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GNU Image Manipulation Program/Image maps

Imagemaps are created on GIMP using one of the extensions. Beside learning the GIMP way of creating ImageMaps the learning resource introduces into technical aspects imagemaps in Wikiversity.

Explore an ImageMap in Wikiversity
The image on the right contains an image map. Click on the people in the painting to get more information and answer the question "who is this?"

Learning Resources

 * (Basics about ImageMaps) Learn about the basic principles of Image Maps and explore the learning environment about ImageMaps on W3School.
 * (ImageMaps with GIMP) Learn about Creating and ImageMap with GIMP with a Youtube video by TanUv90. Download GIMP
 * (Javascript ImageMap Editor) A Javascript ImageMap Editor allows a direct generation of imagemap in the mediawiki format (see https://niebert.github.io/imgmap/ Online Example).

Imagemaps from GIMP to Mediawiki

 * If you want to create an imagemap, you need
 * an image and
 * a list of clickable areas in the image that includes a reference to another web resource,
 * to understand the concept of imagemaps try a clickable image on the right first.
 * then try to create an image for yourself with Online image map editor. Use e.g. the the following image on wikimedia.
 * https://upload.wikimedia.org/wikipedia/commons/5/5e/Annweiler_Queich_05.JPG
 * insert the link to the image in Online image map editor and press the small arrow on the right.
 * mark the water wheel with (e.g. you get the coordinates 500,145,630,525, see image on the right)
 * one click on the top left of corner
 * one click on the bottom right of corner
 * you will see a rectangle, and analyse the HTML code

 * Watch the Youtube Video about ImageMaps with GIMP by Youtube user TanUv90
 * install the GIMP's plug-in for image maps. When you use the plug-in:
 * it has simple selection tools and when an outline of an area is closed (or double mouse clicked) a dialog shows up that has text boxes for the different image table attributes and instructions.
 * Use the "alt" attribute for what the name for the object that is highlighted on a mouse-over event.
 * Use the "href" attribute to create a link, that is displayed when you click on the area.
 * The GIMP output and the output of the Online image map editor looks like commons:Image_talk:Afrotropic-Ecozone-Biocountries-IM.svg and to make the wiki markup from that it is necessary
 * to search and replace "<area shape="" just gets removed,
 * "" coords="" replaced with a space,
 * "," replaced also with a space (in the list of coordinates/numbers),
 * " href=" is replaced with " " or " [[:Category:" and " />" with "".

Web-based ImageMap converter
The MediaWiki needs the imagemap in a specific format mentioned above. The imagemap editor by Adam Maschek was forked to change the export format directly to MediaWiki. No application of regular expressions for replacement the standard image map format for HTML is necessary anymore.

 * Online MediaWiki ImageMap Creator

Note: Image Maps are create in the browser no images are uploaded to any server.
 * (Select Image) Select the image for which you want to create an image map
 * (Upload) upload the image to the Online Image Editor that you have already uploaded to WikiMedia
 * (WikiMedia) If you want to use a image from wikimedia use the full size URL to the image
 * (Sample Image) Use one of the 4 sample images to explore the features of the Online Image Editor
 * (Add Links on Image) Clickable areas on the image can be created as
 * rectangle,
 * circle or
 * polygone (finsh the selection of point list for the polygone by a SHIFT-click)

Manual Replacement of Coordinates
From GIMP or the web-based tool you will get an image map syntax like this.  After converting the coordinates the code for the imagemap will look like this: poly 203 193 202 222 195 238 202 268 239 249 264 284 281 289 333 288 329 322 370 277 377 243 346 250 322 221 306 181 256 186 254 194 248 199 226 198 226 205 205 205 Category:Northeast Tropical Africa View source code of this page to analyse the image map used for the painting.


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User:Eas4200c.f08/HW report table/The best of HW4

 Under construction; not final (but close to being stable after a couple of weeks). The intention here is to document the best features in any HW report for the readers (including you); if you see excellent features in any HW report (including your team's) that I may have missed noticing, don't hesitate to let me know. (I don't have time to read all HW reports in detail.) Eml4500.f08 13:46, 28 October 2008 (UTC)

 * classnotes graded over 100%
 * idealized single-cell airfoil (semi-circular leading edge, straight-line body), find $$\displaystyle J$$; need to complete the solution to find the numerical value of $$\displaystyle J$$. Some team went well beyond expectation and solved this problem by coding it up with matlab; see Team Aero6.
 * analytical result: $$\displaystyle J = 0.100 m^4$$ Team Aero, $$\displaystyle J = 0.1001 m^4$$ Team VQCrew
 * numerical result: $$\displaystyle J = 0.125 m^4$$ Team Aero6; the error compared to the analytical result was large; perhaps there was not enough accuracy in the quadrature.
 * idealized 2-cell airfoil (rectangular body), find $$\displaystyle J$$; need to complete the solution to find the numerical value of $$\displaystyle J$$ Some team went well beyond expectation and solved this problem by coding it up with matlab; see Team Aero6.
 * analytical result: $$\displaystyle J = 0.855 \times 10^{-4} m^4$$ Team VQCrew
 * numerical result: $$\displaystyle J = 0.855 \times 10^{+4} cm^4 = 0.855 \times 10^{-4} m^4$$ Team Aero6
 * engineering (ad-hoc) derivation of expression for $$\displaystyle \theta$$ in terms of $$\displaystyle (G, \bar{A}, q, t)$$. Best explanation: Team VQCrew
 * figure for the derivation. The figure in Team VQCrew is not exactly correct since the line $$\displaystyle \overline{PP^{\prime\prime}}$$ is not distinguishable from the tangent to the contour; see the more correct figure in Exam 2.
 * 3 ad-hoc aspects of engineering derivation. Best description: Team VQCrew
 * torsional analysis by elasticity theory (cont'd)
 * kinematic assumption (review from previous lecture)
 * 4 zero strain components for torsion problem
 * strain-stress relation for isotropic elasticity
 * inversion for stress-strain relation for isotropic elasticity
 * 4 zero stress components for torsion problem

 * matlab problem graded over 100%
 * problem description (a repeat of my wiki page Airfoil with figure describing the various variables defining the NACA airfoil). Best description: Team ZYX had matlab figure of the NACA airfoil with all of its geometric quantities presented.
 * figure showing various points on the NACA airfoil and method of area integration by triangles for each cell. Best figure: Team Aero_(Eelman) Team Aero
 * those who did not finish their matlab problem in HW3 should finish it in HW4:
 * computation of centroid coordinates and area $$\displaystyle \bar{A}$$ for single-cell airfoil
 * validate their code with circular airfoil with convergence plot(s)
 * validate their code with various observer points $$\displaystyle P_0$$
 * plot of circular airfoil
 * apply their code to NACA 2415 airfoil with $$\displaystyle c=0.5m$$ to find centroid location and area $$\displaystyle \bar{A}$$ with convergence plot(s)
 * plot of NACA airfoil and centroid at crosshair
 * develop matlab code to find torsional constant $$\displaystyle J$$ of NACA airfoil
 * single-cell NACA 2415 airfoil, compute $$\displaystyle J$$
 * 3-cell NACA 2415 airfoil, compute $$\displaystyle J$$
 * quadrature by triangles at different points in the 3-cell NACA airfoil.
 * compute the area $$\displaystyle \bar{A}_i$$ for cell $$\displaystyle C_i$$, and sum them up to get total area $$\displaystyle \bar{A}$$; error must be within 1% of the area $$\displaystyle \bar{A}$$ for the single-cell NACA airfoil.
 * compare the numerical values of $$\displaystyle J$$ and draw some conclusion
 * best solution: Team Aero Team Aero_(Eelman) Team Aero6, figure showing the quadrature by triangles and selection of observer points $$\displaystyle P_0$$ in each cell; computed $$\displaystyle J_{1C}$$ for the single-cell NACA 2415 airfoil, and $$\displaystyle J_{3C}$$ for the 3-cell NACA 2415 airfoil, showing that $$\displaystyle J_{3C} \approx J_{1C}$$ (as noted in the textbook) even though they did not make this remark; Team VQCrew did make this remark, but they initially obtained $$\displaystyle J_{3C} \approx 2 J_{1C}$$, clearly not consistent with the remark; later, they updated their work to provide the correct result; see Team VQCrew Updated. See also Team ZYX, which did not have a figure to describe the quadrature by triangles and observer points $$\displaystyle P_0$$, but had a section describing the NACA airfoil geometry.
 * Team Aero: $$\displaystyle J_{1C} = 5.129 \times 10^{-6} m^{4}$$, $$\displaystyle J_{3C} = 5.6884 \times 10^{-6} m^{4}$$
 * Team Aero_(Eelman): $$\displaystyle J_{1C} = 5.1163 \times 10^{-6} m^{4}$$, $$\displaystyle J_{3C} = 5.5358 \times 10^{-6} m^{4}$$, relative difference $$\displaystyle 7\%$$.
 * Team Aero6: $$\displaystyle J_{1C} = 4.9978 \times 10^{-6} m^{4}$$, $$\displaystyle J_{3C} = 5.4653 \times 10^{-6} m^{4}$$
 * Team ZYX: $$\displaystyle J_{1C} = 4.970 \times 10^{-6} m^{4}$$, $$\displaystyle J_{3C} = 5.276 \times 10^{-6} m^{4}$$
 * Team VQCrew: $$\displaystyle J_{1C} = 4.7543 \times 10^{-6} m^{4}$$, $$\displaystyle J_{3C} = 8.6086 \times 10^{-6} m^{4}$$, later corrected to $$\displaystyle J_{3C} = 5.250 \times 10^{-6} m^{4}$$ in Team VQCrew Updated.
 * the above 5 teams agreed more or less with the result for $$\displaystyle J_{1C}$$ and for $$\displaystyle J_{3C}$$, except for the value of $$\displaystyle J_{3C}$$ from Team VQCrew; as a result, it is likely that the value of $$\displaystyle J_{3C}$$ from Teams Aero, Aero_(Eelman), Aero6, ZYX is the correct one. After Team VQCrew updated their work, all 5 teams agreed on both $$\displaystyle J_{1C}$$ and $$\displaystyle J_{3C}$$.

 * matlab-code certification: some students signed this section, after having claimed not knowing how to code with matlab but did not report their matlab tutorial work.

EML 4500, The best of HW4
 * contributing team members

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


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User talk:Google~enwikiversity

 Hello Google, 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 14:46, 28 October 2008 (UTC)

Please choose a different username and create a new account because it is inapproriate to use the name of a major website/company for your own use. --mikeu talk 14:46, 28 October 2008 (UTC)

Your account will be renamed
Hello,

The developer team at Wikimedia is making some changes to how accounts work, as part of our on-going efforts to provide new and better tools for our users like cross-wiki notifications. These changes will mean you have the same account name everywhere. This will let us give you new features that will help you edit and discuss better, and allow more flexible user permissions for tools. One of the side-effects of this is that user accounts will now have to be unique across all 900 Wikimedia wikis. See the announcement for more information.

Unfortunately, your account clashes with another account also called Google. To make sure that both of you can use all Wikimedia projects in future, we have reserved the name Google~enwikiversity that only you will have. If you like it, you don't have to do anything. If you do not like it, you can pick out a different name.

Your account will still work as before, and you will be credited for all your edits made so far, but you will have to use the new account name when you log in.

Sorry for the inconvenience.

Yours, Keegan Peterzell Community Liaison, Wikimedia Foundation 23:37, 17 March 2015 (UTC)

Renamed
 This account has been renamed as part of single-user login finalisation. If you own this account you can |log in using your previous username and password for more information. If you do not like this account's new name, you can choose your own using this form after logging in: . -- Keegan (WMF) (talk) 06:09, 19 April 2015 (UTC)

title: User talk:Google~enwikiversity


text: Category:Consensus in a wiki environment


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User:Eml4500.f08.Ateam.paquette/Homework5

 Lecture notes 24 October 20, 2008

Justification of eliminating rows 1,2,5,6.
The goal of eliminating rows 1,2,5 and 6 in a stiffness matrix is to find $$\mathbf {K_{2x2}}$$ in the original 2 bar truss shown below:

The Force Displacement relationship can be described as: $$\mathbf {K_{6x6}d_{6x1} = F_{6x1}}$$

This can be rearranged into the equation $$\mathbf {K_{6x6}d_{6x1} - F_{6x1} = 0_{6x1}}$$ (1)

For the Principle of Virtual Work (PVW):

$$\mathbf {W_{6x1}\cdot (K_{6x6}d_{6x1} - F_{6x1}) = 0_{1x1}}$$ (2)

This is for all $$\mathbf {W_{6x1}}$$, also known as the Weighting Matrix.

As seen above, it is trivial when going from equation 1 to equation 2, but it is very important to understand how to get from equation 2 to equation 1. The next part will be verifying the steps in going from equation 2 to equation 1.

Eq. 1 $$\Leftrightarrow $$ Eq. 2
When proving Eq. 2 $$\Leftrightarrow $$ Eq. 1, choices have to be made for what the Weighting Coefficient will equal.

*Choice 1

Let $$\mathbf {W_1 = 1}$$ and $$\mathbf {W_2 = W_3 = W_4 = W_5 = W_6 = 0}$$

This gives the Weighting Matrix $$ W = \begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \end{bmatrix}_{6x1}$$

$$(\mathbf{{W})\cdot({K} {d} - {F})=}$$

$$1\cdot \left[ \sum_{j=1}^{6}{k_{1j}d_{j}-F_{1}} \right]$$ $$+ 0\cdot\left[ \sum_{j=1}^{6}{k_{2j}d_{j}-F_{2}} \right]$$

$$+ 0\cdot\left[ \sum_{j=1}^{6}{k_{3j}d_{j}-F_{3}}\right]$$ $$+ 0\cdot\left[ \sum_{j=1}^{6}{k_{4j}d_{j}-F_{4}}\right]$$

$$+ 0\cdot\left[ \sum_{j=1}^{6}{k_{5j}d_{j}-F_{5}}\right]$$ $$+ 0\cdot\left[ \sum_{j=1}^{6}{k_{6j}d_{j}-F_{6}}\right]$$

The only equation that is used is the first one since the rest become zero.

$$ 1\cdot\left[ \sum_{j=1}^{6}{k_{1j}d_{j}-F_{1}}\right]$$

or

$$ \sum_{j=1}^{6}{k_{1j}d_{j}-F_{1}}$$

*Choice 2

Next $$\mathbf W_2$$ will be equal to 1 while the rest of the weighting matrix is equal to 0.

$$ W = \begin{bmatrix} 0 \\ 1 \\ 0 \\ 0 \\ 0 \\ 0 \end{bmatrix}_{6x1}$$

The same process will be taken to find the next part of the equation.

$$(\mathbf{{W})\cdot({K} {d} - {F})=}$$

$$0\cdot \left[ \sum_{j=1}^{6}{k_{1j}d_{j}-F_{1}} \right]$$ $$+ 1\cdot\left[ \sum_{j=1}^{6}{k_{2j}d_{j}-F_{2}} \right]$$

$$+ 0\cdot\left[ \sum_{j=1}^{6}{k_{3j}d_{j}-F_{3}}\right]$$ $$+ 0\cdot\left[ \sum_{j=1}^{6}{k_{4j}d_{j}-F_{4}}\right]$$

$$+ 0\cdot\left[ \sum_{j=1}^{6}{k_{5j}d_{j}-F_{5}}\right]$$ $$+ 0\cdot\left[ \sum_{j=1}^{6}{k_{6j}d_{j}-F_{6}}\right]$$

Again all of the equations are going to become zero, but this time the second equation will be the only one that matters since does not have a zero coefficient. The final equation for choice 2 is:

$$ \sum_{j=1}^{6}{k_{2j}d_{j}-F_{2}}$$

*Choice 3

$$\mathbf W_3$$ for this choice will be the factor equal to 1.

$$ W = \begin{bmatrix} 0 \\ 0 \\ 1 \\ 0 \\ 0 \\ 0 \end{bmatrix}_{6x1}$$

$$(\mathbf{{W})\cdot({K} {d} - {F})=}$$

$$0\cdot \left[ \sum_{j=1}^{6}{k_{1j}d_{j}-F_{1}} \right]$$ $$+ 0\cdot\left[ \sum_{j=1}^{6}{k_{2j}d_{j}-F_{2}} \right]$$

$$+ 1\cdot\left[ \sum_{j=1}^{6}{k_{3j}d_{j}-F_{3}}\right]$$ $$+ 0\cdot\left[ \sum_{j=1}^{6}{k_{4j}d_{j}-F_{4}}\right]$$

$$+ 0\cdot\left[ \sum_{j=1}^{6}{k_{5j}d_{j}-F_{5}}\right]$$ $$+ 0\cdot\left[ \sum_{j=1}^{6}{k_{6j}d_{j}-F_{6}}\right]$$

The final equation for choice 3 becomes:

$$ \sum_{j=1}^{6}{k_{3j}d_{j}-F_{3}}$$

*Choice 4

$$ W = \begin{bmatrix} 0 \\ 0 \\ 0 \\ 1 \\ 0 \\ 0 \end{bmatrix}_{6x1}$$

$$(\mathbf{{W})\cdot({K} {d} - {F})=}$$

$$0\cdot \left[ \sum_{j=1}^{6}{k_{1j}d_{j}-F_{1}} \right]$$ $$+ 0\cdot\left[ \sum_{j=1}^{6}{k_{2j}d_{j}-F_{2}} \right]$$

$$+ 0\cdot\left[ \sum_{j=1}^{6}{k_{3j}d_{j}-F_{3}}\right]$$ $$+ 1\cdot\left[ \sum_{j=1}^{6}{k_{4j}d_{j}-F_{4}}\right]$$

$$+ 0\cdot\left[ \sum_{j=1}^{6}{k_{5j}d_{j}-F_{5}}\right]$$ $$+ 0\cdot\left[ \sum_{j=1}^{6}{k_{6j}d_{j}-F_{6}}\right]$$

The final equation for choice 4 becomes:

$$ \sum_{j=1}^{6}{k_{4j}d_{j}-F_{4}}$$

*Choice 5

$$ W = \begin{bmatrix} 0 \\ 0 \\ 0 \\ 0 \\ 1 \\ 0 \end{bmatrix}_{6x1}$$

$$(\mathbf{{W})\cdot({K} {d} - {F})=}$$

$$0\cdot \left[ \sum_{j=1}^{6}{k_{1j}d_{j}-F_{1}} \right]$$ $$+ 0\cdot\left[ \sum_{j=1}^{6}{k_{2j}d_{j}-F_{2}} \right]$$

$$+ 0\cdot\left[ \sum_{j=1}^{6}{k_{3j}d_{j}-F_{3}}\right]$$ $$+ 0\cdot\left[ \sum_{j=1}^{6}{k_{4j}d_{j}-F_{4}}\right]$$

$$+ 1\cdot\left[ \sum_{j=1}^{6}{k_{5j}d_{j}-F_{5}}\right]$$ $$+ 0\cdot\left[ \sum_{j=1}^{6}{k_{6j}d_{j}-F_{6}}\right]$$

The final equation for choice 5 becomes:

$$ \sum_{j=1}^{6}{k_{5j}d_{j}-F_{5}}$$

*Choice 6

Choice 6 is not any different and it ends the process of trying to prove that going from the equation with a weighting coefficient to the force displacement relationship is not trivial.

$$ W = \begin{bmatrix} 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 1 \end{bmatrix}_{6x1}$$

$$(\mathbf{{W})\cdot({K} {d} - {F})=}$$

$$0\cdot \left[ \sum_{j=1}^{6}{k_{1j}d_{j}-F_{1}} \right]$$ $$+ 0\cdot\left[ \sum_{j=1}^{6}{k_{2j}d_{j}-F_{2}} \right]$$

$$+ 0\cdot\left[ \sum_{j=1}^{6}{k_{3j}d_{j}-F_{3}}\right]$$ $$+ 0\cdot\left[ \sum_{j=1}^{6}{k_{4j}d_{j}-F_{4}}\right]$$

$$+ 0\cdot\left[ \sum_{j=1}^{6}{k_{5j}d_{j}-F_{5}}\right]$$ $$+ 1\cdot\left[ \sum_{j=1}^{6}{k_{6j}d_{j}-F_{6}}\right]$$

The final equation for choice 6 becomes:

$$ \sum_{j=1}^{6}{k_{6j}d_{j}-F_{6}}$$

Boundary Conditions
For the Principle of Virtual Work, the boundary conditions need to be accounted for. In the 2-bar truss, $$\mathbf {d_1 = d_2 = d_5 = d_6 = 0 }$$. The only two displacements the matter are $$\mathbf d_3$$ and $$\mathbf d_4$$ in the two bar truss. Weighting coefficients need to be kinematically admissible, which means they cannot violate the boundary conditions. This concludes that $$ \mathbf {W_1 = W_2 = W_5 = W_6 = 0 }$$. According to the calculus of variations, the weighting coefficients are equivalent to the virtual displacements.

The original equation $$\mathbf({W)\cdot({K} {d} - {F}) = 0}$$ can now be expressed in matrix form and can be further reduced.

$$ (W)_{6x1}K_{6x6} \begin{bmatrix} d_1=0 \\ d_2=0 \\ d_3=x  \\ d_4=y \\ d_5=0 \\ d_6=0  \end{bmatrix}_{6x1}- F_{6x1} = 0$$

Note that x and y are arbitrary values not equal to zero in this problem. The principle of virtual work allows the equation to be condensed to only what is needed and all of the zeros to be removed. This new equations will be noted as equation 3 and is shown below.

$$(W)_{6x1}\cdot \begin{bmatrix} K_{13} & K_{14} \\ K_{23} & K_{24} \\ K_{33} & K_{34} \\ K_{43} & K_{44} \\ K_{53} & K_{54} \\ K_{63} & K_{64} \end{bmatrix}_{6x2} \begin{bmatrix} d_3 \\ d_4 \end{bmatrix}_{2x1} - F_{2x1} = 0 $$

Or: $$ \mathbf{W \cdot(Kd - F)}$$

This equation can be even further reduced to

$$\begin{Bmatrix} W_3\\W_4 \end{Bmatrix}\cdot\left(\begin{bmatrix} K_{33} &K_{34} \\ K_{43}& K_{44}\end{bmatrix}_{2x2} \begin{Bmatrix} d_3\\d_4 \end{Bmatrix}_{2x1} - \begin{Bmatrix} F_3\\F_4 \end{Bmatrix}_{2x1}\right) = 0 $$

This equation can be rewritten as

$$ \mathbf {\overline{K} \overline{d} = \overline{F}}$$

Where

$$\mathbf{\overline{K}}=\begin{bmatrix} K_{33} && K_{34} \\ K_{43} && K_{44}\end{bmatrix} $$

$$\mathbf{\overline{d}}= \begin{Bmatrix}d_{3} \\ d_{4} \end{Bmatrix} $$

$$\overline{F } = \begin{Bmatrix}F_{3} \\ F_{4} \end{Bmatrix} $$

for any $$\begin{Bmatrix}W_3\\W_4 \end{Bmatrix}$$

 Lecture notes 26 October 27, 2008

Matlab Homework
In Homework 5, the 5-bar truss code needs to be reworked into a 6-bar truss. Also the 2-bar code needs to be corrected. The 6-bar truss will have the same E and A as the 5-bar truss. Note that the 2-bar truss has different E and A. $${d_{4x1}^{(i)}}$$ are the element degrees of freedom in the global (x,y) coordinate system for element i. This can be related to the axial displacement in the equation $$ {q}_{2x1}^{(i)} = {T}_{2x4} {d_{4x1}^{(i)}}$$. Also for the 2-bar truss, the complete degrees of freedom in global coordinate system $${d_{6x1}}$$ needs to be changed to d(lmm(i.;)).

Back to Principle of Virtual Work
An example everyone can understand for PVW is the equation for our course grade.

Course Grade = $$\alpha_{o}\,$$ x (Homework grade) + $$\sum_{i=1}^3 \alpha_{i}\,$$ x Exami

where $$\alpha_{o}\,$$ and $$\alpha_{i}\,$$ are the weighting coefficients.

For deriving $$ {k}^{(e)} = {T}^{(e)T}\quad\hat{k}^{(e)} {T^{(e)}}$$ we need to recall the force displacement relationship with the axial degrees of freedoms $$ q^{(e)} $$.

$$\quad\hat{k}_{2x2}^{(e)} {q_{2x1}^{(e)}}=  {p}_{2x1}^{(e)}$$  can be rearranged to the formula

$$\quad\hat{k}_{2x2}^{(e)} {q_{2x1}^{(e)}} - {p}_{2x1}^{(e)} = 0_{2x1} $$ (1) Note that 0 is not a scalar but the 2x1 matrix $$ 0 $$.

With the weighting coefficient added, the PVW equation becomes $$W_{2x1}\cdot \quad\hat{k}_{2x2}^{(e)} {q_{2x1}^{(e)}} - {p}_{2x1}^{(e)} = 0 $$.

Note that since $$W$$ is a 2x1 matrix, the 0 is a scalar. This works for all $$W_{2x1}$$ because equation 1 is equal to zero. Now we have again shown that equation 1 is equivalent to equation 2, like from slide 24-1.

Again recall $$ {q}_{2x1}^{(e)} = {T}_{2x4}^{(e)} {d_{4x1}^{(e)}}$$. (3) Real displacement is represented in this equation.

$$\quad\hat{W}_{2x1} = {T_{2x4}^{(e)}} {W}_{4x1}$$ (4)

The equation above shows that $$\quad\hat{W}$$ relates to the axial $$q$$, and it is also the virtual displacement.

 Lecture notes 27 October 29, 2008

For the next lectures, Chapter 2 sections 1,2,6 and 7 need to be read for a better understanding of what we will be taught.

As a reminder, $$\quad\hat{W}_{2x1}$$ is the virtual axial displacement corresponding to $$ {q}_{2x1}^{(e)} $$. $$ {W}_{4x1} $$ is the virtual displacement in the global coordinate system corresponding to $$ {d}_{4x1}^{(e)} $$.

Our goal is to replace equations 3 and 4 into equation 2. The result is shown below.

$$ (T^{(e)}W)\cdot [\quad\hat{k}^{(e)}(T^{(e)}d^{(e)}) - p^{(e)}] = 0 $$ for all $$W_{4x1}$$. (5)

Transposing the first matrix and multiplying it by the second matrix is equivalent to taking the dot product of two matrices.

Recall that Recall $$\left(AB \right)^{T}=B^{T}A^{T}$$ (6)

Homework: Proof $$\left(AB \right)^{T}=B^{T}A^{T}$$

$${A}_{2x3} = \begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6 \end{bmatrix}$$

$${B}_{3x3} = \begin{bmatrix} 7 & 8 & 9\\1 & 2 & 3 \\4 & 5 & 6 \end{bmatrix}$$

$${AB}=\begin{bmatrix} 21 & 27 & 33\\ 57 & 72 & 87 \end{bmatrix} $$

$${AB}^T=\begin{bmatrix} 21 & 57 \\ 27 & 72 \\33 & 87 \\ \end{bmatrix} $$

$${B}^T_{3x3}=\begin{bmatrix} 7 & 1 & 4\\ 8 & 2 & 5\\ 9 & 3 & 6\end{bmatrix}$$

$${A}^T_{2x3}=\begin{bmatrix} 1 & 4 \\ 2 & 5 \\ 3 & 6 \\ \end{bmatrix}$$

$${B}^T{A}^T=\begin{bmatrix} 21 & 57 \\ 27 & 72 \\ 33 & 87 \\ \end{bmatrix} $$

Therefore, $$\mathbf{{B}^T{A}^T = {(AB)}^T}$$

Also recall that $${a_{nx1} \cdot b_{nx1}=a^T_{1xn}b_{nx1}=S_{1x1}}$$ (7) Now equations 7 and 6 need to be applied into equation 5.

$${\left( T^{(e)}w \right)^T} \left[ {\hat{k}^{(e)}\left(T^{(e)}d^{(e)} \right)-p^{(e)}} \right]=0_{1x1}$$

$${ w^TT^{(e)T}} \left[ {\hat{k}^{(e)}\left(T^{(e)}d^{(e)} \right)-p^{(e)}} \right]=0_{1x1}$$ Next the transposed T matrix is distributed throughout the rest of the equation.

$${ w^T}\cdot \left[ {\left(T^{(e)T}\hat{k}^{(e)}T^{(e)}\right)d^{(e)} - T^{(e)T}p^{(e)}} \right]=0_{1x1}$$

We can set $$\mathbf{(T^{(e)T}\hat{k}^{(e)}T^{(e)}) = k^{(e)}}$$ and $$\mathbf {T^{(e)T}p^{(e)} = f^{(e)}} $$ to shorten the equation to:

$${ w^T}\cdot \left[{{k}^{(e)}d^{(e)} - f^{(e)}} \right]=0_{1x1}$$

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

As shown before, it is important to find the steps going from the equation with the weighting coefficient to the one basic equation.

So far we have only done the discrete case, or the non continuous case, and now we need to learn about the continuous case, or Partial Differential Equations (PDEs). A motivational model problem is an elastic bar with varying A(x) and E(x) subject to varying axial load (distributed) and concentrated loads. The axial and the concentrated loads are both time dependent.


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User:Eml4500.f08.delta 6.castrillon/3-bar truss MATLAB simulation

3-bar Truss Eigenvalues
We have two truss systems, one 3-bar and 5-bar. They are illustrated below:

Where $$E^{(1)}=E^{(2)}=E^{(3)}=1$$; $$A^{(1)}=A^{(2)}=A^{(3)}=1$$; $$L^{(1)}=L^{(2)}=L^{(3)}=1$$. It is possible to analyze the eigenvalues and eigen vectors according to $$\mathbf{K}v=\mathbf{\lambda} v$$ for the 3-bar truss case.

We construct the stffness matrix from each individual element stiffness matrix. It was found to be:
 * K=$$\begin{bmatrix}

0&0&0&0&0&0&0&0\\ 0&1&0&-1&0&0&0&0\\ 0&0&1&0&-1&0&0&0\\ 0&-1&0&2&0&0&0&-1\\ 0&0&-1&0&1&0&0&0\\ 0&0&0&0&0&0&0&0\\ 0&0&0&0&0&0&0&0\\ 0&0&0&-1&0&0&0&1 \end{bmatrix}$$ This matrix was then inputed into MATLAB and its eigenvalues and eigenvectors were determined. The MATLAB results are as follow:

V = -0.0000   0.0000    0.0000   -0.4740    0.7103   -0.2504    0.4082    0.0000   -0.7071   -0.4674   -0.1769    0.2838   -0.0000   -0.7071         0   -0.2136   -0.3585   -0.5702   -0.8165   -0.0000    0.0000   -0.4674   -0.1769    0.2838    0.0000    0.7071    0.0000   -0.2136   -0.3585   -0.5702   -0.0000    0.0000   -0.0000   -0.0556    0.1402   -0.1014   -0.0000    0.0000    0.0000   -0.1604    0.3534   -0.1873    0.4082    0.0000    0.7071   -0.4674   -0.1769    0.2838 D = 3.0000        0         0         0         0         0         0    2.0000         0         0         0         0         0         0    1.0000         0         0         0         0         0         0    0.0000         0         0         0         0         0         0   -0.0000         0         0         0         0         0         0   -0.0000


text:
User:Acey~enwikiversity

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Portal:Taoist Studies

Welcome to the Wikiversity Department of Taoist Studies.

Introduction
The Wikiversity Department of Taoist Studies is a content development project for learning resources related to the Way, for those who wish to follow as well as the curious.

Introduction to Taoism

 * Daoism
 * Introduction to Taoism - An introduction to the major tenets, figures, writings, and schools of historical Taoism.
 * A Brief History of the Way - A more in-depth study of the history of the Taoist faith.

Text Studies

 * Dao De Jing (Tao Te Ching) - The 5,000 character classic considered by most academics to be the founding document of Taoism.
 * Chuang Tzu - A longer and more complex enumeration of Taoist philosophy and cosmology, just as essential to comprehending Taoism as the Tao Te Ching.
 * Lieh-tzu - The third Taoist classic, more down-to-earth and practical than either the Tao Te Ching or the Chuang Tzu.
 * I Ching oracle - A guide to use and interpret the Oracle.

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 * Dao - "The Way that can be followed is not the true Way, the Name that can be named is not the true Name." Doing the impossible, and describing the Dao.
 * De - The virtue of the Way, and how to cultivate it.
 * Wu Wei - Actionless action, explored in detail, including an essay from William S. Burroughs.
 * Po - The "uncarved block" of Lao Tzu.
 * Qi - The breath of life.
 * Eight Trigrams - The binary basis of the universe.
 * Five Elements - The life-cycles in classical Taoist cosmology.
 * Taoist Ethics: Walking the Way - A short course on the ethics and morality of Taoism.

Applied Taoism

 * Taoist Sexual Practices - Immortality through sex.
 * Feng Shui - A layman's course to Taoist geomancy.
 * Tai Chi and Other Taoist Martial Arts - A brief introduction to Taoist philosophy in the martial arts, focusing especially but not exclusively on the "Taoist" arts of Tai Chi, Bagua Zheng, and Xing-yi (Hsing I).
 * Introduction to Traditional Chinese Medicine - What they're doing when they prick you with those needles.
 * Qi - The breath of life.
 * Shen - The holy spirit.
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 * The Flavor That Can Be Tasted... - Taoist dietary practices, focused on modern application.

Social Studies

 * The Peach Wine Cauldron - Where the complex interrelationships between Taoism, Buddhism, and Confucianism will be examined.
 * Modern Taoism: Asia - Exploring the role of Taoism and the Way in modern Chinese and other Asian societies.
 * Modern Taoism: The West - A discussion of Taoism's rise and future in Western countries.
 * Rough Notes on Classical Chinese - Classical Chinese, n., a verb with optional tense, mood, aspect, or nouns.
 * Taoism and Science: Towards a Rational Taoism - Reconciling traditional Chinese beliefs with an Enlightened scientific viewpoint.
 * Missing The Point: The Philosophical/Religious Schism - Discussing the perceived divide between "philosophical" Taoism and "religious" Taoism, and why it doesn't matter after a few cups of peach wine.

title: Portal:Taoist Studies


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User:DarkangelDani

Dani

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User:Stello2008

Hi, I'm Stephen. I have studied maths to 1st year undergraduate level in the past and am looking to revise my previous studies and move onto more advanced material.

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Communities of Practice

Communities of Practice (CoP) are members of an organization who work together to diseminate knowledge throughout their community. They are considered great vehicles that not only promote knowledge reuse but also prmote creation of new knowledge for greater innovation and efficency of the organization. CoP's can be instrumental in the development and documentation of "Best Practices and Leasons Learned"

All CoP's share some basic characteristics. They are made up of members who share common goals. These members agree to the rules of the community and to carry out certain roles in order to achieve the communities goals. They also share a common workspace where members can communicate, store and share knowledge products.

The common workspace of a CoP usually has a virtual aspect in that the CoP will make use of an organization Intranet or the internet, so that real time exchanges and asynchronous discussions can take place.

There are many types of CoP's and they are typically defined by a common focal point, such as members or a profession or industry, work related functions such as production or operations, or topics such as technology or knowledge retention.

CoP's resemble professional organizations. They have a business case, code of ethics, and mission statement.

title: Communities of Practice


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Template:Tutorial 1 - what is Social Psychology?

Some brief definitions about what Social Psychology is:

 * According to Baumeister & Bushman (2008) "Social Psychology is a branch of psychology that seeks a broad understanding of how human beings think, act, and feel"
 * Social psychology is the study of how people and groups interact - Wikipedia
 * Social Psychology examines interactions of individual psychology and group phenomena; examines the influence of real or imagined others on the way people behave (Westen, Burton & Kowalski, 2008).
 * Answers.com defines social psychology as "The branch of human psychology that deals with the behavior of groups and the influence of social factors on the individual."
 * Thoughts, feelings and behaviors of individuals are influenced by actual imagined or implied presence of others (Gordon Allport)

I believe social psychology incorporates aspects of sociology as well as psychology. It is a complex field because of the broad variety of aspects it involves. Social psychology examines the individual, as well as groups, in terms of how they interact with society. It is a dynamic relationship between an individual/group and society each facet influencing one another. Some interesting ideas explored in the lecture and tutorial were:
 * ABC the triad in social psychology stands for: Affect=how people feel inside, Behavior=peoples actions and Cognition= what people think.
 * Social Influence which is behaviors and attitudes brought about by others
 * Social Interaction: How people interact with one another in the social world.
 * Social Perception: Hoe we interpret social objects.
 * The Person vs. Situation debate: is a person influenced primarily by the situation/environment or their biology (NB: this is very similar to the nature vs. nurture debate)

The Person vs. Situation debate is about what influences behavior. In the tutorial we participated in an exercise placing students along a spectrum to see whether they were more inclined to either side. I personally sit somewhere in the middle. Although i believe certain characteristics, such as personality traits, can be genetic (This has been illustrated by Twins studies exclusively). Although there has been no conclusive evidence for a 100% match in personality traits in identical twins. This leaves room for an environmental influence. Although their are a lot of studies regarding this issue, i think that more mapping of the genome (a DNA map) which identifies specific genes for specific things could shed more light on this issue. Although humans are not pre-programmed zombies. Experience and our environment also shape our behavior and thoughts significantly. I don't think this debate can ever be fully resolved, although it proper gates interesting research which i think can be more rewarding than a resolution.

This weeks topics also included research method and history, although there is no need to regurgitate boring and mundane information....i'll save that for the exam. A more interesting aspect of this weeks topic was about culture and nature. The key premise of social psychology is to seek answers for the question how do people think, feel and act? In regard to nature approach to answering these questions scientists examine genes, hormones, brain structure and physiology. Scientists and psychologists from this persuasion generally use the evolutionary theory to explain behavior. Darwin's theory of 'natural selection' investigates traits that increase an animals survival. Culture on the other hand focuses on how environmental factors, such as society and community, can shape peoples thoughts, feelings and actions. ‘ Culture is an information based system, involving both shared understandings and praxis (practical way of doing things), that allows groups of people to live together in an organised fashion and to satisfy their biological and other needs.’ (Baumeister & Bushman, 2008) Cultural influences can be:
 * Parents, society/community and experience
 * Shared things/beliefs i.e. food, language, style of government/shared ideas and meanings which can be defined in language.

In this weeks tutorial we did a number of exercises which classified people into groups. To begin with it was generally factual categorization (eye colour, where you were born etc). Then we formed groups based of beliefs i.e. political or religious. It was interesting to see how the groups were formed and what demographic was in each group. I found it easy to get along with people in each group, probably because of our similarity that put us there in the first place. I am interested to learn more about what drives humans towards certain beliefs and behaviors, how people form groups and make social connections.


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User:Eml4500.f08.lulz.strack/hw5

MATLAB Code
The first part of the required MATLAB code debugs part of the two-bar truss system from Homework 2. The code is presented below for the corrected system representation and is called "TwoBarTrussSystemDebugged.m". It uses the same three functions (PlaneTrussElement.m, NodalSoln.m, PlaneTrussResults.m) defined in Homework 2. The only difference between "TwoBarTrussSystem.m" and "TwoBarTrussSystemDebugged.m" is Line 40 of code reproduced below the program.

The first part of the program defines the specifications of each bar and node including modulus of elasticity (e), cross-sectional area (A), and loading values (P). The "nodes" matrix defines the absolute locations of the three nodes. The "conn" matrix defines the global node numbers at the ends of each bar or element. Each row represents an element. As shown in the code, there are 2 sets of 2 numbers for the 2 nodes corresponding to each element. The "lmm" matrix represents the global degrees of freedom (dof) for each element. Each node has 2 dof's; therefore, each element has 4 corresponding dof's. Degrees of freedom are numbered from 1 to 6 where the first 2 correspond to the x and y direction of Global Node 1, the second 2 correspond to the x and y direction of Global Node 2, and the third 2 correspond to the x and y direction of Global Node 3. From all this data, the global stiffness matrix (K) is created. The displacements of the nodes are calculated by isolating the relevant parts of the global stiffness matrix, inverting them, and multiplying by the load array (R). Finally, the reactions (reactions) at each node are calculated by multiplying the displacement array (d) by the global stiffness matrix.

TwoBarTrussSystemDebugged.m
Line 40

Before

After

Command Window Output

>> TwoBarTrussSystemDebugged

k_local =

0.75       -0.75        -0.75         0.75

k =

0.5625     0.32476      -0.5625     -0.32476      0.32476       0.1875     -0.32476      -0.1875      -0.5625     -0.32476       0.5625      0.32476     -0.32476      -0.1875      0.32476       0.1875

k_local =

5   -5    -5     5

k =

2.5        -2.5         -2.5          2.5         -2.5          2.5          2.5         -2.5         -2.5          2.5          2.5         -2.5          2.5         -2.5         -2.5          2.5

K =

0.5625     0.32476      -0.5625     -0.32476            0            0      0.32476       0.1875     -0.32476      -0.1875            0            0      -0.5625     -0.32476       3.0625      -2.1752         -2.5          2.5     -0.32476      -0.1875      -2.1752       2.6875          2.5         -2.5            0            0         -2.5          2.5          2.5         -2.5            0            0          2.5         -2.5         -2.5          2.5

R =

0    0     0     7     0     0

d =

0           0        4.352       6.1271            0            0

reactions =

-4.4378     -2.5622       4.4378      -4.4378

results =

1.7081      5.1244       5.1244       0.6276        3.138        6.276

The displacement array (d) shows that Global Node 2 was displaced because it was the only unfixed node. The output also displays the reactions at the fixed nodes in the "reactions" matrix. The results matrix (results) shows the strain, stress, and axial force in each element. Homework 2, the results matrix changed from a 3x5 to a 3x3 matrix, showing that the bug was removed. Below are the tabular results of the output above. These values match the values computed using the statics method from Homework 2.

The next part of the MATLAB code constructs and loads a truss system composed of six bars. The code below represents Example 4.1 (SixBarTrussEx.m) on page 226 of the textbook. The code for this program is structured similarly to the code for the two-bar truss system. However, there are six elements and five nodes instead of two elements and three nodes.

There are two cases for this example. The original system (SixBarTrussEx.m), Case 1, is the code provided by textbook. Case 2 (SixBarTrussExModified.m) is the same system with a different modulus of elasticity for each element.

SixBarTrussEx.m (Case 1)
Command Window Output

>> SixBarTrussEx

K =

Columns 1 through 5

85355       35355       -50000            0            0        35355        35355            0            0            0       -50000            0        85355       -35355            0            0            0       -35355  1.0202e+005            0 0           0            0            0        71554            0            0            0            0       -35777            0            0            0            0            0            0            0            0       -66667            0       -35355       -35355       -35355        35355       -71554       -35355       -35355        35355       -35355        35777

Columns 6 through 10

0           0            0       -35355       -35355            0            0            0       -35355       -35355            0            0            0       -35355        35355            0            0       -66667        35355       -35355       -35777            0            0       -71554        35777        17889            0            0        35777       -17889            0        71554        35777       -71554       -35777            0        35777        84555       -35777       -17889        35777       -71554       -35777  2.1382e+005            0 -17889      -35777       -17889            0  1.0649e+005

R =

0           0        10000        17321            0            0            0            0            0            0

d =

0           0      0.21311      0.24998            0            0            0            0   -0.0060971     0.012242

reactions =

-10873     -217.27       874.27      -437.13      -1.7279       -16666

results =

5.3276e-005      10.655        10655 -4.6334e-006    -0.92669      -926.69 -4.8873e-006    -0.97746      -977.46 -8.3326e-005     -16.665       -16665 1.5363e-006     0.30727       307.27 -9.659e-009  -0.0019318      -1.9318

The displacement array (d) shows that Global Nodes 2 and 5 were displaced because these were the only unfixed nodes. The output also displays the reactions at the fixed nodes in the "reactions" matrix. The results matrix (results) show the strain, stress, and axial force in each element. Below are the tabular results of the output above.

To visualize the loaded and unloaded system, the following code graphs the system in its loaded and unloaded state. In order to be able to realize the direction of node movement, the unfixed node displacements have been magnified by the displacement magnification factor (m) of 3000. As shown, the original values from the displacement array (d(3), d(4), d(9), d(10)) above were used below to get the right proportions of the nodal displacements.

SixBarTrussDisplay.m
Command Window Output

>> SixBarTrussDisplay

The dashed red line represents the undeformed system shape. The solid blue line represents the deformed system shape. The abbreviations "GN" and "E" stand for Global Node and Element, respectively. Note only nodes 2 and 5 are displaced because they are unfixed.

The code below is the code for Case 2. Below the program are the line changes to the program for Case 1.

SixBarTrussExModified.m (Case 2)
Lines 2, 22, and 31

Before

After

Command Window Output

>> SixBarTrussExModified

K =

Columns 1 through 5

76391       38891       -37500            0            0        38891        38891            0            0            0       -37500            0        69320       -31820            0            0            0       -31820        98486            0            0            0            0            0        71554            0            0            0            0       -35777            0            0            0            0            0            0            0            0       -66667            0       -38891       -38891       -31820        31820       -71554       -38891       -38891        31820       -31820        35777

Columns 6 through 10

0           0            0       -38891       -38891            0            0            0       -38891       -38891            0            0            0       -31820        31820            0            0       -66667        31820       -31820       -35777            0            0       -71554        35777        17889            0            0        35777       -17889            0        89443        44721       -89443       -44721            0        44721        89027       -44721       -22361        35777       -89443       -44721  2.3171e+005        16015 -17889      -44721       -22361        16015  1.1096e+005

R =

0           0        10000        17321            0            0            0            0            0            0

d =

0           0      0.26485      0.26083            0            0            0            0   0.00063864    -0.001246

reactions =

-9908.3      23.619      -90.274       45.137      -1.4008       -17389

results =

6.6213e-005      9.9319       9931.9 5.347e-007    0.096245       96.245 5.0465e-007     0.10093       100.93 -8.6943e-005     -17.389       -17389 -1.5183e-007   -0.033402      -33.402 -6.2645e-009  -0.0015661      -1.5661

The displacement array (d) shows that Global Nodes 2 and 5 were displaced because these were the only unfixed nodes. The output also displays the reactions at the fixed nodes in the "reactions" matrix. The results matrix (results) show the strain, stress, and axial force in each element. Below are the tabular results of the output above.

To visualize the loaded and unloaded system, the following code graphs the system in its unloaded state and loaded states for Cases 1 and 2. For this display, the magnification factor was changed to 10000. Also, the labels have been removed for the global nodes and elements. The displacement array for Case 1 was changed from "d" to "d1". For Case 2, a new array was created called "d2" using the values in the displacement array from the output above.

SixBarTrussDisplayModified.m
Command Window Output

>> SixBarTrussDisplayModified

The dashed red line represents the undeformed system shape. The solid blue line represents the deformed system shape for Case 1. The dotted black line represents the deformed system shape for Case 2.

The next part of the MATLAB code constructs and loads a three dimensional truss system composed of three bars. The code below represents Example 4.2 (ThreeBarSpaceTrussEx.m) on page 230 of the textbook. The code for this program is structured similarly to the code for the two previous systems. However, this system is a space system where displacement and reactions are measured in three dimensions.

The format of the array "conn" doesn't change. On the other hand, the number of columns in "nodes" and "lmm" increase because the matrices are accommodating for movement in the Z-direction.

This program utilizes two new functions: "SpaceTrussElement.m" and "SpaceTrussResults.m". These are displayed below the program.

SpaceTrussResults.m
Command Window Output

>> ThreeBarSpaceTrussEx

K =

Columns 1 through 6

1459.7      2919.3      -3040.9            0            0            0       2919.3       5838.6      -6081.9            0            0            0      -3040.9      -6081.9       6335.3            0            0            0            0            0            0       3566.7      -3566.7       4953.8            0            0            0      -3566.7       3566.7      -4953.8            0            0            0       4953.8      -4953.8       6880.3            0            0            0            0            0            0            0            0            0            0            0            0            0            0            0            0            0            0      -1459.7      -2919.3       3040.9      -3566.7       3566.7      -4953.8      -2919.3      -5838.6       6081.9       3566.7      -3566.7       4953.8       3040.9       6081.9      -6335.3      -4953.8       4953.8      -6880.3

Columns 7 through 12

0           0            0      -1459.7      -2919.3       3040.9            0            0            0      -2919.3      -5838.6       6081.9            0            0            0       3040.9       6081.9      -6335.3            0            0            0      -3566.7       3566.7      -4953.8            0            0            0       3566.7      -3566.7       4953.8            0            0            0      -4953.8       4953.8      -6880.3            0            0            0            0            0            0            0            0            0            0            0            0            0            0        60000            0            0       -60000            0            0            0       5026.4      -647.45       1912.9            0            0            0      -647.45       9405.4       -11036            0            0       -60000       1912.9       -11036        73216

d =

0           0            0            0            0            0            0            0            0     -0.18705       -2.592      -0.3858

reactions =

6666.7       13333       -13889      -6666.7       6666.7      -9259.3            0            0        23148

results =

0.00050936      101.87        20375   0.00033036       66.072        13214   -0.0001929       -38.58       -23148

The displacement array (d) shows that Global Node 4 was displaced because it was the only unfixed node. The output also displays the reactions at the fixed nodes in the "reactions" matrix. The results matrix (results) show the strain, stress, and axial force in each element. Below are the tabular results of the output above.

This problem is statically determinate. Because the system occupies a three-dimensional space, there exists three equations for equilibrium in the x, y, and z directions. There are also three unknown variables which are the axial forces reacting at Global Node 4 to accommodate the loading force (P).

The relationship between axial forces and directional forces is governed by the following equations.

The place holder "n" represents the nth Global Node or Element. The three equilibrium equations are shown below using the directional forces derived above.

When the equations for the directional forces are substituted into the above equations, the new equations only have the three axial forces as unknowns. The coefficients of these axial forces make up the matrix "F" in the following code. The "L" array defines the lengths of the three elements. The "R" array is the loading at Global Node 4 by force P in the negative Y-direction. The program below simultaneously solves the system of equations to determine the axial forces in the elements.

ThreeBarSpaceTrussStatics.m
Command Window Output

>> ThreeBarSpaceTrussStatics

L =

2.9339      2.8543            2

R =

0      20000           0

F =

0.32721     -0.5045            0      0.65441       0.5045            0     -0.68168     -0.70069           -1

axial_forces =

20375       13214       -23148

The axial forces produced using the statics method matches exactly to the values obtained in the table above for the FEA method. This proves that the system is statically determinate.

To visualize the loaded and unloaded system, the following code graphs the system in its loaded and unloaded state. In order to be able to realize the direction of node movement, the unfixed node displacements have been magnified by the displacement magnification factor (m) of 300. As shown, the original values from the displacement array (d(10), d(11), d(12)) above were used below to get the right proportions of the nodal displacements.

ThreeBarSpaceTrussDisplay.m
Command Window Output

>> ThreeBarSpaceTrussDisplay

The dashed red line represents the undeformed system shape. The solid blue line represents the deformed system shape. The abbreviations "GN" and "E" stand for Global Node and Element, respectively. Note only node 4 is displaced because it is unfixed. This view is a perspective view as if the observer was looking at the system from the global location (-2,-2,3).

The next three view are views along each axis (X, Y, Z), respectively. Line 27 from "ThreeBarTrussDisplay.m" was modified as shown below for each plot. The only other changes to the code were locations of the "GN" and "E" labels which are trivial.

Line 27

Before

After View along X-axis:

View along Y-axis:

View along Z-axis:

title: User:Eml4500.f08.lulz.strack/hw5


text:
User:Emesee/Policy proposals

Reverts
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Red links
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title: User:Emesee/Policy proposals


text:
User:Eml4500.f08.Lulz.Layton.Eric/Hw5

=Homework 5=

Justification of Reduction of Stiffness Matrix for 2-bar Truss Problem
From the general force displacement relationship, the following relation can be derived:

 * $$\mathbf{kd}= \mathbf{F}$$ => $$\mathbf{kd - F}= \mathbf{0}\;\;(1)$$

Using this relation, the Principle of Virtual Work can be expressed as follows:

 * $$\mathbf{W}\cdot\mathbf{(kd - F)} = \mathbf{0}$$ for all $$\mathbf{W}\;\;(2)$$

Where $$\mathbf{W}\;$$ refers to the weighting matrix. It is useful to show the equivalency of the above two expressions. While is is trivial to show how equation 2 comes from equation 1, it is more challenging to show how equation 1 comes from equation 2. In order to prove that this is the case, consider the following weighting coefficients:

 * $$W_{1} = 1\;$$, $$\;W_{2}=...=W_{6} = 0$$

Which can be expressed in transposed form as:

 * $$\mathbf{W^{T}} = \begin{bmatrix} 1&0&0&0&0&0\end{bmatrix}$$

Applying this choice for $$\mathbf{W}$$ to equation 2 yields the following result:

 * $$\mathbf{W}\cdot\mathbf{(kd - F)} = [1\cdot\sum_{j=1}^{6} (k_{1j}d_{j} - F_{1}) + 0\cdot\sum_{j=1}^{6} (k_{2j}d_{j} - F_{2}) + ... + 0\cdot\sum_{j=1}^{6} (k_{6j}d_{j} - F_{6})] = 0$$

It is easily shown that the result of the above expression is:

 * $$\sum_{j=1}^{6} k_{1j}d_{j} = F_{1}$$

Consider another set of weighting coefficients and resulting transposed form:

 * $$W_{1} = 0\;$$, $$W_{2} = 0\;$$ , $$\;W_{3}=...=W_{6} = 0$$

 * $$\mathbf{W^{T}} = \begin{bmatrix} 0&1&0&0&0&0\end{bmatrix}$$

As before, applying this matrix to equation 2 yields a result of $$\sum_{j=1}^{6} k_{2j}d_{j} - F_{2} = 0$$. Thus, it is apparent that for any choice of weighting matrix $$\mathbf{W}$$, the result is equivalent to $$\mathbf{kd - F}= \mathbf{0}$$, which is equation 1.

Using this method, the rest of the force-displacement relations can be found.

Choice 3:

 * $$W_1=0\;, W_2=0, W_3=1, W_4=0, W_5=0, W_6=0$$

 * $$\mathbf{W^T}=\begin{bmatrix} 0&0&1&0&0&0 \end{bmatrix}$$

 * $$\sum_{j=1}^{6} k_{3j}d_{j} - F_{3} = 0$$

 * $$\sum_{j=1}^{6} k_{3j}d_{j} = F_{3}$$

Choice 4:

 * $$W_1=0\;, W_2=0, W_3=0, W_4=1, W_5=0, W_6=0$$

 * $$\mathbf{W^T}=\begin{bmatrix} 0&0&0&1&0&0 \end{bmatrix}$$

 * $$\sum_{j=1}^{6} k_{4j}d_{j} - F_{4} = 0$$

 * $$\sum_{j=1}^{6} k_{4j}d_{j} = F_{4}$$

Choice 5:

 * $$W_1=0\;, W_2=0, W_3=0, W_4=0, W_5=1, W_6=0$$

 * $$\mathbf{W^T}=\begin{bmatrix} 0&0&0&0&1&0 \end{bmatrix}$$

 * $$\sum_{j=1}^{6} k_{5j}d_{j} - F_{5} = 0$$

 * $$\sum_{j=1}^{6} k_{5j}d_{j} = F_{5}$$

Choice 6:

 * $$W_1=0\;, W_2=0, W_3=0, W_4=0, W_5=0, W_6=1$$

 * $$\mathbf{W^T}=\begin{bmatrix} 0&0&0&0&0&1 \end{bmatrix}$$

 * $$\sum_{j=1}^{6} k_{6j}d_{j} - F_{6} = 0$$

 * $$\sum_{j=1}^{6} k_{6j}d_{j} = F_{6}$$

Returning to the specific 2-bar truss problem, accounting for the boundary conditions allows the removal of the following global displacements:

 * $$d_{1}=d_{2}=d_{5}=d_{6}=0\;$$

The weighting coefficients must be "kinematically admissable" - this means that it must not violate the boundary conditions. Thus, the following is a result of the above expression:

 * $$W_{1}=W_{2}=W_{5}=W_{6}=0\;$$

This leads to the conclusion that the weighting coefficients are essentially virtual displacements. As a result of the previous 2 relations, equation 2 can now be rewritten as:

 * $$\begin{Bmatrix}W_{3}\\W_{4}\end{Bmatrix}\cdot\mathbf{(\bar{k}\bar{d} - \bar{F})} = 0$$ for all $$\begin{Bmatrix}W_{3}\\W_{4}\end{Bmatrix}$$

Where $$\mathbf{\bar{k}}\;$$, $$\mathbf{\bar{d}}\;$$, and $$\mathbf{\bar{F}}$$ are defined as follows:

 * $$\mathbf{\bar{k}} = \begin{bmatrix}k_{33}&k_{34}\\k_{43}&k_{44}\end{bmatrix}$$

 * $$\mathbf{\bar{d}} = \begin{Bmatrix}d_{3}\\d_{4}\end{Bmatrix}$$

 * $$\mathbf{\bar{F}} = \begin{Bmatrix}F_{3}\\F_{4}\end{Bmatrix}$$

An example of weighting is the calculation of a student's final grade in this course, which is given as follows: $$course\;grade = \alpha_{0} + HW Grade + \Sigma(\alpha Exam Grade)$$

Prove Kd=F is equivalent to W(kd-F)=0
$$\mathbf{W}\cdot(\mathbf{K}\mathbf{d}-\mathbf{F})=\begin{Bmatrix} w_3\\ w_4\\ \end{Bmatrix}\cdot(\mathbf{\bar{K}}\mathbf{\bar{d}}-\mathbf{\bar{F}})=0 $$ for all $$ \begin{Bmatrix} w_3\\ w_4\\ \end{Bmatrix}$$

Remember and apply our previous derivations for K, d, and F:

k_{33} & k_{34}\\ k_{43} & k_{44}\\ \end{bmatrix}$$ d_3\\ d_4\\ \end{bmatrix}$$ F_3\\ F_4\\ \end{bmatrix}$$
 * $$\mathbf{K}=\begin{bmatrix}
 * $$\mathbf{d}=\begin{bmatrix}
 * $$\mathbf{F}=\begin{bmatrix}

$$\begin{Bmatrix} w_3\\ w_4\\ \end{Bmatrix}\cdot(\begin{bmatrix} k_{33} & k_{34}\\ k_{43} & k_{44}\\ \end{bmatrix}\begin{bmatrix} d_{3}\\ d_{4} \\ \end{bmatrix}-\begin{bmatrix} F_{3}\\ F_{4} \\ \end{bmatrix})=0$$

This matrix can be converted by the dot product and simple matrix math to a single algebraic expression:

$$w_3(k_{33}d_3+k_{34}d_4-F_3)+w_4(k_{43}d_3+k_{44}d_4-F_4)=0$$

This equation will have multiple solutions depending on $$w_3$$ and $$w_4$$. Since by definition the relationships apply for any $$w_3$$ and $$w_4$$ then any arbitrary value can be applied to solve the expression created above. Picking different sets of values will create a system of equations that, when written in matrix form, are identical to $$\mathbf{W}\cdot(\mathbf{K}\mathbf{d}-\mathbf{F})=0$$ and other values will also prove identity.

If we pick 0 for our values of w:

$$w_3=0, w_4=0$$

$$(0)(k_{33}d_3+k_{34}d_4-F_3+(0)(k_{43}d_3+k_{44}d_4-F_4)=0$$

$$0=0$$

If we pick 0 for one value of w and 1 for the other:

$$w_3=0, w_4=1$$

$$(0)(k_{33}d_3+k_{34}d_4-F_3+(1)(k_{43}d_3+k_{44}d_4-F_4)=0$$

$$k_{43}d_3+k_{44}d_4-F_4=0$$

$$F_4=k_{43}d_3 + k_{44}d_4$$

If we pick 1 for our values of w:

$$w_3=1, w_4=1$$

$$(1)(k_{33}d_3+k_{34}d_4-F_3+(1)(k_{43}d_3+k_{44}d_4-F_4)=0$$

$$k_{33}d_3+k_{34}d_4+k_{43}d_3+k_{44}d_4=F_3+F_4$$

$$\begin{bmatrix} k_{33}d_3 + k_{34}d_4\\ k_{43}d_3 + k_{44}d_4\\ \end{bmatrix}=\begin{bmatrix} F_3\\ F_4\\ \end{bmatrix}$$

These three examples each bring us back to our proof that:

$$\mathbf{Kd}=\mathbf{F}$$

Derivation of $$\mathbf{k^{(e)}} = \mathbf{T^{(e)^{T}}}\mathbf{\hat{k}^{(e)}}\mathbf{T^{(e)}}$$
Recall the force-displacement relationship adjusted for axial degrees of freedom $$\mathbf{q^{(e)}}$$:

 * $$\mathbf{\hat{k}^{(e)}}\mathbf{q^{(e)}} = \mathbf{P^{(e)}}$$

This relationship can be rearranged to the following:

 * $$\mathbf{\hat{k}^{(e)}}\mathbf{q^{(e)}} - \mathbf{P^{(e)}} = \mathbf{0}\;\;(1)$$

Applying the Principle of Virtual Work results in the following:

 * $$\mathbf{\hat{W}}\cdot(\mathbf{\hat{k}^{(e)}}\mathbf{q^{(e)}} - \mathbf{P^{(e)}}) = \mathbf{0}$$ for all $$\mathbf{\hat{W}}$$

It has previously been shown that equations 1 and 2 are equivalent. Recalling the following two relations:

 * $$\mathbf{q^{(e)}} = \mathbf{T^{(e)}}\mathbf{d^{(e)}}\;\;(3)$$

 * $$\mathbf{\hat{W}^{(e)}} = \mathbf{T^{(e)}}\mathbf{W^{(e)}}\;\;(4)$$

$$\mathbf{\hat{W}}$$ refers to virtual axis displacements, corresponding to $$\mathbf{q^{(e)}}$$.

$$\mathbf{W}$$ refers to virtual displacements in global coordinate system, corresponding to $$\mathbf{d^{(e)}}$$.

Placing equations 3 and 4 into equation 2 yields the following:

 * $$(\mathbf{T^{(e)}}\mathbf{W})\cdot[\mathbf{\hat{k}^{(e)}}(\mathbf{T^{(e)}}\mathbf{d^{(e)}}) - \mathbf{P^{(e)}}] = \mathbf{0}$$ for all $$\mathbf{W}\;\;(5)$$

Recall the following two vector identities:

 * $$(\mathbf{A}\cdot\mathbf{B})^{T} = \mathbf{B}^{T}\mathbf{A}^{T}\;\;(6)$$

Let matrix A and matrix B be as follows:

$$A = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{bmatrix}$$

$$B = \begin{bmatrix} 7 & 8 & 9 \\ 1 & 2 & 3 \\ 4 & 5 & 6 \end{bmatrix}$$

Then matrix $$AB$$ and matrix $$AB^T$$ are as follows:

$$AB = \begin{bmatrix} 21 & 27 & 33 \\ 57 & 72 & 87 \end{bmatrix}$$

$$AB^T = \begin{bmatrix} 21 & 57 \\ 27 & 72 \\ 33 & 87 \end{bmatrix}$$

The matrix $$A^T, B^T and A^TB^T$$ are as follows:

$$A^T = \begin{bmatrix} 1 & 4 \\ 2 & 5 \\ 3 & 6 \end{bmatrix}$$

$$B^T = \begin{bmatrix} 7 & 1 & 4 \\ 8 & 2 & 5 \\ 9 & 3 & 6 \end{bmatrix}$$

$$A^TB^T = \begin{bmatrix} 21 & 57 \\ 27 & 72 \\ 33 & 87 \end{bmatrix}$$

Therefore, $$A^TB^T = (AB)^T$$

 * $$\mathbf{a}\cdot\mathbf{b} = \mathbf{a}^{T}\mathbf{b}\;\;(7)$$

Applying equations 7 and 8 into equation 5 yields the following derivation:

$$(\mathbf{T^{(e)}}\mathbf{W})^{T}[\mathbf{\hat{k}^{(e)}}(\mathbf{T^{(e)}}\mathbf{d^{(e)}}) - \mathbf{P^{(e)}}] = \mathbf{0}$$ for all $$\mathbf{W}$$

$$\mathbf{T^{(e)}}^{T}\mathbf{W}^{T}[\mathbf{\hat{k}^{(e)}}(\mathbf{T^{(e)}}\mathbf{d^{(e)}}) - \mathbf{P^{(e)}}] = \mathbf{0}$$ for all $$\mathbf{W}$$

$$\mathbf{W}^{T}[\left(\mathbf{T^{(e)}}^{T}\mathbf{\hat{k}^{(e)}}\mathbf{T^{(e)}}\right)\mathbf{d^{(e)}} - \left\{\mathbf{T^{(e)}}^{T}\mathbf{P^{(e)}}\right\}] = \mathbf{0}$$

In the last expression, the term inside the parentheses is equivalent to $$\mathbf{k^{(e)}}$$, and the term inside the curly brackets is equivalent to $$\mathbf{f^{(e)}}$$. Making these substitutions yields the following result:

$$\mathbf{W^{T}}[\mathbf{k^{(e)}}\mathbf{d^{(e)}} - \mathbf{f^{(e)}}] = \mathbf{0}$$ for all $$\mathbf{W}$$.

Because the previous equation has been proven for all values of $$W$$ we can choose $$W$$ to be one and hence the equation becomes:

$$k^{(e)}d^{(e)} = f^{(e)}$$

Continuous Case - Motivational Model Problem

Consider the elastic bar shown in Figure 1. The bar has a varying cross-section and Young's modulus, and is subject to a time-dependent loading condition consisting of a distributed, varying axial load as well as a concentrated load, signified by $$f(x,t)\;$$ and $$P(x)\;$$ respectively. A section of the bar is shown in Figure 2.

The product $$A(x)\rho (x)\;$$ can be expressed as $$m(x)\cdot dx\;$$, where $$m(x)\;$$ is the mass per unit length. The term $$\frac{\partial{^{2}u}}{\partial{x^{2}}}$$ refers to the acceleration of the deforming bar, and can be represented as $$\ddot{u}$$.

The sum of forces on the section can be written as follows:

 * $$\Sigma F_{x} = 0 = -N{x,t} + N(x+dx,t) +f(x,t)dx - m(x)\ddot{u}dx$$

 * $$\Sigma F_{x} = 0 = \frac{\partial{N(x,t)}}{\partial x}dx + f(x,t)dx - m(x)\ddot{u}dx + h.o.t\;\;(1)$$

"h.o.t" refers to the higher order terms.

Recall the general definition of the Taylor Series expansion, $$f(x+dx) = f(x) + \frac{\partial{f(x)}}{\partial x}dx + \frac{1}{2}\frac{\partial{^{2}x}}{\partial{x}^{2}} + ...$$, where $$\frac{1}{2}\frac{\partial{^{2}x}}{\partial{x}^{2}} + ...$$ are the higher order terms and are typically neglected. Dividing out the $$dx\;$$ terms and applying this concept to equation 1 yields equation 2, the equation of motion for the elastic bar:

 * $$\frac{\partial{N}}{\partial x} + f = m\ddot{u}\;\;(2)$$

The constitutive relation for the bar can be given as equation 3:

 * $$N(x,t) = A(x)\sigma (x,t)\;\;(3)$$

The $$\sigma (x,t)\;$$ term can be broken down as follows:

 * $$\sigma (x,t)= E(x)\epsilon (x,t)\;$$

 * $$\epsilon (x,t)= \frac{\partial{u}}{\partial{x}}(x,t)\;$$

The partial differential equation of motion for the elastic bar can be given as equation 4, which is the result of substituting equation 3 and resulting breakdown of terms into equation 2:

 * $$\frac{\partial}{\partial x}[A(x)E(x)\frac{\partial{u}}{\partial{x}}] + f(X,t) = m(x)\ddot{u}\;\;(4)$$

In order to manipulate this equation, boundary and initial conditions are required. Since the equation is of 2nd order in both x and t, two boundary and two initial conditions are necessary. The initial conditions consist of an initial displacement and velocity, given as $$[init. vel., init. disp]\;$$. The boundary conditions depend on the type of system being analyzed. Figure 3 shows a bar of length L with both ends constrained. The boundary conditions for this system are:

 * $$u(0,t) = 0 = u(L,t)\;$$

Figure 4 again shows a bar of length L, but with one end free and subject to force F. The first boundary condition, as with the constrained case, is:

 * $$u(0,t) = 0\;\;(bc\;1)$$

The second boundary condition can be derived using the following line of thought, similar to that used to derive equation 4:

 * $$N(L,t) = F(t)\;$$

 * $$N(L,t) = A(L)\sigma (L,t)\;$$

 * $$\sigma (L,t) = E(L)\epsilon (L,t)\;$$

 * $$\epsilon (L,t) = \frac{\partial u}{\partial x} (L,t)$$

Using the above relations, the second boundary condition for this case can be written as follows:

 * $$\frac{\partial u(L,t)}{\partial x} = \frac{F(t)}{A(L)E(L)}\;\;(bc\;2)$$

Composite Materials Research
Composite materials are materials made from at least two separate materials. The materials used in a composite material are of differing physical/chemical properties so that the final composite material has several material properties. Typical composites include carbon fiber, plywood, fiberglass, concrete, and some types of Kevlar reinforcement. Composites exist as a matrix element and a reinforcement element, both of which are important factors in composite creation. The reinforcement material combines with the matrix materials to create a finite combination of two or more materials, which creates the special composite material.

Composites are arranged so that any kind of loading is applied parallel to their fiber arrangement axis. They are not isotropic materials, so any loading that is applied to perpendicular to the fiber arrangement axis could result in failure of the composite material. If any of the material of the composite becomes damaged or separated from the matrix, imminent failure can occur.

An excellent source for composite material information is on Wikipedia at the following link: http://en.wikipedia.org/wiki/Composite_materials

MATLAB Code
The first part of the required MATLAB code debugs part of the two-bar truss system from Homework 2. The code is presented below for the corrected system representation and is called "TwoBarTrussSystemDebugged.m". It uses the same three functions (PlaneTrussElement.m, NodalSoln.m, PlaneTrussResults.m) defined in Homework 2. The only difference between "TwoBarTrussSystem.m" and "TwoBarTrussSystemDebugged.m" is Line 40 of code reproduced below the program.

The first part of the program defines the specifications of each bar and node including modulus of elasticity (e), cross-sectional area (A), and loading values (P). The "nodes" matrix defines the absolute locations of the three nodes. The "conn" matrix defines the global node numbers at the ends of each bar or element. Each row represents an element. As shown in the code, there are 2 sets of 2 numbers for the 2 nodes corresponding to each element. The "lmm" matrix represents the global degrees of freedom (dof) for each element. Each node has 2 dof's; therefore, each element has 4 corresponding dof's. Degrees of freedom are numbered from 1 to 6 where the first 2 correspond to the x and y direction of Global Node 1, the second 2 correspond to the x and y direction of Global Node 2, and the third 2 correspond to the x and y direction of Global Node 3. From all this data, the global stiffness matrix (K) is created. The displacements of the nodes are calculated by isolating the relevant parts of the global stiffness matrix, inverting them, and multiplying by the load array (R). Finally, the reactions (reactions) at each node are calculated by multiplying the displacement array (d) by the global stiffness matrix.

TwoBarTrussSystemDebugged.m
Line 40

Before

After

Command Window Output

>> TwoBarTrussSystemDebugged

k_local =

0.75       -0.75        -0.75         0.75

k =

0.5625     0.32476      -0.5625     -0.32476      0.32476       0.1875     -0.32476      -0.1875      -0.5625     -0.32476       0.5625      0.32476     -0.32476      -0.1875      0.32476       0.1875

k_local =

5   -5    -5     5

k =

2.5        -2.5         -2.5          2.5         -2.5          2.5          2.5         -2.5         -2.5          2.5          2.5         -2.5          2.5         -2.5         -2.5          2.5

K =

0.5625     0.32476      -0.5625     -0.32476            0            0      0.32476       0.1875     -0.32476      -0.1875            0            0      -0.5625     -0.32476       3.0625      -2.1752         -2.5          2.5     -0.32476      -0.1875      -2.1752       2.6875          2.5         -2.5            0            0         -2.5          2.5          2.5         -2.5            0            0          2.5         -2.5         -2.5          2.5

R =

0    0     0     7     0     0

d =

0           0        4.352       6.1271            0            0

reactions =

-4.4378     -2.5622       4.4378      -4.4378

results =

1.7081      5.1244       5.1244       0.6276        3.138        6.276

The displacement array (d) shows that Global Node 2 was displaced because it was the only unfixed node. The output also displays the reactions at the fixed nodes in the "reactions" matrix. The results matrix (results) shows the strain, stress, and axial force in each element. Homework 2, the results matrix changed from a 3x5 to a 3x3 matrix, showing that the bug was removed. Below are the tabular results of the output above. These values match the values computed using the statics method from Homework 2.

The next part of the MATLAB code constructs and loads a truss system composed of six bars. The code below represents Example 4.1 (SixBarTrussEx.m) on page 226 of the textbook. The code for this program is structured similarly to the code for the two-bar truss system. However, there are six elements and five nodes instead of two elements and three nodes.

There are two cases for this example. The original system (SixBarTrussEx.m), Case 1, is the code provided by textbook. Case 2 (SixBarTrussExModified.m) is the same system with a different modulus of elasticity for each element.

SixBarTrussEx.m (Case 1)
Command Window Output

>> SixBarTrussEx

K =

Columns 1 through 5

85355       35355       -50000            0            0        35355        35355            0            0            0       -50000            0        85355       -35355            0            0            0       -35355  1.0202e+005            0 0           0            0            0        71554            0            0            0            0       -35777            0            0            0            0            0            0            0            0       -66667            0       -35355       -35355       -35355        35355       -71554       -35355       -35355        35355       -35355        35777

Columns 6 through 10

0           0            0       -35355       -35355            0            0            0       -35355       -35355            0            0            0       -35355        35355            0            0       -66667        35355       -35355       -35777            0            0       -71554        35777        17889            0            0        35777       -17889            0        71554        35777       -71554       -35777            0        35777        84555       -35777       -17889        35777       -71554       -35777  2.1382e+005            0 -17889      -35777       -17889            0  1.0649e+005

R =

0           0        10000        17321            0            0            0            0            0            0

d =

0           0      0.21311      0.24998            0            0            0            0   -0.0060971     0.012242

reactions =

-10873     -217.27       874.27      -437.13      -1.7279       -16666

results =

5.3276e-005      10.655        10655 -4.6334e-006    -0.92669      -926.69 -4.8873e-006    -0.97746      -977.46 -8.3326e-005     -16.665       -16665 1.5363e-006     0.30727       307.27 -9.659e-009  -0.0019318      -1.9318

The displacement array (d) shows that Global Nodes 2 and 5 were displaced because these were the only unfixed nodes. The output also displays the reactions at the fixed nodes in the "reactions" matrix. The results matrix (results) show the strain, stress, and axial force in each element. Below are the tabular results of the output above.

To visualize the loaded and unloaded system, the following code graphs the system in its loaded and unloaded state. In order to be able to realize the direction of node movement, the unfixed node displacements have been magnified by the displacement magnification factor (m) of 3000. As shown, the original values from the displacement array (d(3), d(4), d(9), d(10)) above were used below to get the right proportions of the nodal displacements.

SixBarTrussDisplay.m
Command Window Output

>> SixBarTrussDisplay

The dashed red line represents the undeformed system shape. The solid blue line represents the deformed system shape. The abbreviations "GN" and "E" stand for Global Node and Element, respectively. Note only nodes 2 and 5 are displaced because they are unfixed.

The code below is the code for Case 2. Below the program are the line changes to the program for Case 1.

SixBarTrussExModified.m (Case 2)
Lines 2, 22, and 31

Before

After

Command Window Output

>> SixBarTrussExModified

K =

Columns 1 through 5

76391       38891       -37500            0            0        38891        38891            0            0            0       -37500            0        69320       -31820            0            0            0       -31820        98486            0            0            0            0            0        71554            0            0            0            0       -35777            0            0            0            0            0            0            0            0       -66667            0       -38891       -38891       -31820        31820       -71554       -38891       -38891        31820       -31820        35777

Columns 6 through 10

0           0            0       -38891       -38891            0            0            0       -38891       -38891            0            0            0       -31820        31820            0            0       -66667        31820       -31820       -35777            0            0       -71554        35777        17889            0            0        35777       -17889            0        89443        44721       -89443       -44721            0        44721        89027       -44721       -22361        35777       -89443       -44721  2.3171e+005        16015 -17889      -44721       -22361        16015  1.1096e+005

R =

0           0        10000        17321            0            0            0            0            0            0

d =

0           0      0.26485      0.26083            0            0            0            0   0.00063864    -0.001246

reactions =

-9908.3      23.619      -90.274       45.137      -1.4008       -17389

results =

6.6213e-005      9.9319       9931.9 5.347e-007    0.096245       96.245 5.0465e-007     0.10093       100.93 -8.6943e-005     -17.389       -17389 -1.5183e-007   -0.033402      -33.402 -6.2645e-009  -0.0015661      -1.5661

The displacement array (d) shows that Global Nodes 2 and 5 were displaced because these were the only unfixed nodes. The output also displays the reactions at the fixed nodes in the "reactions" matrix. The results matrix (results) show the strain, stress, and axial force in each element. Below are the tabular results of the output above.

To visualize the loaded and unloaded system, the following code graphs the system in its unloaded state and loaded states for Cases 1 and 2. For this display, the magnification factor was changed to 10000. Also, the labels have been removed for the global nodes and elements. The displacement array for Case 1 was changed from "d" to "d1". For Case 2, a new array was created called "d2" using the values in the displacement array from the output above.

SixBarTrussDisplayModified.m
Command Window Output

>> SixBarTrussDisplayModified

The dashed red line represents the undeformed system shape. The solid blue line represents the deformed system shape for Case 1. The dotted black line represents the deformed system shape for Case 2.

The next part of the MATLAB code constructs and loads a three dimensional truss system composed of three bars. The code below represents Example 4.2 (ThreeBarSpaceTrussEx.m) on page 230 of the textbook. The code for this program is structured similarly to the code for the two previous systems. However, this system is a space system where displacement and reactions are measured in three dimensions.

The format of the array "conn" doesn't change. On the other hand, the number of columns in "nodes" and "lmm" increase because the matrices are accommodating for movement in the Z-direction.

This program utilizes two new functions: "SpaceTrussElement.m" and "SpaceTrussResults.m". These are displayed below the program.

SpaceTrussResults.m
Command Window Output

>> ThreeBarSpaceTrussEx

K =

Columns 1 through 6

1459.7      2919.3      -3040.9            0            0            0       2919.3       5838.6      -6081.9            0            0            0      -3040.9      -6081.9       6335.3            0            0            0            0            0            0       3566.7      -3566.7       4953.8            0            0            0      -3566.7       3566.7      -4953.8            0            0            0       4953.8      -4953.8       6880.3            0            0            0            0            0            0            0            0            0            0            0            0            0            0            0            0            0            0      -1459.7      -2919.3       3040.9      -3566.7       3566.7      -4953.8      -2919.3      -5838.6       6081.9       3566.7      -3566.7       4953.8       3040.9       6081.9      -6335.3      -4953.8       4953.8      -6880.3

Columns 7 through 12

0           0            0      -1459.7      -2919.3       3040.9            0            0            0      -2919.3      -5838.6       6081.9            0            0            0       3040.9       6081.9      -6335.3            0            0            0      -3566.7       3566.7      -4953.8            0            0            0       3566.7      -3566.7       4953.8            0            0            0      -4953.8       4953.8      -6880.3            0            0            0            0            0            0            0            0            0            0            0            0            0            0        60000            0            0       -60000            0            0            0       5026.4      -647.45       1912.9            0            0            0      -647.45       9405.4       -11036            0            0       -60000       1912.9       -11036        73216

d =

0           0            0            0            0            0            0            0            0     -0.18705       -2.592      -0.3858

reactions =

6666.7       13333       -13889      -6666.7       6666.7      -9259.3            0            0        23148

results =

0.00050936      101.87        20375   0.00033036       66.072        13214   -0.0001929       -38.58       -23148

The displacement array (d) shows that Global Node 4 was displaced because it was the only unfixed node. The output also displays the reactions at the fixed nodes in the "reactions" matrix. The results matrix (results) show the strain, stress, and axial force in each element. Below are the tabular results of the output above.

This problem is statically determinate. Because the system occupies a three-dimensional space, there exists three equations for equilibrium in the x, y, and z directions. There are also three unknown variables which are the axial forces reacting at Global Node 4 to accommodate the loading force (P).

The relationship between axial forces and directional forces is governed by the following equations.

The place holder "n" represents the nth Global Node or Element. The three equilibrium equations are shown below using the directional forces derived above.

When the equations for the directional forces are substituted into the above equations, the new equations only have the three axial forces as unknowns. The coefficients of these axial forces make up the matrix "F" in the following code. The "L" array defines the lengths of the three elements. The "R" array is the loading at Global Node 4 by force P in the negative Y-direction. The program below simultaneously solves the system of equations to determine the axial forces in the elements.

ThreeBarSpaceTrussStatics.m
Command Window Output

>> ThreeBarSpaceTrussStatics

L =

2.9339      2.8543            2

R =

0      20000           0

F =

0.32721     -0.5045            0      0.65441       0.5045            0     -0.68168     -0.70069           -1

axial_forces =

20375       13214       -23148

The axial forces produced using the statics method matches exactly to the values obtained in the table above for the FEA method. This proves that the system is statically determinate.

To visualize the loaded and unloaded system, the following code graphs the system in its loaded and unloaded state. In order to be able to realize the direction of node movement, the unfixed node displacements have been magnified by the displacement magnification factor (m) of 300. As shown, the original values from the displacement array (d(10), d(11), d(12)) above were used below to get the right proportions of the nodal displacements.

ThreeBarSpaceTrussDisplay.m
Command Window Output

>> ThreeBarSpaceTrussDisplay

The dashed red line represents the undeformed system shape. The solid blue line represents the deformed system shape. The abbreviations "GN" and "E" stand for Global Node and Element, respectively. Note only node 4 is displaced because it is unfixed. This view is a perspective view as if the observer was looking at the system from the global location (-2,-2,3).

The next three view are views along each axis (X, Y, Z), respectively. Line 27 from "ThreeBarTrussDisplay.m" was modified as shown below for each plot. The only other changes to the code were locations of the "GN" and "E" labels which are trivial.

Line 27

Before

After View along X-axis:

View along Y-axis:

View along Z-axis:

Contributing Team Members
Aaron Fisher - Eml4500.f08.Lulz.fisher 21:00, 7 November 2008 (UTC)

Eric Layton - Eml4500.f08.Lulz.Layton.Eric 06:39, 7 November 2008 (UTC)

Sam Miorelli - Eml4500.f08.lulz.abcd 20:48, 7 November 2008 (UTC)

Benjamin Mitchell - Eml4500.f08.Lulz.mitchell.bm 16:59, 7 November 2008 (UTC)

John Saxon - Eml4500.f08.Lulz.js 05:28, 6 November 2008 (UTC)

Andrew Strack - Eml4500.f08.lulz.strack 05:06, 29 October 2008 (UTC)

title: User:Eml4500.f08.Lulz.Layton.Eric/Hw5


text: User:Fatka

title: User:Fatka


text:
User:Stevenarntson/Analytical Writing/Camp

Susan Sontag was one of the first authors to think deeply about cultural effluvia that we might otherwise dismiss as trash. Her essay Notes On Camp is an amazing discussion of those forms of culture that people call "So bad it's good."
 * Today in class we'll take a look at some trash culture and see if a close reading of it might become interesting. In particular, we'll take a look at the comic strips in today's newspaper.


text:
User:TheFearow

Hi All,

I'm a MediaWiki Developer, and keen to contribute to "small" projects (compared to enwiki at least), mainly on the technical side, as they can have a tendency to be left behind.

If anyone has any ideas/suggestions for me, feel free to message me here, although i'm more likely to be reached on. I can also be reached on IRC with the nick MattJohnston.

title: User:TheFearow


text:
User talk:TheFearow

Welcome
Hello TheFearow, 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 06:06, 29 October 2008 (UTC)

title: User talk:TheFearow


text:
Internet and Project Competence/group1

open education

members

 * Earnhart, Lee W. lwearnha@uncc.edu
 * Nowlan, Mary K. mknowlan@uncc.edu
 * St.Charles, Angela R. arstcha@uncc.edu
 * Marlowe, Ashley N.
 * Gilmore, Ashley L.
 * Stewart, Michelle R. mstewa53@uncc.edu
 * Maultsby, Sarah A.  smaults3@uncc.edu
 * Trotter, Allison L. altrotte@uncc.edu
 * Daniela Schweigl   daniela.schweigl@gmx.de
 * Veronica Belcheva

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

Outline for Project

I.  What is Open Education?

Open education does not have a general definition. Perceptions of open education vary. Some say it is an approach to learning that gives students flexibility and choice over what, when, at what pace, where, and how they learn. Open learning often includes aspects of e-learning. Open education can be linked to humanism

II. Humanism “a school of thought that believes human beings are different from other species and possess capacities not found in animals (Edwords, 1989). Humanists, therefore, give primacy to the study of human needs and interests. A central assumption is that human beings behave out of intentionality and values (Kurtz, 2000).” also believe that it is necessary to study the person as a whole, especially as an individual grows and develops over the lifespan. The study of the self, motivation, and goal-setting are also areas of special interest.

http://chiron.valdosta.edu/whuitt/col/affsys/humed.html

Humanists make sense of the world using reason, experience and shared human values. Humanists see no convincing evidence for gods, the supernatural, or life after death. Humanists believe that moral values are properly founded on human empathy and scientific understanding. Humanists believe we must live this life on the basis that it is the only life we'll have -- that, therefore, we must make the most of it for ourselves, each other, and our world.

III. History of Open Education- Lee Earnhart

Open Education in the US came through around the 1960's. They went over to Great Britian and observed their ideas (which were many of what US already had). They put open education into effect so knowledge could be free and open, children could work together, etc.

IV. Key Components in Open Education:

Promotes individual growth, critical thinking, self-reliance, co- operation, commitment to lifelong learning. Most important person is the student not teacher. Not curriculum bound Not age/grade locked Student- centered learning that is intensive, but with relaxed teacher/pupil contact. Needs low teacher/pupil ratio. De-emphasizes schedules	Little control or competition issues between studentsDifficult to draw the line between chaos and order, rebelliousness and expression of rights. Students tend to have better self-concepts and are more creative and co- operative, but academic achievements are lacking

V.  Role of Teacher:- Allison Trotter

Teachers in an open education environment do: Respond to students’ feelings Act as a facilitator for group discussions Use students’ ideas in ongoing discussions Tweak explanations to meet the needs of the students Help students create realistic goals

Teachers in an open education environment do not: Stand up and lecture students Create a negative learning environment for the students Teach to pass a test See learning as limited to one style; there are many different learning styles among students

VI. Role of Student- Angela St. Charles

VII. Challenges-Lee Earnhart

Re-Use - open education formats are open in theory but closed in the practice and re-use. Intrastructure Cost - those putting effot to develop new OERS don't have the chance to make it accessible publicly. Intellectual Property - whether or not open education materials should be commercially usable. Quality Control - ensuring that high quality materials are accessible. Sustainabililty - planning long-term stability.

VIII. Lesson Plans- Mary Katherine Nowlan

VII. Our Thoughts on Open Education everyone needs to input their opinion on this!

Allison Trotter “I like the idea of open education, but I don’t think I would like to implement it all year long in my classroom. Open education is a much more free environment then a regular education classroom. It is something that I think would work for some students, but not for others. “

Daniela Schweigl : Open education don`t has to be good but it can be good. In open education kids can deicide for theirselves what when and with whom they learn. I think open education is good for a lot of children but not for all. My opinion is that good education is a good composition between open education and frontal teaching

Sarah Maultsby's PowerPoint http://upload.wikimedia.org/wikiversity/en/0/02/Open_Education.pdf

Sarah Maultsby Thoughts of Education http://upload.wikimedia.org/wikiversity/en/b/b1/Thoughts_on_Open_Education.pdf

Open Education Powerpoint -

title: Internet and Project Competence/group1


text:
Internet and Project Competence/group2

elementary education program requirements

group members

 * Christiane Rohr
 * Martin, Amanda R.
 * Mclain, Laura K.
 * Reese, Ashley D.
 * Novak, Ashley B.
 * Hemraj, Michelle T.
 * Ledford, Jessica L.
 * Sos, Ashley L.

Power Point Presentation Link
[[media:InternationalProject_Group02.pdf]]

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: Internet and Project Competence/group2


text:
Cooking/Food & Beverage Control

Food and beverage control is vital to an efficient business. While success is not made on controlling costs, many restaurants may fail—not because of poor service, or because of bad food, but because the costs involved ran away from the owner.

Factors to be included in the overall consideration of food and beverage control are:
 * Cost of labor
 * Menu pricing
 * Standards determination
 * Budget
 * Cost-volume-profit analysis
 * Food and beverage equipment management and control.

title: Cooking/Food & Beverage Control


text:
User:Ilaria

it:utente:ilaria

title: User:Ilaria


text:
User talk:Ilaria

it:discussioni utente:ilaria

title: User talk:Ilaria


text:
Food Service Sanitation/Microorganisms

Microorganisms, including virii and bacteria, can range from the relatively harmless to the deadly. In culinary preparation, everything must be sanitized to prevent illness.

Key points to keep in mind:
 * Wash surfaces with hot and soapy water, then sanitizer, before and after each use
 * Be cognizant of the temperature danger zone (41°–140° F, or 5°–60° C)
 * Keep hot foods hot and cold foods cold; bacteria thrive in the mid-range temperatures
 * Cook meats to their "well done" temperatures unless otherwise requested by the customer (be sure to put a notice on the menu that this will increase the likelihood of illness for them; though it may or may not convince the customer to stay away from dangerous circumstances, it will at least protect the establishment from plausible lawsuit for negligence—which is all too common)
 * Steak should be cooked to a core temperature of 145 °F
 * Fish should be cooked to a core temperature of 145 °F
 * Ground beef should be cooked to a core temperature of 160–165 °F
 * Pork should be cooked to a core temperature of 160 °F
 * Eggs should be cooked to a core temperature of 160 °F
 * Chicken should be cooked to a core temperature of 165 °F

title: Food Service Sanitation/Microorganisms


text:
Food Service Sanitation/Personal Hygiene

Personal hygiene is a crucial part of sanitary control in food preparation.

 * Hands should be washed thoroughly under hot, soapy water for thirty seconds before coming into the kitchen
 * Beards should be trimmed closely, or kept under a net if they are long
 * Hair should be short or under a cap or hairnet
 * Clothes should be washed after every work day in the kitchen
 * Aprons can be worn to prevent stains on pants or coats
 * It is a good idea to carry around a fresh, clean towel at all times, to have easy access to it when spills or other detritus need cleaned
 * Employees must shower daily
 * Cuts must be disinfected and covered with a band-aid, and then a glove over that
 * Employees must have no open sores
 * After touching any part of the face, hair, etc., hands must be washed again


text:
User:Yasirniazkhan

TU Kaiserslautern, Germany.

title: User:Yasirniazkhan