\

occurs with four skew, but linearly dependent lines.

It is clear that any axis which meets three of these

forces: [F1c]= o, [F2c]= o, [F3c]::o, then [F4c]= o and the axis also meets the fourth force. The four

lines form part of a ruled quadratic surface - a regulus.

3.8 Resolving forces through a single bar. The set up

for a single bar is, happily, the same in space as in the

plane. For two equilibrium forces F at a and G at b

P + G '= 0 implies G :: -F. We know the two forces share

the line avb. Thus F=Aab and G = -F ::. = Aba. The resolution is given by assigning the scalar -A to the bar, representing the tension or compression.

As in the plane A >O represents compression, A<'O represents

tension.

3.9 Resolving forces on a triangle. Given a set of

three equilibrium rorces F, G, and Hat the verticies

of a non-degenerate triangle abc, we know immediately

that the three forces are concurrent and coplanar with

the triangle. The resolution is virtually identical to

the plane resolution.

'F 't ac = 0 G+pbc..,o(ba= 0 H +'t ca +J3cb = 0

Taking an arbitrary point d not in this plane:

[acbd]=O and [Hbd]+ }{ LCab<0-1- ,B[cbbd]= 0

Thus 0: [Fbd] [abed]

and '( _ - -_lll_b<tl - c a bd - [i'b"Cdj'"

However F + G + H = 0, and Hb:O, so {!bd] = -[Jibd] and

the eqautions are identical. Similarly and abed .

This produces a perfect resolution, provided

If the triangle is

collinear, then it fails to resolve many external

force systems, and has an internal stress. We say the non-degenerate triangle is statically rigid in

space because it resolves all possible external force systems.

Resolving equilibria. farces on a tetrahedron. For four equilibrium forces E, F, G, and Hat the

four corners of a non-degenerate tetrahedron abed, we can seek a resolution:

= 0

= 0

= 0

.. = 0 Joining the first two equations with cd, we find

[Ecd] 0 [Fcdl + )1fi,acd] = 0 However since EiF'-tGlH:.O, we also have [Ecd].._ [Fcd]:O

and the two equations are identical. This gives a

unique solution for the resolution and we can obtain

similar resolutions for the other We conclude that

the non•degenerate tetrahedron is statically rigid, and

since the resolutions are unique, there can be no

internally resolved stress.

On the otherhand a tetrahedron which is coplanar

has both a stress ( as in section 2.13) and is not

statically rigid. The pattern continues into space -

q

(

c/

c

l=,ju.re3./ <f-s,-ny;ftx in 3-spt:trt>.

--------

(

there is a relationship between the number of stresses

and the static rigidity of an object.

3.11 The projected 4-simplex. Consider a complete

5-graph in space, with no 4 verticies coplanar (Fig 3.1).

We can single out one say e, and then Yiew the __ ,... ..- .. .. . . . .

picture as tour lines from e attacking the rigid tetrahedon

abed. Since we have four lines through a single point

there must be a linear dependence of the extensors:

).1ae+ ) 2be + + ) 4de = 0

This gives an equilibrium system which can be resolved

through the tetrahedron abed. When these for .. ces , are

reYersed for application at e, we have an internally

resolved stress on the larger structure. This single

stress, and its multiples, does not change the rigidity

of the structure but represents the necessary static

consequence of adding an- brace beyond rigidity.

4 STATIC RIGIDITY AND STRESS§S 4.1 Basic Concepts. We will present some general definitions and properties which apply to bar and joint structures in the plane and in space. The methods and approach will be generalized in other papers to cover

additional types of structures, but historical patterns as well as mathematical simplicity urge us to study

bar and joint structures as the primary example. We work within the Cayley algebra to present the

basic static problems: the application or forces and the resolution of these forces by the structure. In any dimension a force can be represented as an oriented line

segment along a particular line of application. If the Euclidean force is given with the vector Cr1 ,r2 , ••• ,tn) and point of application (p1 , ••• ,pD) then the force will be

written in the Cayley algebra as the two-extensor:

• • • • r P

a weighted line segment joining the points (p1 , ••• ,p0 ,1)

and (p1+r1 , ••• or equivalently the finite point

(p1 , ••• ,pn,l) and the infinite point <r1 , ••• ,rn,o). This extensor has (n+l)n/2 coordinates - the 2x2 minors -

n of which represent the free vector r 1 , ••• ,fn, and the

remaining n(n-1)/2 represent the moments of the force

about the various coordinate axis. Composition of

(

forces is achieved by adding the free vectors and the moments,

so it will be represented by addition of the two-extensors ( •addition of corresponding minors).

Definition 4.1 Given k forces (2-extensors) Fi i=l, ••• ,k the composit-ion F is defined by the sum being Cayley algebra addition ( or equivalently addition of the corresponding minors)

We note that, in general for n>2, the sum of forces is

not a new single force or extensor but is a dyname or screw. However if all the forces are on lines passing through the

same point Pi:: f i"P , then F: F i r i vp) : (21' i )vp and the composite force is a new force through this point.

Definition 4.2 k forces F i are in equilibrium iff %F i = o. A system of forces in equilibrium will have the

same effect on a rigid body as the 0 force - i.e. it has no effect. This occurs because a rigid body resolves these

forces back into the zero force by setting up an internal transmission of the effects: - some tensions and compressions

which pass the resultants between the various points of application of the forces F1 • The existence of such resolutions becomes the definition of static rigidity for

all types of structures. What distinguishes the various possible kinds of

structures is the points where the external forces can be

applied, and which internal tensions and compressions are available to produce the resolution. Typically we think

(

ot any structure as composed or of material - the members, with connections between two or more members - the articulations. Both the material and the articulations can assume a yariety or forms, each with its own static properties. In the real world materials are elastic and

never quite transmit forces either in tension or compression without some change in length or deformation in shape. However as a first approximation to the problems of statics

we will assume that the material is "perfect" and the members transait the forces with no change in size.

Definition 4.3 A bar and joint structure in n-1 space is

S=<J,B..> where J is a collection of. points<:'al•••••&v> in an n-dim projective apace ( Cayley algebra) - the joints, and B is a collection of unordered pairs < .•• ,(i,j), ••• > 1< i;lj<n - the bars. We assume that at#aj if (i,j )EB.

We assume, as in chapters 2 and 3, that the bars are rigid straight line segments and thus will resolve any system or forces in equilibrium which pass through this line.

Now any force Fk which strikes the line ai bi can be split and treated as two forces Fk at ai and at bi. This split is not unique but uses any appropriate resolution in the plane

For convenience, we will always assume

that the forces which are given are applied directly to

the joints.

Definition 4.4 An equilibrium force system on a bar and joint structure S:<J,B)is an assignment of a force Fi to

(

each joint ai, with F 1"ai:: 0 ( the force passes through the joint) such that ::iF i ::r o. The collection of all possible

called tbe equilibrium force space for the structure.

The resolutions ot forces on this structure must be by tensions or compressions within the bars, which can be

recorded by a scalar At,j for bar (i,j). We write Aij = .Aji since the bars are unordered pairs. The force at a1 is

then while the force at aj is

::-A.l.jajvaj.

Definition 4.5 An internal resolution of the equilibrium force system < ••• ,F1 ,. , .>. on the bar and joint structure <J,B> is an assignment of scalars Aij to the bars

sucb that:

Aijaivaj + Fi:O ( sum over j such that (i,j)eB) The collection of all possible assignments is called the

resolution space. An internally resolved stress is a resolution

of the zero force system <o, ••• ,O}; i.e. an assignment of

scalars such that'1:Aijaivaj:: 0. A non-trivial internally resolved stress is simply called a stress.

It is obvious from the definitions that we have

described several vector indexed by the joints and the bars. We will return to the analysis or these vector

spaces in sections 4.3 and 4.4. We conclude this section

With the basic definition for bar and joint structures.

. I..:.

Definition 4.6 A bar and joint structure is

!tatically rigid iff every equilibrium torce system has

an internal resolution.

4.2 Projectively equivalent structures. It is obvious

from our experience, that if we take a bar and joint

structures and apply a Euclidean transformation ( a

rotation or translation) then the structure has the

same static properties as the original structure. 1he fact that our definitions are stated in the Cayley algebra

indicates a much wider invariance. In fact any general

collineation of the space ( or linear transformation of

the underlying projective coordinates) will preserve the

basic linear equations needed to express an ·equilibrium

force system or a resolution.

It is also traditional in projective geometry to use various weights on the points of a projective space.

These maps or congruences which produce no change in the

position of the structure would be expected to have no

effect on the statics either. In this section we will

confirm that these two sets of maps maintain the static

properties.

Definition 4.7 Two bar and joint structures and

<J•,B•) are projectively equivalent iff there is a

non-singular linear transformation of the underlying

n-space, T, such that T(J)=J' is an isomorphism and

(i,j)EB iff , and (m,k)EB'.

(

Two structures S=-<J ,B> and S'=<J' are congruent iff there is an isomorphism of the joints: a6J

1J(ai) :=. = biEJ', with 0 and B: B•.

Two structures S and S' are equivalent iff there is a structure sa such that S is projectively equivalent to sn and sn is congruent to s•.

Clearly each of these provides an true equivalence

relation on the collection or all bar and joint structures.

Theorem 4.1 If two structures are equivalent then there

are induced non-singular linear transformations between their spaces of equilibrium force systems and between

their resolution spaces. These transformations preserve

static resolutions, internal stresses and static rigidity.

Proof. For a projective equivalence induced by the linear

transformation T we have the induced map of the line segments

T(avb) = T(a)vT'(b). Using this map on the forces we have

T(< ••• ,Pi, ••• >)::: < ..• , T(F i), ••• ) . with the properties that

and F.fO since T is also a linear map on the t 2-extensors. Thus

we haYe a map between the equilibrium force spaces.

For the resolution space we define T(< ••• ,Aijt•••>)

••• ,Aij, ••• ) the identity map. We notice that

).1ja1vaj:-:.r 1 )'::LT(-F i )..., (T(ai )vT(aj) )=. -T(F i) so a resolution of a force system is transformed to a

resolution of the transformed force system. -·t.

' ·[. I

(

For a congruence map we define <{J(< ••• ,Fi, ••• >):< ••• ,Fi, ••• > -the identity map. This is clearly a linear transformation between equilibrium force systems. For the resolution space we define

.. ·•Atj, ... "?) :<. .. .... > ' J This is also a non-singular linear transformation and:

>ftai )vtftaj )=-4>(F 1 ) ' il since the Cayley algebra join is billinear. This map

preserves the static resolutions. We conclude that a composition or the two types or

maps preserves static resolutions and all the related properties. •

A bar and joint structure has an underlying abstract graph or verticies and edges corresponding to the joints and bars. We call a particular structure a realization or the graph G.:<¥ ,E> itt there is, a map <R: 'R<v )=J and E:B.

Within the space of all realizations of a given graph we have this relationship or equivalent realizations which necessarily have the same static properties.

There is a more subtle, unsolved question. We

can define two structures as statically equivalent if they have the !!.same static properties". As we proceed we shall

see which static properties are but the statically

equivalent structures form a broader class than the

equivalence relation defined above. The obvious question is:

what class of transformations, in addition to the projective

• • _J

(

and congruence maps, is needed to generate this larger class?

We know of no reasonable answer but the experience ot strange unexpected relationships between statics and

pro j actions, polarities, :x:-sec.tions and other geometric

constructions can stimulate a variety of strange possibilities.

In the meantime we work within the structural geometry of weighted points in the Cayley algebra and the maps defined here.

4.3 Expected rigidity in the plane. Among the realizations of a given graph we1Nill naturally ·· distinguish between the rigid and the non-rigid realizations. The examples of

section 2 have illustrated the problem and the types of

projective conditions which produce non-rigid placements.

Still there is a problem - some graphs are not rigid

in any realization. An example is the simple quadrilateral.

There simply are not enough resolutions available to match the external equilibrium forces. Enough can be counted

exactly and will be.

For these properties which depend on the underlying

graph we will refer to verticies instead of joints and to

edges instead of bars. Consider a structure in the plane

with V verticies. At each vertex ai we can apply any force Fi, such that which yields a 2-dim vector space or

forces. The space of all possible applied forces

< F 1 , ••• ,F 1 , ••• ,F v )' has dimension 2V and the equilibrium

force systems form a subspace defined by the equation

(

Z: FrO. In the plane this is really 3 independent equations in the three coordinates ( assuming so the equilibrium force systems form a vector space of dimension 2V-3.

The resolutions on the edges <•••t)ij••••> form a vector apace of dimension E. We have the basic linear transformation )<: S 7= J<< < •• • tAij, ••• >) ::: <•. •;f>' ijaivaj, •• •>

The codomain is correct since:

and the resolution goes to an equilibrium force system.

The kernel of this map is the set of internally resolved stresses aud the structure is rigid iff the •ap covers the codomain or bas rank 2V-3J V;,-2.

Proposition 4.2 If a structure is rigid then E42V-3 If a structure has E 2V-3 then it is

rigid iff the dimension of the space of stresses is E-(2V-3).

If a structure has E::: 2V-3 then it is rigid iff there is no non-trivial stress.

Proof. These are obvious results from linear algebra using the equation rank(P{ )+ nullity().{) :w E

along with the fact that the structure is rigid iff

rank ( ,():::: 2V -3. • As we shall see shortly (section 5.4), the condition

that is essentially sufficient for static rigidity

of most realizations of a graph.

.... . ,

(

It is relevant to mention here some common terms from structural engineering. A structure is called

statically determined if each external load (external equilibrium force system) has a unique resolution and statically indeterminant ( or overdetermined) otherwise.

Clearly any structure with with a non-trivial stress is

indeterminfnt and every external force system with one resolution will have many resolutions.

A structure with E:2V-3 which is rigid will then be

statically determined. On the other hand a structure with E:2V-3 which is non-rigid ( and indeterminant) is called a critical form. In general if some realizations of a graph are rigid then the non-rigid realizations are called critical forms. Thus critical forms are always

structures with an extra stress. So far we have emphasized stresses in structures

with However structures are often built up by progressively adding one bar at a time and it is usefull

to isolate the appearance of a stress at any stage. Put another way, it is usefull to look for those subcollections

of the rows of the matrix for which first cause the nullity to reach 1. Definition 4.8 A subset of bars in a structure is

dependent iff there is a non-trivial internally resolved

stress which is zero except on B'. A set of bars is

""" . - ·

independent iff there is only the trivial resolution of the zero force system.

The use of the words dependent and independent is meant to evoke the idea of a combinatorial geometry or a matroid (Crapo-Rota). Given a placement tor the verticies we have an independence structure on the set or

all possible bars - forming a combinatorial geometry or rank

rank 2V-J. In tact it is rank 2V-J unless all the verticies are positioned on a single line, in whivh case it has rank V.

This can be expressed as a matrix equation: bar (i,j)

vertex 1 v 0 ... ... ' .. . • • • • • 0

• • vertex .... ajaj • • • ai ak •. ).ij • -• - • ••• • •

• 0 vertex j ..:, , .. ajai • •• Mk

• . .. . . -. ... • •

where each row records ' an equation f a j-: 0 and the entry is 0 unless the vertex or this row occurs in the

edge of the column. The dependence structure we have defined is the linear

dependence among the ,olut'n'IS- this matrix ( hence the term matroid}. For a particular bar and joint structure on these

joints we simply delete certain bars, which corresponds to

deleting certain columns and this gives a subgeometry or

- l ' . ,

the larger structure geometry. The essential aim or all our

studies is to characterize this geometry by seeking useful!

algorithms for deciding the dependence or independence of a set of bars.

In the vocabulary or combinatorial geometry the rigid

structures in the plane are those containing a basis

(maximal independent set) for the geometry, provided the

joints are not collinear. The minimal dependent sets ( or minimal sets of bars supporting a stress) are called

the circuits or the combinatorial geometry. These circuits

will play an important role in our analysis.

4.4 Expected rigidity in space. We can now restate

all of the definitions and counting arguments for statics

on bar structures in 3-spade. These results can, of course, be stated for n dimensions, but we are still groping for the

exact significance of resolutions of forces in 4 and higher dimensions. While we will find a surprising relationship

to four dimensions in later work, we leave such definitions

as an exercise for any inter•ated readers. P.or a structure in space with V verticies and E edges

the space of possible applied forces has dimension )V. The equilibrium equation corresponds to 6 linear

equations, provided the verticies are not all collinear,

and the space of equilibrium force systems has dimension

3V-6. As before the space of internal resolutions

(

<•••t)ij'•••l bas dimension E and we have the connecting linear J<:

=<···rDtjaiaj,•••> which maps each resolution onto the equilibrium force system which it resolves.

Proposition 4.3 A bar and joint structure in space with V 3 is rigid iff E -6 and the stress space has

dimension -6). Proof. If V>J and the joints are not collinear, then

it is clear that we need so the nullity must be by elementary linear algebra.

If the structure is all collinear with V?3, we claim it is non-rigid. It is true that, for a collinear structureJthe set external forces bas dimension 3V-5.

However at each vertex the net resolution must be another force along the same old line. With the equilibrium ( now a single equation) we know that the rank orl{,V-1. Thus the structure cannot be rigid and the dimension of the stress space is

E-(V-1) )E-(3V-6) since • Once more we have a combinatorial geometry - the

structure geometry - formed by the dependence structure

of the columns matrix for the o. We cal1 a structure dependent iff its bars hold a

non-trivial stress and independent if they hold no stress.

(

Corollary 4.4 A structure with which is collinear is dependent if , and cannot be rigid in space.

: ··· .A 'ttucture with iesdependent. if and is never rigid in space. Proof. For collinear structures the dependence follows

from the proof of proposition 4.3. We note that in collinear

structures the minimal dependent sets - the circuits - are simple polygons on the line and thus the structure geometry

is the usual combinitorial geometry or the graph. Y.or the dependence follows from

proposition 4.2 and the fact that any stress within the plane is still a stress when the structure is embedded into

3-space. The non-rigidity also follows because the maximal

rank or 'k for a coplanar structure is still 2V -3 <3V -6. •

We now know that the structure geometry in the plane

can sneak into our geometry at any time. The ultimate interest of engineers and architects centers on

spacial structures, but a thorough understanding or the structural geometry in the plane remains indispensible for

the spacial work. It is even true that a series of plane

rigid constructions can be used to build space rigid

structures {Whiteley-Porhedra, Intro III, Intro IV).

I '

1¥·5 The ba:sic tasks of statics. We recall that there

are two fundamental types of problems in statics, which

have been emphasised in our preliminary discusions:

1 Decide whether a given framework is rigid

2 Find the internal resolution of a

external load on a given framework

These two kinds of problems fill chapters of statics

texts with particular methods which provide practical

answers for certain classes of structures.

The usual underlying approach to problem 1 is to

find a just-rigid substructure: a framework with the

proper number of bars ( or and then

confirm the rigidity by showing that there is no stress.

Thus we need algorithms for checking the presence of a

stress. The underlying approach to problem 2 usually

assumes there is at most a single answer ( the structure

is statically determined and thus the structure has no

stress) and then seeks to generate the resolution by

some inductive processes.

However the two problems can be solved by essentially

methods. A structure has a stress < ... ,Aij'•••> with\m10 iff the structure with bar (k,m) removed has

a resolution of the external load produced by akam and

Thus the search for a stress is the search for

a resolution of a particular external load. Conversely,

as we shall see in the section on funicular polygons,

53

' I

the resolution of an external equilibrium load is a

particular stress on a derived larger framework. An

algorithm for one problem will provide an algorithm

for the other problem.

Such methods will also provide solutions for

special related problems such as finding the resultant

of a set of forces, or finding the reaction from the

ground to an external load which is not yet in

equilibrium. In the following chapters we will touch on some of the more widely known geometric and

combinatorial methods which are used to solve these

problems. We cannot present a single "best" algorithm,

because such does not exist. Short of massive algebraic

or numerical calculations, of a size which is often

inpractical even on computers, the search for algorithms

remains the unsolved and important problem. In terms

of our mathematical vocabulary we can rephrase the

find an algorithm to identify all dependent

structures, and give the coefficients of

dependence on any minimal dependent structure.

5i

EQUIVALENCE OF STATICS AND INFINITESIMAL KIN!MA1JCS There is a fundamental result which is basic to the

study of any set of structures: provided the correct

definitions are given, a structure is statically rigid

iff it is infinitesimally rigid. This chapter will

present the general equivalence for bar and joint structures

as well as some immediate corollaries.

5.1 Infinitesimal rigidity. In the previous paper in

this series we have given an extensive introduction to

the corresponding projective ana structural theory of

infinitesimal motions and infinitesimal rigidity. We

cannot duplicate all of the motivation here, so we will

simply present a quick summary of the definitions and

basic facts.

Definition 5.1 A motion of a bar and joint structure

is an assignment to the joints { ••• ,Mi••••) where M1 is

a hyperplane segment (n-1 extensor) through the joint ai

(Mi" a1 :. 0). A compatible motion of the bar and joint

structure is a motion < ••• such that

[MiajJ-+ [Mjai] = 0 for each bar (i,j )eB.

Briefly the hyperplane Mi represents the velocity of

the joint a1 by giving.Qhoriented plane area perpendicular

the the Euclidean vector velocity. If the velocity was

••• ,vn-l) in Euclidean terms then the motion is the

hyperplane with coordinates (v1 , ••• ,v0 _ 1 ,-v•a) -precisely

"• ' y

the coefficients ot the usual equation for such a plane.

A motion is compatible it the velocities preserve the lengths of all the bars. This preservation or distance is

implicit):y expressed in the equation [Miajlt£Mja£1: 0 · In our previous analysis we expressed this situation

as a linear transformation !: !.<< ••• ,Mi•••• ,Mj••••">):: •• , (MiajJ+Ji.tjai'] , ••• >

with the compatible motions forming the kernel of .f. Among the compatible motions we always find the

Euclidean or rigid motions of the entire space. These

rigid motions are induced by a single center or n-2

extensor of the algebra C such that M1= Cvai for all ai. A structure is infinitesinally rigid iff

every compatible motion is a rigid motion.

Very clearly this rigidity depends on the nullity of

and thus depends on the rank, and row dependence, of the

matrix In particular the rigidity depends on the

solutions to the equations

vertex vertex 1 j

• • • • •• ••• edge (i,j) o •.• aj ••• a1 o •••

••• ••• • ••

• • Mi

)C • • • Mj • •

0 0 • • • 0

where the ai are written as row vectors and the M1 as columhs.

Finally the degree of freedom of a structure is the dimension of the space of compatible motions, and the

degree of internal freedom is:the degree of freedom)-(the dimension or the set of Euclidean motions of the structurd.

5.2 The fundamental equivalence of statics and infinitesimal Kinematics. What we wish to show is that the dependence structure of the rows of the matrix for 3.- the structure

geometry for infinitesimal motions - is identical to the structural geometry of the statk resolutions. This equivalence has been known for a long time, and is, in fact, much more obvious when the two theories are each expressed

in Euclidean terms (Henneberg,p576-579). However we have extended our theories to include points at infinity and it is important to give a direct proof of the projective theories.

Theorem 5.1 For any bar and joint structure the resolution transformation- •k and the motlon transformation J. ;-:.:'bave

the same ·rank. Proof. While it is simple to confirm this relationship in Euclidean terms, it is disguised in our projective

setting. This is because 'F, the codomain of J:(, is only isomorphic to, but not identical to the domain or t-the space of motions h1, and is a restricted

subset of the hyperplanes satisfying CMia}=o. As an

exercise in the direct linear algebra, and to avoid other

indirect and more obscure forms of reasoning, we will

(

the two matricies. Fork'

Rows for ai

Rows for aj

Column for edge (i,j)

.... 0 •

• ••• .. ., . -.. •

..... u • •

•••

where is written as a column or coordinates.

J'or J..

Row for edge 1.i,j)

Columns for a1

• • • •••

Columns for aj

• •• 0 ••• aJ o •••• 0 a 1 0 •••

••• • • • • •• ... where aj is written as a row of coordinates.

•

•

•

Because is an extra requirement on the

solutions Mi' any dependence of the equations represented

by the rows or! is an assignment of scalars to the rows such that:

•·· 'l .... ( •

sum over j with (i,j)eB By the basic exterior algebra for the operation vwe

Thus the row dependencies for the equations of l are

exactly the column dependencies of t< • Since the number

(

rows of 1-:.JBI ::. the number of columns of t( , we conclude

that the row rank of i. = the colwin rank of)(. 'ftle matricies are essentially transposes of one another - the

elementary condition for adjoint transformations. II

As an abuse of terminology, we will say· #stresses

instead of the dimension of the vector space of internally resolved stresses. Corollary 5.2 For a bar and joint structure in n-space with E bars and V joints;

E - 1i stresses = nV - degree of freedom In the plane, for a structure not all at ,

one point: { 2V-3) -E :: Lr1ternal degree of freedom - # stresses In space, for a non-collinear structure:

(3V-6) - E = internal degree of - # stresses Proof. For the transformation! rankUt+-nullity(.f.):: nV

so rank{!)= nV - degree of freedom For the transformationk rank(k) ::: E

so . r.ank : E - # stresses By the previous theorem rank(K) so we have:

E - I stresses :. nV - degree of freedom. The remaining particular cases follow by simple

arithmetic when we recall that the internal degree of freedom is the degree of freedom less the Euclidean motions ••

Corollary 5.3 Por any bar and joint structure the

structure geometry for infinitesimal motions is identical

to the structure geometry for statics. In particular,

the structure is infinitesimally rigid iff it is statically

rigid.

Proof. The rank of the bar and joint structure, or of

any of its substructures, is the same whether the independence of bars is measured using infinitesimal motions ( the

transformation/) or using static resolutions ( the

Thus the same combinatorial geometry

occurs in each case. Since rigidity is defined by the same number for the rank in each case, we are finished. II

We can use this equivalence to quickly translate

all of the results given in the previous paper on

infinitesimal kinematics, presenting these translations

we will provide related versions which are peculiar to the

vocabulary of stresses and statics. It remains a non-trivial

question to decide which approach and vocabulary will

be most usefull in attacking any particular problem.

5.3 Isostatic linkages and the cut theorem. The problem

of linking two rigid bodies in the plane and in space was

completely described in I-5. We translate the simpleat

form of the result.

Proposition 5.4 Given two full rigid ·bodies inn-space

joined by n(n 1) bars, there is a stress on these linking 2

bars iff the lines lie on a line complex.

Proof. This is a direct consequence of corollary 5.3 and theorem I-5.1. In particular, we note that the

linear dependence of the joining bars guaranteed

j

(

by the line complex, gives the coefficients or the linking bars in the stress. We say a stress is on these bars it

it is non-zero on at least one or them. •

The real cobtent of this result is indicated when we

that the bars joining the two bodies form a

cut set of the larger underlying graph. The result then states that the net force through this cut set is zero

and the lines are dependent. This result holds in general.

Proposition 5.5 Given any collection of bars CCB, any complete component D of the structure with these bars

removed, and any stress on the entire structure then:

sum over (i,j}eC

Proof. The proof is by induction on the number of verticies in D. We first take any single vertex a1• Then for any

stress: = 0 sum oYer (l,j )EB Assume the result is true for any component of n-1

verticies. Given a component of n verticies a1 , ••• ,an we remove one vertex a and all its edges E. By the n induction hypothesis:

=0 sum over l.ti$D-l, (i,j)fCUE

and S:.Au•nvak = 0 sum over We add the two equations to obtain the desired result

•

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This suggests an appropriate generalization of proposition 5.4.

Proposition 5.6 Given two rigid bodies A and B linked

by k bars there is a stress which is non-zero on at least one of these bars iff the lines are

dependent as lines in the space.

In particular, for 5 lines in 3-space the lines

must lie on a line series, and 4 lines in 3-space must be on a line congruence.

Proof. The necessity follows from the cut theorem,

with'% bfOgiving the desired dependence of the lines. Conversely, if the lines are dependent, then

for some Since the two bodies are rigid,

there is a resolution in A of the force system and a resolution in B of the system ) 1b1vai• These

two resolutions, plus give the required stress ••

5.4 Almost-always rigid systems. It is natural to

ask whether the count E:2V-J is a sufficient condition for rigidity. The answer has been developed in work

on infinitesimal

Proposition 5.7 Given .a graph with such that

every subgraph has E'-'2V'-J, then almost every realization

of the graph in the paane produces a rigid structure.

This is a direct translation of theorem I-6.1. We can present an adequate direct proof using stresses

rather than motions, but such a proof seems to loose the

(

detailed information about the poeitionlof the "bad"

realisations, which are located during the original induction. A vtry .-..ice cH\cl-e Kfrnsiot11 hQS prPsen-k.J

(As fMow &.R.o+h II),. There is an interesting corollary to this result

which characterises the rigid circuits of the structure geometry. These structures are important since they form the minimal rigid tensegrity structures (chapter 6). Extending our abuse or language, we say

a nas one stress if it bas a !-dimensional space or stresses, and we refer to any basis of this space as "the stre·ss". Since all of the non-trivial stresses will be scalar multiples of the basis, this should cause

no difficulty. Corollary 5.8 Given a graph with E:2V-2, and for every subgraph, then in almost-all realizations in the plane the structure is rigid and has exactly one stress, which must be non-zero on all the edges.

If some realization of a graph in the plane is rigid, has exactly one stress, and this stress is

non-zero on all the edges, then E:2V-2, for all proper subgraphs, and the rigidity and one stress

occur in almost-all realizations.

Proof. Assume E:r2V- 2 and E '-3. By a direct count of the edges there must be at least one non-trivial stress

Since removing any single bar creates a graph satisfying

proposition 5.7, the structure is rigid in almost-all

realizations in the plane. This is true for the removal

( '

of any single bar, and these rigid realizations with one bar

deleted contain no stress. Intersecting these large classes

of rigid realizations over all the single bars, we still

have a large ( almost-always) class of realizations which

have no stress on any proper subset ot the original graph.

Thus the single stress must be non-zero on all the edges. Conversely, if some realization is rigid and has

exactly one stress, then by prop.4.2. Since

the stress is non-zero on every edge, then removing any

set of edges gives a subgrapb with no stress -We conclude that the graph satisfies the original assumtions

For structures in space there are no comparable

complete characterizations. We can summarize the known

necessary conditions.

Proposition 5.9 If some realization of a graph in 3-space is rigid and has exactly one stress, non-zero on every

edge, then -5, for every subgraph E '-6, the graph is 3-connected in a vertex sense and almost-all

realizations in space are rigid with exactly one stress

non-zero on every edge C ass ul\t i"' 9 II> 3) · Proof. The count of the edges follows directly from

prop. 4.3. The 3-connectivity of rigid structures in

space has been checked using motions. Since one realization

has the required property, this can be treated as a .

in the minors of the matrix for R. Using indeterminants

'4 0 •

\

for the positions of the verticies, we know that for at

least one value all the 3V-6 minors are non-zero. By elementary algebra, this must be true for almost-all values. It is in this way that we use the phrase almost-always. •

The only known converse result is for structures related to convex polyhedra. Proposition 5.10 Given a planar graph with almost-all realizations in space are rigid iff E=3V-6

Proof. If the structure is rigid in any realization

then E:)V-6. Since the graph is planar, it can be drawn on the surface of a sphere, and Euler's formula applies

V- E + F = 2. The largest possible number of edges is

for a complete triangulation, when F: t• and thus 3V -6 E. We conclude that a rigid graph has exactly 3V-6.

If the planar graph has then it completely triangulated• If we lay· this graph on the sphere then

forms a convex polyhedron, since all the faces are triangles, and therefore plane. By Cauchy's theorem

such a polyhedron is rfgid, and thus must be rigid in almost-all realizations.(Gluck, Whitel,y-Polyhedra). I

, and projections. There are intimate connections between rigid and dependent structures in

. .. ' .. • .

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neighboring dimensions. We first recall a relationship examined in the previous paper. Definition 5.3 If S: <J ,B) is a bar and joint

structure and p is any point then the ... cane .. . of S from p, written Syp, is the structure with joints JU.{p) and bars BIIJ>e{p}.

A structure S 'vp is the apparent

__ ·_.. of a structure S which lies in a hyperplane H of the space iff the projection from p of the structure 5' is a structure congruent to s. (In the Cayley algebra

the projection of a point into the hyperplane H is the point (pva}AH. ) Proposition 5.11 If a .. struc:ture S•vp is the apparent

of a structure S in the hyperplane H, then

the structure S'vp has the same #stresses as the structure S.

Proof. By the result of theorem we know that S and S'vp have the same internal degree of freedom d.

If S bas E edges and V verticies, then S vp has V+l

verticies and ViE edges. If the hyperplane has dim k-1 then the space baa dimension k. Assuming the structure . S at least spans a k-2 subspace, then the equations from

corollary 5.2 give: I stresses of S = E - (k-l)V 4-(k-l)k+ d

2 ; EtV - k(V ... l) -rk(k+l)+ d = # stresses of S'vp

2 Similar counting arguments work for smaller structures. II

We know of proof for this propos! tion that

uses direct information about the stresses without

reference to the degrees of freedom. On the other hand the proof using motions was direct and pretty.

There is a simpler relationship for stresses

which generalizes one direction of this result.

Proposition 5.12 If a structure S has an n-dim. stress space, then any structure s•, which is a

projection of S from any point p not on the-line of a ot s; bas an •.

stress space. (If the projection is along some bar of s, then this degenerate bar is deleted from S•). Proof. Take any dependence on S:

<···•Aij••••> such that After projection, these equilibrium equations become

Z >.ij7f(ai)v1T(aj)= 0 \ja!Vaj:::'O since the projection is a linear transformation in the Cayley algebra. If any bar is projected to a single point

then ai \1 a j=O the equation remains true with these

terms If two different stresses were identified

by this projection then thetr difference is sent to zero by the projection. Thus all the edges were collapsed to

points and the projection hit at least one collinear

polygon, contrary to the assumption. •

r'

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It is clear that the converse is not true. Since

any complete 4-grapb has a stress in the plane, but is independent in any non-planar realization in space, we must look elsewhere for a projective source for

general stresses. A more subtle source bas been

found - the projections of oriented polytopes.

These results, extending the work of Maxwell and

Cremona, are discussed further in sectien There is further speculation that even these strange results do not exhaust the connections between

statics and projections from other dimensions.

5.6 Joints at infinity. There is a form of

articulation, common in mechanics, which is the

kinematic equiyalent of a universal joint at infinity - the slide joint (chapter I-9). Since these joints

appear in the algebra for infinitesimal motions, our

basic equivalence already indicates that they will

behave statically like joints at infinity as well. Given a slide joint in the plane which is the

equivalent of the infinite joint (a1 ,a2 ,o), then it permits only the translation (-a2,a1 ) as the relative

motion, in Euclidean terms. Clearly the force which

will pass from a body through this joint is a force

perpendicular to the translation - and thus through

\

the infinite point (a1,a2,o). There is a similar occurance in space where the

slip joints defined in the previous paper represent the joints at infinity.

It is difficult to imagine how a scalar multiple of

an infinite bar can represent a finite static force.

the forces are still written as and if one of the joints is finite then this extensor will

be equivalent to a finite line segment - and thus a finite

length. If both the joints are infinite, then the force

is a couple - which bas no "length" in the ordinary sense but rather a twisting effect which is measured by

its •size" at infinity. In general there is no difficulty posed by having

some joints or bars at_infinity. Such slide joints

are not common in structural engineering and architecture, but they do occur in mechanics and will be implicitly

allowed in all our analysis. They will crop up in

certain reductions of even ordinary structures and thus

are essential to the completeness of our system.

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6 STRESS MODELS AND TENSEGRITY STRUCTiffiES

We have found in our work with infinitesimal rigidity that it is useful. to build models and watch

for movement • . While such motions indicate an unresolved force system, it is desirable to be able to see a stress

in a model. How can a stress be made visible? The answer

is simple. If there is a stress <• •• '"ij, •• with then we can replace the bar (h,k) by a tight

string of the same length and the structure will be

in equilibrium - i.e. will not move from the internal forces. The tension on the string will correspond to

and thus the entire structure can have the

stress «<(. •• and be in equilibrium. It is also clear that if we replace any bar of

a structure by such a tense string and the structure is

in equilibrium, then the structure bas a stress. Since the remaining members resolve the forces from the string

the entire resolution plus the tension in the string

form a stress. Thus models with at least one string

provide a necessary and sufficient condition for a

stress. More sophisticated models to illustrate

these tensions, or corresponding compressions, have

also been developed using less elastic plastic members

with re.stridted· _ slides in the middle. These avoid the

flexibility which comes from the stretching of normal

thread or other simpe tension members.