1. DEPARTMENT OF MECHANICAL AND AEROSPACE ENGINEERING UNIVERSITY OF FLORIDA Ph.D. DISSERTATION Presented by: Jahan B Bayat Summer, 2006

2. 2 Agenda Objective Literature review Analysis of planar tensegrity mechanisms –2-spring planar tensegrity –3-spring planar tensegrity –4-spring planar tensegrity (added to proposal) Future work, Summary and Conclusion

3. 3 Dissertation Objective determine all equilibrium poses for 2-spring, 3-spring, and 4-spring planar tensegrity mechanisms spatial tensegrity struts ties planar tensegrity

4. 4 Contribution analysis of these mechanisms provides a first insight into this class of mechanisms the knowledge gained here may assist in the analysis of more complex structures

5. 5 Definitions Tensegrity is an abbreviation of tension and integrity. Tensegrity structures are formed by a combination of rigid elements in compression called struts and connecting elements that are in tension called ties. In three dimensional tensegrity structures no pair of struts touches and the end of each strut is connected to non- coplanar ties, which are in tension. In two dimensional tensegrity structures, struts still do not touch. A tensegrity structure stands by itself in its equilibrium position and maintains its form solely because of its arrangement of its struts and ties. The potential energy of the system stored in the springs is a minimum in the equilibrium position.

6. 6 Tools for Analysis Linear Algebra with Matrix manipulation Polynomials Sylvester method Maple AutoCAD

7. 7 Basic 3D and 2D Tensegrity Structures triangl e quadrilat eral pentagon hexagon Figure 1: Family of Tensegrity Structures struts ties Figure 2: Planar Tensegrity Structure 1 4 23 a 23 a 41 a 12 a 34

8. 8 Literature Overview of some basic definitions, geometries and applications. An example of practical applications is deployable structures such as Self deployed Space Antenna. Tensegrity is a new science (about 25 years).

9. 9 A few of Literatures: Roth, B., Whiteley, W., “Tensegrity framework,” Transactions American Mathematics Society, Vol.265, 1981. Skelton, R. E., Williamson, D., and Han, J. H., Equilibrium Conditions of Tensegrity Structure, Proceedings of the Third World Conference on Structural Control (3WCSC) Como, Italy, April 7-12, 2002. Duffy, J., and Crane, C., Knight, B., Zhang, "On the Line Geometry of a Class of Tensegrity Structures“. Ian P. Stern, “Development of Design Equations for Self Deployable N-Struts Tensegrity Systems”.

10. 10 Continue Ian P. Stern, “Development of Design Equations for Self Deployable N-Struts Tensegrity Systems”. Yin, J. P., Marsh, D., Duffy, J.“Catastrophe Analysis of Planar Three-Spring Systems”. Carl D. Crane, Joseph Duffy, Kinematics Analysis of Robot Manipulators. Jahan B. Bayat, Carl D. Crane, “Closed-Form Equilibrium Analysis of a Planar Tensegrity Structure”. Etc.

11. 11 Agenda Objective Literature review Analysis of planar tensegrity mechanisms –2-spring planar tensegrity –3-spring planar tensegrity –4-spring planar tensegrity (added to proposal) Future work, Summary and Conclusion

12. 12 Two-Spring Tensegrity System given: –a 12, a 34 strut lengths –a 23, a 41 non-compliant tie lengths –k 1, L 01 k 2, L 02 spring parameters solution 1, find: –L 1, L 2 at equilibrium solution 2, find: –c 4, c 1 at equilibrium 1 L2L2 L1L1 a 41 a 12 a 23 a 34 2 3 4

13. 13 Two-Spring Tensegrity, Solution 1 obtain geometric equation f 1 (L 1, L 2 ) = 0 write potential energy equation U = ½ k 1 (L 1 -L 01 ) 2 + ½ k 2 (L 2 -L 02 ) 2 evaluate this may be written as f 2 (L 1, L 2 ) = 0 solve the two equations for all sets of L 1, L 2 resulted in 28 th degree polynomial in L 1

14. 14 Approach 1 given: –a 12, a 34 strut lengths –a 23, a 41 non-compliant tie lengths –k 1, L 01 k 2, L 02 spring parameters find: –L 1, L 2 at equilibrium 1 L2L2 L1L1 a 41 a 12 a 23 a 34 2 3 4

15. 15 Figure 3: Planar Tensegrity StructureFigure 4: Triangle 4-3-2 a 34 L1L1 a 23 '4'4 2 3 4 1 L2L2 L1L1 a 41 a 12 a 23 a 34 2 3 4 '4'4 "4"4 44 Geometric Constraint  4 +  4 ' =  +  4 " (2-3)(2-1)

16. 16 L1L1 a 12 a 41 "4"4 1 4 2 Figure 5: Triangle 4-1-2 Figure 6: Triangle 4-1-3 4 3 1 44 L2L2 a 41 a 34 Geometric Constraint – Cont. (2-7)(2-5)

17. 17 Geometric Constraint – Cont. cos (  4 +  4 ') = cos (  +  4 ") cos  4 cos  4 ' – sin  4 sin  4 ' = - cos  4 " cos  4 cos  4 ' + cos  4 " = sin  4 sin  4 '. (cos  4 ) 2 (cos  4 ') 2 + 2 cos  4 cos  4 ' cos  4 " + (cos  4 ") 2 = (sin  4 ) 2 (sin  4 ') 2.  4 +  4 ' =  +  4 " (cos  4 ) 2 (cos  ’ 4 ) 2 + 2 cos  4 cos  ’ 4 cos  ” 4 + (cos  ” 4 ) 2 = (1-cos 2  4 ) (1-cos 2  ’ 4 ) (2-12)

18. 18 Geometric Constraint – Cont. substituting (2-3), (2-5), and (2-7) into (2-12) and gives geometry equation. where and B 2, B 0, C 2, and C 0 are expressed in terms of known quantities A L 2 4 + B L 2 2 + C = 0 (2-13) A = L 1 2, B = L 1 4 + B 2 L 1 2 + B 0, C = C 2 L 1 2 + C 0

19. 19 Potential Energy Constraint at equilibrium, the potential energy in the springs will be a minimum U = ½ k 1 (L 1 -L 01 ) 2 + ½ k 2 (L 2 -L 02 ) 2 the mechanism being considered is a one degree of freedom device one parameter can be selected as the generalized coordinate for the problem; L 1

20. 20 Potential E. Constraint – Cont. at a minimum potential energy state, dL 2 /dL 1 can be obtained via implicit differentiation of the geometry constraint as (2-19) (2-20)

21. 21 Potential E. Constraint – Cont. substituting (2-20) into (2-19) gives D L 2 5 + E L 2 4 + F L 2 3 + G L 2 2 + H L 2 + J = 0 (2-21) where the coefficients D through J are polynomials in L 1 equations (2-13) and (2-21) represent two equations in the two unknowns L 1 and L 2 Sylvester’s elimination method is used to obtain values for these parameters that simultaneously satisfy both equations

22. 22 Sylvester’s elimination determinant of coefficient matrix must equal zero which yields a 28 th degree polynomial in L 1

23. 23 Symbolic Expansion determinant of coefficient matrix must equal zero which yields a 28th degree polynomial in L 1 Maple program used to obtain all coefficients symbolically corresponding values of L 2 for each value of L 1 can be readily obtained from solving (2-13) and then (2-21)

24. 24 2.3.4 Numerical Example given: –a 12 = 3 in.a 34 = 3.5 in. –a 41 = 4 in.a 23 = 2 in. –L 01 = 0.5 in.k 1 = 4 lbf/in. –L 02 = 1 in.k 2 = 2.5 lbf/in. find L 1 and L 2 at equilibrium

25. 25 Numerical Example results –coefficients of 28 th degree polynomial in L 1 obtained –8 real roots for L 1 with corresponding values for L 2 –4 cases correspond to minimum potential energy CaseL 1, in.L 2, in. 1-5.48542.3333 2-5.3222-2.9009 3-1.7406-1.4952 4-1.57601.8699 51.62801.7089 61.8628-1.3544 75.1289-3.2880 85.47592.3938

26. 26 spring in compression with a negative spring length spring in tension 4 44 41 11 1 2 2 2 2 3 3 3 3 Case 3Case 4 Case 5Case 6 Numerical Example Cont.

27. 27 Force Balance Verification Case Force in Spring 1, lab Force in Spring 2, lab Force in Strut a 12, lab Force in Strut a 34, lab Force in Tie a 41, lab Force in Tie a 23, lab 3+ 8.9624+ 6.2379-13.4186-16.80979.162318.7570 4+ 8.30402.1749-5.4965-11.70694.103111.9485 54.51201.7722-4.2448-7.10953.08837.4454 65.4514+ 5.8860-11.4149-11.69837.206414.1181

28. 28 Conclusion for Approach 1 closed-form solution to 2 strut, 2 spring tensegrity system presented geometric and potential energy constraints gave two equations in the spring lengths L 1 and L 2 elimination of L 2 resulted in a 28 th degree polynomial in the single variable L 1 numerical example presented showing 4 real solutions other examples have been investigated, but all gave 4 real solutions

29. 29 Approach 2 Determine (c 4 and c 1 ) to Minimize Potential Energy. The objective of this approach is to again investigate, in closed-form, the planar 2-spring tensegrity system.

30. 30 Geometry 4 3 2 1 44 a 41 a 12 a 34 L1L1 L2L2 a 23 11

31. 31 Approach 2 The problem statement is written as: given:a 41, a 12, a 23, a 34 k 1, k 2, L 01, L 02 find:cos  4 (and corresponding value of cos  1 ) when the system is in equilibrium

32. 32 Two-Spring Tensegrity, Solution 2 obtain geometric equations f 1 (c 4, c 1 ) = 0, f 2 (c 4, L 2 ) = 0, f 3 (c 1, L 1 ) = 0, write potential energy equation U = ½ k 1 (L 1 -L 01 ) 2 + ½ k 2 (L 2 -L 02 ) 2 evaluate this may be written as f 4 (c 4, c 1, L 1, L 2 ) = 0 use f 2 and f 3 to eliminate L 1 and L 2 from f 4 solve this equation and f 1 for all sets of c 4, c 1 resulted in 32 nd degree polynomial in c 4

33. 33 Solution Approach 2 from cosine law for planar quadrilateral which can be factored as A c 1 2 + B c 1 + D = 0(2.37) A = A 1 c 4 + A 2 B = B 1 c 4 2 + B 2 c 4 + B 3 where

34. 34 Solution Approach 2 from derivative of potential energy C 10 c 1 10 + C 9 c 1 9 + C 8 c 1 8 + C 7 c 1 7 + C 6 c 1 6 + C 5 c 1 5 + C 4 c 1 4 + C 3 c 1 3 + C 2 c 1 2 + C 1 c 1 + C 0 =0 where the coefficients C i are functions of c 4 (2-55)

35. 35 Solution Approach 2

36. 36 Solution Approach 2 Expansion of the 12×12 determinant yields a 32nd degree polynomial in the parameter c 4. Using earlier numerical values, there are 8 real values and 20 complex values for c 4. Four of real values are identical to real values in approach 1. Table on next page presents real values of second approach.

37. 37 Solution Approach 2 Casec 4 (radian)c 1 (radian)L 1 (inches)L 2 (inches) 1-0.81449029000.2120649698-5.48539508842.3332963547 2-0.70839001240.1385596475-5.3221641782-2.9008756697 3-0.9290901467-0.9154296793-1.7405998094-1.4951507917 4-0.8840460933-0.9381755562-1.57600337871.8699490332 5-0.9046343262-0.93123326021.62800545271.7088706403 6-0.9434161072-0.89707545371.8628443599-1.3543814070 7-0.62281595430.05441345195.1289299905-3.2880318246 8-0.80427672240.20771777655.47587679152.3937944296

38. 38 Agenda Objective Literature review Analysis of planar tensegrity mechanisms –2-spring planar tensegrity –3-spring planar tensegrity –4-spring planar tensegrity (added to proposal) Future work, Summary and Conclusion

39. 39 3-spring planar tensegrity closed-form analysis of a three spring, two strut tensegrity system with one non- compliant element 4 3 2 1 44 a 41 a 12 a 34 L 24 L 31 L 23 11

40. 40 3-S: 1 st approach The problem statement can be explicitly written as: Given:a 12, a 34 lengths of struts, a 41 lengths of non-compliant tie k 1, L 01 spring constant and free length, point 4 to 2 k 2, L 02 spring constant and free length, point 3 to 1 k 3, L 03 spring constant and free length, point 2 to 3 Find: L 1 L 2 L 3 length of springs at equilibrium position,

41. 41 3-S: 1 st approach cont.

42. 42 3-S: 1 st approach cont. Cosine law for triangles 2-4-3 and 4-1-2 :  4 +  4 ' =  +  4 " (3-1) (3-3) (3-7)

43. 43 3-S: 1 st approach cont. 3.2.1 Development of Geometric Equation, G 1 L 3 4 + (G 2 L 2 2 + G 3 ) L 3 2 + (G 4 L 2 4 + G 5 L 2 2 + G 6 ) = 0 3.2.2 Development of Potential Energy Equations (3-13) (3-16)

44. 44 3-S: 1 st approach cont. At equilibrium, the potential energy will be a minimum. (3-17) (3-18)

45. 45 3-S: 1 st approach cont. Substituting, results (3-21) and (3-24) : (D 1 L 2 2 +D 2 ) L 3 3 + (D 3 L 2 2 +D 4 ) L 3 2 + (D 5 L 2 4 +D 6 L 2 2 +D 7 ) L 3 + (D 8 L 2 4 +D 9 L 2 2 +D 10 ) = 0 (E 1 L 2 + E 2 ) L 3 3 + (E 3 L 2 ) L 3 2 + (E 4 L 2 3 + E 5 L 2 2 + E 6 L 2 + E 7 ) L 3 + (E 8 L 2 3 + E 9 L 2 ) = 0 (3-21) (3-24)

46. 46 3-S: 1 st approach cont. Creating solution matrix M 52x52 : M λ = 0 M = λ = [L 2 7 L 3 3, L 2 5 L 3 5, L 2 3 L 3 7, L 2 7 L 3 2, L 2 6 L 3 3, L 2 5 L 3 4, L 2 4 L 3 5, L 2 3 L 3 6, L 2 2 L 3 7, L 2 7 L 3, L 2 6 L 3 2, L 2 5 L 3 3, L 2 4 L 3 4, L 2 3 L 3 5, L 2 2 L 3 6, L 2 L 3 7, L 2 7, L 2 6 L 3, L 2 5 L 3 2, L 2 4 L 3 3, L 2 3 L 3 4, L 2 2 L 3 5, L 2 L 3 6, L 3 7, L 2 6, L 2 5 L 3, L 2 4 L 3 2, L 2 3 L 3 3, L 2 2 L 3 4, L 2 L 3 5, L 3 6, L 2 5, L 2 4 L 3, L 2 3 L 3 2, L 2 2 L 3 3, L 2 L 3 4, L 3 5, L 2 4, L 2 3 L 3, L 2 2 L 3 2, L 2 L 3 3, L 3 4, L 2 3, L 2 2 L 3, L 2 L 3 2, L 3 3, L 2 2, L 2 L 3, L 3 2, L 2, L 3, 1] T.

47. 47 3-S: 1 st approach cont.

48. 48 3-S: 1 st approach cont.

49. 49 3-S: 1 st approach cont.

50. 50 3-S: 1 st approach cont.

51. 51 3-S: 1 st approach cont. In order for a solution for the vector λ 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 expand the determinant symbolically. A numerical case was analyzed and a polynomial of degree 158 in the variable L 1 was obtained. A numerical example is presented next.

52. 52 3-S: 1 st approach cont. Numerical Example : strut lengths: a12 = 14 in.a34 = 12 in. non-compliant tie lengths: a41 = 10 in. spring 1,2 and 3 free length & spring constant: L01 = 8 in. k1 = 1 lbf/in. L02 = 2 in. k2 = 2.687 lbf/in. L03 = 2.5 in. k3 = 3.465 lbf/in.

53. 53 3-S: 1 st approach cont. Solution : (3-13), (3-21), and (3-24) 100 L 3 4 + [(-L 1 2 + 96) L 2 2 + 44 L 1 2 – 52224] L 3 2 + L 1 2 L 2 4 + (L 1 4 – 440 L 1 2 -13824) L 2 2 – 8624 L 1 2 + 6773760 = 0 (3-33) (2.5 L 1 L 2 2 + 90 L 1 – 1600) L 3 3 + (-8.663 L 1 L 2 2 + 381.165 L 1 ) L 3 2 + [-2.5 L 1 L 2 4 + (-6 L 1 3 + 8 L 1 2 + 1196 L 1 – 768) L 2 2 + 44 L 1 3 – 352 L 1 2 – 30664 L 1 + 417792] L 3 + 8.663 L 1 L 2 4 + (17.326 L 1 3 – 3811.651 L 1 ) L 2 2 – 74708.354 L 1 = 0 (3-34) [(2.5 L 1 2 + 160) L 2 – 1074.637] L 3 3 + (831.633 – 8.663 L 1 2 ) L 2 L 3 2 + [(-7 L 1 2 + 192) L 2 3 + (-515.826 + 5.373 L 1 2 ) L 2 2 + (-2.5 L 1 4 + 1188 L 1 2 – 69888) L 2 – 236.420 L 1 2 + 280609.161] L 3 + 17.326 L 1 2 L 2 3 + (- 119755.135 + 8.663 L 1 4 – 3811.651 L 1 2 ) L 2 = 0. (3-35)

54. 54 3-S: 1 st approach cont. Continuation method is applied and solutions were obtained. Only one of the solutions was acceptable (Force Balance). Result: L1 = 13.0 L2 = 8.0 L3 = 7.017 Next, 2 nd approach applied to verify the results.

55. 55 3-S: 2 nd approach Problem Statement : The problem statement is presented bellow: given: a 41, a 12, a 34 k 1, k 2, k 3, L 01, L 02, L 03 find: cos  4, cos  1 when the system is in equilibrium

56. 56 3-S: 2 nd approach, cont. Solution : It is a two d.o.f. system obtain expressions for L 1, L 2, and L 3 in terms of cos  4 and cos  1 write the potential energy equation determine values of cos  4 and cos  1 such that dU/dcos  4 = dU/d cos  1 = 0

57. 57 3-S: 2 nd approach, cont. Geometry equation : quadrilateral 1-2-3-4 can be written as: Where, X 4 = a 34 s 4 Y 4 = -(a 41 + a 34 c 4 ) (3.36) (3.37) (3.38) (3.39)

58. 58 3-S: 2 nd approach, cont. Rewrite the quadrilateral cosine law as: Square both sides and substitute cosine for sines (3.40) (3.41)

59. 59 3-S: 2 nd approach, cont. Obtain an equation w.r.t L23. L 3 4 + A L 3 2 + B = 0 (3.42) A = A 1 c 4 + A 2 c 1 + A 3 c 1 c 4 + A 4 (3.43) Where, B = B 1 c 12 + B 2 c 12 c 4 + B 3 c 1 + B 4 c 1 c 4 + B 5 c 1 c 42 + B 6 c 42 + B 7 c 4 + B 8 (3.44)

60. 60 3-S: 2 nd approach, cont. A cosine law for triangle 4-1-2 may be written as: A cosine law for the triangle 3-4-1 may be written as: (3.46) (3.47)

61. 61 3-S: 2 nd approach, cont. The total potential energy stored in all three springs is given by: U = ½ k 24 (L 24 -L 024 ) 2 + ½ k 23 (L 23 -L 023 ) 2 + ½ k 31 (L 31 -L 031 ) 2 (3.48)

62. 62 3-S: 2 nd approach, cont. The differentiation of U with respect to c1 and c4 gives: dU/dc 4 = k 24 (L 24 – L 024 ) dL 24 /dc 4 + k 23 (L 23 – L 023 ) dL 23 /dc 4 + k 31 (L 31 – L 031 ) (3.49) dU/dc 1 = k 24 (L 24 – L 024 ) dL 24 /dc 1 + k 23 (L 23 – L 023 )dL 23 /dc 1 + k 31 (L 31 – L 031 ) (3.50)

63. 63 3-S: 2 nd approach, cont. dU/dc 4 = 0, This equation does not contain L 24. dU/dc 1 = 0, This equation does not contain L 31. From (3.46) and (3.47), it is apparent that dL 1 /dc 4 and dL 2 /dc 1 are zero. So,

64. 64 3-S: 2 nd approach, cont. (p 1 c 1 2 + p 2 c 1 + p 3 ) c 4 3 + (p 4 c 1 3 + p 5 c 1 2 + p 6 c 1 + p 7 ) c 4 2 + (p 8 c 1 4 + p 9 c 1 3 + p 10 c 1 2 + p 11 c 1 + p 12 ) c 4 + (p 13 c 1 4 + p 14 c 1 3 + p 15 c 1 2 + p 16 c 1 + p 17 ) = 0 (3.65) (q 1 c 1 + q 2 ) c 4 4 + (q 3 c 1 2 + q 4 c 1 + q 5 ) c 4 3 + (q 6 c 1 3 + q 7 c 1 2 + q 8 c 1 + q 9 ) c 4 2 + (q 10 c 1 3 + q 11 c 1 2 + q 12 c 1 + q 13 ) c 4 + (q 14 c 1 3 + q 15 c 1 2 + q 16 c 1 + q 17 ) = 0 (3.66)

65. 65 3-S: 2 nd approach, cont. Next, factor (3.7) into the following form, (r 1 c 1 + r 2 ) c 4 2 + (r 3 c 1 2 + r 4 c 1 + r 5 ) c 4 + (r 6 c 1 2 + r 7 c 1 + r 8 ) = 0 (3.69) Note: Equations (3.65), (3.66), and (3.69) represent three equations in the three unknowns L 3, c 1, and c 4.

66. 66 3-S: 2 nd approach, cont. To apply the Sylvester method, equations 3.65, 3.66, and 3.69 are multiplied by powers of c 1 and c 4. This step is to create a sufficient equation set in the variable c 4 and c 1 where the parameter L 3 is embedded in the coefficients.

67. 67 3-S: 2 nd approach, cont. Equation (3.65) is multiplied by c 1, c 1 2, c 4, c 4 2, c 4 3, c 1 4, c 1 2 c 4, c 1 c 4 2, and c 1 2 c 4 2 to obtain 10 equations including it. Equation (3.69) is multiplied by c 1, c 1 2, c 1 3, c 1 4, c 4, c 4 2, c 4 3, c 4 4, c 1 c 4, c 1 2 c 4, c 1 3 c 4, c 1 4 c 4, c 1 c 4 2, c 1 2 c 4 2, c 1 3 c 4 2, c 1 4 c 4 2, c 1 c 4 3, c 1 2 c 4 3, c 1 3 c 1 4, c 1 c 4 4 and c 1 2 c 4 4 to obtain 22 equations including it. Equation (3.66) is multiplied by c 1, c 1 2, c 1 3, c 4, c 4 2, c 1 c 4, c 1 2 c 4, c 1 3 c 4, c 1 c 4 2 and c 1 2 c 4 2 to obtain 11 equations including it.

68. 68 3-S: 2 nd approach, cont. To show this 43 by 43 coefficient matrix, M, it is divided as a combination of sub-matrices Mij. where,

69. 69 3-S: 2 nd approach, cont. M11, M12, M13, M21, M22, M23, M31, M32, M33 are 11 by 11 matrices. M14, M24, M34 are 11 by 10 matrices. M41, M42, M43 are 10 by 11 matrices. M44 is a 10 by 10 matrix.

70. 70 3-S: 2 nd approach, cont. And all sub-matrices are shown bellow:

71. 71 3-S: 2 nd approach, cont.

72. 72 3-S: 2 nd approach, cont.

73. 73 3-S: 2 nd approach, cont.

74. 74 3-S: 2 nd approach, cont.

75. 75 3-S: 2 nd approach, cont.

76. 76 3-S: 2 nd approach, cont.

77. 77 3-S: 2 nd approach, cont.

78. 78 3-S: 2 nd approach, cont.

79. 79 3-S: 2 nd approach, cont.

80. 80 3-S: 2 nd approach, cont.

81. 81 3-S: 2 nd approach, cont.

82. 82 3-S: 2 nd approach, cont.

83. 83 3-S: 2 nd approach, cont.

84. 84 3-S: 2 nd approach, cont.

85. 85 3-S: 2 nd approach, cont.

86. 86 3-S: 2 nd approach, cont. Equating the determinant of the coefficient matrix Mij to zero will yield a polynomial in the single variable L 23. A numerical example is presented next.

87. 87 3-S: 2 nd approach, cont. The following parameters were selected in analysis of numerical example 2: strut lengths: a34 = 12.0 in.a12 = 14.0 in. non-compliant tie lengths: a41 = 10.0 in. spring 1 free length & spring constant: L024 = 8.0 in.k1 = 1.0 lbf/in. spring 2 free length & spring constant: L031 = 2.6865 in. k2 = 2.0 bf/in. spring 3 free length & spring constant: L023 = 3.4651 in. k2 = 2.5 lbf/in.

88. 88 3-S: 2 nd approach, cont. Find: L 3, c 1, c 4. Solution: The result was 136 solutions for L 3 which 7 were real.

89. 89 3-S: 2 nd approach, cont. Case 1 is presented here (7 real cases): Cosine θ 1 : -0.453571523 Cosine θ 4 : -0.749999925 L 3 : 7.01658755

90. 90 3-S: 2 nd approach, cont. The results of Case 1 is used to calculate L 1, L 2, and L 3 presented here: L 1 : 13.0000000 L 2 : 7.99999999 L 3 : 7.01658830

91. 91 3-S: 2 nd approach, cont. Force balance is applied to each case to verify if the structure is in equilibrium. The summation of forces at point 2 and 3 is expected to become zero if the structure is at equilibrium. The force in each spring is calculated using F = k (Lfinal – Linitial).

92. 92 3-S: 2 nd approach, cont.

93. 93 3-S: 2 nd approach, cont. Testing of real values in second differentiation of energy equation to determine if potential energy of the structure is minimum or maximum. The structure with real values, pass force balance test and minimum potential energy is considered an acceptable solution.

94. 94 3-S: 2 nd approach, cont. In these numerical examples : Obtained same results using two different approaches Satisfied force balance equilibrium evaluation Was at minimum potential energy Presents a stable structure.

95. 95 3-S: 2 nd approach, cont. End of 3 Spring.

96. 96 Agenda Objective Literature review Analysis of planar tensegrity mechanisms –2-spring planar tensegrity –3-spring planar tensegrity –4-spring planar tensegrity (added to proposal) Future work, Summary and Conclusion

97. 97 4 spring planar tensegrity Analysis of a four elastic ties planar tensegrity mechanism to determine equilibrium configurations when no external forces or moments are applied. The equilibrium position is the potential energy stored in four springs is a minimum. A polynomial expressed in terms of the length of one of the springs is developed.

98. 98 4-S given:a 12, a 34 lengths of struts, find: At equilibrium position L 1, L 2, L 3, and L 4 length of spring 1, 2, 3 and 4. k 1, L 01 spring between points 4 and 2, k 2, L 02 spring between points 3 and 1, k 3, L 03 spring between points 3 and 2, k 4, L 04 spring between points 4 and 1.

99. 99 4-S: cont. 4 3 2 1 L1L1 L2L2 L3L3 L4L4 a 12 a 34

100. 100 4-S: cont. Geometry equation : G 1 (L 1 4 L 2 4 ) + G 2 (L 1 4 L 2 2 ) + G 3 (L 3 2 L 1 2 L 2 2 ) + G 4 (L 1 2 L 2 2 ) + G 5 (L 3 2 L 2 2 ) + G 6 (L 2 2 ) + G 7 ( L 3 2 L 1 2 ) + G 8 ( L 1 2 ) + G 9 ( L 3 4 ) + G 10 ( L 3 2 ) + G 1 1 = 0 (4-1)

101. 101 4-S: cont. The potential energy of the system can be evaluated as : At equilibrium, the potential energy will be a minimum.

102. 102 4-S: cont.

103. 103 4-S: cont.

104. 104 4-S: cont. Energy equations : D 1 (L 1 L 2 4 ) + D 2 (L 1 3 L 2 2 ) + D 3 (L 1 2 L 2 2 ) + D 4 (L 3 2 L 1 L 2 2 ) + D 5 (L 1 L 2 2 ) + D 6 (L 3 2 L 2 2 ) + D 7 (L 2 2 ) + D 8 (L 3 4 ) + D 9 (L 3 2 ) + D 10 + D 11 (L 3 2 L 1 3 ) + D 1 2 (L 1 3 ) + D 1 3 (L 3 2 L 1 2 ) + D 1 4 (L 1 2 ) + D 1 5 (L 3 4 L 1 ) + D 1 6 (L 3 2 L 1 ) + D 1 7 (L 1 ) = 0 (4-10) E 1 (L 1 2 L 2 3 ) + E 2 (L 3 2 L 2 3 ) + E 3 (L 2 3 ) + E 4 (L 1 2 L 2 2 ) + E 5 (L 3 2 L 2 2 ) + E 6 (L 2 2 ) + E 7 (L 1 4 L 2 )+ E 8 (L 3 2 L 2 ) + E 9 (L 1 2 L 2 ) + E 10 (L 3 4 L 2 ) + E 11 (L 3 2 L 1 2 L 2 ) + E 12 (L 2 ) + E 13 (L 3 2 )+ E 14 (L 1 2 ) + E 15 (L 3 4 ) + E 16 = 0 (4-12) F 1 (L 1 2 L 2 2 ) + F 2 (L 3 L 1 2 L 2 2 ) + F 3 (L 3 3 L 2 2 ) + F 4 (L 3 2 L 2 2 ) + F 5 (L 3 L 2 2 ) + F 6 (L 2 2 ) + F 7 (L 3 3 L 1 2 ) + F 8 (L 3 L 1 2 ) + F 9 (L 1 2 ) + F 10 (L 3 5 ) + F 11 (L 3 4 ) + F 12 (L 3 3 ) + F 13 (L 3 2 )+ F 14 (L 3 ) + F 15 = 0 (4-14)

105. 105 4-S: cont. Table 4-1 presents coeffecient G, D, E, and F used in the matrix. A numerical example is presented next.

106. 106 4-S: cont. Numerical Example: strut lengths: a12 = 14 in.a34 = 12 in. spring 1 free length & spring constant: L01 = 8 in.k1 = 1 lbf/in. L02 = 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.

107. 107 4-S: cont. Eighteen real roots have been found using Polynomial Continuation Method. The eighteen real roots for this example are presented in Table (4-2) of Chapter 4. Values in Table (4-2) can be either for maximum or minimum energy. The second derivative of a polynomial determines its curvature at specified point in Cartesian coordinate system. These values are tested in second differentiation of energy equation (4-20) with respect to L 1, L 2, and L 3.

108. 108 4-S: cont. The values of L 1, L 2, L 3, and L 4 listed in Table (4-2) satisfy the geometric constraints defined by equation (4-1). Each case of Table (4-2) must be analyzed to determine if the mechanism is in a maximum or minimum potential energy state. This is readily accomplished by evaluating the value of the second derivative of potential energy taken with respect to a change in length of spring 1, 2, and 3. 2 nd derivatives are next.

109. 109 4-S: cont.

110. 110 4-S: cont. Table (4-3) of Chapter 4 shows the value of the potential energy and the second derivative for each of the eighteen real cases. The only acceptable cases are: 2, 3, 13, 15, and 17. These cases are shown in Figures 8 through 12, bellow:

111. 111 4-S: cont. Each of these five configurations was evaluated to determine if it was in equilibrium by determining if forces in the two struts and four elastic ties could be calculated such that the sum of forces at each of the four node points was zero. Table (4-4) presents four equilibrium cases of a tensegrity structure consists of two struts and four elastic ties for the given initial values. Acceptable geometry is next.

112. 112 4-S: cont.

113. 113 4-S: cont. End of Chapter 4.

114. 114 Agenda Objective Literature review Analysis of planar tensegrity mechanisms –2-spring planar tensegrity –3-spring planar tensegrity –4-spring planar tensegrity (added to proposal) Future work, Summary and Conclusion

115. 115 3-spring spatial tensegrity system

116. 116 3-spring spatial tensegrity system The problem statement is given as follows: Given: The length of non-compliant ties, struts, and initial length of compliant ties. Find:The final length of compliant ties.

117. 117 3-spring spatial tensegrity system The following three approaches were considered: minimum potential energy analysis force balance analysis linear dependence of 6 connector lines. Mathematical solution became very complicated.

118. 118 3-spring spatial tensegrity system 5.1.1: 3D Platform Tensegrity: Future suggested research work is a position analysis of a general three dimensional parallel platform device consisting of 3 struts, 3 springs, and 6 ties. 5.1.2: 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 in presented research. Other mathematical methods might exist to solve these problems with less complexity.

119. 119 Conclusion 5.2: Summery and Conclusion: In this research, all possible combination of two dimensional tensegrity structures consists of struts, springs, and non compliance members were reviewed. A closed form solution could is obtained for case of 2 spring, 2 struts, and 2 non compliance members.

120. 120 Continue Finding a simple formula to represents a closed form solution to the other 2 dimensional tensegrity structures consisting of either 3 or 4 springs were not possible due to the complexity of mathematical solutions. A combination of mathematical, numerical and use of engineering software namely Maple, AutoCAD are used to suggest a procedure to solve these problems and verify the solutions.

121. 121 Continue The benefit of this research will be to set a basic mathematical foundation to the closed form solutions of tensegrity structures. The contribution of this research will be to provide a mathematical solution and/or analytical procedure for analysis of more complex structures in tensegrity structures hence robotic.

122. 122 Regards The author expressly thanks his committee members for their valuable suggestions on proposal, their time and effort to review this dissertation.

123. 123 End of presentation Thank you.