More detailed and accessible derivations of the equations in section 2 of the paper "Michelle Trusses and Lines of Principal Action" are provided here. See Stress and Measure Theory modules for background material.
Michelle trusses
Unidirectional curve stress
Let
$C$ be a simple curve in
${\mathbb{R}}^{d}$ . In other words
$C$ is the image of a map
$r\in {C}^{1,1}([0,1],{\mathbb{R}}^{d})$ such that
$\dot{r}\left(t\right)\ne 0$ for each
$t\in [0,1]$ and
$r$ is injective.
We will also require without loss of generality that
$r\left(t\right)$ trace out the curve
$C$ at a constant speed, making
$|\dot{r}\left(t\right)|$ constant on
$[0,1]$ . Therefore
$|\dot{r}\left(t\right)|=\frac{{\int}_{0}^{1}\left|\dot{r}\left(s\right)\right|\phantom{\rule{0.166667em}{0ex}}ds}{1-0}={\mathcal{H}}^{1}\left(C\right)$ for each
$t\in [0,1]$ and the unit tangent vector to
$C$ at time
$t$ will be
$$\tau \left(t\right)=\frac{\dot{r}\left(t\right)}{|\dot{r}\left(t\right)|}=\frac{\dot{r}\left(t\right)}{{\mathcal{H}}^{1}\left(C\right)}.$$
The curvature
$\kappa $ at time
$t$ is then given by
$$\kappa \left(t\right)=\frac{d\tau /dt}{|dr/dt|}=\frac{\ddot{r}\left(t\right)/{\mathcal{H}}^{1}\left(C\right)}{{\mathcal{H}}^{1}\left(C\right)}=\frac{\ddot{r}\left(t\right)}{{\left({\mathcal{H}}^{1}\left(C\right)\right)}^{2}}.$$
We now define a measure
${\sigma}^{C}$ that is proportional to the stress along the curve
$C$ .
$${\sigma}^{C}:=(\tau \otimes \tau ){\mathcal{H}}^{1}{\upharpoonright}_{C}\phantom{\rule{0.166667em}{0ex}}=\frac{1}{{\mathcal{H}}^{1}{\left(C\right)}^{2}}(\dot{r}\otimes \dot{r}){\mathcal{H}}^{1}{\upharpoonright}_{C}$$
Thus
${\sigma}^{C}$ is a rank 1 symmetric matrix measure.
We will now treat
${\sigma}^{C}$ as the functional on
${C}_{c}({\mathbb{R}}^{d},{\mathbb{R}}^{d\times d})$ that corresponds to the measure
${\sigma}^{C}$ .
For any
$\xi \in {C}_{c}({\mathbb{R}}^{d},{\mathbb{R}}^{d\times d})$ ,
$$\begin{array}{ccc}\hfill {\sigma}^{C}\left[\xi \right]& =& \int \u27e8\xi ;\tau \otimes \tau \u27e9\phantom{\rule{0.166667em}{0ex}}d{\mathcal{H}}^{1}{\upharpoonright}_{C}\hfill \\ & =& \frac{1}{{\mathcal{H}}^{1}{\left(C\right)}^{2}}{\int}_{C}\u27e8\xi \left(r\right);\dot{r}\otimes \dot{r}\u27e9\phantom{\rule{0.166667em}{0ex}}d\left|r\right|\hfill \\ & =& \frac{1}{{\mathcal{H}}^{1}\left(C\right)}{\int}_{0}^{1}\u27e8\xi \left(r\left(t\right)\right);\dot{r}\left(t\right)\otimes \dot{r}\left(t\right)\u27e9\phantom{\rule{0.166667em}{0ex}}dt.\hfill \end{array}$$
To find the divergence of the measure
${\sigma}^{C}$ , we let an arbitrary vector-valued function
$u\in {C}_{c}^{1}({\mathbb{R}}^{d},{\mathbb{R}}^{d\times d})$ be given and we evaluate the functional
$-\text{div}\phantom{\rule{4.pt}{0ex}}{\sigma}^{C}\left[u\right]$ .
$$\begin{array}{ccc}\hfill (-\text{div}\phantom{\rule{4.pt}{0ex}}{\sigma}^{C})\left[u\right]& =& {\sigma}^{C}\left[\nabla u\right]\hfill \\ & =& {\int}_{{\mathcal{R}}^{d}}\nabla u\phantom{\rule{0.166667em}{0ex}}d{\sigma}^{C}\hfill \\ & =& \frac{1}{{\mathcal{H}}^{1}\left(C\right)}{\int}_{0}^{1}\u27e8\nabla u\left(r\left(t\right)\right);\dot{r}\left(t\right)\otimes \dot{r}\left(t\right)\u27e9\phantom{\rule{0.166667em}{0ex}}dt\hfill \end{array}$$
Using the definitions of the inner product and the tensor product, we can convert the matrix product
$\u27e8\nabla u\left(r\left(t\right)\right);\dot{r}\left(t\right)\otimes \dot{r}\left(t\right)\u27e9$ into a vector product:
$$\begin{array}{ccc}\hfill \u27e8\nabla u\left(r\left(t\right)\right);\dot{r}\left(t\right)\otimes \dot{r}\left(t\right)\u27e9& =& \u27e8\nabla u\left(r\left(t\right)\right);\dot{r}\left(t\right)\phantom{\rule{0.166667em}{0ex}}\dot{r}{\left(t\right)}^{T}\u27e9\hfill \\ & =& tr\left[\nabla u\left(r\left(t\right)\right)\phantom{\rule{0.166667em}{0ex}}{\left(\dot{r}\left(t\right)\phantom{\rule{0.166667em}{0ex}}\dot{r}{\left(t\right)}^{T}\right)}^{T}\right]\hfill \\ & =& tr\left[\nabla u\left(r\left(t\right)\right)\phantom{\rule{0.166667em}{0ex}}\dot{r}\left(t\right)\phantom{\rule{0.166667em}{0ex}}\dot{r}{\left(t\right)}^{T}\right]\hfill \\ & =& \u27e8\nabla u\left(r\left(t\right)\right)\phantom{\rule{0.166667em}{0ex}}\dot{r}\left(t\right);\dot{r}\left(t\right)\u27e9\hfill \\ & =& \u2329\frac{d}{dt},\left[u\left(r\left(t\right)\right)\right],;,\dot{r},\left(t\right)\u232a.\hfill \end{array}$$
Therefore
$$\begin{array}{ccc}\hfill (-\text{div}\phantom{\rule{4.pt}{0ex}}{\sigma}^{C})\left[u\right]& =& \frac{1}{{\mathcal{H}}^{1}\left(C\right)}{\int}_{0}^{1}\u2329\frac{d}{dt},\left[u\left(r\left(t\right)\right)\right],;,\dot{r},\left(t\right)\u232a\phantom{\rule{0.166667em}{0ex}}dt\hfill \\ & =& {\left(\u2329u,\left(r\left(t\right)\right),;,\frac{\dot{r}\left(t\right)}{{\mathcal{H}}^{1}\left(C\right)}\u232a\right|}_{t=0}^{1}-\frac{1}{{\mathcal{H}}^{1}\left(C\right)}{\int}_{0}^{1}\u27e8u\left(r\left(t\right)\right);\ddot{r}\left(t\right)\u27e9\phantom{\rule{0.166667em}{0ex}}dt\hfill \\ & =& \u2329u,\left(r\left(1\right)\right),;,\frac{\dot{r}\left(1\right)}{{\mathcal{H}}^{1}\left(C\right)}\u232a-\u2329u,\left(r\left(0\right)\right),;,\frac{\dot{r}\left(0\right)}{{\mathcal{H}}^{1}\left(C\right)}\u232a\hfill \\ & & -\frac{1}{{\mathcal{H}}^{1}{\left(C\right)}^{2}}{\int}_{C}\u27e8u\left(r\right);\ddot{r}\u27e9\phantom{\rule{0.166667em}{0ex}}dr\hfill \\ & =& {\int}_{{\mathcal{R}}^{d}}\u27e8u;\tau \u27e9\phantom{\rule{0.166667em}{0ex}}d{\delta}_{B}-{\int}_{{\mathcal{R}}^{d}}\u27e8u;\tau \u27e9\phantom{\rule{0.166667em}{0ex}}d{\delta}_{A}-{\int}_{{\mathcal{R}}^{d}}\u27e8u;\kappa \u27e9\phantom{\rule{0.166667em}{0ex}}d{\mathcal{H}}^{1}{\upharpoonright}_{C}\hfill \\ & =& \tau {\delta}_{B}\left[u\right]-\tau {\delta}_{A}\left[u\right]-\kappa {\mathcal{H}}^{1}{\upharpoonright}_{C}\left[u\right]\hfill \\ & =& (\tau {\delta}_{B}-\tau {\delta}_{A}-\kappa {\mathcal{H}}^{1}{\upharpoonright}_{C})\left[u\right],\hfill \end{array}$$
where
$A=r\left(0\right)$ and
$B=r\left(1\right)$ . Since this equation holds for any
$u\in {C}_{c}^{1}({\mathbb{R}}^{d},{\mathbb{R}}^{d\times d})$ , we say that the measure
$-\text{div}\phantom{\rule{4.pt}{0ex}}{\sigma}^{C}$ is given by
$$-\text{div}\phantom{\rule{4.pt}{0ex}}{\sigma}^{C}=\tau {\delta}_{B}-\tau {\delta}_{A}-\kappa {\mathcal{H}}^{1}{\upharpoonright}_{C}.$$
When the curve
$C$ is just equal to the line segment
$[A,B]$ ,
$\tau =\frac{B-A}{|B-A|}$ and
$\kappa =0$ . Therefore our equation becomes
$$-\text{div}\phantom{\rule{4.pt}{0ex}}{\sigma}^{[A,B]}=({\delta}_{B}-{\delta}_{A})\frac{B-A}{|B-A|}.$$
Stress in a truss
The stress in a beam is given by
$\frac{1}{2}\lambda \phantom{\rule{0.166667em}{0ex}}{\sigma}^{[A,B]}$ for some
$\lambda $ . A finite collection of such beams is a
truss . Therefore the stress in a truss that consists of
$l$ nodes at
$\{{A}_{1},{A}_{2},\cdots ,{A}_{l}\}$ is given by
$$\sigma =\sum _{m,n=1}^{l}-{\lambda}_{m,n}\phantom{\rule{0.166667em}{0ex}}{\sigma}^{[{A}_{m},{A}_{n}]},$$
where we require that
${\lambda}_{m,m}=0$ for each
$m$ .
Recall that the stress
$\sigma $ of a structure that is in equilibrium must satisfy the restrictions
$-\text{div}\phantom{\rule{4.pt}{0ex}}{\sigma}^{T}=F$ and
$\sigma ={\sigma}^{T}$ . It is easy to see that
$\sigma $ will be symmetric by definition. Therefore, if our truss is in equilibrium, there must be a force distribution
$F$ such that
$$F=-\text{div}\phantom{\rule{4.pt}{0ex}}\sigma =-\text{div}\phantom{\rule{4.pt}{0ex}}\left(\sum _{m,n=1}^{l},-,{\lambda}_{m,n},\phantom{\rule{0.166667em}{0ex}},{\sigma}^{[{A}_{m},{A}_{n}]}\right).$$
Since the operator
$-\text{div}$ is linear, we get
$$\begin{array}{ccc}\hfill F& =& \sum _{m,n=1}^{l}{\lambda}_{m,n}\phantom{\rule{0.166667em}{0ex}}\text{div}\phantom{\rule{4.pt}{0ex}}{\sigma}^{[{A}_{m},{A}_{n}]}\hfill \\ & =& \sum _{m,n=1}^{l}{\lambda}_{m,n}\phantom{\rule{0.166667em}{0ex}}({\delta}_{{A}_{m}}-{\delta}_{{A}_{n}})\frac{{A}_{m}-{A}_{n}}{|{A}_{m}-{A}_{n}|},\hfill \end{array}$$
which will be our new requirement for a truss to be balanced
[link] .
Cost of a truss
If we are given a force distribution
$F$ , a set of points
$\mathcal{A}=\{{A}_{1},{A}_{2},\cdots ,{A}_{l}\}$ , and a set of weights
$\Lambda =\{{\lambda}_{m,n}:m,n=1,\cdots l\}$ such that
$F={\sum}_{m,n=1}^{l}{\lambda}_{m,n}\phantom{\rule{0.166667em}{0ex}}({\delta}_{{A}_{m}}-{\delta}_{{A}_{n}})\frac{{A}_{m}-{A}_{n}}{|{A}_{m}-{A}_{n}|}$ , then the cost of the truss structure will be the "total mass" of the structure, which is given by
${\int}_{{\mathcal{R}}^{d}}\left|\sigma \right|$ .
$$\begin{array}{ccc}\hfill {\int}_{{\mathcal{R}}^{d}}\left|\sigma \right|& =& \left|\sigma \right|\left({\mathcal{R}}^{d}\right)\hfill \\ & =& \left|\sigma \right|\left(\bigcup _{1\le m<n\le l},[{A}_{m},{A}_{n}]\right)\hfill \\ & =& \sum _{1\le m<n\le l}\left|\sigma \right|\left([{A}_{m},{A}_{n}]\right)\hfill \\ & =& \sum _{1\le m<n\le l}\left|\sum _{p,q=1}^{l},{\lambda}_{p,q},\phantom{\rule{0.166667em}{0ex}},{\sigma}^{[{A}_{p},{A}_{q}]}\right|\left([{A}_{m},{A}_{n}]\right)\hfill \\ & =& \sum _{1\le m<n\le l}\left|{\lambda}_{m,n},\phantom{\rule{0.166667em}{0ex}},{\sigma}^{[{A}_{m},{A}_{n}]},+,{\lambda}_{n,m},\phantom{\rule{0.166667em}{0ex}},{\sigma}^{[{A}_{n},{A}_{m}]}\right|\left([{A}_{m},{A}_{n}]\right)\hfill \\ & =& 2\sum _{1\le m<n\le l}\left|{\lambda}_{m,n},\phantom{\rule{0.166667em}{0ex}},\frac{{A}_{m}-{A}_{n}}{|{A}_{m}-{A}_{n}|},\otimes ,\frac{{A}_{m}-{A}_{n}}{|{A}_{m}-{A}_{n}|},{\mathcal{H}}^{1},{\upharpoonright}_{[{A}_{m},{A}_{n}]}\right|\left([{A}_{m},{A}_{n}]\right)\hfill \\ & =& \sum _{m,n=1}^{l}\left|{\lambda}_{m,n}\right|\frac{|({A}_{m}-{A}_{n})\otimes ({A}_{m}-{A}_{n})|}{|{A}_{m}-{A}_{n}|}\left|{\mathcal{H}}^{1},{\upharpoonright}_{[{A}_{m},{A}_{n}]}\right|\left([{A}_{m},{A}_{n}]\right)\hfill \\ & =& \sum _{m,n=1}^{l}|{\lambda}_{m,n}\left|\right|{A}_{m}-{A}_{n}|\hfill \end{array}$$
Cost minimization
Given a truss
$T$ , which consists of a set of points
$\mathcal{A}=\{{A}_{1},{A}_{2},\cdots ,{A}_{l}\}$ and a set of weights
$\Lambda =\{{\lambda}_{m,n}:m,n=1,\cdots l\}$ , and given a force distribution
$F$ of finite support, we define a set
${\Sigma}_{F}^{T}\left(\overline{\Omega}\right)$ of all admissible stress measures for the truss T.
$${\Sigma}_{F}^{T}\left(\overline{\Omega}\right)=\left\{\sigma ,\in ,\mathcal{M},(\overline{\Omega},{S}^{d\times d}),:,-,\text{div},\phantom{\rule{4.pt}{0ex}},\sigma ,=,F,,,\phantom{\rule{4.pt}{0ex}},\text{and},\phantom{\rule{4.pt}{0ex}},\sigma ,=,\sum _{m,n=1}^{l},-,{\lambda}_{m,n},\phantom{\rule{0.166667em}{0ex}},{\sigma}^{[{A}_{m},{A}_{n}]}\right\}$$
The requirement
$\sigma ={\sum}_{m,n=1}^{l}-{\lambda}_{m,n}\phantom{\rule{0.166667em}{0ex}}{\sigma}^{[{A}_{m},{A}_{n}]}$ is equivalent to the restriction that
$\sigma $ be rank 1 and supported on a finite collection of simple curves.
We also define the set
$${\Sigma}_{F}\left(\overline{\Omega}\right)=\left\{\sigma ,\in ,\mathcal{M},(,\overline{\Omega},,,{S}^{d\times d},),:,-,\text{div},\phantom{\rule{4.pt}{0ex}},\sigma ,=,F\right\},$$
of all stress measures for structures that balance the force distribution
$F$ . The stresses in this set do not necessarily correspond to trusses, since we have dropped the requirement that the stress be rank 1 and concentrated on curves in
${\mathbb{R}}^{d}$ . In fact, the stress measures in
${\Sigma}_{F}\left(\overline{\Omega}\right)$ may be spread out over a region of
${\mathbb{R}}^{d}$ . Also, we may allow the given force
$F$ to be diffuse.
Our original optimization problem was to find a truss
${T}^{0}$ such that the cost of
${T}^{0}$ is equal to
${inf}_{T}\left\{\phantom{\rule{0.166667em}{0ex}},inf,\left\{\int \left|\sigma \right|:\sigma ,\in ,{\Sigma}_{F}^{T},(,\overline{\Omega}\right\}\right\}$ for a given force distribution
$F$ of finite support. This is equivalent to finding a stress measure
${\sigma}^{0}\in {\bigcup}_{T}\left({\Sigma}_{F}^{T},\left(\overline{\Omega}\right)\right)$ such that
$$\int |{\sigma}^{0}|=inf\left\{\int \left|\sigma \right|:\sigma ,\in ,\bigcup _{T},\left({\Sigma}_{F}^{T},\left(\overline{\Omega}\right)\right)\right\}.$$
Such a
${\sigma}^{0}$ , however, does not always exist
[link] .