DEPARTMENT OF MECHANICAL AND AEROSPACE ENGINEERING UNIVERSITY OF FLORIDA

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

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

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

7. triangle quadrilateral pentagon a23 ties 2 3 struts hexagon a34 a12 Figure 1: Family of Tensegrity Structures a41 4 1 Figure 2: Planar Tensegrity Structure Basic 3D and 2D Tensegrity Structures

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. 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. 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. 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. Two-Spring Tensegrity System • given: • a12, a34 strut lengths • a23, a41 non-compliant tie lengths • k1, L01 k2, L02 spring parameters • solution 1, find: • L1, L2 at equilibrium • solution 2, find: • c4, c1 at equilibrium 3 a23 2 L2 L1 a12 a34 4 a41 1

13. Two-Spring Tensegrity, Solution 1 • obtain geometric equation f1(L1, L2) = 0 • write potential energy equation U = ½ k1 (L1-L01)2 + ½ k2 (L2-L02)2 • evaluate this may be written as f2(L1, L2) = 0 • solve the two equations for all sets of L1, L2 • resulted in 28th degree polynomial in L1

14. Approach 1 • given: • a12, a34 strut lengths • a23, a41 non-compliant tie lengths • k1, L01 k2, L02 spring parameters • find: • L1, L2 at equilibrium 3 a23 2 L2 L1 a12 a34 4 a41 1

15. 3 3 a23 a23 2 2 L2 L1 L1 a12 a34 a34 '4 4 4 a41 1 '4 4 "4 Figure 3: Planar Tensegrity Structure Figure 4: Triangle 4-3-2 Geometric Constraint 4 + 4' =  + 4" (2-1) (2-3)

16. 2 a12 L1 3 4 a41 1 "4 a34 Figure 5: Triangle 4-1-2 L2 4 1 a41 4 Figure 6: Triangle 4-1-3 Geometric Constraint – Cont. (2-5) (2-7)

17. Geometric Constraint – Cont. 4 + 4' =  + 4" 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 . (cos4)2 (cos’4)2 + 2 cos4 cos’4 cos”4 + (cos”4)2 = (1-cos24) (1-cos2’4) (2-12)

18. Geometric Constraint – Cont. • substituting (2-3), (2-5), and (2-7) into (2-12) and gives geometry equation. where and B2, B0, C2, and C0 are expressed in terms of known quantities A L24 + B L22 + C = 0 (2-13) A = L12 , B = L14 + B2 L12 + B0, C = C2 L12 + C0

19. Potential Energy Constraint • at equilibrium, the potential energy in the springs will be a minimum U = ½ k1 (L1-L01)2 + ½ k2 (L2-L02)2 • the mechanism being considered is a one degree of freedom device • one parameter can be selected as the generalized coordinate for the problem; L1

20. Potential E. Constraint – Cont. • at a minimum potential energy state, • dL2/dL1 can be obtained via implicit differentiation of the geometry constraint as (2-19) (2-20)

21. Potential E. Constraint – Cont. • substituting (2-20) into (2-19) gives D L25 + E L24 + F L23 + G L22 + H L2 + J = 0 (2-21) where the coefficients D through J are polynomials in L1 • equations (2-13) and (2-21) represent two equations in the two unknowns L1 and L2 • Sylvester’s elimination method is used to obtain values for these parameters that simultaneously satisfy both equations

22. Sylvester’s elimination • determinant of coefficient matrix must equal zero which yields a 28th degree polynomial in L1

23. Symbolic Expansion • determinant of coefficient matrix must equal zero which yields a 28th degree polynomial in L1 • Maple program used to obtain all coefficients symbolically • corresponding values of L2 for each value of L1 can be readily obtained from solving (2-13) and then (2-21)

24. 2.3.4 Numerical Example • given: • a12 = 3 in. a34 = 3.5 in. • a41 = 4 in. a23 = 2 in. • L01 = 0.5 in. k1 = 4 lbf/in. • L02 = 1 in. k2 = 2.5 lbf/in. • find L1 and L2 at equilibrium

25. Numerical Example • results • coefficients of 28th degree polynomial in L1 obtained • 8 real roots for L1 with corresponding values for L2 • 4 cases correspond to minimum potential energy

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

27. Force Balance Verification

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 L1 and L2 • elimination of L2 resulted in a 28th degree polynomial in the single variable L1 • numerical example presented showing 4 real solutions • other examples have been investigated, but all gave 4 real solutions

29. Approach 2 • The objective of this approach is to again investigate, in closed-form, the planar 2-spring tensegrity system. • Determine (c4 and c1) to Minimize Potential Energy.

30. a23 3 2 L2 L1 a34 a12 1 4 a41 1 4 Geometry

31. Approach 2 The problem statement is written as: given: a41, a12, a23, a34 k1, k2, L01, L02 find: cos4 (and corresponding value of cos1) when the system is in equilibrium

32. Two-Spring Tensegrity, Solution 2 • obtain geometric equations f1(c4, c1) = 0, f2(c4, L2) = 0, f3(c1, L1) = 0, • write potential energy equation U = ½ k1 (L1-L01)2 + ½ k2 (L2-L02)2 • evaluate this may be written as f4(c4, c1, L1, L2) = 0 • use f2 and f3 to eliminate L1 and L2 from f4 • solve this equation and f1 for all sets of c4, c1 • resulted in 32nd degree polynomial in c4

33. Solution Approach 2 • from cosine law for planar quadrilateral which can be factored as A c12 + B c1 + D = 0 (2.37) where A = A1 c4 + A2 B = B1 c42 + B2 c4 + B3

34. Solution Approach 2 • from derivative of potential energy C10 c110 + C9 c19 + C8 c18 + C7 c17 + C6 c16 + C5 c15 + C4 c14 + C3 c13 + C2 c12 + C1 c1 + C0 =0 (2-55) where the coefficients Ci are functions of c4

35. Solution Approach 2

36. Solution Approach 2 • Expansion of the 12×12 determinant yields a 32nd degree polynomial in the parameter c4. • Using earlier numerical values, there are 8 real values and 20 complex values for c4. • Four of real values are identical to real values in approach 1. • Table on next page presents real values of second approach.

37. Solution Approach 2

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. L23 3 2 L31 L24 a34 a12 1 4 a41 1 4 3-spring planar tensegrity • closed-form analysis of a three spring, two strut tensegrity system with one non-compliant element

40. 3-S: 1st approach The problem statement can be explicitly written as: Given: a12, a34 lengths of struts, a41 lengths of non-compliant tie k1, L01 spring constant and free length, point 4 to 2 k2, L02 spring constant and free length, point 3 to 1 k3, L03 spring constant and free length, point 2 to 3 Find: L1 L2 L3 length of springs at equilibrium position,

41. 3-S: 1st approach cont.

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

43. 3-S: 1st approach cont. 3.2.1 Development of Geometric Equation, G1 L34 + (G2 L22 + G3) L32 + (G4 L24 + G5 L22 + G6) = 0 (3-13) 3.2.2 Development of Potential Energy Equations (3-16)

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

45. 3-S: 1st approach cont. Substituting, results (3-21) and (3-24) : (D1 L22+D2) L33 + (D3 L22+D4) L32 + (D5 L24+D6 L22+D7) L3 + (D8 L24+D9 L22+D10) = 0 (3-21) (E1 L2 + E2) L33 + (E3 L2) L32 + (E4 L23 + E5 L22 + E6 L2 + E7) L3 + (E8 L23 + E9 L2) = 0 (3-24)

46. 3-S: 1st approach cont. Creating solution matrix M52x52 : Mλ = 0 M = λ = [L27L33, L25L35, L23L37, L27L32, L26L33, L25L34, L24L35, L23L36, L22L37, L27L3, L26L32, L25L33, L24L34, L23L35, L22L36, L2L37, L27, L26L3, L25L32, L24L33, L23L34, L22L35, L2L36, L37, L26, L25L3, L24L32, L23L33, L22L34, L2L35, L36, L25, L24L3, L23L32, L22L33, L2L34, L35, L24, L23L3, L22L32, L2L33, L34, L23, L22L3, L2L32, L33, L22, L2L3, L32, L2, L3, 1]T .

47. 3-S: 1st approach cont.

48. 3-S: 1st approach cont.

49. 3-S: 1st approach cont.

50. 3-S: 1st approach cont.