# 4.1 Michell trusses  (Page 4/4)

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To prove the first statement, note that

$\begin{array}{cc}\hfill \left(\delta T,x\right)& =\sum _{B\in T}\left(\delta B,x\right)=\sum _{B\in T}\omega \frac{a-b}{|a-b|}·\left(a-b\right)=\sum _{B\in T}\omega |a-b|\hfill \\ & =\sum _{B\in T}\mathrm{Cost}\left(B\right)=\mathrm{Cost}\left(T\right).\hfill \end{array}$

if $\omega >0$ for each $B\in T.$ On the other hand, if $S$ is a truss with $\delta S=\delta T,$ then

$\mathrm{Cost}\left(T\right)=\left(\delta T,x\right)=\left(\delta S,x\right)=\sum _{B\in S}\omega |a-b|\le \mathrm{Cost}\left(S\right).$

Hence $T$ is economical.

To prove the latter statement, let $K$ be the closure of the convex hull of the support of the point forces equilibrated by $T.$ Then $K$ is a convex polyhedron; let $H$ be the hyper-plane passing through one of its sides. Without loss of generality, assume that $H$ is the $xy$ -plane and ${H}_{+}=\left\{x\in {\mathbb{R}}^{3}:{x}_{1}={x}_{2}=0,{x}_{3}>0\right\}$ is the upper half space, and $\mathrm{Support}\left(\delta T\right)\subset {\mathrm{R}}^{3}\setminus {H}_{+}.$ Let $\phi \left(x\right)={e}_{3}\otimes {e}_{3}x$ for ${x}_{3}>0$ and 0 otherwise where ${e}_{3}$ is the unit basis vector $\left(0,0,1\right)$ and ${e}_{3}\otimes {e}_{3}$ is the tensor product

${e}_{3}\otimes {e}_{3}=\left(\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ 0& 0& 1\end{array}\right).$

Then,

$0=\left(\delta T,\phi \right)=\sum _{B\in T}\omega \frac{\left(a-b\right)}{|a-b|}·{e}_{3}\otimes {e}_{3}\left(a-b\right)=\sum _{B\in T}\omega \frac{|{a}_{3}-{b}_{3}{|}^{2}}{|a-b|}$

which implies that $\omega =0$ if it corresponds to a beam lying in ${H}_{+}.$

## 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 analyzeda 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 smallso 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 setof positive two dimensional area.

We define a corner to be the union of three beams $\left\{{B}_{1},{B}_{2},{B}_{3}\right\}\subset T$ which share an endpoint $p,$ lie in a halfplane about $p$ and $\left(\delta {B}_{1}+\delta {B}_{2}+\delta {B}_{3},\phi \right)=0$ for all $\phi$ with $\mathrm{Support}\left(\phi \right)\subset U$ for some neighborhood $U$ of $p.$

By rescaling, translating and rotating, we may assume that ${B}_{1}$ and ${B}_{2}$ form two sides of an isosceles triangle and one endpoint of ${B}_{3}$ lies in the base of this triangle. The base of the triangle and ${B}_{1}$ form an angle of $\theta$ and the base of the triangle and ${B}_{3}$ form an angle of $\phi .$ The height of the triangle is $l.$ We will need the relation

$\frac{\pi }{2}>\phi >\theta >0.$

The sum of ${B}_{1},{B}_{2}$ and ${B}_{3}$ will be called the four point truss, ${T}_{4}$ . The four point truss equilibrates the system of forces ${f}_{1},{f}_{2}$ and ${f}_{3}$ with points of application the intersection of ${B}_{1},{B}_{2}$ and ${B}_{3}$ resp. with the base of the triangle. We require

$\begin{array}{c}{f}_{1}+{f}_{2}+{f}_{3}=0\\ \text{or}\\ |{f}_{1}|+|{f}_{2}|=|{f}_{3}|sin\phi csc\theta ,\phantom{\rule{1.em}{0ex}}|{f}_{1}|-|{f}_{2}|=|{f}_{3}|cos\phi sec\theta .\end{array}$

The simplicity of this truss allows an for a straight forward calculation of the cost. For $i=1,2,3$ let ${\omega }_{i}$ and ${\lambda }_{i}$ denote the strength and length resp. of beam ${B}_{i}.$ 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:

${\omega }_{1}=|{f}_{1}|,\phantom{\rule{1.em}{0ex}}{\omega }_{2}=|{f}_{2}|,\phantom{\rule{1.em}{0ex}}{\omega }_{3}=-|{f}_{3}|.$

The lengths of the the ${T}_{4}$ are also easily calculated. ${B}_{1}$ and ${B}_{2}$ being the equal sides of an isosceles triangle it might be superfluous to state that:

${\lambda }_{1}={\lambda }_{2}$

With the use of trigonometry it is clear that

${\lambda }_{1}={\lambda }_{2}=l\left|csc,\theta \right|$

and

${\lambda }_{3}=l\left|csc,\phi \right|$

The cost of ${T}_{4}$ is given by the formula

$\begin{array}{cc}\hfill \mathrm{Cost}\left({T}_{4}\right)& =\sum _{i=1}^{3}|{\omega }_{i}|{\lambda }_{i}\hfill \\ & =|{f}_{1}|lcsc\theta +|{f}_{2}|lcsc\theta +|{f}_{3}|lcsc\phi \hfill \\ & =l|{f}_{3}|\left(sin\phi {csc}^{2}\theta +csc\phi \right).\hfill \end{array}$

We have used ( ) in relating $|{f}_{1}|$ and $|{f}_{2}|$ to $|{f}_{3}|.$ Note that the cost is linear in $l$ and diverges as $\theta \to 0$ or $\phi \to 0.$

Having calculated the cost of the ${T}_{4}$ we perturb the corner in order to produce a structure that may yield a lower cost compared to that of the ${T}_{4}$ . Our main objective being that of investigation of the consequences or eliminatingcorners from an already existing truss.

The modified truss will be obtained by removing the corner of the ${T}_{4}.$ A horizontal line parallel to the triangle base line a distance $l-h$ from the base of the isosceles traingle is added.The beams ${B}_{1}^{\text{'}}$ and ${B}_{2}^{\text{'}}$ are formed by shortening ${B}_{1}$ and ${B}_{2}$ to where they intersect this added line. To mantain a truss structure ${B}_{3}$ must be replaced by three new beams. ${B}_{4}$ connects ${B}_{1}^{\text{'}}$ and ${B}_{2}$ in the place where the cut of the corner was made. ${B}_{3}^{\text{'}}$ shares a connections point with ${B}_{4}$ and ${B}_{1}^{\text{'}}$ while ${B}_{3}^{\text{'}\text{'}}$ is connected to the node that is shared with both ${B}_{4}$ and ${B}_{2}^{\text{'}}$ . The shape of the new truss that is obtained is that of a trapezoid, the applied forces remain same as for ${T}_{4}$ We shall hence forth refer to this truss structure and the five point tent truss, ${T}_{5,h}.$

The complicated nature of the ${T}_{5,h}$ makes the calculation of the lengths ${\lambda }_{i}$ 's and strengths ${\omega }_{i}$ 's slightly less obvious than they were for the original ${T}_{4}.$ 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 $\phi$ , $\theta$ and $l.$ However, we need three more parameters to calculate the cost. These parameters are the following: $h$ which describes the height of the cut and ${\phi }_{+}$ and ${\phi }_{2}$ represent the angle between the base and ${B}_{3}^{\text{'}}$ and ${B}_{3}^{\text{'}\text{'}}$ resp.

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

$\begin{array}{c}{\lambda }_{1}^{\text{'}}=\left(l-h\right)csc\theta \\ {\lambda }_{2}^{\text{'}}=\left(l-h\right)csc\theta \\ {\lambda }_{3}^{\text{'}}=\sqrt{{\left(lcot\phi -hcot\theta \right)}^{2}+{\left(l-h\right)}^{2}}\\ {\lambda }_{3}^{\text{'}\text{'}}=\sqrt{{\left(lcot\phi +hcot\theta \right)}^{2}+{\left(l-h\right)}^{2}}\\ {\lambda }_{4}=2hcot\theta \end{array}$

The magnitude of strengths are similarly calculated to be:

$\begin{array}{c}|{\omega }_{1}^{\text{'}}|=\left|{f}_{1}\right|\\ |{\omega }_{2}^{\text{'}}|=\left|{f}_{2}\right|\\ |{\omega }_{3}^{\text{'}}|=\left|{f}_{1}\right|sin\theta csc{\phi }_{+}\\ |{\omega }_{3}^{\text{'}\text{'}}|=\left|{f}_{2}\right|sin\theta csc{\phi }_{-}\\ |{\omega }_{4}|=\left|{f}_{1}\right|cos\theta +\left|{f}_{1}\right|sin\theta cot{\phi }_{+}\end{array}$

Having acquired all the information we require we can now calculate the cost of the ${T}_{5,h}$ :

$\begin{array}{cc}\hfill \mathrm{Cost}\left({T}_{5,h}\right)& =\sum _{i=1}^{\nu }{\omega }_{i}{\lambda }_{i}\hfill \\ & =\left|{f}_{1}\right|\left(l-h\right)csc\theta +\left|{f}_{2}\right|\left(l-h\right)csc\theta \hfill \\ & +\left|{f}_{1}\right|sin\theta csc{\phi }_{+}\sqrt{{\left(lcot\phi -hcot\theta \right)}^{2}+{\left(l-h\right)}^{2}}\hfill \\ & +\left|{f}_{2}\right|sin\theta csc{\phi }_{-}\sqrt{{\left(lcot\phi +hcot\theta \right)}^{2}+{\left(l-h\right)}^{2}}\hfill \\ & +\left|{f}_{1}\right|cos\theta +\left|{f}_{1}\right|sin\theta cot{\phi }_{+}\left(2hcot\theta \right)\hfill \end{array}$

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

$\begin{array}{cc}\hfill \frac{d}{dh}\mathrm{Cost}\left({T}_{5,h}\right){|}_{h=0}& =\left|{f}_{1}\right|\left[-csc\theta +sin\theta {cos}^{2}\phi \left(cot\theta -cot\phi \right)\hfill \\ & -sin\theta \left(cot\phi cot\theta +1\right)+2cot\theta \left(cos\theta -sin\theta cot\phi \right)\right]\hfill \\ & +\left|{f}_{2}\right|\left[-csc\theta -sin\theta {cos}^{2}\phi \left(cot\phi +cot\theta \right)\hfill \\ & +sin\theta \left(cot\phi cot\theta -1\right)\right]\hfill \end{array}$

With the help of ( )-( ) we have shown that

$\frac{d}{dh}\mathrm{Cost}\left({T}_{5,h}\right){|}_{h=0}<0$

independent of $l,\theta$ and $\phi .$ 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 ${B}_{1},\cdots ,{B}_{m}$ are beams meeting at a point $p.$ Suppose we shorten each beam by introducing a polyhedron about $p$ whose $m$ vertices meet the beams ${B}_{1},\cdots ,{B}_{m}$ and each edge represents the location of a new beam. From lemma above we see that the cost does not change with the perturbation if the junction consists of only cables of only bars. This studywould 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-triangleswith common vertex in the center. Choose a base and let ${A}_{\rho }$ be the affine transformation of ${\mathbb{R}}^{2}$ which maps points $x$ to points ${A}_{\rho }\left(x\right)$ whose distance to the line passing through the base is scaled by $\rho .$ The affine transformations in the remaining triangles is uniquely determined by continuity. Thus, ${\eta }_{\rho }\left(x\right)$ be the unique piecewise affine transformation of ${\mathbb{R}}^{2}$ with ${\eta }_{\rho }\left(x\right)=x$ for $x$ not in the triangle and ${\eta }_{\rho }\left(x\right)={A}_{\rho }\left(x\right)$ for $x$ in the first sub-triangle. Applying ${\eta }_{\rho }$ to a truss and then adding the necessary beams to ensure that is equilibrated then yields the perturbation.

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