text:
User talk:Mlbolick11

 Hello Mlbolick11, 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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And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! --mikeu talk 03:47, 20 October 2008 (UTC)

HI
Hi! My name is Desi Maneva and I have chosen the thema e-learning. You can visit my wiki site and let a massage for me, if you are interested in this thema and want to discuss some aspects with me. We can make together something like plan and begin work :) Please be my friend--Desi maneva 14:53, 22 October 2008 (UTC)

Upload
Hi Megan, you can upload that file via the Upload File in the bottom left of the side-bar. Hope that helps. -- Jtneill - Talk - c 02:57, 3 December 2008 (UTC)
 * Oh - I see you already did this - so a link to it could be [[Image:International.pdf]] or more directly [[Media:International.pdf]]. Interesting presentation!!! -- Jtneill - Talk - c 03:00, 3 December 2008 (UTC)

title: User talk:Mlbolick11


text:
User:Eas4200c.f08.blue.a/Lecture 18

Quadrature allows an observer outside of a NACA airfoil to find the cross sectional area. An example of this is shown below in Figure 1.

Figure 1

$$\bar{A}=\bar{A_{1}}+\bar{A_{2}} =\bar{A_{1}}\left|\bar{A_{2}} \right|$$

In this close up, $$d\vec{A}$$ is defined as:

$$d\vec{A}=\frac{1}{2}\vec{r}+\vec{PQ}=-\left|dA \right|\vec{i}$$

Dividing the area into trapezoids and adding them is another method used to find the area under a curve called the (Trapezium rule), but it is not as elegant or effective as triangular quadrature in more complicated applications such as finding the area of the oddly shaped airfoil in Figure 2 below.

Figure 2

Returning to the idealized single cell airfoil example from above, where the shear flow is constant, the twist angle can be found using the formula found in step J of the road-map for torsional analysis of an aircraft wing.

$$\Theta =\frac{1}{2G\bar{A}}\oint_{}^{}{\frac{q}{t}ds}$$

because $$\left(q \right)$$ is constant,

$$\left(q \right)=\left( q_{1}\right)=\left( q_{2}\right)=\left( q_{3}\right)$$

$$\Theta =\frac{1}{2G\bar{A}}q\sum_{j=1}^{3}{\frac{l_{j}}{t_{j}}}$$ $$=\frac{1}{2G\bar{A}}q\left[\frac{2\pi \left(\frac{b}{2} \right)}{t_{1}}+\frac{a}{t_{2}}+\frac{\sqrt{a^{2}+b^{2}}}{t_{3}} \right]=q\left(\frac{1}{2\left(5.57m^{2} \right)G} \right)\left(\frac{\pi \left( 1m\right)}{0.008m}+\frac{4m}{0.01m} +\frac{\sqrt{16m^{2}+4m^{2}}}{0.01m} \right)=\frac{q}{\left( 11.14m^{2}\right)G}\left(1239.91 \right)=111.24m^{-2}\frac{q}{G}$$

The Max Shear Stress $$\left( \tau _{max}\right)$$ can also be found now using the formula:

$$ \tau _{max}=\frac{q}{min\left\{t_{1},t_{2},t_{3} \right\}}$$

If the maximum shear stress is equal to the allowable shear stress and from the formula to find shear flow, the allowable torque $$\left[ T_{allow}\right]$$ can be found.

$$\tau _{max}=\left( \tau _{allow}\right)$$

$$q=\frac{T}{2\bar{A}}$$

$$T_{allow}=2\bar{A}\tau _{allow}\left(min\left\{ t_{1},t_{2},t_{3}\right\} \right)$$

Therefore, if $$\tau _{allow}=100GPa$$, and because $$t_{1}=0.008m$$, $$t_{2}=t_{3}=0.01m$$

$$T_{allow}=2\bar{A}\tau _{allow}t_{1}=2\left(5.57m^{2} \right)\left(100GPa \right)\left(0.008m \right)=8.91\left(10^{9} \right)Nm$$

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

title: User:Eas4200c.f08.blue.a/Lecture 18


text:
User:Mrhoden

Hey, I'm Mark Rhoden. I have lived in Monroe, North Carolina all my life. This is my second year at UNCC I transferred in after attending Appalchian State and SPCC each for a semester. I enjoy playing baseball, and I coach a summer baseball team. Last summer I traveled to Alabama with my team, and have been to numerous states along the east coast. I spend any free time and extra money I have on my car. I want to teach forth or fifth grade, and also coach high school baseball.

title: User:Mrhoden


text: Category:Boundary value problems

title: Category:Boundary value problems


text: Category:Differential equations

title: Category:Differential equations


text: Category:Mathematical analysis

title: Category:Mathematical analysis


text:
C Sharp/Math

Lesson Goals
This lesson will focus on basic math functionality in C# as well as user input via the console. At the end of this lesson you should be able to:
 * Understand how to use basic math operations.
 * Explain the order in which math operations are completed.
 * Use the console to read in user input.
 * Understand how to access the more powerful functionality of the math namespace.

Basic Math
Until now we've used hardcoded values to output information to the screen. Now we'll use math on those values to force the computer to perform some mathematical grunt work. A computer uses the same laws for math as we do. The C# language follows the rule of PEDMAS (or BEDMAS depending on your high school teacher). PEDMAS stands for Parenthesis, Exponents, Division, Multiplication, Addition and Subtraction. The operations on the left are executed before the operations to the right. Example: 3 + (4 + 5) * 2 First step (Parenthesis): 3 + 9 * 2 Second step (Multiplication): 3 + 18 Last step (Addition): 21 Following the law of PEDMAS ensures that every computer (and human) will get a consistent answer. With that in mind, it is now possible to unleash mathematical operations on the computer.

Output
The area of a circle with radius 2.5 is 19.6349375 Hit any key to end...

Remarks
Here we've assigned a value for the constant Pi. Pi is required for finding the area of a circle. We've also assigned a constant value for the radius of the circle. Here is where the computer gets to work number crunching. The area of a circle is given as A = pi * r * r. The result of the calculation will be stored in a variable named area. We now output the result of the calculations to the console. As a bonus, we've also output the radius on the circle. These lines of code pause the execution of the program. Without these lines the program would terminate very quickly, and we wouldn't be able to view the program output. Console.ReadyKey waits until the user presses a key.

User Input
A programmer can request the user to input a value in the console. The easiest method to read from the console is to read the console input until the user hits the enter button. A line must be stored in a C# type called a string. A string contains a list of chars. This string must then be converted to the type that was requested by the program. The best way to learn is by example and by doing, so here are some examples of reading a string and then converting that string to another data type.

Each C# data type has a method to parse a string, and convert that string into the corresponding data type. The problem however is that the user might make a mistake an enter a string that cannot be parsed. An example would be if the program requested an integer value, but the user entered a floating point value. The program will then throw an error that the Input string was not in a correct format. For now we will assume that the user will always enter the correct values, and the error that was generated will be handled in a later lesson on try, catch and finally.

Output
Please enter a value for the radius: 2.5 The area of a circle with radius 2.5 is 19.6349375 Hit any key to end...

Remarks
First we prompt the user for a value. This ensures that the user realizes that the program is waiting for input. Then we read a double into the variable named radius. This radius is used to compute the area of the circle. Note that this program will not deal with any errors that could be generated by the Parse method.

Math Namespace
C# has many more math functions stored in the Math namespace. Exploration of the entire Math namespace is beyond the scope of this lesson, but many online references are available. We're going to modify the circle program to use the Math namespace where appropriate.

Output
Please enter a value for the radius: 2.5 The area of a circle with radius 2.5 is 19.6349540849362 Hit any key to end...

Remarks
Notice that there is no longer a line to store a value of pi. The Math namespace already contains a definition of PI (which is much more accurate than the previous version). The Math namespace also has a method for computing powers. radius^2 is represented quite a bit easier with the Math namespace. It would also become very unwieldy to write something like radius^20 with the method that we were using in the previous examples.

Summary

 * Math operations are computed in the order of PEDMAS.
 * User input from the console is first read as a string, and then converted to the correct data type.
 * The conversion method (Parse) can throw an error if the string is incorrectly formatted.
 * The Math namespace contains many methods that make complex mathematical operations much easier to perform.

Practice Exercises

 * Use user input and the Math namespace to generate the sin and cos of a user assigned value.
 * Use user input and the Math namespace to find the maximum value of two user assigned values.

title: C Sharp/Math


text:
User talk:Carlcanary

 Hello Carlcanary, 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:Carlcanary


text:
User talk:Mrhoden

 Hello Mrhoden, 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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And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! --JWSchmidt 14:11, 20 October 2008 (UTC)

title: User talk:Mrhoden


text:
User talk:Every IP in New England

Every IP in New England is welcome to participate at Wikiversity. --JWSchmidt 14:16, 20 October 2008 (UTC)

title: User talk:Every IP in New England


text:
User:Sonet Briggs

- 42 Year old guy who still thirst for knowledge. - have experience programming with xbase (foxpro, clipper, dbase III+) languages

title: User:Sonet Briggs


text:
User:La comadreja/Meetings

All official chats are broadcast on Google Talk and are open to all. Contact reseal05@gmail.com for copies of the publications discussed, or to enter the chat room. Times are Eastern Standard Time (USA).

For now, no chats are scheduled. See User talk:AFriedman for our current lab discussions.

title: User:La comadreja/Meetings


text:
User:Smaults3

Hey! My name is Sarah Maultsby. I am a junior and UNC Charlotte. I have attended UNC Charlotte for two years. I went to another University for my Freshman year. It was a smaller college about an hour away from UNCC called Wingate University. I live in Belmont which is a small town right outside of Charlotte. I enjoy shopping, traveling and swimming. I have a younger brother who is a senior in high school. He wants to attend UNC Chapel Hill in the fall. I have 3 dog's and 1 cat. My dogs names are Carmel, Corkie and Dooney and my cats name is Smokey. I love animals. I have traveled to Europe 3 times. I have been to England, Italy, France, Germany, Switzerland, and Austria. I hope to teach first grade when I am finished with school. I enjoy working with kids that is why I decided to teach.

title: User:Smaults3


text: File:Blah.jpg

title: File:Blah.jpg


text:
User:Caytonpr

Hey everyone!! I am very excited about working with everyone on this international project. As I said on the discussion board, I went to Auburn University for two years, and now I am back in Charlotte finishing up my Education degree. I hope to graduate in December of '09. I love working with kids and can not wait to teach them the fundamentals they will use for their entire lives. I currently babysit regularly as well as work at a barn riding horses for people. I look forward to getting to know everyone through this project. If you would like to know anything else about me, just let me know!!!

KWL CHART
Please click on "edit" (here above), copy the code, paste it on your personal user site and insert Your text in the columns.

title: User:Caytonpr


text:
User:Kathrin L.

Hello,

my name is Kathrin Lehnert and I am studying at the Catholic University Eichstaett. I want to become a primary teacher and thus, I joined this course. I'm curious about what this workshop will be about so time will show ;) For so far, I hope we will have fun, too :)

Greetings,

Kathrin

Discussion about the issue "The teacher's emotional influence on first graders":
User:Daniela_Widdmann/project

title: User:Kathrin L.


text:
User talk:Kathrin L.

 Hello Kathrin L., 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

 * Take a guided tour and learn to edit.
 * Visit a (kind of) random random project.
 * Browse everything on Wikiversity, or visit a portal corresponding to your educational level: pre-school, primary, secondary, tertiary, non-formal education.
 * Find out about research activities on Wikiversity.

 * Read an introduction for teachers and find out how to write an educational resource for Wikiversity.
 * Give feedback about your initial observations
 * Discuss Wikiversity issues or ask questions at the colloquium.
 * Chat with other Wikiversitans on #wikiversity-en.

And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! --mikeu talk 16:33, 20 October 2008 (UTC)

title: User talk:Kathrin L.


text: File:Foto wikiversity.jpg

title: File:Foto wikiversity.jpg


text:
User:Eml4500.f08.ateam.rivero/w4

Meeting 23: October 17, 2008
Recall Equations:

f_{3}^{(1)} + f_{1}^{(2)} = 0 $$
 * Equation 1: $$

$$ [k_{31}^{(1)}*d_{1}^{(1)} + k_{32}^{(1)}*d_{2}^{(1)} + k_{33}^{(1)}*d_{3}^{(1)} + k_{34}^{(1)}*d_{4}^{(1)}] +         [k_{11}^{(1)}*d_{1}^{(2)} +            k_{12}^{(1)}*d_{2}^{(2)} + k_{13}^{(1)}*d_{3}^{(2)} + k_{14}^{(1)}*d_{4}^{(2)}] = 0 $$
 * Equation 1: Expanded

$$ [k_{31}^{(1)}*d_{1} + k_{32}^{(1)}*d_{2} + k_{33}^{(1)}*d_{3} + k_{34}^{(1)}*d_{4}] + [k_{11}^{(1)}*d_{3} + k_{12}^{(1)}*d_{4} + k_{13}^{(1)}*d_{5} + k_{14}^{(1)}*d_{6}] = 0 $$
 * Equation 1: Expanded using global degrees of freedom

f_{4}^{(1)} + f_{2}^{(2)} = P $$
 * Equation 2: $$

$$ [k_{41}^{(1)}*d_{1}^{(1)} + k_{42}^{(1)}*d_{2}^{(1)} + k_{43}^{(1)}*d_{3}^{(1)} + k_{44}^{(1)}*d_{4}^{(1)}] + [k_{21}^{(1)}*d_{1}^{(2)} +           k_{22}^{(1)}*d_{2}^{(2)} + k_{23}^{(1)}*d_{3}^{(2)} + k_{24}^{(1)}*d_{4}^{(2)}] = P $$
 * Equation 2: Expanded

$$ [k_{41}^{(1)}*d_{1} + k_{42}^{(1)}*d_{2} + k_{43}^{(1)}*d_{3} + k_{44}^{(1)}*d_{4}] + [k_{21}^{(1)}*d_{3} + k_{22}^{(1)}*d_{4} + k_{23}^{(1)}*d_{5} + k_{24}^{(1)}*d_{6}] = P $$
 * Equation 2: Expanded using global degrees of freedom

Assembly of   $$  \displaystyle \mathbf     k^{(e)} $$, e = 1,. . ., nel (number of elements) into the global stiffness matrix

$$\displaystyle \mathbf K = \overset{e=nel}{\underset{e=1}{\mathbb A}} \mathbf k^{(e)}$$ $$\displaystyle \overset{e=nel}{\underset{e=1}{\mathbb A}} $$ = Assembly Operator n : total number of global degrees of freedom before eliminating the boundry conditions ned : number of element degrees of freedom ned << n : ned is very small when compared to n

Principle of Virtual Work
Motivation:

 * The principle of virtual work allows us to eliminate the corresponding rows to the boundry conditions to obtain the global stiffness matrix K. Reducing the global stiffness matrix from a (6x2) matrix to a (2x2) matrix.

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

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

 * The ultimate goal is that the principal of virtual work allows us to derive the Finite Element Method for Partial Differential Equations which are heavily used in the world of Heat Transfer, Vibrations, Finite Element Analysis Commercial Codes and many more.

Force Displacement Relation for a bar or spring is expressed in the form $$ kd=F $$, which implies

$$ kd-F=0 $$ Equation (3) $$ w(kd-F)=0 $$ for all w Equation (4) Proof: (3) => (4) Trivial (4) => (3) Not Trivial The reason why going from equation 3 to 4 is trivial is because since equation 4 is valid for all w, select w = 1 and then equation 4 becomes.... $$ 1(kd-F) = 0 $$ which is equal to equation 3

title: User:Eml4500.f08.ateam.rivero/w4


text:
User:Eas4200c.f08.wiki.c/homework report 4

The expression:

$$\theta = \frac{\partial \alpha }{\partial x}$$

can be used to express the rate of twist of a cross section.

Hooke's law can be represented as:

$$\tau = G\gamma = G\rho \theta$$

 $$\tau (s)= G\gamma = G\rho(s) \theta (x)$$

With variables as defined in the previous diagram.

Integrate along $$\textit{C}$$

$$ \oint_\textit{C}{}^{}{\tau (s)ds} = G\theta (x)\oint_{\textit{C}}^{}{\rho (s)ds}$$

where

$$ \tau (s)ds = \frac{q(s)}{t(s)}$$

$$\rho (s) = 2\bar{A}$$

To get an expression for $$\theta$$ as discussed in the previous lecture.

The question, 'What is ad-hoc about the above derivation of $$\theta$$ expression on p 20.2 and the derivation of $$T = 2q\bar{A}$$?' was posed in the discussion of the equation. The answer can be provided in two parts:

1) Strain ($$\gamma$$) must be obtained using the displacement of $$P$$ in the direction tangent to $$\textit{C}$$ at $$P$$, but $$PP''$$ in the previous diagram is not necessarily tangent to $$\textit{C}$$ (but is actually close, due to the small angles involved).

2) $$\tau = \frac{q}{t}$$ obtained from ad-hoc assumption that $$\tau$$ was uniform across wall thickness, this is assuming that the wall thickness is very thin, or else this assumption breaks down.

Now formal justification (derivation) of elasticity theory.

This section refers back to the road map that was previously discussed.

A. Kinematic Assumption

$$u_{x}(y, z) = \theta \Psi (y,z)$$

Where $$\theta$$ is considered constant with respect to x (due to it being a uniform bar).

$$u_{y}(x, z) = -\theta xz$$  $$u_{z}(x, y) = \theta xy$$

To transform equations in Sun[2006] to those using our unified notation, we use cyclic permutation

 $$\varepsilon _{xx} = \varepsilon _{yy} = \varepsilon _{zz} = \gamma _{yz} = 0$$

$$\varepsilon _{xx} = \frac{\partial u_{x}}{\partial x}(y,z) = 0$$

$$\gamma _{yz} = \frac{\partial u_{y}}{\partial z}(x,z) + \frac{\partial u_{z}}{\partial y}(x,y) = -\theta x + \theta x = 0$$ HW:

$$\varepsilon _{yy} = \frac{\partial u_{y}}{\partial y}(x,z) = 0$$

$$\varepsilon _{zz} = \frac{\partial u_{z}}{\partial z}(x,y) = 0$$

These results are obtained due to the fact that both derivatives are taken with respect to a variable that does not show up in either function being differentiated.


text:
User:Eml4500.f08.jamama.justin/HW4

Note: Consider the case: $$\tilde{d}{_{4}}^{(e)}\neq 0$$

$$\tilde{d}{_{1}}^{(e)}=\tilde{d}{_{2}}^{(e)}=\tilde{d}{_{3}}^{(e)}=0$$

$$\tilde{f}^{(e)}=\tilde{k}^{(e)}\tilde{d}^{(e)}=0$$ 4X1  4X4  4X1   4X1

Interp of transverse DOF's.

Recall from Meeting 19: $$\tilde{d}^{(e)}=\tilde{T}^{(e)}d^{(e)}$$

similarly $$\tilde{f}^{(e)}=\tilde{T}^{(e)}f^{(e)}$$

Also: $$\tilde{k}^{(e)}\tilde{d}^{(e)}=\tilde{f}^{(e)}$$

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

If $$\tilde{T}^{(e)}$$ is invertible then:

$$\left[ \tilde{T}^{(e)-1}\hat{k}^{(e)}\tilde{T}^{(e)}\right]d^{(e)}=f^{(e)}$$ (K matrix)

$$\tilde{T}^{(e)}$$=block diagonal matrix (P.19.3)

Consider a general block-diagram matrix:

$$A=\begin{bmatrix} D1 & &  &  &0 \\ & ..&  &  & \\  &  &  ..&  & \\  &  &  &  ..& \\  0&  &  &  & D5 \end{bmatrix}$$

Question: What is $$A^{-1}$$

Simpler example: Diagonal matrix

$$B=\begin{bmatrix} d_{11} & &  &  &0 \\ & d_{22}&  &  & \\ & &  d_{33}&  & \\ & &  &  d_{44}& \\ 0& &  &  &d_{mn} \end{bmatrix}= diag\left[d_{11},d_{22},d_{33}....d_{mn} \right]$$

assuming that $$d_{ij}\neq 0 $$ for i=1,....n then $$B^{-1}=diag\left[\frac{1}{d_{11}},\frac{1}{d_{22}},\frac{1}{d_{33}}....\frac{1}{d_{mn}} \right]$$

For a block diagonal matrix A:

$$A=diag\left[D_{1},....D_{5} \right]$$

$$A^{-1}=diag\left[D_{1}^{-1},....D_{5}^{-1} \right]$$

$$\tilde{T}^{(e)-1}=diag\left[R^{(e)-1},R^{(e)-1} \right]$$

P.19-2 $$R^{(e)T}=\begin{bmatrix} l^{(e)} & -m^{(e)}\\ m^{(e)}&l^{(e)} \end{bmatrix}$$

$$R^{(e)T}R^{(e)}=\begin{bmatrix} l^{(e)} & -m^{(e)}\\ m^{(e)}&l^{(e)} \end{bmatrix} \begin{bmatrix} l^{(e)} & m^{(e)}\\ -m^{(e)}&l^{(e)} \end{bmatrix}= \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}=I $$ (2x2 identity matrix) ... $$\Rightarrow R^{(e)-1}=R^{(e)T}$$

$$\Rightarrow \tilde{T}^{(e)-1}=diag\left[R^{(e)T},R^{(e)T} \right] =\left( diag\left[R^{(e)},R^{(e)} \right]\right)^{T}$$ ...  $$\tilde{T}^{(e)-1}=\tilde{T}^{(e)T}$$

so from earlier it can be shown that: $$\left[ \tilde{T}^{(e)T}\hat{k}^{(e)}\tilde{T}^{(e)}\right]d^{(e)}=f^{(e)}$$

title: User:Eml4500.f08.jamama.justin/HW4


text:
User:EML4500.f08.A-team.VandenBerg/HW4

 Comment: This is HW4 located in the user namespace of EML4500.f08.A-team.VandenBerg 20:28, 24 October 2008 (UTC)

HW4
18-1

Connecting Array
Considering the two-bar truss system from the the earlier page 5-6 for connecting array "conn".

conn: $$ \begin{bmatrix} 1     & 2 \\  2     & 3 \end{bmatrix}$$

Where:
 * Row one represents element one
 * Row 2 represents element two
 * The numbers within the matrix represent the global node numbers

conn(e,j) = This is the global node number of local node "j" of element "e".

Location Matrix Master Array
The following location matrix master is for "1mm":

1mm = $$\begin{vmatrix} 1 & 2 & 3 & 4 \\ 3 & 4 & 5 & 6 \end{vmatrix}$$

Where:
 * Row one represents element one
 * Column one represents local degree of freedom number one
 * Column two represents local degree of freedom number two
 * Column three represents local degree of freedom number three
 * Column four represents local degree of freedom number four
 * The numbers within the matrix represent the global degree of freedom number (or eq. number) in K

18-2 1mm(i,j) = This represents the eq. number (global degree of freedom number) to the j th local degree of freedom number for the element stiffness coefficient.

HW

 * The homework for this meeting is to solve the three bar truss system from page 16-4 by modifying the Matlab code for the two bar truss system.
 * Also, make a transformation matrix that is a 4x4 and (hopefully) invertible by transforming a four degree of freedom system to a four degree of freedom system from page 12-1.

Meeting 19: Wednesday, 8 October 2008. EML4500

 * Goal: We want to find the transformation matrix; $${\underset{4 \times 4}\mathbf \tilde T^{(e)}}$$, that transforms the set of local elemental degrees of freedom; $${\underset{4 \times 1}\mathbf d^{(e)}}$$ which is set on the cartesian cordinate system as seen on the figure below, to the another set of local elemental degrees of freedom $${\underset{4 \times 1}\mathbf \tilde d^{(e)}}$$, such that $${\underset{4 \times 4}\mathbf \tilde T^{(e)}}$$ is inversible.

$$\underset{4 \times 1}\tilde \mathbf d^{(e)}=\underset{4 \times 4}\mathbf \tilde T^{(e)}\underset{4 \times 1} \mathbf d^{(e)} $$

$$\underset{4 \times 1}\tilde \mathbf d^{(e)}=\underset{4 \times 4}\mathbf \tilde T^{(e)}\underset{4 \times 1} \mathbf d^{(e)} $$

$$\underbrace_{q_1^{(e)}} = \begin{bmatrix} l^{(e)} & m^{(e)} \\ \end{bmatrix}\begin{pmatrix} d^{(e)}_1 \\ d^{(e)}_2 \end{pmatrix}$$ {1}

$${\underset{4 \times 1}\mathbf \tilde {d^{(e)}_2}} = \begin{bmatrix} -m^{(e)} & l^{(e)} \\ \end{bmatrix}\begin{pmatrix} d^{(e)}_1 \\ d^{(e)}_2 \end{pmatrix}$$ {2}

$$\mathbf \tilde d^{(e)}_2 = \begin{bmatrix} -sin \theta^{(e)} & cos \theta^{(e)} \\ \end{bmatrix}\begin{pmatrix} d^{(e)}_1 \\ d^{(e)}_2 \end{pmatrix} = \begin{bmatrix} -m^{(e)} & l^{(e)} \\ \end{bmatrix}\begin{pmatrix} d^{(e)}_1 \\ d^{(e)}_2 \end{pmatrix}$$

Put {1} and {2} in matrix form:

$$\begin{pmatrix} \tilde d^{(e)}_1 \\ \tilde d^{(e)}_2 \end{pmatrix} = \underbrace{\begin{bmatrix} l^{(e)} & m^{(e)} \\ -m^{(e)} & l^{(e)} \\ \end{bmatrix}}_{\mathbf R^{(e)}}\begin{pmatrix} d^{(e)}_1 \\ d^{(e)}_2 \end{pmatrix}$$

$$\underset{4 \times 1}\tilde \mathbf d^{(e)}=\underset{4 \times 4}\mathbf \tilde T^{(e)}\underset{4 \times 1} \mathbf d^{(e)} \Rightarrow \underbrace{\begin{pmatrix} \tilde d^{(e)}_1 \\ \tilde d^{(e)}_2 \\ \tilde d^{(e)}_3 \\ \tilde d^{(e)}_4 \end{pmatrix}}_{\underset{4 \times 1}\tilde \mathbf d^{(e)}} = \underbrace{\begin{bmatrix} {\underset{2 \times 2}\mathbf R^{(e)}} & {\underset{2 \times 2}\mathbf 0^{(e)}} \\ {\underset{2 \times 2}\mathbf 0^{(e)}} & {\underset{2 \times 2}\mathbf R^{(e)}} \\ \end{bmatrix}}_{\underset{4 \times 4}\mathbf \tilde T^{(e)}}\underbrace{\begin{pmatrix} d^{(e)}_1 \\ d^{(e)}_2 \\ d^{(e)}_3 \\ d^{(e)}_4 \end{pmatrix}}_{\underset{4 \times 1} \mathbf d^{(e)}}$$

 * Shown in the figure below, the element is represented in matrix form below:

$$ \tilde \mathbf f^{(e)} = k^{(e)}\begin{bmatrix} 1 & 0 & -1 & 0 \\ 0 & 0 & 0 & 0 \\ -1 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \\ \end{bmatrix} \tilde \mathbf d^{(e)}$$

 * $${\underset{4 \times 1} \tilde \mathbf f^{(e)}} ={\underset{4 \times 4}\tilde \mathbf k^{(e)}}{\underset{4 \times 1} \tilde \mathbf d^{(e)}}$$

Meeting 20: Friday, 10 October 2008. EML4500
Note: consider case: $$\tilde{d}^{(e)}_4 \neq 0$$

$$ \tilde{d}^{(e)}_1 {=} \tilde{d}^{(e)}_2 {=} \tilde{d}^{(e)}_3 {=} 0$$

$$ \underline{\tilde{f}}^{(e)}_{4x1} {=} \underline{\tilde{k}}^{(e)}_{4x4} \cdot \underline{\tilde{d}}^{(e)}_{4x1} {=} \underline{0}_{4x1} \leftarrow 4^{th}$$ column of $$\underline{\tilde{k}}^{(e)}$$

Interpretation of trans. dofs

p. 19-3: $$ \underline{\tilde{d}}^{(e)} {=} \underline{\tilde{T}}^{(e)} \cdot \underline{d}^{(e)}$$

Similarly  $$ \underline{\tilde{f}}^{(e)} {=} \underline{\tilde{T}}^{(e)} \cdot \underline{f}^{(e)}$$

Also:  $$ \underline{\tilde{k}}^{(e)} \cdot \underline{\tilde{d}}^{(e)} {=} \underline{\tilde{f}}^{(e)} $$

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

If $$\underline{\tilde{T}}$$ is invertible, then:

$$ [\underline{\tilde{T}}^{(e)-1} \cdot \underline{\tilde{k}}^{(e)} \cdot \underline{\tilde{T}}^{(e)}] \cdot \underline{d}^{(e)} {=} \underline{f}^{(e)} $$

$$ \underline{\tilde{T}}^{(e)}$$ block diag. matrix (p. 19-3)

Consider a general block-diag. matrix:

Question: What is A-1

Simpler example: Diag. matrix:

$$ {=} diag[d_{11}, d_{22} , d_{33} , ... , d_{nn}] $$

$$ \underline{B}^{-1} = Diag[\frac{1}{d_{11}}, \frac{1}{d_{22}} , ... , \frac{1}{d_{nn}}]$$

Assuming $$ d_{ii} \neq 0 $$ for $$ i {=} 1, ... , n$$

For a block-diag matrix A:

$$ \underline{A} {=} Diag[\underline{D}_1, ... , \underline{D}_s]$$

$$ \underline{A}^{-1} {=} Diag[\underline{D}_1^{-1}, ... , \underline{D}^{-1}_{s}]$$

$$ \underline{\tilde{T}}^{(e)-1} {=} Diag[\underline{R}^{(e) -1}, ... , \underline{R}^{(e) -1}] $$

p. 19-2: $$ \underline{R}^{(e) T} {=} \begin{vmatrix} l^{(e)} & -m^{(e)} \\ m^{(e)} & l^{(e)} \end{vmatrix} $$

$$ \underline{R}^{(e)T}_{2x2} \cdot \underline{R}^{(e)}_{2x2} {=} \begin{vmatrix} 1 & 0 \\ 0 & 1 \end{vmatrix}_{2x2} {=} \underline{I}_{2x2} $$

$$ \rightarrow \underline{R}^{(e)-1} = \underline{R}^{(e)T} $$

$$ \rightarrow \underline{\tilde{T}}^{(e)-1} = Diag[\underline{R}^{(e)T}, \underline{R}^{(e)T}] $$

$$ {=} \underbrace{ (Diag[ \underline{R}^{(e)}, \underline{R}^{(e)}]) }_{\underline{\tilde{T}}^{(e)}} $$

$$ \underline{\tilde{T}}^{(e)-1} {=} \underline{\tilde{T}}^{(e)T} $$

p. 20-1: $$ \underbrace{ [\underline{\tilde{T}}^{(e)T} \cdot \underline{\tilde{k}}^{(e)} \cdot \underline{\tilde{T}}^{(e)}]}_{\underline{k}^{(e)}} \cdot \underline{d}^{(e)} {=} \underline{f}^{(e)} $$

HW: Verify: $$ \underline{k}^{(e)} {=} \underline{\tilde{T}}^{(e)T} \cdot \underline{\tilde{k}}^{(e)} \cdot \underline{\tilde{T}}^{(e)} $$

EML4500.f08.A-team.kirley 03:46, 21 October 2008 (UTC) 23-1

Meeting 21: Monday, 13 October 2008. EML4500
Eigenvalue problem: $$\mathbf{kv} = \lambda \mathbf{v}$$

Let $$\left\{\mathbf{u_1}, \mathbf{u_2}, \mathbf{u_3}, \mathbf{u_4} \right\}$$ be the pure eigenvectors corresponding to the 4 zero eigenvalues:

$$\mathbf{ku_i} = 0\mathbf{u_i} = \mathbf{0}\left(i = 1,2,3,4 \right)$$

Linear combination of $$\left\{u_i, i = 1,2,3,4 \right\} $$

$$\sum_{i=1}^{4}{\alpha_{i_{1x1}}\mathbf{u}_{i_{6x1}}=:\mathbf{W}_{_{6x1}}}=\mathbf{0}_{_{1x1}}\cdot \mathbf{W}_{_{6x1}}$$

Where $$=:$$ means "equal by definition"

$$\sum_{i=1}^{4}{\alpha_i\mathbf{u}_i} \equiv \mathbf{W} $$ (equal by definition)

$$ \alpha_i \Rightarrow $$(real numbers)

W is also an eigenvector corresponding to a zero eigenvalue:

$$ \mathbf{kW} = \mathbf{k}\left(\sum_{i=1}^{4}{\alpha_i\mathbf{u}_i} \right) $$

$$= \sum_{i=1}^{4}{\alpha_i\left(\mathbf{ku_i} \right)} = \mathbf{0} = 0 \times \mathbf{W}$$

Eml4500.f08.a-team.robinson 02:58, 24 October 2008 (UTC)

Meeting 22: Wednesday, 15 October 2008. EML4500
Justification of assembly of $$ \underbrace{ element \ stiffness \ matrix }_{\underline{k}^{(e)} } $$ into  $$ \underbrace{ global \ stiffness \ matrix }_{\underline{K}} $$

e = 1,...,n (# of elements)

Consider the example of a two-bar truss.

Recall element FD relation: k(e) d(e) = f(e)

p.11-3: Euler cut principle, method 2 (equilibrium of global node 2)

p.4-2, 4-3: FBDs of element 1 and element 2, element dofs. d(e)

p.8-3: For node 2, identify global dofs to element dofs for both element 1 and 2.

$$ \sum Fx = 0 $$ = -f3(1) - f1(2) = 0

$$ \sum Fy = 0 $$ = P - f4(1) - f2(2) = 0

Next, use element FD relation k(e) d(e) = f(e)

Meeting 23: Friday, 17 October 2008. EML4500
From page 22-2:
 * Equation (1) $$ \rightarrow f_3^{1} + f_1^(2) $$ = 0
 * Equation (2) $$ \rightarrow f_4^{1} + f_2^(2) $$ = P

(1): $$ \underbrace {k_{31}^{(1)} d_1^{(1)} + k_{32}^{(1)} d_2^{(1)} + k_{33}^{(1)} d_3^{(1)} + k_{34}^{(1)} d_4^{(1)}}_{f_3^{(1)}} $$ + $$ \underbrace {k_{11}^{(2)} d_1^{(2)} + k_{12}^{(2)} d_2^{(2)} + k_{13}^{(2)} d_3^{(2)} + k_{14}^{(2)} d_4^{(2)}}_{f_1^{(2)}} $$

For the conversion between global and local nodes please refer to page 8-3 and 8-4: e.g.... $$ k_{31}^{(1)} d_1 $$

From equation (1), using the third and fourth values from the first set of brackets and the first and second values from the second set of brackets you can obtain the third row of K as seen on page 9-1. For applications of K see page 8-4 for the equation:  $$ k_{6x6}d_{6x1} = F_{6x1} $$ 23-2

Assembly of k(e), e = 1,...., ned, into global stiffness matrix K:

$$ \underset {nxn}{\mathbf K} $$ = $$ \overset {ned} {\underset {e=1}{\mathbf A}} $$ $$ {\mathbf {k}}_{nel \ x \ nel}^{(e)} $$

where:
 * "n" is the total number of global degrees of freedom before the boundary conditions are eliminated.
 * "ned" is the number of element degrees of freedom where ned << n (where "<<" means "very small compared to).
 * "A" is the assembly operator

Principle of Virtual Work
The principle of virtual work is also known as $$ \mathbf{P} \mathbf{V} \mathbf{W} $$.

Page 10-1 states that to obtain $$ \overline {\mathbf{K}_{2 \ x \ 2}} $$, the corresponding rows of boundary conditions need to be eliminated. 23-3 Page 12-2: $$ \mathbf{q}_{2x1}^{(e)} $$ = $$ \mathbf{T}_{2x4}^{(e)} \ $$ $$ \mathbf{d}_{4x1}^{(e)} $$

 * $$ \mathbf{k}^{(e)} $$ = $$ \mathbf{T}^{(e)^{T}} \ $$ $$ \hat{\mathbf{k}}^{(e)} \ $$ $$ \mathbf{T}^{(e)} $$: which is from page 14-3

Deriving FEM for PDE's
Where:
 * FEM stands for Finite Element Method
 * PDE stands for Partial Differential Equation
 * FD stands for Force Displacement

The force displacement for a spring or bar is known as kd = F

 * $$ \Rightarrow kd - F = 0 \qquad $$ - (3)

 * $$ \ \Leftrightarrow w(kd - F) = 0 \qquad $$ - for all w  - (4)

Where:
 * "$$ \Rightarrow $$" stands for implies
 * "$$ \Leftrightarrow $$" stands for equivalent

Proof:

$$ \qquad (3) \ \Rightarrow \ (4) \qquad $$ - trivial

$$ \qquad (4) \ \Rightarrow \ (3) \qquad $$ - not trivial

Since (4) is valid for all w, we can select any arbitrary w (i.e. w = 1), and the (4) becomes:

 * $$ \qquad 1.(kd - F) = 0 \ \Rightarrow \ (3) $$

MatLab Problems:

Three Bar Truss System:

The program for the graph is as follows:

Five Bar Truss Evaluation:

Eigen plot of 2-bar truss:

Contributions
--EML4500.f08.A-team.rieth 18:31, 20 October 2008 (UTC)

--EML4500.f08.A-team.kirley 03:47, 21 October 2008 (UTC)

--Eml4500.f08.a-team.robinson 01:03, 24 October 2008 (

--EML4500.f08.A-team.melvin 18:12, 24 October 2008 (UTC)

--EML4500.f08.A-team.VandenBerg 20:28, 24 October 2008 (UTC)

--EML4500.f08.A-team.morford 20:33, 24 October 2008 (UTC)

title: User:EML4500.f08.A-team.VandenBerg/HW4


text: File:Amgraham.jpg

title: File:Amgraham.jpg


title: File:Conn1.jpg


title: File:Lmm1.jpg


text:
User:Stevenarntson/Freshman Seminar/Misology

Today in class I'll lecture on the broad subject of "taste" in aesthetics, and talk about two related terms: "Misology" and "Negative Capability."

Misology
This isn't a term much used in English, but I find it in my translation of the Phaedo. Here is a very loose modernization:

Socrates
 * But first, let us be careful.

Phaedo
 * Of what?

Socrates
 * Lest we become misologists, for nothing worse can happen to a person. As there are misanthropists (people haters), so there are misologists--idea haters. Both spring from the same cause, which is ignorance of the world.  Misanthropy arises out of the confidence of inexperience;--you trust someone because they seem trustworthy, but then they betray you. Then you trust someone else, and they betray you, too. Finally, you end up hating everyone, thinking that people as a whole are untrustworthy. You've probably seen this happen?

Phaedo
 * I have.

Socrates
 * It's unfortunate, because that person is inexperienced. If they took the time to meet a few more people, it would have become apparent that the world is filled with all kinds--some evil, some good, and most somewhere in the middle.

Phaedo
 * What do you mean?

Socrates
 * As with large and small things, most fall inbetween: great and tiny, swift and slow, fair and foul: in all cases, there are only a few extremes, and many in the middle.  Did you never observe this?

Phaedo
 * Yes, I have.

Socrates
 * Yes, and how melancholy, if there is indeed such a thing as truth or certainty or possibility of knowledge. Perhaps a person discovers some argument that seems true at first, but then seems false, and again, and again--perhaps they then decide that there's no such thing as truth, and close themselves off forever to the possibility of finding it.

Phaedo
 * Yes, that's very melancholy.

Socrates
 * Let us beware of it. Let us not believe that there is no such thing as truth. Rather say that we have not yet attained perfection ourselves, and so we must struggle towards it. And we must not be partisan in our approach. Some people take a side just because they think they can win; some people become good arguers just because they enjoy victory; and others are trying desperately to convince themselves of something. Therefore I ask this of you: take care for the truth, but not for Socrates. Agree with me if I seem to speak the truth; if not, then resist me fully, that I not deceive us both.

Negative Capability
Negative capability is a term coined by the poet John Keats in a letter he wrote to his brother. It is his attempt to explain to himself what it takes to be a great artist. Since he wrote it, many have expanded upon and appreciated its simplicity and power.

...several things dovetailed in my mind, & at once it struck me, what quality went to form a Man of Achievement especially in literature & which Shakespeare possessed so enormously - I mean Negative Capability, that is when man is capable of being in uncertainties, Mysteries, doubts without any irritable reaching after fact & reason.

Ellsworth Kelly
In class I'll tell a somewhat lengthy story about my own experience of thinking about aesthetics, which involves some details about the life and work of the painter Ellsworth Kelly. We'll look at some images in class, and have a discussion:

 * Blue, Green, Red II
 * Red Yellow Blue
 * Spectrum V
 * Blue Green Black Red
 * Kelly


text:
User:Gergana

Gergana Georgieva
Hi, I am from SU "Kliment Ohridski", Bulgaria. I am in my 5th semester.

research interests

 * teachers’ professionalism
 * Internet- and Project Competence

Theme of research
http://www.zum.de/Foren/internet/threads/thread419.php
 * Interaction between children and nature

Contacts

 * my e-mail: frany@dir.bg

title: User:Gergana


text:
Talk:Earth-impact events

What's needed?
I was thinking about this project today at work, and wondering what sorts of things the project might need as an initiative in the Think Tank. I was thinking maybe some feedback pages or surveys? For example, an "introduction circle" page where people first arriving on the project could say how much they know about NEOs, where they first learned about them, whether they believe we're capable of doing something to prevent impacts (and/or if an "Earth defense system" should be created), etc. It might also be interesting to have an "ideas section" where people can use their imaginations to come up with ways to prevent an impact, and have the "experts" give feedback on why an idea might or might not work. There have been a number of movies and books about the topic that could also be "fact-checked" as part of the project.

I'm happy to set up a feedback system if you have a list of questions. --SB_Johnny talk 19:14, 20 October 2008 (UTC)

 * Yes, feel free to start a feedback system, or jump in with other ideas. I'll draft a list of questions.  I've also started Earth-impact events/Discussion but that is just a stub at the moment.   --mikeu talk 20:17, 20 October 2008 (UTC)

Dead link
During several automated bot runs the following external link was found to be unavailable. Please check if the link is in fact down and fix or remove it in that case!

 * http://www.slooh.com/blog/2008-tc3-a-serendipitous-event/
 * In Earth-impact events on 2009-11-15 07:32:15, 404 Not Found
 * In Earth-impact events on 2009-11-22 07:32:20, 404 Not Found
 * In Earth-impact events on 2009-11-29 07:32:23, 404 Not Found
 * In Earth-impact events on 2009-12-06 07:32:23, 404 Not Found

--Mu301Bot 07:32, 6 December 2009 (UTC)

Removed broken link. --mikeu talk 14:08, 18 February 2010 (UTC)

Dead link 2
During several automated bot runs the following external link was found to be unavailable. Please check if the link is in fact down and fix or remove it in that case!

 * http://www.slooh.com/blog/suv-sized-asteroid-makes-surprise-pass-between-the-earth-moon/
 * In Earth-impact events on 2009-11-15 07:32:15, 404 Not Found
 * In Earth-impact events on 2009-11-22 07:32:20, 404 Not Found
 * In Earth-impact events on 2009-11-29 07:32:20, 404 Not Found
 * In Earth-impact events on 2009-12-06 07:32:24, 404 Not Found

--Mu301Bot 07:32, 6 December 2009 (UTC)

Removed broken link. --mikeu talk 14:08, 18 February 2010 (UTC)

Dead link 3
During several automated bot runs the following external link was found to be unavailable. Please check if the link is in fact down and fix or remove it in that case!

 * http://www.azstarnet.com/news/264185
 * In Earth-impact events on 2010-01-24 07:32:17, 404 Not Found
 * In Earth-impact events on 2010-01-31 07:32:16, 404 Not Found
 * In Earth-impact events on 2010-02-07 07:32:18, 404 Not Found
 * In Earth-impact events on 2010-02-14 07:32:18, 404 Not Found

--Mu301Bot 07:32, 14 February 2010 (UTC)

Removed broken link. --mikeu talk 14:08, 18 February 2010 (UTC)

Dead link 4
During several automated bot runs the following external link was found to be unavailable. Please check if the link is in fact down and fix or remove it in that case!

 * http://www.theaustralian.news.com.au/story/0,25197,23100798-30417,00.html
 * In Earth-impact events on 2010-03-21 06:32:22, 404 Not Found
 * In Earth-impact events on 2010-03-28 06:32:38, 404 Not Found
 * In Earth-impact events on 2010-04-11 06:32:34, 404 Not Found

--Mu301Bot 06:32, 11 April 2010 (UTC)

title: Talk:Earth-impact events


text:
Wikiversity:Editing disagreements

Practical options when editing material during a disagreement
When using, researching or creating content on Wikiversity it is inevitable that you will find material that you do not agree with.

while philosophical attitude is very important, we need to understand the possible options for dealing with this. There are 8 and they are listed roughly ordered from least to most aggresive.

 * 1) Topic:Editing disagreements/Defer Abandon creating material of that topic and discuss a related topic and/or abandon the topic entirely.
 * 2) Topic:Editing disagreements/Share Place your thoughts on the associated talk page and take action only once the author has agreed to it.
 * 3) Topic:Editing disagreements/Compromise Create a text which compromises between your different viewpoints. Post in the discussion page and wait for feedback.
 * 4) Topic:Editing disagreements/Convince Study the Authors ideas and offer a lengthy cited discussion on the talk page to convince him/her
 * 5) Topic:Editing disagreements/Rewrite Place your own thoughts or draft of the material on the talk page.  Either seek to get your own material posted OR try to get the author to revise his material based upon it.
 * 6) Topic:Editing disagreements/NewForum Create a New forum entirely with your own thoughts.  Relatedly, slightly alter the reported purpose or focus to allow for two conflicting accounts of the same topic.
 * 7) Topic:Editing disagreements/Edit, and alter the project page.  Add conflicting information; create disclaimers.
 * 8) Topic:Editing disagreements/Replace Entirely destroy the authors material, either writing over his thoughts with your own, deleting his material, and removing links to this authors thoughts.

one option is NOT right for all situation and there are several signficant factors.
 * 1) How abandoned is the work?
 * 2) Does your material completely conflict or is more of an update or addition to the authors work?
 * 3) How much consensus do you have the community?
 * 4) What is your expertise and what is the author's expertise?

Please feel free to discuss each of these points. I would like to achieve some consensus about each of the proposed actions.

a link back to the project is here

title: Wikiversity:Editing disagreements


text:
User:Stevenarntson/Analytical Writing/Popular Culture

this page is under construction

Notes on Camp Criticism quotes from Wilde


text:
Wikiversity:Editing Applied ethics

EDITING FOR CONSENSUS, an exercise in applied ethics
I propose that a group form to define what is appropriate behavior when a person, whom is creating content finds An author who gives conflicting information on a page.

Rather this person, called the editor has a variety of possible reactions, given in this page Topic:Editing disagreements

No one option of course in right for every circumstance. and it may be that each of those options may be given a situation, the correct course of action.

Would you visit each action, noting when it is the right thing to do and noting when it is not?

Thank you, comment on the talk page.

title: Wikiversity:Editing Applied ethics


text:
User:EML4500.f08.delta 6.ramirez/101708

Homework from 10/17/08 1. Complete f1(2)

Thus,

$$f_1^{(2)} = [k_{11}^{(2)}d_1^{(2)} + k_{12}^{(2)}d_2^{(2)} + k_{13}^{(2)}d_3^{(2)} + k_{14}^{(2)}d_4^{(2)}]$$

2. Derive the details of the following equation.

$$f_4^{(1)} + f_2^{(2)} = P$$

Where,

1. $$f_4^{(1)} = [k_{41}^{(1)}d_1^{(1)} + k_{42}^{(1)}d_2^{(1)} + k_{43}^{(1)}d_3^{(1)} + k_{44}^{(1)}d_4^{(1)}]$$

2. $$f_2^{(2)} = [k_{21}^{(2)}d_1^{(2)} + k_{22}^{(2)}d_2^{(2)} + k_{23}^{(2)}d_3^{(2)} + k_{24}^{(2)}d_4^{(2)}]$$

Thus,

$$[k_{41}^{(1)}d_1^{(1)} + k_{42}^{(1)}d_2^{(1)} + k_{43}^{(1)}d_3^{(1)} + k_{44}^{(1)}d_4^{(1)}]+ [k_{21}^{(2)}d_1^{(2)} + k_{22}^{(2)}d_2^{(2)} + k_{23}^{(2)}d_3^{(2)} + k_{24}^{(2)}d_4^{(2)}] = P$$

Observation: Last two items in 1. & first two items in 2. correspond to the 4th row in the global stiffness matrix K6x6.

Hence,

$$[(k_{43}^{(1)} + k_{21}^{(2)})(k_{44}^{(1)} + k_{22}^{(2)})]$$

title: User:EML4500.f08.delta 6.ramirez/101708


text:
User:BriannaBowden

The Girl Behind the Name:
Age: 26

Location: Seattle, WA

Hometown: San Antonio, TX

Interests: Art, roller derby, all forms of games (PC, console, board, etc.), watching movies, reading, almost all genres of music, trying new things, etc.

title: User:BriannaBowden


text:
Wikiversity talk:Editing Applied ethics

"The purpose of wikiversity is to create a learning exchange between participants and and is neither served by completely deferring nor replacing the material."
 * I disagree with this. While it frequently is not served by doing either of those things, I think there are times when either could be beneficial to Wikiversity.  The Jade Knight (d'viser) 21:57, 20 October 2008 (UTC)

 * Perhaps then not what I meant.
 * I could envision circumstances where each of the actions were the "right" thing to do.
 * Unfortunately, these two actions are extreme and don't normally foster the learning 'exchange' we would want to have in wikiversity.

 * I will edit it.
 * As always Jade, your careful insightful comments are very welcome.--Jolie 13:02, 21 October 2008 (UTC)

 * People can also learn from adding what they know or think they know and seeing it corrected. Different people learn from different learning styles. --dark[[Image:Yin yang.svg|12px]]lama 19:25, 21 October 2008 (UTC)

title: Wikiversity talk:Editing Applied ethics


text:
User talk:Eml4500.f08.FEABBQ/HW5

 * HW: 88

 * I'll probably have part 8 done before 5, I'm in Reitz working if you need me call.William Kurth 21:01, 7 November 2008 (UTC)
 * For the sake of people's sanity, I'd like to remind you all that (as long as my memory serves me correct), the notes portion of the hw is 80%, whereas all the matlab stuff combined is only a total of 20%. So I guess it may not be the end of the world if we don't get everything done.
 * I don't think it's really fair for Esmail and Andy and them to have to be doing this last minute when they've already done their parts earlier on. Let's try to do as much as we can.  I'm working at the reitz bowling alley...call me if you need anything.William Kurth 17:57, 7 November 2008 (UTC)
 * Billy, do number 8 on the outline I made. Jose and I will do numbers 2,3, and 4. Then Esmail, Andy, and Grant can do numbers 5,6,7. --Eml4500.f08.FEABBQ.vitt 15:59, 7 November 2008 (UTC)

 * Jose/Nik please let me know which part of the hw you want me to work on. Sooner rather than later please.  Call me at 239.410.6903 if any questions. ~Billy
 * Hey guys, I'm really sorry but for some reason I thought the hw was due next week. I sprained my neck last week and was therefore in bed for days, not able to move my neck.  I'm better now but haven't been to class in a while.  I had done notes for the friday right before the week of the test, so I figured I was ok as far as the class notes went.  I didn't expect he'd give so much extra homework.  I'll check again, but I don't remember seeing all these emails about the matlab parts of the homework.  I'm not 100% sure what's left to do, I'll try to cipher through what is up there now and see what I can work on.  Someone call me as soon as they see this to let me know what I should work on.  William Kurth 08:48, 5 November 2008 (UTC)

 * I can meet on Tuesday early since I have another meeting with my design group at 10am, ...Esmail

 * Do you want to meet up on Tuesday morning and knock it off then? I am free until 11:40 - Andy

 * So I just went over both emails Vu-Quoc sent out about HW 5 and unfortunately there is a lot of HW to do besides the notes. Therefore it won't be fair for me and Billy to do it all ourselves. I made an outline of the emails to save everyone the time of having to decipher them. Instead of typing it up I decided to just scan and upload it. We need the book for some of this HW, so I reserved it at the library but there's no way I will get it in time for this HW since there is one person in front of me on the reserve list. Therefore who all has the book so we can get the parts that require it done? Anyways, here's the outline if you want to have a look. The best way to see it is to download it and open with windows viewer and then zoom in a little.

Image:FEA_Outline_1.jpg

Image:FEA_Outline_2.jpg

Lets figure out who's doing what sooner rather than later. --Eml4500.f08.FEABBQ.vitt 17:22, 2 November 2008 (UTC)

 * HWk due Nov. 05

 * I Have posted the notes and the HWk for the 20th, for those who dont have days to cover, they might be a help with the matlab for this homework. as far as I know Jose will be working on it!!!. Esmail Hadman 14:09, 1 November 2008 (UTC)

 * Check this out http://www.codecogs.com/components/equationeditor/equationeditor.php.Stark 19:12, 29 October 2008 (UTC)

title: User talk:Eml4500.f08.FEABBQ/HW5


text:
User talk:BriannaBowden

Welcome
Hello BriannaBowden, 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
 * Take a guided tour and learn to edit;
 * Explore our learning projects;
 * Browse our portals, schools, and research activities;
 * [ Ask] a question, discuss Wikiversity, etc. at the Colloquium;
 * Read and help develop our community policies, or
 * Chat with other Wikiversians on #wikiversity-en.

And don't forget to explore Wikiversity with the links to your left. Be bold, and see you around Wikiversity! --Emesee 22:03, 20 October 2008 (UTC)

Drawing
Good luck on your project! Drawing might have some useful resources, as might Human Anatomy. The Jade Knight (d'viser) 10:25, 30 October 2008 (UTC)

title: User talk:BriannaBowden


text:
Talk:Advanced ANOVA

I think this page needs to be moved; the title implies that it's a generalist project on Research methods and ethics over a variety of fields. However, it is clear that it is limited entirely to the field of Psychology, and I don't know enough about Psychological methods to know whether or not it could be generally useful in psychology, or whether its utility is largely limited to the UCanberra course it was created for. The Jade Knight (d'viser) 22:05, 20 October 2008 (UTC)
 * Mind if we fix the links and put the redirect up for speedy delete? The Jade Knight (d'viser) 00:07, 3 December 2008 (UTC)
 * That's fine, JK, thanks for your help. -- Jtneill - Talk - c 05:05, 3 December 2008 (UTC)

title: Talk:Advanced ANOVA


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Topic talk:Art history

Is this meant to be a History department? The Jade Knight (d'viser) 23:13, 20 October 2008 (UTC)

title: Topic talk:Art history


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User:Uf.team.aero/HW5

= Stress - Strain Relation =

To obtain engineering stress results based on an assessment of the engineering strains, we analyze the previously given strain-stress relation equations represented in matrix form as:

\begin{Bmatrix} \epsilon _{xx}\\ \epsilon _{yy}\\ \epsilon _{zz}\\ \gamma _{yz}\\ \gamma _{zx}\\ \gamma _{xy}\\ \end{Bmatrix}=

\begin{bmatrix} \frac{1}{E}& \frac{-\nu }{E} & \frac{-\nu }{E} & 0 & 0 &0 \\ \frac{-\nu }{E}& \frac{1}{E} & \frac{-\nu }{E} & 0 &  0&0 \\ \frac{-\nu }{E}& \frac{-\nu }{E} &   \frac{1}{E}& 0 & 0 &0 \\ 0 & 0 & 0 & \frac{1}{G} & 0 & 0\\ 0 & 0 & 0 & 0 & \frac{1}{G}& 0\\ 0& 0 &0 &0 & 0 & \frac{1}{G} \end{bmatrix} \begin{Bmatrix} \sigma _{xx}\\ \sigma _{yy}\\ \sigma _{zz}\\ \sigma _{yz}\\ \sigma _{zx}\\ \sigma _{xy}\\ \end{Bmatrix} $$

The above matrices can be rearranged to solve directly for stress in terms of strain by inverting the 6x6 coefficient matrix. To do so, consider the 6x6 matrix as a composition of 3x3 matrices, with the upper-right and lower-left quadrants identical to the zero matrix. Such a composition could be represented as:

\begin{Bmatrix} \epsilon _{xx}\\ \epsilon _{yy}\\ \epsilon _{zz}\\ \gamma _{yz}\\ \gamma _{zx}\\ \gamma _{xy}\\ \end{Bmatrix}=

\begin{bmatrix} \bar{A} & \bar{0} \\ \bar{0} & \bar{B} \\ \end{bmatrix} \begin{Bmatrix} \sigma _{xx}\\ \sigma _{yy}\\ \sigma _{zz}\\ \sigma _{yz}\\ \sigma _{zx}\\ \sigma _{xy}\\ \end{Bmatrix} $$

where the matrices $$\bar{A}$$ and $$\bar{B}$$ are given as

\begin{Bmatrix} \bar{A}\\ \end{Bmatrix}=

\begin{bmatrix} \frac{1}{E}& \frac{-\nu }{E} & \frac{-\nu }{E} \\ \frac{-\nu }{E}& \frac{1}{E} & \frac{-\nu }{E} \\ \frac{-\nu }{E}& \frac{-\nu }{E} &  \frac{1}{E} \\ \end{bmatrix}

\begin{Bmatrix} \bar{B}\\ \end{Bmatrix}=

\begin{bmatrix} \frac{1}{G} & 0 & 0\\ 0 & \frac{1}{G}& 0\\ 0 & 0 & \frac{1}{G} \end{bmatrix} $$

Then, the stress-strain relation can be determined by inverting the coefficient matrix.

\begin{Bmatrix} \sigma _{xx}\\ \sigma _{yy}\\ \sigma _{zz}\\ \sigma _{yz}\\ \sigma _{zx}\\ \sigma _{xy}\\ \end{Bmatrix}=

\begin{bmatrix} \bar{A}^{-1} & \bar{0} \\ \bar{0} & \bar{B}^{-1} \\ \end{bmatrix} \begin{Bmatrix} \epsilon _{xx}\\ \epsilon _{yy}\\ \epsilon _{zz}\\ \gamma _{yz}\\ \gamma _{zx}\\ \gamma _{xy}\\ \end{Bmatrix} $$

This solution is possible because of the structure of the coefficient matrix. The inversion of the coefficient matrix is similar to that of a diagonal matrix, whose inverse is solved by inverting each component of the diagonal (assuming nonzero components) while leaving the zero terms off-diagonal ignored. A confirmation of this solution can be found by premultiplying the $$ \bar{A} $$ and $$ \bar{B} $$ matrices by their inverses within the coefficient matrix and obtaining the Identity Matrix, $$ \bar{I} $$.

\begin{bmatrix} \bar{A}^{-1}\bar{A} & \bar{0} \\ \bar{0} & \bar{B}^{-1}\bar{B} \\ \end{bmatrix}=

\begin{Bmatrix} \bar{I}\\ \end{Bmatrix} $$

It is seen by this solution, since $$ \bar{B} $$ is a diagonal matrix, that for strains $$ \epsilon_{xx}, \epsilon_{yy}, \epsilon_{zz}, \gamma_{yz} $$, the corresponding stress components (i.e. $$ \sigma_{xx}, \sigma_{yy}, \sigma_{zz}, \sigma_{yz} $$) are identically zero. Thus the stress-strain relation reduces to

\begin{Bmatrix} \sigma _{zx}\\ \sigma _{xy}\\ \end{Bmatrix}=

\begin{bmatrix} G & 0 \\ 0 & G \\ \end{bmatrix} \begin{Bmatrix} \gamma _{zx}\\ \gamma _{xy}\\ \end{Bmatrix} $$

= Equilibrium Equation for Stresses in a Non-uniform Stress Field =

Consider a 1-D case as a model.

It is apparent from the uniform case that the axial loading is constant along the length of the member. Conversely, in the non-uniform case, the axial loading is non-constant along the length of the member. This results in a non-uniform stress and strain response of the member.

= Bidirectional Bending Recipe = The image below depicts a non-uniform bar under bidirectional bending

Looking at Figure 2, for this case, the bending moments about the y and z axis, respectively, can be expressed as follows:

 * $$M_y = \int_A z\sigma_{xx}\, dA \, \, =\, \int\int z \sigma_{xx} dydz$$

 * $$M_z = \int_A y\sigma_{xx}\, dA \, \, =\, \int\int y \sigma_{xx} dydz$$

where $$\sigma_{xx}$$ is the axial stress due to the bending moments about the y and z axes at the point (x,y).

Likewise, the moment of inertia tensor - $$I_y, I_z, I_{yz}$$, (which can be equivalently denoted as $$ I_{yy}, I_{zz}, I_{xy}$$ or $$I_{22}, I_{33}, I_{23}$$, respectively) - are defined by the expressions

 * $$\qquad I_y=\int_A z^2dA$$

 * $$\qquad I_z=\int_A y^2dA$$

 * $$\qquad I_{yz}= \int_A yzdA$$

Solving these two groups of expressions yields an expression for the axial due to the bending moments about y and z as

 * $$\sigma_{xx}=E\epsilon_{xx}=\frac{I_yM_z-I_{yz}M_y}{I_yI_z-(I_{yz})^2}y+\frac{I_zM_y-I_{yz}M_z}{I_yI_z-(I_{yz})^2}z$$

If one recalls that

 * $$\underline{I} = \begin{bmatrix} I_{11} & I_{12} & I_{13} \\ & I_{22} & I_{23}\\sym.& & I_{33} \end{bmatrix}$$

One can see that denominator of the $$\sigma_{xx}$$ expression is numerically the determinate of the lower right quadrant of the $$\underline{I}$$ matrix. Also, the symmetry of the $$\underline{I}$$ matrix can be seen by comparing the $$ I_{23}$$ and $$I_{32}$$ terms of the matrix.

 * $$I_{23} = I_{yz} = \int_A yz\ dA$$

 * $$I_{32} = I_{zy} = \int_A zy\ dA$$

since

 * $$\int_A yz\ dA = \int_A zy\ dA$$

 * $$\Rightarrow I_{23} = I_{32}$$

The neutral axis is an axis along which the axial stress due to bending moments is zero, i.e.

 * $$\qquad \sigma_{xx}=0$$

For the case when the y or z axis is an axis of symmetry, and hence $$ I_{yz} = 0$$ the above expression for the axial stress can be reduced to

 * $$\sigma_{xx} = M_yy+M_zz = 0$$

an expression for the neutral axis can be taken from this by solving for z in terms of y:

 * $$z=\left (\frac{-M_y}{M_z} \right ) y = \tan\beta\ y$$

where $$\beta$$ is the angle between the neutral and y axes as shown below in Figure 3.

= Equations of Equilibrium in Terms of $$\sigma$$=

The goal is to obtain an equilibrium equation such that $$ \frac{\partial \sigma_{yx}}{\partial y} + \frac{\partial \sigma_{zx}}{\partial z} = 0 $$, or using indicial notataion: $$ \frac{\partial \sigma_{21}}{\partial x_{2}} + \frac{\partial \sigma_{31}}{\partial x_{3}} = 0 $$.

Recall from previous material [Section 2.4, Sun],

 * $$ \frac{\partial \sigma_{xx}}{\partial x} + \frac{\partial \sigma_{yx}}{\partial y} + \frac{\partial \sigma_{zx}}{\partial z}= 0 $$

 * $$ \frac{\partial \sigma_{yy}}{\partial y} + \frac{\partial \sigma_{zy}}{\partial z} + \frac{\partial \sigma_{xy}}{\partial x}= 0 $$

 * $$ \frac{\partial \sigma_{zz}}{\partial z} + \frac{\partial \sigma_{xz}}{\partial x} + \frac{\partial \sigma_{yz}}{\partial y}= 0 $$

It is seen that a cyclic permutation is possible such that

 * $$ \sum_{i=1}^{3}{\frac{\partial \sigma_{ij}}{\partial x_i}} = 0 $$

This notation is particularly powerful. So powerful in fact, that Einstein used it to formulate the Theory of Relativity.

= Using Taylor Series With Infinitesimal Elements = For the purpose of simplicity, we look at the derivation of equation (4) on pg 27-2 for the one-dimensional model first. Looking at the equilibrium of the infinitesimal figure:

$$\sum\ F_x = 0 = -\sigma_{(x)}\ A + \sigma_{(x + dx)}\ A + f_{(x)} dx\,$$

$$ 0 = A [ \sigma_{(x + dx)}\ - \sigma_{(x)}\ ] + f_{(x)} dx\,$$

Next, we apply Taylor series to the term, $$\sigma_{(x + dx)}\ - \sigma_{(x)}$$:

$$\frac{\partial\sigma_{x}}{\partial\ x} dx + higher~order~terms\,$$

Knowing that the general Taylor Series Approximation for the first three series of terms is given as:

$$F_{(x + dx)} = F_{(x)} + \frac{\partial\ F_{(x)}}{\partial\ dx} dx + \frac{1}{2} \frac{\partial ^2 F_{(x)}}{\partial x^2} (dx)^2 + .~.~.~higher~order~terms\,$$

The higher order terms are neglected and yields the final equation of interest shown below:

$$\frac{\partial\sigma}{\partial\ x} + \frac{f_{(x)}}{A} = 0\,$$

 * where $$f_(x)$$ is a $$\frac{force}{length}\,$$ and has dimensional units of $$\frac{N}{m}\,$$

In the equation above, the applied load $$\frac{f_{(x)}}{A}\,$$ is a Body Force and has dimensional units of $$\frac{N}{m^3}\,$$

Now looking at the nonuniform stress field in 3D and without any applied load, we focus on the x-direction in an attempt to view a better representation of the stress componenents without any cluttering.

In Figure 4 above shows an infinitesimal elemental cube where the stress is not completely uniform. Since the infinitesimal cube is in equilibrium, summing the forces of the x-direction is observed to be the sum of stress components of each facet area multiplied by the respective infinitesimal area. Therefore, the sum of the forces in the x-direction is:

$$\sum\ F_x = 0 = dy dz [ -\sigma_{xx}\ (x,y,z) + \sigma_{xx}\ (x+dx,y,z)]~+~ dz dx [ -\sigma_{yx}\ (x,y,z) + \sigma_{yx}\ (x,y+dy,z)]~+~dx dy [ -\sigma_{zx}\ (x,y,z) + \sigma_{zx}\ (x,y,z+dz)]\,$$

Here the terms can be separated and rewritten in terms of their respective Facets as:

Facets with normal X- $$ dy dz [ -\sigma_{xx}\ (x,y,z) + \sigma_{xx}\ (x+dx,y,z)]\,$$

Facets with normal Y- $$ dz dx [ -\sigma_{yx}\ (x,y,z) + \sigma_{yx}\ (x,y+dy,z)]\,$$

Facets with normal Z- $$ dx dy [ -\sigma_{zx}\ (x,y,z) + \sigma_{zx}\ (x,y,z+dz)]\,$$

The next step after is to take the Taylor Series expansion while neglecting the higher order terms.

Combining the terms from all facets, we derive the following equation:

$$0 = (dx dy dz) \left [ \frac{\partial\sigma_{xx}}{\partial\ x} + \frac{\partial\sigma_{yx}}{\partial\ y} + \frac{\partial\sigma_{zx}}{\partial\ z} \right ]\,$$

= Stringers =

In flight, an aircraft wing is subject to bending. This bending moment must be absorbed and transferred by some structure. It is known that the wing skin, because of its relative thinness, is prone to buckling when subjected to compressive loading. The spars, due to a small moment of inertia, are also relatively ineffective at carrying bending loads. Thus, it is desired to add an additional element to the wing structure.

Stringers are placed at the junction of the skin and spars in "nearly" L-shaped brackets. The stringers run the spanwise length of the wing along with the spars. By placing the stringers at the top and bottom of the wing, their moment of inertia about the wing centroid is maximized, allowing for a greater bending moment to be applied to the wing. The figure below provides a generic schematic of stringer placement for one spar - skin joint location.

= NACA Airfoil Matlab Project Continued =

The purpose of this project was to take the previously created Matlab code and add the option to include stringers to the modeled NACA airfoil. Results will show that it is sufficient to consider only the stringers when analyzing the bending moment effects on a wing.

= Contributing team members =

 * Jared Lee--Eas4200c.f08.aero.lee 18:05, 7 November 2008 (UTC)
 * Oliver Oyama--Eas4200c.fo8.aero.oyama 18:55, 7 November 2008 (UTC)
 * Gonzalo Barcia--Eas4200c.f08.aero.barcia 19:25, 7 November 2008 (UTC)
 * William Kurth--Eas4200c.f08.aero.kurth 20:37, 7 November 2008 (UTC)
 * Ray Strods-- Eas4200c.f08.aero.strods 21:01, 7 November 2008 (UTC)

title: User:Uf.team.aero/HW5


text:
User talk:S&F Publishing

Welcome
Hello S&F Publishing, 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
 * Take a guided tour and learn to edit;
 * Explore our learning projects;
 * Browse our portals, schools, and research activities;
 * [ Ask] a question, discuss Wikiversity, etc. at the Colloquium;
 * Read and help develop our community policies, or
 * Chat with other Wikiversians on #wikiversity-en.

And don't forget to explore Wikiversity with the links to your left. Be bold, and see you around Wikiversity! --Emesee 01:04, 21 October 2008 (UTC)

title: User talk:S&F Publishing


text:
Time breach

Time Breach is a communitarian project for a massively multi-player online game. This game is based on Open Source software.

(Time Breach Project Web Site)

Project presentation
The main idea of the project is to create a MMOG (Massively Multiplayer On-line Game) free game. It runs like that :

 * Each player can create his own character and make it evolve in the time Breach concept (usual feature)
 * Each player can create a world in the Time Breach universe (usually done by the project development team)

This principles already exist in some video games like Neverwinter Nights, which allows the player to create his own worlds. Unfortunately, the worlds in this game haven’t relation between them.

To be part of the MMOG game category, all the servers need to have a coherence between the rules, the story and the game events in their globality.

A central server is thus necessary to unify the worlds from the ones to the worlds of the others and so create an harmony to present one and only one universe to the player. We have wished to create a project that doesn’t and won’t need money. The only central server type which is easily and freely hostable on the net is actually a web site.

Our goal is to propose a game with 3 modes :

 * Playing all together (classic MMOG)
 * Play a character and make it evolve (Role Playing Game)
 * Playing and making business strategy inside his own world (Real Time Strategy)

In the beginning...
In the beginning, the authors wanted to create a free MMOG (i.e. a free World of Warcraft). They contacted people around them to be part of the adventure. Eight people said there were interested and the first appointment took place in June 2005.

The game universe
The specifications of the game universe are essential and sensitive : How to create a game universe that everybody will like ? One solution is the time breach ! This way, players that play with lasers and robots evolves aside with magus, barbarian and monsters without problem.

How to integrate the player in the heart of the story ? Some idea proposal systems with polling exist but the player will not find back his customization in this kind of management. Of course, it must be taken into consideration. One solution is that the players host themselves their worlds. By creating a common centralized library of monsters, building, game elements, … the player uses a common universe and may build his world staying in the universe harmony.

Technical organization
The technical structure of this project must take in account that :
 * It’s a free and open source project (development progress will not be uniform)
 * All the members wanted to be active in the project as soon as possible
 * Team coordination is difficult due to the distance between the project members

In those conditions, the project must be structured in blocks. Each block must respond to one specification and provide an user’s guide. This type of development is quite classic. But, general blocks can’t be written before the basics blocks are defined. Defining a global architecture with a content standard organization of all blocks allow to start the development of all blocks in the same time. Now, for the Time Breach project, the general organization is ready. However, to set up a good block standardization, you need a good knowledge of the used libraries. The goal of video game development is not to reinvent the wheel but to build a given universe. The time breach system is like Russian headstocks : boxes that contains basic libraries : - Central server (to centralize and to coordinate the game universe) - World Server (to host a world) * Graphic engine * Video game Engine * Network engine - Client (to play a character in a world) * Sound engine * Graphic engine * Video game Engine * Network engine - Editor (to create a world) * Sound engine * Graphic engine * Video game Engine * Network engine So we find common blocks in each block. Each of those blocks (called engine) contains themselves other blocks : Game engine *Audiere Graphic engine *Ogre 3D Video game engine *Artificial Intelligence *Script interpreter *Physical engine(surely Open Dynamic Engine) *Common repository of all blocks Network engine * Not yet defined. Using Jabber is evaluated. Some middleware also exists which groups all the elements like massiv allowing to be concentrated on the game only.

The open source tools
An open source project doesn’t have generally the money to buy proprietary and expensive tools. It’s the case with Time Breach. Here are the tools that Time Breach uses for his development
 * blender : 3D creation of mesh
 * gimp 2D creation : texture, images, etc.
 * Eclipse IDE : for programming using Ant and the CDT module
 * Concurrent versions system to organize the development work of the team
 * Xoops as the main communication tool . It’s a CMS
 * Mailing list in case of Xoops failure

Social
The main problems of a community : - There is no obligation to participate to the project - Everybody can left the project when he wants - Everybody work in the project at his own rhythm - There is no constraints if you fail to complete you work In an open source project, you must take in account those problems.

One of the solutions is to foreseen a substitute to the main jobs for the following reasons :
 * If somebody left his job, the substitute can educate a new one.
 * Decisions taken by 2 persons are always better thought
 * If somebody is unavailable, the other can replace him (alternate work)
 * Working together is highly stimulating
 * Work can be done more rapidly with 2 persons (parallel work)

Only the main jobs are fixed (in fact the managers of each project section) : Only tasks limited in duration are to be assigned to the project members This set of rules allows the Time Breach project to keep the stimulation at a high degree and to have a continuous work flow. By the way, Time Breach propose to realize supervised and evaluated projects to schools like IUT of Montpellier (France). This type of partnership allows to speed up the project. Speeding up the project shows new actions. New actions imply a better stimulation. A better stimulation give a advanced product and so on.
 * This lets the freedom to the members to participate according to their free time
 * This allows the manager to select the teams for each work
 * if a member hasn’t do his work in time, the manager can assigned it to someone else.

License
- Artist would that they productions been recognized like artistic productions and not a simple modifiable content. - linux communities are based around GNU GPL that don't recognized arts productions - Propose Time Breach to Linux distributions need to have a license friendly GNU GPL.
 * Choice of license is very important for this reason:

- A Creative Common By Sa 2.5 license for all arts productions (Texts, stories, sounds, graphics etc...) - A GNU GPL license for all programs ans sources code.
 * According to this elements Time Breach use a double license:

This choice able to recognize artist productions and to be compatible with free community.

title: Time breach


text:
Talk:Time breach

* (cur) (last) 01:41, 20 October 2008 Garrett (Talk | contribs) (7,355 bytes) (nominated for deletion) (undo) * (cur) (last) 20:24, 27 July 2008 Dallas1278 (Talk | contribs) m (7,347 bytes) (undo) * (cur) (last) 05:46, 28 June 2006 Derbeth (Talk | contribs) m (Reverted edit of 193.226.19.84, changed back to last version by Kalden) (undo) * (cur) (last) 04:38, 28 June 2006 193.226.19.84 (Talk) (→The game universe) (undo) * (cur) (last) 14:41, 10 June 2006 Kalden (Talk | contribs) (undo) * (cur) (last) 15:59, 2 May 2006 Hagindaz (Talk | contribs) m (Remove cat) (undo) * (cur) (last) 21:40, 22 April 2006 Kalden (Talk | contribs) (undo) * (cur) (last) 21:40, 22 April 2006 Kalden (Talk | contribs) (undo) * (cur) (last) 21:32, 22 April 2006 Kalden (Talk | contribs) (undo) * (cur) (last) 21:30, 22 April 2006 Kalden (Talk | contribs) (undo) * (cur) (last) 21:30, 22 April 2006 Kalden (Talk | contribs) (links corrected) (undo) * (cur) (last) 19:30, 13 April 2006 Hagindaz (Talk | contribs) m (+ ) (undo) * (cur) (last) 07:16, 12 March 2006 83.154.155.227 (Talk) (undo) * (cur) (last) 06:38, 21 February 2006 Kalden (Talk | contribs) (Presentation of Time Breach project (a way to create a massive open source game))
 * (cur) (last) 01:42, 20 October 2008 Garrett (Talk | contribs) m (7,338 bytes) (rm template--only used here and serves no real purpose (the link it contains is now dead)) (undo)

title: Talk:Time breach


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User:Eml4500.f08.gravy.mmm/HW4

21
Closing the loop, infinitesimal displacement, eigenvalues problems, zero eigenvalues, rigid body mechanisms.

HW: Plot the eigenvectors corresponding to zero eigenvalue of 2. bar truss system. - Brian
Eval pb. $$\ \mathbf{kv} = \boldsymbol{\lambda}  \mathbf{v}\,$$

Let {u1,u2,u3,u4} be the pure eigenvalues corresponding to four zero eigenvalues.

$$\ \mathbf{k}_{6x6} \mathbf{u}_i =0 \cdot \mathbf{u}_i \,$$, i=1,2,3,4

where $$ 0 \cdot u_i $$ is the zero vector of size 6x1.

Linear combination of {u_i, i=1,2,3,4}

$$ \sum_{i=1}^4 \alpha_i u_i =: \mathbf{W}_{6x1}$$

=: means equal by definition and $$\alpha$$ is a real number so it is 1x1 and u_1 is 6x1.

W is also an eigenvector corresponding to a zero eigenvalue:

$$\ \mathbf{KW}= \mathbf{K}( \sum_{i=1}^4 \boldsymbol{\alpha}_i \mathbf{W}_i ) =  \sum_{i=1}^4 \boldsymbol{\alpha}_i (\mathbf{Ku}_i) \,$$ where $$(\mathbf{Ku}_i)$$ is a zero matriz of size 6x6. $$\ =\mathbf{0}_{6x1} = 0 \cdot \mathbf{W}_{6x1} \,$$

HW: matlab- Sean

$$\ \mathbf{kv} = \boldsymbol{\lambda}  \,$$

Plot evect. evct. corresp. zero eval for case a use matlab.

a=b=1, E=2,  A=3

22
Justication of assembly of element stiffness matrix $$k^{(e)} e=1, number of elements$$ into global stiffness matrix K

consider example of 2 bar truss from prevous lecture.

Recall element force displacement relation

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

from prevous lecture: Euler cut principal method 2 (equilbrium of global node 2) Euler cut principal method

from prevous report: Free body digram of element 1 and element 2 element degress of freedom d(e) Trusses, matrix method

from prevous lecture: for node 2, identify global degrees of freedom to element degrees of freedom for both element 1 and 2. Global_F-D_Relationship

$$\ \sum F_x= 0-f_3^{(1)}- f_1^{(2)}= 0 \,$$  Equation (1)

$$ \sum F_y= P-f_4^{(1)}- f_2^{(2)}= 0$$  Equation (2)

Next use element Force Displacement relation $$\mathbf{k}^{(e)} \mathbf{d}^{(e)}=\mathbf{f}^{(e)}$$

23
Equation 1 becomes $$\ f_3^{(1)} + f_1^{(2)}=0 \,$$

Equation 2 becomes $$\ f_4^{(1)} + f_2^{(2)}=P \,$$

So substituting for the fs in equation 1 gives

$$\ \underbrace{[ k_{31}^{(1)} d_{1}^{(1)} + k_{32}^{(1)} d_{2}^{(1)} + k_{33}^{(1)} d_{3}^{(1)} + k_{34}^{(1)} d_{4}^{(1)} ]}_{f_3^{(1)}} + \underbrace{[k_{11}^{(2)} d_{1}^{(2)} + k_{12}^{(2)} d_{2}^{(2)} + k_{13}^{(2)} d_{3}^{(2)}+ k_{14}^{(2)} d_{4}^{(2)}]}_{f_1^{(2)}}=0 \,$$

Replacing the local ds with the global ds as explained Global_F-D_Relationship You end up with

$$[ k_{31}^{(1)} d_{1} + k_{32}^{(1)} d_{2} + k_{33}^{(1)} d_{3} + k_{34}^{(1)} d_{4} ]

+ [k_{11}^{(2)} d_{3} + k_{12}^{(2)} d_{4} + k_{13}^{(2)} d_{5}+ k_{14}^{(2)} d_{6}  ]=0 $$

Which is the 3rd row of the K matrix in Kd=F

Hw. Derive details of Eq 2 on 23.1 and the assembly of row 4. - Patrick
The assembly of K(e) e=1 to the number of elements, into the global stiffness matrix K can be represented as

$$\ K_{nxn}=Ak^{(e)}_{mxm}\,$$

Where A is the Assembly operator, n is the total number of global degrees of freedom, m is the number of element degrees of freedom and m is much less than n. Which is the principal of virtual work insert link here

Elimination of rows corresponding to the bountary conitions lets you obtain $$\mathbf{\bar{q}}_{2x2}$$

Remeber that:

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

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

Deriving FEM of Partial differential Equations

Force Displacement relation of a bar or spring: Kd=F implies that Kd-F=0  Equation 3

which is equivalent to w(Kd-F)=0 for all w  Equation 4

Equation 4 is a "weak form" which is Principal of virtual work

Prove:

A) Equation 3 implies equation 4 which is a trivial conclusion

B) Equation 4 implies equation 3 which is not a trivial conclusion

Since equation 4 is valid for all w, first select w=1 and equation 4 becomes 1(Kd-F)=0 which is equation 3

title: User:Eml4500.f08.gravy.mmm/HW4


text:
User:Eml4500.f08.wiki1.aguilar/hw4

Continued Assembly Instructions for a Three-Bar Truss
To continue, we're going to need to be familiar with a "connectivity array" and a "location matrix master array."

A Connectivity array, "conn" is shown below. (Consider the two-bar truss system from your class notes, p.5-6)

Conn(e,j) = global node number of local node j of element e

We also want to establish a Location Matrix Master array: "lmm".

lmm (i,j) = equation number (global degree of freedom number) for element stiffness coefficient corresponding to the jth local degree of freedom number.

Going back to p.12-1, Method 2 to derive $$\mathbf{k}^{(e)}_{4x4} $$

Finding a $$\tilde {\mathbf{T}}^{(e)}$$ that is Invertible
Goal: Our goal is to find a $$\mathbf{\tilde{T}}^{(e)}_{4x4}$$ that can transform the set of local element degrees of freedom, $$\mathbf{d}^{(e)}_{4x1} $$ to another set of local element degrees of freedom, $$\mathbf{\tilde{d}}^{(e)}_{4x1} $$, such that $$\mathbf{\tilde{T}}^{(e)}_{4x4}$$ is invertible.

$${\tilde{d}^{(e)}}_{1}= \begin{bmatrix} l^{(e)} &m^{(e)} \end{bmatrix}\begin{Bmatrix} d^{(e)}_{1}\\ d^{(e)}_{2}\end{Bmatrix}(1)$$

We illustrate this below:

$${\tilde{d}^{(e)}}_{1}={\vec{d}^{(e)}}_{[1]}\cdot \vec{\tilde{j}}$$ This is the component of vector $${\vec{d}^{(e)}}_{[1]}$$ along $$\vec{\tilde{j}}$$, i.e. $$\tilde{y}$$ axis. $${\tilde{d}^{(e)}}_{1}=-\sin \theta ^{(e)} d^{(e)}_{1}+\cos\theta ^{(e)} d^{(e)}_{2}$$

This is also done for $${\tilde{d}^{(e)}}_{2}$$. $${\tilde{d}^{(e)}}_{2}= \begin{bmatrix} l^{(e)} &m^{(e)} \end{bmatrix}\begin{Bmatrix} d^{(e)}_{1}\\ d^{(e)}_{2}\end{Bmatrix}(2)$$

$${\tilde{d}^{(e)}}_{2}={\vec{d}^{(e)}}_{[2]}\cdot \vec{\tilde{j}}$$ This is the component of vector $${\vec{d}^{(e)}}_{[2]}$$ along $$\vec{\tilde{j}}$$, i.e. $$\tilde{y}$$ axis. $${\tilde{d}^{(e)}}_{2}=-\sin \theta ^{(e)} d^{(e)}_{1}+\cos\theta ^{(e)} d^{(e)}_{2}$$

If we put equations (1) and (2) in matrix form:

 $$ \begin{Bmatrix}{\tilde{d}^{(e)}}_{1} \\ {\tilde{d}^{(e)}}_{2} \\ {\tilde{d}^{(e)}}_{3} \\ {\tilde{d}^{(e)}}_{4}\end{Bmatrix}= \begin{bmatrix} \mathbf{R}^{(e)}_{2x2} & 0 \\ 0 & \mathbf{R}^{(e)}_{2x2} \end{bmatrix} \begin{Bmatrix}d^{(e)}_{1} \\ d^{(e)}_{2} \\ d^{(e)}_{3} \\ d^{(e)}_{4}\end{Bmatrix} $$

$${\tilde{\mathbf{d}}^{(e)}}_{4x1}={\tilde{\mathbf{T}}^{(e)}}_{4x4} {\mathbf{d}}^{(e)}_{4x1}$$

$$\mathbf{\tilde f}^{(e)}=k^{(e)}\begin{bmatrix} 1 & 0 & -1 & 0\\ 0 & 0 & 0 & 0\\ -1 & 0 & 1 & 0\\ 0 & 0 & 0 & 0 \end{bmatrix}\mathbf{\tilde d}^{(e)}$$ Note: the zeroes arise because there is no stretching in the transverse direction.

$$\mathbf{\tilde f}^{(e)}_{4x1}=\mathbf{\tilde k}^{(e)}_{4x4} \mathbf{\tilde d}^{(e)}_{4x1}$$

Note: Consider the case: $${\tilde{d}^{(e)}}_{4}\neq 0$$ $${\tilde{d}^{(e)}}_{1}={\tilde{d}^{(e)}}_{2}={\tilde{d}^{(e)}}_{3}=0$$ $$\mathbf{\tilde f}^{(e)}_{4x1}=\mathbf{\tilde k}^{(e)}_{4x4} \mathbf{\tilde d}^{(e)}_{4x1}=\mathbf{0}$$ Here, $$\mathbf{0}_{4x1}$$ corresponds to the 4th column of $$\mathbf{\tilde k}^{(e)}$$ This is the interpolation of transverse degrees of freedom.

From our notes on p.19-3, we recall that:

$${\tilde{\mathbf{d}}^{(e)}}={\tilde{\mathbf{T}}^{(e)}} {\mathbf{d}}^{(e)}$$

We use the same argument for:

$${\tilde{\mathbf{f}}^{(e)}}={\tilde{\mathbf{T}}^{(e)}} {\mathbf{f}}^{(e)}$$

Also:

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

Which leads to:

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

If $${\tilde{\mathbf{T}}^{(e)}}$$ is invertible, then:

$$[{\tilde{\mathbf{T}}^{(e)^{-1}}}{\tilde{\mathbf{k}}^{(e)}}{\tilde{\mathbf{T}}^{(e)}}]{\mathbf{d}}^{(e)}={\mathbf{f}}^{(e)}$$

$${\tilde{\mathbf{T}}^{(e)}}$$ is a block diagonal matrix, as you can see by referring back to your notes on p. 19-3.

Consider a general block diagonal matrix:

Q: What is $$\mathbf{A}^{-1}$$?

Simple example:

$$\mathbf{B}=Diag[d_{11},d_{22},...,d_{nn}]$$

So,
$$\mathbf{B}^{-1}=Diag[\frac{1}{d_{11}},\frac{1}{d_{22}},...,\frac{1}{d_{nn}}]$$ Assuming $$d_{ii}\neq 0$$ for i=1,...,n.

For a block diagonal matrix A:

$$\mathbf{A}=Diag[\mathbf{D}_{1},\mathbf{D}_{2},...,\mathbf{D}_{s}]$$

So,
$$\mathbf{A}^{-1}=Diag[\mathbf{D}_{1}^{-1},\mathbf{D}_{2}^{-1},...,\mathbf{D}_{s}^{-1}]$$

$$\tilde{\mathbf{T}}^{(e)^{-1}} = Diag[\mathbf{R}^{(e)^{-1}} ,\mathbf{R}^{(e)^{-1}}  ]$$ And from p19-2,
$$\tilde{\mathbf{R}}^{(e)^{T}}=\begin{bmatrix}l^{(e)} &-m^{(e)} \\ m^{(e)} &l^{(e)} \end{bmatrix}$$

And,
$$\tilde{\mathbf{R}}^{(e)^{T}}_{2x2}\tilde{\mathbf{R}}^{(e)}_{2x2}=\begin{bmatrix}l^{(e)} &-m^{(e)} \\ m^{(e)} &l^{(e)} \end{bmatrix}\begin{bmatrix}l^{(e)} &m^{(e)} \\ -m^{(e)} &l^{(e)} \end{bmatrix}$$ Thus:$$\tilde{\mathbf{R}}^{(e)^{T}}_{2x2}\tilde{\mathbf{R}}^{(e)}_{2x2}=\begin{bmatrix}l^{(e)^{2}} m^{(e)^{2}} &0 \\ 0 &l^{(e)^{2}} m^{(e)^{2}} \end{bmatrix}$$

Which simplifies to:
$$\tilde{\mathbf{R}}^{(e)^{T}}_{2x2}\tilde{\mathbf{R}}^{(e)}_{2x2}=\begin{bmatrix}1&0 \\ 0 &1\end{bmatrix}$$

Which gives us $$\mathbf{I}_{2x2}$$, an identity matrix.

We can see that:
$$ \mathbf{R}^{(e)^{-1}} = \tilde{\mathbf{R}}^{(e)^{T}}$$

$$\tilde{\mathbf{T}}^{(e)^{-1}} = Diag[\mathbf{R}^{(e)^{T}} ,\mathbf{R}^{(e)^{T}} ]$$

And then,
$$\tilde{\mathbf{T}}^{(e)^{-1}} = (Diag[\mathbf{R}^{(e)} ,\mathbf{R}^{(e)} ])^{T}$$

And then we get:

$$\tilde{\mathbf{T}}^{(e)^{-1}} = \tilde{\mathbf{T}}^{(e)^{T}}$$

Going back to our notes on p.20-1: $${\tilde{\mathbf{k}}^{(e)}}{\tilde{\mathbf{T}}^{(e)}} {\mathbf{d}}^{(e)}= {\tilde{\mathbf{T}}^{(e)}} {\mathbf{f}}^{(e)}$$ which is equal to :$$[{\tilde{\mathbf{T}}^{(e)^{T}}}{\tilde{\mathbf{k}}^{(e)}}{\tilde{\mathbf{T}}^{(e)}}]{\mathbf{d}}^{(e)}={\mathbf{f}}^{(e)}$$

and comparing it with $${\tilde{\mathbf{k}}^{(e)}} {\tilde{\mathbf{d}}^{(e)}} ={\tilde{\mathbf{f}}^{(e)}}$$,

we finally get:

$$\mathbf{k}^{e}=\tilde{\mathbf{T}}^{(e)^{T}} \tilde{\mathbf{k}}^{(e)} \tilde{\mathbf{T}}^{(e)}$$

Meeting 21 Lecture Notes (13 Oct 08)
Comments on HW 3:
 * closing the loop, infinitesimal displacement
 * eigenvalue problem, zero eigenvalues, rigid body motion, mechanisms

 HW : Plot the eigenvectors corresponding to the zero eigenvalues of the 2-bar truss system (Mtg. 11) and interpret the results: mode shapes

These mode shapes (corresponding to the zero eigenvalues) may come out as a linear combination of the pure mode shapes (pure rigid body motions (3) and 1 pure mechanism).

Eigenvalue problem: $$\mathbf{kv} = \lambda \mathbf{v}$$

Let $$\left\{\mathbf{u_1}, \mathbf{u_2}, \mathbf{u_3}, \mathbf{u_4} \right\}$$ be the pure eigenvectors corresponding to the 4 zero eigenvalues:

$$\mathbf{ku_i} = 0\mathbf{u_i} = \mathbf{0}\left(i = 1,2,3,4 \right)$$

Linear combination of $$\left\{u_i, i = 1,2,3,4 \right\} $$

 = \sum_{i=1}^{4}{\alpha_i\left(\mathbf{ku_i} \right)} = \mathbf{0} = 0 \times \mathbf{W}
$$ \sum_{i=1}^{4}{\alpha_i\mathbf{u}_i} \equiv \mathbf{W} $$ (equal by definition)

$$ \alpha_i \Rightarrow $$(real numbers)

W is also an eigenvector corresponding to a zero eigenvalue:

$$ \mathbf{kW} = \mathbf{k}\left(\sum_{i=1}^{4}{\alpha_i\mathbf{u}_i} \right) $$

$$= \sum_{i=1}^{4}{\alpha_i\left(\mathbf{ku_i} \right)} = \mathbf{0} = 0 \times \mathbf{W}$$

 HW : INSERT FIGURE

$$\mathbf{\overline{k}v} = \lambda \mathbf{v}$$

$$\Rightarrow \mathbf{\overline{k}}$$ is stiffness matrix for constrained system

Plot eigenvectors corresponding to zero eigenvalues for case (a).

Matlab code: Modify 2-bar truss code

$$ a = b = 1m$$

$$E = 2$$

$$A = 3$$

Eml4500.f08.wiki1.oatley 20:36, 21 October 2008 (UTC)

Meeting 22 Lecture Notes (15 Oct 08)
ADINA Models and Animations (FEA eigen frequency animation)

Justification of assembly of element stiffness matrix $$\left(\mathbf{k^{(e)}}, e = 1,..., nel \right)\Rightarrow nel =$$ number of element

into global stiffness matrix $$\left(\mathbf{K}\right)$$

Consider example of 2-bar truss, (Mtg. 9):

Recall elemental FD relation: $$\mathbf{k}^{(e)}_{4\times 4}\mathbf{d}^{(e)}_{4\times 1} = \mathbf{f}^{(e)}_{4\times 1}$$

Mtg. 11: Euler cut principle, Method 2 (equilibrium of global node 2)

Mtg. 4: Free Body Diagrams of element 1 and element 2 with element degrees of freedom $$\mathbf{d}^{(e)}_{4\times 1}$$

Mtg. 8: For node 2, identify the relationships between the global degrees of freedom and the element degrees of freedom for elements 1 and 2

Equilibrium of node 2:

$$(1) \sum{F_x} = 0 = -f^{(1)}_3 - f^{(2)}_1 = 0$$

$$(2) \sum{F_y} = 0 = P - f^{(1)}_4 - f^{(2)}_2 = 0$$

Next we use the FD relation: $$\mathbf{k}^{(e)}_{4\times 4}\mathbf{d}^{(e)}_{4\times 1} = \mathbf{f}^{(e)}_{4\times 1}$$

Eml4500.f08.wiki1.oatley 20:37, 21 October 2008 (UTC)

Meeting 23 Lecture Notes (17 Oct 08)
Mtg. 22:
$$\mathbf{Eq. (1) \ }\Rightarrow f^{(1)}_3 + f^{(2)}_1 = 0  \mathbf{\left( 1\right)}$$

$$\mathbf{Eq. (2) \ }\Rightarrow f^{(1)}_4 + f^{(2)}_2 = P \mathbf{\left( 2\right)}$$

$$\mathbf{(1):}\left[k^{(1)}_{31}d^{(1)}_1 + k^{(1)}_{32}d^{(1)}_2 + ... + k^{(1)}_{34}d^{(1)}_4\right] \equiv f^{(1)}_3$$
$$+ \left[k^{(1)}_{11}d^{(2)}_1 + ... + k^{(2)}_{14}d^{(2)}_4\right] \equiv f^{(2)}_1 = 0$$

 HW:  Fill in for missing forces

Mtg. 8: Rewrite local degrees of freedom to global degrees of freedom:

$$\mathbf{\left(1 \right):} \left[k^{(1)}_{31}d_1 + k^{(1)}_{32}d_2 + k^{(1)}_{33}d_3 + k^{(1)}_{34}d_4\right]$$

$$+ \left[k^{(2)}_{11}d_3 + k^{(2)}_{12}d_4 + k^{(2)}_{13}d_5 + k^{(2)}_{14}d_6\right] = 0$$

$$\Rightarrow $$ Thus obtain the 3rd row of $$\mathbf{k}$$ (Mtg. 9) (See also Mtg. 8)

 HW : Derive the details of Eq. (2) and the assembly of row 4 in $$\mathbf{Kd = F} $$

Assembly of $$\mathbf{k^{(e)}}, e = 1,...,nel $$, into global stiffness matrix $$\mathbf{K}$$:

$$\mathbf{K_{n\times n}= }A^{nel}_{e=1}\mathbf{k^{(e)}_{ned\times ned}}$$

A: assembly operator
n: total number of global dof's before eliminating boundary conditions
ned: number of element dof's
(ned<<n)

Principle of Virtual Work (PVW)

Mtg. 10: Elimination of rows corresponding to the boundary conditions to obtain $$\mathbf{\overline{k}_{2\times 2}} $$

Mtg. 12: $$\mathbf{q^{(e)}_{2\times 1}=T^{(e)}_{2\times 4}d^{(e)}_{4\times 1}}$$

Mtg. 14: $$\mathbf{k^{(e)}=T^{(e)^T}\hat{k}^{(e)}T^{(e)}}$$

Deriving FEM for partial differential equations (PDE's):

FD relation for a bar or a spring: $$kd = F$$

$$\Rightarrow kd - F = 0\;\;\;\;\; (3)$$

$$\Leftrightarrow W(kd - F) = 0\;\;\;\left(for \, all\, W\, \right) \;\;\;\;\;(4)$$

W: weighting coefficient
"weak form" $$\equiv $$ PVW

 Proof :

$$A) \; \left(3 \right)\Rightarrow (4): trivial$$

$$B) \; \left(4 \right)\Rightarrow (3): \mathbf{NOT\; TRIVIAL}$$

Since $$\left(4 \right)$$ is valid for all $$W$$, select $$W=1$$ Then $$\left(4 \right)$$ becomes:

$$1\times (kd - F) = 0 \Rightarrow \left(3 \right)$$

Eml4500.f08.wiki1.oatley 20:37, 21 October 2008 (UTC)

title: User:Eml4500.f08.wiki1.aguilar/hw4


title: File:Hw4matlabfig3.png


title: File:Bdm1.jpg


title: File:Dm1.jpg


title: File:Hw4matlabfig4.png


text:
User:Lfavaloro~enwikiversity

L Favaloro Sydney, Australia

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User:Apletters

Hello friends, I’m Anil Prasad P, from Kerala India. I have been a Wikipedian since June 2007. I always love to write online on free and democratic pages like wikipedia that are easily accessible and useful to mankind all over the world.

I also serve the Free and Open Education movement as a Volunteer Ambassaor for WikiEducator project - http://www.wikieducator.org

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User:Raia

Raina Stoikova
Hello, I am from SU " St. Kliment Ohridski", Sofia, Bulgaria. I am in my 3rd year at university, studing Pedagogy with English language.

research interests

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Contacts

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title: User:Raia


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User talk:Raia

 Hello Raia, 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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And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! Jtneill - Talk - c 09:31, 21 October 2008 (UTC)

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Well done
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User:EML4500.f08.FEABBQ.Jayma/HW4

Notes for October 13, 2008
The four zero-eigenvalues of the $$ \underline{K}$$ matrix correspond to
 * 3 Rigid Body Modes
 * 1 Mechanism

HW 1
Plot the eigenvectors corresponding to the zero eigenvalues of the 2-bar truss system; interpret the results.

Before we do this assignment, it is important to understand the nature of these mode shapes. These mode shapes (that correspond to the zero eigenvalues) may come out as a linear combination of the pure mode shapes (pure rigid body motions and pure mechanism(s)). For example, take a body that has only two rigid body motions: horizontal and directly vertical. The linear combination of these two pure modes would be a diagonal motion. ''In Structural Mechanics, the eigenvalues represent the frequencies while the eigenvectors represent the mode shapes during free vibration. The lower frequencies (eigenvalues) tend to have more prominent vibration.''

Eigenvalue Problem: $$ \underline{K}\underline{v}=\lambda\underline{v} $$

Let {$$ \underline{u}_{1}, \underline{u}_{2}, \underline{u}_{3}, \underline{u}_{4} $$} be the pure eigenvectors corresponding to the four zero eigenvalues.

$$ \underline{K}_{_{6x6}}\underline{u}_{i_{6x1}}=0\cdot\underline{u}_{i_{6x1}}=\underline{0}_{_{6x1}} $$, where i=1,...,4.

A linear combination of {$$ \underline{u}_{i}, i=1,...,4$$}

$$ \sum_{i=1}^4 \alpha_{i_{1x1}} \underline{u}_{i_{6x1}} =: \underline{W}_{_{6x1}} $$, where $$ \alpha_{i} $$ are real numbers.

Note: the $$=:$$ means "equality definition."

$$ \underline{W} $$ is also an eigenvector corresponding to a zero eigenvalue: $$ \underline{K}\underline{W}=\underline{K}(\sum_{i=1}^4\alpha_{i} \underline{u}_{i})=\sum_{i=1}^4 \alpha_{i}(\underline{K}\underline{u}_{i})=\underline{0}=0\cdot\underline{W} $$

From the two-bar truss, we have the following output of $$ \underline{K} $$

And the following is the are the eigenvectors and eigenvalues, using the notation [V,D], where V are the eigenvectors and D are the eigenvalues.

The first four column vectors of V, which correspond to the four zero eigenvalues, are the eigenvectors for those zero eigenvalues. The matrix V is called the modal matrix.

The following are the four column vectors (eigenvectors) plotted as mode shapes. Note: they are shown in the order from the first column to the fourth column of the modal matrix.

HW2
Solve for $$ \underline{\overline{K}}\underline{v}=\lambda\underline{v} $$, the stiffness matrix for the constrained system below. Plot the eigenvectors corresponding to the zero eigenvalues for case (a).

The following is the Matlab code which solves for the global stiffness matrix, $$ \underline{K} $$, for the three bar truss in case (a) above.

The following is the output for the stiffness matrix of the three bar truss in case (a)

Using the method to solve for the eigenvectors of case(a), we get the following. Note: the first five column vectors correspond to zero eigenvalues.

The first five eigenvectors above correspond to zero eigenvalues, therefore their mode shapes are plotted below in order.

The Matlab code which solves for case (a) can be modified to solve for the stiffness matrix, $$ \underline{K} $$ for the four bar truss, case (b)

The following is the output for the stiffness matrix, $$ \underline{K} $$, for the four bar truss system.

title: User:EML4500.f08.FEABBQ.Jayma/HW4


text:
User:64ByteKiller

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64ByteKiller 14:32, 21 October 2008 (UTC)

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title: File:3Bar4Bar.jpg


text: File:Truss for conn & lmm matrices.JPG

title: File:Truss for conn & lmm matrices.JPG


text: File:Truss for conn & lmm matrices - elements.JPG


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User:Lilia~enwikiversity

INTRODUCTION

Hi!My name is Lilia Popova and i'm a 3rd year student of Primary School Pedagogics in SU "St. Kliment Ohridski" in Sofia, Bulgaria.I love reading,watching movies and hanging out with my friends!

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