The Dynamics of the Class 1 Shell Tensegrity Structuremaeresearch.ucsd.edu/.../pinuad_dynamics_CRC.pdf · 2004-04-05 · dynamics of all components (structural components, controls,

18

The Dynamics of theClass 1 Shell

Tensegrity Structure

18.1 Introduction18.2 Tensegrity Definitions

A Typical Element • Rules of Closure for the Shell Class

18.3 Dynamics of a Two-Rod Element18.4 Choice of Independent Variables and

Coordinate Transformations18.5 Tendon Forces18.6 ConclusionAppendix 18.A Proof of Theorem 18.1Appendix 18.B Algebraic Inversion of the Q MatrixAppendix 18.C General Case for (n, m) = (i, 1)Appendix 18.D Example Case (n,m) = (3,1)Appendix 18.E Nodal Forces

Abstract

A tensegrity structure is a special truss structure in a stable equilibrium with selected membersdesignated for only tension loading, and the members in tension forming a continuous network ofcables separated by a set of compressive members. This chapter develops an explicit analyticalmodel of the nonlinear dynamics of a large class of tensegrity structures constructed of rigid rodsconnected by a continuous network of elastic cables. The kinematics are described by positionsand velocities of the ends of the rigid rods; hence, the use of angular velocities of each rod is avoided.

The model yields an analytical expression for accelerations of all rods, making the model efficientfor simulation, because the update and inversion of a nonlinear mass matrix are not required. Themodel is intended for shape control and design of deployable structures. Indeed, the explicitanalytical expressions are provided herein for the study of stable equilibria and controllability, butcontrol issues are not treated.

18.1 Introduction

The history of structural design can be divided into four eras classified by design objectives. In theprehistoric era, which produced such structures as Stonehenge, the objective was simply to opposegravity, to take static loads. The classical era, considered the dynamic

response and placed designconstraints on the eigenvectors as well as eigenvalues. In the modern era, design constraints couldbe so demanding that the dynamic response objectives require feedback control. In this era, the

Robert E. Skelton

University of California, San Diego

Jean-Paul Pinaud

University of California, San Diego

D. L. Mingori

University of California, Los Angeles

8596Ch18Frame Page 389 Wednesday, November 7, 2001 12:18 AM

© 2002 by CRC Press LLC

control discipline followed the classical structure design, where the structure and control disciplineswere ingredients in a multidisciplinary system design, but no interdisciplinary tools were developedto integrate the design of the structure and the control. Hence, in this modern era, the dynamics ofthe structure and control were not cooperating to the fullest extent possible. The post-modern eraof structural systems is identified by attempts to unify the structure and control design for a commonobjective.

The ultimate performance capability of many new products and systems cannot be achieved untilmathematical tools exist that can extract the full measure of cooperation possible between thedynamics of all components (structural components, controls, sensors, actuators, etc.). This requiresnew research. Control theory describes how the design of one component (the controller) shouldbe influenced by the (given) dynamics of all other components. However, in systems design, wheremore than one component remains to be designed, there is inadequate theory to suggest how thedynamics of two or more components should influence each other at the design stage. In the future,controlled structures will not be conceived merely as multidisciplinary design steps, where a plate,beam, or shell is first designed, followed by the addition of control actuation. Rather, controlledstructures will be conceived as an interdisciplinary process in which both material architecture andfeedback information architecture will be jointly determined. New paradigms for material andstructure design might be found to help unify the disciplines. Such a search motivates this work.Preliminary work on the integration of structure and control design appears in Skelton

1,2

andGrigoriadis et al.

3

Bendsoe and others

4-7

optimize structures by beginning with a solid brick and deleting finiteelements until minimal mass or other objective functions are extremized. But, a very importantfactor in determining performance is the paradigm used for structure design. This chapter describesthe dynamics of a structural system composed of axially loaded compression members and tendonmembers that easily allow the unification of structure and control functions. Sensing and actuatingfunctions can sense or control the tension or the length of tension members. Under the assumptionthat the axial loads are much smaller than the buckling loads, we treat the rods as rigid bodies.Because all members experience only axial loads, the mathematical model is more accurate thanmodels of systems with members in bending. This unidirectional loading of members is a distinctadvantage of our paradigm, since it eliminates many nonlinearities that plague other controlledstructural concepts: hysteresis, friction, deadzones, and backlash.

It has been known since the middle of the 20th century that continua cannot explain the strengthof materials. While science can now observe at the nanoscale to witness the architecture of materialspreferred by nature, we cannot yet design or manufacture manmade materials that duplicate theincredible structural efficiencies of natural systems. Nature’s strongest fiber, the spider fiber,arranges simple nontoxic materials (amino acids) into a microstructure that contains a continuousnetwork of members in tension (amorphous strains) and a discontinuous set of members in com-pression (the

β

-pleated sheets in Figure 18.1).

8,9

This class of structure, with a continuous network of tension members and a discontinuousnetwork of compression members, will be called a Class 1 tensegrity structure. The importantlessons learned from the tensegrity structure of the spider fiber are that

1. Structural members never reverse their role. The compressive members never take tensionand, of course, tension members never take compression.

2. Compressive members do not touch (there are no joints in the structure).3. Tensile strength is largely determined by the local topology of tension and compressive

members.

Another example from nature, with important lessons for our new paradigms is the carbonnanotube often called the Fullerene (or Buckytube), which is a derivative of the Buckyball. Imagine



a 1-atom thick sheet of a graphene, which has hexagonal holes due to the arrangements of materialat the atomic level (see Figure 18.2). Now imagine that the flat sheet is closed into a tube bychoosing an axis about which the sheet is closed to form a tube. A specific set of rules must definethis closure which takes the sheet to a tube, and the electrical and mechanical properties of theresulting tube depend on the rules of closure (axis of wrap, relative to the local hexagonal topol-ogy).

10

Smalley won the Nobel Prize in 1996 for these insights into the Fullerenes. The spider fiberand the Fullerene provide the motivation to construct manmade materials whose overall mechanical,thermal, and electrical properties can be predetermined by choosing the local topology and therules of closure which generate the three-dimensional structure from a given local topology. Bycombining these motivations from Fullerenes with the tensegrity architecture of the spider fiber,this chapter derives the static and dynamic models of a shell class of tensegrity structures. Futurepapers will exploit the control advantages of such structures. The existing literature on tensegritydeals mainly

11-23

with some elementary work on dynamics in Skelton and Sultan,

24

Skelton andHe,

25

and Murakami et al.

26

FIGURE 18.1

Nature’s strongest fiber: the Spider Fiber. (From Termonia, Y.,

Macromolecules

, 27, 7378–7381,1994. Reprinted with permission from the American Chemical Society.)

FIGURE 18.2

Buckytubes.

amorphouschain

β-pleated sheet

entanglement

hydrogen bond

yz

x6nm



18.2 Tensegrity Definitions

Kenneth Snelson built the first tensegrity structure in 1948 (Figure 18.3) and Buckminster Fullercoined the word “tensegrity.” For 50 years tensegrity has existed as an art form with some archi-tectural appeal, but engineering use has been hampered by the lack of models for the dynamics.In fact, engineering use of tensegrity was doubted by the inventor himself. Kenneth Snelson in aletter to R. Motro said, “As I see it, this type of structure, at least in its purest form, is not likelyto prove highly efficient or utilitarian.” This statement might partially explain why no one botheredto develop math models to convert the art form into engineering practice. We seek to use scienceto prove the artist wrong, that his invention is indeed more valuable than the artistic scope that heascribed to it. Mathematical models are essential design tools to make engineered products. Thischapter provides a dynamical model of a class of tensegrity structures that is appropriate for spacestructures.

We derive the nonlinear equations of motion for space structures that can be deployed or heldto a precise shape by feedback control, although control is beyond the scope of this chapter. Forengineering purposes, more precise definitions of tensegrity are needed.

One can imagine a truss as a structure whose compressive members are all connected with balljoints so that no torques can be transmitted. Of course, tension members connected to compressivemembers do not transmit torques, so that our truss is composed of members experiencing nomoments. The following definitions are useful.

Definition 18.1

A given configuration of a structure is in a

stable equilibrium

if, in the absenceof external forces, an arbitrarily small initial deformation returns to the given configuration.

Definition 18.2

A tensegrity structure is a stable system of axially loaded members.

Definition 18.3

A stable structure is said to be a “Class 1” tensegrity structure if the membersin tension form a continuous network, and the members in compression form a discontinuous setof members.

FIGURE 18.3

Needle Tower of Kenneth Snelson, Class 1 tensegrity. Kröller Müller Museum, The Netherlands.(From Connelly, R. and Beck, A.,

American Scientist

, 86(2), 143, 1998. With permissions.)



Definition 18.4

A stable structure is said to be a “Class 2” tense grity structure if the membersin tension form a continuous set of members, and there are at most tw o members in compressionconnected to each node.

Figure 18.4 illustrates Class 1 and Class 2 tensegrity structures.Consider the topology of structural members given in Figure 18.5, where thick lines indicate

rigid rods which tak e compressi ve loads and the thin lines represent tendons. This is a Class 1tense grity structure.

Definition 18.5

Let the topology of Figure 18.5 describe a three-dimensional structure by con-necting points A to A, B to B, C to C,…, I to I. This constitutes a “Class 1 tense grity shell” if thereexists a set of tensions in all tendons (

α

= 1

→

10,

β

= 1

→

n,

γ

= 1

→

m) such that thestructure is in a stable equilibrium.

FIGURE 18.4

Class 1 and Class 2 tense grity structures.

FIGURE 18.5

Topology of an (8,4) Class 1 tense grity shell.

1 2

tαβγ ,



18.2.1 A Typical Element

The axial members in Figure 18.5 illustrate only the pattern of member connections and not theactual loaded configuration. The purpose of this section is two-fold: (i) to define a typical “element”which can be repeated to generate all elements, and (ii) to define rules of closure that will generatea “shell” type of structure.

Consider the members that make the typical

ij

element where

i

= 1, 2, …, n indexes the elementto the left, and

j

= 1, 2, …, m indexes the element up the page in Figure 18.5. We describe theaxial elements by vectors. That is, the vectors describing the

ij

element, are

t

1

ij

,

t

2

ij

, …

t

10

ij

and

r

1

ij

,

r

2

ij

, where, within the

ij

element,

t

α

ij

is a vector whose tail is fixed at the specified end of tendonnumber

α

, and the head of the vector is fixed at the other end of tendon number

α

as shown inFigure 18.6 where

α

= 1, 2, …, 10. The

ij

element has two compressive members we call “rods,”shaded in Figure 18.6. Within the

ij

element the vector

r

1

ij

lies along the rod r

1

ij

and the vector

r

2

ij

lies along the rod

r

2

ij

. The first goal of this chapter is to derive the equations of motion for thedynamics of the two rods in the

ij

element. The second goal is to write the dynamics for the entiresystem composed of

nm

elements. Figures 18.5 and 18.7 illustrate these closure rules for the case(

n,

m

) = (8,4) and (

n, m

) = (3,1).

Lemma 18.1

Consider the structure of Figure 18.5 with elements defined by Figure 18.6. A Class 2tensegrity shell is formed by adding constraints such that for all i =

1, 2, …

,

n, and for m > j >

1,

FIGURE 18.6

A typical

ij

element.



(18.1)

This closes nodes n

2ij

and n

1(i+1)(j+1)

to a single node, and closes nodes n

3(i–1)j

and n

4i(j–1)

to a singlenode (with ball joints). The nodes are closed outside the rod, so that all tension elements are onthe exterior of the tensegrity structure and the rods are in the interior.

The point here is that a Class 2 shell can be obtained as a special case of the Class 1 shell, byimposing constraints (18.1). To create a tensegrity structure not all tendons in Figure 18.5 arenecessary. The following definition eliminates tendons

t

9

ij

and

t

10

ij

, (i

= 1

→

n, j = 1

→

m).

Definition 18.6

Consider the shell of Figures 18.4. and 18.5, which may be Class 1 or Class 2depending on whether constraints (18.1) are applied. In the absence of dotted tendons (labeled t

9

and t

10

), this is called a primal tensegrity shell. When all tendons t

9

, t

10

are present in Figure 18.5,it is called simply a Class 1 or Class 2 tensegrity shell.

The remainder of this chapter focuses on the general Class 1 shell of Figures 18.5 and 18.6.

18.2.2 Rules of Closure for the Shell Class

Each tendon exerts a positive force away from a node and

f

αβγ

is the force exerted by tendon

t

αβγ

and denotes the force vector acting on the node

n

α

ij

. All

f

α

ij

forces are postive in the directionof the arrows in Figure 18.6, where

w

α

ij

is the external applied force at node

n

α

ij

,

α

= 1, 2, 3, 4. Atthe base, the rules of closure, from Figures 18.5 and 18.6, are

t

9

i

1

= –

t

1

i

1

,

i

= 1, 2, …,

n

(18.2)

t

6

i

0

=

0

(18.3)

t

600

= –

t2n1 (18.4)

t901 = t9n1 = –t1n1 (18.5)

0 = t10(i–1)0 = t5i0 = t7i0 = t7(i–1)0, i = 1, 2, …, n. (18.6)

FIGURE 18.7 Class 1 shell: (n,m) = (3,1).

− + =+ =+ =+ =

t t

t t

t t

t t

1 4

2 3

5 6

7 8

ij ij

ij ij

ij ij

ij ij

0

0

0

0

,

,

,

.

fαij



At the top, the closure rules are

t10im = –t7im (18.7)

t100m = –t70m = –t7nm (18.8)

t2i(m+1) = 0 (18.9)

0 = t1i(m+1) = t9i(m+1) = t3(i+1)(m+1)

= t1(i+1)(m+1) = t2(i+1)(m+1). (18.10)

At the closure of the circumference (where i = 1):

t90j = t9nj, t60(j–1) = t6n(j–1), t70(j–1) = t7n(j–1) (18.11)

t80j = t8nj, t70j = t7nj, t100(j–1) = t10n(j–1). (18.12)

From Figures 18.5 and 18.6, when j = 1, then

0 = f7i(j–1) = f7(i–1)(j–1) = f5i(j–1) = f10(i–1)(j–1), (18.13)

and for j = m where,

0 = f1i(m+1) = f9i(m+1) = f3(i+1)(m+1) = f1(i+1)(m+1). (18.14)

Nodes n11j, n3nj, n41j for j = 1, 2, …, m are involved in the longitudinal “zipper” that closes thestructure in circumference. The forces at these nodes are written explicitly to illustrate the closurerules.

In 18.4, rod dynamics will be expressed in terms of sums and differences of the nodal forces,so the forces acting on each node are presented in the following form, convenient for later use.The definitions of the matrices Bi are found in Appendix 18.E.

The forces acting on the nodes can be written in vector form:

f = Bd f d + Bo f o + Wow (18.15)

where

Wo = BlockDiag [ ,W1, W1, ],

f

f

f

f

f

f

f

f

f

f

w

w

w

=

=

=

=

11

21 1

MM

M M

m

d

d

d

md

o

o

mo

m

, , , ,

L L

B

B B

B B

B

B B

B B

B

B B

B

B

B

d o=

=

3 4

5 6

5

6 4

5 8

1 2

7

2

7

0 0

0 0

0 0

0 0

0

0

0 0

L

O O M

O O

M O O

K

L

O O M

M O O O

M O O

L L

,



and

(18.16)

Now that we have an expression for the forces, let us write the dynamics.

18.3 Dynamics of a Two-Rod Element

Any discussion of rigid body dynamics should properly begin with some decision on how themotion of each body is to be described. A common way to describe rigid body orientation is touse three successive angular rotations to define the orientation of three mutually orthogonal axesfixed in the body. The measure numbers of the angular velocity of the body may then be expressedin terms of these angles and their time derivatives.

This approach must be reconsidered when the body of interest is idealized as a rod. The reasonis that the concept of “body fixed axes” becomes ambiguous. Two different sets of axes with acommon axis along the rod can be considered equally “body fixed” in the sense that all massparticles of the rod have zero velocity in both sets. This remains true even if relative rotation isallowed along the common axis. The angular velocity of the rod is also ill defined because thecomponent of angular velocity along the rod axis is arbitrary. For these reasons, we are motivatedto seek a kinematical description which avoids introducing “body-fixed” reference frames andangular velocity. This objective may be accomplished by describing the configuration of the systemin terms of vectors located only the end points of the rods. In this case, no angles are used.

We will use the following notational conventions. Lower case, bold-faced symbols with anunderline indicate vector quantities with magnitude and direction in three-dimensional space. Theseare the usual vector quantities we are familiar with from elementary dynamics. The same bold-faced symbols without an underline indicate a matrix whose elements are scalars. Sometimes wealso need to introduce matrices whose elements are vectors. These quantities are indicated with anupper case symbol that is both bold faced and underlined.

As an example of this notation, a position vector can be expressed as

In this expression, pi is a column matrix whose elements are the measure numbers of for the mutuallyorthogonal inertial unit vectors e1, e2, and e3. Similarly, we may represent a force vector as

Matrix notation will be used in most of the development to follow.

ff

f f

f

f

f

f

f

f

f

f

w

w

w

w

w

ijo

ijijd

ij

ij

ij

=

=

=

5

1

2

3

4

6

7

8

9

10

1

2

3

4

, ,

p e e e Epi

i

i

i

i

p

p

p

=

=[ ] .1 2 3

1

2

3

pi

fi

ˆ ˆ .f Efi i=



We now consider a single rod as shown in Figure 18.8 with nodal forces and applied tothe ends of the rod.

The following theorem will be fundamental to our development.

Theorem 18.1 Given a rigid rod of constant mass m and constant length L, the governingequations may be described as:

(18.17)

where

The notation denotes the skew symmetric matrix formed from the elements of r:

and the square of this matrix is

The matrix elements r1, r2, r3, q1, q2, q3, etc. are to be interpreted as the measure numbers of thecorresponding vectors for an orthogonal set of inertially fixed unit vectors e1, e2, and e3. Thus,using the convention introduced earlier,

r = Er, = Eq, etc.

FIGURE 18.8 A single rigid rod.

f1 f 2

˙ ˜q Kq Hf+ =

qq

q

p p

p p=

=

+−

1

2

1 2

2 1

ff f

f f H =

I

q

~

~=

+−

=

−

ˆ ˆ

ˆ ˆ , ,˙ ˙

.1 2

1 2

3

322 2

2 2 3

2

2m LL

T

0

0K

0 0

0 q q I

r~

r r = ~ =−

−−

0

0

0

3 2

3 1

2 1

1

2

3

r r

r r

r r

r

r

r

,

r~2 =− −

− −− −

r r r r r r

r r r r r r

r r r r r r

22

32

1 2 1 3

2 1 12

32

2 3

3 1 3 2 12

22

.

q



The proof of Theorem 18.1 is given in Appendix 18.A. This theorem provides the basis of ourdynamic model for the shell class of tensegrity structures.

Now consider the dynamics of the two-rod element of the Class 1 tensegrity shell in Figure 18.5.Here, we assume the lengths of the rods are constant. From Theorem 18.1 and Appendix 18.A, themotion equations for the ij unit can be described as

(18.18)

(18.19)

where the mass of the rod αij is mαij and rαij = Lαij. As before, we refer everything to a commoninertial reference frame (E). Hence,

and the force vectors appear in the form

.

m

ij ij ij

ijij ij ij ij ij

ij ij ij ij

ij ij ij

ij

m

L

12 1 1 2

12 2 2 2 1

2 2 2 2

2 2 12

6

q f f

q q q f f

q q q q

q q

..

.. ^ ^

. . ..

= +

× = × −

+ =

=

^ ^

( ) ( )

. .

.

,

0

m

ij ij ij

m

ij ij ij ij ij

ij ij ij ij

ij ij ij

ij

ij

L

2

2

2

6

22

˙ ˆ ˆ

( ˙ ) (ˆ ˆ )

˙ . ˙ . ˙

.

,

q f f

q q q f f

q q q q

q q

3 3 4

4 4 4 4 3

4 4 4 4

4 4

= +

× = × −

+ =

=

0

q q q q1

11

12

13

2

21

22

23

3

31

32

33

4

41

42

43

ij

ij

ij

ij

ij

ij

ij

ij

ij

ij

ij

ij

ij

ij

ij

ij

q

q

q

q

q

q

q

q

q

q

q

q

∆ ∆ ∆ ∆

, , , ,

q q q q q1 1 2 3 4ij ijT

ijT

ijT

ijT

T∆ , , , ,[ ]

HI

q HI

q211

3

322

2

3

342

2 2

12

22

ijij L ij

ijij L ijm m

ij ij

=

=

0

0

0

0˜ , ˜ ,

HH

H f

f f

f f

f f

f f

ij

ij

ijij

ij ij

ij ij

ij ij

ij ij

=

+

−

+

−

1

2

1 2

1 2

3 4

3 4

0

0,

ˆ ˆ

ˆ ˆ

ˆ ˆ

ˆ ˆ

∆



Using Theorem 18.1, the dynamics for the ij unit can be expressed as follows:

where

The shell system dynamics are given by

(18.20)

where f is defined in (18.15) and

18.4 Choice of Independent Variables and Coordinate Transformations

Tendon vectors tαβγ are needed to express the forces. Hence, the dynamical model will be completedby expressing the tendon forces, f, in terms of variables q. From Figures 18.6 and 18.9, it followsthat vectors and ij can be described by

(18.21)

(18.22)

To describe the geometry, we choose the independent vectors r1ij, r2ij, t5ij, for i = 1, 2, …, n, j =1, 2, …, m and ρρρρ11, t1ij, for i = 1, 2, …, n, j = 1, 2, …, m, and i < n when j = 1.

This section discusses the relationship between the q variables and the string and rod vectorstαβγ and rβij. From Figures 18.5 and 18.6, the position vectors from the origin of the reference frame,E, to the nodal points, p1ij, p2ij, p3ij, and p4ij, can be described as follows:

˙ ,q q H fij ij ij ij ij+ =ΩΩ

ΩΩ ΩΩ11

22 2 3

222

4 4 3ij

ij ijT

ijij

ij ijT

ijL L

=

=

− −

0 0

0

0 0

0˙ ˙ , ˙ ˙ ,q q I

q q I

ΩΩΩΩ

ΩΩij

ij

ij

=

1

2

0

0,

q q q q q q q= [ ]11 1 12 2 1T

nT T

nT

mT

nmT T

,..., , ,..., ,..., ,..., .

˙ ,q K q Hf+ =r

q q q q q q q

K

H = H H H H H H

= [ ]= [ ]

[ ]

11 1 12 2 1

11 1 12 2 1

11 1 12 2 1

TnT T

nT

mT

nmT T

r n n m nm

n n m nm

BlockDiag

BlockDiag

,..., , ,..., ,..., ,..., ,

,..., , ,..., ,..., ,..., ,

,..., , ,..., ,..., ,..., .

ΩΩ ΩΩ ΩΩ ΩΩ ΩΩ ΩΩ

ij

ρρ ρρij k

k

i

k

k

i

ik ik

k

j

ij

k

j

= + − + + −= =

−

=

−

=∑ ∑ ∑∑11 1 1

1

1 1

1

1

1 5

1

1

1

2

r t t t r

ˆ .ρρ ρρij ij ij ij ij = + r + t r 1 5 2−



(18.23)

We define

(18.24)

Then,

(18.25)

FIGURE 18.9 Choice of independent variables.

p

p r

p

p r

1

2 1

3

4 2

ij ij

ij ij ij

ij ij

ij ij ij

=

= +

=

= +

ρρ

ρρ

ρρ

ρρ

ˆ

ˆ

q p p

q p p

q p p

q p p

1 2 1 1

2 2 1 1

3 4 3 1 2

4 4 3 2

2

2

ij ij ij ij ij

ij ij ij ij

ij ij ij ij ij

ij ij ij ij

= + = +

= − =

= + = +

= − =

∆∆

∆∆

∆∆

∆∆

ρρ

ρρ

r

r

r

r

ˆ

q

q

q

q

q

I I

I I

I I

I I

p

p

p

p

I I

I

I I

I

ij

ij ij

=

=−

−

=

∆∆

1

2

3

4

3 3

3 3

3 3

3 3

1

2

3

4

3 3

3

3 3

3

2

2

0 0

0 0

0 0

0 0

0 0

0 0 0

0 0

0 0 0

ij ij

ρρ

ρρr

r

1

2

ˆ



In shape control, we will later be interested in the p vector to describe all nodal points of thestructure. This relation is

p = Pq P = BlockDiag […,P1, …,P1,…] (18.26)

The equations of motion will be written in the q coordinates. Substitution of (18.21) and (18.22)into (18.24) yields the relationship between q and the independent variables t5, t1, r1, r2 as follows:

(18.27)

To put (18.27) in a matrix form, define the matrices:

and

P

I I

I I

I I

I I

I I

I I

I I

I I

1

3 3

3 3

3 3

3 3

13 3

3 3

3 3

3 3

12

=−

−

=

−

−

−

∆∆

0 0

0 0

0 0

0 0

0 0

0 0

0 0

0 0

q r t t t r

q r

q r t t t

1 11 1 1

1

1 1 1

2

5

1

1

1

1

1

2 1

3 11 1 1

1

1 1 1

2

5

11

1

2

2

ij k

k

i

k ik

k

j

ik

k

j

k

i

ij

ij ij

ij k

k

i

k ik

k

j

ik

k

j

k

i

= + − + +

−

=

= + − + +

= = =

−

=

−

= = ==

−

∑ ∑ ∑∑

∑ ∑ ∑∑

ρρ

ρρ

−

=

r

q r

2

4 2

ij

ij ij

l

r

r

t

t

ij

ij

ij

ij

ij

j m=

=

1

2

5

1

2 3for , ,..., ,

lr

r

t

l

t

r

r

t

( - )

11

11

111

211

511

1

1 1 1

1 1

2 1

5 1

2=

=

=

ρρ

, ,... , ,i

i

i

i

i

i nfor

l l l l l l l l= [ ]11 21 1 12 2 1

T TnT T

nT

mT

nmT T

, , , , , , , , , , ,K K K K

A

I I

I

I I I I

I

B

I I

I

I I I I

I

=−

=

−

− −

2

2 2 2

2

2 2 2

3 3

3

3 3 3 3

3

3 3

3

3 3 3 3

3

0 0

0 0 0

0 0 0

0 0

0 0 0

0 0 0

,



Then (18.27) can be written simply

q = Ql, (18.28)

where the 12nm × 12nm matrix Q is composed of the 12 × 12 matrices A–H as follows:

(18.29)

n × n blocks of 12 × 12 matrices,

12n × 12n matrix,

Q22 = BlockDiag […,C, …,C],

Q32 = BlockDiag […,J, …,J],

C

I I

I

I I I

I

D

I I

I I=

−

−

=

3 3

3

3 3 3

3

3 3

3 3

2

2 2

2 2

2 2

0 0

0 0 0

0

0 0 0

0 0

0 0 0 0

0 0

0 0 0 0

,

E

I I

I I F

I I I

I I I=

−

−

=

2 2

2 2

2 2 2

2 2 2

3 3

3 3

3 3 3

3 3 3

0 0

0 0 0 0

0 0

0 0 0 0

0

0 0 0 0

0

0 0 0 0

,

J

I I

I I G

I I I

I I I=

=

−

0 0

0 0 0 0

0 0

0 0 0 0

0

0 0 0 0

0

0 0 0 0

2 2

2 2

2 2 2

2 2 2

3 3

3 3

3 3 3

3 3 3

,

Q

Q

Q Q

Q Q Q

Q Q Q Q

Q Q Q Q Q

=

11

21 22

21 32 22

21 32 32 22

21 32 32 32 22

0 0

0

L L

O M

O M

M M M M O O

,

Q

A

D B

D E B

D E E B

D E E E B

11

0

=

0 0L L L

O M

O M

O M

M M M O O

L

Q

F

D G

D E G

D E E

D E E E E G

21 =

0 0

0

L L L

O M

O M

O O M

M M M O O



where each Qij is 12n × 12n and there are m row blocks and m column blocks in Q. Appendix18.B provides an explicit expression for the inverse matrix Q, which will be needed later to expressthe tendon forces in terms of q.

Equation (18.28) provides the relationship between the selected generalized coordinates and anindependent set of the tendon and rod vectors forming l. All remaining tendon vectors may bewritten as a linear combination of l. This relation will now be established. The following equationsare written by inspection of Figures 18.5, 18.6, and 18.7 where

(18.30)

and for i = 1, 2, …, n, j = 1, 2, …,m we have

(18.31)

For j = 1 we replace t2ij with

For j = m we replace t6ij and t7ij with

where and i + n = i. Equation (18.31) has the matrix form,

t r1 1 1 1 1 11n n n= + −ρρ ρρ

t = + r

t

t t + r = r

t r

2 1 2 1

3 1

4 3 1 1 1

6 1 1 2

1ij ij i j i j

ij i j ij

ij ij ij ij ij i j

ij i j ij ij

j

j

ρρ ρρ

ρρ ρρ

ρρ ρρ

ρρ ρρ

− >

= −

= − + −

= − + <

− −

−

−

+ +

(ˆ ) , ( )

ˆ

ˆ

(ˆ ) , (

( ) ( )

( )

( )

( )( ) mm

j mij ij i j

ij ij ij ij ij ij ij

ij i j ij ij

ij i j i j ij

)

ˆ , ( )

ˆ

( )

ˆ ˆ .

( )( )

( )

( ) ( )

t

t r r t r

t r

t r

7 1 1

8 1 5 2

9 1 1

10 1 2 1

= − <

= + − = − − +

= − +

= + −

+ +

+

+ +

ρρ ρρ

ρρ ρρ

ρρ ρρ

ρρ ρρ

1

t2 1 1 1 1i i i= − +ρρ ρρ( ) .

t r r

t r

6 1 2 1 2

7 1 2 1

im i m i m im im

im im i m i m

= + − +

= − +

+ +

+ +

ˆ (ˆ )

ˆ (ˆ ).

( ) ( )

( ) ( )

ρρ ρρ

ρρ ρρ

ρρ ρρ ρρ ρρ0 j nj oj nj∆∆ ∆∆, ˆ ˆ ,

t

I I

r

ijd

ij

=

=

− −

∆∆

t

t

t

t

t

t

t

t

0 0

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

2

3

4

6

7

8

9

10

3 3

1

ρρ

ρρrr

I

I

r

r2 1

3

3

1

2 1

+

−

− −i j i j( ) ( )

ˆ

0 0 0 0

0 0 0

0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

ρρ

ρρ



(18.32)

+

−

− −

−− −

−

+

I

I

I I

I I

I

I I I

I I

I

r

r

3

3

3 3

3 3

3

3 3 3

3 3

3

1

2

0 0 0

0 0 0

0 0

0 0

0 0 0

0

0 0

0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

0

ρρ

ρρ

ij

00 0 0

0 0 0 0

0 0 0

0 0

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0

0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

I

I I

r

r

+I

I

3

3 3

1

2 1

3

3

−

+

ρρ

ρρ

( )i j

+ +

ρρ

ρρr

r

1

2 1 1

ˆ ,

( )( )i j

t

t

t

t

t

t

t

t

t

I

I

r

r

id

i

1

2

3

4

6

7

8

9

10 1

3

3

1∆∆

=

−

0 0 0 0

0 0 0

0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

ρρ

ρρ

22 1 1

3

3

3

3 3

3

3 3 3

3 3

3

1

2

+

−

− −

−− −

−

−( )

ˆ

i

I

I

I I

I I

I

I I I

I I

I

r

r

0 0 0

0 0 0

0 0

0 0

0 0 0

0

0 0

0 0 0

3 ρρ

ρρ

+

−

+−

+

i

i

1

3

3

3 3

1

2 1 1

3

3

I

I

I I

r

r

I

I

0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0

0 0

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0

0 0 0

ρρ

ρρ

( )00 0 0 0

0 0 0 0

0 0 0 0

+

ρρ

ρρr

r

1

2 1 2

ˆ ,

( )i



Equation (18.25) yields

(18.33)

Hence, (18.32) and (18.33) yield

0 0

=

− −

t

t

t

t

t

t

t

t

t

I2

3

4

6

7

8

9

10

3

imd

im

II

r

r

I

I

3

1

2 1

3

3

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0

0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0

+

−

−

ρρ

ρρ

( )i m

00

0 0 0

0 0 0

0 0

0 0

0 0 0

0

0 0

0 0 0

+

−

− −

−− −

−

−

ρρ

ρρr

r

I

I

I I

I I

I

I I I

I I

I

1

2 1

3

3

3

3 3

3

3 3

3 3

3

ˆ

( )i m

3

3

+− −

ρρ

ρρ

ρρ

ρρr

r

I I

I I

I

I I

r

r

1

2

3 3

3 3

3

3 3

1

2

ˆ ˆ

im

0 0 0 0

0 0 0 0

0 0 0 0

0 0

0 0

0 0 0 0

0 0 0

0 0

+( )

.

i m1

ρρ

ρρr

r

I I

I

I I

I

q1

2

12 3

12 3

3

12 3

12 3

3

ˆ

=

−

−

ij

ij

0 0

0 0 0

0 0

0 0 0

t

t

t

t

t

t

t

t

t

qijd

ij

i j=

=

∆∆

2

3

4

6

7

8

9

10

12

0 0 I I

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

3 3– –

( −− −+

1 1

12) ( )

–

–

0 0 0 0

0 0 I I

0 0 I I

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

3 3

3 3

q i j

∆∆



(18.34)

+ 12

–

–

– –

–

I I 0 0

I I 0 0

I I 0 0

0 0 I I

0 0 I I

I

3 3

3 3

3 3

3 3

3 3

33 3 3 3

3 3

3 3

3 3

3 3

I I I

I I 0 0

0 0 I I

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

I I 0 0

0 0 I I

–

– –

–

–

+

q ij

12

+

+

+ +

q

q

( )

( )( ),

–

–

i j

i j

1

1 1

12

0 0 0 0

0 0 0 0

0 0 0 0

I I 0 0

I I 0 0

0 0 0 0

0 0 0 0

0 0 0 0

3 3

3 3

t

t

t

t

t

t

t

t

t

qid

i

1

2

3

4

6

7

8

9

10 1

12

∆∆

=

−−

0 0 0 0

0 0 I I

0 0 I I

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

3 3

3 3

(( )i i− +

−−

− −−

−− −

−

+

−

1 1 1

12

12

I I 0 0

I I 0 0

I I 0 0

0 0 I I

0 0 I I

I I I I

I I 0 0

0 0 I I

I I 0 0

0 0 0 0

0 0 0 0

0

3 3

3 3

3 3

3 3

3 3

3 3 3 3

3 3

3 3

3 3

q

00 0 0

0 0 0 0

0 0 0 0

I I 0 0

0 0 I I

0 0 0 0

0 0 0 0

0 0 0 0

I I 0 0

I I 0 0

0 0 0 0

0 0 0 0

0 0 0 03 3

3 3

3 3

3 3

−

+−

−

+q( )i 1 1

12

+q( ) ,i 1 2



Also, from (18.30) and (18.32)

(18.35)

With the obvious definitions of the 24 × 12 matrices E1, E2, E3, E4, Ê4, , E5, equations in(18.34) are written in the form, where q01 = qn1, q(n+1)j = qij,

(18.36)

=

− −

t

t

t

t

t

t

t

t

t

imd

im

2

3

4

6

7

8

9

10

12

∆∆

0 0 I I

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

3 3

+

−−

+

−−

− −q qi m i m( ) ( )1 1

12

12

0 0 0 0

0 0 I I

0 0 I I

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

I I 0 0

I I 0 0

I I 0 0

3 3

3 3

3 3

3 3

3 3

00 0 I I

0 0 I I

I I I I

I I 0 0

0 0 I I

0 0 0 0

0 0 0 0

0 0 0 0

0 0 I I

0 0 I I

0 0 0 0

I I 0 0

0 0 I I

3 3

3 3

3 3 3 3

3 3

3 3

3 3

3 3

3 3

3 3

− −−

−− −

−

+− −

−

q im

12

+q( ) .i m1

t Ir

r

I Ir

r

t I I q I I q

E q

1 1 31

2 11

3 31

2 1

1 112 3

12 3 11

12 3

12 3 1

6 11

n

n

n n

= −[ ]

+ [ ]

= −[ ] + [ ]=

, , ,

,

, , ˆ , ˆ

, , , , , ,

0 0 0 0 0

0 0 0 0

ρρ

ρρ

ρρ

ρρ

++

= [ ]

∈ ∈

= = [ ] ∈

× ×

×

E q

R

t R q R q R R

7 1

6 7

11

21

1

63 12

73 12

1 1 0 1 0 03 12

n

n

nn

,

, , , , , , ,

, .,

E 0 0 E

q

q

q

E E

0

LM

E4

t E q E q E q E q

t E q E q E q E q E q

t E q E q

ild

i i i i

ijd

i j i j ij i j i j

imd

i m i m

= + + +

= + + + +

= + +

− + +

− − + + +

− −

2 1 1 3 1 4 1 1 5 1 2

1 1 2 1 3 4 1 5 1 1

1 1 2 1

( ) ( ) ( )

( ) ( ) ( ) ( )( )

( ) ( )

ˆ ,

,

EE q E q3 4 1im i m+ +( ) .



Now from (18.34) and (18.35), define

to get

(18.37)

and have the same structure as R11 except E4 is replaced by , and , respectively.Equation (18.37) will be needed to express the tendon forces in terms of q. Equations (18.28) and(18.37) yield the dependent vectors (t1n1, t2, t3, t4, t6, t7, t9, t10) in terms of the independent vectors(t5, t1, r1, r2). Therefore,

(18.38)

18.5 Tendon Forces

Let the tendon forces be described by

(18.39)

l t t t t t t t

t t t t

dn

dT dT dTndT dT

ndT

nmdT

T

ndT dT dT

ndT T

= [ ]= [ ]

1 1 11 21 1 12 2

1 1 1 2

, , , , , , ,

, , , , ,

K K K

K

l ld nm nm nm d nm= ∈ ∈ ∈+ ×Rq R R q R R( + ), , , ,( )24 3 12 12 24 3

R

R

R R

R R R

R R R

R R

R

R R

R R R R

R

E E E

E E E

=

∈ ∈

=

× ×

0

11 12

21 11 12

21 11 12

21 11

12

21 11

24 120

3 12

11

3 4 2

2 3 4

0 0

0

0

0 0

0

L L L

O M

O M

O M

M O O

M O O O

L L

L L

O

ˆ

, , ,ijn n n

MM

O M

M O

O O O

L

L L

O M

M O O O O M

M O O O

O O

L L

0

0

0

0

0 0 0

0 0

0

0

0 0 0

E E E

E E E

E

E 0 E E

R

E

E

E

E

R

2 3 4

2 3 4

4

4 2 3

12

5

5

5

5

=

,

i(( ) ),

, , ,

, , , , , , , ,

, , , .

i k i k i

i

k k

BlockDiag i

+ +

×

= > = >

= [ ] ∈ = →

= [ ] = −[ ]

= [ ]

0 0

0 0 0 0

0 0

if if

,

1 1

1 5

12

12

21 1 124 12

0 6 7 6 3 3

7 3 3

R

R E E E R

R E E E I I

E I I

(

L L

L

R11 R11 E4 E4

l RQld = .

ft

tα αα

αij ij

ij

ij

F=



For tensegrity structures with some slack strings, the magnitude of the force Fαij can be zero, fortaut strings Fαij > 0. Because tendons cannnot compress, Fαij cannot be negative. Hence, themagnitude of the force is

(18.40)

where

(18.41)

where is the rest length of tendon tαij before any control is applied, and the control is uαij,the change in the rest length. The control shortens or lengthens the tendon, so uαij can be positiveor negative, but . So uαij must obey the constraint (18.41), and

(18.42)

Note that for t1n1 and for α = 2, 3, 4, 6, 7, 8, 9, 10 the vectors tαij appear in the vector ld relatedto q from (4.7) by ld = Rq, and for α = 5, 1 the vectors tαij appear in the vector l related to q from(18.28), by l = Q–1 q. Let Pαij denote the selected row of R associated with tαij for αij = 1n1 andfor α = 2, 3, 4, 6, 7, 8, 9, 10. Let Pαij also denote the selected row of Q–1 when α = 5, 1. Then,

(18.43)

(18.44)

From (18.39) and (18.40),

fαij = – Kαij (q)q + bαij (q)uαij

where

(18.45)

(18.46)

Hence,

F k Lij ij ij ijα α α α= −( )t

kif L

k if Lij

ij ij

ij ij ij

αααα αα

αα αα αα

∆∆0

0

,

,

t

t

>

> ≤

L u Lij ij ijo

α α α− + ≥ 0

L ijoα > 0

L ijoα > 0

u Lij ijo

α α≤ > 0.

t qαα αα ααij ij ij

nm= ∈ ×R R, 3 12

t q qαα αα ααij

Tij

Tij

2= R R

K q q q Kαα αα αα αα αα αα ααij ij ij

o Tij

Tij ij ij

nmk L( ) ,= ( ) −

∈

− ×∆∆ R R R12 1 3 12R

b q q q q bαij ij

Tij

Tij ij ijk( ) ( ) , = ∈− ×∆∆

αα αα αα αα ααR R R R12 3 1

f

f

f

f

f

f

f

f

f

K

K

K

K

K

K

K

K

qijd

ij

ij

ij

ij

ij

ij

ij

ij

ij

ij

ij

ij

ij

ij

ij

ij

=

= −

2

3

4

6

7

8

9

10

2

3

4

6

7

8

9

10



or

(18.47)

and

or

(18.48)

Now substitute (18.47) and (18.48) into

Hence, in general,

or by defining

(18.49)

+

b

b

b

b

b

b

b

b

2

3

4

6

7

8

9

10

2

3

4

6

7

8

9

10

ij

ij

ij

ij

ij

ij

ij

ij

ij

ij

ij

ij

ij

ij

ij

ij

u

u

u

u

u

u

u

u

f K q P uijd

ijd

ijd

ijd= − + ,

ff

f

K

Kq

bbij

o ij

ij

ij

ij

ij

ij

ij

ij

u

u=

= −

+

5

1

5

1

5

1

5

10

0

f K q P uijo

ijo

ijo

ijo= − + .

f

f

f

f

f

K

K

K

K

q

P

P

P

P

u

u

u

1

1 1

11

21

1

1 1

11

21

1

1 1

11

21

1

1 1

11

21

1

d

nd

d

nd

nd

d

nd

nd

d

nd

nd

d

nd

u

=

= −

+

M M O M

= − +K q P u1 1 1d d d

f

f

f

f

K

K

K

q

P

P

P

u

u

u

K q P2

12

22

2

12

22

2

12

22

2

12

22

2

2 2d

d

d

nd

d

d

nd

d

d

nd

d

d

nd

d=

= −

+

= − +M M O M 22 2

d du .

f K q P ujd

jd

jd

jd= − +

K

K

K

K

P

P

P

P

d

d

d

md

d

d

d

md

=

=

1

2

1

2

M M,



fd = –Kdq + Pdud.

Likewise, for forces (18.48),

(18.50)

fo = – Koq + Pouo.

Substituting (18.49) and (18.50) into (18.E.21) yields

f = –(BdKd + BoKo)q + BdPdud + BoPouo + Wow, (18.51)

which is written simply as

(18.52)

by defining,

f11o

f

f

f

f

K

K

K

q

P

P

P

u

u

u

1

11

21

1

11

21

1

11

21

1

11

21

1

o

o

o

no

o

o

no

o

o

no

o

o

no

=

= −

+

M M O M

f

f

f

f

K

K

K

q

P

P

P

u

u

u

jo

jo

jo

njo

jo

jo

njo

jo

jo

njo

jo

jo

njo

=

= −

+

1

2

1

2

1

2

1

2

M M O M

f K q P ujo

jo

jo

jo= − +

f Kq Bu W w= − + +˜ ˜ o ,

˜ ,K B K B K= +∆ d d o o

˜ , ,B B P B P=[ ]∆ d d o o

B P

B P B P

B P B P B P

B P B P

B P

B P

B P B P

d d

d d

d d d

d d

d

md

md

md

=

−

3 1 4 2

5 1 6 2 4 3

5 2 6 3

5 3

4

5 1 8

0 0

0

0

0 0

L L

O M

O O M

M O O O

M O O O

L L

,



(18.53)

(18.54)

In vector in (18.54), u1n1 appears twice (for notational convenience u1n1 appears in and in. From the rules of closure, t9i1 = – t1i1 and t7im = – t10im, i = 1, 2, …, n, but t1i1, t7im, t9i1, t10im all

appear in (18.54). Hence, the rules of closure leave only n(10m – 2) tendons in the structure, but(18.54) contains 10nm + 1 tendons. To eliminate the redundant variables in (18.54) define =Tu, where u is the independent set u , and is given by (18.54). We chooseto keep t7im in u and delete t10im by setting t10im = – t7im. We choose to keep t1i1 and delete t9i1 bysetting t9i1 = – t1i1, i = 1, 2, …, n. This requires new definitions of certain subvectors as follows in(18.57) and (18.58). The vector is now defined in (18.54). We have reduced the vector by2n + 1 scalars to u. The T matrix is formed by the following blocks,

B P

B P B P

B P B P

B P

B P

B P

o o

o o

o o

o

mo

mo

=

1 1 2 2

7 2 2 3

7 3

2

7

0 0

0

0

0 0

L

O M

M O O

M O O

L L

K B K B K

B K B K B K B K

B K B K B K B K B K

B K B K B K B K B K

B K B K B K B K= + =

+ + ++ + + ++ + + ++ + +d d o o

d o d o

d d d o o

d d d o o

d d d o

3 1 1 1 4 2 2 2

5 1 6 2 4 3 7 2 2 3

5 2 6 3 4 4 7 3 2 4

5 3 5 4 4 5 7 4 ++

+ + + ++ +

− − −

−

B K

B K B K B K B K B K

B K B K B K

2 5

5 2 6 1 4 7 1 2

5 1 6 7

o

md

md

md

mo

mo

md

md

mo

M

˜ ,

ˆ

ˆu

u

u

u

u

u

u

u

u

u

u

u

u

u

u

u

u

u

u

u

u

u

=

=

1

2

3

4

1

2

3

4

1

2

3

4

1

2

3

4

d

d

d

d

md

o

o

o

o

mo

d

d

d

d

md

o

o

o

o

mo

M

M

M

M

u u1d

u1o )

u∈ −n m( )10 2 u ∈ +R10 1nm

u u



(18.55)

where

(18.56)

There are n blocks labeled T1, n(m – 2) blocks labeled I8 (for m ≤ 2 no I8 blocks needed, seeappendix 18.D), n blocks labeled T2, nm blocks labeled I2 blocks, and n blocks labeled S.

The block becomes

(18.57)

T

T S

T S

I

I

T

T

I

I

I

I

=

∈ +( )× −( )( )

0 0 0 0 0L

O O

O

O

O

O

1

1

1

8

8

2

2

2

2

2

2

10 1 10 2R nm n m

T

I

1

6 6 1

1 6

1 6

8 70

1

=

∈×

×

×

×

0

0

0

R

TI

27 8 7

0 0 0 0 1 0 0=

−

∈ ×R

S

= −

∈×

×0

0

6 28 20 1

0R .

u1d

ˆ

ˆ

ˆ

ˆ

ˆ

, ˆ , , , , ...u

u

u

u

u

u 1

11

21

31

1

1

2 1

3 1

4 1

6 1

7 1

8 1

10 1

7 1 1 2 3d

d

d

d

nd

id

n

n

n

n

n

n

n

u

u

u

u

u

u

u

i n=

=

∈ =×∆∆

M

R ,, j = 1



The block becomes

(18.58)

The block is the block with the first element u1n1 removed, because it is included in From (18.17) and (18.52),

(18.59)

where,

Kp = H(q) (q),

B = H(q) (q)T,

D = H(q)Wo.

The nodal points of the structure are located by the vector p. Suppose that a selected set of nodalpoints are chosen as outputs of interest. Then

yp = Cp = CPq (18.60)

where P is defined by (18.26). The length of tendon vector tαij = is given from (18.44).Therefore, the output vector yl describing all tendon lengths, is

.

Another ouput of interest might be tension, so from (18.40) and (18.44)

.

The static equilibria can be studied from the equations

Kp(q)q = B(q)u + D(q)w, yp = CPq. (18.61)

umd

ˆ

ˆ

ˆ

ˆ

ˆ

, ˆ , , , , ... , .u

u

u

u

u

u md

md

md

md

nmd

imd

nm

nm

nm

nm

nm

nm

nm

u

u

u

u

u

u

u

i n j m=

=

∈ = =×∆∆

1

2

3

2

3

4

6

7

8

9

7 1 1 2 3

M

R

u1d u1

d uno1.

˙ ( ( ˙ ) ( )) ( ) ( ) ,q K q K q q B q u D q w+ + = +r p

K

B

ααij q

y yl ij=

M

Mαα ,

y q qαα ααααij

T Tijij

= ( ) 12

y yf ij ij ij ij ijF F k L=

= −M

Mαα αα αα αα αα, ( )



Of course, one way to generate equilibria is by simulation from arbitrary initial conditions andrecord the steady-state value of q. The exhaustive definitive study of the stable equilibria is in aseparate paper.27

Damping strategies for controlled tensegrity structures are a subject of further research. Theexample case given in Appendix 18.D was coded in Matlab and simulated. Artificial critical dampingwas included in the simulation below. The simulation does not include external disturbances orcontrol inputs. All nodes of the structure were placed symmetrically around the surface of a cylinder,as seen in Figure 18.10. Spring constants and natural rest lengths were specified equally for alltendons in the structure. One would expect the structure to collapse in on itself with this giveninitial condition. A plot of steady-state equilibrium is given in Figure 18.11 and string lengths inFigure 18.12.

FIGURE 18.10 Initial conditions with nodal points on cylinder surface.

FIGURE 18.11 Steady-state equilibrium.

FIGURE 18.12 Tendon dynamics.

10 5 0 5 10

105051002468

101214161820

Initial Conditions Perspective View

10 8 6 4 2 0 2 4 6 8 1010

8

6

4

2

0

2

4

6

8

10 Initial Conditions Top View

10 5 0 5 10

105051002468

101214161820

Steady State Equilibrium Perspective View

10 8 6 4 2 0 2 4 6 8 10-10

-8

-6

-4

-2

0

2

4

6

8

10Steady State Equilibrium Top View

0 5 10 15 20 25 30 35 40 45 503

4

5

6

7

8

9

10

Ten

don

Leng

th

time

Tendon Dynamicst1, t7, t9, t10, (Diagonal Strings)

t4, t8, (Saddle Strings)

t5, t3, (Vertical Strings)

t2,t6, (Base, Top Strings)



18.6 Conclusion

This chapter developed the exact nonlinear equations for a Class 1 tensegrity shell, having nm rigidrods and n(10m – 2) tendons, subject to the assumption that the tendons are linear-elastic, and therods are rigid rods of constant length. The equations are described in terms of 6nm degrees offreedom, and the accelerations are given explicitly. Hence, no inversion of the mass matrix isrequired. For large systems this greatly improves the accuracy of simulations.

Tensegrity systems of four classes are characterized by these models. Class 2 includes rods thatare in contact at nodal points, with a ball joint, transmitting no torques. In Class 1 the rods do nottouch and a stable equilibrium must be achieved by pretension in the tendons. The primal shellclass contains the minimum number of tendons (8nm) for which stability is possible.

Tensegrity structures offer some potential advantages over classical structural systems composedof continua (such as columns, beams, plates, and shells). The overall structure can bend but allelements of the structure experience only axial loads, so no member bending. The absence ofbending in the members promises more precise models (and hopefully more precise control).Prestress allows members to be uni-directionally loaded, meaning that no member experiencesreversal in the direction of the load carried by the member. This eliminates a host of nonlinearproblems known to create difficulties in control (hysteresis, dead zones, and friction).

Acknowledgment

The authors recognize the valuable efforts of T. Yamashita in the first draft of this chapter.



Appendix 18.A Proof of Theorem 18.1

Refer to Figure 18.8 and define

using the vectors and which locate the end points of the rod. The rod mass center is locatedby the vector,

(18.A.1)

Hence, the translation equation of motion for the mass center of the rod is

(18.A.2)

where a dot over a vector is a time derivative with respect to the inertial reference frame. A vector locating a mass element, dm, along the centerline of the rod is

(18.A.3)

and the velocity of the mass dm, v, is

(18.A.4)

The angular momentum for the rod about the mass center, hc, is

(18.A.5)

where the mass dm can be described using dx = L dρ as

(18.A.6)

Hence, (18.A.2) can be rewritten as follows:

(18.A.7)

where (18.A.1) and (18.A.3) yield

(18.A.8)

(18.A.4)–(18.A.8) yield

q p p q p p1 2 1 2 2 1

= + = −, ,

p1

p2

p qc

= 12 1

.

mm

c˙ ˙ (ˆ ˆ ),p q f f= = +

2 1 1 2

p

p p (p p ) q ( q= + − = + − ≤ ≤ =1 2 1 1 2

12

12

0 1ρ ρ ρ ρ) , , ,xL

v p q q= = + −˙ ˙ ( ) ˙ .12

121 2

ρ

h p p pc cm

dm= − − ×∫ ( ) ˙

dmmL

L d m d=

=( ) . ρ ρ

h p p p( c cm d= − ×∫ ( ) ˙ )

0

1

ρ

p p q− = −c

( ) .ρ 12 2



(18.A.9)

The applied torque about the mass center, is

Then, substituting hc and from (18.A.9) into Euler’s equations, we obtain

or

(18.A.10)

Hence, (18.A.2) and (18.A.10) yield the motion equations for the rod:

(18.A.11)

We have assumed that the rod length L is constant. Hence, the following constraints for hold:

Collecting (18.A.11) and the constraint equations we have

(18.A.12)

h q q q

q q q

q q

c m d

m d d

m

= − × + −

= × − + −

= × −

+ −

∫

∫∫

( ) ˙ ( ) ˙

˙ ( ) ˙ ( )

˙ ( )

ρ ρ ρ

ρ ρ ρ ρ

ρ ρ ρ

12

12

12

12

12

12

12

12

12

13

12

2 1 20

1

2 1 2

2

0

1

0

1

2

2

0

1

1

33

0

1

2

2 212

= ×

˙

˙

q

q qm

ττc,

ττc = × −12 2 2 1q f f(ˆ ˆ )

ττc

hc c=

˙ ( ˙ ˙ ˙ )

˙ (ˆ ˆ )

h q q q q

q q q f f

c

m

m

= × + ×

= × = × −

12

1212

2 2 2 2

2 2 2 2 1

m

m

2 1 1 2

6 2 2 2 2 1

˙ ˆ ˆ

( ˙ ) (ˆ ˆ )

q f f

q q q f f

= +× = × −

q2

q q

q q q q q q q q

q q q q q q

2 2

2

2 2 2 2 2 2 2 2

2 2 2 2 2 2

2 0

0

⋅ =

⋅ = ⋅ + ⋅ = ⋅ =

⋅ = ⋅ + ⋅ =

L

d

dt

d

dt

( ) ˙ ˙ ˙

( ˙ ) ˙ ˙ ˙

m

m

L

2 1 1 2

6 2 2 2 2 1

2 2 2 2

2 2

2

0

˙ ˆ ˆ

( ˙ ) (ˆ ˆ )

˙ ˙ ˙

q f f

q q q f f

q q q q

q q

= +× = × −

⋅ + ⋅ =⋅ =



We now develop the matrix version of (18.A.12). Recall that

Also note that ET⋅E = 3 × 3 identity. After some manipulation, (18.A.12) can be written as:

(18.A.13)

Introduce scaled force vectors by dividing the applied forces by m and mL

Then, (18.A.13) can be rewritten as

(18.A.14)

Solving for requires,

(18.A.15)

Lemma For any vector q, such that qTq = L2,

Proof:

Since the coefficient of in (18.A.15) has linearly independent columns by virtue of the Lemma,the unique solution for is

q E f Efqi i i i= =; ˆ ˆ

m

m

T T

T L

2 1 1 2

6 2 2 2 2 1

2 2 2 2

2 22

˙ ˆ ˆ

˜ ˙ ˜ (ˆ ˆ )

˙ ˙ ˙

.

q f f

q q q f f

q q q q

q q

= +

= −

= −

=

g f f g f f1 1 2 2 1 2 2

2 6= + = −∆ ∆(ˆ ˆ ) , (ˆ ˆ ) .

m mL

˙

˜ ˙ ˜ ( )

˙ ˙ ˙

.

q g

q q q g

q q q q

q q

1 1

2 2 2 22

2 2 2 2

2 22

== −= −=

L

L

T T

T

˙q2

˜˙

˙ ˙

˜q

qq

q qq

g2

22

2 2

22

2T T L

=

−

−

0

0

˜ ˜q

q

q

qI

T

T

TL

= 2 3

0

0

0

0

0

0

3 2 1

3 1 2

2 1 3

3 2

3 1

2 1

1 2 3

23

q q q

q q q

q q q

q q

q q

q q

q q q

L

−−

−

−−

−

= I ∆.

˙q2˙q2



(18.A.16)

where the pseudo inverse is uniquely given by

It is easily verified that the existence condition for in (18.A.15) is satisfied since

Hence, (18.A.16) yields

(18.A.17)

Bringing the first equation of (18.A.14) together with (18.A.17) leads to

(18.A.18)

Recalling the definition of g1 and g2 we obtain

(18.A.19)

where we clarify

Equation (18.A.19) is identical to (18.17), so this completes the proof of Theorem 18.1.

Example 1

The generalized forces are now defined as

˙˜

˙ ˙

˜,q

q

q q qq

g22

2 2 2

22

2=

−

−

+

T T L0

0

˜ ˜ ˜ ˜ ˜q

q

q

q

q

q

q

q

q

q2

2

2

2

2

2

1

2

2

2 2

2T T

T

T T

T

T

T

L

=

=

+ −

−

˙q2

Iq

q

q

q q qq

g−

−

−

=

+˜ ˜

˙ ˙

˜,2

2

2

2 2 2

22

2 0T T T L0

0

˙˙ ˙

˜qq q

q q g22 2

2 2 22

2= − +T

L

˙

˙ ˜˙ ˙q

q I

q

q

I

q

g

gq q1

2 3

1

2

3

22

1

22 2

2

+

=

0 0

0

0

0T

L

˙

˙ ˜˙ ˙q

q I

q

q

I

q

f

fq q1

2 3

1

2

3

322

1

22 2

2 2

2

+

=

0 0

0

0

0T

L Lm

f

ff f

f f1

2

1 2

1 2

=

+−

ˆ ˆ

ˆ ˆ .

q p p1 1 211 21

12 22

11

12

= + =++

=

p p

p p

q

q,

q p p2 2 121 11

22 12

21

22

= − =−−

=

p p

p p

q

q,



From (18.A.16),

Example 2

Using the formulation developed in 18.A.3, 18.A.4, and 18.A.5 we derive the dynamics of a planartensegrity. The rules of closure become:

We define the independent vectors lo and ld:

The nodal forces are

FIGURE 18.A.1 A rigid bar of the length L and mass m.

g f f g f f1 1 211 21

12 222 2 2 1 2

21 11

22 12

2 2 6 6= + =++

= − =

−−

m m

f f

f f mL mL

f f

f f(ˆ ˆ ) , (ˆ ˆ )

˙

˙

˙

˙

˙ ˙

˙ ˙

q

q

q

q

q

q

q

q

qq q

Lq q

L

11

12

21

22

11

12

21

22

222

0 0 0 0

0 0 0 0

0 0 0

0 0 0

1 0 0 0

0 1 0 0

0 0212

222

2

212

222

2

= −

−

+−+

+

qq q

q q q21 22

21 22 2120 0

1

2

−

g

g

t t

t t

t t

t t

5 4

8 1

7 2

6 3

= −

=

= −

= −

lr

r

t

l

t

t

t

o d=

=

ρ

1

2

5

1

2

3

,

f

f f

f f

f f

f f

f f w f f w

f f w f f w

f f w f f w

f f w

=

+−+−

=

− + + − +− + − − +

+ + + − − ++ +

ˆ ˆ

ˆ ˆ

ˆ ˆ

ˆ ˆ

( ) ( )

( ) ( )

( ) ( )

(

1 2

1 2

3 4

3 4

3 2 1 5 1 2

3 2 1 5 1 2

1 2 3 3 5 4

1 2 33 3 5 4) ( )− − − +

f f w



We can write

where

Or, with the obvious definitions for Bo, Bd, and W1, in matrix notation:

Bo fo + Bd fd + W1w. (18.A.20)

The nodal vectors are defined as follows:

and

We define

The relation between q and p can be written as follows:

f

I

I

I

I

f

I I I

I I I

I I I

I I I

f

I I

I I

I I

I I

w=−−

+

− −−

−

+−

−

2

2

2

2

2 2 2

2 2 2

2 2 2

2 2 2

2 2

2 2

2 2

2 2

o d

0 0

0 0

0 0

0 0

f f f

f

f

f

w

w

w

w

w

o d= [ ] =

=

5

1

2

3

1

2

3

4

, , .

f =

p

p r

p

p r

1

2 1

3

4 2

=

= +

=

= +

ρρ

ρρ

ρρ

ρρ

ˆ

ˆ

,

ˆ .ρρ ρρ= + + −r t r1 5 2

q p p r

q p p r

q p p r r t r

q p p r

1 2 1 1

2 2 1 1

3 4 3 2 1 5 2

4 4 3 2

2

2 2

= + = +

= − =

= + = + = + + −

= − =

∆∆

∆∆

∆∆

∆∆

ρρ

ρρ ρρˆ ( )

q

q

q

q

q

I I

I I

I I

I I

p

p

p

p

=

=−

−

1

2

3

4

2 2

2 2

2 2

2 2

1

2

3

4

0 0

0 0

0 0

0 0

,



and

(18.A.21)

We can now write the dependent variables ld in terms of independent variables lo. From (18.30)and (18.31):

t1 = ρρρρ + r1 –(ρρρρ + r1 + t5 – r2),

t2 = ρρρρ – (ρρρρ + r1 + t5 – r2).

By inspection of Figure 18.A.2, (18.30) and (18.31) reduce to:

t1 = r2 – t5

t2 = –r1 + r2 – t5

t3 = r1 + t5, (18.A.22)

or,

Equation (18.A.21) yields

FIGURE 18.A.2 A planar tensegrity.

q

q

q

q

q

I I

I

I I I I

I

r

r

t

Q=

=−

=

1

2

3

4

2 2

2

2 2 2 2

2

1

2

5

2

2 2 2

0 0

0 0 0

0 0 0

l

ρρ

o.

I

t

t

t

I I

I I I

I I

r

r

t

d =

=−

− −

1

2

3

2 2

2 2 2

2 2

1

2

5

0 0

0

0 0

ρρ



Hence,

We can now write out the tendon forces as follows:

or

fd = –Kdq + Pdud

and

fo = [f5] = –[K5] q + [b5] [u5],

or

fo = – Koq + Pouo,

using the same definitions for K and b as found in (18.45) and (18.46), simply by removing the ijelement indices. Substitution into (18.A.20) yields:

With the matrices derived in this section, we can express the dynamics in the form of (18.20):

l

I I

I

I

I I I I

Q qo =

=

−

− −

=

= −

ρρr

r

t

0 0

0 0 0

0 0 0

q

q

q

q

1

2

5

2 2

2

2

2 2 2 2

1

2

3

4

112

2

2.

l

I I I I

I I I I q Rqd =−

− −−

=12

2 2 2 2

2 2 2 2

2 2 2 2I I I I

f

f

f

f

K

K

K

q

b

b

b

d

u

u

u

=

−

1

2

3

1

2

3

1

2

3

1

2

3

= + ,

f B K q P u B K q P u

B K B K q B P u B P u

= − + + − +

= − + + +

o o o o

o o o o o

( ) ( )

( ) ( ).

d d d d

d d d d d

˙ ( ) ,q K K q B Dw+ + = + r p u

Kr = 1,

K HKp = ˜ ,

B HB = ˜ ,



,

where

and

D HW H W = =o1 1

H H

I 0 0 0

0 q 0 0

0 0 I 0

0 0 0 q2

= =

1

2

322

322

2 12

22

mL

L

˜

˜

,

K

0 0 0 0

0 q q I 0 0

0 0 0 0

0 0 0 q q I

K B K B K

B B P B P

r

T

T

d d o o

d d o o

L

L

=

+

−

−

11

22 2 2

22

4 4 4

˙ ˙

˙ ˙

,

˜ ,

˜ [ , ],

∆∆

∆∆

˜ , ˜ ,

˙ ˙ ˙ ˙ , ˙ ˙ ˙ ˙ .

q q

q q q q

22 22

221 22

21 22 212 4

2 422

41 42

41 42 412

2 2 212

222

4 4 412

422

=−

−

=

−−

= + = +

q q q

q q q

q q q

q q q

q q q qT T



Appendix 18.B Algebraic Inversion of the Q Matrix

This appendix will algebraically invert a 5 × 5 block Q matrix. Given Q in the form:

(18.B.1)

we define x and y matrices so that

Qx = y (18.B.2)

where

(18.B.3)

Solving (18.B.2) for x we obtain

x = Q–1y (18.B.4)

Substituting (18.B.1) and (18.B.3) into (18.B.2) and carrying out the matrix operations we obtain

(18.B.5)

Solving this system of equations for x will give us the desired Q–1 matrix. Solving each equationfor x we have

(18.B.6)

Elimination of x on the right side of (18.B.6) by substitution yields

Q

Q

Q Q

Q Q Q

Q Q Q Q

Q Q Q Q Q

=

11

21 22

21 32 22

21 32 32 22

21 32 32 32 22

0 0 0 0

0 0 0

0 0

0

x

x

x

x

x

x

y

y

y

y

y

y

=

=

1

2

3

4

5

1

2

3

4

5

,

Q x y

Q x Q x y

Q x Q x Q x y

Q x Q x Q x Q x y

Q x Q x Q x Q x Q x y

11 1 1

21 1 22 2 2

21 1 32 2 22 3 3

21 1 32 2 32 3 22 4 4

21 1 32 2 32 3 32 4 22 5 5

=+ =+ + =+ + =+ + + =

+

+

x Q y

x Q Q x y

x Q Q x Q x y

x Q Q x Q x Q x y

x Q Q x Q x Q x Q x

1 11 1

2 221

21 1 2

3 22 21 1 32 2 3

4 22 21 1 32 2 32 3 4

5 22 21 1 32 2 32 3 32 4

( +

( +

( +

(

1

1

1

1

== −= − −= − − −= − − − −

−

−

−

−

−

)

)

)

+ y5)



(18.B.7)

Or, in matrix form,

(18.B.8)

where we have used the notation with the following definitions

Using repeated patterns, the inverse may be computed by

(18.B.9)

x y

x y y

x y y y

x y y y y

x y y y y y

1 11 1

2 21 1 22 2

3 31 1 32 2 33 3

4 41 1 42 2 43 3 44 4

5 51 1 52 2 53 3 54 4 55 5

== += + += + + += + + + +

ΛΛΛΛ ΛΛΛΛ ΛΛ ΛΛΛΛ ΛΛ ΛΛ ΛΛΛΛ ΛΛ ΛΛ ΛΛ ΛΛ

Q−

1 =

ΛΛΛΛ ΛΛΛΛ ΛΛ ΛΛΛΛ ΛΛ ΛΛ ΛΛΛΛ ΛΛ ΛΛ ΛΛ ΛΛ

11

21 22

31 32 33

41 42 43 44

51 52 53 54 55

0 0 0 0

0 0 0

0 0

0

Λrow column,

ΛΛ

ΛΛ ΛΛ ΛΛ ΛΛ

ΛΛ

ΛΛ

11 111

21 221

21 111

221

21 11 22 21 11

31 221

32 221

21 111

221

21 111

41 221

32 221

32 221

21 111

221

32 221

21 1112

=

= − = − = −

= −

= − + −

−

− − −

− − − − −

− − − − − −

Q

Q Q Q Q Q Q

Q Q Q Q Q Q Q Q

Q Q Q Q Q Q Q Q Q Q Q Q Q–2222

121 11

1

51 221

32 221

32 221

21 111

221

32 221

21 111

221

32 221

32 221

32 221

21 111

221

21 111

22 33 44 55 221

32

3 3

− −

− − − − − − −

− − − − − − −

−

= − +

+ −

= = = =

=

Q Q

Q Q Q Q Q Q Q Q Q Q Q Q

Q Q Q Q Q Q Q Q Q Q Q Q

Q

ΛΛ

ΛΛ ΛΛ ΛΛ ΛΛ

ΛΛ ΛΛ4343 54 221

32 221= = − − −ΛΛ Q Q Q

ΛΛ ΛΛ

ΛΛ

42 53 221

32 221

32 221

221

32 221

52 221

32 221

32 221

32 221

221

32 221

32 221

221

32 2212

= = −

= − + −

− − − − −

− − − − − − − − −

Q Q Q Q Q Q Q Q

Q Q Q Q Q Q Q Q Q Q Q Q Q Q Q

Q− =

1

11

21 22

21 32 222

21 32 32 223

212

31 32 32 224

213

312

31 32 32 22

ΛΛΛΛ ΛΛΘΘΛΛ ΛΛ ΛΛΘΘ ΛΛ ΘΘΛΛ ΛΛ ΛΛΘΘ ΛΛ ΘΘ ΛΛ ΘΘΛΛ ΛΛ ΛΛΘΘ ΛΛ ΘΘ ΛΛ ΘΘ ΛΛ ΘΘΛΛ ΛΛ ΛΛM M O O O O O



using

Only and powers of Θ need to be calculated to obtain for any (n, m).The only matrix inversion that needs to be computed to obtain is and , substantiallyreducing computer processing time for computer simulations.

ΘΘ ΛΛ

ΛΛ

ΛΛ ΛΛ ΛΛ

ΛΛ

ΛΛ ΛΛ ΛΛ

=

=

=

=

=

( )

.

I Q

Q

Q

Q

Q

−

−

−

−

−

22 32

11 111

21 22 21 11

22 221

32 22 32 22

ΛΛ ΛΛ ΛΛ ΛΛ11 22 21 32, , , , Q−1

Q−1 Q111− Q22

1−



Appendix 18.C General Case for (n, m) = (i, 1)

In Appendix E the forces acting on each node are presented, making special exceptions for thecase when (j = m = 1). The exceptions arise for (j = m = 1) because one stage now contains bothclosure rules for the base and the top of the structure. In the following synthesis we use and

from (18.E.1) and and from (18.E.4). At the right, where (i = 1, j = 1):

At the center, where (1 < i < n, j = 1):

At the left end of the base in Figure 18.5, where (i = n, j = 1):

(18.C.1)

Using,

f111

f211 f311 f411

f

f f

f f

f f

f f

f f f f w f f f f w

f f f f w f f f

11

+

=

+−+−

=

− + + + + − − − +− + + + + − − −

ˆ ˆ

ˆ ˆ

ˆ ˆ

ˆ ˆ

( ) ( )

( ) (

111 211

111 211

311 411

311 411

211 311 1 1 2 1 111 511 111 411 811 211

211 311 1 1 2 1 111 511 111 411

n n

n n −− +− − + + − − + +− − + − − − + +

f w

f f f f w f f f f w

f f f f w f f f f w

811 211

811 711 321 421 311 611 511 6 1 7 1 411

811 711 321 421 311 611 511 6 1 7 1 411

)

( ) ( )

( ) ( )

. +

+n n

n n

f

f f

f f

f f

f f

f f f f w f f f f

i

i i

i i

i i

i i

i i i i i i i i

1

1 1 2 1

1 1 2 1

3 1 4 1

3 1 4 1

1 1 1 2 1 1 2 1 3 1 1 1 1 1 4 1 5 1 8

=

+−+−

=

+ − + + + − − + −− −

ˆ ˆ

ˆ ˆ

ˆ ˆ

ˆ ˆ

( ) (( ) ( ) ii i

i i i i i i i i i i

i i i i i i i

1 2 1

1 1 1 2 1 1 2 1 3 1 1 1 1 1 4 1 5 1 8 1 2 1

8 1 7 1 3 1 1 4 1 1 3 1 5 1 6 1

++ − + + − − − + − +

− − + + + − +− −

w

f f f f w f f f f w

f f f f w f f

)

( ) ( )

( ) (( ) ( )

( + ) ( + ) −− +− − + + − − + − +

− −

− −

f f w

f f f f w f f f f w6 1 7 1 1 4 1

8 1 7 1 3 1 1 4 1 1 3 1 5 1 6 1 6 1 7 1 1 4 1

( ) ( )

( + ) ( + ) ( ) ( )

+

+i i i

i i i i i i i i i i

)

( ) ( )

.

f

f f

f f

f f

f f

f f f f w f f f f

n

n n

n n

n n

n n

n n n n n n n n

1

1 1 2 1

1 1 2 1

3 1 4 1

3 1 4 1

1 1 1 2 1 1 3 1 2 1 1 1 1 1 4 1 5 1 8

=

+−+−

=

+ + − + + − − + −− −

ˆ ˆ

ˆ ˆ

ˆ ˆ

ˆ ˆ

( ) (( ) ( ) nn n

n n n n n n n n n n

n n n n n n

1 2 1

1 1 1 2 1 1 3 1 2 1 1 1 1 1 4 1 5 1 8 1 2 1

7 1 8 1 311 411 3 1 6 1 1 5 1 6 1

++ + − + − − − + − +

− + − + + + − − + +− −

−

w

f f f f w f f f f w

f f f f w f f f f

)

( ) ( )

( ) (( ) ( )

( ) 77 1 1 4 1

7 1 8 1 311 411 3 1 6 1 1 5 1 6 1 7 1 1 4 1

( )

( ) ( )

n n

n n n n n n n n

−

− −

+− + − + + − − − + + +

w

f f f f w f f f f w

)

( ) ( )

.



Or, in matrix notation, with the implied matrix definitions,

(18.C.2)

ff

ff

f

f

f

f

f

f

f

f

w

w

w

w

w

ijo

ijijd

ij

ij

ij

=

=

=

5

1

2

3

4

6

7

8

9

10

1

2

3

4

, , ,

f

I 0 0 0 0 0 0 0

I 0 0 0 0 0 0 0

0 0 0 I I 0 0 0

0 0 0 I I 0 0 0

f

I I I 0 0 I 0 0

I I I 0 0 I 0 0

0 0 0 I I I 0 0

0 0 0 I I I 0 0

3

3

3 3

3 3

3 3 3 3

3 3 3 3

3 3 3

3 3 3

11 1=−

−

+

− − −−

−− −

nd

f11d

+−−

+

−−−

+

+

0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0

0 I I 0 0 0 0 0

0 I I 0 0 0 0 0

f

I I

I I

I 0

I 0

f

I

I

0

0

f

I I 0 0

3 3

3 3

21

3 3

3 3

3

3

11

3

31 1

3 3

d on

II I 0 0

0 0 I I

0 0 I I

w3 3

3 3

3 3

11

−

−

.

f B f B f B f B f B f Ww11 1 1 11 11 21 21 11 11 1 1 1 1 11= + + + + +ˆ .nd

nd d d d d o o

n n

f

0 0 0 0 0 0 0

0 0 0 0 0 0 0

0 0 0 0 0 0

0 0 0 0 0 0

f

0 0 0 0

0 0 0 0

0 0 0 0 0

0 0

i id

1 1 1=

−−

+

− − −−

−−( )

I

I

I I

I I

I I I I

I I I I

I I I

3

3

3 3

3 3

3 3 3 3

3 3 3 3

3 3 3

00 0 0

f

0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0

0 0 0 0 0 0

0 0 0 0 0 0

f

0

0

0 0

0 0

f

− −

+−−

+

+( ) −(

I I I

I I

I I

I

I

3 3 3

3 3

3 3

3

3

id

id

i

1

1 1 1))

+

−−−

+−

−

1

1 1

o

io

i

I I

I I

I

I

I I

I I

I I

I I

3 3

3 3

3

3

3 3

3 3

3 3

3 3

0

0

f

0 0

0 0

0 0

0 0

w .



Or, in matrix form,

(18.C.3)

Or, in matrix form,

(18.C.4)

Now assemble (18.C.2)–(18.C.4) into the form

(18.C.5)

where

Or, simply, (18.C.5) has the form (18.E.21), where

f B f B f B f B f B f Wwi

di

d did d

id o

io o

io

i1 01 1 1 11 1 21 1 1 01 1 1 11 1 1= + + + +− −ˆ .( ) ( + ) ( ) +

f

I 0 0 0 0 0 0 0

I 0 0 0 0 0 0 0

0 0 0 I I 0 0 0

0 0 0 I I 0 0 0

f

I I I 0 0 I 0 0

I I I 0 0 I 0 0

0 0 0 I I I 0 0

0

n nd

1

3

3

3 3

3 3

1 1

3 3 3 3

3 3 3 3

3 3 3

=−

−

+

− − −−

−−( )

00 0 I I I 0 0

f

0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0

0 I I 0 0 0 0 0

0 I I 0 0 0 0 0

f

0 I

0 I

0 0

0 0

f

− −

+−−

+

−( )

3 3 3

1

3 3

3 3

11

3

31 1

nd

dnoo

nlo

n n

+−−

+

−

+−

−

I 0

I 0

I 0

I 0

f

I

I

0

0

f

I I 0 0

I I 0 0

0 0 I I

0 0 I I

w

3

3

3

3

3

31 1

3 3

3 3

3 3

3 3

1 .

f B f B f B f B f B f B f Wwnd

nd d

nd d d o

no

no

no

n n n1 01 1 1 11 1 21 11 01 1 1 1 1 1 1 1 1 1= + + + + + +−( ) −( )ˆ .

f B f B f W w1 3 1 1 1 1 1= + +d o ,

f

f

f

f

ff

ff

f

f

f

w

w

w

w

j

j

j

nj

d nd j

o

jo

jo

njo

j

j

j

nj

=

=

=

=

1

21

1 1

1

1

2

1

2

M M M, , ,

B

B B B 0 0 B

0 B 0

0 0

0 B

B B 0 0 B B

B

B 0 0

B

0

B 03

1 1 11 21 1

01

21

1 1 21 01 11

1

11

01

11

=

=

nd d

nd

d

d

nd d d

o

o

o

L

O O O

M O O O O O M

O O O O

O O O O

L

L L

O O M

O O O M

M O O

ˆ

ˆ

ˆ

,

00 0 B BL 01 1o

no

,

W W W1 = [ ]BlockDiag K K, , , .



The next set of necessary exceptions that apply to the model (i,1) arises in the form of the Rmatrix that relates the dependent tendons set to the generalized coordinates . For any(i,1) case R takes the form following the same procedure as in (18.32) and (18.33).

(18.C.6)

where

The transformation matrix T that is applied to the control inputs takes the following form. Theonly exception to (18.55) is that there are no I8 blocks due to the fact that there are no stagesbetween the boundary conditions at the base and the top of the structure. The second set of I2

blocks is also not needed since m = 1. Hence, the appropriate T matrix for = Tu is

where S is defined by (18.56), and

f f f f f f w w W W = = 1 1 1 1 1, , , , ,d d o o o= = =

B B B Bd o= =3 1, .

(ld = Rq)

R =R

R 0

11˜ ,

˜

˜

˜

˜

˜

˜

˜

R

E E 0 E

E E E 0

0 E E E

E E E 0

0 E

E 0 0 E E

11

3 4 2

2 3 4

2 3 4

2 3 4

4

4 2 3

=

L L

O

O M

M O

O O O

L

˜ .E

I I

I I

I I

I I

I I

4

3 3

3 3

3 3

3 3

3 3

12

=

−

− −

−

0 0

0 0 0 0

0 0 0 0

0 0

0 0

0 0 0 0

0 0

0 0

u

T =

S

S

I

I

0 01

1

1

2

2

10 1 8

L

O O

O

T

T

′

′

∈ +( )×R n n,



There are n blocks, n S blocks, and n I2 blocks. The control inputs are now defined as

The block becomes

Appendix 18.D will explicitly show all matrix forms for the specific example (n,m) = (3,1).

′ =

−

∈ ×T

16

1 0 0 0 0 0

0 1 0 0 0 0

0 0 1 0 0 0

0 0 0 1 0 0

0 0 0 0 1 0

0 0 0 0 0 1

0 0 0 0 0 0

0 0 0 0 1 0

8R .

′T1

u

u

u = 1

1

d

o

.

u1d

u

u

u

u

u

u1

11

21

31

1

1

2 1

3 1

4 1

6 1

7 1

8 1

6 1 1 2 3 1′

′

′

′

′

′=

=

∈ = = =×d

d

d

d

nd

id

n

n

n

n

n

n

u

u

u

u

u

u

i n j m

M

L, , , , , ,R



Appendix 18.D Example Case (n,m) = (3,1)

Given the equation for the dynamics of the shell class of tensegrity structures:

We explicitly write out the matrices that define the problem:

,

where Q11 is (36 × 36). Furthermore,

which is also a (36 × 36) matrix.

(36 × 36) = (36 × 36) [(36 × 75) ∗ (75 × 36) + (36 × 18) ∗ (18 × 36)]

In order to form , R is needed. In order to form Q–1 is needed. Therefore, we obtainR as follows:

˙ ( ( ˙ ) ( )) ( ) ( ) .q K q K q q B q u D q w+ + = +r p

q q

q

q

q

q

q

q

q

q

q

q

q

q

q

q

q

Q l

A 0 0

D B 0

D E B

l

l

l

= =

=

= =

1

11

21

31

111

211

311

411

121

221

321

421

131

231

331

431

11 1

11

21

31

=

A 0 0

D B 0

D E B

r

r

r

r

r

r

r

r

r

r

r

ρρ11

111

211

511

111

121

221

521

121

131

321

531

K q

r ( ˙ ) =

ΩΩΩΩ

ΩΩ

11

21

31

0 0

0 0

0 0

K q HK H B K B K H B K B Kpd d o o d o( ) ˜ ( ) ( )= = + = +1 3 1 1 1

0

K

H

H

H

B B B B

B B B

B B B B

K

K

K

K

B

p

nd d

nd

d d d

nd d d

d

d

d

o

=

+

11

21

31

1 1 11 21 1

01 11 21

1 1 21 01 11

131

11

21

31

11

0 0

0 0

0 0

0

ˆ

ˆ

ˆ

00 0

0

0

B B

B B

K

K

K01 11

01 1

11

21

31

o o

ono

o

o

o

K11d K1

o,



where the matrix is dimension (75 × 36).

where the dimensions are (36 × 25) = (36 × 75) ∗ (75 × 25).

and the dimensions are (36 × 6) = (36 × 18) ∗ (18 × 6).

and the dimensions are (36 × 24) = (36 × 36) ∗ (36 × 31) ∗ (31 × 24).The control inputs uαij are defined as

l lt

tRq

R

Rq

t

t

t

t

E E

E E E

E E E

E E E

q

q

q

d dd

d

d

d

d

d

= =

=

=

1131

1

0

111

131

11

21

31

6 7

3 4 2

2 3 4

4 2 3

11

21

31

= ˜ [ ],

˜

˜

˜

0

,

B q HBT H B P B P T( ) ˜ [ , ]= = d d o o

B P B P

B B B B

B B B

B B B B

P

P

P

P

d d dn

d dnd

d d d

nd d d

nd

d

d

d

[ ] [ ]

= = 3 1

1 1 11 21 1

01 11 21

1 1 21 01 11

1 1

11

21

31

ˆ

ˆ

ˆ0

0 0 0

0 0 0

0 0 0

0 0 0

B P B P

B

B B

B B

P

P

P

o o o

o

o o

ono

o

o

o

[ ] [ ]

= = 1 1

11

01 11

01 1

11

21

31

0 0

0

0

0 0

0 0

0 0

B =

H

H

H

B P B P

T S

T S

T S

I

I

I

11

21

31

3 1 1 1

1

1

1

2

2

2

0 0 01

0 0

0 0

0 0

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0 0

0 0 0 0 0

0 0 0 0 0

[ ]

′′

′

d o,

L L

uu

u

u

u

u

u

u

u

u

u

=

=

′

=

′′′

×d

o

d

o

d

d

d

o

o

o

1

1

11

21

31

11

21

31

24 1( )



where

The external forces applied to the nodes arise in the product Dw, where

with dimensions: (36 × 36) = (36 × 36) ∗ (36 × 36).

so

′ =

∈ = =×u id

n

n

n

n

n

n

u

u

u

u

u

u

i j1

2 1

3 1

4 1

6 1

7 1

8 1

6 1 1 2 3 1R , , , .

Bu∈ ×R( ) .36 1

D q HW H W

H

H

H

W

W

W

( ) = o = [ ]=

1

11

21

31

0 0

0 0

0 0

0 0

0 0

0 0

w w

w

w

w

= 1

11

21

31

36 1=

×( )

Dw ∈ ×R (36 1).



Appendix 18.E Nodal Forces

At the base, right end of Figure 18.5, where (i = 1, j = 1):

At the center of the base, where (1 < i < n, j = 1):

At the left end of the base in Figure 18.5, where (i = n, j = 1):

(18.E.1)

At the second stage, where (1 ≤ i ≤ n, j = 2):

f

f f

f f

f + f

f f

f f f f w f f f f w

f f f f w f f f

11

111 211

111 211

311 411

311 411

211 311 1 1 2 1 111 511 111 411 811 211

211 311 1 1 2 1 111 511 111

=

+−

−

=

− + + + + + − − − +− + + + + − − −

ˆ ˆ

ˆ ˆ

ˆ ˆ

ˆ ˆ

( ) ( )

( ) (n n

n n 411411 811 211

811 711 1011 321 42 1 311 611 511 212 112 411

811 711 1011 321 42 1 311 611 511 212 112 411

− +− + − + + + − + + +− + − + + − − + + +

f w

f f f f f w f f f f w

f f f f f w f f f f w

)

( ) ( )

( ) ( )

f

f f

f f

f + f

f f

f f f f w f f f f

i

i i

i i

i i

i i

i i i i i i i i

1

1 1 2 1

1 1 2 1

3 1 4 1

3 1 4 1

1 1 1 2 1 1 2 1 3 1 1 1 1 1 4 1 5 1

=

+−

−

=

+ − + + + − − + −− −

ˆ ˆ

ˆ ˆ

ˆ ˆ

ˆ ˆ

( ) (( ) ( ) 88 1 2 1

1 1 1 2 1 1 2 1 3 1 1 1 1 1 4 1 5 1 8 1 2 1

8 1 7 1 10 1 3 1 1 4 1 1 3 1 5

i i

i i i i i i i i i i

i i i i i i i

++ − + + − − − − − +

− + − + + + −− −

+ +

w

f f f f w f f f f w

f f f f f w f

)

( ) ( )

( ) (( ) ( )

( ) ( ) 11 6 1 10 1 1 2 2 2 4 1

8 1 7 1 10 1 3 1 1 4 1 1 3 1 5 1 6 1 10 1 1 2 2 2 4 1

+ − + +− + − + + − − + − + +

−

+ + −

f f f f + w

f f f f f w f f f f f + wi i i i i

i i i i i i i i i i i i

( )

( ) ( ) ( )

)

( ) ( )

f

f f

f f

f + f

f f

f f f f w f f f f

n

n n

n n

n n

n n

n n n n n n n n

1

1 1 2 1

1 1 2 1

3 1 4 1

3 1 4 1

1 1 1 2 1 1 3 1 2 1 1 1 1 1 4 1 5 1

=

+−

−

=

+ + − + + − − + −− −

ˆ ˆ

ˆ ˆ

ˆ ˆ

ˆ ˆ

( ) (( ) ( ) 88 1 2 1

1 1 1 2 1 1 3 1 2 1 1 1 1 1 4 1 5 1 8 1 2 1

7 1 8 1 10 1 311 411 3 1 10 1 1 5 1

n n

n n n n n n n n n n

n n n n n n

+

+ + − + − − − + − +

− + + − + + + − − +

− −

−

w

f f f f w f f f f w

f f f f f w f f

)

( ) ( )

( ) (

( ) ( )

( ) ff f f w

f f f f f w f f f f f w

6 1 1 2 2 2 4 1

7 1 8 1 10 1 311 411 3 1 10 1 1 5 1 6 1 1 2 2 2 4 1

n n n n

n n n n n n n n n n

+ + +

− + + − + + − − − + + + +

−

)

( ) ( )( )



(18.E.2)

At the typical stage (1 ≤ j < m, 1 ≤ i ≤ n):

f

f f

f f

f f

f f


f f f f f

12

112 212

112 212

312 412

312 412

6 1 7 1 9 2 212 312 112 112 412 512 812 912 212

6 1 7 1 9 2 212

=

+−+−

=

− + − − + +( ) + − − + − + +( )− + − − +

ˆ ˆ

ˆ ˆ

ˆ ˆ

ˆ ˆ

n n n

n n n 312312 112 112 412 512 812 912 212

712 812 1012 322 422 312 512 612 10 2 113 213 412

712 812 1012 322 422 312 512 612 10 2 113 213 412

+( ) − − − + − + +( )− + + − + +( ) + − + − + + +( )− + + − + +( ) − − + − + + +

w f f f f f w


f f f f f w f f f f f wn

n(( )

=

+−+−

=

− + − − +− − −

f

f f

f f

f f

f f

f f f f f

i

i i

i i

i i

i i

i i i i i

2

1 2 2 2

1 2 2 2

3 2 4 2

3 2 4 2

6 1 1 7 1 1 9 1 2 2 2 3

ˆ ˆ

ˆ ˆ

ˆ ˆ

ˆ ˆ

( ) ( ) ( ) 22 1 2 1 2 4 2 5 2 8 2 9 2 2 2

6 1 1 7 1 1 9 1 2 2 2 3 2 1 2 1 2 4 2 5 2 8 2 9 2 2

+( ) + − − + − + +( )− + − − + +( ) − − − + − + +− − −

w f f f f f w

f f f f f w f f f f f wi i i i i i i

i i i i i i i i i i i i( ) ( ) ( ) 22

7 2 8 2 10 2 3 1 2 4 1 2 3 2 5 2 6 2 10 1 2 1 3 2 3 4 2

7 2 8 2 10 2 3 1 2 4 1 2

( )− + + − + +( ) + − + − + + +( )− + + − + +

+ + −

+ +


f f f f f wi i i i i i i i i i i i

i i i i i

( ) ( ) ( )

( ) ( ) 33 2 5 2 6 2 10 1 2 1 3 2 3 4 2

2

1 2 2 2

1 2 2 2

3 2 4 2

3 2 4 2

i i i i i i i

n

n n

n n

n n

n n

( ) − − + − + + +( )

=

+−+−

−f f f f f w

f

f f

f f

f f

f f

( )

ˆ ˆ

ˆ ˆ

ˆ ˆ

ˆ ˆ

=

− + − − + +( ) + − − + − + +( )− + − − +

− − −

− − −


f f f f6 1 1 7 1 1 9 1 2 2 2 3 2 1 2 1 2 4 2 5 2 8 2 9 2 2 2

6 1 1 7 1 1 9 1 2 2 2

( ) ( ) ( )

( ) ( ) ( )

n n n n n n n n n n n n

n n n n ff w f f f f f w


f

3 2 1 2 1 2 4 2 5 2 8 2 9 2 2 2

7 2 8 2 10 2 312 412 3 2 5 2 6 2 10 1 2 1 3 2 3 4 2

7 2

n n n n n n n n

n n n n n n n n n n

n

+( ) − − − + − + +( )− + + − + +( ) + − + − + + +( )− +

−( )

ff f f f w f f f f f w 8 2 10 2 312 412 3 2 5 2 6 2 10 1 2 1 3 2 3 4 2n n n n n n n n n+ − + +( ) − − + − + + +( )

−( )

f

f f

f f

f f

f f


1

11 21

11 21

31 41

31 41

6 1 7 1 9 21 31 11 11 41 51 81 91 21

j

j j

j j

j j

j j

n j n j nj j j j j j j j j

=

+−+−

=

− + − − + +( ) + − − + − + +− −

ˆ ˆ

ˆ ˆ

ˆ ˆ

ˆ ˆ

( ) ( ) (( )− + − − + +( ) − − − + − + +( )

− + + − + +( ) + − + − +− −

+


f f f f f w f f f f6 1 7 1 9 21 31 11 11 41 51 81 91 21

71 81 101 32 42 31 51 61 10 11 1

n j n j nj j j j j j j j j j

j j j j j j j j nj j

( ) ( )

( ) ++ +( )− + + − + +( ) − − + − + + +( )

+

+ +

f w

f f f f f w f f f f f w21 1 41

71 81 101 32 42 31 51 61 10 11 1 21 1 41

( )

( ) ( )

j j

j j j j j j j j nj j j j



(18.E.3)

(1 ≤ i ≤ n, j = m)

(18.E.4)

f

f f

f f

f f

f f

f f f f f w f f f

ij

ij ij

ij ij

ij ij

ij ij

i j i j i j ij ij ij ij ij ij

=

+−+−

=

− + − − + + + − − + −− − − − −

ˆ ˆ

ˆ ˆ

ˆ ˆ

ˆ ˆ

( ) (( )( ) ( )( ) ( )

1 2

1 2

3 4

3 4

6 1 1 7 1 1 9 1 2 3 1 1 4 5 ff f w


f f f f f

8 9 2

6 1 1 7 1 1 9 1 2 3 1 1 4 5 8 9 2

7 8 10 3 1 4 1

ij ij ij

i j i j i j ij ij ij ij ij ij ij ij ij

ij ij ij i j i

+ +− + − − + + − − − + − + +− + + − +

− − − − −

+ +

)

( ) ( )

(( )( ) ( )( ) ( )

( ) ( )) ( ) ( ) ( )

( ) ( ) ( ) ( )

) ( )

( ) (j ij ij ij i j i j i j ij

ij ij ij i j i j ij ij ij i j i j

+ + − + − + + +− + + − + + − − + − + +

− + +

+ + − +

w f f f f f w

f f f f f w f f f f f3 5 6 10 1 1 1 2 1 4

7 8 10 3 1 4 1 3 5 6 10 1 1 1 22 1 4

3 4

3 4

6 1 1 7 1 1 9 1

i j ij

nj

nj nj

nj nj

nj nj

nj nj

n j n j n j

( )

( )( ) ( )( ) ( )

)

ˆ ˆ

ˆ ˆ

ˆ ˆ

ˆ ˆ

(

+

− − − − −

+

=

+−+−

=

− + − −

w

f

f f

f f

f f

f f

f f f

1 2

1 2

ff f w f f f f f w

f f f f f w f f f f f w2 3 1 1 4 5 8 9 2

6 1 1 7 1 1 9 1 2 3 1 1 4 5 8 9 2

nj nj nj nj nj nj nj nj nj

n j n j n j nj nj nj nj nj nj nj nj nj

+ + + − − + − + +− + − − + + − − − + − + +−

− − − − −

) ( )

( ) ( )

(( )( ) ( )( ) ( )

ff f f f f w f f f f f w

f f f f f w f f f f(

7 8 10 31 41 3 5 6 10 1 1 1 2 1 4

7 8 10 31 41 3 5 6 10 1 1

nj nj nj j j nj nj nj n j n j n j nj

nj nj nj j j nj nj nj n j n

+ + − + + + − + − + + +− + + − + + − − + − +

− + +

−

) ( )

( ) (( ) ( ) ( )

) (( ) ( ) )j n j nj+ ++ +

1 2 1 4f w

f

f f

f f

f f

f f


1m

m m

m m

m m

m m

n m n m m m nm nm m m m m m

=

+−+−

=

− + − + − + + − + − − + +− −

ˆ ˆ

ˆ ˆ

ˆ ˆ

ˆ ˆ

( ) (( ) ( )

11 21

11 21

31 41

31 41

6 1 7 1 21 31 9 1 41 51 81 11 91 2nmnm

n m n m m m nm nm m m m m m nm

m m m m m m m nm nm m

)

( ) ( )

( ) ( )

(

( ) ( )− + − + − + − − + − − + +− + − + + + − + − + +−

− −f f f f f w f f f f f w

f f f f w f f f f w 6 1 7 1 21 31 9 1 41 51 81 11 91 2

71 81 32 42 31 51 61 6 7 41

ff f f f w f f f f w

f

f f

f f

f f

f f

f

71 81 32 42 31 51 61 6 7 41

1 2

1 2

3 4

3 4

m m m m m m m nm nm m

im

im im

im im

im im

im im

+ − + + − − + − + +

=

+−+−

=

−

) ( )

ˆ ˆ

ˆ ˆ

ˆ ˆ

ˆ ˆ

( 66 1 1 7 1 1 9 1 2 3 1 1 4 5 8 9 2

6 1 1 7 1 1 9 1 2 3

( )( ) ( )( ) ( )

( )( ) ( )( ) ( )

) ( )

(i m i m i m im im im im im im im im im

i m i m i m im im

− − − − −

− − − − −

+ − − + + + − − + − + +− + − − + +

f f f f w f f f f f w

f f f f f w11 1 4 5 8 9 2

7 8 3 1 4 1 3 5 6 1 6 7 1 4

7 8 3 1

im im m im im im im

im im i m i m im im i m im i m im

im im i

) ( )

( ) ( )

(( ) ( ) ( ) ( )

( )

− − − + − + +− + − + + + − − + + +− + −

+ + − −

+

f f f f f w

f f f f w f f f f w

f f f

i

mm i m im im i m im i m im

nm

nm nm

nm nm

nm nm

nm nm

+ + − − − + + +

=

+−+−

+ − −f w f f f f w

f

f f

f f

f f

f f

4 1 3 5 6 1 6 7 1 4

1 2

1 2

3 4

3 4

( ) ( ) ( )) ( )

ˆ ˆ

ˆ ˆ

ˆ ˆ

ˆ ˆ

=

− + − − + + + − − + − + +− + − −

− − − − −

− − − − −

( ) ( )

(( )( ) ( )( ) ( )

( )( ) ( )( ) ( )


f f f f6 1 1 7 1 1 9 1 2 3 1 1 4 5 8 9 2

6 1 1 7 1 1 9 1 2

n m n m n m nm nm nm nm nm nm nm nm nm

n m n m n m nm ++ + − − − + − + +− + − + + + − + − + +− + − + +

− −

f w f f f f f w

f f f f w f f f f w

f f f f w

3 1 1 4 5 8 9 2

7 8 31 41 3 5 6 6 1 7 1 4

7 8 31 41 3

nm nm nm nm nm nm nm nm

nm nm m m nm nm nm n m n m nm

nm nm m m nm

) ( )

( ) ( )

(( ) ( )

)) ( )( ) ( )− − + − + +

− −f f f f w 5 6 6 1 7 1 4nm nm n m n m nm



Or, in matrix notation, with the implied matrix definitions,

(18.E.5)

ff

ff

f

f

ff

f

ff

f

w

w

w

w

w

ijo

ijijd

ij

ij

ij

=

=

=

5

1

2

3

4

6

7

8

9

10

1

2

3

4

, ,

f f

I I I

I I I

I

I

I

I

I I I

I I

11

3

31

3 3 3

3 3 3

3

3

3

3

3 3 3

3 3

=

+

− −−

−

−

−−

I 0 0 0

I 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

0

0

0 0 0

0 0 0

0 0 0

0 0 0

0 n

d

00

0 0 0 0

0 0 0 0

I 0 0 0

I 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

0 I I 0

0 I I 0

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

I

f

f

3

11

3

3

123 3

3 3

+

−

+−−

d

d

+−−

−

+

−

+

+−

f

I

I

I

I

I

If

I

I

f

I

If

I I

I I

21

3

3

3

3

3

311

3

3

12

3

31 1

3 3

3 3

d

o on0

0

0

0

0

0

0

0

0

0

0 0

0 0

0 0 II I

I I

w3 3

3 3

11

0 0 −

.

f B f B f B f B f B f B f B f Ww11 1 1 11 11 12 12 21 21 11 11 12 12 1 1 1 1 11= + + + + + + +nd

nd d d d d d d o o o o

n n .

f

I

I

I

I

f

I I I

I I I

I

I

I

I

I I I

I I I

i id

1

3

3

3

3

1 1

3 3 3

3 3 3

3

3

3

3

3 3 3

3 3 3

=

+

− −−

−

−

−−

−

0 0 0

0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0

0 0 0

0

0

0 0 0

0 0 0

0 0 0

0 0 0

0

0

( )

fid1

+

−

+−−

+

+

0 0 0 0

0 0 0 0

0 0 0

0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

0 0

0 0

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

0

0

I

I

fI I

I I

f

3

3

23 3

3 3

1 1 id

id( )

00

0

0

0

0

0

0

0

0

0

0

0

0 0

0 0

0 0

I

If

I

I

I

I

I

If

I

I

f

I I

I I

I I

3

31 1

3

3

3

3

3

31

3

3

2

3 3

3 3

3 3

+−−

−

+

−

+−

− ( )io

io

io

00 0 I I

w

3 3

1

−

i



Or, in matrix form,

(18.E.6)

Or, in matrix form,

(18.E.7)

f B f B f B f B f B f

B f B f Ww

id

id d

id d

id d

id o

io

oio o

io

i

1 01 1 1 11 1 12 2 21 1 1 01 1 1

11 1 12 2 1

= + + + +

+ + +

− + −( ) ( ) ( )

.

f

I

I

I

I

f

I I I

I I I

I

I

I

I

I I I

I I I

n nd

1

3

3

3

3

1 1

3 3 3

3 3 3

3

3

3

3

3 3 3

3 3 3

=

+

− −−

−

−

−−

−

0 0 0

0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0

0 0 0

0

0

0 0 0

0 0 0

0 0 0

0 0 0

0

0

( )

fnd1

+

−

+−−

+

0 0 0 0

0 0 0 0

I 0 0 0

I 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

f f

0

0

0

0

I

I

3

3

nd d2

3 3

3 3

11

3

0 0 0 0

0 0 0 0

0 I I 0

0 I I 0

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

331 1

3

3

3

3

13

3

2

3

31 1

3 3

0

0

f

I

I

I

I

0

0

0

0

f

0

0

0

0

0

0

I

I

f

I

I

0

0

f

I I 0 0

+−−

+

−

+

−

+

−( )no

no

no

n

II I 0 0

0 0 I I

0 0 I I

w3 3

3 3

3 3

1

−

−

n .


B f B f B f Ww

nd

nd d

nd d

nd d d o

no

no

no o

no

n n n

1 01 1 1 11 1 12 2 21 11 01 1 1

1 1 12 2 1 1 1 1 1

= + + + +

+ + + +

− −( ) ( )

.

f

I I I

I I I

I

I

I I

I I

I I I

I I I

f12

3 3 3

3 3 3

3

3

3 3

3 3

3 3 3

3 3 3

123 3

3 3

=

− −−

−

−−

−−

+−−

0

0

0 0 0

0 0 0

0 0

0 0

0

0

0 0 0 0

0 0 0 0

0 I I 0

0 I I 0

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

d

+

−

+

−−

f

f

22

3

3

13

3

3

3

3

d

d

0 0 0 0

0 0 0 0

I 0 0 0

I 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 I

0 0 0 I

0 0 0 0

0 0 0 0

I 0 0 0

I 0 0 0

0 0 0 0

0 0 0 0

fnd1



Or, in matrix form,

(18.E.8)

Or, in matrix form,

(18.E.9)

f

I

I

I

I

I

If

I

I

f+

−−

−

+−−

−

+

−

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

0 0 I 0

0 0 I 0

0 0 0 I

0 0 0 I

0

0

0

0

0

0

0

03

3

3

3

2

3

3

3

3

3

312

3

3

nd o

1313

3 3

3 3

3 3

3 3

12

o

I I

I I

I I

I I

w+−

−

0 0

0 0

0 0

0 0

.


B f B f Ww

12 12 12 21 22 12 13 1 1 2 2

11 12 12 13 12

= + + + +

+ + +

d d d d d dnd

nd

nd

nd

o o o o .

f f

I I I

I I I

I

I

I I

I I

I I I

I

i id

2

3

3

3

31 1

3 3 3

3 3 3

3

3

3 3

3 3

3 3 3

3

=

−−

+

− −−

−

−−

−−

−

0 0 0 I

0 0 0 I

0 0 0 0

0 0 0 0

I 0 0 0

I 0 0 0

0 0 0 0

0 0 0 0

0

0

0 0 0

0 0 0

0 0

0 0

0 ( )

II I

f

f

3 3

2

3

3

3

3

3

3

3

0

0 0 0 0

0 0 0 0

I 0 0 0

I 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

0 0 I 0

0 0 I 0

0 0 0 I

0 0 0 I

+

−

+

−−

−

id

id

+−−

+−−

−

+

−

+

f

f

I

I

I

I

I

If

( )

( )

id

id

io

1 2

3 3

3 3

1 2

3

3

3

3

3

32

0 0 0 0

0 0 0 0

0 I I 0

0 I I 0

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

0

0

00

0

0

0

0

0

0 0

0 0

0 0

0 0

I

I

f

I I

I I

I I

I I

w3

3

3

3 3

3 3

3 3

3 3

2

−

+−

−

io

i .


B f B f Ww

i nd

id d

id d

id

nd

id d

id

oio o

io

i

2 1 1 1 12 2 12 3 2 1 2 21 1 2

11 2 12 3 2

= + + + +

+ + +

− − +( ) ( ) ( )

.

f f fnd

nd

23 3

3 3

12

3

3

3

31 1=

−−

+

−−

−

0 0 0 0

0 0 0 0

0 I I 0

0 I I 0

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 I

0 0 0 I

0 0 0 0

0 0 0 0

I 0 0 0

I 0 0 0

0 0 0 0

0 0 0 0

( )



Or, in matrix form,

(18.E.10)

Or, in matrix form,

(18.E.11)

+

−−

−

+

− −−

−

−−

−−

−

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

0 0 I 0

0 0 I 0

0 0 0 I

0 0 0 I

0

0

0 0 0

0 0 0

0 0

0 0

0

3

3

3

3

1 2

3 3 3

3 3 3

3

3

3 3

3 3

3 3 3

3 3

f

I I I

I I I

I

I

I I

I I

I I I

I I

( )nd

00

0 0 0 0

0 0 0 0

I 0 0 0

I 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

0

0

0

0

0

0

I

f

f I

I

f

I

I

I

I

I

I

3

2

3

3

33

3

3

3

3

3

3

3

+

−

+

−

+−−

−

nd

nd

no 33

2

3 3

3 3

3 3

3 3

2

0

0

0 0

0 0

0 0

0 0

+−

−

f

I I

I I

I I

I I

w

no

n .

f B f B f B f B f B f B f

B f Ww

nd d

nd

nd

nd

nd d

nd d

nd o

no

ono

n

2 21 12 1 1 1 2 1 2 12 2 12 3 12 3

11 2 2

= + + + + +

+ +

− −( ) ( )

.

f f

I I I

I I I

I

I

I I

I I

I I I

I I

1

3

3

3

31

3 3 3

3 3 3

3

3

3 3

3 3

3 3 3

3

j n jd=

−−

+

− −−

−

−−

−−

−

0 0 0 I

0 0 0 I

0 0 0 0

0 0 0 0

I 0 0 0

I 0 0 0

0 0 0 0

0 0 0 0

0

0

0 0 0

0 0 0

0 0

0 0

0( )

33 3

1

3

3

3

3

3 3

3 3

0

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

0 0 I 0

0 0 I 0

0 0 0 I

0 0 0 I

0 0 0 0

0 0 0 0

0 I I 0

0 I I 0

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

I

f

f

+

−−

−

+−−

jd

njd

+

−

+−−

−

+

+

f

f

I

I

I

I

I

If

2

3

3

1 1

3

3

3

3

3

31

jd

jd

jo

0 0 0 0

0 0 0 0

I 0 0 0

I 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

0

0

0

0

0

0

0

0

( )

II

I

f

I I

I I

I I

I I

w3

3

1 1

3 3

3 3

3 3

3 3

1

−

+−

−

+( ) .jo

j

0 0

0 0

0 0

0 0


B f B f Ww

1 1 1 12 1 2 21 2 12 1 1

11 1 12 1 1 1

j nd

n jd d

jd

nd

njd d

jd d

jd

oj

o oj

oj

= + + + +

+ + +

− +

+

( ) ( )

( ) .



Or, in matrix form,

(18.E.12)

f f

I I I

I I I

I

I

I I

I I

I I Iij i jd=

−−

+

− −−

−

−−

−−

− −

0 0 0 I

0 0 0 I

0 0 0 0

0 0 0 0

I 0 0 0

I 0 0 0

0 0 0 0

0 0 0 0

0

0

0 0 0

0 0 0

0 0

0 0

0

3

3

3

31 1

3 3 3

3 3 3

3

3

3 3

3 3

3 3 3

( )( )

II I I

f

3 3 30

ijd

+

−−

−

+−−

− +

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

0 0 I 0

0 0 I 0

0 0 0 I

0 0 0 I

0 0 0 0

0 0 0 0

0 I I 0

0 I I 0

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

3

3

3

3

13 3

3 3

1 f f( ) (i jd

i ))

( )

jd

i jd

ijo+

−

+−−

−

+

−

+

0 0 0 0

0 0 0 0

I 0 0 0

I 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

0

0

0

0

0

0

0

0

3

3

1

3

3

3

3

3

3

3

3

f

I

I

I

I

I

If

I

I

+−

−

+f

I I

I I

I I

I I

wi jo

ij( ) .1

3 3

3 3

3 3

3 3

0 0

0 0

0 0

0 0


B f B f Ww

ij nd

i jd d

ijd

nd

i jd d

i jd d

i jd

oijo o

i jo

ij

= + + + +

+ + +

− − − + +

+

1 1 1 12 2 1 21 1 12 1

11 12 1

( )( ) ( ) ( ) ( )

( ) .

f f

I I I

I I I

I

I

I I

I I

I I Inj n jd=

−−

+

− −−

−

−−

−−

− −

0 0 0 I

0 0 0 I

0 0 0 0

0 0 0 0

I 0 0 0

I 0 0 0

0 0 0 0

0 0 0 0

0

0

0 0 0

0 0 0

0 0

0 0

0

3

3

3

31 1

3 3 3

3 3 3

3

3

3 3

3 3

3 3 3

( )( )

II I I

f

f

3 3 3

3

3

3

3

13 3

3 3

0

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

0 0 I 0

0 0 I 0

0 0 0 I

0 0 0 I

0 0 0 0

0 0 0 0

0 I I 0

0 I I 0

0 0 0 0

0 0 0 0

0 0

+

−−

−

+−−

−

njd

n jd ( ) 00 0

0 0 0 0

0 0 0 0

0 0 0 0

I 0 0 0

I 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

0

0

0

0

+

−

+−−

−

++

f

f

I

I

I

I

I

If

1

3

3

1

3

3

3

3

3

3

jd

n jd

njo

( ) 00

0

0

0

0 0

0 0

0 0

0 0

I

I

f

I I

I I

I I

I I

w

3

3

1

3 3

3 3

3 3

3 3

−

+−

−

+n jo

nj

( )

.



Or, in matrix form,

(18.E.13)

Or, in matrix form,

(18.E.14)

Or, in matrix notation,

(18.E.15)


B f B f Ww

nj nd

n jd d

njd

nd

n jd d

jd d

n jd

onjo o

n jo

nj

= + + + +

+ + +

− − − +

+

1 1 1 12 2 1 21 1 12 1

11 12 1

( )( ) ( ) ( )

( ) .

f

I I I

I I I

I

I

I I

I I

I I

I I

f1

3 3 3

3 3 3

3

3

3 3

3 3

3 3

3 3

13 3

3 3

m md=

− −−

−

−−

−−

+−−

0

0

0 0 0

0 0 0

0 0

0 0

0 0

0 0

0 0 0 0

0 0 0 0

0 I I 0

0 I I 0

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

f2md

+−

−

+

−−

+−

−( )

0 0 0 0

0 0 0 0

0 0 0 I

0 0 0 I

0 0 0 0

0 0 0 0

I 0 0 0

I 0 0 0

0 0 0 I

0 0 0 I

0 0 0 0

0 0 0 0

I 0 0 0

I 0 0 0

0 0 0 0

0 0 0 03

3

3

3

3

3

3

31

3

f f

I

nmd

n md

II

I

I

I

If

I I

I I

I I

I I

W3

3

3

3

31

3 3

3 3

3 3

3 3

1−

−

+−

−

0

0

0 0

0 0

0 0

0 0

mo

m .

f B f B f B f B f B f Ww1 1 1 21 2 1 1 11 1 1m md

md d

md

nmd

nmd

nd

n md o

mo

m= + + + + +−( ) .

f fim i md=

−−

+−

−−

−

−( ) −( )

0 0 0 I

0 0 0 I

0 0 0 0

0 0 0 0

I 0 0 0

I 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 I

0 0 0 I

0 0 I 0

0 0 I 0

I 0 0 0

I 0 0 0

3

3

3

31 1

3

3

3

3

3

3

− −−

−

−−

−−

+−−

−( )f

+

I I I

I I I

I

I

I I

I I

I I

I I

f

i md

md

1

3 3 3

3 3 3

3

3

3 3

3 3

3 3

3 3

13 3

3 3

0

0

0 0 0

0 0 0

0 0

0 0

0 0

0 0

0 0 0 0

0 0 0 0

0 I I 0

0 I I 0

0 0 0 0

0 0 0 0

00 0 0 0

0 0 0 0

0

0

0 0

0 0

0 0

0 0

−−

−

+−

−

+( )f

+

I

I

I

I

I

If

I I

I I

I I

I I

w

i md

imo

im

1

3

3

3

3

3

3

3 3

3 3

3 3

3 3

.

f B f B f B f B f

+ B f Ww

im nd

i md

nmd

i md

md

imd d

i md

oimo

im

= + + +

+

−( ) −( ) −( ) +( )1 1 1 1 1 21 1

11 .



Or, in matrix notation,

(18.E.16)

Now assemble (18.E.5)–(18.E.16) into the form

(18.E.17)

where

f fnm n md=

−−

+−

−−

−

− −

0 0 0 I

0 0 0 I

0 0 0 0

0 0 0 0

I 0 0 0

I 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 I

0 0 0 I

0 0 I 0

0 0 I 0

I 0 0 0

I 0 0 0

3

3

3

31 1

3

3

3

3

3

3

( )( )

+

− −−

−

−−

−−

+−−

−f

I I I

I I I

I

I

I I

I I

I I

I I

f

( )n md

nmd

1

3 3 3

3 3 3

3

3

3 3

3 3

3 3

3 3

3 3

3 3

0

0

0 0 0

0 0 0

0 0

0 0

0 0

0 0

0 0 0 0

0 0 0 0

0 I I 0

0 I I 0

0 0 0 0

0 0 0 0

0 0 00 0

0 0 0 0

0

0

0 0

0 0

0 0

0 0

+−−

−

+−

−

f

I

I

I

I

I

If

I I

I I

I I

I I

w

1

3

3

3

3

3

3

3 3

3 3

3 3

3 3

md

nmo

nm .

f B f B f B B f

B f Ww

nm nd

n md

nmd

n md

md

nmd d

md

onmo

nm

= 1 1 1 1 1 1 21 1

11

( )( ) ( )

.

− − −+ + +

+ +

f

f B f B f B f B f W w1 3 1 4 2 1 1 2 2 1 1 = d d o o+ + + +

ff

ff

f

f

f

f

f

f

f

w

w

w

w

f

f

f1

1 1

1

1

2

1

2

1

2

1

2d nd j

d

jd

jd

njd

jo

jo

jo

njo

j

j

j

nj

j

j

j=

=

=

=

=, , , ,M M M M

ffnj

B

B B B 0 0 B

0 B 0

0 0

0 B

B B 0 0 B B

3

1 1 11 21 1

01

21

1 1 21 01 11

=

nd d

nd

d

d

nd d d

L

O O O

M O O O O O M

O O O O

O O O O

L

B B B

B B B

4 12 12

2 12 12

= … …

= … …

BlockDiag

BlockDiag

d d

o d

[ , , , ],

[ , , , ],



W1 = BlockDiag […, W, W, …].

Also, from (18.E.5)–(18.E.16)

(18.E.18)where

Also from (18.E.5)–(18.E.16)

(18.E.19)

(18.E.20)

Or, simply, the vector form of (18.E.17)–(18.E.19) is

(18.E.21)

where

B

B

B

B

B B

1

11

01

11

01 1

=

o

o

o

ono

0 0

0

0

0 0

L L

O O M

O O O M

M O O

L

,

f = B f B f B f B f B f W w 2 5 2 6 2 4 3 7 2 2 3 1 2d d d o o+ + + + + ,

B 0, B B B B5 5 7 11 11= [ ] = [ ], , , , ,BlockDiag o oL L

B

B

B

B

B

B B B

B

B

B B B

5

1

1

1

6

12 21 2

2

21

21 2 12

=

=

0 0 0

0

0

0

0 0 0

0 0

0

0

0

0

0 0

L

O O

O O O M

M O O O

L

L

O O O

O O O O M

M O O O O

O O O

L

nd

nd

nd

d dnd

nd

d

dnd d

,

.

f B f B f B f B f B f W w 3 5 2 6 3 4 4 7 3 2 4 1 3= + + + + +d d d o o ,

f = B f B f B f W w , m md

md

mo

m5 1 8 7 1( )− + + +

B

B B 0 0 B

B 0

0

0

0 B

B 0 0 B B

8

1 21

21

21 1

=

md d

nmd

nm

d

dnmd

md

L

O O O

O O O O M

M O O O O

O O O

f = B f B f W w d d o o o+ + ,

f

f

f

f

f

f

f

f

f

f

w

w

w

=1

1

21 1

MM

M M

m

d

d

d

md

o

o

mo

m

=

=

=

, , , ,



References

1. R. E. Skelton, Systems design, the absentee in system theory, American Automatic ControlConference, San Diego, CA, June 1999.

2. R. E. Skelton, A systems approach to integrate structure and control design, SPIE Conference onMathematics for Smart Structures, Newport Beach, CA, March 2000.

3. K. M. Grigoriadis, G. Zhu, and R. E. Skelton, Optimal redesign of linear systems, Journal ofDynamic Systems, Measurement and Control, 118(3), 598–605, September 1996.

4. M. P. Bendsoe, Optimal shape design as a material distribution problem, Structural Optimization,1, 193–202, 1989.

5. M. P. Bendsoe, Optimization of Structural Topology, Shape, and Material, Springer, Berlin, 1995.6. M. P. Bendsoe and N. Kikuchi, Generating optimal topologies in structural design using a

homozenization method, Computer Methods in Applied Mechanics and Engineering, 71, 197–224,1988.

7. A. R. Diaz and M. P. Bendsoe, Shape optimization of structures for multiple loading conditionsusing a homogenization method, Structural Optimization, 4, 17–22, 1992.

8. A. H. Simmons, C. A. Michal, and L. W. Jelinski, Molecular orientation and two-component natureof the crystalline fraction of dragline silk, Science, 271, 84–87, 1996.

9. Y. Termonia, Molecular modeling of spider silk elasticity, Macromolecules, 27, 7378–7381, 1994.10. B. I. Yakobson and R.E. Srnalley, Fullerene nanotubes beyond, American Scientist, 85, 324–337,

1997.11. R. Chu, Tensegrity, Journal of Synergetics, 2(l), 1988.12. R. B. Fuller and R. Marks, The Dymaxion World of Buckminister Fuller, Anchor Books Edition,

Garden City, NY, 1973.13. R. Buckminister Fuller, Tensile-integrity structures, US Patent 3,063,521, 1962.14. R. Buckminister Fuller, Synergetics Explorations in the Geometry of Thinking, Collier Macmillan

Publishers, London, 1975.15. H. Furuya, Concept of deplorable tensegrity structures in space applications, International Journal

of Space Structures, 7(2),143–151, 1992.16. D. E. Ingber, Cellular tensegrity: Defining new rules of biological design that govern the cyto-

skeleton, Journal of Cell Science, 104(3), 613–627, 1993.17. D. E. Ingber, Tensegrity: The architectural basis of cellular mechanotransduction, Annual Review

of Physiology, 59, 575–599, 1997.18. D. E. Ingber, Architecture of life, Scientific American, 48–57, January 1998.19. H. Kenner, Geodesic Math and How to Use It, University of California Press, Berkeley, 1976.20. R. Motro, Tensegrity systems: The state of the art, International Journal of Space Structures, 7(2),

75–83, 1992.21. I. J. Oppenheim and W. J. Williams, Tensegrity prisms as adaptive structures, ASME Annual

meeting, Dallas, November 1997.22. A. Pugh, An Introduction to Tensegrity, University of California Press, Berkeley, 1976.23. R. Adhikari, R. E. Skelton, and W. J. Helton, Mechanics of Tensegrity Beams, UCSD, Structural

System and Control Lab., Report No. 1998-1, 1998.

W W Wo BlockDiag= [ , , , ],L L1 1

B

B B 0 0

B B

0 B 0

B B

0 0 B B

B

B B 0 0

0 B

0

B

0 0 B

d o=

3 4

5 6

5

6 4

5 8

1 2

7

2

7

L

O O M

O O

M O O

L

L

O O M

M O O O

M O O

L L

=

, .



24. R. E. Skelton and C. Sultan, Controllable tensegrity, a new class of smart structures, SPIEConference, Mathematics and Control in Smart Structures, San Diego, March 1997.

25. R. E. Skelton and M. He., Smart tensegrity structure for nestor, SPIE Conference, Smart Structuresand Integrated Systems, San Diego, CA, March 1997.

26. H. Murakami, Y. Nishimura, T. J. Impesullo, and R. E. Skelton. A virtual reality environment fortensegrity structures, In Proceedings of the 12th ASCE Engineering Mechanics Conference, 1998.

27. D. Williamson and R. E. Skelton. A general class of tensegrity systems: Geometric definition,ASCE Conference Engineering Mechanics for the 21st Century, pp. 164–169, La Jolla, CA, May1998.



Documents

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