut lengths: a12 = 14 in., a34 = 12 in. Spring free lengths and spring constants: Loi = 8 in. ki = 1 lbf/in. Lo2 = 2.68659245 in. k2 = 2.0 lbf/in. L03 = 3.46513678 in. k3 = 2.5 lbf/in. L04 = 7.3082878 in. k4 = 1.5 lbf/in. Equations (41), (410), (412), and (414) can now be expressed respectively as L24L1 + (L4 L32L12 440.0 L12 + 96.0 L32 13824.0) L22 8424.0 L12 8624.0 L12 + 44.0 L32 Li2 + 6773760.0 52224.0 L32 + L34 = 0 (416)  4.03756830 L24 Li + (11520.0 5.96243170 L I2 Li + 80.0 Li2 18.07513660.0 Li3 + 3216.53005200 Li + 80.0 L32) L22 + 11200.0 L32 + 1960.0 Li3 10 L32 Li3 _ 1577.65300520 L32 Li + 10.0 Li L34 80.0 L34 + 0.2257920 107 15680.0 Li2 247420.01098080 Li + 80. L32 Li2 = 0 (417) (28.07513660 Li2 + 2880.0 20.0 L32) L23 + (7737.38625600 + 53.73184900 L32 + 53.73184900 Li2) L22 + (3187.60655680 L32 + 5696.53005200 Li2 15.96243170 L32 Li2 + 20.0 L34 4.03756830 Li4 508664.65582080) L2 + 7522.45886000 L32 + 0.151652770617600 10 53.73184900 L34 10531.44240400 Li2 + 53.73184900 L32 Li = 0 (418)  494742.03310080 L3 + 12127.978730000 L32 16979.170222000 Li2 + 86.628419500 L32 Li2 86.628419500 L34 4307.51366000 L33 + 4722.34699480 L3 Li2 + 25.0 L35  25.0 L33 Li2 + (20.96243170 L3 Li2 + 86.628419500 Li2 + 3212.39344320 L3 25.0 L33  12474.492408000 + 86.628419500 L32) L22 + 0.2445000511968000 107 = 0 (419) The Polynomial Continuation Method was used to find simultaneous roots for the four equations. Eighteen real solutions and 490 complex solutions were found. All the real and complex solutions were shown to satisfy Equations (416) through (419) with neglible residuals. The real solutions are listed in Table 42. Table 42. Eighteen real solutions. (Solutions that are geometrically impossible are shaded.) Li, in. L2, in. L3, in. L4, in. 1 19.028528 3.624423 3.108440 4.306682 2 13.000001 7.999996 7.016593 9.999998 3 11.632986 4.613617 9.950038 12.269992 Table 42. Continued. L1, in. 4 3.141567 5 7.985150 6 0.398136 7 6.868035 8 4.200027 9 19.507910 10 16.748403 11 4.520731 12 4.077402 13 13.473294 14 22.800126 15 14.616270 16 20.778750 17 12.774217 18 3.806954 L2, in. 2.156404 10.049547 4.497517 1.105366 3.538652 4.034579 8.590494 0.603385 4.708715 11.823371 6.225147 9.580282 0.084697 12.496006 2.744884 L3, in. 5.241983 6.186358 10.533965 3.486019 10.462012 3.169571 2.764444 7.518882 9.316941 2.235364 13.541713 4.547034 4.369638 1.650677 8.098941 L4, in. 18192258 14.286022 15.669134 20.971818 15.411513 5.347832 2.329743 18.852623 16.646605 7.821817 5.861200 8.863331 6.733159 8.309365 17.877213 Cases 1, 4, 7, 9, 10, 11, 14, 16, and 18 are not physically realizable. For example, Li in case 1 is greater than sum of a12 and L4, and hence point 2 cannot be constructed. The solutions in Table 42 can correspond to either maximum or minimum potential energy configurations. The second derivative of the potential energy with respect to the variables L1, L2, and L3 are now evaluated to identity the minimal potential energy cases. Derivatives of Equations (44) through (46) with respect to L1, L2, and L3, respectively can be written as d2U dLi2 d2U dL,2 d2U dL32 k+k k4(dL4 )2 +k4(L4 dL, k +k4(d4 )2 +k4(L4 dL2 k,3 +k4 (L4 )2 +k4(L4 dL,3 k, +k4 4 2+k4(L, dL, dL, dL 1 dL L dL32 d2L dL 3 (420) Table 43 shows the value of the potential energy and the second derivative for each of the remaining nine cases. It is seen that cases 2, 3, 13, 15, and 17 have all positive second derivative values and represent possible equilibrium solutions. Each of these solutions was analyzed to determine if it truly corresponds to an equilibrium configuration and only cases 2, 3, 13, and 15 were indeed in equilibrium. Table 43. Identification of minimum potential energy state. (Solutions that do not have all positive second derivatives are shaded.) 02U 02U 02U Case L1, in L2, in L3, in L4, in U, inlbf L a12 OL2 2 L4 2 lbf/in lbf/in Ibf/in 2 13.000 7.999 7.016 9.999 61.932 1.146 3.052 4.569 3 11.632 4.613 9.950 12.269 130.923 0.740 0.566 4.065 5 7.985 10.049 6.186 14.286 334.931 0.470 1.241 1.594 6 0.398 4.497 10.533 15.669 153.432 3.597 1.889 0.750 8 4.200 3.538 10.462 15.411 156.415 0.659 16.134 2494.334 12 4.077 4.708 9.316 16.646 119.990 0.221 15.521 379.009 13 13.473 11.823 2.235 7.821 139.276 4.967 51.239 56.665 15 14.616 9.580 4.547 8.863 25.420 2.405 25.608 32.573 17 12.774 12.496 1.650 8.309 345.474 10.529 50.466 39.862 Figure 42 shows the four feasible solutions that are in stable equilibrium drawn to scale. Dashed lines are used to identify the springs that have negative spring lengths. It must be noted that when a spring has a negative spring length, the two points at the end of the spring want to move towards one another as the spring attempts to return to its free length. 2 2 2 2 4 1 4 1 4 1 4 1_ 1 case 2 case 3 case 13 case 15 Figure 42. Four equilibrium solutions A force analysis was conducted for each of the four cases and these cases were found to be in static equilibrium. Table 44 shows the calculated force in each of the four springs and the two struts for the equilibrium cases. Negative values in the table correspond to the springs which have a negative spring length. Table 44. Calculated forces in struts, and elastic ties. Force in Force in Force in Force in Force in Force in Case Spring Spring L2, Spring L3, Spring L4, Strut a12, Strut a34, L1, lbf lbf lbf lbf lbf lbf 2 5.000 10.625 8.877 4.036 11.828 7.220 3 3.632 14.599 16.212 7.441 17.942 9.253 13 5.473 18.273 14.250 0.769 18.5294 5.8228 15 6.616 24.533 20.030 2.332 25.005 7.6394 4.6 Conclusions The problem of determining the equilibrium configurations of a fourspring planar tensegrity system was formulated by developing four equations in terms of the four spring lengths. It was not practical to use Sylvester's method to solve this set of equations. Rather, the continuation method was used for a particular numerical example and four equilibrium configurations were obtained. Cases may exist that would yield more than four realizable configurations, and this could be investigated in future work. The objective here, however, was to determine if it was tractable at all to solve the equilibrium problem for multiple (if not all) solutions, either symbolically or numerically, and this goal was met. CHAPTER 5 FUTURE WORK, SUMMARY, AND CONCLUSION 5.1 Future Work 5.1.1 The 3D Tensegrity Platform As part of future suggested research work, an introduction to the position analysis of a general three dimensional parallel platform device consisting of 3 struts, 3 springs, and 6 ties (Figure 51) is presented here. The problem statement is given as follows: Given: length of noncompliant ties, struts, and free lengths and spring constants of the compliant ties. Find: final length of compliant ties at all equilibrium configurations. 21 Figure 51. Threedimensional parallel platform of 3 struts, 3 springs, and 6 ties. Three approaches to this problem are presented here: 1. Minimum potential energy analysis. 2. Force balance analysis. 3. Linear dependence of 6 connector lines. The unexpected complexity, i.e. high degree polynomials, encountered in the planar analysis may indicate that the analysis of the general three strut tensegrity system will be very complex. This is a three degree of freedom system. One approach is to define the system by six angles, subject to three constraints. That is, the coordinates of the tip of each strut (points 4, 5, and 6) can be defined by angles ai and Pi, i = 1..3, which define the orientation of each of the three struts relative to the base triangle. The three constraints are that the distances between points 4, 5, and 6 are given. The potential energy can be written in terms of these six angles and three equations can be generated by equating the partial derivative of the potential energy with respect to three of the angles to zero. These three equations, together with the three distance constraint equations would give a set of six equations in the six angles. In the second approach, force balance equations at points 4, 5, and 6 could be written as functions of the lengths of the three springs. This is complicated in that the positions of points 4, 5, and 6 would have to be determined from a forward analysis of the device, i.e. the device is equivalent to a 33 parallel mechanism, whose forward analysis was solved by Griffis and Duffy [27]. The solution of the forward analysis of the 33 parallel mechanism requires rooting a fourth degree polynomial in the square of one parameter, resulting in eight possible configurations for the device. The nature of the forward analysis makes it infeasible to obtain a symbolic solution to the equilibrium configuration problem. A numerical solution may be possible. The third approach utilizes the fact that if the device is in equilibrium with no external forces applied, then the Plucker coordinates of the lines along the three struts and the three springs must be linearly dependent. For this solution, it is then necessary to obtain the Pluicker coordinates of these lines as functions of the three spring lengths. This again requires a forward analysis of the 33 parallel mechanism. Further, only a single equation results from equating the determinant of the 6x6 matrix of line coordinates to zero. This leads one to believe that the fact that the lines must be linearly dependent is a necessary, but not sufficient condition for equilibrium. Further analysis is necessary. 5.1.2 The 2D Tensegrity structures Cases of two dimensional tensegrity structures proved to be mathematically challenging in spite of their simple configurations. Different methods were tried to solve these problems. It was expected the final closed form solution be a simple mathematical expression. In each case the intermediate mathematical expressions become complicated to solve and the relevant matrices become large to manipulate. Other mathematical procedures might exist to solve these problems with less complexity. 5.2: Summary and Conclusion In this research, all possible combinations of two dimensional tensegrity structures consisting of two struts, springs, and noncompliant members were analyzed. A closed form solution could only be obtained for case of 2 spring, 2 struts, and 2 noncompliant members. Finding a simple solution approach for a closed form solution to the other two dimensional tensegrity structures consisting of either 3 or 4 springs was not possible due 86 to the complexity of mathematical solutions. A numerical solution using the Continuity Method was used to solve individual numerical examples. The contribution of this research was to set a basic mathematical foundation for the closed form solutions of tensegrity structures and also to present numerical approaches where a closed form solution could not be obtained. The contribution of this research leads to a better understanding of the complexity of these devices and hopes to lay a ground work for future analyses of more complex systems. APPENDIX SAMPLE OUTPUT The following coefficients belong to the numerical example of a Maple sessions programming. (pi ci2 + p2 C1 + p3) C43 + (p4 C13 + p5 C12 + p6 C1 + p7) C42 + (p8 C14 + p9 C13 + p1o C12 + P11 ci + pi2) c4 + (p13 c14 + P14 c13 + P15 c12 + P16 1 + p17) = 0 (365) (qi ci + q2) c44 + (q3 c12 + q4 c1 + q5) c43 + (q6 c13 + q7 c12 + q8 ci + q9) c42 + (q1o c13 + qi Ci2 + q12 c1 + q13) C4 + (q14 C13 + q15 C12 + q16 1 + q17) = 0 (366) (ri ci + r2) C42 + (r3 C12 + r4 c + r5) 4 + (r6 12 + r7 c + r8) = 0 (369) > # ># Numerical example 14 a34:= 12 a41 := 10 8 k:= 1 2 k2 := 2.687 2.5 k3 := 3.465 0.5773663961 1015 L32 0.1110100300 1016 L32 0.5335964670 1015 L32 0.4853912146 1015 L32 0.2080248063 1016 L3 + 0.1873780685 1016 0.4198962128 1016 L3 + 0.3961707733 1016 0.2114119475 1016 L3 + 0.2094045517 1016 0.2087909024 1016 L3 + 0.2186077465 1016 al2:z L01 : L02 : L03 : p2: p3: p4: p5 := 0.1733540051 1013 L34 + 0.7456817937 1013 L33 + 0.2423387462 1016 L32 0.9462608682 1016 L3 + 0.9212755033 1016 p6:= 0.3864997467 1013 L34 + 0.1452389258 1014 L33 + 0.3406073884 1016 L32 0.1266504024 1017 L3 + 0.1176377359 1017 p7:= 0.2113487353 1013 L34 + 0.7162795261 1013 L33 + 0.1466145513 1016 L32 0.5300984714 1016 L3 + 0.4722921577 1016 p8:= 0.1020169484 1015 L32 0.5100847420 1015 L3 + 0.6376059274 1015 p9 := 0.7286924884 1012 L34 + 0.3643462442 1013 L33 + 0.1068305616 1016 L32 0.4737062676 1016 L3 + 0.5137282043 1016 plO := 0.1301236586 1010 L36 0.6506182936 1010 L35 + 0.2052962276 1014 L33 + 0.3082075504 1016 L32 + 0.1219206631 1017 pll := 0.3300435984 + 0.2941695331 +0.1113665191 pl2 := 0.2092793462 + 0.1246857458 + 0.3440522646 1010 L36 0.1289836349 1011 L35 1014 L33 + 0.3455740168 1016 L32 1017 1010 L36 0.5893635686 1010 L35 1014 L33 + 0.1339547765 1016 L32 1016 0.4747632005 0.1236996613 0.7989672476 0.1248534533 0.3983027155 0.4345164946 pl3 := 0.1037172309 1015 L32 0.5185861544 1015 L3 + 0.6482326928 1015 pl4 := 0.7408373632 1012 L34 + 0.3704186816 1013 L33 + 0.5844049572 1015 L32 0.2657927757 1016 L3 + 0.2963349453 1016 pl5 := 0.1322923863 + 0.1316433327 + 0.4841901337 pl6:= 0.3355443251 + 0.1489517912 + 0.3326208569 pl7 := 0.2121019430 + 0.5272839394 + 0.8167029969 10 10L36 0.6614619316 1010 L35 1014 L33 + 0.1235296500 1016 L32 1016 1010 L36 0.1311333621 1011 L35 1014 L33 + 0.1159476235 1016 L32 1016 1010 L36 0.5991862948 1010 L35 1013 L33 + 0.4069170754 1015 L32 1015 0.3034952946 0.4981478398 0.4128749122 0.4015136920 1013 L34 1016 L3 1013 L34 1016 L3 0.1862278592 1013 L34 0.1160024667 1016 L3 1013 L34 1017 L3 1013 L34 1017 L3 1013 L34 1016 L3 ql := 0.8744309862 1014 L32 0.4372154932 1015 L3 + 0.5465193664 1015 q2 := 0.9243984712 1014 L32 0.4621992356 1015 L3 + 0.5777490445 1015 q3 := 0.4669520350 1015 L32 0.2187549572 1016 L3 + 0.2550423709 1016 q4 := 0.7286924880 1012 L34 + 0.3643462442 1013 L33 + 0.9975040455 1015 L32 0.4749518798 1016 L3 + 0.5610932160 1016 q5 := 0.7703320590 1012 L34 + 0.3851660295 1013 L33 + 0.5326615535 1015 L32 0.2576221616 1016 L3 + 0.3081328237 1016 q6:= 0.6233888280 1015 L32 0.2723886871 1016 L3 + 0.2975494328 1016 q7:= 0.1945633479 1013 L34 + 0.9114789880 1013 L33 + 0.2467565040 1016 L32 0.1115537420 1017 L3 + 0.1253958324 1017 q8 := 0.1518109351 1010 L36 0.7590546756 1010 L35 0.4948249499 1013 L34 + 0.2326400716 1014 L33 + 0.2978451536 1016 L32 0.1366694272 1017 L3 + 0.1548714436 1017 q9 := 0.1604858457 1010 L36 0.8024292286 1010 L35 0.3056661932 1013 L34 + 0.1440713627 1014 L33 + 0.1127231047 1016 L32 0.5199241185 1016 L3 + 0.5873781950 1016 qlO := 0.1238733084 1016 L32 0.5475596549 1016 L3 + 0.6050171798 1016 qll :=0.4022366239 1013 L34 + 0.1777220459 1014 L33 + 0.3554116422 1016 L32 0.1568854809 1017 L3 + 0.1720685863 1017 ql2 := 0.3260393580 1010 L36 0.1447643958 1011 L35 0.7932329962 1013 L34 + 0.3478741954 1014 L33 + 0.3352073482 1016 L32 0.1470298991 1017 L3 + 0.1588185279 1017 ql3 := 0.3446701784 1010 L36 0.1530366470 1011 L35 0.3895030685 1013 L34 + 0.1691392672 1014 L33 + 0.1035832980 1016 L32 0.4479338270 1016 L3 + 0.4707584806 1016 ql4:= 0.6153702088 1015 L32 0.2751416753 1016 L3 + 0.3075503998 1016 ql5:= 0.2075805867 1013 L34 + 0.8667876209 1013 L33 + 0.1553312052 1016 L32 0.6722506098 1016 L3 + 0.7212669096 1016 ql6:= 0.1750560045 1010 L36 0.6792486624 1010 L35 0.3717087746 1013 L34 + 0.1514057165 1014 L33 + 0.1284193314 1016 L32 0.5346826116 1016 L3 + 0.5463422536 1016 90 ql7 := 0.7180628716 1010 L35 0.1603893518 103 L34 + 0.6318953271 1013 L33 + 0.1830521647 1010 L36 0.1390169719 1016 L3 + 0.1348526898 1016 + 0.3482593233 1015 L32 rl := 8 al2 a41 a342 r2 := 4 a412 a342 + 4 a342 al22 r3 := 8 a34 a41 al22 r4 := 12 al2 a412 a34 4 al2 a34 L32 + 4 al2 a343 + 4 al23 a34 r5 := 4 a34 a41 L32 + 4 a413 a34 + 4 a34 a41 al22 + 4 a343 a41 r6 := 4 a412 al22 + 4 a342 al22 r7 := 4 al23 a41 4 al2 a41 L32 + 4 al2 a41 a342 + 4 al2 a413 r8 := L34 + (2 a342 2 a412 2 al22) L32 + a414 + a344 + al24 + 2 a412 a342 + 2 a412 al22 2 a342 al22 The following portion of Maple program presents the three equations used in continuation method. (371)= eqnlforcont (372) = eqn3_16_sq (373) = eqn3_17_sq > # 3 springs cosine of thetal, cosine of theta2 and L23, Continuation method >a41:= 10 ; al2:= 14.0 ; a34:= 12 ; a41 := 10 al2 := 14.0 a34 := 12 >k24:= 1 ; k31:= 2.0 ; k23:= 2.5 ; k24 := 1 k31 := 2.0 