text:
User talk:MarissaRicchetti

Marissa, Welcome to the Wikiversity! The link you posted on your project is quite amazing!
 * Stevenarntson 21:59, 15 October 2008 (UTC)

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

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

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

And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! --The Jade Knight (d'viser) 09:17, 30 October 2008 (UTC)

You may find Embroidery helpful. We currently have no project on Embroidery here at Wikiversity; consider sharing when you've learned more about it. Good luck! The Jade Knight (d'viser) 09:17, 30 October 2008 (UTC)

title: User talk:MarissaRicchetti


text:
User:Elizabethnguyen

Born and raised in Seattle, WA in March 14th. Attending the Art Institute of Seattle for Bachelor's in Fashion Design.

=Humanities I/The Commons=

Midterm Research Project
U District's Tubs, located on the corner of Northeast 50th Street and Roosevelt Way http://www.youtube.com/watch?v=Im1Wbkdp6eQ

Once known as Tubs, a place where people came to relax in hot tubs was assumed to be a sleezy place where people came to do drugs and have sex. But, it was actually a spa like it said it was. Rumors were possibly started because of the intimacy of each room with a hot tub along with tanning beds and a sauna. It was opened in 1982 and closed in 2007.

The graffiti was done by the Free Sheep Foundation and is unfortunately a temporary display of graffiti style murals covering the building. The graffiti itself was done by 20 artists and some of them have messages regarding what's going on in the world. When the murals were complete, there were some mixed reviews about it. Some people that live there believe that it will only encourage more graffiti artists to tag in the U District while others say that it's amazing and it will hopefully attract graffiti artist and keep them in one area. When the building is taken down, it will be replaced by a 60 unit apartment building with ground floor retail.

http://www.flickr.com/groups/1062435@N21/

http://www.flickr.com/photos/9586804@N08/3499264669/sizes/o/in/pool-1062435@N21/

http://www.flickr.com/photos/slightlynorth/3476313847/sizes/l/in/pool-1062435@N21/

http://www.flickr.com/photos/slightlynorth/3481047938/sizes/l/in/pool-1062435@N21/

http://www.flickr.com/photos/russ3ll/4041184096/sizes/l/in/pool-1062435@N21/

http://www.flickr.com/photos/liquidnight/3499498167/in/pool-1062435@N21

http://www.flickr.com/photos/allisonoconnor/3821115574/sizes/l/in/pool-1062435@N21/

http://www.flickr.com/photos/allisonoconnor/3820312119/sizes/l/in/pool-1062435@N21/

http://www.flickr.com/photos/scarequotes/3600018710/sizes/l/in/pool-1062435@N21/

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

title: User:Elizabethnguyen


text:
User:Mattwentz

My name is Matt Wentz

I am 21 years of age

I attend the Art Institute of Seattle in the Audio Production program

I enjoy music, movies, family, friends, my puppy, and snowboarding

My brithday is in January and mustard is fantastic

title: User:Mattwentz


text:
User talk:Mattwentz

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

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

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

And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! Jtneill - Talk - c 23:50, 15 October 2008 (UTC)

 Hi , I am Percy. Nice to meet you. I love the beach, swimming, fishing, and I am currently doing some research at Wikiversity on oceanography. I am also learning to play piano. What do you like? Can I be your mascot?

title: User talk:Mattwentz


text:
User:Eml4500.f08.ateam.boggs.t/week 7

Lecture 19

title: User:Eml4500.f08.ateam.boggs.t/week 7


text:
User:Eml4500.f08.ateam.boggs.t/Lecture 19

The goal is to find $$ \tilde{T}^{(e)}$$, a 4x4 matrix that transforms the set of local element degrees of freedom $$ d^{(e)} $$ a 4x1 matrix to another set of local, element, degrees of freedom $$ \tilde{d}^{(e)}$$ such that $$ \tilde{T}^{(e)}$$ is invertible.

$$ \tilde{d}^{(e)}$$ is the rotated degrees of freedom matrix corresponding to the $$\tilde{x}$$ coordinate system. The relationship is described below,

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

The components that make up the $$\tilde{d}^{(e)}$$ matrix are as follows,

 * $$ \tilde{d_{1}}^{(e)} =

\begin{bmatrix} l^{(e)} & m^{(e)} \end{bmatrix} \begin{bmatrix} d_{1}^{(e)}\\ d_{2}^{(e)} \end{bmatrix} $$

Here $$\tilde{d_{1}}^{(e)}$$ directly corresponds to the axial displacement $$q_{1}^{(e)}$$. And,

 * $$ \tilde{d_{2}}^{(e)} = \overrightarrow{d_{[1]}}^{(e)} \cdot \overrightarrow{\tilde{j}}$$

where $$ \tilde{d_{2}}^{(e)}$$ is a component of $$\overrightarrow{d_{[1]}}^{(e)}$$ along the unit vector $$\overrightarrow{\tilde{j}}$$, i.e., along the $$\tilde{y}$$-axis.

Here,

 * $$ \overrightarrow{i} \cdot \overrightarrow{\tilde{j}} = \cos{(\theta + \frac{\pi}{2})} = -\sin\theta$$

 * $$ \overrightarrow{j} \cdot \overrightarrow{\tilde{j}} = \cos\theta $$

This becomes,

 * $$\tilde{d_{2}}^{(e)} = -d_{1}^{(e)}\sin\theta + d_{2}^{(e)}\cos\theta$$

Where,

 * $$-\sin\theta^{(e)} = -m^{(e)}$$
 * $$ \cos\theta^{(e)} = l^{(e)}$$

So,

\tilde{d_{2}}^{(e)} = \begin{bmatrix} -m^{(e)} & l^{(e)} \end{bmatrix} \begin{bmatrix} d_{1}^{(e)}\\ d_{2}^{(e)} \end{bmatrix} $$

Combining these two equations equates to

\begin{bmatrix} \tilde{d_{1}}^{(e)}\\ \tilde{d_{2}}^{(e)} \end{bmatrix} = \begin{bmatrix} l^{(e)} & m^{(e)}\\ -m^{(e)} & l^{(e)} \end{bmatrix} \begin{bmatrix} d_{1}^{(e)}\\ d_{2}^{(e)} \end{bmatrix} $$

Here,

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

Looking at the entire element e and combining the local degrees of freedom one gets,

\begin{bmatrix} \tilde{d_{1}}^{(e)}\\ \tilde{d_{2}}^{(e)}\\ \tilde{d_{3}}^{(e)}\\ \tilde{d_{4}}^{(e)} \end{bmatrix} = \begin{bmatrix} l^{(e)} & m^{(e)} & 0 & 0\\ -m^{(e)} & l^{(e)} & 0 & 0\\ 0 & 0 & l^{(e)} & m^{(e)}\\ 0 & 0 & -m^{(e)} & l^{(e)} \end{bmatrix} \begin{bmatrix} d_{1}^{(e)}\\ d_{2}^{(e)}\\ d_{3}^{(e)}\\ d_{1}^{(e)} \end{bmatrix} $$

The first matrix is the local element degrees of freedom corresponding to the $$\tilde{x}$$ coordinate system, the second is the transformation matrix, and the third is the matrix of the local element degrees of freedom.

The transformation matrix can be broken down as follows,

\begin{bmatrix} l^{(e)} & m^{(e)} & 0 & 0\\ -m^{(e)} & l^{(e)} & 0 & 0\\ 0 & 0 & l^{(e)} & m^{(e)}\\ 0 & 0 & -m^{(e)} & l^{(e)} \end{bmatrix} = \begin{bmatrix} R^{(e)} & 0\\ 0 & R^{(e)} \end{bmatrix} $$

If the transverse displacement is taken away, the force-displacement relationship will look the same as in previous lectures.

 * $$ k^{(e)}

\begin{bmatrix} 1 & -1\\     -1 & 1 \end{bmatrix} \begin{bmatrix} q_{1}^{(e)}\\ q_{2}^{(e)} \end{bmatrix} = \begin{bmatrix} p_{1}^{(e)}\\ p_{2}^{(e)} \end{bmatrix} $$

Looking at the force-displacement relationship,

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

Here the axial displacements are present and the transverse displacements are taken to be zero because if one acts along the transverse the axial displacements are zero and the spring has no effect.

This now becomes

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

In this equation force is represented by a 4x1 matrix, the spring stiffness is represented by a 4x4 matrix, and the displcement is a 4x1 matrix.


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

HW 4
The code used to solve the two-bar truss system was altered to solve the three-bar truss system. The altered code is displayed below, followed by the functions required to solve the system, along with a diagram near the end portraying the deformation in the system. % Three bar truss example taken from 2 bar truss example % modified for three bars instead clear all; % Matrices e, A, and L represent Young's Modulus, cross sectional area, and % length of the three elements e = [2 4 3]; A = [3 1 2]; L=[5 5 10]; % Verical force exerted on node 2 P = 30; % Angles between each element and the horizontal alpha = pi/6; beta = pi/6; ceta = pi/4;  % note: "ceta" is not actually a greek letter % Coordinates of each of the 4 nodes calculated with trigonometry nodes = [0, 0; L(1)*cos(alpha), L(1)*sin(alpha); 2*L(1)*cos(alpha), 0; L(1)*cos(alpha)-L(3)*cos(ceta),L(1)*sin(alpha)-L(3)*sin(ceta)]; % Since the problem contains 4 nodes, there are 8 global degrees of freedom dof=2*length(nodes); % The connectivity matrix describes which nodes are connected to one % another by listing each row as an element and the global nodes it % connects conn=[1,2; 2,3; 2,4]; % The location matrix master array details which global degrees of freedom % are related to matching local degrees of freedom. lmm = [1, 2, 3, 4; 3, 4, 5, 6; 3, 4, 7, 8]; % The number of elements is determined by the size of the lmm matrix elems=size(lmm,1); % The K (square) and R (columnar) matrices are initialized with zeros 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 The above code uses the following functions obtained from the website highlighted in the Fundamental Finite Element Analysis and Applications textbook by Bhatti. function k = PlaneTrussElement(e, A, coord) % PlaneTrussElement(e, A, coord) % Generates stiffness matrix for a plane truss element % e = modulus of elasticity % A = area of cross-section % coord = coordinates at the element ends x1=coord(1,1); y1=coord(1,2); x2=coord(2,1); y2=coord(2,2); L=sqrt((x2-x1)^2+(y2-y1)^2); ls=(x2-x1)/L; ms=(y2-y1)/L; k = e*A/L*[ls^2, ls*ms,-ls^2,-ls*ms; ls*ms, ms^2,-ls*ms,-ms^2; -ls^2,-ls*ms,ls^2,ls*ms; -ls*ms,-ms^2,ls*ms,ms^2]; function [d, rf] = NodalSoln(K, R, debc, ebcVals) % [nd, rf] = NodalSoln(K, R, debc, ebcVals) % Computes nodal solution and reactions % K = global coefficient matrix % R = global right hand side vector % debc = list of degrees of freedom with specified values % ebcVals = specified values dof = length(R); df = setdiff(1:dof, debc); Kf = K(df, df); Rf = R(df) - K(df, debc)*ebcVals; dfVals = Kf\Rf; d = zeros(dof,1); d(debc) = ebcVals; d(df) = dfVals; rf = K(debc,:)*d - R(debc); function [eps, sigma, force] = PlaneTrussResults(e, A, coord, disps) % [eps, sigma, force] = PlaneTrussResults(e, A, coord, disps) % Compute plane truss element results % e = modulus of elasticity % A = Area of cross-section % coord = coordinates at the element ends % disps = displacements at element ends % The output quantities are eps = axial strain % sigma = axial stress and force = axial force. x1=coord(1,1); y1=coord(1,2); x2=coord(2,1); y2=coord(2,2); L=sqrt((x2-x1)^2+(y2-y1)^2); ls=(x2-x1)/L; ms=(y2-y1)/L; T=[ls,ms,0,0; 0,0,ls,ms]; d = T*disps; eps= (d(2)-d(1))/L; sigma = e.*eps; force = sigma.*A; Results of the solved 3-bar truss problem are as follows: k_local = 1.2        -1.2         -1.2          1.2 k = 0.9     0.51962         -0.9     -0.51962      0.51962          0.3     -0.51962         -0.3         -0.9     -0.51962          0.9      0.51962     -0.51962         -0.3      0.51962          0.3 k_local = 0.8        -0.8         -0.8          0.8 k = 0.6    -0.34641         -0.6      0.34641     -0.34641          0.2      0.34641         -0.2         -0.6      0.34641          0.6     -0.34641      0.34641         -0.2     -0.34641          0.2 k_local = 0.6        -0.6         -0.6          0.6 k = 0.3         0.3         -0.3         -0.3          0.3          0.3         -0.3         -0.3         -0.3         -0.3          0.3          0.3         -0.3         -0.3          0.3          0.3 K = 0.9     0.51962         -0.9     -0.51962            0            0            0            0      0.51962          0.3     -0.51962         -0.3            0            0            0            0         -0.9     -0.51962          1.8      0.47321         -0.6      0.34641         -0.3         -0.3     -0.51962         -0.3      0.47321          0.8      0.34641         -0.2         -0.3         -0.3            0            0         -0.6      0.34641          0.6     -0.34641            0            0            0            0      0.34641         -0.2     -0.34641          0.2            0            0            0            0         -0.3         -0.3            0            0          0.3          0.3            0            0         -0.3         -0.3            0            0          0.3          0.3 R = 0    0     0    30     0     0     0     0 d = 0           0      -11.674       44.405            0            0            0            0 reactions = -12.567     -7.2557       22.387      -12.925      -9.8194      -9.8194 results = 2.4186      6.4625       2.3145

Displaying the three-bar truss problem can be done with the code written below. clear; close; % The 3-bar truss problem has 8 connection points between lines that will % be displayed n_node = 8; % there are 9 dotted lines to be displayed, showing the 3-bar truss system % before deformation and showing the lines AB, AC, AD n_elem = 9; % P values calculated from the results above based on the equation % P = sqrt[(f1_1)^2 + (f2_1)^2] P2_1 = 14.512; P1_2 = 25.849; P1_3 = 13.88676; k_1 = 1.2; k_2 = 0.8; k_3 = 0.6; % Deformation in each bar calculated by P(e)/k(e) AB = P2_1/k_1; AC = P1_2/k_2; AD = P1_3/k_3; % Positions of all 8 nodes based on the origin at global node 4 position(:,1) = [10*cos(pi/4)-5*cos(pi/6); 10*sin(pi/4)-5*sin(pi/6); 0];  % global node 1 position(:,2) = [10*cos(pi/4); 10*sin(pi/4); 0];    % Point A, global node 2 position(:,3) = [10*cos(pi/4)+5*cos(pi/6); 10*sin(pi/4)-5*sin(pi/6); 0];     % global node 3 position(:,4) = [0; 0; 0]; % global node 4 position(:,5) = [10*cos(pi/4)+AB*cos(pi/6); 10*sin(pi/4)+AB*sin(pi/6); 0]; % point B position(:,6) = [10*cos(pi/4)-AC*cos(pi/6); 10*sin(pi/4)+AC*sin(pi/6); 0];  % point C position(:,7) = [10*cos(pi/4)+AD*cos(pi/4); 10*sin(pi/4)+AD*sin(pi/4); 0];  % point D position(:,8) = [-4.60; 51.48; 0];  % point E, or global node 2 after deformation % Creates x,y,z coordinates for the position of each node above for i = 1 : n_node; x(i) = position(1,i); y(i) = position(2,i); z(i) = position(3,i); end % node_connect(local node number, element number) = global node number node_connect(1,1) = 1; node_connect(2,1) = 2; node_connect(1,2) = 2; node_connect(2,2) = 3; node_connect(1,3) = 2; node_connect(2,3) = 4; node_connect(1,4) = 2; node_connect(2,4) = 5; node_connect(1,5) = 2; node_connect(2,5) = 6; node_connect(1,6) = 2; node_connect(2,6) = 7; node_connect(1,7) = 5; node_connect(2,7) = 8; node_connect(1,8) = 6; node_connect(2,8) = 8; node_connect(1,9) = 7; node_connect(2,9) = 8; % Plotting a dotted line between nodes for i = 1 : n_elem; node_1 = node_connect(1,i); node_2 = node_connect(2,i); xx = [x(node_1),x(node_2)]; yy = [y(node_1),y(node_2)]; zz = [z(node_1),z(node_2)]; plot3(xx,yy,zz,':') hold on end % Plotting a solid line to represent the deformed 3-bar truss system % Element 1 node_1 = 1; node_2 = 8; xx = [x(node_1),x(node_2)]; yy = [y(node_1),y(node_2)]; zz = [z(node_1),z(node_2)]; % Element 2 node_1 = 3; node_2 = 8; xxx = [x(node_1),x(node_2)]; yyy = [y(node_1),y(node_2)]; zzz = [z(node_1),z(node_2)]; % Element 3 node_1 = 4; node_2 = 8; xxxx = [x(node_1),x(node_2)]; yyyy = [y(node_1),y(node_2)]; zzzz = [z(node_1),z(node_2)]; plot3(xx,yy,zz,'-') plot3(xxx,yyy,zzz,'-') plot3(xxxx,yyyy,zzzz,'-') title('Deformed and undeformed three-bar truss system') xlabel('x') ylabel('y') zlabel('z') view([0 0 1])     % xy plane view

''Note: this MATLAB code was created using the following codes as examples:

- The MATLAB script for analyzing the Two Bar Truss System was obtained from Professor Vu-Quoc's website http://clesm.mae.ufl.edu/~vql/courses/fead/2008.fall/codes/twoBarTrussEx.txt

- The MATLAB script for the plot of a crab-leg structure created by X.G. Tan (2003) and obtained from Professor Vu-Quoc's website http://apollo.mae.ufl.edu/~vql/courses/fead/2006.fall/codes/matlab_plot_example2.m

- The MATLAB script for the plot of an electric pylon (undeformed five-bar truss structure) created by Team MeshWorks (2003) and obtained from Professor Vu-Quoc's website http://apollo.mae.ufl.edu/~vql/courses/fead/2006.fall/code.web.pages/electric_pylon.m''

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


text:
User:Eas4200c.f08.aeris.guan/hw4

Derivation of the Twist Angle in the Case of a Uniform, Non-Circular Cross Sectional Bar with Varying Shear Flow
Regarding the previous lecture, once $$\theta$$ is found in terms of T, use $$T=GJ\theta$$ or $$J=\frac{T}{G\theta}$$ to find the torsional constant J.  For this homework assignment, the NACA 2415 airfoil, referring to lecture 16.2, is partitioned into three cells. The cells are separated by two walls of assigned thickness. To perform the Matlab analysis similar to the homework 3, more information must be provided in order to properly calculate the "average" areas $$\bar{A}_1, \bar{A}_2, and \bar{A}_3$$ corresponding to the three cells of the airfoil and along with the torque (not indicated but will be treated as a constant), thickness of the walls and shell, contour length and shear modulus will allow us to calculate the shear flow on the airfoil. First, one wall is .25 times the chord length away from the leading edge, and a second wall is .75 times the chord length away from the leading edge. The shear modulus is not needed because the equations used will cancel out these terms. There will be three unknowns $$q_1,q_2, and q_3$$ and using the three following equations, these unknowns will be solved for. $$T=2\sum_{i=1}^3{q_i\bar{A}_i}$$ $$\displaystyle \theta_1=\theta_2$$ $$ \displaystyle \theta_2=\theta_3$$ Then, the torsional constant J will be found by the equation mentioned above. As we have already finished deriving the follow equations, $$T=GJ\theta, T=2q\bar{A}$$ Now we will derive the twist angle equation for a uniform, non-circular cross sectional with an engineering (ad hoc) method. $$\theta=\frac{1}{2G\bar{A}}\oint\frac{q}{t}ds$$ Let's consider this uniform, non-circular cross sectional bar subject to a torque. The bottom two pictures describe this type of bar with infinitesimally small axial length dx and the cross section is shown in the (y,z) plane.  Displacement PP' is due to alpha, and approximating alpha as very small, we have the following expression for PP' and OP in a simple geometric triangular relation. $$\frac{PP'}{OP}=tan(\alpha)$$ Now, project the displacement PP' in the direction orthogonal to OP'. In other words, $$\displaystyle PP' = PP'' cos( \alpha)$$ $$\displaystyle PP=(OPtan(\alpha))cos(\alpha)=(OPcos(\alpha))tan(\alpha)=OPtan(\alpha)$$ Recall, that OP=R, the radial coordinate of the point P and thus $$\rho$$ must be the perpendicular distance from the origin to P which equals, in this case, OP. Knowing that, the following expression can be written. $$\displaystyle PP''=(Rcos(\alpha))tan(\alpha)=\rho \alpha$$ Here, PP" is the displacement of P in the direction "tangent" to the lateral surface of the bar. Strain is $$\gamma=\frac{PP}{dx}$$ where dx is the distance between very close planes on the bar. $$\gamma=\frac{PP}{dx}=\frac{\rho \alpha}{dx}=\rho \theta$$ Here theta is the rate of twist angle.

title: User:Eas4200c.f08.aeris.guan/hw4


title: File:Naca3jin1.JPG


text:
User:Eas4200c.f08.blue.a/Lecture 17

Cubature - a method used for integrating volumes by breaking up the object into cubes and adding the individual cubes to find the total volume. To make the method more accurate, use smaller cubes to better approximate the volume.

Squaring (quadrature) of the circle - wikipedia.

Algebraic sums of the areas as the airfoil is swept counterclockwise is the NACA airfoil area.

A = A1 + A2 = A1 - |A2|

NOTE: When writing the code plotting the NACA arifoil, we do not need if statements to calculate the area if the arifoil. As long as the cross product is taken between the 2 sweeping vectors, the signs will add up correctly to give the appropriate area.

How about using trapezoids to integrate??

The method of trapezoids is easier to integrate over the surface and gives a more precise answer than Riemann Sums however it has a distinct disadvantage over the method of quadrature. The disadvantage being, changes in curvature are not accurately taken into account. For example, a segment with a lot of curvature is not approximated very well with a straight line. By using the method of quadrature, as long as there are a large number of segments, the area can be approximated better. Not to mention, THE TRAPEZOID METHOD IS NOT AS ELEGANT AS THE METHOD OF QUADRATURE (TRIANGLES)!!!!!!!

Returning to the single cell airfoil:

Shear flow along the airfoil is constant: q = q1 = q2 = q3

The angular rate of twist is:

$$ \theta =\frac{1}{2GA}q\sum_{j=1}^3 \frac{l}{t} $$

$$ \theta =\frac{1}{2GA}q [\frac{\pi b}{t1} + \frac{a}{t2} + \frac{\sqrt{a^2 + b^2}}{t3}] $$

$$ \theta =(factor)q $$

Homework: Computing the factor for the rate of twist.

Max shear stress = tau_max

If tau_max = tau_all (which is given), and since q = T/2A then,

Tall = 2Atauall[min{t1,t2,t3}]

Homework: If tau_all = 100 Gpa, fint T_all.

title: User:Eas4200c.f08.blue.a/Lecture 17


text:
User:Countrymike/Masters Thesis

I am currently writing a Masters Thesis under the working title "Community Radio in New Zealand" at the Dept. of Film, Television, and Media Studies at the University of Auckland. I will be using this as a space to take notes, etc.

Thesis
/Literature Review/

Books
Renni, Ellie. (2006). Community Media: A Global Introduction. Lanham, Maryland, USA. Rowman and Littlefield.

Articles
Mouffe, Chantal (1992). Democratic Citizenship and the Political Community. /Mouffe Notes/

Web Pages
Communitarianism - a group of related but distinct philosophies opposing exalted forms of individualism while advocating phenomena such as civil society. Communitarianism emphasizes the need to balance individual rights and interests with that of the community as a whole, and that autonomous selves are shaped by the culture and values of the community. See also: http://plato.stanford.edu/entries/communitarianism/

title: User:Countrymike/Masters Thesis


text:
User:Countrymike/Masters Thesis/Mouffe Notes

Mouffe, Chantal (1992). Democratic Citizenship and the Political Community. In Dimensions of Radical Democracy.

Liberalism vs Civic Republicanism

Liberal view is that 'citizenship is the capacity for each person to form, revise and rationally pursue his/her definition of the good.'

Communitarians object to this on the grounds that it seems to privilege self-interest rather than a "natural"(?) desire to join with others to pursue common action towards a common good. Does not allow for the notion that community might constitute the identity of individuals; only allows for 'instrumental' community.

Communitarians then believe in the revival of civic republican view of politics that puts an emphasis on the notion of a public good, prior to the individuals desires and interests. Notion mostly replaced by Liberalism

Republicanism and Liberalism

Pluralism, individual liberty, separation of church and state, development of civil society are all constitutive of modern democracies. Require distinction between the private and the public domain, the realm of morality and the realm of politics. Mouffe suggests that a a modern democratic political community cannot be organized around a single idea of a common good. Individual liberty cannot be sacraficed. Communitarian critique of liberalism becomes dangerously conservative.

ideas of the 'common good' are argued by liberals to have totalitarian implications. These were enjoyed by pre-modern society and are 'nostalgic relics' which should be discarded.

Rawls argues that individual rights can't be sacraficed for the sake of general welfare, as is the case with Utilitarianism. Justice should impose restrictions on what 'goods' individuals are allowed to pursue and such principles need respect plurality of competing conceptions of the good.

modern democracy is characterized by absense of substantive common good.

"On the one side, liberalism's exclusive concern with individuals and their rights has not provided content and guidance for the exercise of those rights. This has led to the devaluation of civic action, of common concern, which has caused an increasing lack of social cohesion in democratic societies." Mouffe, p.230 <== GOOD QUOTE

what makes us citizens of liberal democratic society is not a idea of the good, but a set of political principles specific to that tradition: the principles of freedom and equality for all. There will always as well be competing interpretations of these principles.

"discursive surfaces" - political community, community, state, etc.

encounters between the "private" and the "public". <-- might work for interpretation of "music" in community radio. The projection of private likes/pleasures into the public arena.

radical democratic citizen concern with equality and liberty should inform actions in all areas of social life. No sphere is immune from those concerns.

private/public individual/citizen


text: Nanotechnology readings

title: Nanotechnology readings


text:
Introduction to US History/Foreign Exploration

INTRODUCTION TO US HISTORY LECTURER:GEOFFREY D. PLOURDE FOREIGN EXPLORATION

Previous Lecture / Course Page / Next Lecture Welcome back to the spectacular field of US History! Last week, we learned about the cultures present in America before the Europeans arrived. This week's lecture moves along in history and is on the exploration of America by Europe.

As was previously mentioned, Columbus was not the first explorer to visit the Americas. The honor goes to the Vikings. The Vikings are most commonly known for pillaging. They were excellent sailors and put these skills to use in seeking out villages to attack. In the process, they explored much of the world around them. Sometime around the year 1000 AD, a Viking named Leif Ericson left Greenland and sailed west with a crew. He discovered a green fruitful land that he named Vinland. Today this is known as Newfoundland in Canada. Due to internal conflict, the Vikings did not leave a lasting presence in America. In addition to the Vikings, the Chinese, Egyptians, Phoenicians and Romans are also claimed to have reached America. There is no proof of any of these groups reaching America at this time. This does not necessarily mean that they didn't reach America, just that no evidence has been discovered. Therefore the Vikings can safely be considered the first explorers to America for whom we have proof of their visits.

America became unknown until 1492. Europe needed resources to fund its wars, forcing it to look outside of its borders. Spain was desperate for a new route to the Indies and Christopher Columbus brought forth the ambitious and logical plan of sailing west to India. King Ferdinand and Queen Isabella financed three ships, the Nina, the Pinta, and the Santa Maria. When he landed on Cuba, he was able to acquire gold, reinforcing his belief that he had arrived in India. Needless to say, the Spanish sent more expeditions to America. In 1519, Hernan Cortes arrived in Mexico. Through superior firepower, he was able to conquer the Aztecs, giving the Spanish most of Mexico. From 1522 to 1532, Francisco Pizarro fought the inhabitants of South America. His territorial gains dramatically expanded the Spanish empire. From these Spanish possessions, the Spaniards extracted much gold. which assisted them with their plethora of wars in Europe.

While the Spanish were invading South America, the English were focusing on North America. They heard about the discovery of a new route to "India" and promptly attempted to send ships themselves. Operating under the English flag, John Cabot landed in Canada during 1497. Needless to say, the English and other European colonies sent more explorers. Unfortunately, there was not much gold in the areas that the English landed in. Instead they came across a variety of climates, some suitable for farming, others suitable for grazing. They were able to make this up by raiding Spanish galleons and disrupting the flow of gold and silver to Spain.

The lighter and more agile English ships allowed England to wrest naval dominance from Spain, an action that would have many important effects down the line. This concludes Lecture 2. Please work on the section from the US History Wikibook

When the Europeans and Indians met
Although horses benefitted the Indians and their livelihoods, it wasn't enough to stop conflict. The Europeans' biggest weapon was the diseases they brought, including smallpox, syphillis, and measles. The Tainos were wiped out because of diseases. Columbus paved the way for European expansion (including Hernan Cortez and Francisco Coronado) and significantly altered the livelihood of the Native Americans due to the epidemics that were taking place. As for the New World, motivation, glory and gold all were reasons to continue expansion. Gold was taken out of South America and shipped back to Spain, Catholicism was mingled with the natives, and "mestizos", byproduct of the Spanish settles and Mexican natives, were created.

Commerce
Mercantilism is an old philosophy which argued that there are only so many minerals in this world, so it is up to a nation to mine and attain those minerals to accumulate wealth. Nations do not trade freely with other nations and nations controlled trade in order to attain wealth.

Commodification turned every-day items for the Natives (chica (corn beer) and tobacco) into trading items.

Columbian Exchange
The Columbian Exchange are the exchanges that took place between the Old and New World. Cattle, diseases and wheat were sent to the Old World from the New World - and vice versa: tobacco, corn, potatoes and beans. Sugar was the most valuable item of the Columbian Exchange with tobacco coming second. Illnesses and ideas were spread. Shifts from a religious to a secular point of view were becoming normalized. "Mourning wars", where the natives kidnapped a European and "rebirthed" them into the identity of the dead native [due to illness], was practiced by the Iroquois.

First Settlements
The first settlements in the US were:

 * St. Augustine, Florida [1565]
 * Roanoke Island, North Carolina [1585]
 * Jamestown, VA [1607]
 * Plymouth, MA [1620]

The first major explorations were:

 * 1492 - Columbus claims West Indies for Spain
 * 1497 - John Cabot claims North America for England
 * 1513 - Vasco Balboa claimed everything that touched the Pacific Ocean for Spain
 * 1520 - The "Marching God" Hernan Cortez invades the Aztecs.
 * 1585 - Roanake Colony by the British disappears. John White, leader of the colony, had a granddaughter: Virginia Dare - the first white settler to be born on US soil. The colony disappeared when White returned to it in 1590.
 * 1607 - Successful British settlement named after King James I, becoming Jamestown.

England vs. Natives

 * England - Saw colonies as "business oppertunities", believed in land ownership, and wanted to make investors back home filthy rich. They also wanted to use North America as a platform to spy on the Spanish.
 * Natives - Semi-normadic, hunter and gatherers; took care of land for next gen., lost most wars

Impact of Elizabeth I

 * Gave Sir Walter Raleigh the go to establish a colony (NC) in the 1500s.
 * Wanted England to come out supreme in terms of wealth and reputation and hated the Spanish

Why come to America?
Many reasons were out there for Europeans to come to America. There was a massive population boom in the colonies that attracted viewers. The law of primogeniture, where a person passes their property to their first legitimate son once they die, was active in the colonies as land was abundant. Fame, fortune, religious freedom, lack of social classes and economic improvement were other reasons for Europeans to travel to the New World. Various people from all backgrounds came to the New World.

Other oppertunities for children to gain wealth were to become a soldier, become a religious figure or learn a trade in an apprenticeship.

Tasks

 * Joint-Stock Company - investors, not the royal family, bankroll the colony.
 * Looking for Gold, not God.
 * Needed to find a route to the West Indies/China, find the missing Roanake Colony and spy on the Spanish.

Not Easy!

 * May 1607: 107 colonists go down to 34 after their first winter.
 * 1609: Starving time, 400 colonists --> 60 colonists.
 * 1625: 8k colonists arrived, but only half survived.

Chaos

 * The Indians were friendly at first, but Captain John Smith was a poor leader. In 1610, the colony was abandoned - but Lord Del La Ware convinced the colonists to come back to Jamestown and declared war on the Indians.
 * Pocahontas married John Rolfe in 1614 in order to solidify peace between the Indians and the English. She eventually converted to Christianity and moved to London, where she eventually died of disease. Rolfe came back to Jamestown with only words and no proof, leading to massive, violent clashes. The Indian massacre of 1622 was proof of these deteriorating relationships.

1619
In Jamestown, the House of Burgesses was created as a form of local government, tobacco became the crop of trade, indentured servants and slaves were presented as a means of labor and women were sold for 125lb of tobacco. After many years of no set regulation, a local government was set in place for many reasons. The distance to get to and from England was about 2 months, four months of total traveling time - thereby failing to give fair, speedy decisions. Witnesses weren't of much value if traveling distances was this long. Finally, life in Jamestown was different from life in England. The penalty for stealing a horse is severely different due to each land's circumstances.

Tobacco, with major cultivations taking place by Rolfe from 1612-1618, was a good and bad thing. It was a great tool for trade and gave the colony a financial standing, but it required great labor which promoted the New World's slavery culture. Thanks to a Dutch ship that came by Jamestown in 1619 selling slaves, people, some reluctantly, accepted the need of permanent human labor - soon becoming a port for slavery. Indentured servants were people who worked on whatever task for a master for about 7 years in repayment for bringing them to the colonies and sheltering/feeding them.

James I wasn't happy with this and eventually declared Virginia a royal colony after revoking their Company's charter. This was because the government stood by principles that he went against, such as slavery, tobacco and "abuse" of his royal power.


text:
Third Reich

HITLER'S THIRD REICH
Course Description This course will allow the student to examine the various facets of fascism in Nazi Germany. Topics will include: post-Treaty of Versailles, fall of the Weimar Republic, Rise of the Nazis, Nazi Germany, World War II, and the Fall of the Third Reich. Students will be expected to complete readings on a nightly basis, respond to them in the forms of discussion questions or essays, and tests/examinations.

Grading Scheme This course will follow the following system of weights and measures for grading:
 * 50%	 Tests (approximately four)
 * 25%	Writing assignments/essays (unlimited number)
 * 25%	Reading quizzes (LOWEST two scores will be dropped)

Attendance Since this is an online course, attendance records will ONLY be kept in the form of extra credit. Contributing to online discussions posted on CourseZone will allow the student to accumulate up to 5 POINTS worth of extra credit, which can be applied to tests and other examinations.

Syllabus

 * Week 1... Post-World War I Germany and the Lure of Fascism
 * Week 2... Hitler and the Early Nazi Party
 * Week 3... The Enabling Act and Complete Control over the Third Reich
 * Week 4-5... Development of the Nazi State (1933-1939)
 * Weeks 5-6... Dreams of an Aryan Empire
 * Weeks 7-8... The Holocaust
 * Week 9... German Involvement in World War II
 * Week 10... Fall of the Third Reich and Final Examination

FIRST ASSIGNMENT
Please read the following articles regarding German reparation payments, the Treaty of Versailles, and the weak Weimar Republic.

 * Treaty_of_Versailles Concerning the Treaty of Versailles
 * World_War_I_reparations Concerning German reparation payments
 * World_War_I_reparations the Weimar Republic

1. What was the ultimate goal of the Allied powers during negotiations held at the Paris Peace Conference?

2. What were the conditions of Article 231/the War Guilt Clause contained within the Treaty of Versailles?

3. How was the map of Europe essentially redrawn following changes made by the treaty?

4. Why was the French occupation of the Ruhr Valley a disaster? What prompted the French to occupy the region in the first place? What eventually happened to the value of the German mark?

5. How did the Dawes Plan attempt to reduce reparations and stabilize Germany's payment system?

6. What were the conditions of the Treaty of Locarno?

7. How did the withdraw of American funds from German banks lead to international banking faulures?

8. Describe a few of the difficulties the Weimar Republic faced during its early years.

title: Third Reich


text:
Portal:Preparatory Academy/Courses

Courses offered by Wikiversity that can assist you in high school will be linked here. Wherever possible, it is recommended that you take the most rigorous course possible.

Social Sciences

 * AP US History - Offered through the Department of North American History

title: Portal:Preparatory Academy/Courses


text:
User talk:Vishnumane

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

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

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

And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! --mikeu talk 11:17, 16 October 2008 (UTC)

title: User talk:Vishnumane


text:
Internet radio

What is Internet Radio?

Internet Radio lets you listen to the radio via the Internet. Dedicated device or computer software radios pick up stations and on-demand content which are streamed on the Internet via a broadband connection.

Internet radio can easily be broadcasted globally, with the exception of per basis blocks to certain regions or countries. The number of available Internet radio stations is always rising but at the last count there are over 10,000 stations you can listen to, meaning access to up-to-date news and music from all over the world. And, since lots of stations store their back catalog of shows, you can even listen on-demand if you miss something you love.

Open-source computer programs
Shell-fm is a text based internet radio player, that subscribes to Last-fm.

Box devices
Roku - can be hooked to a sound system The Intempo GX was a good example of an internet radio and has been hailed as one of the easiest Internet Radios to use.

Internet streamers
Tune-in has select local radio stations and popular podcasts. Last-fm is owned by cbs, and it contains a music stream of by most artists based on user choices. Yahoo music MOG - paid service but plays the users' choice of music

Red Bull Radio

NTS.live

Open-source
Using open source computer programs may facilitate the easiest way to create and broadcast an internet radio station.

title: Internet radio


text:
BCP/Chrysanthemum 'Sheffield Pink'

Chrysanthemum 'Sheffield Pink'

title: BCP/Chrysanthemum 'Sheffield Pink'


text:
BCP/Chrysanthemum 'Sheffield Pink'/Logs

2008 Logs

 * (no archived logs yet)

Recent Logs

 * --SB_Johnny talk 11:54, 16 October 2008 (UTC)
 * --SB_Johnny talk 10:45, 3 November 2008 (UTC)

title: BCP/Chrysanthemum 'Sheffield Pink'/Logs


text:
User:Fkatzdobler

My name is Franz-Josef Katzdobler and I'm currently studying Computer Science at the Vienna University of Technology.

title: User:Fkatzdobler


text:
User talk:Fkatzdobler

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

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

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

And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! Jtneill - Talk - c 12:59, 16 October 2008 (UTC)

 Woof! My name is Jack(Russell). I am a dog and a Wikiversity mascot. I am pretty new around Wikiversity. Perhaps we can learn together! Tail wag...

title: User talk:Fkatzdobler


text:
User:Eml4500.f08.ramrod.D/HW4

October 10th, 2008
Note: Consider case where $$\tilde{d}_{4}^{(e)} \ne 0$$

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

$$\tilde{\mathbf{f}}_{4x1}^{(e)}=\tilde{\mathbf{k}}_{4x4}^{(e)}\tilde{\mathbf{d}}_{4x1}^{(e)}=\mathbf{0}_{4x1}$$

The $$\mathbf{0}_{4x1}$$ is the 4th column of $$\mathbf{k}^{(e)}$$

Interpretation of the the transverse degrees of freedom

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

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

Also:

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

Therefore:

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

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

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

$$\tilde{\mathbf{T}}^{(e)} $$ block diagonal matrix: $$\begin{bmatrix} \mathbf{R}_{2x2}^{(e)} & \mathbf{0}_{2x2} \\ \mathbf{0}_{2x2} & \mathbf{R}_{2x2}^{(e)} \end{bmatrix}$$

Consider a general block diagonal matrix: $$ \mathbf{A}=\begin{bmatrix} \mathbf{D}_{1}& \mathbf{0} \\ \mathbf{0} & \mathbf{D}_{s} \end{bmatrix}$$

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

Here is a simple example

$$ \mathbf{B}=\begin{bmatrix} \mathbf{d}_{11}& \mathbf{0}& \mathbf{0} \\ \mathbf{0} & \mathbf{d}_{22} & \mathbf{0} \\ \mathbf{0} & \mathbf{0} & \mathbf{d}_{nn} \end{bmatrix}$$

$$\mathbf{B}=Diag[d_{11}, d_{22}, ..., d_{nn}]$$ (This is the shortened notation for $$ \mathbf{B} $$)

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

Assumming $$d_{ii}\ne 0$$ for $$ i=1, ..., n$$

For a block diagonal matrix A:

$$\mathbf{A}=Diag[d_{1} ..., d_{s}]$$

$$\mathbf{A}^{-1}=Diag[d_{1}^{-1} ..., d_{s}^{-1}]$$

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

We know:

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

Also

$$\mathbf{R}_{2x2}^{(e)}\mathbf{R}_{2x2}^{(e)T}=\begin{bmatrix} 1& 0 \\ 0 & 1 \end{bmatrix}_{2x2}=\mathbf{I}_{2x2}$$ (This is the Identity matrix)

In conclusion: $$\mathbf{R}^{(e)-1}=\mathbf{R}^{(e)T}$$

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

This shows that: $$\tilde{\mathbf{T}}^{(e)-1}=\tilde{\mathbf{T}}^{(e)T}$$

5 Truss MATLAB code
The 5 bar truss system in example 1.4 in the book can be solved by the below MATLAB code. analysis and explanation of the MATLAB code is shown below. This code and the the functions were downloaded from the books accompanying website:

http://bcs.wiley.com/he-bcs/Books?action=index&itemId=0471648086&bcsId=2256

Functions
The functions PlaneTrussElement.m, NodalSoln.m, PlaneTrussResults.m were all called in the code for the 5 truss system. They are all necessary for the completion of the truss problem.

PlaneTrussElement.m

This function take the Young's Modulus (e), the Area of the element (A) and the coordinates of the element ends (coord) and generates the stiffness matrix for a plane truss element. The function first calculates the lengths of the the elements. Once those are calculated the stiffness matrix is calculated.

NodalSoln.m

The function takes the global coefficient matrix (K), the global right hand side vector (R), the list of degrees of freedom with specified values, and the specified values to determine the displacements and reactions at each node. The dof of the the system is first found by using the  command which finds the longest dimension of R. df then finds the difference between the dofs that have known values (a value of zero) and the dof that were found in the previous line. The displacements and the reactions are then calculated.

PlaneTrussResults.m

This function computes the plane truss element results. It takes in the modulus of elasticity (e), the area of the cross-section (A), the coordinates at the element ends (coord), and the displacements at the elemtent ends (disps) and calculates the axial strain (eps), stress(sigma) and force (force). The results are stored in the variable  and sent back to the main program.

Results from the 5 Truss MATLAB example
The results from the 5 Truss System MATLAB code is shown below. As stated in the PlaneTrussResults.m function the  columns go in the order of axial strain, axial stress, and last axial force.

Code for the plot of the undeformed and deformed five truss system
Below is the code for the plot of the five bar system. Using the results that were obtained from the above MATLAB solution the deformed plot was created. The dotted lines are the undeformed system and the solid lines are the deformed system. The the displacements were multiplied by a factor of 500 so that they would appear on the plot.

Three Bar Truss System
A program was also written to solve the three bar truss system. The code follows the model code that was given to us in HW 2. The functions PlaneTrussElement.m, NodalSoln.m, and PlaneTrussResults.m were once again used in the code to obtain the results. Descriptions of these functions can be found in the section for the Five bar truss system.

Below are the results from the above code.

Plot of undeformed and deformed three bar truss system
Below is the code and the plot of the three bar truss system. The plot was made from the results that were found in the above code. The results demonstrate that only node two had a displacement of (-15.167, 48.231). The undeformed system is plotted with the dotted lines and the deformed system is plotted with solid lines. The deformed displacements have been scaled down by a tenth in order to enhance in the plot.

title: User:Eml4500.f08.ramrod.D/HW4


text:
User:Tinucherian

I am usually active in English Wikipedia Tinucherian 15:32, 16 October 2008 (UTC)

title: User:Tinucherian


text:
User talk:Tinucherian

I am usually active in English Wikipedia Tinucherian 15:32, 16 October 2008 (UTC)

title: User talk:Tinucherian


text:
School:History/Reorg

Welcome to the School of History!

A complete listing of history-related projects may be found at the History Portal.

title: School:History/Reorg


text:
Template:History/Reorg/Departments

General Divisions and Departments:
 * African History
 * Chinese History  [ Help out! ]
 * Diplomatic History
 * East Asian History
 * European History [ Help out! ]
 * Historiography [ Help out! ]
 * Indic History
 * Medieval History
 * North American History [ Help out! ]
 * History of Racism
 * Southeast Asian History

Integrated Divisions and Departments:
 * Board Game Studies

title: Template:History/Reorg/Departments


text:
School:History/Reorg/Introduction

Are you new to Wikiversity or the School of History, and wondering what you can do to get involved or help out? We keep a special page of things that need to be done for people who just want to jump in!
 * Get involved!

Have any questions? Leave them on the School's talk page, or contact Michael-Forest with your question.

title: School:History/Reorg/Introduction


text:
User:Daniela Widdmann

{| cellspacing="0" cellpadding="0" style="margin:0em 0em 1em 0em; width:100%" Daniela Widdmann
 * colspan="1" style="width:100%; vertical-align:top; border:1px solid Sienna; background-color:Cornsilk;" |

About me

 * Hello everybody!!

My name is Daniela Widdmann and I live in Germany (most time in Eichstaett but at the weekend in Munich). I want to become a primary school teacher and also a school psychologist and in around 2 years my dream will come true :-)

KWL Chart

 * }

title: User:Daniela Widdmann


text:
User:Eml4500.f08.qwiki.bishop/Lecture 20

Note

 * Consider case: $$\tilde{d}^{(e)}_{4}\neq 0$$
 * $$\tilde{d}^{(e)}_{1}=\tilde{d}^{(e)}_{2}=\tilde{d}^{(3)}_{4}=0$$
 * $$\tilde{f}^{(e)}=\tilde{k}^{(e)}\tilde{d}^{(e)}$$=0
 * (4x1)=(4x4)(4x1)=(4x1)
 * The zero matrix represents the fourth column of the elemental stiffness matrix, k

T matrix application to the elemental Force matrix
\tilde{f}^{(e)}_{1}\\ \tilde{f}^{(e)}_{2} \end{Bmatrix}=\begin{bmatrix} l^{(e)} & m^{(e)} \\ -m_{(e)}&l^{(e)} \end{bmatrix}\begin{Bmatrix} {f}^{(e)}_{1}\\ {f}^{e}_{2}\\ \end{Bmatrix}=\textbf{R}^{(e)}\begin{Bmatrix} {f}^{(e)}_{1}\\ {f}^{(e)}_{2}\\ \end{Bmatrix}$$ \tilde{f}^{(e)}_{1}\\ \tilde{f}^{(e)}_{2}\\ \tilde{f}^{(e)}_{3}\\ \tilde{f}^{(e)}_{4} \end{Bmatrix}=\begin{bmatrix} \mathbf{R}^{(e)}_{2x2} & \mathbf{0}_{2x2}\\ \mathbf{0}_{2x2} & \mathbf{R}^{(e)}_{2x2} \end{bmatrix}\begin{Bmatrix} {f}^{(e)}_{1}\\ {f}^{(e)}_{2}\\ {f}^{(e)}_{3}\\ {f}^{(e)}_{4} \end{Bmatrix}\Rightarrow \tilde{\mathbf{f}}^{(e)}_{4x1}=\tilde{\mathbf{T}}^{(e)}_{4x4}\mathbf{f}^{(e)}_{4x1}$$
 * p.19-3, $$\tilde{d}^{(e)}=\tilde{T}^{(e)}d^{(e)}$$
 * Similarly, $$\tilde{f}^{(e)}=\tilde{T}^{(e)}f^{(e)}$$
 * $$\begin{Bmatrix}
 * $$\begin{Bmatrix}
 * The set of R and zero matrices can be represented by the T matrix.
 * If we look at the force-displacement relationship, we can apply the T matrix conversion between global and axial coordiantes.
 * $$\tilde{\mathbf{k}}^{(e)}\tilde{\mathbf{d}}^{(e)}=\tilde{\mathbf{f}}^{(e)}\Rightarrow \tilde{\mathbf{k}}^{(e)}\tilde{\mathbf{T}}^{(e)}\mathbf{d}^{(e)}=\tilde{\mathbf{T}}^{(e)}\mathbf{f}^{(e)}$$
 * But if the T matrix can be inverted, then:
 * $$[\tilde{\mathbf{T}}^{(e) -1}\tilde{\mathbf{k}}^{(e)}\tilde{\mathbf{T}}^{(e)}]\mathbf{d}^{(e)}=\mathbf{f}^{(e)}$$

General block diagonal matrix
d_{11}&0 &0 \\ 0& d_{22}&0 \\ 0& 0 &d_{33} \end{bmatrix}=diag[d_{11}, d_{22}, d_{33}]$$ l^{(e)} & -m^{(e)} \\ m_{(e)}&l^{(e)} \\ \end{bmatrix}$$ 1 & 0 \\ 0&1 \\ \end{bmatrix}=\mathbf{I}$$
 * $$\tilde{\mathbf{T}}^{(e)}$$ is a block diagonal matrix.
 * To denote a block diagonal matrix in shorthand use the following notation:
 * $$\mathbf{B}=\begin{bmatrix}
 * The inverse of B is:
 * $$\mathbf{B^{(-1)}}=diag[\frac{1}{d_{11}},\frac{1}{d_{22}}, \frac{1}{d_{33}}]=diag[d_{11}^{(-1)}, d_{22}^{(-1)}, d_{33}^{(-1)}]$$
 * Assuming that no value of d is zero
 * In applying this to the T-matrix:
 * $$\tilde{\mathbf{T}}^{(e)-1}=diag[\mathbf{R}^{(e)-1},\mathbf{R}^{(e)-1}]$$
 * In reference to p.19-2:
 * $$\mathbf{R}^{(e)T}=\begin{bmatrix}
 * In order to prove that the transpose of R is equivalent to the inverse of R:
 * $$\mathbf{R}^{(e)T}\mathbf{R}^{(e)}=\begin{bmatrix}
 * So, $$\mathbf{R}^{(e)T}=\mathbf{R}^{(e)-1}$$
 * The inverse of the T-matrix can be written as:
 * $$\tilde{\mathbf{T}}^{(e)-1}=diag[\mathbf{R}^{(e)T},\mathbf{R}^{(e)T}]=\tilde{\mathbf{T}}^{(e)T}$$

Force-Displacement Relationship with T

 * $$[\tilde{\mathbf{T}}^{(e)T}\tilde{\mathbf{k}}^{(e)}\tilde{\mathbf{T}}^{(e)}]\mathbf{d}^{(e)}=\mathbf{f}^{(e)}$$
 * To show that $$[\tilde{\mathbf{T}}^{(e)T}\tilde{\mathbf{k}}^{(e)}\tilde{\mathbf{T}}^{(e)}]=\mathbf{k}^{(e)}$$
 * $$\mathbf{k}^{(e)}=\begin{bmatrix}

l^{(e)} &-m^{(e)} & 0 &0 \\ m^{(e)}& l^{(e)} & 0 &0 \\ 0 & 0 & l^{(e)}&-m^{(e)} \\ 0 & 0 & m^{(e)} & l^{(e)} \end{bmatrix}\begin{bmatrix} 1 & 0 &-1 &0 \\  0& 0 & 0 & 0\\  -1&0  &1  &0 \\  0&  0&  0& 0 \end{bmatrix}\begin{bmatrix} l^{(e)} &m^{(e)} & 0 &0 \\ -m^{(e)}& l^{(e)} & 0 &0 \\ 0 & 0 & l^{(e)}&m^{(e)} \\ 0 & 0 & -m^{(e)} & l^{(e)} \end{bmatrix}$$
 * $$\mathbf{k}^{(e)}=\begin{bmatrix}

l &0 &-l  & 0\\ m& 0& -m & 0\\ -l & 0& l & 0\\ -m& 0 & m&0 \end{bmatrix}\begin{bmatrix} l^{(e)} &m^{(e)} & 0 &0 \\ -m^{(e)}& l^{(e)} & 0 &0 \\ 0 & 0 & l^{(e)}&m^{(e)} \\ 0 & 0 & -m^{(e)} & l^{(e)} \end{bmatrix}$$
 * $$\mathbf{k}^{(e)}=\begin{bmatrix}

l^2 &lm &-l^2  & -ml\\ ml& m^2& -ml & -m^2\\ -l^2 & -lm& l^2 & lm\\ -ml& -m^2 & ml&m^2 \end{bmatrix}$$
 * So, the relationship between the elemental k matrix in global coordinates and the k matrix in axial coordinates using the T matrix is confirmed.


text:
User talk:Daniela Widdmann

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

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

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

And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! --mikeu talk 17:44, 16 October 2008 (UTC)

title: User talk:Daniela Widdmann


text:
User:Eas4200c.f08.aero.e/Week 7

=Monday=

The quadrature of a NACA airfoil is shown below.

The shear flow is q around the airfoil. A is the total area of the airfoil. This is found by subtracting the area under the bottom of the airfoil from P to Q, represented by the green A, and the area under the top airfoil from P to Q, represented by the blue A.

How to find the areas are shown below.

$$ d \bar{A} = \frac{1}{2}( \bar{r} \times \overline{PQ}) = -\left | dA \right \vert \bar{i} $$

A question was asked, "Why no use the trapezoidal method to integrate?"

The answer is, with the change in curvature that method has disadvantages and is less elegant.

Going back to the single cell airfoil.

The shear flow across the foil above is equal over the entire foil.

$$ q = q_1 = q_2 = q_3 $$

The angle of twist, θ, is represented by the below.

$$ \theta = \frac{1}{2G \bar{A}} q \sum_{j=1}^3 \frac{l_i}{t_j} $$

$$ \theta = \frac{1}{2G \bar{A}} (q) \left [ \frac{2 \pi (b/2)}{2t_1} + \frac{a}{t_2} + \frac{\sqrt{a^2 + b^2}}{t_3} \right ] $$

Knowing that

$$ a = 4m  \qquad\quad\quad   q = \frac{T}{2\bar{A}} $$ $$ b = 2m  \quad\quad\quad   \bar{A} = \frac{\pi (b/2)^2}{2} + \frac{1}{2}ba  $$ $$ t_1 = 0.08m  \qquad\quad\quad   t_2 = t_3 = 0.01m $$

we can find $$\theta$$ in terms of T and G.

$$ \theta = 7.141 \frac{T}{G} rad/m $$

Max shear stress - $$ \tau_{max} = \frac{q}{[min(t_1, t_2, t_3)]} = \frac{T}{t_{min}\bar{A}} $$.

If $$ \tau_{max} = \tau_{all}$$ (given) and since $$ q = \frac{T}{2 \bar{A}}$$, then $$ T_{all} = 2 \bar{A} \tau_{all} [min(t_1, t_2, t_3)] $$

Letting $$ \tau_{all} = 100 GPa$$, and knowing that t2 and t3 are the tmin,

$$ T_{all} = 2 \tau_{max} t_2 \bar{A} = 1.114 \times 10^10 Nm $$

=Wednesday= Multi Cell Airfoil (Sec. 3.6)

Recall:

$$ T = 2q\bar A $$

$$\theta = \frac{1}{G\bar A} \oint\frac{{q}_{i}}{{t}_{i}}ds $$

Specific example first (generalization later)

Compare this example with page 94 in textbook.

$$ T = 2\sum_{i=1}^1 {q}_{i} \bar {A}_{i}  $$

Find theta as a function of T and J(torsional constant)

$$T = {T}_{1} + {T}_{2} = 2{q}_{1}\bar {A}_{1} + 2{q}_{2} \bar {A}_{2} $$   (1)

$$\bar {A}_{1} = ac, \bar {A}_{1} = bc$$

$${\theta}_{1} = \frac{1}{G\bar {A}_{1}} \oint\frac{{q}_{1}}{{t}_{1}}ds $$

$$\oint\frac{{q}_{1}}{{t}_{1}}ds$$ q and t can be thought of as functions of (s), their curvilinear coordinates

therefore we can replace by discrete summation,

$${\theta}_{1} = \frac{1}{G\bar {A}_{1}} 2\frac{{q}_{1}a}{{t}_{1}} + \frac{{q}_{1}c}{{t}_{1}} + \frac{({q}_{1} - {q}_{2})c}{{t}_{12}}$$  (3)

$${\theta}_{2} = \frac{1}{G\bar {A}_{2}} 2\frac{{q}_{2}a}{{t}_{2}} + \frac{{q}_{2}c}{{t}_{2}} + \frac{({q}_{2} - {q}_{1})c}{{t}_{12}}$$  (4)

Since they are rigidly connect cells 1 and 2 must have the same rate of twist angle.

Therefore,

$${\theta}_{1} = {\theta}_{2}$$  (2)

Think of q1 and q2 as 2 unknowns. Equations (1) and (2) are the 2 equations for the 2 unknowns (q1 and q2) with T being a variable. Use (1) and (2) to find an expression for q1 and q2 in terms of T.

$${q}_{1} = {\beta}_{1}T {q}_{2} = {\beta}_{2}T$$

Next use the experssion (3) or (4) to find the expression between theta and T

$$\theta = {\theta}_{1} = {\theta}_{2} = \frac{T}{GJ} $$

Finally from this equation the torsional constant J can be deduced.

=Friday=

Information about 3-cell NACA 2415

The airfoil has two partition walls:

1st wall at ¼ C from leading edge.

2nd wall at ¾ C from leading edge.

3 Unknowns: q1, q2, q3.

1. $$T = 2\sum_{i=1}^{3}{}q_{i}\bar{A_{i}}$$

2.$$\theta _{1} = \theta _{2}$$

3.$$\theta _{2} = \theta _{3}$$

-

Now, we must recall from previous classes that we derived the equations:

$$T = GJ\theta $$ and $$T = 2q\bar{A}$$

Engineering (ad-hoc) derivation of:

$$\theta = \frac{1}{2gJ}\oint_{}^{}{}\frac{q}{t}ds$$

We refer to the next figure to carry out such derivation.

Then we find the displacement PP' due to α:

$$\frac{PP'}{OP} = (tan\alpha) \approx \alpha    $$  for small α

Then, we project the displacement of PP' on the direction orthogonal to OP'

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

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

And we know that:

$$OP''=(OP)(cos\alpha )$$

And we also must recall that:

$$OP = r$$ and $$OP'' = \rho $$

Then we get:

$$PP''= (rcos\alpha )(tan\alpha )$$

Where

$$\rho = (r)(cos\alpha )$$  and   $$\alpha \cong  (tan\alpha )$$

PP" = Displacement of point P in direction tangent to the lateral surface of the bar.

The the strain will be;

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

Finally, we must recall that:

Rate of twist angle = $$\theta = \frac{\alpha }{dx}$$ and we must notice that α is very small ad it can be denoted by dα

So, finally we obtain:

$$\theta = \frac{d\alpha }{dx}$$

(The next step is to integrate over the contour of the bar. Such derivation will be presented in the next day of homework)

title: User:Eas4200c.f08.aero.e/Week 7


text:
User:Eas4200c.f08.aero.e/Week 8

=Monday=

Hooke's Law: $$\tau (s)=G\gamma =G\rho \theta $$

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

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

If we now integrate along the contour $$\varsigma$$:

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

$$\oint_{\varsigma }^{}{}\rho (s)dA=2\bar{A}$$

$$\theta =\frac{d\alpha }{dx}$$

What is Ad hoc about the above expression $$(\theta )$$ and the derivative for $$T=2q\bar{A}$$ ?

1) Strain $$(\gamma)$$ must be obtained using the displacement of $$(\rho)$$ in the direction tangent to $$ (\varsigma)$$ at (roe),          but PP’’ is not necessarily tangent to $$(\varsigma)$$.  (However, it is close)

2) $$\tau=q/t$$ obtained from the ad-hoc assumption that $$\tau$$ was uniform across the wall thickness.

Now lets visit a formal justification by the Elasticity Theory:

A. Kinematic assumption.

$$U_{x}(y,z) = \theta \psi (y,z)$$

$$U_{z}(x,y) = \theta x Y$$

$$U_{y}(x,z) = \theta x Z$$

It is important to mention that there is a difference in notation with respect to the text Sun (2006). In order to Rectify the difference we change to the unified notation using cyclic permutation.

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

$$\varepsilon _{xx}=\frac{dU_{x}}{dx}=0$$

$$\gamma _{yz}=\frac{dU_{y}}{dz} + \frac{dU_{z}}{dy} = 0$$

$$\frac{dU_{y}}{dz} = -\theta x $$

$$\frac{dU_{z}}{dy} = \theta x $$

Homework assignment for this lecture was to perform the above operation for $$\varepsilon _{yy}$$ &  $$\varepsilon _{zz}$$

$$\varepsilon _{yy}=\frac{dU_{y}}{dy}=0$$

$$\varepsilon _{zz}=\frac{dU_{z}}{dz}=0$$

=Wednesday=

As a continuation of the previous lecture here is a third Ad Hoc assumption for the engineering derivative.

3) There is an inconsistancy in the assumption on the size of $$\alpha$$.               *To get PP' assume a small $$\alpha$$               *To get PP" assume $$\alpha$$ is infinite and $$\rho=r\cos\alpha$$                *Then reintroduce a small $$\alpha$$ again

The following question was posed in class: How many strain components exist in three dimensions?

answer = 6

$$\varepsilon = \begin{bmatrix} \varepsilon _{xx} & \varepsilon _{xy} & \varepsilon _{xz}\\ \varepsilon _{yx} & \varepsilon _{yy} & \varepsilon _{yz}\\ \varepsilon _{zx} & \varepsilon _{zy} & \varepsilon _{zz} \end{bmatrix} = \begin{bmatrix} \varepsilon _{11} & \varepsilon _{12} & \varepsilon _{13}\\ \varepsilon _{21} & \varepsilon _{22} & \varepsilon _{23}\\ \varepsilon _{31} & \varepsilon _{32} & \varepsilon _{33} \end{bmatrix}$$

The matrix transformation shown above was changed to indicial notation where x=1, y=2 and z=3.

$$\varepsilon = \left[\varepsilon _{ij} \right]$$ i,j = 1,2,3

* i = Row * j = Colum

tensorial rotation;

Symmetry of $$\varepsilon$$ : $$\varepsilon _{ij} = \frac{1}{2}(\frac{du_{i}}{dx_{j}} + \frac{du_{j}}{dx_{i}})$$

Coord.

$$x = x_{1}$$

$$y = x_{2}$$

$$z = x_{3}$$

$$\varepsilon _{11} = \varepsilon _{yx} = \frac{1}{2}(\frac{du_{1}}{dx_{1}} + \frac{du_{1}}{dx_{1}}) = \frac{du_{1}}{dx_{1}} = \frac{du_{x}}{dx} $$

$$\varepsilon _{12}=\varepsilon _{21} \Leftrightarrow \varepsilon _{xy}=\varepsilon _{yx}$$ *Only six independent components of $$\varepsilon$$ *Similarly for the stress tensor $$\sigma$$

$$\sigma =\left[\sigma _{ij} \right]_{3x3}$$ *Also only six independent components

$$\sigma _{xx}=\sigma _{yy}=\sigma _{zz}=\tau _{yz}=0$$ *Due to the stress strain relation *See page 70 in Sun(2006) *Isotropic elastic material

MatLab Guide

1) Set $$P_{o}=D$$, then the sweep area from B to L and then from L to E. (Area $$A_{1}$$)

1) Set $$P_{o}=E$$, then the sweep area from F to B. (Area $$A_{21}$$)

1) Set $$P_{o}=F$$, then the sweep area from E to H. (Area $$A_{22}$$)

1) Set $$P_{o}=G$$, then the sweep area from H to T and then from T to F. (Area $$A_{3}$$)

$$\bar{A}=\bar{A_{1}}+\bar{A_{21}}+\bar{A_{22}}+\bar{A_{3}}$$

=Friday=

Normal Strains:

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

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

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

 Shear Strains:

$$\gamma_{xy}=2\epsilon_{xy}=\frac{\tau_{xy}} {G}$$

$$\gamma_{yz}=2\epsilon_{yz}=\frac{\tau_{yz}} {G}$$

$$\gamma_{xz}=2\epsilon_{xz}=\frac{\tau_{xz}} {G}$$

Due to symmetry of strain tensor and stress tensor, we can write the relations in column matrix form:
 * $$\underline{\underline{\epsilon}} = \begin{bmatrix}

\epsilon_{11}\\ \epsilon_{22}\\ \epsilon_{33} \\ \epsilon_{23} \\ \epsilon_{31} \\ \epsilon_{12}\end{bmatrix}$$

 * $$\underline{\underline{\sigma}} = \begin{bmatrix}

\sigma_{11}\\ \sigma_{22}\\ \sigma_{33} \\ \sigma_{23} \\ \sigma_{31} \\ \sigma_{12}\end{bmatrix}$$

Hooke's Law for isotropic elasticity is as follows:

 * $$\underline{\underline{\epsilon}} = \begin{bmatrix}

\epsilon_{11}\\ \epsilon_{22}\\ \epsilon_{33} \\ \epsilon_{23} \\ \epsilon_{31} \\ \epsilon_{12}\end{bmatrix}=\begin{bmatrix} \frac{1}{E} & \frac{-\nu}{E} & \frac{-\nu}{E} & 0 & 0 & 0\\ \frac{-\nu}{E} & \frac{1}{E} & \frac{-\nu}{E} & 0 & 0 & 0\\ \frac{-\nu}{E} & \frac{-\nu}{E} & \frac{1}{E} & 0 & 0 & 0\\ 0 & 0 & 0 & \frac{1}{2G} & 0 & 0\\ 0 & 0 & 0 & 0 & \frac{1}{2G} & 0\\ 0 & 0 & 0 & 0 & 0 & \frac{1}{2G}\end{bmatrix}\begin{bmatrix} \sigma_{11}\\ \sigma_{22}\\ \sigma_{33} \\ \sigma_{23} \\ \sigma_{31} \\ \sigma_{12}\end{bmatrix}$$

Note: The Poisson ratio is determined by the type of material. The only material with a virtually zero Poisson ratio is cork, used for closing wine bottles. As the material is stretched or compressed in on direction, the other directions do not experience a strain. Most materials have Poisson ratios between 0 and 0.5. Steel has a Poisson ratio of 0.3, while rubber has a 0.5 Poisson ratio.

title: User:Eas4200c.f08.aero.e/Week 8


text:
User:Eml4500.f08.bottle.butler:Homework/Week 1

Finite Element Analysis is a method of solving partial differential equations (PDE’s) and integral equations by approximating solutions. The PDE’s and integrals are reduced into less complex equations and solved using common methods. It is typically used in engineering to perform complex computer analysis on materials and designs. The item being analyzed is broken down into a series of matrices and analyzed on a smaller scale. It was first applied to vibration systems in 1943 by R. Courant (http://www.sv.vt.edu/classes/MSE2094_NoteBook/97ClassProj/num/widas). Since then it has been used in a wide variety of fields and industries from manufacturing to biomedical research.

The Finite Element Analysis and Design (FEAD) class at the University of Florida, EML 4500, is offered on Monday, Wednesday, and Friday’s this fall. It meets from 3:00-3:50 in the Florida gym beginning on August 25th and ending on December 10th. The class is composed of lectures, group projects, and three exams. These exams are scheduled for Wednesday September 24th, Wednesday October 22nd, and Wednesday December 3rd. These dates are none negotiable and the policy is no make-up exams. A wide variety of material will be covered in class, including but not limited to trusses, beams, solid structures, linear stability, mixed algebra, and design analysis.

The book used in the current fall semester is Finite Element Analysis With Mathematica and Matlab Computations and Practical Applications: Fundamental Concepts by M. Asghar Bhatti and MATLAB software will be used extensively.

MATLAB is an integral part of this course. The first assignment for the class is to explore and attempt the University of Florida’s MATLAB primer tutorial, as seen below. It is imperative that students begin the class with a solid foundation in the software that will be used in later analyses.

A large portion of the course will be used to encourage group work and utilize cooperative learning techniques. Students will submit reports containing a summarization of the week’s lecture notes and a problem or two exemplifying the theories explained in lecture. Wikiversity software will be used to submit the assignment. This is advantageous to the student in several ways. First, it will allow the student to learn a new form of communication /idea sharing or allow a student already familiar with Wikiversity to gain a better understanding of the software. A second advantage is that is accessible from any place that has an internet connection. This allows students to contribute to the group project at any time and allows all group members to oversee the final report. It also allows the instructor to see the individual contributions of each group member and ensures that all members are participating.

To submit a report each group member must create an individual Wikipedia account. From there work can be uploaded to individual pages or the group page and can be accessed by all.


text:
User:Eml4500.f08.bottle.butler:Homework/2

9/3 Given: Truss with elastic bars When faces with a problem as presented above in Statics, the boundary conditions are removes and replaced with forces. Certain problems are left statically indeterminate. The flowing steps are used to solve a statically indeterminate problem using matrices.

First, draw a global free body diagram. The axis for the FBD are chosen based on the direction of the reaction forces. The FBD is then labeled using the following notation: 1. Nodes are numbered within circles 2. Elements are then numbered within triangles 3. Unknown reaction forces are labeled 4. The three equations of quality are written.

Then, a FBD for each element is drawn. The following notation is used: 1. Each node is labeled within a box 2. Element is numbered within a triangle 3. The reaction at each node is labeled in the following way a. f degree of freedomelement #, d degree of freedomelement # numbering begins with 1 on X axis and then moves to Y axis, then proceeds to next node.

Once the elements are labeled the force displacement relationship is written. An example used to clarify the method is a spring example where f=kd Forces are arranged in a matrix:

where k is the stiffness matrix

9/5 The big picture: Steps to solve a simple truss problem: 1. Global Picture (Description)

a. At structure level i.    Global degrees of freedom (displacement, dof’s) ii. Global forces

2. Displacement degrees of freedom are partitioned into

a. A known part (e.g. fixed dof, constraint/applied force) b.     An unknown part (solve using FEM/resultant forces)

3. Element picture

a. Element DOF’s b.     Element faces

4. Global Force Displacement relationship

a. Element stiffness matrices in global coordinates b.     Element force matrices in global c.     Assembly of element stiffness matrix and element force matrix in global FD rel i.    Kd=F

5. Elimination of known dof’s to reduce global relationship

a. Stiffness matrix is non singular

6. Compute forces from known d= element stress 7. compute reactions

Specific example:

Element Length L(1)= 4 L(2)=2 Youngs modulus E(1)= 3 E(2)=5

(1)=30 (2)=-45

Matrices: {F}=[k]{d}

~Remember to use convenient coord system and apply load to only one node.

k(e) = K

where l(e)= cosθ(e) and m(e)= sinθ(e)

9/10

The focus of the lesson is to continue with the computation of the simple truss problem presented in lecture 4. We know that θ(1) = 30°. Therefore, l(1)= lnθ(1)= cos 30° = And m(1) = sinθ(1) = sin30° = Also, K(1) = = =

Therfore, we can compute the stiffness matrix k(1), which is calculated in the following form:

K matrix with super and sub scripts

Using the numbers computed above we calculated k(1) to be the following:

Image of K stiffness matrix completed

When looking at the stiffness matrix we are able to observe that only 3 values need to be calculated: l(e)m(e), m(e), and l(e). The absolute values of these numbers form the stiffness matrix with only the sign changing. We are also able to observe that the matrix is symmetric, or the transpose of k(e) is equal to k(e).

We can use the same method described above to find the stiffness matrix for element 2. θ(2)= l(2)= cos – 45° =

m(2)= sin -45° =

A brief description of the force-displacement relationship:

Elemental FD: k(e)d(e)= f(e) d(e) = and f (e)=

Global FD: Kd = F

9/12

The global notation can be simplified into the form [Kij]{dj} = {Fi} where K= [Kij] is the global stiffness matrix, d= {dj} is the global displacement matrix and F= {Fi} is the global force matrix. The elemental notation can be simplified into the form k(e)d(e)=f (e), where k= [k(e)ij] is the element stiffness matrix, d= {d (e)j} is the element displacement matrix and f= {f (e)i} is the element force matrix.

One can move from the element to the global matrices through an assembly process. The correspondence between the element degrees of freedom and the global displacement degrees of freedom can be seen below for element 1 and 2:

For node 1: d1 = d(1)1 d2 = d(1)2

For node 2: d3 = d(1)3= d(2)1 d4= d(1)4= d(2)2

For node 3: d5 = d(3)4 d6 = d(2)4

When fully completed we can see that the topography of K resembles the figure below.

Picture of stiffness matrix with K1 and K2 blocks

9/15

Therefore, for example 1, K 6x6 =

Picture of K

Where K11= k(1)11 = ; K33= k(1)33 = k(2)11 = 3.0625; etc. At this step we can eliminate the known DOF’s. Because the truss example we are calculating is static, we know d1=d2=d5=d6= 0. When we apply these boundary conditions we can delete the corresponding columns in the stiffness matrix and by using the Principle of Virtual Work we can also eliminate the corresponding rows. We are left with the following equation:

Kd=


text:
User:Eml4500.f08.bottle.butler:Homework/3

For this homework assignment I typed all of the formulas in the notes in Matlab code and added to the notes with Jonathon Hipps. I have included the entire note section here in hopes that it is easier for the reader to follow.

Initially in the previous lectures, we derived an element force diagram with $$f \! $$ representing element force and $$ d \! $$ representing element displacement on a truss element. We replace these unknowns with $$ P \! $$ representing the element force and $$ q \! $$ representing the element displacement through the axis of the truss element.

With:

1.   $$q_i^{(e)}$$=axial displacement of element e at local node i   2.    $$P_i^{(e)}$$=axial force of element e at local node i

To derive $$ k \!$$ $$* \!$$$$d \!$$$$= \!$$$$f \!$$ from $$k \!$$$$* \!$$$$q \!$$$$=$$$$P \!$$, we need to find the relation between $$q\!$$ and $$d\!$$, and $$P\!$$ and $$f\!$$. For displacement, we derive that:

$$ q_{2X1}^{(e)}$$ $$ = \!$$ $$T_{2X4}^{(e)}$$ $$ * \!$$ $$d_{4X1}^{(e)}$$

In order to derive $$T \!$$, we must consider the displacement vector of the nodes in each truss element. Considering the previous truss system derived in the lecture notes, we derive vector $$ d \! $$ of node 1 on the axis x of element e, giving us:

$$q_1^{(e)}$$ $$=$$ $$d_1^{(e)}$$ $$ *$$ $$ \bar{i} $$ $$ = $$ $$d_1^{(e)}$$ $$ \tilde{i} $$ $$ + $$ $$ d_2^{(e)} \tilde {j} $$ $$ $$ = $$ $$d_1^{(e)}$$ $$ ($$ $$\tilde{i}$$$$*$$ $$\bar{i}$$$$)$$ $$ + $$ $$ d_2^{(e)}$$ $$ ( $$ $$ \tilde{j}$$ $$*$$$$ \bar{i}) $$ $$=$$$$ d_1^{(e)}$$$$*$$ $$\cos (\theta^{(e)}) \! $$ $$ + $$ $$ d_2^{(e)}$$$$ * $$ $$ \sin (\theta^{(e)}) \!$$ $$q_1^{(e)}$$$$= \!$$$$ d_1^{(e)}$$$$ * \!$$ $$ l^{(e)} \!$$ $$+ \!$$ $$d_2^{(e)} $$ $$ * \!$$ $$ m^{(e)} \!$$
 * $$ $$ \bar {i} $$

Homework Problem Using the same procedure as before for element 2 of the truss system gives us:

$$q^{(e)}_2= (d^{(e)}_3\bar{i}+d^{(e)}_4\bar{j}) \cdot \bar{\tilde{i}}$$

$$= d^{(e)}_3(\bar{i}\cdot \bar{\tilde{i}})+d^{(e)}_4(\bar{j} \cdot\bar{\tilde{i}}) $$

$$= d_3^{(e)}\cdot cos(\theta^{(e)})+d_4^{(e)}\cdot sin(\theta^{(e)})$$

$$q_2^{(e)}$$ $$ = \!$$ $$d_3^{(e)}$$$$ *\! $$$$ l^{(e)}\!$$ $$ + \!$$ $$d_4^{(e)}$$ $$ *\!$$ $$ m^{(e)}\!$$

Combining these two equations gives us the result for the $$ T \! $$ matrix:

$$ \begin{Bmatrix} q_1^{(e)} \\ q_2^{(e)} \end{Bmatrix} $$ $$ = \! $$$$ \begin{bmatrix} l^{(e)} & m^{(e)} & 0 & 0 \\  0  &  0  & l^{(e)} &  m^{(e)} \end{bmatrix} $$ $$ * \! $$ $$ \begin{Bmatrix} d_1^{(e)} \\d_2^{(e)} \\d_3^{(e)} \\ d_4^{(e)}\end{Bmatrix} $$

The same argument can also be applied to the axial force on the truss system, using the same procedure used to find the axial displacement of the truss system:

$$P^{(e)} \!$$ $$ = \!$$ $$T^{(e)}\!$$ $$* \!$$ $$f^{(e)}\!$$

Gives:

$$ \begin{Bmatrix} P_1^{(e)} \\ P_2^{(e)} \end{Bmatrix} $$ $$ = \! $$$$ T_{2X4}^{(e)} $$ $$ * \! $$ $$ \begin{Bmatrix} f_1^{(e)} \\f_2^{(e)} \\f_3^{(e)} \\ f_4^{(e)}\end{Bmatrix} $$

We can then replace the axial forces for the equations derive, representing the element forces:

$$k_{2X1}^{(e)}$$ $$*$$ $$ q_{2X1}^{(e)}$$$$=\!$$$$ P_{2x1}^{(e)}$$ and $$k^{(e)} \!$$$$ * \!$$ $$\underbrace{(T^{(e)}\! d^{(e)}) \!}_{q^{(e)}}=\!$$ $$ \underbrace{(T^{(e)} \!  f^{(e)})\!}_{P^{(e)}}$$ Using the inverse of $$ T \! $$, we can get the following solution for the element forces: $$ \begin{bmatrix} (T^{(e)})^{T}k^{(e)}I^{(e)} \end{bmatrix}*d^{(e)} = f^{(e)}$$ However, we cannot invert $$ T \!$$ with itself since it is a rectangular matrix with a two by two matrix $$ k \!$$; we instead use the Principal of Virtual Work (PVW) to find $$f_1 \! $$, $$ f_2 \!$$, $$ f_5 \!$$, and $$ f_6 \!$$. There is no need to find $$ f_3 \! $$ and $$ f_4 \!$$ since they are the applied load on the truss system, which is already known.

The following loop used for the two bar truss system to compare between the Finite Element Method (FEM) and the statics method went as follows, starting with the FEM method:

FEM Compute displacements Compute reactions

Then the loop goes back using the statics method:

Statics Compute reactions Compute displacements By using the statics method, the reactions $$ f_1 \! $$, $$ f_2 \! $$, $$ f_5 \! $$, and $$ f_6 \! $$ are known and with it, the forces $$ P \! $$ are then known. This knowledge can then be used to find the axial displacement $$ q \! $$ at each node on the truss system:

$$q_2^{(1)} $$ $$ =\frac{P_{2}^{(1)}}{k^{(1)}} \! $$$$ = \!$$ $$AC \!$$

$$q_1^{(1)} $$$$ = \!$$ $$0 \!$$ ,fixed node $$1 \!$$

$$q_1^{(2)}$$ $$ =\frac{P_{2}^{(2)}}{k^{(2)}} $$ $$= \!$$ $$ -AB \! $$

$$q_2^{(2)} $$$$=\! $$ $$0 \! $$ ,fixed node $$ 2 \! $$

Another method used to find the displacement of global node 2 of the two body truss system is to find the deformed shape of the truss system itself.

The first part is to find the deformed lengths of $$ AC \! $$ and $$ AB \! $$:

$$ AC = \! $$ $$ \frac{P_{2}^{(1)}}{k^{(1)}} = \frac{5.1243}{.75} = 6.8324$$

$$ AB = \! $$ $$\frac{P_{1}^{(2)}}{k^{(2)}}= \frac{6.2760}{5} = 1.255 \!$$

Then the x-y coordinates of points B and C are determined, leaving only two unknowns $$ x_D \! $$ and $$ y_D \! $$. To find this displacement, the formula used to determine said point is:

$$ x-x_{p}= (PQ)cos \theta \!$$ $$ y-y_{p}= (PQ)sin\theta \!$$

$$ y-y_{p}= tan(\theta + \frac{\pi }{2})(x-x_{p})$$

with $$ x_p \!$$ and $$ y_p \! $$ representing points B and C on the displacement diagram. However, we can simply use point A as our reference point since it is already on the point of origin on the x-y axis.

$$AD = (x_{d}-x_{a})\bar{i} + (y_{d}-y_{a})\hat{j} \!$$ $$AD = (x_{d} -0)\bar{i} + (y_{d} -0)\hat{j} \! $$

With the resulting definition becoming:

$$AD = d_{3}*\bar{i} + d_{4}*\hat{j} \! $$

This ultimately closes the loop for the two bar truss system. The same procedure can be used to find the displacement for a three bar truss system


text:
User:Eml4500.f08.bottle.butler:Homework/4

Our goal is to find $$ \tilde{T}^{(e)}_{4X4} $$ that transforms the set of local element dof's $$ d^{(e)}_{4X1} $$ to another set of local element $$ \tilde{d}^{(e)}_{4X1}$$ such that $$ \tilde{T}^{(e)} $$ is invertible.

image

From previous lectures we know that:

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

and $$ \tilde{d}_{2}^{(e)}= \vec{d}^{(e)}_{1}\circ \vec{\tilde{j}} $$ $$= -\sin \theta ^{(e)}d_{1}^{(e)}+\cos \theta ^{(e)}d_{2}^{(e)}$$

therefore:

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

If we combine the two we get the following equation in matrix form: $$ \begin{Bmatrix} \tilde{d}_{1}^{(e)}\\ \tilde{d}_{2}^{(e)} \end{Bmatrix} = \begin{bmatrix} l^{(e)} & m^{(e)}\\ -m^{(e)} & l^{(e)} \end{bmatrix}\begin{Bmatrix} d_{1}^{(e)}\\ d_{2}^{(e)} \end{Bmatrix} $$

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


text:
User:Eml4500.f08.bottle.butler/HW 1

Finite Element Analysis is a method of solving partial differential equations (PDE’s) and integral equations by approximating solutions. The PDE’s and integrals are reduced into less complex equations and solved using common methods. It is typically used in engineering to perform complex computer analysis on materials and designs. The item being analyzed is broken down into a series of matrices and analyzed on a smaller scale. It was first applied to vibration systems in 1943 by R. Courant (http://www.sv.vt.edu/classes/MSE2094_NoteBook/97ClassProj/num/widas). Since then it has been used in a wide variety of fields and industries from manufacturing to biomedical research.

The Finite Element Analysis and Design (FEAD) class at the University of Florida, EML 4500, is offered on Monday, Wednesday, and Friday’s this fall. It meets from 3:00-3:50 in the Florida gym beginning on August 25th and ending on December 10th. The class is composed of lectures, group projects, and three exams. These exams are scheduled for Wednesday September 24th, Wednesday October 22nd, and Wednesday December 3rd. These dates are none negotiable and the policy is no make-up exams. A wide variety of material will be covered in class, including but not limited to trusses, beams, solid structures, linear stability, mixed algebra, and design analysis.

The book used in the current fall semester is Finite Element Analysis With Mathematica and Matlab Computations and Practical Applications: Fundamental Concepts by M. Asghar Bhatti and MATLAB software will be used extensively.

MATLAB is an integral part of this course. The first assignment for the class is to explore and attempt the University of Florida’s MATLAB primer tutorial, as seen below. It is imperative that students begin the class with a solid foundation in the software that will be used in later analyses.

A large portion of the course will be used to encourage group work and utilize cooperative learning techniques. Students will submit reports containing a summarization of the week’s lecture notes and a problem or two exemplifying the theories explained in lecture. Wikiversity software will be used to submit the assignment. This is advantageous to the student in several ways. First, it will allow the student to learn a new form of communication /idea sharing or allow a student already familiar with Wikiversity to gain a better understanding of the software. A second advantage is that is accessible from any place that has an internet connection. This allows students to contribute to the group project at any time and allows all group members to oversee the final report. It also allows the instructor to see the individual contributions of each group member and ensures that all members are participating.

To submit a report each group member must create an individual Wikipedia account. From there work can be uploaded to individual pages or the group page and can be accessed by all.

title: User:Eml4500.f08.bottle.butler/HW 1


text:
User:Eml4500.f08.bottle.butler/HW 2

9/3 Given: Truss with elastic bars When faces with a problem as presented above in Statics, the boundary conditions are removes and replaced with forces. Certain problems are left statically indeterminate. The flowing steps are used to solve a statically indeterminate problem using matrices.

First, draw a global free body diagram. The axis for the FBD are chosen based on the direction of the reaction forces. The FBD is then labeled using the following notation: 1.	Nodes are numbered within circles 2.	Elements are then numbered within triangles 3.	Unknown reaction forces are labeled 4.	The three equations of quality are written.

Then, a FBD for each element is drawn. The following notation is used: 1.	Each node is labeled within a box 2.	Element is numbered within a triangle 3.	The reaction at each node is labeled in the following way a.	f degree of freedomelement #, d degree of freedomelement # numbering begins with 1 on X axis and then moves to Y axis, then proceeds to next node.

Once the elements are labeled the force displacement relationship is written. An example used to clarify the method is a spring example where f=kd Forces are arranged in a matrix: where k is the stiffness matrix

9/5 The big picture: Steps to solve a simple truss problem: 1.	Global Picture (Description) a.	At structure level i.	Global degrees of freedom (displacement, dof’s) ii. Global forces 2.	Displacement degrees of freedom are partitioned into a.	A known part (e.g. fixed dof, constraint/applied force) b.	An unknown part (solve using FEM/resultant forces) 3.	Element picture a.	Element DOF’s b.	Element faces

4.	Global Force Displacement relationship a.	Element stiffness matrices in global coordinates b.	Element force matrices in global c.	Assembly of element stiffness matrix and element force matrix in global FD rel i.	Kd=F 5.	Elimination of known dof’s to reduce global relationship a.	Stiffness matrix is non singular 6.	Compute forces from known d= element stress 7.	compute reactions

Specific example:

Element Length L(1)= 4 L(2)=2 Youngs modulus E(1)= 3 E(2)=5

(1)=30 (2)=-45

Matrices: {F}=[k]{d}

~Remember to use convenient coord system and apply load to only one node.

k(e) = K

where l(e)= cosθ(e) and m(e)= sinθ(e)

9/10

The focus of the lesson is to continue with the computation of the simple truss problem presented in lecture 4. We know that θ(1) = 30°. Therefore, l(1)= lnθ(1)= cos 30° = And m(1) = sinθ(1) = sin30° = Also, K(1) = =   =

Therfore, we can compute the stiffness matrix k(1), which is calculated in the following form:

K matrix with super and sub scripts

Using the numbers computed above we calculated k(1) to be the following:

Image of K stiffness matrix completed

When looking at the stiffness matrix we are able to observe that only 3 values need to be calculated: l(e)m(e), m(e), and l(e). The absolute values of these numbers form the stiffness matrix with only the sign changing. We are also able to observe that the matrix is symmetric, or the transpose of k(e) is equal to k(e).

We can use the same method described above to find the stiffness matrix for element 2. θ(2)=   l(2)= cos – 45° =

m(2)= sin -45° =

A brief description of the force-displacement relationship:

Elemental FD: k(e)d(e)= f(e) d(e) =  and f (e)=

Global FD: Kd = F

9/12

The global notation can be simplified into the form [Kij]{dj} = {Fi} where K= [Kij] is the global stiffness matrix, d= {dj} is the global displacement matrix and F= {Fi} is the global force matrix. The elemental notation can be simplified into the form k(e)d(e)=f (e), where k= [k(e)ij] is the element stiffness matrix, d= {d (e)j} is the element displacement matrix and f= {f (e)i} is the element force matrix.

One can move from the element to the global matrices through an assembly process. The correspondence between the element degrees of freedom and the global displacement degrees of freedom can be seen below for element 1 and 2:

For node 1: d1 = d(1)1 d2 = d(1)2

For node 2: d3 = d(1)3= d(2)1 d4= d(1)4= d(2)2

For node 3: d5 = d(3)4 d6 = d(2)4

When fully completed we can see that the topography of K resembles the figure below.

Picture of stiffness matrix with K1 and K2 blocks

9/15

Therefore, for example 1, K 6x6 =

Picture of K

Where K11= k(1)11 = ;  K33= k(1)33 = k(2)11 = 3.0625; etc. At this step we can eliminate the known DOF’s. Because the truss example we are calculating is static, we know d1=d2=d5=d6= 0. When we apply these boundary conditions we can delete the corresponding columns in the stiffness matrix and by using the Principle of Virtual Work we can also eliminate the corresponding rows. We are left with the following equation:

Kd=

title: User:Eml4500.f08.bottle.butler/HW 2


text:
User:Eml4500.f08.bottle.butler/HW 3

For this homework assignment I typed all of the formulas in the notes in Matlab code and added to the notes with Jonathon Hipps. I have included the entire note section here in hopes that it is easier for the reader to follow.

Initially in the previous lectures, we derived an element force diagram with $$f \! $$ representing element force and $$ d \! $$ representing element displacement on a truss element. We replace these unknowns with $$ P \! $$ representing the element force and $$ q \! $$ representing the element displacement through the axis of the truss element.

With:

1.   $$q_i^{(e)}$$=axial displacement of element e at local node i   2.    $$P_i^{(e)}$$=axial force of element e at local node i

To derive $$ k \!$$ $$* \!$$$$d \!$$$$= \!$$$$f \!$$ from $$k \!$$$$* \!$$$$q \!$$$$=$$$$P \!$$, we need to find the relation between $$q\!$$ and $$d\!$$, and $$P\!$$ and $$f\!$$. For displacement, we derive that:

$$ q_{2X1}^{(e)}$$ $$ = \!$$ $$T_{2X4}^{(e)}$$ $$ * \!$$ $$d_{4X1}^{(e)}$$

In order to derive $$T \!$$, we must consider the displacement vector of the nodes in each truss element. Considering the previous truss system derived in the lecture notes, we derive vector $$ d \! $$ of node 1 on the axis x of element e, giving us:

$$q_1^{(e)}$$ $$=$$ $$d_1^{(e)}$$ $$ *$$ $$ \bar{i} $$ $$ = $$ $$d_1^{(e)}$$ $$ \tilde{i} $$ $$ + $$ $$ d_2^{(e)} \tilde {j} $$ $$ $$ = $$ $$d_1^{(e)}$$ $$ ($$ $$\tilde{i}$$$$*$$ $$\bar{i}$$$$)$$ $$ + $$ $$ d_2^{(e)}$$ $$ ( $$ $$ \tilde{j}$$ $$*$$$$ \bar{i}) $$ $$=$$$$ d_1^{(e)}$$$$*$$ $$\cos (\theta^{(e)}) \! $$ $$ + $$ $$ d_2^{(e)}$$$$ * $$ $$ \sin (\theta^{(e)}) \!$$ $$q_1^{(e)}$$$$= \!$$$$ d_1^{(e)}$$$$ * \!$$ $$ l^{(e)} \!$$ $$+ \!$$ $$d_2^{(e)} $$ $$ * \!$$ $$ m^{(e)} \!$$
 * $$ $$ \bar {i} $$

Homework Problem Using the same procedure as before for element 2 of the truss system gives us:

$$q^{(e)}_2= (d^{(e)}_3\bar{i}+d^{(e)}_4\bar{j}) \cdot \bar{\tilde{i}}$$

$$= d^{(e)}_3(\bar{i}\cdot \bar{\tilde{i}})+d^{(e)}_4(\bar{j} \cdot\bar{\tilde{i}}) $$

$$= d_3^{(e)}\cdot cos(\theta^{(e)})+d_4^{(e)}\cdot sin(\theta^{(e)})$$

$$q_2^{(e)}$$ $$ = \!$$ $$d_3^{(e)}$$$$ *\! $$$$ l^{(e)}\!$$ $$ + \!$$ $$d_4^{(e)}$$ $$ *\!$$ $$ m^{(e)}\!$$

Combining these two equations gives us the result for the $$ T \! $$ matrix:

$$ \begin{Bmatrix} q_1^{(e)} \\ q_2^{(e)} \end{Bmatrix} $$ $$ = \! $$$$ \begin{bmatrix} l^{(e)} & m^{(e)} & 0 & 0 \\  0  &  0  & l^{(e)} &  m^{(e)} \end{bmatrix} $$ $$ * \! $$ $$ \begin{Bmatrix} d_1^{(e)} \\d_2^{(e)} \\d_3^{(e)} \\ d_4^{(e)}\end{Bmatrix} $$

The same argument can also be applied to the axial force on the truss system, using the same procedure used to find the axial displacement of the truss system:

$$P^{(e)} \!$$ $$ = \!$$ $$T^{(e)}\!$$ $$* \!$$ $$f^{(e)}\!$$

Gives:

$$ \begin{Bmatrix} P_1^{(e)} \\ P_2^{(e)} \end{Bmatrix} $$ $$ = \! $$$$ T_{2X4}^{(e)} $$ $$ * \! $$ $$ \begin{Bmatrix} f_1^{(e)} \\f_2^{(e)} \\f_3^{(e)} \\ f_4^{(e)}\end{Bmatrix} $$

We can then replace the axial forces for the equations derive, representing the element forces:

$$k_{2X1}^{(e)}$$ $$*$$ $$ q_{2X1}^{(e)}$$$$=\!$$$$ P_{2x1}^{(e)}$$ and $$k^{(e)} \!$$$$ * \!$$ $$\underbrace{(T^{(e)}\! d^{(e)}) \!}_{q^{(e)}}=\!$$ $$ \underbrace{(T^{(e)} \!  f^{(e)})\!}_{P^{(e)}}$$ Using the inverse of $$ T \! $$, we can get the following solution for the element forces: $$ \begin{bmatrix} (T^{(e)})^{T}k^{(e)}I^{(e)} \end{bmatrix}*d^{(e)} = f^{(e)}$$ However, we cannot invert $$ T \!$$ with itself since it is a rectangular matrix with a two by two matrix $$ k \!$$; we instead use the Principal of Virtual Work (PVW) to find $$f_1 \! $$, $$ f_2 \!$$, $$ f_5 \!$$, and $$ f_6 \!$$. There is no need to find $$ f_3 \! $$ and $$ f_4 \!$$ since they are the applied load on the truss system, which is already known.

The following loop used for the two bar truss system to compare between the Finite Element Method (FEM) and the statics method went as follows, starting with the FEM method:

FEM Compute displacements Compute reactions

Then the loop goes back using the statics method:

Statics Compute reactions Compute displacements By using the statics method, the reactions $$ f_1 \! $$, $$ f_2 \! $$, $$ f_5 \! $$, and $$ f_6 \! $$ are known and with it, the forces $$ P \! $$ are then known. This knowledge can then be used to find the axial displacement $$ q \! $$ at each node on the truss system:

$$q_2^{(1)} $$ $$ =\frac{P_{2}^{(1)}}{k^{(1)}} \! $$$$ = \!$$ $$AC \!$$

$$q_1^{(1)} $$$$ = \!$$ $$0 \!$$ ,fixed node $$1 \!$$

$$q_1^{(2)}$$ $$ =\frac{P_{2}^{(2)}}{k^{(2)}} $$ $$= \!$$ $$ -AB \! $$

$$q_2^{(2)} $$$$=\! $$ $$0 \! $$ ,fixed node $$ 2 \! $$

Another method used to find the displacement of global node 2 of the two body truss system is to find the deformed shape of the truss system itself.

The first part is to find the deformed lengths of $$ AC \! $$ and $$ AB \! $$:

$$ AC = \! $$ $$ \frac{P_{2}^{(1)}}{k^{(1)}} = \frac{5.1243}{.75} = 6.8324$$

$$ AB = \! $$ $$\frac{P_{1}^{(2)}}{k^{(2)}}= \frac{6.2760}{5} = 1.255 \!$$

Then the x-y coordinates of points B and C are determined, leaving only two unknowns $$ x_D \! $$ and $$ y_D \! $$. To find this displacement, the formula used to determine said point is:

$$ x-x_{p}= (PQ)cos \theta \!$$ $$ y-y_{p}= (PQ)sin\theta \!$$

$$ y-y_{p}= tan(\theta + \frac{\pi }{2})(x-x_{p})$$

with $$ x_p \!$$ and $$ y_p \! $$ representing points B and C on the displacement diagram. However, we can simply use point A as our reference point since it is already on the point of origin on the x-y axis.

$$AD = (x_{d}-x_{a})\bar{i} + (y_{d}-y_{a})\hat{j} \!$$ $$AD = (x_{d} -0)\bar{i} + (y_{d} -0)\hat{j} \! $$

With the resulting definition becoming:

$$AD = d_{3}*\bar{i} + d_{4}*\hat{j} \! $$

This ultimately closes the loop for the two bar truss system. The same procedure can be used to find the displacement for a three bar truss system

title: User:Eml4500.f08.bottle.butler/HW 3


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

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

Figure 1

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

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

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

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

Figure 2

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

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

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

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

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

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

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

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

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

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

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

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

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

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