Page 1

1 Copyright © 2007 byASME

Proceedings of the ASME 2007 International Design Engineering Technical Conferences & Computers and

Information in Engineering Conference

IDETC/CIE 2007

September 4-7, 2007, Las Vegas, Nevada, USA

DETC2007-34249

CLOSED-FORM EQUILIBRIUM ANALYSIS

OF PLANAR TENSEGRITY STRUCTURES

Jahan Bayat

Center for Intelligent Machines and Robotics

Department of Mechanical andAerospace Engineering

University of Florida, Gainesville, Florida 32611, USA

Carl D. Crane III

Professor and author of correspondence, Phone: (352) 392-9461, Fax: (352) 392-1071, Email: ccrane@ufl.edu.

ABSTRACT

This paper presents a closed-form analysis of a series of

planar tensegrity structures

equilibrium configurations for each device when no external

forces or moments are applied.

determined by identifying the configurations at which the

potential energy stored in the springs is a minimum. The degree

of complexity associated with the solution was far greater than

expected. For a two-spring system, a 28th degree polynomial

expressed in terms of the length of one of the springs is

developed where this polynomial identifies the cases where the

change in potential energy with respect to an infinitesimal

change in the spring length is zero.

systems are also analyzed. These more complex systems were

solved using the Continuation Method.

are presented.

to determineall possible

The equilibrium position is

Three and four spring

Numerical examples

1. INTRODUCTION

The word tensegrity is a combination of the words tension

and integrity (Edmondson, 1987 and Fuller, 1975). Tensegrity

structures are spatial structures formed by a combination of

rigid elements in compression (struts) and connecting elements

that are in tension (ties). In a classic definition, no pair of struts

touch and the end of each strut is connected to three non-

coplanar ties (Yin et al, 2002). The entire configuration stands

by itself and maintains its form solely because of the internal

arrangement of the struts and ties (Tobie, 1976).

The development of tensegrity structures is relatively new

and the works related have only existed for approximately

twenty five years. Kenner, 1976, established the relation

between the rotation of the top and bottom ties. Tobie, 1976,

presented procedures for the generation of tensile structures by

physical and graphical means. Yin, 2002, obtained Kenner’s

results using energy considerations and found the equilibrium

position for unloaded tensegrity prisms. Stern, 1999, developed

generic design equations to find the lengths of the struts and

elastic ties needed to create a desired geometry for a symmetric

case. Knight, 2000, addressed the problem of stability of

tensegrity structures for the design of deployable antennae.

Duffy, 2000 and 2002, presents static analysis and equilibrium

condition for selected tensigrity structures. Roth, 1981,

establishes some basis for flexibility and rigidity of tensigrity

frameworks. Skelton, 2002, reduces the static model of some

tensigritystructuresto linear

characterizing the problems in vector space. Tibert, 2002, uses

the tensegrity structure in deployable space applications. Festl,

2003, explains the adjustable tensigrity structures and their

applications.

algebra equations by

2. TWO-SPRING SYSTEM

A planar tensegrity system is shown in Figure 1.

device consists of two rigid struts and four ties. For the two-

spring system, two of the ties are compliant (the ties between

points 4 and 2 and points 1 and 3) and two of the ties are non-

compliant. The objective is to determine all equilibrium

configurations for the system when given the lengths of the

struts and non-compliant ties as well as the free length and

spring constant for each of the compliant ties.

The specific problem statement for the two-spring system is

written as follows:

given: a12, a34

lengths of struts,

a23, a41

lengths of non-compliant ties,

k1, L01

spring constant and free length of compliant

tie between points 4 and 2,

The

Page 2

2 Copyright © 2007 byASME

k2, L02

spring constant and free length of compliant tie

between points 3 and 1.

length of spring 1 at equilibrium position,

length of spring 2 at

corresponding to length of spring 1, i.e. L1.

It should be noted that the problem statement could be

formulated in a variety of ways, i.e. a different variable (such as

the relative angle between strut a34and tie a41) could have been

selected as the generalized parameter for this problem.

Attempts at using this alternate approach did not successfully

yield a closed form solution for the equilibrium positions.

find: L1

L2

equilibrium position

2.1 Development of Geometry and Energy Equations

Figure 2 shows the nomenclature that is used. L1and L2

are the extended lengths of the compliant ties between points 4

and 2 and points 3 and 1. A cosine law for the triangle with

sides a34, a23, and L1can be written as

a

2

Solving for cos4' yields

2

23

4

L2

A cosine law for the triangle with sides a41, a12, and L1can

be written as

a

2

Solving for cos4" yields

2

12

4

2

A cosine law for the triangle with sides a41, a34, and L2can

be written as

a

2

Solving for cos4yields

2

a

' cosθ

aL

2

L

2

23

4 341

2

34

2

1

. (1)

341

2

34

2

1

a

aLa

'

θ

cos

. (2)

2

a

" cosθ

aL

2

L

2

12

4 411

2

41

2

1

. (3)

411

2

41

2

1

aL

aLa

" cosθ

. (4)

2

L

cosθ

aa

2

a

2

2

4 4134

2

41

2

34

. (5)

4134

2

41

2

34

2

2

4

aa2

aaL

cosθ

. (6)

From Figure 2 it is apparent that

4+ 4' = + 4" (7)

Equating the cosine of the left and right sides of (7) yields

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

and expanding this equation yields

cos4cos4' – sin4sin4' = - cos4" .

Rearranging (9) yields

cos4cos4' + cos4" = sin4sin4' .

Squaring both sides of (10) gives

(cos4)2(cos4')2+ 2 cos4cos4' cos4" + (cos4")2

= (sin4)2(sin4')2.

Substituting for (sin4)2and (sin4')2in terms of cos4 and

cos4' gives

(cos4)2(cos4')2+ 2 cos4cos4' cos4" + (cos4")2

= (1-cos24) (1-cos24') .

Equations (2), (4), and (6) are substituted into (12) to yield

a single equation in the parameters L1and L2which can be

written as

AL2

where

A= L1

B = L1

and where

B2= - (a23

B0= (a12– a41) (a12+ a41) (a23– a34) (a23+ a34) ,

C2= (a34– a41) (a34+ a41) (a23– a12) (a23+ a12) ,

C0= (a41a23+ a34a12) (a41a23- a34a12) (a41

(8)

(9)

(10)

(11)

(12)

4+ B L2

2+ C = 0(13)

2,

4+ B2L1

2+ B0, C = C2L1

2+ C0

(14)

2+ a34

2+ a41

2+ a12

2),

2+ a23

2– a12

2– a34

2).

(15)

The potential energy of the system can be evaluated as

1

U

2

0222

2

0111

)L (Lk

2

1

)L (Lk

2

. (16)

Figure 1: Planar Tensegrity System

4

a41

a12

a34

2

3

a23

struts

ties

1

Figure 2: Two-Spring Planar Tensegrity Structure

4

3

2

1

θ4

θ'4

θ"4

a41

a23

a34

a12

L1

L2

Page 3

3 Copyright © 2007 byASME

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

condition can be determined as the configuration of the

structure whereby the derivative of the potential energy taken

with respect to the length L1equals zero, i.e.

dU

2 0111

1

The derivative dL2/dL1 can be determined via implicit

differentiation from equation (13) as

aaL2L(L[L

dL

41232112

1

0

dL

dL

)L (Lk)L (Lk

dL

1

2

022

. (17)

))]

(18)

aa )(a a ()aaaaL2L(L[L

)]aa )(a a ()

aa dL

2

34

2

23

2

41

2

12

2

12

2

34

22222

2

34

2

41

2

23

2

12

2

12

2

34

2

41

2

23

2

1

2

2

2

21

2

.

Substituting (18) into (17) and regrouping gives

D L2

where

D = D1L1,

E = E1L1,

F = F3L1

G = G3L1

H = H5L1

J = J1L1

and where,

D1= k2, E1= -k2L02,

F3= 2 (k2– k1), F2= 2 k1L01,

F1= - k2(a12

G3= -2 k2L02, G1= k2L02(a12

H5= - k1, H4= k1L01, H3= k1(a12

H2= - k1L01(a12

H1= - k1(a34

(a23

H0= k1L01(a34

J1= k2L02(a34

5+ E L2

4+ F L2

3+ G L2

2+ H L2+ J = 0(19)

3+ F2L1

3+ G1L1,

5+ H4L1

2+ F1L1,

4+ H3L1

3+ H2L1

2+ H1L1+ H0,

(20)

2+ a23

2+ a34

2+ a41

2),

2+ a23

2+ a34

2+ a23

2),

2) + k2(a34

2+ a41

2+ a34

2),

2+ a41

2),

2+ a23

2) (a41

2),

2– a23

2– a41

2+ a34

2– a12

2+ a41

2– a23

2– a41

2)

2– a12

2) (a41

2) (a12

2– a12

2– a23

2),

2) . (21)

2.2 Solution of Geometry and Energy Equations

Equations (19) and (13) represent two equations in the two

unknowns L1and L2. These equations can be solved by using

Sylvester’s variable elimination procedure by multiplying

equation (13) by L2, L2

L2

matrix form as

B0A0000

2, L2

3, and L2

4and equation (19) by L2,

2, L2

3to yield a total of nine equations that can be written in

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

C0

JHGFED000

2

2

2

3

2

4

2

5

2

6

2

7

2

8

2

. (22)

A solution to this set of equations can only occur if the

determinant of the 9×9 coefficient matrix is equal to zero.

Expansion of this determinant yields a 30thdegree 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 L1

coefficients of the 28thdegree polynomial were obtained

symbolically in terms of the given quantities, but are not

presented here due to their length and complexity.

Values for L2that correspond to each value of L1can be

determined by first solving (13) for four possible values of L2.

Only one of these four values also satisfies equation (19).

2to yield a 28thdegree polynomial.The

2.3 Numerical Example

The following parameters were selected to show the results

of a numerical example:

strut lengths:

a12= 3 in.a34= 3.5 in.

non-compliant tie lengths:

a41= 4 in.a23= 2 in.

spring 1 free length & spring constant:

L01= 0.5 in.k1= 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 L1.

The real values for L1and the corresponding values of L2are

shown in Table 1.

Table 1: Eight Real Solutions

CaseL1, in.

1 -5.485

2 -5.322

3 -1.741

4 -1.576

51.628

6 1.863

75.129

8 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 1 satisfy the

geometric constraints defined by equation (13) and the energy

condition defined by equation (19). 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. Afree body analysis of struts

a12and a34was performed to show that these bodies were indeed

in equilibrium.Figure 3 shows the four equilibrium

configurations for the numerical case under consideration.

Efforts were undertaken to obtain a numerical example that

would yield more than eight real roots. One example of each

type of Grashof and non-Grashof four-bar mechanism was

analyzed, yet for all these cases a maximum of eight real roots

were found.The complex values of L1 were analyzed and

Page 4

4 Copyright © 2007 byASME

corresponding complex values of L2were determined. In every

case it was possible to obtain complex pairs of L1and L2that

satisfied the geometric constraint equation, (13), and the

derivative of potential energy equation (19).

extraneousroots wereintroduced

elimination procedure.

Thus no

variable during the

3. THREE-SPRING SYSTEM

Figure 4 shows a three-spring tensegrity system.

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.

descriptive parameters are the angles θ4 and θ1. Considering the

non-compliantmembera41

specification of θ4will define the location of point 3. Similarly,

specification of θ1will define the location of point 2.

Two approaches to solve this problem have been analyzed.

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 L1and L2are chosen as the

descriptive parameters.Derivatives of the potential energy

equation are obtained with respect to L1and L2and 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 θ4 and θ1were 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 of lesser degree in that, for example, a single value of

cosθ4accounts for the obvious symmetry in solutions that will

occur with respect to the fixed member a41.

Two

One obvious set of

asbeing fixed toground,

3.1 Approach 1 – Descriptive Parameters L1and L2

3.1.1 Problem Formulation

The problem statement can be explicitly written as:

given: length of struts (a12, a34) ; length of the non-

compliant tie (a41) ; spring constants and free lengths

of three springs (k1, L01; k2, L02; k3, L03)

find:length of springs 1 and 2 (L1, L2) and corresponding

length of spring 3 (L3) at equilibrium

The analysis for this case can proceed in a manner similar

to that presented for the two-spring system. Specifically, the

term a23in (15) can be replaced by L3and equation (13) can be

factored into the form

G1L3

where

G1= a41

G2= G2aL1

G3= G3aL1

G4= G4aL1

G5= G5aL1

G6= G6aL1

and where the remaining coefficients are written in terms of the

constant mechanism parameters as

G2a= -1,

G2b= a12

G3a= (a34

G3b= a41

G4a= 1,

G5a= 1,

G5b= (-a12

G5c= a34

G6a= a12

4+ (G2L2

2+ G3) L3

2+ (G4L2

4+ G5L2

2+ G6) = 0 (23)

2,

2+ G2b,

2+ G3b,

2,

4+ G5bL1

2+ G6b

2+ G5c,

(24)

2– a41

2– a41

2(a41

2,

2),

2– a12

2– a34

2) – a12

2a34

2,

2– a34

2(a41

2(a41

2– a41

2– a12

2– a34

2),

2),

2),G6b= a12

2a34

2(a12

2+ a34

2– a41

2) .

(25)

The potential energy of the system can be written as

1

)L (Lk

2

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

condition can be determined as the configuration of the

2

0333

2

0222

2

0111

)L (Lk

2

1

)L (Lk

2

1

U

.(26)

4

44

41

11

1

2

2

2

2

3

3

3

3

Case3 Case 4

Case 5 Case 6

spring in compression with

a negative spring length

spring in tension

Figure 3: Four Equilibrium Configurations

for Two-Spring Tensegrity System

4

3

2

1

4

a41

a12

a34

L1

L2

L3

1

Figure 4: Three-Spring Tensegrity System

Page 5

5 Copyright © 2007 byASME

structure whereby the derivative of the potential energy taken

with respect to the descriptive parameters L1and L2both equal

zero. The geometric constraint equation, equation (23),

contains three unknown terms, L1, L2, and L3.

equation, L3can be considered as a dependent variable of L1

and L2. The following two expressions may be written:

U

3 0111

1

U

3 0222

2

The derivatives δL3/δL1and δL3/δL2can be determined via

implicit differentiation from Equation 23 as

LLGLL2[L

L

21a2313

1

LLGLL [2L

L

21a2313

2

Substituting (29) into (27) and rearranging gives

(D1L2

(D5L2

where

D1= D1aL1,

D2= D2aL1+ D2b,

D3= D3aL1,

D4= D4aL1,

D5= D5aL1,

D6= D6aL1

D7= D7aL1

D8= D8aL1,

D9= D9aL1

D10= D10aL1

and

D1a= - G2ak3,

D2a= 2 G1k1– G3ak3, D2b= - 2 G1k1L01,

D3a= G2ak3L03,

D4a= G3ak3L03,

D5a= - k3,

D6a= G2ak1– 2 k3,

D6b= -G2ak1L01, D6c= G2bk1– G5bk3, D6d= -G2bk1L01,

D7a= G3ak1, D7b= -G3ak1L01, D7c= G3bk1– G6ak3,

D7d= -G3bk1L01,

D8a= k3L03,

D9a= 2 k3L03, D9b= G5bk3L03

D10a= G6ak3L03

Substituting (30) into (28) and rearranging gives

(E1L2+ E2) L3

+ E5L2

where

E1= E1aL1

E2= - 2 G1k2L02,

E3= E3aL1

E4= E4aL1

E5= E5aL1

From this

0

L

L

)L (Lk)L (Lk

L

1

3

033

, (27)

0

L

L

)L (Lk)L (Lk

L

2

3

033

. (28)

]GLGLGLLGLG [2L

]GL

GL

LG

L

b3

2

1a3

2

2b

L

2

222

a6

2

2b5

4

2

2

3a3

2

3

2

2a2

2

2

2

113

(29)

]GLGLGLLGLG [2L

]GL

GL

G

L

b3

2

1a3

2

2b2

222

c5

2

1b5

4

1

2

3b2

2

3

2

1a2

2

2

2

123

(30)

2+D2) L3

2+D7) L3+ (D8L2

3+ (D3L2

2+D4) L3

4+D9L2

2+

2+D10) = 0

4+D6L2

(31)

3+ D6bL1

3+ D7bL1

2+ D6cL1+ D6d,

2+ D7cL1+ D7d,

3+ D9bL1

(32)

(33)

3+ (E3L2) L3

2+ (E4L2

3+ E9L2) = 0

3

2+ E6L2+ E7) L3+ (E8L2

(34)

2+ E1b,

2+ E3b,

2+ E4b,

2+ E5b,

E6= E6aL1

E7= E7aL1

E8= E8aL1

E9= E9aL1

4+ E6bL1

2+ E7b,

2,

4+ E9bL1

2+ E6c,

2+ E9c

(35)

and

E1a= -G2ak3,

E3a= G2ak3L03,

E4a= G2ak2– 2 k3,

E5a= -G2ak2L02,

E6a= - k3, E6b= G3ak2– G5bk3,E6c= G3bk2– G5ck3,

E7a= -G3ak2L02,E7b= -G3bk2L02,

E8a= 2 k3L03,

E9a= k3L03, E9b= G5bk3L03, E9c= G5ck3L03.

E1b= -G2bk3+ 2 G1k2,

E3b= G2bk3L03,

E4b= G2bk2,

E5b= -G2bk2L02,

(36)

3.1.2Solution of Three Simultaneous Equations in Three

Unknowns – Sylvester’s Method

Equations (23), (31), and (34) are three equations in the

three unknowns L1, L2, and L3.

applied in order to obtain sets of values for these parameters

that simultaneously satisfy all three equations. In this solution,

the parameter L1is 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 L1) must satisfy such that the three equations can

have common roots for L2and L3will yield a single polynomial

in L1.

Equation (23) was multiplied by L2, L3, L2L3, L3

L2L3

L2

L2

Equation (34) was multiplied by L2, L3, L2L3, L3

L2

L2

equations that can be written in matrix for as

M λ = 0 .

The vector λ is written as

λ = [L2

L2

L2L3

L2

L2

L2

The coefficient matrix M is a 52×52 matrix whose

elements are the coefficients G1through G6, D1through D10,

and E1 through E9 which are polynomials in terms of the

variable L1. This matrix is not presented here due to its size.

Since the set of 52 simultaneous equations represented by (37)

must be linearly dependent, the determinant of the matrix M

must equal zero. This will yield an equation in terms of the

single variable L1.

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.

possible to numerically solve this high degree polynomial for

Sylvester’s method can be

2, L2

2, and

2, L3

2,

2, L2

3L3

2, L2L3

2L3, L2

2L3

2, L3

3, L2

3, L2L3

3, L2

2L3

3, L2

3L3, L2

2, L2L3

3L3

2, L2

3L3, L2

3L3

3. Equation (31) was multiplied by L3, L2, L3

3, L2

3,

4.

2,

4,

2L3, L2

2L3

2, L2

3, L3

4, L2

3L3, L2

2, and L2L3

2, L2L3

3L3

2L3, L2

4L3, L2

2L3

4L3

2, L3

2, L2L3

3, L2

4, and L2

3, L2L3

3, L2

2L3

2L3

3, L2

2, L2

4, L3

4. This resulted in a set of 52

(37)

7L3

3L3

3, L2

6, L2

7, L2

6, L2

3L3

3, L2

5L3

2L3

7, L2

5, L2

7, L2

6L3, L2

4L3

3, L2L3

2, L3

3L3

7L3, L2

5L3

2, L2

4, L3

3, L2

7, L2

7L3

6L3

2, L2

3L3

5, L2

2, L2L3, L3

2, L2

2, L2

4L3

2L3

4, L2

6L3

5L3

3, L2

4, L2L3

3L3, L2

2, L2, L3, 1]T.

3, L2

3, L2

3L3

5L3

4L3

4, L2

5, L3

2L3

4, L2

4, L2

2L3

6, L2

2, L2L3

4L3

3L3

5, L2L3

5, L2

5,

5, L2

2L3

6,

7,

6, L3

4L3,

3, L3

5L3, L2

2, L2

2L3, L2L3

3, L2

2L3

4,

(38)

It was not