department of mechanical and aerospace engineering university of florida

DEPARTMENT OF MECHANICAL AND AEROSPACE ENGINEERING

UNIVERSITY 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

Dissertation Objective

• determine all equilibrium poses for 2-spring, 3-spring, and 4-spring planar tensegrity mechanisms

spatial tensegrity

struts

ties

planar 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

Basic 3D and 2D Tensegrity Structures

triangle

quadrilateral

pentagon

hexagon

Figure 1: Family of Tensegrity Structures

struts

ties

Figure 2: Planar Tensegrity Structure

1

4

2 3a23

a41

a12

a34

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




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

1

L2L1

a41

a12

a23

a34

2

3

4

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

0dL

dU

1

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

1

L2L1

a41

a12

a23

a34

2

3

4

15

Figure 3: Planar Tensegrity Structure Figure 4: Triangle 4-3-2

a34

L1

a23

'4

2

3

41

L2L1

a41

a12

a23

a34

2

3

4'4

"44

Geometric Constraint

341

234

21

223

4 aL2

aLa'cos

4 + 4' = + 4" (2-3)(2-1)

16

L1

a12

a41

"4

1

4

2

Figure 5: Triangle 4-1-2 Figure 6: Triangle 4-1-3

4

3

1

4

L2

a41

a34

Geometric Constraint – Cont.

411

241

21

212

4 aL2

aLa"cos

4134

241

234

22

4 aa2

aaLcos

(2-7)(2-5)

17


cos (4 + 4') = cos ( + 4")

cos4 cos4' – sin4 sin4' = - cos4"

cos4 cos4' + cos4" = sin4 sin4' .

(cos4)2 (cos4')

2 + 2 cos4 cos4' cos4" + (cos4")2 = (sin4)

2 (sin4')2 .

4 + 4' = + 4"

(cos4)2 (cos’4)2 + 2 cos4 cos’4 cos”4 + (cos”4)2 = (1-cos24) (1-cos2’4)

(2-12)

18


• 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 ofknown quantities

A L24 + B L2

2 + C = 0 (2-13)

A = L12 , B = L1

4 + B2 L12 + B0, C = C2 L1

2 + 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

0dL

dL)LL(k)LL(k

dL

dU

1

202220111

1

))]aa)(aa()aaaaL2L(L[L

)]aa)(aa()aaaaL2L(L[L

dL

dL2

342

232

412

122

122

342

412

232

22

12

12

234

241

223

212

212

234

241

223

21

22

221

1

2

(2-19)

(2-20)

21

Potential E. Constraint – Cont.

• substituting (2-20) into (2-19) gives

D L25 + E L2

4 + F L23 + G L2

2 + 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

0

0

0

0

0

0

0

0

0

1

L

L

L

L

L

L

L

L

0000C0B0A

000JHGFED

000C0B0A0

00JHGFED0

00C0B0A00

0JHGFED00

0C0B0A000

C0B0A0000

JHGFED000

2

22

32

42

52

62

72

82

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

Case L1, in. L2, in.

1 -5.4854 2.3333

2 -5.3222 -2.9009

3 -1.7406 -1.4952

4 -1.5760 1.8699

5 1.6280 1.7089

6 1.8628 -1.3544

7 5.1289 -3.2880

8 5.4759 2.3938

26

spring in compression with a negative spring length

spring in tension

4

44

41

1 1

1

2

2

2

2

3

33

3

Case 3 Case 4

Case 5 Case 6

Numerical Example Cont.

27

Force Balance Verification

CaseForce in Spring 1,

labForce in Spring 2,

labForce in Strut a12,

lab

Force in Strut a34,

lab

Force in Tie a41,

lab

Force in Tie a23,

lab

3 + 8.9624 + 6.2379 -13.4186 -16.8097 9.1623 18.7570

4 + 8.3040 2.1749 -5.4965 -11.7069 4.1031 11.9485

5 4.5120 1.7722 -4.2448 -7.1095 3.0883 7.4454

6 5.4514 + 5.8860 -11.4149 -11.6983 7.2064 14.1181

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

• Determine (c4 and c1) to Minimize Potential Energy.

• The objective of this approach is to again investigate, in closed-form, the planar 2-spring tensegrity system.

30

Geometry

4

32

1

4

a41

a12a34L1

L2

a23

1

31

Approach 2

The problem statement is written as:

given: a41, a12, a23, a34

k1, k2, L01, L02

find: cos4 (and corresponding value of cos1) when the system is in equilibrium

32

Two-Spring Tensegrity, Solution 2

• obtain geometric equationsf1(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

0dc

dU

4

33

Solution Approach 2

• from cosine law for planar quadrilateral

which can be factored as2

aZ

223

41

A c12 + B c1 + D = 0 (2.37)

A = A1 c4 + A2

B = B1 c42 + B2 c4 + B3

where

34

Solution Approach 2

• from derivative of potential energy

C10 c110 + C9 c1

9 + C8 c18 + C7 c1

7 + C6 c16 + C5 c1

5 + C4 c1

4 + C3 c13 + C2 c1

2 + C1 c1 + C0 =0

where the coefficients Ci are functions of c4

(2-55)

35

Solution Approach 2

0

0

0

0

0

0

0

0

0

0

0

0

1

c

c

c

c

c

c

c

c

c

c

c

0CCCCCCCCCCC

CCCCCCCCCCC0

000000000DBA

00000000DBA0

0000000DBA00

000000DBA000

00000DBA0000

0000DBA00000

000DBA000000

00DBA0000000

0DBA00000000

DBA000000000

1

21

31

41

51

61

71

81

91

101

111

012345678910

012345678910

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

Case c4 (radian) c1 (radian) L1 (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

38

Agenda




39

3-spring planar tensegrity

• closed-form analysis of a three spring, two strut tensegrity system with one non-compliant element

4

32

1

4

a41

a12a34L24

L31

L23

1

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



Find: L1 L2 L3

length of springs at equilibrium position,

41

3-S: 1st approach cont.

42


Cosine law for triangles 2-4-3 and 4-1-2 :

2

L'cosaL

2

a

2

L 23

4341

234

21

2

a"cosaL

2

a

2

L 212

4411

241

21

4 + 4' = + 4"

(3-1)

(3-3)

(3-7)

43


3.2.1 Development of Geometric Equation,

G1 L34 + (G2 L2

2 + G3) L32 + (G4 L2

4 + G5 L22 + G6) = 0

3.2.2 Development of Potential Energy Equations

20333

20222

20111 )LL(k

2

1)LL(k

2

1)LL(k

2

1U

(3-13)

(3-16)

44


• At equilibrium, the potential energy will be a minimum.

0L

L)LL(k)LL(k

L

U

1

303330111

1

0L

L)LL(k)LL(k

L

U

2

303330222

2

]GLGLGLLGLG2[L

]GLGLLGLLGLL2[L

L

L

b32

1a32

2b22

22

1a22

313

a62

2b54

22

3a32

32

2a22

22

11

1

3

]GLGLGLLGLG2[L

]GLGLLGLLGLL2[L

L

L

b32

1a32

2b22

22

1a22

313

c52

1b54

12

3b22

32

1a22

22

12

2

3

(3-17)

(3-18)

45


Substituting, results (3-21) and (3-24) :

(D1 L22+D2) L3

3 + (D3 L22+D4) L3

2 + (D5 L2

4+D6 L22+D7) L3 + (D8 L2

4+D9 L22+D10) = 0

(E1 L2 + E2) L33 + (E3 L2) L3

2 + (E4 L2

3 + E5 L22 + E6 L2 + E7) L3 + (E8 L2

3 + E9 L2) = 0

(3-21)

(3-24)

46


Creating solution matrix M52x52 :

M λ = 0

M =

2221

1211

MM

MM

λ = [L27L3

3, L25L3

5, L23L3

7, L27L3

2, L26L3

3, L25L3

4, L24L3

5, L2

3L36, L2

2L37, L2

7L3, L26L3

2, L25L3

3, L24L3

4, L23L3

5, L22L3

6, L2L3

7, L27, L2

6L3, L25L3

2, L24L3

3, L23L3

4, L22L3

5, L2L36, L3

7, L2

6, L25L3, L2

4L32, L2

3L33, L2

2L34, L2L3

5, L36, L2

5, L24L3,

L23L3

2, L22L3

3, L2L34, L3

5, L24, L2

3L3, L22L3

2, L2L33, L3

4, L2

3, L22L3, L2L3

2, L33, L2

2, L2L3, L32, L2, L3, 1]T .

47


000D0D000000D0D000D0D00000

0D0000D0D00000000000000000

0000D0D0000D0D000000000000

00000000000G0G0G0000000000

0000G0G0G00000000000000000

000G0G0G000000000000000000

0G000000000000000000000000

00000000000000000000000000

0000EEEE0000E0E00000000000

E0000E0E000000000000000000

0000E0E0000000000000000000

E0000000000000000000000000

00000000000000000000000000

00000D0D0000D0D00000000000

0000D0D0000000000000000000

D0000000000000000000000000

G0000000000000000000000000

00000000000000000000000000

00000000000000000000000000

00000000000000000000000000

00000000000000000000000000

00000000000000000000000000

00000000000000000000000000

00000000000000000000000000

00000000000000000000000000

00000000000000000000000000

263815

815

3815

124

124

124

4

235814

814

14

4

3815

15

5

4

11M

48


0000EE000000E0000EEEE00EE0

0000E00000EEEE000E0E000000

000000EE000000E0000EEEE0EE

E000000E00000EEEE000E0E000

00EEEE0000E0E0000000000000

E00000EEEE0000E0E000000000

000D0D0000D0D00000D0D00DD0

00000D0D0000D0D00000D0D0DD

D0000D0D000000D0D000D0D000

00D0D000000D0D000D0D000000

D000000D0D0000D0D000000000

000000000000G0G00000000GGG

00000G0G0000000000G0G0G000

G00000000000G0G0G000000000

0000G0G0000000000G0G0G0000

0000000000G0G0G00000000000

00000G0G0G0000000000000000

00G0G0G0000000000000000000

000000E00000EEEE000E0E0000

00000EEEE0000E0E0000000000

00000E00000EEEE000E0E00000

000EEEE0000E0E000000000000

EE0000E0E00000000000000000

000E0E00000000000000000000

0000D0D000000D0D000D0D0000

000000D0D0000D0D0000000000

796235814

6235814

796335814

96235814

235814

6235814

49263815

49263815

9263815

263815

63815

35124

35124

5124

35124

124

124

124

6235814

235814

6235814

235814

5814

14

263815

3815

21M

49


00000000000D0000D00000D0D0

00000D00D000D0D00D0D00000D

000000D000D0000D0D000D0D00

000000000000G0000000000G0G

00000000G00000000G0G000000

0000000G00000000G0G0000000

00000G000000G0G00000000G0G

000G000000G0G00000000G0G0G

00000000000000000EE00000E0

000000000000EE0000E0000EEE

00000000000EE0000E0000EEEE

00000000EE000E000EEEE000E0

000000EE000E000EEEE000E0E0

0000000D000D0000D0D000D0D0

000D00D000D0D00D0D00000D0D

00D0D00D0D0D0D0000D0D000D0

0000G000000G0G00000000G0G0

0G0000G0G000000G0G0G000000

00G0000G0G000000G0G0G00000

0000000EE000E000EEEE000E0E

000EE00E00EEEE00E0E0000000

0000EE00E00EEEE00E0E000000

0D0D00D0D0D0D0000D0D000D0D

0EE0E0EEEE0E0E000000000000

DD0D0DD0D000D0D00D0D000000

G00G0G0000G0G0G00000000000

10749

10749263

74926

635

635

635

63512

635124

796

796235

7962358

79623581

96235814

1074926

107492638

1074926381

63512

635124

635124

796235814

796235814

796235814

10749263815

796235814

10749263815

635124

12

10

7

M

50


00000000000000000000000000

0000000000000000000000EE00

00000000000000000000000000

0000000000000000000000000E

000000000000000EE00000E000

0000000000000000000EE00000

0000000000000000D00000D000

000000000000000000D00000D0

0000000000000D0000D00000D0

0000000000D0000D00000D0D00

000000000D000D0000D0D000D0

000000000000000000000000G0

000000000000000000G0000000

0000000000000G0000000000G0

00000000000000000G00000000

00000000000G0000000000G0G0

000000000G00000000G0G00000

000000G00000000G0G00000000

000000000000000000000000EE

000000000000000000EE00000E

00000000000000000000000EE0

0000000000000000EE00000E00

0000000000000EE0000E0000EE

0000000000EE0000E0000EEEE0

000000000000D0000D00000D0D

00000000D000D0000D0D000D0D

79

7

796

79

107

107

1074

10749

107492

6

6

63

6

635

635

635

79

796

79

796

79623

7962358

10749

1074926

22M

51


• 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 L1 was obtained.

• A numerical example is presented next .

52


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


Solution : (3-13), (3-21), and (3-24)

• 100 L34 + [(-L1

2 + 96) L22 + 44 L1

2 – 52224] L32 + L1

2 L24 + (L1

4 – 440 L1

2 -13824) L22 – 8624 L1

2 + 6773760 = 0 (3-33)

• (2.5 L1 L22 + 90 L1 – 1600) L3

3 + (-8.663 L1 L22 + 381.165 L1) L3

2 +

[-2.5 L1 L24 + (-6 L1

3 + 8 L12 + 1196 L1 – 768) L2

2 + 44 L13 – 352 L1

2 – 30664 L1 + 417792] L3 + 8.663 L1 L2

4 + (17.326 L13 – 3811.651 L1) L2

2 – 74708.354 L1 = 0 (3-34)

• [(2.5 L12 + 160) L2 – 1074.637] L3

3 + (831.633 – 8.663 L12) L2 L3

2

+ [(-7 L12 + 192) L2

3 + (-515.826 + 5.373 L12) L2

2 + (-2.5 L14 + 1188 L1

2 – 69888) L2 – 236.420 L1

2 + 280609.161] L3 + 17.326 L12 L2

3+ (-119755.135 + 8.663 L1

4 – 3811.651 L12) L2 = 0 . (3-35)

54


• 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, 2nd approach applied to verify the results.

55

3-S: 2nd approach

Problem Statement :

• The problem statement is presented bellow:

• given: a41, a12, a34

k1, k2, k3, L01, L02, L03

• find: cos4, cos1 when the system is in equilibrium

56

3-S: 2nd approach, cont.

Solution :• It is a two d.o.f. system

• obtain expressions for L1, L2, and L3 in terms of cos4 and cos1

• write the potential energy equation

• determine values of cos4 and cos1 such that dU/dcos4 = dU/d cos1= 0

57


• Geometry equation :

quadrilateral 1-2-3-4 can be written as:

Where,

2)(

212

414141241

aZcYsXaZ

X4 = a34 s4

Y4 = -(a41 + a34 c4)

(3.36)

(3.37)

(3.38)

(3.39)

2

LZ

23

41

58


44134

241

234

4 caa2

a

2

aZ

Rewrite the quadrilateral cosine law as:

Square both sides and substitute cosine for sines

(3.40)

(3.41)2

L

2

aZcYasXa

23

212

414121412

59


Obtain an equation w.r.t L23.

L34+ A L3

2 + B = 0 (3.42)

A = A1 c4 + A2 c1 + A3 c1 c4 + A4 (3.43)

Where,

B = B1 c12 + B2 c12 c4 + B3 c1 + B4 c1 c4 + B5 c1 c42 + B6 c42 + B7 c4 + B8

(3.44)

60


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)

2

Lcaa

2

a

2

a 21

14112

241

212

2

Lcaa

2

a

2

a 22

44134

241

234

61


The total potential energy stored in all three springs is given by:

U = ½ k24 (L24-L024)2 + ½ k23 (L23-L023)2 +

½ k31 (L31-L031)2 (3.48)

62


The differentiation of U with respect to c1 and c4 gives:

dU/dc4 = k24 (L24 – L024) dL24/dc4 + k23 (L23 – L023) dL23/dc4 +

k31 (L31 – L031) (3.49)

dU/dc1 = k24 (L24 – L024) dL24/dc1 + k23 (L23 – L023)dL23/dc1 + k31 (L31 – L031) (3.50)

63


dU/dc4 = 0, This equation does not contain L24.

dU/dc1 = 0, This equation does not contain L31.

From (3.46) and (3.47), it is apparent that dL1/dc4 and dL2/dc1 are zero. So,

64


(p1 c12 + p2 c1 + p3) c4

3 + (p4 c1

3 + p5 c12 + p6 c1 + p7) c4

2 + (p8 c1

4 + p9 c13 + p10 c1

2 + p11 c1 + p12) c4 +(p13 c1

4 + p14 c13 + p15 c1

2 + p16 c1 + p17) = 0 (3.65)

(q1 c1 + q2) c44 +

(q3 c12 + q4 c1 + q5) c4

3 + (q6 c1

3 + q7 c12 + q8 c1 + q9) c4

2 + (q10 c1

3 + q11 c12 + q12 c1 + q13) c4 +

(q14 c13 + q15 c1

2 + q16 c1 + q17) = 0 (3.66)

65


Next, factor (3.7) into the following form,

(r1 c1 + r2) c42 +

(r3 c12 + r4 c1 + r5) c4 +

(r6 c12 + r7 c1 + r8) = 0 (3.69)

Note: Equations (3.65), (3.66), and (3.69) represent three equations in the three unknowns L3, c1, and c4.

66


•To apply the Sylvester method, equations 3.65, 3.66, and 3.69 are multiplied by powers of c1 and c4.

•This step is to create a sufficient equation set in the variable c4 and c1 where the parameter L3 is embedded in the coefficients.

67


• Equation (3.65) is multiplied by c1, c12, c4, c4

2, c43, c1

4, c1

2c4, c1c42, and c1

2c42 to obtain 10 equations including it.

• Equation (3.69) is multiplied by c1, c12, c1

3, c14, c4, c4

2, c43,

c44, c1c4, c1

2c4, c13c4, c1

4c4, c1c42, c1

2c42, c1

3c42, c1

4c42, c1c4

3, c1

2c43, c1

3c14, c1c4

4 and c12c4

4 to obtain 22 equations including it.

• Equation (3.66) is multiplied by c1, c12, c1

3, c4, c42, c1c4,

c12c4, c1

3c4, c1c42 and c1

2c42 to obtain 11 equations including

it.

68


To show this 43 by 43 coefficient matrix, M, it is divided as a combination of sub-matrices Mij.

:= Mij

M11 M12 M13 M14

M21 M22 M23 M24

M31 M32 M33 M34

M41 M42 M43 M44

where,

69


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


And all sub-matrices are shown bellow:

:= M11

r8 r7 r6 0 0 0 0 r5 r4 r3 0

0 r8 r7 r6 0 0 0 0 r5 r4 r3

0 0 r8 r7 r6 0 0 0 0 r5 r4

0 0 0 r8 r7 r6 0 0 0 0 r5

0 0 0 0 r8 r7 r6 0 0 0 0

0 0 0 0 0 0 0 r8 r7 r6 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 r8 r7 r6

0 0 0 0 0 0 0 0 0 r8 r7

71


:= M12

0 0 0 r2 r1 0 0 0 0 0 0

0 0 0 0 r2 r1 0 0 0 0 0

r3 0 0 0 0 r2 r1 0 0 0 0

r4 r3 0 0 0 0 r2 r1 0 0 0

r5 r4 r3 0 0 0 0 r2 r1 0 0

0 0 0 r5 r4 r3 0 0 0 0 r2

0 0 0 r8 r7 r6 0 0 0 0 r5

0 0 0 0 0 0 0 0 0 0 r8

0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 r5 r4 r3 0 0 0 0

r6 0 0 0 0 r5 r4 r3 0 0 0

72


:= M13

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

r1 0 0 0 0 0 0 0 0 0 0

r4 r3 0 0 0 0 r2 r1 0 0 0

r7 r6 0 0 0 0 r5 r4 r3 0 0

0 0 0 0 0 0 r8 r7 r6 0 0

r2 r1 0 0 0 0 0 0 0 0 0

0 r2 r1 0 0 0 0 0 0 0 0

73


:= M14

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 r2 r1 0 0 0 0 0 0

0 r5 r4 r3 0 0 r2 r1 0 0

0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0

74


:= M21

0 0 0 0 0 0 0 0 0 0 r8

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 r8 r7

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

75


:= M21

0 0 0 0 0 0 0 0 0 0 r8

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 r8 r7

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

76


:= M23

0 0 r2 r1 0 0 0 0 0 0 0

0 0 0 r2 r1 0 0 0 0 0 0

r5 r4 r3 0 0 0 0 r2 r1 0 0

0 r2 r1 0 0 0 0 0 0 0 0

0 0 r5 r4 r3 0 0 0 0 r2 r1

0 0 0 r5 r4 r3 0 0 0 0 r2

r8 r7 r6 0 0 0 0 r5 r4 r3 0

0 r8 r7 r6 0 0 0 r5 r4 r3 0

0 0 r8 r7 r6 0 0 0 0 r5 r4

0 0 0 0 0 0 0 r8 r7 r6 0

0 0 0 0 0 0 0 0 r8 r7 r6

77


:= M24

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

r1 0 0 0 0 0 0 0 0 0

0 0 r2 r1 0 0 0 0 0 0

0 0 0 r2 r1 0 0 0 0 0

r3 0 0 0 r2 r1 0 0 0 0

0 0 r5 r4 r3 0 0 r2 r1 0

0 0 0 r5 r4 r3 0 0 r2 r1

78


:= M31

q17 q16 q15 q14 0 0 0 q13 q12 q11 q10

0 q17 q16 q15 q14 0 0 0 q13 q12 q11

0 0 q17 q16 q15 q14 0 0 0 q13 q12

0 0 0 q17 q16 q15 q14 0 0 0 q13

0 0 0 0 0 0 0 q17 q16 q15 q14

0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 q17 q16 q15

0 0 0 0 0 0 0 0 0 q17 q16

0 0 0 0 0 0 0 0 0 0 q17

0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0

79


:= M32

0 0 0 q9 q8 q7 q6 0 0 0 q5

q10 0 0 0 q9 q8 q7 q6 0 0 0

q11 q10 0 0 0 q9 q8 q7 q6 0 0

q12 q11 q10 0 0 0 q9 q8 q7 q6 0

0 0 0 q13 q12 q11 q10 0 0 0 q9

0 0 0 q17 q16 q15 q14 0 0 0 q13

q14 0 0 0 q13 q12 q11 q10 0 0 0

q15 q14 0 0 0 q13 q12 q11 q10 0 0

q16 q15 q14 0 0 0 q13 q12 q11 q10 0

0 0 0 0 q17 q16 q15 q14 0 0 0

0 0 0 0 0 q17 q16 q15 q14 0 0

80


:= M33

q4 q3 0 0 0 0 q2 q1 0 0 0

q5 q4 q3 0 0 0 0 q2 q1 0 0

0 q5 q4 q3 0 0 0 q2 q1 0 0

0 0 q5 q4 q3 0 0 0 q2 q1 0

q8 q7 q6 0 0 0 q5 q4 q3 0 0

q12 q11 q10 0 0 0 q9 q8 q7 q6 0

q9 q8 q7 q6 0 0 0 q5 q4 q3 0

0 q9 q8 q7 q6 0 0 0 q5 q4 q3

0 0 q9 q8 q7 q6 0 0 0 q5 q4

q13 q12 q11 q10 0 0 0 q9 q8 q7 q6

0 q13 q12 q11 q10 0 0 0 q9 q8 q7

81


:= M34

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 q2 q1 0 0 0 0 0 0 0

0 q5 q4 q3 0 0 q2 q1 0 0

0 0 q2 q1 0 0 0 0 0 0

0 0 0 q2 q1 0 0 0 0 0

q3 0 0 0 q2 q1 0 0 0 0

0 0 q5 q4 q3 0 0 q2 q1 0

q6 0 0 q5 q4 q3 0 0 q2 q1

82


:= M41

p17 p16 p15 p14 p13 0 0 q12 q11 q10 q9

0 p17 p16 p15 p14 p13 0 0 q12 q11 q10

0 0 p17 p16 p15 p14 p13 0 0 q12 q11

0 0 0 0 0 0 0 p17 p16 p15 p14

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 p17 p16 p15

0 0 0 0 0 0 0 0 0 p17 p16

0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0

83


:= M42

p8 0 0 p7 p6 p5 p4 0 0 0 p3

q9 p8 0 0 p7 p6 p5 p4 0 0 0

q10 q9 p8 0 0 p7 p6 p5 p4 0 0

p13 0 0 q12 q11 q10 q9 p8 0 0 p7

0 0 0 p17 p16 p15 p14 p13 0 0 q12

0 0 0 0 0 0 0 0 0 0 p17

p14 p13 0 0 q12 q11 q10 q9 p8 0 0

p15 p14 p13 0 0 q12 q11 q10 q9 p8 0

0 0 0 0 p17 p16 p15 p14 p13 0 0

0 0 0 0 0 p17 p16 p15 p14 p13 0

84


:= M43

p2 p1 0 0 0 0 0 0 0 0 0

p3 p2 p1 0 0 0 0 0 0 0 0

0 p3 p2 p1 0 0 0 0 0 0 0

p6 p5 p4 0 0 0 0 p3 p2 p1 0

q11 q10 q9 p8 0 0 p7 p6 p5 p4 0

p16 p15 p14 p13 0 0 q12 q11 q10 q9 p8

p7 p6 p5 p4 0 0 0 p3 p2 p1 0

0 p7 p6 p5 p4 0 0 0 p3 p2 p1

q12 q11 q10 q9 p8 0 0 p7 p6 p5 p4

0 q12 q11 q10 q9 p8 0 0 p7 p6 p5

85


:= M44

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 p3 p2 p1 0 0 0 0 0 0

0 p7 p6 p5 p4 0 p3 p2 p1 0

0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0

0 0 p3 p2 p1 0 0 0 0 0

p4 0 0 p3 p2 p1 0 0 0 0

86


• Equating the determinant of the coefficient matrix Mij to zero will yield a polynomial in the single variable L23.

• A numerical example is presented next.

87


• 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


Find:L3, c1, c4.

Solution:

The result was 136 solutions for L3 which 7 were real.

89


• Case 1 is presented here (7 real cases):

• Cosine θ1: -0.453571523

• Cosine θ4: -0.749999925

• L3: 7.01658755

90


• The results of Case 1 is used to calculate L1, L2, and L3 presented here:

• L1: 13.0000000

• L2: 7.99999999

• L3: 7.01658830

91


• 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


93


• 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


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


End of 3 Spring.

96

Agenda




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

4-S

• given: a12, a34lengths of struts,

• find: At equilibrium position

• L1, L2, L3, and L4 length of spring 1, 2, 3 and 4.

•

• k1, L01 spring between points 4 and 2,



• k4, L04 spring between points 4 and 1.

99

4-S: cont.

4

3

2

1

L1

L2

L3

L4

a12

a34

100

4-S: cont.

Geometry equation :

• G1 (L14 L2

4) + G2 (L14 L2

2) + G3 (L32 L1

2 L22) + G4 (L1

2 L22) + G5

(L32 L2

2) + G6 (L22) + G7( L3

2 L12 ) + G8( L1

2 ) + G9( L34 ) +

G10( L32 ) + G1

1 = 0 (4-1)

101

4-S: cont.

• The potential energy of the system can be evaluated as :

20444

20333

20222

20111 )(

2

1)(

2

1)(

2

1)(

2

1LLkLLkLLkLLkU

At equilibrium, the potential energy will be a minimum.

102

4-S: cont.

0)()(1

404440111

1

dL

dLLLkLLk

dL

dU

0)()(2

404440222

2

dL

dLLLkLLk

dL

dU

0)()(3

404440333

3

dL

dLLLkLLk

dL

dU

103

4-S: cont.

212

23

234

23

43

23

24

212

234

21

23

21

212

23

22

21

22

234

224

234

23

212

234

23

24

24

212

24

22

212

22

23

22

21

22

234

22

421

1

4

2(

)]2[

aLaLLLLaaLLLaLLLLaLL

aLaaLLLaLLaLLLLLaLLL

dL

dL

)2(

)]2[2

122

32

342

34

32

32

42

122

342

12

32

12

122

32

22

12

22

342

24

212

23

23

24

212

234

21

24

234

24

21

212

21

23

41

21

234

21

222

2

4


aLLLaaLLaLLaLLLLaLLL

dL

dL

)2(

)]2[2

122

32

342

34

32

32

42

122

342

12

32

12

122

32

22

12

22

342

24

21

234

212

234

24

212

234

24

23

24

44

21

24

212

22

24

22

21

223

3

4


LaaaLaaLLLLLLaLLLLLL

dL

dL

104

4-S: cont.

Energy equations :

• D1(L1 L24) + D2(L1

3 L22) + D3(L1

2 L22) + D4(L3

2 L1 L22) + D5(L1 L2

2) + D6(L3

2 L22) + D7(L2

2) + D8(L34) + D9(L3

2) + D10 + D11(L32 L1

3) + D12(L1

3) + D1

3(L32 L1

2) + D14(L1

2) + D15(L3

4 L1) + D16(L3

2 L1) + D17 (L1) = 0

(4-10)

• E1(L12L2

3) + E2(L32L2

3) + E3(L23) + E4(L1

2L22) + E5(L3

2L22) + E6(L2

2) + E7(L1

4L2)+ E8(L32L2) + E9(L1

2L2) + E10(L34L2) + E11(L3

2L12L2) + E12(L2) +

E13(L32)+ E14(L1

2) + E15(L34) + E16 = 0 (4-12)

• F1(L12L2

2) + F2(L3L12L2

2) + F3(L33L2

2) + F4(L32L2

2) + F5(L3L22) + F6(L2

2) + F7(L3

3L12) + F8(L3 L1

2) + F9(L12) + F10(L3

5) + F11(L34) + F12(L3

3) + F13(L3

2)+ F14(L3) + F15 = 0 (4-14)

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

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

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 L1, L2, and L3.

108

4-S: cont.

• The values of L1, L2, L3, and L4 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.

• 2nd derivatives are next.

109

4-S: cont.

12

42

04442

1

4412

1

2

)()(dL

LdLLk

dL

dLkk

dL

Ud

22

42

04442

2

4422

2

2

)()(dL

LdLLk

dL

dLkk

dL

Ud

32

42

04442

3

4432

3

2

)()(dL

LdLLk

dL

dLkk

dL

Ud

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

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

4-S: cont.

113

4-S: cont.

• End of Chapter 4.

114

Agenda




115

3-spring spatial tensegrity system

116


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


• 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


• 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

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

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

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

Regards

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

123

End of presentation

Thank you.

department of mechanical and aerospace engineering university of florida

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