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User:Eml4500.f08.group/HW6

=Homework 6=

 Can someone do this section of the MATLAB part? It doesn't require any coding, simply discussion of statically determination.

 * is the problem statically determinate?

 * If yes, provide an argument to support your answer, and compute the reactions. Extend your matlab code to compute the reactions, and compare the computed reactions to the reactions obtained from statics.

 * If no, provide an argument to support your answer.

http://en.wikiversity.org/wiki/User:Eml4500.f08.group/HW6/Matlab#Statics

 * Lecture Notes (current version).Lecture Notes (archived version)
 * MATLAB Code (current version). MATLAB Code (archived version)
 * HW6 Contributors (current version). Contributors (archived version)

 * Note - the links to the archived version will need to be corrected to our archived version that we turn in - these are just placeholders... Eml4500.f08.group.taylor 20:20, 16 November 2008 (UTC)

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

title: User:Eml4500.f08.group/HW6


text:
User:Eml4500.f08.group/HW6/Contributors

 Comment: This page is identical to the previous link, except that Julia Setlak Eml4500.f08.group.setlak was added as a contributing member. --Eml4500.f08.group.setlak 19:33, 9 December 2008 (UTC) *
 * The other members of Team Group agree with this update. Eml4500.f08.group.cady 19:38, 9 December 2008 (UTC) (Group Coordinator)

Brian Taylor Eml4500.f08.group.taylor 20:22, 16 November 2008 (UTC)

Tom Cady Eml4500.f08.group.cady 19:56, 20 November 2008 (UTC)

Jeff Wright Eml4500.f08.group.w 07:19, 21 November 2008 (UTC)

Aaron Clausen Eas4200c.f08.carbon.clausen 18:57, 21 November 2008 (UTC)

Melisa Gaar Eml4500.f08.group.gaar 21:06, 21 November 2008 (UTC)

Julia Setlak Eml4500.f08.group.setlak 19:33, 9 December 2008 (UTC)

title: User:Eml4500.f08.group/HW6/Contributors


text:
User:EML4500.f08.JAMAMA/Homework6

Meeting 29
Initial Conditions  (2 conditions) at t = 0, prescribe:

Condition 1: $$ u(x, t=0) = \bar{u}(x) $$
 * $$ \uparrow $$
 * known function displacement

Condition 2: $$ \frac{\partial u}{\partial t}(x, t=0) = \dot {u}(x, t=0) = \bar{v}(x)$$
 * $$ \uparrow $$
 * known function velocity

PVW (continuous) of dynamics of elastic bar

PDE (Partial Differential Equation):

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

Discrete EOM (Equation of Motion)
 * $$\downarrow $$

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

 * $$\Downarrow $$

$$ \mathbf{M\ddot d} + \mathbf{Kd} = \mathbf{F} $$
 * $$ \nwarrow $$
 * Multiple Degree of Freedom System (MDOF)

Single Degree of Freedom System (SDOF)

$$ \Rightarrow \ \mathbf{m\ddot d} + \mathbf{kd} = \mathbf{f} $$

Derive (2) from (1)

$$ \int \limits_0^L ( \frac{\partial}{\partial x}[EA \frac{\partial u}{\partial x}] + f - m \ddot {u} ) dx = 0,$$ $$ \qquad $$   for all possible  $$\mathbf{w(x)} \qquad \qquad \qquad (3)$$
 * $$ \uparrow $$
 * weighting function

$$ (1) \Rightarrow (3) $$  >>  trivial

$$ (3) \Rightarrow (1) $$  >>  not trivial

(3) rewritten as : $$ \int w(x) g(x) dx, \qquad $$ for all $$ \mathbf{w(x)}$$

Since (3) holds for all $$ \mathbf{w(x)}$$, select $$ \mathbf{w(x)} = \mathbf{g(x)}$$, then (3) becomes $$\int \underbrace{g^2(x)}_{\ge 0} dx = 0 \Rightarrow g(x) = 0 $$

Meeting 30
Integration by parts:$$ r(x), s(x) $$

$$ (rs)' = r's + rs'$$

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

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

$$ => \underline{\int{r's} = rs - \int{rs'}}$$

We can recall cont. PVW (Principal of Virtual Work) Eq. (3) p. 29-1

First term: $$ r(x)=(EA)\frac{\partial{u}}{\partial{x}}  ;   s(x) = W(x)$$

By integrating by parts we can get:

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

$$ = W(L)(EA)(L)\frac{\partial{u}}{\partial{x}}(L,t) - W(0)(EA)(0)\frac{\partial{u}}{\partial{x}}(0,t) - \int_{0}^{L}{\frac{dw}{dx}(EA)\frac{\partial{u}}{\partial{x}}dx}$$

Now lets consider the following model problem: p.28-3  2 b.c.'s

At x=o, select W(x) so that W(0)=0 (i.e. Kinematically admissible)

Motivation : Discrete PVW applied to Equation on  p. 10-1

$$\mathbf{W}_{6x1}. \underbrace{\left(\left[      \right]_{6x2}\begin{Bmatrix}d3\\d4\end{Bmatrix}_{2x1} - \mathbf{F}_{6x1}\right)}_{6x1} = 0 $$     for all W

$$ \mathbf{F}^T = \begin{bmatrix} F_1&F_2&F_3&F_4&F_5&F_6\end{bmatrix}$$  with $$F_1, F_2, F_5, F_6$$: unknown reactions

Since W can be selected arbitrarily, select W so that:

$$ W_1 = W_2 = W_5 = W_6 = 0$$; so to eliminate equations involving unknown reactions => eliminate rows 1,2,5,6;

Let equation (1) be: $$ \mathbf{K}_{2x2} \mathbf{d}_{2x1} = \mathbf{F}_{2x1}$$

with: $$ \mathbf{d}_{2x1} = \begin{Bmatrix} d_3 \\ d_4 \end{Bmatrix}$$ and: $$ \mathbf{F}_{2x1} = \begin{Bmatrix} F_3 \\ F_4 \end{Bmatrix}$$

Let us note that for the step before eq.(1)

 * $$ \mathbf{W}.\left(\mathbf{K}\mathbf{d}-\mathbf{F}\right) = 0$$ with $$ \mathbf{W} = \begin{Bmatrix} W_3 \\ W_4 \end{Bmatrix}$$

Now back to the PVW:

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

Cont. PVW :

$$W(L)f(t)- \int_0^L{\frac{dw}{dx}(EA)\frac{\partial{u}}{\partial{x}}dx} + \int_0^L{W(x)\left[f - m\ddot{u}\right]dx} = 0$$

So for all W(x) so that W(0) = 0,

$$=> \int_0^L{W(m\ddot{u})}dx + \int_0^L{\frac{dW}{dx}(EA)\frac{\partial{u}}{\partial{x}}}dx = W(L)f(t) + \int_0^L{Wf}dx$$
 * for all W so that W(0) = 0.

Meeting 31

Assumption: 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 $$u(x)$$

2-bar truss:

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
Rube Goldberg Device: This is a device that accomplishes a simple task through many unnecessary steps, and in a very indirect fashion.

An example of this type of device can be seen in the following youtube link: []

NOTE: Honesty, imagination, ethics. It is always very important to do your own work, and not plagiarize other people's work. Honesty is the best policy.

Cont PVW to discrete PVW.

Lagrangian Interpolation.

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

1) Mi(x) and Mi+1(x) are linear (straight lines), thus any linear combination of Ni and Ni+1 are also linear and in the particular expression for U(x) on Page 31-3.

$$N_{i}(x)=\alpha _{i}+\beta _{i}x $$ with $$(\alpha _{i},\beta _{i})$$ being numbers.

$$N_{i+1}(x)=\alpha _{i+1}+\beta _{i+1}x$$ Linear combo of Ni and $$(\alpha _{i},\beta _{i})$$ numbers.

$$N_{i+1}$$: $$N_{i}d_{i}+N_{i+1}d_{i+1}=(\alpha _{i}+\beta _{i}x)d_{i}+(\alpha _{i+1}+\beta _{i+1}x)d_{i+1}$$ $$=(\alpha _{i}d_{i}+\alpha _{i+1}d_{i+1})+(\beta _{i}d_{i}+\beta _{i+1}d_{i+1})x$$ is clearly a function of x.

2) Recall equation for U(x) (interp of U(x)) on Page 31-3:

$$U(x_{i})=N_{i}(x_{i})d_{i}+N_{i+1}(x_{i})d_{i+1}=d_{i}$$

=1 =0

Meeting 33
FEM (Finite Element Method) via (Principles of Virtual Work) Continued

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

Applying the same interpolation for W(x) as for u(x)

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

The element stiffness matrix then becomes

$$\beta =\int_{x_{i}}^{x_{i+1}}{\left[\frac{dN_{i}}{dx}(x)W_{i}+\frac{dN_{i+1}}{dx}(x)W_{i+1} \right]\left[EA \right]\left[\frac{dN_{i}}{dx}(x)d_{i}+\frac{dN_{i+1}}{dx}(x)d_{i+1} \right]}dx$$

$$u(x)=\begin{bmatrix} N_{i}(x) & N_{i+1}(x) \end{bmatrix} \begin{bmatrix} d_{i}\\ d_{i+1} \end{bmatrix} \Rightarrow u(x)=\textbf{N}(x) \begin{bmatrix} d_{i}\\ d_{i+1} \end{bmatrix}$$

$$\frac{du(x)}{dx}=\begin{bmatrix} N^{'}_{i}(x) & N^{'}_{i+1}(x) \end{bmatrix} \begin{bmatrix} d_{i}\\ d_{i+1} \end{bmatrix} \Rightarrow \frac{du(x)}{dx}=\textbf{B}(x) \begin{bmatrix} d_{i}\\ d_{i+1} \end{bmatrix}$$

Just like u(x), W(x) is represented the same way

$$W(x)=\textbf{N}(x) \begin{bmatrix} d_{i}\\ d_{i+1} \end{bmatrix}$$

$$\frac{dW(x)}{dx}=\textbf{B}(x) \begin{bmatrix} d_{i}\\ d_{i+1} \end{bmatrix}$$

Now recall the element degrees of freedom as shown in the picture below

$$\begin{bmatrix} d_{i}\\ d_{i+1} \end{bmatrix}= \begin{bmatrix} d_{1}^{(i)}\\ d_{2}^{(i)} \end{bmatrix}=\textbf{d}^{(i)}$$

$$\begin{bmatrix} W_{i}\\ W_{i+1} \end{bmatrix}= \begin{bmatrix} W_{1}^{(i)}\\ W_{2}^{(i)} \end{bmatrix}=\textbf{W}^{(i)}$$

Now the element stiffness matrix becomes

$$\beta =\int_{x_{i}}^{x_{i+1}}{\left[\textbf{BW}^{(i)}\right]\left[EA \right]\left[\textbf{Bd}^{(i)}\right]}dx=\textbf{W}^{(i)}\bullet \left(\textbf{k}^{(i)} \textbf{d}^{(i)}\right)$$

Each set of matrices and scalars multiplied inside the brackets when evaluated come out to be scalars so the bracketed items can be multiplied in any order to come out to the same result.

$$\beta =\int_{x_{i}}^{x_{i+1}}{\left[EA \right]\left[\textbf{BW}^{(i)}\right]\left[\textbf{Bd}^{(i)}\right]}dx$$

Since the the quantity of $$\left[\textbf{BW}^{(i)}\right]$$ is a scalar value the transpose of the value is the same. The transpose of the scalar value $$\left[\textbf{BW}^{(i)}\right]^{T}$$ is also equal to $$\left[\textbf{W}^{(i)T} \textbf{B}^{(i)T}\right]$$

$$\beta =\textbf{W}^{(i)}\left( \int_{x_{i}}^{x_{i+1}}{\textbf{B}^{T}\left[EA \right]\textbf{B}dx}\right) \textbf{d}^{(i)}$$

So the stiffness matrix is equal to

$$\textbf{k}^{(i)}=\int_{x_{i}}^{x_{i+1}}{\textbf{B}^{T}\left(EA \right)\textbf{B}dx}$$

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

$$\displaystyle L^{(i)}=x_{i+1}-x_{i}$$

Transformation of variable from $$\displaystyle x$$ to $$\displaystyle \tilde{x}$$

$$\tilde{x}=x-x_{i}$$

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

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

Find expression for $$\textbf{k}^{(i)}$$ using the equation above.

$$A(\tilde{x})=N^{(i)}_{1}(\tilde{x})A_{1}+N^{(i)}_{2}(\tilde{x})A_{2}$$

$$E(\tilde{x})=N^{(i)}_{1}(\tilde{x})E_{1}+N^{(i)}_{2}(\tilde{x})E_{2}$$

$$\textbf{k}^{(i)}=\int_{\tilde{x}=0}^{\tilde{x}=L^{(i)}}{\textbf{B}(\tilde{x})^{T}\left(\left(N^{(i)}_{1}(\tilde{x})E_{1}+N^{(i)}_{2}(\tilde{x})E_{2}\right)\left(N^{(i)}_{1}(\tilde{x})A_{1}+N^{(i)}_{2}(\tilde{x})A_{2}\right) \right)\textbf{B}(\tilde{x})d\tilde{x}}$$

Meeting 34
$$N_{1}^{(i)}(\tilde{x})=\tilde{x}=\left\{ \begin{align} & 1\text{ at }\tilde{x}=0 \\ & 1\text{ at }\tilde{x}=L^{(i)} \\ \end{align} \right.$$

$$N_{2}^{(i)}=\frac{L^{(i)}}=\left\{ \begin{align} & 0\text{ at }\tilde{x}=0 \\ & 1\text{ at }\tilde{x}=L^{(i)} \\ \end{align} \right.$$

Shape functions (basis): $$N_{1}^{(i)},N_{2}^{(i)}$$

$$K^{(i)}=\frac{(E_{1}+\frac{x}{L}(E_{2}-E_{1}))(A_{1}+\frac{x}{L}(A_{2}-A_{1}))}{L}\begin{bmatrix} 1 &-1 \\ -1 &1 \end{bmatrix}$$
 * HW 6: Book p. 159:

$$K^{(i)}=\frac{(E)(A_{1}+\frac{x}{L}(A_{2}-A_{1}))}{L}\begin{bmatrix} 1 &-1 \\ -1 &1 \end{bmatrix}$$

$$\int_{0}^{L}{(A_{1}+\frac{x}{L}(A_{2}-A_{1}))dx}=(A_{1}x+\frac{x^{2}}{2L}(A_{2}-A_{1}))\mid ^{L}_{0}$$

$$A_{1}L+\frac{L^{2}}{2L}(A_{2}-A_{1})=A_{1}(L-\frac{L}{2})+\frac{L}{2}A_{2}=\frac{A_{1}+A_{2}}{2}$$

$$A_{1}=A_{L}$$...$$A_{2}=A_{R}$$

so:

$$K^{(i)}=\frac{E(A_{1}+A_{2})}{2L}\begin{bmatrix} 1 & -1\\ -1 & 1 \end{bmatrix}=K^{(i)}=\frac{E(A_{L}+A_{R})}{2L}\begin{bmatrix} 1 & -1\\ -1 & 1 \end{bmatrix}$$

Set $$E_{1}=E_{2}=E$$ and let $$A(\tilde{x})$$ be linear as on p. 33-5. Obtain $$k^{(i)}$$ from the previous problem (p. 33-5) and compare it to the expression given in the book.

$$\frac{E(A_{1}+A_{2})}{L^{(i)}2}\left[ \begin{matrix} 1 & -1 \\   -1 & 1  \\ \end{matrix} \right]=K^{(i)}$$

Next, compare the general $$k^{(i)}$$ on p. 33-5 to the stiffness matrix obtained by using $$\frac{1}{2}(A_{1}+A_{2})$$ and $$\frac{1}{2}(E_{1}+E_{2})$$.

NOTE: $$E_{1}\ne E_{2}$$

$$\frac{(E_{1}+E_{2})(A_{1}+A_{2})}{4L^{(i)}}\left[ \begin{matrix} 1 & -1 \\   -1 & 1  \\ \end{matrix} \right]=k_{ave}^{(i)}$$

find $$k^{(i)}-k_{ave}^{(i)}$$

Recall the mean value theorem (MVT) and its relation to the centroid.

MVT: $$\int_{x=a}^{x=b}{f(x)dx=f(\bar{x})\left[ b-a \right]}$$ for $$\bar{x}\in [a,b]\text{ or }a\le \bar{x}\le b$$

$$Centroid\int_{A}{xdA}=\bar{x}\int_{A}{dA}=\bar{x}A$$

$$\int_{x=a}^{x=b}{f(x)g(x)dx=f(\bar{x})g(\bar{x})[b-a]\text{ for }a\le \bar{x}\le b}$$

but: $$f(\bar{x})\ne \underbrace{\frac{1}{b-a}\int_{a}^{b}{f(x)dx}}_{average\text{ }f}$$

$$g(\bar{x})\ne \underbrace{\frac{1}{b-a}\int_{a}^{b}{g(x)dx}}_{average\text{ g}}$$

The following electric pylon is statically indeterminate because it has nodes with multiple (more than two) elements running to them. Because of this fact, there are too many unknowns for the amount of equations that can be used to solve it statically, therefore it is statically indeterminate.

Electric Pylon
{| class="toccolours collapsible collapsed" style="width:100%" ! Matlab Code For Electric Pylon

Matlab Code

results=[]; for i=1:n_elem lm=lmm(i,:); location = [x(node_connect(1,i)),y(node_connect(1,i)); x(node_connect(2,i)),y(node_connect(2,i))]; results = [results; PlaneTrussResults(E,A,location,d(lm))]; end results

 * }

Contributing Members
William Mueller Eml4500.f08.jamama.mueller 22:19, 16 November 2008 (UTC)

Justin McIntire --Eml4500.f08.jamama.justin 18:38, 20 November 2008 (UTC)

Cedric Adam--Eml4500.f08.jamama.adam 08:35, 21 November 2008 (UTC)

Arunas Janulevicius --Eml4500.f08.jamama.jan 10:13, 21 November 2008 (UTC)

Megan Alvarez EML4500.f08.jamama.megan 17:10, 21 November 2008 (UTC)

title: User:EML4500.f08.JAMAMA/Homework6


text:
User:Eml4500.f08.group/HW6/Lecture Notes

=Lecture Notes=

Continuous PVW for Elastic Bar
Principle of Virtual Work for dynamics of elastic bar

PDE:

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

Recall:

$$-kd + f=m\ddot{d}$$

$$M\ddot{d}+kd=F$$  Equation (2)

Derive (2) from (1):

$$\int_{x=0}^{x=L}W(x)\left\{\frac{\partial }{\partial x} \left[EA\frac{\partial u}{\partial x} \right] + f-m\ddot{u}\right\}dx$$  Equation (3) = 0 for all possible W(x)-->Weighting Function

(1) => (3) trivial

(3) => (1) not trivial

(3) rewritten as:

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

Since(3) holds for all W(x), you can select W(x) to be equal to g(x)

$$W(x)=g(x)$$, then (3) becomes:

$$\int g^{2}dx=0$$, where $$g^{2}\geq 0$$ => $$g(x)=0$$

Integration by Parts with Application
Review of Integration by Parts:

Decompose the function into one composed of two separate functions, r(x) and s(x).

$$(rs)' = r's + rs'$$, where $$r' = \frac{dr}{dx}, s' = \frac {ds}{dx}$$

$$\int (rs)' = \int r's + \int rs' = > \int r's = rs - \int rs'$$

Recall the Continuous Principle of Virtual Work and Equation 3 from Lecture 29-1:

The first term: $$r(x) = (EA) \frac{\partial u}{\partial x}$$, where s(x) = w(x)

From integration by parts we see:

$$\int_{x=0}^{x=L} w(x) \frac{\partial}{\partial x}\left[(EA) \frac {\partial u}{\partial x}\right] dx = \left[w(EA) \frac {\partial u}{\partial x}\right]_{x=0}^{x=L} - \int_0^L \frac{\partial w}{\partial x}(EA)\frac{\partial u}{\partial x} dx$$

$$= w(L)(EA)(L)\frac{\partial u}{\partial x}(L,t) - w(0)(EA)(0)\frac{\partial u}{\partial x}(0,t) - \int_0^L \frac{\partial w}{\partial x}(EA)\frac{\partial u}{\partial x} dx$$

Model Problem and PVW
Now consider the following model problem:

At x=0, select w(x) such that w(0)=0 (i.e. the problem is kinematically admissible)

Motivation for the problem: The Discrete Principle of Virtual Work as applied in Lecture 10-1:

$$\boldsymbol w_{6x1} \left ([k]_{6x2} \begin{Bmatrix} d_{3}\\ d_{4}\end{Bmatrix}_{2x1}- \boldsymbol F \right) = 0_{1x1} $$

$$\boldsymbol {F^T} = [F_1 F_2 F_3 F_4 F_5 F_6]$$, where $$F_1, F_2, F_5, F_6$$ are the unknown reactions.

Since w can be selected arbitrarily, select w such that $$w_1 = w_2 = w_5 = w_6$$

So, to eliminate equations involving unknown reactions we need to eliminate rows 1,2,5, and 6, this leaves: $$\boldsymbol k_{2x2} \boldsymbol d_{2x1} = \boldsymbol F_{2x1}$$

 * Note that $$\boldsymbol {w(kd-F)} = 0 $$ for all w.

Back to continuous PVW...

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

From continuous PVW: $$w(L)F(t) - \int_0^L \frac{\partial w}{\partial x}(EA) \frac{\partial u}{\partial x}dx + \int_0^L w(x)[f-m\ddot{u}]dx = 0$$ for all w(x) such that w(0)=0.

$$=> \int_0^L w(m\ddot{u})dx + \int_0^L \frac{\partial u}{\partial x}(EA) \frac{\partial u}{\partial x}dx = w(L)F(t) + \int_0^L wfdx$$ for all w(x) such that w(0)=0.

Interpolation of Deformed Bar
Stiffness term:

In the figure above, assuming the displacement u(x) for $$x_{i} \leq x \leq x_{i+1}$$ (for example, $$x\in [x_{i}, x_{i+1}]$$, where the symbol $$\in $$ means "belongs to". The figure shows the interpolation of u(x) where there is only axial displacement (or, no transverse displacement)

The motivation for linear interpolation of u(x)
In a 2-bar truss the deformation can be shown as:

The deformed shape is a straight line, meaning that there is an implicit asumption of linear interpolation of displacement between two nodes.

u(x) can be expressed in terms of $$d_{i} = u(x_{i})$$ and $$d_{i+1} = u(x_{i+1})$$ as a linear function in x (for example, linear interpolation) as follows:

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

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

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

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

Therefore, $$u(x) = \frac {x-x_{i+1}}{x_{i} - x_{i+1}}d_{i} + \frac {x - x_{i}}{x_{x+1} - x_{i}}d_{i+1}$$

Lagrangian Interpretation
Motivation for form of $$\displaystyle N_i(x)$$ and $$\displaystyle N_{i + 1}(x)$$

1) $$\displaystyle N_i(x)$$ and $$\displaystyle N_{i + 1}(x)$$ are linear (straight lines), thus any linear combination of $$\displaystyle N_i$$ and $$\displaystyle N_{i + 1}$$ is also linear, and in particular the expression for $$\displaystyle u(x)$$ on p. 31-3: $$\displaystyle u(x) = N_i (x) d_i + N_{i + 1} (x) d_{i + 1}$$

$$\displaystyle N_i(x) = \alpha_i + \beta_i x$$ $$\displaystyle N_{i + 1}(x) = \alpha_{i + 1} + \beta_{i + 1} x$$ with $$\displaystyle \alpha_i$$ and $$\displaystyle \beta_i$$ as numbers

Linear combo of $$\displaystyle N_i(x)$$ and $$\displaystyle N_{i + 1}(x)$$: $$\displaystyle N_i(x) d_i + N_{i + 1} d_{i +1}(x) = (\alpha_i + \beta_i x) d_i + (\alpha_{i +1} + \beta_{i + 1} x) d_{i + 1}$$ $$\displaystyle => N_i(x) d_i + N_{i + 1} d_{i +1}(x) = (\alpha_i d_i + \alpha_{i +1} d_{i +1}) + (\beta_i d_i+ \beta_{i + 1}d_{i +1})x$$

2) Recall equation for $$\displaystyle u(x)$$ on p. 31-3: $$\displaystyle u(x_i) = N_i (x_i) d_i + N_{i + 1} (x_i) d_{i + 1}$$

set $$\displaystyle N_i (x_i) = 1$$ and $$\displaystyle N_{i + 1} (x_i) = 0$$

therefore $$\displaystyle u(x_i) = d_i$$

FEM via PVW
Finite Element Method via the Principal of Virtual Work. (cont'd)

Our goal here is to calculate the Element stiffness matrix $$ \beta $$, which is defined as follows.

By examining the figure drawn during lecture 31, we can interpolate to find the values of u(x) and w(x). u(x) = Ni(x)*di+Ni+1(x)*di+1 w(x) = Ni(x)*wi+Ni+1(x)*wi+1

After differentiating both u(x) and w(x), we can plug into the above equation to yield the following element stiffness matrix for a generic element i.

$$\frac{\partial u(x)}{\partial x}=\begin{bmatrix} N^{'}_{i}(x) & N^{'}_{i+1}(x) \end{bmatrix} \begin{Bmatrix} d_{i}\\ d_{i+1} \end{Bmatrix} $$

$$B(x)=\begin{bmatrix} N^{'}_{i}(x) & N^{'}_{i+1}(x) \end{bmatrix}$$

Similarly: $$W(x)=N(x)\begin{Bmatrix} w_{i}\\ w_{i+1}

\end{Bmatrix}$$

$$\frac{\partial W}{\partial x}(x)=B(x)\begin{Bmatrix} W_{i}\\ W_{i+1}

\end{Bmatrix}$$

Recall the element dof(S):

$$\begin{Bmatrix} d_{i}\\ d_{i+1}

\end{Bmatrix} = \begin{Bmatrix} d^{(i)}_{1}\\ d^{(i)}_{2}

\end{Bmatrix} =d^{(i)}$$

$$\begin{Bmatrix} w_{i}\\ w_{i+1}\\ \end{Bmatrix} = \begin{Bmatrix} w^{(i)}_{1}\\ w^{(i)}_{2}\\ \end{Bmatrix} =w^{(i)}$$

$$\displaystyle \beta = \int_{x_i}^{x_{i+1}} (\boldsymbol B w^{(i)})_{1x1} (EA)_{1x1} (\boldsymbol B d^i)_{1x1} dx$$ $$\displaystyle => \beta = \boldsymbol w^{(1)} * (k^{(1)} d^{(1)})$$ This is the Goal from p. 31-3

$$\displaystyle \beta = \int_{x_i}^{x_{i+1}} (EA) (\boldsymbol B w^{(i)})_{1x1} * (\boldsymbol B d^i)_{1x1} dx$$

$$\displaystyle (\boldsymbol B w^{(i)})_{1x1} * (\boldsymbol B d^i)_{1x1} = (\boldsymbol B w^{(i)})^T (\boldsymbol B d^i)$$

where $$\displaystyle (\boldsymbol B w^{(i)})^T = w^{(i) T} \boldsymbol B^T = w^{(i)} * \boldsymbol B^T$$

$$\displaystyle => \beta = w^{(i)} * (\int B^T (EA) B dx) d^{(i)}$$

$$\boldsymbol k^{(i)} = \int_{x_i}^{x_{i+1}} \boldsymbol B(x) (EA) B(x) dx$$

$$\boldsymbol B(x)=\begin{bmatrix} \frac{-1}{L^{(i)}} & \frac{1}{L^{(i)}} \end{bmatrix}$$, where $$L^{(i)}=x_{i+1}-x_i$$

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

Transformation of variables (coordinates) from $$x$$ to $$\tilde x$$.

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

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

 * {| class="toccolours collapsible collapsed" width="60%" style="text-align:left"

!HW Expression for $$\mathbf{k^{(i)}}$$, from team group
 * The expression for $$\mathbf{k^{(i)}}$$ can be solved as such:
 * The expression for $$\mathbf{k^{(i)}}$$ can be solved as such:

$$\mathbf{k^{(i)}} = \int_{x=0}^{\tilde{x} = L^{(i)}}{\mathbf{B^{T}}(\tilde{x})(EA)(\tilde{x})\mathbf{B}(\tilde{x})d\tilde{x}} = \frac {EA}{\tilde{x}}\begin{bmatrix} 1 & -1\\ -1 & 1 \end{bmatrix}^{L^{(i)}}_{0} = \frac {EA}{L^{(i)}}\begin{bmatrix} 1 & -1\\ -1 & 1 \end{bmatrix}$$

The equation of the area is: $$A(\tilde{x}) = N_1^{(i)}(\tilde{x})A_1 + N_2^{(i)}(\tilde{x})A_2$$

The equation of the modulus of elasticity is: $$E(\tilde{x}) = N_1^{(i)}(\tilde{x})E_1 + N_2^{(i)}(\tilde{x})E_2$$

Substituting the equations above into the stiffness matrix, is as follows:

$$\mathbf{k^{(i)}} = \frac {[N_1^{(i)}(\tilde{x})E_1 + N_2^{(i)}(\tilde{x})E_2][N_1^{(i)}(\tilde{x})A_1 + N_2^{(i)}(\tilde{x})A_2]}{L^{(i)}}\begin{bmatrix} 1 & -1\\ -1 & 1 \end{bmatrix}$$

contribution by carbon.w
 * }

Stiffness Matrices Application
Refering to the figures drawn during lecture 31, we find the shape functions N(i)1 and N(i)2. N(i)2 was given in class as the following.

HW: Find an expression for N(i)1

Noting that the value for N(i)1 has to equal 1 at x = 0 and it must also equal 0 at x = 1, we find the linearized value for N(i)1 to be:

HW: obtain expressions for k(i) using the average values for A and E as well as Linearized values for A and E

Using the average values for A $$\frac{A_{1} + A_{2}}{2}$$ into the equation $$ k = \frac {EA} {L} $$, we find the element stiffness matrix for element i to be:

Next, we use the following linearized equation for A to solve for the same stiffness matrix.

HW: Compare stiffness matrices obtained by using average values and linearized values for both E and A

Average vales of A and E,, yield the following stiffness matrix.

While the linearized values of A and E,, yield this stiffness matrix.

$$k^{(i)} = \frac {[(L^{(i)} - \tilde{x}) E_1 + (\tilde{x}) E_2] [(L^{(i)} - \tilde{x}) A_1 + (\tilde{x}) A_2]} {L^{(i)^3}} \begin{pmatrix} 1 & -1 \\ -1 & 1 \end{pmatrix} $$

Next, the mean value theorem (MVT) is recalled. This theorem is to help determine the location of a centroid of an area. For example, take the area shown below.

Recall that to find the centroid we use the equation

In our case, when calculating a stiffness matrix, there could be two different functions of x in the same equation. In that case, the mean value theorem yields.

where it is important to note that, in general:


text: File:Elem dofs.JPG

title: File:Elem dofs.JPG


text:
User:Eml4500.f08.delta 6.lopez/HW6

=FEM via PVW (cont.)=

Apply same interpolation for W(X). For example:

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

=Elemental Stiffness Matrix for Element i=

Now the element stiffness matrix can be written as:

$$ \beta = \int_{x_{i}}^{x_{i+1}}{[N_{i}'W_{i}+N_{i+1}'W_{i+1}](EA)}[N_i'd_i+N_{i+1}'d_{i+1}]dx$$ where $$N_{i}'W_{i}+N_{i+1}'W_{i+1}=W'(x)$$ and $$N_i'd_i+N_{i+1}'d_{i+1}=U'(x)$$

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

Likewise, the relation is the same for Ni+1':

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

 * NOTE

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

$$ \frac{du(x)}{dx}=\begin{bmatrix} N_i'(x) & N_{i+1}'(x) \end{bmatrix} \begin{Bmatrix} d_i\\ d_{i+1} \end{Bmatrix} =\mathbf{B(x)} \begin{Bmatrix} d_i\\ d_{i+1} \end{Bmatrix} $$

Similarly:

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

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

Now, let us recall the element degrees of freedom:

$$ \begin{Bmatrix} d_i\\ d_{i+1}

\end{Bmatrix} = \begin{Bmatrix} d_1^{(i)}\\ d_2^{(i)} \end{Bmatrix} = \mathbf{d^{(i)}} $$

$$ \begin{Bmatrix} W_i\\ W_{i+1}

\end{Bmatrix} = \begin{Bmatrix} W_1^{(i)}\\ W_2^{(i)} \end{Bmatrix} = \mathbf{W^{(i)}} $$

The element stiffness matrix now becomes: $$ \beta =\int_{x_{i}}^{x_{i+1}}{\left(\textbf{BW}^{(i)}\right)\left(EA \right)\left(\textbf{Bd}^{(i)}\right)}dx$$

$$ =\textbf{W}^{(i)}* \left(\textbf{k}^{(i)} \textbf{d}^{(i)}\right) $$

$$ \beta =\int_{x_i}^{X_{i+1}}(EA)(\mathbf{BW^{(i)}})*(\mathbf{Bd^{(i)}})dx-(\mathbf{BW^{(i)}})^T(\mathbf{Bd^{(i)}}) $$

where

$$ \mathbf{BW^{(i)}}= \mathbf{W^{(i)T}B^T} = \mathbf{W^{(i)}*B^T} $$

$$ \beta= \mathbf{W^{(i)}}*\int \mathbf{B^T}EA\mathbf{B}dx\mathbf{d^{(i)}} $$

$$ \mathbf{k^{(i)}}=\int_{x_i}^{x^{i+1}}{\mathbf{B(x)^T}}(EA)\mathbf{B(x)}dx $$

=Transformation of variable coordinate system from $$x$$ to $$ \tilde{x} $$=

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

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

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

=Homework=

Show that, if EA is constant,

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

Knowing that

$$ \beta = \int_{0}^{L}{\frac{dw}{dx}EA\frac{du}{dx}dx}\; \; \; \; \; \; \; \; \; \; Equation (1) $$

$$ \frac{dW}{dx}=[N_i'W_i + N_{i+1}'W_{i+1}] $$

$$ \frac{du}{dx}=[N_i'd_i + N_{i+1}'d_{i+1}] $$

Equation 1 reduces to

$$ \beta = \int_{x_i}^{x_{i+1}}{\mathbf{B}W^{(i)}EA\mathbf{B}d^{(i)}dx}\; \; \; \; \; \; \; \; \; \; Equation (2) $$

where

$$ W^{(i)} = \begin{bmatrix} W_i\\ W_{i+1} \end{bmatrix},

d^{(i)} = \begin{bmatrix} d_i\\ d_{i+1} \end{bmatrix},

\mathbf{B} = \begin{bmatrix} N_i' &N_{i+1}' \end{bmatrix} $$

Manipulating Equation (2) with transpose definitions, we determine:

$$ \mathbf{B} = W^{(i)}*\left(\int_{x_i}^{x_{i+1}}{\mathbf{B^T}EA\mathbf{B}dx} \right)d^{(i)}\; \; \; \; \; \; \; \; \; \; Equation (3) $$

Equation (3) shows that

$$ \mathbf{k^{(i)}}= \int_{x_i}^{x_{i+1}}{\mathbf{B^T}EA\mathbf{B}dx} $$

Since EA remains constant, this reduces to

$$ \mathbf{k^{(i)}}= EA\int_{x_i}^{x_{i+1}}{\mathbf{B^T}\mathbf{B}dx} $$

Now $$\mathbf{B(x)}$$ must be analyzed.

$$ \mathbf{B(x)}= \begin{bmatrix} N_i'(x) & N_{i+1}'(x) \end{bmatrix} = \begin{bmatrix} \frac{1}{x_i-x_{i+1}} & \frac{1}{x_{i+1}-x_i} \end{bmatrix}= \begin{bmatrix} \frac{1}{-L^{(i)}} & \frac{1}{L^{(i)}} \end{bmatrix} $$

$$ \mathbf{B^T}=\begin{bmatrix} \frac{1}{-L^i}\\ \frac{1}{L^i} \end{bmatrix} $$

Hence,

$$ \mathbf{k^{(i)}}= EA\int_{x_i}^{x_{i+1}}{\begin{bmatrix} \frac{1}{-L^i}\\ \frac{1}{L^i} \end{bmatrix}\begin{bmatrix} \frac{1}{-L^{(i)}} & \frac{1}{L^{(i)}} \end{bmatrix}dx} $$

Multiplying matrices gives

$$ \mathbf{k^{(i)}} = EA\begin{bmatrix} \int_{x_i}^{x_{i+1}}{\frac{1}{L^2}dx} & \int_{x_i}^{x_{i+1}}{\frac{-1}{L^2}dx}\\ \int_{x_i}^{x_{i+1}}{\frac{-1}{L^2}dx} & \int_{x_i}^{x_{i+1}}{\frac{1}{L^2}dx} \end{bmatrix} $$

 * NOTE

One integer will be used for example. Others differ by a positive or negative sign.

$$ \int_{x_i}^{x_{i+1}}{\frac{1}{L^2}dx} = \frac{x_{i+1}-x_i}{L^2} = \frac{L}{L^2} = \frac{1}{L} $$

Thus,

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

Extracting $$\frac{1}{L}$$ gives us

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

=Homework 2=

where

$$ A(\tilde{x})=N^{(i)}_{1}(\tilde{x})A_{1}+N^{(i)}_{2}(\tilde{x})A_{2} $$

and

$$ E(\tilde{x})=N^{(i)}_{1}(\tilde{x})E_{1}+N^{(i)}_{2}(\tilde{x})E_{2} $$

We will find an expression for $$\mathbf{k^{(i)}}$$ using

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

By plugging the above equations in for $$E(\tilde{x})$$ and $$A(\tilde{x})$$, we get

$$ \mathbf{k^{(i)}}=\int_{\tilde{x}=0}^{\tilde{x}=L^{(i)}}\mathbf{B^T(\tilde{x})}(N^{(i)}_{1}(\tilde{x})E_{1}+N^{(i)}_{2}(\tilde{x})E_{2})(N^{(i)}_{1}(\tilde{x})A_{1}+N^{(i)}_{2}(\tilde{x})A_{2})\mathbf{B(\tilde{x})}d\tilde{x} $$

title: User:Eml4500.f08.delta 6.lopez/HW6


text:
Template:Course Invite

This template can be placed on a Wikiversity page containing an active course. Please feel free to include the course on the Wikiversity_Courses page.

title: Template:Course Invite


text: File:FEA Nov 7 Pic 5.jpg

title: File:FEA Nov 7 Pic 5.jpg


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User:Eml4500.f08.delta 6.guzman/Lecture Notes 11/07

Stiffness Term

Note: Nodes do not have to be equidistant.

Plot

Note: Since you are drawing a straight line between nodes you are assuming the their displacement was linear.

Assume displacement $$ u\bigl(\begin{matrix}x\end{matrix}\bigr)$$ for $$ x_{i}\leq x\leq x_{i+1} $$ (i.e. $$x \epsilon \left [ x_i, x_{i+1} \right ]$$)

Motivation for linear interpolation of $$ u\bigl(\begin{matrix}x\end{matrix}\bigr)$$

In this plot can be easily observed that the deformed shape is a straight line confirming that there was an implicit assumption of the linear interpolation of the displacement between the two nodes. Consider the case where these are only axial displacements (i.e. 0 translational displacement).

Question: Express $$ u\bigl(\begin{matrix}x\end{matrix}\bigr)$$ in terms of $$d_{i}=u\bigl(\begin{matrix}x_{i}\end{matrix}\bigr)$$ and $$d_{i+1}=u\bigl(\begin{matrix}x_{i+1}\end{matrix}\bigr)$$ as a linear function in x (i.e. linear interpolation)

$$u\bigl(\begin{matrix}x_{i}\end{matrix}\bigr)=N_{i}(x)d_{i}+N_{i+1}(x)d_{i+1}$$

where $$N_{i}\bigl(\begin{matrix}x\end{matrix}\bigr)$$ and $$N_{i+1}\bigl(\begin{matrix}x\end{matrix}\bigr)$$ are linear functions of x.

Note: From this plot it can be easily observed that $$ N_{i+1}(x)=\frac{x-x_{i}}{x_{i+1}-x_{i}} $$.


text:
The US Presidents/George Washington

For an in depth look at the 1st US President, follow this link: George Washington.

Elected: 1st Term in 1789 (only US election to occur in an odd-numbered year)

Elected: 2nd Term in 1792

Presidency: 1789 to 1797

Major US Laws
The following is a list of major laws in US History signed by George Washington during his presidency:
 * Residence Act of 1790, allowing the President to select a location for the future capital of the country.
 * The first 10 Amendments were added to the Constitution as the Bill of Rights in 1791.
 * He also gave his famous Farewell Address in 1796 before retiring from the Presidency

Major US Events
The following is a list of major events in US History during the presidency of George Washington:
 * US governmental operations started under the new US Constitution on March 4, 1789.
 * The first session of the Supreme Court convened on February 2, 1790.
 * The Whiskey Rebellion was quelled by state militia in 1794, proving the strength of the new Constitution.
 * Vermont (1791), Kentucky (1792), and Tennessee (1796) joined the Union.

Foreign Affairs
The following is a list of events in US Foriegn Affairs during the presidency of George Washington:
 * Jay Treaty

Major World Events
The following is a list of major events in World History during the presidency of George Washington:
 * The French Revolution began, establishing the First French Republic.
 * George Washington had nothing to with it the first 3 years.

Interesting Facts

 * George Washington was the only US president ever to have been elected unanimously by the Electoral College, which he was during both terms.
 * Washington refused to run for a third term once his second term was done, setting a precedent for future presidents, a tradition which held until Franklin Delano Roosevelt's third election in 1940.

title: The US Presidents/George Washington


text: File:Elastic Bar divided with X nodes.jpg


text:
A course in troll sockpuppets

Introduction
Trolls are creatures of myth and also real denizens of the Internet. Here the two concepts meet in a series of puppet designs.

Requisite skills:
 * Good hand sewing
 * Basic wireworking
 * Basic fabric painting
 * Soft sculpture

Recommended skills:
 * Basic crochet
 * Tambour embroidery
 * Buttonhole stitch embroidery
 * Latch hook rugmaking
 * Polymer clay sculpture

Substitution suggestions are provided for instructions that use recommended skills.

Part 1: Introduction to trolling
Instructions for creating a troll sockpuppet.
 * 1) Basic trolling
 * 2) Feeding a troll

Part 2: How to build an army of sockpuppet trolls
Wardrobe and accessory designs for a rogues' gallery of sockpuppet trolls.
 * 1) Sgt. Troll
 * 2) Insecure troll
 * 3) Hacker troll
 * 4) Conspiracy theorist troll
 * 5) Pseudointellectual troll
 * 6) Pseudoscience troll
 * 7) Fergus MacTroll
 * 8) Hamlet, Prince of Trollmark
 * 9) Interview with a female troll

Part 3: How to kid around with a troll and feed him to the sharks
A guide to creating a sockpuppet theater production of "The Three Billy Goats Gruff", supplementing the troll with three goats and a set design.
 * 1) The Three Billy Goats Gruff
 * 2) Billy the Kid
 * 3) Medium sized billy goat
 * 4) Great big billy goat
 * 5) Set design

Part 4: Advanced trolling
Design advice for inventing additional trolls, puppets, and sets.
 * 1) Advanced trolling

title: A course in troll sockpuppets


text:
A course in troll sockpuppets/Introduction to trolling

When beginning any creative endeavor, one of the most important things is to choose a clear inspiration. John Bauer illustrated Norwegian folk tales during the early twentieth century. The example at left is representative of his work and contains several trolls in various poses.

Although a sockpuppet cannot reproduce all the elements of a detailed illustration, important preparatory work to designing a new sockpuppet includes analysis of distinctive features. The head dominates a sockpuppet so particular interest goes to upper features. The hand is also of interest for this course because several of the puppets carry a prop.

This lesson designs a basic troll with head and neck. Lesson 2 builds a hand for the puppet.

Noteworthy features

 * Large bulbous nose
 * Closely set eyes
 * Marks underneath the eyes
 * Long thin ears that curl outward at the top
 * Low forehead
 * Long stringy hair
 * Lower jaw that meets the underside of the nose
 * Muted colors
 * Thick hands
 * Pointed fingernails

Puppet construction
Basic material for the troll is a brown cotton men's sock. Exact measurement depends on the size of the puppeteer's hand. The end of the sock may project beyond the puppeteer's fingers, but the mouth itself should not curl with the puppeteer's hand. The cut for the mouth begins from the toe seam.

Hemming is necessary in order to prevent knitted fabric from unraveling. To hem the sock, roll fabric inward and secure with an overhand stitch. The completed mouth should resemble the mouth of a Bauer troll: shorter than the nose yet closing directly into the nose. The hemmed cut may curl naturally. Flatten sock when sizing black felt for the mouth liner. Cut slightly large, then trim as appropriate. Liner should fit within the hem and bend at the base of the thumb. Secure with overhand stitches. After completing the mouth, begin work on the eyes. Preparatory work with buttons has the advantage of stabilizing googly eyes. A button base provides a more stable surface for plastic eyes than knitted sock fabric, since the eyes can only be secured with glue. Stitch the buttons close together high on the fingers, leaving a shallow forehead for troll hair. Glue eyes on afterward. Allow at least one hour for glue to dry. Ears are the next important feature. Ears should go at the outer edge of the head slightly forward of the bend of the hand and above the mouth crease. A single green pipe cleaner cut in two and folded double is sufficient for both ears. Twist the halves for structural integrity and splay the open ends. Then crimp the folded end with a pair of pliers. un each ear up the body of the sock and into position, then pull through fabric. Curl the ends after both ears are in place. Bend the remaining base beneath the sock into a triangular or es-shaped configuration. In order to recreate the bulbous appearance of the noses in the original artwork, insert synthetic pillow filler into the end of the nose. Then, using the puppeteer's fingers as a sizing indicator, create a set of loose quilting stitches within the upper forelip to delineate a nose. Then add semicircles of green fabric paint beneath the eyes. Wait for paint to dry.

The trolls in Bauer's artwork appear to have a part running down the middle of their foreheads. To imitate that effect, select an appropriate yarn and cut a number of equal lengths to approximate the over-the-shoulders effect of Bauer's trolls. Then lay across the puppet head behind the ears and secure with tight backstitches. Trim ends of hair as appropriate. To complete the basic puppet, cut small sections of felt in a neutral color and glue to the base of the ears. Allow plenty of time to dry.


title: File:HW61.JPG


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User:Eml4500.f08.jamama.jan/Meeting29

Initial Conditions  (2 conditions) at t = 0, prescribe:

Condition 1: $$ u(x, t=0) = \bar{u}(x) $$
 * $$ \uparrow $$
 * known function displacement

Condition 2: $$ \frac{\partial u}{\partial t}(x, t=0) = \dot {u}(x, t=0) = \bar{v}(x)$$
 * $$ \uparrow $$
 * known function velocity

PVW (continuous) of dynamics of elastic bar

PDE (Partial Differential Equation):

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

Discrete EOM (Equation of Motion)
 * $$\downarrow $$

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

 * $$\Downarrow $$

$$ \mathbf{M\ddot d} + \mathbf{Kd} = \mathbf{F} $$
 * $$ \nwarrow $$
 * Multiple Degree of Freedom System (MDOF)

Single Degree of Freedom System (SDOF)

$$ \Rightarrow \ \mathbf{m\ddot d} + \mathbf{kd} = \mathbf{f} $$

Derive (2) from (1)

$$ \int \limits_0^L ( \frac{\partial}{\partial x}[EA \frac{\partial u}{\partial x}] + f - m \ddot {u} ) dx = 0,$$ $$ \qquad $$   for all possible  $$\mathbf{w(x)} \qquad \qquad \qquad (3)$$
 * $$ \uparrow $$
 * weighting function

$$ (1) \Rightarrow (3) $$  >>  trivial

$$ (3) \Rightarrow (1) $$  >>  not trivial

(3) rewritten as : $$ \int w(x) g(x) dx, \qquad $$ for all $$ \mathbf{w(x)}$$

Since (3) holds for all $$ \mathbf{w(x)}$$, select $$ \mathbf{w(x)} = \mathbf{g(x)}$$, then (3) becomes $$\int \underbrace{g^2(x)}_{\ge 0} dx = 0 \Rightarrow g(x) = 0 $$

{| class="toccolours collapsible collapsed" style="width:100%" ! Matlab Code For Electric Pylon

Matlab Code

results=[]; for i=1:n_elem lm=lmm(i,:); location = [x(node_connect(1,i)),y(node_connect(1,i)); x(node_connect(2,i)),y(node_connect(2,i))]; results = [results; PlaneTrussResults(E,A,location,d(lm))]; end results

title: User:Eml4500.f08.jamama.jan/Meeting29


text: File:Plot of Displacement in relation to X.jpg


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text: File:Plot of 2 Bar Truss system.jpg

title: File:Plot of 2 Bar Truss system.jpg


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User:Eas4200c.f08.blue.b/Lecture 30

Continuing Torsional Analysis:

Dimensional analysis, from lecture 29, is applied to the differential stress formula from lecture 27 to show that the differential of stress over distance x has the dimensions of force over volume.

$$\left[\frac{\delta \sigma _{ij}}{\delta x_{i}} \right]=\frac{F}{L^{3}}=\frac{force}{volume}$$

In equation 3.14 in Mechanics of aircraft structures by C.T. Sun, it is shown that the of partial derivatives of shear force in the x and y directions is equal to zero.

$$\frac{\partial \tau _{xz}}{\partial x}+\frac{\partial \tau _{yz}}{\partial y}=\frac{\partial \sigma _{xz}}{\partial x}+\frac{\partial \sigma _{yz}}{\partial y}=0$$

From the roadmap described in lecture 16, the Prandtl stress function $$\phi $$ can be found using the following formulas (Sun 3.15):

$$\sigma _{yx}=\frac{\delta \phi }{\delta z}$$ $$\sigma _{zx}=\frac{-\delta \phi }{\delta y}$$

As can be seen, the Prandtl stress function plays the role of a potential function. $$\left(\sigma _{yx},\sigma _{zx} \right)$$ is a component of the "gradient" of the Prandtl function with respect to y and z.

Recall that the gradient of a scalar function f(x,y,z) is simply the partial differential of f in each direction (x, y, and z) or (i, j, and k).

$$\vec{\bigtriangledown} f(x,y,z)=\frac{\partial f}{\partial x}\hat{i}+\frac{\partial f}{\partial y}\hat{j}+\frac{\partial f}{\partial z}\hat{k} $$

In lecture 27, the class found that the sum of the partial derivative with respect to y of stress in the yx direction and the partial derivative with respect to z of the stress in the zx direction results in zero. Now we can replace the stresses with the Prandtl stress function equivalents.

$$\frac{\partial \sigma _{yx}}{\partial y}+\frac{\partial \sigma _{xz}}{\partial z}=0$$

$$\frac{\partial }{\partial y}\left(\frac{\partial \phi }{\partial z} \right)+\frac{\partial }{\partial z}\left(\frac{-\partial \phi }{\partial y} \right)=0$$

$$\frac{\partial ^{2}\phi }{\partial y\partial z}-\frac{\partial ^{2}\phi }{\partial z\partial y}=0$$

Since $$\phi$$ is continuous and smooth then the second mixed derivative is interchangable.

Now the Prandtl stress function $$\phi (x,y)$$ can be introduced in terms of the shear stress as in equation 3.15 (Sun, 70).

$$\tau _{xz}=\frac{\partial \phi }{\partial y}$$ $$\tau _{yz}=-\frac{\partial \phi }{\partial x}$$

It follows that:

$$\gamma _{xz}=\frac{\partial w}{\partial x}+\frac{\partial u}{\partial z}=\frac{\partial w}{\partial x}-\theta y$$

$$\gamma _{yz}=\frac{\partial w}{\partial y}+\frac{\partial v}{\partial z}=\frac{\partial w}{\partial y}+\theta x$$

$$\frac{\partial \gamma _{yz}}{\partial x}-\frac{\partial \gamma _{xz}}{\partial y}=2\theta $$

$$\gamma _{yz}=\frac{1}{G}\tau _{yz}$$

$$\gamma _{xz}=\frac{1}{G}\tau _{xz}$$

$$\frac{\partial \tau _{yz}}{\partial x}-\frac{\partial \tau _{xz}}{\partial y}=2G\theta $$

$$\frac{\partial ^{2}\phi }{\partial x^{2}}+\frac{\partial ^{2}\phi }{\partial y^{2}}=-2G\theta $$

$$\frac{\partial ^{2}\phi }{\partial y^{2}}-\frac{\partial ^{2}\phi }{\partial z^{2}}=-2G\theta=\bigtriangledown ^{2}\phi $$

This equation for the Prandtl Equation is equivalent to step F of the Torsional analysis Roadmap.

To find a function for the Prandtl Function, a two dimensional case is first looked at:

In the above image, a cross section of a non-uniform bar is shown with a force (t) pu.

This image represents the stress vector (t) on an infinitesimal surface of a non-uniform beam.

title: User:Eas4200c.f08.blue.b/Lecture 30


text: File:Linear Interpolation example.jpg

title: File:Linear Interpolation example.jpg


text:
PCP HIV AIDS Toolkit/Summary of RA 8054

RA 8504 - The Philippine AIDS Law
(This is a condensed version of the RA 8504)

Article 1
The Government must promote education and information campaigns.
 * Schools and non-formal education programs must include HIV education.
 * The DOH must conduct public health education campaigns.
 * The government must provide education for Filipinos going abroad.
 * Employers, working with DOLE must develop workplace education and safety.
 * The Department of Tourism must provide education for tourists and transients.
 * LGUs must develop local HIV prevention and education efforts.
 * Accurate information about prophylactics will be provided.
 * It is against the law to give false or intentionally misleading information on HIV/AIDS.

Article 2
The Government will develop and promote safe medical practices
 * All blood, organs, and tissue for donation will be tested. If it tests positive, it will be disposed of safely, and not used.
 * The DOH will provide guidelines on safe surgery and medical procedures.
 * Willfully or negligently dangerous practices will be against the law.

Article 3
The Government will provide appropriate confidential HIV testing.
 * HIV testing can only be done with the consent of the person tested.
 * The DOH will provide anonymous HIV testing.
 * Counseling will be conducted before and after testing.
 * The DOH will make adequate and affordable HIV testing available in all of the Philippines.

Article 4
The Government will ensure adequate health and support services for people with HIV.
 * Hospitals will provide adequate care for persons with HIV.
 * LGUs must provide community-based prevention and care efforts.
 * Livelihood efforts will be made available for people with HIV.
 * The DOH is to do STD prevention and control efforts.
 * The DOH is to make sure that health insurance is available to people with HIV.

Article 5
The Government will monitor HIV/AIDS
 * The Government will establish an AIDSWATCH council.
 * Reporting procedures will be developed to track HIV rates, while respecting client confidentiality.
 * Contract tracing is permitted provided confidentiality is not breached.

Article 6
Confidentiality will be protected on HIV status.
 * All healthcare workers, and anyone handling health records will strictly preserve patient confidentiality on HIV status, and the identity of people with HIV.
 * HIV status can be shared by health officials in three circumstances; officials of the AIDSWATCH council, Health workers who are involved in treatment and need to know for their own safety, and a judge if under subpoena for an official court case.
 * HIV results may be given to the person tested, an official of the AIDSWATCH council, and a parent or guardian.  It cannot be given to anyone else.
 * Legal penalties can be provided for breaching confidentiality.
 * People with HIV must inform spouses and sex partners as soon as reasonably possible.

Article 7
Discrimination against people with HIV is illegal.
 * Employers cannot discriminate in hiring, firing, promotion, or assignment based on actual or suspected HIV status.
 * Schools are not allowed to refuse admission, punish students, or deny participation in activities based on real or perceived HIV status.
 * A person with HIV has the legal right to travel, live, and lodge with the same freedom as any other citizen.  Quarantines and other restrictions are illegal.
 * A person with HIV has every legal right to seek pubic office.
 * Credit services cannot be denied based on HIV status.  Insurance cannot be refused based on HIV status, provided the person does not lie about their HIV status.
 * Hospitals can heath services cannot refuse treatment or discriminate based on HIV status.
 * Decent burials cannot be denied based on HIV status.
 * Legal penalties for discrimination will be enforced.

Article 8
The government will establish a Philippine National AIDS Council
 * This council will be responsible for implementing the act and conducting all national AIDS performances.

title: PCP HIV AIDS Toolkit/Summary of RA 8054


text:
A course in troll sockpuppets/Feeding a troll

Adding a hand to a puppet increases its expressive range and gives the puppet a way to carry props. Here we build a hand that several of the advanced puppets will use. It makes puppeteering a two handed endeavor, with the puppeteer holding a wooden dowel. A sleeve conceals the puppeteer and connects the hand to the puppet. As a starter prop this lesson also creates a puppet-sized pizza, since in order for a community to develop a really serious trolling problem a troll has to get fed.

As noted in Lesson 1, Bauer's troll hands are thick and sharp nailed. The puppet version of the hand reduces these details to a thumb and three sharp green fingernails. A large palm is needed to provide an attachment base for props.

Instructions
Rather than beginning with the hand, this construction begins with the prop and then sketches a hand with sufficient proportions to support the prop. The pizza needs a structural base that begins with cardboard. Ordinary cereal box cardboard was used for this puppet. Cardboard from shirt packaging is also good for puppet props. The appropriate size of the pizza depends on the sock and the puppeteer's hand.

Hold the sock for comparison and choose a circle approximately the same diameter as the width of the puppet's head. For precise sizing a compass is ideal, but tracing a cylindrical household object is a functional substitute. It is strongly recommended to use different pairs of scissors for cutting cardboard and for cutting fabric. Cardboard dulls the cutting edge and makes it difficult to manipulate a pair of blades over softer materials. Once the cardboard base is ready, cut matching felt sections in light brown. Trim into matching circles and glue to the cardboard. Wait to dry. Add a slightly smaller circle in yellow to resemble pizza. Center within the pizza base and glue into place. Add red fabric paint circles to represent pepperoni. Pepperoni may be added gently while glue is wet. To create a crust, use wire cutters to trim a brown pipe cleaner slightly longer than the circumference of the pizza. Wait to dry completely before proceeding to next step. Wrap the pipe cleaner around a cylindrical object to bend it into a circular shape the same size as the diameter of the pizza. Bend ends with small pliers and make short wraps to complete a stable circle. Flatten any bends that may result in the wire. Glue pipe cleaner to pizza and wait for glue to dry. Sketch the outline for a troll hand on another piece of cardboard, using the pizza and the sockpuppet to estimate sizing. Allow a moderate length past the hand itself for a wrist and part of a forearm. Unlike the pizza base, the cardboard troll hand will be used only as a tracing model and serves no structural purpose. Cut out the tracing model with cheap scissors, then select felt in a color that resembles the sockpuppet color and make two hand tracings onto felt. Cut the felt with good scissors.

Use 22 gauge wire and/or pipe cleaners for structural support when sewing together the hand. Bend with pliers to reinforce the thumb, palm, and wrist. Synthetic pillow filling will buffer the wires and fill out a soft thick hand.

Begin sewing with tight overhand stitches from the wrist upward away from the thumb side, adding wire and pillow stuffing while stitching the fingers. At least two wires should extend downward from the wrist beyond the length of the felt. Be sure all wire and hand stuffing is secure before closing the natural base of the hand. Insert a wooden dowel up the arm to the wrist. Complete the stitching, closing as much of the arm base as feasible. Upon completion of initial stitching the troll hand is ambidextrous. Before adding finishing touches, decide on the puppet's hand preference. Note that the puppet's hand preference is opposite of the puppeteer's hand preference. Puppeteers manipulate the mouth and face with their dominant hand, using the non-dominant hand for the dowel that supports the puppet's hand. So a right-handed puppeteer will control a left-handed puppet.

After selecting a hand preference, bend wrist and paint green fingernails. After paint dries sew pizza to the palm. Soft knitted fabric such as an old t-shirt is ideal for a puppet sleeve. The sleeve must be long enough to wrap around the arm and conceal the puppeteer's hand, allow a naturalistic distance to the puppet body, and wrap around the puppet body like a shirt. Vertical fabric edges should be concealed. Cut and hem by hand or machine. Hand stitch the sleeve to the puppet wrist, then wrap around the puppet body and secure with a few stitches. Attachment of the sleeve to the wrist is permanent, but attachment to the body needs to be removable if the puppet will be wearing troll costumes that do not use a hand. Wrap extending wires around dowel and protect hands from scratching by bending and crimping the ends of the wires with a small pair of pliers. Then enjoy the pizza: you've just fed a troll.


text:
User:Eas4200c.f08.blue.a/Lecture 32

Fig.1 to the right shows an unsymmetric thin-walled section. $$\displaystyle y_c$$ and $$\displaystyle z_c$$ are measured from the centroid of the shaded area. As labeled in the figure, the thickness $$\displaystyle t_s$$ can be a function of s, the curvilinear coordinate.

A general equation for thin-walled sections is the following (5.1):
 * $$\iint_{A_s} \frac{d\sigma_{xx}}{dx} dA = -q_s$$

A comparison of equations for symmetric and unsymmetric sections: (Note: in the following symmetric cases, the section is symmetric about the y-axis)

Symmetric: $$\displaystyle \sigma_{xx} = \frac{M_yz}{I_y}$$

Unsymmetric: $$\displaystyle \sigma_{xx} = (k_yM_z - k_{yz}M_y)y + (k_zM_y - k_{yz}M_z)z$$

where
 * $$\displaystyle k_y = \frac {I_y}{D}$$
 * $$\displaystyle k_{yz} = \frac {I_{yz}}{D}$$
 * $$\displaystyle k_z = \frac {I_z}{D}$$
 * $$\displaystyle D = I_yI_z - I_{yz}^2$$

The above equation for an unsymmetric section can be rewritten in matrix form to become
 * $$\displaystyle \sigma_{xx} = \begin{bmatrix}

y & z \end{bmatrix} \begin{bmatrix} k_y & -k_{yz} \\ -k_{yz} & k_z \end{bmatrix} \begin{Bmatrix} M_z \\ M_y \end{Bmatrix} = \begin{bmatrix} z & y \end{bmatrix} \begin{bmatrix} k_z & -k_{yz} \\ -k_{yz} & k_y \end{bmatrix} \begin{Bmatrix} M_y \\ M_z \end{Bmatrix} $$

The equation for an unsymmetric section can also be particularized to become the equation for a symmetric section (considering $$\displaystyle M_z$$ to be equal to zero). With a symmetric section, $$\displaystyle I_{yz}$$ is equal to zero.

 * $$\displaystyle M_z = 0 $$
 * $$\displaystyle I_{yz} = 0 $$
 * $$\displaystyle\therefore D = I_yI_z$$
 * $$\displaystyle k_y = \frac {I_y}{I_yI_z} = \frac {1}{I_z}$$
 * $$\displaystyle k_{yz} = \frac {I_{yz}}{I_yI_z} = 0$$
 * $$\displaystyle k_z = \frac {I_z}{I_yI_z} = \frac {1}{I_y}$$
 * $$\displaystyle\therefore \sigma_{xx} = (k_yM_z - k_{yz}M_y)y + (k_zM_y - k_{yz}M_z)z = (0-0)y + (k_zM_y - 0)z = \frac{M_yz}{I_y}$$

Continuing with our comparisons of equations between symmetric and unsymmetric sections:

Symmetric: $$\displaystyle q(s) = -\frac {V_zQ_y}{I_y}$$
 * $$Q_y = \int_{A_s} z\,dA = z_cA_s$$

Unsymmetric: $$\displaystyle q(s) = -(k_yV_y - k_{yz}V_z)Q_z - (k_zV_z - k_{yz}V_y)Q_y$$
 * $$Q_y = \int_{A_s} z\,dA$$
 * $$Q_z = \int_{A_s} y\,dA$$

The above equation for an unsymmetric section can be rewritten in matrix form to become
 * $$\displaystyle q(s) = -\begin{bmatrix}

Q_z & Q_y \end{bmatrix} \begin{bmatrix} k_y & -k_{yz} \\ -k_{yz} & k_z \end{bmatrix} \begin{Bmatrix} V_y \\ V_z \end{Bmatrix} = -\begin{bmatrix} Q_y & Q_z \end{bmatrix} \begin{bmatrix} k_z & -k_{yz} \\ -k_{yz} & k_y \end{bmatrix} \begin{Bmatrix} V_z \\ V_y \end{Bmatrix} $$

As with the equation for $$\displaystyle \sigma_{xx}$$, the equation for $$\displaystyle q(s)$$ with an unsymmetric section can be particularized to become the equation with a symmetric section. Similar to the case for $$\displaystyle I_{yz}$$ proven above, $$\displaystyle Q_z$$ is equal to zero due to symmetry about the y-axis. Again, $$\displaystyle I_{yz}$$ is also equal to zero.
 * $$\displaystyle Q_z = \int_{A_s} y\,dA = 0$$
 * $$\displaystyle k_y = \frac {I_y}{I_yI_z} = \frac {1}{I_z}$$
 * $$\displaystyle k_{yz} = \frac {I_{yz}}{I_yI_z} = 0$$
 * $$\displaystyle k_z = \frac {I_z}{I_yI_z} = \frac {1}{I_y}$$
 * $$\displaystyle \therefore q(s) = -(k_yV_y - k_{yz}V_z)Q_z - (k_zV_z - k_{yz}V_y)Q_y = -(k_yV_y-0)0 - (k_zV_z - 0)Q_y = -\frac {V_zQ_y}{I_y}$$

When dealing with stringer-web sections, it is assumed that the thickness $$\displaystyle t$$ of the skin and spar web is very small, therefore their areas are neglected in the computation of $$\displaystyle I_y, I_z, I_{yz}, Q_y, Q_z$$. The only areas used are the areas of the stringers (as shown in Fig. 2 to the left).

With a section that is symmetric about the y-axis, the areas will be set such that $$\displaystyle A_3 = A_2$$ and $$\displaystyle A_4 = A_1$$. However, for the unsymmetric case, the areas will be set such that $$\displaystyle A_1 = A,\,\, A_2 = 2A,\,\, A_3 = 3A,\,\, A_4 = 4A$$.

title: User:Eas4200c.f08.blue.a/Lecture 32


text: A course in troll sockpuppets/Pseudoscience troll


text:
A course in troll sockpuppets/Sgt. Troll

Every troll army needs an efficient noncommissioned officer. He's the fellow who sets up the mailing list and spots opportunities for mass trolling. Sgt. Troll noticed that a troll got fed over here and told everyone to come join the fun. Soon the site will get inundated with other trolls. But first, a close look at Sgt. Troll and his uniform.

Instructions
Sgt. Troll's distinctive elements are really just two features: a helmet and rank insignia. Any country's gear would do. The U.S. Army M1 helmet, used from World War II through Vietnam, became the model for Sgt. Troll's helmet.

To make the helmet, begin with a square of household tinfoil. Fold several times into multiple thicknesses and shape gently into a rounded dome. Should fit snugly between the puppet ears and flatten at the edges to resemble the M1 helmet. Cut a section of dark green knit fabric and stretch across helmet. Glue into place.

For insignia, cut a section of yellow felt. Use fabric paint for outline and chevrons. Wait to dry. Stitch into place.


text:
A course in troll sockpuppets/Pseudointellectual troll

The pseudointellectual troll thinks the ability to compose sentences in polysyllables is a substitute for reasoned discussion.

Some pseudointellectual trolls think that a degree makes them all important. We have a copy of one such degree.


text:
A course in troll sockpuppets/Hamlet, Prince of Trollmark

The drama sock.


text:
Talk:A course in troll sockpuppets

Wow. A (semi-)troll course about troll sockpuppets. Brilliant. --93.73.49.245 16:10, 30 January 2010 (UTC)

Troll University - not sure yet what it is really about :-) Erkan Yilmaz uses the Chat 16:48, 9 January 2011 (UTC)

title: Talk:A course in troll sockpuppets


text:
Talk:Object Oriented Enterprise Modelling

Is this or will this be part of a wikiversity course?.... if so

Please note that wikiversity courses are quite different from a regular university course for some advice please feel free to review Wikiversity_Courses


text:
User:Eml4500.f08.delta 6.castrillon/HW 6

Integration by Parts Recalling the method of integration by pats for integration, it is described below. We have two functions, $$r(x)$$ and $$s(x)$$. When we want to determine the derivative of the product of these two functions, we have: $$(rs)'=r's+rs'$$ where, $$r'=\frac{dr}{dx}$$ and $$s'=\frac{ds}{dx}$$ Then, $$(rs)=\int (rs)'dx=\int rs'dx+\int r'sdx$$ $$\int r'sdx=rs-\int rs'dx$$

Recalling the Principle of Virtual Work, we can apply this method of integration to this principle.

Let $$r(x)=EA \frac{\partial u}{\partial x}$$ and $$s(x)=w(x)$$

Now applying Integration by Parts,

 * $$\int _{x=0}^{L}w(x) \frac{\partial }{\partial x} \left[ EA \frac {\partial u}{\partial x} \right] dx$$
 * $$\left[w(EA)\frac {\partial u}{\partial x} \right]_{0}^{L}-\int_{0}^{L}\frac{dw}{dx}EA\frac{\partial u}{\partial x}dx$$
 * $$w(L)EA(L) \frac{\partial u(L,t)}{\partial x}-w(0)EA(0) \frac{\partial u(0,t)}{\partial x}- \int_{0}^{L} \frac{dw}{dx}EA\frac{\partial u}{\partial x}dx$$

It is noticeable that
 * $$N(L,t)=EA(L) \frac{\partial u(L,t)}{\partial x}$$

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

Now we need to account for the boundary conditions (BCs). We use the BC defined in the diagram below:

At the wall, (x=0), we have to select w(x) such that w(0)=0. This requirement would be kinematically admissible as it would adhere to the constraints.

Descrete PVW applied to the reduction of the global FD rel

 * $$w$$ $$_{6x1}$$•$$\left( \begin{bmatrix}

K \end{bmatrix}_{6x2}

\begin{pmatrix} d \end{pmatrix}_{2x1}-F_{6x1} \right)=0_{1x1}$$; for all $$w$$

We have the transpose of F:

 * $$F^T$$$$=\begin{bmatrix}

F_1 & F_2 & F_5 & F_6 \end{bmatrix}$$

where F_1, F_2, F_5, and F_6 are all unknown reactions.

Since $$w$$ can be selected arbitrarily, we say that $$w$$ is such that:
 * $$w_1=w_2=w_5=w_6=0$$

Therefore, to eliminate the equation involving unknown reactions (element rows 1, 2, 5, and 6), we say that:
 * $$\bar{w}_{2x1}$$•$$\left( \mathbf{\bar{K} \bar{d} \bar{F}} \right)=0$$; for all $$w$$.

This yields to,
 * $$\bar{K}_{2x2} \bar{d}_{2x1}$$ = $$\bar{F}_{2x1}$$

where,
 * $$\bar{d}_{2x1}$$=$$\begin{pmatrix}

d_3\\ d_4 \end{pmatrix}$$ and
 * $$\bar{F}_{2x1}$$=$$

\begin{pmatrix} F_3\\ F_4 \end{pmatrix}$$

Turning our attention back to the Principle of Virtual Work, and recalling the reaction terms,

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

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

we can build the following expression:

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

Rewriting in row form,

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


text:
User:Eas4200c.f08.gator.edwards/Wk6 Class Notes

Notes for Meeting 29: November 3rd
Eas4200c.f08.gator.reger 05:08, 19 November 2008 (UTC)

Equation of Equilibrium:

Recall 1-D case (from meeting 28): Dimensional Analysis

$$\left[f\right] = \frac{f}{L}$$ $$\left[A\right] = L^2\,$$ Therefore: $$ \left[ \frac{f}{A}\right] = \frac{F}{L^3}$$ Where $$f$$ is 'dimension of', $$F\,$$ is force, $$L\,$$ is length, and $$A\,$$ is area.

Also, $$ \left[dx\right] = L \,$$ And, $$ \left[\sigma\right] = \frac{F}{L^2}\,$$ Therefore: $$ \left[\frac{d \sigma }{dx}\right] = \frac{F}{L^3}\,$$

It can be shown that:

$$\left[ \epsilon \right] = \left[\frac{du}{dx}\right] = \frac{L}{L} = 1  (Non-dimensional)\,$$ And, $$\left[ \mu \right]=\left[\frac{ \epsilon _{yy} }{ \epsilon _{xx}}\right] = 1  (Non-dimensional)\,$$

Torsion Analysis Continued
Case 1: Thin walled cross section (closed)

Outer Surface $$\phi=C_{0}$$ on $$S_{0}$$

Inner Surface $$\phi=C_{1}$$ on $$S_{1}$$

Case 2: Solid cross section

Outer Surface $$\phi=0$$ on $$S_{0}$$

Uniform bar with solid circular cross section

$$\phi (y,z)=C\left(\frac{y^{2}}{a^{2}}+\frac{z^{2}}{b^{2}}-1 \right)$$

Because the cross section is a circle $$a=b$$.

$$\phi=0$$ on $$S_{0}$$ because the cross section is solid.

$$\bigtriangledown^{2} \phi = -2G \theta $$

$$C=-\frac{1}{2}a^{2}G\theta $$

$$T = 2 \int_{A}^{} \phi \partial A = 2 C(\frac{J}{a^{2}} - A)$$

rearrange and substitute the area of a circle for A to get,

$$J = \int_{A}^{} r^{2} \partial A = \frac{1}{2} \pi a^{4}$$

Traction Vector
Relate ds to dz and dy via θ.

$$\partial y = \partial s*sin(\theta)$$

$$\partial z = \partial s*cos(\theta)$$

$$n_{y} = cos(\theta)$$

$$n_{z} = sin(\theta)$$

Now, sum forces in all directions:

$$\sum_{}^{} F_{y} = 0 = \sum_{}^{} \sigma A = \sum_{}^{} \sigma \ell$$

$$ = -\sigma_{yy} \partial z - \sigma_{yz} \partial y + t_{y} \partial s$$

$$ = -\sigma_{yy} \partial s *cos(\theta) - \sigma_{yz} \partial s* sin(\theta) + t_{y} \partial s$$

$$ = -\sigma_{yy}*cos(\theta) - \sigma_{yz}*sin(\theta) + t_{y}$$

$$t_{y} = \sigma_{yy}*n_{y} + \sigma_{yz}*n_{z}$$

$$\sum_{}^{} F_{z} = 0 = \sum_{}^{} \sigma A = \sum_{}^{} \sigma \ell$$

$$ = -\sigma_{zz} \partial y - \sigma_{yz} \partial z + t_{z} \partial s$$

$$ = -\sigma_{zz} \partial s *sin(\theta) - \sigma_{yz} \partial s* cos(\theta) + t_{z} \partial s$$

$$ = -\sigma_{zz}*sin(\theta) - \sigma_{yz}*cos(\theta) + t_{z}$$

$$t_{z} = \sigma_{zz}*n_{z} + \sigma_{yz}*n_{y}$$

The values for ty and tz are pressures and hence have units of pascals or psi. These pressures combine to form the traction vector, which is a surface force.

Matrix Notation of t The components of the traction vector can be displayed in matrix notation as such:

$$\begin{pmatrix} {t_{y}}\\ {t_{z}} \end{pmatrix} = \begin{pmatrix} {\sigma_{yy}}&{\sigma_{yz}}\\ {\sigma_{zy}}&{\sigma_{zz}} \end{pmatrix}

\begin{pmatrix} {n_{y}}\\ {n_{z}} \end{pmatrix}$$

For the three dimensional consideration with axes 1, 2, and 3:

$$\begin{pmatrix} {t_{1}}\\ {t_{2}}\\ {t_{3}} \end{pmatrix} = \begin{pmatrix} {\sigma_{11}}&{\sigma_{12}}&{\sigma_{13}}\\ {\sigma_{21}}&{\sigma_{22}}&{\sigma_{23}}\\ {\sigma_{31}}&{\sigma_{32}}&{\sigma_{33}} \end{pmatrix} \begin{pmatrix} {n_{1}}\\ {n_{2}}\\ {n_{3}} \end{pmatrix}$$

or in a more condensed form,

$$\begin{pmatrix} {t_{i}} \end{pmatrix}_{3x1} = \begin{pmatrix} {\sigma_{ij}} \end{pmatrix}_{3x3} \begin{pmatrix} {n_{j}} \end{pmatrix}_{3x1}$$

Notes from meeting 34, November 14, 2008
Unsymmetric thin walled cross sections

General equation for Unsymmetric cross sections $$\int_{A_{s}}^{} \frac{\partial \sigma_{xx}}{\partial x} \partial A = -q_{s} = -q(s)$$

Generalized nonsymmetric case:

$$\sigma_{xx} = (k_{y} M_{z} - k_{yz} M_{y}) y + (k_{z} M_{y} - k_{yz} M_{z}) z$$

$$k_{y} = \frac{I_{y}}{det I}$$ $$k_{z} = \frac{I_{z}}{det I}$$ $$k_{yz} = \frac{I_{yz}}{det I}$$

$$I = \begin{pmatrix} {I_{11}}&{I_{12}}&{I_{13}}\\ {I_{21}}&{I_{22}}&{I_{23}}\\ {I_{31}}&{I_{32}}&{I_{33}} \end{pmatrix} $$

$$\sigma_{xx} = \begin{pmatrix} {y}&{z} \end{pmatrix} \begin{pmatrix} {k_{y}}&{-k_{yz}}\\ {-k_{yz}}&{k_{z}} \end{pmatrix} \begin{pmatrix} {M_{z}}\\ {M_{y}} \end{pmatrix}$$

$$\sigma_{xx} = \begin{pmatrix} {y}&{z} \end{pmatrix} \begin{pmatrix} {k_{y} M_{z} - k_{yz} M_{y}}\\ {-k_{yz} M_{z} - k_{z} M_{y}} \end{pmatrix} = (k_{y} M_{z} - k_{yz} M_{y}) y + (-k_{yz} M_{z} + k_{z} M_{y}) z$$

Consider $$M_{z} = 0$$

$$I_{yz} = 0 \Rightarrow det\ I = I_{y} - I_{z}$$

$$k_{y} = \frac{1}{I_{z}},\qquad k_{z} = \frac{1}{I_{y}},\qquad k_{yz} = 0$$

$$q(s) = -(k_{y} V_{y} - k_{yz} V_{z}) Q_{z} - (k_{z} V_{z} - k_{yz} V_{y}) Q_{y}$$

$$Q_{z} = \int_{A_{s}}^{} y \partial A$$ $$Q_{y} = \int_{A_{s}}^{} z \partial A$$

For the case of a cross section that is symmetric about the y-axis:

$$\sigma_{xx} = \frac{M_{y} z}{I_{y}}$$

$$q(s) = \frac{-V_{z} Q_{y}}{I_{y}}$$

$$Q_{y} = \int_{A_{s}}^{} z \partial A = z_{c} A_{s}$$

Example 5.2 From Textbook
(The beginning of Example 5.2) Symmetric w.r.t. the y-axis:

$$A_3=A_2$$

$$A_4=A_1$$

Non-Symmetric:

$$A_1=A$$

$$A_2=2A$$

$$A_3=3A$$

$$A_4=4A$$

A Note on Matrix Brackets
Column matrices use curly brackets, { and }

Row matrices use "l" brackets

Larger matrices use the straight brackets, [ and ]

Contributing Members
Eas4200c.f08.spars.prey 07:27, 17 November 2008 (UTC)

William Mueller Eas4200c.f08.gator.mueller 18:33, 17 November 2008 (UTC)

Jacob Papp Eas4200c.f08.gator.papp 23:55, 18 November 2008 (UTC)

Robert Reger Eas4200c.f08.gator.reger 05:09, 19 November 2008 (UTC)

Eas4200c.f08.spars.prey 20:53, 19 November 2008 (UTC)

Justin Dickinson User:Eas4200c.f08.gator.dickinson 05:34, 19 November 2008 (UTC)

Greg Edwards Eas4200c.f08.gator.edwards 18:28, 21 November 2008 (UTC)


text:
User:Jtneill/Marking dispute process

Marking dispute process

From time to time emerging scholars receive a grade/mark and/or feedback on an assessment piece which either they don't fully understand or which doesn't seem to be fair. Don't panic! - Ideally this wouldn't happen, but markers make mistakes too.

In such situations, it is recommended that you read over the feedback provided, take a bit of time to cool off, and then if you are still unclear or have concerns, seek clarification and request further explanation. This is best done by drawing attention to a list of specific concerns in writing.

The steps below offer a guide to a suggested process. Based on past experience, most issues that tend to come up with marking and feedback can be resolved quickly, simply, and satisfactorily via this suggested process.

Step 1: Email your case

 * 1) Email the unit convener the list of questions or comments you have. Very important: be as specific as possible . Some hints:
 * 2) The more precise and specific your comments, the greater the likelihood of you presenting evidence/argument that may lead to a different decision on review.
 * 3) The closer the correspondence between your comments, the marking criteria, and feedback that was already provided, the more likely it is that you will present evidence/argument that may lead to a different decision on review.
 * 4) The more honest you are, the more likely it is that a fair and accurate assessment can be made on review.
 * 5) Claims such as "I put in a lot of effort, therefore I deserve at least an X grade" or "I don't usually get a mark like this!" or "I am really keen to get into 4th year" or "I really want to get an HD" are very unlikely to support your case. Instead, explain your view about how the submitted work met the requirements listed in the guidelines and marking criteria.
 * 6) Each of the points raised will be reviewed and you will receive a response with additional feedback and a decision (if needed) which could be any of:
 * 7) Disagree and provide further explanation/feedback
 * 8) Agree and possibly make a corresponding adjustment to the marks
 * 9) Other proposal - e.g., request for more information

Step 2: A face to face meeting

 * 1) If Step 1 does not lead to a satisfactory outcome, then a face-to-face meeting with the unit convener may be suggested to discuss and resolve points of difference.

Step 3: Informal appeal

 * If, after Steps 1 and 2, a student is still not satisfied with the assessment and grading, the next step is to appeal the decision in writing to the course convener(s) or head of discipline.

Step 4: Formal appeal

 * If, after Steps 1, 2, and 3, a student is still not satisfied with the assessment and grading, the next step is to lodge a formal appeal with the Associate Dean of Education for the Faculty. Students are encouraged (but not required) to try to resolve matters informally (via Steps 1, 2, and 3) before writing to lodge a formal appeal. For more information, see University of Canberra's student grievance policy and procedure.
 * 1) A student can appeal any grade if there is some breach of appropriate policy or procedure. However, a student cannot, as a general rule, appeal academic judgement (there would have to have been some breach of policy that led to impaired academic judgement). The student needs to be clear about what the grounds for appeal.

Support

 * 1) If a student would like to discuss the matter with someone else at UC, contact Support at UC.

title: User:Jtneill/Marking dispute process


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Template:Bloom clock plant hardiness

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This template does not work yet.

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 * h = hardy
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title: Template:Bloom clock plant hardiness


title: Template:Bcp-8.11.17


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Talk:Amateur astronomy/Initial experiences

This should be renamed to something with the word "Astronomy" or the like in it, or moved to be a subpage—"Initial experiences" could apply to a number of different fields, so some specificity is necessary here. The Jade Knight (d'viser) 07:18, 17 November 2008 (UTC)
 * Yes, agree. I'd suggest moving to something like Amateur astronomy/Initial experiences. -- Jtneill - Talk - c 08:48, 17 November 2008 (UTC)

Anyone mind if I delete this page, now? The Jade Knight (d'viser) 20:50, 18 November 2008 (UTC)
 * Redirect page deleted. --mikeu talk 19:22, 30 November 2008 (UTC)


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User:Eml4500.f08.bottle.butler/HW 6

Integration by Parts (IBP)
Assume we have the following functions: $$r(x)\! $$ and $$ s(x)\! $$ $$ (rs)' = r's + rs' \! $$ where $$ r'=\frac{dr}{dx} \! $$ and $$ s'=\frac{ds}{dx} \! $$ Therefore, $$ \underbrace{\int (rs)'}_{rs}=\int r's + \int rs' \! $$  $$ \int r's = rs-\int rs' \! $$

Applying IBP to PVW
Returning to the continuous PVW, $$ (3)\; \int_{x=0}^{x=L}{W(x)}(\frac{\partial}{\partial x} [EA\frac{\partial u}{\partial x}]+f-m\ddot{u})dx =0 \! $$ for all possible $$ W(x) \! $$ If r(x) = EA\frac{\partial u}{\partial x}\! and $$s(x)=W(x) \!$$ and we integrate by parts we find the following: $$ \int_{x=0}^{x=L}\underbrace_{s(x)} \frac{\partial}{\partial x} \underbrace{[EA\frac{\partial u}{\partial x}]}_{r(x)}dx = [W(EA)\frac{\partial u}{\partial x}]^{x=L}_{x=0}-\int_{x=0}^{x=L}{\frac{dW}{dx}(EA)\frac{\partial u}{\partial x}dx} \! $$  $$ = W(L)\underbrace{(EA(L))\frac{\partial u (L,x)}{\partial x}}_{N(L,t)} - \!$$$$\underbrace{W(0)(EA(0))\frac{\partial u (0,t)}{\partial x}}_{N(0,t)} \!$$$$ - \int_{0}^{L}{\frac{dw}{dx}(EA)\frac{\partial u}{\partial x}}dx \! $$

Boundary Conditions
We now account for boundary conditions. We consider the model problem: image At $$ x=0 \! $$ select $$ W(x) \! $$ such that $$ W(0)=0 \! $$ (i.e. kinematically adminisble).

Motivation
Let's reexamine the discrete PVW as applied to the equation on page 10-1. $$w_{6X1}\circ ([]_{6X2}\begin{Bmatrix} d_{3}\\ d_{4}

\end{Bmatrix}_{2X1}-F_{6X1})=0_{1X1} \! $$ $$F^{T}=\begin{bmatrix} F_{1} & F_{2} & F_{3} & F_{4} & F_{5} & F_{6} \end{bmatrix} \! $$  where $$F_{2} \! $$ and $$ F_{3} \! $$ are known and $$ F_{1}, F_{2},  F_{5}, F_{6} \! $$ are unknown.  Since $$ W \! $$ can be selected arbitrary, then we select $$ W \! $$ such that $$ W_{1}=W_{2}=W_{5}=W_{6}=) \! $$ in order to eliminate equations involving unknown reactions. We are then left with : $$ K_{2X2}d_{2X1}=F_{2X1} = K\begin{Bmatrix} d_{3}\\ d_{4}

\end{Bmatrix}=\begin{Bmatrix} F_{3}\\ F_{4}

\end{Bmatrix} \! $$

Continuous case PVW cont.
The unknown reaction $$N(0,t)- EA(0)\frac{\partial u (0,t)}{\partial x} \! $$ where $$ N= \! $$ the normal force. $$ W(L)F(t)-\int_{0}^{L}{\frac{dW}{dx}(EA)\frac{\partial u}{\partial x}}dx+\int_{0}^{L}{W(x)[f-m\ddot{u}]dx}=0 \! $$ for all $$ W(x) \! $$ such that $$ W(0)=0 \! $$ $$\Rightarrow \int_{0}^{L}{W(m\ddot{u})dx}+\int_{0}^{L}{\frac{dW}{dx}(EA)\frac{\partial u}{\partial x}dx} \! $$ $$W(L)F(t)+\int_{0}^{L}{wfdx} \! $$ for all $$ w(x) \! $$ such that $$ w(0)=0 \! $$

Connecting continuous PVW to the discrete PVW
To connect the two theories we use a Lagrangian interpretation. We use the form of $$ N_{i}(x) \! $$ and $$N_{i+1}(x) \! $$ 1) $$ N_{i}(x) \! $$ and $$N_{i+1}(x) \! $$ are linear, thus any linear combination of $$ N_{i}(x) \! $$ and $$N_{i+1}(x) \! $$ are also linear, specifically $$u(x)= N_{i}(x)d_{i}+ N_{i+1}(x) d_{i+1} \! $$ $$N_{i}(x)=\alpha _{i}+\beta _{i} \! $$ $$N_{i+1}(x)=\alpha _{i+1}+\beta _{i+1} \! $$ where $$ \alpha \! $$ and $$\beta \! $$ are real numbers. Therefore we get the linear combination: $$ N_{i}(x)d_{i}+ N_{i+1}(x) d_{i+1} = (\alpha _{i}+\beta _{i})d_{i} + (\alpha _{i+1}+\beta _{i+1})_{i+1} \! $$   $$ = (\alpha _{i}d_{i} +\alpha _{i+1}d_{i+1})+(\beta _{i}d_{i}+\beta _{i+1}d_{i+1})x \! $$  2) Recall the equation  for $$ u(x)\! $$ (the interpolation of $$ u(x) \! $$ consider: $$ \underbrace {N_{i}(x)d_{i}}_{1}+ \underbrace {N_{i+1}(x) d_{i+1}}_{0} \! $$  therefore $$ u(x_{i})=d_{i} \! $$  and $$u(x_{i+1})=d_{i+1} \! $$  We apply the same theory to  W(x) \!  $$ W(x)=N_{i}(x)W_{i}+N_{i+1}W_{i+1} \! $$

Element Stiffness Matrix for Element i
$$ (\beta ) \! \;$$ $$\int_{x_{i}}^{x_{i+1}}{\underbrace{[N_{i}'W_{i}+N_{i+1}'W_{i+1}]}_{W'(x)}}(EA(x))\underbrace{[N_{i}'d_{i}+N_{i+1}'d_{i+1}]}_{u'(x)}dx \! $$ where $$ N'_{i}=\frac{dN_{i}(x)}{dx} \! $$ and $$N'_{i+1}=\frac{dN_{i+1}(x)}{dx} \! $$ Note: $$ u(x) \! $$ can be written as $$ \underbrace{[N_{i}'N_{i+1}']}_{N(x)_{1X2}} \! $$ Therefore,  \frac{du(x)}{dx}=\underbrace{[N_{i}'(x)N_{i+1}'(x)]}_{B(x)_{2X1}}\begin{Bmatrix} d_{1}\\d_{i+1}

\end{Bmatrix}_{2X1} \! Similarly: $$ W(x) = N(x)\begin{Bmatrix} W_{i}\\W_{i+1}

\end{Bmatrix} \! $$  \frac{dW(x)}{dx}=B(x)_{2X1}\begin{Bmatrix} W_{1}\\W_{i+1}

\end{Bmatrix}_{2X1} \! Recall the element dof's: image $$\begin{Bmatrix} d_{i}\\d_{i+1}

\end{Bmatrix}= \begin{Bmatrix} d^{i}_{1}\\d^{1}_{2}

\end{Bmatrix}=d^{i} \! $$ and $$\begin{Bmatrix} W_{i}\\W_{i+1}

\end{Bmatrix}= \begin{Bmatrix} W^{i}_{1}\\W^{1}_{2}

\end{Bmatrix}=W^{i} \! $$  $$ (\beta ) = \int_{x_{i}}^{x_{i+1}}{\underbrace{(B(x)W^{(i)})}_{1X1}}\underbrace{(EA)}_{1X1} \underbrace{(\beta d^{(i)})}_{1X1}dx \! $$ $$ = W^{(i)}\circ (k^{(i)}d^{(i)}) \! $$ where our goal is to find $$ k \! $$ We can rewrite $$ (\beta )=\int_{x_{i}}^{x_{i+1}}{(EA) \underbrace{(B(x)W^{(i)})}_{1X1}} \circ \underbrace{(B d^{(i)})}_{1X1}dx  \! $$ as: $$(BW^{(i)})^{T}(Bd^{(i)}) \! $$ Therefore $$ W^{(i)}\circ (\underbrace{\int B^{T}(EA)Bdx)}_{k^{(i)}}d^{(i)} \! $$  $$ k^{(i)}_{2X2}=\int_{x_{i}}^{x_{i+1}}{\underbrace{B(x)^{T}}}_{2X1}\underbrace{(EA)}_{1X1}\underbrace{B(x)}_{1X2}dx  \! $$ We can then transform the coordinantes from $$ x \! $$ to $$ \tilde{x} \! $$ where $$ \tilde{x}= x-x_{i} \! $$ and $$ d\tilde{x} = dx \! $$  $$ k^{(i)}=\int_{\tilde{x}=0}^{\tilde{x}=L^{(i)}}{B^{T}(\tilde{x})(EA(\tilde{x})B(\tilde{x}) d\tilde{x}} \! $$

title: User:Eml4500.f08.bottle.butler/HW 6


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User talk:Mike.lifeguard/BCP update to versions 8.11.17

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User:Eml4500.f08.ateam.rivero/w6

Honesty, Imagination, Ethics
The following quotes are excerpts from Dr. L. Vu-Quoc main homework page for the Finite Element Analysis: EML 4500 course at the University of Floida.

Cont. PVW to Discrete PVW (Cont'd)
Langrangian Interpolation

Motivation for the form of $$ N_{i}(x) $$ and $$ N_{i+1}(x) $$

1. $$ N_{i}(x) $$ and $$ N_{i+1}(x) $$ are linear (straight line) thus any linear combo of $$ N_{i} $$ and $$ N_{i+1} $$ is also linear, and in particular expression for $$ u(x) $$ on p.31-3.

$$ N_{i}(x) = \alpha_{i} + \beta_{i}(x) $$, where $$ \alpha_{i} , \beta_{i} $$, are real numbers $$ N_{i+1}(x) = \alpha_{i+1} + \beta_{i+1}(x) $$, where $$ \alpha_{i+1} , \beta_{i+1} $$, are real numbers $$ N_{i}*d_{i}+N_{i+1}*d_{i+1} = (\alpha_{i}+\beta_{i}x)d_{i}+(\alpha_{i+1}+\beta_{i+1}x)d_{i+1} $$ ;is clearly a linear function in x 2. Recall equation for $$ u(x) $$ $$ u(xi) = N_{i}(xi)d_i+N_{i+1}(xi)d_{i+1} $$ Where N_{i}(xi)d_i = 1 and N_{i+1}(xi)d_{i+1} = 0

title: User:Eml4500.f08.ateam.rivero/w6


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User:Stevenarntson/Freshman Seminar/Education

Today in class, we'll have a discussion about education, beginning with experiences you've had:
 * 1 good learning experience you had outside of school.
 * 1 bad learning experience you had outside of school.
 * 1 good learning experience you had in school.
 * 1 bad learning experience you had in school.

Additionally, we'll talk about what resources are required for a school, and possibly spend a little time in a Utopic enterprise.
 * Curriculum (required classes?)
 * Assessment (grades?)
 * Residency (dorms?)
 * Admissions (standards?)


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User:Stevenarntson/English Composition/example papers

Yours for the Revolution

 * John Benedict

Sheets of snow melted instantly on my steaming face, only to freeze again leaving trails of ice daggers down from my eyes and across my nose in this sun forgotten hell I call home. I am alive here. With half a ton on my shoulders, I run as fast as I can. The other dogs beside me huff from the weight. But, as the lead sled dog, I am steady and craving more with each stride. “Mush!” the sleigh man yells, and with even more words of encouragement the whip speaks, ripping the air in two, forging a hole big enough for an explosion to escape. The snow crunched beneath us, though barely absorbing the displacement of the entire team of dogs. I truly am alive here. My name is Jack London. I am a writer in the time where nature, it seems, is merely a concept in my pen. The year is barely 1900 and man has forgotten the womb whence he came and the milk that has sustained him. My writings are a metaphor of my own life’s saga. Just like the main character, which happens to be a dog, in The Call of the Wild; I too, was taken from my home early in life to endure a brief time of extreme hardship and unknown fate. I chose this fate. I was 16 years old. Long strands of straw pricked any exposed skin as I rolled with my dreams. The air shifted from feeling comfortable to a nervous chill and the dark air became light as the giant wooden boxcar door slid open. A figure, no smaller than an thirty year old oak tree, quickly blocked the morning sunlight that poured in through the large square opening. My squinty eyes grew heavy again. I was still drunk with sleep. The light disappeared again as if the new day came and went, ready for night. I fell back asleep. My dreams to follow began normal. I imagined myself back home again entangled in another argument with my mother. “If you leave this house again, you’re not welcome back.” “I’m not coming back.” “You’re just like your father. He never amounted to anything and neither will you. You both run away when you get scared.” “I never had a chance to know what dad was like and have you ever thought that just maybe we’re running away from you? I won’t be back. I promise.” I was about to close the front door when I heard a tapping sound coming from above my head inside the house. For some reason I stopped. It was steady. It spoke to me. I walked up the staircase trying not to confuse the creaking steps with this alluring sound. It had a thudding beat. Something driven by force was connecting with a giving solid mass, a noise that maybe only a professional boxer would know too familiarly. I got to the top of the stairs. The sound was emanating from a bedroom around the corner. I crept right up to the door. It controlled me at this moment, wanting me to discover it. The door swung open spewing the most brilliant colors at my eyes, almost blinding me. And there I saw it on the floor before me. The beating continued. Before my eyes, I watched my biological father being beaten to death by a group of men. The lights grew brighter and even wilder. At this point I couldn’t see my father any more. I could feel his pain though. The blood soaked floor remained where my father had been. Yet, I still heard the beatings. My body began to tremble. “What’s happening?” I asked myself. Lightning rattled through my bones. I’m in such pain. The figure from the train car stood over me clutching an iron bar. Between swings, I could glimpse the bright morning sky. My body was no longer on the boxcar. It had been dragged out into the open air. With each blow from the iron bar, I was quickly forgetting the weird dream I had dreamt and was now quickly feeling the regret from ever hopping a train. This figure was the railroad bull. A man can not fight back in this situation. Here, the railroad is king. Growing up in San Francisco and around Oakland in the late 1800’s, it was necessary to know how to fight in order to survive in the gangs we were in. But, this wasn’t a rival gang. This was the law. “Jump my train again. I beg you.” I lay there next to nothingness. A life lesson just realized. Despite my close encounter with death on the railroad, I continued to hop trains throughout my teenage years. It seemed the only way to find myself was to get lost first. I travelled from my hometown of San Francisco to New York. And when I grew restless, I hopped a train again back west, this time in the neighboring Canada. I wasn’t alone on these journeys, though. Other bums and vagrants must have been searching for their lost souls as well, for there were herds of them each boxcar I jumped. At times they were so plentiful that tiny skirmishes would unveil to establish ownership of a train car or even food. Having a rough boyhood, or no real boyhood for that matter, callused me so that I stood my ground like a mountain. The only battles I lost were those against the Bulls, but I quickly learned never again to engage in their kind of war. Those few months were the most exciting thus far. I felt more confident with my position in life. And yet I yearned for more, thirsty for my life-flavored water. I returned home. As soon as I was back, however, I was forced to leave again by this gravity force-like drive that fueled my body and mind. The next thing I knew, I was in motion again. My mode of travel this time wasn’t steam powered, but wind powered. The ropes were thrown and left to the mercy of the land from where we once were moored. Men ran about and around me in a hurried calm. Jumping and running for and aft, from helm to mast and port to starboard. These nautical terms I quickly learned in order to be part of the crew. We were sailing west. I had joined an American expedition to Japan. The prize, seals. This was a hunting party in search of their fortune. I was still in search for my inner being. All the same, it seemed to work in both favors. The massive canvas sheet inhaled what seemed like a timeless breath of wind and propelled the vessel smoothly and effortlessly through the marble polished sea water, as if sharp shears were tearing through silk fabric. The weather in the waters of Japan, this time of year, was ice worthy. “Pull it over the side! Don’t let it go!” My jagged harpoon spear was thrashing in and out of the water stuck in the back of a female seal. The frozen sweat on my palms was not a reassuring factor that I was definitely going to be reeling in this so-called money pot. The seal moaned in pain turning her head back and up staring in my eyes asking me to kill her quickly. With all my might and the help of two other deck hands, we hauled the seal’s one hundred pound body into the ship. Blood sprayed out of her back and sides where she was stabbed as she blindly lunged at her attackers in pure desperation to some how salvage her life. “Hit her over the head. Jackie! Use your spear! Jack!” “What? Oh.” Locked into the eyes of my victim, I couldn’t help but experience her pain. I was enthralled in the seal’s death ballet. I’ve never killed anything before. How could I destroy a creature in its own waters, on its own time? What once was a graceful swimmer exploring the sea in search of food, is now this monstering* blood and flesh mess clinging to its last breath. I performed the deed and the dance was done. This was the beginning of many more dances to come. After this, I am done. No more death. I am Seawolf. We sailed home wealthy from our three month seal onslaught. I was ready to move on to new frontiers with the same prospects, discovering myself. The amazing opportunities I have encountered in these few years since leaving school and home has inspired me to theme them in stories featuring me as the main character in one form or another. I don’t want to tell my stories, I want to live my stories. The waning day exhausts a final encore of colors. The trees speak to each other in short whispers about the sun and sky show they are receiving, while the tree birds roost in preparation of the upcoming day. Triplet candles light my words. A glass of cognac aids my inspiration. This pen bleeds from my hand and speaks from my mind. “Look at my life,” I say to myself. This belongs in a book.
 * monstering: to act/move in a monster-like manner.

Sources
 * 1) London, Jack. The Call of the Wild and White Fang . Harcourt: New York, New York, 1903.
 * 2) "Jack London." Great Writers . Kultur: New York, 1995

A Simple Peace
Jonny Darko My small hut was filled with the aroma of rootleaf stew. Vile it is, a putrid smell that came from the swamps, but learned to love it I did. The smell of home, it is. Better than most, this simple meal suits me. Much like my stew, my exiled life on Degobah is. Simple it is. Simple, but rougher now that closer to the dark cave I have moved. After 837 years, I feel at home again, but not at peace. Wait for my peace I will. Haunted by the decisions of my past I am, haunted by my thoughts and feelings. How could I have missed so much darkness? Felt it growing I should have, but the dark side clouds everything, even the purest thoughts of a master jedi. Now in exile, to reflection my thoughts must turn.

Failed my fellow jedi, I have. I see it all so clearly now but clouded, my judgment was when I needed clarity most. Now, slow and broken am I, like an old man at the end of his life. “More spice my stew needs.” Made the same mistakes, would my master have?

Centuries ago it was, on an uncharted planet when first I met N’Kata Del Gormo. A couple of days it was since the crash, when approached I was by master Gormo. Knew nothing of the force I did, but it matters not. Feel the force surging through me he did. Strong and unfocused, more natural power than any other jedi did I have. In the mighty age of the Galactic Republic it was, an age of enlightenment and power. The jedi searched the entire galaxy for new members, those who were strong in the force. In me did he find his new apprentice.

“Why are you here?” Feel the force in him I could, but know what it was I did not. “I have come for you, my young padawan. You will become my apprentice.” Never met any of his species had I.   Much fear did I have when saw him I did. Coiled was his tail as if preparing to attack, but his presence was calm and confident. Much taller was he and carried strange weapons and equipment he did.

A few short weeks did my training last. Learn much from my new master I did, much about the light and dark sides of the force. “You are unlike any padawan I have ever heard of. The force flows through you freely, and you have almost mastered your jedi skills. There is very little for me to teach you my young apprentice.” Much confusion did I sense. See through him I could. Strong I felt with my new skills. “Be mindful of your thoughts young padawan. You may have the knowledge of the force now, but knowledge is nothing without proper direction.”

True were his words. Never again did I see him. Left to seek the jedi council did I, as I was instructed. The lesson about knowledge, the most important training it was. Knowledge is power and power had I, but what now.

“Tea! Something to go with the stew.”  Helped us build the republic did this knowledge and power. “Yarum seed and Gimer bark will do, yes.” It was knowledge and power that clouded our thoughts. Without balance, even the strongest structure will fall. Fall we did, at the peak of the jedi council’s power and knowledge, because of our arrogance it was.

At the head of the council did I sit, looking deep into the boy. “The chosen one you say, hmm?” All of the masters on the council knew of the prophecy. Bring balance to the force the chosen one will. “Think on this we must.” Obvious it was that many council members had already decided. Flourished did the council in the waning days of the republic and corrupted by power were they. Comfortable, arrogant, and lazy many of the jedi had become. “What say you, master Windu?” Wait I did, until the chamber was empty. Wanted our opinions to shake the council, I did not. “The boy does possess a strong force presence, but he is too old to begin the training.” Clear it was even the most knowledgeable of masters, the dark side can blind. “Perhaps, and perhaps the chosen one he is. Meditate on this, I must.”

How could I not have seen the truth? “Almost ready my stew is. Where did I put that bowl?” The chosen one the boy was, born to bring balance to the force. As jedi, our mission it was to purge the galaxy of those who would live in the dark side of the force. “Bowl, spoon, cup. Good for the stew, this tea will be, mmm. Now eat can I.” Successful, we had been and shifted was now the balance. Many jedi had we, dedicated to the light side of the force. Find a way in this boy, the force did, the chosen one, mmm. Power would corrupt him but balance to the force would he bring.

On Kashyyyk was I, helping the wookiees in their great battle. Through battle meditation, able was I to provide focus and clarity to the wookiees as they waged a war for their planet. “Much pain and suffering do I sense. Pain and death.” A powerful dark sith lord had the emperor become. The order, he gave, to kill all jedi throughout the galaxy. Strong was he in the force. All the knowledge and power of a jedi master did he possess. Corrupt him it did. A powerful agent of the dark side did he become. Unnoticed, all control he took. Dead was the old republic. Give birth to the age of the Galactic Empire it did. “Into exile I must go. Failed, I have.”

A simple satisfaction had I, from the food and tea. Balanced and pleasant, simple was the meal. “Meditate now, must I.” Searched my thoughts I did. Search the past I did, many times for more answers. Clear are my memories, but the simple truth eludes me still. How did I miss this? Fix this, can I? “Simple is my life. Simple is the way of the jedi. Perhaps simple my thoughts must be, yes.” Knowledge. Power. Corruption. Balance. Say these simple words I did, again in my head. Easy are they to understand and hard at the same time. Thinking too much I am. Dwelling on the past can lead to sadness. Sadness leads to fear. Fear is the path to the dark side. “Tired am I. Rest now, must I get. Meditate on this tomorrow I will.”

Crawling into bed, consuming me, my thoughts are. “Time for rest now, it is.” Knowledge. Power. Corruption. Balance. Simple are these words, silent they will not be. With knowledge comes power. Power to learn more. Power to teach. Power to take and do harm. Power to give and do good. Knowledge is power. “Comforting is this thought, yes.” Proud would my master be. Chose I did, to use my power for good. To teach and to give service to the republic. Comfort me, does the simple warmth of my bed.

Power. Without direction, power can corrupt. Corrupted was the council. Corrupted was the emperor. Power caused this corruption. Power corrupts. “Mindful of my thought must I remain.” Hide I cannot from the simple truth. Knowledge is power. Power corrupts. How did knowledge become corruption? Knowledge is corruption. So simple and true it is, but so wrong. For good, knowledge can be used, and for bad. Lack of direction it must be, and lack of balance. “Yes, lack of balance in the force.” Created a lack of balance in the force by hunting agents of the dark side, we did. Try to make the galaxy perfect we did. “Balance is the key. The good without the bad, you cannot have.” Tired am I now, too tired to continue this train of thought. Sleep now.

Give me energy, the morning sounds and smells do. Old am I now and soon, go into the force I will, but not until balance I bring to the force again. “More jedi will there be. Teach them I will, to use the force for good.” Hope there is in the younglings. Feel the force in them I can, and come to me they will. Teach them I must, to use the force. Teach them I will, the lessons of knowledge, power, and corruption. Teach them the lesson of balance. How it is the key. Live through them I will, and bring balance to the force.

“Breakfast, I must have.” Breakfast I must have, and live must I.  Prepare myself to teach the younglings, I must. Knowledge and power. Now direction have I, yes. Only after I teach them can I rest. At peace I will be, only after we bring balance to the force again. Only then can I die. Only then will I take my place in the force. “Only then can I be at peace.”


text:
A course in troll sockpuppets/Insecure troll

One of the easiest trolls to create is the insecure troll. He's desperate to compensate for something and doesn't have a girlfriend. So in between rounds of downloading pornography he tells everyone what a great big man he is. Nobody's fooled but that doesn't stop him from trying.

Instructions
To build the insecure troll, mark off two inches from a regular ruler onto a piece of cardboard in quarter inch increments. Trim and attach to the troll hand with basting stitches. Or substitute a magnifying glass.

Poor little troll...


text:
User:Eml4500.f08.team.foskey.ckf/teamHW6tosubmit

Team "TEAM"

HW6 Wiki Report

Lecture Notes

Nov 3 (Meeting 29) - Finite Element Analysis Using Partial Differential Equations

Nov 5 (Meeting 30) - The Principle of Virtual Work Using Integration by Parts

Nov 7 (Meeting 31) - Linear Interpolations of Displacements

Nov 10 (Meeting 32) - Continuous and Discrete Principles of Virtual Work

Nov 12 (Meeting 33) - Finite Element Analysis via Principles of Virtual Work

Nov 14 (Meeting 34) - Using Properties that Vary with Distance

Determining if the Electric Pylon is Statically Indeterminate

MATLAB Project

The Electric Pylon Problem

The Lecture Notes are summaries of the notes provided in class by Dr. Vu-Quoc for EML4500

Contributing Group Members Cory Foskey Eml4500.f08.team.foskey.ckf 17:38, 17 November 2008 (UTC) Jin Yu Guan Eml4500.f08.team.guan 21:35, 17 November 2008 (UTC) Joseph Figuerrez Eml4500.f08.team.figuerrez 23:38, 18 November 2008 (UTC) James Dwyer Eml4500.f08.team.dwyer.jd 19:14, 21 November 2008 (UTC) Timothy Bengtson Eml4500.f08.team.Bengtson 19:19, 21 November 2008 (UTC)