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Template talk:Note

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- just checking about the changes to this template in Jan 2018 - the template no longer wraps text? e.g., Survey research and design in psychology/Assessment/Survey/TSQFUS1 -- Jtneill - Talk - c 22:23, 14 March 2018 (UTC)

 * Better now. Thanks! -- Dave Braunschweig (discuss • contribs) 23:18, 14 March 2018 (UTC)

title: Template talk:Note


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User:Eas4200c.f08.vqcrew.f

$$\frac{W_f}{W_o} = \frac{T}{W_o}\frac{t}{Isp}$$

User:Eas4200c.f08.vqcrew.f/hw3

User:Eas4200c.f08.vqcrew.f/hw4

User:Eas4200c.f08.vqcrew.f/hw5

User:Eas4200c.f08.vqcrew.f/hw6

Non-Symmetric Thin Walled Cross-Section

Now that the formulas have been laid out for non-symmetric case, as seen in HW6, it is now time to look at a problem utilizing these equations. The non-symmetric case can be seen in Fig #. This case is vary similar to the symmetrical cross-section in example problem 5.2 in the book, however the areas will vary by stringer for the non-symmetrical case. The figure also shows the location of the centroid which was calculated in the HW section below, and the shear flow across each skin panel.

 * $$A_1 = A \;$$

 * $$A_2 = 2A \;$$

 * $$A_3 = 3A \;$$

 * $$A_4 = 4A \;$$

In evaluating Fig #., the mean value theorem will be used in analyzing the first moment of the area, Q_y and Q_z, so the equations can be modified to the following:

 * $$ Q_z = \int_A \bar{y} \,dA = \bar{y} A $$

 * $$ Q_y = \int_A \bar{z} \,dA = \bar{z} A $$

The Area to be used for the overall moments will have to be the sum of the area of the components. Neglecting the skin and spar webs the equation for the overall area the sum of the area of $$A_i\;$$

 * $$A = \sum_{i=1}^{4}{A_i}\;$$

From the first moment of area equations and the known area it is now possible to examine the shear flow along the path s. Looking at Fig #., V_z is zero for this case so the shear flow equations becomes the following:

 * $$ q(s) = (k_{yz}Q_z^{(s)} - k_z Q_y^{(s)}) V_z $$

Now if we break down the shear flow equation and examine each variable it can be determined the each variable is actually independent of the past s.

 * $$\left.\begin{matrix}

1) & V_z & \\ 2) & k_{yz} & k_z\\ 3) & Q_z & Q_y \end{matrix}\right\}$$ independent of s

From the two first moment of inertia equations it can also be seen the they are constant between two stringers. This is due to all areas are concentrated of stringers, thus constant and independent of s between the stringers.

These relations yield the final assumption that the shear flow q(s) is constant between two stringers, but q(s) would increment (jump) when crossing over a stringer.

Now that we have the following relation with the stringers, it is now time to lay out a game plan to solve for the shear flows along the path s.

Step 1: Find $$ (\bar{y}, \bar{z})$$

Step 2: Find $$ I_y, I_z, I_{yz}\;$$

Step 3: Find $$ k_y, k_z, and k_yz \;$$

Step 4: Follow path "s" to find $$q_{1-2}, q_{2-3}, q_{3-4}\;$$

$$ q_{1-2}\;$$ represents the shear flow in skin panel 1-2. This can be represented in the same form as the shear flow along any path.

 * $$ q_{12} = (k_{yz}Q_z^{12} - k_z Q_y^{12}) V_z $$

Where the first moment area of inertia is local around each panel. Where $$y_1$$ is the y component of the string with respect to the origin at the centroid, and similarly for $$z_1$$

 * $$ Q_z^{12} = y_1 A_1 $$

 * $$ Q_y^{12} = z_1 A_1 $$

A quick look for reference of the following skin panel 2-3 shows what the first moment of inertia would resemble in that case.

 * $$ Q_z^{23} = y_1 A_1 + y_2 A_2 $$

A look at a different type of non-symmetrical case might help better explain different approaches that could be taken when looking for the shear flow. Fig #. shows different paths and areas for each path that could be used for solving such a shape.

The area for this case that was split into two section is the sum of the two sections:

 * $$ A = A_{\hat{s}} + A_{\hat{s}, L} $$

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

title: User:Eas4200c.f08.vqcrew.f


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Antisocial behavior

 is doing something “bad” to someone or society which results in damaging relationships and interfering with a society’s functioning. The effect of antisocial behavior can be seen as a reduction in social capital. is contrasted with prosocial behavior.

Examples of behaviours commonly considered to be antisocial are hurting others, disobeying rules, conflicting with others, and other socially unacceptable behaviours.

title: Antisocial behavior


text:
User talk:Setsuri

 Hello Setsuri, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may

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title: User talk:Setsuri


text: File:Cylindrical cross sections.jpg

title: File:Cylindrical cross sections.jpg


text: File:Torsion.JPG

title: File:Torsion.JPG


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User:Fabianarocha

Internet nada mais é que uma rede de computadores interligados entre si, que compartilham arquivos, informações e periféricos em tempo real. As redes de computadores tem como objetivos principais: administração centralizada e suporte, compartilhamento de arquivos, hardware e software. O controle de envio e de recebimento de informações na Internet é feito a partir de protocolos. Pode-se dizer que um protocolo é a realização de certas ações e troca de mensagens entre componentes, ou seja, são regras que definem o que deve ser feito entre as máquinas receptoras e máquinas transmissoras. O processamento de informações é feito de duas formas: centralizada e distribuída. Na forma, centralizada existe um computador central com alto poder de processamento que é responsável por processar informações aos outros computadores, com alto custo de investimento e físico, já na forma distribuída o processamento de informações passou a ser distribuído entre os terminais, isto é, computadores menores passaram a dividir as tarefas com o computador central. Existem dois tipos de configuração de redes: redes ponto-a-ponto: esse tipo de redes são mais utilizados com redes que possuam no máximo até 10 computadores, sem que haja hierarquia entre os computadores redes baseadas em servidor: são redes utilizadas para redes com mais de 10 computadores e com a presença de um ou mais servidores. Uma topologia de rede refere-se ao formato físico de como a rede é ligada, isto é, como computadores, cabos, e outros componentes são interligados. As topologias de redes são: Existem cinco camadas de protocolos: - HTTP(realiza a transferência de documentos pela Web) - SMTP(realiza a transferência de mensagens pelo correio eletrônico) - FTP(realiza a transferência de arquivos entre dois hospedeiros) - TCP: provê o serviço orientado para conexão - UDP: provê o serviço não orientado para conexão.
 * barramento: computadores ligados em série por meio de um único cabo, considerada uma topologia rudimentar.
 * estrela: os computadores são ligados através de vários cabos a um único componente central.
 * anel: os computadores são ligados por meio de um único cabo, só que em círculo.
 * malha: os computadores são ligados por meio de vários cabos entre si.
 * camada de aplicação: é responsável pelas aplicações de rede juntamente com seus protocolos. Ela possui os seguintes protocolos:
 * camada de transporte: é responsável pelo transporte de mensagens entre a camada de aplicação do cliente e do servidor. Ela possui dois protocolos:
 * camada de rede: é responsável pela movimentação de pacotes de camada de rede. Utiliza dois protocolos: o protocolo IP e o protocolo de roteamento.
 * camada de enlace: é responsável pelos erros de tansmissão, bem como pelo roteio de um datagrama.
 * camada física: é responsável pelo movimento de bits no quadro de um nó.

Tanenbaum, Andrew S. 2003 KUROSE, James. F. and ROSS, Keith W.,2005

title: User:Fabianarocha


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User:Eml4500.f08.bike.viale/homework report 3

=October 1=

"Closing the Loop" continued
The method of infinitesimal displacement, which is related to virtual displacement, is used to analyze the two bar truss system.

Some sample calculations for the lengths of AC and AB were performed as follows:

$$AC = \frac{|P_{2}^{(1)}|}{k^{(1)}} = \frac{5.1243}{.75} = 6.8324$$

$$ AB = \frac{|P_{1}^{(2)}|}{k^{(2)}} = \frac{6.276}{5} = 1.255$$

The (x,y) coordinates $$(x_{B}, y_{B})$$ and $$(x_{C}, y_{C})$$ must then be found. This can be done using simple trigonometry and the values of the magnitude of the line and the values of theta obtained in HW 2. This method is further explained using the figure on the left, along with the intermediate calculation steps shown below.

2 unknowns $$(x_{D}, y_{D})$$

The equations for lines AB and BC are then needed.

$$ \vec{PQ} = (PQ)\vec{\tilde{i}} = (PQ)[cos\Theta \vec{i} + sin\theta \vec{j}]$$

$$= (x - x_{p})\vec{i} + (y - y_{p})\vec{j}$$

$$\Rightarrow x - x_{p} = (PQ)cos\Theta$$

and $$y - y_{p} = (PQ)sin\Theta$$

$$\Rightarrow \frac{y - y_{p}}{x - x_{p}} = tan\Theta$$

$$y - y_{p} = (tan\Theta )(x - x_{p})$$

Using this equation along with $$\Theta^{1} = 30^{o}$$, it can be found that $$(x_{C}, y_{C})$$ = $$(5.92, 3.416)$$

Also, the equation for a line perpendicular to the above line, passing P is as follows:

$$y - y_{p} = (tan(\Theta + \frac{\pi}{2}))(x - x_{p})$$

This is useful in finding the value of $$(x_{B}, y_{B})$$, and using the value of $$\Theta^{2} = 135^{o}$$ it can be found that $$(x_{B}, y_{B})$$ = $$(-.877, .877)$$

To find the value of $$(x_{D}, y_{D})$$ one can use the following equation:

$$ \vec{AD} = (x_{D} - x_{A})\vec{i} + (y_{D} - y_{A})\vec{j}$$

Where the value of $$(x_{A}, y_{A})$$ are zero due to choosing the axis' origin to be point A.

By definition:

$$\vec{AD} = d_{3}\vec{i} + d_{4}\vec{j}$$

This means that, using the values obtained in HW 2, the line AD can be expressed as:

$$\vec{AD} = 4.35\vec{i} + 6.127\vec{j}$$

Which is the displacement vector of A from FEM. This ends closing the loop.

3-Bar Truss System
The three bar truss system shown above will now be analyzed using the values below as parameters.

$$ E^{(1)} = 2$$ $$ E^{(2)} = 4$$ $$ E^{(3)} = 3$$

$$ A^{(1)} = 3$$ $$ A^{(2)} = 1$$ $$ A^{(3)} = 2$$

$$ L^{(1)} = 5$$ $$ L^{(2)} = 5$$ $$ L^{(3)} = 10$$

$$P = 30$$ $$\Theta ^{(1)} = 30^{o}$$ $$\Theta ^{(2)} = 30^{o}$$ $$\Theta ^{(3)} = 45^{o}$$

Care should be taken when numbering the local nodes in order to make the assembly of the $$K$$ matrix convenient.


text:
User talk:Eas4200C.f08.spars.mcgilvray/homework3

I waded through all the info from vu-quoc's messages and posted some of the relevant stuff to help stay organized. Lecture dates: 22 sep to 3 oct. Summary of required problems: Prove that A = BH/2 for a triangle ...Done. --Eas4200C.f08.spars.mcgilvray 00:21, 5 October 2008 (UTC) If a = radius of circular x-section, prove J = πa4/2 from the integral of r2dA ...I'll do this one as well. (Thomas) Compare J of solid cross section to that of thin cross section. Read and discuss relevant material from Chapter 3. Compute Abar (centroid?) if Abar=πr2/2 + ba/2 Matlab code. Do we post the code on wiki or send in the program? Delvis, do you need help with this?

Rondon's Figures

Equation Editor

Tables

Best hw1 report

--Eas4200C.f08.spars.mcgilvray 23:20, 4 October 2008 (UTC)

Hey, it's Eddie. Delvis is working on the airfoil, I'll take care of chapter three I suppose. If you need any figures scanned and put on the page, draw them out and give them to me tomorrow. I'll scan them between class and 4pm I guess. --Eas4200c.f08.spars.rondon 21:45, 7 October 2008 (UTC)

--- um, whomever's been reworking the entries i've made for chapter 3, can you maybe hold off on that? please and thanks. --Eas4200c.f08.spars.prey 16:42, 8 October 2008 (UTC)

Images
I'll need one more image uploaded than I mentioned in class. It's the airfoil on page 90 with the following changes the height is b=2m the length is a=4m t1=0.008m and t2=t3=0.01m

Reply
Hey, this is eddie, damn, i didn't know you needed this scanned. I have class from 4:05 to 5:00. If anyone can go scan this at one of the libraries... that'd be great, cause I won't have a chance to do it before 5.

TURN IN TIME
Guys, I'll be turning this in as late as possible, pretty close to 5 pm. You can call me if you need to (321)543-5268 --Thomas TURNED IN --Eas4200C.f08.spars.mcgilvray 20:17, 8 October 2008 (UTC)


text:
User:Eas4200C.f08.spars.mcgilvray/homework3

=Homework #3=

Big Picture Road Map
In order to build up to the torsional analysis of multi-cell structures Dr. Vu-Quoc has suggested the following set of steps. The points in bold represent areas of particular importance.

 * Kinematic assumptions
 * Strain displacement relationship
 * Equilibrium equation for stresses
 * Prondtl stress function $$\phi \,\! (x,y)$$
 * Strain compatibility equation
 * Equation for $$\phi \,\!$$
 * Boundary conditions for $$\phi \,\!$$
 * Relationship between $$t$$ and $$\phi \,\!$$
 * Thin-walled cross section
 * Twist angle $$\theta \,\!$$: method one
 * Section 3.6 multi-cell thin-walled cross section

Meeting 1
A shear panel (fig 15) is a thin sheet of material that is intended to withstand in-plane shear forces. The shear force in the panel is distributed across the depth of the panel, but because the panel is assumed to be very thin this force may be viewed as being constant across the depth. With this assumption the internal forces can be seen as a shear flow (q) represented by the equation
 * $$ q=\tau \!t$$

Where $$\tau \!$$ is the uniform shear stress and $$\!t$$ is the thickness of the panel.

With this in mind the panel may be viewed as a line in the (Z,Y) plane (fig 16). The shear force acting in the panel can therefore be broken up into an infintessimal force ($$\overrightarrow{dF}$$) acting on an infintessimal length ($$\overrightarrow{dF}$$) represented by the following equation.

 * $$ \overrightarrow{dF} = q\overrightarrow{dL} = q(dL_y\overrightarrow{j}+dL_z\overrightarrow{k})$$

Zooming in on an infintessimal length (fig. 18) allows this equation to be taken further.

 * $$\overrightarrow{dF} = q(dL cos \theta \overrightarrow{j} + dL sin\theta\overrightarrow{k}) = q(dy\overrightarrow{j} + dz\overrightarrow{k})$$

The resultant shear force vector can then be determined by integrating this infintessimal force along the line representing the panel as follows.

 * $$\overrightarrow{F}=\int_A^B\overrightarrow{dF}=q[(\int_A^Bdy)\overrightarrow{j} + (\int_A^Bdz)\overrightarrow{k}]=q(a\overrightarrow{j} + b\overrightarrow{k})$$

Breaking this force into components

 * $$\overrightarrow{F}=F_y\overrightarrow{j} + F_z\overrightarrow{k}$$ → $$F_y=qa\!$$ $$F_z=qb\!$$

Recalling that $$ q=\tau \!t$$ this equation is the same as equations (1.4) and (1.5) (Sun, pg.5). Calculating the resultant shear force magnitude

 * ||$$\overrightarrow{F}$$|| $$=[(F_y)^2 + (F_z)^2]^{1/2}= q[a^2+b^2]^{1/2} = qd \!$$

Where d is the length of a straight line from point A to point B and does not depend of the curve. This equation is the same as equation (3.49a) (sun, pg.85)

Relating the force (||$$\overrightarrow{F}$$||) to torque (T)

 * $$T=2q\overline{A}$$

Which is proven by considering a thin walled cross-section with a shear flow along that wall.

 * $$T=\oint \rho q ds = \int \int_{\overline{A}} 2q dA = 2q\overline{A}$$ (equation 3.48 Sun, pg 85)

Meeting 2

The equations previously derived for a closed thin walled cross-section are also valid for an open thin walled cross-section (fig 19).

 * $$T_0=2q\overline{A}$$ → $$Re=T=2q\overline{A}$$

In considering a uniform bar of circular cross section assume the cross-section behaves like a rigid disc without warping(as it turns out this is the first point on Dr. Vu-Quoc's "road map", the kinematic assumptions). In other words after a torque is applied the planal section of the shaft both remains in the plane and circular, also the diameter in the plane remains straight. From this assumption it can be derived that shear strain is a linear function of the radial distance between a point and the center of the circular section (figs. 20 and 21). Note that these assumptions are only valid for circular cross-sections.

 * $$T=\int\int_A rG\gamma dA = \int\int_A r\tau dA $$ (Hooke's Law)
 * $$\gamma=r\frac{d\alpha}{dx}$$, $$\frac{d\alpha}{dx}=: \theta$$ rate of twist
 * $$T=\int_A rG(r\theta)dA=G\theta\int_Ar^2dA\!$$
 * $$\int_Ar^2dA= J$$

Where $$ J $$ is the second polar area moment of inertia.

It can then be shown that $$J$$ is proportional to $$\overline{A}^{\frac{3}{2}}$$ with a proportionality factor of $$2\pi^{-\frac{1}{2}}t$$.

Meeting 3
In discussing the plotting of an airfoil Dr. Vu-Quoc recommended taking the quadrature (dividing an area into a number of rectangles). In this manner the area of a complex shape can be reduced down to a sum of a number of rectangles that fit into that shape(fig. 22). Of course the more rectangles the area is divided into the more accurate the approximation becomes. From this approximation
 * $$\overrightarrow{dT}=\overrightarrow{r}\times \overrightarrow{dF}=\overrightarrow{r} \times(q\overrightarrow{PQ})$$
 * $$\overrightarrow{dT}=q(\overrightarrow{r}\times \overrightarrow{PQ})=q(2dA\overrightarrow{i})$$

As opposed to a uniform, circular bar under torsion a uniform, non-circular bar experiences what is known as warping. Warping is the axial displacement of a point on the deformed cross section along the length of the bar.

Before continuing with warping it is helpful to make a few notes on small angle approximations.

Assuming the angle $$\alpha\!$$ is very small (fig. 24)

 * $$\overline{PP'}\perp\overline{OP}$$
 * $$\overline{OP'}>\overline{OP}$$
 * $$v=-(\overline{PP'})sin(\beta)=-\alpha(\overline{OP})sin(\beta)$$

Recalling that

 * $$Rsin(\alpha)\!$$≈$$R\alpha\!$$
 * $$Rcos(\alpha)\!$$≈$$R\!$$

Meeting 4
Continuing with the torsion of a uniform bar with a non-circular cross-section (fig.25), The rate of twist ($$\theta\!$$) can then be given by the total angle of rotation over the length in the direction of the axis normal to the cross-section.

 * $$\theta\!=\frac{\alpha}{x}$$

For small displacement angles the components of the displacement can be expressed as follows

 * $$v=u_y=-\overline{PP'}sin(\beta)=-\overline{OP}\alpha sin(\beta)=-\alpha x_p=-\theta xz$$
 * $$w=u_z= \overline{PP'}cos(\beta)= \overline{OP}\alpha cos(\beta)= \alpha y_p= \theta xy$$

The displacement in the x-direction is independent of x, therefore

 * $$u=u_x=\theta\psi(y,z)\!$$

Note that the displacement in the (y,z) plane is the same as if it were a rigid disc, while the displacement in the x-direction varies with the warping function $$\psi(y,z)$$. These equations illustrate the kinematic assumptions in the first point of the previously mentioned "road map".

Meeting 5
Two equations form the basis of the torsional analysis of a multi-cell thin walled cross-section.

 * $$T=2q\overline{A}\!$$
 * $$\theta\!=\frac{1}{2G\overline{A}}\oint\frac{q}{t}ds$$

Noting that $$T=\sum_{i=1}^n T_i$$ and that $$\theta=\sum_{i=1}^{n}\theta_i$$ these equations become

 * $$T=2\sum_{i=1}^nq_i\overline{A_i}$$
 * $$\theta_i=\frac{1}{2G_i\overline{A_i}}\oint\frac{q_i}{t_i}ds$$

Where $$q_i$$ is the shear flow in cell i, $$\overline{A_i}$$ is the average area in the cell,$$G_i$$ is the shear modulus of cell i, and $$t_i$$ is the thickness of cell i.

Note that $$t_i\!$$ can be viewed as a function of the curvilinear coordinates along the cell wall.

Chapter 3: Torsion
Chapter 3 begins the in-depth coverage of torsion, and it's effect on aircraft structures. The mechanisms that are covered in the chapter are as follows

Matlab Project
=Contributing Team Members= --Eduardo Rondon Eas4200c.f08.spars.rondon 11:52, 8 October 2008 (UTC)

--Eas4200c.f08.spars.prey 16:48, 8 October 2008 (UTC)

-- Thomas McGilvray --Eas4200C.f08.spars.mcgilvray 18:39, 8 October 2008 (UTC)

--Michael Lee Eas4200c.f08.spars.lee 19:15, 8 October 2008 (UTC)

=Placeholder for Figures=

=References=

2. Sun, C.T. (2006). Mechanics of Aircraft Structures John Wiley & Sons, New York. ISBN 0471699667.

3. Vu-Quoc, Loc (2008) Lecture Notes University of Florida


title: User:A3pbe


text:
Biomedical engineering

Clinical engineering
Clinical engineering is a branch of biomedical engineering for professionals responsible for the management of medical equipment in a hospital. The tasks of a clinical engineer are typically the acquisition and management of medical device inventory, supervising biomedical engineering technicians (BMETs), ensuring that safety and regulatory issues are taken into consideration and serving as a technological consultant for any issues in a hospital where medical devices are concerned. Clinical engineers work closely with the IT department and medical physicists.

A typical biomedical engineering department does the corrective and preventive maintenance on the medical devices used by the hospital, except for those covered by a warantee or maintenance agreement with an external company. All newly acquired equipment is also fully tested. That is, every line of software is executed, or every possible setting is exercised and verified. Most devices are intentionally simplified in some way to make the testing process less expensive, yet accurate. Many biomedical devices need to be sterilized. This creates a unique set of problems, since most sterilization techniques can cause damage to machinery and materials. Most medical devices are either inherently safe, or have added devices and systems so that they can sense their failure and shut down into an unusable, thus very safe state. A typical, basic requirement is that no single failure should cause the therapy to become unsafe at any point during its life-cycle. See safety engineering for a discussion of the procedures used to design safe systems.

Medical devices
A medical device is intended for use in:

 * the diagnosis of disease or other conditions, or
 * in the cure, mitigation, treatment, or prevention of disease,
 * intended to affect the structure or any function of the body of man or other animals, and which does not achieve any of its primary intended purposes through chemical action and which is not dependent upon being metabolized for the achievement of any of its primary intended purposes.

Some examples include pacemakers, infusion pumps, the heart-lung machine, dialysis machines, artificial organs, implants, artificial limbs, corrective lenses, cochlear implants, ocular prosthetics, facial prosthetics, somato prosthetics, and dental implants.

Medical devices can be regulated and classified (in the US) as shown below:

 * 1) Class I devices present minimal potential for harm to the user and are often simpler in design than Class II or Class III devices. Devices in this category include tongue depressors, bedpans, elastic bandages, examination gloves, and hand-held surgical instruments and other similar types of common equipment.
 * 2) Class II devices are subject to special controls in addition to the general controls of Class I devices. Special controls may include special labelling requirements, mandatory performance standards, and post market surveillance. Devices in this class are typically non-invasive and include x-ray machines, PACS, powered wheelchairs, infusion pumps, and surgical drapes.
 * 3) Class III devices require premarket approval, a scientific review to ensure the device's safety and effectiveness, in addition to the general controls of Class I. Examples include replacement heart valves, silicone gel-filled breast implants, implanted cerebellar simulators, implantable pacemaker pulse generators and endosseous (intra-bone) implants.

Medical Imaging
Imaging technologies are often essential to medical diagnosis, and are typically the most complex equipment found in a hospital including:

 * Fluoroscopy
 * Magnetic resonance imaging (MRI)
 * Nuclear Medicine
 * Positron Emission Tomography (PET) PET scansPET-CT scans
 * Projection Radiography such as X-rays and CT scans
 * Tomography
 * Ultrasound
 * Electron Microscopy

Tissue engineering
One of the goals of tissue engineering is to create artificial organs for patients that need organ transplants. Biomedical engineers are currently researching methods of creating such organs. In one case bladders have been grown in lab and transplanted successfully into patients[1]. Bioartificial organs, which utilize both synthetic and biological components, are also a focus area in research, such as with hepatic assist devices that utilize liver cells within an artificial bioreactor construct[2].

Regulatory Issues
Regulatory issues are never far from the mind of a biomedical engineer. To satisfy safety regulations, most biomedical systems must have documentation to show that they were managed, designed, built, tested, delivered, and used according to a planned, approved process. This is thought to increase the quality and safety of diagnostics and therapies by reducing the likelihood that needed steps can be accidentally omitted again.

In the United States, biomedical engineers may operate under two different regulatory frameworks. Clinical devices and technologies are generally governed by the Food and Drug Administration (FDA) in a similar fashion to pharmaceuticals. Biomedical engineers may also develop devices and technologies for consumer use, such as physical therapy devices, which may be governed by the Consumer Product Safety Commission. See US FDA 510(k) documentation process for the US government registry of biomedical devices.

Other countries typically have their own mechanisms for regulation. In Europe, for example, the actual decision about whether a device is suitable is made by the prescribing doctor, and the regulations are to assure that the device operates as expected. Thus in Europe, the governments license certifying agencies, which are for-profit. Technical committees of leading engineers write recommendations which incorporate public comments and are adopted as regulations by the European Union. These recommendations vary by the type of device, and specify tests for safety and efficacy. Once a prototype has passed the tests at a certification lab, and that model is being constructed under the control of a certified quality system, the device is entitled to bear a CE mark, indicating that the device is believed to be safe and reliable when used as directed.

The different regulatory arrangements sometimes result in technologies being developed first for either the U.S. or in Europe depending on the more favourable form of regulation. Most safety-certification systems give equivalent results when applied diligently. Frequently, once one such system is satisfied, satisfying the other requires only paperwork.

Training
Biomedical engineers combine sound knowledge of engineering and biological science, and therefore tend to have a bachelors of science and advanced degrees from major universities, who are now improving their biomedical engineering curriculum because interest in the field is increasing. Many colleges of engineering now have a biomedical engineering program or department from the undergraduate to the doctoral level. Traditionally, biomedical engineering has been an interdisciplinary field to specialize in after completing an undergraduate degree in a more traditional discipline of engineering or science, the reason for this being the requirement for biomedical engineers to be equally knowledgeable in engineering and the biological sciences. However, undergraduate programs of study combining these two fields of knowledge are becoming more widespread. As such, many students also pursue an undergraduate degree in biomedical engineering as a foundation for a continuing education in medical school.

Though the number of biomedical engineers is currently low (under 10,000), the number is expected to rise as modern medicine improves. Currently, according to U.S. News & World Report, the program at Johns Hopkins University is ranked first among American universities in the category of bioengineering/biomedical engineering. At the undergraduate level, an increasing number of programs are also becoming recognized by ABET as accredited bioengineering/biomedical engineering programs in the United States. Duke University, ranked second in the U.S. by U.S. News, was the first program accredited by the Engineering Council for Profession Development (now ABET) in September of 1972.

Subdivisions and Departments
Like divisions, subdivisions and departments are pages in the Topic namespace, and their names start with the "Topic:" prefix. Individual departments can be used by multiple schools. Schools that use (link to) the same department should cooperate to develop the department. Subdivisions are used by large divisions to help organize related departments. Not all divisions have subdivisions. If you need a subdivision in your division, you can use the Template:Subdivision boilerplate template to start a subdivision.

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Research Projects
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Related news

 * March 17, 2007 - New propulsion system using diodes may allow novel applications. (See also...)

title: Biomedical engineering


text:
Talk:Biomedical engineering

just pasted from topic Emesee 00:45, 5 October 2008 (UTC)

diff copied from Wikipedia article, as of 20:08, October 3, 2008 pasted 00:44, 5 October 2008 (UTC). - Sidelight12 Talk 04:44, 24 June 2013 (UTC)

title: Talk:Biomedical engineering


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User:Plonk420

hoping wikiversity is the proper place for an "educational article" on (how to use x264, esp it's switches)...

template for the chart

This article is a sortable list of x264-to-FFmpeg options and how they map to each other

http://en.wikiversity.org/w/index.php?title=X264_FFmpeg_option_mapping&action=submit

http://en.wikiversity.org/wiki/User:Plonk420/testarticle

title: User:Plonk420


text:
User:Eml4500.f08.gravy.mmm/HW3

12.1
Derivation of element Force Displacement with respect to global coordinate system.

$$\ k^{(e)}_{4x4} d^{(e)}_{4x1}=f^{(e)}_{4x1} \,$$ equation 1

12.2
The force displacement can be written as

$$ \ K^{(e)} * \begin{bmatrix} 1 & -1 \\ -1 &1 \end{bmatrix}_{\hat{K}_{2x2}^{(e)}}* \begin{Bmatrix} q_1^{(e)} \\ q_2^{(e)} \end{Bmatrix}_{q_{2x1}^{(e)}} = \begin{Bmatrix} p_1^{(e)} \\ p_2^{(e)} \end{Bmatrix}_{p_{2x1}^{(e)}} \,$$ equation 2

$$\ q_i^{(e)} = \,$$ axial disp.of element at local node i

$$\ p_i^{(e)} = \,$$ axial disp.of element at local node i

Goal: Derive Equation 1 from Equation 2

Want the relationship between $$\underline {q^{(e)}}$$ and $$\underline {d^{(e)}}$$ also $$\underline {p^{(e)}}$$ and $$\underline {f^{(e)}}$$

12.3
These relationships can be expressed in the form:

$$\ \underline {q^{(e)}_{2x1}}= \underline { T^{(e)}_{2x4}} * \underline {d^{(e)}_{4x1}} \,$$

Consider the displacement vector of local node 1 denoted by

$$ \vec{d^{(e)}_{1}} $$

$$ \vec{d^{(e)}_{1}}= \vec{d^{(e)}_{1}}* \vec{i} + \vec{d^{(e)}_{2}} *\vec{j} $$

$$\ q_i^{(e)} = \,$$ axial displacement of node 1 is the orthogonal projection of the displacement.

12.4
Vector $$ \vec{d^{(e)}_{1}} $$ of node 1 on the axis $$\tilde{x}$$ of element $$e^{(1)}$$

$$\ \vec{d^{(e)}_{1}}* \vec{\tilde{i}} \,$$

$$\ = ( d_1^{(e)}*\vec{i} + d_2^{(e)}*\vec{j}) \cdot \vec{\tilde{i}} \,$$

$$\ = d_1^{(e)}* (\vec{i} \cdot \vec{\tilde{i}}) + d_2^{(e)}*(\vec{j} \cdot \vec{\tilde{i}} ) \,$$

$$\ (\vec{i} \cdot \vec{\tilde{i}} )= cos \theta^{(e)} = l^{(e)} \,$$

$$\ (\vec{j} \cdot \vec{\tilde{i}} )=sin \theta^{(e)} = m^{(e)} \,$$

$$\ q_1^{(e)}=l^{(e)}*d_1^{(e)} + m^{(e)}* d_2^{(e)} \,$$

$$\ = \begin{bmatrix} l^{(e)} & m^{(e)} \end{bmatrix}_{1x2} * \begin{Bmatrix} d_1^{(e)} \\ d_2^{(e)} \end{Bmatrix}_{2x1} \,$$

12.5
Similarly for node 2 : HW.

$$\ q_2^{(e)}=l^{(e)}*d_1^{(e)} + m^{(e)}* d_2^{(e)}= \begin{bmatrix} l^{(e)} & m^{(e)} \end{bmatrix}_{1x2} * \begin{Bmatrix} d_3^{(e)} \\ d_4^{(e)} \end{Bmatrix} \,$$

$$ \begin{Bmatrix}q_1^{(e)} \\ q_2^{(e)} \end{Bmatrix}_{\underline{q_{2x1}^{(e)}}} = \begin{bmatrix} l^{(e)} & m^{(e)} & 0& 0\\0 & 0 & l^{(e)} & m^{(e)} \end{bmatrix}_{\underline{T_{2x4}^{(e)}}}* \begin{Bmatrix} d_1^{(e)} \\ d_2^{(e)} \\d_3^{(e)} \\ d_4^{(e)} \end{Bmatrix}_{\underline{d_{4x1}^{(e)}}}$$

$$\ \underline{ q_{2x1}^{(e)} }= \underline{ T_{2x4}^{(e)} }* \underline{ d_{4x1}^{(e)} } \,$$

14.1
Similarly (same argument) :

$$\ \begin{Bmatrix} P_1^{(e)}\\P_2^{(e)} \end{Bmatrix}_{\underline{ P_{2x1}^{(e)} } } = \underline{ T_{2x4}^{(e)} } * \begin{Bmatrix} f_1^{(e)} \\ f_2^{(e)} \\ f_3^{(e)} \\ f_4^{(e)} \end{Bmatrix} _{\underline{ f_{4x1}^{(e)} } } \,$$

$$\underline{ P^{(e)} } = \underline{ T^{(e)}* f^{(e)} }$$

Recall element axial Force Displacement relation:

$$\ \underline{\hat{k}_{2x2}^{(e)}} * \underline{q_{2x1}^{(e)}}= \underline{P_{2x1}^{(e)}} \,$$

14.2
$$\underline{\hat{k}^{(e)}} *( \underline{ T^{(e)} } * \underline{ d^{(e)} })_{\underline{q^{(e)}}} = ( \underline{ T^{(e)} } * \underline{ f^{(e)} })_{\underline{P^{(e)}}}$$

Goal: Want to have $$\ \underline{k^{(e)}} * \underline{d^{(e)}}= \underline{f^{(e)}} \,$$

So "move " $$\underline{ T^{(e)}}$$ from the right hand side to the left hand side by multiplying the equation by $$ \underline{ T^{(e) -1}}$$ (inverse of $$\underline{ T^{(e)}} $$)

Unfortunately, $$\underline{ T^{(e)}}$$ is a rectangular matrix (2x4) so you can not invert it.

Ans: $$[\underline{ T_{4x2}^{(e)T} } * \underline{\hat{k}_{2x2}^{(e)}} * \underline{ T_{2x4}^{(e)}} ]_{is the matrix we are looking for} * d_{4x1}^{(e)}= f_{4x1}^{(e)}$$

14.3
$$\ \underline{k^{(e)} } * \underline{d^{(e)} }= \underline{f^{(e)} } \,$$

1) $$\ \underline{k^{(e)} } = \underline{T^{(e)T}} * \underline{\hat{k}^{(e)} }* \underline{T^{(e)} } \,$$

HW: verify above.

Justification for 1) : PVW (later see 10-1 for fist approach of PVW Reduction of Global Force Displacement relation)

$$ \underline{K_{6x6}}* \underline{d_{6x1}}= \underline{F_{6x1}} which reduces to \underline{K_{2x2}}* \underline{d_{2x2}}= \underline{F_{2x2}}$$

Remark: Why not solve as follows

$$ \underline{d} = \underline{K^{-1}}* \underline{F} $$ ?

The reason is that K can't be unversed because it is singular

14.4
Singular matrix has a determinate of 0 so it is not invertible

Recall: need to compute $$\frac{1}{det(K)}$$ to find K-1

Why? For an unconstrained structure system there are three possible rigid body motion in 2-D (2 translation, and 1 rotation)

HW: compute eigenvalues values of K6x6 and make observation about the number of zero eigenvalues

Dyn. eval Problem : $$\underline{K}_{stiff. mat} *v= \lambda_{related to vibration frequency}*M_{mass mat. }*v$$

14.5
Zeros corresponds to zero stored elastic energy or rigid body modes. modes=mode shapes=eigenvectors

//(end Remark)

17.1

$$ \sum{F_x} =0$$

$$ \sum{F_y} =0$$

$$ \sum{M_A} =0$$ trivial

Question: How about $$\sum{M_B}?$$ It still equals 0

$$ \sum{M_B} = \vec{BA'}\times \vec{F}$$

for A' on line of action of $$\vec{F}$$

$$\vec{BA'}= \vec{BA} + \vec{AA'}$$

17.2
$$ \sum{M_B}=(\vec{BA}+ \vec{AA'})\times \vec{F}$$ $$ = \vec{BA}\times F + \vec{AA'}\times \vec{F}= 0 $$

Back to 3-bar truss

Node A is in equilibrium : $$\sum_{i=0}{\vec{F_i}} =\vec{0}$$

$$ \sum_i{M_{B,i}}=\sum_i{\vec{BA'_i}}\times \vec{F_i}$$

A'i= any point on line of action of $$\vec{F_i}$$

$$ \sum_i{M_{B,i}}=\sum_i{\vec{BA}}\times \vec{F_i} \vec{BA} \times \sum_i{\vec{F_i}} = 0 $$

Where $$K_{33}= k^{(1)}_{33}+ k^{(2)}_{11}+k^{(3)}_{11} $$ and

$$K_{34}= k^{(1)}_{34}+ k^{(2)}_{13}+k^{(3)}_{12} $$

HW: fill out the rest

hw
From previous lecture

Element Length

$$\ L^{(1)}=4 \,$$

$$\ L^{(2)}=2 \,$$

Young's Modules

$$\ E^{(1)}=3 \,$$

$$\ E^{(2)}=5 \,$$

Cross-sectional Area

$$\ A^{(1)}=1 \,$$

$$\ A^{(2)}=2 \,$$

Inclination angle

 * $$\ \theta^{(1)} = 30^o \,$$

 * $$\ \theta^{(2)} = -45^o \,$$

 * $$\ \theta^{(1)} = 30^o \,$$

Since $$ \ P_{(e)} = T_{(e)} * f_{(e)}\,$$

And the T matrix is

$$\ T^{(e)} = \begin{bmatrix} l^{(e)} & m^{(e)}& 0 & 0 \\ 0 & 0 & l^{(e)} & m^{(e)}\end{bmatrix} \,$$

we can solve for P

The f matrix was solved in a previous lecture and is:

$$\ F= \begin{bmatrix} -4.4378 \\ -2.5622 \\ +4.4378 \\ +2.5622 \end{bmatrix}_{4X1}\,$$

l m and k were also solved in a previous lecture and are:

$$\ l^{(1)}=cos\theta^{(1)}= cos 30^o= \frac{\sqrt{3}}{2} \,$$

$$\ m^{(1)}=sin\theta^{(1)}= sin 30^o= \frac{1}{2} \,$$

$$\ k^{(1)}= \frac{E^{(1)}A^1}{L^1} \frac{\left(3\right ) \left(1\right )}{4}= \frac {3}{4}\,$$

$$\ l^{(2)}=cos\theta^{(2)}= cos -45^o= .707 \,$$

$$\ m^{(2)}=sin\theta^{(2)}= sin -45^o= -.707 \,$$

$$\ k^{(1)}= \frac{E^{(5)}A^1}{L^1} \frac{\left(2\right ) \left(1\right )}{2}= 5\,$$

$$T^{(1)}$$ and $$T^{(2)}$$ are solved as the following:

$$\ T^{(1)} = \begin{bmatrix} \frac{1}{2} & \frac{1}{2} & 0 & 0 \\ 0 & 0 & \frac{1}{2} & \frac{1}{2}\end{bmatrix} \,$$

$$\ T^{(2)} = \begin{bmatrix} .707 & -.707 & 0 & 0 \\ 0 & 0 & .707 & -.707\end{bmatrix} \,$$

With T and f the P s were found as being,

$$P_{2}^{(1)}$$= $$P_{1}^{(2)}$$=5.1243

so

title: User:Eml4500.f08.gravy.mmm/HW3


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text:
User talk:Plonk420

It probably could be the proper place... as long as "educational" does in fact mean educational. I would encourage you to explore our policies or simply to be bold. Welcome. Emesee 05:27, 5 October 2008 (UTC)

 Hello Plonk420, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may

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title: User talk:Plonk420


title: File:Localdiagramhw3.jpg


text:
Portal:Dance

< School of Music and Dance

Welcome to the Dance department, which is part of the School of music and dance.

Courses

 * The history of dance
 * Introduction to coupled dancing
 * Types of dance
 * Geography of dance
 * Dance technique
 * Dance culture
 * Dance science
 * Celtic Dance

On this wiki

 * Ten stupid things beginners do to mess up their contradance experience
 * African music and dance

Active participants

 * (if you contribute to a course or resource, please add your name here)

 * Since 1 September 2017 with Compositions across disciplines --Marshallsumter (discuss • contribs) 03:27, 17 April 2018 (UTC)

title: Portal:Dance


text:
User:Fievarsty

Wikipedia:ja:user:Fievarsty ja:user:Fievarsty

title: User:Fievarsty


text:
The History of Dancing

Lessons

 * /Ancient dances/
 * Medieval dances
 * 19:th century dancing
 * 20:th century dancing
 * 21:st century dancing

There are a number of different types of dancing that have first occurred during the 20th and 21st centuries. They include:
 * Freestyle dance;
 * Slow Dance;
 * Street Dance; etc

title: The History of Dancing


text:
Introduction to coupled dancing

to School:Music and Dance ?

Lessons

 * Introduction
 * Catching the beat
 * Basic steps
 * Position
 * Leading and following
 * Connection
 * Rotation and spinning
 * Musicality
 * Syncopated music
 * Air steps

title: Introduction to coupled dancing


text:
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title: User talk:Niteshbhatia008~enwikiversity


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User talk:Fievarsty

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text: User:Vandalismterminator

title: User:Vandalismterminator


text:
User:Eml4500.f08.ateam.shah/Week 3

Homework Report 3 Friday September 25, 2009

\begin{bmatrix} P_1^{(e)}\\ P_2^{(e)}\\ \end{bmatrix} = T^{(e)} \cdot \begin{bmatrix} f_1^{e}\\ f_2^{e}\\ f_3^{e}\\ f_4^{e}\\ \end{bmatrix} $$

Where T(e) is a 2x4 transformation matrix which can be used to transform distance and force from one coordinate system to another.

q^{(e)} = T^{(e)} \cdot d^{(e)} $$

P^{(e)} = T^{(e)} \cdot f^{(e)} $$

Recall

P^{(e)} = q^{(e)} \cdot k^{(e)} $$

Where q and P are axial components with k being a 2x2 overlap stiffness matrix and q being 2x1 distance matrix.

Thus

T^{(e)} \cdot d^{(e)} = q^{(e)} $$

T^{(e)} \cdot f^{(e)} = P^{(e)} $$

k^{(e)} \cdot (T^{(e)} \cdot d^{(e)}) = (T^{(e)} \cdot f^{(e)}) $$

The goal is to have a FD distance relationship as described by:

k(e) * d(e) = f(e)

So you move the transformation matrix from the RHS to the LHS by premultiplying equation by Y(e)-1. (inverse of the T matrix)

However the T matrix is a rectangular 2x4 matrix, thus it cannot be inverted. However the T matrix can be transposed to achieve similar results:

[T^{(e)T} \cdot k^{(e)} \cdot T^{(e)}] \cdot d^{(e)} = (T^{(e)} \cdot f^{(e)}) $$

Where transpose of T is a 4x2 matrix and the overall resultant matrix of the transpose of TT * k * T leads to a 4x4 matrix.

The following problem has been worked out to verify the above statement. $$     \begin{bmatrix} &  &   &   &   \\          &   &   &   &   \\          &   & K &   &   \\ &  &   &   &   \\          &   &   &   &      \end{bmatrix} = \begin{bmatrix} l^{(e)} & m^{(e)} & 0 & 0\\ 0 & 0 & l^{(e)} & m^{(e)} \end{bmatrix} \cdot \begin{bmatrix} &  &   &   &   \\          &   &   &   &   \\          &   & K &   &   \\ &  &   &   &   \\          &   &   &   &      \end{bmatrix} \cdot \begin{bmatrix} l^{(e)} & m^{(e)} & 0 & 0\\ 0 & 0 & l^{(e)} & m^{(e)} \end{bmatrix}^T $$

The justification of this concept can be provided using the principal virtual work theorem introduced later in the semester.

 * See P 10 for the 1st example of PVW/Reduction of global FD relationship.

REMARK The k equation on the exam to determine the displacement could not be solved using: k-1 * F = d because  k  is a singular matrix.

 K  is a singular matrix, which means the determinant of  k  is 0 and thus  k  cannot be inverted.

RECALL: You need to computed (det)-1 to find the inverse of the  k  matrix.

 * A famous quote narrated by the professor:

" Ask what not your country can do for you               ask what you can do for your country." JFK For an unconstrained structure system, there are 3 possible rigid body motions in 2-D (2 truss - 1 rotation).

HW- Find the eigenvalues of the K matrix, and make observation about the number of 0 eigenvalues.

A =

0.5625   0.3248   -0.5625   -0.3248         0         0    0.3248    0.2500   -0.3248   -0.2500         0         0   -0.5625   -0.3248    3.0625   -2.1752   -2.5000    2.5000   -0.3248   -0.2500    2.8248    1.9375    2.5000   -2.5000         0         0   -2.5000    2.5000    2.5000   -2.5000         0         0    2.5000   -2.5000   -2.5000    2.5000

>> eig(A)

ans =

8.4194   2.5879   -0.1729   -0.0219   -0.0000   -0.0000

Dynamic Evaluation Problem:
 * k * v = λ * M * V

Where k*v is the stiffness matrix, λ is the eigenvalue related to the vibrational frequency, M is the mass of the system and V is the velocity matrix. Thus zero eigenvalues lead to zero stored elastic energy. This means the system is rigid.

title: User:Eml4500.f08.ateam.shah/Week 3


text:
User talk:Vandalismterminator

 Hello Vandalismterminator, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may

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Custodian tools
Generally you need to develop a relationship with a mentor first :-). The IRC channel (#wikiversity-en on freenode) is often a good place to start. --SB_Johnny talk 14:58, 5 October 2008 (UTC)

Unblock
Eh boy :-(. Well, I thought you "looked" familiar... I don't know why you're so interested in sysop tools, but here's a bit of advice: you get them when you prove yourself to be a trustworthy and interested contributor (and don't run alternate accounts that do the opposite). It takes time and patience, but it's worth it, and I hope you'll put in the effort. I will not unblock that terminal. --SB_Johnny talk 19:14, 8 October 2008 (UTC)

title: User talk:Vandalismterminator


text:
Combining general relativity and quantum theory

Quantum theory wishes to simplify micro behavioral movements in any of earth and cosmos gravitational matter. From meter to highest level of energy all that can be linearly described, are levels of critical masses and the definition of forces binding them?

Electromagnetic, magnetic field and electrical current are bind in presence, there for in meter. Gravitational forces, the truth is that the amount of matter is relatively small and the most of earth space, or our cosmic creation, is hollow.

General relativity theory - implies to behavioral issues in a vertical approach means a two dimensional axis model. There for it cannot fully describe three simultaneous forces; electromagnetic, strong nuclear, weak nuclear forces.

Quantum theory general relativity in motion - Usually described as radial sourced forces binding critical amounts of mass, expanding in space along time.

The most successful numerical approach to quantum field theory begins with a formulation of quantum mechanics developed by Feynman in which a quantum amplitude is described as a weighted integral over all possible paths (not necessarily obeying the classical equations) which start at the system's initial state and end at the final state. For single particle quantum mechanics the quantum amplitude  for a transition from position qi at time ti to position qf at time tf is written as:
 * Numerical treatment

 = òdq[t]eiA[q]

where A[q(t)] is the classical action for the path q(t) given by

A = òtitf {½ m q'2 - V(q(t))}dt

This is a sophisticated Wiener integration over function space and is typically an awkward formalism for analytic calculation. However, it is nicely suited for numerical work since it replaces the normal operator/Hilbert space formalism of quantum mechanics with an explicit integral.

The path integral appropriate for quantum field theory is similar to the equation above except that the integration must be performed over all possible time evolutions of field configurations rather than particle trajectories. In our physical problem, a field configuration specifies both the quark and gluon fields as particular functions of space. A particular time evolution then specifies these fields as functions of space and time. This problem is easily put in a numerically tractable from by replacing the space-time continuum by a grid or lattice of points, conventionally a uniform, four-dimension mesh.

The field theory analogue of the single-particle action given in the equation above is a similar polynomial in the field variables and their derivatives, integrated over space-time. Thus, the corresponding discrete field theory action will be a four-dimensional sum of a local density which depends on the lattice field variables at a specific lattice site and its nearest neighbors.

The actual integration appropriate for the lattice QCD evaluation of an observable O is typically performed as a Monte Carlo average,

<0| O |0> = 1/N Sn=1N O({U}n),

over an ensemble of configurations for the gluon fields {U}n, N ³ n ³ 1. Each configuration assign a specific 3x3 complex matrix U to each link connecting neighboring sites in the lattice. The ensemble used above is generated by a Metropolis or molecular dynamics algorithm so as to be distributed according to the positive definite statistic weight:

e{b/6SP tr UP} det(D+m)

The sum is over all elementary squares or plaquettes, P, that can be constructed out of four lattice links and UP is the ordered product of the corresponding U matrices associated with those links. The quark fields correspond to anti-commuting classical variable and cannot be treated numerically as an integral but instead are represented by the determinant above. Here D is a nearest neighbor difference operator and 'm', the quark mass matrix. Typically, the force generated by the determinant is computed using a noisy estimator which can be done using, for example, a conjugate gradient method for computing the inverse of the sparse matrix D+m. In addition to the quark mass matrix 'm', the coupling strength, related to the parameter b, is the only other free parameter in the calculation.

This is an ideal formulation for massively parallel computing. A typical large-scale lattice calculation might work with 324 hypercubic lattice. If each processor in a parallel machine is assigned a 44 subvolume, 4K processors would be required. The most computationally demanding part of a conjugate-gradient iteration requires about 500 floating point operations per lattice site or 128K flops/processor. A 3-component complex vector must be transferred both in and out of the processor for each link that joins the 44 subcube to its neighbors, or 43 · 8 · 6 · 2 or 6K words total. This suggests a reasonably favorable 20:1 computation to communications load for each processor.


text:
User:Eml4500.f08.bottle.vitello/HW 3

Lecture Notes
This part of the lecture concerns the derivation of the element force diagram with respect to the global coordinate system.

Initially in the previous lectures, we derived an element force diagram with $$f \! $$ representing element force and $$ d \! $$ representing element displacement on a truss element. We replace these unknowns with $$ P \! $$ representing the element force and $$ q \! $$ representing the element displacement through the axis of the truss element.

With:

1.   $$q_i^{(e)}$$=axial displacement of element e at local node i   2.    $$P_i^{(e)}$$=axial force of element e at local node i

To derive $$ k \!$$ $$* \!$$$$d \!$$$$= \!$$$$f \!$$ from $$k \!$$$$* \!$$$$q \!$$$$=$$$$P \!$$, we need to find the relation between $$q\!$$ and $$d\!$$, and $$P\!$$ and $$f\!$$. For displacement, we derive that:

$$ q_{2X1}^{(e)}$$ $$ = \!$$ $$T_{2X4}^{(e)}$$ $$ * \!$$ $$d_{4X1}^{(e)}$$

In order to derive $$T \!$$, we must consider the displacement vector of the nodes in each truss element. Considering the previous truss system derived in the lecture notes, we derive vector $$ d \! $$ of node 1 on the axis x of element e, giving us:

$$q_1^{(e)}$$ $$=$$ $$d_1^{(e)}$$ $$ *$$ $$ \bar{i} $$ $$ = $$ $$d_1^{(e)}$$ $$ \tilde{i} $$ $$ + $$ $$ d_2^{(e)} \tilde {j} $$ $$ $$ = $$ $$d_1^{(e)}$$ $$ ($$ $$\tilde{i}$$$$*$$ $$\bar{i}$$$$)$$ $$ + $$ $$ d_2^{(e)}$$ $$ ( $$ $$ \tilde{j}$$ $$*$$$$ \bar{i}) $$ $$=$$$$ d_1^{(e)}$$$$*$$ $$\cos (\theta^{(e)}) \! $$ $$ + $$ $$ d_2^{(e)}$$$$ * $$ $$ \sin (\theta^{(e)}) \!$$ $$q_1^{(e)}$$$$= \!$$$$ d_1^{(e)}$$$$ * \!$$ $$ l^{(e)} \!$$ $$+ \!$$ $$d_2^{(e)} $$ $$ * \!$$ $$ m^{(e)} \!$$
 * $$ $$ \bar {i} $$

Using the same procedure as before for element 2 of the truss system gives us:

$$q^{(e)}_2= (d^{(e)}_3\bar{i}+d^{(e)}_4\bar{j}) \cdot \bar{\tilde{i}}$$

$$= d^{(e)}_3(\bar{i}\cdot \bar{\tilde{i}})+d^{(e)}_4(\bar{j} \cdot\bar{\tilde{i}}) $$

$$= d_3^{(e)}\cdot cos(\theta^{(e)})+d_4^{(e)}\cdot sin(\theta^{(e)})$$

$$q_2^{(e)}$$ $$ = \!$$ $$d_3^{(e)}$$$$ *\! $$$$ l^{(e)}\!$$ $$ + \!$$ $$d_4^{(e)}$$ $$ *\!$$ $$ m^{(e)}\!$$

Combining these two equations gives us the result for the $$ T \! $$ matrix:

$$ \begin{Bmatrix} q_1^{(e)} \\ q_2^{(e)} \end{Bmatrix} $$ $$ = \! $$$$ \begin{bmatrix} l^{(e)} & m^{(e)} & 0 & 0 \\  0  &  0  & l^{(e)} &  m^{(e)} \end{bmatrix} $$ $$ * \! $$ $$ \begin{Bmatrix} d_1^{(e)} \\d_2^{(e)} \\d_3^{(e)} \\ d_4^{(e)}\end{Bmatrix} $$

The same argument can also be applied to the axial force on the truss system, using the same procedure used to find the axial displacement of the truss system:

$$P^{(e)} \!$$ $$ = \!$$ $$T^{(e)}\!$$ $$* \!$$ $$f^{(e)}\!$$

Gives:

$$ \begin{Bmatrix} P_1^{(e)} \\ P_2^{(e)} \end{Bmatrix} $$ $$ = \! $$$$ T_{2X4}^{(e)} $$ $$ * \! $$ $$ \begin{Bmatrix} f_1^{(e)} \\f_2^{(e)} \\f_3^{(e)} \\ f_4^{(e)}\end{Bmatrix} $$

We can then replace the axial forces for the equations derive, representing the element forces:

$$k_{2X1}^{(e)}$$ $$*$$ $$ q_{2X1}^{(e)}$$$$=\!$$$$ P_{2x1}^{(e)}$$ and $$k^{(e)} \!$$$$ * \!$$ $$\underbrace{(T^{(e)}\! d^{(e)}) \!}_{q^{(e)}}=\!$$ $$ \underbrace{(T^{(e)} \!  f^{(e)})\!}_{P^{(e)}}$$ Using the inverse of $$ T \! $$, we can get the following solution for the element forces: $$ \begin{bmatrix} (T^{(e)})^{T}k^{(e)}I^{(e)} \end{bmatrix}*d^{(e)} = f^{(e)}$$ However, we cannot invert $$ T \!$$ with itself since it is a rectangular matrix with a two by two matrix $$ k \!$$; we instead use the Principal of Virtual Work (PVW) to find $$f_1 \! $$, $$ f_2 \!$$, $$ f_5 \!$$, and $$ f_6 \!$$. There is no need to find $$ f_3 \! $$ and $$ f_4 \!$$ since they are the applied load on the truss system, which is already known.

The following loop used for the two bar truss system to compare between the Finite Element Method (FEM) and the statics method went as follows, starting with the FEM method:

FEM Compute displacements Compute reactions

Then the loop goes back using the statics method:

Statics Compute reactions Compute displacements

By using the statics method, the reactions $$ f_1 \! $$, $$ f_2 \! $$, $$ f_5 \! $$, and $$ f_6 \! $$ are known and with it, the forces $$ P \! $$ are then known. This knowledge can then be used to find the axial displacement $$ q \! $$ at each node on the truss system:

$$q_2^{(1)} $$ $$ =\frac{P_{2}^{(1)}}{k^{(1)}} \! $$$$ = \!$$ $$AC \!$$

$$q_1^{(1)} $$$$ = \!$$ $$0 \!$$ ,fixed node $$1 \!$$

$$q_1^{(2)}$$ $$ =\frac{P_{2}^{(2)}}{k^{(2)}} $$ $$= \!$$ $$ -AB \! $$

$$q_2^{(2)} $$$$=\! $$ $$0 \! $$ ,fixed node $$ 2 \! $$

Another method used to find the displacement of global node 2 of the two body truss system is to find the deformed shape of the truss system itself.

The first part is to find the deformed lengths of $$ AC \! $$ and $$ AB \! $$:

$$ AC = \! $$ $$ \frac{P_{2}^{(1)}}{k^{(1)}} = \frac{5.1243}{.75} = 6.8324$$

$$ AB = \! $$ $$\frac{P_{1}^{(2)}}{k^{(2)}}= \frac{6.2760}{5} = 1.255 \!$$

Then the x-y coordinates of points B and C are determined, leaving only two unknowns $$ x_D \! $$ and $$ y_D \! $$. To find this displacement, the formula used to determine said point is:

$$ x-x_{p}= (PQ)cos \theta \!$$ $$ y-y_{p}= (PQ)sin\theta \!$$

$$ y-y_{p}= tan(\theta + \frac{\pi }{2})(x-x_{p})$$

with $$ x_p \!$$ and $$ y_p \! $$ representing points B and C on the displacement diagram. However, we can simply use point A as our reference point since it is already on the point of origin on the x-y axis.

$$AD = (x_{d}-x_{a})\bar{i} + (y_{d}-y_{a})\hat{j} \!$$ $$AD = (x_{d} -0)\bar{i} + (y_{d} -0)\hat{j} \! $$

With the resulting definition becoming:

$$AD = d_{3}*\bar{i} + d_{4}*\hat{j} \! $$

This ultimately closes the loop for the two bar truss system. The same procedure can be used to find the displacement for a three bar truss system

The bars are broken up into their respected elements:

The overall momentum at point A is equal to zero and therefore, point A is in equilibrium, with the summation of forces of x and y equal to zero. Momentum along any other point near point A will also equal to zero. The sample approach to find the stiffness matrix for a two body system is also applied to a three body system.

Derivation of the Node 2 Axial to Global Displacement Matrix
For Node 1, it was found that

$$q^{(e)}_1 = (d^{(e)}_1\vec{i}+d^{(e)}_2\vec{j}) \cdot \vec{\tilde{i}}$$

For Node 2

$$q^{(e)}_2 = (d^{(e)}_3\vec{i}+d^{(e)}_4\vec{j}) \cdot \vec{\tilde{i}}$$

Given the definitions of

$$ l^{(e)} $$ and  $$ m^{(e)} $$

from the notes above, it follows that

$$q_2^{(e)}=l^{(e)}d_3^{(e)}+m^{(e)}d_4^{(e)}$$

and hence

$$q_2^{(e)}=\begin{bmatrix}l^{(e)} & m^{(e)} \end{bmatrix}\begin{Bmatrix}d_3^{(e)}\\ d_4^{(e)}\end{Bmatrix}$$

Using the Global FD Relation
Using the global FD relation, where displacements $$\displaystyle d_1, d_2, d_5, d_6 $$ all equal zero, the first two columns and the last two columns for the $$\displaystyle K $$ matrix can be ignored, yielding the following relation:

 * $$ \begin{bmatrix} K_{13} & K_{14}

\\ K_{23} & K_{24} \\ K_{33} & K_{34} \\ K_{43} & K_{44} \\ K_{53} & K_{54} \\ K_{63} & K_{64} \end{bmatrix} \begin{Bmatrix} d_3 \\d_4 \\ \end{Bmatrix} = \begin{Bmatrix} F_1 \\F_2 \\ F_3 \\ F_4 \\ F_5 \\ F_6  \end{Bmatrix}$$

which can also be expressed in terms of the local $$\displaystyle k^{(e)} $$ as shown

 * $$ \begin{bmatrix} k_{13}^{(1)} & k_{14}^{(1)}

\\ k_{23}^{(1)} & k_{24}^{(1)} \\ k_{33}^{(1)} + k_{11}^{(2)} & k_{34}^{(1)} + k_{12}^{(2)} \\ k_{43}^{(1)} + k_{21}^{(2)} & k_{44}^{(1)} + k_{22}^{(2)} \\ k_{31}^{(2)} & k_{32}^{(2)} \\ k_{41}^{(2)} & k_{42}^{(2)} \end{bmatrix} \begin{Bmatrix} d_3 \\d_4 \\ \end{Bmatrix} = \begin{Bmatrix} F_1 \\F_2 \\ F_3 \\ F_4 \\ F_5 \\ F_6 \end{Bmatrix} $$

Inputting the values for each $$\displaystyle K $$ and $$\displaystyle d $$ yields

 * $$ \begin{bmatrix} -{9 \over 16} & -{3 \sqrt 3 \over 16}

\\ -{3 \sqrt 3 \over 16} & -{3 \over 16} \\ {9 \over 16} + {5 \over 2} & {3 \sqrt 3 \over 16} + -{5 \over 2} \\ {3 \sqrt 3 \over 16} + -{5 \over 2} & {3 \sqrt 3 \over 16} + {5 \over 2} \\ -{5 \over 2} & {5 \over 2} \\ {5 \over 2} & -{5 \over 2} \end{bmatrix} \begin{Bmatrix} 4.352 \\ 6.1271 \\ \end{Bmatrix} = \begin{Bmatrix} F_1 \\F_2 \\ 0 \\ 7 \\ F_5 \\ F_6 \end{Bmatrix} $$

Note that since $$\displaystyle F_3 $$ and $$\displaystyle F_4 $$ are known (because they are the applied loads, 0 and 7 respectively) rows three and four do not need to be computed.

Thus, the computed solution is

 * $$ \textbf{F} = \begin{Bmatrix} -4.4378 \\ -2.5622 \\ 0 \\ 7 \\ 4.4378 \\ -4.4378 \end{Bmatrix} $$

Notice that these results are the same as those determined from Method 1.

Multiplication with the Transpose
$$\ \mathbf{k}^{(e)} = \mathbf{T}^{(e)T} \mathbf{\hat{k}}^{(e)} \mathbf{T}^{(e)}  $$

Prove that the above is true by multiplying right hand side of the equation.

$$\ \mathbf{T}^{(e)} $$ is defined in class as $$\ \begin{bmatrix} l^{(e)}&m^{(e)}&0&0\\0&0&l^{(e)}&m^{(e)} \end{bmatrix}_{2x4} $$

$$\ \mathbf{\hat{k}}^{(e)}$$ is defined in class as $$\ \begin{Bmatrix} k&-k\\-k&k\end{Bmatrix}_{2x2}$$

$$\ \mathbf{T}^{(e)T} $$ is the transpose of $$\ \mathbf{T}^{(e)}_, $$ where $$\ \mathbf{T}^{(e)T}_{ij} = \mathbf{T}^{(e)}_{ji} $$ from $$\ 1 \le i \le n_, \ 1 \le j \le m $$ for matrix $$\ \mathbf{T}^{(e)}$$  of dimensions $$\ n * m $$

Therefore, $$\ \mathbf{T}^{(e)T} = \begin{bmatrix} l^{(e)}&0\\m^{(e)}&0\\0&l^{(e)}\\0&m^{(e)} \end{bmatrix}_{4x2} $$

$$\ \mathbf{T}^{(e)T}*\mathbf{\hat{k}}^{(e)}= \begin{bmatrix} l^{(e)}*k^{(e)}&-l^{(e)}*k^{(e)}\\m^{(e)}*k^{(e)}&-m^{(e)}*k^{(e)}\\-l^{(e)}*k^{(e)}&l^{(e)}*k^{(e)}\\-m^{(e)}*k^{(e)}&m^{(e)}*k^{(e)} \end{bmatrix}_{4x2} $$

$$\ \mathbf{T}^{(e)T}*\mathbf{\hat{k}}^{(e)}*\mathbf{T}^{(e)}= \begin{bmatrix} l^{(e)}*l^{(e)}*k^{(e)}&l^{(e)}*m^{(e)}*k^{(e)}&-l^{(e)}*l^{(e)}*k^{(e)}&-l^{(e)}*m^{(e)}*k^{(e)}\\l^{(e)}*m^{(e)}*k^{(e)}&m^{(e)}*m^{(e)}*k^{(e)}&-l^{(e)}*m^{(e)}*k^{(e)}&-m^{(e)}*m^{(e)}*k^{(e)}\\-l^{(e)}*l^{(e)}*k^{(e)}&-l^{(e)}*m^{(e)}*k^{(e)}&l^{(e)}*l^{(e)}*k^{(e)}&l^{(e)}*m^{(e)}*k^{(e)}\\-l^{(e)}*m^{(e)}*k^{(e)}&-m^{(e)}*m^{(e)}*k^{(e)}&l^{(e)}*m^{(e)}*k^{(e)}&m^{(e)}*m^{(e)}*k^{(e)} \end{bmatrix}_{4x4} $$

Bringing out the $$\ k^{(e)} $$ and combining common terms gives $$\ \mathbf{T}^{(e)T}*\mathbf{\hat{k}}^{(e)}*\mathbf{T}^{(e)}= k^{(e)}\begin{bmatrix} (l^{(e)})^2&l^{(e)}*m^{(e)}&-(l^{(e)})^2&-l^{(e)}*m^{(e)}\\l^{(e)}*m^{(e)}&(m^{(e)})^2&-l^{(e)}*m^{(e)}&-(m^{(e)})^2\\-(l^{(e)})^2&-l^{(e)}*m^{(e)}&(l^{(e)})^2&l^{(e)}*m^{(e)}\\-l^{(e)}*m^{(e)}&-(m^{(e)})^2&l^{(e)}*m^{(e)}&(m^{(e)})^2 \end{bmatrix}_{4x4} $$

This is equivalent to the $$\ \mathbf{k}^{(e)} $$ given in class.

Therefore, $$\ \mathbf{f}^{(e)}=\mathbf{k}^{(e)}\mathbf{d}^{(e)} $$ can be written as

$$\ \mathbf{f}^{(e)}=\mathbf{T}^{(e)T}\mathbf{\hat{k}}^{(e)}\mathbf{T}^{(e)}\mathbf{d}^{(e)} $$

Closing the Loop and Associated Problems
From above, the following was established:

$$ AC = \! $$ $$ \frac{P_{2}^{(1)}}{k^{(1)}} = \frac{5.1243}{.75} = 6.8324$$

$$ AB = \! $$ $$\frac{P_{1}^{(2)}}{k^{(2)}}= \frac{6.2760}{5} = 1.255 \!$$

To find the coordinates of (xb, yb) and (xc, yc) of the two body truss system example, it is necessary to compute the equations for lines AB and BC.

This is found by computing the vector PQ as

$$ \vec{PQ} = (PQ)\vec{\tilde{i}} = (PQ)(cos\theta \vec{i} + sin\theta \vec{j})$$

$$ \vec{PQ} = (x - x_{p})\vec{i} + (y - y_{p})\vec{j}$$

The corresponding expressions thus become

$$ x - x_{p} = (PQ)cos\theta$$ $$ y - y_{p} = (PQ)sin\theta$$

or through rearrangement

$$ \frac{y - y_{p}}{x - x_{p}} = tan\theta$$

The expression for a line perpendicular to the above line and passing through P can be expressed as

$$ y - y_{p} = tan(\theta + \frac{\pi}{2})(x - x_{p})$$

From here, knowing that the angle for finding the b coordinates is 30 degrees and the angle for finding the c coordinates is -135 degrees, substitution provides

$$ (x_{b}, y_{b}) = (6.8324*cos(30), 6.8324*sin(30)) = (5.917, 3.416)$$

$$ (x_{c}, y_{c}) = (1.255*cos(-135), 1.255*sin(-135)) = (-0.887, -0.887)$$

Finally, to find (xD, yD) it necessary to understand the following expression

$$ \vec{AD} = (x_{D} - x_{A})\vec{i} + (y_{D} - y_{A})\vec{j}$$

Which can reduce to

$$ \vec{AD} = (x_{D})\vec{i} + (y_{D})\vec{j}$$

based on the choice of the origin.

By definition

$$ \vec{AD} = d_{3}\vec{i} + d_{4}\vec{j}$$

and knowing that d3=4.352 and d4=6.1271, vector AD becomes

$$ \vec{AD} = 4.352\vec{i} + 6.1271\vec{j}$$

such that

$$ (x_{D}, y_{D}) = (4.352, 6.1271)$$

Stiffness Matrix K and its Corresponding Eigenvalues
Given that the stiffness matrix K is defined as

$$\textbf{K}= \begin{bmatrix} K_{11}^{(1)} & K_{12}^{(1)} & K_{13}^{(1)} & K_{14}^{(1)} & 0 & 0  \\ K_{21}^{(1)}& K_{22}^{(1)} & K_{23}^{(1)} & K_{24}^{(1)} & 0 & 0 \\ K_{31}^{(1)} & K_{32}^{(1)} & K_{33}^{(1)} + K_{11}^{(2)} & K_{34}^{(1)} + K_{12}^{(2)} & K_{13}^{(2)} & K_{14}^{(2)} \\ K_{41}^{(1)}& K_{42}^{(1)} & K_{43}^{(1)} + K_{21}^{(2)} & K_{44}^{(1)} + K_{22}^{(2)} & K_{23}^{(2)} & K_{24}^{(2)} \\ 0 & 0 & K_{31}^{(2)} & K_{32}^{(2)} & K_{33}^{(2)} & K_{34}^{(2)} \\ 0 & 0 & K_{41}^{(2)} & K_{42}^{(2)} & K_{43}^{(2)} & K_{44}^{(2)}  \end{bmatrix}\,$$

with values determined in HW 2, K then becomes

$$\textbf{K}= \begin{bmatrix} 0.5625 & 0.3248 & -0.5625 & -0.3248 & 0 & 0 \\ 0.3248 & 0.1875 & -0.3248 & -0.1875 & 0 & 0 \\ -0.5625 & -0.3248 & 3.0625 & -2.1752 & -2.5 & 2.5 \\ -0.3248 & -0.1875 & -2.1752 & 2.6875 & 2.5 & -2.5 \\ 0 & 0 & -2.5 & 2.5 & 2.5 & -2.5 \\ 0 & 0 & 2.5 & -2.5 & -2.5 & 2.5  \end{bmatrix}\,$$

Solving for the eigenvalues of K using MATLAB results in

K =

0.5625   0.3248   -0.5625   -0.3248         0         0    0.3248    0.1875   -0.3248   -0.1875         0         0   -0.5625   -0.3248    3.0625   -2.1752   -2.5000    2.5000   -0.3248   -0.1875   -2.1752    2.6875    2.5000   -2.5000         0         0   -2.5000    2.5000    2.5000   -2.5000         0         0    2.5000   -2.5000   -2.5000    2.5000

EDU>> eig(K)

ans =

-0.0001  -0.0000    0.0000    0.0000    1.4706   10.0294

Due to round-off error with MATLAB, the solution to the eigenvalues of the K matrix are 1.4706, 10.0294, and four values of zero.

Two Bar Stress MATLAB Problem
% Matlab coding for HW 3 problem % For HW3, write a matlab code to plot the initial underformed % configuration of the two-bar truss system solved in class in % dotted line, and then superpose on the same figure the deformed % configuration in solid line. clear; close; % L(1) = 4 % theta(1) = 30 % k(1) = 3/4 % AC = 6.8324 % L(2) = 2 % theta(2) = -45 % k(2) = 5 % AB = 1.255 n_node = 8; n_elem = 6; P1_1 = 5.1243; P1_2 = 6.276; k_1 = .75; k_2 = 5; AC = P1_1/k_1; AB = P1_2/k_2; position(:,1) = [0; 0; 0]; position(:,2) = [4*cos(.5236); 4*sin(.5236); 0]; position(:,3) = [4*cos(.5236)+2*cos(.7854); 4*sin(.5236)-2*sin(.7854); 0]; position(:,4) = [4*cos(.5236)-1.255*cos(.7854); 4*sin(.5236)+1.255*sin(.7854); 0]; position(:,5) = [4*cos(.5236)+6.8324*cos(.5236); 4*sin(.5236)+6.8324*sin(.5236); 0]; position(:,6) = [4*cos(.5236)+6.8324*cos(.5236)-4*cos(1.047); 4*sin(.5236)+6.8324*sin(.5236)+4*sin(1.047); 0]; position(:,7) = [4*cos(.5236)-1.255*cos(.7854)+8*cos(.7854); 4*sin(.5236)+1.255*sin(.7854)+8*cos(.7854); 0]; position(:,8) = [7.8157; 8.1264; 0];   % intersection between AC and AB                                         % determined by zooming on graph % intersection for i = 1 : n_node; x(i) = position(1,i); y(i) = position(2,i); z(i) = position(3,i); end node_connect(1,1) = 1; node_connect(2,1) = 2; node_connect(1,2) = 2; node_connect(2,2) = 3; node_connect(1,3) = 2; node_connect(2,3) = 4; node_connect(1,4) = 2; node_connect(2,4) = 5; node_connect(1,5) = 4; node_connect(2,5) = 7; node_connect(1,6) = 5; node_connect(2,6) = 6; node_connect(1,7) = 1; node_connect(2,7) = 8; node_connect(1,8) = 3; node_connect(2,8) = 8; for i = 1 : n_elem; node_1 = node_connect(1,i); node_2 = node_connect(2,i); xx = [x(node_1),x(node_2)]; yy = [y(node_1),y(node_2)]; zz = [z(node_1),z(node_2)]; axis([-1 10 -1 10]) plot3(xx,yy,zz,':') hold on end for i = 7 : 8; node_1 = node_connect(1,i); node_2 = node_connect(2,i); xx = [x(node_1),x(node_2)]; yy = [y(node_1),y(node_2)]; zz = [z(node_1),z(node_2)]; plot3(xx,yy,zz,'-') hold on end title('Deformed and undeformed two-bar truss system') xlabel('x') ylabel('y') zlabel('z') view([0 0 1])     % xy plane view

The image below represents the two-bar truss system plotted in Matlab. The dotted line corresponds to the two-bar truss system before deformation, and the solid line includes deformation. The arrow pointing from Global Node 2 to Global Node 6 (which is the same as Global Node 2 for the deformed case) is the deformation vector.

Individual Contributions
Eml4500.f08.bottle.vitello 15:50, 5 October 2008 (UTC)

Eml4500.f08.bottle.hipps 18:38, 5 October 2008 (UTC)

Eml4500.f08.bottle.butler 23:23, 5 October 2008 (UTC)

Eml4500.f08.bottle.barnes 21:59, 7 October 2008 (UTC)

Eml4500.f08.bottle.brockmiller 00:30, 8 October 2008 (UTC)

Eml4500.f08.bottle.ranto 03:07, 8 October 2008 (UTC)

Eml4500.f08.bottle.loschak 04:38, 8 October 2008 (UTC)

title: User:Eml4500.f08.bottle.vitello/HW 3


text:
Combinatorics/Graph & Ramsey Theory

Welcome to the Lesson of Graph & Ramsey Theory In mathematics and computer science, graph theory is the study of graphs: mathematical structures used to model pairwise relations between objects from a certain collection. A "graph" in this context refers to a collection of 'vertices' or 'nodes' and a collection of edges that connect pairs of vertices. A graph may be undirected, meaning that there is no distinction between the two vertices associated with each edge, or its edges may be directed from one vertex to another. The graphs studied in graph theory should not be confused with "graphs of functions" and other kinds of graphs.

Theorems
Ramsey's Theorem is the solution to the Party Planner Problem

Schur's Theorem is a central theorem in Ramsey theory and combinatorial number theory that is concerned with arithmetic progressions.

Van der Waerden's Theorem is another theorem that is concerned with arithmetic progressions on the integers.

title: Combinatorics/Graph & Ramsey Theory


text:
User:Eml4500.f08.qwiki.bishop/HW3

 Comment: The pictures that were previously used in this homework submission have been changed. The previous images ( and ) were based on images that were not originally authored by Team Qwiki, however I did not intentionally plagiarize Team Echo's work. As Lee Harris, fellow Team Qwiki member, also stated in Homework 4, while working on Homework 2, both Team Qwiki and Team Echo created two images, with the same name: and . Due to the nature of MediaWiki, if two different teams create an image with the same name, only the latest version is displayed when called upon in an article. Therefore, even though Team Qwiki uploaded two of our own images (Element1.jpg and Element2.jpg) during homework 2, Team Echo's images were displayed because their upload was the most recent. In the future, we will make sure to upload files that have a unique name. For Homework 3, I created and  on October 5 because it was the two bar truss used in Homework 2, and to keep the images showing the same two bar truss with identical angles and lengths, I based the mentioned images on them. I was under the impression that Element1.jpg was the work of Team Qwiki. The problem has been resolved and Team Qwiki files are now displayed here. In the future Team Qwiki and myself will be sure to: A comparison between the current version of Homework 3 and our prior submission can be found here. Andrea Booher EML4500.f08.qwiki.booher 00:17, 13 November 2008 (UTC)
 * Create unique names when uploading files
 * Check the file history of all images before I right click and save to my computer

=Lecture 12 - Monday, September 22, 2008=

For a more thorough understanding of the Finite Element Method, it is wise to derive the element force displacement with respect to the global coordinate system.

Recall from Page 6-1, $$k^{(e)}d^{(e)}=f^{(e)}$$ (Equation 1)

Therefore, the element force displacement matrix can be written as follows:

$$\mathbf{k^{(e)}}\begin{bmatrix} 1 & -1 \\ -1 & 1\\ \end{bmatrix}\begin{pmatrix} q_{1}^{(e)}\\ q_{2}^{(e)}\\ \end{pmatrix}=\begin{pmatrix} P_{1}^{(e)}\\ P_{2}^{(e)}\\ \end{pmatrix} $$

$$q^{(e)}_{i}$$=axial displacement of element e at local node $$i$$

$$P^{(e)}_{i}$$=axial force of element e at local node $$i$$

The overall goal is to derive equation 1 from equation 2 (already derived in Meeting 4). We want to find the relationship between: The relationships can be expressed in the form: $$q^{(e)}_{2x1}=T^{(e)}_{2x4}d^{(e)}_{4x1}$$
 * $$q^{(e)}_{2x1}$$ and $$d^{(e)}_{4x1}$$
 * $$P^{(e)}_{2x1}$$ and $$f^{(e)}_{4x1}$$

Consider the displacement vector of local node i, denoted by $$d^{(e)}_{i}$$:

 * $$\vec{d^{(e)}_{[i]}}=d^{(e)}_{1}\vec{i}+d^{(e)}_{2}\vec{j}$$

 * $$q^{(e)}_{i}$$=axial displacement of node [1] is the orthogonal projection of the displacement vector $$\vec{d_{[1]}^{(e)}}$$ of node [1] on the $$\tilde{x}$$ axis of element e

Therefore, for node [1], $$q_1^{(e)}$$ can be derived using the following steps.

$$q_{1}^{(e)}=d_{[1]}^{(e)}*\vec{\tilde{i}}$$ $$q_{1}^{(e)}=(d_{1}^{(e)}\vec{i}+d_{2}^{(e)}\vec{j})\vec{\tilde{i}}$$ $$q_{1}^{(e)}=d_{1}^{(e)}(\vec{i}*\vec{\tilde{i}})+d_{2}^{(e)}(\vec{j}*\vec{\tilde{i}})$$ $$q_{1}^{(e)}=l^{(e)}d_{1}^{(e)}+m^{(e)}d_{2}^{(e)}$$ $$q_{1}^{(e)}=\begin{bmatrix} l^{(e)} & m^{(e)}\\ \end{bmatrix}\begin{pmatrix} d_{1}^{(e)}\\ d_{2}^{(e)}\\ \end{pmatrix}$$

The values for $$q_2^{(e)}$$ and $$q_2^{(e)}$$ can now be substituted into the following matrix:

$$\begin{pmatrix} q_{1}^{(e)}\\ q_{2}^{(e)}\\ \end{pmatrix}=\begin{bmatrix} l^{(e)} & m^{(e)} & 0 & 0\\ 0 & 0 & l^{(e)} & m^{(e)}\\ \end{bmatrix}\begin{pmatrix} d_{1}^{(e)}\\ d_{2}^{(e)}\\ d_{3}^{(e)}\\ d_{4}^{(e)}\\ \end{pmatrix}$$

=Lecture 13 - Wednesday, September 24, 2008=

Test day, no lecture notes given. Exam 1 Solutions

=Lecture 14 - Friday, September 26, 2008=

Similar to our objective in Meeting 12, today we would like to transform from a 4x2 matrix to a 2x2. The transformation relationship

\begin{bmatrix} P_1^{(e)}\\ P_2^{(e)}\\ \end{bmatrix} =T_{(2x4)}^{(e)} \begin{bmatrix} f_1^{(e)}\\ f_2^{(e)}\\ f_3^{(e)}\\ f_4^{(e)}\\ \end{bmatrix} $$ derived in Meeting 12 can also be expressed in a more general form:
 * $$\mathbf{P}^{(e)}_i=\mathbf{T}^{(e)}\mathbf{f}^{(e)}_i$$, where

 * $$\mathbf{P}^{(e)}_{i}$$ is the axial force of element e at local node $$i$$
 * $$\mathbf{T}^{(e)}_{(2x4)}$$ is the matrix used to transform force components to an axial force
 * $$\mathbf{f}^{(e)}_{i}$$ is the $$i$$th force component of element e

The overall goal is to find the relationship between: Let us consider the force transformation relationship:
 * $$\mathbf{q}^{(e)}_{(2x2)}$$ and $$\mathbf{d}^{(e)}_{(4x2)}$$, (axial displacements of each element node and their respective displacement component)
 * $$\mathbf{P}^{(e)}_{(2x2)}$$ and $$\mathbf{f}^{(e)}_{(4x2)}$$, (axial forces at each element node and their respective force components)
 * $$\mathbf{P}^{(e)}_{(2x2)}=\mathbf{T}^{(e)}_{(2x4)}\mathbf{f}^{(e)}_{(4x2)}$$

Recall that the element axial FD relationship is given by:
 * $$\mathbf{\hat{k}}^{(e)}_{(2x2)}\mathbf{q}^{(e)}_{(2x2)}=\mathbf{P}_{(2x2)}^{(e)}$$

By introducing a transformation matrix into the FD relationship, we obtain the following relationship:
 * $$\mathbf{\hat{k}}^{(e)}\mathbf{T}^{(e)}\mathbf{d}^{(e)}=\mathbf{T}^{(e)}\mathbf{f}^{(e)}$$

The ultimate goal is to express the above equation in the form $$\mathbf{k}^{(e)}\mathbf{d}^{(e)}=\mathbf{f}^{(e)}$$. If $$\mathbf{T}^{(e)}$$ were an nxn matrix, then one could simply "move" $$\mathbf{T}^{(e)}$$ from the right hand side to the left hand side by premultiplying the entire equation by the inverse $$\mathbf{T^{(e)}}^{-1}$$. Unfortunately, $$\mathbf{T}^{(e)}$$ is a rectangular 2x4 matrix, which means it is non-invertible.

Since we cannot solve for the inverse of the transformation ($$\mathbf{T}^{(e)}$$ is not an nxn matrix), we must instead make use of the transpose:
 * $$\mathbf{T}^{(e)T}_{(4x2)}\hat{k}^{(e)}_{(2x2)}\mathbf{T}^{(e)}_{(2x4)}\mathbf{d}^{(e)}_{(4x1)}=\mathbf{f}^{(e)}_{(4x1)}$$

Note that $$\mathbf{k}^{(e)}=\mathbf{T}^{(e)T}_{(4x2)}\mathbf{\hat{k}}^{(e)}_{(2x2)}\mathbf{T}^{(e)}_{(2x4)}$$ Therefore, $$\mathbf{k}^{(e)}\mathbf{d}^{(e)}=\mathbf{f}^{(e)}$$

The relationship $$\mathbf{k}^{(e)}=\mathbf{T}^{(e)T}\mathbf{\hat{k}}^{(e)}\mathbf{T}^{(e)}$$ can be justified by the principle of virtual work, which was first introduced in Meeting 10. Additionally, applying the principle of virtual work to the relationship $$K_{(6x6)}d_{(6x1)}=F_{(6x1)}$$ reduces it to $$K_{(2x2)}d_{(2x1)}=F_{(2x1)}$$.

One may wonder why we cannot simply solve $$d=K^{(-1)}F$$ Answer : The equation cannot be solved in the current form due to the singularity of $$K$$, i.e, $$detK=0$$ and thus $$K$$ is not invertible. (See above for definition of interbile matrix) Recall that in order to find $$K^{(-1)}$$ we must first compute $$1/{(detK)}$$. If a matrix is singular then $$1/{(detK)}$$ is undefined and cannot be solved. For our case of an unconstrained structural system, there are three possible rigid body motions in 2-dimensions (2 translations and 1 rotation) for each node.

References

The blue solid line represents the undeformed truss system, while the green dashed line represents the deformed truss system.

=Contributing Team Members= The following students contributed to this report
 * Michael Berry Eml4500.f08.qwiki.berry 10:41, 8 October 2008 (UTC)
 * Andrea Booher Eml4500.f08.qwiki.booher 11:32, 8 October 2008 (UTC)
 * Chris Bishop Eml4500.f08.qwiki.bishop 13:02, 8 October 2008 (UTC)
 * David Nobles Eml4500.f08.qwiki.nobles 15:29, 8 October 2008 (UTC)
 * Eric Gatch Eml4500.f08.qwiki.gatch 12:15, 8 October 2008 (UTC)
 * Lee HarrisEml4500.f08.qwiki.harris 17:16, 8 October 2008 (UTC)

title: User:Eml4500.f08.qwiki.bishop/HW3


text:
User:Carlesmari

My user page in Catalan-Valencian

title: User:Carlesmari


text:
User:Eml4500.f08.gravy.jad/Notes3

Lecture 15
HW problem goes here. (closing the loop between Finite Element Method and statics: virtual displacement.)

Two-Bar Truss System

There are two different ways on solving these types of problems: using FEM, or using statics.

(insert pictures)

These pictures are FBD's of 2 force bodies of element 1 & 2. By Statics reactions are known and therefore the member forces $$\ [P_{1}^{(1)}, P_{2}^{(2)}] \,$$ are equal and opposite in direction. We need to compute the axial displacement dofs ( amount of extension of bars):

$$\ q_{2}^{(1)}= P^{(1)}/k^{(1)}= \underline{AC} \,$$ . ( $$\ q_{1}^{(1)}=0 \,$$ because it's fixed at global node 1.

$$\ q_{1}^{(2)}= -P^{(2)}/k^{(2)}= -\underline{AC} \,$$ . ( $$\ q_{2}^{(2)}=0 \,$$ because it's fixed at global node 3.

Q : How to back out from above results of displacement of dofs of global node 2?

insert picture here triangle thing

insert matlab portion

$$\ u_{y}=R*sin(\alpha) \simeq R*\alpha \,$$

$$\ u_{x}=R*(1-cos(\alpha)) \simeq 0 \,$$ $$\ (cos(\alpha) \simeq 1) \,$$

hence

Lecture 16
Closing the loop continued Infinitesimal displacement (relationship to virtual displacement)

$$\ AC= \frac{|P_{2}^{(1)}|}{k^{(1)}} = \frac{5.1243}{3/4} = 6.8324 \,$$

insert AB HW

insert xb,yb HW

- P is fixed, Q is "sliding" on the line

- the two lines are parallel $$\ \bar{PQ} =(PQ)*\vec {\tilde i} = (PQ)[cos\theta \vec {i} + sin\theta  \vec {j}] = (x-x_p)\vec i + (y-y_p)\vec j \,$$

$$\ \Rightarrow x-x_p=PQ cos\theta \,$$

$$\ \Rightarrow y-y_p=PQ sin\theta \,$$

$$\ \frac{y-y_p}{x-x_p}= tan\theta \,$$

$$\ (y-y_p)= (tan\theta)(x-x_p) \,$$

eq for a line perpendicular to the line, passing P: $$\ (y-y_p)= \tan(\theta+\frac{\pi}{2})(x-x_p) \,$$

insert HW xD,yD

since $$\ x_A=0 and y_A=0 \,$$ then $$\ \vec {AD} =(x_D-x_A)\vec i + (y_D - y_A)\vec j = x_D-\vec i + y_D\vec j \,$$

By definition: $$\ \vec{AD} = d_3 \vec i + d_4 \vec j \,$$ (these are displacement vectors of A from FEM

///end closing loop

3-Bar Truss Syst:

$$\ E^{(1)} =2, E^{(2)} =4, E^{(3)} =3 \,$$

$$\ A^{(1)} =3, A^{(2)} =1, A^{(3)} =2 \,$$

$$\ L^{(1)} =5, L^{(2)} =5, L^{(3)} =10 \,$$

$$\ P=30, \theta^{(1)}=30^\circ = -\theta^{(2)}, \theta^{(3)}=45^\circ \,$$

Local node numbering is equivalent to the convenience in assembly k.

Convient local node numbering

title: User:Eml4500.f08.gravy.jad/Notes3


text: File:FEA loop.JPG

title: File:FEA loop.JPG


text:
User:Eml4500.f08.delta 6.ramirez/matrix proof

Note: Since the inverse of T(e) cannot be used the transpose is utilized. The following demonstrates that both sides are equal.

$$k^{(e)}= T^{(e)T}\hat{k}^{(e)} T^{(e)}$$

$$k^{(e)} = k_{(e)}\begin{bmatrix} l^{(e)2} & l^{(e)}m^{(e)} & -l^{(e)2} & -l^{(e)}m^{(e)}\\ l^{(e)}m^{(e)} & m^{(e)2} & -l^{(e)}m^{(e)} & -m^{(e)2} \\ -l^{(e)2} & -l^{(e)}m^{(e)} & l^{(e)2} & l^{(e)}m^{(e)}\\ -l^{(e)}m^{(e)} & -m^{(e)2} & l^{(e)}m^{(e)} & m^{(e)2} \end{bmatrix}$$ where,

$$k_{(e)} = \frac{E^{(e)}A^{(e)}}{L^{(e)}}$$

thus,

$$k_{(e)}\begin{bmatrix} l^{(e)2} & l^{(e)}m^{(e)} & -l^{(e)2} & -l^{(e)}m^{(e)}\\ l^{(e)}m^{(e)} & m^{(e)2} & -l^{(e)}m^{(e)} & -m^{(e)2} \\ -l^{(e)2} & -l^{(e)}m^{(e)} & l^{(e)2} & l^{(e)}m^{(e)}\\ -l^{(e)}m^{(e)} & -m^{(e)2} & l^{(e)}m^{(e)} & m^{(e)2} \end{bmatrix} = T^{(e)T}\hat{k}^{(e)} T^{(e)}$$

To prove that both sides are equal, the matrices must be multiplied in the correct order. The matrices are given as:

$$ T^{(e)T} = \begin{bmatrix} l^{(e)} & 0 \\ m^{(e)} & 0\\ 0 & l^{(e)}\\ 0 & m^{(e)} \end{bmatrix} $$

$$\hat{k}^{(e)} = k_{(e)}\begin{bmatrix} 1 & -1\\ -1 & 1 \end{bmatrix}$$

$$T^{(e)} = \begin{bmatrix} l^{(e)} & m^{(e)} & 0 & 0\\ 0 & 0 & l^{(e)} & m^{(e)} \end{bmatrix}$$

Multiplying out the first two matrices gives:

$$T^{(e)T}\hat{k}^{(e)} = \begin{bmatrix} l^{(e)} & 0 \\ m^{(e)} & 0\\ 0 & l^{(e)}\\ 0 & m^{(e)} \end{bmatrix} k_{(e)}\begin{bmatrix} 1 & -1\\ -1 & 1 \end{bmatrix} = k_{(e)} \begin{bmatrix} l^{(e)} & -l^{(e)}\\ m^{(e)} & -m^{(e)}\\ -l^{(e)} & l^{(e)}\\ -m^{(e)} & m^{(e)} \end{bmatrix}$$

Taking the result from this and now multiplying it by the third matrix T(e) gives:

$$k_{(e)} \begin{bmatrix} l^{(e)} & -l^{(e)}\\ m^{(e)} & -m^{(e)}\\ -l^{(e)} & l^{(e)}\\ -m^{(e)} & m^{(e)} \end{bmatrix} T^{(e)} = k_{(e)} \begin{bmatrix} l^{(e)} & -l^{(e)}\\ m^{(e)} & -m^{(e)}\\ -l^{(e)} & l^{(e)}\\ -m^{(e)} & m^{(e)} \end{bmatrix} \begin{bmatrix} l^{(e)} & m^{(e)} & 0 & 0\\ 0 & 0 & l^{(e)} & m^{(e)} \end{bmatrix} = k_{(e)}\begin{bmatrix} l^{(e)2} & l^{(e)}m^{(e)} & -l^{(e)2} & -l^{(e)}m^{(e)}\\ l^{(e)}m^{(e)} & m^{(e)2} & -l^{(e)}m^{(e)} & -m^{(e)2} \\ -l^{(e)2} & -l^{(e)}m^{(e)} & l^{(e)2} & l^{(e)}m^{(e)}\\ -l^{(e)}m^{(e)} & -m^{(e)2} & l^{(e)}m^{(e)} & m^{(e)2} \end{bmatrix} $$

Observation: both sides are equal.