text:
User:Eml4500.f08.Lulz.mitchell.bm

BEN IS FREAKING AWESWOME
Just kidding

HW1 Notes
8-25: Introduced website and went over syllabus.

8-27: Set the deadline for submitting the team structure for Wed, 3 Sep, 2008.

Teams should consist of 5 or 6 students. Teams are to submit a paper at the end of class on 3 Sep with the following information:
 * Team name; should be 4-6 characters
 * Member names and UF-ID numbers
 * Wiki usernames

8-29: The class namespace is announced as being Eml4500.fo8.

9-3: Began lecturing on "Trusses, Matrix Method" from chapter 4 of the book.

Notation:
 * Global Node Number - a number inside a circle; designates a node for the whole structure.
 * Local Node Number - a number inside a square; designates a node for a specific element.
 * Element Number - a number inside a triangle; designates a specific element of the entire structure.

Here we see a truss with 2 elastic, deformable bars

Here we see the global free body diagram for the truss system.

The R's represent unknown reaction forces. The P is a known applied force. This FBD has 4 unknowns and 3 equations of equilibrium which makes it statically indeterminate.

2 FBDs of 2 bar elements:

Notation: fi(e) Ex: f2(1) represents the 2nd internal force in the 1st element.
 * i - ith internal force
 * (e) - element number
 * displacements are written in the same manner: di(e)

Since corresponding forces and displacements act along the same line, they can be represented by the same arrow.

Ex: f1(2),d1(2)

The next big step is to come up with Force-Displacement (FD) relationships.

Below we see the FD relationship of a 1 dimensional spring element with one end fixed followed by the FD relationship for a 1 dimensional spring element with both ends free.

Note: An elastic bar inclined in a plane is a 2 dimensional spring.

The FD relationship for the system can be written in matrix form with forces and displacements arranged as column matrices. The k matrix is called the stiffness matrix.

$$\binom {f1}{f2}=\begin{vmatrix} k & -k\\ -k &k \end{vmatrix} \binom{d1}{d2}$$

Some special cases to consider:

Case 1: An observer sits on node 1.

The observer would move with node 1 and the spring would seem to stretch by d2-d1.

f2=k(d1-d2)

Case 2: An observer sits on node 2.

The observer would move with node 2.

f1=-f2=-k(d2-d1)

9-5: Reading assignment: Ch 4 (Trusses, Beams, Frame) Ch 1 (Big Picture): 1.1 Discretization 1.1.1 Plane Truss elements 1.2 Assembly of Element equations (Ex: 1.4 Five-bar trusses) 1.4 Element solution of model ralidity 1.4.1 Plane Truss elemnts

Steps to solve a simple truss:

1. global picture (description); at structure level
 * global dofs (disp. dofs); disp. unknowns
 * global forces; usually known except for reactions

disp. dofs are partitioned into:
 * a known part; e.g. fixed dofs, constraints
 * an unknown part; solved using FEM

similarly for global forces:
 * a known part; applied forces
 * an unknown part; reaction forces

2. element picture
 * element dofs (either local or global coordinates)
 * element forces (either local or global coordinates)

3. global force-displacement
 * element stiffness matrices in global coordinates
 * element force matrices in global coordinates
 * assembly of element stiffness matrices and element force matrices into global force-displacement relationship: kd=F
 * k is NxN;d is Nx1;F is Nx1
 * N = # of known and unknown displacement dofs
 * "free-free" system, unconstrained, singular, determinate is 0

4. elimination of known dops to reduce the global force-displacement relationship
 * stiffness matrix is non-singular; therefore invertible
 * can solve the system of equations - kd=F
 * only known dofs left
 * k is MxM; d is Mx1;F is Mx1
 * M = # of unknown displacement dofs
 * k is non-singular; therefore k-1 exists
 * therefore d=k-1F

5.compute element forces from now known d
 * gives element stresses

6.compute reactions (unknown forces)

Specific example (to see how the recipe works)

(pic)

Element Lenght:
 * L(1)=4
 * L(2)=2

Young's Modulus:
 * E(1)=3
 * E(2)=5

X-sectional Area:
 * A(1)=1
 * A(2)=2

Inclination Angle:
 * o (1)=30 deg
 * o (2)=-45 deg

1)Global Picture:

-global dofs

(pic)

-numbering the disp. dofs and global forces
 * follow order of nodes
 * for each node: label x then y

-it is possible to arrange global forces and disp. dofs in matrix form

(matrix)

HW 2 Notes
From where we left off.

The second step is to draw the element pictures. In these pictures you should label the element dofs and the element forces in either global or local coordinates.

(Pic)

The third step is to develop the global force-displacement relationship at the element level. As we saw earlier, for a specified element the force-displacement relationship is: $$\textbf{k}^{(e)}*\textbf{d}^{(e)}=\textbf{f}^{(e)}$$

Where k(e) is the element stiffness matrix, d(e) is the element displacement matrix and f(e) the element force matrix.

The axial stiffness of a specified element is determined by the equation:

$$k=\frac{E^{(e)}A^{(e)}}{L^{(e)}}$$

Where E is the Young's modulus for the element, A is the cross-sectional area for the element and L is the length if the element.

Now we need to find the director cosines (l(e) and m(e)) for each element using the following equations:

$$l^{(e)}=\textbf{i}^{\sim}\cdot\textbf{i}=cos\theta^{(e)}$$

$$m^{(e)}=\textbf{j}^{\sim}\cdot\textbf{j}=sin\theta^{(e)}$$

(pic)

Using the director cosines, axial stiffness and the following formula we can determine the element stiffness matrix.

$$\textbf{k}^{(e)}=k^{(e)}\left[\begin{array}{c c c c} (l^{(e)})^2 & l^{(e)}m^{(e)} & -(l^{(e)})^2 & -l^{(e)}m^{(e)}\\ l^{(e)}m^{(e)} & (m^{(e)})^2 & -l^{(e)}m^{(e)} & -(m^{(e)})^2\\ (l^{(e)})^2 & -l^{(e)}m^{(e)} & (l^{(e)})^2 & l^{(e)}m^{(e)}\\ -l^{(e)}m^{(e)} & -(m^{(e)})^2 & l^{(e)}m^{(e)} & (m^{(e)})^2 \end{array}\right ]$$

For element 1:

$$\theta=30$$

$$l=cos\theta=\frac{\sqrt{3}}2$$

$$m=sin\theta=\frac{1}2$$

$$k=\frac{EA}L=\frac{(3)(1)}4=\frac{3}4$$

This gives us the stiffness matrix for element 1:

$$k^{(1)}=\left[\begin{array}{c c c c} \frac{9}{16} & \frac{3\sqrt{3}}{16} & -\frac{9}{16} & -\frac{3\sqrt{3}}{16}\\ \frac{3\sqrt{3}}{16} & \frac{3}{16} & -\frac{3\sqrt{3}}{16} & -\frac{3}{16}\\ -\frac{9}{16} & -\frac{3\sqrt{3}}{16} & \frac{9}{16} & \frac{3\sqrt{3}}{16}\\ -\frac{3\sqrt{3}}{16} & -\frac{3}{16} & \frac{3\sqrt{3}}{16} & \frac{3}{16} \end{array}\right]$$

Notice that the stiffness matrix for element 1 is symmetric.

For element 2:

$$\theta=-45$$

$$l=cos(-45)=\frac{\sqrt{2}}2$$

$$m=sim(-45)=\frac{\sqrt{2}}2$$

$$k=\frac{EA}L=\frac{(5)(2)}2=5$$

This gives us the stiffness matrix for element 2:

$$k^{(2)}=\left[\begin{array}{c c c c} \frac{5}2 & -\frac{5}2 & \frac{5}2 & -\frac{5}2\\ -\frac{5}2 & \frac{5}2 & -\frac{5}2 & \frac{5}2\\ -\frac{5}2 & \frac{5}2 & -\frac{5}2 & \frac{5}2\\ \frac{5}2 & -\frac{5}2 & \frac{5}2 & -\frac{5}2 \end{array}\right]$$

Notice that the stiffness matrix for element 2 is also symmetric.

Now we have to assemble the global stiffness matrix. This is illustrated in the picture below where the shaded area represents the addition of overlapping parts of $$\textbf{k}^{(1)}$$ and $$\textbf{k}^{(2)}$$.

$$K_{33}=k_{33}^{(1)} + k_{11}^{(2)}$$

$$K_{34}=k_{34}^{(1)} + k_{12}^{(2)}$$

$$K_{43}=k_{43}^{(1)} + k_{21}^{(2)}$$

$$K_{44}=k_{44}^{(1)} + k_{22}^{(2)}$$

For this example we get:

$$\textbf{K}=\left[\begin{array}{c c c c c c} \frac{9}{16} & \frac{3\sqrt{3}}{16} & -\frac{9}{16} & -\frac{3\sqrt{3}}{16} & 0 & 0\\ \frac{3\sqrt{3}}{16} & \frac{3}{16} & -\frac{3\sqrt{3}}{16} & -\frac{3}{16} & 0 & 0\\ -\frac{9}{16} & -\frac{3\sqrt{3}}{16} & \frac{49}{16} & \frac{-40+3\sqrt{3}}{16} & -\frac{5}2 & -\frac{5}2\\ -\frac{3\sqrt{3}}{16} & -\frac{3}{16} & \frac{-40+3\sqrt{3}}{16} & \frac{43}{16} & -\frac{5}2 & -\frac{5}2\\ 0 & 0 & -\frac{5}2 & -\frac{5}2 & \frac{5}2 & \frac{5}2\\ 0 & 0 & -\frac{5}2 & -\frac{5}2 & \frac{5}2 & \frac{5}2 \end{array}\right]$$

Now on the fourth step we eliminate known dofs to reduce the global force-displacement relationship.

Due to the applied constraints, $$d_1=d_2=d_5=d_6=0$$. This allows us to eliminate columns 1,2,5 and 6 due to the principal of virtual work. We also note that $$F_1=F_2=F_3=F_5=F_6=0$$. This allows us to eliminate rows 1,2,5 and 6 by the same principal. The resulting, simplified force-displacement relationship follows:

$$\left[\begin{array}{c c} K_{33} & K_{34}\\ K_{34} & K_{44} \end{array}\right] \begin{Bmatrix} d_3\\ d_4 \end{Bmatrix}=\begin{Bmatrix} F_3\\ F_4 \end{Bmatrix}$$

Let $$F_3=0$$ and $$F_4=10$$.Now plugging in numbers we get:

$$\left[\begin{array}{c c} \frac{49}{16} & \frac{-40+3\sqrt{3}}{16}\\ \frac{-40+3\sqrt{3}}{16} & \frac{43}{16} \end{array}\right] \begin{Bmatrix} d_3\\ d_4 \end{Bmatrix}= \begin{Bmatrix} 0\\ 10 \end{Bmatrix}$$

Now we can solve for the unknown displacement dofs by multiplying both sides by:

$$\textbf{K}^{-1}=\frac{1}{det\textbf{K}} \left[\begin{array}{c c} K_{44} & -K_{34}\\ -K_{43} & K_{33} \end{array}\right]$$.

Plugging in numbers we get:

$$\textbf{K}^{-1}=\frac{1}{3.4988} \left[\begin{array}{c c} \frac{49}{16} & \frac{40-3\sqrt{3}}{16}\\ \frac{40-3\sqrt{3}}{16} & \frac{43}{16} \end{array}\right]$$

It is important to note that: $$\textbf{K}^{-1}\neq\frac{1}{det\textbf{K}}\textbf{K}^T$$

Following through with this, we get:

$$\begin{Bmatrix} d_3\\ d_4\end{Bmatrix}= \textbf{K}^{-1}\begin{Bmatrix} F_3\\ F_4\end{Bmatrix}$$

Plugging in the numbers now gives:

$$\begin{Bmatrix} d_3\\ d_4\end{Bmatrix}= \left[\begin{array} {c c} .8753 & .6217\\ .627 & .7681 \end{array}\right] \begin{Bmatrix} 0\\ 10\end{Bmatrix}$$

Solving this system gives us:

$$d_3=6.217$$

$$d_4=7.681$$

Now in step five we can compute reactions by one of two methods.

The first method is to use the element force-displacement relationship.

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

For element 1 we would have:

$$\left[\begin{array}{c c c c} \frac{9}{16} & \frac{3\sqrt{3}}{16} & -\frac{9}{16} & -\frac{3\sqrt{3}}{16}\\ \frac{3\sqrt{3}}{16} & \frac{3}{16} & -\frac{3\sqrt{3}}{16} & -\frac{3}{16}\\ -\frac{9}{16} & -\frac{3\sqrt{3}}{16} & \frac{9}{16} & \frac{3\sqrt{3}}{16}\\ -\frac{3\sqrt{3}}{16} & -\frac{3}{16} & \frac{3\sqrt{3}}{16} & \frac{3}{16} \end{array}\right]\begin{Bmatrix} 0\\ 0\\ 6.217\\ 7.681\end{Bmatrix}= \begin{Bmatrix} f_1^{(1)}\\ f_2^{(1)}\\ f_3^{(1)}\\ f_4^{(1)}\end{Bmatrix}$$

Solving this system we get:

$$f_1^{(1)}=5.992$$

$$f_2^{(1)}=3.459$$

$$f_3^{(1)}=-5.992$$

$$f_4^{(1)}=-3.459$$

For element 2 we have:

$$\left[\begin{array}{c c c c} \frac{5}2 & -\frac{5}2 & \frac{5}2 & -\frac{5}2\\ -\frac{5}2 & \frac{5}2 & -\frac{5}2 & \frac{5}2\\ -\frac{5}2 & \frac{5}2 & -\frac{5}2 & \frac{5}2\\ \frac{5}2 & -\frac{5}2 & -\frac{5}2 & \frac{5}2 \end{array}\right]\begin{Bmatrix} 6.217\\ 7.681\\ 0\\ 0\end{Bmatrix}= \begin{Bmatrix} f_1^{(2)}\\ f_2^{(2)}\\ f_3^{(2)}\\ f_4^{(2)}\end{Bmatrix}$$

Solving this system we get:

$$f_1^{(2)}=-3.660$$

$$f_2^{(2)}=3.660$$

$$f_3^{(2)}=3.660$$

$$f_4^{(2)}=-3.660$$

The second method is to solve by statics. Since this system is statically indeterminate we must use the Euler cut method.

(pic)

HW 3 Matlab, Problem, and Latex
Matlab-

% Matlab Script %***************************************************** % % Purpose: %  Sketch the undeformed and deformed %  two-bar truss system from the clas %  example problem. % % Author: Benjamin Mitchell % % Created: 10-07-2008 00:25:38 EST % %***************************************************** clear; close; dof = 6; %dof per node d1 = 0;        %disp of node 1 in the x direction d2 = 0;        %disp of node 1 in the y direction d3 = 4.3520;   %disp of node 2 in the x direction d4 = 6.1271;   %disp of node 2 in the y direction d5 = 0;        %disp of node 3 in the x direction d6 = 0;        %disp of node 3 in the y direction %input nodal coordinates num_nodes = 3;              %total number of nodes num_elems = 2;              %total number of elements tot_dofs = 2 * num_nodes;   %total dofs of the system %element 1 details L1 = 4; E1 = 3; A1 = 1; theta1 = 30; %deg thetar1 = (pi * theta1)/180; %element 2 details L2 = 2; E2 = 5; A2 = 2; theta2 = -45; %deg thetar2 = (pi * theta2)/180; %undeformed positions pos(1,:)=[0,L1*cos(thetar1),L1*cos(thetar1) + L2*cos(thetar2)]; pos(2,:)=[0,L1*sin(thetar1),0]; %endeformed nodal coordinate arrays for i=1:num_nodes x(i)=pos(1,i); y(i)=pos(2,i); end %element connectivity array %follows: %node_conn(local node number, elemnt number) = global node number node_conn(1,1)=1;    %element 1 node_conn(2,1)=2; node_conn(1,2)=2;    %element 2 node_conn(2,2)=3; %plot the undeformed 2-bar truss system for i=1:num_elems node_1=node_conn(1,i); node_2=node_conn(2,i); xx=[x(node_1),x(node_2)]; yy=[y(node_1),y(node_2)]; axis([-4 10 -4 10]) plot(xx,yy,'--rs') hold on end text(x(node_conn(1,1)),y(node_conn(1,1))-.5,'Global Node 1','HorizontalAlignment','center') text(x(node_conn(2,1)),y(node_conn(2,1))+.5,'Global Node 2','HorizontalAlignment','center') text(x(node_conn(2,2)),y(node_conn(2,2))-.5,'Global Node 3','HorizontalAlignment','center') text(x(node_conn(2,1))/2,y(node_conn(2,1))/2,'Element 1','HorizontalAlignment','center') text(x(node_conn(2,1))+(x(node_conn(2,2))-x(node_conn(2,1)))/2,(y(node_conn(1,2)))/2,'Element 2','HorizontalAlignment','center') %deformed positions pos(1,:)=[0 + d1,L1*cos(thetar1) + d3,L1*cos(thetar1) + L2*cos(thetar2) + d5]; pos(2,:)=[0 + d2,L1*sin(thetar1) + d4,0 + d6]; %deformed nodal coordinate arrays for i=1:num_nodes x(i)=pos(1,i); y(i)=pos(2,i); end %plot the deformed 2-bar truss system for i=1:num_elems hold on    node_1=node_conn(1,i); node_2=node_conn(2,i); xx=[x(node_1),x(node_2)]; yy=[y(node_1),y(node_2)]; plot(xx,yy,'-s') end text(x(node_conn(2,1)),y(node_conn(2,1))-.5,'Global Node 2','HorizontalAlignment','center') text(x(node_conn(2,1))/2,y(node_conn(2,1))/2,'Element 1','HorizontalAlignment','center') text(x(node_conn(2,1))+(x(node_conn(2,2))-x(node_conn(2,1)))/2,(y(node_conn(1,2)))/2,'Element 2','HorizontalAlignment','center') title('Deformed and Undeformed Two-bar Truss System') xlabel('X') ylabel('Y')

Latex-

--12-1 eq 1--

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

--12-1 eq 2--

$$k^{(e)}\hat{\textbf{k}}^{(e)}\textbf{q}^{(e)}=\textbf{p}^{(e)}$$

$$k^{(e)}\left[\begin{array}{c c} 1 & -1\\ -1 & 1 \end{array}\right] \begin{Bmatrix} q_1^{(e)}\\ q_2^{(e)} \end{Bmatrix}= \begin{Bmatrix} P_1^{(e)}\\ P_2^{(e)} \end{Bmatrix}$$

--12-2 relationship--

$$\textbf{q}^{(e)}=\textbf{T}^{(e)}\textbf{f}^{(e)}$$

--12-3 q director cosines--

$$q_1^{(e)}=\vec{d_1^{(e)}}\cdot\vec{\tilde{i}}$$

$$q_1^{(e)}=(d_1^{(e)}\vec{i}+d_2^{(e)}\vec{j})\cdot\vec{\tilde{i}}$$

$$q_1^{(e)}=d_1^{(e)}(\vec{i}\cdot\vec{\tilde{i}})+d_2^{(e)}(\vec{j}\cdot\vec{\tilde{i}})$$

$$(\vec{i}\cdot\vec{\tilde{i}})$$ and $$(\vec{j}\cdot\vec{\tilde{i}})$$ are director cosines

$$q_1^{(e)}=l^{(e)}d_1^{(e)}+m^{(e)}d_2^{(e)}$$

$$q_1^{(e)}=\lfloor\begin{array}{c c} l^{(e)} & m^{(e)} \end{array}\rfloor \begin{Bmatrix} d_1^{(e)}\\ d_2^{(e)} \end{Bmatrix}$$

similarly for local node 2: $$q_2^{(e)}=\lfloor\begin{array}{c c} l^{(e)} & m^{(e)} \end{array}\rfloor \begin{Bmatrix} d_3^{(e)}\\ d_4^{(e)} \end{Bmatrix}$$

$$\begin{Bmatrix} q_1^{(e)}\\ q_2^{(e)} \end{Bmatrix}= \left[\begin{array}{c c c c} l^{(e)} & m^{(e)} & 0 & 0\\ 0 & 0 & l^{(e)} & m^{(e)} \end{array}\right] \left\{\begin{array}{c} d_1^{(e)}\\ d_2^{(e)}\\ d_3^{(e)}\\ d_4^{(e)} \end{array}\right\}$$

--14-1 similarly--

$$\begin{Bmatrix} P_1^{(e)}\\ P_2^{(e)} \end{Bmatrix}=\textbf{T}^{(e)} \left\{\begin{array}{c} f_1^{(e)}\\ f_2^{(e)}\\ f_3^{(e)}\\ f_4^{(e)} \end{array}\right\}$$

elem axial FD rel

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

--14-2--

$$\hat{\textbf{k}}^{(e)}(\textbf{T}^{(e)}\textbf{d}^{(e)})=\textbf{T}^{(e)}\textbf{f}^{(e)}$$

want: $$\hat{\textbf{k}}^{(e)}\textbf{d}^{(e)}=\textbf{f}^{(e)}$$

so move $$\textbf{T}^{(e)}$$ to left hand side.

$$\textbf{T}^{(e)}$$ is rectangular so we can't invert

Ans: $$[\textbf{T}^{(e)T}\hat{\textbf{k}}^{(e)}\textbf{T}^{(e)}]\text{d}^{(e)}=\textbf{f}^{(e)}$$

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

--14-3--

$$\textbf{k}^{(e)}=\textbf{T}^{(e)T}\hat{\textbf{k}}^{(e)}\textbf{T}^{(e)}$$

--15-2--

$$q_2^{(1)}=\frac{P_1^{(1)}}{k^{(1)}}=\frac{P_2^{(1)}}{k^{(1)}}$$

--15-3--

$$q_1^{(1)}=0$$

$$q_1^{(2)}=\frac{-P_2^{(2)}}{k^{(2)}}$$

$$\textbf{AC}=\frac{P_2^{(1)}}{k^{(1)}}$$

$$-\textbf{AB}=\frac{-P_2^{(2)}}{k^{(2)}}$$

--15-4--

$$u_y=Rsin(\alpha)$$

$$u_x=R[1-cos(\alpha)]$$

--16-1--

$$\textbf{AC}=\frac{|P_2^{(2)}|}{k^{(1)}}$$

Problem 9
$$\left[\begin{array}{c c c c c c c c} k_{11}^{(1)}&k_{12}^{(1)}&k_{13}^{(1)}&k_{14}^{(1)}&0&0&0&0\\ k_{21}^{(1)}&k_{22}^{(1)}&k_{23}^{(1)}&k_{24}^{(1)}&0&0&0&0\\ k_{31}^{(1)}&k_{32}^{(1)}&(k_{33}^{(1)}+k_{11}^{(2)}+k_{11}^{(3)})& (k_{34}^{(1)}+k_{12}^{(2)}+k_{12}^{(3)})&k_{13}^{(2)}&k_{14}^{(2)}& k_{13}^{(3)}&k_{14}^{(3)}\\ k_{41}^{(1)}&k_{42}^{(1)}&(k_{43}^{(1)}+k_{21}^{(2)}+k_{21}^{(3)})& (k_{44}^{(1)}+k_{22}^{(2)}+k_{22}^{(3)})&k_{23}^{(2)}&k_{24}^{(2)}& k_{23}^{(3)}&k_{24}^{(3)}\\ 0&0&k_{31}^{(2)}&k_{32}^{(2)}&k_{33}^{(2)}&k_{34}^{(2)}&0&0\\ 0&0&k_{41}^{(2)}&k_{42}^{(2)}&k_{43}^{(2)}&k_{44}^{(2)}&0&0\\ 0&0&k_{31}^{(3)}&k_{32}^{(3)}&0&0&k_{33}^{(3)}&k_{34}^{(3)}\\ 0&0&k_{41}^{(3)}&k_{42}^{(3)}&0&0&k_{43}^{(3)}&k_{44}^{(3)} \end{array}\right]$$

Prob. 5 Eigenvalues
Now we want to plot the eigenvectors corresponding to the 0 eigenvalues of the 2-bar truss system that we have been looking at in class for the past few weeks. We want to assume no constraints on the 2-bar truss system so we will use the full 6x6 global stiffness matrix.

We know that the global stiffness matrix K is:

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

If we use the eig function in Matlab we get that the eigenvalues are as follows:

Eval=eig(K) Eval= -0.0001        -0.0000         -0.0000          0.0000          1.4706         10.0294

Next we can use the function [V,D] = EIG(X)which gives us a diagonal matrix D of eigenvalues and a full matrix V whose columns contain the corresponding eigenvectors that corresponds to the equation:

$$\textbf{X}\times\textbf{V}=\textbf{D}\times\textbf{V}$$

For our particular case, we set X equivalent to K and the diagonal matrix D equivalent to $$\lambda$$ to get the following:

$$\textbf{K}\times\textbf{V}=\lambda\times\textbf{V})$$

Therefore, using Matlab we get:

[V,Lam]=eig(K) Lam = -0.0001        0         0         0         0         0         0   -0.0000         0         0         0         0         0         0   -0.0000         0         0         0         0         0         0    0.0000         0         0         0         0         0         0    1.4706         0         0         0         0         0         0   10.0294  V = -0.4494   0.3877    0.5158    0.0160    0.6174   -0.0139    0.6754   -0.4281    0.4825    0.0244    0.3565   -0.0080    0.1682    0.3877    0.5158    0.0160   -0.5409    0.5123   -0.3942   -0.4281    0.4825    0.0244   -0.4330   -0.4904    0.2812    0.4163   -0.0116    0.7023   -0.0765   -0.4984   -0.2812   -0.3995   -0.0449    0.7107    0.0765    0.4984

Where Lam contains the previously found eigenvalues and V contains the corresponding eigenvectors. It is important to note now that while Matlab calculated the eigenvalues to precise decimals, we can assume that the first four values in the diagonal Lam matrix are 0.

Now we would like to plot the eigenvectors that correspond to these four 0 eigenvalues so we may get a look at their mode shapes. Using the following Matlab function allows us to plot the four mode shapes at the same time.

for i=1:4 figure plot(V(:,i)); title(['Mode Shape ',num2str(i)]) end

Now let us consider the following three-bar truss system.

Assume that all three bars have the same material properties:

L=1    E=2      A=3

$$k=\frac{EA}L=\frac{(2)(3)}1=6$$

For element 1 and element 3:

$$l^{(1,3)}=0$$

$$m^{(1,3)}=1$$

For elemnt 2:

$$l^{(2)}=1$$

$$m^{(2)}=0$$

$$\textbf{k}^{(1)}=\textbf{k}^{(3)}= \left [\begin{array}{c c c c} 0 & 0 & 0 & 0\\ 0 & 36 & 0 & -36\\ 0 & 0 & 0 & 0\\ 0 & -36 & 0 &-36 \end{array}\right]$$

$$\textbf{k}^{(2)}= \left [\begin{array}{c c c c} 36 & 0 & -36 & 0\\ 0 & 0 & 0 & 0\\ -36 & 0 & 36 & 0\\ 0 & 0 & 0 & 0 \end{array} \right]$$

$$\textbf{K}= \left [\begin{array}{c c c c c c c c} 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 36 & 0 & -36 & 0 & 0 & 0 & 0\\ 0 & 0 & 36 & 0 & -36 & 0 & 0 & 0\\ 0 & -36 & 0 & 36 & 0 & 0 & 0 & 0\\ 0 & 0 & -36 & 0 & 36 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 36 & 0 & -36\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & -36 & 0 & 36 \end{array} \right] $$

In this example we are assuming that global nodes 1 and 4 are constrained so that they don't move. This means that $$d_1=d_2=d_7=d_8=0$$. Therefore we only have to consider the following modified global stiffness matrix.

$$\overline{\textbf{K}}= \left[ \begin{array} {c c c c} 36 & 0 & -36 & 0\\ 0 & 36 & 0 & 0\\ -36 & 0 & 36 & 0\\ 0 & 0 & 0 & 36 \end{array} \right]$$

If we follow the same steps as above we get:

Kbar = 36    0   -36     0     0    36     0     0   -36     0    36     0     0     0     0    36 Eval=eig(Kbar) Eval = 0   36    36    72 [V,Lam]=eig(Kbar) V = -0.7071        0         0   -0.7071         0    1.0000         0         0   -0.7071         0         0    0.7071         0         0    1.0000         0 Lam = 0    0     0     0     0    36     0     0     0     0    36     0     0     0     0    72

In this example we have only one zero eigenvalue and it can be plotted as follows:

plot(V(:,1)); title(['Mode Shape',num2str(1)])

HW 5
Choice 3:

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

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

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

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

Choice 4:

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

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

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

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

Choice 5:

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

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

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

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

Choice 6:

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

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

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

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

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User:Eas4200c.f08.wiki.b

=Homework 1=

User:Eas4200c.f08.wiki.b/HW1

=Homework 2=

User:Eas4200c.f08.wiki.b/HW2

=Homework 3=

User:Eas4200c.f08.wiki.b/HW3

=Homework 4=

User:Eas4200c.f08.wiki.b/HW4

=Homework 5=

User:Eas4200c.f08.wiki.b/HW5

=Homework 6=

User:Eas4200c.f08.wiki.b/HW6

=Homework 7=

User:Eas4200c.f08.wiki.b/HW7

title: User:Eas4200c.f08.wiki.b


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User:Eas4200c.f08.wiki.b/HW1

=Basics of Aerospace Structures=

The Finite Element method (FEM) can be used in a wide variety of fields, from various types of engineering, to applied math, and even geology. Aerospace structures is one of the fields where FEM can be used to solve the partial differential equations that govern this particular field.

The structure of an aircraft must be designed with two main goals in mind: it must be light, and it must be strong. More specifically, the structure of an aircraft should have a high stiffness, high strength, and be composed of light weight materials.

Stiffness refers to a Young’s modulus $$\displaystyle E$$, or the slope, in a linear relation between stress $$\displaystyle \sigma$$ and strain $$\displaystyle \epsilon$$, or $$\displaystyle \sigma = E \epsilon$$. Strength refers to a particular stress, such as the yield stress $$\displaystyle \sigma_Y$$, where a material begins to weaken, and the ultimate stress $$\displaystyle \sigma_U$$, where a material fails. Toughness is simply the capacity to endure fracture or damage.

Examples of materials with high stiffness and high strength include steel and titanium alloys, since they have a large Young’s modulus and large ultimate and yield strengths. Glass is an example of a material with high stiffness, but low toughness. It does not withstand much plastic deformation and does not resist fracture well. Plastics and nylon are materials with a low stiffness, and high toughness, as they are they behave oppositely to glass.

Aircraft structures typically have monocoque or semi-monocoque designs to them. The term monocoque refers to a “single shell” design, while semi-monocoque is a shell that is stiffened by a supporting structure. These designs allow aircraft to have low weight, while also remaining strong and resisting bending. The geometry of an aircraft is, of course, also restricted by aerodynamics such as lift and drag on an airfoil.

title: User:Eas4200c.f08.wiki.b/HW1


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User:Eml4500.f08.ateam.nobrega/Homework Report One

Lecture Notes Through September 5th
Lecture Notes

What is Matlab?
MATLAB is short for "matrix laboratory". It is a program software that allows users to solve numerical problems using a programming language similar to, yet simpler and faster than, Fortran or C++. It is a matrix based system that allows for quick computation and visualization of scientific and engineering problems such as, but not limited to, plotting data and functions, implementing algorithms, solving algebraic expressions and creating programs.

An Introduction to the Matlab Primer
The MATLAB Primer Tutorial was written based on the 4.0 and 4.1 version of MATLAB. However, it can be used in updated versions of MATLAB as well. The Primer was written by Kermit Sigmon, Department of Mathematics at the University of Florida. It is designed to help one use MATLAB. It is encouraged that the Primer be used hands on with the reader sitting at their computer going through the examples. Users should type the commands into the MATLAB command window and view the MATLAB generated output for each topic.

If the user is unsure of how to perform a task or if an error message arises, he or she should call on the help function. This function can be called by typing the word help into the command window. Doing this will display a list of topics for the user to choose from. To call for a specific help topic type in the word help followed by a space and the name of the topic one is searching for. Another helpful command is to type in the word demo into the command space, allowing the user to preview some of the MATLAB functions.

The MATLAB Primer is a set of instructions for how to perform basic MATLAB functions. The MATLAB User's Guide and Reference Guide will need to be accessed for more advanced applications. To find these documents consult your instructor or visit your local library or computer lab.

"MATLAB is available for a number of environments: Sun/Apollo/VAXstation/HPworkstations, VAX, MicroVAX, Gould, PC and AT compatibles, 80386 and 80486 computers, Apple Macintosh, and several parallel machines. There is a relatively inexpensive Student Edition available from Prentice Hall publishers."

Accessing MATLAB
After logging on to a computer, the easiest way to access MATLAB is to double-click the MATLAB icon on the desktop or search for MATLAB in your program list. The MATLAB icon is featured below:

Entering Matrices
Matrices in MATLAB can be entered in several different ways. Below are two examples that output the same matrix.

>> A=[1,2,3;4,5,6;7,8,9]

A = 1    2     3     4     5     6     7     8     9

>> A = [ 1 2 3 4 5 6 7 8 9 ]

A = 1    2     3     4     5     6     7     8     9

Entering Matrices with complex numbers. Below are two different ways to enter complex matrices.

>> B = [1 2;3 4] + i*[5 6;7 8]

B =

1.0000 + 5.0000i  2.0000 + 6.0000i 3.0000 + 7.0000i  4.0000 + 8.0000i

>> B = [1+5i 2+6i;3+7i 4+8i]

B =

1.0000 + 5.0000i  2.0000 + 6.0000i 3.0000 + 7.0000i  4.0000 + 8.0000i

Referencing an element from a matrix. For example in the matrix A above, if you want to reference the value of the element located in row 2, column 3 type A(2,3). View example below.

>> A(2,3)

ans =

6

Matrix and Array Operations
MATLAB incorporates the following matrix operations: Care must be taken to ensure that sizes of the matrices are compatible or an error message will result, and the computation will not be performed."It is important to observe that these other operations, *, ^, \, and /, can be made to operate entry-wise by preceding them by a period." See the following example for further clarification.
 * Addition +
 * Subtraction –
 * Multiplication *
 * Exponential Power ^
 * Conjugate Transpose ‘
 * Left Division \
 * Right Division /

>>[6,7,8,9].*[6 7 8 9]

ans =

36   49    64    81 Notice that the operation performed was not matrix multiplication but rather an entry by entry multiplication. See the following example for matrix multiplication.

>>A=[1 2;3 4]

A =

1    2     3     4

>> B=[1 2;3 4]

B =

1    2     3     4

>> A*B

ans =

7   10    15    22

Statements, Expressions and Variables
"MATLAB is an expression language. Expressions you type are interpreted and evaluated." Examples of statement formatting:

$$variable = expression$$ or    $$expression$$

Expressions may consist of functions, operators, and names of variables. The return key executes your typed command. MATLAB is case sensitive and therefore care must be exercised when entering variable names. For example:

$$plot(x,Y) \ne plot(X, y)$$

The command "who" or "whois" displays variables currently stored in MATLAB. The command " clear variablename " deletes a specific variable.

The command "eps" prints the machine round off point useful for iterative convergence.

If a run away computation is experienced "CTRL-C" or "CTRL-BREAK" will break the calculation.

Saving the Workspace
When MATLAB is exited, all variables are lost unless previously saved. The command "save" will generate a file entitled "matlab.mat". When re-entering the program, the command "load" will reopen the saved workspace.

Matrix Building Function
Matrix building functions are

eye	       identity matrix zeros          matrix of zero ones           matrix of ones diag           create or extract diagonals triu           upper triangular part of the matrix tril           lower triangular part of the matrix rand           randomly generated matrix hilb           Hilbert Matrix magic          magic square topelitz       see help topelitz

Functions like zeros, eye, and ones can be used to create matrices. The functions are followed by the size of the matrix.

Ex:

$$>> zeros(2,3)$$ $$ans =$$ 0    0     0     0     0     0 Functions such as diagonals vectors require a vector as an input.

Ex:

$$>>A =$$ 2    2     2     2

$$>> diag(A)$$ $$ans = $$    2     2

You can combine and create new matrices using multiple functions.

Ex: This function will create a 5x5 matrix shown below.

$$>> 	B= [A, zeros(3,2); zeros(2,3), eye(2)]$$

$$B =$$

0    0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     1     0     0     0     0     0     1

For, While, If - and Relations
For.

Using the For. statement to produce a vector. Example for n = 3

>> x = []; for i = 1:n, x=[x,i^2], end

or

>> x = []; >> for i = 1:n x = [x,i^2] end

Will result in:

x =

1

x =

1    4

x =

1    4     9

Producing the same vector as shown above, but in reverse order. Example is shown below.

>> x = []; for i = n:-1:1, x=[x,i^2],end

x =

9

x =

9    4

x =

9    4     1

While.

A while loop allows for the statement to be executed continuosly aslong as the relation defined remains true.

General Form:

while relation statement end Example:

>> a =30; >> n = 0; >> while 2^n < a    n = n + 1; end

Will result in:

>> n

n = 5

If.

An if statement is only executed if the relation is true.

General Form:

if relation statement end

Example:

>> if n < 0 parity = 0; elseif rem(n,2) == 0 parity = 2; else parity = 1; end

Relations

MATLAB Relational Operators:

MATLAB Logical Operators:

If the statement is true it is a value of 1. If the statement is false it is a value of 0. See the examples below.

>> 3 < 5

ans =

1

>> 3 > 5

ans =

0

>> 3 == 5

ans =

0

>> 3 == 3

ans =

1

>> a = rand(5)

a =

0.9501   0.7621    0.6154    0.4057    0.0579    0.2311    0.4565    0.7919    0.9355    0.3529    0.6068    0.0185    0.9218    0.9169    0.8132    0.4860    0.8214    0.7382    0.4103    0.0099    0.8913    0.4447    0.1763    0.8936    0.1389

>> b = triu(a)

b =

0.9501   0.7621    0.6154    0.4057    0.0579         0    0.4565    0.7919    0.9355    0.3529         0         0    0.9218    0.9169    0.8132         0         0         0    0.4103    0.0099         0         0         0         0    0.1389

>> a == b

ans =

1    1     1     1     1     0     1     1     1     1     0     0     1     1     1     0     0     0     1     1     0     0     0     0     1

Scalar Functions
MATLAB has built in functions that operate on scalars, but can also operate on individual elements of a matrix. The most common of these functions are as follows:

See the following examples:

>>sin(3*pi/2)

ans =

-1

>> log(1)

ans =

0

Vector Functions
Some MATLAB functions perform actions on vectors in a column by column fashion and produce a row vector with the results. If one wants to operate on a row by row basis, the transpose can be used.

max min sort sum prod median mean std any all

For example, to calculate on a row by row basis and find the mean:

mean(A')'

For example, the max value in a matrix B is determined by:

max(max(B))

The command example above is nested within itself. The inner most "max" function produces a single row of the max values for each column of the original matrix. The outer function produces the maximum value of these remaining row vector values.

Matrix Functions
eig       eigenvalues and eigenvectors chol      cholesky factorization svd       singular value decomposition inv       inverse lu        LU factorization qr	       QR factorization hess      hessenberg form schur     schur decomposition rref      reduced row echelon form expm      matrix exponential sqrtm     matrix square root poly      characteristic polynomial det       determinant size      size norm      1-norm, 2-norm F-norm cond      condition number in the 2-form rank      rank

Above is a list of matrix functions that can be used. These functions can be called using a variety of commands.

Ex:

$$Y=eig(A)$$

Or

$$eig(A)$$

Command Line Editing and Recall
Keys and their functions to edit in MATLAB

Submatrices and Colon Notaion
"Vectors and submatrices are often used in MATLAB to achieve fairly complex data manipulation effects. "Colon notation" ( which is used both to generate vectors and reference submatrices) and subscripting by integral vectors are keys to effcient manipulation of these objects." A row vector of length n may be defined as 1:n where n represents the length of the vector. The default increment is taken to be 1 unless otherwise specified. See the following examples for further clarification. A row vector of length n may be defined as 1:n where n represents the length of the vector. The default increment is taken to be 1 unless otherwise specified. See the following examples for further clarification.

>>1:5

ans =

1    2     3     4     5

>> 1:2:5

ans =

1    3     5 Both vectors are row vectors from 1 to 5, but the second vector is from 1 to 5 with an increment of 2."

Colon Notation" may also be used to access submatrices of a matrix, and to replace elements within matrices. See the following example for further clarification.

>>A =

1    2     3     4     5     6     7     8     9    10    11    12    13    14    15    16

>> A(1:2,3)

ans =

3    7 This creates a column vector consisting of the first 2 entries of the 3rd column of matrix A.

M-files
M-files are a specific type of file MATLAB is capable of executing. A M-file consists of a series of sequential commands and are characterized by the ".m" file extension. MATLAB M-files fall into two categories: script files and function files.

Script Files
Script files are composed of normal MATLAB commands. To execute the commands inside of a script file, simply type the file name. For example, if the file name is "transpose.m", the command "transpose" will execute all the statements to be performed.

Details of script files:
 * 1) Can reference other M-files or itself
 * 2) Variables are global and will overwrite current session variables
 * 3) Useful to avoid repetitious data entry

To create a .m script file, simply use the built in editor of MATLAB and save the file with a .m extension. For example if the following M-file entitled "test.m" was executed by using the command "test":

B=[1,2,3;4,5,6;7,8,9]

A=[9,8,7;6,5,4;3,2,1]

C=A+B

The current workspace will now have three variables imported into memory: A, B, and the result, C.

Function Files
The second type of M-file is the function file. This tool extends the convenience of built in functions but allows for customization to specific tasks. Note that variables in a function file are local by default. See the following example from the MATLAB Primer document.

function a = randint(m,n) %RANDINT Randomly generated integral matrix. % randint(m,n) returns an m-by-n such matrix with entries % between 0 and 9 a = floor(10*rand(m,n));

Note the following procedures when creating a function file:
 * 1) First line must begin with "function" and define the input variables
 * 2) "%"-Allows for comments
 * 3) Last line defines output variables

As an example to run the above program, "k = randint(4,3)" could be used.

This command passes 4 and 3 to the program and stores the result in a new matrix, k. Note that variable names are not carried over outside of the function M-file.

Text Strings, Error Message, Input
Text strings can be entered into Matlab using single quotations. These strings can be displayed by using the disp command.

Ex: s= ‘peter pan had an accident’

disp(‘peter pan had an accident’)

Error messages that are placed in M-files can abort the program.

Ex:

error(‘Sorry, this matrix is wrong’)

Strings can also be added into an M-file to make the program interactive.

Ex:

iter = input(‘Enter the number of times you want to enter a number:’)

Managing M-Files
It is recommended that if your system allows you to run multiple processes that you keep both the MATLAB Command Window and your m-file editor open simultaneously. At any point if you would like to edit your m-file use the function !ed. For example you would type into the command window as follows: >>!ed filename.m

Note: Make sure to replace filename with the file you are trying to edit.

Commands useful for managing m-files.

Comparing Efficency of Algorithms Flops, Tic and Toc
Older versions of MATLAB had the function flops which kept a running total of flops performed to solve and algorithm. In more resent versions of MATLAB the command function flops is unavailable and an error message will appear if the command is entered. The command function “tic, any statement, toc” can be used to obtain the elapsed time (in seconds) it takes MATLAB to solve the algorithm. This enables the user to compare the efficiency of his / her algorithms. See the example below for further explanation.

>> A=[ 1 2 3; 7 8 9; 4 5 6]

A =

1    2     3     7     8     9     4     5     6

>> b= [1 2; 4 5; 8 9]

b =

1    2     4     5     8     9 >> x=A*b

x =

33   39   111   135    72    87 >> tic, x=A*b; toc

Elapsed time is 0.000000 seconds.

Output Format
MATLAB calculations are performed in double precision. However, output formatting can be changed with the following commands:

format short      fixed point with 4 decimal places (default) format long       fixed point with 14 decimal places format short e    scientific notation with 4 decimal places format long e     scientific notation with 15 decimal places format rat        approximation by ratio of small integers format hex        hexadecimal format format bank       fixed dollars and cents format +          +, -, blank

Hardcopy
The easiest way to safe your command file in Matlab is the diary command. The diary command, when followed by a file name will save the data under the new name. However if no file name is provided, the file will be saved under a generic name of diary. The file can be opened later for further editing or printing.

$$diary$$ $$filename$$

Planar Plots
The plot command creates linear x-y plots; if x and y are vectors of the same length, the command plot(x,y) opens a graphics window and draws an x-y plot of the elements of the x versus the elements of y.

For example command: >> x = -4:.01:4; y=sin(x); plot(x,y)

Results in:

One can also refrence a function from an m-file in the given format: >>fplot('expnormal',[-1.5,1.5])

Results in:

Plots of parametrically defined curves can also be made. For example:

Commands for plots:

Manualing Scaling Axes:

Multiple Plots on a Single Graph: Two possible ways

1. Example: Listing them within the same command line >> x=0:.01:2*pi; Y=[sin(x)', sin(2*x)', sin(4*x)']; plot(x,Y)

Result in:

2. Example: Using the hold function >> x=0:1:10; y=x; plot(x,y) >> hold on >> x=0:1:10; y=2*x; plot(x,y) >> x=0:1:10; y=3*x; plot(x,y) >> hold off

Result in:

Editing Linetypes, Marktypes, and Colors of plots:

Commands:

 * Linetypes: solid(-), dashed(--), dotted, dashdot(-.)
 * Marktypes: point(.), plus(+), star(*), circle(o), x-marks(x)
 * Colors: yellow(y), magenta(m), cyan(c), red(r), green(g), blue(b), white(w), black(k)

Example Code: >> x=0:.01:2*pi; y1=sin(x); y2=sin(2*x); y3=sin(4*x); plot(x,y1,'--',x,y2,':',x,y3,'+')

Graphics Hardcopy
To obtain a hardcopy of your plots you would use the commmand: >>print

3-D Line Plots
To plot a 3-D line defined by vectors x, y, z of the same size, you would use the command plot3(x,y,z). Below is an example:

>> t=.01:.01:20*pi; x=cos(t); y=sin(t); z=t.^3; plot3(x,y,z)

Result in:

3-D Mesh and Surface Plots
3-D Surface Mesh plots are drawn with the command mesh

Example Code:

>> mesh(eye(10))

Result in:

Faceted surface plots are drawn with the command surf

Example Code:

>> surf(eye(10))

Result in:

Plotting 3-D Functions: Function:

Code:

>> xx = -2:.2:2; >> yy = xx; >> [x,y] = meshgrid(xx,yy); >> z = exp(-x.^2 - y.^2); >> mesh(z)

Resulting in:

One could also use the surf command for the same function resulting in: Code:

>> surf(z)

Shading and Color Profiles

Commands:

*shading faceted *shading interp *shading flat

*colormap(hsv) Note: Default *colormap(hot) *colormap(cool) *colormap(jet) *colormap(pink) *colormap(copper) *colormap(flag)

Other Useful Commands: Can further explore under the help command

*meshz *surfc *surfl *contour *pcolor

*colormap(gray) *colormap(bone)

Sparse Matrix Computations
A great deal on computation time and memory storage space can saved when performing calculations on a matrix that has many zero entries by using the command sparse. “Matlab has two storage modes, full and sparse with full the default. The functions full and sparse convert between the two modes.” "A sparse matrix is stored as a linear array of it’s nonzero elements along with their row and column indices.” See the following example for further explanation.

F =

9    4     0     0     0     0     2     0     7     0     0     0     0     8     1     0     0     0     0     0     4     3     6     0     0     0     0     8     2     4     0     0     0     0     1     4

>> S=sparse(F)

S =

(1,1)       9   (2,1)        2   (1,2)        4   (3,2)        8   (2,3)        7   (3,3)        1   (4,3)        4   (4,4)        3   (5,4)        8   (4,5)        6   (5,5)        2   (6,5)        1   (5,6)        4   (6,6)        4

>> F=full(S)

F =

9    4     0     0     0     0     2     0     7     0     0     0     0     8     1     0     0     0     0     0     4     3     6     0     0     0     0     8     2     4     0     0     0     0     1     4

The command S=sparse(F) creates the sparse matrix S from the matrix F, and the command F=full(S) returns matrix F back to its original form.

Contributing Team Members
The list below is in no particular order.

Pedro Rivero User:Eml4500.f08.ateam.rivero 18:14, 17 September 2008 (UTC)

Akash Shah Eml4500.f08.ateam.shah 14:49, 18 September 2008 (UTC)

Tatiana Nobrega User:Eml4500.f08.ateam.nobrega 23:03, 18 September 2008 (UTC)

Michael Carr Eml4500.f08.ateam.carr 23:13, 18 September 2008 (UTC)

Taylor Boggs Eml4500.f08.ateam.boggs.t 04:34, 19 September 2008 (UTC)

Josh McNally Eml4500.f08.ateam.mcnally 03:41, 19 September 2008 (UTC)


text:
User:Eas4200c.f08.radsam.s

1.4 - Aircraft Materials
Material selection has always been a driving force throughout the history of aeronautics. The technology involved with developing new materials has pushed next generation aircraft to be lighter, stronger, and more resistant to nature's elements. From the wood and canvas structures of early aircraft to modern-day titanium alloys and composites, aircraft have come a long way with the help of advances in material.

Currently, the common materials used in aircraft structures are aluminum, titanium, and steel alloys. The advantages and disadvantages of each are listed below:

*Aluminum is relatively light weight, and has good fatigue life. It is not as strong as other materials and cannot stand high temperatures. *Steel Alloys are most used for their high strength. They are heavy and strongly affected by corrosion. *Titanium can withstand very high temperatures, and has good resistance to corrosion. It is, however, difficult to machine which makes its manufacturing cost higher.

Another material being used more frequently is fiber-reinforced composites. These composites are manufactured in a careful manner in order to provide maximum strength without being too brittle. Different types of composites exist, such as glass-fiber and carbon-fiber. The mechanical properties for some of these materials yield great strength to weight ratios, which make them an attractive option for aerospace applications.

Look at how far aircraft structures have come due to material influences in the span of a few decades: *This is a picture of an early Wright Brothers' airplane made of wood and canvas: http://www.edunetconnect.com/TimeMachine/contentimages/flyght1.gif *This is a picture of a modern-day airplane constructed mostly of composite materials: http://franklin-innovation.com/wp-content/uploads/2007/01/carbonf.jpg

title: User:Eas4200c.f08.radsam.s


text:
User:Eml4500.f08.team.allen.oca

 * homework 1
 * homework 2-1
 * homework 2-2
 * Test Page
 * homework 3
 * homework 4, 10-13-08

title: User:Eml4500.f08.team.allen.oca


text:
Template:Taxobox

it:Template:Tassobox

title: Template:Taxobox


text:
The Impact of Web-Based Lecture Technologies on Current and Future Practices in Learning and Teaching

Introduction
This page contains an extract of a research study's executive summary with the intention of providing a point of departure for further consideration and exploration of web-based additions or alternatives to the traditional lecture. A key finding relates to an apparent disconnection between staff and student about the perceived value of web-based learning technologies (WBLT). Feel free to discuss on the accompanying talk page or over at Lecture 2.0.

Method
A survey-based study of staff and students at four Australian universities, 2006 to 2008:
 * 1) * Macquarie University
 * 2) * Murdoch University
 * 3) * Flinders University
 * 4) * University of Newcastle
 * 5) Staff (N = 155 responded out of 676; 22.9%)
 * 6) Students (N = 815 responded out of 13728; 5.9%)

Licensing of this extract
This work was originally published under the terms of the Creative Commons Attribution-Noncommercial-ShareAlike 2.5 Australia Licence. Under this Licence you are free to copy, distribute, display and perform the work and to make derivative works for non-commercial purposes.

Attribution: You must attribute the work to the original authors (Maree Gosper, David Green, Margot McNeill, Rob Phillips, Greg Preston, and Karen Woo) and include the following statement: Support for the original work was provided by The Carrick Institute for Learning and Teaching in Higher Education Ltd (now known as the Australian Learning and Teaching Council), an initiative of the Australian Government Department of Education, Employment and Workplace Relations

Noncommercial: You may not use this work for commercial purposes.

Share Alike. If you alter, transform, or build on this work, you may distribute the resulting work only under a license identical to this one. For any reuse or distribution, you must make clear to others the license terms of this work.

Any of these conditions can be waived if you get permission from the copyright holder. Permission is being sought for a WMF-compatible license from the authors - see talk page.

Key extracts from the Executive Summary
The three key outcome measures used in the surveys were:
 * 1) positive experience with Web-based Learning Technologies (WBLT);
 * 2) perceptions of benefits for learning; and
 * 3) perceptions of achievement of better results.

Regardless of age, gender, enrolment mode or attendance pattern, 76% of students reported positive experiences with WBLT almost always or frequently. Staff experiences, on the other hand, were more varied with 54% of respondents finding use of WBLT to be generally positive, while another 26% found the experience to be negative.

Overall, there was a clear mis-match between staff and student views on learning and achievement of better results. Sixty seven percent (67%) of students compared with 30% of staff agreed that WBLT helped students achieve better results. In addition, 80% of students compared with 49% of staff agreed that WBLT made it easier for students to learn.

This mis-match between student and staff perceptions is one of several key themes that have emerged. Insights into the range of issues at the heart of this mismatch were explored more thoroughly through qualitative comments on the surveys, interviews and case studies. They are as follows:

Students appreciate the flexibility in access and support for learning - staff have concerns
Although it has long been acknowledged that external students need flexibility, the data indicates that students enrolled in internal mode also appreciate this aspect of WBLT. From the survey responses, 56.2% of students indicated that they didn’t attend at least some of the face-to-face lectures available. Of these students who listened to WBLT rather than attending face-to-face lectures, 75.3% indicated this was because they ‘couldn’t attend’.

Responses from staff supported this view with a high proportion of respondents (81.9%) introducing WBLT to support students who could not attend. Staff also recognised the value of the technology as a study tool with 64.5% of staff providing it to support students in their learning. While many staff agreed that WBLT was a useful resource for external students, some were concerned that on-campus students were choosing not to attend lectures as a result of using the technologies and this was perceived as having a negative impact on their learning.

WBLT have contributed to a blurring of the boundaries between internal and external students
Many programs and individual units are being offered both internally and externally, with the same lectures being delivered to both cohorts. The data suggest that staff perceive access to recorded lectures is beneficial for external students but use of WBLT can be disadvantageous to internal students if they use them as a replacement for attending lectures. Students, on the other hand, do not perceive the difference. Moreover, internal students exhibit strong similarities to their external counterparts in the way they use WBLT as a study tool - to revise for exams, review complex materials and take comprehensive notes. This raises the question: Is there any difference between the learning needs of an internal student who cannot attend and an external student who is not expected to attend? More generally, where WBLT are used in combination with other eLearning technologies to access and interact with content, communicate and collaborate, we need to question whether the distinction between external and internal modes of enrolment is of relevance to an increasing number of students.

Introducing WBLT will change lecture attendance patterns and may raise questions about the role of lectures
While staff seem to understand the need of flexibility for their students, they are, nevertheless, concerned about falling lecture attendance.

Students appreciated the flexibility offered by WBLT with 75% using the technology because they couldn’t come to class and another 69% because it was the only class they had on that day. Importantly, they also viewed lectures as important to their learning. They found lectures motivating, they valued contact with the lecturers and their peers and they found the visual aids helpful. Importantly, the use of WBLT did not necessarily exclude lecture attendance and students often ‘double up’ by attending lectures and listening to the recordings.

Our findings indicated that students are quite strategic about the choices they make, basing decisions on lecture attendance around three types of factors: educational value; convenience and flexibility; and social opportunities to meet other students, exchange ideas and make new friendships. With students being offered the technologies and choosing not to attend, some academics have begun questioning the role of lectures. At least 80% of the staff surveyed use lectures to inspire and motivate students; build conceptual frameworks; establish connections with students; use multimedia content; provide structured experiences for students; impart information and make announcements. This raise the question of whether there are more effective ways of achieving these functions.

Using WBLT demands changes in the way students learn and teachers teach
The statistics are compelling - 68% of students using WBLT believe they can learn just as well using WBLT as they can face-to-face. They use the tools to help revise for exams, review complex materials, work at their own pace and place of convenience, pick up on things that they missed in class, go back and take comprehensive notes after the lecture so they can concentrate on what is happening in the lecture, and check what was said before approaching their lecturer for clarification of issues, ideas or misunderstandings.

Aside from some concerns about IP around the re-use of lectures and copyright issues associated with using visual aids such as videos on WBLT, staff were most concerned about WBLT reducing two-way communication with their students and their ability to inspire and motivate students. On the other hand, there was recognition that the technology could help to provide a structured experience for students and facilitate information exchange.

From the findings it is clear that many staff recognise the strengths and limitations of WBLT and are concerned about the impact these technologies may have on learning. Nevertheless, there has been a mixed response to dealing with the changing context. Approximately one third have not made substantial changes to what they do in lectures. A common approach has been to maintain the status quo by re-emphasising the importance of lectures and the need for students to attend. In managing the limitation of the technology some have reduced their movements around the lecture theatre and reduced multimedia content due to copyright restrictions.

On the other hand, another third of lecturers have taken a more proactive approach and made changes to cater for students who are present as well as those using WBLT. Many of the changes are reflective of sound inclusive practice for example changing teaching strategies to accommodate students not present by explaining the actions in the class and by repeating students’ questions when they are being recorded, scripting the lecture more tightly to provide a more controlled presentation, and using discussion forums and other activities to extend communication and interactive opportunities beyond the lecture experience. .

Introducing WBLT is more than a teaching issue – it will affect the design of the whole curriculum
The introduction of any new technology is not an isolated experience and it impacts the entire teaching and learning context: including the ways in which students and staff communicate and the relationship between other elements of the curriculum. Despite this, our study clearly showed 75% of staff reported they had not changed the structure of their unit.

Rather than focussing on the lecture alone, a shift is needed for staff to consider the whole curriculum, taking into account the learning outcomes and needs of students and using a range of different activities and technologies (tutorials, workshops, online communication, etc) to provide stimulating and engaging learning environments and experiences.

Introducing WBLT has professional and organisational development implications
In addition to strategies for successful implementation at a curriculum level, the project also highlighted several professional and organisational implications. For staff, a correlation between choice in the use of WBLT and a positive experience with WBLT sends a strong message that policies enforcing the uptake of technologies may be counterproductive. Empowering academics by encouraging a culture of innovation and experimentation with new technologies and enabling them to make informed decisions about the appropriateness of technologies in their own context may be more effective and sustainable in the longer term. Professional development is an essential ongoing requirement to enable staff to implement new technologies into their curricula. Similarly, students need support to use them effectively and the technologies themselves need to be embedded in a robust infrastructure and technical support network.


text:
User talk:EAS4200c.f08.Blue.E

Assignments

Lecture 4

Lecture 10

Lecture 16

Lecture 22

Oct31st

nov12th

dec1st

title: User talk:EAS4200c.f08.Blue.E


text:
Wikimania/2009

Wikimania 2009 will be held in Buenos Aires, Argentina.

Present your project
Hello,

do anyone of you folks is interested in make a presentation of your project on the next Wikimania? The next Wikimania would probably held in July in Buenos Aires, Argentina. The actual date is not confirmed yet. Traditionally in February there would be a Call for Participation, where you can submit your talks and about in April you can request for a scholarship to finantial your flight and accomodation as a speaker on the conference. I would like to see someone of you folks to make a presentation of your project there. Is anyone interested in this and can prepare for a presentation?--Wing 08:56, 3 September 2008 (UTC)
 * I'd be interested, though the only significant project I've started here is the History of Ireland through Song, and that hasn't really taken off yet. Mostly I've been busy being a bureaucrat for the School of History, English Language Division, and French Department.  The Jade Knight 12:29, 3 September 2008 (UTC)
 * I think User:Cormaggio plans to attend Wikimania 2009.
 * Interview (alt. mp3) with Cormac Lawler from Wikimania. (2008) (13m:24s)
 * --JWSchmidt 14:51, 3 September 2008 (UTC)

title: Wikimania/2009


text:
Talk:The Impact of Web-Based Lecture Technologies on Current and Future Practices in Learning and Teaching

Licensing
I am conscious of a potential licensing issue in uploading this extract from the reports' Executive Summary, because it is cc-by-nc licensed, which is more restrictive than WMF-approved licenses. I have contacted the authors to request WMF-compatible licensing, but I'd also be interested in hearing other perspectives on this potential issue. -- Jtneill - Talk - c 17:42, 9 September 2008 (UTC)
 * The lead author has responded via email and is discussing the request for a WMF-compatible license to be applied to this extract with her co-authors. -- Jtneill - Talk - c 18:47, 12 September 2008 (UTC)


text:
User talk:Eas4200c.f08.radsam.s

 Hello Eas4200c.f08.radsam.s, 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! --JWSchmidt 17:53, 9 September 2008 (UTC)

title: User talk:Eas4200c.f08.radsam.s


text:
User:EML4500.f08.A-team.melvin

Lecture 1

Lecture 10

Lecture 15

Lecture 17

Lecture 22

Lecture 28

Lecture 34

Lecture 38

title: User:EML4500.f08.A-team.melvin


text:
User:Jcarroll

Learning goals (as encouraged by the student union: Feel free to contact me through the university email system.
 * Become an expert in Open Hardware (design)
 * Gain an education sufficient to provide for my family.
 * Serve my community through teaching

Cheers! --Jcarroll 19:14, 9 September 2008 (UTC)

title: User:Jcarroll


text:
Introduction to Diplomatic History

DIPL 1001: Introduction to Diplomatic History is a survey of the history of diplomacy throughout recorded history. This course is meant to introduce new students to the study of diplomatic history by explaining the concept of diplomacy, terms that the student will use often in the Department of Diplomatic History, and a broad history of international relations on a global scale. This is the first course students new to the department should complete as it will be the foundation for successive courses.

Courses within the Department of Diplomatic History will be primarily self-study, placing the educational responsibility on the student rather than the professor. Although the professor will provide a plan for study and general information, it is the responsibility of the student to fulfil the goals and expectations of the course by completing the readings, studying the provided lesson, and performing independent research to prove comprehension. This course is also designed for homeschooling use, with information for the homeschool instructor to aid the student in successful completion. Recommendations for further study will be provided with each homeschooling lesson to expand upon the basic course.

Denelder Homeschooling Series on History
As denoted by the infobox above, this course is a lesson created by Dr. Jason Denelder as a part of his lesson series for homeschooled students. This lesson was designed as an aid for homeschoolers to further their knowledge on a specific topic, not to replace regular homeschooling lesson plans and resources. The Introduction to Diplomatic History is intended as the first lesson in the Department of Diplomatic History and is classified as HS-DH001 in the Homeschooling Series.

Suggested Readings

 * Diplomacy: An Historical Perspective Simon Szykman. Spring 1995. Carnegie Mellon University.
 * A Brief History of Diplomacy eDiplomat.
 * History and the Evolution of Diplomacy Richard Langhorne.
 * Diplomacy as an Instrument of Good Governance Vladimir Petrovsky.

Goals and Expectations
That students will understand the concept of diplomacy in a historical sense, will understand the basic terms of diplomatic research, and will gain an understanding of international relations on a global scale throughout history. That students will be able to implement their newly gained knowledge and successfully demonstrate their understanding by developing a researched essay on a general topic of diplomatic history.

Essay
The purpose of the essay is to demonstrate an understanding of the basics of diplomacy in the context of diplomatic history. As this is considered a university first-year course, grammar, spelling, and rhetoric should be correct and consistent throughout. Substance, however, is the most important component of a successful essay, demonstrating comprehension of the idea of diplomacy and its practical uses. Students have open control over the topic of their choice as long as it relates to the course material. Essays should be at least two pages long, although this should not be looked upon as a maximum. As the purpose is to demonstrate comprehension, the student should not write a descriptive essay, but rather an analytical essay. This means that the student should analyze a topic in diplomatic history, discussing the consequences of such an action and possible alternate paths and their consequences, using terms and concepts from the lesson.

Since this is an introductory course that encompasses the entire history of international relations, possible topics are endless in scope. In the essay, it is better to narrow in on a particular incident rather than discuss a broad topic. Do not analyze the diplomacy of Europe during the First World War, rather, analyze the effects of the War Guilt Clause on relations between Germany and the United Kingdom. Be specific in your analysis.

For those students wishing for constructive criticism of their essays may email them to Dr. Denelder at jason.denelder@yahoo.com. He will read your essay, making comments and noting corrections where necessary. If submitting the essay, please allow up to two weeks to accommodate Dr. Denelder's busy schedule.

Course Completion
Successful completion of DIPL 1001 is dependent on the student's understanding of the concept of diplomacy and the broad history of international relations demonstrated by the satisfactory completion of an essay on a selected topic of diplomatic history. DIPL 1001 serves as the prerequisite course for continuation into more detailed courses of diplomatic history and should be completed first.

Terms

 * Diplomacy: the art and practice of conducting negotiations between representatives of groups or states; also, the employment of tact to gain strategic advantage or to find mutually acceptable solutions to a common challenge
 * International Diplomacy: the conduct of international relations through the intercession of professional diplomats with regard to issues of peace-making, trade, war, economics, and culture
 * Diplomatic Immunity: a form of legal immunity and a policy held between governments, which ensures that diplomats are given safe passage and are considered not susceptible to lawsuit or prosecution under the host country's laws
 * Diplomatic Recognition: a unilateral political act, with domestic and international legal consequences, whereby a state acknowledges an act or status of another state or government
 * Paradiplomacy: the international relations conducted by subnational, regional, local or non-central governments

Course Roster

 * --Jolie 22:22, 9 October 2008 (UTC)

Professorship

 * Dr. Jason Denelder - For questions, concerns, discussion, or comment, please use my talk page or email me at jason.denelder@yahoo.com.

title: Introduction to Diplomatic History


text:
MediaWiki:Longpagewarning

 This page is $1 kilobytes long; please consider breaking this big page into smaller pages (such as subpages) so that dialup users, and those using mobile devices or old ancient browsers can edit and access pages without difficulties or delays.

title: MediaWiki:Longpagewarning


text:
User:Eml4500.f08.ateam.boggs.t/Class Notes

Week 1

Week 2

Week 3

Week 5

Week 7

Week 9

Week 11

Week 13

Week 15


text:
User:Tcady63

SOLVE USING FEA

For the first part of our homework, the 2-truss system shown in the figure below had to be analyzed and solved using the F.E.A. procedure. Where P is an applied downward force of 7, R1x, R1y, R3x, and R3y are the reactions in the x and y directions at nodes 1 and 3 respectively.

The element 1 is inclined at an angle of 30 degrees relative to the x axis while element 2 is declined at an angle of 45 degrees.

L1 = 4 L2 = 2 A1 = 1 A2 = 2 E1 = 3 E2 = 5

The next step is to draw the Free Body Diagram for the 1st element and determine its stiffness matrix. This Free Body Diagram is shown in the following figure and the stiffness matrix is defined as follows.

where k(e) = k(1) = E(1)*A(1)/L(e), l(e) = cos(θ(e)), m(e) = sin(θ(e))

In the case of element 1,

k(1) = 3*1/4 = 3/4 = 0.75

l(1) = cos(30) = 0.866

m(1) = sin(30) = 0.500

The resulting stiffness matrix for element 1 is, k(1) $$ = \begin{bmatrix} 0.5625&0.32476&-0.5625&-0.32476 \\ 0.32476&0.1875&-0.32476&-0.1875\\ -0.5625&-0.32476 & 0.5625&0.32476\\ -0.32476&-0.1875&0.32476&0.1875 \end{bmatrix} $$

Next it is necessary to calculate the stiffness matrix for the second element.

k(2) = 5*2/2 = 5

l(2) = cos(-45) = 0.500

m(2) = sin(-45) = -0.500

The resulting stiffness matrix for element 2 is, k(2) $$ = \begin{bmatrix} 2.5&-2.5&-2.5&2.5 \\ -2.5&2.5&2.5&-2.5\\ -2.5&2.5&2.5&-2.5\\ 2.5&-2.5&-2.5&2.5 \end{bmatrix} $$

Given that each element stiffness matrix is a 4x4 (because there are 4 unknown forces), it is next necessary to assemble a global (6x6) force-displacement matrix. This is done by recognizing that

F(1)1 = R(1)x

F(1)2 = R(1)y

F(2)3 = R(3)x

F(2)4 = R(3)y

F(1)3 = F(2)1 = Px

F(1)4 = F(2)2 = Py

Applying the above relationships to the global (6x6) matrix yields the following

K = $$ \begin{bmatrix} 0.5625&0.32476&-0.5625&-0.32476&0&0 \\ 0.32476&0.1875&-0.32476&-0.1875&0&0 \\ -0.5625&-0.32476&3.0625&-2.1752&-2.5&2.5 \\ -0.32476&-0.1875&-2.1752&2.6875&2.5&-2.5 \\ 0&0&-2.5&2.5&2.5&-2.5 \\ 0&0&2.5&-2.5&-2.5&2.5 \end{bmatrix} $$

Next, we observe that since node 1 and node 3 are pinned and fixed. This means that no displacement can occur at these points and thus d11, d12, d23, and d24 = 0. We have also been given that Py = 7 and is oriented vertically Px = 0. Applying these observations into the equation F = KD yields the following system.

$$ \begin{bmatrix} 0.5625&0.32476&-0.5625&-0.32476&0&0 \\ 0.32476&0.1875&-0.32476&-0.1875&0&0 \\ -0.5625&-0.32476&3.0625&-2.1752&-2.5&2.5 \\ -0.32476&-0.1875&-2.1752&2.6875&2.5&-2.5 \\ 0&0&-2.5&2.5&2.5&-2.5 \\ 0&0&2.5&-2.5&-2.5&2.5 \end{bmatrix} $$ * $$\begin{bmatrix} 0\\ 0\\ D3\\ D4\\ 0\\ 0 \end{bmatrix} $$ = $$ \begin{bmatrix} Rx1\\ Ry1\\ 0\\ 7\\ Rx3\\ Ry3 \end{bmatrix}$$

These equations can be used to solve for D3 and D4 (the x and y displacement of node 2).

D3 = 4.353

D4 = 6.1271

We can now plug these values back into the matrix equation and solve for the unknown reactions at nodes 1 and 2. When we do this, we find:

Rx1 = -4.4378

Ry1 = -2.5622

Rx3 = 4.4378

Ry3 = -4.4378

SOLVE USING STATICS

To verify these results, it is possible to solve the same problem using statics. Thus can been done by realize that in a truss, any reaction is directed along the axis of the truss. When the reactions at nodes 1 and 3 are combined into a single reaction forces (R1 and R3)we can solve the problem as follows.

ΣFy = R1sin(30) + R2sin(45) = 7

ΣFx = -R1cos(30) + R2cos(45) =

$$\begin{bmatrix} sin(30)&sin(45) \\ -cos(30)&cos(45) \end{bmatrix}$$ * $$\begin{bmatrix} R1\\ R3 \end{bmatrix}$$ = $$\begin{bmatrix} 7\\ 0 \end{bmatrix}$$

R1 = 5.12436

R3 = 6.27603

Plugging these values into the above equations, we find that

Rx1 = -5.12436*cos(30) = -4.4378

Ry1 = -5.12436*sin(30) = -2.562

Rx3 = 6.27603*cos(45) = 4.4378

Ry3 = -6.27603*sin(45) = -4.4378

These values agree perfectly with the ones attained using the Finite Element method thus confirming the results.

PROVE THE EXAMPLE IS A 2-FORCE BODY

The above statics solutions assumes a 2-force body and that the solutions attained via this assumption match those of the solutions attained using Finite Element Analysis. Given that the two solutions sets match, this proves that the trusses in the example problem are indeed 2-force bodies.

NODE 2 IN EQUILIBRIUM

In order for node 2 to be in equilibrium, the sum of the forces in the y direction have to equal to 0 as to the sum of the forces in the x direction. That is

R3sin(45) + R1sin(30) = 7

R3cos(45) - R1 cos(30) = 0

(6.27603)(0.707107) + (5.12436)(0.5) = 7

(6.27603)(0.707107) - (5.12436)(0.866025) = 0

Thus the sum of the forces in both the x and y directions equal 0, i.e. the node does not move and is in equilibrium.

title: User:Tcady63


text:
Wikiversity:Appropriate Venues

Wikiversity hosts a number of of independent projects on a number of topics, and each of these are contributed to by an independent segment of the larger community. These various communities should be "self-organizing and self-governing" to the extent that they are responsible for setting goals, methods, and acceptable practices within individual projects. The main discussion fora of Wikiversity should not be used as subpages of a particular project.

 * 1) In most cases, the talk page of the top project page will be the best place to discuss issues pertinent to a particular project.
 * 2) If discussing a topic that should be considered by several projects within a particular topic, a Topic: talk page should be used.
 * 3) The general fora (discussion pages within the Wikiversity namespace, including the Colloquium) should only be used for introducing a new project, asking advice on how to start a new project, or calling for intervention from the larger community in cases where communication in the local community breaks down.
 * 4) If the use of custodian tools is needed to handle a topic, use WV:RFC
 * 5) If review is needed on the behavior of another user, use WV:CR

title: Wikiversity:Appropriate Venues


text:
User talk:Eas4200c.f08.vqcrew.b

 Hello Eas4200c.f08.vqcrew.b, 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! --JWSchmidt 20:49, 9 September 2008 (UTC)

title: User talk:Eas4200c.f08.vqcrew.b


text:
User talk:Tcady63

 Hello Tcady63, 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! --JWSchmidt 20:50, 9 September 2008 (UTC)

title: User talk:Tcady63


text:
User talk:EML4500.f08.A-team.melvin

 Hello EML4500.f08.A-team.melvin, 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
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User:Eml4500.f08.ateam.nobrega/Lecture Notes Homework 1

Introduction to EML4500 Finite Element Analysis
The start of EML4500 was an introduction of the course detailing the schedule, the grading system, and the method for submitting work in the class. Groups of five or six students were to be created to work on homework and submit homework reports. These reports were originally going to be submitted on Wikipedia, but are now to be submitted on Wikiversity. Dr. Vu-Quoc also introduced the concept of collaborative learning and its benefits when juxtaposed with traditional teaching/learning methods. Each team is responsible for summarizing each lecture and identifying confusing or ambiguous points discussed. All of the teams' input will be used by Dr. Vo-Quoc as feedback and to address any underlying issues if necessary. Using Wikiversity would enable a sharing of knowledge between different teams enabling the class to be better prepared for the exams and encouraging discussions of course content. This would be accomplished by the posting of the top three homework reports on the class main page, making the correct answers to the homework available to the whole class with the material presented in a clear and concise manner. The top three teams would be given extra credit, but if any flaws were discovered in their reports after posting some of the extra credit could be taken away and given to those who found the flaws.

Using Wikiversity
Instructions were given on how to register and use Wikipedia (Wikiversity). It was explained to the class how to create an account and access one's user page as well as the class page. To do this the students were instructed to go to the Wikiversity main page and select

log in/create account

Next the user would need to type in the appropriate user name and create a password. The user names were to be created as follows:

eml4500.f08.teamname.lastname.first initial

Adding the first initial at the end of the user name was optional. We were then taught how to create links in our user page. Links are created by typing user:eml4500.f08.teamname.lastname.first initial/link|Link Name into ______________. This allows the user to create multiple pages within their user page or to link to a group member's user page or the class page.

Following this discussion it was explained in more detail what is expected of our homework reports, including content and what is expected of each group member. As a group member one is expected to work on his or her part and post in his or her user page. Once one finishes their contribution they are to post it on the group leader's user page where it will be compiled in the final report. This allows the Teaching Assistants and the Professor to possibly view each group members contribution to the homework report as well as the times of each submission. To account for any disparity between individual group member's contribution the group will rate each member's productivity on a scale from 0 to 8. The ratings will be tallied and a percentage will be taken based on the best possible rating one can receive. This percentage is then multiplied with the group's homework report grade and that member's individual homework grade is determined.

The advantages of using Wikiversity and Wikimedia followed, ending this series of notes with a comparison to the older more traditional method of structuring a course. It was explained to the class that at any point in the semester, if the majority of the class was in agreement, it would be possible to switch back to the more traditional method and grading scale.

Trusses, Beams and Frames (Ch4)
Trusses are structural systems arranged to resist primarily axial forces. Beams are long slender members that are commonly subjected to normal loading with respect to their longitudinal axis; as a result, they must withstand shear and bending forces. Frames can be viewed as a hybrid between trusses and beams because they are meant to resist bending and axial forces. Consider a truss composed of two elastic (deformable) bars as shown in Figure 1. The bars are fixed (constrained to zero displacement) at the two ends shown and a known force P is applied at the joint of the two bars. By inspection one realizes that there are four reaction forces, one in the x-direction and one in the y-direction of each fixed end.

Another method may be appropriate to solve this problem. First, inspect the system as a whole by drawing a global force body diagram (FBD). This diagram is featured below in Figure 2 and the FBDs of bar elements one and two are represented in Figures 3 and 4 respectively.

Then analyze each individual component by drawing the individual bar element FBD. In the Global FBD each node number is placed inside of a circle, for the FBD of the individual bar elements the node number is distinguished by a local node number that is placed inside of a square. The number may or may not directly correspond to the Global node number that it is referencing. For example, with bar element one in figure 3 below the Global node number one is also the local node number one, but depending on the user's preference the local node number for global node one could be two and the local node number for two could be one. This is also apparent with bar element two in figure 4 below as Global node number two is local node number one and Global node number three is local node number two. For the forces shown in the FBD's of the two bar elements the correct nomenclature is a capital F with a superscript of the bar element number in parenthesis and a subscript degree of freedom. It is correct to start with the x-direction and then continue with the y-direction at each node when labeling the degree of freedom.

For bar elements one and two the displacements directly correspond to the forces and are to follow the same strategies listed above when labeling the diagram. Instead of a capital F being used, for the displacements a lower case d will occupy its space.

The next big step is determining the force displacement (FD) relationship.

If you recall from previous classes the spring force equation is:

$$F=kd$$

This equation represents the force-distance relationship for a one-dimensional spring.

For a one-dimensional spring with 2 free ends the force-displacement relationship is more complex than the previous equation and can be determined with the use of matrices.

The force and displacement of the spring is shown in the image below:

When placing the equation for the spring above into matrix form one gets,

The equation above is a simple spring force equation for a spring with two nodes. The left side of the equation represents the spring forces at node one and two in a 2x1 matrix. The right hand side of the equation consists of the spring constant and the displacement of each node. The spring constant is listed in a 2x2 matrix while the displacement belongs in a 2x1 matrix.

Case 1: For the figure above with the observer is seated at node one

$$F$$2$$= k*$$$$(d$$2 $$- d$$1$$)$$

Case 2: For the figure above with the observer seated at node two or at equilibrium where F1 + F2 = 0

$$F$$1$$ = -F$$2$$= k*(d$$1$$ - d$$2$$)$$

Instructions (Recipe): Simple Trusses
This lecture given by Professor Vu-Quoc was a “recipe” for the steps to solve a simple truss system. The steps for solving a simple truss system are as follows:

Step 1 describes the Global Picture at the structure level. Identify the Global Degrees of Freedom (dof). (The displacement dofs) The displacement dofs are divided into a know part which is the fixed dof constraint, and unknown part which is solved for using the Finite Element Method. Next Identify the Global Forces. Similar the Global dofs the Global Forces are also divided into two parts. The know parts are the applied forces, and the unknown are the reaction forces.

Step 2 describes the Element Picture. First break the truss down into individual elements and identify element dofs, and element forces. When identifying the element dofs and forces you may either use the global or a local element coordinate system to make computation easier.

Step 3 defines the Global Force Displacement Relation. Define the element stiffness matrix K in the global coordinate system. Define The element force matrix F in the global coordinate system. The final step is the assembly of the element stiffness matrix and element force matrix into the global force displacement relation. The global force displacement relation is given by the equation: Kd=F where K is a(nxn) matrix, d is a (nx1) matrix, and F is a (nx1) matrix.It is assumed that the stiffness matrix is non-singular and invertible.

Step 4 the elimination of the known dofs to reduce the global Force Displacement Relation. Because the stiffness matrix K is non-singular and invertible the number of unknown displacement dofs are less than the number of know displacement dofs. The equation above in Step 3 is then solved for d by inverting K and multiplying by F resulting in d=K(inverse)F

Step 5 compute the element Forces from the now known ‘’’d’’’ element stresses.

Step 6 compute the unknown reaction Forces from the now known element forces and element stresses.

For further explanation please see the illustrated example below.

Example Problem:

The example below will showcase the steps described in the recipe above to solving simple truss problems. Trusses are composed of elements that only allow for axial loads. The elements are connected by nodes or joints. This problem has two elements and three nodes where each node has two degrees of freedom. The problem indicates an applied load P is present at node two. After solving the problem it would be possible to quantify the deflections and force reactions at each node.

Data:

Element Length: *L(1) = 4 *L(2) = 2

Young's Modulus: *E(1) = 3 *E(2) = 5

Cross Sectional Area: *A(1) = 1 *A(2) = 2

Inclination Angle: *Θ(1) = 30o *Θ(2) = 45o

Step 1: Global Picture

Labeling : Global degrees of freedom (dofs) and forces

The numbers inscribed in the green circles refer to a node value n. In the example described above the node values range from one to three. When numbering the displacement degrees of freedom (d1, . . . d6), you want to start at node one and with the x displacement. This displacement degree of freedom would be labeled ( d1 ). This would be followed by the y displacement at node one; labeled ( d2 ). Next would be node two, again starting with the x displacement; labeled ( d3 ). Followed again by the y displacement; labeled ( d4 ). One would continue this process until all the nodes are labeled with their corresponding degrees of freedom. The same systematic order would be followed to label the global forces ( F1, . . . F6 ). Refer to the image above for further explanation on systematic labeling. These displacements and forces are then put into a matrix format as seen in the image below.


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User:Eml4500.f08.team.foskey.ckf/HW1b

The Matrix Method

The Global Picture

To demonstrate using the Matrix Method to solve a problem, we will examine a truss with two elastic, or deformable, bars.

[Truss with two elastic bars]

A force P is applied at the top of the truss where the two bars connect, and each bar is constrained to the ground. Due to these constraints, the displacement at these locations will be zero, as we will show later.

A global free-body diagram is created showing all of the forces acting upon the structure as a whole. These include the force P, as well as the horizontal and vertical reaction forces on each of the fixed locations on the bars. It is also important to label the parts of the truss. Here each node is numbered and circled. Each bar represents an element of the whole, and is numbered and enclosed by a triangle. From this point forward, the bars will be referred to as elements.

[Global FBD of Truss]

The Element Picture

Each element of the structure is now looked at as an independent part. The nodes are renumbered and enclosed by a box. The triangle enclosed element number remains. The forces acting on the element are shown and labeled using the following format: fi(e) where e is the element number and i is the ith internal force of element e. Numbering of i is done in order of the nodes, beginning with the x-direction, and then the y-direction.

Displacements are also shown and labeled for each element. They are labeled similarly to the forces, except they are denoted by the letter d, as shown: di(e)

[Element 1]

[Element 2]

The Force Displacement Relation

There is a relationship between displacement and force. A simple example would be a spring. Force is proportional to the displacement of the spring. The coefficient for the relationship is k.

[Spring Example Diagram]

[Force-displacement relationship for spring example]

This concept can be applied to the elements by relating their displacements, forces, and the axial stiffness coefficient k. The matrices for the displacements, forces, and axial stiffenesses are assembled into the force-displacement relation equation: kd=F

where k is an (n x n) matrix, and d and F are (n x 1) matrices.

Elimination of knowns and computations

Once the force-displacement equation is created, knowns can be inserted to reduce the amount of unkown values than must be solved for. These knowns include the known applied forces (force P), and any constraints that have known displacements (the constrained ends of the elements is zero because they cannot move, as stated earlier). From there, the unknown displacements and the unknown forces can be calculated using matrix algebra.

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User:Eml4500.f08.ateam.boggs.t/team members

Tatiana

Mike

Akash

Pedro

Josh


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SPICE

SPICE (Simulation Program with Integrated Circuit Emphasis) is an open-source, circuit simulation program used to analyze and predict the behavior of analog and digital electronic circuits.

Readings

 * SPICE

Parts

 * 2 Emitter NPN
 * 3 Emitter NPN
 * 4 Emitter NPN
 * VCC 5V

Schematics

 * 1 Resistor Circuit
 * 2 Resistor Parallel Circuit
 * 2 Resistor Series Circuit
 * 3 Resistor SPP Circuit
 * 4 Resistor SPSP Circuit
 * NAND 2 Input TTL
 * NOT TTL

title: SPICE


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User:Eml4500.f08.ateam.boggs.t/Homework One

MATLAB Introduction


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User:Eml4500.f08.ateam.boggs.t/week 1

Lecture 1

Lecture 2

Lecture 3

title: User:Eml4500.f08.ateam.boggs.t/week 1


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User:Eml4500.f08.ateam.boggs.t/week 2

Lecture 4

Lecture 5

title: User:Eml4500.f08.ateam.boggs.t/week 2


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