# 0.5 Michell truss stress measure

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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.

## Unidirectional curve stress

Let $C$ be a simple curve in ${\mathbb{R}}^{d}$ . In other words $C$ is the image of a map $r\in {C}^{1,1}\left(\left[0,1\right],{\mathbb{R}}^{d}\right)$ such that $\stackrel{˙}{r}\left(t\right)\ne 0$ for each $t\in \left[0,1\right]$ 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 $|\stackrel{˙}{r}\left(t\right)|$ constant on $\left[0,1\right]$ . Therefore $|\stackrel{˙}{r}\left(t\right)|=\frac{{\int }_{0}^{1}|\stackrel{˙}{r}\left(s\right)|\phantom{\rule{0.166667em}{0ex}}ds}{1-0}={\mathcal{H}}^{1}\left(C\right)$ for each $t\in \left[0,1\right]$ and the unit tangent vector to $C$ at time $t$ will be

$\tau \left(t\right)=\frac{\stackrel{˙}{r}\left(t\right)}{|\stackrel{˙}{r}\left(t\right)|}=\frac{\stackrel{˙}{r}\left(t\right)}{{\mathcal{H}}^{1}\left(C\right)}.$

The curvature $\kappa$ at time $t$ is then given by

$\kappa \left(t\right)=\frac{d\tau /dt}{|dr/dt|}=\frac{\stackrel{¨}{r}\left(t\right)/{\mathcal{H}}^{1}\left(C\right)}{{\mathcal{H}}^{1}\left(C\right)}=\frac{\stackrel{¨}{r}\left(t\right)}{{\left({\mathcal{H}}^{1}\left(C\right)\right)}^{2}}.$

We now define a measure ${\sigma }^{C}$ that is proportional to the stress along the curve $C$ .

${\sigma }^{C}:=\left(\tau \otimes \tau \right){\mathcal{H}}^{1}{↾}_{C}\phantom{\rule{0.166667em}{0ex}}=\frac{1}{{\mathcal{H}}^{1}{\left(C\right)}^{2}}\left(\stackrel{˙}{r}\otimes \stackrel{˙}{r}\right){\mathcal{H}}^{1}{↾}_{C}$

Thus ${\sigma }^{C}$ is a rank 1 symmetric matrix measure.

We will now treat ${\sigma }^{C}$ as the functional on ${C}_{c}\left({\mathbb{R}}^{d},{\mathbb{R}}^{d×d}\right)$ that corresponds to the measure ${\sigma }^{C}$ . For any $\xi \in {C}_{c}\left({\mathbb{R}}^{d},{\mathbb{R}}^{d×d}\right)$ ,

$\begin{array}{ccc}\hfill {\sigma }^{C}\left[\xi \right]& =& \int ⟨\xi ;\tau \otimes \tau ⟩\phantom{\rule{0.166667em}{0ex}}d{\mathcal{H}}^{1}{↾}_{C}\hfill \\ & =& \frac{1}{{\mathcal{H}}^{1}{\left(C\right)}^{2}}{\int }_{C}⟨\xi \left(r\right);\stackrel{˙}{r}\otimes \stackrel{˙}{r}⟩\phantom{\rule{0.166667em}{0ex}}d|r|\hfill \\ & =& \frac{1}{{\mathcal{H}}^{1}\left(C\right)}{\int }_{0}^{1}⟨\xi \left(r\left(t\right)\right);\stackrel{˙}{r}\left(t\right)\otimes \stackrel{˙}{r}\left(t\right)⟩\phantom{\rule{0.166667em}{0ex}}dt.\hfill \end{array}$

To find the divergence of the measure ${\sigma }^{C}$ , we let an arbitrary vector-valued function $u\in {C}_{c}^{1}\left({\mathbb{R}}^{d},{\mathbb{R}}^{d×d}\right)$ be given and we evaluate the functional $-\text{div}\phantom{\rule{4.pt}{0ex}}{\sigma }^{C}\left[u\right]$ .

$\begin{array}{ccc}\hfill \left(-\text{div}\phantom{\rule{4.pt}{0ex}}{\sigma }^{C}\right)\left[u\right]& =& {\sigma }^{C}\left[\nabla u\right]\hfill \\ & =& {\int }_{{\mathcal{R}}^{d}}\nabla u\phantom{\rule{0.166667em}{0ex}}d{\sigma }^{C}\hfill \\ & =& \frac{1}{{\mathcal{H}}^{1}\left(C\right)}{\int }_{0}^{1}⟨\nabla u\left(r\left(t\right)\right);\stackrel{˙}{r}\left(t\right)\otimes \stackrel{˙}{r}\left(t\right)⟩\phantom{\rule{0.166667em}{0ex}}dt\hfill \end{array}$

Using the definitions of the inner product and the tensor product, we can convert the matrix product $⟨\nabla u\left(r\left(t\right)\right);\stackrel{˙}{r}\left(t\right)\otimes \stackrel{˙}{r}\left(t\right)⟩$ into a vector product:

$\begin{array}{ccc}\hfill ⟨\nabla u\left(r\left(t\right)\right);\stackrel{˙}{r}\left(t\right)\otimes \stackrel{˙}{r}\left(t\right)⟩& =& ⟨\nabla u\left(r\left(t\right)\right);\stackrel{˙}{r}\left(t\right)\phantom{\rule{0.166667em}{0ex}}\stackrel{˙}{r}{\left(t\right)}^{T}⟩\hfill \\ & =& tr\left[\nabla u\left(r\left(t\right)\right)\phantom{\rule{0.166667em}{0ex}}{\left(\stackrel{˙}{r}\left(t\right)\phantom{\rule{0.166667em}{0ex}}\stackrel{˙}{r}{\left(t\right)}^{T}\right)}^{T}\right]\hfill \\ & =& tr\left[\nabla u\left(r\left(t\right)\right)\phantom{\rule{0.166667em}{0ex}}\stackrel{˙}{r}\left(t\right)\phantom{\rule{0.166667em}{0ex}}\stackrel{˙}{r}{\left(t\right)}^{T}\right]\hfill \\ & =& ⟨\nabla u\left(r\left(t\right)\right)\phantom{\rule{0.166667em}{0ex}}\stackrel{˙}{r}\left(t\right);\stackrel{˙}{r}\left(t\right)⟩\hfill \\ & =& 〈\frac{d}{dt},\left[u\left(r\left(t\right)\right)\right],;,\stackrel{˙}{r},\left(t\right)〉.\hfill \end{array}$

Therefore

$\begin{array}{ccc}\hfill \left(-\text{div}\phantom{\rule{4.pt}{0ex}}{\sigma }^{C}\right)\left[u\right]& =& \frac{1}{{\mathcal{H}}^{1}\left(C\right)}{\int }_{0}^{1}〈\frac{d}{dt},\left[u\left(r\left(t\right)\right)\right],;,\stackrel{˙}{r},\left(t\right)〉\phantom{\rule{0.166667em}{0ex}}dt\hfill \\ & =& {\left(〈u,\left(r\left(t\right)\right),;,\frac{\stackrel{˙}{r}\left(t\right)}{{\mathcal{H}}^{1}\left(C\right)}〉|}_{t=0}^{1}-\frac{1}{{\mathcal{H}}^{1}\left(C\right)}{\int }_{0}^{1}⟨u\left(r\left(t\right)\right);\stackrel{¨}{r}\left(t\right)⟩\phantom{\rule{0.166667em}{0ex}}dt\hfill \\ & =& 〈u,\left(r\left(1\right)\right),;,\frac{\stackrel{˙}{r}\left(1\right)}{{\mathcal{H}}^{1}\left(C\right)}〉-〈u,\left(r\left(0\right)\right),;,\frac{\stackrel{˙}{r}\left(0\right)}{{\mathcal{H}}^{1}\left(C\right)}〉\hfill \\ & & -\frac{1}{{\mathcal{H}}^{1}{\left(C\right)}^{2}}{\int }_{C}⟨u\left(r\right);\stackrel{¨}{r}⟩\phantom{\rule{0.166667em}{0ex}}dr\hfill \\ & =& {\int }_{{\mathcal{R}}^{d}}⟨u;\tau ⟩\phantom{\rule{0.166667em}{0ex}}d{\delta }_{B}-{\int }_{{\mathcal{R}}^{d}}⟨u;\tau ⟩\phantom{\rule{0.166667em}{0ex}}d{\delta }_{A}-{\int }_{{\mathcal{R}}^{d}}⟨u;\kappa ⟩\phantom{\rule{0.166667em}{0ex}}d{\mathcal{H}}^{1}{↾}_{C}\hfill \\ & =& \tau {\delta }_{B}\left[u\right]-\tau {\delta }_{A}\left[u\right]-\kappa {\mathcal{H}}^{1}{↾}_{C}\left[u\right]\hfill \\ & =& \left(\tau {\delta }_{B}-\tau {\delta }_{A}-\kappa {\mathcal{H}}^{1}{↾}_{C}\right)\left[u\right],\hfill \end{array}$

where $A=r\left(0\right)$ and $B=r\left(1\right)$ . Since this equation holds for any $u\in {C}_{c}^{1}\left({\mathbb{R}}^{d},{\mathbb{R}}^{d×d}\right)$ , we say that the measure $-\text{div}\phantom{\rule{4.pt}{0ex}}{\sigma }^{C}$ is given by

$-\text{div}\phantom{\rule{4.pt}{0ex}}{\sigma }^{C}=\tau {\delta }_{B}-\tau {\delta }_{A}-\kappa {\mathcal{H}}^{1}{↾}_{C}.$

When the curve $C$ is just equal to the line segment $\left[A,B\right]$ , $\tau =\frac{B-A}{|B-A|}$ and $\kappa =0$ . Therefore our equation becomes

$-\text{div}\phantom{\rule{4.pt}{0ex}}{\sigma }^{\left[A,B\right]}=\left({\delta }_{B}-{\delta }_{A}\right)\frac{B-A}{|B-A|}.$

## Stress in a truss

The stress in a beam is given by $\frac{1}{2}\lambda \phantom{\rule{0.166667em}{0ex}}{\sigma }^{\left[A,B\right]}$ for some $\lambda$ . A finite collection of such beams is a truss . Therefore the stress in a truss that consists of $l$ nodes at $\left\{{A}_{1},{A}_{2},\cdots ,{A}_{l}\right\}$ is given by

$\sigma =\sum _{m,n=1}^{l}-{\lambda }_{m,n}\phantom{\rule{0.166667em}{0ex}}{\sigma }^{\left[{A}_{m},{A}_{n}\right]},$

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

$F=-\text{div}\phantom{\rule{4.pt}{0ex}}\sigma =-\text{div}\phantom{\rule{4.pt}{0ex}}\left(\sum _{m,n=1}^{l},-,{\lambda }_{m,n},\phantom{\rule{0.166667em}{0ex}},{\sigma }^{\left[{A}_{m},{A}_{n}\right]}\right).$

Since the operator $-\text{div}$ is linear, we get

$\begin{array}{ccc}\hfill F& =& \sum _{m,n=1}^{l}{\lambda }_{m,n}\phantom{\rule{0.166667em}{0ex}}\text{div}\phantom{\rule{4.pt}{0ex}}{\sigma }^{\left[{A}_{m},{A}_{n}\right]}\hfill \\ & =& \sum _{m,n=1}^{l}{\lambda }_{m,n}\phantom{\rule{0.166667em}{0ex}}\left({\delta }_{{A}_{m}}-{\delta }_{{A}_{n}}\right)\frac{{A}_{m}-{A}_{n}}{|{A}_{m}-{A}_{n}|},\hfill \end{array}$

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 $\mathcal{A}=\left\{{A}_{1},{A}_{2},\cdots ,{A}_{l}\right\}$ , and a set of weights $\Lambda =\left\{{\lambda }_{m,n}:m,n=1,\cdots l\right\}$ such that $F={\sum }_{m,n=1}^{l}{\lambda }_{m,n}\phantom{\rule{0.166667em}{0ex}}\left({\delta }_{{A}_{m}}-{\delta }_{{A}_{n}}\right)\frac{{A}_{m}-{A}_{n}}{|{A}_{m}-{A}_{n}|}$ , then the cost of the truss structure will be the "total mass" of the structure, which is given by ${\int }_{{\mathcal{R}}^{d}}|\sigma |$ .

$\begin{array}{ccc}\hfill {\int }_{{\mathcal{R}}^{d}}|\sigma |& =& |\sigma |\left({\mathcal{R}}^{d}\right)\hfill \\ & =& |\sigma |\left(\bigcup _{1\le m

## Cost minimization

Given a truss $T$ , which consists of a set of points $\mathcal{A}=\left\{{A}_{1},{A}_{2},\cdots ,{A}_{l}\right\}$ and a set of weights $\Lambda =\left\{{\lambda }_{m,n}:m,n=1,\cdots l\right\}$ , 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.

${\Sigma }_{F}^{T}\left(\overline{\Omega }\right)=\left\{\sigma ,\in ,\mathcal{M},\left(\overline{\Omega },{S}^{d×d}\right),:,-,\text{div},\phantom{\rule{4.pt}{0ex}},\sigma ,=,F,,,\phantom{\rule{4.pt}{0ex}},\text{and},\phantom{\rule{4.pt}{0ex}},\sigma ,=,\sum _{m,n=1}^{l},-,{\lambda }_{m,n},\phantom{\rule{0.166667em}{0ex}},{\sigma }^{\left[{A}_{m},{A}_{n}\right]}\right\}$

The requirement $\sigma ={\sum }_{m,n=1}^{l}-{\lambda }_{m,n}\phantom{\rule{0.166667em}{0ex}}{\sigma }^{\left[{A}_{m},{A}_{n}\right]}$ is equivalent to the restriction that $\sigma$ be rank 1 and supported on a finite collection of simple curves.

We also define the set

${\Sigma }_{F}\left(\overline{\Omega }\right)=\left\{\sigma ,\in ,\mathcal{M},\left(,\overline{\Omega },,,{S}^{d×d},\right),:,-,\text{div},\phantom{\rule{4.pt}{0ex}},\sigma ,=,F\right\},$

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 ${inf}_{T}\left\{\phantom{\rule{0.166667em}{0ex}},inf,\left\{\int |\sigma |:\sigma ,\in ,{\Sigma }_{F}^{T},\left(,\overline{\Omega }\right\}\right\}$ 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

$\int |{\sigma }^{0}|=inf\left\{\int |\sigma |:\sigma ,\in ,\bigcup _{T},\left({\Sigma }_{F}^{T},\left(\overline{\Omega }\right)\right)\right\}.$

Such a ${\sigma }^{0}$ , however, does not always exist [link] .

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