0.1 Derived copy of michell trusses  (Page 4/5)

Let $b=a_{\nu +1}+\rho f_{\nu +1}$ and let

Define the point force field $\mathbf{g}=g_1\delta \left(b_1\right)+\cdots +g_{\nu }\delta \left(b_{\nu }\right)$ by

We claim that $\mathbf{g}$ is balanced; by [link]

and by [link]

By induction, there exists a truss $R$ equilibrating $\mathbf{g}.$ Let $S$ be the truss consisting of the collection of beams $a_1-b,a_2-b,a_3-b$ and $b-a_{\nu +1}$ with weights ${\Omega }_1,{\Omega }_2,{\Omega }_3$ and $|f_{\nu }|$ resp. Arguing as in Lemma 2, we find that $T=R+S$ equilibrates $\mathbf{f}.$

Some economical trusses

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

whenever $\delta S=\delta T.$ That is to say, the cost of $T$ is less than or equal to that of any truss which equilibrates the same force system as $T.$ We begin by describing some global statements about economical trusses that can proven by choosing special test vector fields $\phi$ in [link] . Then we make local perturbations on trusses with corners to find a necessary conditionfor 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

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

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 $H_{+}=\{x\in {\mathbb{R}}^3:x_1=x_2=0,x_3>0\}$ 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 $(0,0,1)$ and $e_3\otimes e_3$ is the tensor product

Then,

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 $\{B_1,B_2,B_3\}\subset T$ which share an endpoint $p,$ lie in a halfplane about $p$ and $(\delta B_1+\delta B_2+\delta B_3,\phi )=0$ 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