# 0.5 Michell truss stress measure

 Page 1 / 1
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] .

what is variations in raman spectra for nanomaterials
I only see partial conversation and what's the question here!
what about nanotechnology for water purification
please someone correct me if I'm wrong but I think one can use nanoparticles, specially silver nanoparticles for water treatment.
Damian
yes that's correct
Professor
I think
Professor
what is the stm
is there industrial application of fullrenes. What is the method to prepare fullrene on large scale.?
Rafiq
industrial application...? mmm I think on the medical side as drug carrier, but you should go deeper on your research, I may be wrong
Damian
How we are making nano material?
what is a peer
What is meant by 'nano scale'?
What is STMs full form?
LITNING
scanning tunneling microscope
Sahil
how nano science is used for hydrophobicity
Santosh
Do u think that Graphene and Fullrene fiber can be used to make Air Plane body structure the lightest and strongest. Rafiq
Rafiq
what is differents between GO and RGO?
Mahi
what is simplest way to understand the applications of nano robots used to detect the cancer affected cell of human body.? How this robot is carried to required site of body cell.? what will be the carrier material and how can be detected that correct delivery of drug is done Rafiq
Rafiq
what is Nano technology ?
write examples of Nano molecule?
Bob
The nanotechnology is as new science, to scale nanometric
brayan
nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale
Damian
Is there any normative that regulates the use of silver nanoparticles?
what king of growth are you checking .?
Renato
What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ?
why we need to study biomolecules, molecular biology in nanotechnology?
?
Kyle
yes I'm doing my masters in nanotechnology, we are being studying all these domains as well..
why?
what school?
Kyle
biomolecules are e building blocks of every organics and inorganic materials.
Joe
anyone know any internet site where one can find nanotechnology papers?
research.net
kanaga
sciencedirect big data base
Ernesto
Introduction about quantum dots in nanotechnology
what does nano mean?
nano basically means 10^(-9). nanometer is a unit to measure length.
Bharti
do you think it's worthwhile in the long term to study the effects and possibilities of nanotechnology on viral treatment?
absolutely yes
Daniel
how did you get the value of 2000N.What calculations are needed to arrive at it
Privacy Information Security Software Version 1.1a
Good
Got questions? Join the online conversation and get instant answers!  By By By By  By By By By Rhodes