0.1 Derived copy of michell trusses  (Page 2/5)

Notation

We will restrict ourselves to points and vectors in three dimensional Euclidean space, ${\mathbb{R}}^3.$ A point force is a vector valued measure $v\delta \left(a\right)$ where $v$ is a vector and $a$ is a point in Euclidean space and $\delta (·)$ is the Kronecker delta mass. If $\phi$ is a continuous vector field on Euclidean space we define

and extend this definition linearly to a sum of point forces. A beam $B$ is a pair of distinct points $a$ and $b$ in Euclidean space with a weight $\omega \in \mathbb{R}.$ We can associate with $B$ a mass

and point forces field

$\delta B$ corresponds to the reaction force of a cable if $\omega <0$ and of a bar if $\omega >0$ with endpoints at $a$ and $b.$ Note that $(\delta B,\phi )=0$ if $\phi$ is a constant vector field. To see this, note that

because $\phi \left(a\right)=\phi \left(b\right)$ since $\phi$ is a constant vector field. We choose the notation $\delta B$ to be consistent with the notion of first variation of mass, although in this case it differs by the sign of $\omega .$ A truss $T$ is a finite collection of beams and we define $\mathrm{Cost}\left(T\right)$ and $\delta T$ by extending the definitions [link] and [link] linearly;

and

If $\mathbf{f}$ is a point force field (a sum of point forces) and $T$ is a truss, then $T$ is said to equilibrate $\mathbf{f}$ if $\delta T=\mathbf{f}$ in the sense that

for all continuous vector fields $\phi .$ $\mathbf{f}=f_1\delta \left(a_1\right)+\cdots +f_{\mu }\delta \left(a_{\mu }\right)$ is said to be balanced if

Existence of equilibrating trusses

One may easily check that $\delta B$ is equilibrated if $B$ is a beam and by linearity $\delta T$ is equilibrated if $T$ is a truss. The first natural question we deal with is the converse question; is every balanced point force field equal to $\delta T$ for some truss $T?$ 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 $\mathbf{f}=f\delta \left(a\right)+g\delta \left(b\right)$ be balanced. Then $\mathbf{f}$ is equilibrated by a truss $T.$

Let $T$ consist of the single beam $(a,b)$ with $\omega =g·(a-b)/|a-b|.$ Then [link] implies $f=-g$ and then [link] implies that $f_2$ is parallel to $a-b,$ because $f\times (a-b)=0$ . $\omega =\frac{g·(a-b)}{|a-b|}=\frac{-f·(a-b)}{|a-b|}=\frac{|-f||a-b|}{|a-b|}=\left|f\right|$ because $f$ is parallel to $a-b$ . Therefore $f=\left|f\right|\frac{a-b}{|a-b|}$ because $\frac{a-b}{|a-b|}$ is a unit vector parallel to $f$ . Thus $f=\omega \frac{a-b}{|a-b|}$ . By definition

Clearly then [link] holds.

Lemma 2 Let $\mathbf{f}=f\delta \left(a\right)+g\delta \left(b\right)+h\delta \left(c\right)$ be balanced with $a,b$ and $c$ not lying on the same line. Then $\mathbf{f}$ is equilibrated by a truss $T.$

Without loss of generality, $a,b,$ and $c$ lie in the $xy$ -plane and $a=0.$ $f,g$ , and $h$ must lie in the plane, because otherwise it would be impossible to balance the forces. Because $a,b,$ and $c$ do not all lie on the same line, two of the points, say $b$ and $c$ , are linearly independent; dotting [link] with $b$ and then $c$ implies that $f·e_3=g·e_3=h·e_3=0$ where $e_3$ is the unit basis vector parallel with the $z$ -axis. Hence $f$ can be expressed as a linear combination of $b$ and $c$

Consider the point force field $\mathbf{e}=e_b\delta \left(b\right)+e_c\delta \left(c\right)$ where $e_b=g+{\omega }_b\frac{b}{\left|b\right|}$ and $e_c=h+\omega \frac{c}{\left|c\right|}.$ We claim $\mathbf{e}$ is balanced;