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# Michell Trusses

Module by: Rolf Ryham. E-mail the author

Summary: This report summarizes work done as part of the Michell Trusses PFUG under Rice University's VIGRE program. VIGRE is a program of Vertically Integrated Grants for Research and Education in the Mathematical Sciences under the direction of the National Science Foundation. A PFUG is a group of Postdocs, Faculty, Undergraduates and Graduate students formed around the study of a common problem. In this PFUG, we study cost minimizing trusses. We ask what is the structure, constructed of bars and cables, of least cost that will withstand a set of point forces. This work was studied in the Rice University VIGRE class MATH499 in the Spring and Fall of 2008.

## Introduction

The Michell Truss PFUG studies a variational problem posed by mechanical engineer Anthony George Maldon Michell in the early part of the last century; what configuration of bars and cables is needed to withstand a system of balanced point forces is most economical? Suppose that some collection of forces with their points of application (point forces) is given and you are asked to build a structure which can withstand these forces. Your choice of materials are bars and cables. Bars can withstand compression (forces parallel and pointing inward) but will break under extension (forces parallel and pointing outward). Cables can withstand extension but will buckle under compression. You may also choose the strength of the bar or cable; the strength is the largest compression or extension the bar or cable respectively can withstand. A structure built of such cables and bars will be called a truss.

The first question the PFUG has addressed is whether one can build such a truss at all. Notice that a bar or cable by itself can withstand only one set of point forces; that of two opposite forces applied to the endpoints of the bar or cable. These point forces have the property that if the points of application were attached to a stationary body, the center of mass would not accelerate and the body would not rotate. This property is called balanced. Since point forces can be linearly superimposed, a necessary condition that a set of point forces be withstood by a truss is that it be balanced. Below we show that being balanced is also a sufficient condition; if the set of point forces is balanced then there is a truss withstanding it.

Next, the PFUG formulated necessary conditions for a truss to be economical. The cost of a bar or cable is its strength times its length. The cost of a truss is defined to be the sum of the cost of bars and cables constituted by the truss. In general, there are many trusses which withstand a given a set of point forces. The question is, which is most economical, i.e. which truss costs the least? In general this is a very difficult question to answer because a minimizing sequence, which we know to exist from the first part, may converge in a larger class of measures than described here and it is difficult to formulate any class of perturbations of a truss. Below we describe one class of perturbations on trusses with corners which yields a surprising necessary condition for economy. A set of perturbations to be studied in the future are also described.

## Notation

We will restrict ourselves to points and vectors in three dimensional Euclidean space, R3.R3. A point force is a vector valued measure vδ(a)vδ(a) where vv is a vector and aa is a point in Euclidean space and δ(·)δ(·) is the Kronecker delta mass. If φφ is a continuous vector field on Euclidean space we define

( v δ ( a ) , φ ) = φ ( a ) · v ( v δ ( a ) , φ ) = φ ( a ) · v
(1)

and extend this definition linearly to a sum of point forces. A beam BB is a pair of distinct points aa and bb in Euclidean space with a weight ωR.ωR. We can associate with BB a mass

Cost ( B ) = | ω | | a - b | Cost ( B ) = | ω | | a - b |
(2)

and point forces field

δ B = ω a - b | a - b | ( δ ( a ) - δ ( b ) ) . δ B = ω a - b | a - b | ( δ ( a ) - δ ( b ) ) .
(3)

δBδB corresponds to the reaction force of a cable if ω<0ω<0 and of a bar if ω>0ω>0 with endpoints at aa and b.b. Note that (δB,φ)=0(δB,φ)=0 if φφ is a constant vector field. We choose the notation δBδB to be consistent with the notion of first variation of mass, although in this case it differs by the sign of ω.ω. A truss TT is a finite collection of beams and we define Cost (T) Cost (T) and δTδT by extending the definitions (Equation 2) and (Equation 3) linearly;

Cost ( T ) = B T Cost ( B ) Cost ( T ) = B T Cost ( B )
(4)

and

δ T = B T δ B . δ T = B T δ B .
(5)

If ff is a point force field (a sum of point forces) and TT is a truss, then TT is said to equilibrateff if δT=fδT=f in the sense that

( δ T , φ ) = ( f , φ ) ( δ T , φ ) = ( f , φ )
(6)

for all continuous vector fields φ.φ.f=f1δ(a1)++fνδ(aν)f=f1δ(a1)++fνδ(aν) is said to be balanced if

i = 1 ν f i = 0 i = 1 ν f i × a i = 0 . i = 1 ν f i = 0 i = 1 ν f i × a i = 0 .
(7)

## Existence of equilibrating trusses

One may easily check that δBδB is equilibrated if BB is a beam and by linearity δTδT is equilibrated if TT is a truss. The first natural question we deal with is the converse question; is every balanced point force field equal to δTδT for some truss T?T? In other words, can any balanced point force field be equilibrated by a truss? This section answers this question in the affirmative. The sufficiency will follow from a proof by induction.

Lemma 1 Let f=fδ(a)+gδ(b)f=fδ(a)+gδ(b) be balanced. Then ff is equilibrated by a truss T.T.

Let TT consist of the single beam (a,b)(a,b) with ω=g·(a-b)/|a-b|.ω=g·(a-b)/|a-b|. Then (Equation 7) implies f=-gf=-g and then ((Reference)) implies that f2f2 is parallel to a-b,a-b, i.e. g=ω(a-b)/|a-b|.g=ω(a-b)/|a-b|. By definition

δ T = ω a - b | a - b | ( δ ( a ) - δ ( b ) ) = f δ ( a ) - f δ ( b ) = f δ ( a ) + g δ ( b ) = f . δ T = ω a - b | a - b | ( δ ( a ) - δ ( b ) ) = f δ ( a ) - f δ ( b ) = f δ ( a ) + g δ ( b ) = f .
(8)

Clearly then (Equation 6) holds.

Lemma 2 Let f=fδ(a)+gδ(b)+hδ(c)f=fδ(a)+gδ(b)+hδ(c) be balanced with a,ba,b and cc not lying on the same line. Then ff is equilibrated by a truss T.T.

Without loss of generality, a,b,ca,b,c lie in the xyxy-plane and a=0.a=0. Since a,ba,b and cc do not all lie on the same line, bb and cc are linearly independent; dotting ((Reference)) with bb and then cc implies that f·e3=g·e3=h·e3=0f·e3=g·e3=h·e3=0 where e3e3 is the unit basis vector parallel with the zz-axis. Hence ff can be expressed as a linear combination of bb and cc

f = ω b b + ω c c . f = ω b b + ω c c .
(9)

Consider the point force field e=ebδ(b)+ecδ(c)e=ebδ(b)+ecδ(c) where eb=g+ωbbeb=g+ωbb and ec=h+ωc.ec=h+ωc. We claim ee is balanced;

e b + e c = g + h + ω b b + ω c c = f + g + h = 0 . e b + e c = g + h + ω b b + ω c c = f + g + h = 0 .
(10)

and (a=0a=0)

e b × b + e c × c = g × b + h × c = 0 . e b × b + e c × c = g × b + h × c = 0 .
(11)

According to lemma (Reference), there is a truss RR with equilibrating e.e. Let SS be the truss consisting of the collection of beams (0,b)(0,b) and (0,c)(0,c) with weights ωbωb and ωcωc resp. We claim T=R+ST=R+S equilibrates f.f. Let φφ be a continuous vector field. Then

δ T = δ R + δ S = ω b b ( δ ( 0 ) - δ ( b ) ) + ω c c ( δ ( 0 ) - δ ( c ) ) + e = f δ ( 0 ) - ω b b δ ( b ) - ω c c δ ( c ) + ( g + ω b b ) δ ( b ) + ( h + ω c c ) δ ( c ) = f δ T = δ R + δ S = ω b b ( δ ( 0 ) - δ ( b ) ) + ω c c ( δ ( 0 ) - δ ( c ) ) + e = f δ ( 0 ) - ω b b δ ( b ) - ω c c δ ( c ) + ( g + ω b b ) δ ( b ) + ( h + ω c c ) δ ( c ) = f
(12)

which then implies (Equation 6).

The following lemma shows when the volume of the parallelpiped spanned by three vectors can be made to be nonzero by sheering the parallelpiped in a fixed direction by an appropriate amount. An example in two dimensions is drawn below.

Lemma 3 Let a,b,ca,b,c be distinct, nonzero points in Euclidean space and vv a nonzero vector. Then there exists tRtR such that v Span {a+tv,b+tv,c+tv}.v Span {a+tv,b+tv,c+tv}.

The scalar triple product

D ( t ) = ( ( a + t v ) × ( b + t v ) ) · ( c + t v ) = ( a × b ) · c + t ( ( a - b ) × ( b - c ) ) · v D ( t ) = ( ( a + t v ) × ( b + t v ) ) · ( c + t v ) = ( a × b ) · c + t ( ( a - b ) × ( b - c ) ) · v
(13)

is constant in tt if v Span {a,b,c}v Span {a,b,c} and is nonzero for all but one tt otherwise. In the former case, setting t=0t=0 while in the latter choosing tt so that D(t)0D(t)0 and consequently Span {a+tv,b+tv,c+tv}=R3 Span {a+tv,b+tv,c+tv}=R3 then implies the lemma.

Combining the above lemmas, we have

Theorem 1 A necessary and sufficient condition that a point force field f=f1δ(a1)++fνδ(aν)f=f1δ(a1)++fνδ(aν) be balanced is that it is equilibrated by a truss T.T.

(Necessity) Suppose ff is balanced by the truss T.T. Let vR3vR3 and φ(x)=vφ(x)=v for all xR3.xR3. Then (Equation 6) implies

0 = ( δ T , φ ) = ( f , φ ) = i = 1 μ f i · v = 0 . 0 = ( δ T , φ ) = ( f , φ ) = i = 1 μ f i · v = 0 .
(14)

Since vv was arbitrary, we infer (Equation 7). For wR3,wR3, define the skew symmetric matrix

w ^ = 0 - w 3 w 2 w 3 0 - w 1 - w 2 - w 1 0 w ^ = 0 - w 3 w 2 w 3 0 - w 1 - w 2 - w 1 0
(15)

One then checks that w^v=w×v.w^v=w×v. Let vR3vR3 and φ(x)=x^v.φ(x)=x^v. Then

( δ T , φ ) = i = 1 μ ω i ( φ ( a i ) - φ ( b i ) ) · a i - b i | a i - b i | = i = 1 μ ω i ( a ^ i - b ^ i ) v · a i - b i | a i - b i | = i = 1 μ ω i ( a i - b i ) × v · a i - b i | a i - b i | = 0 . ( δ T , φ ) = i = 1 μ ω i ( φ ( a i ) - φ ( b i ) ) · a i - b i | a i - b i | = i = 1 μ ω i ( a ^ i - b ^ i ) v · a i - b i | a i - b i | = i = 1 μ ω i ( a i - b i ) × v · a i - b i | a i - b i | = 0 .
(16)

By (Equation 6), this implies

0 = ( δ T , φ ) = ( f , φ ) = i = 1 ν f i · φ ( a i ) = i = 1 ν f i · a ^ i v = i = 1 ν f i · a i × v = i = 1 ν f i × a i · v . 0 = ( δ T , φ ) = ( f , φ ) = i = 1 ν f i · φ ( a i ) = i = 1 ν f i · a ^ i v = i = 1 ν f i · a i × v = i = 1 ν f i × a i · v .
(17)

Since vv is arbitrary, we infer ((Reference)). Thus ff is balanced.

(Sufficiency) We proceed by induction on ν.ν. If ν=2,ν=2, then ff is equilibrated by the truss TT guaranteed in lemma (Reference). If ν=3ν=3 and a1,a2a1,a2 and a3a3 do not all lie on a line, then ff is equilibrated by the truss TT guaranteed in lemma (Reference). If ν=3ν=3 and a1,a2a1,a2 and a3a3 lie on a line, consider the equivalent point force field g=f1δ(a1)+f2δ(a2)+f3δ(a3)+0δ(a4)g=f1δ(a1)+f2δ(a2)+f3δ(a3)+0δ(a4) where a4a4 is a arbitrary point chosen off the line containing a1,a2a1,a2 and a3.a3. The result for ν=4ν=4 proves that gg is equilibrated by a truss T.T. However, since (g,φ)=(f,φ)(g,φ)=(f,φ) for all continuous vector fields φ,φ, this implies that TT also equilibrates f.f. Assume that sufficiency holds for ν3.ν3. If fν+1=0,fν+1=0, then we are done. Otherwise, according to lemma (Reference), there is ρRρR so that

f ν + 1 Span ( a 1 - ( a ν + 1 + ρ f ν + 1 ) , a 2 - ( a ν + 1 + ρ f ν + 1 ) , a 3 - ( a ν + 1 + ρ f ν + 1 ) ) . f ν + 1 Span ( a 1 - ( a ν + 1 + ρ f ν + 1 ) , a 2 - ( a ν + 1 + ρ f ν + 1 ) , a 3 - ( a ν + 1 + ρ f ν + 1 ) ) .
(18)

Let b=aν+1+ρfν+1b=aν+1+ρfν+1 and let

f ν + 1 = Ω 1 ( a 1 - b ) + Ω 2 ( a 2 - b ) + Ω 3 ( a 3 - b ) . f ν + 1 = Ω 1 ( a 1 - b ) + Ω 2 ( a 2 - b ) + Ω 3 ( a 3 - b ) .
(19)

Define the point force field g=g1δ(b1)++gνδ(bν)g=g1δ(b1)++gνδ(bν) by

g i = f i + ( a i - b ) Ω i , i = 1 , 2 , 3 , g i = f i , i = 4 , , ν , b i = a i , i = 1 , , ν . g i = f i + ( a i - b ) Ω i , i = 1 , 2 , 3 , g i = f i , i = 4 , , ν , b i = a i , i = 1 , , ν .
(20)

We claim that gg is balanced; by (Equation 7)

i = 1 ν g i = i = 1 ν f i + i = 1 3 ( a i - b ) Ω i = i = 1 ν f i + f ν + 1 = 0 i = 1 ν g i = i = 1 ν f i + i = 1 3 ( a i - b ) Ω i = i = 1 ν f i + f ν + 1 = 0
(21)

and by ((Reference))

i = 1 ν g i × b i = i = 1 ν f i × a i + i = 1 3 Ω i ( a i - b ) × a i = i = 1 ν f i × a i + i = 1 3 Ω i ( a i - b ) × ( a i - b ) + i = 1 3 Ω i ( a i - b ) × b = i = 1 ν f i × a i + f ν + 1 × ( a ν + 1 + ρ f ν + 1 ) = 0 . i = 1 ν g i × b i = i = 1 ν f i × a i + i = 1 3 Ω i ( a i - b ) × a i = i = 1 ν f i × a i + i = 1 3 Ω i ( a i - b ) × ( a i - b ) + i = 1 3 Ω i ( a i - b ) × b = i = 1 ν f i × a i + f ν + 1 × ( a ν + 1 + ρ f ν + 1 ) = 0 .
(22)

By induction, there exists a truss RR equilibrating g.g. Let SS be the truss consisting of the collection of beams a1-b,a2-b,a3-ba1-b,a2-b,a3-b and b-aν+1b-aν+1 with weights Ω1,Ω2,Ω3Ω1,Ω2,Ω3 and |fν||fν| resp. Arguing as in lemma (Reference), we find that T=R+ST=R+S equilibrates f.f.

## Some Economical Trusses

After proving the question of existence we turn to the question of economy. We say that a truss TT is economical if it satisfies

Cost ( T ) Cost ( S ) Cost ( T ) Cost ( S )
(23)

whenever δS=δT.δS=δT. That is to say, the cost of TT is less than or equal to that of any truss which equilibrates the same force system as T.T. We begin by describing some global statements about economical trusses that can proven by choosing special test vector fields φφ in (Equation 6). Then we make local perturbations on trusses with corners to find a necessary condition for economy.

### Some Global Properties

A surprising and easily proven fact is

Lemma 4 A truss is economical if the weights of the beams are all of the same sign. Such a truss lies in the closed convex hull of the support of the point forces it equilibrates.

To prove the first statement, note that

( δ T , x ) = B T ( δ B , x ) = B T ω a - b | a - b | · ( a - b ) = B T ω | a - b | = B T Cost ( B ) = Cost ( T ) . ( δ T , x ) = B T ( δ B , x ) = B T ω a - b | a - b | · ( a - b ) = B T ω | a - b | = B T Cost ( B ) = Cost ( T ) .
(24)

if ω>0ω>0 for each BT.BT. On the other hand, if SS is a truss with δS=δT,δS=δT, then

Cost ( T ) = ( δ T , x ) = ( δ S , x ) = B S ω | a - b | Cost ( S ) . Cost ( T ) = ( δ T , x ) = ( δ S , x ) = B S ω | a - b | Cost ( S ) .
(25)

Hence TT is economical.

To prove the latter statement, let KK be the closure of the convex hull of the support of the point forces equilibrated by T.T. Then KK is a convex polyhedron; let HH be the hyper-plane passing through one of its sides. Without loss of generality, assume that HH is the xyxy-plane and H+={xR3:x1=x2=0,x3>0}H+={xR3:x1=x2=0,x3>0} is the upper half space, and Support (δT)R3H+. Support (δT)R3H+. Let φ(x)=e3e3xφ(x)=e3e3x for x3>0x3>0 and 0 otherwise where e3e3 is the unit basis vector (0,0,1)(0,0,1) and e3e3e3e3 is the tensor product

e 3 e 3 = 0 0 0 0 0 0 0 0 1 . e 3 e 3 = 0 0 0 0 0 0 0 0 1 .
(26)

Then,

0 = ( δ T , φ ) = B T ω ( a - b ) | a - b | · e 3 e 3 ( a - b ) = B T ω | a 3 - b 3 | 2 | a - b | 0 = ( δ T , φ ) = B T ω ( a - b ) | a - b | · e 3 e 3 ( a - b ) = B T ω | a 3 - b 3 | 2 | a - b |
(27)

which implies that ω=0ω=0 if it corresponds to a beam lying in H+.H+.

### Some Local Properties: Cutting Corners

Our object of interest are two dimensional trusses with corners. We have shown that an economical truss cannot have corners. To show this, we analyzed a perturbation of a truss with corners which consists of cutting the corner and replacing it with a flat top. Specifically, we showed that for any corner, the cut can be made sufficiently small so that the perturbed truss costs less. This surprising result suggests that any economical truss, if made of both cables and bars, has as boundary a differentiable curve and is supported on a set of positive two dimensional area.

We define a corner to be the union of three beams {B1,B2,B3}T{B1,B2,B3}T which share an endpoint p,p, lie in a halfplane about pp and (δB1+δB2+δB3,φ)=0(δB1+δB2+δB3,φ)=0 for all φφ with Support (φ)U Support (φ)U for some neighborhood UU of p.p.

By rescaling, translating and rotating, we may assume that B1B1 and B2B2 form two sides of an isosceles triangle and one endpoint of B3B3 lies in the base of this triangle. The base of the triangle and B1B1 form an angle of θθ and the base of the triangle and B3B3 form an angle of φ.φ. The height of the triangle is l.l. We will need the relation

π 2 > φ > θ > 0 . π 2 > φ > θ > 0 .
(28)

The sum of B1,B2B1,B2 and B3B3 will be called the four point truss, T4T4. The four point truss equilibrates the system of forces f1,f2f1,f2 and f3f3 with points of application the intersection of B1,B2B1,B2 and B3B3 resp. with the base of the triangle. We require

f 1 + f 2 + f 3 = 0 or | f 1 | + | f 2 | = | f 3 | sin φ csc θ , | f 1 | - | f 2 | = | f 3 | cos φ sec θ . f 1 + f 2 + f 3 = 0 or | f 1 | + | f 2 | = | f 3 | sin φ csc θ , | f 1 | - | f 2 | = | f 3 | cos φ sec θ .
(29)

The simplicity of this truss allows an for a straight forward calculation of the cost. For i=1,2,3i=1,2,3 let ωiωi and λiλi denote the strength and length resp. of beam Bi.Bi. It is easy to see that for this truss to equilibrate the the strengths of each of individual beam must be equal to the force that is based on a point of the beam and aligned to it, thus:

ω 1 = | f 1 | , ω 2 = | f 2 | , ω 3 = - | f 3 | . ω 1 = | f 1 | , ω 2 = | f 2 | , ω 3 = - | f 3 | .
(30)

The lengths of the the T4T4 are also easily calculated. B1B1 and B2B2 being the equal sides of an isosceles triangle it might be superfluous to state that:

λ 1 = λ 2 λ 1 = λ 2
(31)

With the use of trigonometry it is clear that

λ 1 = λ 2 = l csc θ λ 1 = λ 2 = l csc θ
(32)

and

λ 3 = l csc φ λ 3 = l csc φ
(33)

The cost of T4T4 is given by the formula

Cost ( T 4 ) = i = 1 3 | ω i | λ i = | f 1 | l csc θ + | f 2 | l csc θ + | f 3 | l csc φ = l | f 3 | ( sin φ csc 2 θ + csc φ ) . Cost ( T 4 ) = i = 1 3 | ω i | λ i = | f 1 | l csc θ + | f 2 | l csc θ + | f 3 | l csc φ = l | f 3 | ( sin φ csc 2 θ + csc φ ) .
(34)

We have used ((Reference)) in relating |f1||f1| and |f2||f2| to |f3|.|f3|. Note that the cost is linear in ll and diverges as θ0θ0 or φ0.φ0.

Having calculated the cost of the T4T4 we perturb the corner in order to produce a structure that may yield a lower cost compared to that of the T4T4. Our main objective being that of investigation of the consequences or eliminating corners from an already existing truss.

The modified truss will be obtained by removing the corner of the T4.T4. A horizontal line parallel to the triangle base line a distance l-hl-h from the base of the isosceles traingle is added. The beams B1'B1' and B2'B2' are formed by shortening B1B1 and B2B2 to where they intersect this added line. To mantain a truss structure B3B3 must be replaced by three new beams. B4B4 connects B1'B1' and B2B2 in the place where the cut of the corner was made. B3'B3' shares a connections point with B4B4 and B1'B1' while B3''B3'' is connected to the node that is shared with both B4B4 and B2'B2'. The shape of the new truss that is obtained is that of a trapezoid, the applied forces remain same as for T4T4 We shall hence forth refer to this truss structure and the five point tent truss, T5,h.T5,h.

The complicated nature of the T5,hT5,h makes the calculation of the lengths λiλi's and strengths ωiωi's slightly less obvious than they were for the original T4.T4. However, easy trigonometry and some calculations yield both the length's and the strengths of each individual beam. We use the same parameters to describe this truss as before, namely φφ, θθ and l.l. However, we need three more parameters to calculate the cost. These parameters are the following: hh which describes the height of the cut and φ+φ+ and φ2φ2 represent the angle between the base and B3'B3' and B3''B3'' resp.

Keeping in mind some of the properties that govern the truss we have calculated the length's of each individual beams to be:

λ 1 ' = ( l - h ) csc θ λ 2 ' = ( l - h ) csc θ λ 3 ' = ( l cot φ - h cot θ ) 2 + ( l - h ) 2 λ 3 ' ' = ( l cot φ + h cot θ ) 2 + ( l - h ) 2 λ 4 = 2 h cot θ λ 1 ' = ( l - h ) csc θ λ 2 ' = ( l - h ) csc θ λ 3 ' = ( l cot φ - h cot θ ) 2 + ( l - h ) 2 λ 3 ' ' = ( l cot φ + h cot θ ) 2 + ( l - h ) 2 λ 4 = 2 h cot θ
(35)

The magnitude of strengths are similarly calculated to be:

| ω 1 ' | = f 1 | ω 2 ' | = f 2 | ω 3 ' | = f 1 sin θ csc φ + | ω 3 ' ' | = f 2 sin θ csc φ - | ω 4 | = f 1 cos θ + f 1 sin θ cot φ + | ω 1 ' | = f 1 | ω 2 ' | = f 2 | ω 3 ' | = f 1 sin θ csc φ + | ω 3 ' ' | = f 2 sin θ csc φ - | ω 4 | = f 1 cos θ + f 1 sin θ cot φ +
(36)

Having acquired all the information we require we can now calculate the cost of the T5,hT5,h:

Cost ( T 5 , h ) = i = 1 ν ω i λ i = f 1 ( l - h ) csc θ + f 2 ( l - h ) csc θ + f 1 sin θ csc φ + ( l cot φ - h cot θ ) 2 + ( l - h ) 2 + f 2 sin θ csc φ - ( l cot φ + h cot θ ) 2 + ( l - h ) 2 + f 1 cos θ + f 1 sin θ cot φ + ( 2 h cot θ ) Cost ( T 5 , h ) = i = 1 ν ω i λ i = f 1 ( l - h ) csc θ + f 2 ( l - h ) csc θ + f 1 sin θ csc φ + ( l cot φ - h cot θ ) 2 + ( l - h ) 2 + f 2 sin θ csc φ - ( l cot φ + h cot θ ) 2 + ( l - h ) 2 + f 1 cos θ + f 1 sin θ cot φ + ( 2 h cot θ )
(37)

In order to check that the perturbation lowers the cost for some sufficiently small h,h, we differentiate this expression and evaluate the derivate at h=0.h=0. Even further simplification reduces the formula to the following somewhat complicated expression:

d d h Cost ( T 5 , h ) | h = 0 = f 1 [ - csc θ + sin θ cos 2 φ ( cot θ - cot φ ) - sin θ ( cot φ cot θ + 1 ) + 2 cot θ ( cos θ - sin θ cot φ ) ] + f 2 [ - csc θ - sin θ cos 2 φ ( cot φ + cot θ ) + sin θ ( cot φ cot θ - 1 ) ] d d h Cost ( T 5 , h ) | h = 0 = f 1 [ - csc θ + sin θ cos 2 φ ( cot θ - cot φ ) - sin θ ( cot φ cot θ + 1 ) + 2 cot θ ( cos θ - sin θ cot φ ) ] + f 2 [ - csc θ - sin θ cos 2 φ ( cot φ + cot θ ) + sin θ ( cot φ cot θ - 1 ) ]
(38)

With the help of (Equation 36)-((Reference)) we have shown that

d d h Cost ( T 5 , h ) | h = 0 < 0 d d h Cost ( T 5 , h ) | h = 0 < 0
(39)

independent of l,θl,θ and φ.φ. This proves the claim that two dimensional trusses cannot have corners.

### Other Deformations

Other deformations of trusses worth considering are perturbations of other junctions. Consider the case when B1,,BmB1,,Bm are beams meeting at a point p.p. Suppose we shorten each beam by introducing a polyhedron about pp whose mm vertices meet the beams B1,,BmB1,,Bm and each edge represents the location of a new beam. From lemma (Reference) above we see that the cost does not change with the perturbation if the junction consists of only cables of only bars. This study would help rule out certain cases when both cables and bars meet at a junction.

A second deformation consists of making piecewise affine deformations within a triangle as follows; divide the triangle into three sub-triangles with common vertex in the center. Choose a base and let AρAρ be the affine transformation of R2R2 which maps points xx to points Aρ(x)Aρ(x) whose distance to the line passing through the base is scaled by ρ.ρ. The affine transformations in the remaining triangles is uniquely determined by continuity. Thus, ηρ(x)ηρ(x) be the unique piecewise affine transformation of R2R2 with ηρ(x)=xηρ(x)=x for xx not in the triangle and ηρ(x)=Aρ(x)ηρ(x)=Aρ(x) for xx in the first sub-triangle. Applying ηρηρ to a truss and then adding the necessary beams to ensure that is equilibrated then yields the perturbation.

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