text:
User:Sltruesd

Hello,

My name is Sidney Truesdale. I am a junior at UNC-Charlotte. My major is Elementary Education with my concentration in Social Studies. I love working with children and can't wait to start teaching. Before moving to Charlotte I lived in a very small town. There were only 170 people in my graduating class and now I go to school with over 20,000. I hadn't decided on coming to Charlotte until May my senior year and I just thought I would give it a try and if I didn't like it I would transfer, now I am a junior!

title: User:Sltruesd


text:
Template:History of Haiti/Intro

hello

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

title: Template:History of Haiti/Intro


text:
History of Haiti/Catalog

 * Introduction to Haitian History
 * The Natives and Spanish Rule
 * Colonial Haiti
 * Haitian Revolution

title: History of Haiti/Catalog


text: File:FINAL YEAR PROJEC1.pdf

title: File:FINAL YEAR PROJEC1.pdf


text:
User:Mahendaran/Virtual Matchmaking Gateway

[[media:FINAL_YEAR_PROJEC1.pdf]]


text:
User talk:A. Morphous

Account blocked for vandalism. Yann 10:09, 10 November 2008 (UTC)

title: User talk:A. Morphous


text:
User:A. Morphous

Account blocked for vandalism. Yann 10:10, 10 November 2008 (UTC)

title: User:A. Morphous


text:
User:Darkoneko

fr:user:Darkoneko

This account tend to be used for steward actions, such as vandalism fight

title: User:Darkoneko


text:
User talk:Darkoneko

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

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What is History

Time Estimation
Time estimation for the completion of the required material in this section.
 * Lecture = 30 min.
 * Required Reading & Review Questions = 30 min.
 * Test Yourself Questions = 30 min.

Lecture
Coming Soon

Outline
 * A definition of history as a discipline
 * The birth of history by Herodotus
 * Thucydides, the founder of scientifc history

Required Reading

 * History. Read the introductory part above the Contents and Etymology and Description.
 * Herodotus, Histories. Read the introduction and the first three paragraphs from History of Herodotus/Book 1.
 * Thucidides, History of the Peloponnesian War. Read paragraphs 20-23 from History of the Peloponnesian War/Book 1.

Review Questions

 * Where does the word ‘history’ come from?
 * What were the uses of the word ‘history’ in ancient and modern times?
 * Give a definition of history as a discipline.
 * Name some of the complementary disciplines that historians use in order to understand the past?
 * Why did Herodotus write his Histories?
 * How did Herodotus learn about the events he refers to?
 * Identify the part(s) of Herodotus writing in the first three paragraphs that he makes reference(s) to different versions of the same event.
 * Identify the part(s) of Herodotus writing in the first three paragraphs that he makes reference(s) to stories he heard.
 * Should historians, according to Thucydides, take for granted whatever they are told? Give examples of his opinion from the assigned paragraphs.
 * What is the method that Thucydides used in order to find the true course of events?
 * Why does Thucydides believe that his account may not attract much interest?
 * Why has he written his account?
 * Does Thucydides make any differentiation among the causes of the Peloponnesian War he is going to describe?

Test Yourself Questions

 * Write a short paragraph explaining what history is.
 * In a short essay compare and contrast the methods that Herodotus and Thucydides used in order to write their accounts.

Additional Information
(Help for playing wikipedia media files)
 * Herodotus biography
 * Thucydides biography and comparision with Herodotus
 * [[Image:Speaker Icon.svg|24px]] Herodotus speaking. (6 min.)
 * [[Image:Speaker Icon.svg|24px]] Thucydides speaking. (5 min.)

title: What is History


text:
User:Uf.team.wiki/HW6

Principle of Virtual Work (continuous) of dynamics of the elastic bar problem
Partial Differential Equation:

$$ \frac{\partial }{\partial x} \left [(EA) \frac{\partial u}{\partial x}\right ] + f = m \ddot u \qquad \qquad (1)$$

Equation (1) is equivalent to the multiple degree of freedom system (MDOF):

$$- \mathbf {Kd} + \mathbf F = \mathbf {M \ddot d} \Rightarrow $$

$$ \mathbf {M \ddot d} + \mathbf {Kd} = \mathbf F \qquad \qquad (2)$$

Derive (2) from (1):

$$ \int\limits_{x=0}^{x=L} W(x) \left \{ \frac{\partial }{\partial x}\left [(EA) \frac{\partial u}{\partial x}\right ] + f - m \ddot u \right \} dx = 0 \qquad \qquad (3) \qquad$$

 * NOTE: Equation (3) applies to all possible W(x), where W(x) is the weighting function

 * $$ (1) \Rightarrow (3)$$ : trivial
 * $$ (3) \Rightarrow (1)$$ : not trivial

 * (3) written as $$ \int W(x)g(x)dx = 0 $$ [for all W(x)]

Since (3) holds for all W(x), select W(x)=g(x), then (3) becomes:

$$ \int \underbrace{g^2}_{\ge 0} dx = 0 \quad \Rightarrow \quad g(x)=0$$

Integration by parts
Recall: Continuous PVW {Equation (3), above}

 * 1st term: $$ \qquad r(x) = (EA)\frac{\partial u}{\partial x} $$
 * 2nd term: $$ \qquad s(x) = W(x) $$

Integration by parts:

$$ \int\limits_{x=0}^{x=L} \underbrace{W(x)}_{s} \frac{\partial }{\partial x} \underbrace{\left [(EA) \frac{\partial u}{\partial x}\right ]}_{r} dx = 0 \left [ W (EA) \frac{\partial u}{\partial x}  \right ] - \int\limits_{0}^{L}\frac{dW}{dx}(EA)\frac{\partial u}{\partial x}dx $$

$$ \Rightarrow \underbrace{W(L)(EA)(L)\frac{\partial u}{\partial x}(L,t)}_{F=N(L,t)} - \underbrace{W(0)(EA)(0)\frac{\partial u}{\partial x}(0,t)}_{N(0,t)}- \int\limits_{0}^{L}\frac{dW}{dx}(EA)\frac{\partial u}{\partial x}dx $$

Model Problem

Looking at the figure above, it is shown that there are two Boundary Conditions
 * @ x=0, select W(x) such that W(0)=0 (ie: kinematically admissible)

Motivation: Apply the discrete Principle of Virtual Work:

$$ \underset{6 \times 1}{\mathbf W } \cdot \underbrace {\left (\underset{6 \times 2}{[\qquad] }\underset{2 \times 1}{\begin{Bmatrix} d_{3}\\ d_{4}\\ \end{Bmatrix}} - \underset{2 \times 1}{\mathbf F} \right )}_{6 \times 1} = \underset{1 \times 1}{0}$$

$$ \mathbf F^T = \begin{bmatrix} F_1 &  F_2  &  F_3  &  F_4  &  F_5  &  F_6 \\ \end{bmatrix} $$

(where F1, F2, F5, F6 are known reactions)

Since W can be selected arbitrarily, select W such that W1 = W2, W5 = W6 so to eliminate equations involving unknown reactions → eliminate rows 1,2,5,6:

$$ \underset{2 \times 2}{\overline \mathbf K} \underset{2 \times 1}{\overline \mathbf d} = \underset{2 \times 1}{\overline \mathbf F} $$

NOTE: Recall that:

$$\overline \mathbf W \cdot (\mathbf {Kd}- \mathbf F) = 0 \hookrightarrow for all \quad{\underset{6 \times 1}\mathbf W} \qquad (2)$$

Where: W_{3}\\ W_{4}\\ \end{Bmatrix} $$ d_{3}\\ d_{4}\\ \end{Bmatrix} $$ F_{3}\\ F_{4}\\ \end{Bmatrix} $$
 * $$\underset{2 \times 1}{\overline \mathbf W} \begin{Bmatrix}
 * $$\underset{2 \times 1}{\overline \mathbf d} \begin{Bmatrix}
 * $$\underset{2 \times 1}{\overline \mathbf F} \begin{Bmatrix}

Unknown reaction: $$ N(0,t) = (EA)(0) \frac {\partial u}{\partial x} (0,t) $$

Continuous PVW: $$ W(L)F(t) - \int\limits_{0}^{L} \frac {\partial W}{\partial x}(EA)\frac {\partial u}{\partial x}dx + \int W(x)[f-m \ddot u ] dx = 0 \qquad \qquad $$

(for all W(x) such that W(0)=0)

$$ \Rightarrow \int\limits_{0}^{L} W (m \ddot u )dx + \int \limits_{0}^{L} \frac {\partial W}{\partial x}(EA)\frac {\partial u}{\partial x}dx = W(L)F(t) + \int \limits_{0}^{L} W f dx \qquad $$

(for all W(x) such that W(0)=0)

Sample table for comparisons
When looking at the Principles of Virtual Work, a comparison can be made between the equations for the continuous setting and the discrete setting. A summary of this comparison is made below and shown in the table.

Looking at the stiffness term, and focusing on the general element i with the use of of a dotted area shown in Figure 2 below, represents an example of interpolation using an elastic bar.

Focusing in on a general element, i in the interpolation process, we may look at the figure below to aid in the calculation and assuming that the displacement u(x) is for $$x_i\leq x\leq x_{i+1}$$.

Figure 4 shown below is an example of a 2 - Bar truss example that serves as motivation when calculating the interpolation of u(x). Here there is an assumed linear interpolation between both nodes 1 and 2. Here ther deformed shape is a straight line, that was part of the implicit assumption made in the linear interpolation.

Continuation: Using the Principle of Virtual Work to obtain the Discrete Principal of Virtual Work
Rube Goldberg device

Walter Benjamin

- "honesty, imagination, ethics"

Going from continuous PVW to Discrete PVW
Lagrangian interpolation

motivation for form of $$\mathbf{N_{i}(x)}$$ and $$\mathbf{N_{i+1}(x)}$$

1. It is assumed that $$\mathbf{N_{i}(x)}$$ and $$\mathbf{N_{i+1}(x)}$$ are straight lines, and thus any linear combination of $$\mathbf{N_{i}(x)}$$ and $$\mathbf{N_{i+1}(x)}$$ is also linear, and in particular expression for $$\mathbf{u(x)}$$:

$$\mathbf{u(x)=N_i(x)d_i+N_{i+1}(x)d_{i+1}}$$

where $$N_i(x) and N_{i+1}(x)$$ are linear functions in x and equations are shown below.

 * $$\mathbf{N_{i}(x)={\alpha_i} + {\beta_i}x}$$

 * $$\mathbf{N_{i+1}(x)={\alpha_{i+1}} + {\beta_{i+1}}x}$$

where ($$\mathbf{\alpha_{i}, \beta_i, \alpha_{i+1}, \beta_{i+1}}$$) are all real numbers

Looking at the linear combination of $$\mathbf{N_{i}}$$ and $$\mathbf{N_{i+1}(x)}$$ :

$$\mathbf{N_{i+1}(x)}$$ : $$\mathbf{N_{i}d_{i} + N_{i+1}d_{i+1} = (\alpha_i + \beta_{i}x)d_i + (\alpha_{i+1} + \beta_{i+1}x)d_{i+1} }$$

$$ =\mathbf{(\alpha_{i}d_{i} + \alpha_{i+1}d_{i+1}) +(\beta_{i}d_{i} + \beta_{i+1}d_{i+1})x} $$

Looking at the solution above, one can see that it is clearly a linear function of x

2) For the second case, we can recall equation for $$u(x)$$ (interpolation of $$u(x))$$

$$ \mathbf{u(x_{i})= N_{i}(x_{i})d_{i}+ N_{i+1}(x_{i})d_{i+1}}$$

 * where $$ \mathbf{u(x_{i})= d_{i}}$$ and $$\mathbf{ N_{i+1}(x_{i})d_{i+1} }= 0$$

Use of Interpolation
Looking at the interpolation result for u(x), the same interpolation is applied to form W(x) shown below for an approximation of the weighting coefficient:

$$\ W(x)= N_{i}(x)W_{i}+ N_{i+1}(x)W_{i+1}$$

Next, we are ready to construct the element stiffness matrix for any standard element i, from the elemental stiffness matrix:

$$\beta = \int_{ x }^{{x}_{ i+1}}  [{ N}_{ \acute{i}}{w}_{i}+ {N }_{\acute{i}+1 }{w}_{i+1}](EA) [{ N}_{\acute{i}}{d}_{i}+ {N }_{\acute{i}+1 }{d}_{i+1}]dx$$

Where : $${N}_{\acute{i}}=\frac{d{N}_{i}(x)}{dx}$$ and $${N}_{\acute{i}+1}=\frac{d{N}_{i+1}(x)}{dx}$$

From β:

$$\acute{w}(x)=[{ N}_{ \acute{i}}{w}_{i}+ {N }_{\acute{i}+1 }{w}_{i+1}]$$

$$\acute{u}(x)=[{ N}_{ \acute{i}}{d}_{i}+ {N }_{\acute{i}+1 }{d}_{i+1}]$$

It is important to note:

$$u(x) = {\begin{bmatrix}N_i(x) & N_{i+1}(x)\end{bmatrix}}\begin{Bmatrix}d_i \\ d_{i+1}\end{Bmatrix}$$

where N(x) can be approximated as:

$$\overline \underset{1\times 2}{\mathbf N(x)}=[{N}_{i}(x) {N}_{{i}+1}(x)]$$

Next, we can take the derivative of u with respect to x:

$$\frac{du}{dx} = {\begin{bmatrix}N'_i(x) & N'_{i+1}(x)\end{bmatrix}}\begin{Bmatrix}d_i \\ d_{i+1}\end{Bmatrix}$$

where B(x) can be approximated as:

$$\overline \underset{1\times 2}{\mathbf B(x)}=[{N}_{\acute{i}}(x) {N}_{{\acute{i}}+1}(x)]$$

Similarly the weighting coefficient, W(x) can be approximated as:

$$W(x)=\underline N(x) \begin{Bmatrix}W_i \\ W_{i+1}\end{Bmatrix}$$

where taking the derivative with respect to x yields:

$$\frac{dw }{ dx}=\underline B(x) \begin{Bmatrix}W_i \\ W_{i+1}\end{Bmatrix}$$

Recall: Use of element with degrees of freedom
In the figure below, both the global and local degrees of freedom are shown on each of the axial displacement vectors

From the above figure, one may truly appreciate the power of simplifying both notations for global and local degree of freedoms by using matricies to express the displacements and weighting coefficients

$$\begin{Bmatrix} d_i \\ d_{i+1} \end{Bmatrix} = \begin{Bmatrix} d_1^{(i)} \\ d_2^{(i)} \end{Bmatrix} = \underline\textbf{d}^{(i)} $$

$$\begin{Bmatrix} W_i \\ W_{i+1} \end{Bmatrix} = \begin{Bmatrix} W_1^{(i)} \\ W_2^{(i)} \end{Bmatrix} = \underline\textbf{W}^{(i)} $$

$$\beta =\int_{0}^{L}{\frac{dW}{dx}(EA)\frac{\delta u}{\delta x}dx}=\int_{x_{i}}^{x_{i+1}}{(\underline B \times \underline W^i)(EA)(\underline B \times \underline d^i)dx } $$

$$=w^i\cdot (\underline k ^i \underline d ^i)$$

The above equation for β is rearranged:

$$\beta = \underline w^i \cdot (\int \underline B^T (EA)\underline B dx)\underline d^i$$

The elemental stiffness matrix can be rewritten as seen below:

$$\underline k^i=\int_{x_{i}}^{X_{i+1}}{\underline B(x)^T (EA)\underline B(x) dx)\underline dx}$$

For this assignment EA is taken to be constant, then the equivalent value for K becomes:

$$\textbf{k}^{(i)} = \frac{EA}{L^{(i)}}\begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix}$$

Coordinate Transformation
The first step is to transform the variable coordinate system from $$x$$ to $$\tilde{x}$$

Therefore, the notation can be abbreviated as:

$$\tilde{x} := x - x_i $$

$$d\tilde{x} = dx$$

Using this relationship, the equation below for $$\textbf{k}^{(i)}:$$

$$\textbf{k}^{(i)} = \int_{\tilde{x} = 0}^{\tilde{x} = L^{(i)}}\underline\textbf{B}^T(\tilde{x})(EA)(\tilde{x})\underline\textbf{B}(\tilde{x})d\tilde{x}$$

Below is a figure of an element with varying areas, and moduli of elasticity and will be solved.

Associated with the figure above, the equations similarly for the area and the modulus of elasticities are shown below for the element i.

$$A(\tilde{x}) = N_1^{(i)}(\tilde{x})A_1 + N_2^{(i)}(\tilde{x})A_2$$

$$E(\tilde{x}) = N_1^{(i)}(\tilde{x})E_1 + N_2^{(i)}(\tilde{x})E_2$$

Incorporation of the Mean Value Theorm (MVT)

$$N_{2}^{(i)}(\tilde{x})=\frac{\tilde{x}}{L^{(i)}}=\begin{Bmatrix} 0 & at & \tilde{x}=0\\ 1 & at & \tilde{x}=L \end{Bmatrix}$$

 * Shape functions $$N_{1}^{(i)},N_{2}^{(i)} $$ for the Basis

 Comparing Stiffness Matices

 * $$k^{i}$$ to the matrix obtained from:

$$\frac{1}{2}(A_{1}+A_{2})$$ and $$\frac{1}{2}(E_{1}+E_{2})$$     (where $$E_{1} \neq E_{2} $$)

Provides:

$$\frac{(E_{1}+E_2)(A_{1}+A_{2})}{4L^{i}}\begin{bmatrix} 1 & -1\\ -1 & 1 \end{bmatrix}= k_{ave}^{(i)}$$

 Find:  k (i) - k (i)ave

 * Recall the Mean Value Theorem (MVT) related to a centroid:

$$MVT: \int_{x=a}^{x=b}{f(x)dx=f(\bar{x})[b-a]}$$ $$\bar{x} \epsilon [a,b] $$ where $$a\leq \bar{x}\leq b$$
 * For:

$$\int_{a}^{}{X dA }=\bar{X}\int_{a}^{}{dA}=\bar{X}A$$

$$\int_{x=a}^{x=b}{f(x)g(x)dx=f(\bar{x}g(\bar{x})[b-a]}$$ (where $$a\leq \bar{x}\leq b$$)

 * In most cases

 * The average value of  f :

$$f(\bar{x)}\neq \frac{1}{b-a}\int_{a}^{b}{f(x)dx}$$

 * The average value of  g :

$$g(\bar{x)}\neq \frac{1}{b-a}\int_{a}^{b}{g(x)dx}$$

Analysis of the Electric Pylon
This section analyzes an Electric Pylon with:

Number of Elements = 91

Material = 300M Steel

Young's Modulus of Each Element = 200GPa

Cross Sectional Area of Each Element = 4$${cm}^{2}$$

Required MATLAB Functions
Before running the 2-Bar Truss System program, the following three MATLAB program functions must be in the same directory. They are obtained from the textbook website (See HW1 for textbook information). Matlab Truss Files

PlaneTrussElement.m
This function takes the Modulus of Elasticity (e), the Cross-Sectional Area (A) and the Nodal Coordinates (coord) of the element and generates the local stiffness matrix.

NodalSoln.m
This function takes the Global Stiffness Matrix (K), the Global Force vector (R), and two inputs that are the boundary conditions. The output is an array of the displacement and reaction forces.

PlaneTrussResults.m
This function inputs the Modulus of Elasticity (e), Cross-Sectional Area (A), Nodes of Each Element (coord), and the Displacement calculated in the previous function (disps). The output are the Strain, Stress, and Axial Force of each element.

Main MATLAB Function
The following MATLAB program (electric.m) is the main function to break down the 6-Bar Truss System.

Results
The following are the results to the Electric Pylon

The displacement matrix of the Electric Pylon is as follows (Units are in millimeters (mm)):

$$\begin{Bmatrix} d_{1} \\ d_{2} \\ d_{3} \\ d_{4} \\ d_{5} \\ d_{6} \\ d_{7} \\ d_{8} \\ d_{9} \\ d_{10} \\ d_{11} \\ d_{12} \\ d_{13} \\ d_{14} \\ d_{15} \\ d_{16} \\ d_{17} \\ d_{18} \\ d_{19} \\ d_{20} \\ d_{21} \\ d_{22} \\ d_{23} \\ d_{24} \\ d_{25} \\ d_{26} \\ d_{27} \\ d_{28} \\ d_{29} \\ d_{30} \\ d_{31} \\ d_{32} \\ d_{33} \\ d_{34} \\ d_{35} \\ d_{36} \\ d_{37} \\ d_{38} \\ d_{39} \\ d_{40} \\ d_{41} \\ d_{42} \\ d_{43} \\ d_{44} \\ d_{45} \\ d_{46} \\ d_{47} \\ d_{48} \\ d_{49} \\ d_{50} \\ d_{51} \\ d_{52} \\ d_{53} \\ d_{54} \\ d_{55} \\ d_{56} \\ d_{57} \\ d_{58} \\ d_{59} \\ d_{60} \\ d_{61} \\ d_{62} \\ d_{63} \\ d_{64} \\ d_{65} \\ d_{66} \\ d_{67} \\ d_{68} \\ d_{69} \\ d_{70} \\ d_{71} \\ d_{72} \\ d_{73} \\ d_{74} \\ d_{75} \\ d_{76} \\ d_{77} \\ d_{78} \\ d_{79} \\ d_{80} \\ d_{81} \\ d_{82} \\ d_{83} \\ d_{84} \\ d_{85} \\ d_{86} \\ d_{87} \\ d_{88} \\ d_{89} \\ d_{90} \\ d_{91} \\ d_{92} \\ \end{Bmatrix}$$ = $$\begin{Bmatrix} 0 \\            0  \\            0  \\            0  \\     -0.24614  \\      0.21895  \\     -0.15704  \\   -0.0027926  \\     -0.21101  \\      -0.3432  \\     -0.21576  \\      0.38648  \\     -0.21763  \\     -0.14083  \\     -0.21754  \\     -0.62552  \\      0.30459  \\      0.35724  \\      0.33289  \\      -0.6888  \\      0.84792  \\      0.74729  \\      0.87754  \\     -0.24461  \\      0.91043  \\      -1.1024  \\       1.5418  \\       1.2099  \\       1.5409  \\      0.23911  \\         1.48  \\     -0.75546  \\       1.4796  \\      -1.4891  \\        2.174  \\      0.72439  \\       1.9197  \\      -1.1672  \\       4.5898  \\       2.8773  \\       3.4349  \\        -2.63  \\       6.0714  \\       6.5191  \\       6.0714  \\       4.4011  \\       6.0714  \\        3.522  \\       6.0738  \\       2.8633  \\       6.0862  \\       2.0639  \\       6.1061  \\       1.2663  \\       6.0417  \\      0.11623  \\        5.919  \\      -1.1413  \\       5.7852  \\      -2.4771  \\        5.643  \\      -3.7848  \\       5.4056  \\      -6.3546  \\       4.8496  \\      -18.249  \\       6.4456  \\       4.9206  \\        6.754  \\      -8.1536  \\        6.649  \\       4.0415  \\       7.1973  \\      -5.2394  \\       6.7724  \\       3.5162  \\       6.7648  \\       2.4998  \\       6.7485  \\       1.6336  \\       6.7999  \\      0.71154  \\       6.9414  \\     -0.44367  \\       7.1235  \\      -1.8243  \\       7.3567  \\      -3.8516  \\         8.42  \\       3.5162  \\       10.425  \\      -3.8516  \\ \end{Bmatrix}$$

The reactions at the legs of the Electric Pylons are as follows (Units are in Newtons (N)):

$$\begin{Bmatrix} R_1 \\ R_2 \\ R_3 \\ R_4 \\ \end{Bmatrix}$$ = $$\begin{Bmatrix} 183.78 \\     -501.65 \\      -183.78 \\       1501.7 \\ \end{Bmatrix}$$

In the results array, the left column is the Axial Strain, the middle column is the Axial Stress (MPa), and the right column is the Axial Force (N) for each element.

Looking at the results array, the element with the highest tensile and compressive stress are as follows

$$\sigma_{81}=10.981MPa$$

$$\sigma_{55}=-10.679MPa$$

Plot
The following MATLAB function is used to plot the Electric Pylon. It is a modified version of the function provided here.There are comments embedded in the function. The elements of highest compression and tension stresses are shown by arrows.

The following plot shows the undeformed and deformed shape of the Electric Pylon. The deformed displacement is magnified by a factor of 1000 to get a better look.

Statics Method
The Electric Pylon is Statically Determinate. Unlike previous 2-D problems, taking the moments about a point does not yield a 0 = 0. The first two equations are the sum of the forces in the x and y directions

$$\sum{F_{x}}=R_{1}+R_{3}=0$$

$$\sum{F_{y}}=R_{2}+R_{4}-1000=0$$

The third equations is the sum of the moments taken about Node 1. The convention is that positive are moments pointing into the pylon. The distances are calculated from the MATLAB plot.

$$\sum{M_{1}}=1000(6-1.488)+R_{4}(4.493-1.488)=0$$

From this:

$$\R_{4}=1501.5$$

$$\R_{2}=-501.5$$

Using the nodal locations of Element 1 for relations of the X and Y component of the force, it is calculated that:

$$\R_{1}=197.9$$

And since:

$$\R_{1}=-\R_{3}$$

$$\R_{3}=-197.9$$

The solution of the Statics Method only differ by a small margin versus the FEA Method.

Contributing Members
The following Team Wiki members contributed to this report.

Renee Hood --Eml4500.f08.wiki.hood 00:35, 18 November 2008 (UTC)

Rodney Dagulo --Eml4500.f08.wiki.dagulo.r 19:45, 20 November 2008 (UTC)

Gonzalo Barcia --Eml4500.f08.wiki.barcia 10:45, 21 November 2008 (UTC)

Thomas Kehoe --Eml4500.f08.wiki.Kehoe 12:12, 21 November 2008 (UTC)

Cortland Glowacki --Eml4500.f08.wiki.cort 12:57, 21 November 2008 (UTC)

Ricardo Lopez --Eml4500.f08.wiki.Lopez 4:12, 21 November 2008 (UTC)

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User:EML4500.f08.A-team.VandenBerg/HW6

 Comment: This is HW6 located in the user namespace of EML4500.f08.A-team.VandenBerg 16:08, 12 November 2008 (UTC)

Meeting 29: Monday, 3 November 2008. EML4500
At the beginning of class comments were made on mechanisms and eigenvalues from HW4, and then HW6 was talked about and how to redo the vector truss.

PVW (continuous) of Dynamics of Elastic Bar
PDE:

 * (1) $$ \frac{\partial}{\partial x}$$$$\begin{bmatrix}

(EA) \frac{\partial u}{\partial x} \end{bmatrix} + f = m\ddot{u} $$

Discrete EOM:

 * - Kd + F = M \underline{\ddot{u}}

 * $$ \rightarrow (2) \qquad \underline{M}\underline{\ddot{u}} + \underline{K}\underline{d} = \underline{F} $$

 * This is a multiple degree of freedom system, M.D.O.F.
 * A single degree of freedom system, S.D.O.F., is a system that involves something like a mass and spring (This is an identical representation of what is being taught in the University of Florida Vibrations course).

deriving the equation (2) from equation (1), the following equation (3) is retrieved:

 * (3) $$ \int\limits_{0}^{\alpha = L}w(x)$$

$$ \{ $$ $$ \frac{\partial}{\partial x} $$ $$ \begin{bmatrix} (EA) \frac{\partial u}{\partial x} \end{bmatrix}$$ $$ + f - m\ddot{u} \} = 0 \rightarrow $$ for all possible weighting functions, w(x).

(1) \rightarrow </math (3) trivial (3) \rightarrow </math (1) not trivial

 * Equation (3) can be rewritten as:

 * $$ \int w(x)g(x)dx = 0$$ for all w(x)

Since equation (3) holds true for all w(x), select W(x) = g(x), then equation (3) would become:

 * $$ \int g^2, dx = 0 \rightarrow g(x) = 0 $$

where: $$ g^2 is \geqq 0 $$

Integration by Parts: r(x), s(x)
$$ (rs)' = r's + rs' $$

$$ r' = \frac{dr}{dx} \quad, \quad s' = \frac{ds}{dx}$$

$$ \underbrace{\int(rs)'}_{rs} = \int r's + \int rs' $$

$$\Rightarrow \int r's = rs - \int rs' $$

Recall continuous PVW (Eqn. (3) from meeting 29)

 * 1st term: $$ \qquad r(x) = (EA)\frac{\partial u}{\partial x} $$
 * 2nd term: $$ \qquad s(x) = W(x) $$

By integration by parts:

$$ \int\limits_{x=0}^{x=L} \underbrace{W(x)}_{s} \frac{\partial }{\partial x} \underbrace{\left [(EA) \frac{\partial u}{\partial x}\right ]}_{r} dx = \left [W (EA) \frac{\partial u}{\partial x}  \right]_{x=0}^{x=L} - \int\limits_{0}^{L}\frac{dW}{dx}(EA)\frac{\partial u}{\partial x}dx $$

$$ \Rightarrow W(L) \underbrace{(EA)(L)\frac{\partial u}{\partial x}(L,t)}_{F=N(L,t)} - W(0) \underbrace{(EA)(0)\frac{\partial u}{\partial x}(0,t)}_{N(0,t)}- \int\limits_{0}^{L}\frac{dW}{dx}(EA)\frac{\partial u}{\partial x}dx $$

Model Problem
Image

At x = 0, select W(x) so that W(0) = 0, meaning the system is kinematically admissible.

Motivation: Discrete PVW applied to the equation shown below:

$$ \mathbf{W}_{6x1}\left(\left[ \mathbf{K} \right]_{6x2}\begin{Bmatrix} d_{3} \\ d_{4}\end{Bmatrix}_{2x1}-\mathbf{F}_{6x1}\right)=\mathbf{0}_{1x1} $$ for all $$ \mathbf{W} $$

where $$ \mathbf{F}^{T}=\left[F_{1 }F_{2} F_{3} F_{4} F_{5} F_{6}\right] $$

$$ F_{3}, F_{4} $$ are known reactions and $$ F_{1}, F_{2}, F_{5}, F_{6} $$ are unknown reactions.

Since $$ \mathbf{W} $$ can be selected arbitrarily, select $$ \mathbf{W} $$ such that $$ W_{1}=W_{2}=W_{5}=W_{6}=0 $$ so to eliminate equations involving unknown reactions $$\Rightarrow$$ This eliminates rows 1, 2, 5, and 6. The result is shown below:

$$ \mathbf{K}_{2x2}\mathbf{d}_{2x1}=\mathbf{F}_{2x1} \Rightarrow$$ Equation (1)

Note: $$ \mathbf{W}\left(\mathbf{K}\mathbf{d}-\mathbf{F}\right)=0 $$ for all $$ \mathbf{W} $$

Back to the continuous principle of virtual work

The unknown reaction is:

$$ N(0,t)=(EA)(0)\frac{\partial u}{\partial x}(0,t) $$

Due to continuous PVW, the equation then becomes:

$$ W\left(L\right)F(t)-\int_{0}^{L}\frac{\partial W}{\partial x}(EA)\frac{\partial u}{\partial x}dx+\int W(x)\left[f-m\ddot{u}\right]dx=0 $$ for all $$ W\left(x\right) $$ such that $$ W\left(0\right)=0 $$

The final equation can be written as:

$$ \int_{0}^{L}W\left(m\ddot{u}\right)dx+\int_{0}^{L} \frac{\partial W}{\partial x}(EA)\frac{\partial u}{\partial x}dx=W(L)F(t)+\int_{0}^{L}Wfdx $$ for all $$ W\left(x\right) $$ such that $$ W\left(0\right)=0 $$

Meeting 31: Friday, 7 November 2008. EML4500
Displacement within element $$i$$ is linear displacement $$u(x)$$ for $$x_{i}\le x\le x_{i+1}$$, (i.e., $$x\in \left[ x_{i},x_{i+1} \right]$$)

Motivation for linear interpolation of $$u(x)$$

Deformed shape is a straight line (linear interpolation) of the displacement of 2-nodes.

Consider the case where there are only axial displacements:

Q: Express $$x$$ in terms of $$u_{i}$$ $$(u(x_{i})\text{ and }d_{i+1}=u(x_{i+1}))$$ as a linear function of $$x$$ (i.e. linear interpolation)

$$u(x)=N_{i}(x)d_{i}+N_{i+1}(x)d_{i+1}$$

express $$N_{i}(x)$$ and $$N_{i+1}(x)$$ as linear functions of $$x$$.

$$N_{i}(x)=N_{1}^{(i)}(\tilde{x})$$

$$N_{i+1}(x)=\frac{x-x_{i}}{x_{i+1}-x_{i}}$$

This linear interpolation utilizes the geometry properties of similar triangles in order to provide the unknown value, which lies in between two known values.

Meeting 32: Monday, 10 November 2008. EML4500
32-1 In the beginning of class, a video was watched about a Rude-Goldberg machine as well as Walter Benjamin (Pronounced Valter Benyamin in German; another word like this is Volkswagen, which is pronounced like Folksvagen).

After this, there was a long discussion of "Honesty, imagination, and ethics" as well as narration in homework reports.

After this, previous slides from meeting 31 were shown and ta;led about

Cont. of PVW to Discrete PVW (continued)
Legrangian Interpolation

The equation: u(x) = Ni(x)di + Ni+1(x)di+1 was talked about from page 31-3.

Motivation
Motivation for the form of Ni(x) and Ni+1(x)

(1) Ni(x) and Ni+1(x) are linear (slight lines), thus any linear combination of Ni and Ni+1 is also linear, and in particular for the expression of u(x) on page 31-3.

Ni(x) = $$\alpha$$i + $$\beta$$i

Ni+1(x) = $$\alpha$$i+1 + $$\beta$$i+1  (Numbers: $$\alpha$$i+1, $$\beta$$i+1 32-2

Linear Combinations
Linear combinations of Ni and Ni+1:

 * Nidi + Ni+1di+1 = ($$\alpha$$i + $$\beta$$ix)di + ($$\alpha$$i+1 + $$\beta$$i+1x)di+1

 * ($$\alpha$$idi + $$\alpha$$i+1di+1) + $$\beta$$idi + $$\beta$$i+1di+1)x

as you can see from this result, this is clearly a linear function in x.

(2) If you recall, the interpretation of the equation of u(x) from page 31-2:

 * u(xi) = Ni(xi)di + Ni+1(xi)di+1 = di

because:

 * Ni(xi) = 1

 * Ni+1(xi) = 0

Meeting 33: Wednesday, 12 November 2008. EML4500
FEM vs. PVW (cont.)

$$ u(x_{i+1}) {=} d_{i+1} $$

p. 31-1: interpolation for u(x)

Apply same interpolation for w(x), i.e.

$$ w(x) {=} N_i(x)w_i + N_{i+1}(x)w_{i+1} $$

Element stiffness matrix for element i:

$$ \int_{x_i}^{x_{i+1}} [N_i^'w_i + N_{i+1}^'w_{i+1}](EA)[N_i^'d_i + N_{i+1}^'d_{i+1}]\, dx $$

$$ N_i^' :{=} \frac{dN_i(x)}{dx} $$

Likewise for $$ N_{i+1}^' $$:

$$ N_{i+1}^' :{=} \frac{dN_{i+1}(x)}{dx} $$

Note: $$ u(x) {=} \underbrace{ \begin{vmatrix} N_i(x) & N_{i+1}(x) \end{vmatrix}}_{N(x)_{1x2}} \begin{Bmatrix} d_i \\ d_{i+1} \end{Bmatrix}_{2x1} $$

$$ \frac{du(x)}{dx} = \underbrace{ \begin{vmatrix} N_i^'(x) & N_{i+1}^'(x) \end{vmatrix}}_{\underline{B}(x)_{1x2}} \begin{Bmatrix} d_i \\ d_{i+1} \end{Bmatrix}_{2x1} $$

Similarly: $$ w(x) {=} \underline{N}(x) \begin{Bmatrix} W_i \\ W_{i+1} \end{Bmatrix} $$

$$ \frac{dW(x)}{dx} {=} \underline{B}(x) \begin{Bmatrix} W_i \\ W_{i+1} \end{Bmatrix} $$

Recall the element degree of freedom conventions:

$$ \begin{Bmatrix} d_i \\ d_{i+1} \end{Bmatrix} {=} \begin{Bmatrix} d_1^{(i)} \\ d_2^{(i)} \end{Bmatrix} {=} \underline{d}^{(i)} $$

$$ \begin{Bmatrix} w_i \\ w_{i+1} \end{Bmatrix} {=} \begin{Bmatrix} w_1^{(i)} \\ w_2^{(i)} \end{Bmatrix} {=} \underline{w}^{(i)} $$

$$ \beta {=} \int_{x^i}^{x^{i+1}} \underbrace{(\underline{B} \underline{w}^{(i)})}_{1x1} \underbrace{ (EA)}_{1x1} \underbrace{(\underline{B}\underline{d}^{(i)})}_{1x1}, dx (scalar) $$

$$ {=} \underline{w}^i \cdot (\underline{k}^{(i)} \underline{d}^{(i)}) $$

$$ \beta {=} \int_{x^i}^{x^{i+1}} (EA) \underbrace{(\underline{B} \underline{w}^{(i)})}_{1x1} \cdot \underbrace{(\underline{B}\underline{d}^{(i)})}_{1x1} dx $$

$$ (\underline{B}\underline{w}^{(i)})^T(\underline{B}\underline{d}^{(i)}) $$

$$ (\underline{B}\underline{w}^{(i)})^T {=} \underline{w}^{(i)T} \underline{B}^T {=} \underline{w}^{(i)} \cdot \underline{B}^T $$

\beta {=} \underline{w}^{(i)} \cdot (\int \underline{B}^T(EA)\underline{B} dx )d^{(i)}

$$ \underline{k}^{(i)}_{2x2} {=} \int_{x_i}^{x_{i+1}} \underbrace{\underline{B}^T(x)}_{2x1} \underbrace{(EA)}_{\underbrace{(x)}_{1x1}} \underbrace{\underline{B}(x)}_{1x2} dx $$

$$ B(x) {=} \begin{vmatrix} HW & \frac{1}{L^{(i)}} \end{vmatrix} $$

$$ L^{(i)} {=} x_{i+1} - x_i $$ (Length of element i)

HW 6: Consider EA = const.

$$ k^{(i)} {=} \frac{EA}{L^{(i)}} \begin{vmatrix} 1 & -1 \\ -1 & 1 \end{vmatrix} $$

Transfer of variable coordinates from x to $$ \tilde{x} $$

$$ \tilde{x} :{=} x - x_i $$

$$ d\tilde{x} {=} dx $$

$$ k^{(i)} {=} \int^{\tilde{x} {=} L^{(i)}}_{\tilde{x} {=} 0} \underline{B}^T(\tilde{x})(EA)(\tilde{x})(\underline{B}(\tilde{x}) d\tilde{x} $$

HW6: Find expression for $$ \underline{k}^{(i)} $$ using above:

$$ A(\tilde{x}) {=} N_1^{(i)}(\tilde{x})A_1 + N_2^{(i)}(\tilde{x})A_2 $$

$$ E(\tilde{x}) {=} N_1^{(i)}(\tilde{x})E_1 + N_2^{(i)}(\tilde{x})E_2 $$

$$ \underline{k}^{(i)} {=} ? $$

Meeting 34: Friday, 14 November 2008. EML4500
p.31-4: N1(1) $$ (\tilde{x}) $$ = HW6

N2(1) $$ (\tilde{x}) = \frac{\tilde{x}}{L^{(i)}} = \begin{cases} 0, & \mbox{at }\tilde{x}\mbox{ = 0} \\ 1, & \mbox{at }\tilde{x}\mbox{ = L} \end{cases} $$

shape function N1(i), N2(i)

(basis)

HW6 : book p.159

set E1 = E2 = E

Let A($$ \tilde{x} $$) be linear as on p.33-5

Obtain k (i) from previous problem (p.33-5) and compare to expression given in book.

$$ \frac{E}{L^{(i)}} \underbrace{\frac{(A_1 + A_2}{2}}_{Ave. \ Area} \begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix} $$ = k (i)

Next, compare the general k (i) on p.33-5 to the stiffness matrix obtained by using $$ \frac{1}{2} (A_1 + A_2) $$. Note: $$ E_1 \ne E_2 $$

$$ \frac{(E_1 + E_2)(A_1 + A_2)}{4L^{(i)}} \begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix} $$ = k (i)ave.

Find k (i) - k (i)ave.

Rem : Recall the Mean Value Theorem (MVT) and its relation to centroid:

MVT: $$ \int_{x=a}^{x=b} f(x)\, dx = f(\overline{x})[b-a] $$

for $$ \overline{x} \ \epsilon \ [b-a] $$

$$ \epsilon \ $$= "belongs to"

$$ a \le \overline{x} \le b $$

$$ \int_{A}^{} x\, dx = \overline{x} \int_{A}^{} \,dA = \overline{x} A $$

$$ \int_{x=a}^{x=b} f(x)g(x)\,dx = f(\overline{x})g(\overline{x}) [b-a] $$

$$ a \le \overline{x} \le b $$

But $$ \ f(\overline{x}) \ \ne \ \underbrace{\frac{1}{b-a} \ \int_{a}^{b} f(x) \, dx}_{Ave. \ value \ of \ f} $$

In general

$$ \ g(\overline{x}) \ \ne \ \underbrace{\frac{1}{b-a} \ \int_{a}^{b} g(x) \, dx}_{Ave. \ value \ of \ g} $$

Modify 2-bar truss code to accommodate general k (i) on p.33-5

Three Bar Truss Redo
Below are the plots of the original undeformed 3 bar truss followed by an individual plot of each of the five zero eigenvectors.

Here is the code for the Unstable 3 bar truss:

And here are the results showing the stiffness matrix K, and the eigenvalues and eigenvectors:

K =

0    0     0     0     0     0     0     0     0     6     0    -6     0     0     0     0     0     0     6     0    -6     0     0     0     0    -6     0     6     0     0     0     0     0     0    -6     0     6     0     0     0     0     0     0     0     0     6     0    -6     0     0     0     0     0     0     0     0     0     0     0     0     0    -6     0     6

V =

Columns 1 through 7

1.0000        0         0         0         0         0         0         0   -0.7071         0         0         0         0         0         0         0   -0.7071         0         0   -0.7071         0         0   -0.7071         0         0         0         0         0         0         0   -0.7071         0         0    0.7071         0         0         0         0         0   -0.7071         0   -0.7071         0         0         0    1.0000         0         0         0         0         0         0         0   -0.7071         0    0.7071

Column 8

0  -0.7071         0    0.7071         0         0         0         0

D =

0    0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0    12     0     0     0     0     0     0     0     0    12     0     0     0     0     0     0     0     0    12

3 Bar Truss Stabalized by Addition of Another Bar Element Redo
Below are the plots of the original Stabalized 4 bar truss, along with individual plots of each of the zero eigenvectors.

Here is the code for the Stable 4 Bar Truss

Below once again are the stiffness matrix and the eigenvectors and eigenvalues

K =

Columns 1 through 7

2.1213   2.1213         0         0   -2.1213   -2.1213         0    2.1213    8.1213         0   -6.0000   -2.1213   -2.1213         0         0         0    6.0000         0   -6.0000         0         0         0   -6.0000         0    6.0000         0         0         0   -2.1213   -2.1213   -6.0000         0    8.1213    2.1213         0   -2.1213   -2.1213         0         0    2.1213    8.1213         0         0         0         0         0         0         0         0         0         0         0         0         0   -6.0000         0

Column 8

0        0         0         0         0   -6.0000         0    6.0000

V =

Columns 1 through 7

0.2372   0.1828   -0.6190   -0.3565    0.5948   -0.0000    0.0000    0.2512   -0.5293    0.1627    0.1742    0.1586    0.2656    0.5127    0.5834    0.0231    0.0449   -0.2390   -0.4362   -0.5767   -0.0263    0.2512   -0.5293    0.1627    0.1742    0.4362   -0.2656   -0.5127    0.5834    0.0231    0.0449   -0.2390   -0.1586    0.5767    0.0263   -0.0950   -0.3695   -0.5012    0.0566   -0.1586   -0.3112    0.4863    0.3447    0.3636   -0.2394    0.8317    0.0000    0.0000    0.0000   -0.0950   -0.3695   -0.5012    0.0566   -0.4362    0.3112   -0.4863

Column 8

0.2150   0.4914    0.2764   -0.2764   -0.4914   -0.4914    0.0000    0.2764

D =

Columns 1 through 7

-0.0000        0         0         0         0         0         0         0   -0.0000         0         0         0         0         0         0         0   -0.0000         0         0         0         0         0         0         0    0.0000         0         0         0         0         0         0         0    3.8183         0         0         0         0         0         0         0   12.0000         0         0         0         0         0         0         0   12.0000         0         0         0         0         0         0         0

Column 8

0        0         0         0         0         0         0   16.6670

EDU>>

91 Bar Electric Pylon (Superman)
Below are the plots of the undeformed and deformed electric pylon, along with plots of the three lowest eigenvectors

Here is the code used to solve for the reactions of the Electric Pylon

Below are the results as caluclated by MatLab

d =

0        0         0         0   -0.0002    0.0002   -0.0001   -0.0000   -0.0002   -0.0003   -0.0001    0.0003   -0.0001   -0.0001   -0.0001   -0.0005    0.0004    0.0003    0.0005   -0.0006    0.0010    0.0006    0.0010   -0.0002    0.0011   -0.0010    0.0017    0.0010    0.0017    0.0002    0.0017   -0.0006    0.0017   -0.0013    0.0023    0.0006    0.0022   -0.0010    0.0047    0.0024    0.0040   -0.0024    0.0064    0.0057    0.0064    0.0038    0.0064    0.0030    0.0064    0.0024    0.0064    0.0016    0.0064    0.0009    0.0063   -0.0002    0.0062   -0.0013    0.0061   -0.0024    0.0060   -0.0033    0.0059   -0.0051    0.0055   -0.0126    0.0068    0.0043    0.0070   -0.0063    0.0070    0.0035    0.0074   -0.0044    0.0072    0.0030    0.0071    0.0021    0.0071    0.0012    0.0072    0.0003    0.0073   -0.0007    0.0074   -0.0019    0.0076   -0.0034    0.0090    0.0030    0.0105   -0.0034

reactions =

1.0e+003 *

0.1496  -0.5017   -0.1496    1.5017

results =

1.0781e-005 2.1562e+006       862.48 -6.5356e-006 -1.3071e+006     -522.85 6.7339e-006 1.3468e+006       538.71 6.4556e-006 1.2911e+006       516.45 -3.5152e-006 -7.0304e+005     -281.22 -2.3798e-005 -4.7596e+006     -1903.8 -1.2325e-005 -2.465e+006      -986.01 7.1182e-006 1.4236e+006       569.46 1.9269e-005 3.8538e+006       1541.5 -2.8828e-005 -5.7655e+006     -2306.2 -1.8941e-007      -37882      -15.153 8.778e-009      1755.6      0.70224 1.9328e-005 3.8655e+006       1546.2 1.4698e-005 2.9395e+006       1175.8 -1.8494e-005 -3.6988e+006     -1479.5 -2.883e-005 -5.7661e+006     -2306.4 1.7948e-006 3.5896e+005       143.58 2.621e-005  5.242e+006       2096.8 9.8986e-006 1.9797e+006       791.89 -1.8874e-005 -3.7749e+006       -1510 -3.258e-005 -6.5159e+006     -2606.4 2.7892e-006 5.5784e+005       223.14 3.2194e-006 6.4388e+005       257.55 2.9828e-005 5.9656e+006       2386.3 -4.0852e-006 -8.1704e+005     -326.82 -1.9808e-005 -3.9616e+006     -1584.6 1.16e-005   2.32e+006       928.02 -4.0653e-006 -8.1307e+005     -325.23 -2.8778e-005 -5.7556e+006     -2302.2 -1.4237e-007      -28474       -11.39 -4.9677e-006 -9.9353e+005     -397.41 -7.6072e-008      -15214      -6.0857 2.9608e-005 5.9216e+006       2368.6 2.7907e-007       55813       22.325 -2.4358e-005 -4.8717e+006     -1948.7 6.9573e-006 1.3915e+006       556.59 -1.5116e-008     -3023.3      -1.2093 -2.8794e-005 -5.7588e+006     -2303.5 -2.427e-005 -4.854e+006      -1941.6 6.9522e-006 1.3904e+006       556.17 -9.4154e-007 -1.8831e+005     -75.323 1.129e-005  2.258e+006       903.22 -3.16e-006  -6.32e+005       -252.8 -2.0051e-005 -4.0102e+006     -1604.1 -2.0266e-019 -4.0531e-008 -1.6213e-011 7.3233e-019 1.4647e-007  5.8586e-011 3.2073e-007       64147       25.659 1.3501e-006 2.7001e+005       108.01 1.3575e-006 2.7149e+005        108.6 -1.0389e-005 -2.0778e+006     -831.13 -2.2136e-005 -4.4271e+006     -1770.9 -2.9388e-005 -5.8777e+006     -2351.1 -3.6774e-005 -7.3549e+006     -2941.9 -4.331e-005 -8.662e+006      -3464.8 -4.3478e-005 -8.6957e+006     -3478.3 -1.2903e-019 -2.5807e-008 -1.0323e-011 1.5471e-019 3.0942e-008  1.2377e-011 0           0            0  1.2205e-019  2.4409e-008  9.7638e-012 -8.8522e-007 -1.7704e+005     -70.818 1.1034e-006 2.2069e+005       88.276 -9.5965e-007 -1.9193e+005     -76.772 9.9436e-007 1.9887e+005       79.549 1.0732e-005 2.1464e+006       858.56 -1.0415e-005 -2.083e+006      -833.19 1.1448e-005 2.2895e+006       915.81 -1.1321e-005 -2.2643e+006     -905.71 1.0517e-005 2.1035e+006       841.38 -1.0732e-005 -2.1464e+006     -858.56 7.0689e-006 1.4138e+006       565.51 -6.8859e-006 -1.3772e+006     -550.87 7.9177e-006 1.5835e+006       633.42 -1.8464e-005 -3.6928e+006     -1477.1 -4.8696e-007      -97393      -38.957 -1.1845e-007      -23690      -9.4762 1.5604e-007       31208       12.483 4.5239e-005 9.0479e+006       3619.2 0           0            0  4.5158e-005  9.0317e+006       3612.7 -1.5877e-018 -3.1754e-007 -1.2702e-010 4.5256e-005 9.0511e+006       3620.5 -6.5877e-007 -1.3175e+005     -52.702 -1.4822e-006 -2.9645e+005     -118.58 8.9829e-006 1.7966e+006       718.63 2.2438e-005 4.4876e+006         1795 3.3117e-005 6.6233e+006       2649.3 3.8878e-005 7.7755e+006       3110.2 0           0            0  2.4017e-020  4.8034e-009  1.9214e-012 -1.9214e-019 -3.8428e-008 -1.5371e-011 1.0741e-019 2.1482e-008  8.5927e-012

Element_Stress =

2.1562e+006 -1.3071e+006 1.3468e+006 1.2911e+006 -7.0304e+005 -4.7596e+006 -2.465e+006 1.4236e+006 3.8538e+006 -5.7655e+006 -37882      1755.6  3.8655e+006 2.9395e+006 -3.6988e+006 -5.7661e+006 3.5896e+005 5.242e+006 1.9797e+006 -3.7749e+006 -6.5159e+006 5.5784e+005 6.4388e+005 5.9656e+006 -8.1704e+005 -3.9616e+006 2.32e+006 -8.1307e+005 -5.7556e+006 -28474 -9.9353e+005 -15214 5.9216e+006 55813 -4.8717e+006 1.3915e+006 -3023.3 -5.7588e+006 -4.854e+006 1.3904e+006 -1.8831e+005 2.258e+006 -6.32e+005 -4.0102e+006 -4.0531e-008 1.4647e-007 64147 2.7001e+005 2.7149e+005 -2.0778e+006 -4.4271e+006 -5.8777e+006 -7.3549e+006 -8.662e+006 -8.6957e+006 -2.5807e-008 3.0942e-008 0 2.4409e-008 -1.7704e+005 2.2069e+005 -1.9193e+005 1.9887e+005 2.1464e+006 -2.083e+006 2.2895e+006 -2.2643e+006 2.1035e+006 -2.1464e+006 1.4138e+006 -1.3772e+006 1.5835e+006 -3.6928e+006 -97393      -23690        31208  9.0479e+006 0 9.0317e+006 -3.1754e-007 9.0511e+006 -1.3175e+005 -2.9645e+005 1.7966e+006 4.4876e+006 6.6233e+006 7.7755e+006 0 4.8034e-009 -3.8428e-008 2.1482e-008

Tstress =

9.0511e+006

Tstress_Element =

81

Cstress =

-8.6957e+006

Cstress_Element =

55

lamda =

132.16      2468.4       2940.3

column =

79   82    83

T =

0.54654     0.12647      0.11587

The first column of results d, is the displacement experienced by the structure before a magnifying factor was applied. The next set is the reaction forces. Next is all of the strains and stresses applied to each element of the structure, it was observed that the highest tensile stress occurred in element number 81 and was equal to about 9.05e6 Pa. The highest compressive stress observed was in element 55 and was equal to about -8.7e6 Pa.

The electric pylon is not statically determinant since there are not enough equations to satisfy all of the unknowns.

Below are the three lowest eigenvalues of the structure as determined by MatLab:

132.16

2468.4

2940.3

Finally the three lowest vibrational periods are as follows:

0.54654

0.12647

0.11587

Below are the additional MatLab m-files created to assist in the analysis of the electric pylon:

ReducedMassM.m

and

ReducedStiffM.m

Contributions
--EML4500.f08.A-team.rieth 21:24, 10 November 2008 (UTC)

--EML4500.f08.A-team.melvin 23:04, 14 November 2008 (UTC)

--Eml4500.f08.a-team.robinson 22:17, 20 November 2008 (UTC)

--EML4500.f08.A-team.VandenBerg 22:32, 20 November 2008 (UTC)

--EML4500.f08.A-team.morford 21:05, 21 November 2008 (UTC)

EML4500.f08.A-team.kirley 23:20, 22 November 2008 (UTC)

title: User:EML4500.f08.A-team.VandenBerg/HW6


text:
Portal:Haitian history

School:History/Resources

title: Portal:Haitian history


text:
Portal:Haitian history/Introduction

Haiti! Many people, upon mentioning this name, will show complete ignorance. Those who've actually paid attention in 5th Grade Geography will recall that it's an island on the Caribbean Sea--quite poor and in such a despereate condition. There are much more to Haiti than meets the eye, and what little speculations one might have heard. This is thus the purpose of this department, to educate people on this subject.

Haiti shares the Island of Hispaniola with the Dominican Republic. It occupies the western third of the island, with a population of over seven million. It is true that a vast majority of the population is under poverty, but modern shame is found sometimes consolidated by past prides and glories, which Haiti has seen many of. It's one of the most interesting and compelling histories ever recorded by man.

title: Portal:Haitian history/Introduction


text: Portal:Haitian history/box-header

title: Portal:Haitian history/box-header


text: Portal:Haitian history/box-footer

title: Portal:Haitian history/box-footer


text: Talk:Educational Media Awareness Campaign/Music/POTD 8


text: File:Radsam hmwrk5 Part2a.png

title: File:Radsam hmwrk5 Part2a.png


text:
User:Eas4200c.f08.aeris.guan/hw6/117

Deriving Traction Force Equation cont...
Recall that the equation we are trying to derive is the following. $$\displaystyle [t]_{3x1}=[\sigma]_{3x3}[n]_{3x1}$$ Traction force is the stress applied to a lateral surface. For example, stresses due to aerodynamic forces on a wing such as drag or even friction can be a traction force. The goal is to essentially derive the three components of the traction force matrix from the stress tensor and the surface's normal vector. This expression will be obtained by geometrical manipulation. Let's consider the following figure to illustrate an infinitesimally small surface element.  In the diagram, ds is the length of the lateral surface, n is the unit normal vector of the lateral surface, and the base and height of this triangular element is dy and dz respectively. Now, in order to relate the traction force to the stress tensor components, the angle of the lateral surface with respect to the coordinate system must be utilized. First, the base and height of the triangle will be written in terms of the length of the lateral surface. Thus, it is simple to see the following by simple geometry.  $$\displaystyle dz=dscos(\theta)$$ $$\displaystyle dy=dssin(\theta)$$ Similarly, the normal vector can be written in terms of the angle $$\theta$$. Realize that the magnitude of the normal vector is one, the following can be written.  $$||\vec{n}||=1$$ $$\displaystyle n_y=cos(\theta)$$ $$\displaystyle n_z=sin(\theta)$$ To continue with the derivation, the next logical step is to write the equilibrium equations of the forces in the z and y directions. To simplify things, the depth in and out of the page will be assumed to be one (the area of the base and the height will be the base and the height respectively). Although this is the derivation for the 2D case, the process here can be extrapolated with another axis and equations for the 3D case can also be derived. Let's equate forces in the z and y directions.  $$\sum {F}_y=0=-\sigma_{yy}dz-\sigma_{yz}dy +t_yd_s$$ Substituting the base and height will the lateral surface length and its respect direction cosine, we get the following equation. $$\displaystyle =-\sigma_{yy}ds n_y-\sigma_{yz}ds n_z +t_yd_s$$ It is possible to divide out the ds terms resulting in... $$\displaystyle t_y= \sigma_{yy}n_y+\sigma_{yz}n_z ... (1)$$ The sum of the forces in the z direction can be found in the below collapsible box. Once the equations 1 and 2 are obtained, this can be arranged into a matrix notation. In book the subscripts are different in that it uses the (x,y) coordinate system. Using cyclic permutation, we have the following equation.  $$\begin{bmatrix} t_y\\ t_z\\ \end{bmatrix} = \begin{bmatrix} \sigma_{yy} & \sigma_{yz}\\ \sigma_{zy} & \sigma_{zz}\\ \end{bmatrix} \begin{bmatrix} n_y\\ n_z\\ \end{bmatrix}...(3) $$

The above equation is true because $$\sigma_{yz}=\sigma_{zy}$$. Equation 3 is written this way so that the first subscript indicates the row of the equation. In indicial notation, the equation is rewritten as the following.  $$\begin{bmatrix} t_1\\ t_2\\ \end{bmatrix} = \begin{bmatrix} \sigma_{11} & \sigma_{12}\\ \sigma_{21} & \sigma_{22}\\ \end{bmatrix} \begin{bmatrix} n_1\\ n_2\\ \end{bmatrix}...(3') $$

This can be projected into the third dimension. Generalizing this equation into the 3D case in indicial notation yields the following. $$ \begin{bmatrix} t_1\\ t_2\\ t_3\\ \end{bmatrix} = \begin{bmatrix} \sigma_{11} & \sigma_{12}& \sigma_{13}\\ \sigma_{21} & \sigma_{22}& \sigma_{23}\\ \sigma_{31} & \sigma_{32}& \sigma_{33}\\ \end{bmatrix} \begin{bmatrix} n_1\\ n_2\\ n_3\\ \end{bmatrix}...(4) $$ The advantage of using the indicial notation is that it can be written in a compact way that still provides all the information of the matrix notation.  $$t_i=\sum_{j=1}^3{\sigma_{ij}n_j},i=1,2,3$$ Now, the original equation is derived. $$\displaystyle [t_i]_{3x1}=[\sigma_{ij}]_{3x3}[n_i]_{3x1} $$     From the road map, we have the following three equations. $$\displaystyle T=2\int\int_{A}^{}{}\phi dA$$ $$\displaystyle T = GJ\theta $$ $$\displaystyle J=\frac{-4}{\nabla^{2}\phi }\int \int _{A}\phi dA$$ Combining these three equations and solving for the Lagrangian of phi, we have the following equation. $$ \frac{\partial^2{\phi}}{\partial{y}^2}+\frac{\partial^2{\phi}}{\partial{z}^2}=-2G\theta $$ where $$\phi$$ is constant on the lateral surface of the bar.

title: User:Eas4200c.f08.aeris.guan/hw6/117


title: File:Tracforcedia1.JPG


title: File:Tracforcedia2.JPG


text:
Portal:Haitian history/CURRICULUM

Courses Under Construction

 * HIST-HT-1101 Introduction to Haitian History
 * HIST-HT-1102 The Natives and Spanish Rule
 * HIST-HT-1103 Colonial Haiti
 * HIST-HT-1104 Haitian Revolution

Proposed Courses

 * Pirates of Hispaniola
 * Haiti: A Young Republic--Expansion, and Decline
 * U.S. Occupation

title: Portal:Haitian history/CURRICULUM


text:
Template:Haitian history/COURSE INSTRUCTORS

Course Instructors

 * If you possess intermediate knowledge of this subject, and are interested in teaching a course, please contact Franco Lucien. After you are qualified to administer a class, your name will then be added down here, next to the class you are assigned to. Please click here for more information.


text:
Portal:Haitian history/DEPARTMENT NEWS

 * November 8, 2008 - Department of Haitian History Founded!

title: Portal:Haitian history/DEPARTMENT NEWS


text:
Managing a private wiki/Adding extensions

Getting the .php file(s)
First, one must download the  files. This can be done in several ways.
 * SVN
 * MediaWiki code is kept in a subversion repository (SVN repo). Download the files using an SVN client to the appropriate location ( usually)

 * Add either a  or   near the bottom of your LocalSettings.php file, and check Special:Version on your wiki to see if the extension is installed properly. See the examples.

Setting up the database

 * Some extensions require database tables - one of the files in the extension folder will be an SQL query to create the table. phpMyAdmin is a useful tool for altering the database.

Troubleshooting

 * If the extension fails to install properly, check the relevant page on MediaWiki.org if there are specialized installation requirements. Some extensions may have dependencies requiring installation of additional software.


text:
Managing a private wiki/Upgrading

 * Main page: Manual:Upgrading to 1.13

Subversion is a version control software that allows users to download the very latest version of a branch without having to wait for someone to get around to packaging it. Advantages to using Subversion include the latest version, vastly simplified updating, the ability to roll back an upgrade, the ability to create and submit patches. See http://www.mediawiki.org/wiki/Subversion for more information.

If you use SVN, upgrading is comparatively easy. Simple SVN up to the latest revision of the code you wish to use, and run. (You are supposed to backup your files, but it is just as easy to revert using SVN)

For those not using SVN, you can download a zipped package. Simply unzip it and replace your files with the contents.

title: Managing a private wiki/Upgrading


text:
User:Eas4200c.f08.aeris.guan/hw6/1110

Bending and Torsional Analysis Continued
To continue torsional analysis, let us consider two cases of different cross sections. Case 1 is a thin-walled cross section that is closed. An example of this kind of cross section would be a NACA airfoil. It is two lateral surfaces denoted by S with corresponding subscripts as shown in the figure below. Case 2 is a solid cross section with only one lateral surface. Both cross sections are uniform and non circular.  In case 2, solid sections with a single contour boundary have phi as an arbitrary constant and thus can be chosen to be zero. This is due to a traction free boundary condition meaning that the traction force t_z is equal to zero. In case 1, using the Prandtl stress function phi, the stress-free boundary condition says that the derivative of phi with respect to the contour length is zero on lateral surfaces o and i. This means that phi on S_o and S_i are different constants. Differing from case 2, these two constants cannot be set to equal zero. Now consider a uniform bar with circular, solid cross section. The Prandtl stress function of this geometry is as follows.  $$ \phi(y,z)=c\left(\frac{y^2}{a^2}+\frac{z^2}{b^2}-1\right) $$ where a=b=radius and thus constituting a circle. If a did not equal b, the shape of the cross section would be an ellipse. Applying equation 3.19 in the book, it is possible to solve for the factor C. $$
 * }

\nabla^2{\phi}=\frac{\partial^2{\phi}}{\partial{y}^2}+\frac{\partial^2{\phi}}{\partial{z}^2}=-2G\theta$$ $$\nabla^2{\left[C(\frac{y^2}{a^2}+\frac{z^2}{a^2}-1)\right]}={C}\nabla^2{\left[(\frac{y^2}{a^2}+\frac{z^2}{a^2}-1)\right]}$$ $$=C\nabla{(\frac{2y}{a^2}+\frac{2z}{a^2})}=\frac{2C}{a^2}\nabla{(y+z)}=\frac{4C}{a^2}$$ Applying the compatibility equation yields the following $$C=-\frac{a^2}{2}G\theta$$

title: User:Eas4200c.f08.aeris.guan/hw6/1110


text:
User:Timh2870

... nothing here yet

title: User:Timh2870


text:
Portal:Haitian history/MISCELLANEOUS

Citadelle Henri Christophe
It lies approximately 17 miles south of Cap-Haïtien, and it's the largest fortress in the Western Hemisphere. The Citadel, one of Haiti's most valuable piece of history, was built between 1805 and 1820 as part of a system of fortications for retaliation against French forces, in an attempt to protect the nation's newly gained independence. However many lives and amount of blood the cost was, the fortress never saw use for it's original intents. Today, it is a major tourist destination, and features on many tourist ministry posters and on stamps and currency.

title: Portal:Haitian history/MISCELLANEOUS


text:
Portal:Haitian history/DIVERSE FACTS

Citadelle Henri Christophe

 * Also known as Citadelle Laferrière (in English: the Citadel), lies approximately 17 miles south of Cap-Haïtien, and it's the largest fortress in the Western Hemisphere. The Citadel, one of Haiti's most valuable piece of history, was built between 1805 and 1820 as part of a system of fortifications for retaliation against French forces, in an attempt to protect the nation's newly gained independence. However many lives and amount of blood the cost was, the fortress never saw use for it's original intents. Today, it is a major tourist destination, and features on many tourist ministry posters and on stamps and currency.

title: Portal:Haitian history/DIVERSE FACTS


text:
User:Melancholie/mmA frame

 * To make this list more useful, import this JavaScript code into MediaWiki:Common.js!

Search requests/traffic (source / blacklist)

 * Most often requested nonexistent articles per day (based on ' days in year ').
 *  Note : List contains misspellings, based on Squid server requests; filtered case-insensitively!

en:Wikiversity:Most_missed_articles

title: User:Melancholie/mmA frame


text:
Talk:High school physics

Under the heading "Advanced Topics for Exploration", the first topic listed is 'Cold Fusion'. Claims of Cold Fusion are controversial and poorly documented; most physicists doubt that the field is valid. This subject may not be appropriate for study, or alternately, if it was listed in "Advanced Topics for Exploration" so that students might evaluate the evidence themselves, it would probably be advisable to add a note to that effect, or perhaps a disclaimer. Thundragon 07:50, 11 November 2008 (UTC)

title: Talk:High school physics


text:
Colonial Haiti

Return to Department

Class Objective
This course will take a total of 6 weeks to complete. Using the resources, information, and the materials provided, the students will be able complete the final exam, an assessment which measures the knowlege of the learners, with an acceptable performance, as described in the class rubric.

Class Rubric

 * 89%-100%--A
 * 79%-88%---B
 * 69%-78%---C
 * 59%-69%---D
 * Below 59%-F

Course Instructor's Letter
Dear students,

My name is Franco L. Lucien, and I will be your instructor for this course. If you've wondered about what qualifies me to teach this subject, it's years of personal studies and living experience. I've lived in Haiti for over seven years, but once I got back to the States, I just had to go back and take a year of college level courses on the history of the island. Before, I've just read and heard stories about it's past, hence I took such an obsessive interest on the subject and began three years of personal studies.

I will be available to answer your questions. You can either go to my user page, and send me a quick email, or you can drop a note on either the course talk page or mine.

Class Material

 * Unit 1-Age of Discovery
 * Unit 2-Spanish Colonization of Haiti
 * Unit 3-French Colonization of Haiti
 * Unit 4-Piracy and the Caribbean
 * Unit 5-Revolutions

Class Syllabus

 * Students will turn in assignments at the end of every week (Friday)
 * There will be a set of vocabs, including terms and concepts at the beginning of each unit; questions and essays; and check-point quizes at the end of each unit.
 * The learner may send assignments electronically through email, except for the check-point quizes, which are simply as means of evaluating the understanding of each unit, and therefore will not count towards the students accumulative grade.
 * Week 1
 * Age of Discovery (reading)
 * Vocabs
 * Essays
 * Chk-pt quiz
 * Week 2
 * Week 3
 * Week 4
 * Week 5
 * Week 6

Questions
If you've got any questions, edit this space here and post your questions. Alternatively, you can leave you messages either on my talk page, or the course talk page. Either way, I will do my best to answer you every questions within a timely manner.

Student Registration
Please, include your signatures here if you are interested in following through with this course. After doing so, you will automatically become active, thus this course begins thereafter.

title: Colonial Haiti


text:
User talk:Timh2870

 Hello Timh2870, 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! Jtneill - Talk - c 10:20, 11 November 2008 (UTC)

 Hi! Our name is Vicky-Verity. We are twins and are Wikiversity mascots. We think Wikiversity is a great place to learn and make friends (when we aren't sleeping). Leave us a message!

title: User talk:Timh2870


text: Category:Haiti

title: Category:Haiti


text:
Talk:Managing a private wiki/Upgrading

What's SVN? (I actually want to do an upgrade on my wiki, but putting it off for now). --SB_Johnny talk 12:33, 11 November 2008 (UTC)
 * SVN is a client software wich work by an update of only changed file until the installed current version in the client computer. A modification of a code source take an ID (ask as version) in the SVNadmin (software master) and when you ak an update ( by the command svn up) the SVN client tell to the master its current version, and the master send the update needing to reach the current version. This system allows to know all the source code changes.
 * This system work in the upgrade, but it's the same if you want to downgrade, you can ask to the SVN client to use an older version by the ID if the newest do not work well.
 * to use the update via SVN, requires that the installation is made also under SVN Crochet.david 13:07, 11 November 2008 (UTC)

title: Talk:Managing a private wiki/Upgrading


text:
User:La comadreja/Funding

At present, the lab has no steady source of funding. Email Anne Friedman reseal05@gmail.com if you would like to donate.

Budget

title: User:La comadreja/Funding


text:
User:La comadreja/Funding/Budget

11/10/08: -$14.08 on CD-ROMs for data storage (record expense) 12/16/08: +$10.00 received from Columbia University's Computer Graphics and User Interfaces Laboratory, for participating in a study (record amount received) 12/20/08: +25.00 for Hanukkah (record amount received) 1/5/09: +10.00 in change 1/21/09: +0.95 found 2/24/09: $10.10 found 4/6/09: $5.00 earned in survey 5/25/2009: $500.10 from fellowship and found money (record amount received) 7/10/2009: $21.05 found

Balance: +$568.12

title: User:La comadreja/Funding/Budget


text:
User:Stevenarntson/The Commons/Collage

Today in class we'll talk about materials and ideas in terms of questions surrounding "the commons"--what are the materials one can use in artwork, and what materials are excluded? We'll look at a number of artists who use a variety of kinds of materials; some of whom have gotten into trouble over their choices.

collage
Collage in Western art came from the concerns of an art movement known as cubism, which was created by Pablo Picasso and George Braque in the early 1900s.
 * Les Demoiselles D’Avignon
 * Three Musicians
 * Guitar, Sheet Music, and Wine Glass

juxtaposition
"What goes next to what" is one of the primary modes in which collage functions aesthetically.
 * The Interaction of Color by Josef Albers. (text. cover image.)
 * Koyaanisqatsi by Godfrey Reggio.

control

 * "Accident In Art" by Richard Hovey.
 * William S. Burroughs's "cutup" technique is a method of willfully introducing random events into an artwork, in the hope of creating some productive accidents. Burroughs and his collaborators experimented with writing, audio, and filmic versions of the technique.

freedom
Artist Kurt Schwitters described a desire for "Freedom from All Fetters"--a wish that an artist might use any material that presented itself, whether it were paint or a paint can; canvas or cardboard. Many artists took up this call in the decades that followed, using an amazing variety of kinds of materials in their work.
 * Robert Rauschenberg's Bed and Monogram are what he called "combines."
 * Musician John Cage used a wide variety of unusual objects in his music. Here's his "Water Walk"

appropriation
Appropriation is another word for stealing, more or less--when an artist borrows from the work of another artist without asking permission. There are many issues relating to ownership of cultural products. Earlier this quarter, we watched a short film called The Amen Break by Nate Harrison that described, in miniature, a number of interesting issues. Two good resources for news and information are the Stanford Copyright and Fair Use Center and Copyright Website.
 * Wizard People, Dear Reader is a project by comic book artist Brad Neely. Some people have put two and two together on this youtube video.
 * Danger Mouse's The Grey Album is another famous example of appropriation.
 * Generally, Illegal-Art.org is an excellent resource for such work.

public domain
Nobody owns it if it's in the public domain, but is it in the public domain, or not?
 * Public Domain Day
 * Happy Birthday!
 * Digital Copyright Slider
 * Copyright Term Extension Act
 * You Tomb, maintained by MIT.

fair use
Fair Use laws determine if your unpermissioned borrowings from another artist's work are legitimate or not. There are 4 questions generally asked, but each is a gray area--there are no hard and fast rules in fair use law. Fair Use is as robust as our society decides to make it.

 * An excellent summary of these factors is provided here by the Stanford University Library.
 * World Fair Use Day is on January 12.
 * DJ Drama's Lawsuit

copyleft
There are a number of reforms right now trying to define more nuanced application of copyright--places between "all rights reserved" and "public domain."
 * Creative Commons
 * Electronic Frontier Foundation
 * My Online Resources page contains other interesting links.
 * Here's a talk by Lawrence Lessig, founder of Creative Commons.

title: User:Stevenarntson/The Commons/Collage


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User:Eas4200c.f08.wiki.f/11-3-08

Recall the 1-D case from earlier: Dimensional Analysis
$$\left. [f]=\frac{F}{L} \right \} \;\left [ \frac{f}{A} \right] = \frac{f}{L^3}$$

Notes :

 * The square brackets mean "dimension of"

 * F = Force

 * L = Length

$$[A]=L^2$$

$$[\sigma] = \frac{F}{L^2} \Rightarrow \left [ \frac{d \sigma}{dx} \right] = \frac{[d \sigma]}{[dx]} = \frac{\frac{F}{L^2}}{L} = \frac{F}{L^3}$$

 * F/L3 = Force/Volume or Body Force

$$[dx] = L$$

$$\left. \begin{matrix} \varepsilon = \frac{du}{dx}\\ E = \frac{\Delta L}{L} \end{matrix}  \right \} [\varepsilon] = \frac{[du]}{[dx]} = \frac{L}{L} = 1$$

$$\nu =\frac{\varepsilon_{yy}}{\varepsilon_{xx}} \Rightarrow [\nu] = \frac{[\varepsilon_{yy}]}{[\varepsilon_{xx}]} = 1$$

title: User:Eas4200c.f08.wiki.f/11-3-08


text:
Learn SignWriting for American Sign Language

There is a proposal to open a Wikipedia in American Sign Language. At present, it looks like SignWriting will be the dominant mode for content, potentially supplemented with video.

SignWriting is a system for representing the movements of visuospatial signed languages. This project will focus on writing and reading American Sign Language, with a particular focus on use within a Wikipedia or Wiktionary project.

 * /Introduction/
 * /Handshapes/
 * /Contact/
 * /Movement/
 * /Expression/


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Colonial Haiti/Unit 1

Course / Next Unit

Hispaniola
The recorded history of Haiti began on December 5, 1492 when the European navigator Christopher Columbus happened upon a large island in the region of the western Atlantic Ocean that later came to be known as the Caribbean Sea. It was inhabited by the Taíno, an Arawakan people, who variously called their island Ayiti, Bohio, or Kiskeya. Columbus promptly claimed the island for the Spanish Crown, and renamed it La Isla Española ("the Spanish Island"), or Hispañola (later Anglicized as Hispaniola).

Portugal Leads Exploration

The exploration age begun as a result of trade, need for fresh conquests, and technological advances witnessed in Portugal. 15th century Portugal had sean many breakthroughs in the fields of mathematics and naval technology, thus making it the scientific center of that time. In light of their location on the European map, the Portuguese were a seafaring people. It's no wonder that most trade routes were by sea, delivering goods to England, Flanders, Italy, and the Hanseantic league towns. By land, trade would be more of a challenge for them, as they'd face the task of crossing Castile and Aragon territory (now modern-day Spain).

Columbus and New World
Christopher Columbus, Italian Cristoforo Colombo, was a Genoese navigator. He made his fist voyage to the New World in August 3, 1492 with three ships: Santa Maria, Pinta, and Nina. In December 5, 1492, he arrived in northern Hispaniola, and later had to abandon the Santa Maria since she crashed ashore. He returned back to Spain, since his voyage was charted by the monarch thereof, with twenty-five natives (only seven survived the voyage) and left 39 men who founded the settlement of La Navidad.

La Navidad
Christopher, after one of his ships have wrecked, ordered his men to have the ship dismantled providing building materials for the construction of a small fortress. This would be the very first European settlement in the New Word. Christopher, however, returned to Spain, leaving Deiogo de Arana as governor.

When Columbus came back from Spain during his second voyage, on November 27, 1493 he arrived hoping to see a bustling village. To his surprise, only thing that was left of the village were corpse of his men. Nearby Tainos told him that the Natives retaliated against them because they mistreated the Tainos. Comlumbus did not give up, thus went and built another settlement further east called La Islabella, after his queen.

Unit 1 Vocabulary

 * Taíno
 * Bohio, Kiskeya
 * Hispaniola
 * Genoese
 * La Navidad & La Islabella

title: Colonial Haiti/Unit 1


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Colonial Haiti/Unit 2

Course / Previous Unit / Next Unit

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Colonial Haiti/Unit 3

Course / Previous Unit / Next Unit

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Colonial Haiti/Unit 4

Course / Previous Unit / Next Unit

title: Colonial Haiti/Unit 4


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Colonial Haiti/Unit 5

Course / Previous Unit

title: Colonial Haiti/Unit 5


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Primary mathematics:Introduction to variables

A variable is any part of an equasion that replaces a number with something that can represent a number. For example, $$5+E=14$$. In this case, $$E$$ is the variable. You may have seen this before as $$5+$$__$$=14$$. To solve a variable including problem with one step, you have to get rid of the first number. So for example, take $$x+36=45$$. To get rid of 36, we need to create $$-36$$. We also need to balance the problem on both sides. Therefore, the problem looks somewhat like this:. Now you solve the problem vertically. So $$36+(-36)$$ equals 0, so we can get rid of that. Then we take $$45-36$$ which equals $$9$$. Therefore, $$x=9$$.

The same overall process goes for other operations, just remember: Addition is the opposite of subtraction and multiplication is the opposite of division. By doing the opposite of the operation in the equation, you eliminate the variable and come to an answer.


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User:EAS4200C.F08.WIKI.A/My Contribution: 6

=Torsional Analysis Continued=

Two Cases of Torsional Analysis
There are two specific cases that we will concern ourselves with. The first being the closed thin walled cross-section and the second being the solid cross section. the two are depicted below.

Case 1: Closed Thin Walled Cross-Section (compare equations)
This case can be compared to the NACA Airfoil that is continuously being analyzed. It is important to note that the Prandtl stress function $$\displaystyle \Phi$$ is constant on the outer surface $$\displaystyle S_o$$ and the inner surface $$\displaystyle S_i$$. $$\displaystyle S_i$$ can also be represented by $$\displaystyle S_1$$ as well. The image below represents this case. It is comparable to Figure 3.11 in the text (Sun 2006).

Case 2: Solid Cross-Section
The one thing that is important to note about this case is that the Prandtl stress function $$\displaystyle \Phi$$ is zero on the outter surface $$\displaystyle S_o$$. This case is depicted below.

Uniform Bar With Solid Circular Cross-Section
This section coincides with section 3.3 in the text (Sun 2006). The following figure depicts the cross section of a solid circular bar and we will be determining the expression for the torque and polar moment of inertia in terms of the the Prandtl stress function.

Torque In Terms of $$\displaystyle \phi$$
The axis of the bar coincides with the origin of the coordinate system. The boundary of the bar can be expressed as the following equation;

$$\displaystyle x^2 +y^2 = a^2$$ where $$\displaystyle a^2$$ is the radius of the cross section

It is important to note that this is equivalent to the equation of a circle. We will assume that the stress function equals the following;

 $$\displaystyle \phi = C \left(\frac{x^2}{a^2} + \frac{y^2}{a^2} -1 \right)$$

With this information we are able to compute the torque. This is done below.

$$\displaystyle T = 2\int \int_{A}^{}{\phi dxdy} =2 C \int \int_{A}^{}{\left(\frac{x^2}{a^2} + \frac{y^2}{a^2} -1 \right) dxdy} = 2C\int \int_{A}^{}{\left(\frac{r^2}{a^2} -1 \right) dA} = 2C \left(\frac{{\pi} a^2}{2} - ({\pi}a^2) \right) = -C \pi a^2$$

We must now solve for the variable $$\displaystyle C$$. We know that the Prandtl stress function is equal to zero on the outer surface. We also that the gradient of the Prandtl stress function is equal to the following;

$$\displaystyle \nabla^2 \phi = -2G\theta = \frac{\partial^2 \phi }{\partial x^2} + \frac{\partial^2 \phi}{\partial y^2} = C\frac{2}{a^2} + C\frac{2}{a^2} = C\frac{4}{a^2}$$

Solving for $$\displaystyle C$$ yields;

 $$\displaystyle C = -\frac{a^2G\theta}{2}$$

Plugging this back into the equation for torque we can obtain the final expression.

$$\displaystyle T = \frac {\theta G \pi a^4}{2}$$

Polar Moment of Inertia In Terms of $$\displaystyle \phi$$
We know that the polar moment of inertia $$\displaystyle J$$ equals the following;

 $$\displaystyle J = \frac {T}{G\theta}$$

where

$$\displaystyle T = 2\int \int_{A}^{}{\phi dxdy}= -C\pi a^2$$  $$\displaystyle G \theta = -\frac{\nabla^2 \phi}{2} = \frac{2C}{a^2}$$

The polar moment of inertia then equals the following;

$$\displaystyle J = \frac {T}{G\theta} = --\frac{C\pi a^2}{\frac{2C}{a^2}} = \frac{\pi a^4}{2}$$

This is the exact equation that has been derived multiple time before using other methods.


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User talk:Goodsonj

group 6a: The teacher's (emotional) influence on first graders
Due to some of my clinicals I have learned first hand the essentialness of the teacher's emotional influence on first graders. Not to disclude first graders, seeing as that i what our discussion is on, but this influence relates to all elementary grades to provide an opportunity of success. I have witnessed the emotionality of different types of teachers, all in which display unique emotions to allow their students to strive in the classroom. Some of the best things I've seen thus far. I've seen a male elementary teacher, working in a title 1 school really connect with his students. Some of these students do not have the best home lives, so he makes it the best experience possible for them while they are there. The thing is, kids respect him and that is why they look up to him so highly and commence doing such great work within his class. My favorite line that he says to them is, "if you all do not listen, I'm going to have to give you a knuckle sandwhich." Some of us would look at that and say-a teacher is not supposed to say that, but it makes his children laugh, and I've learned very fast that if you do not have laughter in the classroom, especially at young ages you do not have the willingness or fun in children to want to learn. I have also observed a teacher that allows her children to roam aroud her classroom and workon their work, or explor different types of work. Her whole philosophy is that as long as they are learning something, then I am doing my job. I have to say I totally agree. They way she praises them when they get their work done, is amazing. You can tell that the kids want to get their work done and want to impress her. So in my first responce to our group disucssion I'd say that it takes a very emotional responce in teachers to produce a successful child.

title: User talk:Goodsonj


text:
The Early Humans and The Neolithic Revolution

Time Estimation
Time estimation for the completion of the required material in this section.
 * Lecture = 30 min.
 * Required Reading & Review Questions = 40-50 min.
 * Test Yourself Questions = 30-40 min.

Lecture
Coming Soon

Outline
 * The Evolution of Human
 * The Paleolithic (Old Stone) Age
 * The Neolithic (New Stone) Age
 * The Primitive Society

Abbreviations
 * m.y.a. stands for million years ago
 * B.C., Before Christ
 * A.D. "anno Domini"

Required Reading

 * Prehistory. Read the main essay only not the linked essays.

Review Questions

 * What was the human evolution.
 * What are the main differences between the Old Stone (Paleolithic) and the New Stone (Neolithic) Age?

Test Yourself Questions

 * Describe changes in human life from the Paleolithic to the Neolithic Age.

Additional Information

 * Human evolution
 * Homo sapiens sapiens (Modern Human)


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Computer Architecture Lab/WS2008/AUA

Another Useless Architecture (AUA) is a RISC microprocessor architecture developed by Jakob Wilhelm, Stefan Rottensteiner and Stefan Tauner as a course project for the Computer Architecture Lab at the Vienna University of Technology in winter 2008/2009. It has been implemented in VHDL and tested in a Cyclone II FPGA from Altera on a DE2 development board from Terasic Technology.

Description
AUA is a 16bit RISC architecture providing 32 registers apparently having similarities to MIPS. E.g. MIPS branches are supported by individual compare and branch instructions (i.e. you have to compare two values/registers and use the result as input for the branch decision). The branch delay is 0 for untaken and 1 cycle for taken branches. The branch decision and branch address calculation is done in the ID stage. An additional cycle is lost if the branch uses the result of the preceding instruction. It is not possible to circumvent this stall with forwarding as this would introduce a new longest path (from the ID/EX pipeline registers through the alu, through the branch decision unit in ID, to the PC register in IF). The EX stage forwards its results to resolve data hazards in all other instruction sequences though (i.e. if instruction $$n+1$$ uses the (EX) result of instruction $$n$$ as (EX) input). We used on-chip memory of the FPGA (M4K blocks) for the register file. As the memory requires registered inputs, this registers can be regarded as pipeline registers for the WB stage. To be able to read two registers (the two operands) and write one (the result of the previous instruction), we had to duplicate the register file into two "simple dual-port mode" memories. Another possibility would be to clock the memory twice as fast, but we did prefer a single clock domain. In dual-port mode the on-chip memory has a latency of 1.5 cycles, which made another forwarding necessary if instr. $$n+2$$ uses the result of instr. $$n$$. Beside the branch stall, where IF and (partially) ID get locked, the EX stage needs to lock all stages, when a load/store operation takes more than one cycle.

Memory and I/O
Instructions, data and memory-mapped I/O devices all share the same address space. The MMU is responsible for handling all memory transactions and transfers the data to the instruction cache and from/to the load/store unit in the EX stage. LD/ST transactions have priority over instruction fetches, so that the pipeline can progress normally. IF will schedule a nop if it is blocked because the current instr. is not available.

The MMU can access on-board SRAM, on-chip ROM (e.g. for a bootloader) and an unlimited number of SimpCon devices. They all share the same address space, which is divided in parts by the MMU. The SimpCon bus is specified as a point-to-point connection, but it is possible to create a shared SimpCon bus by connecting every SimpCon slave to an independant 'RD' and 'WR' line. In the opposite direction (i.e. from the SC slave to the master) it is necessary to multiplex 'WR_DATA' and 'RDY_CNT', because there are no tri-state buses in FPGAs (see Tri-State Buses in Altera Devices). Both tasks are fullfilled by a multiplexer process in the top entity. It matches the 'ADDRESS' similar to IP network masks and selects the corresponding signals.

Pipeline stages
AUA has three stages (four if you include write-back): IF, ID, EX/MEM. IF requests the instruction at the current address/PC from the cache/MMU and extracts the various fields (opcode, immediate, operand and destinstation register addresses). These fields get registered and feed into ID.

The immediate field gets expanded and sign extended before a logic block controls a mux according to the current opcode to choose the correct form of immediate or drop it altogether in favor of a value read from the register file. ID also decides, if a branch is taken and if so, which address is next to be fetched by IF. It is not possible to read the PC directly, but to support returning from functions, ID can store the PC into a register by scheduling an ordinary MOV and overriding the source operand with the current PC value.

EX contains the ALU and a small control curcuit to process LD/ST instructions in cooperation with the MMU. The result of either the ALU or the MMU is selected and transfered to the register file (through the ID unit). If the destination is the zero register, EX enforces a value of 0 to not alter the register.

Instruction Cache
It is possible to plug in different instruction cache implementations between the IF stage and the MMU. Currently a dummy cache (that caches nothing and just connects the signals between IF and the MMU) and a generic direct-mapped cache are provided.

The number of lines cached by the direct-mapped cache can be configured. Currently only on word per cache line is working, because there is no prefetching done by the MMU. It is planned to change this soon. It is also planned to use on-chip memory instead of dedicated logic registers for the cache.

Instruction Set Architecture
The ISA of AUA is typical for a RISC. It was designed with the decoding in mind: All types of fields (opcode, immediates etc.) have a fixed location in the fixed-sized instruction word of 16b. One attribute of our architecture particularly constrained the design of the ISA: We decided that we want 32 registers. This means that out of the 16 bits of one instruction word, 5 bits are needed to address a register.

Most commands need two input registers and one destination register. We use 2 address encoding, where one source register is used as the destination too. This leaves us with at most 6 bits for the opcode, or a maximum of 64 instructions. This number gets reduced by the ldi instruction, because 8b are needed for the immediate itself plus 5b for the destination register, leaving 3b for the ldi opcode. Or in other words $$6-3b$$ are wasted for a part of the immediate value. These 3 bits can't be used for other instructions. Since we could not think of 56 useful instructions anyway, we introduced another type of instruction, that uses immediates.(addi,muli) These use 7 bits to encode the immediate, wasting 2b of opcode space each. So the total number of instructions is limited to $$2^6 - (2^3-1) - 4*(2^2-1) = 45$$.

Instructions
The following table lists all instructions currently implemented in hardware (see the pseudo instruction section below for the rest). There are still some white spots, which are reserved for later use. Cells filled with a s form an encoded source register. d, i likewise for destination registers and immediates. a denotes that the memory address (e.g. for a branch) is taken from the register encoded by these cells. The Imm. column indicates how immediates are interpreted (e.g. the values used with branches get sign-extended and can represent forward as well as backward (=negative) jumps).

ldi – Load Immediate (signed, byte)

Load an 8 bit signed immediate to the Low Byte of Register Rd

jmpl – Jump and Link

Jump to the Addr

brez – Branch if Equal Zero

Branch to the Addr if the content of Rs is equal to zero

brnez – Branch if Not Equal Zero

Branch to the Addr if the content of Rs is not equal to zero

brezi – Branch if Equal Zero Immediate

Branch to the Addr if the imm is equal to zero

brnezi – Branch if Not Equal Zero Immediate

Branch to the Addr if the imm is not equal to zero

Addi – Add Immediate

Add an 7 bit signed Immediate to Rd and wirite the result back to Rd

Muli – Multiply Immediate

Multiply an 7 bit signed Immediate with Rd and wirite the result back to Rd After execution, Rd holds the low word of the multiplication.

Add – Add without carry

Add the content of register Rs to register Rd and write the result back to Rd Carrybit is set, if an overflow occurred

Addc – Add with Carry

Add Carrybit and content of Rs to Rd and write the result back to Rd Carrybit is set if result greater than max. value otherwise cleared.

Sub – Sub without carry

Subtract the content of register Rs from register Rd and write the result back to Rd Carrybit is set, if an overflow occurred

Subc – Sub with Carry

Subtract Carrybit and content of Rs from Rd and write the result back to Rd Carrybit is set if result less than min. value, otherwise cleared.

Mul – Multiply low word signed

Multiply the content of Rs and Rd signe-interpreted and write the result back to Rd. After execution, Rd holds the low word of the multiplication.

Mulu – Multiply low word Unsigned

Multiply content of Rs and Rd unsigne-interpreted and write the result back to Rd. After execution, Rd holds the low word of the multiplication.

Mulh – Multiply High word signed

Multiply the content of Rs and Rd signe-interpreted and write the result back to Rd. After execution, Rd holds the high word of the multiplication.

Mulhu – Multiply High word Unsigned

Multiply content of Rs and Rd unsigne-interpreted and write the result back to Rd. After execution, Rd holds the high word of the multiplication.

Or – Logical Or

Bitwise or combination of the two registers content, and write result back to Rd

And – Logical And

Bitwise and combination of the two registers content, and write result back to Rd

Xor – Logical Xor

Bitwise xor combination of the two registers content, and write result back to Rd

Not – Invert register

Invert the bits of Rs, and write the result to Rd.

Neg – Negate (2er-Complement)

Calculate the Two'ers Complement of Rs and write the result to Rd. For the max. negative value, the result is the max. negative value again

Asr – Aritmethic Shift Right

Hold MSB and shift all bits in Rd one bit to the right. LSB is shifted out.

Lsl – Logical Shift Left

Shift all bits in Rd to the left. MSB is shifted out, LSB is cleared.

Lsr – Logical Shift Right

Shift all bits in Rd to the right. MSB is cleared, LSB is shifted out.

Lsli – Logical ShiftLeft Immediate

Shift all bits about the given value to the left. MSB is shifted out, LSB is cleared. Immediate is 4 bits long and interpreted as unsigned

Lsri – Logical Shift Right Immediate

Shift all bits about the given value to theright. MSB is cleared, LSB is shifted out. Immediate is 4 bits long and interpreted as unsigned

Scb – Set/Clear Bit

Bit 3 to bit 0 of immediate indicates which bit in Rd should be set or cleared If bit 4 is 1, bit is set otherwise cleared.

Roti – Rotate right/left about an Immediate

Bit 3 to bit 0 of immediate indicates the length to rotate bits in Rd. if bit4 is 1, rotate right, otherwise rotate left.

Cmplt – Compare Less Than

Test if Rd < Rs If true write 1 to Rd otherwise 0

Cmpltu – Compare Less Than Unsigned

Test if Rd < Rs If true write 1 to Rd otherwise 0

Cmplte – Compare Less Than Equal

Test if Rd <= Rs If true write 1 to Rd otherwise 0

Cmplteu – Compare Less Than Equal Unsigned

Test if Rd <= Rs If true write 1 to Rd otherwise 0

Cmpe – Compare Equal

Test if Rd = Rs If true write 1 to Rd otherwise 0

Cmpei – Compare Equal Immediate

Test if Rd = Imm If true write 1 to Rd otherwise 0

Mov – Copy Registercontent

Copy content of Rs to Rd

Ld – Load word

Load a word from Addr. to Rd

Ldb – Load Byte

Load a byte from Addr. to the lower byte of Rd

St – Store word

Store content of Rs to the Addr.

Stb – Store Byte

Store the lower byte of Rs to the Addr.

Assembler
Before we wrote our own assembler, we tried to use an assembler generator that takes an architecture definition (how many registers; definition and encoding of the instructions etc) and outputs an assembler, a disassembler, a high level simulator etc. We tried ArchC, a brazilian project that patches the GNU binutils according to your architecture. Sounds great? Theoretically, yes, but ArchC needs SystemC and both take quite some time to setup, if you don't know exactly what you need to do (e.g. ArchC requires an closely matched version of the binutil sources, or patching fails). When we finally got it working, we found some bugs in the generated assembler (e.g. labels were not handled correctly all the time). We had spent a few hours digging binutils code, when we decided that it's not worth the trouble and started work on our own tools.

The assembler is coded in c++ with help of the boost library. It supports various pseudo instructions (see below), can link/include multiple files and is able to write constants (words, strings, arrays of words) into a "data section".

Pseudo Instructions
Nop – No Operation

Do nothing Implemented as a ldi 0 to zero-register Example: ldi $0, 0

ret – Return

Implemented as a branch to Addr. in R31 if R0 equals zero(always true) Example: brez $0, $31

Jmp – Jump

Implemented as a branch to Addr. in Rs if R0 equals zero(always true) Example: brez $0, Rs

Rjmpi – Relative Jump Immediate

Implemented as a branch to Addr represented by Imm if R0 equals zero(always true) Example: brezi $0,Imm

Cmpgt – Compare Greater Than

Test if Rd > Rs If true write 1 to Rd otherwise 0 Implemented as a cmplte between Rs and Rd Example: CMPGT R1, R2 maps to CMPLTE R2, R1

Cmpgt – Compare Greater than Unsigned

Test if Rd > Rs If true write 1 to Rd otherwise 0 Implemented as a cmplteu between Rs and Rd Example: CMPGTU R1, R2 maps to CMPLTEU R2, R1

Cmpgte – Compare Greater Than Equal

Test if Rd >= Rs If true write 1 to Rd otherwise 0 Implemented as a cmplt between Rs and Rd Example: CMPGTE R1, R2 maps to CMPLT R2, R1

Cmpgteu – Compare Greater Than Equal Unsigned

Test if Rd >= Rs If true write 1 to Rd otherwise 0 Implemented as a cmpltu between Rs and Rd Example: CMPGTEU R1, R2 maps to CMPLTU R2, R1

Swpb – Swap Bytes

Change the byte order in the register Rs and write the result to Rd Implemented as a roti in register rd, about 8 bits Example: SWPB R1 maps to ROTI R1,8

Set – Set all bits to 1

Set all bits in Rd to 1 Implemented as a not Rd and zero-register Example: Set R1 maps to NOT R1, R0

Clr – Clear all bits to 0

Set all bits in Rd to 0 Implemented as a mov Rd and zero-register Example: CLR R1 maps to MOV R1, R0

Inc – Increment without carry

Increment the content of Rd about 1. Implemented as a addi with immediate 1 Example: INC R1 maps to ADDI R1, 1

Dec – Decrement without carry

Decrement the content of Rd about 1. Implemented as a addi with immediate -1 Example: DEC R1 maps to ADDI R1, -1

ldiw – Load Immediate Word

Load a word to the Rd Implemented as: load immediate to register shift bits in register about 8 digits load immediate to register Example: LDIW R1, 0x3F1A maps to LDI R1, 3F LSLI R1,8 LDI R1, 1A

setb - Set Bit

Set the corresponding bit in Rd Implemented as scb where the 4th bit is 1. Example: SETB R10, 0xF maps to SCB R10, 0x1F

clrb - Clear Bit

Clear the corresponding bit in Rd Implemented as scb where the 4th bit is 0. Example: CLRB R10, 0xF maps to SCB R10, 0x0F

Build Instructions
Get the source at https://github.com/stefanct/aua/.

VHDL (hw directory)
In the project folder is an aua.qsf file, that contains all necessary settings for Altera's Quartus II. The most important .vhd file is hw/src/aua_types_de2.vhd, which holds a bunch of important constants (e.g. CPU and UART frequencies) and defines various data types used all over in the source files. Another important file, especially if you wanna port AUA to another fpga, is src/aua_top.vhd, where the top entity and the configuration reside. In the configuration section one can choose between different caches (see above) and two ALUs.

Included are also two Modelsim scripts for RTL and gate level simulation. They can be found in hw/sim. They use the testbenches in the same directory.

Assembler (as)
A makefile is included in the as directory. The assembler links with boost_regex, so be sure to have boost installed. To assemble a program and embed the binary into ../hw/mmu/rom.vhd execute "as -r .asm" or to get the binary alone (useable with the simulator or for bootstrapping) call "as .asm ".

High Level Simulator (sim)
The simulator is written in the scripting language python, so no compilation is necessary. When called with "pyhton sim.py" (in sim/src), it reads the binary file from ../../as/boot. See the online help (command "h") for further instructions.

Example program
Hello World

title: Computer Architecture Lab/WS2008/AUA


text:
Talk:Computer logic

give on step in program development life cycle?

Question
In the arithmetic introduction the article says 'those whose MSB is 1 are negative' however, later, in the addition section, it add 1111 (15 decimal) to 0101 (5 decimal). Should not 1111 be -15 decimal? Tum Dushul 15:45, 19 November 2009 (UTC)

title: Talk:Computer logic