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.

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

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 ω∈R.ω∈R. We can associate with BB a mass

and point forces field

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;

and

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

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

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

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

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;

and (a=0a=0)

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

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.

Figure 1

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

The scalar triple product

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 v∈R3v∈R3 and φ(x)=vφ(x)=v for all x∈R3.x∈R3. Then (Equation 6) implies

Since vv was arbitrary, we infer (Equation 7). For w∈R3,w∈R3, define the skew symmetric matrix

One then checks that w^v=w×v.w^v=w×v. Let v∈R3v∈R3 and φ(x)=x^v.φ(x)=x^v. Then

By (Equation 6), this implies

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 ν≥3.ν≥3. If fν+1=0,fν+1=0, then we are done. Otherwise, according to lemma (Reference), there is ρ∈Rρ∈R so that

Let b=aν+1+ρfν+1b=aν+1+ρfν+1 and let

Figure 2

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

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

and by ((Reference))

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.

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

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 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 .

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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 .

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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 | .

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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

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With the use of trigonometry it is clear that

λ 1 = λ 2 = l csc θ λ 1 = λ 2 = l csc θ

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and

λ 3 = l csc φ λ 3 = l csc φ

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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 ) .

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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 θ→0θ→0 or φ→0.φ→0.

Figure 3

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

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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 +

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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 )

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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 ) ]

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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

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