Kinestatic Analyses of Mechanisms with Compliant Elements

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1 Kinestatic Analyses of Mechanisms with Compliant Elements Carl Crane How can such a simple mechanism have such a high order solution?

2 Tensegrity structures comprised of struts in compression and ties in tension

3 Self-deployable tensegrity structures certain ties replaced by elastic members

4 Can we solve the basic problem? determine in closed-form all equilibrium configurations of a self-deployable tensegrity structure given: strut lengths tie lengths free lengths and spring constant of elastic members any applied loads Stern [1999] performed closed-form analysis of unloaded symmetric systems solutions, n=..6 Correa [001] obtained numerical solution for general loaded systems numerical convergence to a solution

5 Let s start with a warm up problem. Planar -strut -spring tensegrity structure given: strut lengths a 1, a tie lengths a 1, a spring parameters k 1, L 01, k, L 0 determine: all equilibrium poses tie a L 1 a struts a 1 a 1 1 L elastic tie

6 Planar -strut -spring tensegrity structure the struts and non-elastic ties form a simple -bar mechanism pose can be defined by one parameter several descriptive parameters tried analysis was performed using an energy method and using L 1 as the descriptive parameter L 1 a a 1 L a a 1 1

7 Geometric constraints A L + B L + C = 0 where A = L 1, B = L 1 + B L 1 + B 0, C = C L 1 + C 0 and where B, B 0, C, and C 0 are expressed in terms of known quantities

8 Verification of geometric constraint equation L 1 L L 1 L L 1 L 1 L L 1 Figure 7: Possible Configurations for Numerical Example

9 Potential energy constraint at equilibrium, the potential energy in the springs will be a minimum U = ½ k 1 (L 1 -L 01 ) + ½ k (L -L 0 ) at a minimum potential energy state, (17) dl /dl 1 can be obtained via implicit differentiation of the geometry constraint (1) as (18)

10 Geometry and Potential energy constraints geometry constraint A L + B L + C = 0 (1) potential energy constraint D L 5 + E L + F L + G L + H L + J = 0 (19) where the coefficients D through J are polynomials in L 1 Sylvester s elimination method can be used to identify the condition that the coefficients must satisfy in order for (1) and (19) to have common roots for L multiply (1) by L, L, L, L multiply (19) by L, L, L obtain 9 homogeneous equations in 9 unknowns

11 Sylvester s elimination tie a struts L 1 elastic tie determinant of coefficient matrix must equal zero which yields a 8 th degree polynomial in L 1 a a 1 L a 1 1

12 Numerical example given: a 1 = in. a =.5 in. a 1 = in. a = in. L 01 = 0.5 in. k 1 = lbf/in. L 0 = 1 in. k =.5 lbf/in. find L 1 and L at equilibrium a tie struts L 1 elastic tie a a 1 L a 1 1

13 Numerical Example results coefficients of 8 th degree polynomial in L 1 obtained 8 real roots for L 1 with corresponding values for L all 0 complex solution pairs (L 1, L ) satisfied equations (1) and (19), i.e. geometry constraint and du/dl 1 = 0 cases correspond to minimum potential energy Case L 1, in. L, in

14 Numerical Example 1 1 Case Case spring in compression with a negative spring length spring in tension 1 Case 5 Case 6 1

15 Let s look at a second simple problem. In many papers involving compliant elements, researchers assume that their springs have a free length of zero. How much more complicated does the problem become if the spring free lengths are not zero? x 5 6 y 1 γ 1 γ L L 1 L x 1 y 1

16 Problem Statement given: L 1, p x, p y L 5, p 6x, p 6y L k 1, L 01 k, L 0 find: γ 1 and γ for all equilibrium configurations x x 5 6 y 1 γ 1 γ L L 1 L y 1

17 Case 1: Numerical Example L 1 = 6 m, p x = -1.5 m, p y = m, L 5 = 5.5 m, p 6x = -1.1 m, p 6y = -.1 m L = 10 m k 1 = N/m, L 01 = 0, k =.5 N/m, L 0 = 0 Solution # γ 1, radians γ, radians i i i i case 1 case case case

18 Case : L 0 = 0, L 01 0 (E 1 x + E x + E ) d 1 + E x + E 5 x + E 6 = 0 (F 1 x + F x + F ) d 1 + F x + F 5 x + F 6 = 0 (G 1 x + G x + G ) d 1 + G x + G 5 x + G 6 = 0 Sylvester s Solution method First two equations are multiplied by x, d 1, and d 1 x, d 1 and d 1 x. Third equation multiplied by x, d 1, and d 1 x. Results in 16 homogeneous equations in 16 unknowns. The determinant of the coefficient matrix must equal zero. Resulted in a nd degree polynomial in x 1. Divided by (1+x 1 ) and by square of the nd degree polynomial corresponding to d = 0. 0 th degree solution 18

19 Case : Numerical Example L 01 =. m, 8 real solutions case 1 case case case case 5 case 6 case 7 case 8 19

20 Case : L 0 0, L 01 0 Will obtain equations of the form (C 1 x + C x + C ) + (C x + C 5 x + C 6 ) d i + (C 7 x + C 8 x + C 9 ) d 1i = 0 (D 1 x + D x + D ) + (D x + D 5 x + D 6 ) d i + (D 7 x + D 8 x + D 9 ) d 1i = 0 (1) () (M 1 x + M x + M ) + (M x + M 5 x + M 6 ) d 1i = 0 () (N 1 x + N x + N ) + (N x + N 5 x + N 6 ) d i = 0 () where the coefficients are functions of x 1 0

21 Case : L 0 0, L 01 0 Multiply Equations (1), () by {1, d 1i, d i, d 1i, d i, d 1i d i, d 1i d i, d 1i d i } *{1,x } Multiply Equation () by {1, d 1i, d i, d 1i d i, d i } *{1,x } Multiply Equation () by {1, d 1i, d i, d 1i d i, d 1i } *{1,x } Total Equations = 5 Unknowns {1, d 1i, d i, d 1i, d i, d 1i d i, d 1i d i, d 1i d i, d 1i, d i, d 1i d i, d 1i d i, d 1i d i }*{x, x, x, 1} 1

22 Case : L 0 0, L 01 0 Expansion of M = 0 yielded a 10 th degree polynomial in x 1. This can be divided by (1+x 1 ) 1 to get 78 th degree polynomial in x 1. Of the 78 solutions 16 were extraneous. 6 solutions were obtained. Numerical continuation method * found 88 solutions (6 + 6 circular points). * PHCpack software, Jan Verschelde, U. Illinois, Chicago

23 Case : L 0 0, L 01 0 L 01 = 5.1 m, L 0 = m Out of the 6 solutions 8 were complex and were real All the solutions satisfy the four equations

24 Case : Real Solutions - I

25 Case : Real Solutions - II

26 Conclusion Case 1, L 01 = L 0 = 0 6 solutions Case : L 0 = 0, L solutions x 5 6 y γ L L 1 L Case : L 0 0, L γ 1 6 solutions y 1 x 1 6

27 given: Back to the primary problem. strut lengths top tie lengths bottom tie lengths spring constant and free length of side ties find: all equilibrium configurations

28 How to structure the problem? First, attempt to solve the problem when all the spring free lengths equal zero.

29 Problem Formulation given: A j, a j, k j find: X j, Y j, Z j, x j, y j, z j such that is an extremum subject to for j = 1..n. This formulation was selected to avoid any need to use tan half-angle substitutions to convert trigonometric functions into polynomials. 9

30 Example Three strut tensegrity where spring free lengths equal zero. There are 9 unknowns, i.e. X j, Y j, Z j, j=1... There are 9 constraint equations: 0

31 Numerical Example Three strut tensegrity a 1 =10, a =1., a = 15 cm s 1 = 0, s =, s = 10.5 cm k 1 =.8, k =, k =. N/cm The homotopy continuation method* was used to solve the set of 9 equations in 9 unknowns. 10 real configurations were obtained which satisfy the 9 equations (160 complex solutions obtained) * PHCpack, Jan Verschelde, Univ. of Illinois at Chicago 1

32 Stability Analysis nd Order Analysis A nd order analysis was conducted to classify the real solutions as stable equilibrium unstable equilibrium neutral equilibrium a small perturbation will continuously deform the structure to another neutral equilibrium state have a statically balanced mechanism The equilibrium study becomes the examination of the positive definiteness of the Hessian matrix of the Lagrangian function w.

33 Numerical Example (cont) stability analysis of the 10 real solutions 1 is unstable (negative definite Hessian) 7 are directionally stable (indefinite Hessian) are stable (positive definite Hessian) unstable case

34 Numerical Example (cont) stability analysis of the 10 real solutions 1 is unstable (negative definite Hessian) 7 are directionally stable (indefinite Hessian) are stable (positive definite Hessian) stable cases

35 x 5 6 y 1 γ 1 γ L L 1 L y 1 8 th degree univariate polynomial numerical example had 8 real roots, of which were stable equilibrium 1 of the stable equilibrium had both spring lengths > 0 x 1 Case 1, L 01 = L 0 = 0 6 solutions, real Case : L 0 = 0, L solutions, 8 real Case : L 0 0, L solutions, real problem formulation has extraneous roots free lengths equal zero 9 equations in 9 unknowns continuation method yielded 10 real and 160 complex solutions cases were in stable equilibrium

Kinetostatic Analysis and Solution Classification of a Planar Tensegrity Mechanism

Kinetostatic Analysis and Solution Classification of a Planar Tensegrity Mechanism P. Wenger 1 and D. Chablat 1 1 Laboratoire des Sciences du Numérique de Nantes, UMR CNRS 6004 Ecole Centrale de Nantes,