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User:EAS4200C.F08.WIKI.A/My Contribution: 4

=Multicell Airfoil=

In this section we will examin a multicel airfoil. We are interested to find the rate of twist as a function of the torque $$\displaystyle T$$ and the torsional constant $$\displaystyle J$$. The figure depicted below is the object that will be in concideration.

We first have to express the torque of the while object. To do so we will can use the principal of superposition. In other words the torque of the whole object is the summation of the torques from each cell, $$\displaystyle C_1$$ and $$\displaystyle C_2$$.

$$\displaystyle \sum_{i=1}^{N_c}{2q_i\bar{A_i}}= 2q_1\bar{A_1} + 2q_2\bar{A_2}$$

We now must find the rate of twist for each cell. The formulation for cell one is shown below.

$$\displaystyle \theta_1 = \frac{1}{2G\bar{A_1}}\oint_{a}^{a}{\frac{q_1}{t_1} ds} = \frac{1}{2G\bar{A_1}}[\frac{2q_1a}{t_1} + \frac{q_1c}{t_1} + \frac{(q_1-q_2)c}{t_{12}}]$$

When all known variables are put in and noting that $$\displaystyle \bar{A_1} = ac = (30cm)(40cm) = 1200cm^2$$ equals. the expression for the rate of twist for cell one then becomes;

$$\theta_1 = \frac{1}{Gcm^2}[.181q_1 - .042q_2]$$

The same is done for cell two and is performed below.

$$\displaystyle \theta_2 = \frac{1}{2G\bar{A_2}}\oint_{a}^{a}{\frac{q_2}{t_2} ds} = \frac{1}{2G\bar{A_2}}[\frac{2q_2b}{t_2} + \frac{q_2c}{t_2} + \frac{(q_2-q_1)c}{t_{12}}]$$

Substituting values and noting that $$\displaystyle \bar{A_2} = bc = (60cm)(40cm) = 2400cm^2$$, the expression reduces to;

$$\theta_2 = \frac{1}{Gcm^2}[-.021q_1 + .062q_2]$$

Now it is imperative to notice that the rates of twist for each cell are equivalent. Setting them equal to each other and solving for either $$\displaystyle q_1$$ or $$\displaystyle q_2$$ is the next step in solving this problem. the result is as follows. We chose to solve for $$\displaystyle q_1$$.

$$ \frac{1}{Gcm^2}[.181q_1 - .042q_2] = \frac{1}{Gcm^2}[-.021q_1 + .062q_2]$$  $$\displaystyle q_1= .65 q_2$$

Now we pace this expression for $$\displaystyle q_1$$ into the torque equation that was first derived in order to obtain the rate of twist for cell two $$\displaystyle q_2$$ in terms of the torque $$\displaystyle T$$. It is then possible to obtain the rate of twist for cell one $$\displaystyle q_1$$ in terms of the torque.

$$\displaystyle T = 2q_2(.65)(1200cm^2) + 2q_2(2400cm^2) $$  $$\displaystyle q_2 = \frac{1.57x10^{-4}T}{cm^2}$$  $$\displaystyle q_1 = \frac{1.02x10^{-4}T}{cm^2}$$

Now we must put these expressions back into either one of the rates of twist equations. Doing so will put the rate of twist in terms of the torque. This is done below.

$$\theta_1 = \frac{1}{Gcm^2}[.181q_1 - .042q_2] = \frac{1}{Gcm^2}[.181(\frac{1.02x10^{-4}T}{cm^2}) - .042(\frac{1.57x10^{-4}T}{cm^2})]$$
$$\theta_1 = \frac{1.18x10^{-5}T}{Gcm^4} = \theta_2$$

Now we must relate the angle of twist to the torsional constant. We know that the angle of twist is equal to the following.

$$\displaystyle \theta = \frac{T}{GJ}$$

We can conclude that

 $$\displaystyle \frac{1}{J} = \frac{1.18x10^{-6}}{cm^4}$$

And that

$$\displaystyle J = \frac{1}{1.18x10^{-6}} = 84745.8cm^4$$


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

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User:EML4500.f08.RAMROD.E/HW4

HW: Eigenvectors for the zero Eigenvalues of K
Using Matlab, the eigenvalues for K were found (see HW 3) and the eigenvectors were plotted for the zero values. The eigenvectors were found to be:

$$E_1 = \begin{Bmatrix} -0.4494 & 0.6754 & 0.1682 & -0.3942 & 0.2812 & -0.2812 \end{Bmatrix}$$

$$E_2 = \begin{Bmatrix} 0.3877&-0.4281&0.3877&-0.4281&0.4163&-0.3995 \end{Bmatrix}$$

$$E_3 = \begin{Bmatrix} 0.5158&0.4825&0.5158&0.4825&-0.0116&-0.0449 \end{Bmatrix}$$

$$E_4 = \begin{Bmatrix} 0.0160&0.0244&0.0160&0.0244&0.7023&0.7107 \end{Bmatrix}$$

The plots are shown below.

These plots show the mode shapes corresponding to each of the zero eigenvalues and are representative of a linear combination of the four pure mode shapes (x-translation, y-translation, rotational, and mechanism).

For the eigenvalue/eigenvector problem:

$$Kv = \lambda v$$

We can let $$\begin{Bmatrix} u_1 & u_2 & u_3 & u_4 \end{Bmatrix}$$ be the pure eigenvetors corresponding to the four zero eigenvalues. Assuming this tells us that:

$$Ku_i = 0u_i$$ for i = 1,2,3,4.

To get the true eigenvector, w, for each zero eigenvalue, we take the sum of the four contributors, each multiplied by a constant $$\alpha _{1}$$, giving us:

$$ w = \alpha_1 u_1 + \alpha_2 u_2 + \alpha_3 u_3 + \alpha_4 u_4$$

Because w consists of the combination of eigenvectors for zero eigenvalues, w too represents an eigenvector corresponding to a zero eigenvalue. That is that:

"$\bar{K}\bar{w} = K(\Sigma \alpha _iu_i) = \bar{0} = 0*/bar{w}$"

title: User:EML4500.f08.RAMROD.E/HW4


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User talk:JWSchmidt/Blog/19 October 2008

Why do something that sickens you? Your responses are, at this point, so verbose (and generally unstructured), and community interest in the conflict is currently so low, that I expect most people will simply say (or think) "TLDR" and move on. Why not simply say "I absolutely disagree with the way my actions have been characterized, and I have responded to several of the accusations and can respond to others if requested", and let the whole thing slide into history? The Jade Knight (d'viser) 21:25, 23 October 2008 (UTC)


text: File:Ideal airfoil.JPG

title: File:Ideal airfoil.JPG


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User:Eml4500.f08.bottle.ranto/HW4

HW Problem. Prove that $$ k^{(e)} = T^{(e)^{T}} \hat{k}^{(e)} T^{(e)} $$

Where

$$ \textbf{k}^{(e)}= \left[ \begin{array}{cccc} l_{s}^{2} & l_{s}m_{s} & -l_{s}^{2} & -l_{s}m_{s} \\ l_{s}m_{s} & m_{s}^{2} & l_{s}m_{s} & -m_{s}^{2} \\ -l_{s}^{2} & -l_{s}m_{s} & l_{s}^{2} & l_{s}m_{s} \\ -l_{s}m_{s} & -m_{s}^{2} & l_{s}m_{s} & m_{s}^{2} \end{array} \right]$$

As given in the notes above

$$ \textbf{T}^{(e)^{T}} = \left [ \begin{array}{cccc} l_{s}^{(e)} & 0 \\ m_{s}^{(e)} & 0 \\ 0 & l_{s}^{(e)} \\ 0 & m_{s}^{(e)} \end{array} \right] $$

and

$$ \hat\textbf{k}^{(e)} = \left [ \begin{array}{cccc} 1 & -1 \\ -1 & 1 \end{array} \right] $$

Therefore

$$ \hat\textbf{k}^{(e)}\textbf{T}^{(e)} = \left [ \begin{array}{cccc} l_{s}^{(e)} & m_{s}^{(e)} & -l_{s}^{(e)} & -m_{s}^{(e)} \\ -l_{s}^{(e)} & -m_{s}^{(e)} & l_{s}^{(e)} & m_{s}^{(e)} \end{array} \right] $$

Finally, multiplying by $$ T^{(e)^{T}} $$ gives

$$ \left[ \begin{array}{cccc} l_{s}^{2} & l_{s}m_{s} & -l_{s}^{2} & -l_{s}m_{s} \\ l_{s}m_{s} & m_{s}^{2} & l_{s}m_{s} & -m_{s}^{2} \\ -l_{s}^{2} & -l_{s}m_{s} & l_{s}^{2} & l_{s}m_{s} \\ -l_{s}m_{s} & -m_{s}^{2} & l_{s}m_{s} & m_{s}^{2} \end{array} \right]$$

title: User:Eml4500.f08.bottle.ranto/HW4


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User:EAS4200C.F08.WIKI.A/Homework4

=Quadrature Method=

Quadratures can be used to find the average area of foils. Below is an image of the quadrature of a foil.

$$\overline{A} = \overline{A}_1 + \overline{A}_2 = \overline{A}_1 - \begin{vmatrix}\overline{A}_2\end{vmatrix}$$
 * $$ \overline{A}_1 > 0 $$
 * $$ \overline{A}_2 > 0 $$

 * $$\overrightarrow{r}x\overrightarrow{PQ}\; goes\; in\; negative\; x\; direction\; $$

Using Trapezoidals to Integrate
There are problems with using this method because of the curvature of the foil. When you split up the area of the foil, not every section will end up looking like a trapezoid. This method should not be used for foils.

Shear Flow in a Single Cell Airfoil
Below is an image of a single cell airfoil.

The shear flow is constant, therefore:

 * $$ q=q_1=q_2=q_3$$

Rate of Twist Angle θ

$$\theta = \frac{1}{2G\overline{A}}\; q\sum_{j=1}^{}{\frac{l_j}{t_j}}$$

$$\theta = \frac{1}{2G\overline{A}}\; q[\frac{\pi (\frac{b}{2})}{t_1} + \frac{a}{t_2} + \frac{\sqrt{a^2 + b^2}}{t_2}]$$

$$\theta = \frac{1}{2G\overline{A}}\; q[\frac{\pi bt_2t_3 + 2at_1t_3 + 2t_1t_2\sqrt{a^2 + b^2}}{t_1t_2t_3}]$$

Shear Stress
Max Shear Stress τmax

$$\tau_{max} = \frac{q}{min(t_1, t_2, t_3)}$$

 * $$if\; \tau_{max} = \tau_{all}$$

 * $$and\; since\; q = \frac{T}{2\overline{A}} \; then$$

$$T_{all} = 2\bar{A}\tau_{all}\;min(t_1, t_2, t_3)$$

=Multicell Airfoil=

In this section we will examin a multicel airfoil. We are interested to find the rate of twist as a function of the torque $$\displaystyle T$$ and the torsional constant $$\displaystyle J$$. The figure depicted below is the object that will be in concideration.

We first have to express the torque of the while object. To do so we will can use the principal of superposition. In other words the torque of the whole object is the summation of the torques from each cell, $$\displaystyle C_1$$ and $$\displaystyle C_2$$.

$$\displaystyle \sum_{i=1}^{N_c}{2q_i\bar{A_i}}= 2q_1\bar{A_1} + 2q_2\bar{A_2}$$

We now must find the rate of twist for each cell. The formulation for cell one is shown below.

$$\displaystyle \theta_1 = \frac{1}{2G\bar{A_1}}\oint_{a}^{a}{\frac{q_1}{t_1} ds} = \frac{1}{2G\bar{A_1}}[\frac{2q_1a}{t_1} + \frac{q_1c}{t_1} + \frac{(q_1-q_2)c}{t_{12}}]$$

When all known variables are put in and noting that $$\displaystyle \bar{A_1} = ac = (30cm)(40cm) = 1200cm^2$$ equals. the expression for the rate of twist for cell one then becomes;

$$\theta_1 = \frac{1}{Gcm^2}[.181q_1 - .042q_2]$$

The same is done for cell two and is performed below.

$$\displaystyle \theta_2 = \frac{1}{2G\bar{A_2}}\oint_{a}^{a}{\frac{q_2}{t_2} ds} = \frac{1}{2G\bar{A_2}}[\frac{2q_2b}{t_2} + \frac{q_2c}{t_2} + \frac{(q_2-q_1)c}{t_{12}}]$$

Substituting values and noting that $$\displaystyle \bar{A_2} = bc = (60cm)(40cm) = 2400cm^2$$, the expression reduces to;

$$\theta_2 = \frac{1}{Gcm^2}[-.021q_1 + .062q_2]$$

Now it is imperative to notice that the rates of twist for each cell are equivalent. Setting them equal to each other and solving for either $$\displaystyle q_1$$ or $$\displaystyle q_2$$ is the next step in solving this problem. the result is as follows. We chose to solve for $$\displaystyle q_1$$.

$$ \frac{1}{Gcm^2}[.181q_1 - .042q_2] = \frac{1}{Gcm^2}[-.021q_1 + .062q_2]$$  $$\displaystyle q_1= .65 q_2$$

Now we pace this expression for $$\displaystyle q_1$$ into the torque equation that was first derived in order to obtain the rate of twist for cell two $$\displaystyle q_2$$ in terms of the torque $$\displaystyle T$$. It is then possible to obtain the rate of twist for cell one $$\displaystyle q_1$$ in terms of the torque.

$$\displaystyle T = 2q_2(.65)(1200cm^2) + 2q_2(2400cm^2) $$  $$\displaystyle q_2 = \frac{1.57x10^{-4}T}{cm^2}$$  $$\displaystyle q_1 = \frac{1.02x10^{-4}T}{cm^2}$$

Now we must put these expressions back into either one of the rates of twist equations. Doing so will put the rate of twist in terms of the torque. This is done below.

$$\theta_1 = \frac{1}{Gcm^2}[.181q_1 - .042q_2] = \frac{1}{Gcm^2}[.181(\frac{1.02x10^{-4}T}{cm^2}) - .042(\frac{1.57x10^{-4}T}{cm^2})]$$
$$\theta_1 = \frac{1.18x10^{-5}T}{Gcm^4} = \theta_2$$

Now we must relate the angle of twist to the torsional constant. We know that the angle of twist is equal to the following.

$$\displaystyle \theta = \frac{T}{GJ}$$

We can conclude that

 $$\displaystyle \frac{1}{J} = \frac{1.18x10^{-6}}{cm^4}$$

And that

$$\displaystyle J = \frac{1}{1.18x10^{-6}} = 84745.8cm^4$$

=Ad-Hoc Derivation=

The engineering, or ad-hoc, derivation of the rate of twist of the following equation

 $$\displaystyle \theta = {1 \over 2G \bar A} \oint {q \over t} ds $$

applies to a uniform bar with a non-circular cross section.

The displacement of segment $$\displaystyle PP' $$ due to an angle $$\displaystyle \alpha $$ is defined as

 $$\displaystyle {PP' \over OP} = \tan \alpha $$

But for a small angle $$\displaystyle \alpha $$, the following approximation is appropriate:

 $$\displaystyle \tan \alpha \approx \alpha $$

Hence,

 $$\displaystyle {PP' \over OP} \approx \alpha $$

The projected displacement of $$\displaystyle PP' $$ in the direction perpendicular to $$\displaystyle OP' $$ can be written

 $$\displaystyle PP'' = PP' \cos \alpha $$

By substitution

 $$\displaystyle PP'' = (OP \tan \alpha) \cos \alpha $$

By the associative property of multiplication

 $$\displaystyle PP'' = (OP \cos \alpha) \tan \alpha $$

Recall that $$\displaystyle OP = r $$ and $$\displaystyle OP'' = \rho $$, thus we can express it as

 $$\displaystyle PP'' = (r \cos \alpha) \tan \alpha $$

Strain, $$\displaystyle \gamma $$ can be written as

 $$\displaystyle \gamma = {PP' \over dx} = {\rho \alpha \over dx} = \rho \theta $$

where, as previously stated, $$\displaystyle \theta $$ is the rate of twist.

The expression:

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

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

Hooke's law can be represented as:

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

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

With variables as defined in the previous diagram.

Integrate along $$\textit{C}$$

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

where

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

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

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

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

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

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

Now formal justification (derivation) of elasticity theory.

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

A. Kinematic Assumption

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

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

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

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

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

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

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

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

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

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

=Strain=

Normal strains

$$\varepsilon _{xx}=\frac{\sigma _{xx}}{E}-\frac{\nu \sigma _{yy}}{E}-\frac{\nu\sigma _{zz} }{E} $$

$$\varepsilon _{yy}=-\frac{\nu\sigma _{xx}}{E}+\frac{ \sigma _{yy}}{E}-\frac{\nu\sigma _{zz} }{E} $$

$$\varepsilon _{zz}=-\frac{\nu\sigma _{xx}}{E}-\frac{\nu \sigma _{yy}}{E}+\frac{\sigma _{zz} }{E} $$

Shear Strains

$$\gamma _{xy} =2 \zeta _{xy}= \frac{\tau_{xy}}{G}=\frac{\sigma_{xy}}{G}$$

$$\gamma _{yz} =2 \zeta _{yz}= \frac{\tau_{yz}}{G}=\frac{\sigma_{yz}}{G}$$

$$\gamma _{zx} =2 \zeta _{zx}= \frac{\tau_{zx}}{G}=\frac{\sigma_{zx}}{G}$$

Unroll this matrix along the diagonal and then up and around counterclockwise.

$$\epsilon = \begin{vmatrix} \varepsilon_ {11} & \varepsilon_{12} & \varepsilon _{13}\\ \varepsilon _{21} &\varepsilon _{22} &\varepsilon _{23} \\ \varepsilon _{31} &\varepsilon _{32} &\varepsilon _{33} \end{vmatrix}$$

Due to the symmetry, you may rewrite this unrolled matrix into a $$6_x 1$$ with a Voigt Notation:

$$ \left\{\varepsilon _{ij} \right\}_{6x1}= \begin{Bmatrix} \varepsilon _{11}\\ \varepsilon _{22}\\ \varepsilon _{33}\\ \varepsilon _{23}\\ \varepsilon _{13}\\ \varepsilon _{12} \end{Bmatrix}$$

$$\left\{\sigma _{ij} \right\}_{6x1}= \begin{Bmatrix} \sigma _{11}\\ \sigma _{22}\\ \sigma _{33}\\ \sigma _{23}\\ \sigma _{13}\\ \sigma _{12} \end{Bmatrix}$$

Hooke's Law for isotropic elasticity:

$$\begin{Bmatrix} \varepsilon _{11}\\ \varepsilon _{22}\\ \varepsilon _{33}\\ \gamma _{23}\\ \gamma _{13}\\ \gamma _{12} \end{Bmatrix}=

\begin{vmatrix} \frac{1}{E}& \frac{-\nu }{E} & \frac{-\nu }{E} & 0 &0 &0 \\ \frac{-\nu }{E}& \frac{1}{E} &\frac{-\nu }{E}  & 0 & 0 & 0\\ \frac{-\nu }{E} & \frac{-\nu }{E} & \frac{1}{E} & 0 &0  &0 \\ 0& 0 & 0 & \frac{1}{G} & 0 & 0\\ 0& 0 & 0 & 0 & \frac{1}{G} &0 \\ 0& 0 & 0 & 0 & 0 & \frac{1}{G} \end{vmatrix} \begin{Bmatrix} \sigma _{11}\\ \sigma _{22}\\ \sigma _{33}\\ \sigma _{23}\\ \sigma _{13}\\ \sigma _{12} \end{Bmatrix}$$

It can also be written as

$$\begin{Bmatrix} \varepsilon _{11}\\ \varepsilon _{22}\\ \varepsilon _{33}\\ \varepsilon _{23}\\ \varepsilon _{13}\\ \varepsilon _{12} \end{Bmatrix}=

\begin{vmatrix} \frac{1}{E}& \frac{-\nu }{E} & \frac{-\nu }{E} & 0 &0 &0 \\ \frac{-\nu }{E}& \frac{1}{E} &\frac{-\nu }{E}  & 0 & 0 & 0\\ \frac{-\nu }{E} & \frac{-\nu }{E} & \frac{1}{E} & 0 &0  &0 \\ 0& 0 & 0 & \frac{1}{2G} & 0 & 0\\ 0& 0 & 0 & 0 & \frac{1}{2G} &0 \\ 0& 0 & 0 & 0 & 0 & \frac{1}{2G} \end{vmatrix} \begin{Bmatrix} \sigma _{11}\\ \sigma _{22}\\ \sigma _{33}\\ \sigma _{23}\\ \sigma _{13}\\ \sigma _{12} \end{Bmatrix}$$

Poisson's ratio values for different materials
Poisson's Ratio table []

=MATLAB=

Idealized Single Cell Airfoil
function IdealizedSingleCell

a = 4; b = 2; r = b/2; t1 = .008; t2 = .01; t3 = .01;

y = linspace(0, -r, 25);

y(26:50) = linspace(-r, 0, 25);

y(51) = a; y(52) = 0;

z = sqrt(r^2 - y(1:25).^2);

z(26:50) = -sqrt(r^2 - y(26:50).^2);

z(51) = -r; z(52) = r;

plot(y,z)

A_bar = pi/2*(b/2)^2 + .5*b*a;

ThetaG = 1/(2*A_bar)*(pi*(b/2)/t1 + a/t2 + sqrt(a^2+b^2)/t3); % *q

J = 2*A_bar/ThetaG

EDU>> IdealizedSingleCell

J =

0.1001

Rectangular 2-Cell Airfoil
function Rectangle2Cell

a = .3; b = .6; c = .4;

t1 = .003; t2 = .005; t12 = .004;

A1_bar = a*c; A2_bar = b*c;

X = 2*a/t1 + c/t1 + c/t12;

Y = 2*b/t2 + c/t2 + c/t12;

Z = -c/t12;

alpha = (A2_bar*X - A1_bar*Z)/(A1_bar*Y - A2_bar*Z);

J = (4*A1_bar^2 + 4*alpha*A1_bar*A2_bar)/(X + Z*alpha)

EDU>> Rectangle2Cell

J =

8.5507e-004

Circle Validation
function Circle(c,P_0)

R = c/2; check = 1; j = 1;

while check == 1 ns = 2*j; y = linspace(0,c,ns); for i = 1:ns z_u(i) = sqrt(R^2-(y(i)-R)^2); z_l(i) = -sqrt(R^2-(y(i)-R)^2); end

plot(y,z_u,y,z_l)

A_bar = 0; y_bar = 0; z_bar = 0; s = 0; y_u = y;   y_l = y;

for i = 1:ns-1 PQ = [0 y(ns-i)-y(ns-i+1) z_u(ns-i)-z_u(ns-i+1)]; r = [0 y(ns-i+1)-P_0(2) z_u(ns-i+1)-P_0(3)]; a = cross(r,PQ)/2; A_bar = A_bar + a;       y_bar = y_bar + (y_u(i)+y_u(i+1))/2*norm([y_u(i+1) z_u(i+1)]-[y_u(i) z_u(i)]); z_bar = z_bar + (z_u(i)+z_u(i+1))/2*norm([y_u(i+1) z_u(i+1)]-[y_u(i) z_u(i)]); s = s + norm([y_u(i+1) z_u(i+1)]-[y_u(i) z_u(i)]); end

for i = 1:ns-1 PQ = [0 y(i+1)-y(i) z_l(i+1)-z_l(i)]; r = [0 y(i)-P_0(2) z_l(i)-P_0(3)]; a = cross(r,PQ)/2; A_bar = A_bar + a;       y_bar = y_bar + (y_l(i)+y_l(i+1))/2*norm([y_l(i+1) z_l(i+1)]-[y_l(i) z_l(i)]); z_bar = z_bar + (z_l(i)+z_l(i+1))/2*norm([y_l(i+1) z_l(i+1)]-[y_l(i) z_l(i)]); s = s + norm([y_u(i+1) z_u(i+1)]-[y_u(i) z_u(i)]);

end

NS(j) = ns; A_BAR(j) = A_bar(1);

if abs(pi*R^2-A_bar(1))/(pi*R^2)<.01 check = 0; else j = j+1; end end

y_bar = y_bar/s; z_bar = z_bar/s;

A_bar = A_BAR(j)

subplot(2,1,1), plot(y_u,z_u,y_l,z_l,y_bar,z_bar,'x'), xlabel('y'), ylabel('z'), title('NACA 2415 Airfoil')

subplot(2,1,2), plot(NS,A_BAR), xlabel('ns'), ylabel('A_b_a_r'), title('ns vs. A_b_a_r')

EDU>> Circle(1,[0 0 0])

A_bar =

0.7779

NACA 2415 Single Cell Airfoil
function Airfoil(NACA,c,P_0)

m = floor(NACA/1000)/100;

p = (floor(NACA/100)-floor(NACA/1000)*10)/10;

ta = (NACA-floor(NACA/100)*100)/100;

ts = .002;

check = 1; j = 1;

while check ==1 ns = 5*j; y = linspace(0,1,ns);

for i = 1:ns if y(i)<=p z_c(i) = (m/p^2)*(2*p*y(i)-y(i)^2); theta(i) = atan((m/p^2)*(2*p-2*y(i))); else z_c(i) = (m/(1-p)^2)*((1-2*p)+2*p*y(i)-y(i)^2); theta(i) = atan((m/(1-p)^2)*(2*p-2*y(i))); end z_t(i) = 5*ta*(.2969*y(i)^.5-.1260*y(i)-.3516*y(i)^2+.2843*y(i)^3-.1015*y(i)^4);

y_u(i) = (y(i) - z_t(i)*sin(theta(i)))*c; z_u(i) = (z_c(i) + z_t(i)*cos(theta(i)))*c;

y_l(i) = (y(i) + z_t(i)*sin(theta(i)))*c; z_l(i) = (z_c(i) - z_t(i)*cos(theta(i)))*c; end

Area = 0; y_bar = 0; z_bar = 0; s = 0;

for i = 1:ns-1 PQ = [0 y_u(ns-i)-y_u(ns-i+1) z_u(ns-i)-z_u(ns-i+1)]; r = [0 y_u(ns-i+1)-P_0(2) z_u(ns-i+1)-P_0(3)]; a = cross(r,PQ)/2; Area = Area + a;       y_bar = y_bar + (y_u(i)+y_u(i+1))/2*norm([y_u(i+1) z_u(i+1)]-[y_u(i) z_u(i)]); z_bar = z_bar + (z_u(i)+z_u(i+1))/2*norm([y_u(i+1) z_u(i+1)]-[y_u(i) z_u(i)]); s = s + norm([y_u(i+1) z_u(i+1)]-[y_u(i) z_u(i)]); end

for i = 1:ns-1 PQ = [0 y_l(i+1)-y_l(i) z_l(i+1)-z_l(i)]; r = [0 y_l(i)-P_0(2) z_l(i)-P_0(3)]; a = cross(r,PQ)/2; Area = Area + a;       y_bar = y_bar + (y_l(i)+y_l(i+1))/2*norm([y_l(i+1) z_l(i+1)]-[y_l(i) z_l(i)]); z_bar = z_bar + (z_l(i)+z_l(i+1))/2*norm([y_l(i+1) z_l(i+1)]-[y_l(i) z_l(i)]); s = s + norm([y_u(i+1) z_u(i+1)]-[y_u(i) z_u(i)]); end

NS(j) = ns; A_BAR(j) = Area(1); J(j) = 4*Area(1)^2*ts/s;

if j ~= 1 if abs((J(j)-J(j-1))/J(j-1))*100<1 check = 0; end end j = j+1; end

y_bar = y_bar/s; z_bar = z_bar/s;

A_bar = A_BAR(j-1)

subplot(2,1,1), plot(y_u,z_u,y_l,z_l,y_bar,z_bar,'x'), xlabel('y'), ylabel('z'), title('NACA 2415 Airfoil')

subplot(2,1,2), plot(NS,J), xlabel('ns'), ylabel('Torsional Constant, J'), title('ns vs. J')

NACA 2415 3-Cell Airfoil
function Airfoil3Cell(NACA,c,P_0)

m = floor(NACA/1000)/100;

p = (floor(NACA/100)-floor(NACA/1000)*10)/10;

ta = (NACA-floor(NACA/100)*100)/100;

ts = .002; tp = .003;

check = 1; j = 1;

while check ==1 ns = 20*j; y = linspace(0,1,ns);

for i = 1:ns if y(i)<=p z_c(i) = (m/p^2)*(2*p*y(i)-y(i)^2); theta(i) = atan((m/p^2)*(2*p-2*y(i))); else z_c(i) = (m/(1-p)^2)*((1-2*p)+2*p*y(i)-y(i)^2); theta(i) = atan((m/(1-p)^2)*(2*p-2*y(i))); end z_t(i) = 5*ta*(.2969*y(i)^.5-.1260*y(i)-.3516*y(i)^2+.2843*y(i)^3-.1015*y(i)^4);

y_u(i) = (y(i) - z_t(i)*sin(theta(i)))*c; z_u(i) = (z_c(i) + z_t(i)*cos(theta(i)))*c;

y_l(i) = (y(i) + z_t(i)*sin(theta(i)))*c; z_l(i) = (z_c(i) - z_t(i)*cos(theta(i)))*c; end

for i = 1:ns if y_u(i) > c/4 P1UpperIndex = i;           break end end

for i = 1:ns if y_l(i) > c/4 P1LowerIndex = i;           break end end

for i = P1UpperIndex:ns if y_u(i) > 3*c/4 P2UpperIndex = i;           break end end

for i = P1LowerIndex:ns if y_l(i) > 3*c/4; P2LowerIndex = i;           break end end

%Cell 1 C1y = (y_u(P1UpperIndex)+y_l(P1LowerIndex))/2; for i = 1:P1UpperIndex C1y(i+1) = y_u(P1UpperIndex-i+1); end for i = 1:P1LowerIndex C1y = [C1y, y_l(i)]; end C1y = [C1y, (y_u(P1UpperIndex)+y_l(P1LowerIndex))/2];

C1z = (z_u(P1UpperIndex)+z_l(P1LowerIndex))/2; for i = 1:P1UpperIndex C1z(i+1) = z_u(P1UpperIndex-i+1); end for i = 1:P1LowerIndex C1z = [C1z, z_l(i)]; end C1z = [C1z, (z_u(P1UpperIndex)+z_l(P1LowerIndex))/2];

%Cell 2 for i = 1:P2UpperIndex-P1UpperIndex C21y(i) = y_u(P2UpperIndex-i+1); end C21y = [C21y, y_l(P1LowerIndex), y_u(P2UpperIndex);

for i = 1:P2UpperIndex-P1UpperIndex C21z(i) = z_u(P2UpperIndex-i+1); end C21z = [C21z, z_l(P1LowerIndex), z_u(P2UpperIndex);

for i=1:P2LowerIndex-P1LowerIndex C22y(i) = y_l(P1LowerIndex+i-1); end C22y = [C22y, y_u(P2UpperIndex), y_l(P1LowerIndex)]

for i=1:P2LowerIndex-P1LowerIndex C22z(i) = z_l(P1LowerIndex+i-1); end C22z = [C22z, z_u(P2UpperIndex), z_l(P1LowerIndex)]

%Cell 3 C3y = (y_u(P2UpperIndex)+y_l(P2LowerIndex))/2; for i = 1:ns-P2LowerIndex C3y(i+1) = y_l(P2LowerIndex+i-1) end for i = 1:ns-P2UpperIndex C3y'(i) = y_u(ns-i+1) end C3y = [C3y, C3y', (y_u(P2UpperIndex)+y_l(P2LowerIndex))/2];

C3z = (z_u(P2UpperIndex)+z_l(P2LowerIndex))/2; for i = 1:ns-P2LowerIndex C3z(i+1) = z_l(P2LowerIndex+i-1) end for i = 1:ns-P2UpperIndex C3z'(i) = z_u(ns-i+1) end C3z = [C3z, C3z', (z_u(P2UpperIndex)+z_l(P2LowerIndex))/2];

%Area & Torsional Constant Area1 = 0; Area21 = 0; Area22 = 0; Area3 = 0;

for i = 1:size(C1y) PQ = [0 C1y(ns-i)-C1y(ns-i+1) C1z(ns-i)-C1z(ns-i+1)]; r = [0 C1y(ns-i+1)-(y_u(P1UpperIndex)+y_l(P1LowerIndex))/2 C1z(ns-i+1)-0]; a = cross(r,PQ)/2; Area1 = Area1 + a;

end

for i = 1:size(C21y) PQ = [0 C21y(i+1)-C21y(i) C21z(i+1)-C21z(i)]; r = [0 C21y(i)-y_l(P1LowerIndex) C21z(i)-z_l(P1LowerIndex)]; a = cross(r,PQ)/2; Area21 = Area21 + a;

end

for i = 1:size(C22y) PQ = [0 C22y(i+1)-C22y(i) C22z(i+1)-C22z(i)]; r = [0 C22y(i)-y_u(P2UpperIndex) C22z(i)-z_u(P2UpperIndex)]; a = cross(r,PQ)/2; Area22 = Area22 + a;

end

for i = 1:size(C3y) PQ = [0 C3y(i+1)-C3y(i) C3z(i+1)-C3z(i)]; r = [0 C3y(i)-(y_u(P2UpperIndex)+y_l(P2LowerIndex))/2 C3z(i)-0]; a = cross(r,PQ)/2; Area3 = Area3 + a;

end

NS(j) = ns; A_BAR(j) = Area(1);

if j ~= 1 if abs((A_BAR(j)-A_BAR(j-1))/A_BAR(j-1))*100<.01 check = 0; end end j = j+1; end

y_bar = y_bar/s; z_bar = z_bar/s;

A_bar = A_BAR(j-1)

P1y = [y_u(P1UpperIndex) y_l(P1LowerIndex)];

P1z = [z_u(P1UpperIndex) z_l(P1LowerIndex)];

P2y = [y_u(P2UpperIndex) y_l(P2LowerIndex)];

P2z = [z_u(P2UpperIndex) z_l(P2LowerIndex)];

subplot(2,1,1), plot(y_u,z_u,y_l,z_l,y_bar,z_bar,'x',P1y,P1z,P2y,P2z), xlabel('y'), ylabel('z'), title('NACA 2415 Airfoil')

subplot(2,1,2), plot(NS,A_BAR), xlabel('ns'), ylabel('A_b_a_r'), title('ns vs. A_b_a_r')

=Individual Contributions=

Christopher Pergola EAS4200C.F08.WIKI.A 20:59, 24 October 2008 (UTC)

Braden Barnes Eas4200c.f08.wiki.b 21:54, 23 October 2008 (UTC)

Chris Fontana Eas4200c.f08.WIKI.E 18:55, 24 October 2008 (UTC)

Eric Viale Eas4200c.f08.wiki.c 19:51, 24 October 2008 (UTC)

Jeff Huenink Eas4200c.f08.wiki.f 20:07, 24 October 2008 (UTC)

Michael Rolle Eas4200c.f08.wiki.d 20:59, 24 October 2008 (UTC)

title: User:EAS4200C.F08.WIKI.A/Homework4


text:
User:Eas4200c.f08.vqcrew.d/hw4

Ad-hoc (Engineering) Derivation of Rate of Twist Formula
To derive the twist angle formula $$\theta = \frac{1}{2G\bar{A}}\oint_{C}^{}{\frac{q}{t}ds}$$ we will consider a uniform bar with non-circular cross section subjected to twisting, as shown in Figure XX

To begin, the displacement $$PP'$$ is found to be

$$PP'=OPtan \mathbf{\alpha}$$ Then, project $$PP'$$ into the direction perpendicular to $$OP'$$ : $$PP''=PP'cos \mathbf{\alpha}$$

Now the displacement of $$P$$ in the direction tangent to the lateral surface of the beam, $$PP$$, can be expressed in terms of $$OP$$ : $$PP=(OPtan \mathbf{\alpha})cos \mathbf{\alpha} = OPcos \mathbf{\alpha}tan \mathbf{\alpha} = OPtan \mathbf{\alpha}$$ As in Figure XX, $$OP''$$ is defined as the perpendicular distance to the surface, $$\mathbf{\rho}$$. If we assume that the twist angle, $$\mathbf{\alpha}$$, is a small angle, $$PP''$$ simplifies to: $$PP''=\mathbf{\rho}\mathbf{\alpha}$$

The shear strain, $$\mathbf{\gamma}$$, can be defined as the ratio of this displacement distance, $$PP''$$, to a small displacement in the axial (x) direction: $$\gamma = \frac{PP''}{dx}$$ We can then substitute for $$PP''$$, and since we assumed $$\mathbf{\alpha}$$ is very small it can be replaced by it's differential element $$d\mathbf{\alpha}$$. We can also define the rate of twist, $$\mathbf{\theta}$$, as the differential twist angle $$d\mathbf{\alpha}$$ per differential change in the axial direction $$dx$$ : $$\gamma = \frac{\rho d\alpha}{dx} = \rho \theta$$

We can use Hooke's Law to relate the shear stress to the shear strain: $$\mathbf{\tau =G \gamma =G \rho \theta}$$ $$\mathbf{\tau(s) =G \rho (s) \theta (x)}$$

If we take the contour integral along C, the equation becomes: $$\oint_{C}^{}{\tau(s)ds} = G \theta(x)\oint_{C}^{}{\rho (s)ds}$$ Next we can substitute in values from previous derivations to arrive at the final form for $$\mathbf{\theta}$$: $$\oint_{C}^{}{\frac{q(s)}{t(s)}ds} = G \theta(x)2\bar{A}$$ $$\Rightarrow \theta = \frac{1}{2G\bar{A}}\oint_{C}^{}{\frac{q}{t}ds}$$

What makes this derivation "ad hoc"?

 * 1) The shear strain $$\mathbf{\gamma}$$ must be obtained using the displacement of $$P$$ in the direction tangent to $$C$$ at $$P$$, but $$PP''$$ is not necessarily tangent to C (but is close).
 * 2) The equation for the shear stress, $$\mathbf{\tau} =\frac{q}{t}$$, was obtained from an ad hoc assumption that $$\mathbf{\tau}$$ is uniform across the wall thickness.
 * 3) There is an inconsistency in the assumption on the size of $$\mathbf{\alpha}$$. To get line $$PP'$$, we assumed $$\mathbf{\alpha}$$ was small, but to get $$PP''$$ we assumed a finite $$\mathbf{\alpha}$$ to use the relation $$\mathbf{\rho}=OPcos \mathbf{\alpha}$$. We then reintroduced the small angle approximation for $$\mathbf{\alpha}$$ in the rest of the derivation.

title: User:Eas4200c.f08.vqcrew.d/hw4


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text:
User:Eml4500.f08.ateam.mcnally/Week 6

Further Evaluation and Understanding of the Zero Eigen Values
Meeting 21 on Monday October 13th 2008 was a review and further understanding of a previously assigned homework problem dealing with the Eigen values of the Global K Matrix for a two bar truss system. The solution to the Eigen value problem actually yields four zero Eigen which corresponds to rigid body motion. The four zero values correspond to the global dof constraints: two translations, one rotation, and one mechanism which relates the idea of zero stored energy in the truss system if you remove both supports from the two bar truss system.

The next section is a plot of the Eigen vectors which correspond to the zero Eigen values of the two bar truss system.

K =

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

The corresponding Eigen values in a column vector are as follows:

eig(K)

ans =

-0.0000  -0.0000    0.0000    0.0000    1.4706   10.0294

The command [U,D]=eig(K) produces a matrix U whose columns are the eigenvectors of K and a diagonal matrix D with the eigenvalues of K on its diagonal

[U,D]=eig(K)

U =

-0.1118   0.5043   -0.0000   -0.5931    0.6174   -0.0139   -0.0814   -0.8634    0.0000   -0.3476    0.3565   -0.0080   -0.4628    0.0089    0.0000   -0.4803   -0.5409    0.5123    0.5266   -0.0053   -0.0000   -0.5429   -0.4330   -0.4904   -0.4947    0.0071   -0.7071    0.0313   -0.0765   -0.4984    0.4947   -0.0071   -0.7071   -0.0313    0.0765    0.4984

D =

-0.0000        0         0         0         0         0         0   -0.0000         0         0         0         0         0         0    0.0000         0         0         0         0         0         0    0.0000         0         0         0         0         0         0    1.4706         0         0         0         0         0         0   10.0294

From these matrices (U and D) the four Eigenvectors corresponding to the zero Eigen values can be pulled from matrix U and are as follows:

E1 =

-0.1118  -0.0814   -0.4628    0.5266   -0.4947    0.4947

E2 =

0.5043  -0.8634    0.0089   -0.0053    0.0071   -0.0071

E3 =

0        0         0         0   -0.7071   -0.7071

E4 =

-0.5931  -0.3476   -0.4803   -0.5429    0.0313   -0.0313

The results of the plots are the Mode Shapes which are a linear combination of the pure mode shapes. (pure rigid body motion two translations, one rotation, and one pure mechanism). The results explained above can also be represented mathematically with the following arguments.

$$Kv = \lambda v$$

Let $$\begin{Bmatrix} U_1 & U_2 & U_3 & U_4 \end{Bmatrix}$$ be the pure eigenvetors corresponding to the four zero eigenvalues. $$K*U_i = 0*U_i$$ for i = 1,...,4.

To get W for each zero eigenvalue, we take the sum of the four contributors, each multiplied by a constant $$\alpha _{1}$$, which gives us the following relation:

$$ W = \alpha_1 u_1 + \alpha_2 u_2 + \alpha_3 u_3 + \alpha_4 u_4$$

W consists of the combination of eigenvectors for zero eigenvalues, therefore W also represents an eigenvector corresponding to a zero eigenvalue.

title: User:Eml4500.f08.ateam.mcnally/Week 6


text: File:Eigenvector 1.JPG

title: File:Eigenvector 1.JPG


text: File:Eigenvector 3.JPG

title: File:Eigenvector 3.JPG


text: File:Eigenvector 4.JPG

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text:
User:Eml4500.f08.bottle.barnes/HW4

2-Bar Truss Eigenvectors
The eigenvalues of the global stiffness matrix are as follows:

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

To find the eigenvectors corresponding to the eigenvalues of the global stiffness matrix, first the matrix of eigenvalues is defined in MATLAB:

k =

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

Then, the following command is implemented:

EDU>> E = eig(k)

E =

-7.0650  -0.0000   -0.0000    0.0000    1.4401    7.1249

Plotting these values in yields the following graph:

title: User:Eml4500.f08.bottle.barnes/HW4


text:
The Web Economy/32

IPhone ERP
This is a lesson part of the module The Web Economy out of the Open Source ERP/Executive Diploma course conducted by a private university.

Topics

 * IPhone delighted consumers with its highly natural and intuitive style.
 * As a killer application, it can be integrated into an ERP System where our voice through the phone can call up the CRM and ERP information without keypad or text.
 * Many accompanying technologies such as voice synthesis and recognition, VOIP and high storage and data carrier are matured to be integrated to give a seamless experience.
 * The User Experience will be as if s/he is interacting via phone to a real human operator that is highly knowledgable and skilled in handling any operational tasks and transactions. For example:
 * "Please raise me a PO to Mr ABC". Machine will ask "For what Product? Please state the first letter, category or Product Name".
 * "P, Category Car, Proton Perdana". Machine will answer, "Product order for Proton Perdana which is a car is been placed with Mr. ABC. Same payment terms?"
 * "No. I want to pay in 30 days". Machine will answer, "PO issued via email and fax. You shall receive details copy via sms".
 * System then updates the CRM Request module, Inventory module and Commitment Accounting values.

Assignment Tasks

 * If Apple or anyone else is to incorporate the idea above, what will happen to conventional ERP or even open source ones?
 * Will there be a market for such a service or application?
 * How would you make it better?
 * What other telephony projects out there that are similar or as creative as this?
 * What other sensory-based projects are there out there? Can you suggest any that uses the other senses of a human?

Activities

 * Publish your works in an attractive manner in your user page, stating your own views and findings providing links to your sources. Use the talk page here to score marks.
 * Create or edit sub-pages of course materials within wikiversity.
 * Discuss in the forum (link shall be provided) by offering your ideas and answering or comment on others' postings.

Links to Student Notes
(Provided by Students - subject to edit ranking by tutor)

Notable Links to Resources
(Provided by Students - subject to edit ranking by tutor)

Sub-Pages
(Done in wikiversity as course material by the students under the guidance of the tutor)

Course Navigation
Next - Universal Supply Chain >> The Web Economy/33

title: The Web Economy/32


title: File:2bartruss.svg


text: File:A-forces.svg

title: File:A-forces.svg


text: File:Node2 2BarTruss.jpg

title: File:Node2 2BarTruss.jpg


title: File:L21Fig1.SVG


text:
Consensus in a wiki environment

Welcome to the Wikiversity Project discussing 'Consensus in a wiki environment'.

Intended participants are anyone with an interest in the way the 'Wiki-verse' treats the concept of 'consensus', from the micro-practical level in determining the wording of a subject heading, to the broad philosophical issues.

We may develop areas including;
 * /An introduction to Consensus/, there are some particular nuances to 'consensus' in a wiki environment, this will get you started.
 * /Measuring Consensus/, how to avoid tyranny of the majority, creeping towards voting etc.
 * /Managing Consensus/, how to deal with change within a consensus environment
 * /Establishing consensus/
 * /Practical consensus/ the difference between a normative definition and an empirical definition and the natural consequence of this
 * Your topic here! - if you're interested in discussing any aspect of 'Consensus', feel free to suggest or develop a new topic

title: Consensus in a wiki environment


text:
The Web Economy/33

Universal Supply Chain
This is a lesson part of the module The Web Economy out of the Open Source ERP/Executive Diploma course conducted by a private university.

Topics

 * Companies or projects are trying to come up with a killer app that encompasses the whole world into a single system.
 * ERP can be one such system where the world supply chain of producers and consumers are all connected via the web.
 * For maximum connectivity, users should all rely on one single standard and protocol of how things should be done.
 * Either all software vendors agree on such standards or there emerged an ubiquitous free software for easiest adoption.
 * There has to be much lowering of prices to remove barriers of user entry.
 * One target business model of SaaS providers will be to aggregate from a high ocean of traffic and provide cheaper sources of suppliers and consumers.

Assignment Tasks

 * How much has the price of ERP software changed since the last 10 years?
 * What are the open source ERP out there that has the highest number of adoption? What are their trends of growth and what are their factors?
 * Which kind of traders or producers/consumers will want to rely on a global supply chain? What about privacy and trade secrets?
 * Wouldnt over dependency on the Web be its biggest risk? What if suddenly the whole Web crashed due to some virus? Can business be normal again?

Activities

 * Publish your works in an attractive manner in your user page, stating your own views and findings providing links to your sources. Use the talk page here to score marks.
 * Create or edit sub-pages of course materials within wikiversity.
 * Discuss in the forum (link shall be provided) by offering your ideas and answering or comment on others' postings.

Links to Student Notes
(Provided by Students - subject to edit ranking by tutor)

Notable Links to Resources
(Provided by Students - subject to edit ranking by tutor)

Sub-Pages
(Done in wikiversity as course material by the students under the guidance of the tutor)

Course Navigation
Return to Course or Main Module: Course - The Web Economy Course Page Module - Open Source ERP / Executive Diploma Module Page

title: The Web Economy/33


text:
Consensus in a wiki environment/An introduction to Consensus

What is consensus?

Wikipedia defines consensus as "having two common meanings. One is a general agreement among the members of a given group or community, each of which exercises some discretion in decision making and follow-up action. The other is as a theory and practice of getting such agreements."

Wikipedia defines consensus decision-making as "[a] process that not only seeks the agreement of most participants, but also the resolution or mitigation of minority objections."

A census is a process to count living things and also to gain individual information from them. The prefix con- denotes that the census information is put together. For example, if a census finds 2 red ducks, 5 orange ducks, and 3 gray ducks, then a possible consensus is to say there are 10 ducks. Another possible consensus is to say there are red, orange, and gray ducks. The agenda of a consensus relates what information is put together from a census.

Meta-Wikimedia says that '"consensus can only work among reasonable editors who are making a good faith effort to work together to accurately and appropriately describe the different views on the subject."

Consensus is always flexible within its agenda, and consensus may be amended whenever someone has something new and reasonable to say.

The role of consensus
Many well established wikis, such as Wikiversity, use consensus as a guiding tool for many, if not all, project decisions. In the case of Wikimedia projects, consensus is used to decide whether material is suitable for inclusion, whether or not to grant certain 'editing rights' to individual editors, as well as to resolve disagreements between editors. In short, consensus' role in wiki environments is hugely important, and ubiquitous.

How is consensus used to make decisions?
Essentially consensus is interpreted following discussion, hence the importance of understanding and experience on the part of the interpreter. We would like to expand this section to detail formal and informal consensus discussions - can you help?


text: File:Three bar truss with cross U and D.jpg


text: File:Plot E1.jpg

title: File:Plot E1.jpg


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title: File:Plot E3.jpg


text: File:Plot E4.jpg

title: File:Plot E4.jpg


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title: File:Trusscasea.JPG


text: File:Consensus example1.svg

title: File:Consensus example1.svg


text: File:A-matrix.svg

title: File:A-matrix.svg


text: File:B-matrix.svg

title: File:B-matrix.svg


text: File:Unstable 3 bar truss setup.jpg

title: File:Unstable 3 bar truss setup.jpg


text:
User:Eml4500.f08.delta 6.lopez/HW4

=Justification of assembly of elemental stiffness matrix into global stiffness matrix= Consider the example of the 2-bar truss system discussed earlier. If we recall the elemental Force-Discplacement relationship:

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

The equilibrium of node 2 can be found using three methods we have discussed earlier this semester. 1. The Euler Cut Principle 2. The Free Body Diagrams of each element 3. The Identity of Global D.O.F.'s to Element D.O.F.'s

At node 2, the elquilibrium can be found by simple summing the forces in each direction.

$$ \sum{F_{x}} = 0 = -f_{3}^{(1)} - f_{1}^{(2)} = 0 $$ $$ \sum{F_{y}} = 0 = P -f_{4}^{(1)} - f_{2}^{(2)} = 0 $$

=Homework: Plot the eigenvectors corresponding to the zero eigenvalues, and interpret the results=

Using MATLAB, the output of the Global Stiffness Matrix is:

The V and D matrices below correspond to the eigenvectors and eignevalues, respectively.

As can be seen by the D matrix, the first four columns of the V matrix correspond to the eigenvectors of the zero eigenvalues.

Columns 1 & 2

Columns 3 & 4

title: User:Eml4500.f08.delta 6.lopez/HW4


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User talk:Scottmccaffree

Scott, thanks for your help with the "online resources" page; though I generally prefer if people make adds to the discussion tabs, I thought your edits were very helpful!
 * Stevenarntson 01:39, 24 October 2008 (UTC)

Oh, ok. Sorry, I didn't know that there was a difference on that page. I'll look closer. I also tried to clean up (a lot actually) the discussion for everyone's projects. It was non-linear and unorganized so I tried to make it look better and easier to fallow. I also put the main code in its own cell so people didn't get confused about how to do it, as some had when I went through to fix it. (I only changed formatting, I didn't touch anything that was below their title, it's all as they wrote it.)

I'll be weary about if I'm in discussion or not next time, I don't want to change your project or take it over, so if I step over, please let me know.

Thanks for the feedback

 * Scottmccaffree 05:19, 24 October 2008 (UTC)

I much prefer people stepping over than not stepping up at all! :) I looked over the code changes, and they seemed very productive--thanks for taking the time to do it!
 * Stevenarntson 16:35, 24 October 2008 (UTC)

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

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And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! --The Jade Knight (d'viser) 10:02, 30 October 2008 (UTC)

That's quite an ambitious project you've got in mind, Scott. I will say, though, I do enjoy Daft Punk. I'd love to see you post some of your progress here to Wikiversity. The Jade Knight (d'viser) 10:02, 30 October 2008 (UTC)

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User talk:Stevenarntson/English Composition/Vispo

 * VisPo for Kids!
 * Stevenarntson 01:40, 24 October 2008 (UTC)


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User:Eml4500.f08.gravy.jad/Notes4

$$\ \,$$ $$\  \,$$ $$\  \,$$

'''Connectivity Array '''

Here we are using a connectivity array "conn" (from p.5-6 2-bar truss system)

$$\ conn = \begin{vmatrix} 1 & 2 \\ 2 & 3 \end{vmatrix} \,$$

Where the columns represent the local node number, but the numbers inside the matrix state the global node number. The rows tell the element number. (ie. row 1 represents the first element, row 2 represents the second element, ect.)

conn(e,j)= global node # of local node j of element e.

 Location Matrix Master Array

Here we are using a location matrix master array "lmm" (from p.5-6 2-bar truss system)

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

Where the columns represent the local dof's, but the numbers inside the matrix state the global dof's or equation number in matrix k. The rows tell the element number. (ie. row 1 represents the first element, row 2 represents the second element, ect.)

lmm(i,j)= eq # (global dof's #) for the elem. stiffness coeff. corresponding to the jth local dof #.

Insert HW 18-2

Back to p.12-1: Method 2 to desire k(e)4x4 : transform a system with 4 dofs into a system with also 4 dofs (instead of only 2 dofs) so that the transformation matrix is now 4x4 and inverted. We want a system of equations that follows a coordinate system which correlates to the axial direction of the bar.

Goal : Want to find $$\ T^{(e)}_{4x4} \,$$ that transform the set of local (elem) dofs $$\ d^{(e)}_{4x4} \,$$ to another set of local (elem) dofs $$\ \underline{\tilde{d}}^{(e)}_{4x4} \,$$ such that $$\ \underline{\tilde{T}}^{(e)} \,$$  is invertible.

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

Continuing the example form p.12-4

$$\ \tilde{d}^{(e)}_{1} = \left \lfloor \ l^{e} m^{e} \right \rfloor \begin{Bmatrix} d^{(e)}_{1} \\ d^{(e)}_{2} \end{Bmatrix} \,$$  where $$\ \underline{\tilde{ d }}^{(e)}_{1} \,$$ also equals $$\ q^{(e)}_{1} \,$$ (Equation 1)

$$\ \tilde{d}^{(e)}_{2} = \tilde{d}^{(e)}_{1} = \vec{d}^{(e)}_{1} \cdot \vec{\tilde{j}} \,$$ (component of $$\ \tilde{d}^{(e)}_{1} \,$$ along $$\ \vec{\tilde{j}} \,$$ j, ie. $$\ \tilde{y}-axis \,$$) (of the local nodes)

$$\ = - \sin\theta^{(e)} d^{(e)}_{1} + \cos \theta^{(e)} d^{(e)}_{2} \,$$

$$\ = \left \lfloor \ -m^{e} l^{e} \right \rfloor \begin{Bmatrix} d^{(e)}_{1} \\ d^{(e)}_{2} \end{Bmatrix} \,$$ (Equation 2)

Insert HW 19-2

Put (EQ 1) and (EQ 2) in matrix form:

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

where $$\ \underline{R}^{(e)} = \begin{bmatrix} l^{(e)} & m^{(e)} \\ -m^{(e)} & l^{(e)} \end{bmatrix} \,$$

A complete matrix form of all dofs

$$\ \underbrace{ \begin{Bmatrix} \tilde{d}^{(e)}_{1} \\ \tilde{d}^{(e)}_{2} \\ \tilde{d}^{(e)}_{3} \\ \tilde{d}^{(e)}_{4} \end{Bmatrix} }_{\underline{\tilde{d}}^{(e)}_{4x1}} = \underbrace{ \begin{bmatrix} \underline{R}^{(e)}_{2x2} & \underline{0}_{2x2} \\ \underline{0}_{2x2} & \underline{R}^{(e)}_{2x2} \end{bmatrix} }_{\underline{\tilde{T}}^{(e)}_{4x4}} \underbrace{ \begin{Bmatrix} d^{(e)}_{1} \\ d^{(e)}_{2} \\d^{(e)}_{3} \\d^{(e)}_{4} \end{Bmatrix} }_{\underline{d}^{(e)}_{4x1}} \,$$

$$\ \underline {\tilde{f}}^{(e)}_{4x1}= k^{(e)} \begin{bmatrix} 1 & 0 & -1 & 0 \\ 0 & 0 & 0 & 0 \\ -1 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix} \underline{\tilde{d}}^{(e)} \,$$

In conclusion:

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

Special Case

$$\ \tilde{d}^{(e)}_{4} \ne 0 \,$$, $$\ \tilde{d}^{(e)}_{1} =\tilde{d}^{(e)}_{2} =\tilde{d}^{(e)}_{3} = 0 \,$$

Therefore: $$\ \underline{\tilde{f}}^{(e)}_{4x1} = \underline{\tilde{k}}^{(e)}_{4x4} \underline{\tilde{d}}^{(e)}_{4x1} = \underline{0}_{4x1} \,$$ where the zero matrix is the 4th column of $$\ \underline{\tilde{k}}^{(e)} \,$$

Now interpret the transverse of the dofs: $$\ \underline{\tilde{d}}^{(e)} = \underline{\tilde{T}}^{(e)} \tilde{d}^{(e)} \,$$

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

Insert HW 20-1

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

Hence, if $$\ \underline{\tilde{T}}^{(e)} \,$$ is invertible then: $$\ \left [ {\underline{\tilde{T}}^{(e)}}^{-1} \underline{\tilde{k}}^{(e)} \underline{\tilde{T}}^{(e)} \right ] \underline{\tilde{d}}^{(e)} = \underline{f}^{(e)} \,$$

$$\ \underline{\tilde{T}}^{(e)} \,$$ is a block diagonal matrix (from p.19-3).

Now consider the general block-diagonal matrix: $$\ A = \begin{bmatrix} \underline{D}_{1} & & \underline{0} \\ & \ddots &  \\ \underline{0} &  & \underline{D}_{s} \end{bmatrix} \,$$

Question: Then what is $$\ A^{-1} \,$$?

Answer: Lets make a simpler case- $$\ B = \begin{bmatrix} d_{11} & & & \underline{0} \\ & d_{22} & &  \\ & & \ddots & \\ \underline{0} &  & & D_{s} \end{bmatrix} = Diag \left [ d_{11},d_{22}, \ldots, d_{nn} \right ] \,$$

$$\ B^{-1}= Diag \left [ \frac{1}{d_{11}},\frac{1}{d_{22}}, \ldots, \frac{1}{d_{nn}} \right ] \,$$ assuming that $$\ d_{ij} \ne 0 \,$$ for $$\ i=1 \ldots n \,$$.

$$\ B \cdot B^{-1} = I \,$$

Now going back to the A-matrix:

$$\ A= Diag \left [ \underline{D}_{1},\underline{D}_{2}, \ldots, \underline{D}_{s} \right ] \,$$

$$\ A^{-1}= Diag \left [ \frac{1}{\underline{D}_{1}},\frac{1}{\underline{D}_{2}}, \ldots, \frac{1}{\underline{D}_{s}} \right ] = Diag \left [ \underline{D}^{-1}_{1},\underline{D}^{-1}_{2}, \ldots, \underline{D}^{-1}_{s} \right ] \,$$

Therefore: $$\ {\underline{\tilde{T}}^{(e)}}^{-1} = Diag \left [ {\underline{R}^{(e)}}^{-1}, {\underline{R}^{(e)}}^{-1} \right ] \,$$

Recall from p.19-2: $$\ \underline{R}^{(e)} = \begin{bmatrix} l^{(e)} & m^{(e)} \\ -m^{(e)} & l^{(e)} \end{bmatrix} \,$$

Now: $$\ {\underline{R}^{(e)}}^{T} = \begin{bmatrix} l^{(e)} & -m^{(e)} \\ m^{(e)} & l^{(e)} \end{bmatrix} \,$$

Therefore: $$\ {\underline{R}^{(e)}}^{T} \underline{R}^{(e)} = \begin{bmatrix} l^{(e)} & -m^{(e)} \\ m^{(e)} & l^{(e)} \end{bmatrix}_{2x2} \begin{bmatrix} l^{(e)} & m^{(e)} \\ -m^{(e)} & l^{(e)} \end{bmatrix}_{2x2} = \begin{bmatrix} {l^{(e)}}^{2} + {m^{(e)}}^{2} & 0 \\ 0 & {l^{(e)}}^{2} + {m^{(e)}}^{2} \end{bmatrix} \,$$

since $$\ {l^{(e)}}^{2} + {m^{(e)}}^{2} = 1 \,$$, then $$\ \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I_{2x2} \,$$ the Identity matrix.

In conclusion: $$\ \Rightarrow {R^{(e)}}^{-1} = {R^{(e)}}^{T} \,$$ and $$\ \Rightarrow {\underline{\tilde{T}}^{(e)}}^{-1} = {\underline{\tilde{T}}^{(e)}}^{T} \,$$

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User talk:Ivanceras

Welcome
Hello Ivanceras, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may
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And don't forget to explore Wikiversity with the links to your left. Be bold, and see you around Wikiversity! --Emesee 03:21, 24 October 2008 (UTC)

title: User talk:Ivanceras


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Terraforming/Mars

Learning Tasks

 * (Duration) Terraforming takes time and activities and interventions of today will have an impact many years later. Explain the scientific workflow, how the duration of terraforming can be calculated and estimated. Explain similarities and differences between terraforming and forrest management according to time and duration of intervention! e.g. plant trees today might result in harvesting the wood in 100 years. What is the motivation of people to plan so many years ahead?
 * (Human Interventions on Earth) Interventions on earth of mankind formed the earth in way like no other species did before. What are the similarities an differences between the way mankind changes the global ecosystem on earth (e.g. climate change) and terraforming of other planets? Is there anything we can learn from terraforming to keep our earth "in good shape" and visualized our impact on earth, (e.g. Earth Overshot Day) or vice versa what are the lessons learnt from climate change that can be used in the context of terraforming.
 * (Drivers and Activities) What kind of activities or drivers could you identify beyond commercial objectives and exploitations of resources for application on earth. What are potential obstacles and requirements and constraints that can be challenge?
 * (Transport and Resource Management) Transportation of resources or tools to other planets for terraforming is necessary and expensive. Explain the role of resource management for terraforming to use existing resource on the planet to perform terraforming. What are steps before terraforming that are required prior to any interventions planned for other planets.

Wikipedia

 * Terraforming of Mars
 * Mars
 * Colonization of Mars

Wikibooks

 * Colonising Mars

title: Terraforming/Mars


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User:Eml4500.f08.delta 6.guzman/Lecture Notes 10/10

$$f^{(e)}_{4x1} = k_{4x4} ^{(e)} * d_{4x1}^{(e)}$$ Note: Consider the case where $$ d_{4} \neq 0 $$ then $$ d_1^{(e)} = d_2^{(e)} = d_3^{(e)} = 0 $$

you will get the following relationship $$ f_{4x1} ^{(e)} = k_{4x4} ^{(e)} * d_{4x1} ^{(e)} = 0_{4x1} $$ where the zero matrix is the 4th column of matrix $$ k^{(e)} $$ (interpretation of the transverse dofs)

Recall from 19-3 that $$ d^{(e)} = T^{(e)} * d^{(e)} $$ similarly there is the same relationship for f where $$f^{(e)} = T^{(e)} * f^{(e)} $$

Also: $$ k^{(e)} = T^{(e)} * d^{(e)} $$

The lesson here is that we should use the transverse dofs instead of just using the inverse dofs therefore finding the relationship for the global matrix $$ k=d*f $$ $$k^{(e)} *T^{(e)} * d^{(e)}= T^{(e)} * f^{(e)}$$

If $$ T^{(e)}$$ is inversible, then the matrix that we are looking for is:

$$f^{(e)} = d^{(e)} * \begin{bmatrix} T^{(e)^{-1}} * k^{(e)} * T^{(e)} \end{bmatrix}$$ where $$T^{(e)}$$ is the block diagram matrix.

Considering a general block-diagram matrix A:

$$ A=

\begin{bmatrix}

D_{1} & 0 & 0 & 0\\

0 & . & 0 & 0\\

0 & 0 & . & 0\\

0 & 0 & 0 & D_{s}

\end{bmatrix} $$

so what will be the inverse of matrix A? This can be answered by observing the simple example of the diagonal matrix B:

$$ B=

\begin{bmatrix}

d_{11} & 0 & 0 & 0\\

0 & d_{22} & 0 & 0\\

0 & 0 & . & 0\\

0 & 0 & 0 & d_{mm}

\end{bmatrix} $$

$$ B= Diag \begin{bmatrix} d_{11}, d_{22}, ... , d_{mm} \end{bmatrix} $$

$$ B^{-1} = Diag\begin{bmatrix} \frac{1}{d_{11}}, \frac{1}{d_{22}}, ... , \frac{1}{d_{mm}} \end{bmatrix} $$

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

For a block diagonal matrix A

$$ A = Diag\begin{bmatrix} D_{11}, D_{22}, ... , D_{s} \end{bmatrix} $$

$$ A^{-1} = Diag\begin{bmatrix} D_{11}^{-1}, D_{22}^{-1}, ... , D_{s}^{-1} \end{bmatrix} $$

$$T^{(e)^{-1}} = Diag \begin{bmatrix} R^{(e)^{-1}}, ... , R^{(e)^{-1}} \end{bmatrix}$$

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

Since $$l^{2}*m^{2} = 1 $$ the matrix becomes where $$ I_{2x2} $$ is the identity matrix.

$$R_{2x2}^{(e)^{T}}*R_{2x2}^{(e)} = \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}_{2x2} = I_{2x2} $$

$$ R^{(e)^{T}}= R^{(e)^{-1}} $$ $$ T^{(e)^{-1}} = Diag\begin{bmatrix} R^{(e)^{T}}, R^{(e)^{T}} \end{bmatrix}$$ where $$ T^{(e)} = \begin{pmatrix} Diag \begin{bmatrix} R^{(e)}, R^{(e)} \end{bmatrix} \end{pmatrix}^{T} $$ $$ T^{(e)^{-1}} = T^{(e)^{T}} $$ $$ \begin{bmatrix} T^{(e)^{-T}} * k^{(e)}* T^{(e)} \end{bmatrix} *d^{(e)}= f^{(e)} $$ where $$ k^{e} = \begin{bmatrix} T^{(e)^{-T}} * k^{(e)}* T^{(e)} \end{bmatrix} $$


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User:Eas4200c.f08.gator.edwards/Matlab Program Continued

The end goal of the MATLAB program for this homework is to compute the torsion constant J for the three-cell airfoil. The first program shown is the validation for the method of area computation. This is done by using the code on a circle.

Next it needs to be shown that the area computed will turn out the same regardless of where the point P0 is placed. This will be done with the original program for finding the area and centroid of a NACA 4-digit airfoil.

Now we can move on to computing the torsion constant J. The first program is a modified version of the above airfoil program. It calculates J for a single-cell airfoil. It also plots the airfoild but this plot is irrelevant so it will be left out (the plot can be seen in Homework 3). The second program is a further modified version of the above program, this time modified to calculate J for a three-cell airfoil. The plot from this program is included since this is the first time addressing the three-cell airfoil.

The value of J from the three-cell airfoil is seen to be three orders of magnitude higher than that of the single-cell airfoil. It is desirable to have a high J value so a three-cell airfoil is clearly preferable to a single-cell airfoil.


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Consensus in a wiki environment/Measuring Consensus

On this page we will look at various ways of measuring consensus, pitfalls to avoid, and some examples of 'best practice'. Please feel free to contribute to building this resource!

Consensus without majority rule
Parliamentary procedure contrasts with consensus by broad support. In parliamentary procedure, a majority rule determines the outcome, which may lead to stakeholders disappointed if there input is shorted or excluded. Those that operate by broad support feel they can achieve greater consensus than what simple majority rule achieves. To measure consensus without majority rule, the participants attempt to find unanimity, by broad support, among the stakeholders, which does not mean the stakeholders must compromise what is important to them.

Consensus differs from a vote, like a majority rule. In a consensus the viewpoints are significant, but in a vote the individual tally from each person is significant.

How to balance a consensus
Consensus can balance the needs of the majority with the needs of the minority, so it is important to understand everyone's viewpoint.

Examples:
 * 1) If a majority of the people have the same viewpoint, then that is one viewpoint in the consensus. If the minority of the people have the same viewpoint, then that is also another viewpoint in the consensus. If there is only a minority viewpoint and majority viewpoint, then there are only two viewpoints among all the stakeholders.
 * 2) If 4 people share a common view, 2 people share a different common view, and 5 more people share another common view, than 3 viewpoints are said to exist. People who agree, but agree for different reasons, may still have their views counted as separate viewpoints.

Does a new stakeholder represent an existing or new viewpoint?
Measuring consensus involves determining whether or not concerns were relevant and considers whether concerns brought up were reasonably addressed or satisfied by discussion or changes. Sometimes the determined outcome of consensus can also be questioned, questionable or members of the community may question the motives of the person who made the consensus determination, which may lead to further discussion that may result in a different outcome.

Who measures the consensus?
Community participants may sometimes question how consensus was measured, or feel that the motivation of the person who made the determination is questionable, which may lead to further discussion with a different outcome.


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Consensus in a wiki environment/Managing Consensus

Consensus evolves over time, on this page we will look at how that is managed.


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Pakistan

Pakistan is an Islamic Republic

Geography
Pakistan is a country located in South Asia, Southwest Asia, Middle East and converges with Central Asia and the Middle East.It has a border with Afghanistan, Iran, India and the People's Republic of China. It's total area is about 803,940 km².

Punjab
It has the second largest city in Pakistan, Lahore. It is ranked 22nd in the Largest Cities of the World. It is the agriculture centre of the country.

Gilgit Baltistan
In Gilgit Baltistan there is the 2nd biggest mountain in the world called "Mt. Godwin Austin" or "K2".

Sindh
It has the largest city in Pakistan, Sindh. It is ranked 12th in the Largest Cities of the World.

Balochistan
It's biggest city is Quetta.

Independence from the United Kingdom
=== Presidents ===

Facts
Pakistan has a census and an election every 5 years.

Maps of Pakistan

Basic Map with Close Countries

Learning resources

 * Pakistan
 * Pakistan a Brief History
 * Islamabad - Capital of Pakistan
 * Provinces of Pakistan
 * Cities of Pakistan
 * /Natural Resources/
 * /Marriage/

Universities

 * Universities in Pakistan
 * Pakistan Academy of Sciences
 * Pakistan Educational Research Network
 * Higher Education Commission of Pakistan

title: Pakistan


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File:5-beam truss def.jpg

Matlab simulated 5-beam deformed (solid line) vs. undeformed (dashed line) truss

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User:Eas4200c.f08.vqcrew.c/Homework 4/my

Torsional Analysis of Idealized Single-Cell Airfoil
It was determined in an earlier section that the following relationship is true for the idealized airfoil given in Figure ##:

$$\theta= \frac{111.29 q}{G}$$

The two relationships for the torque are also known:

$$T = GJ\theta\;$$

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

Combining the above two equations and solving for $$J\;$$ yields:

$$J = \frac{2q\bar{A}}{G \theta}$$

Substituting in the expression for $$\theta\;$$ results in the following expression:

$$J = \frac{2\bar{A}}{111.29}$$

The area $$\bar{A}\;$$ was previously calculated in HW3. Plugging in this value into the above yields the following value for $$J\;$$.

$$J = .1001\; m^{4}$$

Torsional Analysis of Two-Celled Rectangular Airfoil
Figure ## depicts a rectangular airfoil of two cells. It was found in an earlier section that the following three relationships are true:

$$T = T_1 + T_2 = 2\ (q_1\ \bar A_1 + q_2\ \bar A_2)$$

$$\theta_1 = \frac{1}{2\ G\ \bar A_1}\ (\frac{2\ q_1\ a}{t_1} + \frac{q_1\ c}{t_1} + \frac{(q_1 - q_2)\ c}{t_{12}})$$

$$\theta_2 = \frac{1}{2\ G\ \bar A_2}\ (\frac{2\ q_2\ a}{t_2} + \frac{q_2\ c}{t_2} + \frac{(q_2 - q_1)\ c}{t_{12}})$$

Plugging in the known variables available in the earlier section, and solve for $$q_{1}\;$$ and $$q_{2}\;$$ in terms of T, the following relationships can be determined:

$$q_{1} = 1.012 T\;$$

$$q_{2} = 1.577 T\;$$

Noting that, by definition, $$\theta = \theta_{1} = \theta_{2}\;$$, plugging the above into the equation for $$\theta_{1}$$ yields:

$$\theta = 1170.125 \frac{T}{G}$$

It is also known that $$J\;$$ in terms of $$\theta\;$$ can be expressed as:

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

Which leades directly to solving for J:

$$J = \frac{1}{1170.125} = 8.55x10^{-4} m^{4}$$

MATLAB Airfoil Code Validation
Part of the first version of the MATLAB problem was to verify that the code was working properly by testing it both against a known area and also for replication of results. In the collapsible table below there is a "validation code" which calls part of the HW3 MATLAB code as a subfunction.

The resulting output from running the validation code is provided below.

The first section of the validation involved a "circular" airfoil. The first task was to determine the area of a circle of radius 1 meter at various numbers of contour segments. A table summarizing the above results is as follows:

 * {| border=1

 * bgcolor="red"|Number of Segments
 * bgcolor="#99ccff" style="text-align: center;" |20
 * bgcolor="#99ccff" style="text-align: center;" |40
 * bgcolor="#99ccff" style="text-align: center;" |60
 * bgcolor="#99ccff" style="text-align: center;" |80
 * bgcolor="#99ccff" style="text-align: center;" |100
 * bgcolor="red"|Calculated Area
 * bgcolor="#99ccff" |3.1045
 * bgcolor="#99ccff" |3.1285
 * bgcolor="#99ccff" |3.1344
 * bgcolor="#99ccff" |3.1369
 * bgcolor="#99ccff" |3.1383
 * }
 * }

The second task was to determine the $$ns_{max}\;$$, the number of segments at which the calculated area is within 1% of the actual area, which in this case is equal to 3.1416 square meters. This was done by calculating the area at various values of n and then comparing them. A plot of the results is given in Figure ##, where $$ns_{max}\;$$ and its corresponding area are denoted by a black square. For reference, the calculated value for $$ns_{max}\;$$ was found to be 23 segments, and the calculated area at this point was 3.1115 square meters, which differs from the actual value by .96%.

The final task for this section was to calculate the circle area at various locations of the origin point $$P_{0}\;$$. The three locations used were the coordinate system origin $$[0,0]\;$$, the circle center $$[1,0]\;$$, and an arbitrary external point $$[1,-4]\;$$. A table summarizing the results is shown below. As expected, the area is identical in all three cases.

 * {| border=1

 * bgcolor="red"|Coordinate of Origin Point
 * bgcolor="#99ccff" style="text-align: center;" |[0,0]
 * bgcolor="#99ccff" style="text-align: center;" |[1,0]
 * bgcolor="#99ccff" style="text-align: center;" |[1,-4]
 * bgcolor="red"|Calculated Area (square meters)
 * bgcolor="#99ccff" |3.1115
 * bgcolor="#99ccff" |3.1115
 * bgcolor="#99ccff" |3.1115
 * }
 * }

The second section of the validation involved a NACA 2415 airfoil of chord length c = 1 meter. The first task was to find $$ns_{max}\;$$ as with the circular airfoil. However, instead of comparing against a known area, the calculated area at a given number of segments is compared with the calculated area using one less segment, such that $$ns_{max}\;$$ is located when the difference between two successive area calculations is less than 1% of the current area calculation. A plot of the results is given in Figure ##. For reference, the calculated value for $$ns_{max}\;$$ was found to be 9 segments, and the calculated area at this point was 0.0244 square meters.

The second task was to show that the area calculation is valid at any choice of the origin point $$P_{0}\;$$. The three locations used were the coordinate system origin $$[0,0]\;$$, the airfoil centroid $$[.210,.008]\;$$ (previously calculated in HW3), and an arbitrary external point $$[.25,-1]\;$$. A table summarizing the results is shown below. Unlike with the circular case, the calculated areas are not identical for each case. However, a quick relative error calculation shows that the calculated values are all within 1% of each other, and such a small difference can be easily explained by the small numerical errors inherent in this kind of analysis.

 * {| border=1

 * bgcolor="red"|Coordinate of Origin Point
 * bgcolor="#99ccff" style="text-align: center;" |[0,0]
 * bgcolor="#99ccff" style="text-align: center;" |[.210,.008]
 * bgcolor="#99ccff" style="text-align: center;" |[.25,-1]
 * bgcolor="red"|Calculated Area (square meters)
 * bgcolor="#99ccff" style="text-align: center;" |.0244
 * bgcolor="#99ccff" style="text-align: center;" |.0246
 * bgcolor="#99ccff" style="text-align: center;" |.0247
 * }
 * }

Description of Airfoil Geometry and Equations

Figure ## shows the geometry of a generic NACA 4-digit airfoil. The various quantities are defined as follows:

$$LE\;$$ - The point corresponding to the leading edge of the airfoil. $$TE\;$$ - The point corresponding to the trailing edge of the airfoil. $$c\;$$ - The chord length, describing the length of the straight line connecting the leading edge to the trailing edge. $$m\;$$ - The maximum camber, corresponding to the first digit of the 4-digit series, and given in percentage of chord. $$p\;$$ - The location of maximum camber, corresponding to the second digit of the 4-digit series, and given in tenths of chord. $$t_{a}\;$$ - The maximum thickness, corresponding to the third and fourth digits of the 4-digit series, and given in percentage of chord. $$y_{u}\;$$ - The contour defining the airfoil's upper surface. $$y_{l}\;$$ - The contour defining the airfoil's lower surface. $$MCL\;$$ - The contour defining the airfoil's mean camber line. Referred to as $$z_{c}\;$$ in the airfoil equations.

A set of general equations used to generate the various contours for an arbitrary NACA 4-digit airfoil can be found here.

Torsional Constant Comparison Between Single- and Multi-celled Airfoils
The new facet to the MATLAB assignment deals with a comparison of single and multi-celled airfoils, in terms of the torsional constant $$J\;$$. The code for the single-celled airfoil is in the table below:

The first portion of the assignment was to determine $$ns_{max}\;$$, as in previous assignments. The result was a $$ns_{max}\;$$ value of 12 segments and a $$J\;$$ value of $$4.7543x10^{-6} m^{4}\;$$. The second portion involved generating a graph of torsional constant vs. number of segments. The result is shown in Figure ##.

The second portion of the assignment was to determine $$J\;$$ for a NACA2415 airfoil with 3 cells. The MATLAB code used to carry out this task is given in a table below.

The plotted airfoil, with partitions at quarter and three-quarter chord, is shown in Figure ##. The calculated $$J\;$$ for this case is $$8.6086x10^{-6} m^{4}\;$$.

Comparison:

The two calculated values for $$J\;$$ are very close to each other. The difference is likely a result of numerical errors during the generation of the torsional constant in the above MATLAB codes. Accounting for this effect, the overall result echoes the conclusion given in Chapter 3 of Sun: the torsional rigidity of a thin-walled, closed cross-section cannot be appreciably improved by compartmentalization


text:
User:Eml4500.f08.lulz.strack/hw4

MATLAB Code: Five-bar and Three-bar Trusses
The first part of the required MATLAB code constructs and loads a truss system composed of five bars. The code below represents Example 1.4 (FiveBarTrussEx.m) on page 25 of the textbook. The first part defines the specifications of each bar and node including modulus of elasticity (e), cross-sectional area (a), and loading values (P). The "nodes" matrix defines the absolute locations of the four nodes. The "conn" matrix defines the global node numbers at the ends of each bar or element. Each row represents an element and nodes are numbered by the lowest first. As shown in the code, there are 5 sets of 2 numbers for the 2 nodes corresponding to each element. The "lmm" matrix represents the global degrees of freedom (dof) for each element. Each node has 2 dof's; therefore, each element has 4 corresponding dof's. Degrees of freedom are numbered from 1 to 2*(number of nodes) where the first 2 correspond to Global Node 1, the second 2 correspond to the Global Node 2, etc. From all this data, the global stiffness matrix (K) is created. The displacement of the loaded node is calculated using the relevant parts of the global stiffness matrix, inverting them, and multiplying by the load array (R). Finally, the reactions (reactions) at each node are calculated by multiplying the displacement array (d) by the global stiffness matrix. The functions "PlaneTrussElement.m", "NodalSoln.m", and "PlaneTrussResults.m" are used again from Section 6 of Homework 2.

FiveBarTrussEx.m
% Five bar truss example e1=200*10^3; e2=70*10^3; a1=40*100; a2=30*100; a3=20*100; P = 150*10^3; nodes = 1000*[0, 0; 1.5, 3.5; 0, 5; 5, 5]; conn = [1,2; 2,4; 1,3; 3,4; 2,3]; lmm = [1,2,3,4; 3, 4, 7, 8; 1, 2, 5, 6; 5, 6, 7, 8; 3, 4, 5, 6]; K=zeros(8); % Generate stiffness matrix for each element and assemble it. for i=1:2 lm=lmm(i,:); con=conn(i,:); k=PlaneTrussElement(e1, a1, nodes(con,:)); K(lm, lm) = K(lm, lm) + k; end for i=3:4 lm=lmm(i,:); con=conn(i,:); k=PlaneTrussElement(e1, a2, nodes(con,:)); K(lm, lm) = K(lm, lm) + k; end

lm=lmm(5,:); con=conn(5,:); k=PlaneTrussElement(e2, a3, nodes(con,:)); K(lm, lm) = K(lm, lm) + k

% Define the load vector R = zeros(8,1); R(4)=-P

% Nodal solution and reactions [d, reactions] = NodalSoln(K, R, [1,2,7,8], zeros(4,1)) results=[]; for i=1:2 results = [results; PlaneTrussResults(e1, a1, ...           nodes(conn(i,:),:), d(lmm(i,:)))]; end for i=3:4 results = [results; PlaneTrussResults(e1, a2, ...           nodes(conn(i,:),:), d(lmm(i,:)))]; end format short g results = [results; PlaneTrussResults(e2, a3, ...       nodes(conn(5,:),:), d(lmm(5,:)))]

Output to the Command Window

>> FiveBarTrussEx

K =

32600       76067       -32600       -76067            0            0            0            0        76067  2.9749e+005       -76067 -1.7749e+005            0    -1.2e+005            0            0 -32600      -76067  2.4309e+005  1.1914e+005       -32998        32998 -1.7749e+005       -76067 -76067 -1.7749e+005 1.1914e+005  2.4309e+005        32998       -32998       -76067       -32600 0           0       -32998        32998    1.53e+005       -32998    -1.2e+005            0 0   -1.2e+005        32998       -32998       -32998    1.53e+005            0            0 0           0 -1.7749e+005       -76067    -1.2e+005            0  2.9749e+005        76067 0           0       -76067       -32600            0            0        76067        32600

R =

0          0           0     -150000           0           0           0           0

d =

0           0      0.53895     -0.95306       0.2647      -0.2647            0            0

reactions =

54927 1.5993e+005 -54927     -9926.7

results =

-0.0001743     -34.859 -1.3944e+005 -3.15e-005     -6.2999       -25200 -5.2941e-005     -10.588       -31764 -5.2941e-005     -10.588       -31764 0.00032087      22.461        44922

The results show the strain, stress, and axial force in each element corresponding to the three columns in the "results" matrix.

To visualize the loaded and unloaded system, the following code graphs the system in its loaded and unloaded state. In order to be able to realize the direction of node movement, the unfixed node displacements have been magnified by a factor of two (using the variable "m"). As shown, the original values from the displacement array (d3-d6) above were used below to get the right proportions of the nodal displacements.

FiveBarTrussDisplay.m
%Displaying the five bar truss clear; close;

m = 2;

d1 = 0;         %disp of node 1 in the x direction d2 = 0;         %disp of node 1 in the y direction d3 = 0.53895*m; %disp of node 2 in the x direction d4 = -0.95306*m; %disp of node 2 in the y direction d5 = 0.2647*m;  %disp of node 3 in the x direction d6 = -0.2647*m; %disp of node 3 in the y direction d7 = 0;         %disp of node 4 in the x direction d8 = 0;         %disp of node 4 in the y direction

%input nodal coordinates num_nodes = 4;              %total number of nodes num_elems = 5;              %total number of elements tot_dofs = 2 * num_nodes;   %total dofs of the system

%undeformed positions pos(1,:)=[0 1.5 0 5]; pos(2,:)=[0 3.5 5 5];

%undeformed nodal coordinate arrays for i=1:num_nodes x(i)=pos(1,i); y(i)=pos(2,i); end

%element connectivity array %follows: %node_conn(local node number, element number) = global node number

node_conn(1,1)=1;    %element 1 node_conn(2,1)=2;

node_conn(1,2)=2;    %element 2 node_conn(2,2)=4;

node_conn(1,3)=1;    %element 3 node_conn(2,3)=3;

node_conn(1,4)=3;    %element 4 node_conn(2,4)=4;

node_conn(1,5)=2;    %element 5 node_conn(2,5)=3;

%plot the undeformed 2-bar truss system for i=1:num_elems node_1=node_conn(1,i); node_2=node_conn(2,i); xx=[x(node_1),x(node_2)]; yy=[y(node_1),y(node_2)]; axis([-1 6 -1 6]) plot(xx,yy,'--rs') hold on end

n = -0.5; o = -0.25; text(0+n,0,'GN1','HorizontalAlignment','center') text(1.5+n,3.5,'GN2','HorizontalAlignment','center') text(0+n,5,'GN3','HorizontalAlignment','center') text(5+n,5,'GN4','HorizontalAlignment','center') text(0.75+o,1.75,'E1','HorizontalAlignment','center') text(3.25+o,4.25,'E2','HorizontalAlignment','center') text(0+o,2.5,'E3','HorizontalAlignment','center') text(2.5+o,5,'E4','HorizontalAlignment','center') text(0.75+o,4.25,'E5','HorizontalAlignment','center')

%deformed positions pos(1,:)=[0+d1 1.5+d3 0+d5 5+d7]; pos(2,:)=[0+d2 3.5+d4 5+d6 5+d8];

%deformed nodal coordinate arrays for i=1:num_nodes x(i)=pos(1,i); y(i)=pos(2,i); end

text(1.5+d3+n,3.5+d4,'GN2','HorizontalAlignment','center') text(0+d5+n,5+d6,'GN3','HorizontalAlignment','center') text(pos(1,1)+(pos(1,2)-pos(1,1))/2+o,pos(2,1)+(pos(2,2)-pos(2,1))/2,'E1','HorizontalAlignment','center') text(pos(1,2)+(pos(1,4)-pos(1,2))/2+o,pos(2,2)+(pos(2,4)-pos(2,2))/2,'E2','HorizontalAlignment','center') text(pos(1,1)+(pos(1,3)-pos(1,1))/2+o,pos(2,1)+(pos(2,3)-pos(2,1))/2,'E3','HorizontalAlignment','center') text(pos(1,3)+(pos(1,4)-pos(1,3))/2+o,pos(2,3)+(pos(2,4)-pos(2,3))/2,'E4','HorizontalAlignment','center') text(pos(1,2)+(pos(1,3)-pos(1,2))/2+o,pos(2,2)+(pos(2,3)-pos(2,2))/2,'E5','HorizontalAlignment','center')

%plot the deformed 2-bar truss system for i=1:num_elems hold on   node_1=node_conn(1,i); node_2=node_conn(2,i); xx=[x(node_1),x(node_2)]; yy=[y(node_1),y(node_2)]; plot(xx,yy,'-s') end

title('Deformed and Undeformed Five-bar Truss System') xlabel('X') ylabel('Y')

The dotted line represents the undeformed system shape. The solid line represents the deformed system shape. The abbreviation "GN" and "E" stand for Global Node and Element, respectively. Note only nodes 2 and 3 are displaced because they are unfixed.

For the three bar truss created in the class notes, the two bar truss system code (Two Bar Truss System.m) was modified from Section 6 of Homework 2. One node and element each were added to the system, and e, A, P, and L values were changed to coincide with the problem.

ThreeBarTrussEx.m
% Three bar truss example clear all; e = [2 4 3]; A = [3 1 2]; P = 30; L = [5 5 10]; alpha = pi/6; beta = -pi/6; gamma = -3*pi/4; nodes = [0, 0; L(1)*cos(alpha), L(1)*sin(alpha); L(1)*cos(alpha) + L(2)*cos(beta), L(1)*sin(alpha) + L(2)*sin(beta); L(1)*cos(alpha) + L(3)*cos(gamma), L(1)*sin(alpha) + L(3)*sin(gamma)]; dof = 2*length(nodes); conn = [1 2; 2 3; 2 4]; lmm = [1 2 3 4; 3 4 5 6; 3 4 7 8]; elems = size(lmm,1); K = zeros(dof); R = zeros(dof,1); debc = [1 2 5 6 7 8]; ebcVals = zeros(length(debc),1);

%load vector R = zeros(dof,1); R(4) = P;

% Assemble global stiffness matrix K = zeros(dof); for i = 1:elems lm = lmm(i,:); con = conn(i,:); k_local = e(i)*A(i)/L(i)*[1 -1; -1 1] k = PlaneTrussElement(e(i), A(i), nodes(con,:)) K(lm, lm) = K(lm, lm) + k; end K R % Nodal solution and reactions [d, reactions] = NodalSoln(K, R, debc, ebcVals) results = []; for i = 1:elems results = [results; PlaneTrussResults(e, A, nodes(conn(i,:),:), d(lmm(i,:)))]; end format short g results

Output to the Command Window

>> ThreeBarTrussEx

k_local =

1.2000  -1.2000   -1.2000    1.2000

k =

0.9000   0.5196   -0.9000   -0.5196    0.5196    0.3000   -0.5196   -0.3000   -0.9000   -0.5196    0.9000    0.5196   -0.5196   -0.3000    0.5196    0.3000

k_local =

0.8000  -0.8000   -0.8000    0.8000

k =

0.6000  -0.3464   -0.6000    0.3464   -0.3464    0.2000    0.3464   -0.2000   -0.6000    0.3464    0.6000   -0.3464    0.3464   -0.2000   -0.3464    0.2000

k_local =

0.6000  -0.6000   -0.6000    0.6000

k =

0.3000   0.3000   -0.3000   -0.3000    0.3000    0.3000   -0.3000   -0.3000   -0.3000   -0.3000    0.3000    0.3000   -0.3000   -0.3000    0.3000    0.3000

K =

0.9000   0.5196   -0.9000   -0.5196         0         0         0         0    0.5196    0.3000   -0.5196   -0.3000         0         0         0         0   -0.9000   -0.5196    1.8000    0.4732   -0.6000    0.3464   -0.3000   -0.3000   -0.5196   -0.3000    0.4732    0.8000    0.3464   -0.2000   -0.3000   -0.3000         0         0   -0.6000    0.3464    0.6000   -0.3464         0         0         0         0    0.3464   -0.2000   -0.3464    0.2000         0         0         0         0   -0.3000   -0.3000         0         0    0.3000    0.3000         0         0   -0.3000   -0.3000         0         0    0.3000    0.3000

R =

0    0     0    30     0     0     0     0

d =

0        0  -11.6737   44.4051         0         0         0         0

reactions =

-12.5672  -7.2557   22.3866  -12.9249   -9.8194   -9.8194

results =

2.4186      4.8371       9.6742       7.2557       14.511       9.6742       14.511       6.4625       12.925        25.85       19.387       38.775        25.85       38.775       2.3145       4.6289       9.2578       6.9434       13.887       9.2578       13.887

After running this code, the output produces displacement of node 2 to be -11.6737 in the x-direction and 44.4051 in the y-direction. On the other hand, the "results" matrix has 7 columns instead of 4. This is due to the fact that vectors for the modulus of elasticity (e), cross-sectional area (A), and element length (L) were passed as variables instead of the scalar values. The first column is the strain array for elements 1-3. This is followed by two 3x3 matrices created for stresses and axial forces of the elements instead of two arrays of length 3 each. The real values for stress and axial force lie along the diagonals of each of these matrices.

The code for displaying the three bar truss was similar to the code displaying the five bar truss. However, a displacement magnification factor (m) of 0.25 was used to ease visualization of the deformation.

ThreeBarTrussDisplay.m
%Displaying the three bar truss clear; close;

m = 0.25;

d1 = 0;         %disp of node 1 in the x direction d2 = 0;         %disp of node 1 in the y direction d3 = -11.6737*m; %disp of node 2 in the x direction d4 = 44.4051*m; %disp of node 2 in the y direction d5 = 0;         %disp of node 3 in the x direction d6 = 0;         %disp of node 3 in the y direction d7 = 0;         %disp of node 4 in the x direction d8 = 0;         %disp of node 4 in the y direction

%input nodal coordinates num_nodes = 4;              %total number of nodes num_elems = 3;              %total number of elements tot_dofs = 2 * num_nodes;   %total dofs of the system

%element 1 details L1 = 5; E1 = 2; A1 = 3; theta1 = 30; %deg thetar1 = (pi * theta1)/180;

%element 2 details L2 = 5; E2 = 4; A2 = 1; theta2 = -30; %deg thetar2 = (pi * theta2)/180;

%element 3 details L3 = 10; E3 = 3; A3 = 2; theta3 = -135; %deg thetar3 = (pi * theta3)/180;

%undeformed positions pos(1,:)=[0,L1*cos(thetar1),L1*cos(thetar1)+L2*cos(thetar2),L1*cos(thetar1)+L3*cos(thetar3)]; pos(2,:)=[0,L1*sin(thetar1),L1*sin(thetar1)+L2*sin(thetar2),L1*sin(thetar1)+L3*sin(thetar3)];

%undeformed nodal coordinate arrays for i=1:num_nodes x(i)=pos(1,i); y(i)=pos(2,i); end

%element connectivity array %follows: %node_conn(local node number, element number) = global node number

node_conn(1,1)=1;    %element 1 node_conn(2,1)=2;

node_conn(1,2)=2;    %element 2 node_conn(2,2)=3;

node_conn(1,3)=2;    %element 3 node_conn(2,3)=4;

%plot the undeformed 2-bar truss system for i=1:num_elems node_1=node_conn(1,i); node_2=node_conn(2,i); xx=[x(node_1),x(node_2)]; yy=[y(node_1),y(node_2)]; axis([-5 10 -10 20]) plot(xx,yy,'--rs') hold on end

n = -1; o = -0.5; text(x(node_conn(1,1))+n,y(node_conn(1,1)),'GN1','HorizontalAlignment','center') text(x(node_conn(2,1))+n,y(node_conn(2,1)),'GN2','HorizontalAlignment','center') text(x(node_conn(2,2))+n,y(node_conn(2,2)),'GN3','HorizontalAlignment','center') text(x(node_conn(2,3))+n,y(node_conn(2,3)),'GN4','HorizontalAlignment','center') text(x(node_conn(2,1))/2+o,y(node_conn(2,1))/2,'E1','HorizontalAlignment','center') text(x(node_conn(2,1))+(x(node_conn(2,2))-x(node_conn(2,1)))/2+o,y(node_conn(2,1))+(y(node_conn(2,2))-y(node_conn(2,1)))/2,... 'E2','HorizontalAlignment','center') text(x(node_conn(2,1))+(x(node_conn(2,3))-x(node_conn(2,1)))/2+o,y(node_conn(2,1))+(y(node_conn(2,3))-y(node_conn(2,1)))/2,... 'E3','HorizontalAlignment','center')

%deformed positions pos(1,:)=[0+d1,L1*cos(thetar1)+d3,L1*cos(thetar1)+L2*cos(thetar2)+d5,L1*cos(thetar1)+L3*cos(thetar3)+d7]; pos(2,:)=[0+d2,L1*sin(thetar1)+d4,L1*sin(thetar1)+L2*sin(thetar2)+d6,L1*sin(thetar1)+L3*sin(thetar3)+d8];

%deformed nodal coordinate arrays for i=1:num_nodes x(i)=pos(1,i); y(i)=pos(2,i); end

text(x(node_conn(2,1))+n,y(node_conn(2,1)),'GN2','HorizontalAlignment','center') text(x(node_conn(2,1))/2+o,y(node_conn(2,1))/2,'E1','HorizontalAlignment','center') text(x(node_conn(2,1))+(x(node_conn(2,2))-x(node_conn(2,1)))/2+o,y(node_conn(2,1))+(y(node_conn(2,2))-y(node_conn(2,1)))/2,... 'E2','HorizontalAlignment','center') text(x(node_conn(2,1))+(x(node_conn(2,3))-x(node_conn(2,1)))/2+o,y(node_conn(2,1))+(y(node_conn(2,3))-y(node_conn(2,1)))/2,... 'E3','HorizontalAlignment','center')

%plot the deformed 2-bar truss system for i=1:num_elems hold on   node_1=node_conn(1,i); node_2=node_conn(2,i); xx=[x(node_1),x(node_2)]; yy=[y(node_1),y(node_2)]; plot(xx,yy,'-s') end

title('Deformed and Undeformed Three-bar Truss System') xlabel('X') ylabel('Y')

The dotted line represents the undeformed system shape. The solid line represents the deformed system shape. The abbreviation "GN" and "E" stand for Global Node and Element, respectively. Note only node 2 is displaced because it is unfixed.

title: User:Eml4500.f08.lulz.strack/hw4


text: File:Circleplot.png

title: File:Circleplot.png


text: File:Airfoilplot.png

title: File:Airfoilplot.png


text:
User:Mariana~enwikiversity

.

Mariana Igova
Hello to all, I'm 3rd year student of Primary School of Pedagogics in SU "St. Kliment Ohridski", Bulgaria.It will be a pleasure for me to meet and work all of you.

research interests

 * Connection, interference and behaviour between teachers and students
 * Internet- and Project Competence

Theme of research

 * The importance of teachers in children's education in Primary School

Contacts

 * my e-mail: lortela@yahoo.com

title: User:Mariana~enwikiversity


text:
Autism spectrum/Dealing with outbursts

Some people with autism have outbursts. Children diagnosed with Asperger syndrome (AS; see Note) present a special challenge in the educational milieu. Typically viewed as eccentric and peculiar by classmates, their inept social skills often cause them to be made victims of scapegoating. Clumsiness and an obsessive interest in obscure subjects add to their "odd" presentation. Children with AS lack understanding of human relationships and the rules of social convention; they are naive and conspicuously lacking in common sense. Their inflexibility and inability to cope with change causes these individuals to be easily stressed and emotionally vulnerable. At the same time, children with AS (the majority of whom are boys) are often of average to above-average intelligence and have superior rote memories. Their single-minded pursuit of their interests can lead to great achievements later in life.

Asperger syndrome is considered a disorder at the higher end of the autistic continuum. Comparing individuals within this continuum, Van Krevelen (cited in Wing, l99l) noted that the low-functioning child with autism "lives in a world of his own," whereas the higher functioning child with autism "lives in our world but in his own way" (p.99).

Naturally, not all children with AS are alike. Just as each child with AS has his or her own unique personality, "typical" AS symptoms are manifested in ways specific to each individual. As a result, there is no exact recipe for classroom approaches that can be provided for every youngster with AS, just as no one educational method fits the needs of all children not afflicted with AS.

However, that is just one individual's suggestion. There are many ways to deal with outbursts.

title: Autism spectrum/Dealing with outbursts


text:
Physical exercises

Recommended Exercises for the Elderly-JUST MOVE!!!

Endurance Activities : swimming, bicycling, walking briskly, golf w/o a cart, dancing, etc. Vigorous Activities : climbing stairs, jogging, brisk bicycling up hills, swimming laps, etc. Rotating with overload by a senior.

Exercises to Improve Strength/Balance

Biceps Curl Strengthens upper-arm muscles.
 * 1) Sit in armless chair with your back supported by back of chair and keep feet flat on the floor.
 * 2) Hold hand weights straight down at your sides, with palms facing inward.
 * 3) Slowly bend one elbow, lifting weight toward chest and hold for 1 second. (Rotate palm to face shoulder while lifting weight.)
 * 4) Slowly lower arm to starting position. Pause. Repeat with other arm.
 * 5) Alternate arms until you have done 8 to 15 repetitions with each arm. Rest; then do another set of 8 to 15 alternating repetitions.

Knee Flexion Strengthens muscles in back of thigh. Use ankle weights, if you are ready.
 * 1) Stand straight holding onto a table or chair for balance and slowly bend knee as far as possible. Don’t move your upper leg at all; bend your knee only.
 * 2) Hold position for 1 second.  Slowly lower foot all the way back down. Pause.
 * 3) Repeat with other leg. Alternate legs until you have done 8 to 15 repetitions with each leg. Rest; then do another set of 8 to 15 alternating repetitions.

Hip Flexion
 * 1) Stand straight to the side or behind a chair or table, holding on for balance.
 * 2) Slowly bend one knee toward chest, without bending waist or hips.
 * 3) Hold position for 1 second.
 * 4) Slowly lower leg all the way down. Pause. Repeat with other leg. Alternate legs until you have done 8 to 15 repetitions with each leg. Rest; then do another set of 8 to 15 alternating repetitions.

Side Leg Raise Strengthens muscles at sides of hips and thighs. Use ankle weights, if you are ready.
 * 1) Stand straight, directly behind table or chair, feet slightly apart.
 * 2) Hold onto a table or chair for balance.
 * 3) Slowly lift one leg 6-12 inches out to side. Keep your back and both legs straight. Don’t point your toes outward; keep them facing forward.
 * 4) Hold position for 1 second.  Slowly lower leg. Pause. Repeat with other leg.
 * 5) Alternate legs until you have done 8 to 15 repetitions with each leg. Rest; then do another set of 8 to 15 alternating repetitions.

Stretching Exercises

Alternative Hamstrings Stretch Stretches muscles in the back of the thigh.
 * 1) Stand behind chair, holding the back of it with both hands.
 * 2) Bend forward from the hips (not waist), keeping back and shoulders straight at all times.
 * 3) When upper body is parallel to floor, hold position for 10 to 30 seconds. You should feel a stretch in the backs of your thighs.
 * 4) Repeat 3 to 5 times.

Ankles Stretches front ankle muscles. lean back, using pillows to support your back.
 * 1) Remove your shoes. Sit toward the front edge of a chair and
 * 1) Stretch legs out in front of you.
 * 2) With your heels still on the floor, bend ankles to point feet toward you. Bend ankles to point feet away from you.
 * 3) If you don’t feel the stretch, repeat with your feet slightly off the floor. Hold the position for 1 second. Repeat 3 to 5 times.

Triceps Stretch Stretches muscles in back of upper arm.
 * 1) Hold one end of a towel in right hand.
 * 2) Raise and bend right arm to drape towel down back. Keep your right arm in this position, and continue holding onto the towel.
 * 3) Reach behind your lower back and grasp bottom end of towel with left hand.
 * 4) Climb left hand progressively higher up towel, which also pulls your right arm down. Continue until your hands touch, or as close to that as you can comfortably go. Reverse positions. Repeat each position 3 to 5 times.

Shoulder Rotation Stretches shoulder muscles. back bothers you, place a rolled towel under your knees.
 * 1) Lie flat on floor, pillow under head, legs straight. If your
 * 1) Stretch arms straight out to side. Your shoulders and upper arms will remain flat on the floor throughout this exercise.
 * 2) Bend elbows so that your hands are pointing toward the ceiling. Let your arms slowly roll backwards from the elbow. Stop when you feel a stretch or slight discomfort, and stop immediately if you feel a pinching sensation or a sharp pain.
 * 3) Hold position for 10 to 30 seconds.
 * 4) Slowly raise your arms, still bent at the elbow, to point toward the ceiling again. Then let your arms slowly roll forward, remaining bent at the elbow, to point toward your hips. Stop when you feel a stretch or slight discomfort. Hold position for 10 to 30 seconds.
 * 5) Alternate pointing above head, then toward ceiling, then toward hips. Begin and end with pointing-above-head position. Repeat 3 to 5 times.

Single Hip Rotation Stretches muscles of pelvis and inner thigh. Don’t do this exercise if you have had a hip replacement, unless your surgeon approves.
 * 1) Lie on your back on floor, knees bent and feet flat on the floor.
 * 2) Keep shoulders on floor throughout exercise.
 * 3) Lower one knee slowly to side, keeping the other leg and your pelvis in place.
 * 4) Hold position for 10 to 30 seconds. Bring knee back up slowly. Repeat with other knee.  Repeat 3 to 5 times on each side.

	Citation for material discussed: Exercise: a guide from the national institute on aging. NIH Publication Number 01-4258. Reprinted April 2004

title: Physical exercises


text: File:Airfoil-geometry.png

title: File:Airfoil-geometry.png


text:
User:Eml4500.f08.group/Homework4/Matlab

= Matlab Code =

Results

Kg =

0.9000   0.5196   -0.9000   -0.5196         0         0         0         0    0.5196    0.3000   -0.5196   -0.3000         0         0         0         0   -0.9000   -0.5196    1.8000    0.4732   -0.6000    0.3464   -0.3000   -0.3000   -0.5196   -0.3000    0.4732    0.8000    0.3464   -0.2000   -0.3000   -0.3000         0         0   -0.6000    0.3464    0.6000   -0.3464         0         0         0         0    0.3464   -0.2000   -0.3464    0.2000         0         0         0         0   -0.3000   -0.3000         0         0    0.3000    0.3000         0         0   -0.3000   -0.3000         0         0    0.3000    0.3000

Fg =

-12.5672  -7.2557         0   30.0000   22.3866  -12.9249   -9.8194   -9.8194

Dg =

0        0  -11.6737   44.4051         0         0         0         0

title: User:Eml4500.f08.group/Homework4/Matlab


title: File:GroupFEA3bartrussjpg.jpg