# 10.6 Networks of uniform strings  (Page 3/3)

 Page 3 / 3

## Example #1: the tritar

Depicted above is a network of three strings called a tritar. We are interested in how the eigenvalues of this simple network vary with changes in the transverse stiffness k i of each string. We assume that the longitudinal stiffnesses σ i are 1 for each string, and we also assume that the lengths of the strings are all 1 for conveninence.

We define our orientation vectors by

$\begin{array}{c}\hfill \stackrel{^}{{v}_{1}}=\left[\begin{array}{c}1\\ 0\end{array}\right]\phantom{\rule{1.em}{0ex}}\phantom{\rule{1.em}{0ex}}\stackrel{^}{{v}_{2}}=\frac{1}{\sqrt{2}}\left[\begin{array}{c}1\\ 1\end{array}\right]\phantom{\rule{1.em}{0ex}}\phantom{\rule{1.em}{0ex}}\stackrel{^}{{v}_{3}}=\frac{1}{\sqrt{2}}\left[\begin{array}{c}1\\ -1\end{array}\right]\end{array}$

First, we compute the stiffness matrices. For the first string, we obtain

$\begin{array}{ccc}\hfill {P}_{1}& =& {K}_{1}\left(I-\stackrel{^}{{v}_{1}}{\stackrel{^}{{v}_{1}}}^{T}\right)+{\sigma }_{1}\stackrel{^}{{v}_{1}}{\stackrel{^}{{v}_{1}}}^{T}\hfill \\ & =& k\left(\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right],-,\left[\begin{array}{cc}1& 0\\ 0& 0\end{array}\right]\right)+\left[\begin{array}{cc}1& 0\\ 0& 0\end{array}\right]\hfill \\ & =& \left[\begin{array}{cc}1& 0\\ 0& k\end{array}\right].\hfill \end{array}$

For the second string, we get

$\begin{array}{ccc}\hfill {P}_{2}& =& {K}_{2}\left(I-\stackrel{^}{{v}_{2}}{\stackrel{^}{{v}_{2}}}^{T}\right)+{\sigma }_{2}\stackrel{^}{{v}_{2}}{\stackrel{^}{{v}_{2}}}^{T}\hfill \\ & =& k\left(\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right],-,\frac{1}{2},\left[\begin{array}{cc}1& 1\\ 1& 1\end{array}\right]\right)+\frac{1}{2}\left[\begin{array}{cc}1& 1\\ 1& 1\end{array}\right]\hfill \\ & =& \frac{1}{2}\left[\begin{array}{cc}k+1& -k+1\\ -k+1& k+1\end{array}\right],\hfill \end{array}$

and for string number three, we have

$\begin{array}{ccc}\hfill {P}_{3}& =& {K}_{3}\left(I-\stackrel{^}{{v}_{3}}{\stackrel{^}{{v}_{3}}}^{T}\right)+{\sigma }_{3}\stackrel{^}{{v}_{3}}{\stackrel{^}{{v}_{3}}}^{T}\hfill \\ & =& k\left(\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right],-,\frac{1}{2},\left[\begin{array}{cc}1& -1\\ -1& 1\end{array}\right]\right)+\frac{1}{2}\left[\begin{array}{cc}1& -1\\ -1& 1\end{array}\right]\hfill \\ & =& \frac{1}{2}\left[\begin{array}{cc}k+1& k-1\\ k-1& k+1\end{array}\right].\hfill \end{array}$

Letting the vector ${u}_{i}={\left[{u}_{i1}\phantom{\rule{0.222222em}{0ex}}{u}_{i2}\right]}^{T}$ represent the displacements of string i , we obtain the following system of differential equations:

$\begin{array}{ccc}\hfill {P}_{1}{u}_{1}^{\text{'}\text{'}}& =& -{\lambda }^{2}{u}_{1}\hfill \\ \hfill {P}_{2}{u}_{2}^{\text{'}\text{'}}& =& -{\lambda }^{2}{u}_{2}\hfill \\ \hfill {P}_{3}{u}_{3}^{\text{'}\text{'}}& =& -{\lambda }^{2}{u}_{3}.\hfill \end{array}$

Expanding via matrix multiplication, we see that, component-wise, this translates into

$\begin{array}{ccc}\hfill {u}_{11}^{\text{'}\text{'}}& =& -{\lambda }^{2}{u}_{11}\hfill \\ \hfill k{u}_{12}^{\text{'}\text{'}}& =& -{\lambda }^{2}{u}_{12}\hfill \\ \hfill {s}_{2}^{\text{'}\text{'}}& =& -{\lambda }^{2}{s}_{2}\hfill \\ \hfill k{d}_{2}^{\text{'}\text{'}}& =& -{\lambda }^{2}{d}_{2}\hfill \\ \hfill k{s}_{3}^{\text{'}\text{'}}& =& -{\lambda }^{2}{s}_{3}\hfill \\ \hfill {d}_{3}^{\text{'}\text{'}}& =& -{\lambda }^{2}{d}_{3},\hfill \end{array}$

where

$\begin{array}{c}\hfill {s}_{i}={u}_{i1}+{u}_{i2}\\ \hfill {d}_{i}={u}_{i2}-{u}_{i2}.\end{array}$

for $i=1,2$ . For boundary conditions, we require that the ends be clamped:

$\begin{array}{ccc}\hfill {u}_{1}\left(0\right)& =& 0\hfill \\ \hfill {u}_{2}\left(1\right)& =& 0\hfill \\ \hfill {u}_{3}\left(1\right)& =& 0,\hfill \end{array}$

that there be continuity at the central node:

$\begin{array}{c}\hfill {u}_{1}\left(1\right)={u}_{2}\left(0\right)={u}_{3}\left(0\right),\end{array}$

and that the forces balance at the central node:

$\begin{array}{c}\hfill {P}_{1}{u}_{1}^{\text{'}}\left(1\right)-{P}_{2}{u}_{2}^{\text{'}}\left(0\right)-{P}_{3}{u}_{3}^{\text{'}}\left(0\right)=0.\end{array}$

As with many systems of differential equations, this one can be solved via the time-honored method of guessing. Noting that the differential equations of this form equate the second derivative of a function with a constant multiple of itself, wehypothesize that the solution for each component of displacement is some linear combination of sines and cosines:

$\begin{array}{ccc}\hfill {u}_{11}& =& {a}_{1,1}cos\lambda x+{b}_{1,1}sin\lambda x\hfill \\ \hfill {u}_{12}& =& {a}_{1,2}cos\frac{\lambda }{\sqrt{k}}x+{b}_{1,2}sin\frac{\lambda }{\sqrt{k}}x\hfill \\ \hfill {u}_{21}& =& {a}_{2}cos\lambda x+{b}_{2}sin\lambda x+{\alpha }_{2}cos\frac{\lambda }{\sqrt{k}}x+{\beta }_{2}sin\frac{\lambda }{\sqrt{k}}x\hfill \\ \hfill {u}_{22}& =& {a}_{2}cos\lambda x+{b}_{2}sin\lambda x-{\alpha }_{2}cos\frac{\lambda }{\sqrt{k}}x-{\beta }_{2}sin\frac{\lambda }{\sqrt{k}}x\hfill \\ \hfill {u}_{31}& =& {a}_{3}cos\frac{\lambda }{\sqrt{k}}x+{b}_{3}sin\frac{\lambda }{\sqrt{k}}x+{\alpha }_{3}cos\lambda x+{\beta }_{3}sin\lambda x\hfill \\ \hfill {u}_{32}& =& {a}_{3}cos\frac{\lambda }{\sqrt{k}}x+{b}_{3}sin\frac{\lambda }{\sqrt{k}}x-{\alpha }_{3}cos\lambda x-{\beta }_{3}sin\lambda x.\hfill \end{array}$

where we have guessed

$\begin{array}{ccc}\hfill {s}_{2}& =& 2{a}_{2}cos\lambda x+2{b}_{2}sin\lambda x\hfill \\ \hfill {d}_{2}& =& 2{\alpha }_{2}cos\frac{\lambda }{\sqrt{k}}x+2{\beta }_{2}sin\frac{\lambda }{\sqrt{k}}x\hfill \\ \hfill {s}_{3}& =& 2{a}_{3}cos\frac{\lambda }{\sqrt{k}}x+2{b}_{3}sin\frac{\lambda }{\sqrt{k}}x\hfill \\ \hfill {d}_{3}& =& 2{\alpha }_{3}cos\lambda x+2{\beta }_{3}sin\lambda x\hfill \end{array}$

and have used the above to translate these into u 21 , u 22 , u 31 , and u 32 . We need to determine the coefficients. Applying the boundary condition that ${u}_{1}\left(0\right)=0$ , we get

$\begin{array}{ccc}\hfill {a}_{1,1}& =& 0\hfill \\ \hfill {a}_{1,2}& =& 0.\hfill \end{array}$

The fact that ${u}_{2}\left(1\right)=0$ implies that

$\begin{array}{ccc}\hfill {a}_{2}cos\lambda +{b}_{2}sin\lambda & =& 0\hfill \\ \hfill {\alpha }_{2}cos\frac{\lambda }{\sqrt{k}}+{\beta }_{2}sin\frac{\lambda }{\sqrt{k}}& =& 0.\hfill \end{array}$

Likewise, for ${u}_{3}\left(1\right)=0$ , we get

$\begin{array}{ccc}\hfill {a}_{3}cos\frac{\lambda }{\sqrt{k}}+{b}_{3}sin\frac{\lambda }{\sqrt{k}}& =& 0\hfill \\ \hfill {\alpha }_{3}cos\lambda +{\beta }_{3}sin\lambda & =& 0.\hfill \end{array}$

Next, we apply the continuity conditions. Since ${u}_{1}\left(1\right)={u}_{2}\left(0\right)$ , we see that

$\begin{array}{ccc}\hfill {a}_{1,1}cos\lambda +{b}_{1,1}sin\lambda -{a}_{2}-{\alpha }_{2}& =& 0\hfill \\ \hfill {a}_{1,2}cos\frac{\lambda }{\sqrt{k}}+{b}_{1,2}sin\frac{\lambda }{\sqrt{k}}-{a}_{2}+{\alpha }_{2}& =& 0,\hfill \end{array}$

and because ${u}_{2}\left(0\right)={u}_{3}\left(0\right)$ we have

$\begin{array}{ccc}\hfill {a}_{2}-{a}_{3}& =& 0\hfill \\ \hfill {\alpha }_{2}-{\alpha }_{3}& =& 0.\hfill \end{array}$

Finally, from the force-balance equation, we have

$\begin{array}{ccc}\hfill -{a}_{1,1}\lambda sin\lambda & +& {b}_{1,1}\lambda cos\lambda -\frac{1}{2}\left(k+1\right)\left({b}_{2},\lambda ,+,{\beta }_{2},\frac{\lambda }{\sqrt{k}}\right)-\frac{1}{2}\left(k+1\right)\left({b}_{2},\lambda ,-,{\beta }_{2},\frac{\lambda }{\sqrt{k}}\right)-\frac{1}{2}\left(k-1\right)\left({b}_{3},\frac{\lambda }{\sqrt{k}},+,{\beta }_{3},\lambda \right)\hfill \\ & -& \frac{1}{2}\left(k+1\right)\left({b}_{3},\frac{\lambda }{\sqrt{k}},-,{\beta }_{3},\lambda \right)=0\hfill \end{array}$

and

$\begin{array}{ccc}\hfill -{a}_{1,2}\frac{\lambda }{\sqrt{k}}sin\frac{\lambda }{\sqrt{k}}& +& {b}_{1,2}\frac{\lambda }{\sqrt{k}}cos\frac{\lambda }{\sqrt{k}}-\frac{1}{2}\left(1-k\right)\left({b}_{2},\lambda ,+,{\beta }_{2},\frac{\lambda }{\sqrt{k}}\right)-\frac{1}{2}\left(k+1\right)\left({b}_{2},\lambda ,-,{\beta }_{2},\frac{\lambda }{\sqrt{k}}\right)-\frac{1}{2}\left(k-1\right)\left({b}_{3},\frac{\lambda }{\sqrt{k}},+,{\beta }_{3},\lambda \right)\hfill \\ & -& \frac{1}{2}k+1\left({b}_{3},\frac{\lambda }{\sqrt{k}},-,{\beta }_{3},\lambda \right)=0.\hfill \end{array}$

These twelve equations can be represented by a single matrix equation with the following coefficient matrix:

$\begin{array}{c}\hfill \left[\begin{array}{cccccccccccc}1& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0\\ 0& 1& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& cos\lambda & sin\lambda & 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& cos\frac{\lambda }{\sqrt{k}}& sin\frac{\lambda }{\sqrt{k}}& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 0& 0& cos\frac{\lambda }{\sqrt{k}}& sin\frac{\lambda }{\sqrt{k}}& 0& 0\\ 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& cos\lambda & sin\lambda \\ cos\lambda & 0& sin\lambda & 0& -1& 0& -1& 0& 0& 0& 0& 0\\ 0& cos\frac{\lambda }{\sqrt{k}}& 0& sin\frac{\lambda }{\sqrt{k}}& -1& 0& 1& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 1& 0& 0& 0& -1& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 1& 0& 0& 0& -1& 0\\ -\lambda sin\lambda & 0& \lambda cos\lambda & 0& 0& -\lambda & 0& \frac{\lambda }{\sqrt{k}}& 0& \left(1-k\right)\frac{\lambda }{\sqrt{k}}& 0& -\lambda \\ 0& -\frac{\lambda }{\sqrt{k}}sin\frac{\lambda }{\sqrt{k}}& 0& \frac{\lambda }{\sqrt{k}}cos\frac{\lambda }{\sqrt{k}}& 0& -\lambda & 0& \frac{\lambda }{\sqrt{k}}& 0& -\frac{\lambda }{\sqrt{k}}& 0& \lambda \end{array}\right].\end{array}$

The determinant of this matrix is

$\begin{array}{ccc}& & -sin\left(\frac{\lambda }{\sqrt{k}}\right)sin\left(\lambda \right){l}^{2}\left(6,\phantom{\rule{0.166667em}{0ex}},cos,\left(\lambda \right),sin,\left(\frac{\lambda }{\sqrt{k}}\right),{k}^{2},cos,\left(\frac{\lambda }{\sqrt{k}}\right),sin,\left(\lambda \right),+,4,\phantom{\rule{0.166667em}{0ex}},{\left(cos,\left(\lambda \right)\right)}^{2},{\left(sin,\left(\frac{\lambda }{\sqrt{k}}\right)\right)}^{2},{k}^{3/2}\right)\hfill \\ & & \phantom{\rule{1.em}{0ex}}\left(+,3,\phantom{\rule{0.166667em}{0ex}},cos,\left(\lambda \right),sin,\left(\frac{\lambda }{\sqrt{k}}\right),k,cos,\left(\frac{\lambda }{\sqrt{k}}\right),sin,\left(\lambda \right),+,{\left(sin,\left(\lambda \right)\right)}^{2},{\left(cos,\left(\frac{\lambda }{\sqrt{k}}\right)\right)}^{2},{k}^{3/2},-,{\left(sin,\left(\lambda \right)\right)}^{2},{\left(cos,\left(\frac{\lambda }{\sqrt{k}}\right)\right)}^{2},\sqrt{k}\right)\hfill \\ & & \phantom{\rule{1.em}{0ex}}\left(+,2,\phantom{\rule{0.166667em}{0ex}},{\left(sin,\left(\lambda \right)\right)}^{2},{\left(cos,\left(\frac{\lambda }{\sqrt{k}}\right)\right)}^{2},{k}^{5/2}){k}^{-3/2},\hfill \end{array}$

and by substituting in the desired value of k and setting this determiniant to zero we can then solve for the eigenvalues λ of our tritar net. A plot of the first seven eigenvalues as a function of k is displayed below: First seven planar eigenvalues for a tritar net, computed with fzero().

As expected, the eigenvalues increase as k increases; however for eigenvalues beyond the first, we observe some rather strange behavior, which suggests that something has gone wrong in the above process. The eigenvalues in the above plot werecomputed using MATLAB's fzero() function at a tolerance of 1e-10. Using a more naive bisection method (which is less likely to lock onto the wrong root) at the same tolerance, we obtain the following plot: First seven planar eigenvalues for a tritar net, computed with the bisection method.

This seems to have fixed some of the erratic behavior, but neither tightening the tolerance nor increasing the fine-ness of the mesh along which the determinant is evaluated provides much further improvement. On the other hand, it is apparent thatthe solver's structure and parameters impacts the shape of the plots. Perhaps a better solver of some sort (e.g. Newton's method, but adapted to search only in a given interval) can fix more of the problem.

## Example #2: the quintar A depiction of the quintar, with the end angles at 90 degrees.

Rather than develop all of the mathematical relations as in the previous example, it should suffice to say that the same procedure is followed. The solutions to the differential equations are still sums of sines and cosines, but more equations have been added to the system. The function obtained by setting the determinant equal to zero is not enlightening and longer than that for the quintar, and so only the final plots will be presented here. By iteratively increasing the angle at the ends of the network, a plot of the angle versus the eigenvalues is obtained in which traces are formed as the eigenvalues change. It is interesting to note how some of the eigenvalues increase in magnitude while others decrease.

A second plot is presented in which the first nine eigenvalues are plotted versus the transverse stiffness parameter for our analytic model. A region of stiffnesses was chosen where the root-finding algorithm is able to successfully lock onto the zeroes of the determinant as they change. It is clear from this plot that for this range of stiffnesses, increasing stiffness results in the modes of vibration increasing in frequency. This result is expected, since in general stiffer members possess higher vibrational frequencies. A plot of the opening angle at the ends versus the eigenvalues. A plot of the first nine eigenvalues for planar modes versus the transverse stiffness.

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
Berger describes sociologists as concerned with
Got questions? Join the online conversation and get instant answers!   By Mistry Bhavesh  By  By   By Rhodes